Embed Size (px)
Citation preview
Lecture Notes on Functional Analysis
and
Linear Partial Differential Equations
Alberto Bressan
August 17, 2010
Contents
1 Introduction 5
0.1 Linear systems of algebraic equations . . . . . . . . . . . . . . . . . . . 5
0.2 Linear systems of Ordinary Differential Equations . . . . . . . . . . . . 7
1 Summary of contents . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . 8
2 Banach Spaces 11
1 Basic definitions . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . 11
1.1 Examples of normed and Banach spaces . . . . . . . . . . . . . . . . . . 13
2 Linear operators . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . 15
2.1 Examples of linear operators . . . . . . . . . . . . . . . . . . . . . . . . 16
3 Finite dimensional spaces . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . 18
4 Extension theorems . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . 20
5 Dual spaces and weak convergence . . . . . . . . . . . . . . . . . . . . . . . . . 25
6 Problems . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . 26
3 Spaces of Continuous Functions 34
1 Bounded continuous functions . . . . . . . . . . . . . . . . . . . . . . . . . . . . 34
2 The Stone-Weierstrass approximation theorem . . . . . . . . . . . . . . . . . . 35
3 Ascoli’s compactness theorem . . . . . . . . . . . . . . . . . . . . . . . . . . . . 39
4 Spaces of Holder continuous functions . . . . . . . . . . . . . . . . . . . . . . . 40
5 Problems . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . 41
4 Linear Operators 44
1 The open mapping theorem . . . . . . . . . . . . . . . . . . . . . . . . . . . . . 44
1
2 The closed graph theorem . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . 46
3 Adjoint operators . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . 47
4 Compact operators . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . 48
5 Problems . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . 50
5 Hilbert Spaces 52
1 Spaces with an inner product . . . . . . . . . . . . . . . . . . . . . . . . . . . . 52
2 Orthogonal projections . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . 54
3 Linear functionals on a Hilbert space . . . . . . . . . . . . . . . . . . . . . . . . 56
4 Gram-Schmidt orthogonalization . . . . . . . . . . . . . . . . . . . . . . . . . . 57
5 Orthonormal sets . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . 58
5.1 Fourier series . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . 61
6 Positive definite operators . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . 61
7 Weak convergence . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . 64
8 Problems . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . 66
6 Compact Operators on a Hilbert Space 70
1 Fredholm theory . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . 70
2 Spectrum of a compact operator . . . . . . . . . . . . . . . . . . . . . . . . . . 74
3 Symmetric operators . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . 75
4 Problems . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . 78
7 Linear Semigroups 80
1 O.D.E’s in a Banach space . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . 80
1.1 Linear homogeneous ODEs . . . . . . . . . . . . . . . . . . . . . . . . . 82
2 Semigroups of linear operators . . . . . . . . . . . . . . . . . . . . . . . . . . . 84
2.1 Definition and basic properties of semigroups . . . . . . . . . . . . . . . 86
3 Resolvents . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . 88
4 Generation of a semigroup . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . 92
5 Problems . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . 96
8 Sobolev Spaces 100
2
1 Distributions and weak derivatives . . . . . . . . . . . . . . . . . . . . . . . . . 100
1.1 Distributions . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . 102
2 Mollifications . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . 105
3 Sobolev spaces . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . 107
4 Approximations and extensions of Sobolev functions . . . . . . . . . . . . . . . 113
5 Embedding theorems . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . 118
5.1 Morrey’s inequality . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . 118
5.2 The Gagliardo-Nirenberg inequality . . . . . . . . . . . . . . . . . . . . 122
5.3 High order Sobolev estimates . . . . . . . . . . . . . . . . . . . . . . . . 125
6 Compact embeddings . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . 127
7 Differentiability properties . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . 130
8 Problems . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . 131
9 Linear Partial Differential Equations 135
1 Second order elliptic equations . . . . . . . . . . . . . . . . . . . . . . . . . . . 135
1.1 Existence of solutions . . . . . . . . . . . . . . . . . . . . . . . . . . . . 137
1.2 Eigenfunction representation . . . . . . . . . . . . . . . . . . . . . . . . 139
1.3 General linear elliptic operators . . . . . . . . . . . . . . . . . . . . . . . 140
2 Parabolic equations . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . 142
2.1 Representation of solutions in terms of eigenfunctions . . . . . . . . . . 144
2.2 More general operators . . . . . . . . . . . . . . . . . . . . . . . . . . . 146
3 Hyperbolic equations . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . 150
4 Problems . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . 154
10 Appendix 159
1 Metric and Topological Spaces . . . . . . . . . . . . . . . . . . . . . . . . . . . 159
2 The Contraction Mapping Theorem . . . . . . . . . . . . . . . . . . . . . . . . 161
3 The Baire Category Theorem . . . . . . . . . . . . . . . . . . . . . . . . . . . . 162
4 Review of Lebesgue measure theory . . . . . . . . . . . . . . . . . . . . . . . . 163
4.1 Measurable sets . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . 163
4.2 Lebesgue integration . . . . . . . . . . . . . . . . . . . . . . . . . . . . . 164
3
5 Mollifications . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . 167
5.1 Partitions of unity . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . 170
6 Integrals of functions taking values in a Banach space . . . . . . . . . . . . . . 171
7 Inequalities . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . 172
7.1 Convex sets and convex functions . . . . . . . . . . . . . . . . . . . . . . 172
7.2 Basic inequalities . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . 174
7.3 A differential inequality . . . . . . . . . . . . . . . . . . . . . . . . . . . 176
8 Some technical points . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . 177
9 Problems . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . 178
10 Hints to selected problems . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . 180
10.1 Chapter 2 . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . 180
10.2 Chapter 3 . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . 181
10.3 Chapter 4 . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . 181
10.4 Chapter 5 . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . 182
10.5 Chapter 6 . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . 183
10.6 Chapter 7 . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . 183
10.7 Chapter 8 . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . 183
10.8 Chapter 9 . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . 185
10.9 Chapter 10 . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . 186
4
Chapter 1
Introduction
These lecture notes provide an introduction to linear functional analysis, extending techniquesand results of classical linear algebra to infinite dimensional spaces. With the development ofa theory of function spaces, these techniques provide a solution to basic problems in ordinaryand partial differential equations.
The following remarks highlight some key results in linear algebra, and their infinite-dimensional counterparts.
0.1 Linear systems of algebraic equations
Let A be an n × n matrix. Given a vector b ∈ IRn, a fundamental problem in linear algebraconsists in finding a vector x ∈ IRn such that
Ax = b . (0.1)
In the theory of linear PDEs, an analogous problem is the following. Consider a boundedopen set Ω ⊂ IRn, and a linear partial differential operator
Lu.= −
n∑
i,j=1
(aij(x)uxi)xj +n∑
i=1
bi(x)uxi + c(x)u . (0.2)
Given a function f : Ω 7→ IR, we seek a function u, vanishing on the boundary of Ω, such that
Lu = f . (0.3)
There are fundamental differences between the problems (0.1) and (0.3). The matrix A yieldsa continuous linear transformation on the finite dimensional space IRn. On the other hand, thedifferential operator L can be regarded as an unbounded (hence discontinuous) linear operatoron the infinite dimensional function space L2(Ω). In particular, the domain of L is not theentire space L2(Ω) but only on a suitable subspace.
In spite of these differences, since both (0.1) and (0.3) are linear problems, there are a numberof techniques from linear algebra that remain valid for the problem (0.3).
5
1 - POSITIVITY
Assume the matrix A is positive definite, i.e., there exists a constant β > 0 such that
〈Ax, x〉 ≥ β |x|2 for all x ∈ IRn.
Then the equation (0.1) has a unique solution for every b ∈ IRn.
This result has a direct counterpart for the elliptic PDEs: Namely, assume that the operatorL is positive definite, in the sense that (after a formal integration by parts)
(Lu, u)L2 =
∫
Ω
(aij(x)uxiuxj +
n∑
i=1
bi(x)uxiu+ c(x)u2)dx ≥ β ‖u‖2H1
0 (Ω) (0.4)
for some constant β > 0 and all u ∈ H10 (Ω). Here H
10 (Ω) is the space of Lebesgue measurable
functions which vanish on the boundary of Ω and such that
‖u‖H10 (Ω)
.=
(∫
Ω|u|2 + |∇u|2 dx
)1/2
< ∞ .
If (0.4) holds, one can then prove that the problem (0.3) has a unique solution u ∈ H10 (Ω).
for every given f ∈ L2(Ω).
A key assumption, in order for the inequality (0.4) to hold, is that the operator L should beelliptic. Namely, at each point x ∈ Ω the n × n matrix (aij(x)) should be strictly positivedefinite.
2 - FREDHOLM ALTERNATIVE
A well known criterion in linear algebra states that the equation (0.1) has a unique solutionfor every given b ∈ IRn if and only if the homogeneous equation
Ax = 0
has only the solution x = 0. Of course, this holds if and only if the matrix A is invertible.
In general, continuous linear operators on an infinite dimensional space X do not share thisproperty. In other words, one may construct a bounded linear operator Λ : X 7→ X which isone-to-one but not onto, or viceversa.
Still, the finite dimensional theory carries over to an important class of operators, having theform Λ = I − K where I is the identity and K is a compact operator. If Λ is in this class,then one can still prove the equivalence
Λ is one-to-one ⇐⇒ Λ is onto.
By an application of this theory it follows that, for a linear elliptic operator, the equation(0.3) has a unique solution u ∈ H1
0 (Ω) for every f ∈ L2(Ω) if and only if the homogeneousequation
Lu = 0
6
has only the zero solution.
3 - DIAGONALIZATION
Let A be an n × n symmetric matrix. Then one can find an orthonormal basis v1, . . . ,vnof the Euclidean space IRn consisting of eigenvectors of A. Moreover, the correspondingeigenvalues λ1, . . . , λn are all real numbers.
In this case, the solution x of (0.1) can be more easily computed in terms of the orthonormalbasis:
x =n∑
i=1
〈x,vi〉vi, Ax =n∑
i=1
λi 〈x,vi〉vi = b =n∑
i=1
〈b,vi〉vi .
Hence, if all eigenvalues are non-zero, we have the explicit formula
x =n∑
i=1
1
λi〈b,vi〉vi .
The above results remain valid for compact symmetric operators on an infinite dimensionalHilbert space H. Saying that K : H 7→ H is symmetric means that 〈Ku , v〉 = 〈u , v〉 forevery u, v ∈ H, where 〈·, ·〉 is now the inner product on the Hilbert space.
This theory can be used in the analysis of the elliptic operator L in (0.2), whenever bi(x) ≡ 0.In this case, one can find a countable orthonormal basis φ1, φ2, . . . of the space L2(Ω)consisting of functions φk ∈ H1
0 (Ω) such that
Lφk = λkφk (0.5)
for a suitable sequence of eigenvalues λk → +∞. Assuming that λk 6= 0 for all k, the uniquesolution of (0.3) can now be written as
u =∞∑
k=1
1
λk〈f, φk〉φk . (0.6)
0.2 Linear systems of Ordinary Differential Equations
Let A be an n× n matrix. For a given initial state x ∈ IRn, consider the Cauchy problem
d
dtx(t) = Ax(t), x(0) = b . (0.7)
According to linear ODE theory, this problem has a unique solution, given by
x(t) = etAb =
( ∞∑
k=0
tkAk
k!
)b. (0.8)
Notice that the right hand side of (0.6) is defined as an absolutely convergent series of n × nmatrices. The family of matrices etA ; t ∈ IR has the “group property”, namely
e0A = I , etAesA = e(t+s)A for all t, s ∈ IR .
7
In the special case where A is symmetric and admits an orthonormal basis of eigenvectorsv1, . . . ,vn, with the corresponding eigenvalues λ1, . . . , λn, this solution can be written moreexplicitly as
etAb =n∑
k=1
etλk〈b, vk〉vk .
The theory of linear semigroups provides an extension of these results to unbounded linear op-erators in infinite dimensional spaces. In particular, it applies to parabolic evolution equationsof the form
d
dtu(t) = − Lu(t), u(0) = g ∈ L2(Ω) , (0.9)
where L is the partial differential operator in (0.2). In the case where bi(x) ≡ 0 and theelliptic operator L is symmetric, the solution can be decomposed along the orthonormal basisφ1, φ2, . . . of the space L2(Ω), considered at (0.5). This yields the representation
u(t) = e−Ltg =∞∑
k=1
e−tλk〈g, φk〉φk t ≥ 0 . (0.10)
Notice that, while the operator L is unbounded (its eigenvalues satisfy λk → +∞ as k →∞),the exponential operators e−L t in (0.10) are bounded for every t ≥ 0 (but not for t < 0).
1 Summary of contents
After the present Introduction, these lecture notes contain eight main chapters.
• Chapter 1 starts with the definition and basic properties of Banach spaces, with sev-eral examples. We then introduce bounded linear operators, defining their norms anddiscussing various examples. The following section contains an analysis of finite dimen-sional subspaces, including a proof that a locally compact Banach space must be finitedimensional. The last sections are devoted to he Hahn-Banach extension theorem, withapplications to separation of convex sets.
• Chapter 2 is devoted to spaces of continuous functions. After the basic definitions, someuseful convergence theorems are proved. Section 2 is devoted to the Stone-Weierstrassapproximation theorem, while Section 3 contains a proof of the Ascoli-Arzela compact-ness theorem. The chapter is concluded with a few remarks on spaces of Holder contin-uous functions.
• Chapter 3 contains some basic theory of abstract linear operators. The first sectioncovers the open mapping and the closed graph theorem. The second section presentsa few facts about adjoint operators. The chapter is concluded by proving a few factsabout compact operators.
• Chapter 4 discusses Hilbert spaces. After the basic definitions, Section 2 establishes themain properties of orthogonal projections. Riesz’ theorem on representation of contin-uous linear functions is proved in Section 3. The following sections describe the Gram-Schmidt orthogonalization technique, and the construction of orthonormal Hilbert bases.
8
This method is applied, in particular, to the representation of a function as sum of aFourier series. The following sections introduce the notion of positive definite operators,and prove the Lax-Milgram theorem on unique solvability of a linear equation definedby a strictly positive operator. The chapter is concluded by the analysis of weak conver-gence of a sequence of elements in a Hilbert space. Relations with strong convergence,and compositions with compact operators, are also discussed.
• Chapter 5 is concerned with compact operators in a Hilbert space. The first sectionpresents the Fredholm theory of operators of the form I+K, obtained as the sum of theidentity plus a compact operator. Many of the results proved in classical linear algebrafor linear operators on the Euclidean space IRn remain valid in this more general setting.The second section discusses eigenvalues and eigenvectors of compact operators. Finally,the last section contains a proof of the fundamental Hilbert-Schmidt theorem: given acompact self adjoint operator K on a Hilbert space H, one can select an orthonormalbasis of H consisting of eigenvectors of K.
• Chapter 6 contains an introduction to the theory of linear semigroups, extending thedefinition of the exponential function etA to a suitable class of (possibly unbounded)linear operators. The first sections review the theory of ODE’s, and examine the con-struction of forward and backward Euler approximate solutions. Section 4 describes themain properties of linear semigroups and their generators. Section 5 covers the definitionand properties of resolvent operators, and clarifies their connection with backward Eulerapproximations. Finally, the fundamental theorem on existence and uniqueness of thesemigroup generated by a linear operator is proved in the Section 6.
• Chapter 7 introduces Sobolev spaces and discusses their basic properties. These providea key tool, in order to apply abstract functional analysis toward the solution of prob-lems in ordinary and partial differential equations. The first sections introduce the basicdefinitions of distribution and weak derivatives. The link between weak derivatives andpartial derivatives in the classical sense is here established by a mollification technique,described in Section 3. Section 4 introduces the definitions of Sobolev spaces and theirelementary properties. Approximation and extensions of Sobolev functions are discussedin Section 5, while Sections 6 to 8 work out proofs of the Morrey and the Gagliardo-Nirenberg inequalities, as well as more general Sobolev embedding theorems. Section 9is devoted to the Rellich-Kondrachov compact embedding theorem, which plays a keyrole in many existence proofs in PDE’s and in the Calculus of Variations. As a first ap-plication, we prove an inequality by Poincare, relating the Lp norm of a function (havingzero average) with the Lp norm of its gradient. The final section establishes an extensionof Rademacher’s theorem, proving that certain Sobolev functions are a.e. differentiable.
• Chapter 8 covers applications of the previous theory to problems in linear PDE’s. Sec-tion 1 introduces the definition of uniformly elliptic operators, and the concept of weaksolution to an elliptic boundary value problem. Using the Riesz representation theorem,or the Lax-Milgram theorem, some results on the existence and uniqueness of weak solu-tions are proved in Section 2. Section 3 covers the case of a self-adjoint operators, whichcan be diagonalized in terms of an orthonormal basis of the underlying L2 space. In thiscase, the solution admits a simple representation as sum of a series of eigenfunctions.Section 4 shows that, for more general elliptic operators, the Fredholm theory applies:the existence and uniqueness of solutions to a boundary value problem with general righthand side can be determined by looking at the corresponding homogeneous problem.
9
The following sections cover initial-boundary value problems of parabolic type. Thefirst result proves the existence of a semigroup of solutions, in the case where the ellipticoperator is strictly positive definite. In this operator is also self-adjoint, we obtain arepresentation of the form (0.10), in terms of an orthonormal basis of eigenfunctions. Afurther result establishes the existence of the semigroup of solutions, for general ellipticoperators. This section is concluded by discussing the properties of semigroup trajecto-ries, and in which sense these trajectories provide solutions to the parabolic evolutionproblem.
The final section is concerned with second order hyperbolic equations. Here the analysisis restricted to the case of a self-adjoint elliptic operator L, which can thus be diagonal-ized in terms of an orthonormal basis of eigenfunctions. We show that the second orderevolution equation can be reduced to a first order system. A semigroup of boundedlinear operators is then constructed, using the basis of eigenfunctions. The section isconcluded by discussing in which sense the semigroup trajectories provide solutions tothe hyperbolic problem.
• At the end, an Appendix collects various background material. This includes: definitionand properties of metric spaces, the contraction mapping theorem, the Baire categorytheorem, a review of Lebesgue measure theory, mollification techniques and partitions ofunity, integrals of functions taking values in a Banach space, a collection of inequalities,and a version of Gronwall’s lemma.
These notes are illustrated by 30 figures, while 130 homework problems, together with somehints for the solutions, are collected at the end of the various chapters.
10
Chapter 2
Banach Spaces
Given a vector space X over the real (or complex) numbers, we wish to introduce a distanced(·, ·) among points of X. This will allow us to define limits, convergent sequences andseries, and continuous mappings. In particular, we shall be able to construct solutions tomathematical equations by taking limits of a sequence of approximations.
Since X has the algebraic structure of a vector space, the distance d should be consistent withthis structure. It is thus natural to require the following properties:
• The distance d is positively homogeneous. Namely: d(λx, λy) = |λ| d(x, y), for any scalarλ ∈ IR and any x, y ∈ X.
• The distance d is invariant under translations. In particular, d(x, y) = d(x− y, 0).
• Every ball B(x0, r).= x ∈ X ; d(x, x0) < r is a convex set.
The invariance under translations implies that the distance d(·, ·) is entirely determined assoon as we specify the function x 7→ ‖x‖ = d(x, 0), i.e. the distance of a point x from theorigin. This is what we call the norm of a vector x ∈ X. It can be taken as starting pointfor the entire theory.
1 Basic definitions
Let X be a (possibly infinite dimensional) vector space, over the field IK of numbers. Weshall always assume that either IK = IR is the field of real numbers or IK = C is the fieldof complex numbers. A norm on X is a map x 7→ ‖x‖, from X into IR, with the followingproperties.
(N1) For every x ∈ X one has ‖x‖ ≥ 0, with equality holding if and only if x = 0.
(N2) For every x ∈ X and λ ∈ IK one has ‖λx‖ = |λ| ‖x‖.
11
(N3) For every x, y ∈ X there holds ‖x+ y‖ ≤ ‖x‖+ ‖y‖.
One can think of ‖x‖ as measuring the size of the vector x.
A vector space X with a norm ‖ · ‖ satisfying (N1)–(N3) is called a normed space.
A norm determines a distance between vectors, namely
d(x, y).= ‖x− y‖. (1.1)
We check that, for all x, y, z ∈ X, the three basic properties of a distance are satisfied:
(D1) d(x, y) ≥ 0, d(x, y) = 0 if and only if x = y,
(D2) d(x, y) = d(y, x),
(D3) d(x, z) ≤ d(x, y) + d(y, z).
Indeed, (D1) is an immediate consequence of (N1). To prove (D2) we write
d(y, x) = ‖y − x‖ = ‖(−1)(x − y)‖ = | − 1| ‖x − y‖ = ‖x− y‖.
The triangular inequality follows from (N3), replacing x, y with x− y and y − z, respectively:
d(x, z) = ‖x− z‖ = ‖(x− y) + (y − z)‖ ≤ ‖x− y‖+ ‖y − z‖ = d(x, y) + d(y, z).
In turn, a distance determines a topology on the vector space X. We can thus talk aboutopen sets, closed sets, convergent sequences, continuous mappings.
The open ball centered at a point x with radius r > 0 will be denoted as
B(x, r).= y ∈ X ; ‖y − x‖ < r.
If B(x, r) ⊆ V for some r > 0, we say that the set V is a neighborhood of x. Equivalently,x is an interior point of V .
A set V ⊆ X is open if it is a neighborhood of each of its points. In other words, for everyx ∈ V , there exists r > 0 such that B(x, r) ⊆ V .
A set U ⊆ X is closed if its complement X \ U is open.
We say that sequence (xn)n≥1 converges to a point x ∈ X if
limn→∞
‖xn − x‖ = 0.
12
Given two normed spaces X,Y , we say that a map f : X 7→ Y is continuous if, for everyx ∈ X and ε > 0 there exists δ > 0 such that
‖f(x′)− f(x)‖ < ε whenever x′ ∈ X, ‖x′ − x‖ < δ.
A sequence (xn)n≥1 is a Cauchy sequence if for every ε > 0 one can find an integer N largeenough so that
‖xm − xn‖ ≤ ε whenever m,n ≥ N.
A normed space X is complete if every Cauchy sequence converges to some limit point. Acomplete normed space is called a Banach space.
1.1 Examples of normed and Banach spaces
2x
x1
2x
1x x1
x2
−1 1
−1
1
Figure 2.1: The unit balls in IR2 with norms ‖ · ‖1 (left), ‖ · ‖2 (center), and ‖ · ‖∞ (right).
1. The finite dimensional space IRn = x = (x1, . . . , xn) , xi ∈ IR with Euclidean norm
‖x‖2 .=
(x21 + · · ·+ x2n
)1/2(1.2)
is a Banach space over the real numbers.
2. On the space IRn one can consider the alternative norms
‖x‖p .=(|x1|p + · · ·+ |xn|p
)1/p, ‖x‖∞ .
= max1≤i≤n
‖xi|.
Here 1 ≤ p <∞. Each of these norms also makes IRn into a Banach space.
3. For a given interval [a, b], the space of continuous functions f : [a, b] 7→ IR with norm
‖f‖C0.= max
x∈[a,b]|f(x)| (1.3)
is a Banach space.
13
4. Let Ω be an open subset of IRn. For every 1 ≤ p < ∞, the space Lp(Ω) of all Lebesguemeasurable functions f : Ω 7→ IR, with norm
‖f‖Lp.=
(∫
Ω|f(x)|p dx
)1/p
(1.4)
is a Banach space. Two functions f, f are here regarded as the same element of Lp(Ω) if they
coincide almost everywhere, i.e. if meas(x ∈ Ω ; f(x) 6= f(x)
)= 0.
Similarly, the space L∞(Ω) of all essentially bounded, measurable functions on Ω is a Banachspace with norm
‖f‖L∞.= ess- sup
x∈Ω|f(x)|. (1.5)
5. For a fixed p ≥ 1, consider the space of all sequences of real numbers whose p-th powersare summable:
ℓp.=x = (x1, x2, . . .) ;
∞∑
k=1
|xk|p <∞
(1.6)
This is a Banach space with norm
‖x‖p .=
( ∞∑
k=1
|xk|p)1/p
. (1.7)
6. The space ℓ∞ of all bounded sequences of real numbers, with norm
‖x‖∞ .= sup
k≥1|xk|, (1.8)
is a Banach space. Within this space, we can consider the subspace c0 of sequences (xk)k≥1
that converge to zero, as k →∞. This is also a Banach space, for the norm (1.8).
Remark 1. Within the space ℓp, 1 ≤ p <∞, consider the family of unit vectors
e1 = (1, 0, 0, 0, . . .), e2 = (0, 1, 0, 0, . . .), e3 = (0, 0, 1, 0, . . .), . . . (1.9)
These are linearly independent. The set of all linear combinations
spanek ; k ≥ 1 = N∑
k=1
θkek ; N ≥ 1, θk ∈ IR
(1.10)
does not coincide with the entire space ℓp, but is a dense subset of ℓp. (Note: as linearcombination we always mean a finite sum, not a series.) Indeed, it consists of all vectors ofthe form x = (x1, x2, . . . , xN , 0, 0, 0, . . .) whose coordinates are all equal to 0 beyond a certainindex.
The set ek ; k ≥ 1 is not an algebraic basis for ℓp, but it provides a topological basis. Inother words, every element x ∈ ℓp can be obtained as a sum of a convergent series
x =∞∑
k=1
xk ek .
Namely, defining the partial sums yn.=∑n
k=1 xk ek, one has
limn→∞ ‖yn − x‖p = 0. (1.11)
14
2 Linear operators
Let X,Y be normed spaces on the same field IK of scalar numbers. A linear operator is amapping Λ from a subspace Dom(Λ) ⊆ X into Y such that
Λ(c1x1 + c2x2) = c1Λx1 + c2Λx2 for all x1, x2 ∈ X c1, c2 ∈ IK.
Here Dom(Λ) is the domain of Λ. The range of Λ is the subspace
Range(Λ).= Λx ; x ∈ Dom(Λ) ⊂ Y.
The null space, or kernel of Λ is the subspace
Ker(Λ).= x ∈ X ; Λx = 0 ⊆ X.
Notice that Λ is one-to-one if and only if Ker(Λ) = 0. The operator Λ is densely definedif its domain is dense in X, i.e. if Dom(Λ) = X.
A mapping φ : X 7→ Y is bounded if the image of a bounded subset of X is a bounded subsetof Y . If Λ : X 7→ Y is a linear operator defined on the entire space X (i.e., with Dom(Λ) = X),then Λ is bounded if and only if
‖Λ‖ .= sup
‖x‖≤1‖Λx‖ = sup
‖x‖=1‖Λx‖ = sup
x 6=0
‖Λx‖‖x‖ < ∞. (2.1)
A bounded linear operator Λ : X 7→ Y is compact if, for every bounded sequence (xn)n≥1 ofpoints in X, there exists a subsequence (xnj)j≥1 such that Λxnj converges to some limit pointy ∈ Y .
Theorem 1.1 (continuity of bounded operators). A linear operator Λ : X 7→ Y isbounded if and only if it is continuous.
Proof. Assume that Λ is continuous at the origin. Then, choosing ε = 1, there exists δ > 0such that ‖x‖ ≤ δ implies ‖Λx‖ ≤ 1. By linearity, this implies
‖Λx‖ ≤ 1
δwhenever ‖x‖ ≤ 1 .
Hence Λ is bounded.
Viceversa, let Λ be bounded, so that (2.1) holds. Then, by linearity, ‖Λx1−Λx2‖ ≤ ‖Λ‖ ‖x1−x2‖, showing that Λ is uniformly Lipschitz continuous, with Lipschitz constant ‖Λ‖.
If X,Y are normed spaces over the same field of scalars, we denote by B(X; Y ) the space ofbounded linear operators from X into Y . Notice that here we require that the domain ofthese operators should be the entire space X.
Theorem 1.2 (the space of bounded linear operators). The space B(X; Y ) of allbounded linear operators from X into Y is a normed space, with norm defined at (2.1). If Yis a Banach space, then B(X; Y ) is a Banach space.
15
Proof. If Λ1,Λ2 are linear operators, and c1, c2 ∈ IK, then the linear combination is of coursedefined as
(c1Λ1 + c2Λ2)(x).= c1Λ1x+ c2Λ2x .
We now check that the properties (N1)–(N3) of a norm are satisfied.
1. If Λ = 0 is the zero operator, then Λx = 0 for all x ∈ X, and ‖Λ‖ = 0. On the other hand,if Λ is not the zero operator, then Λx 6= 0 for some x and the supremum in (2.1) is strictlypositive. This proves (N1).
2. If α ∈ IK, then (N2) is proved by
‖αΛ‖ = sup‖x‖≤1
‖αΛx‖ = |α| sup‖x‖≤1
‖Λx‖ = ‖α‖Λ‖ .
3. To check the triangle inequality (N3), for every x ∈ X with ‖x‖ ≤ 1 we write
‖(Λ2 + Λ2)x‖ = ‖Λ1x+ Λ2x‖ ≤ ‖Λ1x‖+ ‖Λ2x‖ ≤ ‖Λ1‖+ ‖Λ2‖.
Taking the supremum over all x ∈ X with ‖x‖ ≤ 1 we obtain (N3).
Next, assume that Y is a Banach space. We need to show that B(X; Y ) is complete. Let(Λn)n≥1 be a Cauchy sequence of bounded linear operators. For every x ∈ X, this implies
lim supm,n→∞
‖Λmx− Λnx‖ ≤ lim supm,n→∞
‖Λm − Λn‖ ‖x‖ = 0.
Therefore the sequence of points (Λnx)n≥1 is Cauchy in Y . Since Y is complete, this sequencehas a unique limit, which we call Λx.
It is clear that Λ is linear. We claim that Λ is also bounded (and hence continuous). Byassumption, there exists N large enough such that
‖Λk − ΛN‖ ≤ 1 for all k ≥ N.
Therefore, for any x ∈ X with ‖x‖ ≤ 1 one has
‖Λx‖ = limn→∞
‖Λkx‖ ≤ ‖ΛNx‖+ lim supn→∞
‖Λk − ΛN‖‖x‖ ≤ ‖ΛN‖+ 1 .
Since x was an arbitrary point in the unit ball, this proves the boundedness of the limitoperator Λ.
2.1 Examples of linear operators
Example 1 (diagonal operators on a space of sequences). Let 1 ≤ p ≤ ∞, and considerthe space X = ℓp of all sequences x = (x1, x2, . . .) of real numbers, with norm defined at (1.7)or (1.8). Let (λ1, λ2, . . .) be an arbitrary sequence of real numbers, and define the operatorΛ : X 7→ X as
Λ(x1, x2, x3, . . .).= (λ1x1 , λ2x2 , λ3x3 , . . . ) (2.2)
16
With reference to the basis of unit vectors e1, e2, . . . in (1.9), we can think of Λ as an infinitematrix:
λ1 00 λ2
λ3. . .
with λ1, λ2, . . . along the diagonal and 0 everywhere else. In this case we have
(i) If the sequence (λk)k≥1 is bounded, then the operator Λ is bounded, and its norm is
‖Λ‖ = supk|λk| . (2.3)
(ii) If the sequence (λk)k≥1 is unbounded, then the operator Λ is not bounded. Its domain
Dom(Λ) =x ∈ ℓp ; Λx ∈ ℓp
is a vector subspace, strictly contained in ℓp.
(iii) If λk → 0 as k →∞, then the operator Λ is compact.
In spaces of functions, as a “rule of thumb” one might say that integral operators are com-pact, shift operators and multiplications operators (by a bounded function) are bounded -hence continuous, while differential operators are unbounded - hence discontinuous. The nextexamples will make these claims a bit more precise.
Example 2 (differentiation). Let I = ]0, π[ and let X = C(I; IR) be the space of bounded,continuous real valued functions on I. Then the differential operator Λf = f ′ is a linear,unbounded operator of X into itself. Indeed, consider the sequence of functions fk = sin kx.Then f ′k = k cos kx, hence
‖fk‖ = 1 , ‖Λfk‖ = k for all k ≥ 1.
The domain Dom(Λ) of this differential operator is the space of functions f ∈ C(I ; IR) whichare everywhere differentiable and have a bounded, continuous derivative. This is a proper,dense subspace of C(I ; IR).
Example 3 (shift operator on Lp(IR)). Let 1 ≤ p ≤ ∞. Fix any a ∈ IR. Given a functionf ∈ Lp(IR), define (Λaf)(x)
.= f(x− a). Clearly ‖Λf‖Lp = ‖f‖Lp . Therefore, Λa : Lp 7→ Lp
is a bounded linear operator, with norm ‖Λa‖ = 1. Notice that the operator Λa is one-to-oneand onto.
Example 4 (shift operators on ℓp). Let 1 ≤ p ≤ ∞. Define the operators Λ+ : ℓp 7→ ℓp
and Λ− : ℓp 7→ ℓp asΛ+(x1, x2, x3, . . .)
.= (0, x1, x2, . . .) ,
Λ−(x1, x2, x3, . . .).= (x2, x3, x4, . . .) .
Observe that these are linear continuous operators, with ‖Λ+‖ = ‖Λ−‖ = 1. Moreover, Λ+
is one-to-one but not onto, while Λ− is onto, but not one-to-one.
17
Example 5 (multiplication operator). Let Ω ⊂ IRn and let g : Ω 7→ IR be a bounded,measurable function. On the space Lp(Ω), consider the multiplication operator (Λf)(x)
.=
g(x) f(x). This is a continuous operator, with norm
‖Λ‖ .= sup
f 6=0
‖gf‖Lp
‖f‖Lp= ‖g‖L∞ . (2.4)
Example 6 (integral operator). Let a < b and consider the space X = C([a, b] ; IR) of realvalued continuous functions defined on the interval [a, b]. Consider the integral operator
(Λf)(x).=
∫ x
af(y) dy .
Then Λ : X 7→ X is a bounded operator. Indeed,
∣∣∣(Λf)(x)∣∣∣ =
∣∣∣∣∫ x
af(y) dy
∣∣∣∣ ≤∫ x
a|f(y)| dy ≤ (b− a) ‖f‖
Hence ‖Λf‖ ≤ (b− a)‖f‖ and ‖Λ‖ ≤ (b− a).
In this case, the operator Λ is actually compact. Indeed, let (fn)n≥1 be a bounded sequenceof continuous functions, say
|fn(x)| ≤M for all n ≥ 1, x ∈ [a, b].
Then, for any x, x′ ∈ [a, b], say with x < x′, we have
∣∣∣(Λfn)(x)− (Λfn)(x′)∣∣∣ ≤
∣∣∣∣∣
∫ x′
xfn(y) dy
∣∣∣∣∣ ≤∫ x′
x|fn(y)| dy ≤ M |x′ − x| .
Hence the integral functions Λfn are uniformly bounded and uniformly Lipschitz continuousof constant M . By Ascoli’s compactness theorem, the sequence Λfn admits a subsequencethat converges uniformly on the whole interval [a, b].
3 Finite dimensional spaces
We now show that every finite dimensional normed space is equivalent to the standardEuclidean space IKN , with the standard Euclidean norm. If α = (α1, . . . , αN ), then‖α‖ .=
√∑k |αk|2.
Theorem 1.3 (a finite dimensional normed space is homeomorphic to IKN). LetX be a finite dimensional normed space, over the field IK of real or complex numbers. LetB = u1, u2, . . . , uN be a basis of X. Then
(i) X is complete, and hence a Banach space.
(ii) For every α = (α1, α2, . . . , αN ) ∈ IKN , define Λα as the linear combination
Λα = α1u1 + α2u2 + · · ·+ αNuN .
18
Then the linear operator Λ : IKN 7→ X is bijective and bounded. Moreover, its inverse Λ−1 :X 7→ IKN is also bounded.
Proof. 1. The fact that u1, . . . , uN is a basis implies that Λ is one-to-one and onto. Hencethe inverse operator Λ−1 : X 7→ IKN is well defined.
2. Writing
‖Λα‖ ≤N∑
k=1
‖αkuk‖ ≤ ‖α‖N∑
k=1
‖uk‖ ,
we see that Λ : IK 7→ X is a bounded linear operator, hence continuous.
3. We claim that Λ−1 is also bounded. Otherwise, there exists a bounded sequence (xn)n≥1
in X such that‖Λ−1xn‖ → ∞ as n→∞.
Consider the normalized vectors
βn.=
Λ−1xn‖Λ−1xn‖
∈ IKN .
Then ‖βn‖ = 1 for every n and Λβn → 0 as n→∞. Since (βn)n≥1 is a bounded sequence inthe Euclidean space IKn, by the Heine-Borel theorem there exists a convergent subsequence,say αnk
→ α ∈ IKN . Clearly
‖α‖ = limk→∞
‖αnk‖ = 1, Λα = lim
k→∞Λαnk
= 0 ,
because Λ is continuous. This contradict the fact that Λ is one-to-one. We thus conclude thatΛ−1 is bounded, hence continuous. This proves (ii).
4. To prove that X with norm ‖ · ‖ is complete, let (xn)n≥1 be a Cauchy sequence in X. ThenΛ−1xn defines a Cauchy sequence in IKN , which converges to some point β ∈ IKN , becauseIKN is complete. Since Λ is bounded, the sequence xn converges to Λβ.
Theorem 1.4 (locally compact normed spaces are finite dimensional). Let X be anormed space such that the closed unit ball B1
.= B(0, 1) = x ∈ X ; ‖x‖ ≤ 1 is compact.
Then X is finite dimensional.
Proof. 1. By assumption, the closed unit ball B1 is compact, hence it is pre-compact andcan be covered with finitely many open balls B(pi, 1/2) centered at points pi, i = 1, . . . , n andwith radius 1/2.
2. Consider the finite dimensional subspace V = spanp1, . . . , pn. Observe that V is closedin X, because by the previous theorem every finite-dimensional normed space is complete.
We claim that V = X, hence X itself is finite-dimensional. If not, we could find a pointx ∈ X \ V . Let ρ
.= d(x, V )
.= infy∈V ‖y − x‖ . Notice that ρ > 0 because V is closed. Then
19
Vp
0
z
i
x
v
Figure 2.2: The construction used to prove Theorem .
there exists a point v ∈ V such that
ρ ≤ ‖x− v‖ ≤ 3
2ρ . (3.1)
Consider the unit vector
z.=
x− v‖x− v‖ ∈ B1 .
By construction, there exists a point pi ∈ B1 such that ‖z − pi‖ < 12 . We thus have
x = v + ‖x− v‖ z = v + ‖x− v‖pi + ‖x− v‖(z − pi) .
Since v + ‖x− v‖pi ∈ V , we must have
‖x− v‖ ‖z − pi‖ ≥ d(x, V ) ≥ ρ .
Hence ‖x − v‖ ≥ 2ρ, in contradiction with (3.1). This shows X = V , completing the proof.
4 Extension theorems
In the following, we consider a vector spaceX over the real numbers, and a function p : X 7→ IRsuch that
p(x+ y) ≤ p(x) + p(y) , p(tx) = t p(x) for all x, y ∈ X, t ≥ 0. (4.1)
Notice that, if X is a normed space, for any κ > 0 the function p(x) = κ ‖x‖ satisfies theabove properties.
In general, (4.1) implies that the function p is convex. Indeed
p(θx+(1−θ)y) ≤ p(θx)+p((1−θ)y) = θp(x)+(1−θ)p(y) for all x, y ∈ X , θ ∈ [0, 1] .
However, compared with a norm, for x 6= 0 here we also allow p(x) ≤ 0. Moreover, we do notrequire p to be symmetric w.r.t. the origin. In other words, one may well have p(x) 6= p(−x).
20
Example 7. Let X be a normed space and let Ω ⊂ X be an open, convex set containing theorigin. Then the functional
p(x).= inf λ ≥ 0 ; x ∈ λΩ (4.2)
satisfies the assumptions (4.1).
Theorem 1.5 (Hahn-Banach extension theorem). Let V be a subspace of a vector spaceX over the reals. Let p : X 7→ IR be a map with the properties (4.1). Let f : V 7→ IR be alinear functional such that
f(x) ≤ p(x) for all x ∈ V (4.3)
Then there exists a linear functional F : X 7→ IR such that F (x) = f(x) for all x ∈ V and
−p(−x) ≤ F (x) ≤ p(x) for all x ∈ X . (4.4)
Proof. If V = X there is nothing to prove. Otherwise, choose any vector x0 /∈ V and considerthe strictly bigger subspace
V0.= x+ tx0 ; x ∈ V, t ∈ IR.
For every x, y ∈ V , the bound on f yields
f(x) + f(y) = f(x+ y) ≤ p(x+ y) ≤ p(x− x0) + p(x0 + y).
Thereforef(x)− p(x− x0) ≤ p(y + x0)− f(y) for all x, y ∈ V.
Choosing β.= supx∈V
f(x)− p(x− x0)
, we have
f(x)− p(x− x0) ≤ β ≤ p(y + x0)− f(y) for all x, y ∈ V . (4.5)
2. We now extend f to a linear functional defined on the larger space V0, by setting
f(x+ tx0).= f(x) + β t x ∈ V, t ∈ IR .
We claim that this extension still satisfies
f(x+ tx0) ≤ p(x+ tx0) for all x ∈ V0 , t ∈ IR . (4.6)
Indeed, if t = 0, the above inequality follows from our initial assumptions. If t > 0, replacingboth x and y by x/t in (4.5) we obtain
t
[f(xt
)− p
(x
t− x0
)]≤ tβ ≤ t
[p
(x
t+ x0
)− f
(xt
)],
f(x)− p(x− tx0) ≤ tβ ≤ p(x+ tx0)− f(x) .
21
Therefore, for x ∈ V and t ≥ 0 we have
f(x− tx0) = f(x)− βt ≤ p(x− tx0) , f(x+ tx0) = f(x) + βt ≤ p(x+ tx0),
proving (4.6).
3. The previous two steps show that every bounded linear functional f defined on a propersubspace of V ⊂ X can be extended to a strictly larger subspace. To complete the proof, weinvoke a general maximality principle.
Namely, let F be the family of all couples (W,φ), whereW is a subspace of X and φ :W 7→ IRis a linear functional such that
φ(x) ≤ p(x) for all x ∈W.This family can be partially ordered by setting
(W,φ) ≺ (W , φ) if and only if W ⊆ W and φ coincides with the restriction of φ to W .
By the Hausdorff maximality principle (equivalent to Zorn’s lemma, and to the axiom ofchoice), the partially ordered family F contains a maximal element, say (W , F ). If W 6=X, then by the previous step the linear functional F could be extended to a strictly largersubspace, against the maximality assumption. Hence W = X and F : X 7→ IR provides alinear functional such that
F (x) ≤ p(x) for all x ∈ X.By linearity, this implies
−p(x) ≤ − F (−x) = F (x) ,
completing the proof.
V
y
F(y)
f(x)
x 0
p
Figure 2.3: By the Hahn-Banach theorem, a linear functional f : V 7→ IR defined on a subspaceV ⊂ X and such that f(x) ≤ p(x) for all x ∈ V can be extended to a linear functional F : X 7→ IRsatisfying F (y) ≤ p(y) for all y ∈ X .
The previous theorem has a natural application to the case where p(x) = ‖x‖ is a norm.
22
Theorem 1.6 (extension theorem for bounded linear functionals). Let X be a vectorspace over the field IK of real or complex numbers. Let f : V 7→ IK be a bounded linearfunctional defined on a subspace V ⊆ X. Then f can be extended to a bounded linear functionalF : X 7→ IK having the same norm:
‖F‖ .= sup
x∈X,‖x‖≤1|F (x)| = sup
x∈V,‖x‖≤1|f(x)| .
= ‖f‖.
Proof. 1. First assume IK = IR, so that f is real valued. Set κ.= ‖f‖, and define p(x)
.= κ‖x‖.
Then the result follows immediately from the previous theorem.
2. Next, we consider the case where IK is the field of complex numbers. Notice that in thiscase, V and X are also vector spaces over the real numbers.
For x ∈ V , define u(x).= Ref(x). This is a real valued linear functional on V with norm
‖u‖ ≤ ‖f‖ .= κ. Hence by the previous steps it admits an extension U : X 7→ IR with norm
‖U‖ ≤ κ. We claim that the map
F (x).= U(x)− iU(ix)
satisfies all requirements. Indeed, for x ∈ V ,
F (x) = Ref(x)− iRef(ix) = Ref(x) + iImf(x) = f(x).
Moreover, let α ∈ C be such that |α| = 1, αF (x) = |F (x)|. Then
|F (x)| = αF (x) = U(αx) ≤ κ‖αx‖ = κ‖x‖
Hence ‖F‖ ≤ κ .= ‖f‖.
Corollary 1.1. Let X be a Banach space. For every couple of vectors x, x′ ∈ X with x 6= x′,there exists a continuous linear functional φ ∈ X∗ such that φ(x) 6= φ(x′).
Corollary 1.2. Let X be a Banach space. For every vector x ∈ X, there exists a continuouslinear functional φ ∈ X∗ such that φ(x) = ‖x‖ and ‖φ‖ = 1.
Theorem 1.7 (separation of convex sets). Let X be a normed space over the reals, andlet A,B be nonempty, disjoint convex subsets of X.
(i) If A is open, then there exists a bounded linear functional Λ : X 7→ IR and a number c ∈ IRsuch that
Λa < c ≤ Λb for all a ∈ A , b ∈ B . (4.7)
23
B
AA
B
Λ(x) = c
A ρ
Figure 2.4: Left: if A is open, the disjoint convex sets A,B can be separated. Right: If A is compactand B is closed, the the disjoint convex sets A,B can be strictly separated.
(ii) If A is compact and B is closed, then there exists a bounded linear functional Λ : X 7→ IRand numbers c1, c2 ∈ IR such that
Λa ≤ c1 < c2 ≤ Λb for all a ∈ A , b ∈ B . (4.8)
Proof. 1. Choose points a0 ∈ A and b0 ∈ B and set x0.= b0 − a0. Consider the open set
Ω.= A−B + x0
.=(a− a0) + (b0 − b) ; a ∈ A, b ∈ B
.
Since A,B are convex and A is open, it is clear that Ω is an open, convex neighborhood ofthe origin. Moreover, x0 /∈ Ω, because otherwise
x0 = a− b+ x0 , a− b = 0 for some a ∈ A, b ∈ B.
This is a contradiction because A ∩B = ∅.
2. Consider the functional
p(x).= inf λ ≥ 0 ; x ∈ λΩ (4.9)
Since Ω is a neighborhood of the origin, we have B(0, ρ) ⊆ Ω for some ρ > 0. Hence
p(x) ≤ ‖x‖ρ
for all x ∈ X. (4.10)
Moreover, the convexity of Ω implies
p(x+ y) ≤ p(x) + p(y) , p(tx) = t p(x) for all x, y ∈ X , t ≥ 0. (4.11)
Notice that p(x0) ≥ 1 because x0 /∈ Ω.
3. On the one-dimensional subspace V.= tx0 ; t ∈ IR, define the linear functional f by
setting f(tx0).= t. Observe that
f(x0) = 1 , f(tx0) = t ≤ tp(x0) = p(tx0) .
24
By the Hahn-Banach extension theorem, there exists a linear functional Λ : X 7→ IR such that
−p(−x) ≤ Λx ≤ p(x) for all x ∈ X .
