Linear Analysis: An Introductory Course, Second edition

Bela Bollobás II an introductory course SECOND EDITION CAMBRIDGE MATHEMATICAL TEXTBOOKS LINEAR ANALYSIS An Introduc...

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Bela Bollobás

II an introductory course SECOND EDITION

CAMBRIDGE MATHEMATICAL TEXTBOOKS

LINEAR ANALYSIS

An Introductory Course Bela Bollobás University of Cambridge

CAMBRIDGE UNIVERSITY PRESS

PUBLISHED BY THE PRESS SYNDICATE OF THE UNIVERSITY OF CAMBRIDGE The Pitt Building, Trumpington Street, Cambridge CB2 1RP, United Kingdom CAMBRIDGE UNIVERSITY PRESS The Edinburgh Building, Cambridge CB2 2RU, UK http://www.cup.cam.ac.uk

40 West 20th Street, New York, NY 10011-4211. USA http://www.cup.org 10 Stamford Road. Oakleigh, Melbourne 3166, Australia

First edition © Cambridge University Press 1990 Second edition © Cambridge University Press 1999

This book is in copyright. Subject to statutory exception and to the provisions of relevant collective licensing agreements. no reproduction of any part may take place without the written permission of Cambridge University Press. First published 1990 Second edition 1999

Printed in the United Kingdom at the University Press, Cambridge Typeset in Times 10/l3pt A catalogue record for this book is available from the British Library ISBN 0 521 65577 3 paperback

To Mark

Qui cupit, capit omnia

CONTENTS

Preface

ix

1.

Basic inequalities

2.

Normed spaces and bounded linear operators

18

3.

Linear functionals and the Hahn—Banach theorem

45

4.

Finite-dimensional normed spaces

60

5.

The Baire category theorem and the closed-graph theorem

75

6.

Continuous functions on compact spaces and the Stone— Weierstrass theorem

85

I

7.

The contraction-mapping theorem

101

8.

Weak topologies and duality

114

9.

Euclidean spaces and Hubert spaces

130

10.

Orthonormal systems

141

11.

Adjoint operators

155

12.

The algebra of bounded linear operators

167

Contents 13.

Compact operators on Banach spaces

186

14.

Compact normal operators

198

15.

Fixed-point theorems

213

16.

invariant subspaces

226

Index of notation

233

index of terms

235

PREFACE

This book has grown out of the Linear Analysis course given in Cambridge on numerous occasions for the third-year undergraduates reading mathematics. It is intended to be a fairly concise, yet readable and down-

to-earth, introduction to functional analysis, with plenty of challenging exercises. In common with many authors, I have tried to write the kind of book that I would have liked to have learned from as an undergraduate. I am convinced that functional analysis is a particularly beautiful and elegant area of mathematics, and I have tried to convey my enthusiasm to the reader. In most universities, the courses covering the contents of this book are given under the heading of Functional Analysis; the name Linear Analysis has been chosen to emphasize that most of the material in on linear functional analysis. Functional Analysis, in its wide sense, includes partial differential equations, stochastic theory and non-commutative harmonic analysis, but its core is the study of normed spaces, together with linear functionals and operators on them. That core is the principal topic of this volume. Functional analysis was born around the turn of the century, and within

a few years, after an amazing burst of development, it was a wellestablished major branch of mathematics. The early growth of functional analysis was based on 19th century Italian function theory, and was given

a great impetus by the birth of Lebesgue's theory of integration. The subject provided (and provides) a unifying framework for many areas: Fourier Analysis, Differential Equations, Integral Equations, Approximation Theory, Complex Function Theory, Analytic Number Theory, Measure Theory, Stochastic Theory, and so on.

ix

Preface

x

From the very beginning, functional analysis was an international sub-

ject, with the major contributions coming from Germany, Hungary, Poland, England and Russia: Fisher, Hahn, Hilbert, Minkowski and Radon from Germany, Fejér, Haar, von Neumann, Frigyes Riesz and Marcel Riesz from Hungary, Banach, Mazur, Orlicz, Schauder, SierpiElski and Steinhaus from Poland, Hardy and Littlewood from England, Gelfand, Krein and Milman from Russia. The abstract theory of normed

spaces was developed in the 1920s by Banach and others, and was presented as a fully fledged theory in Banach's epoch-making monograph, published in 1932.

The subject of Banach's classic is at the heart of our course; this material is supplemented with a body of other fundamental results and some pointers to more recent developments. The theory presented in this book is best considered as the natural continuation of a sound basic course in general topology. The reader would benefit from familiarity with measure theory, but he will not be at a great disadvantage if his knowledge of measure theory is somewhat shaky or even non-existent. However, in order to fully appreciate the power of the results, and, even more, the power of the point of view, it is advisable to look at the connections with integration theory, differential equations, harmonic analysis, approximation theory, and so on. Our aim is to give a fast introduction to the core of linear analysis, with emphasis on the many beautiful general results concerning abstract spaces. An important feature of the book is the large collection of exercises, many

of which are testing, and some of which are quite difficult. An exercise which is marked with a plus is thought to be particularly difficult. (Needless to say, the reader may not always agree with this value judgement.) Anyone willing to attempt a fair number of the exercises should obtain a thorough grounding in linear analysis. To help the reader, definitions are occasionally repeated, various basic

facts are recalled, and there are reminders of the notation in several places.

The third-year course in Cambridge contains well over half of the contents of this book, but a lecturer wishing to go at a leisurely pace will find enough material for two terms (or semesters). The exercises should certainly provide enough work for two busy terms. There are many people who deserve my thanks in connection with this book. Undergraduates over the years helped to shape the course; numerous misprints were found by many undergraduates, including John Longley, Gábor Megyesi, Anthony Quas, Alex Scott and Alan Stacey.

Preface

xi

I am grateful to Dr Pete Casazza for his comments on the completed manuscript. Finally, I am greatly indebted to Dr Imre Leader for having suggested many improvements to the presentation. Cambridge, May 1990

Bela Bollobás

For this second edition, I have taken the opportunity to correct a number of errors and oversights. I am especially grateful to R. B. Burckel for providing me with a list of errata. B. B.

1. BASIC INEQUALITIES

The arsenal of an analyst is stocked with inequalities. In this chapter we present briefly some of the simplest and most useful of these. It is an

indication of the size of the subject that, although our aims are very modest, this chapter is rather long. Perhaps the most basic inequality in analysis concerns the arithmetic and geometric means; it is sometimes called the AM-GM inequality. of n reals is The arithmetic mean of a sequence a = (a1,. . , .

A(a) = a is non-negative then the geometric mean is

/n

G(a) =

(II

\1/fl

a.) /

where the non-negative nth root is taken.

Theorem 1. The geometric mean of n non-negative reals does not exceed their arithmetic mean: if a = (a1,. .. , G(a)

Equality holds if a1 =

then

A(a).

(1)

=

Proof. This inequality has many simple proofs; the witty proof we shall present was given by Augustin-Louis Cauchy in his Cours d'Analyse (1821). (See Exercise I for another proof.) Let us note first that the theorem holds for n = 2. Indeed, (a1 —a2)2 = a?—2a1a2+a?

0; 1

Chapter 1: Basic inequalities

2

so

(a1+a2)2

with equality iff a1 =

4a1a2

a2.

Suppose now that the theorem holds for n = m. We shall show that it holds for n = 2m. Let a1 ,. . ,a,,,, b1 ,. . . be non-negative reals. .

Then —

.

jj

.am,

m

m —

ai+...+am+bi+...+bm 2m

If equality holds then, by the induction hypothesis, we have a1 = = tZm = b1 = = bm. This implies that the theorem holds whenever n is a power of 2. Finally, suppose n is an arbitrary integer. Let

n<2k= N

and

a=

Set a1 =

=

a1

a1

a't,

= aN, in other words if a1 =

with equality if a1 =

=

0

In 1906 Jensen obtained some considerable extensions of the AM-GM inequality. These extensions were based on the theory of convex functions, founded by Jensen himself. A subset D of a real vector space is convex if every convex linear

combination of a pair of points of D is in D, i.e. if x,y E D and 0 < t < I imply that tx + (1 — t)y E D. Note that if D is convex, x1 ,... ,

E D,

,

> 0 and

t1

=

1

then

E D.

Chapter 1: Basic inequalities

3

Indeed, assuming that a convex linear combination of n — is in D, we find that

1

points of D

and so

ED.

=

is

Given a convex subset D of a real vector space, a function f: D —p said to be convex if

f(tx+(1—t)y)

tf(x)+(1—t)f(y)

(2)

x,y E D and 0 < t < 1. We call f strictly convex if it is convex and, moreover, f(tx+(1—t)y) = tf(x)+(1—t)f(y) and 0< t < I imply that x = y. Thus f is strictly convex if strict inequality holds in (2) whenever x y and 0 < I < 1. A function f is concave if —f is convex and it is strictly concave if —f is strictly convex. Clearly, f is convex 1ff the set {(x,y) E Dxli: y f(x)} is convex. Furthermore, a function f: D R is convex (concave, ...) iff its restriction to every interval [a,b} = {ta+(1—t)b:0 t 1) in D is convex (concave, ...). Rolle's theorem implies that if f: (a, b) —' R is differentiable then f is convex if f' is increasing and f is concave if f' is decreasing. In particular, if f is twice differentiable and f" 0 then f is convex, while if f" 0 then j is concave. Also, if f" > 0 then f is whenever

strictly convex and if f" < 0 then f is strictly concave. The following simple result is often called Jensen's theorem; in spite of its straightforward proof, the result has a great many applications. Theorem 2. Let f: D —

be a concave function. Then tf(x1)

(3)

E (0, 1) and t, = 1. Furtherwhenever x1,... , E D, t1 ,. . more, if f is strictly concave then equality holds in (3) if x1 = = .

Proof. Let us apply induction on n. As for n = there is nothing to prove and for n = 2 the assertions are immediate from the definitions, 1

let us assume that n

3 and the assertions hold for smaller values of n.

Chapter 1: Basic inequalities

4

Suppose first that f is concave, and let E (0,1)

2,...,n, set

with

that 1 = 1. Then, by applying the induction hypothesis twice, first for n — 1 and then for 2, we find that i

=

t = t1/(1—t1), so

t,f(x1) =

t1f(x1) + (1—t1) taxi)

f(t1x1 +

)

1=2

=I

If f is strictly concave, n

3 and not all x, are equal then we may

assume that not all of x2,. . . ,x,, are equal. But then < (1

(1 —t1)

t;xI);

1=2

so

0

the inequality in (3) is strict.

It is very easy to recover the AM-GM inequality from Jensen's theorem: logx is a strictly concave function from (0, to IL so for a1,... , —

loga1

which is equivalent to (1). In fact, if t1 ,. . t,logx1

with equality if x1 =

=

a

—,

log

> 0 and

.

tx1,

log giving

t, = 1 then (4)

the following extension of

Theorem 1. Theorem 3. Let a1,.. Then

0 and ps,...

> 0 with

P1

=

1.

Chapter 1: Basic inequalities

af' with equality if a1 =

5

p1a,,

(5)

=

Proof. The assertion is trivial if some a, is 0; if each a is positive, the 0 assertion follows from (4).

The two sides of (5) can be viewed as two different means of the a generalized geometric mean sequence a1,... , and the right-hand side is a generalized arithmetic mean, with the various terms and factors taken with different weights. In fact, it is rather natural to define a further extension of these notions. will play the role of Let us fix P = 1: the weights or probabilities. Given a continuous and strictly monotonic (0, — R, function (a1 > 0) is defined as

the c-mean of a sequence a =

(a1,.

. .

,

a,,)

Mç(a) = Note that M, need not be rearrangement invariant: for a permutation ir

the c-mean of a sequence a1,... ,a,, need not equal the c-mean of the sequence .. , a,,.(.,). Of course, if = p,, = 1/n then every = c-mean is rearrangement invariant. It is clear that mm

a

M (a)

max a,.

In particular, the mean of a constant sequence (a0,..

.

,

a0)

is precisely

a0.

For which pairs and ifs are the means Mq, and M,,,, comparable? More precisely, for which pairs q and i/i is it true that Mç(a) for every sequence a = (a1,... ,a,,) (a1 > 0)? It may seem a little surprising that Jensen's theorem enables us to give an exact answer to these questions (see Exercise 31). Theorem 4. Let Pi ,.. . ,p,, > 0 be fixed weights with p, = 1 and let ifs: (0, P be continuous and strictly monotone functions, such is concave if ç is increasing and convex if is decreasing. that Then Mq,(a)

Chapter 1: Basic inequalities

6

(a1> 0). If for every sequence a = (a1,.. is strictly concave (respectively, strictly convex) then equality holds 1ff a1 = = is concave. Set b• =

Proof. Suppose that is increasing and and note that, by Jensen's theorem,

=

=

=

\i=1

I

If

- is strictly concave and not all a, are equal then the inequality above is strict since not all b• are equal. The case when is decreasing and is convex is proved analo-

0

gously.

When studying the various means of positive sequences, it is convcnicnt to usc thc convention that a stands for a sequence (a1,... and so on; furthermore, b for a sequence (b1 ,. .. =

ab = and so

a+x =

= (a1b1

(XE

abc =

on.

If p(t) = r (—cx 0 we define the mean M, for all non-negative sequences: if a = (a1,... (a1 0) then

in =

=l

\1/r

p,af

Note that if Pi = = p,, = 1/n then M1 is the usual arithmetic mean A, M2 is the quadratic mean and M_1 is the harmonic mean. As an

immediate consequence of Theorem 4, we shall see that Mr is a continuous monotone increasing function of r.

a natural extension from to the whole of the extended real line [—oo, such that Mr(a) is a continuous monotone increasing function. To be precise, put In fact, Mr(a)

has

Chapter 1: Basic inequalities

a,

= max Thus M0(a)

7

M0(a) = H

af.

is the weighted geometric mean of the a1. It is easily for all r

checked that we have M,(a) =

r

Theorem 5. Let a = (a1,... be a sequence of positive numbers, not all equal. Then Mr(a) is a continuous and strictly increasing function of r r on the extended real line Proof. It is clear that M,(a) is continuous on (—x,O)U(O,x). To

show that it is strictly increasing on this set, let us fix r and s, with < r <s r 0 and s # 0. If 0 < r then ( is an increasing function of t > 0, and is a concave function, and if r < 0 then t' is (i'S convex. is Hence, by Theorem 4, we have decreasing and <M5(a). Let us write A(a) and G(a) for the weighted arithmetic and geometric i.e. set means of a =

A(a) =

M1(a)

pa,

=

G(a) = M0(a) = II

and

af.

To complete the proof of the theorem, all we have to do is to show

that M,0(a) = lim Mr(a),

G(a) = limM,(a).

= urn Mr(a),

The proofs of the first two assertions are straightforward. Indeed, let m n be such that am = Then for r > 0 we have 1 =

Mr(a) so

am = =

we have lim,....,.

=

pi/ra.

him Mr(a

final assertion, G(a) = lim,....0 keeping with our conventions, for ar = (ar,. . Then, clearly, The

Mr(a) =

1)}

I

= lim

Mr(a).

requires a little care. In < r < (r 0) let us write

.

Also, it is immediate that

for every r,

Since Mr(a)

as required. Also,

A(a')".

Chapter 1: Basic inequalities

8

= log a

(a[ — 1) =

urn

r—.oT

r

and so

lim!{A(a')_l} r—.O

r

= logG(a).

(6)

Since

logt

(—1

for every t> 0, if r> 0 then log G(a) =

Letting r and so

0,

log A(a')

log G(a')

{A(a') — 1}.

we see from (6) that the right-hand side tends to log G(a) lim logM,.(a) = urn

r-4.Q+

r

logA(a') = logG(a),

implying

lim Mr(a) = G(a).

r-40+

Finally,

lim M,(a) =

urn

=

G(a'Y'

= G(a).

0

The most frequently used inequalities in functional analysis are due to Holder, Minkowski, Cauchy and Schwarz. Recall that a hermitian form on a complex vector space V is a function p: Vx V C = Aç(x,z)+A.ç(y,z) and q'(y,x) = for all such that = A x,y,z E V and A,p E C. (Thus hermitian form is said to be positive if ç(x, x) is a positive real number

for alix E V(x

0).

Let ç(•,) Then,

be a positive hermitian form on a complex vector space V. given x,y E V, the value

ço(Ax+y,Ax+y) = is

real and non-negative for we

all A E C.

For x

0, setting A =

find that ço(x,x)p(y,y)

and

the same inequality holds, trivially, for x =

0

as weH. This is the

Chapter 1: Basic inequalities

9

Cauchy—Schwarz inequality. In particular, as n

ç(x,y) 1=1 is

a positive hermitian form on C'1, /

\1/2/

\I/2

x,p, 1=1

i—I

i—I

and so

\1/2/

/ xy, i—i

(

,,

=

I

\i=i

/

\:=1

1y112

\i—1

/

/

\i—1 (7)

Our next aim is to prove an extension of (7), namely Holder's inequality.

Theorem 6. Suppose

p,q>1 Then for complex numbers a1,. . / bk k—i

.

, a,,, I ak

(

b1,. . . , b,, we have \i/,O/ \1/q I

'

I

\k—1

I

with equality if all ak are 0 or all k and some t and 0. Proof.

11

—+—=1.

and

(

(8)

I

\k—1

I

and akbk =

=

and b, set x1

Given non-negative reals a

e

Iakbkl for

= a1', x2 =

Pt = I/p and P2 = 1/q. Then, by Theorem 3,

—+—, with equality if a1' = HOlder's inequality is a short step away from here. Indeed, if i)

(±

I

i)

then by homogeneity we may assume that IakI1' k—i

But then, by (9),

=

= k—i

1.

(9)

Chapter 1: Basic inequalities

10

)=_+._=1. 1

+

k=I

q

1

Furthermore, if equality holds then

Iak!° = IbkI"

and

= k=I

Iakbkl,

Conversely, it is immediate that under these implying akbk = e Iakbk conditions we have equality in (8). 0 .

Note that if Mr denotes the rth mean with weights p, = and b = we put 1,...,n) and for a = and lbl = ab = al = (I =

then HOlder's inequality states that if p1 + q -' = 1 with p, q >

1,

then

Mp(lal)Mq(lbt).

M1(labl)

A minor change in the second half of the proof implies that (8) can be and Mq with extended to an inequality concerning the means M1, arbitrary weights (see Exercise 8).

Thc numbers p and q appearing in Holder's inequality are said to be conjugate exponents (or conjugate indices). It is worth remembering that = is the same as the condition p 1

(p—1)qp

(p—1)(q—1)= 1,

or

(q—1)p=q.

Note that 2 is the only exponent which is its own conjugate. As we remarked earlier, the special case p = q = 2 of Holder's inequality is called the Cauchy—Schwarz inequality. In fact, one calls I and conjugate exponents as well. HOlder's inequality is essentially trivial for the pair M1(labl)

with equality iff there is a 0 such that bk I = I) and akbk = 0. whenever ak elakbkl The next result, Minkowski's inequality, is also of fundamental importance: in chapter 2 we shall use it to define the classical I,, spaces.

Theorem 7. Suppose 1 numbers. Then

p

fn

\l/P

\k=I

/

and a1,...

<

/n I

\k=1

\l/P /

are complex

Chapter 1: Basic inequalities

11

with equality 1ff one of the following holds:

(1) allak are 0; (ii) bk = tak for all k and some t (iii) p = 1 and, for each k, either

0; = 0 or bk = Ikak for some

k

0.

if p = 1 so let us suppose that not all ak are 0 and not all bk are 0. Let q be the conjugate of p: p' + q -' = 1. Note that Proof. The assertion is obvious

1
ak+bkV

k=1

Iak+bkV'Iakl

Iak+bkV'Ibk(.

+

Applying (8), HOlder's inequality, to the two sums on the right-hand

side with exponents q and p, we find that

1/q

up

up

IakV)

\k=1

1\k=i

=(

\k=1

\h/P

n

/

Iak+bkV)

/

n

+(

/

\k=1

/

we obtain (8). The case of equality follows from that in Holder's ine-

0

quality.

Minkowski's inequality is also essentially trivial for p = M,, mean:

with

equality if there

is

an index k

such that

ak I

i.e. for the

=

a I),

and Iak+bkl = Iakl+IbkI. The last two theorems are easily carried over from sequences to integrable functions, either by rewriting the proofs, almost word for word, or by approximating the functions by suitable step functions. IbkI

Readers unfamiliar with Lebesgue measure will lose nothing if they take f and g to be piecewise continuous functions on [0, 1].

Chapter 1: Basic inequalities

12

Theorem 8. (HOlder's inequality for functions) Let p and q be conjugate indices and let f and g be measurable complex-valued functions on a measure space such that fl" and are integrable. Then fg is integrable and

o

(f IV' d,.L)"'(f

Jfg

9. (Minkowski's inequality for functions) Let I p < and let f and g be measurable complex-valued functions on a measure space and IgI" are integrable. Then (X, such that is integrable and Theorem

(Jlfv th)"'+(f

(I

IgI" th)11"

Exercises All analysts spend half their time hunting through the literature for inequalities which they want to use but cannot prove.

Harald Bohr

1.

Let

= fx = (x1)7:

x1

n and

0 for every

c

is bounded on and attains its E supremum at some point z = A and x1 = minx, <x2 = maxx1. Set Yi = i n. Show that y = and = x1 for 3 Y2 = and g(y) > g(x). Deduce that z = 1 for all i. E (iii) Deduce the AM-GM inequality. are convex func2. Show that if c(': (a,b) — (c,d) and cc: (c,d) (i) Show that g(x) =

x,

tions and cc is increasing then (coiIi)(x) =

3. Suppose that f: (a, b) —' that f is convex.

4. Let!: (a,b)

(0,

(c,b) and cc:

is convex.

is such that log! is convex. Prove be such that

and ç1of

convex. Show that f is convex. 5. Let {f,,: y E 1) be a family of convex functions on (a, b) such that are

f(x) =

f7(x)

for every x E (a,b). Show that f(x) is

also convex.

6. Suppose that f: (0, 1)

R

is an infinitely differentiable strictly

convex function. Is it true that f"(x) > 0 for every x E (0, 1)?

Chapter 1: Basic inequalities

13

7. Let p, q> I be conjugate exponents. By considering the areas of the domains = {(x,y): 0

a and 0

x

y

and

prove inequality (8) and hence deduce HOlder's theorem. that (p—1)(q—1) = 1, so if y = xe'' thenx = 8.

[Note

Prove the following form of HOlder's theorem for the means

In M,((a1fl) =

Mr(a)

(

\i=I

\1/r

p,a[)

(r > 0).

/

If p,q> 1 are conjugate exponents, that is (a1

,..

•

and (b1 ,..

,

. ,

=

1,

and

are complex sequences then Mp(IaI)Mq(IbI),

with equality as in Theorem 6.

9. Deduce from the inequality in the previous exercise that if a = (a1,. . , a,,) is a positive sequence, that is, a sequence of posi.

tive reals, not all equal, then Mr(a) is a strictly log-convex function is a strictly convex function of r. of r, i.e. 10. Show that for a fixed positive sequence a, log Mi,r(a) is a mono= tone decreasing convex function of r > 0, that is, if then {Mp(a)Mq(a)}"2.

11. Show that Mi,r(a)

is

a monotone decreasing convex function of r.

12. Show thatif 0
I,

(

\(r—p)/(r—q)/ ,,

\(p—q)/(r—q)

/

for all positive reals

13. Deduce from the previous exercise that if 0 < q

0 is continuous on (a, b) (or just measurable) then b

ía

b

(r—p)/fr—q)

b

(L

frth)

(p—q)/fr—q)

Chapter 1: Basic inequalities

14

14. Show that if p,q,r >0 are such that

=

then

Mp(a)Mq(b)

and b = (b1,.. . + q1 + r1 = also that if p. q, r, s > 0 are such that

for all positive sequences a = (a1,... Prove

then M5(abc)

M,,(a) Mq(b) Mr(C)

for all non-negative sequences a, b and c. State and prove the analogous inequality for k sequences.

15. Let a and b be positive sequences. Show that if r < 0 <s < f and then

=

M,(ab)

Mjrfr4) M5(b).

Deduce that if 0 < r < 1 then + b)

16. Let Pp,q(x,y) =

p,q

I and (p— 1)(q— 1) (p—1)(q—1) 1.

Mr(a) + Mr(b).

(x,y 0). Show that Ppq is concave 1ff 1 and that it is convex iffp,q 1 and

17. Deduce from the result in the previous exercise that M1(ab) and 1. To be precise, if Mq(b) are comparable if (p — 1)(q — 1) 1 then p,q >1 and (p—1)(q—I) Mp(a)Mq(b)

and if p,q < I and (p—1)(q—1) M1(a)

Mp(a)Mq(b).

isa decreasing func-

=

18.

tion of x 19.

1 then

0 and tends to M1(a) as x

Show

r 1 then M,(a + x) — x is an increasing function of x also tends to Mi(a) as x —* Prove that if r> 1, a > 0 and a 1 then

aT—1 > r(a—1) and

a—i r

Deduce that if a, b > 0 and a

b then

also that if 0 and it

Chapter 1: Basic inequalities

15

<ar_br < ra''(a—b) if

r<

or r> 1, and that

0

0
20. Prove Chebyshev's inequality:

the reverse inequalities hold if

if r> 0, 0 <

a2

and Mr(0)

unless all the a1 or all the b are equal. Prove also that the mequality is reversed if a = is monotone increasing and b = is monotone decreasing. [HINT: Note first that it suffices to prove the result for r = 1. For r = 1 the difference of the two sides is

—

21. For ar,... ,a,, >

0

of the form that the sequence ir0, products

for

all k (1

a2) let Irk be the arithmetic mean of the a (1 i1 < n). Show <1k .

.

, ii,, is strictly log-concave:

n—i). [HINT: Apply induction on n. For the assertion is just the AM-GM inequality for two terms.

n= 2 Now let n Ot,...

k

3 and denote by ir the appropriate average for Check that

= and

deduce that for =

A 22.

<

B

-ire

Deduce from the inequality in the previous exercise the following considerable extension of the AM-GM inequality, already known to Newton: if a1,..., a,, are positive, n 2 and a1 a2 then >

23.

C<

<

Let f: that if a1 ,.

R .

. , a,,,

... > ir,1l/n.

be a strictly convex function with f(0) 0. Prove 0 and at least two a are non-zero then f(a,)

as).

Chapter 1: Basic inequalities

16

24.

Let f: (0,

—

(0,

be a monotone increasing function such that

f(x) /x2 is monotone decreasing. For a, b >

f(a,b) Prove

=

that if a,, b, >

and 0

/

\i=1

set

g(a,b) =

then

\2

ab,)

0

\fn

\

b.))( / \i=i

(

\i=i

g(a1,

b,))

b,2

(

/

\i=1

/ \i=i

satisfy (*) for all a,b1 > Prove that if f,g: then they are of the form given in the previous exercise. 26. Show that 25.

abi)

+

b.2) a,2+b12

0

(± a12)(± b?)

for all real a , b with a? + b? > 0. be a rearrangement of the positive numbers 27. Let b1 , b2,. .. , Prove that a1,a2,. . .

n

28.

Let f(x)

0 be a convex function. Prove that

fdx. Show that is best possible. 29. Let f: [0, a] —+ R be a continuously differentiable function satisfying f(0) = 0. Prove the following inequality due to G. H. Hardy: {f'(x)}2 dx.

[HINT: Note that — 1/211(x) \' f(x)—x

f(x)

and so

(f(x))2(f(x)yf(x)]

Chapter 1: Basic inequalities

17

30. Show that Theorem 4 characterizes comparable means, i.e. if M4,(a) for all a = (a1)7 (a, > 0) then is concave if q is decreasing. is increasing and convex if 31. In a paper the authors claimed that if 0 for k = 1,2,. x," x1 (2x2 — x1) (3x3 — 2x2).

.

(n

.

Show that this is indeed true if 2x2 x1, 3x3 x1_ 1' and equality holds if and only ifx1 = ...

inequality need not hold if any of the n

—1

—

1)

.

., n then

i).

2x2,. . ., (n — 1) Show also that the inequalities kxk (k — 1)

=

Xk_I fails.

Notes

The foundation of the theory of convex functions is due to J. L. W. V. Jensen, Sur les fonctions convexes et les inégalites entre les valeurs moyennes, Acta Mathematica, 30 (1906), 175—93. Much of this chapter is

based on the famous book of G. H. Hardy, J. E. Littlewood and G. Polya, Inequalities, Cambridge University Press, First edition 1934, Second edition 1952, reprinted 1978, xii + 324 pp. This classic is still in print, and although its notation is slightly old-fashioned, it is well worth reading. Other good books on inequalities are D. S. Mitrinovic, Analytic mequalities, Springer-Verlag, Berlin and New York, 1970, xii + 400 pp., and A. W. Marshall and I. 01km, inequalities: Theory of Majorization and Its Applications, Academic Press, New York, 1979, xx + 569 pp.

2. NORMED SPACES AND BOUNDED LINEAR OPERATORS

In this long chapter we shall introduce the main objects studied in linear analysis: normed spaces and linear operators. Many of the normed spaces encountered in practice are spaces of functions (in particular, functions on i.e. sequences), and the operators are often defined in terms of derivatives and integrals, but we shall concentrate on the notions defined in abstract terms.

As so often happens when starting a new area in mathematics, the ratio of theorems to definitions is rather low in this chapter. However, the reader familiar with elementary linear algebra and the rudiments of the theory of metric spaces is unlikely to find it heavy going because the concepts to be introduced here are only slight extensions of various con-

cepts arising in those areas. Moreover, the relatively barren patch is rather small: as we shall see, even the basic definitions lead to fascinating questions.

A normed space is a pair (V, fl), where V is a vector space over or C and is a function from V to = {r E r O} satisfying (I) Dxli = 0 iff x = 0; (ii) (iii)

liAxD

= iA

for all x E V and scalar A; for all x,y E V. lxii + lxii

We call lixii the norm of the vector x: it is the natural generalization of the length of a vector in the Euclidean spaces or C's. Condition

(iii) is the triangle inequality: in a triangle a side is no longer than the sum of the lengths of the other two sides. In most cases the scalar field may be taken to be either R or C, even when, for the sake of simplicity, we specify one or the other. If we want to emphasize that the ground field is C, say, then we write 'complex normed space', 'complex 1,, space', 'complex Banach space', etc. 18

Chapter 2: Normed spaces and bounded linear operators

Furthermore, unless there is some danger of confusion, we shall identify with its underlying vector space V, and a normed space X = (V. call the vectors in V the points or vectors of X. Thus x E X means that x is a point of X, i.e. a vector in V. We also say that is a norm on

x. Every normed space is a metric space and so a topological space, and

we shall often make use of some basic results of general topology. Although this book is aimed at the reader who has encountered metric

spaces and topological spaces before, we shall review some of the basic concepts of general topology. A metric space is a pair (X, d), where X is a set and d is a function from Xx X into = [0, oo) such that (i) d(x,y) = 0 if x = y, (ii) d(x,y) = d(y,x) for all x,y C X and (iii) d(x, z) d(x, y) + d(y, z) for all x, y, z C X. We call d(x, y) the distance between x and y; the function d is a metric on X. Condition (iii) is again the triangle Inequality.

A topology r on a set X is a collection of subsets of X such that (i) 0 C r and X E r, (ii) is closed under arbitrary unions: if U,, E i for v C I' then U,, U,, C i, and (iii) r is closed under finite intersecr then tion: if U1 ,..., U1,, U1 C r. The elements of the collection r are said to be open (in the topology r). A topological space is a pair (X, 'r), where X is a set and r is a topology on X. If it is clear that the topology we take is r then we do not mention r explicitly and we call X a topological space. If Y is a subset (also called a subspace) of a topological space (X, r) then {Yfl U: U C r} is a topology on Y, called the subspace topology or the topology induced by r. In most cases every subset Y is considered to be endowed with the subspace topology. Given a topological space X, a set N C X is said to be a neighbourhood of a point x E X if there is an open set U such that x C U C N. A subset of X is closed if its complement is open. Since the intersection of a collection of closed sets is closed, every subset A of X is contained in a unique minimal closed set A = {x C X: every neighbourhood of x meets A}, called the closure of A. It is often convenient to specify a topology by giving a basis for it. Given a topological space (X, r), a basis for r is a collection a of subsets

of X such that a C r and every set in r is a union of sets from a. Clearly, if a C i.e. is a family of subsets of X, then a is a basis for a topology if (i) every point of X is in some element of a; (ii) if B1 , B2 E a then B1 fl B2 is a union of some sets from a.

Chapter 2: Normed spaces and bounded linear operators

20

A neighbourhood base at a point x0 is a collection v of neighourhoods of x0 such that every neighbourhood of x0 contains a member of v. There are numerous ways of constructing new topological spaces from

old ones; let us mention here the possibility of taking products, to be studied in some detail in Chapter 8. Let (X,o) and (Y,i) be topological spaces. The product topology on Xx Y = {(x, y): x E X, y E Y} is the topology with basis {Ux V: U E o, V r}. Thus a set W C Xx Y is

open if for every (x,y) E W there are open sets U C X and V C Y such that (x,y) E UXV C W. If d is a metric on X then the open balls D(x, r) = [y E X: d(x, y)
(x E X, r> 0)

form a basis for a topology. This topology is said to be defined or induced by the metric d; we also call it the topology of the metric space. Not every topology is induced by a metric; for example, = {U

C R: U = 0 or the complement R\U of U is countable}

is a topology on IR and it is easily seen that it is not induced by any metric.

