real and functional analysis: part b: functional analysis

Second Edition

Real and Functional Analysis PARTB FUNCTIONAL ANALYSIS

MA THEMATICAL CONCEPTS AND METHODS IN SCIENCE AND ENGINEERING

Series Editor: Angelo Miele Mechanical Engineering and Mathematical Sciences Rice University

Recent volumes in this series:

22 APPLICATIONS OF FUNCTIONAL ANALYSIS IN ENGINEERING • J. L. Nowinski

23 APPLIED PROBABILITY. Frank A. Haight

24 THE CALCULUS OF VARIATIONS AND OPTIMAL CONTROL: An Introduction • George Leitmann

25 CONTROL, IDENTIFICATION, AND INPUT OPTIMIZATION • Robert Kalaba and Karl Spingarn

26 PROBLEMS AND METHODS OF OPTIMAL STRUCTURAL DESIGN • N. V. Banichuk

27 REAL AND FUNCTIONAL ANALYSIS, Second Edition Part A: Real Analysis • A. Mukherjea and K. Pothoven

28 REAL AND FUNCTIONAL ANALYSIS, Second Edition Part B: Functional Analysis. A. Mukherjea and K. Pothoven

29 AN INTRODUCTION TO PROBABILITY THEORY WITH STATISTICAL APPLICATIONS. Michael A. Golberg

30 MULTIPLE-CRITERIA DECISION MAKING: Concepts, Techniques, and Extensions. Po-Lung Yu

31 NUMERICAL DERIVATIVES AND NONLINEAR ANALYSIS • Harriet Kagiwada, Robert Kalaba, Nima Rasakhoo, and Karl Spingarn

32 PRINCIPLES OF ENGINEERING MECHANICS Volume 1: Kinematics- The Geometry of Motion • Millard F. Beatty, Jr.

33 PRINCIPLES OF ENGINEERING MECHANICS Volume 2: Dynamics- The Analysis of Motion. Millard F. Beatty, Jr.

34 STRUCTURAL OPTIMIZATION Volume 1: Optimality Criteria. Edited by M. Save and W. Prager

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Second Edition

Real and Functional Analysis PART B FUNCTIONAL ANALYSIS

A. Mukherjea and K. Pothoven University of South Florida Tampa, Florida

Springer Science+Business Media, LLC

Library of Congress Cataloging in Publication Data

Mukherjea, Arunava, 1941-Real and functional analysis.

(Mathematical concepts and methods in science and engineering; 27-28) Includes bibliographical references and indexes. Contents: pt. A. Real analysis-pt. B. Functional analysis. 1. Functions of real variables. 2. Functional analysis. I. Pothoven, K. II. Title. III.

Series. QA331.5.M84 1984 515.8 84-8363 ISBN 978-1-4899-4560-0 ISBN 978-1-4899-4558-7 (eBook)

© 1986 Springer Science+Business Media New York Originally published by Plenum Press, New York in 1986

Softcover reprint of the hardcover 2nd edition 1986

All rights reserved

No part of this book may be reproduced, stored in a retrieval system, or transmitted, in any form or by any means, electronic, mechanical, photocopying, microfilming,

recording, or otherwise, without written permission from the Publisher

DOI 10.1007/978-1-4899-4558-7

Preface to the Second Edition

The second edition is composed of two volumes. The first volume, Part A, is entitled Real Analysis and contains Chapters 1, 2, 3, 4, parts of Chapter 5, and the first five sections of Chapter 7 of the first edition, together with additional material added to each of these chapters. The second volume, Part B, is entitled Functional Analysis and contains Chapters 5 and 6 and the Appendix of the first edition together with additional topics in functional analysis including a new section on topological vector spaces, a complete chapter on spectral theory, and an appendix on invariant subspaces. Included in this edition are many new problems, new proofs of theorems, and additional material. Our goal has been, as before, to present the essentials of analysis as well as to include in the book many interesting, useful, and relevant results (usually not available in other books) so that the book can be useful as a reference for the student of analysis.

As in the first edition, certain portions of the text designated by (.) can be omitted. In this volume, problems that are designated by (X) are an integral part of the text and should be worked by the student. Problems that are difficult are starred (*).

We are again grateful to friends and colleagues who have pointed out errors in the first edition and given suggestions for improving the text. Particularly, we thank Professors G. Hognas, R. A. Johnson, and B. Schreiber.

Tampa, Florida

v

A. Mukherjea K. Pothoven

Contents

6. Banach Spaces . . . . . . . . . . . . . . . . . . . . . . . 1

6.1. Basic Concepts and Definitions. . . . . . . . . . . . . . 2

6.2. Bounded Linear Functionals and the Hahn-Banach Theorem. 14

6.3. The Open Mapping Theorem, the Closed Graph Theorem, and the Principle of Uniform Boundedness ..... 33

6.4. Reflexive Banach Spaces and the Weak Topology. 48

6.5. Compact Operators and Spectral Notions . . . . 70

6.6. Topological Vector Spaces. . . . . . . . . . . 93

6.7. The Kakutani Fixed Point Theorem and the Haar Measure on a Compact Group . . . . . . . . . . . . . . . . . . . .. 111

7. Hilbert Spaces . . . . . . . . . . . . . . . . . . . . . 121

7.1. The Geometry of Hilbert Space ........... 122

7.2. Subspaces, Bases, and Characterizations of Hilbert Spaces 131

7.3. The Dual Space and Adjoint Operators . . . . . . . . 142

7.4. The Algebra of Operators. The Spectral Theorem and the Approximation Theorem for Compact Operators .

7.5. Spectral Decomposition of Self-Adjoint Operators

8. Spectral Theory

152

177

195 8.1. Spectral Theory for Bounded Operators Revisited 195

8.2. Unbounded Operators and Spectral Theorems for Unbounded

Appendix

C.

Self-Adjoint Operators ................. 219

Invariant and Hyperinvariant Subspaces .

vii

255 255

viii Contents

BIBLIOGRAPHY . . . . . . . . . . . . . . . . . . . . . . . . . . 265

DEFINITION, THEOREM, PROPOSITION, LEMMA, AND COROLLARY INDEX 269

SYMBOL AND NOTATION INDEX 271

SUBJECT INDEX. . .

ERRATA FOR PART A

273

277

6

Banach Spaces

Integral equations occur in a natural way in numerous physical problems and have attracted the attention of many mathematicians including Volterra, Fredholm, Hilbert, Schmidt, F. Riesz, and others. The works of Volterra and Fredholm on integral equations emphasized the usefulness of the techniques of the integral operators. Soon it was realized that many problems in analysis could be attacked with greater ease if placed under a suitably chosen axiomatic framework. Axioms closely related to those of a Banach space were introduced by Bennett. t Using the axioms of a Banach space, F. Riesz; extended much of the Fredholm theory of integral equations. In 1922 using similar sets of axioms for such spaces, Banach,§ Wiener, II and Hahn· all independently published papers. But it was Banach who continued making extensive and fundamental contributions in the development of the theory of these spaces, now well known as Banach spaces. Banach-space techniques are widely known now and applied in numerous physical and abstract problems. For example, using the Hahn-Banach Theorem (asserting the existence of nontrivial continuous linear real-valued functions on Banach spaces), one can show the existence of a translation-invariant, finitely additive measure on the class of all bounded subsets of the reals such that the measure of an interval is its length.

The purpose of this chapter is to present some of the basic properties and principles of Banach spaces. In Section 6. I we introduce the basic

t A. A. Bennett, Proc. Nat. Acad. Sci. U.S.A. 2, 592-598 (1916). ; F. Riesz, Acta Math. 41, 71-98 (1918). § S. Banach, Fund. Math. 3, 133-181 (1922). II N. Wiener, Bull. Soc. ·Math. France 50, 119-134 (1922). [ H. Hahn, Monatsh. Math. Phys. 32, 3-88 (1922).

1

2 Chap. 6 • Banach Spaces

concepts and definitions. In Sections 6.2 and 6.3 we present what are acknowledged as the four most important theorems in Banach spaces-the Hahn-Banach Theorem, the Open Mapping Theorem, the Closed Graph Theorem, and the Principle of Uniform Boundedness. In Section 6.4 we introduce the reflexive spaces, derive representation theorems for the duals of various important Banach spaces, and present in detail the interplay between reflexivity and weak topology. In Section 6.5 we introduce compact operators, present the classical Fredholm alternative theory, and discuss spectral concepts for such operators. In Section 6.6, topological vector spaces, locally convex spaces, and the Krein-Milman theorem are introduced. We present also Liapounoff's convexity theorem, as an application of the Krein-Milman theorem, in this section. The final section contains the Kakutani fixed point theorem and its application showing the existence of an invariant measure on a compact Hausdorff topological group.

For the sake of completeness and continuity, a slight overlap between parts of Sections 3.5 and 4.3 in Part A and parts of Sections 6.1 and 6.2 in this chapter has been unavoidable. The reader who is already familiar with those sections in Part A may, of course, skip this material in this chapter.

Throughout this chapter, F will denote either the real numbers R or the complex numbers C.

6.1. Basic Concepts and Definitions

We begin with several fundamental definitions.

Definition 6.1. A linear space X over a field F is an Abelian group under addition (+), together with a scalar multiplication from Fx X into X such that

(i) a(x + y) = ax + ay,

(ii) (a + P)x = ax + px, (iii) (ap)x = a(px),

(iv) Ix = x,

for all a, p E F, x and Y E X. (Here 1 denotes the multiplicative identity in F and 0 will denote the additive identity in X.)

Definition 6.2. A linear space X over a field F is called a normed linear space if to each x E X is associated a nonnegative real number II x II,

Sec. 6.1 • Basic Concepts and Definitions

called the norm of x such that

(i) II x II = 0,

(ii) II ax II = I a I II x II,

(iii) II x + y II < II x II + II y II,

if and only if x = 0;

for all a E F;

for all x, y E X.

3

Defining d(x, y) = II x - y II, it can easily be verified that d defines a metric in X. A normed linear space X is called a Banach space if it is complete in this metric. The topology induced by d will usually be referred to as the topology mX.

A linear space is also called a vector space. Let X be a vector space over a field F. (The elements of X are often called vectors and those of F scalars.) Let A C X. If the subset A is also a vector space over F with respect to the same operations, then A is called a linear subspace (or simply, a subspace). The subset A is said to span X if any element of X can be expressed as a finite linear combination of elements from A, that is, for each x E X there exist scalars aI' a2, ... , an in F, and vectors Xl, X2' ... , Xn in X such that x = alXI + a2X2 + ... + anxn. Note that in a vector space, for any given subset B there is always a subspace (denoted by (B» spanned by B, namely, the set of all finite linear combinations of elements from B. The subspace (B) is the smallest subspace of X containing B. In other words, if S is a subspace of X containing B, then S contains the subspace (B).

A finite subset {Xl' X2' ... , xn} of a vector space X over F is called linearly independent if any relation of the form

ai E F (i = I, 2, ... , n)

implies that al = a2 = ... = an = 0. A subset (not necessarily finite) of a vector space is called linearly independent if every finite subset of it is linearly independent. Notice that if (Ak)kEK is a family (possibly infinite) of linearly independent subsets in a vector space such that for kl' k2 in K (kl o:F k 2), either Akl C Ak2 or Ak2 C Akl , then U {Ak : k E K} is also a linearly independent subset. Thus, considering the partial order "inclusion" in the family !T of all linearly independent subsets in a vector space X

(X o:F {O}), it is clear that !T is nonempty (since it contains all nonzero singleton subsets), and each nonempty totally (or linearly) ordered subset of!T has an upper bound. It follows by using Zorn's lemma (Part A, Theorem 1.2) that X has a maximal linearly independent set H. Such a set is called a Hamel basis for X. Equivalently, a subset H of X is a Hamel basis

for X if every vector x in X can be expressed uniquely as a finite linear combination of vectors from H. Note that any linearly independent subset of X is contained in a Hamel basis for X. (This is proven in the same manner as before using Zorn's lemma in the family of all linearly independent subsets of X which contain as a subset the given linearly independent set.) The cardinality of a Hamel basis for a vector space X is a number independent of the Hamel basis, and is called the dimension of X. (See Problem 6.1.28 and the hint therein.)

Now we consider some examples of normed linear spaces.

Examples

6.1. Rn = {(a1 , a2 , ••• , an): ai E R, 1 <i< n} is a Banach space over R if we define addition and scalar multiplication in the natural way and

n II (a1 , a2 , ••• , an) II = L I ai I·

i=l

6.2. e1[0, 1], the usual linear space of complex-valued continuous functions over [0, 1], is a Banach space over F if we define

11/11= sup I/(x) I· 0$:1;$1

(See Problem 6.1.5.) This norm will be referred to as the uniform or "sup" norm.

6.3. The set of all polynomials on [0, 1] as a subspace of qo, I] in Example 6.2 is a normed linear space, but not a Banach space. The reason is that nonpolynomial continuous functions can be uniformly approximated by polynomials on [0, I], by Weierstrass' theorem.

6.4. Let (X, ~ f1) be a measure space. For real number p > I, let Lp be the linear space of all scalar-valued measurable functions I such that 1 I IP is integrable. If functions that are equal a.e. are identified, then Lp becomes a Banach space with the norm

II flip = (f 1 I IP df1 ) lip .

These spaces were introduced in Section 3.5 as were the spaces L=. L= is the space of all measurable functions that are bounded except possibly on a set of measure zero. Again, if we identify functions in L= that are equal a.e., L= is a Banach space with the norm

II I II = ess sup 1 I I,

Sec. 6.1 • Basic Concepts and Definitions 5

where ess sup I f I = inf{M: .u{x: I f(x) I > M} = O}.

When the measure space X is the set of positive integers with each integer having measure I, the Lp spaces are called Ip spaces. Thus Ip(l < p < 00) is the set of all sequences x = (Xn)~~l' Xn E F with L::'1 I Xn IP < 00. Ip, like L p , is a Banach space under the norm

( 00 ) lip II x lip = ];1 I xn IP .

The 100 space is the set of all bounded sequences x = (Xn)~~l' Xn E F. Again, 100 , like Leo, is a Banach space under the norm II x 1100 = supn I Xn ,.

Definition 6.3. Suppose X is a linear space over F, and II . 111 and II . 112 are two norms defined on it. Then these norms are called equivalent if there exist positive numbers a and b such that

a . II x 111 < II X 112 < b . II X 111

for all x EX. I

Equivalence of norms is clearly reflexive, symmetric, and transitive. In an infinite-dimensional normed linear space, there are always two norms that are not equivalent (Problem 6.1.12). But the situation is much nicer in the finite-dimensional case, as the following theorem shows.

Theorem 6.1. In a finite-dimensional linear space X, all norms are equivalent. I

Proof. Let {ZI' Z2' ... , zn} be a basis for X. Let x E X and x =

L:7~I XiZi' We define: II x II = SUPl5isn I Xi I. Then II . II defines a norm in X, and X is complete in this norm (see Problem 6.1.2). We show that any given norm II . ilIon X is equivalent to this sup-norm.

Let x = L:7~I Xizi and b = L7~1 II Zi Ill' Then b > 0 and

II X III < b II x II for all x E X.

To find a similar inequality in the other direction, we use induction on n. When n = I,

II X 111 = I Xl I II ZI 111 = II X II II Zl lit-

Thus, the theorem is true for n = I. Suppose that the theorem is true for all spaces of dimension less than n, n > I. Let Xi be the subspace spanned by {ZI' ... , Zi-I, Zi+I' ... , zn}. Then dim Xi < n. Since Xi is complete in II . II and since the norms II . II and II . III are equivalent in Xi by the induction hypothesis, the subspace Xi is also complete in II . III and is therefore a closed subspace of (X, II . Ill). Since Zi (/: Xi, there is a positive number di

such that

This means that for all scalars aj E F,

We can repeat the process for each i, 1 O. Then

II f aizi II > sup I ai I . min I di I, i=I 1 Isisn Isisn

which means that a II x II < II x Ill, for all x E X,

where a = minISiSnl di I > O. I

Corollary 6.1. Every finite-dimensional normed linear space is com-plete. I

Corollary 6.2. Let (Xl' II . Ill) and (X2' II . 112) be any two finitedimensional normed linear spaces of the same dimension over F. Then they are topologically isomorphic, i.e., there is a mapping from one onto the other, which is an algebraic isomorphism as well as a topological homeomorphism. I

The proof is an easy consequence of Theorem 6.1 and is left to the reader. Next, we present a result due to Riesz, often useful in proving various results in the theory of normed linear spaces, besides being of independent geometric interest.

Proposition 6.1. Let Y be a proper closed linear subspace of a normed linear space X over F. Let 0 < a <: 1. Then there exists some xa E X such that II xa II = 1 and inf1lEy II Xa - Y II > a. I

Proof. Let x E X - Y and d = inf1ley II x - y II. Then d > 0, since Y is closed. Now there exists Yo E Y such that ° < II x - Yo II < dla. Let Xa = [x - Yo]1 II x - Yo II. Now the reader can easily check that Xa satisfies the requirements of the propositon. I

Note that in the above proposition, the proper subspace Y has to be necessarily closed. For instance, if X is C[O,I] with the uniformt norm and Y is the subspace of all polynomials on [0, I], then Y = X and therefore the proposition fails to work in this case.

Also, one cannot generally take a = 1 in Proposition 6.1. For example, let X be the real-valued continuous functions on [0, 1] which vanish at 0, with the uniform norm, and Y = {f E X: Hf(x) dx = O}. Then Y is a closed proper subspace of X. Suppose there exists hEX - Y such that inf/eyll h - f II > 1, where for g E X, II g II = sUPo:s:x:S:ll g(x) I. If g EX - Y and a(g) = [H hex) dx]/[H g(x) dx], then

f: [hex) - a(g) . g(x)] dx = ° or h - a(g) . g E Y;

therefore, II h - [h - a(g) . g] II = II a(g) . g II > 1. Let gn(x) = Xlln,

I < n < 00. Then gn E X - Y and hence II a(gn) . gn II > I. But a(gn) = [en + 1)ln] H hex) dx. Hence, since II gn II = 1, we have

I J: hex) dx I > n: I '

for each positive integer n. This means that I H hex) dx I > 1. But since h(O) = ° and II h II = 1, I f~ hex) dx I < 1, which is a contradiction.

Now we show how Proposition 6.1 can help us understand the notion of compactness in a normed linear space. We know that a set in Rn or in any finite-dimensional normed linear space is compact if and only if it is closed and bounded. But if X is an infinite-dimensional normed linear space, then by Proposition 6.1 we can find (xn) with II Xn II = I, I < n < 00 and for each n, II xn - Xi II > t, for I < i < n; clearly these xn's cannot have a limit point and therefore the closed unit ball in X is not compact. Thus we have the following theorem.

Theorem 6.2. Let X be a normed linear space over F. Then its closed unit ball is compact if and only if X is finite dimensional. I

t The same as the "sup"-norm.

Definition 6.4. A series L:k=IX/; in a normed linear space X over F is called summable if /I L:Z-IX/; - X " - 0 as n - 00 for some x E X. For a summable series, we writeL::'lx/; = limn-+ooL:Z=lx/;. The series L:k:1X/; is called absolutely summable if L:k:lll x/; II < 00. I

We know that an absolutely summable series of real numbers is summabIe. This is a consequence of the completeness of the real numbers. In fact, we have the foHowing theorem, which is often useful in establishing the completeness of a normed linear space.

Theorem 6.3. Every absolutely sum mabIe series in a normed linear space X is summable if and only if X is complete. I

Proof. For the "if" part, let X be complete and for each positive integer n let Xn be an element of X such that L::'111 Xn II < 00. Let Y/; = L:~=IXn' Then

II Yk+p - Yk II = /I n%1 xn II < n%1 II xn II,

which converges to zero as k - 00. Hence the sequence (Y/;)k=1 is Cauchy in X. Therefore since X is complete, there exists x E X such that x = Iim/;->-ooL:~=lxn' This proves the "if" part.

For the "only if" part, suppose every absolutely summable series in X is summable. Let (xn) be a Cauchy sequence in X. For each positive integer k, there is a positive integer n/; such that II Xn - Xm II < 1/2k for all nand m greater than or equal to n/;o We choose n/;H > n/;o Let Yl = xn1 and Y/;H = xnk+l - xnk ' k > 1. Then L:k:lll y/; II < 00. Therefore, there exists y E X such that

m y = lim L Y/; = lim xnm '

m-+oo /;=1 m-+oo

Since (xn) is Cauchy, limn-+ooxn is also y. I

Finally, in this section we introduce the concept of quotient spaces. Quotient spaces are often useful in tackling certain problems in normed linear spaces, as will be seen in the later sections of this chapter. See also Problem 6.1.10.

Suppose M is a closed subspace of a normed linear space X over a field F. Let [x] = {y EX: Y""" X}, where x E X and Y""" x if and only if y - x E M. Clearly, ",...,," defines an equivalence relation on X. Let XIM

denote the set of all equivalence classes. For a E F and [x], [y] in XI M, we define

[x] + [y] = [x + y] and a[x] = [ax].

These operations are well defined and make XI M a vector space. Let II [x] 111 = infYEM11 x - y II. One can easily check that (XIM, II . 111) is a normed linear space, usually called the quotient of X by M. If rJ> is the natural map from X onto XIM defined by rJ>(x) = [x], then rJ> is a continuous, linear, and open mapping. The linearity of rJ> is trivial. The continuity of rJ> follows from the fact that II [xn - x] 111 < II Xn - x II. Also, if V = {y E X: II y - x II < r}, then rJ>(V) = {[y] E XIM: II [y] - [x] 111 < r}. This implies that rJ> is open.

In many cases, several properties of the quotient space XI M are strongly related to similar properties of the space X. Problem 6.1.9 and the next proposition will illustrate two of them.

Proposition 6.2. Let M be a closed linear subspace of a normed linear space X. Then X is complete if and only if M and XI M are complete. I

Proof. For the "if" part, let M and XIM be complete. Let (Xn)~=l be a Cauchy sequence in X. Then ([Xn]):'l is Cauchy in XI M and therefore there exists [y] E XIM such that [xn] -+ [y] in XIM as n -+ 00. This means that infzEM IIxn - y - z II -+ 0 as n -+ 00. Hence there exists a subsequence (nk) of positive integers and a sequence (Zk) in M such that xnk - y - Zk -+ 0 as k -+ 00. This means that the sequence (Zk):'l is Cauchy in M and therefore there exists Z E M such that Zk -->- Z as k -->- 00 so that Xn -+ Y + Z as n -+ 00. The "if" part is proved.

For the "only if" part, we will use Theorem 6.3. Let X be complete. Then M, being a closed subspace, is also complete. To prove the completeness of XIM, let 1::'111 [Xn] III < 00. We are finished if we can show that there exists [y] E XI M such that [y] = limk-+ooL~=l [xn]. Let Yn E M be chosen such that II Xn + Yn II < II [Xn] 111 + 1/2n, for each positive integer n. Then L:'lll Xn + Yn II < 00. Since X is complete, there exists Y E X such that Y = limk-+ooL~=l(Xn + Yn). Since the natural map rJ> is continuous and linear,

k k [y] = lim L [xn + Yn] = lim L [xnl

k-""'oo n .... l k -+00 n= 1 I

Problems

X 6.1.1. Let X be a normed linear space over a field F. Let y E X and a E F, a # 0. Show that the mappings x -+ x + y and x -+ a . x are homeomorphisms of X onto itself. X 6.1.2. Let X be a linear space (of dimension n) over a field F. Let {Zl' Z2' ... , zn} be a basis of X. If x = L?=la.jZi, ai E F and II x II = SUPl$i$n I aj I, then show that (X, II . II) is a Banach space. X 6.1.3. Let X be a normed linear space. If Sir) = {y E x: II y - x II < r}, then show that Sx(r) = {y E X: II y - x II < r}. X 6.1.4. Let A and B be two subsets of a normed linear space X. Let

A + B = {x + y: x E A, y E B}.

Show that (a) A + B is open whenever either A or B is open; and (b) A + B is closed whenever A.is compact and B is closed. (Note that A + B need not be closed even if A and B are both closed.) X 6.1.5. Prove that CI[O, I], the linear space of complex-valued continuous functions on [0, 11, is a Banach space under the uniform norm. X 6.1.6. Show that a finite-dimensional subspace of an infinite-dimensional normed linear space X is nowhere dense in X. X 6.1.7. Use Problem 6.1.6 and the Baire Category Theorem to prove that an infinite-dimensional Banach space cannot have a countable Hamel basis. X 6.1.8. Let X be a normed linear space and f be the mapping defined by f(x) = rx/(I + II x 11). Show that f is a homeomorphism from X onto {x: II x II < r}. [Hint: f-l(y) = y/(r - II y II).] X 6.1.9. Let M be a closed subspace of a normed linear space X. Show that X is separable if and only if M and X/Mare both separable. X 6.1.10. (i) Show that the sum of two closed subspaces of a normed linear space is closed whenever one of the subs paces is finite dimensional. [Hint: Let A be a closed subspace and B be a finite-dimensional subspace of a normed linear space X. If (/J is the natural map from X onto X/A, then (/J-l«(/J(B») = A + B.]

(ii) Let M be a closed subspace of a normed linear space X. Show that M + N is closed for every subspace N if dim(X/M) is finite.

(iii) Let X = 12 and for each k, let ek = (0, ... ,0, 1,0,0, ... ), where the only nonzero entry is I and at the kth position. Let M be the closed subspace of X spanned by the elements (e2k)~1 and N be the closed subspace spanned by the elements (e2k + (l/k)e2k-l)~I' Show that M + N is dense in X, but does not contain the element (1,0,1,0, t, ... ).

X 6.1.11. Show that a normed linear space whose closed unit ball is totally bounded is finite dimensional.

6.1.12. Let X be an infinite-dimensional linear space over F. Let (xn) be an infinite set contained in H, a Hamel basis for X. If x E X and x = I: a"x" , x" E H and a" E F, then define II x III = sup I a" I and II x 112 = I: I a" I. Show that these two norms are not equivalent. [Hint: Let xn E H for n > I and Yn = (lIn) (XI + X 2 + ... + xn). Then II Yn III ~ ° as n ~oo; II Yn 112 = I for each n.] See Example 6.12 later for II • 112'

X 6.1.13. Prove that Lp(X, ~ fl), p > I, is a Banach space under the norm II I lip = (f I I IP dfl)IIP. X 6.1.14. Prove that LcyAX, ~ fl) is a Banach space under the norm

II I 1100 = ess sup I I I· 6.1.15. Prove that a subset A C lp, I X 2 , ... ) EA.

6.1.16. A Schauder basis for a normed linear space X is a sequence (un) in X such that for every x in X, there is a unique sequence of scalars (an) such that x = I::'lanUn , i.e., limn ..... oo II x - I:£=laiui II = 0. Prove t the following assertions:

(i) A normed linear space with a Schauder basis is separable. (ii) Let en denote the element (Xi) in lp, where Xi = ° for i 7'=- nand

Xn = 1. Then the sequence (en) is a Schauder basis for lp, I o} is not a Schauder basis for C[O, I] with the "sup" norm. (Hint: See if a continuous function, not differentiable at 0, can be represented by a uniformly convergent power series on [0, 1])

(iv) Consider C[O, I] with the "sup" norm. Define a sequence (fn) in C[O, I] as follows: Let (t i ) be the sequence of dyadic points in [0, I] given, respectively, by

0, I, t , ! , i, 1, i, i, i, l6 , l6' ...

Let h(t) = I, 12(t) = t for each t in [0, I]. For n > 2, let In(t i) = ° for i < n, = I for i = n, and In be linear between any two consecutive points among the first n dyadic points. Let g E C[O, I]. Then the sequence I:':..laJi, where the a/s are given by

n-I

an = g(tn) - I aJi(tn), a l = g(O), i=1

converges uniformly to g on [0, I]. [Thus, the sequence (fn) is a Schauder basis for C[O, I].]

t For these and other information on Schauder bases, the serious reader should consult R. C. James, Amer. Math. Monthly, Nov. 1982, pp. 625-640.

(v) Let X be a Banach space and (en) be a Schauder basis of X.

Define III x III by III x III = sup {II L~~lxiei II: n > I} if x = L:IXie·i' Then this defines a norm in X. The space X is complete in this new norm, and the identity mapping from (X, III . III) to (X, II . II) is continuous. X 6.1.17. Let Y be a finite-dimensional subspace of a normed linear space X. Let x E X - Y. Show that there exists y E Y such that

IIx-yll=inf{llx-zll:zE V}.

(The element y is called a best approximation of x in Y.) Also, show that the set of best approximations of x in Y is convex. X 6.1.18. A normed linear space X is called strictly convex if for any x, y in X such that II x II = II y II = I and x of::- y, we have II !(x + y) II < I. Prove that X is strictly convex if and only if for any x, y in X, II x + y II = II x II + II y II implies that x = ay or y = ax with a > O. X 6.1.19. Let X and Y be as in Problem 6.1.17. Suppose also that X is strictly convex. Show that any element x in X has a unique best approximation in Y.

6.1.20. Show that the Lp spaces, I < p < 00, are strictly convex under usual norm, whereas Ll and Loo (with usual norms) are not. (Use the remark on the validity of equality in the Holder's Inequality for equality in the Minkowski's Inequality.)

{Remark: Notice that C[a, b], the real-valued continuous functions with the "sup" norm is not strictly convex. However, we may remark that for each positive integer n, every element in C[a, b] has a unique best approximation in the finite-dimensional subspace of polynomials of degree not exceeding n. In fact, this assertion remains true for any finite-dimensional subspace Y satisfying the following condition (known as the Haar condition): each nonzero y in Y has at most m - I zeros in [a, b], where m = dim Y. Also, the Haar condition, it turns out, is necessary and sufficient for the uniqueness of the best approximation in Y.}

6.1.21. Let X be a strictly convex normed linear space and Y be a finite-dimensional subspace of X. For each x in X, let T(x) be the best approximation of x in Y. Prove that T is continuous. (Note that T need not be linear.)

6.1.22. Uniformly Convex Normed Linear Spaces. A normed linear space X is called uniformly convex if for sequences (xn) and (Yn) in X such that II Xn II < 1, II Yn II < 1, and II HXn + Yn) II ->- I as n ->- 00, we have limn-+ooll xn - Yn II = O. [Notice that if the norm satisfies the parallelogram law, i.e., II x + y 112 + II x - Y 112 = 211 X 112 + 211 y 112 for every x, y in X, then X is uniformly convex.] Prove that uniformly convex spaces are strictly

convex. Also prove that a normed linear space X is uniformly convex if and only if for every e, 0 < e < 2, there exists a (lee) > 0 such that II x II < I, II Y II < I, and II x - y II > e imply II Hx + y) II < I - (lee).

6.1.23. Best approximations in normed linear spaces. Let X be a uniformly convex normed linear space and A be a nonempty convex subset of X such that A is complete in the norm of X. Show that for any x in X, there exists a unique y in A such that

II x - y II = inf {II x - Z II: Z E A}.

[Hint: Let Yn E A be such that II x - Yn 11-+ a(x) == inf {II x - Z II: Z E A}. Let a(x) > O. Write Zn for x - Yn' Then, limn-+ooll Zn II = limn,m-+ooll i(zn + zm) II = a(x). Let zn' = zn/II Zn II. Use uniform convexity to show that limn,m-+ooll zn' - zm' II = 0.]

6.1.24. Prove that the Lp , I 2, use the inequality

where a and b are any two complex numbers, to prove that for J, g in Lp ,

For I < p < 2, the proof of uniform convexity is less trivial. The reader is referred to the paper of J. A. Clarkson [Trans. Am. Math. Soc. 40 (1936)].}

6.1.25. Prove that a normed linear space is not separable if and only if it contains an uncountable set of pairwise disjoint balls of radius 1. [Hint for the "only if" part: For each positive integer n, let f?;. be a maximal (with respect to inclusion) family of pairwise disjoint balls of radius I /n; f?;. exists by Zorn's Lemma. If f?;. is countable for each n, then the space is separable.]

6.1.26. Let X and Y be two normed linear spaces and T: X -+ Y be a linear operator such that Ker T is closed in X. For [x] in X/Ker T, define To([x]) = T(x). Prove the following assertions:

(i) To is an isomorphism of X/Ker Tonto T(X) C Y. (ii) To is continuous if and only if T is continuous, and in this case,

II To II = 11TII· (iii) T is open if and only if To is open. 6.1.27. Prove that every separable Banach space X is a quotient space

of 11' [Hint: Define p: 11 -+ X by ({l1' (l2' ... ) -+ L~l{lnxn' where Xn is a countable dense subset of the unit ball of X. Then p is onto, continuous, and linear.] (See Proposition 6.13 in Section 6.3.)

6.1.28. Let X be a vector space, and let A and B be any two Hamel bases for X. Prove that

(i) For each x E A, there exists y E B such that the set (A - {x}) U

{y} is linearly independent; (ii) Card A = Card B.

[Hint for (ii): It is enough to show that there are 1 - 1 functions from A intoB and from B into A. Consider the nonempty family!?"' = {f: f is 1 - 1, Df C A, Rf C B, and Rf U (A - Df ) is linearly independent}, where Df = the domain of J, and Rf = the range of f Partially order!T by <, where f < g if Df C Dg , and g(x) = f(x) for x E Df . By Zorn's lemma, !T has a maximal function m. If Dm = A, we are done. Suppose that Dm -# A. Then Rm -# B since Rm U (A - Dm) is linearly independent. Let zE A - Dm and wE B - Rm. The set Rm U {w} U (A - Dm) cannot be linearly independent since otherwise m can be extended by defining m(z) = w. Thus, there exist scalars aI, a2 , ••• , an and vectors Xl, X2' ... , xn in Rm U (A - Dm) such that for some k, ak -# 0 and Xk E A - Dm and w = alxl + ... + anxn. For a similar reason, the set Rm U {w} U

(A - (Dm U {Xk}») cannot be linearly independent. This contradicts the linear independence of Rm U (A - Dm).]

6.2. Bounded Linear Functionals and the Hahn-Banach Theorem

The aim of this section is to prove the Hahn-Banach Theorem. This theorem is one of the most fundamental theorems in functional analysis. It yields the existence of nontrivial continuous linear functionals on a normed linear space, a basic fact necessary for the development of a large portion of functional analysis. Moreover, it is an indispensible tool in the proofs of many important theorems in analysis. (For example, see Proposition 6.6.)

This theorem was first proved by Hahn in 1927 for a normed linear space over the reals, and then by Banach in 1929 for a real linear space (in the absence of any topology). The complex version of this theorem is due to Bohnenblust and Sobczyk in 1938 and, independently, to Soukhomlinoff, also in 1938.

We start this section by introducing the concept of a bounded linear operator.

Definition 6.S. Let X and Y be vector spaces over the same scalar field. Then a mapping T from X into Y is called a linear operator if for all

Sec. 6.2 • Bounded Linear Functionals; Hahn-Banach Theorem 15

Xl' X 2 E X and scalars a, (J,

I

Examples 6.5. Let X be the real-valued continuous functions defined on [0, 1],

under the uniform norm and Y be the reals. Let

T(f) = f: f(x) dx, fE X.

Then T is a continuous linear operator from X into Y. 6.6. Let X be the class of real-valued continuously differentiable

functions on [0, 1] and Y be the class of real-valued continuous functions on [0, 1], both under the uniform norm. Let

T(f) = df dx'

fE X.

Then T is a linear operator. But T is not continuous, since the sequence xn/n -+ 0 in X, but the sequence T(xn/n) = x n- l does not converge to 0 in Y.

6.7. Let X be a n-dimensional normed linear space over F and let Y be any normed linear space over F. Let

where a/s are scalars, {Xl' x 2 , ... , xn } is a basis of X, and YI, Y2, ... , Yn

are arbitrarily chosen, but fixed elements of Y. Then T is a linear bperator. t

But T is also continuous, since for X = 2::f-Iaixi,

n II T(x) II < L II Yi II • sup I ai I

i-I ISiSn

n < L II Yi II • K· II X II ,

i-I

where K is a constant such that

t It is also clear that every linear operator from X into Y must be of this form.

for all n-tuples {aI' a2, ... , an}. Such a K can always be found since in a finite-dimensional space any norm is equivalent to the sup-norm. (See Theorem 6.1.) Note that this example shows that a linear operator from a finitedimensional normed linear space into any normed linear space is continuous.

6.S. Let T be a mapping from 11 into F defined by 00

T(x) = LXi' i-I

Then T is linear, but not continuous if we define a new norm in 11 by considering it as a subspace of 100 , T is not continuous since T(zn) = I for all n, where

Zn = (_1_, _1_, ... , _1_,0,0, ... ) , n n n

, , v

n terms

and II Zn 1100 ~ 0 as n ~ 00. Note that II X 1100 = sup I Xi I· Is:i<oo

Definition 6.6. A linear operator T from a normed linear space X into a normed linear space Y is called bounded if there is a positive constant M such that

II T(x) II < M II X II, for all X EX. I

Proposition 6.3. Let T be a linear operator from a normed linear space X into a normed linear space Y. Then the following are equivalent:

(a) T is continuous at a point. (b) T is uniformly continuous on X. (c) T is bounded. I

Proof. (a)=> (b). Suppose Tis continuous at a point Xo. Then given e > 0, there exists b > 0 such that II x - XO II II Tx - Txo II < e. Now let y and Z be elements in X with II y - Z II < b. Then II (y - Z + xo) - XO II (c). Suppose T is uniformly continuous on X and not bounded. Hence for each positive integer n there exists Xn E X such that II T(xn) II > n . II Xn II. This means that II T(xn/n II Xn II) II > 1. But this contradicts the continuity of T at the origin since II xn/n II Xn II II ~ ° as n ~ 00.

(c) => (a) Boundedness of T trivially implies the continuity of T at the origin. I

The bounded linear operators from a normed linear space X into a normed linear space Y, denoted by L(X, Y). form a vector space where

addition of vectors and scalar multiplication of vectors are defined by

(aT)(x) = a . T(x).

Let us define on this vector space

II Til = sup II T(x) /I x*o II x II

Equivalently,

II Til = sup II T(x) II = sup II T(x) II = sup II T(x) II. IIXII:51 IIxlI=l II xII <1

This defines a norm on L(X, Y) and L(X, Y) becomes a normed linear space. The completeness of L(X, Y) in this norm depends upon that of Y. More precisely, we have the following proposition.

Proposition 6.4. L(X, Y) is a Banach space if and only if Y is complete. (We assume, of course, that X*-{ O}.) I

Proof. For the "if" part, let (Tn) be a Cauchy sequence in L(X, Y).

Then for each x E X, II Tn(x) - T m(x) II < II Tn - T m II II x II, so that limn-+ooTn(x) exists in Y, Ybeing complete. Let us define T(x) = limn--~=Tn(x).

Then Tis a linear operator from Xinto Y. We wish to show that Tis bounded and II Tn - T II converges to 0 as n -+ 00. To do this, let e > O. There exists N such that for n, m > N, we have II Tn - T m II < e or II Tn II < II TN II + e. Therefore II T(x) II = limHoo II Tn(x) II < (II TN II + e) • II X II. Hence T is bounded. Now, for x E X and n > N,

II Tix) - T(x) II = lim II Tn(x) - Tm(x) II m-+oo

< lim II Tn - Tm II· II x 11< e II x II· m-+oo

Therefore,

II Tn - Til = sup II Tn(x) - T(x) II < e, IIxll,;l

if n > N. The "if" part of the proof follows. The proof of the "only if" part is an application of the Hahn-Banach theorem and is left to the reader as a problem with hint. See Problem 6.2.4. I

Before we present the Hahn-Banach theorem we state a useful proposition providing a criterion for the existence of a bounded inverse of a bounded linear operator.

Proposition 6.5. Suppose T is a linear operator from a normed linear space X into a normed linear space Y. Then the inverse mapping T-I exists and is a bounded linear operator from T(X) into X if and only if there is some k> 0 such that k II x II < II T(x) II, for all x EX. I

The proof of this proposition is routine and left to the reader.

Definition 6.7. When Y is the scalar field F, which is a Banach space over itself under the absolute-value norm, the elements of L(X, Y) are called the bounded linear functionals on X. The class L(X, F) is denoted by X*. (Hence X* is also a Banach space.) X* is called the dual of X. I

Example 6.9. Suppose X is a n-dimensional normed linear space under the "sup"-norm over the real numbers R. Let T be a bounded linear functional on X. Then II Til = sUPllzlI_II T(x) I. To compute II T II, let {Xl' x2 , ••• ,xn} be a basis of X and T(Xi) = ri' I 0, and bi = - I if ri < 0, then, for X = l:i=Ib.;Xi,

II X II = sup I bi I = 1 Isisn

and

From this equality and the above inequality, it follows that n

II Til = L Ird· i-I

We now state and prove the main theorem of this section, which will show, among other things, the nontriviality of X*, the dual (also called adjoint) of a normed linear space X.

Theorem 6.4. The Hahn-Banach Theorem (Real Version). Suppose X is a linear space over the reals R. Let S be a subspace of X and p be a real-valued function on X with the following properties:

(i) p(x + y) < p(x) + p(y);

(ii) p(ax) = apex), if a> O.

If/is a linear functional on S (i.e., a linear mapping from S into R) such that /(s) < pes) for all s E S, then there exists a linear functional cP on X such that CP(x) < p(x) for all x E X and CP(s) = /(s) for all s E S. I

Proof. Let !?" be the set of all linear functionals g defined on a subspace of X containing S, such that g(s) = /(s) for all s E Sand g(x) < p(x), whenever g is defined. Clearly / E!?". We partially order!?" by requiring g < h if and only if h is a linear extension of g. By the Hausdorff Maximal Principle, there is a maximal chain.ro. We define a functional cP by setting

and

domain cP = U domain g gEYo

CP(x) = g(x), if x E domain g.

It is easy to check· that domain cP is a subspace of X and cP is a linear extension of f. Moreover, cP is a maximal extension, since if G is a proper extension of CP, then .ro u {G} will be a chain in !l'; contradicting the maximality of .ro.

We are finished if we can show that domain cP = X. This will be shown by showing that any g E fr, with its domain a proper subspace of X, has a proper extension.

Let y EX - domain g. We wish to extend g to the subspace spanned by y and domain g. Thus we need to define a functional h by

h(ay + x) = ah(y) + g(x),

x E domain g and a, a scalar where h(y) is chosen so that h(ay + x) < p(ay + x). In particular, we need to choose h(y) so that

hey) < p(y + x) - g(x)

and

-h(y) < p( -y -x) + g(x)

or

-pc -y -z) - g(z) < h(y) < p(y + x) - g(x),

where x and z are eiements from domain g. Since, for x, z E domain g,

p(y + x) - g(x) + p( - y - z) + g(z) > p(x - z) - g(x - z) > 0,

sup [- p(- y - z) - g(z)] < inf[p(y + x) - g(x)]. Z II:

Hence we choose h(y) = inf{p(y + x) - g(x): x E domain g}. Now the theorem will be proved if we only check that h( ay + x) < p( ay + X).

To do this, let a > O. Then

h(ay + x) = ah(y) + g(x)

< ap(y + x/a) - ag(x/a) + g(x)

=p(ay + x).

The cases a = 0 and a < 0 can similarly be taken care of. I

Theorem 6.5. The Hahn-Banach Theorem (Complex Version). Let X be a linear space over the complex numbers, S a linear subspace, and p, a real-valued function on X such that p(x + y) < p(x) + p(y) and p(ax) = 1 a 1 p(x). Let f be a linear functional on S such that 1 f(s) 1 <p(s) for all s E S. Then there is a linear functional tP defined on X such that tP(s) = f(s) for all s E Sand 1 tP(x) 1 <p(x) for all x E X. I

Proof. Let us define the mappings g and h on S, by taking g(s) to be the real part of f(s) and h(s) its imaginary part. Then g and h are linear in the real sense, by considering X (and hence S) as a vector space over the reals.

Now f = g + ih. Since, for s E S,

g(is) + ih(is) = f(is) = if(s) = ig(s) - h(s),

we have h(s) = -g(is). Since g(s) < If(s) 1 <p(s), by Theorem 6.4 we can extend g to a real-valued linear functional G on X considered as a real vector space such that G(x) < p(x) for all x E X. Let C/>(x) = G(x) - iG(ix). Then for s E S, tP(s) = g(s) + ih(s) = f(s). It is now easy to check that tP is linear in the complex sense. Finally for x E X, if a is a complex number with 1 a 1 = I and atP(x) = 1 C/>(x) I, then 1 tP(x) 1 = tP(ax) = G(ax) < p(ax) = 1 a 1 p(x) = p(x). I

In the rest of the section, we will consider some of the important consequences and a few applications of the Hahn-Banach theorem.

Theorem 6.6. The Hahn-Banach Theorem (Normed Linear Space Version). Let X be a normed linear space over a field F and let Y be a subspace of X. If y* E Y*, then there exists some x* E X* such that "y* " = " x* " and y*(x) = x*(x) for all x E Y. I

Proof. Let us define p(x) = II y* 1111 x II, x E X. Then p(x + y) < p(x) + p(y) and p(ax) = I a I p(x). Also I y*(x) I <p(x), since II y* II = suPx,,",o I y*(x) 1/11 x II, for x E Y. Hence by Theorem 6.5, there exists a linear functional x* on X such that (i) x*(x) = y*(x) for all x E Y, and (ii) I x*(x) I <p(x)t for all x E X. This means that I x*(x) I < II y* 1111 x II, x E X or II x* II < II y* II. But then (i) implies that II x* II = II y* II· I

Corollary 6.3. Let Y be a subspace of a normed linear space X over a field F. Suppose x E X and infyEy l1 x - y II = d> O. Then there exists x* E X* such that II x* II = I, x*(x) = d, and x*(y) = 0 for all y E Y. I

Proof. Let Z be the subspace spanned by Y and x. We define a linear functional z* on Z by

z*(ax + y) = ad, yE Yand aE F.

Thenz*(y) = Oforally E Yandz*(x) = d. To show that z* E Z*,foraoF 0, a E Fand y E Y,

II ax + y II = I a I II x + y la II > I a I d

or I z*(ax + y) I < II ax + y II for all y in Y. Hence II z* II < 1. But there exists a sequence Yk E Y such that II x - Yk II - d as k - 00. Therefore given e> 0, I z*(ax - aYk) I = I a I d> II ax - aYk II - e for sufficiently large k. This means that II z* II > 1. Hence II z* II = 1. The corollary now follows by extending z* by Theorem 6.6. I

The next corollary shows that there are sufficiently many bounded linear functionals on a normed linear space to separate points of the space.

Corollary 6.4. Given x E X (oF {O}), a normed linear space over F, thele exists x* E X* such that II x* II = I and x*(x) = II x II. In particular, if x OF y, there exists x* E X* such that x*(x) - x*(y) = II x - y II oF o. I

Proof. The proof follows immediately from Corollary 6.3, by taking y= {O}. I

t Note that p(x) = p( -x). Thus, for real scalars, x*(x) :::; p(x) implies that I x*(x) I :::; p(x).

Corollary 6.5. For any x in a normed linear space X over F,

1/ x II = sup i x*(x) i· I x·eX·

""'·"-1 We leave the proof to the reader. Corollary 6.4 above is an important consequence of the Hahn-Banach

Theorem. There are many applications of this outstanding theorem. One interesting application is the existence of a finitely additive measure, defined on the class of all bounded subsets of R, which is translation invariant and an extension of the Lebesgue measure. We will present this application in this section.

To do this, we first need an extension of the Hahn-Banach Theorem.

Theorem 6.7. An Extension of the Hahn-Banach Theorem. Let p be a real-valued function of the linear space X over the reals such that p(x + y) < p(x) + p(y) and p(ax) = apex), if a >0. Supposefis a linear functional on a subspace S such that f(s) < pes) for all s E S. Suppose also that .r is an Abelian semigroup of linear operators on X (that is TI , T2 E ~ implies TIT2 = T2TI E ~) such that if T EfT, then p(T(x» < p(x) for all x E X andf(T(s» = f(s) for all s E S. Then there is an extension c/> off to a linear functional on X such that C/>(x) < p(x) and C/> (T(x) ) = C/>(x) for all x E X. I

Proof. The proof will be an application of the Hahn-Banach Theorem. First we need to choose a new subadditive function (like p). Let us define

q(x) = inf(1/n)p(T1(x) + ... + Tn(x», XE X,

where the infimum is taken over all possible finite subsets {Tl' T2 , ••• ,

Tn} E fT. To show that this infimum is a real number, we have q(x) < p(x). Also since

and

we have

and therefore

- p( - x) < q(x) < p(x).

Moreover, q(ax;) = aq(x), a >0. Now we wish to show that q(x + y) < q(x) + q(y). Soletx,y E XandE > O. Then there exist {T1 • T2 • ...• Tn} and {SI' S2 • .. , , Sm} E!!T such that

1 E n p(T1(x) + ... + Tn(x») < q(x) + 2

and

Then

q(x + y} < n~ p( tl %1 TiS;(x + y»)

< n~ pC~ SjC~ Ti(X»)) + n~ p( tl TiCtl Sj(y»))

< + p( tl Ti(X») + ~ p( jtl Sj(y»)

< q(X) + q(y) + E.

Since E is arbitrary, q(x + y) < q(x) + q(y). Also, since for S E S

f(s) = _1 f(T1(s) + ... + Tn(s») < _1 P(Tl(S) + '" + Tn(s»). n n

f(s) < q(s). Therefore, by Theorem 6.4, there exists a linear extension tP of Ito all of X such that tP(x) < q(x) < p(x) for all x EX. The proof will be complete if we can show that tP(T(x») = tP(x), x E X and TEfl'. To do this, let x E X, T E f!'; and n be any positive integer. Then

q(x - T(x») < + p( tl Ti(x - T(X)))

1 = n p(T(x) - Tn+l(x»)

1 < n [p(x) + p( - x)].

Letting n approach 00, q(x - T(x») < O. Since (/>(x - T(x») < q(x -T(x»), we have tP(x) < (/>(T(x»). Applying this to -x, we get (/>(x) =

tP(T(x»). I

[Note that this theorem yields the Hahn-Banach Theorem (real version) when!?, consists of the identity operator alone. The complex version of this theorem can also be formulated like that of the Hahn-Banach Theorem and proved.]

We proved in Chapter 3 that it is impossible to define a translationinvariant countably additive measure on the class of all bounded subsets of R, such that the measure of an interval is its length. But interestingly enough, an application of Theorem 6.7 will show that there exists a finitely additive measure having the above properties.

Proposition 6.6. There is a finitely additive measure /-l defined for all bounded subsets of R such that

(i) /-l(A + t) = /-leA), A C Rand t E R; (ii) if A C R is Lebesgue measurable, then /-leA) is the Lebesgue mea-

sure of A. I

Proof. Let X be the linear space over R of all real bounded functions defined on [0, I), under the natural operations of pointwise addition and scalar multiplication. Let Y be the subspace of all bounded Lebesguemeasurable functions of [0, I). For J E X, let us define

p(f) = l.u.b. {J(x): x E [0, I)}.

Then

Juh EX,

and

(ii) for a >0, pea!) = ap(f) , for all JE X.

Let rp be the linear functional defined on Y by rp(f) = H J(x) dx. Let f!' be the Abelian semi group of linear operators on X defined by

:tr = {Tt: ° < t I}.

Now for ° < t < 1, an easy computation shows that for any A C [0, 1),

x + tEA ¢> x E A + (1 - t).

Since the Lebesgue measure of a Lebesgue-measurable set A is the same

Sec. 6.2 • Bounded Linear Functionals; Hahn-Banach Theorem 2S

as that of A-+-(1 - t), we have

f: XA(X) dx = f: XA(X -+- t) dx.

Hence for fe Yand ° < t < 1, t1>(f) = t1>(Tt[f]). Also by the definition of p and Tt , we have

p(Tt[f]) < p(f) , for all f e X, ° < t < 1.

Hence by Theorem 6.7, there exists a linear extension 1p of t1> defined on X such that 1p(f) < p(f) and 1p(Tt[f)) = 1p(f) for all f e X.

For A C [0, I), let us define f1(A) = 1p(XA). Then since 1p is linear, f1 is finitely additive. Since 1p( - XA) < p( - XA) < 0, - 1p(XA) < ° or f1(A) > 0. Also since for ° <t < 1, XA(X -+- t) = XA+U-t)(X), we have

f1(A) = 1p(XA) = 1p(XA(X -+- t») = f1(A -+- (l - t»).

If t > t, then 1 - t < t. Hence, for A c [0, t),

f1(A) = f1(A + s), for all s e [0, t).

Now to extend f1 to the class of all bounded subsets on R, consider any Be [nI2, (n + 1)/2), where n is any integer. Then B - (nI2) C [0, 1). It is the value f1(B - n12) we will take as f1(B). Thus for any bounded set A C [-m, m],

2m-l ( [n n + 1 )) f1(A) = n=~mf1 AnT' -2- .

This f1 will satisfy the requirements of the theorem. I

Before we close this section, we briefly discuss another useful concept -that of separability. A set A in a normed linear space X is called separable if there is a countable subset of A that is dense in A. qo, 1] is separable under the supremum norm. lp for 1 0, 0, ... )} of all sequences where the a/s are rational is countable and dense in Ip. Also the Lp space of Lebesgue-measurable functions (1 <p < 00) is separable since the polynomials with rational coefficients (being dense in qo, I]) are dense in Lp. But 100 , as well as Loo , is not separable. (See Problem 6.2.7.) Also the Lp (I < P < 00) space over an arbitrary measure space is not, in general, separable. For example, if S is an uncountable set with the counting measure, then for I <p < 00

Lp = {f: S -+ R: L I f(s) IP.< oo}; .8ES

and for IE L p ,

[ ] lip

II I lip = L I I(s) IP . seS

If Sl and S2 are in S, then

This means that Lp(S) is not separable. It so happens that h* = 100 (equality means the existence of a linear

isometry onto). This will be discussed in the next section. Therefore 11* is not separable, despite the separability of II. But the converse situation is different, and once again the Hahn-Banach Theorem helps us clarify the converse.

Proposition 6.7. If the dual X* of a normed linear space X is separable, then X is also separable. I

Proof. Let (Xk*)k-l be a countable dense set in X*. Let (Xk)k=l be elements of X such that for all k, II Xk II = 1 and I Xk*(Xk) I > ! II Xk* II. Let Y be the closed subspace spanned by the Xk'S. Then Y is separable. We are finished if Y = X. If Y:f=. X, by Corollary 6.3 there exists x* E X* such that II X* II > 0 and x*(y) = 0 for all y E Y. Since (Xk*)k=l is dense in X*, there is a subsequence (xl,) such that II xl, - X* II - 0 as i - 00. But

II Xli - x* II > I Xl,(Xk,) - X*(Xk,) I = I Xl,(Xk,) I > ! II xl, II·

Hence II xt, II - 0 as i - 00. This means that II x* II = 0, which is a contradiction. I

Problems

X 6.2.1. Let X be a finite-dimensional space with basis {Xl' X2 , ••• ,

xn }, such that II l:i'=IU.jXi II = SUPlsisn I Ui I, ui E F. Define the linear operator A from X into X by

Then find II A II.

n A(xj) = L UijXj,

j=1 I <i< n.

X 6.2.2. Let X be as in Problem 6.2.1, with II L:f-1aixi II = (L:i-11 ai 12)112.

Then if IE X*, find II I II. X 6.2.3. Let X be a normed linear space over R and I: X ->- R.

(a) Suppose that I is continuous and for all x, y in X,/(x) + I(y) =/(x + y). Prove that/E X*.

(b) Suppose that I is linear, but discontinuous. Show that 1-1(0) is dense in X. (Hint: Note that Y = 1-1(0) is not closed since I is discontinuous. Let y be a limit point of Y but not in Y. Then for any x in X, x =

{x - [/(x)ll(y)]y} + [/(x)ll(y)]y.) (c) Let I be linear. Show that I is continuous if and only if 1-1(0) is

closed. (d) Let I be linear and 1-1(0) *- X. Then /-1(0) is a maxi.mallinear

subspace of X, that is, a linear subspace W such that for any x in X - W,

{x} U W spans all of X. (For this result to hold, X needs to be only a linear space.)

[REMARK: Note that if T: X ->- Y is a linear operator between two normed linear spaces X and Y and if T(X) is finite dimensional, then T is continuous if and only if Ker T is closed.]

6.2.4. Show that if L(X, Y) is complete, where X and Yare any two normed linear spaces over F and X*- {O}, then Y is complete. {Hint: Let (Yn) be a Cauchy sequence in Y. Then define Tn E L(X, Y) by Tn(x) =

x*(x)Yn, where x*(*- 0) E X*. Show that (Tn) is Cauchy in L(X, Y) and hence limn~=yn = [llx*(x)] limn~ooTn(x), where x*(x) *- O.} X 6.2.5. Let I <p< = and lip + llq = I. For each gE Lq , define Tg E Lp * by Tg(f) = fig dft. Show that for p > 1, II Tg II = II g Ilq, and that for p = 1 this equality holds for all g E Loo if and only if the measure is semifinite. [Hint: If I < p < = and I = I g Iq,P sgn g,t then Til) =

II g Ilq II I lip·] X 6.2.6. For any fixed l(t) E qo, 1] (under the uniform norm), let cP be the linear functional on C[O, I] defined by

CP(g(t») = J:/(t)g(t) dt.

Then show that cP is bounded and find II cP II. X 6.2.7. Prove that 100 as well as Loo[O, I] is not separable. [Hint: Let Xk E 100 and Xk = (Xkl, Xk2, ... ). Define x = (aI' a2, ... ) by ak = 0, if I xli > I; = I + I XkX: I, if I Xkk I < I. Then x E 100 and II x - Xx: 1100 > I for all k.]

t By definition, sgn IX = 0 if IX = 0 and ~n IX = IX/I IX I if IX * O.

X 6.2.8. If S is a linear subspace of a Banach space X, the annihilator So of S is defined to be So = {x* E X*: x*(s) = ° for all s E S}. t If Tis a subspace of X*, then °T = {x EX: x*(x) = ° for all X* E T}. Show that

(i) So is a closed subspace of X*; (ii) O(SO) = S;

(iii) if S is closed, then S* is linearly isometric to X* I So and So is linearly isometric to (XIS)*. X 6.2.9. For I < p < 00, show that Ip * is linearly isometric to Iq, lip + Ilq = I. [Hint: Let

ek = (0, 0, ... , 0, I, 0, 0, ... ).

k

If x* E Ip*, let x*(ek) = bk. Show that b = (bI , b2, ... ) E Iq and II x* II = II b Il q ·]

X 6.2.10. Show that 11 * is linearly isometric to lco. X 6.2.11. Let c be the space of all sequences of complex numbers (Xl' X2' ... ) such that limn-+ooxn exists with natural addition and scalar multiplication. For an element X = (Xl' X2, ... ) E c, let II X II = supn I Xn I. Then show that

(i) c is a Banach space; (ii) c* is linearly isometric to h.

[Hint: Let ek = (0,0, .,. ,0, 1,0, ... ) as in Problem 6.2.9 for I <k < 00, eo = (I, I, I, ... ), x*(ek) = bb I <k < 00, and x*(eo) = m, where x* E c*. Show that (dI , d2, ... ) E 11, where dl = m - l:.';;Ibb di+1 = bi , I < i < 00 and II x* II = l:.::1 I di I.]

(iii) A subset A C C is relatively compact if and only if A is bounded and sUPi,j2':n I Xi - Xj I -+ ° as n -+ 00, uniformly for all X == (Xi) E A. X 6.2.12. Let Co be the subspace of C (see Problem 6.2.11 above) such that limn-+coxn = 0, if (Xl' X2, ... ) E Co. Show that Co * is linearly isometric to 11'

Further show that a subset A C Co is relatively compact if and only if A is bounded and sUPi2':n I Xi I -+ ° as n -+ 00, uniformly for all X == (Xi) EA.

6.2.13. Consider the linear space L p , ° < p < I, in the Lebesgue measure space on [0, I], with the pseudo metric

d(f, g) = f: I f(t) - get) IP dt.

t Although AO is also used to denote the interior of a set A in this text, the meaning of the symbol should be clear from the context.

Then show the following: (i) (Lp , d), 0 < p < I, is a pseudometric linear space [that is, a

vector space with metric (pseudo) topology where the vector addition and scalar multiplication are continuous].

(ii) J can be written as g + h, d(g, 0) = d(h, 0) = M(f, 0). {Hint: Let gx = J. XEO,x] , hx = J. X[x,I]' Then d(gx, 0) + d(hx, 0) = d(f, 0). But d(gx, 0) is a continuous function from [0, I] onto [0, d(f, 0)]. Use the Intermediate Value theorem.}

(iii) The only continuous linear functional on Lp is the zero functional. [Hint: Let t/J E Lp * with t/J(f) = I. Then, by part (ii), there exists g E Lp with t/J(g) > to d(g, 0) = td(f, 0). Let gI = 2g. Then t/J(gl) > I, d(gl' 0) = 2P - Id(f, 0). Continue the process to get gl, g2, ... with t/J(gn) > I and d(gn' 0) = 2n(P-1ld(f, 0).]

6.2.14. The Volterra Fixed Point Theorem. Let X be a Banach space and T be a bounded linear operator on X such that :L:'I II Tn II < 00. Prove that the transformation S defined by

SeX) = y + T(x),

has a unique fixed point given by

y a fixed element in X,

= Xo = Y + L Tn(y).

n~1

[Hint: For any Z E X, consider Xl = Z and Xn+I = T(xn); show that (xn) is Cauchy in X and limH~n = 0.]

6.2.15. An Application oj Problem 6.2.14. Show that the Volterra equation

J(t) = get) + f: K(s, t)J(s) ds,

where g E qo, 1] and K E C([O, 1] x [0, 1]) are given functions, has a unique solution JE qo, 1]. [Hint: Define T on the Banach space qo, 1] with "sup" norm by

T(f)(t) = f: K(s, t)J(s) ds.

Then

II T" < M = sup I K(s, t) I and "T"" < Mn/n!. O$8 t 1Sl

Now use Problem 6.2.14.] 6.2.16. Existence oj Banach Limits. (An Application oj the Hahn

Banach Theorem). Banach limits are linear functionals f/J on the space 1=

bounded sequences of real numbers (xn):'o satisfying these conditions:

(i) <P[(xn)] > 0 if Xn > 0 for 0 <n < 00;

(ii) (/)[(xn+1)] = (/)[(xn )];

(iii) (/)[(1)] = 1, (I) = (I, 1, 1, ... ).

Prove the following assertions:

(1) ( 1 n-1 ) p[(xn)] = lim sup - L xi+j

n-+oo j n i~O

exists and is a sublinear functional on 100 . [Hint: If

then akm < am and (r + km )ar+km < rar + kmam or limk-+oo sup ar+km < am for r = 1,2, ... , m. Hence lim sup an < am for each m or lim an exists.]

(2) For (xn) E c( C 100), define

f[(xn)] = lim Xn < p[(xl1)]·

n-+oo

(3) By Theorem 6.4, there is an extension <P of f to 100 and (/) is a Banach limit. {Hint: I p[(xn+1) - (xn)] I = I limn[supjn-1(xj+n - Xj)] I < limn2n-1 supjl Xj 1= 0; also I P[(Xn) - (Xn+1)] 1= o.}

(4) The maximal value of Banach limits on (xn) is p(xn). (Hint: First, for any Banach limit L, L[(xn)] < p[(xn)]. To see this, by (i) L[(supj Zj - zn)] > 0, and therefore SUPjZj > L[(zn)]; then if Zn = (1Im)"f.:::,;:;/ Xi+n' L[(xn)] = L[(zn)] < SUPjZj for each m and therefore letting m -+00, < p[(xn)]. Conversely, given (x,,), there exists a Banach limit L such that L[(xn») = p[(xn)]. Indeed, if (xn) $ c, define L on c as f [in (2)], and as in the proof of Theorem 6.4: L[(xn)] = inf(Yn)6c{p[(xn + Yn)] - lim Yn}; then as in Theorem 6.4, L can be extended to 100 and L[(xn)J = p[(xn)], since p[(xn + Yn)] = p[(xn)] + lim Yn.)

(5) The minimal value of Banach limits on (xn) is

( 1 n-1 ) lim inf - L xi+j . n-+oo j n i~O

(6) A necessary and sufficient condition in order that all Banach limits on (xn) agree and be equal is that limn-+oo(l In )Lf':-~ Xi+j = s uniformly in j. Hence on convergent sequences, Banach limits agree with limits of the sequences.

The formulation of the above results in the form they appear is due to L. Sucheston [40].

6.2.17. Let (X, .s¥', fl) be a finite measure space and let S be the linear space of all real-valued measurable functions (where two functions agreeing a.e. are equal) with the topology of convergence in measure {equivalently, with the metric d(/, g) = f [I f - g 1/(1 + If - g I)] dfl}. Prove the following assertions:

(i) If A is an atom in.s¥', then t/>.ttCf) = fA f dfl defines a continuous linear functional on S.

(ii) If t/> is a continuous linear functional on S, then t/>(fXA) = 0 for all f E S if A contains no atoms.

(iii) Let fT be an (at most countable) collection of pairwise disjoint atoms of .Ji/ and T = {t/> A: A E fT}. Then the class S* of all continuous linear functionals on S is precisely the linear span of T. [This result is due to T. K. Mukherjee and W. H. Summers.]

(iv) S* contains a nonzero element if and only if .Ji/ contains an atom.

6.2.18. Another Application of the Hahn-Banach Theorem. The classical moment problem can be stated as follows: Given a sequence of real numbers (an), when does there exist a real-valued function g of bounded variation on [0, 1] such that H xn dg(x) = an, n = 0, 1, 2, ... ? Show that this problem can be answered in an abstract setup as follows:

Let X be a normed linear space, (XA)A€A be elements in X, and (aA)A€A be scalars. Then the following are equivalent:

(i) There exists x* E X* such that X*(XA) = aA for each A. EA.

(ii) There exists a positive number M such that

I L aAbA I < M . "L bAxA II,

for every subset (bA)'\EA of scalars with all but finitely many of the b/s zero.

[Hint: (ii) => (i). Let Y be the subspace spanned by (X,\),\€A' Define x* E y* by X*(L bAxA) = L bAaA; show that x* is well defined because of (ii). Then use the Hahn-Banach Theorem.]

