MAT 578 Functional Analysis Lecture Notesquigg/teach/courses/578/2019fall... · 2020. 3. 17. · 6 Locally convex spaces 36 7 Weak topologies 43 8 Extreme points 50 Introduction Most

MAT 578 Functional AnalysisLecture Notes

Professor: John Quigg

Semester: Fall 2019

Revised February 23, 2020

c©2020 Arizona State University School of Mathematical & Statistical Sciences

Contents

1 Banach spaces 2

2 Bounded linear maps 7

3 Hahn-Banach Theorem 14

4 Hilbert spaces 20

5 Baire Category consequences 31

6 Locally convex spaces 36

7 Weak topologies 43

8 Extreme points 50

Introduction

Most of these notes follow the books by Conway, Folland (particularly Chapter 5), and Rudin[2, 3, 4].

All vector spaces will have scalar field F = R or C. We say that X is a vector space overF, and we say that X is real or complex if F = R or C, respectively.

1

1 BANACH SPACES 2

1 Banach spaces

Definition 1.1. A norm on a vector space X is a function ‖ · ‖ : X → R such that:

(1) ‖x‖ ≥ 0 for all x ∈ X, and ‖x‖ = 0 if and only if x = 0;

(2) ‖cx‖ = |c|‖x‖ for all c ∈ F and x ∈ X;

(3) ‖x+ y‖ ≤ ‖x‖+ ‖y‖ for all x, y ∈ X.

When ‖ · ‖ is a norm on X we call the pair (X, ‖ · ‖) a normed space.

Property (3) is called the triangle inequality.

If we weaken Property (1) to allow ‖x‖ = 0 with x 6= 0, then we have a seminorm. Ifwe go further, eliminating (1) and also weakening Property (2) to only be for c ≥ 0, thenwe have a sublinear functional. An element x in a normed space is called a unit vector if‖x‖ = 1. The open ball of radius r > 0 centered at x ∈ X is Br(x) = y ∈ X : ‖y−x‖ < r,and we just write Br for Br(0). The closed unit ball of X is

BX = x ∈ X : ‖x‖ ≤ 1.

In practice, we frequently just say “X is a normed space” and use the notation ‖ · ‖ forthe norm without comment.

Lemma 1.2. A normed space X becomes a metric space with d(x, y) = ‖x− y‖.

Definition 1.3. A Banach space is a complete normed space.

We’ll see later that the completion of a normed space is a Banach space.

Definition 1.4. Two norms ‖ · ‖ and ‖ · ‖′ on X are called equivalent if there are constantsc, d > 0 such that

c‖x‖ ≤ ‖x‖′ ≤ d‖x‖ for all x ∈ X.

Lemma 1.5. Equivalent norms determine equivalent metrics, and in particular the sameconvergent sequences.

Example 1.6. The 1-norm on C[0, 1] is not equivalent to the sup norm, because the sequence(xn) converges to 0 in the 1-norm but not in the sup norm.

We’ll see later that all norms on a finite-dimensional vector space are equivalent.

Definition 1.7. A vector subspace Y of a normed space X becomes a normed space withthe restriction of the norm, and is just called a subspace of X.

1 BANACH SPACES 3

Lemma 1.8. A closed subspace of a Banach space is a Banach space.

Recall from linear algebra that the span of a subset E of a vector space X is the subspace

spanE =

n∑i=1

cixi : ci ∈ F, xi ∈ X

,

and satisfies: spanE is the intersection of all the subspaces of X containing E, and if Y isany subspace of X containing E then spanE ⊆ Y , so that spanE is in fact the smallestsubspace of X containing E.

Definition 1.9. If X is a normed space and E ⊆ X, then the closed span of E is

spanE = spanE.

Lemma 1.10. If X is a normed space and E ⊆ X, then spanE is the intersection of allthe closed subspaces of X containing E, and if Y is any closed subspace of X containing Ethen spanE ⊆ Y .

Thus spanE is in fact the smallest closed subspace containing E.

Definition 1.11. The direct sum of normed spaces X and Y is the vector-space direct sumX ⊕ Y with the product norm

‖(x, y)‖ = max‖x‖, ‖y‖.

We sometimes call this the external direct sum, to distinguish between this and the construc-tion in Definition 1.13 below.

Lemma 1.12. The direct sum of two Banach spaces is a Banach space.

Definition 1.13. Let X be a normed space.

(1) If Y, Z are closed subspaces with X = Y +Z and Y ∩Z = 0, we say X is the internaldirect sum of Y and Z, written X = Y ⊕ Z.

(2) A closed subspace Y of X is complemented in X if there exists a closed subspace Z ofX such that X = Y ⊕ Z.

We will see later that if X is complete then the internal and external direct sums areisomorphic as Banach spaces.

Definition 1.14. If Y is a closed subspace of a normed space X, then the quotient space isthe quotient vector space X/Y with the quotient norm

‖x+ Y ‖ = infy∈Y‖x+ y‖.

1 BANACH SPACES 4

Lemma 1.15. The quotient norm really is a norm.

Proof. The only nonobvious part is the triangle inequality: let x, y ∈ X. Then for anyz, w ∈ Y we have

‖(x+ Y ) + (y + Y )‖ = ‖(x+ y) + Y ‖≤ ‖x+ y + z + w‖≤ ‖x+ z‖+ ‖y + w‖.

Taking the inf over z, w gives

‖(x+ Y ) + (y + Y )‖ ≤ ‖x+ Y ‖+ ‖y + Y ‖.

We will wait until the next section to prove that the quotient of a Banach space bya closed subspace is a Banach space, because the most efficient proof uses bounded linearmaps.

Exercises

1. Prove that if p is a seminorm on X, then

|p(x)− p(y)| ≤ p(x− y) for all x, y ∈ X.

2. If X and Y are normed spaces, then 0 ⊕ Y is a closed subspace of X ⊕ Y , and (X ⊕Y )/(0 ⊕ Y ) is isometrically isomorphic to X.

3. The closure of any subspace of a normed space X is a subspace of X.

4. If X is a topological space, then the set Cb(X) of continuous bounded (scalar-valued) func-tions on X becomes a Banach space with the sup norm (also called the uniform norm)

‖f‖u = supx∈X|f(x)|.

If X is compact we just write C(X), since boundedness follows from continuity.

5. If X is a locally compact Hausdorff space, then the set C0(X) of continuous functions f onX that vanish at infinity, in the sense that for all ε > 0 the set

|f | ≥ ε = x ∈ X : |f(x)| ≥ ε

is compact, is a closed subspace of Cb(X), and hence a Banach space. If X has the discretetopology (i.e., is just a set), we write c0(X) for C0(X).

1 BANACH SPACES 5

6. (a) If X is a locally compact Hausdorff space, then for all f ∈ C0(X) the set

f 6= 0 = x ∈ X : f(x) 6= 0

is σ-compact, i.e., is a countable union of compact sets.

(b) If X is any set then for all f ∈ c0(X) the set f 6= 0 is countable.

7. If (X,M, µ) is a measure space and 1 ≤ p ≤ ∞, then the set Lp(X,µ) is a Banach spacewith the p-norm ‖ · ‖p. If M = P(X) and µ is counting measure, we write `p(X). Notethat `∞(X) = Cb(X) when X is given the discrete topology. We write `p if X = N and `pn ifX = 1, . . . , n. We also write Fn for the particular case `2n. Also, we write c0 for C0(N).

8. The set c = x ∈ `∞ : limn→∞ xn exists of convergent sequences is a Banach subspace.

9. If X is a locally compact Hausdorff space, then the set Cc(X) of continuous functions f onX whose support

supp f = f 6= 0

is compact is a dense subspace of C0(X).

10. If a < b in R then the set C1[a, b] of continuously differentiable functions f in [a, b] is a densesubspace of C[a, b], but becomes a Banach space with the norm

‖f‖ = ‖f‖u + ‖f ′‖u.

11. If X is a normed space, then

(a) the vector space operations

• (c, x) 7→ cx : F×X → X and

• (x, y) 7→ x+ y : X ×X → X

are continuous, and

(b) the norm ‖ · ‖ : X → R is continuous.

12. The direct sum of a family Xii∈S of normed spaces, written⊕∞

i∈S Xi, is the set of elementsx = (xi) in the Cartesian product set

∏i∈S Xi whose product norm

‖x‖∞ = supi∈S‖xi‖

is finite.⊕∞

i∈S Xi is a normed space, and is a Banach space if and only if every Xi is.

Instructor comment : There are other possible `p choices for the norm, but we won’t needthem.

1 BANACH SPACES 6

13. Let p be a seminorm on a vector X, and let Z = x ∈ X : p(x) = 0. Then Z is a vectorsubspace of X, and the quotient vector space X/Z becomes a normed space with

‖x+ Z‖ = p(x).

14. (Riesz Lemma) Let Y be a proper closed subspace of a normed space X. Then for all0 < a < 1 there exists x ∈ X such that ‖x‖ = 1 and ‖x− y‖ > a for all y ∈ Y . (Hint: firstuse the quotient norm to find x with ‖x‖ < 1, then scale it.)

15. The Banach spaces c0 and `p for p <∞ are separable, but `∞ is not separable.

16. C[0, 1] is separable.

17. (Series in normed spaces) Let (xn) be a sequence in a normed space X. The series with termsxn, written

∑∞n=1 xn, is the sequence (sk =

∑kn=1 xn)∞k=1 of partial sums. The series converges

to the sum∑∞

n=1 xn if the sequence (sk) of partial sums does, and converges absolutely if theseries

∑∞n=1 ‖xn‖ converges in R. Prove that X is complete if and only if every absolutely

convergent series in X converges in X. (Hint for the hard direction: let (xn) be a Cauchysequence in X, and find a subsequence (yk) such that

‖yk − yk+1‖ < 2−k for all k ∈ N,

and then consider the series whose partial sums are (yk).)

18. Every absolutely convergent series in a Banach space converges unconditionally, i.e., iff : N→ N is a bijection, then

∞∑n=1

xn =∞∑n=1

xf(n).

19. (Nets in normed spaces) Since a normed space X is a metric space, sequence are sufficient todetermine topological properties. However, sometimes nets arise anyhow, so we must knowhow to deal with them. Let S be a directed set, and let (xi)i∈S be a net in X. From yourexperience with topological spaces, you are not at all surprised by the definition that (xi)converges to x, written xi → x, if for all ε > 0 there exists t ∈ S such that

‖xi − x‖ < ε for all i ≥ t.

Similarly, you are probably not surprised by: (xi) is Cauchy if for all ε > 0 there exists t ∈ Ssuch that

‖xi − xj‖ < ε for all i, j ≥ t.

What you must do in this problem is prove the following: if X is a Banach space, then everyCauchy net converges. (Of course, the converse it trivial, since sequences are special casesof nets.)

2 BOUNDED LINEAR MAPS 7

20. (Generalized series) We will occasionally need a more general notion of series than in Prob-lem 17. Let X be a normed space, let S be a set, and let (xi)i∈S be an S-tuple in X. Let Fdenote the family of finite subsets of S, and for each F ∈ F let

yF =∑i∈F

xi.

Then the series with terms xii∈S, written∑

i∈S xi, is the net (yF )F∈F , where F is orderedby inclusion. Each yF is called a partial sum of the series. If yF → x we say x is the sum ofthe series

∑i∈S xi, and we indicated this by writing x =

∑i∈S xi. What you must do in this

problem is prove that the series is Cauchy (meaning that the net of partial sums is Cauchy)if and only if for all ε > 0 there exists a finite subset F ⊆ S such that∥∥∥∥∥∑

i∈E

xi

∥∥∥∥∥ < ε for all finite E ⊆ S \ F.

21. Let∑

i∈S xi be a Cauchy series in a normed space X. Define f : S → F by

f(i) = ‖xi‖.

(a) f ∈ c0(S).

(b) A = i ∈ S : xi 6= 0 is countable.

(c) Assuming that the set A defined in the preceding part is infinite, choose any bijection

n 7→ in : N→ S.

Then the series∑∞

n=1 xin is Cauchy.