Given topological spaces (Xj,r1) and (X2,r2), a map f: X1 —+ X2 is said to be continuous if C for every U E r2, i.e. if the inverse image of every open set is open. are continuous A bijection f from X1 to X2 such that both f and is said to be a homeomorphism; furthermore, (X1 ,r1) and (X2,r2) are said to be homeomorphic if there is a homeomorphism from X1 to X2. A sequence in a topological space (X, r) is said to be convergent to a point x0 C X, denoted —p x0 or = x0, if for every neighbourhood N of x0 there is an n0 such that C N whenever n0. Writing S for the subspace {n': n = 1,2,.. .}u{0} of R with n the Euclidean topology, we see that x,, = x0 iff the map f: S —' X, given by f(n = and f(0) = x0 is continuous.

The topology of a metric space is determined by its convergent sequences. Indeed, a subset of a metric space is closed iff it contains the limits of its convergent sequences.

If a and i

are

topologies on a set X and a C r then a is said to be

weaker (or coarser) than and r is said to be stronger (or finer) than a. Thus a- is weaker than i- iff the formal identity map (X, r) (X, a) is continuous. The topological spaces occuring in linear analysis are almost always Hausdorff spaces, and we often consider compact Hausdorif spaces. A

Chapter 2: Normed spaces and bounded linear operators

21

topology r on a set X is a Hausdorff topology if for any two points and U,, such that x E U, and x,y E X there are disjoint open sets y E Ui,. A topological space (X, r) is compact if every open cover has a where each U,, is an finite subcover, i.e. if whenever X = U,,E:f. open set, then X U,,EF U,, for some finite subset F of r. A subset A of a topological space (X, r) is said to be compact if the topology on A induced by r is compact. Every closed subset of a compact space is compact, and in a compact Hausdorff space a set is compact 1ff it is closed.

It is immediate that if K is a compact space and f: K —÷

R

is continu-

ous then f is bounded and attains its supremum on K. Indeed, if we hadf(x) <s = sup{f(y): yE K}for every XE K then E

U

K: f(x)
r<s

would be an open cover of K without a finite subcover.

Every normed space X is a metric space with the induced metric d(x,y) = lix—yll. Conversely, given a metric d on a vector space X, setting lixil = d(x,O) defines a norm on X 1ff d(x,y) = d(x+z,y+z) and d(Ax,Ay) = AId(x,y) for all x,y,z X and scalar A. The induced metric in turn, defines a topology on X, the norm topology. We shall always freely consider a normed space as a metric space with the induced metric and a topological space with the induced topology, and we shall use the corresponding terminology. Let X be a normed space. By a subspace of X we mean a linear subspace Y of the underlying vector space, endowed with the norm on X (to be pedantic, with the restriction of the norm to Y). A subspace is closed if it is closed in the norm topology. Given a set Z C X, the subspace spanned by Z is

linZ

:

Zk

C Z, Ak scalar, n = 1,2,..

=

it is called the linear span of Z and it is the minimal subspace containing

z. A normed space is complete if it is complete as a metric space, i.e. if every Cauchy sequence is convergent: if C X is such that —* 0 as min{n,m} then converges to d(xn,xm) = some point x0 in X (i.e. —' 0). A complete = normed space is called a Banach space. It is easily seen that a subset of a complete metric space is complete 1ff it is closed; thus a subspace of a Banach space is complete if it is closed.

Chapter 2: Normed spaces and bounded linear operators

22

A metric space is separable if it contains a countable dense set, i.e. a countable set whose closure is the whole space. A normed space is separable if as a metric space it is separable. Most normed spaces we

shall consider are separable. Given a normed space X, the unit sphere of X is S(X) = {x E X: lixil =

1}

and the (closed) unit ball is B(X) = {x E X: lxii

IL

More generally, the sphere of radius r about a point x0 (or centre x0) is

= S(x0,r) = {x E X: iix—xoiI = r} and

the (closed) ball of radius r about x0 (or centre x0) is = B(x(}. r) = {x E X: lix —xoli

r}.

Occasionally we shall need the open ball of radius r and centre x0.

= D(x0, r) = {x E X: lix —xoii < r}.

Note that the sets x + tB(X) (t > 0) form a neighbourhood base at the point x. is The definition of the norm implies that Br(X0) is closed, is IfltBr(Xü) = Dr(X0). The open, and for r> 0 the interior of closure of äBr(Xø) of

Dr(Xø) = Br(XO). the sphere Sr(X0). is

is

Furthermore, the boundary

The definition also implies that if X is a normed space then the map Xx X —p X given by (x, y) '—f x + y is uniformly continuous. Similarly, Ax is continuous. X) given by (A,x) —, X (or CXX the map is also Continuous. from X to Note that the norm function fi Let us give a host of examples of normed spaces. Most of these are important spaces, whilc some othcrs are presented only to illustrate the definitions. The vector spaces we take are vector spaces of sequences or and functions with pointwise addition and multiplication: if x = funcg(t) are = f(t) and then Ax +/i.y = g = (Ax, if f y= = Af(t) tions then .

the examples below, and throughout the book, various sets are often assumed to be non-empty when the definitions would not make sense otherwise. Thus S 0 in (ii), T 0 in (iii), and so on. In

Chapter 2: Normed spaces and bounded linear operators

23

Examples 1. (i) The n-dimensional Euclidean space: the vector space is

or C" and the norm is lixil =

where

x=

(x1

In

\1/2

\=i

/

(

,.. . ,x,,). The former is a real Euclidean space, the latter

is a complex one.

(ii) Let S be any set and let be the vector space of all bounded let scalar-valued functions on S. For f E = suplf(s)I.

Ilfil

sES

This norm is the uniform or supremum norm.

(iii) Let L be a topological space, let X = C(L) be the vector space of all bounded continuous functions on L and set, as in example (ii), = supjf(OI.

Ilfil

tEL

(iv) This is a special, but very important, case of the previous example. Let K be a compact Hausdorff space and let C(K) be the space of continuous functions on K, with the supremum norm IlfO

=

=

:

x E K}.

K is compact, If(x) is bounded on K and attains its supremum. (v) Let X be R" or C" and set

Since

IIxII

=

Ix,

This is the space li"; the norm is the l1-norm.

Also, = max IxaI is

a norm, the (vi) Let 1 p

the space it gives is lx".

For x = (x1,. . . ,x,,) IIXIIp

1k"

(or C") put

kkl.0). =

(real or complex); the norm is the The notation is consistent with that in example (v) for li" and also, in a natural way, with that for (see Exercise 6). Note that I"' is exactly This defines the space l,'

the n-dimensional Euclidean space.

Chapter 2: Normed spaces and bounded linear operators

24

(vii) Let X consist of all Continuous real-valued functions f(t) on that vanish outside a finite interval, and put If(t)I dt.

j-

=

(viii) Let X consist of all continuous complex-valued functions on

[0,1] and forfE Xput

/ rl 11f112 = I j

\0

(ix) For

1

p<

\1/2 If(t)12 dt

the space i,,

consists of all scalar sequences

x = (x1,x2,...) for which \I/P /

\i=1

The norm of an element x

is

i/p IIXIIp

The space

=

consists of all bounded scalar sequences with = sup 1x11

and

Co

is

the space of all scalar sequences tending to 0, with the same As remarked earlier, we have both real and complex forms

norm of these spaces. the space (x) For I p measurable functions on [0, 1] for which

/

= (j

1)

consists of those Lebesgue

\iIP d:}

/

Note that L1(0, 1) is precisely the space of integrable functions. Strictly speaking, a point of 1) is an equivalence class of functions, two functions being equivalent if they agree almost everywhere.

Putting it another way, fi and f2 are equivalent (and so are considered = 0. to be identical) if (xi) By analogy with the previous examples, the astute reader will guess that L(0, 1) Consists of all essentially bounded Lebesgue measurable functions on [0, 11, i.e. those functions f for which

Chapter 2: Normed spaces and bounded linear operators

25

= esssuplf(t)I Recall that the essential supremum esssuplf(r)I is defined as

S C [0, 1] and [0, 1]\S has measure 0} = inf{a: the set {t: 11(1)1 > a) has measure 0}, where f IS

is

the restriction off to S and so =

5).

s

1) consists of equivalence classes of

Once again, strictly speaking

functions bounded on [0, 1].

Lc,(a,co), etc., are defined similarly to

(xii)

the spaces in (x) and (xi). Of these, once again, the Hubert spaces etc., are the most important. 1) consist of the functions f(t) on (0, 1) having n con(xiii) Let tinuous and bounded derivatives, with norm iiiii =

0<

sup{±

t

<

(xiv) Let X consist of all polynomials f(t) k=O

of degree at most n, with norm Ilfil

(k+1)lckI.

=

(xv) Let X be the space of bounded continuous functions on (—1, 1) which are differentiable at 0, with norm If'(O)I

+ 'gIfWI.

(xvi) Let X consist of all finite trigonometric polynomials

f(t)

= k—n

(n = 1,2,...)

cke,

with norm 1/2

ICkI

11111 = ( —n

Chapter 2: Normed spaces and bounded linear operators

26

(xvii) Let V be a real vector space with basis (e1,e2), and

for x E V

define

lixil = inf1iai + lbI + id:

(xviii) Let V be

as

a,b,c

E

and x =

ae1+be2+

+ e2)

in (xvii) and set

114

andx (xix) V1 ,

L'2,. .

Let .

,

V

be

an

= ae1+be2+

c(e1 +e2)+d(e1 —e2)

n-dimensional real vector space and let

be vectors spanning V. For x E V define lxii

=

Ic1 I: c,

E

and x =

c.vl}.

Let us prove that the examples above are indeed normed spaces, i.e. that the underlying spaces are vector spaces and the functions U ii satisfy conditions (i)—(iii). It is obvious that conditions (i) and (ii) are satisfied in each example so we have to check only (iii), the triangle inequality. Then it will also be clear that the underlying space is a vector space. Furthermore, the triangle inequality is obvious in examples (ii)—(v), (vii), (xi) and (xii)—(xv). In the rest, with the exception of the last three examples, the triangle inequality is precisely Minkowski's inequality in one of its many guises. Thus Theorem 1.7 (Minkowski's inequality for sequences) is just the triangle inequality in 1: ilx+ylip

IIxIlp+Ilyllp

(1)

x,y 1,?; this clearly implies the analogous inequality in Theorem 1.9 (Minkowski's inequality for functions) is precisely the tri1): angle inequality in whenever

llflip +

(2)

1), etc. In turn, the triangle inequality implies that the underlying spaces are indeed vector spaces. Examples (xvii)—(xix) are rather similar, with (xix) being the most general case; we leave the proof to the reader (Exercise 7). 0 whenever f, g E

The spaces I,, (1 p oo) are the simplest classical sequence spaces, 1) (1 and the spaces C(K) and are the simplest classical p

Chapter 2: Normed spaces and bounded linear operators

27

function spaces. These spaces have been extensively studied for over

eighty years: much is known about them but, in spite of all this attention, many important questions concerning them are waiting to be settled.

In many ways, the most pleasant and most important of all

infinite-dimensional Banach spaces is '2' the space of square-summable is the canonical example of a Hilberi space. sequences. This space Similarly, of the finite-dimensional normed spaces, the space 1$ is central: this is the n-dimensional Euclidean space. The spaces in Examples (i)—(vi), (ix)—(xiv) and (xvii)—(xix) are com-

plete, the others are incomplete. Some of these are easily seen, we shall see the others later. Let us remark that if X is a normed space then its completion X as a

metric space has a natural vector space structure and a natural norm. Thus every normed space is a dense subspace of a Banach space. We shall expand on this later in this chapter. Having mentioned the more concise formulations of Minkowski's inequality (inequalities (1) and (2)), let us draw attention to the analogous formulations of Holder's inequality. Suppose p and q are conjugate 1 indices, with p E then Y=

and q =

k=1

Similarly, 1ff E

permitted.

kkYkl

1) and g E ilfgiii

=

If x =

E I,,

and

iiXiipiiYiiq.

Lq(O,

lfgj th

1) then fg is integrable and lifilpiigiiq.

It is easy to describe the general form of a norm on a vector space V in terms of its open unit ball. Let X = (V, fi) be a normed space and let D = {x E V: lxii < 1) be the open unit ball. Clearly is determined by D: if x 0 then lxii = inf{t: t > 0, x E :D}. The set D has the following properties; (1) if x,y E D and IAI+ E (ii) if x C D then x+€D C D for some e = E(x) > 0; (iii) for x V (x 0) there are non-zero scalars A and such that fi

Ii

Ax C D and D satisfying (i) is said to be absolutely convex: this property is a consequence of the triangle inequality. Property (ii) follows from the fact that D is an open ball, while (iii) holds since Dxli < for every x and = 0 1ff y = 0.

Chapter 2: Normed spaces and bounded linear operators

28

Conversely, if D C V satisfies (i)—(iii) then

q(x) = inf{z: :> 0, x E :D}

defines a norm on V, and in this norm D is the open unit ball. The function q(x) is the Minkowski functional determined by D. Note that in conditions (i)—(iii) we do not assume any topology on V. When studying normed spaces, it is often useful to adopt a geometric point of view and examine the geometry, i.e. the 'shape', of the unit ball. Much of the material presented in this book concerns linear function-

als and linear operators. Let X and V be normed spaces over the same ground field. A linear operator from X to Y is a linear map between the underlying vector spaces, i.e. a map T: X Y such that T(A1x1+A2x2) = A1T(x1)+A2T(x2)

for all x1,x2 E X and scalars A1 and A2. The vector space of linear

operators from X to Y is denoted by

Y).

The image of T is

Im T = {Tx: x E X} and the kernel of T is Ker T = {x: Tx = 0}. Clearly Ker T is a subspace of X and Im T is a subspace of V. Furthermore, T is a vector-space isomorphism if Ker T = {0} and Im T = V.

A linear operator T E £e(X, Y) is bounded if there is an N> 0 such that IITxII

We shall write

NIIxII

for all x E X.

Y) for the set of bounded linear operators from X

X) is the set of bounded linear operators on X. Y) is a vector space. The operators in Y) are said to be unbounded. A linear functional on X is a linear operator from X into the scalar field. We write X' for the space of linear functo Y;

=

Clearly

tionals on X, and X* for the vector space of bounded linear functionals on X. It is often convenient to use the bracket notation for the value of a functional on an element: for x EX and f€X' we set = (x,f) = f(x). This is in keeping with the fact that the map XXX' R (or C) given by (x, f) f(x) is bilinear.

Theorem 2. Let X and Y be normed spaces and let T: X—+ V be a linear operator. Then the following conditions are equivalent: (1) T is continuous (as a map of the topological space X into the topological space Y);

(ii) T is continuous at some point x0 E X; (iii) T is bounded.

Chapter 2: Normed spaces and bounded linear operators Proof. The implication (1)

29

(ii) is trivial.

(iii). Suppose T is continuous at x0. Since Tx0+ B(Y) is a (ii) neighbourhood of T(x0), there is a 6 > 0 such that

x0+y

x

Tx =

x0 + ÔB(X)

Hence

& implies IITyII (iii) (i). Suppose lITxII we have IITx— Tyll <€.

Tx0 +

Ty E Tx0+ B(Y).

and so IITzD for all z. NIIxII for all x E X. Then if lix—yll < 1

0

Two normed spaces X and Y are said to be isomorphic if there is a Linear map T: X Y which is a topological isomorphism (i.e. a

bomeomorphism). We call X and Y isometrically isomorphic if there is

a linear isometry from X to Y, i.e. if there is a bijective operator

Y) such that T' E X. TE and IITxII = IIxII for all x Two norms, and 112' on the same vector space V are said to be equivalent if they induce the same topology on V, i.e. if the formal identity map from X1 = (V, to X2 = (V, 1112) is a topological isomor11

phism. As an immediate consequence of Theorem 2, we see that if a linear map is a topological isomorphism then both the map and its inverse are bounded.

Corollary 3. Let X and Y be normed spaces and let T E Y). Then X). T is a topological isomorphism if T E Y) and T' E Two norms, on the same vector space V are equivalent and V if there are constants c, d> 0 such that dlxii

dIixiI

i

for all x E V.

It is immediate that equivalence of norms is indeed an equivalence relation, and Corollary 3 implies that if a normed space is complete then it is also complete in every equivalent norm. The equivalence of norms has an intuitive geometrical interpretation

in terms of the (open or closed) unit balls of the normed spaces. Let and lb be norms on V, with closed unit balls B1 and B2. Then II. and 112 are equivalent if and only if B2 C cB1 and B1 C dB2 for some constants c and d. Indeed, these relations hold if and only if dlixiJj for all x E V. lxiii ciIxiI2 and iixIi2 It is clear that Y) is a subY) and Xt are vector spaces: R). In fact, they are also space of Y) and X* is a subspace of normed spaces with a natural norm. The operator norm or simply norm fl

Chapter 2: Normed spaces and bounded linear operators

30

Y) is given by

on 11111

= inf{N>O:

Niixii for alix E X} = sup{iiTxIi:

liTx!i

1}.

ilxii

Although this gives us the definition of the norm of a bounded linear functional as well, let us spell it out: the norm of f E is 11111

= inf{N >

f(x)I

0:

NIIXII

for

all

xE X} = sup{If(x)l:

IIxU

1).

Note that in these definitions the infimum is attained so Ii

Txfl

lxii

II

and

11(x) I

for all x.

lifli lixil

The terminology is justified by the following simple result. defined above, is a norm on Y). In particular, is

Theorem 4. The function fl

Y is complete then so is

Y). If

a complete

normed space.

Proof. Only the second assertion needs proof. Let be a Cauchy sequence in Y). Then is a Cauchy sequence in Y for every xE Xand so there is a uniquey E Ysuch that T,1x—*y. Set Tx = y. To complete the proof, all we have to check is that T E Y) and Given x1 , x2 E X and scalars

and A2, we have

T(A1x1+A2x2) = tim

= lim

TE

lim

Y).

Furthermore, given e > 0, there is an n0 such that n,m Then for x C X and m n0 we have

ii

=

= iihm(Tn—Tm)xIi = eiixiI.

Theref ore

—

ii <€ if

Chapter 2: Normed spaces and bounded linear operators

(C + TmII)

€IIxIl

and so T E

Y).

31

Finally, from the same inequality we find that

0 If we extend the operator norm to the whole of

Y) by putting unbounded operator, then = sup{I!TxII: lixil Y) having finite norm. Y) consists of the operators in is called the dual of X. For T E The Banach space Y) and g E r define a function rg: X (or C) by for

1} =

an

(rg) (x) =

g(Tx).

Then rg is a linear functional on X and r: Y* X* is easily seen to be a linear map, Furthermore, T* is not only in but is, fact, a bounded linear operator since = so

that

Irgil The

iirii

and

IITU

liii.

operator T* is the adjoin: of T. In fact, as we shall see later

(Theorem 3.9),

ijrii = liii.

definition of the adjoint looks even more natural in the bracket notation: rg is the linear functional satisfying The

(rg,x) =

(g,Tx).

Given maps T: X Y and S: Y —' Z, we can compose them: the map ST: X—' Z is given by (ST)(x) = (SOT)(x) = S(T(x)). IfS and T are linear then so is ST; if they are bounded linear operators then so is ST.

Theorem 5. Let X, V S

(F',

and

Z

be

normed spaces and let T€

(X,

F),

Z). Then ST: X —' Z is bounded linear operator and 1ISTII

ilSPlilllI.

= X) is closed under multiplication (i.e. under the composition of operators). In particular,

Proof. Clearly ll(ST)(x) II = IIS(Tx)

ilTxlI

ilSllil TII xli.

0

Chapter 2: Normed spaces and bounded linear operators

32

The last result shows that if X is a Banach space then is a unital Banach algebra: it is an algebra with a unit (identity element) which is also a Banach space, such that the identity has norm 1 and the norm of the product of two elements is at most the product of their norms.

Examples 6. (i) Define T:

by putting where 1 = min{n,m}.

Then T E and TO = 1. (ii) If A is any endomorphism of R" then A E q (1 p,q (iii) Define S E by

1) for all p and

Sx = (O,x1,x2,...). This is the right shift operator. The left shift T E is defined by Tx = (x2,x3,...).

Clearly S is an injection but not a surjection, T is a surjection but not an injection, IISII = 11Th = 1, and TS = 1 (the identity operator) but 1: KerST= {(x1,O,...):x1 E R}. = (iv) Let 1) be the normed space of functions on (0, 1) with k continuous and bounded derivatives, as in Examples 1 (xiii). Let D: —p C°" be the differentiation operator: Df = f'. Then ODD

= 1.

(v) Let T:

be

the formal identity map Tf = f. Then

11111 = 1. (vi) In order to define linear functionals on function spaces, let us

introduce the following notation. Given sequences x = (Xk)° and let (x,y) = and let y=

xy E then (x,y) is well defined. Inequality (3), i.e. HOlder's inequality for sequences, can now be restated once again in the following concise form: if p and q are conjugate indices, with 1

E I,,, andy E 'q' thenxy El1 and (5)

Inequality (5) implies that for y E bounded linear functional on 1,, and and for 1

identifies

the function hIYIIq•

In fact,

= (x,y) is a =

is a linear isometry which p < x the correspondence y with i (see Exercise 1 of the next chapter). The dual of

Chapter 2: Normed spaces and bounded linear operators

33

contains as a rather small subspace (L and are not separable but is); however, is the dual of c0, the closed subspace of L consisting of sequences tending to 0 (again, see Exercise I of the next chapter). (vii) Let p and q be conjugate indices and g E Lq(O, 1). Define, for

fE

40gW

fg

=

is a bounded linear functional on

Then, by Holder's inequality (4),

In fact, 1I40g11 = IIgIIq' and for 1 the is a linear isometry which identifies Lq(O, 1) With

and 1140gI1

IIgIIq.

correspondence g 1)*.

(viii) Let C[O, 1] be

the space of continuous functions with the

supremum norm: Ilfil = 1 = max{If(t)I: 0 t 1}. Let 0 and define CEO, 1] R by is a bounded = 1(b). Then linear functional of norm 1. (ix) Let X be the subspace of CEO, 1] consisting of differentiable functions With continuous derivative. Then D: X C[O, 1], defined by Df = f', is an unbounded linear operator.

(x) Let V be the vector space of all scalar sequences x = = (V, where finitely many non-zero terms. Set \l/P

(Xk

with

/ and IIxIL,

= max{IxkI: 1

k <

r,s

be the formal identity map: TrsX = let Trs: Xr is a linear operator for all r and s. If 1 r s then T,5 is bounded: in fact, = 1, but if r > s then is unbounded

For

1

x. Then

(see Exercise 27).

0

As promised earlier in the chapter, let us say a few words about completions. Every metric space has a unique completion, and if the metric is induced by a norm then this completion is a normed space. Before examining the completion of a normed space, let us review briefly the basic facts about the completion of a metric space. A metric space X is said to be the completion of a metric space X if X is complete and X is a dense subset of X. (A subset A of a topological

space T is dense if its closure is the whole of T.) The completion is

Chapter 2: Normed spaces and bounded linear operators

34

unique in the sense that if X is a dense subset of the complete metric spaces Y and Z then there is a unique continuous map q: Y Z whose restriction to X is the identity, and this map is an isometry onto Z. This is easily seen since if y E Y then y is the limit (in Y) of a sequence C X. Then is a Cauchy sequence in Z and so it has a limit = z z Z. Since ç is required to be continuous, we must have and so ç, if it exists, is unique. On the other hand, if we define a map Y Z by setting ç(y) = z, then this map is clearly an isometry from Y onto Z.

The completion of a metric space is defined by taking equivalence classes of Cauchy sequences. To be precise, given a metric space X, let [X] be the set of Cauchy sequences of points of X: [X] =

E X, lim sup d(Xn,Xm) = fl—.x m

on [X] by setting

Define an equivalence relation 0,

O}.

—

and let X = [X]/— be the collection of equivalence

classes with respect to —. For 1,9 E X set

d(i,j) = is E I and [0, E 9. It is immediate that d: XxX well defined, i.e. that is independent of the representatives chosen. Furthermore, if C I, i then 9 and

where

d(i, 1) = lim

lim

,

+

,

=

= d(i,9)+d(9,i). Finally,

x = y.

= 0, and if

d(i,9)

= 0 then

so

that

-

metric space (X,d) is complete. Indeed, let 1(k) (k = 1,2,...). Let sequence in Xand let The

<2-k

be a Cauchy be such that

if n,m

It is easily checked that C [X] and that 1(k) tends to the equivalence class of tbe sequence (Xk Writing [x] for the equivalence class of the constant sequence x,x,..., we find that the map X—* X, given by x [x], is an isometry. Set Xk =

Thus, with a slight abuse of terminology, X can be considered to be a subset of X. Clearly, X is a dense subset of X, since if is a

Chapter 2: Normed spaces and bounded linear operators

n0 then

<E whenever n, m

representative of I E X and d(x,, , d(i,x,,0) = d(i,

35

If X is not only a metric space but also a normed space then its completion has a natural Banach-space structure. Theorem 7. For every normed space X there is a Banach space X such that X is dense in X. This space X is unique in the sense that if X is a dense subspace of a Banach space X then there is a unique continuous map X X such that the restriction of to Xis the identity; furthermore, this map ç is a linear isometry from X to X. Proof. Considering X as a metric space, with the metric induced by the

norm, i.e. with d(x,y) = lIx—yiI, let X be its completion. Given x,y X, let x,, x andy,, —' y, where x,,,y,, E X. Then, for scalars A is a Cauchy sequence. Now, if Xis to be given a normed-space structure such that the norm on X induces the metric on X, then tim = A limx,,+p limy,, = and

the sequence (Ax,,

Ax

+.&y must be defined to be the limit of the Cauchy sequence

It is easily checked that with this definition of addition and scalar multiplication X becomes a vector space. Furthermore, is a norm on this vector space, turning X lull = d(x,O) = (Ax,,

into a Banach space. The remaining assertions are clear.

0

In fact, Theorem 7 is also an immediate consequence of Theorem 3.10. The space X in this result is called the completion of the normed space X. In view of

Theorem 7, if X is a dense subspace of a Banach space Y then Y is often regarded as the completion of X. For example, the space C[0, I] of continuous functions on [0, 1] is the completion of the space of piecewise linear functions

on [0, 1J with supremum norm =

:0

x

1} = max{lf(x)I :0

x

1}.

The same space CEO, 1] is also the completion of the space of polynomi-

als and of the space of infinitely differentiable functions. Similarly, for 1) is the completion of the space of polynomials with 1 p< the norm = and

(L'

dx)

also of the space of piecewise linear functions with the same norm.

Chapter 2: Normed spaces and bounded linear operators

36

When working in Banach spaces, one often considers series. Given a normed space X, a series

(x,1.€X)

n1 is said

to the convergent to x E X if N

n1

x,—x

i.e. if N

=0.

urn x—

N-."

ii=1

We also say that x is the sum of the x,, and write

xn.

n1 A series is said to the absolutely convergent if In a Banach space X every absolutely convergent series

< x,, is

be the Nth partial sum. We convergent. Indeed, let YN = have to show that is a Cauchy sequence. Given e > 0, choose an <€. Then for n0 N < M we have n0 such that n—no M Ii

=

M IIXnII

<6.

nN+1 nN+1 is indeed a Cauchy sequence and so converges to some Hence x E X, so that x = In an incomplete space X the assertion above always fails for some absolutely convergent series x,: this is often useful in checking whether a space is complete or not. Before showing this, let us mention a very useful trick when dealing with Cauchy sequences in metric spaces: we may always assume that we are dealing with a 'fast' Cauchy sequence. To be precise, given a Cau)°, a priori all we know is that there is a sequence chy sequence = 2k, such that d(Xn,Xm) < 1/k if n,m However, say we may always assume that our sequence is much 'better' than this: fl, Oi d(xn,xm) <2-2 if m n, or whatever d(Xn,Xm) <2's if m = f(1) and suits us. Indeed, let f(n) > 0 be an arbitary function. Set choose an n1 such that d(xn,xm) <Ej whenever n,m n1. Then set €2 = f(2) and choose an n2 > n1 such that ,Xm) <€2 whenever

Chapter 2: Normed n, m

spaces

and bounded linear operators

37

n2. Continuing in this way, we find a sequence n1
x,)
we find that the is such that d(yk ,y,)
n

,

k. Setting Yk =

no we have ,

y) =

since Yk = x,1 and

that a Cauchy sequence indeed assume that ,

urn Yk)

k—.oo

urn sup

,

Yk)

k is sufficiently large. Thus, in proving in a metric space is convergent, we may say, if m n.

Theorem 8. A normed space is complete if and only if every absolutely convergent series in it is convergent.

Proof. We have already seen that in a Banach space every absolutely convergent series is convergent. Suppose then that in our space X every absolutely convergent series is convergent. Our aim is to show that every Cauchy sequence is convergent. Let be a Cauchy sequence in X. As we have just seen, we may for n m. Set x0 = 0 and assume that d(x1 , = 111n — Xml (k 1). Then Yk = XkXk_1 Yk and IJYkfl <2_k41 There=

fore the series x1 ,

Yk is absolutely convergent and has partial sums By assumption, is convergent, say to a vector x,

0 Absolutely convergent series in Banach spaces have many properties analogous to those of absolutely convergent series in For example, if x,, is absolutely convergent to x and n1 , n2,... is a permutation of 1,2,... then x,,. is also absolutely convergent to x. However, as we shall see in a moment, unlike in this property does not characterize absolutely convergent series in a Banach space. The use of series often enables one to identify a Banach space with a

sequence space endowed with a particular norm. This identification is made with the aid of a basis, provided that the space does have a basis. Given a Banach space X, a sequence (e1)' is said to be a (Schauder) basis of X if every x E X can be represented in the form x = A-e1 (A1 scalar) and this representation is unique. For 1 p
Chapter 2: Normed spaces and bounded linear operators

38

ith term is

and the others are 0. It

1

(x1 , x2,...)

I,,

only if

lxi

is

easily seen that if x =

then x =

xe1 and this representation is unique (see Exercise 11). In particular, x, e. is convergent (in I,,) if and This implies that in 12 every rearrangement of e,/n is convergent (to the same sum), but the series is not abso-

lutely convergent.

To conclude this chapter, let us say a few words about defining new spaces from old. We have already seen the simplest way: every (algebraic) subspace Y of a normed space X is a normed space, with the restriction of the norm of X to Y as the norm. If X is complete then Y is complete if and only if it is closed. As the intersection of a family of subspaces is again a subspace, for every set S C X there is a unique smallest subspace containing S, called the linear span of S and denoted by tin 5: it is the intersection of all subspaces containing S and also the set of all (finite) linear combinations of elements of S.

linS = fl{W: WisasubspaceofXandSC W} s1,. ..,s,, =

€5; n =

the closed linear span of S, denoted lin 5, is the unique smallest closed subspace containing S; it is the intersection of all closed subspaces containing S, and is also the closure of tinS, the linear span of S. Let us turn now to quotient spaces. Given a vector space X and a subspace Z, define an equivalence relation — on X by setting x y if Similarly,

y E Z. For x C X, let [xl be the equivalence class of x: putting it another way, [x] = x+Z. Then X/— = {[x]: x E X} is a vector space, with vector-space operations induced by those on X: A[x] +p.[y] = [Ax + ky]. Note that [x] = 0 if x E Z. Now if X is a normed space and Z is a closed subspace of X then we can define a norm fi on X/Z x—

lb

by setting Il[x]ilo = inf{IIyii : y

x} = inf{lIx+zlI : z E Z}.

It is easily checked that 11110 is indeed a norm on X/Z: the homogeneity and the triangle inequality are obvious and Ii [0] lb = 0. All that remains is to show that if [x] 1 = 0 then [x] = 0. To see this, note that if [xlii = 0 then lix — fi —' 0 for some sequence (zn) C Z. Hence z,,

x and so, as Z is closed, x C Z.

We call X/Z, endowed with the quotient normed space of X by Z, and call the quotient norm. Throughout this book, a quotient 11

lb

Chapter 2: Normed spaces and bounded linear operators

39

space of a normed spaces will always be endowed with the quotient norm.

Given normed spaces X and Y, and a. bounded linear operator T: X Y, the kernel Z = Ker T = T'(O) of T is a closed subspace of X, and T induces a linear operator T0: X/Z —+ Y. Analogously to many standard results in algebra, we have the following theorem.

Theorem 9. Let X and Y be normed spaces, T E Y) and Z = Ker T. Let T0: X/Z Y be the linear map induced by T. Then T0 is a bounded linear operator from the quotient space X/Z to Y, and its norm is precisely 11711. Proof. H

To[x] II

> 11111

II

1171111 [xJ

let e > 0 and choose an x E X such that

IITxIl > 11711—c.

Hence

x} =

11111.

II

Conversely,

:y

inf{II ill

Then

II[x}H

1

and

lixil

= 1 and

IIT0[x]Il = IIT1V > 11711—c.

0

—

In fact, the quotient norm II 110 on X/Z is the minimal norm on X/Z such that if T E Y) and Ker T 3 Z, then the operator T0: X/Z Y induced by T has norm at most 11711 (see Exercise 13).