6.2.19. Let X and Y (=1= {O}) be normed linear spaces, and dim X =

00. Show that there is at least one and therefore infinitely many unbounded linear operators T: X -+ Y. (Hint: Use the fact that X has a Hamel basis, i.e., a linearly independent subset that spans X, and a linear operator on X is uniquely determined by its values on a Hamel basis.)

6.2.20. Let X be a linear space over Rand Y C X. Show that the following are equivalent:

(a) Y is a hyperplane (that is, Y = Xo + Z, where Xo E X and Z is a maximalt linear subspace of X).

(b) There exists a linear functional f : X -+ R, f * 0, and a real n umber r such that Y = {x E X: f(x) = r}.

[Hint: For (a) => (b), define f(ax + z) = a, where x E X - Z and z E Z. Note that {x} U Z spans X. Then,J(xo) = r. For (b) => (a), let y be such thatf(y) = I; then, take Xo = ry and Z = f-1(0). Use 6.2.3(d).]

X 6.2.21. Let X be a linear space over F and let K C X be a convex absorbing set. ("Absorbing" means that for each x in X, there exists a > 0 such that x E aK.) For x E X, write

PK(X) = inf{a: a> 0 and x E aK}.

Show that for x, y in X

(i) PK(X) > 0; (ii) PK(O) = 0;

(iii) PK(X + y) < PK(X) + PK(y); (iv) PK(ax) = apK(x) for a > O.

The functional P K is called the Minkowski functional of K.

6.2.22. Let X be a linear space over R, K C X a convex absorbing set, and L a hyperplane such that K n L = 0. Show that there exists a linear functional x* on X such that

(i) L = {x: x*(x) = I}; (ii) K C {x: x*(x) < I};

(iii) for x EX, x*(x) < PK(X), where PK is as in Problem 6.2.21. [Hint: Use Problem 6.2.20 to establish (i). For (ii), let x, y E K such that x*(x) < 1 < x*(y). Then for s in [0, 1], sx + (I - s)y E K and sx*(x) + (I - s)x*(y), as a function of s, attains the value I for some So. Note that

soX + (I - so)Y E K n L.] 6.2.23. A Geometric Form of the Hahn-Banach Theorem. Let X be a

linear space over Rand K C X be a convex absorbing set. If Lo C X is such that Lo = Xo + Wand K n Lo = 0, where W is a linear subspace, then there exists a hyperplane L and a linear functional x* on X such that

(i) L = {x: x*(x) = I}; (ii) Lo C L;

(iii) x*(x) < PK(X), x E X; (iv) K C {x: x*(x) < I}.

t That is, a proper subspace Z such that for any z E X - Z, the subspace spanned by Z u {z} is X.

Sec. 6.3 • Open Mapping, Closed Graph Theorems; Uniform Boundedness 33

(Hint: Note that Xo (/; W, since 0 E K. Let {xo} U W span the subspace V. Then Lo is a hyperplane in V and K n V n Lo = 0. Use Problem 6.2.22 and Theorem 6.4.)

6.2.24. Let X be. a normed linear space over Rand LeX be a hyperplane such that L = {x: x*(x) = I} for some linear functional x* on X. Let G be an open subset of X such that G C {x: x*(x) < I}. Show that (i) L is closed, (ii) G C {x: x*(x) < I} and x* is continuous, and (iii) G C {x: x*(x) < I}. (We remark that the results outlined in this and the next problem hold in the more general context of topological vector spaces. See Problem 6.7.29.) [Hint: If L is not closed, then L is dense and therefore {x: x*(x) = 2} is also dense and must intersect G. For (ii), use Problem 6.2.3(c).J

6.2.25. Let X be a normed linear space over Rand K C X be an open convex set containing O. Suppose also that K n Lo = 0, where Lo = Xo + Wand W is a linear subspace. Show that there exists a closed hyperplane L and a continuous linear functional x* on X such that

(i) L = {x: x*(x) = I}; (ii) Lo C L;

(iii) K C {x: x*(x) < I}. (Hint: Use Problems 6.2.23 and 6.2.24.)

6.3. The Open Mapping Theorem, the Closed Graph Theorem, and the Principle of Uniform Boundedness

In this section we will consider three basic principles of functional analysis, which rank in importance with the Hahn-Banach Theorem of the previous section. They provide the foundation for many far-reaching modem results in diverse disciplines of analysis such as ergodic theory, the theory of differential equations, integration theory, and so on. The first, called the Open Mapping Theorem, asserts that certain continuous linear mappings between Banach spaces map open sets into open sets. The second is called the Closed Graph Theorem and asserts that a linear map between two Banach spaces that has a closed graph is continuous. The third is the Principle of Uniform Boundedness, which asserts that a pointwise bounded family of continuous linear mappings from one Banach space to another is uniformly bounded. One form of the Open Mapping and the Closed Graph Theorems was first proved by Banach in 1929; a more general form is due to Schauder in 1930. The principle of uniform boundedness was proved for


bounded linear functionals on a Banach space by Hahn in 1922, for continuous linear mappings between Banach spaces by Hildebrandt in 1923, and also by Banach and Steinhaus (in a more general case) in 1927. The Principle of Uniform boundedness is sometimes called the Banach-Steinhaus Theorem.

This section is based on the concept of a closed linear operator. We start with its definition.

Let X and Y be normed linear spaces over the same scalars. Then X X Y is the normed linear space of all ordered pairs (x, y), x E X and y E Y, with the usual definitions of addition and scalar multiplication and norm defined by II (x, y) II = max{ II x II, II y II}.

Definition 6.S. Let T: D ---+ Y be a linear operator, where D is a subspace of X. Then T is called closed if its graph GT = {(x, T(x)): xED} is a closed subspace of X X Y. Equivalently, T is closed if and only if the following condition holds: Whenever xn E D, xn ---+ x E X, and T(xn ) ---+ y E Y, then xE D and y = T(x). I

Examples 6.10. A Linear Operator That Is Closed But Not Continuous. Let

X = Y = C[O, 1] and D = {x EX: d[x(t)]fdt E C[O, 1 n. Define T: D ---+ Y

by T(x) = dx/dt. Then if xn(t) = tn, II xn(t) II = I; but

II T(xn ) II = II n . tn- 1 II = n ---+ 00.

Hence T is not bounded. The reader can easily check that T is closed. Note that D is not complete.

6.11. A Linear Operator That Is Continuous But Not Closed. Let D be a nonclosed subspace of any normed linear space X. Let Y = X and i: D ---+ Y be the identity map. Then i is clearly continuous but not closed.

Clearly we have the following propositiqn.

Proposition 6.S. Let T: D ---+ Y, D C X, be a continuous linear operator. Then if D is a closed subspace of X, T is closed. Conversely, if Y is complete and T is closed, then D is a closed subspace of X. I

We leave the proof to the reader.

Example 6.12. A Closed Linear Operator from a Banach Space into a Normed Linear Space That Is Not Continuous. Let X be any infinitedimensional Banach space and let H be a Hamel basis for X. We may and do assume that the elements in H have norm 1. Let Y be the vector space X with a new norm II • 111 given by

Sec. 6.3 • Open Mapping, Closed Graph Theorems; Uniform Boundedness 3S

For each x E X, II X III > II x II. Let (Yn) be an infinite sequence in H. If

zn = tl ( ~2 ) . Yk>

then for n < m, m 1

II Zn - Zm 111 = L F ~ 0, k~n+l

as n, m ~ 00.

Hence (zn) is Cauchy in Yas well as in X. For any Z in Y, Z = 2:f-Iaihi for some positive integer p with hi E H, for i = 1,2, ... ,p. Hence for n > k, II Zn - Z III > Ijk2 if k is large enough for Yk *- hi' i =1, 2, ... ,po This means that (zn) does not converge to any Z in Y, so that Y is not complete. Now if we consider the identity map i: X ~ Y, then i is clearly closed, one-to-one, onto, and has a continuous inverse. However, i is not continuous, since in that case Y has to be complete.

Example 6.13. A Closed Linear Operator from a Normed Linear Space onto a Banach Space That Is Not an Open Map. Consider Example 6.12 and the identity map i: Y ~ X. Then i is a closed linear map that is not open.

Now we state and prove the Open Mapping Theorem.

Theorem 6.8. The Open Mapping Theorem. Let X be a Banach space and Y be a normed linear space of the second category. If T is a closed linear operator from a linear subspace D in X onto Y, then T is an open map. I

Before we prove this theorem, we prove the following lemma.

Lemma 6.1. Let T be a closed linear operator from a linear subspace D of a Banach space X into a normed linear space Y. Suppose that T(G) has a non empty interior whenever G is a relatively open subset of D containing O. Then T is open and onto. I

Proof. Let G be open in D and 0 E G. Then there is an open ball Vr = {x ED: II x II < r} such that Vr - Vr C G. Therefore

T(Vr) - T(Vr) C T(Vr) - T(Vr) C T(Vr - Vr) C T(G),

so 0 is an interior point of T(G), by the assumptions in the lemma. Notice that if x E G, then for some s > 0, x + V2• C G, so that T(x) + T(V2.) C

T(G). Thus, it is enough to prove that T(Vr) C T(V2r) for r > O. Let r > 0 and kn -4- 0 such that for each n > 1

Let Y E T(Vr)' There exists YI E T(Vr) such that II Y - YI II < kl . Then

Y - YI E T( Vr/2 ), and there exists Y2 E T( Vr12 ) such that II Y - YI - Y2 II < k2. It is clear that there exist (Yn)::l such that Yn E T(V(r/2n'-.1» and II Y - :Li~IYi II < kn· Let Yn = T(xn) and II Xn II < r/2n- 1 so that :L:III Xi II < 2r. By Theorem 6.3, there exists x in X such that :Li~IXi -4- x. Since T is closed, Y = Tx and II x II < 2r. I

Proof of Theorem 6.S. Let r > O. Then D = U::lnVr and Y =

U::lnT(Vr). (Vr is the open r-ball in D.) Since Y is of the second category, there is a positive integer n such that nT(Vr) is not nowhere dense.

Thus T( Vr) has a non empty interior, and T is open by Lemma 6.1. I

For bounded linear maps, the property of being open is equivalent to the continuity or bounded ness of the inverse map when the inverse exists. In general, we have the following result.

Proposition 6.9. Let T be a bounded linear map from a normed linear space X onto a normed linear space Y. Then T is open if and only if there exists k > 0 such that for each Y E Y, there exists x E X with T(x) = Y and II x II < kll Y II· I

Proof. Recall from Section 6.1 that the map 1>: X -4- X/K, where K = T-I(O), is continuous and open, so all open sets in the quotient space X/K are of the form 1>(U), U being an open subset of X. Then the map To: y -4- X/K defined by To(Y) = x + K, T(x) = y, is well-defined and linear. Since TO-I (1)(U) ) = T(U), To is continuous if and only if T is open. Recalling the norm in X/K, it follows that To is continuous if and only if there exists a > 0 such that

inf II x - z II < a II Y II, Y = T(x). ZEK

The rest of the proof is left to the reader. I

Proposition 6.9 leads to a partial converse of Theorem 6.8.

Proposition 6.10. If T is a bounded linear operator from a Banach space X onto a normed linear space Y, then the "openness" of T implies the completeness of Y. I

Proof. Let (Yn) be a Cauchy sequence in Y. Then we can find a sequence (nk) of positive integers such that nk < nk+l and for each k, II Yn1:+1 - Yn1: \I < 1/2k. If T is open, by Proposition 6.9 there exists N > 0 such that we can find Xk with T(Xk) = .Jlnlc+1 - Ynlc and II Xk II < N II Yn.l:+1 - Ynlc II. Then L:'1 II Xk II < 00. Since X is complete, there exists x E X such that x = limn~L~_IXk. Since T is continuous and linear, L~_IT(Xk) - T(x), which meansthatYn.l:+1-Ynl + T(x). Since (Yn) is CauchY,Yn - Ynl + T(x).1

Now as applications of the Open Mapping Theorem, we present the next three propositions and also the Closed Graph Theorem. Proposition 6.12 is also needed for Theorem 7.10.

Proposition 6.11. Let X be a vector space that is complete in each of the norms II • " and II • Ill. Suppose there exists k> 0 such that

II x II <k II X III

for each x E X. Then the norms are equivalent. I

Proof. If i is the identity map from (X, II . Ill) onto (X, II • II), then i is clearly closed and linear. By the Open Mapping Theorem, i is also open and so i-I is continuous. I

Proposition 6.12. Let X and Y be Banach spaces. Then the set of all surjective maps in L(X, Y) is open in L(X, Y). I

Proof. Let T be a surjective map in L(X, Y). Then T is open by Theorem 6.8. Let S E L(X, Y) and "T - S " < I 12k, where k is a real number for the map T with the property as stated in Proposition 6.9. We claim that S is surjective. To prove this, let Y E Y and II Y II < I. Then by Proposition 6.9, there exists x EX, T(x) = Y and II x II <k. Let Yl = T(x) - Sex). Then, II Yl " < l and there exists Xl EX, T(xl ) = Yl and II Xl II < k12. LetY2 = T(xl ) - S(Xl). Then, II Y2 II < 1/22. Continuing inductively, we find Xn such that T(xn} = Yn and

"Yn II < 1/211.,

Then, Y = Sex) + S(xl ) + ... + S(Xn) + Yn+1' Write z = x + L:.IXn. Then S(z) = Y and our claim is proven. I

Proposition 6.13. Let X be a separable Banach space. Then there exists a closed linear subspace L of the Banach space 11 such that X is topologically isomorphic to the quotient space III L. I

Proof. Let (xn) be a dense sequence in the unit ball of X. We define the mapping T: 11 - X by T(y) = L anxn , y = (an)~l' Clearly, II T(y) II < II y Ill' Let L = T-l{O}. Now the mapping T defined by T(y + L) = T(y) is a well-defined continuous linear one-to-one mapping from III L into X. If T is surjective, then T is surjective, in which case an application of the Open Mapping Theorem will finish the proof. So it suffices to show that T is surjective.

Suppose x E X and II x II < 1. By an induction argument, we can find a subsequence (x",) such that x = L':.oxnJ2i. [Choose xno such that II x - xno II < I; then, choose n1 > no such that II 2(x - xno) - x n1 II < I and so on.] Let y = (ak) be defined so that ak = 1/(2i ) if k = nit = 0 if k =1= ni for all i. Then YEll and T(y) = x. It follows that T is surjective. I

Theorem 6.9. The Closed Graph Theorem. A closed linear operator T mapping a Banach space X into a Banach space Y is continuous. I

Proof. Since Tis closed, GT = {(x, T(x)): x E X} is a closed subspace of the Banach space Xx Y. We define P1[(x, T(x))] = x and P2[(x, T(x))] = T(x). Then P1 and P2 are both continuous maps from GT onto X and into Y, respectively. By the Open Mapping Theorem, PI is open. Since P1 is oneto-one, p1-l is continuous and therefore, P2P1-I = T is continuous. I

Actually, we could state the Closed Graph Theorem in a more general form.

Theorem 6.10. A closed linear operator T mapping a normed linear space X of the second category into a Banach space Y is continuous. I

Proof. We will briefly outline the proof. Let M = {x E X I T(x) = O}. Then M is a closed subspace of X, since T is closed. If X = M, the theorem is trivial. Suppose X =1= M. Then the quotient space XI M is of the second category (see Problem 6.3.3). Define To: XIM - Y by To((/>(x)) = T(x), where (/> is the natural map from X onto XI M. Then To is well defined, oneto-one, closed, and linear. (See Problem 6.3.4.) Therefore, the mapping To-I: T(X)(C Y) - XIM is again closed. By the Open Mapping Theorem, To -1 is open and therefore To is continuous. This means that T is continuous. I

By means of quotient space arguments as used in the proof of Theorem 6.10, it is possible to give a proof of the Open Mapping Theorem for the case when D is closed, using Theorem 6.10 as a starting point. This means that the Open Mapping and the Closed Graph Theorems are two different forms of the same theorem.

Next, we present an incomplete normed linear space of the second category.

Example 6.14. Let X be the Lebesgue-integrable functions on [0, 1] with the Ll norm. Let (rn) be the rationals in (0, 1], and Xn be the characteristic function of [0, r n]. Then A, the set containing the xn's, is linearly independent and hence can be extended to a Hamel basis H of X. Since X is complete, H is uncountable. (See Problem 6.1.7.) Let (Yn) be a sequence of elements in H - A. Let Yn be the subspace spanned by H - {Yn' Yn+1' ... }. Then since X = U:'l Yn, for some n = p, Yp is of the second category. Also Yp is not complete, since Yp contains A and the span of A is dense in X (the step functions being dense in Ll).

Finally, we come to the Principle of Uniform Boundedness. We need first a lemma.

Lemma 6.2. Let S be a nonempty set and Y be a normed linear space. Let B(S, Y) = {f: S -+ Y I SUPSES II f(s) II < oo}. If Y is complete and II f II = sUPsEs11 f(s) II, then B(S, Y) is a Banach space with this norm. I

The proof of this lemma is left as Problem 6.3.5.

Theorem 6.11. The Principle of Uniform Boundedness. Let S be a family of bounded linear operators from a Banach space X into a Banach t space Y. Suppose that for x E X there is a constant M(x) such that sUPTEsl1 T(x) II < M(x) < 00. Then there exists M > ° such that

sup{11 Til: TE S} < M. I

Proof. We define the linear operator A from Xinto B(S, Y) by A(x)[T] = T(x). Then A is well defined since sUPTEsl1 T(x) II < 00. A is also closed. (Why?) By the Closed Graph Theorem, A is continuous. Therefore,

t Completeness of Y is not needed in this theorem. One can consider B(S, y), where Y is the completion of Y. See Proposition 6.15. Also it suffices to let X be a normed linear space of the second category.

sUPliziisll1 A(x) II < M, for some M> 0. But since II A (x) II = sUPPesll T(x) II we have sUPllzlI:s;tsuPPesll T(x) II < M or SUPPeS II Til < M. I

Remarks on Theorem 6.11 and Some Applications

6.1. Theorem 6.11 need not be true if X is not complete. Let X = {(ai) E 12 : ai = ° for all but finitely many i}. Then XC 12 • We define Tn: X ~ 12 linearly by

T. (e.) = {O, n t nen ,

i-::j:.n, i= n,

where ei = (0, 0, ... , 0, 1, 0, 0, ... ).

Then for x = Lt=laiei, Tn(x) = ° for n > k and therefore sUPlsn<ooll Tix) U

is finite. But II Tn II = nand sUPlsn<ooll Tn II = 00.

6.2. From Theorem 6.11, it follows that for S C X*, X a Banach space, sUPzooesl1 x* II < 00 whenever for each x EX, sUPzooesl x*(x) I is finite.

6.3. If seX a normed linear space such that SUpzeS I x*(x) I < 00

for each x* E X*, then sUPzesll x II < 00. This immediately follows by applying Theorem 6.11 to the family of operators [J(x)]zeS from X* into F (the scalar field), J being the natural map from X into X**, that is [J(x)](x*) = x*(x), x* E X*.

6.4. Analyticity of a Banach Space Valued Function of a Complex Variable. Let X be a complex Banach space, G an open set in the complex plane, and f a function from G into X. Then f is called analytic on G iff is differentiable at each point Ao E G, that is,

II f(~ =1Ao) - J'(Ao) II ~ 0,

as A ~ Ao for some function J': G ~ X. We will apply the principle of uniform boundedness (actually Remark 6.3 above) to show that a necessary and sufficient condition for f to be differentiable on G is that x*(J(A» be differentiable on G for each x* E X*.

Since the "necessary" part is trivial, we show only the "sufficient" part. Let x* (J(A) be differentiable on G and Ao E G. There exists r > ° such that

I A - Ao I < r ==-- A E G.

By Cauchy's integral formula (in elementary complex analysis), we have

x*(J(A) = _1_. f x*(J(z)) dz, 2m 0 z-A

(6.1)

where I A - Ao I < rand C is the positively oriented circle I z - Ao I = r. It follows from equation (6.1) after a simple calculation that for A#: 1', I A - Ao I <ir and II' - Ao I <ir (and therefore, I z - A I >ir and I z - I' I > ir for z E C),

X*([/(A) - I(Ao) _ IC!') - I(Ao) ]/A _ 1') A - Ao I' - Ao

= -1_·f x*(J(z)) dz. 2ni c (z - A)(Z -I')(z - Ao)

(6.2)

Since M = sup{11 x*(J(z)) II: z E C} < 00, the absolute value of the lefthand side of equation (6.2) does not exceed 4M/r2. Applying Remark 6.3 above, we have

II I(A) - I(Ao) _ IC!') - I(Ao) II < K . I A - I' I ' A - Ao f-l - Ao

for some constant K, whenever 0 < I A - Ao I <r/2 and 0 < I f-l - Ao I < r/2. Since X is complete, it follows that I is differentiable at Ao. • 6.5. Divergence 01 the Fourier Series. By the Fourier series of a function I in L 1[ -n, n], we mean the series

f /(k)eikt, k=-oo

where the Fourier transform / of I is defined by

A I f'" ·k I(k) = 2n _"I(s)e-' 8 ds, k E Z.

Such series were invented originally to serve as tools to solve problems in heat conduction, the theory of oscillation, and various other fields. The many problems that arose to determine whether the Fourier series of I converges to I or whether I is determined by its Fourier series gave rise to an important branch of analysis, known as harmonic analysis. We will not consider the convergence of the Fourier series here. In the next chapter (Remark 7.4), we shall show that the Fourier series of I in L 2[ -n, n] converges to I in L 2-norm. Actually a much more nontrivial result holds, namely, that the Fourier series of such a function converges almost everywhere. This result was first conjectured by N. N. Lusin in 1951 and then proven in 1966 by L. Carleson [8]. Later on, in 1968, the result was extended to Lp[ -n, n}, 1 < p < 00, by R. A. Hunt [17}. We will not go into the details of these results here. Here we will consider the divergence of the Fourier series to show another application of Theorem 6.11.

We prove the following: There exists a continuous function I on [-1£, 1£] such that the Fourier series of I diverges at O. The set of all such functions is of the second category in Cl [-1£, 1£]. t

Proof. We write Snl(x) = L~ __ n!(k)eikz. Then we have, for any I in Cl [ -1£, 1£],

where

1 f" S,J(x) = 21£ _" l(t)Dix - t) dt,

n Dn(t) = L eikt =

k=-n

= 2n + 1,

sin(n + !)t sin(lt)

if eit = 1.

if eit =I=- 1,

For each nonnegative integer n, we define

Then xn* is a bounded linear functional on CI[-n, 1£] with norm (I/2n) II Dn 111' (The reader should verify this.) Now, suppose that the Fourier series of every IE Cl[--n, 1£] converges at x = O. Then for each/E Cl[-n, 1£], supn I xn*(f) 1< =. By Theorem 6.11, it follows that

sup II xn * II = -21 sup II D" 111 < 00. n 1£ n

But we have

II Dn 111 = f"--I sin~n + !)t 1 dt = 4 f"/21 sin(2~ + l)t 1 dt .. sm(!t) 0 sm t

f"/21 sin(2n + 1)1 1 8 2n 1 >4 dl>- L --. - 0 1 -nk=ok+l

Thus, sup" II D" 111 = 00, a contradiction. To complete the proof, we assume that the set XC Cl[ -1£,1£] offunctions

f such that sUPnl r.~=-n!(k) I < 00 is of the second category. Noting that Theorem 6.11 holds even when X is a normed linear space of the second category, it follows as before that sup" II x" * II < 00, which is a contradiction. The proof is complete. I

t The complex-valued continuous functions on [-:re,:re] with the sup-norm.

Before we close this section, we present a theorem on projection, an application of the Closed Graph Theorem. The reader will find projection operators extremely useful' in Hilbert space theory.

Definition 6.9. A bounded linear opt:rator P from a normed linear space X onto a subspace M of X is called a projection if p2 = P.t I

Remarks

6.6. If P is a projection and P #- 0, then "P " > I since

" P " = "P2 " < " P " . "P II. In fact, "P "can be greater than I. Let X = R2 with " (x, y) " = I x I + I y I and M = {(x, x) I x E R}, a closed subspace of X. We define P(x, y») = (y, y). Then P is a projection onto M. But p(O, I») = (I, I) and therefore, II P II > 1.

6.7. There is always a projection from a normed linear space X onto any of its finite-dimensional subspaces M. To see this, let Xl' x2 , ..• , Xn

be the basis of M eX. Define Xi* E X* by

*( ) _ { I, Xi Xj - 0, i =j, i#-j.

I..et P(x) = Lf=IXi*(X)Xi' Then P is a projection.

Proposition 6.14. Let M be a closed subspace of a Banach space X. Then there exists a projection P from X onto M if and only if there is a closed subspace N of X such that X = M + N, M n N = {O}. I

Proof. For the "only if" part, let N = P-l( {O}), where P is a projection of X onto M. Since x = Px + (x - Px) and x - Px EN, X = M+N. Also Z EM n N implies P(z) = 0 and z = P(x) for some x E X, which means z = O.

For the "if" part, let X = M + N, M n N = {O}. Then for each x E X, there exist unique mE M and n EN such that x = m + n. We define P(x) = m. Then P is linear and p2 = P. To show that P is bounded, it is sufficient to show that it is closed (because of the Closed Graph Theorem). Let Xk ---+ x and P(Xk) ---+ y. Let Xk = mk + nk, mk E M and

t In a Hilbert space, projection operators, by definition, need to satisfy an additional condition. See Section 7.4.

nk E N. Also let x = m + n, m E M and n E N. Then P(Xk) = mk ---+ Y E M. Hence nk = (mk + nk) - mk ---+ m + n - y. This means that m + n - y EN or m - y EM n N = {o}. Therefore y = m = P(x) and P is closed .•

Problems

6.3.1. The notations are as in Lemma 6.2. Show that S,k) C TSx (1), if Sy(r) C TSx(l/(l - e» for all e E (0, I). X 6.3.2. Prove Proposition 6.8 and Proposition 6.9.

6.3.3. Let X be a normed linear space of the second category and M be a closed proper subspace of X. Show that XIM is of the second category. X 6.3.4. Let T be a closed linear operator from a normed linear space X into a normed linear space Y. Let M = T-I{O}. Show that there is a unique one-to-one, closed, and linear operator To: XIM ---+ Y such that To 0 rp = T, where rp is the natural map from X onto XIM. Also show that TO-I: T(X)(C Y) ---+ XIM is a closed linear operator.

X 6.3.5. Prove Lemma 6.2. X 6.3.6. Let Tn E L(X, Y), where X is a Banach space and Y is any normed linear space. If for each x E X, T(x) = limn-+ooTn(x), then show that T E L(X, Y).

X 6.3.7. Let A be a linear operator from X into Y such that y* 0 A is continuous for each y* E Y*. Show that A E L(X, Y). (X and Yare normed linear spaces which are not necessarily complete.)

X 6.3.8. If A is a bounded linear operator from D (a dense subspace of a normed linear space X) into a Banach space Y, then show that there exists a unique bounded linear operator B from X into Y with II A II = II B II and B(x) = A(x), xED. [Note that completeness is essential here; for consider D = Y = the polynomials on [0, I], X = crO, I] and i: D ---+ D, the identity map.]

X 6.3.9. Let A E L(X, Y). The adjoint of A is an operator A*: y* ---+ X* defined by A*y*(x) = y*(A(x», x E X. Show the following:

(i) II A* II = II A II· (ii) (aA + (JB)* = aA* + (JB· for all a, (J E F and A, B E L(X, Y).

(iii) If A E L(X, y), BE L(Y, Z), then (BA)* = A*B*. (iv) If A E L(X, Y), A is onto and A-I exists and belongs to L(Y, X),

then (A-I)* = (A*)-I.

Note that if X = Y = Ip, I <p < 00, then a bounded linear operator A on Ip can be represented by an infinite matrix (aij). If ei is that element in lp where the ith entry is I and every other entry is 0, and if ei* E Ip * such that

Sec. 6.3 • Open Mapping, Closed Graph Theorems; Uniform Boundedness 4S

ei*(ej) = 1 if i = j, = 0 if i =I=- j, then aij = e,* (A (ej) ). If A(x) = y and x = L Xiei, then e,*(y) = LjaijXj' Since Ip * can be identified with 1'1' where pq = p + q, the adjoint A* can be easily verified to be represented by the transpose of the infinite matrix (aij) in the same sense as above.

In the next few problems and later on, the range of A is denoted by R(A) or RA •

X 6.3.10. Let A E L(X, Y), A be one-to-one and X, Y both Banach spaces. Then R(A) is closed if and only if there exists C> 0 such that II x II < C II A(x) II for each x E X.

6.3.11. If X and Yare Banach spaces and A E L(X, Y), then R(A) is closed in Y if and only if there exists C > 0 such that inf{ II x - y II: A(y) = O} < C II A(x) II for each x E X.

X 6.3.12. Let sex. Then So (the annihilator of S) = {x* E X*: x*(Y) = 0 for each YES}. So = X* when S is empty. Similarly, for E C X*, °E = {x EX: x*(x) = 0 for each x* E E}. °E = X, when E is empty. Show the following:

(i) So and ° E are both closed subspaces. (ii) For sex, O(SO) is the closed subspace spanned by S.

(iii) [R(AW = N(A*), the null space of A*

(iv) R(A) = O[N(A*)]. (v) O[R(A*)] = N(A).

(vi) R(A*) C [N(AW.

X 6.3.13. Let A E L(X, y), X and Y both Banach spaces. If R(A) is closed in Y, then show that R(A*) = [N(A)r and hence is closed in X*. [Hint: Use Problem 6.3.12 (vi) and the following. For x* E [N(A)]O, define f: R(A) ~ F by f(A(x» = x*(x). Show, by Problem 6.3.11, I f(A(x») I < C II x* II· II A(x) II, C > O. Extendfto y* E y* so that A*(y*) = x*.]

6.3.14. Let A E L(X, Y). Then show that R(A*) = X* if and only if A-I exists and is continuous.

6.3.15. Let A E L(X, Y), Y complete and R(A) = Y. Then show that A* has a continuous inverse.

6.3.16. Prove that L 2[0, 1] is of the first category in L 1[0, 1]. (Hint: The identity map from L2 into Ll is continuous, but not onto. Use the Open Mapping Theorem.)

6.3.17. Joint Continuity of a Separately Continuous Bilinear Function. Suppose X and Yare Banach spaces and T: Xx Y ~ R is a mapping such

that for each x E X and each y E Y, the functions y -->- T(x, y) and x -->- T(x, y) are bounded linear functionals. Show that T is continuous on Xx Y. (Hint: Use the Principle of Uniform Boundedness.)

6.3.18. Convex Functions and the Principle of Uniform Boundedness. Let :tr be the smallest a-algebra containing the open sets of a Banach space X. Then every real-valued convex function defined on X and measurable with respect to :tr is continuous. Using this result, the Principle of Uniform Boundedness can be proven as follows. The function p(x) = sup{ II T;.(x) II: A. E A}, where {T;.: A. E A} is a family of bounded linear operators from X into a normed linear space Y such that p(x) is a realvalued function on X, is a lower semicontinuous convex function and therefore continuous. Then there exists 15 > 0 such that II x II < 15 => I p(x) - p(O) I < 1. This means that for each A. E A II T;. II < 1/15.

6.3.19. Another Application of the Principle of Uniform Boundedness. Let X be the Banach space (with "sup" norm) of periodic continuous functionsf on R with period 2n. For fE X, let Tnf(t) = n(J(t + lin) - f(t)). Then by Corollary 1.2 (Chapter 1), limn-+ooTnf(t) =f'(t) exists for all fin a dense subset of X. However, the set of nondifferentiable functions is of the second category in X. [Notice that II Tn II = 2n. If DC X is the set of functions differentiable at 0 in Rand D is of the second category, then the Principle of Uniform Boundedness applies to the sequence x* 0 Tn' where x*(f) = f(O).]

6.3.20. Equivalent Norms in qo, I]. Any complete norm II . II in qo, I], where limn-+ooll fn - f II = 0 => limn-+oofn(t) = f(t) for all t E [0, I], is equivalent to the usual "sup" norm. {Here the Closed Graph Theorem can be used to show that the identity map from qo, I] with "sup" norm into (qo, I], II . II) is continuous; then Proposition 6.11 applies.} * 6.3.21. Necessary and Sufficient Condition for a Sequence to be a Schauder+ Basis in a Banach Space (An Application of the Open Mapping Theorem). Let X be a Banach space and (en) a sequence in X such that the linear span of (en) is dense in X and no en is O. Prove that (en) is a Schauder basis for X if and only if there is a positive ko such that for all positive integers nand p and scalars (ai),

(**)

[Hint: Use Problem 6.1.I6(v) and the norm III . III defined there. By the

+ See Problem 6.1.16 for definition.

open mapping theorem, T-l is continuous, where T is the identity map from (X, III . III) to (X, II . II). This proves the "only if" part. For the "if" part, assume (**) and let Y be the linear span of (en). Define e/ on Y by

ei*( ~ ajej) = ai j=1

=0

if n > i,

otherwise.

By (**), ei* is continuous, and can be extended by continuity to all of X. Given any subsequence (mk), there exists (nk) C (mk) such that Ynk =

Li'!lai,kei and II x - Ynk II < IJ(ko2k+2). For i < nk, ei*(Ynk) = ai,k' Use (**) to show that

so that for each p > 1,

Now let p go to infinity, and then, let k go to infinity.]

* 6.3.22. A Schauder Basis for Lp[O, 1], 1 1 <j< 2'. (Thus, for r = 2, 11 ,2 = [0, t), 12 ,2 = [!, t), 13 ,2 = [t, !), and 14 ,2 = [!, 1 ].) Define the sequence of functions (gi) on [0, 1] by

g2T-1+k = 1,

= -1,

= 0,

on 12k-l,r,

on 12k,r>

otherwise,

where 1 < k < 2r - l • Notice that all characteristic functions of the intervals Ij,r are contained in the linear span of the g/s, which is, therefore, dense in Lp[O, 1]. Use condition (**) in Problem 6.3.21 to show that (gi) is a Schauder basis for Lp[O, 1], 1 <p < 00.

6.3.23. Continuation of Problem 6.3.20. Prove the following assertions in qo, 1]:

(i) The norm II f II = L::l(l J2n) I f(tn) I, where (t,,) is a dense seq uence in (0, 1), is incomplete. (Does convergence in this norm imply pointwise convergence?)

(ii) The norm II f II = SUPtE[O,lJ I tf(t) I + I f(O) I is incomplete.

(iii) There are at least two complete norms in C[O, 1] that are not equivalent. What if qo, I] is replaced by an infinite-dimensional vector space? [Hint for (iii): qo, I], as well as L2 [0, 1], has a Hamel basis with cardinality c. In fact, a Hamel basis of any separable Banach space has the cardinality c (under the assumption of the continuum hypothesis). Let (fn) be a uniformly bounded linearly independent sequence in qo, I] such that II In 112 -+ 0, but In -;. ° in qo, I]. Extend (fn) to a Hamel basis HI of qo, I] and also to a Hamel basis H2 of LAO, I]. It is now easy to transport the L 2-norm to qo, I] via a linear bijection.]

6.3.24. Let S be a closed subspace of LI [0, I]. If S C Lp[O, I] for some p > I, then what can be said about S?

6.4. Reflexive Banach Spaces and the Weak Topology

In the theory of Banach spaces an often useful concept is that of reflexivity, which is based upon a characterization of a class of bounded linear functionals. The notion and properties of reflexive Banach spaces, along with some representation theorems for bounded linear functionals on certain well-known normed linear spaces, is the subject of this section. One reason for the study of such spaces is that a large number of useful results that are not true in general Banach spaces are valid in such spaces; also, one comes across a wide class of such spaces in theory as well as in practice. For instance, the Lp (I < p < 00) spacest will be shown to belong to this class.

Among other things, we also show in this section the interplay between reflexivity and a basic convergence concept, the concept of weak convergence. The topics of weak convergence and weak topology are essential in functional analysis. They find an immense number of applications in various contexts in the theory of differential equations and in the calculus of variations.

Definition 6.10. The natural map J of a normed linear space X into its second conjugate space X**[= L(X*, F)] is defined by

[J(x)](x*) = x*(x), x* E X*.

If the range of J is all of X**, then X is called reflexive. I

t In this section, for convenience the Lp spaces are taken over the reals.

Sec. 6.4 • Reflexive Banach Spaces and the Weak Topology 49

Remark 6.S. A reflexive normed linear space is complete. The reason is that

II J(x) II = sup I x*(x) I = II x II, IIX'II-1

by Corollary 6.5; and therefore J is a linear isometry from X onto X**, which is complete. However, the existence of a linear isometry from X onto X** does not guarantee the reflexivity of X. For an example demonstrating this fact, the serious reader is referred to R. C. James. t

Remark 6.9. A finite-dimensional normed linear space X is reflexive. The reason is that dim X = dim X* = dim X**, and a one-to-one linear operator between finite-dimensional spaces of the same dimension is also onto.

Proposition 6.15. Every normed linear space X is a dense subspace of a Banach space. I

Proof. The idea is simple. Simply identify X with J(X) and note that

J(X) is a Banach space. The details are slightly messy. Let X = X u (J(X) -J(X»). Denoting vector addition in X** by +, we define addition in X as follows:

(i) x + y is the same as it is in X, if x E X and y EX;

(ii) x + y = J(x) + y, if x E X and y E X - X;

(iii) x + y = x + y, if x EX - X, Y EX - X and x + y $ J(X); (iv) x + y = J-l(X + y), if x EX - X, Y EX - X and x + y E J(X).

We define scalar multiplication in X similarly so that X is a vector space with X as a subspace. For x E X, let its norm in X be its norm in X. For x EX - X, we define its norm to be its norm in X**. It follows easily that

X is a normed linear space with X as a dense subspace. The map (j>: X --+ J(X) defined by

(j>(x) = J(x), if x E X;

= x, if x E X - X,

is a surjective linear isometry. This means that X is complete. I Before we get involved with the properties of reflexive spaces, we should

consider some interesting examples of such spaces. To this end, we study

t R. C. James, A nonreflexive Banach space isometric with its second conjugate, Proc. Nat. Acad. Sci. U.S.A. 37, 174-177 (1951).

first the conjugate (or dual) spaces of some important Banach spaces. Let us recall the following four results, which were proven in Chapter 4 of Part A and appeared there as Theorem 4.8, Corollaries 4.4, 4.5, and 4.6. These results involve Lp , I <p < 00, spaces. For simplicity, we consider only real-valued Lp functions.

Theorem 6.12. Riesz Representation Theorem. Let (X, .JJf, p) be a a-finite measure space and (/J be a bounded linear functional on L p ,

I <p < 00. If lip + Ilq = I, then there is a unique element g E Lq such that

(/J(f) = f fg dp,

and I

Corollary 6.6. Let (X,.JJf, p) be any measure space and (/J be a bounded linear functional on L p , I < p < 00. Then there is a unique element g E Lq , lip + Ilq = I such that

(/J(f) = f fg dp, for all fE L p ,

and II (/J II = II g Ilq· I

Corollary 6.7. For I < P < 00, Lp * is linearly isometric to Lq, IIp+llq=1. I

Corollary 6.8. For I <p < 00, 1/ is linearly isometric to Iq, lip + Ilq = 1. I

Corollary 6.9. For I < p < 00, Lp is reflexive. I

Proof. Let J be the natural map from Lp into Lp **. To show that J is surjective,letx** E Lp**. Let X be the map from Lq into Lp*, lip + Ilq = I, defined by

[x(g)](f) = f.rg dp,

Then by Corollary 6.6 X is a surjective linear isometry. Let us define the map

x* = x** 0 x.

Then x* E 4*. Hence by Corollary 6.6, there exists hELp such that x*(g) = J gh df-', g E Lq. Let y* E Lp*. Then there exists go E Lq such that X(go) = y*. Now clearly

x**(y*) = x*(go) = f goh df-' = [x(go)](h) = y*(h) = [J(h)](y*).

Hence J is surjective. I

Corollary 6.10. For 1 < p < 00, Ip is reflexive. However, It is not reflexive. I

Proof. The first part follows from Corollary 6.9. For the second part suppose 11 is reflexive. Then 11 **, being homeomorphic to 11, is separable and therefore, by Proposition 6.7, 11* (and hence 100) is separable, which is a contradiction. I

We have seen above that the dual of Lp , 1 <p < 00, is (linearly isometric to) Lq in a l1-finite measure space. Unfortunately, such a representation does not hold for the bounded linear functionals on Looo Consider the Lebesgue measure space on [0, 1]. Let (/> be the linear functional on qo, 1] C Loo defined by (/>(f) = 1(0). Let (/>0 be the continuous extension of (/> (possible by the Hahn-Banach Theorem) to Looo Then II (/>0 II > II (/> II = 1. Suppose there is agE Ll such that (/>0(1) = J~/(t)g(t) dt for all IE Looo Let us define In E qo, 1] by

{ 0 Ifn<t< I, In(t)= 1'-nt, O<t<l/n.

Then H In(t)g(t) dt --+ 0 as n --+ 00, but (/>o(Jn) = InCO) = 1, which is a contradiction.

From the above results it is clear that there are plenty of infinite-dimensional reflexive spaces and therefore development of a theory for such spaces is in order and naturally will be very useful.

So far we have explored to some extent the class of all bounded linear functionals on a normed linear space. The concept of reflexivity is based upon characterizing a class of bounded linear functionals. Why is it that we do not considel with equal interest the class of all linear (not necessarily bounded) functionals on a normed linear space? The next result sheds some light on this question. First we need a definition.

Definition 6.11. Let X be a vector space over F, and let X' be the vector space (under natural operations) of all linear functionals on X-called


the algebraic dual of X. LetJbe the natural map from Xinto X" (the algebraic dual of X') defined by

[J(x)](x') = x'(x), X'E X'.

Then if J is onto, X is called algebraically reflexive. I

Proposition 6.16. A vector space X over F is algebraically reflexive if and only if X is finite-dimensional. I

Proof. We will only prove the "only if" part. Let H be the Hamel basis of X. Suppose X is infinite dimensional. Then H is infinite and let H = {Xi: i E I}, I being an infinite indexed set.

Let us define x/ E X' by

and

Then A' = {x/: i E I} is linearly independent in X'. (Why?). Let H' be a Hamel basis of X' containing A'. Let us define x" E X" such that

x"(x/) = Pi' x"(x') = 0,

i E I,

x' E H' - A',

where infinitely many of the P/s are nonzero. If J is onto, then there exists x E X such that [J(x)]x' = x'(x) = x"(x') for all x' E X'. However x/(x) = x"(xn for all i E I. If i E I is such that the term Xi is missing in the unique representation of x as a linear combination of elements of H, and x"(x/) = Pi:i=- 0, then x/(x) = 0:i=- Pi = x"(x/), which is a contradic-tion. I

We see therefore from Proposition 6.16 that the concept of algebraic reflexivity is not very useful since the study of algebraically reflexive spaces is nothing more than the study of finite-dimensional spaces.

Now let us reconsider the space c (see Problem 6.2. II ), a subspace of 100- By Problem 6.2.11, .c* is linearly isometric to 11 and therefore c** is

linearly isometric to 11*. (Why?) If c is reflexive, then c** is separable since c is separable. But this contradicts the nonseparability of 11* (or 100)' Hence c as well as c* (or 11) is not reflexive. Here arises a natural question: Does


the reflexivity of a Banach space X imply that of X*? The following proposition answers this.

Proposition 6.17. A Banach space X is reflexive if and only if X* is reflexive. I

Proof. Suppose X is reflexive. To prove that the natural map J: X* -+ X*** is onto, let x*** E X***. We define

x* = x*** 0 Jx ,

where Jx is the natural map from X onto X**. Then x* E X* and J(x*)[Jx(x») = Jx(x)[x*) = x*(x) = x***[Jx(x»). Since Jx(X) = X**, x*** = J(x*) and J is onto.

To prove the converse, let X* be reflexive. If X is not reflexive, then Jx(X) is a closed proper subspace of X**. By Corollary 6.3 there exists x*** E X*** such that x***(x**) *- 0 for some x** E X** - Jx(X) and x***[Jx(x») = 0 for each x EX. Since J(X*) = X***, there exists x* E X* such that J(x*) = x***. Therefore, 0 = J(x*)[Jx(x») = Jx(x)[x*) = x*(x) for each x E X, which means that x* = 0; and therefore x*** = 0, which is a contradiction. I

We will not try to determine the space 100* in this book, since this determination is nontrivial; and we will not need it in our discussion. However, knowing that c is not reflexive, we can assert that 100 is also not reflexive. The following proposition makes it possible.

Proposition 6.18. Every closed subspace of a reflexive Banach space is reflexive. I

Proof. Let Y be a closed subspace of a reflexive Banach space X. Let Jy : Y -+ y** and Jx : X -+ X** be the natural maps. Given y** E Y**, we define x** E X** by x**(x*) = y**(x*ly). Since Jx is onto, there exists x E X such that Jx(x) = x**. We claim that x E Yand Jy(x) = y**. If x if; Y, then by Corollary 6.3, there exists x* E X* such that x*(x) *- 0 and x*(y) = 0 for each y E Y. Then x**(x*) = y**(x*ly) = 0 or Jx(x)[x*] = x*(x) = 0, which is a contradiction. Hence x E Y. The rest is left to the ~~ I

Next we consider a very important nonreflexive Banach space-the space eta, b) of (real valued) continuous functions under the uniform

(sup) norm. It is clear that for any function g of bounded variation on [a, b]

f{J(f) = J: J(t) dg(t)

defines a bounded linear functional f{J on qa, b). Actually we will see in what follows that every bounded linear functional on qa, b) has the above form. This fact was first discovered by F. Riesz in 1909. Later, in 1937, it was extended by Banach to the case of a compact metric space (instead of [a, b]) and then by Kakutani in 1941 to the case of a compact Hausdorff space (instead of [a, b]). Kakutani considered signed measures instead offunctions of bounded variation. (See Chapter 5.)

Let us denote by BV [a, b) the Banach space of real functions of bounded variation on [a, b) with the norm

II g II = V(g) + I g(a) I,

where V(g) denotes the total variation of g on [a, b). (The reader should convince him- or herself that BV[a, b) is a Banach space-a fact that is neither difficult nor trivial to verify.) We will denote by B[a, b) the Banach space of all real bounded functions on [a, b) with the usual supremum norm.

Theorem 6.13. (Riesz).+ For every bounded linear functional if> on C[a, b) there is a function g of bounded variation such that, for each JE C[a, b),

if>(!) = J: J(t) dg(t)

and II if> II = V(g). I Proof. Since qa, b) C B[a, b), by the Hahn-Banach Theorem there

exists a continuous linear functional if>o on B[a, b) extending <P such that II <Po II = II <P II. We define for s E (a, b)

zs(t) = { ~: za(t) == O.

a<t<s, s < t < b,

Define g(s) = <Po(zs). We claim that V(g) < II <Po II < 00. Clearly if

t The reader may note that a similar result holds in ella, b] when the scalars are complex numbers.

Sec. 6.4 • ReOexive Banach Spaces and the Weak Topology 55

and 1 <i< n,

then we have

since the function inside the parenthesis has norm 1. Hence V(g) < II tPo II. Now to complete the proof of the theorem, letfE qa, b]. For a = 10 < II

< ... < tn- 1 < tn = b, let h(t) = L:-t!(ti)[Zt,(t) - Zt;_l(t)]. Then we have

1 h(t) - f(t) 1 = I f(t i ) - f(t) I, ti- 1 < t < ti ,

= 1 f(tl) - f(t) I, t = a ,

and n

tPo(h) = L f(li)[g(t i ) - g(ti - 1)]. i-I

It is clear now that in the limit when n -+ 00 and maxlSiSnl Ii - ti - 1 1 -+ 0,

tP(f) = tPo(f) = J: f(t) dg(t).

Since 1 J! f(t) dg(t) 1 < II f II V(g), II tP II < V(g). The proof is complete. I

The above theorem does not provide a one-to-one correspondence between bounded linear functionals tP on qa, b] and functions of bounded variation on [a, b], as the following lemma shows.

Lemma 6.3. Let g E BV[a, b]. Let h be defined by

{ g(t + 0) - g(a),

h(t) = g(b) - g(a), 0,

Then h E BV[a, b] and for each f E qa, b]

a < t < b, t = b, t = a.

f: f(t) dg(t) = f: f(t) dh(t)

and V(h) < V(g). I

The proof of this lemma is left to the reader. To provide a one-to-one correspondence between C*[a, b] and a suit

able subspace of BV[a, b], we need the following definition.

Definition 6.12. A function g E BV[a, b] is called normalized if g(a) = 0 and get + 0) = get), a < t < b. I

The collection of normalized functions of bounded variation on [a, b] is denoted by NBV[a, b]. The next two results will show that NBV[a, b] will provide us with the one-to-one correspondence desired above. We need another lemma.

Lemma 6.4. Let g E BV[a, b] such that J~ f(t) dg(t) = 0 whenever fE C[a, b]. Then g(a) = g(b) and for a < t < b, get - 0) = get + 0) =~ I

Proof. Clearly if f= 1, then J~ dg(t) = 0 and therefore g(a) = g(b). For a < e < band 0 < h < b - e, let us define

{I,

f(t) = 1 - (t - e)jh, 0,

ThenfE C[a, b] and

a<t<e, e<t<e+h, e + h < t< b.

fb f~ o = a f(t) dg(t) = gee) - g(a) + c f(t) dg(t).

Integrating by parts and simplifying,

1 fC+h o = - g(a) + h C get) dt, 0< h < b - e,

and therefore by letting h - 0, g(e + 0) = g(a). Similarly, g(e - 0) = g(a) for a < e < b. I

Theorem 6.14. The dual space of qa, b] is linearly isometric to NBV(a, b]. I

Proof. For r.p E C*[a, b], by Theorem 6.13 we can find g E BV[a, b] such that for each f E C[a, b] we have

r.pU) = f: f(t) dg(t),

and II rp II = V(g). Let h be as defined in Lemma 6.3. Then hE NBV[a, b], V(h) < V(g) and for IE C[a, b]

rp{f) = f: I(t) dh(t).

Then since II rp II < V(h), V(h) = II rp II. If we define T(rp) = h, then Tis a linear isometry from C*[a, b] onto NBV[a, b]. That T is well defined (that is, there is a unique h E NBV[a, b] with the above properties) is guaranteed ~~~M I

Finally in this section, we wish to consider a new notion of convergence of a sequence of vectors in a normed linear space-called weak convergence. The reason for considering this topic is that a very important interplay exists between reflexivity and weak convergence, as will be clear in what follows.

Definition 6.13. A sequence (xn) in a normed linear space X is said to converge weakly to x E X if for every x* E X*, x*(xn) - x*(x) as n-oo. I

Remarks

6.10. If Xn - x in X, then Xn --'!. x, i.e., Xn converges weakly to x. 6.11. The converse to Remark 6.10 need not be true. For instance, if

X = lp (1 < p < 00) and Xn = (0, 0, ... ,0, I, 0, ... ), then II xn lip = I n

for each n while Xn ~ 0. [Recall that for each x* E X* there is b = (bl , b2 , ••• ) E Iq, lip + Ilq = I such that x*(xn) = bn.]

6.12. If Xn ~ x, then sUPIs:n<ooll Xn II < 00. The reason is that if J(xn)[x*] = x*(xn), then [J(Xn)]:'1 becomes a pointwise bounded family of bounded linear operators from X* into F; and therefore by the Principle of Uniform Boundedness SUPI s:n<oo II J(xn) II = SUPI s:n <00 II Xn II < 00.

6.13. If X is finite dimensional, then Xn -! x if and only if Xn - x. See Problem 6.4.3.

6.14. If a sequence (In) in qa, b] converges weakly to IE qa, h], then the sequence is uniformly bounded and for a < t < h, limn-+ooln(t) = J(t). This is because of Remark 6.12 above and because Yt*(fn) - y'*(f), where y,* E C*[a, h} is defined by y,*(h) = h(t). The converse is also true by Theorem 6.13.

6.15. If Xn = (Xnl' Xn2, ... ) E lp, Z = (Zl' Z2, ... ) E lp with I < p

< 00, then Xn --'!. Z if and only if IiIDn~ni = Zi' I < i < 00, and SUPls:n<oo II Xn lip < 00. See Problem 6.4.4.

6.16. A normed linear space X is called weakly sequentially complete if every weak Cauchy sequence (xn) E X [that is, (x*(xn» is Cauchy for each x* E X*] converges weakly to some element x E X. A reflexive Banach space X is weakly sequentially complete. To prove this, let (xn) be a weak Cauchy sequence in X. Let T(x*) = limn->-oox*(xn). Then T is linear; and by Remark 6. I 2above, II Til < sUPl$n<ooll xn II < 00 and therefore T E X**. If X is reflexive, there exists x E X such that J(x) = T, J being the usual natural map from X onto X**. Hence limn->-oox*(xn) = x*(x).

6.17. C[O, I] is not weakly sequentially complete and hence not reflexive. Consider xn(t) = (1 - t)n. Then xn(t) cannot converge pointwise to a continuous function and hence by Remark 6.14 cannot converge weakly. But for y* E C* [0, 1], there exists g E NBV[O, 1] such that for

fl ftl fl y*(zmn) = zmn(t) dg(t) = zmn(t) dg*(t) + zmn(t) dg(t), t o 0 11

where g*(t) = g(t + 0), ° < t < I. The first integral can be made arbitrarily small by taking II (>0) sufficiently close to 0, g*(t) being continuous from the right; and the second integral can be made arbitrarily small {since xn(t) -+ 0 uniformly in [11, I]} by taking nand m sufficiently large. This means that (xn) is weakly Cauchy in C[O, t].

6.18. In a reflexive Banach space X, each bounded (that is, norm bounded) sequence has a weakly convergent subsequence.

To prove this, let (xn) be a sequence in X such that SUPI$n<oo II Xn II < 00.

Consider the closed linear subspace Y spanned by the xn's. Then Y is separable. Y is also reflexive by Proposition 6.18. Therefore y** is separable, and by Proposition 6.7 y* is also separable. Let (Ym*)meN be dense in Y*. Since the sequence Yl * (xn) is bounded, there exists a subsequence (Xl n)neN such that Yl *(xI,n) converges as n -+ 00. Similarly, there exists a subsequence (x2,n) C (Xl,n) such that Y2*(X2,n) converges as n -+ 00. Using induction we can find for each positive integer k a subsequence (Xk+1,n) C (Xk,n) such that for each i, I < i < k + 1, Yi*(Xk+1,n) converges as n -+ 00. If (Xnn):'1 is the diagonal sequence, then it is easy to show that for each i, y/(xnn) converges; and therefore for each y* E Y*, y*(xnn) converges. We define

y**(y*) = lim y*(xnn ). n->-oo

t Note that for 1 E C[O, 1] with 1(0) = 0, f: 1 dg = f: 1 dg* for all ( in (0, 1].

Sec. 6.4 • Reflexive Banach Spaces and the Weak Topology S9

Then y** E Y**, since SUPI:5n<ooll Xn II < 00. Since Yis reflexive, there exists

Y E Y such that y*(y) = y**(y*) whenever y* E Y*. Hence xnn -'!. y. 6.19. It is clear from the previous remark that in a reflexive Banach

space if the weak convergence of a sequence implies its convergence, then its closed unit ball must be compact, and consequently the space must be finite dimensional. This is not true in a nonreflexive space, as the following remark shows.

6.20. In II, a nonreflexive space, the weak convergence of a sequence implies its (strong) convergence. If this were not the case, there would be an 8 > 0 and a sequence of elements Xn = (xl n, x 2n , xan, ... ) in II such that for each (WI' W2 , ••• ) E 100 , we have

and

00

L Wjx/--O, j-I

00

L I x/ I> 8, j=1

as n--oo,

(6.3)

I <n < 00.

Choosing the Wk'S properly, it follows easily that for each k, limn-+ooxkn = O. Let mo = no = 1. Then the sequence (mk' nk) is defined inductively as follows: nk is the smallest integer n > nk-I such that

mk-l 8

L IX/kl <-, j=1 5

(6.4)

and mk is the smallest integer m > mk-I such that

(6.5)

We define W = (WI' W2, ... ) by

mk-I <j < mk,

mk <j < mk+1' etc.

Then by inequalities (6.4) and (6.5) we have

I 00 00 I mk-l 00 48 ~ WjX/k - ~ I X/i: I <2 ~ I X/k 1+ 2.L I X/k 1< S· }-I J=I J=I }=mk

Therefore it follows from (6.3) that for each k,

I f W·X·nl; I >~ j_1 }) - 5 '

which contradicts the first statement in (6.3).

From the above remarks the reader has some ideas about weak convergence in a normed linear space and its connection with reflexivity of the space. This connection will be more distinct if we study what is called the weak topology of a normed linear space X.

The weak topology of X is the weakest topology on X such that every element in X* is continuous. Clearly a basis for the weak topology on X consists of sets of the form

{x: lfi(x) - fi(xo) 1< e, i = 1,2, ... , n},

where Xo EX, e > 0, and Ii E X*. Then it is easily proven that the weak topology is contained in the metric topology of X so that every weakly closed set is strongly closed. (The reader should verify this.)

That the converse is not true is clear from Remark 6.11. However, the following proposition holds.

Proposition 6.19. A linear subspace Y of a normed linear space X is weakly closed if and only if it is strongly closed. I

Proof. Suppose Y is strongly closed and y $ Y. Then infzeyll y - z II > () > O. Hence by Corollary 6.3 there is x* E X* such that x*(y) * 0 and x*(z) = 0 for each z E Y. This means that A = {w E X: I x*(w) I > O} is an open set in the weak topology, but AnY is empty. Hence y is not a weak-closure point of Y. The rest is clear. I

In Remark 6.20 we have seen that in 11 weak convergence of a sequence is equivalent to its strong convergence. However, the topology of a topological space is not determined by the concept of convergence of a sequence unless the space is first countable. We will see that in 11, as well as in any infinitedimensional normed linear space, the weak topology is properly contained in the strong topology. The following theorem demonstrates this.

Theorem 6.15. The weak topology of a normed linear space coincides with its strong topology if and only if the space is finite dimensional. I

Proof. We prove only the "only if" part. We will prove that if the open unit ball in X is weakly open, then X* (and therefore X) is finite dimensional.

Suppose S = {x E X: II x II < I} is weakly open. Then there exist Xl*' ... , xn* in X* and positive real numbers r1 , ••• , rn such that

{XE X: IXi*(X) 1< ri' I <i<n} C S.

Ifni~l{XEX: Xi*(X)=O} (=A, say) contains xo, then for every real number r, rxo E A C S. This implies that Xo = 0, so A = {OJ. We claim that {Xl*' '" , xn*} spans X*. To prove this let x* E X*. We define for x E X, T(x) = (xl*(x), ... , Xn * (x» C Fn, where F is the scalar field. We also define h from T(X) into F by h(T(x)) = x*(x). Then h is well defined since T(x) = T(y) implies x - YEA = {OJ or x = y. Since h is a linear functional on T(X) C Fn and T(X) [and therefore T(X)*] is finite dimensional, we may assume that there exist hI> h2' ... ,hm (l < m < n) in T(X)* such that hi (11' 12 , ••• , In) = Ii' I < i < m and h = 2::::'laihi, ai E F. Then x* = L:laixi*' Hence X* is finite dimensional. I

Before we close this section, we will present characterizations of a reflexive Banach space in terms of the weak compactness and also weak sequential compactness of its closed unit ball. We will do this by introducing another useful concept called the weak* topology, a topology of X*.

We know that the weak topology in X* is the weakest topology in X* such that each element in X** is continuous. However, this topology turns out to be less useful than the topology in X* generated by the elements in J(X), J being the natural map from X into X**. This latter topology is called the weak* topology for X* and is clearly weaker than its weak topology. A base for the weak* topology is given by the sets of the form

{IE X*: I/(xi) - lo(xi) 1< e, i = I, ... , n},

where Xl' X2 , ••• , Xn EX, e> 0, and 10 E X*. If X is reflexive, then J(X) = X** and therefore the weak topology for

X* and its weak* topology coincide. The usefulness of the weak* topologyt stems mainly from the following basic theorem. A sequential form of this important theorem for separable Banach spaces was proved by Banach in 1932. Alaoglu proved the theorem in the following general form in 1940.

Theorem 6.16. The Banach-Alaoglu Theorem. The closed unit ball in X* is compact in its weak* topology. I

Proof. Let S* = {I E X*: Ilf II < I}. If I E S*. then I(x) E {c E F: I c I < " X "} = Ix. say. Then we can think of S* as a subset of P = ILexIx, which is the set of all functions I on X with I(x) E Ix, given the usual product topology. The topology which S* inherits as a subset of P is the weak*

t The reader should verify by using the Hahn-Banach Theorem that the weak and the weak*·topology are both Hausdorff.

topology of S*. Since P is compact by Tychonoff's theorem (Theorem 1.5), S* will be compact if it is closed as a subset of P.

Let f be a point of closure of S* in P. Then f: X --+ F and I f(x) I < II x II. Now for x, y E X and a, fl E F, the set

v = {g E P: I g(x) - f(x) I < B, I g(y) - fey) I < B,

I g(ax + fly) -f(ax + fly) 1< B}

is an open subset of P containing f and hence V II S* *- 0. Since for gE V II S*, g is linear, I f(ax + fly) - af(x) - flf(y) 1 0, f is linear and therefore fE S*. I

If X is reflexive, then X and X** can be identified, and therefore the weak topology on X can be regarded as the weak* topology of X**. Hence by Theorem 6.16 the closed unit ball in X is weakly compact. The converse is also true. To prove this we need the following lemma.

Lemma 6.5. Let sex and S** C X** be defined as follows:

S = {x: II x II <l} and S** = {x**: II x** II < l}.

Then J(S) is dense in S** with the weak* topology of X**, where J is the natural map from X into X**. I

Proof. Let xo** E S**, B > 0, and XI*' ... , Xn * E X*. The lemma will be proved if we can find Xo E S such that

i = 1,2, ... , n. (6.6)

Let Y = II {x EX: Xi*(X) = 0, i = 1,2, ... , n}. Then Y is a closed subspace of X. Let yo = {x* E X*: x*(y) = ° for each y E V}. Then by an argument similar to that used in the proof of Theorem 6.15 (see also Problem 6.4.7), yo is the closed subspace of X* spanned by XI*' ... ,Xn *. Consider the mapping T: (X/Y)* --+ yo by T(V*) = x*, where x*(x) = V*([xD. T is well defined and an onto linear isometry. (The reader can easily verify this.) Hence (X/Y)* (and therefore X/Y) is finite dimensional. Define V** E (X/Y)** by V**(V*) = xo** a T(V*). Then II V** II < I since T is an isometry. Since X/ Y is reflexive (being finite dimensional), there exists V E X/V such that V**(V*) = V*(V) for each V* E (X/Y)* and II V II = II V** II < 1. Let us choose k such that sUPIsisnil Xi* II < k. Then there

exists x E V such that

II x II < II V II + elk < I + elk,

and V*(V) = x*(x), where T(V*) = x*. This means that for each x* E yo, x*(x) = xo**(x*). If Xo = [kl(k + e)]x, then II XO II < I; and for x* E yo,

I x*(xo) - xo**(x*) I = I x*(xo) - x*(x) I < II x* IIII XO - x II < (elk) II x* II·

It follows that Xo satisfies the inequalities (6.6). I

Theorem 6.17. A normed linear space X is reflexive if and only if the closed unit ball in X is weakly compact. I

Proof. The "only if" part follows easily from the Banach-Alaoglu Theorem. To prove the "if" part we see that the natural map J from X (with weak topology) into X** (with its weak* topology) is a linear homeomorphism onto J(X) C X**. Hence if the closed unit ball S of X is weakly compact, J(S) is compact (and hence closed) in the weak* topology of X**. By Lemma 6.5, J(S) = S* * = {x** E X* *: II x** II < I}. Since J is linear, J(X) = X**. I

Our next theorem in this section is the well-known Eberlein-Smulian Theorem, one of the most remarkable results in Banach spaces. In 1940, v. L. Smulian proved that the weak countable compactness of the weak closure of a subset A of a Banach space X implies its weak sequential compactness. W. F. Eberlein, in 1947, proved that the subset A is weakly compact if and only if it is weakly closed and weakly sequentially compact. Note that these results are nontrivial since the weak topology need not be even first countable. It is relevant here only to mention (without proof) at this point that the weak topology of the closed unit ball of a Banach space X is a metric topology if and only if X* is separable.

Definition 6.14. A set A C X*, where X is a. normed linear space, is called total if whenever x*(x) = 0 for every x* E A, then x = O. I

If X is a separable Banach space, then X* contains a countable total set; for if (xn) is a dense sequence in {x: II x II = I} and Xn * E X* with

Xn * (xn) = II Xn * II = I, then (xn *) is a total set and for all x in X, II x II = sUPnl Xn *(x) I.


The proof of the Eberlein-Smulian Theorem that we present here is due to R. Whitley. First we need a lemma.

Lemma 6.6. Let X be a normed linear space such that X* contains a countable total set. Then the weak topology on a weakly compact subset of X is metrizable. I

Proof. Let (xn *) be total and " Xn * II = I. Let

00 I d(x, y) = n~l 2n I xn*(x - y) I

and let A be a weakly compact subset of X. By Remark 6.3, A is bounded since x*(A) is compact for x* E X*. Let i be the identity map from A (with weak topology) onto A (with the metric topology induced by d). i is clearly continuous and therefore a homeomorphism; for if B is a weakly closed subset of A, then B is weakly compact and i(B) is compact in the metric d and therefore closed, in this metric. I

Corollary 6.11. Let A C X, a normed linear space. Let B, the weak closure of A, be compact in the weak topology. Then B is also sequentially compact in the weak topology. I

Proof. Let (an) be a sequence from B, and let sp(an) be the (norm) closure of the linear subspace spanned t by (an). By Proposition 6.19 sp(an) is weakly closed and therefore B () sp(an ) is a weakly compact subset of the separable normed linear space sp(an ). By Lemma 6.6, the weak topology on B () sp(an ) is metrizable and therefore sequentially compact by Proposition 1.22 of Chapter I. The rest is clear. I

Theorem 6.18. The Eberlein-Smulian Theorem. The following are equivalent for any subset A of a Banach space X:

(a) The weak closure of A is weakly compact. (b) Any sequence in A has a weakly convergent (in X) subsequence. (c) Every countable infinite subset of A has a weak-limit point in X.I

Proof. (a) =:> (b) by Corollary 6.11, and (b) =:> (c) trivially. We establish only (c) =:> (a). So we assume (c). Since x*(A) is a bounded set of

t Sometimes the subspace spanned by a set E is also denoted by IE].

