Chalmers not… · Contents vii Chapter6. HilbertSpaces 85 x6.1. HermitianFormsandInnerProducts ::::::::::::::::::::::::: 85 x6.2. Orthoprojections

Contents

Preface to the English Translation v

Preface to the First Russian Edition vi

Chapter 1. An Excursus into Set Theory 1

§ 1.1. Correspondences . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . 1

§ 1.2. Ordered Sets . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . 3

§ 1.3. Filters . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . 6

Exercises . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . 8

Chapter 2. Vector Spaces 10

§ 2.1. Spaces and Subspaces . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . 10

§ 2.2. Linear Operators . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . 13

§ 2.3. Equations in Operators . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . 15

Exercises . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . 19

vi Contents

Chapter 3. Convex Analysis 21

§ 3.1. Sets in Vector Spaces . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . 21

§ 3.2. Ordered Vector Spaces . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . 23

§ 3.3. Extension of Positive Functionals and Operators . . . . . . . . . . . . . 26

§ 3.4. Convex Functions and Sublinear Functionals . . . . . . . . . . . . . . . . . 27

§ 3.5. The Hahn–Banach Theorem . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . 30

§ 3.6. The Kre–ın–Milman Theorem . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . 33

§ 3.7. The Balanced Hahn–Banach Theorem . . . . . . . . . . . . . . . . . . . . . . . . 35

§ 3.8. The Minkowski Functional and Separation . . . . . . . . . . . . . . . . . . . 37

Exercises . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . 40

Chapter 4. An Excursus into Metric Spaces 42

§ 4.1. The Uniformity and Topology of a Metric Space . . . . . . . . . . . . . . 42

§ 4.2. Continuity and Uniform Continuity . . . . . . . . . . . . . . . . . . . . . . . . . . 45

§ 4.3. Semicontinuity . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . 47

§ 4.4. Compactness . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . 49

§ 4.5. Completeness Completeness . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . 50

§ 4.6. Compactness and Completeness . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . 52

§ 4.7. Baire Spaces . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . 55

§ 4.8. The Jordan Curve Theorem and Rough Drafts . . . . . . . . . . . . . . . 57

Exercises . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . 58

Chapter 5. Multinormed and Banach Spaces 60

§ 5.1. Seminorms and Multinorms . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . 60

§ 5.2. The Uniformity and Topology of a Multinormed Space . . . . . . . 64

§ 5.3. Comparison Between Topologies . . . . . . . . . . . . . . . . . . . . . . . . . . . . . 66

§ 5.4. Metrizable and Normable Spaces . . . . . . . . . . . . . . . . . . . . . . . . . . . . . 68

§ 5.5. Banach Spaces . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . 71

§ 5.6. The Algebra of Bounded Operators . . . . . . . . . . . . . . . . . . . . . . . . . . 77

Exercises . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . 83

Contents vii

Chapter 6. Hilbert Spaces 85

§ 6.1. Hermitian Forms and Inner Products . . . . . . . . . . . . . . . . . . . . . . . . . 85

§ 6.2. Orthoprojections . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . 89

§ 6.3. A Hilbert Basis . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . 92

§ 6.4. The Adjoint of an Operator . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . 95

§ 6.5. Hermitian Operators . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . 98

§ 6.6. Compact Hermitian Operators . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . 100

Exercises . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . 103

Chapter 7. Principles of Banach Spaces 105

§ 7.1. Banach’s Fundamental Principle . . . . . . . . . . . . . . . . . . . . . . . . . . . . . 105

§ 7.2. Boundedness Principles . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . 107

§ 7.3. The Ideal Correspondence Principle . . . . . . . . . . . . . . . . . . . . . . . . . . 110

§ 7.4. Open Mapping and Closed Graph Theorems . . . . . . . . . . . . . . . . . 113

§ 7.5. The Automatic Continuity Principle . . . . . . . . . . . . . . . . . . . . . . . . . 117

§ 7.6. Prime Principles . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . 119

Exercises . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . 124

Chapter 8. Operators in Banach Spaces 126

§ 8.1. Holomorphic Functions and Contour Integrals . . . . . . . . . . . . . . . . 126

§ 8.2. The Holomorphic Functional Calculus . . . . . . . . . . . . . . . . . . . . . . . . 132

§ 8.3. The Approximation Property . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . 138

§ 8.4. The Riesz–Schauder Theory . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . 140

§ 8.5. Fredholm Operators . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . 144

Exercises . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . 151

Chapter 9. An Excursus into General Topology 153

§ 9.1. Pretopologies and Topologies . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . 153

§ 9.2. Continuity . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . 155

§ 9.3. Types of Topological Spaces . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . 158

§ 9.4. Compactness . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . 162

§ 9.5. Uniform and Multimetric Spaces . . . . . . . . . . . . . . . . . . . . . . . . . . . . . 166

§ 9.6. Covers, and Partitions of Unity . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . 172

Exercises . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . 175

viii Contents

Chapter 10. Duality and Its Applications 177

§ 10.1. Vector Topologies . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . 177

§ 10.2. Locally Convex Topologies . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . 179

§ 10.3. Duality Between Vector Spaces . . . . . . . . . . . . . . . . . . . . . . . . . . . . . 181

§ 10.4. Topologies Compatible with Duality . . . . . . . . . . . . . . . . . . . . . . . . 184

§ 10.5. Polars . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . 185

§ 10.6. Weakly Compact Convex Sets . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . 187

§ 10.8. Reflexive Spaces . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . 189

§ 10.8. The Space C(Q, R) . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . 189

§ 10.9. Radon Measures . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . 195

§ 10.10. The Spaces D(�) and D ′(�) . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . 202

§ 10.11. The Fourier Transform of a Distribution . . . . . . . . . . . . . . . . . . . 209

Exercises . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . 220

Chapter 11. Banach Algebras 221

§ 11.1. The Canonical Operator Representation . . . . . . . . . . . . . . . . . . . . 221

§ 11.2. The Spectrum of an Element of an Algebra . . . . . . . . . . . . . . . . . 223

§ 11.3. The Holomorphic Functional Calculus in Algebras . . . . . . . . . . . 224

§ 11.4. Ideals of Commutative Algebras . . . . . . . . . . . . . . . . . . . . . . . . . . . . 226

§ 11.5. Ideals of the Algebra C(Q,C) . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . 227

§ 11.6. The Gelfand Transform . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . 228

§ 11.7. The Spectrum of an Element of a C∗-Algebra . . . . . . . . . . . . . . . 232

§ 11.8. The Commutative Gelfand–Na–ımark Theorem . . . . . . . . . . . . . . . 234

§ 11.9. Operator ∗-Representations of a C∗-Algebra . . . . . . . . . . . . . . . . 237

Exercises . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . 243

References 237

Notation Index 255

Subject Index 259

Prefaceto the English Translation

This is a concise guide to basic sections of modern functional analysis. Includedare such topics as the principles of Banach and Hilbert spaces, the theory of multi-normed and uniform spaces, the Riesz–Dunford holomorphic functional calculus,the Fredholm index theory, convex analysis and duality theory for locally convexspaces.

With standard provisos the presentation is self-sufficient, whereas exposingabout a hundred of celebrated “named” theorems furnished with complete proofsand culminating in the Gelfand–Naımark–Segal construction for C∗-algebras.

The first Russian edition was printed by the Siberian Division of the “Nauka”Publishers in 1983. Since then the monograph has served as the standard textbookon functional analysis at the University of Novosibirsk.

This volume is translated from the second Russian edition printed by theSobolev Institute of Mathematics of the Siberian Division of the Russian Acad-emy of Sciences in 1995. It incorporates new sections on Radon measures, theSchwartz spaces of distributions, and a supplementary list of theoretical exercisesand problems for drill.

This edition was typeset using AMS-TEX, the American Mathematical Soci-ety’s TEX system.

To clear my conscience completely, I also confess that := stands for the definor,the assignment operator, / marks the beginning of a (possibly empty) proof, and .signifies the end of the proof.

S. Kutateladze

Prefaceto the First Russian Edition

As the title implies, this book treats of functional analysis. At the turn of thecentury the term “functional analysis” was coined by J. Hadamard who is famousamong mathematicians for the formula of the radius of convergence of a power se-ries. The term “functional analysis” was universally accepted then as related to thecalculus of variations, standing for a new direction of analysis which was intensivelydeveloped by V. Volterra, C. Arzela, S. Pincherle, P. Levy, and other representa-tives of the French and Italian mathematical schools. J. Hadamard’s contributionto the recent discipline should not be reduced to the invention of the word “func-tional” (or more precisely to the transformation of the adjective into a proper noun).J. Hadamard was fully aware of the relevance of the rising subject. Working hard,he constantly advertised problems, ideas, and methods just evolved. In particular,to one of his students, M. Frechet, he suggested the problem of inventing somethingthat is now generally acclaimed as the theory of metric spaces. In this connectionit is worthy to indicate that neighborhoods pertinent to functional analysis in thesense of Hadamard and Volterra served as precursors to Hausdorff’s famous researchheralded the birth of general topology.

For the sequel it is essential to emphasize that one of the most attractive,difficult, and important sections of classical analysis, the calculus of variations,became the first source of functional analysis.

The second source of functional analysis was provided by the study directed tocreating some algebraic theory for functional equations or, stated strictly, to sim-plifying and formalizing the manipulations of “equations in functions” and, in par-ticular, linear integral equations. Ascending to H. Abel and J. Liouville, the theoryof the latter was considerably expanded by works of I. Fredholm, K. Neumann,F. Noether, A. Poincare, et al. The efforts of these mathematicians fertilized soilfor D. Hilbert’s celebrated research into quadratic forms in infinitely many variables.His ideas, developed further by F. Riesz, E. Schmidt, et al., were the immediatepredecessors of the axiomatic presentation of Hilbert space theory which was un-

Preface xi

dertaken and implemented by J. von Neumann and M. Stone. The resulting sectionof mathematics has vigorously influenced theoretical physics, first of all, quantummechanics. In this regard it is instructive as well as entertaining to mention thatboth terms, “quantum” and “functional,” originated in the same year, 1900.

The third major source of functional analysis encompassed Minkowski’s geo-metric ideas. His invention, the apparatus for the finite-dimensional geometryof convex bodies, prepared the bulk of spatial notions ensuring the modern devel-opment of analysis. Elaborated by E. Helly, H. Hahn, C. Caratheodory, I. Radon,et al., the idea of convexity has eventually shaped the fundamentals of the theoryof locally convex spaces. In turn, the latter has facilitated the spread of distribu-tions and weak derivatives which were recognized by S. L. Sobolev as drasticallychanging all tools of mathematical physics. In the postwar years the geometric no-tion of convexity has conquered a new sphere of application for mathematics, viz.,social sciences and especially economics. An exceptional role in this process wasperformed by linear programming discovered by L. V. Kantorovich.

The above synopsis of the strands of functional analysis is schematic, incom-plete, and approximate (for instance, it casts aside the line of D. Bernoulli’s super-position principle, the line of set functions and integration theory, the line of opera-tional calculus, the line of finite differences and fractional derivation, the line of gen-eral analysis, and many others). These demerits notwithstanding, the three sourceslisted above reflect the main, and most principal, regularity: functional analysis hassynthesized and promoted ideas, concepts, and methods from the classical sectionsof mathematics: algebra, geometry, and analysis. Therefore, although functionalanalysis verbatim means analysis of functions and functionals, even a superficialglance at its history gives grounds to claim that functional analysis is algebra,geometry, and analysis of functions and functionals.

A more viable and penetrating explanation for the notion of functional analysisis given by the Soviet Encyclopedic Dictionary: “Functional analysis is one of theprincipal branches of modern mathematics. It resulted from mutual interaction,unification, and generalization of the ideas and methods stemming from all partsof classical mathematical analysis. It is characterized by the use of concepts per-taining to various abstract spaces such as vector spaces, Hilbert spaces, etc. It findsdiverse applications in modern physics, especially in quantum mechanics.”

The S. Banach treatise Theorie des Operationes Lineares, printed half a cen-tury ago, inaugurated functional analysis as an essential activity in mathematics.Its influence on the development of mathematics is seminal: Ubiquitous and om-nipresent, Banach’s ideas, propounded in the book, captivate the realm of modernmathematics.

Outstanding contribution toward progress in functional analysis belongs tothe renowned Soviet scientists: I. M. Gelfand, L. V. Kantorovich, M. V. Keldysh,

xii Preface

A. N. Kolmogorov, M. G. Kreın, L. A. Lyusternik, and S. L. Sobolev. The char-acteristic feature of the Soviet school is that its research on functional analysis isalways conducted in connection with profound applied problems. The research hasexpanded the scope of functional analysis which becomes the prevailing languageof the applications of mathematics.

The next fact is demonstrative: In 1948 considered as paradoxical was even thetitle of Kantorovich’s insightful article Functional Analysis and Applied Mathemat-ics which provided a basis for numerical mathematics of today. Whereas in 1974S. L. Sobolev stated that “to conceive the theory of calculations without Banachspaces is as impossible as without electronic computers.”

The exponential accumulation of knowledge within functional analysis is nowobserved alongside a sharp rise in demand for the tools and concepts of the disci-pline. A thus resulting conspicuous gap widens permanently between the state-ofthe art of analysis as such and as reflected in the literature accessible to the readingcommunity. To alter this ominous trend is the purpose of the present book.

Prefaceto the Second Russian Edition

More than a decade the monograph serves as a reference book for the compulsoryand optional courses in functional analysis which are delivered at Novosibirsk StateUniversity. The span of time proves that the principles of compiling the book arelegitimate. The present edition is enlarged with sections addressing fundamentalsof distribution theory. Theoretical exercises are supplemented and the list of refer-ences is updated. Also, the inaccuracies are improved that were mostly indicatedby my colleagues.

I seize the opportunity to express my gratitude to all those who helped mein preparation of the book. My pleasant debt is to acknowledge the financial supportof the Sobolev Institute of Mathematics of the Siberian Division of the RussianAcademy of Sciences, the Russian Foundation for Basic Research, the InternationalScience Foundation and the American Mathematical Society during the compilationof the second edition.

S. KutateladzeMarch, 1995

Chapter 1

An Excursus into Set Theory

1.1. Correspondences

1.1.1. Definition. Let A and B be sets and let F be a subset of the productA × B := {(a, b) : a ∈ A, b ∈ B}. Then F is a correspondence with the setof departure A and the set of arrival B or just a correspondence from A (in)to B.

1.1.2. Definition. For a correspondence F ⊂ A×B the set

domF := D(F ) := {a ∈ A : (∃ b ∈ B) (a, b) ∈ F}

is the domain (of definition) of F and the set

imF := R(F ) := {b ∈ B : (∃ a ∈ A) (a, b) ∈ F}

is the codomain of F , or the range of F , or the image of F .

1.1.3. Examples.(1) If F is a correspondence from A into B then

F−1 := {(b, a) ∈ B ×A : (a, b) ∈ F}

is a correspondence from B into A which is called inverse to F or the inverse of F .It is obvious that F is the inverse of F−1.

(2) A relation F on A is by definition a subset of A2, i.e. a correspon-dence from A to A (in words: “F acts in A”).

(3) Let F ⊂ A×B. Then F is a single-valued correspondence if for alla ∈ A the containments (a, b1) ∈ F and (a, b2) ∈ F imply b1 = b2. In particular,if U ⊂ A and IU := {(a, a) ∈ A2 : a ∈ U}, then IU is a single-valued correspon-dence acting in A and called the identity relation (over U on A). A single-valued

2 Chapter 1

correspondence F ⊂ A × B with domF = A is a mapping of A into B or a map-ping from A (in)to B. The terms “function” and “map” are also in current usage.A mapping F ⊂ A×B is denoted by F : A→ B. Observe that here domF alwayscoincides with A whereas imF may differ from B. The identity relation IU on A isa mapping if and only if A = U and in this case IU is called the identity mappingin U or the diagonal of U2. The set IU may be treated as a subset of U × A. Theresulting mapping is usually denoted by ι : U0 = 0 OO and is called the identicalembedding of U into A. It is said that “F is a correspondence from A onto B”if imF = B. Finally, a correspondence F ⊂ X × Y is one-to-one whenever thecorrespondence F−1 ⊂ B ×A is single-valued.

(4) Occasionally the term “family” is used instead of “mapping.” Na-mely, a mapping F : A0 = 0 OO is a family in B (indexed in A by F ), also denotedby (ba)a∈A or a 7→ ba (a ∈ A) or even (ba). Specifically, (a, b) ∈ F if and onlyif b = ba. In the sequel, a subset U of A is often treated as indexed in itselfby the identical embedding of U into A. It is worth recalling that in set theory ais an element or a member of A whenever a ∈ A. In this connection a family in Bis also called a family of elements of B or a family of members of B. By way ofexpressiveness a family or a set of numbers is often addressed as numeric. Also,common abusage practices the identification of a family and its range. This sin isvery enticing.

(5) Let F ⊂ A × B be a correspondence and U ⊂ A. The restrictionof F to U , denoted by F |U , is the set F ∩(U×B) ⊂ U×B. The set F (U) := imF |Uis the image of U under F .

If a and b are elements of A and B then F (a) = b is usually written insteadof F ({a}) = {b}. Often the parentheses in the symbol F (a) are omitted or replacedwith other symbols. For a subset U of B the image F−1(U) of U under F−1 isthe inverse image of U or the preimage of U under F . So, inverse images are justimages of inverses.

(6) Given a correspondence F ⊂ A×B, assume that A is the productof A1 and A2, i.e. A = A1 ×A2. Fixing a1 in A1 and a2 in A2, consider the sets

F (a1, · ) := {(a2, b) ∈ A2 ×B : ((a1, a2), b) ∈ F};F ( · , a2) := {(a1, b) ∈ A1 ×B : ((a1, a2), b) ∈ F}.

These are the partial correspondences of F . In this regard F itself is often sym-bolized as F ( · , · ) and referred to as a correspondence in two arguments. Thisbeneficial agreement is effective in similar events.

1.1.4. Definition. The composite correspondence or composition of corre-spondences F ⊂ A×B and G ⊂ C ×D is the set

G ◦ F := {(a, d) ∈ A×D : (∃ b) (a, b) ∈ F & (b, d) ∈ G}.

An Excursus into Set Theory 3

The correspondence G ◦ F is considered as acting from A into D.

1.1.5. Remark. The scope of the concept of composition does not diminishif it is assumed in 1.1.4 from the very beginning that B = C.

1.1.6. Let F be a correspondence. Then F ◦ F−1 ⊃ I imF . Moreover, theequality F ◦ F−1 = I imF holds if and only if F |domF is a mapping. CB

1.1.7. Let F ⊂ A × B and G ⊂ B × C. Further assume U ⊂ A. Then thecorrespondence G ◦ F ⊂ A× C satisfies the equality G ◦ F (U) = G(F (U)). CB

1.1.8. Let F ⊂ A×B, G ⊂ B×C, and H ⊂ C×D. Then the correspondencesH ◦ (G ◦ F ) ⊂ A×D and (H ◦G) ◦ F ⊂ A×D coincide. CB

1.1.9. Remark. By virtue of 1.1.8, the symbol H ◦ G ◦ F and the like aredefined soundly.

1.1.10. Let F, G, and H be correspondences. Then

H ◦G ◦ F =⋃

(b,c)∈G

F−1(b)×H(c).

C (a, d) ∈ H ◦ G ◦ F ⇔ (∃ (b, c) ∈ G) (c, d) ∈ H & (a, b) ∈ F ⇔(∃ (b, c) ∈ G) a ∈ F−1(b) & d ∈ H(c) B

1.1.11. Remark. The claim of 1.1.10, together with the calculation intendedas its proof, is blatantly illegitimate from a formalistic point of view as based onambiguous or imprecise information (in particular, on Definition 1.1.1!). Experiencejustifies treating such criticism as petty. In the sequel, analogous convenient (and,in fact, inevitable) violations of formal purity are mercilessly exercised with nocircumlocution.

1.1.12. Let G and F be correspondences. Then

G ◦ F =⋃

b∈ imF

F−1(b)×G(b).

C Insert H := G, G := I imF and F := F into 1.1.10. B

1.2. Ordered Sets

1.2.1. Definition. Let σ be a relation on a set X, i.e. σ ⊂ X2. Reflexivityfor σ means the inclusion σ ⊃ IX ; transitivity, the inclusion σ◦σ ⊂ σ; antisymmetry,the inclusion σ ∩ σ−1 ⊂ IX ; and, finally, symmetry, the equality σ = σ−1.

4 Chapter 1

1.2.2. Definition. A preorder is a reflexive and transitive relation. A sym-metric preorder is an equivalence. An order (partial order, ordering, etc.) is an an-tisymmetric preorder. For a set X, the pair (X, σ), with σ an order on X, isan ordered set or rarely a poset. The notation x ≤σ y is used instead of y ∈ σ(x).The terminology and notation are often simplified and even abused in commonparlance: The underlying set X itself is called an ordered set, a partially orderedset, or even a poset. It is said that “x is less than y,” or “y is greater than x,” or“x ≤ y,” or “y ≥ x,” or “x is in relation σ to y,” or “x and y belong to σ,” etc. Anal-ogous agreements apply customarily to a preordered set, i.e. to a set furnished witha preorder. The convention is very propitious and extends often to an arbitraryrelation. However, an equivalence is usually denoted by the signs like ∼.

1.2.3. Examples.(1) The identity relation; each subset X0 of a set X bearing a relation

σ is endowed with the (induced) relation σ0 := σ ∩X0 ×X0.(2) If σ is an order (preorder) in X, then σ−1 is also an order (preorder)

which is called reverse to σ.(3) Let f : X0 = 0 OO be a mapping and let τ be a relation on Y .

Consider the relation f−1 ◦ τ ◦ f appearing on X. By 1.1.10,

f−1 ◦ τ ◦ f =⋃

(y1,y2)∈τ

f−1(y1)× f−1(y2).

Hence it follows that (x1, x2) ∈ f−1 ◦ τ ◦ f ⇔ (f(x1), f(x2)) ∈ τ. Thus, if τis a preorder then f−1 ◦ τ ◦ f itself is a preorder called the preimage or inverseimage of τ under f . It is clear that the inverse image of an equivalence is alsoan equivalence. Whereas the preimage of an order is not always antisymmetric.In particular, this relates to the equivalence f−1 ◦ f (= f−1 ◦ IY ◦ f).

(4) Let X be an arbitrary set and let ω be an equivalence on X. Definea mapping ϕ : X0 = 0 OO P(X) by ϕ(x) := ω(x). Recall that P(X) stands forthe powerset of X comprising all subsets of X and also denoted by 2X . Let X :=X/ω := imϕ be the quotient set or factor set of X by ω or modulo ω. A memberof X/ω is usually referred to as a coset or equivalence class. The mapping ϕ isthe coset mapping (canonical projection, quotient mapping, etc.). Note that ϕ istreated as acting onto X. Observe that

ω = ϕ−1 ◦ ϕ =⋃x∈X

ϕ−1(x)× ϕ−1(x).

Now let f : X0 = 0 OO be a mapping. Then f admits factorization through X; i.e.,there is a mapping f : X0 = 0 OO (called a quotient of f by ω) such that f ◦ ϕ = fif and only if ω ⊂ f−1 ◦ f . CB


(5) Let (X, σ) and (Y, τ) be two preordered sets. A mapping f : X0 =0 OO is increasing or isotone (i.e., x ≤σ y ⇒ f(x) ≤τ f(y)) whenever σ ⊂ f−1 ◦ τ ◦ f .That f decreases or is antitone means σ ⊂ f−1 ◦ τ−1 ◦ f . CB

1.2.4.Definition. Let (X, σ) be an ordered set and let U ⊂ X. An element xof X is an upper bound of U (write x ≥ U) if U ⊂ σ−1(x). In particular, x ≥ ∅.An element x of X is a lower bound of U (write x ≤ U) if x is an upper bound of Uin the reverse order σ−1. In particular, x ≤ ∅.

1.2.5. Remark. Throughout this book liberties are taken with introducingconcepts which arise from those stated by reversal, i.e. by transition from a (pre)ord-er to the reverse (pre)order. Note also that the definitions of upper and lower boundsmake sense in a preordered set.

1.2.6. Definition. An element x of U is greatest or last if x ≥ U and x ∈ U .When existent, such element is unique and is thus often referred to as the greatestelement of U . A least or first element is defined by reversal.

1.2.7. Let U be a subset of an ordered set (X, σ) and let πσ(U) be thecollection of all upper bounds of U . Suppose that a member x of X is the greatestelement of U . Then, first, x is the least element of πσ(U); second, σ(x) ∩ U ={x}. CB

1.2.8. Remark. The claim of 1.2.7 gives rise to two generalizations of theconcept of greatest element.

1.2.9. Definition. Let X be a (preordered) set and let U ⊂ X. A supremumof U in X is a least upper bound of U , i.e. a least element of the set of all upperbounds of U . This element is denoted by supX U or in short supU . Certainly,in a poset an existing supremum of U is unique and so it is in fact the supremumof U . An infimum, a greatest lower bound inf U or infX U is defined by reversal.

1.2.10. Definition. Let U be a subset of an ordered set (X, σ). A member xof X is a maximal element of U if σ(x) ∩ U = {x}. A minimal element is againdefined by reversal.

1.2.11. Remark. It is important to make clear the common properties anddistinctions of the concepts of greatest element, maximal element, and supremum.In particular, it is worth demonstrating that a “typical” set has no greatest elementwhile possibly possessing a maximal element.

1.2.12. Definition. A lattice is an ordered set with the following property:each pair (x1, x2) of elements of X has a least upper bound, x1∨x2 := sup{x1, x2},the join of x1 and x2, and a greatest lower bound, x1 ∧ x2 := inf{x1, x2}, the meetof x1 and x2.

6 Chapter 1

1.2.13. Definition. A lattice X is complete if each subset of X has a supre-mum and an infimum in X.

1.2.14. An ordered set X is a complete lattice if and only if each subset of Xhas a least upper bound. CB

1.2.15. Definition. An ordered set (X, σ) is filtered upward provided thatX2 = σ−1 ◦ σ. A downward-filtered set is defined by reversal. A nonempty poset isa directed set or simply a direction if it is filtered upward.

1.2.16. Definition. Let X be a set. A net or a (generalized) sequence in Xis a mapping of a direction into X. A mapping of the set N of natural numbers,N = {1, 2, 3 . . . }, furnished with the conventional order, is a (countable) sequence.

1.2.17. A lattice X is complete if and only if each upward-filtered subset of Xhas a least upper bound. CB

1.2.18. Remark. The claim of 1.2.17 implies that for calculating a supremumof each subset it suffices to find suprema of pairs and increasing nets.

1.2.19. Definition. Let (X, σ) be an ordered set. It is said that X is orderedlinearly whenever X2 = σ ∪ σ−1. A nonempty linearly-ordered subset of X isa chain in X. A nonempty ordered set is called inductive whenever its every chainis bounded above (i.e., has an upper bound).

1.2.20. Kuratowski–Zorn Lemma. Each inductive set contains a maximalelement.

1.2.21. Remark. The Kuratowski–Zorn Lemma is equivalent to the axiomof choice which is accepted in set theory.

1.3. Filters

1.3.1. Definition. Let X be a set and let B, a nonempty subset of P(X),consist of nonempty elements. Such B is said to be a filterbase (onX) if B is filtereddownward. Recall that P(X) is ordered by inclusion. It means that a greater subsetincludes a smaller subset by definition; this order is always presumed in P(X).

1.3.2. A subset B of P(X) is a filterbase if and only if1 B 6= ∅ and ∅ 6∈ B;(2) B1, B2 ∈ B ⇒ (∃B ∈ B) B ⊂ B1 ∩B2.

1.3.3. Definition. A subset F of P(X) is a filter (on X) if there is a filter-base B such that F is the set of all supersets of B; i.e.,

F = filB := {C ∈P(X) : (∃B ∈ B) B ⊂ C}.

In this case B is said to be a base for the filter F (so each filterbase is a filter base).


1.3.4. A subset F in P(X) is a filter if and only if(1) F 6= ∅ and ∅ 6∈ F ;(2) (A ∈ F & A ⊂ B ⊂ X)⇒ B ∈ F ;(3) A1, A2 ∈ F ⇒ A1 ∩A2 ∈ F . CB

1.3.5. Examples.(1) Let F ⊂ X × Y be a correspondence and let B be a downward-

filtered subset of P(X). Put F (B) := {F (B) : B ∈ B}. It is easy to see thatF (B) is filtered downward. The notation is alleviated by putting F (B) := filF (B).If F is a filter on X and B ∩ domF 6= ∅ for all B ∈ F , then F (F ) is a filter(on Y ) called the image of F under F . In particular, if F is a mapping then theimage of a filter on X is a filter on Y .

(2) Let (X, σ) be a direction. Clearly, B := {σ(x) : x ∈ X} isa filterbase. For a net F : X0 = 0 OO , the filter filF (B) is the tail filter of F . Let(X, σ) and F : X → Y be another direction and another net in Y . If the tail filterof F includes the tail filter of F then F is a subnet (in a broad sense) of the netF . If there is a subnet (in a broad sense) G : X → X of the identity net ((x)x∈Xin the direction (X, σ)) such that F = F ◦G, then F is a subnet of F (sometimesF is addressed as a Moore subnet or a strict subnet of F ). Every subnet is a subnetin a broad sense. It is customary to speak of a net having or lacking a subnet withsome property.

1.3.6. Definition. Let F (X) be the collection of all filters on X. TakeF1, F2 ∈ F (X). Say that F1 is finer than F2 or F2 refines F1 (in other words,F1 is coarser than F2 or F1 coarsens F2) whenever F1 ⊃ F2.

1.3.7. The set F (X) with the relation “to be finer” is a poset. CB

1.3.8. Let N be a direction in F (X). Then N has a supremum F0 :=sup N . Moreover, F0 = ∪{F : F ∈ N }.

C To prove this, it is necessary to show that F0 is a filter. Since N is notempty it is clear that F0 6= ∅ and ∅ /∈ F0. If A ∈ F0 and B ⊃ A then, choosingF in N for which A ∈ F , conclude that B ∈ F ⊂ F0. Given A1, A2 ∈ F0,find an element F of N satisfying A1, A2 ∈ F , which is possible because N isa direction. By 1.3.4, A1 ∩A2 ∈ F ⊂ F0. B

1.3.9. Definition. An ultrafilter is a maximal element of the ordered setF (X) of all filters on X.

1.3.10. Each filter is coarser than an ultrafilter.C By 1.3.8, the set of filters finer than a given filter is inductive. Recalling the

Kuratowski–Zorn Lemma completes the proof. B

8 Chapter 1

1.3.11. A filter F is an ultrafilter if and only if for all A ⊂ X either A ∈ For X \A ∈ F .

C ⇒: Suppose that A 6∈ F and B := X \ A 6∈ F . Note that A 6= ∅ andB 6= ∅. Put F1 := {C ∈ P(X) : A ∪ C ∈ F}. Then A 6∈ F ⇒ ∅ 6∈ F1 andB ∈ F1 ⇒ F1 6= ∅. The checking of 1.3.4 (2) and 1.3.4 (3) is similar. Hence, F1is an ultrafilter. By definition, F1 ⊃ F . In addition, F is an ultrafilter and soF1 = F . Observe that B 6∈ F and B ∈ F , a contradiction.

⇐: Take F1 ∈ F (X) and let F1 ⊃ F . If A ∈ F1 and A 6∈ F then X \A ∈ Fby hypothesis. Hence, X\A ∈ F1; i.e., ∅ = A∩(X\A) ∈ F1, which is impossible. B

1.3.12. If f is a mapping from X into Y and F is an ultrafilter on X thenf(F ) is an ultrafilter on Y . CB

1.3.13. Let X := XF0 := {F ∈ F (X) : F ⊂ F0} for F0 ∈ F (X). Then Xis a complete lattice.

C It is obvious that F0 is the greatest element of X and {X} is the leastelement of X . Therefore, the empty set has a supremum and an infimum in X :in fact, sup∅ = inf X = {X} and inf ∅ = supX = F0. By 1.2.17 and 1.3.8,it suffices to show that the join F1 ∨F2 is available for all F1, F2 ∈X . ConsiderF := {A1 ∩ A2 : A1 ∈ F1, A2 ∈ F2}. Clearly, F ⊂ F0 while F ⊃ F1 andF ⊃ F2. Thus to verify the equality F = F1 ∨F2, it is necessary to establishthat F is a filter.

Plainly, F 6= ∅ and ∅ 6∈ F . It is also immediate that (B1, B2 ∈ F ⇒B1 ∩ B2 ∈ F ). Moreover, if C ⊃ A1 ∩ A2 where A1 ∈ F1 and A2 ∈ F2, thenC = {A1 ∩A2} ∪ C = (A1 ∪ C) ∩ (A2 ∪ C). Since A1 ∪ C ∈ F1 and A2 ∪ C ∈ F2,conclude that C ∈ F . Appealing to 1.3.4, complete the proof. B

Exercises

1.1. Give examples of sets and nonsets as well as set-theoretic properties and non-set-theoretic properties.

1.2. Is it possible for the interval [0, 1] to be a member of the interval [0, 1]? For theinterval [0, 2]?

1.3. Find compositions of the simplest correspondences and relations: squares, disks andcircles with coincident or distinct centers in RM × RN for all feasible values of M and N .

1.4. Given correspondences R, S, and T , demonstrate that

(R ∪ S)−1 = R−1 ∪ S−1; (R ∩ S)−1 = R−1 ∩ S−1;(R ∪ S) ◦ T = (R ◦ T ) ∪ (S ◦ T ); R ◦ (S ∪ T ) = (R ◦ S) ∪ (R ◦ T );(R ∩ S) ◦ T ⊂ (R ◦ T ) ∩ (S ◦ T ); R ◦ (S ∩ T ) ⊂ (R ◦ S) ∩ (R ◦ T ).

1.5. Assume X ⊂ X ×X. Prove that X = ∅.1.6. Find conditions for the equations XA = B and AX = B to be solvable for X in cor-

respondences or in functions.


1.7. Find the number of equivalences on a finite set.1.8. Is the intersection of equivalences also an equivalence? And the union of equivalences?1.9. Find conditions for commutativity of equivalences (with respect to composition).1.10. How many orders and preorders are there on two-element and three-element sets? List

all of them. What can you say about the number of preorders on a finite set?

1.11. Let F be an increasing idempotent mapping of a set X into itself. Assume that Fdominates the identity mapping: F ≥ IX . Such an F is an abstract closure operator or, briefly,an (upper) envelope. Study fixed points of a closure operator. (Recall that an element x is a fixedpoint of F if F (x) = x.)

1.12. Let X and Y be ordered sets and M(X, Y ), the set of increasing mappings from Xto Y with the natural order (specify the latter). Prove that

(1) (M(X, Y ) is a lattice) ⇔ (Y is a lattice);(2) (M(X, Y ) is a complete lattice) ⇔ (Y is a complete lattice).

1.13. Given ordered sets X, Y , and Z, demonstrate that(1) M(X, Y × Z) is isomorphic with M(X, Y )×M(Y, Z);(2) M(X × Y, Z) is isomorphic with M(X, M(Y, Z)).

1.14. How many filters are there on a finite set?1.15. How do the least upper and greatest lower bounds of a set of filters look like?1.16. Let f be a mapping from X onto Y . Prove that each ultrafilter on Y is the image

of some ultrafilter on X under f .

1.17. Prove that an ultrafilter refining the intersection of two filters is finer than eitherof them.

1.18. Prove that each filter is the intersection of all ultrafilters finer than it.1.19. Let A be an ultrafilter on N containing cofinite subsets (a cofinite subset is a subset

with finite complement). Given x, y ∈ s := RN, put x ∼A y := (∃A ∈ A ) x|A = y|A. Denote∗R := RN/∼A. For t ∈ R the notation ∗t symbolizes the coset with the constant sequence t definedas t(n) := t (n ∈ N). Prove that ∗R \ {∗t : t ∈ R} 6= ∅. Furnish ∗R with algebraic and orderstructures. How are the properties of R and ∗R related to each other?

Chapter 2

Vector Spaces

2.1. Spaces and Subspaces

2.1.1. Remark. In algebra, in particular, modules over rings are studied.A module X over a ring A is defined by an abelian group (X, +) and a repre-sentation of the ring A in the endomorphism ring of X which is considered as leftmultiplication · : A × X0 = 0 OO by elements of A. Moreover, a natural agreementis presumed between addition and multiplication. With this in mind, the followingphrase is interpreted: “A module X over a ring A is described by the quadruple(X, A, +, · ).” Note also that A is referred to as the ground ring of X.

2.1.2. Definition. A basic field is the field R of real numbers or the field C ofcomplex numbers. The symbol F stands for a basic field. Observe that R is treatedas embedded into C in a standard (and well-known) fashion so that the operationRe of taking the real part of a number sends C onto the real axis, R.

2.1.3. Definition. Let F be a field. A module X over F is a vector space(over F). An element of the ground field F is a scalar in X and an element of Xis a vector in X or a point in X. So, X is a vector space with scalar field F.The operation + : X × X0 = 0 OO is addition in X and · : F × X0 = 0 OO is scalarmultiplication in X. We refer to X as a real vector space in case F = R and asa complex vector space, in case F = C. A more complete nomenclature consistsof (X, F, +, · ), (X, R, +, · ), and (X, C, +, · ). Neglecting these subtleties,allow X to stand for every vector space associated with the underlying set X.

2.1.4. Examples.(1) A field F is a vector space over F.(2) Let (X, F, +, · ) be a vector space. Consider (X, F, +, ·∗), where

·∗ : (λ, x) 7→ λ∗x for λ ∈ F and x ∈ X, the symbol λ∗ standing for the conventionalcomplex conjugate of λ. The so-defined vector space is the twin of X denoted by X∗.If F := R then the space X and the twin of X, the space X∗, coincide.

Vector Spaces 11

(3) A vector space (X0, F, +, · ) is called a subspace of a vector space(X, F, +, · ), if X0 is a subgroup of X and scalar multiplication in X0 is the re-striction of that in X to F × X0. Such a set X0 is a linear set in X, whereas Xis referred to as an ambient space (for X0). It is convenient although not perfectlypuristic to treat X0 itself as a subspace of X. Observe the important particularityof terminology: a linear set is a subset of a vector space (whose obsolete title isa linear space). To call a subset of whatever space X a set in X is a mathematicalidiom of long standing. The same applies to calling a member of X a point in X.Furthermore, the neutral element of X, the zero vector of X, or simply zero of X,is considered as a subspace of X and is denoted by 0. Since 0 is not explicitlyrelated to X, all vector spaces including the basic fields may seem to have a pointin common, zero.

(4) Take (Xξ)ξ∈�, a family of vector spaces over F, and let X :=∏ξ∈�Xξ be the product of the underlying sets, i.e. the collection of mappings

x : �0 = 0 OOξ∈�Xξ such that xξ := x(ξ) ∈ Xξ as ξ ∈ � (of course, here � is not

empty). Endow X with the coordinatewise or pointwise operations of addition andscalar multiplication:

(x1 + x2)(ξ) := x1(ξ) + x2(ξ) (x1, x2 ∈X , ξ ∈ �);(λ · x)(ξ) := λ · x(ξ) (x ∈X , λ ∈ F, ξ ∈ �)

(below, as a rule, we write λx and sometimes xλ rather than λ · x). The so-constructed vector space X over F is the product of (Xξ)ξ∈�. If � := {1, 2, . . . , N}then X1 ×X2 × . . .×XN := X . In the case Xξ = X for all ξ ∈ �, the designationX� := X is used. Given � := {1, 2, . . . , N}, put XN := X .

(5) Let (Xξ)ξ∈� be a family of vector spaces over F. Consider theirdirect sum X0 :=

∑ξ∈�Xξ. By definition, X0 is the subset of X :=

∏ξ∈�Xξ

which comprises all x0 such that x0(�\�0) ⊂ 0 for a finite subset �0 ⊂ � (routinelyspeaking, �0 is dependent on x0). It is easily seen that X0 is a linear set in X .The vector space associated with X0 presents a subspace of the product of (Xξ)ξ∈�and is the direct sum of (Xξ)ξ∈�.

(6) Given a subspace (X, F, +, · ) of a vector space (X0, F, +, · ),introduce

∼X0 := {(x1, x2) ∈ X2 : x1 − x2 ∈ X0}.

Then ∼X0 is an equivalence on X. Denote X := X/∼X0 and let ϕ : X0 = 0 OO X bethe coset mapping. Define operations on X by letting

x1 + x2 := ϕ(ϕ−1(x1) + ϕ−1(x2)) (x1, x2 ∈X );λx := ϕ(λϕ−1(x)) (x ∈X , λ ∈ F).

12 Chapter 2

Here, as usual, for subsets S1 and S2 of X, a subset � of F, and a scalar λ, a memberof F, it is assumed that

S1 + S2 := +{S1 × S2};�S1 := · (�× S1); λS1 := {λ}S1.

Thus X is furnished with the structure of a vector space. This space, denotedby X/X0, is the quotient (space) of X by X0 or the factor space of X modulo X0.

2.1.5. Let X be a vector space and let Lat(X) stand for the collection of allsubspaces of X. Ordered by inclusion, Lat(X) presents a complete lattice.

C It is clear that inf Lat(X) = 0 and supLat(X) = X. Further, the intersectionof a nonempty set of subspaces is also a subspace. By 1.2.17, the proof is complete.B

2.1.6. Remark. With X1, X2 ∈ Lat(X), the equality X1 ∨ X2 = X1 + X2holds. It is evident that inf E = ∩{X0 : X0 ∈ E } for a nonempty subset Eof Lat(X). Provided that E is filtered upward, supE = ∪{X0 : X0 ∈ E }. CB

2.1.7. Definition. Subspaces X1 and X2 of a vector space X split X into(algebraic) direct sum decomposition (in symbols, X = X1 ⊕X2), if X1 ∧X2 = 0and X1 ∨X2 = X. In this case X2 is an (algebraic) complement of X1 to X, andX1 is an (algebraic) complement of X2 to X. It is also said that X1 and X2 are(algebraically) complementary to one another.

2.1.8. Each subspace of a vector space has an algebraic complement.

C Take a subspace X1 of X. Put

E := {X0 ∈ Lat(X) : X0 ∧X1 = 0}.

Obviously, 0 ∈ E . Given a chain E i0 in E , from 2.1.6 infer that X1 ∧ supE i0 = 0,i.e. supE i0 ∈ E . Thus E is inductive and, by 1.2.20, E has a maximal element,say, X2. If x ∈ X \ (X1 +X2) then

(X2 + {λx : λ ∈ F}) ∧X1 = 0.

Indeed, if x2 + λx = x1 with x1 ∈ X1, x2 ∈ X2 and λ ∈ F, then λx ∈ X1 +X2 andso λ = 0. Hence, x1 = x2 = 0 as X1 ∧X2 = 0. Therefore, X2 + {λx : λ ∈ F} = X2because X2 is maximal. It follows that x = 0. At the same time, it is clear thatx 6= 0. Finally, X1 ∨X2 = X1 +X2 = X. B

Vector Spaces 13

2.2. Linear Operators

2.2.1. Definition. Let X and Y be vector spaces over F. A correspondenceT ⊂ X × Y is linear, if T is a linear set in X × Y . A linear operator on X (orsimply an operator, with linearity apparent from the context) is a mapping of Xand a linear correspondence simultaneously. If need be, we distinguish such T froma linear single-valued correspondence S with domS 6= X and say that T is givenon X or T is defined everywhere or even T is a total operator, whereas S is referredto as not-everywhere-defined or partially-defined operator or even a partial operator.In the case X = Y , a linear operator from X to Y is also called an operator in Xor an endomorphism of X.

2.2.2. A correspondence T ⊂ X × Y is a linear operator from X to Y if andonly if domT = X and

T (λ1x1 + λ2x2) = λ1Tx1 + λ2Tx2 (λ1, λ2 ∈ F; x1, x2 ∈ X). CB

2.2.3. The set L (X, Y ) of all linear operators carrying X into Y constitutesa vector space, a subspace of Y X . CB

2.2.4.Definition. Amember of L (X, F) is a linear functional onX, and thespace X# := L (X, F) is the (algebraic) dual of X. A linear functional on X∗ isa ∗-linear or conjugate-linear functional on X. If the nature of F needs specifying,then we speak of real linear functionals, complex duals, etc. Evidently, when F = Rthe term “∗-linear functional” is used rarely, if ever.

2.2.5. Definition. A linear operator T , a member of L (X, Y ), is an (alge-braic) isomorphism (of X and Y , or between X and Y ) if the correspondence T−1

is a linear operator, a member of L (Y, X).

2.2.6. Definition. Vector spaces X and Y are (algebraically) isomorphic,in symbols X ' Y , provided that there is an isomorphism between X and Y .

2.2.7. Vector spaces X and Y are isomorphic if and only if there are operatorsT ∈ L (X, Y ) and S ∈ L (Y, X) such that S ◦ T = IX and T ◦ S = IY (in thisevent S = T−1 and T = S−1). CB

2.2.8. Remark. Given vector spaces X, Y , and Z, take T ∈ L (X, Y ) andS ∈ L (Y, Z). The correspondence S◦T is undoubtedly a member of L (X, Z). Forsimplicity every composite operator S ◦T is denoted by juxtaposition ST . Observealso that the taking of composition (S, T ) 7→ ST is usually treated as the mapping◦ : L (Y, Z) × L (X, Y )0 = 0 OO L(X, Z). In particular, if E ⊂ L (Y, Z) andT ∈ L (X, Y ) then we let E ◦ T := ◦(E × {T}). One of the reasons behind theconvention is that juxtaposition in the endomorphism space L (X) := L (X, X)of X which comprises all endomorphisms of X transforms L (X) into a ring (andeven into an algebra, the endomorphism algebra of X, cf. 5.6.2).

14 Chapter 2

2.2.9. Examples.(1) If T is a linear correspondence then T−1 is also a linear correspon-

dence.(2) If X1 is a subspace of a vector space X and X2 is an algebraic

complement of X1 then X2 is isomorphic with X/X1. Indeed, if ϕ : X0 = 0 OO /X1is the coset mapping then its restriction to X2, i.e. the operator x2 7→ ϕ(x2) withx2 ∈ X2, implements a desired isomorphism.CB

(3) Consider X :=∏ξ∈�Xξ, the product of vector spaces (Xξ)ξ∈�.

Take a coordinate projection, i.e. a mapping Prξ : X 0 = 0 OOξ defined by Prξ x := xξ.

Clearly, Prξ is a linear operator, Prξ ∈ L (X , Xξ). Such an operator is oftentreated as an endomorphism of X , a member of L (X ), on implying a naturalisomorphism between Xξ and Xξ, where Xξ :=

∏η∈�Xη with Xη := 0 if η 6= ξ and

Xξ := Xξ.(4) Let X := X1 ⊕ X2. Since +−1 is an isomorphism between X and

X1 × X2, we may define P1, P2 ∈ L (X) as P1 := PX1||X2 := Pr1 ◦(+−1) andP2 := PX2||X1 := Pr2 ◦(+−1). The operator P1 is the projection of X onto X1 alongX2 and P2 is the complementary projection to P1 or the complement of P1 (insymbols, P2 = P d1 ). In turn, P1 is complementary to P2, and P2 projects X ontoX2 along X1. Observe also that P1 +P2 = IX . Moreover, P 2

1 := P1P1 = P1, and soa projection is an idempotent operator. Conversely, every idempotent P belongingto L (X) projects X onto P (X) along P−1(0).

For T ∈ L (X) and P a projection, the equality PTP = TP holds if and onlyif T (X0) ⊂ X0 with X0 = imP (read: X0 is invariant under T ). CB

The equality TPX1||X2 = PX1||X2T holds whenever both X1 and X2 are invari-ant under T , and in this case the direct sum decomposition X = X1⊕X2 reduces T .The restriction of T to X1 is acknowledged as an element T1 of L (X1) which iscalled the part of T in X1. If T2 ∈ L (X2) is the part of T in X2, then T isexpressible in matrix form

T ∼(T1 00 T2

).

Namely, an element x of X1 ⊕ X2 is regarded as a “column vector” with compo-nents x1 and x2, where x1 = PX1||X2x and x2 = PX1||X2x; matrices are multipliedaccording to the usual rule, “rows by columns.” The product of T and the columnvector x, i.e. the vector with components T1x1 and T2x2, is certainly consideredas Tx (in this case, we also write Tx1 and Tx2). In other words, T is identifiedwith the mapping from X1 ×X2 to X1 ×X2 acting as(

x1x2

)7→(T1 00 T2

)(x1x2

).

Vector Spaces 15

In a similar way we can introduce matrix presentation for general operators con-tained in L (X1 ⊕X2, Y1 ⊕ Y2). CB

(5) A finite subset E of X is linearly independent (in X) provided that∑e∈E λee = 0, with λe ∈ F (e ∈ E ), implies λe = 0 for all e ∈ E . An arbitrary

subset E of X is linearly independent if every finite subset of E is linearly inde-pendent. A Hamel basis (or an algebraic basis) for X is a linearly independent setin X maximal by inclusion. Each linearly independent set is contained in a Hamelbasis. All Hamel bases have the same cardinality called the dimension of X anddenoted by dimX. Every vector space is isomorphic to the direct sum of a family(F)ξ∈� with � of cardinality dimX. Suppose that X1 is a subspace of X. The codi-mension of X1 is the dimension of X/X1, with codimX1 standing for the former.If X = X1 ⊕X2 then codimX1 = dimX2 and dimX = dimX1 + codimX1. CB

2.3. Equations in Operators

2.3.1. Definition. Given T ∈ L (X, Y ), define the kernel of T as kerT :=T−1(0), the cokernel of T as cokerT := Y/ imT , and the coimage of T as coimT :=X/ kerT . Agree that an operator T is an monomorphism whenever kerT = 0.An operator T is an epimorphism in the case of the equality imT = Y .

2.3.2. An operator is an isomorphism if and only if it is a monomorphism andan epimorphism simultaneously. CB

2.3.3. Remark. Below use is made of the concept of commutative diagram.So the phrase, “The following diagram commutes,”

V W

X Yα1

α2α3α4

α5

-

?

@@@R-

�

encodes the containments α1 ∈ L (X, Y ), α2 ∈ L (Y, W ), α3 ∈ L (X, W ),α4 ∈ L (V, Y ) and α5 ∈ L (V, W ), as well as the equalities α2α1 = α3 andα5 = α2α4.

2.3.4. Definition. A diagram XT−→ Y

S−→ Z is an exact sequence (at theterm Y ), if kerS = imT . A sequence . . . 0 = 0 OO

k−10 = 0 OOk0 = 0 OO

k+10 = 0 OO is exactat Xk, if for all k the subsequence Xk−10 = 0 OO

k0 = 0 OOk+1 (symbols of operators

are omitted), and is exact if it is exact at every term (except the first and the last,if any).

16 Chapter 2

2.3.5. Examples.(1) An exact sequence X T−→ Y

S−→ Z is semi-exact, i.e. ST = 0. Theconverse is not true.

(2) A sequence 00 = 0 OO T−→ Y is exact if and only if T is a monomor-phism. (Throughout the book 00 = 0 OO certainly denotes the sole element of L (0, X),zero (cf. 2.1.4 (3)).)

(3) A sequence X T−→ Y 0 = 0 OO is exact if and only if T is an epi-morphism. (Plainly, the symbol Y 0 = 0 OO again stands for zero, the single elementof L (Y, 0).)

(4) An operator T from X to Y , a member of L (X, Y ), is an isomor-phism if and only if 00 = 0 OO T−→ Y 0 = 0 OO is exact.

(5) Suppose that X0 is a subspace of X. Let ι : X00 = 0 OO stand forthe identical embedding of X0 into X. Consider the quotient space X/X0 and letϕ : X0 = 0 OO /X0 be the corresponding coset mapping. Then the sequence

00 = 0 OO 0ι−→ X

ϕ−→ X/X00 = 0 OO

is exact. (On similar occasions the letters ι and ϕ are usually omitted.) In a sense,this sequence is unique. Namely, consider a so-called short sequence

00 = 0 OO T−→ YS−→ Z0 = 0 OO

and assume that it is exact. Putting Y0 := imT , arrange the following commutativediagram

0

0

Y0

X

Y

Y

Y/Y0

Z

0

0

α

T

β

S

γ

-

-

-

-

-

-

-

-

? ? ?

where α, β, and γ are some isomorphisms. In other words, a short exact sequenceactually presents a subspace and the corresponding quotient space. CB

(6) Every T in L (X, Y ) generates the exact sequence

00 = 0 OO T0 = 0 OO T−→ Y 0 = 0 OO T0 = 0 OO

which is called the canonical exact sequence for T .

2.3.6. Definition. Given T0 ∈ L (X0, Y ), with X0 a subspace of X, callan operator T from X to Y an extension of T0 (onto X, in symbols, T ⊃ T0),provided that T0 = Tι, where ι : X00 = 0 OO is the identical embedding of X0into X.

Vector Spaces 17

2.3.7. Let X and Y be vector spaces and let X0 be a subspace of X. Theneach T0 in L (X0, Y ) has an extension T in L (X, Y ).

C Putting T := T0PX0 , where PX0 is a projection onto X0, settles the claim. B

2.3.8. Theorem. Let X, Y , and Z be vector spaces. Take A ∈ L (X, Y ) andB ∈ L (X, Z). The diagram

Z

X YA

XB

-

?

@@@R

is commutative for some X in L (Y, Z) if and only if kerA ⊂ kerB.

C ⇒: It is evident that kerA ⊂ kerB in case B = X A.⇐: Set X := B◦A−1. Clearly, X ◦A(x) = B◦(A−1◦A)x = B(x+kerA) = Bx.

Show that X0 := X |imA is a linear operator. It suffices to check that X is single-valued. Suppose that y ∈ imA and z1, z2 ∈X (y). Then z1 = Bx1, z2 = Bx2 andAx1 = Ax2 = y. By hypothesis B(x1− x2) = 0; therefore, z1 = z2. Applying 2.3.7,find an extension X of X0 to Y . B

2.3.9. Remark. Provided that the operator A is an epimorphism, there isa unique solution X in 2.3.8. It is the right place to emphasize that the phrase,“There is a unique X ,” implies that X is available as well as unique.

2.3.10. Every linear operator T admits a unique factorization through itscoimage; i.e., there is a unique quotient T of T by the equivalence ∼codimT .

C Immediate from 2.3.8 and 2.3.9. B

2.3.11. Remark. The operator T is sometimes called the monoquotient of Tand treated as acting onto imT . In this connection, observe that T is expressibleas the composition T = ιTϕ, with ϕ an epimorphism, T an isomorphism, and ιa monomorphism; i.e., the following diagram commutes:

coimT

X

imT

Y

ϕ

T

Tι6

-

-

?

2.3.12. Let X be a vector space and let f0, f1, . . . , fN belong to X#. Thefunctional f0 is a linear combination of f1, . . . , fN (i.e., f0 =

∑Nj=1 λjfj with λj in

F ) if and only if ker f0 ⊃ ∩Nj=1 ker fj .

18 Chapter 2

C Define the linear operator (f1, . . . , fN ) : X0 = 0 OO F N by (f1, . . . , fN )x :=(f1(x), . . . , fN (x)). Obviously, ker(f1, . . . , fN ) = ∩Nj=1 ker fj . Now apply 2.3.8 tothe problem

F

X FN(f1,... ,fN )

f0

-

?

@@@@@R

on recalling what FN# is. B

2.3.13. Theorem. Let X, Y , and Z be vector spaces. Take A ∈ L (Y, X)and B ∈ L (Z, X). The diagram

Z

X YA

XB

�

6

@@@I

is commutative for some X in L (Z, Y ) if and only if imA ⊃ imB.

C ⇒: imB = B(Z) = A(X (Z)) ⊂ A(Y ) = imA.

⇐: Let Y0 be an algebraic complement of kerA to Y and A0 := A|Y0 . Then A0

is an isomorphism between Y0 and imA. The operator X := ιA−10 B is obviously

a sought solution, with ι the identical embedding of Y0 into Y . B

2.3.14. Remark. Provided that the operator A is a monomorphism, there isa unique operator X in 2.3.13. CB

2.3.15. Remark. Theorems 2.3.8 and 2.3.13 are in “formal duality.” Oneresults from the other by “reversing arrows,” “interchanging kernels and images,”and “passing to reverse inclusion.”

2.3.16. Snowflake Lemma. Let S ∈ L (Y, Z) and T ∈ L (X, Y ). Thereare unique operators α1, . . . , α6 such that the following diagram commutes:

Vector Spaces 19

0 0

kerST kerS

0 kerT X Y cokerT 0

Z

cokerS cokerST

0 0

α2

α1 α3

T

ST S

α6 α4

α5

-

��/

SSSw

��/

-

SSSw

��/

-

SSSw

SSSSSSSw

Sw �/

-

Sw

-

SSSw

��/

�

��7

�/

-

Moreover, the sequence

00 = 0 OO Tα1−→ kerST α2−→ kerS α3−→ cokerT α4−→ cokerST α5−→ cokerS0 = 0 OO

is exact. CB

Exercises

2.1. Give examples of vector spaces and nonvector spaces. Which constructions lead tovector spaces?

2.2. Study vector spaces over the two-element field Z2.2.3. Describe vector spaces with a countable Hamel basis.2.4. Prove that there is a discontinuous solution f : R0 = 0 OO R to the function equation

f(x+ y) = f(x) + f(y) (x, y ∈ R).

How to visualize such an f graphically?

2.5. Prove that the algebraic dual to the direct sum of (Xξ) is presentable as the productof the algebraic duals of (Xξ).

2.6. Let X ⊃ X0 ⊃ X00. Prove that the spaces X/X00 and (X/X0)/(X00/X0) are isomor-phic.

2.7. Define the “double sharp” mapping by the rule

x## : x# 7→ 〈x |x#〉 (x ∈ X, x# ∈ X#).

Show that this mapping embeds a vector space X into the second dual X##.

20 Chapter 2

2.8. Prove that finite-dimensional spaces and only such spaces are algebraically reflexive;i.e.,

##(X) = X## ⇔ dimX < +∞.

2.9. Are there any analogs for a Hamel basis in general modules?2.10. When does a sum of projections present a projection itself?2.11. Let T be an endomorphism of some vector space which satisfies the conditions Tn−1 6=

0 and Tn = 0 for a natural n. Prove that the operators T 0, T, . . . , Tn−1 are linearly independent.

2.12. Describe the structure of a linear operator defined on the direct sum of spaces andacting into the product of spaces.

2.13. Find conditions for unique solvability of the following equations in operators XA = Band AX = B (here the operator X is unknown).

2.14. What is the structure of the spaces of bilinear operators?2.15. Characterize the real vector space that results from neglecting multiplication by imag-

inary scalars in a complex vector space (cf. 3.7.1).

2.16. Given a family of linearly independent vectors (xe)e∈E, find a family of functionals(x#e )e∈E satisfying the following conditions:

〈xe |x#e 〉 = 1 (e ∈ E );

〈xe |x#e 〉 = 0 (e, e′ ∈ E, e 6= e′).

2.17. Given a family of linearly independent functionals (x#e )e∈E, find a family of vectors

(xe)e∈E satisfying the following conditions:

〈xe |x#e 〉 = 1 (e ∈ E );

〈xe |x#e 〉 = 0 (e, e′ ∈ E, e 6= e′).

2.18. Find compatibility conditions for simultaneous linear equations and linear inequalitiesin real vector spaces.

2.19. Consider the commutative diagram

W−→ XT−→ Y−→ Z

α ↓ β ↓ γ ↓ δ ↓

W−→ XT−→ Y−→ Z

with exact rows and such that α is an epimorphism, and δ is a monomorphism. Prove thatker γ = T (kerβ) and T−1(im γ) = imβ.

Chapter 3

Convex Analysis

3.1. Sets in Vector Spaces

3.1.1. Definition. Let � be a subset of F2. A subset U of a vector space isa � -set in this space (in symbols, U ∈ (� )) if (λ1, λ2) ∈ � ⇒ λ1U + λ2U ⊂ U.

3.1.2. Examples.(1) Every set is in (∅). (Hence, (∅) is not a set.)(2) If � := F2 then a nonempty � -set is precisely a linear set in a vector

space.(3) If � := R2 then a nonempty � -set in a vector space X is a real

subspace of X.(4) By definition, a cone is a nonempty � -set with � := R2

+. In otherwords, a nonempty set K is a cone if and only if K +K ⊂ K and αK ⊂ K for allα ∈ R+. A nonempty R2

+ \ 0-set is sometimes referred to as a nonpointed cone; anda nonempty R+× 0-set, as a nonconvex cone. (From now on we use the convenientnotation R+ := {t ∈ R : t ≥ 0}.)

(5) A nonempty � -set, for � := {(λ1, λ2) ∈ F2 : λ1 + λ2 = 1}, isan affine variety or a flat. IfX0 is a subspace ofX and x ∈ X then x+X0 := {x}+X0is an affine variety in X. Conversely, if L is an affine variety in X and x ∈ L thenL− x := L+ {−x} is a linear set in X. CB

(6) If � := {(λ1, λ2) ∈ F2 : |λ1|+ |λ2| ≤ 1} then a nonempty � -set isabsolutely convex.

(7) If � := {(λ, 0) ∈ F2 : |λ| ≤ 1} then a nonempty � -set is balanced.(In the case F := R the term “star-shaped” is occasionally employed; the word“symmetric” can also be found.)

(8) A set is convex if it is a � -set for � := {(λ1, λ2) ∈ R2 : λ1 ≥ 0, λ2 ≥0, λ1 + λ2 = 1}.

22 Chapter 3

(9) A conical segment or conical slice is by definition a nonempty � -setwith � := {(λ1, λ2) ∈ R2

+ : λ1 + λ2 ≤ 1}. A set is a conical segment if and onlyif it is a convex set containing zero. CB

(10) Given � ⊂ F2, observe that X ∈ (� ) for every vector space Xover F. Note also that in 3.1.2 (1)–3.1.2 (9) the set � is itself a � -set.

3.1.3. Let X be a vector space and let E be a family of � -sets in X. Then∩{U : U ∈ imE } ∈ (� ). Provided that imE is filtered upward (by inclusion),∪{U : U ∈ imE } ∈ (� ). CB

3.1.4. Remark. The claim of 3.1.3 means in particular that the collectionof � -sets of a vector space, ordered by inclusion, presents a complete lattice.

3.1.5. Let X and Y be vector spaces and let U and V be � -sets, with U ⊂ Xand V ⊂ Y . Then U × V ∈ (� ).C If U or V is nonempty then U × V = ∅, and there is nothing to prove.

Now take u1, u2 ∈ U , v1, v2 ∈ V , and (λ1, λ2) ∈ � . Find λ1u1 + λ2u2 ∈ U andλ1v2 + λ2v1 ∈ V . Hence, (λ1u1 + λ2u2, λ1v1 + λ2v2) ∈ U × V . B

3.1.6. Definition. Let X and Y be vector spaces and � ⊂ F2. A correspon-dence T ⊂ X × Y is a � -correspondence provided that T ∈ (� ).

3.1.7. Remark. When a � -set bears a specific attribute, the latter is pre-served for naming a � -correspondence. With this in mind, we speak about linearand convex correspondences, affine mappings, etc. The next particularity of thenomenclature is worth memorizing: a convex function of one variable is not a con-vex correspondence, save trivial cases (cf. 3.4.2).

3.1.8. Let T ⊂ X × Y be a �1-correspondence and let U ⊂ X be a �2-set.If �2 ⊂ �1 then T (U) ∈ (�2).

C Take y1, y2 ∈ T (U). Then (x1, y1) ∈ T and (x2, y2) ∈ T with somex1, x2 ∈ U . Given (λ1, λ2) ∈ �2, observe that λ1(x1, y1) + λ2(x2, y2) ∈ T ,because by hypothesis (λ1, λ2) ∈ �1. Finally, infer that λ1y1 + λ2y2 ∈ T (U). B

3.1.9. The composition of � -correspondences is also a � -correspondence.C Let F ⊂ X × V and G ⊂W × Y , with F, G ∈ (� ). Note that

(x1, y1) ∈ G ◦ F ⇔ (∃ v1) (x1, v1) ∈ F & (v1, y1) ∈ G;(x2, y2) ∈ G ◦ F ⇔ (∃ v2) (x2, v2) ∈ F & (v2, y2) ∈ G.

To complete the proof, “multiply the first row by λ1; the second, by λ2, where(λ1, λ2) ∈ � ; and sum the results.” B

Convex Analysis 23

3.1.10. If subsets U and V of a vector space are � -sets for some � ⊂ F2 thenαU + βV ∈ (� ) for all α, β ∈ F.

C The claim is immediate from 3.1.5, 3.1.8, and 3.1.9. B

3.1.11. Definition. Let X be a vector space and let U be a subset of X. For� ⊂ F2 the � -hull of U is the set

H� (U) := ∩{V ⊂ X : V ∈ (� ), V ⊃ U}.

3.1.12. The following statements are valid:(1) H� (U) ∈ (� );(2) H� (U) is the least � -set including U ;(3) U1 ⊂ U2 ⇒ H� (U1) ⊂ H� (U2);(4) U ∈ (� )⇔ U = H� (U);(5) H� (H� (U)) = H� (U). CB

3.1.13. The Motzkin formula holds:

H� (U) = ∪{H� (U0) : U0 is a finite subset of U}.

C Denote the right side of the Motzkin formula by V . Since U0 ⊂ U ; applying3.1.12 (3), deduce that H� (U0) ⊂ H� (U); and, hence, H� (U) ⊃ V . By 3.1.12 (2),it is necessary (and, surely, sufficient) to verify that V ∈ (� ). The last follows from3.1.3 and the inclusion H� (U0) ∪H� (U1) ⊂ H� (U0 ∪ U1). B

3.1.14. Remark. The Motzkin formula reduces the problem of describingarbitrary � -hulls to calculating � -hulls of finite sets. Observe also that � -hulls forconcrete � s have special (but natural) designations. For instance, if � := {(λ1, λ2) ∈R2

+ : λ1+λ2 = 1}, then the term “convex hull” is used and co(U) stands for H� (U).If U 6= ∅, the notation for HF2(U) is lin(U); moreover, it is natural and convenientto put lin(∅) := 0. The subspace lin(U) is called the linear span of U . The conceptsof affine hull, conical hull, etc. are introduced in a similar fashion. Note also thatthe convex hull of a finite set of points comprises their convex combinations; i.e.,

co({x1, . . . , xN}) =

{N∑k=1

λkxk : λk ≥ 0, λ1 + . . .+ λN = 1

}. /.

3.2. Ordered Vector Spaces

3.2.1. Definition. Let (X, R, +, · ) be a vector space. A preorder σ on Xis compatible with vector structure if σ is a cone in X2; in this case X is an orderedvector space. (It is more precise to call (X, R, +, · , σ) a preordered vector space,reserving the term “ordered vector space” for the case in which σ is an order.)

24 Chapter 3

3.2.2. If X is an ordered vector space and σ is the corresponding preorderthen σ(0) is a cone and σ(x) = x+ σ(0) for all x ∈ X.

C By 3.1.3, σ(0) is a cone. The equality (x, y) = (x, x) + (0, y − x) yieldsthe equivalence (x, y) ∈ σ ⇔ y − x ∈ σ(0). B

3.2.3. Let K be a cone in a vector space X. Denote

σ := {(x, y) ∈ X2 : y − x ∈ K}.

Then σ is a preorder compatible with vector structure and K coincides with thecone σ(0) of positive elements of X. The relation σ is an order if and onlyif K ∩ (−K) = 0.C It is clear that 0 ∈ K ⇒ IX ⊂ σ and K+K ⊂ K ⇒ σ ◦σ ⊂ σ. Furthermore,

σ−1 = {(x, y) ∈ X2 : x − y ∈ K}. Therefore, σ ∩ σ−1 ⊂ IX ⇔ K ∩ (−K) = 0.To show that σ is a cone, take (x1, y1), (x2, y2) ∈ σ and α1, α2 ∈ R+. Findα1y1 + α2y2 − (α1x1 + α2x2) = α1(y1 − x1) + α2(y2 − x2) ∈ α1K + α2K ⊂ K. B

3.2.4. Definition. A cone K is an ordering cone or a salient cone providedthat K ∩ (−K) = 0.

3.2.5. Remark. By virtue of 3.2.2 and 3.2.3, assigning a preorder to X isequivalent to distinguishing some cone of positive elements in it which is the positivecone ofX. The structure of an ordered vector space arises from selecting an orderingcone. Keeping this in mind, we customarily call a pair (X, X+) with positive coneX+, as well as X itself, a (pre)ordered vector space.

3.2.6. Examples.(1) The space of real-valued functions R� with the positive cone R�+ :=

(R+)� constituted by the functions assuming only positive values has the “naturalorder.”

(2) Let X be an ordered vector space with positive cone X+. If X0is a subspace of X then the order induced in X0 is defined by the cone X0 ∩X+.In this way X0 is considered as an ordered vector space, a subspace of X.

(3) IfX and Y are preordered vector spaces then T ∈ L (X, Y ) is a pos-itive operator (in symbols, T ≥ 0) whenever T (X+) ⊂ Y+. The set L+(X, Y ) of allpositive operators is a cone. The linear span of L+(X, Y ) is denoted by Lr(X, Y ),and a member of Lr(X, Y ) is called a regular operator.

3.2.7. Definition. An ordered vector space is a vector lattice or a Riesz spaceif the ordered set of its vectors presents a lattice.

3.2.8. Definition. A vector lattice X is called a Kantorovich space or brieflya K-space if X is boundedly order complete or Dedekind complete, which meansthat each nonempty bounded above subset of X has a least upper bound.

Convex Analysis 25

3.2.9. Each nonempty bounded below subset of a Kantorovich space has agreatest lower bound.

C Let U be bounded below: x ≤ U for some x. So, −x ≥ −U . Applying 3.2.8,find sup(−U). Obviously, − sup(−U) = inf U . B

3.2.10. If U and V are nonempty bounded above subsets of a K-space then

sup(U + V ) = supU + supV.

C If U or V is a singleton then the equality is plain. The general case followsfrom the associativity of least upper bounds. Namely,

sup(U + V ) = sup{sup(u+ V ) : u ∈ U}= sup{u+ supV : u ∈ U} = supV + sup{u : u ∈ U}

= supV + supU. .

3.2.11. Remark. The derivation of 3.2.10 is valid for an arbitrary orderedvector space provided that the given sets have least upper bounds. The equalitysupλU = λ supU for λ ∈ R+ is comprehended by analogy.

3.2.12. Definition. For an element x of a vector lattice, the positive partof x is x+ := x ∨ 0, the negative part of x is x− := (−x)+, and the modulus of x is|x| := x ∨ (−x).

3.2.13. If x and y are elements of a vector lattice then

x+ y = x ∨ y + x ∧ y.C x+ y − x ∧ y = x+ y + (−x) ∨ (−y) = y ∨ x B3.2.14. x = x+ − x−; |x| = x+ + x−.C The first equality follows from 3.2.13 on letting y := 0. Furthermore, |x| =

x ∨ (−x) = −x+ (2x) ∨ 0 = −x+ 2x+ = (x+ − x−) + 2x+ = x+ + x−. B

3.2.15. Interval Addition Lemma. If x and y are positive elements of a vec-tor lattice X then

[0, x+ y] = [0, x] + [0, y].(As usual, [u, v] := σ(u) ∩ σ−1(v) is the (order) interval with endpoints u and v.)C The inclusion [0, x]+[0, y] ⊂ [0, x+y] is obvious. Assume that 0 ≤ z ≤ x+y

and put z1 := z ∧ x. It is easy that z1 ∈ [0, x]. Now if z2 := z − z1 then z2 ≥ 0 andz2 = z − z ∧ x = z + (−z) ∨ (−x) = 0 ∨ (z − x) ≤ 0 ∨ (x+ y − x) = 0 ∨ y = y. B

3.2.16. Remark. The conclusion of the Interval Addition Lemma is oftenreferred to as the Riesz Decomposition Property.

3.2.17. Riesz–Kantorovich Theorem. Let X be a vector space and let Ybe a K-space. The space of regular operators Lr(X, Y ), ordered by the coneof positive operators L+(X, Y ), is a K-space. CB

26 Chapter 3

3.3. Extension of Positive Functionals and Operators

3.3.1. Counterexamples.(1) Let X be B([0, 1], R), the space of the bounded real-valued func-

tions given on [0, 1]; and let X0 be C([0, 1], R), the subspace of X comprising allcontinuous functions. Put Y := X0 and equip X0, X, and Y with the natural order(cf. 3.2.6 (1) and 3.2.6 (2)). Consider the problem of extending the identical em-bedding T0 : X00 = 0 OO to a positive operator T in L+(X, Y ). If the problem hada solution T , then each nonempty bounded set E in X0 would have a least upperbound supX0

E calculated in X0. Namely, supX0E = T supX E , where supX E is

the least upper bound of E in X; whereas, undoubtedly, Y fails to be a K-space.(2) Denote by s := RN the sequence space and furnish s with the natural

order. Let c be the subspace of s comprising all convergent sequences, the convergentsequence space. Demonstrate that the positive functional f0 : c0 = 0 OO R definedby f0(x) := limx(n) has no positive extension to s. Indeed, assume that f ∈s#, f ≥ 0 and f ⊃ f0. Put x0(n) := n and xk(n) := k ∧ n for k, n ∈ N. Plainly,f0(xk) = k; moreover, f(x0) ≥ f(xk) ≥ 0, since x0 ≥ xk ≥ 0, a contradiction.

3.3.2. Definition. A subspace X0 of an ordered vector space X with positivecone X+ is massive (in X) if X0 +X+ = X. The terms “coinitial” or “minorizing”are also in common parlance.

3.3.3. A subspace X0 is massive in X if and only if for all x ∈ X there areelements x0 and x0 in X0 such that x0 ≤ x ≤ x0. CB

3.3.4.Kantorovich Theorem. IfX is an ordered vector space, X0 is massivein X, and Y is a K-space; then each positive operator T0, a member of L+(X0, Y ),has a positive extension T , a member of L+(X, Y ).C Step I. First, let X := X0 ⊕X1, where X1 is a one-dimensional subspace,

X1 := {αx : α ∈ R}. Since X0 is massive and T0 is positive, the set U := {T0x0 :x0 ∈ X0, x0 ≥ x} is nonempty and bounded below. Consequently, there is some ysuch that y := inf U . Assign Tx := {T0x0+αy : x = x0+αx, x0 ∈ X0, α ∈ R}. It isclear that T is a single-valued correspondence. Further, T ⊃ T0 and domT = X.So, only the positivity of T needs checking. If x = x0 + αx and x ≥ 0, thenthe case of α equal to 0 is trivial. If α > 0 then x ≥ −x0/α. This implies that−T0x0/α ≤ y, i.e., Tx ∈ Y+. In a similar way for α < 0 observe that x ≤ −x0/α.Thus, y ≤ −T0x0/α and, finally, Tx = T0x0 + αy ∈ Y+.

Step II. Now let E be the collection of single-valued correspondences S ⊂X × Y such that S ⊃ T0 and S(X+) ⊂ Y+. By 3.1.3, E is inductive in orderby inclusion and so, by the Kuratowski–Zorn Lemma, E has a maximal element T .If x ∈ X \ domT , apply the result of Step I with X := domT ⊕ X1, X0 :=domT, T0 := T and X1 := {αx : α ∈ R} to obtain an extension of T . But T ismaximal, a contradiction. Thus, T is a sought operator. B

Convex Analysis 27

3.3.5. Remark. When Y := R, Theorem 3.3.4 is sometimes referred to asthe Kreın–Rutman Theorem.

3.3.6. Definition. A positive element x is discrete whenever [0, x] = [0, 1]x.3.3.7. If there is a discrete functional on (X, X+) then X = X+ −X+.

C Let T be such functional and := X+ −X+. Take f ∈ X#. It suffices tocheck that ker f ⊃⇒ = . Evidently, T + f ∈ [0, T ]; i.e., T + f = αT for someα ∈ [0, 1]. If T | = 0 then 2T ∈ [0, T ]. Hence, T = 0 and f = 0. Now if T (x0) 6= 0for some x0 ∈, then α = 1 and f = 0. B

3.3.8. Discrete Kreın–Rutman Let X be a massive subspace of an orderedvector space X and let T0 be a discrete functional on X0. Then there is a discretefunctional T on X extending T0.

C Adjust the proof of 3.3.4.Step I. Observe that the exhibited functional T is discrete. For, if T ′ ∈ [0, T ]

then there is some α ∈ [0, 1] such that T ′(x0) = αT (x0) for all x0 ∈ X0 and so(T − T ′)(x0) = (1− α)T (x0). Estimating, find

T ′(x) ≤ inf{T ′(x0) : x0 ≥ x, x0 ∈ X0} = αT (x);(T − T ′)(x) ≤ inf{(T − T ′)(x0) : x0 ≥ x, x0 ∈ X0} = (1− α)T (x).

Therefore, T ′ = αT and [0, T ] ⊂ [0, 1]T . The reverse inclusion is always true.Thus, T is discrete.

Step II. Let E be the same as in the proof of 3.3.4. Consider Ed, the setcomprising all S in E such that the restriction S|domS is a discrete functionalon domS. Show that Ed is inductive. To this end, take a chain E0 in Ed and putS := ∪{S0 : S0 ∈ E0}; obviously, S ∈ E . Verifying the discreteness of S willcomplete the proof.

Suppose that 0 ≤ S′(x0) ≤ S(x0) for all x0 ∈ (domS)+ and S′ ∈ (domS)#.If S(x0) = 0 for all x0 then S′ = 0S, as was needed. In the case S(x0) 6= 0 forsome x0 ∈ (domS)+ choose S0 ∈ E0 such that S0(x0) = S(x0). Since S0 is discrete,deduce that S′(x′) = αS(x′) for all x′ ∈ domS0. Furthermore, α = S′(x0)/S(x0);i.e., α does not depend on the choice of S0. Since E0 is a chain, infer that S′ = αS. B

3.4. Convex Functions and Sublinear Functionals

3.4.1. Definition. The semi-extended real line R· is the set R· with somegreatest element +∞ adjoined formally. The following agreements are effective:α(+∞) := +∞ (α ∈ R+) and +∞+ x := x+ (+∞) := +∞ (x ∈ R·).

28 Chapter 3

3.4.2. Definition. Let f : X0 = 0 OO R· be a mapping (also called an extendedfunction). The epigraph of f is the set

0 = 0 OOOO := {(x, t) ∈ X × R : t ≥ f(x)}.

The effective domain of definition or simply the domain of f is the set

dom f := {x ∈ X : f(x) < +∞}.

3.4.3. Remark. Inconsistency in overusing the symbol dom f is illusory. Na-mely, the effective domain of definition of f : X0 = 0 OO R· coincides with the domainof definition of the single-valued correspondence f ∩ X × R. We thus continue towrite f : X0 = 0 OO R, omitting the dot in the symbol R· whenever dom f = X.

3.4.4. Definition. If X is a real vector space then a mapping f : X0 = 0 OO R·

is a convex function provided that the epigraph 0 = 0 OOOO is convex.

3.4.5. A function f : X0 = 0 OO R· is convex if and only if the Jensen inequalityholds:

f(α1x1 + α2x2) ≤ α1f(x1) + α2f(x2)

for all α1, α2 ≥ 0, α1 + α2 = 1 and x1, x2 ∈ X.C ⇒: Take α1, α2 ≥ 0, α1 + α2 = 1. If either of x1 and x2 fails to be-

long to dom f then the Jensen inequality is evident. Let x1, x2 ∈ dom f . Then(x1, f(x1)) ∈ 0 = 0 OOOO and (x2, f(x2)) ∈ 0 = 0 OOOO . Appealing to 3.1.2 (8), findα1(x1, f(x1)) + α2(x2, f(x2)) ∈ 0 = 0 OOOO .⇐: Take (x1, t1) ∈ 0 = 0 OOOO and (x2, t2) ∈ 0 = 0 OOOO , i.e. t1 ≥ f(x1) and

t2 ≥ f(x2) (if dom f = ∅ then f(x) = +∞ (x ∈ X) and 0 = 0 OOOO = ∅). Applyingthe Jensen inequality, observe the containment (α1x1+α2x2, α1t1+α2t2) ∈ 0 = 0 OOOO

for all α1, α2 ≥ 0, α1 + α2 = 1. B

3.4.6. Definition. A mapping p : X0 = 0 OO R· is a sublinear functional pro-vided that 0 = 0 OOOO is a cone.

3.4.7. If dom p 6= 0 then the following statements are equivalent:(1) p is a sublinear functional;(2) p is a convex function that is positively homogeneous:

p(αx) = αp(x) for all α ≥ 0 and x ∈ dom p;(3) if α1, α2 ∈ R+ and x1, x2 ∈ X, then p(α1x1 + α2x2) ≤

α1p(x1) + α2p(x2);(4) p is a positively homogeneous functional that is subadditive:

p(x1 + x2) ≤ p(x1) + p(x2) (x1, x2 ∈ X). CB

Convex Analysis 29

3.4.8. Examples.(1) A linear functional is sublinear; an affine functional is a convex

function.(2) Let U be a convex set in X. Define the indicator function δ(U) :

X0 = 0 OO R· of U as the mapping

δ(U)(x) :={

0, if x ∈ U+∞, if x 6∈ U .

It is clear that δ(U) is a convex function. If U is a cone then δ(U) is a sublinearfunction. If U is an affine set then δ(U) is an affine function.

(3) The sum of finitely many convex functions and the supremum (orupper envelope) of a family of convex functions (calculated pointwise, i.e. in (R·)X)are convex functionals. Sublinear functionals have analogous properties.

(4) The composition of a convex function with an affine operator, i.e.an everywhere-defined single-valued affine correspondence, is a convex function.The composition of a sublinear functional with a linear operator is sublinear.

3.4.9. Definition. If U and V are subsets of a vector space X then U absorbsV if V ⊂ nU for some n ∈ N. A set U is absorbing (in X) if it absorbs every pointof X; i.e., X = ∪n∈N nU .

3.4.10. Let T ⊂ X × Y be a linear correspondence with imT = Y . If U isabsorbing (in X) then T (U) is absorbing (in Y ).

/ Y = T (X) = T (∪n∈N nU) = ∪n∈NT (nU) = ∪n∈N nT (U) .

3.4.11. Definition. Let U be a subset of a vector space X. An element xbelongs to the core of U (or x is an algebraically interior point of U) if U − x isabsorbing in X.

3.4.12. If f : X0 = 0 OO R· is a convex function and x ∈ core dom f then for allh ∈ X there is a limit

f ′(x)(h) := limα↓0

f(x+ αh)− f(x)α

= infα>0

f(x+ αh)− f(x)α

.

Moreover, the mapping f ′(x) : h 7→ f ′(x)h is a sublinear functional f ′(x) : X0 =0 OO R, the directional derivative of f at x.C Set ϕ(α) := f(x + αh). By 3.4.8 (4), ϕ : R0 = 0 OO R· is a convex function

and 0 ∈ core domϕ. The mapping α 7→ (ϕ(α)−ϕ(0))/α (α > 0) is increasing andbounded above; i.e., ϕ′(0)(1) is available. By definition, f ′(x)(h) = ϕ′(0)(1).

30 Chapter 3

Given β > 0 and h ∈ H, successively find

f ′(x)(βh) = inff(x+ αβh)− f(x)

α= β inf

f(x+ αβh)− f(x)αβ

= βf ′(x)(h).

Moreover, using the above result, for h1, h2 ∈ X infer that

f ′(x)(h1 + h2) = 2 limα↓0

f(x+ 1/2α(h1 + h2)

)− f(x)

α

= 2 limα↓0

f(1/2(x+ αh1) + 1/2(x+ αh2)

)− f(x)

α

≤ limα↓0

f(x+ αh1)− f(x)α

+ limα↓0

f(x+ αh2)− f(x)α

= f ′(x)(h1) + f ′(x)(h2).

Appealing to 3.4.7 ends the proof. B

3.5. The Hahn–Banach Theorem

3.5.1. Definition. Let X be a real vector space and let f : X0 = 0 OO R· bea convex function. The subdifferential of f at a point x in dom f is the set

∂x(f) := {l ∈ X# : (∀ y ∈ X) l(y)− l(x) ≤ f(y)− f(x)}.