By (4.10), it is clear that Λ is bounded. Indeed, ‖Λ‖ ≤ ρ−1.
4. If now a ∈ A and b ∈ B, we have
Λa− Λb+ 1 = Λ(a− b+ x0) ≤ p(a− b+ x0) < 1
because Λx0 = f(x0) = 1 while a− b+ x0 ∈ Ω and Ω is open. Therefore
Λa < Λb for all a ∈ A , b ∈ B .
The sets ΛA and ΛB are non-empty, disjoint convex sets of IR, with ΛA open. Takingc.= supa∈A Λa, the conclusion (4.7) is satisfied. This proves (i).
5. To prove (ii), observe that the assumptions on A,B imply
d(A,B).= inf ‖a − b‖ ; a ∈ A, b ∈ B > 0 .
If we choose ρ.= d(A,B), then the open neighborhood Aρ
.= x ∈ X ; d(x,A) < ρ of radius
ρ around A does not intersect B. We can thus apply part (i) the the disjoint convex sets Aρ
and B. This yields a continuous linear functional Λ : X 7→ IR and a constant c2 such that
Λx < c2 ≤ Λy for all x ∈ Aρ , y ∈ B .
We now observe that the set Λ(A) is compact, being the image of a compact set by a continuousmap. Hence c1
.= supx∈a Λx < c2. This achieves the proof of (ii).
0
0
a0
b0
A B
b − a
A − B + b − a
= 1Λ
0 0
0
Figure 2.5: The construction used in the proof of the separation theorem.
5 Dual spaces and weak convergence
LetX be a Banach space over the field IK of real or complex numbers. The set of all continuouslinear functionals ϕ : X 7→ IK is called the dual space of X, and denoted by X∗. We observethat X∗ is a Banach space with norm
‖ϕ‖ .= sup
x 6=0
|ϕ(x)|‖x‖ . (5.1)
25
It is often useful to adopt the more symmetric notation
〈x∗, x〉 .= x∗(x) x∗ ∈ X∗, x ∈ X
to denote the bilinear mapping between a Banach space X and its dual X∗.
Given a set S ⊆ X, we denote by
S⊥ = φ ∈ X∗ ; φ(s) = 0 for all s ∈ S ⊆ X∗
the family of all bounded operators that vanish on S. This is a closed subspace of X∗.Similarly, for S ⊆ X∗, we denote by
S⊥ = x ∈ X ; φ(x) = 0 for all φ ∈ S ⊆ X .
This is a closed subspace of X.
A sequence x1, x2, . . . in a Banach space X is called weakly convergent if there exists x ∈ Xsuch that
limn→∞
ϕ(xn) = ϕ(x) for all ϕ ∈ X∗
In this case, x is called the weak limit of the sequence xn, and we write xn x. We recallthat the sequence xn converges strongly to x if ‖xn−x‖ → 0. Since by definition every ϕ ∈ X∗
is continuous, the strong convergence xn → x clearly implies the weak convergence xn x.
We observe that, if a weak limit exists, then it is necessarily unique: if xn x and xn y,then x = y. Indeed, assume y 6= x. Then by the Hahn-Banach theorem there exists acontinuous linear functional φ ∈ X∗ such that φ(x) 6= φ(y). This leads to a contradiction,because
φ(x) = limn→∞
φ(xn) = φ(y)
6 Problems
1. Check if the following are normed spaces. In the negative case, identify which of theproperties (N1)–(N3) fails. In the positive case, decide if they are Banach spaces.
(i) Let X = IR, with
‖x‖ =
x if x ≥ 0,−2x if x < 0.
(ii) Let X be the vector space of all sequences of real numbers x = (x1, x2, x3, . . .) such thatxk = 0 for all except finitely many k. On X consider the norm (1.8).
26
(iii) Let X be the space of all polynomials (of any degree), with norm (1.3).
(iv) Let X be the space of all polynomials of degree ≤ 2, with norm
‖p‖ .= |p(0)|+ |p′(0)|+ |p′′(0)|.
(v) Let X be the space of all continuous functions on the interval [0, 1], with
‖f‖ .=
∫ 1
0|f(x)| dx.
(vi) Let X = R2. Given x = (x1, x2), for a fixed p > 0 define
‖x‖ .=(|x1|p + |x2|p
)1/p.
Consider the cases 0 < p < 1 and 1 ≤ p <∞ separately.
(vii) Fix κ ∈ IR and let X be the space of all continuous functions f : [0,∞[ 7→ IR such that
‖f‖ .= sup
t≥0eκt |f(t)| <∞.
2. Let X,Y be Banach spaces. Prove that the cartesian product
X × Y =(x, y) ; x ∈ X, y ∈ Y
is also a Banach space, with norm
‖(x, y)‖ .= max‖x‖, ‖y‖. (6.2)
3. Let X be a Banach space over the field IK of real or complex numbers. Notice that X ×Xand IK ×X are Banach spaces, with product norms as in (6.2). Prove:
(i) The mapping (x, y) 7→ x+ y from X ×X into X is continuous.
(ii) The mapping (α, x) 7→ αx from IK ×X into X is continuous.
4. Let X be a finite dimensional space. let ‖ · ‖ and ‖ · ‖† be two norms on X. Prove thatthey are equivalent, namely there exists a constant C > 1 such that
1
C‖x‖ ≤ ‖x‖† ≤ C‖x‖ for all x ∈ X .
27
5. Let X be a vector space. The convex hull of a set A ⊂ X is defined as the set of allconvex combinations of elements of A, namely
coA.=
N∑
k=1
θkak ; N ≥ 1, ak ∈ A , θk ∈ [0, 1],N∑
k=1
θk = 1
.
Prove that (i) coA is convex, and (ii) coA is the intersection of all convex sets that contain A.
6. Let X be a normed space with norm ‖ · ‖. Prove that every subspace V ⊂ X is also anormed space, with the same norm. If X is a Banach space and V is closed, then V is also aBanach space.
7. Let X be a Banach space and let A,B ⊂ X. Prove the following statements.
(i) If A is open, then coA is open as well.
(ii) If A is bounded, then coA is bounded.
(iii) If A,B are bounded, then the set A+B.= a+ b ; a ∈ A, b ∈ B is bounded as well.
(iv) If A is closed and B is compact, then A+B is closed.
(v) The sum of two closed sets may not be closed.
(vi) If A is convex, then x+ y ; x ∈ A , y ∈ A .= A+A = 2A
.= 2x ; x ∈ A.
(vii) If A is closed and A+A = 2A, then A must be convex.
Hint for (vii): for any two points x, y ∈ A, the assumption implies x+ y = 2z for some z ∈ A,hence x
2 + y2 ∈ A. By induction, show that θx + (1 − θ)y ∈ A for every dyadic coefficient
θ =∑n
i=1 ci2−i, ci ∈ 0, 1. Then use the closure of A. Notice that the result can fail if A is
not closed. For example, let A be the set of all rational numbers.
8. Let X be a normed space. We say that a subset S ⊆ X is symmetric if a ∈ S implies−a ∈ S. Prove that
(i) If S is convex, then its closure is convex as well.
(ii) If S is symmetric, then its closure is symmetric as well.
9. Let X,Y be Banach spaces over the real numbers, and let Λ : X 7→ Y be a bounded linearoperator. Prove that
(i) If S is convex, then its image Λ(S) is convex.
(ii) If S is symmetric, then its image Λ(S) is symmetric.
10. Let Λ : X 7→ Y be a linear operator. Assume that, for every sequence xn → 0, the
28
sequence (Txn)n≥1 is bounded. Prove that T is continuous.
11. Let X,Y be Banach spaces, and let T : X 7→ Y be a linear map. Prove that the threeways of defining the norm ‖T‖ in (2.1) are the same.
12. Let X be an infinite dimensional Banach space, and let S be a set of linearly independentvectors. Prove that
(i) If S = v1, . . . , vN is a finite set, then span(S) is a closed subspace of X.
(ii) If S = vk ; k ≥ 1 is an infinite sequence, then the vector space span(S) cannot be closedin X.
13. (spaces over the reals and over the complex numbers) Let X be a vector spaceover the complex numbers. Then X is also a vector space over the real numbers. If Φ : X 7→ Cis a complex linear functional and φ(x) = ReΦ(x) is its real part, prove that φ is a real linearfunctional, and Φ(x) can be reconstructed from φ as
Φ(x) = φ(x)− iφ(ix).
14. Consider the spaces ℓp of sequences of real numbers, defined at (1.6)–(1.8).
If 1 ≤ p ≤ q ≤ ∞, prove that ℓp ⊆ ℓq, with equality holding only if p = q. Moreover, provethat the identity operator Λ : ℓp 7→ ℓq defined as Λx = x is continuous, for every p ≤ q.
15. Prove that the for 1 ≤ p < ∞, the closure of the subspace spanek ; k ≥ 1 in (1.10) isthe entire space ℓp. For p = ∞, show that this closure coincides with the subspace c0 of allsequences that converge to zero.
16 (Properties of diagonal operators). For 1 ≤ p ≤ ∞, consider the operator Λ : ℓp 7→ ℓp
defined at (2.2). Prove the claims (i)-(iii) in Example 1.
17. Prove that the norm of the multiplication operator in Example 4 is indeed given by (2.4).
18. Fix 1 ≤ p ≤ ∞. Let (fn)n≥1 be a sequence of functions in Lp(IR), converging weakly toa function f ∈ Lp(IR). Prove that
limn→∞
∫ b
afn(x) dx =
∫ b
af(x) dx for all a < b.
29
19. Consider the sequence of functions
fn(x) =
1/n if x ∈ [0, n]0 otherwise
(i) Prove that fn → 0 strongly, in every space Lp(IR) with 1 < p ≤ ∞.
(ii) On the other hand, show that in the space L1(IR) this sequence is not strongly convergent.In fact, this sequence does not even admit any weakly convergent subsequence.
20. Let Λ1 : X 7→ Y and Λ2 : Y 7→ Z be a bounded linear operators. If either Λ1 or Λ2 iscompact, prove that the composition Λ2 Λ1 : X 7→ Z is a compact operator as well.
21. Let φ : IR2 7→ IR be a linear functional, say φ(x1, x2) = ax1 + bx2. Give a direct proofthat
(i) If IR2 is endowed with the norm ‖x‖1 = |x1| + |x2|, then the corresponding norm (5.1) is‖φ‖∞ = max|a|, |b|.
(ii) If IR2 is endowed with the norm ‖x‖∞ = max|x1|, |x2|, then the corresponding norm(5.1) is ‖φ‖1 = |a|+ |b|.
(iii) If IR2 is endowed with the norm ‖x‖p = (|x1|p + |x2|p)1/p, with 1 < p < ∞, then thecorresponding norm (5.1) is ‖φ‖q = (|a|q + |b|q)1/q, with 1
p + 1q = 1.
22. Let X be a Banach space over the reals, and let X∗ be its dual space. Prove Corollary 2.As a consequence, show that for every x ∈ X one has
‖x‖ = sup〈x∗ , x〉 ; ‖x∗‖ ≤ 1
.
23. Let X be a vector space. Let B ⊂ X be a convex subset such that, for every nonzerovector x ∈ X, there exists a positive number θx > 0 such that
αx ∈ B if and only if |α| ≤ θx .
For r ≥ 0, call rB.= rb ; b ∈ B. Prove that
‖x‖ .= min r ≥ 0 ; x ∈ rB
is a norm on X, and B = x ∈ X ; ‖x‖ ≤ 1 is the unit ball in this norm.
24. Let X be a separable Banach space. Show that in this case the Hahn-Banach theoremcan be proved without using transfinite induction (i.e. the Hausdorff maximality principle).
30
25. Let X be a Banach space. The sum of a series is defined as
∞∑
k=1
xk.= lim
n→∞
n∑
k=1
xk . (6.3)
Prove that, if∑∞
k=1 ‖xk‖ = M < ∞, then the series converges, i.e. the limit in (6.3) is welldefined.
26. Let X be a Banach space. Consider any set S ⊂ X and assume x ∈ span(S). Prove thatthere exist points xj ∈ S and coefficients cj ∈ IK, j = 1, . . . , N , such that
x =N∑
j=1
cjxj ,
∥∥∥∥∥∥
k∑
j=1
cjxj
∥∥∥∥∥∥≤ 2‖x‖ for all k = 1, . . . , N . (6.4)
27. Let S be a subset of a Banach spaceX. Prove that the following statements are equivalent.
(i) x ∈ span(S)
(ii) x =∑∞
j=1 cjxj , with xj ∈ S and cj ∈ IK for every j.
28. Consider the Banach space ℓ∞ consisting of all bounded sequences x = (x1, x2, x3, . . .) ofreal numbers. Prove that there exists a linear functional F : ℓ∞ 7→ IR such that
|F (x)| ≤ supn≥1|xn| ,
F (x) = limn→∞
xn if the limit exists.
20. For every bounded sequence of real numbers x = (x1, x2, x3, . . .), define
Λx.=
1
2
(lim infn→∞
xn)+
1
2
(lim supn→∞
xn
).
Is Λ a bounded linear functional on the Banach space ℓ∞ ?
30. On the Banach space of all bounded sequences of real numbers x = (x1, x2, x3, . . .), definethe operator Λ : ℓ∞ 7→ ℓ∞ by setting Λx = y = (y1, y2, y3, . . .), where
yn.=
1
n
n∑
i=1
xi . (6.5)
(i) Prove that Λ a bounded linear operator. Compute its norm. Is Λ a compact operator ?
31
(ii) Consider the operator Λ defined as in (6.5), but on the space ℓ1 of absolutely summablesequences. Is Λ a bounded linear operator ? Is it compact ?
31. In the Banach space X = L∞(IR), consider the subspace V consisting of all boundedcontinuous functions.
Prove that there exists a bounded linear functional Λ : L∞(IR) 7→ IR with ‖Λ‖ = 1 suchthat Λf
.= f(0) for every bounded continuous function f . However, show that there exists no
function g ∈ L1(IR) such that Λf =∫fg dx for every f ∈ L∞(IR). Note: this shows that
the dual space of L∞(IR) cannot be identified with L1(IR).
32. Let X be a normed space and let Ω ⊂ X be an open, convex set containing the origin.Consider the functional
p(x).= inf λ ≥ 0 ; x ∈ λΩ (6.6)
(i) Prove that p(·) satisfies the conditions
p(x+ y) ≤ p(x) + p(y) , p(tx) = t p(x) for all x, y ∈ X, t ≥ 0 .
(ii) Assuming that Br.= x ∈ X ; ‖x‖ < r ⊆ Ω, prove that p(x) ≤ ‖x‖/r.
(iii) Assuming that Ω = x ∈ X ; ‖x‖ < 1 is the open unit ball, prove that p(x) = ‖x‖.
33. Let C0([0, 1]) be the Banach space of all real valued continuous functions f : [0, 1] 7→ IR,with norm ‖f‖ = maxx∈[0,1] |f(x)|.
(i) Show that X = f ∈ C0([0, 1]) ; f(0) = 0 is a closed subspace of C0, hence a Banachspace.
(ii) Prove that the map f 7→ Λf.=∫ 10 f(x) dx is a continuous linear functional on X. Compute
its norm ‖Λ‖ .= sup‖f‖≤1 |Λf |. Is this supremum over the closed unit ball actually attained
as a maximum ?
34. Let X be a real Banach space. By a hyperplane we mean a set of the form H = x ∈X ; Λx = α, for some linear (possibly discontinuous) mapping Λ : X 7→ IR and a numberα ∈ IR.
Prove that every hyperplane is either closed or dense in H.
35. Let X be a Banach space and let X∗ be its dual. Let Ω ⊂ X be a convex set containingthe origin. Define
Ω∗ .= x∗ ∈ X∗ ; 〈x∗, x〉 ≤ 1 for all x ∈ Ω ,
Ω∗∗ .= x ∈ X ; 〈x∗, x〉 ≤ 1 for all x∗ ∈ Ω∗ .
Prove that Ω∗∗ is the closure of Ω.
36. Let φ be a non-zero vector in a Banach space X. Call U = spanφ = λφ ; λ ∈ IR.
32
Prove that there exists a closed subspace V ⊂ X such that X = U⊕V . Namely, every elementx ∈ X can be written uniquely as a sum
x = u+ v with u ∈ U , v ∈ V .
Moreover, show that the projections x 7→ u = πU x and x 7→ v = πV x are continuouslinear operators.
37. Let X be an infinite dimensional Banach space and let (xn)n≥1 be a sequence of linearlyindependent vectors. Define the subspaces Vn = spanx1, . . . , xn. Prove that the union
V =⋃
n≥1
Vn
is a subspace of X which is not closed.
33
Chapter 3
Spaces of Continuous Functions
1 Bounded continuous functions
Let E be a metric space. Let IK be either the field of real numbers, or of complex numbers.By B(E, IK) we denote we denote the space of all bounded maps f : E 7→ IK, with norm
‖f‖∞ .= sup
x∈E|f(x)| . (1.1)
By C(E; IK) denotes the space of all bounded continuous functions f : E 7→ IK, with the samenorm (1.1).
Lemma 1. C(E) is closed of B(E).
Proof. we need to show that if a sequence of continuous functions (fn)n≥1 converges to somefunction f uniformly on E, then f is continuous as well. Fix any x ∈ E and ε > 0.
By uniform convergence, there exists an integer N large enough such that |fN (x)−f(x)| < ε/3for every x ∈ E.
Since fN is continuous, there exists δ > 0 such that |fN (y)−fN (x)| < ε/3 whenever d(y, x) < δ.
Putting together the above inequalities, when d(y, x) < δ we have
|f(y)− f(x)| ≤ |f(y)− fN (y)|+ |fN (y)− fN (x)|+ |fN (x)− f(x)| < ε
3+ε
3+ε
3= ε ,
proving that f is continuous at the point x.
Remark 1. In the above lemma, the assumption of uniform convergence is essential. Forexample, on the interval E = [0, 1], the sequence of functions fn(t) = tn converges pointwiseto the discontinuous function
f(t) =
0 if 0 ≤ t < 1 ,1 if t = 1 .
Clearly, here the convergence is not uniform on the whole interval [0, 1].
34
The following theorem shows how, for an increasing sequence of functions, the pointwiseconvergence can imply uniform convergence.
Theorem 2.1 (Dini). Let E be a compact metric space. If (fn)n≥1 is an increasing sequenceof functions in CIR(E), converging pointwise to a continuous limit function g, then fn → guniformly on E.
Proof. Fix any ε > 0.
For every x ∈ E, there exists an integer N(x) such that |fN(x)(x)− g(x)| < ε/3.
Since fN(x) and g are continuous, there exists an open neighborhood Vx of x such that|fN(x)(y)− fN(x)(x)| < ε/3 and |g(y) − g(x)| < ε/3 for every y ∈ Vx.
Since E is compact, we can cover E with finitely many of these neighborhoods, say E ⊆Vx1 ∪ · · · ∪ Vxm .
Choose the integer N = max N(x1), . . . , N(xm). For every n ≥ N and y ∈ E, assumingy ∈ Vxi we have the estimate
|fn(y)−g(y)| ≤ |fN(xi)(y)−g(y)| ≤ |fN(xi)(y)−fN(xi)(xi)|+|fN(xi)(xi)−g(xi)|+|g(xi)−g(y)| < ε .
Since ε > 0 was arbitrary, this establishes the uniform convergence fn → g.
2 The Stone-Weierstrass approximation theorem
In many applications, one would like to approximate a general continuous function f ∈ C(E)with special functions: for example polynomials, or exponential functions, or trigonometricpolynomials. It is thus important to know whether any function f ∈ C(E) can be uniformlyapproximated by these functions. The following theorem is a key result in this direction.
We observe that the space C(E) is an algebra, in the sense that
if f, g ∈ C(E) then also fg ∈ C(E)
Moreover, the norm of the product satisfies
‖fg‖ ≤ ‖f‖ ‖g‖ .
We say that a subspace A ⊆ C(E) is a subalgebra if f, g ∈ A implies fg ∈ A.
Lemma 2. If A ⊆ C(E) is a subalgebra, then its closure A is a subalgebra as well.
Indeed, assume f, g ∈ A. Then there exist uniformly convergent sequences fn, gn ∈ A withfn → f and gn → g. One has
‖fg − fngn‖ ≤ ‖fg − fng‖ + ‖fng − fngn‖ ≤ ‖f‖ ‖g − gn‖+ ‖fn‖ ‖g − gn‖ .
Since the sequence (gn)n≥1 is uniformly bounded, as n → ∞ the right hand side approacheszero. This shows the convergence fngn → fg. Since A is an algebra, fngn ∈ A for every n.Hence fg ∈ A.
35
We say that a subset A ⊆ C(E) separates points if, for every couple of distinct pointsx, y ∈ E, there exists a function f ∈ A such that f(x) 6= f(y)
Theorem 2.2 (Stone-Weierstrass). Let E be a compact metric space. If A is a subalgebraof CIR(E; IR) that separates points and contains all constant functions, then A = CIR(E; IR).
Proof. 1. There exists a sequence of polynomials (pn)n≥1 such that pn(t) →√t uniformly
for t ∈ [0, 1].
The underlying idea is to construct approximate solutions to the equation t − p2(t) = 0, byiteration. We thus set p0(t) ≡ 0 and, by induction on n = 1, 2, 3, . . .
pn+1(t) = pn(t) +1
2(t− p2n(t)). (2.1)
By induction, one checks that pn(t) ≤ pn+1(t) ≤√t for every t ∈ [0, 1]. Indeed,
√t− pn+1(t) =
√t− pn(t)−
1
2(t− p2n(t)) = (
√t− pn(t))
(1− 1
2(√t+ pn(t))
).
For every fixed t ∈ [0, 1], the sequence pn(t) is increasing and bounded above. Hence it has aunique limit, say g(t). By (2.1), this limit satisfies t− g2(t) = 0. Since g(t) ≥ 0 we concludethat g(t) =
√t.
Finally, by Dini’s theorem, the convergence is uniform for t ∈ [0, 1].
2. For every continuous function f ∈ A, one has |f | ∈ A.
Indeed, let κ = supx |f(x)|. Then all functions fn(x) = pn(f2(x)/κ2) lie in A, because A is an
algebra. Since f2(x)/κ2 ∈ [0, 1], the previous step yields the convergence
fn(x) →√f2(x)/κ2 =
|f(x)|κ
,
uniformly for x ∈ E. Therefore |f |κ ∈ A, and hence also |f | ∈ A.
3. We now apply the previous step to the subalgebra A and conclude that, if f, g ∈ A, thenthe functions
maxf, g =1
2(f + g + |f − g|) , minf, g =
1
2(f + g − |f − g|)
also lie in A.
4. For any two distinct points y1, y2 ∈ E and any couple of real numbers a1, a2, there existsa function f ∈ A such that f(y1) = a1 and f(y2) = a2.
Indeed, by assumption there exists a continuous function g ∈ A such that g(y1) 6= g(y2). SinceA contains all constant functions, we can define
f(x).= a1 + (a2 − a1)
g(x)− g(y1)g(y2)− g(y1)
.
36
y
z zy zz
yy
z1 2 3 5
y1
1 2
4
z4hz1
h
y3
gy2
g
f
fG
g
Figure 3.1: Steps 5 and 6 in the proof of the Stone-Weierstrass theorem.
5. Given any continuous function f , a point y ∈ E, and ε > 0, there exists a function gy ∈ Asuch that
gy(y) = f(y), g(x) ≤ f(x) + ε for every x ∈ E . (2.2)
Indeed, for every point z ∈ E, there exists a function hz ∈ A such that hz(y) = f(y) andhz(z) = f(z).
Since f and hz are both continuous, there exists an open neighborhood Vz of z such thathz(x) < f(x) + ε for every x ∈ Vz.
Cover the compact set E with finitely many neighborhoods: E ⊆ Vz1 ∪ · · · ∪ Vzm . Then thefunction
gy(x).= minhz1(x), . . . , hzm(x)
lies in A and satisfies the conditions in (2.2).
6. The closure of A is the entire space CIR(E).
Let f ∈ CIR(E) be any continuous function and let ε > 0 be given. For each y ∈ E, by theprevious step there exists a function gy ∈ A such that
g(y) = f(y), gy(x) ≤ f(x) + ε for every x ∈ E .
By continuity of f and gy, there exists a neighborhood Uy of y such that
gy(x) ≥ f(x)− ε for all x ∈ Uy .
37
We now cover the compact set E with finitely many neighborhoods: E ⊆ Uy1∪· · ·∪Uyν . Thenthe function
G(x).= max gy1(x), . . . , gyν (x)
lies in A and satisfies
|G(x)− f(x)| ≤ ε for all x ∈ E .
Corollary 1. Let E be any compact subset of IRn. Let F be the family of all real-valuedpolynomials in the variables (x1, . . . , xn). Then F is dense on C(E; IR). In other words, everycontinuous function f : E 7→ IR can be uniformly approximated by polynomials.
As stated, the Stone-Weierstrass approximation theorem is NOT valid for complex-valuedfunctions. However, the following holds.
Theorem 2.3. Let E be a compact metric space. Let A be a subalgebra of CIR(E; C) thatseparates points and contains all constant functions. Moreover, assume that if f : E 7→ C liesin A, then also the complex conjugate function f lies in A. Then A = CIR(E; C).
Proof. By the assumptions, if f ∈ A, then its real and its imaginary parts
Re(f) =f + f
2, Im(f) =
f − f2
also lie in A. Let A0 be the sub-algebra of A (on the real numbers). consisting of all functionsf ∈ A with real values. Applying the Stone-Weierstrass theorem to A0, we see that A0 isdense on C(E; IR).
Given any f ∈ C(E; C), we write f as a sum of its real and imaginary part f = Re(f)+i Im(f).By the previous step, there exist two sequences of real-valued functions gn, hn ∈ A0 such that
gn → Re(f) hn → Im(f)
as n→∞, uniformly on E.
We now define fn.= gn+ ihn ∈ A0+ iA0 = A. This achieves the uniform convergence fn → f .
Example 1 (trigonometric polynomials). To see an application of the previous theorem,let E be the unit circle x2+y2 = 1 in IR2. Points on E will be parameterized with the angleθ ∈ [0, 2π]. Let A be the algebra of all trigonometric polynomials:
p(θ) =∑
−N≤n≤N
cneinθ.
where N ≥ 0 is any integer and the cn are complex-valued coefficients. It is clear that A is analgebra, contains the constant functions, separates points, and moreover p ∈ A implies p ∈ Aas well. By Theorem 2.3, A = C(E; C).
This result can be reformulated in terms of the space of continuous functions f : IR 7→ C whichare periodic of period 2π, i.e. f(t) = f(t+ 2π) for every t ∈ IR.
Corollary. Let f : IR 7→ C be a continuous function, periodic of period 2π. Then there existsa sequence of trigonometric polynomials (pk)k≥1 that converges to f uniformly on IR.
38
3 Ascoli’s compactness theorem
Let E be a metric space. A family of bounded continuous functions F ⊂ C(E) is calledequicontinuous if, for every x ∈ E and ε > 0, there exists δ > 0 such that
d(y, x) < δ implies |f(y)− f(x)| < ε (3.1)
for all functions f ∈ F . Here the point is that δ depends on x and ε, but not on f .
Lemma 3. Let F ⊂ C(E) be equicontinuous, and assume that the metric space E is compact.Then F is uniformly equicontinuous. Namely, for every ε > 0 there exists δ > 0 such that
d(x, y) < δ implies |f(x)− f(y)| < ε for all x, y ∈ E, f ∈ F . (3.2)
Proof. Let ε > 0 be given. For each x ∈ E, choose δ(x) > 0 such that (3.1) holds,simultaneously for all functions f ∈ F . We now cover the compact space E with finitely manyballs:
E ⊆ B(x1, δ1) ∪ · · · ∪B(xn, δn).
By Lemma 1 in the Appendix, there exists ρ > 0 so small that every ball B(x, ρ) is entirelycontained inside one of the balls B(xk, δk).
Now assume d(x, y) < ρ. Then there exists an index k ∈ 1, . . . , n such that x, y ∈ B(xj , δj).This implies
|f(x)− f(y)| ≤ |f(x)− f(xj)|+ |f(y)− f(xj)| ≤ ε+ ε.
for every function f ∈ F . Since ε > 0 was arbitrary, this proves the lemma.
In many applications, we construct a sequence of approximate solutions to a given problem,and ask if there exists at least a convergent subsequence. In several cases, a positive answeris provided by the following compactness theorem.
Theorem 2.4 (Ascoli). Let E be a compact metric space. Let F ⊂ C(E) be an equicontinuousfamily of functions, such that
supf∈F
|f(x)| < +∞ for every x ∈ E. (3.3)
Then F is a relatively compact subset of C(E).
Proof. Consider any sequence (fk)k≥1 of functions in F . We need to prove that there existsa subsequence that converges to some limit function f , uniformly on E. This will be provedin several steps.
1. Since the metric E is compact, we can find a sequence of points (yi)i≥1 dense in E.
2. Since the sequence of numbers (fk(y1))k≥1 is bounded, we can extract a first subsequence,say
f11 , f12 , f13, . . . (3.4)
39
which converges at the point y1. Say, f1k(y1)→ f(y1) as k →∞, for some value f(y1).
From the first subsequence (f1k)k≥1 we then extract a second subsequence, say (f2k)k≥1, whichconverges at the point y2. Namely, f1k(y2)→ f(y2) as k →∞, for some value f(y2).
By induction, for every integer i we extract a further subsequence (fik)k≥1, which convergesat the point yi. Say, fik(yj)→ f(yi) as k →∞, for some value f(yi).
3. We claim that the diagonal subsequence f11, f22, f33, . . . is a Cauchy sequence, and henceit converges uniformly on E to some continuous function f ∈ C(E).
Let ε > 0 be given. By Lemma 3, the family F is uniformly equicontinuous, hence there existsδ > 0 such that (3.2) holds.
Since E is compact and the sequence of points (yi)i≥1 is dense in E, there exists an integer msuch that
E ⊆ B(y1, δ) ∪ · · · ∪B(ym, δ) .
Since the m sequences
f11(yi) , f22(yi) , f33(yi), . . . i ∈ 1, . . . ,mare all convergent, they are all Cauchy sequences. Hence we can choose an integer N so largethat, for every i = 1, . . . ,m,
|fjj(yi)− fkk(yi)| < ε whenever j, k ≥ N . (3.5)
Now assume j, k ≥ N and consider any point x ∈ E, say x ∈ B(yi, δ). Using (3.2) and (3.5),we obtain the estimate
|fjj(x)− fkk(x)| ≤ |fjj(x)− fjj(yi)|+ |fjj(yi)− fkk(yi)|+ |fkk(yi)− fkk(x)| ≤ ε+ ε+ ε .
Since ε > 0 was arbitrary, this proves that the diagonal sequence (fkk)k≥1 is Cauchy.
4 Spaces of Holder continuous functions
Let Ω ⊂ IRn be an open set, and 0 < γ ≤ 1. We say that a function f : Ω 7→ IR is Holdercontinuous with exponent γ if there exists a constant C such that
|f(x)− f(y) ≤ C |x− y|γ for all x, y ∈ Ω .
We denote by C0,γ(Ω) the space of all bounded Holder continuous functions on Ω, with norm
‖f‖C0,γ(Ω).= sup
x∈Ω|f(x)| + sup
x,y∈Ω, x 6=y
|f(x)− f(y)||x− y|γ . (4.1)
More generally, given an integer k ≥ 0, we denote by Ck,γ(Ω) the space of all continuousfunctions with Holder continuous partial derivatives up to order k. This space is endowedwith the norm
‖f‖Ck,γ(Ω).=
∑
|α|≤k
(supx∈Ω
|Dαf(x)|)
+∑
|α|=k
(sup
x,y∈Ω, x 6=y
|Dαf(x)−Dαf(y)||x− y|γ
). (4.2)
40
Theorem 2.5 (Holder spaces are complete). Let Ω ⊆ IRn be an open set. For everynon-negative integer k and any 0 < γ ≤ 1, the space Ck,γ(Ω) is a Banach space.
Proof. The fact that (4.2) defines a norm is clear. To prove that the space Ck,γ(Ω) is complete,let (fm)m≥1 be a Cauchy sequence w.r.t. the norm (4.2). Then, for every x ∈ Ω, the sequencefm(x) is Cauchy and converges to some value f(x) uniformly on Ω.
The assumption also imply that, for every |α| ≤ k, the sequence of partial derivatives Dαfmis Cauchy, hence converges to some continuous function vα(x) = Dαf(x) uniformly on Ω.
Finally we show that, for any |α| = k,
limm→∞ sup
x,y∈Ω, x 6=y
∣∣∣Dα(fm − f)(x)−Dα(fm − f)(y)∣∣∣
|x− y|γ = 0 . (4.3)
By assumption,
limm,n→∞
supx,y∈Ω, x 6=y
∣∣∣Dα(fm − fn)(x) −Dα(fm − fn)(y)∣∣∣
|x− y|γ = 0 .
Hence, for any ε > 0, there exists N large enough such that
supx,y∈Ω, x 6=y
∣∣∣Dα(fm − fn)(x)−Dα(fm − fn)(y)∣∣∣
|x− y|γ ≤ ε for all m,n ≥ N .
Keeping m fixed and letting n→∞ we obtain
supx,y∈Ω, x 6=y
∣∣∣Dα(fm − f)(x)−Dα(fm − f)(y)∣∣∣
|x− y|γ ≤ ε for all m ≥ N .
Since ε > 0 was arbitrary, this proves (4.3).
5 Problems
1. Let E be a compact metric space. Assume that a family of real-valued continuous functionsF ⊂ C(E; IR) satisfies the following two conditions:
(i) For every x, y ∈ E and a, b ∈ IR, there exists a function f ∈ F such that f(x) = a andf(y) = b.
(ii) If f, g ∈ F , then the functions maxf, g and minf, g lie in the closure F .
Prove that F is dense in C(E; IR).
2. Prove or disprove the following statements.
(i) For every continuous function f : IR 7→ IR (possibly unbounded), there exists a sequenceof polynomials (pn)n≥1 such that pn(x) → f(x) uniformly on every bounded interval [a, b].(Note: here the sequence of polynomials should be independent of the interval [a, b] )
41
(ii) There exists a sequence of polynomials pn that converges to the function f(x) = e−x2
uniformly on IR.
3. Let f : IR 7→ IR be a real-valued continuous function, periodic of period 2π. Prove that,for any ε > 0, there exists a trigonometric polynomial of the form
p(x) =N∑
k=1
ak sin kx+N∑
k=0
bk cos kx
such that |p(x)− f(x)| < ε for every x ∈ IR.
If f is an even function, i.e. f(x) = f(−x), show that in the above approximation one cantake ak = 0 for every k ≥ 1. If f is an odd function, i.e. f(x) = −f(−x), show that one cantake bk = 0 for every k ≥ 0.
4. Let g : [0, π] 7→ IR be a continuous function.
(i) Prove that, for any ε > 0, there exists a trigonometric polynomial of the form
p(x) =N∑
k=0
bk cos kx
such that |p(x)− f(x)| < ε for every x ∈ [0, π].
(ii) If g(0) = g(π) = 0, prove that, for any ε > 0, there exists a trigonometric polynomial ofthe form
p(x) =N∑
k=1
ak sin kx
such that |p(x)− f(x)| < ε for every x ∈ [0, π].
5. Let f : [a, b] 7→ IR be continuously differentiable. Show that there exists a sequence ofpolynomials pn such that pn → f and p′n → f ′, uniformly on [a, b].
6. Let Ω ⊂ IRn be a bounded open set, and 0 < γ ≤ 1. Prove that the embedding C0,γ(Ω) ⊂⊂C0(Ω) is compact. In other words, if (fn)n≥1 is a bounded sequence in C0,γ(Ω), then it admitsa subsequence that converges in C0(Ω).
7. Let E be the unit circle x2 + y2 = 1 in IR2. Points on E will be parameterized with theangle θ ∈ [0, 2π]. Let A be the family of all trigonometric polynomials of the form
p(θ) =N∑
n=0
cneinθ.
where N ≥ 0 is any integer and the cn are complex-valued coefficients.
(i) Is A is an algebra ?
(ii) Does A contains the constant functions, and separate points ?
42
(iii) Is A dense on the space of all continuous, complex-valued functions f : E 7→ C ?
8. Let Ω ⊂ IRn be a bounded open set. Prove that the family of all polynomials p =p(x1, x2, . . . , xn) in n variables is dense on the space Lp(Ω), for every 1 ≤ p <∞.
9. Consider the rectangle Q = (x, y) ; x ∈ [0, a] , y ∈ [0, b].
(i) Given any continuous function f = f(x, y) on Q and any ε > 0, construct a finite numberof continuous functions g1, . . . , gN : [0, a] 7→ IR and h1, . . . , hn : [0, b] 7→ IR such that
∣∣∣∣∣f(x, y)−N∑
0=1
gi(x)hi(y)
∣∣∣∣∣ ≤ ε for all (x, y) ∈ Q.
(ii) Given any continuous function f : Q 7→ IR which vanishes on the boundary of Q, showthat one can choose an integer M ≥ 1 and finitely many coefficients cmn, 1 ≤ m,n ≤M , suchthat ∣∣∣∣∣∣
f(x, y)−M∑
m,n=1
cmn sinπmx
asin
πn y
b
∣∣∣∣∣∣≤ ε for all (x, y) ∈ Q.
43
Chapter 4
Linear Operators
1 The open mapping theorem
In this chapter we look in more detail at bounded linear operators on a normed space. Webegin with some results based on the Baire category theorem.
Theorem 4.1 (Uniform Boundedness Principle). Let X,Y be Banach spaces. Let F ⊂B(X,Y ) be any family of bounded linear operators. Then either A is uniformly bounded, sothat
supT∈F
‖T‖ < ∞,
or else there exists a dense set S ⊂ X such that
supT∈F
‖Tx‖ = ∞ for all x ∈ S. (1.1)
Proof. For every integer n, consider the set
Sn.= x ∈ X ; ‖Tx‖ > n for some T ∈ F.
If one of these sets, say Sn, is not dense in X, then there exists x0 ∈ X and a radius r > 0such that the closed ball B(x0, r) does not intersect Sn. In other words,
‖Tx‖ ≤ n for all T ∈ F , x ∈ B(x0, r).
If now ‖x‖ < r, then
‖Tx‖ = ‖T (x0 + x)− Tx0‖ ≤ ‖T (x0 + x)‖+ ‖Tx0‖ ≤ 2n for all T ∈ F .
Hence ‖T‖ ≤ 2n/r for every T ∈ F . In this case, the family of operators F is uniformlybounded.
The other possibility is that the set Sn is dense in X, for every n ≥ 1. We observe that eachSn is open. By Baire’s theorem, the intersection S
.=⋃
n≥1 Sn of countably many open densesubsets of the complete metric space X is dense in X. By construction, for each x ∈ S thereexists an operator T ∈ F such that ‖Tx‖ > n. Hence (1.1) holds.
44
Corollary 4.1. Let X,Y be Banach spaces. Let (Tn)n≥1 be a sequence of bounded linearoperators in B(X; Y ). Assume that
Tx.= lim
n→∞Tnx (1.2)
exists for every x ∈ X. Then the map T defined by (1.2) is a bounded linear operator.
Proof. For every x ∈ X, the sequence (Tnx)n≥1 is bounded. By the previous theorem thesequence of operators Tn is uniformly bounded. Hence
‖Tx‖ = limn→∞
‖Tnx‖ ≤ ‖x‖ · supn≥1‖Tn‖
is globally bounded on the unit ball x ∈ X ; ‖x‖ ≤ 1.
If X,Y are metric spaces, we say that a map f : X 7→ Y is open if, for every open subsetU ⊆ X, the image f(U) is open subset of Y . This is the case iff, for every x ∈ X and r > 0,the image f(B(x, r)) of the open ball centered at x with radius r > 0 contains a ball centeredat f(x).
Theorem 4.2 (open mapping). Let X,Y be Banach spaces. Let T : X 7→ Y be a bounded,surjective linear operator. Then T is open.
Proof. 1. By linearity, the image of an open ball B(x, r) can be written as
T (B(x, r)) = T (x) + T (B(0, r)) = T (x) + r T (B(0, 1)).
Therefore, calling Br the open ball centered at the origin with radius r, the theorem will beproved if we show that the image T (B1) contains an open ball around the origin in Y .
2. Since T is surjective, Y =⋃∞
n=1 T (Bn). Since Y is a complete metric space, Baire’s categorytheorem implies that at least one of the closures T (Bn) ⊂ Y has non-empty interior.
Since Br is a homeomorphic image of B1, this implies that already T (B1) has nonemptyinterior. In other words, there exists y0 ∈ Y and r > 0 such that B(y0, r) ⊂ T (B1).
Observing that the unit ball B1 is convex and symmetric, the same is true of its image T (B1)and of the closure T (B1). In particular, by symmetry B(−y0, r) ⊆ T (B1), while convexityimplies that the ball B(0, r) ⊂ Y satisfies
B(0, r) =1
2B(y0, r) +
1
2B(−y0, r) ⊆ T (B1).
Using again the linearity of T , by rescaling we obtain
B(0, 2−nr) ⊆ T (B2−n) for all n ≥ 1 . (1.3)
3. We conclude the proof by showing that B(0, r/2) ⊆ T (B1). Indeed, assume y ∈ B(0, r/2).We proceed by induction.
- By density, we can find x1 ∈ B2−1 such that ‖y − Tx1‖ < 2−2r.
45
- In turn, we can find x2 ∈ B2−2 such that ‖(y − Tx1)− Tx2‖ < 2−3r.
- Continuing by induction, for each n we have
y −n−1∑
j=1
Txj ∈ B(0, 2−nr) ⊆ T (B2−n).
Therefore we can select a point xn ∈ B2−n such that∥∥∥∥∥∥
(y −
n−1∑
j=1
Txj)− Txn
∥∥∥∥∥∥< 2−n−1r .
It is now clear that the series∑
n xn converges, say∑∞
n=1 xn = x. We observe that
‖x‖ ≤∞∑
n=1
‖xn‖ <∞∑
n=1
2−n = 1, Tx = limn→∞
n∑
j=1
Txj = y .
Hence the image T (B1) contains all points y ∈ Y with ‖y‖ < r/2.
Corollary 4.2. If X,Y are Banach spaces and T : X 7→ Y is a continuous bijection, then itsinverse T−1 : Y 7→ X is continuous as well.
Indeed, if T is a bijection, saying that T is open is the same as saying that T−1 is continuous.
2 The closed graph theorem
Let X,Y be Banach spaces. The product space X × Y is the set of all couples (x, y) withx ∈ X and y ∈ Y . This is a Banach space with norm
‖(x, y)‖ .= ‖x‖+ ‖y‖ . (2.4)
Next, let Λ : X 7→ Y be a linear operator. We say that Λ is closed if its graph
Graph(Λ).= (x, y) ; x ∈ X , y = Λx
is a closed subset of the product space X × Y . In other words, Λ is closed provided that thefollowing holds:
If xn ∈ X and yn = Λxn ∈ Y are sequences such that xn → x and yn → y, then Tx = y.
Notice that every continuous operator is closed. Indeed, the assumption that xn → x alonesuffices to conclude that Tx = y. The next result shows that a converse is also true.
Theorem 4.3 (closed graph). Let X,Y be Banach spaces, and let Λ : X 7→ Y be a closedlinear operator. Then Λ is continuous.
Proof. Call Γ.= Graph(Λ). By assumption, Γ is a closed subspace of the Banach space
X × Y , hence it is a Banach space as well.
46
Consider the projections π1 : Γ 7→ X and π2 : Γ 7→ Y , defined as
π1(x,Λx) = x , π2(x,Λx) = Λx .
The map π1 is a bijection between Γ and X, hence by the Corollary, its inverse π−11 is contin-
uous. Therefore Λ = π2 π−11 is the composition of two continuous maps, hence continuous.
Λ
X
Y
2π
π1
Graph( )
Figure 4.1: Proving the closed graph theorem.
Example 1. Consider the space X = C0(]a, b[) of all bounded continuous functions on theopen interval ]a, b[ , with norm ‖f‖C0
.= supa<x<b |f(x)|. Let Λ be the differentiation operator
defined by Λf = f ′. Its domain is the subspace
Dom(A) = f ∈ C0(]a, b[) ; f ′ ∈ C0(]a, b[) = C1(]a, b[)
consisting of all continuously differentiable functions with bounded derivative. Clearly, Λ isnot bounded (hence not continuous). However, its graph is closed. Indeed, consider a sequencefn ∈ Dom(Λ) such that, for some functions f, g ∈ C0, one has
‖fn − f‖C0 → 0 , ‖f ′n − g‖C0 → 0 .
Then f is continuously differentiable and f ′ = g. Hence the point (f, g) ∈ X ×X lies in thegraph of Λ.
Notice that this example does not contradict the closed graph theorem, because Λ is notdefined on the entire space X.
3 Adjoint operators
Let X,Y be Banach spaces on the field IK, let X∗, Y ∗ be their duals. Let Λ : X 7→ Y be abounded linear operator. Then, for every continuous functional y∗ : Y 7→ IK, the composedmap x∗ : X 7→ IK defined as x∗(x) = y∗(Λx) is a continuous functional on X. The mapy∗ 7→ Λ∗y∗
.= y∗ Λ is a bounded continuous operator from Y ∗ into X∗, which we call the
dual of Λ. In other words,
〈Λ∗y∗, x〉 = 〈y∗ , Λx〉 for all x ∈ X .
Theorem 4.4. (i) ‖Λ∗‖ = ‖Λ‖.
47
R
X Y
YX
X YΛ Λ
Λ ***
y*Λ y* *
Figure 4.2: The maps involved in the definition of the adjoint operators
(ii) Ker(Λ) = [Range(Λ∗)]⊥ and Ker(Λ∗) = [Range(Λ)]⊥.
Proof. The statement (i) follows from
‖Λ‖ = sup ‖Λx‖ ; ‖x‖ ≤ 1
= sup |〈y∗, Λx〉| ; ‖x‖ ≤ 1 , ‖y∗‖ ≤ 1
= sup |〈Λ∗y∗, x〉| ; ‖x‖ ≤ 1 , ‖y∗‖ ≤ 1
= sup ‖Λ∗y∗‖ ; ‖y∗‖ ≤ 1 = ‖Λ∗‖ .
To prove (ii), we observe that the following statements are all equivalent:
x ∈ Ker(Λ)Λx = 0〈Λx, y∗〉 = 0 for all y∗ ∈ Y ∗
〈x, Λ∗y∗〉 = 0 for all y∗ ∈ Y ∗
x ∈ [Range(Λ∗)]⊥.
Similarly, the following statements are all equivalent:
y∗ ∈ Ker(Λ∗)Λ∗y∗ = 0〈x, Λ∗y∗〉 = 0 for all x ∈ X〈Λx, y∗〉 = 0 for all x ∈ Xy∗ ∈ [Range(Λ)]⊥.
4 Compact operators
Let X,Y be Banach spaces. A bounded operator Λ : X 7→ Y is compact if, for every boundedsequence (xn)n≥1 of points in X, there exists a subsequence (xnj ) such that Λxnj converges.Equivalently, Λ is compact if and only if, for any bounded set U ⊂ X, the image Λ(U) ⊂ Yhas compact closure.