Sec. 6.4 • Reflexive Banach Spaces and the Weak Topology 6S

scalars for each x* E X*, by Remark 6.3 A is bounded. If J is the natural map of X into X**, J(A) is bounded; and therefore by Theorem 6.16, w* (J(A», the weak * closure of J(A), is compact in the weak * topology of X** (i.e., the topology induced by X*). Since J is a homeomorphism from X (with the weak topology) onto J(X) (with the weak* topology), it is sufficient to show that w* {l(A» C J(X).

To show this, let x** E w*(j(A». We will use induction. Let Xl* E X* with II Xl* II = 1. Now there is a l E A with I (x** - J(al»(xl*) I < 1. Let E2 be the finite-dimensional subspace spanned by x** and x** - J(al ). Since the surface of the closed unit ball in E2 is compact, there are (yi**)f!2 E E2 with II Yi** II = I such that for any y** E E2 and II y** II = I, II y** - Yi** II < ! for some i. Let Xi* E X*, II Xi* II = I be such that Yi**(Xi*) > t, 2 1 II y** II· Again there exists a2 E A so that

Then we consider the space E3 spanned by x**, x** - J(al), and x** - J(a2);

we find (Xi*X!!ns+1 and then choose a3 E A as before so that for every y** E E3 , we have

max{1 Y**(Xi*) I: n2 1 II y** II

and

In this way, we continue to construct the sequence an' By (c), the sequence (an) obtained above has a weak-limit point x.

Clearly x E sp(an) and so x** - J(x) is in the space sp(x**, x** - J(lln) for I < n < 00). Therefore by the construction of (an) above, we have

Also

sup {I(x** - J(x»(xi*)I} > 1 II x** - J(x) II. (6.7) lsi <00

I [x** - J(X)](Xi*) I < I [X** - J(ap)](Xi*) I + I Xi*(ap - x) I < lip + I Xi*(flp - x) I , for i < np'

Since x is a weak-limit point of (an)' it follows that [x** - J(x)](x,*) = 0 for all i. By inequality (6.7), x** =-J(x). The proof is complete. I

Remark 6.21. By Theorems 6.17 and 6.18, a Banach space X is reflexive if and only if its closed unit ball is weakly sequentially compact. This is true even for any normed linear space (Problem 6.4.4).

An application of the Eberlein-Smulian Theorem characterizes weakly compact subsets of qo, I J. (See Problem 6.4.13.) The Banach-Alaoglu Theorem has already been utilized in obtaining Theorems 6.17 and 6.18. Another application of this theorem is outlined in Problem 6.4.14. In what follows, we present still another application of this theorem in the context ofweak*-sequential compactness and then use this result to solve a problem in harmonic analysis.

First, we prove the following useful result.

• Theorem 6.19. Let X be a normed linear space and S* = {IE X*: II I II < I}. Then X is separable if and only if the weak* topology on X* restricted to S* is a metric topology. I

Proof. Suppose X is separable. Let A C X be a countable dense subset of X. Then the weak* topology restricted to S* is the topology on S* that is induced by A; i.e., as in the proof of Theorem 6.16, the topology induced on S* is a subset of the set PA = ILeAlz , where Iz is the set of all scalars c with I c I < II x II, with product topology. Since PA , as a countable product of metric spaces, is metrizable, the "only if" part of the theorem follows.

To prove the "if" part, we assume that the weak* topology on S* is metrizable and therefore has a countable local base at O. Then there exist real numbers r n and finite subsets An C X such that

00

{O} = n Wn , Wn = {x* E S*: I x*(x) 1< rn' x E An}. n-l

Let A = U:'lAn. It is clear that x* = 0 whenever x* E S* and x*(x) = 0 for every x E A. Hence, by an application of the Hahn-Banach Theorem, it follows that the closed linear subspace spanned by A is X. Since A is countable, X is separable. I

In a metric space, compactness and sequential compactness are equivalent and therefore an application of the Banach-Alaoglu Theorem leads to the following

• Corollary 6.12. In a separable normed linear space X, the closed unit ball in X* is weak*-sequentially compact. I

5«. 6.4 • Reflexive Banach Spaces and tbe Weak Topology 67

We will now close this section with an application to a problem of harmonic analysis. The problem is to characterize all those operators in L(X, y), X = L1(R) and Y = L,.(R) where I < p < 00, which commute with convolution. Note that we have already introduced the notion of convolution in Problem 4.3.21. We recall thatfor/e LI andge L,. (1 < p $. oo),I*g is the convolution of I and g, and

f· g(x) ~ J f(x - y)g(y) dy .

• Theorem 6.20. Let Te L(X, Y), where X = L1(R), y = 4(R) and I (i) by Problem4.3.21, we prove only that (i) => (ii). We assume (i) and define the sequence

u,,(x) = (n /2)XI_I/".I/"I(X), n = I, 2, 3, .. ..

Then ~ U" ~ l = I and for any Ie Ce(R) (and so for Ie L1),

lim ~ u" * I - I ~ l = O . . -Hence

lim ~ T(f) - T(ulI).f ~" = lim ~ T(f) - T(u".f) V" .-< ! T II . lim ~ f - u.' fl. ~ o . . -

Since the sequence T(u,,) is bounded in L,.-norm, it follows (after considering L" as Lq*, pq = p + q) by Corollary 6.12 that some subsequence T(ullj) converges to some h in L" in the weak· topology of L". This means that for Ie LI and g e 4,. we have

J T(f)(x)g( - x)dx ~!'::: J T(u,).j(x)g( - x)dx

~ lim ([T(u.,)'fJ.g)(O) .-~ lim [T(u. ,) ' (f. g»)(O) .-~ lim J T(u,)(x)f'g(- x) dx .-

68

= f h(x)f*g(-x)dx

= [(h * f) * g](O)

Chap. 6 • Banach Spaces

= f (h*f)(x)g(- x)dx.

It follows easily that T(f) = h * f whenever fELl. We leave the details to the reader. I

Problems

X 6.4.1. Let X = {Xl' X2} and f-l be a measure on 2x such that f-l( {Xl}) = 1 and f-l({x2}) = 00. Show that dim LI{f-l) = dim LI*{f-l) = 1, whereas dim Loo{f-l) = dim Loo *{f-l) = 2. X 6.4.2. Prove that BV[a, b] is a Banach space under the norm II gil = V(g) + I g(a) I. Prove Lemma 6.3. X 6.4.3. Show that in a finite-dimensional normed linear space a sequence is convergent if and only if it is weakly convergent. X 6.4.4. Prove Remarks 6.15 and 6.21. X 6.4.5. Letfbe a real-valued measurable function in a a-finite measure space such that for all g in Lp (1 < P < 00), fg ELI. Show that fE Lq where lip + llq = I. What happens when the measure is semifinite? (Hint: Write X = u Xn, Xn C Xn+l' and f-l(Xn) < 00. Let fn(x) =

Xxn(x) inf{lf(x) I, n}. Define Tn(g) = f fng df-l. Use the uniform boundedness principle.)

6.4.6. Let S be a linear subspace of C[O, 1] which is closed as a subspace of L 2 [O, 1]. Show that S is finite-dimensional. (Hint: Show that S is closed as a subspace of qo, 1] and that therefore there exists k> 0 such that II f 1100 < k II! 112 for all fin S. Use this to show that the closed unit ball of S in L2 is compact.) X 6.4.7. Let g,fl>f2' ... ,fn be linear functionals on a vector space X such that nf=I{X: heX) = O} C {x: g(x) = O}. Show that g is a linear combination ofthef.. [Hint: Consider the mapping (g(x),h(x), ... ,fn(x»- (h(x), ... ,fn(x», which is injective.] X 6.4.8. Show that if X is reflexive and separable, then so is X*.

6.4.9. Show that if

fn(t) = nt,

= 2 - nt,

o <t< lin,

lin < t < 21n,

= 0, 21n < t < 1,

then j~ ~ ° in C[O, 1]; but In -f+ ° in C[O, 1]. 6.4.10. Let Y be a closed subspace of X. Show that (XI Y)* is linearly

isometric onto yo = {x* E X*: x*(y) = ° for all y in Y}. {Hint: Consider the mapping T(V*) = x* where x*(x) = V*([x]).}

6.4.11. Let Y be a closed subspace of X. Show that X is reflexive if and only if Yand XIY are reflexive. (Use Problem 6.4.10 for the "only if" part.) X 6.4.12. (i) If f is a linear functional on X, then show that f is continuous relative to the weak topology if and only if f E X*.

(ii) If g is a linear functional on X*, then show that g is continuous relative to the weak*-topology if and only if there is x E X such that g(x*) = x*(x) for every x* E X*.

6.4.13. Weak Compactness in qo, 1]. Prove that a subset E of qo, 1] is weakly compact if and only if E is weakly closed, norm bounded, and every sequence (fn) in E has a subsequence (fn,.) such that limHoofn,t(x) = f(x) for some f in E and all x E [0, 1]. Also, is a pointwise convergent and uniformly bounded sequence in qo, 1] weakly convergent?

6.4.14. Banach Spaces as Spaces of Continuous Functions. Show that given a real Banach space X there exists a compact Hausdorff space S such that X is linearly isometric onto a closed linear subspace of C(S) with uniform norm. (Hint: Take S = {x* E X*: II x* II < 1}, which is compact in the relative weak*-topology. Let f:r:(x*) = x*(x) for x E X and x* E S. Consider e(x) =h.)

6.4.15. Let X be an infinite-dimensional Banach space. Prove that the weak closure of {x E X: II x II = I} is the unit ball {x E X: II x II < I}. [Hint: Suppose that II y II < 1 and there is no x with norm 1 such that I Xi*(X - y) I >r > ° for i = 1,2, ... , n. Since for any nonzero x, there is a real t such that II y + Ix II = 1, the mapping x ~ (X1*(X), ... , xn*(x» is injective.] X 6.4.16. Weak Convergence in Lp, 1 < p < 00. Show that a norm bounded sequence Un) in Lp converges weakly to f in Lp if fn ~ f in measure.

6.4.17. Prove the following result due to E. Hewitt: Letfbe a realvalued measurable function in a semifinite measure space such thatf f/: LpCp) for some p > 1. Then the set {g E Lq : fg E L1 and J fg dp = O} is dense in Lq, where lip + llq = 1. (Hint: The set E = {g E Lq:fg E L 1} is dense in Lq since it contains all functions X..4., where

,u(A) < 00 and A C {x: n < I f(x) I < n + I}.

Then T, where T(g) = Ilg df-l, is not continuous, but linear on E. Now use Problem 6.2.3(b).)

6.4.18. Let X be a compact Hausdorff space. Let e: X ~ C(X)* be defined by e(x)(f) = I(x). Show that e is a homeomorphism from X onto e(X) C C(X)* (with weak*-topology). (Since X need not be sequentially compact, this shows that the Eberlein-Smulian theorem is false for weak*compact subsets.)

6.4.19. Show that a Banach space X is reflexive if and only if every total subspace of X* is dense in X*. (Hint: For the "if" part, suppose x** E X** - J(X). Then the subspace {x* E X*: x**(x*) = O} is not dense in X*. Show that it is total.)

6.4.20. Banach-Saks Theorem. Every weakly convergent sequence (In) in L2 [O, 1] has a subsequence (Ink = gk) such that the sequence (hn), hn = (l/nn:~~lgk> converges in L2 norm. (Hint: Consider a subsequence (Ink) such that for j> niH and 1 < k < i, II jj(x)lnk(x) dx I < 1/2i+l .)

This result remains true in L p , p > 1. The reader can later observe that the proof in L2 extends easily to any Hilbert space.

6.4.21. Prove that Ll(f-l) is weakly sequentially complete if f-l is a-finite. (Hint: If (In) is a weak Cauchy sequence in L l , then for each measurable set E, limn-+oo IE In df-l exists. Use Problem 4.3.16 to show that veE) = limn-+oo IE In df-l defines a bounded signed measure absolutely continuous with respect to f-l. Now apply the Radon-Nikodym theorem.)

* 6.4.22. Weak Convergence in Ll . Prove that in a a-finite measure space, a sequence (In) in Ll converges weakly to I in Ll if and only if sUPnllln 111 < 00 and for each measurable set E, the sequence IE In df-l converges.

* 6.4.23. Prove that X* is weak*-sequentially complete if X is a Banach space. [It is relevant to mention here that though Ll as well as any reflexive Banach space is weakly sequentially complete, a normed linear space X is weakly complete (that is, every weak Cauchy net converges weakly) if and only if X is finite-dimensional. A similar statement holds for the weak* topology of X*.]

6.5. Compact Operators and Spectral Notions

To prove some of Fredholm's results on integral equations, F. Riesz devised vector space techniques which easily extend and can be applied to a special class of linear operators called compact operators. These operators are very useful .and often find applications in classical integral equations as well as in nonsingular problems of mathematical physics.

Sec. 6.5 • Compact Operators and Spectral Notions 71

In this section we will derive basic properties of compact operators and then consider the Riesz-Schauder theory of such operators. The connection between the classical approximation problem for compact operators (by finite-dimensional operators) and the Schauder-basis problem in Banach spaces will be briefly discussed. We will finally introduce the spectral notions for a bounded linear operator on a Banach space and then consider briefly the spectral theory of compact operators.

Let X and Y be normed linear spaces over the same scalars.

Definition 6.15. A linear operator A from X into Y is called compact (or completely continuous) if A maps bounded sets of X into relatively compact (that is, having compact closure) sets of Y. I

Remarks

6.22. If A is compact then A is continuous. 6.23. If A E L(X, Y) and A (X) is finite-dimensional, then A is compact. 6.24. An operator A E L(X, Y) need not be compact. For example,

the identity operator on an infinite-dimensional normed linear space is not compact. (See Theorem 6.2.)

Example 6.15. Let A: C[O, I] ->- C[O, I] be defined by

Af(x) = J: k(x, y)f(y) dy,

where k(x, y) is a continuous function of (x, y) on [0, 1] x [0, 1]. The reader can easily check that if S = {f E qo, 1]: II f 1100 < I}, then A (S) is uniformly bounded and equicontinuous. By the Arzela-Ascoli Theorem A(S) is relatively compact and therefore A is compact.

The next few results give some basic properties of compact operators.

Proposition 6.20. Any finite linear combination of compact operators is compact. I

The proof is left to the reader.

Proposition 6.21. Let A and B be in L(X, X) with A compact. Then AB and BA are both compact. I

Proof. If S is a bounded set in X, then AB(S) = A (B(S)) is relatively compact since B(S) is bounded and A is compact. Also BA(S) C B(A(S)), which is compact since B is continuous and A(S) is compact. I

Proposition 6.22. Let A E L(X, Y) and (An)N be a sequence of compact operators from X into Y, where Y is complete and limn-+oo/l An - A /I = 0. Then A is compact. I

Proof. We will show that A (S), for S bounded in X, is totally bounded and hence relatively compact in Y, a complete metric space. Let 8 > 0. Then there is a positive integer n such that for each XES, /I An(x) - A(x) II < 8. Since An is compact, An(S) is compact and hence totally bounded. Therefore there exist Xl, X2 , ••• ,Xm in S such that, for each xE S,

inf II Aix) - An{xJ II < 8.

This means that for each XES

inf II A{x) - A(Xi) II < 38.

Hence A{S) is totally bounded. I

The following example shows that completeness of Y is essential in Proposition 6.22.

Example 6.16. Let B be the operator from co{ C Icc,) into 12 defined by

Let X = Co, Y = the range of B, and A E L(X, Y) be defined by A(x) = B{x). However, A is not compact. To see this, let

Xn = (1, 1, 1,: .. ,1,0,0, ... ) E co; n

then the sequence A (xn) = (l, 1/2, ... , l/n, 0, 0, ... ) converges to (l / k )kEN in 12 • However, (l/k)kEN $ Y, since if B(ak» = (l /k)then (1, 1, 1, ... ) E co, which is a contradiction. Hence the sequence A(xn) cannot have a convergent subsequence in Y or A is not compact. Nevertheless if we define An E L(X, Y) by

where 1 <k<n, k >n,

then the An's are compact and limn-+ooll An - A II = 0.

Proposition 6.22 above shows that the compact operators form a closed linear subspace of L(X, Y), when Y is complete. Also every operator in L(X, Y) with finite-dimensional range is compact. It is therefore natural to ask the following:

(A) Is every compact operator Tin L(X, Y) a limit in the norm of operators with finite-dimensional range?

This is the famous approximation problem in Banach spaces and formerly was one of the most widely known unsolved problems in the theory of Banach spaces. The approximation problem was studied in detail by Grothendieck. t He conjectured that (A) is not true in general. Only recently has it been solved in the negative by Per Enflo. t

This problem is really a problem of the structure of Banach spaces. To clarify this let us say that a Banach space Y is said to have the approximation property if for every compact set KeY and every e > 0, there is P E L( Y, Y), depending on K and e, with finite-dimensional range such that II P(x) - x II < e for every x E K. Now if Y has the approximation property, then the answer to (A) is affirmative. Indeed, if T is compact and K is the compact set T(Sx), where Sx is the closed unit ball of X, then dim PT(X) < 00 and II PT - T II < e.

Also it is clear that a Banach space has the approximation property if every separable subspace of it has this property. To see this, one has to consider the closed subspace (separable) YK spanned by the compact set K in the definition above and then use the Hahn-Banach Theorem to extend P from an operator in L(YK , YK ) to one in L(Y, Y). It can be easily verified that this is possible since the range of P is finite dimensional. This shows that (A) is related to the classical basis problem:

(B) Does every separable Banach space Y have a Schauder basis (Xi)~l (that is, can every y E Y be written uniquely in the form L~lAiXi)?

Actually, if Y has a Schauder basis, then Y has the approximation property. To see this, let (Xi)~l be the Schauder basis of Y and Pn E L(Y, Y) be defined by PnCL~lAiXi) = Lf=lAiXi' Then II Pn(x) - X 11-+ 0 uniformly as n -+ 00 on every compact subset of Y (see Problem 6.5.2), and, consequently, Y has the approximation property. Hence a positive answer to (B) must give a positive answer to (A). Per Enflo (in the paper mentioned above) solved (A) in the negative [and therefore also (B) in the negative]

t A. Grothendieck, Can. J. Math. 7, 552-561 (1955). t Per Enflo, Acta Math. 130, 3-4, 309-317 (t973),

by giving an example of a separable reflexive Banach space that does not have the approximation property.

In the next chapter, we will see that every separable subspace of a Hilbert space has a Schauder basis, and as such Problem (A) has an affirmative solution in a Hilbert space. A different proof of this fact will also be considered there.

We now leave Problem (A) and consider other basic properties of a compact operator.

Proposition 6.23. If A is a compact operator, then the range of A is separable. I

Proof. The range of A is contained in U:'lA(Sn), where Sn = {x: 1/ x II < n}. Since this is a countable union of compact sets, and a compact metric space is separable, the result follows. I

Proposition 6.24. Let A E L(X, y), where X is infinite dimensional and A is compact. Then A-I, if it exists, is not bounded. I

The proof is left to the reader. Let us now recall the definition of an adjoint operator (Problems 6.3.9,

6.3.12). If A E L(X, Y), then A* E L(Y*, X*) is defined by

A*(y*) = y* 0 A.

Then II A* II = II A II. The concept of the adjoint of a bounded linear operator A is useful in obtaining information about the range and inverse of A, as was outlined in Problems 6.3.12-6.3.15. There is a further duality between A and A* as the following theorem shows.

Theorem 6.21. Let A E L(X, Y). If A is compact, then A* is compact. Conversely, if A* is compact and Y complete, then A is compact. I

Proof. Let A be compact. It suffices to show that A*(W), for W bounded in Y*, is totally bounded in X*. Let e> 0 and S = {x EX: II x II < I}. Since A is compact, there exist Xl, X 2 , ... , Xn E X such that, for each XES,

inf II A(x) - A(Xi) II < e. (6.8) l:5i:5n

Let B be defined from y* into P by

B(y*) = (y* 0 A (XI), ... , y* 0 A(xn)),

then B E L( Y*, En). Hence B( W) is totally bounded. Thus there exist Yl*' ... , Ym* E W such that for each y* E Wand each i, 1 < i < n,

inf I y* 0 A(Xi) - y/ 0 A(Xi) I < e. (6.9) Isjsm

Now let y* E Wand let y/ for 1 <j< m be such that for each i, 1 <i< n,

(6.10)

Then

II A*(y*) - A*(y/) " = sup I y* 0 A(x) - y/ 0 A(x) I, IIxliSI

which is, by inequalities (6.8) and (6.9), less than or equal to (2M + 1 )e, where M is an upper bound for II y* II, y* E W. This proves that A*(W) is totally bounded and hence A* is compact.

Conversely, let A* be compact. By what we have just proved, AU is compact. Let (xn) be a bounded sequence in X. Then the sequence (J(xn)) is also bounded in X**, J being the natural mapping from X into X**. Since A** E L(X**, Y**) and A** is compact, there exists a subsequence [A**(J(Xnk))]kEN which converges in Y**. Now J1 (A(xnk)) = A**(J(xnk)), where J1 is the natural mapping from Y into Y**. Since J 1 is an isometry, (A(Xnk))kEN is a Cauchy sequence in Y and therefore must converge since Y is complete. This means that A is compact. I

Corollary 6.13. t If the range of a compact operator A is complete, then the range of A is finite dimensional. I

Proof. Let A E L(X, Y) be compact and let Z be the range of A. Then if Z is complete, A as an operator in L(X, Z) is compact, and therefore, by Theorem 6.21, A* is compact. Also, by Problem 6.3.15, A* has a bounded inverse. Hence by Proposition 6.24 Z* (and therefore Z) is finite dimensional. I

t When X is complete, one can prove this noting that A is open onto A(X) (by Theorem 6.8) and that the unit ball in A(X) is compact (by Proposition 6.9).

Next we present the famous Fredholm Alternative Theorem, which was developed by F. Riesz as a tool for the study of linear integral equations.

The proof of this theorem uses the properties of So (the annihilator of S) as given in Problems 6.3.12 and 6.3.13.

Theorem 6.22. The Fredholm Alternative Theorem. Let X be complete and K E L(X, X) be compact. Then R{I - K) is closed and dim N(I - K) = dim N(I - K*) < 00, where N denotes the null space, R the range, and I the identity operator. In particular, either R(I - K) = X and N(I - K) = {O}, or R(/- K)=I= X and N(/- K)=I= {O}. I

Proof. We will prove the theorem in several steps. We will write A for 1- K.

Step I. In this step we will show that R(A) is closed [or equivalently, there exists k > 0 such that for every x E X, d(x, N(A)) < k II A(x) II, where d denotes the usual distance]. (See Problem 6.3.11.)

Suppose R(A) is not closed. Then there exists a sequence (Xn)neN E X such that d(xn' N(A» = 1 and II A(xn) II ~ 0 as n ~ 00. Therefore, there also exists a sequence (zn) such that d(zn' N(A» = 1, II A(zn) II ~ 0 as n ~ 00, and I < II zn II < 2. Since K is compact, there exists a subsequence (Znt)ieN such that K(zn,) converges to some Z E X. Then Znj' which is A(zn) + K(znt)' also converges to z; or A(zn,) converges to A (z). Hence A(z) = 0 or Z E N(A). This is a contradiction, since for each i, d(zn, , N(A)) = 1 and zn, ~ Z as i ~ 00.

Step II. In this step we will show that dim N(A) < 00 and dim N(A*) < 00.

Consider a sequence (Xn)N in N(A) with II xn II < 1. Then for each n, A(xn) = 0 or K(xn) = Xn. Since K is compact, there exists a subsequence (Xn)ieN such that K(xn) = xn, is convergent. This means that the unit ball (closed) of N(A) (considered as a normed linear space) is compact, and therefore dim N(A) < 00. Similarly, dim N(A*) < 00.

Step Ill. In this step we will prove that dim N(A) = 0 if and only if dim N(A*) = O.

Suppose that dim N(A*) = o. Since R(A) is closed by Step I, R(A) = °N(A*) = X. (See Problem 6.3.12.) Suppose dim N(A) > O. Then there exists XI"# 0 such that A(XI) = O. Since R(A) = X, there exists a sequence (Xn)N such that A(xn+1) = xn. Now An+1(xn+1) = An(xn) = ... = A(xl ) = 0, and An(XnH) = An-l(xn) = ... = A(x2) = Xl =1= O. This means that for each positive integer n, N(An) isa proper closed subspace of N(An+I).

By Riesz's lemma (Proposition 6.1), there exists Zn E N(An) such that II Zn II = I and d(zn' N(An-l)) > 1/2. Since n > m implies

II K(zn} - K(zm) II = )) Zn - A(zn) - Zm + A(zm) II = II Zn - [zm - A(zm) + A(zn)] II

> d(zn' N(An-I»> 1/2,

we have a contradiction to the fact that K is compact. Hence dim N(A) = o. Conversely, suppose that dim N{A) = O. Then R{A*) = N(A)O = X*

(see Problem 6.3.13). Since A* = 1- K* and K* is also compact, by a similar argument to that above, dim N(A*) = o.

Step IV. In this step we will show that dim N(A) = dim N(A*), and this will complete the proof of the theorem.

Suppose dim N(A) = n > 0 and dim N(A*) = m > O. Let {Xl' X2' ... , xn} be a basis for N(A) and {XI*' X2*' ... , xm*} be a basis for N(A*). Then we claim the following:

(i) There exists xo* E X* such that XO*(Xi) = 0 for 1 < i < n, and

Xo*(Xn) * o. (ii) There exists Xo E X such that Xi*(XO) = 0 for 1 < i < m, and

xm*(xo)-=I= O.

(iii) If Al = A - Ko where Ko{x) = xo*(x) . Xo, then dim N{AI) = n - 1 and dim N(AI *) = m - 1.

Statement (i) follows easily. We prove (ii) by an inductive argument. Let p(k) be the statement "There exist {Zj: j = I, 2, ... , k} such that for 1 < t, j < k, Xi*(Zj) is equal to zero if i -=1= j and is equal to one if i = j." Clearly p(k) holds for k = 1. Suppose thatp(k) holds for k = m - 1. Then for 1 < i < m - I, we have

for each X EX. If for each X EX, Xm*(X - Li=""iIX;*(X)Zj) = 0, then Xm * = Li=~l xm * (zJx;* , a contradiction. Hence there exists zm' such that Xm *(zm') = 1 and for 1 < i < m - 1, Xi*(Zm') = O. Now letting z/ = Zj

- Xm * (Zj)zm' for 1 <j< m - 1, we have for 1 < i, j < m, Xi*(Z/) equals zero if i -=1= j and equals one if i = j. The argument is complete and (ji) is established.

To prove (iii), let x E N(Al). Then A(x) = Ko(x) = xo*(x)xo. Since Xo rf O[N(A*)] = R(A) (see Problem 6.3.12), xo*(x) = 0 and therefore x E N(A). So we can write x = 'LJ-lajXj for some scalars aj, and since

n 0= xo*(x) = L ajxo*(xj) = an . xo*(xn),

j-l

we have an = 0 or x = 'Lj:~ajxj. This means that dim N(A l ) = n - 1. To prove that dim N(Al*) = m - 1, let x* E N(Al*) or A*x* = Ko*x* = x*(xo) . xo*. Since xo* rf N(A)O = R(A*) (see Problem 6.3.13), x*(xo) = 0 or x* E N(A*). Therefore, we can write x* = 'L';lPjX/ for some scalars Pj. Then since x*(xo) = 0 and x*(xo) = Pmxm * (xo), Pm = O. This means that dim N(Al*) = m - 1.

Now the proof of the theorem will follow easily. If n < m, then dim N(Al) *- dim N(AI *). Repeating the above process a finite number of times, we end up with an operator An = 1- (K + Ko + ... + Kn- l ) = I - (a compact operator) such that dim N(An) = 0 and dim N(An *) > O. But this contradicts the result in Step III. Hence n > m. Similarly, n < m. This proves the theorem. I

At the end of this section, we will outline some applications of the above theorem in linear integral equations.

Now we consider briefly what is called the spectral theory of linear operators, which is the systematic study of various connections between T - AI and (T - A1)-1, where T E L(X, X), A is a scalar, and I is the identity operator. A large part of the theory of bounded linear operators is centered around their spectral theory. The most highly developed spectral theory is that for a class of operators called self-adjoint operators on Hilbert spaces (this will be discussed in depth in the next chapter).

If T is a linear operator on en, then the eigenvalues of T are complex numbers A such that (T - A1)X = 0 for some nonzero x. The set of all such A is called the spectrum of T. If A is not an eigenvalue, then T - AI has an inverse. The study of the spectrum of T often leads to a better understanding of the operator T. For example, if A = (aij) is the n X n matrix that represents T with respect to some basis in en and if A is Hermitian (that is, aij = aji), then it is well known from linear algebra that T can be represented with respect to some orthonormal basis {Xl' X2' ••• , xn} by a diagonal matrix so that T(Xi) = AiXi, i = 1, 2, ... ,n, for some scalars AI' A2 , ••• , An' This gives a very simple representation for such operators in

Sec. 6.S • Compact Operators and Spectral Notions 79

terms of their spectrum. The spectral theory of linear operators in infinitedimensional spaces is much more involved and extremely important in the study of these operators. (See, for example, for a generalization of the discussion above to the infinite-dimensional situation, the statement of Theorem 7.8 in the next chapter.) In what follows, we d.efine the spectrum of T and show that the spectrum of a bounded linear operator on a Banach space is nonempty, and for compact such operators, the spectrum is at most countable. At the end of this section, we give an application of these results to integral equations.

Definition 6.16. Let T: X ->- X be a linear operator on a nonzero complex normed linear space X. The resolvent set rAT) of T is the set

{A E C: R(T - AI) = X and T - AI has a continuous inverse}.

The complement in the complex plane of the resolvent set of T is called the spectrum of T and is denoted by a(T). The point spectrum Pa(T) consists of all scalars A in a(T) such that T - AI is not one-to-one. If A E Pa(T) then A is also called an eigenvalue of T and a nonzero x E X such that Tx = AX is called a corresponding eigenvector to A. The continuous spectrum Ca(T) consists of all scalars A in a(T) such that T - AI is one-to-one and R(T - AI) is dense in X but (T - AI)-l is not continuous as an operator from R(T -AI) onto X. The residual spectrum Ra(T) of T consists of scalars A in a(T) such that T - AI has an inverse (T - AI)-l which mayor may not be continuous but its domain R(T - AI) is not dense in X. I

We will deal with the continuous and residual spectrums when we study operators on Hilbert spaces later in the text.

Observe that if X is complete and T is closed, then when A E e(T) the operator (T - AI)-l is in L(X, X). This follows from Proposition 6.10 since (T - AI)-l is in this case closed and continuous and has therefore a closed domain equal to X.

We denote (T - AI)-l by R()., T) whenever (T - ).1)-1 E L(X, X).

Example 6.17. Let (X, d, ft) be a semifinite measure space. If G E Loo(ft) and TG E L(L2 , L 2 ) is given by TGJ = GJ, then

a(Ta) = {A I for all e>O,ft({x:IG(x)-).1 <e}»O},

the so-called essential range of G. (See Problem 6.5.21.)

Remarks

6.25. We note that when X is finite dimensional, A E aCT) if and only if T - U is not one-to-one, which is true if and only if A is an eigenvalue of T. Since the eigenvalues of T - U are the solutions of the equation

det (Tm - AI) = 0,

where I is the identity matrix and T m is the matrix representing T (with respect to some fixed basis of X), it follows that aCT) is nonempty. (Note that every nonconstant polynomial with complex coefficients has a complex root.)

6.26. aCT) can also be defined in real normed linear spaces X, but the difficulty is that aCT) can be empty even when X is finite dimensional. For example, let T be defined from R2 into R2 linearly by T( (I, 0») = (0, -I) and T(O, I» = (1,0); then aCT) is empty.

6.27. When X is infinite dimensional, there can be elements in aCT) that are not eigenvalues. For example, let X = 12 and T be defined on 12 by

Since for x E 12 II T(x) II = II x II, T is one-to-one and bounded so that ° is not an eigenvalue of T. But since (1,0, 0, ... ) rf the range of T, T-I is is not defined on X and therefore ° E aCT).

6.28. When X is infinite dimensional, aCT) can be uncountably infinite. For example, let X = 12 and let T be defined on 12 by

T(Xl' X2, ... » = (X2' X a , ... ).

Then T E L(X, X). If A is a complex number with I A I < I, then x = (I, A, A2, •.. ) E 12 and T(x) = (A, A2, .•• ) = AX. This means that a(T) :) {A.: I A I < I}. (The operator T is called the shift on 12.)

Next we present a basic result concerning a(T): for complex scalars, it is always nonempty. First we need the following result.

Proposition 6.25. Let X be complete and T E L(X, X). Then the resolvent set e(T) is open and if A, p, E e(T), then [writing RI' = R(p" T)]

Moreover, RA as a function from e(T) to L(X, X) has derivatives of all orders. (See Remark 6.4 for definition.) I

Proof. Suppose A E e(T) and I f-l - A I < 1 I II R;. II. Then T - f-lI = (T - AI)[I - (p, - A)R;.]. Now L::'o(p, - A)nR;.n is convergent in L(X, X), and it follows easily that [I - (p, - A)R;.]-1 = L::'o(p, - A)nR/'. Hence (T - f-l1)-1 E L(X, X) and f-l E e(T); consequently, e(T) is open. Also for A, f-l E e(T) we have

Rp - R;. = R;.[R,'l-1 - Rp-l]Rp

= R;.[T - AI - (T - f.11)]Rp

= (f-l - A)R;.Rw

Hence (Rp - R;.)/(p, - A) = R;.Rp or as f-l- A, (R;. - Rp)j(A - f-l) - Rl, in the L(X, X) norm. [Note that II Rp II < II R;. II (I - I A - f-ll II R;. 11)-\ when I A - f-l I is sufficiently small.] By induction, it can be shown that R .. has derivatives of all higher orders. I

Proposition 6.26. Let X be complete and T E L(X, X). Then I). I > II T II implies that A E e(T) and R;. = - L:=IA-nTn-l. Hence aCT) is a compact subset of the complex plane. I

We leave the proof to the reader. We use Proposition 6.25 and 6.26 to prove that a(T) is nonempty.

Theorem 6.23. If X is complete and T E L(X, X), then aCT) is not empty. I

Proof. We use Liouville's theorem from complex analysis and the Hahn-Banach Theorem. Let x E X and x* E X*. Then the complex-valued function x* (R;.(x») is,by Proposition 6.25, an analytic (or differentiable) function on e(T). By Proposition 6.26 for I A I > II T II, we have

00 II T Iln-l 1 II R;. II <,{;1 I). In = 1).1 (1 _ II T 111I ).1) - ° ,

as I A I - 00. This means that x* (R;. (x)) is a bounded function on the entire complex plane if aCT) is empty, since then e(T) is the entire complex plane. By Liouville's theorem, x*(R;.(x», being a bounded entire function. must be a constant (= 0, in this case). An application of Corollary 6.4 then asserts that for all x E X, R;,(x) = 0, which is a contradiction. I

Now we show that limn400 II Tn lI I1n exists and equals sup {I A I: A E a(T)}. Let C be any circle with origin as center and radius greater than

II T II. For x* E X* and x E X, we consider the complex-valued analytic function x* (Rix)) on e(T). Using the line integral over C taken in a suitable direction, we have from Proposition 6.26

f An. x* (Rix)) dA = - f X*(Tk-I(X)) . f An- k dA = 2ni· x*(P(x)). o k-I 0

It follows that if Co is any circle containing aCT) in its interior, then

x*(rn(x)) = + f ).n. x*(Rix )) dA, nl 0 0

where the line integral is taken in a suitable direction. This leads to the following

Proposition 6.27. Let X be complete and T E L(X, X). Let r(T) = SUPAEa(Tll A I. Then limn~ooll rn 111,n exists and is equal to r(T). [Here r(T) is called the spectral radius of T.] I

Proof. Let 10 > 0, C be the circle 1 A 1 = r(T) + e. Then for x* E X* with II x* II = 1 and x E X with II x II = I, we have

1 x*(rn(x)) 1 = I 2~i f 0 An . x*(Rix )) dAI

1 < 2n . [r(T) + e]n . 2n(r(T) + e) . M(C),

where M(C) = sUPAEcl1 RA II < =, since RA is a continuous function of A. This means that

lim sup II rn 111,n < r(T) + e. n-+oo

Since 10 > 0 is arbitrary, we have

lim sup II Tn 111,n < r(T). n~oo

We will now complete the proof by showing that

r(T) < lim inf II rn IIl/n. n~oo

To show this, we note that A E a(T) => An E a(Tn). The reason is that

Tn - An/= (T- U)A = A(T- U),

Sec. 6.S • Compact Operators and Spectral Notions

where n-l

A = )' AkT"-k-l, t=o

83

so that An E e(T") ==> T - A./ is bijective ==> A E e(T), by the Open Mapping Theorem. Hence, if A E a(T), by Proposition 6.26

I An I < II T" II, for each positive integer n. It follows that

r(T) < lim inf II T" pIn. n~

This completes the proof. I

Our next result is what is called the Spectral Mapping Theorem. This theorem answers a natural question, namely, when the equationp(T)(x) = y nas a unique solution for each y in X, where p is a polynomial, T E L(X, X), and peT) = 'L,7r-oakTk, when p(x) = I:7r-oakxk. Here TO == I. It is clear that the equation is solvable if ° is not in the spectrum of p(T). The Spectral Mapping Theorem answers the question more precisely: The above equation can be solved uniquely for each y in X if and only if no A in a(T) is a root of p. Here X is a Banach space.

Theorem 6.24. The Spectral Mapping Theorem. For T E L(X, X) and any polynomial p,p(a(T)) = G(p(T)). [Here p(G(T)) = {peA): A E G(T)}.] I

Proof. Let A E G(T). Since A is a root of the polynomial pet) - peA), we can write

pet) -- peA) = q(t) . (t - A) and

peT) - p(A)I = q(T)(T - A./) = (T - A./)q(T),

for some polynomial q(t). Now if peA) E e(p(T)), then T - A./ is bijective and therefore, by the Open Mapping Theorem, A E e(T). This proves that p(G(T)) C G(p(T». For the opposite inclusion, let A E G(p(T)). Suppose that AI' A2 , ••• , An are the complex roots of pet) - A. Then

c=i=o.

If each Ai is in e(T), then for each i, (T - Al)-1 E L(X, X) and therefore, by the above, A E e (p(T)), which is a contradiction. Thus, one of the Ai must be in G(T). Since this Ai is a root of pet) - A, it follows that A E p( G(T)). The proof is complete. I

Our last theorem in this section (preceding the applications) concerns the spectrum of a compact operator T on a Banach space X. The spectrum of such operators is at most countable and contains 0 when X is infinite dimensional; this follows from Proposition 6.24 and the next result.

Theorem 6.25. Let T be a compact operator on a Banach space X. If A "* 0, then A E (l(T) or A is an eigenvalue of T. Moreover, a(T) is at most countable and 0 is its only possible limit point. I

Proof. Suppose A "* 0 is in the spectrum of T. If T - AI is not oneto-one, then clearly). is an eigenvalue of T. If (T - ).1)-1 is bounded, then by Theorem 6.22 R(T - AI) = X and A E (l(T). On the other hand, if (T - AI)-l exists as a function but is not bounded, then by Proposition 6.5, for each positive integer n, there exists an Xn in X with II Xn II = I and with II (T - AI)xn II < lin. This means TXn - Axn ~ 0 in X. Since T is compact, TXn has a convergent subsequence (TXn')kEN converging to Y in X. Since also TXnk - AXnk ~ 0, it is clear that ).xnk ~ y. Since T is continuous,

Ty = lim T(Axnk) = ). lim T(xnk) = ).y. k-+oo k-+oo

Since y"* 0 as II y II = limHoo II AXnk II = I A Ilimk-+oo II xnk II = 1 ). I, A is an eigenvalue of T.

To prove the rest of the theorem, it is sufficient to prove that for any

B > 0, the set Pe = {).: I A I > B and A E aCT)}

is finite. Suppose this is false. Then there exists B > 0 such that p. is infinite. Let (AiX:l be a sequence of distinct eigenvalues in p., with (Xi)~l the corresponding eigenvectors. Since eigenvectors corresponding to distinct eigenvalues are linearly independent, t the subspace Xn spanned by {Xl' X2, ... , Xn} is properly contained in Xn+1 spanned by {Xl' X2, ... ,Xn+1}' By Riesz's result (Proposition 6.1), there exist Yn E Xn with II Yn II = I and infxExn_lll Yn - X II > 1/2. We write Yn = :Li:.laixi; then

n n AnYn - T(Yn) = L ai).nXi - L aiAixi

i=l i=I

n-l

= L ai(An - ).i)Xi E Xn- l · i-I

t ~::'lCiXI = 0 => ~ CiA-IX, = 0 = T(~ C,Xi)' The linear independence of the XI'S follows by induction.

Therefore, for n > m,

which is a contradiction to the compactness of T. The theorem follows. I

We conclude this section with some applications of the preceding theory to the study of integral equations.

• Remark 6.29. Applications to Integral Equations: The Dirichlet Problem. A number of problems in applied mathematics and mathematical physics can be reduced to equations of the type

f(s) - ;. f: K(s, t)f(t) dt = g(s), (6.11 )

where K(s, t) is a complex-valued Lebesgue-measurable function on [a, b) X [a, b) such that

II K 1122 = f: f: 1 K(s, t) 12 ds dt < 00,

J, g E L2 [a, b). Here;' is a nonzero complex number and f is the unknown function. These equations are usually called Fredholm equations of the second kind and the function K is called the kernel of the equation.

In what follows, we will apply Theorems 6.22 and 6.25 to solve the problem of existence of solutions of equation (6.11); the results on integral equations will then be useful in studying a fundamental problem of mathematical physics-the Dirichlet problem.

First, we define the operator T by

Tf(s) = f: K(s, t)f(t) dt, (6.12)

It follows from Fubini's Theorem (Theorem 3.7) and the Holder Inequality (Proposition 3.13) that, for almost all s,

1 Tf(s) 12 < f: 1 K(s, t) 12 dt . f: 1 f(t) 12 dt

and therefore

(6.13)

Thus, T is a bounded linear operator on L2 • We claim that T is compact.


To prove our claim, we assume with no loss of generality that K is a continuous function of (s, t). This is possible since by LlJsin's Theorem (see Problem 3.l.l3) we can approximate the kernel by a continuous kernel in L 2-norm and then Proposition 6.22 applies. Now by Theorem 1.26, a continuous kernel K(s, t) can be approximated uniformly by kernels Kn(s, t) of the form Lf-1Ui(S)Vi(t). If we define

Tnf(s) = f: Kn(s, t)f(t) dt,

then

n fb Tnf(s) = &;:1 uM) a vi(t)f(t) dt.

This means that Tn is finite dimensional and therefore, by Remark 6.23 Tn is compact. By the same argument as used in obtaining (6.13), we have

It follows that limn-+= II T - Tn II = O. By Proposition 6.22, T is compact. Now we write equation (6.11) as

(I - ).T)f= g (6.14)

or, equivalently, (T - ).-lI)f = _ ).-lg. (6.15)

Taking ).T as the operator K in Theorem 6.22, we obtain easily the following.

Theorem A. Either the equation (6.14) has a unique solution

f = (I - )'T)-lg

for each g E L2 or the homogeneous equation

f(s) - ). f: K(s, t)f(t) dt = 0 (6.16)

has a nonzero solution fin L 2 • In the latter case, the number of linearly independent solutions of equation (6.16) is finite. I

Now an application of Theorem 6.25 gives us immediately the following


Theorem B. The equation (6.16) can have nonzero solutions for at most countably many values of A. If there is an infinite sequence (An) of such values, then I An 1-->-00. I

To find some more information on equation (6.11), we need to find the adjoint of T. For h E L2, let h* denote the linear functional on L2 defined by

h*(f) = f: f(t)h(t) dt,

Then it can be verified by a simple computation that T*h* = g*, where g* is the linear functional induced by g as above and g is given by

g(s) = f: K(t, s)h(t) dt. (6.17)

Another application of Theorem 6.22 leads to our next result.

Theorem C. The equation

f(s) - ;: f: K(t, s)f(t) dt = 0 (6.18)

and equation (6. 1 6) have the same number oflinearly independent solutions. Moreover, if A-I is an eigenvalue of T, then equation (6.11) has a solution in L2 for a given g in L2 if and only if

J: g(t)f(t) dt = 0,

whenever f is a solution of equation (6.18). I

Note that this last result follows since (J - AT)(L2 ) = O[N(l- AT*)].

We remark that the preceding Fredholm theory is also valid if the kernel K(s, t) is a continuous function on G X G, where G is a compact set in Rn, and the operator T acts on C(G).

We now consider the Dirichlet Problem, a fundamental problem in mathematical physics and one of the oldest problems in potential theory.

The Dirichlet Problem. This is the first boundary-value problem of potential theory. The problem is to find a harmonic function on an open connected set E in Rn, which is continuous on E and coincides with a given continuous function g on the boundary of E. The problem originates in the


study of various physical phenomena from electrostatics, fluid dynamics, heat conduction, and other areas of physics. During the last hundred years or so, this problem has been studied by many celebrated mathematicians including Dirichlet, Poincare, Lebesgue, Hilbert, and Fredholm, and many different methods have been discovered for solving this problem. Though this problem is not solvable for all domains E, the existence of solutions has been proven in many important cases. In what follows, we shall consider only the two-dimensional case and show how the Fredholm theory can be applied in showing the existence of a solution under certain general assumptions. We shall assume (without proving) several facts from potential theory, our intent being to give the reader only an idea of the applicability of the Fredholm theory. For a detailed discussion of the Dirichlet Problem, the reader can consult Partial Differential Equations. t

Let E be an open connected set in R2 bounded by and in the interior of a simple closed curve C with continuous curvature [i.e., points of C have rectangular coordinates xes), y(s) (in terms of arc length s), possessing continuous second derivatives]. A function u(x, y) on E with continuous second-order derivatives and satisfying the equation

a2u a2u Ju = ax2 + ay2 = 0

in E is called a harmonic function on E. Let f be a continuous function on C. Then it is a fact from potential

theory that the function

v(p) = fof(t) a:, 10gCP~tl)dt (6.19)

is a harmonic function in E as well as in (E)c. Here a/ant represents the derivative in the direction of the interior normal nt at t. For SEC, let us write

and

v-(s) = lim vet) t-+. teB

v+(s) = lim vet). t-+, t~j

It is known that these limits exist and the following equalities are valid. [Note that the integral in equation (6.19) defines V(p) even when p E C.]

t P. R. Garabedian, Partial Differential Equations, John Wiley, New York (1967).

Sec. 6.S • Compact Operators and Spectral Notions 89

(i) v-(s) = v(s) + nf(s);

(ii) v+(s) = v(s) - nf(s); (6.20)

(iii) the normal derivative of v is continuous on C.

The reader can find detailed proofs of similar equalities in Garabedian's book.

It is clear from equations (6.19) and (6.20) that the function u(t), given by

u(t) = { vet), v-(t),

tEE, tEe,

will be a solution of the Dirichlet problem if we find a solution f of the equation

_I g(s) = f(s) + I K(s, t)f(t) dt, n 0

(6.21 )

where

1 fJ ( 1 ) K(s, t) = - ~ log I I n une s - t

and g is defined as the given continuous function on the boundary. Here one can show by straightforward computations and by using the continuous curvature of C that K(s, t) is a continuous function of (s, t), even when s = t. To apply the Fredholm theory, we consider the homogeneous equation

f(s) + Io K(s, t)f(t) dt = 0 (6.22)

and show that this equation does not have any nonzero continuous solution. To show this, we note that any continuous solution f of equation (6.22) will define, as in equation (6.19), a function F harmonic in E as well as in (E)e such that by equation (6.20), F-(s) = F(s) + f(s) = 0, sEC. Since a harmonic function is known to assume its maximum and minimum values on the boundary, the function F(t) = 0 for all tEE. This means that (fJF/fJn)- = 0 on C, and by equation (6.20) (iii), (f}F/f}n)+ = 0 on C. Now using the harmonic property of Fin (E)e, it can be proven by using the classical divergence theorem (or Green's first identity) that

II [(:~r+ (:;rJ dxdy = - Io F+(t) (::r dt = o. Ii?)C

90

This means that

of =0= of ax oy

Cbap. 6 • Banach Spaces

and therefore F is constant in (BY. Since F, because of its representation as in equation (6.19), is known to be zero at infinity, F+(s) = 0, SEC and by equation (6.20)

f(s) = ![v-(s) - v+(s)] = 0, sE C.

This proves that equation (6.22) cannot have any nonzero solutions. Since an analog of Theorem A holds also for continuous kernels and for operators T acting on C(B), the following result is immediate.

Theorem D. For every continuous function g given on the boundary C, th; Dirichlet problem has a solution. I

Problems

X 6.5.1. Prove Proposition 6.20. X 6.5.2. Suppose (Xi):l is a Schauder basis for a Banach space X. Let

Show that II Pn{x) - X II - ° uniformly on every compact set K as n - 00.

X 6.5.3. Prove Proposition 6.24. X 6.5.4. Let T be the operator

T(f)(x) = f: K(x, y)f(y) dy

from L1I [0, I] into Lq[O, I], where lip + Ilq = I, 1 < p < 00, and K{x, y) E Lq([O, I] X [0, I]). Show that T is compact. (Hint: First prove the result when K is continuous; then approximate K by continuous functions and use Proposition 6.22.) X 6.5.5. Show that a compact operator T E L(X, Y) maps weakly convergent sequences onto convergent sequences. X 6.5.6. (i) Let T be a bounded linear operator on a reflexive Banach space. If T maps weakly convergent sequences onto convergent sequences, then show T is compact.

(ii) Show that the result of (i) may not be true in a nonreflexive space. (Hint: Look at 11')

6.5.7. Consider the following Fredholm integral equation:

f(x) = g(x) + ..t J: K(x, y)f(y) dy,

where g E L 2 [0, I] and K E L 2([0, 1] X [0, I)). Prove that if g = ° implies f = 0, then there exists a unique solution of the equation for any g E L 2 [0, 1].

6.5.8. Let X be a compact metric space and f.l be a finite measure on it. Let K(x, y) be continuous on Xx X, and suppose the only continuous solution of

f(x) =..t f K(x, y)f(y) df.l(Y)

is f = 0. Prove that for every continuous function g(x) on X, there exists a unique continuous solutionf(x) of the integral equation in Problem 6.5.7.

6.5.9. Consider the Volterra integral equation

f(x) = g(x) + J: K(x, t)f(t) dt, ° <x< 1,

where K(x, t) is continuous on [0 1] x [0, 1]. Prove that for any continuous function g, there exists a unique continuous solution f of the Volterra equation.

6.5.10. Let T E L(X, X), X a complex Banach space. Show that e(T) = e(T*) and R(A, T*) = [R(A, T)]*.

6.5.11. Let X, Y, and Z be Banach spaces, K E L(X, Y) and T E L(Z, Y). If K is compact and T(Z) C K(X), then prove that T is compact. (Hint: Let N = K-I({O}). Then Ko, defined by Ko(x + N) = K(x), is a compact operator from XI N into Y.)

6.5.12. Weakly Compact Operators. A linear operator mapping bounded sequences onto sequences having a weakly convergent subsequence is called weakly compact. Prove the following: (i) Weakly compact operators are continuous. (ii) If T E L(X, Y) and either X or Y is reflexive, then Tis weakly compact. (iii) If T is the operator from LI[O, 1] into Lp[O, 1], 1 <p < 00, defined by T(f)(x) = J~ K(x, y)f(y) dy, where K(x, y) is a bounded measurable function on [0, I] x [0, I], then T is weakly compact. (Hint: Use Problem 6.4.22 for p = I.)

6.5.13. Let TE L(X, Y) be compact and Z = T(X). Define To: X ---+ Z by To(x) = T(x). Is To compact? What if X is reflexive? What if

Z is closed in Y? What happens if "compact" is replaced by "weakly compact"?

6.5.14. If T E L(X, X) is compact and T + U is invertible in L(X, X), prove there exists a scalar Ao and a compact operator To such that (T + Al)-I = To + Ao/.

6.5.15. Prove that the set of invertible elements of L(X, X), X a Banach space, is open. [Hint: If T is invertible let S E L(X, X) have II S II < II T-I 11-1. Show that T + S = T[I + T-IS] and that (I + T-IS)-I = L:'o( -1 )n(T-Is)n.]

6.5.16. Banach Algebras. A Banach algebra is a complex Banach space A on which an operation of multiplication is defined satisfying for all a, b, and c in A and A E C the properties (ab)c = a(bc); a(b + c) = ab + ac; (b + c)a = ba + bc; A(ab) = (Aa)b = a(Ab); and II ab II < II a 1111 b II. If there is an element e in A such that ae = ea = a for all a in A and II e II = 1 then A is said to have a unit e. If ab = ba for all a, b in A, then A is said to be commutative.

(i) Show L(X, X), X a complex Banach space, is a commutative Banach algebra with a unit.

(ii) Show CI(X), the space of continuous complex valued functions on a compact Hausdorff space, is a commutative Banach algebra with a unit.

(iii) If A is a Banach algebra with a unit, define for each a E A, the mapping Ma in L(A, A) by the rule Mib) = abo Show that the mapping a ~ Ma is a linear isometry from A into L(A, A) which also preserves multiplication.

(iv) If A is an in (iii), the spectrum set a(a) of a in A is {A I a - Ae has no inverse in A}. [a - Ae has an inverse if there exists b in A such that b(a - Ae) = (a - Ae)b = e.] Prove a(a) = a(Ma)' Conclude that a(a) is nonempty and compact.

(v) Gelfand-Mazur Theorem. If A is a Banach algebra with unit and if every nonzero element in A has an inverse, then A is isometrically isomorphic to the algebra of complex numbers. Prove this. (Hint: Show that for each a in A there is exactly one Aa such that a - Aae = 0.) Conclude that A is commutative.

6.5.17. (i) If X is a complex Banach space and if T E L(X, X), prove that there is a maximal closed subspace M of L(X, X) that contains T, contains I, is closed with respect to composition, and on which composition is commutative. (M is an example of a commutative Banach algebra with unit.)

(ii) Prove a(T) = aM(T), which is defined to be {A I T - U has no inverse in M}.

Sec. 6.6 • Topological Vector Spaces 93

6.5.1S. This is a continuation of Problem 6.5.17. (i) If W is a proper ideal in M (that is, a proper subspace of M such

that ABE W whenever A E M and B E W), prove that W is contained in a maximal ideal 2 which is closed (topologically) and 2"* M. [Hint: Use Zorn's lemma to get a maximal ideal 2 which does not contain I. Since Z is also an ideal and the set of invertible elements in L(X, X) is open (6.5.15), 2 = Z.]

(ii) Prove M /2 (see Section 6.1) is an algebra in which every nonzero element has an inverse. [Hint: Let (/): M --+ M/2 be the natural map (/)(A) = [A]. Let Vo be the ideal {ATo + S , A E M, S E 2} for some To E M"-..2. Then Vo = 2 so (/)(ATo + S) = (/)(J) for some A and S or [A ][Tol = [I].

(iii) Prove M /2 is isometrically isomorphic to the algebra of complex numbers. [Hint: Gelfand-Mazur, 6.S.16(v).]

(iv) Prove aCT) = {h(T) ,h is a homomorphism (preserving scalar multiplication, addition, and multiplication) from M onto C}. [Hint: If A E aCT) = aM(T), then {(T - AI)S ,·S EM} is a proper ideal W of M contained in a maximal proper ideal 2 of M. Let h = i 0 (/): M --+ M/2 --+ C. Then h(T - AI) = O. Conversely, if A tf; aCT), then for some S E M, (T - AI)S = I. Hence for every homomorphism h from M onto C, h(T - AI) ::;t: O. Here i is the map from (iii).]

6.5.19. Using (ii) of Problem 6.5.17 and (iv) of Problem 6.5.18, prove the following spectral mapping theorems.

(i) If T E L(X, X) and p is any polynomial of one complex variable, then a(p(T») = p(a(T»).

(ii) If T and S are commuting operators in L(X, X) and p(x, y) is a complex polynomial in two variables, then a(p(S, T)) C p(a(S), aCT)) == {pea, fJ): a E a(S), fJ E a(T)}.

6.5.20. Let T be a linear operator on a normed linear space X. If A is an eigenvalue of T, let M). = {x E X: Tx = Ax}, the eigenspace of T corresponding to A. Show that if T is closed, then M). is a closed linear subspace of X which is invariant under T, that is, T(M)) eM)..

6.5.21. Verify Example 6.17. [Hint: It is sufficient to prove To is not invertible if and only if ,u {x " G(x) , < E} > 0 for all E > 0.]

6.6. Topological Vector Spaces

A natural extension of normed linear spaces are vector spaces where the topology is induced by a family of norms, instead of one norm. Such generalizations include many important examples such as the weak topology in a normed linear space X and the weak *-topology in X*, and provide

proper framework for the consideration of various applications. Vector spaces with such a topology are special cases of a more general class of spaces, called topological vector spaces. The main purpose of this section is to give a brief introduction of topological vector spaces, leading to the basic properties of locally convex spaces and a discussion of the KreinMilman theorem and one of its applications in measure theory.

Let X be a vector space over F (the field of real or complex numbers) with a topology T such that the maps

It: XxX--+X and 12: FxX--+X,

defined by

It(x, y) = x + y and 12({J, x) = {Jx,

are continuous, where X X X and F x X have the usual product topologies. For convenience, 0 will denote both the zero vector and the scalar zero. The meaning will be clear from the context. The vector space X with topology T is called a topological vector space or a TVS, in short. In a TVS, the translation map x --+ x + Xo is a homeomorphism, and thus, a local base t at 0 for T gives automatically a local base at any other point. The following proposition gives properties that characterize a base for a topology for a TVS.

Proposition 6.28. Let X be a TVS. Then there is a local base 9J at 0 having the following properties:

(i) If G and H are in c95', then there is A E c95' such that A e G n H. (ii) If G E.'-J5' and x E G, then there exists H E.~ such that x +

He G. (iii) If G E .Sil, then there exists HE cY.1 such that H + He G. (iv) If x E X and G E .Y.1, then there exists {J E F such that x E {JG. (v) If G E.YiJ and 0 < I t I < 1, then tG e G and tG E 9J.

Conversely, a given collection.Sf1 of subsets of a vector space X containing o and satisfying the above properties is a local base at 0 for a topology T

that makes X a TVS. The topology T is Hausdorff if and only if

(vi) n {G: G E .Sf1} = {O}. I

Proof. Let X be a TVS with topology T. Let VET and 0 E V. Since 0.0 = 0 and scalar multiplication is continuous, there exists u > 0 and

t A local base at 0 is a family of open subsets containing 0 such that each open set containing 0 contains a member of the family.

WE T, 0 E W such that IW C V, whenever I I I <u. If U = uW, then o E U, U E T and IU C V for I I I < 1. (Note that we have used the fact that multiplication by a nonzero scalar is a homeomorphism so that U E T.)

Let G = u {tU: I t I < I}. Then, 0 E G C V, GET and IG C G for I t I < 1. Let!?4 be the collection of all such G. Then the reader can verify that properties (ii), (iii), and (iv) follow from the continuity of addition and scalar multiplication and the observations that x + 0 = x,O + 0 = 0, and Ox = o. Property (i) follows from the definition of topology and property (v) from the definition of !?4.

Conversely, let!?4 be a collection of subsets of a vector space X containing 0 and satisfying properties (i)-(v). Define T to be the collection of arbitrary union of sets from the family

!T = {x + G: x E X and G E !?4}.

Note that if Z Ex + G E ?; then by (ii) there exists U E!?4 such that Z + U C x + G. Thus, if Z E (x + G) n (y + H), G and H in !?4, then there exists [by (i) and (ii)] Uo E!?4 such that Z + Uo C (x + G) n (y + H). Thus, !T fulfills the conditions to be a base for a topology T. We need to show that in this topology addition and scalar multiplication are continuous.

If x + y = z and V E!?4, then by (iii) there exists U E !?4 such that (x + U) + (y + U) C z + V; thus, addition is continuous in T. To prove the continuity of (f3, x) ---+ f3x, let us take first f3 = O. Given U E !?4, there exists V E!?4 such that V + V C U. By (iv) and (v), there exists 0 < u < 1 such that for I I I < u, Ix E V and therefore, I(X + V) C V + tV C V + V C U. Now let f3 * O. Let WE!?4. By (iii), there exists W' E !?4 satisfying (v) such that W' + W' C W. Let Wo = (1/1 + I (31)W'. Then Wo E!?4 and there exists 0 < u < 1 such that for I I I <u, Ix E W' and (f3 + t)Wo C W'. Thus, for I t I < u, we have

(f3 + t)(x + Wo) = (f3 + t)x + (f3 + I) WO C f3x + W' + W' C f3x + W.

Finally, we consider the Hausdorff property of T. Suppose that (vi) holds and x * y. Then there exist WI and W2 in !?4 such that x - y f/; W2

and WI + WI C W2 • Since -WI = WI' (x + WI) n (y + WI) = 0. Thus, T is T2. The converse is immediate, since x * 0 and the T2 property imply that there exists V' E!?4 such that x f/; V'. I

For further discussions on topological vector spaces, we need a definition.

Definition 6.17. Let X be a vector space over F and A C X. The set A is called balanced if tA C A for 1 t 1 < I. The set A is called absorbing if given x E X, there exists u > 0 such that tx E A, whenever 1 t 1 < u. I

Notice that in a TVS X, if V is a balanced open set and if given x E X, there exists some nonzero 10 E F such that loX E V, then for 1 t 1 < 1 10 I, Ix = (t Ilo)tox E (t Ito) V C V, and consequently, V is absorbing. It follows from Proposition 6.28 that in a TVS, 0 has a local base consisting of balanced and absorbing sets.

Let us now consider two important examples of topological vector spaces which will also lead us to the study of a special important class of these spaces, called locally convex spaces. First, let !'T be a family of linear functionals on a vector space X. Then, the weakest topology on X which makes each member in !'T continuous has a local base at 0 consisting of sets of the form

{x: 1.Ii(x) 1< r, i = 1,2, ... , n} (6.23)

where r > 0 and each.li belongs to !T. This base has the properties (i)-(v) of Proposition 6.28 and therefore, induces a topology that makes X a TVS. Note that the sets in (6.23) are convex [that is, for all IE (0, I), tx + (I -t)y are in the set whenever x and yare]. This topology is called the weak topology on X induced by !T. As for the second example, let C1(S) be the complex-valued continuous functions on some Hausdorff topological space S. Let us define the seminorm t PK on C1(S) for each compact subset K C S by

PK(X) = sup {I x(s) I: S E K}.

Then the collection of sets of the form

{x E C1(S): PKi(X) < r, i = 1,2, ... , n} (6.24)

induces a Hausdorff topology on C1(S) making it a TVS. The sets in (6.24) are again convex. This topology is called the topology of uniform convergence on compact sets.

Definition 6.18. A TVS X with topology T is called locally convex if T has a local base at 0 consisting of convex sets. I

t A seminorm n on a vector space X over F is a nonnegative real function such that for x, y in X and fl E F, n(flx) = I fll n(x) and n(x + y) ::; n(x) + n(y). Thus, a seminorm is a norm if and only if n(x) = 0 implies that x = o.

A normed linear space with its norm topology is locally convex. It follows from the proof of Proposition 6.28 that in a locally convex space (that is, a locally convex TVS), 0 has a local base consisting of balanced, convex, and absorbing sets. The reason is that in this proof the convex hull of the set G, when V is convex and open, is open, balanced, absorbing, and contained in V. In the second example above, the family of seminorms {PK: K is a compact subset of S} induced a topology [meaning the topology induced by the sets in (6.24)] making C1(S) a locally convex space. The same is the case for all locally convex spaces, as the following theorem shows.

Theorem 6.26. Let X be a TVS. Then X is locally convex if and only if the topology T of X has a local base at 0 given by the sets

Nfh ,{J2, ... ,{3n;r = {x: P{J/x) < r, i = 1,2, . , . ,n},

where (P{J)' {J E A (A an indexed set), is a family of seminorms and r is a positive real number. I

Proof. The "if" part is obvious since the sets N{Jl,{32'''' ,{3n; r are convex. For the "only if" part, let.9iJ be the local base at 0 consisting of balanced, convex, and absorbing sets. For each U E .9iJ, let us define the Minkowski functional Pu : X -+ [0,00) by

Pu(x) = inf{{J: (J > 0 and x E {JU}.

Then Pu can be verified to have the following properties:

(i) Pu({Jx) = I (J I Pu(x) for (J E F.

(ii) Pu(x + y) < Pu(x) + Pu(Y),

(iii) {x: Pu(x) < r} C rU and belongs to T, r > O.

(6.25)

(6.26)

(6.27)

(6.28)

Thus, Pu is a seminorm by (i) and (ii). It is clear from (iii) that the topology induced by the family of seminorms {Pu: U E 93'} is the same as the original locally convex topology. I

It is always important to know if a particular topology is metrizable, since metric spaces have many nice properties. The next theorem considers this question for a locally convex space.

Theorem 6.27. Let X be a locally convex space. Then the following are equivalent:

(i) The topology of X is given by a pseudometric. (ii) 0 has a countable local base.

(iii) The topology of X is generated by an at most countable family of seminorms.