3.5.2. Examples.(1) Let p : X0 = 0 OO R· be a sublinear functional. Put ∂ (p) := ∂0(p).

Then

∂ (p) = {l ∈ X# : (∀x ∈ X) l(x) ≤ p(x)};∂x(p) = {l ∈ ∂ (p) : l(x) = p(x)}.

(2) If l ∈ X# then ∂ (l) = ∂x(l) = {l}.(3) Let X0 be a subspace of X. Then

∂ (δ(X0)) = {l ∈ X# : ker l ⊃ X0}.

(4) If f : X0 = 0 OO R· is a convex function then

∂x(f) = ∂ (f ′(x))

whenever x ∈ core dom f . CB

Convex Analysis 31

3.5.3.Hahn–Banach Theorem. Let T ∈ L (X, Y ) be a linear operator andlet f : Y 0 = 0 OO R· be a convex function. If x ∈ X and Tx ∈ core dom f then

∂x(f ◦ T ) = ∂Tx(f) ◦ T.

C By 3.4.10 it follows that x ∈ core dom f . From 3.5.2 (4) obtain ∂x(f ◦ T ) =∂ ((f ◦ T )′(x)). Moreover,

(f ◦ T )′(x)(h) = limα↓0

(f ◦ T )(x+ αh)− (f ◦ T )(x)α

= limα↓0

f(Tx+ αTh)− f(Tx)α

= f ′(Tx)(Th)

for h ∈ X. Put p := f ′(Tx). By 3.4.12, p is a sublinear functional. Whence, onappealing again to 3.5.2 (4), infer that

∂ (p) = ∂ (f ′(Tx)) = ∂ Tx(f);∂ (p ◦ T ) = ∂ ((f ◦ T )′(x)) = ∂x(f ◦ T ).

Thereby, the only claim left unproven is the equality

∂ (p ◦ T ) = ∂ (p) ◦ T.

If l ∈ ∂ (p) ◦T (i.e., l = l1 ◦T , where l1 ∈ ∂ (p)) then l1(y) ≤ p(y) for all y ∈ Y .In particular, l(x) ∈ l1(Tx) ≤ p(Tx) = p ◦ T (x) for all x ∈ X and so l ∈ ∂ (p ◦ T ).This argument yields the inclusion ∂ (p) ◦ T ⊂ ∂ (p ◦ T ).

Now take l ∈ ∂ (p ◦ T ). If Tx = 0 then l(x) ≤ p(Tx) = p(0) = 0; i.e., l(x) ≤ 0.The same holds for −x. Therefore, l(x) = 0; in other words, ker l ⊃ kerT . Hence,by 2.3.8, l = l1 ◦ T for some l1 ∈ Y #. Putting Y0 := T (X), observe that thefunctional l1 ◦ ι belongs to ∂ (p ◦ ι), with ι the identical embedding of Y0 into Y .With ∂ (p ◦ ι) ⊂ ∂ (p) ◦ ι proven, observe that l1 ◦ ι = l2 ◦ ι for some l2 ∈ ∂ (p), whichconsequently yields l = l1 ◦ T = l1 ◦ ι ◦ T = l2 ◦ ι ◦ T = l2 ◦ T ; i.e., l ∈ ∂ (p) ◦ T .

Thus, to complete the proof of the Hahn–Banach Theorem, showing that ∂ (p◦ι) ⊂ ∂ (p) ◦ ι is in order.

Take an element l0 in ∂ (p◦ι) and consider the functional T0 : (y0, t) 7→ t−l0(y0)given on the subspace Y0 := Y0 × R of Y := Y × R. Equip Y with the coneY+ := 0 = 0 OOOO . Note that, first, Y0 is massive since

(y, t) = (0, t− p(y)) + (y, p(y)) (y ∈ Y, t ∈ R).

Second, by 3.4.2, t ≥ p(y0) for (y0, t) ∈ Y0 ∩Y+, and so T0(y0, t) = t− l0(y0) ≥ 0;i.e., T0 is a positive functional on Y0. By 3.3.4, there is a positive functional Tdefined on Y which is an extension of T0. Put l(y) := T (−y, 0) for y ∈ Y . It isclear that l ◦ ι = l0. Furthermore, T (0, t) = T0(0, t) = t. Hence, 0 ≤ T (y, p(y)) =p(y)− l(y), i.e., l ∈ ∂ (p). B

32 Chapter 3

3.5.4. Remark. The claim of Theorem 3.5.3 is also referred to as the formulafor a linear change-of-variable under the subdifferential sign or the Hahn–BanachTheorem in subdifferential form. Observe that the inclusion ∂ (p ◦ ι) ⊂ ∂ (p) ◦ ι isoften referred to as the Hahn–Banach Theorem in analytical form or the DominatedExtension Theorem and verbalized as follows: “A linear functional given on a sub-space of a vector space and dominated by a sublinear functional on this subspacehas an extension also dominated by the same sublinear functional.”

3.5.5. Corollary. If X0 is a subspace of a vector space X and p : X0 = 0 OO Ris a sublinear functional then the (asymmetric) Hahn–Banach formula holds:

∂ (p+ δ(X0)) = ∂ (p) + ∂ (δ(X0)).

C It is obvious that the left side includes the right side. To prove the reverseinclusion, take l ∈ ∂ (p + δ(X0)). Then l ◦ ι ∈ ∂ (p ◦ ι), where ι is the identicalembedding of X0 into X. By 3.5.3, l ◦ ι ∈ ∂ (p) ◦ ι, i.e., l ◦ ι = l1 ◦ ι for somel1 ∈ ∂ (p). Put l2 := l − l1. By definition, find l2 ◦ ι = (l − l1) ◦ ι = l ◦ ι− l1 ◦ ι = 0,i.e., ker l2 ⊃ X0. As was mentioned in 3.5.2 (3), this means l2 ∈ ∂ (δ(X0)). B

3.5.6. Corollary. If f : X0 = 0 OO R· is a linear functional and x ∈ core dom fthen ∂x(f) 6= ∅.C Let p := f ′(x) and let ι : 00 = 0 OO be the identical embedding of 0 to X.

Plainly, 0 ∈ ∂ (p ◦ ι), i.e., ∂ (p ◦ ι) 6= ∅. By the Hahn–Banach Theorem, ∂ (p) 6= ∅(otherwise, ∅ = ∂ (p) ◦ ι = ∂ (p ◦ ι)). To complete the proof, apply 3.5.2 (4). B

3.5.7. Corollary. Let f1, f2 : X0 = 0 OO R· be convex functions. Assumefurther that x ∈ core dom f1 ∩ core dom f2. Then

∂x(f1 + f2) = ∂x(f1) + ∂x(f2).

C Let p1 := f ′1(x) and p2 := f ′2(x). Given x1, x2 ∈ X, define p(x1, x2) :=p1(x1) + p2(x2) and ι(x1) := (x1, x1). Using 3.5.2 (4) and 3.5.3, infer that

∂x(f1 + f2) = ∂ (p1 + p2) = ∂ (p ◦ ι)= ∂ (p) ◦ ι = ∂ (p1) + ∂ (p2) = ∂x(f1) + ∂x(f2). .

3.5.8. Remark. Corollary 3.5.6 is sometimes called the Nonempty Subdiffer-ential Theorem. On the one hand, it is straightforward from the Kuratowski–Zorn Lemma. On the other hand, with Corollary 3.5.6 available, demonstrate∂ (p ◦ T ) = ∂ (p) ◦ T as follows: Define p T (y) := inf{p(y + Tx) − l(x) : x ∈ X},where l ∈ ∂ (p) and the notation of 3.5.3 is employed. Check that pT is sublinearand every l1 in ∂ (p T ) satisfies the equality l = l1 ◦T . Thus, the Nonempty Subdif-ferential Theorem and the Hahn–Banach Theorem in subdifferential form constitutea precious (rather than vicious) circle.

Convex Analysis 33

3.6. The Kreın–Milman Theorem

3.6.1. Definition. Let X be a real vector space. Putting

seg(x1, x2) := {α1x1 + α2x2 : α1, α2 > 0, α1 + α2 = 1},

define the correspondence seg ⊂ X2×X that assigns to each pair of points the opensegment joining them. If U is a convex set in X and segU is the restriction of segto U2, then a convex subset V of U is extreme in U if seg−1

U (V ) ⊂ V 2. An extremesubset of U is sometimes called a face in U . A member x of U is an extreme pointin U if {x} is an extreme subset of U . The set of extreme points of U is usuallydenoted by extU .

3.6.2. A convex subset V is extreme in U if and only if for all u1, u2 ∈ U andα1, α2 > 0, α1 + α2 = 1, the containment α1u1 + α2u2 ∈ V implies that u1 ∈ Vand u2 ∈ V . CB

3.6.3. Examples.(1) Let p : X0 = 0 OO R· be a sublinear functional and let a point x of X

belong to dom p. Then ∂x(p) is an extreme subset of ∂ (p).C For, if α1l1 +α2l2 ∈ ∂x(p) and l1, l2 ∈ ∂ (p), where α1, α2 > 0, α1 +α2 = 1;

then 0 = p(x) − (α1l1(x) + α2l2(x)) = α1(p(x) − l1(x)) + α2(p(x) − l2(x)) ≥ 0.Moreover, p(x)−l1(x) ≥ 0 and p(x)−l2(x) ≥ 0. Hence, l1 ∈ ∂x(p) and l2 ∈ ∂x(p). B

(2) Let W be an extreme set in V and let V be in turn an extreme setin U . Then W is an extreme set in U . CB

(3) If X is an ordered vector space then a positive element x of X isdiscrete if and only if the cone {αx : α ∈ R+} is an extreme subset of X+.C ⇐: If 0 ≤ y ≤ x then x = 1/2(2y) + 1/2(2(x − y)). Therefore, by 3.6.2,

2y = αx and 2(x − y) = βx for some α, β ∈ R+. Thus, 2x = (α + β)x. The casex = 0 is trivial. Now if x 6= 0 then α/2 ∈ [0, 1] and, consequently, [0, x] ⊂ [0, 1]x.The reverse inclusion is evident.

⇒: Let [0, x] = [0, 1]x, and suppose that αx = α1y1+α2y2 for α ≥ 0; α1, α2 >0, α1 + α2 = 1 and y1, y2 ∈ X+. If α = 0 then α1y1 ∈ [0, x] and α2y2 ∈ [0, x];hence, y1 is a positive multiple of x; the same is valid for y2. In the case α > 0observe that (α1/α)y1 = tx for some t ∈ [0, 1]. Finally, (α2/α)y2 = (1− t)x. B

(4) Let U be a convex set. A convex subset V of U is a cap of U ,if U \ V is convex. A point x in U is extreme if and only if {x} is a cap of U . CB

3.6.4. Extreme and Discrete Lemma. Let p : X0 = 0 OO R be a sublinearfunctional and l ∈ ∂ (p). Assign := X ×R, + := = OOOO , and Tl : (x, t) 7→ t− l(x) (x ∈X, t ∈ R). Then the functional l is an extreme point of ∂ (p) if and only if Tl isa discrete functional.

34 Chapter 3

C ⇒: Consider T ′ ∈# such that T ′ ∈ [0, Tl]. Put

t1 := T ′(0, 1), l1(x) := T ′(−x, 0);t2 := (Tl − T ′)(0, 1), l2(x) := (Tl − T ′)(−x, 0).

It is clear that t1 ≥ 0, t2 ≥ 0, t1 + t2 = 1; l1 ∈ ∂ (t1p), l2 ∈ ∂ (t2p), and l1 + l2 = l.If t1 = 0 then l1 = 0; i.e., T ′ = 0 and T ′ ∈ [0, 1]Tl. Now if t2 = 0 then t1 = 1; i.e.,T ′ = Tl, and so T ′ ∈ [0, 1]Tl. Assume that t1, t2 > 0. In this case 1/t1 l1 ∈ ∂ (p)and 1/t2 l2 ∈ ∂ (p); moreover, l = t1

(1/t1 l1

)+ t2

(1/t2 l2

). Since, by hypothesis,

l ∈ ext ∂ (p), from 3.6.2 it follows that l1 = t1l; i.e., T ′ = t1Tl. B⇐: Let l = α1l1 +α2l2, where l1, l2 ∈ ∂ (p) and α1, α2 > 0, α1 + α2 = 1. The

functionals T ′ := α1Tl1 and T ′′ := α2Tl2 are positive; moreover, T ′ ∈ [0, Tl], sinceT ′ + T ′′ = Tl. Therefore, there is some β ∈ [0, 1] such that T ′ = βTl. At the point(0, 1), find α1 = β. Hence, l1 = l. By analogy, l2 = l. CB

3.6.5. Kreın–Milman Let T ∈ L (X, Y ) be a linear operator and let f :Y 0 = 0 OO R· be a convex function. If x ∈ X and Tx ∈ core dom f then

ext ∂x(f ◦ T ) ⊂ (ext ∂ Tx(f)) ◦ T.

C Start arguing as in the proof of the Hahn–Banach Theorem: Put p := f ′(Tx)and observe that the only claim left to checking is the inclusion ext ∂ (p ◦ T ) ⊂(ext ∂ (p)) ◦ T. Take l ∈ ext ∂ (p ◦ T ). Since ext ∂ (p ◦ T ) ⊂ ∂ (p ◦ T ) ⊂ ∂ (p) ◦ T ,there is some f in ∂ (p) such that l = f ◦ T . Denote by f0 the restriction of f toY0 := imT and notice that f0 ∈ ext ∂ (p ◦ ι), with ι the identical imbedding of Y0into Y .

Now in Y := Y ×R consider Y+ := 0 = 0 OOOO and introduce the space Y0 := Y0×R.Note that Y+ ∩ Y0 = 0 = 0 OOOO p ◦ ι). Applying 3.6.4 with X := Y0, l := f0, andp := p ◦ ι, observe that Tf0 is a discrete functional on Y0; moreover, Y0 is massivein Y (cf. the proof of 3.5.3). By 3.3.8, find a discrete extension S ∈ Y# of Tf0 .Evidently, S = Tg, where g(y) := S(−y, 0) for y ∈ Y . Appealing again to 3.6.4,infer that g ∈ ext ∂ (p). By construction, l(x) = f(Tx) = f0(Tx) = g(Tx) for allx ∈ X. Finally, l ∈ (ext ∂ (p)) ◦ T . B

3.6.6. Corollary. If p : X0 = 0 OO R is a sublinear functional then for every xin X there is an extreme functional l, a member of ext ∂ (p), such that l(x) = p(x).

C From 3.6.5 it is easy that ext ∂ (p) 6= ∅ for every p (cf. 3.5.6). Usingthis and 3.4.12, choose l in ext ∂x(p′(x)). Applying 3.5.2 (2) and 3.5.2 (4), obtainl ∈ ext ∂x(p). By 3.6.3 (1), ∂x(p) is extreme in ∂ (p). Finally, 3.6.3 (2) implies thatl is an extreme point of ∂ (p). B

Convex Analysis 35

3.7. The Balanced Hahn–Banach Theorem

3.7.1. Definition. Let (X, F, +, · ) be a vector space over a basic field F.The vector space (X, R, +, · |R×X) is called the real carrier or realificationof (X, F, +, · ) and is denoted by XR.

3.7.2. Definition. Given a linear functional f on a vector space X, definethe mapping Re f : x 7→ Re f(x) (x ∈ X). The real part map or realifier is themapping Re : (X#)R0 = 0 OO XR)#.

3.7.3. The real part map Re is a linear operator and, moreover, an isomor-phism of (X#)R onto (XR)#.

C Only the case F := C needs verifying, because Re is the identity mappingwhen F := R.

Undoubtedly, Re is a linear operator. Check that Re is a monomorphism andan epimorphism simultaneously (cf. 2.3.2).

If Re f = 0 then 0 = Re f(ix) = Re(if(x)) = Re(i(Re f(x) + i Im f(x))) =− Im f(x). Hence, f = 0 and so Re is a monomorphism.

Now take g ∈ (XR)# and put f(x) := g(x)−ig(ix). Evidently, f ∈ L (XR, CR)and Re f(x) = g(x) for x ∈ X. It is sufficient to show that f(ix) = if(x), becausethis equality implies f ∈ X#. Straightforward calculation shows f(ix) = g(ix) +ig(x) = i(g(x) − ig(ix)) = if(x), which enables us to conclude that Re is alsoan epimorphism. B

3.7.4. Definition. The lear trap map or complexifier is the inverse Re−1 :(XR)#0 = 0 OO X#)R of the real part map.

3.7.5. Remark. By 3.7.3, for the complex scalar field

Re−1g : x 7→ g(x)− ig(ix) (g ∈ (XR)#, x ∈ X).

In the case of the reals, F := R, the complexifier Re−1 is the identity operator.

3.7.6. Definition. Let (X, F, +, · ) be a vector space over F. A seminormon X is a function p : X0 = 0 OO R· such that dom p 6= ∅ and

p(λ1x1 + λ2x2) ≤ |λ1|p(x1) + |λ2|p(x2)

for x1, x2 ∈ X and λ1, λ2 ∈ F.

3.7.7. Remark. A seminorm is a sublinear functional (on the real carrierof the space in question).

36 Chapter 3

3.7.8. Definition. If p : X0 = 0 OO R· is a seminorm then the balanced subdif-ferential of p is the set

|∂ |(p) := {l ∈ X# : |l(x)| ≤ p(x) for all x ∈ X}.

3.7.9. Balanced Subdifferential Lemma. Let p : X0 = 0 OO R· be a semi-norm. Then

|∂ |(p) = Re−1(∂ (p)); Re (|∂ |(p)) = ∂ (p)

for the subdifferentials |∂ |(p) and ∂ (p) of p.C If F := R then the equality |∂ |(p) = ∂ (p) is easy. Furthermore, in this case

Re is the identity operator.Let F := C. If l ∈ |∂ |(p) then (Re l)(x) = Re l(x) ≤ |l(x)| ≤ p(x) for all x ∈ X;

i.e., Re (|∂ |(p)) ⊂ ∂ (p). Take g ∈ ∂ (p) and put f := Re−1g. If f(x) = 0 then|f(x)| ≤ p(x); for f(x) 6= 0 set θ := |f(x)|/f(x). Thereby |f(x)| = θf(x) = f(θx) =Re f(θx) = g(θx) ≤ p(θx) = |θ|p(x) = p(x), since |θ| = 1. Finally, f ∈ |∂ |(p). B

3.7.10. Let X be a vector space, let p : X0 = 0 OO R be a seminorm, and let X0be a subspace of X. The asymmetric balanced Hahn–Banach formula holds:

|∂ |(p+ δ(X0)) = |∂ |(p) + |∂ |(δ(X0)).

C From 3.7.9, 3.5.5, and the results of Section 3.1 obtain

|∂ |(p+ δ(X0)) = Re−1(∂ (p+ δ(X0))) = Re−1(∂ (p) + ∂ (δ(X0)))= Re−1(∂ (p)) + Re−1(∂ (δ(X0))) = |∂ |(p) + |∂ |(δ(X0)). .

3.7.11. If X and Y are vector spaces, T ∈ L (X, Y ) is a linear operator, andp : Y 0 = 0 OO R is a seminorm; then p ◦ T is also a seminorm and, moreover,

|∂ |(p ◦ T ) = |∂ |(p) ◦ T.

C Applying 2.3.8 and 3.7.10, successively infer that

|∂ |(p ◦ T ) = |∂ |(p+ δ(imT )) ◦ T = (|∂ |(p) + |∂ |(δ(imT ))) ◦ T

= |∂ |(p) ◦ T + |∂ |(δ(imT )) ◦ T = |∂ |(p) ◦ T. .3.7.12. Remark. If T is the identical embedding of a subspace and the ground

field is C then 3.7.11 is referred to as the Sukhomlinov–Bohnenblust–Sobczyk Theo-rem.

3.7.13. Balanced Hahn–Banach Let X be a vector space. Assume furtherthat p : X0 = 0 OO R is a seminorm and X0 is a subspace of X. If l0 is a linearfunctional given on X0 such that |l0(x0)| ≤ p(x0) for x0 ∈ X0, then there is a linearfunctional l on X such that l(x0) = l0(x0) whenever x0 ∈ X0 and |l(x)| ≤ p(x) forall x ∈ X. CB

Convex Analysis 37

3.8. The Minkowski Functional and Separation

3.8.1. Definition. Let R stand for the extended real axis or extended reals(i.e., R denotes R· with the least element −∞ adjoined formally). If X is an arbi-trary set and f : X0 = 0 OO R is a mapping; then, given t ∈ R, put

{f ≤ t} := {x ∈ X : f(x) ≤ t};{f = t} := f−1(t);

{f < t} := {f ≤ t} \ {f = t}.Every set of the form {f ≤ t}, {f = t}, and {f < t} is a level set or Lebesgue setof f .

3.8.2. Function Recovery Lemma. Let T ⊂ R and let t 7→ Ut (t ∈ T ) bea family of subsets of X. There is a function f : X0 = 0 OO R such that

{f < t} ⊂ Ut ⊂ {f ≤ t} (t ∈ T )if and only if the mapping t 7→ Ut increases.

C⇒: Suppose that T contains at least two elements s and t. (Otherwise thereis nothing left to proof.) If s < t then

Us ⊂ {f ≤ s} ⊂ {f < t} ⊂ Ut.⇐: Define a mapping f : X0 = 0 OO R by setting f(x) := inf{t ∈ T : x ∈ Ut}.

If {f < t} is empty for t ∈ T then all is clear. If x ∈ {f < t} then f(x) < +∞,and so there is some s ∈ T such that x ∈ Us and s < t. Thus, {f < t} ⊂ Us ⊂ Ut.Continuing, for x ∈ Ut find f(x) ≤ t; i.e., Ut ⊂ {f ≤ t}. B

3.8.3. Function Comparison Lemma. Let f, g : X0 = 0 OO R be functionsdefined by (Ut)t∈T and (Vt)t∈T as follows:

{f < t} ⊂ Ut ⊂ {f ≤ t};{g < t} ⊂ Vt ⊂ {g ≤ t} (t ∈ T ).

If T is dense in R (i.e., (∀ r, t ∈ R, r < t) (∃ s ∈ T ) (r < s < t)) then the inequalityf ≤ g (in RX ; i.e., f(x) ≤ g(x) for x ∈ X) holds if and only if

(t1, t2 ∈ T & t1 < t2)⇒ Vt1 ⊂ Ut2 .C ⇒: This is immediate from the inclusions

Vt1 ⊂ {g ≤ t1} ⊂ {f ≤ t1} ⊂ {f < t2} ⊂ Ut2 .⇐: Assume that g(x) 6= +∞ (if not, f(x) ≤ g(x) for obvious reasons). Given

t ∈ R such that g(x) < t < +∞, choose t1, t2 ∈ T so as to satisfy the conditionsg(x) < t1 < t2 < t. Now

x ∈ {g < t1} ⊂ Vt1 ⊂ Ut2 ⊂ {f ≤ t2} ⊂ {f < t}.Thus, f(x) < t. Since t is arbitrary, obtain f(x) ≤ g(x). B

38 Chapter 3

3.8.4. Corollary. If T is dense in R and the mapping t 7→ Ut (t ∈ T ) increasesthen there is a unique function f : X0 = 0 OO R such that

{f < t} ⊂ Ut ⊂ {f ≤ t} (t ∈ T ).

Moreover, the level sets of f may be presented as follows:

{f < t} = ∪ {Us : s < t, s ∈ T};

{f ≤ t} = ∩{Ur : t < r, r ∈ T} (t ∈ R).

C It is immediate from 3.8.2 and 3.8.3 that f exists and is unique. If s < t ands ∈ T then Us ⊂ {f ≤ s} ⊂ {f < t}. If now f(x) < t then, since T is dense, thereis an element s in T such that f(x) < s < t. Therefore, f ∈ {f < s} ⊂ Us, whichproves the formula for {f < t}.

Suppose that r > t, r ∈ T . Then {f ≤ t} ⊂ {f < r} ⊂ Ur. In turn, if x ∈ Urfor r ∈ T, r > t; then f(x) ≤ r for all r > t. Hence, f(x) ≤ t. B

3.8.5. Let X be a vector space and let S be a conical segment in X. Givent ∈ R, put Ut := ∅ if t < 0 and Ut := tS if t ≥ 0. The mapping t 7→ Ut (t ∈ R)increases.

C If 0 ≤ t1 < t2 and x ∈ t1S then x ∈ (t1/t2) t2S. Hence, x ∈ t2S. B

3.8.6. Definition. The Minkowski functional or the gauge function or simplythe gauge of a conical segment S is a functional pS : X0 = 0 OO R such that

{pS < t} ⊂ tS ⊂ {pS ≤ t} (t ∈ R+)

and {p < 0} = ∅. (Such a functional exists and is unique by 3.8.2, 3.8.4, and 3.8.5.)In other words,

pS(x) = inf{t > 0 : x ∈ tS} (x ∈ X).

3.8.7. Gauge Theorem. The Minkowski functional of a conical segment isa sublinear functional assuming positive values. Conversely, if p is a sublinearfunctional assuming positive values then the sets {p < 1} and {p ≤ 1} are conicalsegments. Moreover, p is the Minkowski functional of each conical segment S suchthat {p < 1} ⊂ S ⊂ {p ≤ 1}.C Consider a conical segment S and its Minkowski functional pS . Let x ∈ X.

The inequality pS(x) ≥ 0 is evident. Take α > 0. Then

pS(αx) = inf{t > 0 : αx ∈ tS} = inf{t > 0 : x ∈ t/αS

}= inf{αβ > 0 : x ∈ βS, β > 0}

= α inf{β > 0 : x ∈ βS} = αpS(x).

Convex Analysis 39

To start checking that pS is subadditive, take x1, x2 ∈ X. Noting the inclusiont1S + t2S ⊂ (t1 + t2)S for t1, t2 > 0, in view of the identity

t1x1 + t2x2 = (t1 + t2)(

t1t1 + t2

x1 +t2

t1 + t2x2

),

successively infer that

pS(x1 + x2) = inf{t > 0 : x1 + x2 ∈ tS}≤ inf{t : t = t1 + t2; t1, t2 > 0, x1 ∈ t1S, x2 ∈ t2S}

= inf{t1 > 0 : x1 ∈ t1S}+ inf{t2 > 0 : x2 ∈ t2S} = pS(x1) + pS(x2).

Let p : X0 = 0 OO R· be an arbitrary sublinear functional with positive valuesand let {p < 1} ⊂ S ⊂ {p ≤ 1}. Given t ∈ R+, put Vt := {p < t} and Ut := tS;given t < 0, put Vt := Ut := ∅. Plainly,

{pS < t} ⊂ Ut ⊂ {pS ≤ t}; {p < t} ⊂ Vt ⊂ {p ≤ t}

for t ∈ R. If 0 ≤ t1 < t2 then Vt1 = {p < t1} = t1{p < 1} ⊂ t1S = Ut1 ⊂ Ut2 .Moreover, Ut1 ⊂ t1{p ≤ 1} ⊂ {p ≤ t1} ⊂ {p < t2} ⊂ Vt2 . Therefore, by 3.8.3 and3.8.4, p = pS . B

3.8.8. Remark. A conical segment S in X is absorbing in X if and onlyif dom pS = X. Also, pS is a seminorm whenever S is absolutely convex. Conversely,for every seminorm p the sets {p < 1} and {p ≤ 1} are absolutely convex. CB

3.8.9. Definition. A subspace H of a vector space X is a hypersubspacein X if X/H is isomorphic to the ground field of X. An element of X/H is calleda hyperplane in X parallel to H. By a hyperplane in X we mean an affine varietyparallel to some hypersubspace of X. An affine variety H is a hyperplane if H−h isa hypersubspace for some (and, hence, for every) h in H. If necessary, a hyperplanein the real carrier of X is referred to as a real hyperplane in X.

3.8.10. Hyperplanes inX are exactly level sets of nonzero elements ofX#. CB

3.8.11. Separation Theorem. Let X be a vector space. Assume furtherthat U is a nonempty convex set in X and L is an affine variety in X. If L∩U = ∅then there is a hyperplane H in X such that H ⊃ L and H ∩ coreU = ∅.

C Without loss of generality, it may be supposed that coreU 6= ∅ (otherwisethere is nothing left unproven) and, moreover, 0 ∈ coreU . Take x ∈ L and putX0 :=L − x. Consider the quotient space X/X0 and the corresponding coset mappingϕ : X0 = 0 OO /X0. Applying 3.1.8 and 3.4.10, observe that ϕ(U) is an absorbing

40 Chapter 3

conical segment. Hence, by 3.8.7 and 3.8.8, the domain of the Minkowski functionalp := pϕ(U) is the quotient X/X0; moreover,

ϕ(coreU) ⊂ coreϕ(U) ⊂ {p < 1} ⊂ ϕ(U).

In particular, this entails the inequality p(ϕ(x)) ≥ 1, since ϕ(x) 6∈ ϕ(U).Using 3.5.6, find a functional f in ∂x(p ◦ ϕ); now the Hahn–Banach Theorem

impliesf ∈ ∂x(p ◦ ϕ) = ∂ϕ(x)(p) ◦ ϕ.

Put H := {f = p ◦ ϕ(x)}. It is clear that H is a real hyperplane and, undoubtedly,H ⊃ L. Appealing to 3.5.2 (1), conclude that H ∩ coreU = ∅. Now let f := Re−1fand H := {f = f(x)}. There is no denying that L ⊂ H ⊂ H. Thus, the hyperplaneH provides us with what was required. B

3.8.12. Remark. Under the hypotheses of the Separation Theorem, it may beassumed that coreU ∩ L = ∅. Note also that Theorem 3.8.11 is often referred to asthe Hahn–Banach Theorem in geometric form or as the Minkowski–Ascoli–MazurTheorem.

3.8.13. Definition. Let U and V be sets in X. A real hyperplane H in Xseparates U and V if these sets lie in the different halfspaces defined by H; i.e.,if there is a presentation H = {f = t}, where f ∈ (XR)# and t ∈ R, such thatV ⊂ {f ≤ t} and U ⊂ {f ≥ t} := {−f ≤ −t}.

3.8.14. Eidelheit Separation Theorem. If U and V are convex sets suchthat coreV 6= ∅ and U ∩ coreV = ∅, then there is a hyperplane separating Uand V and disjoint from coreV . CB

Exercises

3.1. Show that a hyperplane is precisely an affine set maximal by inclusion and other thanthe whole space.

3.2. Prove that every affine set is an intersection of hyperplanes.3.3. Prove that the complement of a hyperplane to a real vector space consists of two convex

sets each of which coincides with its own core. The sets are named open halfspaces. The unionof an open halfspace with the corresponding hyperplane is called a closed halfspace. Find out howa halfspace can be prescribed.

3.4. Find possible presentations of the elements of the convex hull of a finite set. What usecan be made of finite dimensions?

3.5. Given sets S1 and S2, let S :=⋃

0≤λ≤1 λS1∩(1−λ)S2. Prove that S is convex wheneverso are S1 and S2.

3.6. Calculate the Minkowski functional of a halfspace or a cone and of the convex hullof the union or intersection of conical segments.

Convex Analysis 41

3.7. Let S := {p + q ≤ 1}, where p and q are the Minkowski functionals of the conicalsegments Sp and Sq . Express S via Sp and Sq .

3.8. Describe sublinear functionals with domain RN .3.9. Calculate the subdifferential of the upper envelope of a finite set of linear or sublinear

functionals.

3.10. Let p and q be sublinear functionals in general position, i.e. such that dom p−dom q =dom q − dom p. Prove the symmetric Hahn–Banach formula (cf. 3.5.7): ∂ (p+ q) = ∂ (p) + ∂ (q).

3.11. Let p, q : X0 = 0 OO R be total (= everywhere-defined) sublinear functionals on X.Then the equality holds: ∂ (p ∨ q) = co(∂ (p) ∪ ∂ (q)).

3.12. Find the Minkowski functional of a ball in a Hilbert space whose center of symmetryis not necessarily the zero of the space.

3.13. A symmetric square 2 × 2-matrix is called positive, provided that its eigenvalues arepositive. Does the resulting order in the space of such matrices agree with vector structure? Doesit define the structure of a Kantorovich space?

3.14. Each ordered vector space admits a nonzero positive linear functional, doesn’t it?3.15. What are the means for transforming RN into a Kantorovich space?3.16. Under what conditions does the claim of the Hahn–Banach Theorem in analytical

form hold for a partial (= not-everywhere-defined) sublinear functional?

3.17. Find the extreme points of the subdifferential of the conventional norm on l∞.3.18. Find possible generalizations of the Hahn–Banach Theorem for a mapping acting into

a Kantorovich space.

3.19. Given a set C in a space X, define the Hormander transform H(C) of C as

H(C) = {(x, t) ∈ X × R : x ∈ tC}.

Study the properties of the Hormander transform on the collection of all convex sets.

Chapter 4

An Excursus into Metric Spaces

4.1. The Uniformity and Topology of a Metric Space

4.1.1. Definition. A mapping d : X20 = 0 OO R+ is a metric on X if

(1) d(x, y) = 0⇔ x = y;

(2) d(x, y) = d(y, x) (x, y ∈ X);

(3) d(x, y) ≤ d(x, z) + d(z, y) (x, y, z ∈ X).The real d(x, y) is usually referred to as the distance between x and y. The pair(X, d) is a metric space. In this situation, it is convenient to take the libertyof calling the underlying set X a metric space. An element of a metric space X isalso called a point of X.

4.1.2. A mapping d : X20 = 0 OO R+ is a metric if and only if

(1) {d ≤ 0} = IX ;

(2) {d ≤ t} = {d ≤ t}−1 (t ∈ R+);

(3) {d ≤ t1} ◦ {d ≤ t2} ⊂ {d ≤ t1 + t2} (t1, t2 ∈ R+).

C Items 4.1.2 (1)–4.1.2 (3) reformulate 4.1.1 (1)–4.1.1 (3). B

4.1.3. Definition. Let (X, d) be a metric space and take ε ∈ R+\0, a strictlypositive real. The relation Bε := Bd,ε := {d ≤ ε} is the closed cylinder of size ε. The

set◦Bε :=

◦Bd,ε := {d < ε} is the open cylinder of size ε. The image Bε(x) of a point x

under the relation Bε is called the closed ball with center x and radius ε. By analogy,

the set◦Bε(x) is the open ball with center x and radius ε.

4.1.4. In a nonempty metric space open cylinders as well as closed cylindersform bases for the same filter. CB

An Excursus into Metric Spaces 43

4.1.5. Definition. The filter on X2 with the filterbase of all cylinders ofa nonempty metric space (X, d) is the metric uniformity on X. It is denotedby UX , Ud, or even U , if the space under consideration is implied. Given X := ∅,put UX := {∅}. An element of UX is an entourage on X.

4.1.6. If U is a metric uniformity then(1) U ⊂ fil {IX};(2) U ∈ U ⇒ U−1 ∈ U ;(3) (∀U ∈ U ) (∃V ∈ U ) V ◦ V ⊂ U ;(4) ∩{U : U ∈ U } = IX . CB

4.1.7. Remark. Property 4.1.6 (4) reflecting 4.1.1 (1) is often expressed asfollows: “U is a Hausdorff or separated uniformity.”

4.1.8. Given a space X with uniformity UX , put

τ(x) := {U(x) : U ∈ U }.

Then τ(x) is a filter for every x in X. Moreover,(1) τ(x) ⊂ fil {x};(2) (∀U ∈ τ(x)) (∃V ∈ τ(x) & V ⊂ U) (∀ y ∈ V ) V ∈ τ(y). CB

4.1.9. Definition. The mapping τ : x 7→ τ(x) is the metric topology on X.An element of τ(x) is a neighborhood of x or about x. More complete designationsfor the topology are also in use: τX , τ(U ), etc.

4.1.10. Remark. All closed balls centered at x form a base for the neighbor-hood filter of x. The same is true of open balls. Note also that there are disjoint(= nonintersecting) neighborhoods of different points in X. This property, cipheredin 4.1.6 (4), reads: “τX is a Hausdorff or separated topology.”

4.1.11. Definition. A subset G of X is an open set in X whenever G isa neighborhood of its every point (i.e., G ∈ Op(τ) ⇔ ((∀x ∈ G) G ∈ τ(x))).A subset F of X is a closed set in X whenever its complement to X is open(in symbols, F ∈ Cl(τ)⇔ (X \ F ∈ Op(τ))).

4.1.12. The union of a family of open sets and the intersection of a finitefamily of open sets are open. The intersection of a family of closed sets and theunion of a finite family of closed sets are closed. CB

4.1.13. Definition. Given a subset U of X, put

intU :=◦U := ∪{G ∈ Op(τX) : G ⊂ U};

clU := U := ∩{F ∈ Cl(τX) : F ⊃ U}.

44 Chapter 4

The set intU is the interior of U and its elements are interior points of U . Theset clU is the closure of U and its elements are adherent to U . The exterior of Uis the interior of X \ U ; the elements of the former are exterior to U . A boundarypoint of U is by agreement a point of X neither interior nor exterior to U . Thecollection of all boundary points of U is called the boundary of U or the frontierof U and denoted by frU or ∂U .

4.1.14. A set U is a neighborhood about x if and only if x is an interior pointof U . CB


4.1.15. Remark. In connection with 4.1.14, the set Op(τX) is also referredto as the topology of the space X under study, since τX is uniquely determinedfrom Op(τX). The same relates to Cl(τX), the collection of all closed sets in X.

4.1.16. Definition. A filterbase B on X converges to x in X or x is a limitof B (in symbols, B0 = 0 OO ) if filB is finer than the neighborhood filter of x; i.e.,filB ⊃ τ(x).

4.1.17. Definition. A net or (generalized) sequence (xξ)ξ∈� converges to x(in symbols, xξ0 = 0 OO ) if the tail filter of (xξ) converges to x. Other familiar termsand designations are freely employed. For instance, x = limξ xξ and x is a limitof (xξ) as ξ ranges over �.

4.1.18. Remark. A limit of a filter, as well as a limit of a net, is uniquein a metric space X. This is another way of expressing the fact that the topologyof X is separated. CB

4.1.19. For a nonempty set U and a point x the following statements areequivalent:

(1) x is an adherent point of U ;

(2) there is a filter F such that F0 = 0 OO and U ∈ F ;

(3) there is a sequence (xξ)ξ∈� whose entries are in U and which con-verges to x.

C (1)⇒ (2): Since x is not exterior to U , the join F := τ(x)∨fil {U} is availableof the pair of the filters τ(x) and fil {U}.

(2) ⇒ (3): Let F0 = 0 OO and U ∈ F . Direct F by reverse inclusion. TakexV ∈ V ∩ U for V ∈ F . It is clear that xV 0 = 0 OO .

(3) ⇒ (1): Let V be a closed set. Take a sequence (xξ)ξ∈� in V such thatxξ0 = 0 OO . In this case it suffices to show that x ∈ V , which is happily evident.Indeed, were x in X \ V we would find at least one ξ ∈ � such that xξ ∈ X \ V . B

4.1.20. Remark. It may be assumed that F has a countable base in 4.1.19(2), and � := N in 4.1.19 (3). This property is sometimes formulated as follows:“In metric spaces the first axiom of countability is fulfilled.”

4.2. Continuity and Uniform Continuity

4.2.1. If f : X0 = 0 OO and τX and τY are topologies on X and Y then thefollowing conditions are equivalent:

46 Chapter 4

(1) G ∈ Op(τY )⇒ f−1(G) ∈ Op(τX);

(2) F ∈ Cl(τY )⇒ f−1(F ) ∈ Cl(τX);

(3) f(τX(x)) ⊃ τY (f(x)) for all x ∈ X;

(4) (x ∈ X & F0 = 0 OO )⇒ (f(F )0 = 0 OO (x)) for a filter F ;

(5) f(xξ)0 = 0 OO (x) for every point x and every sequence (xξ) conver-gent to x.

C The equivalence (1) ⇔ (2) follows from 4.1.11. It remains to demonstratethat (1) ⇒ (3) ⇒ (4) ⇒ (5) ⇒ (2).

(1) ⇒ (3): If V ∈ τY (f(x)) then W := intV ∈ Op(τY ) and f(x) ∈ W . Hence,f−1(W ) ∈ Op(τX) and x ∈ f−1(W ). In other words, f−1(W ) ∈ τX(x) (see 4.1.14).Moreover, f−1(V ) ⊃ f−1(W ) and, consequently, f−1(V ) ∈ τX(x). Finally, V ⊃f(f−1(V )).

(3) ⇒ (4): Given F0 = 0 OO , by Definition 4.1.16 filF ⊃ τX(x). From thehypothesis infer that f(F ) ⊃ f(τX(x)) ⊃ τY (f(x)). Appealing to 4.1.16, againreveal the sought instance of convergence, f(F )0 = 0 OO (x).

(4) ⇒ (5): The image of the tail filter of (xξ)ξ∈� under f is coarser than thetail filter of (f(xξ))ξ∈�.

(5) ⇒ (2): Take a closed set F in Y . If F = ∅ then f−1(F ) is also emptyand, hence, closed. Assume F nonempty and let x be an adherent point of f−1(F ).Consider a sequence (xξ)ξ∈� converging to x and consisting of points in f−1(F ) (theclaim of 4.1.19 yields existence). Then f(xξ) ∈ F and f(xξ)0 = 0 OO (x). Anothercitation of 4.1.19 guarantees that f(x) ∈ F and, consequently, x ∈ f−1(F ). B

4.2.2. Definition. A mapping satisfying one (and hence all) of the equivalentconditions 4.2.1 (1)–4.2.1 (5) is continuous. If 4.2.1 (5) holds at a fixed point x thenf is said to be continuous at x. Thus, f is continuous onX whenever f is continuousat every point of X.

4.2.3. Every composition of continuous mappings is continuous.

C Apply 4.2.1 (5) thrice. B

4.2.4. Let f : X0 = 0 OO and let UX and UY be uniformities on X and Y . Thefollowing statements are equivalent:

(1) (∀V ∈ UY ) (∃U ∈ UX) ((∀x, y)(x, y) ∈ U ⇒ (f(x), f(y)) ∈ V );

(2) (∀V ∈ UY ) f−1 ◦ V ◦ f ∈ UX ;

(3) f×(UX) ⊃ UY , with f× : X20 = 0 OO defined as f× : (x, y) 7→(f(x), f(y));

(4) (∀V ∈ UY ) f×−1(V ) ∈ UX ; i.e., f×−1(UY ) ⊂ UX .


C By 1.1.10, given U ⊂ X2 and V ⊂ Y 2, observe that

f−1 ◦ V ◦ f =⋃

(v1,v2)∈V

f−1(v1)× f−1(v2)

= {(x, y) ∈ X2 : (f(x), f(y)) ∈ V } = f×−1(V );

f ◦ U ◦ f−1 =⋃

(u1,u2)∈U

f(u1)× f(u2)

= {(f(u1), f(u2)) : (u1, u2) ∈ U} = f×(U). .

4.2.5. Definition. A mapping f : X0 = 0 OO satisfying one (and hence all)of the equivalent conditions 4.2.4 (1)–4.2.4 (4) is called uniformly continuous (theterm “uniform continuous” is also in common parlance).

4.2.6. Every composition of uniformly continuous mappings is uniformly con-tinuous.

C Consider f : X0 = 0 OO , g : Y 0 = 0 OO and h := g ◦ f : X0 = 0 OO . Plainly,

h×(x, y) = (h(x), h(y)) = (g(f(x)), g(f(y))) = g×(f(x), f(y)) = g× ◦ f×(x, y)

for all x, y ∈ X. Hence, by 4.2.4 (3) h×(UX) = g×(f×(UX)) ⊃ g×(UY ) ⊃ UZ ,meaning that h is uniformly continuous. B

4.2.7. Every uniformly continuous mapping is continuous. CB

4.2.8. Definition. Let E be a set of mappings from X into Y and let UX

and UY be the uniformities in X and Y . The set E is equicontinuous if

(∀V ∈ UY )⋂f∈E

f−1 ◦ V ◦ f ∈ UX .

4.2.9. Every equicontinuous set consists of uniformly continuous mappings.Every finite set of uniformly continuous mappings is equicontinuous. CB

4.3. Semicontinuity

4.3.1. Let (X1, d1) and (X2, d2) be metric spaces, and := X ×X. Givenx := (x1, x2) and y := (y1, y2), put

d(x, y) := d1(x1, y1) + d2(x2, y2).

Then d is a metric on . Moreover, for every x := (x1, x2) in the presentation holds:

τ(x) = fil{U1 × U2 : U1 ∈ τX1(x1), U2 ∈ τX2(x2)}. /.

4.3.2. Definition. The topology τ is called the product of τX1 and τX2 or theproduct topology of X1 ×X2. This topology on X1 ×X2 is denoted by τX1 × τX2 .

48 Chapter 4

4.3.3. Definition. A function f : X0 = 0 OO R· is lower semicontinuous if itsepigraph 0 = 0 OOOO is closed in the product topology of X × R.

4.3.4. Examples.(1) A continuous real-valued function f : X0 = 0 OO R is lower semicon-

tinuous.(2) If fξ : X0 = 0 OO R· is lower semicontinuous for all ξ ∈ �, then the

upper envelope f(x) := sup{fξ(x) : ξ ∈ �} (x ∈ X) is also lower semicontinuous.A simple reason behind this is the equality 0 = 0 OOOO = ∩ξ∈�0 = 0 OOOO ξ.

4.3.5. A function f : X0 = 0 OO R· is lower semicontinuous if and only if

x ∈ X ⇒ f(x) = limy0=0 OO inf f(y).

Here, as usual,

limy0=0 OO inf f(y) := lim

y0=0 OOf(y) := sup

U∈τ(x)inf f(U)

is the lower limit of f at x (with respect to τ(x)).C ⇒: If x 6∈ dom f then (x, t) 6∈ 0 = 0 OOOO for all t ∈ R. Hence, there is a neigh-

borhood Ut of x such that inf f(Ut) > t. This implies that limy0=0 OO inf f(y) =+∞ = f(x). Suppose that x ∈ dom f . Then inf f(V ) > −∞ for some neighbor-hood V of x. Choose ε > 0 and for an arbitrary U in τ(x) included in V findxU in U so that inf f(U) ≥ f(xU ) − ε. By construction, xU ∈ dom f and, more-over, xU0 = 0 OO (by implication, the set of neighborhoods of x is endowed withthe conventional order by inclusion, cf. 1.3.1). Put tU := inf f(U) + ε. It is clearthat tU0 = 0 OO := limy0=0 OO inf f(y) + ε. Since (xU , tU ) ∈ 0 = 0 OOOO , from the closureproperty of 0 = 0 OOOO obtain (x, t) ∈ 0 = 0 OOOO . Thus,

limy0=0 OO inf f(y) + ε ≥ f(x) ≥ lim

y0=0 OO inf f(y).

⇐: If (x, t) 6∈ 0 = 0 OOOO then

t < limy0=0 OO inf f(y) = sup inf

U∈τ(x)f(U).

Therefore, inf f(U) > t for some neighborhood U of x. It follows that the comple-ment of 0 = 0 OOOO to X × R is open. B

4.3.6. Remark. The property, stated in 4.3.5, may be accepted as an initialdefinition of lower semicontinuity at a point.


4.3.7. A function f : X0 = 0 OO R is continuous if and only if both f and −fare lower semicontinuous. CB

4.3.8. A function f : X0 = 0 OO R· is lower semicontinuous if and only if forevery t ∈ R the level set {f ≤ t} is closed.

C ⇒: If x 6∈ {f ≤ t} then t < f(x). By 4.3.5, t < inf f(U) in a suitableneighborhood U about x. In other words, the complement of {f ≤ t} to X is open.⇐: Given limy0=0 OO inf f(y) ≤ t < f(x) for some x ∈ X and t ∈ R, choose

ε > 0 such that t+ ε < f(x). Repeating the argument of 4.3.5, given U ∈ τ(x) takea point xU in U ∩{f ≤ inf f(U)+ε}. Undoubtedly xU ∈ {f ≤ t+ε} and xU0 = 0 OO ,a contradiction. B

4.4. Compactness

4.4.1. Definition. A subset C of X is called a compact set whenever forevery subset E of Op(τX) with the property C ⊂ ∪{G : G ∈ E } there is a finitesubset E0 in E such that still C ⊂ ∪{G : G ∈ E0}.

4.4.2. Remark. Definition 4.4.1 is verbalized as follows: “A set is compactif its every open cover has a finite subcover.” The terms “covering” and “subcovering”are also in current usage.

4.4.3. Every closed subset of a compact set is also compact. Every compactset is closed. CB

4.4.4. Remark. With regard to 4.4.3, it stands to reason to use the term“relatively compact set” for a set whose closure is compact.

4.4.5. Weierstrass Theorem. The image of a compact set under a continu-ous mapping is compact.

C The inverse images of sets in an open cover of the image compose an opencover of the original set. B

4.4.6. Each lower semicontinuous real-valued function, defined on a nonemptycompact set, assumes its least value; i.e., the image of a compact domain has a leastelement.

C Suppose that f : X0 = 0 OO R· and X is compact. Let t0 := inf f(X). In thecase t0 = +∞ there is nothing left to proof. If t0 < +∞ then put T := {t ∈ R :t > t0}. The set Ut := {f ≤ t} with t ∈ T is nonempty and closed. Check that∩{Ut : t ∈ T} is not empty (then every element x of the intersection meets theclaim: f(x) = inf f(X)). Suppose the contrary. Then the sets {X \ Ut : t ∈ T}compose an open cover of X. Refining a finite subcover, deduce ∩{Ut : t ∈ T0} = ∅.However, this equality is false, since Ut1 ∩ Ut2 = Ut1∧t2 for t1, t2 ∈ T . B

50 Chapter 4

4.4.7. Bourbaki Criterion. A space is compact if and only if every ultrafilteron it converges (cf. 9.4.4).

4.4.8. The product of compact sets is compact.C It suffices to apply the Bourbaki Criterion twice. B

4.4.9. Cantor Theorem. Every continuous mapping on a compact set isuniformly continuous. CB

4.5. Completeness Completeness

4.5.1. If B is a filterbase on X then {B2 : B ∈ B} is a filterbase (and a basefor the filter B×) on X2.

C (B1 ×B1) ∩ (B2 ×B2) ⊃ (B1 ∩B2)× (B1 ∩B2) B

4.5.2. Definition. Let UX be a uniformity on X. A filter F is calleda Cauchy filter or even Cauchy (the latter might seem preposterous) if F× ⊃ UX .A net in X is a Cauchy net or a fundamental net if its tail filter is a Cauchy filter.The term “fundamental sequence” is treated in a similar fashion.

4.5.3. Remark. If V is an entourage on X and U is a subset of X then U isV -small whenever U2 ⊂ V . In particular, U is Bε-small (or simply ε-small) if andonly if the diameter of U , diamU := sup(U2), is less than or equal to ε. In thisconnection, the definition of a Cauchy filter is expressed as follows: “A filter isa Cauchy filter if and only if it contains arbitrarily small sets.”

4.5.4. In a metric space the following conditions are equivalent:(1) every Cauchy filter converges;(2) each Cauchy net has a limit;(3) every fundamental sequence converges.

C The implications (1) ⇒ (2) ⇒ (3) are obvious; therefore, we are left withestablishing (3) ⇒ (1).

Given a Cauchy filter F , let Un in F be aB1/n-small set. Put Vn := U1∩. . .∩Unand take xn ∈ Vn. Observe that V1 ⊃ V2 ⊃ . . . and diamVn ≤ 1/n. Hence, (xn)is a fundamental sequence. By hypothesis it has a limit, x := limxn. Check thatF0 = 0 OO . To this end, choose n0 ∈ N such that d(xm, x) ≤ 1/2n as m ≥ n0. Thenfor all n ∈ N deduce that d(xp, y) ≤ diamVp ≤ 1/2n and d(xp, x) ≤ 1/2n wheneverp := n0∨2n and y ∈ Vp. It follows that y ∈ Vp ⇒ d(x, y) ≤ 1/n; i.e., Vp ⊂ B1/n(x).In conclusion, F ⊃ τ(x). B

4.5.5. Definition. A metric space satisfying one (and hence all) of the equiv-alent conditions 4.5.4 (1)–4.5.4 (3) is called complete.


4.5.6. Cantor Criterion. A metric space X is complete if and only if ev-ery nonempty downward-filtered family of nonempty closed subsets of X whosediameters tend to zero has a point of intersection.

C ⇒: If B is such family then by Definition 1.3.1 B is a filterbase. By hy-pothesis, B is a base for a Cauchy filter. Therefore, there is a limit: B0 = 0 OO . Thepoint x meets the claim.

⇐: Let F be a Cauchy filter. Put B := {clV : V ∈ F}. The diameters of thesets in B tend to zero. Hence, there is a point x such that x ∈ clV for all V ∈ F .Plainly, F0 = 0 OO . Indeed, let V be an ε/2-small member of F and y ∈ V . Somey ′ in V is such that d(x, y ′) ≤ ε/2. Therefore, d(x, y) ≤ d(x, y ′) + d(y ′, y) ≤ ε.Consequently, V ⊂ Bε(x) and so Bε(x) ∈ F . B

4.5.7.Nested Ball Theorem. A metric space is complete if and only if everynested (= decreasing by inclusion) sequence of balls whose radii tend to zero hasa unique point of intersection. CB

4.5.8. The image of a Cauchy filter under a uniformly continuous mapping isa Cauchy filter.

C Let a mapping f act from a space X with uniformity UX into a space Y withuniformity UY and let F be a Cauchy filter on X. If V ∈ UY then, by Definition4.2.5, f−1 ◦V ◦ f ∈ UX (cf. 4.2.4 (2)). Since F is a Cauchy filter, U2 ⊂ f−1 ◦V ◦ ffor some U ∈ F . It turns out that f(U) is V -small. Indeed,

f(U)2 =⋃

(u1,u2)∈U2

f(u1)× f(u2)

= f ◦ U2 ◦ f−1 ⊂ f ◦ (f−1 ◦ V ◦ f) ◦ f−1 = (f ◦ f−1) ◦ V ◦ (f ◦ f−1) ⊂ V,

because, by 1.1.6, f ◦ f−1 = Iim f ⊂ IY . B

4.5.9. The product of complete spaces is complete.C The claim is immediate from 4.5.8 and 4.5.4. B

4.5.10. Let X0 be dense in X (i.e., clX0 = X). Assume further that f0 :X00 = 0 OO is a uniformly continuous mapping from X0 into some complete space Y .Then there is a unique uniformly continuous mapping f : X0 = 0 OO extending f0;i.e., f |X0 = f0.C For x ∈ X, the filter Fx := {U ∩X0 : U ∈ τX(x)} is a Cauchy filter on X0.

Therefore, 4.5.8 implies that f0(FX) is a Cauchy filter on Y . By the completenessof Y , there is a limit y ∈ Y ; that is, f0(Fx)0 = 0 OO . Moreover, this limit is unique(cf. 4.1.18). Define f(x) := y. Checking uniform continuity for f readily completesthe proof. B

52 Chapter 4

4.5.11. Definition. An isometry or isometric embedding or isometric map-ping of X into X is a mapping f : (X, d)0 = 0 OO X, d ) such that d = d ◦ f×.A mapping f is an isometry onto X if f is an isometry of X into X and, moreover,im f = X. The expressions, “an isometry of X and X” or “an isometry between Xand X” and the like, are also in common parlance.

4.5.12. Hausdorff Completion Theorem. If (X, d) is a metric space thenthere are a complete metric space (X, d ) and an isometry ι : (X, d)0 = 0 OO X, d )onto a dense subspace of (X, d ). The space (X, d ) is unique to within isometryin the following sense: The diagram

(X1, d1)

(X, d) (X, d )ι

�ι1

-

?

@@@R

commutes for some isometry � : (X, d )0 = 0 OO X1, d1), where ι1 : (X, d)0 =0 OO X1, d1) is an isometry of X onto a dense subspace of a complete space (X1, d1).C Uniqueness up to isometry follows from 4.5.10. For, if �0 := ι1 ◦ ι−1 then

�0 is an isometry of the dense subspace ι(X) of X onto the dense subspace ι1(X)of X1. Let � be the unique extension of �0 to the whole of X. It is sufficientto show that � acts onto X1. Take x1 in X1. This element is the limit of somesequence (ι1(xn)), with xn ∈ X. Clearly, the sequence (xn) is fundamental. Thus,(ι(xn)) is a fundamental sequence in X. Let x := lim ι(xn), x ∈ X. Proceed asfollows: �(x) = lim�0(ι(xn)) = lim ι1 ◦ ι−1(ι(xn)) = lim ι1(xn) = x1.

We now sketch out the proof that X exists. Consider the set of all fundamentalsequences in X. Define some equivalence relation in X by putting x1 ∼ x2 ⇔d(x1(n), x2(n))0 = 0 OO . Assign X := /∼ and d(ϕ(x1), ϕ(x2)) := lim d(x1(n), x2(n)),where ϕ : = OO X is the coset mapping. An isometry ι : (X, d)0 = 0 OO X, d ) isimmediate: ι(x) := ϕ(n 7→ x (n ∈ N)). B

4.5.13. Definition. The space (X, d ) introduced in Theorem 4.5.12, as wellas each space isomorphic to it, is called a completion of (X, d).

4.5.14. Definition. A set X0 in a metric space (X, d) is said to be completeif the metric space (X0, d|X2

0), a subspace of (X, d), is complete.

4.5.15. Every closed subset of a complete space is complete. Every completeset is closed. CB

4.5.16. If X0 is a subspace of a complete metric space X then a completionof X0 is isometric to the closure of X0 in X.


C Let X := clX0 and let ι : X00 = 0 OO X be the identical embedding. It isevident that ι is an isometry onto a dense subspace. Moreover, by 4.5.15 X iscomplete. Appealing to 4.5.12 ends the proof. B

4.6. Compactness and Completeness

4.6.1. A compact space is complete. CB

4.6.2. Definition. Let U be a subset of X and V ∈ UX . A set E in X isa V -net for U if U ⊂ V (E).

4.6.3.Definition. A subset U ofX is a totally bounded set inX if for every Vin UX there is a finite V -net for U .

4.6.4. If for every V in U a set U in X has a totally bounded V -net then Uis totally bounded.

C Let V ∈ UX andW ∈ UX be such thatW ◦W ⊂ V . Take a totally boundedW -net F for U ; i.e., U ⊂W (F ). Since F is totally bounded, there is a finite W -netE for F ; that is, F ⊂W (E). Finally,

U ⊂W (F ) ⊂W (W (E)) =W ◦W (E) ⊂ V (E);

i.e., E is a finite V -net for U . B

4.6.5. A subset U of X is totally bounded if and only if for every V in UX

there is a family U1, . . . , Un of subsets of U such that U = U1 ∪ . . . ∪ Un and eachof the sets U1, . . . , Un is V -small. CB

4.6.6. Remark. The claim of 4.6.5 is verbalized as follows: “A set is totallybounded if and only if it has finite covers consisting of arbitrarily small sets.”

4.6.7. Hausdorff Criterion. A set is compact if and only if it is completeand totally bounded. CB

4.6.8. Let C(Q, F) be the space of continuous functions with domain a com-pact set Q and range a subset of F. Furnish this space with the Chebyshev metric

d(f, g) := supx∈Q

dF(f(x), g(x)) := supx∈Q|f(x)− g(x)| (f, g ∈ C(Q, F));

and, given θ ∈ UF, put

Uθ :={(f, g) ∈ C(Q, F)2 : g ◦ f−1 ⊂ θ

}.

Then Ud = fil {Uθ : θ ∈ UF}. CB

54 Chapter 4

4.6.9. The space C(Q, F) is complete. CB

4.6.10. Ascoli–Arzela Theorem. A subset E of C(Q, F) is relatively com-pact if and only if E is equicontinuous and the set ∪{g(Q) : g ∈ E } is totallybounded in F.

C ⇒: It is beyond a doubt that ∪{g(Q) : g ∈ E } is totally bounded. Toshow equicontinuity for E take θ ∈ UF and choose a symmetric entourage θ′ suchthat θ′ ◦ θ′ ◦ θ′ ⊂ θ. By the Hausdorff Criterion, there is a finite Uθ′ -net E ′ for E .Consider the entourage U ∈ UQ defined as

U :=⋂f∈E ′

f−1 ◦ θ′ ◦ f

(cf. 4.2.9). Given g ∈ E and f ∈ E ′ such that g ◦ f−1 ⊂ θ′, observe that

θ′ = θ′−1 ⊃ (g ◦ f−1)−1 = (f−1)−1 ◦ g−1 = f ◦ g−1.

Moreover, the composition rules for correspondences and 4.6.8 imply

g×(U) = g ◦ U ◦ g−1 ⊂ g ◦ (f−1 ◦ θ′ ◦ f) ◦ g−1

⊂ (g ◦ f−1) ◦ θ′ ◦ (f ◦ g−1) ⊂ θ′ ◦ θ′ ◦ θ′ ⊂ θ.

Since g is arbitrary, the resulting inclusion guarantees that E is equicontinuous.⇐: By 4.5.15, 4.6.7, 4.6.8, and 4.6.9, it is sufficient to construct a finite Uθ-net

for E given θ ∈ UF. Choose θ′ ∈ UF such that θ′ ◦ θ′ ◦ θ′ ⊂ θ and find an opensymmetric entourage U ∈ UQ from the condition

U ⊂⋂g∈E

g−1 ◦ θ′ ◦ g

(by the equicontinuity property of E , such an U is available). The family {U(x) :x ∈ Q} clearly forms an open cover of Q. By the compactness of Q, refine a finitesubcover {U(x0) : x0 ∈ Q0}. In particular, from 1.1.10 derive

IQ ⊂⋃

x0∈Q0

U(x0)× U(x0)

=⋃

(x0,x0)∈IQ0

U−1(x0)× U(x0) = U ◦ IQ0 ◦ U.

The set {g|Q0 : g ∈ E } is totally bounded in FQ0 . Consequently, there is a finiteθ′-net for this set. Speaking more precisely, there is a finite subset E ′ of E with thefollowing property:

g ◦ IQ0 ◦ f−1 ⊂ θ′


for every g in E and some f in E ′. Using the estimates, successively infer that

g ◦ f−1 = g ◦ IQ ◦ f−1 ⊂ g ◦ (U ◦ IQ0 ◦ U) ◦ f−1

⊂ g ◦ (g−1 ◦ θ′ ◦ g) ◦ IQ0 ◦ (f−1 ◦ θ′ ◦ f) ◦ f−1

= (g ◦ g−1) ◦ θ′ ◦ (g ◦ IQ0 ◦ f−1) ◦ θ′ ◦ (f ◦ f−1)= Iimg ◦ θ′ ◦ (g ◦ IQ0 ◦ f−1) ◦ θ′ ◦ Iimf

⊂ θ′ ◦ θ′ ◦ θ′ ⊂ θ.

Thus, by 4.6.8, E ′ is a finite Uθ-net for E . B

4.6.11. Remark. It is an enlightening exercise to translate the proof of theAscoli–Arzela Theorem into the “ε-δ” language. The necessary vocabulary is in hand:“θ and Uθ stand for ε,” “θ′ is ε/3,” and “δ is U .” It is also rewarding and instructiveto find generalizations of the Ascoli–Arzela Theorem for mappings acting into moreabstract spaces.

4.7. Baire Spaces

4.7.1. Definition. A set U is said to be nowhere dense or rare whenever itsclosure lacks interior points; i.e., int clU = ∅. A set U is meager or a set of firstcategory if U is included into a countable union of rare sets; i.e., U ⊂ ∪n∈NUn withint clUn = ∅. A nonmeager set (which is not of first category by common parlance)is also referred to as a set of second category.

4.7.2. Definition. A space is a Baire space if its every nonempty open set isnonmeager.

4.7.3. The following statements are equivalent:(1) X is a Baire space;(2) every countable union of closed rare sets in X lacks interior points;(3) the intersection of a countable family of open everywhere dense sets

(i.e., dense in X) is everywhere dense;(4) the complement of each meager set to X is everywhere dense.

C (1) ⇒ (2): If U := ∪n∈NUn, Un = clUn, and intUn = ∅ then U is meager.Observe that intU ⊂ U and intU is open; hence, intU as a meager set is necessarilyempty, for X is a Baire space.

(2) ⇒ (3): Let U := ∩n∈NGn, where Gn’s are open and clGn = X. ThenX\U = X\∩n∈NGn = ∪n∈N (X\Gn). Moreover, X\Gn is closed and int(X\Gn) =∅ (since clGn = X). Therefore, int(X \ U) = ∅, which implies that the exteriorof U is empty; i.e., U is everywhere dense.

56 Chapter 4

(3) ⇒ (4): Let U be a meager set in X; i.e., U ⊂ ∪n∈NUn and int clUn = ∅.It may be assumed that Un = clUn. Then Gn := X \ Un is open and everywheredense. By hypothesis, ∩n∈NGn = X \ ∪n∈NUn is everywhere dense. Moreover, theset is included into X \ U , and so X \ U is everywhere dense.

(4)⇒ (1): If U is nonempty open set in X then X \U is not everywhere dense.Consequently, U is nonmeager. B

4.7.4. Remark. In connection with 4.7.3 (4), observe that the complementof a meager set is (sometimes) termed a residual or comeager set. A residual setin a Baire space is nonmeager.

4.7.5.Osgood Theorem. LetX be a Baire space and let (fξ : X0 = 0 OO R)ξ∈�be a family of lower semicontinuous functions such that sup{fξ(x) : ξ ∈ �} < +∞for all x ∈ X. Then each nonempty open set G in X includes a nonempty opensubset G0 on which (fξ)ξ∈� is uniformly bounded above; i.e., supx∈G0

sup{fξ(x) :ξ ∈ �} ≤ +∞. CB

4.7.6. Baire Category Theorem. A complete metric space is a Baire space.

C Let G be a nonempty open set and x0 ∈ G. Suppose by way of contradictionthat G is meager; i.e., G ⊂ ∪n∈NUn, where intUn = ∅ and Un = clUn. Takeε0 > 0 satisfying Bε0(x0) ⊂ G. It is obvious that U1 is not included into Bε0/2(x0).Consequently, there is an element x1 in Bε0/2(x0) \ U1. Since U1 is closed, find ε1satisfying 0 < ε1 ≤ ε0/2 and Bε1(x1) ∩ U1 = ∅. Check that Bε1(x1) ⊂ Bε0(x0).For, if d(x1, y1) ≤ ε1, then d(y1, x0) ≤ d(y1, x1) + d(x1, x0) ≤ ε1 + ε0/2, becaused(x1, x0) ≤ ε0/2. The ball Bε1/2(x1) does not lie in U2 entirely. It is thus possibleto find x2 ∈ Bε1/2(x1) \ U2 and 0 < ε2 ≤ ε1/2 satisfying Bε2(x2) ∩ U2 = ∅. It iseasy that again Bε2(x2) ⊂ Bε1(x1). Proceeding by induction, obtain the sequenceof nested balls Bε0(x0) ⊃ Bε1(x1) ⊃ Bε2(x2) ⊃ . . . ; moreover, εn+1 ≤ εn/2 andBεn(xn) ∩ Un = ∅. By the Nested Ball Theorem, the balls have a common point,x := limxn. Further, x 6= ∪n∈NUn; and, hence, x 6∈ G. On the other hand,x ∈ Bε0(x0) ⊂ G, a contradiction. B

4.7.7. Remark. The Baire Category Theorem is often used as a “pure exis-tence theorem.” As a classical example, consider the existence problem for contin-uous nowhere differentiable functions.

Given f : [0, 1]0 = 0 OO R and x ∈ [0, 1), put

D+f(x) := limh↓0

inff(x+ h)− f(x)

h;

D+f(x) := limh↓0

supf(x+ h)− f(x)

h.


The elements D+f(x) and D+f(x) of the extended axis R are the lower right andupper right Dini derivatives of f at x. Further, let D stand for the set of functionsf in C([0, 1], R) such that for some x ∈ [0, 1) the elements D+f(x) and D+f(x)belong to R; i.e., they are finite. Then D is a meager set. Hence, the functionslacking derivatives at every point of (0, 1) are everywhere dense in C([0, 1], R).However, the explicit examples of such functions are not at all easy to find andgrasp. Below a few of the most popular are listed:

van der Waerden’s function:∞∑n=0

〈〈4nx〉〉4n

,

with 〈〈x〉〉 := (x − [x]) ∧ (1 + [x] − x), the distance from x to the whole numbernearest to x;

Riemann’s function:+∞∑n=0

1n2 sin (n2πx);

and, finally, the historically first

Weierstrass’s function:∞∑n=0

bn cos (anπx),

with a an odd positive integer, 0 < b < 1 and ab > 1 + 3π/2.

4.8. The Jordan Curve Theorem and Rough Drafts

4.8.1. Remark. In topology, in particular, many significant and curious factsof the metric space R2 are scrutinized. Here we recall those of them which are ofuse in the sequel and whose role is known from complex analysis.

4.8.2. Definition. A (Jordan) arc is a homeomorphic image of a (nonde-generate) interval of the real axis. Recall that a homeomorphism or a topologicalmapping is by definition a one-to-one continuous mapping whose inverse is alsocontinuous. A (simple Jordan) loop is a homeomorphic image of a circle. Conceptslike “smooth loop” are understood naturally.

4.8.3. Jordan Curve Theorem. If γ is a simple loop in R2 then there areopen sets G1 and G2 such that

G1 ∪G2 = R2 \ γ; γ = ∂G1 = ∂G2. /.

4.8.4. Remark. Either G1 or G2 in 4.8.3 is bounded. Moreover, each of thetwo sets is connected; i.e., it cannot be presented as the union of two nonemptydisjoint open sets. In this regard the Jordan Curve Theorem is often read as follows:“A simple loop divides the plane into two domains and serves as their mutualboundary.”

58 Chapter 4

4.8.5. Definition. Let D1, . . . , Dn and D be closed disks (= closed balls)in the plane which satisfy D1, . . . , Dn ⊂ intD and Dm ∩Dk = ∅ as m 6= k. Theset

D \n⋃k=1

intDk

is a holey disk or, more formally, a perforated disk. A subset of the plane which isdiffeomorphic (= “smoothly homeomorphic”) to a holey disk is called a connectedelementary compactum. The union of a finite family of pairwise disjoint connectedelementary compacta is an elementary compactum.

4.8.6. Remark. The boundary ∂F of an elementary compactum F comprisesfinitely many disjoint smooth loops. Furthermore, the embedding of F into the(oriented) plane R2 induces on F the structure of an (oriented) manifold with (ori-ented) boundary ∂F . Observe also that, by 4.8.3, it makes sense to specify thepositive orientation of a smooth loop. This is done by indicating the orientationinduced on the boundary of the compact set surrounded by the loop.

4.8.7. If K is a compact subset of the plane and G is a nonempty open setthat includes K then there is a nonempty elementary compactum F such that

K ⊂ intF ⊂ F ⊂ G. /.

4.8.8. Definition. Every set F appearing in 4.8.7 is referred to as a roughdraft for the pair (K, G) or an oriented envelope of K in G.

Exercises

4.1. Give examples of metric spaces. Find methods for producing new metric spaces.4.2. Which filter on X2 coincides with some metric uniformity on X?4.3. Let S be the space of measurable functions on [0, 1] endowed with the metric

d(f, g) :=

1∫0

|f(t)− g(t)|1 + |f(t)− g(t)|

dt (f, g ∈ S)

with some natural identification implied (specify it!). Find the meaning of convergence in thespace.

4.4. Given α, β ∈ NN, put

d(α, β) := 1/min {k ∈ N : αk 6= βk}.

Check that d is a metric and the space NN is homeomorphic with the set of irrational numbers.


4.5. Is it possible to metrize pointwise convergence in the sequence space? In the functionspace F[0,1]?

4.6. How to introduce a reasonable metric into the countable product of metric spaces? Intoan arbitrary product of metric spaces?

4.7. Describe the function classes distinguishable by erroneous definitions of continuity anduniform continuity.

4.8. Given nonempty compact subsets A and B of the spaces RN, define

d(A, B) :=

(supx∈A

infy∈B

|x− y|)∨(supy∈B

infx∈A

|x− y|).

Show that d is a metric. The metric is called the Hausdorff metric. What is the meaningof convergence in this metric?

4.9. Prove that nonempty compact convex subsets of a compact convex set in RN constitutea compact set with respect to the Hausdorff metric. How does this claim relate to the Ascoli–ArzelaTheorem?

4.10. Prove that each lower semicontinuous function on RN is the upper envelope of somefamily of continuous functions.

4.11. Explicate the interplay between continuous and closed mappings (in the product topol-ogy) of metric spaces.

4.12. Find out when a continuous mapping of a metric space into a complete metric spacesis extendible onto a completion of the initial space.

4.13. Describe compact sets in the product of metric spaces.4.14. Let (Y, d) be a complete metric spaces. A mapping F : Y 0 = 0 OO is called expanding

whenever d(F (x), F (y)) ≥ βd(x, y) for some β > 1 and all x, y ∈ Y . Assume that an expandingmapping F : Y 0 = 0 OO acts onto Y . Prove that F is one-to-one and possesses a sole fixed point.

4.15. Prove that no compact set can be mapped isometrically onto a proper part of it.4.16. Show normality of an arbitrary metric space (see 9.3.11).4.17. Under what conditions a countable subset of a complete metric spaces is nonmeager?4.18. Is it possible to characterize uniform continuity in terms of convergent sequences?4.19. In which metric spaces does each continuous real-valued function attain the supremum

and infimum of its range? When is it bounded?

Chapter 5

Multinormed and Banach Spaces

5.1. Seminorms and Multinorms

5.1.1. Let X be a vector space over a basic field F and let p : X0 = 0 OO R· bea seminorm. Then

(1) dom p is a subspace of X;(2) p(x) ≥ 0 for all x ∈ X;(3) the kernel ker p := {p = 0} is a subspace in X;

(4) the sets◦Bp := {p < 1} and Bp := {p ≤ 1} are absolutely convex;

moreover, p is the Minkowski functional of every set B such that◦Bp ⊂ B ⊂ Bp;

(5) X = dom p if and only if◦Bp is absorbing.

C If x1, x2 ∈ dom p and α1, α2 ∈ F then by 3.7.6

p(α1x1 + α2x2) ≤ |α1|p(x1) + |α2|p(x2) < +∞+ (+∞) = +∞.

Hence, (1) holds. Suppose to the contrary that (2) is false; i.e., p(x) < 0 for somex ∈ X. Observe that 0 ≤ p(x) + p(−x) < p(−x) = p(x) < 0, a contradiction. Theclaim of (3) is immediate from (2) and the subadditivity of p. The validity of (4)and (5) has been examined in part (cf. 3.8.8). What was left unproven follows fromthe Gauge Theorem. B

5.1.2. If p, q : X0 = 0 OO R· are two seminorms then the inequality p ≤ q(in (R·)X) holds if and only if Bp ⊃ Bq.

C ⇒: Evidently, {q ≤ 1} ⊂ {p ≤ 1}.⇐: In view of 5.1.1 (4), observe that p = pBp and q = pBq . Take t1, t2 ∈ R

such that t1 < t2. If t1 < 0 then {q ≤ t1} = ∅, and so {q ≤ t1} ⊂ {p ≤ t2}.If t1 ≥ 0 then t1Bq ⊂ t1Bp ⊂ t2Bp. Thus, by 3.8.3, p ≤ q. B

Multinormed and Banach Spaces 61

5.1.3. LetX and Y be vector spaces, let T ⊂ X×Y be a linear correspondence,and let p : Y 0 = 0 OO R· be a seminorm. If pT (x) := inf p ◦ T (x) for x ∈ X thenpT : X0 = 0 OO R· is a seminorm and the set BT := T−1(Bp) is absolutely convex.In addition, pT = pBT .

C Given x1, x2 ∈ X and α1, α2 ∈ F, infer that

pT (α1x1 + α2x2) = inf p(T (α1x1 + α2x2))

≤ inf p(α1T (x1) + α2T (x2)) ≤ inf(|α1|p(T (x1)) + |α2|p(T (x2)))

= |α1|pT (x1) + |α2|pT (x2);

i.e., pT is a seminorm.The absolute convexity of BT is a consequence of 5.1.1 (4) and 3.1.8. If x ∈ BT

then (x, y) ∈ T for some y ∈ Bp. Hence, pT (x) ≤ p(y) ≤ 1; that is, BT ⊂ BpT .

If in turn x ∈◦BpT then pT (x) = inf{p(y) : (x, y) ∈ T} < 1. Thus, there is some

y ∈ T (x) such that p(y) < 1. Therefore, x ∈ T−1(◦Bp) ⊂ T−1(Bp) = BT . Finally,

◦BpT ⊂ BT ⊂ BpT . Referring to 5.1.1 (4), conclude that pBT = pT . B

5.1.4. Definition. The seminorm pT , constructed in 5.1.3, is the inverseimage or preimage of p under T .

5.1.5. Definition. Let p : X0 = 0 OO R be a seminorm (by 3.4.3, this impliesthat dom p = X). A pair (X, p) is referred to as a seminormed space. It isconvenient to take the liberty of calling X itself a seminormed space.

5.1.6. Definition. A multinorm on X is a nonempty set (a subset of RX)of everywhere-defined seminorms. Such a multinorm is denoted by MX or simplyby M, if the underlying vector space is clear from the context. A pair (X, MX),as well as X itself, is called a multinormed space.

5.1.7. A set M in (R·)X is a multinorm if and only if (X, p) is a seminormedspace for every p ∈M. CB

5.1.8. Definition. A multinorm MX is a Hausdorff or separated multinormwhenever for all x ∈ X, x 6= 0, there is a seminorm p ∈ MX such that p(x) 6= 0.In this case X is called a Hausdorff or separated multinormed space.

5.1.9. Definition. A norm is a Hausdorff multinorm presenting a singleton.The sole element of a norm on a vector space X is also referred to as the normon X and is denoted by ‖ · ‖ or (rarely) by ‖ · ‖X or even ‖ · |X‖ if it is necessaryto indicate the space X. A pair (X, ‖ · ‖) is called a normed space; as a rule, thesame term applies to X.

62 Chapter 5

5.1.10. Examples.

(1) A seminormed space (X, p) can be treated as a multinormed space(X, {p}). The same relates to a normed space.

(2) If M is the set of all (everywhere-defined) seminorms on a space Xthen M is a Hausdorff multinorm called the finest multinorm on X.

(3) Let (Y, N) be a multinormed space, and let T ⊂ X ×Y be a linearcorrespondence such that domT = X. By 3.4.10 and 5.1.1 (5), for every p in N theseminorm pT is defined everywhere, and hence M := {pT : p ∈ N} is a multinormon X. The multinorm N is called the inverse image or preimage of N under thecorrespondence T and is (sometimes) denoted by NT . Given T ∈ L (X, Y ), setM := {p ◦ T : p ∈ N} and use the natural notation N ◦ T := M. Observein particular the case in which X is a subspace Y0 of Y and T is the identicalembedding ι : Y00 = 0 OO . At this juncture Y0 is treated as a multinormed spacewith multinorm N ◦ ι. Moreover, the abuse of the phrase “N is a multinorm on Y0”is very convenient.

(4) Each basic field F is endowed, as is well known, with the standardnorm | · | : F0 = 0 OO R, the modulus of a scalar. Consider a vector space X andf ∈ X#. Since f : X0 = 0 OO F , it is possible to define the inverse image of the normon F as pf (x) := |f(x)| (x ∈ X). If is some subspace of X# then the multinormσ(X, ) := { : ∈} is the weak multinorm on X induced by .

(5) Let (X, p) be a seminormed space. Assume further thatX0 is a sub-space of X and ϕ : X0 = 0 OO /X0 is the coset mapping. The linear correspondenceϕ−1 is defined on the whole of X/X0. Hence, the seminorm pϕ−1 appears, called thequotient seminorm of p by X0 and denoted by pX/X0 . The space (X/X0, pX/X0)is called the quotient space of (X, p) by X0. The definition of quotient space foran arbitrary multinormed space requires some subtlety and is introduced in 5.3.11.

(6) Let X be a vector space and let M ⊂ (R·)X be a set of seminormson X. In this case M can be treated as a multinorm on the space X0 := ∩{dom p :p ∈ M}. More precisely, thinking of the multinormed space (X0, {pι : p ∈ M}),where ι is the identical embedding of X0 into X, we say: “M is a multinorm,” or“Consider the (multinormed) space specified by M.” The next example is typical:The family of seminorms{

pα,β(f) := supx∈RN

|xα∂βf(x)| : α and β are multi-indices}

specifies the (multinormed) space of infinitely differentiable functions on RN de-creasing rapidly at infinity (such functions are often called tempered, cf. 10.11.6).


(7) Let (X, ‖·‖) and (Y, ‖·‖) be normed spaces (over the same groundfield F). Given T ∈ L (X, Y ), consider the operator norm of T , i.e. the quantity

‖T‖ := sup{‖Tx‖ : x ∈ X, ‖x‖ ≤ 1} = supx∈X

‖Tx‖‖x‖

.

(From now on, in analogous situations we presume 0/0 := 0.)It is easily seen that ‖ · ‖ : L (X, Y )0 = 0 OO R· is a seminorm. Indeed, putting

BX := {‖ · ‖X ≤ 1} for T1, T2 ∈ L (X, Y ) and α1, α2 ∈ F, deduce that

‖α1T1 + α2T2‖ = sup ‖ · ‖α1T1+α2T2(BX)= sup ‖ · ‖((α1T1 + α2T2)(BX)) ≤ sup ‖α1T1(BX) + α2T2(BX)‖≤ |α1| sup ‖ · ‖T1(BX) + |α2| sup ‖ · ‖T2(BX) = |α1| ‖T1‖+ |α2| ‖T2‖.

The subspace B(X, Y ), the effective domain of definition of the above semi-norm, is the space of bounded operators; and an element of B(X, Y ) is a boundedoperator. Observe that a shorter term “operator” customarily implies a boundedoperator. It is clear that B(X, Y ) is a normed space (under the operator norm).Note also that T in L (X, Y ) is bounded if and only if T maintains the normativeinequality; i.e., there is a strictly positive number K such that

‖Tx‖Y ≤ K ‖x‖X (x ∈ X).

Moreover, ‖T‖ is the greatest lower bound of the set of Ks appearing in the nor-mative inequality. CB

(8) Let X be a vector space over F and let ‖·‖ be a norm on X. Assumefurther that X ′ := B(X, F) is the (normed) dual of X, i.e. the space of boundedfunctionals fs with the dual norm

‖f‖ = sup{|f(x)| : ‖x‖ ≤ 1} = supx∈X

|f(x)|‖x‖

.

Consider X ′′ := (X ′)′ := B(X ′, F), the second dual of X. Given x ∈ X andf ∈ X ′, put x′′ := ι(x) : f 7→ f(x). Undoubtedly, ι(x) ∈ (X ′)# = L (X ′, F).In addition,

‖x′′‖ = ‖ι(x)‖ = sup{|ι(x)(f)| : ‖f | X ′ ≤ 1}= sup{|f(x)| : (∀x ∈ X) |f(x)| ≤ ‖x‖X} = sup{|f(x)| : f ∈ |∂ |(‖ · ‖X)} = ‖x‖X .

The final equality follows for instance from Theorem 3.6.5 and Lemma 3.7.9.Thus, ι(x) ∈ X ′′ for every x in X. It is plain that the operator ι : X0 = 0 OO ′′,

64 Chapter 5

defined as ι : x 7→ ι(x), is linear and bounded; moreover, ι is a monomorphism and‖ιx‖ = ‖x‖ for all x ∈ X. The operator ι is referred to as the canonical embeddingof X into the second dual or more suggestively the double prime mapping. As a rule,it is convenient to treat x and x′′ := ιx as the same element; i.e., to consider X asa subspace of X ′′. A normed space X is reflexive if X and X ′′ coincide (under theindicated embedding!). Reflexive spaces possess many advantages. Evidently, notall spaces are reflexive. Unfortunately, such is C([0, 1], F) which is irreflexive (theterm “nonreflexive” is also is common parlance). CB

5.1.11. Remark. The constructions, carried out in 5.1.10 (8), show somesymmetry or duality between X and X ′. In this regard, the notation (x, f) :=〈x | f〉 := f(x) symbolizes the action of x ∈ X on f ∈ X ′ (or the action of f on x).To achieve and ensure the utmost conformity, it is a common practice to denoteelements of X ′ by symbols like x′; for instance, 〈x |x′〉 = (x, x′) = x′(x).