Suppose that X and Y are closed subspaces of a normed space Z, with Xfl Y = {0} and X+ Y = Z. If the projections Px: Z —* X, and py: Z —p Y, given by = x and py(x,y) = y, are bounded (i.e. continuous) then we call Z a direct sum of the subspaces X and V. It is easily seen that Z is a direct sum of its subspaces X and Y if the topology on Z (identified with Y = {(x, y): x E X, y E Y}) is precisely the product of the topologies on X and V. Note that, if Z is a direct sum of X and V. Z' is a direct sum of X' and V'. and X is isomorphic to

X' and Y is isomorphic to Y', then Z is isomorphic to Z'. If Z is a direct sum of X and Y then the projection Px: Z —. X induces an isomorphism between Z/Y and X. Conversely, given normed spaces X and V. there are various natural ways of turning Y, the algebraic direct sum of the underlying vector spaces, into a normed space. For example, for 1 p we may take the norm Thus for 1 p < we take = Il(lIxII, =

and

for p =

we

define

=

It is easily seen that all these norms are equivalent; indeed, each induces the product topology on XEPJ V. The normed max{IlxII, IlylI}.

Chapter 2: Normed spaces and bounded linear operators

40

space (X$ Y,

usually denoted by X Y. Considering X and V Y is a direct sum of X and V. as subspaces of X$,, Y, we see that Finally, given a family {II II,.: y C fl of norms on a vector space V. if is

IIxII

=

a norm on V. Note that the analogous assertion about the infimum of norms does not hold in general (see for every x C V. then

fl

is

Exercise 15).

Having got a good many of the basic definitions under our belts, we are ready to examine the concepts in some detail. In the next chapter we shall study continuous linear functionals. Exercises

1. Show that in a normed space, the closure of the open ball Dr(X0) (r > 0) is the closed ball B,(xo) and the boundary c3Br(XO) of B,(x0) Do these statements hold in a general metric is the sphere space as well? 2. Let B1 D B2 J ... be closed balls in a normed space X, where

B,, =

B(x1, , r1,)

= {x C X: 1k1, — xli

(r1, > r > 0).

r1,}

Does

n1 B1,? Is there a ball B(x,r) r> 0, contained in 3. Prove or disprove each of the following four statements. In a comspace every nested sequence of closed has a plete

hold?

non-empty intersection.

4. Let X = (V, il) be a normed space and W a subspace of V. Supis a norm on W which is equivalent to the restriction of pose W. Show that there is a norm II on V that is equivalent to II H

ii

and whose restriction to W is precisely I. to be two norms on a vector space Vand let Wbe a and 5. Let subspace of V that is Il-dense in V. Suppose that the restrictions necessarily to W are equivalent. Are . II and of . II and equivalent? on R1, Show that if let be the 6. For 1 p then I p< r llXIlr. For which points x do we have equality? I

I

I

Chapter 2: Normed spaces and bounded linear operators

41

Prove that for every €> 0 there is an N such that if N

7. Show that the

space defined in Examples 1 (xix)

is indeed a

nonned space.

8. Let 1

p,q,r

co

be

such that p1+q1+r1 =

defined to be 0. Show that for x, y, z

1,

with

we have

IIXIIpIIYIIqIIZIIr.

State and prove the analogous inequality for s vectors from C's. 9. Show that I,, is a Banach space for every p, (1 p and that c0, the set of all sequences tending to 0, is a closed subspace of Show also that 4 (1 p
10. Let p, q and r be positive reals satisfying p1 + q' = r1. Show that for f E 1) and g E Lq(O, 1) the function fg belongs to Lr(0,1) and IIfIIpIl8iIq.

<x. 11. Let e, = (O,...,0,1,0,...) = €4, where 1 Show that (e1,e2,...) is a basis of 4, called the standard basis, i.e. every x 4 has a unique representation in the form x=

A•e1.

For x = (x1)r E li', the support of x is suppx = 0}. Let
norm of the induced operator T0: X/Z —. Y is at most 11711. 14.

A seminorm on a vector space V is a function p: V —' such that p(Ax) = IA Ip(x) and p(x +y) p(x) +p(y) for all vectors x,y E V and scalar A. [Thus a seminorm p is a norm if p(x) = 0 implies x = O.J Let {p,.: y I) be a family of seminorms on a vector space V such that

42

Chapter 2: Normed spaces and bounded linear operators 0

sup{p7(x):

y E fl <

0). Show that p() is a norm on V. for every x E V(x 15. Let lixili and 0x112 be norms on a vector space V. Is jixil = min{11x111, 11x112} necessarily a norm? tend16. Consider the vector space c0 of complex sequences x = E c0, let be the decreasing ing to 0. For a sequence Formally, if x rearrangement of I )°. I{k: IxkI >x}j Let x =

I{k: IXkI

>0 be such that yc1b1=oo. Define, for E c0,

: m = 1,2,.

=

IIxII'

.

.

Let d0

Show that

17. For x E

is

a norm on d0. Is this space complete?

set IIxII'

=

and that 11 is complete in this II' is a norm on norm. Is IIxII' equivalent to the l1-norm 11x111 = Show that on every infinite-dimensional normed space X there is a discontinuous (unbounded) linear functional. [By Zorn's lemma X has a Hamel basis, i.e. a set {x7: y E fl C X such that every Show that

18.

= {x E c0: IIxII' < rx}

II

x E X has a unique representation in the form x =

Ax7.

19. Prove that if two norms on the same vector space are not equivalent then at least one of them is discontinuous on the unit sphere in the other norm. Can each norm be discontinuous when restricted to the unit sphere of the other? 20. Give two norms on a vector space such that one is complete and the other is incomplete. 21. Find two inequivalent norms and 1116 on a vector space V such that (V, fl fly) and (V,

•112)

are

isometric normed spaces.

22. Let Y be a closed subspace of a Banach space X and let x be an element of X. Is the distance of x from Y attained? (Is there a pOint Yo E Y such that

= inf{IIx—yII: y E Y}?)

Chapter 2: Normed spaces and bounded linear operators

43

'12'. . . ,f,, be linear functionals on a vector space V. Show that there is a norm liii on V such that each f, is continuous on (V, II•II). Can this be done for infinitely many linear functionals? And what about infinitely many linearly independent linear func-

23. Let

tionals?

)° C X is a sequence such 24k. Let X be a Banach space. Suppose that every x X has a unique representation in the form x = consists of isolated points. Prove that the set 25. Check the assertion in Examples 6(x). 26. Let Y be a closed subspace of a normed space X. Show that if X/Y and Y are separable then so is X. 27. Let Y be a closed subspace of a normed space X. Show that if any two of the three spaces I, Y and X/Y are complete then so is the third.

28. Let Y be a subspace of a normed space X. Show that Y ii a closed subspace if and only if its unit ball, B(Y), is closed in X. Show also that if Y is complete then Y is closed.

29. Prove the converse of one of the assertions of Theorem 4: if Y) is complete then so is Y.

[Thus every x E V Let V be a vector space with basis (x 0) has a unique expression in the form x =
Let

C be a closed convex set in a normed space X such that

C+B(X) D Bl+E(X) for some e > 0. Does it follow that IntC 0, i.e. that C contains a ball of positive radius? Notes

Abstract normed spaces were first defined and investigated by Stefan Banach in 1920 in his Ph.D. thesis at the University of Léopol (i.e. the Polish town of Lwôw, now in the Soviet Union). Much of this thesis was published as an article: Sur les operations dans les ensembles abstra its et leur applications aux equations intégrales, Fundamenta Mathematica, 3 (1922), 133—81. A few years later Banach wrote the

first book wholly devoted to normed spaces and linear operators: Théorie des Operations Linéaires, Warsaw, 1932, vii + 254 pp. Banach gave an elegant account of the work of many mathematicians involved in the creation of functional analysis, including Frédéric (Frigyes) Riesz,

whom he quoted most frequently, Just ahead of himself, Maurice

Chapter 2: Normed spaces and bounded linear operators

44

Fréchet, Alfred Haar, Henri Lebesgue, Stanislaw Mazur, Juliusz Schauder and Hugo Steinhaus. It is perhaps amusing to note that, when

writing about Banach spaces, Banach used the term 'espace du type (B)'.

This beautiful book of Banach has had a tremendous influence on functional analysis; it is well worth reading even today, expecially in its English translation: Theory of Linear Operators (translated by F. Jellett), North-Holland, Amsterdam, 1987, ix + 237 pp. This edition is particularly valuable because the second part, by A. and Cz.

Bessaga (Some aspects of the present theory of Banach spaces, pp. 161—237), brings the subject up to date, with many recent results and references.

Another classic on functional analysis is F. Riesz and B. Sz.-Nagy, Functional Analysis (translated from the 2nd French edition by L. F. Boron), Blackie and Son Ltd., London and Glasgow, 1956, xii + 468 pp. This volume concentrates on the function-theoretic and measuretheoretic aspects of functional analysis, so it does not have too much in common with our treatment of linear analysis.

There are a good many monographs on normed spaces and linear operators, including the massive treatise by N. Dunford and J.

Schwartz, Linear Operators, in three parts; Interscience, New York; Part I: General Theory, 1958, xiv + 858 pp.; Part II: Spectral Theory, Self Adjoint Operators in Hubert Space, 1963, ix ÷ 859—1923 pp. + 7 pp. Errata; Part LII: Spectral Operators, 1971, xix + 1925—2592 pp., L. V. Kantorovich and G. P. Akilov, Functional Analysis in Normed Spaces, Pergamon Press, International Series of Monographs on Pure and Applied Mathematics, vol. 45, Oxford, 1964, xiii + 771 pp., A. N. Kolmogorov and S. V. Fomin, Introductory Real Analysis (translated and edited by R. A. Silverman), Prentice-Hall, Inc., Englewood Cliffs, N. J., 1970, xii + 403 pp., Mahlon M. Day, Normed Linear Spaces, Third Edition, Ergebnisse der Mathematik und Ihrer Grenzgebiete, vol. 21, Springer-Verlag, Berlin, 1973, viii + 211 pp. and J. B. Conway, A Course in Functional Analysis, Graduate Texts in Mathematics, vol. 96, Springer-Verlag, New York, 1985, xiv + 404 pp.

3. LINEAR FUNCTIONALS AND THE HAIIN-BANACH THEOREM

let us write X' for the algebraic Given a normed space X = (V, dual of X, i.e. for the vector space V' of linear functionals on V. Thus X*, the space of bounded linear functionals on X, is a subspace of the II

vector space X'. We know from the standard theory of vector spaces that every independent set of vectors is contained in a (Hamel) basis (see Exercise 18 of Chapter 2). In particular, for every non-zero vector u E V there is a linear functional fE V withf(u) 0. Equivalently, V is a large enough to distinguish the elements of Even V: for all x, yE V(x y) there is a functionalfE V such thatf(x) more, the dual V' of V is large enough to accommodate V: there is a natural embedding of V into V' which is an isomorphism if V is finite-dimensional.

But what happens if we restrict our attention to bounded linear func-

tionals on a normed space X? Are there sufficiently many bounded linear functionals to distinguish the elements of X? In other words, given an element x E X (x 0) is there a functional f E r such that f(x) 0? As X is a normed space, one would like to use X* to obtain some information about the norm on X. So can we estimate lixII by f(x) for some f E r with lifli = 1? To be more precise, we know that for every x E X, IIxll

sup{jf(x)I: f E

But is the right-hand-side comparable to lix II? As a matter of fact, so far we do not even know that for every non-

zero normed space there is at least one non-zero linear functional, i.e.

we have not even ruled Out the utter indignity that X* = {0} while X = (V, fi) is large, say V is infinite-dimensional. The main aim of this chapter is to show that, as far as the questions fi

above are concerned, Candide and Pangloss were right, tout est pour le mieu.x dans le meilleur des mondes possibles; indeed, everything is for 45

46

Chapter 3: Linear functionals and the Hahn—Banach theorem

the best in the world of bounded linear functionals. Before we present the result implying this, namely the Hahn—Banach theorem, we shall point out some elementary facts concerning linear functionals.

Let us show that f E X' is bounded if and only if f(B) is not the entire ground field. Let B = B(X) = B(O, 1) be the closed unit ball of 1 then AB = B(O, Al) C B. Hence for f X' we have X. If Al where llfll = if f is suplf(B)I = sup{If(x)l: lixil 1} = unbounded, and Af(B) = f(AB) C f(B). Consequently f(B) is either {A : IAI < lIfIl} or {A : IAI llfO}. Similarly, if D = D(X) = D(O, 1) is

{A: IAI < llfll} for all f E X' Hence f E X' is bounded 1ff f(B) is not the entire ground

the open unit ball of X then f(D) = (f

0).

field, as claimed.

it is often useful to think of a (non-zero) linear functional as a hyperplane in our vector space. An affine hyperplane or simply a hyperplane H in X is a set

H = {x0}+ Y = {x0+y: y

Y},

where x0 X and Y C X is a subspace of codimension 1, i.e. a subspace with dim X/Y = 1. We say that H is a translate of Y. Given a non-zero functional f E X', let K(f) = f'(O) = {x E X: f(x) = 0} be

the null space, i.e. the kernel, of f and let

1(f) =

= (XE X: f(x) =

1}.

Let us recall the following simple facts from elementary linear algebra.

Theorem 1. Let X be a (real or complex) vector space. (a) If f E X' and f(x0) 0, then K(f) is a subspace of codimension 0 then every vector x E X has a unique 1. Moreover, if f(x0) representation in the form x = y + Ax0, where y K(f) and A is a scalar.

Furthermore, 1(f) is a hyperplane not containing 0. (b) If f,g E X'—(O} thenf = Ag 1ff K(f) = K(S). (c) The map f '—* 1(f) gives a 1—1 correspondence between non-zero

0

linear functionals and hyperplanes not containing 0.

Continuous (i.e. bounded) linear functionals are easily characterized in terms of K(f) or 1(f). Note that Ilfil 111 and only if lf(x) < 1 for all x with 11111 < 1, i.e. if and only if 1(f) is disjoint from the open unit ball D(0,1) = {x X: llxll <1). I

Chapter 3: Linear functionals and the Hahn—Banach theorem

47

A subset A of a topological space T is nowhere dense if its closure has empty interior.

Theorem 2. Let X be a (real or complex) nonned space.

(a) Let f E X', (f

0).

If f is continuous (i.e. f E

then K(f)

and 1(f) are closed and nowhere dense in X. If f is discontinuous (i.e. unbounded) then K(f) and 1(f) are dense in X. (b) The map f 1(f) gives a 1—1 correspondence between non-zero bounded linear functionals and closed hyperplanes not containing 0.

Proof. (a) Suppose that f is continuous.

Then K(f) and 1(f) are

closed, since they are inverse images of closed sets. If f(x0) 0 then 0. Hence K(f) and 1(f) have empty interiors, f(x) f(x + fx0) for and so are nowhere dense. Suppose now that K(f) is not dense in X, say B(x0, r) fl K(f) = 0 for some x0 E X and r> 0. Then f(B(xo, r)) = f(x0) +rf(B(X)) does not contain 0, so f(B(X)) is not the entire ground field. Hence, as we have seen, f is bounded.

Since 1(f) is a translate of K(f), it is dense if K(f) is dense. (b) This is immediate from (a) and Theorem 1(c).

In fact, B(x0, r) fl K(f) = namely

the

I IJxII/r

bound

0 implies a bound on the norm of f,

f(xo)I/r. for some x E X and so Ilfil

=

0

—

Indeed, otherwise

If(x)I >

r)

and f(y) = 0, contradicting our assumption. Now we turn to one of the cornerstones of elementary functional analysis, the Hahn—Banach theorem which guarantees that functionals can be extended from subspaces without increasing their norms. This means that all the questions posed at the beginning of the chapter have reassuring answers. Although the proof of the general form of the Hahn—Banach theorem uses Zorn's lemma, the essential part of the proof is completely elementary and very useful in itself.

Let YCX be vector spaces and let f€X' and gE 1". Iff(y) = for all y E Y (i.e. flY, the restriction off to Y, is g) then f is an extension of g. We express this by writing g C f. A function on a real vector space X is said to be a convex p: X-+ = functional if it is positive homogeneous, i.e. p(zx) = tp(x) for all t 0 g(y)

Chapter 3: Linear functionals and the Hahn—Banach theorem

48

and x E X, and is a convex function (as used in Chapter 1), i.e. if x,y E X, and 0 t 1 then p(tx+(1—Oy) tp(x)+(1—Op(y). By the positive homogeneity of p, the second condition is equivalent to p(x+y) p(x) +p(y) for all x,y E X, i.e. to the subadditivity of p. As customary for the operations on R = we use the convention =

=

that

for all

s

ER;

= 0; and

=

fort>0.

Note that a norm is a convex functional, as is every linear functional. Furthermore, if X = (V. II' fi) is a normed space then a linear functional f E X* and f E X' is dominated by the convex functional NIIxII S —+ R is said to dominate a function N. As usual, a function liffi

if

i/r: S

R if i/i(s)

—+

q'(s) for all S E S.

Lemma 3. Let p be a convex functional on a real vector space X and let fo be a linear functional on a 1-codimensional subspace Y of X. Suppose that fo is dominated by p, i.e. p(y)

fo(Y)

for all y E Y.

Then fo can be extended to a linear functional f E X' dominated by p: f(x)

p(x)

for all x E X.

Proof. Fix z E X (z 1') so that every x X has a unique representation in the form x = y + tz, where y E Y and t E R. The functional f we are looking for is determined by its value on z, say f(z) = c. To prove (1), we have to show that for some choice of c we have

f(y+tz) in other words

f0(y)+tc

p(y+iz)

for all y E Y and t E R.

For t> 0 inequality t=

—s

(2)

gives an upper bound on c, and for

<0 it gives a lower bound. Indeed, for t> 0, (2) becomes

for all y

Y. For s > 0 we have fo(Y) —Sc

C> for all y E Y. The former holds if

p(y—sz) and so

Chapter 3: Linear functionals and the Hahn —Banach theorem c

49

p(y' +z)—f0(y')

for all y' E Y, and the latter holds iff c

—p(y" — z) + fo(Y")

for all y" E Y. Hence there is an appropriate c 1ff

p(y'+z)—f0(y')

(3)

for all y',y" E Y. But (3) does hold since f0(y')+f0(y") =

p(y'+y")

p(y'+z)+p(y"—z),

0

completing the proof. The following theorem is a slight strengthening of Lemma 3.

Theorem 4. Let Y be a subspace of a real vector space X such that X is the linear span of Y and a sequence z1 , Suppose fo E Y' is dom-

mated by a convex functional p on X. Then fo can be extended to a linear functional f X' dominated by p. If X is a real normed space and Jo E r then to has an extension to a functional f on X such that 11111 = Ilfoll. Proof. Set K, = lin{Y,z1,.. . ,z,,}. By Lemma such that C 12 C fUflCtioflalS to C

inated by p. Define f: X

3 we can define linear and each f,, is dom-

by setting f(x) = f X' extends Jo and it is dominated by p. The second part is immediate from the first. Indeed, fo is dominated where N = Ilfoll. Hence there by the convex functional p(x) = R

is an f E Xt extending to and dominated by p. But then f(x) N = Iltoll, implying NfIxII for all x E X, so that Ilfif

p(x) =

11111

IltoIl.

=

0

restriction on Y in Theorem 4 is, in fact, unnecessary. As we shall see, this is an easy consequence of Zorn's lemma, the standard weapon of an analyst which ensures the existence of maximal objects. For the sake of completeness, we shall state Zorn's lemma, but before The

doing so we have to define the terms needed in the statement. A partial order or simply order on a set P is a binary relation

that (i) a

a for every a

P. (ii) if a

b and b

such

c for a,b,c E P

forsomea,bE Pthena = Briefly,

b.

is a transitive and reflexive binary relation on P. We call the

50

Chapter 3: Linear functionals and the Hahn—Banach theorem

pair (P. a partially ordered set; in keeping with our custom concerning normed spaces and topological spaces, (P, is often abbreviated to P. A subset C of P is a chain or a totally ordered set if for all a, b E C we have a orb a. Anelementm E Pisamaximalelejnentof P if m a implies that a = m; furthermore, we say that b is an upper foralisES. Itcan beshownthatthe bound axiom of choice is equivalent to the following assertion. Zorn's lemma. If every chain in a non-empty partially ordered set P has an upper bound, then P has at least one maximal element. 0

The fact that Theorem 4 holds for any subspace Y of X is the celebrated Hahn—Banach extension theorem.

Theorem 5. Let Y be a subspace of a real vector space X and let fo E Y'. Let p be a convex functional on X. If fo is dominated by p on Y, i.e. fo(y) p(y) for every y E Y, then Jo can be extended to a linear functional f E IC dominated by p.

If X is a real normed space and Jo E r then fo has a normpreserving extension to the whole of X: there is a functional f E X* such that fo C f and 11111 = Ilfoll.

= {f,,: y E 1) of all extensions of fo dominated by p: for each y there is a subspace V,, and a linear functional E such that Y C Y,,, fo C and f., is dominated by p. Clearly is 'less than or equal to' fo if the relation 'C' is a partial order on is a non-empty chain (i.e. - a totally = if,,: y E C fe). If Proof. Consider the set

ordered set) then it has an upper bound, namely f E Y', where Y,, and f(y) = f,,(y) if y E Y, (y E Fe). Therefore, by = Zorn's lemma, there is a maximal extension. But by Lemma 3 every maximal extension is defined on the whole of X. The second part follows as before. 0 With a little work one can show that norm-preserving extensions can be guaranteed in complex normed spaces as well. A complex normed space X can be considered as a real normed space; as such, we denote it by XR. We write for the dual of Xft. It is easily checked that the

mapping r: r

defined by r(f) = Ref (i.e. r(J)(x) = Ref(x) for

r

x E X) is a one-to-one norm-preserving map onto

The inverse of r —' is the map c: defined by c(S) (x) = g(x) — ig(ix). This enables us to deduce the complex form of the Hahn—Banach extension theorem.

Chapter 3: Linear functionals and the Hahn—Banach theorem

51

Theorem 6. Let Y be a subspace of a complex normed space X and let Then fo has a norm-preserving extension to the whole of X: fo such that fo C and 11111 = 111011. there is a functional f E

r

Proof.

f

By Theorem 5, we can extend r(f0) to a functional g on XR The complex functional f = c(g)

satisfying lid = IIr(fo)II =

Er 0

extends fo and satisfies IlfIl = IIfo II.

The Hahn—Banach theorem has many important consequences; we give some of them here. Corollary 7. Let X be a normed space, and let x0 X. Then there is a such that f(x0) = IIxoII. In particular, Iixoii C 1ff functional f C C for all g 5(r). ig(xo)P

Proof. We may assume that x0

Let Y be the 1-dimensional sub-

0.

space lin{x0} and define to E VS by f0(Ax0) =

A IixoII.

Then

and

1

its extension f, guaranteed by the& Hahn—Banach theorem, has the

0

required properties.

Corollary 8. Let X be a normed space, and let x0 C X. If f(x0) =

allfErthenx0=0.

0

for

0

The functional f whose existence is guaranteed by Corollary 7 is said to be a support functional at x0. Note that if x0 C 5(X) and f is a support functional at x0 then the hyperplane 1(f) is a support plane of the convex body B(X) at x0; in other words: x0 C B(X) flI(f) and 1(f) contains no interior point of B(X). The norm on X is said to be smooth if every x0 C S(X) has a unique support functional. T

Corollary 7 implies that the map given by Y) —' is an isometry, as remarked after Theorem 2.4, when we

r,

defined the adjoint.

Theorem 9. If X and Y are normed spaces and T C

r

and itrii

Y) then

= 11711.

Proof. As usual, we may and shall assume that X and Y are non-trivial spaces: X {0} and V {0}. We know that fi r 0, liii. Given there is an x0 S(X) such that if Tx011 11111 — e. Let g C S(VS) be a support functional at Tx0: g(Txo) = iITxoll. Then

(Tg)(x0) = g(Tx0) = so that

lIrgIl

and iirii

IlTxoII

IIi1I—€.

11711

52

Chapter 3: Linear functionals and the Hahn—Banach theorem

Given a vector space V with dual V' and second dual V" = (Vt)', there is a natural embedding V V" defined by v v", where v" is defined by v"(f) = f(v) for f C V. Rather trivially, this embedding is an isomorphism if V is finite dimensional. If X is a normed space with dual r, second dual Xt. and x C X, then we write i for the restriction of x" to X*: I is the linear functional on given by 1(f) = f(x) for In other words, with the bracket notation, fC

(I,f) = (f,x) for all f C X*. Since Ii(f)I = If(x)I IlfIllIxIl, we have I C (not just I E (X*)I), and moreover Dxli. The Hahn—Banach theorem implies that, in fact, we have equality here. Theorem 10. The natural map x I is a norm-preserving isomorphism (embedding) of a normed space X into its second dual X**.

Proof. For x C X (x 0), let f be a support functional at x: lifil = 1 and f(x) = ilxII. Then ii(f)i = If(x)i = iixli and 11(1)1 IIIIiiifiI = so

that iIxli

hID.

0

In view of Theorem 10 it is natural to consider X as a subspace of

X the whole of X**, i.e. then X is said to be reflexive. We know that X* * is complete

X= even when X is not, so a reflexive space is necessarily complete. However, a Banach space need not be reflexive. For example, 1, is reflexive for 1

every Banach space X is a closed subspace of C(L) for some metric space L. Indeed, put L = B(X*), and for x C X define = IlL. Then, by Theorem 10, the map X —' C(L), given by x is a linear isometry onto a closed subspace of C(L). We shall see in chapter 8 that considerably more is true: instead of C(L) we may take C(K), the space of all continuous functions on a compact Hausdorff space K with the uniform norm.

To conclude this chapter, let us present a strengthening of the Hahn—Banach theorem. This time we wish to impose not only an upper bound but also a lower bound on our linear functional to be found, the upper bound being a convex functional, as before, and the lower bound a concave functional.

Chapter 3: Linear fwzc:ionals and the Hahn—Banach theorem

53

Given a real vector space X, a function q: R. = is said to be a concave functional if —q(x) is a convex functional, i.e. if q is positive homogeneous and q(x + y) q(x) + q(y), i.e., superadditive. Given a convex functional p and a concave functional q, our aim is to find a linear functional f E X' such that q(x)

f(x)

p(x)

(4)

for all x E X. What condition does f have to satisfy on a subspace Y in order for f to be extendable? One's first guess is surely that q(y)

f(y)

p(y)

for all y E Y. While this condition is undoubtedly necessary, it need not be the whole story. Indeed, as f(y) = f(x+y) —f(x) and —f(x) —q(x), we must have

f(y)

(5)

for all y E Y and x E X. Inequality (5) is stronger than (4): putting x = 0 in (5) we find that f(y) p(y) for all y E Y, and setting x = —y we see that f(y) —q(—y), i.e. q(—y) f(—y) for all y Y. Of course, if Y = X then conditions (4) and (5) are equivalent. The following strengthening of the Hahn—Banach theorem shows that (5) is sufficient to guarantee the existence of an extension from Y to the whole of X.

Theorem 11. Let p be a convex functional and q a concave functional on a real vector space X, let Y be a subspace of X and let fo E Y' be such that fo(Y)

for all y E Y and x

f

X. Then q(x)

for every x

p(x+y) —q(x)

f(x)

X' such that

p(x)

X.

Proof. The heart of the matter is the analogue of Lemma 3: once we have managed to extend to a slightly larger (i.e. one dimension larger) subspace, the rest follows as before. Pick a vector z X (z Y) and set Z = lin{Y,z}. Let us show that fo has an extension f1 to Z satisfying f1(u)

Chapter 3: Linear functionaLs and the Hahn-Banach theorem

54

for all u E Z and x E X. As in the proof of Lemma 3, we have to show that there is a suitable choice c for f1(z), i.e. that there is a c such that

R

f1(y+z) = and f1(y'—z)—— f0(y')—c

for all y,y'

Y and x,x'

X. Such a c

—p(x' -I-y' — z) +

exists

if and only if

+p(x +y + z) — q(x)

fo(Y)

for all y,y' E Y and x,x' E X. But this inequality does hold, since

fo(y)±fo(y') =

to Z. fo has a suitable extension This assertion implies the analogue of Theorem 4, and an application of Zorn's lemma gives our theorem in its full generality. 0

Corollary 12. Let p be a convex functional and q a concave functional on a real vector space X such that q is dominated by p: for all x X we have q(x) p(x). Then there is a linear functional f E V' such that q(x)

Proof. Let Y =

(0)

f(x)

p(x)

for

all x

X.

C X. Then the trivial linear functional fo on Y

satisfies the condition in Theorem 11, namely p(x)

h(O)

—

q(x)

for all x X. Hence fo has an extension to a linear functional f E X' such that q(x) f(x) p(x) for all x X. 0 The following separation theorem is an easy consequence of Corollary 12.

Theorem 13. Let A and B be disjoint non-empty convex subsets of a real vector space X. Suppose that for some ccE A and every xE X there is an

e(x)>

0

such that

+ tE A for all

t, fri

(x). Then

A and B can be

separated by a hyperplane, i.e. there is a non-zero linear functionalfE X' and

a real number c such thatf(x)

c

for all x€ A and yE B.

Chapter 3: Linear functionals and the Hahn—Banach theorem

55

for every xE X there is an i =e(x)>Osuchthat[—€x,exJCA.DefinefunctionspandqonXby Proof. We may assume that a =

0, i.e.,

setting, for XE X,

0: xE tA};

p(x) = inf{t It is

0: XE tB}.

q(x) = sup{t

easily checked that p: X

R is

a convex functional and

is a concave functional. Furthermore, as tA fl tB = 0 for t> 0, we have q(x) p(x). Hence, by Corollary 12, there is a non-zero linear functional f E X' such that q(x) f(x) p(x) for all x E X. To complete the proof, note that if x E A and y B then

q: X

f(x)

Hence we may take c =

p(x)

1

q(y)

f(y).

0

1.

As the last result of this chapter, we shall show that the separation theorem gives a pleasant description of the closed convex hull of a set in a normed space. The convex hull co S of a set S in a vector space X is

the intersection of all convex subsets of X containing S, so it is the unique smallest convex set containing S. Clearly,

coS

t1x1

:

S, t,

0 (i = 1,...,n),

t1

=

1

(n =

=

if X is a normed space then the closed convex hull S of S is the intersection of all closed convex subsets of X containing S, so it is the unique smallest closed convex set containing S. As the closure of a convex set S is the closure of co S. is convex,

The following immediate consequence of the separation theorem is the intersection of all closed half-spaces containing S. shows that It is, of course, trivial that S is contained in this intersection.

Theorem 14. Let S be a non-empty subset of a real normed space X. = {x E X: f(x) su2f(s) for allf E

Then

SE.)

S. Then B(x0, r) fl S = 0 for some 0) separate the convex sets B(x0, r) and 5:

Proof. Suppose that x0

r> 0. Let f E X' (f

f(x) for all x E B(x0, r) and y

bounded above,f€ X* and

c

coS. Since the restriction of f to B(xo, r) is O,f(x0) > c.

0

Chapter 3: Linear functionaLc and the Hahn—Banach theorem

56

Throughout the book, we shall encounter many applications of the Hahn—Banach theorem and its variants. For example, the last result will be used in chapter 8. Exercises

1. Let p and q be conjugate indices, with 1 p < Prove that Show also that [Note that this gives a quick = = proof of the fact that for 1

the space is complete.] p 2. Let c be the subspace of consisting of all convergent sequences. What is the general form of a bounded linear functional on c? 3. Let p and q be conjugate indices, with 1 p < Prove that the dual of 1) is Lq(O. 1). is reflexive. Check also 4. Check that for 1

Let X1 X2,... be normed spaces and let X = *

space of all sequences x

and x1 = 0

x,, E

if n

X, be

the

(x1,x2,...) which are eventually zero:

is

sufficiently large, with pointwise

opera-

tions and norm \1/2 lxii

What 7.

Let X be

a

=1

Banach space. Show that if X*** = X* then

X is

= X.

i.e.

A linear functional f E 1,. is positive if f(x) 0 when every 0 for all n. Show that every positive x= E 1,. with x, linear functional

9.

(

is X*?

reflexive,

8.

=

Show Ofl

icc.

such

on ic,.

and deduce

is bounded.

= lim

that p0(x) =

is a convex functional

the existence of a linear functional f:

that

f(x) for

every x =

E

10. Let X be a complex normed space and let Y be a subspace of XR. (Thus if Yi Y2 E Y and r1 , r2 C R then r1y1 + r2y2 C Y.) Let (i.e. let Jo be a bounded real linear functional on Y) such Jo C

Chapter 3: Linear funcrionals and the Hahn—Banach theorem

57

that if y1,y2,A1y1+A2y2 E Y for some A1,A2 E C and Y1'Y2 E Y then f0(A1y1 +A2y2) = A1f0(y1) +A2h(y2). Show that fo need not

have an extension to a bounded complex linear functional on the whole of X, i.e. it need not have an extension to a functional

fEr.

11k. Let X be a Banach space, Y a 1-codimensional subspace of X and X1 and X2 dense subspaces of X. Is X1 nA'2 dense in X? Is X1 fl Y dense in Y? What are the answers if A'1 and X2 have codimension 1?