Moreover, the topology of X is given by a seminorm if and only if the family of semi norms determining the topology of X has a finite subfamily (i.e., finitely many semi norms from the same family) determining the same

~~~. I

Proof. (i) ~ (ii) by Remark 1.79. Also, (ii) ~ (iii), since if ,~ is a countable local base at 0 consisting of convex, balanced and absorbing sets, then the seminorms {Pu: U E ,Y6'} [defined as in (6.25)] will generate the same topology as ,9,5'.

Now let us assume that (Pn) is a countable family of seminorms generating the topology of X. Define

(6.29)

Then P is a pseudometric. Notice that for x, Y in X,

Pn(X + y) --'-"-'-----..,.---"-'--,- 0, a> fJ if and only if a/(\ + a) > fJ/(\ + fJ), and therefore, it follows that IimH.=p(xi, y) = 0 if and only if for each n, limH.=Pn(xi -y) = O. Thus, P induces the same topology on X as that induced by

(pnr:~I. For the second part of the theorem, we observe that the topology in

duced by the seminorms {PI' P2, ... , Pn} is the same as that induced by the seminorm L~~IPi. Now suppose that the topology of X is generated by the family of semi norms {Pi3: fJ E A}, and that it is also generated by the seminorm p. Then there exists u > 0 and fJI, fJ2, ... , fJm in A such that

{x: ppJx) < u, i = 1,2, ... , m} C {x: p{x) < 1}. (6.30)

We claim that, for all x in X,

(6.31 )

If the claim is false, there exists z in X such that

1 m p(z) > r > - L P{3t(z),

U i-I (6.32)

for some real number r. Thenp{3j(z/r) < U, i = 1,2, ... ,m. Butp(z/r) > I. This contradicts (6.30) and the claim (6.3\) follows. It follows that the seminorms {P{3t: i = I, 2, ... , m} generate the same topology as does p. I

We now turn to the three basic principles of functional analysis discussed in Section 6.3 in the context of topological vector spaces. We need a definition.

Definition 6.19. A complete metrizable TVS is called a Frechet space. I

In Frechet spaces, the open mapping theorem,. the closed graph theorem, and the principle of uniform boundedness hold in the following forms.

Theorem 6.28. Let X and Y be Frechet spaces and T: X -+ Y be a closed linear onto map. Then T is open. I

(As before, the map T above is called closed if its graph is a closed subset of Xx Y with usual product topology.)

Theorem 6.29. A closed linear map from a Frechet space into anoth-er is continuous. I

Theorem 6.30. Let X and Y be two complete metrizable locally convex topological vector spaces. Let fl' be a family of continuous linear maps from X into Y such that for every continuous seminorm p on Yand every x in X, the set {p(T(x»): TEfl'} is bounded. Then for each such p, there is a constant C > 0 and a continuous seminorm q on X such that for all x in X and T E !l':

p(T(x») < Cq(x). I

The proofs of Theorems 6.28, 6.29 and 6.30 are analogous to those of Theorems 6.8, 6.9, and 6.11, respectively, and will not be given here. However, we prove below two other forms of uniform boundedness principles in topological vector spaces. We need another definition.


Definition 6.20. A set A in a TVS X is called bounded if given any open neighborhood U of 0, there exists u > 0 such that uA C U. I

Note that in a normed linear space, a set is bounded if and only if it is bounded in the norm.

Theorem 6.31. Let X be a complete metrizable TVS and Y be a TVS. Let?' be a pointwise bounded family of continuous linear maps from X into Y. Then?' is equicontinuous. t I

Proof. Let V C Y be an open neighborhood of O. Then there is an open and balanced neighborhood U of 0 such that U + U + U + U C V. Let W be the set defined by

W = {x E X: T(x) E [j for all T in !l7}. Then W is closed and balanced. Let Z E X. The set {T(z): T E!17} is a bounded subset of Y and therefore, there exists {3 > 0 such that

{3 {T(z): T E!17} C U.

Hence, {3z E Wand therefore, W is absorbing. Thus, X = U;;"~!nW. Since X is a complete metric space, there is a positive integer n such that n W is not nowhere dense. It follows that W contains an open set G. Let x, YEW

and T EfT Then T(x - y) E 0 - [j C U + U + U + U C V, since [j C U + U. Thus, the set H = G - G is an open neighborhood of 0 and for all T E :?: T(H) C V. I

Theorem 6.32. Let X be a complete metrizable TVS and Y be a TVS. Suppose that (Tn) is a sequence of continuous linear maps from X into Y such that limn-+ooTn(x) exists for all x E X. If T(x) = limn-+ooTn(x),

then T is continuous. I

Proof. By Theorem 6.31, the family (Tn);:! is equicontinuous. Let W be an open subset of Y containing O. Let V be an open neighborhood of o in Y such that V + V C W. By Theorem 6.31, there is an open neighborhood U of 0 in X such that for each n, Tn(U) C V. Then for any x in U,

T(x) = T(x) - Tn(x) + Tn(x) E V + V C W for sufficiently large n, since T(x) - Tn(x) --+- 0 as n --+- CXJ. I

t ~ is called equicontinuous if given any open set V c Yand 0 E V, there exists an open set U c X, 0 E U, and T(U) c V for every TE c'iT.

An important feature of locally convex spaces is that in such spaces there exist many continuous linear functionals, and consequently many useful results hold in such spaces. One such result is the famous KreinMilman theorem. In what follows, we will present first some useful results concerning the existence of continuous linear functionals on locally convex spaces, and then, to conclude this section we will present the Krein-Milman theorem and one of its many applications.

We start with an extension of the Hahn-Banach theorem to locally convex spaces.

Theorem 6.33. Let X be a locally convex space and Y C X be a linear subspace. Let f be a continuous linear functional on Y. Then there is a continuous linear functional g on X such that for y E Y, g(y) = fey)· I

Proof. Let (P{J){JEA be a family of seminorms determining the topology of X. The relative topology on Y is given by the restrictions of these seminorms to Y. Thus, there exists u > 0 and fJI, fJ2' ... , fJn in A such that

{YE Y:P{J/Y) <u,i= 1,2, ... ,n}C {yE Y:If(y) I < I}.

Let P = (l/u) L~IP{Ji. Then the above set inclusion implies that for each y E Y, I fey) I <p(y). By Theorems 6.4 and 6.5, there exists a linear functional g on X such that g extendsfand for alI x E X, I g(x) I <p(x). Clearly g is continuous. I

Definition 6.21. Let X be a vector space over F. Iff is a nonzero real linear functional on X regarded as a vector space over the reals, then the set {x E X:f(x) = r}, where r is a real number, is called a hyperplane. I

Thus, a subset L in X, a vector space over the reals, is a hyperplane if and only if L = x + W, where x E X and W is a maximal linear subspace (that is, a proper subspace W such that the subspace spanned by W U {z}, z EX - W, is X). (See Problem 6.2.20.)

Definition 6.22. In a TVS X, the sets U and V are said to be separated by a hyperplane if for some continuous real-valued functional f on X (regarded as a vector space over the reals) and some real number r,

UC {XE X: f(x) < r} and VC {XE X: I(x) > r}.

In the case

U C {x E X: f(x) < r} and V C {x E X: f(x) > r},

we say that U and V are strictly separated. I

The above two definitions will now allow us to present a theorem which shows, among other things, that in a locally convex Hausdorff space for any two distinct points x and y there always exists a continuous linear functional f such thatf(x) * fey).

Theorem 6.34. Let U and V be disjoint convex sets in a locally convex space X. Then the following assertions hold:

(a) If U is open, then U and V can be separated by a hyperplane. (b) If U and V are both open, then they can be strictly separated by

a hyperplane. (c) If U is compact and Vis closed, then they can be strictly separated

by a hyperplane. I

Proof. We regard X as a locally convex space over the reals. Let U be open, x E U and y E V. Let W = Wo - Wo, where Wo = U - V + {y - x}. Then W is open (and therefore, absorbing), convex, and symmetric. Since ° E W, W is also balanced. Consider the Minkowski functional Pw as defined in (6.25). Let Y = {a(y - x): a E R} and f(a(y - x)) = a for a in R. Thenfis a linear functional on the subspace Yand for each z E Y, fez) < Pw(z) since y - x rt Wand Pw(y - x) > I. By Theorem 6.4, f can be extended to a linear functional g on X such that for each z E X, g(z) < Pw(z). Since W is an open neighborhood of ° and for each z E W, Pw(z) < I, it is clear that W C g-l [-1, 1] and g is continuous. Notice that for x' E U and y' E V, x' - y' + y - X E Wo C W, and therefore, since g(y - x) = 1 and g(x' - y' + y - x) < 1, we have g(x') < g(y'). Thus,

sup {g(x'): x' E U} < inf{g(y'): y' E V}, (6.33)

and the proof of part (a) is complete. To prove (b), let U and V be both open. Consider the nonzero functional

g above. Then g( U) is open. To see this, let x E U. If g(x) = 0, then since g is nonzero, there exists y E Uo C U such that g(y) * 0, where Uo is balanced. It follows that the interval (-g(y),g(y))C g(Uo)C g(U). If g(x) * 0, then there exists {J > ° such that {(I + t)x: I t I < {J} C U so that the interval ({I - {J)g(x), (I + {J)g(x)) C g( U). This proves that g( U) is


open. Similarly, g(V) is also open. By (6.33), g(U) n g(V) is a singleton, if nonempty. Since g(U) n g(V) is open, it must be empty.

To prove (c), let U be compact and V be closed. Then for each x E U, there is a convex open set Gx such that 0 E Gx and (x + Gx) n V = 0. Since U is compact, there exist Xl' X 2 , ••. , Xn in U such that U C U7~1 (Xi

+ iGxJ If G = n?~l iGxt , then (U + G) n V = 0. This means that (U + iG) n (V - iG) = 0. Since U + iG and V - iG are disjoint open convex sets containing U and V respectively, the proof of (b) now applies. I

We now present the celebrated Krein-Milman theoremt and give J. L. Kelly's proof of it. The Krein-Milman theorem is an abstract analog for compact convex sets in locally convex spaces of the fact that in two dimensions any point inside or on a triangle determined by three vertices can be expressed as a convex linear combination of these vertices. For its proof, we need a definition and a proposition.

Definition 6.23. Let X be a TVS over F and K C X. A non empty subset A C K is called an extremal subset of K if the following holds: If X

and yare both points of K and the point fJx + (1 - fJ)y is in A for some fJ in (0, I), then X and y must be points of A. An extremal subset of K consisting of exactly one point is called an extreme point of K. I

The vertices of a triangle in R2 are the extreme points of the triangle, whereas its sides are examples of its other extremal subsets. Note that an open triangle in R2 or an open solid sphere in R3 has no extreme points, whereas for a closed solid sphere, its surface is an extremal subset and every point on it is an extreme point. It follows immediately from the definition that if in a TVS X, Be A C K, B is an extremal subset of A and A is an extremal subset of K, then B is an extremal subset of K.

Proposition 6.29. Let X be a locally convex Hausdorff space over F and K C X be a nonempty compact subset. Then K has extreme points. I

Proof. Let!;T be the family of closed extremal subsets of K. Then !;T is nonempty since K E!;T. Partially order !;T by inclusion. Consider a linearly ordered subfamily {Ap: Ap E :r. fJ E B}. Then the set n {Ap: fJ E B} is nonempty (and closed) since K is compact and the Ap's have finite intersection property. (See Remark 1.31.) This intersection is also an ex-

t M. Krein and D. Milman, Studia Math. 9, (1940).

tremal subset of K. Thus, it is an upper bound for the subfamily. By Zorn's Lemma, !7 contains a maximal element A. Suppose that A has two distinct points z and w. By Theorem 6.34, there exists a nonzero real-valued continuous linear functional 1 on X (as a TVS over the reals) such that I(z) </(w). This means that the set D defined by D = {XE A:/(x) = SUPyeA I(y)} is a nonempty closed proper subset of A. (Note that A is compact and therefore 1 attains its supremum in A.) Also, it follows from the definition of D that D is an extremal subset of A, and therefore, an extremal subset of K. This contradicts the maximality of A. Hence, A must be a singleton, and therefore, is an extreme point of K. I

We need one final definition in this section.

Definition 6.24. Let X be a TVS over F and E C X. Then the convex hull of E is defined as

co(E) = { t aixi: ai > 0, t ai = I; Xi E E, i = I, 2, ... , n, I < n < oo}. \=1 \-1

The closure of co(E) is called the closed convex hull of E, and denoted by oo~. I

We may remark that co(E) = () {A: A J E, A convex} and co(E) = () {A: A J E, A closed convex}, since co(E) (co(E), respectively) is a convex (closed convex, respectively) set containing E.

Theorem 6.35. The Krein-Milman Theorem. Let X be a locally convex Hausdorff space and K a compact subset of X. Let E be the set of its extreme points. Then K C co(E). If K is also convex, K = co (E). I

Proof. Let X E K - co(E). By Theorem 6.34, there is a nonzero real continuous linear functional 1 on X (as a TVS over the reals) and real numbers T, s such that

I(x) < r < s < inf{f(y): y E co(E)}.

Let Ko = {z E K: I(z) = infyeKI(y)}. Then Ko is a closed extremal subset of K and Ko () E = 0. Since an extreme point of Ko is an extreme point of K, Ko has no extreme points, and this contradicts Proposition 6.29. I

Finally, we give an application of the Krein-Milman theorem to measure theory. We present below J. Lindenstrauss' prooft of the Liapounoff convexity theorem using Theorem 6.35.

Theorem 6.36. Let fll' fl2' ... ,fln be finite nonatomic (positive) measures on some measurable space (X, .JY"). Then the set of points in Rn

of the form (fll(A), fl2(A), ... , fln(A»), A Ed, is a closed and convex subset of Rn. I

Proof. We use induction on n. Before we give the proof for n = I, it is helpful to carry out the induction step. (As we will see, the proof for n = I will be clear from that of the induction step.) Write fl = fll + fl2

+ ... + fln' and let W = {g E L=(fl): 0 <g < I}. Define T: W -+ Rn by T(g) = (f g dfll , f g dfl2, ... , f g dfln). Let W be given the weak *-topology of Loo(fl) as the dual of LI(fl). Then W is compact by Theorem 6.16, Tis continuous, and T( W) is compact and convex. We claim that T( W) =

((fll(A), fl2(A), ... , fln(A»): A E L()I'}. To prove this, let 13 = (131,132, ... , f3n) E T(W). Then the set Wo = T-I(f3) is a weak*-compact convex subset of We Loo(fl). Since Loo(fl), with the weak*-topology, is a locally convex Hausdorff space, it follows by Theorem 6.35 that Wo has extreme points. Let g be an extreme point of WOo Suppose that there exists u > 0 and a subset A E.JY" such that fll(A) > 0 and u < g(x) < I - u for x E A. We will show that this will contradict the fact that g is an extreme point of Wo, thus proving that g is of the form XlI, HE JV:

Since fll is non atomic, there exists BEd, Be A such that fll(B) > 0 and fll(A - B) > O. By induction hypothesis, replacing X by B, we see that there exists C C B, C E d, such that

Similarly, there exists DE.JY" such that DCA - Band

Now we choose real numbers sand t such that

Is I <u, I t I <u, and

t J. Lindenstrauss, J. Math. Meeh. 15 (6), (1966).

Let h = s(2Xc - XII) + t(XA -II - 2Xn). Then h ¥= 0 in Loo(f-l), since f-l(C) > 0 and on C, h = s. [If s = 0, then h = -Ion D, f-l(D) > 0 and 1 of:. 0.] Notice that

Ihl <u<g< I -u< l-Ihl, on A,

and h = 0 on X-A. This means that g + hand g - h are both in W. Since f h df-l'i = 0 for i = I, 2, ... ,n, the functions g + hand g - h are both in Wo. This contradicts the assertion that g is an extreme point of Wo, since

g = (g + h)j2 + (g - h)j2.

This means that g is a measurable function of the form XH and T(g) = f3 = (f-lI(H), f-l2(H), ... ,f-ln(H)). This proves the claim and completes the induction step. The reader can now furnish the proof for n = I (following the induction step). I

Problems

X 6.6.1. Let T: X ~ Y be a linear map from a TVS X into a locally convex TVS Y. Suppose that the topology of Y is given by a family {pp: f3 E B} of semi norms. Prove that T is continuous if and only if Pp 0 T is continuous for each f3 E B.

6.6.2. Give examples of two distinct locally convex topologies on a vector space X with the same class of continuous linear functionals. X 6.6.3. Show that a locally convex TVS is Hausdorff if and only if the family of semi norms {pp: f3 E B} generating its topology is total, i.e., x E X is 0 whenever pp(x) = 0 for each f3 E B.

X 6.6.4. Prove that a maximal subspace of a TVS is either closed or dense.

X 6.6.5. Prove that a TVS is regular. [Hint: Observe that for any subset B, B = n {B + U: U is open and 0 E U}.] X 6.6.6. Let X be a TVS. Show that X is Ta if and only if {O} is closed in X if and only if given x of:. 0, there exists an open set U such that 0 E U and x ft U.

X 6.6.7. Bounded sels in a TVS. A set S in a TVS X is called bounded if for every open set U containing 0, there exists e > 0 such that eS C U. Prove the following assertions:

(i) A compact set in a TVS is bounded. (ii) The closure of a bounded set in a TVS is bounded.

(iii) In a locally convex TVS, the convex hull of a bounded set is bounded.

X 6.6.8. Prove that if a TVS has a bounded convex open set containing 0, then its topology is given by a seminorm. (Hint: Use the seminorm Pv, where V is a bounded, balanced, and convex open set containing 0.) X 6.6.9. Let X be a locally convex TVS and {Pfl: fJ E B} is a family of seminorms generating its topology. Prove that a subset seX is bounded if and only if Pj3(S) is bounded for each fJ E B.

6;6.10. Consider the space [p, 0 < p < I. Notice that II x II = (L:I / Xi /P)I1P does not define a norm, since for 0 < p < I, II x + Y II may exceed II x II + II Y II. However, if we define

00

d(x, y) = L / Xi - Yi /P, i~I

where x = (Xi) and Y = (Yi) are in [p, then (/p, d) becomes a complete metric TVS. Show that a set in lp is bounded as a bounded set in a TVS if and only if it is bounded in the metric d. Also show that the convex hull of {x: d(O, x) < I} is not bounded in d. [Hint: The sequence (lIn, lin, ... , lin, 0, 0, ... ) = xn , the first n entries of Xn being lin, is not bounded in d. By Problem 6.6.7, (/p, d), 0 < p < I, is not locally convex.]

6.6.11. Let X be a vector space. Suppose that dI and dz are two complete metrics such that (X, dI ) and (X, dz) are topological vector spaces. If dl is stronger than d2 , then show that they are equivalent. (Hint: Use the open mapping theorem.)

6.6.12. Show that the product of an arbitrary family of topological vector spaces, equipped with the product topology, is a TVS.

6.6.13. Show that an infinite product of Hausdorff topological vector spaces (each of positive dimension) cannot have a bounded absorbing set. [Hint: Let X = TIj3EB Xj3 and nj3: X -+ Xfl be the projection on Xj3' If A C X is absorbing, then nj3(A) = Xfl for all but finitely many fJ.]

6.6.14. Let X be a vector space over F and p a seminorm on X. Prove the following assertions:

(i) The sets A = {x: p(x) < I} and B = {x: p(x) < I} are balanced and convex.

(ii) If p' is another seminorm on X and {x: p(x) < I} = {x: p'(X)

< I}, then p = p'. {Hint: Let e > 0 and Y = [p(x) + e]-Ix .}

(iii) If X is a TVS and p is continuous at 0, then B is the interior of A and A = E.

X 6.6.15. Let T: X -+ Y be a linear map from a TVS X into a TVS Y. Show that if T( V) is bounded for some open set V containing 0, then T is continuous. Further show that if Y has an open and bounded set containing 0, then T( V) is bounded for some open set V containing 0, whenever T is continuous.

6.6.16. Prove that the product of infinitely many normed linear spaces is not a normed linear space.

6.6.17. Let X be a TVS. Prove the following assertions: (i) A semi norm P on X is continuous if and only if {x: p(x) < I}

is open. (ii) Suppose that a family !7 of semi norms generate the topology

of X. Then a seminorm p on X is continuous if and only if there exists

M > ° and PI, P2' ... , Pn in!7 such that

p(x) < M max{pi(x): 1 <i< n},

for all x EX. (iii) Let X be as in (ii). Then the family of continuous semi norms on

X also generates the topology of X.

6.6.18. Let X be a TVS and M a linear subspace of X. Then the set XjM = {x + M: x E X} of co sets of M is a vector space over F if we define vector addition and scalar multiplication by

(x + M) + (y + M) = x + y + M, fJ(x + M) = fJx + M.

Let n: X -+ XjM be defined by n(x) = x + M. Then n is linear and onto. Let XjM be given the quotient topology, i.e., the largest topology on XjM such that n is continuous. Prove the following assertions:

(i) Xj M is a TVS. (ii) Xj M is Hausdorff if and only if M is closed.

(iii) n is open. (iv) A local base ,~ of ° in X generates a local base {n( U): U E ,'%'}

of ° in XjM.

6.6.19. Let X be a complete metrizable locally convex TVS and !7 be a pointwise bounded family of continuous semi norms on X. Show that there exists an open neighborhood U of ° such that U C {x: p(x) sup {p(x): P E !7}. Therefore, V cannot be nowhere dense. Note that tv - tv = V.]

6.6.20. Prove that the product of an arbitrary number of locally convex spaces is locally convex.

6.6.21. Let X be a TVS and Y, Z linear subspaces of X such that X = Y + Z and Y n Z = {o}. Note that each x in X has a unique representation x = y + z, where y E Y and Z E Z. Prove that X is the topological direct sum of Y and Z (that is, the mappings (y, z) -+ y + z and x = y + z -+ (y, z), where y E Y and Z E Z, are continuous when Y x Z is given the product topology) if and only if the mapping y + Z -+ Y is continuous, and in this case, Z is called the topological supplement of Y.

6.6.22. If X is a Hausdorff TVS over F with finite dimension d, then show that every vector space isomorphism I: Fd -+ X is also a homeomorphism. X 6.6.23. Let X be a Hausdorff TVS over F. Show that X is finite dimensional if and only if X is locally compact.

6.6.24. Let X be a locally convex TVS over F. Consider the product space F' = TIpEBFp, where each Fp is F. Show that every continuous linear mapping I: Y -+ F', where Y is a linear subspace of X, can be extended to a continuous linear mapping from X into F'. (Hint: Let :rt:p: F' -+ Fp be the coordinate projection so that :rt:{3 01: Y -+ F is a continuous linear functional on Y.)

6.6.25. Let Y be a linear subspace of a TVS X. Prove that Y possesses a topological supplement (see Problem 6.6.21) if anlonly if there is a continuous linear map p: X -+ X such that pop = p and p(X) = Y.

6.6.26. Let X be a locally convex Hausdorff TVS over F. Prove that every finite-dimensional linear subspace Y of X has a topological supplement. (Hint: Let I: Y -+ Fn be a linear homeomorphism. Use Problem 6.6.24 to extend I to a continuous linear map 10 on X. Then p = 1-1 010: X -+ X is a continuous linear map with range Y such that pop = p.) X 6.6.27. Let I: X -+ F be a nonzero linear functional on a TVS X over F. Prove that I is continuous if and only if 1-1(0) is closed if and only if 1-1(0) is not dense.

6.6.28. Let X be a TVS over the reals and U, V be subsets such that U has nonempty interior. Suppose that U and V are separated by a nonzero linear functional I on X. Prove that I is continuous. [Hint: Use Problem 6.6.27. If U C {x: f(x) < r} and V C {x: I(x) > r}, then I-1(r) n (Int U) = 0.]

6.6.29. Do Problems 6.2.24 and 6.2.25 when X is a TVS over the reals.

6.6.30. Prove that the set of extreme points of a compact convex set in R2 is closed. Show that this is not always true in R3. * 6.6.31. Show that in a uniformly convex normed linear space the set of extreme points of the closed ball {x: II x - XO II < r} is the set {x: II x - XO II = r}, which is its surface.

6.6.32. Let X be a vector space over the reals and let.9J be defined by

.9J = {V C X: 0 E V, V is convex such that for any x EX,

there exists fJ > 0 such that fJx E V}.

Prove the following assertions:

(i) ,';[iJ is a local base for a topology on X, making X a locally convex TVS.

(ii) The topology induced by ,qg is stronger than any other locally convex topology on X.

6.6.33. Recall that c is the space of all real sequences (xn) such that lim Xn exists, endowed with the norm II Xn II = sup I Xn I, and Co is the subspace of sequences converging to zero. Show that the unit ball of Co does not have extreme points, whereas the unit ball of c has exactly two extreme points, namely, (I, I, I, ... ) and (-I, -I, -I, ., .).

6.6.34. Use the Krein-Milman theorem to show that neither c nor Co can be isometrically isomorphic to the dual of any normed linear space.

6.6.35. Let I < P < 00. Show that the set of extreme points of the closed unit ball of Lp[O, I] is {f: II I lip = I}. (See Problem 6.6.31.)

6.6.36. Show that the closed unit ball of Ll [0, I] has no extreme points. [Hint: If I has zero norm, then I is not an extreme point. Since I = t(2 -II I Ill)! + i II I Ih f, the norm of an extreme point must be I. Let II I 111 = I and 1= IXA + IXB, where m(A) > 0, m(B) > 0, and A U B = {x: I(x) -::j::. O}. Notice that I = fJlt + (I - fJ)J;., where fJ = II IXA 111 ,It = IXA/II IXA 111

and 12 = IXB/II IXB 111'] 6.6.37. Prove that the set of extreme points of the closed unit ball of

Loo[O, I] is the set {f: I I(x) I = I a.e. in [0, I]}. 6.6.38. Find the extreme points of the closed unit ball of lp, I <p

<00. * 6.6.39. Find the extreme points of the closed unit ball of C(S), the Banach space of real continuous functions with supremum norm, where S is a compact Hausdorff space.

6.6.40. Totally Bounded Subsets in a TVS. Let X be a Hausdorff TVS. A subset A is called totally bounded if for every open neighborhood V of 0, there exists a finite subset {Xl' X 2 , ••• ,xn } of A such that A C Uf=l(Xi

Sec. 6.7 • Kakutani Fixed Point Theorem, Haar Measure 111

+ V). Prove the following assertions: (i) A totally bounded subset of X is bounded.

(ii) Let Y be a TVS, Z a proper closed subspace of Y, and U a bounded open neighborhood of O. Let V be a balanced open neighborhood of 0 such that V + V C U. Then U is not contained in Z + V. {Hint: Suppose that U C Z + V. Then V + V CUe Z + V implies that 2n V C Z + V for each positive integer n. Then VC nm21[Z + (ljm)V] = Z.}

(iii) Let 0 E U C A, where U is open and A is a totally bounded subset of X. Then X is finite dimensional. [Hint: Let X be infinite dimensional. Use (ii) to construct an infinite sequence (xn) in A such that Xn+l if'. V + Wn , where V is a balanced open subset, 0 E V, V + V C A, and Wn is the subspace spanned by {Xl' X2 , ••• , xn}. Then Xn - Xm if'. V if n # m. Let V' be a balanced open neighborhood of 0 such that V' + V' C V. There exist Yt, Y2, ... ,Yk such that A C U~~I(Yi + V'). But then there are m, n such that xn - Xm E V' + V' C V.]

(iv) Every subset of a totally bounded subset A of X is totally bounded.

(v) Let AI, A 2 , ••• , An be a finite number of compact convex subsets of X. Then the convex hull of A = Ur~l Ai is also compact. [Hint: Let Pj > 0, Lj~]Pj = 1, and Xi E Ai for 1 < i < n. Consider the mapping (p], ... , Pn, Xl' ... , Xn) --+ Lr~IPiXi' which is continuous as a map from a compact space into a Hausdorff space.]

(vi) Let X be a locally convex Hausdorff TVS and A a totally bounded subset. Then the convex hull of A is also totally bounded. [Hint: Let 0 E V, V open and convex. Let A C Ur~](Xi + V). If Y E coCA), then there exists z E co {Xl' X2 , ••• , Xn} == K, say, such that Y - Z E V. Thus, coCA) C K + V, and by (v), K is compact.]

(vii) Let A and X be as in (vi). Then the balanced hull of A, i.e., the set {px: X E A and I P I < I } is also totally bounded.

6.7. The Kakutani Fixed Point Theorem and the Haar Measure on a Compact Group

The theory of Haar measure is an important branch of measure theory and constitutes an extremely useful generalization of the theory of Lebesgue measure.

The Haar measure is a translation-invariant measure on a locally compact topological group [i.e., an algebraic group with locally compact


topology where the mappings (x, y) -+- x . y and x -+- X-I are continuous]. The foundations of the theory of topological groups were laid around 1926-1927 by O. Schreier and F. Leja. To study the structure of certain topological groups, D. Hilbert in 1900 posed the following problem (now famous as Hilbert's fifth problem-fifth in the list of 23 problems he posed at the International Congress of Mathematics): Is every topological group that is locally Euclidean (i.e., every point has an open neighborhood homeomorphic to an open subset of Rn) necessarily a Lie group [i.e., a manifold that is a group and where the mappings (x, y) -+- xy and x -+- X-I

are analytic]? In 1933 A. Haar took a fundamental step towards the solution of this problem. He established the existence of a translationinvariant measure (now known as the Haar measure) on a second countable locally compact topological group. Soon after, in the same year, von Neumann utilized Haar's result and solved Hilbert's fifth problem in the affirmative for compact locally Euclidean groups. He also proved the uniqueness of the Haar measure in 1934, and later on in 1940 A. Wei! extended Haar's result to all locally compact topological groups. Hilbert's fifth problem was solved completely (in 1952) in the affirmative by A. Gleason, D. Montgomery, and L. Zippin.

In this section we will present a fixed point theorem of S. Kakutani t and utilize it to prove the existence of the Haar measure on a compact topological group.

First we need a definition.

Definition 6.25. A family !T of linear operators on a topological vector space X is called equicontinuous on a subset K of X if for every open set V with 0 E V there is an open U with 0 E U such that if kl , k2 E K and kl - k2 E U, then !T(k1 - k 2) C V, i.e., T(kl - k 2) E V for every T E!iT. I

Theorem 6.37. The Kakutani Fixed Point Theorem. Let K be a nonempty compact convex subset of a locally convex Hausdorff space X, and let!T be a group of linear operators on X. Suppose!T is equicontinuous on K and !T(K) C K.t Then there is x E X such that T(x) = x for every TE!T. I

Proof. By Zorn's Lemma there is a minimal nonempty compact convex set Kl such that.r(K1) C K 1 • If Kl is a singleton, there is nothing to

t S. Kakutani, Proc. Imp. Acad. Tokyo 14, 242-245 (1938). t ."r(K) = u {T(K): Te .?"'}.

prove. If Kl is not a singleton, we will reach a contradiction to the minimality of K1 , and the theorem will follow.

Suppose there are ZI #- Z2, ZI, Z2 E K1 • Let y = Zl - Z2' Then there are open sets U, V with 0 E U n V such that V is convex and balanced,

y$ P, (6.34)

and

Let Uo = co(7(U)). Then Uo is open. Using the group property of .r, the reader can verify that

7(Uo) = Uo and 7(Do) = Do·

Since 7 contains the identity operator, U C Uo and inf{t: t > 0 and Kl - Kl C tUo} = to < 00. Let W = toUo. Then it is clear that, for each e in (0, 1),

(6.36)

and (6.37)

Since 0 E W, Kl C UkeK1(k + ! W); and therefore by the compactness of K1 , there exist k i E Kl with 1 < j < n such that

n Kl C U (k; + ! W).

;=1

Now we claim that

is a nonempty compact convex set satisfying

(i) 7(K2) C K2 ,

(ii) K2 # K1 •

(6.38)

This claim, once verified, will prove the theorem since K2 will contradict the minimality of K1 • We prove the "non empty" assertion last.

To prove (i), let Z E K2 and k E K1 • Since for T E 7, T-I(K1) C K1 ,

then k = T(ko) for some ko E K1 • Also Z E ko + (I - 1/4n)W or T(z) E k + (1 - 1/4n) . W, thus proving (i). Now to prove (ii), by (6.36) th~re are x, y E KI such that x - y $ (I - I/4n)W. This means that x rIO y +

(I - 1/4n)W or x I/: K2, thus proving (ii). To prove K2 c:F- 0, we show that p = (lin) L?~Iki [k;'s as in (6.38)] belongs to K2 • Clearly p E KI , since KI is convex. Let y E KI • Then by (6.38)

(6.39)

for some j, where 1 <j< n. Also, by (6.37), for each i c:F- j where 1 <i< n,

Y E k i + (1 + 1/4n)W.

By (6.39) and (6.40), we have

P E + {(y - lW) + (n - 1)[Y - (1 + ;n )W])

=y_ (1 __ 1 __ 1 )w 4n 4n2 ,

(6.40)

or p E Y + (1 - 1 14n) W (since W = - W), proving that p E K2 • The proof of the theorem is now complete. I

Before we go into the existence of the Haar measure, let us consider a few examples of topological groups. First, a definition and a few basic remarks.

Definition 6.26. A topological group is a group G with a topology such that the mappings (x, y) -+ xy from G X G into G and x -+ X-I from G into G are continuous. Clearly, the continuity requirements in a topological group are equivalent to the requirement that (x, y) -+ xy-I is continuous. I

Remarks

6.30. In a topological group the mappings x -+ zx, x -+ XZ, and x -+ X-I are all homeomorphisms.

6.31. Suppose a topological group G has the property (To): if x =F y and x, y E G, then there is an open set V containing one of them but excluding the other. Note that (TI ) =:;. (To) (see Chapter 1, Section 1.4). But curiously enough, the following are equivalent in G:

(a) G is a To-space. (b) G is a Trspace. (c) G is a T2-space. (d) n {U: U open and e E U} = {e}, where e is the identity of G.

Proof. (a) =- (b). Let XoF y and x E V open and y i V. Then e EX-I V = W [open by (i)]. Now if U = WnW-I, then y E yU (open). Now if x E yU; then X-I E Uy-l (since U = U-l), X-l E X-IVy-I, or y E V (which is a contradiction). Hence x E yU. Hence (b) is true.

(b) =- (c). Let x oF y. Then by (b), y E V = G - {x} (open). Hence e E y-l V (open). Since G is a topological group, there is open W with

e E Wand WW-l C y-l V. Let U = G - y W. If x E y W, then x W n y W

oF 0 and so x E yWW-l C V, which a contradiction. Hence xi yW and x E U. Hence (c).

The proofs of (c) =- (d) => (a) are easy and are left to the reader. I

6.32. A Hausdorff topological group is completely regular. A first countable topological group is metrizable. The proofs of these facts are somewhat involved and are omitted. The reader might consult Pontrjagin's text. t

6.33. Every real-valued continuous function f with compact support defined on a topological group G is uniformly continuous [i.e., given e > 0, there exists an open set V with e E V such that If(x) - fey) I < e whenever x-ly E V].

Proof. Suppose K is the compact support of f By using the compactness of K and the continuity of (x, y) --+ xy, we can find for e > 0 an open set VI with e E VI such that

If(x) -f(xy) 1< e, (6.41 )

whenever x E K and y E VI. Now for each x E A, where

A = {y: If(y) I > e} C the interior of K,

let Vx be an open set with e E Vx such that x Vx C K. Let e E Wx (open) such that Wx Wx C VX • By the compactness of A there exist Xl' X2' ... , Xn E A such that A C U~lXi WXj or AWe K, where W = U?~l WXi • Therefore if x i K and yEW-I, then

I f(x) - f(xy) I = I f(xy) I < e. (6.42)

From (6.41) and (6.42), taking V = VI n W-l, we have for every x E G and yE V, I f(x) - f(xy) I < e.

The proof of Remark 6.33 is now clear. I

t L. Pontrjagin, Topological Groups, Princeton University Press, Princeton, New Jersey (1939).

6.34. Let G be a compact topological group and C{G) be the realvalued continuous functions on G with "sup" norm. Let

!?!={ f aJ(six): O<ai< I, f ai = I, SiE G, and nE N}. i-I i-I

In other words, !?! is the convex hull of the set of all left translates of I [in C(G)]. Then ?i is a compact subset of C(G).

Proof. Suffice it to show that !?! is equicontinuous, since then the compactness of !?i will follow by Arzela-Ascoli's Theorem (Chapter I, Section 1.6). Since G is compact, by Remark 6.33, there is an open V with e E V and x-Iy E V such that I/(sx) - I(sy) I < e for every s E G. The equicontinuity of !?! now follows. I

Examples

6.18. R - {O} with the usual relative topology of R is a locally compact topological group under multiplication. R itself is also so under addition, and the Lebesgue measure is a translation-invariant measure on R.

6.19. The complex numbers of absolute value 1 is a compact topo-logical group under multiplication and usual topology of R2.

6.20. Let S be the set of all matrices of the form (~ n, where x > 0 and y is any real number. Then S is a locally compact topological group with the usual matrix multiplication and with the topology induced by the usual topology of R2 (in an obvious manner).

Theorem 6.38. There exists a unique regular probability measure f-l [i.e., f-l(G) = 1] on every compact Hausdorff topological group G such that for every S E G,

(i) f!s df-l = f I df-l,

(ii) f 18 dp, = f I dp"

(iii) f 1-1 dp, = f I dp"

where IE C(G), !sex) = I(sx), 1 8(x) = I(xs), and 1-1 (x) = I(x-I). This f-l is called the Haar measure on G. I

Proof. If L8: 1--+ Is, then for s E G, L8 is an isometry from C(G) (with "sup" norm) into itself. Therefore, {L8: s E G} is an equicontinuous

group of linear operators on the Banach space C(G). From Remark 6.34, ?j is a compact convex subset of C(G), and for s E G, Ls(?j) C?j. By Theorem 6.37, there exists fo E ?j such that Jo(xs) = fo(x) for every x, s E G; and thereforeJo(x) = fo(e) or fo is a constant c which can be uniformly approximated by convex combinations of left translates off If c' is a similar constant corresponding to the right translates of J, and c" is a similar constant corresponding to the left translates of f-l' we claim that c = c' = e". To prove this, let E > o. Then there are 0 < ai' bj , ek < 1 with L:f-lai = L:i=lbj = L~=lek = 1 such that

Ie - t ad(six ) I < E, (6.43)

~-1

Ie' - f b;f(xtj ) I < E, }=1

(6.44)

(6.45)

for all x E G and some (Si), (lj), and (Wk) E G. In the inequality (6.43), writing tj for x, multiplying by bj , and then summing over j, we have

I e - ~ aibj f(Sitj) I < E. (6.46)

Similarly, inequalities (6.44) and (6.45) can be written as

I c' - ~. aibjf(sitJ I < E, ''',}

(6.47)

and

(6.48)

From inequalities (6.46) and (6.47), e = c'; also by writing (6.43) in a form similar to (6.48), we get e = e". This argument also shows that there can be only one constant obtained as above corresponding to the translates of f or f-l. If we call this constant 1(1), then we have

(A) l(l) = 1,

(8) I(af) = al(f) , for all reals a,

(e) l(f) > 0, for f> 0, (D) 1(1s) = 1(f8) = l(f-l), SE G, (E) 1(f + g) = 1(f) + leg).

We establish only (E). Let e > 0.. Then, as in (6.43), we get

XE G, (6.49)

where Li~lai = I, 0 <ai' and Si E G. Let hex) = L~la~(six), Then h E ~ and !?i. C ~; therefore, since each of these sets contains a unique constant function, I(g) = I(h). Hence there are bj > 0, LJ..1bj = I, and tj E G such that

I leg) - j~l bjh(tjx) I < e.

By replacing h with g, we have

I leg) - ~. aibjg(SitjX) I < e. '.J

Writing (6.49) as

I l(f) - ~. aibd(SitjX) I < e, '.J

we have by inequalities (6.51) and (6.52)

I l(.f) + I(g) - t a,bif + g)(SitjX) I < 2e,

thus proving (E).

(6.50)

(6.51)

(6.52)

(6.53)

By the Riesz Representation Theorem (Theorem 5.10), there is a unique regular Borel measure fl satisfying l(f) = f f dfl for all f E C(G). Since 1(1) = 1, fl(G) = 1. All the properties of fl in the theorem now follow from property (D) above. I

Problems

6.7.1. Suppose fl is the Haar measure in a compact group G. Show that

(i) fl( V) > 0 for every open set V( =r'= 0), and (ii) fl(Bx) = fl(xB) = fl(B-I) for every Borel set B and x E G.

6.7.2. Suppose fl is a weakly Borel measure (possibly infinite) on a Hausdorff topological group G such that

(i) fl(xK) = fl(K) for all compact sets K and x E G, and (ii) 0 < fl(V) < = for some open set V with compact closure.

Prove that (a) G is locally compact, and (b) G is compact if fl(G) < =.

(This means that it is impossible to have any meaningful translation-invariant measure on a non-locally-compact topological group.)

6.7.3. Let H be the component containing the identity of a topological group G. Prove that H is a subgroup of G such that X-I Hx = H for alI xE G.

6.7.4. Prove that every open subgroup of a topological group is closed. 6.7.5. Suppose G is a group with first countable topology such that

(i) (x, y) --+ xy is separately continuous, and (ii) for any two compact sets A and B the set AB-I = UXEB{Y:

YX E A} is compact.

Prove that G is a topological group. 6.7.6. Suppose that G is a group with a metric topology with property

(i) of Problem 6.7.5 and the property d(x, y) = d(xz, yz) for any x, y, and z E G. Then prove that G is a topological group. [Here it is relevant to mention a beautiful result of R. Ellis: Suppose G is a group with locally compact Hausdorff topology such that (x, y) --+ xy is separately continuous. Then G is a topological group. For a proof, see his paper.t]

6.7.7. Let G be a Hausdorff topological group and let f1 be a weakly Borel measure such that

(i) f1(Kx) = f1(K) for every compact set K and x E G,

(ii) 0 < f1(V) < 00 for some open set V, and (iii) f1( {y}) > 0 for some y E G.

Prove that G is discrete.

6.7.8. Subsemigroups with Nonempty Interiors in a Compact Topological Group. Let H be a subsemigroup of a compact topological group G such that H has a nonempty interior. Show that H is a compact subgroup of G. [Hint: Let S be the interior of H. Then S is a subsemigroup. If f1 is the Haar measure of G, then for XES and any open set V, f1(x- I V n S) = f1(V n S) = f1(Vx-I n S). This means that Sx = xS = S, or S is a compact subgroup. Observe now that S = S, since for YES, yS-I n S i= 0.]

6.7.9. Use Problem 6.7.8 to prove that every locally compact subsemigroup of positive Haar measure in a compact topological group is a compact subgroup. [Note: This result remains true without the requirement of "positive Haar measure." Actually, using an important result of K. Numakura (that a cancellative semigroup with compact topology and

t R. Ellis, Duke Math. J. 24, 119-125 (1957).

jointly continuous multiplication is a topological group), it is very easy to do Problems 6.7.8 and 6.7.9. The reader should try this.]

6.7.10. A Fixed Point Theorem/or Affine Maps. Let Tbe a continuous map from a compact convex subset K of a normed linear space X into itself. Suppose that T is affine, i.e., for 0 < a <1 and x, y E K, T(ax + (I - a)y) = aT(x) + (1 - a)T(y). Then T has a fixed point. [Hint: For x E K and Xn = (I/n)LZ:JTk(x), observe that II T(xn) - Xn II ->- 0 as n ->- 00.]

6.7.11. The Markov-Kakutani Fixed Point Theorem. Let~be a family of continuous maps from a compact convex subset K of a normed linear space X into itself such that for any T and S in this family, T is affine and T(S(x» = S(T(x» for all x E K. Then ~ has a common fixed point. [Hint: For T E~, the set KT = {x E K: T(x) = x} is nonempty by Problem 6.7.10. If S E~, then S: KT ->- KT and there exists x E KT such that Sex) = x. An induction argument shows that every finite subfamily of ~ has a common fixed point. Now use the finite intersection property of the compact sets of common fixed points of the finite subfamilies of ~.] This theorem was first proven by A. Markov in 1936. In 1938, S. Kakutani gave an alternative proof of this theorem, along with an extension in the noncommutative case.

Apply the above fixed point theorem to show the existence of Banach limits on 100 , For a discussion of Banach limits, see Problem 6.2.16.

6.7.12. The Schauder Fixed Point Theorem. It T is a continuous map from a compact convex subset of a Banach space into itself, then T has a fixed point.

The proof of this theorem is difficult and depends on the Brouwer theorem in Chapter 1. For a proof, the reader might consult [II]. Around 1930 J. Schauder first proved this theorem. In 1935, A. Tychonoff proved this theorem in the more general context of locally convex Hausdorff spaces.

This theorem is useful for various applications in differential and integral equations. Using this theorem, the existence part of the initialvalue problem considered in Chapter 1, Theorem 1.24 can be proven easily assuming only the continuity of the function / and requiring no Lipschitz condition. Demonstrate this.

6.7.13. (Johnson). Let Y = [0,.0] with the order topology and v be the nonregular measure on Y as in Problem 5.3.13 or Example 5. II. Let A be a discrete group with two elements and X = X {X .. : A. E [O,.o)}, where X A == A V A, with the product topology. Let ft be the Haar measure on the compact group X (with coordinatewise multiplication). Prove that M = {(x, y): x EX, Y E Yand XA = e if A. > y} is a compact set such that v(Mx) is not measurable with respect to the completion of ft.

7

Hilbert Spaces

In this chapter we will study aspects of the theory of Hilbert spaces. Roughly we may say that a Hilbert space is a Banach space whose norm is defined in a particular manner. We shall give a characterization in terms of the norm of those Banach spaces that are actually Hilbert spaces. This well-known result (Proposition 7.2) is due to Jordan and von Neumann.

Infinite-dimensional Hilbert spaces are natural generalizations of the finite-dimensional spaces Rn and en with the usual "Euclidean norms." Their study was initiated in the early 1900's by Hilbert, who studied the particular spaces 12 and L 2 • The abstract axiomatization of Hilbert space was later given by von Neumann in the separable case in the 1920's, t and in general by U:iwigt and Rellich,§ Many others have made significant contributions.

Our aim in this chapter is to study Hilbert spaces starting with very basic properties of the structure of Hilbert spaces and ending with a brief exposition of some essential data concerning the spectral theory of self-adjoint operators. Primarily our aim is to prove the spectral theorem for bounded self-adjoint operators-an important tool in the further study of bounded linear operators in Hilbert space theory in that self-adjoint operators are represented as a sum (integral) of projection operators.

t J. von Neumann, Allgemeine Eigenwerttheorie Hermitescher Functionaloperen, Math. Ann. 102, 49-131 (1929-1930); Mathematische Begriindung der Quantenmechanik, Nachr. Ges. Wiss. Gottingen Math.-Phys. Kl., 1-57 (1927).

t H. Lowig, Komplexe euklidische Riiume von beliebiger endlicher oder unendlicher DimensionzahI, Acta Sci. Math. (Szeged.) 7, 1-33 (1934). F. ReIIich, Spectraltheorie in nichtseparabeln Riiumen, Math. Ann. 110,342-356 (1935).

121

122 Chap. 7 • Hilbert Spaces

7.1. The Geometry of Hilbert Space

In this section Vand W will denote vector spaces over the field F of real or complex numbers. ii will denote the complex conjugate of the complex number a.

Definition 7.1. A sesquilinear form B on Vx W is a mapping B: V X W -+ F such that, for all a and fJ in F, x and y in V, and wand z in W,

(i) B(ax + fJy, z) = aB(x, z) + fJB(y, z)

and

(ii) B(x, aw + fJz) = iiB(x, w) + PB(x, z). I

In case V = W, a sesquilinear form on Vx W is referred to as a sesquilinear form on V. A sesquilinear form on V is called Hermitian if B(x, y)

= B(y, x) for all x and y in V. Since B(x, x) is necessarily a real number if B is Hermitian, we say that a Hermitian form on V is positive if B(x, x) > 0 for all x in V and strictly positive if B(x, x) > 0 when x*- O.

Sometimes when F is the field of real numbers so that ii = a for all scalars, a sesquilinear form on V is called a bilinear form and a Hermitian form is called a symmetric form since B(x, y) = B(y, x) for all x and y in V. Since the development to follow is true-unless specifically indicated-for real and complex vector spaces, we will continue to use the terms sesquilinear and Hermitian regardless of whether F = R.

The following proposition gives in a nutshell some facts we will find extremely useful regarding sesquilinear forms on V.

Proposition 7.1. Let B be a sesquilinear form on V.

(i) Polarization Identity. If V is a complex vector space, then for all x and y in V

B(x, y) = HB(x + y, x + y) - B(x - y, x - y)

+ iBex + ;y, x + iy) - iBex - ;y, x - iy)]. (7.1)

If V is a real vector space, then for all x and y

B(x, y) = t[B(x + y, x + y) - B(x - y, x - y)], (7.2)

provided B is Hermitian.

Sec. 7.1 • The Geometry of Hilbert Space 123

(ii) Parallelogram Law. For all x and y in V

B(x + y, x + y) + B(x - y, x - y) = 2B(x, x) + 2B(y, y). (7.3)

(iii) Cauchy-Schwarz Inequality. If B is a positive Hermitian sesquilinear form on V, then for all x and y in V

I B(x, y) 12 < B(x, x)B(y, y). (7.4)

(iv) If B is a positive Hermitian sesquilinear form on V, then for all x and y in V,

[B(x + y, x + y)]112 < [B(x, X)]1I2 + [B(y, y)]1I2. I

Proof. Statements (i) and (ii) are verified by direct computation and the verifications are left to the reader. Assuming momentarily that (iii) has been verified, we can easily prove (iv). Indeed using (iii)

B(x + y, x + y) = B(x, x) + B(x, y) + B(y, x) + B(y, y)

= B(x, x) + 2Re[B(x, y)] + B(y, y)

< B{x, x) + 2 I B{x, y) I + B(y, y)

< B(x, x) + 2[B(x, X)]1I2[B(y, y)]112 + B(y, y)

= ([B(x, X)]1/2 + [B(y, y)]1I2}2.

It remains therefore to establish (iii). For all real numbers r and for a E F with I a I = I,

o < B{rax + y, rax + y) = r2B(x, x) + raB(x, y) + raBey, x) + B(y, y)

= r2B(x, x) + 2r Re[aB(x, y)] + B(y, y). (7.5)

Since equation (7.5) holds for all real numbers r, the quadratic function fer) = B(x, x)r2 + 2Re[aB(x, y)]r + B(y, y) has at most one distinct real root. Hence its discriminant must be nonpositive, that is,

{Re[aB(x, y)]}2 < B(x, x)B(y, y), (7.6)

for all a with 1 a 1 = I. Choose a so that aB(x, y) = 1 B(x, y) I. Then inequality (7.6) yields

I B(x, y) 12 < B(x, x)B(y, y). I

Remark 7.1. Clearly if B is a strictly positive sesquilinear form on V

and B(x, y) = B(x, z) for all x in V, then y = z. Indeed

B(y - z, y - z) = B(y - z, y) - B(y - z, z) = O.

Other interesting and useful facts are given in the following corollary of Proposition 7.1 (i).

Corollary 7.1. Assume V is a complex vector space.

(i) If B: Vx V -+ C and B': Vx V -+ Care sesquilinear forms such that B(x, x) = B'(x, x) for all x, then B = B'.

(ii) A sesquilinear form B: Vx V -+ C is Hermitian if and only if B(x, x) is real for all x. I

Proof. The proof of (i) is readily seen by examining equation (7.1).

To prove (ii) note that B(x, x) is real if B is Hermitian since B(x, x) = B(x, x). Conversely, if B(x, x) is real for all x, the sesquilinear form B'(x, y)

= B(y, x) is such that B'(x, x) = B(x, x) for all x. By (i), B = B' or B is Hermitian. I

With the information given in Proposition 7.1 we are in a good position to begin our study of Hilbert and pre-Hilbert spaces.

Definition 7.2. A pre-Hilbert space P over the field F is a vector space P over F together with a strictly positive Hermitian sesquilinear form on P. I

The sesquilinear form in a pre-Hilbert space is often called an inner product and a pre-Hilbert space is accordingly called an inner product space. The image in F of the ordered pair (x, y) in P x P by the inner product B on P will be denoted by (x 1 y) instead of B(x, y). In what follows, P, unless otherwise mentioned, will denote a pre-Hilbert space.

Examples. Here are some simple yet important examples of inner product spaces:

7.1. For any positive integer n the space cn(Rn) of ordered n-tuples x = (Xl' ... , xn) of complex (real) numbers with inner product given by

n _

(x 1 y) = L XiYi' i=l

7.2. The space 12 of all complex (real) sequences x = (Xi)N such that L~ll Xi 12 < 00 with inner product given by

00

(x I y) = L XiYi' i=l

7.3. For any measure space (X, d, ft), the space L 2{ft) of all measurable functions f for which J If 12 dft < 00 with inner product given by

(II g) = f fg d,u.

(Note that Examples 7.1 and 7.2 are special cases of Example 7.3 if X is chosen to be {1,2, ... ,n} and N, respectively-each with the counting measure.)

7.4. The vector space of continuous functions f on an interval [a, b] with inner product

(I I g) = f: f(t)g(t) dt.

Any pre-Hilbert space P is a normed linear space by virtue of the following definition: If x E P, define the norm of x by

II x II = (x I X)1!2. (7.7)

Since an inner product is a positive sesquilinear form, we have

II x II > 0 and II x II = 0 , if and only if x = O.

Also,

II ax II = I a I II x II, since II ax 112 = (ax I ax)

= aa(x I x) = I a 12 II x 112.

Finally, Proposition 7.1 (iv) becomes the triangle inequality

II x + Y II < II x II + II Y II· (7.8)

One should also note that in any inner product space the Parallelogram Law and the Cauchy-Schwarz Inequality, respectively, now have the following forms:

II x + Y 112 + II x - Y 112 = 2 II X 112 + 2 II Y 11 2,

I (x I y) I < II x II II Y II·

(7.9)

(7.10)

Examining the geometrical meaning of the Parallelogram Law in R2 demonstrates the aptness of its title: The sum of the squares of the diagonals in a parallelogram is equal to the sum of the squares of the four sides.

Remark 7.2. If xn ---+ x and Yn ---+ Y in P, then (xn I Yn) ---+ (x I y) in F. This follows from the inequality

I (x IY) - (xn IYn) I = I (x IY) - (x IYn) + (x IYn) - (xn IYn) I < II x II IIY - Yn II + II x - Xn II II Yn II·


Definition 7.3. A Hilbert space is a complete pre-Hilbert space with norm II x II = (x I X)I/2. I

Not all pre-Hilbert spaces are Hilbert spaces. For example the subspace of 12 (Example 7.2) consisting of finitely nonzero sequences x = (Xi)N [a sequence x = (Xi)N is finitely nonzero if there exists some positive integer M such that Xi = 0 for all i > M] is a pre-Hilbert space that is not complete. Also Example 7.4 is not complete. The completion (see Problem 7.1.3) of this space is the space L2([a, b D. The verifications of these statements are left as exercises (Problem 7.1.4).

Briefly we can say that a Hilbert space is a Banach space with the norm defined by an inner product as in equation (7.7). When is a Banach space a Hilbert space? The Paralleiogram Law gives us one characterization. t Precisely, we have the following characterization, whose proof we have outlined in the Problems (Problem 7.1.5).

Proposition 7.2. A Banach space is a Hilbert space with its norm given by an inner product if and only if its norm satisfies the parallelogram identity (7.9). I

Definition 7.4. Two vectors x and y in a pre-Hilbert space P are said to be orthogonal (or perpendicular), written x -.l y, if (x I y) = O. If E and F are subsets of P, then E and F are said to be orthogonal (to each other), written E -.l F, if x -.l y for each x in E and y in F. A subset E of P is said to be an orthogonal set if x -.l y for each nonequal pair of vectors x and y in E. If in addition 1/ x II = 1 for each x in E, then E is said to be orthonormal. I

Remark 7.3. Any orthogonal set E in a pre-Hilbert space P that does not contain the zero vector is linearly independent. Indeed, if {Xl' X2 , .•. ,

xn} is a finite subset of E and (lIXl + (l2X2 + .,. + (lnXn = 0, then (li II Xj 112 = (Lf-l(liXi I Xj) = 0 so that (lj = 0 for each j.

In the space 12 the countable set of vectors like (0, 0, ... , 1, 0, 0, ... ) where 1 is the ith coordinate for i = 1, 2, .. , is an orthonormal set. In R3, the set {(I, 1,0), (0, -1,0)} is independent but not orthogonal. In L 2([0, 2nD the set {eint I n = 0, ± 1, ± 2, ... } is orthogonal, but not orthonormal.

Definition 7.5. A family of vectors (Xi)ieI in a normed linear space is

t For other characterizations, see [22].

called summable to x, written LIXi = x, if for each e > 0 there exists a finite subset F(e) of I such that if J is a finite subset of I containing F(e) then

II LieJXi - x II < e. I

It can be shown by the reader that in a Banach space a family (Xi)I is summable (to some x) if and only if for each number e > 0 there exists a finite subset F(e) of I such that if J is a finite subset of I with J n F(e) = 0, then II LieJXi II < e. From this criterion it follows that if (Xi)I is summable, the set of indices i for which Xi *- 0 is at most countable. Indeed for each positive integer n, let F{1ln) be the finite subset of I such that II LJXi II < lin if J n F{1ln) = ° and J is finite. If x ft U:"lF(1ln), a countable set, II x II < I In for all n.

It is easy to verify the following rules in any pre-Hilbert space:

(i) If LIXi = x, then LIaxi = ax for any scalar a.

(ii) If LIXi = x and LIYi = y, then LIXi + Yi = X + y.

(iii) If LIXi = x, then LI(Xi I y) = (x I y) and

LI(Y I Xi) = (y I x) for every vector y.

(7.11)

To verify (i) for instance, let e > 0 be arbitrary. Then there is a finite subset F(e) such that if Jis finite and J J F(e), then II LJaxi - ax II = I a I II LJXi - x II < I a I e. Hence LIaxi = ax.

Proposition 7.3. Pythagorean Theorem.

(i) If {Xl' X 2 , ••• ,xn } is any orthogonal family of vectors in a Hilbert space H, then

(ii) Any orthogonal family (Xi)I of vectors in His summable if and only if (II Xi 112)1 is summable. If x = LIXi, then II x 112 = LIII Xi 112. I

Proof. An inductive argument proves (i). To prove (ii) note that (Xi)] is summable if and only if for each e > 0 there exists a finite subset F(e) of I such that if J is a finite subset of I with J n F(e) = 0, then

(7.12)

By virtue of the equality in (7.12), this condition is also necessary and sufficient for (II Xi 112)z to be summable.

If X = LIXi, then by equation (7.11) we have

II x 112 = (x Ix) = (f Xii x)

= L (Xi I x) = L (Xi I L Xj) = L L (Xi I Xj) = L II Xi 112. I I iE! jEl i ... , xn} is any finite family of orthonormal vectors in P and x is any vector in P, then

n L I (x I Xi) 12 < Ii X 1!2. i-I

(ii) If (Xi)I is any orthonormal family of vectors in P and x is any vector in P, then

L I (x I Xi) 12 < II X 112. I I

Proof. (i)

o < II x - tl (x I Xi)Xi 112 = II X 112 - tl (x I Xi)(X I Xi)

n n __ n

- L (x I Xi)(Xi I x) + L (x I Xi)(X I Xi) = II x 112 - L I (x I Xi) 12, i-I i-I i-I

from which (i) follows.

(ii) By (i), LiEF I (x I Xi) 12 < II X 112 for any finite subset F of I. tHence by the definition of summability the inequality must hold for I. I

Using the concept of summability of an arbitrary family of vectors or scalars, we can give an example of a class of Hilbert spaces which we shall see later represents all Hilbert spaces.

Example 7.5. For any non empty set I let CI(RI) be the vector space of all complex- (real-) valued functions on I, that is, the set of all families of elements (X;).iEI, where Xi is a scalar. Let 12(1) be the vector subspace of CI(RI) of all families (Xi)iEI such that LI I Xi 12 is sum mabie (written LII Xi 12 < 00). 12(/) is a Hilbert space with inner product given by (x I y)

t Note that {j E I: I (x I XI) I > O} is at most countable.

= LIXiYi for x = (Xi)I and Y = (Yi)I. Using the Holder Inequality (Proposition 3.13 in Chapter 3), the reader can verify that this does in fact define an inner product, and in particular that (XiYi)I is summable. We here establish the completeness of 12(1). [The completeness also follows from that of L2 (see Theorem 3.12 in Chapter 3), where the measure is the counting measure; but we here give a different proof.]

To this end, let Xk = (XniEI, k = 1,2, ... be a Cauchy sequence in 12(/). Since for each i

II Xjn - Xim 112 < L 1 X/I - X/" 1 2 = II Xn -- Xm !12, ieI

(Xnk=1.2 •... is a Cauchy sequence of scalars for each i in I. Hence there exists for each i in 1 an Xi such that limk-+ooxl = Xi. Let X = (Xi)iEI. We wish to show X E 12(/) and xk - x in 12(1).

Let J be any finite subset of 1 and 8 > 0 be arbitrary. Then there exists N > 0 such that if n, m > N, then

L I Xin - Xjm 12 < L 1 Xi n - Xim 12 < 82• ;.EJ iEI

Letting n - 00, then for m > N

L 1 Xi - Xim 12 < 82• iEJ

Since J is an arbitrary finite subset of I, this means LII Xi - Xim 12 < 82 and II (Xi - Xim)I II - 0 as m - 00. In particular (Xi - Xr)I is in 12(/), so that (Xi)I = (Xi - Xim)I + (Xr)I is in 12(/).

Problems

X 7.1.1. (i) If X (# 0) and yare any vectors in a pre-Hilbert space, prove I (x I y) I = II xliii Y II if and only if Y = AX for some A E F. [Hint: Look at the proof of Proposition 7.1 (iii).]

(ii) If X and yare nonzero vectors in a pre-Hilbert space, prove II X + Y II = II X II + II y II if and only if y = Ax for some A > O.

(iii) Prove in a pre-Hilbert space, II X - z II = II X - y II + II y - z II if and only if y = ax + (1 - a)z for some a in [0,1].

7.1.2. (i) Prove that if (xn) is an orthogonal sequence of vectors in a pre-Hilbert space such that L:llI Xi 112 < 00, then the sequence (Lf-lXi)nEN