5.2. The Uniformity and Topology of a MultinormedSpace

5.2.1. Let (X, p) be a seminormed space. Given x1, x2 ∈ X, put dp(x1, x2):= p(x1 − x2). Then

(1) dp(X2) ⊂ R+ and {d ≤ 0} ⊃ IX ;(2) {dp ≤ t} = {dp ≤ t}−1 and {dp ≤ t} = t{dp ≤ 1} (t ∈ R+ \ 0);(3) {dp ≤ t1} ◦ {dp ≤ t2} ⊂ {dp ≤ t1 + t2} (t1, t2 ∈ R+);(4) {dp ≤ t1} ∩ {dp ≤ t2} ⊃ {dp ≤ t1 ∧ t2} (t1, t2 ∈ R+);(5) p is a norm ⇔ dp is a metric. CB

5.2.2. Definition. The uniformity of a seminormed space (X, p) is the filterUp := fil {{dp ≤ t} : t ∈ R+ \ 0}.

5.2.3. If Up is the uniformity of a seminormed space (X, p) then(1) Up ⊂ fil {IX};(2) U ∈ Up ⇒ U−1 ∈ Up;(3) (∀U ∈ Up) (∃ V ∈ Up) V ◦ V ⊂ U . CB

5.2.4. Definition. Let (X, M) be a multinormed space. The filter U :=sup{Up : p ∈M} is called the uniformity of X (the other designations are UM, UX ,etc.). (By virtue of 5.2.3 (1) and 1.3.13, the definition is sound.)

5.2.5. If (X, M) is a multinormed space and U is the uniformity of X then(1) U ⊂ fil {IX};(2) U ∈ U ⇒ U−1 ∈ U ;(3) (∀U ∈ U ) (∃V ∈ U ) V ◦ V ⊂ U .


C Examine (3). Given U ∈ U , by 1.2.18 and 1.3.8 there are seminormsp1, . . . , pn ∈ M such that U = U{p1,... ,pn} = Up1 ∨ . . . ∨ Upn . Using 1.3.13, findsets Uk ∈ Upk satisfying U ⊃ U1 ∩ . . .∩Un. Applying 5.2.3 (3), choose Vk ∈ Upk soas to have Vk ◦ Vk ⊂ Uk. It is clear that

(V1 ∩ . . . ∩ Vn) ◦ (V1 ∩ . . . ∩ Vn) ⊂ V1 ◦ V1 ∩ . . . ∩ Vn ◦ Vn⊂ U1 ∩ . . . ∩ Un.

Moreover, V1 ∩ . . . ∩ Vn ∈ Up1 ∨ . . . ∨Upn ⊂ U . B

5.2.6. A multinorm M on X is separated if and only if so is the uniformityUM; i.e., ∩{V : V ∈ UM} = IX .

C ⇒: Let (x, y) 6∈ IX ; i.e., x 6= y. Then p(x − y) > 0 for some seminorm pin M. Hence, (x, y) 6∈ {dp ≤ 1/2 p(x − y)}. But the last set is included in Up,and thus in UM. Consequently, X2 \ IX ⊂ X2 \ ∩{V : V ∈ UM}. Furthermore,IX ⊂ ∩{V : V ∈ UM}.

⇐: If p(x) = 0 for all p ∈ M then (x, 0) ∈ V for every V in UM. Hence,(x, 0) ∈ IX by hypothesis. Therefore, x = 0. B

5.2.7. Given a space X with uniformity UX , define

τ(x) := {U(x) : U ∈ UX} (x ∈ X).

Then τ(x) is a filter for every x ∈ X. Moreover,(1) τ(x) ⊂ fil {x};(2) (∀U ∈ τ(x)) (∃V ∈ τ(x) & V ⊂ U) (∀ y ∈ V ) V ∈ τ(y).

C All is evident (cf. 4.1.8). B

5.2.8.Definition. The mapping τ : x 7→ τ(x) is called the topology of a multi-normed space (X, M); a member of τ(x) is a neighborhood about x. The designationfor the topology can be more detailed: τX , τM, τ(UM), etc.

5.2.9. The following presentation holds:

τX(x) = sup{τp(x) : p ∈MX}

for all x ∈ X. CB

5.2.10. If X is a multinormed space then

U ∈ τ(x)⇔ U − x ∈ τX(0)

for all x ∈ X.C By 5.2.9 and 1.3.13, it suffices to consider a seminormed space (X, p). In this

case for every ε > 0 the equality holds: {dp ≤ ε}(x) = εBp+x, where Bp := {p ≤ 1}.Indeed, if p(y − x) ≤ ε then y = ε(ε−1(y − x)) + x and ε−1(y − x) ∈ Bp. In turn,if y ∈ εBp + x, then p(y − x) = inf{t > 0 : y − x ∈ tBp} ≤ ε. B

66 Chapter 5

5.2.11. Remark. The proof of 5.2.10 demonstrates that in a seminormedspace (X, p) a key role is performed by the ball with radius 1 and centered at zero.The ball bears the name of the unit ball of X and is denoted by Bp, BX , etc.

5.2.12. A multinorm MX is separated if and only if so is the topology τX ;i.e., given distinct x1 and x2 in X, there are neighborhoods U1 in τX(x1) and U2in τX(x2) such that U1 ∩ U2 = ∅.

C⇒: Let x1 6= x2 and let ε := p(x1−x2) > 0 for p ∈MX . Put U1 := x1+ε/3Bpand U2 := x2 + ε/3Bp. By 5.2.10, Uk ∈ τX(xk). Verify that U1 ∩ U2 = ∅. Indeed,if y ∈ U1 ∩ U2 then p(x1 − y) ≤ ε/3 and p(x2 − y) ≤ ε/3. Therefore, p(x1 − x2) ≤2/3 ε < ε = p(x1 − x2), which is impossible.⇐: If (x1, x2) ∈ ∩{V : V ∈ UX} then x2 ∈ ∩{V (x1) : V ∈ UX}. Thus,

x1 = x2 and, consequently, MX is separated by 5.2.6. B

5.2.13. Remark. The presence of a uniformity and the corresponding topol-ogy in a multinormed space readily justifies using uniform and topological conceptssuch as uniform continuity, completeness, continuity, openness, closure, etc.

5.2.14. Let (X, p) be a seminormed space and let X0 be a subspace of X.The quotient space (X/X0, pX/X0) is separated if and only if X0 is closed.

C ⇒: If x 6∈ X0 then ϕ(x) 6= 0 where, as usual, ϕ : X0 = 0 OO /X0 is the cosetmapping. By hypothesis, 0 6= ε := pX/X0(ϕ(x)) = pϕ−1(ϕ(x)) = inf{p(x + x0) :x0 ∈ X0}. Hence, the ball x+ ε/2Bp does not meet X0 and x is an exterior pointof X0. Thus, X0 is closed.⇐: Suppose that x is a nonzero point of X/X0 and x = ϕ(x) for some x in X.

If pX/X0(x) = 0 then 0 = inf{p(x − x0) : x0 ∈ X0}. In other words, there isa sequence (xn) in X0 converging to x. Consequently, by 4.1.19, x ∈ X0 and x = 0,a contradiction. B

5.2.15. The closure of a � -set is a � -set.

C Given U ∈ (� ), suppose that U 6= ∅ (otherwise there is nothing worthyof proving). By 4.1.9, for x, y ∈ clU there are nets (xγ) and (yγ) in U such thatxγ0 = 0 OO and yγ0 = 0 OO . If (α, β) ∈ � then αxγ + βyγ ∈ U . Appealing to 4.1.19again, infer αx+ βy = lim(αxγ + βyγ) ∈ clU . B

5.3. Comparison Between Topologies

5.3.1. Definition. If M and N are two multinorms on a vector space then M

is said to be finer or stronger than N (in symbols, M � N) if UM ⊃ UN. If M � N

and N � M simultaneously, then M and N are said to be equivalent (in symbols,M ∼ N).


5.3.2. For multinorms M and N on a vector space X the following statementsare equivalent:

(1) M � N;(2) the inclusion τM(x) ⊃ τN(x) holds for all x ∈ X;(3) τM(0) ⊃ τN(0);(4) (∀ q ∈ N) (∃ p1, . . . , pn ∈M) (∃ ε1, . . . , εn ∈ R+ \ 0)

Bq ⊃ ε1Bp1 ∩ . . . ∩ εnBpn ;(5) (∀ q ∈ N) (∃ p1, . . . , pn ∈ M) (∃ t > 0) q ≤ t(p1 ∨ . . . ∨ pn) (with

respect to the order of the K-space RX).C The implications (1) ⇒ (2) ⇒ (3) ⇒ (4) are evident.(4) ⇒ (5): Applying the Gauge Theorem (cf. 5.1.2), infer that

q ≤ pBp1/ε1 ∨ . . . ∨ pBpn/εn =(1/ε1p1

)∨ . . . ∨

(1/εnpn)≤(1/ε1) ∨ . . . ∨ (1/εn) p1 ∨ . . . ∨ pn.

(5) ⇒ (1): It is sufficient to check that M � {q} for a seminorm q in N.If V ∈ Uq then V ⊃ {dq ≤ ε} for some ε > 0. By hypothesis

{dq ≤ ε} ⊃ {dp1 ≤ ε/t} ∩ . . . ∩ {dpn ≤ ε/t}with suitable p1, . . . , pn ∈M and t > 0. The right side of this inclusion is an elementof Up1 ∨ . . . ∨Upn = U{p1,... ,pn} ⊂ UM. Hence, V is also a member of UM. B

5.3.3. Definition. Let p, q : X0 = 0 OO R be two seminorms on X. Say that pis finer or stronger than q and write p � q whenever {p} � {q}. The equivalenceof seminorms p ∼ q is understood in a routine fashion.

5.3.4. p � q ⇔ (∃ t > 0) q ≤ tp⇔ (∃ t ≥ 0) Bq ⊃ tBp;p ∼ q ⇔ (∃ t1, t2 > 0) t2p ≤ q ≤ t1p⇔ (∃ t1, t2 > 0) t1Bp ⊂ Bq ⊂ t2Bp.C Everything follows from 5.3.2 and 5.1.2. B

5.3.5. Riesz Theorem. If p, q : FN0 = 0 OO R are seminorms on the finite-dimensional space FN then p � q ⇔ ker p ⊂ ker q. CB

5.3.6. Corollary. All norms in finite dimensions are equivalent. CB

5.3.7. Let (X, M) and (Y, N) be multinormed spaces, and let T be a linearoperator, a member of L (X, Y ). The following statements are equivalent:

(1) N ◦ T ≺M;(2) T×(UX) ⊃ UY and T×−1(UY ) ⊂ UX ;(3) x ∈ X ⇒ T (τX(x)) ⊃ τY (Tx);(4) T (τX(0)) ⊃ τY (0) and τX(0) ⊃ T−1(τY (0));(5) (∀ q ∈ N) (∃ p1, . . . , pn ∈M) q ◦ T ≺ p1 ∨ . . . ∨ pn. CB

68 Chapter 5

5.3.8. Let (X, ‖ · ‖X) and (Y, ‖ · ‖Y ) be normed spaces and let T be a linearoperator, a member of L (X, Y ). The following statements are equivalent:

(1) T is bounded (that is, T ∈ B(X, Y ));(2) ‖ · ‖X � ‖ · ‖Y ◦ T ;(3) T is uniformly continuous;(4) T is continuous;(5) T is continuous at zero.

C Each of the claims is a particular case of 5.3.7. B

5.3.9. Remark. The message of 5.3.7 shows that it is sometimes convenientto substitute for M a multinorm equivalent to M but filtered upward (with respectto the relation ≥ or �). For example, we may take the multinorm M := {supM0 :M0 is a nonempty finite subset of M}. Observe that unfiltered multinorms shouldbe treated with due precaution.

5.3.10. Counterexample. Let X := F�; and let X0 comprise all constantfunctions; i.e., X0 := F1, where 1 : ξ 7→ 1 (ξ ∈ �). Set M := {pξ : ξ ∈ �},with pξ(x) := |x(ξ)| (x ∈ F�). It is clear that M is a multinorm on X. Now letϕ : X0 = 0 OO /X0 stand for the coset mapping. Undoubtedly, Mϕ−1 consists of thesole element, zero. At the same time Mϕ−1 is separated.

5.3.11. Definition. Let (X, M) be a multinormed space and let X0 be a sub-space of X. The multinorm Mϕ−1 , with ϕ : X0 = 0 OO /X0 the coset mapping,is referred to as the quotient multinorm and is denoted by MX/X0 . The space(X/X0, MX/X0) is called the quotient space of X by X0.

5.3.12. The quotient space X/X0 is separated if and only if X0 is closed. CB

5.4. Metrizable and Normable Spaces

5.4.1. Definition. A multinormed space (X, M) is metrizable if there isa metric d on X such that UM = Ud. Say that X is normable if there is a normon X equivalent to the initial multinorm M. Say that X is countably normableif there is a countable multinorm on X equivalent to the initial.

5.4.2. Metrizability Crtiterion. A multinormed space is metrizable if andonly if it is countably normable and separated.

C ⇒: Let UM = Ud. Passing if necessary to the multinorm M, assume thatfor every n in N it is possible to find a seminorm pn in M and tn > 0 such that{d ≤ 1/n} ⊃ {dpn ≤ tn}. Put N := {pn : n ∈ N}. Clearly, M � N. If V ∈ UM

then V ⊃ {d ≤ 1/n} for some n ∈ N by the definition of metric uniformity. Hence,by construction, V ∈ Upn ⊂ UM, i.e., M ≺ N. Thus, M ∼ N. The uniformity Ud


is separated, as indicated in 4.1.7. Applying 5.2.6, observe that UM and UN areboth separated.

⇐: Passing if necessary to an equivalent multinorm, suppose that X, the spacein question, is countably normed and separated; that is, M := {pn : n ∈ N} and Mis a Hausdorff multinorm on X. Given x1, x2 ∈ X, define

d(x1, x2) :=∞∑k=1

12k

pk(x1 − x2)1 + pk(x1 − x2)

(the series on the right side of the above formula is dominated by the convergentseries

∑∞k=1

1/2k , and so d is defined soundly).Check that d is a metric. It suffices to validate the triangle inequality. For

a start, put α(t) := t(1 + t)−1 (t ∈ R+). It is evident that α′(t) = (1 + t)−2 > 0.Therefore, α increases. Furthermore, α is subadditive:

α(t1 + t2) = (t1 + t2)(1 + t1 + t2)−1

= t1(1 + t1 + t2)−1 + t2(1 + t1 + t2)−1 ≤ t1(1 + t1)−1 + t2(1 + t2)−1

= α(t1) + α(t2).

Thus, given x, y, z ∈ X, infer that

d(x, y) =∞∑k=1

12kα(pk(x− y)) ≤

∞∑k=1

12kα(pk(x− z) + pk(z − y))

≤∞∑k=1

12k

(α(pk(x− z)) + α(pk(z − y))) = d(x, z) + d(z, y).

It remains to established that Ud and UM coincide.First, show that Ud ⊂ UM. Take a cylinder, say, {d ≤ ε}; and let (x, y) ∈

{dp1 ≤ t} ∩ . . . ∩ {dpn ≤ t}. Since α is an increasing function, deduce that

d(x, y) =n∑k=1

12k

pk(x− y)1 + pk(x− y)

+∞∑

k=n+1

12k

pk(x− y)1 + pk(x− y)

≤ t

1 + t

n∑k=1

12k

+∞∑

k=n+1

12k≤ t

1 + t+

12n.

Since t(1 + t)−1 + 2−n tends to zero as n0 = 0 OO and t0 = 0 OO , for appropriate tand n observe that (x, y) ∈ {d ≤ ε}. Hence, {d ≤ ε} ∈ UM, which is required.

70 Chapter 5

Now establish that UM ⊂ Ud. To demonstrate the inclusion, given pn ∈ Mand t > 0, find ε > 0 such that {dpn ≤ t} ⊃ {d ≤ ε}. For this purpose, take

ε :=12n

t

1 + t,

which suffices, since from the relations

12n

pn(x− y)1 + pn(x− y)

≤ d(x, y) ≤ ε = 12n

t

1 + t

holding for all x, y ∈ X it follows that pn(x− y) ≤ t. B

5.4.3. Definition. A subset V of a multinormed space (X, M) is a boundedset in X if sup p(V ) < +∞ for all p ∈M which means that the set p(V ) is boundedabove in R for every seminorm p in M.

5.4.4. For a set V in (X, M) the following statements are equivalent:(1) V is bounded;(2) for every sequence (xn)n∈N in V and every sequence (λn)n∈N in F

such that λn0 = 0 OO , the sequence (λnxn) vanishes: λnxn0 = 0 OO

(i.e., p(λnxn)0 = 0 OO for each seminorm p in M);(3) every neighborhood of zero absorbs V .

C (1) ⇒ (2): p(λnxn) ≤ |λn|p(xn) ≤ |λn| sup p(V )0 = 0 OO .(2)⇒ (3): Let U ∈ τX(0) and suppose that U fails to absorb V . From Definition

3.4.9, it follows that (∀n ∈ N) (∃xn ∈ V ) xn 6∈ nU . Thus, 1/n xn 6∈ U for all n ∈ N;i.e., (1/n xn) does not converge to zero.

(3)⇒ (1): Given p ∈M, find n ∈ N satisfying V ⊂ nBp. Obviously, sup p(V ) ≤sup p(nBp) = n < +∞. B

5.4.5.Kolmogorov Normability Criterion. Amultinormed space is normableif and only if it is separated and has a bounded neighborhood about zero.

C ⇒: It is clear.⇐: Let V be a bounded neighborhood of zero. Without loss of generality,

it may be assumed that V = Bp for some seminorm p in the given multinorm M.Undoubtedly, p ≺ M. Now if U ∈ τM(0) then nU ⊃ V for some n ∈ N. Hence,U ∈ τp(0). Using Theorem 5.3.2, observe that p �M. Thus, p ∼M; and, therefore,p is also separated by 5.2.12. This means that p is a norm. B

5.4.6. Remark. Incidentally, 5.4.5 shows that the presence of a boundedneighborhood of zero in a multinormed space X amounts to the “seminormabil-ity” of X.


5.5. Banach Spaces

5.5.1. Definition. A Banach space is a complete normed space.5.5.2. Remark. The concept of Frechet space, complete metrizable multi-

normed space, serves as a natural abstraction of Banach space. It may be shownthat the class of Frechet spaces is the least among those containing all Banachspaces and closed under the taking of countable products. CB

5.5.3. A normed space X is a Banach space if and only if every norm conver-gent (= absolutely convergent) series in X converges.

C ⇒: Let∑∞n=1 ‖xn‖ < +∞ for some (countable) sequence (xn). Then the

sequence of partial sums sn := x1 + · · ·+ xn is fundamental because

‖sm − sk‖ = ‖m∑

n=k+1

xn‖ ≤m∑

n=k+1

‖xn‖0 = 0 OO

for m > k.⇐: Given a fundamental sequence (xn), choose an increasing sequence (nk)k∈N

such that ‖xn−xm‖≤ 2−k as n, m ≥ nk. Then the series xn1 +(xn2−xn1)+(xn3−xn2) + · · · converges in norm to some x; i.e., xnk0 = 0 OO . Observe simultaneouslythat xn0 = 0 OO . B

5.5.4. If X is a Banach space and X0 is a closed subspace of X then thequotient space X/X0 is also a Banach space.C Let ϕ : X0 = 0 OO := X/X0 be the coset mapping. Undoubtedly, for every

x ∈ there is some x ∈ ϕ−1(x) such that 2‖x‖ ≥ ‖x‖ ≥ ‖x‖. Hence, given∑∞n=1 xn,

a norm convergent series in , it is possible to choose xn ∈ ϕ−1(xn) so that thenorm series

∑∞n=1 ‖xn‖ be convergent. According to 5.5.3, the sum x :=

∑∞n=1 xn

is available. If x := ϕ(x) then

‖x−n∑k=1

xk‖ ≤ ‖x−n∑k=1

xk‖0 = 0 OO .

Appealing to 5.5.3 again, conclude that is a Banach space. B

5.5.5.Remark. The claim of 5.5.3 may be naturally translated to seminormedspaces. In particular, if (X, p) is a complete seminormed space then the quotientspace X/ ker p is a Banach space. CB

5.5.6. Theorem. If X and Y are normed spaces and X 6= 0 then B(X, Y ),the space of bounded operators, is a Banach space if and only if so is Y .

72 Chapter 5

C⇐: Consider a Cauchy sequence (Tn) in B(X, Y ). By the normative inequal-ity, ‖Tmx− Tkx‖ ≤ ‖Tm − Tk‖ ‖x‖0 = 0 OO for all x ∈ X; i.e., (Tnx) is fundamentalin Y . Thus, there is a limit Tx := limTnx. Plainly, the so-defined operator T islinear. By virtue of the estimate |‖Tm‖ − ‖Tk‖| ≤ ‖Tm−Tk‖ the sequence (‖Tn‖) isfundamental in R and, hence, bounded; that is, supn ‖Tn‖ < +∞. Therefore, pass-ing to the limit in ‖Tnx‖ ≤ supn ‖Tn‖ ‖x‖, obtain ‖T‖ < +∞. It remains to checkthat ‖Tn−T‖0 = 0 OO . Given ε > 0, choose a number n0 such that ‖Tm−Tn‖ ≤ ε/2as m, n ≥ n0. Further, for x ∈ BX find m ≥ n0 satisfying ‖Tmx − Tx‖ ≤ ε/2.Then ‖Tnx− Tx‖ ≤ ‖Tnx− Tmx‖+ ‖Tmx− Tx‖ ≤ ‖Tn − Tm‖+ ‖Tmx− Tx‖ ≤ εas n ≥ n0. In other words, ‖Tn − T‖ = sup{‖Tnx − Tx‖ : x ∈ BX} ≤ ε for allsufficiently large n.

⇒: Let (yn) be a Cauchy sequence in Y . By hypothesis, there is a norm-oneelement x in X; i.e., x has norm 1: ‖x‖ = 1. Applying 3.5.6 and 3.5.2 (1), findan element x′ ∈ |∂ |(‖ · ‖) satisfying (x, x′) = ‖x‖ = 1. Obviously, the rank-one operator (with range of dimension 1) Tn := x′ ⊗ yn : x 7→ (x, x′)yn belongsto B(X, Y ), since ‖Tn‖ = ‖x′‖ ‖yn‖. Hence, ‖Tm − Tk‖ = ‖x′ ⊗ (ym − yk)‖=‖x′‖ ‖ym − yk‖ = ‖ym − yk‖, i.e., (Tn) is fundamental in B(X, Y ). Assign T :=limTn. Then ‖Tx− Tnx‖ = ‖Tx− yn‖ ≤ ‖T − Tn‖ ‖x‖0 = 0 OO . In other words, Txis the limit of (yn) in Y . B

5.5.7. Corollary. The dual of a normed space (furnished with the dual norm)is a Banach space. CB

5.5.8. Corollary. Let X be a normed space; and let ι : X0 = 0 OO ′′, the doubleprime mapping, be the canonical embedding of X into the second dual X ′′. Thenthe closure cl ι(X) is a completion of X.

C By virtue of 5.5.7, X ′′ is a Banach space. By 5.1.10 (8), ι is an isometryfrom X into X ′′. Appealing to 4.5.16 ends the proof. B

5.5.9. Examples.(1) “Abstract” examples: a basic field, a closed subspace of a Banach

space, the product of Banach spaces, and 5.5.4–5.5.8.(2) Let E be a nonempty set. Given x ∈ F E , put ‖x‖∞ := sup |x(E )|.

The space l∞(E ) := l∞(E , F) := dom ‖ · ‖∞ is called the space of bounded functionson E . The designations B(E ) and B(E , F) are also used. For E := N, it iscustomary to put m := l∞ := l∞(E ).

(3) Let a set E be infinite, i.e. not finite, and let F stand for a filteron E . By definition, x ∈ c(E , F )⇔ (x ∈ l∞(E ) and x(F ) is a Cauchy filter on F).In the case E := N and F is the finite complement filter (comprising all cofinite setseach of which is the complement of a finite subset) of N, the notation c := c(E , F )is employed, and c is called the space of convergent sequences. In c(E , F ) the


subspace c0(E , F ) := {x ∈ c(E , F ) : x(F )0 = 0 OO } is distinguished. If F isthe finite complement filter then the shorter notation c0(E ) is used and we speakof the space of functions vanishing at infinity. Given E := N, write c0 := c0(E ).The space c0 is referred to as the space of vanishing sequences. It is worth keepingin mind that each of these spaces without further specification is endowed with thenorm taken from the corresponding space l∞(E , F ).

(4) Let S := (E , X,∫) be a system with integration. This means that

X is a vector sublattice of RE , with the lattice operations in X coincident withthose in RE , and

∫: X0 = 0 OO R is a (pre)integral; i.e.,

∫∈ X#

+ and∫xn ↓ 0

whenever xn ∈ X and xn(e) ↓ 0 for e ∈ E . Moreover, let f ∈ F E be a measurablemapping (with respect to S) (as usual, we may speak of almost everywhere finiteand almost everywhere defined measurable functions).

Denote Np(f) := (∫|f |p)1/p for p ≥ 1, where

∫is the corresponding Lebesgue

extension of the initial integral∫. (The traditional liberty is taken of using the

same symbol for the original and its successor.)An element of domN1 is an integrable or summable function. The integrability

of f ∈ F E is equivalent to the integrability of its real part Re f and imaginary partIm f , both members of RE . For the sake of completeness, recall the definition

N(g) := inf{sup

∫xn : (xn) ⊂ X, xn ≤ xn+1, (∀ e ∈ E ) |g(e)| = lim

nxn(e)

}for an arbitrary g in F E . If F = R then domN1 obviously presents the closure of Xin the normed space (domN, N).

The Holder inequality is valid:

N1(fg) ≤ Np(f)Np′(g),

with p′ the conjugate exponent of p, i.e. 1/p + 1/p′ = 1 for p > 1.C This is a consequence of the Young inequality, xy − xp/p ≤ yp

′

/p′ for allx, y ∈ R+, applied to |f |/Np(f) and |g|/Np′(g) when Np(f) and Np′(g) are bothnonzero. If Np(f)Np′(g) = 0 then the Holder inequality is beyond a doubt. B

The set Lp := domNp is a vector space.

C |f + g|p ≤ (|f |+ |g|)p ≤ 2p(|f | ∨ |g|)p = 2p(|f |p ∨ |g|p) ≤ 2p(|f |p + |g|p) .

The function Np is a seminorm, satisfying the Minkowski inequality:

Np(f + g) ≤ Np(f) + Np(g).

74 Chapter 5

C For p = 1, this is trivial. For p > 1 the Minkowski inequality follows fromthe presentation

Np(f) = sup{N1(fg)/Np′(g) : 0 < Np′(g) < +∞} (f ∈ Lp)

whose right side is the upper envelope of a family of seminorms. To prove the abovepresentation, using the Holder inequality, note that g := |f |

p/p′ lies in Lq whenNp(f) > 0; furthermore, Np(f) = N1(fg)/Np′(g). Indeed, N1(fg) =

∫|f |

p/p′+1 =Np(f)p, because p/p′+1 = p

(1− 1/p

)+1 = p. Continue arguing to find Np′(g)p

′=∫

|g|p′ =∫|f |p = Np(f)p, and so Np′(g) = Np(f)

p/p′. Finally,

N1(fg)/Np′(g) = Np(f)p/Np(f)p/p′ = Np(f)

p−p/p′ = Np(f)p(1−1/p′) = Np(f). .

The quotient space Lp/ kerNp, together with the corresponding quotient norm‖ · ‖p, is called the space of p-summable functions or Lp space with more completedesignations Lp(S), Lp(E , X,

∫), etc.

Finally, if a system with integration S arises from inspection of measurable stepfunctions on a measure space (�, A , µ) then it is customary to write Lp(�, A , µ),Lp(�, µ) and even Lp(µ), with the unspecified parameters clear from the context.

Riesz–Fisher Completeness Theorem. Each Lp space is a Banach space.C We sketch out the proof. Consider t :=

∑∞k=1 Np(fk), where fk ∈ Lp.

Put σn :=∑nk=1 fk and sn :=

∑nk=1 |fk|. It is seen that (sn) has positive entries

and increases. The same is true of (spn). Furthermore,∫spn ≤ tp < +∞. Hence,

by the Levy Monotone Convergence Theorem, for almost all e ∈ E there existsa limit g(e) := lim spn(e), with the resulting function g a member of L1. Puttingh(e) := g

1/p(e), observe that h ∈ Lp and sn(e)0 = 0 OO (e) for almost all e ∈ E . Theinequalities |σn| ≤ sn ≤ h imply that for almost all e ∈ E the series

∑∞k=1 fk(e)

converges. For the sum f0(e) the estimate holds: |f0(e)| ≤ h(e). Hence, it maybe assumed that f0 ∈ Lp. Appealing to the Lebesgue Dominated Convergence

Theorem, conclude that Np(σn − f0) =(∫|σn − f0|p

)1/p 0 = 0 OO . Thus, in theseminormed space under consideration each seminorm convergent series converges.To complete the proof, apply 5.5.3–5.5.5. B

If S is the system of conventional summation on E ; i.e., X :=∑e∈E R is the

direct sum of suitably many copies of the ground field R and∫x :=

∑e∈E x(e), then

Lp comprises all p-summable families. This space is denoted by lp(E ). Further,

‖x‖p :=(∑

e∈E |x(e)|p)1/p . In the case E := N the notation lp is used and lp is

referred to as the space of p-summable sequences.(5) Define L∞ as follows: Let X be an ordered vector space and let

e ∈ X+ be a positive element. The seminorm pe associated with e is the Minkowskifunctional of the order interval [−e, e], i.e.,

pe(x) := inf{t > 0 : −te ≤ x ≤ te}.


The effective domain of definition of pe is the space of bounded elements (withrespect to e); the element e itself is referred to as the strong order-unit in Xe.An element of ker pe is said to be nonarchimedean (with respect to e). The quotientspace Xe/ ker pe furnished with the corresponding quotient seminorm is called thenormed space of bounded elements (generated by e in X). For example, C(Q, R),the space of continuous real-valued functions on a nonempty compact setQ, presentsthe normed space of bounded elements with respect to 1 := 1Q : q 7→ 1 (q ∈ Q)(in itself). In RE the same element 1 generates the space l∞(E ).

Given a system with integration S :=(E , X,

∫ ), assume that 1 is measurable

and consider the space of functions acting from E into F and satisfying

N∞(f) := inf{t > 0 : |f | ≤ t1} < +∞,

where ≤ means “less almost everywhere than.” This space is called the space of es-sentially bounded functions and is labelled with L∞. To denote the quotient spaceL∞/ kerN∞ and its norm the symbols L∞ and ‖ · ‖∞ are in use.

It is in common parlance to call the elements of L∞ (like the elements of L∞)essentially bounded functions. The space L∞ presents a Banach space. CB

The space L∞, as well as the spaces C(Q, F), lp(E ), c0(E ), c, lp, and Lp (p ≥1), also bears the unifying title “classical Banach space.” Nowadays a Lindenstraussspace which is a space whose dual is isometric to L1 (with respect to some systemwith integration) is also regarded as classical. It can be shown that a Banachspace X is classical if and only if the dual X ′ is isomorphic to one of the Lp spaceswith p ≥ 1.

(6) Consider a system with integration S :=(E , X,

∫ )and let p ≥ 1.

Suppose that for every e in E there is a Banach space (Ye, ‖ · ‖Ye). Given an arbi-trary element f in

∏e∈E Ye, define |||f ||| : e 7→ ‖f(e)‖Ye . Put Np(f) := inf{Np(g) :

g ∈ Lp, g ≥ |||f |||}. It is clear that domNp is a vector space equipped with theseminorm Np. The sum of the family in the sense of Lp or simply the p-sumof (Ye)e∈E (with respect to the system with integration S) is the quotient spacedomNp/ kerNp under the corresponding (quotient) norm |||·|||p.

The p-sum of a family of Banach spaces is a Banach space.C If

∑∞k=1Np(fk) < +∞ then the sequence (sn :=

∑nk=1|||fk|||) tends to some

almost everywhere finite positive function g and Np(g) < +∞. It follows that foralmost all e ∈ E the sequence (sn(e)) (i.e., the series

∑∞k=1 ‖fk(e)‖Ye) converges.

By the completeness of Ye, the series∑∞k=1 fk(e) converges to some sum f0(e) in Ye

with ‖f0(e)‖Ye ≤ g(e) for almost every e ∈ E . Therefore, it may be assumed thatf0 ∈ domNp. Finally, Np (

∑nk=1 fk − f0) ≤

∑∞k=n+1Np(fk)0 = 0 OO . .

In the case when E := N with conventional summation, for the sum Y of a se-quence of Banach spaces (Yn)n∈N (in the sense of Lp) the following notation is oftenemployed:

Y := (Y1 ⊕ Y2 ⊕ · · · )p,

76 Chapter 5

with p the type of summation. An element y in Y presents a sequence (yn)n∈N suchthat yn ∈ Yn and

|||y|||p :=

( ∞∑k=1

‖yn‖pYn

)1/p

< +∞.

In the case Ye := X for all e ∈ E , where X is some Banach space over F, put Fp :=domNp and Fp := Fp/ kerNp. An element of the so-constructed space is a vectorfield or a X-valued function on E (having a p-summable norm). Undoubtedly, Fpis a Banach space. At the same time, if the initial system with integration containsa nonmeasurable set then extraordinary elements are plentiful in Fp (in particular,for the usual Lebesgue system with integration Fp 6= Lp). In this connection thefunctions with finite range, assuming each value on a measurable set, are selectedin Fp. Such a function, as well as the corresponding coset in Fp, is a simple, finite-valued or step function. The closure in Fp of the set of simple functions is denotedby Lp (or more completely Lp(X), Lp(S, X), Lp(�, A , µ), Lp(�, µ), etc.) and isthe space of X-valued p-summable functions. Evidently, Lp(X) is a Banach space.

It is in order to illustrate one of the advantages of these spaces for p = 1. First,notice that a simple function f can be written as a finite combination of character-istic functions:

f =∑x∈imf

χf−1(x)x,

where f−1(x) is a measurable set as x ∈ im f , with χE(e) = 1 for e ∈ E andχE(e) = 0 otherwise. Moreover,∫

|||f ||| =∫ ∑

x∈imf

‖χf−1(x)x‖

=∫ ∑

x∈imf

χf−1(x)‖x‖ =∑x∈imf

‖x‖∫χf−1(x) < +∞.

Next, associate with each simple function f some element in X by the rule∫f :=

∑x∈imf

∫χf−1(x)x.

Straightforward calculation shows that the integral∫defined on the subspace of sim-

ple functions is linear. Furthermore, it is bounded because

‖∫f‖ = ‖

∑x∈imf

∫χf−1(x)x‖ ≤

∑x∈imf

∫χf−1(x)‖x‖

=∫ ∑

x∈imf

‖x‖χf−1(x) =∫|||f |||.


By virtue of 4.5.10 and 5.3.8, the operator∫has a unique extension to an element

of B(L1(X), X). This element is denoted by the same symbol,∫(or

∫E , etc.), and

is referred to as the Bochner integral.(7) In the case of conventional summation, the usage of the scalar the-

ory is preserved for the Bochner integral. Namely, the common parlance favoursthe term “sum of a family” rather than “integral of a summable function,” and thesymbols pertinent to summation are perfectly welcome. What is more important,infinite dimensions bring about significant complications.

Let (xn) be a family of elements of a Banach space. Its summability (in thesense of the Bochner integral) means the summability of the numeric family (‖xn‖),i.e. the norm convergence of (xn) as a series. Consequently, (xn) has at mostcountably many nonzero elements and may thus be treated as a (countable) se-quence. Moreover,

∑∞n=1 ‖xn‖ < +∞; i.e. the series x1 + x2 + · · · converges in

norm. By 5.5.3, for the series sum x =∑∞n=1 xn observe that x = limθ sθ, where

sθ :=∑n∈θ xn is a partial sum and θ ranges over the direction of all finite sub-

sets of N. In this case, the resulting x is sometimes called the unordered sum of(xn), whereas the sequence (xn) is called unconditionally or unorderly summableto x (in symbols, x =

∑n∈N xn). Using these terms, observe that summability in

norm implies unconditional summability (to the same sum). If dimX < +∞ thenthe converse holds which is the Riemann Theorem on Series. The general case isexplained by the following deep and profound assertion:

Dvoretzky–Rogers Theorem. In an arbitrary infinite-dimensional Banachspace X, for every sequence of positive numbers (tn) such that

∑∞n=1 t

2n < +∞

there is an unconditionally summable sequence of elements (xn) with ‖xn‖ = tn forall n ∈ N.

In this regard for a family of elements (xe)e∈E of an arbitrary multinormedspace (X, M) the following terminology is accepted: Say that (xe)e∈E is summableor unconditionally summable (to a sum x) and write x :=

∑e∈E xe whenever x is

the limit in (X, M) of the corresponding net of partial sums sθ, with θ a finitesubset of E ; i.e., sθ0 = 0 OO in (X, M). If for every p there is a sum

∑e∈E p(xe)

then the family (xe)e∈E is said to be multinorm summable (or, what is more exact,fundamentally summable, or even absolutely fundamental).

In conclusion, consider a Banach space Y and T ∈ B(X, Y). The operator Tcan be uniquely extended to an operator from L1(X) into L1(Y) by putting Tf :e 7→ Tf(e) (e ∈ E ) for an arbitrary simple X-valued function f . Then, givenf ∈ L1(X), observe that Tf ∈ L1(Y) and

∫E Tf = T

∫E f . This fact is verbalized

as follows: “The Bochner integral commutes with every bounded operator.” CB

5.6. The Algebra of Bounded Operators

5.6.1. Let X, Y , and Z be normed spaces. If T ∈ L (X, Y ) and S ∈ L (Y, Z)

78 Chapter 5

are linear operators then ‖ST‖ ≤ ‖S‖ ‖T‖; i.e., the operator norm is submultiplica-tive.

C Given x ∈ X and using the normative inequality twice, infer that

‖STx‖ ≤ ‖S‖ ‖Tx‖ ≤ ‖S‖ ‖T‖ ‖x‖. .

5.6.2. Remark. In algebra, in particular, (associative) algebras are studied.An algebra (over F) is a vector space A (over F) together with some associativemultiplication ◦ : (a, b) 7→ ab (a, b ∈ A). This multiplication must be distributivewith respect to addition (i.e., (A, +, ◦) is an (associative) ring) and, moreover, theoperation ◦ must agree with scalar multiplication in the following sense: λ(ab) =(λa)b = a(λb) for all a, b ∈ A and λ ∈ F. A displayed notation for an algebra is(A, F, +, · , ◦). However, just like on the other occasions it is customary to usethe term “algebra” simply for A.

5.6.3. Definition. A normed algebra (over a ground field) is an associativealgebra (over this field) together with a submultiplicative norm. A Banach algebrais a complete normed algebra.

5.6.4. Let B(X) := B(X, X) be the space of bounded endomorphisms ofa normed space X. The space B(X) with the operator norm and compositionas multiplication presents a normed algebra. If X 6= 0 then B(X) has a neutralelement (with respect to multiplication), the identity operator IX ; i.e., B(X) isan algebra with unity. Moreover, ‖IX‖ = 1. The algebra B(X) is a Banach algebraif and only if X is a Banach space.

C If X = 0 then there is nothing left to proof. Given X 6= 0, apply 5.5.6. B

5.6.5. Remark. It is usual to refer to B(X) as the algebra of bounded opera-tors in X or even as the (bounded) endomorphism algebra of X. In connection with5.6.4, given λ ∈ F, it is convenient to retain the same symbol λ for λIX . (In partic-ular, 1 = I0 = 0!) For X 6= 0 this procedure may be thought as identification of Fwith FIX .

5.6.6. Definition. If X is a normed space and T ∈ B(X) then the spectralradius of T is the number r(T ) := inf

{‖Tn‖1/n : n ∈ N

}. (The rationale of this

term will transpire later (cf. 8.1.12).)

5.6.7. The norm of T is greater than the spectral radius of T .C Indeed, by 5.6.1, the inequality ‖Tn‖ ≤ ‖T‖n is valid. B

5.6.8. The Gelfand formula holds:

r(T ) = lim n√‖Tn‖.


C Take ε > 0 and let s ∈ N satisfy ‖T s‖1/s ≤ r(T )+ε. Given n ∈ N with n ≥ s,observe the presentation n = k(n)s+ l(n) with k(n), l(n) ∈ N and 0 ≤ l(n) ≤ s−1.Hence,

‖Tn‖ = ‖T k(n)sT l(n)‖ ≤ ‖T s‖k(n) ‖T l(n)‖≤(1 ∨ ‖T‖ ∨ . . . ∨ ‖T s−1‖

)‖T s‖k(n) =M ‖T s‖k(n).

Consequently,

r(T ) ≤ ‖Tn‖1/n ≤M

1/n‖T s‖k(n)/n

≤M1/n(r(T ) + ε)

k(n)s/n =M1/n(r(T ) + ε)

(n−l(n))/n .

Since M1/n0 = 0 OO and (n−l(n))/n0 = 0 OO , find r(T ) ≤ lim sup‖Tn‖1/n ≤ r(T ) + ε.

The inequality lim inf ‖Tn‖1/n ≥ r(T ) is evident. Recall that ε is arbitrary, thuscompleting the proof. B

5.6.9. Neumann Series Expansion Theorem. With X a Banach spaceand T ∈ B(X), the following statements are equivalent:

(1) the Neumann series 1+T+T 2+· · · converges in the operator normof B(X);

(2) ‖T k‖ < 1 for some k and N;

(3) r(T ) < 1.If one of the conditions (1)–(3) holds then

∑∞k=0 T

k = (1− T )−1.

C (1)⇒ (2): With the Neumann series convergent, the general term (T k) tendsto zero.

(2) ⇒ (3): This is evident.(3) ⇒ (1): According to 5.6.8, given a suitable ε > 0 and a sufficiently large

k ∈ N, observe that r(T ) ≤ ‖T k‖1/k ≤ r(T ) + ε < 1. In other words, some tailof the series

∑∞k=0 ‖T k‖ is dominated by a convergent series. The completeness

of B(X) and 5.5.3 imply that∑∞k=0 T

k converges in B(X).

Now let S :=∑∞k=0 T

k and Sn :=∑nk=0 T

k. Then

S(1− T ) = limSn(1− T ) = lim (1 + T + · · ·+ Tn) (1− T ) = lim(1− Tn+1) = 1;(1− T )S = lim(1− T )Sn = lim(1− T )(1 + T + · · ·+ Tn) = lim

(1− Tn+1) = 1,

because Tn0 = 0 OO . Thus, by 2.2.7 S = (1− T )−1. B

80 Chapter 5

5.6.10. Corollary. If ‖T‖ < 1 then (1 − T ) is invertible (= has a boundedinverse); i.e., the inverse correspondence (1 − T )−1is a bounded linear operator.Moreover, ‖(1− T )−1‖ ≤ (1− ‖T‖)−1.

C The Neumann series converges and

‖(1− T )−1‖ ≤∞∑k=0

‖T k‖ ≤∞∑k=0

‖T‖k = (1− ‖T‖)−1. .

5.6.11. Corollary. If ‖1− T‖ < 1 then T is invertible and

‖1− T−1‖ ≤ ‖1− T‖1− ‖1− T‖

.

C By Theorem 5.6.9,

1 +∞∑k=1

(1− T )k =∞∑k=0

(1− T )k = (1− (1− T ))−1 = T−1.

Hence,

‖T−1 − 1‖ = ‖∞∑k=1

(1− T )k‖ ≤∞∑k=1

‖(1− T )k‖ ≤∞∑k=1

‖1− T‖k. .

5.6.12. Banach Inversion Stability Theorem. Let X and Y be Banachspaces. The set of invertible operators Inv(X, Y ) is open. Moreover, the inversionT 7→ T−1 acting from Inv(X, Y ) to Inv(Y, X) is continuous.C Take S, T ∈ B(X, Y ) such that T−1 ∈ B(Y, X) and ‖T−1‖ ‖S−T‖ ≤ 1/2.

Consider the operator T−1S in B(X). Observe that‖1− T−1S‖ = ‖T−1T − T−1S‖ ≤ ‖T−1‖ ‖T − S‖ ≤ 1/2 < 1.

Hence, by 5.6.11, (T−1S)−1 belongs to B(X). Put R := (T−1S)−1T−1. Clearly,R ∈ B(Y, X) and, moreover, R = S−1(T−1)−1T−1 = S−1. Further,

‖S−1‖ − ‖T−1‖ ≤ ‖S−1 − T−1‖= ‖S−1(T − S)T−1‖ ≤ ‖S−1‖ ‖T − S‖ ‖T−1‖ ≤ 1/2‖S−1‖.

This implies ‖S−1‖ ≤ 2‖T−1‖, yielding the inequalities‖S−1 − T−1‖ ≤ ‖S−1‖ ‖T − S‖ ‖T−1‖ ≤ 2‖T−1‖2‖T − S‖. .

5.6.13.Definition. IfX is a Banach space over F and T ∈ B(X) then a scalarλ ∈ F is a regular or resolvent value of T whenever (λ− T )−1 ∈ B(X). In this caseput R(T, λ) := (λ− T )−1 and say that R(T, λ) is the resolvent of T at λ. The setof the resolvent values of T is denoted by res(T ) and called the resolvent set of T .The mapping λ 7→ R(T, λ) from res(T ) into B(X) is naturally called the resolventof T . The set F\ res(T ) is referred to as the spectrum of T and is denoted by Sp(T )or σ(T ). A member of Sp(T ) is said to be a spectral value of T (which is enigmaticfor the time being).


5.6.14. Remark. If X = 0 then the spectrum of the only operator T = 0in B(X) is the empty set. In this regard, in spectral analysis it is silently presumedthat X 6= 0. In the case X 6= 0 for F := R the spectra of some operators can alsobe void, whereas for F := C it is impossible (cf. 8.1.11). CB

5.6.15. The set res(T ) is open. If λ0 ∈ res(T ) then

R(T, λ) =∞∑k=0

(−1)k(λ− λ0)kR(T, λ0)k+1

in some neighborhood of λ0. If |λ| > ‖T‖ then λ ∈ res(T ) and the expansion

R(T, λ) =1λ

∞∑k=0

T k

λk

holds. Moreover, ‖R(T, λ)‖0 = 0 OO as |λ|0 = 0 OO ∞.C Since ‖(λ−T )−(λ0−T )‖ = |λ−λ0|, the openness property of res(T ) follows

from 5.6.12. Proceed along the lines

λ− T = (λ− λ0) + (λ0 − T ) = (λ0 − T )R(T, λ0)(λ− λ0) + (λ0 − T )= (λ0 − T )((λ− λ0)R(T, λ0) + 1) = (λ0 − T )(1− ((−1)(λ− λ0)R(T, λ0))).

In a suitable neighborhood of λ0, from 5.6.9 derive

R(T, λ) = (λ− T )−1

= (1− ((−1)(λ− λ0)R(T, λ0)))−1(λ0 − T )−1 =∞∑k=0

(−1)k(λ− λ0)kR(T, λ0)k+1.

According to 5.6.9, for |λ| > ‖T‖ there is an operator(1− T /λ

)−1 presentingthe sum of the Neumann series; i.e.,

R(T, λ) =1λ

(1− T /λ

)−1=

1λ

∞∑k=0

T k

λk.

It is clear that‖R(T, λ)‖ ≤ 1

|λ|· 11− ‖T‖/|λ|

. .

5.6.16. The spectrum of every operator is compact. CB

82 Chapter 5

5.6.17. Remark. It is worth keeping in mind that the inequality |λ| > r(T )is a necessary and sufficient condition for the convergence of the Laurent series,R(T, λ) =

∑∞k=0 T

k/λk+1, which expands the resolvent of T at infinity.

5.6.18. An operator S commutes with an operator T if and only if S commuteswith the resolvent of T .

C ⇒: ST = TS ⇒ S(λ − T ) = λS − ST = λS − TS = (λ − T )S ⇒R(T, λ)S(λ− T ) = S ⇒ R(T, λ)S = S R(T, λ) (λ ∈ res(T )).⇐: SR(T, λ0) = R(T, λ0)S ⇒ S = R(T, λ0)S(λ0 − T ) ⇒ (λ0 − T )S =

S(λ0 − T )⇒ TS = ST . B

5.6.19. If λ, µ ∈ res(T ) then the first resolvent equation, the Hilbert identity,holds:

R(T, λ)−R(T, µ) = (µ− λ)R(T, µ)R(T, λ).

C “Multiplying the equality µ − λ = (µ − T ) − (λ − T ), first, by R(T, λ)from the right and, second, by R(T, µ) from the left,” successively infer the soughtidentity. B

5.6.20. If λ, µ ∈ res(T ) then R(T, λ)R(T, µ) = R(T, µ)R(T, λ). CB

5.6.21. For λ ∈ res(T ) the equality holds:

dk

dλkR(T, λ) = (−1)kk!R(T, λ)k+1. /.

5.6.22. Composition Spectrum Theorem. The spectra Sp(ST ) andSp(TS) may differ only by zero.

C It suffices to establish that 1 6∈ Sp(ST ) ⇒ 1 6∈ Sp(TS). Indeed, fromλ 6∈ Sp(ST ) and λ 6= 0 it will follow that

1 6∈ 1/λ Sp(ST )⇒ 1 6∈ Sp(1/λST )⇒ 1 6∈ Sp

(1/λTS)⇒ λ 6∈ Sp(TS).

Therefore, consider the case 1 6∈ Sp(ST ). The formal Neumann series expan-sions

(1− ST )−1 ∼ 1 + ST + (ST )(ST ) + (ST )(ST )(ST ) + · · · ,T (1− ST )−1S ∼ TS + TSTS + TSTSTS + · · · ∼ (1− TS)−1 − 1

lead to conjecturing that the presentation is valid:

(1− TS)−1 = 1 + T (1− ST )−1S


(which in turn means 1 6∈ Sp(TS)). Straightforward calculation demonstrates theabove presentation, thus completing the entire proof:

(1 + T (1− ST )−1S)(1− TS) = 1 + T (1− ST )−1S − TS + T (1− ST )−1(−ST )S= 1 + T (1− ST )−1S − TS + T (1− ST )−1(1− ST − 1)S= 1 + T (1− ST )−1S − TS + TS − T (1− ST )−1S = 1;

(1− TS)(1 + T (1− ST )−1S) = 1− TS + T (1− ST )−1S + T (−ST )(1− ST )−1S

= 1− TS + T (1− ST )−1S + T (1− ST − 1)(1− ST )−1S

= 1− TS + T (1− ST )−1S + TS − T (1− ST )−1S = 1. .

Exercises

5.1. Prove that a normed space is finite-dimensional if and only if every linear functionalon the space is bounded.

5.2. Demonstrate that it is possible to introduce a norm into each vector space.5.3. Show that a vector space X is finite-dimensional if and only if all norms on X are

equivalent to each other.

5.4. Demonstrate that all separated multinorms introduce the same topology in a finite-dimensional space.

5.5. Each norm on RN is appropriate for norming the product of finitely many normedspaces, isn’t it?

5.6. Find conditions for continuity of an operator acting between multinormed spaces andhaving finite-dimensional range.

5.7. Describe the operator norms in the space of square matrices. When are such normscomparable?

5.8. Calculate the distance between hyperplanes in a normed space.5.9. Find the general form of a continuous linear functional on a classical Banach space.5.10. Study the question of reflexivity for classical Banach spaces.5.11. Find the mutual disposition of the spaces lp and lp′ as well as Lp and Lp′ . When is

the complement of one element of every pair is dense in the other?

5.12. Find the spectrum and resolvent of the Volterra operator (the taking of a primitive),a projection, and a rank-one operator.

5.13. Construct an operator whose spectrum is a prescribed nonempty compact set in C.5.14. Prove that the identity operator (in a nonzero space) is never the commutator of any

pair of operators.

5.15. Is it possible to define some reasonable spectrum for an operator in a multinormedspace?

5.16. Does every Banach space over F admit an isometric embedding into the space C(Q, F),with Q a compact space?

84 Chapter 5

5.17. Find out when Lp(X)′ = Lp′ (X′), with X a Banach space.5.18. Let (Xn) be a sequence of normed spaces and let

X0 =

{x ∈

∏n∈N

Xn : ‖xn‖0 = 0 OO}

be their c0-sum (with the norm ‖x‖ = sup{ ‖xn‖ : n ∈ N} induced from the l∞-sum). Prove thatX0 is separable if and only if so is each of the spaces Xn.

5.19. Prove that the space C(p)[0, 1] presents the sum of a finite-dimensional subspace anda space isomorphic to C[0, 1].

Chapter 6

Hilbert Spaces

6.1. Hermitian Forms and Inner Products

1. Definition. Let H be a vector space over a basic field F. A mappingf : H20 = 0 OO F is a hermitian form on H provided that

(1) the mapping f( · , y) : x 7→ f(x, y) belongs to H# for every y in Y ;(2) f(x, y) = f(y, x)∗ for all x, y ∈ H, where λ 7→ λ∗ is the natural

involution in F; that is, the taking of the complex conjugate of a complex number.

2. Remark. It is easy to see that, for a hermitian form f and each x in H themapping f(x, · ) : y 7→ (x, y) lies in H#

∗ , where H∗ is the twin of H (see 2.1.4 (2)).Consequently, in case F := R every hermitian form is bilinear, i.e., linear in eachargument; and in case F := C, sesquilinear, i.e., linear in the first argument and∗-linear in the second.

3. Every hermitian form f satisfies the polarization identity:

f(x+ y, x+ y)− f(x− y, x− y) = 4Re f(x, y) (x, y ∈ H).

C f(x+ y, x+ y) = f(x, x) + f(x, y) + f(y, x) + f(y, y)−f(x− y, x− y) = f(x, x)− f(x, y)− f(y, x) + f(y, y)

2(f(x, y) + f(y, x)) B

4.Definition. A hermitian form f is positive or positive semidefinite providedthat f(x, x) ≥ 0 for all x ∈ H. In this event, write (x, y) := 〈x | y〉 := f(x, y) (x, y ∈H). A positive hermitian form is usually referred to as a semi-inner product onH. A semi-inner product on H is an inner product or a (positive definite) scalarproduct whenever (x, x) = 0⇒ x = 0 with x ∈ H.

5. The Cauchy–Bunyakovskiı–Schwarz inequality holds:

|(x, y)|2 ≤ (x, x)(y, y) (x, y ∈ H).

86 Chapter 6

C If (x, x) = (y, y) = 0 then 0 ≤ (x+ty, x+ty) = t(x, y)∗+t∗(x, y). Lettingt := −(x, y), find that −2|(x, y)|2 ≥ 0; i.e., in this case the claim is established.

If, for definiteness, (y, y) 6= 0; then in view of the estimate

0 ≤ (x+ ty, x+ ty) = (x, x) + 2tRe(x, y) + t2(y, y) (t ∈ R)

conclude that Re(x, y)2 ≤ (x, x)(y, y).If (x, y) = 0 then nothing is left to proof. If (x, y) 6= 0 then let θ :=

|(x, y)| (x, y)−1 and x := θx. Now |θ| = 1 and, furthermore,

(x, x) = (θx, θx) = θθ∗(x, x) = |θ|2(x, x) = (x, x);|(x, y)| = θ(x, y) = (θx, y) = (x, y) = Re(x, y).

Consequently, |(x, y)|2 = Re(x, y)2 ≤ (x, x)(y, y). .

6. If ( · , · ) is a semi-inner product on H then the mapping ‖·‖ : x 7→ (x, x)1/2

is a seminorm on H.C It suffices to prove the triangle inequality. Applying the Cauchy–Bunyakov-

skiı–Schwarz inequality, observe that

‖x+ y‖2 = (x, x) + (y, y) + 2Re(x, y)≤ (x, x) + (y, y) + 2‖x‖ ‖y‖ = (‖x‖+ ‖y‖)2. .

7. Definition. A space H endowed with a semi-inner product ( · , · ) and theassociate seminorm ‖ · ‖ is a pre-Hilbert space. A pre-Hilbert space H is a Hilbertspace provided that the seminormed space (H, ‖ · ‖) is a Banach space.

8. In a pre-Hilbert space H, the Parallelogram Law is effective

‖x+ y‖2 + ‖x− y‖2 = 2(‖x‖2 + ‖y‖2) (x, y ∈ H)

which reads: the sum of the squares of the lengths of the diagonals equals the sumof the squares of the lengths of all sides.

/ ‖x+ y‖2 = (x+ y, x+ y) = ‖x‖2 + 2Re(x, y) + ‖y‖2;‖x− y‖2 = (x− y, x− y) = ‖x‖2 − 2Re(x, y) + ‖y‖2 .

9.Von Neumann–Jordan Theorem. If a seminormed space (H, ‖ · ‖) obeysthe Parallelogram Law then H is a pre-Hilbert space; i.e., there is a unique semi-inner product ( · , · ) on H such that ‖x‖ = (x, x)

1/2 for all x ∈ H.

Hilbert Spaces 87

C Considering the real carrier HR of H and x, y ∈ HR, put

(x, y)R := 1/4(‖x+ y‖2 − ‖x− y‖2

).

Given the mapping ( · , y)R, from the Parallelogram Law successively derive

(x1, y)R + (x2, y)R

= 1/4(‖x1 + y‖2 − ‖x1 − y‖2 + ‖x2 + y‖2 − ‖x2 − y‖2

)= 1/4

((‖x1 + y‖2 + ‖x2 + y‖2

)−(‖x1 − y‖2 + ‖x2 − y‖2

))= 1/4

(1/2(‖(x1 + y) + (x2 + y)‖2 + ‖x1 − x2‖2)

−1/2(‖(x1 − y) + (x2 − y)‖2 + ‖x1 − x2‖2)

)= 1/4

(1/2‖x1 + x2 + 2y‖2 − 1/2‖x1 + x2 − 2y‖2)

= 1/2

(∥∥∥(x1+x2)/2 + y∥∥∥2−∥∥∥(x1−x2)/2 − y

∥∥∥2)

= 2(

(x1+x2)/2, y)

R.

In particular, (x2, y)R = 0 in case x2 := 0, i.e. 1/2(x1, y)R =(1/2x1, y

)R. Analo-

gously, given x1 := 2x1 and x2 := 2x2, infer that

(x1 + x2, y)R = (x1, y)R + (x2, y)R.

Since the mapping ( · , y)R is continuous for obvious reasons, conclude that( · , y)R ∈ (HR)#. Put

(x, y) := (Re−1( · , y)R)(x),

where Re−1 is the complexifier (see 3.7.5).In case F := R it is clear that (x, y) = (x, y)R = (y, x) and (x, x) = ‖x‖2;

i.e., nothing is to be proven. On the other hand, if F := C then

(x, y) = (x, y)R − i(ix, y)R.

This entails sesquilinearity:

(y, x) = (y, x)R − i(iy, x)R = (x, y)R − i(x, iy)R

= (x, y)R + i(ix, y)R = (x, y)∗,

since

(x, iy)R = 1/4(‖x+ iy‖2 − ‖x− iy‖2

)= 1/4

(|i| ‖y − ix‖2 − | − i| ‖ix+ y‖2

)= −(ix, y)R.

88 Chapter 6

Furthermore,

(x, x) = (x, x)R − i(ix, x)R

= ‖x2‖ − i/4(‖ix+ x‖2 − ‖ix− x‖2

)= ‖x‖2

(1− i/4

(|1 + i|2 − |1− i|2

))= ‖x‖2.

The claim of uniqueness follows from 6.1.3. B

10. Examples.(1) A Hilbert space is exemplified by the L2 space (over some system

with integration), the inner product introduced as follows (f, g) :=∫fg∗ for f, g ∈

L2. In particular, (x, y) :=∑e∈E xey∗e for x, y ∈ l2(E ).

(2) Assume that H is a pre-Hilbert space and ( · , · ) : H20 = 0 OO Fis a semi-inner product on H. It is clear that the real carrier HR with the semi-inner product ( · , · )R : (x, y) 7→ Re(x, y) presents a pre-Hilbert space with thenorm of an element of H independent of whether it is calculated in H or in HR.The pre-Hilbert space (HR, ( · , · )R) is the realification or decomplexification of(H, ( · , · )). In turn, if the real carrier of a seminormed space is a pre-Hilbert spacethen the process of complexification leads to some natural pre-Hilbert structureof the original space.

(3) Assume that H is a pre-Hilbert space and H∗ is the twin vectorspace of H. Given x, y ∈ H∗, put (x, y)∗ := (x, y)∗. Clearly, ( · , · )∗ is a semi-inner product on H∗. The resulting pre-Hilbert space is the twin of H, with thedenotation H∗ preserved.

(4) Let H be a pre-Hilbert space and let H0 := ker ‖ · ‖ be the kernelof the seminorm ‖ · ‖ on H. Using the Cauchy–Bunyakovskiı–Schwarz inequality,Theorem 2.3.8 and 6.1.10 (3), observe that there is a natural inner product onthe quotient space H/H0: If x1 := ϕ(x1) and x2 := ϕ(x2), with x1, x2 ∈ H andϕ : H0 = 0 OO /H0 the coset mapping, then (x1, x2) := (x1, x2). Moreover, the pre-Hilbert space H/H0 may be considered as the quotient space of the seminormedspace (H, ‖·‖) by the kernel of the seminorm ‖·‖. Thus, H/H0 is a Hausdorff spacereferred to as the Hausdorff pre-Hilbert space associated with H. Completing thenormed space H/H0, obtain a Hilbert space (for instance, by the von Neumann–Jordan Theorem). The so-constructed Hilbert space is called associated with theoriginal pre-Hilbert space.

(5) Assume that (He)e∈E is a family of Hilbert spaces and H is the2-sum of the family; i.e., h ∈ H if and only if h := (he)e∈E , where he ∈ He for e ∈ Eand

‖h‖ :=

(∑e∈E

‖he‖2)1/2

< +∞.

Hilbert Spaces 89

By 5.5.9 (6), H is a Banach space. Given f, g ∈ H, on successively applying theParallelogram Law, deduce that

1/2(‖f + g‖2 + ‖f − g‖2

)= 1/2

(∑e∈E

‖fe + ge‖2 ++∑e∈E

‖fe − ge‖2)

=∑e∈E

1/2(‖fe + ge‖2 + ‖fe − ge‖2

)=∑e∈E

(‖fe‖2 + ‖ge‖2

)= ‖f‖2 + ‖g‖2.

Consequently, H is a Hilbert space by the von Neumann–Jordan Theorem. Thespace H, the Hilbert sum of the family (He)e∈E , is denoted by ⊕e∈EHe. WithE := N, it is customary to use the symbol H1 ⊕H2 ⊕ . . . for H.

(6) Let H be a Hilbert space and let S be a system with integration.The space L2(S, H) comprising all H-valued square-integrable functions is alsoa Hilbert space. CB

6.2. Orthoprojections

6.2.1. Let Uε be a convex subset of the spherical layer (r+ ε)BH \ rBH withr, ε > 0 in a Hilbert space H. Then the diameter of Uε vanishes as ε tends to 0.

C Given x, y ∈ Uε, on considering that 1/2(x + y) ∈ Uε and applying theParallelogram Law, for ε ≤ r infer that

‖x− y‖2 = 2(‖x‖2 + ‖y‖2

)− 4

∥∥∥(x+y)/2

∥∥∥2

≤ 4(r + ε)2 − 4r2 = 8rε+ 4ε2 ≤ 12rε. .

6.2.2. Levy Projection Theorem. Let U be a nonempty closed convex setin a Hilbert space H and x ∈ H \ U . Then there is a unique element u0 of U suchthat

‖x− u0‖ = inf{‖x− u‖ : u ∈ U}.

C Put Uε := {u ∈ U : ‖x−u‖ ≤ inf ‖U −x‖+ ε}. By 6.2.1, the family (Uε)ε>0constitutes a base for a Cauchy filter in U . B

6.2.3. Definition. The element u0 appearing in 6.2.2 is the best approxima-tion to x in U or the projection of x to U .

6.2.4. Let H0 be a closed subspace of a Hilbert space H and x ∈ H \ H0.An element x0 of H0 is the projection of x to H0 if and only if (x− x0, h0) = 0 forevery h0 in H0.

90 Chapter 6

C It suffices to consider the real carrier (H0)R of H0. The convex functionf(h0) := (h0 − x, h0 − x) is defined on (H0)R. Further, x0 in H0 serves as theprojection of x to H0 if and only if 0 ∈ ∂x0(f). In view of 3.5.2 (4) this containmentmeans that (x− x0, h0) = 0 for every h0 in H0, because f ′(x0) = 2(x0 − x, · ). B

6.2.5. Definition. Elements x and y of H are orthogonal, in symbols x ⊥ y,if (x, y) = 0. By U⊥ we denote the subset of H that comprises all elementsorthogonal to every point of a given subset U ; i.e.,

U⊥ := {y ∈ H : (∀ x ∈ U) x ⊥ y}.

The set U⊥ is the orthogonal complement or orthocomplement of U (to H).

6.2.6. Let H0 be a closed subspace of a Hilbert space H. The orthogonalcomplement of H0, the set H⊥0 , is a closed subspace and H = H0 ⊕H⊥0 .

C The closure property of H⊥0 in H is evident. It is also clear that H0 ∧H⊥0 =H0 ∩H⊥0 = 0. We are left with showing only that H0 ∨H⊥0 = H0 +H⊥0 = H. Takean element h of H \H0. In virtue of 6.2.2 the projection h0 of H to H0 is availableand, by 6.2.4, h− h0 ∈ H⊥0 . Finally, h = h0 + (h− h0) ∈ H0 +H⊥0 . .

6.2.7. Definition. The projection onto a (closed) subspace H0 along H⊥0 isthe orthoprojection onto H0, denoted by PH0 .

6.2.8. Pythagoras Lemma. x ⊥ y ⇒ ‖x+ y‖2 = ‖x‖2 + ‖y‖2. /.

6.2.9. Corollary. The norm of an orthoprojection is at most one: (H 6= 0 &H0 6= 0)⇒ ‖PH0‖ = 1. /.

6.2.10. Orthoprojection Theorem. For an operator P in L (H) such thatP 2 = P , the following statements are equivalent:

(1) P is the orthoprojection onto H0 := imP ;(2) ‖h‖ ≤ 1⇒ ‖Ph‖ ≤ 1;(3) (Px, P dy) = 0 for all x, y ∈ H, with P d the complement of P , i.e.

P d= IH − P ;(4) (Px, y) = (x, Py) for x, y ∈ H.

C (1) ⇒ (2): This is observed in 6.2.9.(2) ⇒ (3): Let H1 := kerP = imP d. Take x ∈ H⊥1 . Since x = Px+ P dx and

x ⊥ P dx; therefore, ‖x‖2 ≥ ‖Px‖2 = (x−P dx, x−P dx) = (x, x)−2Re(x, P dx)+(P dx, P dx) = ‖x‖2 + ‖P dx‖2. Whence P dx = 0; i.e., x ∈ imP . Considering 6.2.6,from H1 = kerP and H⊥1 ⊂ imP deduce the equalities H⊥1 = imP = H0. Thus(Px, P dy) = 0 for all x, y ∈ H, since Px ∈ H0 and P dy ∈ H1.

(3)⇒ (4): (Px, y) = (Px, Py+P dy) = (Px, Py) = (Px, Py)+(P dx, Py) =(x, Py).

Hilbert Spaces 91

(4) ⇒ (1): Show first that H0 is a closed subspace. Let h0 := limhn withhn ∈ H0, i.e. Phn = hn. For every x in H, from the continuity property of thefunctionals ( · , x) and ( · , Px) successively derive

(h0, x) = lim (hn, x) = lim (Phn, x) = lim (hn, Px) = (Ph0, x).

Whence (h0 − Ph0, h0 − Ph0) = 0; i.e., h0 ∈ imP .Given x ∈ H and h0 ∈ H0, now infer that (x − Px, h0) = (x − Px, Ph0) =

(P (x − Px), h0) = (Px − P 2x, h0) = (Px − Px, h0) = 0. Therefore, from 6.2.4obtain Px = PH0x. .

6.2.11. Let P1, and P2 be orthoprojections with P1P2 = 0. Then P2P1 = 0./ P1P2 = 0 ⇒ imP2 ⊂ kerP1 ⇒ imP1 = (kerP1)⊥ ⊂ (imP2)⊥ = kerP2 ⇒

P2P1 = 0 .

6.2.12. Definition. Orthoprojections P1 and P2 are orthogonal, in symbols,P1 ⊥ P2 or P2 ⊥ P1, provided that P1P2 = 0.

6.2.13. Theorem. Let P1, . . . , Pn be orthoprojections. The operator P :=P1 + . . .+ Pn is an orthoprojection if and only if Pl ⊥ Pm for l 6= m.C ⇒: First, given an orthoprojection P0, observe that ‖P0x‖2 = (P0x, P0x) =

(P 20 x, x) = (P0x, x) by Theorem 6.2.10. Consequently,

‖Plx‖2 + ‖Pmx‖2

≤n∑k=1

‖Pkx‖2 =n∑k=1

(Pkx, x) = (Px, x) = ‖Px‖2 ≤ ‖x‖2

for x ∈ H and l 6= m.In particular, putting x := Plx, observe that

‖Plx‖2 + ‖PmPlx‖2‖Plx‖2 ⇒ ‖PmPl‖ = 0.

⇐: Straightforward calculation shows that P is an idempotent operator. Indeed,

P 2 =

(n∑k=1

Pk

)2

=n∑l=1

n∑m=1

PlPm =n∑k=1

P 2k = P.

Furthermore, in virtue of 6.2.10 (4), (Pkx, y) = (x, Pky) and so (Px, y) = (x, Py).It suffices to appeal to 6.2.10 (4) once again. B

6.2.14. Remark. Theorem 6.2.13 is usually referred to as the pairwise orthog-onality criterion for finitely many orthoprojections.

92 Chapter 6

6.3. A Hilbert Basis

6.3.1. Definition. A family (xe)e∈E of elements of a Hilbert space H is or-thogonal, if e1 6= e2 ⇒ xe1 ⊥ xe2 . By specification, a subset E of a Hilbert space His orthogonal if so is the family (e)e∈E .

6.3.2. Pythagoras Theorem. An orthogonal family (xe)e∈E of elements ofa Hilbert space is (unconditionally) summable if and only if the numeric family(‖xe‖2)e∈E is summable. Moreover,∥∥∥∑

e∈E

xe

∥∥∥2=∑e∈E

‖xe‖2.

C Let sθ :=∑e∈E xe, where θ is a finite subset of E . By 6.2.8, ‖sθ‖2 =∑

e∈θ ‖xe‖2. Given a finite subset θ′ of E which includes θ, thus observe that

‖sθ′ − sθ‖2 = ‖sθ′\θ‖2 =∑e∈θ′\θ

‖xe‖2.

In other words, the fundamentalness of (sθ) amounts to the fundamentalness of thenet of partial sums of the family (‖xe‖2)e∈E . On using 4.5.4, complete the proof. B

6.3.3. Orthoprojection Summation Theorem. Let (Pe)e∈E be a familyof pairwise orthogonal orthoprojections in a Hilbert space H. Then for every xin H the family (Pex)e∈E is (unconditionally) summable. Moreover, the operatorPx :=

∑e∈E Pex is the orthoprojection onto the subspace

H :=

{∑e∈E

xe : xe ∈ He := imPe,∑e∈E

‖xe‖2 < +∞

}.

C Given a finite subset θ of E , put sθ :=∑e∈θ Pe. By Theorem 6.2.13, sθ is

an orthoprojection. Hence, in view of 6.2.8, ‖sθx‖2 =∑e∈θ ‖Pex‖2 ≤ ‖x‖2

for every x in H. Consequently, the family (‖Pex‖2)e∈E is summable (the netof partial sums is increasing and bounded). By the Pythagoras Theorem, thereis a sum Px :=

∑e∈E Pex; i.e., Px = limθ sθx. Whence P 2x = limθ sθPx =

limθ sθ limθ′ sθ′x = limθ limθ′ sθsθ′x = limθ limθ′ sθ∩θ′x = limθ sθx = Px. Finally,‖Px‖ = ‖ limθ sθx‖ = limθ ‖sθx‖ ≤ ‖x‖ and, moreover, P 2 = P . Appealing to6.2.10, conclude that P is the orthoprojection onto imP .

If x ∈ imP , i.e., Px = x; then x =∑e∈E Pex and by the Pythagoras Theorem∑

e∈E ‖Pex‖2 = ‖x‖2 = ‖Px‖2 < +∞. Since Pex ∈ He (e ∈ E ); therefore, x ∈H .If xe ∈ He and

∑e∈E ‖xe‖2 < +∞ then for x :=

∑e∈E xe (existence follows from

the same Pythagoras Theorem) observe that x =∑e∈E xe =

∑e∈E Pexe = Px;

i.e., x ∈ imP . Thus, imP = H . .

Hilbert Spaces 93

6.3.4. Remark. This theorem may be treated as asserting that the space Hand the Hilbert sum of the family (He)e∈E are isomorphic. The identification isclearly accomplished by the Bochner integral presenting the process of summationin this case.

6.3.5. Remark. Let h in H be a normalized or unit or norm-one element;i.e., ‖h‖ = 1. Assume further that H0 := Fh is a one-dimensional subspace of Hspanned over h0. For every element x of H and every scalar λ, a member of F,observe that

(x− (x, h)h, λh) = λ∗((x, h)− (x, h))(h, h) = 0.

Therefore, by 6.2.4, PH0 = ( · , h)⊗ h. To denote this orthoprojection, it is conve-nient to use the symbol 〈h〉. Thus, 〈h〉 : x 7→ (x, h)h (x ∈ H).

6.3.6. Definition. A family of elements of a Hilbert space is called orthonor-mal (or orthonormalized) if, first, the family is orthogonal and, second, the norm ofeach member of the family equals one. Orthonormal sets are defined by specifica-tion.

6.3.7. For every orthonormal subset E of H and every element x of H, thefamily (〈e〉x)e∈E is (unconditionally) summable. Moreover, the Bessel inequalityholds:

‖x‖2 ≥∑e∈E

|(x, e)|2.

C It suffices to refer to the Orthoprojection Summation Theorem, for

‖x‖2 ≥

∥∥∥∥∥∑e∈E

〈e〉x

∥∥∥∥∥2

=

∥∥∥∥∥∑e∈E

(x, e)e

∥∥∥∥∥2

=∑e∈E

‖(x, e)e‖2 . .

6.3.8. Definition. An orthonormal set E in a Hilbert space H is a Hilbertbasis (for H) if x =

∑e∈E 〈e〉x for every x in H. An orthonormal family of elements

of a Hilbert space is a Hilbert basis if the range of the family is a Hilbert basis.

6.3.9. An orthonormal set E is a Hilbert basis for H if and only if lin(E ), thelinear span of E , is dense in H. CB

6.3.10. Definition. A subset E of a Hilbert space is said to meet the Steklovcondition if E⊥ = 0.

6.3.11. Steklov Theorem. An orthonormal set E is a Hilbert basis if andonly if E meets the Steklov condition.C ⇒: Let h ∈ E⊥. Then h =

∑e∈E 〈e〉h =

∑e∈E (h, e)e =

∑e∈E 0 = 0.

⇐: For x ∈ H, in virtue of 6.3.3 and 6.2.4, x−∑e∈E 〈e〉x ∈ E⊥. .

94 Chapter 6

6.3.12. Theorem. Each Hilbert space has a Hilbert basis.C By the Kuratowski–Zorn Lemma, each Hilbert space H has an orthonormal

set E maximal by inclusion. If there were some h in H \H0, with H0 := cl lin(E );then the element h1 := h−PH0h would be orthogonal to every element in E . Thus,for H 6= 0 we would have E ∪ {‖h1‖−1h1} = E . A contradiction. In case H = 0there is nothing left to proof. B

6.3.13. Remark. It is possible to show that two Hilbert bases for a Hilbertspace H have the same cardinality. This cardinality is the Hilbert dimension of H.

6.3.14.Remark. Let (xn)n∈N be a countable sequence of linearly independentelements of a Hilbert space H. Put x0 := 0, e0 := 0 and

yn := xn −n−1∑k=0

〈ek〉xn, en :=yn‖yn‖

(n ∈ N).

Evidently, (yn, ek) = 0 for 0 ≤ k ≤ n − 1 (for instance, by 6.2.13). Also, yn 6= 0,since H is infinite-dimensional. Say that the orthonormal sequence (en)n∈N resultsfrom the sequence (xn)n∈N by the Gram–Schmidt orthogonalization process. Usingthe process, it is easy to prove that a Hilbert space has a countable Hilbert basisif and only if the space has a countable dense subset; i.e., whenever the space isseparable. CB

6.3.15. Definition. Let E be a Hilbert basis for a space H and x ∈ H. Thenumeric family x := (xe)e∈E in F E , given by the identity xe := (x, e), is the Fouriercoefficient family of x with respect to E or the Fourier transform of X (relativeto E ).

6.3.16. Riesz–Fisher Isomorphism Theorem. Let E be a Hilbert basisforH. The Fourier transform F : x 7→ x (relative to E ) is an isometric isomorphismof H onto l2(E ). The inverse Fourier transform, the Fourier summation F−1 :l2(E )0 = 0 OO , acts by the rule F−1(x) :=

∑e∈E xee for x := (xe)e∈E ∈ l2(E ).

Moreover, for all x, y ∈ H the Parseval identity holds:

(x, y) =∑e∈E

xey∗e .

C By the Pythagoras Theorem, the Fourier transform acts in l2(E ). By The-orem 6.3.3, is an epimorphism. By the Steklov Theorem, is a monomorphism.It is beyond a doubt that F−1x = x for x ∈ H and F−1(x) = x for x ∈ l2(E ).The equality

‖x‖2 =∑e∈E

‖xe‖2 = ‖x‖22 (x ∈ H)

Hilbert Spaces 95

follows from the Pythagoras Theorem. At the same time

(x, y) =

(∑e∈E

xee,∑e∈E

yee

)=∑e,e′∈E

xey∗e′(e, e

′) =∑e∈E

xey∗e . .

6.3.17. Remark. The Parseval identity shows that the Fourier transform pre-serves inner products. Therefore, the Fourier transform is a unitary operator ora Hilbert-space isomorphism; i.e., an isomorphism preserving inner products. Thisis why the Riesz–Fisher Theorem is sometimes referred to as the theorem on Hilbertisomorphy between Hilbert spaces (of the same Hilbert dimension).

6.4. The Adjoint of an Operator

6.4.1. Riesz Prime Theorem. Let H be a Hilbert space. Given x ∈ H, putx′ := ( · , x). Then the prime mapping x 7→ x′ presents an isometric isomorphismof H∗ onto H ′.

C It is clear that x = 0⇒ x′ = 0. If x 6= 0 then

‖x′‖H′ = sup‖y‖≤1

|(y, x)| ≤ sup‖y‖≤1

‖y‖ ‖x‖ ≤ ‖x‖;

‖x′‖H′ = sup‖y‖≤1

|(y, x)| ≥ |(x/‖x‖, x)| = ‖x‖.

Therefore, x 7→ x′ is an isometry of H∗ into H ′. Check that this mapping isan epimorphism.

Let l ∈ H ′ and H0 := ker l 6= H (if there no such l then nothing is to beproven). Choose a norm-one element e in H⊥0 and put grad l := l(e)∗e. If x ∈ H0then

(grad l)′(x) = (x, grad l) = (x, l(e)∗e) = l(e)∗∗(x, e) = 0.

Consequently, for some α in F and all x ∈ H in virtue of 2.3.12 (grad l)′(x) = αl(x).In particular, letting x := e, find

(grad l)′(e) = (e, grad l) = l(e)(e, e) = αl(e);

i.e., α = 1. B

6.4.2. Remark. From the Riesz Prime Theorem it follows that the dualspace H ′ possesses a natural structure of a Hilbert space and the prime mappingx 7→ x′ implements a Hilbert space isomorphism from H∗ onto H ′. The inversemapping now coincides with the gradient mapping l 7→ grad l constructed in theproof of the theorem. Implying this, the claim of 6.4.1 is referred to as the theoremon the general form of a linear functional in a Hilbert space.

96 Chapter 6

6.4.3. Each Hilbert space is reflexive.C Let ι : H0 = 0 OO ′′ be the double prime mapping; i.e. the canonical embedding

of H into the second dual H ′′ which is determined by the rule x′′(l) = ι(x)(l) = l(x),where x ∈ H and l ∈ H ′ (see 5.1.10 (8)). Check that ι is an epimorphism. Letf ∈ H ′′. Consider the mapping y 7→ f(y ′) for y ∈ H. It is clear that this mapping isa linear functional over H∗ and so by the Riesz Prime Theorem there is an elementx ∈ H = H∗∗ such that (y, x)∗ = (x, y) = f(y ′) for every y in H. Observe thatι(x)(y ′) = y ′(x) = (x, y) = f(y ′) for all y ∈ H. Since by the Riesz Prime Theoremy 7→ y ′ is a mapping onto H ′, conclude that ι(x) = f. .

6.4.4. Let H1 and H2 be Hilbert spaces and T ∈ B(H1, H2). Then there isa unique mapping T ∗ : H20 = 0 OO 1 such that

(Tx, y) = (x, T ∗y)

for all x ∈ H1 and y ∈ H2. Moreover, T ∗ ∈ B(H2, H1) and ‖T ∗‖ = ‖T‖.C Let y ∈ H2. The mapping x 7→ (Tx, y) is the composition y ′ ◦ T ; i.e.,

it presents a continuous linear functional over H1. By the Riesz Prime Theoremthere is precisely one element x of H1 for which x′ = y ′ ◦ T . Put T ∗y := x.It is clear that T ∗ ∈ L (H2, H1). Furthermore, using the Cauchy–Bunyakovskiı–Schwarz inequality and the normative inequality, infer that

|(T ∗y, T ∗y)| = |(TT ∗y, y)| ≤ ‖TT ∗y‖ ‖y‖ ≤ ‖T‖ ‖T ∗y‖ ‖y‖.

Hence, ‖T ∗y‖ ≤ ‖T‖ ‖y‖ for all y ∈ H2; i.e., ‖T ∗‖ ≤ ‖T‖. At the same timeT = T ∗∗ := (T ∗)∗; i.e., ‖T‖ = ‖T ∗∗‖ ≤ ‖T ∗‖. .

6.4.5. Definition. For T ∈ B(H1, H2), the operator T ∗, the member ofB(H2, H1) constructed in 6.4.4, is the adjoint of T . The terms like “hermitian-conjugate” and “Hilbert-space adjoint” are also in current usage.

6.4.6. LetH1 andH2 be Hilbert spaces. Assume further that S, T ∈ B(H1, H2)and λ ∈ F. Then

(1) T ∗∗ = T ;(2) (S + T )∗ = S∗ + T ∗;(3) (λT )∗ = λ∗T ∗;(4) ‖T ∗T‖ = ‖T‖2.

C (1)–(3) are obvious properties. If ‖x‖ ≤ 1 then

‖Tx‖2 = (Tx, Tx) = |(Tx, Tx)| = |(T ∗Tx, x)| ≤ ‖T ∗Tx‖ ‖x‖ ≤ ‖T ∗T‖.

Furthermore, using the submultiplicativity of the operator norm and 6.4.4, infer‖T ∗T‖ ≤ ‖T ∗‖ ‖T‖ = ‖T‖2, which proves (4). B

Hilbert Spaces 97

6.4.7. Let H1, H2, and H3 be three Hilbert spaces. Assume further thatT ∈ B(H1, H2) and S ∈ B(H2, H3). Then (ST )∗ = T ∗S∗.C (STx, z) = (Tx, S∗z) = (x, T ∗S∗z) (x ∈ H1, z ∈ H3) .

6.4.8.Definition. Consider an elementary diagramH1T−→ H2. The diagram

H1T∗←− H2 is the adjoint of the initial elementary diagram. Given an arbitrary

diagram composed of bounded linear mappings between Hilbert spaces, assumethat each elementary subdiagram is replaced with its adjoint. Then the resultingdiagram is the adjoint or, for suggestiveness, the diagram star of the initial diagram.