(d) If∑

i∈S xi converges to x then so does∑∞

n=1 xin .

(e) The converse of part (d) is false.

2 Bounded linear maps

Definition 2.1. If X and Y are normed spaces, a linear map T : X → Y is called boundedif

sup‖Tx‖ : ‖x‖ = 1 <∞.

The above sup is called the operator norm of T and is written ‖T‖, and we write B(X, Y )for the set of all bounded linear maps X → Y . We write B(X) = B(X,X), and call linearmaps from X to itself operators on X.


It is an exercise that the boundedness condition is equivalent to

supx∈BX

‖Tx‖ <∞.

Lemma 2.2. B(X, Y ) is a normed space, and is a Banach space if Y is.

Proof. We only prove the triangle inequality and completeness; the rest is an exercise. First,if S, T ∈ B(X, Y ) and ‖x‖ = 1, then

‖(S + T )x‖ = ‖Sx+ Tx‖≤ ‖Sx‖+ ‖Tx‖≤ ‖S‖+ ‖T‖.

Taking the sup over x gives‖S + T‖ ≤ ‖S‖+ ‖T‖.

For the other part, let (Tn) be a Cauchy sequence in B(X, Y ), and assume that Y is complete.For each x ∈ X the sequence (Txn) is Cauchy in Y , since

‖Txn − Txk‖ = ‖T (xn − xk)‖ ≤ ‖T‖‖xn − xk‖.

Thus (Txn) converges in Y , and we define T : X → Y by

Tx = limTxn.

Obvious computations show that T is linear; we will show that it is bounded. Note that sincethe sequence (Tn) is Cauchy it is bounded, so we can choose M > 0 such that ‖Tn‖ ≤M forall n. Let ‖x‖ = 1. For each n we have ‖Tnx‖ ≤M , so

‖Tx‖ = lim ‖Tnx‖ ≤M.

It remains to show that Tn → T in B(X). Let ε > 0, and choose p ∈ N such that‖Tn − Tk‖ < ε for all n, k ≥ p. Let n ≥ p and ‖x‖ = 1. Then

‖Tnx− Tkx‖ < ε for all k ≥ p,

so, letting k →∞, we get‖Tnx− Tx‖ ≤ ε.

Taking the sup over x gives ‖Tn−T‖ ≤ ε for all n ≥ p, and we conclude that ‖Tn−T‖ → 0,i.e., Tn → T in the normed space B(X).

Lemma 2.3. If X and Y are normed spaces and T : X → Y is linear, then the followingare equivalent:

(1) T is bounded;


(2) T is continuous;

(3) T is continuous at 0.

Proof. Assume (1). Since

‖Tx− Ty‖ = ‖T (x− y)‖ ≤ ‖T‖‖x− y‖,

T is continuous (in fact Lipschitz).

(2)⇒(3) is trivial. Assume (3). Choose δ > 0 such that

‖Tx‖ < 1 whenever ‖x‖ < δ.

Then ‖Tx‖ ≤ 1 whenever ‖x‖ ≤ δ, so ‖x‖ = 1 implies

‖Tx‖ =

∥∥∥∥1

δT (δx)

∥∥∥∥ =1

δ‖T (δx)‖ ≤ 1

δ,

and therefore T is bounded.

As our first application of bounded linear maps, we complete (sorry!) the basic theoryof quotient spaces:

Corollary 2.4. The quotient of a Banach space by a closed subspace is a Banach space.

Proof. Let X be a Banach space and Y ⊆ X a closed subspace, and let∑∞

n=1(xn + Y ) bean absolutely convergent series in X/Y . For each n choose yn ∈ Y such that

‖xn + yn‖ ≤ ‖xn + Y ‖+ 2−n.

Then ∑n

‖xn + yn‖ ≤∑n

‖xn + Y ‖+∑n

2−n <∞.

Since X is complete, the series∑

n(xn + yn) converges to a vector z. Since the quotient mapQ : X → X/Y is linear and bounded, we have

Qz =∑n

Q(xn + yn) =∑n

(xn + Y ).

Thus the series∑

n(xn + Y ) converges in X/Y .

Example 2.5. Let (X,M, µ) be a σ-finite measure space and g ∈ L∞(X,µ). For f ∈Lp(X,µ) define Mgf by

(Mgf)(x) = g(x)f(x).

Then Mg ∈ B(Lp(X,µ)) and ‖Mg‖ = ‖g‖∞. We call Mg a multiplication operator.


Example 2.6. If k ∈ C([0, 1]×[0, 1]), then there is a bounded linear operator K on L2([0, 1])defined by

(Kf)(x) =

∫k(x, y)f(y) dy.

K is called a kernel operator, and k is called its kernel

Example 2.7. The Volterra operator V ∈ B(L2[0, 1]) is defined by

(V f)(x) =

∫ x

0

f(y) dy.

Lemma 2.8. If T ∈ B(X, Y ) and S ∈ B(Y, Z) then the composition ST = S T is bounded,and

‖ST‖ ≤ ‖S‖‖T‖.

Definition 2.9. T ∈ B(X, Y ) is:

(1) an isomorphism, or invertible, if it is bijective and T−1 : Y → X is bounded;

(2) isometric, or an isometry, if ‖Tx‖ = ‖x‖ for all x ∈ X;

(3) an isometric isomorphism if it is both isometric and an isomorphism.

Sometimes we say linear isometry for emphasis, because “isometry” technically couldjust mean a metric thing.

Example 2.10. If Y is a subspace of a normed space X, then the inclusion map T : Y → Xdefined by Ty = y is an isometry.

Example 2.11. The forward shift on `2, defined by

S(x1, x2, . . . ) = (0, x1, x2, . . . )

is an isometry that is not invertible.

Definition 2.12. Let X amd Y be Banach spaces. A linear map T : X → Y is calledcompact if T (BX) has compact closure in Y . K(X, Y ) denotes the set of all compact linearmaps.

Although it would make sense to define compact linear maps between possibly incompletenormed spaces, we will have no use for this, and it would significantly alter the development ofthe theory. In particular, the proof of (1) of the following lemma depends upon completenessof Y .

Lemma 2.13. Let X and Y be Banach spaces.

(1) K(X, Y ) is a closed subspace of B(X, Y ).


(2) For every Banach space Z,

K(Y, Z)B(X, Y ) ∪B(Y, Z)K(X, Y ) ⊆ K(X,Z).

Proof. (1) Let (Tn) be a sequence in K(X, Y ), and suppose that Tn → T in B(X, Y ). Letε > 0. Choose k ∈ N such that ‖Tk − T‖ < ε. Then choose x1, . . . , xp ∈ BX such that

Tk(BX) ⊆p⋃i=1

Bε(Tkxi).

Let y ∈ BX . Choose i such that Tky ∈ Bε(Tkxi). Then

‖Ty − Tkxi‖ ≤ ‖Ty − Tky‖+ ‖Tiy − Tkxi‖< ‖T − Tk‖‖y‖+ ε

≤ 2ε.

Thus

T (BX) ⊆p⋃i=1

B2ε(Tkxi),

and we have shown that T (BX) is totally bounded. Since Y is complete, T (BX) is compact.

(2) Let T ∈ B(X, Y ) and S ∈ B(Y, Z). If T is compact then

ST (BX) = S(T (BX)) ⊆ S(T (BX)

),

which is compact because S is continuous and T (BX) is compact. On the other hand, if Sis compact then

ST (BX) = S(T (BX))

⊆ S (‖T‖BY )

= ‖T‖S(BY ),

which has compact closure because S(BY ) does.

Exercises

1. If T ∈ B(X, Y ), then

‖T‖ = supx∈BX

‖Tx‖

= sup

‖Tx‖‖x‖

: x 6= 0

.

= infM ≥ 0 : ‖Tx‖ ≤M‖x‖ for all x ∈ X= minM ≥ 0 : ‖Tx‖ ≤M‖x‖ for all x ∈ X.


2. The maps

• (T, x) 7→ Tx : B(X, Y )×X → Y and

• (T, S) 7→ ST : B(X, Y )×B(Y, Z)→ B(X,Z)

are continuous.

3. If T : X → Y is a linear bijection, then T−1 is bounded if and only if T is bounded below inthe sense that there exists M > 0 such that

‖Tx‖ ≥M‖x‖ for all x ∈ X.

4. Prove that all norms on a finite-dimensional vector space are equivalent by verifying all thestatements in the following outline:

(a) Let X be a vector space with basis e1, . . . , en. The linear map T : Fn → X by definedby

T (x1, · · · , xn) =n∑i=1

xiei.

is bounded.

(b) T takes the unit sphereS = x ∈ Fn : ‖x‖ = 1

to a compact subset of X not containing 0. It follows that the inverse T−1 : X → Fn isbounded.

(c) Parts (a) and (b) together imply the statement we are trying to prove.

5. Every finite-dimensional subspace of a normed space is closed.

6. Let Y and Z be subspaces of a normed space X. If Y is closed and Z is finite-dimensional,then Y + Z is closed.

7. Every subspace of X that has either finite dimension or finite codimension is complemented.

8. If X and Y are normed spaces and X is finite-dimensional then every linear map T : X → Yis bounded.

9. For a nonzero linear functional f on a normed space X, the following are equivalent:

(i) f is bounded;

(ii) ker f is closed;

(iii) ker f is not dense.


10. Every infinite-dimensional Banach space has an unbounded linear functional. (Hint: choosea Hamel basis of unit vectors. Note: a Hamel basis is a basis in the sense of linear algebra,i.e., a maximal linearly independent subset.)

11. If X is an infinite-dimensional Banach space, then BX is not compact. (Hint: use the RieszLemma to inductively construct a sequence (xn) in BX such that ‖xn − xk‖ ≥ 1/2 for alln 6= k.)

12. The identity operator I on a Banach space X is compact if and only if X is finite-dimensional.

13. Let Y be a closed subspace of X. The quotient map Q : X → X/Y is defined by Qx = x+Y .Suppose that Y is a proper subspace — i.e., 6= X.

(a) ‖Q‖ = 1.

(b) Q is open in the sense that Q(U) is open whenever U is.

(c) X/Y has the quotient topology in the sense that Q−1(V ) is open if and only if V is.

14. Let T ∈ B(X, Y ). Put K = kerT , and let Q : X → X/K be the quotient map.

(a) There is a unique linear map T such that the diagram

XT //

Q

K

X/KT

!

<<

commutes.

(b) T is injective, ran T = ranT , and ‖T‖ = ‖T‖.

15. If X =⊕∞

i Xi is a product of normed spaces, then for each i the coordinate projectionπi : X → Xi defined by

πi(x) = xi

is linear and has norm 1.

16. Let X and Y be topological spaces, and let f : X → Y be continuous. Define a mapf ∗ : Cb(Y )→ Cb(X) by f ∗(g) = g f .

(a) f ∗ is a bounded linear map.

(b) If f is a homeomorphism then f ∗ is an isometric isomorphism.

3 HAHN-BANACH THEOREM 14

17. If 1 ≤ p ≤ ∞ then `p(2, 3, 4, . . . ) is isometrically isomorphic to `p.

18. Let C1[0, 1] have the norm ‖f‖ = ‖f‖u + ‖f ′‖u.

(a) The map D : C1[0, 1]→ C[0, 1] defined by Df = f ′ is bounded. What is its norm?

(b) The map D : C1[0, 1]→ C[0, 1] defined by Df = f is compact.

19. Let k ∈ C([0, 1]× [0, 1]).

(a) The operator T on C[0, 1] defined by

(Tf)(x) =

∫ 1

0

k(x, y)f(y) dy

is compact.

(b) The kernel operator K on L2[0, 1] with kernel k is compact.

20. The Volterra operator on L2[0, 1] is compact.

3 Hahn-Banach Theorem

Throughout this section X will be a normed space (except in examples where it may be ameasure space or a topological space).

Definition 3.1. The dual space of X is the Banach space X∗ = B(X,F) of bounded linearfunctionals on X.