is a Cauchy sequence.

(ii) Give an example where the conclusion of (i) may fail if (xn ) is not orthogonal.

X 7.1.3. If P is a pre-Hilbert space with inner product (x I y), prove that there is a Hilbert space H with inner product B(x, y) and a linear map T: P -+ H such that B(Tx, Ty) = (x I y) for all x and y in P and T(P) is dense in H. Prove that if (H', B') is another Hilbert space satisfying these criteria, then Hand H' are linearly isometric-that is, there is a linear map S from H onto H' such that B'(Sx, Sy) = B(x, y) for all x and y in H. H is called the completion of P.

X 7.1.4. (i) Prove that the subspace of 12 (see Example 7.2) consisting of finitely nonzero sequences is not complete, but its completion is 12 ,

(ii) Prove that the space of Example 7.4 is not complete, but its completion is L 2([a, bD. {Hint: Look at the sequencefn(t} = 0 if a < t < (a + b)/2; = n[t - (a + b}/2] if (a + b)/2 < t < (a + b)/2 + lin; = I otherwise.}

X 7.1.5. (i) Prove Proposition 7.2 by showing that each Banach space whose norm satisfies the Parallelogram Law, equation (7.9), is a Hilbert space. [Hint: If B is a real Banach space, define (x I y) as in equation (7.2) by t {II x + y 112 - II x - y 112} while if B is a complex Banach space define (x I y) as in equation (7.1) by HII x + y 112 - II x - y 112 + i II x + iy 112 - i II x - iy 112}. In the real case show (x I y) + (z I y) = (x + z I y), (xn I y) ~ (x I y) if Xn ~ x in B, and conclude (ax I y) = a(x I y) for all real a. Note that in the complex case Im(x I y) = Re(x I iy).]

(ii) Prove LI[O, I] is not a Hilbert space by showing that the Parallelogram Law is not satisfied.

7.1.6. Show that the result in Problem 7.1.5 (i) can be extended as follows: Let V be a real vector space and II . II: V -+ R be a function satisfying the parallelogram law [equation (7.9)] and the following property: For every x E V, the function a -+ II ax lion R is continuous at O. Then (x I y) = t {II x + y 112 - II x - y 112} defines a nonnegative Hermitian bilinear form on V. (This extension is due to D. Fearnley-Sander and J. Symons.)

7.1.7. State and prove a complex version of the result outlined in Problem 7.1.6.

7.1.8. Define on 12 the norm

II f II = (~ I fen) 12t2 + ~?E I fen) I·

Show that this norm is equivalent to the usual norm in 12 but it does not come from an inner product.

Sec. 7.2 • Subspaces, Bases, and Characterization 131

7.2. Subspaces, Bases, and Characterizations of Hilbert Spaces

We will now turn our attention to subspaces of pre-Hilbert and Hilbert spaces. It is clear that any vector subspace of a pre-Hilbert space is a preHilbert space with the restricted inner product.

Crucial to the study of the structure of Hilbert spaces and subspaces is the following result, not valid in every normed linear space.

Theorem 7.1. Let S be a complete and convex (x, yES implies ax + (\ - a)y E S for all a E [0, 1]) subset of a pre-Hilbert space P. Given any vector x in P there exists one and only one vector Yo E S such that

II x - Yo II < II x - Y II for all Y in S. I

(In regard to this theorem see Problems 7.2.1 and 7.2.2.)

Proof. Let Yi be a sequence of vectors in S such that II x - Yi II converges to 15, the inf of {II x - Y II: Y in S}. We will show that Yi is a Cauchy sequence in S converging to the desired vector Yo. Using the Parallelogram Law, equation (7.9) of Section 7.1,

or

Since t(Yi + Yj) E s,

Hence,

(7.13)

As i, j~oo, the right-hand side of equation (7.13) goes to zero so that (Yi)N is a Cauchy sequence in S. Since S is complete, Yi converges to some Yo in S. Since II Yi - x 1/ ~ II Yo - x II, II Yo - x II = 15.

If Yo' E S also satisfies II Yo' - x II = 15, then using the Parallelogram Law again

II Yo - Yo' 112 = 2 II Yo - x 1/2 + 2 1/ x - Yo' 1/ 2 - 4 1/ i(yo + Yo') - X 1/ 2

< 2 II Yo - x 112 + 2 II x - Yo' 112 - 4152 = 4152 - 4152 = 0. I

Lemma 7.1. If S is a proper complete subspace of pre-Hilbert space P, then there exists x in P - S such that {x} -.l S. I

Proof. By Theorem 7.1 for any vector z in P - S there exists a unique vector yo(z) in S such that II z - yo(z) II < liz - y II for all y in S. Let x = z - yo(z). We will show x E P - S and {x} -.l S. Clearly x if; S since z if; S. Since for every scalar a, Yotz) + ay E S for every y in S,

II ay - x 112 = II [Yo(z) + ay] - z 112 > II z - Yo(z) 112 = II X 112.

Hence

o < II x - ay 112 - II X 112 = - a(y I x) - a(x I y) + aa II y 112.

Letting a = - fJ(x I y) for any real fJ we get

o < fJ(x I y)(y I x) + fJ(x I y)(x I y) + fJ2(X I y)(x I y) II Y 112

= 2fJ I (x I y) 12 + fJ2 I (x I y) 12 II Y 112

= fJ I (x I y) 12[2 + fJ II y 1[2]. (7.14)

If fJ is chosen to be a negative number such that fJ >- 2/11 Y 11 2, (7.14) forces I (x I y) I = 0 or x -.l y. I

Definition 7.6. If S is any subset of a pre-Hilbert space P, the annihilator or orthogonal complement of S is the set

S1- = {x E P: x -.l y, for all y in S}. I

The orthogonal complement of any set S always contains the zero element of P. Clearly S n S1- C {O} and S C (S1- )1-. More can be said.

Lemma 7.2. If S is any subset of P, then S1- is a closed subspace of P.I

The proof is easy using Remark 7.2. Note that in Theorem 7.1 and Lemma 7.1 if P is a Hilbert space,

the word "complete" may be replaced by "closed" since in any complete metric space a subset is closed if and only if it is complete. In particular, Lemma 7.2 assures us that S1- is complete in a Hilbert space.

If M and N are subspaces of a pre-Hilbert space P, then M + N is the subspace defined as {m + n: m E M, n EN}. If M -.l N, then each element of M + N has a unique representation as m + n with m E M and n E N. Indeed, if m + n = m' + n' with m, m' E M and n, n' E N, then

Sec. 7.2 • Subspaces, Bases, and Chamcterization 133

m - m' = n' - n. Hence m - m' E M n N so that (m - m' I m - m') = O. Hence m = m'. Similarly n = n'. In this case we write M + N as MEElN.

Theorem 7.2. If M is a complete linear subspace of pre-Hilbert space P, then P = M EEl M1. and M = (M1.)1.. (See also Problem 7.2.6.) I

Proof. Let Z be any vector in P. By Theorem 6.1, there exists a unique vector Yo(z) in M such that

II Z - Yo(z) II < II Z - Y II, for all Y in M.

As shown in the proof of Lemma 7.1, x(z) = z - Yo(z) is in M1.. Hence z = Yo(z) + x(z) E M + M1.. Since M ~ M1., P = M + M1. = M EEl M1..

Clearly for any set S, S C (S1. )1.. If Z E (M 1. )1., then Z = Y + x with Y EM and x E M1.. Since also Y E (M1.)1., X = Z - Y E (M1.)1.. But (M1.) n (M1.)1. = 0, so x = O. Thus Z = Y EM. Hence (M1.)l = M. I

If S is a set in a pre-Hilbert space P, we denote by [S] the smallest subspace of P containing S. It is easy to see that [S] is the vector space of all finite linear combinations of elements of S.

Proposition 7.5. The Gram-Schmidt Orthonormalization Process. If {Xi: i = I, 2, ... , N} for I < N < 00 is a linearly independent set in a pre-Hilbert space P, then there is an orthonormal set {Zi: i = 2, ... , N} such that [{Zi: i = 1,2, ... , n}] = [{Xi: i = 1,2, ... , n}] for each n = 1,2, ... , N. I

Proof. For N = 00 we proceed by induction. Let Zl = II Xl II-lxl . Clearly [Xl] = [Zl]. Assume an orthonormal set {Zi: i = 1,2, ... , n - I} exists such that [{Zi: i = 1, ... , n - I}] = [{Xi: i = 1, ... , n - I}]. Define Zn = II Yn II-lYn, where Yn = Xn - Li:f(xn I Zi)Zi' This equation and the inductive hypothesis guarantee that each Zi for i = 1, 2, ... , n is a linear combination of the set {Xi: i = 1, ... , n} and each Xi for i = 1,2, ... , n is a linear combination of the set {Zi: i = 1,2, ... , n}. Also, for each j = I, 2, ... , n - I,

= !I Yn II-l{(xn I Zj) -- (xn I Z¥Zj I Zj)}

=0.

The modifications for the case when N is finite are obvious. I

If {Xi: i = 1,2, ... , N} for some N in 1 < N < 00 is a vector space basis for a pre-Hilbert space P, then by Proposition 7.5, P has a basis {z,: i = 1, 2, ... , N} which is an orthonormal set. This basis {Zi: i = 1, 2, ... , N} has the following property: If X in P is orthonormal to {Zi: i = 1,2, ... , N}, then x = O. Indeed if x is in P, x = Lf-1aizi for some scalars ai' However, for each j = 1, 2, ... , N ,

N

aj = aj II Zj II = (ajzj I Zj) = C~ aizi I Zj) = (x I Zj) = 0,

so that x = O. Because the set {Zi: i = 1,2, ... , N} has this property, it is an example of a "complete" orthonormal set. Precisely, we have the following definition.

Definition 7.7. An orthonormal set S in a pre-Hilbert space is complete if whenever {x}.l S then x = O. A complete orthonormal set in a pre-Hilbert space is called a (pre-Hilbert space) basis. I

A word of caution and explanation is in order. The word "basis" is used with two different meanings. In one sense it means an algebraic or Hamel basis, that is, a linearly independent set in a vector space which spans the vector space. In the other sense it means a Hilbert space basis as defined in Definition 7.7.

Subsequently, whenever we speak of a basis we mean a pre-Hilbert space basis unless otherwise indicated.

Proposition 7.6. Let (Xi)/ be an orthonormal family in a Hilbert space H. The following statements are equivalent.

(i) The family (Xi)l is a basis for H.

(ii) The family (Xi)l is a maximal orthonormal family in H.

(iii) If x E H, then x = L/(X I Xi)Xi (the Fourier expansion of x).

(iv) If x and yare in H, then

(Parseval's identity).

(v) If x is in H, then II x 112 = L/ I (x I Xi) 12. I

Proof. (i) => (ii). Suppose S is an orthonormal family containing

Sec. 7.2 • Subspaces, Bases, and Characterization 13S

(Xi)[ and S E S - (Xi)[' Then s ..1 Xi for each i E I implies s = O. However, this contradicts the fact that II s II = 1.

(ii) ~ (iii). By Bessel's Inequality (Proposition 7.4) the family

[I (x I Xi) 12][ has a convergent sum since for any finite subset A of I,

L I (x 1 Xi) 12 < II X 112. ieA

Hence LI(X I Xi)Xi converges since for each e > 0 there exists a finite subset F(e) of I so that if J is finite and J n F(e) = 0. then (using Proposition 7.3)

Let y = LI(X I Xi)xi' It remains to show x = y. For each j E I,

(x - y 1 Xj) = {x - L (x 1 X;)Xi I Xj) = (x 1 Xj) - (x 1 Xj) = o. [

If x - y * 0, then {II x - y II-I(x - y)} U {Xi}ie[ is an orthonormalfamily properly containing {x;}iE[. Since this is impossible, x = y.

(iii) ~ (iv).

(x 1 y) = {L (x I Xi)Xi I L ()' 1 Xj)xJ = L «(x 1 Xi)Xi I L (y I Xj)Xj) iei jeI ieI .leI

= L (X 1 X;)(L (y 1 X)(Xi 1 Xj» = L (X 1 Xi)(Xi 1 y). ieI JEI 'ieI

(iv) ~ (V). Take x = y in (iv).

(v) ~ (i). If (x 1 Xi) = 0 for each i, then II x 112 = L(x I Xi)2 = O. I Does each pre-Hilbert space have a basis? It is sufficient to ask whether

each pre-Hilbert space has a maximal orthonormal set. Consider the collection of all orthonormal sets in a given pre-Hilbert space. Order this collection by set inclusion. Since the union of an increasing family of orthonormal sets is orthonormal, Zorn's Lemma guarantees the existence of a maximal orthonormal set. We have proved the following theorem.

Theorem 7.3. Every pre-Hilbert space has a basis. I

Proposition 7.7. Any two bases of a pre-Hilbert space H have the same cardinality. I

Proof. If H is finitely generated, the result follows from the theory of finite-dimensional vector spaces. So assume that {Xdl and {Yj}J are two bases of infinite cardinality. Since for each k E I, Xk = LAxk I Yj)Yj by Proposition 6.6 (iii), the set Jk of those indices j for which (Xk I Yj) 0:/=. 0 is at most countable. Since {xih is a basis, no Yj can be orthogonal to each Xk' This means J C Ukel Jk. Hence

card J < ~o . card I = card I. A symmetrical argument gives card I < card J. I

Proposition 7.7 gives meaning to the following definition.

Definition 7.8. The dimension of a pre-Hilbert space H is the cardi-nality of any basis for H. I

It is an interesting fact that in a pre-Hilbert space H the "distance" between any two distinct elements Xi and Xj of a basis {xih is 2112. Indeed,

II Xi - Xj 112 = (Xi - Xj I Xi - Xj) = (Xi I Xi) + (Xj I Xj) = 2.

It follows that an open neighborhood N(Xi' 2112/2) = {X E H: II X - Xi II < 2112/2} of Xi contains no other element of the basis {XJI except Xi' In fact the collection of such neighborhoods is pairwise disjoint. If S is a dense subset of H, each N(Xi' 2112/2) for i E I must contain a point of S. This means that the cardinality of I is no greater than that of S. In other words if H is a separable Hilbert (metric) space so that S is countable, card I < ~o. We have partly proved the following proposition.

Proposition 7.8. The dimension of a Hilbert space H is less than or equal to ~o if and only if H is separable. I

The converse is left to the reader. A linear isometry T from a Hilbert space H into a Hilbert space K

is a linear mapping from H into K such that II Tx II = II X II for all X in H. If T: H - K is a linear isometry, then it follows from the polarization identity (Proposition 7.1) and the equation (Tx I Tx) = II Tx 112 = II X 112 = (x I x) that (Tx I Ty) = (x I y) for all x and Y in H. Hence T: H - K is a linear isometry if and only if T is linear and (Tx I Ty) = (x I y) for all x and Y in H. Two Hilbert spaces Hand K are said to be linearly isometric if there is a linear isometry from H onto K. t

t In this chapter we write Tx, rather than T(x), as was done in Chapter 6, for ease of notation in the context of inner products.

Proposition 7.9. Two Hilbert spaces Hand K are linearly isometric if and only if they have the same dimension. I

Proof. If Hand K are linearly isometric, let T: H -4- K be an isometry from H onto K. If{hi : i E I} is a basis of H, then the set {Thi : i E I} is a basis in K of the same cardinality.

Conversely, if the dimension of K equals the dimension of H, let {hi: i E I} and {ki : i E I} be bases of Hand K, respectively, indexed by the same set I. If h = LI(h I hi)hi is in H, define Th as LI(h I hi)ki . Clearly T is a linear mapping from H onto K and II Th 112 = L I (h I hi) 12 = II h 112. I

Theorem 7.4

(i) A Hilbert space of finite dimension n is linearly isometric to 12(/), where 1 = {I, 2, ... , n}.

(ii) A separable Hilbert space of infinite dimension is linearly isometric to 12 •

(iii) A Hilbert space of dimension u is linearly isometric to 12(/),

where 1 is a set of cardinality u. I

Proof. For any set I, a basis for 12(/) is the set B = {fi E M/): i E 1

andfi(j) = <5ij }, where <5ij is the Kronecker delta. The dimension of 12(/) is then the cardinality of B, which is the cardinality of I. The result follows from Proposition 6.9. I

• Remark 7.4. Applications of Proposition 7.6: Fourier Analysis in the Hilbert Space L 2 , Recall the definition of Fourier series from Chapter 6. Remark 6.5. The Fourier series of a function fELl [- n, n] is sometimes written as the trigonometric series

I 00

2:" ao + ,{;l [an cos nt + bn sin nt], (7.15)

where

1 f" an = n _" f(s) cos ns ds, bn = - f(s) sin ns ds. 1 f" n _"

Note that the series (7.15) can be easily derived from the series

00

L /(k)eikt, k--oo

by using the formula eins = cos ns + i sin ns. By the Riemann-Lebesgue

theorem (see Problem 3.2.18), we have

lim len) = 0, IE L)[- n, n]. (7.16) 1"1""*00

This fact also follows from Theorem A below. It can be proven (see Problem 7.2.8) that the family {(1 /2n )1/2eikt : k = 0, ± 1, ±2, ... } is a basis for L 2[ - n, n]. Using this fact, Theorem A below follows immediately from Proposition 7.6.

Theorem A. Let IE L 2[ - n, n]. Then

(7.17)

and

(7.18)

(7.19)

and if (ak):'-oo is a sequence of scalars such that

00

L 1 ak 12 < 00, (7.20) k~-oo

then there exists a unique IE L 2[ - n, n] such that 1= 2:.';.._ooakeikt and Ok = l(k). I

If we take g = X[a,b] , - n < a < b < n, in equation (7.19) above, we obtain the following striking result.

Theorem B. For IE L 2 [ - n, n],

fb 00 fb l(t) dt = L j(k)eikt dt.

a k--oo a I

This theorem is surprising since we have obtained above the integral of I by integrating the terms of the Fourier series. Note that usually term-byterm integration of an infinite series is only possible by an assumption of

uniform convergence, and in the case of the Fourier series above we have not assumed even pointwise convergence of the series.

Now we present a useful relationship between the convolution products of functions in L2 and their Fourier coefficients. Let f, g E L2 [-n, n]. We continue these functions on R with period 2n. For convenience, we define f * g as

1 f" f * get) = 2n _!(t - s)g(s) ds. (7.21)

Then f * g is a continuous function on [- n, n], since

2n If* g(t2) - f* g(tl) I < Ilf(t2 - s) - f(tl - s) 112 II g(s) 112 ---+- 0

as I tl - t2 I---+- O.

(Note that this convergence to zero is trivially justified when f is continuous, and therefore it follows for f E L2 since the continuous functions are dense in L2.)

Theorem C. The Fourier series for f* g [as defined in equation (7.21) above] is absolutely convergent (uniformly in t) and given by

00

f * get) = L /(k)g(k)ei1ct• k--oo

(7.22)

I

Proof. By using the periodicity of the functions, substitution, and Fubini's Theorem one can easily verify that --(f* g)(k) = /(k)g(k);

and thus the series in equation (7.22) is the Fourier series of f * g. Since (/(n» and (g(n» are both in 12 by equation (7.18), it follows that the series in equation (7.22) is absolutely convergent, uniformly in t, and thus has a continuous sum function, say F(t). Since the series is uniformly convergent, term-by-term integration is possible and it follows that the series is the Fourier series of F. Now it is clear from equation (7.17) that F = f * g

almost everywhere. Since these functions are continuous, F = f * g. I

Problems

X 7.2.1. Show that Theorem 7.1 may be false if S is not complete.


[Hint: Consider the set S of all real sequences (Xi) such that L~lXi = 1 in the pre-Hilbert space of finitely nonzero sequences of real numbers. S contains no vector of minimum norm.]

7.2.2. (i) Show that the uniqueness conclusion of Theorem 7.1 is not valid in RxR with norm II (x, y) II = max{11 x II, II y II}.

(ii) Show that the existence conclusion of Theorem 7.1 is not valid in some normed linear spaces.

X 7.2.3. In the space L 2 [0, I], let {fl>h, ... } be the orthonormal set obtained from the set {I, x, X2, .•. } by the Gram-Schmidt orthonormalization process. Using the fact that the set of polynomials is dense in L 2[0, 1], prove that {It, h, ... } is a basis of L 2 [0, I].

X 7.2.4. (i) Show that on any real or complex vector space an inner product can be defined.

(ii) Give an example of a vector space (necessarily infinite dimensional) that is a pre-Hilbert space under two different inner products but the completions of each of the pre-Hilbert spaces are not linearly isometric Hilbert spaces. [Hint: Consider /2 and form a new inner product by considering a Hamel basis (Xi)iel such that (Xi I Xj) = t5ij.]

7.2.5. As in Example 7.4, let P be the pre-Hilbert space of continuous real-valued functions on [- 1, 1]. Show that the set of odd functions 0 is orthogonal to the set of even functions E. Show P = E EB O. Is E = 01.?

X 7.2.6. Let P be the subspace of Example 7.2 of finitely nonzero real sequences. Le~ M be the subspace of P of all sequences x = (Xk) such that Lk=lXk/k = 0. Prove that M is closed but M =I=- (M1. )1.. Compare Theorem 7.2. (Hint: Show M1. = {a} since {l, 0, 0, ... , -n, 0, ... } is in M where -n is the nth coordinate.)

7.2.7. Prove Theorem 7.3 for separable Hilbert spaces without using Zorn's Lemma (or an equivalent).

7.2.S. Prove that S = {eikt /(2n)1I2: k = 0, ±l, ±2, ... } is a basis for L 2([0, 2nD. {Hint: Show S is complete by showing f = ° a.e. if H" f(t)e- ikt dt = 0 and f E L 2([0, 2nD. To accomplish this let F(t) be the periodic absolutely continuous function F(t) = f~f(u) duo By partial integration show that f~" [F(t) - C]e-ikt dt = 0 for k = ± I, ±2, .... Pick c so that this is true also if k = 0. Since F - c has period 2n, approximate F - c uniformly on [0, 2n] by a trigonometric polynomial T(t) = 'L':.nCkeikt

(see Corollary 1.2 in Chapter 1). Show that f~ 1 F(t) - C 12 dt is arbitrarily small. Hence F(t) = c a.e.}

X 7.2.9. Show that (iii)-(v) of Proposition 7.6 are equivalent in any pre-Hilbert space P and so are (i) and (ii).

7.2.10. Show that a maximal orthonormal set {x",} in a pre-Hilbert space P need not be a basis for P, in the sense that it may not be possible to write each x E P uniquely as LC",X",. (Hint: Let {eJ be an orthonormal basis for a Hilbert space Hand P be the linear subspace spanned by {L:'ln-Ien, e2, ea, ... }. Then {e2, ea, ... } is a maximal orthonormal set in P, though not a basis for P.)

* 7.2.11. Let HI and H2 be Hilbert spaces over the same scalars, and let T: HI -4- H2 be a linear operator such that T(B) = 0, where B is an orthonormal basis for HI. Suppose that T(HI) = H2, G is the graph of T, and H = HI EB H2 [the-direct sum {(x, y): x E HI' Y E H2} with inner product (Xl' YI) 1 (x2, h») = (xII X2) + (YI 1 Y2)]· Then show that G = Hand dim G = dim HI. (This result is taken from S. Gudder. t)

* 7.2.12. Is the dimension of a pre-Hilbert space the same as that of its completion? [Hint: Use Problem 7.2.11 to find the answer in the negative. Take HI = 12, dim H2 = C, B an orthonormal basis for HI' and D("J B) a Hamel basis for HI. Then dim D = c. Let T: HI -4- H2 so that T(B) = 0 and T(D - B) is an orthonormal basis for H2. Then dim G (the graph of T) = ~o, but dim G = c.]

7.2.13. Use Problem 7.2.12 to show that the "necessity" part of Proposition 6.8 is not necessarily true in a pre-Hilbert space.

7.2.14. Measure in a Hilbert Space. Let (X, 5$, p,) be a measure space where 5$ is the smallest a-algebra containing the open sets of a Hilbert space X. Suppose p,(G) > 0 for every open set G, p,(B + x) = p,(B) for every B E 5$, X E X, and X is infinite dimensional. Then prove that p,(G) = 00 for every nonempty open set G. [Hint: Let (Xi) be an infinite orthonormal set in X. Let G = {x: II x II < r} and Gn = {x: II x - (r/2) . Xn II < r/4}. Then each Gn C G, and Gn's are pairwise disjoint with the same positive measure.]

7.2.15. Prove that the linear subspace spanned by the set {Xne-x"/2:

n = 0,1,2, ... } is dense in L 2( -00,00). [Hint: Assume for some

t S. Gudder, Am. Math. Mon. 81(1), 29-36 (1974).

fe L2(-00, 00),

f:f(t)e-tl/2tn dt = 0, for n = 0, 1,2, ....

Let

for complex numbers z. Show F(kl(Z) = 0 for k = 1,2, ... implying that F(z) is identically zero. Therefore

if -00 < x < 00. Multiply this equality by e-iZ71, where y is real and integrate with respect to x from -w to w to get

f oo f(t)e-tB/2 sin w(t - y) dt = 0, -00 t-y

for every wand y. Conclude that f(t) = 0 a.e.]

7.3. The Dual Space and Adjoint Operators

In this section we will consider the dual of a Hilbert space and characterize each element of the dual space. Leading up to our study of operators in the next section, we will also define the adjoint of an operator and give its basic properties.

The reader will recall from Chapter 6 that the dual space E* of a normed linear space E is the Banach space of all bounded linear functionals on E. In a pre-Hilbert space P each element y in P gives rise to a special element y* of P* defined by

y*(x) = (x I y), for all x in P. (7.23)

It is easy to verify using the Cauchy-Schwarz Inequality [Proposition 7.1 (iii)] that y* is a bounded linear functional with II y* II = II y II. Thus

Sec. 7.3 • The Dual Space and Adjoint Operators 143

we can define a mapping y -- y* of Pinto P* that is an isometry and conjugate linear, that is,

(y + z)* = y* + z*, for all y, z E P

(ay)* = ay* , for all scalars a and all y in P.

Since p* is always complete, P will be complete-a Hilbert space-if this mapping is surjective. The converse is also true, as shown in the next theorem.

Theorem 7.5. Let H be a Hilbert space. For each continuous linear functional fin H*, there exists a unique y in H such that f(x) = (x 1 y) for all x in H. Thus the conjugate linear isometry y -- y* [given by equation (7.23)] of a pre-Hilbert space P into its dual P* is surjective if and only if P is actually a Hilbert space. I

Proof. Let N be the null space {x E H: f(x) = O} of f, a closed linear subspace of H. If N = H, f = 0 and y = O. If N oj::. H, since (Nly = N by Theorem 7.2, Nl. does not equal the zero space. Let z E Nl. with z oj::. O. Since N (') Nl. = {O}, fez) oj::. O. Replacing z by [f(Z)]-lZ we may assume f(z) = 1.

Now for x E H,f(x - f(x)z) = f(x) - f(x) = 0 so that x - f(x)z EN. Since z E Nl.,

0= (x - f(x)z 1 z) = (x 1 z) - f(x)(z 1 z)

or

. (x 1 z) (I z ) j(x) = (z 1 z) = x --r;r .

Letting y = z/II Z 11 2, f(x) = (x 1 y), where y is independent of x. If also w is in H so that f(x) = (x 1 w) for all x in H, then in par-

ticular f( w - y) = (w - y 1 w) = (w - y 1 y) or w = y. I

Remark 7.5. The dual H* of a Hilbert space H is a Hilbert space with inner product given by

(x* 1 y*) = (y 1 X)H, (7.24)

for each x* and y* in H* where x -- x* and y -- y* as in equation (7.23) under mapping H -- H*. [Here (I)H denotes the inner product in H.]

The inner product given by equation (7.24) is compatible with the already existent norm on H* as II y* 112 = II Y 112 = (y I Y)H = (y* I y*). We may s:ry therefore that Hand H* are "conjugate" isomorphic since the isometric surjection y ---+ y* from H to H* is conjugate linear and (x* I y*) = (x I Y)H .

Corollary 7.2. Each Hilbert space is reflexive. I

Proof. The composition of the conjugate isometric surjections H ---+ H* and H* ---+ (H*)* is easily seen to be the natural mapping J: H ---+ H** given by J(h) = h, where h(h*) = h*(h) for all h* in H*. I

Definition 7.9. If X and Yare normed linear spaces, a sesquilinear form B on Xx Y is said to be bounded if there exists a constant M such that I R(x, y) I < M II x II II y II for all x in X and yin Y. The norm II B II of B is given by

II B II = inf{M: I B(x, y) I < Mil x II II y II for all x in X, y in Y}. I

Proposition 7.10.

(i) If B is a bounded sesquilinear form on Xx Y, where X and Y are normed linear spaces, then

II B II = sup{1 B(x, y) I: II x II < I and II y II < I}

= sup{1 B(x, y) I: II x II < I and II y II < l} = sup{1 B(x, y) I: II x II = I and II y II = I},

and I B(x, y) I < II B II II x II II y II for all x in X and y in Y.

(ii) If P is a pre-Hilbert space and B is a bounded Hermitian sesquilinear form on P then

II B II = sup{1 B(x, x) I: II x II < l}.

[Problem 7.3.6 shows that a converse of (ii) is not always true.] I

Proof. The proof of part (i) is analogous to the proof of the corresponding equalities for norms of linear operators on normed linear spaces and is left to the reader.

To prove (ii) we first note that by assumption B(x, x) is real since B is Hermitian. If II x II < I, then by part (i), I B(x, x) I < II B II so that

S < II B II if S equals the sup{1 B{x, x) I: II x II < I}. It suffices to show that for any x and Y in P with II x II < 1 and II y II < 1 we have I B{x, y) I < S. We look at two cases.

Case 1. Suppose B(x, y) is real. By Proposition 7.1 (i)t we have

B{x, y) = !{B(x + y, x + y) - B{x - y, x - y)}.

Using the Parallelogram Law,

I B(x, y) I < HI B(x + y, x + y) I + I B(x - y, x - y) I} < l{S II x + y 112 + S II x - y 112} = tS{211 x 112 + 211 y 112} < S.

Case 2. In general, write I B(x, y) I = aB(x, y), where a is a complex number of norm l. Then B(ax, y), equal to aB{x, y), is a real number. Hence by case I,

I B(x, y) I = I B{ax, y) I < S. I If Hand K are Hilbert spaces each bounded linear operator T: H -+ K

generates a bounded sesquilinear form BT on Hx K by the formula

BT{x, y) = (Tx I Y)K, (7.25)

where (I )K is the inner product of K. lt is easy to verify that BT is sesquilinear and that II BT II < II T II using the Cauchy-Schwarz Inequality, Proposition 7.1. If II x II < I and II y II < I, then by definition of II BT II,

I BT(x, y) I = I (Tx I y)K I = I (y I TX)K I < II BT II.

Fixing x and taking the supremum over II y II < I, we get II Tx II < " BT " since (y I TX)K for fixed x is a continuous linear functional on K of norm II Tx II. Now taking the supremum over II x II < I, we get II Til < II BT II. Hence II Til = II BT II.

lt is interesting that every bounded sesquilinear form B on Hx K is equal to BT for some bounded linear operator T from H to K. This is the content of the next theorem.

Theorem 7.6. If Hand K are Hilbert spaces, then for each bounded sesquilinear form Bon Hx K there exists a unique bounded linear operator T: H -+ K such that B{x, y) = (Tx I y) for all x and y. Moreover II B II = II Til. I t The other two terms on the right side of Proposition 7.1 (i) cancel out, in this case,

after simplification.

Proof. For each x in H, define Ix: K -+ F (the scalar field) by the formula

(7.26)

Then Ix is linear (easily checked) and

11x(y) I = I B(x, y) I < II B II II x II II y II.

So Ix is bounded with II Ix II < II B II II x II. By Theorem 7.5, there exists for each x in H a unique element Zx in K such that II Ix II = II ZX II and Ix(y) = (y I zx) for all y in K. Define T: H -+ K by Tx = Zx. Then

B(x, y) = Ix(y) = (y I zx) = (y I Tx),

so that (Tx I y) = B(x, y). We now assert that T is linear, bounded with norm II B II, and is the unique operator satisfying B(x, y) = (Tx I y):

T is linear: for all y, (T(a1x1 + a2x2) I y) = (a1Txl + a2Tx2 I y). T is bounded: II Tx II = IIIx II < II B II II x II. (Clearly now II Til

= II B II by our remarks before the theorem.) T is unique: If S is also a bounded linear operator satisfying (Sx I y)

= B(x, y), then the equality (Sx - Tx I y) = 0 for all y implies Sx = Tx ~d~ I

Given a bounded linear operator T: H -+ K, we have shown there is a unique bounded sesquilinear form BT on Hx K satisfying the formula BT(x, y) = (Tx I y). Moreover, II BT II = II T II. In the same manner

B*(y, x) = BT(x, y) = (Tx I y) = (y I Tx)

defines a bounded sesquilinear form B* on Kx H with norm II B* II satisfying

II B* II = II BT II = II Til·

However, by Theorem 7.6 we know that to B* there corresponds a unique operator S: K -+ H satisfying B*(y, x) = (Sy I x) and with II S II = II B* II. Hence corresponding to T: H -+ Kthere corresponds a (necessarily unique) bounded linear operator S: K -+ H satisfying

(Tx I y) = BT(x, y) = B*(y, x) = (Sy I x) = (x I Sy)

and

II Til = II BT II = II B* II = II S II·


The operator S is called the adjoint of T and is generally denoted by T*. Formally we have proved the following result.

Theorem 7.7. To each bounded linear operator T: H -- K there corresponds one and only one linear operator T*: K -- H, called the adjoint of T satisfying (Tx I Y)K = (x I T*Y)H for all x in Hand Y in K. Moreover II Til = II T* II· I

Problem 7.3.7 gives some examples of adjoints of operators.

Remark 7.6. In Chapter 6, the adjoint T*: y* -- X* of a continuous linear operator T: X -- Y was defined for normed space X and Y by the equ~tion T*(f*)(x) = f*(T(x)) for f* E y* and x EX. Denoting this adjoint momentarily by T' instead of T*, we remark that T' is not the same operator defined in Theorem 7.7 when X and Yare Hilbert spaces. Indeed the domains are different-that of T' being the dual space Y*. To emphasize this distinction the T* of Theorem 7.7 is sometimes called the Hilbert space adjoint. The relationship between T* and T' is examined in Problem 7.3.8.

We conclude this section with several easily verified results giving some properties of bounded linear operators and their adjoints on Hilbert spaces.

Proposition 7.11. For any bounded linear operators T: H -- K and S: H -- K and scalar a we have the following:

(i) (y I TX)K = (T*y I X)H for all x E Hand Y E K;

(ii) (S + T)* = (T* + S*);

(iii) (aT)* = liT*;

(iv) (T*)* = T;

(v) II T*T II = II TT* II = II T 112;

(vi) T*T = 0 if and only if T = o.

Proof

(i) Left to the reader.

(ii) (x I (S + T)*Y)H = «S + T)x I Y)K = (Sx I Y)K + (Tx I Y)K

I

= (x I S*Y)H + (x I T*Y)H = (x I S*y + T*Y)H

for all x in Hand y in K. Hence

(S + T)*y = (S* + T*)y for all y.

(iv) (T*)*x I y)K = (y I (T*)*X)K = (T*y I X)K = (Tx I y)K

for all x in Hand y in K. Hence (T*)* = T.

(v) For II x II < 1,

II Tx 112 = (Tx I TX)K = (T*Tx I X)H < II T*Tx II II x II < II T*T II.

Hence II T 112 < II T*T II. However, II T*T II < II T* II II Til = II T 112 so II T*T II = II T 112. Replacing T by T* and T* by (T*)* or T gives II TT* II = II T* 112 = II T 112.

(vi) follows from (v) and the proof of (iii) is similar to that of (ii). I

Proposition 7.12. If T: H -+ K and S: K -+ L are continuous linear operators (H, K, and L are Hilbert spaces), then (ST)* = T*S*. I

Proof. For all x in Hand y in L,

(ST)*y I X)H = (y I STxh = (S*y I TX)K = (T*S*y I X)H' I

Proposition 7.13. If T: H -+ K is a continuous linear mapping and Me Hand N C K with T(M) C N, then T*(N.l) C M.l. I

Proof. If y E N.l and m E M,

(T*y I m)H = (y I Tm)K = 0 so that T*y E M.l. I A stronger conclusion in Proposition 7.13 can be obtained if N is a

closed subspace, as Proposition 7.14 shows.

Proposition 7.14. If T: H -+ K is a continuous linear mapping, Mis a linear subspace of H, and N is a closed linear subspace of K, then TM C N if and only if T*(NJ.) C M.l. I

Proof. The necessity is proved by Proposition 7.13. If T*(N.l) C M.l, then also T**(MH) C NH by Proposition 7.13. Since MH) M, NH = N by Theorem 7.2 and T** = T, T(M) C N. I

Proposition 7.15. If T: H -+ K is a continuous linear mapping, then

(i) {x: Tx = O} = [T*(K)].l;

Sec. 7.3 • The Dual Space and Adjoint Operators

(ii) {x: Tx = O}.l = T*(K);

(iii) {y: T*y = O} = [T(H}].L;

(iv) {y: T*y = O}.l = T(H).

Proof. For a fixed x E H,

(Tx I y)K = 0 for any y E K,

if and only if

(x I T*Y)H = 0 for any y E K.

This proves (i). From (i),

{x: Tx = O} = [T*(K)].1 = [T*(K)].l,

so that

{x: Tx = O}.l = [T*(K)].l.l = T*(K).

149

I

(7.27)

(7.28)

(7.29)

(7.30)

This proves (ii). The other two parts follow similarly if we work with T* instead of T and note that T** = T. I

Having introduced the idea of the adjoint in this section and having given some of its essential properties, we will use it in the next section to define special classes of operators. In particular we will show in Section 7.5 that each bounded linear operator T: H -+ H which equals its adjoint T* has a neat representation.

Problems X 7.3.1. If P is a pre-Hilbert space such that N.l.l = N for every closed linear subspace N of P, show P is a Hilbert space by showing that every continuous linear functional y* in P* is of the form y*(x) = (x I y) for some y in P. X 7.3.2. Prove that if I is a linear functional on a Hilbert space H, then the null space N = {x: I(x) = O} is dense in H if I is not continuous. X 7.3.3. (i) Prove that in a Hilbert space H, a sequence (xn) converges weakly to x in H if and only if the sequence (xn I y» converges to (x I y) for each y in H.

(ii) Give an example of a sequence (xn) in the infinite-dimensional Hilbert space H that converges weakly but not strongly in H. [Hint: Use (i) and Bessel's Inequality.]

7.3.4. (i) Let Px Q be the direct product of pre-Hilbert spaces P and Q. For x = (PI' ql) and y = (P2' q2) in Px Q, define (x I y) as (PI I P2)P + (ql I q2)Q' Show that P x Q becomes a pre-Hilbert space with this inner product and that II (x I y) 112 = II X 112 + II y 112. Show that Px Q equals Po EB Qo, where Po is the range of the mapping from Pinto P x Q given by P --+ (p, 0) and similarly Qo is the range of the mapping from Q into P x Q. Show that P x Q is a Hilbert space if and only if P and Q are Hilbert spaces.

(ii) Generalize (i) to an arbitrary collection (P ,,) of pre-Hilbert spaces.

7.3.5. If X and Yare normed linear spaces, prove that the set B(X, Y) of bounded sesquilinear forms on Xx Y is a normed linear space linearly isometric to L(X, (Y)*), where Y is the complex conjugate of Y -that is, the same space as Y except that addition EB and scalar multiplication 8 in Y is given by a 8 x EB {3 8 y = ii . x + P . y for all scalars a, (3 and vectors x and y, where + and . represent the operations in Y. Conclude B(X, Y) is a Banach space. Compare Theorem 7.6.

7.3.6. (i) Let P be a complex pre-Hilbert space and define B(x, y)

= i(x I y). Prove B is a bounded sesquilinear form on P with II B II = sup {I H(x, x) I: II x II < I} but B is not Hermitian. Compare Proposition 7.1O(ii).

(ii) Let P be any pre-Hilbert space. Suppose T: P --+ P is a continuous linear mapping with II T II = 1 and Tp = P for some nonzero P in P. Define BT(x, y) = (Tx I y). Show that BT is sesquilinear with II BT II = sup{1 BT(x, x) I: II x II < I}, but BT is not necessarily Hermitian.

X 7.3.7. (i) Let T be a mapping from a separable Hilbert space H into itself and let (ei) be a basis of H. Then Tej = Li-laijei as in Proposition 6.6. The collection (aij) for i = 1,2, ... and j = 1,2, ... is called the matrix of T with respect to the basis (ei)' Show (Tej I ek) = akj and that the matrix of T* with respect to (ei) is the collection of scalars ({3ij) for i = 1,2, ... , and j = 1,2, ... where {3ij = (iji'

(ii) What is the adjoint of the shift operator T: 12 --+ 12 given by T(al , a2 , ••• ) = (0, aI' a2 , ••• )1

X 7.3.8. If Hand K are Hilbert spaces, and U: H --+ H* and V: K --+ K* are the conjugate isometries discussed in Theorem 7.5, prove that T* = U-IT'V, where T* is the Hilbert space adjoint of T, and T' is the Banach space adjoint from K* to H* given by T'(k*)h = k*(Th).

7.3.9. Let Hand K be Hilbert spaces and {hJiEI and {kj}jEJ be bases for Hand K, respectively. If T: H --+ K is a bounded linear mapping, LIII Thi 112 converges if and only if LJII T*kj 112 converges and in this case both sums equal Li,il (Thi I k j ) 12.

7.3.10. A bounded linear operator T: H -+ K is said to be a HilbertSchmidt operator if there exists a basis {hi} in H such that LIII Thi 112 < 00.

Prove the following using Problem 7.3.9: (i) If T is a Hilbert-Schmidt operator, show LIII Tgi 112 < 00 for

every basis {gih of H; (ii) T is a Hilbert-Schmidt operator if and only if T* is a Hilbert

Schmidt operator. (iii) The class of Hilbert-Schmidt operators is a pre-Hilbert space

with the inner product (S I T) = LI(Shi I Th i ) where {hih is a basis of H. (iv) The pre-Hilbert space of (iii) is complete. Prove this by showing

that it is isomorphic to '2(1 x J) via T -+ (Xij)IXJ, where Xij = (Thi I k j) and where {hJI and {kj}J are bases of Hand K, respectively.

(v) Show that a Hilbert-Schmidt operator is compact. (vi) If (X,3iJ, /1) is a a-finite measure space and Sf I K(x, y) 12

d/1 (x) d/1 (y) < 00, prove that T: L 2(/1) -+ L 2(/1) given by (Tf)(x) =

f f(y)K(xy) d/1 (y) is a Hilbert-Schmidt operator.

7.3.11. Prove that the following statements are equivalent in a pre-Hilbert space P.

(i) P is complete. (ii) If M is a closed subspace of P, then M EB Ml. = P.

(iii) If M is a closed subspace of P, then M = MH.

(iv) If M is a proper closed subspace of P, then Ml. oj::. {O}. (v) If f is a continuous linear functional on P, then there exists

x E P such that fey) = (y I x) for all YEP.

X 7.3.12. Suppose that S C R and that for each s in S there is a bounded linear operator T(s) on H. An operator A on H is the weak limit of the function T: S -+ L(H, H) as s -+ so, written A = w-lims-+soT(s), if for any pair hand k in H

(Ah I k) = li~ (T(s)h I k).

An operator A on H is the strong limit of T(s) as s -+ so, written A = s-lims ..... oT(s) if for all h in H

lim II (T(s) -- A)h 11-+ O. 8-+80

An operator A on H is the uniform limit of T(s) as s -+ So if

lim II T(s) - A 11-+ O. 8"~80

Show that if A is the uniform limit of T(s) as s ->- So, then A is the strong limit of T(s) as s ->- So. Show also that if A is the strong limit of T(s) as s ->- so, then A is the weak limit of T(s) as s ->- so. (Note here that So may be ± 00 and S may be the natural numbers N.)

X 7.3.13. Let A be a linear operator from a Hilbert space H into itself with (x lAy) = (Ax I y) for all x and y E H. Prove that A is bounded. (Hint: Use the Closed Graph Theorem.)

7.3.14. If T E L(H, H), define weT) = sup {I (Tx I x) I: II x II = I}. Prove that

(i) w(cT) = I c I weT) for all scalars c; (ii) weT} + Tz) < weT}) + w(Tz) for T}, T2 E L(H, H);

(iii) weT) < II T II < 2w(T). Conclude that w defines a norm on L(H, H) equivalent to the usual operator norm. [Hint: To prove II T II < 2w(T), let B(x, y) = (Tx I y) and prove I B(x, y) I < 2w(T) for all II x II = II y II = I.]

7.4. The Algebra of Operators. The Spectral Theorem and the Approximation Theorem for Compact Operators

In this section we will first examine some special classes of operators on a Hilbert space H-that is, special classes of operators from H into itself. These special classes of operators can all be defined by use of the adjoint. Secondly, we wi11 prove a spectral theorem for compact operators which the spectral theorem of the next section wi11 generalize. Finally, we will show how each completely continuous operator can be approximated by operators with finite-dimensional ranges.

Let us first give in capsule form the definition of all types of operators we will consider in this section.

Definition 7.10. Let T: H ->- H be a continuous linear operator with adjoint T*.

(i) T is isometric if and only if T*T = I, the identity of H.

(ii) T is unitary if and only if T*T = TT* = I. (iii) T is self-adjoint (or Hermitian) if and only if T = T*.

(iv) T is a projection if and only if T2 = T and T* = T.

(v) T is normal if and only if T*T = TT*. I

Sec. 7.4 • The Algebra of Operators 153

Note that when we speak of T as being anyone of the types of operators of our Definition 7.10, T is understood to be continuous and linear. The reader should consult Problem 7.4.1 for various examples of these types of operators.

Clearly, every unitary, every self-adjoint, and every projection operator is normal. The following four propositions give equivalences of isometric, unitary, self-adjoint, and normal operators, respectively. We first prove the following lemma.

Lemma 7.3. If H is a complex Hilbert space and S, T: H ~ Hare bounded linear operators such that (Sx I x) = (Tx I x) for all x, then S=~ I

Proof. The sesquilinear forms Bs(x, y) = (Sx I y) and BT(x, y) = (Tx I y) are such that Bs(x, x) = BT(x, x) for all x in H. By the polarization identity [Proposition 7.1(i)], Bs(x, y) = B7,(x, y) for all x and y in H. By Remark 7.1, S = T. I

Proposition 7.16. The following conditions on T: H -+ H are equivalent.

(i) T is isometric.

(ii) (Tx I Ty) = (x I y) for all x and y in H.

(iii) II Tx II = II x II for all x in H. I

Proof. (i) => (ii) since (Tx I Ty) = (x I T*Ty) = (x I y). Trivially (ii) implies (iii). If we assume (ii), then (T*Tx I y) = (x I y) for all x and y so that T*Tx = x for all x, or T*T = I. Hence (ii) => (i). If H is a real Hilbert space, then by Proposition 7.1 (i)

(Tx I Ty) = HII Tx + Ty 112 - II Tx - Ty Jj2}.

Clearly (iii) implies (ii) if H is real. Similarly if H is complex Proposition 6.1 (i) gives that (iii) => (ii). I

Proposition 7.17. The following conditions on T: H ~ H are equivalent.

(i) T is unitary.

(ii) T* is unitary.

154

(iii) T and T* are isometric.

(iv) T is isometric and T* is injective.

(v) T is isometric and surjective.

(vi) T is bijective and T-l = T*.

Chap. 7 • Hilbert Spaces

I

Proof. Since T** = T, the equivalences of (i), (ii), and (iii) are trivial, as are the implications (iii):= (iv) and (vi):= (i). The proof is completed by showing (iv):= (v) and (v):= (vi).

(iv):= (v): Since T is isometric, T(H) is closed. Hence by Proposition 6.15,

H = {O}.L = {x: T*x = O}.L = T(H) = T(H).

(v):= (vi): Let S = T-I. To show T* = S. Since T is isometric,

T* = T*I = T*(TS) = (T*T)S = IS = S. I

Proposition 7.18. Let T: H --+ H be a continuous linear operator. Let (i), (ii), (iii), and (iv) represent the following statements:

(i) T is self-adjoint. (ii) (Tx 1 y) = (x 1 Ty) for all x and y in H.

(iii) (Tx 1 x) = (x 1 Tx) for all x in H.

(iv) (Tx 1 x) is real for all x in H.

Then (i) is equivalent to (ii), (iii) is equivalent to (iv), and if H is a complex Hilbert space, all the statements are equivalent. I

Proof. All the implications are trivial except perhaps (iv) ~ (i) in the complex case. Suppose (iv) holds. Then

(T*x 1 x) = (x 1 Tx) = (Tx 1 x)

for each x in H. By Lemma 7.3, T = T*. I

Proposition 7.19. If H is a Hilbert space, then T: H --+ H is normal if and only if II T*x II = II Tx II for each x in H. I

Proof. If T is normal,

II T*x 112 = (T*x 1 T*x) = (TT*x 1 x) = (T*Tx 1 x) = (Tx 1 Tx) = II Tx112. (7.31 )

Sec. 7.4 • The Algebra of Operators ISS

Conversely, if II T*x II = II Tx II, equation (7.31) shows that T is normal using Lemma 7.3 in the case H is complex. However, in either case the Hermitian sesquilinear forms Band B', given by B(x, y) = (Tx I Ty) and B'(x, y) = (T*x I T*y), respectively, are such that

B(x, x) = (Tx I Tx) = II Tx 112 = II T*x 112 = (T*x I T*x) = B'(x, x),

for all x in H. By the polarization identity, Proposition 7.1 (i), B(x, y) = B'(x, y) for all x and y in H. Hence

(T*Tx I y) == (Tx I Ty) = (T*x I T*y) = (IT*x I y).,

for all x and y in H. Hence T*T = IT* by Remark 7.1. I

We will take a closer look at self-adjoint operators in what follows. Before doing so, let us examine a few pertinent theorems regarding some special self-adjoint operators-the projection operators.

Recall that if N is a closed linear subspace of H, then H = N EEl Nl.. We can define P: H ~ H by the following rule: If x = y + z with YEN and Z E Nl., let Px = y. P is easily checked to be a bounded linear operator, but also satisfies the equations p2 = P and P* = P. Indeed, Px E N implies P(Px) = Px and

where Xl = YI + Zl and X 2 = Y2 + Z2 with YI' Y2 E Nand Zl' Z2 E Nl.. P is called the projection t operator associated with N and is denoted by P N'

Proposition 7.20. If T: H ~ H is a projection operator, then there is one and only one closed linear subspace N such that T = PN' In fact N = T(H) and Nl. = Ker T. I

Proof. LetN = {x: (T - I)x = O} = {x: Tx = x}, a closed subspace. If y is in the range R(T) of T, then for some x, y = Tx = T(Tx) = Ty so that y is in N. Conversely, if YEN, y = Ty E R(T). Therefore N = R(T). Now by Proposition 7.15

{x: Tx = O} = [T*(H)]1. = [T(H)]l.

t P is also called the orthogonal projection on N.

or N1- = {x: Tx = O}. Let x E H.Then x = y + z with YEN and z E N1-. Hence Tx = T(y + z) = Ty + Tz = y. Hence T = PN •

If T = PN = PM for some subspaces Nand M then N = PN(H)

=~~=M I

• An Application of Proposition 7.20: The Mean Ergodic Theorem. In Chapter 3, we discussed measure-preserving mappings and the Individual Ergodic Theorem showing the existence of an a.e. pointwise limit for Ll functions. Here we show how the subject of unitary operators comes up naturally in the context of ergodic theory. We present below the Mean Ergodic Theorem, proven by von Neumann in 1933, which will consider L2 functions and convergence in the L2 norm. We will use Proposition 6.20.

First, we notice that if cJ> is a measure-preserving mapping on a measure space (X, d, p,) and if

T(f)(x) = f(cJ>(x» , fE Lip),

then T is a unitaryt operator on L 2 • Thus the Mean Ergodic Theorem in L2 can be presented in the following Hilbert space setup.

The Mean Ergodic Theorem. Let T be a unitary operator on a Hilbert space H. Let P be the orthogonal projection onto the closed linear subspace N = {x: Tx = x}. Then, for any x E H,

as n --+ 00.

Proof. First, we notice that by Proposition 7.15,

[(I - T)(H)]1- = Ker(I - T*) = {x: T*x = x} = {x: T-1x = x}

=N(= P(H».

Now if x = y - Ty, then since II T II = 1 [by Proposition 7.11 (v)],

II Tnx II = II {ljn)[y - Tny] II < (2jn) II y 11--+ 0, as n --+ =.

By Theorem 7.2 and Proposition 7.20,

Ker P = N1- = (I - T)(H).

t Note that T*(f)(x) = f«(P-l(X».

I

Therefore, for x E NJ., limn-+ooTnxt = 0 = Px. Also, for x E N = P(H), limn-+=Tnx = Tx = x = Px. Since H = NEB Nl., the theorem now follows easily. I

Having established a one-to~one correspondence between projections on H and closed linear subspaces of H, it is interesting to investigate the preservation properties of this correspondence. Before giving these properties we define a partial ordering on the class of self-adjoint operators.

Definition 7.11. A self-adjoint operator T is said to be positive in case (Tx I x) > 0 for all x E H. If Sand T are self-adjoint operators, S < T if T - S is positive, written T - S > o. I

Proposition 7.21. Suppose MI and M2 are closed linear subspaces corresponding, respectively, to projections PI and P2 on H. Then the following hold:

(i) The assertions MI 1- M2 , P1P2 = 0, P2P1 = 0, P1(M2) = {O}. and P2(M1) = {O} are equivalent.

(ii) P1P2 is a projection if and only if P1P2 = P2P1 • If this condition is satisfied the range of P1P2 is MI () M 2 •

(iii) The assertions PI < P2 , "PIX II < "P2x " for all x, MI C Ma, P2P1 = PI' and P1P2 = PI are equivalent. I

Proof. (i) MI 1- M2 implies M2 C MIl.. As shown in the proof of Proposition 7.20, M2 = {x: P2x = x} = R(P2) and MIl. = {x: PIX = O}. Since M2 C MIl. and P2x E M2 for any x in H, P1P2 = O. Now suppose P1P2 = O. Then if x E M2, P2x = x and therefore PIX = P1P2X = O. Hence P1(M2) = {O}. Now suppose P1(M2) = {O}. Then M2 C MIl. and therefore M2 1- MI' The other equivalences of (i) follow by symmetry.

(ii) P1P2 a projection implies

P1P2 = (P1P2)* = P2*P1* = P2P1 •

Conversely, P1P2 = P2P1 implies

t Note that

" Tnx " ~ " Tnz " + " Tn(x - z) " ~ " Tnz" + " x - Z "

for Z E (I - T)(H).

and (PIP2)* = P2*PI* = P2PI = PIP2·

Finally if PIP2 is a projection, then the range M of P = PIP2 is contained in MI and M2 since PI and P2 commute. Also if x E MI n M 2, then PIX

= X = P2x or Px = X so that x EM. Hence M = MI n M 2·

(iii) If PI < Pz, then II PIX liZ = (PIX I x) < (Pzx I x) = II P2x 112 for all x. If II PIX II < II P2x II for all x, then for X E M I ,

so that II P2x II > II X II. Since II X 112 = II P 2x 112 + II X - P 2x 11 2, P2x = X and X E M 2 • If MI C M 2, then for all x, PIX E M 2 • Hence P2PIX = PIX.

If P2PI = PI' then PIP2 = (P2PI )* = PI* = Pl' If PIP2 = PI' then

We have therefore seen that the one-to-one correspondence between closed subspaces of H and projections on H preserves order: MeN if and only if PM < PN • Thus the set of projections is a partially ordered set such that for every family of projections {Pih there exists a "greatest" projection "smaller than" each Pi and a "smallest" projection "greater than" each Pi'

Self-adjoint operators are also called Hermitian operators. This terminology is first of all consistent with that used for square complex matrices (aij) where aij = iiij' Indeed if T is a self-adjoint operator on the Hilbert space en, let (aij) i = I, ... , nand j = I, ... , n be the matrix that represents T relative to some orthonormal basis {cI , ... , cn} in en-that is, TCj = L~~laijCi' Problem 7.3.7 (i) shows that the matrix of T* is (bij),

where bij = iiji . Thus T is Hermitian if and only if (aij) is a Hermitian matrix. In particular we know from the theory of diagonalization of Hermitian matrices that T can be represented with respect to some orthonormal basis, say {Xl, ... , x n }, by a diagonal matrix so that TXi = AjXi,

i = I, ... , n, for some scalars AI,"" An' Hence if X E en, X =

Lb.I(X I Xi)Xi by Proposition 7.6 so that

n Tx = L Ai(X I Xi)Xi'

i~1

(7.32)

t For a projection operator Q"* 0, 1/ Q /I = 1 since Q = Q2 and /I Qx II" :s; II Qx II" + II x - Qx 112 = II X 11 2. See 7.4.19.

It is formula (7.32) we wish to generalize to compact self-adjoint operators in this section. We return to it in Theorem 7.8.

The term "Hermitian" for self-adjoint operators is also consistent with that used for sesquilinear forms. Indeed. if BT is defined on Hx H by BT(x, y) = (Tx I y), then BT(x, y) = (Tx I y) = (y I Tx) = (T*y I x) so that BT(x, y) = BT(y, x) if and only if T = T*, by Remark 7.1. Hence if T is self-adjoint, then by Proposition 7.10 (ii),

II Til = II BT II = sup{l (Tx I x) I: II x II < I}.

These remarks enable us to prove the next results.

Lemma 7.4. Let T be a self-adjoint operator on H. Let mT = inf{(Tx I x): II x II = I} and MT = sup{(Tx I x): II x II = I}. Then mTI< T< MTI and II T 1/ = max{MT' - mT}. I

Proof. For all x in H,

mT(x I x) = mT11 x 112 < (Tx I x) < MTII X 112 = MT(x I x),

which shows mTI < T < MTI. The rest is clear. I

Lemma 7.5. Let T be a positive self-adjoint operator on H. Then

I (Tx I y) 12 < (Tx I x)(Ty I y).

for all x, y E H. I

Proof. Apply the Cauchy-Schwarz inequality, Proposition 7.1 (iii). to the positive Hermitian sesquilinear form B(x, y) = (Tx I y). I

Lemma 7.6. Suppose T is a self-adjoint operator, on a Hilbert space H.

(i) If ft is an eigenvalue of T, then ft is real and mT < ft < M T,

mT = inf{(Tx I x): II x II = I}, MT = sup{(Tx I x): II x II = l}.

Eigenvectors corresponding to distinct eigenvalues are orthogonal.

(ii) a(T) C [mT' MT] and the endpoints mT and MT are both in aCT). I

Proof. (i) Since (Tx I x) is real and equals /-lex I x) with (x I x) > 0, /-l must be real. Also if x is an eigenvector with II x II = 1,

/-l = /-l II X 112 = (/-lxl x) = (Tx I x)

so that mT < /-l < MT·

Now suppose /-lo is an eigenvalue with /-lo * /-l and let Xo and x be eigenvectors corresponding to /-lo and /-l, respectively. Since /-lo is real,

Thus (x I xo) = 0 since /-l * /-lo· (ii) Suppose A E F - [mT' Md. Then t5 = dCA, [mT' MTD> 0 and

for every x in H with II x II = 1,

II (T - U)x II > I (T - U)x I x) I = I (Tx I x) - A I > t5. (7.33)

Hence by Proposition 6.5 in Chapter 6, T - )J is a bijection onto the range R(T - AI) of T - )J with a bounded inverse on R(T - AI). In particular R(T - AI) is complete and closed. Suppose R(T - AI)1- * {O}. Then for y E R(T - AJ)1- with II y II = I, we have (T - AI)(y) I y) = 0, which contradicts (7.33). Hence R(T - AI) = H and A rf. aCT).

Finally we show that m1' and M l' are in aCT). Since by Lemma 7.4 M TI - T > 0, we have, using Lemma 7.5 that for II x II = 1

II (MTI- T)x 114 = I (MTI - T)x I (MTI - T)x) 12

< (MTI - T)x I x)(MTI - T)2X I (MTI - T)x)

< II MTI - T 11 3 (MT - (Tx I x».

By virtue of the definition of M T,

inf{11 (MTI - T)x 114: II x II = l} < inf{MT - (Tx I x): II x II = I}

. II M TI - T 11 3 = 0.

Thus (MTI - T)-l does not exist as a bounded operator, so MT E aCT). Similarly mT E aCT). I

In particular if T is a nonzero compact, self-adjoint operator on H, then Lemmas 7.4, 7.6 and Theorem 6.25 show us that either II T II or - II T II is an eigenvalue of T. Hence there is a real number A such that I A I = II T II and a vector x in H with II x II = 1 and Tx = AX. We have proved the following result.

Lemma 7.7. Suppose T is a nonzero compact, self-adjoint operator on H. Then either II T II or -II T II is an eigenvalue of T and there is

a corresponding eigenvector x such that II x II = 1 and I (Tx I x) I = II T II. I

Before proving the spectral theorem for compact self-adjoint operators, let us establish two results concerning self-adjoint operators which will be needed in the next section.

Lemma 7.S. If T: H -+ H is a self-adjoint operator, then T is positive if and only if aCT) C [0,00). I

Proof. Let m7' = inf{(Tx I x): II x II = I}. If T is posItIve, m7' > ° and so by Lemma 7.6, a(T) C [0,00). On the other hand, if aCT) C [0,00) then m1' > ° by Lemma 7.6 and (Tx I x) > ° for all x. I

Lemma 7.9. Suppose (Tn) is a bounded sequence of self-adjoint operators with Tn < Tn+! (as in Definition 7.11) for n = 1,2, .... Then there is a self-adjoint operator T such that Tn converges strongly to T (that is, Tnx -+ Tx for each x in H). I

Proof. By assumption there exists a positive number M such that II Tn II < M or -MI < Tn < Ml. If tM-l(Tn + MI) converges strongly to a self-adjoint operator S, then Tn converges strongly to the self-adjoint operator 2MS - Ml. Thus, replacing Tn by tM-l(Tn + MI), we may suppose ° < Tn n, then ° < T mn < I if T mn = Tm - Tn, and II Tmn II < I by Lemma 7.4. Using Lemma 7.5 and Proposition 7.1 (iii)

II Tmx - Tnx 114 = II Tmnx 114 = II (Tmnx I Tmnx) 112

< (Tmnx I X)(r;.nX I Tmnx )

< (Tmnx I x) II T;'nx II II Tmnx II < II x 112[(Tmx I x) - (Tnx I x)]. (7.34)

Since ° < Tn < Tn+! < I, (Tnx I x) is a bounded increasing sequence of real numbers which converges so that (Tnx) is a Cauchy sequence by (7.34). Let Tx = limn-+ooTnx. It is easy to check that the operator T thus defined is a bounded self-adjoint operator. I

We have now come to a main objective of this section.

Theorem 7.S. Suppose T: H -+ H is a nonzero, compact, self-adjoint operator.


(i) If T(H) is finite-dimensional, then the nonzero eigenvalues of T

form a real finite sequence AI, A2 , ••• , An and Tx = L:f~lA/X I Xi)Xi for all x in H, where Xl, X2 , ••• ,Xn is a corresponding orthonormal set of eigenvectors.

(ii) If T does not have a finite-dimensional range, then the nonzero eigenvalues of T form a real infinite sequence AI, A2 , • •• with I Ai I --+ 0 such that Tx = L::I Ai(x I Xi)Xi for each X in H where Xl' X2 , • •• is an orthonormal set of eigenvectors with Xi corresponding to Ai' I

Proof. Suppose first that T is not of finite rank. We show first that for each n, there exists a nonzero closed subspace Xn of H, an eigenvalue An of T with I An I = /I T IXn /I, and a corresponding eigenvector Xn for An with /I Xn /I = 1 such that

I An I > I An+! I, Xn+I = {X: (X I Xi) = 0, for i = 1,2, ... ,n}, (7.35)

Xn E Xn , for n = 1,2, ....

Let Xl = Hand TI = T. By Lemma 7.7, there exists a vector Xl and a scalar Al such that /I Xl /I = 1, I Al I = /I TI /I, and TXI = AIXI ·

Proceeding inductively, suppose that for each i = 1, 2, ... , n - 1, Xi' Ai' and Xi have been chosen satisfying equation (7.35). Define

Xn = {x: (x I Xi) = 0, for i = 1,2, ... , n - I}.

Xn"# 0, for if Xn = ° then the set {Xl' X2 , ••• , xn- l } forms a basis for H, which contradicts the fact that T is of infinite rank. Clearly, Xn C Xn- l and T: Xn --+ Xn since for X E Xn

for j = I, 2, ... , n - 1. Letting Tn = T Ix , Tn is compact and self-adjoint. fI

Now if Tn = 0, then letting Yn-l = X - L:i.:-f(x I Xi)Xi we see that Yn-l E Xn so that TYn-1 = ° or Tx = L:i.:-IAi(x I Xi)Xi for each X in H. This is impossible since T has infinite rank. Hence Tn is nonzero. By Lemma 7.7 again, there is a vector xn E Xn and a scalar An such that /I xn II = I, I An I = /I Tn /I, and TXn = Anxn- Since X" ~ Xn- 1 ,

The sequence (Ai) of eigenvalues thus formed converges to zero. If not, Xn = T(xnIAn) would have a convergent subsequence x7I /c since (xnIAn)

is bounded and T is compact. However, since the xnk are orthonormal, II xnk - X nk' 112 = 2 if k:j::. k' and the sequence (xnk) can never converge.

Setting again Yn = X - L~~I(X I Xk)Xk, we have

n II Yn !12 = II X 112 - L I (x I Xk) 12 < II X 112.

k-l

Since Yn E Xn+! and I An+! I = II T IXn+l II,

II TYn II < I An+! I II Yn II < I An+! I II x II,

which means TYn ~ O. Hence 00

Tx = L Aix I Xk)Xk' k~l

Finally, if A is a nonzero eigenvalue excluded from (Ak), then Tx = AX for some vector x with II x II = 1 and (x I Xk) = 0 for all k. Hence Ax = Tx = 0, which is a contradiction. We have proved (ii).

Suppose now T has finite rank. Clearly T has only a finite set of distinct eigenvalues, for if AI' A2 , ••• is an infinite set of distinct nonzero eigenvalues with Xl' X 2 , •• 0 a corresponding orthogonal set of eigenvectors, then T(H) is infinite dimensional as each Xk is in T(H) since Xk = T(Ak-IXk)'

Just as in expression (7.35), using the same procedure, let (Ak) be the finite set of all eigenvalues with corresponding eigenvectors (Xk) such that I Ak+! I < I Ak I = II T IXk II. If m + I is the smallest integer such that Am+! = 0, then T IXm+l = 0, and, as before, Tx = Li!.lAi(X I Xi)Xi' In the other case, there is a smallest integer m such that Xm+! = {O}; then {Xl' X2'

o • 0 ,Xm } is a basis for H. I

At this juncture we can prove another significant result using the theorem just proved. In short, this result says that each Hilbert space satisfies the approximation property (see Chapter 6). More precisely, we have the following theorem.