6.4.9. Diagram Star Principle. A diagram is commutative if and only if sois its adjoint diagram.

C Follows from 6.4.7 and 6.4.6 (1). B

6.4.10. Corollary. An operator T is invertible if and only if T ∗ is invertible.Moreover, T ∗−1 = T−1∗. CB

6.4.11. Corollary. If T ∈ B(H) then λ ∈ Sp(T )⇔ λ∗ ∈ Sp(T ∗). CB

6.4.12. Sequence Star Principle (cf. 7.6.13). A sequence

. . . 0 = 0 OOk−1

Tk−−−−→ HkTk+1−−−−→ Hk+10 = 0 OO

is exact if and only if so is the sequence star

. . .← Hk−1T∗k←−−−− Hk

T∗k+1←−−−− Hk+1 ← . . . . /.

6.4.13. Definition. An involutive algebra or ∗-algebra A (over a groundfield F) is an algebra with an involution ∗, i.e. with a mapping a 7→ a∗ from Ato A such that

(1) a∗∗ = a (a ∈ A);(2) (a+ b)∗ = a∗ + b∗ (a, b ∈ A);(3) (λa)∗ = λ∗a∗ (λ ∈ F, a ∈ A);(4) (ab)∗ = b∗a∗ (a, b ∈ A).

A Banach algebra A with involution ∗ satisfying ‖a∗a‖ = ‖a‖2 for all a ∈ A isa C∗-algebra.

6.4.14. The endomorphism space B(H) of a Hilbert space H is a C∗-algebra(with the composition of operators as multiplication and the taking of the adjointof an endomorphism as involution). CB

98 Chapter 6

6.5. Hermitian Operators

6.5.1. Definition. Let H be a Hilbert space over a ground field F. An ele-ment T of B(H) is a hermitian operator or selfadjoint operator in H provided thatT = T ∗.

6.5.2. Rayleigh Theorem. For a hermitian operator T the equality holds:

‖T‖ = sup‖x‖≤1

|(Tx, x)|.

C Put t := sup{|(Tx, x)| : ‖x‖ ≤ 1}. It is clear that |(Tx, x)| ≤ ‖Tx‖ ‖x‖ ≤‖T‖ provided ‖x‖ ≤ 1. Thus, t ≤ ‖T‖.

Since T = T ∗; therefore, (Tx, y) = (x, Ty) = (Ty, x)∗ = (y, Tx)∗; i.e.,(x, y) 7→ (Tx, y) is a hermitian form. Consequently, in virtue of 6.1.3 and 6.1.8

4Re(Tx, y) = (T (x+ y), x+ y)− (T (x− y), x− y)≤ t(‖x+ y‖2 + ‖x− y‖2) = 2t(‖x‖2 + ‖y‖2).

If Tx = 0 then it is plain that ‖Tx‖ ≤ t. Assume Tx 6= 0. Given ‖x‖ ≤ 1 andputting y := ‖Tx‖−1Tx, infer that

‖Tx‖ = ‖Tx‖(

Tx

‖Tx‖,

Tx

‖Tx‖

)= (Tx, y) = Re(Tx, y) ≤ 1/2 t

(‖x‖2 +

∥∥Tx/‖Tx‖ ∥∥2)≤ t;

i.e., ‖T‖ = sup{‖Tx‖ : ‖x‖ ≤ 1} ≤ t. .

6.5.3. Remark. As mentioned in the proof of 6.5.2, each hermitian operator Tin a Hilbert space H generates the hermitian form fT (x, y) := (Tx, y). Conversely,let f be a hermitian form, with the functional f( · , y) continuous for every yin H. Then by the Riesz Prime Theorem there is an element Ty of H such thatf( · , y) = (Ty)′. Evidently, T ∈ L (H) and (x, Ty) = f(x, y) = f(y, x)∗ =(y, Tx)∗ = (Tx, y). It is possible to show that in this case T ∈ B(H) and T = T ∗.In addition, f = fT . Therefore, the condition T ∈ B(H) in Definition 6.5.1 can bereplaced with the condition T ∈ L (H) (the Hellinger–Toeplitz Theorem).

6.5.4. Weyl Criterion. A scalar λ belongs to the spectrum of a hermitianoperator T if and only if

inf‖x‖=1

‖λx− Tx‖ = 0.

C ⇒: Put t := inf{‖λx − Tx‖ : x ∈ H, ‖x‖ = 1} > 0. Demonstrate thatλ /∈ Sp(T ). Given an x in H, observe that ‖λx− Tx‖ ≥ t‖x‖. Thus, first, (λ− T )

Hilbert Spaces 99

is a monomorphism; second, H0 := im(λ − T ) is a closed subspace (because ‖(λ −T )xm − (λ − T )xk‖ ≥ t‖xm − xk‖; i.e., the inverse image of a Cauchy sequenceis a Cauchy sequence); and, third, which is final, (λ − T )−1 ∈ B(H) wheneverH = H0 (in such situation ‖R(T, λ)‖ ≤ t−1). Suppose to the contrary thatH 6= H0. Then there is some y in H⊥0 satisfying ‖y‖ = 1. For all x ∈ H, note that0 = (λx − Tx, y) = (x, λ∗y − Ty); i.e., λ∗y = Ty. Further, λ∗ = (Ty, y)/(y, y)and the hermiticity of T guarantees λ∗ ∈ R. Whence λ∗ = λ and y ∈ ker(λ − T ).We arrive at a contradiction: 1 = ‖y‖ = ‖0‖ = 0.⇐: If λ /∈ Sp(T ) then the resolvent of T at λ, the member R(T, λ) of B(H),

is available. Hence, inf{‖λx− Tx‖ : ‖x‖ = 1} ≥ ‖R(T, λ)‖−1. .

6.5.5. Spectral Endpoint Theorem. Let T be a hermitian operator ina Hilbert space. Put

mT := inf‖x‖=1

(Tx, x), MT := sup‖x‖=1

(Tx, x).

Then Sp(T ) ⊂ [mT , MT ] and mT , MT ∈ Sp(T ).C Considering that the operator T − Reλ is hermitian in the space H under

study, from the identity

‖λx− Tx‖2 = | Imλ|2‖x‖2 + ‖Tx− Reλx‖2

infer the inclusion Sp(T ) ⊂ R by 6.5.4. Given a norm-one element x of H andinvoking the Cauchy–Bunyakovskiı–Schwarz inequality, in case λ < mT deducethat

‖λx− Tx‖ = ‖λx− Tx‖ ‖x‖ ≥ |(λx− Tx, x)|= |λ− (Tx, x)| = (Tx, x)− λ ≥ mT − λ > 0.

On appealing to 6.5.4, find λ ∈ res(T ). In case λ > MT , similarly infer that

‖λx− Tx‖ ≥ |(λx− Tx, x)| = |λ− (Tx, x)| = λ− (Tx, x) ≥ λ−MT > 0.

Once again λ ∈ res(T ). Finally, Sp(T ) ⊂ [mT , MT ].Since (Tx, x) ∈ R for x ∈ H; therefore, in virtue of 6.5.2

‖T‖ = sup{|(Tx, x)| : ‖x‖ ≤ 1}= sup{(Tx, x) ∨ (−(Tx, x)) : ‖x‖ ≤ 1} =MT ∨ (−mT ).

Assume first that λ := ‖T‖ =MT . If ‖x‖ = 1 then

‖λx− Tx‖2 = λ2 − 2λ(Tx, x) + ‖Tx‖2 ≤ 2‖T‖2 − 2‖T‖(Tx, x).

100 Chapter 6

In other words, the next estimate holds:inf‖x‖=1

‖λx− Tx‖2 ≤ 2‖T‖ inf‖x‖=1

(‖T‖ − (Tx, x)) = 0.

Using 6.5.4, conclude that λ ∈ Sp(T ).Now consider the operator S := T−mT . It is clear thatMS =MT−mT ≥ 0 and

mS = mT −mT = 0. Therefore, ‖S‖ =MS and in view of the above MS ∈ Sp(S).Whence it follows thatMT belongs to Sp(T ), since T = S+mT andMT =MS+mT .It suffices to observe that mT = −M−T and Sp(T ) = − Sp(−T ). .

6.5.6. Corollary. The norm of a hermitian operator equals the radius of itsspectrum (and the spectral radius). CB

6.5.7. Corollary. A hermitian operator is zero if and only if its spectrumconsists of zero. CB

6.6. Compact Hermitian Operators

6.6.1. Definition. Let X and Y be Banach spaces. An operator T , a memberof L (X, Y ), is called compact (in symbols, T ∈ K (X, Y )) if the image T (BX) ofthe unit ball BX of X is relatively compact in Y .

6.6.2. Remark. Detailed study of compact operators in Banach spaces is thepurpose of the Riesz–Schauder theory to be exposed in Chapter 8.

6.6.3. Let T be a compact hermitian operator. If 0 6= λ ∈ Sp(T ) then λ isan eigenvalue of T ; i.e., ker(λ− T ) 6= 0.

C By the Weyl Criterion, λxn − Txn0 = 0 OO for some sequence (xn) such that‖xn‖ = 1. Without loss of generality, assume that the sequence (Txn) converges toy := limTxn. Then from the identity λxn = (λxn−Txn)+Txn obtain that there isa limit (λxn) and y = limλxn. Consequently, Ty = T (limλxn) = λ limTxn = λy.Since ‖y‖ = |λ|, conclude that y is an eigenvector of T , i.e. y ∈ ker(λ− T ). B

6.6.4. Let λ1 and λ2 be distinct eigenvalues of a hermitian operator T . Assumefurther that x1 and x2 are eigenvectors with eigenvalues λ1 and λ2 (i.e., xs ∈ker(λs − T ), s := 1, 2). Then x1 and x2 are orthogonal.

/ (x1, x2) = 1λ1(Tx1, x2) = 1

λ1(x1, Tx2) = λ2

λ1(x1, x2) .

6.6.5. For whatever strictly positive ε, there are only finitely many eigenvaluesof a compact hermitian operator beyond the interval [−ε, ε].

C Let (λn)n∈N be a sequence of pairwise distinct eigenvalues of T satisfying|λn| > ε. Further, let xn be an eigenvector corresponding to λn and such that‖xn‖ = 1. By virtue of 6.6.4 (xk, xm) = 0 for m 6= k. Consequently,

‖Txm − Txk‖2 = ‖Txm‖2 + ‖Txk‖2 = λ2m + λ2

k ≥ 2ε2;i.e., the sequence (Txn)n∈N is relatively compact. We arrive at a contradiction tothe compactness property of T. .

Hilbert Spaces 101

6.6.6. Spectral Decomposition Lemma. Let T be a compact hermitianoperator in a Hilbert space H and 0 6= λ ∈ Sp(T ). Put Hλ := ker(λ−T ). Then Hλ

is finite-dimensional and the decomposition H = Hλ ⊕ H⊥λ reduces T . Moreover,the matrix presentation holds

T ∼(λ 00 Tλ

),

where the operator Tλ, the part of T in H⊥λ , is hermitian and compact, withSp(Tλ) = Sp(T ) \ {λ}.C The subspace Hλ is finite-dimensional in view of the compactness of T .

Furthermore, Hλ is invariant under T . Consequently, the orthogonal complementH⊥λ of Hλ is an invariant subspace of T ∗ (coincident with T ), since h ∈ H⊥λ ⇒(∀x ∈ Hλ) x ⊥ h⇒ (∀x ∈ Hλ) 0 = (h, Tx) = (T ∗h, x)⇒ T ∗h ∈ H⊥λ .

The part of T in Hλ is clearly λ. The part Tλ of T in H⊥λ is undoubtedlycompact and hermitian. Obviously, for µ 6= λ, the operator

µ− T ∼(µ− λ 00 µ− Tλ

)is invertible if and only if so is µ − Tλ. It is also clear that λ is not an eigenvalueof Tλ. .

6.6.7. Hilbert–Schmidt Theorem. Let H be a Hilbert space and let T bea compact hermitian operator in H. Assume further that Pλ is the orthoprojectiononto ker(λ− T ) for λ ∈ Sp(T ). Then

T =∑

λ∈Sp(T )

λPλ.

C Using 6.5.6 and 6.6.6 as many times as need be, for every finite subset θof Sp(T ) obtain the equality∥∥∥∥∥T −∑

λ∈θ

λPλ

∥∥∥∥∥ = sup{|λ| : λ ∈ (Sp(T ) ∪ 0) \ θ}.

It suffices to refer to 6.6.5. B

6.6.8. Remark. The Hilbert–Schmidt Theorem provides essentially new in-formation, as compared with the case of finite dimensions, only if the operator Thas infinite-rank, that is, the dimension of its range is infinite or, which is the same,H⊥0 is an infinite-dimensional space, where H0 := kerT . In fact, if the operator T

102 Chapter 6

has finite rank (i.e., its range is finite-dimensional) then, since the subspace H⊥0 isisomorphic with the range of T , observe that

T =n∑k=1

λk〈ek〉 =n∑k=1

λke′k ⊗ ek,

where λ1, . . . , λn are nonzero points of Sp(T ) counted with multiplicity, and {e1, . . . , en}is a properly-chosen orthonormal basis for H⊥0 .

The Hilbert–Schmidt Theorem shows that, to within substitution of series forsum, an infinite-rank compact hermitian operator looks like a finite-rank operator.Indeed, for λ 6= µ, where λ and µ are nonzero points of Sp(T ), the eigenspacesHλ and Hµ are finite-dimensional and orthogonal. Moreover, the Hilbert sum⊕λ∈Sp(T )\0Hλ equals H⊥0 = cl imT , because H0 = (imT )⊥. Successively se-lecting a basis for each finite-dimensional space Hλ by enumerating the eigen-values in decreasing order of magnitudes with multiplicity counted; i.e., puttingλ1 := λ2 := . . . := λdimHλ1

:= λ1, λdimHλ1+1 := . . . := λdimHλ1+dimHλ2:= λ2, etc.,

obtain the decomposition H = H0 ⊕Hλ1 ⊕Hλ2 ⊕ . . . and the presentation

T =∞∑k=1

λk〈ek〉 =∞∑k=1

λke′k ⊗ ek,

where the series is summed in operator norm. CB

6.6.9. Theorem. Let T in K (H1, H2) be an infinite-rank compact operatorfrom a Hilbert space H1 to a Hilbert space H2. There are orthonormal families(ek)k∈N in H1, (fk)k∈N in H2, and a numeric family (µk)k∈N in R+ \0, µk ↓ 0, suchthat the following presentation holds:

T =∞∑k=1

µke′k ⊗ fk.

C Put S := T ∗T . It is clear that S ∈ B(H1) and S is compact. Furthermore,(Sx, x) = (T ∗Tx, x) = (Tx, Tx) = ‖Tx‖2. Consequently, in virtue of 6.4.6, Sis hermitian and H0 := kerS = kerT . Observe also that Sp(S) ⊂ R+ by Theorem6.5.5.

Let (ek)k∈N be an orthonormal basis forH⊥0 comprising all eigenvalues of S andlet (λk)k∈N be a corresponding decreasing sequence of strictly positive eigenvaluesλk > 0, k ∈ N (cf. 6.6.8). Then every element x ∈ H1 expands into the Fourierseries

x− PH0x =∞∑k=1

(x, ek)ek.

Hilbert Spaces 103

Therefore, considering that TPH0 = 0 and assigning µk :=√λk and fk := µ−1

k Tek,find

Tx =∞∑k=1

(x, ek)Tek =∞∑k=1

(x, ek)µkµkTek =

∞∑k=1

µk(x, ek)fk.

The family (fk)k∈N is orthonormal because

(fn, fm) =(Tenµn

,T emµm

)=

1µnµm

(Ten, T em)

=1

µnµm(T ∗Ten, em) =

1µnµm

(Sen, em)

=1

µn, µm(λnen, em) =

µnµm

(en, em).

Successively using the Pythagoras Theorem and the Bessel inequality, argue asfollows: ∥∥∥∥∥

(T −

n∑k=1

µke′k ⊗ fk

)x

∥∥∥∥∥2

=

∥∥∥∥∥∞∑

k=n+1

µk(x, ek)fk

∥∥∥∥∥2

=∞∑

k=n+1

µ2k|(x, ek)|2 ≤ λn+1

∞∑k=n+1

|(x, ek)|2 ≤ λn+1‖x‖2.

Since λk ↓ 0, finally deduce that∥∥∥∥∥T −n∑k=1

µke′k ⊗ fk

∥∥∥∥∥ ≤ µn+10 = 0 OO . .

6.6.10. Remark. Theorem 6.6.9 means in particular that a compact operator(and only a compact operator) is an adherent point of the set of finite-rank op-erators. This fact is also expressed as follows: “Every Hilbert space possesses theapproximation property.”

Exercises

6.1. Describe the extreme points of the unit ball of a Hilbert space.6.2. Find out which classical Banach spaces are Hilbert spaces and which are not.6.3. Is a quotient space of a Hilbert space also a Hilbert space?6.4. Is it true that each Banach space may be embedded into a Hilbert space?6.5. Is it possible that the (bounded) endomorphism space of a Hilbert space presents

a Hilbert space?

104 Chapter 6

6.6. Describe the second (= repeated) orthogonal complement of a set in a Hilbert space.6.7. Prove that no Hilbert basis for an infinite-dimensional Hilbert space is a Hamel basis.6.8. Find the best approximation to a polynomial of degree n+ 1 by polynomials of degree

at most n in the L2 space on an interval.

6.9. Prove that x ⊥ y if and only if ‖x+ y‖2 = ‖x‖2 + ‖y‖2 and ‖x+ iy‖2 = ‖x‖2 + ‖y‖2.6.10. Given a bounded operator T , prove that

(kerT )⊥ = cl imT ∗, (imT )⊥ = kerT ∗.

6.11. Reveal the interplay between hermitian forms and hermitian operators (cf. 6.5.3).6.12. Find the adjoint of a shift operator, a multiplier, and a finite-rank operator.6.13. Prove that an operator between Hilbert spaces is compact if and only if so is its adjoint.6.14. Assume that an operator T is an isometry. Is T ∗ an isometry too?6.15. A partial isometry is an operator isometric on the orthogonal complement of its kernel.

What is the structure of the adjoint of a partial isometry?

6.16. What are the extreme points of the unit ball of the endomorphism space of a Hilbertspace?

6.17. Prove that the weak topology of a separable Hilbert space becomes metrizable if re-stricted onto the unit ball.

6.18. Show that an idempotent operator P in a Hilbert space is an orthoprojection if andonly if P commutes with P ∗.

6.19. Let (akl)k,l∈N be an infinite matrix such that akl ≥ 0 for all k and l. Assume furtherthat there are also pk and β, γ > 0 satisfying

∞∑k=1

aklpk ≤ βpl;∞∑l=1

aklpl ≤ γpk (k, l ∈ N).

Then there is some T in B(l2) such that (ek, el) = akl and ‖T‖ =√βγ, where ek is the

characteristic function of k, a member of N.

Chapter 7

Principles of Banach Spaces

7.1. Banach’s Fundamental Principle

7.1.1. Lemma. Let U be a convex set with nonempty interior in a multi-normed space: intU 6= ∅. Then

(1) 0 ≤ α < 1⇒ α cl U + (1− α) intU ⊂ intU ;(2) coreU = intU ;(3) clU = cl intU ;(4) int clU = intU .

C (1): For u0 ∈ intU , the set intU − u0 is an open neighborhood of zeroin virtue of 5.2.10. Whence, given 0 ≤ α < 1, obtain

α clU = clαU ⊂ αU + (1− α)(intU − u0)= αU + (1− α) intU − (1− α)u0

⊂ αU + (1− α)U − (1− α)u0 ⊂ U − (1− α)u0.

Thus, (1− α)u0 + α clU ⊂ U and so U includes (1− α) intU + α clU . The last setis open as presenting the sum of α clU and (1− α) intU , an open set.

(2): Undoubtedly, intU ⊂ coreU . If u0 ∈ intU and u ∈ coreU then u =αu0 + (1− α)u1 for some u1 in U and 0 < α < 1. Since u1 ∈ clU , from (1) deducethat u ∈ intU .

(3): Clearly, cl intU ⊂ clU for intU ⊂ U . If, in turn, u ∈ clU ; then, choosingu0 in the set intU and putting uα := αu0+(1−α)u, infer that uα0 = 0 OO as α0 = 0 OO

and uα ∈ intU when 0 < α < 1. Thus, by construction u ∈ cl intU .(4): From the inclusions intU ⊂ U ⊂ clU it follows that intU ⊂ int clU .

If now u ∈ int clU then, in virtue of (2), u ∈ core clU . Consequently, taking u0in the set intU again, find u1 ∈ clU and 0 < α < 1 satisfying u = αu0 + (1−α)u1.Using (1), finally infer that u ∈ intU . B

106 Chapter 7

7.1.2. Remark. In the case of finite dimensions, the condition intU 6= ∅ maybe omitted in 7.1.1 (2) and 7.1.1 (4). In the opposite case, the presence of an interiorpoint is an essential requirement, as shown by numerous examples. For instance,take U := Bc0 ∩ X, where c0 is the space of vanishing sequences and X is thesubspace of terminating sequences in c0, the direct sum of countably many copiesof a basic field. Evidently, coreU = ∅ and at the same time clU = Bc0 . CB

7.1.3. Definition. A subset U of a (multi)normed space X is an ideally con-vex set in X, if U is closed under the taking of countable convex combinations.More precisely, U is ideally convex if for all sequences (αn)n∈N and (un)n∈N, withαn ∈ R+,

∑∞n=1 αn = 1 and un ∈ U such that the series

∑∞n=1 αnun converges

in X to some u, the containment holds: u ∈ U .

7.1.4. Examples.(1) Translation (by a vector) preserves ideal convexity. CB(2) Every closed convex set is ideally convex. CB(3) Every open convex set is ideally convex.

C Take an open and convex U . If U = ∅ then nothing is left to proof. If U 6= ∅then by 7.1.4 (1) it may be assumed that 0 ∈ U and, consequently, U = {pU < 1},where pU is the Minkowski functional of U . Let (un)n∈N and (αn)n∈N be sequencesin U and in R+ such that

∑∞n=1 αn = 1, with the element u :=

∑∞n=1 αnun failing

to lie in U . By virtue of 7.1.4 (2), u belongs to clU = {pU ≤ 1} and so pU (u) = 1.On the other hand, it is clear that pU (u) ≤

∑∞n=1 αnpU (un) ≤ 1 =

∑∞n=1 αn

(cf. 7.2.1). Thus, 0 =∑∞n=1(αn − αnpU (un)) =

∑∞n=1 αn(1 − pU (un)). Whence

αn = 0 for all n ∈ N. We arrive at a contradiction. B(4) The intersection of a family of ideally convex sets is ideally con-

vex. CB(5) Every convex subset of a finite-dimensional space is ideally con-

vex. CB

7.1.5. Banach’s Fundamental Principle. In a Banach space, each ideallyconvex set with absorbing closure is a neighborhood of zero.

C Let U be such set in a Banach space X. By hypothesis X = ∪n∈N n clU . Bythe Baire Category Theorem, X is nonmeager and so there is some n in N such thatintn clU 6= ∅. Therefore, int clU = 1/n intn clU 6= ∅. By hypothesis 0 ∈ core clU .Consequently, from 7.1.1 it follows that 0 ∈ int clU . In other words, there is a δ > 0such that clU ⊃ δBX . Consequently,

ε > 0⇒ cl 1/εU ⊃ δ/εBX .

Using the above implication, show that U ⊃ δ/2BX .

Principles of Banach Spaces 107

Let x0 ∈ δ/2BX . Putting ε := 2, choose y1 ∈ 1/εU from the condition ‖y1 −x0‖ ≤ 1/2εδ. Thus obtain an element u1 of U such that

∥∥1/2u1 − x0∥∥ ≤ 1/2εδ =

1/4δ.Now putting x0 := −1/2u1 + x0 and ε := 4 and applying the argument of the

preceding paragraph, find an element u2 of U satisfying∥∥1/4u2 + 1/2u1 − x0

∥∥ ≤1/2εδ = 1/8δ. Proceeding by induction, construct a sequence (un)n∈N of the pointsof U which possesses the property that the series

∑∞n=1

1/2nun converges to x0.Since

∑∞n=1

1/2n = 1 and the set U is ideally convex, deduce x0 ∈ U. .7.1.6. For every ideally convex set U in a Banach space the following four

sets coincide: the core of U , the interior of U , the core of the closure of U and theinterior of the closure of U .C It is clear that intU ⊂ coreU ⊂ core clU . If u ∈ core clU then cl(U − u)

equal to clU−u is an absorbing set. An ideally convex set translates into an ideallyconvex set (cf. 7.1.4 (1)). Consequently, U−u is a neighborhood of zero by Banach’sFundamental Principle. By virtue of 5.2.10, u belongs to intU . Thus, intU =coreU = core clU . Using 7.1.1, conclude that int clU = intU. .

7.1.7. The core and the interior of a closed convex set in a Banach spacecoincide.

C A closed convex set is ideally convex. B

7.1.8. Remark. Inspection of the proof of 7.1.5 shows that the condition forthe ambient space to be a Banach space in 7.1.7 is not utilized to a full extent.There are examples of incomplete normed spaces in which the core and interior ofeach closed convex set coincide. A space with this property is called barreled. Theconcept of barreledness is seen to make sense also in multinormed spaces. Barreledmultinormed spaces are plentiful. In particular, such are all Frechet spaces.

7.1.9. Counterexample. Each infinite-dimensional Banach space containsabsolutely convex, absorbing and not ideally convex sets.

C Using, for instance, a Hamel basis, take a discontinuous linear functional f .Then the set {|f | ≤ 1} is what was sought. B

7.2. Boundedness Principles

7.2.1. Let p : X0 = 0 OO R be a sublinear functional on a normed space (X, ‖·‖).The following conditions are equivalent:

(1) p is uniformly continuous;(2) p is continuous;(3) p is continuous at zero;(4) {p ≤ 1} is a neighborhood of zero;(5) ‖p‖ := sup{|p(x)| : ‖x‖ ≤ 1} < +∞; i.e., p is bounded.

108 Chapter 7

C The implications (1) ⇒ (2) ⇒ (3) ⇒ (4) are immediate.(4) ⇒ (5): There is some t > 0 such that t−1BX ⊂ {p ≤ 1}. Given ‖x‖ ≤ 1,

thus find p(x) ≤ t. In addition, the inequality −p(−x) ≤ p(x) implies that −p(x) ≤t for x ∈ BX . Finally, ‖p‖ ≤ t < +∞.

(5) ⇒ (1): From the subadditivity of p, given x, y ∈ X, observe that

p(x)− p(y) ≤ p(x− y); p(y)− p(x) ≤ p(y − x).

Whence |p(x)− p(y)| ≤ p(x− y) ∨ p(y − x) ≤ ‖p‖ ‖x− y‖. .

7.2.2. Gelfand Theorem. Every lower semicontinuous sublinear functionalwith domain a Banach space is continuous.

C Let p be such functional. Then the set {p ≤ 1} is closed (cf. 4.3.8). Sincedom p is the whole space; therefore, by 3.8.8, {p ≤ 1} is an absorbing set. ByBanach’s Fundamental Principle {p ≤ 1} is a neighborhood of zero. Applicationto 7.2.1 completes the proof. B

7.2.3. Remark. The Gelfand Theorem is stated amply as follows: “If X isa Banach space then each of the equivalent conditions 7.2.1 (1)–7.2.1 (5) amountsto the statement: ‘p is lower semicontinuous on X.’ ” Observe immediately thatthe requirement dom p = X may be slightly relaxed by assuming dom p to bea nonmeager linear set and withdrawing the condition for X to be a Banach space.

7.2.4. Equicontinuity Principle. Suppose that X is a Banach space and Yis a (semi)normed space. For every nonempty set E of continuous linear operatorsfrom X to Y the following statements are equivalent:

(1) E is pointwise bounded; i.e., for all x ∈ X the set {Tx : T ∈ E } isbounded in Y ;

(2) E is equicontinuous.C (1) ⇒ (2): Put q(x) := sup{p(Tx) : T ∈ E }, with p the (semi)norm of Y .

Evidently, q is a lower semicontinuous sublinear functional and so by the GelfandTheorem ‖q‖ < +∞; i.e., p(T (x − y)) ≤ ‖q‖ ‖x − y‖ for all T ∈ E . Consequently,T×−1({dp ≤ ε}) ⊃ {d‖·‖ ≤ ε/‖q‖} for every T in E , where ε > 0 is taken arbitrarily.This means the equicontinuity property of E .

(2) ⇒ (1): Straightforward. B

7.2.5. Uniform Boundedness Pronciple. Let X be a Banach space andlet Y be a normed space. For every nonempty family (Tξ)ξ∈� of bounded operatorsthe following statements are equivalent:

(1) x ∈ X ⇒ supξ∈� ‖Tξx‖ < +∞;(2) supξ∈� ‖Tξ‖ < +∞.

C It suffices to observe that 7.2.5 (2) is another expression for 7.2.4 (2). B


7.2.6. Let X be a Banach space and let U be a subset of X ′. Then thefollowing statements are equivalent:

(1) U is bounded in X ′;(2) for every x in X the numeric set {〈x |x′〉 : x′ ∈ U} is bounded

in F.C This is a particular case of 7.2.5. B

7.2.7. LetX be a normed space and let U be a subset ofX. Then the followingstatements are equivalent:

(1) U is bounded in X;(2) for every x′ in X ′ the numeric set {〈x |x′〉 : x ∈ U} is bounded

in F.C Only (2) ⇒ (1) needs examining. Observe that X ′ is a Banach space

(cf. 5.5.7) and X is isometrically embedded into X ′′ by the double prime map-ping (cf. 5.1.10 (8)). So, the claim follows from 7.2.6. B

7.2.8.Remark. The message of 7.2.7 (2) may be reformulated as “U is boundedin the space (X, σ(X, X ′))” or, in view of 5.1.10 (4), as “U is weakly bounded.”The duality between 7.2.6 and 7.2.7 is perfectly revealed in 10.4.6.

7.2.9. Banach–Steinhaus Theorem. Let X and Y be Banach spaces. As-sume further that (Tn)n∈N, Tn ∈ B(X, Y ), is a sequence of bounded operators.Put E := {x ∈ X : ∃ limTnx}. The following conditions are equivalent:

(1) E = X;(2) supn∈N ‖Tn‖ < +∞ and E is dense in X.

Under either (and, hence, both) of the conditions (1) and (2) the mapping T0 :X0 = 0 OO , defined as T0x := limTnx, presents a bounded linear operator and ‖T0‖ ≤lim inf ‖Tn‖.

C If E = X then, of course, clE = X. In addition, for every x in X the se-quence (Tnx)n∈N is bounded in Y (for, it converges). Consequently, by the UniformBoundedness Principle supn∈N ‖Tn‖ < +∞ and (1) ⇒ (2) is proven.

If (2) holds and x ∈ X then, given x ∈ E and m, k ∈ N, infer that

‖Tmx− Tkx‖ = ‖Tmx− Tmx+ Tmx− Tkx+ Tkx− Tkx‖≤ ‖Tmx− Tmx‖+ ‖Tmx− Tkx‖+ ‖Tkx− Tkx‖≤ ‖Tm‖ ‖x− x‖+ ‖Tmx− Tkx‖+ ‖Tk‖ ‖x− x‖

≤ 2 supn∈N‖Tn‖ ‖x− x‖+ ‖Tmx− Tkx‖.

Take ε > 0 and choose, first, x ∈ E such that 2 supn ‖Tn‖ ‖x − x‖ ≤ ε/2, and,second, n ∈ N such that ‖Tmx− Tkx‖ ≤ ε/2 for m, k ≥ n. By virtue of what was

110 Chapter 7

proven ‖Tmx − Tkx‖ ≤ ε; i.e., (Tnx)n∈N is a Cauchy sequence in Y . Since Y isa Banach space, conclude that x ∈ E. Thus, (2) ⇒ (1) is proven.

To complete the proof it suffices to observe that

‖T0x‖ = lim ‖Tnx‖ ≤ lim inf ‖Tn‖ ‖x‖

for all x ∈ X, because every norm is a continuous function. B

7.2.10. Remark. Under the hypotheses of the Banach–Steinhaus Theorem,the validity of either of the equivalent items 7.2.9 (1) and 7.2.9 (2) implies that(Tn) converges to T0 compactly on X (= uniformly on every compact subset of X).In other words,

supx∈Q

‖Tnx− T0x‖0 = 0 OO

for every (nonempty) compact set Q in X.C Indeed, it follows from the Gelfand Theorem that the sublinear functional

pn(x) := sup{‖Tmx−T0x‖ : m ≥ n} is continuous. Moreover, pn(x) ≥ pn+1(x) andpn(x)0 = 0 OO for all x ∈ X. Consequently, the claim follows from the Dini Theorem:“Each decreasing sequence of continuous real functions which converges pointwiseto a continuous function on a compact set converges uniformly.” B

7.2.11. Singularity Fixation Principle. Let X be a Banach space and letY be a normed space. If (Tn)n∈N is a sequence of operators, Tn ∈ B(X, Y )and supn ‖Tn‖ = +∞ then there is a point x of X satisfying supn ‖Tnx‖ = +∞.The set of such points “fixing a singularity” is residual.C The first part of the assertion is contained in the Uniform Boundedness

Principle. The second requires referring to 7.2.3 and 4.7.4. B

7.2.12. Singularity Condensation Principle. Let X be a Banach spaceand let Y be a normed space. If (Tn,m)n,m∈N is a family of operators, Tn ∈ B(X, Y ),such that supn ‖Tn,m‖ = +∞ for everym ∈ N then there is a point x of X satisfyingsupn ‖Tn,mx‖ = +∞ for all m ∈ N. CB

7.3. The Ideal Correspondence Principle

7.3.1. Let X and Y be vector spaces. A correspondence F ⊂ X ×Y is convexif and only if for x1, x2 ∈ X and α1, α2 ∈ R+ such that α1 +α2 = 1, the inclusionholds:

F (α1x1 + α2x2) ⊃ α1F (x1) + α2F (x2).

C⇐: If (x1, y1), (x2, y2) ∈ F and α1, α2 ≥ 0, α1+α2 = 1, then α1y+α2y2 ∈F (α1x1 + α2x2) since y1 ∈ F (x1) and y2 ∈ F (x2).⇒: If either x1 or x2 fails to enter in domF then there is nothing to prove. If

y1 ∈ F (x1) and y2 ∈ F (x2) with x1, x2 ∈ domF then α1(x1, y1) +α2(x2, y2) ∈ Ffor α1, α2 ≥ 0, α1 + α2 = 1 (cf. 3.1.2 (8)). B


7.3.2. Remark. Let X and Y be Banach spaces. It is clear that there aremany ways for furnishing the space X × Y with a norm so that the norm topologybe coincident with the product of the topologies τX and τY . For instance, it ispossible to put ‖(x, y)‖ := ‖x‖X + ‖y‖Y ; i.e., to define the norm on X × Y as thatof the 1-sum of X and Y . Observe immediately that the concept of ideally convexset has a linear topological character: the class of objects distinguished in a spaceis independent of the way of introducing the space topology; in particular, the classremains invariant under passage to an equivalent (multi)norm. In this connectionthe next definition is sound.

7.3.3. Definition. A correspondence F ⊂ X × Y , with X and Y Banachspaces, is called ideally convex or, briefly, ideal if F is an ideally convex set.

7.3.4. Ideal Correspondence Lemma. The image of a bounded ideallyconvex set under an ideal correspondence is an ideally convex set.

C Let F ⊂ X × Y be an ideal correspondence and let U be a bounded ideallyconvex set in X. If U ∩ domF = ∅ then F (U) = ∅ and nothing is left unproven.Let now (yn)n∈N ⊂ F (U); i.e., yn ∈ F (xn), where xn ∈ U and n ∈ N. Let, finally,(αn) be a sequence of positive numbers such that

∑∞n=1 αn = 1 and, moreover,

there is a sum of the series y :=∑∞n=1 αnyn in Y . It is beyond a doubt that

∞∑n=1

‖αnxn‖ =∞∑n=1

αn‖xn‖ ≤∞∑n=1

αn sup ‖U‖ = sup ‖U‖ < +∞

in view of the boundedness property of U . Since X is complete, from 5.5.3 itfollows that X contains the element x :=

∑∞n=1 αnxn. Consequently, (x, y) =∑∞

n=1 αn(xn, yn) in the space X × Y . Successively using the ideal convexity of Fand U , infer that (x, y) ∈ F and x ∈ U . Thus, y ∈ F (U). .

7.3.5. Ideal Correspondence Principle. Let X and Y be Banach spaces.Assume further that F ⊂ X × Y is an ideal correspondence and (x, y) ∈ F .A correspondence F carries each neighborhood of x onto a neighborhood about yif and only if y ∈ coreF (X).

C ⇒: This is obvious.⇐: On account of 7.1.4 it may be assumed that x = 0 and y = 0. Since each

neighborhood of zero U includes εBX for some ε > 0, it suffices to settle the caseU := BX . Since U is bounded; by 7.3.4, F (U) is ideally convex. To complete theproof, it suffices to show that F (U) is an absorbing set and to cite 7.1.6.

Take an arbitrary element y of Y . Since by hypothesis 0 ∈ coreF (X), there isa real α in R+ such that αy ∈ F (X). In other words, αy ∈ F (X) for some x in X.

112 Chapter 7

If ‖x‖ ≤ 1 then there is nothing to prove. If ‖x‖ > 1 then λ := ‖x‖−1 < 1. Whence,using 7.3.1, infer that

αλy = (1− λ)0 + λαy ∈ (1− λ)F (0) + λF (x)⊂ F ((1− λ)0 + λx) = F (λx) ⊂ F (BX) = F (U).

Here use was made of the fact that ‖λx‖ = 1; i.e., λx ∈ BX . B

7.3.6. Remark. The property of F , described in 7.3.5, is referred to as theopenness of F at (x, y).

7.3.7. Remark. The Ideal Correspondence Principle is formally weaker thatBanach’s Fundamental Principle. Nevertheless, the gap is tiny and can be easilyfilled in. Namely, the conclusion of 7.3.5 remains valid if we suppose that y ∈core clF (X), on additionally requiring ideal convexity from F (X). The requirementis not too stringent and certainly valid provided that the domain of F is boundedin virtue of 7.3.4. As a result of this slight modification, 7.1.5 becomes a particularcase of 7.3.5. In this connection the claim of 7.3.5 is often referred to as Banach’sFundamental Principle for a Correspondence.

7.3.8. Definition. Let X and Y be Banach spaces and let F ⊂ X × Y bea correspondence. Then F is called closed if F is a closed set in X × Y .

7.3.9. Remark. For obvious reasons, a closed correspondence is often referredto as a closed-graph correspondence.

7.3.10. A correspondence F is closed if and only if for all sequences (xn) in Xand (yn) in Y such that xn ∈ domF, yn ∈ F (xn) and xn0 = 0 OO , yn0 = 0 OO , itfollows that x ∈ domF and y ∈ F (x). CB

7.3.11. Assume that X and Y are Banach spaces and F ⊂ X × Y is a closedconvex correspondence. Further, let (x, y) ∈ F and y ∈ core imF . Then F carrieseach neighborhood of x onto a neighborhood about y.

C A closed convex set is ideally convex and so all follows from 7.3.5. B

7.3.12. Definition. A correspondence F ⊂ X×Y is called open if the imageof each open set in X is an open set in Y .

7.3.13. Open Correspondence Principle. Let X and Y be Banach spacesand let F ⊂ X × Y be an ideal correspondence, with imF an open set. Then F isan open correspondence.

C Let U be an open set in X. If y ∈ F (U) then there is some x in U suchthat (x, y) ∈ F . It is clear that y ∈ core imF . By the criterion of 7.3.5 F (U) isa neighborhood of y because U is a neighborhood of x. This means that F (U) isan open set. B


7.4. Open Mapping and Closed Graph Theorems

7.4.1. Definition. A member T of L (X, Y ) is a homomorphism, if T ∈B(X, Y ) and T is an open correspondence.

7.4.2. Assume that X is a Banach space, Y is a normed space and T is a ho-momorphism from X to Y . Then imT = Y and Y is a Banach space.

C It is obvious that imT = Y . Presuming T to be a monomorphism, observethat T−1 ∈ L (Y, X). Since T is open, T−1 belongs to B(Y, X), which ensures thecompleteness of Y (the inverse image of a Cauchy sequence in a subset is a Cauchysequence in the inverse image of the subset). In the general case, consider thecoimage coimT := X/ kerT endowed with the quotient norm. In virtue of 5.5.4,coimT is a Banach space. In addition, by 2.3.11 there is a unique quotient T of Tby coimT , the monoquotient of T . Taking account of the definition of quotient normand 5.1.3, conclude that T is a homomorphism. Furthermore, T is a monomorphismby definition. It remains to observe that imT = imT = Y. .

7.4.3. Remark. As regards the monoquotient T : coimT0 = 0 OO of T , it maybe asserted that ‖T‖ = ‖T‖. CB

7.4.4. Banach Homomorphism Theorem. Every bounded epimorphismfrom one Banach space onto the other is a homomorphism.

C Let T ∈ B(X, Y ) and imT = Y . On applying the Open CorrespondencePrinciple to T , complete the proof. B

7.4.5. Banach Isomorphism Theorem. Let X and Y be Banach spacesand T ∈ B(X, Y ). If T is an isomorphism of the vector spaces X and Y , i.e.kerT = 0 and imT = Y ; then T−1 ∈ B(Y, X).

C A particular case of 7.4.4. B

7.4.6. Remark. The Banach Homomorphism Theorem is often referred to asthe Open Mapping Theorem for understandable reasons. Theorem 7.4.5 is brieflyformulated as follows: “A continuous (algebraic) isomorphism of Banach spaces isa topological isomorphism.” It is also worth observing that the theorem is sometimesreferred to as theWell-Posedness Principle and verbalized as follows: “If an equationTx = y, with T ∈ B(X, Y ) and X and Y Banach spaces, is uniquely solvable givenan arbitrary right side; then the solution x depends continuously on the right sidey.”

7.4.7. Banach Closed Graph Theorem. Let X and Y be Banach spacesand let T in L (X, Y ) be a closed linear operator. Then T is continuous.

C The correspondence T−1 is ideal and T−1(Y ) = X. B

114 Chapter 7

7.4.8. Corollary. Suppose that X and Y are Banach spaces and T is a linearoperator from X to Y . The following conditions are equivalent:

(1) T ∈ B(X, Y );(2) for every sequence (xn)n∈N in X, together with some x in X and y

in Y satisfying xn0 = 0 OO and Txn0 = 0 OO , it happens that y = Tx.C (2): This is a reformulation of the closure property of T . B

7.4.9. Definition. A subspace X1 of a Banach space X is complemented(rarely, a topologically complemented), ifX1 is closed and, moreover, there is a closedsubspace X2 such that X = X1 ⊕X2 (i.e., X1 ∧X2 = 0 and X1 ∨X2 = X). TheseX1 and X2 are called complementary to one another.

7.4.10. Complementation Principle. For a subspace X1 of some Banachspace X one of the following conditions amounts to the other:

(1) X1 is complemented;(2) X1 is the range of a bounded projection; i.e., there is a member P

of B(X) such that P 2 = P and imP = X1.C (1) ⇒ (2): Let P be the projection of X onto X1 along X2 (cf. 2.2.9 (4)).

Let (xn)n∈N be a sequence in X with xn0 = 0 OO and Pxn0 = 0 OO . It is clear thatPxn ∈ X1 for n ∈ N. Since X1 is closed, by 4.1.19 y ∈ X1. Similarly, the condition(xn − Pxn ∈ X2 for n ∈ N) implies that x− y ∈ X2. Consequently, P (x− y) = 0.Furthermore, y = Py; i.e., y = Px. It remains to refer to 7.4.8.

(2) ⇒ (1): It needs showing only that X1 equal to imP is closed. Take a se-quence (xn)n∈N in X1 such that xn0 = 0 OO in X. Then Pxn0 = 0 OO x in view of theboundedness of P . Obtain Pxn = xn, because xn ∈ imP and P is an idempotent.Finally, x = Px, i.e. x ∈ X1, what was required. B

7.4.11. Examples.(1) Every finite-dimensional subspace is complemented. CB(2) The space c0 is not complemented in l∞.

C It turns out more convenient to work with X := l∞(Q) and Y := c0(Q),where Q is the set of rational numbers. Given t ∈ R, choose a sequence (tn)of pairwise distinct rational numbers other than t and such that tn0 = 0 OO . LetQt := {tn : n ∈ N}. Observe that Qt′ ∩Qt′′ is a finite set if t′ 6= t′′.

Let χt be the coset containing the characteristic function of Qt in the quotientspace X/Y and V := {χt : t ∈ R}. Since χt′ 6= χt′′ for t′ 6= t′′, the set V isuncountable. Take f ∈ (X/Y )′ and put Vf := {v ∈ V : f(v) 6= 0}. It is evidentthat Vf = ∪n∈NVf (n), where Vf (n) := {v ∈ V : |f(v)| ≥ 1/n}. Given m ∈ N andpairwise distinct v1, . . . , vm in Vf (n), put αk := |f(vk)|/f(vk) and x :=

∑nk=1 αkvk

to find ‖x‖ ≤ 1 and ‖f‖ ≥ |f(x)| = |∑mk=1 αkf(vk)| = |

∑mk=1 |f(vk)|| ≥ m/n.


Hence, Vf (n) is a finite set. Consequently, Vf is countable. Whence it followsthat for every countable subset F of (X/Y )′ there is an element v of V satisfying(∀ f ∈ F ) f(v) = 0. At the same time the countable set of the coordinate projectionsδq : x 7→ x(q) (q ∈ Q) is total over l∞(Q); i.e., (∀ q ∈ Q) δq(x) = 0 ⇒ x = 0 forx ∈ l∞(Q). It remains to compare the above observations. B

(3) Every closed subspace of a Hilbert space is complemented in virtueof 6.2.6. Conversely, if, in an arbitrary Banach space X with dimX ≥ 3, each closedsubspace is the range of some projection P with ‖P‖ ≤ 1; then X is isometricallyisomorphic to a Hilbert space (this is the Kakutani Theorem). The next fact ismuch deeper:

Lindenstrauss–TzafririTheorem. A Banach space having every closed sub-space complemented is topologically isomorphic to a Hilbert space.

7.4.12. Sard Theorem. Suppose that X, Y, and Z are Banach spaces. TakeA ∈ B(X, Y ) and B ∈ B(Y, Z). Suppose further that imA is a complementedsubspace in Y. The diagram

Z

X YA

XB

-

?

@@@R

is commutative for some X in B(Y, Z) if and only if kerA ⊂ kerB.C Only ⇐ needs examining. Moreover, in the case imA = Y the sole member

X0 of L (Y, Z) such that X0A = B is continuous. Indeed, X −10 (U) = A(B−1(U))

for every open set U in Z. The set B−1(U) is open in virtue of the boundednessproperty of B, and A(B−1(U)) is open by the Banach Homomorphism Theorem.In the general case, construct X0 ∈ B(imA, Z) and take as X the operator X0P ,where P is some continuous projection of Y onto imA. (Such a projection isavailable by the Complementation Principle.) B

7.4.13. Phillips Theorem. Suppose that X, Y, and Z are Banach spaces.Take A ∈ B(Y, X) and B ∈ B(Z, X). Suppose further that kerA is a comple-mented subspace of Y . The diagram

Z

X YA

XB

�

6

@@@I

is commutative for some X in B(Z, Y ) if and only if imA ⊃ imB.

116 Chapter 7

C Once again only ⇐ needs examining. Using the definition of complementedsubspace, express Y as the direct sum of kerA and Y0, where Y0 is a closed subspace.By 5.5.9 (1), Y0 is a Banach space. Consider the part A0 of A in Y0. Undoubtedly,imA0 = imA ⊃ imB. Consequently, by 2.3.13 and 2.3.14 the equation A0X0 = Bhas a unique solution X0 := A−1

0 . It suffices to prove that the operator X0, treatedas a member of L (Z, Y0), is bounded.

The operator X0 is closed. Indeed (cf. 7.4.8), if zn0 = 0 OO and A−10 Bzn0 = 0 OO

then Bzn0 = 0 OO z, since B is bounded. In addition, by the continuity of A0,the correspondence A−1

0 ⊂ X × Y0 is closed; and so 7.3.10 yields the equalityy = A−1

0 Bz. B

7.4.14. Remark. We use neither the completeness of Z in proving the SardTheorem nor the completeness of X in proving the Phillips Theorem.

7.4.15. Remark. The Sard Theorem and the Phillips Theorem are in “formalduality”; i.e., one results from the other by reversing arrows and inclusions andsubstituting ranges for kernels (cf. 2.3.15).

7.4.16. Two Norm Principle. Let a vector space be complete in each of twocomparable norms. Then the norms are equivalent.

C For definiteness, assume that ‖ · ‖2 � ‖ · ‖1 in X. Consider the diagram

(X, ‖ · ‖1)

(X, ‖ · ‖1) (X, ‖ · ‖2)IX

XIX

�

6

@@@I

By the Phillips Theorem some continuous operator X makes the diagram commu-tative. Such an operator is unique: it is IX . B

7.4.17. Graph Norm Principle. Let X and Y be Banach spaces and let Tin L (X, Y ) be a closed operator. Given x ∈ X, define the graph norm of x as‖x‖gr T := ‖x‖X + ‖Tx‖Y . Then ‖ · ‖gr T ∼ ‖ · ‖X .

C Observe that the space (X, ‖ · ‖gr T ) is complete. Further, ‖ · ‖gr T ≥ ‖ · ‖X .It remains to refer to the Two Norm Principle. B

7.4.18. Definition. A normed space X is a Banach range, if X is the rangeof some bounded operator given on some Banach space.

7.4.19. Kato Criterion. Let X be a Banach space and X = X1⊕X2, whereX1, X2 ∈ Lat(X). The subspaces X1 and X2 are closed if and only if each of themis a Banach range.


C ⇒: A corollary to the Complementation Principle.⇐: Let Z be a some Banach range, i.e. Z = T (Y ) for some Banach space Y

and T ∈ B(Y, Z). Passing, if need be, to the monoquotient of T , we may assumethat T is an isomorphism. Put ‖z‖0 := ‖T−1z‖Y . It is clear that (Z, ‖ · ‖0) isa Banach space and ‖z‖ = ‖TT−1z‖ ≤ ‖T‖‖T−1z‖ = ‖T‖‖z‖0; i.e., ‖ · ‖0 � ‖ · ‖Z .Applying this construction to X1 and X2, obtain Banach spaces (X1, ‖ · ‖1) and(X2, ‖ · ‖2). Now ‖ · ‖k � ‖ · ‖X on Xk for k := 1, 2.

Given x1 ∈ X1 and x2 ∈ X2, put ‖x1 + x2‖0 := ‖x1‖1 + ‖x2‖2. Thereby weintroduce in X some norm ‖·‖ that is stronger than the initial norm ‖·‖X . By con-struction (X, ‖ · ‖0) is a Banach space. It remains to refer to 7.4.16. B

7.5. The Automatic Continuity Principle

7.5.1. Lemma. Let f : X0 = 0 OO R· be a convex function on a (multi)normedspace X. The following statements are equivalent:

(1) U := int dom f 6= ∅ and f |U is a continuous function;(2) there is a nonempty open set V such that sup f(V ) < +∞.

C (1) ⇒ (2): This is obvious.(2) ⇒ (1): It is clear that U 6= ∅. Using 7.1.1, observe that each point u of U

has a neighborhood W in which f is bounded above, i.e. t := sup f(W ) < +∞.Without loss of generality, it may be assumed that u := 0, f(u) := 0 and W isan absolutely convex set. From the convexity of f , for every α ∈ R+ such thatα ≤ 1 and for an arbitrary v in W , obtain

f(αv) = f(αv + (1− α)0) ≤ αf(v) + (1− α)f(0) = αf(v);f(αv) + αf(−v) ≥ f(αv) + f(α(−v))

= 2(1/2f(αv) + 1/2f(−αv)

)≥ 2f(0) = 0.

Therefore, |f(αW )| ≤ αt, which implies that f is continuous at zero. B

7.5.2. Corollary. If x ∈ int dom f and f is continuous at x then the sub-differential ∂x(f) contains only continuous functionals.

C If l ∈ ∂x(f) then (∀x ∈ X) l(x) ≤ l(x) + f(x) − f(x) and so l is boundedabove on some neighborhood about x. Consequently, l is continuous at x by 7.5.1.From 5.3.7 derive that l is continuous. B

7.5.3. Corollary. Every convex function on a finite-dimensional space is con-tinuous on the interior of its domain. CB

7.5.4.Definition. A function f : X0 = 0 OO R· is called ideally convex if 0 = 0 OOOO

is an ideal correspondence.

118 Chapter 7

7.5.5.Automatic Continuity Principle Every ideally convex function on a Ba-nach space is continuous on the core of its domain.

C Let f be such function. If core dom f = ∅ then there is nothing to prove.If x ∈ core dom f then put t := f(x) and F := (0 = 0 OOOO )−1 ⊂ R × X. Applyingthe Ideal Correspondence Principle, find a δ > 0 from the condition F (t + BR) ⊃x+ δBX . Whence, in particular, infer the estimate f(x+ δBX) ≤ t+ 1. In virtueof 7.5.1, f is continuous on int dom f . Since x ∈ int dom f ; therefore, by Lemma7.1.1, core dom f = int dom f . B

7.5.6. Remark. Using 7.3.6, it is possible to prove that an ideally convexfunction f , defined in a Banach space on a subset with nonempty core, is locallyLipschitz on int dom f . In other words, given x0 ∈ int dom f , there are a positivenumber L and a neighborhood U about x0 such that ‖f(x)− f(x0)‖ ≤ L‖x− x0‖whenever x ∈ U . CB

7.5.7. Corollary. Let f : X0 = 0 OO R· be an ideally convex function on a Ba-nach space X and x ∈ core dom f . Then the directional derivative f ′(x) is a con-tinuous sublinear functional and ∂x(f) ⊂ X ′.

C Apply the Automatic Continuity Principle twice. B

7.5.8. Remark. In view of 7.5.7, in studying a Banach space X, only contin-uous functionals on X are usually admitted into the subdifferential of a functionf : X0 = 0 OO R· at a point x; i.e., we agree to define

∂x(f) := ∂x(f) ∩X ′.

Proceed likewise in (multi)normed spaces. If a need is felt to distinguish the “old”(wider) subdifferential, a subset of X#, from the “new” (narrower) subdifferential,a subset of X ′; then the first is called algebraic, whereas the second is called topo-logical. With this in mind, we refer to the facts indicated in 7.5.2 and 7.5.7 as to theCoincidence Principle for algebraic and topological subdifferentials. Observe finallythat if f := p is a seminorm on X then, for similar reasons, it is customary to put|∂|(p) := |∂|(p) ∩X ′.

7.5.9. Ideal Hahn–Banach Theorem. Let f : Y 0 = 0 OO R· be an ideallyconvex function on a Banach space Y . Further, let X be a normed space andT ∈ B(X, Y ). If a point x in X is such that Tx ∈ core dom f then

∂x(f ◦ T ) = ∂ Tx(f) ◦ T.

C The right side of the sought formula is included into its left side for obviousreasons. If l in X ′ belongs to ∂x(f ◦ T ), then by the Hahn–Banach Theorem thereis an element l1 of the algebraic subdifferential of f at Tx for which l = l1 ◦ T .It suffices to observe that, in virtue of 7.5.7, l1 is an element of Y ′ and so it isa member of the topological subdifferential ∂ Tx(f). B


7.5.10. Balanced Hahn–Banach Theorem. Suppose that X and Y arenormed spaces. Given T ∈ B(X, Y ), let p : Y 0 = 0 OO R be a continuous seminorm.Then

|∂|(p ◦ T ) = |∂|(p) ◦ T.

C If l ∈ |∂|(p ◦ T ) then l = l1 ◦ T for some l1 in the algebraic balancedsubdifferential of p (cf. 3.7.11). From 7.5.2 it follows that l1 is continuous. Thus,|∂|(p ◦ T ) ⊂ |∂|(p) ◦ T . The reverse inclusion raises no doubts. B

7.5.11. Continuous Extension Principle. Let X0 be a subspace of X andlet l0 be a continuous linear functional on X0. Then there is a continuous linearfunctional l on X extending l0 and such that ‖l‖ = ‖l0‖.C Take p := ‖l0‖ ‖ · ‖, and consider the identical embedding ι : X00 = 0 OO .

On account of 7.5.10, l0 ∈ |∂|(p ◦ ι) = |∂|(p) ◦ ι = ‖l0‖ |∂|(‖ · ‖) ◦ ι. It suffices toobserve that |∂|(‖ · ‖X) = BX′ . B

7.5.12. Topological Separation Theorem. Let U be a convex set withnonempty interior in a space X. If L is an affine variety in X and L ∩ intU = ∅then there is a closed hyperplane H in X such that H ⊃ L and H ∩ intU = ∅. CB

7.5.13. Remark. When applying Theorem 7.5.12, it is useful to bear in mindthat a closed hyperplane is precisely a level set of a nonzero continuous linearfunctional. CB

7.5.14. Corollary. Let X0 be a subspace of X. Then

clX0 = ∩{ker f : f ∈ X ′, ker f ⊃ X0}.

C It is clear that (f ∈ X ′ & ker f ⊃ X0) ⇒ ker f ⊃ clX0. If x0 /∈ clX0then there is an open convex neighborhood about x0 disjoint from clX0. In virtueof 7.5.12 and 7.5.13 there is a functional f0, a member of (XR)′, such that ker f0 ⊃clX0 and f0(x0) = 1. From the properties of the complexifier infer that the func-tional Re−1f0 vanishes on X0 and differs from zero at the point x0. It is also beyonda doubt that the functional is continuous. B

7.6. Prime Principles

7.6.1. Let X and Y be (multi)normed vector spaces (over the same groundfield F). Assume further that X ′ and Y ′ are the duals of X and Y respectively.Take a continuous linear operator T from X to Y . Then y ′ ◦ T ∈ X ′ for y ′ ∈ Y ′and the mapping y ′ 7→ y ′ ◦ T is a linear operator. CB

7.6.2. Definition. The operator T ′ : Y ′0 = 0 OO ′, constructed in 7.6.1, is thedual or transpose of T : X0 = 0 OO .

120 Chapter 7

7.6.3. Theorem. The prime mapping T 7→ T ′ implements a linear isometryof the space B(X, Y ) into the space B(Y ′, X ′).C The prime mapping is clearly a linear operator from B(X, Y ) to L (Y ′, X ′).

Furthermore, since ‖y‖ = sup{|l(y) : l ∈ |∂|(‖ · ‖)}; therefore,

‖T ′‖ = sup{‖T ′y ′‖ : ‖y ′‖ ≤ 1}= sup{|y ′(Tx)| : ‖y ′‖ ≤ 1, ‖x‖ ≤ 1} = sup{‖Tx‖ : ‖x‖ ≤ 1} = ‖T‖,

what was required. B

7.6.4. Examples.(1) Let X and Y be Hilbert spaces. Take T ∈ B(X, Y ). Observe first

that, in a plain sense, T ∈ B(X, Y )⇔ T ∈ B(X∗, Y∗). Denote the prime mappingof X by (·)′X : X∗0 = 0 OO ′, i.e., x 7→ x′ := ( · , x); and denote the prime mappingof Y by (·)′Y : Y∗0 = 0 OO ′, i.e., y 7→ y ′ := ( · , y).

The adjoint of T , the member T ∗ of B(Y, X), and the dual of T , the mem-ber T ′ of B(Y ′, X ′), are related by the commutative diagram:

X∗T∗←−Y∗

(·)′X ↓ ↓ (·)′YX ′

T ′←−Y ′

C Indeed, it is necessary to show the equality T ′y ′ = (T ∗y)′ for y ∈ Y . Givenx ∈ X, by definition observe that

T ′y ′(x) = y ′(Tx) = (Tx, y) = (x, T ∗y) = (T ∗y)′(x).

Since x is arbitrary, the proof is complete. B(2) Let ι : X00 = 0 OO be the identical embedding of X0 into X. Then

ι′ : X0 = 0 OO ′0. Moreover, ι′(x′)(x0) = x′(x0) for all x0 ∈ X0 and x′ ∈ X ′ and ι′ is

an epimorphism; i.e., X ′ ι′−→ X ′00 = 0 OO is an exact sequence. CB

7.6.5. Definition. Let an elementary diagram XT−→ Y be given. The dia-

gram Y ′T ′−→ X ′ is referred to as resulting from setting primes or as the diagram

prime of the original diagram or as the dual diagram. If primes are set in everyelementary subdiagram in an arbitrary diagram composed of bounded linear op-erators in Banach spaces, then the so-obtained diagram is referred to as dual orresulting from setting primes in the original diagram. The term “diagram prime” isused for suggestiveness.


7.6.6. Double Prime Lemma. Let X ′′ T′′

−→ Y ′′ be the diagram that resultsfrom setting primes in the diagram X

T−→ Y twice. Then the following diagramcommutes:

XT−→Y

′′ ↓ ↓′′

X ′′T ′′−→Y ′′

Here ′′ : X0 = 0 OO ′′ and ′′ : Y 0 = 0 OO ′′ are the respective double prime mappings;i.e. the canonical embeddings of X into X ′′ and of Y into Y ′′ (cf. 5.1.10 (8)).C Let x ∈ X. We have to prove that T ′′x′′ = (Tx)′′. Take y ′ ∈ Y ′. Then

T ′′x′′(y ′) = x′′(T ′y ′) = T ′y ′(x) = y ′(Tx) = (Tx)′′(y ′).

Since y ′ ∈ Y ′ is arbitrary, the proof is complete. B

7.6.7. Diagram Prime Principle. A diagram is commutative if and onlyif so is its diagram prime.

C It suffices to convince oneself that the triangles

XT−→ Y

R↘ ↙ S

Z

X ′T ′←− Y ′

R′ ↖ ↗ S′

Z ′

are commutative or not simultaneously. Since R = ST ⇒ R′ = (ST )′ = T ′S′;therefore, the commutativity of the triangle on the left entails the commutativityof the triangle on the right. If the latter commutes then by what was alreadyproven R′′ = S′′T ′′. Using 7.6.6, argue as follows: (Rx)′′ = R′′x′′ = S′′T ′′x′′ =S′′(T ′′x′′) = S′′(Tx)′′ = (STx)′′ for all x ∈ X. Consequently, R = ST . B

7.6.8. Definition. Let X0 be a subspace of X and let X0 be a subspaceof X ′. Put

X⊥0 := {f ∈ X ′ : ker f ⊃ X0} = |∂|(δ(X0));⊥X0 := {x ∈ X : f ∈X0 ⇒ f(x) = 0} = ∩{ker f : f ∈X0}.

The subspace X⊥0 is the (direct) polar of X0, and the subspace ⊥X0 is the reversepolar of X0. A less exact term “annihilator” is also in use.

7.6.9. Definition. Let X and Y be Banach spaces. An arbitrary element Tof B(X, Y ) is a normally solvable operator, if imT is a closed subspace of Y . Thenatural term “closed range operator” is also in common parlance.

122 Chapter 7

7.6.10. An operator T , a member of B(X, Y ), is normally solvable if and onlyif T is a homomorphism, when regarded as acting from X to imT .C ⇒: The Banach Homomorphism Theorem.⇐: Refer to 7.4.2. B

7.6.11. Polar Lemma. Let T ∈ B(X, Y ). Then(1) (imT )⊥ = ker(T ′);(2) if T is normally solvable then

imT = ⊥ ker(T ′), (kerT )⊥ = im(T ′).

C (1): y ′ ∈ ker(T ′)⇔ T ′y ′ = 0⇔ (∀x ∈ X) T ′y ′(x) = 0⇔ (∀x ∈ X) y ′(Tx) = 0⇔ y ′ ∈ (imT )⊥.

(2): The equality cl imT = ⊥ ker(T ′) follows from 7.5.13. Furthermore, by hy-pothesis imT is closed.

If x′ = T ′y ′ and Tx = 0 then x′(x) = T ′y ′(x) = y ′(Tx) = 0, which meansthat x′ ∈ (kerT )⊥. Consequently, im(T ′) ⊂ (kerT )⊥. Now take x′ ∈ (kerT )⊥.Considering the operator T acting onto imT , apply the Sard Theorem to the leftside of the diagram

XT−→ imT −→ Y

x′ ↘ ↓ y′0 ↙ y′

F

As a result of this, obtain y ′0 in (imT )′ such that y ′0◦T = x′. By the ContinuousExtension Principle there is an element y ′ of Y ′ satisfying y ′ ⊃ y ′0. Thus, x′ = T ′y ′,i.e. x′ ∈ im(T ′). B

7.6.12. Hausdorff Theorem. Let X and Y be Banach spaces. Assume fur-ther that T ∈ B(X, Y ). Then T is normally solvable if and only if T ′ is normallysolvable.

C ⇒: In virtue of 7.6.11 (2), im(T ′) = (kerT )⊥. Evidently, the subspace(kerT )⊥ is closed.

⇐: To begin with, suppose cl imT = Y . It is clear that 0 = Y ⊥ = (cl im T )⊥= (imT )⊥ = ker(T ′) in virtue of 7.6.11. By the Banach Isomorphism Theoremthere is some S ∈ B(im(T ′), Y ′) such that ST ′ = IY ′ . The case r := ‖S‖ = 0 istrivial. Therefore, it may be assumed that ‖T ′y ′‖ ≥ 1/r‖y ′‖ for all y ′ ∈ Y ′.

Show now that clT (BX) ⊃ 1/2rBY . With this at hand, from the ideal con-vexity of T (BX) it is possible to infer that T (BX) ⊃ 1/4rBY . The last inclusionimplies that T is a homomorphism.

Let y /∈ clT (BX). Then y does not belong to some open convex set includingT (BX). Passing, if need be, to the real carriers of X and Y , assume that F := R.


Applying the Topological Separation Theorem, find some nonzero y ′ in Y ′ satisfying‖y ′‖ ‖y‖ ≥ y ′(y) ≥ sup

‖x‖≤1y ′(Tx) = ‖T ′y ′‖ ≥ 1/r‖y ′‖.

Whence ‖y‖ ≥ 1/r > 1/2r. Therefore, the sought inclusion is established and theoperator T is normally solvable under the above supposition.

Finally, address the general case. Put Y0 := cl imT and let ι : Y00 = 0 OO bethe identical embedding. Then T = ιT , where T : X0 = 0 OO 0 is the operator actingby the rule Tx = Tx for x ∈ X. In addition, im(T ′) = im(T ′ι′) = T ′(im(ι′)) =T ′(Y ′0), because ι′(Y ′) = Y ′0 (cf. 7.6.4 (2)). Thus, T ′ is a normally solvable operator.By what was proven, T is normally solvable. It remains to observe that imT =imT . B

7.6.13. Sequence Prime Principle. A sequence

. . . 0 = 0 OOk−1

Tk−−−−→ XkTk+1−−−−→ Xk+10 = 0 OO

is exact if and only if so is the sequence prime

. . .← X ′k−1T ′k←−−−− X ′k

T ′k+1←−−−− X ′k+1 ← . . . .

C ⇒: Since imTk+1 = kerTk+2; therefore, Tk+1 is normally solvable. Usingthe Polar Lemma, conclude that

ker(T ′k) = (imTk)⊥ = (kerTk+1)⊥ = im(T ′k+1).⇒: By the Hausdorff Theorem Tk+1 is normally solvable. Once again appealing

to 7.6.11 (2), infer that(imTk)⊥ = ker(T ′k) = im(T ′k+1) = (kerTk+1)⊥.

Since Tk is normally solvable by Theorem 7.6.12; therefore, imTk is a closed sub-space. Using 7.5.14, observe that

imT k = ⊥((imTk)⊥) = ⊥((kerTk+1)⊥) = kerTk+1.

Here account was taken of the fact that kerTk+1 itself is a closed subspace. B

7.6.14. Corollary. For a normally solvable operator T , the following isomor-phisms hold: (kerT )′ ' coker(T ′) and (cokerT )′ ' ker(T ′).C By virtue of 2.3.5 (6) the sequence

00 = 0 OO T0 = 0 OO T−→ Y 0 = 0 OO T0 = 0 OO

is exact. From 7.6.13 obtain that the sequence

00 = 0 OO cokerT )′0 = 0 OO ′ T ′−→ X ′0 = 0 OO kerT )′0 = 0 OO

is exact. B

7.6.15. Corollary. T is an isomorphism ⇔ T ′ is an isomorphism. CB

7.6.16. Corollary. Sp(T ) = Sp(T ′). CB

124 Chapter 7

Exercises

7.1. Find out which linear operators are ideal.7.2. Establish that a separately continuous bilinear form on a Banach space is jointly con-

tinuous.

7.3. Is a family of lower semicontinuous sublinear functionals on a Banach space uniformlybounded on the unit ball?

7.4. Let X and Y be Banach spaces and T : X0 = 0 OO . Prove that ‖Tx‖Y ≥ t‖x‖X for somestrictly positive t and all x ∈ X if and only if kerT = 0 and imT is a complete set.

7.5. Find conditions for normal solvability of the operator of multiplication by a functionin the space of continuous functions on a compact set.

7.6. Let T be a bounded epimorphism of a Banach space X onto l1(E ). Show that kerT iscomplemented.

7.7. Establish that a uniformly closed subspace of C([a, b]) composed of continuously dif-ferentiable functions (i.e., elements of C(1)([a, b])) is finite-dimensional.

7.8. Let X and Y be different Banach spaces, with X continuously embedded into Y . Es-tablish that X is a meager subset of Y .

7.9. Let X1 and X2 be nonzero closed subspaces of a Banach space and X1∩X2 = 0. Provethat the sum X1 +X2 is closed if and only if the next quantity

inf{‖x1 − x2‖/‖x1‖ : x1 6= 0, x1 ∈ X1, x2 ∈ X2}

is strictly positive.

7.10. Let (amn) be a double countable sequence such that there is a sequence (x(m)) ofelements of l1 for which all the series

∑∞n=1 amnx

(m)n fail to converge in norm. Prove that there

is a sequence x in l1 such that the series∑∞

n=1 amnxn fail to converge in norm for all m ∈ N.

7.11. Let T be an endomorphism of a Hilbert space H which satisfies 〈Tx | y〉 = 〈x |Ty〉 forall x, y ∈ H. Establish that T is bounded.

7.12. Let a closed cone X+ in a Banach space X be reproducing: X = X+ − X+. Provethat there is a constant t > 0 such that for all x ∈ X and each presentation x = x1 − x2 withx1, x2 ∈ X+, the estimates hold: ‖x1‖ ≤ t‖x‖ and ‖x2‖ ≤ t‖x‖.

7.13. Let lower semicontinuous sublinear functionals p and q on a Banach space X be suchthat the cones dom p and dom q are closed and the subspace dom p − dom q = dom q − dom p iscomplemented in X. Prove that

∂ (p+ q) = ∂ (p) + ∂ (q)

in the case of topological subdifferentials (cf. Exercise 3.10).

7.14. Let p be a continuous sublinear functional defined on a normed space X and let T bea continuous endomorphism of X. Assume further that the dual T ′ of T takes the subdifferential∂ (p) into itself. Establish that ∂ (p) contains a fixed point of T ′.

7.15. Given a function f : X0 = 0 OO R· on a (multi)normed space X, put

f∗(x′) := sup{〈x |x′〉 − f(x) : x ∈ dom f} (x′ ∈ X′);

f∗∗(x) := sup{〈x |x′〉 − f∗(x′) : x′ ∈ dom(f∗)} (x ∈ X).

Find conditions for f to satisfy f = f∗∗.


7.16. Establish that l∞ is complemented in each ambient Banach space.7.17. A Banach space X is called primary, if each of its infinite-dimensional complemented

subspaces is isomorphic to X. Verify that c0 and lp (1 ≤ p ≤ +∞) are primary.

7.18. Let X and Y be Banach spaces. Take an operator T , a member of B(X, Y ), suchthat imT is nonmeager. Prove that T is normally solvable.

7.19. Let X0 be a closed subspace of a normed space X. Assume further that X0 and X/X0are Banach spaces. Show that X itself is a Banach space.

Chapter 8

Operators in Banach Spaces

8.1. Holomorphic Functions and Contour Integrals

8.1.1. Definition. Let X be a Banach space. A subset � of the ball BX′in the dual space X ′ is called norming (for X) if ‖x‖ = sup{|l(x)| : l ∈ �}for all x ∈ X. If each subset U of X satisfies sup ‖U‖ < +∞ on condition thatsup{|l(u)| : u ∈ U} < +∞ for all l ∈ �, then � is a fully norming set.

8.1.2. Examples.(1) The ball BX′ is a fully norming set in virtue of 5.1.10 (8) and 7.2.7.(2) If �0 is a (fully) norming set and �0 ⊂ �1 ⊂ BX′ then �1 itself is

a (fully) norming set.(3) The set extBX′ of the extreme points of BX′ is norming in virtue

of the Kreın–Milman Theorem in subdifferential form and the obvious equalityBX′ = |∂|(‖ · ‖X) which has already been used many times. However, extBX′can fail to be fully norming (in particular, the possibility is realized in the spaceC([0, 1], R)). CB

(4) Let X and Y be Banach spaces (over the same ground field F) andlet �Y be a norming set for Y . Put

�B := {δ(y′,x) : y ′ ∈ �Y , x ∈ BX},

where δ(y′,x)(T ) := y ′(Tx) for y ′ ∈ Y, x ∈ X and T ∈ B(X, Y ). It is clear that

|δ(y′,x)(T )| = |y ′(Tx)| ≤ ‖y ′‖ ‖Tx‖ ≤ ‖y ′‖ ‖T‖ ‖x‖;

i.e., δ(y′,x) ∈ B(X, Y )′. Furthermore, given T ∈ B(X, Y ), infer that

‖T‖ = sup{‖Tx‖ : ‖x‖ ≤ 1} = sup{|y ′(Tx)| : y ′ ∈ �Y , ‖x‖ ≤ 1}= sup{|δ(y′,x)(T )| : δ(y′,x) ∈ �B}.

Operators in Banach Spaces 127

Therefore, �B is a norming set for B(X, Y ). If �Y is a fully norming set then �Bis also a fully norming set. Indeed, if U is such that the numeric set {|y ′(Tx)| :T ∈ U} is bounded in R for all x ∈ BX and y ′ ∈ �Y , then by hypothesis theset {Tx : T ∈ U} is bounded in Y for every x in X. By virtue of the UniformBoundedness Principle it means that sup ‖U‖ < +∞.

8.1.3. Dunford–Hille Theorem. Let X be a complex Banach space and let� be a fully norming set for X. Assume further that f : D0 = 0 OO is an X-va-lued function with domain D an open (in C R ' R2) subset of C. The followingstatements are equivalent:

(1) for every z0 in D there is a limit

limz0=0 OO

0

f(z)− f(z0)z − z0

;

(2) for all z0 ∈ D and l ∈ � there is a limit

limz0=0 OO

0

l ◦ f(z)− l ◦ f(z0)z − z0

;

i.e., the function l ◦ f : D0 = 0 OO C is holomorphic for l ∈ �.C (1) ⇒ (2): This is obvious.(2) ⇒ (1): For the sake of simplicity assume that z0 = 0 and f(z0) = 0. Con-

sider the disk of radius 2ε centered at zero and included in D ; i.e., 2εD ⊂ D , whereD := BC := {z ∈ C : |z| ≤ 1} is the unit disk. As is customary in complex analysis,treat the disk εD as an (oriented) compact manifold with boundary εT, where T isthe (properly oriented) unit circle T := {z ∈ C : |z| = 1} in the complex plane CR.Take z1, z2 ∈ εD \ 0 and the holomorphic function l ◦ f (the functional l lies in �).Specifying the Cauchy Integral Formula, observe that

l ◦ f(zk)zk

=12πi

∫2εT

l ◦ f(z)z(z − zk)

dz (k := 1, 2).

If now z1 6= z2 then, using the condition |z − zk| ≥ ε (k := 1, 2) for z ∈ 2εT andthe continuity property of the function l ◦ f on D , find∣∣∣∣l( 1

z1 − z2

(f(z1)z1− f(z2)

z2

))∣∣∣∣=

∣∣∣∣∣∣ 1z1 − z2

· 12πi

∫2εT

l ◦ f(z)(

1z(z − z1)

− 1z(z − z2)

)dz

∣∣∣∣∣∣=

12π

∣∣∣∣∣∣∫

2εT

l ◦ f(z) 1z(z − z1)(z − z2)

dz

∣∣∣∣∣∣ ≤M supz∈2εT

|l ◦ f(z)| < +∞

128 Chapter 8

for a suitable M > 0. Since � is a fully norming set, conclude that

supz1 6=z2;z1,z2 6=0|z1|≤ε,|z2|≤ε

1|z1 − z2|

∥∥∥∥f(z1)z1− f(z2)

z2

∥∥∥∥ < +∞.

This final inequality ensures that the sought limit exists. B

8.1.4. Definition. A mapping f : D0 = 0 OO satisfying 8.1.3 (1) (or, which isthe same, 8.1.3 (2) for some fully norming set �) is called holomorphic.

8.1.5. Remark. A meticulous terminology is used sometimes. Namely, if fsatisfies 8.1.3 (1) then f is called a strongly holomorphic function. If f satisfies8.1.3 (2) with � := BX′ then f is called weakly holomorphic. Under the hypothesesof 8.1.3 (2) and 8.1.2 (4), i.e. for f : D0 = 0 OO (X, Y ), �Y := BY ′ and the corre-sponding � := �B , the expression, “f is weakly operator holomorphic,” is employed.With regard to this terminology, the Dunford–Hille Theorem is often referred toas the Holomorphy Theorem and verbalized as follows: “A weakly holomorphicfunction is strongly holomorphic.”

8.1.6. Remark. It is convenient in the sequel to use the integrals of thesimplest smooth X-valued forms f(z)dz over the simplest oriented manifolds, theboundaries of elementary planar compacta (cf. 4.8.5) which are composed of finitelymany disjoint simple loops. An obvious meaning is ascribed to the integrals:Namely, given a loop γ, choose an appropriate (smooth) parametrization � : T0 =0 OO (with orientation accounted for) and put∫

γ

f(z)dz :=∫T

f ◦ �d�,

with the integral treated for instance as a suitable Bochner integral (cf. 5.5.9 (6)).The soundness of the definition is beyond a doubt, since the needed Bochner integralexists independently of the choice of the parametrization �.

8.1.7. Cauchy–Wiener Integral Theorem. Let D be an nonempty opensubset of the complex plane and let f : D0 = 0 OO be a holomorphic X-valuedfunction, with X a Banach space. Assume further that F is a rough draft for thepair (∅, D). Then ∫

∂F

f(z)dz = 0.

Moreover,

f(z0) =12πi

∫∂F

f(z)z − z0

dz

for z0 ∈ intF .


C By virtue of 8.1.3 the Bochner integrals exist. The sought equalities followreadily from the validity of their scalar versions basing on the Cauchy IntegralFormula, the message of 8.1.2 (1) and the fact that the Bochner integral commuteswith every bounded functional as mentioned in 5.5.9 (6). B

8.1.8. Remark. The Cauchy–Wiener Integral Theorem enables us to inferanalogs of the theorems of classical complex analysis for X-valued holomorphicfunctions on following the familiar patterns.

8.1.9. Taylor Series Expansion Theorem. Let f : D0 = 0 OO be a holomor-phic X-valued function, with X a Banach space, and take z0 ∈ D . In every opendisk U := {z ∈ C : |z−z0| < ε} such that clU lies in D , the Taylor series expansionholds (in a compactly convergent power series, cf. 7.2.10):

f(z) =∞∑n=0

cn(z − z0)n,

where the coefficients cn, members of X, are calculated by the formulas:

cn =12πi

∫∂U

f(z)(z − z0)n+1 dz =

1n!dnf

dzn(z0).

C The proof results from a standard argument: Expand the kernel u 7→(u− z)−1 of the formula

f(z) =12πi

∫∂U ′

f(u)u− z

du (z ∈ clU)

in the powers of z − z0; i.e.,

1u− z

=1

(u− z0)(1− z−z0

u−z0

) =∞∑n=0

(z − z0)n

(u− z0)n+1 .

The last series converges uniformly in u ∈ ∂U ′. (Here U ′ = U + qD for someq > 0 such that clU ′ ⊂ D .) Taking it into account that sup ‖f(∂U ′)‖ < +∞and integrating, arrive at the sought presentation of f(z) for z ∈ clU . Applyingthis to U ′ and using 8.1.7, observe that the power series under study convergesin norm at every point of U ′. This yields uniform convergence on every compactsubset of U ′, and so on U . B

130 Chapter 8

8.1.10. Liouville Theorem. If an X-valued function f : C0 = 0 OO , with Xa Banach space, is holomorphic and sup ‖f(C)‖ < +∞ then f is a constant map-ping.

C Considering the disk εD with ε > 0 and taking note of 8.1.9, infer that

‖cn‖ ≤ supz∈εT‖f(z)‖ · ε−n ≤ sup ‖f(C)‖ · ε−n

for all n ∈ N and ε > 0. Therefore, cn = 0 for n ∈ N. B

8.1.11. Each bounded endomorphism of a nonzero complex Banach space hasa nonempty spectrum.C Let T be such an endomorphism. If Sp(T ) = ∅ then the resolvent R(T, · ) is

holomorphic on the entire complex plane C, for instance, by 5.6.21. Furthermore,by 5.6.15, ‖R(T, λ)‖0 = 0 OO as |λ|0 = 0 OO ∞. By virtue of 8.1.10 conclude thatR(T, · ) = 0. At the same time, using 5.6.15, observe that R(T, λ)(λ− T ) = 1 for|λ| > ‖T‖. A contradiction. B

8.1.12. The Beurling–Gelfand formula holds:

r(T ) = sup{|λ| : λ ∈ Sp(T )}

for all T ∈ B(X), with X a complex Banach space; i.e., the spectral radius ofan operator T coincides with the radius of the spectrum of T .C It is an easy matter that the spectral radius r(T ) is greater than the radius

of the spectrum of T . So, there is nothing to prove if r(T ) = 0. Assume now thatr(T ) > 0. Take λ ∈ C so that |λ| > sup{|µ| : µ ∈ Sp(T )}. Then the disk of radius|λ|−1 lies entirely in the domain of the holomorphic function (cf. 5.6.15)

f(z) :={R(T, z−1), for z 6= 0 and z−1 ∈ res(T )0, for z = 0.

Using 8.1.9 and 5.6.17, conclude that |λ|−1 < r(T )−1. Consequently, |λ| > r(T ). B

8.1.13. LetK be a nonempty compact subset of C. Denote byH(K) the set ofall functions holomorphic in a neighborhood of K (i.e., f ∈ H(K)⇔ f : dom f0 =0 OO C is a holomorphic function with dom f ⊃ K). Given f1, f2 ∈ H(K), let thenotation f1 ∼ f2 mean that it is possible to find an open subset D of dom f1∩dom f2satisfying K ⊂ D and f1|D = f2|D . Then ∼ is an equivalence in H(K). CB

8.1.14. Definition. Under the hypotheses of 8.1.13, put H (K) := H(K)/∼.The element f in H (K) containing a function f in H(K) is the germ of f on K.


8.1.15. Let f, g ∈H (K). Take f1, f2 ∈ f and g1, g2 ∈ g. Put

x ∈ dom f1 ∩ dom g1 ⇒ ϕ1(x) := f1(x) + g1(x),x ∈ dom f2 ∩ dom g2 ⇒ ϕ2(x) := f2(x) + g2(x).

Then ϕ1, ϕ2 ∈ H(K) and ϕ1 = ϕ2.C Choose open sets D1 and D2 such that K ⊂ D1 ⊂ dom f1 ∩ dom f2 and

K ⊂ D2 ⊂ dom g1 ∩ dom g2, with f1|D1 = f2|D1 and g1|D2 = g2|D2 . Observe nowthat ϕ1 and ϕ2 agree on D1 ∩D2. B

8.1.16. Definition. The coset, introduced in 8.1.15, is the sum of the germsf1 and f2. It is denoted by f1 + f2. The product of germs and multiplication ofa germ by a complex number are introduced by analogy.