Theorem 4.5. (i) Let X,Y be Banach spaces, and let Λ : X 7→ Y be a bounded linearoperator. If the range of Λ is finite-dimensional, then Λ is compact.
48
(ii) Let Λn : X 7→ Y be a compact operator, for each n ≥ 1. Assume limn→∞ ‖Λn −Λ‖ = 0.Then the operator Λ is compact as well.
Proof. (i) A bounded linear operator Λ : X 7→ Y is compact if and only the unit ball B1 ⊂ Xhas image Λ(B1) ⊂ Y whose closure is compact. If Range(Λ) is finite dimensional and Λ isbounded, then the closure Λ(B1) is a closed bounded subset of a finite dimensional space,hence compact.
2. To prove (ii), we observe that, since Y is complete, the closure Λ(B1) is compact if andonly if Λ(B1) is precompact. This means: for every ε > 0, the set Λ(B1) can be covered byfinitely many balls of radius ε.
Under the assumptions (ii), let ε > 0 be given. Choose k such that ‖Λ−Λk‖ < ε/2. Since Λk
is compact, we can select finitely many elements y1, . . . , yN ∈ Y such that
Λk(B1) ⊆N⋃
i=1
B(yi,
ε
2
). (4.1)
If ‖x‖ ≤ 1, then ‖Λx − Λkx‖ < ε/2. By (4.1) there exists a point yi with ‖Λkx − yi‖ < ε/2.By the triangle inequality, ‖Λx − yi‖ < ε. This proves that the finitely many balls B(yi, ε)cover Λ(B1).
Theorem 4.6 (adjoint of a compact operator). Let X,Y be Banach spaces, and letΛ : X 7→ Y be a bounded linear operator. Then Λ is compact if and only if its adjointΛ∗ : Y ∗ 7→ X∗ is compact.
Proof. Assume that Λ is compact. Call B1.= x ∈ X ; ‖x‖ ≤ ρ the unit ball in X. By
assumption, the image Λ(B1) has compact closure in Y .
Let (y∗n)n≥1 be a sequence in Y ∗, with ‖y∗n‖ ≤ 1 for every n. We need to show that thesequence Λ∗y∗n admits a convergent subsequence.
Define the sequence of linear functionals
φn(y).= 〈y, y∗n〉 y ∈ Y.
These functions are uniformly Lipschitz continuous. Indeed
|φn(y)− φn(y′)| ≤ ‖y∗n‖ ‖y − y′‖ ≤ ‖y − y′‖ y, y′ ∈ Y .
By Ascoli’s theorem, there exists a subsequence φnj which converges to a function φ, uniformly
on the compact set Λ(B1).
We now observe that
‖Λ∗y∗ni− Λ∗y∗nj
‖ = sup‖x‖≤1
|〈x , Λ∗y∗ni− Λ∗y∗nj
〉| = sup‖x‖≤1
|〈Λx , y∗ni− y∗nj
〉|
= sup‖x‖≤1
|φni(Λx)− φnj(Λx)| → 0 as i, j →∞ .
49
This shows that the subsequence Λ∗y∗njis Cauchy, hence converges to a limit x∗ ∈ X∗. There-
fore Λ∗ is compact.
The converse implication can be proved by the same arguments.
5 Problems
1. Prove that L2([0, 1]) is a vector subspace of L1([0, 1]) of first category. Indeed, for eachn ≥ 1, the set En
.= f : [0, 1] 7→ IR ;
∫ 10 |f |2 dx ≤ n is a closed subset of L1 with empty
interior.
2. Let X,Y,Z be Banach spaces. Let B : X × Y 7→ Z a bilinear map which is continuous atthe origin. Prove that B is bounded, namely there exists a constant C such that
‖B(x, y)‖ ≤ C ‖x‖ ‖y‖ for all x ∈ X , y ∈ Y.
3. Let X be the vector space of all polynomials of one real variable, with norm
‖p‖ .=∫ 1
0|p(t)| dt .
Consider the bilinear functional B : X ×X 7→ IR defined as
B(p, q) =
∫ 1
0p(t)q(t) dt.
Show that, for each fixed p ∈ X, the map q 7→ B(p, q) is a continuous linear functional on X.Similarly, for q ∈ X fixed, the map p 7→ B(p, q) is a continuous linear functional. However,prove that B is not continuous from the product space X ×X into IR.
4. Let X,Y be Banach spaces. Let (Λn)n≥1 be a sequence of linear functionals from X intoY which are equibounded, i.e. ‖Λn‖ ≤ M for some constant M and all n ≥ 1. Consider theset
V.=x ∈ X ; lim
n→∞Λnx exists in Y
.
(i) Prove that V is a closed subspace of X.
(ii) If V = X, prove that Λx.= limn→∞Λnx is a bounded linear operator.
5. Let X be a Banach space. Let Ω ⊂ X be a closed, convex set such that⋃
n≥1 nΩ = X.Prove that Ω contains a neighborhood of the origin.
50
6. Let X,Y be infinite dimensional Banach spaces. IfK : X 7→ Y is a compact linear operator,prove that the K(X) 6= Y , i.e. K cannot be surjective.
As an example, consider the map K : ℓ1 7→ ℓ1 defined as follows. If x = (x1, x2, x3, . . .) with
‖x‖ℓ1 .=∑
n≥1 |xn| <∞, then Kx =(x1,
x22 ,
x33 , . . . ,
xnn , . . .). Find a point y ∈ ℓ1 \K(ℓ1).
7. Let f : IRm 7→ IRn be a bounded function with closed graph. Prove that f is continuous.
On the other hand, construct a function g : IR 7→ IR with closed graph which is not continuous.
8. Let X,Y be Banach spaces. Prove that every continuous function f : X 7→ Y has closedgraph.
Construct a bounded continuous function g : IR 7→ ℓ1 which has closed graph but is notcontinuous.
9. Let X,Y be Banach spaces. Prove that, within the space B(X;Y ) of bounded linearoperators from X into Y , the family of compact operators is a closed subspace.
51
Chapter 5
Hilbert Spaces
1 Spaces with an inner product
Let H be a vector space over the field IK of real or complex numbers. An inner producton H is a map (·, ·) that, to each couple of elements x, y ∈ H, associates a number (x, y) ∈ IKwith the following properties. For every x, y, z ∈ H and α ∈ IK one has
(i) (x, y) = (y, x) where the upper bar denotes complex conjugation,
(ii) (x+ y, z) = (x, z) + (y, z),
(iii) (αx, z) = α(x, z),
(iv) (x, x) ≥ 0, with equality holding if and only if x = 0.
Notice that the above properties also imply
(x, y + z) = (x, y) + (x, z), (x, αy) = α(x, y). (1.1)
In the case where IK = IR, the properties (i)–(iii) simply say that an inner product on a realvector space is a symmetric bilinear mapping. In connection with the above inner product,we also define
‖x‖ .=√(x, x) . (1.2)
The Minkowski inequality, proved below, shows that this is indeed a norm on the vectorspace H.
Theorem 4.1 (two basic inequalities). Let H be a vector space with inner product (·, ·).Then
(i) |(x, y)| ≤ ‖x‖ ‖y‖ (Cauchy-Schwarz inequality)
(ii) ‖x+ y‖ ≤ ‖x‖+ ‖y‖ (Minkowski inequality)
52
Proof. (i) If y = 0, the first inequality is trivial. To cover the general case, set
a.= (x, x), b
.= (x, y), c
.= (y, y).
For every scalar λ ∈ IK one has
0 ≤ (x+ λy, x+ λy) = a+ bλ+ bλ+ cλλ.
Choosing λ = −b/c we obtain
0 ≤ a− bb
c.
Since c = (y, y) > 0, multiplying both sides by c we obtain
0 ≤ ac− |b|2 = ‖x‖2 ‖y‖2 − |(x, y)|2,
proving (i).
(ii) By the Schwarz inequality,
Re(x, y) ≤ |(x, y)| ≤ ‖x‖ ‖y‖ .
Therefore
‖x+y‖2 = (x+y, x+y) = ‖x‖2+‖y‖2+2Re(x, y) ≤ ‖x‖2+‖y‖2+2‖x‖ ‖y‖ =(‖x‖+‖y‖
)2.
Taking square roots we obtain (ii).
A vector spaceH with an inner product (·, ·), which is complete w.r.t. the norm ‖x‖ .=√(x, x),
is called a Hilbert space.
Example 1. The Euclidean space IRn with inner product (x, y) = x1y1 + · · · + xnyn is aHilbert space over the real numbers.
Example 2. The space ℓ2 of all infinite sequences of complex numbers x = (x1, x2, . . .) suchthat
‖x‖ =∞∑
k=1
|xk|2 <∞
is a Hilbert space over the complex numbers, with inner product
(x, y).=
∞∑
k=1
xk yk .
Example 3. Let Ω ⊆ IRn be any open set. The space L2(Ω; IR) of square integrable mapsf : Ω 7→ IR is a Hilbert space, with
(f, g).=
∫
Ωf(x) g(x) dx , ‖f‖L2
.=
∫
Ω|f(x)|2 dx .
53
2 Orthogonal projections
We recall that span(S) is the set of all finite linear combination of elements of the set S,namely
span(S).=
N∑
i=1
ciai ; N ≥ 1, ci ∈ IK, ai ∈ S. (2.1)
In general, span(S) is a subspace of H, possibly not closed. The closure V.= span(S) is called
the space generated by S. We say that the set S is total if it generates the whole space H.In other words, S is total if, for every x ∈ H, there exists a sequence of elements xn ∈ span(S)such that ‖xn − x‖ → 0.
Two elements x, y in a Hilbert space H are said to be orthogonal if (x, y) = 0.
Given any subset S ⊆ H, its orthogonal subspace is defined as
S⊥ .= x ∈ H ; (x, s) = 0 for all s ∈ S.
Notice that S⊥ is always a closed subspace of H.
Theorem 4.2 (perpendicular projections). Let H be a Hilbert space and let V ⊆ H be aclosed subspace. Then
(i) H = V ⊕ V ⊥, in the sense that each x ∈ H can be uniquely written as x = y + z, wherey ∈ V and z ∈ V ⊥.
(ii) y = PV (x) is the unique point in V having minimal distance from x, while z = PV ⊥(x)isthe unique point in V ⊥ having minimal distance from x.
(iii) The perpendicular projections x 7→ y.= PV (x) and x 7→ PV ⊥(x) are linear continuous
operators, with norm ≤ 1.
n V
V ⊥P x
PV
⊥V
z
yy
Figure 5.1: Constructing the perpendicular projections on the subspace V and on the orthogonalsubspace V ⊥.
Proof. 1. Given x ∈ H, we begin by showing that there exists a unique point y ∈ V havingminimal distance from x. Let α
.= d(x, V ). Then there exists a sequence of points (yn)n≥1
such that limn→∞
‖x− yn‖ = α. We claim that (yn)n≥1 is a Cauchy sequence. Indeed, for any
two points u, v ∈ H one has
‖u+ v‖2 + ‖u− v‖2 = 2 ‖u‖2 + 2 ‖v‖2,
54
Applying this equality to u = x− ym and v = x− yn we obtain
‖ym − yn‖2 = 2 ‖x− ym‖2 + 2 ‖x− yn‖2 − 4∥∥∥x− ym + yn
2
∥∥∥2
(2.2)
Since ym+yn2 ∈ V , we have
∥∥∥x− ym+yn2
∥∥∥2≥ α2. Therefore
lim supm,n→∞
‖ym − yn‖2 ≤ 2 lim supm→∞
‖x− ym‖2 + 2 lim supn→∞
‖x− yn‖2 − 4 lim infm,n→∞
∥∥∥x− ym + yn2
∥∥∥2
≤ 2α2 + 2α2 − 4α2 = 0,
proving our claim.
2. Since V is closed (and hence complete), the sequence (yn)n≥1 converges to a unique limity such that ‖x − y‖ = d(x, V ). We claim that this point is unique: if ‖x − y′‖ = d(x, V ) forsome other point y′, then the same argument used in (2.2) yields
‖y − y′‖2 = 2 ‖x− y‖2 + 2 ‖x− y′‖2 − 4∥∥∥x− y + y′
2
∥∥∥2≤ 2α2 + 2α2 − 4α2,
because y+y′
2 ∈ V . Hence y′ = y. This proves that the map x 7→ PV (x) is well defined.
3. We now show that PV x can also be characterized as the unique point y ∈ V such that
x− y ∈ V ⊥. (2.3)
Choose any vector v ∈ V . By step 1, for λ ∈ IR, the real-valued map
λ 7→ ‖x− (y + λv)‖2 = ‖x− y‖2 + |λ|2‖v‖2 + 2Re(x− y, λv)
attains its unique global minimum at λ = 0. Therefore, its derivative at λ = 0 must be zero.This already proves that Re(x− y, v) = 0 for every v ∈ V . If H is a complex Hilbert space,we can replace v with −iv and conclude that Im(x− y, v) = Re(x− y, −iv) = 0 as well. Thisproves (2.3).
4. Next, we prove that the point y ∈ V such that x− y ∈ V ⊥ is unique. Indeed, if y′ ∈ V isanother point such that x− y′ ∈ V ⊥, then
‖y − y′‖2 = (y − y′, y − y′) = (y − y′, x− y′)− (y − y′, x− y) = 0
because y − y′ ∈ V while x− y′ ∈ V ⊥ and x− y ∈ V ⊥.
5. Finally, we show that the perpendicular projection PV : H 7→ V is a linear operator withunit norm.
If y = PV x and y′ = PV x′, then for any scalars α,α′ ∈ IK there holds
αy + α′y′ ∈ V, αx+ α′x′ − (αy + α′y′) ∈ V ⊥.
By step 3, this suffices to conclude PV (αx + α′x′) = αy + α′y′, proving that the projectionoperator PV is linear. Similarly, the operator PV ⊥ = I − PV is linear.
55
Since the vectors PV (x) and PV ⊥(x) = x−PV (x) are perpendicular, by Pythagoras’ theoremwe have
‖PV (x)‖2 ≤ ‖PV (x)‖2 + ‖x− PV (x)‖2 = ‖x‖2.This proves that ‖PV ‖ ≤ 1 and ‖PV ⊥‖ ≤ 1.
Remark. In the above theorem, if V 6= 0, then ‖PV ‖ = 1. If V 6= H, then ‖PV ⊥‖ = 1.
3 Linear functionals on a Hilbert space
The next theorem shows that a Hilbert space coincides with its dual. In other words, everyelement x ∈ H determines a bounded linear functional φx : H 7→ IK defined as phix(y) =(y, x). Conversely, every bounded linear functional on H is of the form y 7→ (y, x), for somex ∈ H.
Theorem 4.3 (Riesz representation of linear functionals). Let H be a Hilbert space.
(i) For every x ∈ H, the map y 7→ (y, x) is a continuous linear functional on H.
(ii) Let y 7→ Ay be a continuous linear functional on H. Then there exists a unique elementa ∈ H such that Ay = (y,a) for every y ∈ H.
Ker (A)
0
a
b
Figure 5.2: If Ay = (y, a) for every y, the vector a must be perpendicular to Ker(A). Since [Ker(A)]⊥
is one-dimensional, given any nonzero vector b ∈ [Ker(A)]⊥, we must have a = tb, where the numbert is determined by the identity Ab = (b, tb).
Proof. (i) Let x ∈ H be given. From the definition of inner product, we already know thatthe map φx defined as φx(y)
.= (y, x) is a linear map. To prove its continuity, it suffices to
check that this map is bounded. This follows from the Cauchy-Schwarz inequality:
‖φx‖ .= sup
‖y‖≤1|(y, x)| ≤ sup
‖y‖≤1‖y‖ ‖x‖ = ‖x‖ .
(ii) Viceversa, let a linear continuous functional y 7→ Ay be given. If Ay ≡ 0 for all y ∈ Hthe conclusion is obvious. Otherwise, V
.= Ker(A) is a closed hyperplane. The orthogonal
complement V ⊥ is a subspace of dimension one. Choose any non-zero vector b ∈ V ⊥ anddefine κ = Ab/‖b‖2, a .
= κb. This choice yields
0 6= Ab = κ(b,b) = (b, κb) = (b,a).
56
Given any vector y ∈ H, we can decompose y as a sum of a vector in V and a vector in V ⊥:
y = PV y + αb
for some α ∈ IK. Then
Ay = A(PV y) +A(αb) = 0 + αAb = 0 + α(b,a) = (PV y, a) + (αb,a) = (y,a).
Remark. By the previous theorem, the map x 7→ φx is an isometric isomorphism between Hand its dual space H∗ (= the space of bounded linear functionals on H). We can thus identifythe two spaces H and H∗.
If now Λ : H 7→ H is a bounded linear operator, its adjoint Λ∗ : H∗ 7→ H∗ can be identifiedwith an operator Λ∗ : H 7→ H. This adjoint is characterized by the identities
(x , Λ∗y) = (Λx , y) for all x, y ∈ H .
4 Gram-Schmidt orthogonalization
Let H be a Hilbert space. We say that x ∈ H is normalized if ‖x‖ = 1. A subset E ⊂ H isorthonormal if every vector in E has unit norm and any two vectors in E ore orthogonal toeach other.
Next, consider a set S = v1, v2, . . . , vn of finitely many linearly independent vectors. Assumex ∈ spanv1, . . . , vn, so that
x =n∑
k=1
θkvk
How do we compute the coefficients θk ?
Observe that, for every j = 1, . . . , n we must have
(x, vj) =n∑
k=1
θk(vk, vj)
Therefore the coefficients θ1, . . . , θn are obtained by solving the system of n linear equations
(v1, v1) · · · (vn, v1)...
. . ....
(v1, vn) · · · (vn, vn)
θ1...θn
=
(x, v1)...
(x, vn)
(4.1)
This system is much easier to solve when the matrix is diagonal, namely (vi, vj) = 0 for i 6= j.This is the case where the vectors v1, v2, . . . , vn are orthogonal to each other. The explicitsolution is then computed as
θk =(x, vk)
(vk, vk).
57
~
v1
v2
e1
e2
v2
~
2
0
v −2v
Figure 5.3: The Gram-Schmidt orthonormalization procedure, applied to the two vectors v1, v2 .
In the special case where the set v1, . . . , vn is orthonormal, so that (vk, vk) = 1 for everyk = 1 . . . , n, the above formula simplifies further to
θk = (x, vk).
The previous analysis shows that using an orthonormal basis makes computations much eas-ier. Given any linearly independent set S = v1, v2, . . ., we now describe a general pro-cedure to generate an orthonormal set e1, e2, . . . in such a way that spane1, . . . , en =spanv1, . . . , vn for every n ≥ 1. This is achieved by induction.
(i) Start by setting e1.= v1
‖v1‖ .
(ii) If e1, . . . , en−1 have been constructed, let
vn be the perpendicular projection of vn on the subspace spanv1, . . . , vn−1 =spane1, . . . , en−1. Then set
en.=
vn − vn‖vn − vn‖
Observe that vn 6= vn because vn /∈ spanv1, . . . , vn−1. Hence en is well defined and has normone. Moreover, en is perpendicular to all vectors e1, . . . , en−1.
Notice that
vn =n−1∑
k=1
(vn, ek)ek ,
hence
en =vn −
∑n−1k=1(vn, ek)ek∥∥∥vn −
∑n−1k=1(vn, ek)ek
∥∥∥.
5 Orthonormal sets
If e1, e2, . . . , en is an orthonormal basis of IRn, then every vector x ∈ IRn can be written asa linear combination
x =n∑
k=1
(x, ek)ek .
58
In an infinite dimensional Hilbert space H, the above finite sum should be replaced by aninfinite series. We now examine in which cases the corresponding series converges, and wehave the equality
x =∞∑
k=1
(x, ek)ek .
Lemma 1. Let H be a Hilbert space. For every S ⊆ H the orthogonal subspace S⊥ is a closedsubspace of H. Moreover, the following are equivalent:
(i) The set S is total, i.e. span(S) is dense in H.
(ii) S⊥ = 0.
Proof. 1. The fact that S⊥ is a subspace of H is clear. Now assume that xn ∈ S⊥ andxn → x. Then for every a ∈ S we have
(x, a) = limn→∞
(xn, a) = 0.
Hence a ∈ S⊥, showing that this subspace is closed.
2. To prove that (i) implies (ii), assume that the set S is total and that x ∈ S⊥, so that(x, a) = 0 for every a ∈ S. Then there exists a sequence of linear combinations
xn =n∑
k=1
θn,kak
such that xn → x as n→∞. This implies
(x, x) = limn→∞(x, xn) = lim
n→∞
(x ,
n∑
k=1
θn,kak)
=n∑
k=1
θn,k (x, ak) = 0
Hence x = 0.
3. Finally, assume that (ii) holds, and let V be the closure of the set of all linear combinationsof elements of S. If V 6= H, then there exists an element y /∈ V . Consider the perpendicularprojection PV (y). Then w
.= y−PV y is a non-zero vector perpendicular to V , hence (y, a) = 0
for all a ∈ S, against the assumption. Hence (i) holds as well.
Theorem 4.4. Let S = e1, e2, . . . be an orthonormal subset of a Hilbert space H. Let V bethe closed subspace generated by S and call PV : H 7→ V the perpendicular projection. Then,for every x ∈ H, one has the Bessel inequality
∞∑
k=1
|(x, ek)|2 = ‖PV x‖2 ≤ ‖x‖2. (5.1)
Moreover, the following series converges:
∞∑
k=1
(x, ek)ek = PV x . (5.2)
59
Proof. For any n ≥ 1, call Vn.= spane1, . . . , en. Then
PVnx =n∑
k=1
(x, ek)ek .
and hence, by orthonormality,
‖PVnx‖2 =( n∑
j=1
(x, ej)ej ,n∑
k=1
(x, ek)ek)
=n∑
k,j=1
(x, ej)(x, ek)(ej , ek) =n∑
j=1
|(x, ej)|2
Since ‖PVnx‖ ≤ ‖x‖ for every n, the first part of the theorem is proved.
Next, we show that the sequence of partial sums
xn.=
n∑
k=1
(x, ek)ek
is Cauchy. Observe that all terms in the series (5.2) are orthogonal to each other. For m < n,using Pithagoras’ theorem we thus obtain
‖xn − xm‖2 =n∑
k=m+1
|(x, ek)|2 → 0 as m,n→∞.
Since H is complete, the series has a well defined sum, say x. To complete the proof, we needto show that x = PV x.
Since xn ∈ V for each n and xn → x, it is clear that x lies in the closed subspace V .
Moreover,(x− x , ek) = lim
n→∞(x− xn, ek) = 0
because (x−xn, ek)=0 as soon as n ≥ k. This proves that (x−x) is perpendicular to all vectorsek, and hence to every linear combination of these vectors. By density, x− x is perpendicularto every vector v ∈ V .
We say that an orthonormal set S ⊂ H is complete if the set of all linear combinations
N∑
k=1
αkxk with N ≥ 1, αk ∈ IK, xk ∈ S
is dense in H. In this case, the closed subspace generated by S is V = H. Hence PV x = x forevery x ∈ H, and (5.1)-(5.2) yield
∞∑
k=1
|(x, ek)|2 = ‖x‖2,∞∑
k=1
(x, ek)ek = x . (5.3)
60
5.1 Fourier series
As an application of the previous theory, consider the Hilbert space of complex valued functionsL2([−π, π] ; C), with inner product
(f, g).=
∫ π
−πf(x) g(x) dx . (5.1)
Within this space, the set of functions
ϕn(x).=
einx√2π
n ∈ ZZ
is orthonormal. Indeed,
(ϕm, ϕn) =
∫ π
−πeimx einx dx =
∫ π
−πei(m−n)x dx =
0 if m 6= n ,
2π if m = n .
We claim that the countable set S.= einx ; n ∈ ZZ is total in L2([−π, π]). Indeed, consider
any function f ∈ L2([−π, π] ; C). We can extend f by periodicity to the entire real line.Taking mollifications fε
.= Jε ∗f , we obtain a family of smooth functions fε, periodic of period
2π, such that‖fε − f‖L2([−π, π]) → 0 as ε→ 0 .
In turn, by the Stone-Weierstrass theorem, each continuous function fε : IR 7→ C, periodicof period 2π, can be uniformly approximated by a trigonometric polynomial of the formp(x) =
∑Nk=−N αke
ikx. Observing that
‖g‖L2([−π, π]) =
(∫ π
−π|g(x)|2 dx
)1/2
≤√2π ‖g‖C0([−π, π]) ,
it is clear that each continuous functions fε can be approximated by trigonometric polynomialsp(·) also in the (weaker) norm of L2([−π, π]). Hence span(S) = L2([−π, π]).
We now consider the trigonometric series
∞∑
k=−∞ak
eikx√2π
ak.= (f, ϕk) =
∫ π
−πf(x)
e−ikx
√2π
dx . (5.2)
By the previous theorems, this series converges to f in L2([−π, π]). More precisely
limN→∞
∥∥∥∥∥∥f −
N∑
k=−N
akϕk
∥∥∥∥∥∥L2([−π, π])
= 0 . (5.3)
6 Positive definite operators
A basic problem of linear algebra is to solve the system of linear equations
Ax = b ,
61
where A is an n × n matrix and b is a vector in IRn. One condition which guarantees theexistence and uniqueness of solutions is that A be strictly positive definite. Indeed, (denotingby x† the transpose of the column vector x), if x†Ax > 0 for every x 6= 0, then A must have fullrank. Hence the above system of linear equations will have a unique solution. In this sectionwe show that this result can be extended to the setting of (possibly infinite dimensional)Hilbert spaces.
Let H be a Hilbert space over the reals. We say that a bounded linear operator A : H 7→ His positive definite if there exists β > 0 such that
(Au, u) ≥ β ‖u‖2 for all u ∈ H. (6.1)
Theorem 4.5 (positive definite operators). Let H be a Hilbert space over the reals. LetA : H 7→ H be a bounded linear operator which is positive definite, so that (6.1) holds. Then,for every f ∈ H, there exists a unique u
.= A−1f ∈ H such that
Au = f . (6.2)
The inverse operator A−1 satisfies
‖A−1‖ ≤ 1
β. (6.3)
Proof. We need to show that, under the assumption (6.2), the continuous map A is one-to-oneand onto.
1. From (6.1) it followsβ‖u‖2 ≤ (Au, u) ≤ ‖Au‖ ‖u‖
Henceβ‖u‖ ≤ ‖Au‖. (6.4)
If Au = 0, then u = 0, proving that Ker(A) = 0 and A is one-to-one.
2. Next, we claim that Range(A) is closed. Consider any sequence of points vn ∈ Range(A),such that vn → v. We need to show that v = Au for some u ∈ H.
By assumption, vn = Aun for some un ∈ H. Using (6.4) we obtain
lim supm,n→∞
‖um − un‖ ≤ lim supm,n→∞
1
β‖Aum −Aun‖ = lim sup
m,n→∞
1
β‖vm − vn‖ = 0.
Hence the sequence (un)n≥1 is Cauchy, and converges to some limit u ∈ H. By continuity,Au = v, proving our claim.
3. We now claim that Range(A) = H. If not, since Range(A) is closed, we could find anonzero vector w ∈ [Range(A)]⊥. But this would imply
β‖w‖2 ≤ (Aw,w) = 0.
62
reaching a contradiction.
4. By the previous steps, A : H 7→ H is a bijection, hence the equation (6.2) has a uniquesolution u
.= A−1f . By (6.4) it follows
‖A−1f‖ = ‖u‖ ≤ ‖f‖β
for all f ∈ H .
This yields (6.3).
The above result can also be conveniently formulated in terms of bilinear forms.
Theorem 4.6 (Lax-Milgram). Let H be a Hilbert space over the reals and let B : H×H 7→IR be a continuous bilinear functional. This means
B[au+ bu′ , v] = aB[u, v] + bB[u′, v], B[u, av + bv′] = aB[u, v] + bB[u, v′],∣∣∣B[u, v]
∣∣∣ ≤ C ‖u‖ ‖v‖for some constant C and all u, u′, v, v′ ∈ H, a, b ∈ IR. In addition, assume that B is strictlypositive definite, i.e. there exists a constant β > 0 such that
B[u, u] ≥ β ‖u‖2 for all u ∈ H .
Then, for every f ∈ H, there exists a unique u ∈ H such that
B[u, v] = (f, v) for all v ∈ H . (6.5)
Moreover,‖u‖ ≤ β−1‖f‖ . (6.6)
Proof. For every fixed u ∈ H the map v 7→ B[u, v] is a continuous linear functional on H.By the Riesz representation theorem, there exists a unique vector which we call Au ∈ H suchthat
B[u, v] = (Au, v) for all v ∈ H .
We claim that A is a bounded, positive definite linear operator.
The linearity of A is easy to check. To prove that A is bounded we observe that, for everyu ∈ H,
‖Au‖ .= sup
‖v‖=1|(Au, v)| = sup
‖v‖=1|B[u, v]| ≤ C ‖u‖ .
Hence ‖A‖ ≤ C.
Moreover,(Au, u)
.= B[u, u] ≥ β ‖u‖2 ,
proving that A is strictly positive definite.
By the previous theorem the equation Au = f has a unique solution u = A−1f , satisfying‖u‖ ≤ β−1‖f‖. By the definition of A, this provides a solution to (6.5).
63
7 Weak convergence
Let H be a Hilbert space. We say that a sequence of points xn ∈ H converges weakly to apoint x ∈ H, and write xn x, if
limn→∞
(y, xn) = (y, x) for all y ∈ H.
We recall that, if H is infinite dimensional, then the closed unit ball B1.= x ∈ H ; ‖x‖ ≤ 1
is not compact (w.r.t. the topology determined by the norm). In particular, B1 contains aninfinite orthonormal sequence, which does not admit any convergent subsequence.
On the other hand, the following theorem shows that every closed ball in a Hilbert space issequentially weakly compact.
Theorem 4.7 (weakly convergent sequences). Let H be a Hilbert space.
(i) Every weakly convergent sequence is bounded. Namely, if xn x, then ‖xn‖ ≤ C for someconstant C and all n ≥ 1.
(ii) Every bounded sequence of points xn ∈ H admits a weakly convergent subsequence:xnj x for some x ∈ H.
Proof. 1. Every xn can be regarded as a continuous linear map, namely y 7→ (xn, y) from Hinto IR. The Hilbert norm of xn coincides with the norm of xn as a linear functional on H,namely
‖xn‖ = sup‖y‖≤1
(xn, y) .
By assumption, for every y ∈ H, the set (xn, y) ; n ≥ 1 is bounded. By the UniformBoundedness Principle, the countable set of linear functionals xn ; n ≥ 1 is uniformlybounded. This establishes (i).
2. To prove (ii), consider the vector space V.= spanxn ; n ≥ 1, i.e. the closure of the set
of all linear combinations of the points x1, x2, . . . Call V⊥ the orthogonal subspace. Clearly,
V is separable. Hence we can construct an orthonormal basis e1, e2, . . . of V . This can beobtained from the xn by the Gram-Schmidt orthogonalization procedure.
3. The sequence of numbers (e1, xn)n≥1 is bounded, hence we can find an infinite set of indicesI1 ⊂ N such that the subsequence (e1, xn)n∈I1 converges.
Similarly, the sequence (e2, xn)n∈I1 is bounded, hence we can find an infinite set of indicesI2 ⊂ I1 such that the subsequence (e2, xn)n∈I2 converges.
By induction, for each m we can find an infinite set of indices Im ⊂ Im−1 such that thesubsequence (em, xn)n∈Im converges.
We now choose a subsequence xn1 < xn2 < xn3 < · · · with xnk∈ Ik for every k. This
guarantees the convergence limk→∞(em, xnk) → αm ∈ IK, for every m ≥ 1.
64
4. Consider the point
z.=
∞∑
m=1
αmem . (7.1)
We first show that this is well defined, i.e. the series is convergent.
By assumption, ‖xn‖ ≤ C for some constant C and all n ≥ 1. Hence, for every integer N ≥ 1we have
N∑
m=1
α2m = lim
k→∞
N∑
m=1
|(em, xnk)|2 ≤ lim sup
k→∞
N∑
m=1
‖xnk‖2 ≤ C2 . (7.2)
This proves that the series∑
m α2m is convergent.
Since the sequence ek ; k ≥ 1 is orthonormal, for any n < m we have∥∥∥∥
m∑
k=n+1
αkek
∥∥∥∥2
=m∑
k=n+1
α2k → 0 as n,m→∞ .
Hence the sequence of partial sums is Cauchy, and the series in (7.1) has a well defined sum.Moreover, (7.2) yields
‖z‖2 = limN→∞
N∑
m=1
α2m ≤ C2.
5. We claim that the subsequence (xnk)k≥1 converges weakly to z. Indeed, consider any vector
y ∈ H. We need to show thatlimk→∞
(y, xnk) = (y, z) . (7.3)
Write y = y1 + y2, with
y1 =∞∑
m=1
bmem ∈ V , y2 ∈ V ⊥.
Let ε > 0 be given, and choose N large enough so that∑
m>N
|bm|2 < ε2 .
Then
(y , xnk−z) = (y1 , xnk
−z) =( ∑
m≤N
bmem , xnk−z)+( ∑
m>N
bmem , xnk−z) .
= Ak+Bk .
From the inequalities
‖xn‖ ≤ C , ‖z‖ ≤ C ,∥∥∥∑
m>N
bmem∥∥∥2
=∑
m>N
|bm|2 < ε2 ,
it follows
limk→∞
Ak = 0 , |Bk| ≤∥∥∥∥∥∑
m>N
bmem
∥∥∥∥∥ ‖xnk− z‖ ≤ ε (‖xn‖+ ‖z‖) ≤ 2C ε .
Thereforelim supk→∞
|(y , xnk− z)| ≤ 2C ε.
Since ε > 0 was arbitrary, this completes the proof.
65
The following result shows that a compact operator maps weakly convergent sequences intostrongly convergent ones.
Theorem 4.8. In a Hilbert space H, consider a weakly convergence sequence xn x. LetΛ : H 7→ H be a compact operator. Then one has the strong convergence
‖Λxn − Λx‖ → 0 . (7.4)
Proof. To prove (7.4), it suffices to show that, from any subsequence (xn)n∈I1 one can extracta further subsequence (xn)n∈I2 , with I2 ⊆ I1 ⊆ N , such that
limn→∞, n∈I2
‖Λxn − Λx‖ = 0 .
Let a subsequence (xn)n∈I1 be given. Since this sequence is weakly convergent, by the previoustheorem it is globally bounded, say ‖xn‖ ≤ C for some every n ∈ I1 .
Since Λ is compact, form this bounded sequence we can extract a further subsequence (xn)n∈I2 ,with I2 ⊆ I1, such that the images converge strongly:
limn→∞, n∈I2
‖Λxn − y‖ = 0
for some y ∈ H. It remains to show that y = Λx.
Calling Λ∗ the adjoint operator, we now have
(Λxn − Λx , v) = (xn − x , Λ∗v) → 0 for all v ∈ H.
This proves the weak convergence Λxn Λx. Since the weak limit is unique, this yieldsΛx = y, completing the proof.
8 Problems
1. In a Hilbert space H, prove Pithagoras’ theorem: if two vectors x, y ∈ H are orthogonalto each other, then ‖x‖2 + ‖y‖2 = ‖x+ y‖2.
2. Let H be a vector space with inner product (·, ·). Prove that, as the inner product is acontinuous map from the product space H×H into K is continuous. In other words, if xk → xand yk → y w.r.t. the norm (1.2), then the sequence of inner products (xk, yk) converges to(x, y).
3. Let X be a Banach space over the reals, with norm ‖ · ‖. Does there exist an inner product(·, ·) on X such that ‖x‖ =
√(x, x) for every x ∈ X? To answer, follow the two steps below.
66
(i) Let H be a real Hilbert space. Prove that its norm ‖x‖ =√(x, x) satisfies the parallelogram
identity‖x+ y‖2 + ‖x− y‖2 = 2(‖x‖2 + ‖y2‖) for all x, y ∈ X . (8.5)
(ii) Viceversa, Let X be a Banach space over the real numbers, whose norm satisfies theparallelogram identity (8.5). Show that
(x, y).=
1
2
(‖x+ y‖2 − ‖x‖2 − ‖y‖2
).
is an inner product on X, which yields exactly the same norm ‖ · ‖.
4. On the vector space IR2, prove that the norm ‖x‖p .= (|x1|p + |x2|p)1/p does NOT satisfy
the parallelogram identity (8.5) for all p ≥ 1, p 6= 2.
5. Consider the Hilbert space H = L2([−1, 1]), with inner product (f, g) =∫ 1−1 fg dx. Show
that the sequence of polynomials 1, x, x2, x3, . . . is linearly independent, and total in thespace H. Applying the Gram-Schmidt orthogonalization procedure, construct an orthonor-mal sequence of polynomials p0, p1, p2, p3, . . . (proportional to the Legendre polynomials).Explicitly compute the first three polynomials.
6. Let (en)n≥1 be an orthonormal sequence in a Hilbert space H. Prove the weak convergenceen 0. In other words show that, for every x ∈ H, one has limn→∞(x, en) = 0.
7. Let H be Hilbert space and let any vector x ∈ H be given, with ‖x‖ ≤ 1. Construct asequence of vectors xn with ‖xn‖ = 1 for every n ≥ 1, such that xn x.
8. Let (vk)k≥1 be a sequence of unit vectors in a real Hilbert space H. Let (αk)k≥1 be asequence of real numbers.
(i) If∑∞
k=1 |αk| <∞, prove that the series∑∞
k=1 αkvk converges.
(ii) Assume that the sequence (ek)k≥1 is orthonormal. In this case, prove that the series∑∞k=1 αkek converges if and only if
∑∞k=1 |αk|2 < ∞.
9. Let S = e1, e2, e3, . . . an orthonormal sequence in a Hilbert space H. Prove that thefollowing are equivalent.
(i) span(S) is dense in H.
(ii) For every x ∈ H, one has∑
n≥1 |(x, en)|2 = ‖x‖2.
(iii) For every x ∈ H, one has∑
n≥1(x, en)en = x.
10. Consider the Hilbert space ℓ2 of all sequences of real numbers x = (x1, x2, . . .) such that∑∞k=1 |xk|2 < ∞, with inner product (x,y)
.=∑∞
k=1 xkyk . Define the operator Λ : H 7→ H
67
by setting Λ(x1, x2, x3, . . .).= (x2, x3, x4, . . .). Compute the adjoint operator Λ∗. Are the
operators Λ,Λ∗ surjective ? One-to-one ?
11. Let S be a convex set. We say that x ∈ S is an extreme point of S if x cannot beexpressed as a convex combination of distinct points of S. In other words,
x 6= θx1 + (1− θ)x2 whenever 0 < θ < 1, x1, x2 ∈ S , x1 6= x2 .
Prove the following.
(i) If S is the closed unit ball in a Hilbert space H, then every point x ∈ S with ‖x‖ = 1 is anextreme point of S. This is true, in particular, for the space H = L2([0, 1]).
(ii) In contrast, consider the unit ball in L1([0, 1]), i.e.
B.=f : [0, 1] 7→ IR ;
∫ 1
0|f(t)| dt ≤ 1
.
Prove that B does not contain any extreme point.
12. Let (xn)n≥1 be a sequence of points in a Hilbert spaceH such that C.= lim infn→∞ ‖xn‖ <
∞. Prove that there exists a weakly convergent subsequence xnj x for some point x ∈ Hsatisfying ‖x‖ ≤ C.
13. Let H be an infinite dimensional, separable Hilbert space over the reals, and let ℓ2 be the
space of all sequences of real numbers a = (a1, a2, . . .) such that ‖a‖ℓ2 .=(∑∞
k=1 a2k
)1/2< ∞.
Construct a linear bijection Λ : H 7→ ℓ2 which preserves distances, i.e. such that
‖Λx‖ℓ2 = ‖x‖H for all x ∈ H .
14. Given n vectors v1, v2, . . . , vn in a real Hilbert space H, define their Gram determinantas the determinant of the n× n symmetric matrix in (4.1):
G(v1, . . . , vm).= det
(v1, v1) · · · (vn, v1)...
. . ....
(v1, vn) · · · (vn, vn)
.
(i) Prove that the vectors v1, . . . , vn are linearly independent if and only if G(v1, . . . , vn) 6= 0.
(ii) Assuming that the vectors are linearly independent, consider the subspace V.=
spanv1, . . . , vn. For any vector x ∈ H, show that the distance of x to V is
d(x, V ) = ‖x− PV (x)‖ =
√G(x, v1, v2, . . . , vn)
G(v1, v2, . . . , vn).
68
(iii) How is G(v1, v2, . . . , vn) related to the n-dimensional volume of the parallelepiped withedges v1, . . . , vn?
15. Let f ∈ L2(IR). Prove that there exists a unique even function g0 such that
‖f − g0‖L2 = ming∈L2(IR), g even
‖f − g‖L2 .
Explicitly determine the function g0. (Note: g even means g(x) = g(−x) for all x ∈ IR.)
16. Let e1, e2, . . . be an orthonormal basis of a Hilbert space H. For each n ≥ 1, letKn : H 7→ H be a bounded linear operator whose range is contained in spane1, e2, . . . en.Assuming that ‖Kn −K‖ → 0, prove that the limit operator K is compact.
17. Let un∞n=1 and vn∞n=1 be two orthonormal sets in a Hilbert space H. Assume that
∞∑
n=1
‖un − vn‖ < 1.
Show that if one set is complete, the other set is also complete.
18. Let un ; n ≥ 1 be a countable set of linearly independent unit vectors in a Hilbertspace H. Consider the vector y
.=∑∞
n=1 2−nun .
(i) Assuming that all vectors un are mutually orthogonal, prove that the set S.=
y, u1, u2, u3, . . . is linearly independent.
(ii) Show by an example that, if the vectors un are not mutually orthogonal, the above set Scan be linearly dependent.
69
Chapter 6
Compact Operators on a Hilbert
Space
1 Fredholm theory
Let H be a Hilbert space. We recall that a bounded linear operator K : H 7→ H is compactif for every bounded sequence of points un ∈ H one can extract a subsequence (unj )j≥1 suchthat the images converge: Kunj → v for some v ∈ H.
The next theorem shows that several properties of finite-dimensional operators A : IRn 7→ IRn
can be extended to a Hilbert space setting. Indeed, these properties remain valid for operatorsof the form I −K, where I is the identity and K is a compact operator.
Theorem 5.1 (Fredholm). Let H be a Hilbert space over the reals and let K : H 7→ H be acompact linear operator. Then
(i) Ker(I −K) is finite dimensional,
(ii) Range(I −K) is closed,
(iii) Range(I −K) = Ker(I −K∗)⊥
(iv) Ker(I −K) = 0 if and only if Range(I −K) = H.
(v) Ker(I −K) and Ker(I −K∗) have the same dimension.
Proof. 1. If the kernel of (I − K) is infinite-dimensional, one can find an orthonormalsequence (en)n≥1 contained in Ker(I −K). In this case Ken = en for every n. Moreover, byPithagoras’theorem, for m 6= n one has
‖em − en‖2 = ‖em‖2 + ‖en‖2 = 2.
Therefore ‖Kem −Ken‖ = ‖em − en‖ =√2 for every m 6= n. Therefore, from the sequence
(Ken)n≥1 one cannot extract any convergent subsequence. This contradiction establishes (i).
70
2. Toward the proof of (ii), we first show that there exists β > 0 such that
‖u−Ku‖ ≥ β‖u‖ for all u ∈ Ker(I −K)⊥. (1.1)
Indeed, if (1.1) fails, we could find a sequence of points un ∈ Ker(I −K)⊥ such that ‖u‖ = 1and ‖un −Kun‖ < 1/n.
Since the sequence (un)n≥1 is bounded, by extracting a subsequence and relabelling, we canassume that this sequence converges weakly, say un u for some u ∈ H.
Since K is a compact operator, this implies the strong convergence Kun → Ku.
We now have
‖un −Ku‖ ≤ ‖un −Kun‖+ ‖Kun −Ku‖ → 0 asn→∞.
This implies the strong convergence un → Ku. Recalling the weak limit un u, we concludethat u = Ku and un → u strongly.
By construction, we now have
‖u‖ = limn→∞ ‖un‖ = 1, u ∈ Ker(I −K)⊥
while, at the same time, u−Ku = 0 hence u ∈ Ker(I−K). We thus reached a contradiction,proving (1.1).
3. We now prove that Range(I−K) is closed. Consider a sequence of points vn ∈ Range(I −K), with vn → v as n→∞. We need to find some u such that v = u−Ku. By assumption,for each n ≥ 1, there exists un such that vn = un − Kun. Notice that, if un → u for someu ∈ H, then by continuity we could immediately conclude that v = u − Ku. In general,however, there is no guarantee that the sequence (un)n≥1 converges.
To fix this problem, let un to be the perpendicular projection of un on Ker(I −K), and letzn
.= un − un These definitions yield
zn.= un − un ∈ Ker(I −K)⊥, vn = un −Kun = zn −Kzn .
Using (1.1), for every pair of indices m,n we obtain
‖vm − vn‖ ≥ β‖zm − zn‖.
Since the sequence (vn)n≥1 is Cauchy, this proves that the sequence (zn)n≥1 is a Cauchysequence as well. Therefore there exists u ∈ H such that zn → u, and hence
u−Ku = limn→∞ zn −Kzn = lim
n→∞ vn = v.
4. Since Range(I − K) and Ker(I − K∗)⊥ are closed subspaces, the assertion (iii) holds ifand only if
Range(I −K)⊥ = Ker(I −K∗). (1.2)
71
3
Ker (I−K) ⊥
zn
nu
0
~
Ker (I−K)
un
H
H
H
H12
Figure 6.1: Left: by choosing zn ∈ Ker(I −K)⊥ we achieve the inequality β‖zn‖ ≤ ‖vn‖, used in theproof of (ii). Right: if I −K is one-to-one but not onto, the sequence of subspaces Hn
.= (I −K)n(H)
is strictly decreasing. This leads to a contradiction, used in the proof of (iv).
This is proved by observing that the following statements are all equivalent.
x ∈ Ker(I −K∗)(I −K∗)x = 0(y , (I −K∗)x) = 0 for all y ∈ H((I −K)y , x) = 0 for all y ∈ Hx ∈ Range(I −K)⊥.
5. Concerning (iv), assume that Ker(I −K) = 0, so that the operator I −K is one-to-one.If Range(I −K) 6= H, we shall derive a contradiction.
Indeed, assume H1.= (I −K)(H) 6= H. By (ii), H1 is a closed subspace of H. Since I −K is
one-to-one, we must haveH2
.= (I −K)(H1) ⊂ H1 .
We can continue this process by induction: for every n, we set
Hn.= (I −K)n(H)
By induction, we see that each Hn is a closed subspace of H, and
H ⊃ H1 ⊃ H2 ⊃ · · ·
For each n ≥ 1 we now choose a vector en ∈ Hn ∩ H⊥n+1 with ‖en‖ = 1. Observe that, if
m < n, then
Kem −Ken = − (em −Kem) + (en −Ken) + (em − en) = em + zm
with zm.= − (em −Kem) + (en −Ken)− en ∈ Hm+1.