Theorem 7.9. Let T: H ~ H be a compact operator. Then there is a sequence Tn of finite-rank linear operators converging to T in the norm (uniformly). I

Proof. First assume H is separable. Then H contains a complete orthonormal sequence (Xk)' Define Tn by

n Tnx = L {x I Xk)Xk o

k~l

Clearly (by Proposition 7.6) II Tn II < I, III - Tn II < I, and Tnx -+ x as n -+ 00 for each x in H. If the conclusion of the theorem is false, then there exists a 150 > 0 such that II T - A II > 150 for all finite-rank linear operators A. Since TnT has finite rank, there exists for each n a Yn in H with II Yn II = 1 and II (T -. TIIT)Yn II > 150 /2. Since the sequence (Yn) is bounded, there exists a subsequence (Yni) of (Yn) such that TYni converges to some z in H. However,

II (T - Tn,T)Yn, II = II (I - Tn,)TYnj II < II (l- Tn)(TYnj - z) II + II {l- Tn,)z II, (7.36)

and the left side of (7.36) goes to zero as i -+ 00. Since this is a contradiction, the theorem is proved when H is separable.

Now let H be any Hilbert space. Let S = T*T. Then S* = S, so S is self-adjoint and compact. t By Theorem 7.8, Sx = L%:1Aix I Xk)Xk for each x in H, where (Xk) is some orthonormal sequence of eigenvectors with corresponding eigenvalues (Ak)' Let Ho be the subspace of H spanned by {PXk: n = 0, I, ... , and k = 1,2, ... }. Then Ho is a separable Hilbert space and T: Ho -+ Ho. However, we know by what we have just proved that there is a sequence {Fn} of finite-rank operators converging to THo in norm. Now H = Ho EEl Ho.L. Therefore for each x in H, x = Y + z with Y E Ho and z E Ho.L. Since z 1. Xk for all k = I, 2, ... , Sz = O. Therefore T*Tz = 0 so that Tz = 0 as (Tz I Tz) = (T*Tz, z) = O.

Define Gn on H by Gnx = FnY for n = 1, 2, .... Then Gn are finiterank operators, and since II x 112 = II Y 112 + II Z 11 2,

II T - Gn II = sup II Tx - Gnx II = sup II Ty - FnY II < II THo - Fn II· I IIXII=1 IIxll=1

• Application of Theorem 7.8: Fredholm Integral Equations and the Sturm

Liouville Problem

Let us first consider here solving what is known as a Fredholm integral equation with a Hilbert-Schmidt-type kernel. It is an equation of the form

f(s) '- A f: K(s, t)f(t) dt = g(s), (7.37)

where K(s, t) is a complex-valued measurable function on [a, b] X [a, b]

t By Proposition 6.21 and Theorem 6.21.

such that

II K 112 = f: J: I K(s, t) 12 ds dt < 00,

fE L2[a, b] and g E L2[a, b]. The function K is called the kernel of equation (7.37). A is a nonzero complex parameter. Solving equation (7.37) means finding a function fin L 2 [a, b] that satisfies equation (7.37).

As shown in Remark 6.29 of Chapter 6, the operator T given by

Tf(s) = f: K(s, t)f(t) dt

is a compact bounded linear operator from L2 [a, b) to L2[a, b] with norm II T II satisfying II T II < II K 112' Moreover T is a self-adjoint operator if and only if K(s, t) = K(t, s) a.e. with respect to (s, t). In this case the kernel K is called symmetric.

If we assume that the kernel K is symmetric, the equation (7.37) takes the form

f= ATf+g

or

Tx - p.x = y, (7.38)

where p. = l/A, x = J, y = -(l/A)g, and T is a compact self-adjoint operator. Clearly y = 0 if and only if g = O. For g = 0, the values of A for which equation (7.37) has a nontrivial solution are called characteristic values of equation (7.37). For y = 0, the values of p. for which equation (7.38) has a nonzero solution are the eigenvalues of T. Clearly characteristic values and nonzero eigenvalues are reciprocally related.

We can consider the equation (7.38) as an equation in any Hilbert space H where T is a self-adjoint compact operator on H and x and yare elements of H. Since T is compact, the nonzero values of p. are either regular values of T or eigenvalues of T; that is, T - p.I either has a bounded inverse on H or p. is an eigenvalue of T. If (T - p.1)-1 exists, then equation (7.38) has a unique solution for each y in H, and in particular 0 is the only solution of Tx - p.x = O.

If P. is an eigenvalue of T, let Hp. be the eigenspace of T, the subspace of H consisting of all eigenvectors corresponding to p.. Let Hp.l. b~ the orthogonal complement of Hp.' Note that T(Hp.) C Hp. and T(Hp.l.) C Hp.l.. If we consider T restricted to Hp.l., then p. is not an eigenvalue for this restricted operator. Consequently as an operator from Hp.l. to Hp.l.,


T - pJ has a bounded inverse on H/ and equation (7.38) has a unique solution for any y in H/. Since any x in H can be written uniquely as Xl + X2 with Xl E HI' and X2 E H/ and (T - P1)XI = 0 we have

(T - pI)x = (T - PI)X2 E H/.

Hence equation (7.38) has a solution only for y in H",J-. If X 2 is any solution

of (7.38) and Xl is any element of HI" then Xl + X2 is also a solution. Consequently, as in the case when P was not an eigenvalue, we can say that (7.38) has a unique solution for y in H if and only if the homogeneous equation (T - pJ)x = 0 has a unique solution. Moreover if p of::. 0 is an eigenvalue, then (7.38) has a solution if and only if y is in the subspace HJ1..l of vectors orthogonal to the space of eigenvectors.

Theorem 7.8 furnishes a technique for solving equation (7.38) and of course equation (7.37). First of all, suppose p is not an eigenvalue. By Theorem 7.8, for each X in H

Tx = L Pi(X I Xi)Xi· i

From (7.38), Tx = y + px or

I I X = - (Tx - y) = -L Pi(X I Xi)Xi - y.

P P i (7.39)

For any n, by taking scalar products of both sides of this equation, we have

from which

Pn(X I xn) = ~ ([ ~ Pi(X I Xi)Xi - Y] I Xn)

= ~n [,un(x I xn) - (y I xn)] ,

(X I X ) = (f-tnlp)(Y I xn) = Pn(y I xn) • Pn n Pnl P - I Pn - P

Hence from equation (7.39) the solution X is given by

(7.40)

Secondly if P is an eigenvalue, then equation (7.39) is still valid and as before we can solve for Pn(x I xn) as in equation (7.40) if P of::. Pn· If P = Pn' the coefficient of eigenvector Xn can be taken as arbitrary since this term


represents a vector 10 HI"" Consequently, a solution of equation (7.38) is given by

where Cn = ftn(y I xn)/{ftn - ft) if ftn 7'=- ft and Cn is arbitrary if ftn = ft· These results are summarized in the following theorem.

Theorem A. Suppose T is a compact self-adjoint operator on H. If ft 7'=- 0 is not an eigenvalue of T and y is an element of H, then the solution of (T - ftI)x = y exists and is the unique vector

_ 1 [". fti(Y I Xi)Xi ] x-- .t.. -y , ft i fti - ft

where the fti are the eigenvalues of T and the Xi are corresponding eigenvectors. If ft 7'=- 0 is an eigenvalue of T and y E Hi, then a solution of (T - ftI)x = y exists and is given by

X = -;- (~ CiXi - y) , where Ci = fti(Y I Xi)/{fti - ft) if fti 7'=- ft and Ci is arbitrary if fti = ft· I

An equation which is solved by means of the theory for solving Fredholm integral equations is the so-called StUrm-Liouville equation. It is an equation of the form

-[V'(X)f'(X)]' + q;(x)f(x) - ftf(x) + g(x) = 0, (7.41 )

where V'(x), V"(x), and q;(x) are real continuous functions on [a, b], g(x) is a complex-valued continuous function on [a, b], and V'(x) > 0 for x in [a, b]. Equations of this sort arise in the study of vibrating strings and membranes, transmission lines, and resonance in a cavity. Solving the SturmLiouville problem means finding a function fin L 2 [a, b] satisfying equation (7.41) such that

ad(a) + ad'(a) = 0,

bd(b) + bd'(b) = 0,

f" exists and is continuous on [a, b],

(7.42)

(7.43)

(7.44)

where ai' a2 , b1 and b2 are real constants such that I a1 I + I a2 I > 0 and I b1 I + I b2 I > O.


Together with equation (7.41) the associated homogeneous equation (7.45) in which g(x) = 0 can also be considered:

-[1p(x)f'(x)]' + q;(x)f(x) - pf(x) = O. (7.45)

In consideration of this equation, two possibilities arise. Either fl = 0 admits the existence of a nontrivial solution to equation (7.45) that satisfies equations (7.42)-(7.44) or fl = 0 admits only the trivial solution f = O. Alternatively, we say, respectively, that either 0 is an eigenvalue of equation (7.45) or 0 is not an eigenvalue. We deal with each of these cases below.

Let us observe first that if fll and fl2 are two distinct eigenvalues of equation (7.45), that is, two scalars that admit nontrivial solutions It and f2' respectively, of equation (7.45), then It and h. are orthogonal elements of L 2 [a, b]. Indeed, since

[1p(x)It'(x)]' - q;(x)];.(x) + fldl(X) = 0,

[1p(X)f2'(X)]' - q;(x)h.(x) + ild;(x) = 0,

it follows that upon multiplying the first equation by .h and the second by It and then subtracting,

(7.46)

Since a1 ];.(a) + ad/(a) = 0, ad2(a) + ad2'(a) = 0, and I a1 I + I a2 I > 0, the determinant ];.(a)N(a) - !t'(a)f2(a) is zero. Similarly f2(b)ft'(b) - It(b)j/(b) = O. Upon integrating both sides of equation (7.46), we find

therefore that f~fl(X)h(X) dx = O. Since L2 [a, b] is separable, it contains at most a countable set of or

thogonal elements. Consequently there are at most countably many distinct eigenvalues of equation (7.45) and uncountably many real numbers that are not eigenvalues of equation (7.45).

Let us first consider now the case where 0 is an eigenvalue of equation (7.45) and show that this case can be reduced to the case where 0 is not an eigenvalue. Suppose Ao is a real number that is not an eigenvalue of equation (7.45), that is,

-[1p(x)f'(x)]' + q;(x)f(x) - Aof(x) = 0

has only the trivial solution that satisfies equations (7.42) and (7.43). Letting y(x) = q;(x) - Ao, this equation has the form

[-1p(x)f'(x)]' + y(x)f(x) = 0,


and equation (7.41) has the form

-[1p(x)f'(x)]' + y(x)f(x) - (fl - "'o)f(x) + g(x) = o. (7.47)

By assumption the homogeneous counterpart of equation (7.47), i.e., with g(x) = 0, has only the trivial solution when fl - "'0 = 0, which is precisely the assumption made when the case in which 0 is not an eigenvalue of equation (7.45) is considered.

Let us then assume the latter case: fl = 0 admits only the trivial solution f = 0 to equation (7.45) that satisfies equations (7.42)-(7.44). Let fl = O. From the theory of elementary differential equations, a nontrivial real solution U] of equation (7.45) exists that satisfies equation (7.42) and a nontrivial solution U2 of equation (7.45) exists that satisfies equation (7.43). Moreover, the Wronskian of U1 and U2 is given by c/1p{x) for each x in [a, b), where c is a constant. Solutions U1 and U2 are linearly independent, for if one were a constant multiple of the other, each would satisfy equation (7.45) and conditions (7.42)-(7.44). This means that c #- O. By the proper choice of U] and U 2 we can assume that c = - I, whereby Ul and U 2

satisfy the equation

u](x)u2'(x) - u]'(x)u2(x) = -l/1p(x), x E [a, b). (7.48)

Using the method of variation of parameters from elementary differential equations, a solution of equation (7.41) with fl = 0 is given by

u(x) = c](x)u](x) + C2(X)U2(x),

where C1 and C2 are functions on [a, b) satisfying

and

From equation (7.48), the solution of these equations is given by

or

Hence

c/(x) = -U2(X)g(X) and

and

U(x) = Ul(X) J: u2(y)g(y) dy + u2(x) f: u](y)g(y) dy

= f: G(x, y)g(y) dy,

170

where

Chap. 7 • Hilbert Spaces

ifx<y, if y<x.

(7.49)

Observe that the function u satisfies the boundary conditions (7.42) and (7.43). Indeed,

and

u'(a) = ul'(a) f: u2{y)g(y) dy,

so that

alu(a) + a2u'(a) = [a1ul(a) + a2u/(a)] f: u2(y)g(y) dy = O.

Similarly blu(b) + b2u'(b) = O. Observe also that u is the only solution of equation (7.41) satisfying equations (7.42) and (7.43), for if v is any other solution, then u - v = 0 as u - v satisfies equation (7.45) with f1, = O.

The function G(x, y) is called Green's function. It is easily seen to be continuous on [a, b] X [a, b] and to satisfy the symmetry property G(x, y) = G(y, x). The solution of equation (7.41) is seen therefore to be the operator value of the compact self-adjoint linear operator G: L2 [a, b] -- L2 [a, b] given by

(Gg)(x) = f: G(x, y)g(y) dy, (7.50)

corresponding to the continuous function g in L 2 [a, b]. Specifically we have the following theorem relating the Sttirm~Liouville equation to an integral equation.

Theorem B. (i) Assume f1, = 0 is not an eigenvalue of equation (7.45). Then there is a real continuous function G defined on [a, b] X [a, b] and satisfying G(x, y) = G(y, x) such that f(x) is a solution of

-[tp(x)f'(x»)' + fP(x)f(x) + g(x) = 0

satisfying equations (7.42) and (7.43) if and only if

f(x) = f: G(x, y)g(v) dy.


In particular, there is a real symmetric function G on [a, b] X [a, b] such that f is a solution of

[tp(x)f'(x)]' - q;(x)f(x) = Af(x) (7.51)

satisfying equations (7.42) and (7.43) if and only if f is a solution of the integral equation

f(x) = A J: G(x, y)f(y) dy. (7.52)

(ii) Assume 0 is an eigenvalue of equation (7.45). If Ao is not an eigenvalue, there is a real symmetric function G(x, y, Ao) such that f(x) is a solution of

-[tp(x)f'(x)]' + q;(x)f(x) - Aof(x) + g(x) = 0

if and only if

f(x) = J: G(x, y, Ao)g(y) dy.

In particular, G(x, y, Ao) exists such that f is a solution of

-[tp(x)f'(x)]' + q;(x)f(x) = Af(x)

if and only if f is a solution of

f(x) = (Ao - A) f: G(x, y, Ao)f(y) dy. I

In the foregoing discussion of the Sturm-Liouville problem, the first two terms of equation (7.41) can be considered as the values of an operator S in L 2[a, b) whose values are given by

Sf(x) = -[tp(x)f'(x)]' + q;(x)f(x).

S is actually an unbounded self-adjoint operator on the domain of functions in L2 [a, b] satisfying equations (7.42), (7.43), and (7.44). (This type of operator and its eigenvalues are studied in Chapter 8.) Equation (7.45) thus has the form

(S - ft)f= O. (7.53)

We have seen in Theorem B that if 0 is not an eigenvalue thenj is a solution of equation (7.53) if and only iff is a solution of the equation

(G - llft)f = 0, (7.54)


where G is a compact operator on L2 [a, b). Hence we see that the values of /-t for which equation (7.53) has a nontrivial solution (the eigenvalues of S) are the reciprocals of the nonzero eigenvalues of the integral operator G. In other words, the eigenvalues of S are the characteristic values of the Fredholm integral equation (7.52). When zero is not an eigenvalue of G, the set of eigenvalues of G form a bounded real sequence converging to zero by Theorem 7.8, whereby the eigenvalues of S then form a real sequence (/-t;)N such that I /-ti I -->- 00.

Problems

X 7.4.1. In this problem A, S, T, U, and Q are the continuous linear operators on 12 given by the following rules:

where (ai) is a bounded sequence of scalars;

S(XI' X2 , .•. ) = (X2' X3' ... );

T(XI' X 2 , ... ) = (0, Xl' X2, X3 , ... );

U(XI' X2' ..• ) = (Xil' Xi., Xis' ... ),

where the set {iI, i2 , ••• } is a permutation of {I, 2, ... }; and

Q(XI' X2 , ... ) = (0, X 2 , X3' ... ).

(i) Show that A is always normal but A is self-adjoint if and only if ai = ai for each i, and A is unitary if and only if I ai I = I for all i.

(ii) Show that S = T* and S* = T. Show that T is isometric but not unitary.

(iii) Show that U is unitary, but not a projection. (iv) Show that Q is a projection but not isometric. (Hint: See Problem

7.3.7.) . X 7.4.2. Prove Lemma 7.3 is false if H is a real Hilbert space. (Hint: Consider a rotation in R2.) Show that the conclusion of Lemma 7.3 is true for all Hilb~rt spaces if T and S are self-adjoint.

7.4.3. (i) Prove that the productt (that is, composition) AB of two normal operators A and B is normal if and only if II A * Bx II = II BA * X II for each xE H.

t A normal, B normal and AB normal need not imply BA normal.

(ii) Prove that the product of two self-adjoint operators is self-adjoint if and only if the given operators commute.

7.4.4. Prove that every continuous linear operator T: H ---+ H, where H is a complex Hilbert space, can be expressed uniquely as A + iB, where A and B are self-adjoint, and that T is normal if and only if A and B commute. X 7.4.5. Let H be any Hilbert space and (Xi)ie[ be a basis for H. Let T: H ---+ H be a bounded linear operator.

(i) Prove that T is isometric if and only if {TXi: i E I} is an orthonormal set.

(ii) Prove that Tis unitary if and only if {TXi: i E I} is a basis for H. X 7.4.6. (i) Prove that if T: H ---+ H is any bounded linear operator, then II T*T II = II TT* II = II T 112.

(ii) Prove that if T is self-adjoint or normal, then for any positive integer n, II rn II = II T lin. X 7.4.7. If A and B are linear operators from H into H such that (Ax I y) = (x I By) for all x and y in H, show that A is continuous and that A * = B. [Hint: Show that A has a closed graph and use the Closed Graph Theorem (Theorem 6.9 in Chapter 6).]

7.4.8. If TI and T2 are self-adjoint operators on H such that Tl < T2 and T2 < TI , show that TI = T2 if H is a complex Hilbert space. Is this true if H is a real Hilbert space? (See Problem 7.4.2.) X 7.4.9. Suppose T is a linear operator on a separable Hilbert space H with orthonormal basis (Xi)N such that Tx = L::Imi(X I Xi)Xi when x =

L:I (x I Xi)X for a sequence (mi)N. Prove the following:

(i) T is bounded if and only if (mi)N E 100 •

(ii) T is compact if and only if (mi)N E co.

(iii) Suppose mi ---+ 1, mi are distinct, and mi #- 1 for all i. Then each mi is an eigenvalue of T, 1 is in the continuous spectrum, and all other complex numbers are regular values of T.

7.4.10. If M is a closed subspace of an infinite-dimensional Hilbert space, prove that PM, the projection on M, is compact jf and only if M is finite dimensional.

7.4.11. If T: H ---+ K is a bounded linear operator, show that T is compact if and only if T* is compact.

7.4.12. Let The the shift operator of Problem 7.4.1 and let Tn = (T)n. Let Vi be the operator on the subspace N of 12 of finitely nonzero sequences given by Vi «XI , X2 , ••• » = (lXI' 2X2' ... , iXi' Xi+!' Xi+2, ... ). Finally

let Ui be the operator given by Ui (Xl' X 2 , ••• » = (0, 0, 0, ... , Xi'

Xi+!, •.. ). Prove the following: (i) The sequence (Ui ) converges strongly to the zero operator (that

is, Uix -+ ° for each X in H), but (Ui) does not converge uniformly to ° (that is, II Ui II -f 0).

(ii) (Vi) converges strongly to the operator V on N given by V(XI' X2 , ••• ) = (Xl' 2x2 , 3xa, ... ), but V is not bounded.

(iii) (Tn) converges neither strongly nor uniformly to the zero operator but (Tn) converges weakly to ° [that is, h*(Tnf) -+ ° for all h* E 12 * and fin 12],

X 7.4.13. Let T be the shift operator of Problem 7.4.1. Prove the following:

(i) T has no eigenvalue, but ° E aCT). (Hint: If Tx = I-'x, show that X = 0.)

(ii) ° is an eigenvalue of T*. (iii) I-' is an eigenvalue for T* <0> II-' I < 1.

X 7.4.14. (i) If T is a normal operator, prove that I-' is an eigenvalue of T if and only if fi is an eigenvalue of T*.

(ii) If A is the operator of Problem 7.4.1, show {ai: i EN} are the only eigenvalues of A. Conclude that {iii: i E N} are the only eigenvalues of A*.

(iii) Prove a(A) is the closure of {ai: i EN}. 7.4.15. Let T be the operator on the complex Hilbert space L 2[0, 1]

given by Tf(x) = xf(x). Prove the following:

(i) T is self-adjoint. (ii) II T II = 1.

(iii) T has no eigenvalue. (iv) aCT) = [0, 1].

7.4.16. Let T be a linear operator on H. A complex number A is ~alled an approximate eigenvalue of T if there exists a sequence (hn) of unit vectors in H (i.e., II hn II = 1) such that limn-+oo(T - A)hn = 0.

(i) Prove that every approximate eigenvalue belongs to the spectrum aCT) of T.

(ii) Prove that if A is a bounded normal operator, then every point of the spectrum is an approximate eigenvalue.

X 7.4.17. Let P be a nontrivial projection on the Hilbert space H. Prove that the spectrum of P consists of two eigenvalues, 0 and I. X 7.4.18. Let P E L(H, H) and suppose p2 = P. Prove Ker P -.l Range P if and only if P is a projection (that is, P* = P).

X 7.4.19. If P E L(H, H), P *- 0, and p2 = P, prove II P II > I. Prove moreover that II P II = I if and only if P is a projection.

7.4.20. Let PI, ... , Pn be projections onto M1 , ••• , Mn , respectively. Prove P == PI + P2 + ... + Pn is a projection if and only if Mi -.l Mj

when i *- j. In this case P is the projection onto Ml ffi M2 ffi ... ffi Mn.

X 7.4.21. If T E L(H, H) and P = I, prove that T = T* if and only if II T II = I.

X 7.4.22. Let H be a separable Hilbert space and let S, T E L(H, H). Let (Xi)N be an orthonormal basis for H.

(i) Show that with respect to (Xi)N, Tcan be represented by a unique matrix M = (m;j) with a countable number of rows and columns in the sense that if x = L (x / Xi)X; then Tx = Li(Ljmij(X / Xj»)x;.

(ii) If (aj)N is a sequence of scalars such that Ll=l I aj /2 < 00, prove that

(iii) Show that T* is represented by the matrix M* = (mt) where

mlj = mji' (iv) If S is represented by A = (nij) with nij = mij' prove that

T* = S. (v) Prove T is unitary if and only if

00 00

L mijmik = L miimki = bjk' i=1 i=1

7.4.23. Let T be a linear operator on a complex Hilbert space H. Prove that T is a bounded self-adjoint operator if and only if (x I Tx) is real for all x E H.

7.4.24. Suppose (Tn) is a sequence of bounded self-adjoint operators on H converging weakly [that is, h*(Tnf) -+ h*(Tf) for all h* E H* and f E H) to a bounded linear operator T on If. Prove T is self-adjoint.

X 7.4.25. Let T be a bounded normal (unitary) operator on H. Let M). = {x I Tx = Ax} be the eigenspace of T corresponding to the eigenvalue A. Prove

(i) if A is eigenvalue of T then l is an eigenvalue of T*(T-l); (ii) M). is also the eigenspace of T* corresponding to l; and

(iii) M). reduces T, i.e., T(M).) C M). and T(M;.1-) C M;.1-.


7.4.26. Prove that if A"* 0 is an eigenvalue of a compact operator Ton H, then the eigenspace MA is finite dimensional.

7.4.27. For f E L 2 [0, I], let (Tf)(x) = J~f(t) dt for x E [0, ]]. Show T is a compact operator on L 2 [0, I] with no eigenvalues. Find a(T). [Hint: Compute reT), the spectral radius.]

7.4.28. Let T be a bounded normal operator on Hilbert space H.

(i) Prove H = Ker TEB R(T). (ii) Prove A E e(T) if R(T - AI) = H.

(iii) Prove A E Ca(T) if R(T - AI) "* H but is dense in H. (iv) Prove A is an eigenvalue if R(T - AI) is not dense in H. (v) Prove the residual spectrum is empty.

X 7.4.29. If A: H ~ H is a compact self-adjoint operator, prove that a set of orthonormal eigenvectors (Xi) corresponding to the set of eigenvalues of A is a complete orthonormal set if 0 is not an eigenvalue of A. (Hint: Use Theorem 7.8.) X 7.4.30. Let T be a compact self-adjoint operator on H.

(i) Prove that T can be written as L AkPko where the Ak are the nonzero eigenvalues of T, Pk is the projection onto the eigenspace MAk corresponding to Ak, and the convergence is uniform.

(ii) Show that if 0 is not an eigenvalue of T, T can be represented by a diagonal matrix with respect to a basis of orthonormal eigenvectors.

7.4.31. Generalization of Theorem 7.8 to Normal Operators. If T: H ~ H is a compact, normal operator, prove the following:

(i) H has an orthonormal basis (Xi) such that each Xi is an eigenvector of T.

(ii) T can be written as Tx = Li:lAi(X I Xi)Xi, where (Ai) is a sequence of complex numbers such that Ai ~ 0 as i ~ 00.

[Hint: To prove (i), use Zorn's Lemma to get a maximal orthonormal set of {xJ of eigenvectors of T. Let M be the closed linear span of {xd. Show M reduces T. If MJ. #- {O}, show TMJ. is compact, normal, and nonzero with an eigenvector in M, contradicting the maximality of {xd. You might need the fact that yeT) = r II T II if T is normal and the Theorem 6.25.]

7.4.32. (M. Rosenblum). (i) If S and Tare bounded linear operators on Hilbert spaces Hand K, respectively, and if !T E L( L(K, H), L(K, H») is given by!T(X) = sx - XT then prove a(!T) C a(S) - a(T). [Hint: If .9' and ~ are the operators on L(K, H) given by .9'(X) = SX and ~(X) = XT, respectively, then show that if A$. a(S) then A f/; a(.9'), and so a(.9') C a(S). Likewise a(~) C a(T). Since .9' and ~ commute and !T = .9' - ~, a(!T) C a(.9') - a(~) by 7.5.l9(ii).]

Sec. 7.5 • Spectral Decomposition of Self-Adjoint Operators 177

(ii) If a(S) n aCT) = 0 then prove that for each operator Y from K to H, there is a unique operator X from K to H such that SX - XT = Y.

X 7.4.33. Let (Xk)k':l b~ a sequence of nonzero vectors in a Hilbert space Hand ({Jk)k':l a sequence of nonzero real scalars. Prove the following assertions:

(i) If for x E H, Tx = Lk.:l{Jk(X I Xk)Xk> where n is a fixed positive integer, then T is a compact self-adjoint operator on H.

(ii) Suppose that {x" X2' ... ,xn} is a linearly independent set and let T be as in (i). Then 0 is an eigenvalue of T if and only if dim H > n.

(iii) Suppose that {Xl' X2, ... , xn} is an orthonormal set. Let T be as in (i). Then the set of all nonzero eigenvalues of T is the set {{Jl, {J2, ... , {In}.

(iv) Let (Xk)k':l be an orthonormal sequence and G be the closed subspace of H spanned by (Xk)k':l' Then (Xk)k':l is an orthonormal basis of G.

(v) Let the Xk be as in (iv) and limk-+oo I fh I = O. Then the operator T defined by

00

Tx = L {Jk(X I Xk)Xk, XEH, k=l

is a compact self-adjoint operator and the set of all nonzero eigenvalues of Tis {{Jl, {J2, ... }. [Hint: Suppose that {J =F 0, x =F 0 and Tx = {Jx. Write x = y + z, Y E G and Z E GJ., G as in (iv). Then Tx = Ty = (Jy.]

(vi) Let T be a nonzero compact self-adjoint operator on an infinitedimensional Hilbert space. Then M > 0, where M = sup{(Tx I x): II x II = I}. [Hint: Use Theorem 7.8 to see that T is as in (ii) when M < 0.]

7.4.34. Let X = Y + Z, where Y and Z are subspaces of a Banach space X and Y n Z = {O}. If x = y + z, then define P(x) = y. Prove that P is bounded if and only if both Y and Z are closed.

7.4.35. Let T E L(H, H). Prove that aCT) C the closure of N7" where the numerical range NT = {(Tx I x): II x II = l}. [Hint: If il is not an approximate eigenvalue of T (see Problem 7.4.16) and il E aCT), then R(T - ill) is closed, but not all of H. Then there exists x E H such that II x II = 1 and (T - ill)x I x) = 0.]

7.5. Spectral Decomposition of Self-Adjoint Operators

Our goal in this section is simply stated: to prove what is known as the Spectral Theorem for bounded self-adjoint operators. It can be said that all our preceding work in Hilbert space theory-although important in


itself-has been preparatory to proving this outstanding theorem. As seen earlier, self-adjoint operators are generalizations of Hermitian matrices. The Spectral Theorem is a generalization of the diagonalization theorem for such matrices.

This theorem was originally proved by Hilbert between 1904 and 1910. t Proofs of this result were also given by F. Riesz in 1910 and 19l3. t

Since that time other proofs of spectral theorems have been given by many others for self-adjoint, unitary, and normal operators both in the bounded and unbounded cases. The interested reader may also consult for further study the works of Dunford and Schwartz [17], Halmos [23], Riesz and Sz-Nagy [48], Stone [59], Prugovecki [46], among others, as well as the Appendix.

The spectral theory of certain classes of operators has been studied extensively since it was initiated by Hilbert in the early 1900's. The theory has profound applications in the study of operators on Hilbert space and in areas of classical analysis such as differential equations. See for example the work of Dunford and Schwartz [17].

Although the Spectral Theorem can be generalized to broader classes of operators, as mentioned above, we consider here only self-adjoint bounded operators. In Chapter 8 other formulations of the Spectral Theorem for bounded and unbounded operators will be given-the proofs of which involve measure theoretic techniques. We will here prove the Spectral Theorem using an approach that makes use of what is known as the "functional" or "operational" calculus and involves no measure theory. After proving this theorem, some applications of this key result will be outlined in the Problems.

Recall from Theorem 7.8 that if T: H -+ H is a nonzero compact, self-adjoint operator, then for each x in H,

00

Tx = L Ak(X I Xk)Xk, (7.55) k=l

where the Ak are eigenvalues of T and the Xk form a corresponding set of eigenvectors. After Definition 7.13 we will show how the Spectral Theorem generalizes the formula (7.55) to arbitrary self-adjoint operators.

There is one stipUlation we must make, however, before proceeding. All Hilbert spaces considered in this section are complex Hilbert spaces.

t D. Hilbert, Grundzlige einer allgemeinen Theorie der !inearen Integralgleichungen, Nachr. Akad. Wiss. Gottingen Math.-Phys. IV, Kl. 157-227 (1906).

t F. Riesz, Ober quadratische Formen von unendlich vielen Veranderlichen, Nachr. Akad. Wiss. Gottingen Math.-Phys. Kl. 190-195 (1910); Les systemes d'equations lineaires a une infinite d'inconnues, Gauthier-Villars, Paris (1913).

Sec. 7.S • Spectral Decomposition of Self-Adjoint Operators 179

The need for this stipulation arises from the fact that the complex numbers are algebraically closed while the real numbers are not-meaning every real polynomial has complex roots but not necessarily real roots. The necessity for using complex Hilbert spaces becomes already evident in the proof of Lemma 7.10.

In this section T will denote an arbitrary self-adjoint (bounded linear) operator on H, and m and M will denote real numbers such that ml < T < M I (see Lemma 7.4).

To prove a form of the Spectral Theorem for bounded self-adjoint operators, we first prove a theorem that describes the so-called "Continuous Functional Calculus." It gives us much more information than actually needed to prove the Spectral Theorem for a bounded self-adjoint operator, but we prove it here since we will have need of it in our more general considerations of spectral theory in Chapter 8. We first must present two lemmas.

Lemma 7.10. Suppose p(x) = L:=4l~k is a polynomial with real or complex coefficients. If T is a bounded self-adjoint operator, let p(T) = L:_oa~.Tk. Then

<1(p(T») = {p().); ). E <1(T)} == p(<1(T»). I

For proof, see Theorem 6.24 in Chapter 6.

Lemma 7.11. Let T be a bounded self-adjoint operator. Then if p is as in Lemma 7.10,

II p(T) II = sup I p(J..) I· I Aea(T)

Proof. II p(T) 112 = II p(T) *p(T) II = II (Pp)T II sup I ). 1 , by Lemma 7.6 (ii)

Aea(jp(T»

sup 1 Pp().) I, by Lemma 7.10 Aea(T)

= (sup I p().) 1)2. Aea(T)

[Here P(x) = p(x), where p(x) is the complex conjugate of p(x).] I

If p is a polynomial and T is a bounded (self-adjoint) operator, p(T) is defined as in Lemma 7.10.

Theorem 7.10. Continuous Functional Calculus. Let T be a selfadjoint operator on a Hilbert space H. There exists a unique map cp:

Cda(Dl ~ L(H, H) with the following properties:

(i) q;(af + pg) = aq;(f) + pq;(g), q;(fg) = q;(f)q;(g), q;(I) = I H ,

q;U) = q;(f)*, for all f, gin C1[a(T)] and scalars a and p.

(ii) II q;(f) II < C II f 1100 ' for all f, for some constant C.

(iii) If f(x) = x, q;(f) = T.

Moreover, q; satisfies the following additional properties.

(iv) If Th = Ah for all h, q;(f)h = f(A)h.

(v) a(q;(f» = f(a(T»).

(vi) Iff> 0, q;(f) > O.

(vii) II q;(f) II = Ilflloo == sup{lf(A) I: A E aCT)}.

[Part (v) is called the Spectral Mapping Theorem.] I

Proof. Define q;(p) to be peT) for each polynomial p in C1 (a(T»). By Lemma 7.11, II q;(p) II = II p 1100 so that q; has a unique continuous linear extension to the closure of the set of polynomials in C1 (a(D). By the StoneWeierstrass Theorem (Theorem 1.26), this closure is all of C1 (a(T»). Clearly, the extension of q; to C1 (a(T» satisfies (i), (ii), and (iii) and is the unique function satisfying these properties.

Since (iv) and (vii) are valid for polynomials, they are valid by the continuity of q; for all continuous functions. To prove (vi), observe that if f> 0, then f = g2, where g is real-valued and g E C1(a(D). Hence, q;(f) = q;(g2) = [q;(g)]2, where q;(g) is self-adjoint so that q;(f) > O.

It remains to show (v). To this end suppose A is a scalar and A oF feu) for any u in aCT). Let g = (f - A)-I. Then

q;(f-A)q;(g) = q;((f - A)g) = q;(l) = IH = q;(g(f-A» = q;(g)q;(f - A),

so that q;(g) = [q;(f) - A]-I. This means Ai a(q;(f»). Hence a(q;(f»

C f( a(T»). Conversely, suppose A = f(fl), fl E aCT) and A i a (J(T) ). Then f(D - ).J is surjective. By Proposition 6.12, there exists b > 0 such that for A E L(H, H),

II A - [J(T) - AI] II < b =? A is surjective.

Let 0 < e < 15. Let P be a polynonlial such that II p - f 1100 < teo Then, 1/ p(T) - p{p)!- [f(T) - AI] 1/ < e. Consequently, p(T) - p(ft)! is surjective. Since p{P) E p(a(T») = a(p(T»), it follows that either p(T) - p(ft)! is not one-to-one or [p(T) - p{P)!]-l, when it exists, is not bounded. In either case, by Proposition 6.5 there exists h E H such that II h II = 1 and II [p(T) - p{P)!]h II < leo But then II [f(T) - U]h II < 2e. This contradicts our original assumption that A $ a (J(T»). I

It should be emphasized that any real-valued continuous function is mapped by cp to a self-adjoint operator, as (i) shows; and, in particular, a nonnegative continuous function is mapped to a positive self-adjoint operator, as shown by (vi).

An immediate consequence of Theorem 7.10 is the following corollary.

Corollary 7.3. If T > 0, then there exists a positive operator S such that S2 = T. (S is called the square root of T.)t I

Proof. If T> 0, then a(T) C [0,00) by Lemma 7.8. Let f(x) = Xl/I

on [0, 00). Then if S = f(T), then S2 = T. I

We wish now to "extend" (see Problem 7.5.2) the mapping cp to certain nonnegative discontinuous functions defined below. To do so we will need the following lemma.

Let P+ denote the class of all (real-valued) nonnegative polynomials defined on a(T).

Lemma 7.12.

(i) If P E P+, peT) is a positive self-adjoint operator.

(ii) If (Pn) is a sequence in P+ with Pn+I < Pn for n = 1,2, ... , then Pn(T) converges strongly to a positive self-adjoint operator.

(iii) Let (Pn) and (qn) be sequences in P+ with Pn+I < Pn and qn+I < qn for n = 1, 2, ... , and let Sp and Sq be the strong limits of the sequences (Pn(T») and (qn(T»), respectively. If limn-+ooPn(t) < limn-+ooqn(t) for all tin aCT), then Sp < Sq. I

t In Problem8.1.7 it is established that there is exactly one positive operator S such that 52 = T.

Proof.

(i) By (i) and (vi) of Theorem 7.10.

(ii) As in Lemma 7.9, the sequence (Pn(T)) converges strongly to a self-adjoint operator Sp. Since Pn(T) > 0,

(Spx I x) = lim (Pn(T)x I x» 0 or Sp> O. n~oo

(iii) For each t in aCT), the sequences (Pn(t» and (qn(t))-as nonnegative, nonincreasing sequences-must converge. Let k be a fixed positive integer. For each t in aCT),

(7.56)

Let mn(t) = max{{Pn - qk)(t), O} for each t and n = 1,2, .... From equation (7.56) it follows that limn~oomn(t) = 0 for each t in a(T). Clearly, mn+l < mn. Hence, by Dini's Theorem, Problem 1.5.18, the sequence mn converges uniformly to 0 in aCT). This means that for any e > 0 there is a positive integer N so that whenever n > N,

Pn(t) - qit) < e,

for any t in aCT). Hence, by (vi) of Theorem 7.10,

for n > N. Hence Sp < eI + qk(T) and since k is arbitrary Sp < eI + Sq. The fact that e is arbitrary means Sp < Sq. I

Now let L+ represent the class of all real-valued functions J on aCT) for which there exists a sequence (Pn) of nonnegative polynomials defined on aCT) such that

(i) 0 <Pn+l(t) < Pn(t) , for n = I, 2, ... ,

and (7.57)

(ii) lim pit) = J(t) , for each t in aCT).

The preceding lemma enables us to define J(T) for each J in L + as the strong limit of (Pn(T)) , where (Pn) is a sequence as in condition (7.57). Indeed, if (qn) is another sequence of nonnegative polynomials satisfying condition (7.57), then Lemma 7.12 assures us that the strong limit of

(Pn(T») and (qn(T)) are the same positive self-adjoint operator. Hence, J(T) is well-defined as this positive self-adjoint operator.

Lemma 7.13. The mapping J ->-J(T) of L + into the class of positive self-adjoint operators satisfies the following properties:

(i) (f + g)T = J(T) + geT) for all f, g in L+.

(ii) (af)T = aJ(T) for all J in L+ and a > O.

(iii) (fg)T = J(T)g(T) for all f, g in L+.

(iv) If J < g then J(T) < geT) for all f, gEL +. I

Proof. The proofs of (i) and (ii) follow from the corresponding statements for P and q in P+. Statement (iv) is an easy consequence of Lemma 7.12 (iii). We here prove (iii). Choose any nonincreasing sequences (Pn) and (qn) in P+ with J(t) = limnPit) and get) = limnqn(t). Then, Pn+1qn+1 < Pnqn and limn(Pnqn)(t) = Jg(t). For all x and y in H

(J(T)g(T)x I y) = (g(T)x I J(T)y) = lim (qn(T)x I Pn(T)y) n-+oo

= lim (Pn(T)qn(T)x I y) = lim (Pnqn(T)x I y)= ((fg)Tx I y). n-+C"Q n-+oo

Hence J(T)g(T) = (fg)T by Remark 7.1. I

Let L be the set of all bounded real-valued functions on a(T) of the formJ - g withf, g E L+. Clearly L is a subspace of the real linear space of all bounded real-valued functions on aCT). If P is any real-valued polynomial on aCT), then pEL since P + al is in P+ for some positive a and P = (p + al) - al.

Our development has led us to the point where we can extend the mapping P ->- peT) of real-valued polynomials to L. Let J ELand choose g and h in L + so that J = g - h. As expected define J(T) as geT) - h(T). Notice that if also J = g' - h', then g + h' = h + g' so that by Lemma 6.13 (i), geT) + h'(T) = h(T) + g'(T). Hence geT) - h(T) = g'(T) - h'(T) and J(T) is well defined.

Proposition 7.22. The mappingJ ->- J(T) on L is a linear mapping into the class of self-adjoint operators, and for all J and g in L,

(i) J < g implies J(T) < geT),

(ii) (fg)T = J(T)g(T). I

Proof. The proof follows readily from Lemma 7.13. I

There are special functions in L + that will be used to prove the Spectral Theorem. For each real number s define ea on aCT) as follows: If m < s < M put

es(t) = { ~: for t E [m, s) n aCT) for t E (s, M) n aCT).

If s < m, set es = 0; if s > M, set es = 1.

(7.58)

It is easy to see that es is in L+ if s < m or s > M. For s E [m, M), we appeal to the Weierstrass approximation theorem (Corollary l.l, in Part A). For s E [m, M) let N be the least positive integer such that s + IIN<M. For n>Nlet

for t E [m, s) n a(T) , {

I.

in(t) = -nt + ns + I ' 0,

for t E (s, s + lin) n aCT), (7.59) for t E [s + lin, M) n aCT).

For each such n, in is a teal-valued continuous function on aCT) with o <in+! <in for n = N, N + I, ... ell) is the limit ofin(t) for each tin [m, M). Utilizing the Weierstrass approximation theorem, we can find a sequence (Pn) in P with

es(t) < in(t) + 2-n- 1 < Pn(t) < in(t) + 2-n, (7.60)

for each t in aCT) and n = N, N + 1, .... From inequalities (7.60) it follows that Pn E P+ and PnH < Pn for n = N, N + 1, .... Also es(t) =

limnPn(t). Thus, es E L+.

Definition 7.12. For every real number s, define E(s) to be the positive self-adjoint operator given by E(s) = es(T), where es is the element of L+ given by equation (7.58). I

Proposition 7.23. Let T be a self-adjoint operator on H and suppose ml < T < MI. For each real number s there is a projection E(s) on H such that

(i) E(s)T = TE(s),

(ii) E(s) < E(s') , for s < s',

(iii) E(s) = 0, for s < m,

(iv) E(s) = I, for s> M,

(v) limh-+O+E(s + h) = E(s) (in the strong sense). I

Sec. 7.S • Spectral Decomposition of Self-Adjoint Operators 185

Proof. For each s, E(s) is the positive self-adjoint operator es(T). Since T = J(T), where J(x) = x on u(T), (i) follows from Proposition 7.22 (ii). Since es2 = eSt es < es" for s < Sf, e. = 0 for s < m, and e. =.1 for s > M, it is clear that E(s) is a projection and that statements (ii), (iii), and (iv) are true. To prove (v) we note that analogous to (7.60) we can construct a sequence (Pn) in P+ with Pn+1 < Pn and eS+lIn -< Pn such that liffin-+ooPn(t) = e.(t). Thls means Pn(T) > E(s + lIn) > E(s). Since limnPn(T) = E(s), we have E(s + lIn) -+ E(s) as n -+ 00. Using (ii) we see that limh-+o+ E(s + h) = E(s). I

Any projection-valued function E satisfying the conditions of Proposition 7.23 is called a resolution oj the identity associated with T.

Definition 7.13. Suppose E is a function on R that assigns to each real number s a self-adjoint operator E(s) on H. Let a and b be real numbers with a 0 there exists a ~ > 0 such that, for all partitions {a = So < Sl < ... < Sn = b} of [a, b] with Sk - Sk-l < ~ for k = 1, ... , n and for all numbers tl , t 2 , ••. , tn with Sk-l < tk < Sk

for k = 1, ... , n,

The operator S is called the integral oj J with respect to E and is denoted by J~J(t) dE(t). I

Example 7.6. A Concrete Example oj the Resolution oj the Identity Jor a Compact Self-Adjoint Operator. Let us pause a moment in our development and see that the functionJ(t) = tis E-integrable with respect to some operator-valued function E on R and its integral J~ t dE(t) is equal to a given compact self-adjoint operator T. To this end let T be a fixed nonzero compact, self-adjoint operator on H, where dim H = 00. Then, by Problem 7.4.33(vi), M > O. Recall from Theorem 7.8 that, for each x in H,

Tx = L Aix I Xi)Xi, i

(7.61)

where the Ai are the eigenvalues of T and the Xi form an orthonormal

set of eigenvectors with Xi corresponding to Ai' For each real number S define E(s) by

for S < 0,

for s::::: o. (7.62)

It is understood that if there are no Ak < S for S < 0 or no Ak > S for S > 0, then the respective sums in equation (7.62) are zero. If B is the orthonormal set {Xi: i = 1,2, ... } and G is the closed linear span of B, then B is an orthonormal basis+ for G and each element X in H can be written in the form

(7.63)

where Yx E G.l by Theorem 7.2. Clearly for s < 0, E(s) is the projection onto the closed linear span G. of {Xk I Ak < s}, and for s > 0, E(s) is the projection on G.l EB G •. It is left as an exercise (Problem 7.5.1) for the reader to show that E is a projection-valued function on R that satisfies the criteria of Proposition 7.23. In short, we may say that the values of E "increase" from the zero projection when s < m to the identity projection when s> M where mI< T< MI.

Now let a and b be real numbers with a < m and b > M. If P = {a = So < S1 < ... < Sn = b} is any partition of [a, b], using equation (7.63) we easily see that

[E(s.j) - E(Si-1)](X) = L (x I Xk)Xb if Si < 0, (7.64) 81_1 <.lk $' i

[E(Si) - E(Si_1)](X) = Yx + L (x I Xk)Xk, if Si-1 < 0 <Sj, (7.65) 81-1 <.lk $81

[E(Si) - E(Si-1)](X) = L (x I Xk)Xb if 0 <Si-1 < Si' (7.66) 81_1 <.lk$Sj

We show now that the compact operator in equation (7.61) is the integral J~ t dE(t) as in Definition 7.13, where E is as in equation (7.62). Let e > 0 be arbitrary. With c5 = e, let P = {a = So < S1 < S2 . .. < Sn = b} be any partition of [a, b] with Si - S;_1 < c5 for i = 1, 2, ... ,n. Let io

be the index with Sio-1 < 0 < Sio' If then ti is arbitrary in [Si-1, Si], we have, using equation (7.61) and the orthonormality of the set {Xl' X2 , •.. }, that

II Tx - tl tj[E(Si) - E(Si_l)](X) W

=11 t [ L (A'k-ti)(XIXk)Xk]+ L (A'k-·tio)(XIXk)Xk+tio·YxI12 ~~1 '1_1 <Ak"'1 'lo-1<Ak""0 '*'0

= .~I Li-l~k9i (Ak - t.J2 I (X I Xk) 12] + SiO-1l;k 9 iO

(Ak - t io )2 I (X I Xk) 12 "*'0 + I tio 12 II Yx 112

n < e2 L L I (x I Xk) 12 + e2 L I (x I Xk) 12 + [-:2 II Yx 112

i=1 'j-1"Ak"'1 Sjo-1"Ak"Slo iot-io

Hence,

provided Si - Si-1 < (J and ti E [Si-l, s;]. Hence T = f~ t dE(t). Using this example as a motivation, we now come to the goal-and

the most outstanding theorem-of this section.

Theorem 7.11. The Spectral Theorem (Resolution of the Identity Formulation). Let T be a self-adjoint operator, and m and M be real numbers with ml < T < MI. Then there exists a unique resolution of the identity E on R associated with T such that if a and b are real numbers with a < m and b > M, then the mapping t -- t of [a, b] into R is E-integrable and

T = f: t dE(t). (7.67)

I

Proof. If es represents the function defined in equation (7.58), it is easy to verify that if S < u then

euCt) - esCt) = { ~: for t E (s, u] () a(T) , for t E a(T) - (s, u].

Hence for all t E aCT),

s[eit) - e.(t)] < t[eit) - e.(t)] < u[eit) - e.(t)].

Defining E by Definition 7.12, and using Proposition 7.22, we obtain

s[E(u) - E(s)] < T[E(u) - E(s)] < u[E(u) - E(s)]. (7.68)

We know from the proof of Proposition 7.23 that E satisfies conditions (i)-(v) of Proposition 7.23. We must show that the mapping t ---* t of [a, b] into R is E-integrable. To this end, let e > 0 be arbitrary and let {a = So < S1 < '" < Sn = b} be a partition of [a, b] with Sk - Sk-1 < e for k = 1,2, ... , n. Then from (7.68) we get the inequality

n n n L Sk-1[E(Sk) - E(Sk-1)] < T L [E(Sk) - E(Sk-1)] < L sk[E(Sk) - E(Sk-l)]' k~ k~ k~

However, since a < m and b > M, Proposition 6.23 tells us that

n L [E(Sk) - E(Sk-1)] = E(b) - E(a) = I. (7.69) k=1

Hence, n n L Sk-1[E(Sk) - E(Sk_l)] < T < L sk[E(Sk) - E(Sk_l)]' k-l k=l

Now let Ik be a real number with Sk-l < Ik < Sk for k = 1,2, ... , n. Then

n n f;l (Sk-l - Ik)[E(Sk) -- E(Sk_I)] < T - k'id-1

tk[E(Sk) - E(Sk_l)]

n < L (Sk - Ik)[E(Sk) - E(Sk-I)]' (7.70)

k=1

However, Sk-l - Ik > -e, Sk - Ik < e, and E(Sk) - E(Sk-l) > 0 so that using equation (7.69), the inequalities (7.70) reduce to

so that from Lemma 7.4

Hence T = J: I dE(t).

It remains to show the uniqueness conclusion of the theorem. The proof of this is outlined in Problems 7.5.4 and 7.5.5. I

Two applications of the preceding results are outlined in Problems 7.5.6 and 7.5.7. For a further study of the spectral theory of self-adjoint operators in Hilbert space along with applications of this theory, the reader may wish to consult the exhaustive work of Dunford and Schwartz [12].

To conclude this section, we present an example to illustrate much of the preceding theory.

Example 7.7. The Spectrum, the Eigenvalues, and the Resolution of Identity of a Bounded Self-Adjoint Operator. Let H = L2([a, P], ft) where -00 < a < P < 00 and ft is the Lebesgue measure of [a, Pl. Define T: H ~ H by Tf =, Af(A). T is called the multiplication operator. The equation

# f# fP f I Tfl2 dft = I Af(A) 12 dft(A) < m2 If(A) 12 d,I{A) = m2 11f 112 < 00,

"" " (7.71)

where m = max{1 a I, I P I}, shows that TfE L 2[a, P] for every f and that T is a bounded operator. T is also self-adjoint, since

(Tfl g) = f: Af(A)g(A) dft(A) = f:f(A)Ag(A) dft(A) = UI Tg). (7.72)

Let Ao be any scalar and let f be an element of L 2 [a, Pl such that (T - Aol)f = O. Then f(A) = 0 almost everywhere or f = O. Hence T has no eigenvalues.

Next suppose that Ao E [a, Pl. For each positive integer n, letfn be the characteristic function of [Ao - lin, Ao + lin] n [a, Pl. Noting that Ilfn II > 0, let gn be the function fnl II fn II. If Ao (/; aCT), then

I = II g" 112 = II (T - Aol)-l(T - Aol)gn 112 < II (T - Aol)-l 112 II (T - Aol)gn 112

= II (T - Aol)-l 112 f: I A - Ao 121 gn(A) 12 dft{A}

< II (T - Aol)-l 112 ~ fP I gn(A) 12 dft(A) n "

This contradiction shows that Ao E aCT) and that [a, P] C aCT).

Finally, let ~o be any scalar outside of [a, Pl. We want to show ~o f/; a(T) and conclude [a, P] = aCT). Let 15 = inf{1 ~ - ~o I: ~ E [a, Pl} > O. For any g e L2 [a, P], define Sg to be (~ - ~O)-lg(~). Since

Sg E L 2[a, P] for each g. Moreover,

which means that (T - ~oI)S = J. In addition SeT - ~oI) = J, as is easily verified. Moreover, the inequality (7.73) shows that II S II < 1/15. We conclude that ~o f/; aCT).

To complete this example, let us calculate the unique resolution of the identity associated with Tsatisfying equation (7.67). Since [a, P1 = aCT) C[mT, M T ] and mT and MT are in aCT), a = mT and P = MT. Define E: R -->- L(H, H) by

E(A.) = { 0, J,

and by

E(A.) f(~) = { f(~), 0,

if A. < a, if A. > p,

ifa<~<A.,

if A. < ~ < p.

It is easy to verify that E is a resolution of the identity associated with T. It remains to show that T = f~ t dE(t) if a < a and b > p.

Let e > 0 be arbitrary and let {a = So < SI < ... < Sn = b} be a partition of [a, b) with Si - Si-l < e. For each i, let Si-l < ti < Si' Now

II Tf - ~ ti[E(Si) - E(Si_l)]fIl 2= II A.f(A.) - tl tJ (A.)X[a,/l)()[s'_l'B,)(A.)1I 2

= f f I A. - ti 12If(A.) 12 dp(A.) < e2 f If(A.) 12 dp(A.) = e211f112. i-I [«,/l)()[BI_l>BI)

Hence T = f~ A. dE(A.).

Problems

X 7.5.1. Prove that the function of R defined by equation (7.62) is a resolution of the identity associated with the compact self-adjoint operator

T. [Hint: To prove (v) of Proposition 7.23 recall from Theorem 7.8 that I Al I > I A2 I > ... -->- 0.]

In the following problems T is a self-adjoint operator on Hand mI<T<MI. X 7.5.2. (i) Prove that each real-valued continuous function I on a(T) is an element of L, that is,f = g - h, where g and h are in L+. [Hint: If I> 0, then by the Weierstrass approximation theorem, there exist polynomials Pn(t) such that I(t) + l/(n + 1) < pit) </(t) + lin for all I in a(T).]

(ii) Prove that the mapping in Proposition 7.22 of L into the class of self-adjoint operators is an extension of the mapping rp in Theorem 7.10 on the class of continuous real-valued functions.

(iii) Let {E(t): IE R} be a family of projections in L(H, H) such that S < I=> E(s) < E(/). Let hn -->- 0+ and kn -->- 0+. By Lemma 7.9, there are At and Bt in L(H, H) such that E(t + hn) -)0 At and E(t + kn )

-->- B t strongly. Show that At = Bt . Thus, the strong limit limh-+o+E(1 + h) = E(t+) exists and equals At. Prove that E(t-) also exists, and that E(/+) and E(t-) are both projections. X 7.5.3. Let E' be any projection-valued function on R satisfying the conditions (i), (ii), (iii), and (iv) of Proposition 7.23. If a < m and b > M, suppose T = J~ I dE'(t). (This can be proved as in Theorem 7.11.) Prove the following:

(i) Tm = J~ tm dE' (I) for every nonnegative integer m. {Hint: If i < j, [E'(Si) - E'(Si_l)][E'(sj) - E'(Sj_l)] = O. Hence

Show that

where K(m) is a constant dependent on m.}

(ii) Po(T) = J:p(/) dE'V) for every real polynomial P on [a, b], where Po = P la(T).

(iii) 10(T) = J~/(t) dE'(t) for every continuous real-valued function Ion [a, b], where 10 = I la(7'). [Note that 10(T) is defined by Proposition 7.22 since 10 E L by Problem 7.5.2.] {Hint: There exists a real polynomial p so that -e/3 < pet) - J(t) < e/3 for all t in [a, b]. By Lemma 7.4,

II poeT) - !o{n II < 813. Also

-(8/3)1 < L[P(ti) - f(1;)][E'(s;) - E'(Si-I)] < (813)1,

so that

Use part (ii).} X 7.5.4. This problem outlines a proof of the uniqueness conclusion of the Spectral Theorem.

(i) Suppose E' is a projection-valued function on R satisfying conditions (i}-(iv) of Proposition (7.23), and T = J~ I dE'(/) (a < m < M < b). Suppose m < s < M andfn are the continuous functions on [a, b] given by

{I,

fn(x) = n(s - x) + 1, 0,

if a <x< s, if s < x < s + lin, if s + lin < x < b,

for n = N + 1, N + 2, ... , where N is the least positive integer so that s + liN < M. Note that fnla(T)(T) is equal to J~ fn(t) dE'(t) by (iii) of Problem 7.5.3. Prove that J! fn(t) dE'(t) converges strongly to limh-+O+ E'(s + h).

(ii) Establish the uniqueness conclusion of the Spectral Theorem.

7.5.5. (i) If E is any projection-valued function satisfying conditions (ii), (iii), and (iv) of Proposition 7.23, show that the function I-+- (E(t)x I y) is a bounded variation function on [a, b], where x and yare fixed in H. [Hint: (E(/)x I x) is a monotonic function; for any t, (E{t)x I y) is a sesquilinear form on H. Use the polarization identity.]

(ii) If the assumptions are as in Problem 7.5.3 (using E instead of E'), prove that for all x and y in H

(!o(nx I y) = f: f(/) d(E(t)x I y)

in the Riemann-Stieltjes sense. (iii) Give another proof of the uniqueness conclusion of Theorem

7.11 using part (ii).

7.5.6. Let E be the resolution of the identity satisfying Theorem 7. I I. (i) Prove that a real number A. is not in the spectrum a(n of T if and

only if there exists a positive number 8 such that E(s) = E(t) whenever A. - 8 <s < t < A. + 8. {Hint: To prove the sufficiency, define a continuous functionfon [a, b] so thatf(t) = (t - A.)-l for t $ [a, b] n [A. - E,

1.+ c]. Let get) = t - A for all t. Show (T - A)J(T) = g(T)J(T) =

f~J(t)g(t) dE(t) = I whenever I.E [m, M]. To prove the necessity, assume that for each c > 0, sand t exist in [A - c, 1.+ c] such that E(s) -=F- E(t). Exhibit a y such that E(s)y = 0 and E(t)y = Y if s < t. Using Problem 7.5.5 (ii), show II (T - AI)(y) 112 < c2 II Y 11 2.}

(ii) Deduce from (i) that if 1.1= aCT), (T - 1.)-1 = f~J(t) dE(t), where J(t) is a continuous function on [a, b] equal to (t - 1.)-1 for t 1= [a, b] n [A - c, 1.+ c].

7.5.7. Prove that a real number A is an eigenvalue of T if and only if E(A) -=F- E(A- ), where E is the resolution of the identity satisfying Theorem 7.11. {Hint: To prove the sufficiency, by Problem 7.5.5 (ii) II (T - A)X 112

= f~ (t - 1.)2 d II E(t)x 112 for any x in H. Apply this to a particular x, where E(t)x = x if t > A and E(t)x = 0 if t < A. To prove the necessity, if A is an eigenvalue, suppose Txo - AXo = 0, where Xo -=F- o. By Problem 7.5.5 (iii),

«T - )·)2.~o I xo) = J: (t - J.)2d(E(t)xo I xo) = O.

Since A E (a, b], show E(A + c)xo = Xo and E(A - c)xo = 0 by considering the integrals

fM+. fA-. (t - A)2d(E(t)xo I xo) and (t - A)2d(E(t)xo I xO).}

A+. m

7.5.S. An operator T E L(H, H) is a partial isometry if there exists a closed subspace Min H such that T 1M is isometric (that is II Tm II = II m II for all mE M) and T IM.L = o. Prove

(i) Ker T = M1-. (ii) If T is a partial isometry, so is T*. (Hint: Show T*IT(H) is iso

metric.) (iii) T is a partial isometry if and only if T*T is a projection. In this

case T*T is a projection on (Ker T)1-. (iv) T is a partial isometry if and only if T = TT*T. 7.5.9. Polar Decomposition. (i) Prove that if TE L(H, H), then there

exists a unique positive operator P and a unique partial isometry U such that T = UP and Ker U = Ker P = Ker T. This is called the polar decomposition of T. [Hint: Let P = (T*T)1I2 and U(PJ) = TJ on the range of P.]

(ii) Show also that T = QV, where Q = (TT*)1I2 and V is a partial isometry.

(iii) Prove that if T is normal then UP = PU.

7.5.10. Let T E L(H, H). (i) Prove T is compact if and only if (T*T)1I2 is compact. [Hint: If T = UP is the polar decomposition of T, U*T = (T*T)1I2.]

(ii) Prove that if T is compact, then T is a Hilbert-Schmidt operator if and only if the eigenvalues of (T*T)1I2 are in 12.

7.5.11. The Square Root of a Positive Operator. Let T be a bounded linear operator on a complex Hilbert space H such that (Tx I x) > 0 for every x E H. Note that T is then self-adjoint (via a polarization identity argument). Let P E L(H, H), P > 0, such that p2 = T. Let Q be the square root of T, that is, Q = f(T), where f(x) = x1l2. Since PT = TP, P commutes with Q. Since (P - Q)P(P - Q) and (P - Q)Q(P - Q) are both positive and their sum is zero, both are zero operators. Then (P - Q)3, their difference, is also zero. Use Problem 7.4.6(ii) to show that P = Q. Prove also the following assertions:

(i) If T > 0 and T is compact, then T1I2 is compact. (ii) If 0 < Tl < T2 and T2 is compact, then Tl is compact.

[Hint: By Theorem 7.8, Tx = L%:lAk(X I Xk)Xb Ak > O. Note that T1I2X = L%:lA/12(X I xk)xd

8

Spectral Theory

8.1. Spectral Theory for Bounded Operators Revislted

In this section we utilize the knowledge of measure theory at our disposal to prove other versions of the Spectral Theorem for bounded self-adjoint operators-other than the resolution of the identity version given in Chapter 7. Although it is quite possible to generate a spectral measure from the resolution of the identity corresponding to a given self-adjoint operator, we prefer here to prove a more sophisticated spectral measure version of the Spectral Theorem making use of an elegant functional calculus version of the Spectral Theorem. Many of the results of this section will be utilized and duplicated in the next section, where we deal with unbounded operators.

Throughout this section, H denotes a complex Hilbert space.

Definition S.l. If X is a set and gp is a ring of subsets of X, then a positive-operator-valued measure E is a function E: gp -+ L(H, H) such that

(i) E(M) > 0 for all M in gp,

(u) E(U~lMi) = li1llnlL~=lE(Mi)] (in the strong convergence sense) whenever (Mi) is a disjoint sequence of measurable sets whose union is also in gp.

If the values of E are projections, then E is called a spectral measure. If X E gp and E(X) = I, then spectral measure E is called normalized. I

Remark S.l. If E is a positive-operator-valued measure, then

(i) E(0) = 0;

195

196 Chap. 8 • Spectral Theory

(ii) E is finitely additive;

(iii) if MeN, then E(M) < E(N);

(iv) E(M u N) + E(M n N) = E(M) + E(N);

(v) E(M n N) = E(M)E(N) if and only if E is a spectral measure.

Theorem 8.1. Let E: .9l ---+ L(H, H) be a function whose values are positive operators. Then E is a positive-operator-valued measure if and only if, for each h in H, the formula fliM) = (E(M)h I h) defines a measure on.9l. I

Proof. The necessity is clear. For the converse, let flh be a measure on .9l for each h and let (Mi) be a disjoint sequence of sets in .9l with U~lMi in.9l. Now

If

then II An II is bounded t by II E(U~lMi) II, and we have by Lemma 7.5 in Chapter 7 that

so that by equation (8.1), Anh ---+ O. Hence E(U~lMi) is the strong limit of Lf-1E(Mi ) as n ---+ 00. I

Proposition 8.1. Let 92 be a ring of subsets of a set X, and suppose that for each vector h in H there is given a finite measure flh on .9l. There exists a unique positive-operator-valued measure E on .9l such that flh(M) = (E(M)h I h) for all h in H and for all M in 92 if and only if for all vectors hand k and for each M in .9l

(i) Luh+iM)]1I2 < Luh(M)]1I2 + [,uk(M)]1I2,

(ii) flch(M) = I c 12flh(M) for each scalar c,

t Since Ph is a measure, An is a positive operator and II An II = suP.h,:S;l(Anh I h).

Sec. 8.1 • Spectral Theory for Bounded Operators 197

(iii) p.h+iM) + P.h-k(M) = 2P.h(M) + 2P.k(M), (iv) for each M in gp there exists a constant kM such that

I Proof. Let M be arbitrary in gp. Define the real-valued function

II 11M on H by II h 11M = Luh(M)]1/2.

As in the proof of Proposition 7.2 the conditions (i}-(iii) are equivalent to saying that II 11M is a pseudonorm (II h 11M = 0 does not necessarily mean that h = 0) satisfying the Parallelogram Law, which in turn is equivalent to saying that

BM(h, k) = HII h + k 11M2 - II h - k 11M2 + i II h + ik 11M2 - ill h - ik 11M2)

defines a Hermitian sesquilinear form on H such that BM(h, h) = II h 11M2. The boundedness of BM is equivalent to (iv). Hence by Theorem 7.6, conditions (i}-(iv) are equivalent to the existence for each M of a bounded self-adjoint operator E(M) with (E(M)h I h) = BM(h, h) = p.h(M). By Theorem 8.1, E(M) defines a positive-operator-valued measure. I

The next result is easily verified.

Proposition 8.2. If E is a positive-operator-valued measure whose domain is a O'-ring, then

sup{ II E(M) II: M E gp} < 00. I From this point on through Theorem 8.2, E will denote a positive

operator-valued measure on a O'-ring gpo If P.h is the measure

p.h(M) = (E(M)h I h) ,

then by Proposition 8.2 there exists a positive number K such that P.h(M) < K II h 112 for all M and for all h.

If 1= g + ih is a complex-valued function on X, then I is measurable in case g and h are measurable. If p. is a measure on gp and g and hare p.-integrable, then we say lis p.-integrable and define (as in Problem 3.2.24)

f I d" = f g dp. + i f h dp..

I is p.-integrable if and only if I I I is p.-integrable, and in this case

\ f I dp.\ < f I I I dp..

Definition 8.2. Suppose!T == {,uhheH is a family of measures on a O'-ring.9P. [In particular!T could be generated by E in the sense that ,uh(M) = (E(M)h I h) for each h and each M.] A measurable complex function on (X,.9P) is !T-integrable (in particular E-integrable if !T is generated by E) if it is ,uh-integrable for each h. If I is !T-integrable then for any ordered pair of vectors in H, we write

f I d,uh.k = i(f I d,uh+k - f I d,uh-k + i f fdph+ik - i f I d,uh-ik)' (8.2) I

Lemma 8.1. For each pair of vectors hand k in H, the complexvalued function Lh.k given by

Lh.k(f) = f I d,uh.k

is a linear functional on the complex vector space of !T-integrable functions. Moreover if !T is generated by E, then