8.1.17. The set H (K) with operations defined in 8.1.16 is an algebra. CB

8.1.18. Definition. The algebra H (K) is the algebra of germs of holomor-phic functions on a compact set K.

8.1.19. Let K be a compact subset of C, and let R : C \ K0 = 0 OO be anX-valued holomorphic function, with X a Banach space. Further, take f ∈H (K)and f1, f2 ∈ f . If F1 is a rough draft for the pair (K, dom f1) and F2 is a roughdraft for the pair (K, dom f2) then∫

∂F1

f1(z)R(z)dz =∫∂F2

f2(z)R(z)dz.

C Let K ⊂ D ⊂ intF1∩ intF2, with D open and f1|D = f2|D . Choose a roughdraft F for the pair (K, D). Since f1R is holomorphic on dom f1 \K and f2R isholomorphic on dom f2 \K, infer the equalities

∫∂F


f1(z)R(z)dz,

∫∂F


f2(z)R(z)dz

(from the nontrivial fact of the validity of their scalar analogs). Since f1 and f2agree on D , the proof is complete. B

132 Chapter 8

8.1.20. Definition. Under the hypotheses of 8.1.19, given an element hin H (K), define the contour integral of h with kernel R as the element∮

h(z)R(z)dz :=∫∂F

f(z)R(z)dz,

where h = f and F is a rough draft for the pair (K, dom f).

8.1.21. Remark. The notation h(z) in 8.1.20 is far from being ad hoc. It iswell justified by the fact that w := f1(z) = f2(z) for every point z in K and everytwo members f1 and f2 of a germ h. In this connection the element w is said tobe the value of h at z, which is expressed in writing as h(z) = w. It is also worthnoting that the function R in 8.1.2 may be assumed to be given only in U \ K,where intU ⊃ K.

8.2. The Holomorphic Functional Calculus

8.2.1. Definition. Let X be a (nonzero) complex Banach space and let T bea bounded endomorphism of X; i.e., T ∈ B(X). For h ∈ H (Sp(T )), the contourintegral with kernel the resolvent R(T, · ) of T is denoted by

RTh :=12πi

∮h(z)R(T, z)dz

and called the Riesz–Dunford integral (of the germ h). If f is a function holomorphicin a neighborhood about Sp(T ) then put f(T ) := RT f := RT f . We also use moresuggestive designations like

f(T ) =12πi

∮f(z)z − T

dz.

8.2.2. Remark. In algebra, in particular, various representations of mathe-matical objects are under research. It is convenient to use the primary notions ofrepresentation theory for the most “algebraic” objects, namely, algebras. Recall thesimplest of them.

Let A1 and A2 be two algebras (over the same field). A morphism from A1to A2 or a representation of A1 in A2 (rarely, over A2) is a multiplicative linearoperator R, i.e. a member R of L (A1, A2) such that R(ab) = R(a)R(b) for alla, b ∈ A1. The expression, “T represents A1 in A2,” is also in common parlance.A representation R is called faithful if kerR = 0. The presence of a faithful repre-sentation R : A10 = 0 OO 2 makes it possible to treat A1 as a subalgebra of A2.


If A2 is a (sub)algebra of the endomorphism algebra L (X) of some vectorspace X (over the same field), then a morphism of A1 in A2 is referred to asa (linear) representation of A1 on X or as an operator representation of A1. Thespace X is then called the representation space for the algebra A1.

Given a representation R, suppose that the representation space X for A hasa subspace X1 invariant under all operators R(a), a ∈ A. Then the representa-tion R1 : A0 = 0 OO L(X1) arises naturally, acting by the rule R1(a)x1 = R(a)x1

for x1 ∈ X1 and a ∈ A and called a subrepresentation of R (induced in X1).If X = X1⊕X2 and the decomposition reduces each operator R(a) for a ∈ A, thenit is said that the representation R reduces to the direct sum of subrepresentationsR1 and R2 (induced in X1 and X2). We mention in passing the significance ofstudying arbitrary irreducible representations, each of which has only trivial sub-representations by definition.

8.2.3. Gelfand–Dunford Theorem. Let T be a bounded endomorphismof a Banach space X. The Riesz–Dunford integral RT represents the algebra ofgerms of holomorphic functions on the spectrum of T on the space X. Moreover,if f(z) =

∑∞n=0 cnz

n (in a neighborhood about Sp(T )) then f(T ) =∑∞n=0 cnT

n

(summation is understood in the operator norm of B(X)).

C There is no doubt that RT is a linear operator. Check the multiplicativityproperty of RT . To this end, take f1, f2 ∈H (Sp(T )) and choose rough drafts F1

and F2 such that Sp(T ) ⊂ intF1 ⊂ F1 ⊂ intF2 ⊂ F2 ⊂ D , with the functions f1 inf1 and f2 in f2 holomorphic on D .

Using the obvious properties of the Bochner integral, the Cauchy Integral For-

134 Chapter 8

mula and the Hilbert identity, successively infer the chain of equalities

RT f1 ◦RT f2 = f1(T )f2(T ) =12πi

12πi

∫∂F1

f1(z1)z1 − T

dz1 ◦∫∂F2

f2(z2)z2 − T

dz2

=12πi

12πi

∫∂F2

∫∂F1

f1(z1)R(T, z1)dz1

f2(z2)R(T, z2)dz2

=12πi

12πi

∫∂F1

∫∂F2

f1(z1)f2(z2)R(T, z1)R(T, z2)dz2dz1

=12πi

12πi

∫∂F1

∫∂F2

f1(z1)f2(z2)R(T, z1)−R(T, z2)

z2 − z1dz2dz1

=12πi

∫∂F1

f1(z1)

12πi

∫∂F2

f2(z2)z2 − z1

dz2

R(T, z1)dz1

− 12πi

∫∂F2

f2(z2)

12πi

∫∂F1

f1(z1)z2 − z1

dz1

R(T, z2)dz2

=12πi

∫γ

f1(z1)f2(z1)R(T, z1)dz1 − 0 = f1f2(T ) = RT (f1f2).

Choose a circle γ := εT that lies in res(T ) as well as in the (open) disk ofconvergence of the series f(z) =

∑∞n=0 cnz

n. From 5.6.16 and 5.5.9 (6) derive

f(T ) =12πi

∫γ

f(z)∞∑n=0

z−n−1Tndz

=12πi

∞∑n=0

∫γ

f(z)z−n−1Tndz =∞∑n=0

12πi

∫γ

f(z)zn+1 dz

Tn =∞∑n=0

cnTn

in virtue of 8.1.9. B

8.2.4. Remark. Theorem 8.2.3 is often referred to as the Principal Theoremof the holomorphic functional calculus.

8.2.5. Spectral Mapping Theorem. For every function f holomorphic in a neigh-borhood about the spectrum of an operator T in B(X), the equality holds:

f(Sp(T )) = Sp(f(T )).


C Assume first that λ ∈ Sp(f(T )) and f−1(λ) ∩ Sp(T ) = ∅. Given z ∈(C \ f−1(λ)) ∩ dom f , put g(z) := (λ− f(z))−1. Then g is a holomorphic functionon a neighborhood of Sp(T ), satisfying g(λ − f) = (λ − f)g = 1C. Using 8.2.3,observe that λ ∈ res(f(T )), a contradiction. Consequently, f−1(λ) ∩ Sp(T ) 6= ∅;i.e., Sp(f(T )) ⊂ f(Sp(T )).

Now take λ ∈ Sp(T ). Put

λ 6= z ⇒ g(z) :=f(λ)− f(z)

λ− z; g(λ) := f ′(λ).

Clearly, g is a holomorphic function (the singularity is “removed”). From 8.2.3obtain

g(T )(λ− T ) = (λ− T )g(T ) = f(λ)− f(T ).Consequently, if f(λ) ∈ res(f(T )) then the operator R(f(T ), f(λ))g(T ) is inverseto λ− T . In other words, λ ∈ res(T ), which is a contradiction. Thus,

f(λ) ∈ C \ res(f(T )) = Sp(f(T ));

i.e., f(Sp(T )) ⊂ Sp(f(T )). B

8.2.6. Let K be a nonempty compact subset of C and let g : dom g0 = 0 OO Cbe a holomorphic function with dom g ⊃ K. Given f ∈ H(g(K)), put

◦g(f) := f ◦ g.

Then◦g is a representation of the algebra H (g(K)) in the algebra H (K). CB

8.2.7.Dunford Theorem. For every function g : dom g0 = 0 OO C holomorphicin the neighborhood dom g about the spectrum Sp(T ) of an endomorphism T ofa Banach space X, the following diagram of representations commutes:

B(X)

H (Sp(T )) H (Sp(g(T )))

Rg(T )

◦g

RT

�

?

@@@@@R

C Let f ∈ H (g(Sp(T ))) with f : D0 = 0 OO C such that f ∈ f and D ⊃g(Sp(T )) = Sp(g(T )). Let F1 be a rough draft for the pair (Sp(g(T )), D) andlet F2 be a rough draft for the pair (Sp(T ), g−1(intF1)). It is clear that nowg(∂F2) ⊂ intF1 and, moreover, the function z2 7→ (z1 − g(z2))−1 is defined andholomorphic on intF2 for z1 ∈ ∂F1. Therefore, by 8.2.3

R(g(T ), z1) =12πi

∫∂F2

R(T, z2)z1 − g(z2)

dz2 (z1 ∈ ∂F1).

136 Chapter 8

From this equality, successively derive

Rg(T )f =12πi

∫∂F1

f(z1)z1 − g(T )

dz1 =12πi

12πi

∫∂F1

f(z1)

∫∂F2

R(T, z2)z1 − g(z2)

dz2

dz1

=12πi

12πi

∫∂F2

∫∂F1

f(z1)z1 − g(z2)

dz1

R(T, z2)dz2.

Since g(z2) ∈ intF1 for z2 ∈ ∂F2 by construction; therefore, the Cauchy IntegralFormula yields the equality

f(g(z2)) =12πi

∫∂F1

f(z1)z1 − g(z2)

dz1.

Consequently,

Rg(T )f =12πi

∫∂F2

f(g(z2))R(T, z2)dz2 = RT◦g(f).

8.2.8. Remark. The Dunford Theorem is often referred to as the CompositeFunction Theorem and written down symbolically as f ◦ g(T ) = f(g(T )).

8.2.9. Definition. A subset σ of Sp(T ) is a clopen or isolated part or rarelyan exclave of Sp(T ), if σ and its complement σ′ := Sp(T ) \ σ are closed.

8.2.10. Let σ be a clopen part of Sp(T ) and let κσ be some function thatequals 1 in an open neighborhood of σ and 0 in an open neighborhood of σ′. Further,assign

Pσ := κσ(T ) :=12πi

∮κσ(z)z − T

dz.

Then Pσ is a projection in X and the (closed) subspace Xσ := imPσ is invariantunder T .

C Since κ2σ = κσ, it follows from 8.2.3 that κσ(T )2 = κσ(T ). Furthermore,

T = RT IC, where IC : z 7→ z. Hence TPσ = PσT (because ICκσ = κσIC).Consequently, in virtue of 2.2.9 (4), Xσ is invariant under T . B

8.2.11. Definition. The projection Pσ is the Riesz projection or the Rieszidempotent corresponding to σ.


8.2.12. Spectral Decomposition Theorem. Let σ be a clopen part of thespectrum of an operator T in B(X). Then X splits into the direct sum decompo-sition of the invariant subspaces X = Xσ ⊕Xσ′ which reduces T to matrix form

T ∼(Tσ 00 Tσ′

),

with the part Tσ of T in Xσ and the part Tσ′ of T in Xσ′ satisfying

Sp(Tσ) = σ, Sp(Tσ′) = σ′.

C Since κσ + κσ′ = κSp(T ) = 1C, in view of 8.2.3 and 8.2.10 it suffices to es-tablish the claim about the spectrum of Tσ.

From 8.2.5 and 8.2.3 obtain

σ ∪ 0 = κσIC(Sp(T )) = Sp(κσIC(T )) = Sp(RT (κσIC))= Sp(RTκσ ◦RT IC) = Sp(PσT ).

Moreover, in matrix form

PσT ∼(Tσ 00 0

).

Let λ be a nonzero complex number. Then

λ− PσT ∼(λ− Tσ 0

0 λ

);

i.e., the operator λ − PσT is not invertible if and only if the same is true of theoperator λ− Tσ. Thus,

Sp(Tσ) \ 0 ⊂ Sp(PσT ) \ 0 = (σ ∪ 0) \ 0 ⊂ σ.Suppose that 0 ∈ Sp(Tσ) and 0 /∈ σ. Choose disjoint open sets Dσ and Dσ′ so

that σ ⊂ Dσ, 0 /∈ Dσ and σ′ ⊂ Dσ′ , and put

z ∈ Dσ ⇒ h(z) :=1z;

z ∈ Dσ′ ⇒ h(z) := 0.

By 8.2.3, h(T )T = Th(T ) = Pσ. Moreover, since hκσ = κσh, the decompositionX = Xσ⊕Xσ′ reduces h(T ) and h(T )σTσ = Tσh(T )σ = 1 for the part h(T )σ of h(T )in Xσ. So, Tσ is invertible; i.e., 0 /∈ Sp(Tσ). We thus arrive at a contradiction whichimplies that 0 ∈ σ. In other words, Sp(Tσ) ⊂ σ.

Observe now that res(T ) = res(Tσ) ∩ res(Tσ′). Consequently, by the above

Sp(T ) = C \ res(T ) = C \ (res(Tσ) ∩ res(Tσ′))= (C \ res(Tσ)) ∪ (C \ res(Tσ′)) = Sp(Tσ) ∪ Sp(Tσ′) ⊂ σ ∪ σ′ = Sp(T ).

Considering that σ ∩ σ′ = ∅, complete the proof. B

138 Chapter 8

8.2.13. Riesz–Dunford Integral Decomposition Let σ bea clopen part of Sp(T ) for an endomorphism T of a Banach space X. The directsum decomposition X = Xσ ⊕ Xσ′ reduces the representation RT of the algebraH (Sp(T )) in X to the direct sum of the representations Rσ and Rσ′ . Moreover,the following diagrams of representations commute:

B(Xσ)

H (Sp(T )) H (σ)

RTσ

κσ

Rσ

-

?

@@@@@R

B(Xσ′)

H (Sp(T )) H (σ′)

RTσ′

κσ′

Rσ′

-

?

@@@@@R

Here κσ(f) := κσf and κσ′(f) := κσ′f for f ∈ H(Sp(T )) are the representationsinduced by restricting f onto σ and σ′. CB

8.3. The Approximation Property

8.3.1. Let X and Y be Banach spaces. For K ∈ L (X, Y ) the followingstatements are equivalent:

(1) the operator K is compact: K ∈ K (X, Y );

(2) there are a neighborhood of zero U in X and a compact subset Vof Y such that K(U) ⊂ V ;

(3) the image under K of every bounded set in X is relatively compactin Y ;

(4) the image under K of every bounded set in X is totally boundedin Y ;

(5) for each sequence (xn)n∈N of points of the unit ball BX , the se-quence (Kxn)n∈N has a Cauchy subsequence. CB

8.3.2. Theorem. Let X and Y be Banach spaces over a basic field F. Then(1) K (X, Y ) is a closed subspace of B(X, Y );

(2) for all Banach spaces W and Z, it holds that

B(Y, Z) ◦K (X, Y ) ◦B(W, X) ⊂ K (W, Z);

i.e., if S ∈ B(W, X), T ∈ B(Y, Z) and K ∈ K (X, Y ) then TKS ∈ K (W, Z);

(3) IF ∈ K (F) := K (F, F).


C That K (X, Y ) is a subspace of B(X, Y ) follows from 8.3.1. If Kn ∈K (X, Y ) and Kn0 = 0 OO ; then, given ε > 0, for n sufficiently large observe that‖Kx −Knx‖ ≤ ‖K −Kn‖ ‖x‖ ≤ ε whenever x ∈ BX . Therefore, Kn(BX) servesas an ε-net (= Bε-net) for K(BX). It remains to refer to 4.6.4 and thus completethe proof of the closure property of K (X, Y ). The other claims are evident. B

8.3.3. Remark. Theorem 8.3.2 is often verbalized as follows: “The class ofall compact operators is an operator ideal.” Behind this lies a conspicuous analogywith the fact that K (X) := K (X, X) presents a closed bilateral (two-sided) idealin the (bounded) endomorphism algebra B(X); i.e., K (X) ◦ B(X) ⊂ K (X) andB(X) ◦K (X) ⊂ K (X).

8.3.4. Calkin Theorem. The ideals 0, K (l2), and B(l2) exhaust the list ofclosed bilateral ideals in the endomorphism algebra B(l2) of the Hilbert space l2.

8.3.5. Remark. In view of 8.3.4 it is clear that a distinguishable role in op-erator theory should be performed by the algebra B(X)/K (X) called the Calkinalgebra (on X). The performance is partly delivered in 8.5.

8.3.6. Definition. An operator T , a member of L (X, Y ), is called a finite-rank operator provided that T ∈ B(X, Y ) and imT is a finite-dimensional subspaceof Y . In notation: T ∈ F (X, Y ). A hasty term “finite-dimensional operator” wouldabuse consistency since T as a subspace of X × Y is usually infinite-dimensional.

8.3.7. The linear span of the set of (bounded) rank-one operators comprisesall finite-rank operators:

T ∈ F (X, Y )

⇔ (∃x′1, . . . , x′n ∈ X ′) (∃ y1, . . . , yn ∈ Y ) T =n∑k=1

x′k ⊗ yk. /.

8.3.8. Definition. Let Q be a (nonempty) compact set in X. Given T ∈B(X, Y ), put

‖T‖Q := sup ‖T (Q)‖.

The collection of all seminorms B(X, Y ) of type ‖ · ‖Q is the Arens multinormin B(X, Y ), denoted by κB(X,Y ). The corresponding topology is the topology ofuniform convergence on compact sets or the compact-open topology (cf. 7.2.10).

8.3.9. Grothendieck Theorem. Let X be a Banach space. The followingconditions are equivalent:

(1) for every ε > 0 and every compact set Q in X there is a finite-rankendomorphism T of X, a member of F (X) := F (X, X), such that ‖Tx − x‖ ≤ εfor all x ∈ Q;

140 Chapter 8

(2) for every Banach space W , the subspace F (W, X) is dense in thespace B(W, X) with respect to the Arens multinorm κB(W,X);

(3) for every Banach space Y , the subspace F (X, Y ) is dense in thespace B(X, Y ) with respect to the Arens multinorm κB(X,Y ).

C It is clear that (2) ⇒ (1) and (3) ⇒ (1). Therefore, we are to show onlythat (1) ⇒ (2) and (1) ⇒ (3).

(1) ⇒ (2): If T ∈ B(W, X) and Q is a nonempty compact set in W then,in view of the Weierstrass Theorem, T (Q) is a nonempty compact set in X. So,for ε > 0, by hypothesis there is a member T0 of F (X) such that ‖T0 − IX‖T (Q) =‖T0T − T‖Q ≤ ε. Undoubtedly, T0T ∈ F (W, X).

(1) ⇒ (3): Let T ∈ B(X, Y ). If T = 0 then there is nothing to be proven.Let T 6= 0, ε > 0 and Q be a nonempty compact set in X. By hypothesis thereis a member T0 of F (X) such that ‖T0 − IX‖Q ≤ ε‖T‖−1. Then ‖TT0 − T‖Q ≤‖T‖ ‖T0 − IX‖Q ≤ ε. Furthermore, TT0 ∈ F (X, Y ). B

8.3.10. Definition. A Banach space satisfying one (and hence all) of theequivalent conditions 8.3.9 (1)–8.3.9 (3) is said to possess the approximation prop-erty.

8.3.11. Grothendieck Criterion. A Banach space X possesses the approx-imation property if and only if, for every Banach space W , the equality holds:clF (W, X) = K (W, X), with the closure taken in operator norm.

8.3.12. Remark. For a long time there was an unwavering (and yet unprov-able) belief that every Banach space possesses the approximation property. There-fore, P. Enflo’s rather sophisticated example of a Banach space lacking the approx-imation property was acclaimed as sensational in the late seventies of the currentcentury. Now similar counterexamples are in plenty:

8.3.13. Szankowski Counerexample. The space B(l2) lacks the approxi-mation property.

8.3.14.Davis–Figiel–Szankowski Couterexample. The spaces c0 and lpwith p 6= 2 have closed subspaces lacking the approximation property.

8.4. The Riesz–Schauder Theory

8.4.1. ε-Perpendicular Lemma. Let X0 be a closed subspace of a Banachspace X and X 6= X0. Given an ε > 0, there is an ε-perpendicular to X0 in X; i.e.,an element xε in X such that ‖xε‖ = 1 and d(xε, X0) := inf d‖·‖({xε}×X0) ≥ 1−ε.C Take 1 > ε and x ∈ X \X0. It is clear that d := d(x, X0) > 0. Find x′ in the

subspace X0 satisfying ‖x−x′‖ ≤ d/(1−ε), which is possible because d/(1−ε) > d.


Put xε := (x− x′)‖x− x′‖−1. Then ‖xε‖ = 1. Finally, for x0 ∈ X0, observe that

‖x0 − xε‖ =∥∥∥∥x0 −

x− x′

‖x− x′‖

∥∥∥∥=

1‖x′ − x‖

‖(‖x− x′‖x0 + x′)− x‖ ≥ d(x, X0)‖x′ − x‖

≥ 1− ε. .

8.4.2. Riesz Criterion. Let X be a Banach space. The identity operatorin X is compact if and only if X is finite-dimensional.

C Only the implication ⇒ needs proving. If X fails to be finite-dimensional,then select a sequence of finite-dimensional subspaces X1 ⊂ X2 ⊂ . . . in X suchthat Xn+1 6= Xn for all n ∈ N. In virtue of 8.4.1 there is a sequence (xn) satisfyingxn+1 ∈ Xn+1, ‖xn+1‖ = 1 and d(xn+1, Xn) ≥ 1/2, namely, some sequence of 1/2-perpendiculars to Xn in Xn+1. It is clear that d(xm, xk) ≥ 1/2 for m 6= k. In otherwords, the sequence (xn) lacks Cauchy subsequences. Consequently, by 8.3.1 theoperator IX is not compact. B

8.4.3. Let T ∈ K (X, Y ), with X and Y Banach spaces. The operator T isnormally solvable if and only if T has finite rank.

C Only the implication ⇒ needs examining.Let Y0 := imT be a closed subspace in Y . By the Banach Homomorphism

Theorem, the image T (BX) of the unit ball of X is a neighborhood of zero in Y0.Furthermore, in virtue of the compactness property of T , the set T (BX) is relativelycompact in Y0. It remains to apply 8.4.2 to Y0. B

8.4.4. Let X be a Banach space and K ∈ K (X). Then the operator 1−K isnormally solvable.

C Put T := 1 − K and X1 := kerT . Undoubtedly, X1 is finite-dimensionalby 8.4.2. In accordance with 7.4.11 (1) a finite-dimensional subspace is comple-mented. Denote a topological complement of X1 to X by X2. Considering that X2is a Banach space and T (X) = T (X2), it suffices to verify that ‖Tx‖ ≥ t‖x‖ forsome t > 0 and all x ∈ X2. In the opposite case, there is a sequence (xn) such that‖xn‖ = 1, xn ∈ X2 and Txn0 = 0 OO . Using the compactness property of K, we mayassume that (Kxn) converges. Put y := limKxn. Then the sequence (xn) convergesto y, because y = lim(Txn + Kxn) = limxn. Moreover, Ty = limTxn = 0; i.e.,y ∈ X1. It is beyond a doubt that y ∈ X2. Thus, y ∈ X1 ∩ X2; i.e., y = 0. Wearrive at a contradiction: ‖y‖ = lim ‖xn‖ = 1. B

8.4.5. For whatever strictly positive ε, there are only finitely many eigenvaluesof a compact operator beyond the disk centered at zero and having radius ε.

142 Chapter 8

C Suppose by way of contradiction that there is a sequence (λn)n∈N of pair-wise distinct eigenvalues of a compact operator K, with |λn| ≥ ε for all n ∈ N andε > 0. Suppose further that xn satisfying 0 6= xn ∈ ker(λn −K) is an eigenvectorwith eigenvalue λn. Establish first that the set {xn : n ∈ N} is linearly inde-pendent. To this end, assume the set {x1, . . . , xn} linearly independent. In casexn+1 =

∑nk=1 αkxk, we would have 0 = (λn+1−K)xn+1 =

∑nk=1 αk(λn+1−λk)xk.

Consequently, αk = 0 for k := 1, . . . , n. Whence the false equality xn+1 = 0 wouldensue.

Put Xn := lin({x1, . . . , xn}). By definition X1 ⊂ X2 ⊂ . . . ; moreover, as wasproven, Xn+1 6= Xn for n ∈ N. By virtue of 8.4.1 there is a sequence (xn) such thatxn+1 ∈ Xn+1, ‖xn+1‖ = 1 and d(xn+1, Xn) ≥ 1/2. For m > k, straightforwardcalculation shows that z := (λm+1−K)xm+1 ∈ Xm and z+Kxk ∈ Xm+Xk ⊂ Xm.Consequently,

‖Kxm+1 −Kxk‖ = ‖ − λm+1xm+1 +Kxm+1 + λm+1xm+1 −Kxk‖= ‖λm+1xm+1 − (z +Kxk)‖ ≥ |λm+1|d(xm+1, Xm) ≥ ε/2.

In other words, the sequence (Kxn) has no Cauchy subsequences. B

8.4.6. Schauder Theorem. Let X and Y be Banach spaces (over the sameground field F). Then

K ∈ K (X, Y )⇔ K ′ ∈ K (Y ′, X ′).

C ⇒: Observe first of all that the restriction mapping x′ 7→ x′|BX implementsan isometry of X ′ into l∞(BX). Therefore, to check that K ′(BY ′) is relativelycompact we are to show the same for the set V := {K ′y ′|BX : y ′ ∈ BY ′}. SinceK ′y ′|BX (x) = y ′ ◦ K|BX (x) = y ′(Kx) for x ∈ BX and y ′ ∈ BY ′ , consider the

compact set Q := clK(BX) and the mapping◦K : C(Q, F)0 = 0 OO ∞(BX) defined

by the rule◦Kg : x 7→ g(Kx). Undoubtedly, the operator

◦K is bounded and, hence,

continuous. Now put S := {y ′|Q : y ′ ∈ BY ′}. It is clear that S is simultaneouslyan equicontinuous and bounded subset of C(Q, F). Consequently, by the Ascoli–Arzela Theorem, S is relatively compact. From the Weierstrass Theorem derive that◦K(S) is a relatively compact set too. It remains to observe that

◦Ky ′|Q = K ′y ′|BX

for y ′ ∈ BY ′ ; i.e.,◦K(S) = V .

⇐: If K ′ ∈ K (Y ′, X ′) then, as is proven, K ′′ ∈ K (X ′′, Y ′′). By the DoublePrime Lemma, K ′′|X = K. Whence it follows that the operator K is compact. B

8.4.7. Every nonzero point of the spectrum of a compact operator is isolated(i.e., such point constitutes a clopen part of the spectrum).


C Taking note of 8.4.4 and the Sequence Prime Principle, observe that eachnonzero point of the spectrum of a compact operator K is either an eigenvalue of Kor an eigenvalue of the dual of K. Using 8.4.5 and 8.4.6, conclude that, for eachstrictly positive ε, there are only finitely many points of Sp(K) beyond the diskcentered at zero and having radius ε. B

8.4.8. Riesz–Schauder Theorem. The spectrum of a compact operator Kin an infinite-dimensional space contains zero. Each nonzero point of the spectrumof K is isolated and presents an eigenvalue of K with the corresponding eigenspacefinite-dimensional.

C Considering K, a compact endomorphism of a Banach space X, we mustonly demonstrate the implication

0 6= λ ∈ Sp(K)⇒ ker(λ−K) 6= 0.

First, settle the case F := C. Note that {λ} is a clopen part of Sp(K). Puttingg(z) := 1/z in some neighborhood about λ and g(z) := 0 for z in a suitableneighborhood about {λ}′, observe that κ{λ} = gIC. Thus, by 8.2.3 and 8.2.10,P{λ} = g(K)K. By virtue of 8.3.2 (2), P{λ} ∈ K (X). From 8.4.3 it follows thatimP{λ} is a finite-dimensional space. It remains to invoke the Spectral Decompo-sition Theorem.

In the case of the reals, F := R, implement the process of complexification.Namely, furnish the space X2 with multiplication by an element of C which isintroduced by the rule i(x, y) := (−y, x). The resulting complex vector spaceis denoted by X ⊕ iX. Define the operator K(x, y) := (Kx, Ky) in the spaceX ⊕ iX. Equipping X ⊕ iX with an appropriate norm (cf. 7.3.2), observe thatthe operator K is compact and λ ∈ Sp(K). Consequently, λ is an eigenvalue of Kby what was proven. Whence it follows that λ is an eigenvalue of K. B

8.4.9. Theorem. Let X be a complex Banach space. Given T ∈ B(X),assume further that f : C0 = 0 OO C is a holomorphic function vanishing only at zeroand such that f(T ) ∈ K (X). Then every nonzero point λ of the spectrum of T isisolated and the Riesz projection P{λ} is compact.C Suppose the contrary; i.e., find a sequence (λn)n∈N of distinct points of Sp(T )

such that λn0 = 0 OO 6= 0 (in particular, X is infinite-dimensional). Then f(λn)0 =0 OO (λ) and f(λ) 6= 0 by hypothesis. By the Spectral Mapping Theorem, Sp(f(T )) =f(Sp(T )). Thus, by 8.4.8, f(λn) = f(λ) for all sufficiently large n. Whence itfollows that f(z) = f(λ) for all z ∈ C and so f(T ) = f(λ). By the Riesz Criterionin this case X is finite-dimensional. We come to a contradiction meaning that λis an isolated point of Sp(T ). Letting g(z) := f(z)−1 in some neighborhood aboutλ disjoint from zero, infer that gf = κ{λ}. Consequently, by the Gelfand–DunfordTheorem, P{λ} = g(T )f(T ); i.e., in virtue of 8.3.2 (2) the Riesz projection P{λ} iscompact. B

144 Chapter 8

8.4.10. Remark. Theorem 8.4.9 is sometimes referred to as the GeneralizedRiesz–Schauder Theorem.

8.5. Fredholm Operators

8.5.1. Definition. Let X and Y be Banach spaces (over the same groundfield F). An operator T , a member of B(X, Y ), is a Fredholm operator (in symbols,T ∈ F r(X, Y )) if kerT := T−1(0) and cokerT := Y/ imT are finite-dimensional;i.e., if the following quantities, called the nullity and the deficiency of T , are finite:

α(T ) := nul T := dim kerT ; β(T ) := def T := dim cokerT.

The integer ind T := α(T )−β(T ), a member of Z, is the index or, fully, the Fredholmindex of T .

8.5.2. Remark. In the Russian literature, a Fredholm operator is usuallycalled a Noether operator, whereas the term “Fredholm operator” is applied only toan index-zero Fredholm operator.

8.5.3. Every Fredholm operator is normally solvable.C Immediate from the Kato Criterion. B

8.5.4. For T ∈ B(X, Y ), the equivalence holds:

T ∈ F r(X, Y )⇔ T ′ ∈ F r(Y ′, X ′).

Moreover, ind T = − ind T ′.C By virtue of 2.3.5 (6), 8.5.3, 5.5.4 and the Sequence Prime Principle, the

next pairs of sequences are exact simultaneously:

0→ kerT0 = 0 OO T−→ Y → cokerT → 0;

0← (kerT )′ ← X ′T ′←− Y ′ ← (cokerT )′ ← 0;

0→ ker(T ′)→ Y ′T ′−→ X ′ → coker(T ′)→ 0;

0← (ker(T ′))′ ← YT←− X ← (coker(T ′))′ ← 0.

Moreover, α(T ) = β(T ′) and β(T ) = α(T ′) (cf. 7.6.14). B

8.5.5. An operator T is an index-zero Fredholm operator if and only if so isthe dual of T .C This is a particular case of 8.5.4. B


8.5.6. Fredholm Alternative. For an index-zero Fredholm operator Teither of the following mutually exclusive events takes place:

(1) The homogeneous equation Tx = 0 has a sole solution, zero. Thehomogeneous conjugate equation T ′y ′ = 0 has a sole solution, zero. The equationTx = y is solvable and has a unique solution given an arbitrary right side. Theconjugate equation T ′y ′ = x′ is solvable and has a unique solution given an arbitraryright side.

(2) The homogeneous equation Tx = 0 has a nonzero solution. The ho-mogeneous conjugate equation T ′y ′ = 0 has a nonzero solution. The homogeneousequation Tx = 0 has finitely many linearly independent solutions x1, . . . , xn. Thehomogeneous conjugate equation T ′y ′ = 0 has finitely many linearly independentsolutions y ′1, . . . , y ′n.

The equation Tx = y is solvable if and only if y ′1(y) = . . . = y ′n(y) = 0.Moreover, the general solution x is the sum of a particular solution x0 and thegeneral solution of the homogeneous equation; i.e., it has the form

x = x0 +n∑k=1

λkxk (λk ∈ F ).

The conjugate equation T ′y ′ = x′ is solvable if and only if x′(x1) = . . . =x′(xn) = 0. Moreover, the general solution y ′ is the sum of a particular solution y ′0and the general solution of the homogeneous equation; i.e., it has the form

y ′ = y ′0 +n∑k=1

µky′k (µk ∈ F ).

C This is a reformulation of 8.5.5 with account taken of the Polar Lemma. B

8.5.7. Examples.(1) If T is invertible then T is an index-zero Fredholm operator.(2) Let T ∈ L (F n, F m). Let rankT := dim imT be the rank of T .

Then α(T ) = n− rankT and β(T ) = m− rankT . Consequently, T ∈ F r(F n, F m)and ind T = n−m.

(3) Let T ∈ B(X) and X = X1 ⊕ X2. Assume that this direct sumdecomposition of X reduces T to matrix form

T ∼(T1 00 T2

).

Undoubtedly, T is a Fredholm operator if and only if its parts are Fredholmoperators. Moreover, α(T ) = α(T1)+α(T2) and β(T ) = β(T1)+β(T2); i.e., ind T =ind T1 + ind T2. CB

146 Chapter 8

8.5.8. Fredholm Theorem. Let K ∈ K (X). Then 1 −K is an index-zeroFredholm operator.

C First, settle the case F := C. If 1 /∈ Sp(K) then 1 − K is invertible andind (1 − K) = 0. If 1 ∈ Sp(K) then in virtue of the Riesz–Schauder Theoremand the Spectral Decomposition Theorem there is a decomposition X = X1 ⊕X2such that X1 is finite-dimensional and 1 /∈ Sp(K2), with K2 the part of K in X2.Furthermore,

1−K ∼(1−K1 0

0 1−K2

).

By 8.5.7 (2), ind (1−K1) = 0 and, by 8.5.7 (3), ind (1−K) = ind (1−K1)+ind (1−K2) = 0.

In the case of the reals, F := R, proceed by way of complexification as inthe proof of 8.4.8. Namely, consider the operator K(x, y) := (Kx, Ky) in thespace X ⊕ iX. By above, ind (1 −K) = 0. Considering the difference between Rand C, observe that α(1 − K) = α(1 − K) and β(1 − K) = β(1 − K). Finally,ind (1−K) = 0. B

8.5.9. Definition. Let T ∈ B(X, Y ). An operator L, a member of B(Y, X),is a left approximate inverse of T if LT − 1 ∈ K (X). An operator R, a member ofB(Y, X), is a right approximate inverse of T if TR − 1 ∈ K (Y ). An operator S,a member of B(Y, X), is an approximate inverse of T if S is simultaneously a leftand right approximate inverse of T . If an operator T has an approximate inverse Sthen T is called approximately invertible. The terms “regularizer” and “parametrix”are all current in this context with regard to S.

8.5.10. Let L and R be a left approximate inverse and a right approximateinverse of T , respectively. Then L−R ∈ K (Y, X).

/ LT = 1 +KX (KX ∈ K (X))⇒ LTR = R+KXR;TR = 1 +KY (KY ∈ K (Y ))⇒ LTR = L+ LKY .

8.5.11. If L is a left approximate inverse of T and K ∈ K (Y, X) then L+Kis also a left approximate inverse of T .

/ (L+K)T − 1 = (LT − 1) +KT ∈ K (X) .

8.5.12. An operator is approximately invertible if and only if it has a rightapproximate inverse and a left approximate inverse.C Only the implication⇐ needs examining. Let L and R be a left approximate

inverse and a right approximate inverse of T , respectively. By 8.5.10, K := L−R ∈K (Y, X). Consequently, by 8.5.11, R = L−K is a left approximate inverse of T .Thus, R is an approximate inverse of T . B


8.5.13. Remark. The above shows that, in the case X = Y , an operator Sis an approximate inverse of T if and only if ϕ(S)ϕ(T ) = ϕ(T )ϕ(S) = 1, whereϕ : B(X)0 = 0 OO (X)/K (X) is the coset mapping to the Calkin algebra. In otherwords, a left approximate inverse is the inverse image of a left inverse in the Calkinalgebra, etc.

8.5.14. Noether Criterion. An operator is a Fredholm operator if and onlyif it is approximately invertible.

C ⇒: Let T ∈ F r(X, Y ). Using the Complementation Principle, consider thedecompositions X = kerT ⊕X1 and Y = imT ⊕ Y1 and the respective finite-rankprojections P which carries X onto kerT along X1 and Q which carries Y ontoY1 along imT . It is clear that the restriction T1 := T |X1 is an invertible operatorT1 : X10 = 0 OO T . Put S := T−1

1 (1−Q). The operator S may be viewed as a memberof B(Y, X). Moreover, it is beyond a doubt that ST + P = 1 and TS +Q = 1.⇐: Let S be an approximate inverse of T ; i.e., ST = 1+KX and TS = 1+KY

for appropriate compact operatorsKX andKY . Consequently, kerT ⊂ ker(1+KX);i.e., kerT is finite-dimensional since so is ker(1+KX) in virtue of 8.5.8. Furthermore,imT ⊃ im(1 +KY ); and so the range of T is of finite codimension because 1 +KY

is an index-zero Fredholm operator. B

8.5.15. Corollary. If T ∈ F r(X, Y ) and S is an approximate inverse of Tthen S ∈ F r(Y, X). CB

8.5.16. Corollary. The product of Fredholm operators is itself a Fredholmoperator.

C The composition of approximate inverses (taken in due succession) is an ap-proximate inverse to the composition of the originals. B

8.5.17. Consider an exact sequence

00 = 0 OO 10 = 0 OO 20 = 0 OO 0 = 0 OOn−10 = 0 OO

n0 = 0 OO

of finite-dimensional vector spaces. Then the Euler identity holds:

n∑k=1

(−1)k dimXk = 0.

C For n = 1 the exactness of the sequence 00 = 0 OO 10 = 0 OO means that X1 = 0,and for n = 2 the exactness of 00 = 0 OO 10 = 0 OO 20 = 0 OO amounts to isomorphybetween X1 and X2 (cf. 2.3.5 (4)). Therefore, the Euler identity is beyond a doubtfor n := 1, 2.

148 Chapter 8

Suppose now that for m ≤ n− 1, where n > 2, the desired identity is alreadyestablished. The exact sequence

0→ X10 = 0 OO 20 = 0 OO → Xn−2Tn−2−−−−→ Xn−1

Tn−1−−−−→ Xn → 0

reduces to the exact sequence

0→ X1 → X2 → . . .→ Xn−2Tn−2−−−→ kerTn−1 → 0.

By hypothesis,

n−2∑k=1

(−1)k dimXk + (−1)n−1 dimkerTn−1 = 0.

Furthermore, since Tn−1 is an epimorphism,

dimXn−1 = dimkerTn−1 + dimXn.

Finally,

0 =n−2∑k=1

(−1)k dimXk + (−1)n−1(dimXn−1 − dimXn)

=n∑k=1

(−1)k dimXk. .

8.5.18.Atkinson Theorem. The index of the product of Fredholm operatorsequals the sum of the indices of the factors.

C Let T ∈ F r(X, Y ) and S ∈ F r(Y, Z). By virtue of 8.5.16, ST ∈ F r(X, Z).Using the Snowflake Lemma, obtain the exact sequence of finite-dimensional spaces

00 = 0 OO T0 = 0 OO ST0 = 0 OO S0 = 0 OO T0 = 0 OO ST0 = 0 OO S0 = 0 OO .

Applying 8.5.17, infer that

α(T )− α(ST ) + α(S)− β(T ) + β(ST )− β(S) = 0;

whence ind (ST ) = ind S + ind T . B

8.5.19.Corollary. Let T be a Fredholm operator and let S be an approximateinverse of T . Then ind T = − ind S.

C ind (ST ) = ind (1 +K) for some compact operator K. By Theorem 8.5.8,1 +K is an index-zero Fredholm operator. B


8.5.20. Compact Index Stability Theorem. The property of beinga Fredholm operator and the index of a Fredholm operator are preserved under com-pact perturbations: if T ∈ F r(X, Y ) and K ∈ K (X, Y ) then T +K ∈ F r(X, Y )and ind (T +K) = ind T .

C Let S be an approximate inverse to T ; i.e.,

ST = 1 +KX ; TS = 1 +KY

with some KX ∈ K (X) and KY ∈ K (Y ) (that S exists is ensured by 8.5.14). It isclear that

S(T +K) = ST + SK = 1 +KX + SK ∈ 1 + K (X);(T +K)S = TS +KS = 1 +KY +KS ∈ 1 + K (Y );

i.e., S is an approximate inverse of T +K. By virtue of 8.5.14, T +K ∈ F r(X, Y ).Finally, from 8.5.19 infer the equalities ind (T+K) = − ind S and ind T = − ind S.B

8.5.21. Bounded Index Stability Theorem. The property of beinga Fredholm operator and the index of a Fredholm operator are preserved undersufficiently small bounded perturbations: the set F r(X, Y ) is open in the space ofbounded operators, and the index of a Fredholm operator ind : F r(X, Y )0 = 0 OO Zis a continuous function.

C Let T ∈ F r(X, Y ). By 8.5.14 there are operators S ∈ B(Y, X), KX ∈K (X) and KY ∈ K (Y ) such that

ST = 1 +KX ; TS = 1 +KY .

If S = 0 then the spaces X and Y are finite-dimensional by the Riesz Criterion,i.e., nothing is left to proof: it suffices to refer to 8.5.7 (2). If S 6= 0 then for allV ∈ B(X, Y ) with ‖V ‖ < 1/‖S‖, from the inequality of 5.6.1 it follows: ‖SV ‖ < 1and ‖V S‖ < 1. Consequently, in virtue of 5.6.10 the operators 1 + SV and 1+ V Sare invertible in B(X) and in B(Y ), respectively.

Observe that

(1 + SV )−1S(T + V ) = (1 + SV )−1(1 +KX + SV )= 1 + (1 + SV )−1KX ∈ 1 + K (X);

i.e., (1 + SV )−1S is a left approximate inverse of T + V . By analogy, show thatS(1 + V S)−1 is a right approximate inverse of T + V . Indeed,

(T + V )S(1 + V S)−1 = (1 +KY + V S)(1 + V S)−1

= 1 +KY (1 + V S)−1 ∈ 1 + K (Y ).

150 Chapter 8

By 8.5.12, T+V is approximately invertible. In virtue of 8.5.14, T+V ∈ F r(X, Y ).This proves the openness property of F r(X, Y ). When left and right approximateinverses of a Fredholm operator W exist, each of them is an approximate inverseto W (cf. 8.5.12), Therefore, from 8.5.19 and 8.5.18 obtain

ind (T + V ) = − ind ((1 + SV )−1S)= − ind (1 + SV )−1 − ind S = − ind S = ind T

(because (1 + SV )−1 is a Fredholm operator by 8.5.7 (1)). This means that theFredholm index is continuous. B

8.5.22. Nikol ′skiı Criterion. An operator is an index-zero Fredholm opera-tor if and only if it is the sum of an invertible operator and a compact operator.

C ⇒: Let T ∈ F r(X, Y ) and ind T = 0. Consider the direct sum decom-positions X = X1 ⊕ kerT and Y = imT ⊕ Y1. It is beyond a doubt that theoperator T1, the restriction of T to X1, implements an isomorphism between X1and imT . Furthermore, in virtue of 8.5.5, dimY1 = β(T ) = α(T ); i.e., there isa natural isomorphism Id : kerT0 = 0 OO 1. Therefore, T admits the matrix presen-tation

T ∼(T1 00 0

)=(T1 00 Id

)+(0 00 − Id

).

⇐: If T := S + K with K ∈ K (X, Y ) and S−1 ∈ B(Y, X) then, by 8.5.20and 8.5.7 (1), ind T = ind (S +K) = ind S = 0. B

8.5.23. Remark. Let Inv(X, Y ) stand as before for the set of all invertibleoperators from X to Y (this set is open by Theorem 5.6.12). Denote by F (X, Z)the set of all Fredholm operators acting from X to Y and having index zero. TheNikol′skiı Criterion may now be written down as

F (X, Y ) = Inv(X, Y ) + K (X, Y ).

As is seen from the proof of 8.5.22, it may also be asserted that

F (X, Y ) = Inv(X, Y ) + F (X, Y ),

where, as usual, F (X, Y ) is the subspace of B(X, Y ) comprising all finite-rankoperators. CB

Exercises

8.1. Study the Riesz–Dunford integral in finite dimensions.8.2. Describe the kernel of the Riesz–Dunford integral.8.3. Given n ∈ N, let fn be a function holomorphic in a neighborhood U about the spectrum

of an operator T . Prove that the uniform convergence of (fn) to zero on U follows from theconvergence of (fn(T )) to zero in operator norm.


8.4. Let σ be an isolated part of the spectrum of an operator T . Assume that the partσ′ := Sp(T ) \ σ is separated from σ by some open disk with center a and radius r so thatσ ⊂ {z ∈ C : |z − a| < r}. Considering the Riesz projection Pσ , prove that

Pσ = limn

(1− z−n(T − a)n)−1;

x ∈ im(Pσ)⇔ lim supn‖(a− T )nx‖

1/n < r.

8.5. Find conditions for a projection to be a compact operator.8.6. Prove that every closed subspace lying in the range of a compact operator in a Banach

space is finite-dimensional.

8.7. Prove that a linear operator carries each closed linear subspace onto a closed set if andonly if the operator is normally solvable and its kernel is finite-dimensional or finite-codimensional(the latter means that the kernel has a finite-dimensional algebraic complement).

8.8. Let 1 ≤ p < r < +∞. Prove that every bounded operator from lr to lp or from c0 to lpis compact.

8.9. Let H be a separable Hilbert space. Given an operator T in B(H) and a Hilbert basis(en) for H, define the Hilbert–Schmidt norm as

‖T‖2 :=

(∞∑n=1

‖Ten‖2)1/2

.

(Examine soundness!) An operator with finite Hilbert–Schmidt norm is a Hilbert–Schmidt oper-ator. Demonstrate that an operator T is a Hilbert–Schmidt operator if and only if T is compactand

∑∞n=1 λ

2n < +∞, where (λn) ranges over the eigenvalues of some operator (T ∗T )

1/2 (definethe latter!).

8.10. Let T be an endomorphism. Then

im(T 0) ⊃ im(T 1) ⊃ im(T 2) ⊃ . . . .

If there is a number n satisfying im(Tn) = im(Tn+1) then say that T has finite descent. Theleast number n with which stabilization begins is the descent of T , denoted by d(T ). By analogy,considering the kernels

ker(T 0) ⊂ ker(T 1) ⊂ ker(T 2) ⊂ . . . ,

introduce the concept of ascent and the denotation a(T ). Demonstrate that, for an operator Twith finite descent and finite ascent, the two quantities, a(T ) and d(T ), coincide.

8.11. An operator T is a Riesz–Schauder operator, if T is a Fredholm operator and hasfinite descent and finite ascent. Prove that an operator T is a Riesz–Schauder operator if and onlyif T is of the form T = U + V , where U is invertible and V is of finite rank (or compact) andcommutes with U .

8.12. Let T be a bounded endomorphism of a Banach space X which has finite descent andfinite ascent, r := a(T ) = d(T ). Prove that the subspaces im(T r) and ker(T r) are closed, thedecomposition X = ker(T r)⊕ im(T r) reduces T , and the restriction of T to im(T r) is invertible.

152 Chapter 8

8.13. Let T be a normally solvable operator. If either of the next quantities is finite

α(T ) := dim kerT, β(T ) := dim cokerT,

then T is called semi-Fredholm. Put

Ф+(X) := {T ∈ B(X) : imT ∈ Cl(X), α(T ) < +∞};

Ф−(X) := {T ∈ B(X) : imT ∈ Cl(X), β(T ) < +∞}.

Prove that

T ∈ Ф+(X)⇔ T ′ ∈ Ф−(X′);

T ∈ Ф−(X)⇔ T ′ ∈ Ф+(X′).

8.14. Let T be a bounded endomorphism. Prove that T belongs to Ф+(X) if and only if forevery bounded but not totally bounded set U , the image T (U) is not a totally bounded set in X.

8.15. A bounded endomorphism T in a Banach space is a Riesz operator, if for every nonzerocomplex λ the operator (λ − T ) is a Fredholm operator. Prove that T is a Riesz operator if andonly if for all λ ∈ C, λ 6= 0, the following conditions are fulfilled:

(1) the operator (λ− T ) has finite descent and finite ascent;(2) the kernel of (λ− T )k is finite-dimensional for every k ∈ N;(3) the range of (λ− T )k has finite deficiency for k ∈ N,

and, moreover, all nonzero points of the spectrum of T are eigenvalues, with zero serving as theonly admissible limit point (that is, for whatever strictly positive ε, there are only finitely manypoints of Sp(T ) beyond the disk centered at zero with radius ε).

8.16. Establish the isometric isomorphisms: (X/Y )′ ' Y ⊥ and X′/Y ⊥ ' Y ′ for Banachspaces X and Y such that Y is embedded into X.

8.17. Prove that for a normal operator T in a Hilbert space and a holomorphic function f ,a member of H(Sp(T )), the operator f(T ) is normal. (An operator is normal if it commutes withits adjoint, cf. 11.7.1.)

8.18. Show that a continuous endomorphism T of a Hilbert space is a Riesz operator if andonly if T is the sum of a compact operator and a quasinilpotent operator. (Quasinilpotency ofan operator means triviality of its spectral radius.)

8.19. Given two Fredholm operators S and T , members of Fr(X, Y ), with ind S = ind T ,demonstrate that there is a Jordan arc joining S and T within Fr(X, Y ).

Chapter 9

An Excursus into General Topology

9.1. Pretopologies and Topologies

9.1.1. Definition. Let X be a set. A mapping τ : X0 = 0 OO P (P(X)) isa pretopology on X if

(1) x ∈ X ⇒ τ(x) is a filter on X;(2) x ∈ X ⇒ τ(x) ⊂ fil{x}.

A member of τ(x) is a (pre)neighborhood about x or of x. The pair (X, τ), as wellas the set X itself, is called a pretopological space.

9.1.2. Definition. Let T (X) be the collection of all pretopologies on X.If τ1, τ2 ∈ T (X) then τ1 is said to be stronger than τ2 or finer than τ2 (in symbols,τ1 ≥ τ2) provided that x ∈ X ⇒ τ1(x) ⊃ τ2(x). Of course, τ2 is weaker or coarserthan τ1.

9.1.3. The set T (X) with the relation “to be stronger” presents a completelattice.C If X = ∅ then T (X) = {∅} and there is nothing to be proven. If X 6= ∅

then refer to 1.3.13. B

9.1.4. Definition. A subset G of X is an open set in X, if G is a (pre)neigh-borhood of its every point (in symbols, G ∈ Op(τ) ⇔ (∀x ∈ G) (G ∈ τ(x))).A subset F of X is a closed set in X if the complement of F to X is open; that is,F ∈ Cl(τ)⇔ X \ F ∈ Op(τ).

9.1.5. The union of a family of open sets and the intersection of a finite familyof open sets are open. The intersection of a family of closed sets and the unionof a finite family of closed sets are closed. CB

9.1.6. Let (X, τ) be a pretopological space. Given x ∈ X, put

U ∈ t(τ)(x)⇔ (∃V ∈ Op(τ)) x ∈ V & U ⊃ V.The mapping t(τ) : x 7→ t(τ)(x) is a pretopology on X. CB

154 Chapter 9

9.1.7. Definition. A pretopology τ on X is a topology if τ = t(τ). The pair(X, τ), as well as the underlying set X itself, is then called a topological space. Theset of all topologies on X is denoted by the symbol T(X).

9.1.8. Examples.(1) A metric topology.(2) The topology of a multinormed space.(3) Let τ◦ := inf T (X). It is clear that τ◦(x) = {X} for x ∈ X.

Consequently, Op(τ◦) = {∅, X} and so τ◦ = t(τ◦); i.e., τ◦ is a topology. Thistopology is called trivial, or antidiscrete, or even indiscrete.

(4) Let τ◦ := supT (X). It is clear that τ◦(x) = fil {x} for x ∈ X.Consequently, Op(τ◦) = 2X and so τ◦ = t(τ◦); i.e., τ◦ is a topology. This topologyis called discrete.

(5) Let Op be a collection of subsets in X which is stable under thetaking of the union of each of its subfamilies and the intersection of each of its finitesubfamilies. Then there is a unique topology τ on X such that Op(τ) = Op.C Put τ(x) := fil {V ∈ Op : x ∈ V } for x ∈ X (in case X = ∅ there is nothing

to prove). Observe that τ(x) 6= ∅ since the intersection of the empty family equalsX (cf. inf ∅ = +∞). From the construction derive that t(τ) = τ and Op ⊂ Op(τ).If G ∈ Op(τ) then G = ∪{V : V ∈ Op, V ⊂ G} and so G ∈ Op by hypothesis.The claim of uniqueness raises no doubts. B

9.1.9. Let the mapping t : T (X)0 = 0 OO T (X) act by the rule t : τ 7→ t(τ).Then

(1) im t = T(X); i.e., τ ∈ T (X)⇒ t(τ) ∈ T(X);(2) τ1 ≤ τ2 ⇒ t(τ1) ≤ t(τ2) (τ1, τ2 ∈ T (X));(3) t ◦ t = t;(4) τ ∈ T (X)⇒ t(τ) ≤ τ ;(5) Op(τ) = Op(t(τ)) (τ ∈ T (X)).

C The inclusion Op(τ) ⊃ Op(t(τ)) holds because it is easier to be open in τ .The reverse inclusion Op(τ) ⊂ Op(t(τ)) follows from the definition of t(τ). Theequality Op(τ) = Op(t(τ)) makes everything evident. B

9.1.10. A pretopology τ on X is a topology if and only if

(∀U ∈ τ(x))(∃V ∈ τ(x) & V ⊂ U)(∀ y) (y ∈ V ⇒ V ∈ τ(y))

for x ∈ X.C Straightforward from 9.1.9 (5). B

An Excursus into General Topology 155

9.1.11. Let τ1, τ2 ∈ T(X). The following statements are equivalent:(1) τ1 ≥ τ2;(2) Op(τ1) ⊃ Op(τ2);(3) Cl(τ1) ⊃ Cl(τ2). CB

9.1.12. Remark. As follows from 9.1.8 (5) and 9.1.11, the topology of a spaceis uniquely determined from the collection of its open sets. Therefore, the set Op(X)itself is legitimately called the topology of the space X. In particular, the collectionof open sets of a pretopological space (X, τ) makes X into the topological space(X, t(τ)) with the same open sets in stock. Therefore, given a pretopology τ , thetopology t(τ) is usually called the topology associated with τ .

9.1.13. Theorem. The set T(X) of all topologies on X with the relation“to be stronger” presents a complete lattice. Moreover, for every subset E of T(X)the equality holds:

supT(X) E = supT (X) E .

C Evidently, t(supT (X) E ) ≥ supT (X) t(E ) ≥ supT (X) E ≥ t(supT (X) E ).Thus, τ := supT (X) E belongs to T(X). It is clear that τ ≥ E . Furthermore,if τ0 ≥ E and τ0 ∈ T(X) then τ0 ≥ τ and so τ = supT(X) E . It remains to referto 1.2.14. B

9.1.14. Remark. The explicit formula for the greatest lower bound of E ismore involved:

infT(X) E = t(infT (X) E ).

However, the matter becomes simpler when the topologies are given by means oftheir collections of open sets in accordance with 9.1.12. Namely,

U ∈ Op(infT(X) E )⇔ (∀ τ ∈ E ) U ∈ Op(τ).

In other words,Op(infT(X) E ) =

⋂τ∈E

Op(τ).

In this connection it is in common parlance to speak of the intersection of the set Eof topologies (rather than of the greatest lower bound of E ). CB

9.2. Continuity

9.2.1. Remark. The presence of a topology on a set obviously makes it pos-sible to deal with such things as the interior and closure of a subset, convergenceof filters and nets, etc. We have already made use of this circumstance while in-troducing multinormed spaces. Observe for the sake of completeness that in everytopological space the following analogs of 4.1.19 and 4.2.1 are valid:

156 Chapter 9

9.2.2. Birkhoff Theorem. For a nonempty subset U and a point x of a topo-logical space the following statements are equivalent:

(1) x is an adherent point of U ;(2) there is some filter containing U and converging to x;(3) there is a net of elements of U which converges to x. CB

9.2.3. For a mapping f between topological spaces the following conditionsare equivalent:

(1) the inverse image under f of an open set is open;(2) the inverse image under f of a closed set is closed;(3) the image under f of the neighborhood filter of an arbitrary point

x is coarser than the neighborhood filter of f(x);(4) for all x, the mapping f transforms each filter convergent to x into

a filter convergent to f(x);(5) for all x, the mapping f sends a net that converges to x to a net

that converges to f(x). CB

9.2.4. Definition. A mapping acting between topological spaces X and Yand satisfying one (and hence all) of the equivalent conditions 9.2.3 (1)–9.2.3 (5)is called continuous. A continuous one-to-one mapping f from X onto Y whoseinverse f−1 acts continuously from Y to X is a homeomorphism or a topologicalmapping or a topological isomorphism between X and Y .

9.2.5. Remark. If f : (X, τX)→ (Y, τY ) meets 9.2.3 (5) at some point x inXthen it is customary to say that f is continuous at x (cf. 4.2.2). Observe that thedifference is immaterial between the definitions of the continuity property at a pointof X and the general continuity property (on X). Indeed, if we let τx(x) := τX(x)and τx(x) := fil {x} for x ∈ X, x 6= x, then the continuity property of f at x (withrespect to the topology τX in X) amounts to that of f : (X, τx)→ (Y, τY ) (at everypoint of the space X with topology τx).

9.2.6. Let τ1, τ2 ∈ T(X). Then τ1 ≥ τ2 if and only if IX : (X, τ1)→ (X, τ2)is continuous. CB

9.2.7. Let f : (X, τ) → (Y, ω) be a continuous mapping and let τ1 ∈ T(X)and ω1 ∈ T(Y ) be such that τ1 ≥ τ and ω ≥ ω1. Then f : (X, τ1) → (Y, ω1) iscontinuous.

C By hypothesis the following diagram commutes:

(X, τ) f−→ (Y, ω)IX ↑ ↓ IY(X, τ1)

f−→ (Y, ω1)


It suffices to observe that every composition of continuous mappings is continu-ous. B

9.2.8. Inverse Image Topology Theorem. Let f : X → (Y, ω). Put

T0 := {τ ∈ T(X) : f : (X, τ)→ (Y, ω) is continuous}.

Then the topology f−1(ω) := inf T0 belongs to T0.C From 9.2.3 (1) it follows that

τ ∈ T0 ⇔ (x ∈ X ⇒ f−1(ω(f(x))) ⊂ τ(x)).

Let τ(x) := f−1(ω(f(x))). Undoubtedly, t(τ) = τ . Furthermore, f(τ(x)) =f(f−1(ω(f(x)))) ⊃ ω(f(x)); i.e., τ ∈ T0 by 9.2.3 (3). Thus, f−1(ω) = τ . B

9.2.9. Definition. The topology f−1(ω) is the inverse image of ω undera mapping f or simply the inverse image topology under f .

9.2.10. Remark. Theorem 9.2.8 is often verbalized as follows: “The inverseimage topology under a mapping is the weakest topology on the set of departurein which the mapping is continuous.” Moreover, it is easy for instance from 9.1.14that the open sets of the inverse image topology are precisely the inverse imagesof open sets. In particular, (xξ → x in f−1(ω)) ⇔ (f(xξ) → f(x) in ω); likewise,(F → x in f−1(ω))⇔ (f(F )→ f(x) in ω) for a filter F . CB

9.2.11. Image Topology Let f : (X, τ)→ Y . Put

�0 := {ω ∈ T(Y ) : f : (X, τ)→ (Y, ω) is continuous}.

Then the topology f(τ) := sup�0 belongs to �0.C Appealing to 9.1.13, observe that

f(τ)(y) = (supT(Y ) �0)(y) = (supT (Y ) �0)(y) = sup{ω(y) : ω ∈ �0}

for y ∈ Y . By virtue of 9.2.3 (3),

ω ∈ �0 ⇔ (x ∈ X ⇒ f(τ(x)) ⊃ ω(f(x))).

Comparing the formulas, infer that f(τ) ∈ �0. B

9.2.12. Definition. The topology f(τ) is the image of τ under a mapping for simply the image topology under f .

158 Chapter 9

9.2.13. Remark. Theorem 9.2.11 is often verbalized as follows: “The imageof a topology under a mapping is the strongest topology on the set of arrival in whichthe mapping is continuous.”

9.2.14. Theorem. Let (fξ : X → (Yξ, ωξ))ξ∈� be a family of mappings.Further, put τ := supξ∈� f

−1ξ (ωξ). Then τ is the weakest (= least) topology on X

making all the mappings fξ (ξ ∈ �) continuous.C Using 9.2.8, note that

(fξ : (X, τ)→ (Yξ, ωξ) is continuous) ⇔ τ ≥ f−1ξ (ωξ). .

9.2.15. Theorem. Let (fξ : (Xξ, τξ) → Y )ξ∈� be a family of mappings.Further, assign ω := infξ∈� fξ(τξ). Then ω is the strongest (= greatest) topologyon Y making all the mappings fξ (ξ ∈ �) continuous.

C Appealing to 9.2.11, conclude that

(fξ : (Xξ, τξ)→ (Y, ω) is continuous) ⇔ ω ≤ fξ(τξ). .

9.2.16. Remark. The messages of 9.2.14 and 9.2.15 are often referred to asthe theorems on topologizing by a family of mappings.

9.2.17. Examples.(1) Let (X, τ) be a topological space and let X0 be a subset of X.

Denote the identical embedding of X0 into X by ι : X0 → X. Put τ0 := ι−1(τ).The topology τ0 is the induced topology (by τ in X0), or the relative or subspacetopology; and the space (X0, τ0) is a subspace of (X, τ).

(2) Let (Xξ, τξ)ξ∈� be a family of topological spaces and let X :=∏ξ∈�Xξ be the product of (Xξ)ξ∈�. Put τ := supξ∈� Pr

−1ξ (τξ), where Prξ : X→ Xξ

is the coordinate projection (onto Xξ); i.e., Prξ x = xξ (ξ ∈ �). The topology τis the product topology or the product of the topologies (τξ)ξ∈�, or the Tychonofftopology of X. The space (X, τ) is the Tychonoff product of the topological spacesunder study. In particular, if Xξ := [0, 1] for all ξ ∈ � then X := [0, 1]� (with theTychonoff topology) is a Tychonoff cube. When � := N, the term “Hilbert cube” isapplied.

9.3. Types of Topological Spaces

9.3.1. For a topological space the following conditions are equivalent:(1) every singleton of the space is closed;(2) the intersection of all neighborhoods of each point in the space

consists solely of the point;


(3) each one of any two points in the space has a neighborhood disjointfrom the other.

C To prove, it suffices to observe that

y ∈ cl{x} ⇔ (∀V ∈ τ(y)) x ∈ V ⇔ x ∈ ∩{V : V ∈ τ(y)},

where x and y are points of a space with topology τ . B

9.3.2. Definition. A topological space satisfying one (and hence all) of theequivalent conditions 9.3.1 (1)–9.3.1 (3), is called a separated space or a T1-space.The topology of a T1-space is called a separated topology or (rarely) a T1-topology.

9.3.3. Remark. By way of expressiveness, one often says: “A T1-space isa space with closed points.”

9.3.4. For a topological space the following conditions are equivalent:(1) each filter has at most one limit;(2) the intersection of all closed neighborhoods of a point in the space

consists of the sole point;(3) each one of any two points of the space has a neighborhood disjoint

from some neighborhood of the other point.C (1) ⇒ (2): If y ∈ ∩U∈τ(x) clU then U ∩ V 6= ∅ for all V ∈ τ(y), provided

that U ∈ τ(x). Therefore, the join F := τ(x) ∨ τ(y) is available. Clearly, F → xand F → y. By hypothesis, x = y.

(2)⇒ (3): Let x, y ∈ X, x 6= y (if such points are absent then either X = ∅ orX is a singleton and nothing is left unproven). There is an neighborhood U in τ(x)such that U = clU and y /∈ U . Consequently, the complement V of U to X is open.Furthermore, U ∩ V = ∅.

(3) ⇒ (1): Let F be a filter on X. If F → x and F → y then F ⊃ τ(x)and F ⊃ τ(y). Thus, U ∩ V 6= ∅ for U ∈ τ(x) and V ∈ τ(y), which means thatx = y. B

9.3.5. Definition. A topological space satisfying one (and hence all) of theequivalent conditions 9.3.4 (1)–9.3.4 (3) is a Hausdorff space or a T2-space. A nat-ural meaning is ascribed to the term “Hausdorff topology.”

9.3.6. Remark. By way of expressiveness, one often says: “A T2-space isa space with unique limits.”

9.3.7. Definition. Let U and V be subsets of a topological space. It is saidthat V is a neighborhood of U or about U , provided intV ⊃ U . If U is nonemptythen all neighborhoods of U constitute some filter that is the neighborhood filterof U .

160 Chapter 9

9.3.8. For a topological space the following conditions are equivalent:(1) the intersection of all closed neighborhoods of an arbitrary closed

set consists only of the members of the set;(2) the neighborhood filter of each point has a base of closed sets;(3) if F is a closed set and x is a point not in F then there are disjoint

neighborhoods of F and x, respectively.C (1) ⇒ (2): If x ∈ X and U ∈ τ(x) then V := X \ intU is closed and x /∈ V .

By hypothesis there is a set F in Cl(τ) such that x /∈ F and intF ⊃ V . PutG := X \F . Clearly, G ∈ τ(x). Moreover, G ⊂ X \ intF = cl(X \ intF ) ⊂ X \V ⊂intU ⊂ U . Consequently, clG ⊂ U .

(2) ⇒ (3): If x ∈ X and F ∈ Cl(τ) with x /∈ F then X \F ∈ τ(x). Thus, thereis a closed neighborhood U in τ(x) lying in X \ F . Thus, X \ U is a neighborhoodof F disjoint from U .

(3) ⇒ (1): If F ∈ Cl(τ) and intG ⊃ F ⇒ y ∈ clG, then U ∩G 6= ∅ for everyU in τ(y) and every neighborhood G of F . This means that y ∈ F . B

9.3.9. Definition. A T3-space is a topological space satisfying one (and henceall) of the equivalent conditions 9.3.8 (1)–9.3.8 (3). A separated T3-space is calledregular.

9.3.10. Urysohn Little Lemma. For a topological space the following con-ditions are equivalent:

(1) the neighborhood filter of each nonempty closed set has a base ofclosed sets;

(2) if two closed sets are disjoint then they have disjoint neighborhoods.C (1) ⇒ (2): Let F1 and F2 be closed sets in some space X with F1 ∩F2 = ∅.

Put G := X \ F1. Obviously, G is open and G ⊃ F2. If F2 = ∅, then there isnothing to be proven. It may be assumed consequently that F2 6= ∅. Then there isa closed set V2 such that G ⊃ V2 ⊃ intV2 ⊃ F2. Put V1 := X \ V2. It is clear thatV1 is open, and V1 ∩ V2 = ∅. Moreover, V1 ⊃ X \G = X \ (X \ F1) = F1.

(2) ⇒ (1): Let F = clF , G = intG and G ⊃ F . Put F1 := X \ G. ThenF1 = clF1 and so there are open sets U and U1 satisfying U ∩U1 = ∅, with F ⊂ Uand F1 ⊂ U1. Finally, clU ⊂ X \ U1 ⊂ X \ F1 = G. B

9.3.11. Definition. A T4-space is a topological space meeting one (and henceboth) of the equivalent conditions 9.3.10 (1) and 9.3.10 (2). A separated T4-spaceis called normal.

9.3.12. Continuous Function Recovery Lemma. Let a subset T be densein R and let t 7→ Ut (t ∈ T ) be a family of subsets of a topological space X. Thereis a unique continuous function f : X → R such that

{f < t} ⊂ Ut ⊂ {f ≤ t} (t ∈ T )


if and only if(t, s ∈ T & t < s)⇒ clUt ⊂ intUs.

C ⇒: Take t < s. Since {f ≤ t} is closed and {f < s} is open, the inclusionshold:

clUt ⊂ {f ≤ t} ⊂ {f < s} ⊂ intUs.

⇐: Since Ut ⊂ clUt ⊂ intUs ⊂ Us for t < s, the family t 7→ Ut (t ∈ T ) increasesby inclusion. Therefore, f exists by 3.8.2 and is unique by 3.8.4. Consider thefamilies t 7→ Vt := clUt and t 7→ Wt := intUt. These families increase by inclusion.Consequently, on applying 3.8.2 once again, find functions g, h : X → R satisfying

{g < t} ⊂ Vt ⊂ {g ≤ t}, {h < t} ⊂Wt ⊂ {h ≤ t}

for all t ∈ T . If t, s ∈ T and t < s, then in view of 3.8.3

Wt = intUt ⊂ Ut ⊂ Us ⇒ f ≤ h;Vt = clUt ⊂ intUs =Ws ⇒ h ≤ g;Ut ⊂ Us ⊂ clUs = Vs ⇒ g ≤ f.

So, f = g = h. Taking account of 3.8.4 and 9.1.5 and given t ∈ R, find

{f < t} = {h < t} = ∪{Ws : s < t, s ∈ T} ∈ Op(τX);{f ≤ t} = {g ≤ t} = ∩{Vs : t < s, s ∈ T} ∈ Cl(τX).

These inclusions readily provide continuity for f . B

9.3.13. Urysohn Great Lemma. Let X be a T4-space. Assume further thatF is a closed set in X and G is a neighborhood of F . Then there is a continuousfunction f : X → [0, 1] such that f(x) = 0 for x ∈ F and f(x) = 1 for x /∈ G.C Put Ut := ∅ for t < 0 and Ut := X for t > 1. Consider the set T of the

dyadic-rational points of the interval [0, 1]; i.e., T := ∪n∈NTn with Tn := {k2−n+1 :k := 0, 1, . . . , 2n−1}. It suffices to define Ut for all t in T so that the family t 7→ Ut(t ∈ T := T ∪ (R \ [0, 1])) satisfy the criterion of 9.3.12. This is done by wayof induction.

If t ∈ T1, i.e., t ∈ {0, 1}; then put U0 := F and U1 := G. Assume nowthat, for t ∈ Tn and n ≥ 1, some set Ut has already been constructed, satisfyingclUt ⊂ intUs whenever t, s ∈ Tn and t < s. Take t ∈ Tn+1 and find the two pointstl and tr in Tn nearest to t:

tl := sup{s ∈ Tn : s ≤ t};tr := inf{s ∈ Tn : t ≤ s}.

162 Chapter 9

If t = tl or t = tr, then Ut exists by the induction hypothesis. If t 6= tl andt 6= tr, then tl < t < tr and again by the induction hypothesis clUtl ⊂ intUtr .By virtue of 9.3.11 there is a closed set Ut such that

clUtl ⊂ intUt ⊂ Ut = clUt ⊂ intUtr .

It remains to show that the resulting family satisfies the criterion of 9.3.12.To this end, take t, s ∈ Tn+1 with t < s. If tr = sl, then for s > sl by con-

structionclUt ⊂ clUtr = clUsl ⊂ intUs.

For t < tr = sl similarly deduce the following:

clUt ⊂ intUtr = inf Usl ⊂ intUs.

If tr < sl then, on using the induction hypothesis, infer that

clUt ⊂ clUtr ⊂ intUsl ⊂ intUs,


9.3.14. Urysohn Theorem. A topological space X is a T4-space if and onlyif to every pair of disjoint closed sets F1 and F2 in X there corresponds a continuousfunction f : X → [0, 1] such that f(x) = 0 for x ∈ F1 and f(x) = 1 for x ∈ F2.C ⇒: It suffices to apply 9.3.13 with F := F1 and G := X \ F2.⇐: If F1 ∩ F2 = ∅ and F1 and F2 are closed sets, then for a corresponding

function f the sets G1 := {f < 1/2} and G2 := {f > 1/2} are open and disjoint.Moreover, G1 ⊃ F1 and G2 ⊃ F2. B

9.3.15.Definition. A topological spaceX is a T31/2-space, if to a closed set Fin X and a point x not in F there corresponds a continuous function f : X → [0, 1]such that f(x) = 1 and y ∈ F ⇒ f(y) = 0. A separated T31/2 -space is a Tychonoffspace or a completely regular space.

9.3.16. Every normal space is a Tychonoff space.C Straightforward from 9.3.1 and 9.3.14. B

9.4. Compactness

9.4.1. Let B be a filterbase on a topological space and let

clB := ∩{clB : B ∈ B}

be the set of adherent points of B (also called the adherence of B). Then


(1) clB = cl filB;(2) B → x⇒ x ∈ clB;(3) (B is an ultrafilter and x ∈ clB)⇒ B → x.

C Only (3) needs demonstrating, since (1) and (2) are evident. Given U ∈ τ(x)and B ∈ B, observe that U ∩ B 6= ∅. In other words, the join F := τ(x) ∨B is available. It is clear that F → x. Furthermore, F = B, because B isan ultrafilter. B

9.4.2. Definition. A subset C of a topological space X is a compact set in Xif each open cover of C has a finite subcover (cf. 4.4.1).

9.4.3. Theorem. Let X be a topological space and let C be a subset of X.The following statements are equivalent:

(1) C is compact;(2) if a filterbase B lacks adherent points in C then there is a member

B of B such that B ∩ C = ∅;(3) each filterbase containing C has an adherent point in C;(4) each ultrafilter containing C has a limit in C.

C (1) ⇒ (2): Since clB ∩ C = ∅; therefore, C ⊂ X \ clB. Thus, C ⊂X \ ∩{clB : B ∈ B} = ∪{X \ clB : B ∈ B}. Consequently, there is a finitesubset B0 of B such that C ⊂ ∪{X \ clB0 : B0 ∈ B0} = X \∩{clB0 : B0 ∈ B0}.Let B ∈ B satisfy B ⊂ ∩{B0 : B0 ∈ B0} ⊂ ∩{clB0 : B0 ∈ B0}. ThenC ∩B ⊂ C ∩ (∩{clB0 : B0 ∈ B0}) = ∅.

(2) ⇒ (3): In case C = ∅, there is nothing to be proven. If C 6= ∅, then forB ∈ B by hypothesis B ∩ C 6= ∅ because C ∈ B. Thus, clB ∩ C 6= ∅.

(3) ⇒ (4): It suffices to appeal to 9.4.1.(4) ⇒ (1): Assume that C 6= ∅ (otherwise, nothing is left to proof).Suppose that C is not compact. Then there is a set E of open sets such that

C ⊂ ∪{G : G ∈ E } and at the same time, for every finite subset E0 of E , theinclusion C ⊂ ∪{G : G ∈ E0} fails. Put

B :=

{ ⋂G∈E0

X \G : E0 is a finite subset of E

}.

It is clear that B is a filterbase. Furthermore,

clB = ∩{clB : B ∈ B} = ∩{X \G : G ∈ E }= X \ ∪{G : G ∈ E } ⊂ X \ C.

Now choose an ultrafilter F that is coarser than B, which is guaranteed by 1.3.10.By supposition each member of B meets C. We may thus assume that C ∈ F(adjusting the choice of F , if need be). Then F → x for some x in C and so, by 9.4.1(2), clF ∩ C 6= ∅. At the same time clF ⊂ clB. We arrive at a contradiction. B

164 Chapter 9

9.4.4. Remark. The equivalence (1) ⇔ (4) in Theorem 9.4.3 is called theBourbaki Criterion and verbalized for X = C as follows: “A space is compact if andonly if every ultrafilter on it converges” (cf. 4.4.7). An ultranet is a net whosetail filter is an ultrafilter. The Bourbaki Criterion can be expressed as follows:“Compactness amounts to convergence of ultranets.” Many convenient tests forcompactness are formulated in the language of nets. For instance: “A space X iscompact if and only if each net in X has a convergent subnet.”

9.4.5. Weierstrass Theorem. The image of a compact set under a continu-ous mapping is compact (cf. 4.4.5). CB

9.4.6. Let X0 be a subspace of a topological space X and let C be a subsetof X0. Then C is compact in X0 if and only if C is compact in X.

C ⇒: Immediate from 9.4.5 and 9.2.17 (1).⇐: Let B be a filterbase on X0. Further, let V := clX0 B stand for the

adherence of B relative to X0. Suppose that V ∩C = ∅. Since B is also a filterbaseon X, it makes sense to speak of the adherence W := clX B of B relative to X.It is clear that V = W ∩X0 and, consequently, W ∩ C = ∅. Since C is compactin X, by 9.4.3 there is some B in B such that B ∩C = ∅. Using 9.4.3 once again,infer that C is compact in X0. B

9.4.7. Remark. The claim of 9.4.6 is often expressed as follows: “Compact-ness is an absolute concept.” It means that for C to be or not to be compactdepends on the topology induced in C rather than on the ambient space inducingthe topology. For that reason, it is customary to confine study to compact spaces,i.e. to sets “compact in themselves.” A topology τ on a set C, making C intoa compact space, is usually called a compact topology on C. Also, such C is referredto as “compact with respect to τ .”

9.4.8. Tychonoff Theorem. The Tychonoff product of compact spaces iscompact.

C Let X :=∏ξ∈�Xξ be the product of such spaces. If at least one of the

spaces Xξ is nonempty then X = ∅ and nothing is left to proof. Let X 6= ∅ andlet F be an ultrafilter on X. By 1.3.12, given ξ ∈ � and considering the coordinateprojection Prξ : X→ Xξ, observe that Prξ(F ) is an ultrafilter on Xξ. Consequently,in virtue of 9.4.3 there is some xξ in Xξ such that Prξ(F ) → xξ. Let x : ξ 7→ xξ.It is clear that F → x (cf. 9.2.10). Appealing to 9.4.3 once more, infer that X iscompact. B

9.4.9. Every closed subset of a compact space is compact.C Let X be compact and C ∈ Cl(X). Assume further that F is an ultrafilter

on X and C ∈ F . By Theorem 9.4.3, F has a limit x in X: that is, F → x. By theBirkhoff Theorem, x ∈ clC = C. Using 9.4.3 again, conclude that C is compact. B


9.4.10. Every compact subset of a Hausdorff space is closed.C Let C be compact in a Hausdorff space X. If C = ∅ then there is nothing

to prove. Let C 6= ∅ and x ∈ clC. By virtue of 9.2.2 there is a filter F0 on X suchthat C ∈ F0 and F0 → x. Let F be an ultrafilter finer than F0. Then F → xand C ∈ F . By 9.4.3, F has a limit in C. By 9.3.4 every limit in X is unique.Consequently, x ∈ C. B

9.4.11. Let f : (X, τ) → (Y, ω) be a continuous one-to-one mapping withf(X) = Y . If τ is a compact topology and ω is a Hausdorff topology, then f isa homomorphism.

C It suffices to establish that f−1 is continuous. To this end, we are to demon-strate that F ∈ Cl(τ) ⇒ f(F ) ∈ Cl(ω). Take F ∈ Cl(τ). Then F is compactby 9.4.9. Successively applying 9.4.5 and 9.4.10, infer that f(F ) is closed. B

9.4.12. Let τ1 and τ2 be two topologies on a set X. If (X, τ1) is a compactspace and (X, τ2) is a Hausdorff space with τ1 ≥ τ2, then τ1 = τ2. CB

9.4.13. Remark. The message of 9.4.12 is customarily verbalized as follows:“A compact topology is minimal among Hausdorff topologies.”

9.4.14. Theorem. Every Hausdorff compact space is normal.C Let X be the space under study and let B be some filterbase on X. Assume

further that U is a neighborhood of clB. It is clear that X \ intU is compact(cf. 9.4.9) and clB ∩ (X \ intU) = ∅. By Theorem 9.4.3 there is a member B of Bsuch that B∩(X\intU) = ∅; i.e., B ⊂ U . Putting, if need be, B := {clB : B ∈ B},we may assert that clB ⊂ U .

To begin with, take x ∈ X and put B := τ(x). By virtue of 9.3.4, clB = {x}and, consequently, the filter τ(x) has a base of closed sets. Thus, X is regular.

Now take nonempty closed subset F of X. Take as B the neighborhood filterof F . By 9.3.8, clB = F , and, as is already established, B has a base of closedsets. In accordance with 9.3.9, X is a normal space. B

9.4.15. Corollary. Each Hausdorff compact space is (to within a homeomor-phism) a closed subset of a Tychonoff cube.

C The compactness property of a closed subset of a Tychonoff cube followsfrom 9.4.8 and 9.4.9. Moreover, every cube is a Hausdorff space and so such is eachof its subspaces.

Now take some Hausdorff compact space X. Let Q be the collection of allcontinuous function from X to [0, 1]. Define the mapping � : X → [0, 1]Q as�(x)(f) := f(x) where x ∈ X and f ∈ Q. From 9.4.14 and 9.3.14 infer that� carries X onto �(X) in a one-to-one fashion. Furthermore, � is continuous.Application to 9.4.11 completes the proof. B

166 Chapter 9

9.4.16. Remark. Corollary 9.4.15 presents a part of a more general assertion.Namely, a Tychonoff space is (to within a homeomorphism) a subspace of a Ty-chonoff cube. CB

9.4.17. Remark. Sometimes a Hausdorff compact space is also called a com-pactum (cf. 4.5 and 4.6).

9.4.18. Dieudonne Lemma. Let F be a closed set and let G1, . . . , Gn beopen sets in a normal topological space, with F ⊂ G1 ∪ . . . ∪Gn. There are closedsets F1, . . . , Fn such that F = F1 ∪ . . . ∪ Fn and Fk ⊂ Gk (k := 1, . . . , n).

C It suffices to settle the case n := 2. For k := 1, 2 the set Uk := F \ Gkis closed. Moreover, U1 ∩ U2 = ∅. By 9.3.10 there are open V1 and V2 such thatU1 ⊂ V1, U2 ⊂ V2 and V1∩V2 = ∅. Put Fk := F \Vk. It is clear that Fk is closed andFk ⊂ F \Uk = F \(F \Gk) ⊂ Gk for k := 1, 2. Finally, F1∪F2 = F \(V1∩V2) = F . B

9.4.19. Remark. From 9.3.14 we deduce that under the hypotheses of 9.4.18there are continuous functions h1, . . . , hn : X → [0, 1] such that hk|G′

k= 0 and∑n

k=1 hk(x) = 1 for every point x in some neighborhood about F . (As usual,G′k := X \Gk.)

9.4.20. Definition. A topology is called locally compact if each point pos-sesses a compact neighborhood. A locally compact space is a set furnished witha Hausdorff locally compact topology.

9.4.21. A topological space is locally compact if and only if it is homeomor-phic with a punctured compactum (= a compactum with a deleted point), i.e. thecomplement of a singleton to a compactum.

C⇐: In virtue of the Weierstrass Theorem it suffices to observe that each pointof a punctured compactum possesses a closed neighborhood (since every compactumis regular). It remains to make use of 9.4.9 and 9.4.6.