Since em ∈ H⊥m+1, by Pythagoras’ theorem this implies
‖Kem −Ken‖ ≥ ‖em‖ = 1.
Therefore, the sequence (Ken)n≥1 cannot have any strongly convergent subsequence, contra-dicting the compactness of K.
72
6. To prove the converse implication in (iv), we use a duality theorem. Assume that Range(I−K) = H. Then Ker(I −K∗) = Range(I −K)⊥ = H⊥ = 0. Since K∗ is compact, by theprevious step we have Range(I − K∗) = H. Using again the duality theorem, we obtainKer(I −K) = Range(I −K∗)⊥ = H⊥ = 0, as claimed.
7. Toward a proof of (v), we first show that the dimension of Ker(I −K) is greater or equalto the dimension of Range(I −K)⊥. Indeed, suppose on the contrary that
dim Ker(I −K) < dim Range(I −K)⊥. (1.3)
Then there exists a linear map A : Ker(I−K) < dim Range(I−K)⊥ which is one-to-one butnot onto. We extend A to a linear map A : H 7→ Range(I −K)⊥ defined on the whole spaceH, by requiring that Au = 0 if u ∈ Ker(I −K)⊥. Since the range of A is finite dimensional,the operator A is compact, and so is K +A.
We claim that Ker(I − (K +A)) = 0. Indeed, consider any vector u ∈ H and write
u = u1 + u2 u1 ∈ Ker(I −K), u2 ∈ Ker(I −K)⊥.
Then
(I −K −A)(u1 + u2) = (I −K)u2 −Au1 ∈ Range(I −K)⊕Range(I −K)⊥. (1.4)
Since (I−K)u2 is orthogonal to Au1 the sum (I−K)u2+Au1 can vanish only if (I−K)u2 = 0and Au1 = 0. Recalling that the operator I −K is one-to-one on Ker(I −K)⊥ and A is one-to-one on Ker(I −K), we conclude that u1 = u2 = 0.
Applying (iv) to the compact operator K + A, we conclude that Range(I − (K + A)) = H.However, this is impossible: by construction, there exists a vector v ∈ Range(I −K)⊥ withv /∈ Range(A). By (1.4), the equation
u−Ku−Au = v
has no solution. This contradiction shows that (1.3) cannot hold.
8. Recalling that Range(I −K∗)⊥ = Ker(I −K), from step 7. we deduce
dimKer(I −K∗) ≥ dimRange(I −K∗)⊥ = dimKer(I −K).
Interchanging the roles of K and K∗ we obtain the opposite inequality.
Remark 1. When K is a compact operator, the above theorem provides information aboutexistence and uniqueness of solutions to the linear equation
u−Ku = f. (1.5)
Namely, two cases can arise.
CASE 1: Ker(I − K) = 0. Then the operator I − K is one-to-one and onto. For everyf ∈ H the equation (1.5) has exactly one solution.
73
CASE 2: Ker(I − K) 6= 0. This means that the homogeneous equation u − Ku = 0has a nontrivial solution. In this case, the equation (1.5) has solutions if and only if f ∈Ker(I −K∗)⊥.
The above dichotomy is known as the Fredholm alternative.
2 Spectrum of a compact operator
Let H be a Hilbert space over the reals, and let Λ : H 7→ H be a bounded linear operator.
• The resolvent set of Λ, denoted as ρ(Λ), is the set of numbers η ∈ IR such that ηI − Λis one-to-one and onto. Notice that in this case, by the open mapping theorem, the inverseoperator (ηI − Λ)−1 is continuous.
• The complement of the resolvent set: σ(K).= IR \ ρ(Λ) is called the spectrum of Λ.
• The point spectrum of Λ, denoted as σp(Λ), is the set of numbers η ∈ IR such that ηI −Λis not one-to-one. Equivalently, η ∈ σp(Λ) if there exists a non-zero vector w ∈ H such that
Λw = ηw.
In this case, η is called an eigenvalue of Λ and w is an associated eigenvector.
• The essential spectrum of Λ, denoted as σe(Λ) = σ(Λ) \ σp(Λ), is the set of numbersη ∈ IR such that ηI − Λ is one-to-one but not onto.
Theorem 5.2 (spectrum of a compact operator). Let H be an infinite dimensionalHilbert space, and let K : H 7→ H be a compact linear operator. Then
(i) 0 ∈ σ(K).
(ii) σ(K) = σp(K) ∪ 0.
(iii) Either σp is finite, or else σp(K) = λk ; k ≥ 1 with limk→∞ λk = 0.
Proof. 1. To prove (i) we argue by contradiction. If 0 /∈ σ(K), then by definition K has acontinuous inverse K−1 : H 7→ H. We thus have I = K K−1. Since the composition of acontinuous operator with a compact operator is compact, this implies that the identity is acompact operator. But this is false, because H is an infinite dimensional space and the closedunit ball in H is not compact.
2. To prove (ii), assume that λ ∈ σ(K), with λ 6= 0. If Ker(λI −K) = 0, the Fredholmalternative would imply Range(λI−K) = H. By the open mapping theorem, (λI−K) wouldhave a bounded inverse, against the assumptions.
3. Assume that (λn)n≥1 is a sequence of eigenvalues of K, with λn → λ. We claim that λ = 0.
Indeed, since λn ∈ σp(K), there exists an eigenvector wn such that Kwn = λnwn. Call
74
Hn.= spanw1, . . . , wn. Since eigenvectors corresponding to distinct eigenvalues are linearly
independent, Hn ⊂ Hn+1.
For every n ≥ 2 we also have (K − λnI)Hn ⊆ Hn−1.
For each n choose an element en ∈ Hn ∩H⊥n−1 with ‖en‖ = 1. If m < n, then
(Ken − λnen) ∈ Hn−1 , (Kem − λmem) ∈ Hm−1 ⊂ Hn−1 , em ∈ Hm ⊆ Hn−1 ,
while en ∈ H⊥n−1. Hence
‖Ken −Kem‖ =∥∥∥(Ken − λnen)− (Kem − λmem) + λnen − λmem
∥∥∥ ≥ ‖λnen‖ = |λn|
Thereforelim infm,n→∞
‖Ken −Kem‖ ≥ limn→∞
|λn| = |λ| .
This shows that the sequence (Ken)n≥1 cannot have any convergent subsequence, contradictingthe assumption that K is compact.
3 Symmetric operators
Let A = (aij)i,j=1,...,n be a symmetric n×n matrix. Then A determines a linear map x 7→ Axfrom IRn into IRn. It also determines the quadratic form
x 7→ (x, Ax) =n∑
i,j=1
aij xixj .
The quantitiesm
.= min
|x|=1(x, Ax), M
.= max
|x|=1(x, Ax)
provide the smallest and the largest eigenvalue of A, respectively. The next lemma extendsthis result to the case of a symmetric linear operator on a Hilbert space.
Lemma 1 (bounds on the spectrum of a symmetric operator). Let Λ : H 7→ H bea bounded linear symmetric operator on a real Hilbert space H. Define the upper and lowerbounds
m.= inf
u∈H , ‖u‖=1(Λu, u), M
.= sup
u∈H , ‖u‖=1(Λu, u).
Then
(i) The spectrum σ(Λ) is contained in the interval [m,M ].
(ii) m,M ∈ σ(Λ).
(iii) ‖Λ‖ = max−m,M.
Proof. 1. Let η > M . Then
(ηu− Λu , u) ≥ (η −M)‖u‖2 for all u ∈ H.
75
By the Lax-Milgram theorem, the linear continuous operator ηI − Λ is one-to-one and onto.By the open mapping theorem, it has a continuous inverse. This proves that every η > M isin the resolvent set of Λ. Similarly, replacing Λ with −Λ, we see that every η < m lies in theresolvent set. This proves (i).
2. From now on, to fix the ideas, we assume |m| ≤ M . The opposite case, where M < −mcan be handled by entirely similar arguments, replacing Λ by −Λ.
For every u, v ∈ H we have
4(Λu, v) = (Λ(u+ v), u+ v)− (Λ(u− v), u− v)≤ M
(‖u+ v‖2 + ‖u− v‖2
)
= 2M(‖u‖2 + ‖v‖2).
If Λu 6= 0, setting v.= (‖u‖/‖Λu‖)Λu we obtain
2‖u‖ ‖Λu‖ = 2(Λu, v) ≤ M(‖u‖2 + ‖v‖2) = 2M ‖u‖2.
Therefore‖Λu‖ ≤M‖u‖ for all u ∈ H (3.1)
Indeed, (3.1) trivially holds also if Λu = 0.
Since ‖Λ‖ ≥ sup‖u‖=1(Λu, u) =M , from (3.1) it follows ‖Λ‖ =M , proving (iii).
3. Next, we claim that M ∈ σ(Λ). Choose a sequence (un)n≥1 with
(Λun, un)→M , ‖un‖ = 1 for all n.
Then
‖Λun −Mun‖2 = ‖Λun‖2 − 2M(Λun, un) +M2‖un‖2 ≤ 2M2 − 2M(Λun, un) → 0
Therefore the operator Λ−MI cannot have a bounded inverse.
According to a classical theorem in linear algebra, every symmetric n × n matrix A can bereduced to diagonal form by an orthogonal transformation. Equivalently, one can choose anorthonormal basis of IRn consisting of eigenvectors of A. The following theorem shows thatthis result remains valid for compact symmetric operators.
Theorem 5.3 (Hilbert-Schmidt) (eigenvectors of a compact symmetric operator).Let H be a separable Hilbert space, and let K : H 7→ H be a symmetric compact linear operator.Then there exists a countable orthonormal basis of H consisting of eigenvectors of K.
Proof. 1. If H = IRn, this is a classical result in linear algebra. We thus assume that H isinfinite dimensional. Set η0 = 0 and let η1, η2, . . . be the set of all non-zero eigenvalues ofK. Consider the eigenspaces
H0.= Ker(K), H1
.= Ker(K − η1I), H2
.= Ker(K − η2I) . . .
76
Observe that 0 ≤ dimH0 ≤ ∞, while 0 < dim Hk <∞ for every k ≥ 1.
2. We claim that, for m 6= n, the subspaces Hm and Hn are orthogonal. Indeed, assumeu ∈ Hm, v ∈ Hn. Then
ηm(u, v) = (Ku, v) = (u,Kv) = ηn(u, v)
Since ηm 6= ηm, this implies (u, v) = 0.
3. Next, we show that the subspaces Hk generate the entire space H. More precisely, considerthe set of all linear combinations
H.= N∑
k=0
αkuk , N ≥ 1, uk ∈ Hk , αk ∈ IK.
We claim thatH⊥ ⊆ Ker(K)
.= H0 . (3.2)
Indeed, K(H) ⊆ H. Moreover if u ∈ H⊥ and v ∈ H, then Kv ∈ H and hence
(Ku, v) = (u,Kv) = 0.
This shows that K(H⊥) ⊆ H⊥.
Let K be the restriction of K to the subspace H⊥. Clearly K is a compact, symmetricoperator. By Lemma 1 we have
‖K‖ = supu∈H⊥, ‖u‖=1
|(Ku, u)| .= M .
If M 6= 0, then by the lemma either λ =M or λ = −M is in the spectrum of K. In this case,since K is compact, λ is in the point spectrum, and there exists an eigenvector w ∈ H⊥ suchthat
Kw = Kw = λw .
But this is impossible, because all eigenvectors of K corresponding to a nonzero eigenvalue arealready contained in the union of the subspaces Hk, k ≥ 1. We thus conclude that ‖K‖ = 0,proving (3.2).
In turn, (3.2) yieldsH⊥ ⊆ H⊥
0 ∩H0 = 0,proving that H is dense in H.
4. For each k ≥ 1, the finite-dimensional subspace Hk admits an orthonormal basis Bk =ek,1 ek,2 , . . . , ek,N(k). Moreover, since H is separable, the closed subspace H0 = Ker(K)admits a countable orthonormal basis B0 = e0,1 , e0,2 , . . .. The union B .
=⋃
k≥0 Bk is anorthonormal basis of H.
77
Remark 5.2. Given a countable set S = u1, u2, . . . in a Banach space X over the reals, abasic problem is to decide if span(S) is dense on X. Positive answers can be provided in twoimportant cases.
(I) X = C(E) is the space of all continuous real valued functions on a compact metric spaceE. In this case, if span(S) is an algebra that contains the constant functions and separatespoints, then the Stone-Weierstrass theorem yields span(S) = X.
(II) X is a separable Hilbert space, and there exists a compact self-adjoint operator Λ : X 7→X such that
(i) span(S) contains all eigenvectors of Λ, and
(ii) span(S) contains the kernel of Λ.
In this case, the previous theorem yields span(S) = X.
4 Problems
1. Let H be a Hilbert space, and let Λ : H 7→ H be a compact operator. Prove that thereexists a sequence of bounded linear operators (Λn)n≥1 such that
(i) for every n, the image Λn(H) is finite dimensional
(ii) ‖Λn − Λ‖ → 0 as n→∞.
2. (i) Find two linear continuous operators Λ1,Λ2 from L2([0,∞[) into itself such that Λ1Λ2 =I (the identity operator), but Λ2 Λ1 6= I.
(ii) Let H be a real Hilbert space, and let Λ,K : H 7→ H be bounded linear operators, withK compact. Show that in this case
Λ(I −K) = I if and only if (I −K)Λ = I .
3. On the Hilbert space L2([0,∞[), consider the linear operator defined by
(Λf)(x) = f(x+ 1) x ≥ 0 .
(i) Is Λ a bounded operator ? Is it compact ?
(ii) Is Λ one-to-one ? Is it onto ?
(iii) Explicitly determine the adjoint operator Λ∗.
(iv) Describe KerΛ and KerΛ∗.
78
4. Let H be a real Hilbert space, and let Λ : H 7→ H be a bounded linear operator, withnorm ‖Λ‖ =M . Give a direct proof that σ(Λ) ⊆ [−M,M ].
5. On the Hilbert space H = L2([0, 1]), consider the linear operator
(Λf)(x) =
∫ 1
0K(x, y) f(y) dy , (4.3)
where K : [0, 1] × [0, 1] 7→ IR is a continuous map, satisfying
K(x, y) = K(y, x) , |K(x, y)−K(x′, y)| ≤ C|x−x′| for all x, x′, y ∈ [0, 1]. (4.4)
Prove that Λ is a compact self-adjoint linear operator.
As a special case, show that the conditions (4.4) are satisfied by the integral kernel
K(x, y).=
(1− y)x if 0 ≤ x ≤ y ,(1− x)y if y ≤ x ≤ 1 .
6. On the Hilbert space L2([0, ∞) ), consider the linear operator Λu(x) = 2u(x+ 1).
(i) Determine Ker(Λ) and Range(Λ). Is Λ a compact operator ?
(ii) Determine the adjoint operator Λ∗.
(iii) Compute the operator norm ‖Λ‖.
79
Chapter 7
Linear Semigroups
1 O.D.E’s in a Banach space
The standard existence-uniqueness theory for ordinary differential equations can be extendedto Lipschitz continuous ODEs in Banach spaces, without any substantial change.
Let X be a Banach space, and let g : X 7→ X be a Lipschitz continuous map, so that
‖g(x) − g(y)‖ ≤ L ‖x− y‖ (1.1)
for some Lipschitz constant L and every x, y ∈ X.
Given an initial point x ∈ X, we consider the Cauchy problem
x(t).=
d
dtx(t) = g(x(t)), x(0) = x . (1.2)
The global existence and uniqueness of a solution can be obtained from the contraction map-ping principle.
Theorem 6.1 (existence-uniqueness of solutions to a Cauchy problem for an ODEwith Lipschitz continuous right hand side). Let X be a Banach space and assume thatthe vector field g : X 7→ X satisfies the Lipschitz condition (1.1). Then for every x ∈ X theCauchy problem (1.2) has a unique solution t 7→ x(t), defined for all t ∈ IR.
Proof. Fix any T > 0 and consider the Banach space C([0, T ]; X) of all continuous mappingsw : [0, T ] 7→ X, with the equivalent norm
‖w‖† .= max
t∈[0,T ]e−2Lt‖w(t)‖ . (1.3)
Observe that a function x : [0, T ] 7→ X provides a solution to the Cauchy problem (1.2) if andonly if x(·) is a fixed point of the Picard operator
Φ(w(·))(t) = x+
∫ t
0g(w(s)) ds, t ∈ [0, T ]. (1.4)
80
We claim that Φ is a strict contraction, w.r.t. the equivalent norm (1.3). Indeed given anyu, v ∈ C([0, T ]; X), set δ
.= ‖u− v‖†. This yields‖u(s)− v(s)‖ ≤ δe2Ls for all s ∈ [0, T ] .
For every t ∈ [0, T ], the assumption of Lipschitz continuity in (1.1) implies
e−2Lt‖Φ(u)(t) − Φ(v)(t)‖ = e−2Lt
∥∥∥∥∫ t
0g(u(s)) − g(v(s)) ds
∥∥∥∥
≤ e−2Lt∫ t
0
∥∥∥g(u(s))− g(v(s))∥∥∥ ds ≤ e−2Lt
∫ t
0L‖u(s)− v(s)‖ ds
≤ e−2Lt∫ t
0Lδe2Ls ds <
δ
2.
Therefore, ∥∥∥Φ(u)− Φ(v)∥∥∥†≤ 1
2‖u− v‖† . (1.5)
We can now apply the contraction mapping theorem, obtaining the existence of a unique fixedpoint for Φ, i.e. a continuous mapping x(·) such that
x(t) = x+
∫ t
0g(x(s)) ds for all t ∈ [0, T ].
Hence x(·) provides the unique solution to the Cauchy problem. By reversing time, we canconstruct a unique solution on any time interval of the form [−T, 0].
y(t)
_x
z(t)
x(t)
x1
2x
Figure 7.1: Euler approximations to the system x1 = x2, x2 = −x1. Here x(·) is an exact solution,y(·) is a piecewise affine approximation obtained by the forward Euler scheme, while z(·) is obtainedby the backward Euler scheme.
There are two well-known methods for constructing approximate solutions to the ODE (1.2),see fig. 7.1. Fix a time step h > 0, and define the times tj = j · h, j = 0, 1, 2, . . . As initialcondition we take x(t0) = x(0) = x.
• Forward Euler approximations. The values of the approximate solution at the timestj = j · h are defined by induction, according to the formula
x(tj+1) = x(tj) + hF (x(tj)). (1.6)
81
• Backward Euler approximations. The values of the approximate solution at the timestj = j · h are here defined by
x(tj+1) = x(tj) + hF (x(tj+1)). (1.7)
In both cases, as soon as the values x(tj) have been determined, we let x(t) be the piecewiseaffine function that takes the values x(tj) at the times t0, t1, . . .
Forward Euler approximations are easy to construct: during the whole time interval [tj, tj+1]we let the derivative x(t) be constant, equal to the value of F at the initial point:
x(t) = F (x(tj)) t ∈ [tj, tj+1].
On the other hand, in a backward Euler approximation, for t ∈ [tj , tj+1] we let the derivativex(t) be constant, equal to the value of F at the terminal point:
x(t) = F (x(tj+1)) t ∈ [tj, tj+1].
In this case, given x(tj), to find x(tj+1) one needs to solve the implicit equation (1.7). Back-ward Euler approximations thus require more computational work. On the other hand, theyoften have much better stability and convergence properties.
1.1 Linear homogeneous ODEs
Let A : IRn 7→ IRn be a linear operator. Then the solution to the linear Cauchy problem
x = Ax x(0) = x (1.8)
is the map t 7→ etAx, where
etA.=
∞∑
k=0
tkAk
k!. (1.9)
This same formula remains valid for every bounded linear operator A on a Banach space X.We observe that the series in (1.9) is absolutely convergent for every t ∈ IR. Moreover, theexponential map has the following properties:
(i) e0A = I, the identity map.
(ii) For every x ∈ X, the map t 7→ etAx is continuous.
(iii) esAetA = e(s+t)A (semigroup property).
The family of operators etA ; t ≥ 0 is called a linear semigroup.
The theory of linear semigroups studies the correspondence
A ←→ etA ; t ≥ 0between a linear operator and its exponential.
82
When A is a bounded linear operator, its exponential function is computed by the convergentseries (1.9). Viceversa, given the family of operators etA, one can recover A as the limit
A = limt→0+
etA − It
.
There are, however, important cases where the operators etA are all bounded, while A is anunbounded operator. These are indeed the most interesting applications of semigroup theory,which are useful in the analysis of PDEs of parabolic or hyperbolic type.
Example 1. If A : IRn 7→ IRn is a diagonal matrix, then its exponential can be readilycomputed. Indeed
A =
λ1 · · · 0...
. . ....
0 · · · λn
etA =
etλ1 · · · 0...
. . ....
0 · · · etλn
.
Observe that the corresponding operator norms are
‖A‖ = maxk|λk| , ‖etA‖ = max
k|etλk |.
Example 2. Let ℓ1 be the Banach space of all sequences of complex numbers x =(x1, x2, x3, . . .) with norm
‖x‖1 .=
∑
k
|xk| .
Given any sequence of complex numbers (λk)k≥1, consider the linear operator
Ax.= (λ1x1, λ2x2, λ3x3, . . .). (1.10)
The corresponding exponential operators are
etAx = (etλ1x1, etλ2x2, e
tλ3x3, . . .)
It is important to observe that the quantity
‖A‖ = supk|λk|
may well be infinite, but at the same time the norm
‖etA‖ = supk|etλk |
can be bounded, for each time t ≥ 0. Indeed, assume that the real part of all eigenvalues λkis uniformly bounded above, say
λk = αk + iβk , αk ≤ ω for all k ≥ 1 ,
for some constant ω ∈ IR. Then
|etλk | = |etαk | ≤ etω for all t ≥ 0. (1.11)
83
Therefore, for each t ≥ 0 the operator etA is bounded, namely ‖etA‖ ≤ etω.
Two further cases are worth exploring.
CASE 1: Assume that all real parts of the eigenvalues λk are bounded above and below, say
−ω ≤ αk ≤ ω
for some constant ω > 0 and all k ≥ 1. Then, for all t ∈ IR we have the estimate
|etλk | = |etαk | ≤ e|t|ω. (1.12)
Hence the operators etA are all bounded, also for t < 0. This shows that the differentialequation (1.8) can be solved both forward and backward in time. The family of operatorsetA ; t ∈ IR forms a group of bounded linear operators.
CASE 2: Assume that all eigenvalues λk = αk + iβk are contained in a sector of the complexplane. More precisely, assume that there exists ω ∈ IR and an angle 0 < θ < π/2 such that,for every k we can write
αk = ω − r cos θk β = −rk sin θk (1.13)
for some rk ≥ 0 and θk ∈ [−θ, θ].
In this case, since αk ≤ ω, it is clear that (1.11) holds and the operator ‖etA‖ is bounded, forevery t ≥ 0. However, much more is true. Indeed, for every t > 0, we claim that the composedoperator AetA is bounded. Indeed
‖AetA‖ = supk|λketλk | ≤ sup
k(ω + rk)e
(ω−rk cos θk)t
≤ supr≥0, |θ|≤θ
(ω + r)e(ω−r cos θ)t ≤ supr≥0
(ω + r)e(ω−r cos θ)t < +∞,
because cos θ > 0. In this case, A is called a sectorial operator.
Even if the initial condition x(0) = x does not lie in the domain of the operator A, for everyt > 0 we have x(t) = etAx ∈ Dom(A).
2 Semigroups of linear operators
Consider a linear evolution equation in a Banach space X, say
d
dtu(t) = Au(t) u(0) = u ∈ X. (2.1)
For t ≥ 0, would like to express the solution as u(t) = etAu, for some family of linear operatorsetA ; t ≥ 0.
Typically, (2.1) is a PDE of evolution type, for example ut = ux. In this case, the operator Ais a differential operator: Au
.= ux. Regardless of the norm we use on the space of functions
84
ω−ω
iR
θ_
ωθrλ
iR
R
iR
λk
ω
λk
0R R
0
Figure 7.2: Left: when all the eigenvalues λk of the operator A in (1.10) have real part Re(λk) ≤ ω,then for every t ≥ 0 the exponential operator is bounded: ‖etA‖ ≤ etω. Center: if all eigenvalues λksatisfy Re (λk) ∈ [−ω, ω], then for every t ∈ IR the exponential operator is bounded: ‖etA‖ ≤ e|t|ω.Right: if all eigenvalues λk lie in a sector of the complex plane with angle θ < π/2, then for each t > 0the operator AetA is also bounded.
X, differential operators are unbounded. For example, in the space of bounded continuousfunctions u : IR 7→ IR the linear map A : u 7→ ux satisfies
‖A‖ = sup‖u‖≤1
‖ux‖ ≥ supk≥1‖ ddx
sin kx‖ ≥ supk≥1
(supx|k cos kx|
)= +∞
On the other hand the solution of the Cauchy problem
d
dtu = ux u(0) = u
isu(t)(x) = u(x+ t) t ∈ IR .
This indicates that, although the differential operator A = ∂x is unbounded, the correspondingexponential operator etA is precisely the shift operator (by −t)
(etAu)(x) = u(x+ t).
For every t, the linear operator etA is an isometry: ‖etAu‖ = ‖u‖. Hence ‖etA‖ = 1.
In the theory of linear semigroup one considers two main problems:
(1) Given a semigroup of linear operators St ; t ≥ 0, find its generator, i.e. the operator Asuch that St = etA.
(2) Given a linear operator A, construct the corresponding semigroup etA ; t ≥ 0 and studyits properties.
For applications, (2) is clearly more important. However, for the development of the theory,it is convenient to start with (1).
85
2.1 Definition and basic properties of semigroups
Definition. Let X be a Banach space. A strongly continuous semigroup of linearoperators on X is a family of linear maps St ; t ≥ 0 with the following properties.
(i) Each St : X 7→ X is a bounded linear operator.
(ii) For every s, t ≥ 0, the composition satisfies StSs = St+s (semigroup property). MoreoverS0 = I (the identity operator).
(iii) for every u ∈ x, the map t 7→ Stu is continuous from [0,∞[ into X.
We say that St ; t ≥ 0 is a semigroup of type ω if, in addition, the bounded linearoperators St satisfy the bounds
‖St‖ ≤ etω t ≥ 0. (2.2)
A semigroup of type ω = 0 is called a contractive semigroup. Indeed, in this case ‖St‖ ≤ 1for every t ≥ 0, hence
‖Stu− Stv‖ ≤ ‖u− v‖ for all u, v ∈ X, t ≥ 0.
The operator
Au.= lim
t→0+
Stu− ut
(2.3)
is called the generator of the semigroup St ; t ≥ 0. This is clearly a linear operator.Its domain Dom(A) is the set of all u ∈ X for which the limit in (2.3) exists.
For a given u ∈ X, we regard the map t 7→ Stu as the solution to a linear ODE, namely
d
dtu(t) = Au(t) u(0) = u . (2.4)
Notice that here, in a sense, we are approaching the problem backwards: given the solutionu(t) = Stu, find the equation, i.e. the operator A.
Some elementary properties of the semigroup S and its generator A are now examined.
Theorem 6.2 (properties of semigroups). Let St ; t ≥ 0 be a strongly continuoussemigroup and let A be its generator. Assume u ∈ Dom(A). Then (i) For every t ≥ 0 onehas Stu ∈ Dom(A) and AStu = StAu.
(ii) The map t 7→ u(t).= Stu is continuously differentiable and provides a solution to the
Cauchy problem (2.4).
Proof. 1. Let u ∈ Dom(A), so that the limit in (2.3) exists. Then
lims→0+
SsStu− Stus
= lims→0+
StSsu− Stus
= St lims→0+
Ssu− us
= StAu.
Therefore Stu ∈ Dom(A) and AStu = StAu, proving (i).
86
2. Next, assume u ∈ Dom(A), t > 0. Then, by the semigroup property,
limh→0+
Stu− St−hu
h− StAu
= lim
h→0+
St−h
(St−hu− u
h
)− StAu
= limh→0+
St−h
(Shu− u
h−Au
)+ (St−hAu− StAu)
= 0.
Indeed, Shu−uu → Au, while ‖St−h‖ ≤ etω . Moreover, since Au ∈ X is well defined, the map
s 7→ SsAu is continuous. The above computation shows that the map t 7→ u(t) = Stu has aleft derivative:
limh→0+
Stu− St−hu
h= StAu .
The right derivative is computed as
limh→0+
St+hu− Stuh
= St limh→0+
Shu− uh
= StAu .
Therefore, for every t > 0, the map t 7→ Stu is differentiable, with derivative ddtStu = StAu =
AStu. Since Au ∈ X, by the definition of semigroup the map t 7→ St(Au) is continuous. Hencethe map t 7→ Stu is continuously differentiable.
The following theorem collects some properties of the generator A. We recall that a linearoperator is closed if its graph
Graph(A) =(x, y) ∈ X ×X ; x ∈ Dom(A), y = Ax
is a closed subset of the product space X ×X.
Theorem 6.3 (properties of generators). Let St ; t ≥ 0 be a semigroup, and let A beits generator. Then
(i) The domain of A is dense in X.
(ii) The operator A is closed.
Proof. 1. Let any u ∈ X and consider the approximation Uε.= ε−1
∫ ε
0Ssu ds. Since the
map t 7→ Stu is continuous, we have Uε → u as ε → 0+. To prove (i), we now show thatUε ∈ Dom(A) for every ε > 0. Since Dom(A) is a vector subspace, it suffices to show that
uε.= εUε =
∫ ε
0Ssu ds ∈ Dom(A).
For 0 < h < ε we have
Shuε − uεh
=1
h
Sh( ∫ ε
0Ssu ds
)−( ∫ ε
0Ssu ds
)=
1
h
∫ ε
0(Ss+hu− Ssu)ds
=1
h
∫ ε+h
εSsu ds −
1
h
∫ h
0Ssu ds → Sεu− u as h→ 0 + .
This shows that uε ∈ Dom(A) for every ε > 0, proving (i).
87
2. Next, we prove that the graph of A is closed. Let (uk, vk) be a sequence of points on thegraph of A. More precisely, let uk ∈ DomA, vk = Auk, and assume the convergence uk → u,vk → v, for some u, v ∈ X. We need to show that u ∈ Dom(A) and Au = v. For each k ≥ 1,since uk ∈ Dom(A) the previous theorem implies
Shuk − uk =
∫ h
0
(d
dtStuk
)dt =
∫ h
0StAuk dt.
Letting k →∞, we obtain
Shu− u =
∫ h
0Stv dt.
Therefore,
limh→0+
Shu− uh
= limh→0+
1
h
∫ h
0Stv dt = v.
By definition, this means that u ∈ Dom(A) and Au = v.
t
u
S ut
h0 ε ε+h
−E u
h
Figure 7.3: Left: the intervals of integration, in the proof of Theorem 6.3 (i). Right: according to(3.12), the backward Euler approximation with step h > 0 is the weighted average of points on thetrajectory t 7→ Stu, with exponentially decreasing weight w(t) = e−t/h/h.
3 Resolvents
The crucial link between a semigroup St ; t ≥ 0 and its generator A is provided by theso-called “resolvent”. This can be best understood in terms of backward Euler approximations.
Assume that we want to solve (2.4) approximately, by backward Euler approximations. Wethus fix a time step h > 0 and iteratively solve
u(t+ h) = u(t) +Au(t+ h)
In other words, at each step, given a value u(t) ∈ X, we need to compute
u(t+ h) = (I − hA)−1u(t)
The backward Euler operator
E−h
.= (I − hA)−1 (3.1)
with h > 0 small, plays a key role in semigroup theory. Given A, we have two main ways torecover the solution of the Cauchy problem (2.4), in terms of this operator.
88
1. As a limit of backward Euler approximations.
For a fixed τ > 0, we consider the time step h = τ/n. After n steps, the backward Eulerapproximation scheme yields
u(τ) ≈ E−τ/n · · · E−
τ/n =
(I − τ
nA
)−n
u .
Keeping τ fixed and letting n → ∞, we thus expect to recover the exact solution of (2.4) inthe limit:
u(τ) = Sτ u.= lim
n→∞
(I − τ
nA
)−1
u. (3.2)
2. As a limit of solutions to approximate evolution equations.
Namely, fix again a time step h > 0 and set λ = 1/h. We then define the operator Aλ : X 7→ Xas
Aλu.= AE−
h u = A(I − hA)−1u.
In other words, Aλu is the value of A computed not at u but at the nearby point E−h u, i.e. at
the first step in a backward Euler approximation, with h = 1/λ. It turns out that, for h > 0small enough, Aλ = A1/h is a well defined, bounded linear operator. We can thus consider the
exponential operators etAλ =∑
k≥0(tAλ)k/k! and define
u(t) = Stu.= lim
λ→∞etAλ u. (3.3)
In the following, we shall thus study the properties of the operator E−h.= (I − hA)−1.
Example 3. Consider the scalar ODE
x = ax, x(0) = x
Its solution is x(t) = eatx. In this case we trivially have
aλ = a1/h =a
1− ha, eatx = limh→0
eta1/h x = limh→0+
eta/(1−ha)x .
This corresponds to the limit (3.3). On the other hand, the limit (3.2) is provided by theidentity
eta = limn→∞
(1− τ
na
)−n
.
Notice also the important identities
∫ ∞
0
e−t/h
hdt = 1, (1− ha)−1x =
∫ ∞
0
(e−t/h
h
)etax dt (3.4)
The second identity says that the backward Euler operator E−h x = (1−ha)−1x can be obtained
as an average of points on the trajectory t 7→ etax, with an exponentially decreasing weightw(t) = e−t/h/h.
89
Motivated by the previous analysis, we introduce some definitions.
Definition. Let A be a linear operator on a Banach Space X. The resolvent set of A is theset ρ(A) of all real numbers λ such that the operator
λI −A : Dom(A) 7→ X
is one-to-one and onto. If λ ∈ ρ(A), the resolvent operator Rλ : X 7→ X is defined by
Rλu.= (λI −A)−1u.
We remark that, by the closed graph theorem, the operator Rλ : X 7→ Dom(A) ⊆ X is abounded linear operator. Moreover
ARλu = RλAu if u ∈ Dom(A).
There is a close connection between resolvents and backward Euler operators. Namely
λRλ = E−1/λ .
Theorem 6.4 (resolvent identities). Let A be a closed linear operator. If λ, µ ∈ ρ(A),then
Rλ −Rµ = (µ − λ)RλRµ , (3.5)
RλRµ = RµRλ . (3.6)
Proof. For any u ∈ X one has
v.= Rλu−Rµu = (λI −A)−1u− (µI −A)−1u ∈ Dom(A),
(λI −A)v = u− (λI − µI + µI −A)(µI −A)−1u = (λ− µ)(µI −A)−1u .
Applying the operator (λI −A)−1 to both sides of the above identity we obtain (3.5).
Next, using (3.5), for any λ 6= µ in the resolvent set ρ(A), we obtain
RλRµ =Rλ −Rµ
λ− µ =Rµ −Rλ
µ− λ = RµRλ .
Theorem 6.5 (integral formula for the resolvent). Let St ; t ≥ 0 be a semigroup oftype ω, and let A be its generator. Then for every λ > ω one has λ ∈ ρ(A) and
Rλu =
∫ ∞
0e−tλStu dt . (3.7)
Moreover,
‖Rλ‖ ≤1
λ− ω . (3.8)
90
Proof. 1. By assumption, the semigroup satisfies the bounds ‖St‖ ≤ etω . Therefore theintegral in (3.7) is absolutely convergent:
∥∥∥∥∫ ∞
0e−tλStu dt
∥∥∥∥ ≤∫ ∞
0etλ‖St‖ ‖u‖ dt ≤
∫ ∞
0e(ω−λ)t‖u‖ dt ≤ 1
λ− ω ‖u‖. (3.9)
Define
Rλu.=
∫ ∞
0e−tλStu dt .
The above estimate shows that Rλ is a bounded linear operator, with norm
‖R‖ ≤ 1
λ− ω .
2. We now show that
(λI −A)Rλu = u for all u ∈ X . (3.10)
Indeed, for any h > 0 we have
ShRλu− Rλu
h=
1
h
∫ ∞
0e−λt (St+hu− Stu) dt
=1
h
∫ ∞
0
(e−λ(t−h) − e−λt
)Stu dt−
1
h
∫ h
0e−λ(t−h) Stu dt
=(eλh − 1
h
) ∫ ∞
0e−λt Stu dt−
eλh
h
∫ h
0e−λt Stu dt .
Taking the limit as h→ 0+ we obtain
limh→0+
ShRλu− Rλu
h= λRλu− u .
By definition of generator, this means
Rλu ∈ Dom(A), ARλu = λRλu− u.Therefore, (3.10) holds.
3. The identity (3.10) already shows that the map v 7→ (λI − A)v from Dom(A) into X isonto. We now prove that this map is one-to-one. If u ∈ Dom(A), then
ARλu = A
∫ ∞
0e−λtStu dt =
∫ ∞
0e−λtAStu dt =
∫ ∞
0e−λtStAudt = RλAu .
This proves the commutativity relation
Rλ(λI −A)u = (λI −A)Rλu for all u ∈ Dom(A). (3.11)
If now (λI −A)u = (λI −A)v, using (3.11) and (3.10) we obtain
u = Rλ(λI −A)u = Rλ(λI −A)v = v,
proving that the map (λI −A) is one-to-one.
We conclude that λ ∈ ρ(A) and Rλ = (λI −A)−1 = Rλ.
91
Remark 1. According to the integral representation formula (3.7), the resolvent operatorsRλ provide the Laplace transform of the semigroup S. Taking 0 < h = λ−1, this same formulashows that the backward Euler approximations can be obtained as
E−h u
.= (I − hA)−1u =
∫ ∞
0
e−t/h
hStu dt . (3.12)
Here the integral is convergent, provided that h is sufficiently small: if S is a semigroup oftype ω, we need h < ω−1.
Generalizing the identity (3.4), we see that the first step in an Euler backward approximationcoincides with an averaged value of the entire trajectory t 7→ Stu, with the exponentiallydecreasing weight w(t) = e−t/h/h.
4 Generation of a semigroup
In this section we tackle the most important question. Namely, given an operator A fromDom(A) ⊂ X into X, under which conditions does there exist a semigroup St ; t ≥ 0generated by A ?
Theorem 6.6 (existence of the semigroup generated by a linear operator). Let Abe linear operator on a Banach space X. Then the following are equivalent.
(i) A is the generator of a semigroup of linear operators St ; t ≥ 0, of type ω.
(ii) A is a closed, densely defined operator. Moreover, every real number λ > ω is in theresolvent set of A, and
∥∥∥(λI −A)−1∥∥∥ ≤ 1
λ− ω for all λ > ω . (4.1)
Proof. The fact that (i) implies (ii) has already been proved. We shall thus assume that (ii)holds and prove that A generates a semigroup of type ω. The proof will be achieved in severalsteps.
1. By assumption, for each λ > ω the resolvent operator Rλ = (λI−A)−1 is well defined. Wecan thus consider the bounded linear operator
Aλ.= − λI + λ2Rλ = λARλ . (4.2)
Notice that, setting h = 1/λ, we can write
Aλu = A(I − hA)−1u = A(E−h u)
In other words, Aλu is the value of A computed not at the point u but at the point E−h u
obtained by a backward Euler step of size h = 1/λ, starting at u. It is thus expected that Aλ
should be a good approximation for the operator A, at least when λ is large (so that h = 1/λis small).
92
2. Since each Aλ is bounded, we can construct its exponential as
etAλ.=
∞∑
k=0
(tAλ)k
k!= e−λteλ
2tRλ = e−λt∞∑
k=0
(λ2t)k Rkλ
k!.
We observe that, if A is unbounded, then ‖Aλ‖ → ∞ as λ → ∞. However, the norms of theexponential operators remain uniformly bounded as λ → ∞. Indeed, the assumption (4.1)together with the definition (4.2) yields
‖etAλ‖ ≤ e−λt∞∑
k=0
(λ2t)k ‖Rλ‖kk!
≤ e−λteλ2t/(λ−ω) = eλωt/(λ−ω) .
In particular, as soon as λ ≥ 2ω we have
‖etAλ‖ ≤ e2ωt for all t ≥ 0 . (4.3)
3. In this step we establish the limit
limλ→∞
Aλu = Au for all u ∈ Dom(A). (4.4)
To prove (4.4), we start with the identity
λRλu− u = ARλu = RλAu ,
which is valid for all u ∈ Dom(A). This yields
‖λRλu− u‖ = ‖RλAu‖ ≤ ‖Rλ‖ ‖Au‖ ≤1
λ− ω‖Au‖ → 0 as λ→∞.
Hencelimλ→∞
λRλu = u for all u ∈ Dom(A).
Using the fact that A is densely defined, one can prove that the above limit holds for allv ∈ X as well. Indeed, for every v ∈ X and ε > 0, since Dom(A) is dense in X we can findu ∈ Dom(A) such that ‖u− v‖ < ε. Using the triangle inequality, one obtains
lim supλ→∞
‖λRλv − v‖ ≤ lim supλ→∞
‖λRλv − λRλu‖+ lim supλ→∞
‖λRλu− u‖+ ‖u− v‖
≤ lim supλ→∞
‖λRλ‖ ‖v − u‖+ 0 + ‖u− v‖ ≤ lim supλ→∞
λ
λ− ω ε+ ε = 2ε .
Since ε > 0 was arbitrary, this proves
limλ→∞
λRλv = v for all v ∈ X. (4.5)
In terms of backward Euler approximations, as the time step h = 1/λ→ 0+, (4.5) yields
limh→0+
E−h v = v for all v ∈ X .
93
If now u ∈ Dom(A), we can then take v = Au in (4.5) and conclude
limλ→∞
Aλu = limλ→∞
λARλu = limλ→∞
λRλAu = limλ→∞
λRλv = v = Au .
This proves (4.4).
4. We claim that, as λ → ∞, the family of uniformly bounded operators etAλ converges tosome linear operator, which we call St. To prove this, for λ, µ > 2ω we will estimate thedifference etAλ − etAµ .
The commutativity relation RλRµ = RµRλ implies AλAµ = AµAλ. Therefore
AµetAλ = Aµ
∞∑
k=1
(tAλ)k
k!= etAλAµ .
For every u ∈ X we thus have
etAλu− etAµu =
∫ t
0
d
ds
[e(t−s)AµesAλ
]ds
=
∫ t
0e(t−s)Aµ(Aλ −Aµ)e
sAλu ds =
∫ t
0e(t−s)AµesAλ(Aλu−Aµu) ds .
By (4.3), this implies
‖etAλu− etAµu‖ ≤∫ t
0e2(t−s)ωe2sω‖Aλu−Aµu‖ ds = t e2ωt ‖Aλu−Aµu‖ .
If u ∈ Dom(A), then (4.4) yields Aλu→ Au and Aµu→ Au as λ, µ→∞. Therefore
lim supλ,µ→∞
‖etAλu− etAµu‖ ≤ t e2ωt lim supλ,µ→∞
‖Aλu−Aµu‖ = 0.
λ
u
λ usA
e
etA
e
tA µe u
(t−s)AesA
u
λ u
µ
Figure 7.4: Proving the convergence of the approximations etAλ as λ→∞. The difference ‖etAλu −etAµu‖ can be estimated by the length of the curve s 7→ e(t−s)AµesAλu, for s ∈ [0, t].
Using the triangle inequality, we now show that the same limit is valid for all v ∈ X, uniformlyas t ranges in bounded intervals. Indeed, given ε > 0, choose u ∈ Dom(A) with ‖u− v‖ ≤ ε.Then
lim supλ,µ→∞
‖etAλv − etAµv‖
≤ lim supλ→∞
‖etAλv − etAλu‖+ lim supλ,µ→∞
‖etAλu− etAµu‖+ lim supµ→∞
‖etAµu− etAµv‖
≤ 2 lim supλ→∞
∥∥∥etAλ
∥∥∥ ‖v − u‖ ≤ e2ωt ε .
94
Since ε > 0 was arbitrary, our claim is proved.
5. By the previous step, for every t ≥ 0 and u ∈ X the following limits is well defined:
Stu.= lim
λ→∞etAλu .
We claim that St ; t ≥ 0 is a strongly continuous semigroup of type ω.
Indeed, the semigroup property follows from
St Ss u = limλ→∞
etAλesAλu = limλ→∞
e(t+s)Aλu = St+su .
For fixed u ∈ X, the map t 7→ Stu is continuous, being the limit of the continuous mapst 7→ etAλu, uniformly for t in bounded intervals.
Finally, for every t ≥ 0 and u ∈ X with ‖u‖ ≤ 1 we have the estimate
‖Stu‖ = limλ→∞
‖etAλu‖ ≤ lim supλ→∞
‖etAλ‖ ‖u‖ ≤ limλ→∞
etλω/(λ−ω)‖u‖ = etω‖u‖.
This proves that ‖St‖ ≤ etω, i.e. that the semigroup is of type ω.
6. It remains to prove that the linear operator A is indeed the generator of the semigroup.Toward this goal, call B the generator of the semigroup St ; t ≥ 0. By our earlier analysis,we know that B is a linear, closed operator, with domain Dom(B) dense in X. We need toshow that B = A.
Since Aλ is the generator of the semigroup etAλ ; t ≥ 0, for every λ > ω we have
etAλu− u =
∫ t
0esAλAλu ds . (4.6)
Moreover, for u ∈ Dom(A), the triangle inequality yields
‖esAλAλu− SsAu‖ ≤ ‖esAλ‖ ‖Aλu−Au‖+ ‖esAλAu− SsAu‖. (4.7)
As λ → ∞, the right hand side of (4.7) goes to zero, uniformly for s in bounded intervals.Taking the limit as λ→∞ in (4.6) we thus obtain
Stu− u =
∫ t
0SsAuds for all t ≥ 0, u ∈ Dom(A) . (4.8)
As a consequence, Dom(B) ⊃ Dom(A) and
Bu = limh→0+
Stu− ut
= limt→0+
1
t
∫ t
0StAuds = Au for all u ∈ Dom(A) .
To prove that A = B, it remains to show that Dom(B) ⊆ Dom(A). For this purpose, chooseany λ > ω. We know that the operators
λI −A : Dom(A) 7→ X and λI −B : Dom(B) 7→ X
are both one-to-one and surjective. In particular, the restriction of λI−B toDom(A) coincideswith λI − A and is thus surjective. Hence this operator λI − B cannot be extended to anydomain strictly larger than Dom(A), preserving the one-to-one property. This shows thatDom(B) = Dom(A), completing the proof.
95
We now investigate uniqueness. Given a semigroup of linear operators St ; t ≥ 0, itsgenerator is uniquely defined by the limit in (2.3). Viceversa, given a linear operator Asatisfying the assumption of Theorem 6.6, we now show that the semigroup generated by Ais uniquely determined.
Theorem 6.7 (uniqueness of the semigroup). Let St, St be two semigroups ofbounded linear operators, having the same generator A. Then St = St for every t ≥ 0.
Proof. Let u ∈ Dom(A). Then Ssu ∈ Dom(A) and St−sSsu ∈ Dom(A) for every 0 ≤ s ≤ t.We can thus estimate
Stu− Stu =
∫ t
0
d
ds
[St−sSsu
]ds =
∫ t
0
[St−sASsu−ASt−sSsu
]ds = 0.