Example 3.2. Let (X,M, µ) be a σ-finite measure space, 1 ≤ p < ∞, and 1/p + 1/q = 1.For each g ∈ Lq(X,µ) define φg : Lp(X,µ)→ F by

φg(f) =

∫X

fg dµ.

Then g 7→ φg is an isometric isomorphism of Lq onto (Lp)∗, and we frequently abuse notationby writing (Lp)∗ = Lq.

When 1 < p < ∞, the σ-finiteness condition can be removed. In fact, for countingmeasure we even have `1(X)∗ = `∞(X).


Example 3.3. Let X be a locally compact Hausdorff space. Then the set M(X) of regularcomplex Borel measures on X becomes a Banach space with pointwise addition and scalarmultiplication, and norm

‖µ‖ = |µ|(X),

where |µ| is the total variation measure, i.e., the smallest positive Borel measure on X suchthat |µ(E)| ≤ |µ|(E) for every Borel set E. For µ ∈M(X) define φµ : C0(X)→ F by

φµ(f) =

∫X

f dµ.

Then the Riesz Representation Theorem says that µ 7→ φµ is an isometric isomorphism ofM(X) onto C0(X)∗, and we just write C0(X)∗ = M(X).

A fundamental special case is c0(X)∗ = `1(X) for any set X, since M(X) can be identifiedwith `1(X) (since every regular complex Borel measure is absolutely continuous with respectto counting measure).

Theorem 3.4 (Hahn-Banach Theorem). Let Y be a subspace of a normed space X, and letf ∈ Y ∗. Then there exists g ∈ X∗ such that

(1) g|Y = f ;

(2) ‖g‖ = ‖f‖.

The above formulation is the most frequently used. However, we now state a significantlymore general version, from which the above Hahn-Banach Theorem will quickly follow. Firstrecall that a sublinear functional on a vector space X is a map p : X → R such that

• p(x+ y) ≤ p(x) + p(y) for all x, y ∈ X;

• p(cx) = cp(x) for all x ∈ X and c ≥ 0.

Theorem 3.5 (Hahn-Banach Theorem for sublinear functionals). Let Y be a subspace of areal vector space X, let p be a sublinear functional on X, and let f : Y → R be linear withf ≤ p on Y . Then there exists a linear g : X → R such that g|Y = f and g ≤ p.

Once we have this result, we will first of all be able to deduce a version for seminorms:

Corollary 3.6 (Hahn-Banach Theorem for seminorms). Let Y be a subspace of a vectorspace X, let p be a seminorm on X, and let f : Y → F be linear with |f | ≤ p on Y . Thenthere exists a linear g : X → F such that g|Y = f and |g| ≤ p.

Proof. To see this, first assume that F = R, and note that since p is a seminorm, by Problem 1we have f ≤ p if and only if |f | ≤ p, so Theorem 3.5 does the job.

On the other hand, if F = C, by Problem 2 we can regard X as a real vector space, thenapply the above to Re f instead of f . We get a real-linear extension u of Re f to X with|u| ≤ p, and then by Problem 2 again we get a complex-linear g extending f to X such that|g| ≤ p.


Proof of Theorem 3.4. Apply Corollary 3.6 to p(x) = ‖f‖‖x‖,

Now we (finally!) prove the sublinear version of the Hahn-Banach theorem:

Proof of Theorem 3.5. We extend one dimension at a time and apply Zorn’s lemma. Firstlet x ∈ X \ Y . Note that for all y, u ∈ Y ,

f(y) + f(u) = f(y + u)

≤ p(y + u)

= p(y − x+ x+ u)

≤ p(y − x) + p(x+ u),

sof(y)− p(y − x) ≤ p(x+ u)− f(u).

Thussupy∈Y

(f(y)− p(y − x)

)≤ inf

u∈Y

(p(x+ u)− f(u)

),

so we can choose c ∈ R such that

f(y)− p(y − x) ≤ c ≤ p(x+ u)− f(u) for all y, u ∈ Y.

Define g : Y + Rx byg(y + tx) = f(y) + tc.

Then g is a linear extension of f , and we will show that g ≤ p on Y + Rx. If y ∈ Y andt > 0,

g(y + tx) = f(y) + tc

= t(f(yt

)+ c)

≤ t(f(yt

)+ p

(x+

y

t

)− f

(yt

))= p(y + tx),

while

g(y − tx) = f(y)− tc

= t(f(yt

)− c)

≤ t(f(yt

)−(f(yt

)− p

(yt− x)))

= p(y − tx).

Now consider the set E of all linear extensions of f dominated by p, ordered by inclusion(identifying functions with their graphs). Since the union of any totally ordered subset ofE is again in E , by Zorn’s lemma E has a maximal element. That is, there is a subspace Zof X and a linear extension g of f to Z dominated by p such that there is no proper linearextension of g dominated by p. But then we must have Z = X, since we could always extendby one dimension.


Corollary 3.7. For all x ∈ X \ 0 there exists f ∈ X∗ such that ‖f‖ = 1 and f(x) = ‖x‖.

Proof. Define g ∈ (Fx)∗ by g(tx) = t‖x‖. Then |g(tx)| = ‖tx‖ for all tx ∈ Fx, so ‖g‖ = 1.By the Hahn-Banach theorem we can extend g to f ∈ X∗ with ‖f‖ = 1. Since f(x) =g(x) = ‖x‖, the proof is complete.

Corollary 3.8. There is a linear isometry J : X → X∗∗ defined by

(Jx)(f) = f(x) for x ∈ X, f ∈ X∗.

We also writex = Jx.

Proof. Obviously J is a linear map of X into the vector space of all linear functionals onX∗. The point is to show that ‖Jx‖ = ‖x‖. If f ∈ BX∗ we have

|f(x)| ≤ ‖f‖‖x‖ ≤ ‖x‖,

so ‖Jx‖ ≤ ‖x‖. On the other hand, Corollary 3.7 implies that ‖Jx‖ ≥ ‖x‖, so we must have‖Jx‖ = ‖x‖.

Remark 3.9. We frequently identify X with its image J(X) ⊆ X∗∗. Then we can regardthe norm closure X in X∗∗ as a completion of X.

Definition 3.10. In Corollary 3.8, J is called the canonical embedding into the double dual.X is called reflexive if J is surjective (and hence an isomorphism). If X is reflexive, we abusenotation by writing X = X∗∗.

Example 3.11. Every finite-dimensional normed space is reflexive.

Example 3.12. If 1 < p <∞ and (X,µ) is a measure space then Lp(X,µ) is reflexive.

Lemma 3.13. If T ∈ B(X, Y ) and f ∈ Y ∗ define T ∗f = f T . The map T 7→ T ∗ is alinear isometry from B(X, Y ) to B(Y ∗, X∗).

Proof. Since the map

B(Y,F)×B(X, Y ) = Y ∗ ×B(X, Y )→ B(X,F) = X∗

is bilinear and satisfies‖f T‖ ≤ ‖f‖‖T‖,

T ∗ is a bounded linear map from X∗ to Y ∗ of norm at most ‖T‖. On the other hand, ifa < ‖T‖ then we can choose x ∈ X such that ‖x‖ = 1 and ‖Tx‖ > a, and then by theHahn-Banach theorem we can choose f ∈ Y ∗ such that ‖f‖ = 1 and |f(Tx)| > a, so that

a < |T ∗f(x)| ≤ ‖T ∗‖‖f‖‖x‖ = ‖T ∗‖.

Taking the sup over a < ‖T‖, we see that ‖T‖ ≤ ‖T ∗‖, and therefore we must have ‖T ∗‖ =‖T‖. Finally, again by the bilinearity, the map T 7→ T ∗ is linear.

Definition 3.14. With the above notation, T ∗ is the dual map of T .


Exercises

1. If p is a seminorm on X and f : X → R is linear, then f ≤ p if and only if |f | ≤ p.

2. (a) Let X be a complex vector space. Note that X can alternatively be regarded as a realvector space by restricting scalars. Let f : X → C be linear. Then u = Re f : X → R islinear and

f(x) = u(x)− iu(ix) for all x ∈ X. (1)

Conversely, if u : X → R is linear then Equation (1) defines a linear map f : X → C. Thisgives a bijection between C-linear and R-linear functionals on X.

(b) Now suppose in addition that p is a seminorm on X. Then |f | ≤ p if and only if |u| ≤ p.(Hint: if |u(x)| ≤ p(x) and f(x) = α|f(x)| with α ∈ T, then f(αx) = u(αx).)

(c) If X is a normed space, then ‖u‖ = ‖f‖.

3. The annihilator of a subset E ⊆ X is the set

E⊥ = f ∈ X∗ : f(x) = 0 for all x ∈ E,

and the preannihilator of a subset R ⊆ X∗ is the set

⊥R = x ∈ X : f(x) = 0 for all f ∈ R.

(a) E⊥ is a closed subspace of X∗.

(b) ⊥(E⊥) is the closed subspace spanned by E.

(c) If Y is a closed subspace of X then Y = ⊥(Y ⊥).

(d) A subspace Y of X is dense if and only if Y ⊥ = 0.

4. Let X and Y be Banach spaces and T ∈ B(X, Y ). Then:

(a) kerT ∗ = (ranT )⊥;

(b) kerT = ⊥(ranT ∗).

(c) Then T ∗ is injective if and only if ranT is dense in Y .

Instructor comment : The companion result with T and T ∗ switched is more subtle, and willhave to wait until we study weak topologies in Section 7.


5. If Y is a closed subspace of X and x ∈ X \ Y , then there exists f ∈ X∗ such that f(x) 6= 0and Y ⊆ ker f .

6. If Y is a closed subspace of X and Q : X → X/Y is the quotient map, then the dual mapQ∗ : (X/Y )∗ → X∗ is an isometric isomorphism onto Y ⊥. Consequently, for all x ∈ X thereexists f ∈ Y ⊥ such that ‖f‖ = 1 and

f(x) = ‖x+ Y ‖.

7. Let Y be a closed subspace of X and S : Y → X the inclusion map.

(a) The dual map S∗ : X∗ → Y ∗ is given by S∗f = f |X , and is a linear surjection with kernelY ⊥.

(b) The unique linear map R making the diagram

X∗ S∗//

Q

Y ∗

X∗/Y ⊥R

;;

commute, where Q is the quotient map, is an isometric isomorphism.

8. There is a regular Borel measure µ on [0, 1] such that∫p dµ = p′(0)

for every polynomial p of degree at most 100.

9. There is no regular Borel measure µ on [0, 1] such that∫p dµ = p′(0)

for every polynomial p.

10. Let X be a Banach space. Let J : X → X∗∗ and K : X∗ → X∗∗∗ be the canonical embed-dings.

(a) J∗K = IX∗ .

(b) X is reflexive if and only if X∗ is.

11. (a) c0 is not reflexive.

4 HILBERT SPACES 20

(b) `1 is not reflexive.

12. Let X be a Banach space.

(a) If X∗ is separable then so is X. (Hint: let fn be a countable dense subset of the unitsphere of X∗, and for each n choose xn ∈ X such that ‖xn‖ = 1 and |fn(xn)| ≥ 1/2.)

(b) Give an example where X is separable but X∗ is not.

13. (Universal Property) Let X be a normed space, let J : X → X∗∗ be the canonical embedding,and let X = J(X) (norm closure). Then for any Banach space Y and T ∈ B(X, Y ) thereexists a unique T ∈ B(X,Y ) such that T = TJ .

14. (Essential Uniqueness) Let X be a normed space. A completion of X is a pair (Z,L), whereZ is a Banach space and L ∈ B(X,Z) is a linear isometry such that for any Banach space Yand T ∈ B(X, Y ) there exists a unique T ∈ B(Z, Y ) such that T = TL. (For example, thepair (X, J) from the preceding exercise.) Prove that if (Z1, L1) and (Z2, L2) are completionsof X then there is a unique isometric isomorphism U making the diagram

XL1 //

L2

Z1

U!Z2

commute.

4 Hilbert spaces

Definition 4.1. An inner product on a vector space X is a map 〈·, ·〉 : X × X → F suchthat

(1) 〈x, x〉 ≥ 0 for all x ∈ X, and 〈x, x〉 = 0 if and only if x = 0.