⇒: Put the initial space X in X · := X ∪{∞}, adjoining to X a point ∞ takenelsewhere. Take the complements to X · of compact subsets of X as a base for theneighborhood filter about ∞. A neighborhood of a point x of X in X · is declaredto be a superset of a neighborhood of x in X. If A is an ultrafilter in X · and Kis a compactum in X then A converges to a point in K provided that K ∈ A.If A contains the complement of each compactum K in X to X, then A convergesto ∞. B

9.4.22. Remark. If a locally compact space X is not compact in its own rightthen X · of 9.4.20 is the one-point or Alexandroff compactification of X.

9.5. Uniform and Multimetric Spaces

9.5.1. Definition. Let X be a nonempty set and let UX be a filter on X2.The filter UX is a uniformity on X if


(1) UX ⊂ fil {IX};(2) U ∈ UX ⇒ U−1 ∈ UX ;(3) (∀U ∈ UX)(∃V ∈ UX) V ◦ V ⊂ U .

The uniformity of the empty set is by definition {∅}. The pair (X, UX), as wellas the underlying set X, is called a uniform space.

9.5.2. Given a uniform space (X, UX), put

x ∈ X ⇒ τ(x) := {U(x) : U ∈ UX}.

The mapping τ : x 7→ τ(x) is a topology on X.C Clearly, τ is a pretopology. If W ∈ τ(x) then W = U(x) for some U

in UX . Choose a member V of UX so that V ◦ V ⊂ U . If y ∈ V (x) then V (y) ⊂V (V (x)) = V ◦ V (x) ⊂ U(x) ⊂ W . In other words, the set W is a neighborhoodabout y for every y in V (x). Therefore, V (x) lies in intW . Consequently, intW isa neighborhood about x. It remains to refer to 9.1.6. B

9.5.3.Definition. The topology τ appearing in 9.5.2 is the topology of the uni-form space (X, UX) under consideration or the uniform topology on X. It is alsodenoted by τ(UX), τX , etc.

9.5.4. Definition. A topological space (X, τ) is called uniformizable pro-vided that there is a uniformity U on X such that τ coincides with the uniformtopology τ(U ).

9.5.5. Examples.(1) A metric space (with its metric topology) is uniformizable (with its

metric uniformity).(2) A multinormed space (with its topology) is uniformizable (with its

uniformity).(3) Let f : X → (Y, UY ) and f−1(UY ) := f×−1(UY ), where as usual

f×(x1, x2) := (f(x1), f(x2)) for (x1, x2) ∈ X2. Evidently, f−1(UY ) is a uniformityon X. Moreover,

τ(f−1(UY )) = f−1(τ(UY )).

The uniformity f−1(UY ) is the inverse image of UY under f . Therefore, the inverseimage of a uniform topology is uniformizable.

(4) Let (Xξ, Uξ)ξ∈� be a family of uniform spaces. Assume furtherthat X :=

∏ξ∈�Xξ is the product of the family. Put UX := supξ∈� Pr

−1ξ (Uξ). The

uniformity UX is the Tychonoff uniformity. It is beyond a doubt that the uniformtopology τ(UX) is the Tychonoff topology of the product of (Xξ, τ(Uξ))ξ∈�. CB

168 Chapter 9

(5) Each Hausdorff compact space is uniformizable in a unique fashion.C By virtue of 9.4.15 such space X may be treated as a subspace of a Ty-

chonoff cube. From 9.5.5 (3) and 9.5.5 (4) it follows that X is uniformizable. Sinceeach entourage of a uniform space includes a closed entourage; therefore, the com-pactness property of the diagonal IX of X2 implies that every neighborhood of IXbelongs to UX . On the other hand, each entourage is always a neighborhood of thediagonal. B

(6) Assume that X and Y are nonempty sets, UY is a uniformity on Yand B is an upward-filtered subset of P(X). Given B ∈ B and θ ∈ UY , put

UB,θ := {(f, g) ∈ Y X × Y X : g ◦ IB ◦ f−1 ⊂ θ}.

Then U := fil {UB,θ : B ∈ B, θ ∈ UY } is a uniformity on Y X . This uniformityhas a cumbersome (but exact) title, the “uniformity of uniform convergence on themembers of B.” Such is, for instance, the uniformity of the Arens multinorm(cf. 8.3.8). If B is the collection of all finite subsets of X, then U coincides with theTychonoff uniformity on Y X . This uniformity is called weak, and the correspondinguniform topology is called the topology of pointwise convergence or, rarely, thatof simple convergence. If B is a singleton {X}, then the uniformity U is calledstrong and the corresponding topology τ(U ) in Y X is the topology of uniformconvergence on X.

9.5.6. Remark. It is clear that, in a uniform (or uniformizable) space, theconcepts make sense such as uniform continuity, total boundedness, completeness,etc. It is beyond a doubt that in such space the analogs of 4.2.4–4.2.9, 4.5.8, 4.5.9,and 4.6.1–4.6.7 are preserved. It is a rewarding practice to ponder over a possibilityof completing a uniform space, to validate a uniform version of the Hausdorff Cri-terion, to inspect the proof of the Ascoli–Arzela Theorem in an abstract uniformsetting, etc.

9.5.7. Definition. Let X be a set and put R·+ := {x ∈ R· : x ≥ 0}. A map-ping d : X2 → R·+ is called a semimetric or a pseudometric on X, provided that

(1) d(x, x) = 0 (x ∈ X);(2) d(x, y) = d(y, x) (x, y ∈ X);(3) d(x, y) ≤ d(x, z) + d(z, y) (x, y, z ∈ X).

A pair (X, d) is a semimetric space.

9.5.8. Given a semimetric space (X, d), let Ud := fil {{d ≤ ε} : ε > 0}. ThenUd is a uniformity. CB

9.5.9. Definition. Let M be a (nonempty) set of semimetrics on X. Thenthe pair (X, M) is a multimetric space with multimetric M. The multimetricuniformity on X is defined as UM := sup{Ud : d ∈M}.


9.5.10. Definition. A uniform space is called multimetrizable, if its unifor-mity coincides with some multimetric uniformity. A multimetrizable topologicalspace is defined by analogy.

9.5.11. Assume that X, Y , and Z are sets, T is a dense subset of R, and(Ut)t∈T and (Vt)t∈T are increasing families of subsets of X × Z and Z × Y , re-spectively. Then there are unique functions f : X × Z → R, g : Z × Y → R andh : X × Y → R such that

{f < t} ⊂ Ut ⊂ {f ≤ t}, {g < t} ⊂ Vt ⊂ {g ≤ t},{h < t} ⊂ Vt ◦ Ut ⊂ {h ≤ t} (t ∈ T ).

Moreover, the presentation holds:

h(x, y) = inf{f(x, z) ∨ g(z, y) : z ∈ Z}.

C The sought functions exist by 3.8.2. The claim of uniqueness is straightfor-ward from 3.8.4. The presentation of h via f and g raises no doubts. B

9.5.12. Definition. Let f : X ×Z → R and g : Z × Y → R. The function h,given by 9.5.11, is called the ∨-convolution (read: vel-convolution) of f and g andis denoted by

f

9.5.13. Definition. A mapping f : X2 → R·+ is a K-ultrametric withK ∈ R, K ≥ 1, if

(1) f(x, x) = 0 (x ∈ X);

(2) f(x, y) = f(y, x) (x, y ∈ X);

(3) 1/Kf(x, u) ≤ f(x, y) ∨ f(y, z) ∨ f(z, u) (x, y, z, u ∈ X).9.5.14.Remark. Condition 9.5.13 (3) is often referred to as the (strong) ultra-

metric inequality. In virtue of 9.5.12 this inequality may be rewritten as K−1f ≤ f

9.5.15. 2-Ultrametric Lemma. To every 2-ultrametric f : X2 → R·+ therecorresponds a semimetric d such that 1/2f ≤ d ≤ f .

C Let f1 := f and fn+1 := fnWe are left with proving that 1/2f ≤ d. To this end, show that fn ≥ 1/2f for

n ∈ N. Proceed by way of induction.The desired inequalities are obvious when n := 1, 2. Assume now that f ≥ f1 ≥

. . . ≥ fn ≥ 1/2f and at the same time fn+1(x, y) < 1/2f(x, y) for some (x, y)in X2 and n ≥ 2.

170 Chapter 9

For suitable z1, . . . , zn in X by construction

t := f(x, z1) + f(z1, z2) + . . .+ f(zn−1, zn) + f(zn, y)< 1/2f(x, y).

If f(x, z1) ≥ t/2 then t/2 ≥ f(z1, z2) + . . . + f(zn, y) ≥ 1/2f(z1, y). It followsthat t ≥ f(x, z1) and t ≥ f(z1, y). On account of 9.5.13 (3), 1/2f(x, y) ≤f(x, z1)∨ f(z1, y) ≤ t. Whence we come to 1/2f(x, y) > t ≥ 1/2f(x, y), which isfalse.

Thus, f(x, z1) < t/2. Find m ∈ N, m < n, satisfying

f(x, z1) + . . .+ f(zm−1, zm) < t/2;

f(x, z1) + . . .+ f(zm, zm+1) ≥ t/2.

This is possible, since assuming m = n would entail the false inequality f(zn, y) ≥t/2. (We would have t/2 ≥ f(x, z1) + . . . + f(zn−1, zn) ≥ 1/2f(x, zn) and so1/2f(x, y) > t ≥ f(x, zn) ∨ f(zn, y) ≥ 1/2f(x, y).)

We obtain the inequality

f(zm+1, zm+2) + . . .+ f(zn−1, zn) + f(zn, y) < t/2.

Using the induction hypothesis, conclude that

f(x, zm) ≤ 2(f(x, z1) + . . .+ f(zm−1, zm)) ≤ t;f(zm, zm+1) ≤ t;

f(zm+1, y) ≤ 2(f(zm+1, zm+2) + . . .+ f(zn, y)) ≤ t.

Consequently, by the definition of 2-ultrametric

1/2f(x, y) ≤ f(x, zm) ∨ f(zm, zm+1) ∨ f(zm+1, y) ≤ t< 1/2f(x, y).

We arrive at a contradiction, completing the proof. B

9.5.16. Theorem. Every uniform space is multimetrizable.C Let (X, UX) be a uniform space. Take V ∈ UX . Put V1 := V ∩V −1. If now

Vn ∈ UX then find a symmetric entourage V = V−1

, a member of UX , satisfyingV ◦ V ◦ V ⊂ Vn. Define Vn+1 := V . Since by construction

Vn ⊃ Vn+1 ◦ Vn+1 ◦ Vn+1 ⊃ Vn+1 ◦ IX ◦ IX ⊃ Vn+1;


therefore, (Vn)n∈N is a decreasing family.Given t ∈ R, define a set Ut by the rule

Ut :=

∅, t < 0IX , t = 0Vinf{n∈N : t≥2−n}, 0 < t < 1V1, t = 1X2, t > 1.

By definition, the family t 7→ Ut (t ∈ R) increases. Consider a unique functionf : X2 → R satisfying the next conditions (cf. 3.8.2 and 3.8.4)

{f < t} ⊂ Ut ⊂ {f ≤ t} (t ∈ R).

If Wt := U2t for t ∈ R then

Us ◦ Us ◦ Us ⊂Wt

for s < t. Consequently, in virtue of 3.8.3 and 9.2.1 the mapping f is a 2-ultrametric.Using 9.5.15, find a semimetric dV such that 1/2f ≤ dV ≤ f . Clearly, UdV =

fil {Vn : n ∈ N}. It is also beyond a doubt that UM = UX for the multimetricM := {dV : V ∈ UX}. B

9.5.17. Corollary. A topological space is uniformizable if and only if it isa T31/2-space. CB

9.5.18. Corollary. A Tychonoff space is the same as a separated multimetricspace. CB

9.6. Covers, and Partitions of Unity

9.6.1. Definition. Let E and F be covers of a subset of U in X; i.e., E , F ⊂P(X) and U ⊂ (∪E )∩ (∪F ). It is said that F coarsens E or E refines F , if eachmember of E is included in some member of F ; i.e., (∀E ∈ E ) (∃F ∈ F ) E ⊂ F .It is also said that E is a refinement of F . Observe that if F is a subcover of E(i.e., F ⊂ E ) then E refines F .

9.6.2. Definition. A cover E of a set X is called locally finite (with respectto a topology τ on X), if each point in X possesses a neighborhood (in the senseof τ) meeting only finite many members of E . In the case of the discrete topology onX, such cover is called point finite. If X is regarded as furnished with a prescribedtopology τ then, speaking of a locally finite cover of X, we imply the topology τ .

172 Chapter 9

9.6.3. Lefschetz Lemma. Let E be a point finite open cover of a normalspace X. Then there is an open cover {GE : E ∈ E } such that clGE ⊂ E for allE ∈ E .C Let the set S comprise the mappings s : E → Op(X) such that ∪s(E ) = X

and for E ∈ E either s(E) = E or cl s(E) ⊂ E. Given functions s1 and s2, puts1 ≤ s2 := (∀E ∈ E ) (s1(E) 6= E ⇒ s2(E) = s1(E)). It is evident that (S, ≤) isan ordered set and IE ∈ S. Show that S is inductive.

Given a chain S0 in S, for all E ∈ E put s0(E) := ∩{s(E) : s ∈ S0}. If s0(E) =E then s(E) = E for all s ∈ S0. If the case s0(E) 6= E observe that s0(E) =∩{s(E) : s(E) 6= E, s ∈ S0}.

Since the order of S0 is linear, infer that s0(E) = s(E) for s ∈ S0 with s(E) 6=E. Hence, s0(E ) ⊂ Op(X) and s0 ≥ S0. It remains to verify that s0 is a cover of X(and so s0 ∈ S). By the hypothesis of point finiteness, given x ∈ X, there are someE1, . . . , En in E such that x ∈ E1 ∩ . . . ∩ En and x /∈ E for the other members Eof E . If s0(Ek) = Ek for some k, then there is nothing to prove, for x ∈ ∪s0(E ).In the case when s0(Ek) 6= Ek for every k, there are s1, . . . , sn ∈ S0 meeting theconditions sk(Ek) 6= Ek (k := 1, 2 . . . , n). Since S0 is a chain, it may be assumedthat sn ≥ {s1, . . . , sn−1}. Moreover, x ∈ sn(E) ⊂ E for an appropriate E in E .It is clear that E ∈ {E1, . . . , En} (because x /∈ E for the other members E of E ).Since s0(E) = sn(E), it follows that x ∈ s0(E).

By the Kuratowski–Zorn Lemma there is a maximal element s in S. TakeE ∈ E . If F := X \∪ s(E \ {E}), then F is closed and s(E) is a neighborhood of F .For a suitable G in Op(X) by 9.3.10 F ⊂ G ⊂ clG ⊂ s(E). Put s(E) := G ands(E) := s(E) for E 6= E (E ∈ E ). It is clear that s ∈ S. If s(E) = E, then s ≥ sand so s = s. Moreover, s(E) ⊂ clG ⊂ s(E) = E; i.e., cl s(E) ⊂ E. If s(E) 6= E,then cl s(E) ⊂ E by definition. Thus, s is a sought cover. B

9.6.4. Definition. Let f be a numeric or scalar-valued function on a topo-logical space X, i.e. f : X → F. The set supp(f) := cl{x ∈ X : f(x) 6= 0}is the support of f . If supp(f) is a compact set then f is a compactly-supportedfunction or a function of compact support. The designation spt (f) := supp(f) isused sometimes.

9.6.5. Let (fe)e∈E be a family of numeric functions onX and let E := {supp(fe) :e ∈ E } be the family of their supports. If E is a point finite cover of X then thefamily (fe)e∈E is summable pointwise. If in addition E is locally finite and everymember of (fe)e∈E is continuous, then the sum

∑e∈E fe is also continuous.

C It suffices to observe that in a suitable neighborhood about a point in Xonly finitely many members of the family (fe)e∈E are distinct from zero. B

9.6.6. Definition. It is said that a family of functions (f : X → [0, 1])f∈F isa partition of unity on a subset U of X, if the supports of the members of the family


composes a point finite cover of X, and∑f∈F f(x) = 1 for all x ∈ U . The empty

family of functions in this context is treated as summable to unity at each point.The term “continuous partition of unity” and the like are understood naturally.

9.6.7. Definition. Let E be a cover of a subset U of a topological spaceand let F be a continuous partition of unity on U . If the family of supports{supp(f) : f ∈ F} refines E then F is a partition of unity subordinate to E .A possibility of finding such an F for E is also verbalized as follows: “E admitsa partition of unity.”

9.6.8. Each locally finite open cover of a normal space admits a partitionof unity.

C By the Lefschetz Lemma, such cover {Uξ : ξ ∈ �} has an open refinement{Vξ : ξ ∈ �} satisfying the condition clVξ ⊂ Uξ for all ξ ∈ �. By the UrysohnTheorem, there is a continuous function gξ : X → [0, 1] such that gξ(x) = 1 forx ∈ Vξ and gξ(x) = 0 for x ∈ X\Uξ. Consequently, supp(gξ) ⊂ Uξ. In virtue of 9.6.5the family (gξ)ξ∈� is summable pointwise to a continuous function g. Moreover,g(x) > 0 for all x ∈ X by construction. Put fξ := gξ/g (ξ ∈ �). The family (fξ)ξ∈�is what we need. B

9.6.9. Definition. A topological space X is called paracompact, if each coverof X has a locally finite open refinement.

9.6.10.Remark. The theory of paracompactness contains deep and surprisingfacts.

9.6.11. Theorem. Every metric space is paracompact.

9.6.12. Theorem. A Hausdorff topological space is paracompact if and onlyif its every open cover admits a partition of unity.

9.6.13.Remark. The metric space RN possesses a number of additional struc-tures providing a stock of well-behaved, smooth (= infinitely differentiable) func-tions (cf. 4.8.1).

9.6.14. Definition. A mollifier or a mollifying kernel on RN is a real-valuedsmooth function a having unit (Lebesgue) integral and such that a(x) > 0 for|x| < 1 and a(x) = 0 for |x| ≥ 1. In this event, supp(a) = {x ∈ RN : |x| ≤ 1} isthe (unit Euclidean) ball B := BRN .

9.6.15. Definition. A delta-like sequence is a family of real-valued (smooth)functions (bε)ε>0 such that, first, limε→0(sup | supp(bε)|) = 0 and, second, the equal-ity holds

∫RN bε(x) dx = 1 for all ε > 0. The terms “δ-sequence” and “δ-like se-

quence” are also in current usage. Such a sequence is often assumed countablewithout further specification.

174 Chapter 9

9.6.16. Example. The function a(x) := t exp(−(|x|2−1)−1) is taken as a mostpopular mollifier when extended by zero beyond the open ball intB, with the con-stant t determined from the condition

∫RN a(x) dx = 1. Each mollifier generates

the delta-like sequence aε(x) := ε−Na(x/ε) (x ∈ RN ).

9.6.17. Definition. Let f ∈ L1,loc(RN ); i.e., let f be a locally integrablefunction, that is, a function whose restriction to each compact subset of RN isintegrable. For a compactly-supported integrable function g the convolution f ∗ gis defined as

f ∗ g(x) :=∫

RN

f(x− y)g(y) dy (x ∈ RN ).

9.6.18. Remark. The role of a mollifying kernel and the corresponding delta-like sequence (aε)ε>0 becomes clear from inspecting the aftermath of applying thesmoothing process f 7→ (f∗aε)ε>0 to a function f belonging to L1,loc(RN ) (cf. 10.10.7(5)).

9.6.19. The following statements are valid:(1) to every compact set K in the space RN and every neighborhood U

of K there corresponds a truncator (= a bump function) ψ := ψK,U , i.e. a smoothmapping ψ : RN → [0, 1] such that K ⊂ int{ψ = 1} and supp(ψ) ⊂ U ;

(2) assume that U1, . . . , Un ∈ Op(RN ) and U1 ∪ . . . ∪ Un is a neigh-borhood of a compact set K; there are smooth functions ψ1, . . . , ψn : RN → [0, 1]such that supp(ψk) ⊂ Uk and

∑nk=1 ψk(x) = 1 for x in some neighborhood of K.

C (1) Put ε := d(K, RN \ U) := inf{|x − y| : x ∈ K, y /∈ U}. It is clear thatε > 0. Given β > 0, denote the characteristic function of K + εB by χβ . Takea delta-like sequence (bγ)γ>0 of positive functions and put ψ := χβ ∗ bγ . Whenγ ≤ β and β + γ ≤ ε with γ := sup | supp(bγ)|, observe that ψ is a sought function.

(2) By the Diedonne Lemma there are closed sets Fk, with Fk ⊂ Uk, composinga cover of K. Put Kk := Fk ∩K and choose some truncators ψk := ψKk,Uk . Thefunctions ψk/

∑nk=1 ψk (k := 1, . . . , n), defined on {

∑nk=1 ψk > 0}, meet the claim

after extension by zero onto {∑nk=1 ψk = 0} and multiplication by a truncator

corresponding to an appropriate neighborhood of K. B

9.6.20. Countable Partition Theorem. Let E be a family of open setsin RN and � := ∪E . There is a countable partition of unity which is composedof smooth compactly-supported functions on RN and subordinate to the cover Eof �.

C Refine from E a countable locally finite cover A of � with compact sets sothat the family (α := intα)α∈A be also an open cover of �. Choose an open cover(Vα)α∈A of � from the condition clVα ⊂ α for α ∈ A. In virtue of 9.6.19 (1) thereare truncators ψα := ψclVα,α. Putting ψα(x) := ψα(x)/

∑α∈A ψα(x) for x ∈ � and

ψα(x) := 0 for x ∈ RN \ �, arrive to a sought partition. B


9.6.21.Remark. It is worth observing that the so-constructed partition of unity(ψα)α∈A possesses the property that to each compact subset K of � there corre-spond a finite subset A0 of A and a neighborhood U of K such that

∑α∈A0

ψα(x) =1 for all x ∈ U (cf. 9.3.17 and 9.6.19 (2)).

Exercises

9.1. Give examples of pretopological and topological spaces and constructions leading to them.9.2. Is it possible to introduce a topology by indicating convergent filters or sequences?9.3. Establish relations between topologies and preorders on a finite set.9.4. Describe topological spaces in which the union of every family of closed sets is closed.

What are the continuous mappings between such spaces?

9.5. Let (fξ : X0 = 0 OO Yξ, τξ))ξ∈� be a family of mappings. A topology σ on X is calledadmissible (in the case under study), if for every topological space (Z, ω) and every mappingg : Z0 = 0 OO the following statement holds: g : (Z, ω)0 = 0 OO X, σ) is continuous if and onlyif so is each mapping fξ ◦ g (ξ ∈ �). Demonstrate that the weakest topology on X making everyfξ (ξ ∈ �) continuous is the strongest admissible topology (in the case under study).

9.6. Let (fξ : (Xξ, σξ)0 = 0 OO )ξ∈� be a family of mappings. A topology τ on Y is calledadmissible (in the case under study), if for every topological space (Z, ω) and every mappingg : Y 0 = 0 OO the following statement is true: g : (Y, τ)0 = 0 OO Z, ω) is continuous if and onlyif each mapping g ◦ fξ (ξ ∈ �) is continuous. Demonstrate that the strongest topology on Ymaking every fξ (ξ ∈ �) continuous is the weakest admissible topology (in the case under study).

9.7. Prove that in the Tychonoff product of topological spaces, the closure of the productof subsets of the factors is the product of closures:

cl

(∏ξ∈�

Aξ

)=∏ξ∈�

clAξ.

9.8. Show that a Tychonoff product is a Hausdorff space if and only if so is every factor.9.9. Establish compactness criteria for subsets of classical Banach spaces.9.10. A Hausdorff space X is called H-closed, if X is closed in every ambient Hausdorff

space. Prove that a regular H-closed space is compact.

9.11. Study possibilities of compactifying a topological space.9.12. Prove that the Tychonoff product of uncountably many real axes fails to be a normal

space.

9.13. Show that each continuous function on the product of compact spaces depends on atmost countably many coordinates in an evident sense (specify it!).

9.14. Let A be a compact subset and let B be a closed subset of a uniform space, withA ∩B = ∅. Prove that V (A) ∩ V (B) = ∅ for some entourage V .

9.15. Prove that a completion (in an appropriate sense) of the product of uniform spaces isuniformly homeomorphic (specify!) to the product of completions of the factors.

9.16. A subset A of a separated uniform space is called precompact if a completion of A iscompact. Prove that a set is precompact if and only if it is totally bounded.

9.17. Which topological spaces are metrizable?9.18. Given a uniformizable space, describe the strongest uniformity among those inducing

the initial topology.

9.19. Verify that the product of a paracompact space and a compact space is paracompact.Is paracompactness preserved under general products?

Chapter 10

Duality and Its Applications

10.1. Vector Topologies

10.1.1. Definition. Let (X, F, +, · ) be a vector space over a basic field F.A topology τ on X is a topology compatible with vector structure or, briefly, a vectortopology, if the following mappings are continuous:

+ : (X ×X, τ × τ)→ (X, τ),· : (F×X, τF × τ)→ (X, τ).

The space (X, τ) is then referred to as a topological vector space.

10.1.2. Let τX be a vector topology. The mappings

x 7→ x+ x0, x 7→ αx (x0 ∈ X, α ∈ F \ 0)

are topological isomorphisms in (X, τX). CB

10.1.3. Remark. It is beyond a doubt that a vector topology τ on a space Xpossesses the next “linearity” property:

τ(αx+ βy) = ατ(x) + βτ(y) (α, β ∈ F \ 0; x, y ∈ X),

where in accordance with the general agreements (cf. 1.3.5 (1))

Uαx+βy ∈ ατ(x) + βτ(y)⇔ (∃Ux ∈ τ(x) & Uy ∈ τ(y)) αUx + βUy ⊂ Uαx+βy.

In this regard a vector topology is often called a linear topology and a topologicalvector space, a linear topological space. This terminology should be used on theunderstanding that a topology may possess the “linearity” property while failingto be linear. For instance, such is the discrete topology of a nonzero vector space.

Duality and Its Applications 177

10.1.4. Theorem. Let X be a vector space and let N be a filter on X.There is a vector topology τ on X such that N = τ(0) if and only if the followingconditions are fulfilled:

(1) N + N = N ;(2) N consists of absorbing sets;(3) N has a base of balanced sets.

Moreover, τ(x) = x+ N for all x ∈ X.C ⇒: Let τ be a vector topology and N = τ(0). From 10.1.2 infer that

τ(x) = x + N for x ∈ X. It is also clear that (1) reformulates the continuityproperty of addition at zero (of the space X2). Condition (2) may be rewrittenas τF(0)x ⊃ N for every x in X, which is the continuity property of the mappingα 7→ αx at zero (of the space R) for every x in X. Condition (3) with account takenof (2) may in turn be rendered in the form τF(0)N = N , which is the continuityproperty of scalar multiplication at zero (of the space F×X).

⇐: Let N be a filter satisfying (1)–(3). It is evident that N ⊂ fil {0}. Putτ(x) := x+N . Then τ is a pretopology. From the definition of τ and (1) it followsthat τ is a topology, with every translation continuous and addition continuousat zero in X2. Thus, addition is continuous at every point of X2. The validityof (2) and (3) means that the mapping (λ, x) 7→ λx is jointly continuous at zeroand continuous at zero in the first argument given the second argument. By virtueof the identity

λx− λ0x0 = λ0(x− x0) + (λ− λ0)x0 + (λ− λ0)(x− x0),

we are left with examining the continuity property of scalar multiplication at zeroin the second argument given the first argument. In other words, it is necessaryto show that λN ⊃ N for λ ∈ F. With this in mind, find n ∈ N such that |λ| ≤ n.Let V in N and W in N be such that W is balanced and W1 + . . . +Wn ⊂ V ,where Wk :=W . Then λW = n

(λ/nW

)⊂ nW ⊂W1 + . . .+Wn ⊂ V . B

10.1.5. Theorem. The set VT(X) of all vector topologies on X presentsa complete lattice. Moreover,

supVT(X) E = supT(X) E

for every subset E of VT(X).C Let τ := supT(X) E . Since for τ ∈ E each translation by a vector is a topolog-

ical isomorphism in (X, τ); therefore, this mapping is a topological isomorphism in(X, τ). Using 9.1.13, observe that the filter τ(0) meets conditions 10.1.4 (1)–10.1.4(3), since these conditions are fulfilled for every filter τ(0) with τ ∈ E . It remainsto refer to 1.2.14. B

178 Chapter 10

10.1.6. Theorem. The inverse image of a vector topology under a linear op-erator is a vector topology.

C Take T ∈ L (X, Y ) and ω ∈ VT(Y ). Put τ := T−1(ω). If xγ → x andyγ → y in (X, τ) then by 9.2.8 Txγ → Tx and Tyγ → Ty. So T (xγ+yγ)→ T (x+y).This means in virtue of 9.2.10 that xγ+yγ → x+y in (X, τ). Thus, τ(x) = x+τ(0)for all x ∈ X and, moreover, τ(0) + τ(0) = τ(0). Successively applying 3.4.10and 3.1.8 to the linear correspondence T−1, observe that the filter τ(0) = T−1(ω(0))consists of absorbing sets and has a base of balanced sets. The reason is as follows:by 10.1.4 the filter ω(0) possesses these two properties. Once again using 10.1.4,conclude that τ ∈ VT(X). B

10.1.7. The product of vector topologies is a vector topology.


10.1.8. Definition. Let A and B be subsets of a vector space. It is said thatA is B-stable if A+B ⊂ A.

10.1.9. To every vector topology τ onX there corresponds a unique uniformityUτ having a base of IX -stable sets and such that τ = τ(Uτ ).

C Given U ∈ τ(0), put VU := {(x, y) ∈ X2 : y− x ∈ U}. Observe the obviousproperties:

IX ⊂ VU ; VU + IX = VU ; (VU )−1 = V−U ;VU1∩U2 ⊂ VU1 ∩ VU2 ; VU1 ◦ VU2 ⊂ VU1+U2

for all U , U1, U2 ∈ τ(0). Using 10.1.4, infer that Uτ := fil {VU : U ∈ τ(0)} isa uniformity and τ = τ(Uτ ). It is also beyond a doubt that Uτ has a base ofIX -stable sets.

If now U is another uniformity such that τ(U ) = τ , and W is some IX -stableentourage in U ; then W = VW (0). Whence the sought uniqueness follows. B

10.1.10. Definition. Let (X, τ) be a topological vector space. The unifor-mity Uτ , constructed in 10.1.9, is the uniformity of X.

10.1.11. Remark. Considering a topological vector space, we assume it to befurnished with the corresponding uniformity without further specification.

10.2. Locally Convex Topologies

10.2.1. Definition. A vector topology is locally convex if the neighborhoodfilter of each point has a base of convex sets.


10.2.2. Theorem. Let X be a vector space and let N be a filter on X. Thereis a locally convex topology τ on X such that N = τ(0) if and only if

(1) 1/2N = N ;(2) N has a base of absorbing absolutely convex sets.

C ⇒: By virtue of 10.1.2 the mapping x 7→ 2x is a topological isomorphism.This means that 1/2N = N . Now take U ∈ N . By hypothesis there is a convexset V in N such that V ⊂ U . Applying 10.1.4, find a balanced set W satisfyingW ⊂ V . Using the Motzkin formula and 3.1.14, show that the convex hull co(W )is absolutely convex. Moreover, W ⊂ co(W ) ⊂ V ⊂ U .⇐: An absolutely convex set is balanced. Consequently, N satisfies 10.1.4 (2)

and 10.1.4 (3). If V ∈ N and W is a convex set, W ∈ N and W ⊂ V ; then1/2W ∈ N . Furthermore, 1/2W + 1/2W ⊂ W ⊂ V because of the convexityproperty of W . This means that N + N = N . It remains to refer to 10.1.4. B

10.2.3. Corollary. The set LCT (X) of all locally convex topologies on X isa complete lattice. Moreover,

supLCT (X) E = supT(X) E

for every subset E of LCT (X). /.

10.2.4.Corollary. The inverse image of a locally convex topology under a lin-ear operator is a locally convex topology. CB

10.2.5. Corollary. The product of locally convex topologies is a locally con-vex topology. CB

10.2.6. The topology of a multinormed space is locally convex. CB

10.2.7. Definition. Let τ be a locally convex topology on X. The set ofall everywhere-defined continuous seminorms on X is called the mirror (rarely, thespectrum) of τ and is denoted by Mτ . The multinormed space (X, Mτ ) is calledassociated with (X, τ).

10.2.8. Theorem. Each locally convex topology coincides with the topologyof the associated multinormed space.C Let τ be a locally convex topology on X and let ω := τ(Mτ ) be the topology

of the associated space (X, Mτ ). Take V ∈ τ(0). By 10.2.2 there is an absolutelyconvex neighborhood B of zero, B ∈ τ(0), such that B ⊂ V . In virtue of 3.8.7

{pB < 1} ⊂ B ⊂ {pB ≤ 1}.

It is obvious that pB is continuous (cf. 7.5.1); i.e., pB ∈Mτ , and so {pB < 1} ∈ ω(0).Consequently, V ∈ ω(0). Using 5.2.10, infer that ω(x) = x+ω(0) ⊃ x+τ(0) = τ(x);i.e., ω ≥ τ . Furthermore, τ ≥ ω by definition. B

180 Chapter 10

10.2.9. Definition. A vector space, endowed with a separated locally convextopology, is a locally convex space.

10.2.10. Remark. Theorem 10.2.8 in slightly trimmed form is often verbal-ized as follows: “The concept of locally convex space and the concept of sepa-rated multinormed space have the same scope.” For that reason, the terminologyconnected with the associated multinormed space is lavishly applied to studyinga locally convex space (cf. 5.2.13).

10.2.11. Definition. Let τ be a locally convex topology on X. The symbol(X, τ)′ (or, in short, X ′) denotes the subspace of X# that comprises all continuouslinear functionals. The space (X, τ)′ is the dual (or τ -dual) of (X, τ).

10.2.12. (X, τ)′ = ∪{|∂|(p) : p ∈Mτ}. CB10.2.13. Prime Theorem. The prime mapping τ 7→ (X, τ)′ from LCT (X)

to Lat(X#) preserves suprema; i.e.,

(X, supE )′ = sup{(X, τ)′ : τ ∈ E }

for every subset E of LCT (X).C If E = ∅ then supE is the trivial topology τ◦ of X and, consequently,

(X, τ◦)′ = 0 = inf Lat(X#) = supLat(X#) ∅. By virtue of 9.2.7 the prime mappingincreases. Given a nonempty E , from 2.1.5 infer that

(X, supE )′ ≥ sup{(X, τ)′ : τ ∈ E }.

If f ∈ (X, supE )′, then in view of 10.2.12 and 9.1.13 there are topologiesτ1, . . . , τn ∈ E such that f ∈ (X, τ1 ∨ . . . ∨ τn)′. Using 10.2.12 and 5.3.7, findp1 ∈Mτ1 , . . . , pn ∈Mτn satisfying f ∈ |∂|(p1 ∨ . . .∨ pn). Recalling 3.5.7 and 3.7.9,observe that |∂|(p1 + . . .+ pn) = |∂|(p1) + . . .+ |∂|(pn). Finally,

f ∈ (X, τ1)′ + . . .+ (X, τn)′ = (X, τ1)′ ∨ . . . ∨ (X, τn)′. .

10.3. Duality Between Vector Spaces

10.3.1. Definition. Let X and Y be vector spaces over the same ground fieldF. Also, Assume given a bilinear form (or, as it is called sometimes, a bracketing)〈· | ·〉 acting from X ×Y to F, i.e. a mapping linear in each of its arguments. Givenx ∈ X and y ∈ Y , put

〈x | : y 7→ 〈x | y〉, 〈· | : X → F Y, 〈X | ⊂ Y #;| y〉 : x 7→ 〈x | y〉, | ·〉 : Y → FX, |Y 〉 ⊂ X#.

The mappings 〈· | and | ·〉 are the bra-mapping and the ket-mapping of the initialbilinear form. By analogy, a member of 〈X | is a bra-functional on X and a memberof |Y 〉 is a ket-functional on Y .


10.3.2. The bra-mapping and the ket-mapping are linear operators. CB

10.3.3. Definition. A bracketing of some vector spaces X and Y is a pairing,if its bra-mapping and ket-mapping are monomorphisms. In this case we say thatX and Y are (set) in duality, or present a duality pair, or that Y is the pair-dualof X, etc. This is written down as X ↔ Y . Each of the bra-mapping and the ket-mapping is then referred to as dualization. For suggestiveness, the correspondingpairing of some spaces in duality is also called their duality bracket.

10.3.4. Examples.(1) Let X ↔ Y with duality bracket 〈· | ·〉. Given (y, x) ∈ Y × X,

put 〈y |x〉 := 〈x | y〉. It is immediate that the new bracketing is a pairing of Yand X. Moreover, the pairs of dualizations with respect to the old (direct) andnew (reverse) duality brackets are the same. It thus stands to reason to drawno distinction between the two duality brackets unless in case of an emergency(cf. 10.3.3). For instance, Y is the pair-dual of X in the direct duality bracketif and only if X is the pair-dual of Y in the reverse duality bracket. Therefore,the hair-splitting is neglected and a unified term “pair-dual” is applied to eachof the spaces in duality, with duality treated as a whole abstract phenomenon.Observe immediately that the mapping 〈x | y〉R := Re〈x | y〉 sets in duality the realcarriers XR and YR. By way of taking liberties, the previous notation is sometimesreserved for the arising duality XR ↔ YR; i.e., it is assumed that 〈x | y〉 := 〈x | y〉R,on considering x and y as members of the real carriers.

(2) Let H be a Hilbert space. The inner product on H sets H and H∗in duality. The prime mapping is then coincident with the ket-mapping.

(3) Let (X, τ) be a locally convex space and let X ′ be the dual of X.The natural evaluation mapping (x, x′) 7→ x′(x) sets X and X ′ in duality.

(4) Let X be a vector space and let X# := L (X, F) be the (algebraic)dual of X. It is clear that the evaluation mapping (x, x#) 7→ x#(x) sets the spacesin duality.

10.3.5. Definition. Let X ↔ Y . The inverse image in X of the Tychonofftopology on F Y under the bra-mapping, further denoted by σ(X, Y ), is the bra-topology or the weak topology on X induced by Y . The bra-topology σ(Y, X)of Y ↔ X is the ket-topology of X ↔ Y or the weak topology on Y induced by X.

10.3.6. The bra-topology is the weakest topology making every ket-functionalcontinuous. The ket-topology is the weakest topology making every bra-functionalcontinuous.C xγ → x (in σ(X, Y )) ⇔ 〈xγ | → 〈x | (in F Y ) ⇔ (∀ y ∈ Y ) 〈xγ | (y) →

〈x | (y) ⇔ (∀ y ∈ Y ) 〈xγ | y〉 → 〈x | y〉 ⇔ (∀ y ∈ Y ) | y〉(xγ) → | y〉(x) ⇔ (∀ y ∈ Y )xγ → x (in | y〉−1(τF)) B

182 Chapter 10

10.3.7. Remark. The notation σ(X, Y ) agrees perfectly with that of theweak multinorm in 5.1.10 (4). Namely, σ(X, Y ) is the topology of the multinorm{|〈· | y〉| : y ∈ Y }. Likewise, σ(Y, X) is the topology of the multinorm {|〈x | ·〉| :x ∈ X}. CB

10.3.8. The spaces (X, σ(X, Y )) and (Y, σ(Y, X)) are locally convex.


10.3.9. Dualization Theorem. Each dualization is an isomorphism betweenthe pair-dual and the weak dual of the pertinent member of a duality pair.

C Consider a duality pair X ↔ Y . We are to prove exactness for the sequences

0→ X〈·|−→(Y, σ(Y, X))′ → 0; 0→ Y

|·〉−→(X, σ(X, Y ))′ → 0.

Since the ket-mapping of X ↔ Y is the bra-mapping of Y ↔ X, it suffices to showthat the first sequence is exact. The bra-mapping is a monomorphism by definition.Furthermore, from 10.2.13 and 10.3.6 it follows that

(Y, σ(Y, X))′ = (Y, sup{〈x |−1(τF) : x ∈ X})′

= sup{(Y, 〈x |−1(τF))′ : x ∈ X} = lin({(Y, f−1(τF))′ : f ∈ 〈X |}) = 〈X |,

since in view of 5.3.7 and 2.3.12 (Y, f−1(τF))′ = {λf : λ ∈ F} (f ∈ Y #). .

10.3.10. Remark. Theorem 10.3.9 is often referred to as the theorem on thegeneral form of a weakly continuous functional. Here a useful convention revealsitself: apply the base form “weak” when using objects and properties that arerelated to weak topologies. Observe immediately that, in virtue of 10.3.9, Example10.3.4 (3) actually lists all possible duality brackets. That is why in what followswe act in accordance with 5.1.11, continuing the habitual use of the designation(x, y) := 〈x | y〉, since it leads to no misunderstanding. For the same reason, givena vector space X, we draw no distinction between the pair-dual of a spaceX and theweak dual of X. In other words, considering a duality pair X ↔ Y , we sometimesidentify X with (Y, σ(Y, X))′ and Y with (X, σ(X, Y ))′, which justifies writingX ′ = Y and Y ′ = X.

10.3.11. Remark. A somewhat obsolete convention relates to X ↔ X ′ withX a normed space. The ket-topology σ(X ′, X) is customarily called the weak∗

topology (read: weak-star topology) in X ′, which reflects the concurrent notationX∗ for X ′. The term “weak∗” proliferates in a routine fashion elsewhere.


10.4. Topologies Compatible with Duality

10.4.1. Definition. Take a duality pair X ↔ Y and let τ be a locally convextopology on X. It is said that τ is compatible with duality (between X and Yby pairing X ↔ Y ), provided that (X, τ)′ = |Y 〉. A locally convex topology ωon Y is compatible with duality (by pairing X ↔ Y ), if ω is compatible with duality(by pairing Y ↔ X); i.e., if the equality holds: (Y, ω)′ = 〈X |. A unified conciseterm “compatible topology” is also current in each of the above cases.

10.4.2. Weak topologies are compatible.

C Follows from 10.3.9. B

10.4.3. Let τ(X, Y ) stand for the least upper bound of the set of all locallyconvex topologies on X compatible with duality (between X and Y ). Then thetopology τ(X, Y ) is also compatible.

C Denote the set of all compatible topologies on X by E . Theorem 10.2.13readily yields the equalities

(X, τ(X, Y ))′ = sup{(X, τ)′ : τ ∈ E } = sup{|Y 〉 : τ ∈ E } = |Y 〉,

because E is nonempty by 10.4.2. B

10.4.4. Definition. The topology τ(X, Y ), constructed in 10.4.3 (i.e., thefinest locally convex topology on X compatible with duality by pairing X ↔ Y ), isthe Mackey topology (on X induced by X ↔ Y ).

10.4.5.Mackey–Arens Theorem. A locally convex topology τ onX is com-patible with duality between X and Y if and only if

σ(X, Y ) ≤ τ ≤ τ(X, Y ).

C By 10.2.13 the prime mapping τ 7→ (X, τ)′ preserves suprema and, in par-ticular, increases. Therefore, given τ in the interval of topologies, from 10.4.2and 10.4.3 obtain

|Y 〉 = (X, σ(X, Y )) ⊂ (X, τ)′ ⊂ (X, τ(X, Y ))′ = |Y 〉.

The remaining claim is obvious. B

10.4.6. Mackey Theorem. All compatible topologies have the samebounded sets in stock.

184 Chapter 10

C A stronger topology has fewer bounded sets. So, to prove the theoremit suffices to show that if some set U is weakly bounded in X (= is bounded in thebra-topology) then U is bounded in the Mackey topology.

Take a seminorm p from the mirror of the Mackey topology and demonstratethat p(U) is bounded in R. Put X0 := X/ ker p and p0 := pX/ ker p. By 5.2.14, p0 isclearly a norm. Let ϕ : X → X0 be the coset mapping. It is beyond a doubt thatϕ(U) is weakly bounded in (X0, p0). From 7.2.7 it follows that ϕ(U) is boundedin the norm p0. Since p0 ◦ ϕ = p; therefore, U is bounded in (X, p). B

10.4.7. Corollary. Let X be a normed space. Then the Mackey topologyτ(X, X ′) coincides with the initial norm topology on X.

C It suffices to refer to the Kolmogorov Normability Criterion implying that thespace X with the topology τ(X, X ′) finer than the original topology is normable.Appealing to 5.3.4 completes the proof. B

10.4.8. Strict Separation Theorem. Let (X, τ) be a locally convex space.Assume further that K and V are nonempty convex subsets of X with K compact,V closed and K ∩ V = ∅. Then there is a functional f , a member of (X, τ)′, suchthat

supRe f(K) < inf Re f(V ).

C A locally convex space is obviously a regular space. Since K is compact,it thus follows that, for an appropriate convex neighborhood of zero, say W , theset U := K +W does not meet V (it suffices to consider the filterbases comprisingall subsets of the form K +W and V +W , with W a closed neighborhood of zero).By 3.1.10, U is convex. Furthermore, K ⊂ intU = coreU . By the EidelheitSeparation Theorem, there is a functional l, a member of (XR)#, such that thehyperplane {l = 1} in XR separates V from U and does not meet the core of U .Obviously, l is bounded above on W and so l ∈ (XR, τ)′ by 7.5.1. If f := Re−1lthen, of view of 3.7.5, f ∈ (X, τ)′. It is clear that f is a sought functional. B

10.4.9. Mazur Theorem. All compatible topologies have the same closedconvex sets in stock.

C A stronger topology has more closed sets. So, to prove the theorem it sufficesin view of 10.4.5 to show that if U is a convex set closed in the Mackey topologythen U is weakly closed. The last claim is beyond a doubt since by Theorem 10.4.8U is the intersection of weakly closed sets of the form {Re f ≤ t}, with f a (weakly)continuous linear functional and t ∈ R. B

10.5. Polars

10.5.1. Definition. Let X and Y be sets and let F ⊂ X×Y be a correspon-


dence. Given a subset U of X and a subset V of Y, put

π(U) := πF (U) := {y ∈ Y : F−1(y) ⊃ U};π−1(V ) := π−1

F (V ) := {x ∈ X : F (u) ⊃ V }.

The set π(U) is the (direct) polar of U (under F ), and the set π−1(V ) is the (reverse)polar of V (under F ).

10.5.2. The following statements are valid:(1) π(u) := π({u}) = F (u) and π(U) = ∩u∈Uπ(u);(2) π(∪ξ∈�Uξ) = ∩ξ∈�π(Uξ);(3) π−1

F (V ) = πF−1(V );(4) U1 ⊂ U2 ⇒ π(U1) ⊃ π(U2);(5) U × V ⊂ F ⇒ (V ⊂ π(U) & U ⊂ π−1(V );(6) U ⊂ π−1(π(U)). CB

10.5.3. Akilov Criterion. A subset U of X is the polar of some subset of Yif and only if given x ∈ X \ U there is an element y in Y such that

U ⊂ π−1(y), x /∈ π−1(y).

C ⇒: If U = π−1(V ) then U = ∩v∈V π−1(v) by 10.5.2 (1).⇐: The inclusion U ⊂ π−1(y) means that y ∈ π(U). Thus, by hypothesis

U = ∩y∈π(U)π−1(y) = π−1(π(U)). B

10.5.4. Corollary. The set π−1(π(U)) is the (inclusion) least polar greaterthan U . CB

10.5.5. Definition. The set π−1F (πF (U)) is the bipolar of a subset U (under

the correspondence F ).

10.5.6. Examples.(1) Let (X, σ) be an ordered set and let U be a subset of X. Then

πσ(U) is the collection of all upper bounds of U (cf. 1.2.7).(2) Let (H, ( · , · )H) be a Hilbert space and F := {(x, y) ∈ H2 :

(x, y)H = 0}. Then π(U) = π−1(U) = U⊥ for every subset U of H. The bipolarof U in this case coincides with the closed linear span of U , that is, the closure ofthe linear span of U .

(3) Let X be a normed space and let X ′ be the dual of X. ConsiderF := {(x, x′) : x′(x) = 0}. Then π(X0) = X⊥0 and π−1(X0) = ⊥X0 for a subspaceX0 of X and a subspace X0 of X ′ (cf. 7.6.8). Moreover, π−1(π(X0)) = clX0by 7.5.14.

186 Chapter 10

10.5.7. Definition. Let X ↔ Y . Put

pol := {(x, y) ∈ X × Y : Re〈x | y〉 ≤ 1};abs pol := {(x, y) ∈ X × Y : |〈x | y〉| ≤ 1}.

To refer to direct or inverse polars under pol , we use the unified term “polar” (withrespect to X ↔ Y ) and the unified designations π(U) and π(V ). In the case of thecorrespondence abs pol , we speak of absolute polars (with respect to X ↔ Y ) andwrite U◦ and V ◦ (for U ⊂ X and V ⊂ Y ).

10.5.8. Bipolar Theorem. The bipolar π2(U) := π(π(U)) is the (inclusion)least weakly closed conical segment greater than U .

C Straightforward from 10.4.8 and the Akilov Criterion. B

10.5.9. Absolute Bipolar Theorem. The absolute bipolar U◦◦ := (U◦)◦ isthe (inclusion) least weakly closed absolutely convex set greater than U .C It suffices, first, to observe that the polar of a balanced set U coincides with

the absolute polar of U and, second, to apply 10.5.8. B

10.6. Weakly Compact Convex Sets

10.6.1. Let X be a locally convex real vector space and let p : X → R bea continuous sublinear functional on X. Then the (topological) subdifferential ∂(p)is compact in the topology σ(X ′, X).

C Put Q :=∏x∈X [−p(−x), p(x)] and endow Q with the Tychonoff topology.

Evidently ∂(p) ⊂ Q, with the Tychonoff topology on Q and σ(X ′, X) inducing thesame topology on ∂(p). It is beyond a doubt that the set ∂(p) is closed in Q by thecontinuity of p. Taking note of the Tychonoff Theorem and 9.4.9, conclude that∂(p) is a σ(X ′, X)-compact set. B

10.6.2. The balanced subdifferential of each continuous seminorm is weaklycompact. CB

10.6.3. Theorem. Let X be a real vector space. A subset U of X# is thesubdifferential of a (unique total) sublinear functional sU : X → R if and only if Uis nonempty, convex and σ(X#, X)-compact.C ⇒: Let U = ∂(sU ) for some sU . The uniqueness of sU is ensured by 3.6.6.

In view of 10.2.12 it is easy that the mirror of the Mackey topology τ(X, X#) isthe strongest multinorm on X (cf. 5.1.10 (2)). Whence we infer that the functionalsU is continuous with respect to τ(X, X#). In virtue of 10.6.1 the set U is compactin σ(X#, X). The convexity and nonemptiness of U are obvious.

⇐: Put sU (x) := sup{l(x) : l ∈ U}. Undoubtedly, sU is a sublinear functionaland dom sU = X. By definition, U ⊂ ∂(sU ). If l ∈ ∂(sU ) and l /∈ U , then by theStrict Separation Theorem and the Dualization Theorem sU (x) < l(x) for some xin X. This is a contradiction. B


10.6.4. Definition. The sublinear functional sU , constructed in 10.6.3, is thesupporting function of U . The term “support function” is also in current usage.

10.6.5. Kreın–Milman Theorem. Each compact convex set in a locallyconvex space is the closed convex hull (= the closure of the convex hull) of the setof its extreme points.

C Let U be such subset of a space X. It may be assumed that the spaceX is real and U 6= ∅. By virtue of 9.4.12, U is compact with respect to thetopology σ(X, X ′). Since σ(X, X ′) is induced in X by the topology σ(X ′#, X ′)on X ′#; therefore, U = ∂(sU ). Here (cf. 10.6.3) sU : X ′ → R acts by the rulesU (x′) := supx′(U). By the Kreın–Milman Theorem in subdifferential form, theset extU of the extreme points of U is not empty. The closure of the convex hullof extU is a subdifferential by Theorem 10.6.3. Moreover, this set has sU as itssupporting function, thus coinciding with U (cf. 3.6.6). B

10.6.6. Let X ↔ Y and let S be a conical segment in X. Assume further thatpS is the Minkowski functional of S. The polar π(S) is the inverse image of the(algebraic) subdifferential ∂(pS) under the ket-mapping; i.e.,

π(S) = | ∂(pS) 〉−1R .

If S is absolutely convex, then the absolute polar S◦ is the inverse image of the(algebraic) balanced subdifferential |∂|(pS) under the ket-mapping; i.e.,

S◦ = | |∂|(pS) 〉−1.

C If y ∈ YR and y ∈ | ∂(pS) 〉−1R then | y〉R belongs to ∂(pS). Hence, Re〈x | y〉 =

〈x | y〉R = | y〉R(x) ≤ pS(x) ≤ 1 for x ∈ S, because S ⊂ {pS ≤ 1} by the GaugeTheorem. Consequently, y ∈ π(S).

If, in turn, y ∈ π(S) then | y〉R belongs to ∂(pS). Indeed, 1 > pS(α−1x) forall x in XR and α > pS(x); i.e., α−1x ∈ {pS < 1} ⊂ S. Whence 〈α−1x | y〉R =Re〈α−1x | y〉 = α−1 Re〈x | y〉 ≤ 1. Finally, observe that | y〉R(x) ≤ α. Since αis arbitrary, this inequality means that | y〉R(x) ≤ pS(x). In other words, y ∈| ∂(ps)〉−1

R , which implies that π(S) = | ∂(pS)〉−1R . The remaining claim follows

from the properties of the complexifier (cf. 3.7.3 and 3.7.9). B

10.6.7. Alaoglu–Bourbaki Theorem. The polar of a neighborhood of zeroof each compatible topology is a weakly compact convex set.

C Let U be a neighborhood of zero in a space X and let π(U) be the polar of U(with respect to X ↔ X ′). Since U ⊃ {p ≤ 1} for some continuous seminorm p,by 10.5.2 (4), π(U) ⊂ π({p ≤ 1}) = π(Bp) = B◦p . Using 10.6.6 and recalling thatp is the Minkowski functional of Bp, obtain the inclusion π(U) ⊂ |∂|(p). By virtueof 10.6.2 the topological balanced subdifferential |∂|(p) is σ(X ′, X)-compact. By def-inition π(U) is weakly closed. To infer the σ(X ′, X)-compactness of π(U), it re-mains to appeal to 9.4.9. The convexity property of π(U) is beyond a doubt. B

188 Chapter 10

10.7. Reflexive Spaces

10.7.1. Kakutani Criterion. A normed space is reflexive if and only if itsunit ball is weakly compact.

C ⇒: Let X be reflexive, i.e. ′′(X) = X ′′. In other words, the image of Xunder the double prime mapping coincides with X ′′. Since the ball BX′′ is thepolar of the ball BX′ with respect to X ′′ ↔ X ′; therefore, BX′′ is a σ(X ′′, X ′)-compact set by the Alaoglu–Bourbaki Theorem. It remains to observe that BX′′ is(the image under the double prime mapping of) BX , and σ(X, X ′) is (the inverseimage under the double prime mapping of) σ(X ′′, X ′).

⇐: Consider the duality pair X ′′ ↔ X ′. By definition, the ball BX′′ presentsthe bipolar of BX (more precisely, the bipolar of (BX)′′). Using the AbsoluteBipolar Theorem and observing that the weak topology σ(X, X ′) is induced in Xby the topology σ(X ′′, X ′), conclude that BX′′ = BX (because of the obviousconvexity and closure properties of BX , the latter following from compactness sinceX is separated). Thus, X is reflexive. B

10.7.2. Corollary. A space X is reflexive if and only if every bounded closedconvex set in X is weakly compact. CB

10.7.3. Corollary. Every closed subspace of a reflexive space is reflexive.C By the Mazur Theorem, such subspace and, hence, the unit ball of it are

weakly closed. It thus suffices to apply the Kakutani Criterion twice. B

10.7.4. Pettis Theorem. A Banach space and its dual are (or are not) re-flexive simultaneously.

C If X is reflexive then σ(X ′, X) coincides with σ(X ′, X ′′). Therefore, by theAlaoglu–Bourbaki Theorem, BX′ is σ(X ′, X ′′)-compact. Consequently, X ′ is re-flexive. If, in turn, X ′ is reflexive, then so is X ′′ by what was proven. However, X,as a Banach space, is a closed subspace of X ′′. Thus, X is reflexive by 10.7.3. B

10.7.5. James Theorem. A Banach space is reflexive if and only if eachcontinuous (real) linear functional attains its supremum on the unit ball of thespace.

10.8. The Space C(Q, R)

10.8.1. Remark. Throughout Section 10.8 let Q stand for a nonempty Haus-dorff compact space, denoting by C(Q, R) the set of continuous real-valued func-tions on Q. Unless specified otherwise, C(Q, R) is furnished with the natural point-wise algebraic operations and order and equipped with the sup-norm ‖ · ‖ := ‖ · ‖∞related to the Chebyshev metric (cf. 4.6.8). Keeping this in mind, we treat thestatements like “C(Q, R) is a vector lattice,” “C(Q, R) is a Banach algebra,” etc.Other structures, if ever introduced in C(Q, R), are specified deliberately.


10.8.2. Definition. A subset L of C(Q, R) is a sublattice (in C(Q, R)) iff1 ∨ f2 ∈ L and f1 ∧ f2 ∈ L for f1, f2 ∈ L, where, as usual,

f1 ∨ f2(q) := f1(q) ∨ f2(q),f1 ∧ f2(q) := f1(q) ∧ f2(q) (q ∈ Q).

10.8.3. Remark. Observe that to be a sublattice in C(Q, R) means morethan to be a lattice with respect to the order induced from C(Q, R).

10.8.4. Examples.(1) ∅, C(Q, R), and the closure of a sublattice.(2) The intersection of each family of sublattices is also a sublattice.(3) Let L be a sublattice and let Q0 be a subset of Q. Put

LQ0 := {f ∈ C(Q, R) : (∃ g ∈ L) g(q) = f(q) (q ∈ Q0)}.

Then LQ0 is a sublattice. Moreover, L ⊂ LQ0 .(4) Let Q0 be a compact subset of Q. Given a sublattice L in C(Q, R),

putL∣∣Q0

:={f∣∣Q0

: f ∈ L}.

Therefore,LQ0 =

{f ∈ C(Q, R) : f

∣∣Q0∈ L

∣∣Q0

}.

It is clear that L∣∣Q0

is a sublattice of C(Q0, R). Furthermore, if L is a vec-tor sublattice of C(Q, R) (i.e., a vector subspace and simultaneously a sublatticeof C(Q, R)); then L

∣∣Q0

is a vector sublattice of C(Q0, R) (certainly, if Q0 6= ∅).

(5) Let Q := {1, 2}. Then C(Q, R) ' R2. Each nonzero vector sublat-tice of R2 is given as

{(x1, x2) ∈ R2 : α1x1 = α2x2}

for some α1, α2 ∈ R+.(6) Let L be a vector sublattice of C(Q, R). For q ∈ Q, the alternative

is offered: either L{q} = C(Q, R) or L{q} = {f ∈ C(Q, R) : f(q) = 0}. If q1 andq2 are distinct points of Q and L

∣∣{q1,q2}

6= 0, then by 10.8.4 (5) there are some α1,α2 ∈ R+ such that

L{q1,q2} = {f ∈ C(Q, R) : α1f(q1) = α2f(q2)}.

Moreover, if L contains a constant function other than zero (i.e. a nonzero multipleof the constantly-one function 1) then as α1 and α2 in the above presentationof L{q1,q2} the unity, 1, may be taken. CB

190 Chapter 10

10.8.5. Let L be a sublattice of C(Q, R). A function f in C(Q, R) belongsto the closure of L if and only if for all ε > 0 and (x, y) ∈ Q2 there is a functionf := fx,y,ε in L satisfying the conditions

f(x)− f(x) < ε, f(y)− f(y) > −ε.

C ⇒: This is obvious.⇐: Basing it on 3.2.10 and 3.2.11, assume that f = 0. Take ε > 0. Fix x ∈ Q

and consider the function gy := fx,y,ε ∈ L. Let Vy := {q ∈ Q : gy(q) > −ε}. ThenVy is an open set and y ∈ Vy. By a standard compactness argument, there arey1, . . . , yn ∈ Q such that Q = Vy1 ∪ . . . ∪ Vyn . Put fx := gy1 ∨ . . . ∨ gyn . It isclear that fx ∈ L. Furthermore, fx(x) < ε and fx(y) > −ε for all y ∈ Q. Now letUx := {q ∈ Q : fx(q) < ε}. Then Ux is open and x ∈ Ux. Using the compactnessof Q once again, find x1, . . . , xm ∈ Q such that Q = Ux1 ∪ . . . ∪ Uxm . Finally, putl := fx1 ∧ . . . ∧ fxm . It is beyond a doubt that l ∈ L and ‖l‖ < ε. B

10.8.6. Remark. The message of 10.8.5 is often referred to as the GeneralizedDini Theorem. (cf. 7.2.10).

10.8.7. Kakutani Lemma. Every sublattice L of C(Q, R) is expressible as

clL =⋂

(q1,q2)∈Q2

cl(L{q1,q2}

).

C The inclusion of clL into cl(L{q1,q2}

)for all (q1, q2) ∈ Q2 raises no doubts.

If f ∈ cl(L{q1,q2}

)for all such q1 and q2, then by 10.8.5, f ∈ clL. B

10.8.8. Corollary. Every vector sublattice L of C(Q, R) is expressible as

clL =⋂

(q1,q2)∈Q2

L{q1,q2}.

C Observe that every set of the form L{q1,q2} is closed. B

10.8.9. Definition. A subset U of FQ separates the points of Q, if for all q1,q2 ∈ Q such that q1 6= q2 there is a function u ∈ U assuming different values at q1and q2, i.e. u(q1) 6= u(q2).

10.8.10. Stone Theorem. If a vector sublattice of C(Q, R) contains constantfunctions and separates the points of Q, then it is dense in C(Q, R).C Given such sublattice L, observe that

L{q1,q2} = C(Q, R){q1,q2}

for every pair (q1, q2) in Q2 (cf. 10.8.4 (6)). It remains to appeal to 10.8.8. B


10.8.11. Let µ ∈ C(Q, R)′. Put

N (µ) := {f ∈ C(Q, R) : [0, |f |] ⊂ kerµ}.

Then there is a unique closed subset supp(µ) of Q such that

f ∈ N (µ)⇔ f∣∣supp(µ) = 0.

C By the Interval Addition Lemma

[0, |f |] + [0, |g|] = [0, |f |+ |g|].

Thus, f, g ∈ N (µ) ⇒ |f | + |g| ∈ N (µ). Since N (µ) is an order ideal; i.e.,(f ∈ N (µ) & 0 ≤ |g| ≤ |f | ⇒ g ∈ N (µ)), conclude that N (µ) is a linear set.Moreover, N (µ) is closed. Indeed, assuming fn ≥ 0, fn → f and fn ∈ N (µ), forg ∈ [0, f ] find g ∧ fn → g and g ∧ fn ∈ [0, fn]. Whence it follows that µ(g) = 0;i.e., f ∈ N (µ).

Since N (µ) is an order ideal, from 10.8.8 deduce that

N (µ) =⋂q∈Q

N (µ){q}.

Define the set supp(µ) as

q ∈ supp(µ)⇔ N (µ){q} 6= C(Q, R)⇔ (f ∈ N (µ)⇒ f(q) = 0).

It is beyond a doubt that supp(µ) is closed. Moreover, the equalities hold:

N (µ) =⋂

q∈supp(µ)

N (µ){q} ={f ∈ C(Q, R) : f

∣∣supp(µ) = 0

}.

The claim of uniqueness follows from the normality ofQ (cf. 9.4.14) and the UrysohnTheorem. B

10.8.12. Definition. The set supp(µ) under discussion in 10.8.11 is the sup-port of µ (cf. 10.9.4 (5)).

10.8.13. Remark. If µ is positive then

N (µ) = {f ∈ C(Q, R) : µ(|f |) = 0}.

Consequently, when µ(fg) = 0 for all g ∈ C(Q, R), observe that f∣∣supp(µ) = 0.

By analogy supp(µ) = ∅ ⇔ N (µ) = C(Q, R) ⇔ µ = 0. Therefore, it is quiteconvenient to work with the support of a positive functional.

192 Chapter 10

Let F be a closed subset of Q. It is said that F supports or carries µ or thatX \ F lacks µ or is void of µ if µ(|f |) = 0 for every continuous function f withsupp(f) ⊂ Q\F . The support supp(µ) of µ carries µ; moreover, supp(µ) is includedin every closed subset of Q supporting µ. In other words, the support of µ is thecomplement of the greatest open set void of µ (cf. 10.10.5 (6)).

It stands to reason to observe that in virtue of 3.2.14 and 3.2.15 to everybounded functional µ there correspond some positive (and hence bounded) func-tionals µ+, µ−, and |µ| defined as

µ+(f) = supµ[0, f ]; µ−(f) = − inf µ[0, f ]; |µ| = µ+ + µ−,

given f ∈ C(Q, R)+.Moreover, C(Q, R)′ is a Kantorovich space (cf. 3.2.16). CB

10.8.14. The supports of µ and |µ| coincide.C By definition N (µ) = N (|µ|). B

10.8.15. Considering a ∈ C(Q, R) with 0 ≤ a ≤ 1, define aµ : f 7→ µ(af) forf ∈ C(Q, R) and µ ∈ C(Q, R)′. Then |aµ| = a|µ|.

C Given f ∈ C(Q, R)+, infer that

(aµ)+(f) = sup{µ(ag) : 0 ≤ g ≤ f} ≤ supµ[0, af ]= µ+(af) = aµ+(f).

Furthermore,

µ+ = (aµ+ (1− a)µ)+ ≤ (aµ)+ + ((1− a)µ)+ ≤ aµ+ + (1− a)µ+ = µ+.

Consequently, (aµ)+ = aµ+, whence the claim follows. B

10.8.16.De Branges Lemma. Let A be a subalgebra of C(Q, R) containingconstant functions. Take µ ∈ ext(A⊥ ∩ BC(Q,R)′). Then the restriction of eachmember of A to the support of µ is a constant function.

C If µ = 0 then supp(µ) = ∅, and there is nothing to be proven. If µ 6= 0then, certainly, ‖µ‖ = 1. Take a ∈ A. Since the subalgebra A contains constantfunctions, it suffices to settle the case in which 0 ≤ a ≤ 1 and

q ∈ supp(µ)⇒ 0 < a(q) < 1.

Put µ1 := aµ and µ2 := (1 − a)µ. It is clear that µ1 + µ2 = µ and the functionalsµ1 and µ2 are both nonzero. Moreover,

‖µ‖ ≤ ‖µ1‖+ ‖µ2‖


= sup‖f‖≤1

µ(af) + sup‖g‖≤1

µ((1− a)g) = sup‖f‖≤1,‖g‖≤1

µ(af + (1− a)g) ≤ ‖µ‖,

because it is obvious that

aBC(Q,R) + (1− a)BC(Q,R) ⊂ BC(Q,R).

Thus, ‖µ‖ = ‖µ1‖+ ‖µ2‖. Since

µ = ‖µ1‖µ1

‖µ1‖+ ‖µ2‖

µ2

‖µ2‖,

and µ1, µ2 ∈ A⊥, conclude that µ1 = ‖µ1‖µ. By 10.8.15, a|µ| = |aµ| = |µ1| =‖µ1‖ |µ|. Consequently, |µ|((a − ‖µ1‖1)g) = 0 for all g ∈ C(Q, R). Using 10.8.13and 10.8.14, infer that the function a is constant on the support of µ. B

10.8.17. Stone–Weierstrass Theorem. Let A be a subalgebra of the alge-bra C(Q, R). Suppose that A contains constant functions and separates the pointsof Q. Then A is dense in C(Q, R).

C Proceeding by way of contradiction, assume the contrary. By the AbsoluteBipolar Theorem, the subspace A⊥ (coincident with A◦) of C(Q, R)′ is nonzero.Using the Alaoglu–Bourbaki Theorem, observe that A⊥ ∩ BC(Q,R)′ is a nonemptyabsolutely convex weakly compact set. Thus, by the Kreın–Milman Theorem, theset has an extreme point, say, µ.

Undoubtedly, µ is a nonzero functional. By the de Branges Lemma the supportof µ fails to contain two distinct points, since A separates the points of Q. Thesupport of µ is not a singleton, since µ annihilates constant functions. Thus, supp(µ)is empty. But then µ is zero (cf. 10.8.13). We arrive at a contradiction completingthe proof. B

10.8.18. Corollary. The closure of a subalgebra of C(Q, R) is a vector sub-lattice of C(Q, R).

C Using the Stone–Weierstrass Theorem, find a polynomial pn satisfying

|pn(t)− |t| | ≤ 1/2n

for all t ∈ [−1, 1]. Then |pn(0)| ≤ 1/2n. Therefore, the polynomial

pn(t) := pn(t)− pn(0)

maintains the inequality |pn(t)− |t| | ≤ 1/n when −1 ≤ t ≤ 1. By construction, pnlacks the constant term. Now, if a function a lies in a subalgebra A of C(Q, R)and ‖a‖ ≤ 1, then

|pn(a(q))− |a(q)| | ≤ 1/n (q ∈ Q).

Moreover, the function q 7→ pn(a(q)) is clearly a member of A. B

194 Chapter 10

10.8.19. Remark. Corollary 10.8.18 (together with 10.8.8) completely de-scribes all closed subalgebras of C(Q, R). In turn, as the proof prompts, theclaim of 10.8.18 is immediate on providing some sequence of polynomials whichconverges uniformly to the function t 7→ |t| on the interval [−1, 1]. It takes nopains to demonstrate 10.8.17, with 10.8.18 available.

10.8.20. Tietze–Urysohn Theorem. Let Q0 be a compact subset of a com-pact set Q and f0 ∈ C(Q0, R). Then there is a function f in C(Q, R) such thatf∣∣Q0

= f0.

C Suppose that Q0 6= ∅ (otherwise, there is nothing to prove). Consider theidentical embedding ι : Q0 → Q and the bounded linear operator

◦ι : C(Q, R) →

C(Q0, R) acting by the rule◦ιf := f ◦ ι. We have to show that

◦ι is an epimorphism.

It is beyond a doubt that im◦ι is a subalgebra of C(Q0, R) separating the points

of Q0 and containing constant functions. In virtue of 10.8.17 it is thus sufficient(and, clearly, necessary) to examine the closure of im

◦ι.

Consider the monoquotient ι of the operator◦ι and the coset mapping ϕ :

C(Q, R)0 = 0 OO ◦ι. Given f ∈ C(Q, R), put

g := (f ∧ sup |f(Q0)|1) ∨ (− sup |f(Q0)|1).By definition f

∣∣Q0

= g∣∣Q0

; i.e., f := ϕ(f) = ϕ(g). Consequently, ‖g‖ ≥ ‖f‖.Furthermore,

‖f‖ = inf{‖h‖C(Q,R) :

◦ι(h− f) = 0

}= inf

{‖h‖C(Q,R) : h

∣∣Q0

= f∣∣Q0

}≥ inf

{‖h∣∣Q0‖C(Q,R) : h

∣∣Q0

= f∣∣Q0

}= sup |f(Q0)| = ‖g‖ ≥ ‖f‖.

Therefore,

‖ιf‖ = ‖◦ιg‖ = ‖◦ιg‖C(Q0,R)

= ‖g ◦ ι‖C(Q0,R) = sup |g(Q0)| = ‖g‖ = ‖f‖;

i.e., ι is an isometry. Successively applying 5.5.4 and 4.5.15, infer first that coim◦ι is

a Banach space and second that im ι is closed in C(Q0, R). It suffices to observethat im

◦ι = im ι. B

10.9. Radon Measures

10.9.1. Definition. Let � be a locally compact topological space. Put K(�):= K(�, F) := {f ∈ C(�, F) : supp(f) is compact}. If Q is compact in � thenlet K(Q) := K�(Q) := {f ∈ K(�) : supp(f) ⊂ Q}. The space K(Q) is furnishedwith the norm ‖ · ‖∞. Given E ∈ Op (�), put K(E) := ∪{K(Q) : Q b E}. (Thenotation Q b E for a subset E of � means that Q is compact and Q lies in theinterior of E relative to �.)


10.9.2. The following statements are valid:(1) if Q b � and f ∈ C(Q, F) then

f∣∣∂Q

= 0⇔ (∃ g ∈ K(Q)) g∣∣Q= f ;

moreover, K(Q) is a Banach space;(2) letQ, Q1, andQ2 be compact sets andQ b Q1×Q2; the linear span

in the space C(Q, F) of the restrictions to Q of the functions like u1 · u2(q1, q2) :=u1 ⊗ u2(q1, q2) := u1(q1)u2(q2) with us ∈ K(Qs) is dense in C(Q, F);

(3) if � is compact then K(�) = C(�, F); if � fails to be compact andis embedded naturally into C(�·, F), with �· := � ∪ {∞} the Alexandroff com-pactification of �, then the space K(�) is dense in the hyperplane {f ∈ C(�·, F) :f(∞) = 0};

(4) the mapping E ∈ Op (�) 7→ K(E) ∈ Lat (K(�)) preserves suprema;(5) for E′, E′′ ∈ Op (�) the following sequence is exact:

00 = 0 OO (E′ ∩ E′′)ι(E′,E′′)−−−−−→ K(E′)×K(E′′)

σ(E′,E′′)−−−−−→ K(E′ ∪ E′′)0 = 0 OO ,

with ι(E′,E′′)f := (f, −f), and σ(E′,E′′)(f, g) := f + g.C (1): The boundary ∂Q of Q is at the same time the boundary of the exterior

of Q, the set int(� \Q).(2): The set under study is a subalgebra. Apply 9.3.13 and 10.8.17 (cf. 11.8.2).(3): It may be assumed that F = R. The claim will then follow from 10.8.8

since K(�) is an order ideal separating the points of �· (cf. 10.8.11).(4): Clearly, K(sup∅) = K(∅) = 0. If E ⊂ Op (�) and E is filtered upward

then, for f ∈ K(∪E ), observe that supp(f) ⊂ E for some E ∈ E by the compactnessof supp(f). Whence K(∪E ) = ∪{K(E) : E ∈ E }. To conclude, take E1, . . . , En ∈Op (�) and f ∈ K(E1 ∪ . . . ∪ En). In accordance with 9.4.18 there are someψk ∈ K(Ek) such that

∑nk=1 ψk = 1. Moreover, f =

∑nk=1 ψkf and supp(fψk) ⊂

Ek (k := 1, . . . , n).(5): This is straightforward from (4). B

10.9.3. Definition. A functional µ, a member of K(�, F)#, is a measure(or, amply, a Radon F-measure) on �, in symbols, µ ∈ M (�) := M (�, F); ifµ∣∣K(Q) ∈ K(Q)′ whenever Q b �. The following notation is current:∫

�

f dµ :=∫f dµ :=

∫f(x) dµ(x) := µ(f) (f ∈ K(�)).

The scalar µ(f) is the integral of f with respect to µ. In this connection µ is alsocalled an integral.

196 Chapter 10

10.9.4. Examples.(1) For q ∈ � the Dirac measure f 7→ f(q) (f ∈ K(�)) presents

a Radon measure. It is usually denoted by the symbol δq and called the delta-function at q.

Suppose also that � is furnished with group structure so that the takingof an inverse q ∈ � 7→ q−1 ∈ � and the group operation (s, t) ∈ � × � 7→ st ∈ �are continuous; i.e., � is a locally compact group. By δ we denote δe where e is theunity of � (recall the concurrent terms: “identity,” “unit,” and “neutral element,”all meaning the same: es = se = s for all s ∈ �). The nomenclature pertinentto addition is routinely involved in abelian (commutative) groups.

Given a ∈ �, acknowledge the operation of (left or right) translation or shiftby a in K(�) (in fact, every function is shifted in �× F):

(aτf)(q) := af(q) := f(a−1q), (τaf)(q) := fa(q) := f(qa−1)

with f ∈ K(�) and q ∈ �. Clearly, aτ , τa ∈ L (K(�)).A propitious circumstance of paramount importance is the presence of a non-

trivial measure on �, a member of M (�, R), which is invariant under left (or right)translations. All (left)invariant Radon measures are proportional. Each nonzero(left)invariant positive Radon measure is a (left) Haar measure (rarely Haar in-tegral). In the case of right translations, the term “(right) Haar measure” is incommon parlance. In the abelian case, we speak only of a Haar measure and evenof Haar measure, neglecting the necessity of scaling. The familiar Lebesgue measureon RN is Haar measure on the abelian group RN . That is why the conventional no-tation of Lebesgue integration is retained in the case of an abstract Haar measure.In particular, the left-invariance condition is written down as∫

�

f(a−1x) dx =∫�

f(x) dx (f ∈ K(�), a ∈ �).

(2) Let M(�) := (K(�), ‖ · ‖∞)′. A member of M(�) is a boundedRadon measure. It is clear that a bounded measure belongs to the space C(�·, F)′(cf. 10.9.2 (2)).

(3) Given µ ∈ M (�), put µ∗(f) = µ(f∗)∗, where f∗(q) := f(q)∗ forq ∈ � and f ∈ K(�). The measure µ∗ is the conjugate of µ. Distinction be-tween µ∗ and µ is perceptible only if F = C. In case µ = µ∗, we speak of a realC-measure. It is clear that µ = µ1 + iµ2, where µ1 and µ2 are uniquely-determinedreal C-measures. In turn, each real C-measure is generates by two R-measures (alsocalled real measures), members of M (�, R), because K(�, C) coincides with thecomplexification K(�, R) ⊕ iK(�, R) of K(�, R). The real R-measures obvi-ously constitute a Kantorovich space. Moreover, the integral with respect to such


measure serves as (pre)integral. So, an opportunity is offered to deal with the corre-sponding Lebesgue extension and spaces of scalar-valued or vector-valued functions(cf. 5.5.9 (4) and 5.5.9 (5)). We seize the opportunity without much ado and spec-ification.

To every Radon measure µ we assign the positive measure |µ| that is definedfor f ∈ K(�, R), f ≥ 0, by the rule

|µ|(f) := sup{|µ(g)| : g ∈ K(�, F), |g| ≤ f}.

Observe that, in current usage, the word “measure” often means a positive measure,whereas a “general” measure is referred to as a signed measure or a charge.

If µ and ν are measures and |µ| ∧ |ν| = 0 then µ and ν are called disjoint orindependent (of one another). A measure ν is absolutely continuous with respectto µ on condition that ν is independent of every measure independent of µ. Sucha measure ν may be given as ν = fµ, where f ∈ L1,loc(µ) and the measure fµhaving density f with respect to µ acts by the rule (fµ)(g) := µ(fg) (g ∈ K(�))(this is the Radon–Nikodym Theorem).

(4) Given �′ ∈ Op (�) and µ ∈M (�), consider the restriction µ�′ :=µ∣∣K(�′) of µ to K(�′). The restriction operator µ 7→ µ�′ from M (�) to M (�′),

viewed as depending on a subset of �, meets the agreement condition: if �′′ ⊂�′ ⊂ � and µ ∈ M (�) then µ�′′ = (µ�′)�′′ . This situation is verbalized asfollows: “The mapping M : E ∈ Op (�) 7→ M (E) and the restriction operator(referred to jointly as the functor M ) defines a presheaf (of vector spaces over �).”It stands to reason to convince oneself that the (values of the) restriction operatorneed not be an epimorphism.

(5) Let E ∈ Op (�) and µ ∈ M (�). It is said that E lacks µ or isvoid of µ or that � \ E supports or carries µ if µE = 0. By 10.9.2 (4) there isa least closed set supp(µ) supporting µ, the support of µ. It may be shown thatsupp(µ) = supp(|µ|). The above definition agrees with 10.8.12. The Dirac measureδq is a unique Radon measure supported at {q} to within scaling.

(6) Let �k be a locally compact space and µk ∈ M (�k) (k := 1, 2).There is a unique measure µ on the product �1 × �2 such that∫

�1×�2

u1(x)u2(y) dµ(x, y) =∫�1

u1(x) dµ1(x)∫�2

u2(y) dµ2(y)

with uk ∈ K(�k). The next designations are popular: µ1 × µ2 := µ1 ⊗ µ2 := µ.Using 10.9.2 (4), infer that for f ∈ K(�1 × �2) the value µ1 × µ2(f) may becalculated by repeated integration (this is the Fubini Theorem).

198 Chapter 10

(7) Let G be a locally compact group, and µ, ν ∈M(G). For f ∈ K(G)the function _f(s, t) := f(st) is continuous and |(µ × ν)(_f)| ≤ ‖µ‖ ‖ν‖ ‖f‖∞. Thisdefines the Radon measure µ ∗ ν(f) := (µ× ν)(_f) (f ∈ K(G)), the convolution of µand ν. Using vector integrals, obtain the presentations:

µ ∗ ν =∫

G×G

δs ∗ δt dµ(s)dν(t) =∫G

δs ∗ ν dµ(s) =∫G

µ ∗ δt dν(t).

The space M(G) of bounded measures with convolution as multiplication, ex-emplifies a Banach convolution algebra.

This algebra is commutative if and only if G is an abelian group. In that eventthe space L1(G), constructed with respect to Haar measure m, also possesses a nat-ural structure of a convolution algebra (namely, that of a subalgebra ofM(G)). Thealgebra (L1(G), ∗) is the group algebra of G. Thus, for f, g ∈ L1(G), the defini-tions of convolution for functions and measures agree with one another (cf. 9.6.17):(f ∗ g)dm = fdm ∗ gdm. By analogy, the convolution of µ ∈M(G) and f ∈ L1(G)is defined as (µ ∗ f)dm := µ ∗ (fdm); i.e., as the density of the convolution of µand fdm with respect to Haar measure dm. In particular,

f ∗ g =∫G

(δx ∗ g)f(x) dm(x) =∫G

τx(g)f(x) dm(x).

Wendel Theorem. Let T ∈ B(L1(G)). Then the following statements areequivalent:

(i) there is a measure µ ∈M(G) such that Tf = µ ∗ f for f ∈ L1(G);(ii) T commutes with translations: Tτa = τaT for a ∈ G, where τa is

a unique bounded extension to L1(G) of translation by the elementa in K(G);

(iii) T (f ∗ g) = (Tf) ∗ g for f, g ∈ L1(G);(iv) T (f ∗ ν) = (Tf) ∗ ν for ν ∈M(G) and f ∈ L1(G).

10.9.5. Definition. The spaces K(�) and M (�) are set in duality (inducedby the duality bracket K(�) ↔ K(�)#). In this case, the space M (�) is fur-nished with the topology σ(M (�), K(�)), usually called vague. The space K(�)is furnished with the Mackey topology τK(�) := τ(K(�), M (�)) (thereby, in par-ticular, (K(�), τK(�))′ = M (�)). The space of bounded measuresM(�) is usuallyconsidered with the dual norm:

‖µ‖ := sup{|µ(f)| : ‖f‖∞ ≤ 1, f ∈ K(�)} (µ ∈M(�)).

10.9.6. The topology τK(�) is strongest among all locally convex topologiesmaking the identical embedding of K(Q) to K(�) continuous for every Q withQ b � (i.e., τK(�) is the so-called inductive limit topology (cf. 9.2.15)).


C If τ is the inductive limit topology and µ ∈ (K(�), τ)′ then by definitionµ ∈ M (�), because µ ◦ ιK(Q) is continuous for Q b �. In turn, for µ ∈ M (�)the set VQ := {f ∈ K(Q) : |µ(f)| ≤ 1} is a neighborhood of zero in K(Q). Fromthe definition of τ , infer that ∪{VQ : Q b �} = {f ∈ K(�) : |µ(f)| ≤ 1} isa neighborhood of zero in τ . Thus, µ ∈ (K(�), τ)′ and τ is compatible withduality. Therefore, τ ≤ τK(�).

On the other hand, if p is a seminorm in the mirror of the Mackey topologythen p is the supporting function of a subdifferential lying in M (�). Consequently,the restriction q := p ◦ ιK(Q) of p to K(Q) is lower semicontinuous anyway. By theGelfand Theorem (in view of the barreledness of K(Q)) the seminorm q is contin-uous. Consequently, the identical embedding ιK(Q) : K(Q) → (K(�), τK(�)) iscontinuous and τ ≥ τK(�) by the definition of inductive limit topology. B

10.9.7. A subset A of K(RN ) is bounded (in the inductive limit topology),if sup ‖A‖∞ < +∞ and, moreover, the supports of the members of A lie in a com-mon compact set.