Indeed,
d
ds
[St−sSsu
]= lim
h→0
St−s−h(Ss+hu)− St−sSsu
h
= limh→0
St−s−h(Ss+hu− Ssu)h
+ limh→0
St−s−h(Ssu)− St−s(Ssu)
h
= St−s(ASsu)−ASt−s(Ssu) .
This shows that, for every fixed t ≥ 0, the bounded linear operators St and St coincide on thedense subset Dom(A). Hence St = St.
Applications of the theory of semigroups to the solution of partial differential equations ofevolutionary type will be illustrated in Chapter 8.
5 Problems
1. Assume that the operator A in (1.10) is sectorial, so that (1.13) holds, for some ω ∈ IRand some angle θ < π/2.
(i) For t > 0, give an a priori estimate on the norm of the operator ‖AetA‖, depending onlyon ω, θ.
(ii) More generally, for t > 0 and k ≥ 1, prove that the operator AketA is also bounded.Estimate the norm ‖AketA‖.
2. Let A be the generator of a contractive semigroup. Show that, for every λ > 0, the boundedoperator Aλ in (4.2) also generates a contractive semigroup.
96
3. Fix λ > 0 and consider the weight function
w(t) =
λ e−λt if λ ≥ 0,
0 if λ < 0.
By induction on n ≥ 1, check that the n-fold convolution wn.= w ∗ w ∗ · · · ∗ w is given by
wn(t) =
λn
(n− 1)!tn−1 e−λt if t ≥ 0,
0 if t < 0.(5.9)
Show that w(·) is the density of a probability measure with
mean value =
∫ ∞
0t w(t) dt =
1
λ,
variance =
∫ ∞
0
(t− 1
λ
)2
w(t) dt =
∫ ∞
0
(t2 − 1
λ2
)w(t) dt =
1
λ2.
Therefore, wn(·) is the density of a probability measure with mean n/λ and variance n/λ2.
Now fix T > 0 and let λn.= n/T . For every n ≥ 1, set
Wn(t) =
(n/T )n
(n− 1)!tn−1 e−(n/T )t if t ≥ 0,
0 if t < 0.(5.10)
Show that Wn(·) is the density of a probability measure with mean value n/λn = T andvariance n/(n/T )2 = T 2/n. In particular, for every ε > 0, one has
limn→∞
[∫ T−ε
0Wn(t) dt+
∫ ∞
T+εWn(t) dt
]= 0 , lim
n→∞
∫ T+ε
T−εWn(t) dt = 1 . (5.11)
4 (convergence of backward Euler approximations). Let St ; t ≥ 0 be a contractivesemigroup on a Banach space X, generated by the linear operator A. Fix h > 0 and setλ = 1/h. By induction on n ≥ 1, extend the formula (3.12) for backward Euler approximations,showing that
(I − hA)−nu =
∫ ∞
0wn(t)Stu dt ,
where wn is the function defined at (5.9).
Setting h = T/n, so that λn = n/T , prove that(I − T
nA
)−n
u =
∫ ∞
0Wn(t)Stu dt → STu as n→∞.
5 (positively invariant sets). Let A be the generator of a contractive semigroup St ina Banach space X. Let Ω ⊂ X be a closed, convex subset. Prove that the following areequivalent.
97
Figure 7.5: A plot of the weight functionsWn. Here T = 4, while n = 1, 2, 10, 100. All these probabilitydistributions have average value 4, while their variance (= 16/n) decreases to zero as n→∞.
(i) Ω is positively invariant for the semigroup: if u ∈ Ω then Stu ∈ Ω for all t ≥ 0.
(ii) Ω is invariant w.r.t. the backward Euler scheme: if u ∈ Ω, then for every h > 0 one has(I − hA)−1u ∈ Ω.
6. Let St ; t ≥ 0 be a semigroup of type ω, and let A be its generator. Prove that, forevery γ ∈ IR the family of bounded linear operators eγtSt ; t ≥ 0 is a semigroup of typeω + γ, having A+ γI as its generator.
7. On the space Lp(IR), consider the linear operators St defined by
(Stf)(x) = e−2tf(x+ t) .
(i) Prove that the family of linear operators St ; t ≥ 0 is a strongly continuous, contractivesemigroup on Lp(IR), for all 1 ≤ p < ∞. Find the generator A of this semigroup. What isDom(A)?
(ii) Show that the family of operators St ; t ≥ 0 is a NOT a strongly continuous semigroupon L∞(IR).
8 (Semilinear equations). Let A be a linear operator on a Banach space X, generating thecontractive semigroup St ; t ≥ 0. We say that a map u : [0, T ] 7→ X is a mild solution of
98
the semilinear Cauchy problem
d
dtu(t) = Au(t) + f(t, u(t)) u(0) = u (5.12)
if
u(t) = Stu+
∫ t
0St−sf(s, u(s)) ds for all t ∈ [0, T ] . (5.13)
(i) Assuming that f : [0, T ]×X 7→ X is continuous and satisfies the bounds
‖f(t, x)‖ ≤ M , ‖f(t, x)− f(t, y)‖ ≤ L ‖x− y‖ for all t, x, y,
prove that the Cauchy problem (5.12) admits a unique mild solution.
(ii) In the special case where A is a bounded operator, so that St = etA, prove that u(·)satisfies (5.13) if and only if u is a continuously differentiable solution to the Cauchy problem(5.12)
99
Chapter 8
Sobolev Spaces
1 Distributions and weak derivatives
We denote by L1loc(IR) the space of locally integrable functions f : IR 7→ IR. These are the
Lebesgue measurable functions which are integrable restricted to every bounded interval.
The support of a function ϕ, denoted by Supp(φ), is the closure of the set x ; ϕ(x) 6= 0,where ϕ does not vanish. By C∞c (IR) we denote the space of continuous functions with compactsupport, having continuous derivatives of every order.
A locally integrable function f ∈ L1loc(IR) determines a linear functional Λf : C∞c (IR) 7→ IR,
namely
Λf (ϕ).=
∫
IRf(x)ϕ(x) dx . (1.1)
Notice that this integral is well defined for all ϕ ∈ C∞c (IR) , because ϕ vanishes outside acompact set. Moreover, if Supp(ϕ) ⊆ [a, b], we have the estimate
|Λf (ϕ)| ≤(∫ b
a|f(x)| dx
)‖ϕ‖C0 . (1.2)
Now assume that f is continuously differentiable. Then its derivative
f ′(x) = limh→0
f(x+ h)− f(x)h
is continuous, hence locally integrable. In turn, f ′ also determines a linear functional onC∞c (IR), namely
Λf ′(ϕ).=
∫
IRf ′(x)ϕ(x) dx = −
∫
IRf(x)ϕ′(x) dx . (1.3)
At this stage, a key observation is that the first integral in (1.3) is defined only if f ′(x) existsfor a.e. x, and is locally integrable. However, the second integral is well for every locallyintegrable function f , even if f does not have a pointwise derivative at any point. Moreover,if Supp(ϕ) ⊆ [a, b], we have the estimate
|Λf ′(ϕ)| ≤(∫ b
a|f(x)| dx
)‖ϕ‖C1 .
100
This construction can be performed for any derivative. Given an integer k ≥ 1, the distribu-tional derivative of order k of f ∈ L1
loc is the linear functional
ΛDkf (ϕ) = (−1)k∫
IRf(x)Dkϕ(x) dx
If there exists a locally integrable function g such that ΛDkf = Λg, namely
∫
IRg(x)ϕ(x) dx = (−1)k
∫
IRf(x)Dkϕ(x) dx for all ϕ ∈ C∞c (IR),
then we say that g is the weak derivative of order k of f .
Remark 7.1. Classical derivatives are defined pointwise, as limits of difference quotients.Weak derivatives are defined only in an integral sense, up to a set of measure zero. Byarbitrarily changing the function f on a set of measure zero we do not affect its weak derivativesin any way.
Example 7.1. Consider the function
f(x).=
0 if x ≤ 0,x if x > 0 .
Its distributional derivative is the map
Λ(ϕ) = −∫ ∞
0x · ϕ′(x) dx =
∫ ∞
0ϕ(x) dx =
∫
IRH(x)ϕ(x) dx ,
where
H(x) =
0 if x ≤ 0,1 if x > 0 .
(1.4)
In this case, the Heaviside function H in (1.4) is the weak derivative of f .
Example 7.2. The function H in (1.4) is locally integrable. Its distributional derivative isthe linear functional
Λ(ϕ).= −
∫
IRH(x)ϕ′(x) dx = ϕ(0) .
This corresponds to the Dirac measure, concentrating a unit mass at the origin. We observethat the function H does not have any weak derivative. Indeed, assume that, for some locallyintegrable function g, one has
∫g(x)ϕ(x) dx = ϕ(0) for all ϕ ∈ C∞c .
By the Lebesgue Dominated Convergence Theorem,
limh→0
∫ h
−h|g(x)| dx = 0.
101
Hence we can choose δ > 0 so that∫ δ−δ |g(x)| dx ≤ 1/2. Let ϕ : IR 7→ [0, 1] be a smooth
function, with ϕ(0) = 1 and with support contained in the interval [−δ, δ]. We now reach acontradiction by writing
1 = ϕ(0) = Λ(ϕ) =
∫
IRg(x)ϕ(x) dx =
∫ δ
−δg(x)ϕ(x) ≤
∫ δ
−δ|g(x)| dx ·max
x|ϕ(x)| ≤ 1
2.
Example 7.3. Consider the function
f(x).=
0 if x is rational,
2 + sinx if x is irrational .
Clearly, f is discontinuous at every point x. Hence it is not differentiable at any point. On theother hand, the function g(x) = cosx provides a weak derivative for f . Indeed, the behaviorof f on the set of rational points (having measure zero) is irrelevant. We thus have
−∫f(x)ϕ′(x) dx = −
∫(2 + sinx)ϕ′(x) dx =
∫(cos x)ϕ(x) dx.
1.1 Distributions
The construction described in the previous section can be extended to any open domain, ina multi-dimensional space. Let Ω ⊆ IRn be an open set. By L1
loc(Ω) we denote the space ofloccaly integrable functions on Ω. These are the measurable functions f : Ω 7→ IR which areintegrable restricted to every compact subset K ⊂ Ω.
Example 7.4. The functions ex, ln |x|, and |x|−1/2 are in L1loc(IR), while x
−1 /∈ L1loc(IR). On
the other hand, taking Ω = ]0,∞[, the function f(x) = xk is in L1loc(Ω) for every (positive or
negative) integer k.
By C∞c (Ω) we denote the space of continuous functions φ : Ω 7→ IR, having continuous partialderivatives of all orders, and whose support is a compact subset of Ω. Functions ϕ ∈ C∞c (Ω)are usually called “test functions”. We recall that the support of a function φ is the closureof the set where φ does not vanish.
Supp (φ).= x ∈ Ω ; φ(x) 6= 0.
We shall need an efficient way to denote higher order derivatives of a function f . A multi-index α = (α1, α2, . . . , αn) is an n-tuple of non-negative integer numbers. Its length is definedas
|α| .= α1 + α2 + · · · + αn .
Each multi-index α determines a partial differential operator of order |α|, namely
Dαf =
(∂
∂x1
)α1(∂
∂x2
)α2
· · ·(
∂
∂xn
)αn
f .
102
Definition 7.1. By a distribution on the open set Ω ⊆ IRn we mean a linear functionalΛ : C∞c (Ω) 7→ IR such that the following boundedness property holds.
• For every compact K ⊂ Ω there exist an integer N ≥ 0 and a constant C such that
|Λ(φ)| ≤ C ‖φ‖CN for every φ ∈ C∞ with support contained inside K. (1.5)
In other words, for all test functions φ which vanish outside a given compact set K, the valueΛ(φ) should be bounded in terms of the maximum value of derivatives of φ, up to a certainorder N .
Notice that here both N and C depend on the compact subset K. If there exists an integerN ≥ 0 independent of K such that (1.5) holds (with C = CK possibly still depending on K),we say that the distribution has finite order. The smallest such integer N is called the orderof the distribution.
Example 7.5. Let Ω be an open subset of IRn and consider any function f ∈ L1loc(Ω). Then
the linear map Λf : C∞c (Ω) 7→ IR defined by
Λf (φ).=
∫
Ωf φ dx , (1.6)
is a distribution. Indeed, it is clear that Λf is well defined and linear. Given a compact subsetK ⊂ Ω, for every test function φ with Supp(φ) ⊆ K we have the estimate
|Λf (φ)| =
∣∣∣∣∫
Kf φ dx
∣∣∣∣ ≤∫
K|f(x)| dx ·max
x∈K|φ(x)| ≤ C ‖φ‖C0 .
Hence the estimate (1.5) holds with C =∫K |f | dx and N = 0. This provides an example of a
distribution of order zero.
The family of all distributions on Ω is clearly a vector space. A remarkable fact is that,while a function f may not admit a derivative (in the classical sense), for a distribution Λ anappropriate notion of derivative can always be defined.
Definition 7.2. Given a distribution Λ and a multi-index α, we define the distribution DαΛby setting
DαΛ(φ).= (−1)|α|Λ(Dαφ) . (1.7)
It is easy to check that DαΛ is itself a distribution. Indeed, the linearity of the map φ 7→DαΛ(φ) is clear. Next, let K be a compact subset of Ω and let φ is a test function withsupport contained in K. By assumption, there exists a constant C and an integer N ≥ 0 suchthat (1.5) holds. In turn, this implies
|DαΛ(φ)| = |Λ(Dαφ)| ≤ C ‖Dαφ‖CN ≤ C ‖φ‖CN+|α| .
Hence DαΛ also satisfies (1.5), with N replaced by N + |α|.
Notice that, if Λf is the distribution at (1.1) corresponding to a function f which is |α|-timescontinuously differentiable, then we can integrate by parts and obtain
DαΛf (φ) = (−1)|α|Λf (Dαφ) = (−1)|α|
∫f(x)Dαφ(x) dx =
∫Dαf(x)φ(x) dx = ΛDαf (φ) .
103
This justifies the formula (1.7).
More generally, in some cases the distribution DαΛf coincides with a distribution Λg, for somelocally integrable function g. This leads to the concept of weak derivative.
Definition 7.3. Let f ∈ L1loc(Ω) be a locally integrable function on the open set Ω ⊆ IRn and
let Λf be the corresponding distribution, as in (1.1). Given a multi-index α = (α1, . . . , αn), ifthere exists a locally integrable function g ∈ L1
loc(Ω) such that DαΛf = Λg, i.e.
∫f Dαφdx = (−1)|α|
∫g φ dx for all test functions φ ∈ C∞c (Ω), (1.8)
then we say that g is the weak α-th derivative of f .
In general, a weak derivative may not exist. In particular, Example 7.2 shows that theHeaviside function does not admit a weak derivative. Indeed, its distributional derivative is aDirac measure, not a locally integrable function.
On the other hand, if a weak derivative does exist, then it is uniquely determined up to setsof measure zero.
Lemma 7.1 (uniqueness of weak derivatives). Assume f ∈ L1loc(Ω) and let g, g ∈ L1
loc(Ω)be weak α-th derivatives of f , so that
∫f Dαφdx = (−1)|α|
∫g φ dx = (−1)|α|
∫g φ dx
for all test functions φ ∈ C∞c (Ω). Then g(x) = g(x) for a.e. x ∈ Ω.
Proof. Indeed, the assumptions imply that the function (g − g) ∈ L1loc(Ω) satisfies
∫(g − g)φdx = 0 for all test functions φ ∈ C∞c (Ω).
By Corollary A.1 in the Appendix, we thus have g(x)− g(x) = 0 for a.e. x ∈ Ω.
Lemma 7.2 (convergence of weak derivatives). Consider a sequence of functions un ∈L1loc(Ω). For a fixed multi-index α, assume that each un admits an α-th weak derivative Dαun.
If un → u and Dαun → vα in L1loc, then vα = Dαu.
Proof. For every test function φ ∈ C∞c (Ω), a direct computation yields
∫
Ωvα φdx = lim
n→∞
∫
ΩDαun φdx = lim
n→∞(−1)|α|
∫
ΩunD
αφdx = (−1)|α|∫
ΩuDαφdx .
By definition, this means that vα is the α-th weak derivative of u.
104
2 Mollifications
As usual, let Ω ⊆ IRn be an open set. For a given ε > 0, define the subset
Ωε.= x ∈ IRn ;B(x, ε) ⊂ Ω , (2.1)
Then for every u ∈ L1loc(Ω) the mollification
uε(x).= (Jε ∗ u)(x) =
∫
B(x,ε)Jε(x− y)u(y) dy
is well defined for every x ∈ Ωε. In particular, uε ∈ C∞(Ωε). A very useful property of themollification operator is that it commutes with weak differentiation.
Lemma 7.3 (mollifications). Let Ωε ⊂ Ω be as in (2.1). Assume that a function f ∈ L1loc(Ω)
admits a weak derivative Dαu, for some multi-index α. Then the derivative of the mollification(which exists in the classical sense) coincides with the mollification of the weak derivative:
Dα(Jε ∗ u) = Jε ∗Dαu x ∈ Ωε . (2.2)
Proof. Observe that, for each fixed x ∈ Ωε, the function φ(y).= Jε(x− y) is in C∞c (Ω). Hence
we can apply the definition of weak derivative Dαu, using φ as a test function. Writing Dαx
and Dαy to distinguish differentiation w.r.t. the variables x or y, we thus obtain
Dαuε(x) = Dαx
(∫
ΩJε(x− y)u(y) dy
)
=
∫
ΩDα
xJε(x− y)u(y) dy
= (−1)|α|∫
ΩDα
y Jε(x− y)u(y) dy
= (−1)|α|+|α|∫
ΩJε(x− y)Dα
y u(y) dy
=(Jε ∗Dαu
)(x).
The property (2.2) of mollifications provides the key tool to relate weak derivatives withderivatives in the usual sense. As a first application, we prove
Corollary 7.1 (constant functions). Let Ω ⊆ IRn be an open, connected set, and assumeu ∈ L1
loc(Ω). If the first order weak derivatives of u satisfy
Dxiu(x) = 0 for i = 1, 2, . . . , n and a.e. x ∈ Ω ,
105
ΓεΩ
x
y
ΩΩ
Figure 8.1: Left: even if Ω is connected, the subdomain Ωε.= x ∈ Ω ; B(x, ε) ⊆ Ω may not be
connected. Right: any two points x, y ∈ Ω can be connected by a polygonal path Γ remaining insideΩ. Hence, if ε > 0 is sufficiently small, x and y belong to the same connected component of Ωε.
then u coincides a.e. with a constant function.
Proof. 1. For ε > 0, consider the mollified function uε = Jε ∗ u. By the previous analysis,uε : Ωε 7→ IR is a smooth function, whose derivativesDxiuε vanish identically on Ωε. Therefore,uε must be constant on each connected component of Ωε.
2. Now consider any two points x, y ∈ Ω. Since the open set Ω is connected, there exista polygonal path Γ joining x with y, remaining inside Ω. Let δ
.= minz∈Γ d(z, ∂Ω). Then
for every ε < δ the whole polygonal curve Γ is in Ωε. Hence x, y lie in the same connectedcomponent of Ωε. In particular, uε(x) = uε(y).
3. Call u(x).= limε→0 uε(x). By the previous step, u is a constant function on Ω. Moreover,
u(x) = u(x) for every Lebesgue point of u, hence almost everywhere on Ω. This concludes theproof.
Relying again on Lemma 7.3, we now characterize the set of functions of a scalar variablehaving a weak derivative in L1.
Lemma 7.4 (absolutely continuous functions). Let Ω ⊂ IR be an open interval, andassume that u : Ω 7→ IR has a weak derivative v ∈ L1(Ω). Then there exists an absolutelycontinuous function u such that
u(x) = u(x) for a.e. x ∈ Ω , (2.3)
v(x) = limh→0
u(x+ h)− u(x)h
for a.e. x ∈ Ω . (2.4)
Proof. Let x0 ∈ Ω be a Lebesgue point of u, and define
u(x).=
∫ x
x0
v(y) dy
Clearly, u is absolutely continuous and satisfies (2.4).
In order to prove (2.3), let Jε be the standard mollifier, and call uε.= Jε ∗u, vε .
= Jε ∗ v. Then
106
uε, vε ∈ C∞. Moreover, every x ∈ Ω we have
uε(x) = uε(x0) +
∫ x
x0
vε(y) dy . (2.5)
By the properties of mollifiers, as ε→ 0 we have uε(x0) → u(x0) as well as uε(x) → u(x) fora.e. x ∈ Ω. Letting ε→ 0 in (2.5), the left hand side converges to u(x) for every x ∈ Ω, whilethe right hand side converges to u(x) for every Lebesgue point of u. Hence (2.3) holds.
3 Sobolev spaces
Consider an open set Ω ⊆ IRn, fix p ∈ [1,∞] and let k be a non-negative integer. We say thatan open set Ω♯ is compactly contained in Ω if the closure Ω♯ is a compact subset of Ω.
Definition 7.4.
(i) The Sobolev space W k,p(Ω) is the space of all locally summable functions u : Ω 7→ IRsuch that, for every multi-index α with |α| ≤ k, the weak derivative Dαu exists and belongs toLp(Ω).
On W k,p we introduce the norm
‖u‖W k,p.=
∑
|α|≤k
∫
Ω|Dαu|p dx
1/p
if 1 ≤ p <∞, (3.1)
‖u‖W k,∞.=
∑
|α|≤k
ess- supx∈Ω|Dαu| if p =∞ . (3.2)
(ii) The subspace W k,p0 (Ω) ⊆ W k,p(Ω) is defined as the closure of C∞c (Ω) in W k,p(Ω). More
precisely, u ∈ W k,p0 (Ω) if and only if there exists a sequence of functions un ∈ C∞c (Ω) such
that‖u− un‖W k,p → 0.
(iii) By W k,ploc (Ω) we mean the space of functions which are locally in W k,p. These are the
functions u : Ω 7→ IR satisfying the following property. If Ω♯ is an open set compactly containedin Ω, then the restriction of u to Ω♯ is in W k,p(Ω♯).
One can think of the closed subspace W k,p0 (Ω) as the space of all functions u ∈W k,p(Ω) such
that Dαu = 0 along the the boundary of Ω, for every |α| ≤ k − 1.
Definition 7.5. In the special case where p = 2, we define the Hilbert-Sobolev spaceHk(Ω)
.=W k,2(Ω). The space Hk(Ω) is endowed with the inner product
(u, v)Hk.=
∑
|α|≤k
∫
ΩDαuDαv dx . (3.3)
107
Similarly, we define Hk0 (Ω)
.=W k,2
0 (Ω).
Theorem 7.1. (i) Each Sobolev space W k,p(Ω) is a Banach space.
(ii) The space W k,p0 (Ω) is a closed subspace of W k,p(Ω), hence it is a Banach space, with the
same norm.
(iii) The spaces Hk(Ω) and Hk0 (Ω) are Hilbert spaces.
Proof. 1. Let u, v ∈ W k,p(Ω). For |α| ≤ k, call Dαu, Dαv their weak derivatives. Then,for any λ, µ ∈ IR, the linear combination λu+ µv is a locally integrable function. One easilychecks that its weak derivatives are
Dα(λu+ µv) = λDαu+ µDαv . (3.4)
Therefore, Dα(λu + µv) ∈ Lp(Ω) for every |α| ≤ k. This proves that W k,p(Ω) is a vectorspace.
2. Next, we check that (3.1) and (3.2) satisfy the conditions (N1)–(N3) defining a norm.
Indeed, for λ ∈ IR and u ∈W k,p one has
‖λu‖W k,p = |λ| ‖u‖W k,p ,
‖u‖W k,p ≥ ‖u‖Lp ≥ 0
with equality holding if and only if u = 0.
Moreover, if u, v ∈W k,p(Ω), then for 1 ≤ p <∞ Minkowski’s inequality yields
‖u+ v‖W k,p =
∑
|α|≤k
‖Dαu+Dαv‖pLp
1/p
≤ ∑
|α|≤k
(‖Dαu‖Lp + ‖Dαv‖Lp
)p
1/p
≤ ∑
|α|≤k
‖Dαu‖pLp
1/p
+
∑
|α|≤k
‖Dαv‖pLp
1/p
= ‖u‖W k,p + ‖v‖W k,p .
In the case p =∞, the above computation is replaced by
‖u+v‖W k,∞ =∑
|α|≤k
‖Dαu+Dαv‖L∞ ≤∑
|α|≤k
(‖Dαu‖L∞+‖Dαv‖L∞
)= ‖u‖W k,∞+‖v‖W k,∞ .
108
3. To conclude the proof of (i), we show that the space W k,p(Ω) is complete, hence it is aBanach space.
Let (un)n≥1 be a Cauchy sequence in W k,p. Then, for every multi-index α with |α| ≤ k, thesequence of weak derivatives Dαun is Cauchy in Lp(Ω). Since the space Lp(Ω) is complete,there exist functions u and uα, such that
‖un − u‖Lp → 0, ‖Dαun − uα‖Lp → 0 for all |α| ≤ k . (3.5)
By Lemma 7.4, for |α| ≤ k, the limit function uα is precisely the weak derivative Dαu.
4. The fact that W k,p0 (Ω) is a closed subspace of W k,p(Ω) follows immediately from the
definition. The fact that (3.3) is an inner product is also clear.
Example 7.5. Let Ω = ]a, b[ be an open interval. By Lemma 7.4, each element of the spaceW 1,p(Ω) coincides a.e. with an absolutely continuous function f : ]a, b[7→ IR having derivativef ′ ∈ Lp(]a, b[).
−γ
x2
x1
u(x) = |x|
Figure 8.2: For certain values of p, n a function u ∈ W 1,p(IRn) may not be continuous, or bounded.
Example 7.6. Let Ω = B(0, 1) ⊂ IRn be the open ball centered at the origin, with radiusone. Fix γ > 0 and consider the radially symmetric function
u(x).= |x|−γ =
(n∑
i=1
x2i
)−γ/2
0 < |x| < 1.
109
Observe that u ∈ C1(Ω \ 0). Outside the origin, its partial derivatives are computed as
uxi = − γ
2
(n∑
i=1
x2i
)−(γ/2)−1
2xi =−γxi|x|γ+2
. (3.6)
Hence, the gradient ∇u = (ux1 , . . . , uxn) has norm
|∇u(x)| =n∑
i=1
|uxi(x)|2 =γ
|x|γ+1.
On the open set Ω\0, the function u clearly admits weak derivatives of all orders, and thesecoincide with the classical ones.
We wish to understand in which cases the formula (3.6) defines the weak derivatives of u onthe entire domain Ω. This means
∫
Ωuxi φdx = −
∫
Ωuφxi dx
for every smooth function φ ∈ C∞c (Ω) whose support is a compact subset of Ω (and not onlyfor those functions φ whose support is a compact subset of Ω \ 0).
Observe that, for any ε > 0, one has
∫
ε<|x|<1(uφ)xi dx =
∫
|x|=εu(x)φ(x) νi(x) dS
where dS is the (n − 1)-dimensional measure on the surface of the ball B(0, ε), and νi(x) =−xi/|x|, so that ν = (ν1, ν2, . . . , νn) is the unit normal pointing toward the interior of the ballB(0, ε). Since φ is a bounded continuous function, we have
∫
|x|=ε|u(x)φ(x) νi(x)| dS ≤ ‖φ‖L∞ ε−γ σnε
n−1 → 0 as ε→ 0
provided that n− 1 > γ. In other words, if γ < n− 1, then
∫
Ω
−γxi|x|γ+2
φ(x) dx =
∫
Ω
1
|x|γ φxi(x) dx for every test function φ ∈ C∞c (Ω).
and the locally integrable function uxi defined at (3.6) is in fact the weak derivative of u onthe whole domain Ω.
Observe that
∫
Ω
(1
|x|γ+1
)p
dx =
∫
x∈IRn, |x|<1|x|−p(γ+1) dx = σn
∫ 1
0rn−1r−p(γ+1) dr < ∞
if and only if n− 1− p(γ + 1) > −1, i.e. γ < n−pp .
The previous computations show that, if 0 < γ < n−pp , then u ∈W 1,p(Ω).
Notice that u is absolutely continuous (in fact, smooth) on a.e. line parallel to one of thecoordinate axes. However, there is no way to change u on a set of measure zero, so that itbecomes continuous on the whole domain Ω.
110
An important question in the theory of Sobolev spaces is whether one can estimate the normof a function in terms of the norm of its first derivatives. The following result provides anelementary estimate in this direction. It is valid for domains Ω which is contained in a slab,say
Ω ⊆ x = (x1, x2, . . . , xn) ; a < x1 < b . (3.7)
Theorem 7.2 (Poincare’s inequality - I). Let Ω ⊂ IRn be an open set which satisfies (3.7)for some a, b ∈ IR. Then, every u ∈ H1
0 (Ω) satisfies
‖u‖L2(Ω) ≤ 2(b− a) ‖Dx1u‖L2(Ω) . (3.8)
Proof. 1. Assume first that u ∈ C∞c (Ω). We extend u to the whole space IRn by settingu(x) = 0 for x /∈ Ω. Using the variables x = (x1, x
′) with x′ = (x2, . . . , xn), we compute
u(x1, x′) =
∫ x1
a2uux1(t, x
′) dt .
An integration by parts yields
‖u‖2L2 =
∫
IRnu2(x) dx =
∫
IRn−1
∫ b
a1 ·(∫ x1
a2uux1(t, x
′) dt)dx1 dx
′
=
∫
IRn−1
∫ b
a(b− x1) 2uux1(t, x
′) dx1 dx′ ≤ 2(b− a)
∫
IRn|u| |ux1 | dx
≤ 2(b− a)‖u‖L2‖ux1‖L2 .
Dividing both sides by ‖u‖L2 we obtain (3.8), for every u ∈ C∞c (Ω).
2. Now consider any u ∈ H10 (Ω). By assumption there exists a sequence of functions un ∈
C∞c (Ω) such that ‖un − u‖H1 → 0. By the previous step, this implies
‖u‖L2 = limn→∞
‖un‖L2 ≤ limn→∞
2(b− a)‖Dx1un‖L2 = 2(b− a) ‖Dx1u‖L2 .
To proceed in the analysis of Sobolev spaces, we need to derive some more properties of weakderivatives. Given two multi-indices α = (α1, . . . , αn) and β = (β1, . . . , βn), the notationβ ≤ α means βi ≤ αi for every i = 1, . . . , n. Moreover,
(αβ
).=
α!
β! (α− β)!.=
α1!
β1! (α1 − β1)!· α2!
β2! (α2 − β2)!· · · α2!
β2! (α2 − β2)!.
Theorem 7.3 (properties of weak derivatives). Let Ω ⊆ IRn be an open set, p ∈ [1,∞],and |α| ≤ k. If u, v ∈W k,p(Ω), then
111
(i) The restriction of u to any open set Ω ⊂ Ω is in the space W k,p(Ω).
(ii) Dαu ∈W k−|α|,p(Ω). If |α|+ |β| ≤ k, then Dα(Dβu) = Dβ(Dαu) = Dα+βu.
(iii) If η ∈ Ck(Ω), then the product satisfies η u ∈ W k,p(Ω). Its weak derivatives are given bythe Leibniz formula
Dα(ηu) =∑
β≤α
(αβ
)DβηDα−βu . (3.9)
Moreover there exists a constant C depending on Ω and on ‖η‖Ck but not on u, such that
‖ηu‖W k,p(Ω) ≤ C ‖u‖W k,p(Ω) . (3.10)
(iv) Let Ω♯ ⊆ IRn is an open set and let ϕ : Ω♯ 7→ Ω be a Ck diffeomorphism whose Jacobianmatrix has a uniformly bounded inverse. Then the composition satisfies u ϕ ∈ W k,p(Ω♯).Moreover there exists a constant C, depending on Ω♯ and on ‖ϕ‖Ck but not on u, such that
‖u ϕ‖W k,p(Ω♯) ≤ C ‖u‖W k,p(Ω) . (3.11)
Proof. 1. The statement (i) is obvious.
2. To prove (ii), consider any test function φ ∈ C∞c (Ω). Since Dβφ ∈ C∞c (Ω), we have∫
Ω(Dαu) (Dβφ) dx = (−1)|α|
∫
Ωu (Dα+βφ) dx
= (−1)|α|(−1)|α+β|∫
Ω(Dα+βu)φdx
= (−1)|β|∫
Ω(Dα+βu) φdx .
This shows that Dα+βu is the β-th derivative of Dαu, in the weak sense.
3. To prove (iii), let Jε be the standard mollifier and set uε.= Jε∗u. Since the Leibniz formula
holds for the product of smooth functions, for every ε > 0 we obtain
Dα(ηuε) =∑
β≤α
(αβ
)Dβη Dα−βuε . (3.12)
For every test function φ ∈ C∞c (Ω), we thus have
(−1)|α|∫
Ω(ηuε)D
αφdx =∑
β≤α
(−1)|α−β|(αβ
)∫
ΩuεD
α−β(φDβη) dx .
Letting ε→ 0, this yields
(−1)|α|∫
Ω(ηu)Dαφdx =
∑
β≤α
(−1)|α−β|(αβ
)∫
ΩuDα−β(φDβη) dx
=∑
β≤α
(αβ
)∫
Ω(Dα−βu)(Dβη)φdx .
112
Hence (3.9) holds.
Since all derivatives of η are bounded, namely
‖Dβη‖L∞ ≤ ‖η‖Ck for all |β| ≤ k,
the bound (3.10) is an immediate consequence of (3.9).
4. We prove (iv) by induction on k. By assumption, the n×n Jacobian matrix (Dxiϕj)i,j=1,...,n
has a uniformly bounded inverse. Hence the case k = 0 is clear.
Next, assume that the result is true for k = 0, 1, . . . ,m− 1. Setting uε.= Jε ∗ u, one has
Dxi(uε ϕ)(x) =n∑
j=1
(Dxjuε)(ϕ(x)) · (Dxiϕj)(x) .
If u ∈Wm,p(Ω), letting ε→ 0 we obtain
‖Dxi(u ϕ)‖Wm−1,p(Ω♯) ≤ C ′ ‖∇u‖Wm−1,p(Ω) ‖ϕ‖Cm(Ω♯) ≤ C ‖u‖Wm,p(Ω) ,
showing that the result is true also for k = m. By induction, this achieves the proof.
4 Approximations and extensions of Sobolev functions
Theorem 7.4 (approximation with smooth functions). Let Ω ⊆ IRn be an open set.Let u ∈W k,p(Ω) with 1 ≤ p <∞. Then there exists a sequence of functions un ∈ C∞(Ω) suchthat ‖un − u‖W k,p(Ω) → 0 as n→∞.
Proof. 1. Let ε > 0 be given. Consider the following locally finite open covering of the set Ω.
V1.=x ∈ Ω ; d(x, ∂Ω) >
1
2
, Vj
.=x ∈ Ω ;
1
j + 1< d(x, ∂Ω) <
1
j − 1
k = 2, 3, . . .
Let η1, η2, . . . be a smooth partition of unity subordinate to the above covering. By Theo-rem 7.3, for every j ≥ 0 the product ηju is in W k,p(Ω). By construction, it has supportcontained in Vj.
2. Consider the mollifications Jǫ ∗ (ηju). By Lemma 7.3, for every |α| ≤ k we have
Dα(Jǫ ∗ (ηju)) = Jǫ ∗ (Dα(ηju)) → Dα(ηju)
as ǫ → 0. Since each ηj has compact support, here the convergence takes place in Lp(Ω).Therefore, for each j ≥ 0 we can find 0 < ǫj < 2−j small enough so that
‖ηju− Jǫj ∗ (ηju)‖W k,p(Ω) ≤ ε 2−k.
3. Consider the function
U.=
∞∑
j=1
Jεj ∗ (ηju)
113
Notice that the above series may not converge in W k,p. However, it is certainly pointwiseconvergent because every compact setK ⊂ Ω intersects finitely many of the sets Vj. Restrictedto K, the above sum contains only finitely many non-zero terms. Since each term is smooth,this implies U ∈ C∞(Ω).
4. Consider the subdomains
Ω1/n.=x ∈ Ω ; d(x, ∂Ω) >
1
n
.
Recalling that∑
j ηj(x) ≡ 1, for every n ≥ 1 we find
‖U − u‖W k,p(Ω1/n)≤
n+2∑
j=1
‖ηju− Jεj ∗ (ηju)‖W k,p(Ω) ≤∞∑
j=1
ε 2−j = ε.
This yields‖U − u‖W k,p(Ω) = sup
n≥1‖U − u‖W k,p(Ω1/n)
≤ ε .
Since ε > 0 was arbitrary, this proves that the set of C∞ function is dense on W k,p(Ω).
Using the above approximation result, we describe a regularity theorem for Sobolev functions.
Theorem 7.5 (relation between weak and strong derivatives). Let u ∈W 1,1(Ω), whereΩ ⊆ IRn is an open set having the form
Ω =x = (x, x′) ; x′
.= (x2, . . . , xn) ∈ Ω′ , α(x′) < x1 < β(x′)
. (4.1)
Then there exists a function u with u(x) = u(x) for a.e. x ∈ Ω, such that the following holds.For a.e. x′ = (x2, . . . , xn) ∈ Ω′ ⊂ IRn−1 (w.r.t. the (n − 1)-dimensional Lebesgue measure),the map
x1 7→ u(x1, x′)
is absolutely continuous. Its derivative coincides a.e. with the weak derivative Dx1u.
Proof. 1. By the previous theorem, there exists a sequence of functions un ∈ C∞(Ω) suchthat
‖un − u‖W 1,1 < 2−n . (4.2)
We claim that this implies the pointwise convergence
un(x)→ u(x), Dx1un(x)→ Dx1u(x) for a.e. x ∈ Ω .
Indeed, consider the functions
f(x).= |u1(x)|+
∞∑
n=1
|un+1(x)−un(x)| , g(x).= |Dx1u1(x)|+
∞∑
n=1
∣∣∣Dx1un+1(x)−Dx1un(x)∣∣∣ .
(4.3)By (4.2), there holds
‖un − un+1‖W 1,1 ≤ 21−n,
114
x1
Ω
x =1 α (x , x )2 3 x =x
x2
31 β(x , x )
2 3
’Ω
Figure 8.3: The domain Ω at (4.1). If u has a weak derivative Dx1u ∈ L1(Ω), then (by possibly
changing its values on a set of measure zero), the function u is absolutely continuous on almost everysegment parallel to the x1-axis, and its partial derivative ∂u/∂x1 coincides a.e. with the weak derivative.
hence f, g ∈ L1(Ω) and the series in (4.3) are absolutely convergent for a.e. x ∈ Ω. Therefore,they converge pointwise almost everywhere. Moreover, we have the bounds
|un(x)| ≤ f(x) , |Dx1un(x)| ≤ g(x) for all n ≥ 1, x ∈ Ω . (4.4)
2. Since f, g ∈ L1(Ω), by Fubini’s theorem there exists a null set N ⊂ Ω′ (w.r.t. the n − 1dimensional Lebesgue measure) such that, for every x′ ∈ Ω′ \ N one has
∫ β(x′)
α(x′)f(x1, x
′) dx1 < ∞ ,
∫ β(x′)
α(x′)g(x1, x
′) dx1 < ∞. (4.5)
Fix such a point x′ ∈ Ω′ \ N . Choose a point y1 ∈ ]α(x′) , β(x′)[ where the pointwiseconvergence un(y1, x
′)→ u(y1, x′) holds. For every α(x′) < x1 < β(x′), since un is smooth we
have
un(x1, x′) = un(y1, x
′) +∫ x1
y1Dx1un(s, x
′) ds . (4.6)
We now let n→∞ in (4.6). By (4.4) and (4.5), the functions Dx1un(·, x′) are all bounded bythe integrable function g(·, x′) ∈ L1. By the dominated convergence theorem, the right handside of (4.6) thus converges to
u(x1, x′)
.= u(y1, x
′) +∫ x1
y1Dx1u(s, x
′) ds . (4.7)
Clearly, the right hand side of (4.7) is an absolutely continuous function of the scalar variablex1. On the other hand, the left hand side satisfies
u(x1, x′)
.= lim
n→∞un(x1, x′) = u(x1, x
′) for a.e. x1 ∈ [α(x′), β(x′)].
This achieves the proof.
115
Remarks. (i) It is clear that a similar result holds for any other derivative Dxiu, withi = 1, 2, . . . , n.
(ii) If u ∈W k,p(Ω) and Ω ⊂ Ω, then the restriction of u to Ω lies in the space W k,p(Ω).
(iii) If Ω ⊂ IRn is a bounded open set and u ∈ W k,p(Ω), then we also have u ∈ W k,q(Ω) forevery q ∈ [1, p].
The next theorem shows that, given a bounded open domain Ω ⊂ IRn with C1 boundary, eachfunction u ∈W 1,p(Ω) can be extended to a function Eu ∈W 1,p(IRn).
Theorem 7.6 (extension operators). Let Ω ⊂⊂ Ω ⊂ IRn be a bounded open sets, withΩ containing the closure of Ω. Assume that ∂Ω ∈ C1. Then there exists a bounded linearoperator E : W 1,p(Ω) 7→W 1,p(IRn) and a constant C such that
(i) Eu(x) = u(x) for a.e. x ∈ Ω,
(ii) Eu(x) = 0 for x /∈ Ω,
(iii) One has the bound ‖Eu‖W 1,p(IRn) ≤ C‖u‖W 1,p(Ω).
Proof. 1. We first consider the special case where the domain is a half plane: Ω = x =(x1, x2, . . . , xn) ; x1 > 0, and Ω = IRn. In this case, any function u ∈ W 1,p(Ω) can beextended to the whole space IRn by reflection, i.e. by setting
(E♯u)(x1, x2, . . . , xn).= u ( |x1|, x2, x3, . . . , xn) . (4.8)
Observe that, on the region where x1 < 0,the weak derivatives of E♯u are given by
Dx1E♯u(−x1, x2, . . . , xn) = −Dx1u(x1, x2, . . . , xn), for all x1 > 0, x2, . . . , xn ∈ IR.
It is easily checked that this extension operator E♯ : W 1,p(Ω) 7→ W 1,p(IRn) is linear andsatisfies
‖E♯u‖W 1,p(IRn) ≤ 2‖u‖W 1,p(Ω) .
2. To handle the general case, we shall use a partition of unity. For every x ∈ Ω (the closureof Ω), choose an open ball B(x, rx) centered at x with radius rx > 0, so that the followingholds.
- If x ∈ Ω, then B(x, rx) ⊂ Ω.
- If x ∈ ∂Ω, then B(x, rx) ⊂ Ω. Moreover, there exists a C1 bijection ϕx : B(0, 1) 7→ B(x, rx),whose inverse is also C1, which maps the upper half ball
B+(0, 1).=
y = (y1, y2, . . . , yn) ;
n∑
i=1
y2i < 1 , y1 > 0
onto the set B(x, rx) ∩ Ω.
Choosing rx > 0 sufficiently small, the existence of such a bijection follows from the assumptionthat Ω has a C1 boundary.
116
0
xj
x
B
B i
ϕi
Ω~Ω
i
i 1y
Figure 8.4: Left: the open covering of the set Ω. Right: the diffeomorphism mapping the unit ball Bonto the ball Bi = B(xi, rxi
), so that B+ is mapped onto Bi ∩Ω. The shaded areas show the supportof the functions (ηiu) ϕi and ηiu, respectively.
Since Ω is bounded, its closure Ω is compact. Hence it can be covered with finitely many ballsBi = B(xi, rxi), i = 1, . . . , N . Let ϕi : B(0, 1) 7→ Bi be the corresponding bijections. Recallthat ϕi maps B+(0, 1) onto B+
i.= Bi ∩ Ω.
3. Let η1, . . . , ηN be a smooth partition of unity subordinated to the above covering. Forevery x ∈ Ω we thus have
u(x) =N∑
i=1
ηi(x)u(x) .
We split the set of indices as1, 2, . . . , N = I ∪ J ,
where I contains the indices with xi ∈ Ω while J contains the indices with xi ∈ ∂Ω.
For every i ∈ J , we have
ηiu ∈W 1,p(B+i ) , (ηiu) ϕi ∈W 1,p(B+(0, 1)).
Therefore, the extension E♯ constructed at (4.8) satisfies
E♯((ηiu) ϕi
)∈ W 1,p(B+(0, 1)) , E♯
((ηiu) ϕi
) ϕ−1
i ∈ W 1,p(B+i ) .
Summing together all these extensions, we define
Eu.=
∑
i∈Iηiu+
∑
i∈JE♯((ηiu) ϕi
) ϕ−1
i .
It is now clear that the extension operator E satisfies all requirements. Indeed, E is a finitesum of bounded linear operators. Moreover, for every u ∈W 1,p(Ω), the support of the functionEu is contained in ∪Ni=1Bi ⊆ Ω.
117
5 Embedding theorems
In one space dimension, a function u : IR 7→ IR which admits a weak derivative Du ∈ L1(IR)is absolutely continuous (after changing its values on a set of measure zero). In several spacedimensions, however, there exists functions u ∈W 1,p(IRn) which are not continuous, and noteven bounded. This is indeed the case of the function u(x) = |x|−γ considered in Example 7.6,for a suitable γ > 0.
In several applications to PDEs or to the Calculus of Variations, it is important to understandthe degree of regularity enjoyed by functions u ∈W k,p(IRn). We shall prove two basic resultsin this direction.
1. (Morrey). If p > n, then every function u ∈ W 1,p(IRn) is Holder continuous (after amodification on a set of measure zero).
2. (Gagliardo-Nirenberg). If p < n, then every function u ∈ W 1,p(IRn) lies in the space
Lp∗(IRn), with the larger exponent p∗ = p+ p2
n−p .
In both cases, the result can be stated as an embedding theorem: after a modification on a setof measure zero, each function u ∈W 1,p(Ω) also lies in some other Banach space X. TypicallyX = C0,γ, or X = Lq for some q > p. The basic approach is as follows:
I - Prove an a priori inequality valid for all smooth functions. Given any functionu ∈ C∞ ∩W k,p(Ω), one proves that u also lies in another Banach space X, and there exists aconstant C depending on k, p,Ω but not on u, such that
‖u‖X ≤ C ‖u‖W k,p for all u ∈ C∞ ∩W k,p(Ω) . (5.1)
II - Extend the embedding to the entire space, by continuity. Since C∞ is densein W k,p, for every u ∈ W k,p(Ω) we can find a sequence of functions un ∈ C∞ such that‖u− un‖W k,p → 0. By (5.1),
lim supm,n→∞
‖um − un‖X ≤ lim supm,n→∞
C ‖um − un‖W k,p = 0 .
Therefore the functions un form a Cauchy sequence also in the space X. By completeness,un → u for some u ∈ X. Observing that u(x) = u(x) for a.e. x ∈ Ω, we conclude that, up toa modification on a set N ⊂ Ω of measure zero, each function u ∈ W k,p(Ω) lies also in thespace X.