12. Let A' be a normed space, Y a dense subspace of X and Z a closed finite-codimensional subspace of A'. Is Zfl Y dense in Z? 13. Let V be a subspace of a normed space A'. Show that the closure of Y is =

fl{Kerf: f E

Y C Kerf}.

14. Let K be a closed convex set in a real normed space X. Show that

every boundary point of K has a support functional: for every

x0 E ÔK there is an f E r such that f

0 and SUPXEK f(x) =

f(xo).

15. Let A be a set of points in a real normed space X, and let fo: A —÷ R. Show that there is a functional f(a) = f0(a) for all a E A if and only if

A(a)f0(a)

aEF

a€F

f E B(X')

that

A(a)a

for every finite subset F of A and for every function A: F 16.

such

Let V be a subspace of a normed space X and let x

R.

X. Show

that d(x, Y) = inf{IJx—yII: y E Y}

1

if and only if there is a linear functional f E B(X) V C Ken and f(x) =

such

that

1.

The results in the final exercises, all due to Banach, enable us to define finitely additive 'integrals' of large classes of functions and to attach a 'limit' to every bounded sequence.

17. Let T = Fl/i be the circle group, i.e. the additive group of reals modulo the integers. Let X = be the vector space of bounded real-valued functions on T. For

58

Chapter 3: Linear functionals and the Hahn—Banach theorem

f = f(t) E X

a1 ,... ,

and

R

set

=

and define

n = 1,2,...;

p(f) =

a convex functional and deduce that there is a genand such that for f, g E eralized 'integral' 1(f)

A,p.,t9 E R we have = Al(f) + id(s);

(i) 1(Af+

(iii) J(J(t + t0)) (v) 1(1) =

1

1(f)

(ii)

0

if f

0;

(iv) 1(J( —1)) =

= I(J(t));

(i.e. the integral of the identically 1 function is 1).

18. Let X be the vector space of all real-valued functions f(t) on R such that urn SUPt..IOØ

f = f(t)

EX

a1,...

and

ER

set =

limsup!

and define

p(f) =

n = 1,2,...; a convex functional, and deduce that there is a gen-

eralized 'limit' LIM,...,. f(t) on X such that +

(I)

(ii) LIM f(t)

}= 0

(—C

A

LIM f(t) +

if lim inff(r)

(iii) LIMf(t+t0) = LLMf(t);

(iv) LIM1 =

1.

0;

g(t);

Chapter 3: Linear functionals and the Hahn—Banach theorem

19. For x =

4, and n1 <

59

1r(x;nl,...,nk) =

k

and define

±

i—I

p is a convex functional, and deduce the existence of a linear functional L: 4, R such that L(x)

p(x)

for every x 1,,. The value L(x) is said to be a Banach limit or a generalized limit of the sequence and is usually denoted by

LIM

LIM

LIM

LIM Let rn LIM

+ LIM

N be fixed and define

= n/rn —

What is

20. Show that if in Theorem 13 we drop the condition on A then the assertion is no longer true. [Hint. Let X be a vector space with basis e1, e2,. .

.,

and let A = — B

=

x2,:

> 0, n = 1,2,...

Check

}.

that f(A) = f(B) =

R

for all fE X', f

0.]

Notes

The Hahn—Banach theorem for real normed spaces was proved by H. Hahn, Uber lineare Gleichungen in linearen Räumen, J. für die reine und angewandte Mathematik, 157 (1927), 214—29; and by S. Banach, Sur les fonctionnelles linéaires II, Studia Math., 1 (1929), 22 3—39. The complex version, namely Theorem 6, was proved by H. F. Bohnenblust and A. Sobczyk, Extensions of funclionals on complex linear spaces, Bull. Amer. Math. Soc., 44 (1938), 91—3.

Many dual spaces were first identified by F. Riesz, in Sur les operations fonctionnelles Iinéaires, Comptes Rendus, 149 (1909), 974—7, and in Uiuersuchungen über Systeme integrierbarer Funk:ionen, Mathematische Annalen, 69 (1910), 449—97.

4. FINITE-DIMENSIONAL NORMED SPACES

As the next cautious step in our exploration of normed spaces and operators on them, we look at the 'smallest' normed spaces, namely the finite-dimensional ones. As far as the crude classification of norms is concerned, these spaces are very simple indeed: any two norms on a finite-dimensional vector space are equivalent. This can be proved in many different ways; the proof we give here is based on a lemma about the space Lemma 1. The closed unit ball of

Proof. We shall show that the standard basis of so that

is compact.

is sequentially compact. Let (e,)? be

=

A1 I.

with Xk = we have CB= I a bounded sequence of complex numbers has a convergent subsequence, by repeatedly selecting subsequences, we can find a converges to some scalar A for every such that subsequence i (1 i n). Given

Setting x =

A,e1 we find that lim

and so

=0

x E B.

Theorem 2. On a finite-dimensional vector space any two norms are equivalent. 60

Chapter 4: Finite-dimensional normed spaces

61

Proof. Let V be an n-dimensional vector space with basis be

the

Let

-norm on V given by =

i=I

i=1

be an arbitrary norm on V. It suffices to show that and let are equivalent. II

Let X1 = (V, III). f(x) = lxii. The set

S1

= S(X1)

and let f: S1

II

and

be defined by

closed subset of the compact set B(X1) and therefore it is compact. Furthermore 11(x)

—f(y)I

is a

ilx—yll

—

1=1

lx1—y11

= (max Iieali)Ux—yiii I

f is a continuous function on the compact set S1. Hence f attains its infimum m and supremum M on Si. Since f(x) = lixIl > 0 for all so

xE S1. we have m >0. By the definition of f, for any xE V we have llxlI

Mllxflj.

0

This theorem has several easy but important consequences.

Corollary 3. Let X and Y be normed spaces, with X finite-dimensional. Then every linear operator T: X —+ Y is Continuous. In particular, every linear functional on X is continuous. Proof. Note that llxD' = llxll + llTxll is a norm on X; since lll and are equivalent, there is an N such that ilxll' Niixll for all x and so

0 Corollary 4. Any two finite-dimensional spaces of the same dimension are isomorphic. Proof. If dim X = dim Y then there is an invertible operator TC Y). As both T and T1 are bounded, X and Y are isomorphic. 0

Chapter 4: Finite-dimensional normed spaces

Corollary 5. Every finite-dimensional space is complete.

Proof. If a space is complete in one norm then it is complete in every equivalent norm. Since, for example, the space is complete, the

0

assertion follows.

Corollary 6. In a finite-dimensional space a set is compact 1ff it is closed

and bounded. In particular, the closed unit ball and the unit sphere are compact.

Proof. Recall that pact set is compact.

is compact, and that a closed subset of a com-

0

Corollary 7 Every finite-dimensional subspace of a normed space is closed and complete. Proof. The assertion is immediate from Corollary 5.

0

In fact, as proved by Frederic Riesz, the compactness of the unit ball characterises finite-dimensional normed spaces. We shall prove this by making use of the following variant of a lemma also due to Riesz. Theorem 8. Let Y be a proper subspace of a normed space X. S(X) whose 0 there is a point x (a) If Y is closed then for every distance from Y is at least 1 —

d(x,Y) = inf{I)x—yII:y€ Y is finite-dimensional then there is a point x E S(X) whose distance from Y is 1.

Proof. Let z E X\Y and set Z = lin{Y, z}. Define a linear functional f: Z R by f(y + Az) = A for y E Y and A E R. Then f is a bounded linear functional since Kerf = V is a closed subspace of Z. By the Hahn—Banach theorem, f has an extension to a bounded linear functional F E X' with IIFII = 11ff > 0. Note that Y C KerF. (a) Let x E S(X) be such that F(x)

(1—€)IIFII.

Then for y E Y we have IIx—yIj

F(x)

F(x—y) Fl

=

1—c.

Chapter 4: Finite-dimensional normed spaces (b)

63

If Y is finite-dimensional then F attains its supremum on the com-

pact set S(X) so there is an x E S(X) such that Rx) = IIF1I. But then for y E Y we have

>F(x-y)F(x)1 UF1I

—

11111

be finite-dimensional subspaces of a normed space, with all inclusions proper. Then there are unit vectors = for all n 2. x1,x2,... such that x,, E X,, and

Corollary 9. Let X1 C X2 C

1

In particular, an infinite-dimensional normed space contains an infinite of 1-separated unit vectors (i.e. with 1 for sequence n

en).

Proof. To find x,, E

apply Theorem 8(b) to the pair (X, Y) =

0 Theorem 10. A normed space is finite-dimensional if and only if its unit

ball is compact.

Proof. From Corollary 6, all we have to show is that if X is infinitedimensional then its unit ball B(X) is not compact. To see this, simply take an infinite sequence x1 , x2,... E B(X) whose existence is 1 for i j. As this sequence guaranteed by Corollary 9: 11x1 —x1fl has no convergent subsequence, B(X) is not compact. 0

Theorem 10 is often used to prove that a space under consideration is finite-dimensional: the compactness of the unit ball tells us precisely this, without giving any information about the dimension of the space.

The above proof of Theorem 10 is based on the existence of a 1 for all i j. Let sequence of unit vectors such that 11x1 —x,II us show that, in fact, we can do better: we can make sure that the inequalities are strict. All we need is the compactness of the unit ball of a finite-dimensional normed space.

Lemma 11. Let x1,... , x,

linearly independent vectors in a real E S(X) normed space X of dimension n 2. Then there is a vector i < n). such that > 1 for all i (1 — be

Proof. We may assume that dim X = n. Let f S(X) be such that i < n. (In other words, we require

f(x1) = 0 for 1

Chapter 4: Finite-dimensional normed spaces

64

K(f) = lin(x1,..

.

and so we have precisely two choices for f: a functional and its negative.) Furthermore, let g X* be such that g(x1) = (1 i < n). Since S(X) is compact, so is

1

for every i

K={xES(X):f(x)= Let we have

K}. Then for 1

x

—x,) =

1.

Since

the choice of x,, tells us that x,, —x, i < n). IIx,,—x111 > 1 for every 1(1

—x1) =

K, so that

i
—1

—xJ

1.

Hence

0

Theorem 12. Let X1 C X2 C be subspaces of a real normed space, with dim X,, = n. Then there is a sequence x1 , x2,... of unit vectors such that IIx,—x111 > 1 if i j and lin{x1,. .. ,x,j = for n = 1,2

There is another elegant way of finding 1-separated sequences of unit vectors. This time we shall rely on the finite-dimensional form of the Hahn—Banach theorem. Let us choose vectors x1,x2,... E S(X) and support functionals x,x,... as follows. Pick x1 C S(X1) and let E be a support functional at x1 : = 1. Suppose = n, x1,.. . k
x,.. .,4

k

W

k

fl K(x7)= fl

Kerx7

has dimension at least n — k (in fact, precisely n — k) and so we can pick Let a vector Xk+1 be a support functional at Xk+1, i.e. let C and 4+I(Xk+L) = 1. This implies that if X1 C X2 C ... are subspaces of X = n)

then there are sequences lin{x1 ,.. , x,j = X,, = .

lar,

XmII

1 for n

C S(X) and 1

and

=

m.

The canonical bases of 1,, and

C 0

such that

for n <m. In particu-

i,

+ where = 1, are examples of pairs of sequences of this type. In fact, with = (0,..., 0,1,0,...) 6 and = (0,..., 0,1,0,...) where the l's occur in the nth places, we find that = = 1 and = 0 for all n m, not only when n <m. A pair of sequences

=

satisfying these conditions is called a normali.ced biorthogonal system.

To be precise, given a normed space X, a biorthogonal system on X

Chapter 4: Finite-dimensional normed spaces

65

consists of vectors x1 ,.. , x,, and bounded linear functionals = 1). Such a (i.e. x7(x1) = 0 if i j and such that xr(x1) = lixtil = I for all i. system is normalised if lxiii .

Theorem 13. Let X be an n-dimensional real normed space. Then there

x.

is a normalised biorthogonal system (x1 )7, (xr

write [fl) and S = S(X). For u1 ,... , u,, E for the n-dimensional Euclidean volume of the n-simplex Let x1 ,... , x,, E S be such that v(x1,... , x,,) with vertices 0, u1 ,.. ,

Let X =

Proof.

v(uj,.

. .

,

.

is

maximal.

We claim that these vectors will do. What are the functionals (x i. But = I and x7(x1) = 0 for j Clearly xr has to be defined by does x7 have norm 1? Indeed it does, since 11x711 > 1 implies that x7(y1) > I for some y, E S and so D

The sequence

whose existence is guaranteed by Theorem 13 is

called an A uerbach system.

In order to see what the existence of an Auerbach system really means, it is worthwhile to reformulate Theorem 13 in geometric terms. norm and Let x1 ,.. , x,, be a basis of X. Define two norms on X: the norm defined by taking x1,. , x,, as canonical basis. Thus for the A1x1 set x= .

.

and

ku1 =

.

= max

1=1

Let

X1 and X,. be the normed spaces defined by these norms. Then

Theorem 13 claims precisely that x1,. .. ,x,,, can be chosen in such a way that B(X1) C B(X) C B(X,0).

Clearly

and

8(X1) is the convex hull of the 2n vectors ±x1, ±x2,..., ±x, is the n-dimensional parallelepiped whose vertices are (€, E {—1,1}).

Let us return to the opening statement of this chapter: the isomorphic classification of finite-dimensional normed spaces is trivial, with two spaces being isomorphic if and only if they have the same dimension. Based on this, one could come to the hasty verdict that there is nothing to finite-dimensional normed spaces: they are not worth studying. As it

Chapter 4: Finite-dimensional normed spaces

66

happens, this would not only be a hasty verdict but it would also be utterly incorrect. There are a great many important and interesting questions, only the isomorphic classification is not one of them. All these questions, many of which are still open, concern the metric properties of the finite-dimensional normed spaces. Perhaps the most fundamental question is the following: given two ndimensional normed spaces, how close are they to being isometric? Let us formulate this question precisely.

Let X and Y be isomorphic normed spaces, i.e. let X and Y be such

that there

is

a bounded linear operator T C

Y) which has a

bounded inverse. The Banach—Mazur distance between X and Y is d(X, Y) = inf{1171111T'JI: T C

Y), T' C

If X and Y are not isomorphic then one defines their Banach—Mazur distance to be Note that d(X, Y) for any two 1, d(Y, X) = d(X, Y) = d(X*, spaces, and d(X,Z)

d(X,Y)d(Y,Z)

for any three spaces. Thus in some sense it would be more natural to measure the 'distance' between X and Y by log d(X, Y). What does d(X, Y)
!B(Y) C TB(X) C c28(Y) for some c1 and c2 satisfying c1c2 < d, in other words that SB(Y)C B(X)CcSB(Y)

for some S C

and c < d. Equivalently, we may demand that B(Y) C TB(X) C cB(Y)

for some c < d. Thus the distance is less than d if after a linear transformation the unit ball of one of the spaces is sandwiched between the unit ball of the other space and c times that unit ball, where c < d. Corollary 4 tells us that if dim X = dim Y = n then X and Y are isoBut how large can d(X, Y) be? A commorphic and so d(X, Y) pactness argument implies immediately that d(X, Y) f(n) for some R. In fact, the following simple consequence of Theorem 13 gives a bound on d(X, Y).

function f: N

Chapter 4: Finite-dimensional normed spaces Corollary Then

Let X = (V,

14.

H

fi) be

67

an n-dimensional normed space.

n.

be an Auerbach system in X so that, in particular, on X by setting is a basis of V. Define a norm fi Proof. Let

jA1 I.

= Then X1 = (V,

lii)

is

isometric to

1k";

we have to show that

so

n. d(X,X1) We claim that the formal identity J: X3 X is such that 11111 Ax1 we have n. Indeed, forx = IIJ(x)II and so 11111

I

=

1. (As IIJxiII

IIx,II

1 and

A1 I = lixili

=

= lixilli = 1, in fact IIJII = 1.) let (x,' be the other half of the normal-

In order to estimate ised biothogonal system (x1)?, (x7)7; in other words, let 4 be defined Given x = by 4(x1) = Ax1, choose €, E (—1, 1} such that Then Ilfil n eA1 = 1A11 for i = 1,... ,n. Set f =

r.

and

f(x)

AJ = IIxIh,

and so

= IIxIIi = f(x)

nIIxJI,

implying 11J' II The alert reader may have noticed that the formal proof above is, in fact, unnecessary, because the result is a trivial consequence of (1), the relation equivalent to Theorem 13. Indeed, B(X,,,) C nB(X1) and so B(X1) C B(X) C nB(X1). Thus d(X,X1) n. But X1 is isometric to so that d(X, n. An immediate consequence of Corollary 14 is that d(X, Y) n2 for any two n-dimensional spaces, but this trivial estimate is far from being best possible. In 1948 Fritz John proved an essentially best-possible upper bound for d(X, Y) by first bounding the distance of an n-dimensional space from rather than from Before giving this result, let us think a little about the Banach—Mazur distance from a Euclidean space.

Chapter 4: Finite-dimensional normed spaces

An easy compactness argument implies that if dim X = dim Y = n then d(X, Y) is attained: there is an operator S E say, such that SB( Y)

where

d = d(X, Y).

c B(X) c dSB( }')

Now if Y =

= (Rn, II11) then B(Y) is the

Euclidean unit ball

the image SD of the Euclidean ball For a linear operator S E D is an ellipsoid centred at 0; conversely, for every ellipsoid E C with centre at 0, there is a linear map S E R") such that SD = E. d if and only if there is an ellipsoid E, with centre 0, Thus d(X, such that

ECB(X)CdE. The following elegant and important theorem of John shows that one can obtain a good upper bound for d(X, be taking the ellipsoid E of minimal volume containing B(X).

Theorem 15. (John's theorem) Let X =

a normed space with unit ball B = B(X). Then there is a unique ellipsoid D of minimal (Euclidean) volume containing B. Furthermore, (R'1, IFII) be

C BC D. In particular,

d(X,lfl

n1"2.

Proof. A simple compactness argument shows that the infimum of the

volumes of ellipsoids containing B is attained. Furthermore, every ellipsoid of minimal volume is centred at 0.

Let us first prove uniqueness. Suppose that D and D' are ellipsoids of minimal volume containing B. If T E is invertible then T(D) are elliposids of minimal volume containing T(B). Hence we and T(D') may assume that and 1=1

where

>

0

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

1=1

Chapter 4: Finite-dimensional normed spaces

a=

a,, and

the volume of D, we have vol D' = Let E be the ellipsoid

Denoting by so

69

1.

E=

E

Then

BC DflD' CE and

(volE)2 (vol D)2

2

—

2a2

—

2a

—

1 1

a is 1. This contradicts the assumption that D was an ellipsoid of minimal volume containing B. 1

Let us turn to the proof of n 112D C B. Suppose that this is not the case, so that B has a boundary point in the interior of n112D. By taking a support plane of B at such a point and rotating B to make this we may support plane parallel to the plane of the axes x2 , x3,. . . , assume that

BC P =

E

lxii

for some c > For

a>

b

> 0 define an ellipsoid Eab E W':

Ea,b so

that (VOl D)/(vol Ea,b)

+ b2

= ab"'. If x C B then x C DflP and so

+ b2 i=2

x? =

+ b2 a

Hence B C

by

and vol Ea,b

+1,2

+b2.

and

>

Thus, to complete the proof, it suffices to show that these inequalities

Chapter 4: Finite-dimensional normed spaces

70 are

satisfied for some choice of 0 < b
which, unfortunately, looks a little untidy. Put 0< e < a = and b = 1—e. Then ab"1 > 1. Also, 2_ ++ C

2

C

2

\

C

=

<1

=

if e >

is

0

0

is sufficiently small.

Theorem 15 can be reformulated in terms of inscribed ellipsoids: there a unique ellipsoid D of maximal volume contained in B and

if B is not a unit ball but only a bounded convex body in P'1 then some translate B' of B satisfies D C B C n 112D.

Furthermore,

D C B' C nD for the ellipsoid D of maximal volume contained in B'

(see Exercise 18).

Corollary 16. Let X and Y be n-dimensional normed spaces. Then d(X,Y) n "2n"2

= n.

0

Although the estimate in Corollary 16 does seem somewhat crude, since the Banach- Mazur distance between X and Y is estimated by going through it is, in fact, close to being best possible: in 1981 the Russian mathematician E. D. Gluskin proved that there is a constant c > 0 such that for every a there are n-dimensional normed spaces X and Y with d(X, Y) cn. At the time this result was extremely surprising; the proof was based on a probabilistic argument which has become an important tool in the so-called local theory of Banach spaces, the theory concerning finite-dimensional spaces. Exercises

1. Let x1 ,. . A

.

,

be non-zero elements in a normed space. For

= (A1)7 C P'1 set DAlI0

= max{H±

:

€, C {—1, 1} (i =

Chapter 4: Finite-dimensional normed spaces

Show that

II

andj =

a norm on E R' we have

Show also that for A =

Ho is

IlAllo

71

=

= IIiIJomaxIA1I.

2. Let X1 = (V,,1111) (i = 1,...,n) be normed spaces. set E V = p ooandx =

For

1

= that is a norm on V. Show also that any two of these is usually denoted by norms are equivalent. [The space (V, fi it isadirect sum of be a symmetric norm on R", i.e. 3. Let

Show

with

1e11

O,1,O,...,O)I = I

=

=

and let X1,. .. , X,, be normed spaces. Show that II(xI,x2,...

=

11X2112,

. .. ,

. a norm on X = inducing the product topology on this space. Check that each X1 is a closed subspace of X and that

is

X is complete 1ff each X1 is complete. 4. Let x1 ,.. , be unit vectors in a normed space such that .

1=1

for all A,. .. ,

E R. Show that for k2 u =

p.1x1

n there are unit vectors

(i = 1,... ,k)

jE U,

such that U1flU,. =

0 for i

i' and

k

A,u1

c max

1A11

1=1

for all Aj,...,Ak ER. Let X be a finite-dimensional real normed space, x0 E X and TE Can be dense in X? [HINT: First prove it for complex normed spaces.]

Let Y be a proper subspace of a finite-dimensional normed space X.

Can one always find a vector z0 E X (z0

0) such that

Chapter 4: Finite-dimensional normed spaces

72

for all y E Y? Or a subspace Z C X such that Y+Z = Xand Iy+zlj Nyu for all yE Yand z E Z?

7. Let Y be a finite codimensional closed subspace of a normed space

X and let T E

Z), where Z is a normed space. Show that if

the restriction of T to Y is continuous then so is T. 8. Let x1 ,.. , be unit vectors in a normed space. Let 0 <E < and suppose that .

1=1

9.

for all A1,...

C

for all A1,...

C R.

Prove that

Let X and Y be n-dimensional normed spaces. Show that the Banach—Mazur distance between X and Y is attained, i.e. there is an operator T0 E Y) such that E X) and d(.X, Y) = inf{IITIIJIT'lI : T C

10.

Show that if 2

p

Y)} =

11T01111T0111.

then

d(1;,!n 11. Let X be a normed space with unit ball B = B(X) and suppose that B can be covered by a finite number of translates of

there

EXsuch that B C

U

(x1

+

B) = Li B(x1,

By making use of Corollary 7 but no subsequent result, prove that

X is finite-dimensional, and deduce Theorem 10. 12. Let Y be a finite-dimensional subspace of an infinite-dimensional

normed space X. Show that for every

0 there is an x C S(X)

such that y C Y. (1 + €)Itx + 13. A sequence in a normed space is said to be a basic sequence 0 for every n, and there is a constant K such that if AkxkH

for all

I

m

ii and scalars a,.. • ,

AkxkD

The minimal K satisfying

Chapter 4: Finite-dimensional normed spaces

73

Deduce from the result in the this is the basis constant of previous exercise that every infinite-dimensional normed space contains a basic sequence. be a basic sequence in a Banach space. Show that 14. Let is a Schauder basis of its closed linear span, i.e. every x E has a unique representation in the form x = Akxk, with the series convergent in norm. 15. Deduce from Corollary 9 that a Banach space cannot have a countwhere e1 , e2,... ably infinite algebraic basis, i.e. if X = are independent, then X is incomplete. 16. Let X and Y be finite-dimensional normed spaces and set NK

= {T E B(X, Y):

11111,

K}.

Prove that NK is a compact subset of B(X, Y) (with the operator norm). 17k. Let X be an infinite-dimensional separable Banach space such that for every subspace V C X (dim V = n) there is an operator such that Q JiJ K. Prove that there is an E B(Y, operator T E B(X,12) such that K. fl

be a bounded convex body (i.e. IntB 0). Show 18. Let B C that there is an ellipsoid D of centre 0 such that some translate B' of B satisfies D C B' C nD. 19. Prove that if a linear functional is continuous on a closed finitecodimensional subspace then it is a bounded linear functional. (V1, II•IJ) and X2 = (V2, be subspaces of a normed with X1 finite-dimensional. Show that if space X = (V, V= then X is the direct sum of X1 and X2, i.e. the projections X = X1 + X2 X1 (i = 1,2) are continuous.

20. Let X1 =

Notes

Fritz John's theorem was proved in the paper Extremum problems with inequalities as subsidiary conditions, in Courant Anniversary Volume, Interscience, New York, 1948, pp. 187—204. The original paper of E. D. Gluskin, The diameter of the Minkowski compactum is roughly equal to n, Funct. Anal. AppI., 15 (1981), 72—3, is an all-too-concise account of his celebrated results.

Excellent accounts of much of the excitement concerning finitedimensional spaces can be found in the following volumes: V. D. Mi!-

man and 0. Schechtman, Asymptotic Theory of Finite Dimensional

74

Chapter 4: Finite-dimensional normed spaces

Normed Spaces, Lecture Notes in Mathematics, vol. 1200, SpringerVerlag, Berlin, Heidelberg, 1986, viii + 156 pp., 0. Pisier, Volume of Convex Bodies and Geometry of Banach Spaces, Cambridge University Press, 1989, xv + 250 pp., and N. Tomczak-Jaegermann, Banach— Mazur Distance and Finite Dimensional Operator Ideals, Pitman, 1989.

5. THE BAIRE CATEGORY THEOREM AND THE CLOSED-GRAPH THEOREM

In this chapter we shall present several of the most fundamental and useful results in classical functional analysis: the Baire category theorem,

the Banach—Steinhaus theorem, the theorem of the condensation of singularities, the open-mapping theorem and the closed-graph theorem. All these results are very closely related; indeed, they are practically variants of each other. Nevertheless, it is rather useful to emphasize the many facets of the same phenomenon, because even a subtle variation in the formulation can make quite a difference in an application. The simplest member of this group of results is due to Baire and concerns metric spaces. Recall that a set Y in a topological space X is in X if Y, the closure of Y, is X. In other words, Y is dense if Yl) G 0 whenever G is a non-empty open set. We shall give several equivalent formulations of the Baire category theorem: here is the first.

Theorem 1. Let G1, G2,... be a sequence of dense open subsets of a is dense in X. complete metric space X. Then G = Proof. As in a normed space, for x E X and r> 0 we set Dr(X) = {y

Although

X: d(x,y)

and

=

(ye X: d(x,y)

r}.

need not be Br(X), we have D,.(x) C Br(X) C D,÷E(x) for

all 0. Since is a dense open subset of X, for all x E X and in a non-empty open set so there r> 0 the open ball D,(X) meets C C Ii Dr(X). This is the are y E X and s > 0 such that

property we shall exploit.

Let then x E X and r>O. We have to show that D,(x)flG 0. Let us construct sequences of points x1 , x2,... E X and positive numbers r1,r2,... as follows. Choose x1 E X and 0 <

r1

<

1

such that 75

Chapter 5: The Baire category theorem

76

Next choose x2 E X and 0 < r2 < 4 such that then choose x3 E X and 0 < r3 < 4 such that Br2(X2) C —* 0 and B,1(XI) D etc. By construction, Brj(X3) C B,2(x2) J ... and so, by the completeness of X, = {x0} for some x0 E X. As x0 E B,1(x1) C D,(x) and B,1(x1) C G1flD,(x).

0

x0 E D,(X)flG.

of its importance and many applications, we give some other forms of Theorem 1 and thereby explain the name of the result as well. The first is a slightly weaker but very useful variant. Because

Theorem 1'. If a complete metric space is the union of countably many closed sets then at least one of the closed sets has non-empty interior. where X is a complete metric space and each Proof. Let X = (3,, = 0 so that at F,, is closed. Setting G,, = X\F, we find that

least one of the open sets IntF,,=X\G,,.

G,, is

not dense in X: G,,

X.

But

0

Strictly speaking, Theorem 1' says precisely that the set Gin Theorem 1 is not empty. However, replacing X by B,(x), we see that G fl B,(x) is not empty either, and so G is dense. Thus Theorem 1 is an easy consequence of Theorem 1'. A subset Y of a topological space X is said to be nowhere dense in X if the closure of Y has empty interior: hit Y = 0. Note that a set Y is nowhere dense if and gnly if its closure, Y, is nowhere dense. A subset Z C X is meagre in X or of the first category in X if it is the union of countably many nowhere-dense sets. Clearly, the union of countably many meagre sets is meagre. A subset of X is of the second category in

X if it is not meagre in X. Thus U is of the second category in X if, whenever F1,F2,... are closed sets and U C F,, then 0 for some n. Let us use this terminology to give another reformulation of the Baire category theorem. Theorem The complement of a meagre subset of a complete metric space is dense. In particular, a complete metric space is of the second category.

Also, in a complete metric space the complement of a set of the first category (a meagre set) is of the second category.

Chapter 5: The Baire category theorem Proof. Suppose Y =

1',,

77

C X with X complete and each

nowhere dense in X. Then 1',, is nowhere dense in X so G,, =

a dense

open

set and therefore G =

is dense.

is

Clearly

GCX\Y. To see the third assertion, note that the union of two meagre sets is

0

meagre.

The two parts of the last variant are trivially equivalent since if A is a set of the second category in a topological space X and B C X is meagre then A\B is again of the second category.

The intuitive meaning of a set of the second category is that it is a large set. Thus, saying that 'the set of points having a certain property (e.g. where a given function is continuous) is of the second category' is rather similar to saying that 'almost all points have this property', i.e. the set of points not having this property has measure zero. Of course, the second statement makes sense only if we have a measure on the space, while being of the second category is an intrinsic property of a metric space. For example, given a collection of convex bodies in loosely speaking one may say that 'almost all sets in have property P', the set of convex meaning that in the natural (Hausdorif) metric on bodies having P is of the second category. It is perhaps worth pointing out that the two notions are only similar but not comparable, even if we restrict our attention to [0, 1] c It Indeed, [0, c [0, 1] does not have full measure, yet is of the second category, and one can construct a subset of[0, 1] with measure I that is of the first category

(cf. Exercise 23).

in 1897 Osgood proved the following pretty result about continuous Let fj,f2,...: [0,1] R be continuous functions such that is bounded. Then there is an for each t E [0, 1] the sequence functions.

interval [a,bJ C [0,1] (a
is

uni-

formly bounded.

Let us see that the Baire category theorem implies the following extension of this result. Theorem 2. (Principle of uniform boundedness) Let U be a set of the be a family of continuous second category in a metric space X and let .

is bounded for every functions f: X —' R such that {f(u): f E are uniformly bounded in some ball ii E U. Then the elements of n holds for some n and all f E B,(x0) (x0 X; r > 0), i.e. If(x)I and x E B,(xo). In particular, the conclusion holds if X is complete and is bounded for every x X. {f(x): E

f

Chapter 5: The Baire category theorem

78

Proof. For n

1 let

flf'([—n,n]).

I

Then as the intersection of a family of closed sets, is closed. By our assumption,

n1 so

by the definition of a set of the second category, mt

0 for some

The second assertion is immediate from Theorem 1".

0

n.

The next result is only a little more than a reformulation of Theorem 2 in the setting of linear maps. Theorem 3. (Banach—Steinhaus theorem) Let X and Y be normed spaces, let U be a set of the second category in X and let C Y) be a family of bounded linear operators such that sup{IITulJ: T

<

N for some N and all T E the conclusion holds if X is complete and T

In particular,

for all u E U. Then JITII

TE

x

by

that IITxII

is

a ball

n for some n and all T E

proof of Theorem 2.2, this implies that Indeed,

let T E

E

X; r > 0) such

and x E B,.(x0). As in the E N = fir for all T E

and x E B(X). Then 10 + rx

and x0 — rx

are in

Br(XO) and so IITxO

and

therefore

11J

=N

=

0

N for all T E

The last result has the following immediate consequence concerning the condensation of singularities.

Theorem 4. Let X and Y be normed spaces, with X complete, and let E Y) (n,m = 1,2...) be such that urn

=

for all

79

Chapter 5: The Baire category theorem

Then there is a set U C X of the second category in X such that for u E U we have for all n.

IIT,,,,,(u)II =

Proof.

For a fixed n, let V,, C X be the set of vectors v such that Then, by Theorem 3, V,, is of the first category.

and thus by Theorem 1" the set U = X\V is Hence so is V = U..1 0 of the second category in X. The next major result, the open-mapping theorem, is a consequence of the Baire category theorem and the important lemma below. Recall that for a normed space X we write D(X) for the open unit ball and

B(X) for the closed unit ball; furthermore, we use the notation D,(X) = rD(X) and Lemma 5.

= rB(X).