C Suppose to the contrary that, for some Q such that Q b RN , the inclusionA ⊂ K(Q) fails. In other words, for n ∈ N there are some qn ∈ RN and an ∈ Asatisfying an(qn) 6= 0 and |qn| > n. Put B := {n|an(qn)|−1δqn : n ∈ N} and observethat this set of Radon measures is vaguely bounded and so the seminorm p(f) :=sup{|µ|(|f |) : µ ∈ B} is continuous. Moreover, p(an) ≥ n|an(qn)|−1δqn(|an|) = n,which contradicts the boundedness property of A. B

10.9.8. Remark. Let (fn) ⊂ K(RN ). The notation fn �K

0 symbolizes the

proposition (∃Q b RN )(∀n) supp(fn) ⊂ Q & ‖fn‖∞ → 0. From 10.9.7 it isimmediate that µ ∈ K(RN )# is a Radon measure if µ(fn) → 0 whenever fn �

K0.

Observe also that the same holds for every locally compact � countable at infinityor σ-compact, i.e. for � presenting the union of countably many compact sets.

10.9.9. Remark. There is a sequence (pn) of real-valued positive polynomialson R such that the measures pndx converge vaguely to δ as n→ +∞. Consideringproducts of the measures, arrive at some polynomials Pn on the RN , with (Pndx)converging vaguely to δ (here, as usual, dx := dx1 × . . . × dxN is the Lebesguemeasure on RN ). Now, take f ∈ K(RN ) such that f is of class C(m) in someneighborhood of a compact set Q (i.e., f has continuous derivatives up to order mon Q). Arranging the convolutions (f ∗Pn), obtain a sequence of polynomials whichapproximates not only f but also the derivatives of f up to orderm uniformly on Q.The possibility of such reqularization is referred to as the Generalized WeierstrassTheorem in RN (cf. 10.10.2 (4)).

10.9.10. Measure Localization Pronciple. Let E be an open cover of �and let (µE)E∈E be a family of Radon measures, µE ∈ M (E). Assume further

200 Chapter 10

that, for every pair (E′, E′′) of the members of E , the restrictions of µE′ and µE′′to E′ ∩E′′ coincide. Then there is a unique measure µ on � whose restriction to Eequals µE for all E ∈ E .C Using 10.9.2 (5), arrange the sequence∑

{E′,E′′}E′,E′′∈E, E′ 6=E′′

K(E′ ∩ E′′) ι−→∑E∈E

K(E) σ−→ K(�)→ 0,

where ι is generated by summation of the coordinate identical embeddings ι(E′,E′′)and σ is the standard addition. Every direct sum is always assumed to be furnishedwith the inductive limit topology (cf. 10.9.6).

Examine the exactness of the sequence. Since K(�) = ∪Qb�K(Q) and in viewof 10.9.2 (4), it suffices to settle the case of a finite cover by checking exactness at thesecond term. Thus, proceeding by induction, suppose that for every n-element cover{E1, . . . , En} (n ≥ 2) the following sequence is exact:

Knιn−→

n∏k=1

K(Ek)σn−→ K(E1 ∪ . . . ∪ En)→ 0,

where ιn is the “restriction” of ι to Kn, the symbol σn stands for addition and

Kn :=∏k<l

k,l∈{1,... ,n}

K(Ek ∩ El).

By hypothesis, im ιn = kerσn. If σn+1(f , fn+1) = 0 where f := (f1, . . . , fn), thenσnf = −fn+1 and fn+1 ∈ K((E1 ∪ . . . ∪ En) ∩ En+1). Since σn is an epimorphismby 10.9.2 (5), there are some θk ∈ K(Ek ∩ En+1) such that σnθ = −fn+1 forθ := (θ1, . . . , θn). Whence (f − θ) ∈ kerσn, and by hypothesis there is a member κof Kn such that ιnκ = f − θ. Clearly,

Kn+1 = Kn ×n∏k=1

K(Ek ∩ En+1)

(to within isomorphism), κ := (κ, θ1, . . . , θn) ∈ Kn+1, and ιn+1κ = (f , fn+1).Passing to the diagram prime (cf. 7.6.13), arrive at the exact sequence

0→M (�) σ′−→∏E∈E

M (E) ι′−→∏

{E′,E′′}E′,E′′∈E ,E′ 6=E′′

M (E′ ∩ E′′).

The proof is complete. B


10.9.11. Remark. In topology, a presheaf recoverable from its patches or localdata in the above manner is a sheaf. In this regard, the claim of 10.9.10 is verbalizedas follows: the presheaf � 7→M (�) of Radon measures is a sheaf or, putting it morecategorically, the functor M is a sheaf (of vector spaces over �, cf. 10.9.4 (4)).

10.10. The Spaces D() and D ′()

10.10.1. Definition. A compactly-supported smooth function f : RN→ F isa test function; in symbols, f ∈ D(RN ) := D(RN , F). Given Q b RN and � ∈Op (RN ), designate D(Q) :=

{f ∈ D(RN ) : supp(f) ⊂ Q

}and D(�) := ∪{D(Q) :

Q b �}.

10.10.2. The following statements are valid:(1) D(Q) = 0⇔ intQ = ∅;(2) given Q b RN , put

‖f‖n,Q :=∑|α|≤n

‖∂αf‖C(Q) :=∑

α∈(Z+)Nα1+...+αN≤n

sup |(∂α1 . . . ∂αnf)(Q)|

for a function f smooth in a neighborhood of Q (as usual, Z+ := N ∪ {0}); themultinorm MQ := {‖ · ‖n,Q : n ∈ N} makes D(Q) into a Frechet space;

(3) the space of smooth functions C∞(�) := E (�) on � in Op (RN )with the multinorm M� := {‖ · ‖n,Q : n ∈ N, Q b �} is a Frechet space; moreover,D(�) is dense in C∞(�);

(4) if Q1 b RN , Q2 b RM and Q b Q1 × Q2; then the linear spanin D(Q) of the restrictions to Q of the functions like f1f2(q1, q2) := f1⊗f2(q1, q2) :=f1(q1)f2(q2), with qk ∈ Qk and fk ∈ D(Qk), is dense in D(Q);

(5) the mapping E ∈ Op (�) 7→ D(E) ∈ Lat (D(�)) preserves suprema:

D(E′ ∩ E′′) = D(E′) ∩D(E′′), D(E′ ∪ E′′) = D(E′) + D(E′′);D(∪E ) = L (∪{D(E) : E ∈ E }) (E ⊂ Op (�)).

Moreover, the next sequence is exact (cf. 10.9.2 (5)):

00 = 0 OO D(E′ ∩ E′′)ι(E′,E′′)−−−−−→ D(E′)×D(E′′)

σ(E′,E′′)−−−−−→ D(E′ ∪ E′′)0 = 0 OO .

C (1) and (2) are obvious.(3): Choose a sequence (Qm)m∈N such that Qm b �, Qm b Qm+1 and

∪m∈NQm = �. The multinorm {‖ · ‖n,Qm : n ∈ N, m ∈ N} is countable and

202 Chapter 10

equivalent to M�. A reference to 5.4.2 yields the metrizability of C∞(�). Thecompleteness of C∞(�) raises no doubts.

To show that D(�) is dense in C∞(�), consider the truncator set Tr(�) := {ψ ∈D(�) : 0 ≤ ψ ≤ 1}. Make Tr(�) a direction on letting ψ1 ≤ ψ2 ⇔ supp(ψ1) ⊂int{ψ2 = 1}. It is clear that, for f ∈ C∞(�), the net (ψf)ψ∈Tr(�) approximates fas is needed.

(4): Take q′ ∈ RN and q′′ ∈ RM . Let a(q′, q′′) := a′(q′)a′′(q′′), where a′and a′′ are mollifiers on RN and RM , respectively. Given f ∈ D(Q), n ∈ N, andε > 0, choose χ from the condition ‖f − f ∗ aχ‖n,Q ≤ ε/2. Using the equicontinuityproperty of the family F := {∂αf(q)τq(aχ) : |α| ≤ n, q ∈ Q1×Q2}, find finite sets�′ and �′′, with �′ ⊂ Q1 and �′′ ⊂ Q2, so that the integral of each function in Fbe approximated to within 1/2(N +1)−nε by a Riemann sum of it using the pointsof �′×�′′. This yields a function f , a member of the linear span of D(Q1)×D(Q2),which was sought; i.e., ‖f − f‖n,Q ≤ ε.

(5): Check this as in 10.9.2 (4), on replacing 9.4.18 with 9.6.19 (2). B

10.10.3. Remark. The Generalized Weierstrass Theorem may be appliedto the demonstration of 10.10.2 (4), when combined with due truncation providingthat the constructed approximation has compact support.

10.10.4.Definition. A functional u, a member of D(�, F)#, is a distributionor a generalized function whenever u |D(Q) ∈ D ′(Q) := D(Q)′ for all Q b �. This isexpressed in writing as u ∈ D ′(�) := D ′(�, F). Sometimes a reference is appendedto the nature of the ground field F.

The usual designations are as follows: 〈u, f〉 := 〈f |u〉 := u(f). Often we usethe most suggestive and ubiquitous symbol∫

f(x)u(x) dx := u(f) (f ∈ D(�)).

10.10.5. Examples.(1) Let g ∈ L1,loc(RN ) be a locally integrable function. The mapping

ug(f) :=∫f(x)g(x) dx (f ∈ D(�))

determines a distribution. A distribution of this type is regular. To denote a regulardistribution ug, a more convenient symbol g is also employed. In this connection,we write D(�) ⊂ D ′(�) and ug = | g〉.

(2) Every Radon measure is a distribution. Each positive distribution u(i.e., such that f ≥ 0⇒ u(f) ≥ 0) is determined by a positive measure.


(3) A distribution u is said to has order at most m, if to every Q suchthat Q b RN there corresponds a number tQ satisfying

|u(f)| ≤ tQ‖f‖m,Q (f ∈ D(Q)).

The notions of the order of a distribution and of a distribution of finite order areunderstood in a matter-of-fact fashion. Evidently, it is false that every distributionhas finite order.

(4) Let α be a multi-index, α ∈ (Z+)N ; and let u be a distribution,u ∈ D ′(�). Given f ∈ D(�), put (∂αu)(f) := (−1)|α|u(∂αf). The distribution∂αu is the derivative of u (of order α). We also speak of generalized differentiation,of derivatives in the distribution sense, etc. and use the conventional symbolsof differential calculus.

A derivative (of nonzero order) of a Dirac measure is not a measure. At thesame time δ ∈ D ′(R) is the derivative of the Heaviside function δ(−1) := H, whereH : R → R is the characteristic function of R+. If a derivative of a (regular)distribution u is some regular distribution ug, then g is a weak derivative of uor a generalized derivative of u in the Sobolev sense. For a test function, suchderivative coincides with its ordinary counterpart.

(5) Given u ∈ D ′(�), put u∗(f) := u(f∗)∗. The distribution u∗ is theconjugate of u. The presence of the involution ∗ routinely justifies speaking of realdistributions and complex distributions (cf. 10.9.3 (3)).

(6) Let E ∈ Op (�) and u ∈ D ′(�). For f ∈ D(E), the scalar u(f)is easily determined. This gives rise to the distribution uE , a member of D ′(E),called the restriction of u to E. The functor D ′ is clearly a presheaf.

Given u ∈ D ′(�) and E ∈ Op (�), say that E lacks or is void of u, if uE = 0.By 10.10.4 (5), if the members of a family of open subsets of � are void of u thenso is the union of the family. The complement (to RN ) of the greatest open setvoid of u is the support of u, denoted by supp(u). Observe that supp(∂αu) ⊂supp(u). Moreover, a distribution with compact support (= compactly-supporteddistribution) has finite order.

(7) Let u ∈ D ′(�) and f ∈ C∞(�). If g ∈ D(�) then fg ∈ D(�). Put(fu)(g) := u(fg). The resulting distribution fu is the product of f and u. Considerthe truncator direction Tr(�). If there is a limit limψ∈Tr(�) u(fψ) then say that uapplies to f . It is clear that a compactly-supported distribution u applies to everyfunction in C∞(�). Moreover, u ∈ E ′(�) := C∞(�)′. In turn, each element uof E ′(�) (cf. 10.10.2 (3)) uniquely determines a distribution with compact support,which is implicit in the notation u ∈ D ′(�).

If f ∈ C∞(�) and ∂αf∣∣supp(u) = 0 for all α, |α| ≤ m, where u is a compactly-

supported distribution of order at most m, then it is easy that u(f) = 0. In con-

204 Chapter 10

sequence, only linear combinations of a Dirac measure and its derivatives are sup-ported at a singleton. CB

(8) Let �1, �2 ∈ Op (RN ) and uk ∈ D ′(�k). There is a unique distri-bution u on �1 × �2 such that u(f1f2) = u1(f1)u2(f2) for all fk ∈ D(�k). Thedistribution is denoted by u1 × u2 or u1 ⊗ u2. Using 10.10.2 (4), infer that forf ∈ D(�1 ×�2) the value u(f) of u at f appears from successive application of u1and u2. Strictly speaking,

u(f) = u2(y ∈ �2 7→ u1(f( · , y))) = u1(x ∈ �1 7→ u2(f(x, · ))).

More suggestive designations prompt the Fubini Theorem:∫∫�1×�2

f(x, y)(u1 × u2)(x, y) dxdy

=∫�2

∫�1

f(x, y)u1(x) dx

u2(y) dy =∫�1

∫�2

f(x, y)u2(y) dy

u1(x) dx.

It is worth noting that

supp(u1 × u2) = supp(u1)× supp(u2).

(9) Let u, v ∈ D ′(RN ). Given f ∈ D(RN ), put+f := f ◦ +. It is

clear that+f ∈ C∞(RN × RN ). Say that the distributions u and v convolute or

admit convolution or are convolutive provide that the product u × v applies to

the function+f ⊂ C∞(RN × RN ) for each f in D(RN ). It is easy (cf. 10.10.10)

that the resulting linear functional f 7→ (u × v)(+f ) (f ∈ D(RN )) is a distribu-

tion called the convolution of u and v and denoted by u ∗ v. It is beyond a doubtthat the convolutions of functions (cf. 9.6.17) and measures on RN (cf. 10.9.4 (7))are particular cases of the convolution of distributions. In some classes, every twodistributions convolute. For instance, the space E ′(RN ) of compactly-supporteddistributions with convolution as multiplication presents an (associative and com-mutative) algebra with unity the delta-function δ. Further, ∂αu = ∂αδ ∗ u and∂α(u ∗ v) = ∂αu ∗ v = u ∗ ∂αv. Moreover, the following remarkable equality, theLions Theorem of Supports, holds:

co (supp(u ∗ v)) = co (supp(u)) + co (supp(v)).


It is worth emphasizing that the pairwise convolutivity of distributions fails in gen-eral to guarantee the associativity of convolution (for instance, (1 ∗ δ′) ∗ δ(−1) = 0whereas 1 ∗ (δ′ ∗ δ(−1)) = 1, with 1 := 1R).

Each distribution u convolutes with a test function f , yielding some regulardistribution (u ∗ f)(x) = u(τx(f ˜ )), where f ˜:= f is the reflection of f ; i.e.,f (x) := f(−x) (x ∈ RN ). The operator u∗ : f 7→ u ∗ f from D(RN ) to C∞(RN )is continuous and commutes with translations: (u∗)τx = τxu∗ for x ∈ RN . It iseasily seen that the above properties are characteristic of u∗; i.e., if an operator T ,a member of L (D(RN ), C∞(RN )), is continuous and commutes with translations,then there is a unique distribution u such that T = u∗; namely, u(f) := (T ′δ)(f)for f ∈ D(RN ) (cf. the Wendel Theorem).

10.10.6. Definition. The spaces D(�) and D ′(�) are set in duality (in-duced by the duality bracket D(�) ↔ D(�)#). Moreover, D ′(�) is furnishedwith the topology of the distribution space σ(D ′(�), D(�)), and D(�) is furnishedwith the topology of the test function space, the Mackey topology τD := τD(�) :=τ(D(�), D ′(�)).

10.10.7. Let � ∈ Op (RN ). Then(1) τD is the strongest of the locally convex topologies making the

identical embedding of D(Q) into D(�) continuous for all Q b � (i.e., τD is theinductive limit topology);

(2) a subset A of D(�) is bounded if and only if A lies in D(Q) forsome Q such that Q b � and is bounded in D(Q);

(3) a sequence (fn) converges to f in (D(�), τD) if and only if thereis a compact set Q such that Q b �, supp(fn) ⊂ Q, supp(f) ⊂ Q and (∂αfn)converges to ∂αf uniformly on Q for every multi-index α (in symbols, fn � f);

(4) an operator T , a member of L (D(�), Y ) with Y a locally convexspace, is continuous if and only if Tfn → 0 provided that fn � 0;

(5) a delta-like sequence (bn) serves as a (convolution) approximateunity in D(RN ) as well as in D ′(RN ); i.e., bn ∗ f � f (in D(RN )) and bn ∗ u→ u(in D ′(RN )) for f ∈ D(RN ) and u ∈ D ′(RN ).

C (1): This is established in much the same way as 10.9.6; and (2), by analogywith 10.9.7 using the presentation of � as the union � = ∪n∈NQn, whereQn b Qn+1for n ∈ N.

(3): Note that a convergent countable sequence is bounded, and apply 10.10.7(2) (cf. 10.9.8).

(4): In virtue of 10.10.7 (1) the continuity of T amounts to that of the restrictionT∣∣D(Q) for all Q b �. By 10.10.2 (2) the space D(Q) is metrizable. It remains

to refer to 10.10.7 (3).

206 Chapter 10

(5): It is clear that all of the supports supp(bn ∗ f) lie in some compact neigh-borhood about supp(f). Furthermore, for g ∈ C(RN ), it is evident that bn ∗ g → guniformly on compact subsets of RN . Applying this to ∂αf and considering (3),infer that bn ∗ f � f .

On account of 10.10.5 (9), observe the following:

u(f) = (u ∗ f)(0) = limn(u ∗ (bn ∗ f))(0)

= limn((u ∗ bn) ∗ f)(0) = lim

n(bn ∗ u)(f)

for f ∈ D(RN ). B

10.10.8.Remark. In view of 10.10.7 (3), for � ∈ Op (RN ) andm ∈ Z+ it is of-ten convenient to consider the space D (m)(�) := C(m)

0 (�) comprising all compactly-supported functions f whose derivatives ∂αf are continuous for all |α| ≤ m. Thespace D (m)(Q) := {f ∈ D (m)(�) : supp(f) ⊂ Q} for Q b � is furnished with thenorm ‖·‖m,Q making it into a Banach space. In that event, D (m)(�) is endowed withthe inductive limit topology. Thus, D (0)(�) = K(�) and D(�) = ∩m∈ND (m)(�).For a sequence (fn) to converge in D (m)(�) means to converge uniformly with allderivatives up to order m on some Q such that Q b � and supp(fn) ⊂ Q for all suf-ficiently large n. Note that D (m)(�)′ comprises all distributions of order at most m.Correspondingly,

D ′F (�) :=⋃m∈N

D (m)(�)′

is the space of finite-order distributions.

10.10.9. Let � ∈ Op (RN ). Then(1) the space D(�) is barreled; i.e., each barrel, a closed absorbing

absolutely convex subset, is a neighborhood of zero;(2) every bounded closed set in D(�) is compact; i.e., D(�) is a Montel

space;(3) every absolutely convex subset of D(�), absorbing each bounded

set, is a neighborhood of zero; i.e., D(�) is a bornological space;(4) the test functions are dense in the distribution space.

C (1): Given a barrel V in D(�), observe that VQ := V ∩ D(Q) is a barrelin D(Q) for all Q b �. Thus, VQ is a neighborhood of zero in D(Q) (cf. 7.1.8).

(2): Such a set lies in D(Q) for some Q b � by 10.10.7 (2). In virtue of 10.10.2(2), D(Q) is metrizable. Applying 4.6.10 and 4.6.11 proves the claim.

(3): This follows from the barreledness of D(Q) for Q b �.(4): Let g ∈ |D(�)〉◦, with the polar taken with respect to D(�) ↔ D ′(�).

It is clear that uf (g) = 0 for f ∈ D(�); i.e.,∫g(x)f(x) dx = 0. Thus, g = 0.

It remains to refer to 10.5.9. B


10.10.10. Schwartz Theorem. Let (uk)k∈N be a sequence of distributions.Assume that for every f in D(�) there is a sum

u(f) :=∞∑k=1

uk(f).

Then u is a distribution and

∂αu =∞∑k=1

∂αuk

for every multi-index α.C The continuity of u is guaranteed by 10.10.9 (1). Furthermore, for f ∈ D(�)

by definition (cf. 10.10.5 (4))

∂αu(f) = u((−1)|α|∂αf

)=∞∑k=1

uk((−1)|α|∂αf

)=∞∑k=1

∂αuk(f). .

10.10.11. Theorem. The functor D ′ is a sheaf.C It is immediate (cf. 10.9.10 and 10.9.11). B

10.10.12. Remark. The possibility of recovering a distribution from localdata, the Distribution Localization Principle stated in 10.10.11, admits clarificationin view of the paracompactness of RN . Namely, consider an open cover E of � anda distribution u ∈ D ′(�) with local data (uE)E∈E . Take a countable (locally finite)partition of unity (ψk)k∈N subordinate to E . Evidently, u =

∑∞k=1 ψkuk, where

uk := uEk and supp(ψkuk) ⊂ Ek (k ∈ N).

10.10.13. Theorem. Each distribution u on � of order at most m may beexpressed as sum of derivatives of Radon measures:

u =∑|α|≤m

∂αµα,

where µα ∈M (�).C To begin with, assume that u has compact support. Let Q with Q b � be

a compact neighborhood of supp(u). By hypothesis (cf. 10.10.5 (7) and 10.10.8)

|u(f)| ≤ t∑|α|≤m

‖∂αf‖∞ (f ∈ D(Q))

for some t ≥ 0.

208 Chapter 10

Using 3.5.7 and 3.5.3, from 10.9.4 (2) obtain

u = t∑|α|≤m

να ◦ ∂α = t∑|α|≤m

(−1)|α|∂ανα

for a suitable family (να)|α|≤m, where να ∈ |∂|(‖ · ‖∞).Passing to the general case and invoking the Countable Partition Theorem, find

a partition of unity (ψk)k∈N with ψk ∈ D(�) such that some neighborhoods Qkof supp(ψk) compose a locally finite cover of � (cf. 10.10.12). For each of thedistributions (ψku)k∈N it is already proven that

ψku =∑|α|≤m

∂αµk,α,

where every µk,α is a Radon measure on � and supp(µk,α) ⊂ Qk.From the Schwartz Theorem it is immediate that the sum

µα(f) :=∞∑k=1

µk,α(f)

exists for all f ∈ K(�). Moreover, the resulting distribution µα is a Radon measure.Once again appealing to 10.10.10, infer that

u =∞∑k=1

ψku =∞∑k=1

∑|α|≤m

∂αµk,α =∑|α|≤m

∂α( ∞∑

k=1

µk,α

)=∑|α|≤m

∂αµα,

which was required. B

10.10.14. Remark. The claim of 10.10.13 is often referred to as the theoremon the general form of a distribution. Further abstraction and clarification are avail-able. For instance, it may be verified that a compactly-supported Radon measureserves as a derivative (in the distribution sense) of suitable order of some contin-uous function. This enables us to view each distribution as a result of generalizeddifferentiation of a conventional function.

10.11. The Fourier Transform of a Distribution

10.11.1. Let χ be a nonzero functional over the space L1(RN ) := L1(RN , C).The following statements are equivalent:

(1) χ is a character of the group algebra (L1(RN ), ∗); i.e., χ 6= 0,χ ∈ L1(RN )′ and

χ(f ∗ g) = χ(f)χ(g) (f, g ∈ L1(RN ))

(in symbols, χ ∈ X(L1(RN )), cf. 11.6.4);


(2) there is a unique vector t in RN such that

χ(f) = f(t) := (f ∗ et)(0) :=∫

RN

f(x)ei(x,t) dx

for all f ∈ L1(RN ).C (1) ⇒ (2): Suppose that χ(f)χ(g) 6= 0. If x ∈ RN then

χ(δx ∗ f ∗ g) = χ(δx ∗ f)χ(g) = χ(δx ∗ g)χ(f).

Put ψ(x) := χ(f)−1χ(δx ∗ f). This soundly defines some continuous mapping ψ :RN → C. Moreover,

ψ(x+ y)= χ(f ∗ g)−1χ(δx+y ∗ (f ∗ g)) = χ(f)−1χ(g)−1χ(δx ∗ f ∗ δy ∗ g)

= χ(f)−1χ(δx ∗ f)χ(g)−1χ(δy ∗ g) = ψ(x)ψ(y)

for x, y ∈ RN ; i.e., ψ is a (unitary) group character: ψ ∈ X(RN ). Calculus showsthat ψ = et for some (obviously, unique) t ∈ RN . Further, by the properties of theBochner integral

χ(f)χ(g) = χ(f ∗ g) = χ

∫RN

(δx ∗ g)f(x) dx

=∫

RN

χ(δx ∗ g)f(x) dx =∫

RN

f(x)χ(g)ψ(x) dx = χ(g)∫

RN

f(x)ψ(x) dx.

Thus,

χ(f) =∫

RN

f(x)ψ(x) dx (f ∈ L1(RN )).

(2)⇒ (1): Given t ∈ RN and treating f , g and f ∗ g as distributions, infer that

f ∗ g(t) = uf∗g(et)

=∫

RN

∫RN

f(x)g(y)et(x+ y) dxdy =∫

RN

f(x)et(x) dx∫

RN

g(y)et(y) dy

210 Chapter 10

= uf (et)ug(et) = f(t)g(t). .

10.11.2. Remark. The essential steps of the above argument remain validfor every locally compact abelian group G, and so there is a one-to-one correspon-dence between the character space X(L1(G)) of the group algebra and the set X(G)comprising all (unitary) group characters of G. Recall that such character is a con-tinuous mapping ψ : G→ C satisfying

|ψ(x)| = 1, ψ(x+ y) = ψ(x)ψ(y) (x, y ∈ G).

Endowed with pointwise multiplication, the set G := X(G) becomes a commutativegroup. In virtue of the Alaoglu–Bourbaki Theorem, X(L1(G)) is locally compactin the weak topology σ((L1(G))′, L1(G)). So, Gmay be treated as a locally compactabelian group called the character group of G or the dual group of G. Each elementq in G defines the character q : q ∈ G 7→ q(q) ∈ C of the dual group G. The resulting

embedding of G into G is surprisingly an isomorphism of the locally compact abeliangroups G and G (the Pontryagin–van Kampen Duality Theorem).

10.11.3. Definition. For a function f in L1(RN ), the mapping f : RN → C,defined by the rule

f(t) := f(t) := (f ∗ et)(0),

is the Fourier transform of f .

10.11.4. Remark. By way of taking convenient liberties, we customarily usethe term “Fourier transform” expansively. First, it is retained not only for the oper-ator F : L1(RN )→ C RN acting by the rule Ff := f but also for its modifications(cf. 10.11.13). Second, the Fourier transform F is identified with each operatorFθf := f ◦ θ, where θ is an automorphism (= isomorphism with itself) of RN .Among the most popular are the functions: θ(x) := ∼(x) := −x, θ(x) := 2π(x) := 2πxand θ(x) := −2π(x) := −2πx (x ∈ RN ). In other words, the Fourier transform isusually defined by one the following formulas:

F∼f(t) =∫

RN

f(x)e−i(x,t) dx,

F2πf(t) =∫

RN

f(x)e2πi(x,t) dx,

F−2πf(t) =∫

RN

f(x)e−2πi(x,t) dx.


Since the character groups of isomorphic groups are also isomorphic, there aregrounds to using the same notation f for generally distinct functions Ff , F∼f ,and F±2πf . The choice of the symbol for F2π (F−2π) dictates the denotation ∨

for F−2π (F2π) (cf. 10.11.12).

10.11.5. Examples.

(1) Let f : R0 = 0 OO R be the characteristic function of the interval[−1, 1]. Clearly, f(t) = 2t−1 sin t. Observe that if kπ ≥ t0 > 0 then

∫[t0,+∞)

|f(t)| dt ≥∫

[kπ,+∞)

|f(t)| dt =∞∑n=k

∫[nπ,(n+1)π]

|f(t)| dt

≥∞∑n=k

∫[nπ,(n+1)π]

2| sin t|(n+ 1)π

dt = 4∞∑n=k

1(n+ 1)π

= +∞.

Thus, f /∈ L1(R).

(2) For f ∈ L1(RN ) the function f is continuous, with the inequality‖f‖∞ ≤ ‖f‖1 holding.

C The continuity of f follows from the Lebesgue Dominated Convergence The-orem; and the boundedness of f , from the obvious estimate

|f(t)| ≤∫

RN

|f(x)| dx = ‖f‖1 (t ∈ RN ). .

(3) If f ∈ L1(RN ), then |f(t)| → 0 as |t| → +∞ (= the Riemann–Lebesgue Lemma).

/ The claim is obvious for compactly-supported step functions. It sufficesto refer to 5.5.9 (6) and the containment F ∈ B(L1(RN ), l∞(RN )). .

(4) Let f ∈ L1(RN ), ε > 0, and fε(x) := f(εx) (x ∈ RN ). Thenfε(t) = ε−N f (t/ε) (t ∈ RN ).

/ fε(t) =∫

RNf(εx)et(x) dx = ε−N

∫RN

f(εx)et/ε(εx) dεx = ε−N f (t/ε) .

(5) F (f∗) = (F∼f)∗, τxf = exf , and exf = τxf for all f ∈ L1(RN )and x ∈ RN .

212 Chapter 10

C We will only check the first equality. Since a∗b = (ab∗)∗ for a, b ∈ C;on using the properties of conjugation and integration, given t ∈ RN , infer that

F (f∗)(t) =∫

RN

f(x)ei(x,t) dx =

∫RN

f(x)(ei(x,t)

)∗dx

∗

=

∫RN

f(x)e−i(x,t) dx

∗ = (F∼f)∗(t). .

(6) If f, g ∈ L1(RN ) then

f ∗ g = f g;∫

RN

fg =∫

RN

fg.

C The first equality is straightforward from 10.11.1. The second, the multipli-cation formula, follows on applying the Fubini Theorem:∫

RN

fg =∫

RN

∫RN

f(x)et(x) dxg(t) dt

=∫

RN

∫RN

g(t)et(x) dt

f(x) dx =∫

RN

fg. .

(7) If f , f and g belong to L1(RN ) then (fg) = f˜∗ g.C Given x ∈ RN , observe that

(fg)(x) = ∫RN

g(t)f(t)et(x) dt =∫

RN

∫RN

g(t)f(y)et(y)et(x) dydt

=∫

RN

∫RN

f(y)g(t)et(x+ y) dtdy

=∫

RN

f(y)g(x+ y) dy =∫

RN

f(y − x)g(y) dy = f˜∗ g(x). .(8) If f ∈ D(RN ) and α ∈ (Z+)N then

F (∂αf) = i|α|tαFf, ∂α(Ff) = i|α|F (xαf);

F2π(∂αf) = (2πi)|α|tαF2πf, ∂α(F2πf) = (2πi)|α|F2π(xαf)


(these equalities take the rather common liberty of designating xα := tα := (·)α :y ∈ RN 7→ yα1

1 · . . . · yαNN ).

C It suffices (cf. 10.11.4) to examine only the first row. Since ∂αet = i|α|tαet;therefore,

F (∂αf)(t) =(et ∗ ∂αf

)(0)

=(∂αet ∗ f

)(0) = i|α|tα(et ∗ f)(0) = i|α|tαf(t).

By analogy, on differentiating under the integral sign, infer that

∂

∂t1(Ff)(t) =

∂

∂t1

∫RN

f(x)ei(x,t) dx =∫

RN

f(x)ix1ei(x,t) dx = F (ix1f)(t). .

(9) If fN (x) := exp(−1/2|x|2

)for x ∈ RN , then fN = (2π)

N/2fN .

C It is clear that

fN (t) =N∏k=1

∫R

eitkxke−1/2|x|2k dxk (t ∈ RN ).

Consequently, the matter reduces to the case N = 1. Now, given y ∈ R, observethat

f1(y) =∫R

e−1/2x

2eixy dx =

∫R

e−1/2(x−iy)2−1/2y

2dx

= f1(y)∫R

e−1/2(x−iy)2 dx.

To calculate the integral A that we are interested in, consider (concurrently ori-ented) straight lines λ1 and λ2 parallel to the real axis R in the complex planeCR ' R2. Applying the Cauchy Integral Theorem to the holomorphic functionf(z) := exp

(− z2/2

)(z ∈ C) and a rectangular with vertices on λ1 and λ2 and

properly passing to the limit, conclude that∫λ1f(z) dz =

∫λ2f(z) dz.Whence it fol-

lows that

A =∫R

e−1/2(x−iy)2 dx =

∫R

e−1/2x

2dx =

√2π. .

214 Chapter 10

10.11.6. Definition. The Schwartz space is the set of tempered (or rapidlydecreasing) functions, cf. 10.11.17 (2),

S (RN ) :={f ∈ C∞(RN ) : (∀α, β ∈ (Z+)N ) |x| → +∞⇒ xα∂βf(x)→ 0

}with the multinorm {pα,β : α, β ∈ (Z+)N}, where pα,β(f) := ‖xα∂βf‖∞.

This space is treated as a subspace of the space of all functions from RN to C.

10.11.7. The following statements are valid:

(1) S (RN ) is a Frechet space;

(2) the operators of multiplication by a polynomial and differentiationare continuous endomorphisms of S (RN );

(3) the topology of S (RN )may equivalently be given by the multinorm{pn : n ∈ N}, where

pn(f) :=∑|α|≤n

‖(1 + | · |2)n∂αf‖∞ (f ∈ S (RN ))

(as usual, |x| stands for the Euclidean norm of a vector x in RN );

(4) D(RN ) is dense in S (RN ); furthermore, the identical embeddingof D(RN ) into S (RN ) is continuous and S (RN )′ ⊂ D ′(RN );

(5) S (RN ) ⊂ L1(RN ).

C We will check (4), because the other claims are easier.Take f ∈ S (RN ) and let ψ be a truncator in D(RN ) such that B ⊂ {ψ = 1}.

Given x ∈ RN and ξ > 0, put ψξ(x) := ψ(ξx), and fξ = ψξf. Evidently, fξ ∈ D(RN ).Let ε > 0 and α, β ∈ (Z+)N . It is an easy matter to show that sup{‖∂γ(ψξ−1)‖∞ :γ ≤ β, γ ∈ (Z+)N} < +∞ for 0 < ξ ≤ 1. Considering that xα∂βf(x) → 0 as|x| → +∞, find r > 1 satisfying |xα∂β((ψξ(x) − 1)f(x))| < ε whenever |x| > r.Moreover, fξ(x)− f(x) = (ψ(ξx)− 1)f(x) = 0 for |x| ≤ ξ−1. Therefore,

pα,β(fξ − f) = sup|x|>ξ−1

|xα∂β((ψξ(x)− 1)f(x))|

≤ sup|x|>r

|xα∂β((ψξ(x)− 1)f(x))| < ε

for ξ ≤ r−1. Thus, pα,β(fξ−f)→ 0 as ξ → 0; i.e., fξ → f in S (RN ). The requiredcontinuity of the identical embedding raises no doubt. B


10.11.8. The Fourier transform is a continuous endomorphism of S (RN ).C Given f ∈ D(RN ), from 10.11.5 (8), 10.11.5 (2) and the Holder inequality

obtain‖tαf‖∞ = ‖(∂αf)‖∞ ≤ ‖∂αf‖1 ≤ K‖∂αf‖∞.

Thus,‖tα∂β f‖∞ = ‖tα(xβf)‖∞ ≤ K ′‖∂α(xβf)‖∞.

Whence it is easily seen that f ∈ S (RN ) and the restriction of F to D(Q) withQ b RN is continuous. It remains to refer to 10.10.7 (4) and 10.11.7 (4). B

10.11.9. Theorem. The repeated Fourier transform in the space S (RN ) isproportional to the reflection.

C Let f ∈ S (RN ) and g(x) := fN (x) = exp(−1/2|x|2

). From 10.11.8 and

10.11.7 derive that f , f , g ∈ L1(RN ) and so, by 10.11.5 (7), (fg) = f ˜∗ g. Putgε(x) := g(εx) for x ∈ RN and ε > 0. Then for the same x, in view of 10.11.5 (4)find ∫

RN

g(εt)f(t)et(x) dt =1εN

∫RN

f(y − x)g(yε

)dy =

∫RN

f(εy − x)g(y) dy.

Using 10.11.5 (9) and the Lebesgue Dominated Convergence Theorem as ε → 0,infer that

g(0)∫

RN

f(t)et(x) dt = f(−x)∫

RN

g(y) dy

= (2π)N/2f(x)

∫RN

e−1/2|x|2 dx = (2π)Nf(−x).

Finally, F 2f = (2π)Nf . B

10.11.10. Corollary. F 22π is the reflection and (F2π)−1 = F−2π.

C Given f ∈ S (RN ) and t ∈ RN , deduce that

f(−t) = (2π)N∫

RN

ei(x,t)f(x) dx =∫

RN

e2πi(x,t)f(2πx) dx

= (F2π(F2πf))(t).

Since F2πf˜= F−2πf , the proof is complete. B

216 Chapter 10

10.11.11. Corollary. S (RN ) is a convolution algebra (= an algebra withconvolution as multiplication).

C For f , g ∈ S (RN ) the product fg is an element of S (RN ) and so f g ∈S (RN ). From 10.11.5 (6) infer that F2π(f ∗ g) ∈ S (RN ) and, consequently,by 10.11.10, f ∗ g = F−2π(F2π(f ∗ g)) ∈ S (RN ). B

10.11.12. Inversion Theorem. The Fourier transform F := F2π is a topo-logical automorphism of the Schwartz space S (RN ), with convolution carried intopointwise multiplication. The inverse transform F−1 equals F−2π, with pointwisemultiplication carried into convolution. Moreover, the Parseval identity holds:∫

RN

f g∗ =∫

RN

f g ∗ (f, g ∈ S (RN )).

C In view of 10.11.10 and 10.11.5 (5), only the sought identity needs examining.Moreover, given f and g, from 10.11.5 (7) and 10.11.7 (4), obtain (fg)(0) = (f ∗g)(0). Using 10.11.5 (5), conclude that

∫RN

fg∗ = (F(F−1f)g∗)(0) = ((F−1f)˜∗ Fg∗)(0)

=∫

RN

Ff(Fg∗) dx =∫

RN

FfF∼(g∗) dx =∫

RN

Ff(Fg)∗. .

10.11.13. Remark. In view of 10.11.9, the theorem on the repeated Fouriertransform, the following mutually inverse operators

Ff(t) =1

(2π)N/2

∫RN

f(x)ei(x,t) dx;

F−1f(x) =

1(2π)N/2

∫RN

f(t)e−i(x,t) dt

are considered alongside F. In this case, an analog of 10.11.12 is valid on conditionthat convolution is redefined as f ∗ g := (2π)−

N/2f∗g (f, g ∈ L1(RN )). The meritsof F and F

−1are connected with some simplification of 10.11.5 (8). In the case of F,

a similar goal is achieved by introducing the differential operator Dα := (2πi)−|α|∂αwith α ∈ (Z+)N .


10.11.14. Plancherel Theorem. The Fourier transform in the Schwartzspace S (RN ) is uniquely extendible to an isometric automorphism of L2(RN ).

C Immediate from 10.11.12 and 4.5.10 since S (RN ) is dense in L2(RN ). B

10.11.15. Remark. The extension, guaranteed by 10.11.14, retains the pre-vious name and notation. Rarely (in search of emphasizing distinctions and sub-tleties) one speaks of the Fourier–Plancherel transform or the L2-Fourier transform.It that event it stands to reason to specify the understanding of the integral formulasfor Ff and F−1f with f ∈ L2(RN ) which are treated as the results of appropriatepassage to the limit in L2(RN ).

10.11.16. Definition. Let u ∈ S ′(RN ) := S (RN )′. Such a distributionu is referred to as a tempered distribution (variants: a distribution of slow growth,a slowly increasing distribution, etc.). The space S ′(RN ) of tempered distributionsis furnished with the weak topology σ(S ′(RN ), S (RN )) and is sometimes calledthe Schwartz space (as well as S (RN )).

10.11.17. Examples.

(1) Lp(RN ) ⊂ S ′(RN ) for 1 ≤ p ≤ +∞.

C Let f ∈ Lp(RN ), ψ ∈ S (RN ), p < +∞ and 1/p′ + 1/p = 1. Using Holderinequality, for suitable positive K, K ′, and K ′′ successively infer that

‖ψ‖p′ ≤

∫B

|ψ|p

1/p

+

∫RN\B

∣∣(1 + |x|2)N (1 + |x|2)−Nψ(x)∣∣p dx

1/p

≤ K ′‖ψ‖∞ + ‖(1 + | · |2)Nψ‖∞

∫RN\B

dx

(1 + |x|2)Np

1/p

≤ K ′′p1(ψ).

Once again using the Holder inequality, observe that

|uf (ψ)| = |〈ψ | f〉| =

∣∣∣∣∣∣∫

RN

fψ

∣∣∣∣∣∣ ≤ ‖f‖p ‖ψ‖p′ ≤ Kp1(ψ).

The case p = +∞ raises no doubts. B

(2) S (RN ) is dense in S ′(RN ).

C Follows from 10.11.7 (4), 10.11.17 (1), 10.11.7 (5), and 10.10.9 (4). B

218 Chapter 10

(3) Let µ ∈M (RN ) be a tempered Radon measure; i.e.,∫RN

d|µ|(x)(1 + |x|2)n

< +∞

for some n ∈ N. Evidently, µ is a tempered distribution.(4) If u ∈ S ′(RN ), f ∈ S (RN ) and α ∈ (Z+)N then fu ∈ S ′(RN )

and ∂αu ∈ S ′(RN ) in virtue of 10.11.7 (2). By a similar argument, puttingDαu(f) := (−1)|α|uDαf for f ∈ S (RN ), infer that Dαu ∈ S ′(RN ) and Dαu =(2πi)−|α|∂αu.

(5) Every compactly-supported distribution is tempered.C In accordance with 10.10.5 (7) such u in D ′(RN ) may be identified with

a member of E ′(RN ). Since the topology of S (RN ) is stronger than that inducedby the identical embedding in C∞(RN ), conclude that u ∈ S ′(RN ). B

(6) Let u ∈ S ′(RN ). If f ∈ S (RN ) then u convolutes with f andu ∗ f ∈ S (RN ). In may be shown that u also convolutes with every distribution v,a member of E ′(RN ), and u ∗ v ∈ S ′(RN ).

(7) Take u ∈ D ′(RN ) and x ∈ RN . Let τxu := (τ−x)′u = u ◦ τ−x bethe corresponding translation of u. A distribution u is periodic (with period x) ifτxu = u. Every periodic distribution is tempered. Periodicity is preserved underdifferentiation and convolution.

(8) If un ∈ S ′(RN ) (u ∈ N) and for every f ∈ S (RN ) there is a sumu(f) :=

∑∞n=1 un(f), then u ∈ S ′(RN ) and ∂αu =

∑∞n=1 ∂

αun (cf. 10.10.10).

10.11.18. Theorem. Each tempered distribution is the sum of derivativesof tempered measures.

C Let u ∈ S ′(RN ). On account of 10.11.7 (3) and 5.3.7, for some n ∈ Nand K > 0, observe that

|u(f)| ≤ K∑|α|≤n

∥∥(1 + | · |2)n∂αf∥∥∞ (f ∈ S (RN )).

From 3.5.3 and 3.5.7, for some µα ∈M(RN ), obtain

u(f) =∑|α|≤n

µα((1 + | · |2)n∂αf

)(f ∈ S (RN )).

Let να := (−1)|α|(1 + | · |2)nµα. Then να is tempered and u =∑|α|≤n ∂

ανα. .


10.11.19. Definition. For u ∈ S ′(RN ), the distribution Fu acting as

〈f |Fu〉 = 〈Ff |u〉 (f ∈ S (RN ))

is the Fourier transform or, amply, the Fourier–Schwartz transform of u.

10.11.20. Theorem. The Fourier–Schwartz transform F is a unique extensionof the Fourier transform in S (RN ) to a topological automorphism of S ′(RN ). Theinverse F−1 of F is a unique continuous extension of the inverse Fourier transformin S (RN ).

C The Fourier–Schwartz transform in S ′(RN ) is the dual of the Fourier trans-form in S (RN ). It remains only to appeal to 10.11.7 (5), 10.11.12, 10.11.17 (2),and 4.5.10. B

Exercises

10.1. Give examples of linear topological spaces and locally convex spaces as well as con-structions leading to them.

10.2. Prove that a Hausdorff topological vector space is finite-dimensional if and only if it islocally compact.

10.3. Characterize weakly continuous sublinear functionals.10.4. Prove that the weak topology of a locally convex space is normable or metrizable

if and only if the space is finite-dimensional.

10.5. Describe weak convergence in classical Banach spaces.10.6. Prove that a normed space is finite-dimensional if and only if its unit sphere, compris-

ing all norm-one vectors, is weakly closed.

10.7. Assume that an operator T carries each weakly convergent net into a norm convergentnet. Prove that T has finite rank.

10.8. Let X and Y be Banach spaces and let T be a linear operator from X to Y . Provethat T is bounded if and only if T is weakly continuous (i.e., continuous as a mapping from(X, σ(X, X′)) to (Y, σ(Y, Y ′))).

10.9. Let ‖ · ‖1 and ‖ · ‖2 be two norms making X into a Banach space. Assume furtherthat (X, ‖ · ‖1)′ ∩ (X, ‖ · ‖2)′ separates the points of X. Prove that these norms are equivalent.

10.10. Let S act from Y ′ to X′. When does S serve as the dual of some operator from Xto Y ?

10.11. What is the Mackey topology τ(X, X#)?10.12. Let (Xξ)ξ∈� be a family of locally convex spaces, and let X :=

∏ξ∈�Xξ be the prod-

uct of the family. Validate the presentations:

σ(X, X′) =∏ξ∈�

σ(Xξ, X′ξ); τ(X, X′) =∏ξ∈�

τ(Xξ, X′ξ).

10.13. Let X and Y be Banach spaces and let T be a element of B(X, Y ) satisfyingimT = Y . Demonstrate that Y is reflexive provided so is X.

220 Chapter 10

10.14. Show that the spaces (X′)′′ and (X′′)′ coincide.10.15. Prove that the space c0 has no infinite-dimensional reflexive subspaces.10.16. Let p be a continuous sublinear functional on Y , and let T ∈ L (X, Y ) be a contin-

uous linear operator. Establish the following inclusion of the sets of extreme points: extT ′(∂p) ⊂T ′(ext ∂p).

10.17. Let p be a continuous seminorm on X and let X be a subspace of X. Prove thatf ∈ ext(X ◦ ∩ ∂p) if and only if the next equality holds:

X = clX + {p− f ≤ 1} − {p− f ≤ 1}.

10.18. Prove that the absolutely convex hull of a totally bounded subset of a locally convexspace is also totally bounded.

10.19. Establish that barreledness is preserved under passage to the inductive limit. Whathappens with other linear topological properties?

Chapter 11

Banach Algebras

11.1. The Canonical Operator Representation

11.1.1. Definition. An element e of an algebra A is called a unity elementif e 6= 0 and ea = ae = a for all a ∈ A. Such an element is obviously unique and isalso referred to as the unity or the identity or the unit of A. An algebra A is unitalprovided that A has unity.

11.1.2. Remark. Without further specification, we only consider unital alge-bras over a basic field F. Moreover, for simplicity, it is assumed that F := C, unlessstated otherwise. In studying a representation of unital algebras, we naturallypresume that it preserves unity. In other words, given some algebras A1 and A2,by a representation of A1 in A2 we henceforth mean a morphism, a multiplicativelinear operator, from A1 to A2 which sends the unity element of A1 to the unityelement of A2.

For an algebra A without unity, the process of unitization or adjunction of unityis in order. Namely, the vector space Ae := A × C is transformed into an algebraby putting (a, λ)(b, µ) := (ab+µa+λb, λµ), where a, b ∈ A and λ, µ ∈ C. In thenormed case, it is also taken for granted that ‖(a, λ)‖Ae := ‖a‖A + |λ|.

11.1.3. Definition. An element ar in A is a right inverse of a if aar = e.An element al of A is a left inverse of a if ala = e.

11.1.4. If an element has left and right inverses then the latter coincide.

/ ar = (ala)ar = al(aar) = ale = al .

11.1.5. Definition. An element a of an algebra A is called invertible, inwriting a ∈ Inv(A), if a has a left and right inverse. Denote a−1 := ar = al. Theelement a−1 is the inverse of a. A subalgebra (with unity) B of an algebra A iscalled pure or full or inverse-closed in A if Inv(B) = Inv(A) ∩B.

222 Chapter 11

11.1.6. Theorem. Let A be a Banach algebra. Given a ∈ A, put La : x 7→ ax(x ∈ A). Then the mapping

LA := L : a 7→ La (a ∈ A)

is a faithful operator representation. Moreover, L(A) is a pure closed subalgebraof B(A) and L : A→ L(A) is a topological isomorphism.C Clearly,

L(ab) : x 7→ Lab(x) = abx = a(bx) = La(Lbx) = (La)(Lb)x

for x, a, b ∈ A; i.e., L is a representation (because the linearity of L is obvious).If La = 0 then 0 = La(e) = ae = a, so that L is a faithful representation. To provethe closure property of the range L(A), consider the algebra Ar coinciding with Aas a vector space and equipped with the reversed multiplication ab := ba (a, b ∈ A).

Let R := LAr , i.e. Ra := Ra : x 7→ xa for a ∈ A. Check that L(A) is in factthe centralizer of the range R(A), i.e. the closed subalgebra

Z(imR) := {T ∈ B(A) : TRa = RaT (a ∈ A)}.

Indeed, if T ∈ L(A), i.e. T = La for some a ∈ A, then LaRb(x) = axb =Rb(La(x)) = RbLa(x) for all b ∈ A. Hence, T ∈ Z(R(A)). If, in turn, T ∈ Z(R(A))then, putting a := Te, find

Lax = ax = (Te)x = Rx(Te) = (RxT )e = (TRx)e = T (Rxe) = Tx

for all x ∈ A. Consequently, T = La ∈ L(A). Thus, L(A) is a Banach subalgebraof B(A).

For T = La there is a T−1 in B(A). Put b := T−1e and observe that ab =Lab = Tb = TT−1e = e. Moreover, ab = e ⇒ aba = a ⇒ T (ba) = Laba = aba =a = Lae = Te. Whence ba = e, because T is a monomorphism. Thus, L(A) isa subalgebra of A.

By the definition of a Banach algebra, the norm is submultiplicative, providing

‖L‖ = sup{‖La‖ : ‖a‖ ≤ 1} = sup{‖ab‖ : ‖a‖ ≤ 1, ‖b‖ ≤ 1} ≤ 1.

Using the Banach Isomorphism Theorem, conclude that L is a topological isomor-phism (i.e., L−1 is a continuous operator from L(A) onto A). B

11.1.7. Definition. The representation LA, constructed in 11.1.6, is thecanonical operator representation of A.

Banach Algebras 223

11.1.8.Remark. The presence of the canonical operator representation allowsus to confine the subsequent exposition to studying Banach algebras with norm-oneunity.

For such an algebra A the canonical operator representation LA implementsan isometric embedding of A into the endomorphism algebra B(A) or, in short, anisometric representation of A in B(A). In this case, LA implements an isometricisomorphism between the algebras A and L(A). The same natural terminology isused for studying representations of arbitrary Banach algebras. Observe immedi-ately that the canonical operator representation of A, in particular, justifies theuse of the symbol λ in place of λe for λ ∈ C, where e is the unity of A (cf. 5.6.5).In other words, henceforth the isometric representation λ 7→ λe is considered as theidentification of C with the subalgebra Ce of A.

11.2. The Spectrum of an Element of an Algebra

11.2.1. Definition. Let A be a Banach algebra and a ∈ A. A scalar λ in Cis a resolvent value of a, in writing λ ∈ res(a)), if (λ− a) ∈ Inv(A). The resolventR(a, λ) of a at λ is R(a, λ) := 1

λ−a := (λ − a)−1. The set Sp(a) := C \ res(a) isthe spectrum of a, with a point of Sp(a) a spectral value of a. When it is necessary,more detailed designations like SpA(a) are in order.

11.2.2. For a ∈ A the equalities hold:

SpA(a) = SpL(A)(La) = Sp(La);

LR(a, λ) = R(La, λ) (λ ∈ res(a) = res(La)). /.

11.2.3. Gelfand–Mazur Theorem. The field of complex numbers is up toisometric isomorphism the sole Banach division algebra (or skew field); i.e., eachcomplex Banach algebra with norm-one unity and invertible nonzero elements hasan isometric representation on C.

C Consider � : λ 7→ λe, with e the unity of A and λ ∈ C. It is clear that� represents C in A. Take a ∈ A. By virtue of 11.2.2 and 8.1.11, Sp(a) 6= ∅.Consequently, there is λ ∈ C such that the element (λ − a) is not invertible; i.e.,a = λe by hypothesis. Hence, � is an epimorphism. Moreover, ‖�(λ)‖ = ‖λe‖ =|λ| ‖e‖ = |λ| so that � is an isometry. B

11.2.4. Shilov Theorem. Let A be a Banach algebra and let B be a closed(unital) subalgebra of A. Then

SpB(b) ⊃ SpA(b), ∂ SpA(b) ⊃ ∂ SpB(b)

for all b ∈ B.

224 Chapter 11

C If b := λ − b ∈ Inv(B) then surely b ∈ Inv(A). Whence resB(b) ⊂ resA(b);i.e.,

SpB(b) = C \ resB(b) ⊃ C \ resA(b) = SpA(b).

If λ ∈ ∂ SpB(b) then b ∈ ∂ Inv(B). Therefore, there is a sequence (bn), bn ∈Inv(B), convergent to b. Putting t := supn∈N

∥∥b−1n

∥∥, deduce that∥∥b−1n − b−1

m

∥∥ =∥∥b−1n (1− bnb−1

m )∥∥ =

∥∥b−1n (bm − bn)b−1

m

∥∥ ≤ t2∥∥bn − bm∥∥.In other words, if t < +∞ then there is a limit a := lim b−1

n in B. Multiplicationis jointly continuous, and so ab = ba = 1; i.e., b ∈ Inv(B). Since Inv(B) is openby the Banach Inversion Stability Theorem and 11.1.6, arrive at a contradictionto the containment b ∈ ∂ Inv(B).

Therefore, it may be assumed (on dropping to a subsequence, if need be) that∥∥b−1n

∥∥→ +∞. Put an :=∥∥b−1n

∥∥−1b−1n . Then∥∥ban∥∥ =∥∥(b− bn)an + bnan

∥∥≤∥∥b− bn∥∥ ‖an‖+ ∥∥b−1

n

∥∥−1∥∥bnb−1n

∥∥→ 0.

Whence it follows that b is not invertible. Indeed, in the opposite case for a := b−1

it would hold that

1 = ‖an‖ =∥∥aban∥∥ ≤ ‖a‖ ∥∥ban∥∥→ 0.

Finally, conclude that λ− b does not belong to Inv(A); i.e., λ ∈ SpA(b). Sinceλ is a boundary point of a larger set SpB(b); undoubtedly, λ ∈ ∂ SpA(b). B

11.2.5. Corollary. If SpB(b) lacks interior points then SpB(b) = SpA(b)./ SpB(b) = ∂ SpB(b) ⊂ ∂ SpB(b) ⊂ ∂ SpA(b) ⊂ SpA(b) ⊂ SpB(b) .

11.2.6. Remark. The Shilov Theorem is often referred to as the UnremovableSpectral Boundary Theorem or Spectral Permanence Theorem and verbalized asfollows: “A boundary spectral value is unremovable.”

11.3. The Holomorphic Functional Calculus in Algebras

11.3.1. Definition. Let a be an element of a Banach algebra A, and leth ∈H (Sp(a)) be a germ of a holomorphic function on the spectrum of a. Put

Rah :=12πi

∮h(z)z − a

dz.

The element Rah of A is the Riesz–Dunford integral of h. If, in particular, f ∈H(Sp(a)) is a function holomorphic in a neighborhood about the spectrum of a,then f(a) := Raf .

Banach Algebras 225

11.3.2. Gelfand–Dunford The Riesz–Dunford integral Ra represents the al-gebra of germs of holomorphic functions on the spectrum of an element a of a Banachalgebra A in A. Moreover, if f(z) :=

∑∞n=0 cnz

n (in a neighborhood of Sp(a)) thenf(a) :=

∑∞n=0 cna

n.

From 11.2.3, 8.2.1, and 11.2.2 obtain

(LRah)(b) = LRahb = (Rah)b

=12πi

∮h(z)R(a, z) dzb =

12πi

∮h(z)R(a, z) bdz

=12πi

∮h(z)R(La, z) bdz =

12πi

∮h(z)R(La, z) dzb

= RLah(b)for all b ∈ A. In particular, imL includes the range of RLa(H (Sp(a))). Therefore,the already-proven commutativity of the diagram

H (Sp(a))

B(A) A

Ra

L

RLa

�?

@@@@@R

implies that the following diagram also commutes:

H (Sp(a))

L(A) A

Ra

L−1

RLa

-?

@@@@@R

It remains to appeal to 11.1.6 and the Gelfand–Dunford Theorem in an operatorsetting. B

11.3.3. Remark. All that we have established enables us to use in the sequelthe rules of the holomorphic functional calculus which were exposed in 8.2 for theendomorphism algebra B(X), with X a Banach space.

226 Chapter 11

11.4. Ideals of Commutative Algebras

11.4.1. Definition. Let A be a commutative algebra. A subspace J of A isan ideal of A, in writing J C A, provided that AJ ⊂ J .

11.4.2. The set J(A) of all ideals of A, ordered by inclusion, is a completelattice. Moreover,

supJ(A) E = supLat(A) E , infJ(A) E = infLat(A) E ,

for every subset E of J(A); i.e., J(A) is embedded into the complete lattice Lat(A)of all subspaces of A with preservation of suprema and infima of arbitrary subsets.

C Clearly, 0 is the least ideal, whereas A is the greatest ideal. Furthermore,the intersection of ideals and the sum of finitely many ideals are ideals. It remainsto refer to 2.1.5 and 2.1.6. B

11.4.3. Let J0 C A. Assume further that ϕ : A→ A/J0 is the coset mappingof A onto the quotient algebra A := A/J0. Then

J C A⇒ ϕ(J) C A; J C A⇒ ϕ−1(J) C A.

C Since by definition ab := ϕ(ϕ−1(a)ϕ−1(b)) for a, b ∈ A, the operator ϕ ismultiplicative: ϕ(ab) = ϕ(a)ϕ(b) for a, b ∈ A. Whence successively derive

ϕ(J) ⊂ Aϕ(J) = ϕ(A)ϕ(J) ⊂ ϕ(AJ) ⊂ ϕ(J);ϕ−1(J) ⊂ Aϕ−1(J) ⊂ ϕ−1(ϕ(A)J) = ϕ−1(AJ) ⊂ ϕ−1(J). .

11.4.4. Let J C A and J 6= 0. The following conditions are equivalent:

(1) A 6= J ;

(2) 1 /∈ J ;(3) no element of J has a left inverse. CB

11.4.5. Definition. An ideal J of A is called proper if J is other than A.A maximal element of the set of proper ideals (ordered by inclusion) is a maximalideal.

11.4.6. A commutative algebra is a division algebra if and only if it has noproper ideals other than zero. CB

11.4.7. Let J be a proper ideal of A. Then (J is maximal)⇔ (A/J is a field).

Banach Algebras 227

C ⇒: Let J C A/J . Then, by 11.4.3, ϕ−1(J) C A. Since, beyond a doubt,J ⊂ ϕ−1(J); therefore, either J = ϕ−1(J) and 0 = ϕ(J) = ϕ(ϕ−1(J)) = J , orA = ϕ−1(J) and J = ϕ(ϕ−1(J)) = ϕ(A) = A/J in virtue of 1.1.6. Consequently,A/J has no proper ideals other than zero. It remains to refer to 11.4.6.⇐: Let J0 C A and J0 ⊂ J . Then, by 11.4.3, ϕ(J0) C A/J . In virtue of 11.4.6,

either ϕ(J0) = 0 or ϕ(J0) = A/J . In the first case, J0 ⊂ ϕ−1 ◦ϕ(J0) ⊂ ϕ−1(0) = Jand J = J0. In the second case, ϕ(J0) = ϕ(A); i.e., A = J0+J ⊂ J0+J0 = J0 ⊂ A.Thus, J is a maximal ideal. B

11.4.8.Krull Theorem. Each proper ideal is included in some maximal ideal.C Let J0 be a proper ideal of an algebra A. Assume further that E is the set

comprising all proper ideals J of A such that J0 ⊂ J . In virtue of 11.4.2 each chainE0 in E has a least upper bound: supE = ∪{J : J ∈ E0}. By 11.4.4 the ideal supE0is proper. Thus, E is inductive and the claim follows from the Kuratowski–ZornLemma. B

11.5. Ideals of the Algebra C(Q, C)

11.5.1. Minimal Ideal Theorem. Let J be an ideal of the algebra C(Q, C)of complex continuous functions on a compactum Q. Assume further that

Q0 := ∩{f−1(0) : f ∈ J};J0 := {f ∈ C(Q, C) : int f−1(0) ⊃ Q0}.

Then J0 C C(Q, C) and J0 ⊂ J .C Let Q1 := cl(Q \ f−1(0)) for a function f , a member of J0. By hypothesis,

Q1 ∩ Q0 = ∅. To prove the containment f ∈ J it is necessary (and, certainly,sufficient) to find u ∈ J satisfying u(q) = 1 for all q ∈ Q1. Indeed, in that eventuf = f .

With this in mind, observe first that for q ∈ Q1 there is a function fq in J suchthat fq(q) 6= 0. Putting gq := f∗q fq, where as usual f∗q : x 7→ fq(x)∗ is the conjugateof fq, observe that gq ≥ 0 and, moreover, gq(q) > 0. It is also clear that gq ∈ J forq ∈ Q1. The family (Uq)q∈Q1 , with Uq := {x ∈ Q1 : gq(x) > 0}, is an open coverof Q1. Using a standard compactness argument, choose a finite subset {q1, . . . , qn}of Q1 such that Q1 ⊂ Uq1 ∪ . . .∪Uqn . Put g := gq1 + . . .+ gqn . Undoubtedly, g ∈ Jand g(q) > 0 for q ∈ Q1. Let h0(q) := g(q)−1 for q ∈ Q1. By the Tietze–UrysohnTheorem, there is a function h in C(Q, R) satisfying h

∣∣Q1

= h0. Finally, let u := hg.This u is a sought function.

We have thus demonstrated that J0 ⊂ J . Moreover, J0 is an ideal of C(Q, C)for obvious reasons. B

228 Chapter 11

11.5.2. For every closed ideal J of the algebra C(Q, C) there is a uniquecompact subset Q0 of Q such that

J = J(Q0) := {f ∈ C(Q, C) : q ∈ Q0 ⇒ f(q) = 0}.

C Uniqueness follows from the Urysohn Theorem. Define Q0 as in 11.5.1.Then, surely, J ⊂ J(Q0). Take f ∈ J(Q0) and, given n ∈ N, put

Un :={|f | ≤ 1/2n

}, Vn :=

{|f | ≥ 1/n

}.

Once again using the Urysohn Theorem, find hn ∈ C(Q, R) satisfying 0 ≤ hn ≤ 1with hn

∣∣Un

= 0 and hn∣∣Vn

= 1. Consider fn := fhn. Since

int f−1n (0) ⊃ intUn ⊃ Q0,

therefore, from 11.5.1 derive fn ∈ J . It suffices to observe that fn → f by con-struction. B

11.5.3. Maximal Ideal Theorem. A maximal ideal of C(Q, C) has theform

J(q) := J({q}) = {f ∈ C(Q, C) : f(q) = 0},

with q a point of Q.C Follows from 11.5.2, because the closure of an ideal is also an ideal. B

11.6. The Gelfand Transform

11.6.1. Let A be a commutative Banach algebra, and let J be a closed ideal ofA other than A. Then the quotient algebra A/J , endowed with the quotient norm,is a Banach algebra. If ϕ : A → A/J is the coset mapping then ϕ(1) is the unityof A/J , the operator ϕ is multiplicative and ‖ϕ‖ = 1.C Given a, b ∈ A, from 5.1.10 (5) derive

‖ϕ(a)ϕ(b)‖A/J = inf{‖a′b′‖A : ϕ(a′) = ϕ(a), ϕ(b′) = ϕ(b)}≤ inf{‖a′‖A‖b′‖A : ϕ(a′) = ϕ(a), ϕ(b′) = ϕ(b)}

= ‖ϕ(a)‖A/J‖ϕ(b)‖A/J .

In other words, the norm of A/J is submultiplicative. Consequently, ‖ϕ(1)‖ ≥ 1.Furthermore,

‖ϕ(1)‖A/J = inf{‖a‖A : ϕ(a) = ϕ(1)} ≤ ‖1‖A = 1;

i.e., ‖ϕ(1)‖ = 1. Whence, in particular, the equality ‖ϕ‖ = 1 follows. The remainingclaims are evident. B

Banach Algebras 229

11.6.2. Remark. The message of 11.6.1 remains valid for a noncommutativeBanach algebra A under the additional assumption that J is a bilateral ideal of A;i.e., J is a subspace of A satisfying the condition AJA ⊂ J .

11.6.3. Let χ : A → C be a nonzero multiplicative linear functional on A.Then χ is continuous and ‖χ‖ = χ(1) = 1 (in particular, χ is a representation of Ain C).

C Since χ 6= 0; therefore, 0 6= χ(a) = χ(a1) = χ(a)χ(1) for some a in A.Consequently, χ(1) = 1. If now a in A and λ in C are such that |λ| > ‖a‖, thenλ−a ∈ Inv(A) (cf. 5.6.15). So, 1 = χ(1)χ(λ−a)χ((λ−a)−1). Whence χ(λ−a) 6= 0;i.e., χ(a) 6= λ. Thus, |χ(a)| ≤ ‖a‖ and ‖χ‖ ≤ 1. Since ‖χ‖ = ‖χ‖ ‖1‖ ≥ |χ(1)| = 1,conclude that ‖χ‖ = 1. B

11.6.4. Definition. A nonzero multiplicative linear functional on an alge-bra A is a character of A. The set of all characters of A is denoted by X(A), fur-nished with the topology of pointwise convergence (induced in X(A) by the weaktopology σ(A′, A)) and called the character space of A.

11.6.5. The character space is a compactum.

C It is beyond a doubt that X(A) is a Hausdorff space. By virtue of 11.6.3,X(A) is a σ(A′, A)-closed subset of the ball BA′ . The latter is σ(A′, A)-compactby the Alaoglu–Bourbaki Theorem. It remains to refer to 9.4.9. B

11.6.6. Ideal and Character Theorem. Each maximal ideal of a commu-tative Banach algebra A is the kernel of a character of A. Moreover, the mappingχ 7→ kerχ from the character space X(A) onto the set M(A) of all maximal idealsof A is one-to-one.

C Let χ ∈ X(A) be a character of A. Clearly, kerχ C A. From 2.3.11 it followsthat the monoquotient χ : A/ kerχ→ C of χ is a monomorphism. In view of 11.6.1,χ(1) = χ(1) = 1; i.e., χ is an isomorphism of A/ kerχ and C. Consequently,A/ kerχ is a field. Using 11.4.7, infer that kerχ is a maximal ideal; i.e., kerχ ∈M(A). Now, let m ∈ M(A) be some maximal ideal of A. It is clear that m ⊂ clm,clm C A, and 1 /∈ clm (because 1 ∈ Inv(A), and the last set is open by theBanach Inversion Stability Theorem and 11.1.6). Therefore, the ideal m is closed.Consider the quotient algebra A/m and the coset mapping ϕ : A→ A/m. In viewof 11.4.7 and 11.6.1, A/m is a Banach field. By the Gelfand–Mazur Theorem, thereis an isometric representation ψ : A/m → C. Put χ := ψ ◦ ϕ. It is evident thatχ ∈ X(A) and kerχ = χ−1(0) = ϕ−1(ψ−1(0)) = ϕ−1(0) = m.

To complete the proof, it suffices to show that the mapping χ 7→ kerχ is one-to-one. Indeed, let kerχ1 = kerχ2 for χ1, χ2 ∈ X(A). By 2.3.11 χ1 = λχ2 for someλ ∈ C. Furthermore, by 11.6.3, 1 = χ1(1) = λχ2(1) = λ. Finally, χ1 = χ2. B

230 Chapter 11

11.6.7. Remark. Theorem 11.6.6 makes it natural to furnish M(A) with theinverse image topology translated from X(A) to M(A) by the mapping χ 7→ kerχ.In this regard, M(A) is referred to as the compactmaximal ideal space of A. In otherwords, the character space and the maximal ideal space are often identified alongthe lines of 11.6.6.

11.6.8. Definition. Let A be a commutative Banach algebra and let X(A)be the character space of A. Given a ∈ A and χ ∈ X(A), put a(χ) := χ(a). Theresulting function a : χ 7→ a(χ), defined on X(A), is the Gelfand transform of a.The mapping a 7→ a, with a ∈ A is the Gelfand transform of A, denoted by GA(or ).

11.6.9. Gelfand Transform Theorem. The Gelfand transform GA : a 7→ ais a representation of a commutative Banach algebra A in the algebra C(X(A), C).Moreover,

Sp(a) = Sp(a) = a(X(A)),‖a‖ = r(a),

with a ∈ A and r(a) standing for the spectral radius of a.C The implications a ∈ A ⇒ a ∈ C(X(A), C), 1 = 1 and a, b ∈ A ⇒

ab = ab follow from definitions and 11.6.3. The linearity of GA raises no doubts.Consequently, the mapping GA is a representation.

To begin with, take λ ∈ Sp(a). Then λ − a is not invertible, and so the idealJλ−a := A(λ − a) is proper in virtue of 11.4.4. By the Krull Theorem, there isa maximal ideal m of A satisfying the condition m ⊃ Jλ−a. By Theorem 11.6.6,m = kerχ for a suitable character χ. In particular, χ(λ− a) = 0; i.e., λ = λχ(1) =χ(λ) = χ(a) = a(χ). Consequently, λ ∈ Sp(a).

If, in turn, λ ∈ Sp(a) then (λ− a) is not invertible in C(X(A), C); i.e., thereis a character χ ∈ X(A) such that λ = a(χ). In other words, χ(λ − a) = 0. Thus,the assumption λ− a ∈ Inv(A) leads to the following contradiction:

1 = χ(1) = χ((λ− a)−1(λ− a)) = χ((λ− a)−1)χ(λ− a) = 0.

Hence, λ ∈ Sp(a). Finally, Sp(a) = Sp(a).Using the Beurling–Gelfand formula (cf. 11.3.3 and 8.1.12), infer that

r(a) = sup{|λ| : λ ∈ Sp(a)} = sup{|λ| : λ ∈ Sp(a)}= sup{|λ| : λ ∈ a(X(A))} = sup{|a(χ)| : χ ∈ X(A)} = ‖a‖,


Banach Algebras 231

11.6.10. The Gelfand transform of a commutative Banach algebra A is an iso-metric embedding if and only if ‖a2‖ = ‖a‖2 for all a ∈ A.C ⇒: The mapping t 7→ t2, viewed as acting on R+, and the inverse of the

mapping on R+ are both increasing. Therefore, from 10.6.9 obtain

‖a2‖ = ‖a2‖C(X(A),C) = supχ∈X(A)

|a2(χ)| = supχ∈X(A)

|χ(a2)|

= supχ∈X(A)

|χ(a)χ(a)| = supχ∈X(A)

|χ(a)|2

=

(sup

χ∈X(A)|χ(a)|

)2

= ‖a‖2 = ‖a‖2.

⇐: By the Gelfand formula, r(a) = lim ‖an‖1/n . In particular, observe that∥∥a2n∥∥ = ‖a‖2n ; i.e., r(a) = ‖a‖. By 10.6.9, r(a) = ‖a‖, completing the proof. B

11.6.11. Remark. It is interesting sometimes to grasp situations in whichthe Gelfand transform of A is faithful but possibly not isometric. The kernel of theGelfand transform GA is the intersection of all maximal ideals, called the radi-cal of A. Therefore, the condition for GA to be a faithful representation of Ain C(X(A), C) reads: “A is semisimple” or, which is the same, “The radical of A istrivial.”

11.6.12. Theorem. For an element a of a commutative Banach algebra A thefollowing diagram of representations commutes:

H (Sp(a)) = H (Sp(a))

A C(X(A),C)

Ra

GA

Ra

-?

@@@@@R

Moreover, f(a) = f ◦ a = f(a) for f ∈ H(Sp(a)).C Take χ ∈ X(A). Given z ∈ res(a), observe that

χ

(1

z − a(z − a)

)= 1⇒ χ

(1

z − a

)=

1χ(z − a)

=1

z − χ(a).

In other words,

R(a, z)(χ) =1

z − a(χ) =

1z − a(χ)

=1

z − a(χ) = R(a, z)(χ).

232 Chapter 11

Therefore, appealing to the properties of the Bochner integral (cf. 5.5.9 (6)) andgiven f ∈ H(Sp(a)), infer that

f(a) = GA ◦Raf = GA

(12πi

∮f(z)R(a, z) dz

)=

12πi

∮f(z)GA(R(a, z)) dz =

12πi

∮f(z)R(a, z)) dz

=12πi

∮f(z)R(a, z) dz = Ra(f) = f(a).

Furthermore, given χ ∈ X(A) and using the Cauchy Integral Formula, derive thefollowing chain of equalities

f ◦ a(χ) = f(a(χ)) = f(χ(a))

=12πi

∮f(z)

z − χ(a)dz =

12πi

∮χ

(f(z)z − a

)dz

=12πi

χ

(∮f(z)z − a

dz

)= f(a)(χ) = f(a)(χ). .

11.6.13. Remark. The theory of the Gelfand transform may be naturallygeneralized to the case of a commutative Banach algebra A without unity. RetainDefinitions 11.6.4 and 11.6.8 verbatim. A character χ in X(A) generates somecharacter χe in X(Ae) by the rule χe(a, λ) := χ(a) + λ (a ∈ A, λ ∈ C). The setX(Ae)\{χe : χ ∈ X(A)} is a singleton consisting of the sole element χ∞(a, λ) := λ(a ∈ A, λ ∈ C). The space X(A) is locally compact (cf. 9.4.19), because themapping χ ∈ X(A) 7→ χe ∈ X(Ae)\{χ∞} is a homeomorphism. Moreover, kerχ∞ =A × 0. Consequently, the Gelfand transform of a commutative Banach algebrawithout unity represents it in the algebra of continuous complex functions definedon a locally compact space and vanishing at infinity. Given the group algebra(L1(RN ), ∗), observe that by 10.11.1 and 10.11.3 the Fourier transform coincideswith the Gelfand transform, which in turn entails the Riemann–Lebesgue Lemmaas well as the multiplication formula 10.11.5 (6).

11.7. The Spectrum of an Element of a C∗-Algebra

11.7.1. Definition. An element a of an involutive algebra A is called hermit-ian if a∗ = a. An element a of A is called normal if a∗a = aa∗. Finally, an elementa is called unitary if aa∗ = a∗a = 1 (i.e. a, a∗ ∈ Inv(A) with a−1 = a∗ anda∗−1 = a).

Banach Algebras 233

11.7.2. Hermitian elements of an involutive algebra A compose a real subspaceof A. Moreover, for every a in A there are unique hermitian elements x, y ∈ Asuch that a = x+ iy. Namely,

x =12(a+ a∗), y =

12i(a− a∗).

Moreover, a∗ = x− iy.C Only the claim of uniqueness needs examining. If a = x1 + iy1 then, using

the properties of involution (cf. 6.4.13), proceed as follows: a∗ = x∗1 + (iy1)∗ =x∗1 − iy∗1 = x1 − iy1. Thus, x1 = x and y1 = y. B

11.7.3. The unity of A is a hermitian element of A./ 1∗ = 1∗1 = 1∗1∗∗ = (1∗1)∗ = 1∗∗ = 1 .

11.7.4. a ∈ Inv(A) ⇔ a∗ ∈ Inv(A). Moreover, involution and inversion com-mute.

C For a ∈ Inv(A) by definition aa−1 = a−1a = 1. Consequently, a−1∗a∗ =a∗a−1∗ = 1∗. Using 11.7.3, infer that a∗ ∈ Inv(A) and a∗−1 = a−1∗. Repeating thisargument for a := a∗, complete the proof. B

11.7.5. Sp(a∗) = Sp(a)∗. CB11.7.6. The spectrum of a unitary element of a C∗-algebra is a subset of the

unit circle.C By Definition 6.4.13, ‖a2‖ = ‖a∗a‖ ≤ ‖a∗‖ ‖a‖ for an arbitrary a. In other

words, ‖a‖ ≤ ‖a∗‖. Therefore, from a = a∗∗ infer that ‖a‖ = ‖a∗‖. If now a isa unitary element, a∗ = a−1; then ‖a‖2 = ‖a∗a‖ = ‖a−1a‖ = 1. Consequently,‖a‖ = ‖a∗‖ = ‖a−1‖ = 1. Whence it follows that Sp(a) and Sp(a−1) both liewithin the unit disk. Furthermore, Sp(a−1) = Sp(a)−1. B

11.7.7. The spectrum of a hermitian element of a C∗-algebra is real.C Take a in A arbitrarily. From the Gelfand–Dunford Theorem in an algebraic

setting derive

exp(a)∗ =

( ∞∑n=0

an

n!

)∗=∞∑n=0

(an)∗

n!=∞∑n=0

(a∗)n

n!= exp(a∗).

If now h = h∗ is a hermitian element of A then, applying the holomorphicfunctional calculus to a := exp(ih), deduce that

a∗ = exp(ih)∗ = exp((ih)∗) = exp(−ih∗) = exp(−ih) = a−1.

Consequently, a is a unitary element of A, and by 11.7.6 the spectrum Sp(a) of A isa subset of the unit circle T. If λ ∈ Sp(h) then by the Spectral Mapping Theorem(also cf. 11.3.3) exp(iλ) ∈ Sp(a) ⊂ T. Thus, 1 = | exp(iλ)| = | exp(iReλ− Imλ)| =exp(− Imλ). Finally, Imλ = 0; i.e., λ ∈ R. B

234 Chapter 11

11.7.8. Definition. Let A be a C∗-algebra. A subalgebra B of A is calleda C∗-subalgebra of A if b ∈ B ⇒ b∗ ∈ B. In this event, B is considered with thenorm induced from A.

11.7.9. Theorem. Every closed C∗-subalgebra of a C∗-algebra is pure.

C Let B be a closed (unital) C∗-subalgebra of a C∗-algebra A and b ∈ B.If b ∈ Inv(B) then it is easy that b ∈ Inv(A). Let now b ∈ Inv(A). From 11.7.4derive b∗ ∈ Inv(A). Consequently, b∗b ∈ Inv(A) and the element (b∗b)−1b∗ is a leftinverse of b. By virtue of 11.1.4, it means that b−1 = (b∗b)−1b∗. Consequently,to complete the proof it suffices to show only that (b∗b)−1 belongs to B. Since b∗bis hermitian in B, the inclusion holds: SpB(b∗b) ⊂ R (cf. 11.7.7). Using 11.2.5, inferthat SpA(b∗b) = SpB(b∗b). Since 0 /∈ SpA(b∗b); therefore, b∗b ∈ Inv(B). Finally,b ∈ Inv(B). B

11.7.10. Corollary. Let b be an element of a C∗-algebra A and let B be someclosed C∗-subalgebra of A with b ∈ B. Then

SpB(b) = SpA(b). /.

11.7.11. Remark. In view of 11.7.10, Theorem 11.7.9 is often referred to asthe Spectral Purity Theorem for a C∗-algebra. It asserts that the concept of thespectrum of an element a of a C∗-algebra is absolute, i.e. independent of the choiceof a C∗-subalgebra containing a.

11.8. The Commutative Gelfand–Naımark Theorem

11.8.1. The Banach algebra C(Q, C) with the natural involution f 7→ f∗,where f∗(q) := f(q)∗ for q ∈ Q, is a C∗-algebra.

/ ‖f∗f‖ = sup{|f(q)∗f(q)| : q ∈ Q} = sup{|f(q)|2 : q ∈ Q}= (sup |f(Q)|)2= ‖f‖2 .

11.8.2. Stone–Weierstrass Every unital C∗-subalgebra of the C∗-algebraC(Q, C), which separates the points of Q, is dense in C(Q, C).

C Let A be such subalgebra. Since f ∈ A ⇒ f∗ ∈ A; therefore, f ∈ A ⇒Re f ∈ A and so the set ReA := {Re f : f ∈ A} is a real subalgebra of C(Q, R).It is beyond a doubt that ReA contains constant functions and separates the pointsof Q. By the Stone–Weierstrass Theorem for C(Q, R), the subalgebra ReA is densein C(Q, R). It remains to refer to 11.7.2. B

Banach Algebras 235

11.8.3.Definition. A representation of a ∗-algebra agreeing with involution ∗is a ∗-representation. In other words, if (A, ∗) and (B, ∗) are involutive algebras andR : A → B is a multiplicative linear operator, then R is called a ∗-representationof A in B whenever the following diagram commutes:

AR−→B

∗ ↓ ↓ ∗A

R−→B

If R is also an isomorphism then R is a ∗-isomorphism of A and B. In the pres-ence of norms in the algebras, the naturally understood terms “isometric ∗-representation”and “isometric ∗-isomorphism” are in common parlance.

11.8.4. Commutative Gelfand–Naımark Theorem. The Gelfand trans-form of a commutative C∗-algebra A implements an isometric ∗-isomorphism of Aand C(X(A), C).

C Given a ∈ A, observe that

‖a2‖ = ‖(a2)∗a2‖1/2 = ‖a∗aa∗a‖

1/2 = ‖a∗a‖ = ‖a‖2.

In virtue of 11.6.10 the Gelfand transform GA is thus an isometry of A and a closedsubalgebra A of C(X(A), C). Undoubtedly, A separates the points of X(A) andcontains constant functions.

By virtue of 11.6.9 and 11.7.7, h(X(A)) = Sp(h) ⊂ R for every hermitianelement h = h∗ of A. Now, take an arbitrary element a of A. Using 11.7.2, writea = x+ iy, where x and y are hermitian elements. The containments χ(x) ∈ R andχ(y) ∈ R hold for every character χ, a member of X(A).

With this in mind, successively infer that

GA(a)∗(χ) = a∗(χ) = a(χ)∗ = χ(a)∗ = χ(x+ iy)∗

= (χ(x) + iχ(y))∗ = χ(x)− iχ(y) = χ(x− iy) = χ(a∗)

= a∗(χ) = GA(a∗)(χ) (χ ∈ X(A)).

Consequently, the Gelfand transform GA is a ∗-representation and, in particular,A is a C∗-subalgebra of C(X(A), C). It remains to appeal to 11.8.2 and concludethat A = C(X(A), C). B

11.8.5. Assume that R : A → B is a ∗-representation of a C∗-algebra Ain a C∗-algebra B. Then ‖Ra‖ ≤ ‖a‖ for a ∈ A.

236 Chapter 11

C Since R(1) = 1; therefore, R(Inv(A)) ⊂ Inv(B). Hence, SpB(R(a)) ⊂SpA(a) for a ∈ A. Whence it follows from the Beurling–Gelfand formula that theinequality rA(a) ≥ rB(R(a)) holds for the spectral radii. If a is a hermitian elementof A then R(a) is a hermitian element of B, because R(a)∗ = R(a∗) = R(a). If nowA0 is the least closed C∗-subalgebra containing a and B0 is an analogous subalgebracontaining R(a), then A0 and B0 are commutative C∗-algebras. Therefore, fromTheorems 11.8.4 and 11.6.9 obtain

‖R(a)‖ = ‖R(a)‖B0 = ‖GB0(R(a))‖ = rB0(R(a))= rB(R(a)) ≤ rA(a) = rA0(a) = ‖GA0(a)‖ = ‖a‖.