5.1 Morrey’s inequality
In this section we prove that, if u ∈ W 1,p(IRn), where the exponent p is higher than thedimension n of the space, then u essentially coincides with a Holder continuous function
Theorem 7.7 (Morrey’s inequality). Assume n < p < ∞ and set γ.= 1 − n
p > 0. Thenthere exists a constant C, depending only on p and n, such that
‖u‖C0,γ (IRn) ≤ C ‖u‖W 1,p(IRn) (5.2)
118
for every u ∈ C1(IRn) ∩W 1,p(IRn).
z yy ’ ρ ρx1
0 0
Γ
x
1B
1
r
ξ = (ξ , ... , ξ )2H n
Figure 8.5: Proving Morrey’s inequality. Left: the values u(y) and u(y′) are compared with theaverage value of u on the (n − 1)-dimensional ball centered at the mid-point z. Center and right: apoint x in the cone Γ is described in terms of the coordinates (r, ξ) ∈ [0, ρ]×B1 .
Proof. 0. Before giving the actual proof, we outline the underlying idea. From an integralestimate on the gradient of the function u, say
∫
IRn|∇u(x)|p dx ≤ C0, (5.3)
we seek a pointwise estimate of the form
|u(y)− u(y′)| ≤ C1|y − y′|γ for all y, y′ ∈ IRn. (5.4)
To achieve (5.4), a natural attempt is to write
|u(y)−u(y′)| =∣∣∣∣∫ 1
0
[d
dθu(θy + (1− θ)y′)
]dθ
∣∣∣∣ ≤∫ 1
0
∣∣∣∇u(θy+(1−θ)y′)∣∣∣ |y−y′| dθ . (5.5)
However, the integral on the right hand side of (5.5) only involves values of ∇u over thesegment joining y with y′. If the dimension of the space is n > 1, this segment has zeromeasure. Hence the integral in (5.5) can be arbitrarily large, even if the integral in (5.3) issmall. To address this difficulty, we shall compare both values u(y), u(y′) with the average
value uB of the function u over an (n−1)-dimensional ball centered at the midpoint z = y+y′
2 ,as shown in Figure 8.5, left. Notice that the difference |u(y) − uB | can be estimated by anintegral of |∇u| ranging over a cone of dimension n. In this way the bound (5.3) can thus bebrought into play.
1. We now begin the proof, with a preliminary computation. On IRn, consider the cone
Γ.=
x = (x1, x2, . . . , xn) ;
n∑
j=2
x2j ≤ x21 , 0 < x1 < ρ
119
and the function
ψ(x) =1
xn−11
. (5.6)
Let q = pp−1 be the conjugate exponent of p, so that 1
p +1q = 1. We compute
‖ψ‖qLq(Γ) =
∫
Γ
(1
xn−11
)q
dx =
∫ ρ
0cn−1x
n−11
(1
xn−11
)q
dx1 = cn−1
∫ ρ
0s(n−1)(1−q) ds ,
where the constant cn−1 gives the volume of the unit ball in IRn−1. Therefore, ψ ∈ Lq(Γ) ifand only if n < p. In this case,
‖ψ‖Lq(Γ) =
(cn−1
∫ ρ
0s−n−1
p−1 ds
) 1q
= c
(ρ
p−np−1
) p−1p
= c ρp−np , (5.7)
for some constant c depending only on n and p.
2. Consider any two distinct points y, y′ ∈ IRn. Let ρ.= 1
2 |y − y′|. The hyperplane passing
through the midpoint z.= y+y′
2 and perpendicular to the vector y − y′ has equation
H =x ∈ IRn ; 〈x− z , y − y′〉 = 0
.
Inside H, consider the (n− 1)-dimensional ball centered at z with radius ρ,
B.=x ∈ H ; |x− z| < ρ
.
Calling uB the average value of u on the ball B, the difference |u(y)−u(y′)| will be estimatedas
|u(y)− u(y′)| ≤ |u(y)− uB |+∣∣uB − u(y′)
∣∣ . (5.8)
3. By a translation and a rotation of coordinates, we can assume
y = (0, . . . , 0) ∈ IRn, B =x = (x1, x2, . . . , xn) ; x1 = ρ ,
n∑
i=2
x2i ≤ ρ2.
To compute the average value uB , let B1 be the unit ball in IRn−1, and let cn−1 be its (n− 1)-dimensional measure. Points in the cone Γ will be described using an alternative system ofcoordinates. To (r, ξ) = (r, ξ2, . . . , ξn) ∈ [0, ρ] × B1 we associate the point x(r, ξ) ∈ Γ,defined by
(x1, x2, . . . , xn) = (r, rξ) = (r, rξ2, . . . , rξn). (5.9)
Define U(r, ξ) = u(r, rξ), and observe that U(0, ξ) = u(0) for every ξ. Therefore
U(ρ, ξ) = U(0, ξ) +
∫ ρ
0
[∂
∂rU(r, ξ)
]dr ,
uB − u(0) =1
cn−1
∫
B1
(∫ ρ
0
[ ∂∂rU(r, ξ)
]dr
)dξ . (5.10)
We now change variables, transforming the integral (5.10) over [0, ρ]×B1 into an integral overthe cone Γ. The Jacobian of the transformation (5.9) is computed as
dx1 dx2 · · · dxn = rn−1dr dξ2 · · · dξn .
120
Moreover, since |ξ| ≤ 1, the directional derivative of u in the direction of the vector(1, ξ2, . . . , ξn) is estimated by
∣∣∣∣∂
∂rU(r, ξ)
∣∣∣∣ =∣∣∣ux1 +
n∑
i=2
ξiuxi
∣∣∣ ≤ 2|∇u(r, ξ)| . (5.11)
Using (5.11) in (5.10), and the estimate (5.7) on the Lq norm of the function ψ(x).= x1−n
1
introduced at (5.6), we obtain
∣∣∣uB − u(0)∣∣∣ ≤ 2
cn−1
∫
Γ
1
xn−11
|∇u(x)| dx
≤ 2
cn−1‖ψ‖Lq(Γ) ‖∇u‖Lp(Γ)
(q =
p
p− 1
)
≤ C ρp−np ‖u‖W 1,p(IRn) ,
(5.12)
for some constant C. Notice that the last two steps follow from Holder’s inequality and (5.7).
4. Using (5.12) to estimate each term on the right hand side of (5.8), and recalling thatρ = 1
2 |y − y′|, we conclude
|u(y)− u(y′)| ≤ 2C
( |y − y′|2
) p−np
‖u‖W 1,p(IRn). (5.13)
This provides a uniform bound on the Holder norm of u, with Holder constant γ = p−np .
5. To prove an a priori bound on supx |u(x)|, observe that, by (5.13), there exists a constantC1 such that
|u(x)| ≤ |u(y)|+ C1‖u‖W 1,p(IRn) for |y − x| ≤ 1.
Hence, taking the average of the right hand side over the ball centered at x with radius one,
|u(x)| ≤ −∫
B(x,1)|u(y)| dy + C1‖u‖W 1,p(IRn) ≤ C2‖u‖Lp(IRn) + C1‖u‖W 1,p(IRn).
Since C∞ is dense in W 1,p, Morrey’s inequality yields
Corollary 7.2 (embedding). Let Ω ⊂ IRn be a bounded open set with C1 boundary. Assumen < p < ∞ and set γ
.= 1 − n
p > 0. Then every function f ∈ W 1,p(Ω) coincides a.e. with a
function f ∈ C0,γ(Ω). Moreover, there exists a constant C such that
‖f‖C0,γ ≤ C‖f‖W 1,p for all f ∈W 1,p(Ω). (5.14)
Proof. 1. Let Ω.= x ∈ IRn ; d(x,Ω) < 1 be the open neighborhood of radius one around
the set Ω. By Theorem 7.6, there exists a bounded extension operator E, which extends eachfunction f ∈W 1,p(Ω) to a function Ef ∈W 1,p(IRn) with support contained inside Ω.
121
2. Since C1(IRn) is dense in W 1,p(IRn), we can find a sequence of functions gn ∈ C1(IRn)converging to Ef in W 1,p. By Morrey’s inequality
lim supm,n→∞
‖gm − gn‖C0,γ(IRn) ≤ C lim supm,n→∞
‖gm − gn‖W 1,p(IRn) = 0.
This proves that the sequence (gn)n≥1 is a Cauchy sequence also in the space C0,γ . Thereforeit converges to a limit function g ∈ C0,γ(IRn), uniformly for x ∈ IRn.
3. Since gn → Ef in W 1,p(IRn), we also have g(x) = (Ef)(x) for a.e. x ∈ IRn. In particular,g(x) = f(x) for a.e. x ∈ Ω. Since the extension operator E is bounded, from the bound (5.2)we deduce (5.14).
5.2 The Gagliardo-Nirenberg inequality
Next, we study the case 1 ≤ p < n. We define the Sobolev conjugate of p as
p∗ .=
np
n− p > p. (5.15)
Notice that p∗ depends not only on p but also on the dimension n of the space. Indeed,
1
p∗=
1
p− 1
n. (5.16)
As a preliminary, we observe a useful application of the generalized Holder’s inequality. Let
n − 1 non-negative functions g1, g2, . . . , gn−1 ∈ L1(Ω) be given. Then g1
n−1
j ∈ Ln−1 for eachj = 1, . . . , n− 1. The generalized Holder inequality yields
∫
Ωg
1n−1
1 g1
n−1
2 · · · g1
n−1
n−1 ds ≤n−1∏
j=1
‖g1
n−1
j ‖Ln−1 =n−1∏
j=1
‖gj‖1
n−1
L1 . (5.17)
Theorem 7.8 (Gagliardo-Nirenberg inequality). Assume 1 ≤ p < n. Then there existsa constant C, depending only on p and n such that
‖f‖Lp∗(IRn) ≤ C ‖∇f‖Lp(IRn) for all f ∈W 1,p(IRn) . (5.18)
Proof. 1. By an approximation argument, it suffices to prove the inequality for all f ∈C1c (IRn). In this case, for each i ∈ 1, . . . , n and every point x = (x1, . . . , xi, . . . , xn) ∈ IRn,we have
f(x1, . . . , xi, . . . , xn) =
∫ xi
−∞Dxif(x1, . . . , si, . . . , xn) dsi .
In turn, this yields
|f(x1, . . . , xn)| ≤∫ ∞
−∞
∣∣∣Dxif(x1, . . . , si, . . . , xn)∣∣∣ dsi 1 ≤ i ≤ n ,
122
3
(s ,x ,x )(x ,x ,x )1 2 3 1 2 3
x1
x 2
x
Figure 8.6: Proving the Gagliardo-Nirenberg inequality. The integral∫∞
−∞|Dx1
f(s1, x2, x3)| ds1 de-
pends on x2, x3 but not on x1. Similarly, the integral∫∞
−∞|Dx2
f(x1, s2, x3)| ds2 depends on x1, x3 butnot on x2.
|f(x)| nn−1 ≤
n∏
i=1
(∫ ∞
−∞|Dxif(x1, . . . , si, . . . , xn)| dsi
) 1n−1
. (5.19)
We now integrate (5.19) w.r.t. x1. Observe that the first factor on the right hand side doesnot depend on x1. This factor behaves like a constant and can be taken out of the integral.The product of the remaining n− 1 factors is handled using Holder’s inequality, as in (5.17).This yields
∫ ∞
−∞|f | n
n−1 dx1 ≤(∫ ∞
−∞|Dx1f | ds1
) 1n−1
∫ ∞
−∞
n∏
i=2
(∫ ∞
−∞|Dxif | dsi
) 1n−1
dx1
≤(∫ ∞
−∞|Dx1f | ds1
) 1n−1
n∏
i=2
(∫ ∞
−∞
∫ ∞
−∞|Dxif | dsi dx1
) 1n−1
.
(5.20)
We now integrate w.r.t. x2. Observe that one of the n− 1 factors appearing in the product onthe right hand side of (5.20), does not depend on the variable x2 (namely, the one involvingintegration w.r.t. s2). This factor behaves like a constant and can be taken out of the integral.The product of the remaining n− 2 factors is again handled as in (5.17). This yields
∫ ∞
−∞
∫ ∞
−∞|f | n
n−1 dx1 dx2 ≤(∫ ∞
−∞
∫ ∞
−∞|Dx1f | dx1dx2
) 1n−1
(∫ ∞
−∞
∫ ∞
−∞|Dx2f | dx1dx2
) 1n−1
×n∏
i=3
(∫ ∞
−∞
∫ ∞
−∞
∫ ∞
−∞|Dxif | dsi dx1dx2
) 1n−1
.
(5.21)After n integrations we obtain
∫
IRn|f | n
n−1 dx ≤n∏
i=1
(∫ ∞
−∞· · ·∫ ∞
−∞|Dxif |
nn−1 dx1 · · · dxn
) 1n−1
≤(∫
IRn|∇f | dx
) nn−1
.
(5.22)
123
This already implies (∫
IRn|f | n
n−1 dx
)n−1n ≤
∫
IRn|∇f | dx , (5.23)
proving the theorem in the case where p = 1 and p∗ = nn−1 .
2. To cover the general case where 1 < p < n, we apply (5.23) to the function g = |f |β, for asuitable β > 0. Using the standard Holder’s inequality, this yields
(∫
IRn|f |
βnn−1 dx
)n−1n ≤
∫
IRnβ|f |β−1|∇f | dx
≤ β
(∫
IRn|f |
(β−1)pp−1 dx
) p−1p(∫
IRn|∇f |p dx
) 1p
.
(5.24)
If we choose
β.=p(n− 1)
n− p ,
thenβn
n− 1=
(β − 1)p
p− 1=
np
n− p = p∗.
Therefore, (5.24) yields
(∫
IRn|f |p∗ dx
)n−1n ≤ β
(∫
IRn|f |p∗ dx
) p−1p(∫
IRn|∇f |p dx
) 1p
.
Observing that n−1n −
p−1p = n−p
np = 1p∗ , we conclude
(∫
IRn|f |p∗ dx
) 1p∗ ≤ C
(∫
IRn|∇f |p dx
) 1p
.
If the domain Ω ⊂ IRn is bounded, then Lq(Ω) ⊆ Lp∗(Ω) for every q ∈ [1, p∗]. The previoustheorem thus yields
Corollary 7.3 (embedding). Let Ω ⊂ IRn be a bounded open domain with C1 boundary, andassume 1 ≤ p < n. Then, for every q ∈ [1, p∗] with p∗
.= np
n−p , there exists a constant C suchthat
‖f‖Lq(Ω) ≤ C ‖f‖W 1,p(Ω) for all f ∈W 1,p(Ω) . (5.25)
Proof. Let Ω.= x ∈ IRn ; d(x,Ω) < 1 be the neighborhood of radius one around the set
Ω. By Theorem 7.6, there exists a bounded extension operator E : W 1,p(Ω) 7→ W 1,p(IRn),with the property that Ef is supported inside Ω, for every f ∈ W 1,p(Ω). Applying theGagliardo-Nirenberg inequality to Ef , for suitable constants C1, C2, C3 we obtain
‖f‖Lq(Ω) ≤ C1 ‖f‖Lp∗(Ω) ≤ C1 ‖f‖Lp∗(IRn) ≤ C2 ‖Ef‖Lp∗(IRn) ≤ C3‖f‖W 1,p(Ω) .
124
5.3 High order Sobolev estimates
Let Ω ⊂ IRn be a bounded open set with C1 boundary, and let u ∈W k,p(Ω). The number
k − n
p
will be called the net smoothness of u. Let m be the integer part and let 0 ≤ γ < 1 be thefractional part of this number, so that
k − n
p= m+ γ . (5.26)
In the following, we say that a Banach space X is continuously embedded in a Banachspace Y if X ⊆ Y and there exists a constant C such that
‖u‖Y ≤ C ‖u‖X for all u ∈ X .
k
0 1 2 mm+γ
nk − _p
Figure 8.7: Computing the “net smoothness” of a function f ∈ W k,p ⊂ Cm,γ .
Theorem 7.9 (general Sobolev embeddings). Let Ω ⊂ IRn be a bounded open set withC1 boundary, and consider the space W k,p(Ω). Let m,γ be as in (5.26). Then the followingcontinuous embeddings hold.
(i) If k − np < 0 then W k,p(Ω) ⊆ Lq(Ω), with 1
q = 1p − k
n = 1n
(np − k
).
(ii) If k − np = 0, then W k,p(Ω) ⊆ Lq(Ω) for every 1 ≤ q <∞.
(iii) If m ≥ 0 and γ > 0, then W k,p(Ω) ⊆ Cm,γ(Ω).
(iv) If m ≥ 1 and γ = 0, then for every 0 ≤ γ′ < 1 one has W k,p(Ω) ⊆ Cm−1,γ′(Ω) .
Remark. Functions in a Sobolev space are only defined up to a set of measure zero. Bysaying that W k,p(Ω) ⊆ Cm,γ(Ω) one means the following. For every u ∈W k,p(Ω) there existsa function u ∈ Cm,γ(Ω) such that u(x) = u(x) for a.e. x ∈ Ω. Moreover, there exists a constantC, depending on k, p,m, γ but not on u, such that
‖u‖W k,p(Ω) ≤ C ‖u‖Cm,γ(Ω) .
125
Proof. 1. We start by proving (i). Assume k−np < 0 and let u ∈W 1,k(Ω). SinceDαu ∈ Lp(Ω)
for every |α| ≤ k, the Gagliardo-Nirenberg inequality yields
‖Dβu‖Lp∗ (Ω) ≤ C ‖u‖W k,p(Ω) |β| ≤ k − 1 .
Therefore u ∈W k−1,p∗(Ω), where p∗ is the Sobolev conjugate of p.
This argument can be iterated. Set p1 = p∗, p2.= p∗1, . . . , pj
.= p∗j−1, so that by (5.16)
1
p1=
1
p− 1
n, . . .
1
pj=
1
p− j
n.
This is meaningful as long as jp < n. Using the Gagliardo-Nirenberg inequality several times,we obtain
W k,p(Ω) ⊆ W k−1,p1(Ω) ⊆ W k−2,p2(Ω) ⊆ · · · ⊆ W k−j,pj(Ω) . (5.27)
After k steps we find that u ∈W 0,pk(Ω) = Lpk(Ω), with 1pk
= 1p − k
n . Hence pk = q, as statedin (i).
2. In the special case kp = n, repeating the above argument after k − 1 steps we obtain
u ∈W 1,n(Ω) ⊆W 1,n−ε(Ω)
for every ε > 0. Using the Gagliardo-Nirenberg inequality once again, we obtain
u ∈ W 1,n−ε(Ω) ⊆ Lq(Ω) q =n(n− ε)n− (n− ε) =
n2 − εnε
.
Since ε > 0 was arbitrary, this proves (ii).
3. To prove (iii), assume that m ≥ 0 and γ > 0 and let u ∈W k,p(Ω). We use the inequalities(5.27), choosing j to be the smallest integer such that pj > n. We thus have
1
p− j + 1
n<
1
n<
1
pj=
1
p− j
n, u ∈W k−j,pj(Ω) .
Hence, for every multi-index α with |α| ≤ k − j − 1, Morrey’s inequality yields
Dαu ∈ W 1,pj(Ω) ⊆ C0,γ(Ω) with γ = 1− n
pj= 1− n
p+ j .
Since α was any multi-index with length k − j − 1, the above implies
u ∈ Ck−j−1 ,γ(Ω).
To conclude the proof of (iii), it suffices to check that
k − n
p= (k − j − 1) +
(1− n
p+ j
),
so that m = k − j − 1 is the integer part of the number k − np , while γ is its fractional part.
126
4. To prove (iv), assume that m ≥ 1 and j.= n
p is an integer. Let u ∈ W k,p(Ω), and fix anymulti-index with |α| ≤ j−1. Using the Gagliardo-Nirenberg inequality as in step 2, we obtain
Dαu ∈ W k−j,p(Ω) ⊆ W 1,q(Ω)
for every 1 < q <∞. Hence, by Morrey’s inequality
Dαu ∈W 1,q(Ω) ⊆ C0,1−nq (Ω).
Since q can be chosen arbitrarily large, this proves (iv).
6 Compact embeddings
Let Ω ⊂ IRn be a bounded open set with C1 boundary. In this section we study the embeddingW 1,p(Ω) ⊂ Lq(Ω) in greater detail and show that, when 1
q >1p− 1
n , this embedding is compact.
Namely, from any sequence (um)m≥1 which is bounded in W 1,p one can extract a subsequencewhich converges in Lq
As a preliminary we observe that, if p > n, then every function u ∈ W 1,p(Ω) is Holdercontinuous. In particular, if (um)m≥1 is a bounded sequence inW 1,p(Ω) then the functions umare equicontinuous and uniformly bounded. By Ascoli’s compactness theorem we can extracta subsequence (umj )j≥1 which converges to a continuous function u uniformly on Ω. Since Ωis bounded, this implies ‖umj − u‖Lq(Ω) → 0 for every q ∈ [1,∞]. This already shows that theembedding W 1,p(Ω) ⊂ Lq(Ω) is compact whenever p > n and 1 ≤ q ≤ ∞.
In the remainder of this section we thus focus on the case p < n. By the Gagliardo-Nirenberginequality, the space W 1,p(Ω) is continuously embedded in Lp∗(Ω), where p∗ = np
n−p . In turn,
since Ω is bounded, for every 1 ≤ q ≤ p∗ we have the continuous embedding Lp∗(Ω) ⊆ Lq(Ω).
Theorem 7.10 (Rellich-Kondrachov compactness theorem). Let Ω ⊂ IRn be a boundedopen set with C1 boundary. Assume 1 ≤ p < n. Then for each 1 ≤ q < p∗
.= np
n−p one has thecompact embedding
W 1,p(Ω) ⊂⊂ Lq(Ω).
Proof. 1. Using Theorem 7.6 on the extension of Sobolev functions, we can assume thatevery function um is defined on the entire space IRn, and vanishes outside some bounded openset Ω ⊂ IRn (Note: the same set Ω for every um). Let (um)m≥1 be a bounded sequence inW 1,p(Ω). Then, since q < p∗ and Ω is bounded,
‖um‖Lq(IRn) = ‖um‖Lq(Ω)
≤ C ‖um‖Lp∗ (Ω)
≤ C ′ ‖um‖W 1,p(Ω)
for some constants C,C ′. Hence the sequence um is uniformly bounded in Lq.
2. Consider the mollified functions uεm.= Jε ∗ um. We can assume that all these functions are
supported inside Ω. We claim that ‖uεm−um‖Lq(Ω)→ 0 as ε→ 0, uniformly w.r.t. m. Indeed,
127
if um is smooth, then (performing the changes of variable y′ = εy and z = x− εty)
uεm(x)− um(x) =
∫
|y′|<εJε(y
′) [um(x− y′)− um(x)] dy′
=
∫
|y|≤1J(y)[um(x− εy)− um(x)] dy
=
∫
|y|≤1J(y)
(∫ 1
0
d
dt
(um(x− εty)
)dt
)dy
= − ε∫
|y|≤1J(y)
(∫ 1
0∇um(x− εty) · y dt
)dy .
In turn, this yields
∫
Ω|uεm(x)− um(x)| dx ≤ ε
∫
|y|≤1J(y)
(∫ 1
0|∇um(x− εty)| dt
)dy dx
≤ ε
∫
Ω|∇um(z)| dz.
By approximating um in W 1,p with a sequence of smooth functions, we see that the sameestimate remains valid for all functions um ∈W 1,p(Ω). We have thus shown that
‖uεm − um‖L1(Ω)≤ ε‖∇um‖
L1(Ω)≤ εC‖um‖W 1,p(Ω)
, (6.1)
for some constant C.
3. We now use the interpolation inequality for Lp norms. Pick 0 < θ < 1 such that
1
q= θ · 1 + (1− θ) · 1
p∗.
Then‖uεm − um‖Lq(Ω)
≤ ‖uεm − um‖θL1(Ω)
· ‖uεm − um‖1−θ
Lp∗(Ω)≤ C0 ε
θ. (6.2)
for some constant C0 independent of m. Indeed, in the above expression, the L1 norm isbounded by (6.1), while the Lp∗ norm is bounded by a constant, because of the Gagliardo-Nirenberg inequality.
4. Fix any δ > 0, and choose ε > 0 small enough so that (6.2) yields
‖uεm − um‖Lq(Ω)≤ C0 ε
θ ≤ δ
2.
Recalling that uεm = Jε ∗ um, we have
‖uεm‖L∞(Ω)≤ ‖Jε‖L∞ ‖um‖L1 ≤ C1,
‖∇uεm‖L∞(Ω)≤ ‖∇Jε‖L∞ ‖um‖L1 ≤ C2,
128
where C1, C2 are constants depending on ε but not on m. The above shows that, for eachfixed ε > 0, the sequence (uεm)m≥1 is uniformly bounded and equicontinuous. By Ascoli’scompactness theorem, there exists a subsequence (uεmj
) which converges uniformly on Ω tosome continuous function u. We now have
lim supj,k→∞
‖umj − umk‖Lq(Ω)
≤ lim supj,k→∞
(‖umj − uεmj
‖Lq + ‖uεmj− u‖Lq + ‖u− uεmk
‖Lq + ‖uεmk− umk
‖Lq
)
≤ δ2 + 0 + 0 + δ
2 .
(6.3)
5. The proof is now concluded by a standard diagonalization argument. By the previous stepwe can find an infinite set of indices I1 ⊂ N such that the subsequence (um)m∈I1 satifies
lim supℓ,m→∞, ℓ,m∈I1
‖uℓ − um‖Lq(Ω) ≤ 2−1.
By induction on j = 1, 2, . . ., after Ij−1 has been constructed, we choose an infinite set ofindices Ij ⊂ Ij−1 ⊂ N such that the subsequence (um)m∈Ij satisfies
lim supℓ,m→∞, ℓ,m∈Ij
‖uℓ − um‖Lq(Ω) ≤ 2−j .
After the subsets Ij have been constructed for all j ≥ 1, again by induction on j we choosea sequence of integers m1 < m2 < m3 < · · · such that mj ∈ Ij for every j. The subsequence(umj )j≥1 satisfies
lim supj,k→∞
‖umj − umk‖Lq(Ω) = 0 .
Therefore this subsequence is Cauchy, and converges to some limit u ∈ Lq(Ω).
As a first application of the compact embedding theorem, we now prove and estimate on thedifference between a function u and its average value on the domain Ω.
Theorem 7.11 (Poincare’s inequality - II). Let Ω ⊂ IRn be a bounded, connected openset with C1 boundary, and let p ∈ [1,∞]. Then there exists a constant C such that
∥∥∥∥u− −∫
Ωu dx
∥∥∥∥Lp(Ω)
≤ C ‖∇u‖Lp(Ω) . (6.4)
Proof. In the conclusion were false, one could find a sequence of functions uk ∈W 1,p(Ω) with∥∥∥∥u− −
∫
Ωu dx
∥∥∥∥Lp(Ω)
> k ‖∇u‖Lp(Ω)
for every k = 1, 2, . . . Then the renormalized functions
vk.=
u− −∫
Ωu dx
∥∥∥∥u− −∫
Ωu dx
∥∥∥∥Lp(Ω)
129
satisfy
−∫
Ωvk dx = 0 , ‖vk‖Lp(Ω) = 1, ‖Dvk‖Lp(Ω) <
1
kk = 1, 2, . . . (6.5)
By the Rellich-Kondrachov theorem, we have the compact embedding W 1,p(Ω) ⊂⊂ Lp(Ω).Since the sequence (vk)k≥1 is bounded in W 1,p(Ω), it admits a subsequence that converges inLp(Ω) to some function v. By (6.5), the sequence of weak gradients also converges, namely∇vk → 0 in Lp(Ω). By Lemma 7.2 , the zero function is the weak gradient of the limit functionv.
We now have
−∫
Ωv dx = lim
k→∞−∫
Ωvkdx = 0.
Moreover, since ∇v = 0 ∈ Lp(Ω), By Corollary 7.1 the function v must be constant on theconnected set Ω, Hence v(x) = 0 for a.e. x ∈ Ω. But this is in contradiction with
‖v‖Lp(Ω) = limk→∞
‖vk‖Lp(Ω) = 1.
7 Differentiability properties
By Morrey’s inequality, if Ω ⊂ IRn and w ∈W 1,p(Ω) with p > n, then w coincides a.e. with aHolder continuous function. Indeed, after a modification on a set of measure zero, we have
|w(x)− w(y)| ≤ C|x− y|1−np
(∫
B(x, |y−x|)|∇w(z)|p dz
)1/p
. (7.1)
This by itself does not imply that u should be differentiable in a classical sense. Indeed, thereexist Holder continuous functions that are nowhere differentiable. However, for functions in aSobolev space a much stronger differentiability result holds.
Theorem 7.12 (almost everywhere differentiability). Let Ω ⊆ IRn and let u ∈W 1,ploc (Ω)
for some p > n. Then u is differentiable at a.e. point x ∈ Ω, and its gradient equals its weakgradient.
Proof. Let u ∈W 1,ploc (Ω). Since the weak derivatives are in Lp
loc, the same is true of the weakgradient ∇u .
= (Dx1u , . . . ,Dxnu). By the Lebesgue differentiation theorem for a.e. x ∈ Ω wehave
−∫
B(x,r)|∇u(x)−∇u(z)|p dz → 0 as r → 0 . (7.2)
If x is a point for which (7.2) holds, define the remainder
w(y).= u(y)− u(x)−∇u(x) · (y − x). (7.3)
130
Observe that w ∈W 1,ploc (Ω), hence we can apply the estimate (7.1). This yields
|w(y) − w(x)| = |w(y)| = |u(y)− u(x)−∇u(x) · (y − x)|
≤ C|x− y|1−np
(∫
B(x, |y−x|)|∇u(x)−∇u(z)|p dz
)1/p
≤ C|y − x|(−∫
B(x, |y−x|)|∇u(x)−∇u(z)|p dz
)1/p
→ 0 as |y − x| → 0.
By the definition of w at (7.3), this means that u is differentiable at x in the strong sense,and its gradient coincides with its weak gradient.
8 Problems
1. Determine which of the following functionals defines a distribution on Ω ⊆ IR.
(i) Λ(φ) =∞∑
k=1
k!Dkφ(k), with Ω = IR.
(i) Λ(φ) =∞∑
k=1
2−kDkφ(1/k), with Ω = IR.
(iii) Λ(φ) =∞∑
k=1
φ(1/k)
k, with Ω = IR.
(iv) Λ(φ) =
∫ ∞
0
φ(x)
x2dx, with Ω = ]0,∞[ .
2. Give a direct proof that, if f ∈ W 1,p( ]a, b[ ) for some a < b and 1 < p < ∞, then, bypossibly changing f on a set of measure zero, one has
|f(x)− f(y)| ≤ C |x− y|1−1p for all x, y ∈ ]a, b[ .
Compute the best possible constant C.
3. Consider the open square
Q.= (x1, x2) ; 0 < x1 < 1, 0 < x2 < 1 ⊂ IR2.
Let f ∈ W 1,1(Q) be a function whose weak derivative satisfiesDx1f(x) = 0 for a.e. x ∈ Q.Prove that there exists a function g ∈ L1([0, 1]) such that
f(x1, x2) = g(x2) for a.e. (x1, x2) ∈ Q .
131
4. Let Ω ⊆ IRn be an open set and assume f ∈ L1loc(Ω). Let g = Dx1f be the weak derivative
of f w.r.t. x1. If f is C1 restricted to an open subset Ω′ ⊆ Ω, prove that g coincides with thepartial derivative ∂f/∂x1 at a.e. point x ∈ Ω′.
5. (i) Prove that, if u ∈ W 1,∞(Ω) for some open, convex set Ω ⊆ IRn, then u coincidesa.e. with a Lipschitz continuous function.
(ii) Show that there exists a (non-convex), connected open set Ω ⊂ IRn and a function u ∈W 1,∞(Ω) that does not coincide a.e. with a Lipschitz continuous function.
6. Let Ω = B(0, 1) be the open unit ball in IRn, with n ≥ 2. Prove that the unbounded
function ϕ(x) = log log(1 + 1
|x|
)is in W 1,n(Ω).
7. Let Ω = ]0, 1[. Consider the linear map T : C1([0, 1]) 7→ IR defined by Tf = f(0).Show that this map can be continuously extended, in a unique way, to a linear functionalT :W 1,1(Ω) 7→ IR.
8. Let Ω ⊂ IRn be a bounded open set, with smooth boundary. For every continuous functionf : Ω 7→ IR, define the trace
T : C0(Ω) 7→ C0(∂Ω)by letting Tf be the restriction of f to the boundary ∂Ω. Show that the operator T cannot becontinuously extended as a map from Lp(Ω) into Lp(Ω), for any 1 ≤ p < ∞. In other words,a generic function f ∈ Lp(Ω) does not have trace on the boundary ∂Ω.
9. Determine for which values of p ≥ 1 a generic function f ∈ W 1,p(IR3) admits a traceon the x1-axis. In other words, set Γ
.= (t, 0, 0) ; t ∈ IR ⊂ IR3 and consider the map
T : C1c (IR3) 7→ L1loc(Γ), where Tf = f|Γ is the restriction of f to Γ. Find values of p such that
this map admits a continuous extension T :W 1,p(IR3) 7→ Lp(Γ).
10. When k = 0, by definition W 0,p(Ω) = Lp(Ω). If 1 ≤ p <∞ prove that W 0,p0 (Ω) = Lp(Ω)
as well. What is W 0,∞0 (Ω) ?
11. Prove that, for every 1 ≤ p < ∞ and k ≥ 1 one has W k,p0 (IRn) = W k,p(IRn). In other
words, show that C∞c (IRn) is dense in W k,p(IRn).
12. Let Ω ⊂ IRn be an open set. Let R be an n × n rotation matrix, so that R†R = I. Forany u ∈ W 1,p(Ω) show that the composite map x 7→ u(R†x) is in W 1,p(RΩ). Is it true that‖u‖W 1,p(Ω) = ‖u R†‖W 1,p(RΩ) ?
13. Let IR+.= x ∈ IR ; x > 0 and let u ∈W 1,1(IR+). Call v = Dxu the weak derivative of
132
u.
(i) Setting u(x).= u(|x|), prove that u admits a weak derivative on the entire real line, namely
Dxu(x) =
v(x) if x > 0,
−v(−x) if x < 0 .
Hence u ∈W 1,1(IR).
(ii) Next, assume u ∈ W 2,1(IR+) and let w = Dxxu ∈ L1(IR+) be the weak second-orderderivative of u. Give an example showing that, in general, the extension u defined above doesnot admit a second order weak derivative. In particular, the function w(x)
.= w(|x|) is NOT
the second order weak derivative of u on the whole domain IR.
14. Let u ∈ C1c (IRn) and fix p, q ∈ [1,∞[ . For a given λ > 0, consider the rescaled functionuλ(x)
.= u(λx).
(i) Show that there exists an exponent α, depending on n, p, such that
‖uλ‖Lp(IRn) = λα‖u‖Lp(IRn) .
(ii) Show that there exists an exponent β, depending on n, q, such that
‖∇uλ‖Lq(IRn) = λβ‖∇u‖Lq(IRn) .
(iii) Determine for which values of n, p, q one has α = β. Compare with (5.15).
15. Let Ω ⊆ IRn be a bounded open set, with C1 boundary. Let (um)m≥1 be a sequence offunctions which are uniformly bounded in H1(Ω). Assuming that ‖um−u‖L2 → 0, prove thatu ∈ H1(Ω) and
‖u‖H1 ≤ lim infm→∞
‖um‖H1 .
g
x01_n
_1
1
nff
0 1
fn
Figure 8.8: Left: the function f can be approximated in W 1,p with functions fn having compactsupport in Ω. Right: the function f can be approximated in W 1,2 with functions gn having compactsupport in Ω0.
133
16. Let Ω.= (x1, x2) ; x21 + x22 < 1 be the open unit disc in IR2, and let Ω0
.= Ω \ (0, 0)
be the unit disc minus the origin. Consider the function f(x).= 1− |x|. Prove that
f ∈ W 1,p0 (Ω) for 1 ≤ p <∞ ,
f ∈ W 1,p0 (Ω0) for 1 ≤ p ≤ 2 ,
f /∈ W 1,p0 (Ω0) for 2 < p ≤ ∞ .
17. Let Ω = (x1, x2) ; 0 < x1 < 1, 0 < x2 < 1 be the open unit square in IR2.
(i) If u ∈ H1(Ω) satisfies
meas(x ∈ Ω ; u(x) = 0
)> 0 , ∇u(x) = 0 for a.e. x ∈ Ω
prove that u(x) = 0 for a.e. x ∈ Ω.
(ii) For every α > 0, prove that there exists a constant Cα with the following property. Ifu ∈ H1(Ω) is a function such that meas(x ∈ Ω ; u(x) = 0) ≥ α, then
‖u‖L2(Ω) ≤ Cα ‖∇u‖L2(Ω) . (8.4)
18. Let (un)n≥1 be a sequence of functions in the Hilbert space H2(IR3).=W 2,2(IR3). Assume
thatlimn→∞
un(x) = u(x) for all x ∈ IR3, ‖un‖H2 ≤M for all n.
Prove that the limit function u coincides a.e. with a continuous function.
134
Chapter 9
Linear Partial Differential Equations
1 Second order elliptic equations
Let Ω ⊂ IRn be a bounded open set. Given measurable functions aij, bi, c : Ω 7→ IR, considerthe linear, second order differential operator
Lu.= −
n∑
i,j=1
(aij(x)uxi)xj +n∑
i=1
bi(x)uxi + c(x)u . (1.1)
In this chapter, we study solutions to the boundary value problem
Lu = f x ∈ Ω ,u = 0 x ∈ ∂Ω , (1.2)
where f ∈ L2(Ω) is a given function.
Throughout the following we shall assume that the coefficients of the operator L satisfy
aij , bi, c ∈ L∞(Ω) . (1.3)
To study the existence and uniqueness of solutions, an additional key assumption must beintroduced. We say that the partial differential operator L is uniformly elliptic in Ω if thereexists a constant θ > 0 such that
n∑
i,j=1
aij(x)ξiξj ≥ θ|ξ|2 for all x ∈ Ω, ξ ∈ IRn . (1.4)
Otherwise stated, for every x ∈ Ω the n × n symmetric matrix A(x) = (aij(x)) is strictlypositive definite, and its smallest eigenvalue is ≥ θ.
By a classical solution of the boundary value problem (1.2) we mean a function u ∈ C2(Ω)which satisfies the equation and the boundary conditions at every point. In general, due tothe lack of regularity in the coefficients of the equation, we do not expect that (1.2) will haveclassical solutions. A weaker concept of solution is thus needed.
135
Definition 8.1. A weak solution of (1.2) is a function u ∈ H10 (Ω) such that
∫
Ω
( n∑
i,j=1
aijuxivxj +n∑
i=1
biuxiv + cuv
)dx =
∫
Ωfv dx for all v ∈ H1
0 . (1.5)
Remark 8.1. The above equality is formally obtained from∫
Ω(Lu) v dx =
∫
Ωf v dx for all v ∈ H1
0 ,
integrating by parts. Notice that a function u ∈ H10 may not have weak derivatives of second
order. However, the integral in (1.5) is always well defined, for all u, v ∈ H10 .
Remark 8.2. The homogeneous boundary condition u = 0 on ∂Ω is incorporated in therequirement u ∈ H1
0 (Ω). More generally, given a function g ∈ H1(Ω), one can consider thenon-homogeneous boundary value problem
Lu = f x ∈ Ω ,u = g x ∈ ∂Ω , (1.6)
This is reduced to a homogeneous problem for the function u.= u− g, namely
Lu = f − Lg x ∈ Ω ,u = 0 x ∈ ∂Ω . (1.7)
A convenient way to reformulate the concept of weak solution is the following. On the Hilbertspace H1
0 (Ω), consider the bilinear form
B[u, v].=
∫
Ω
n∑
i,j=1
aij uxivxj +n∑
i=1
bi uxiv + c uv
dx . (1.8)
Then a function u ∈ H10 is a weak solution of (1.2) if
B[u, v] = (f, v)L2 for all v ∈ H10 . (1.9)
Here we use the notation
(f, g)L2.=
∫
Ωf g dx (1.10)
for the inner product in L2(Ω), to distinguish it from the inner product in H1(Ω)
(f, g)H1.=
∫
Ωf g dx+
∫
Ω
n∑
i=1
fxigxi dx . (1.11)
Remark 8.3. The minus sign in front of the second order terms in (1.1) disappears in (1.8),after a formal integration by parts. As will become apparent later, this sign is chosen so thatthe corresponding quadratic form B[u, v] can be positive definite.
136
As a first example, consider the boundary value problem
−∆u+ u = f x ∈ Ω ,
u = 0 x ∈ ∂Ω , (1.12)
Clearly, the operator −∆u .= −∑i uxixi is uniformly elliptic, because in this case the n × n
matrix A(x) = (aij(x)) is the identity matrix, for every x ∈ Ω. The existence of solutions to(1.12) is provided by
Lemma 8.1. Let Ω ⊂ IRn be a bounded open set. Then for every f ∈ L2(Ω) the boundaryvalue problem (1.12) has a unique weak solution u ∈ H1
0 (Ω). The corresponding map f 7→ uis a compact linear operator from L2(Ω) into H1
0 (Ω).
Proof. By the Rellich-Kondrachov theorem, the canonical embedding ι : H10 (Ω) 7→ L2(Ω) is
compact. Hence its dual operator ι∗ is also compact. Since H10 and L2 are Hilbert spaces,
they can be identified with their duals. We thus obtain the following diagram:
H10 (Ω)
ι−→ L2(Ω)
H10 (Ω) = [H1
0 (Ω)]∗ ι∗←− [L2(Ω)]∗ = L2(Ω) .
(1.13)
For each f ∈ L2(Ω), the definition of dual operator yields
(ι∗f, v)H1 = (f, ιv)L2 = (f, v)L2 for all f ∈ L2(Ω), v ∈ H10 (Ω) .
By (1.11) this means that ι∗f is a weak solution to (1.12).
1.1 Existence of solutions
As a first step, we study solutions to the elliptic boundary value problem
L0u = f x ∈ Ω ,u = 0 x ∈ ∂Ω , (1.14)
assuming that the differential operator L0 contains only second order terms:
L0u.= −
n∑
i,j=1
(aij(x)uxi)xj . (1.15)
We recall that a weak solution of (1.14) is a function u ∈ H10 (Ω) such that
B0[u, v] = (f, v)L2 = (ι∗f, v)H1 for all v ∈ H10 (Ω) , (1.16)
where B0 : H10 ×H1
0 7→ IR is the continuous bilinear form
B0[u, v].=
∫
Ω
n∑
i,j=1
aijuxivxj dx . (1.17)
137
Theorem 8.1 (solution of a linear elliptic boundary value problem - I). Let Ω ⊂ IRn
be a bounded open set. Let the operator L0 in (1.15) be uniformly elliptic, with coefficientsaij ∈ L∞(Ω). Then, for every f ∈ L2(Ω), the boundary value problem (1.14) has a unique weaksolution u ∈ H1
0 (Ω). The corresponding solution operator, which we denote as L−10 : f 7→ u is
a compact linear operator from L2(Ω) into H10 (Ω).
Proof. The existence and uniqueness of weak solutions to the elliptic boundary value problem(1.14) will be achieved by checking that the bilinear form B0 satisfies all the assumptions ofthe Lax-Milgram theorem.
1. The continuity of B0 is clear. Indeed,
∣∣∣B0[u, v]∣∣∣ ≤
n∑
i,j=1
∫
Ω|aijuxivxj | dx ≤
n∑
i,j=1
‖aij‖L∞‖uxi‖L2‖vxj‖L2 ≤ C ‖u‖H1‖v‖H1 .
2. We claim that B0 is strictly positive definite, i.e. there exists β > 0 such that
B0[u, u] ≥ β ‖u‖2H1(Ω) for all u ∈ H10 (Ω) . (1.18)
Indeed, since Ω is bounded, Poincare’s inequality yields the existence of a constant κ suchthat
‖u‖2L2(Ω) ≤ κ
∫
Ω|∇u|2 dx for all u ∈ H1
0 (Ω) .
On the other hand, the uniform ellipticity condition implies
B0[u, u] =
∫
Ω
n∑
i,j=1
aijuxiuxj dx ≥∫
Ωθ
n∑
i=1
u2xidx = θ
∫
Ω|∇u|2 dx .
Together, the two above inequalities yield
‖u‖2H1(Ω) = ‖u‖2L2(Ω) + ‖∇u‖2L2 ≤ (κ+ 1) ‖∇u‖2
L2 ≤ κ+ 1
θB0[u, u] .
This proves (1.18) with β = θ/(κ+ 1).
3. By the Lax-Milgram theorem, for every f ∈ H10 (Ω) there exists a unique element u ∈ H1
0
such thatB0[u, v] = (f , v)H1 for all v ∈ H1
0 (Ω) . (1.19)
Moreover, the map Λ : f 7→ u is continuous, namely
‖u‖H1 ≤ β−1 ‖f‖H1 .
Choosing f.= ι∗f ∈ H1
0 (Ω), defined at (1.13), we thus achieve
B0[u, v] = (ι∗f, v)H1 = (f, v)L2 for all v ∈ H10 (Ω) . (1.20)
By definition, u is a weak solution of (1.14).
138
4. To prove that the solution operator L−10 : f 7→ u is a compact, consider the the diagram
L2(Ω)ι∗−→ H1
0 (Ω)Λ−→ H1
0 (Ω) .
By Lemma 8.1, the linear operator ι∗ is compact. Moreover, Λ is continuous. Therefore thecomposition L−1
0 = Λ ι∗ is compact.
1.2 Eigenfunction representation
We now study in more detail the structure of the solution operator L−10 for the elliptic boundary
value problem (1.14)-(1.15).
Lemma 8.2. In the setting of Theorem 8.1, the linear operator L−10 : L2(Ω) 7→ L2(Ω) is
compact, one-to-one, and self-adjoint.
Therefore, the space L2(Ω) admits an orthonormal basis φk ; k ≥ 1 consisting of eigenfunc-tions of L−1
0 . Therefore one has the representation
L−10 f =
∞∑
k=1
λk(f, φk)L2 φk . (1.21)
The corresponding eigenvalues satisfy
limk→∞
λk = 0 , λk > 0 for all k ≥ 1. (1.22)
Proof. 1. Because of the compact embedding H10 (Ω) ⊂⊂ L2(Ω), it is clear that L−1
0 is acompact linear operator from L2(Ω) into itself.
2. If u = L−10 f = 0, then
0 = B0[u, v] = (f, v)L2
for every v ∈ H10 (Ω). In particular, for every test function φ ∈ C∞c (Ω) we have
∫
Ωf φ dx = 0 .
This implies f(x) = 0 for a.e. x ∈ Ω. Hence Ker(L−10 ) = 0 and the operator is one-to-one.
3. To prove that L−10 is self-adjoint, assume
f, g ∈ L2(Ω) , u = L−10 f , v = L−1
0 g .
Recalling the definition of weak solution at (1.5) , this implies u, v ∈ H10 (Ω) and
(L−10 f, g)L2 =
∫
Ωu g dx =
∫
Ω
∑
i,j
aijuxivxj dx =
∫
Ωf v dx = (f, L−1
0 g)L2 .