Suppose X and Y are normed spaces, X is complete,

Y) and T(Dr(X)) 3 D3(Y). Then T(D,(X)) 3

T

Proof. We may and shall assume that r = s = 1. Let A = D(Y) fl T(D(X)); then A = B(Y). Given z E D(Y), choose S such that We shall show that and set y = z/(1 —8). IIzII < 1—S < 1 y E T(D(X))/(1 —8) and so z E T(D(X)). Let us define a sequence (y,) C Y. Set Yo = 0 and then choose sucY such that y,, Yn—i E S"'A and fly,, —ylI <8". cessively Yi we can find a suitable y,, since Having chosen Yo'••• and the set S"'A is dense in yE By the definition of A, there exists a sequence

Tx,, = y,,

C X such that

Putting

and

x=

x,,

we have Dxli

ilx,,il

<(1 —Sr'

and

Tx

=

=

y E T(D(X))/(1 —8) and so z E T(D(X)). As z was an arbitrary point of D(Y), this implies that D(Y) C TD(X), as claimed. 0 Consequently

Theorem 6. (Open-mapping theorem) Let X and Y be Banach spaces and let T E Y) be a surjection: T(X) = Y. Then T is an open map, i.e. if U C X is open then so is TU.

Chapter 5: The Baire category theorem

80

Set G = T(D(X)). Since T is a linear map, it suffices to show that G contains a neighbourhood of 0 in Y. Note that = rG and the closure of rG is rG. Since nG, by the Baire category theorem we see that Y = T(X) = 0 for some n and so mt G 0. The set mt nG is convex and symmetric about 0 so mt G 0 implies that G 3 for some r > 0. Indeed, if D(x0, r) C G then D(—xo, r) C G and so the convexity of implies that D(0, r) = DT(Y) C G. But then, by Lemma 5, G = T(D(X)) 3 D,.(Y). 0 Proof.

The final two results are easy but very useful consequences of the open-mapping theorem.

Theorem 7. (Inverse-mapping theorem) If T is a one-to-one bounded linear operator from a Banach space X onto a Banach space Y then T' is also bounded.

Proof. Since T is a bijection, the inverse T' exists and belongs to By Theorem 6, for some r> 0 we have T(D(X)) 3 Thus 11T'II

hr.

0

Theorem 8. (Closed-graph theorem) Let X and Y be Banach spaces and

T E L(X, Y). Then T is bounded if its graph,

1(T) = {(x, Tx) :xE

C Xx Y

is closed in Xx Y in the product topology.

Proof. One of the implications is trivial: the graph of a continuous map into a Hausdorff space is closed, so if T is bounded, 1(T) is closed. Suppose then that 1(T) is closed. Let Z be the direct sum of X and V endowed with the norm II(x,y)II = Dxli + flyll. This norm induces the product topology on Xx V and so, by assumption, 1(T) is a closed subset of Z. In fact, Z is easily seen to be complete, so 1(T) is a closed

subspace of the Banach space Z and therefore it is itself a Banach space. The linear map U: 1(T) —' X, given by U((x,y)) = x, is a norm-decreasing bijection and so, by Theorem 7, the inverse operator is bounded. But then DU'D since IIlxil

iixii —f iilxii = ilL '(x)li

iIU'iillxii.

It is worth pointing out that the closed-graph theorem often makes it much easier to prove that an operator is continuous. When is a linear operator T: X V continuous? If x1 , x2,... E X and —' x0 imply

Chapter 5: The Baire category theorem

81

that —' Tx0. Now if X and Y are Banach spaces, the closed-graph theorem tells us that in order to prove the continuity of a linear operaand Tx,, then tor T: X —, Y, it suffices to show that if x,, —. Tx0, i.e. Tx0 = Yo — a considerable gain! Exercises

1. Let f: R4 0 for all x >

R

0. R

be a continuous function such that Show that

=

= 0.

be infinitely differentiable. Suppose for every Let f: R such that = 0 for all k Prove that x E R there is a f is a polynomial. and, for n 1, set F,, = 3. Let K = [0,1], X =

{f E X: 3 t E K such that

+ h)

—

f(t)

n V h with t + h E

Deduce that the set of continuous nowhere-differentiable real-valued functions on [0,11 is dense in 4. Prove that if a vector space is a Banach space with respect to two Prove that F,, is closed and nowhere dense in CR(K).

norms then the topologies induced by the norms are either equivalent or incomparable (i.e. neither is stronger than the other).

5. Let V be a vector space with algebraic basis

e1 ,

e2,... (so

and every v E V is a unique linear combination of the dim V = e,) and let II be a norm on V. Show that is incomplete.

6. LetXbeanormed space andSCX. Showthatif{f(x):xES}is bounded for every linear functional f E X then the set S is K for some K and all x E S. (Using fancy terminology that will become clear in Chapter 8, a weakly bounded set is norm bounded.) 7. Deduce from the result in the previous exercise that if two norms on a vector space V are not equivalent then there is a linear functional f E V' which is continuous in one of the norms and discontinuous in the other. bounded: lixil

8. Let X be a closed subspace of L1(0, 2). Suppose for every f E L1(0, 1) there is an F C X whose restriction to (0, 1) is f. 9.

Show that there is a constant c such that our function F can always be chosen to satisfy IIF1I clifli. Let 1 p,q and let A = (a11)' be a scalar matrix. Suppose for every x = (x1)° the series is convergent for

Chapter 5: The Baire category theorem

82

Show that the Ely, where y, = map A: I,, —p 'q' defined by x y, isa bounded linear map. [0,1] (n = 1,2,...) be uniformly bounded continu10. Let ous functions such that

every 1, andy =

j

dx

c

0, (n = 1,2,...) and

for some c > 0. Suppose

c,,q,,,(x) n=1

for

everyxE [0,1]. Prove that n1

11.

cn

and y = (y1)° set

For two sequences of scalars x = (x,y) =

x•y,. i=1

I
E

Let

if

K for some K and all n)

is norm bounded (i.e.

and

0 for every i. (This is a characterization of sequences in tending weakly to 0; cf. Exercise 6.) — 0 for 12. Let = E (n = 1,2,...). Show that —* 0 for every i. (Simiis bounded and every y c0 iff

larly to Exercise 11, this is a characterization of sequences in tending weakly to 0.) 13. Show that for I p <

the La-norm I

r'

11111, = (j

'U

\1/P

If(t) "dt

on C[0, ills dominated by the uniform norm Ilfil = and deduce that C[0, 1] is incomplete in the norm

IfO)

I

14. Let P be a projection on a Banach space X, i.e. let P be a linear map of X into itself such that P2 = P. Show that X is the direct and ImP = PX if and only if sum of the subspaces KerP = P is bounded.

15. Let X and Y be normed spaces and T E graph of T is closed 1ff whenever x,,

0

and

Y). Show that the y then y = 0.

Chapter 5: The Baire category theorem

83

As we shall see in the following exercises, the results of this chapter are very useful in proving the fundamental properties of Schauder bases. Recall from Chapter 2 that a sequence is a Schauder basis or simply a basis of a Banach space X if every x X has a unique representawith the series convergent in norm. tion in the form 16. Let

be a basis of a Banach space X. Define P,

by

= Prove that for x E X set Mxlii = is a norm on X and X that is complete in this norm as well. and be as in the previous exercise. Prove that 17. Let X, and [The number is called cf. Exercise 4.13.] the basis constant of 18. Let x1 , x2,... be non-zero vectors in a Banach space X, with is a basis of X if and only if there ) = X. Prove that is a constant K such that and

A,xaO

for all 1 m
,=

[—1, 2k

1] as follows. Set

and define 1

<1+1,

if

(

20.

otherwise.

0

The sequence is called the Haar system. Prove that the Haar system is a basis in 1) for I p Define a sequence of continuous functions ç,1: [0, 1] [0, 1] as follows. Set 1 and for n 2 put =

L

du,

X* is the k th Haar function, defined in the previous exercise. The sequence is called the Schauder system. Prove 1) with the uniform that the Schauder system is a basis in norm. Show also that the basis constant is 1. where

Chapter 5: The Baire category theorem 21.

Let

be a sequence in a normed space such that n=1

for every f E X*. Show that there is a constant M such that

for everyf E X. 22. Use the Baire category theorem to deduce the result in Exercise 4.15: the linear span (not the closed linear span!) of an infinite sequence of linearly independent vectors in a normed space cannot be complete. 23k. Construct a set SC [0, 1] of measure I that is of the first category in [0. 1]. Notes Most of the results in this chapter can be found in Stefan Banach's classic, cited in Chapter 2. The original paper of R. Baire is Sur Ia convergence de variables réelles, Annali Mat. Pura e AppI., 3 (1899), 1—122; Osgood's theorem is in W. F. Osgood, Non-uniform convergence and the integration of series term by term, Amer. J. Math., 19 (1897),

155—90. The Banach—Steinhaus result was proved in S. Banach and H. Steinhaus, Sur le principe de Ia condensation de singularites, Fundamenta Math., 9 (1927), 51—7.

6. CONTINUOUS FUNCTIONS ON COMPACT SPACES AND THE STONE-WEIERSTRASS THEOREM

In this chapter we shall study one of the most important classical Banach spaces, the space C(K) of continuous complex-valued functions on a compact Hausdorif space K, with the uniform (supremum) norm 11111

= sup

z€K

= maxlf(x)I. xEK

In fact, as we shall see in Chapter 8, every Banach space is a closed subspace of C(K) for some compact Hausdorff space K, although this does not really help in the study of a general Banach space. First we shall show that C(K) is large enough, namely that it contains sufficiently many functions; for example, every continuous function on a

dosed subset of K can be extended to a continuous function on the whole of K without increasing its norm. This result, which is a special case of the Tietze—Urysohn extension theorem, is thus reminiscent of the Hahn—Banach theorem: it ensures that the space C(K) is large, just as the Hahn—Banach theorem ensured that the dual space of a normed linear space was large. It is worth emphasizing that the existence of a good stock of continuous functions on a topological space X cannot be taken for granted, not even when X seems to be very pleasant. For example, it can happen that X is a countable Hausdorff space and still every continuous function on X is constant (see Exercise 19). In the second part of the chapter we shall show that C(K) is not too

large in the sense that it is the norm closure (i.e. the closure in the topology induced by the norm) of a good many 'small' subspaces of functions.

We shall start with a standard result in analytic topology which is likely to be familiar to many readers.

Chapter 6: Continuous functions on compact spaces

86

A topological space is said to be normal if every pair of disjoint closed sets can be separated by disjoint open sets: if A and B are disjoint closed sets then there are disjoint open sets U and V such that A C U and B C V. Equivalently, a space is normal if for all A0 C U0, where A0 is closed and U0 is open, one can find a closed set A1 and an open set U1 such that A0 C U1 C A1 C U0. Lemma 1.

(a) In a Hausdorif space any two disjoint compact sets can be separated by open sets. (b) Every compact Hausdorff space is normal.

Proof. (a) Let A and B be non-empty disjoint compact sets in a Hausdorif space. For a E A and b E B, let U(a, b) and V(a, b) be disjoint open neighbourhoods of a and b. Let us fix a point a A. Clearly, V(a, b1) for B C Uh E B V(a, b) and so, as B is compact, B C Set

some

U(a) = fl U(a,b1)

and

V(a) = IJ V(a,b,).

Then U(a) and V(a) are disjoint open sets, U(a) is a neighbourhood of a and V(a) contains B.

Now note that U0 compact, A C

A

U(a,)

is an open cover of A and so, as A for some a1,. ..,am E A. Let

U(a)

U = U U(a1)

and

V=

fl

is

V(a1).

Then U and V are disjoint open sets, A C U and B C V. (b) In a compact space every closed set is compact, so the assertion 0 follows from (a). If A and B arc disjoint closed sets in a metric space X then there is a continuous function f: X [0, 1] such that f is 0 on A and 1 on B; indeed, —

Jmin{d(x,A)/d(x,B),1} i

B

xEB

will do. The next important result tells us that such an f exists not only on metric spaces but also on normal spaces.

Theorem 2. (Urysohn's lemma) Let A and B be disjoint closed subsets of a normal space X. Then there is a continuous function f: X —' [0, 1] and B C f'(l). such that A C

Chapter 6: Continuous functions on compact spaces

87

Proof. Let q0 = 0, q1 = I and let q2,q3.... be an enumeration of the rationals strictly between 0 and 1. Set LI0 = 0. A0 = A, U1 = )AB and A1 = X. Let us construct a sequence of closed sets A0, A1 ,... and a

that U, C A, and if q
U1 ,..

such

(I < n) which is less than q,, and let q1 be the minimal q1 (1 < n) which is greater than q,. Then Ak C U, and so, as X is normal, one can find a closed set A,, and an open set U,, such that Ak C U,, C A,, C U,. Having constructed A0,A1 .... and U0, U1,..., it is easy to define a suitable continuous function f: for x E X set

f(x) =

: x E A,,}.

I and f is 0 on A and I on B. All we have to check
f(x)

are ready to present one of the main results of this section. By Urysohn's lemma, if A and B are disjoint closed (possibly empty) subsets of a normal space X then there is a continuous function g = g(A,B): X—' [—1,1] such that gJA = —I and gIB = 1. Equivalently, if C is a closed subset of X and h is a continuous function from C into 1} then h has an extension to a continuous functhe two-point set tion g: X—' [—1,1]. The Tietze—Urysohn extension theorem below We

claims that every continuous function 1: C [—1, 1] has such an extenThis result is easily proved by using Urysohn's lemma to provide sion. us with continuous functions on X whose restrictions to C are better and better approximations of f. Indeed, setting A0 = {x E C: f(x) 0} and B1) = {x C C: f(x) there is a continuous function G0: [0, fl such that = 0 and Then f' = f—G0 IC maps C into [—I, fl — a very good G01B0 =

start indeed. Next, set A1 = {x C C: f'(x)

—

B1

= {x C C: f'(x)

0)

0] such that HLIAI = and take in a continuous function H1 is a good approximation of f on C, and H21B1 = 0. Then F1 = G0 + and Continuing in this namely f— FIIC maps C into [—

way, we get continuous functions

F2, F3,

... with

2" and

Chapter 6: Continuous functions on compact spaces

88

I

Thus F =

mapping C into the required properties. —

F, will have

For the sake of completeness, in the proof below we shall present a streamlined version of this approach.

Theorem 3. (Tietze—Urysohn extension theorem) Let C be a closed subset of a normal space X and let f: C [—1, 1] be a Continuous function. Then f has a continuous extension to the whole space X, i.e. there is a continuous function F: [—1,1] such that FIC = f. Proof. We shall construct F as the uniform limit of continuous functions on the whole of X whose restrictions to C are better and better approximations of f. To be precise, set fo = A0 = {x

E C: f0(x)

—

and

4}

B0 = {x E C: f0(x)

By Urysohn's lemma there is a continuous function F0: X [—4, 4] such that F0 IA0 = —4 and F0 I = 4. Set fi = fo — F0 IC and note that is a continuous map of C into [—i, In general, having constructed a continuous function : C—p

define

= {x E C:

and

= {x E C:

By Urysohn's lemma there is a continuous function r such

I

2n 12

that =

= i(2)n.

and

Set

and note that Let f1 ,...

in this way. F(x) =

is a continuous map C into .... be the sequences of functions constructed Then for all x X and n 0 and so and F0. F1

is continuous, being a uniform limit of continuous

functions, and IF(x)I

4

=1 ii=0

Chapter 6: Continuous functions on compact spaces

Furthermore, for x E C and n

89

0 we have =

<(4)fl+1

- k=O

0

andsof(x)=F(x).

It may be worth noting that it is trivial to ensure that F(x) F(x) is any continuous extension of f then -

I

1:

if

F(x) = will do.

From now on, let K be a compact Hausdorif space and let us consider the space C(K) of all continuous (complex-valued) functions on K. This is a Banach space in the uniform norm Ilfil = sup{If(x) x E K}. The results above tell us that C(K) is rather rich in functions. Namely, C(K) separates the points of X: for any two distinct points x and y there is a function f E C(K) such that f(x) # f(y). Also, if A and B are disjoint

closed subsets of K then for some f E C(K) we have f(a) =

0 and

f(b)= 1 for all aEA and bEB. Finally, if ACK is closed and f E C(A) then f = FIA for some F E C(K). To see the last statement, simply write f as a sum of its real part and its complex part.

The space C(K) is closed under uniform limits, and this is another source of functions that can be guaranteed to belong to C(K). This leads us to consider relatively compact subsets of C(K), that is subsets in which every sequence has a subsequence convergent to some element of C(K). Recall that a topological space is compact if every open cover has a finite subcover, it is countably compact if every countable open cover has a finite subcover and it is sequentially compact if every sequence has a convergent subsequence. Given a metric space X and N C X, the set N is an €-net if for every x E X there is a pointy EN with d(x,y) <€.

€-nets one tends to consider are finite. A metric space is totally bounded or relatively compact if it contains a finite €-net for every e > 0, or, equivalently, if every sequence has a Cauchy subsequence. For a metric space, the properties of being compact, countably compact and sequentially compact coincide; furthermore, a metric space is compact iff it is complete and relatively compact. In particular, a subset of a complete metric space is relatively compact if its closure is compact; a closed subset of a complete metric space is compact 1ff it is relatively The

compact.

Chapter 6: Continuous functions on compact spaces

90

In order to characterize the relatively compact subsets of the complete

metric space C(K), let us introduce some more terminology. A set S C C(K) is uniformly bounded if it is bounded in the supremum norm, N for some N and all f C S. We say that S is equicontinui.e. if Ilfil 0 there is a neighbourhood U of x ous at a point x C K if for every such that if y E and f E S then 11(x) —f(y)I < €. Furthermore, S is equicontinuous if it is equicontinuous at every point. Theorem 4. (Arzelà—Ascoli theorem) For a compact space K, a set

S C C(K) is relatively compact if and only if

it

is uniformly bounded

and equicontinuous. Proof. (a) Suppose that S is a totally bounded subset of C(K).

Let

0. The set S contains a finite c-net: there is a finite set x E K and f,,} CS such that if fE S then hf—fill
If(x)f(y)l

lf(x) —fa(4l

+ lf(x)—ft(y)l + If,(y)—f(y)I <3€.

(b) Now suppose that S is uniformly bounded and equicontinuous. Given 0, for x E K let If(x)—f(y)I <€ if fE S be

be an open neighbourhood of x such that

and yE

Then

let K = LJEK a finite subcover of K. Since S is uniformly bounded,

the set R = {(f(x1),.

f C S} is a bounded subset of 1,

. .

the

vector space C" endowed with the supremum norm. (As all norms on a finite-dimensional vector space are equivalent, we could have chosen any norm on C".) A bounded set in 1 is totally bounded because a bounded closed set in 1 is compact; therefore R is totally bounded so there are functions ,. , f,,,, E S such that the set .

.

1

is

i

m}

an c-net in R.

We claim that the functions ft,... 'fm form a 3€-net in S. Indeed, given f E S, there is an i (1 I m) such that f(x1) —f,(x1)l < c for n). Let now xE K be an arbitrary point. Then xE all j (I for some 1(1 j n). Consequently, 1(x) —f,(x)h

If(x) —f(x1)I + 1(x1) —f1(x1)I + lf,.(x1) —f1(x)I

<3€.

0

Chapter 6: Continuous functions on compact spaces

91

In conclusion, let us see that CR(K), the space of continuous realvalued functions on K, is closed under monotone pointwise limits. This is Dini's theorem. Theorem

Let K be

5.

CR(K).

a

compact

topological

space

and

let

Suppose that for every x C K the sequence

monotone increasing and tends to f0(x). Then (f,1 ) 0. verges uniformly to Jo' i.e. foli

con-

) i is

such Proof. Let e > 0. For every x E K there is a natural number > f0(x) is a non-negative continuous function, As that x has an open neighbourhood (i, such that if y C U then

<€.

0

and y C U,, then

Note that if n 0

fo(Y)

Jn(Y)

<€.

fo(y)

is certainly a cover of K and K is compact, K is covered

Since

by a finite number of these sets: K =

Il,... ,Xk. The

last inequality implies that if n then IJo(y)fn(y)I <e for ally, and so

for some points n0 = max{nR1,.

. .

0

It is worth noting that all three conditions are needed in Theorem 5, namely that K is compact, the sequence of continuous functions is pointwise increasing and the pointwise limit, Jo is a continuous function (see Exercise 13). Now let us turn to one of the most fundamental results in functional analysis, the Stone—Weierstrass theorem, telling us that the sapce C(K) is not too large: it is the uniform closure of a good many pleasant subspaces, more precisely, it is the uniform closure of a good many subalgebras. Recall that a topological space is said to be locally compact if

every point has a compact neighbourhood. For a locally compact Hausdorif space L, let C(L) =

{f E CL: J is continuous and bounded},

C0(L) = {J

C(L): the set {x: IJ(x)

I

e} is compact for every e > 0}

and

C C(L): suppf is compact}, where suppf, the support of a (not necessarily continuous) function f, is the closure of the set {x: f(x) 0).

Chapter 6: Continuous functions on compact spaces

92

Note that if L is compact then C(L) is just the set of all continuous is the set of all continuous functions with compact support and C0(L) is the set of all continuous function f on L e} is compact for every e > 0. for which {x E L: If(x)l If f and g are functions on the same space, define Af. f+g and fg by pointwise operations. With these operations, C(L) is a commutative functions; similarly.

algebra. This definition,

algebra has a unit, the identically

I

function.

By

C C0(L) C C(L) and CC(L) and C0(L) are subalgebras

of C(L). We know that the vector space C(L) is a Banach space with the uniform (supremum) norm: Ilfil

= sup{lf(x)I: x E L}.

Theorem 6. Let L be a locally compact Hausdorff space. The function

is a norm on C(L) and the space C(L) is complete in this norm; is a dense subspace of C0(L) is a closed subspace of C(L) and C0(L). Furthermore, forf,g E C(L) one has (1)

Ilfil and

I. the identically I function, has norm 1.

is dense in C0(L). Proof. The only assertion needing proof is that e}. Then K0 is = (x E L: If(x)I Given f E C11(L) and >0, set be an open neighbourhood of x with compact. For x E K0 let for Then K0 C compact closure (ix and so K0 C some x1 ,. E K(,. Setting K1 = compact set such that K0 C tnt K1. .

.

we find that K1 is a

By Urysohn's lemma (Theorem 2) there is a continuous function and 0 on K1\lnt K1. Extend h to a funch: K1 [0, 1] that is I on tion g on L by setting it to be 0 outside K1: h(x)

ifxEK1,

0

Then g E

and so gf E

Clearly IIgf—fII

An algebra A which is also a normed space and whose norm satisfies (1) is called a normed algebra. If the algebra has an identity e and the norm of e is I then it is called a unital normed algebra. If the norm is

complete then the algebra is a Banach algebra. We saw in Chapter 2 the set of bounded linear operators on a normed space X, is that is a unital a unital normed algebra and if X is complete then

Chapter 6: Continuous functions on compact spaces

93

Banach algebra. The trivial part of Theorem 6 shows that C(L) is a commutative unital Banach algebra, C0(L) is a commutative Banach is a commutative normed algebra. algebra and are defined analogously; they The spaces CR(L), and consist of the appropriate continuous real-valued functions. These spaces are not only real normed algebras but they are also lattices under the natural lattice operations fvg and fAg. Given real-valued functions f and g on a set S, define new functions fvg and fAg on S by setting, for x E S,

(fvg)(x) = max{f(x),g(x)}

and

We call fvg the join off and g and fAg

(fAg)(x) = min{f(x),g(x)}. the

meet off and g. The join

and the meet are the lattice operations.

Of course, the join of two functions is just their maximum and the and meet is just their minimum. It is clear that CR(L), are all closed under the lattice operations. For a set A of bounded functions on a set S. the uniform closure A of the space of A is the closure of A in the uniform topology on bounded functions on S with the uniform norm (Example 2.1 (ii)). Thus a function f on S belongs to A if for every 0 there exists a g E A such that If(x) <e for all x E S. In this chapter we shall study the uniformly closed subalgebras of C(L), C0(L), CR(L) and so we are particularly interested in uniform approximations by elements from various subsets of these algebras. As the next result shows, the lattice operations are very useful when we look for such approximations. Theorem 7. Let K be a compact Hausdorif space and let A be a subset of which is closed under the lattice operations. Then A, the uniform closure of A, is precisely the set of those continuous (real-valued) functions on K that can be arbitrarily approximated at every pair of points by functions from A.

Remark. Note that A is not assumed to be a subalgebra, not even a subspace, it is merely a set closed under the lattice operations, i.e. it is a sublattice of CR(K).

Proof. It is obvious that the uniform closure A is at most as large as claimed.

Suppose then that f can be approximated by elements of A at every pair of points. We have to show that f E A.

Chapter 6: Continuous functions on compact spaces

94

Let e > 0. The existence of approximations of f means precisely that E A such that for x, yE K there exists a function

<€ = {z E K:

Put

containing

x and y.

the

also belongs

= {y E K: so

K

x1,

•

.

,X,.•,E

K.

is an open set

and so a

x, we find that K = also cover K, say K =

By

minimum of the corresponding functions

>f(x)—E Let

<E•

Fixing

finite number of the sets

assumption, namely =

and

that, Let

to

and

•

fr,,,,,,

A. Clearly for

allyE K.

is open and K = f(y) —e}. Then for some compact, K = be the maximum of the =L1vL2v"

as K is

E A, and

corresponding functions. By assumption,

forallxEK.

f(x)—€
0

0, f E A.

As there is such an fE for every

the space of also closed under the lattice

Lemma 8. A uniformly closed subalgebra A of gb(S), bounded real-valued functions on a set S operations.

is

Since (fvg)(x) = max{f(x),g(x)} = andfAg = f+g—fvg, it suffices to show that ermore, as for n 1 we have nfl = nlfl, andf Proof.

it

suffices

to show that

flEA.

if f

A and

11111

A iff E A. FurthA holds if nf E A,

= supx€s If(x) I <

1

then

3)k be the Tayflffl < 1 and e > 0. Let Ck(t— lor series for about z = 3. As this series converges uniformly on [0, 1], there is a polynomial

Let then f E

A,

P(:)

Ck(t—

3)k

k=O

such that for all 0

t

1.

Hence IP(s2) —(s2+€2)"2J <

for all

—1

s

1

Chapter 6: Continuous functions on compact spaces

95

and so I

Set Q(s) =

P(s2)

P(s2)

—

—

Is II

P(O).

< 2E

for alt —1

s

1.

Then Q is a real polynomial with constant

term 0, and so f E A implies that Q(f) E A. Furthermore, for —1

s

1 we have IQ(s)—lsII

IP(s2)—IsII+IP(O)I

Hence

IQ(f)(x)—If(x)II <4€

foralixEK

and so

IIQ(f) —fill

that IfI can be approximated by the function Q(f) E A within 4€. Since A is uniformly closed, Ill E A. 0 showing

We are ready to prove the first form of the main result of this section. We say that a set A of functions on a set S separates the points of S if for all x,y E S (x y) there is a function f E A with f(x) f(y). Theorem 9. (Stone—Weierstrass theorem) Let K be a compact space

and let A be an algebra of real-valued continuous functions on K that separates the points of K. Then A, the uniform closure of A, is either CR(K) or the algebra of all continuous real-valued functions vanishing at

a single point x. E K.

is no point E K such that f E A. It is easily checked that then for all x y

Proof. (a) Suppose first that there

(x,y E K) there is a function f C A such that f(x) f(y), f(x) 0 and f(y) 0. Then any function on K, say g E fiX, can be approximated at

and y by elements of A. In fact, we can do rather better than only approximate: since (J(x), f(y)) and (f2(x) , f2(y)) are linearly independent vectors in fl2, some linear combination of them is (g(x),g(y)). So for some linear combination h = sf+ jf2 of f and f2 we have h(x) = g(x) and h(y) = g(y). By Theorem 7 and Lemma 8 we have A = (b) Suppose now that = 0 for some C K and all f C A. We x

have to show that if g E CR(K) and g the constant functions to A, i.e. let B C CR(K) be the

algebra of functions of the form f+ c where f C A and c fl Given >0, by part (a) there exists a function h = f+c C B, with! E A and

96

Chapter 6: Continuous fwzc:ions on compact spaces

<€. As E R, such that and so PIf —gil <2€. Hence g E A. C

= Owe have PcI

=

<e

0

Let us see now what Theorem 9 tells us about subalgebras of C(L). For a locally compact Hausdorif space L, the one-point compacuficazion where and a set Sc of L is the topological space on = is open iff(a) SflL is open in Land (b) E S then LIS is compact. It is easily checked that is a compact Hausdorif space inducing the original topology on L. Set =

{f E

= O}.

C,?(L) is a closed subalgebra of CR(L). Furthermore, if L is a locally compact Hausdorif space then f fl L is an isometric isomorand This is why the elements of phism between are said to be the continuous functions on L vanishing at These remarks imply the following reformulation of the Stone— Weierstrass theorem for real functions. We shall say that a set A of functions on a set S separates the points of S strongly if for all x,y S (x y) there are functions,f,g E A such that f(x) f(y),f(x) 0 and g(y) 0. Note that by Theorem 9 if a subalgebra A of CR(K) separates the points of K strongly then A = CR(K). Then

Theorem 9'. Let L be a locally compact Hausdorif space and let A be a strongly separating the points of L. Then A is subalgebra of dense in i.e. A = 0 by CR(L) It is obvious that in Theorem 9' we cannot replace itself is a uniformly closed subalgebra of CR(L). Furthby C0(L) either. This can be seen ermore, we cannot replace from the following example. Let 4 = {z E C: 1} be the closed unit disc in C and let A consist of those continuous functions on 4 that are analytic in the interior of 4. Then A is a closed subalgebra of C0(4) = C(4), it strongly separates the points but does not contain the functionf(z) = C0(4). The example above is, in fact, not an ad hoc example but one that goes to the heart of the matter: what we lack is complex conjugation. This leads us to a form of the Stone—Weierstrass theorem for complex functions, which is an easy consequence of the second variant, Theorem 9'. because

I

Theorem 10. (Stone—Weierstrass theorem for complex functions) Let L

be a locally compact Hausdorif space.

Suppose A is a complex

Chapter 6: Continuous functions on compact spaces

97

subalgebra of C0(L) which is closed under complex conjugation and strongly separates the points of L. Then A = C0(L). the set of real-valued functions in A. Then AR is a real As AR contains subalgebra of A and it is also a subalgebra of Proof. Let

be

(f+ 1)/2

and (f — f) /2i

Theorem

9'. Therefore, C0R

1C0R(L)

Let us theorem;

cA

for every f and

A, it satisfies the conditions of C A and so C0(L)

cA.

=

0

some immediate consequences of the Stone—Weierstrass the first is the original theorem due to Weierstrass. note

Corollary 11. Every continuous real-valued function on a bounded closed subset of IR?* can be uniformly approximated by polynomials.

0

Corollary 12. Every continuous complex-valued function on the circle T = {z E C: IzI = 1} can be uniformly approximated by trigonometric polynomials

o

k = -it

Corollary 13. Let K and L be compact Hausdorff spaces and endow

KxL with the product topology. Then every continuous function f: Kx L C can be uniformly approximated by functions of the form g.(x)h,(y), where g• E C(K) and h,

E C(L) (1

The Stone—Weierstrass theorem is very useful in

i

n).

0

showing that various

of function spaces are dense. As an example, let us look at 1], where 1 p < °o. The subspace of the real space 1] subspaces

consisting of continuous functions is dense. Since every continuous realvalued function on [0,11 can be uniformly approximated by polynomials, and the uniform norm dominates the La-norm, the space of polynomials is also dense in 1]. Exercises

1. Show that a metric space is compact 1ff it is countably compact if it is sequentially compact. 2. Show that a metric space is compact if and only if it is totally bounded and complete. 3. Show that a subset of a complete metric space is totally bounded if and only if its closure is compact.

98

Chapter 6: Continuous functions on compact spaces

4. A topological space X is completely regular if for every closed set A C X and point b not in A there is a continuous function f: [0,1] such that hA = 0 and f(b) = 1. Prove that the halfopen interval topologies of the real line are completely regular.

5. Let A be a subset of a normal topological space X and let f: A [0, 1] be a continuous function. Does it follow that f has a continuous extension to the whole of X? And what is the answer if X is a compact Hausdorff space?

the normed space of bounded continuous realvalued functions on a topological space X, with the uniform

6. Let CR(X)

be

(supremum) norm. Show that CR(X) is a Banach space.

7. Let V be a subspace of CR(X) such that whenever A and B are non-empty disjoint closed subsets of X and a, b E R then there is a a and fIB = b. Prove that V is dense in Ca(X).

function f E V such that f(X) C [a, b], fI A =

8. Let K be a compact metric space and S C C(K). Show that S is equicontinuous if for every e > 0 there is a 8 > 0 such that If(x) — f(y) <e whenever d(x, y) <8. Is this true for every I

metric space K? 9. Prove the following complex version of the Tietze—Urysohn extension theorem. Let A be a closed subset of a normal topological

be a continuous function such that = 1. Show that there is a continuous func-

space X and let f: A —* IlfilA = SUPXEA 1(x) I

C

tion F: = 1. C such that flFflx = 10. Let L be a locally compact Hausdorff space. Show that a set SC is totally bounded if it is bounded, equicontinuous and for every 0 there is a compact set K C L such that jf(x) I <€ for all x E L\K and f E S. D = {zE C 11. Let B, = {zE C Izl < 1) and let F be a M, if IzI = r < 1. set of functions analytic on D such that If(z)l Show that for r < 1 the set FIB, {IIB,: E F} is a relatively compact subset of C(B,). Deduce that every sequence CF which is uniformly convergent on contains a subsequence every B, (r < 1) to a function f analytic on D. 12. Let U be an open subset of C and let '12,...: U —i C be uniformly bounded analytic functions. Prove that there is a subsequence which is uniformly convergent on every compact :

:

f

subset of U.