Given a ∈ A, it is easy to observe that a∗a is a hermitian element. Thus,

‖R(a)‖2 = ‖R(a)∗R(a)‖ = ‖R(a∗a)‖ ≤ ‖a∗a‖ = ‖a‖2. .

11.8.6. Spectral Theorem. Let a be a normal element of a C∗-algebra A,with Sp(a) the spectrum of a. There is a unique isometric ∗-representation Ra

of C(Sp(a), C) in A such that a = Ra(ISp(a)).C Let B be the least closed C∗-subalgebra of A containing a. It is clear

that the algebra B is commutative by the normality of a (this algebra presentsthe closure of the algebra of all polynomials in a and a∗). Moreover, by 11.7.10,Sp(a) = SpA(a) = SpB(a). The Gelfand transform a := GB(a) of a acts from X(B)onto Sp(a) by 11.6.9 and is evidently one-to-one. Since X(B) and Sp(a) are compactsets; on using 9.4.11, conclude that a is a homeomorphism. Whence it is immediate

that the mapping◦R : f 7→ f ◦ a implements an isometric ∗-isomorphism between

C(Sp(a), C) and C(X(B), C).Using Theorem 11.3.2 and the connection between the Gelfand transform and

the Riesz–Dunford integral which is revealed in 11.6.12, for the identity mappingobserve that

a = RaIC = IC ◦ a = IC∣∣a(X(B)) ◦ a = IC

∣∣Sp(a) ◦ a = ISp(a) ◦ a =

◦R(ISp(a)).

Now, put

Ra := G−1B ◦

◦R.

Clearly, Ra is an isometric embedding and a ∗-representation. Moreover,

Ra(ISp(a)) = G−1B ◦

◦R(ISp(a)) = G−1

B (a) = a.

Uniqueness for such representation Ra is guaranteed by 11.8.5 and the Stone–Weierstrass Theorem implies that the C∗-algebra C(Sp(a), C) is its least closed(unital) C∗-subalgebra containing ISp(a). B

Banach Algebras 237

11.8.7. Definition. The representation Ra : C(Sp(a), C) → A of 11.8.6 isthe continuous functional calculus (for a normal element a of A). The elementRa(f) with f ∈ C(Sp(a), C) is usually denoted by f(a).

11.8.8. Remark. Let f be a holomorphic function in a neighborhood aboutthe spectrum of a normal element a of some C∗-algebra A; i.e., f ∈ H(Sp(a)). Thenthe element f(a) of A was defined by the holomorphic functional calculus. Retainthe symbol f for the restriction of f to the set Sp(a). Then, using the continuousfunctional calculus, define the element Ra(f) := Ra

(f∣∣Sp(a)

)of A. This element, as

mentioned in 11.8.7, is also denoted by f(a). The use of the designation is by farnot incidental (and sound in virtue of 11.6.12 and 11.8.6). Indeed, it would beweird to deliberately denote by different symbols one and the same element. Thiscircumstance may be expressed in visual form.

Namely, let ·∣∣Sp(a) stand for the mapping sending a germ h, a member of H (Sp(a)),

to its restriction to Sp(a); i.e., let h∣∣Sp(a) at a point z stand for the value of h at z

(cf. 8.1.21). It is clear that ·∣∣Sp(a) : H (Sp(a)) → C(Sp(a), C). The above con-

nection between the continuous and holomorphic functional calculuses for a normalelement a of the C∗-algebra A may be expressed as follows: The next diagramcommutes

A

H (Sp(a)) C(Sp(a),C)

Ra

·|Sp(a)

Ra

-

?

@@@@@R

11.9. Operator ∗-Representations of a C∗-Algebra

11.9.1. Definition. Let A be a (unital) Banach algebra. An element s in A′is a state of A, in writing s ∈ S(A), if ‖s‖ = s(1) = 1. For a ∈ A, the setN(a) := {s(a) : s ∈ S(A)} is the numeric range of a.

11.9.2. The numeric range of a positive function, a member of C(Q, C), liesin R+.

C Let a ≥ 0 and ‖s‖ = s(1) = 1. We have to prove that s(a) ≥ 0. Take z ∈ Cand ε > 0 such that the disk Bε(z) := z+ εD includes a(Q). Then ‖a− z‖ ≤ ε and,consequently, |s(a−z)| ≤ ε. Hence, |s(a)−z| = |s(a)−s(z)| ≤ ε; i.e., s(a) ∈ Bε(z).

Observe that

∩{Bε(z) : Bε(z) ⊃ a(Q)} = cl co(a(Q)) ⊂ R+.

238 Chapter 11

Thus, s(a) ∈ R+. B

11.9.3. Lemma. Let a be a hermitian element of a C∗-algebra. Then(1) Sp(a) ⊂ N(a);(2) Sp(a) ⊂ R+ ⇔ N(a) ⊂ R+.

C Let B be the least closed C∗-subalgebra, of the algebra A under study, whichcontains a. It is evident that B is a commutative algebra. By virtue of 11.6.9,the Gelfand transform a := GB(a) provides a(X(B)) = SpB(a). In view of 11.7.10,SpB(a) = Sp(a). In other words, for λ ∈ Sp(a) there is a character χ of B satisfyingthe condition χ(a) = λ. By 11.6.3, ‖χ‖ = χ(1) = 1. Using 7.5.11, find a norm-preserving extension s of χ onto A. Then s is a state of A and s(a) = λ. Finally,Sp(a) ⊂ N(a) (in particular, if N(a) ⊂ R+ then Sp(a) ⊂ R+). Now, let s standfor an arbitrary state of A. It is clear that the restriction s

∣∣B

is a state of B.It is an easy matter to show that a maps X(B) onto Sp(a) in a one-to-one fashion.Consequently, B may be treated as the algebra C(Sp(a), C). From 11.9.2 derives(a) = s

∣∣B(a) ≥ 0 for a ≥ 0. Thus, Sp(a) ⊂ R+ ⇒ N(a) ⊂ R+, which ends the

proof. B

11.9.4. Definition. An element a of a C∗-algebra A is called positive if a ishermitian and Sp(a) ⊂ R+. The set of all positive elements of A is denoted by A+.

11.9.5. In each C∗-algebra A the set A+ is an ordering cone.C It is clear that N(a+b) ⊂ N(a)+N(b) and N(αa) = αN(a) for a, b ∈ A and

α ∈ R+. Hence, 11.9.3 ensures the inclusion α1A+ + α2A+ ⊂ A+ for α1, α2 ∈ R+.Thus, A+ is a cone. If a ∈ A+ ∩ (−A+) then Sp(a) = 0. Since a is a hermitianelement, from Theorem 11.8.6 deduce that ‖a‖ = 0. B

11.9.6. To every hermitian element a of a C∗-algebra A there correspond someelements a+ and a− of A+ such that

a = a+ − a−; a+a− = a−a+ = 0.

C Everything is immediate from the continuous functional calculus. B

11.9.7. Kaplansky–Fukamija Lemma. An element a of a C∗-algebra A ispositive if and only if a = b∗b for some b in A.

C ⇒: Let a ∈ A+; i.e., a = a∗ and Sp(a) ⊂ R+. Then (cf. 11.8.6) there isa square root b :=

√a. Moreover, b = b∗ and b∗b = a.

⇐: If a = b∗b then a is hermitian. Therefore, in view of 11.9.6, it may beassumed that b∗b = u− v, where uv = vu = 0 with u ≥ 0 and v ≥ 0 (in the orderedvector space (AR, A+)). Straightforward calculation yields the equalities

(bv)∗bv = v∗b∗bv = vb∗bv = v(u− v)v = (vu− v2)v = −v3.

Banach Algebras 239

Since v ≥ 0, it follows that v3 ≥ 0; i.e., (bv)∗bv ≤ 0. By Theorem 5.6.22, Sp((bv)∗bv)and Sp(bv(bv)∗) may differ only by zero. Therefore, bv(bv)∗ ≤ 0.

In virtue of 11.7.2, bv = a1 + ia2 for suitable hermitian elements a1 and a2.It is evident that a2

1, a22 ∈ A+ and (bv)∗ = a1 − ia2. Using 11.9.5 twice, arrive

at the estimates0 ≥ (bv)∗bv + bv(bv)∗ = 2

(a21 + a2

2)≥ 0.

By 11.9.5, a1 = a2 = 0; i.e., bv = 0. Hence, −v3 = (bv)∗bv = 0. The second appealto 11.9.5 shows v = 0. Finally, a = b∗b = u− v = u ≥ 0; i.e., a ∈ A+. B

11.9.8. Every state s of a C∗-algebra A is hermitian; i.e.,

s(a∗) = s(a)∗ (a ∈ A).

C By Lemmas 11.9.7 and 11.9.3, s(a∗a) ≥ 0 for all a ∈ A. Putting a := a + 1and a := a+ i, successively infer that

0 ≤ s((a+ 1)∗(a+ 1)) = s(a∗a+ a+ a∗ + 1)⇒ s(a) + s(a∗) ∈ R;0 ≤ s((a+ i)∗(a+ i)) = s(a∗a− ia+ ia∗ + 1)⇒ i(−s(a) + s(a∗)) ∈ R.

In other words,

Im s(a) + Im s(a∗) = 0; Re(−s(a)) + Re s(a∗) = 0.

Whence it follows that

s(a∗) = Re s(a∗) + i Im s(a∗) = Re s(a)− i Im s(a) = s(a)∗. .

11.9.9. Let s be a state of a C∗-algebra A. Given a, b ∈ A, denote (a, b)s :=s(b∗a). Then ( · , · )s is a semi-inner product on A.C From 11.9.8 derive

(a, b)s = s(b∗a) = s((a∗b)∗) = s(a∗b)∗ = (b, a)∗s.

Hence, ( · , · )s is a hermitian form. Since a∗a ≥ 0 for a ∈ A in virtue of 11.9.7,(a, a)s = s(a∗a) ≥ 0 by 11.9.3. Consequently, ( · , · )s is a positive-definite hermitianform. B

11.9.10. GNS-Construction Theorem. To every state s of an arbitraryC∗-algebra A there correspond a Hilbert space (Hs, ( · , · )s), an element xs in Hs

and a ∗-representation Rs : A → B(Hs) such that s(a) = (Rs(a)xs, xs)s for alla ∈ A and the set {Rs(a)xs : a ∈ A} is dense in Hs.

240 Chapter 11

C In virtue of 11.9.9, putting (a, b)s := s(b∗a) for a, b ∈ A, obtain a pre-Hilbert space (A, ( · , · )s). Let ps(a) :=

√(a, a)s stand for the seminorm of the

space. Assume that ϕs : A → A/ ker ps is the coset mapping of A onto theHausdorff pre-Hilbert space A/ ker ps associated with A. Assume further thatιs : A/ ker ps → Hs is an embedding (for instance, by the double prime map-ping) of A/ ker ps onto a dense subspace of the Hilbert space Hs associated with(A, ( · , · )s) (cf. 6.1.10 (4)). The inner product in Hs retains the previous notation( · , · )s. Therefore, in particular,

(ιsϕsa, ιsϕsb)s = (a, b)s = s(b∗a) (a, b ∈ A).

Given a ∈ A, consider (the image under the canonical operator representation)La : b 7→ ab (b ∈ A). Demonstrate first that there are unique bounded operatorsLa and Rs(a) making the following diagram commutative:

Aϕs−→A/ ker ps

ιs−→Hs

La ↓ ↓ La ↓ Rs(a)A

ϕs−→A/ ker psιs−→Hs

A sought operator La is a solution to the equation Xϕs = ϕsLa. Using 2.3.8,observe next that the necessary and sufficient condition for solvability of the equa-tion in linear operators consists in invariance of the subspace ker ps under La. Thus,examine the inclusion La(ker ps) ⊂ ker ps. To this end, take an element b of ker ps,i.e. ps(b) = 0. By definition and the Cauchy–Bunyakovskiı–Schwarz inequality,deduce that

0 ≤ (Lab, Lab)s = (ab, ab)s = s((ab)∗ab)= s(b∗a∗ab) = (a∗ab, b)s ≤ ps(b)ps(a∗ab) = 0;

i.e., Lab ∈ ker ps. Uniqueness for La is provided by 2.3.9, since ϕs is an epi-morphism. Observe also that ϕs is an open mapping (cf. 5.1.3). Whence thecontinuity of La is immediate. Therefore, in virtue of 5.3.8 the correspondenceιs ◦ La ◦ (ιs)−1 may be considered as a bounded linear operator from ιs(A/ ker ps)to the Banach spaceHs. By 4.5.10 such an operator extends uniquely to an operatorRs(a) in B(Hs).

Demonstrate now that Rs : a 7→ Rs(a) is a sought representation. By 11.1.6,Lab = LaLb for a, b ∈ A. Consequently,

ϕsLab = ϕsLaLb = LaϕsLb = LaLbϕs.

Since Lab is a unique solution to the equation Xϕs = ϕsLab, infer the equalityLab = LaLb, which guarantees multiplicativity for Rs. The linearity of Rs may beverified likewise. Furthermore,

L1ϕs = ϕsL1 = ϕsIA = ϕs = IA/ ker psϕs = 1ϕs;

Banach Algebras 241

i.e., Rs(1) = 1.For simplicity, put ψs := ιsϕs. Then, on account taken of the definition

of the inner product on Hs (cf. 6.1.10 (4)) and the involution in B(Hs) (cf. 6.4.14and 6.4.5), given elements a, b, and y in A, infer that

(Rs(a∗)ψsx, ψsy)s = (ψsLa∗x, ψsy)s= (La∗x, y)s = (a∗x, y)s = s(y∗a∗x) = s((ay)∗x) = (x, ay)s

= (x, Lay)s = (ψsx, ψsLay)s = (ψsx, Rs(a)ψsy)s = (Rs(a)∗ψsx, ψsy)s.

Now, since imψs is dense in Hs it follows that Rs(a∗) = Rs(a)∗ for all a in A; i.e.,Rs is a ∗-representation.

Let xs := ψs1. Then

Rs(a)xs = Rs(a)ψs1 = ψsLa1 = ψsa (a ∈ A).

Consequently, the set {Rs(a)xs : a ∈ A} is dense in Hs. Furthermore,

(Rs(a)xs, xs)s = (ψsa, ψs1)s = (a, 1)s = s(1∗a) = s(a). .

11.9.11. Remark. The construction, presented in the proof of 11.9.10, iscalled the GNS-construction or, in expanded form, the Gelfand–Naımark–Segal con-struction, which is reflected in the name of 11.9.10.

11.9.12.Gelfand–Naımark Each C∗-algebra has an isometric ∗-representationin the endomorphism algebra of a suitable Hilbert space.

C Let A be a C∗-algebra. We have to find a Hilbert space H and an isomet-ric ∗-representation R of A in the C∗-algebra B(H) of bounded endomorphismsof H. For this purpose, consider the Hilbert sum H of the family of Hilbert spaces(Hs)s∈S(A) which exists in virtue of the GNS-Construction Theorem; i.e.,

H := ⊕s∈S(A)

Hs =

h := (hs)s∈S(A) ∈∏

s∈S(A)

Hs :∑

s∈S(A)

‖hs‖2Hs < +∞

.

Observe that the inner product of h := (hs)s∈S(A) and g := (gs)s∈S(A) is calcu-lated by the rule (cf. 6.1.10 (5) and 6.1.9):

(h, g) =∑

s∈S(A)

(hs, gs)s.

242 Chapter 11

Assume further that Rs is a ∗-representation of A on the space Hs correspond-ing to s in S(A). Since in view of 11.8.5 there is an estimate ‖Rs(a)‖B(Hs) ≤ ‖a‖for a ∈ A; therefore, given h ∈ H, infer that∑

s∈S(A)

‖Rs(a)hs‖2Hs ≤∑

s∈S(A)

‖Rs(a)‖2B(Hs)‖hs‖2Hs ≤ ‖a‖

2∑

s∈S(A)

‖hs‖2Hs .

Whence it follows that the expression R(a)h : s 7→ Rs(a)hs defines an elementR(a)h of H. The resulting operator R(a) : h 7→ R(a)h is a member of B(H).Moreover, the mapping R : a 7→ R(a) (a ∈ A) is a sought isometric ∗-representationof A.

Indeed, from the definition of R and the properties of Rs for s ∈ S(A), it followseasily that R is a ∗-representation of A in B(H). Check for instance that R agreeswith involution. To this end, take a ∈ A and h, g ∈ H. Then

(R(a∗)h, g) =∑

s∈S(A)

(Rs(a∗)hs, gs)s

=∑

s∈S(A)

(Rs(a)∗hs, gs)s =∑

s∈S(A)

(hs, Rs(a)gs)s

= (h, R(a)g) = (R(a)∗h, g).

Since h and g in H are arbitrary, conclude that R(a∗) = R(a)∗.It remains to establish only that the ∗-representation R is an isometry, i.e. the

equalities ‖R(a)‖ = ‖a‖ for all a ∈ A. First, assume a positive. From the SpectralTheorem and the Weierstrass Theorem it follows that ‖a‖ ∈ Sp(a). In virtueof 11.9.3 (1) there is a state s ∈ S(A) such that s(a) = ‖a‖. Using the propertiesof the vector xs corresponding to the ∗-representation Rs (cf. 11.9.10) and applyingthe Cauchy–Bunyakovskiı–Schwarz inequality, infer that

‖a‖ = s(a) = (Rs(a)xs, xs)s ≤ ‖Rs(a)xs‖Hs‖xs‖Hs≤ ‖Rs(a)‖B(Hs)‖xs‖

2Hs = ‖Rs(a)‖B(Hs)(xs, xs)s

= ‖Rs(a)‖B(Hs)(Rs(1)xs, xs)s = ‖Rs(a)‖B(Hs)s(1) = ‖Rs(a)‖B(Hs).

From the estimates ‖R(a)‖ ≥ ‖Rs(a)‖B(Hs) and ‖a‖ ≥ ‖R(a)‖, the formerobvious and the latter indicated in 11.8.5, derive

‖a‖ ≥ ‖R(a)‖ ≥ ‖Rs(a)‖B(Hs) ≥ ‖a‖.

Finally, take a ∈ A. By the Kaplansky–Fukamija Lemma, a∗a is positive. So,

‖R(a)‖2 = ‖R(a)∗R(a)‖ = ‖R(a∗)R(a)‖ = ‖R(a∗a)‖ = ‖a∗a‖ = ‖a‖2.

No further explanation is needed. B

Banach Algebras 243

Exercises

11.1. Give examples of Banach algebras and non-Banach algebras.11.2. Let A be a Banach algebra. Take χ ∈ A# such that χ(1) = 1 and χ(Inv(A)) ⊂ Inv(C).

Prove that χ is multiplicative and continuous.

11.3. Let the spectrum Sp(a) of an element a of a Banach algebra A lie in an open set U .Prove that there is a number ε > 0 such that Sp(a+ b) ⊂ U for all b ∈ A satisfying ‖b‖ ≤ ε.

11.4. Describe the maximal ideal spaces of the algebras C(Q, C) and C(1)([0, 1], C) withpointwise multiplication, and of the algebra of two-way infinite summable sequences l1(Z) withmultiplication

(a ∗ b)(n) :=∞∑

k=−∞

an−kbk.

11.5. Show that a member T of the endomorphism algebra B(X) of a Banach space X hasa left inverse if and only if T is a monomorphism and the range of T is complemented in X.

11.6. Show that a member T of the endomorphism algebra B(X) of a Banach space X hasa right inverse if and only if T is an epimorphism and the kernel of T is complemented in X.

11.7. Assume that a Banach algebra A has an element with disconnected spectrum (havinga proper clopen part). Prove that A has a nontrivial idempotent.

11.8. Let A be a unital commutative Banach algebra and let E be some set of maximalideals of A. Such a set E is a boundary of A if ‖a‖∞ = sup |a(E)| for all a ∈ A. Prove that theintersection of all closed boundaries of A is also a boundary of A. This is the Shilov boundaryof A.

11.9. Let A and B be unital commutative Banach algebras, with B ⊂ A and 1B = 1A.Prove that each maximal ideal of the Shilov boundary of B lies in some maximal ideal of A.

11.10. Let A and B be unital C∗-algebras and let T be a morphism from A to B. Assumefurther that a is a normal element of A and f is a continuous function on SpA(a). Demonstratethat SpB(Ta) ⊂ SpA(a) and Tf(a) = f(Ta).

11.11. Let f ∈ A′, with A a commutative C∗-algebra. Show that f is a positive form (i.e.,f(a∗a) ≥ 0 for a ∈ A) if and only if ‖f‖ = f(1).

11.12. Describe extreme rays of the set of positive forms on a commutative C∗-algebra.11.13. Prove that the algebras C(Q1, C) and C(Q2, C), with Q1 and Q2 compact, are

isomorphic if and only if Q1 and Q2 are homeomorphic.

11.14. Let a normal element a of a C∗-algebra has real spectrum. Prove that a is hermitian.11.15. Using the continuous functional calculus, develop a spectral theory for normal oper-

ators in a Hilbert space. Describe compact normal operators.

11.16. Let T be an algebraic morphism between C∗-algebras, and ‖T‖ ≤ 1. Then T (a∗) =(Ta)∗ for all a.

11.17. Let T be a normal operator in a Hilbert space H. Show that there are a hermitianoperator S in H and a continuous function f : Sp(S)0 = 0 OO C such that T = f(S). Is an analogousassertion valid in C∗-algebras?

11.18. Let A and B be C∗-algebras and let ρ be a ∗-monomorphism from A to B. Provethat ρ is an isometric embedding of A into B.

11.19. Let a and b be hermitian elements of a C∗-algebra A. Assume that ab = ba and,moreover, a ≤ b. Prove that f(a) ≤ f(b) for (suitable restrictions of) every increasing continuousscalar function f over R.

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(1959).

Notation Index

:=, ix/, ., ixAr, 11.1.6A × B, 1.1.1Bp, 5.1.1, 5.2.11◦Bp, 5.1.1BT , 5.1.3BX , 5.1.10, 5.2.11B(X), 5.6.4,B(E , F ), 5.5.9 (2)B(X, Y ), 5.1.10 (7)C(Q, F ), 4.6.8C(m), 10.9.9C∞(�), 10.10.2 (3)Dα, 10.11.13F−1, 1.1.3 (1)F (B), 1.3.5 (1)Fp, 5.5.9 (6)F |U , 1.1.3 (5)F (U), 1.1.3 (5)F (a1, · ), 1.1.3 (6)F ( · , a2), 1.1.3 (6)F ( · , · ), 1.1.3 (6)F (X, Y ), 8.3.6G, 10.11.2G, 10.11.2G ◦ F , 1.1.4H∗, 6.1.10 (3)H(K), 8.1.13H� (U), 3.1.11IC, 8.2.10IU , 1.1.3 (3)J(q), 11.5.3J(Q0), 11.5.2

J C A, 11.4.1K(E), 10.9.1K(Q), 10.9.1K(�), 10.9.1LA, 11.1.6Lp, 5.5.9 (4), 5.5.9 (6)Lp(X), 5.5.9LQ0 , 10.8.4 (3)L∣∣Q0

, 10.8.4 (4)

L∞, 5.5.9 (5)M(�), 10.9.4 (2)N(a), 11.9.1Np, 5.5.9 (6)PH0 , 6.2.7Pσ , 8.2.10PX1||X2 , 2.2.9 (4)P1 ⊥ P2, 6.2.12R (a, λ), 11.2.1R (T, λ), 5.6.13S(A), 11.9.1T ′, 7.6.2T ∗, 6.4.4‖T‖, 5.1.10 (7)U◦, 10.5.7U⊥, 6.2.5U ∈ (� ), 3.1.1〈X |, 10.3.1X′, 5.1.10 (8), 10.2.11X′′, 5.1.10 (8)X∗, 2.1.4 (2)X+, 3.2.5X⊥0 , 7.6.8Xσ , 8.2.10XN , 2.1.4 (4)X#, 2.2.4

Notation Index 257

XR, 3.7.1X�, 2.1.4 (4)X = X1 ⊕ X2, 2.1.7X ⊕ iX, 8.4.8X1 × X2 × . . . × XN , 2.1.4 (4)(X, τ)′, 10.2.11X/X0, 2.1.4 (6)(X/X0, pX/X0 ), 5.1.10 (5)X ↔ Y , 10.3.3X ' Y , 2.2.6|Y 〉, 10.3.1B, 9.6.14C, 2.1.2D, 8.1.3F, 2.1.2N, 1.2.16Q, 7.4.11R, 2.1.2R·, 3.4.1R+, 3.1.2 (4)R, 3.8.1Re, 3.7.3Re−1, 3.7.4T, 8.1.3Z, 8.5.1Z+, 10.10.2 (2)Ae, 11.1.2D(Q), 10.10.1D(�), 10.10.1D ′(�), 10.10.4D ′F (�), 10.10.8D (m)(Q), 10.10.8D (m)(�), 10.10.8D (m)(�)′, 10.10.8E (�), 10.10.2 (3)E ′(RN ), 10.10.5 (9)E ◦ T , 2.2.8F , 10.11.4Fp, 5.5.9 (6)F r(X, Y ), 8.5.1F (X), 1.3.6GA, 11.6.8H (K), 8.1.14K (X), 8.3.3K (X, Y ), 6.6.1L (X), 2.2.8L (X, Y ), 2.2.3Lr(X, Y ), 3.2.6 (3)L∞, 5.5.9 (5)

M (�), 10.9.3N (µ), 10.8.11Np(f), 5.5.9 (4)N∞, 5.5.9 (5)P(X), 1.2.3 (4)RT , 8.2.1Rah, 11.3.1S (RN ), 10.11.6S ′(RN ), 10.11.16T (X), 9.1.2Up, 5.2.2UM , 5.2.4UX , 4.1.5, 5.2.4⊥X0, 7.6.8Fu, 10.11.19M , 5.3.9M ∼ N , 5.3.1M � N , 5.3.1MX , 5.1.6Mτ , 10.2.7N ◦ T , 5.1.10 (3)NT , 5.1.10 (3)Ra, 11.8.7Cl(τ), 4.1.15, 9.1.4Im f , 5.5.9 (4)Inv(A), 11.1.5Inv(X, Y ), 5.6.12�B , 8.1.2 (4)Lat(X), 2.1.5LCT (X), 10.2.3M(A), 11.6.6Op(τ), 4.1.11, 9.1.4Re, 2.1.2Re f , 5.5.9 (4)Sp(a), 11.2.1SpA(a), 11.2.1Sp(T ), 5.6.13T1, 9.3.2T2, 9.3.5T3, 9.3.9T31/2

, 9.3.15T4, 9.3.11T(X), 9.1.7Tr(�), 10.10.2VT(X), 10.1.5X(A), 11.6.4δ, 10.9.4 (1)δ(−1), 10.10.5 (4)δq , 10.9.4 (1)µ∗, 10.9.4 (3)

258 Notation Index

µ+, 10.8.13µ−, 10.8.13|µ|, 10.8.13, 10.9.4 (3)‖µ‖, 10.9.5µ�′ , 10.9.4 (4)µ1 ⊗ µ2, 10.9.4 (6)µ1 × µ2, 10.9.4 (6)µ ∗ f , 10.9.4 (7)µ ∗ ν, 10.9.4 (7)π(U), 10.5.1, 10.5.7π−1(V ), 10.5.1π−1F (πF (U)), 10.5.5

2π , 10.11.4σ′, 8.2.9σ(T ), 5.6.13σ(X, Y ), 10.3.5τ(X, Y ), 10.4.4τaf , 10.9.4 (1)τM , 5.2.8κσ , 8.2.10abs pol , 10.5.7clU , 4.1.13co(U), 3.1.14codimX, 2.2.9 (5)coimT , 2.3.1cokerT , 2.3.1coreU , 3.4.11diamU , 4.5.3dimX, 2.2.9 (5)dom f , 3.4.2domF , 1.1.20 = 0 OOOO , 3.4.2extV , 3.6.1filB, 1.3.3frU , 4.1.13imF , 1.1.2inf U , 1.2.9intU , 4.1.13kerT , 2.3.1lin(U), 3.1.14pol , 10.5.7rankT , 8.5.7 (2)res(a), 11.2.1res(T ), 5.6.13seg, 3.6.1supU , 1.2.9supp(f), 9.6.4supp(µ), 10.8.11, 10.9.4 (5)supp(u), 10.10.5 (6)a, 11.6.8

aµ, 10.8.15aτf , 10.9.4 (1)(a, b)s, 11.9.9c, 3.3.1 (2), 5.5.9 (3)c (E , F ), 5.5.9 (3)c0, 5.5.9 (3)c0(E ), 5.5.9c0(E , F ), 5.5.9 (3)∂αu, 10.10.5 (4)∂ (p), 3.5.2 (1)|∂ |(p), 3.7.8∂U , 4.1.13∂x(f), 3.5.1dp, 5.2.1dx, 10.9.9e, 10.9.4 (1), 11.1.1f , 10.11.3f(a), 11.3.1{f < t}, 3.8.1{f = t}, 3.8.1{f ≤ t}, 3.8.1f(T ), 8.2.1f , 10.10.5 (9)fµ, 10.9.4 (3)f∗, 10.9.4 (3)fu, 10.10.5 (7)fn � f , 10.10.7 (3)fn �

K0, 10.9.8g, 10.11.2

◦g(f), 8.2.6〈h〉, 6.3.5lp, lp(E ), 5.5.9 (4)l∞, l∞(E ), 5.5.9 (2)m, 5.5.9 (2)p � q, 5.3.3pe, 5.5.9 (5)pS , 3.8.6p T , 5.1.3pX/X0 , 5.1.10 (5)r(T ), 5.6.6s, Ex. 1.19tα, 10.11.5 (8)ug , 10.10.5 (1)u∗, 10.10.5 (5)u ∗ f , 10.10.5 (9)u1 ⊗ u2, 10.10.5 (8)u1 × u2, 10.10.5 (8)〈x |, 10.3.1

Notation Index 259

x′, 6.4.1x′′, 5.1.10 (8)xα, 10.11.5 (8)x+, 3.2.12x−, 3.2.12|x|, 3.2.12‖x‖p, 5.5.9 (4)‖x‖∞, 5.5.9 (2)∼(x), 10.11.4∼X0 , 2.1.4 (6)x :=

∑e∈E xe, 5.5.9 (7)

x 7→ x′, 6.4.1x1 ∨ x2, x1 ∧ x2, 1.2.12(x1, x2), 1.2.12〈x | f〉, 5.1.11x ≤σ y, 1.2.2x′ ⊗ y, 5.5.6x ⊥ y, 6.2.5| y〉, 10.3.1

|||y|||p, 5.5.9 (6)‖ · ‖, 5.1.9‖ · ‖n,Q, 10.10.2 (2)‖ · ‖∞, 5.5.9 (5)‖ · ‖X , 5.1.9‖ · |X‖, 5.1.9| ·〉, 10.3.11, 5.3.10, 10.8.4 (6)2X , 1.2.3 (4)∗, 6.4.13b, 10.9.1∫

E, 5.5.9 (6)

〈· | ·〉, 10.3.1〈· |, 10.3.1∼, 1.2.2∑

ξ∈�Xξ, 2.1.4 (5)∏ξ∈�Xξ, 2.1.4 (4)∮h(z)R(z)dz, 8.1.20, 11.6.8

Subject Index

Absolute Bipolar Theorem, 10.5.9, 178absolute concept, 9.4.7, 157absolute polar, 10.5.7, 178absolutely continuous measure,

10.9.4 (3), 190absolutely convex set, 3.1.2 (6), 20absolutely fundamental family

of vectors, 5.5.9 (7), 73absorbing set, 3.4.9, 28addition in a vector space, 2.1.3, 10adherence of a filterbase, 9.4.1, 155adherent point, 4.1.13, 41adherent point of a filterbase, 9.4.1, 155adjoint diagram, 6.4.8, 92adjoint of an operator, 6.4.5, 91adjunction of unity, 11.1.2, 213affine hull, 3.1.14, 22affine mapping, 3.1.7, 21affine operator, 3.4.8 (4), 28affine variety, 3.1.2 (5), 20agreement condition, 10.9.4 (4), 190Akilov Criterion, 10.5.3, 178Alaoglu–Bourbaki Theorem, 10.6.7, 180Alexandroff compactification,

9.4.22, 159algebra, 5.6.2, 73algebra of bounded operators, 5.6.5, 74algebra of germs of holomorphic

functions, 8.1.18, 125algebraic basis, 2.2.9 (5), 14algebraic complement, 2.1.7, 12algebraic dual, 2.2.4, 13algebraic isomorphism, 2.2.5, 13algebraic subdifferential, 7.5.8, 113algebraically complementary subspace,

2.1.7, 12

algebraically interior point, 3.4.11, 28algebraically isomorphic spaces,

2.2.6, 13algebraically reflexive space, Ex. 2.8, 19ambient space, 2.1.4 (3), 11annihilator, 7.6.8, 116antidiscrete topology, 9.1.8 (3), 147antisymmetric relation, 1.2.1, 3antitone mapping, 1.2.3, 4approximate inverse, 8.5.9, 139approximately invertible operator,

8.5.9, 139approximation property, 8.3.10, 133approximation property in Hilbert

space, 6.6.10, 98arc, 4.8.2, 54Arens multinorm, 8.3.8, 133ascent, Ex. 8.10, 144Ascoli–ArzelΓa Theorem, 4.6.10, 50assignment operator, ixassociate seminorm, 6.1.7, 81associated Hausdorff pre-Hilbert

space, 6.1.10 (4), 83associated Hilbert space,

6.1.10 (4), 83associated multinormed space,

10.2.7, 172associated topology, 9.1.12, 148associativity of least upper bounds,

3.2.10, 24asymmetric balanced Hahn–Banach

formula, 3.7.10, 34asymmetric Hahn–Banach formula,

3.5.5, 30Atkinson Theorem, 8.5.18, 141

Subject Index 261

Automatic Continuity Principle,7.5.5, 112

automorphism, 10.11.4, 203

Baire Category Theorem, 4.7.6, 52Baire space, 4.7.2, 52Balanced Hahn–Banach Theorem,

3.7.13, 35Balanced Hahn–Banach Theorem

in a topological setting, 7.5.10, 113balanced set, 3.1.2 (7), 20balanced subdifferential, 3.7.8, 34Balanced Subdifferential Lemma,

3.7.9, 34ball, 9.6.14, 166Banach algebra, 5.6.3, 74Banach Closed Graph Theorem,

7.4.7, 108Banach Homomorphism Theorem,

7.4.4, 108Banach Inversion Stability Theorem,

5.6.12, 76Banach Isomorphism Theorem,

7.4.5, 108Banach range, 7.4.18, 111Banach space, 5.5.1, 66Banach’s Fundamental Principle,

7.1.5, 101Banach’s Fundamental Principle for

a Correspondence, 7.3.7, 107Banach–Steinhaus Theorem, 7.2.9, 104barrel, 10.10.9 (1), 199barreled normed space, 7.1.8, 102barreled space, 10.10.9 (1), 199base for a filter, 1.3.3, 6basic field, 2.1.2, 10Bessel inequality, 6.3.7, 88best approximation, 6.2.3, 84Beurling–Gelfand formula,

8.1.12 (2), 124bilateral ideal, 8.3.3, 132; 11.6.2, 220bilinear form, 6.1.2, 80bipolar, 10.5.5, 178Bipolar Theorem, 10.5.8, 178Birkhoff Theorem, 9.2.2, 148Bochner integral, 5.5.9 (6), 72bornological space, 10.10.9 (3), 199boundary of an algebra, Ex. 11.8, 234boundary of a set, 4.1.13, 41boundary point, 4.1.13, 41bounded above, 1.2.19, 6

bounded below, 3.2.9, 23bounded endomorphism algebra,

5.6.5, 74Bounded Index Stability Theorem,

8.5.21, 142bounded operator, 5.1.10 (7), 59bounded Radon measure,

10.9.4 (2), 189bounded set, 5.4.3, 66boundedly order complete lattice,

3.2.8, 23Bourbaki Criterion, 4.4.7, 46; 9.4.4, 156bracketing of vector spaces, 10.3.1, 173bra-functional, 10.3.1, 173bra-mapping, 10.3.1, 173bra-topology, 10.3.5, 174B-stable, 10.1.8, 171bump function, 9.6.19, 167

Calkin algebra, 8.3.5, 132Calkin Theorem, 8.3.4, 132canonical embedding, 5.1.10 (8), 59canonical exact sequence,

2.3.5 (6), 16canonical operator representation,

11.1.7, 214canonical projection, 1.2.3 (4), 4Cantor Criterion, 4.5.6, 47Cantor Theorem, 4.4.9, 46cap, 3.6.3 (4), 32

Cauchy–BunyakovskiΦß–Schwarzinequality, 6.1.5, 80

Cauchy filter, 4.5.2, 47Cauchy net, 4.5.2, 47Cauchy–Wiener Integral Theorem,

8.1.7, 122centralizer, 11.1.6, 214chain, 1.2.19, 6character group, 10.11.2, 203character of a group algebra,

10.11.1 (1), 201character of an algebra, 11.6.4, 221character space of an algebra,

11.6.4, 221characteristic function, 5.5.9 (6), 72charge, 10.9.4 (3), 190ChebyshΞev metric, 4.6.8, 50classical Banach space, 5.5.9 (5), 71clopen part of a spectrum, 8.2.9, 130closed ball, 4.1.3, 40closed convex hull, 10.6.5, 179

262 Subject Index

closed correspondence, 7.3.8, 107closed cylinder, 4.1.3, 40closed-graph correspondence, 7.3.9, 107closed halfspace, Ex. 3.3, 39closed linear span, 10.5.6, 178closed set, 9.1.4, 146closed set in a metric space, 4.1.11, 41closure of a set, 4.1.13, 41closure operator, Ex. 1.11, 8coarser cover, 9.6.1, 164coarser filter, 1.3.6, 7coarser pretopology, 9.1.2, 146codimension, 2.2.9 (5), 14codomain, 1.1.2, 1cofinite set, Ex. 1.19, 9coimage of an operator, 2.3.1, 15coincidence of the algebraic and

topological subdifferentials,7.5.8, 113

coinitial set, 3.3.2, 25cokernel of an operator, 2.3.1, 15comeager set, 4.7.4, 52commutative diagram, 2.3.3, 15

Commutative Gelfand–NaΦßmarkTheorem, 11.8.4, 227

compact convergence, 7.2.10, 105Compact Index Stability Theorem,

8.5.20, 142compact-open topology, 8.3.8, 133compact operator, 6.6.1, 95compact set, 9.4.2, 155compact set in a metric space, 4.4.1, 46compact space, 9.4.7, 157compact topology, 9.4.7, 157compactly-supported distribution,

10.10.5 (6), 196compactly-supported function,

9.6.4, 165compactum, 9.4.17, 158compatible topology, 10.4.1, 175complementary projection, 2.2.9 (4), 14complementary subspace, 7.4.9, 109Complementation Principle, 7.4.10, 109complemented subspace, 7.4.9, 108complement of an orthoprojection,

6.2.10, 85complement of a projection,

2.2.9 (4), 14complete lattice, 1.2.13, 5complete metric space, 4.5.5, 47complete set, 4.5.14, 49

completely regular space, 9.3.15, 155completion, 4.5.13, 49complex conjugate, 2.1.4 (2), 10complex distribution, 10.10.5 (5), 196complex plane, 8.1.3, 121complex vector space, 2.1.3, 10complexification, 8.4.8, 136complexifier, 3.7.4, 34composite correspondence, 1.1.4, 2Composite Function Theorem,

8.2.8, 129composition, 1.1.4, 2Composition Spectrum Theorem,

5.6.22, 78cone, 3.1.2 (4), 20conical hull, 3.1.14, 22conical segment, 3.1.2 (9), 20conical slice, 3.1.2 (9), 20conjugate distribution, 10.10.5 (5), 196conjugate exponent, 5.5.9 (4), 69conjugate-linear functional, 2.2.4, 13conjugate measure, 10.9.4 (3), 189connected elementary compactum,

4.8.5, 54connected set, 4.8.4, 54constant function, 5.3.10, 64;

10.8.4 (6), 182Continuous Extension Principle,

7.5.11, 113continuous function at a point, 4.2.2,

43; 9.2.5, 149Continuous Function Recovery

Lemma, 9.3.12, 153continuous functional calculus,

11.8.7, 228continuous mapping of a metric

space, 4.2.2, 43continuous mapping of a topological

space, 9.2.4, 149continuous partition of unity, 9.6.6, 166contour integral, 8.1.20, 125conventional summation, 5.5.9 (4), 70convergent filterbase, 4.1.16, 42convergent net, 4.1.17, 42convergent sequence space, 3.3.1 (2), 25convex combination, 3.1.14, 22convex correspondence, 3.1.7, 21convex function, 3.4.4, 27convex hull, 3.1.14, 22convex set, 3.1.2 (8), 20convolution algebra, 10.9.4 (7), 190

Subject Index 263

convolution of a measure anda function, 10.9.4 (7), 191

convolution of distributions,10.10.5 (9), 197

convolution of functions, 9.6.17, 167convolution of measures,

10.9.4 (7), 190convolutive distributions,

10.10.5 (9), 197coordinate projection, 2.2.9 (3), 13coordinatewise operation,

2.1.4 (4), 11core, 3.4.11, 28correspondence, 1.1.1, 1correspondence in two arguments,

1.1.3 (6), 2correspondence onto, 1.1.3 (3), 2coset, 1.2.3 (4), 4coset mapping, 1.2.3 (4), 4countable convex combination,

7.1.3, 101Countable Partition Theorem,

9.6.20, 167countable sequence, 1.2.16, 6countably normable space, 5.4.1, 64cover of a set, 9.6.1, 164C∗-algebra, 6.4.13, 92C∗-subalgebra, 11.7.8, 225

Davis–Figiel–SzankowskiCounterexample, 8.3.14, 134

de Branges Lemma, 10.8.16, 185decomplexification, 6.1.10 (2), 83decomposition reduces an operator,

2.2.9 (4), 14decreasing mapping, 1.2.3, 4Dedekind complete vector lattice,

3.2.8, 23deficiency, 8.5.1, 137definor, ixdelta-function, 10.9.4 (1), 188delta-like sequence, 9.6.15, 166δ-like sequence, 9.6.15, 166δ-sequence, 9.6.15, 166dense set, 4.5.10, 48denseness, 4.5.10, 48density of a measure, 10.9.4 (3), 190derivative in the distribution sense,

10.10.5 (4), 196derivative of a distribution, 10.10.5 (4),

196

descent, Ex. 8.10, 144diagonal, 1.1.3 (3), 2diagram prime, 7.6.5, 115Diagram Prime Principle, 7.6.7, 115diagram star, 6.4.8, 92Diagram Star Principle, 6.4.9, 92diameter, 4.5.3, 47DieudonnΓe Lemma, 9.4.18, 158dimension, 2.2.9 (5), 14Dini Theorem, 7.2.10, 105Dirac measure, 10.9.4 (1), 188direct polar, 7.6.8, 116; 10.5.1, 177direct sum decomposition, 2.1.7, 12direct sum of vector spaces, 2.1.4 (5), 11directed set, 1.2.15, 6direction, 1.2.15, 6directional derivative, 3.4.12, 28discrete element, 3.3.6, 26

Discrete KreΦßn–Rutman Theorem,3.3.8, 26

discrete topology, 9.1.8 (4), 147disjoint measures, 10.9.4 (3), 190disjoint sets, 4.1.10, 41distance, 4.1.1, 40distribution, 10.10.4, 195distribution applies to a function,

10.10.5 (7), 196Distribution Localization Principle,

10.10.12, 200distribution of finite order, 10.10.5 (3),

195distribution size at most m,

10.10.5 (3), 195distribution of slow growth,

10.11.16, 209distributions admitting convolution,

10.10.5 (9), 197distributions convolute, 10.10.5 (9), 197division algebra, 11.2.3, 215domain, 1.1.2, 1Dominated Extension Theorem,

3.5.4, 30Double Prime Lemma, 7.6.6, 115double prime mapping, 5.1.10 (8), 59double sharp, Ex. 2.7, 19downward-filtered set, 1.2.15, 6dual diagram, 7.6.5, 115dual group, 10.11.2, 203dual norm of a functional, 5.1.10 (8), 59dual of a locally convex space,

10.2.11, 173

264 Subject Index

dual of an operator, 7.6.2, 114duality bracket, 10.3.3, 174duality pair, 10.3.3, 174dualization, 10.3.3, 174Dualization Theorem, 10.3.9, 175Dunford–Hille Theorem, 8.1.3, 121Dunford Theorem, 8.2.7 (2), 129Dvoretzky–Rogers Theorem,

5.5.9 (7), 73dyadic-rational point, 9.3.13, 154

effective domain of definition, 3.4.2, 27Eidelheit Separation Theorem,

3.8.14, 39eigenvalue, 6.6.3 (4), 95eigenvector, 6.6.3, 95element of a set, 1.1.3 (4), 2elementary compactum, 4.8.5, 54endomorphism, 2.2.1, 12; 8.2.1, 126endomorphism algebra, 2.2.8, 13;

5.6.5, 74endomorphism space, 2.2.8, 13Enflo counterexample, 8.3.12, 134entourage, 4.1.5, 40envelope, Ex. 1.11, 8epigraph, 3.4.2, 26epimorphism, 2.3.1, 15ε-net, 8.3.2, 132ε-perpendicular, 8.4.1, 134ε-Perpendicular Lemma, 8.4.1, 134Equicontinuity Principle, 7.2.4, 103equicontinuous set, 4.2.8, 44equivalence, 1.2.2, 3equivalence class, 1.2.3 (4), 4equivalent multinorms, 5.3.1, 62equivalent seminorms, 5.3.3, 63estimate for the diameter of a spherical

layer, 6.2.1, 84Euler identity, 8.5.17, 141evaluation mapping, 10.3.4 (3), 174everywhere-defined operator, 2.2.1, 12everywhere dense set, 4.7.3 (3), 52exact sequence, 2.3.4, 15exact sequence at a term, 2.3.4, 15exclave, 8.2.9, 130expanding mapping, Ex. 4.14, 55extended function, 3.4.2, 26extended real axis, 3.8.1, 35extended reals, 3.8.1, 35extension of an operator, 2.3.6, 16exterior of a set, 4.1.13, 41

exterior point, 4.1.13, 41Extreme and Discrete Lemma, 3.6.4, 32extreme point, 3.6.1, 31extreme set, 3.6.1, 31

face, 3.6.1, 31factor set, 1.2.3 (4), 4faithful representation, 8.2.2, 126family, 1.1.3 (4), 2filter, 1.3.3, 6filterbase, 1.3.1, 6finer cover, 9.6.1, 164finer filter, 1.3.6, 7finer multinorm, 5.3.1, 62finer pretopology, 9.1.2, 146finer seminorm, 5.3.3, 63finest multinorm, 5.1.10 (2), 58finite complement filter, 5.5.9 (3), 68finite descent, Ex. 8.10, 144finite-rank operator, 6.6.8, 97; 8.3.6, 132finite-valued function, 5.5.9 (6), 72first category set, 4.7.1, 52first element, 1.2.6, 5fixed point, Ex. 1.11, 8flat, 3.1.2 (5), 20formal duality, 2.3.15, 18Fourier coefficient family, 6.3.15, 89Fourier–Plancherel transform,

10.11.15, 209Fourier–Schwartz transform,

10.11.19, 211Fourier series, 6.3.16, 89Fourier transform of a distribution,

10.11.19, 211Fourier transform of a function,

10.11.3, 203Fourier transform relative to a basis,

6.3.15, 89Fr∆echet space, 5.5.2, 66Fredholm Alternative, 8.5.6, 138Fredholm index, 8.5.1, 137Fredholm operator, 8.5.1, 137Fredholm Theorem, 8.5.8, 139frontier of a set, 4.1.13, 41from A into/to B, 1.1.1, 1Fubini Theorem for distributions,

10.10.5 (8), 197Fubini Theorem for measures,

10.9.4 (6), 190full subalgebra, 11.1.5, 213fully norming set, 8.1.1, 120

Subject Index 265

Function Comparison Lemma, 3.8.3, 36function of class C(m), 10.9.9, 192function of compact support, 9.6.4, 165Function Recovery Lemma, 3.8.2, 35functor, 10.9.4 (4), 190fundamental net, 4.5.2, 47fundamental sequence, 4.5.2, 47fundamentally summable family

of vectors, 5.5.9 (7), 73

gauge, 3.8.6, 37gauge function, 3.8.6, 37Gauge Theorem, 3.8.7, 37-correspondence, 3.1.6, 21-hull, 3.1.11, 21-set, 3.1.1, 20Gelfand–Dunford Theorem in

an operator setting, 8.2.3, 127Gelfand–Dunford Theorem

in an algebraic setting, 11.3.2, 216Gelfand formula, 5.6.8, 74Gelfand–Mazur Theorem, 11.2.3, 215

Gelfand–NaΦßmark–Segal construction,11.9.11, 232

Gelfand Theorem, 7.2.2, 103Gelfand transform of an algebra,

11.6.8, 222Gelfand transform of an element,

11.6.8, 221Gelfand Transform Theorem,

11.6.9, 222general form of a compact operator

in Hilbert space, 6.6.9, 97general form of a linear functional

in Hilbert space, 6.4.2, 90general form of a weakly continuous

functional, 10.3.10, 175general position, Ex. 3.10, 39generalized derivative in the Sobolev

sense, 10.10.5 (4), 196Generalized Dini Theorem, 10.8.6, 183generalized function, 10.10.4, 195Generalized Riesz–Schauder Theorem,

8.4.10, 137generalized sequence, 1.2.16, 6Generalized Weierstrass Theorem,

10.9.9, 192germ, 8.1.14, 124GNS-construction, 11.9.11, 232GNS-Construction Theorem,

11.9.10, 231

gradient mapping, 6.4.2, 90Gram–Schmidt orthogonalization

process, 6.3.14, 89graph norm, 7.4.17, 111Graph Norm Principle, 7.4.17, 111greatest element, 1.2.6, 5greatest lower bound, 1.2.9, 5Grothendieck Criterion, 8.3.11, 133Grothendieck Theorem, 8.3.9, 133ground field, 2.1.3, 10ground ring, 2.1.1, 10group algebra, 10.9.4 (7), 191group character, 10.11.1, 202

Haar integral, 10.9.4 (1), 189Hahn–Banach Theorem, 3.5.3, 29Hahn–Banach Theorem in analytical

form, 3.5.4, 30Hahn–Banach Theorem in geometric

form, 3.8.12, 38Hahn–Banach Theorem in subdifferential

form, 3.5.4, 30Hamel basis, 2.2.9 (5), 14Hausdorff Completion Theorem,

4.5.12, 48Hausdorff Criterion, 4.6.7, 50Hausdorff metric, Ex. 4.8, 55Hausdorff multinorm, 5.1.8, 57Hausdorff multinormed space, 5.1.8, 57Hausdorff space, 9.3.5, 152Hausdorff Theorem, 7.6.12, 117Hausdorff topology, 9.3.5, 152H-closed space, Ex. 9.10, 168Heaviside function, 10.10.5 (4), 196Hellinger–Toeplitz Theorem, 6.5.3, 93hermitian element, 11.7.1, 224hermitian form, 6.1.1, 80hermitian operator, 6.5.1, 93hermitian state, 11.9.8, 230Hilbert basis, 6.3.8, 88Hilbert cube, 9.2.17 (2), 151Hilbert dimension, 6.3.13, 89Hilbert identity, 5.6.19, 78Hilbert isomorphy, 6.3.17, 90Hilbert–Schmidt norm, Ex. 8.9, 144Hilbert–Schmidt operator,

Ex. 8.9, 144Hilbert–Schmidt Theorem, 6.6.7, 96Hilbert space, 6.1.7, 81Hilbert-space isomorphism, 6.3.17, 90Hilbert sum, 6.1.10 (5), 84

266 Subject Index

HΞolder inequality, 5.5.9 (4), 69holey disk, 4.8.5, 54holomorphic function, 8.1.4, 122Holomorphy Theorem, 8.1.5, 122homeomorphism, 9.2.4, 149homomorphism, 7.4.1, 107HΞormander transform, Ex. 3.19, 39hyperplane, 3.8.9, 38hypersubspace, 3.8.9, 38

ideal, 11.4.1, 217Ideal and Character Theorem,

11.6.6, 221ideal correspondence, 7.3.3, 106Ideal Correspondence Lemma,

7.3.4, 106Ideal Correspondence Principle,

7.3.5, 106Ideal Hahn–Banach Theorem, 7.5.9, 113ideally convex function, 7.5.4, 112ideally convex set, 7.1.3, 101idempotent operator, 2.2.9 (4), 14identical embedding, 1.1.3 (3), 2identity, 10.9.4, 188identity element, 11.1.1, 213identity mapping, 1.1.3 (3), 2identity relation, 1.1.3 (3), 1image, 1.1.2, 1image of a filterbase, 1.3.5 (1), 7image of a set, 1.1.3 (5), 2image of a topology, 9.2.12, 150image topology, 9.2.12, 150Image Topology Theorem,

9.2.11, 150imaginary part of a function,

5.5.9 (4), 69increasing mapping, 1.2.3 (5), 4independent measure, 10.9.4 (3), 190index, 8.5.1, 137indicator function, 3.4.8 (2), 27indiscrete topology, 9.1.8 (3), 147induced relation, 1.2.3 (1), 4induced topology, 9.2.17 (1), 151inductive limit topology, 10.9.6, 191inductive set, 1.2.19, 6infimum, 1.2.9, 5infinite-rank operator, 6.6.8, 97infinite set, 5.5.9 (3), 68inner product, 6.1.4, 80integrable function, 5.5.9 (4), 69integral, 5.5.9 (4), 68

integral with respect to a measure,10.9.3, 188

interior of a set, 4.1.13, 41interior point, 4.1.13, 41intersection of topologies, 9.1.14, 148interval, 3.2.15, 24Interval Addition Lemma, 3.2.15, 24invariant subspace, 2.2.9 (4), 14inverse-closed subalgebra, 11.1.5, 213inverse image of a multinorm,

5.1.10 (3), 58inverse image of a preorder, 1.2.3 (3), 4inverse image of a seminorm, 5.1.4, 57inverse image of a set, 1.1.3 (5), 2inverse image of a topology, 9.2.9, 150inverse image of a uniformity,

9.5.5 (3), 160inverse image topology, 9.2.9, 150Inverse Image Topology Theorem,

9.2.8, 149inverse of a correspondence, 1.1.3 (1), 1inverse of an element in an algebra,

11.1.5, 213Inversion Theorem, 10.11.12, 208invertible element, 11.1.5, 213invertible operator, 5.6.10, 75involution, 6.4.13, 92involutive algebra, 6.4.13, 92irreducible representation, 8.2.2, 127irreflexive space, 5.1.10 (8), 59isolated part of a spectrum, 8.2.9, 130isolated point, 8.4.7, 136isometric embedding, 4.5.11, 48isometric isomorphism of algebras,

11.1.8, 215isometric mapping, 4.5.11, 48isometric representation, 11.1.8, 214isometric ∗-isomorphism, 11.8.3, 226isometric ∗-representation, 11.8.3, 226isometry into, 4.5.11, 48isometry onto, 4.5.11, 48isomorphism, 2.2.5, 13isotone mapping, 1.2.3, 4

James Theorem, 10.7.5, 181Jensen inequality, 3.4.5, 27join, 1.2.12, 5Jordan arc, 4.8.2, 54Jordan Curve Theorem, 4.8.3, 54juxtaposition, 2.2.8, 13

Subject Index 267

Kakutani Criterion, 10.7.1, 180Kakutani Lemma, 10.8.7, 183Kakutani Theorem, 7.4.11 (3), 109Kantorovich space, 3.2.8, 23Kantorovich Theorem, 3.3.4, 25Kaplansky–Fukamija Lemma,

11.9.7, 230Kato Criterion, 7.4.19, 111kernel of an operator, 2.3.1, 15ket-mapping, 10.3.1, 173ket-topology, 10.3.5, 174Kolmogorov Normability Criterion,

5.4.5, 66

KreΦßn–Milman Theorem, 10.6.5, 179

KreΦßn–Milman Theorem in subdifferentialform, 3.6.5, 33

KreΦßn–Rutman Theorem, 3.3.5, 26Krull Theorem, 11.4.8, 219Kuratowski–Zorn Lemma, 1.2.20, 6K-space, 3.2.8, 23K-ultrametric, 9.5.13, 162

last element, 1.2.6, 5lattice, 1.2.12, 5lear trap map, 3.7.4, 34least element, 1.2.6, 5Lebesgue measure, 10.9.4 (1), 189Lebesgue set, 3.8.1, 35Lefschetz Lemma, 9.6.3, 165left approximate inverse, 8.5.9, 139left Haar measure, 10.9.4 (1), 189left inverse of an element in an algebra,

11.1.3, 213lemma on continuity of a convex

function, 7.5.1, 112lemma on the numeric range

of a hermitian element, 11.9.3, 229level set, 3.8.1, 35Levy Projection Theorem, 6.2.2, 84limit of a filterbase, 4.1.16, 42Lindenstrauss space, 5.5.9 (5), 71Lindenstrauss–Tzafriri Theorem,

7.4.11 (3), 110linear change of a variable under the

subdifferential sign, 3.5.4, 30linear combination, 2.3.12, 17linear correspondence, 2.2.1, 12;

3.1.7, 21linear functional, 2.2.4, 13linear operator, 2.2.1, 12linear representation, 8.2.2, 126

linear set, 2.1.4 (3), 11linear space, 2.1.4 (3), 11linear span, 3.1.14, 22linear topological space, 10.1.3, 169linear topology, 10.1.3, 169linearly independent set, 2.2.9 (5), 14linearly-ordered set, 1.2.19, 6Lions Theorem of Supports,

10.10.5 (9), 197Liouville Theorem, 8.1.10, 123local data, 10.9.11, 193locally compact group, 10.9.4 (1), 188locally compact space, 9.4.20, 159locally compact topology, 9.4.20, 159locally convex space, 10.2.9, 172locally convex topology, 10.2.1, 171locally finite cover, 9.6.2, 164locally integrable function, 9.6.17, 167locally Lipschitz function, 7.5.6, 112loop, 4.8.2, 54lower bound, 1.2.4, 5lower limit, 4.3.5, 45lower right Dini derivative, 4.7.7, 53lower semicontinuous, 4.3.3, 44L2-Fourier transform, 10.11.15, 209

Mackey–Arens Theorem, 10.4.5, 176Mackey Theorem, 10.4.6, 176Mackey topology, 10.4.4, 176mapping, 1.1.3 (3), 1massive subspace, 3.3.2, 25matrix form, 2.2.9 (4), 14maximal element, 1.2.10, 5maximal ideal, 11.4.5, 218maximal ideal space, 11.6.7, 221Maximal Ideal Theorem, 11.5.3, 220Mazur Theorem, 10.4.9, 177meager set, 4.7.1, 52measure, 10.9.3, 188Measure Localization Principle,

10.9.10, 192measure space, 5.5.9 (4), 69meet, 1.2.12, 5member of a set, 1.1.3 (4), 2metric, 4.1.1, 40metric space, 4.1.1, 40metric topology, 4.1.9, 41metric uniformity, 4.1.5, 40Metrizability Criterion, 5.4.2, 64metrizable multinormed space, 5.4.1, 64minimal element, 1.2.10, 5

268 Subject Index

Minimal Ideal Theorem, 11.5.1, 219Minkowski–Ascoli–Mazur Theorem,

3.8.12, 38Minkowski functional, 3.8.6, 37Minkowski inequality, 5.5.9 (4), 69minorizing set, 3.3.2, 25mirror, 10.2.7, 172module, 2.1.1, 10modulus of a scalar, 5.1.10 (4), 58modulus of a vector, 3.2.12, 24mollifier, 9.6.14, 166mollifying kernel, 9.6.14, 166monomorphism, 2.3.1, 15monoquotient, 2.3.11, 17Montel space, 10.10.9 (2), 199Moore subnet, 1.3.5 (2), 7morphism, 8.2.2, 126; 11.1.2, 213morphism representing an algebra,

8.2.2, 126Motzkin formula, 3.1.13 (5), 22multimetric, 9.5.9, 161multimetric space, 9.5.9, 161multimetric uniformity, 9.5.9, 161multimetrizable topological space,

9.5.10, 161multimetrizable uniform space,

9.5.10, 161multinorm, 5.1.6, 57Multinorm Comparison Theorem,

5.3.2, 62multinorm summable family of vectors,

5.5.9 (7), 73multinormed space, 5.1.6, 57multiplication formula, 10.11.5, 205multiplication of a germ by a complex

number, 8.1.16, 125multiplicative linear operator, 8.2.2, 126

natural order, 3.2.6 (1), 23negative part, 3.2.12, 24neighborhood about a point,

9.1.1 (2), 146neighborhood about a point in a metric

space, 4.1.9, 41neighborhood filter, 4.1.10, 41neighborhood filter of a set, 9.3.7, 152neighborhood of a set, 8.1.13 (2),

124; 9.3.7, 152Nested Ball Theorem, 4.5.7, 47nested sequence, 4.5.7, 47net, 1.2.16, 6

net having a subnet, 1.3.5 (2), 7net lacking a subnet, 1.3.5 (2), 7Neumann series, 5.6.9, 75Neumann Series Expansion Theorem,

5.6.9, 75neutral element, 2.1.4 (3), 11;

10.9.4, 188

Nikol ′skiΦß Criterion, 8.5.22, 143Noether Criterion, 8.5.14, 140nonarchimedean element,

5.5.9 (5), 70nonconvex cone, 3.1.2 (4), 20Nonempty Subdifferential Theorem,

3.5.8, 31non-everywhere-defined operator,

2.2.1, 12nonmeager set, 4.7.1, 52nonpointed cone, 3.1.2 (4), 20nonreflexive space, 5.1.10 (8), 59norm, 5.1.9, 57norm convergence, 5.5.9 (7), 73normable multinormed space, 5.4.1, 64normal element, 11.7.1, 224normal operator, Ex. 8.17, 145normal space, 9.3.11, 153normalized element, 6.3.5, 88normally solvable operator, 7.6.9, 116normative inequality, 5.1.10 (7), 59normed algebra, 5.6.3, 74normed dual, 5.1.10 (8), 59normed space, 5.1.9, 57normed space of bounded elements,

5.5.9 (5), 70norming set, 8.1.1, 120norm-one element, 5.5.6, 68nowhere dense set, 4.7.1, 52nullity, 8.5.1, 137numeric family, 1.1.3 (4), 2numeric function, 9.6.4, 165numeric range, 11.9.1, 229numeric set, 1.1.3 (4), 2

one-point compactification, 9.4.22, 159one-to-one correspondence, 1.1.3 (3), 2open ball, 4.1.3, 40open ball of RN , 9.6.16, 166open correspondence, 7.3.12, 107Open Correspondence Principle,

7.3.13, 107open cylinder, 4.1.3, 40open halfspace, Ex. 3.3, 39

Subject Index 269

Open Mapping Theorem, 7.4.6, 108open segment, 3.6.1, 31open set, 9.1.4, 146open set in a metric space, 4.1.11, 41openness at a point, 7.3.6, 107operator, 2.2.1, 12operator ideal, 8.3.3, 132operator norm, 5.1.10 (7), 58operator representation, 8.2.2, 126order, 1.2.2, 3order by inclusion, 1.3.1, 6order compatible with vector structure,

3.2.1, 22order ideal, 10.8.11, 183order of a distribution, 10.10.5 (3), 195ordered set, 1.2.2, 3ordered vector space, 3.2.1, 22ordering, 1.2.2, 3ordering cone, 3.2.4, 23oriented envelope, 4.8.8, 54orthocomplement, 6.2.5, 85orthogonal complement, 6.2.5, 85orthogonal family, 6.3.1, 87orthogonal orthoprojections, 6.2.12, 86orthogonal set, 6.3.1, 87orthogonal vectors, 6.2.5, 85orthonormal family, 6.3.6, 88orthonormal set, 6.3.6, 88orthonormalized family, 6.3.6, 88orthoprojection, 6.2.7, 85Orthoprojection Summation Theorem,

6.3.3, 87Orthoprojection Theorem, 6.2.10, 85Osgood Theorem, 4.7.5, 52

pair-dual space, 10.3.3, 174pairing, 10.3.3, 174pairwise orthogonality of finitely many

orthoprojections, 6.2.14, 86paracompact space, 9.6.9, 166Parallelogram Law, 6.1.8, 81Parseval identity, 6.3.16, 89;

10.11.12, 208part of an operator, 2.2.9 (4), 14partial correspondence, 1.1.3 (6), 2partial operator, 2.2.1, 12partial order, 1.2.2, 3partial sum, 5.5.9 (7), 73partition of unity, 9.6.6, 165partition of unity subordinate

to a cover, 9.6.7, 166

patch, 10.9.11, 193perforated disk, 4.8.5, 54periodic distribution, 10.11.17 (7), 211Pettis Theorem, 10.7.4, 181Phillips Theorem, 7.4.13, 110Plancherel Theorem, 10.11.14, 209point finite cover, 9.6.2, 164point in a metric space, 4.1.1, 40point in a space, 2.1.4 (3), 11point in a vector space, 2.1.3, 10pointwise convergence, 9.5.5 (6), 161pointwise operation, 2.1.4 (4), 11polar, 7.6.8, 116; 10.5.1, 177Polar Lemma, 7.6.11, 116polarization identity, 6.1.3, 80Pontryagin–van Kampen Duality

Theorem, 10.11.2, 203poset, 1.2.2, 3positive cone, 3.2.5, 23positive definite inner product, 6.1.4, 80positive distribution, 10.10.5 (2), 195positive element of a C∗-algebra,

11.9.4, 230positive form on a C∗-algebra,

Ex. 11.11, 235positive hermitian form, 6.1.4, 80positive matrix, Ex. 3.13, 39positive operator, 3.2.6 (3), 23positive part, 3.2.12, 24positive semidefinite hermitian

form, 6.1.4, 80positively homogeneous functional,

3.4.7 (2), 27powerset, 1.2.3 (4), 4precompact set, Ex. 9.16, 168pre-Hilbert space, 6.1.7, 81preimage of a multinorm,

5.1.10 (3), 58preimage of a seminorm, 5.1.4, 57preimage of a set, 1.1.3 (5), 2preintegral, 5.5.9 (4), 68preneighborhood, 9.1.1 (2), 146preorder, 1.2.2, 3preordered set, 1.2.2, 4preordered vector space, 3.2.1, 22presheaf, 10.9.4 (4), 190pretopological space, 9.1.1 (2), 146pretopology, 9.1.1, 146primary Banach space, Ex. 7.17, 119prime mapping, 6.4.1, 90Prime Theorem, 10.2.13, 173

270 Subject Index

Principal Theorem of the HolomorphicFunctional Calculus, 8.2.4, 128

product, 4.3.2, 44product of a distribution and

a function, 10.10.5 (7), 196product of germs, 8.1.16, 125product of sets, 1.1.1, 1; 2.1.4 (4), 11product of topologies, 9.2.17 (2), 151product of vector spaces, 2.1.4 (4), 11product topology, 4.3.2, 44; 9.2.17 (2),

151projection onto X1 along X2,

2.2.9 (4), 14projection to a set, 6.2.3, 84proper ideal, 11.4.5, 218pseudometric, 9.5.7, 161p-sum, 5.5.9 (6), 71p-summable family, 5.5.9 (4), 70punctured compactum, 9.4.21, 159pure subalgebra, 11.1.5, 213Pythagoras Lemma, 6.2.8, 85Pythagoras Theorem, 6.3.2, 87

quasinilpotent, Ex. 8.18, 145quotient mapping, 1.2.3 (4), 4quotient multinorm, 5.3.11, 64quotient of a mapping, 1.2.3 (4), 4quotient of a seminormed space,

5.1.10 (5), 58quotient seminorm, 5.1.10 (5), 58quotient set, 1.2.3 (4), 4quotient space of a multinormed

space, 5.3.11, 64quotient vector space, 2.1.4 (6), 12

radical, 11.6.11, 223Radon F-measure, 10.9.3, 188Radon–Nikod∆ym Theorem, 10.9.4 (3),

190range of a correspondence, 1.1.2, 1rank, 8.5.7 (2), 139rare set, 4.7.1, 52Rayleigh Theorem, 6.5.2, 93real axis, 2.1.2, 10real carrier, 3.7.1, 33real C-measure, 10.9.4 (3), 189real distribution, 10.10.5 (5), 196real hyperplane, 3.8.9, 38real measure, 10.9.4, 189real part map, 3.7.2, 33real part of a function, 5.5.9 (4), 69real part of a number, 2.1.2, 10

real subspace, 3.1.2 (3), 20real vector space, 2.1.3, 10realification, 3.7.1, 33realification of a pre-Hilbert space,

6.1.10 (2), 83realifier, 3.7.2, 33reducible representation, 8.2.2, 127refinement, 9.6.1, 164reflection of a function, 10.10.5, 197reflexive relation, 1.2.1, 3reflexive space, 5.1.10 (8), 59regular distribution, 10.10.5 (1), 195regular operator, 3.2.6 (3), 23regular space, 9.3.9, 153regular value of an operator, 5.6.13, 76relation, 1.1.3 (2), 1relative topology, 9.2.17 (1), 151relatively compact set, 4.4.4, 46removable singularity, 8.2.5 (2), 128representation, 8.2.2, 126representation space, 8.2.2, 126reproducing cone, Ex. 7.12, 119residual set, 4.7.4, 52resolvent of an element of an algebra,

11.2.1, 215resolvent of an operator, 5.6.13, 76resolvent set of an operator, 5.6.13, 76resolvent value of an element

of an algebra, 11.2.1, 215resolvent value of an operator,

5.6.13, 76restriction, 1.1.3 (5), 2restriction of a distribution,

10.10.5 (6), 196restriction of a measure,

10.9.4 (4), 190restriction operator, 10.9.4 (4), 190reversal, 1.2.5, 5reverse order, 1.2.3 (2), 4reverse polar, 7.6.8, 116; 10.5.1, 177reversed multiplication, 11.1.6, 214Riemann function, 4.7.7, 53Riemann–Lebesgue Lemma,

10.11.5 (3), 204Riemann Theorem on Series,

5.5.9 (7), 73Riesz Criterion, 8.4.2, 134Riesz Decomposition Property,

3.2.16, 24Riesz–Dunford integral, 8.2.1, 126

Subject Index 271

Riesz–Dunford Integral DecompositionTheorem, 8.2.13, 131

Riesz–Dunford integral in an algebraicsetting, 11.3.1, 216

Riesz–Fisher Completeness Theorem,5.5.9 (4), 70

Riesz–Fisher Isomorphism Theorem,6.3.16, 89

Riesz idempotent, 8.2.11, 130Riesz–Kantorovich Theorem, 3.2.17, 24Riesz operator, Ex. 8.15, 145Riesz Prime Theorem, 6.4.1, 90Riesz projection, 8.2.11, 130Riesz–Schauder operator,

Ex. 8.11, 144Riesz–Schauder Theorem, 8.4.8, 136Riesz space, 3.2.7, 23Riesz Theorem, 5.3.5, 63right approximate inverse, 8.5.9, 139right Haar measure, 10.9.4 (1), 189right inverse of an element

in an algebra, 11.1.3, 213R-measure, 10.9.4 (3), 189rough draft, 4.8.8, 54row-by-column rule, 2.2.9 (4), 14

salient cone, 3.2.4, 23Sard Theorem, 7.4.12, 110scalar, 2.1.3, 10scalar field, 2.1.3, 10scalar multiplication, 2.1.3, 10scalar product, 6.1.4, 80scalar-valued function, 9.6.4, 165Schauder Theorem, 8.4.6, 135Schwartz space of distributions,

10.11.16, 209Schwartz space of functions,

10.11.6, 206Schwartz Theorem, 10.10.10, 199second dual, 5.1.10 (8), 59selfadjoint operator, 6.5.1, 93semi-extended real axis, 3.4.1, 26semi-Fredholm operator,

Ex. 8.13, 145semi-inner product, 6.1.4, 80semimetric, 9.5.7, 161semimetric space, 9.5.7, 161seminorm, 3.7.6, 34seminorm associated with a positive

element, 5.5.9 (5), 70seminormable space, 5.4.6, 66

seminormed space, 5.1.5, 57semisimple algebra, 11.6.11, 223separable space, 6.3.14, 89separated multinorm, 5.1.8, 57separated multinormed space, 5.1.8, 57separated topological space, 9.3.2, 151separated topology, 9.3.2, 151separating hyperplane, 3.8.13, 39Separation Theorem, 3.8.11, 38Sequence Prime Principle, 7.6.13, 117sequence space, 3.3.1 (2), 25Sequence Star Principle, 6.4.12, 92series sum, 5.5.9 (7), 73sesquilinear form, 6.1.2, 80set absorbing another set, 3.4.9, 28set in a space, 2.1.4 (3), 11set lacking a distribution,

10.10.5 (6), 196set lacking a functional, 10.8.13, 184set lacking a measure, 10.9.4 (5), 190set of arrival, 1.1.1, 1set of departure, 1.1.1, 1set of second category, 4.7.1, 52set supporting a measure,

10.9.4 (5), 190set that separates the points of another

set, 10.8.9, 183set void of a distribution,

10.10.5 (6), 196set void of a functional, 10.8.13, 184set void of a measure, 10.9.4 (5), 190setting in duality, 10.3.3, 174setting primes, 7.6.5, 115sheaf, 10.9.11, 193shift, 10.9.4 (1), 189Shilov boundary, Ex. 11.8, 234Shilov Theorem, 11.2.4, 215short sequence, 2.3.5, 16σ-compact, 10.9.8, 192signed measure, 10.9.4 (3), 190simple convergence, 9.5.5 (6), 161simple function, 5.5.9 (6), 72simple Jordan loop, 4.8.2, 54single-valued correspondence,

1.1.3 (3), 1Singularity Condensation Principle,

7.2.12, 105Singularity Fixation Principle,

7.2.11, 105skew field, 11.2.3, 215

272 Subject Index

slowly increasing distribution,10.11.16, 209

smooth function, 9.6.13, 166smoothing process, 9.6.18, 167Snowflake Lemma, 2.3.16, 18space countable at infinity, 10.9.8, 192space of bounded elements, 5.5.9 (5), 70space of bounded functions, 5.5.9 (2), 68space of bounded operators,

5.1.10 (7), 59space of compactly-supported

distributions, 10.10.5 (9), 197space of convergent sequences,

5.5.9 (3), 68space of distributions of order at most

m, 10.10.8, 199space of essentially bounded functions,

5.5.9 (5), 70space of finite-order distributions,

10.10.8, 199space of functions vanishing at infinity,

5.5.9 (3), 68space of X-valued p-summable

functions, 5.5.9 (6), 72space of p-summable functions,

5.5.9 (4), 69space of p-summable sequences,

5.5.9 (4), 70space of tempered distributions,

10.11.16, 209space of vanishing sequences,

5.5.9 (3), 68Spectral Decomposition Lemma,

6.6.6, 96Spectral Decomposition Theorem,

8.2.12, 130Spectral Endpoint Theorem, 6.5.5, 94Spectral Mapping Theorem, 8.2.5, 128Spectral Purity Theorem,

11.7.11, 226spectral radius of an operator, 5.6.6, 74Spectral Theorem, 11.8.6, 227spectral value of an element

of an algebra, 11.2.1, 215spectral value of an operator, 5.6.13, 76spectrum, 10.2.7, 172spectrum of an element of an algebra,

11.2.1, 215spectrum of an operator, 5.6.13, 76spherical layer, 6.2.1, 84∗-algebra, 6.4.13, 92

∗-isomorphism, 11.8.3, 226∗-linear functional, 2.2.4, 13∗-representation, 11.8.3, 226star-shaped set, 3.1.2 (7), 20state, 11.9.1, 229Steklov condition, 6.3.10, 88Steklov Theorem, 6.3.11, 88step function, 5.5.9 (6), 72Stone Theorem, 10.8.10, 183Stone–Weierstrass Theorem for

C(Q, C), 11.8.2, 226Stone–Weierstrass Theorem for

C(Q, R), 10.8.17, 186Strict Separation Theorem, 10.4.8, 177strict subnet, 1.3.5 (2), 7strictly positive real, 4.1.3, 40strong order-unit, 5.5.9 (5), 70strong uniformity, 9.5.5 (6), 161stronger multinorm, 5.3.1, 62stronger pretopology, 9.1.2, 146stronger seminorm, 5.3.3, 63strongly holomorphic function,

8.1.5, 122structure of a subdifferential,

10.6.3, 179subadditive functional, 3.4.7 (4), 27subcover, 9.6.1, 164subdifferential, 3.5.1, 29sublattice, 10.8.2, 181sublinear functional, 3.4.6, 27submultiplicative norm, 5.6.1, 73subnet, 1.3.5 (2), 7subnet in a broad sense, 1.3.5 (2), 7subrepresentation, 8.2.2, 126subspace of a metric space, 4.5.14, 49subspace of a topological space,

9.2.17 (1), 151subspace of an ordered vector space,

3.2.6 (2), 23subspace topology, 9.2.17 (1), 151Sukhomlinov–Bohnenblust–Sobczyk

Theorem, 3.7.12, 35sum of a family in the sense of Lp,

5.5.9 (6), 71sum of germs, 8.1.16, 125summable family of vectors,

5.5.9 (7), 73summable function, 5.5.9 (4), 69superset, 1.3.3, 6sup-norm, 10.8.1, 181support function, 10.6.4, 179

Subject Index 273

support of a distribution,10.10.5 (6), 196

support of a functional, 10.8.12, 184support of a measure, 10.9.4 (5), 190supporting function, 10.6.4, 179supremum, 1.2.9, 5symmetric Hahn–Banach formula,

Ex. 3.10, 39symmetric relation, 1.2.1, 3symmetric set, 3.1.2 (7), 20system with integration, 5.5.9 (4), 68Szankowski Counterexample,

8.3.13, 134

tail filter, 1.3.5 (2), 7τ -dual of a locally convex space,

10.2.11, 173Taylor Series Expansion Theorem,

8.1.9, 123tempered distribution, 10.11.16, 209tempered function, 5.1.10 (6), 58;

10.11.6, 206tempered Radon measure, 10.11.17 (3),

210test function, 10.10.1, 194test function space, 10.10.1, 194theorem on Hilbert isomorphy,

6.3.17, 90theorem on the equation AX = B,

2.3.13, 17theorem on the equationXA = B,

2.3.8, 16theorem on the general form

of a distribution, 10.10.14, 201theorem on the inverse image

of a vector topology, 10.1.6, 171theorem on the repeated Fourier

transform, 10.11.13, 209theorem on the structure of a locally

convex topology, 10.2.2, 171theorem on the structure of a vector

topology, 10.1.4, 170theorem on topologizing by a family

of mappings, 9.2.16, 151there is a unique x, 2.3.9, 17Tietze–Urysohn Theorem,

10.8.20, 186topological isomorphism, 9.2.4, 149topological mapping, 9.2.4, 149Topological Separation Theorem,

7.5.12, 114

topological space, 9.1.7, 147topological structure of a convex

set, 7.1.1, 100topological subdifferential, 7.5.8, 113topological vector space, 10.1.1, 169topologically complemented subspace,

7.4.9, 108topology, 9.1.7, 147topology compatible with duality,

10.4.1, 175topology compatible with vector

structure, 10.1.1, 169topology given by open sets, 9.1.12, 148topology of a multinormed space,

5.2.8, 61topology of a uniform space, 9.5.3, 160topology of the distribution space,

10.10.6, 198topology of the test function space,

10.10.6, 198total operator, 2.2.1, 12total set of functionals, 7.4.11 (2), 109totally bounded, 4.6.3, 49transitive relation, 1.2.1, 3translation, 10.9.4 (1), 189translation of a distribution,

10.11.17 (7), 211transpose of an operator, 7.6.2, 114trivial topology, 9.1.8 (3), 147truncator, 9.6.19 (1), 167truncator direction, 10.10.2 (5), 194truncator set, 10.10.2, 194twin of a Hilbert space, 6.1.10 (3), 83twin of a vector space, 2.1.4 (2), 10Two Norm Principle, 7.4.16, 111two-sided ideal, 8.3.3, 132; 11.6.2, 220Tychonoff cube, 9.2.17 (2), 151Tychonoff product, 9.2.17 (2), 151Tychonoff space, 9.3.15, 155Tychonoff Theorem, 9.4.8, 157Tychonoff topology, 9.2.17 (2), 151Tychonoff uniformity, 9.5.5 (4), 160T1-space, 9.3.2, 151T1-topology, 9.3.2, 151T2-space, 9.3.5, 152T3-space, 9.3.9, 153T31/2 -space, 9.3.15, 155T4-space, 9.3.11, 153

ultrafilter, 1.3.9, 7ultrametric inequality, 9.5.14, 162

274 Subject Index

ultranet, 9.4.4, 156unconditionally summable family

of vectors, 5.5.9 (7), 73unconditionally summable sequence,

5.5.9 (7), 73underlying set, 2.1.3, 10Uniform Boundedness Principle,

7.2.5, 103uniform convergence, 7.2.10, 105;

9.5.5 (6), 161uniform space, 9.5.1, 159uniformity, 9.5.1, 159uniformity of a multinormed space,

5.2.4, 60uniformity of a seminormed space,

5.2.2, 60uniformity of a topological vector

space, 10.1.10, 171uniformity of the empty set, 9.5.1, 159uniformizable space, 9.5.4, 160uniformly continuous mapping, 4.2.5, 44unit, 10.9.4, 188unit ball, 5.2.11, 61unit circle, 8.1.3, 121unit disk, 8.1.3, 121unit element, 11.1.1, 213unit sphere, Ex. 10.6, 212unit vector, 6.3.5, 88unital algebra, 11.1.1, 213unitary element, 11.7.1, 224unitary operator, 6.3.17, 90unitization, 11.1.2, 213unity, 11.1.1, 213unity of a group, 10.9.4 (1), 188unity of an algebra, 11.1.1, 213unordered sum, 5.5.9 (7), 73unorderly summable sequence,

5.5.9 (7), 73Unremovable Spectral Boundary

Theorem, 11.2.6, 216upper bound, 1.2.4, 4upper envelope, 3.4.8 (3), 28

upper right Dini derivative, 4.7.7, 53upward-filtered set, 1.2.15, 6Urysohn Great Lemma, 9.3.13, 154Urysohn Little Lemma, 9.3.10, 153Urysohn Theorem, 9.3.14, 1552-Ultrametric Lemma, 9.5.15, 162

vague topology, 10.9.5, 191value of a germ at a point, 8.1.21, 126van der Waerden function, 4.7.7, 53vector, 2.1.3, 10vector addition, 2.1.3, 10vector field, 5.5.9 (6), 71vector lattice, 3.2.7, 23vector space, 2.1.3, 10vector sublattice, 10.8.4 (4), 182vector topology, 10.1.1, 169Volterra operator, Ex. 5.12, 79von Neumann–Jordan Theorem,

6.1.9, 81V -net, 4.6.2, 49V -small, 4.5.3, 47

weak derivative, 10.10.5 (4), 196weak multinorm, 5.1.10 (4), 58weak topology, 10.3.5, 174weak∗ topology, 10.3.11, 175weak uniformity, 9.5.5 (6), 161weaker pretopology, 9.1.2, 146weakly holomorphic function, 8.1.5, 122weakly operator holomorphic function,

8.1.5, 122Weierstrass function, 4.7.7, 53Weierstrass Theorem, 4.4.5, 46;

9.4.5, 157Well-Posedness Principle, 7.4.6, 108Wendel Theorem, 10.9.4 (7), 191Weyl Criterion, 6.5.4, 93

X-valued function, 5.5.9 (6), 71

Young inequality, 5.5.9 (4), 69

zero of a vector space, 2.1.4 (3), 11

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Chalmers not… · Contents vii Chapter6. HilbertSpaces 85 x6.1. HermitianFormsandInnerProducts ::::::::::::::::::::::::: 85 x6.2. Orthoprojections