139
Note that the second inequality follows from the fact the v = L−10 g, using u ∈ H1
0 (Ω) as a testfunction. The third inequality follows from the fact the u = L−1
0 f , using v ∈ H10 (Ω) as a test
function.
4. By the previous steps, we can now apply Theorem 5.3, providing the existence of anorthonormal basis of eigenvectors of L−1
0 . This yields (1.21).
By the compactness of the operator L−10 , the eigenvalues satisfy λk → 0. Finally, choosing
v = φk as test function in (1.16), one obtains
B0[λkφk, φk] = (φk, φk)L2 = 1.
Since the quadratic form B0 is strictly positive definite, we conclude
λk =1
B0[φk, φk]> 0 .
Example 8.1. Let Ω.= ]0, π[⊂ IR and L0u = −uxx. Given f ∈ L2(]0, π[), consider the elliptic
boundary value problem−uxx = f 0 < x < π ,u(0) = u(π) = 0 .
(1.23)
As a first step, we compute the eigenfunctions of L0. Solving the boundary value problem
−uxx = µu , u(0) = u(π) = 0 .
we find the eigenvalues and the normalized eigenfuntions
µk = k2 , φk(x) =
√2
πsin kx .
Of course, the inverse operator L−10 has the same eigenfunctions φk, with eigenvalues λk =
1/k2.
In this special case, the formula (1.21) yields the well known representation of solutions of(1.23) in terms of a Fourier sine series:
u(x) = L−10 f =
∞∑
k=1
λk (f, φk)L2 φk =∞∑
k=1
1
k2
(∫ π
0f(y)
√2
πsin ky dy
)√2
πsin kx
=∞∑
k=1
2
πk2
(∫ π
0f(y) sin ky dy
)sin kx .
1.3 General linear elliptic operators
The existence and uniqueness result stated in Theorem 8.1 relied on the fact that, by Poincare’sinequality, the bilinear form B0 in (1.16) is strictly positive definite on the space H1
0 (Ω).Moreover, the assumption bi ≡ 0 guarantees that B0 is symmetric.
140
These properties no longer hold for the more general bilinear form B in (1.8). For example, ifthe function c(x) is large and negative, one may find some u ∈ H1
0 (Ω) such that B[u, u] < 0.
Example 8.2. Consider the one-dimensional open interval Ω.= ]0, π[ . The operator
Lu.= − uxx − 4u
is uniformly elliptic on Ω. However, the corresponding bilinear form
B[u, v] =
∫ π
0ux vx − 4uv dx
is not positive definite on H10 (Ω). For example, taking u(x) = sinx we find
B[u, u] =
∫ π
0cos2 x− 4 sin2 x dx = − 3π
2.
Observe that, if we take f(x) = sin 2x, then the boundary value problem
−uxx − 4u = sin 2x x ∈ ]0, π[ ,
u(0) = u(π) = 0(1.24)
has no solutions. Indeed, taking v(x) = sin 2x, for every u ∈ H10 (Ω) an integration by parts
yields
B[u, v] =
∫ π
0uxvx − 4uv dx =
∫ π
0
(2ux cos 2x− 4u sin 2x
)dx = 0 6=
∫ π
0sin2 2x dx .
Therefore, it is not possible to satisfy the identity (1.9) for all v ∈ H10 (Ω). In this connection,
one should also observe that the corresponding homogeneous problem
−uxx − 4u = 0 x ∈ ]0, π[ ,
u(0) = u(π) = 0(1.25)
admits infinitely many solutions: u(x) = c sin 2x, where c ∈ IR is any constant.
We now study the existence of weak solutions to the more general boundary value problem(1.2). This can be written as
L0u+ L1u = f x ∈ Ω ,
u = 0 x ∈ ∂Ω , (1.26)
where L0 is defined at (1.15) and
L1u.=
n∑
i=1
bi(x)uxi + c(x)u . (1.27)
Observe that, for every u ∈ H10 (Ω), we have L1u ∈ L2(Ω). A function u ∈ H1
0 (Ω) is a weaksolution to (1.26) if and only if
B0[u, v] = (f, v)L2 − (L1u, v)L2 for all v ∈ H10 (Ω) . (1.28)
141
Therefore, u satisfies (1.28) if and only if
u = L−10 (f − L1u) . (1.29)
We write (1.29) in the form(I +K)u = g (1.30)
whereg
.= L−1
0 f , Ku.= L−1
0 L1u .
Observe that K : H10 (Ω) 7→ H1
0 (Ω) is a compact linear operator, obtained as the compositionof the bounded operator L1 with the compact operator L−1
0 . Namely
H10 (Ω)
L1−→ L2(Ω)L−10−→ H1
0 (Ω).
Therefore, by Theorem 5.1, the Fredholm alternative thus applies to the equation (1.30):
Either
(i) for every h ∈ H10 the equation (I +K)u = h has a unique solution u ∈ H1
0 (Ω),
or else
(ii) the equation u+Ku = 0 has a nontrivial solution u ∈ H10 (Ω).
The identity u + Ku = 0 means that u ∈ H10 (Ω) is a weak solution to the homogeneous
boundary value problem Lu = 0 x ∈ Ω ,u = 0 x ∈ ∂Ω . (1.31)
For the boundary value problem defined by (1.1)-(1.2), we thus have
Theorem 8.2 (solutions of a linear elliptic boundary value problems - II). LetΩ ⊂ IRn be a bounded open set and let the operator L in (1.1) be uniformly elliptic, withaij , bi, c ∈ L∞(Ω). Then
(i) the elliptic boundary value problem (1.2) has a unique solution for every f ∈ L2(Ω)
if and only if
(ii) the homogeneous boundary value problem (1.31) has the only solution u(x) ≡ 0.
2 Parabolic equations
Let Ω ⊂ IRn be a bounded open set and let L be the operator in (1.1). We shall assumethat the coefficients satisfy the uniform ellipticity condition (1.4) and the somewhat strongerregularity conditions
aij ∈W 1,∞(Ω) , bi, c ∈ L∞(Ω) . (2.1)
142
In this section we study the parabolic initial-boundary value problem
ut + Lu = 0 t > 0 , x ∈ Ω ,u(t, x) = 0 t > 0 , x ∈ ∂Ω ,u(0, x) = g(x) x ∈ Ω ,
(2.2)
We reformulate (2.2) as a Cauchy problem in the Hilbert space X = L2(Ω), namely
d
dtu = Au, u(0) = g . (2.3)
for a suitable (unbounded) linear operator A : L2(Ω) 7→ L2(Ω). More precisely
A.= − L , Dom(A) =
u ∈ H1
0 (Ω) ; Lu ∈ L2(Ω). (2.4)
In other words, u ∈ Dom(A) if u is the solution to the elliptic boundary value problem (1.2),for some f ∈ L2(Ω). In this case, Au = −f .
Our eventual goal is to construct solutions of (2.3) using semigroup theory. We first considerthe case where the operator L is strictly positive definite on H1
0 (Ω). More precisely, we assumethat there exists β > 0 such that the bilinear form B : H1
0 (Ω)×H10 (Ω) 7→ IR defined at (1.8)
is strictly positive definite: there exists β > 0 such that
B[u, u] ≥ β‖u‖2H1 for all u ∈ H10 (Ω) , (2.5)
Notice that this is certainly true if bi ≡ 0 and c ≥ 0.
Theorem 8.3 (semigroup of solutions of a parabolic equation - I). Assume that theoperator L in (1.1) satisfies the regularity conditions (2.1) and the uniform ellipticity condition(1.4). Moreover, assume that the corresponding bilinear form B in (1.8) is positive definite,so that (2.5) holds.
Then the operator A = −L generates a contractive semigroup St ; t ≥ 0 of linear operatorson L2(Ω).
Proof. According to Theorem 6.6, to prove that A generates a contraction semigroup onX = L2(Ω) we need to check:
(i) Dom(A) is dense in L2(Ω).
(ii) The graph of A is closed.
(iii) Every real number λ > 0 is in the resolvent set of A, and ‖(λI −A)−1‖ ≤ 1
λ.
1. To prove (i), we observe that, if ϕ ∈ C2c (Ω), then the regularity assumptions (2.1) implyf.= Lϕ ∈ L2(Ω). This proves that Dom(A) contains the subspace C2c (Ω) and is thus dense in
L2(Ω).
2. If (2.5) holds, then, by the Lax-Milgram theorem, for every f ∈ L2(Ω) there exists a uniqueu ∈ H1
0 (Ω) such that
B[u, v] = (f, v)L2 for all v ∈ H10 (Ω) .
143
The map f 7→ u.= L−1f is a bounded linear operator from L2(Ω) into L2(Ω).
We now observe that the pair of functions (u, f) is in the graph of A if and only if (−f, u) liesin the graph L−1. Since L−1 is a continuous operator, its graph is closed. Hence the graph ofA is closed as well.
3. According to the definition of A, to prove (iii) we need to show that, for every λ > 0, theoperator λI−A has a bounded inverse with operator norm ‖(λI−A)−1‖ ≤ 1/λ. Equivalently,for every f ∈ L2(Ω), we need to show that the problem
λu+ Lu = f x ∈ Ω ,
u = 0 x ∈ ∂Ω , (2.6)
has a weak solution satisfying
‖u‖L2 ≤ 1
λ‖f‖L2 . (2.7)
By the Lax-Milgram theorem, there exists a unique u ∈ H10 (Ω) such that
(λu, v)L2 +B[u, v] = (f, v)L2 for all v ∈ H10 (Ω). (2.8)
Taking v = u in (2.8) we obtain
λ‖u‖2L2 +B[u, u] = (f, u)L2 ≤ ‖f‖L2 · ‖u‖L2 .
Since we are assuming B[u, u] ≥ 0, this yields
λ ‖u‖L2 ≤ ‖f‖L2 ,
proving (2.7).
By the generation theorem, the linear operator A generates a contractive semigroup.
2.1 Representation of solutions in terms of eigenfunctions
Consider the special case where L = L0 is the operator in (1.15), containing only second orderterms. In this case, by Poincare’s inequality, the bilinear form B0 in (1.17) is strictly positivedefinite and Theorem 8.3 can be applied.
Relying on Lemma 8.2, one can provide a representation of the semigroup trajectories, interms of the orthonormal basis φk ; k ≥ 1 consisting of eigenfunctions of the compactself-adjoint operator L−1
0 . By construction, for every k ≥ 1 one has
L−10 φk = λkφk ,
where λk > 0 is the corresponding eigenvalue. Therefore
φk ∈ Dom(L0) , L0φk = µk φk µk =1
λk. (2.9)
Notice that λk → 0 and µk → +∞, as k →∞. For every k ≥ 1, the function
u(t).= e−µktφk
144
provides a C1 solution to the Cauchy problem
d
dtu(t) = − L0u(t) , u(0) = φk .
Hence, by the uniqueness of semigroup trajectories, one must have
Stφk = e−µktφk .
By linearity, for any coefficients c1, . . . , cN ∈ IR one has
St
(N∑
k=1
ckφk
)=
N∑
k=1
cke−µktφk .
Since St is a bounded linear operator, decomposing an arbitrary function g ∈ L2(Ω) along theorthonormal basis φk ; k ≥ 1, we thus obtain
Stg =∞∑
k=1
e−µkt(g, φk)L2 φk . (2.10)
The above representation of semigroup trajectories is valid for every g ∈ L2(Ω) and everyt ≥ 0.
Lemma 8.3. Let L = L0 be the operator at (1.15). Then for every g ∈ L2(Ω) the formula(2.10) defines a map t 7→ ut = Stg from [0,∞[ into L2(Ω). This map is continuous fort ∈ [0,∞[ and continuously differentiable for t > 0. Moreover, u(t) ∈ Dom(L0) ⊆ H1
0 (Ω) forevery t > 0 and there holds
d
dtu(t) = L0u(t) for all t > 0 . (2.11)
Proof. 1. Let g ∈ L2(Ω). Since µk > 0 for every k, it is clear that
∣∣∣e−µkt(g, φk)L2
∣∣∣2≤ (g, φk)
2L2 .
Therefore, ∑
k≥1
∣∣∣e−µkt(g, φk)L2
∣∣∣2≤
∑
k≥1
(g, φk)2L2 = ‖g‖2
L2 < ∞ .
Hence the series in (2.10) is convergent, uniformly for t ≥ 0. In particular, since the partialsums are continuous functions of time, the map t 7→ Stg is continuous as well.
2. We claim that, even if g /∈ H10 (Ω), one always has
Stg ∈ Dom(L0) ⊆ H10 (Ω) for all t > 0 . (2.12)
Indeed, a function u =∑
k ckφk lies in Dom(L0) if and only if the coefficients ck satisfy∑
k
(ckµk)2 < ∞ .
145
In the case where ck(t).= e−µkt(g, φk)L2 , we have the estimate
∑
k
(µk ck(t))2 ≤ sup
k
(µk e
−µkt)2·∑
k
(g, φk)2L2 . (2.13)
An elementary calculation now shows that, for ξ ≥ 0 and t fixed, the function ξ 7→ ξ e−tξ
attains its global maximum at ξ = 1/t. In particular,
µk e−µkt ≤ max
ξ≥0ξe−tξ =
1
et.
Using this bound in (2.13) we obtain
∞∑
k=1
(µkck(t)
)2≤ 1
e2 t2‖g‖2
L2 .
Hence the series defining L0u(t) is convergent. This implies u(t) ∈ Dom(L0), for each t > 0.
3. Differentiating the series (2.10) term by term, and observing that the series of derivativesis also convergent, we achieve (2.11).
Example 8.3. As in Example 8.1, let Ω.=]0, π[⊂ IR and L0u = −uxx. Given g ∈ L2(]0, π[),
consider the parabolic initial-boundary value problem
ut = uxx t > 0 , 0 < x < π ,u(0, x) = g(x) 0 < x < π ,u(t, 0) = u(t, π) = 0 .
In this special case, the formula (2.10) yields the solution as the sum of a Fourier sine series:
u(t, x) =∞∑
k=1
2
πe−k2t
(∫ π
0g(y) sin ky dy
)sin kx .
2.2 More general operators
To motivate the following construction, we begin with a finite dimensional example. Let L bean n× n matrix and consider the linear ODE on IRn
d
dtx(t) = − Lx(t) . (2.14)
If L is positive definite, i.e. if 〈Lx, x〉 ≥ 0 for all x ∈ IRn, then −L generates a contractivesemigroup. Indeed
d
dt|x(t)|2 = 2
⟨d
dtu(t) , u(t)
⟩= 2 〈−Lx(t), x(t)〉 ≤ 0 ,
showing that the Euclidean norm of a solution does not increase in time.
146
Next, let L be an arbitrary matrix. We can then find a number γ ≥ 0 large enough so thatthe matrix L+γI is positive definite (and hence it generates a contractive semigroup). In thiscase, if x(t) = e−tLx(0) is a solution to (2.14), writing −L = γI − (L+ γI) one obtains
|x(t)| = |e−Ltx(0)| = |eγI−(L+γI)tx(0)| = eγt|e−(L+γI)tx(0)| ≤ eγt|x(0)|. (2.15)
According to (2.15), the operator −L generates a semigroup of type γ.
We shall work out a similar construction in the case where L is a general elliptic operator, asin (1.1), and the corresponding bilinear form B[u, v] in (1.8) is not necessarily positive definite.As a first step we show that, by choosing a constant γ > 0 large enough, the bilinear form
Bγ [u, v] = B[u, v] + γ(u, v)L2 (2.16)
is strictly positive definite on H10 (Ω).
Remark 8.4. By choosing the constant γ > 0 large enough, it is a completely trivial matterto show that the operator
Bγ.= B[u, v] + γ(u, v)H1
is strictly positive definite on H10 (Ω). Note, however, that in (2.16) we are adding the much
weaker term γ(u, v)L2 .
Lemma 8.4 (estimates on elliptic operators). Let the operator L in (1.1) be uniformlyelliptic, with coefficients aij, bi, c ∈ L∞(Ω). Then there exist constants α, β, γ > 0 such that
|B[u, v]| ≤ α ‖u‖H1‖v‖H1 , (2.17)
β‖u‖2H1 ≤ B[u, u] + γ ‖u‖2L2 . (2.18)
for all u, v ∈ H10 (Ω).
Proof. 1. The boundedness of the bilinear form B : H10 ×H1
0 7→ IR follows from
|B[u, v]| =
∣∣∣∣∣∣
∫
Ω
( n∑
i,j=1
aijuxivxj +n∑
i=1
biuxiv + cuv
)dx
∣∣∣∣∣∣
≤n∑
i,j=1
‖aij‖L∞‖uxi‖L2‖vxj‖L2 +n∑
i=1
‖bi‖L∞‖uxi‖L2‖v‖L2 + ‖c‖L∞‖u‖L2‖v‖L2
≤ α ‖u‖H1 ‖v‖H1 .
2. Toward the second estimate, using the ellipticity condition (1.4) and the elementary in-
147
equality ab ≤ θ2a
2 + 1θ b
2, we obtain
θn∑
i=1
‖uxi‖2L2 = θ
∫
Ω
n∑
i=1
u2xidx ≤
∫
Ω
n∑
i,j=1
aijuxiuxj dx
= B[u, u]−∫
Ω
( n∑
i=1
biuxiu+ cu2)dx
≤ B[u, u] +n∑
i=1
‖bi‖L∞‖uxi‖L2‖u‖L2 + ‖c‖L∞‖u‖2L2
≤ B[u, u] +
( n∑
i=1
θ
2‖uxi‖2L2 +
n∑
i=1
1
θ‖bi‖2L∞‖u‖2
L2
)+ ‖c‖L∞‖u‖2
L2
Therefore,
B[u, u] ≤ θ
2
n∑
i=1
‖uxi‖2L2 − C‖u‖2L2 for all u ∈ H10 (Ω)
for a suitable constant C. Taking β = θ/2 and γ = C + θ, the inequality (2.18) is satisfied.
Motivated by the previous lemma, we define
Lγu.= Lu+ γu , Bγ [u, v]
.= B[u, v] + γ(u, v)L2 .
The parabolic equation in (2.2) can thus be written as
ut = − Lγu+ γu
By the previous analysis, the operator Aγ.= −(L+ γI) generates a contractive semigroup of
linear operators, say S(γ)t ; t ≥ 0. Therefore, the operator A
.= −L = γI −Lγ defined as in
(2.4) generates a semigroup of type γ. Namely St ; t ≥ 0, with
St = eγt S(γ)t t ≥ 0.
Summarizing the above analysis, we have
Theorem 8.4 (semigroup of solutions of a parabolic equation - II). Let Ω ⊂ IRn be abounded open set. Assume that the operator L in (1.1) satisfies the regularity conditions (2.1)and the uniform ellipticity condition (1.4).
Then the operator A defined at (2.4) generates a semigroup St ; t ≥ 0 of linear operatorson L2(Ω).
Having constructed a semigroup St ; t ≥ 0 generated by the operator A, one needs tounderstand in which sense a trajectory of the semigroup t 7→ u(t) = Stf provides a solutionto the parabolic equation (2.2). In the case where L = L0 is the symmetric operator defined
148
at (1.15), the representation (2.10) provides all the needed information. Indeed, according toLemma 8.3, for every initial data g ∈ L2(Ω) the solution t 7→ u(t) = Stg is a C1 map, whichtakes values in Dom(L0) and satisfies (2.11) for every t > 0.
A similar result can be proved for general elliptic operators of the form (1.1). However, thisanalysis is beyond the scope of the present notes. Here we shall only make a few remarks:
(1) - Initial condition. The map t 7→ u(t).= Stg is continuous from [0,∞[ into L2(Ω)
and satisfies u(0) = g. The initial condition in (2.2) is thus satisfied as an identity betweenfunctions in L2(Ω).
(2) - Regular solutions. If g ∈ Dom(A), then u(t) = Stg ∈ Dom(A) for all t ≥ 0. Moreover,the map t 7→ u(t) is continuously differentiable and satisfies the O.D.E. (2.3) at every timet > 0. Since Dom(A) ⊂ H1
0 (Ω), this also implies that u(t) satisfies the correct boundaryconditions, for all t ≥ 0.
(3) - Distributional solutions. Given a general initial condition f ∈ L2(Ω), one canconstruct a sequence of initial data fm ∈ Dom(A) such that ‖fm − f‖L2 → 0 as m→ ∞. Inthis case, if the semigroup is of type γ, we have
‖Stfm − Stf‖L2 ≤ eγt‖fm − f‖L2 .
Therefore the trajectory t 7→ u(t) = Stf is the limit of a sequence of C1 solutions t 7→ um(t).=
Stfm. The convergence is uniform for t in bounded sets.
Relying on these approximations, we now show that the function u = u(t, x) provides a solutionto the parabolic equation
ut =n∑
i,j=1
(aij(x)uxi)xj −n∑
i=1
bi(x)uxi − c(x)u (2.19)
in distributional sense. Namely, for every test function ϕ ∈ C∞c (Ω× ]0,∞[), one has
∫∫ uϕt +
n∑
i,j=1
u(aijϕxj )xi +n∑
i=1
u(biϕ)xi − c uϕdx dt = 0 . (2.20)
To prove (2.20), consider a sequence of initial data fm ∈ Dom(A) such that ‖fm − f‖L2 → 0.Then, for any fixed time interval [0, T ], the trajectories t 7→ um(t)
.= Stfm converge to the
continuous trajectory t 7→ u(t) = Stf in C0([0, T ] ; L2(Ω). Since each um is clearly a solutionin distributional sense, writing
∫∫ umϕt +
n∑
i,j=1
um(aijϕxj )xi +n∑
i=1
um(biϕ)xi − c umϕdx dt = 0 ,
and letting ε→ 0 we obtain (2.20).
149
3 Hyperbolic equations
Consider a linear hyperbolic initial-boundary value problem, of the form
utt + L0u = 0 t ∈ IR , x ∈ Ω ,u(t, x) = 0 t ∈ IR , x ∈ ∂Ω ,u(0, x) = f(x) , ut(0, x) = g(x) x ∈ Ω ,
(3.21)
where L0 is the second order elliptic operator
L0u = −n∑
i,j=1
(aij(x)uxi)xi . (3.22)
We assume that the coefficients aij satisfy
aij = aji ∈W 1,∞(Ω),n∑
i,j=1
aij(x)ξiξj ≥ θ|ξ|2 for all x ∈ Ω, ξ ∈ IRn . (3.23)
According to Lemma 8.2, the space L2(Ω) admits an orthonormal basis φk ; k ≥ 1 consistingof eigenfunctions of L0, so that
φk ∈ Dom(L0) , L0φk = µk φk , (3.24)
for a sequence of eigenvalues µk → +∞ as k →∞.
It is convenient to reformulate (3.21) as a first order system, setting v.= ut. On the product
spaceX
.= H1
0 (Ω)× L2(Ω) , (3.25)
we thus consider the evolution problem
d
dt
(uv
)=
(0 I−L0 0
)(uv
),
(uv
)(0) =
(gh
). (3.26)
A semigroup of solutions of (3.26) will be constructed using the eigenfunctions φk . In thespecial case where
f = akφk , g = bkφk , (3.27)
for a given k ≥ 1 and ak, bk ∈ IR, an explicit solution of (3.26) is found in the form
u(t) = a(t)φk , v(t) = ut(t) = a′(t)φk ,
where the coefficient a(t) satisfies
a′′(t) + µk a(t) = 0 , a(0) = ak , a′(0) = bk .
An elementary computation yields
a(t) = ak cos(õk t) +
bkõk
sin(õk t) .
Hence u(t)
v(t)
=
cos(õk t)
1õk
sin(õk t)
−√µk sin(√µk t) cos(
õk t)
ak φk
bk φk
. (3.28)
150
Observe that t 7→ ( u(t), v(t) ) is a continuously differentiable map from IR into H10 (Ω)×L2(Ω)
which satisfies the initial conditions and the differential equation in (3.26).
By linearity, more general solutions can be obtained by taking linear combinations of solutionsof the form (3.28). This motivates the definition
St
g
h
.
=∞∑
k=1
cos(õk t)
1õk
sin(õk t)
−√µk sin(√µk t) cos(
õk t)
(g, φk)L2 φk
(h, φk)L2 φk
. (3.29)
Theorem 8.5 (solutions of a linear hyperbolic problem). In the above setting, theformula (3.29) defines a strongly continuous group of bounded linear operators St ; t ∈ IRon the space X
.= H1
0 (Ω) × L2(Ω). Each operator St : X 7→ X is an isometry w.r.t. theequivalent norm
‖(u, v)‖X .=(B0[u, u] + ‖v‖2L2
)1/2. (3.30)
Remark 8.5. Consider an elastic membrane which occupies a region Ω in the plane. Letu(t, x) denote the vertical displacement of a point on this membrane, at time t. Then thesquare of the norm ‖(u, ut)‖2X can be interpreted as the total energy of the vibrating membrane.Indeed, the term B0[u, u] describes an elastic potential energy, while ‖ut‖2L2 yields the kineticenergy.
Proof of Theorem 8.5.
1. The equivalence between the norm (3.30) and the standard product norm
‖(u, v)‖H1×L2.=(‖u‖2H1 + ‖v‖2L2
)1/2
is an immediate consequence of (1.18).
2. Let the functions f, g be given by
f =∞∑
k=1
akφk , g =∞∑
k=1
bkφk(ak = (f, φk)L2 , bk = (g, φk)L2
).
Then
B0[f, f ] =∞∑
k=1
(µk akφk , akφk
)L2
=∞∑
k=1
µk a2k , ‖g‖2
L2 =∞∑
k=1
b2k .
Therefore
(fg
)∈ X if and only if
∞∑
k=1
µk a2k < ∞ ,
∞∑
k=1
b2k < ∞ . (3.31)
151
At any time t ∈ IR, we have
∥∥∥∥St(gh
)∥∥∥∥2
X
=∞∑
k=1
µk a2k(t) +
∞∑
k=1
bk(t)2 , (3.32)
where, according to (3.29),
ak(t) = cos(õk t)ak +
1õk
sin(õk t) bk ,
bk(t) = a′k(t) = −√µk sin(√µk t) ak + cos(
õk t) bk .
(3.33)
If
(fg
)∈ X, then for every t ∈ IR the series defining St
(fg
)in (3.29) is convergent. Using
(3.33) inside (3.32), we obtain
∥∥∥∥St(fg
)∥∥∥∥2
X=
∞∑
k=1
µk a2k +
∞∑
k=1
b2k =
∥∥∥∥(fg
)∥∥∥∥2
X. (3.34)
This shows that each linear operator St is an isometry w.r.t. the equivalent norm ‖ · ‖X .
3. It is easy to check that the family of linear operators St ; t ∈ IR satisfies the groupproperties
S0
(fg
)=
(fg
), StSs
(fg
)= St+s
(fg
)t, s ∈ IR .
To complete the proof, we need to show that, for f, g fixed, the map t 7→ St
(fg
)is continuous
from [0,∞[ into X. But this is clear, because the above map is the uniform limit of thecontinuous maps
t 7→ St
(fmgm
), fm
.=
m∑
k=1
(f, φk)L2 φk , gm.=
m∑
k=1
(g, φk)L2 φk , (3.35)
as m→∞.
Having constructed the group of operators St ; t ∈ IR, one needs to understand in whichsense the trajectories
t 7→(u(t)v(t)
)= St
(fg
)(3.36)
provide a solution to the hyperbolic initial-boundary value problem (3.21).
(1) - Initial and boundary conditions. Consider an arbitrary initial data
(fg
)∈ X =
H10 (Ω)× L2(Ω). By the continuity of the map (3.36), it follows that
‖u(t)− f‖H1 → 0 , ‖v(t)− g‖L2 → 0 as t→ 0 .
152
Hence the initial conditions in (3.21) are satisfied.
Moreover, by the definition of the space X, we have u(t) ∈ H10 (Ω) for all t ≥ 0. Hence the
boundary conditions in (3.21) are also satisfied.
(2) - Regular solutions. If both functions f and g are finite linear combinations of theeigenfunctions φk, then the corresponding trajectory (3.36) is a continuously differentiablemap from [0, ∞[ into X, and it satisfies the ODE (3.26) at all times t > 0.
(3) - Distributional solutions. Given general initial data f ∈ H10 (Ω) and g ∈ L2(Ω), one
can construct a sequence of approximations fm, gm as in (3.35). so that ‖fm−f‖H1 → 0, ‖gm−g‖L2 → 0 as m → ∞. The corresponding semigroup trajectories t 7→
(um(t)vm(t)
)= St
(fmgm
)
converge to the trajectory (3.36), uniformly for t ∈ IR.
Relying on these approximations, we now show that the function u = u(t, x) provides a solutionto the parabolic equation
utt =n∑
i,j=1
(aij(x)uxi)xj
in distributional sense. Indeed, consider any test function ϕ ∈ C∞c ( ]0,∞[×Ω ). Since eachfunction um = um(t, x) is a distributional solution of (3.21), there holds
∫∫ um ϕtt +
n∑
i,j=1
aij(x) (um)xiϕxj
dx dt = 0 .
Letting m → ∞ and using the convergence ‖um(t) − u(t)‖H1 → 0, uniformly for t ∈ IR, weobtain ∫∫
uϕtt +n∑
i,j=1
aij(x)uxiϕxj
dx = 0 .
Example 8.4. As in Example 8.1, let Ω.= ]0, π[⊂ IR and L0u = −uxx. Given f ∈ H1
0 (Ω)and g ∈ L2(]0, π[), consider the hyperbolic initial-boundary value problem
ut = uxx t > 0 , 0 < x < π ,u(0, x) = f(x) , ut(0, x) = g(x) 0 < x < π ,u(t, 0) = u(t, π) = 0 .
In this special case, the formula (3.29) yields the solution as the sum of a Fourier sine series:
u(t, x) =∞∑
k=1
2
π
[cos kt
(∫ π
0f(y) sin ky dy
)+
sin kt
k
(∫ π
0g(y) sin ky dy
)]sin kx .
153
4 Problems
1. Let Ω.= (x, y) ; x2+ y2 < 1 be the open unit disc in IR2. Prove that, for every bounded
measurable function f = f(x, y), the problemuxx + xuxy + uyy = f on Ω
u = 0 on ∂Ω
has a unique weak solution.
2. Let Ω ⊂ IRn be a bounded open set, and let the operator L in (1.1) be uniformly elliptic.
(i) Show that a necessary condition in order that the problem (1.2) admits a solution is thefollowing.
If v ∈ H10 (Ω) is a function such that
B[u, v] = 0 for all u ∈ H10 (Ω), (4.37)
then (f, v)L2 = 0.
(ii) By a formal integration by parts, show that the condition (4.37) can be restated by sayingthat v ∈ H1
0 (Ω) provides a weak solution to the homogeneous boundary value problemL∗v = 0 x ∈ Ω ,v = 0 x ∈ ∂Ω , (4.38)
where L∗ is the adjoint operator
L∗v.= −
n∑
i,j=1
(aij(x)vxj )xi −n∑
i=1
(bi(x)v)xi + c(x)v . (4.39)
3. On a bounded open set Ω ⊂ IRn consider an elliptic operator of the form
Lu.= −
n∑
i,j=1
(aij(x)uxi)xj + c(x)u , (4.40)
assuming that the coefficients aij, c satisfy (1.3) and (1.4).
(i) Prove that the space L2(Ω) admits an orthonormal basis φk ; k ≥ 1 consisting ofeigenfunctions of L, so that
φk ∈ H10 (Ω) , Lφk = µkφk
for all k ≥ 1. The corresponding eigenvalues satisfy µk → +∞ as k →∞.
(ii) Show that the general solution to the parabolic initial-boundary value problem (2.2) canbe written as
Stg =∞∑
k=1
e−µkt(g, φk)L2 φk .
154
4. Prove that a representation similar to (3.29) continues to hold, for the solution to the moregeneral hyperbolic initial-boundary value problem
utt + Lu = 0 t ∈ IR , x ∈ Ω ,u(t, x) = 0 t ∈ IR , x ∈ ∂Ω ,u(0, x) = f(x) , ut(0, x) = g(x) x ∈ Ω ,
where L is the elliptic operator in (4.40).
5. Let Ω ⊂ IRn be a bounded open set. Let β > 0 be the constant in Poincare’s inequality:
‖u‖L2(Ω) ≤ β ‖∇u‖L2(Ω) .
(i) Establish a lower bound on the eigenvalues of the operator Lu = −∆u. More precisely, ifφ ∈ H1
0 (Ω) provides a weak solution to
−∆φ = µφ x ∈ Ω ,
φ = 0 x ∈ ∂Ω ,
prove that µ ≥ 1/β2.
(ii) Prove that the solution of the parabolic initial-boundary value problem
ut = ∆u t > 0 , x ∈ Ω ,u(t, x) = 0 t > 0 , x ∈ ∂Ω ,u(0, x) = g(x) x ∈ Ω ,
decays to zero as t→∞. Indeed, ‖u(t)‖L2 ≤ e−t/β2‖g‖L2 for every t ≥ 0.
6. On the interval [0, T ], consider the Sturm-Liouville eigenvalue problem
(p(t)u′)′ + q(t)u = µu 0 < t < T ,u(0) = u(T ) = 0 .
(4.41)
Assume that
p ∈ C1(]0, T [) , q ∈ C0([0, T ]), p(t) ≥ θ > 0 for all t .
Prove that the space L2([0, T ]) admits an orthonormal basis φk ; k ≥ 1 where each φksatisfies (4.41), for a suitable eigenvalue µk. Moreover, µk → −∞ as k →∞.
7. In Lemma 8.2, take L0u = −∆u. Let φk ; k ≥ 1 be an orthonormal basis of L2(Ω)consisting of eigenfunctions of L−1
0 . Show that in this special case the eigenfunctions φk aremutually orthogonal also w.r.t. the inner product in H1, namely (φj , φk)H1 = 0 wheneverj 6= k.
155
8. On a bounded open set Ω ⊂ IRn, let 0 < µ1 ≤ µ2 ≤ · · · be the eigenvalues of the operatorLu = −∆u (with zero boundary conditions).
(i) Given any ϕ ∈ C∞c (Ω), prove that
∫
Ω|∇ϕ|2 dx ≥ µ1
∫
Ω|ϕ|2.
(ii) Prove that the best possible constant in Poincare’s inequality
‖u‖2L2 ≤ C ‖∇u‖2
L2 u ∈ H10 (Ω) (4.42)
is C = 1/µ1.
9. Let Ω ⊆ Ω ⊂ IRn be bounded open sets. Let 0 < µ1 ≤ µ2 ≤ · · · be the eigenvalues of theoperator −∆ on the domain Ω, and let 0 < µ1 ≤ µ2 ≤ · · · be the eigenvalues of the operator−∆ on Ω. Prove that µ1 ≤ µ1.
10. Consider the open rectangle Q.= (x, y) ; 0 < x < a , 0 < y < b. Define the functions
φm,n(x, y).=
√2
absin
mπ x
asin
nπ y
bm, n ≥ 1 .
(i) Check that φm,n ∈ H10 (Q). Moreover, prove that the countable set of functions
φm,n ; m,n ≥ 1 is an orthonormal basis of L2(Q) consisting of eigenfunctions of the ellipticoperator Lu = −∆u. Compute the corresponding eigenvalues µm,n .
(ii) If Ω ⊂ IR2 is an open domain contained inside a rectangle Q with sides a, b, prove that
‖u‖L2(Ω) ≤ab
π√a2 + b2
‖∇u‖L2(Ω) for all u ∈ H10 (Ω) .
11. (Galerkin approximations) Let Ω ⊂ IRn be a bounded open set. Given f ∈ L2(Ω),consider the elliptic problem
−∆ = f x ∈ Ω ,φ = 0 x ∈ ∂Ω ,
Let ϕ1, . . . , ϕm be any set of linearly independent functions in H10 (Ω). Construct an ap-
proximate solution by setting
U =m∑
k=1
ck ϕk
where the coefficients C1, . . . , cm are chosen so that
B[U,ϕj ].=
∫
Ω∇U · ∇ϕj dx = (f, ϕj)L2 j = 1, . . . ,m . (4.43)
156
Show that (4.43) can be written as an algebraic system of m linear equations for the mvariables c1, . . . , cm. Prove that this system has a unique solution.
12. Let Ω = (x, y) ; x2 + y2 < 1 be the open unit disc in IR2, and let u be a smoothsolution to the equation
utt = uxx + xuxy + 3uyy on Ω× [0, T ], (4.44)
u = 0 on ∂Ω× [0, T ].
(i) Write the equation (4.44) in the form utt+Lu = 0, proving that the operator L is uniformlyelliptic on the domain Ω.
(ii) Define a suitable energy e(t) =[kinetic energy] + [elastic potential energy], and check thatit is constant in time.
13. Consider the differential operator on IR2
Lu = −(xux)x − (yuy)y + 2uxy + 3(ux + uy)− 6u.
(i) At which points (x, y) is L elliptic ?
(ii) Determine for which open bounded subsets Ω ⊂ IR2 one can say that operator L isuniformly elliptic on Ω.
14. Let Ω = (x, y) ; x2 + y2 < 1 be the open unit disc in IR2, and let u be a smoothsolution to the equation
utt = uxx + 2xuxy + 3uyy + uy on Ω× [0, T ], (4.45)
u = 0 on ∂Ω× [0, T ].
(i) Write the equation (4.45) in the form utt+Lu = 0, proving that the operator L is uniformlyelliptic on the domain Ω.
(ii) Define a suitable energy E(t) = [kinetic energy] + [elastic potential energy], and checkthat it is constant in time.
15. Let Ω.= (x, y) ; x2+y2 < 1 be the unit disc in IR2. On the space X = H1
0 (Ω), considerthe inner product
〈u, v〉 .=
∫
Ω
[uxvx + 2uyvy + y(uxvy + uyvx)
]dx
(i) Prove that 〈·, ·〉 is indeed an inner product on X, which makes X into a Hilbert space.
157
(ii) Given f ∈ L2(Ω), show that there exists a unique u ∈ X such that
〈u, v〉 =
∫
Ωf v dx for all v ∈ X = H1
0 (Ω) .
What elliptic equation does u solve ?
158
Chapter 10
Appendix
1 Metric and Topological Spaces
A distance on a set X is a map d : X ×X 7→ IR+ satisfying the following three properties:
1 - positivity: d(x, y) ≥ 0, d(x, y) = 0 if and only if x = y,
2 - symmetry: d(x, y) = d(y, x),
3 - triangle inequality: d(x, z) ≤ d(x, y) + d(y, z).
A set X endowed with a distance function d(·, ·) is called a metric space.
The open ball centered at a point x with radius r > 0 will be denoted by
B(x, r).= y ∈ X ; d(y, x) < r.
In turn, a metric d(·, ·) determines a topology on X.
A subset A ⊆ X is open if, for every x ∈ A, there exists a radius r > 0 such that B(x, r) ⊆ A.
A subset C ⊆ A is closed if its complement X \ C = x ∈ X ; x /∈ C is open.
If B(x, r) ⊆ A for some r > 0 we say that A is a neighborhood of the point x, or equivalentlythat x is an interior point of A.
• The union of any family of open sets is open. The intersection of finitely many open sets isopen.
• The intersection of any family of closed sets is closed. The union of finitely many closed setsis closed.
159
The entire space X and the empty set ∅ are always both open and closed. If there exists noother subset S ⊂ X which is at the same time open and closed, we say that the space X isconnected.
The closure of a set A, denoted by A, is the smallest closed set containing A. This is obtainedas the intersection of all closed sets containing A.
A subset S ⊆ X is dense in X if S = X. This is the case if and only if S intersects everynon-empty open subset of X. The space X is separable if it contains a countable densesubset.
We say that sequence (xn)n≥1 converges to a point x ∈ X if limn→∞ d(xn, x) = 0. In thiscase, we also write limn→∞ xn = x or simply xn → x.
• A point x lies in the closure of the set A if and only if there exists a sequence of pointsxn ∈ A that converges to x.
A sequence (xn)n≥1 is a Cauchy sequence if for every ε > 0 one can find an integer N largeenough so that
d(xm, xn) ≤ ε whenever m,n ≥ N.The metric space X is complete if every Cauchy sequence converges to some limit pointx ∈ X.
If X,Y are two metric spaces, a map f : X 7→ Y is continuous if, for every open set A ⊆ Y ,the pre-image f−1(A)
.= x ∈ X ; f(x) ∈ A is an open subset of X.
• A map f : X 7→ Y is continuous if and only if, for every x0 ∈ X and ε > 0, there existsδ > 0 such that
d(x, x0) < δ implies d(f(x), f(x0)) < ε.
We say that a map f : X 7→ Y is Lipschitz continuous if there exists a constant C ≥ 0 suchthat
d(f(x), f(x′)) ≤ C · d(x, x′) for all x, x′ ∈ X .
More generally, we say that f : X 7→ Y is Holder continuous of exponent 0 < α ≤ 1 if thereexists a constant C such that
d(f(x), f(x′)) ≤ C · [d(x, x′)]α for all x, x′ ∈ X .
A collection of open sets Ai ; i ∈ I such that K ⊆ ⋃i∈I Ai is called an open covering ofthe set K. Here the set of indices I may be well be infinite. A set K ⊆ X is compact if, fromevery open covering of K, one can extract a subcovering consisting of finitely many open sets.
160
A set S is relatively compact if its closure S is compact.
A set S is precompact if, for every ε > 0, it can be covered by finitely many balls with radiusε.
Theorem A.1 (compact subsets of IRn). A subset S ⊂ IRn is compact if and only if it isclosed and bounded.
Theorem A.2 (equivalent characterizations of compactness). Let S be a metric space.The following are equivalent:
(i) S is compact.
(ii) S is precompact and complete.
(iii) From every sequence (xk)k≥1 of points in S one can extract a subsequence converging tosome limit point x ∈ S.
2 The Contraction Mapping Theorem
Let φ : X 7→ X be a map from a complete metric space X into itself. A point x∗ such thatφ(x∗) = x∗ is called a fixed point of φ. For strictly contractive maps, the fixed point isunique and can be obtained by a simple iterative procedure.
Theorem A.3 (Contraction Mapping Theorem). Let X be a complete metric space, andlet φ : X 7→ X be a continuous mapping such that, for some κ < 1,
d(φ(x), φ(y)) ≤ κd(x, y) for all x, y ∈ X. (2.1)
Then there exists a unique point x∗ ∈ X such that
x∗ = φ(x∗). (2.2)
Moreover, for any y ∈ X one has
d(y, x∗) ≤ 1
1− κ d(y, φ(y)). (2.3)
*. ..y =y
0
y1
y2 y
3=φ(y)
x
Figure 10.1: Starting from any point y, the iterated images of the function φ converge to the uniquefixed point.
Proof. Fix any point y ∈ X and consider the sequence (see figure 10.1)
y0 = y, y1 = φ(y0), · · · , yn+1 = φ(yn), · · ·
161
By induction, we have
d(y2, y1) ≤ κd(y1, y0),d(y3, y2) ≤ κd(y2, y1) ≤ κ2 d(y1, y0),
· · ·d(yn+1, yn) ≤ κd(yn, yn−1) ≤ κnd(y1, y0) = κnd(φ(y), y).
For m < n we have
d(yn, ym) ≤n−1∑
j=m
d(yj+1, yj) ≤n−1∑
j=m
κj d(y, φ(y)) ≤ κm
1− κ d(y, φ(y)). (2.4)
Since κ < 1, the right hand side of (2.4) approaches zero as m → ∞. Hence the sequence(yn)n≥1 is Cauchy. Since X is complete, this sequence converges to some limit point x∗. Bythe continuity of φ one has
x∗ = limn→∞
yn = limn→∞
φ(yn−1) = φ(limn→∞
yn−1
)= φ(x∗),
hence (2.2) holds. The uniqueness of the fixed point x∗ is proved observing that, if
x1 = φ(x1), x2 = φ(x2),
then by (2.1) it follows
d(x1, x2) = d(φ(x1), φ(x2)) ≤ κd(x1, x2).
The assumption κ < 1 thus implies d(x1, x2) = 0 and hence x1 = x2.
Finally, using (2.4) with m = 0, for every n ≥ 1 we obtain
d(yn, y) ≤1
1− κ d(y, φ(y)).
Letting n→∞ we obtain (2.3).
3 The Baire Category Theorem
Using probability theory, one can introduce a concept of “large sets” and “small sets”. Bigsets are those having probability one, small sets are those with probability zero. Observe that:
(i) The union of countably many small sets is small, and the intersection of countably manylarge sets is big.
(ii) A large set is non-empty.
In a complete metric space, one can still develop a theory of “large sets” and “small sets”,relying exclusively on the topological structure.
162
Theorem A.4 (Baire). Let X be a complete metric space. Let (Vk)k≥1 be a sequence ofopen, dense subsets of X. The the intersection V
.=⋂∞
k=1 Vk is a non-empty, dense subset ofX.
Proof. Let Ω ⊂ X be any open set. We need to show that(⋂∞
k=1 Vk)∩Ω is nonempty.
We begin by choosing a point x0 and a radius r0 < 1 such that B(x0, 3r0) ⊂ Ω.
By induction, for every k ≥ 1 we choose a point xk and a radius rk such that B(xk, 3rk) ⊂Vk ∩B(xk−1, rk−1). This is possible because Vk is open and dense. For each k ≥ 1, the abovechoice implies
rk+1 ≤rk3, d(xk+1, xk) ≤ rk ≤
1
3k.
Therefore, the sequence (xk)k≥1 is Cauchy. Since X is complete, this sequence has a limit:xk → x for some x ∈ X. We now observe that
d(x, xk) ≤∞∑
j=k
d(xj+1, xj) ≤∞∑
j=k
rj ≤∞∑
j=k
3k−jrk =3
2rk .
Therefore, for every k ≥ 1,x ∈ B(xk, 3rk) ⊂ Vk .
When k = 0, this same argument yields x ∈ B(x0, 3r0) ⊂ Ω. Hence x ∈(⋂∞
k=1 Vk)∩ Ω.
Motivated by the above theorem, a set S ⊆ X is said to be of second category (that means:“topologically large”) if S can be obtained as the intersection of countably many open densesets. On the other hand, a set S is said to be of first category, or equivalently meager (i.e.“topologically small”) if S can be obtained as the union of countably many closed sets withempty interior. Notice that these topologically large or small sets satisfy the same properties(i)-(ii) valid for sets of probability 1 or 0, respectively.
4 Review of Lebesgue measure theory
4.1 Measurable sets
A family F of subsets of IRn is called a σ-algebra if
(i) ∅ ∈ F and IRn ∈ F ,
(ii) If A ∈ F then IRn \ A ∈ F ,
(iii) If Ak ∈ F for every k ≥ 1, then⋃∞
k=1Ak ∈ F and⋂∞
k=1Ak ∈ F .
Theorem A.5 (Existence of Lebesgue measure). There exists a σ-algebra F of subsetsof IRn and a mapping A 7→ mn(A) , from F into [0, +∞], with the following properties.
(i) F contains every open subset of IRn and hence also every closed subset of IRn.
163
(ii) If B is a ball in IRn, then mn(B) equals the n-dimensional volume of B.
(iii) If Ak ∈ F for every k ≥ 1 and the sets Ak are pairwise disjoint, then
mn
( ∞⋃
k=1
)=
∞∑
k=1
mn(Ak) (countable additivity)
(iv) If A ⊆ B with B ∈ F and mn(B) = 0, then also A ∈ F and mn(A) = 0.