(2) 〈x, y〉 is linear in x for each fixed y ∈ X.

(3) 〈x, y〉 = 〈y, x〉 for all x, y ∈ X.

An inner product space is a vector space equipped with an inner product.

Condition (3) is written in a form suitable when F = C; when F = R it is intended tomean 〈x, y〉 = 〈y, x〉.

Of course, we could write Condition (2) as: for all c ∈ F, x, y, z ∈ X,

4 HILBERT SPACES 21

• 〈cx, y〉 = c〈x, y〉 and

• 〈x+ y, z〉 = 〈x, z〉+ 〈y, z〉.

Note that if F = R then the inner product is linear in the second variable, too, while if F = Cit is conjugate-linear in the sense that

〈x, y + z〉 = 〈x, y〉+ 〈x, z〉 and 〈x, cy〉 = c〈x, y〉.

The combined properties of linearity in the first variable and conjugate-linearity in the secondare summarized by saying that the map 〈·, ·〉 is a sesquilinear form (“sesqui” means “oneand a half” in Greek).

You saw proofs of the following two results in your linear algebra course.

Lemma 4.2 (Cauchy-Schwartz Inequality). If X is an inner product space, then

|〈x, y〉|2 ≤ 〈x, x〉〈y, y〉 for all x, y ∈ X.

Corollary 4.3. Every inner product space X becomes a normed space with

‖x‖ = 〈x, x〉1/2.

Definition 4.4. A Hilbert space is a complete inner product space.

It’s an exercise that the completion of an inner product space is a Hilbert space in anatural way. We will typically write H for a Hilbert space, and in fact H will typicallydenote a Hilbert space without further comment.

Example 4.5. For any measure space (X,µ), L2(X,µ) becomes a Hilbert space with

〈f, g〉 =

∫X

f(x)g(x) dµ(x).

Definition 4.6. Vectors x and y in a Hilbert space H are called orthogonal, written x ⊥ y,if 〈x, y〉 = 0. The orthogonal complement of a subset E ⊆ H is

E⊥ = y ∈ H : x ⊥ y for all x ∈ E.

More generally, if E,F ⊆ H we sometimes write E ⊥ F to mean x ⊥ y for all x ∈ E, y ∈ F ,or x ⊥ F in the case E = x.

Definition 4.7. A subset E ⊆ H is orthogonal if x ⊥ y for all distinct x, y ∈ E, andorthonormal if it is an orthogonal set of unit vectors.

Note that E is orthonormal if and only if

〈x, y〉 = δx,y for all x, y ∈ E,

where we use the Kronecker delta

δx,y =

1 if x = y

0 if x 6= y.

4 HILBERT SPACES 22

Lemma 4.8. Every orthogonal set of nonzero vectors is linearly independent.

Proof. In your linear algebra course.

Lemma 4.9 (Pythagorean Theorem). If x1, . . . , xn are orthogonal then∥∥∥∥∥n∑i=1

xi

∥∥∥∥∥2

=n∑i=1

‖xi‖2.

Proof. We have ∥∥∥∥∥n∑i=1

xi

∥∥∥∥∥2

=n∑i=1

‖x‖2 +∑i 6=j

〈xi, xj〉,

so the result follows from orthogonality.

Definition 4.10. An orthonormal basis for H is a maximal orthonormal subset of H.

Lemma 4.11. H has an orthonormal basis.

Proof. Routine Zorn’s lemma argument.

Proposition 4.12 (Bessel’s Inequality). If E ⊆ H is orthonormal and x ∈ H then∑e∈E

|〈x, e〉|2 ≤ ‖x‖2.

Proof. By definition of the sum on the left side, it suffices to consider the case when E isfinite. Then Y = spanE is a finite-dimensional (and hence closed) subspace of H, and E isan orthonormal basis for Y . Define

y =∑e∈E

〈x, e〉e.

Then y ∈ Y , and 〈x, u〉 = 〈y, u〉 for all u ∈ E, and hence for all u ∈ Y ; thus z = x− y ∈ Y ⊥,and so

‖x‖2 = ‖y + z‖2

= ‖y‖2 + ‖z‖2 (Pythagorean theorem)

≥ ‖y‖2

=∑e∈E

‖〈x, e〉e‖2 (Pythagoras again)

=∑e∈E

|〈x, e〉|2.

4 HILBERT SPACES 23

Corollary 4.13. Let Y be a closed subspace of H, and let E be an orthonormal basis of Y .

(1) For every x ∈ H, the series∑

e∈E〈x, e〉e converges to a vector y ∈ Y .

(2) z = x− y ∈ Y ⊥.

(3) ‖x− y‖ = d(x, Y ).

(4) H = Y ⊕ Y ⊥.

Proof. (1) By the Pythagorean theorem and Bessel’s inequality, for any finite subset F ⊆ Ewe have ∥∥∥∥∥∑

e∈F

〈x, e〉e

∥∥∥∥∥2

=∑e∈F

|〈x, e〉|2 ≤ ‖x‖2,

so by a routine argument the series is Cauchy, and (1) follows since Y is complete.

(2) By continuity of 〈·, ·〉, we have (similarly to the proof of Bessel’s inequality) 〈x, u〉 =〈y, u〉 for all u ∈ Y , and (2) follows.

(3) For all u ∈ Y ,

‖x− u‖2 = ‖x− y + y − u‖2

= ‖x− y‖2 + ‖y − u‖2 (since y − u ∈ Y )

≥ ‖x− y‖2.

(4) From (1)–(2) we see that H = Y + Y ⊥, and Y ∩ Y ⊥ = 0 since 〈x, x〉 = 0 impliesx = 0.

In Corollary 4.13, we can obviously define a linear operator P on H by

P (y + z) = y for y ∈ Y, z ∈ Y ⊥.

Definition 4.14. With the above notation, P is the orthogonal projection of H on Y .

Corollary 4.15. If P is the orthogonal projection of H on K, then for each x ∈ H, Px isthe best approximation to x from K, and is the unique vector in K such that x− Px ⊥ K.

Proof. This follows quickly from Corollary 4.13.

Corollary 4.16. Let Y be a closed subspace of H, and let E be an orthonormal basis for Y .Define P : H → H by

Px =∑e∈E

〈x, e〉e.

Then P is the orthogonal projection of H on Y .

4 HILBERT SPACES 24

Corollary 4.17. If E is an orthonormal basis for H and x ∈ H, then:

(1) x =∑

e∈E〈x, e〉e;

(2) (Parseval’s Identity) ‖x‖2 =∑

e∈E |〈x, e〉|2.

Corollary 4.18. For an orthonormal set E ⊆ H, the following are equivalent:

(1) E is an orthonormal basis of H;

(2) spanE = H;

(3) E⊥ = 0.

Corollary 4.19. Let H be a Hilbert space and E ⊆ H.

(1) E⊥ is a closed subspace of H.

(2) E⊥⊥ is the closed subspace spanned by E.

(3) If E is a closed subspace then E = E⊥⊥.

Proof. (1) is an exercise, as is the containment E ⊆ E⊥⊥. Then spanE ⊆ E⊥⊥ by (1).To finish, it therefore suffices to note that Corollary 4.13 (4) implies that if E is a closedsubspace then E⊥⊥ ⊆ E.

Proposition 4.20. For each y ∈ H define fy : H → F by

fy(x) = 〈x, y〉.

Then y 7→ fy is a bijection of H onto H∗ that preserves addition and the norm. If F = Cthe map is conjugate-linear, but is an isometric isomorphism if F = R.

Proof. The map is clearly conjugate linear from H to H∗, and

‖y‖ = sup‖x‖=1

|〈x, y〉| = ‖fy‖.

It remains to show that for all f ∈ H∗ there exists y ∈ H such that f = fy. Without loss ofgenerality f 6= 0. Let K = ker f . Since H/K ' F, the subspace K⊥ is 1-dimensional, so hasan orthonormal basis e. Then

ker fe = K = ker f,

so there exists c ∈ F such thatf = cfe = fce.

4 HILBERT SPACES 25

Corollary 4.21. Let T ∈ B(H). Then there is a unique T ∗ ∈ B(H) such that

〈Tx, y〉 = 〈x, T ∗y〉 for all x, y ∈ H.

Proof. Let y ∈ H. Since the map x 7→ 〈Tx, y〉 is linear, by Proposition 4.20, there is aunique T ∗y ∈ H such that 〈Tx, y〉 = 〈x, T ∗y〉 for all x ∈ H. Then routine computationsshow that T ∗ : H → H is linear. We finish by showing it is bounded:

‖T ∗y‖ = sup‖x‖=1

|〈x, T ∗y〉| = sup‖x‖=1

|〈Tx, y〉| ≤ ‖T‖‖y‖.

Definition 4.22. Let T ∈ B(H).

(1) T ∗ is called the adjoint of T .

(2) T is self-adoint if T = T ∗.

(3) T is unitary if T ∗T = TT ∗ = I (the identity operator on H).

Remark 4.23. More generally, if K is another Hilbert space and T ∈ B(H,K), then thereis a unique T ∗ ∈ B(K,H) such that

〈Tx, y〉 = 〈x, T ∗y〉 for all x ∈ H, y ∈ K,

and T is called unitary if T ∗T = IH and TT ∗ = IK , i.e., T is invertible and T ∗ = T−1. Putanother way, a unitary is just an isometric isomorphism between Hilbert spaces.

Note that T is an isometry if and only if 〈Tx, Tx〉 = 〈x, x〉 for all x ∈ H. In fact, a bitmore is true:

Lemma 4.24. T ∈ B(H,K) is an isometry if and only if 〈Tx, Ty〉 = 〈x, y〉 for all x, y ∈ H.Consequently, T is unitary if and only if it is surjective and 〈Tx, Ty〉 = 〈x, y〉 for all x, y ∈ H.

Lemma 4.25. If T ∈ B(H,K) is an isometry then the image under T of any orthonormalset is orthonormal.

Example 4.26. If (X,µ) is a measure space, g ∈ L∞(X,µ), and Mg is the associatedmultiplication operator on L2(X,µ), then

(Mg)∗ = Mg.

Example 4.27. If k ∈ C([0, 1]× [0, 1]) is the kernel of a kernel operator K on L2[0, 1], thenK∗ is a kernel operator whose kernel is the function

(x, y) 7→ k(y, x).

Example 4.28. The adjoint of the forward shift S ∈ B(`2) is the backward shift :

S∗(x1, x2, . . . ) = (x2, x3, . . . ).

4 HILBERT SPACES 26

Corollary 4.29. If P is the orthogonal projection of H on a closed subspace Y , then:

(1) P ∈ B(H) and

(2) P 2 = P ∗ = P .

Conversely, if P ∈ B(H) satisfies (1)–(2), then ranP is closed and P is the orthogonalprojection of H on ranP .

Proof. First assume that P is the orthogonal projection of H on Y . Let x ∈ H, and writex = y + z with y ∈ Y and z ∈ Y ⊥. Then

‖x‖2 = ‖y + z‖2 = ‖y‖2 + ‖z‖2 ≥ ‖y‖2 = ‖Px‖2,

so P ∈ B(H), giving (1).

Next, P 2 = P by construction. To see that P ∗ = P , let x, y ∈ H, and write x = u + v,y = z + w with u, z ∈ Y, v, w ∈ Y ⊥. It suffices to note that

〈Px, y〉 = 〈u, z + w〉 = 〈u, z〉 = 〈u+ v, z〉 = 〈x, Py〉.

Conversely, suppose that P ∈ B(H) satisfies (1)–(2). Since P = P 2, we have ranP =ker(I − P ) by linear algebra. Thus ranP is closed. For the other part,

ranP = ranP = ker(P ∗)⊥ = ker(P )⊥,

and therefore P is the orthogonal projection of H on ranP .

Remark 4.30. Thus, orthogonal projections can be characterized algebraically as self-adjoint idempotents in B(H) (where an operator T is idempotent if T 2 = T ). We haveno real use for general idempotents, and we will use the term projection for a self-adointidempotent.