f XM d,uh.k = (E(M)h I k), (8.3)

for all Min.9P. I Proof. The fact that Lh,k is linear follows from equation (8.2). To

prove the latter part of the lemma, note that if h E H, then

Hence from equation (8.2) and the Polarization Identity (Proposition 7.1 in Chapter 7),

f XM dph.k = H(E(M)(h + k) I h + k) - (E(M)(h - k) I h - k) + i(E(M)(h + ik) I h + ik) - i(E(M)(h - ik) I h - ik)]

= (E(M)h I k). I

Lemma S.2. Let!T = {PhheH be any family of measures on a 0'

ring.9P. Let L be any linear functional on the vector space of !T-integrable functions. Then L = Lh,k for some hand k if and only if

(i) whenever (In) is a sequence of !T-integrable functions such that o <In t f, where I is also !T-integrable, then L(fn) ~ L(f), and

(ii) L(xM) = Lh,k(xM) for all M in .9P. I

Sec. 8.1 • Spectral Theory for Bounded Operators

Proof. To prove the necessity, if 0 <In t I as in (i), then

f Ind/-lh ~ f I d/-lh

for each h by the Monotone Convergence Theorem. Hence

f In d/-lh,k ~ f I d/-lh,k

by equation (8.2).

199

To prove the sufficiency, by (ii), L = Lh,k for all simple functions. If I> 0, choose a sequence of simple functions Un) such that 0 <In t f By (i)

n~oo n-+oo

By linearity LI = Lh,kU) for all ~-integrable functions f I

Lemma 8.3. If for each vector h and for each M in~,

then

f I d/-lh,h = f I d/-lh (8.4)

for each ~-integrable function f In particular, if /-liM) = (E(M)h I h) for each M and h, then equation (8.4) is true for each E-integrable function f I

Proof. Define LI to be f I d/-lh for each f L is linear and satisfies (i) of Lemma 8.2 by the Monotone Convergence Theorem. (ii) is satisfied since

Hence L = Lh,h' I Proposition 8.3. Suppose ~ = (/-lh)heH is a family of measures on a

a-ring ~ such that, for each M in ~,

LIXh,k(XM) = aLh,k(xM),

Lh1+h2 ,k(xM) = Lhlok(xM) + Lh2 ,k(xM),

Lh,IXk(xM) = iiLh,k(xM),

Lh,kl+k2(XM) = Lh,k1(xM) + Lh,k2(xM) ,

(8.5)

(8.6)

(8.7)

(8.8)

for each scalar a and for all vectors in H. Then for each ~-integrable function f, the mapping

(h, k) -+ f I dflh,k == Lh,k(f)

is a sesquilinear form on H. Moreover if for each M in ~

(8.9)

then Lh,k(f) = Lk,h(/) for each ~-integrable function f In addition, if equation (8.4) is true for some bounded measurable function f, and if ~ is a family of measures such that flh(X) < K II h 112 for some constant K independent of h, then the form above is also bounded. I

Proof. By use of Lemma 8.1, statements (8.5)-(8.9) are readily verified to hold for each ~-integrabJe function f For instance, to verify equation (8.9) define L on the class on ~-integrable functions by LU) = Lk,h(I). Then L is linear and satisfies the conditions (i) and (ii) of Lemma 8.1 for the pair (h, k) so that L = Lh,k. The last statement of the proposition follows from the inequality

If Idflh,h I = I f Idflh I <f III dflh < 1I/IIooK II h 112. I

The last four lemmas have been preparatory to proving the next result. It will be most useful in the ensuing work of this section.

Theorem 8.2. Let E be a positive-operator-valued measure on (X, ~). For each bounded measurable complex-valued function I on X, there exists a unique bounded operator Tf on H such that

(Tfh I h) = f I dflh, for all hE H. (8.10)

Moreover,

(Tfh I k) = f I dflh,k, for all hand k in H.

[Here flh is the measure flh(M) = (E(M)h I h ).J I

Proof. Since for each h, flh(M) == (E(M)h I h), each of the equations (8.5), (8.6), (8.7), and (8.8) is true. This can be verified by equation

(8.3). Moreover, by Lemma 8.2, equation (8.4) IS true. Hence corresponding to the bounded sesquilinear form

(h, k) -+ f I dfth,k

is a unique bounded operator Tf such that

(Tfh I k) = f I dfth,k' for all h, k in H.

The following identity shows that equation (8.10) is sufficient for uniqueness. If B is a bounded operator also satisfying equation (8.10), then for any hand k in H,

4 f I dfth,k = f I dfth+k - f I dfth-k + i f I dfth+ik - i f I dfth-ik

= (B(h + k) I h + k) - (B(h - k) I h - k)

+ i(B(h + ik) I h + ik) - i(B(h - ik) I h - ik)

= 4(Bh I k). I

The operator corresponding to I in Theorem 8.2 is denoted by f I dE. Thus for all hand k in H we have

Remark 8.2. Of interest are the following properties of the mapping 1-+ fidE.

(i) f (f + g) dE = fidE + f g dE.

(ii) f aidE = a f I dE for all scalars a.

(iii) fIdE = (f IdE)*.

(iv) f XM dE = E(M) for each M in 92. (v) f Ig dE = fidE f g dE if E is a spectral measure.

(vi) There exists a constant K such that II fidE II < K II 11100 for all f.

Proof. Let us verify (iii), (vi), and (v), respectively. If we define L by LI = f f dfth,k for each E-integrable f~~ction, then the equation

L(XM) = f XM dfth,k = (E(M)h I k) = (E(M)k I h)

and Lemma 8.2 show that L = Lh,k. Hence we have

( (f f dE ) * h I k) = ( (f f dE )k I h) = f f d/Jk,h = f f d/Jh,k = ( (f j dE)h I k); this means (f f dE) * = J IdE.

If f is a real-valued function,

((f f dE)h I h) = Iff d/Jh I < II f II /JiX}.

If f is a complex function k + ig, then

II f f dE II = II f (k + ig) dE II = II f k dE + i f g dE II < 2 II f II /Jh(X),

This equation verifies (vi). Finally since E(M n N) = E(M)E(N) when E is a spectral measure,

(v) is valid if f = lM and g = IN' Iff and g are simple, (v) follows by linearity. If f and g are arbitrary bounded measurable functions, choose simple functions (fn) and (gn) such that II fn - f 1100 --.. 0 and II g - gn ~oo --.. O. Then II fngn - fg 1100 --.. 0 so that

II f fg dE - f f dE f g dE II < II f fg dE - f f~n dE II

+ II ff~ndE- ff~dEII

+IIffng dE - ffg dE II

< 2K{lIfg - fngn 1100 + IIfngn - fng 1100

+ IIfng - fg lloo} --.. O. I

Starting with a positive-operator-valued measure E on a a-ring 9l, we have constructed for each h in Hilbert space H a finite measure /Jh on 9l and then corresponded to E a mappingf --.. Tf on the class of all bounded complex-valued measurable functions on X such that equation (8.10) holds. Our next goal is to start with a fixed bounded self-adjoint operator T on H and associate with it a similar mapping from bounded measurable functions to operators in L(H, H).

Associated with a spectral measure E defined on a measurable space (X, 9l), whe!e X is also a topological space, is the spectrum 1:(E) of E. It is defined as the complement in X of the union of all those open sets

M in g; for which E(M) = o. Obviously .E(E) is a closed set in X. If .E(E) is compact, then E is called a compact spectral measure.

We need the information of the following lemma and proposition. As before, B(R) denotes the a-algebra of Borel sets on R.

Lemma 8.4. If E is a spectral measure on B(R), then

(i) for any M in B(R), E(M) is the "smallest" projection,+ denoted V E(K), "greater than" all E(K), where K is a compact subset of M [in symbols E(M) = V E(K), where K is compact in M];

(ii) E(R - .E(E») = O. I

Proof. (i) Since E is a spectral measure, E(M) > E(K) for all compact subsets K of M; whereby E(M) > V E(K). If E(M) *- V E(K), there exists a nonzero vector h in the range of E(M) that is orthogonal to the range of each E(K). Letting fth be the measure fth(M) = (E(M)h I h), which is regular by Theorem 5.2 in Chapter 5, we have fth(M) = sup{ftiK): K eM, K compact}. Since ftiK) = 0 for each K, fth(M) = 0 or (E(M)h I h) = O. This means h = E(M)h = O.

(ii) In view of (i), it suffices to show E(K) = 0 for K compact in R - .E(E). Since K can be covered by a finite collection of open sets, each of which has zero spectral measure, E(K) = o. I

If E is a compact spectral measure on B(R), let f().) = ).XI(E). Let T = J f dE. The next proposition shows how a(T) is related to .E(E).

Proposition 8.4. If E is a compact normalized spectral measure on B(R) and T = J ).XI(E)().) dE().), then .E(E) = a(T). I

Proof. Since E is normalized, .E(E) *- 0. If Ao E .E(E), then E(M) *- 0 for every open set M containing Ao. Suppose ).0 E .E(E) but Ao ft a(T). If M = {A: I A - Ao I < 1/(2 II (T - Ao)-1 II)}, then there is a unit vector h in the range of E(M) as E(M) *- O. Thus

II Th - Aoh 112 = «T - Ao)*(T - Ao)h 1 h) = f (A - Ao)(A - Ao)XI(E)(A) dfth(A)

= f 1 A - Ao 12 dfth(A) < ( 2 II (T ~ Ao)-l II r t See the remark following Proposition 7.21.

However, I

II h II = II (T - AO)-I(T - Ao)h II < II (T - AO)-1 II 2 II (T _ AO)-1 II = T·

This contradiction shows 1:(E) C aCT). Conversely, if Ao i 1:(E), we show Ao i aCT). If Ao i 1:(E), there is an

open set M containing Ao such that E(M) = O. If <5 = inf{1 A - Ao I: A E Me}, then for every h we have as before that

Hence from Proposition 6.5, (T - ).0)-1 exists and ).0 i aCT). I

Starting with a compact normalized spectral measure E on B(R), we have obtained a bounded self-adjoint operator T = J ).XL'(EP) dE()') on H. Moreover a(T) = 1:(E). We now wish to reverse the process in starting with a given bounded self-adjoint operator and obtaining a spectral (compact, normalized) measure on B(R). By means of the next theorem-sometimes called the functional calculus form of the Spectral Theorem-we will have a neat way of obtaining this spectral measure. It is also a beautiful extension of the Continuous Functional Calculus Theorem (Theorem 7.10) to bounded measurable functions.

Let T be a bounded self-adjoint operator and let aCT) be its spectrum -a compact subset of the real axis. If I E C1 (a(T») and I is real-valued, let I(T) be the bounded linear operator in L(H, H), defined as the image of I under q; in Theorem 7.10. It is easy to verify that if h E H, the function on the real space of real-valued functions in C(a(T)) given by Lh(f) = (J(T)h I h) is a positive linear functional. Hence by Theorem 5.7 there is a unique Borel measure Ph on B( a(T») such that

Lh(f) = (J(T)hlh)= f fdph, (S.ll)

for all fECI (a(T)) that are real-valued. Let ~ now be the particular family of measures {PhhEH satisfying

equation (S.lI). In particular if f is the function Xa(T) in C1 (a(T)), then equation (S.ll) gives for each Ph

Each bounded measurable function on a(T) is thereby integrable for each Ph, and for each such function g we can define J g dPh,k as in equation


(8.2). Since for each continuous function I in C1 (a(T)),

5 f df-lh,h = t[5 I df-lh+h - 5 I df-lh-h + i 5 I df-lh+ih - i 5 I df-lh-ih ]

= H(!(T)(h + h) I h + h) - (!(T)(h - h) I h - h)

+ i(!(T)(h + ih) I h + ih) - i(!(T)(h - ih) I h - ih)]

= (j(T)h I h)

= 5 Idf-lh;

205

equation (8.4) is satisfied for continuous functions. Similarly a simple calculation shows that for each IE C1(a(T»), equations (8.5)-(8.8) are valid (with I replacing XM)' Since each bounded measurable function on a(T) can be approximated in Ll (see Problem 5.3. I 2) by a continuous function on a(T)' the equations (8.4)-(8.8) are seen to hold for all bounded measurable functions. This means (see Lemma 8.3 and Proposition 8.3) that for each bounded measurable function g on aCT) the mapping

(h, k) ---+ 5 gdf-lh,k

is a bounded sesquilinear form on H so that by Theorem 7.6, there exists a unique bounded operator, which we denote by geT), on H such that

(g(T)h I k) = 5 g df-lh,k, for all h, k in H. (8.12)

In particular for any h in H,

(g(T)h I h) = 5 g df-lh,h = 5 g df-lh, (8.13)

and it is this equation for each h which determines geT) uniquely. This is shown by using the same equation employed in the proof of Theorem 8.2.

If I is in C1 (a(T)), I(T) is q;(f), where q; is the function of Theorem 7.10. By the properties of q;, q;(j) = (q;(f»* or in other notation jeT) = (j(T))*. This relation is also true for any bounded measurable function g on aCT). Indeed for IE C1 (a(T)),

5 jdf-lu = (J(T)*h I k) = (j(T)k I h) = 5Idf-lk,h; (8.14)

and since the class C1(0'(T) is dense in Ll> equation (8.14) is also valid for g. This means

(k I g(T)h) = (g(T)h I k) = (g(T)k I h),

so that (g(T)* = g(T). In particular if g is real-valued, geT) is self-adjoint.

To prove our next theorem, we have need of the following lemmas.

Lemma 8.5. If g is a bounded nonnegative measurable function on O'(T) and (gn) is a sequence of measurable functions with 0 < gn < g and fig - gn I dPh -+ 0, then g(T)h = limn-+oo8'n(T)h. I

Proof. From equation (S.13) and the above hypothesis, lim (gn(T)h I h) = (g(T)h I h). Also

II geT) - gn(T) II < II geT) n + II gn(T) II < 2 II geT} II. t

Hence by Lemma 7.5 of Chapter 7,

II [geT) - gn(T)]h 114 = ([g(T) - gn(T)]h I [geT) - gn(T)]h)2

< ([g(T) - gn(T)]h I h )([g(T) - gn(T)]2h I [geT) - gn(T)]h)

< ([g(T) - gn(T}]h I h) II geT) - gn(T) 11 3 II h 112 -+ O. (S.15)

I Lemma 8.6. If g is a nonnegative bounded measurable function and

(In) is a sequence of continuous nonnegative functions such that II In lioo < II g 1100 and f I In - g I dp,h -+ 0 (such a sequence always exists), then g(T}h = lim In(T}h. I

Proof. Let gn = In 1\ g. [gn(x) = min{/n(x), g(x}}.] Since the gn satisfy the conditions of Lemma S.5, g(T}h is the limit of gn(T}h.

Also the sequence (In - gn) is a nonnegative sequence such that f (In - gn) dp,h -+ O. Moreover, the sequence Il/n(T) - gn(T} II is bounded since

Il/n(T} - gn(T} II < Il/n(T} II + II gn(T} II < 2 Illn(T} II,

t II g,,(T) II = SUPIIlIS1(g,,(T)h I h) ~ SUPIIlIS1(g(T)h I h) = II g(T) II·

and (llfnCT) II) is a bounded sequence. [Since fn{T) is self-adjoint,

II fn(T) II = sup (fn(T)h I h) IIhll:S;l

= sup f fn dflh < sup f II fn 1100 dflh < sup f II g 1100 dflh IIhll:S;l IIhII:S;l IIhll:S;l

= II g 1100 sup f dflh = II g 1100·] IIhll:S;l

207

By an inequality similar to (8.15) again, [fn(T) - gn(T)]h converges to 0. We can conclude that fn(T)h converges to g(T)h. I

We have collected sufficient information to prove the following theorem.

Theorem 8.3. Functional Calculus Form of the Spectral Theorem. Let T be a bounded self-adjoint operator on H. There is a unique map pig cP on the class of bounded measurable functions on aCT) into L(H, H) such that

(i) cp(af + fJg) = acp(f) + fJcp(g), cP(fg) = cp(f )cp(g), CPU) = (cp(f»*, cp(l) = I.

(ii) II cp(f) II < II f 1100· (iii) If f().) = )., then cp(f) = T.

(iv) If fn().) -+ f().) for each), and (1lfn 11(0) is bounded, then cp(fn) -+ cp(f) strongly.

In addition, cp satisfies

(v) If Th = )'h, then cp(f)h = f()')h.

(vi) If f> 0, then CPU) > 0.

(vii) If AT = TA, then Acp(f) = cp(f)A. I

Proof. The mapping cp is given by cp(g) = geT), where geT) is the unique bounded operator satisfying equation (8.12). cp is thus an extension of the mapping cp of Theorem 7.10 defined on Cl (a(T») and thereby satisfies (i), (ii), (iii), (v), and (vi) for continuous functions. The linearity of cp is readily verified using equation (8.13); the fact that cp(j) = (cp(f»* and statement (vi) have been verified above.

To prove cp(fg) = cp(f)cp(g), we assume f and g are nonnegative and use linearity to treat the general case. Let (fn) and (gn) be sequences of continuous functions with II fn 1100 < II f 1100, II gn 1100 < II g 1100, gn -+ g in


L1(fth), and fn - fin L1Vtg(T)h + fth) for some h in H. By Lemma S.6,

f(T)[g(T)h] = limfiT)[g(T)h]

and g(T)h = lim gn(T)h.

Hence for any positive integer m,

n-+oo n-+oo

whereby

(fg)(T)h = lim (fng)(T)h = lim [fn(T)g(T)]h = [f(T)g(T)]h. n-+oo n-+oo

The proofs of the other statements-(ii), (v), and (vii)-are similar. The uniqueness of ifJ and statement (iv) remain to be verified. Clearly,

if (fn) converges pointwise to f and (II in 1100) is bounded, then by the Dominated Convergence Theorem (Theorem 3.3),

for each h. Hence for each h, fn(T)h - f(T)h as in equation (S.15), which means fn(T) converges strongly to f(T).

Suppose "P is any mapping satisfying (i)-(iv). By the uniqueness of ffJ of Theorem 7.10, ifJ and "P agree on C1 (a(T»). By linearity, to show"P and ffJ are equal, it suffices to show they agree for real measurable functions. Let h be arbitrary but fixed in H and let VL denote the real vector lattice of real, bounded measurable functions on aCT). If we define I on VL by

I(!) = ("P(f)h I h),

then I is a Daniell integral on VL [here we use (iv) to verify (D) of Definition 5.2]. By Proposition 5.4, there exists a unique (finite) measure 'Ph on a(VL) = B(a(T») such that

("P(f)h I h) = If= f fd'Ph, for all f E VL.

Since fth is the unique measure on B( a(T») such that

(ifJ(f)h I h) = f f dfth


for continuous real functions, and "P and rp agree on continuous functions, it follows that Vh = Ph. As equation (8.13) determines geT) uniquely for nonnegative bounded measurable functions g, "P(g) = geT) = rp(g) for such functions. I

We now consider special bounded measurable functions on aCT) in order to generate a spectral measure on B(R). For each Borel-measurable set Min R consider the characteristic function XM(')a(T). Define P: B(R) ~L(H, H) by

P(M) = rp(XM(')a(T».

In light of the properties of rp, it is easy to convince oneself that P is a normalized spectral measure on B(R).

By Theorem 8.2, corresponding to each bounded measurable function g on aCT) is a unique bounded operator J g dP such that

((f g dP)h I h) = f g dPh ' for all h in H, (8.16)

where Ph(M) = (P(M)h I h) for each h in H and for each Min B(a(T). By equation (8.13), geT) is the unique operator such that

(g(T)h I h) = f g diih , for each h in H, (8.17)

where we use iih momentarily to denote the unique Borel measure on B(a(T» such that

(j(T)h I h) = f f diih'

for all continuous functions f on aCT). Since for every M in B( a(T»

iih = Ph· By the uniqueness of geT) in equation (8.17), we conclude that geT) = J g dP for each bounded measurable function on aCT). In particular

T = f ;, dPo.). (8.18)

We are now ready to prove the following version of the Spectral Theorem.


Theorem 8.4. Spectral Measure Formulation of the Spectral Theorem. There is a one-to-one correspondence between bounded self-adjoint operators T on H and normalized compact spectral measures E on B(R). It is the following correspondence: T = fAXL'(E)(J.) dE(A). I

Proof. If E is a normalized compact spectral measure, let us consider T = fAXL'(E)(A) dE(A) as given in Theorem 8.2. It must be verified that this correspondence is a one-to-one correspondence.

To verify that the association of E ---+ fAXL'(E)(A) dE(A) is onto, let S be any bounded self-adjoint operator. Using Theorem 8.3, let P be the normalized spectral measure on B(R) given by P(M) = rp(XMna(S»' Clearly, P is compact and E(P) = a(S). Moreover, as we have established in equation (8.18)

S = fAXa(S)(J.) dP(A),

so that P maps to S. To verify the assertion E ---+ fAXL'(E)(A) dE(A) is one-to-one, it suffices

to show that when T = f J.XL'(E)(A) dE(A), then E = P, where P is the normalized compact spectral measure given by P(M) = rp(XMna(T»' We know from Proposition 8.4 that E(P) = aCT) = E(E). To show that E = P, it suffices to show that E(M) = P(M) for Me aCT) or that E(M) = rp(XM)' Now by Theorem 8.2, E(M) is the unique operator such that

(E(M)h I h) = f XM dflh; (8.19)

whereas, by Theorem 8.3, rp(XM) is the unique operator such that

(8.20)

where i1h again denotes momentarily the unique Borel measure on aCT) such that

(j(T)h I h) = f f di1h,

for allfE C(a(T». By virtue of Remark 8.2 and the fact that T = f A dE, the unique map of Theorem 7.10 in Chapter 7 is identical to the map of Theorem 8.2 restricted to C1 (a(T»). Hence for every continuous function fin C1(a(T»),

(j(T)h I h) = f fdflh


and flk = ilk' This means, by equations (8. I 9) and (8.20), that E(M)

= tP(xM)· I

Our final consideration in this section is the multiplication operator form of the Spectral Theorem. We will prove it for self-adjoint and normal operators. The proof is relatively easy and depends primarily on the Continuous Functional Calculus Theorem of Chapter 7. We begin with a definition and a preliminary proposition.

Definition 8.3. A vector h in H is called a cyclic vector for T in L(H, H) if the closed linear span of {Tnh: n = 0, 1, ... } is H. I

Proposition 8.5. Let T be a bounded self-adjoint operator with cyclic vector ho . There then exists a unitaryt operator U: H -- L 2 (a(T), flllo)

such that

(UTU-lJ)()") = ).,f().,) a.e.

[Here flko is the unique measure such that U(T)ho I ho) = J f dflko for all fin C(a(T».] I

Proof. Define U on the dense subset

S = {cp(f)ho: fE C1(a(T»} of H (cp is the unique map of Theorem 7.10 in Chapter 7) by U(cp(f)ho) = f To show U is well-defined on S suppose that cp(f)ho = cp(g)ho' Then the equation

° = II cp(f)ho - cp(g)ho 112 = (cp(f - g)*cp(f - g)ho I ho)

= (cp[(f - g)(f - g)]ho I ho) = f If - g 12 dflko (8.21)

implies that f = g a.e. with respect to flko or that f = g in L2 (a( T), flko)·

An equation like (8.21) also shows that U is an isometric map from a dense subset of H onto a dense subset of L 2(M, flho)' namely, C1 (a(T». Hence U can be extended to an isometric map, also designated by U, from H onto L 2(M, flllo)' If fE C1(a(T»,

(UTU-lJ)()") = [UTcp(f)ho](;')

= [Ucp().,)cp(f)ho](;')

= [Ucp().,f)ho](;')

= ).,f(;')·

t By a unitary operator U: H --->- L. we mean a linear isometry from H onto L •.


By continuity, this equation is valid also for any I in L2(a(T), flho) so that the proposition is proved. I

Inasmuch as not all bounded self-adjoint operators have cyclic vectors, the above proposition is not applicable in the general situation. Observe that if h is any vector in H, then the closed linear span M of S = {Ph: n = 0, 1,2, ... } is invariant under T, that is, TM eM. Moreover h is a cyclic vector for M. Since TM C M if and only if T*(MJ.) C MJ. by Proposition 7.14, it is also readily apparent that T(MJ.) C MJ.. To extend the above proposition to the arbitrary case, H must be decomposed into cyclic subspaces as in the next proposition.

If (Hi)iel is an arbitrary collection of Hilbert spaces. each with inner product Wi' the direct sum tHieIHi of (Hi)iel is the Hilbert space H of all families of element (hi)iel with hi E Hi such that the family (II hi 11 2)iel

is summable. Addition, scalar multiplication, and the inner product in H are given by

Proposition 8.6. Let T be a bounded self-adjoint operator on H. Then H = tH1Hi , where (Hi)I is a family of mutually orthogonal closed subspaces of H such that

(i) T leaves Hi invariant, that is, THi C Hi for each i. (ii) For each i there exists a cyclic vector hi for Hi' implying

Hi = {/(T)hi : IE Cl(a(T)} for some vector hi· I

Proof. If h =1= ° is any vector in H, the closed linear span Hl of S = {Ph: n = 0, 1, ... } is invariant with cyclic vector h. If H = H l , the theorem is proved.

If Hl =1= H, there exists a vector h2 orthogonal to Hl and by the same process a subspace H2 of H exists such that H2 is orthogonal to Hl and satisfies (i) and (ii) with h2 the cyclic vector. If H = Hl tH H2, the proposition is proved. If not, we proceed using Zorn's Lemma as follows.

Let?" be the family of all systems (Hi)ieJ consisting of mutually orthogonal subspaces of H satisfying (i) and (ii). By Zorn's Lemma, the

Sec. S.l • Spectral Theory for Bounded Operators 213

family fF", ordered by the inclusion relation C, has a maximal element (Hi)ieI. Now H = EBI Hi' for if not there exists a vector ho orthogonal to EBI Hi and a cyclic subspace Ho satisfying (i) and (ii), contradicting the maximality of (Hi)I· I

Theorem 8.5. Multiplication Operator Form of the Spectral Theorem. Let T be a bounded self-adjoint operator on a Hilbert space H. Then there exists a family of finite measures (Jti)ieI on a(T) and a unitary operator

such that (UTU-1fMA) = AiiO),

I

Proof. By Proposition 8.6, H = EBieI Hi where each Hi is invariant and cyclic with respect to T. By Proposition 8.5, for each i there exists a unitary operator Ui: Hi ----+ L2(a{T), Ili) such that

(UiTIHPi-1fi){A) = Afi{A) ,

for fi E L2{a(T), Ili)' Define U: EBieI Hi ----+ EBieI L 2(a{T), Ili) by U(h) = {UA)ieI. It is easy to see that U satisfies the criteria of the theorem. I

If H is a separable Hilbert space, it is clear that the index set I in Proposition 8.6 and hence in Theorem 8.5 is the finite set {I, 2, ... , n} for some n E N or the set N itself. Using this observation, we have the following important corollary of Theorem 8.5.

Corollary 8.1. Let T be a bounded self-adjoint operator on a Hilbert space H. Then there exists a measure space (X,.N, Il), a bounded function F on X, and a unitary map U: H ----+ ~(X, Il) so that

(UTU-1f)(x) = F{x)f(x) a.e.

If H is separable, then Il can be chosen to be finite. Otherwise Il is semifinite. I

Proof. Let X be the disjoint union of card(I) copies of a(T) and define Il on X by requiring that its restriction to the ith copy of a(T) be Ili' If H is separable, in Proposition 8.6 we can choose the cyclic vectors hi so that II hi II = 2-i . In this case clearly Il(X) = 2::1Ili( a(T») < 00 since Ili( a(T» < 2-i (see Proposition 8.5). Letting U be the composition of the

unitary operator from Theorem 8.5 and the natural unitary operator from EBiEJ L2(a(T), fli) onto L2(X, fl), and letting F be the function whose restriction to the ith copy of a(T) is the identity function, the corollary follows. I

To extend the multiplication operator form of the Spectral Theorem to normal operators, we extend the continuous functional calculus to continuous functions of two variables in the following fashion (see also [25]).

Suppose TI and T2 are two commuting self-adjoint operators. (Actually the argument below applies to any finite collection of commuting selfadjoint operators.) Let S = a(TI) X a(T2) C R x R. Let CPI and rp2 be the unique mappings corresponding to TI and T2, respectively, as in Theorem 8.3 and let PI and P2 be the respective compact spectral measures. Thus for i = I or 2, Pi(M) = rpi(xM) for all measurable M contained in a(Ti). Since by Lemma 8.6, rpi(XM) is the strong limit of a sequence of polynomials in Ti , it is clear that for each measurable set A in a(Tl) and for each measurable set B in a(T2), P1(A) and P2(B) commute.

Let f be any finite linear combination of functions of the form X = XAxB, where A and B are measurable in a(Tl) and a(T2), respectively. Define X(TI' T2) as P1(A)P2(B) and define f(T1, T2) by linearity. It is straightforward to check that f is well defined and also to verify that f can be written as L aiXA/XB" where (Ai x Bi) n (Aj x Bj) = 0 for i *" j. Since f can be so written and

II X(T1, T2) II < sup I X(A.) I, A'S

it is also true that for f

II f(T1 , T2) II < sup I f(A.) I· A.S

The continuous mapping f -+ f(Tl' T2) can thus be extended to uniform limits of functions such as f By virtue of the fact that each function of the form p(x, y) = xiyj on S can be approximated uniformly by functions like f and that polynomials in the variables x and y approximate uniformly continuous functions on S by Corollary I. I the extension to uniform limits includes in particular continuous functions on S.

We thereby have a mapping 'IjJ from C1( a(Tl) X a(T2») -+ L(H, H). This mapping is easily seen to be linear, to preserve multiplication, and to satisfy 'IjJ(j) = ('IjJ(f»)*. Moreover the mapping f(x, y) = x + iy is mapped by 'IjJ to Tl + iT2' as is evident from (iv) of Theorem 8.3 if we consider


sequences of simple functions converging uniformly to fleX, y) = x and f.ix, y) = y, respectively.

Analogous to Definition 8.3 we can consider for ho E H the closed linear span of the set {TliT2jhO: i = 0,1,2, .. . ,j = 0, 1,2, ... }. If this closed span is H, we can define as in Proposition 8.5 a unitary operator U: H -+ L2 (a(TI) X a(T2), f-l) where f-l is the unique measure such that

for all f in Cl ( a(TI) X a(T2», and U is defined by the rule U( tp(f)ho) = f for continuous functions.

Now if T is any normal operator, then T can be written as Tl + iT2' where TI and T2 are the commuting self-adjoint operators given by

Tl = !(T + T*) and T2 = li(T - T*).

Since tp(x + iy) = TI + iT2 = T, the unitary operator U satisfies for f in Cl (a(Tl) X a(T2» the equation

(UTU-lJ)(x, y) = [UTtp(f)ho](x, y)

== [Utp(x + iy)tp(f)ho](x, y)

= [Utp[(x + iy) . f(x, y)]ho](x, y)

= (x + iy)f(x, y).

By continuity, this equation is also valid for any fE L 2(a(TI) X a(T2), f-l). Thus for any normal T, if the closed linear span of {TliT2jho: i = 0,

1, 2, ... , j = 0, 1, 2, ... } is H for some ho in H, then T =, TI + iT2 is unitarily equivalent to a multiplication operator just as for self-adjoint operators. By a similar application of Zorn's Lemma to that used in Proposition 8.6 we arrive at the following generalization of Theorem 8.5.

Theorem 8.6. Let T = Tl + iT2 be a bounded normal operator on H. Then there exists a family of finite measures (f-li)ieI on a(Tl) X a(T2) and a unitary operator

U: H -+ EB L2(a(TI) X a(T2) , f-li), ieI

such that

(UTU-1fMx, y) = (x + iy)fi(x, y) a.e.,

where f = Ui)I is in EBieI L2(a(T), f-li)' I

Immediately we obtain the following corollary of Theorem 8.6, which is analogous to Corollary 8.1.

Corollary S.2. Let T be a bounded normal operator on a Hilbert space H. Then there exists a measure space (X, .J/, p), a bounded complex function G on X, and a unitary map U: H ~ L 2(X, p) so that

(UTU-lj)(A) = G(A)f(A) a.e.

If H is separable, p can be taken to be a finite measure. Otherwise p is semifinite. I

Using this representation of a bounded normal operator we can prove a spectral measure version of the spectral theorem for bounded normal operators similar to Theorem 8.4 for self-adjoint operators. The proof we give here, however, uses a different approach than that used in Theorem 8.4.

Theorem S.7. Spectral Measure Version of the Spectral Theorem for Normal Operators. If T is a bounded normal operator on Hilbert space H, then there exists a unique normalized compact spectral measure P on B(R2), the (weakly) Borel subsets of R2(= R X R), such that T = J AXo-<Tl dP. I

Proof. By Corollary 8.2, we can assume that T is a linear operator on L 2(X, p) given by Tf(x) = G(x)f(x) for some bounded complex function G on X. For each ME B(R2) define P(M) E L(L2' L 2) by P(M)f = la-1(M)f = (XM 0 G)f. Then P is a normalized compact spectral measure on B(R2). [Note: rep) C aCT) since aCT) is the essential range (Example 6.17) of G and so P(O) is 0 for 0 C a(T)c.J We must show that for each fE L 2(X, p), (Tf I f) = fAXI1(T)(A) dp!, where pAM) = (P(M)f I f) for ME B(R2). Theorem 8.2 then gives us T = fAXI1(T)(A) dP.

Fix f in L 2(X, p) and let ME B(R2). Then

J XM dP! = (P(M)f I f) = J (XM 0 G)Jjdp,

whereby, for all simple functions h = L aiXMi' Mi E B(R2),

J h dP! = J (h 0 G)Jjdp.

Therefore we have

fAX,,(7') dftJ = f (AX,,(7') 0 G)fi dft = f Gfi dft

= (Tf I f).

217

To show uniqueness, suppose also E is a normalized compact spectral measure on B(R2) such that T = J AX" (7') (A) dE. Using Remark 8.2 (iii),

J ~X,,(7')(A) dE = T* = f ~X,,(7')(A) dP.

Hence for all polynomials p(A,~) on a(T)

f PXa(7') dE = f PX,,(T) dP,

so that, for any f E L2 ,

f PXa(7') dflJ = f PX,,(7') dftJ

where flJ(M) = (E(M)f I f) and ftiM) = (P(M)f I f» if ME B(R2). Using Theorem 1.27, if g E C1 (a(T),

Since flJ and ftJ are finite regular (weakly) Borel measures on B(R2), flJ = ftJ (See Theorem 5.10 or Theorem 5.11) for each f Therefore P = E. I

Problems

8.1.1. Let T be a bounded linear operator on H. Prove

(i) a(T) C {(Tx I x: II x II = I}, the closure of the numerical range of T;

(ii) if w(T) = sup {I (Tx I x) I: II x II = I}, then r(T) < w(T) < II Til, where r(T) is the spectral radius of T; and

(iii) if T is normal, r(T) = w(T) = II Til. [Hint: Prove r(T) = II T II by showing lim II pn 11 1I2n = II Til.]

8.1.2. A spectral function on R is a function £: R -->- L(H, H) whose values £(J..) are projections and that satisfies

(I) £(.1) < £(A') if A <A', (2) £(A) = (s) lim £(A') (in the strong sense),

A' ... H A';""

(3) £_00 = (s) lim £(A) = ° and £+00 == (s) lim £(A) = I. A-+-OO

A normalized spectral measure on R is a function P from the Borel sets B(R) of R into L(H, H) whose values are projections such that

(a) P(R) = I, (b) P(0) = 0, (c) P(M U N) = P(M) + peN) whenever M n N = 0,

(d) p( 0 Mi) = (s) lim t P(MJ whenever M j n Mj = 0 "~l n "~l for i"* j.

Prove that to every spectral measure P on B(R), there corresponds a unique spectral function £ such that £(A) = p«- 00, An, and conversely.

8.1.3. Let (Xk)N be an orthonormal sequence in a Hilbert space H. Let (Akh be a bounded nondecreasing sequence of real numbers. Define for each real number A the operator £(A) on H by

£(A)h = 0, n

£(J..)h = L (h I Xk)Xk> k=l

£(A)h = h,

(i) Show that £(A) is a spectral function on ~. (ii) Let P be the normalized spectral measure on B(R) corresponding

to £ as in Problem 8.1.2. Show that the measures flk corresponding to P (and hence £) as in Theorem 8.1 are Lebesgue-Stieltjes measures restricted to B(R) and determine a generating function for each flh'

(iii) Calculate the spectrum rep) of P. (iv) Let T = J AXL'(I,)(A) dP(A). Calculate (Th I h) and (Th I k) for

h,k in H.

8.1.4. Let TE L(H, H). Prove T is compact if and only if the range R(T) contains no infinite-dimensional closed subspace. [Hint: For sufficiency, let T = UP be the polar decomposition of T. Assuming R(T) does not contain any infinite-dimensional closed subspaces, prove P is compact by using Corollary 8.1 and proving M p: L 2(X, fl) -->- L 2(X, fl) given by M pf = Ff is compact. To show M p is compact, let e > 0, let X. =

Sec. 8.2 • Unbounded Operators 219

{x EX: F(x) > e} and let Me be the closed subspace {IE L 2(X, fl):/= 0 a.e. on X."}. Show Me is finite dimensional and prove II MF - PeMF II < e, where Pe is the projection onto Me.]

8.1.5. Let (X, d, fl) be a semifinite measure space. Let FE Loo(fl). Define Tp: Lifl) -'-+ L 2(fl) by Tp(g) = Fg. Prove the following:

(i) The mapping F f--+ T p is a one-to-one linear map from Loo into

L(L2' L 2);

(ii) Tp is normal; (iii) T p is self-adjoint if and only if F is real-valued a.e.; (iv) T p is unitary if and only if I F I = I a.e.; (v) II Tp II = II F 1100; (vi) If G is a measurable function, finite a.e., such that Gg E L 2(fl)

for all g E L2(fl), then G E Loo(fl).

8.1.6. Let (X, d, fl) be a finite measure space. Let T E L(L2' L2). If for G E Loo(fl) , To E L(L2' L2) is the operator To! = Gf, prove that if TTa = TaT for all G E Loo(fl), then T = T p for some FE Loo(fl). Deduce that {To: GE Loo(fl)} is a maximal commutative family in L(L2' L2). [Hint: Let F = T(x.d.]

8.1.7. Using Corollary 8.1 show that each positive operator T has a unique positive square root yr. [Hint: We may assume T operates on L 2(X, fl) by the rule T(f) = Ff, where F > O. For existence, let A(f) =

YF f For uniqueness, suppose also T = B2. Show the following: A and B commute with T; A is the uniform limit of a sequence of polynomials in T since the square root function on aCT) is a uniform limit of a sequence of polynomials; A and B commute; Ker T and (Ker T).l are invariant under T, A, and B; Ker T = Ker A = Ker B = Ker(A + B); (A + B)(Ker T).l is dense in (Ker T).l; and (A - B)(A + B) = 0.]

8.1.8. Prove a bounded normal operator on H is (i) self-adjoint, (ii) unitary, (iii) positive, (iv) a projection if and only if its spectrum is respectively (i') rcal, (ii') on the unit circle, (iii') on the nonnegative real axis, or (iv') the set {O, I}.

8.2. Unbounded Operators and Spectral Theorems for Unbounded Self-Adjoint Operators

Most of the operators that occur in applications of the theory of Hilbert space to differential equations and quantum mechanics are unbounded. For this reason, we consider in this section basic definitions and theorems


necessary to deal with unbounded operators and also we prove versions of the spectral theorem for unbounded operators. Let us first define precisely the type of operator we consider in this section.

Definition 8.4. A linear operator T in (in contrast to "on") a Hilbert space H is a linear transformation on a linear subspace DT of H into H. DT is called the domain of T and RT == {Tf: f E DT} is the range of T. I

If Sand T are operators in H, we say T extends S; written SeT, if Ds C DT and Sh = Th for all hinDs. Tis also called an extension of S.

T is bounded if T is bounded as a linear transformation from DT into H.

Example 8.1. Let H = L 2( - 00, 00) and let T be defined in H on

by Tf('),,) = ')"f('),,). Since DT contains all functions in L2( - 00, 00) which vanish outside a finite interval, DT is a dense subset of H and contains moreover the characteristic functions Xln,n+l)' Since II Xln,n+1) II = 1 and

In+!

II TXln,n+U 112 = n x 2 dx > n2 , for n> 0,

T is clearly unbounded.

Analogous to bounded linear operators on H, we can also consider in many cases the adjoint of a linear operator in H whether it be bounded or unbounded by means of the following definition.

Definition 8.5. Let T: DT - H be an operator in H. Let DT* be the possibly empty set given by

DT* = {h E H: corresponding to h is a unique element h * in H such that

(Tk I h) = (k I h*) for all k in DT}.

If DT*:f:. 0, the mapping T*: DT* - H given by T*h = h * is called the adjoint of T. I

The questions that naturally arise are: When does T* exist (DT_ :f:.01) as an operator in H and when is DT* = H1 Obviously, if T is a bounded linear operator on H, then DT_ = Hand T* is the "ordinary"


adjoint defined earlier. The following proposition tells us in general when T* has any meaning at all.

Proposition 8.7. DT."* 0 and T* is a linear mapping in H with domain DT• if and only if DT is dense in H. I

Proof. Suppose DT = Hand hand h * are two vectors satisfying

(Tk I h) = (k I h*), for all k in DT. (8.22)

Then h* is uniquely determined by h and equation (8.22), for if hi * also satisfies equation (8.22), then

(klh*-hl*)=O, forallkinDT

and as DT = H, h* = hi *. This means DT."* 0 since 0 EDT" To show T* is linear, let hi and h2 be in DT_. Then

(k I u1T*h1 + U2T*h2) = o.1(k I T*h1) + o.2(k I T*h2)

= 0.1 (Tk I hi) + o.2(Tk I h2)

= (Tk I u1h1 + U2h2),

so that not only is u1h1 + U2h2 in DT• but T*(u1hl + U2h2) = u1T*h1 + U2 T*h2·

Conversely, suppose DT is not dense in H. Let h be in H - DT. By Theorem 7.2, h = hi + h2' where hi J.. DT and h1"* O. Hence

for all k in DT ,

and corresponding to 0 are two vectors 0 and hi that satisfy equation (8.22). Moreover this means that for every vector x in H for which there exists a vector x* satisfying equation (8.22), there are two vectors x* and x* + hi, that satisfy equation (8.22). This means Dr' = 0 and T* cannot exist. I

The notion of the adjoint of an operator in H gives rise to the idea of self-adjointness of operators in H. This idea generalizes our previous concept of self-adjointness for bounded operators on H.

Definition 8.6. An operator in H with DT = H is symmetric if T* ;2 T. A symmetric operator is self-adjoint if T* = T. I

The next result shows in particular that a self-adjoint operator with DT = H is always bounded.

Proposition 8.8. The adjoint T* of a linear operator on H (Dl' = H) is a bounded operator in H. I

Proof. Assume T* is not bounded. There then exists a sequence (hn )

in DT* such that II hn II = I and II T*hn II -- 00 as n -- 00. Define the functionals f1Jn on H by

The f1Jn are clearly bounded linear functionals on H. Moreover for each h, the sequence (fIJn(h)) is also bounded as shown by

I f1Jn(h) I = I (Th I hn) I < II Th II II hn II = II Th II.

By the Principle of Uniform Boundedness (Theorem 6.11) we have for some constant C,

I f1Jn(h) I < C II h II, n = 1,2, .... (8.23)

Letting h = T*hn in inequality (8.23) we have

which implies II T*(hn) II < C. This is a contradiction to the selection of (hn)· I

Corollary 8.3. Any symmetric operator defined on H is bounded. I

Below are listed a few more important but trivially demonstrable facts.

Remarks

8.3. If Sand T are linear operators in H, SeT and Ds =c H, then T* C S*.

8.4. If T is a linear operator in H with DT = DT* = H, then T C T**.

8.5. Suppose T is a one-to-one operator in H so that T-1 is defined on DT-l == T(DT)' If DT = DT-l = H, then T* is one-to-one and (T*)-l = (T-l)*.


Recall from Chapter 6 that an operator Tin H is closed if whenever hI' h2' ... in DT converges to h in Hand Thl , Th2, ... converges to k in H, then h E DT and Th = k. Equivalently T is closed if its graph GT = {(h, Th): hE DT } is closed in the direct sum H EB H. [H EB H = {(h, k): hE H, k E H} with addition and scalar multiplication defined componentwise and the inner product given by (h, k) I (hI' k l») = (h I hI) + (k I kl)']

Proposition 8.9. The adjoint T* of T in H is closed. I

Proof. Let gi. g2 ... be a sequence in DT* converging to g in Hand suppose T*gl' T*g2' ... converge to h in H. Then for any k in DT ,

(Tk I g) = lim (Tk I gn) = lim (k I T*gn) = (k I h), n n

implying that g E DT* and h = T*g. I

Definition 8.7. An operator T in H is called closable if there is a closed operator S in H which extends T. The closure l' of a closable operator T is the smallest closed operator extending T, that is, any closed operator extending T also extends 1'. I

Obviously the closure exists for every closable operator.

Proposition 8.10. Suppose T is an operator in H with DT = H. Then

(i) if T is closed, DT* = Hand T = T* *,

(ii) DT* = H if and only if T is closable, in which case l' = T* *, and

(iii) if T is closable, (1') * = T*. I

Proof. Observe from the definition ofT* that the set A = {(T*g, -g): g E DT*} is the set of all pairs (g *, -g) with g E DT* such that

(Tk I g) = (k I g*), for all k in DT •

In other words A is the set of points (g*, -g) such that

(k, Tk) I (g*, -g»HeH = 0, for aU k in DT ,

where (I )HeH is the inner product in H EB H. This means A is the

orthogonal complement GTL of the linear space GT. If T is closed then A.l = (Gr).l = GT .

Secondly, observe that if D7,* = H so that T* * exists, then the graph of T* * consists of all the points (f, f*) of H fB H such that

(T*h If) = (h If*) , for all hE DT>,

or for which (T*h, -h) I (f, f*»)HffJH = o.

This implies GT .. = (Gr).l, and in particular GT .. = GT if T is closed. The proofs of (i), (ii), and (iii) follow from these observations. For

(i), assume DT> ~ H and let h be a nonzero element of H such that (h I g) = 0 for all g EDT>. Then

(0, h) I (T*g, -g»)HlffJHa = 0 , for all g in DT>.

Since A.l = GT , (0, h) E GT so that h = O. Hence DT> = H. Observing that always A C A * *, the relation GT» = GT above shows that A = A * *.

In (ii), if T is closable then by (i), D (T» = Hand l' = T* *. However, D(T» eDT> so DT> = H. Conversely, if Dr- = H then T* * exists and T C T* *. If S is any closed extension of T, then G s is closed and contains GT. From the observations above, GT .. = (GTL).l. Hence GT» is the closure of GT and Gs :> GT». Hence l' = T* *.

To prove (iii) notice that, if T is closable,

T* == (T*) = T*** = (1')*. I A symmetric operator is always closable since DT> :> DT and DT is

dense in H. If T is symmetric, T* is a closed extension of T, so the closure T* * of T satisfies T C T* * C T*. If Tis self-adjoint, T = T* * = T*. If T is closed and symmetric, T = T* * C T*. Clearly, a closed symmetric operator is self-adjoint if and only if T* is symmetric.

The following is an illustrative example of three closed operators in H.

Example 8.2. As in Theorem 4.3, a complex-valued function f is absolutely continuous on [a, Pl, where -00 < a < P < 00, if there exists an integrable function g on [a, Pl such that

f(x) = f: get) dt + f(a).

Such a function f is continuous on [a, Pl and differentiable a.e. with

f'(x) = g(x) a.e. on [a, ,8]. Yet its derivative need not belong to L2(a, fJ), as is shown if one considers the function f(x) = X I/2 in L 2 [0, 1].

For the purpose of this example we consider three separate Hilbert spaces as follows:

HI = L2 [a, fJ],

Hz = ~[a, 00),

H3 = L 2(-00, 00).

where -00 < a < fJ < 00,

where -00 < a < 00,

Also we consider three operators TI in HI, Tz in H 2 , and T3 in H3 defined, respectively, on the following three domains:

DI = {g E HI: g = f a.e. where f is absolutely continuous on [a, fJ], f(a) = ° = f(fJ), and f' E L2 [a, fJn,

Dz = {g E Hz: g = f a.e. where f is absolutely continuous on [a, fJ] for each fJ > a, f(a) = 0, and f' E Lz[a, fJn,

D3 = {g E H3: g = f a.e. where f is absolutely continuous on [a, fJ] for each -00 < a < fJ < 00, and f' E H3}'

By definition, TIg = if', T2g = if', and T3g = if' (g, f as above). For each i = 1,2,3, Di = Hi' To show this for DI we recall that

the linear subspace spanned by the set {xn: n = 0, I, 2, ... } is dense in L2 [a, fJ] since the class of all complex polynomials is dense in L 2 [a, fJ]. However, each xn is in DI since each xn can be approximated in L 2 [a, fJ] by a function f in DI as illustrated in Figure 8.1. This means DI = H. To

, x" ,

l Fig. 8.1

l l

prove D2 = Hand D3 = H, it is sufficient to observe analogously that the linear subspace spanned by the set {xne-zll/2 : n = 0, 1, 2 ... } is dense in L2( - 00, 00 ) {and hence their restrictions to [a, 00) are dense in L 2(a, oo)} and then to approximate analogously each X ne-zll/2 by a function I in Dl or D2 •

The important observations to make here about T1 , T2, and T3 are the following: Each is an unbounded symmetric operator in Hi' respectively, and in particular T3 is self-adjoint, whereas Tl and T2 are not self-adjoint. Moreover in each case Ti = Ti * *.

The verification that each Ti is unbounded is accomplished by considering functions in Di of the following form. For a < (3 and n > 2/({3 - a) define In by (see Figure 8.2)

{ n(x - a),

In(x) = 2 - n(x - a), 0,

if x E [a, a + lin] , if x E [a + lin, a + 21n] , if x E [a + 21n, 00).

Clearly In' (X) = n, -n, or ° on the respective intervals and

f<x+2In Il/n 112 = I/n(x) 12 < 21n,

<X

while

f<x+2In II ifn' 112 = <X n2 dx = 2n.

These equations imply that

II DJn II = D.(~) > (2n)1I2 Il/n II ~ Il/n II - (2In)1I2 = n.

The verification that each Ti is symmetric is obtained in the following manner by integrating by parts. For I and g in the domain of Tl

01 0I+2/n {3

Fig. 8.2

and for -00 < a < p < 00,

(if' I g) - (fl ig') = i f: f'(~)g(n d~ + i f: f(~)g'(~) d~ -lfJ = if(~)g(~) ,,= 0, (8.24)

whereby (TIll g) = (fl Tlg). The verifications for T2 and T3 are similar. {Observe that if fE D2 then limx~oof(x) = 0 and likewise if fE D3 then limx~±oof(x) = O. Indeed, for H2 the equation

(flf') + (f' If) = !~~f: [f(~)f'(~) + f'(~)f(~)] d~ = lim [If(x) 12 - I f(a) 12]

X~oo

implies that limx~oolf(x) 12 exists and since fE L 2(a, P), f(x)-+O as x-+oo.}

Let us next calculate the adjoint Tl * of T1 • Let Dl * be the set

Dl* = {g E HI: g = f a.e. where f is absolutely continuous on [a, P], f' E Hd·

Since equation 8.24 still is valid for g in Dl *, the domain of Tl * contains Dl* and Tl*g = if' for g in Dl*' where g and f are as above. We wish to show that the domain of Tl * is Dl *, which will clearly show that Tl C Tl * and Tl =I=- Tl *. To this end, let f be in the domain ofc'1'l *. Let h be the absolutely continuous function given by

where C is a constant chosen so that

f: [f(~) + ih(~) d(~)] = O.

For every g in D1 , an integration by parts gives that

Hence


In particular, letting g be the function in DI given by

g(x) = J: [/(~) + ih(~)] d~ , we obtain that

f: I/(~) + ih(~) la = 0

or that a.e. we have

I(x) = -ih(x) = -i f: TI */(~) d~ - iC

and h is absolutely continuous with h'(x) = TI */(x). Hence I is in DI *. In an almost identical fashion the verification that ~*g = if' on the

domain

Da * = {I E Ha: I is absolutely continuous in each [a, P1 with fJ > a and f' E Ha}

and that Ta *g = if' on the domain

can be carried out. Since DI ~ DI *, Da ~ Da *, and Da = Da *, clearly Ta is self-adjoint and TI * *, Ta * *, and Ta * * are all defined.

It remains to show that Tj = T;* * for j = 1 and 2. In either case since Tj C T;*, we can say that Tj C T;* * C T;*. It suffices to show therefore that Dps •• C Dj • Let IE Dpf*' Then for all g in D;* we have

(T;**/I g) = (II TJ*g).

and moreover since T;* *1 = if' (because T;* * C T;*) we have

0= (if' I g) - (II ig').

If j = I, this means

0= i J: f'(~)g(~) d(~) + i f: I(~)g'(~) d~ = if(~)g(~) I: = i[/(fJ)g(fJ) - I(a)g(a)].

By first letting g(x) = (x - a)/(fJ - a) in DI * and then letting g = (fJ - x)/(fJ - a) in DI*' we obtain/(a) = 0 =1(fJ) implying that IE D1 •

If j = 2, let g(x) = r(z-IX) to obtain I(a) = 0 so that I is again in Da.

Interestingly enough, TI has uncountably many different self-adjoint extensions. Let y E C with I y I = 1 and define Ty in HI on

DTy = {g E L2(a, fJ): g = f a.e. where f is absolutely continuous on [a, fJ], f' E HI' and f(fJ) = yf(a)}

by Tyg = if'. Each Ty is self-adjoint and extends T1 • For each y, we have

TI C Ty C TI*·

The next example presents a type of self-adjoint operator which will be shown in Theorem 8.7 to be the "prototype" of all self-adjoint operators.

Example 8.3. Let (X, JJf, f-l) be a measure space with f-l a finite measure. Suppose that f is an extended real-valued measurable function on X which is finite a.e. Then the operator Tf in L2(X, f-l) defined by Tig) =fg on

is self-adjoint. Indeed, by considering the functions fn defined as

on Nn = {x: I/(x) 1< n}, otherwise,

one can easily verify that gin E Df for each n = 1,2, ... and for each g in L2(X, f-l). Moreover, since the functions gin are dense in L2(X, f-l),

Df = H. Tf is thus clearly symmetric. In addition if h is an element in DT/,

then by the Monotone Convergence Theorem,

II Tf*h II = lim II fn Tf*h II n-+oo

= lim [sup (k Ifn Tf*(h))] n-+oo IIkll-1

= lim [sup (Tikfn) I h)] n-+oo IIkll=1

= lim [sup (k Ilnfh)] = lim IIfnfh II = Ilfh II. n-+oo IIkll-1 n-+oo '

Hence Ih E L2(X, f-l) and hE DTr Therefore, DTf = DTf* and Tf is selfadjoint.

As before for bounded linear operators on H, a scalar A is an eigenvalue of an operator T in H if there exists a nonzero vector h in DT

such that Th = Ah. If T is a symmetric operator, the equation

A(h I h) = (Ah I h) = (Th I h) = (h I Th) = (h I Ah) = X(h I h) shows that eigenvalue A is always real.

Again, the resolvent set e(T) of an operator in H with Dr = H is the set of all scalars A for which R1'-Al is dense in H and for which (T - AI) has a bounded inverse defined on R T - JJ • The spectrum a(T) of T is the complement of e(T) and is clearly decomposed into three disjoint sets: the set of eigenvalues of T (sometimes called the point spectrum Pa(T», the set of scalars A for which RT - U is dense in H but for which (T - 1.1)-1 exists and is not bounded (called the continuous spectrum Ca(T», and the set of scalars A for which (T - 1.1)-1 exists but its domain RT - Al is not dense (called the residual spectrum Ra(T».

Proposition 8.11. If T is a closed linear operator in H with DT = H and I.E e(T), then (T - AI)-1 is a bounded linear operator on (all of) H. I

Proof. R T - U is dense in H and since (T - AI)-1 is bounded on R T - U ,

there exists a positive constant C such that

II h II < C II (T - Al)h II, for all h in D T • (8.25)

If k = limn(T - AI)hn' then by inequality (8.25) the limn~ocA exists, say h. Since T is closed, (T - AI)h = k and k E RT - U • Hence RT - U = H. I

In light of Proposition 8.11, we can see that the resolvent of a closed operator in H is the set of scalars A for which T - AI is a bijection from DT to H and for which (T - AI)-1 is bounded on H.

The next result establishes that the spectrum of a self-adjoint operator in H is contained in the real numbers.

Proposition 8.12. Let T be a self-adjoint operator in H. Then e(T) contains all complex numbers with nonzero imaginary part. Moreover if ImA-=p 0 then

and Im(T - AI)h I h) = Im( -A) II h 11 2 , for all hE DT •

(8.26)

(8.27)

I

Proof. Since T is self-adjoint, (Th I h) is real for all h in DT • Clearly equation (8.27) follows and by the Cauchy-Schwarz inequality,

I Im(A) I II h 112 < I (T - AI)h I h) I < II (T - AI)h II II h II,

which implies that

I Im(A) I "h II < " (T - U)h II. (8.28)

If Im(A) -=1= 0, this inequality implies that (T - AI)-l exists as a linear operator on RT - U since T - U is one-to-one by inequality (8.28). Now RT-U is also dense in H. Indeed, if not, there exists a nonzero vector k in DT (which is dense in H) such that (T - U)h I k) = ° for all h in D7,.

However, then (h I (T* -U)k) = (h I (T - AI)k) = ° for all h in DT; and since DT = H, (T - AI)k = 0. This means Tk = Ak and (Tk I k) = A(k I k), which contradicts the fact that (Tk I k) is real. Therefore RT-J.I is dense in H. Moreover from inequality (8.28), (T - AI)-l is bounded on RT-J.I with bound satisfying inequality (8.26). All this means A E e(T) when 1m A -=1=0. I

The next proposition also shows that the resolvent set is open for a self-adjoint operator.

Proposition 8.13. If T is a closed linear operator in H with DT = H, then the resolvent set is open and if A and p are in (!(T),

RI' - RA = (A - P)RARI"

where, as in Proposition 6.27 of Chapter 6, RA = (T - AI)-l. Moreover R" as a function on e(T) to L(H, H) has derivatives of all orders. I

Proof. Observing that from Proposition 8.11, RA is defined on H for A E e(T), the proof is identical to that of Proposition 6.25 of Chap~~ I

Remarks. Here are some other easy to prove facts concerning the eigenvalues and the spectrum of an operator in H.

8.6. If DT = DT* = Hand T = T**, then A E G(T) if and only if A E G(T*).

8.7. If DT = D7'« = Hand T = T**, then

{h E DTl Th = Ah} = [(T* - U)D7'«ll.

In particular if T is self-adjoint, A is an eigenvalue of T if and only if (T - AI)DT is not dense in H.

8.8. If T is a self-adjoint operator, A E (!(T) if and only if (T- U)DT= H.


Proof. (8.6) If A E e(T), then by Proposition 8.Il D(7'-}'O-1 = H. Since (T - A.J)* = T* - XI, by Remark 8.5 we have (T* - XI)-l = [(T - AI)-I]*. Since [(T - AI)-1 1* is bounded, X E e(T*). Conversely, by reversing the argument, X E e(T*) implies A E e(T).

(8.7) Th - Ah = ° ~ (Th - Ah I k) = 0, ~(h I (T* - X)k) = 0,

~ hE [(T* - XI)DTo].L.

for all k in DTo ,

for all k in DTO ,

(8.8) If A E e(T), we know from Proposition 8.11 that (T - A.J)D7' = H. Conversely, suppose (T - A.J)D7' = H. If A $ R, A E e(T) by Proposition 8.12. If A E R, A is not an eigenvalue by (8.7) and T - A.J is selfadjoint. By Remark 8.5, (T - AI)-l is also self-adjoint and by Proposition 8.8, (T - AI)-l is bounded. Hence A. E e(T)' I

Example 8.4. To illustrate some of the preceding theory let us consider three versions of the differentiation operator idfdt. Let T1 , T2 ,

T3 be the operator idfdt on the respective domains+:

DTI = {fE L2[0, 2n]: f is absolutely continuous on [0, 2nJ and f(O) = O}

DTa = {fE ~[O, 2n]: f is absolutely continuous on [0, 2n] and f(O) =f(2n)}

DTa = {fE L 2[0, 2n]: fis absolutely continuous on [0,2n]}.

From Example 8.2 and the fact that Dl of that example is contained in DT1 , DTa , and DT8 , each of these domains is dense in H. The spectrum of Tl is empty, the spectrum of T2 is the set of integers (which is also the set of eigenvalues of T.), and the spectrum of T3 is the whole complex plane. That the spectrum of Tl is empty follows from the fact that for each A the operator S" given by

(SAK)(t) = J: e-UIt-8)g(s) ds,

is the inverse of Tl - AI. To calculate the spectrum of T2, observe first that a(T2) is a set of real numbers since T2 is self-adjoint as in Example 8.2. For each integer k, the function f{t) = e-ikt is a solution of T2f = kf so that each integer is an eigenvalue. It is easy to verify that for each non-

+ Though not explicitly stated, these domains are here assumed to have the property g = fa.e. and fe DTJ => g e DTJ and TJK = if'·


integer real number A the equation

if'(t) - Af(t) = get)

is solvable for each g in L 2 [0, 211:] so that by Remark 8.8 each such A is in e(T). Finally to calculate the spectrum a(Ta) observe that for each A in C, the function f(A) = e-iA! is a solution of if'(t) - Af(t) = 0 so that each A in C is in fact an eigenvalue of Ta.

Let us now turn our attention to several versions of the Spectral Theorem for unbounded operators. In this case a multiplication operator form of the theorem leads nicely into a functional calculus form and then into a spectral measure version of the Spectral Theorem. The interested reader may also consult references [25] and [31].

The next theorem shows that all self-adjoint operators in Hare unitarily equivalent in the sense of (ii) below to a self-adjoint operator of the type given in Example 8.3 ..

Theorem 8.8. Multiplication Operator Form of the Spectral Theorem. Let T be a self-adjoint operator in a Hilbert space H. There exists then a measure space (X, ..w: 1-'), a unitaryt operator U: H --+ L2(X, 1-'), and a measurable function F on X which is real a.e. such that

(i) hE DT if and only if F(·)Uh(.) is in L 2(X, 1-'),

(ii) if fE U(DT), then (UTU-lJ)(·) = F(· )f(·). I

Proof. To achieve the proof we utilize the multiplication operator form of the Spectral Theorem for bounded normal operators (Corollary 8.2) by applying it to the operator (T + i)-I. Let us first establish that this is a bounded normal operator.

By Propositions 8.11 and 8.12, (T ± i)-I exist as bounded linear operators on H. In particular RT±i = Hand T ± i are one-to-one operators. For any hand k in DT , since T is self-adjoint,

«T - i)h I (T + i)-l(T + i)k) = «T - i)-l(T - i)h I (T + i)k).

This implies that «T + i)-1)* = (T - i)-I. Since (T + i)-I and (T - i)-I commute by Proposition 8.13, we have

(T + i)-l«T + i)-l)* = (T + i)-l(T - i)-I = «T + i)-l)*(T + i)-I,

and (T + i)-I is seen to be a normal operator.

t Bya unitary operator U: H ->- L2 we mean a linear isometry from H onto L2. (See also Proposition 7.17.)


By Corollary 8.2, there is a measure space (X,.J¥; p), a unitary operator U: H --+ L2(X, p), and a bounded, measurable complex function G on X so that

(U(T + i)-IU-lJ)(X) = G(x)f(x) a.e. (8.29)

for all fin L 2(X, p). Since Ker (T + i)-I = {O}, G(x) oF 0 a.e. Therefore if we define F(x)

as G(X)-1 - i for each x in X, I F(x) I is finite a.e. Now iffE U(DT ), then there exists a function g in L2(X, p) such that f(· ) = G(· )g( . ) in L2 • That this is so follows from the inclusions

Observing that U(T + i)-IU-1 is an injection, for any g in the range of U(T + i)-IU-l we have from equation (8.29)

[U(T + i)-IU-l]-lg(X) = [l/G(x)] . g(x) E L 2(X, p).

In particular for f in the set U(DT ),

or

U(T + i)U-lJ(x) = [l/G(x)] . f(x) E L 2(X, p)

or

UTU-lJ(x) = [I/G(x)]f(x) - if(x) = F(x)f(x) E L 2(X, p).

This proves (ii) and the necessity of (i) provided F is real-valued, which we show below. For the converse of (i), if F(x)Uh(x) is in L 2(X, p), then there is a k in H so that Uk = [F(x) + i]Uh(x). Thus G(x)Uk(x) = G(x)[F(x) + i]Uh(x) = Uh(x), so h = (T + i)-lk, whereby hE DT .

To finish the proof it must be established that F is real valued a.e. Observe that the operator in L2(X, p) defined by multiplication by F is self-adjoint since by (ii) it is "unitarily equivalent" to T. Hence for all XM, M a measurable subset of X, (XM I FXM) is real. However, if 1m F> 0 on a set of positive measure, then there exists a bounded set B in the plane so that M = F-l(B) has nonzero measure. Clearly FXM is in L 2(X, p) since B is bounded and Im(XM I FXM) > O. This contradiction shows that 1m F= 0 a.e. I

Using the foregoing theorem we can prove the following result.

Sec. 8.2 • Unbounded Operators 23S

Theorem 8.9. Functional Calculus Form 01 the Spectral Theorem. Let T be a self-adjoint operator in H. There is a unique map !fo from the class, of bounded Borel measurable functions on R into L(H, H) so that

(i) !fo(al + fJg) = a!foU) + fJ!fo(g), !foUg) = !foU)!fo(g), !fo(i) = (!foU»*, !fo(I) = 1.

(ii) II !foU) II < K II 11100 for some K> o. (iii) If Un) is a sequence of bounded Borel functions converging

pointwise to the identity function on Rand Iln(x) I < I x I for all x and n, then for any h E DT , limn!foUn)h = Th.

(iv) If Un) converges pointwise to g and (111n lloo)N is bounded, then ri;Un) - ri;(g) strongly.

In addition:

(v) If Th = Ah, !fo(g)h = g().)h.

(vi) If h > 0, then !fo(h) > o.

Proof. Define tjJ by

tjJ(g) = U-1Tg(FP,

I

where F and U are as in the previous theorem and Tg(FI: L 2(X, p) - L 2(X, p) is given by Tg(FIV' = g(F(. ) )'1'(.). Using the previous theorem, the verification that !fo satisfies the conditions of the theorem is routine but arduous. We illustrate by verifying !foU) = (!foU»* and condition (iii).