The sets contained in the σ-algebra F are called Lebesgue measurable sets, while mn(A)is the n-dimensional Lebesgue measure of the set A ∈ F .
If a property P (x) is true for all points x ∈ IRn, except for those in a measurable set N withmn(N ) = 0, we say that the property P holds almost everywhere (a.e.).
A function f : IRn 7→ IR is measurable if
f−1(U).= x ∈ IRn ; f(x) ∈ U ∈ F
for every open set U ⊆ IR.
Every continuous function is measurable. If f, g are measurable, then f + g and f · g aremeasurable. Given a sequence of measurable functions (fk)k≥1, the functions defined asf∗(x)
.= lim supk→∞ fk(x) and f∗(x)
.= lim infk→∞ fk(x) are both measurable. ????
Theorem A.6 (Egoroff). Let (fk)k≥1 be a sequence of measurable functions, and assumethe pointwise convergence
fk(x)→ f(x) for a.e. x ∈ A,
for some measurable function f and a measurable set A ⊂ IRn with mn(A) < ∞. Then foreach ε > 0 there exists a subset E ⊆ A such that
(i) mn(A \E) ≤ ε
(ii) fk → f uniformly on E.
4.2 Lebesgue integration
In order to define the Lebesgue integral, one begins with a special class of functions. Thecharacteristic function of a set A is
χA(x) =
1 if x ∈ A,0 if x /∈ A.
A function taking finitely many values, having of the form
g(x) =ν∑
i=1
ci χAi(x) (4.1)
164
for some disjoint measurable sets A1, . . . Aν ⊂ IRn and constants c1, . . . , cν ∈ IR is called asimple function.
If g is the simple function in (4.1), its Lebesgue integral is defined as
∫
IRng dx
.=
ν∑
i=1
cimn(Ai).
If f is a non-negative measurable function, its Lebesgue integral is defined as
∫
IRnf dx
.= sup
∫
IRng dx ; g is simple, g ≤ f
.
For any measurable function f , its positive and negative parts are denoted as
f+.= maxf, 0, f−
.= max−f, 0.
We then define the Lebesgue integral of f as∫
IRnf dx
.=
∫
IRnf+ dx−
∫
IRnf− dx (4.2)
provided that at least one of the terms on the right hand side is finite. In this case we saythat f is integrable. Notice that the integral in (4.2) may well be +∞ or −∞.
A measurable function f is summable if∫
IRn|f | dx < ∞.
We say that f is locally summable if the product f · χB
is summable for every ball B.
The essential supremum of a measurable function f is defined as
ess sup f.= inf
α ∈ IR ; f(x) < α for a.e. x ∈ IRn
.
The Lebesgue integration theory has useful convergence theorems:
Theorem A.7 (Fatou’s lemma). Let (fk)k≥1 be a sequence of functions which are non-negative and summable. Then
∫
IRn
(lim infk→∞
fk)dx ≤ lim inf
k→∞
∫
IRnfk dx
Theorem A.8 (monotone convergence). Let (fk)k≥1 be a sequence of measurable functionssuch that f1 ≤ f2 ≤ · · · ≤ fk ≤ fk+1 ≤ · · · Then
∫
IRn
(limk→∞
fk)dx = lim
k→∞
∫
IRnfk dx .
165
Theorem A.9 (dominated convergence). Let (fk)k≥1 be a sequence of measurable func-tions such that
fk(x)→ f(x) for a.e. x ∈ IRn.
Moreover, assume that there exists a summable function g such that
|fk(x)| ≤ g(x) for every k ≥ 1 and a.e. x ∈ IRn.
Then
limk→∞
∫
IRnfk dx =
∫
IRnf dx .
In the following, we denote by
−∫
Af dx
.=
1
meas(A)
∫
Af dx
the average value of f on the set A. Observe that a function f is continuous at the point x0if
limr→0+
supx∈B(x0,r)
|f(x)− f(x0)| = 0. (4.3)
Replacing the supremum with an average, we say that f is quasi-continuous at the pointx0 if
limr→0+
−∫
B(x0,r)|f(x)− f(x0)| dx = 0. (4.4)
Theorem A.10 (Lebesgue differentiation). Let f : IRn 7→ IR be locally summable. Thenf is quasi-continuous at a.e. point y ∈ IRn.
A point y where (4.4) holds is called a Lebesgue point of f .
Next, we consider measurable subsets X ⊆ IRm, Y ⊆ IRn, so that the cartesian product
X × Y .=(x, y) ; x ∈ X, y ∈ Y
is a measurable subset of IRn+m. Given a function of two variables f : X × Y 7→ IR, for eachfixed x we consider the function y 7→ fx(y)
.= f(x, y) of the variable y alone. Similarly, for
each fixed y we consider the function x 7→ f y(x).= f(x, y) of the variable x alone.
Theorem A.11 (Fubini). Let X ⊆ IRm, Y ⊆ IRn be measurable sets, and assume f ∈L1(X × Y ). Then
• f y ∈ L1(X) for a.e. y ∈ Y .
• fx ∈ L1(Y ) for a.e. x ∈ X
• the integral function F (y).=∫X f y(x)dx is in L1(Y )
166
• the integral function G(x).=∫Y f
x(y)dy is in L1(X)
Moreover, one has the identity
∫∫
X×Yf(x, y) dxdy =
∫
Y
[∫
Xf y(x)dx
]dy =
∫
X
[∫
Yfx(y) dy
]dx .
5 Mollifications
In the analysis of Sobolev spaces, an important technique is the approximation of generalfunctions with smooth functions. This can be done by means of mollifications.
The standard mollifier on IRn is defined as
J(x).=
Cn exp 1|x|2−1 if |x| < 1 ,
0 if |x| ≥ 1 ,
(5.1)
where the constant Cn is chosen so that∫IRn J(x) dx = 1 . For each ε > 0 we also define the
rescaled function
Jε(x).=
1
εnJ(xε
).
Notice that Jε ∈ C∞c (IRn), and
∫
IRnJε = 1 , supp(Jε) = |x| ≤ ε.
ε
J
x2
x11ε
J
Figure 10.2: A standard mollifier J and its rescaling Jε.
Let now Ω ⊆ IRn be an open set and f ∈ L1loc(Ω) be a locally integrable function. We then
define the mollification of f in terms of a convolution:
fε.= Jε ∗ f
167
Since f is defined on Ω and Jε has support in the ball centered at the origin with radius ε,this convolution is well defined at all points in the subset
Ωε.=x ∈ Ω ; B(x, ε) ⊆ Ω
.
Indeed
fε(x).=
∫
B(x,ε)Jε(x− y)f(y) dy =
∫
B(0,ε)Jε(z)f(x− z) dz .
Theorem A.12 (properties of mollifiers). Let Ω ⊆ IRn be an open set and let f ∈ L1loc(Ω).
Then
(i) For every ε > 0 one has fε ∈ C∞(Ωε) .
(ii) As ε→ 0 one has the pointwise convergence fε(x)→ f(x) for a.e. x ∈ Ω .
(iii) If f is continuous, then fε → f uniformly on compact subsets of Ω .
(iv) If 1 ≤ p <∞ and f ∈ Lploc(Ω), then fε → f in Lp
loc(Ω) .
Proof. 1. Notice that each Ωε is open. Indeed, if the closed ball B(x0, ε) is entirely containedin the open set Ω, the same holds for the ball B(x, ε), whenever |x− x0| is sufficiently small.
Let e1, . . . , en be the standard ortho-normal basis of IRn. Fix a point x ∈ Ωε and i ∈1, . . . , n. If h is so small that x+ hei ∈ Ωε, then the difference quotient is computed as
fε(x+ hei)− fε(x)h
=1
εn
∫
B(x,ε)
1
h
[J
(x+ hei − y
ε
)− J
(x− yε
)]f(y) dy . (5.2)
Since the closed ball B(x, ε) is entirely contained in Ω, we have f ∈ L1(B(x, ε)). Moreover,since J ∈ C∞, we have the uniform convergence
1
h
[J
(x+ hei − y
ε
)− J
(x− yε
)]→ 1
ε
∂J
∂xi
(x− yε
).
Letting h→ 0 in (5.2) we obtain the existence of the partial derivative
∂fε∂xi
(x) =
∫
B(x,ε)
∂Jε∂xi
(x− y) f(y) dy .
A similar argument shows that, for every multi-index α, the derivative Dαfε exists and
Dαfε(x) =
∫
B(x,ε)DαJε(x− y) f(y) dy .
This proves (i).
2. By the Lebesgue differentiation theorem, for a.e. x ∈ Ω we have
limr→0
−∫
B(x,r)|f(y)− f(x)| dy = 0 . (5.3)
168
If x is a point for which (5.3) holds, then
|fε(x)− f(x)| =
∣∣∣∣∣
∫
B(x,ε)Jε(x− y)[f(y)− f(x)] dy
∣∣∣∣∣
≤ 1
εn
∫
B(x,ε)J
(x− yε
)|f(y)− f(x)| dy ≤ C −
∫
B(x,ε)|f(y)− f(x)| dy .
As ε→ 0, the right hand side goes to zero because of (5.3). This proves (ii).
Ω
εΩ KKρ
Ω
Figure 10.3: The sets Ωε ⊂ Ω and the compact sets K ⊂⊂ Kρ ⊂⊂ Ω, used in the analysis of mollifiers.
3. Assume that f is continuous, and let K ⊂ Ω be a compact subset. Then we can chooseδ > 0 small enough so that the compact neighborhood
Kρ.= x ∈ IRn ; d(x,K) ≤ ρ (5.4)
is still contained inside Ω. Since f is uniformly continuous on the compact set Kρ, the previouscalculations show that fε(x)→ f(x) uniformly for x ∈ K. This proves (iii).
4. To prove (iv), assume 1 ≤ p < ∞ and f ∈ Lploc(Ω). Let K ⊂ Ω be a compact subset and
choose ρ > 0 so that the compact neighborhood Kρ in (5.4) is still contained in Ω. We claimthat, for every 0 < ε < ρ,
‖fε‖Lp(K) ≤ ‖f‖Lp(Kρ) . (5.5)
Indeed, for x ∈ K an application of Holder’s inequality with q = pp−1 yields
|fε(x)| =
∣∣∣∣∣
∫
B(x,ε)Jε(x− y) f(y) dy
∣∣∣∣∣
≤∫
B(x,ε)
(Jε(x− y)
) p−1p(Jε(x− y)
) 1p |f(y)| dy
≤(∫
B(x,ε)Jε(x− y) dy
) p−1p(∫
B(x,ε)Jε(x− y)|f(y)|p dy
) 1p
.
169
Recalling that∫B(x,ε) Jε(x− y) dy = 1, for 0 < ε < ρ the above inequality yields
∫
K|fε(x)|p dx ≤
∫
K
(∫
B(x,ε)Jε(x− y)|f(y)|p dy
)dx
≤∫
Kρ
|f(y)|p(∫
B(y,ε)Jε(x− y) dx
)dy =
∫
Kρ
|f(y)|p dy .
This proves (5.5).
Next, for any δ > 0, we choose g ∈ C(Kρ) such that
‖f − g‖Lp(Kρ) < δ .
Together with (5.5), this yields
‖fε − f‖Lp(K) ≤ ‖fε − gε‖Lp(K) + ‖gε − g‖Lp(K) + ‖g − f‖Lp(K)
≤ ‖f − g‖Lp(Kρ) + ‖gε − g‖Lp(K) + ‖g − f‖Lp(K)
≤ δ + ‖gε − g‖Lp(K) + δ .
Since g is continuous, by (ii) it follows |gε − g| → 0 uniformly on the compact set Kρ. Hencelim supε→0 ‖fε − f‖Lp(K) ≤ 2δ. Since δ > 0 was arbitrary, this proves (iv).
Corollary A.1. Let f ∈ L1loc(Ω) and assume
∫
Ωf φ dx = 0 for every φ ∈ C∞c (Ω).
Then f(x) = 0 for a.e. x ∈ Ω.
Indeed, let x ∈ Ω be a Lebesgue point of f and let Jε be the standard mollifier. Takingφ(y) = Jε(x− y) and letting ε→ 0 one obtains
0 =
∫
ΩJε(x− y)f(y) dy =
∫
B(x,ε)J(x− y)f(y) dy → f(x).
Hence f(x) = 0 for a.e. x ∈ Ω.
5.1 Partitions of unity
Let S ⊆ IRn and let V1, V2, . . . be open sets that cover S, so that
S ⊆⋃
k≥1
Vk .
We say that a family of functions φk ; k ≥ 1 is a smooth partition of unity subordinatedto the sets Vk if
170
(i) For every k ≥ 1, φk : IRn 7→ [0, 1] is a C∞ function with support contained inside theopen set Vk.
(ii) Each point x ∈ S has a neighborhood which intersects the the support of finitely manyfunctions φk. Moreover
∑
k
φk(x) = 1 for all x ∈ S . (5.6)
Note that, by assumption (ii), at each point x ∈ S the summation in (5.6) contains onlyfinitely many non-zero terms.
1
V1V
2
V3
1φ 2
φ3
φ
x
Figure 10.4: A partition of unity subordinate to the sets V1, V2, V3.
Theorem A.13 (existence of a smooth partition of unity). Let S ⊆ IRn and letVk ; k ≥ 1 be a family of open sets covering S. Then there exists a smooth partition ofunity subordinated to the sets Vk.
6 Integrals of functions taking values in a Banach space
Let X be a Banach space with norm ‖ · ‖. Let X∗ be its dual space. Every x∗ ∈ X∗ thusdetermines a linear continuous mapping X 7→ IR.
A function g : [0, T ] 7→ X is simple if it has the form
g(t) =ν∑
i=1
ui χAi(t) t ∈ [0, T ] (6.1)
where, for each i = 1, . . . , ν, the set Ai ⊆ [0, T ] is measurable and ui ∈ X.
A function f : [0, T ] 7→ X is strongly measurable if there exists a sequence of simplefunctions gk : [0, T ] 7→ X such that
gk(t)→ f(t) for a.e. t ∈ [0, T ].
Moreover, we say that f is summable if there exists a sequence of simple functions gk suchthat ∫ T
0‖sk(t)− f(t)‖ dt→ 0. (6.2)
171
A function f : [0, T ] 7→ X is weakly measurable if, for each x∗ ∈ X∗, the scalar functiont 7→ 〈x∗, f(t)〉 is measurable.
A function f : [0, T ] 7→ X is almost separably valued if there exists a subset N ⊂ [0, T ]with zero measure such that the set of images f(t) ; t ∈ [0, T ]\N is separable (i.e., containsa countable dense subset).
Theorem A.14 (Pettis). A map f : [0, T ] 7→ X is strongly measurable if and only if f isweakly measurable and almost separably valued.
The integral of a function f with values in a Banach space is defined in two steps.
If g is the simple function in (6.1), we define
∫ T
0g dt
.=
ν∑
i=1
uim1(Ai).
Here m1 is the one-dimensional Lebesgue measure on the interval [0, T ].
If f is summable, we define ∫ T
0f dt
.= lim
k→∞
∫ T
0gk(t) dt
where (gk)k≥1 is any sequence of simple functions for which (6.2) holds.
Theorem A.15 (Bochner). A strongly measurable function f : [0, T ] 7→ X is summable ifand only if the scalar function t 7→ ‖f(t)‖ is summable. In this case one has
∥∥∥∥∥
∫ T
0f(t) dt
∥∥∥∥∥ ≤∫ T
0‖f(t)‖ dt .
Moreover, for every x∗ ∈ X∗,⟨x∗ ,
∫ T
0f(t) dt
⟩=
∫ T
0〈x∗, f(t)〉 dt .
7 Inequalities
7.1 Convex sets and convex functions
A set Ω ⊆ IRn is convex if
x, y ∈ Ω , θ ∈ [0, 1] =⇒ θx+ (1− θ)y ∈ Ω.
In other words, if Ω contains two points x, y, then it also contains the entire segment joiningx with y.
172
Let Ω ⊂ IRn be a convex set. We say that a function f : Ω 7→ IR is convex if
f(θx+ (1− θ)y) ≤ θ f(x) + (1− θ)f(y) (7.1)
whenever x, y ∈ Ω and θ ∈ [0, 1]. Notice that (7.1) holds if and only if the epigraph of f ,i.e. the set
(x, z) ∈ Ω× IR ; z ≥ f(x),
is a convex subset of IRn × IR.
A twice differentiable function f : IRn 7→ IR is uniformly convex if its Hessian matrix ofsecond derivatives is uniformly positive definite. This means that, for some constant κ > 0,
n∑
i,j=1
fxixj(x) ξiξj ≥ κn∑
i=1
ξ2i for all x, ξ ∈ IRn.
00ω
f(x)z
x0 x
f(x ) + (x−x )
Figure 10.5: A convex function f and its epigraph. A support hyperplane at the point x0.
Theorem A.16 (supporting hyperplanes). Let f : IRn 7→ IRn be a convex function. Thenfor every x0 ∈ IRn there is a (sub-gradient) vector ω such that
f(x) ≥ f(x0) + ω · (x− x0) for all x ∈ IRn.
The hyperplane (x, z) ; z = ω ·(x−x0) ⊂ IRn+1 touches the graph of f at the point x0, andremains below this graph at all points x ∈ IRn. It is thus called a supporting hyperplaneto f at x0.
Theorem A.17 (Jensen’s inequality). Let f : IR 7→ IR be a convex function, and letΩ ⊂ IRn be a bounded open set. If u ∈ L1(Ω) is any integrable function, then
f
(−∫
Ωu dx
)≤ −
∫
Ωf(u) dx . (7.2)
Here −∫Ω u dx.= 1
meas(Ω)
∫Ω udx is the average value of u on the set Ω, while −∫Ω f(u) dx
.=
1meas(Ω)
∫Ω f(u)dx is the average value of f(u). Notice that we do not require that the set Ω
be convex.
173
Proof. Set u0.= −∫Ω u(y) dy. Then there exists a support hyperplane to the graph of f at the
point x0, say f(u) ≥ f(u0) + ω (u− u0) for some constant ω ∈ IR and all u ∈ IR. Hence
f(u(x)) ≥ f
(−∫
Ωu(y) dy
)dx+ ω
(u(x)− −
∫
Ωu(y) dy
).
Taking the average value of both sides on the set Ω, we obtain (7.2).
7.2 Basic inequalities
1. Cauchy’s inequality.
ab ≤ a2 + b2
2(a, b ∈ IR) . (7.3)
Indeed, 0 ≤ (a− b)2 = a2 + b2 − 2ab.
For any ε > 0, replacing a with√2ε a and b with b/
√2ε, from (7.3) we obtain the slightly
more general inequality
ab ≤ εa2 +b2
4ε(a, b ∈ IR , ε > 0) . (7.4)
2. Young’s inequality.
ab ≤ ap
p+bq
q
(a, b > 0 , 1 < p < q <∞ ,
1
p+
1
q= 1
). (7.5)
Indeed, since the map f(u) = eu is convex, one has expup + vq ≤ eu
p + ev
q . Therefore
ab = eln a+ln b = exp
1
pln ap +
1
qln bq
≤ 1
peln ap +
1
qeln bq =
ap
p+bq
q.
For any ε > 0, replacing a with (εp)1/p and b with b/(εp)1/p, from (7.5) one obtains
ab ≤ εap + Cεbq
(a, b, ε > 0 , 1 < p < q <∞ ,
1
p+
1
q= 1
). (7.6)
where the constant Cε = q−1(εp)−q/p is independent of a, b.
3. Holder’s inequality. If f ∈ Lp(Ω), g ∈ Lq(Ω) with 1 ≤ p, q ≤ ∞ and 1p + 1
q = 1, then∫
Ω|fg| dx ≤ ‖f‖Lp(Ω) ‖g‖Lq(Ω) . (7.7)
Indeed, in the special case where ‖f‖Lp(Ω) = ‖g‖Lq(Ω) = 1, Young’s inequality yields∫
Ω|fg| dx ≤ 1
p
∫
Ω|f |p dx1
q
∫
Ω|g|q dx =
1
p+
1
q= 1 = ‖g‖Lp ‖g‖Lq .
To cover the general case, we simply replacing the functions f, g with f = f/‖f‖Lp andg = g/‖g‖Lq , respectively.
174
By induction, one can establish the following more general version of Holder’s inequality. Let1 ≤ p1, . . . , pm ≤ ∞, with 1
p1+ · · ·+ 1
pm= 1. Assume fk ∈ Lpk(Ω) for k = 1, . . . ,m. Then
∫
Ω|f1f2 · · · fm| dx ≤
m∏
k=1
‖fk‖Lpk (Ω) . (7.8)
4. Minkowski’s inequality. For any 1 ≤ p ≤ ∞ and f, g ∈ Lp(Ω), one has
‖f + g‖Lp(Ω) ≤ ‖f‖Lp(Ω) ‖g‖Lp(Ω) . (7.9)
Indeed, in the case p = 1 the result is trivial. If p > 1, applying Holder’s inequality withexponents p and q = p
p−1 , one obtains
‖f + g‖pLp(Ω) =
∫
Ω|f + g|p dx =
∫
Ω|f + g|p−1(|f |+ |g|) dx
≤(∫
Ω|f + g|(p−1)· p
p−1 dx
) p−1p
[(∫
Ω|f |pdx
) 1p(∫
Ω|g|pdx
) 1p
]
= ‖f + g‖p−1Lp(Ω)
(‖f‖Lp(Ω) ‖g‖Lp(Ω)
).
Dividing both sides by ‖f + g‖p−1Lp(Ω) we obtain (7.9).
5. Interpolation inequality. Let 1 ≤ p ≤ r ≤ q ≤ ∞, with
1
r=
θ
p+
1− θq
for some θ ∈ [0, 1] .
If f ∈ Lp(Ω) ∩ Lq(Ω), then we also have f ∈ Lr(Ω) and
‖f‖Lr(Ω) ≤ ‖f‖θLp(Ω) ‖f‖1−θLq(Ω) . (7.10)
Indeed, observing that θrp + (1−θ)r
q = 1, by Holder’s inequality one obtains
∫
Ω|f |rdx =
∫
Ω|f |θr|f |(1−θ)r dx ≤
(∫
Ω|f |θr· p
θr dx
) θrp(∫
Ω|f |(1−θ)r· q
(1−θ)r dx
) (1−θ)rq
.
6. Discrete Holder and Minkowski inequalities. The inequalities (7.7) and (7.9) hold,more generally, when Ω is any measure space. In particular, one can take Ω = 1, . . . , nwith the counting measure. For any collection of numbers a1, . . . , an and b1, . . . , bn, and1 ≤ p, q <∞ with 1
p + 1q = 1, one has the discrete Holder’s inequality
n∑
k=1
|akbk| ≤(
n∑
k=1
|ak|p) 1
p(
n∑
k=1
|bk|q) 1
q
, (7.11)
175
and the discrete Minkowski’s inequality
(n∑
k=1
|ak + bk|p) 1
p
≤(
n∑
k=1
|ak|p) 1
p
+
(n∑
k=1
|bk|p) 1
p
. (7.12)
Given two vectors x = (x1, . . . , xn) and y = (y1, . . . , yn), using the discrete Holder’s inequality(7.11) with p = q = 2, one obtains the Cauchy-Schwarz inequality for the inner product:
|x · y| ≤ |x| |y| . (7.13)
Indeed,
|x · y| =
∣∣∣∣∣n∑
k=1
xk yk
∣∣∣∣∣ ≤n∑
k=1
|xk yk| ≤(
n∑
k=1
|xk|2) 1
2(
n∑
k=1
|yk|q) 1
2
= |x| |y| .
7.3 A differential inequality
Theorem A.18 (Gronwall’s inequality). Let t 7→ z(t) be a non-negative, absolutely con-tinuous function defined for t ∈ [0, T ]. Assume that the time derivative z′ = d
dtz satisfies
z′(t) ≤ φ(t) z(t) + ψ(t) for a.e. t ∈ [0, T ] ,
where φ,ψ ∈ L1([0, T ]) are non-negative functions. Then
z(t) ≤ e∫ t
0φ(s) dsz(0) +
∫ t
0e∫ t
sφ(σ) dσψ(s) ds for all t ∈ [0, T ]. (7.14)
Notice that the right hand side of (7.14) is precisely the solution to the Cauchy problem
Z ′(t) = φ(t)Z(t) + ψ(t) , Z(0) = z(0).
To prove (7.14), we write
d
ds
(e−∫ s
0φ(σ) dσ z(s)
)= e−
∫ s
0φ(σ) dσ
(z′(s)− φ(s)z(s)
)≤ e−
∫ s
0φ(σ) dσψ(s).
Integrating over the interval [0, t], one finds
e−∫ t
0φ(σ) dσz(t) − z(0) ≤
∫ t
0e−∫ s
0φ(σ) dσψ(s) ds .
Multiplying by e∫ t
0φ(σ) dσ we thus obtain (7.14).
176
8 Some technical points
• (proving convergence) To prove that a sequence (yn)n≥1 converges to some point y, itsuffices to show that, from every subsequence (ynj)j≥1, one can extract a further subsequenceconverging to y.
• (ruling out convergence) If a sequence (xn)n≥1 in a normed space X satisfies
‖xm − xn‖ ≥ δ > 0 for all m 6= n,
then it cannot have any convergent subsequence.
• (existence of a convergent subsequence) Consider a sequence (xn)n≥1 in a Banachspace X. Assume that, for every ε > 0, one can extract a subsequence (xnj)j≥1 such that
lim supj,k→∞
‖xnj − xnk‖ < ε.
Then the sequence admits a convergent subsequence.
• (absolutely convergent series) Let (un)n≥1 be Cauchy sequence in a Banach space X.Then we can extract a subsequence (unk
)k≥1 such that
‖unk+1− unk
‖ ≤ 2−k k = 1, 2, . . . (8.15)
In particular, if Y ⊂ X is a dense subspace, then every u ∈ X can be written as the sum ofan absolutely convergent series of elements in Y
u = limn→∞un = lim
k→∞unk
= un1 +∑
k≥1
(unk+1− unk
) .
Note: if ‖un−u‖Lp(Ω) → 0, it may happen that un(x) does not converge to u(x) for any x ∈ Ω.However, if (8.15) holds, then unk
(x)→ u(x) pointwise for a.e. x ∈ Ω.
• (proving that a function vanishes identically) If f : IRn 7→ IR is a locally integrablefunction such that
∫fϕdx = 0 for every ϕ ∈ C∞c (IRn) then, taking limits of mollifications
we findf(x) = lim
ε→0(Jε ∗ f)(x) = 0 for a.e. x ∈ IRn.
Indeed, the above identity holds whenever x is a Lebesgue point of f , i.e.
limε→0
−∫
|y−x|<ε|f(y)− f(x)| dy = 0.
• (approximation with smooth functions) Given ε > 0, every function f ∈ L1(IRn) canbe written as a sum:
f = ϕ+ fε with ϕ ∈ C∞c (IRn) , ‖fε‖L1 < ε .
177
• (L1-convergence implies a.e. pointwise convergence for a subsequence) Considera sequence of functions fn ∈ L1(IR). Even if the sequence is Cauchy, i.e.
limm,n→∞ ‖fm − fn‖L1 → 0 ,
it can happen that the sequence of numbers fn(x) does not converge, for any x ∈ IR. However,if we select a subsequence such that
‖fnk− fnk+1
‖L1 ≤ 2−k .
then we have the pointwise convergence
fnk(x) → f(x) for a.e. x ∈ IR .
• (pointwise convergence and convergence of the integral) If fn(x)→ f(x) for a.e. x ∈IR, in general it is not true that
∫fn dx →
∫f dx. To apply the Dominated Convergence
Theorem we need to find a function g independent of n such that |fn(x)| ≤ g(x) for all nand a.e. x.
If such a function g is not available, a weaker result is provided by Fatou’s lemma: If fn ∈L1(IR), then ∫
|f | dx ≤ lim infn→∞
∫|fn| dx .
• (absolutely continuous functions) A function f : IR 7→ IR with bounded variation isabsolutely continuous if and only if there exists g ∈ L1(IR) such that
f(x) = f(x0) +
∫ x
x0
g(y) dy ,
for some (hence for every) point x0 ∈ IR. In this case, f is differentiable at every Lebesguepoint of g, and f ′(x) = g(x) at such point.
9 Problems
1. Let f ∈ L1loc(IR
n). Assume that
∫
IRnfϕdx = 0 for all ϕ ∈ C∞c (IRn) .
Prove that f(x) = 0 for a.e. x ∈ IRn.
2. Let Ai ; i ∈ I be an open covering of a compact metric space K. prove that there existsρ > 0 such that, for every x ∈ K, the ball B(x, ρ) is entirely contained in one of the sets Ai.
178
3. Consider the function
f(x) =
1
x(lnx)2if 0 < x <
1
2,
0 otherwise.
Let fε.= Jε ∗ f be the corresponding mollifications, and let
F (x).= sup
0<ε<1fε(x).
Prove that f ∈ L1(IR) but F /∈ L1(IR). As a consequence, although fε → f pointwise, onecannot use the Lebesgue dominated convergence theorem to prove that ‖fε − f‖L1(IR) → 0.
4. Let fn : IR 7→ IR, n ≥ 1, be a sequence of absolutely continuous functions such that
(i) At the point x = 0, the sequence fn(0) is bounded.
(ii) There exists a function g ∈ L1(IR) such that the derivatives f ′n satisfy
|f ′n(x)| ≤ g(x) for every n ≥ 1 and a.e. x ∈ IR .
Prove that there exists a subsequence (fnj)j≥1 which converges uniformly on the entire realline.
5. Using a fixed point argument, show that the equation
u(x) = x+1
2
∫ 1
0sin(x+ y)u(y) dy for all x ∈ [0, 1]
has a unique solution u ∈ C([0, 1]).
6. Let f : IR 7→ IR be an absolutely continuous function, with bounded variation. Consider
the sequence of divided differences gn(x) = n[f(x+ 1
n
)− f(x)
]. Prove that gn(x)→ f ′(x)
for a.e. x ∈ IR , and moreover ‖gn − f ′‖L1 → 0.
7. Consider a sequence of functions fn ∈ L1(IR) with ‖fn‖L1 ≤ C for every n ≥ 1. Define
f(x).=
limn→∞
fn(x) if the limit exists,
0 otherwise.
Prove that f is Lebesgue measurable and ‖f‖L1 ≤ C.
179
10 Hints to selected problems
10.1 Chapter 2
12. To prove (ii), consider the finite dimensional subspaces Vn = spanv1, . . . , vn. Use Baire’scategory theorem.
16. To prove (ii), if the sequence (λk)k≥1 is unbounded, then there exists a subsequence with|λnk| ≥ 2k. Consider the vector v =
∑∞k=1 2
−kenk. Show that v ∈ ℓp but v /∈ Dom(Λ).
To prove (iii), for each k ≥ 1 consider the linear functional πk(x).= xk. In other words, πk(x)
is the k-th component of the sequence x = (x1, x2, . . .). Show that this is a continuous linearfunctional with norm ‖πk‖ = 1. Select a subsequence (xnj )j≥1 which converges componentwiseto a vector y
.= (y1, y2, . . .) ∈ ℓp. Namely, limj→∞ πk(xnj ) = yk for each k = 1, 2, 3, . . .
Finally, using the assumption |λk| → 0, prove that ‖Λxnj − Λy‖p → 0 as j →∞.
19. If fn f , then∫ ba f(x) dx = 0 for every bounded interval [a, b]. On the other hand,∫∞
−∞ f(x) dx = limn→∞∫∞−∞ fn(x) dx = 1.
24. Let x1, x2, x3 . . . be a countable dense subset of X. For each N ≥ 1, let fN : span(V ∪x1, . . . , xN) 7→ IR be a linear functional such that
fN (x) = f(x) if x ∈ V , ‖fN‖ ≤ ‖f‖ .
By induction, construct a linear functional f , defined on the dense subset span(V ∪xk ; k ≥ 1). Then extend f to the entire space X by continuity.
26. If x =∑N
i=1 cixi, write
x =
(c1mx1 +
c2mx2 + · · ·+
cNmxN
)+ · · · +
(c1mx1 +
c2mx2 + · · · +
cNmxN
)
(m summands). Choose m large enough.
27. Let x = limn→∞ yn, with yn =∑Nn
i=1 cn,ixn,i ∈ span(S). It is not restrictive to assume‖yn+1 − yn‖ < 2−n for every n. Apply (6.4) with x = yn+1 − yn.
28. Let V ⊂ ℓ∞ be the subspace of all sequences such that limn→∞ xn exists. Use theHahn-Banach extension theorem.
37. Each Vn is a closed subspace with empty interior. Use the Baire category theorem.
180
10.2 Chapter 3
4. For (i), construct an even, periodic function f of period 2π, which coincides with g forx ∈ [0, π]. Then use the result in problem 3. For (ii), construct an odd, periodic function fof period 2π, which coincides with g for x ∈ [0, π]. Then use 3.
6. First show that all functions fn can be extended by continuity to the closure Ω.
8. Given f ∈ Lp(Ω), using Lusin’s theorem show there exists a sequence of continuousfunctions fn : Ω 7→ IR such that ‖fn − f‖Lp(Ω) → 0. Then use the Stone-Weierstrass theorem.
9. For (i), choose δ > 0 such that |f(x, y)− f(x′, y)| ≤ ε whenever |x−x′| ≤ δ. Choose points0 = x0 < x1 < x2 < · · · < xN = a with |xi − xi−1| ≤ δ and define the “tent functions”
ϕi(x).=
xi+1 − xxi+1 − xi
if x ∈ [xi, xi+1]
x− xi−1
xi − xi−1if x ∈ [xi−1, xi]
0 if x /∈ [xi−1, xi+1] .
Estimate the distance ∣∣∣∣∣f(x, y)−N∑
i=0
ϕi(x) f(xi, y)
∣∣∣∣∣ .
For (ii), approximate every function ϕi(·) and f(xi, ·), i = 1, . . . , N − 1 with a sum of sinefunctions, using problem 4.(ii).
10.3 Chapter 4
1. To show that En is closed, assume ‖fk−f‖L1 → 0 and ‖fk‖2L2 ≤ n for every k. By possiblytaking a subsequence we can assume fk(x) → f(x) for a.e. x ∈ [0, 1]. Use Fatou’s lemma toestimate ‖f‖2
L2 .
5. Show that, by possibly replacing Ω with Ω.= Ω ∪ (−Ω), we can assume that Ω = −Ω.
By a Baire category argument, show that Ω has non-empty interior. Then observe that2Ω = Ω + Ω = Ω− Ω.
8. Call ek = (0, . . . , 0, 1, 0, . . .) the sequence of numbers whose k-th element is = 1, while allother elements are = 0. Define g(t) by setting
g(t) =
0 if t ≤ 0,e1 if t ≥ 1,en if t = 1/n, n = 1, 2, 3, . . .
181
and extending g in an affine way on each interval[
1n+1 ,
1n
]. In other words,
g
(θ
n+
1− θn+ 1
)= θen + (1− θ)en+1 for all n ≥ 1, θ ∈ [0, 1].
Show that, for every sequence tk → 0+, the sequence g(tk) has no limit.
10.4 Chapter 5
3. Having proved that (x + x′, y) = (x, y) + (x′, y), conclude that (x + x + · · · + x, y) =(nx, y) = n(x, y) for every positive integer n. Hence also (λx, y) = λ(x, y) for λ rational. Bycontinuity, the same identity must hold for every λ ∈ IR.
7. Hint: consider the set of vectors z ∈ H ; (z, x) = 0 , ‖z‖2 = 1 − ‖x‖2. Then use theresult in the previous problem 6.
12. By taking a subsequence one can assume ‖xn‖ → C. Estimate limn→∞(xn, x).
13. Show that H has an countable orthonormal basis e1, e2, e3, . . .. Then define Λx.= a =
(a1, a2, a3, . . .), with ak.= (x, ek).
v
v2
v1
h
1
v2
v3
h2
3
Figure 10.6: Computing the area of a parallelogram and the volume of a parallelepiped. Here h2 =d(v2 , spanv1) while h3 = d(v3 , spanv1, v2).
14. Let V = spanv1, . . . , vn. To prove (i), let x be the zero vector in (4.1). To prove(ii), observe that, if x ∈ H and y ∈ V then G(x, v1, v2, . . . , vn) = G(x + y, v1, v2, . . . , vn). Inparticular, taking y = PV (x) and z = x− y = PV ⊥(x), we obtain
d(x, V ) = ‖z‖ , G(x, v1, v2, . . . , vn) = G(z, v1, v2, . . . , vn) = (z, z)G(v1, v2, . . . , vn).
To answer (iii), observe that the volume can be computed as a product:
‖vn‖ · d(vn−1 ; spanvn
)· d(vn−2 ; spanvn−1, vn
)· · · d
(v1 ; spanv2, . . . , vn
).
Use (ii) to compute each factor.
17. Let (un)n≥1 be complete. If (vn)n≥1 is not complete, there exists a unit vector x ∈ Hwhich is perpendicular to every vn. A contradiction is obtained from
1 = ‖x‖2 =∞∑
n=1
(x, un)2 =
∞∑
n=1
(x, un − vn)2 ≤∞∑
n=1
‖x‖2 ‖un − vn‖2 ≤∞∑
n=1
‖un − vn‖ .
182
18. (i) Let (un)n≥1 be an orthonormal set. Assume θ0y + θ1u1 + · · · + θNuN = 0, for somecoefficients θj not all zero. If θ0 6= 0, derive a contradiction by taking the inner product ofboth sides with uN+1.
(ii) Let (un)n≥2 be an orthonormal set, and define v.=∑∞
n=2 2−nun, u1
.= v/‖v‖. Observe
that in this case the two vectors y, u1 are parallel.
10.5 Chapter 6
1. Let B1 be the unit ball. By assumption, the image Λ(B1) can be covered with finitelymany balls of radius 2−n, say with centers at y1, . . . , yM(n). Consider the composed op-erators Λn
.= πn Λ, where πn denotes the perpendicular projection on the subspace
Vn.= spany1, . . . , yM(n).
4. If |η| > M , to prove that the Range(ηI − Λ) = H, fix any f ∈ H. Using the contractionmapping theorem, show that the map u 7→ 1
η (f + Λu) has a fixed point.
10.6 Chapter 7
4. To prove convergence, use (5.11) in the previous problem 3, and the continuity of eachtrajectory of the semigroup.
5. Use the result of the previous problem 4.
8. For (i): on the space of all bounded continuous functions w : [0, T ] 7→ X, consider thePicard operator defined as
Φ(w)(t).= Stu+
∫ t
0St−sf(s,w(s)) ds .
Following the proof of Theorem 6.1 in Chapter 7, show that Φ is a strict contraction w.r.t. theequivalent norm
‖w‖† .= max
t∈[0,T ]e−2Lt‖w(t)‖ .
10.7 Chapter 8
3. Consider the smaller square Qε = (x1, x2) ; ε < x1 < 1− ε, ε < x2 < 1 − ε and usemollifications.
5. For (i) use mollifications. For (ii), consider the open set
Ω.= (x1, x2) ; 1 < x21 + x22 < 4 \ (x1, x2) ; x2 = 0 , x1 > 0.
183
Consider the angle function, in polar coordinates.
7. If f ∈W 1,1(Ω), consider a sequence of approximations fn ∈ C∞(Ω) with ‖fn− f‖W 1,1 → 0.Show that the sequence fn(0) is Cauchy, hence it has a unique limit.
8. Construct a sequence of continuous functions fn with ‖fn‖Lp(Ω) → 0 but fn(x) = 1 forevery x ∈ ∂Ω and n ≥ 1.
9. If f ∈W 1,p(IR3), then by Fubini’s theorem for a.e. x1 ∈ IR we have
∫
IR2
(|f(x1, x2, x3)|p + |∇f(x1, x2, x3)|p
)dx2dx3 < ∞ .
If p is suitably large, then f coincides a.e. with a Holder continuous function on the plane(x1, x2, x3) ; x1 = x1 .
10. Consider the subsets Ω1/n.=x ∈ Ω ; |x| ≤ n , B(x, 1/n) ⊂ Ω
. For any f ∈ Lp(Ω),
1 ≤ p <∞, the approximations
fn(x).=
f(x) if x ∈ Ω1/n
0 otherwise
converge to f in Lp. In turn, for ε < 1/n the mollifications Jε ∗ fn lie in C∞c (Ω) and convergeto fn as ε→ 0.
11. Let ϕ : IR 7→ [0, 1] be a C∞ function such that ϕ(r) = 1 for r ≤ 0 and ϕ(r) = 0 for r ≥ 1.Given u ∈ W k,p(IRn), prove that the sequence of functions un(x)
.= ϕ(|x| − n)u(x) converges
to u in W k,p as n→∞. Then take a suitable mollification: u = Jε ∗ un.
12. Consider the cases p = 2 and p 6= 2 separately.
15. Hint: take a weakly converging subsequence umk u in H1(Ω). Show that u = u. Then
let m→∞ in the identity (um − u, um − u)H1 = (um, um)H1 + (u, u)H1 − 2(um, u)H1 .
16. (i) Consider the functions
fn(x).=
1− n+1
n |x| if |x| ≤ nn+1 ,
0 otherwise.
Show that each fn has compact support in Ω. Moreover, ‖fn−f‖W 1,p → 0. Choose a sequenceεn → 0 so that the mollified functions Jεn ∗fn have compact support and converge to f in H1.
(ii) To show that f ∈ W 1,20 (Ω0), let ϕ(x) = log log
(1 + 1
|x|
). Notice that ϕ ∈ W 1,2(Ω) as
claimed in problem 6. For each integer n, define the function
gn(x) = max
0 , min
1− n+ 1
n|x| , n− ϕ(x)
.
184
Show that each gn has compact support in Ω0. Moreover, ‖gn − f‖W 1,p → 0. Choose asequence εn → 0 so that the mollified functions Jεn ∗ fn have compact support contained inΩ0 and converge to f in H1.
(iii) If the functions ϕn ∈ C∞c (Ω0) form a Cauchy sequence w.r.t. the W 1,p norm, with p > 2,then by Morrey’s inequality the sequence (ϕn)n≥1 converges to some continuous function g,uniformly on the closed disc Ω. In particular, g = 0 at the origin. Derive a contradiction,showing that f cannot coincide a.e. with a continuous function that vanishes at the origin.
17. If (8.4) fails, consider a sequence of functions un such that
‖un‖L2 = 1 , ‖∇un‖L2 <1
n,
and meas(x ∈ Ω ; un(x) = 0) ≥ α for every n. Using the fact that H1 is a Hilbert space,select a weakly convergent subsequence so that unj u in H1(Ω). Prove that meas(x ∈Ω ; u(x) = 0) ≥ α and derive a contradiction.
10.8 Chapter 9
1. The matrix
(1 x/2x/2 1
)is strictly positive definite for (x, y) ∈ Ω.
3. For (i), choose γ.= ‖c‖L∞ and write Lu = (Lu + γu) − γu. Show that the operator
Lγ.= L+ γI is strictly positive definite. Its inverse L−1
γ is compact and symmetric.
5. For (i), use the definition (1.5) of weak solution choosing v = u = φ. For (ii), use therepresentation formula (2.10), observing that µk = 1/λk ≥ 1/β2 for every k ≥ 1.
6. Choose γ.= maxt∈[0,T ] q(t), then check that the elliptic operator Lγu
.= −(p(t)u′)′ + (γ −
q(t))u is strictly positive definite on H10 (]0, T [).
8. (i) Let φk ; k ≥ 1 be an orthonormal basis of L2(Ω) consisting of eigenfunctions of theLaplace operator ∆. Since ϕ, ∆ϕ ∈ C∞c (Ω) ⊂ L2(Ω), one can write the convergent series
ϕ =∑
k
(ϕ, φk)L2φk −∆ϕ =∑
k
(−∆ϕ, φk)L2φk =∑
k
µk(ϕ, φk)L2φk . (10.1)
Use (10.1) to compute the inner product
∫
Ω(−∆ϕ)ϕdx =
∫
Ω|∇ϕ|2 dx.
(ii) Given u ∈ H10 (Ω), choose a sequence of functions ϕm ∈ C∞c (Ω) such that ‖u−ϕm‖H1 → 0
as m → ∞. Use (i). Finally, to show that the constant C = 1/µ1 cannot be improved, takeu = φ1.
9. Every function u ∈ H10 (Ω) can be regarded also a function in H1
0 (Ω), equal to zero on
185
Ω \ Ω. Using 8.(ii), prove that
µ1 = minu∈H1
0 (Ω), u 6=0
∫Ω|∇u|2 dx∫Ω|u|2 ≤ min
u∈H10 (Ω), u 6=0
∫Ω |∇u|2 dx∫
Ω |u|2= µ1 . (10.2)
10. (i) By Problem 9 in Chapter 2, every function ϕ ∈ C∞c (Q) can be uniformly approx-imated by a finite linear combination of the functions φm,n. Use this fact to prove thatspanφm,n ; m,n ≥ 1 is dense in L2(Q).
(ii) If µ1 is the smallest eigenvalue of −∆ on Ω, then by (10.2) the assumption Ω ⊆ Q implies
µ1 ≥ µ1,1 = π2
a2 + π2
b2 . Use the result in problem 8.(ii) to conclude.
11. Consider the m ×m matrix with entries Bij =∫Ω∇ϕi · ∇ϕj dx and the m-vector with
components ai =∫Ω f ϕi dx. Use the assumption that the ϕi are linearly independent in order
to show that the matrix (Bij) is strictly positive definite.
10.9 Chapter 10
1. Let x0 be a Lebesgue point for f and choose ϕ(x) = Jε(x − x0), where Jε is the mollifierdefined at (5.1).
2. After a relabelling, we can assume that K ⊆ A1 ∪ · · · ∪ An. Consider the continuousfunction f(x)
.= maxf1(x), . . . , fn(x) , where
fj(x).=
sup
r ; B(x, r) ⊆ Aj
if x ∈ Aj ,
0 if x /∈ Aj .
3. Show that
Jε(x) ≥ φε(x).=
δ
εif 0 < x <
ε
2
0 otherwise
for some constant δ > 0. Taking ε = 2x, notice that F (x) ≥ (φ2x ∗ f)(x).
186
Bibliography
[1] L. C. Evans, Partial Differential Equations. American Mathematical Society, Providence,RI, 1998.
[2] G. B. Folland, Real analysis. Modern techniques and their applications. Second edition.Wiley, New York, 1999.
[3] V. Hutson, J. S. Pym, and M. J. Cloud, Applications of Functional Analysis and OperatorTheory. Second edition. Elsevier, Amsterdam, 2005.
[4] S. Kesavan, Topics in Functional Analysis and Applications. Wiley, New York, 1989.
[5] R. H. Martin, Nonlinear Operators and Differential Equations in Banach Spaces. Wiley,New York, 1976.
[6] M. Miklavcic, Applied Functional Analysis and Partial Differential Equations. World Sci-entific, River Edge, NJ, 1998.
[7] D. Mitrovi and D. Ubrini, Fundamentals of Applied Functional Analysis. Pitman, Long-man, Harlow, 1998.
[8] W. Rudin, Functional analysis. McGraw-Hill,1973.
187