13. Show that all three conditions are needed in Dini's theorem, i.e. that the conclusion 0 in Theorem 5 need not hold if —foil

Chapter 6: Continuous functions on compact spaces

99

only two of the following three conditions hold: (i) K is compact; (ii) the functions are continuous; (ni) is monotonic. 14. Let G be an open subset of P2 and let f: G P be continuous. Use the Arzelà—Ascoli theorem to prove Peano's theorem that for each point (x0,y0) E G, at least one solution of y'(x) = f(x,y) passes

through (x0,y0).

[HINT:

of (x0,y0) and set K = < b such that

Let V be a closed neighbourhood (x,y) 1'). Choose a <

x0

f(x,y):

x

[a,b], x

YYo
The aim is to show that our differential equation has a solution through (x0,y0), defined on [a, b]. Find such a solution by considering piecewise linear approximations, say with division points x0±k/n (k = 15. Let K be a compact metric space. Prove that C(K) is separable. 16. Let K be a compact Hausdorff space. Suppose C(K) = U,,.1 where each C,, is an equicontinuous set of functions. What can you say about K? 17k. We say that e > 0 is a Lebesgue number of a cover U,.EF U,. of a metric space if, for every point x, the ball B(x, €) is contained in some U,.. Show that a metric space (X, d) is compact 1ff for every metric equivalent to d every open cover has a (positive) Lebesgue number. 18. Let K be a compact metric space, L a metric space, and let C(K,L) to be set of continuous mappings from K to L with the uniform metric

d(f,g) =

sup d(f(x),g(x)) xEK

= maxd(J(x),g(x)). xEK

Show that a subset S of C(K,L) is totally bounded if and only if it is equicontinuous (i.e. for all x E K and E > 0 there is a S > 0 such

that if d(x,y) <8 then d(f(x),f(y)) <€ for every f E S) and the set {f(x): f E S} is a totally bounded subset of L for every x K. Construct a countable Hausdorif space on which every continuous function is constant. The first such example was given by Urysohn in 1925.

20. Let K be a compact Haudsorif space. What are the maximal ideals of CR(K)? And the maximal closed subalgebras?

Chapter 6: Continuous functions on compact spaces

100

21. Let K be a compact Hausdorif space. For a continuous surjection

of K onto a compact Hausdorff space K', let be given by ç(f)(x) = (fco)(x) = Show that is a closed unital subalgebra of CR(K) and every closed unital subalgebra of CR(K) can be obtained in this way. Give a similar description of the closed subalgebras of CR(K).

22. Let X be a normal topological space and let U1,. be open sets covering X. Prove that there are continuous functions : X— [0,11 such that fk(x) = 1 for every x and fe,. . .

= 0 for x said

Uk

(k =

1,.

. .

to form a partition of

(The functions f1,. . . , j, are unity subordinate to the cover

,n).

Notes

The relevant results of Urysohn and Tietze are in P. Urysohn, (..Yber die Mdchtigkeit der zusammenhangenden Mengen, Math. Ann., 94 (1925), 262—95 and H. Tietze, Uber Funkrionen, die auf einer abgeschlossenen

Menge stetig sind, J. für die reine und angewandte Mathematik, 145 (1915), 9—14. Heinrich Tietze proved a special case of the extension theorem, while Paul Urysohn (who was Russian but tended to publish in

German) proved Theorems 2 and 3 in the generality we stated them. Various special cases of the extension theorem were proved by a good many people, including Lebesgue, Brouwer, de Ia Vallée Poussin, Bohr and Hausdorif. For the Arzelà—Ascoli theorem see C. Arzelã, Sulk serie di funzioni (paiie prima), Memorie Accad. Sci. Bologna, 8 (1900), 131—86. Of course, there are several excellent books on general topology; J. L. Kelley, General Topology, Van Nostrand, Princeton, New Jersey, 1963, xiv and 298 pp., is particularly recommended.

The original version of Theorem 9 is in K. T. Weierstrass, Uber die analytische Darsteilbarkeit sogenannier willkürlicher Funktionen reeler Argumente, S.-B. Deutsch Akad. Wiss. Berlin, KI. Math. Phys. Tech., 1885, 633—39 and 789—805. The Stone—Weierstrass theorem itself is in

M. H. Stone, Applications of the theory of Boolean rings to general topology, Trans. Amer. Math. Soc., 41(1937), 375—81; an exposition of

the field can be found in M. H. Stone, The generalized Weierstrass approximation theorem, Math. Mag., 21 (1948), 167—84.

7. THE CONTRACTION-MAPPING THEOREM

This chapter is more or less an interlude in our study of normed spaces. S, a point x ES satisfying 1(x) = x Given a set Sand a functionf:

is said to be a fixed point of 1. There is a large and very important body of 'fixed-point theorems' in analysis, that is, results claiming that every function satisfying certain conditions has a fixed point. Theorems

of this kind often enable us to solve equations satisfying rather weak conditions. The aim of this section is to present the most fundamental of these fixed-point theorems and some of its great many applications. In Chapter 15 we shall return to the topic to prove some more sophisticated results. A map f: X —' X of a metric space into itself is said to be a contraction if

d(J(x),f(y)) for some k <

1

kd(x,y)

(1)

and all x,y E X. One also calls f a contraction with

It is immediate from the definition that every contraction map is continuous; in fact, it is uniformly continuous. Although we shall not make much use of this terminology, a function f between

constant k.

metric spaces is said to satisfy the Lipschitz condition with constant k if (1) holds for all x and y. Thus a map f from a metric space into itself is a contraction map if it satisfies a Lipschitz condition with constant less than 1. The result below is usually referred to as Banach's fixed-point theorem or the contraction-mapping theorem. Theorem 1. Let f: X —+ X be a contraction of a (non-empty) complete metric space. Then f has a unique fixed point. 101

102

Chapter 7: The contraction-mapping theorem

Proof. Suppose that k < 1 and f satisfies (1) for all x,y E X. Pick a point x0 E X and set x1 = f(x0), x2 = f(x1), and so on: for n 1 set = f(x,, Writing d for d(x0 , xi), we find that k"d.

Indeed, (2) is trivially true for n =

1;

(2)

assuming that it holds for n, we

have = The triangle inequality

and (2) imply that if n

<m then

rn—n—i

i=O

1—k

Hence is a Cauchy sequence and so it converges to some point x C X. Since f is continuous, the sequence converges to f(x). converges But to x and so f(x) = x. The uniqueness = of the fixed point is even simpler: if f(x) = x and f(y) = y then d(x,y) = d(J(x),f(y)) kd(x,y), and so d(x,y) = 0, i.e. x = y. 0 When looking for a fixed point

of a contraction map f,

it

is often use-

to remember that the fixed point is the limit of every sequence (i, consisting of the iterates of a point x0, i.e. defined by picking a point x0 and setting = for n 1. Let us note the following slight extension of Theorem 1. ful

Theorem 2. Let X be a complete metric space and let f: X X be such that f" is a contraction map for some n 1. Then f has a unique fixed

has a unique fixed point x,

Proof. We know from Theorem 1 that

say. Since f'1(f(x))

= f(f(x))

=

f(x),

is also a fixed point of f and so f(x) = x. As a fixed point of f is also a fixed point of f's, the map f does have a unique fixed point. 0

f(x)

very simple results above and their proofs have a large number of applications. In the rest of this chapter we shall be concerned with two The

Chapter 7: The contraction-mapping theorem

directions: first we shall study maps between Banach spaces and then we shall present some applications to integral equations. The proof of Theorem 1, sometimes called the method of successive approximations, is very important in the study of maps between Banach

spaces. Here is a standard example concerning functions of two vanables which are Lipschitz in their second variable. Recall that for a normed space X we write Dr(X) = {x E X: lixil
F: Dr(X))
Y

be a continuous map such that (3)

and

<s(1—k)

(4)

for all x E Dr(X) and Yi 'Y2 E !)3(Y), where r,s > 0 and 0 < k < Then there is a unique map y: DS(Y) such that y(x) = F(x,y(x))

1.

(5)

for every x E D,(X), and this map y is continuous. Proof. For x E D,(X), define yo(x) = 0, and

d=

Furthermore, if n

= F(x,yo(x)) = F(x,0)

Note that,

11y1(x) — y0(x)

II = IIyi(x) 1 and E DS(Y) then set

by (4), d <s(i — k).

=

We claim that

is

defined for every n, and that

k'd

<s.

(7)

As in the proof of Theorem 1, this follows by induction on n. Indeed, (6) clearly holds for n = 0; assuming that y0(x) , yi(X),. . . , E Y)

and (6) holds for n—i, inequality (3) implies that =

Chapter 7: The contraction-mapping theorem

104

In turn, this implies that (7) holds for n: ILvo(x)II

+

Uy1+i(x)—y1(x)II i—O

< 5,

so, in particular, E As F is a continuous function, each D,(X) D5(Y) is continuous. Inequalities (6) and (7) imply that the sequence is uniformly and

convergent to a continuous function y from Dr(X) to Then (5) holds trivially:

y(x) = lim y(x) is unique, since if yo(x) is a solution of (5) then Ily(x) —yo(x) II

and

= IIF(x,y(x)) — F(x,yo(x))II

—yo(x)

0

so y(x) = y0(x).

The next result claims that a suitable map close to the identity map is, in fact, a homeomorphism.

Theorem 4. Let X be a Banach space and let e: D5(X) traction with constant k < 1 such that

X be a con-

< r, where

s>

0

and r =

4s(1

—k). Define f:

X by

f(x) = x+€(x). Then there is an open neighbourhood U of 0 such that the restriction of

f to U is a homeomorphism of U onto D,(X). Proof. Define F: D,(X) X

X by F(x,y) = x—e(y).

F satisfies the conditions of Theorem 3. Indeed, (3) holds by assumption; furthermore, if x E thefl Then

IIF(x,0)D and

= 11xe(0)II < r+r = s(1—k),

so (4) also holds. Hence, by Theorem 3, there is a unique function

Chapter 7: The contraction-mapping theorem

g: D,.(X)

105

D3(X) such that g(x) = F(x,g(x)) = x—e(g(x)),

i.e. x = g(x) + €(g(x)) = f(g(x))

for all x E Dr(X). Set U = g(D,.(X)). To complete the proof, we have to check that U is an open neighbourhood of 0 and f: U —+ D,(X) is a homeomorphism. Clearly,f is a one-to-one map of U onto D,(X). Since g is unique, the continuity off implies that U =7 '(D,(X)) is open. As g is also continuous,fis indeed a homeomorphism. Finally,f(0) = e (0) E D,.(X) and so

0

OEMU.

From Theorem 4 it is a small step to the inverse-function theorem for Banach spaces.

In defining differentiable maps, it is convenient to use the 'little oh' notation. Let X and Y be Banach spaces. Given a function a: D —+ Y, where D C X, we write

a(h) =

o(h)

if for every e > 0 there is a S > 0 such that Ila(h)II

eflhfl

h E D and lihil <8. Thus 'a(h) tends to 0 faster than h'. Let U be an open subset of X and let f: U Y. We call f differentiY) such that able at x0 E U if there is a linear operator T E whenever

f(x0+h) = f(x0)+Th+o(h).

The operator T is the derivative of f at x0: in notation, f'(x0) = Thus f is differentiable at x0 with derivative T E 0 there is a 8 > 0 such that Ilf(xo+h)—f(xo)--ThH whenever lihIl

T.

Y) 1ff for every

eIIhII

<S.

Note that a linear operator T E Y), x Tx, is differentiable at every point, and T'(x) = T for every x X. We shall need the following simple analogue of the mean-value theorem. Lemma 5. Let X and Y be Banach spaces, U C X an open convex set and f: U Y a differentiable function, with

Chapter

106

7: The contraction-mapping

Ilf'(x)II

theorem

k

for all x E U. Then flf(x)—f(y)II

for all x,y E U. Proof. If the assertion is false then one can find two sequences Yo' and for a 0 either = and C U such that x0 + y,,) and + = + or + = and + =

k' z0,

k —

z0

and all a

the segments

But then there is a z0 E U such that

and IIXn

as

0.

YnII = lix,,

[x,, , y4,,] are

zoil + iIzo—ynli,

nested and

lix,, +

+

iii =

lix,, —

Hence

—f(zo)lI + Ilf(zo) —zo)li +o(x,, —z0)

+ ilf'(zo)(zo—y,,)II +O(Zoy,,) kflx,, —zoli +klizo—y,,Ii +o(x,, —y,,)

= which is a contradiction, as

0

and lix,, —y,,Ii

0.

0

Like Theorem 4, the following theorem ensures that, under suitable circumstances, a function f is a homeomorphism in some neighbourhood.

Theorem 6. Let X and Y be Banach spaces, let U be an open neighY be such that f'(x) exists bourhood of a point x0 E X and let f: U and is continuous in U. Suppose

Then f is a homeomorphism of an open neighbourhood U0 of x0 onto is continuously differentiable an open neighbourhood V0 of f(x0), on V0 and

Chapter 7: The contraction-mapping theorem

107

(f_1)'(y) = fory E V0. Proof. We may assume that x0 =

and f(xo) =

0

0.

Furthermore, setting

X X, we may assume that T0 = f'(O) and replacing f by Y = X and f'(O) is the identity operator I. Set E(x) = f(x) —x. Then e: U —+ Y is continuously differentiable and = 0. Hence there is an open ball D5(X) C U such that Ik'(x)II

4

for x E E)5(X). Then, by Lemma 5, 411x—yII

Ik(x)—e(y)II

for all x,y E J)5(X). Hence, by Theorem 4, f(x) = x+e(x) is a homeomorphism of some open neighbourhood U0 of 0 onto V0 = D514(X). and The rest is plain sailing. For Yo E V0 set x0 = f'(y0) E T = f'(x0). Furthermore, for y0+k E V0 set x0+h = f'(y0+k). As k —* 0, we have h 0, and so k = Th+o(h) and

h= Consequently,

f'(y0+k) = f'(f(xo+h)) proving that

=

= x0+h =

T' =

The continuity of

follows from the fact that the map T

T1 is continuous.

0

The contraction-mapping theorem has numerous applications to differential and integral equations. Here we shall confine ourselves to three rather simple examples.

Theorem 7. Let K(s, t) be a continuous real function on the unit square [0, 1]2, and let v(s) be a continuous real function on [0, 1]. Then there is a unique continuous real function y(s) on [0, 1] such that Cs

y(s) = v(s)

+ I K(s, t)y(t) dz. JO

Proof. We shall consider various operators on the Banach space C[O, 1]

with the uniform (supremum) norm. The linear integral operator L: CEO,!]

C[0,1], defined by Cs

(Lx)(s) = I K(s,t)x(t) dt

Chapter 7: The contraction-mapping theorem

108

is called the Volterra integral operator with kernel K(s, t). In fact, we could assume that K(s, t) is defined only on the closed triangle 0 1, since for s < t we do not use K(s, t). Write K1(s, t) = t s K(s,t) and for n

1 set = K(s, u)Kju,t)du.

t)

Here and in the rest of the proof we take an integral

f(u) du

to be 0 if t > s. Alternatively, we may take each

t) to be 0 outside the closed triangle 0 t s 1. It is easily shown by induction is precisely the Volterra integral operator with kernel on n that Indeed, if this is true for n then +

t)x(t) dt du

f K(s, u)

1x) (s)

=

=

dtdu

f00 J jS

Cu

=J

dux(t) dt

0t J

'-U

dt.

= J0

The function K(s, t) is continuous on the closed unit square and so it is bounded, say IK(s,t)I

for all 0

s, t

M

1. Then, again by induction on n, we see that (n—i)!

for all n

1.

Indeed, if this holds for n then (n—i)!

du

Chapter 7: The contraction-mapping theorem Hence if n is sufficiently large, say n

109

4M, then 1

for ail 0

s,t

1. Therefore

=J

lxii

dt

and so

IILII In particular, L" is a contraction.

It is time to return to our equation. Define a continuous (affine) operator T: CEO, 1] CEO,!] by Tx = v + Lx. The theorem claims that T has precisely one fixed point. But fn—I \ r'x = ( L')v+L"x \i..o / and so T" is a contraction. Hence, by Theorem 2, T has a unique fixed point. 0 The next application concerns a non-linear variant of the Volterra integral operator.

Theorem 8. Let K(s, t) and w(s, t) be continuous real functions on the unit square {0, 1]2, and let v(s) be a continuous real function on [0, 1]. Suppose that Iw(s,t1)—w(s,t2)I for all 0 t1, r2, s 1. Then there is a unique continuous real function y(s) on [0, 1] such that y(s) = v(s)

Proof. Define L: C[0, 1]

+

J

K(s, 0 w(t, y(t)) dt.

C[O, 1] by

(Lx)(s)

=

J

K(s, t) w(t,x(t)) dt.

The theorem claims that the map T: C[0, 1] —* C[0, 1], defined by Tx = v+Lx,

has a unique fixed point.

Chapter 7: The contraction-mapping theorem

110

To show this, we shall turn to trickery. For a > 0, we introduce a new norm II11a on C[O,1}:

=

f

ds.

Jo

Then 1L is indeed a norm on C{0, 1], and it is equivalent to the L1 norm. Set = (C[O, lJ, 11L) and let X0 be the completion of XQ. Clearly, X0 is the vector space L1(O, 1) with the norm a map E: X0 -+ given by the formula for L. Furthermore, with M = max{IK(s,t)I: 0 s, t 1) we have, for = (L1(0, 1), i-I hEx

—

EyhL

J0

= J0 i_i

MN

J

=

J

i_s

—y(t) I

00J

i.1

= MN

K(s, t){w(t, x(t)) — w(t, y(t))} cit ds

dt ds

i_I

OtJ

—y(r) cit dt

e

MNJ

dt

MN This

shows that for a

> MN the map L: Xi., —.

a contraction, and so is T = v + T. It is easily checked that Tniaps into = (C[0, 1], fl so its unique fixed point belongs to C[0, 1], and is also the unique fixed point of T. E

is

In our final example we have to do even less work, since the kernel satisfies a Lipschitz-type condition, just like the function w(s, t) in Theorem 8.

Theorem 9. Let K(s, 1, u) be a continuous function on 0 u 0, such that

IK(x,:,ui) —K(s,I,u2)I where

N(s,

z) is

a continuous function

J

0

satisfying

—u2J,

s, I

1,

Chapter 7: The contraction-mapping theorem

111

for every 0 s 1. Then for every v E CEO, 1] there is a unique function y E CEO, 1] such that

y(s) =

u(s)

di. + 10 K(s,t,y(t))

Proof. Define L: CEO, 1] —' C[O, 1] by

K(s,t,x(t)) di,

(Lx)(s) =

so that a solution of our integral equation is precisely a fixed point of the

x,y

map T: CEO,!] — C[O, 1], C[O,1] we have (Lx — Ly)(s)

=

given

by

Tx = v + Lx.

For

1 {K(s, t,x(t)) — K(s, t,y(t))} 10 N(s, t) Ix(t) —y(t) I d: kflx—yfl.

implying that L is a contraction and thereHence IILx—LyII fore so is T. Consequently, by Theorem 1, T has a unique fixed point.

0 As we remarked earlier, in chapter 15 we shall prove several other fixed-point theorems. But for the time being we return to the study of formed spaces. Exercises

1. Show that a contraction of an incomplete metric space into itself need not have a fixed point. 2. Let A and B be disjoint subsets of a metric space, with A compact and B closed. Show that

d(A,B) = inf{d(x,y): x

A,y E B} >

0.

Show also that this need not be true if we assume only that A and B are closed.

3. Show that if f is a map of a complete metric space X into itself

such that d(f(x),f(y))
X

(x

y) then f

Chapter 7: The contraction-mapping theorem

need not have a fixed point. Show however that if X is compact then f has a unique fixed point. 4. Let f: X —* X be a map from a complete metric space into itself. Suppose that for every

0 there is a

cS

>

such that if

d(x,y) <8 then d(f(x),f(y)) <e. Prove that f has a unique point. 5. Let f be a map of a compact metric space X into itself. Suppose that for every 0 there is a & = 8(€) > 0 such that if d(x,f(x)) <8 then f(B(x,e)) C B(x,€). Let x0 E X and define as in the proof of Theorem 1: = for n 1. Show e

fixed

that

if

,

as n

+ 1) —* 0

then the sequence

con-

verges to a fixed point of f. f be a map of a complete metric space X into itself such that

6. Let

d(J(x),f(y)) for all x,y E

7.

X, where

is a monotone increasing

every t> 0. Deduce from the result in the previous exercise that f has a unique fixed point = a for every x E X. a, and Let f be a continuous map of a (non-empty) compact Hausdorff function

= 0

such that

for

space into itself. Show that there is a (non-empty) compact subset such that f(A) = A.

8. Show that on the space

the norm function x =

c0

differentiable at x if 1111 has a unique maximum (i.e. there is an index j such that 1x11 > I j). 9. Show that on the space the norm function lixil

=

is

i—I

is not differentiable at any point. In

the next five exercises, X, Y

10. Let

and Z are

Banach spaces.

f,g:

X differentiable

an

a

C Y. Suppose

open set (J'(x0))'

E

be at x0 and

an

that f is differentiable at x0 E U and Show that f1: V—p U is differentiable at

Chapter 7: The contraction-mapping theorem

113

Yo = f(x0) and

=

(f'(xo))'.

12. Let U C X and V C Y be open sets, and f: U -+ Y and g: V Z continuous maps with f(U) C V. Suppose that f is differentiable at a point x0 E U and g is differentiable at Yo = f(x0) E V. Show that the map gof: U —' Z is differentiable at x0 and (g°f)'(xo) = g'(y0)of'(x0).

13. Let U be a connected open subset of X and, for n = 1,2,..., let U — Y be differentiable at every point of U. Suppose that U has a neighbourhood N(x0) on which Hfh(x) II is uniformly convergent. Show that if is convergent for some x E U then it is convergent for all x U, the map every x0

f(x) = n—i

is

differentiable at every xEU and

f'(x) =

fh(x). n—i

14.

Let U C X be an open convex set and f: U—' Y differentiable. Show that for all x,y,z E U we have 111(x)

where, 1).

—f(y) —f'(z)(x—y)II

IIx—yII sup If(s) —f'(z)II, .cE(x,y]

as usual, [x, y] is the closed segment {tr + (1 — t) y: 0 Notes

The contraction-mapping theorem is from S. Banach, Sur les operations dans les ensembles abstraits et leur applications aux equations integroJes, Fundamenta Mathematica, 3 (1922), 133—81. Of course, it can be found in most books on general topology and basic functional analysis. A rich

source of applications to differential and integral equations is D. H. Griffel, Applied Functional Analysis, Ellis Harwood, Chichester, 1985, 390 pp.

8. WEAK TOPOLOGIES AND DUALITY

When studying a normed space or various sets of linear operators on a space, it is often useful to consider topologies other than the norm

topology we have introduced so far. Two of the most important of these topologies are the weak topology and the weak-star topology. Before we define these topologies and prove Alaoglu's theorem, the fundamental result concerning the weak-star topology, we shall review some additional concepts in point set topology and present a proof of Tychonov's theorem. Given a topological space (X, i-), a collection a of subsets of X is said to be a sub-basis for i if a C r and every member of r is a union of finite intersections of sets from a. Equivalently, a is a sub-basis for r if a C and for all U E 1 and x E U there are sets S1,. . . ,S,, such that S. C U. More formally, a is a sub-basis for T If xE =

U (1 5:

C

(1) J

is the set of all finite subsets of a. In most cases we may omit the second term on the right, X}, especially if the intersection of an empty collection of subsets of X is taken to be X. Not only do sub-bases tend to be more economical than bases, but they also enable us to use an arbitrary family of sets to define a topology. Indeed, if is any collection of subsets of X then the collection i defined aC by (1) is a topology on X. The possibility of using an arbitrary set system to define a topology enables us to use a set of functions into topological spaces to define a

where

topologyonaset. LetXbeasetandforeachyEl'letf,, beamap of X into a topological space (X,, , r,,). Then there is a unique weakest 114

Chapter 8: Weak topologies and duality

topology r on X such that every f,,: (X, r)

115

(X,, , r,,) is continuous.

Indeed, taking

o= {f'(U,,): U.,.CX,,isopeninr,.(yEfl}, the topology i Is given by (1). Thus UC Xis open in r 1ff for every ..., U, E i,,, such that x E U there are ,. . . , y, E F and U1 E

x E fl

C U.

The topology i defined in this way is called the weak topology generated where y E I). and is denoted by o(X, by = is continuA map! from a topological space (X',T') into (X,o(X, is continuous for every y E I'. l'his soous 1ff f,.of: (X',r') —' called universal property characterizes the weak topology. The product topology is one of the prime examples of a weak topolbe a topological space for each y F and let ogy. Let

X1, be the Cartesian product of the sets X., (y E F). Thus X is the collection of all the functions x: F UYEr X,, such that also, x(y) E X.1,. One usually writes x7 for x(y) and sets x = x,, is the 'y-component of x. Let p.,,: X —' X,, be the projection onto X,,; thus p,,(x) = x,,. The product topology on X is the weak topology generated by the projections p.,. (y F). To spell it out: a set U C X is open in the product topology if for every point x = (X,,),,EF there are such that E indices y,,. . . , y,, and open sets U,,,..., U,. (i = 1,... ,n) and X=

= {y = (y,,) E X: y.,1

(1

(i = 1,. ..,n)} C U.

Thus to guarantee that a point y = (y,.) belongs to a fixed neighbourhood of x = (x,,), it suffices to demand that for finitely many indices each y,, belongs to a certain neighbourhood of x,,. Let us turn to the main topic of this section, the weak and weak-star

topologies of normed spaces. Given a normed space X with dual r, X generated by the

the weak topology a(X,

bounded linear functionals on X. Thus U C X is open in the weak topology (w-open) 1ff for every x E U there are bounded linear funcsuch that tionals and positive reals {y

By replacing

X: byf1/€1,

(i = 1,...,n)} C U. may take every

to be 1.

Chapter 8: Weak topologies and duality

116

The weak-star topology cr(X*, X) on the dual r of a normed space X is the weak topology generated by the elements i E for x X, i.e. by the elements of X acting on as bounded linear functionals (see Theorem 3.10). In more detail: a set G C X is open in the weak-star

if for every g E G there are points x1,. .. , x, E X and positive reals el,... , e,,, such that topology

{f

r; If(x1)—g(x,)j <e, (i = 1,...,n)} C G.

As before, we may take every €, to be 1. The Hahn—Banach theorem implies that if Y is a subspace of a normed space X then the restriction to Y of the weak topology on X is precisely the weak topology on Y, i.e.

u(X,X*)IY = cr(Y, Y*).

(2)

In view of the proof of Tychonov's theorem we shall give, it seems to be appropriate to mention Tukey's lemma, which is equivalent to Zorn's lemma (and so to the axiom of choice). A system of subsets of a set S is said to be of finite character if a subset A of S belongs to if and only if every finite subset of A belongs to

1. (Tukey's lemma) Let be a set system of finite character has a maximal element containing F. and let F E Then Lemma

Proof. Let = {A C F C A}. Order the elements of by inclusion: A B if A C B. Then is a partial order on Also, if C is a chain in then D = C belongs to since all finite subsets of D belong to ?F, and F C D. Clearly, D C is an upper bound for Hence, by Zorn's lemma, has a maximal ele-

0

ment.

of subsets of a set has the finite-intersection property if F, It is easily seen that a topological 0 for all F1,. . , F,, C space X is compact if the intersection of all subsets of a system of closed sets which has the finite-intersection property is non-empty. = {X\F: F E Indeed, is a collection of open sets. Since no finite collection of sets in covers X, the entire system also fails to cover X, so there is a point x C X which belongs to no member of Hence xC F. A trivial reformulation of this characterization of compactness is the following; a topological space is compact if whenever is a system of A system

.

Chapter 8: Weak topologies and duality

subsets with the finite-intersection property then flAE.A A the characterization we shall use in the proof below.

0. This is

Theorem 2. (Tychonov's theorem) The product of a collection of compact sets is compact. Proof. Let {X,,: y

1) be a collection of compact spaces and let

X,, be endowed with the product topology. We have to X show that X is compact, i.e. that if is a system of subsets of X with A the finite-intersection property then be the collection of all set systems on X with the finite-

Let

intersection property. Then is of finite character and so, by Tukey's lemma, there is a maximal set system with the finite-intersection property containing

Since

fl

fl A,

may assume that our set system is itself is maximal for the finiteintersection property. The maximality of implies that the intersection of any two sets in .s4 belongs to as does the intersection of finitely many sets in .si2. Consequently, if B C X meets every set in .sii then B E In particular, if we

and A C B C X then BE So far we have made no use of the fact that X is a product space, let alone a product of compact spaces. Let us do so now. For each y E I',

A

the system {p,,(A): A of subsets of X,, has the finite-intersection property. As X,, is compact, for some point x,, E we have

x,,E fl

p,,(A).

This means precisely that for every neighbourhood U,, of x,,, the set meets every element of

fl whenever

Hence

and so

E

y,,} C 1' and U,,

is a neighbourhood of x7. A. Indeed, for every neighbourhood U of x there are indices y1 ,... , -y,, E I and neighbourhoods U,, of x (i . . , n) such that ,.

.

.

,

Let x = (x,J,,Er

X. We claim that x E

ri

C U.

Chapter 8: Weak topologies and duality

118

Consequently, (3) implies that U E s4, i.e. U intersects every element of Hence every neighbourhood of x intersects every A and so x E A for every A E This proves our claim and completes the proof of our theorem. 0

From here it is a small step to Alaoglu's theorem, the fundamental result concerning the weak-star topology; in view of the fact that both the weak-star topology and the product topology are weak topologies, this is hardly surprising.

Theorem 3. (Alaoglu's theorem) The unit ball

of the dual of a

normed space X is compact in the weak-star topology. = R: IIxII}ifXisarealspaceand IIxII} if Xis a complex space. Furthermore, let D be the product 11xEX endowed with the product topology. Since Proof. =

E C:

a closed ball of the scalar field, is compact, Tychonov's theorem implies that D is a compact space. Let us write B* for endowed with the weak-star topology, and define a map ç: B D by setting, forf C

'p(f) = (J(x))XEX. The

definitions of the product topology and the weak-star topology

imply that is a continuous map of B into D, and it is clear that ç is one-to-one. Furthermore, again by the definitions of the product topology and the weak-star topology, is a continuous map from c(B*)

onto 8*. Thus B* is homeomorphic to the subset of D. Therefore, to complete the proof, it suffices to show that c(B*) is a closed subset of the compact space D. Let then = D be in the closure of

Define a scalar

function f on X by setting f(x) = for x C X. We claim that f is a linear functional, i.e. f E X'. To see this, let x,y C X and let a and be scalars. Since C for every n C N there is a continuous linear functional E B such that 1(x)

As

+ p9y) =

+

+

+

f(ax i.e.f

C X'. Since If(x)I =

<

inequality

(4)

(4) implies that

= af(x) Qxfl for every x

X, f is

not only a

Chapter 8: Weak topologies and duality

linear functional, but it is also continuous and 11111 C Finally, by the definition of f, we have = the proof.

119

1,

i.e. f E B. completing

0

As an immediate consequence of Alaoglu's theorem, we can see that a subspace of the Banach space of continuous functions on a compact Hausdorif space is the most general form of a normed space. Theorem 4. Every normed space X is isometrically isomorphic to a sub-

space of C(K), the Banach space of continuous functions on K, where K endowed with the weak-star topology. In particular, every is

normed space is isometrically isomorphic to a subspace of C(K) for some compact Hausdorff space K.

Proof. Let X and K be as in the theorem; thus K is a compact Hausendowed with the weak-star topology. For dorff space, namely (defined by i(x*) = x(x) for x E X let f, be the restriction of i E to K = B(r). By the definition of the weak-star topology, E f E C(K); indeed, the weak-star topology on r is precisely the weakest topology in which every function i is continuous.

The map x -

of X into C(K) is clearly linear. Furthermore, by

the Hahn—Banach theorem, HfxII

= sup{lfx(x*)I: =

and

so the map X

K}

:fC

C(K) is a linear isometry.

= lixil,

0

It is customary to write w-topology for the weak topology and wtopology for the weak-star topology. Similarly, one may speak of wopen sets, w-neighbourhoods, and so on, all taken in the

appropriate spaces. Thus a w-open subset of X is a o(X, X *)..open of a subset of X* is the closure of the set in subset, and the As before, we consider X as a subspace of the the isometric embedding being given by x i, where (1,x*) = (x,x) for all f C X*. In order to avoid some inessential complications, for the rest of the section we shall consider only real spaces.

The weak topology is weaker than the norm topology, so every w-closed set is also norm closed. The following result claims that for convex sets the converse is also true.