First we show !foU) = (!foU»*. For any hand k in H we have

Similarly,

(!foU)h I k) = (U-lf[F(. )]U(h)(·) I k)

= (U-lf[F(. )]U(h)(·) I U-l[U(k)(·)])

= (J[F(·)]U(h)(·) I U(k)(.»

= f f[F(· )]U(h)(· )U(k)(·) dp(·).

(h I ri;(i)k) = (U-l[U(h)(.)] I U-Y[F(·)]Uk(.»

= f U(h)(·)i[F(.)]U(k)(.)dp(.)

= f f[F(. )U(h)(· )U(k)(. ) dp(·),

whereby (!foU»* = !fo(i)·

Next we verify condition (iii). First observe that since I fn(x) I < I x I and limn-+oofn(x) = x, for any h in DT ,

Ifn(F(· »U(h)(.) - F(· ) U(h) ( .) 12 - 0 (8.31)

and

Ifn(F(. »U(h)(·) - F(· ) U(h) ( .) 12 < 4 I F(· )U(h)(·) 12 , (8.32)

where the right side of equation (8.32) is integrable by (i) of the previous theorem. Hence for any h in DT

f Ifn(F(.»U(h)(.) - F(·)U(h)(·) !2d,u-O. (8.33)

by the Dominated Convergence Theorem. However, for any h in DT ,

II U-I[fn(F(. »]U(h)(·) - Th liB = II U-I[fn(F(. »]U(h)(·) - U-IUTU-IU(h) liB ~ II U-I[fn(F(.»]U(h)(·) - U-IF(·)U(h)(·) liB (b)

= IIfn(F(. »U(h)(·) - F(· )U(h)(·) IIL2(X,I') '

where we have used (ii) of the previous theorem at (a) and the fact that U is unitary at (b). Since the final expression equals the expression in (8.33), rp(fn)h - Th.

The uniqueness of rp must yet be established. The proof is not trivial and we need to do some preliminary work first. This we proceed to do.

Observe that corresponding to each mapping tp from the class of bounded Borel functions on R into L(H, H) that satisfies (i)-(iv) of Theorem 8.9 there is a normalized spectral measure E on B(R) given by E(M) = tp(XM)' In particular, E satisfies the following properties:

(i) E(M) is a projection for each M in B(R),

(ii) E(0) = 0 and E(R) = [, (iii) If M = U:IMi where Mi () M j = 0 for i of::: j, then

n E(M) = lim L E(Mi) (strong sense),

n-+oo i=l

(iv) E(M () N) = E(M)E(N).

The spectral measure which corresponds to rp is denoted by P. By Theorem 8.2, if E is any normalized spectral measure on B(R)

(in particular E could correspond to 'If or be P), then for each bounded complex-valued measurable function f on R, there exists a unique bounded operator Tf such that

(8.34)

where flh is the finite measure flh(M) = (E(M)h 1 h). Tf is denoted by f f dE. Observe that if E is P, then for any bounded Borel function f on R, f f dP = (fU). This is readily verified for characteristic functions XM for M in H(R) since

«(f(XM)h 1 h) = (P(M)h 1 h) = flh(M) = f XM dflh'

From this it follows for simple functions, and from (iv) of the theorem, it follows for nonnegative bounded functions. By additivity it is true for all bounded measurable functions.

If E is any normalized spectral measure on H(R), we can make equation (8.34) be valid for arbitrary measurable functions by the following procedure. If g is any measurable function (I g I finite), then the set

is dense in H. This follows from the fact that, for each h in H,

n h = lim L E(Si)h,

n-1o-oo i=l

where Si = {A. E R: i-I < 1 g(,l) 12 < i} for i = 1,2,3, ... and the fact that for each n, E(Uf-lSi)h is in Dg • Therefore if g is a nonnegative measurable function, defining Tgh for each h in Dg by

Tgh = lim Tgl\n'x[_n,nJ h, n-+oo

where TgAn•x is given by equation (8.34), we define a linear operator [-fl.,n]

Tg on a dense linear subspace of H. By the Monotone Convergence Theorem Tg satisfies for each h in Dg the equation

(8.35)

Now if g is measurable and I g I finite we write g as gl - g2 + i(g3 - g4),

where each gi is nonnegative, measurable, and integrable with respect to P,h for each h in D g. On Dg we define Tg as the linear operator

Tg = Tgl - Tga + i(TgS - T g4)·

By virtue of the fact that each Tg; satisfies equation (8.35), Tg is easily seen to be well-defined and to satisfy equation (8.35).

Equation (8.35) for each h in Dg actually determines Tg uniquely. To see this, we define for each hand k in Dg the sum J g dp,h,k of the integrals as in Definition 8.2. By a straightforward calculation, it is easy to check that

4 f g dp,h,k = f g dp,h+k - f g dp,h-k + i f g dP,h+ik - i f g dp,h-ik = 4(Sh I k) (8.36)

for any operator S satisfying equation (8.35) for all h in Dg • Hence for

any hand k in Dg, (Tgh I k) = (Sh I k) and since Dg = H, Tgh = Sh for any h in H.

Let us summarize this discussion in a lemma.

Lemma 8.7. Let E be a normalized spectral measure on B(R). Corresponding to each complex-valued measurable function g is a unique linear operator Tg in H with domain Dg = {h 1 J':o 1 g 12 dPh < CX>} such that

I

As before, we write Tg = J g dE if Tg is the unique linear operator on Do satisfying equation (8.35).

Observe that if g is a real-valued function, then Tg is a symmetric operator on Dg • Indeed, writing g as gl - g2, where gl and g2 are nonnegative functions, for all hand k in Do,

= lim (TrhAn'x[_n,nlh 1 k) - lim (TY2An'X[_n,nlh 1 k) n~oo n~oo

= lim (h I TOlAn'x[_n,nlk) - lim (h I TY2An'x[_n,nlk) n-+-oo n-+oo

= (h 1 Ty1k) - (h 1 Tgi)

= (h 1 T yk).

If T is a given self-adjoint operator in Hand E is a normalized spectral measure corresponding to Tvia E(M) = V'(XM), where V' is a map from the class of bounded Borel measurable functions R into L(H, H) as in Theorem 8.9, the question arises whether T = J A dE. An affirmative answer is easily obtained from (iii) of Theorem 8.8. To show this we must first establish that DT the domain of T is actually {h I J~oo I AI2 dfth < oo}, which we denote by D)..

First we show DT CD).. Recall that, if In is the function AXI-n,n] , then

Hence if h E D7" then

V'(fn) = f In dE,

[V'(fn)]2 = V'(ln2) = f In2 dE.

(Th I Th) = lim (V'(ln)h I V'(ln)h) n

= lim (V'(ln)*V'(ln)h I h) n

= lim«V'(ln»2h I h)

= lim f A2XI-n,n] dfth n

= f A2 dfth,

which implies J A2 dfth < 00 and hE D).. Now if Tl = J A dE, then Tl is symmetric with domain the set D)..

Also for any h in DT we have by (iii) of Theorem 8.9,

(Th I h) = li~ (V'(fn)h I h) = li~ ( (f In dE)h I h)

= Ii: fAXI-n,n] dfth = f A dfth = (Tlh I h).

Hence T C T1 • This means Tl * C T* = T and since Tl C Tl *, Tl C T. Consequently T = Tl and D7, = D).. More importantly, T = J A dE.

We have proved the following

Lemma 8.8. If T is a self-adjoint operator in Hand E is a normalized spectral measure corresponding to T as in Theorem 8.9 via some map V',

then T = f ).. dE and for every h E DT

II Th 112 = f:oo )..2 dflh ,

with DT = {h I f~oo I ).. 12 dflh < oo}. I In view of the rather lengthy discussion above, it should be clear that

to prove if; is unique it suffices to show that when E is any normalized spectral measure such that

T = f ).. dP = f ).. dE , (8.37)

then P = E. Indeed, if P corresponds to if; and E corresponds to tp, then equation (8.37) is true and if E = P then if; and tp agree on characteristic functions, simple functions, and then all bounded measurable functions. Proof of the uniqueness of if; in this manner will also enable us to quickly prove the final formulation of the spectral theorem-Theorem 8.10 below.

We seek to show then that if E is any normalized spectral measure such that

T = f ).. dP = f ).. dE ,

E = P. First we need some lemmas.

Lemma 8.9. If E is a normalized spectral measure and A is a bounded self-adjoint operator such that

A = f )"dE,

then E is compact. I

Proof. Suppose a(A) C (m, M). Let h be any vector in H and let k = E(-oo, m)h. Then

Hence

E(-oo, )")k = { Ek(,-OO, )")k, if)" < m, if ).. > m.

where flk(M) = (E(M)k I k),

However if k oF 0,

(Ak I k) > inf (Ah I h)(k I k) > m(k I k) , IIAII-1

so that k = O. Hence E(K) = 0 for all K C (-00, m). Similarly E(K) = 0 if K C (M, 00). Hence E is compact. I

Lemma 8.10. Let T be an operator in Hand Q be a projection. If QT C TQ (meaning DT = DQT C DTQ = {h E H: Qh E D T } and QTh = TQh for each h in D T ), then QT = TQ (meaning DQT = D TQ). I

Proof. Since QT C TQ, then QTQ C TQ2 = TQ. However, since DQTQ = D TQ , QTQ = TQ. Inasmuch as (1- Q)T C T(I - Q) also, it

follows in the same way that (I - Q)T(I - Q) = T(I - Q). Moreover

T = QT + (I - Q)T C TQ + T(I - Q) = T ,

since if hE DTQ and hE D TU- Q), then hE DT as h = Qh + (1- Q)h. Since the extremes of this inequality are equal,

QT + (I - Q)T = TQ + T(I - Q).

Applying Q to both sides, one obtains

QT + Q(I - Q)T = QTQ + QT(I - Q)

or QT= TQ + Q(I - Q)T(I - Q) = TQ. I

Lemma 8.11. For any normalized spectral measure E such that

T= f )'dE, (8.38)

ET = TE, that is, E(M)T = TE(M) for all Min B(R). I

Proof. First observe that, for any M, E(M) maps DT into D T • For, if k EDT, then

where f.tB(M)k(N) = (E(N)E(M)k I k) < (E(N)k I k) = f.tk(N) for all N in

B(R). Hence E(M)k EDT by definition of D T . Secondly, for any k in D T ,

we have

(Tk I k) = f )'dp,k

= n~ f ). X[n-l.n) dp,k

= lim (Tok + Tlk + T_lk + ... + Tnk + Lnk I k), (8.39) n-+oo

where Tn = f ).X[n-l.nl dE, a bounded self-adjoint operator on H. Now

so

since k EDT' Hence L Tnk converges in H and from equation (8.39), Tk = L Tnk. Now E(M)Tn = TnE(M) since

E(M)Tn = f XE . ). . X[n-l.n) dE = TnE(M).

Since E(M) is continuous, for k E DT ,

TE(M)k = I TnE(M)k = I E(M)Tnk = E(M)(I Tnk ) = E(M)Tk. ~ ~ n I

Now apply Lemma 8.10.

Lemma 8.12. Let P be the spectral measure corresponding to the selfadjoint operator Tin H via cpo Let A be any operator in L(H, H) such that AT = TA. Then for any integer n, AHn C Hn, where Hn = Pn(H) and Pn = f X[n-l.nl dP. I

Proof. By definition, for any M in B(R) and f E L2(X, p,),

where F, U, and p, are given in Theorem 8.8. Let RM be the range of UP(M)U-l in L 2(X, p,). First let M = [-I, I] CR. Then

F-l(M) = {x E X: -I < F(x) < I}.

Clearly RM = {fE L2(X, p.): f(x) = 0 a.e. if x $ F-l(M)}. It is easy to see that also

RM = {f E ~(p.): II [F(. )]nf(·) liz is bounded for 1 < n < oo}.

[If f(x) = 0 for x $ F-l(M), then

f I [F(x)]nf(x) \Z dp. <.f \ f(x) \2 dp..

Conversely, suppose g E Lz, II Pg 112 is bounded, and g(x) =I=- 0 for x E B for some set B of positive measure on which I F(x) I > 1. Then

f \ F(x)ng(x) pa dp. -- 00 as n -- 00.]

Now suppose f E RM • Then II F(· )nf( .) 112 is bounded and for

fE U(DT ),

II F(· )n(UAU-l)[f(.)] 112 = II (UrnU-l)(UAU-l)f(·) 112

= II uAu-IUrnU-lf(·) 112 < II A IIII F(· )nf(·) 112'

which is bounded. Hence UAU-1(RM) C RM when M = [-I, 1] If M = [a, b] = {A E R: \ A - A.o I <r}, then

F-l(M) = {x EX: \ F~ Ao (x) \ < I}. Since A commutes with T, UAU-l will also commute with multiplication by (F - Ao)/r; again in this case RM will be invariant under UAU-l. If now M = [n - 1, n), then M = U;'lMk when (Mk ) is an increasing sequence of closed intervals. Since RM = {f E ~(X, p.): f(x) = 0 a.e. if x $ F-l(M)} clearly RMl C RMI C ... C RM and by the Dominated Convergence Theo-

r~m, [U;'lRMJ = RM • For any k and fin RMk , UAU-1jE RMk C RM. Smce RM is closed, UAU-l(RM) C RM and RM is invariant under UAU-l. This means

or

I For each integer n, let us continue to let Pn = f X[n-l,nJ dP, that is


Pn = P([n - I, n»), and let Hn = Pn(H). Since

I=P(R) = lim [PO +P1 +P-1 + ... +Pn+P-n ] n-+oo

in the strong sense, for each h in H, h = LneZhn, where hn = Pn(h). This means H = EB Hn, the direct sum of the orthogonal family of subspaces

Hn· As shown above, if E is a normalized spectral measure satisfying

equation (8.38), then TE = ET. From Lemma 8.12, E(M)(Hn ) C Hn for all ME B(R) and each n. Hence each Hn is invariant under P(M) and E(M), ME B(R). Thereby we can define En and Pn by

EiM) = E(M) IH and Pn(M) = P(M) IH' n n

{Note that for each M in B(R), Pn(M) = P([n - I, n) n M).} Since

T= f A. dE,

we have for h E Hn n DT

where vh(M) = (E(M)h I h) = (En(M)h I h), so that T1Hn = f A. dEn. Similarly TIH .. = f A. dPn.

Since Pn is a compact normalized spectral measure, the equality T I H .. = f A. dP n also tells us that T I H.. is a bounded self-adjoint operator defined on Hn. This follows from the one-to-one correspondence established between compact spectral measures and bounded self-adjoint operators established in Theorem 8.4. Since

Pn and En are both compact normalized spectral measures (Lemma 8.9) corresponding to T/Hn' As there can be only one such measure,

E IH = En = Pn = P IH . .. ..

Since H = EB Hn and for each M, E(M) and P(M) are bounded operators,

Sec. 8.2 • Unbounded Operators 24S

if h = LieZhi,

E(M)h = !~ E(M) ( i~n hi)

n n = lim L E(M)hi = lim L 2i(M)hi

n ...... oo i=-n n-+oo i .... -n n

= lim L fti(M)Fii = P(M)h. n-+oo i ... -n

Hence E=P. We have at last completed the proof of Theorem S.9. I

Our next result summarizes our preceding discussion and proof of the uniqueness of ~ in a nutshell.

Theorem 8.10. Spectral Measure Version 01 the Spectral Theorem. There is a one-to-one correspondence between self-adjoint operators T in H and normalized spectral measures P on B(R). The correspondence is given by T = f A dP. Moreover for each real-valued measurable function I [with respect to B(R)] there is a unique self-adjoint operator f(T) given by

I(T) = f IdP,

where T = f A dP. Iff is bounded, f(T) = ~(f), where ~ is given by Theorem S.9. I

The remaining portion of this chapter is devoted to giving some applications of the Spectral Theorem.

Knowledge of the spectral measure corresponding to an operator T can be most valuable in obtaining complete knowledge of the operator T in regard to determining its domain, the value of inner product (Th I k) for hand k in H, the spectrum and eigenspaces of T, and operator functions f(T) of T for f measurable. In Theorem S.II below we show how the spectrum of T is related to the spectral measure corresponding to T.

A few crucial observations should be made regarding the Spectral Theorem. First, for any measurable (real or complex valued) function I on R, the unique operator I(T) satisfying equation (S.35) is defined. Its domain is the set {h I f~oo 1/12 dPh < oo}, dense in H. Secondly, for any h in D/(T) ,

(S.40)

Equation (S.40) is easily verified for the case when f is a nonnegative measurable function. From the discussion preceding equation (S.35), we have

IlJtT)h 112 = lim Ilf 1\ n • X[-n,nl(T)h 112 n-+oo = lim ([f 1\ n . X[_n,nl(T»)2h I h) n-+oo

= ~~oo f~oo [f 1\ n . X[-n,nl().»)2 df-th().)

= f~oo f2 df-th'

In case f is any measurable function, 1 = fl - f2 + i(ga - g4) and 1112 = f12 + j;2 + ga2 + g42. In this situation equation (S.40) can easily be seen to hold.

Recall from Problem 8.1.2 that there is a one-to-one correspondence between spectral functions on R and normalized spectral measures on B(R). It is analogous to the correspondence between Borel measures on R and distribution functions. Given a spectral measure P on B(R), the spectral function Eon R is given by E(x) = P( -00, x). Theorem 8.10 thus implies a one-to-one correspondence between self-adjoint operators T in Hand spectral functions Eon R. Iff is a measurable function, then for any h in H

f f(A) df-th(A) = f f(A) d(E(A)h I h),

which accords with Definition 3.10 in Chapter 3. The applications given below are more easily stated in terms of spectral

functions than spectral measures. Using equation (8.40), the foUowing theorem relating the spectrum

of a self-adjoint operator T to properties of its spectral function is obtained.

Theorem 8.11. Ler T be a self-adjoint operator in H and let E be the spectral function on R corresponding to T. Then we have:

(i) the spectrum aCT) is a subset of the real numbers;

(ii) ).0 is an eigenvalue of T if and only if

E(AO) *- E(Ao-) == lim E().); A<AO A-+AO

moreover the eigenspace of H corresponding to ).0 is REUo>-E<Jlo->;

(iii) Ao is in the continuous spectrum if and only if E(Ao) = E(Ao-), but E(A1) < E(A2) whenever Al < Ao < A2; and

(iv) the residual spectrum of T is empty. I

Proof. (i) has been proved in Proposition 8.12. Clearly

and by (8.40) for h in DT

II (T - AoI)h 112 = f 1 A - Ao 12 d(E(A)h I h). (8.41)

Hence Th = Aoh if and only if E(Ao+) = E(Ao) = E(A) for all A > Ao and E(Ao-) = E(A) for all A < Ao. [Recall E(A)h is right continuous in A for each h, E(A)h ~ 0 as A ~ -00, and E(A)h ~ h as A ~OO.] In other words, Th = Aoh if and only if h = [E(Ao) - E(Ao- )]h. This verifies statement (ii).

We next verify (iv). If the residual spectrum is not empty, then there is a real number Ao for which (T - AoI)-1 exists but its domain RT- AoI is not dense. This means that there is a nonzero ho in H that is orthogonal to RT-}.;JI; that is, (T - AoI)h I ho) = 0 for all h in DT . Hence

(Th I ho) = (Aoh I ho) = (h I Aoho) ,

so that Aoho is in DTo and T*ho = Aoho. Since T is self-adjoint, Tho = Aoho and Ao is an eigenvalue of T. 'Since the point spectrum and residual spectrum are disjoint, this is a contradiction.

It remains to show (iii). From (ii) and (iv), if E(Ao) = E(Ao-) then Ao is either in the resolvent of T or in the continuous spectrum. Now Ao is in the resolvent if and only if there exists a positive constant k such that

II (T - AoI)h II >k II h II, for all h in DT •

In other words from equation (8.41) it is necessary and sufficient that

(8.42)

Now if there exist Al and A2 with Al < Ao < A2 such that Ao - Al = A2 - Ao < k and E(A1) =F E(A2)' then

f:oo (A - Ao)2 d(E(A)h I h) < k 2 f:oo d(E(A)h I h) = k2 II h 1!2,

with h = [£(A2) - £(A1)]X for x in H. Since this contradicts inequality (8.42) if E(A'I) 0:/=- E(A2), Ao is in the continuous spectrum of T. Conversely, if Ao is in the continuous spectrum, £(Ao-) = £(Ao) by (ii). Moreover if there exists Al and A2 with Al < Ao < A2 and £(Al) = E(A2) {implying E(A) is constant on [AI' A2]}, then the function f(A) = I/(A - Ao) is bounded almost everywhere andf(T) = (T - AoI)-1 by Theorem 8.9 is the bounded inverse defined on H of T - Aol. This means Ao is in the resolvent of T, a contradiction. Hence for all Al and A2 with Al < A < A2, E(A1) 0:/=- E(A2)· I

The following example illustrates the use of the preceding theorem.

Example 8.5. Let T be the multiplication operator in L2( - 00, 00) considered in Example 8.1. It is routine to show T is symmetric. To show T is self-adjoint it must be shown that DT* C DT . Suppose g E DT*. Then for every f E DT

J:oo xf(x)g(x) dx = (Tfl g) = (fl T*g) = J:oo f(x)T*g(x) dx ,

whence

J:oo f(x)[g(x)x - T*g(x)] dx = O.

Let [0, b) be a finite interval and define h by hex) = [xg(x) - T*g(x)]X[a,bl' Then J~00[h(x)]2 dx = 0 so that hex) = 0 almost everywhere. Since [0, b] is an arbitrary interval xg(x) = T*g(x) almost everywhere or xg(x) = T*g(x) E L2(-00, 00). Hence g EDT'

Let E be the function on R into L (L2( - 00, 00), L2( - 00, 00» given by

E(A)g = g X(-oo,Al' E is easily checked to be a projection-valued, nondecreasing, right-continuous (in the strong sense) function with E(A) -+ 0 as A -+ -00 and E(A) -+ I as A -+ 00. Moreover, for any f in DT and g in L2( - 00, 00),

(Tf I g) = f:oo xf(x)g(x) dx

= f:oo xd [00 f(t)g(t) dt

= J:oo xd f:oo g(t)E(x)[f(t)] dt

= J:oo x d(E(x)fl g).

This means E is the unique spectral function such that T = J A dE. What is the spectrum of T? Note that for all A, E(A) = E(A-) and

E(A1) < E(A2) if Al < A2 • Hence the continuous spectrum makes up the entire set of real numbers.

It is known that the position operator T of the preceding example is unitarily equivalent to the so-called momentum operator, the operator Ts of Example 8.2. This means there exists a unitary operator F on L 2( - 00,00) such that

F(DTa) = DT and Ts = F-1TF.

The operator F is known as the Fourier-Plancherel operator. The proof of this unitary equivalence is not trivial and not given here. t However, the spectrum of Ts can be easily analyzed to be exactly that of T by means of the following theorem.

Theorem 8.12. If Sand T are unitarily equivalent operators in a Hilbert space H, then the point spectrum, continuous spectrum, and residual spectrum of S are the same as that of T. I

The proof is trivial and is omitted.

Example 8.6. Application of the Spectral Theorem in Solving the SchriJdinger Equation. An equation that occurs in quantum mechanics is the time-dependent SchrOdinger equation given by

i du = Au(t) dt '

where u(t) is an element of a Hilbert space H, A is a self-adjoint operator in H, and t is a time variable with u(t) E DA • An initial condition is u(O) = Uo E DA • The derivative of u is given as the

lim u(t + L1t) - u(t) L1t

in the strong topology of H. The Spectral Theorem enables us to solve the SchrOdinger equation.

Let e-itA be the bounded operator on H given by

t The proof is given in [16], page 135.


where A = J ). dP. We wish to show that

for every h in DA •

~ (e-itAh) = -iA(e-itAh) , dt

To prove this, compute the following limit:

11m + Ie-itA A h . II [ e-i(t+At)A - e-itA .' ] 112 .1t ..... o L1t

foo I e-i(t+.1tlA - e-itA 12 = lim ,1 + ie-itA). d(E()')h 1 h) .1t ..... o -00 t

foo I e-i!LltlA - 1 12 = lim ,1 + i). d(E()')h 1 h) .

.1t ..... o -00 t

(8.43)

Letting M = max of I [e-i(Llt) - l]/L1t + i 12 for L1t E R, the integrand above is bounded by M).2, which is integrable since hE D A' Using the Lebesgue Dominated Convergence Theorem, the limit can be taken inside to the integrand. Since the limit of the integrand is zero, the above limit is zero.

Hence, for every h in D A,

(8.44)

Equation (8.43) follows from (A.44) since, for h in D A,

(8.45)

This follows from the fact that if h is in D A, then e-itAh is in D A since by equation (8.40)

II E(M)e-itAh 112 = f XM 1 e-itA 12 dEh().)

= f XM dEh().) = II E(M)h 112.

Equation (8.45) then follows by an argument similar to that after Lemma 8.4.

The solution u(t) = e-itAuo of the Schrodinger equation is unique. To prove this, suppose v(t) in DA is a solution. Then for any k in H

d. . (ri[t-<S+MI]AV(S + Lfs) I k) - (e-i<t-SIAV(S) I k) - (ro<t-sIAV(S) I k) = lIm -O.-______ ---:;----'_--C... ____ --'-

ds .<Is~O Lfs

( e-i[t-<s+.<Isl)A _ e-i<t-sIA I ) = lim A v(s + Lfs) k

.<I8~O LJS

+ lim (e-i<t-SIA v(s + Lfs) - v(s) I k) .<Is~O Lfs

= (- ~ e-i<t-sIAv(s) I k) + (ri<t-SIA ~~ I k)

= (iri<t-sIAAv(s) I k) + (ri<t-SlA[_ iAv(s)] I k) = O.

Hence, for all k in H,

0= J: ~ (e-i<t-sIAv(s) I k)ds = (riOAv(t) I k) - (e-itAv(O) I k),

and since v(O) = Uo and e-iOA = I, we have

Uniqueness is thus proved.

Problems

8.2.1. Let H be an infinite-dimensional Hilbert space with orthonormal basis (Xitv. Define T: D1' C H -+ H by

where

(i) Show that T is a closed linear operator but T is not bounded on D 1,.

(ii) Prove that T is self-adjoint.

8.2.2. Prove that a closed and unbounded linear operator T in a Hilbert space H cannot have D1' = H.

8.2.3. Prove that every self-adjoint operator T is a maximal sym-


metric operator; that is, there is no symmetric operator extending T with a larger domain.

8.2.4. Let T and S be operators in H with domains DT and Ds , respectively. Suppose U is a unitary operator on H such that U(Ds) = DT and S = U-ITU. (S is said to be unitarily equivalent to T.) Prove the following:

(i) T is bounded if and only if S is bounded. (ii) T is symmetric if and only if S is symmetric.

(iii) T is self-adjoint if and only if S is self-adjoint.

8.2.5. Let M = (mij) be a matrix with rows and columns in MN). Let T be the operator in 12(N) represented by M in the sense that Tf =

Li:l (L;l mij(f I ej) )ei for all f in the domain off = {IE 12(N) I Tf E MN)} [Here ei(j) = (Jij'] Show that

(i) T is a closed operator with dense domain; (ii) if N = (nij), where nij = iiiji and S is the operator represented

by N with domain the set {IE 12(N) I SfE 12}, then S extends T; (iii) S is not necessarily the adjoint of T.

8.2.6. Let T be a symmetric operator in a Hilbert space H. Prove the following statements are equivalent:

(i) T is self-adjoint. (ii) T is closed and the null space of T ± i is {O}.

(iii) R(T ± i) = H. [Hint: Ker (T* ± il) = [Range (T ± il))1-.]

8.2.7. An operator in H is said to be essentially self-adjoint if its closure is self-adjoint. If T is a symmetric operator in H, prove the following are equivalent:

(i) T is essentially self-adjoint; (ii) T* is self-adjoint;

(iii) l' = T*; (iv) Ker(T* ± i) = {O};

(v) R(T± i) = H.

8.2.8. Let M be a closed subspace of H and let P be the projection operator in H onto M. Prove M and M completely reduce T [that is, T(DT n M) C M, T(DT n Ml.) C Ml., and P(DT) eDT] if and only if P(DT) C DT and PT C TP. If DT = H, then T is completely reduced if and only if PT = TP.

8.2.9. If T is a symmetric operator in H and its spectrum consists only of real values, then prove T is either self-adjoint or essentially selfadjoint.


8.2.10. Let T be a symmetric operator in T. Prove the following: (i) All points in the point spectrum and the continuous spectrum

are real whereas the residual spectrum may contain nonreal values.

(ii) If nonreal A = a + bi is in the residual spectrum so is c + di, where d has the same sign as b.

8.2.11. Let T be a symmetric operator in H. Prove the following: (i) the eigenvectors corresponding to distinct eigenvalues are or

thogonal; (ii) the closure All. of MA = {x E H: Tx = Ax} completely reduces T. 8.2.12. Prove that if T is a self-adjoint operator in Hand P is the

normalized spectral measure corresponding to T via T = J A dP, then for each ME B(R), P(M)H and (/ - P)(M)H completely reduce T.

8.2.13. Let T be a closed linear operator in Hilbert space H with D1' = H (e.g., a self-adjoint operator in H). Prove the following:

(i) A E e(T) if and only if X E e(T*). (ii) If A E Ca(T) then X E Ca(T*).

(iii) If A E Ra(T) then X E Pa(T*). (iv) If A E PI1(T) then X E Pa(T*) U Ra(T*).

8.2.14. Let T be an unbounded self-adjoint operator in H. Prove the following:

(i) A E e(T) if R(T - AI) = H. (ii) A E CI1(T) if R(T - AI) * H but R(T - AI) = H.

(iii) A E PaCT) if R(T - AI) eft H. (iv) Ra(T) = 0.

Appendix

C. Invariant and Hyperinvariant Subspaces

One of the best-known unsolved problems in functional analysis is the invariant subspace problem: does every bounded linear operator on a separable, infinite-dimensional, complex Hilbert spacet have a nontrivial closed invariant subspace? The subject of invariant subspaces is very broad and many special cases of this problem have been solved. In this appendix we will restrict ourselves to two main theorems-one for compact operators due to Lomonosov and one for normal operators which is a natural application of the spectral theorems for normal operators. Both of these theorems not only give affirmative partial answers to the invariant subspace problem but also give somewhat stronger conclusions. The reader who wishes to delve more deeply into this problem may consult some of the references listed at the close of the appendix.

The following definition makes precise the concepts with which we deal in this section. From this point H will represent a complex Hilbert space and T a bounded linear operator on H. T is termed nonscalar if it is not a scalar multiple of the identity operator.

Definition C.I. A closed linear subspace M of H is said to be invariant under the operator T if T(M) C M. M is reducing for T if T(M) C M and T(Ml.) C Ml.. M is nontrivial if M * {O} and M * H. I

Note that when we speak of an invariant or reducing subspace we assume the subspace is closed.

t Recently C. J. Read (Bull. Math. Soc. London 16, 337-401 (1984» has exhibited a bounded linear operator, on a separable infinite-dimensional Banach space, which has no nontrivial closed invariant subspace.

255

256 Appendix

The following remarks are easy to verify.

Remark C.I. (i) M is invariant for T if and only if M1. is invariant for T*.

(ii) M is reducing for T if and only if M is invariant under both T and T*.

The basic motivation for the study of invariant subspaces stems from interest in knowing the structure of operators. If M is invariant under T, then T can be written with respect to the decomposition M EB M1. as an operator matrix

( Tn T I2 ),

o T22 where Tij = PiTPj for t, j = 1,2; PI is the projection onto M; and P2 is the projection onto M1.. Tl2 will be the zero operator if and only if in addition T(M 1.) C M 1., in which case T can be represented as a "diagonal" matrix of operators. Further knowledge of T may result if Tn and Tl2 can be similarly further reduced or if properties of T can be deduced from knowledge of the operators Tij , such as knowledge of the spectrum.

If dim H < 00, H can be decomposed into a direct sum of invariant subspaces on each of which T acts in a simple manner (Jordan canonical formt). Likewise, a self-adjoint operator can be written as a direct sum of restrictions to invariant subspaces by means of the spectral theorem (see Problem 8.2.12). Interest in the invariant subspace problem stems partly from a desire to so decompose general operators on infinite-dimensional spaces.

Before we get into the main theorems of this section, let us explain the reason for some of the restrictions-infinite dimensional, complex, separable, closed, complete, and bounded-listed with the invariant subspace problem. We do this with the following remarks and three subsequent propositions.

Remarks C.2. Nonzero linear operators (necessarily continuous) on finite

dimensional (dim> I) complex vector spaces always have nontrivial invariant subspaces-namely, the eigenspace corresponding to a nonzero eigenvalue, whose existence is guarenteed by the Fundamental Theorem of Algebra. This method of generating invariant subspaces does not work in the infinite-dimensional situation, however, since some operators have no

t See, for example, K. Hoffman, and R. Kunze, Linear Algebra, Prentice-Hall, Englewood Cliffs, New Jersey (1971).

C. Invariant and Hyperinvariant Subspaces 257

nonzero eigenvalues. One may consider for example the shift operator T on 12( (Xl' X2, ... ) ---+ (0, Xl, X2, ... » or the operator V on L 2[0, I] given by (Vf)(x) = f~f(t) dt.

C.3. Bounded linear operators on real Hilbert spaces may not have any nontrivial invariant subs paces as a rotation in the plane exemplifies.

C.4. A nonzero linear operator A (bounded or unbounded) on a nonseparable normed linear space X always has nontrivial (closed) invariant subspaces, namely, the closed linear span of {Anx I n = 0, 1, 2, ... } for some x*-O in X, which is separable and, therefore, proper.

The next result is due to Schaefer [35].

Proposition c.t. Nonzero linear operators on real or complex infinite-dimensional vector spaces always have proper (not necessarily closed) invariant subspaces. I

Proof. Let A be a nonzero linear operator on vector space V. It suffices to prove that an infinite-dimensional subspace W with basis {xo, Axo, A2xo, ... } has a proper invariant subspace. Define the linear function Lon Wby

L( to aiAixo) = to ai'

Let S be the subspace {w E WI L(w) = O}. Then {O} ~ S ~ S is invariant under A since L(Aw) = Lw for WE S.

Ws V and

I In contrast, the following result is true if we insist on closed subspaces.

Proposition C.2. A bounded linear operator on a complex pre-Hilbert space may not have a nontrivial (closed) invariant subspace. I

Proof. Let V be the pre-Hilbert space of all polynomials on [0, 1] with complex coefficients and with inner product (p I q) = n pi] dt. Let M: V ---+ V be given by (Mp)(t) = tp(t). Suppose S is a closed subspace of V which is invariant under M and has a nonzero element p. Let e > ° and let q E V. If K is the closed finite set {x E [0, 1] I p(x) = O}, let U be an open set containing K with measure less than e. By Urysohn's Lemma (Lemma 1.1) there exists a continuous function f such that I f I < I,J =- 1 on K, and f = ° on [0, 1] - U. Applying the Stone-Weierstrass Theorem to CI([O, 1] - U) there exists a polynomial q. such that

II q - ~ -: (>q. 1100 < II q - ;P+ ~.f)q. 1100 = II p ! f - q. IL < n ~ 1 '

258 Appendix

where N > II p 1100 and II 1100 is the sup norm over [0, 1 J. This means

II q - pq. 112 < II q - (p + f)q. 112 + II fq. 112

< [f u I q - (p + f)q. 12 dm f'2

+ [f UC I q - (p + f)q.12 dm f'2 + (f u I fq.12 dmY'2

< II q - (p + f)q. lloom(U) + 8 + II fq. 1l00m(U)

< 8(11 q - (p + f)q. 1100 + II fq. 1100 + 1) .

Consequently q E S since S is closed and pq. E S as S is invariant. This means S = V. I

The next proposition is due to Shields [37].

Proposition C.3. There is a linear transformation (not necessarily bounded) on a Hilbert space which has no nontrivial (closed) invariant subspaces. I

Proof. Let H be a (separable) infinite-dimensional Hilbert space. H then has c proper closed infinite-dimensional subspaces. Letting We be the first ordinal number with cardinality c, there exists a one-to-one correspondence a ~ M" between predecessors a of We and the proper closed infinite-dimensional subspaces of H.

For each a < We' let S" be the statement: for each y < a, there exists a pair U;" gy) with /.y E My and gy f/; My and {/.y, gy I y < a} is linearly independent. Clearly SI is valid. Let us assume S{J is valid for all f1 < a. Let V" = span Up, g{J I f1 < a}. Since the algebraic dimension of V" is less than c and the algebraic dimension of M" is not less than c (see Problem 6.1. 7), there is a vector f" E M" - V". The vector subspace V" is not all of H and therefore contains no nonempty open set. Consequently M" U V", ~ H and there exists a vector g" E H - (M" U V,,). Consequently S" is true. By transfinite induction S" is true for all a < We so that there is a linearly independent set .'7 = U;', g" I a < we} such that f" EM" and g" f/; M" for each a.

Extend .'7 to an algebraic basis .'7 U ~ of H. If T is defined on .'7 by setting T(fr.) = g" and T(g,,) = 1a.+1' then no matter how T is defined on ~, T will have no infinite-dimensional (closed) invariant subspaces because Tia. = g" f/; M" for all a < We' If ~ is finite, say {hI, h2' ... ,hn }, define


Thi = hi+! for i = 1, 2, ... , n - 1 and Thn = II. If Wi is infinite, Wi can be well ordered and each element of Wi can be mapped by T to its successor. In either case when T is extended by linearity to H, T has no finitedimensional invariant subs paces either since every linear combination of elements in .9" u Wi is mapped onto a linear combination involving at least one new basis element. I

Even though the invariant subspace question is still unanswered, the analogous question involving reducing subspaces is settled. The following example gives an example of a bounded linear operator on a separable Hilbert space that has invariant subspaces (see [30, p. 45]) but no reducing subspaces.

Example C.I. The unilateral shift T on 12 has no nontrivial reducing subspaces. T is given by T(XI, X2, ... ) = (0, Xl, X2' ... ) or by the rule T(L::IXiei) = L::IXiei+!' (The jth component of ei is bij .) Let xE 12 ,

X * 0, X = L:IXiei' Let no be the smallest i such that Xi * 0. Let y = Tno+I[(T*)no+1(x»). (Recall T* = S where S(XI' X2, ... ) = (xz, Xa, .•. ) by Problem 7.4.1.) Thus Yk = Xk for k * no but YnO = 0. Let M be the smallest subspace containing X that is invariant for T and T*. Then Y EM and so X - Y = (0,0,0, ... ,0, xno ' 0, 0, ... ) E M. By applying T and T* repeatedly to eno ' M is seen to contain all the basis vectors en for n > 1. ~~M=U I

The following proposition gives some essential facts for the study of invariant and reducing subspaces.

Proposition CA. Suppose T E L(H, H) and P is the projection onto the closed subspace M of H. Then (i) M is invariant under T if and only if TP=PTP.

(ii) M is reducing for T if and only if TP = PT. I

Proof. (i) If M is invariant under T and hEM then TPh E T(M) C M and so PTPh = TPh. Conversely, if TP = PTP and if hEM, then Ph = h so that Th = P(Th). Consequently, Th E M and M is invariant under T.

(ii) By definition, M is reducing for T if and only if M and Ml. are invariant under T. By (i) M is reducing if and only if

TP = PTP and T(I - P) = (I - P)T(I - P),

260 Appendix

which is equivalent to

TP = PTP and 0 = -PT + PTP,

which is equivalent to TP = PT. I

A concept related to the notion of invariance is the concept of hyperinvariance.

Definition C.2. A closed subspace M of H is said to be hyperinvariant under the operator T if M is invariant under every operator S for which ~=~ I

The question of whether every bounded linear operator on a Hilbert space has a hyperinvariant subspace is also unsolved. This question seems to be as difficult for nonseparable spaces as for separable.

The next theorem shows that not only do nonzero compact operators have invariant subspaces (a result discovered independently by J. von Neumann (unpublished) and N. Aronszajn and K. T. Smith [2]) but also hyperinvariant subspaces. This remarkable result was proved by V. Lomonosov [21] by using Schauder's fixed point theorem to produce invariant subspaces. The proof presented here of Lomonosov's result is due to H. M. Hilden and does not require any fixed point theorem. It can also be found in [23]. I

Theorem C.I (Lomonosov).+ Every nonzero compact operator K on a complex Banach space X has a nontrivial hyperinvariant subspace.

Proof. We can assume a(K) = {O}. Otherwise X is finite dimensional [if 0 ft a(K)] or K has a nonzero eigenvalue (see Theorem 6.25). In either case K has a hyperinvariant subspace, namely, {x I Kx = AX} for some nonzero eigenvalue A. Therefore assuming a(K) = {O}, the spectral radius formula (Proposition 6.27) implies that II (cK)n 11- 0 as n - 00 for every complex number c.

We can also assume that II K II = 1. Let Xo E X with II Kxo II > 1. Let Sl(XO) = {x I II x - Xo II < I}. Since 1 < II Kxo II < II Xo II, 0 E Sl and

o ft K(Sl)' a compact set.

+ The result stated here is actually a special case of a more general result: If T and K are commuting operators on X with K compact and nonzero and T nonscalar, then T has a hyperinvariant subspace.

c. Invariant and Hyperinvariant Subspaces

For each y E X define My to be the linear subspace given by

My = {A(u); A E L(X, X) and AK = KA}.

261

My is invariant under all operators which commute with K. Hence its closure is a hyperinvariant subspace under K. Since My"# {O} unless y = 0, it remains only to show that for some y "# 0, My"# H.

Suppose on the contrary that for each y "# 0, My is dense in X. Then for each y "# 0, there is an operator A such that AK = AK and" Ay - XO II < l. Let OA = {y; II Ay - XO II < 1}. OA is open since if y E OA with A"# 0 and II x - y II < (I - II Ay - Xo 11)/11 A II then II Ax - XO II < II Ax - Ay II + II Ay - XO II < I. Consequently

K(SI) C U 0A, AEL(X,X)

and by compactness there is a finite set {AI> A 2 , ••• , An} of operators that commute with K such that

n K(SI) C U OAt'

i=1

Since K(xo) E K( SI (xo) ), K(xo) E 0 Ail for some i1 • This means Ai1(K(xo») E SI(XO). Then KAilK(xo) E K(SI(XO»). This means for some i2 , KAi K(xo) E OAi or Ai KAi K(xo) E SI(XO)' Continuing in this manner

1 2 2 1

m times, we get

for some i1 ,i2 , ••• ,im from {1,2, ... ,n}. Let c= max {II Ai II: i= 1, ... , n}. Since each Ai commutes with K,

However II c-1A ij II < I and II (cK)m I! ~ 0 as m ~ 00, so that the sequence

(c-lAim)(c-lAim_) ... (c-1Ai)(cK)m(xo»mEN

converges to O. But 0 (/: SI(XO)' a closed set. This contradiction means My is not dense in X for some y "# 0 and My is a nontrivial invariant subspace under K. I

Our final goal is to prove that every bounded normal operator on a Hilbert space H has a nontrivial invariant subspace. Actually this is a consequence of the Multiplication Operator Form of the Spectral Theorem

262 Appendix

for Normal Operators (Corollary 8.2) since subspaces of L2 functions that vanish on a fixed set of positive measure are invariant for multiplication operators on L 2 • We will, however, prove much more-that every nonscalar bounded normal operator has a nontrivial hyperinvariant subspace. To prove this we will use the Spectral Measure Version of the Spectral Theorem for Normal Operators (Theorem 8.7).

First we prove a preliminary proposition and lemma.

Proposition C.S. Suppose T is a bounded normal operator on the separable Hilbert space Hand T = fAXa(T)O') dP as in Theorem 8.7 (that is, (Th I h) = fAXa(TI dflh for each hE H, where flh(M) = (P(M)h I h) for all M in B(R2) and P is a normalized compact spectral measure on B(R2». Then P(M)H is reducing for T and a(T1P(MIH) C Ai for each M in B(R2). I

Proof. P(M) = f XM dP for each M in B(R2) by Remark 8.2(iv). Therefore by Remark 8.2(v), P(M) commutes with T and consequently P(M)H is reducing for T by Proposition C.4.

To prove a(T1P(MIH) C Ai, by Corollary 8.2 we can assume that T is a linear operator on L 2(X, fl) for some X and fl and that T(J(x») = G(x)f(x) for some bounded complex function G on X. Then T1P(MIH(f) = GXG-l(MI/(see proof of Theorem 8.7) so that a(T1P(MIH) is the essential range of G restricted to G-l(M) (see Example 6.17), which is contained in Ai. I

Lemma C.I. If P is a spectral measure on B(R2), the Borel subsets of R2, then for each M in B(R2), for each e > 0, and for each h in H, there exists a compact set K C M such that II (P(M) - P(K»h II < e. I

Proof. The measure on B(R2) given by flk(M) = (P(M)h I h) is regular by Theorem 5.2. Therefore there exists a compact set K C M such that 0 < fliM) - flh(K) < e. Therefore

II (P(M) - P(K))h II = ([P(M) - P(K)]h I h) < e. I

The following theorem will enable us to prove that every nonscalar normal operator has a nontrivial hyperinvariant subspace. It was proved by B. Fugledet in 1950.

t A commutativity theorem for normal operators, Proc. Nat. Acad. Sci. U.S.A. 36, 35-40 (1950).

Theorem C.2 (Fuglede). If T is a bounded normal operator on the separable Hilbert space H, T = J AXu(Tl dP, and S is any bounded linear operator that commutes with T, then S commutes with P(M) for every Borel set M [equivalently, P(M)H is reducing for S for every Borel set M].

Proof. To show P(M)H is reducing for S, it suffices to show that both P(M)H and P(MC)H are invariant under S. For this it suffices to prove P(M)H is invariant under S or equivalently that P(MC)SP(M) = O. Using the preceding, it suffices to prove that for compact subsets KI and K2 of MC and M, respectively, P(K1)SP(K2) = O.

Let KI and K2 be two such compact sets. Since ST = TS and P(M)H is reducing for T for each M in B(R2), we have

[P(K1)TP(K1)] [P(K1)SP(K2)] = P(K1)TP(K1)SP(K2)

= P(K1)T[I - P(K1C)]SP(K2)

= P(K1)TSP(K2) - P(K1)TP(K{)SP(K2)

= P(K1)TSP(K2) = P(K1)STP(K2)

= [P(K1)SP(K2)][P(K2)TP(K2)].

Since a(P(K1)SP(K1») n a (P(K2)SP(K2) ) C KI n K2 = 0 by Proposition C.S, we have by Problem 7.4.32 that P(K})SP(K2) = o. I

Corollary C.l. Every nonscalar normal operator has a nontrivial hyperinvariant subspace. I

Proof. If T is a non scalar normal operator then aCT) contains at least two distinct points PI and P2' [Otherwise, if we write T as T(f) = Gf as in Corollary 8.2, and the essential range of G (the spectrum of T) contains only one point, then T is a multiple of the identity.] Let P be the compact normalized spectral measure corresponding to T and let

D = {c E R2: I c - Pl I < il PI - P2/}'

Then a(TIP(D» and a(TIP(DC» are proper subsets of aCT) by Proposition C.S. Since P(D) commutes with every operator that commutes with T, P(D)H is hyperinvariant under T. I

The following corollary is actually true for any bounded linear operator T on the Hilbert space H (see [30, p. 32]). The proof, however, depends on developing a functional calculus for general operators as T. Nevertheless, with Fuglede's Theorem we can prove the result for normal operators.

264 Appendix

Corollary C.2. If T is a bounded normal operator with disconnected spectrum, then T has a complementary pair of nontrivial hyperinvariant subspaces. I

Proof. If a(T) is disconnected, then since a(T) is closed there exist disjoint nonempty closed subsets CI and C2 such that a(T) = CI U C2 •

Letting P be the normalized compact spectral measure corresponding to Twe have P([a(T)cJ) = 0 (see proof of Theorem 8.7), I = P(a(T)) = P(CI) + P(C2), and H = P(CI)H EB P(C2)H. Using Proposition C.S, P(CI)H and P(C2)H are reducing under T and a(T,p(Oj)H) C Ci for i = 1,2. Consequently, neither P(CI)H nor P(C2)H is a trivial subspace, and accordingly thus are a nontrivial complementary pair of hyperinvariant subspaces ~~ I

There are many other results giving partial answers to the invariant subspace problem. For example, in [7] the following result is proved:

If T E L(H, H) is subnormal (that is, there exists a Hilbert space HI:) H and a normal operator S E L(HI' HI) such that T(x) = S(x) for all x E H) then T has an invariant subspace.

Likewise, in reference [38] the following result appears:

If T E L(H, H) is hyponormal (that is, (T*T - TT* > 0) and if aD C a(T) C D, where D is the open unit disk in R2, then T has an invariant subspace.

In regard to generalizing Lomonosov's Theorem, Kim, Pearcy, and Shields [18] proved the following result:

If T is a nonscalar operator on a Banach space and if dim(TK - KT)(X) < I for some nonzero compact operator K, then T has a hyperinvariant subspace.

The interested reader who wished to pursue further study into the subspace problem should begin by consulting references [27], [28], [30], [38], [34], and other references cited in these works.

Bibliography

1. Akhiezer, N. I., and Glazman, I. M., Theory of Linear Operators in Hilbert Space, Frederick Ungar, New York (1961, 1963).

2. Aronszajn, N., and Smith, K. T., Invariant subspaces of completely continuous operators, Ann. Math. 60, 345-350 (1954).

3. Bachman, G., and Narici, L., Functional Analysis, Academic, New York (1964). 4. Banach, S., Theorie des Operations Lineaires, Monografje Matematyczne, Warsaw

(1932). 5. Berberian, S. K., Introduction to Hilbert Space, Oxford University Press, New York

(1961). 6. Berberian, S. K., Notes on Spectral Theory, Van Nostrand, Princeton, New Jersey

(1966). 7. Brown, A. L., and Page, A., Elements of Functional Analysis, Van Nostrand Reinhold

Co., London (1970). 8. Carleson, L., On the convergence and growth of partial sums of Fourier series,

Acta Math. 116, 135-157 (1966). 9. Davie, A. M., The Banach approximation problem, J. Approx. Theory 13, 392-394

(1975). 10. Davie, A. M., The approximation problem for Banach spaces, Bull. London Math.

Soc. 5, 261-266 (1973). 11. Dunford, N., and Schwartz, J., Linear Operators. Part I: General Theory, Wiley

(Interscience), New York (1958). 12. Dunford, N., and Schwartz, J., Linear Operators. Part II; Wiley (Interscience), New

York (1964). 13. Fano, G., Mathematical Methods of Quantum Mechanics, McGraw-Hili, New York

(1971). 14. Halmos, P. R., A Hilbert Space Problem Book, Van Nostrand, Princeton, New Jersey

(1967). 15. Halmos, P. R., Introduction to Hilbert Space and the Theory of Spectral Multiplicity,

Chelsea Publishing Co., New York (1951). 16. Helmberg, G., Introduction to Spectral Theory in Hilbert Space, North-Holland,

Amsterdam (1969).

265

266 Bibliography

17. Hunt, R. A., On the convergence of Fourier series, pp. 235-255 in Orthogonal Expansions and Their Continuous Analogs, Southern Illinois University Press, Carbondale, Illinois (1968).

18. Kim, H. W., Pearcy, C., and Shields, A. L., Rank one commutators and hyperinvariant subspaces, Mich. Math. J. 22, 193-194 (1975).

19. Lindenstrauss, J., and Tzafriri, L., Classical Banach Spaces I: Sequence Space, Springer-Verlag, Berlin (1977).

20. Liusternik, L. A., and Sobolev, V. J., Elements of Functional Analysis, Fredrick Ungar, New York (1961).

21. Lomonosov, V., On invariant subspaces of families of operators commuting with a completely continuous operator, Funktsion. Anal. Prilozh. 7, 55-56 (1973). (Russian)

22. Lorch, E. R., On certain implications which characterize Hilbert space, Ann. Math. 49, 523-532 (1948).

23. Michaels, A. J., Hilden's simple proof of Lomonosov's Invariant Subspace Theorem, Adv. Math. 25, 56-58 (1977).

24. Naimark, M. A., Normed Rings (translated from Russian), P. Noordhoof, Groningen, The Netherlands (1964).

25. Nelson, E., Topics in Dynamics I: Flows, Princeton University Press, Princeton, New Jersey (1969).

26. Naylor, A. W., and Sell, G. R., Linear Operator Theory in Engineering and Science, Holt, Rinehart and Winston, New York (1971).

27. Pearcy, C., Some Recent Developments in Operator Theory, CBMS Regional Conference Series in Mathematics No. 36, American Mathematical Society, Providence, Rhode Island (1978).

28. Pearcy, C., and Shields, A. L., A Survey of the Lomonosov Technique in the Theory of Invariant Subs paces (Topics in Operator Theory), American Mathematical Society Surveys No. 13, Providence, Rhode Island (1974), pp. 219-229.

29. Prugoveeki, E., Quantum Mechanics in Hilbert Space, Academic, New York (1971). 30. Radjavi, H., and Rosenthal, P., Invariant Subs paces, Springer Verlag, New York

(1973). 31. Reed, M., and Simon, B., Functional Analysis, Academic, New York (1972). 32. Riesz, F., and Sz-Nagy, B., Functional Analysis, Fredrick Ungar, New York (1955). 33. Roman, P., Some Modern Mathematics for Physicists and Other Outsiders, Vol. 2,

Pergamon, New York (1975). 34. Sarason, D., Invariant Subs paces, Mathematical Survey No. 13, American Mathe

matical Society, Providence, Rhode Island (1974). 35. Schaefer, H. H., Eine Bemerkung zur Existenz invarianter Teilriiume linearer Ab

bildungen, Math. Z. 82 (1963). 36. Shields, A. L., A Survey of Some Results on Invariant Subspaces in Operator Theory

(Linear Spaces and Approximation), International Series of Numerical Mathematics 40 (1977), pp. 641-657.

37. Shields, A. L., A note on invariant subspaces, Mich. Math. J. 17, 231-233 (1970). 38. Stampfli, J. G., Recent developments on the invariant subspace problem (Operator

Theory and Functional Analysis), Res. Notes Math. (Edited by I. Erdelyi) 38, 1-7 (1979).

39. Stone, M. H., Linear Transformations in Hilbert Space, American Mathematical Society, Providence, Rhode Island (1964).

40. Sucheston, L., Banach limits, Am. Math. Mon. 74(3}, 308-311 (1967).

Bibliography 267

41. Taylor, A. E., Introduction to Functional Analysis, John Wiley and Sons, New York (1958).

42. Vulikh, B. Z., Introduction to Functional Analysis for Scientists and Technologists, Pergamon, Oxford (1963).

43. Whitley, R., An elementary proof of the Eberlein-Smulian theorem, Math. Ann. 172, 116--118 (1967).

44. Wilansky, A., Functional Analysis, Blaisdell, New York (1964).

Definition, Theorem, Proposition, Lemma, and Corollary Index

Definitions 6.22 101 8.5 220 6.1 2 6.23 103 8.6 221 6.2 2 6.24 104 8.7 223 6.3 5 6.25 112 C.l 255 6.4 8 6.26 114 C.2 260 6.5 14 7.1 122 6.6 16 7.2 124 6.7 18 7.3 126 Theorems

6.8 34 7.4 126 6.1 5 6.9 43 7.5 126 6.2 7 6.10 48 7.6 132 6.3 8 6.11 51 7.7 134 6.4 18 6.12 56 7.8 136 6.5 20 6.13 57 7.9 144 6.6 20 6.l4 63 7.10 152 6.7 22 6.15 71 7.11 157 6.8 35 6.16 79 7.12 184 6.9 38 6.l7 96 7.13 185 6.10 38 6.l8 96 8.1 195 6.11 39 6.l9 99 8.2 198 6.12 50 6.20 100 8.3 211 6.13 54 6.21 101 8.4 220 6.14 56

269

6.15 60 6.16 61 6.17 63 6.l8 64 6.19 66 6.20 67 6.21 74 6.22 76 6.23 81 6.24 83 6.25 84 6.26 97 6.27 97 6.28 99 6.29 99 6.30 99 6.31 100 6.32 100 6.33 101 6.34 102 6.35 104 6.36 105

270 Definition, Theorem, Proposition, Lemma, and Corollary Index

6.37 112 6.11 37 7.22 183 7.12 181 6.38 116 6.12 37 7.23 184 7.13 183 7.1 131 6.13 38 8.1 196 8.1 198 7.2 133 6.14 43 8.2 197 8.2 198 7.3 135 6.15 49 8.3 199 8.3 199 7.4 137 6.16 52 8.4 203 8.4 203 7.5 143 6.17 53 8.5 211 8.5 206 7.6 145 6.18 53 8.6 212 8.6 206 7.7 147 6.19 60 8.7 221 8.7 238 7.8 161 6.20 71 8.8 222 8.8 239 7.9 163 6.21 71 8.9 223 8.9 240 7.10 179 6.22 72 8.10 223 8.10 241 7.11 187 6.23 74 8.11 230 8.11 241 8.1 196 6.24 74 8.12 230 8.12 242 8.2 200 6.25 80 8.13 231 C.1 262 8.3 207 6.26 81 C.1 257 8.4 210 6.27 82 C.2 257 8.5 213 6.28 94 C.3 258 Corollaries

8.6 215 6.29 103 C.4 259 6.1 6 8.7 216 7.1 122 C.5 262 6.2 6 8.8 233 7.2 126 6.3 21 8.9 235 7.3 127 6.4 21 8.10 245 7.4 128 Lemmas 6.5 22 8.11 246 7.5 133 6.1 35 6.6 50 8.12 249 7.6 134 6.2 39 6.7 50 C.1 260 7.7 135 6.3 55 6.8 50 C.2 263 7.8 136 6.4 56 6.9 50

7.9 137 6.5 62 6.10 51 7.10 144 6.6 64 6.11 64

Propositions 7.11 147 7.1 132 6.12 66 6.1 6 7.12 148 7.2 132 6.13 75 6.2 9 7.13 148 7.3 153 7.1 124 6.3 16 7.14 148 7.4 159 7.2 144 6.4 17 7.15 148 7.5 159 7.3 181 6.5 18 7.16 153 7.6 159 8.1 213 6.6 24 7.17 153 7.7 160 8.2 216 6.7 26 7.18 154 7.8 161 8.3 222 6.8 34 7.19 154 7.9 161 C.1 263 6.9 36 7.20 155 7.10 179 C.2 264 6.10 37 7.21 157 7.11 179

Symbol and Notation Index

Set and element notations X* [= L(X, F)] 18, 143

F 2 X** 48

[x] 8 X' 51

co(E) 104 C 28

<B) 3 °T, So (annihilator) 28 Co 28 BV[a, b] 54

Topological notations NBV[a, b] 56

Sis) 10 sp(A), [A] 64

l:(E) 202 L 2(P), 12(l) 125, 128 M+N, MEBN 132, 133

EBiEI Hi 212 Function space notations Ml.; Sl. 132 C1(X) 96 C1[0, 1] 4 Lp 4, 11 Norm notations

Loo 4, 11 Ilx II 2 B(S, Y) 39 Ilfllp 4, 11 C[a, b] 4, 12 ess sup If I 5 B[a, b] 54 Ilflloo 4, 11

IITII 17

IIBII 144 Normed space notations

R:" 4 Ip 8 Operator notations

X/M 8 R(A), RA 45, 220 L(X, Y) 16 J: X -4-X** 48

271

272 Symbol and Notation Index

a(T) 79 f J(t) dE(t) 185, 201 R(A, T), R). 79, 80 Tf 200 e(T) 79 DT 220 reT) 82 SC T 220 T-l 18 DT• 220 PaCT) 79 l' 223 Ca(T) 79 V E(K) 203 Ra(T) 79 T*, A* 44, 147, 220 Miscellaneous notation PN 155

(x I y) 124 S<T 157 mT, MT 159 x.l y, E.l F 126

peT), J(T), geT) 179, 182, 205 LXi = x 127 I

E(s) 184 w 57 Xn-X

Subject Index

Absolutely summable, 8 Absorbing set, 96 Adjoint of an operator, 44, 147, 220 Affine map, 120

a fixed point theorem for, 120 Algebraic dual, 52 Algebraically reflexive, 52 Annihilator

of a subset, 45, 132 orthogonal complement, 132

Approximate eigenvalue, 174 Approximation problem, 73 Approximation property, 73

Balanced set, 96 Banach algebra, 92 Banach-Alaoglu Theorem, 61 Banach limits, 29 Banach-Saks Theorem, 70 Banach space, 3 Basis

Hamel (algebraic), 3 pre-Hilbert space, 134 problem, 73 Schauder, 11, 73

Bessel's Inequality, 128 Bilinear form (see Sesquilinear form) Bounded linear functional, 18 Bounded linear operator, 16 Bounded sesquilinear form, 144

Cauchy-Schwarz Inequality, 123

273

Characteristic values, 165 Closable operator, 223 Closed Graph Theorem, 38 Closed linear operator, 34, 223 Closure of an operator, 223 Compact linear operator, 71 Compact spectral measure, 203 Complete,

weakly, 70 weakly sequentially, 58

Complete orthonormal set, 134 Completion of a pre-Hilbert space, 130 Conjugate linear, 143 Continuous functional calculus, 179 Continuous spectrum, 79, 230 Convergence

weak (in a normed space), 57 Convex hull, 104 Cyclic vector, 211

Dimension of a pre-Hilbert space, 136 Direct sum (of Hilbert spaces), 212 Dirichlet Problem, 85 Domain of a linear operator, 220 Dual, algebraic, 52 Dual (conjugate) space, 18

Eberlein-Smulian Theorem, 64 Eigenvalue, 79

approximate, 174 Eigenvector, 79

274

E-integrable, 185 Equicontinuity, 100, 112 Equivalent norms, 5

in qO,I], 46 Essential range, 79 Essentially self-adjoint, 252 Extension of a linear operator, 220 Extremal subset, 103

$7-integrable, 198 Fourier expansion, 134 Fourier series, 41, 137-139 Frechet space, 99 Fredholm Alternative Theorem, 76 Fredholm integral equation, 164 Fuglede's Theorem, 263

Gelfand-Mazur Theorem, 92 Gram-Schmidt Orthonormalization

Process, 133 Graph of an operator, 34 Green's function, 170

Haar measure, 111, 116 Hahn-Banach Theorem, 18, 20, 22, 32 Hamel basis, 3 Harmonic function, 88 Hermitian operator, 152 Hermitian sesquilinear form, 122 Hilbert-Schmidt operator, 151 Hilbert space, 126 Hyperinvariant subspace, 260 Hyperplane, 101 Hyponormal operator, 264

Inner product, 124 Inner product space, 124 Integrable (E-integrable), 185, 198 Integral equations, Fredholm, 85, 164 Invariant subspace, 212, 255 Isometric operator, 152 Isomorphic, topologically, 6

Kakutani Fixed Point Theorem, 112 Krein-Milman Theorem, 104

Subject Index

Linear functional, bounded, 18 Linear isometry on a Hilbert space, 136 Linear operator, 14, 220

bounded, 220 closed, 34 compact (completely continuous), 71 weakly compact, 91

Linear space, 2 Locally convex, 96 Lomonosov's Theorem, 260 Lp", lp" spaces, 50

Markov-Kakutani Fixed Point Theorem, 120

Mean Ergodic Theorem, 156 Measure

Haar, 111, 116 in a Hilbert space, 141 operator-valued, 195 spectral, 195

Minkowski functional, 32

Natural map, 48 Norm, 3

uniform or "sup", 4 Normal operator, 152 Normalized bounded variation function,

56 Normalized spectral measure, 195

on R, 218 Normed linear space, 2

Open Mapping Theorem, 35 Operator

adjoint, '44, 220 adjoint (Hilbert), 147 bounded linear, 16, 220 closable, 223 essentially self-adjoint, 252 Hermitian, 152 Hilbert-Schmidt, 151 hyponormal, 264 isometric, 152 linear, 14, 220 nonscalar, 255 normal, 152 projection, 152

Subject Index

Operator (cont.) self-adjoint, 152, 221 shift, 80 subnormal, 264 symmetric, 221 unitary, 152

Orthogonal, 126 Orthogonal complement (annihilator),

132 Orthonormal set, 126

Parallelogram Law, 123 Parseval's Identity, 134 Point spectrum, 79, 230 Polarization Identity, 122 Positive-operator-valued measure, 195 Positive self-adjoint operator, 157 Pre-Hilbert space, 124 Principle of Uniform Boundedness, 39 Projection operator, 43, 152 Pythagorean Theorem, 127

Quotient (of normed spaces), 9

Reducing subspace, 255 Reflexive, 48 Reflexive, algebraically, 52 Residual spectrum, 79, 230 Resolution of the identity, 185 Resolvent set, 79, 230 Riesz Representation Theorem, 50

Schauder basis, II, 73 Schauder Fixed Point Theorem, 120 Schrodinger equation, 249 Self-adjoint operator, 152, 221 Seminorm, 96 Sesquilinear form, 122

bounded, 144 Hermitian, 122 positive, 122

Set absorbing, 32, 96 balanced, 96 bounded, 100 separated (by hyperplane), 101 totally bounded, 110

Shift operator, 80 Space

Banach,3

275

of bounded variation functions, 54 C*[a, b], 56 dual, 18 Frechet, 99 Hilbert, 126 inner product, 124 linear (or vector), 2 I., LiJt), 124, 125 Ip*, Lp*, 50 normed linear, 2 pre-Hilbert, 124 strictly convex, 12 topological vector, 94 uniformly convex, 12

Spectral function on R, 218 Spectral Mapping Theorem, 83, 180 Spectral measure, 195 Spectral radius, 82 Spectral Theorem

Functional Calculus Form, 207, 235 Multiplication Operator Form, 213,

233 Resolution of the Identity

Formulation, 187 Spectral Measure Formulation, 210,

216,245 Spectrum, 79, 230

continuous, 230 residual, 230

of a spectral measure, 202 point, 230

Square-root of an operator, 181, 194 Strictly convex, 12 Strong limit, 151 Sturm-Liouville equation, 167 Sturm-Liouville problem, 164 Subnormal operator, 264 Summable family of vectors, 127 Summable series (absolutely), .8 Sup norm, 4 Symmetric form, 122 Symmetric operator, 221

Topological group, 114 Topologically isomorphic, 6

276

Topological vector space, 94 Topology

locally convex, 96 weak,60 weak*, 61

Total set (in X*), 63 Totally bounded, 110

Uniform Boundedness, Principle of, 39 Uniform limit (of a family of operators),

151 Uniform norm, 4 Uniformly convex space, 12 Unitary operator, 152, 211

Subject Index

Volterra Fixed Point Theorem, 29

Weak Cauchy sequence, 58 Weak compactness in qo, I], 69 Weak convergence (in a normed linear

space), 57 Weak convergence in L p , L 1 , 69, 70 Weak limit (of a family of operators),

151 Weak topology, 60 Weak* topology, 61 Weakly compact linear operator, 91 Weakly complete, 70 Weakly sequentially complete, 58

Errata for Part A

Page 190

In line 5, change "Suppose a > b" to "Suppose aP- l > b." In line 10, change "The case a < b" to "The case aP- l a)."

Page 198, Problem 3.5.20

Change II fill to !ig 111·

Page 230, Problem 4.2.19(iii)

Replace text beginning with "Show that ... " by

"Show that

get) = 1, 1

if < t <-2n + 1 - - 2n ' ° <n < 00,

=0, if 1 < t < ~_-=-2n+1-1 - - 2n +2' n> 1,

and linear on the rest of (0, 1] is an example of such a function."

Page 258, Problem 4.3.12

In this problem, the symbols fl and v are mixed up. The problem should read:

" ... where fll ~ v, ... , and fl3 -.l v and ... [Hint: By Theorem 4.7, fl = flo + fll' fll ~ v, and flo -.l v ... ]"

277

real and functional analysis: part b: functional analysis

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