Theorem 4.31. All orthonormal bases for H have the same cardinality.

Proof. The finite-dimensional case is just linear algebra, so we might as well assume that His infinite-dimensional. Let E and F be orthonormal bases for H. For each e ∈ E let

Fe = f ∈ F : 〈e, f〉 6= 0.

Sincee =

∑f∈F

〈e, f〉f,

Fe is a countable subset of F . Every f ∈ F is contained in some Fe, since E⊥ = 0. Thus

F ⊆⋃e∈E

Fe,

so|F | ≤ ℵ0|E| = |E|.

By symmetry we must have |F | = |E|.

4 HILBERT SPACES 27

Example 4.32. In `2(X), the delta functions δxx∈X give an orthonormal basis. Hereδx : X → C is defined by

δx(y) =

1 if y = x

0 if y 6= x.

Definition 4.33. The dimension of a Hilbert space H, written dimH, is the cardinality ofany orthonormal basis of H.

Example 4.34. The preceding concept of dimension reduces to the usual one from linearalgebra if H is finite-dimensional.

We already had a notion of infinite-dimensional normed space, and in particular Hilbertspace. Our new definition of dimension of a Hilbert space H is consistent with this, in thesense that H is infinite-dimensional if and only if its dimension is an infinite cardinal.

Theorem 4.35. (1) Two Hilbert spaces are isometrically isomorphic if and only if theyhave the same dimension.

(2) An infinite-dimensional Hilbert space is separable if and only if its dimension is ℵ0.

Proof. First let U : H → K be a unitary. Then the image of any orthonormal basis for His an orthonormal set in K whose linear span is dense, and hence is an orthonormal basis ofK. Thus dimH = dimK.

Conversely, let dimH = dimK, and choose orthonormal bases E and F for H and K,respectively. We can choose a bijection from E to F , which then extends uniquely to linearmap U0 : spanE → K. A quick calculation with inner products shows that U0 is isometric,and so extends uniquely to an isometry U : H → K, because spanE = H. Since U(E) = F ,which has dense span in K, the isometry U is surjective, and therefore is unitary.

For (2), first note that `2 is separable and has dimension ℵ0. On the other hand, if Hhas dimension greater than ℵ0, then we can choose an uncountable orthonormal set E ⊆ H,and then B1/

√2(e)e∈E is an uncountable family of nonempty pairwise disjoint open sets,

so H must be nonseparable.

Definition 4.36. Let Hii∈S be a family of Hilbert spaces. The `2-direct sum of the familyis the set ⊕

i∈S

Hi =

x ∈

∏i∈S

Hi :∑i∈S

‖xi‖2 <∞

.

We frequently just say “direct sum”, with the understanding that “`2” is tacitly implied.

Proposition 4.37. The direct sum⊕

i∈S Hi is a Hilbert space, with pointwise vector-spaceoperations and inner product given by

〈x, y〉 =∑i∈S

〈xi, yi〉.

4 HILBERT SPACES 28

Idea of proof. Most of this is left as an exercise. The only hard part is completeness. Up toisometric isomorphism, we can choose pairwise disjoint sets Ωi such that Hi = `2(Ωi). LetΛ =

⋃i∈S Ωi. Then the idea is to verify that⊕

i∈S

Hi ' `2(Λ).

Definition 4.38. If Hii∈S is a pairwise orthogonal family of closed subspaces of a Hilbertspace H, the internal direct sum is the subspace

span⋃i∈S

Hi

of H.

To emphasize the difference, we frequently use the adjective external for the direct sumof Definition 4.36.

For Hilbert spaces, it transpires that the internal direct sum is always isometrically iso-morphic to the external direct sum1. We will state this formally in the following proposition,but first we prepare a bit: Given a pairwise orthogonal family Hi∈S of closed subspaces ofa Hilbert space H, let X be the set of vectors in the external direct sum

⊕i∈S Hi with only

finitely many nonzero coordinates, and note that X is a dense subspace of⊕

Hi.

Proposition 4.39. With the above notation, assume that the Hi’s have dense span in H.Then the map U0 : X → H defined by

U0x =∑i∈S

xi

extends uniquely to a unitary U :⊕

i∈S Hi → H.

Proof. A routine calculation with inner products, using the Pythagorean Theorem, showsthat U0 is isometric. Since U0(X) is dense in H, there is a unique extension to a unitaryU : H → K.

We often blur the distinction between internal and external direct sums of Hilbert spaces,writing

⊕i∈S Hi in both cases.

Exercises

1. (Polarization Identity) Let x and y be vectors in a Hilbert space H.

1this is not the case more generally with Banach spaces

4 HILBERT SPACES 29

(a) If F = R then

〈x, y〉 =1

2

(‖x+ y‖2 − ‖x− y‖2

).

(b) If F = C then

〈x, y〉 =1

4

3∑j=0

ij‖x+ ijy‖2.

2. (Parallelogram Law) For vectors x, y in a Hilbert space,

‖x+ y‖2 + ‖x− y‖2 = 2(‖x‖2 + ‖y‖2

).

3. For all T ∈ B(H):

(a) ‖T ∗‖ = ‖T‖.

(b) (C∗-norm property) ‖T ∗T‖ = ‖T‖2.

4. For all T, S ∈ B(H) and c ∈ F:

(a) (T + S)∗ = T ∗ + S∗;

(b) (cT )∗ = cT ∗;

(c) (TS)∗ = S∗T ∗;

(d) T ∗∗ = T ;

(e) If T is invertible, then so is T ∗, and (T ∗)−1 = (T−1)∗.

5. If T ∈ B(H), then T is an isometry if and only if T ∗T = I.

6. Every Hilbert space is reflexive.

7. The completion of an inner product space is a Hilbert space.

8. Let en be an orthonormal basis for a separable infinite-dimensional Hilbert space H, andlet (cn) be a sequence in C. Then (cn) is bounded if and only if there is a (necessarily unique)T ∈ B(H) such that

Ten = cnen for n ∈ N,

in which case ‖T‖ = supn |cn|. We call such an operator diagonal.

9. For all T ∈ B(H),

4 HILBERT SPACES 30

(a) (ranT )⊥ = kerT ∗;

(b) (ranT ∗)⊥ = kerT .

10. If P is the orthogonal projection of H on K, then:

(a) I − P is the orthogonal projection of H on K⊥.

(b) For all x ∈ H, Px is the closest vector to x from K. (Note: this is part of Corollary 4.15.)

11. (Matrices) Let E = eii∈S be an orthonormal basis of H. For each T ∈ B(H), define anarray (tij)i,j∈S by

tij = 〈Tej, ei〉.We call the array (tij) the matrix of T relative to E.

(a) If x =∑

i∈S xiei, then Tx =∑

i∈S yiei, where

yi =∑j∈S

tijxj for i ∈ S.

(b) Let S ∈ B(H) be another operator, with matrix (sij). Then the matrix (rij) of TS isgiven by

rij =∑k∈S

tikskj for i, j ∈ S.

12. Let X be a Banach space and T ∈ B(X).

(a) Prove that if there are finite-rank operators Tn on X such that ‖T − Tn‖ → 0, then Tis compact. (Here finite-rank means having finite-dimensional range.)

(b) Prove that if X is a Hilbert space and T is compact, then there are finite-rank operatorsTn on X such that ‖T − Tn‖ → 0. (Hint: every compact metric space is separable.)

13. A diagonal operator T ∈ B(`2) determined by the sequence (cn) is compact if and only ifcn → 0.

14. (a) C[0, 1] is dense in L2[0, 1].

(b) L2[0, 1] is separable.

15. Let H be a Hilbert space, x ∈ H, and M ⊆ H nonempty, closed, and convex. Then thereexists a unique y ∈M such that

‖x− y‖ = d(x,M) (= infu∈M‖x− u‖).

(Hint: choose a sequence (yn) in M such that ‖x−yn‖ → d(x,M), and use the parallelogramlaw to show that (yn) is Cauchy.)

5 BAIRE CATEGORY CONSEQUENCES 31

16. In Corollary 4.13, prove directly that (3)⇒(2). Hint: for u ∈ Y , consider the followingquadratic function of the real variable t:

p(t) = ‖z + tu‖2.

17. Let∑∞

n=1 xn be a series of orthogonal terms in a Hilbert space H, and define f : N→ F byf(n) = ‖xn‖.

(a) The series converges if and only if f ∈ `2, in which case∥∥∥∥∥∞∑n=1

xn

∥∥∥∥∥ = ‖f‖2.

(b) If the series converges, it does so unconditionally (recall that this means the convergenceis invariant under rearrangement of the terms).

(c) It is possible for the series to converge but not converge absolutely.

5 Baire Category consequences

Theorem 5.1 (Open Mapping Theorem). Let X and Y be Banach spaces, and let T ∈B(X, Y ) be surjective. Then T is open.

Proof. By linearity, it suffices to show that T (B1) ⊇ Bε for some ε > 0. We have X =⋃∞n=1Bn, so

Y =∞⋃n=1

T (Bn).

By the Baire Category theorem, since Y is complete there exists n ∈ N such that T (Bn) hasnonempty interior, and again by linearity we can assume that n = 1. Thus we can chooset ∈ Y and r > 0 such that

Br(t) = t+Br ⊆ T (B1).

Then we can choose u ∈ B1 such that

Tu ∈ t+Br/2,

and thenTu+Br/2 ⊆ t+Br ⊆ T (B1).

Rearranging,Br/2 ⊆ −Tu+ T (B1) = T (−u) + T (B1) ⊆ T (B2).


Scaling, we see that with s = r/4 we have

Bs ⊆ T (B1).

Let y ∈ Bs/2. We can choose x1 ∈ B1/2 such that

‖y − Tx1‖ <s

4.

Then we can choose x2 ∈ B1/4 such that∥∥∥∥∥y −2∑i=1

Txi

∥∥∥∥∥ < s

8.

Then we can choose x3 ∈ B1/8 such that∥∥∥∥∥y −3∑i=1

Txi

∥∥∥∥∥ < s

16.

Continuing inductively, we get a sequence (xn) in X such that for each n we have ‖xn‖ < 1/2n

and ∥∥∥∥∥y −n∑i=1

Txi

∥∥∥∥∥ < s

2n+1.

Then the series∑∞

n=1 xn converges absolutely, so by completeness it converges to some x ∈ X.We have

‖x‖ ≤∞∑n=1

‖xn‖ <∞∑n=1

1/2n = 1,

so x ∈ B1, and

‖y − Tx‖ = lim

∥∥∥∥∥y −n∑i=1

Txi

∥∥∥∥∥ = 0,

so y = Tx. Thus we can take ε = s/2.

Definition 5.2. Let X and Y be normed spaces, and let T : X → Y be linear. The graphof T is the set

G(T ) = (x, Tx) : x ∈ X ⊆ X × Y.

We say that T has closed graph if G(T ) is closed.

Theorem 5.3 (Closed Graph Theorem). Let X and Y be Banach spaces, and let T : X → Ybe linear. If T has closed graph, then it is bounded.

Proof. Since X and Y are complete, so is X ⊕ Y , and hence so is the closed subspace G(T ).The linear map S : G(T ) → X defined by S(x, Tx) = x is a bounded bijection, so by theOpen Mapping theorem the inverse S−1 : X → G(T ) is bounded. Then T = PY S

−1 isbounded, where PY : X ⊕ Y → Y is the coordinate projection.


Theorem 5.4 (Principle of Uniform Boundedness). Let X and Y be Banach spaces, and letF ⊆ B(X, Y ). If Tx : T ∈ F is bounded in Y for every x ∈ X, then F is bounded inB(X, Y ).

Proof. Consider the Banach space Z =⊕

T∈F Y (with the product norm), and define alinear map S : X → Z by

(Sx)T = Tx for x ∈ X,T ∈ F .

This makes sense by our assumption on F . Claim: S has closed graph. Suppose that(xn, Sxn)→ (0, z) in G(S). Then for each T ∈ F we have

zT = lim(Sxn)T = limTxn = 0.