Chapter 8: Weak topologies and duality

120

Theorem 5. A convex set C of a normed space X is closed if it is wclosed. Proof. Suppose that C is closed and x0 C. Then d = d(x0, C) > 0. Set D = {x E X: d(x0, x)
xEC

Then

U = {x E X: (x*,x) >

s}

is a w-neighbourhood of x0 and Un C = 0. This shows that C is also w-closed. As we remarked earlier, the converse is trivial. fl is the w*.closure of 8(X) in

Theorem 6.

Proof. It is easily checked that the closed unit ball B(X**) is w*.closed. Let B be the of B(X). Since B(X**) is and 8(X) C B(r*), we have B C B(X**). Furthermore, as the of a convex set is convex, B is a convex set which is closed in the norm topology of X Suppose that, contrary to the assertion of the theorem, B

Then there is a point E B(X**)\B so that, by the separation theorem (Theorem 3.13) there is a bounded linear functional 4 * * on *

such

that

(4**,b) for all b E B.

r*. Then

Let

4

be

1<

the restriction of xr* to the subspace X of

(4,x)

1

for all x E 8(X) and so 1140

1, contradicting 1* \X0,X0 1 — \X0 ,X0**

The results above have some beautiful consequences concerning reflexivity. Note that the w-topology on a normed space X is precisely on the restriction to X of the

X is reflexive if B(X) is w-compact.

Proof. By Theorems 3 and 6, 8(X) is a

subset of the w-

compact set B(X**). Since a subset of a compact Hausdorff space is

Chapter 8: Weak topologies and duality

121

if subset of is closed, B(X) is a B(X**). B(X) = But this means precisely that B(X) is w-compact 1ff compact if it

x=x**.

0

The next result can also be deduced from the Hahn—Banach theorem; now we are well equipped to give a very straightforward proof. Corollary 8. A closed subspace of a reflexive space is reflexive. Proof. Let Y be a closed subspace of a reflexive space X. Since B(Y) = B(X) fl Y, the norm-closed convex set B(Y) C X is cr(X,

compact. But, as we remarked in (2), the restriction of o(X, to Y is precisely o-(Y, Hence B(Y) is w-compact and so, by Theorem 7, Y

0

is reflexive.

Theorem 9. Let X be a Banach space. The following assertions are equivalent: (1) X is reflexive;

(ii) o(r,X) = o(r,X**), i.e. on X* the w-topology and the w-

topology coincide; (iii) X* is reflexive.

(ii) is trivial (and so is (1) (iii)). SupProof. The implication (i) is it is also w-compact. pose that (ii) holds. Since (iii). is reflexive. Thus (ii) Hence, by Theorem 7, Finally, suppose that (iii) holds. As X is a Banach space, the ball Therefore, by Theorem 5, B(X) is wB(X) is norm closed in = 1*, this means that closed, i.e. Since B(X) is in But, by Theorem 6, the of B(X) is B(X**), so B(X) = B(X**). Hence (iii) (i). 0

If X if reflexive and f E X* then the function If being continuous on B(X), attains its supremum there: for some x0 = B(X) we have Ilfil

= sup{lf(x)l : x E B(X)} = If(xo)I.

E 8(1) such that 11111 = f(x0). (Of course, 1ff 0 then x0 is not only in the closed unit ball but also in the closed unit sphere S(X).) It is a rather deep theorem of R. C. James that this property characterizes reflexive spaces: a Banach space X is reflexive if every bounded linear functional attains its norm on B(X). Equivalently, there is an

Chapter 8: Weak topologies and duality

122

Our next aim is to show that every Banach space comes rather close to having this property: the functionals which attain their suprema on the unit ball are dense in the dual space. (This property used to be called subreflexivity. As the result claims that every Banach space is subreflexive, the term has gone out of use.) In fact, we shall prove somewhat more. (liven a Banach space X, define 11(X) =

{(x,f): x E S(X), f

S(Xi, f(x) =

1}.

Then 11(X) C XXr and, denoting the projection of Xxr onto X by

Px and onto X. by pr, we

= 5(X). Furthermore,

have

px.(fI(X)) is precisely the set of functionals in S(X*) which attain their norms on S(X). For 8 > 0 let us write 118(X) for the set of pairs (x,f) E S(X)xS(r) such that f is almost 1 at x: 11(x)—li

<6.

dense in 5(r), which is preWe shall show that not only is cisely the subreflexivity of X. but also every point (x, f) of 118(X) is close to some point (y,g) of 11(X): lix—yli

where

(5)

+ Ill—gil
0 as 6 —* 0.

The proof of this theorem is based on the following lemma, whose proof is left to the reader (see Exercise 13).

Lemma 10. Let X be a real normed space and let f,g E

be such

that

whenever x E B(X) and f(x) = 2e or iif±ghi

hf—gil

0,

where 0 <€ <

Then either

0

2e.

Theorem 11. (Bishop—Phelps—Bollobás theorem) Let X be a Banach

5(X) and Ia E

space and let x0

be such that

ifo(xo)11 <8 = where

0<

f1(x1) =

1.

1, hhfo—fiii

Then there exist x1 e and ilxo—xiII

5(X) and

E

such that

Chapter 8: Weak topologies and duality

123

Proof. In the notation introduced before Lemma 10, the theorem claims E 11(X) such that that for every (xo,fo) E 118(X) there is an (x1 ,

IIfofiII

and I)xo—xiII

<€; in particular, (5) holds with

=

say

As linear functionals are determined by their real parts and the map

f '- Ref is an isometry, we may assume that Xis a real space. Let k

on B = B(X) by setting x y if

and define a partial order

lix—yll

kfo(y—x).

This is indeed a partial order, and a simple application of Zorn's lemma (see Exercise 16) shows that the set = {x E B: x0 x} has a maximal element, say x1. Since x0 IIxo—xiII

kfo(xi—xo)

kô

=

C= and let D be the convex hull of B U C:

= {ty+(1—Oz:

yE B, z E C,0

1}.

Then D is a centrally symmetric convex set containing B and C. We claim that x1 IntD. Indeed, otherwise there are 0 <s 0 and x f0(x0)

y implies that f0(x)

fo(y), we see that

= sfo(yi)
and so f0(y1—x1) = (1—s)f0(y1)

Furthermore,

>

>0.

(6)

Chapter 8: Weak topologies and duality

124

y1—x1 = (l—s)y1—(t—s)z1,

and so

<

(7)

Inequalities (6) and (7) give Oyi—xiII

2\ 1+2/€ I (1—s)(1+—J = E/

contradshowing that x1 Yi• But, as x1 is maximal, we have x1 = Int D, as claimed. say. This proves that icting (6), By the separation theorem (Theorem 3.13) there is a linear functional

1 for allx ED. Then such thatf1(x1) = 1 andf1(x) for all x E Bfl and f1(x) 1 Hence, by Lemma 10,

f1

E

1

or But

(f0+f1)(x1) = 1±fo(x1) > 1 > E and so we have ito—fill

e.

0

In Chapter 12 we shall give some applications of Theorem 11 to numerical ranges. Our last aim in this chapter is to prove another major result of functional analysis, the Krein—Milman theorem. Although in its proper generality this result has nothing to do with dual spaces and weak topologies, we present it here since it has some natural applications concerning dual spaces.

Our proof of the Krein—Milman theorem is based on another equivalent form of the axiom of choice, namely on the well-ordering principle, enabling us to apply transfinite induction. An ordered set is a set with a linear (total) order. An ordered set (S. E) is said to be well ordered if every (non-empty) subset T of S has a smallest element, i.e. an element t0 E T such that t0 I for all I E T.

Lemma 12. (Well-ordering principle) Every set can be well ordered.

is said to be an initial segProof. A subset B of an ordered set (A, ment if a E A, b C B and a b imply that a E B. y El) be the set of all well LetS be a set and let P = is a subset of S. For (5,,, in ordered sets such that and (Se,

Chapter 8: Weak topologies and duality

125

Pset 5,, is an initial segment of S8 and on the two orderings coincide. Clearly (P, is a partially ordered set. It is easily seen that in (F, E) every chain has an upper bound. Indeed, let {(S,,, be a chain. Set 5' = S,, and for y if

if T S' T then E 5,, for y E I', and the smallest element of r n 5,, in (5,,, E,,) is also a t'

some

smallest element of 1' in (5, E').

Since every chain has an upper bound, by Zorn's lemma from Chapter 3, (P, has a maximal element (Se, To complete the proof, all we have to show is that S0 = S. Suppose that this is not so, S\S0. Extend the order say s1 to an order E1 on = S0U{s1} by setting s s1 for all s E S0. Then (Si, is a well ordered set and <(S1,

(Se,

0

contradicting the maximality of (S0,

Let K be a convex subset of a vector space. A point in K is said to be an extreme point of K if it is not in the interior of any line segment entirely in K. Thus a E K is an extreme point of K if whenever b,c K, 0< t < 1 and a = tb+(1—t)c, we have a = b = c. The set of extreme points of K is denoted by Ext K.

Recall from chapter 3 that the convex hull of a subset S of a vector space Vis coS

;x1

:xE

0 (i = 1,... ,,z),

5,

=

1 (n

=

=

If S is in a normed space then S, the closed convex hull of S, is the closure of coS; by Theorem 3.14, is the intersection of all closed half-spaces containing S. In its full generality, the Krein—Milman theorem concerns locally con-

vex topological vector spaces. As we do not study these spaces, we shall be satisfied with a somewhat artificial form of this result. Given a vCctor space V with dual V', for F C V' and S C V the F-closed convex hull of S is

COFS= fl {xEV:f(x)Esupf(y)}. IEF

For the rest of the chapter, we say that S is F-convex if

=

S.

By

Chapter 8: Weak topologies and duality

126

Theorem 3.14, if X is a normed space then for all S C X; = a set X is convex and closed (in the norm topology) if it is r-convex. Theorem 13. (Krein—Milman theorem) Let r be a Hausdorif topology

on a vector space V

and let F C V'. Suppose that F separates the points of V(i.e. iff(v) = 0 for alIf C F then v = 0) and eachf C Fis r-continuous. Let K be a non-empty r-compact F-convex subset of V. Then ExtK 0 and K = COFEXtK.

Proof. Let tion s:

y C I') R, y

a well ordering of F. Then there is a funcsuch that for all y C F we have be

s7=max{f.,(x):xCK,f8(x)=s8for6
(8)

This is immediate from the fact that F (i.e. F) is well ordered and if s5 is determined for all 6 E F preceding y F then Kfl

fl

8<7

is a non-empty compact set on which the continuous function 1., attains its supremum, 57•

The existence of the function s implies that there is a point a G K f. This point a is an extreme point of s,, for all y K. Indeed, if a = ib + (1 — t) c for some 0 < t < 1 and b, c C K, then (8) implies that f7(a) = f7(b) = f7(c) for all y F. As F separates the points of V. we have a = b = c. Thus ExtK / 0. Set K' = COFEXt K. Then K' C K since K is F-convex. To complete the proof, we have to show that, in fact, K' K. Suppose that this is such that f7(a) =

not so. Then there is a functional Jo C F such that

maxf0(x) <maxf0(x).

x€K

(9)

xEK

Choosing a well ordering of F in which to is the first element, the point

a produced by the procedure above is such that h(a) = But a

0

K', contradicting (9).

Corollary 14. Let K be a compact convex subset of a Banach space.

0 Corollary 15. Let X be a normed space. Then is By Alaoglu's theorem, X-convex since if C X and Jo x0 C B(X) and f(x0) 1 for all x B(X). Proof.

=

Furthermore, it is then f0(x0) > 1 for some

0

Chapter 8: Weak topologies and duality

127

has 'many" extreme points: The last result implies that sufficiently many to guarantee a 'large" closed convex hull, namely Occasionally this fact is sufficient to ensure that a given space is not a dual space. For example, it is easily seen that Ext B(c0) = 0; so c0 is not a dual space. Indeed, let x = E B(c0). Then IXkI < for n k, Yk = Xk + and for some k and so, with = = we have y = z = E B(c0) and x = 4(y+z), so = that x Ext B(c0). Thus Ext B(c0) is indeed empty. The Krein—Milman theorem is one of the fundamental results of functional analysis, leading to deep theorems in operator theory and abstract harmonic analysis, like the Gelfand—Naimark theorem concerning irreducible *..representations of C-algebras, and the Gelfand—Ralkov theorem about unitary representations of locally compact groups. But these results will not be presented in this book. We shall return to compact convex sets in Chapters 13—16, when we study compact operators and fixed-point theorems. However, our next aim is to study the 'nicest' Banach spaces, namely Hilbert spaces. Exercises

1. Prove identity (2), i.e. show that if Y is a subspace of a normed space X then the weak topology on Y is precisely the topology induced on Y by the weak topology on X. 2. Let Y be a closed subspace of a reflexive space X, and let x0 X. Show that the function x IIx—xoII is weakly lower semicontinu-

ous, i.e. it is lower semicontinuous in the w-topology on X. Deduce that the distance of x0 from Y is attained, i.e. there is a point Yo E Y such that Ilxo—yoIJ

= d(x0,Y) =

Y}.

3. Show that if Y is a closed subspace of a reflexive space X then X/Y is reflexive. 4. Show that

d(x,y) = defines

a metric on

and the restriction of this metric to

induces the weak-star topology on 5. Let X be an infinite-dimensional normed space. Show that the wclosure of S(X) = {x E X: lixil = 1} is B(X) = {x E X: lixil 1}.

Chapter 8: Weak topologies and duality

128

be a sequence in 1,, (1

6. Let x0

7.

1ff

8.

is bounded (i.e.

(Ix,,II

K for some K and all n) and

= 0 for every i, where = , Let C (n = 1,2,...). Show that 0 = for every y C c0 1ff is bounded and = 0 for every i.

9. Formulate and prove assertions analogous to the previous two exercises for the spaces

1) and Lq(O, 1).

10. Let X be an infinite-dimensional normed space. Show that the norm function lxii (x E X) is not weakly continuous at any point of X.

11. Let

be the standard basis of 12 and let

A = {em+men: 1 Prove

m
that 0 is in the w-closure of A but no sequence in A con-

verges weakly to 0.

12. Let X be a Banach space, f C X* and M = Kerf. Prove that for every x C X there is a point of M nearest to x 1ff f attains its norm on B(X). 13. Prove Lemma 10: if X is a real normed space and f,g E are such that (Kerf) fl B(X) C g — ' [ — €, €} for some 0 < then 2€ or i)f+giI 2€. [HINT: ±g then there are If f Ill—gil a,b E X such that f(a) = g(b) = 1 and f(b) = g(a) = 0. Set M = lin{a,b} and N = KerfflKerg, and denote by D the projection of B(X) into M parallel to N. Let i-fl' be the norm on M with unit ball D, and note that with fo = fJM and g0 = giM we have ilafo+Pgoii' = iiaf+flgfl for all E R. This shows that it suffices to prove the lemma in the case when X = M, i.e. when X is 2-dimensional. Use a simple geometric argument to complete the proof.]

14. Show that the assertion of Lemma 10 is best possible: for 0< there are functionals f, g C such that (Kerf)flB(X) C 1 and Ill —gil = 2€. 15. Show that Zorn's lemma indeed implies the existence of a maximal element x1 x0 in the proof of Theorem 11. 16. Check that Theorem 11 can be strengthened a little: if where 0 <e < then we may demand ifo(xo) — fl <

Chapter 8: Weak topologies and duality and IIxo—xtII

129

<€+€2. (This variant of the result is

essentially best possible.)

17. Let K C R" be a compact convex set. Show that if n =

2 then Ext K is compact, but this need not be the case if n 3. 18. Show that B(11) = coExtB(11). 19. What are the extreme points of CR[0, 1]? 20. Let L be a locally compact Hausdorif space, which is not compact. has no extreme points. Show that the closed convex set 21. Let I be an infinite-dimensional Banach space whose unit ball has only finitely many extreme points. Show that X is not a dual space, i.e. there is no normed space Ysuch that X = 22. Show that CR[0, 1] is not a dual space.

r.

Notes

Tychonov's theorem is from A. Tychonoff, Uber em Funktionenraum, Mathematische Annalen, 111 (1935), 762—66; somewhat earlier, an incomplete version of the result appeared in A. Tychonoff, (Jber die topologische Erweilerung von Räumen, Mathematische Annalen, 102 (1929), 544—61. (The modern transliteration of the name 'Tychonov' is closer to the Russian original than the one used in German before the second world war. Similarly, the old-fashioned usage of Markoff, Tschebischeff, Egoroff, etc., is due to German influence.) A proof of the thoerem can be found in almost any good book on general topology, for example, in J. L. Kelley, General Topology, Van Nostrand, Princeton, N.J., 1955, xiv + 298 pp.

Theorem 3 is from L. Alaoglu, Weak topologies of normed linear spaces, Annals of Math., 41(1940), 252—67; in fact, the result was announced two years earlier in Leonidas Alaoglu, Weak convergence of linear functionals, Bull. American Mathematical Society, 44 (1938), p. 196. The theorem of James mentioned after Theorem 9 is from R. C. James, Characterizations of reflexivity, Studia Math., 23 (1964), 205—16. The ori-

ginal form of Theorem 11, stating that for a Banach space X the set of functionals attaining their norms on S(X) is dense in is from E. Bishop and R. R. Phelps, A proof that every Banach space is subreflexive, Bull.

Amer. Math. Soc., 67 (1961), 97—8; the form we gave it in is from B. Bollobás, An extension to the theorem of Bishop and Phelps, Bull. London Math. Soc., 2(1970), 181—2. Our presentation followed that given in F. F. Bonsall and J. Duncan, Numerical Ranges I!, London Math. Soc. Lecture Note Series, vol. 10, Cambridge University Press, 1973, vii + 179 pp.

9. EUCLIDEAN SPACES AND HILBERT SPACES

Having studied general normed spaces and Banach spaces, in the next

two chapters we shall look at the 'nicest' examples of these spaces, namely Euclidean spaces and Hilbert spaces. These spaces are the natural generalizations of the n-dimensional Euclidean space A hermitian form on a complex vector space V is a map f: Vx V C we have such that for all x,y,z V and (I)

=

(ii)

f(y,x) = f(x,y)

for all x,y

C

V.

Similarly, a hermitian form on a real vector space V is a map f: Vx V R satisfying (i) and (ii). In that case (ii) says simply that f is symmetric, so a real hermitian form is just a symmetric bilinear form. If f is a hermitian form on a complex vector space V then g = Ref is a real hermitian form on VR, the space V considered as a real vector space. Conversely, for every real hermitian form g on VR there is precisely one complex hermitian form f on V with g = Ref (see Exercise 1). Let f be a hermitian form on a real or complex vector space V. Note that f(x,x) is real for all x C V and f(Ax,Ax) = IAI2f(x,x) for all x C V and scalar A. We call f positive if, in addition to (i) and (ii), it satisfies (iii)

f(x,x)

0

for all x C V.

Two vectors x and y are said to be orthogonal with respect to I if f(x,y) = 0; orthogonahty is sometimes expressed by writing x .Ly. A vector orthogonal to itself is said to be isotropic. We call f degenerate if some non-zero vector is orthogonal to the entire space; otherwise the form is non-degenerate. 130

Chapter 9: Euclidean spaces and Hubert spaces

131

Theorem 1. Let f be a positive hermitian form on V. Then for all x,y E Vwe have

f(x,x)f(y,y)

If(x,y)12

(1)

and

f(x+y,x+y)"2 Purthermore,

f is

f(x,x)"2+f(y,y)'12.

non-degenerate 1ff 0 is the only

(2)

isotropic vector with

respect to f. Proof. In proving (1), we may replace x by Ax if IA I = 1 so we may assume that f(x,y) is real. Then for all : E R we have 0

f(Lr+y,tx+y) = t2f(x,x)+21f(x,y)+f(y,y)

so (I) follows. To see (2), note that, by (1),

f(x+y,x+y)

f(x,x)+21f(x,y)I+f(y,y) f(x,x) +2[f(x,x)f(y,y)]112+f(y,y)

= Finally, if

vector in V.

x is isotropic then (1) implies that x is orthogonal. to every 0

Inequality (1) is, of course, just the Cauchy—Schwarz inequality; inequality (2) is a variant of Minkowski's inequality for p = 2. In view of Theorem 1, a positive hermitian form is non-degenerate if it is a positive-definite hermitian form, i.e. a hermitian form f such that

ifx

Othenf(x,x) >0.

An inner-product space is a vector space V together with a positive definite hermitian form on V. This positive-definite hermitian form is said to be the inner product or scalar product on V and we shall write it

(.,.) (and, x,y, z

E V and

occasionally, as (,

for

all scalars A

and

= A(x,z)+p.(y,z); (ii) (y,x) = (x,y); (iii) (x,x) 0, with equality if x =

)).

Thus (,) is such that for all

we have

(i)

0.

More often than not, it will not matter whether our inner-product space is a complex or a real inner-product

space.

As the complex case

tends to look a little more complicated, usually we shall work with

132

Chapter 9: Euclidean spaces and Hilbert spaces

complex spaces. By Theorem 1, if (•,•) is an inner product on a vector space V then lxii = (x,x)'12 is a norm on V. A normed space is said to be a Euclidean space or a pre-Hilbert space if its norm can be derived from an inner product. A complete Euclidean space is called a Hubert space. Clearly every subspace of a Euclidean space is a Euclidean space and every closed subspace of a Hilbert space is a Hubert space. The Cauchy—Schwarz inequality states that l(x,y)l lIxIlIlyll; in particular, the inner product is jointly continuous in the induced norm. The inner product defining a norm can easily be recovered from the norm. Indeed, we have the following polarization identities: 4(x,y)

Ox +y112— Ox —y02+ ilIx+ iyll2—illx— iyll2

(3)

if the space is complex, and 2(x,y) =

Ilx+y112—11x112—11y112 =

(4)

the real case. Therefore in a Euclidean space we may, and often shall, use the inner product defining the norm. For this reason, we use the terms 'Euclidean space' and 'inner-product space' interchangeably, and we may talk of orthogonal vectors in a Euclidean space. The complex polarization identity (3) has the following simple extension. If T is a linear operator (not necessarily bounded) on a complex in

Euclidean space then 4(Tx,y) = (T(x+y),x+y)—(T(x—y),x—y)

+i(T(x+iy),x+iy)—i(T(x—iy),x—iy).

(3')

This implies the following result.

Theorem 2. Let E be a complex Euclidean space and let T E such that (Tx,x) = 0 for all x E E. Then T = 0.

Proof. By (3') we have (Tx,y) =

0

for all x,y E E. In particular,

OTxll2=(Tx,Tx)=OforallxeE,soT=0. It

be

0

is worth pointing Out that Theorem 2 cannot be extended to real

Euclidean spaces (see Exercise 2).

In a Euclidean space, the theorem of Pythagoras holds and so does the parallelogram law.

Theorem 3. Let E be a Euclidean space. If x1,...,x,, are pairwise orthogonal vectors then

Chapter 9: Euclidean spaces and Hubert spaces

133

2

xl

=

E then

Furthermore, if x,y

IIx+y112+

= 211x112+211y112.

Proof. Both (5), the Pythagorean theorem, and (6), the parallelogram law, are immediate upon expanding the sides as sums of products. To spell it out, ,

/,,

2

=

\i=1

a

\

a

I

1=1

x.) = =

(x1 , x,)

(x1 ,x1) +

i#j (x1,x,)

11x1112

= and

I!x+y112+ Ix—y112 = (x+y,x+y)—(x—y,x—y) =

211x112+211y112.

In fact, the parallelogram law characterizes Euclidean spaces (see

Exercise 3). This shows, in particular, that a normed space is Euclidean iff all its two-dimensional subspaces are Euclidean. Furthermore, a complex normed space is Euclidean iff it is Euclidean when considered as a real normed space. The last assertion is easily justified without the above characterization of Euclidean spaces. Indeed, if E is an Euclidean space then Re(x.y) is

a real inner product on the underlying real vector space of E and it defines the same norm on E. Conversely, if V is a complex vector space is an inner product on and (.,•) is a real inner product on V (i.e. Vft) such that the induced norm satisfies IIAxII = A IIIxH for all x E V and A E C then

(x,y) = (x,y)—i(ix,y) = (x,y)+i(x,iy)

is a complex inner product on V defining the original norm

and

satisfying Re(x,y) = (x,y). Examples 4. (i) Clearly, (x,y) = is an inner product on 12" and so 12" is an n-dimensional Hubert space.

(ii) Also, (x, ') =

is an inner product on 12 and so 12 is a

separable infinite-dimensional Hilbert space. (iii) Let E be the vector space of all eventually zero sequences of complex numbers (i.e. x = (x,)° belongs to E if x = 0 whenever i is

Chapter 9: Euclidean spaces and Hubert spaces

134

sufficiently large), with inner product (x,y) = dense subspace of 12; it is an incomplete Euclidean space. (iv) It is immediate that

(f,g) is

= Ja

Then E is a

f(t)g(t) dt

an inner product on the vector space C([a, b]); the norm defined is

the 12-norm \1/2

/ 11f112

and

=

(j

If(t)12d1

not the uniform norm. This is an incomplete Euclidean space (see

Exercise 14).

(v) Also,

(f,g) is

= Ja

the inner product on L2(O, 1) =

E

1]: f is measurable and

101 If(i) 12

dt <

defining the L2-norm

\1/2

/ 111112 = (J

\0

If(OI2dt

With this norm L2(O, 1) is a Hubert space.

(vi) Let V = C's, let A = (a11) be an nXn complex matrix and, for in V define x = (x1)? and y =

f(x,y) =

a,1x191. 1,1 = 1

I is a hermitian form if

= a9 for all i, j. Also, every hermitian form on V can be obtained in this way. 0 Then

Theorem 5. The completion of a Euclidean space is a Hilbert space.

Proof. The assertion is that the inner product on a Eucidean space E can be extended to the completion E of E. Let (x,,) and be Cauchy sequences in E and let i and 9 be their equivalence classes in E. Define

Chapter 9: Euclidean spaces and Hubert spaces

135

= Since

i(Xnm,Yn)l+l(Xm,Ynym)I

i(Xn,Yn)(Xm,Ym)i

+ flXmil

—ymIl,

It is easily seen to depend only on I and j and not on the particular representatives (y,,). Furthermore, (, •)is an inner product on E, extending E and defining precisely the norm on the completion E. 0 the limit exists.

Let E be a Euclidean space. For x E E, write x' for the set of vec-

tors orthogonal to x:

x' ={yEE:xJ..y}={yEE:(x,y)=O}. Clearly x' is a closed subspace of E and therefore so is

S' ={y€E:x±yforallxES}= fl x' xES

for every subset S of E. If F is the closed linear span of S. i.e.

F=iiiS,thenS' = F' and(S')' JF.

Call two subspaces F1 and F2 orthogonal if x1 I x2 for all x1 E F, and F2. If F is a subspace of E then F and F' are orthogonal subspaces and Ffl F' = {O} since if x E Ffl F' then x I x and so x =0. If x2

x

F and y E F' then

lix +yii2

= lIxii2+ iiyl12, so the projections

x+y x and x+y '-+ y are both bounded. Thus F+ F' is the orthogonal direct swn of F and F'. In general, F+ F' need not be the entire space, not even when F is closed (see Exercise 6). However, when F is complete, this is the case.

Theorem 6. Let F be a complete subspace of a Euclidean space E. Then E is the orthogonal direct sum of F and every x E E has a unique representation in the form

x=x,+x2 Furthermore, if y E F and y ilx—yil

Proof.

(x1€F,x2EF'). x1 then

> lix—xili = ilx2il.

Let x E E and set d = d(x,F). What characterizes x,? By (7),

it has to be the unique vector in F nearest to x, precisely at distance d.

Chapter 9: Euclidean spaces and Hubert spaces

136

Let then

E F be such that <

d2+!

By identity (6), the parallelogram law, —ymIl2 = 211Xyn112+211Xym112

22 —+—, n

II2XYn —ymII2

m

is a Cauchy sequence. Since F is complete, y,, some x1 E F. Then IIx—xiII = d.

and so

x1

for

Let us show that x2 = x — x1 is orthogonal to F. Suppose this is not so. Then F has an element, say y, such that (x2,y) 0. Replacing y by z = (x2 , y)y we have (x2 , z) E R and (x2, z) > 0. But then for

sufficiently small €> 0 we have IIx_(xi+€z)112 = 11X2EZ112

= d2—2(x2,z)e+11z112€2 < d2, contradicting d(x, F) = d. Inequality (7) is immediate from identity (5), the Pythagorean

theorem. Ify

Fandx1—y

IIx—y112 = flx1

0 then, as (x1—y)J.x2, we have = lxi —y112+ lIx2ll2 >

1lx2112

= d2.

0

In view of Theorem 6, if F is a complete subspace of a Euclidean space

E then we call F' the orthogonal complement of F in E. The map PF: E = F+ F' —. E defined by x = x1 +x2 x1 is called the orthogonal projection onto F; we have just shown that there is such a map. Corollary 7. Let F be a complete subspace of a Euclidean space E. Then there is a unique operator E such that

ix

forxEF,

forxEF'.

Furthermore,

ImPF=F,

KerPF=F1,

(J—PF)2=I—PF

and ifx,y E Ethen (PFx,y)

If F

(0) then

= (PFX,PFY) = (x,PFy).

= 1 and if F

E then

= 1.

0

Chapter 9: Euclidean

spaces

and Hilbert spaces

137

Corollary 8. Let H be a Hubert space, let S C H and let M be the closed linear span of S. Then = (M1)1 = M. Proof.

have

M=

0

=

Given a Hubert space H and a vector x0

H, the function f: H -+

C

defined by f(x) = (x,x0) is a bounded linear functional of norm llxoIl. Indeed, f is clearly linear, and If(x) I = I (x, x0) I lixoll lixil and so HxoII2 and Furthermore, so The ll.xoll. If(xo) I 11111 11111

existence of an orthogonal decomposition implies that all bounded linear functionals on H can be obtained in this way.

Theorem 9. (Riesz representation theorem) Let f be a bounded linear functional on a Hubert space H. Then there is a unique vector x0 E H such that f(x) = (x,x0) for all x H. Furthermore, 11111 = lIxoll.

We may assume that f 0. Then M = Kerf is a closed onecodimensional subspace of H. Consequently is one-dimensional, say M1 = A C}, where x1 H and UxiII = 1. Put x0 = f(x1)x1. Every vector x E H has a unique representation in the form Proof.

x=y+Ax1,

(yEM;AEC).

Then (x,x0) =

= (Ax1,f(x1)x1)

Af(x1) = f(Ax1) = f(y+Ax1) = f(x). To see the uniqueness of x0, note that if (x,x0) = (x,x6) = x E H, then IIxo—x6112

=

0

for all

= 0.

The assertion lit = lixoll was shown immediately before the theorem. V

0 Corollary 10. Let H be a Hilbert space. For y E H let E be H* defined by defined by = (x,y)(x E H). Then the map H y y

f

is an isometric anti-isomorphism between H and H, i.e. if and z g then Ay+pg Af+1g. If H is a real Hubert space

then the map is an isometric isomorphism between H and H.

0

Chapter 9: Euclidean spaces and Hubert spaces

138

The last result enables us to identify a Hilbert space with its dual; this identification is always taken for granted. It is important to remember that the identification is an anti-isomorphism since this will make the

adjoint of a Hubert-space operator seem a little different from the adjoint of a Banach-space operator. Exercises

1. Let f be a hermitian form on a real vector space

U.

Let

V = U+iU and extend f to a function f: VxV—* C by setting

f(x+iy,z+iw)

=

f(x,z)+f(y,w)+if(y,z)—if(x,w).

that I is a hermitian form on the complex vector space V and Il U = f. Show also that f is positive 1ff f is, and that f is Show

degenerate 1ff f is.

2. Give an operator T E andflTIJ=1. 3.

such that (Tx,x) =

0

for all x E

Show that a normed space E is a Euclidean space if and only if the parallelogram law holds in E, i.e. lix +y112 +

4. Let E

be a Euclidean

—yfl2 = 211x112 + 211y112.

space. Show that for x1,.. . ,x,, E Ewe have 11

where the summation is over all

=

i1 sequences (e,)7

(e, =

1

or —1).

5.

F= {x =

=

= Oifkislargeenough}C12.

Show that

F' =

lin{u} = {Au: A is a scalar}.

6. Construct a Euclidean space F and a closed subspace F C E (F E) such that F' = (0). [Note that then, in particular, E F+F'.J 7. Let E be

a Euclidean space and let P E

be

a norm-i projec-

= p and II P11 = 1. Show that P is the orthogonal projection onto F = ImP, i.e. KerP = F1 and E = F+F'. 8. Let A be a non-empty closed convex subset of a Hilbert space H. tion:

Show that the distance from A is always attained: for every b E H

Chapter 9: Euclidean spaces and Hi/bert spaces

139

there is a unique point a = a(b) E A such that Ib—aH

= d(b,A) = inf{IJb—xII: x E A}.

Show also that the map b a(b) is continuous. 9. Let x, y and z be points in a Eucidean space. Prove that IIxIIIly—zH

10.

IIyIlIIz—xli + IIzIIIIx—yU.

Let X be the space of continuously differentiable functions on [0,1], with norm \I/2 / uiu = (j {xf2(x)+21f1(x)12} dx

\0

Show

that X is a Euclidean space. What is the inner product

inducing the norm?

11. Let F be a closed subspace of a Hilbert space H. Show that the quotient space H/F is also a Hilbert space.

12. Let I' be an arbitrary set and let 12(f) be the vector space of all functions f: C such that >.,EI' If(v)I2 In particular, if f E 12(fl then {y E 1': 0} is countable. Show that 12(F) is a Hilbert space with then norm / 11(7)12

11111 = I

13.