Thus z = 0, so G(S) is closed. By the Closed Graph Theorem, S is bounded. Therefore forall x ∈ X we have

‖Tx‖ = ‖(Sx)T‖ ≤ ‖Sx‖ ≤ ‖S‖‖x‖,so ‖T‖ ≤ ‖S‖.

The preceding proof is not the usual one — I found it in [1, Theorem 2.21].

Exercises

1. No infinite-dimensional Banach space has a countable Hamel basis. (Hint: consider finite-dimensional subspaces.)

2. (Inverse Mapping Theorem) Every bijective bounded linear map between Banach spaces isan isomorphism.

3. If X and Y are normed spaces and T : X → Y is linear, then T has closed graph if and onlyif xn → 0 and Txn → y imply y = 0.

4. Let X = C1[0, 1], but with the norm inherited from C[0, 1]. Then the linear map D : X →C[0, 1] defined by Df = f ′ is closed and unbounded.

5. Let X = x ∈ `1 :∑∞

n=1 n|xn| <∞, and define T : X → `1 by (Tx)n = nxn.

(a) T is closed and unbounded.

(b) T is bijective and T−1 is bounded but not open.

6. Let f be an unbounded linear functional on a Banach space X, and define T : X → G(f) byTx = (x, f(x)). Then T is closed and unbounded.


7. If ‖ · ‖1 and ‖ · ‖2 are norms on a vector space X such that X is complete in both norms and‖ · ‖1 ≤ ‖ · ‖2, then the norms are equivalent.

8. Let X and Y be Banach spaces and T : X → Y linear. If f T ∈ X∗ for all f ∈ Y ∗ then Tis bounded.

9. Let X and Y be Banach spaces and (Tn) a sequence in B(X, Y ). Suppose that (Tnx)converges in Y for all x ∈ X, and define Tx = limTnx. Then T ∈ B(X, Y ).

10. If X and Y are Banach spaces and T ∈ B(X, Y ) is injective, then T has closed range if andonly if T is bounded below (recall that this means there exists c > 0 such that ‖Tx‖ ≥ c‖x‖for all x ∈ X).

11. In the hypotheses of the Uniform Boundedness Principle, completeness of X is necessary butcompleteness of Y is not.

12. Let Y and Z be closed subspaces of a Banach space X such that Y ∩Z = 0 and Y +Z = X.Then the map T : Y ⊕ Z → X defined by

T (y, z) = y + z

is an isomorphism. Note: here the notation Y ⊕ Z refers to the external direct sum, withthe product norm.

13. (Fourier transform on the circle) In this problem the scalars are complex. For f ∈ L1(T)

define f : Z→ C by

f(n) =

∫Tf(z)z−n dµ(z),

where µ is normalized arc-length measure on T. (“Normalized” means µ(T) = 1 — this isconvenient in various ways.)

(a) If f(z) = zn then f = δn.

(b) If f is a trig polynomial, meaning f ∈ spanzn : n ∈ Z, then

f ∈ cc(Z) = g ∈ c0(Z) : supp g is finite.

(c) (Riemann-Lebesgue Lemma) f is in c0(Z) for every f ∈ L1(T).

(d) The linear map F : L1(T)→ c0(Z) given by Ff = f is injective.

(e) F has dense range.

(f) F is not surjective. (Hint: use the Open Mapping Theorem and properties of the Banachspaces L1(T) and c0(Z).)


14. The exists f ∈ L2(T) whose Fourier series

∞∑n=−∞

f(n)einx

diverges at x = 0. Here convergence of a series∑

n∈Z cn is defined as convergence of thesequence (

k∑n=−k

cn

)∞k=1

.

Hint: the kth partial sum evaluated at x = 0 is

φk(f) =k∑

n=−k

f(k) =

∫ 2π

0

fDk,

where Dk is the Dirichlet kernel

Dk(x) =k∑

n=−k

einx.

Then φk ∈ L2(T)∗ with ‖φk‖ = ‖Dk‖2. Moreover, ‖Dk‖2k→∞−−−→∞. An appeal to the Uniform

Boundedness Principle now finishes the proof of the result we want.

15. Continuing from the preceding exercise, there exists f ∈ C(T) whose Fourier series divergesat 0. Hint: using the map t 7→ e2πit to replace the measure space (T, µ) with ([0, 1],m/2π),where m denotes Lebesgue measure, we have

Dk(x) =sin(k + 1/2)x

sinx/2,

and φk ∈ C(T)∗ with ‖φk‖ = ‖Dk‖1. We have ‖Dk‖1k→∞−−−→∞, which allows us to finish the

proof of the result we want. (This is not easy.)

16. Here’s an outline of the more common proof of the Uniform Boundedness Principle. Justifyall the steps.

(a) For each n ∈ N define

An = x ∈ X : ‖Tx‖ ≤ n for all T ∈ F.

Then we can choose n ∈ N, u ∈ X, and r > 0 such that

u+Br ⊆ An.

(b) It follows that Br ⊆ A2n.

(c) Let T ∈ F and x ∈ BX . Then ‖Tx‖ ≤ 4n

r.

(d) ‖T‖ ≤ 4n

r.

6 LOCALLY CONVEX SPACES 36

6 Locally convex spaces

Definition 6.1. A topological vector space is a vector space X equipped with a Hausdorfftopology such that addition and scalar multiplication, i.e., the maps

• (x, y) 7→ x+ y : X ×X → X and

• (c, x) 7→ cx : F×X → X,

are continuous.

Definition 6.2. A locally convex space is a topological vector space X for which there is aneighborhood base at 0 consisting of convex sets.

Thus, in a locally convex space the family of all convex neighborhoods of 0 is a neigh-borhood base at 0.

Definition 6.3. A subset U of a vector space X is called:

(1) absorbing if X =⋃t>0 tU ;

(2) balanced if tU ⊆ U whenever |t| ≤ 1;

Lemma 6.4. In a topological vector space, a seminorm p is continuous if and only if p < 1is open.

Proof. For the nontrivial direction, suppose that p is a seminorm and p < 1 is open. Letx ∈ X and ε > 0. Because p < ε = εp < 1 is open, x+ p < ε is a neighborhood of x,so it suffices to note that if y = x+ z with p(z) < ε then

|p(y)− p(x)| ≤ p(y − x) = p(z) < ε.

Lemma 6.5. If p, q are seminorms on X, then p ≤ q if and only if q < 1 ⊆ p < 1.Consequently, if q is continuous and p ≤ q then p is continuous.

Definition 6.6. The Minkowski functional of an absorbing set U is the function pU : X →[0,∞) defined by

pU(x) = inft > 0 : x ∈ tU.

We give two parallel versions inside the following lemma, one for sublinear functionalsand the other for seminorms. In the immediate future we are interested in the seminormversion, but the sublinear version will be useful later when we discuss separation theorems.

Lemma 6.7. Let X be a vector space.


(1) Let U ⊆ X be convex and absorbing. Then the Minkowski functional pU is a sublinearfunctional p on X such that

p < 1 ⊆ U ⊆ p ≤ 1. (2)

Moreover, if U is also balanced then pU is a seminorm, and is the unique seminorm psatisfying (2).

(2) Let p be a sublinear functional on X, and let U = p < 1. Then U is convex andabsorbing. Moreover, if p is a seminorm then U is also balanced, and p is the Minkowskifunctional of U .

Proof. (1) First we show that pU is sublinear. For the subadditivity, let x, y ∈ X. Leta > pU(x) and b > pU(y). Then we can choose t ∈ (0, a) and s ∈ (0, b) and u, v ∈ U suchthat x = tu and y = sv. Then

x+ y = tu+ sv

= (t+ s)

(t

t+ su+

s

t+ sv

)∈ (t+ s)U

since 0 ≤ t/(t+ s), s/(t+ s) and t/(t+ s) + s/(t+ s) = 1. Thus

pU(x+ y) ≤ t+ s < a+ b.

Letting a pU(x) and b pU(y), we get

pU(x+ y) ≤ pU(x) + pU(y).

Now let x ∈ X and t ≥ 0. We must show pU(tx) = tpU(x). Since U is absorbing, we have0 ∈ U , and pU(0) = 0, so without loss of generality t > 0. Then

pU(tx) = infs > 0 : tx ∈ sU

= infs > 0 : x ∈ s

tU

= infs : s > 0, x ∈ s

tU

= t infst

: s > 0, x ∈ stU

= t infr : tr > 0, x ∈ rU= t infr : r > 0, x ∈ rU= tpU(x).

For the inclusions (2), first let pU(x) < 1. Then we can choose c ∈ (0, 1) such thatx ∈ cU . Since U is convex and contains 0, we have cU ⊆ U . Thus x ∈ U . On the otherhand, x ∈ U obviously implies pU(x) ≤ 1.


Now suppose also that U is balanced. Then for all x ∈ X and |a| = 1 we have x ∈ Uif and only if ax ∈ U , so pU(ax) = pU(x) whenever x ∈ X and |a| = 1. Combined withsublinearity, we see that pU is a seminorm.

For the uniqueness, suppose that q and r are seminorms satisfying (2). To show thatq = r, by symmetry it is enough to show q ≤ r, and we argue by contradiction: assume thatx ∈ X and r(x) < q(x). Choose t such that r(x) < t < q(x). Then

r(xt

)< 1,

so x/t ∈ U , and hence q(x/t) ≤ 1. But then

q(x) ≤ t < q(x),

a contradiction.

(2) First let x, y ∈ U . For the convexity it suffices to show that tx + (1 − t)y ∈ U for0 < t < 1:

p(tx+ (1− t)y

)≤ p(tx) + p((1− t)y)

= tp(x) + (1− t)p(y)

< t+ 1− t (because t, 1− t > 0)

= 1.

For the absorbing property, let x ∈ X, and without loss of generality p(x) > 1. Then x ∈ tUwith t = 2p(x).

Now assume that p is in fact a seminorm. Then for all |t| ≤ 1 and x ∈ U ,

p(tx) = |t|p(x) < 1,

so tx ∈ U . Thus U is balanced. Moreover, p trivially satisfies (2), so p = pU .

Definition 6.8. A family P of seminorms on a vector space X is called separating if for allx 6= 0 there exists p ∈ P such that p(x) 6= 0.

Theorem 6.9. A topological vector space X is locally convex if and only if there is a sepa-rating family P of seminorms on X such that the family of sets

p < εp∈P,ε>0

is a neighborhood base at 0.

Proof. First suppose that X is locally convex. Let U be a neighborhood base at 0 consistingof open convex balanced sets. For each U ∈ U , the Minkowski functional pU is a seminormand U = p < 1. Then each pU is continuous, so the set pU < ε is open for everyε > 0, and by construction the sets pU < ε for U ∈ U and ε > 0 form a base of open


neighborhoods of 0. To see that the family pUU∈U of seminorms is separating, let x 6= 0.Then since U is a neighborhood base we can find U ∈ U such that x /∈ U . Then pU(x) ≥ 1,so in particular pU(x) 6= 0.

Conversely, suppose that we have a separating family P of seminorms on X such thatthe family of sets

p < εp∈P,ε>0

is a neighborhood base at 0. Since for each p ∈ P and ε > 0 the set p < ε is convex, X islocally convex.

Remark 6.10. Actually, we could have just used sets of the form p < 1, but in practicewe frequently use p < ε for arbitrarily small ε. For example, in a normed space we useBε = x : ‖x‖ < ε.

Theorem 6.9 has the drawback that to use it we need to know that we have a topologicalvector space. The next theorem fixes this:

Theorem 6.11. Let P be a separating family of seminorms on a vector space X. For eachp ∈ P and ε > 0 define

N(p, ε) = p < ε.

Let B be the family of all finite intersections of the N(p, ε). Then there is a unique locallyconvex topology on X such that B is a neighborhood base at 0.

Proof. Define a set to be open if it is a union of translates of sets in B. This gives atranslation-invariant topology on X such that B is a neighborhood base at 0. We will provethat addition and scalar multiplication are continuous, and that X is T1, so that it is atopological vector space. Then it will in fact be a locally convex space since the sets in Bare convex.

Note that a net (xi) converges to x in X if and only if

p(xi − x)→ 0 for all p ∈ P .

Let xi → x and yi → y, and take p ∈ P . Then

p((xi + yi)− (x+ y)

)≤ p(xi − x) + p(yi − y)→ 0.

If we also have ci → c in F, then

p(cixi − cx) ≤ p(ci(xi − x)

)+ p((ci − c)x

)= |ci|p(xi − x) + |ci − c|p(x)→ 0.

Thus addition and scalar multiplication are continuous.

For the T1 property, by translation-invariance it suffices to note that if x 6= 0 then wecan find p ∈ P such that p(x) > 0, and hence x /∈ 0.


Definition 6.12. In the above theorem, we say that the topology, or the locally convexspace X itself, is generated by P .

Corollary 6.13. If X is a locally convex space generated by a family P of seminorms, thena seminorm q on X is continuous if and only if there exist p1, . . . , pn ∈ P and c > 0 suchthat

q ≤ cmaxipi.

Proof. Note first that for every p ∈ P , the set p < 1 is open, so p is continuous. Thus, byProblem 5 we only need to prove the forward direction: assume that q is continuous. Thenthe neighborhood q < 1 contains

⋂n1pi < εi for some p1, . . . , pn ∈ P and εi > 0. Let

ε = mini εi, so thatpi(x) < ε for all i implies q(x) < 1.

Thusε−1 max

ipi < 1 ⊆ q < 1,

soq ≤ ε−1 max

ipi.

Example 6.14. Every normed space has the topology generated by the norm.

Example 6.15. Let X be a locally compact Hausdorff space, and let C(X) denote the vectorspace of continuous functions on X. The compact-open topology on C(X) is generated bythe seminorms

f 7→ sup|f(x)| : x ∈ K for compact sets K ⊆ X.

A net (fi) in C(X) converges in the compact-open topology if and only if it convergesuniformly on compact sets, and for this reason we also use the term “topology of uniformconvergence on compact sets”.

Example 6.16. If Xii∈S is any family of normed spaces, then the Cartesian product∏

iXi

is a vector space with the coordinate-wise operations, and the product topology is generatedby the seminorms

x 7→ ‖xi‖ for i ∈ S.Theorem 6.17. A locally convex space is metrizable if and only if it is generated by acountable family of seminorms.

Proof. First suppose that X is a metrizable locally convex space. Then it is 1st countable,so there is a countable neighborhood base B at 0, and we can assume that every set inB is open, balanced, and convex. By the proof of Theorem 6.9, the associated family ofseminorms generates X.

Conversely, let (pn) be a sequence of seminorms generating X. Define d : X×X → [0,∞)by

d(x, y) =∞∑n=1

2−npn(x− y)

1 + pn(x− y).

It is an exercise that d is a metric compatible with the topology of X.


Example 6.18. If X is a locally compact Hausdorff space, then C(X) with the compact-open topology is metrizable if and only if X is σ-compact, i.e., is a countable union of compactsets.

Definition 6.19. Let X be a topological vector space. Then B ⊆ X is called bounded if forevery neighborhood U of 0 there exists t > 0 such that B ⊆ tU .

Theorem 6.20. A topological vector space is normable if and only if it has a bounded convexneighborhood of 0.

Proof. For the nontrivial direction, suppose that U is a bounded convex neighborhood of0. By Problem 7, U contains a balanced convex neighborhood V of 0, and of course V isbounded. Let p = pV be the associated Minkowski functional. Since V is bounded and X isT1, p is separating, and hence is a norm. Moreover, if W is any neighborhood of 0, then wecan choose t > 0 such that V ⊆ tW , so that t−1V ⊆ W . Since p < 1 ⊆ V , we see that theopen balls p < ε form a neighborhood base at 0. Therefore the topology of X is generatedby the norm p.

Exercises

1. Let X be a vector space equipped with a topology making addition and scalar multiplicationcontinuous. If points are closed then X is Hausdorff, and hence is a topological vector space.

2. A linear map between topological vector spaces is continuous if and only if it is continuousat 0.

3. If X is a topological vector space then for every y ∈ X and c ∈ F \ 0 the maps

(a) x 7→ x+ y and

(b) x 7→ cx

are homeomorphisms on X.

4. Every absorbing set, and every nonempty balanced set, contains 0.

5. Let X be a topological vector space.

(a) If p and q are continuous seminorms on X, then maxp, q is a continuous seminorm.

(b) If p and q are continuous seminorms on X, then p+ q is a continuous seminorm.

(c) If p is a continuous seminorm on X and c ≥ 0 in F, then cp is a continuous seminorm.


6. In a topological vector space, every neighborhood of 0 is absorbing, and contains a balancedneighborhood of 0.

7. In a topological vector space, every convex neighborhood of 0 contains a balanced convexneighborhood of 0.

8. Every nonzero linear functional on a topological vector space X is an open map.

9. The topology generated by a separating family P of seminorms is the weakest one makingall the seminorms in P continuous. (Hint: Lemma 6.4.)

10. Let X be a locally convex space generated by a set P of seminorms, let (xi) be a net in X,and let x ∈ X. Then xi → x if and only if

p(xi − x)→ 0 for all p ∈ P .

11. A subset A of a vector space X is convex if and only if it is closed under convex combinations,i.e., if x1, . . . , xn ∈ A and c1, . . . , cn ≥ 0 with

∑ni=1 ci = 1, then

∑ni=1 cixi ∈ A.

12. Let A be a subset of a vector space X.

(a) The convex hull of A, written coA, is defined as the intersection of all convex setscontaining X. Prove that coA coincides with the set of all convex combinations of elementsof A, and is the smallest convex set containing A.

(b) The closed convex hull of A, written coA, is defined as the intersection of all closedconvex sets containing A. Prove that coA is the closure of coA, and is the smallest closedconvex set containing A.

13. Let 0 < p < 1, and let `p be the set of all sequences (xn) in F such that

|x|p =∞∑n=1

|xn|p <∞.

Then the map d : `p × `p → R defined by

d(x, y) = |x− y|p

is a metric that makes `p into a topological vector space that is not locally convex.

14. Let 0 < p < 1, and let Lp = Lp[0, 1] be the set of Lebesgue measurable functions on [0, 1]such that

|f |p =

∫ 1

0

|f(x)|p dx <∞.

7 WEAK TOPOLOGIES 43

(a) The map d : Lp × Lp → R defined by

d(f, g) = |f − g|p

is a metric that makes Lp into a topological vector space.

(b) Let V be a convex neighborhood of 0 in Lp. Prove that V = Lp by verifying all thestatements in the following outline: Let f ∈ Lp. We can choose r > 0 such that the openball f : |f |p < r is contained in V . We can choose n ∈ N such that

np−1|f |p < r.

Then we can choose0 < x0 < x1 < · · · < xn = 1

such that ∫ xi

xi−1

|f |p =|f |pn

for i = 1, . . . , n.

Putfi = nfχ(xi−1,xi) for i = 1, . . . , n.

Then fi ∈ V for i = 1, . . . , n, and

f =1

n

n∑i=1

fi.

Therefore f ∈ V .

15. Let X be a topological vector space, let Y be a locally convex space generated by a familyP of seminorms, and let T : X → Y be linear. Then T is continuous if and only if p T iscontinuous for every p ∈ P .

7 Weak topologies

To highlight a certain symmetry between vectors and linear functionals, we will frequentlyuse the following convention: if f is a linear functional on a vector space X, then

〈x, f〉 = f(x) for x ∈ X.

Definition 7.1. A set E of linear functionals on a vector space X is called separating if forall x 6= 0 there exists f ∈ E such that 〈x, f〉 6= 0.

Definition 7.2. Let E be a separating set of linear functionals on a vector space X. Theweak topology on X generated by E, denoted by σ(X,E), is generated by the seminorms

x 7→ |〈x, f〉| for f ∈ E.

If X is given as a topological vector space, we call σ(X,X∗) the weak topology on X.


Note that σ(X,E) is the weakest topology making every functional in E continuous.Obviously, σ(X,E) does not change if we replace E by the vector space it spans, so wemight as well assume that E is already a vector space.

Definition 7.3. Let X be a topological vector space. The weak* topology on the dual X∗

is σ(X∗, X), i.e., the topology generated by the seminorms

f 7→ |〈x, f〉| for x ∈ X.

Note that σ(X∗, X) is the weakest topology on X∗ making every functional x for x ∈ Xcontinuous.

Lemma 7.4. If X is a vector space and Y is a vector space of linear functionals on X, then

(X, σ(X, Y ))∗ = Y.

Proof. Since σ(X, Y ) is the weakest topology on X making every functional in Y continu-ous, certainly Y ⊆ (X, σ(X, Y ))∗. On the other hand, if f ∈ (X, σ(X, Y ))∗, then |f | is acontinuous seminorm on X, so by Theorem 6.13 we can choose g1, . . . , gn ∈ Y and c > 0such that

|f | ≤ cmaxi|gi|.

Then ker f ⊇⋂ni=1 ker gi, so by linear algebra

f ∈ spang1, . . . , gn ⊆ Y.

Let X be a topological vector space, and let Y be a proper dense subspace. ThenX∗ = Y ∗, but the weak topologies σ(X∗, X) and σ(X∗, Y ) are different, by Lemma 7.4.Consequently, the weak* topology on X∗ depends upon the choice of X.

Definition 7.5. Let X and Y be Banach spaces.

(1) The strong operator topology on B(X, Y ) is generated by the seminorms

T 7→ ‖Tx‖ for x ∈ X.

(2) The weak operator topology on B(X, Y ) is generated by the seminorms

T 7→ |〈Tx, f〉| for x ∈ X, f ∈ Y ∗.

Lemma 7.6. For a nonzero linear functional f on a topological vector space X, the followingare equivalent:

(1) f is continuous;

(2) f is continuous at 0;


(3) ker f is closed;

(4) ker f is not dense;

(5) f is bounded on some neighborhood of 0.

Proof. (1)⇔(2) is a special case of Section 6 Problem 2. Clearly (2)⇒(3)⇒(4).

(4)⇒(5). Let x /∈ ker f , and choose a balanced neighborhood V of 0 such that (x+ V )∩ker f = ∅. Then f(x) /∈ f(V ) (since V is balanced), so f(V ) is a balanced proper subset ofF. Thus there exists t ∈ F such that |f | < t on V .

(5)⇒(2). We can choose r > 0 and a balanced neighborhood V of 0 on which |f | < r,and then by scaling we see that for all ε > 0 there is a neighborhood of 0 on which |f | < ε,and hence f is continuous at 0.

Theorem 7.7. Let X be a topological vector space, and let A,B be nonempty disjoint convexsubsets.

(1) If A is open, then there exist f ∈ X∗ and a ∈ R such that

Re f(x) < a ≤ Re f(y) for all x ∈ A, y ∈ B.

(2) If X is locally convex, A is compact, and B is closed there exist f ∈ X∗ and a, b ∈ Rsuch that

Re f(x) ≤ a < b ≤ Re f(y) for all x ∈ A, y ∈ B.

Proof. We might as well assume that F = R, since if we find a suitable real-linear h on Xwe can let f be the unique complex-linear functional with Re f = h.

(1) Let U = A − B. Then U is convex, nonempty, and open. Pick t ∈ U , and letV = U − t. Then V is an open convex neighborhood of 0. Let p = pV be the Minkowskifunctional. Note that −t /∈ V , so p(−t) ≥ 1. Define a linear map g : Rt→ R by g(ct) = −c.Now, g ≤ p|Rt, since if c ≥ 0 then

g(ct) = −c ≤ 0 ≤ p(ct),

while if c < 0 theng(ct) = −c ≤ −cp(−t) = p(ct).

By the Hahn-Banach theorem we can extend g to a linear functional f on X such that f ≤ p.Then f < 1 on V , so f is bounded on a neighborhood of 0, and hence f is continuous byLemma 7.6.

For all x ∈ A and y ∈ B,

f(x− y − t) ≤ p(x− y − t) ≤ 1 since x− y − t ∈ V ,


sof(x)− f(y) = f(x− y) ≤ f(t) + 1 = 0.

Therefore f(x) ≤ f(y). Let a = sup f(A). Then for all x ∈ A and y ∈ B we have

f(x) ≤ a ≤ f(y).

But in fact f(x) < a since f(A) is open (because every nonzero linear functional is an openmap).

(2) We first show that there is a convex neighborhood V of 0 such that

(A+ V ) ∩B = ∅.

Suppose not. Let B be the family of all convex neighborhoods of 0, directed by containment.By assumption, for each V ∈ B we have (A + V ) ∩ B 6= ∅, so we can choose xV ∈ A andyV ∈ V such that

xV + yV ∈ B.

The net (yV ) converges to 0 because B is a neighborhood base at 0 by local convexity. Bycompactness, we can choose a subnet (xVi) of (xV ) that converges to an element x ∈ A.Then also

xVi + yVi → x.

But xVi + yVi ⊆ B for all i, so x ∈ B, contradicting A ∩ B = ∅. Thus the claim is proved,and then A+ V is a convex open set disjoint from B, so by part (1) we can choose f ∈ X∗such that

sup f(A+ V ) ≤ inf f(B).

Since f(A) is a compact convex subset of the convex open set f(A+ V ), we have

max f(A) < sup f(A+ V ),

so we can take a = max f(A) and b = sup f(A+ V ).

Corollary 7.8. If X is a locally convex space then X∗ is separating.

Example 7.9. Corollary 7.8 does not have a converse: if 0 < p < 1 then `p is not locallyconvex, but (`p)∗ = `∞ is separating.

Definition 7.10. Let X be a locally convex space, E ⊆ X, and R ⊆ X∗.

(1) The polar of E is the set

E = f ∈ X∗ : Re〈x, f〉 ≤ 1 for all x ∈ E.

(2) The prepolar of R is the set

R = x ∈ X : Re〈x, f〉 ≤ 1 for all f ∈ R.


Theorem 7.11 (Bipolar Theorem). Let X be a locally convex space.

(1) For E ⊆ X, (E) is the closed convex hull of E ∪ 0.

(2) For R ⊆ X∗, (R) is the weak* closed convex hull of R ∪ 0.

Proof. Let B = co(E ∪ 0). (E) is convex and weakly closed, and hence closed in thegiven topology. Thus B ⊆ (E). Let x /∈ B. It suffices to show that x /∈ (E). Since B isclosed and convex, by Theorem 7.7 we can choose f ∈ X∗ such that

sup Re f(B) < Re f(x).

Since 0 ∈ B, we have 0 ∈ Re f(B), so we can choose t > 0 such that

sup Re f(B) ≤ t < Re f(x).

Now replace f by t−1f , so that for all y ∈ E we have y ∈ B, and hence

Re f(y) ≤ 1 < Re f(x).

Then f ∈ E, and so x /∈ (E).

(2) follows by swapping the roles of X and X∗.

Theorem 7.12 (Alaoglu’s Theorem). If X is a normed space then BX∗ is weak*-compact.

Proof. For each x ∈ X letDx = t ∈ F : |t| ≤ ‖x‖.

Then BX∗ may be regarded as a subset of the Cartesian product

P =∏x∈X

Dx.

Since the weak* topology on BX∗ agrees with the product topology, by Tychonoff’s theoremit suffices to show that BX∗ is closed in P . Let (fi) be a net in BX∗ that converges to g inP . For each x, y ∈ X and t ∈ F we have

g(tx+ y) = lim fi(tx+ y)

= lim(tfi(x) + fi(y)

)= t lim fi(x) + lim fi(y)

= tg(x) + g(y).

Thus g is linear. Also, for each x ∈ X we have |g(x)| ≤ ‖x‖ since g ∈ P , and henceg ∈ BX∗ .


Exercises

1. Let U be a weak neighborhood of 0 in an infinite-dimensional Banach space X. Then Ucontains a subspace of finite codimension.

2. The unit sphere x : ‖x‖ = 1 of an infinite-dimensional Banach space X is weakly dense inthe unit ball.

3. If X is a separable Banach space then BX∗ is weak* metrizable.

4. If (en) is an orthonormal sequence in a Hilbert space then en → 0 weakly.

5. Let X be a Banach space and T ∈ B(X).

(a) If T is compact, then ‖Txn‖ → 0 whenever xn → 0 weakly.

(b) If X is a Hilbert space, and if T has the property that ‖Txn‖ → 0 whenever xn → 0weakly, then T is compact.

6. Let X and Y be Banach spaces, and let (Ti) be a net in B(X, Y ). Then

(a) Ti → 0 in the strong operator topology if and only if Tix→ 0 for every x ∈ X.

(b) Ti → 0 in the weak operator topology if and only if Tix→ 0 weakly for every x ∈ X.

7. Let X and Y be Banach spaces, let (Ti) be a bounded net in B(X, Y ), and let D ⊆ X bedense. Then Ti → 0 in the strong operator topology if and only if Tix→ 0 for every x ∈ D.

8. Let X be a locally compact Hausdorff space. Then a sequence (fn) in C0(X) convergesweakly to 0 if and only if it is bounded and converges pointwise to 0.

9. Let E be an orthonormal basis of a Hilbert space H. Then a sequence (xn) in H convergesweakly to 0 if and only if it is bounded and 〈xn, e〉 → 0 for every e ∈ E.

10. If X is a normed space then(BX) = BX∗ .

11. Let X be a Banach space.

(a) X is weak* dense in X∗∗.

(b) (Goldstine’s Theorem) The unit ball of X is weak* dense in the unit ball of X∗∗.


12. A Banach space is reflexive if and only if its unit ball is weakly compact.

13. Bc0 is not weakly compact.

14. Does zn → 0

(a) weakly in L1(T)?

(b) in norm in L1(T)?

(c) weak* in L∞(T)?

(d) weakly in C(T)?

15. Does δn → 0

(a) weakly in `1?

(b) weak* in `1?

(c) in norm in `1?

(d) weak* in `∞?

(e) weakly in c0?

16. A convex subset of a locally convex space is closed if and only if it is weakly closed.

17. Let X be a locally convex space and Y a subspace of X..

(a) Y is dense if and only if Y ⊥ = 0.

(b) If Y is closed and x ∈ X \ Y , then there exists f ∈ Y ⊥ such that f(x) = 1.

18. Let X be a locally convex space and Y a subspace of X∗.

(a) (⊥Y )⊥ is the weak* closure of Y .

(b) Y is weak* dense if and only if ⊥Y = 0.

19. Let X and Y be locally convex spaces and T : X → Y linear and continuous. Then T isinjective if and only if ranT ∗ is weak* dense in X∗.

20. Let X and Y be Banach spaces and T : X → Y a linear map. Then the following areequivalent:

8 EXTREME POINTS 50

(1) T is continuous.

(2) T is weakly continuous (i.e., continuous from σ(X,X∗) to σ(Y, Y ∗).

(3) For every f ∈ Y ∗ the composition f T is continuous.

Moreover, if these equivalent conditions are satisfied, the dual map T ∗ : Y ∗ → X∗ defined byT ∗f = f T is weak* continuous.

21. Let H be a Hilbert space and T ∈ B(H). Then T (BH) is norm closed.

22. C[0, 1] is weak* dense in L∞[0, 1].

23. L∞[0, 1] is weak* separable.

24. Every normed space is isometrically isomorphic to a subspace of C(X) for some compactHausdorff space X.

25. (Riemann-Lebesgue Lemma) zn → 0 weak* in L∞(T).

26. The delta functions δn converge weakly in `p for 1 < p <∞ but not for p = 1.

27. In a normed space, if xn → x weakly, then ‖x‖ ≤ lim inf ‖xn‖. (Hint: Hahn-Banach Theo-rem)

28. In this problem you will prove a frequently used alternative formulation of the BipolarTheorem. Let X be a locally convex space, E ⊆ X, and R ⊆ X∗. Define

E• = f ∈ X∗ : |〈x, f〉| ≤ 1 for all x ∈ E•R = x ∈ X : |〈x, f〉| ≤ 1 for all f ∈ R.

Then:

(a) •(E•) is the closed convex balanced hull of E.

(b) (•R)• is the weak* closed convex balanced hull of R.

8 Extreme points

Definition 8.1. Let X be a vector space, A ⊆ X, and x ∈ A. Then x is called an extremepoint of A if for all y, z ∈ A and 0 < t < 1, ty + (1− t)z = x implies y = z = x.

Theorem 8.2 (Krein-Milman Theorem). Every convex compact subset of a locally convexspace is the closed convex hull of its extreme points.

8 EXTREME POINTS 51

Proof. Without loss of generality F = R. Let K be a convex compact subset of a locallyconvex space X. Without loss of generality K 6= ∅. Call C ⊆ K a face of K if it isnonempty, convex, compact, and has the property that for all x, y ∈ K and 0 < t < 1, iftx+ (1− t)y ∈ C then x, y ∈ C. It’s routine to verify that a face of a face of K is a face ofK. Consequently, for every face C of K and f ∈ X∗, if a = min f(C) then C ∩ f−1(a) is aface of K, because it is easy to see that it is a face of C.

Let the family of all faces of K be directed by C ≤ D if C ⊇ D. It is easy to see thatthe intersection of any totally ordered family of faces of K is a face of K, because it hasthe finite intersection property, K is compact, and any finite intersection of faces is a face.Thus by Zorn’s lemma K has a minimal face C. We will show that C is a singleton, whichwill give an extreme point of K. Suppose not. Then by the Hahn-Banach theorem we canchoose f ∈ X∗ that is not constant on C. Let a = min f(C). Then C ∩ f−1(a) is a face ofC, and hence of K, and is a proper subset of C since f is nonconstant on C. But this is acontradiction because C is minimal.

Now we have shown that K has at least one extreme point. Let C be the closed convexhull of the set of all extreme points of K. We must show that C = K. Suppose not.Then we can choose x ∈ K \ C, and then by Theorem 7.7 we can find f ∈ X∗ such thatf(x) < min f(C). Let a = min f(K). Then D = f−1(a) is a face of K that is disjoint fromC. By what we already proved, D has an extreme point y, which must also be an extremepoint of K. But y /∈ C, which is a contradiction.

Exercises

1. (a) BL1[0,1] has no extreme points.

(b) L1[0, 1] is not isometrically isomorphic to the dual of a Banach space.

2. Same as Problem 1 but for c0.

3. Prove that every unit vector in a Hilbert space H is an extreme point of BH .

4. For each x ∈ [0, 1] let δx be the point mass at x, and let K = δx : x ∈ [0, 1].

(a) K is weak* compact in M [0, 1].

(b) M [0, 1] is the weak* closed linear span of K.

(c) A probability measure is a positive regular Borel measure of total mass 1. Let P be theset of probability measures on [0, 1]. Then K is the set of extreme points of P .

Hint: for the hard direction, show that every extreme point µ of P has support suppµ = xfor some x ∈ [0, 1], and try proof by contraction.

(d) P is the weak* closed convex hull of K.

REFERENCES 52

References

[1] A. Belton. Functional Analysis. https://www.maths.lancs.ac.uk/~belton/www/

notes/fa_notes.pdf (3/17/2019), 2006. Lancaster University.

[2] J. B. Conway. A course in functional analysis, volume 96 of Graduate Texts in Mathe-matics. Springer-Verlag, New York, second edition, 1990.

[3] G. B. Folland. Real analysis. Pure and Applied Mathematics (New York). John Wiley& Sons, Inc., New York, second edition, 1999.

[4] W. Rudin. Functional analysis. International Series in Pure and Applied Mathematics.McGraw-Hill, Inc., New York, second edition, 1991.

https://www.maths.lancs.ac.uk/~belton/www/notes/fa_notes.pdf

https://www.maths.lancs.ac.uk/~belton/www/notes/fa_notes.pdf

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MAT 578 Functional Analysis Lecture Notesquigg/teach/courses/578/2019fall... · 2020. 3. 17. · 6 Locally convex spaces 36 7 Weak topologies 43 8 Extreme points 50 Introduction Most