Let {H7: y E f) be a family of Hilbert spaces. Let H be the vector space of functions f: F UYEI such that f(y) E and yEr Show that \112

/ 111(7)112

IIfIl = I is

a norm on H and with this norm H is a Eucidean space.

necessarily a Hubert space? 14. Turn CEO, 1] into a Euclidean space by setting =

L

Show that this space is incomplete.

f(t)g(t) dt.

Is

H

Chapter 9: Euclidean

140

15. f: [0,1]

Ft

and Hubert spaces

such thatf' E L2(0,1) and f(0) = 0. Prove that Vis

a real Hubert space with inner product =

16. A real all x,y E

L

f'(t)g'(:) dr.

matrix A is called orthogonal if (Ax,Ay) = (x,y) for

Prove that A is orthogonal 1ff it maps orthogonal

vectors into orthogonal vectors and has norm 1.

17. Let P be a bounded linear projection in a Euclidean space E (i.e. P E B(E) and P2 = P). Show that IIP1I = 1 if P 0 and

ImP±KerP. 18. Let A be a non-empty subset of a Hubert space H and let be such that TH C A and (x— Tx)IA for every XE H. Show that T is a bounded linear operator, A is a closed linear subspace and T is the orthogonal projection onto A: T = Show that the unit ball of '2 contains an infinite set A such that for all x,y E A (x y). Show also that A must conIIx—yII > sist of norm-i vectors. T

19.

Notes

The concept of an abstract Hubert space was introduced by J. von Neumann, Eigenwerz Theorie Hermitescher funktional Operatoren, Math. Ann., 102 (1930), 49—131. Earlier special realizations of a Hubert space had been considered by several people. In particular, from 1904 to 1910 David Hilbert published some fundamental papers on integral equations which led him to consider some 'Hilbert spaces' of functions. It seems that the name 'Hubert space' was first used by F. Riesz in his book, Les systèmes d'equanons a une infinite d'inconnus, Paris, 1913, for what we know as 12: "Considérons l'espace Hilbertien; nous y entendons l'ensemble des systémes (Xk) tels que

12 converge."

JO. ORTHONORMAL SYSTEMS

In this chapter we continue the study of Euclidean spaces and Hubert spaces. As we shall see, every separable Euclidean space contains the This means that exact analogue of the canonical basis of = (Ce', we can use 'coordinates', which in this context we call Fourier coefficients, to identify the points of the space, these coefficients behaving very much like the coordinates in or R". If our space is in fact a Hubert space then the space is thus naturally identified with 12, telling us that all separable Hilbert spaces are isometric. Let E be a Euclidean space and let S C E be a set of vectors. If E is the closed linear span of S, i.e. linS = E, then we call S fundamental or total. We call S orthogonal if it consists of non-zero pairwise orthogonal elements. An orthogonal set of unit vectors is said to be an orthonormal set. An orthogonal set is complete if it is a maximal orthogonal set. A complete orthonormal set in a Hilbert space is said to be an orthonormal basis. We shall see, amonst other results, that every orthonormal basis is

a Schauder basis of the best kind, with basis constant 1 (see Exercises 16—18 of Chapter 5), and every Hilbert space contains an orthonormal basis.

Occasionally, we call a set S of vectors a system of vectors, especially when s is not written in the form of a sequence.

Theorem 1. A fundamental orthogonal set in a Euclidean space is complete. In a Hilbert space every complete orthogonal set is fundamental. Proof. (1) Let S be a fundamental set in a Euclidean space E. Then = = = (0); so if S is orthogonal, then it is complete. (ii) Let S be a complete orthogonal set in a Hilbert space H. Set M = linS. Then Mi = (0) so M = (Mi)i = H. Thus S is a fundamental system. 0 141

Chapter 10: Orthonormal systems

142

The Gram—Schmidt orthogonalization process enables us to replace a sequence of linearly independent vectors by an orthonormal sequence.

Theorem 2. Let x1 , x2,... be linearly independent vectors in a Euclidean space E. Then there exists an orthonormal sequence Yi such that lin{x1,. . .,x,j = lin{y1,. . ,yj for every n. .

Proof. Forn = 0,1,...

set

= where

and

=

isthe orthogonal projection

onto

so that (x —

I M,,

for every x

E E. Then 0 since M, and E Also, Then the sequence Set = has the required properties. Indeed, by the construction, M,, = lin{y1,. . ,yj and so the y are linearly independent. Furthermore, 0 and

In fact, it is easy to define explicitly the orthogonal projection = 0 and for n appearing in the proof of Theorem 2; namely, set

1

define PMX

=

The Gram—Schmidt orthogonahzation process enables us to show that every separable space (i.e. every space containing a countable dense set) contains a fundamental orthonormal sequence. Also, by Zorn's lemma every Eucidean space contains a complete orthonormal system. Theorem 3.

(a) Every separable Euclidean space contains a fundamental orthonormal sequence. (b) In a Euclidean space, every orthonormal system is contained in a complete orthonormal system. In a Hilbert space, every orthonormal system

is contained in an orthonormal

basis.

be a dense sequence in a Euclidean space E. Disif it is in Lin{x1,. . and apply Theorem 2 to the sequence obtained. (b) Let S0 be an orthonormal system in a Euclidean space E. Set Proof. (a) Let card

.

= {S:

S is an

orthonormal

system

in E}.

Then I is partially ordered by inclusion and every totally ordered subset I' of I has an upper bound in I, namely S. Hence, by Zorn's

Chapter 10: Orthonormal systems

143

is a complete orthonorThus lemma, .Z has a maximal element If E is a Hubert space, then the system S1 is mal system containing So. 0 a fundamental system. 4. (1) Let T be the unit circle in the complex plane: {z E C: Iz I = 1}. Equivalently, T is the real line modulo 2ir:

T

T = {e1': t E R} =

0

t

2ir}.

Let E be the Euclidean space CW with inner product i

j2w

(f,g)=y-J IT0

f(t)g7Jjdt,

we have used the second convention: f and g are continuous functions from [O,2ir] to C, with f(0) = f(21T) and g(0) = g(2ir). The = e*hhl (n = 0, ±1, ±2,...) form an orthonormal system in functions E. By the Stone—Weierstrass theorem, the linear span of these functions is dense in C(T) endowed with the uniform norm. Hence n = 0, ±1,.. .} is a fundamental, and so complete, system in E. where

The completion of E is L2(1J. (ii) Let E be the Euclidean space C[— 1, 1] with inner product &.

=

functions 1,:, t2,... are linearly independent in E and, by the Stone—Weierstrass theorem, form a fundamental system in E. Let be the sequence obtained from 1,:,t2,... by the Gram— Schmidt orthogonalization process. Then Q,,(t) is a multiple of the nth

The

Legendre polynomial =

where D is the differentiation operator (see Exercise 1).

(iii) 1,sint,cos:,sin2,cos2t,... is an orthogonal sequence in the Hilbert space 1.2(0, it). Since the subspace of continuous functions is dense in L2(0, it), by the Stone—Weierstrass theorem this sequence is a fundamental orthogonal sequence.

(iv) For n = 0,1,... define the nth Rademacher function r, E by —

1

if

—

—1

if

is even is odd.

1.2(0, 1)

Chapter 10: Orthonormai systems

144

as an element = sign sin 2"irt. (Note that we view is, strictly speaking, an equivalence class of functions of L2(0, 1); so it differing on sets of measure 0. Thus it is also customary to define t t < it is easily an seen to be incomplete. and let ç E C(a,b) be a positive function. a < b (v) Let b) of continuous functions with compact For f and g in the space support, set Equivalently,

rb

(f,g) Then

dt.

= Ja

b); denote its completion by is an inner product on The space L2(ç) consists of all measurable functions f(t) on

[a, bJ such that

dt < = et. Then the functions 1,t,t2, Now let a = 0, b = and form a fundamental system in L2(ç) and the polynomials obtained from them by the orthogonahzation process are the Laguerre polynomials up (n = 0, 1,...) (see Exerto constant factors, i.e. multiples of cise 5).

constructed in Example (v) but

(vi) Consider the space and ç(t) = b = with a =

e'.

Orthogonalizing the sequence 1, t, t2,... we obtain multiples of the Hermite polynomials, i.e. multiples 0 = 0,1,...) (see Exercise 6). If

)7

is an orthonormal basis in an n-dimensional Euclidean space

E, then every vector x E E is X=

a

Furthermore Ck = and =

linear combination of the Qk: Also, if = k=1

dkpk,

then

(x,y)

=

Ckdk

and

k—i

In other words, the correspondence x

11x112

= k—i

ckI2.

identifies E with

Chapter 10: Orthonormal systems

145

The main reason why an orthonormal basis (cok)r in a Hubert space H is very useful is that analogous assertions hold concerning the represen-

tations of vectors in H. As we shall see, for every vector x E H there are unique coefficients (ck)T such that x = ckpk. This sequence satisfies

(ck)r

11x112

and

12

=

every

sequence

satisfying

arises as a sequence of coefficients. Thus an orthonormal basis enables us to identify H with 12, just as any n-dimensional Euclidean space can be identified with 12

I

be an orthonormal sequence in a Hubert space Theorem 5. Let is conH. Then, for a scalar sequence c = (Ck)°, the series vergent iff C C

(i.e.

If the series is convergent then

2

leo k=1

Proof. Set x,, = theorem, we have

=

1ICI12

=

(

Ck(pk. Then, for 1

n

m, by the Pythagorean

lIXn_Xm112

k=n+1

is convergent if

lCk 2 is convergent. Relation (1) holds since, again by the Pythagorean theorem, Hence

k=1

ICkH

A slightly different formulation of Theorem 5 goes as follows. Let

be an orthonormal sequence in a Hilbert space H, and let c = (Ck)° C '2• Then there is a vector x C H such that (X,ck) = Ck for every k. This is usually called the Riesz—Fischer theorem. It looks particularly easy (it is particularly easy) because the space is assumed to be complete. The original form of the theorem concerned L2[O, 11, where the completeness, which is far from trivial, had to be proved. Let be an orthonormal sequence in a Hilbert space H. For x C H, set Ck = (k = 1,2,...). We call c1,c2,... the Fourier coefficients of x with respect to the series k=1

is the Fourier series of x with respect to

Chapter 10: Orthononnal systems

146

It is easily seen that if is an orthonormal basis then every vector is the sum of its Fourier series and so, in particular, it is determined by its Fourier coefficients. In fact, we have the following slightly more general result. be an orthonormal sequence in a Hilbert space H Theorem 6. Let and let M = lin(cok)r. Then, for all x,y E H, we have

= PM(x);

k1 k=1

Rx,ck)12 = IIPM(x)112

(iii)

= (PM(x),PM(y)) = (x,PM(y)) = (PM(x),y). k=1

Also, suppose c =

C 12, i.e.

unique vector u C M with Fourier coefficients

1,2,...), namely u =

if u =

coefficients

Ck(pk.

Then

ICkI

Furthermore,

Ck

=

there is a (k =

v C H has Fourier

PMV.

Proof. Set

= Then, for 1

k=1

n, we have

k

x

a

= 0 and so x = of orthogonal vectors. Therefore, by

the theorem of Pythagoras, 11x112

= IIxnH2+ IIX_xn112

= k=1

Letting n

we

see that

I(x,ck)12 <

so, by Theorem 5,

is convergent, say to x' E M. As for every k we have = = (x,ck)—(x,ck) = 0,

x—x' is orthogonal to M and so x' = PMX. This proves (i); furthermore, (ii) follows from (i) and (1).

Chapter 10: Orthonormal systems

147

To see (iii), set =

Then

= lim

and so

(PMx,y) = (x,PMy) = (PMX,PMY)

=

=

= tim Ic=1

= The proof of the second part is equally easy. By Theorem 5, the series

M. Then

is convergent to some vector u =

fri

c1ço1,

Qk) =

Ck

for every k, and so u does have Fourier coefficients c1 , u = PMV then

c2

Also, if

(V,ck) = (v,PMpk) = (PMV,ck) = (u,ck)

and so u and v have the same Fourier coefficients. Finally, if v E H = 0 for every k and so and = Ck for every k then = M. Therefore u = PMV, as claimed. 0 v—u is orthogonal to The following useful corollary is amply contained in the result above. Corollary 7. is an orthonormal sequence in a (a) (Bessel's inequality) If Euclidean space E and x E E then k=1

(b) (Parseval's identities) If

is a complete orthonormal sequence in a Hubert space H and x, y E H then = 11x112

k—i

= (x,y).

and k—I

0

Chapter JO: Orthonormal syste,ns

148

Parseval's identities imply that every infinite-dimensional Hubert space with an orthonormal basis (i.e. a complete orthonormal sequence) is isometric to 12. Theorem

S. Let

be

an orthonormal basis in a Hubert space H.

For x E H define 1(k) = (x,pk)

Then the map H

I

and

I is a linear isometry of H onto

given by x

Corollary 9. Every n-dimensional Euclidean space is linearly isometric and every separable infinite-dimensional Hubert space is linearly to

0

isometric to '2

Having seen that all separable infinite-dimensional Hilbert spaces are

isometric to 12, what can we say about other Hilbert spaces? As in Exercise 11 of Chapter 9, given a set F, we denote by 12(fl the vector space of complex-valued functions f on F with countable support and Then

such that 11111

is

=

(

lf(y)12

a norm on '2(i) and with this norm 12(F) is a Hilbert space. Thus is precisely '2 and

12({I ,

2,.. . , n}) is

Theorem 10. Every Hubert space is isometrically isomorphic to a space 12(fl.

Proof. Let H be a Hubert space. Then H contains a complete orthonormal system, say {q.,: y E f'}. Then, by Bessel's inequality, for every x E H and for every countable subset of F, we have

Hence the set

= {y: (x,

O} is countable, x—

is orthogonal to every ç,, and so

Chapter 10: Orthonormal systems

the map H

Setting i(y) =

12(fl

given by x

149

i is a linear 0

isometry of H onto l2(fl.

The fact that every separable (infinite-dimensional) Hubert space is isometrically isomorphic to '2 is very important when we are studying the abstract properties of Hubert spaces. Nevertheless, in applications Hilbert spaces often appear as spaces of functions, and then we are interested in the connections between the Hubert space structure and the properties of the functions. the comOne of the most important Hilbert function spaces is pletion of C(T), the space of continuous functions on 'the circle T, ii = defined in Example 4(i). In the standard orthonormal basis 0, ±1, ±2,.. .}, the kth Fourier coefficent Ck of f E L2(T) is Ck

dt

=

and

Sn!

k_n

cket4ldt

is the nth partial sum. Then, by Theorem 8, urn

=0

(2)

with denoting the Hubert space norm, i.e. the for every f E norm in L2(T). Thus the partial sums converge in mean square to f.

For f E C(T) relation

(2) is

an immediate consequence of the

Stone—Weierstrass theorem, which tell us that f can be uniformly approximated by a trigonometric polynomial. (Of course, we used precisely the Stone—Weierstrass theorem to show that {eml: = 0,±1,±2,...} is a complete system.) In particular, for every >0 there is a trigonometric polynomial p such that 11(t) —p(t)I < for all t. But if p has degree n (i.e. p(t) = then = p. Since the projection operator has norm 1, we have

IISnffII

=

<2€.

The very easy relation (2) leaves open the question whether the Fourier series of f E L2(T) tends to f pointwise in some sense. For example, it may not be unreasonable to expect that if f is continuous then tends to f(t) for every t. Sadly, this is not true, as was

Chapter 10: Orthonormai systems

150

first shown by du Bois Reymond in 1876. Nevertheless, Fej6r proved in 1900 that there is a simple way to recover a continuous function from the partial sums of its Fourier series. To be precise, Fej& showed that the Fourier series of a continuous function if Cesdro swnmable and the sum is the function itself: if f is continuous then n+1

Sd, k=O

the average of

the partial sums, tends uniformly to f. Fej6r's was of tremendous importance: it launched the modern theory of Fourier series. Aithougli need not converge to f(t), the set of points at which it fails to tend to f(t) cannot be too large. It was asked by du Bois Reymond whether the partial sums Sn! of a function f C(T) tend to f almost everywhere, and later Lusin conjectured that the answer is in the affirmative for every f L2(T). For many years this was one of the most famous conjectures in analysis; finally, it was proved by Carleson in 1966. The very intricate proof is one of the greatest triumphs of hard analysis; not surprisingly, it is far beyond the scope of this book.

i.e.

theorem

Exercises

1. Show that the Legendre polynomials ngr,2

D —

n

,

RI.'

1J

/

i

—n1

form an orthogonal basis of L2(—1, 1). Deduce that, up to a posi-

tive factor, P,O) is the nth term in the sequence obtained from 1,t,t2,... by the Gram—Schmidt orthogonalization process, counting P0(z) = 1 the 0th term, P1(t) = : the first, P2(t) = 4(3g2_ 1) the second, etc. 2. Show that

=

1

and

= (—1)".

3. Deduce from the previous exercises that

—(2n—

for all n 2. 4. Show that D((:2— 1)Ph(t)) is orthogonal to deduce that

= 0

for = 0

for

every n

1.

k
Chapter 10: Orthonormal systems

5. Check that the Laguerre polynomials etDlt(e_hthl) (n = 0, 1,...) are where ç': [0, oo) —' R is indeed orthogonal polynomials in 6.

e', as in Example 4(v). the analogous assertion about the Hermite polynomials et and weight function = : —' R, as in Example 4 (vi).

7. Let L2(ç) be as defined in Example 4(v), with a positive weight function (a, b) R. Suppose that 1,t, t2,... L2(q). Let be the polynomials obtained by orthogonalizing the sequence 1, t, t2 (We do no: normalize the sequence; so p,,O) is the unique monic polynomial of degree n orthogonal to every

fork
(ii) Show that for n

2 we have = (t —

)p,, —

_2(1)

—

where —

(tPn—i,Pn—i) IIPn _iIl

8. Let x1,. . . ,x, .

,y,, be

and

2

—

IIPn—i112 lIP,, —211

unit vectors in a Euclidean space and let the sequence obtained from it by Gram—Schmidt be

orthogonalization, with =

1, with equality only if y =

Show that

9. For a, = E (i = 1,... ,n) let A = A(a1,...,a,,) = be the nxn complex matrix whose ith row is a,. Prove Hadwnard's inequality, stating that IdetAl

II

with equality if and only if either some a, is 0 or the a are orthogonal.

10. Let x1,. . be linearly independent vectors in a Euclidean space, with N = ("p). Show that there are orthonormal vectors .

Yi ,..

•

, y,, such

that y• =

joint subsets of {1,2,...,N}.

A1x1, where A1 , A2,. . . , A,, are dis-

Chapter 10: Orthonormal systems

152

11. Define the Gram determinant of a sequence x1 , ...... , bert space as

i.e.

in a Hil-

as the determinant of the n xn matrix whose entry in the ith

row and jth column is the inner product (x1,x1). Show that 0, with equality iff the set {x1,.. x,j is linearly G(x1 ,.. , .

.

dependent. Show also that if {x1 ,. then fling a subspace

,

. .

x,,}

is a linearly independent set span-

=

\ G(x1,x2,. . 12. Prove the following weak form of the Riemann—Lebesgue lemma: if f E C(T) then .

dt

—*

0

.1:

13. The aim of this exercise is to prove Fe/Er's theorem. For f set

= (i)

(Skf)(t).

Show that f(x)

t)

—

x)

where Ku(s), the FejEr kernel, is defined as rs

Ku(s)

=

(ii) Prove that if 0 <s < 2ir then Ku(s) =

I —I fl+1\

sInks

and =

= n+1.

E

L2(T)

(iii) Ku(s)

Chapter 10: Orthonormal systems

153

Deduce from (i) and (ii) that Ku(s)

0 for all s E T,

0

uniformly for 0 <

2ir—S < 2ir and

s

1

— 2ir

I

(iv) Deduce Fejér's theorem: if f —*

uniformly in t as n —p (v) Show also that if f E

C(T) then

f(t)

and f is continuous at

then

5

14k. Let H be a Hubert space and let S1 = {y.,: y E fl be such that (Xa

and ôaa =

1).

,

=

(where

y

f) and S2 = = 0 if a

Suppose S1 is a fundamental system, i.e. linS1 =

H. Does it follow that S2 is also fundamental? Notes

The chapter is about the beginning of abstract Fourier analysis. As we have hardly scraped the surface of Fourier analysis proper, the reader is

encouraged to consult a book on the topic; T. W. Körner, Fourier Analysis, Cambridge University Press, 1988, xii + 591 pp., is particularly recommended.

The original Riesz—Fischer theorem was proved in F. Riesz, Sur les systèmes orthogonaux de fonctions, Comptes Rendus, 144 (1907), 615—

19 and 734—36, and E. Fischer, Sur Ia convergence en moyenne, Comptes Rendus, 144 (1907), 1022—4. It was a clear case of independent discovery, precisely described by Fischer in the introduction of his

paper: "Le 11 mars, M. Riesz a présenté a l'Académie une Note sur les systèmes orthogonaux de fonctions (Comptes Rendus, 18 mars 1907). J'étais arrivé au mime résultat et je l'ai démontré dans une conference fait a Ia Société mathématique a Brunn, deja le 5 mars. Ainsi mon indépendance est évidente, mais la priorité de Ia publication revient a M. Riesz." Carleson's theorem, mentioned at the end of the chapter, was proved

in L. Carleson, Convergence and growth of partial sums of Fourier series, Acta Math., 116 (1966), 135—57. Exercise 9 is from J. Hadamard, REsolution d'une question relative aux dEterminants, Bulletin Sd. Math. (2), 17 (1893), 240—348, and Exercise

154

13

Chapter 10: Orthonormal systems

is from L. Fejér, Sur les fonctions bornées et inlegrables, Comptes

Rendus, 131 (1900), 984—7 and Investigations of Fourier series (in Hungarian), Mat. és Fiz. Lapok, 11(1902), 49—68; see also Untersuchungen uber Fouriersche Reihen, Math. Annalen, 58 (1904), 51—69.

11. ADJOINT OPERATORS

In the next four sections we give a brief account of the theory of bounded linear operators on Banach spaces. Our aim is to present several general concepts and prove some of the fundamental results. We are mainly interested in the spectral theory, to be treated in three sections, but before we embark on that, in this chapter we study the basic properties of the adjoint of an operator. Throughout this chapter we shall use the product notation for the value of a linear functional on an element: (x,f) = (f,x) = f(x) for x in a vector space V and f in V', the dual of V. In particular, if X is a normed space and X* is the dual of X, i.e. the Banach space of all

bounded linear functionals on X, then (,) is the bilinear form on Xxr given by (x, f) = f(x). Thus = A(x,f)+!.t(y,f) and

=

(Note the absence of conjugates: (,) is a bilinear form and no: a herinitian form. This is why it is not confusing to have (x,f) = (f,x).) Recall that I(x,f)I lIxIlIlfIl for all x E X andf E r. Furthermore, Oxil

= sup{I(x,f)I: f E

= max{I(x,f)I: f G B(X)}

and 11111

= sup{I(x,f)I : x E B(X)}.

Let X and Y be normed spaces. Recall from Chapter 1 that the adjoin: or dual r of an operator T E Y) is the unique map X such that

r: r

(x,Tg) = (Tx,g)

(1) 155

Chapter 11: Adjoint operators

156

for all x E X and g E r. Indeed, for g E r, we have a function Pg on X defined by (Pg)(x) = (Tx,g); it is easily seen that this function is linear and, in fact, Pg E Y*. In turn, the second dual of T is the

unique map r*: r* ._*

r

such that

= (Q,Pg) for all q E X* * and g E Y*. Let us summarise the basic properties of taking adjoints.

Theorem 1. Let X and Y be normed spaces and let T, T1, T2 E Then

(a) P

Y).

E

and IIP1I = 11711; (b) for scalars A1 and A2, we have (A1 T1+A2T2)* = A17? +A21r;

(c) with the natural inclusions X C

T,i.e. P*x

*

and

YCr*

we

have

Txforallx€X;

(d) if Z is a normed space and S E

(e) if T is invertible, i.e. T' E (T*)1 =

then (ST)* = PS*; then P is also invertible and

Proof. Part (a) is Theorem 3.9, part (b) is immediate from the definition

of the adjoint, and (c) follows from Theorem 3.10, giving the natural embeddings X—* X** and Y—* To see (d), note that for x E X and h

we have

(x,PS*h) = (Tx,S*h) = (STx,h). Finally, (e) follows since T' T = 1x. where

on X and so 'r = Hence (Pr' =

=

is the identity operator Similarly, Iy• =

0

Let now H and K be Hubert spaces, with their inner products written

as (, ). We know from the Riesz representation theorem in Chapter 8 that a Hilbert space is naturally isometric with its dual but the isometry is an anti-isomorphism. Thus x E H can be considered as a functional

on H: it acts on H as multiplication on the right: (,x). Then for we have P E TE Equivalently, P is defined by

(x,Py) =

(Tx,y)

(2)

for all x E H and y E K. Indeed, for a fixed y E K the function

f:

H —p C defined by f(x) = (Tx,y) is a bounded linear functional on H.

Hence, by the Riesz representation theorem, f(x) =

(x,

u) for some

Chapter 11: Adjoin: operators

157

unique vector u E H. We define ry to be this vector u. It ate that 1' is a linear map from K to H. Since l.f(x) I

II TxIl

is immedi-

Dxli

11711

we have Dull

and

= liryli = 11111

IITllIlylI

so r is a bounded linear map and II r fi

Ii TD.

Theorem 2. Let H and K be Hilbert spaces and let T, T2 E for all scalars A1 and A2 we have Then r E IT'll = 11711 and, (3)

(T1T2)* = T17*

(4)

r*T

(5)

and

Furthermore, if K = H then 111112

= IITTII = IIrTIi = IIT'112.

(6)

Proof. We known that I1T'II = 11711 and (T1T2)* = 1?1?; furthermore = T since H is reflexive. To see (3), note that for x E H and

y E K we

have

(x,(A1T1+A2T2)*y)

= ((A1T1+A2T2)x,y) = A1(T1x,y)+A2(T2x,y) = A1(x, 7j'y) +A2(x,

T2y)

=

= Of course, natural

this is also clear from Theorem

1(b)

and the

fact that the

isometry between a Hubert space and its dual is an

anti-

isomorphism.

Finally, let T E 117112

Then

= sup

1x01

IITxlI2

= sup

(Tx, Tx)1

IjxH=1

= sup

i(rTx,x)I

IIrlll

IlrlIlIllI =

0

Chapter 11: Adjoint operators

158

Let us emphasize again that in (3) we have to take the conjugates of the coefficients, unlike in Theorem 1(b), which is the analogous statement for normed spaces. This is because in (3) we take 1? as a map from K to H, rather than from K to H*, as in the normed-space case;

the conjugates appear because H and H* are identified by an antiisomorphism. Theorem 2 is of special interest in the case H = K. An such that (3),

complex Banach algebra is a map x (4) and (5) hold: involution on

a

= x.

(xix2)*

(A1x1+A2x2)

A is a Banach algebra with an involution satisfying (6), i.e. in which IIxxiI = 11x112 (and so 11x112 = Ox*xlI = OxH2). Thus Theorem 2 claims that is a Ce-algebra with involution T Hence every norm-closed subalgebra of which is closed under taking adjoints (i.e. every closed .subalgebra of is also a C*.algebra. What is remarkable is that the converse of this statement is also true: every algebra is isomorphic to a closed *..subalgebra of This is the celebrated Gelfand—Naimark theorem; although it is beyond the scope of this book, we shall present some exercises concerning it at the end of the next chapter. Qiven a normed space X with dual X, for a subspace K C X define the annihilator of K in as

r.

r

K°= {fcr:(x,f) Similarly,

for a subspace L

= OforallxE

C r, the

annihilator

K}.

of L in X (or the

preannihilator of L) is

= {xEX:

=OforaLlfC

L}.

Strictly speaking, K° usually denotes the polar of a set K C X: K° =

{fE r: I(x,f>I

1 for allfE L}

and °L is the prepolar of a set L C X* (or the polar of L in X): = {x E X: I (x, f)

I

1 for all f E }.

Of course, if K and L are subspaces, as we have chosen them, or at least they are unions of one-dimensional subspaces, then the two if

definitions coincide (see Exercise 1). It is clear that for any sets K C X and L C X* the annihilators K° and °L are closed; furthermore,

K° = (lin K)° =

K)°

and

°L = °(lin

L) =

L).

Chapter 11: Adjoint operators

159

Theorem 3. Let X and Y be normed spaces and let T E Ker T = °(Im T)

and

Ker r =

Y). Then

(Im T)°.

0

Proof. Clearly

KerT= {xEr: Tx=O} = {x

E X: (Tx,g) = Ofor allg E

= {xEX: (x,rg) =OforallgE P} = °(Imr). Similarly,

Ker r =

{g

E r: rg = O}

= {g

r: (x, rg)

= {g

Y*: (Tx,g) =

0 for all x E X} 0

for all x

X} = (Im T)°.

Note that if L C X* then °L C X and L° C and so, in general, we cannot expect °L to be equal to L° under the natural inclusion However, if X if reflexive and so X is identified with then for every set L C X* we have °L = L°. If H is a Hubert space then not only is H a reflexive space but also the dual H is identified = LL for every set L C H. with H. With this identification, L° =

XC

Hence Theorem 3 has the following immediate consequence.

Corollary 4. Let H and K be Hubert spaces and let T E Then

Ker T = (Im T)1

and

Ker T = (Im T)

does not hold in general It is worth noting that Im T = °(Ker since Im T need not be closed. However, if Im T is closed then we do have Im T = °(Ker T) (see Exercise 3). is called hermizian Let H be a Hubert space. An operator T or seif-adjoint if r = T. Thus T is self-adjoint if

(Tx,y) = (x, Ty) for all x,y E H.

is self-adjoint if (x,y) = (Tx,y) is a herClearly an operator T E mitian form on H. If S and T are commuting seif-adjoint operators then ST is also hermitian since (STx,y) = (Tx,Sy) = (x,TSy) = (x,STy).

Chapter 11: Adjoint operators

160

In particular, if T is seif-adjoint then so is T" for every n

1.

Note

also that if T is seif-adjoint then by Theorem 2, =

II

etc. Therefore

II

7''

=

IIT2kII 11T2k1I

lii

=

= 117112,

IITII2*.

Also, if 1

IITnT2*_hhII

= II

= II

II

2" then

n

IITfhlIIl112k_,1

and so

T"II = T is__self-adjoint then (Tx,x) is real for every x H since (Tx,x) = (x, Tx). We call a self-adjoint operator T positive if (Tx,x) O for every x E H.

Note that, for T E operator. Indeed,

(rT)' =

the operator rr is a positive self-adjoint

= rT and (rTx,x) = (Tx, Tx)

Replacing T by r, we see that

= IITxlI2

0.

is also a positive self-adjoint opera-

tor.

Theorem 5. Let H be a complex Hilbert space. Then every operator TE has a representation in the form T = + iT2, where T1 and T2 are hermitian, and this representation is unique.

Proof. Set and

T1 =

T2 =

Then T1 and T2 are hermitian and T = T1 + IT2. The uniqueness follows from the fact that if T1 and T2 are hermitian and T1 + IT2 = 0 then T1+iT2 = (T1+iT2)* = T1—iT2

0

andsoT2=OandT1=0.

Examples 6. (I) Let T be the right translation on '2' i.e. let T((x1,x2,...)) = ((0,x1,x2,...)). Then r is the left translation: = ((x2,x3,...)). Clearly 11711 = !lrII = 1, is the identity I but I:

TT*((xi,x2,x3,...)) = ((0,x2,x3,...)). (ii) For

C[O, 1} define TQ: L2(O, 1)

L2(0,

1) as multiplication

by

(Tçf)(t) = ço(t)f(t)

(0

t

1).

Chapter 11: Adjoint operators

161

Then and

I

IIT,Il =

Clearly Tç is a positive hermitian operator 1ff

1}.

is a non-negative real-

valued function.

(iii) Let M be a closed subspace of a Hubert space and let be the is a positive self-adjoint operaorthogonal projection onto M. Then 0 tor and (as every projection) satisfies = It is easily seen that the properties in Example 6(111) characterise the orthogonal projections in a Hubert space.

Theorem 7. Let H be a Hilbert space and let P E

be a self-

P. Then M = ImP is closed and P is the orthogonal projection of H onto M: P = adjoint projection: P2 = P =

Proof. Since P is a projection, we have x— Px E KerP and x = (x — Px) + Px for every x H. So H = Ker P + ImP. By Corollary 4, we have Ker P = (Im = (Im P) and so H is the orthogonal direct

o

sum of KerP and ImP.

In addition to hermitian operators and orthogonal projections, let us Introduce two other important classes of operators. Given a Hilbert is said to be normal if 7T = space H, an operator T E T, and unitary if T is invertible and its inverse is r. Note that every hermitian operator is normal, and so is every unitary operator. In the following

r

results we characterize normal and unitary operators.

Theorem 8. Let H be Hubert space and T E (a) Tis normal 1ff IITxII = IIrxII for all x E H. = for everyn (b) If Tis normal then

KerT =

Kerr

land

= (Im

= (Im

Proof. (a) Clearly, IITxII2—JIT'x112

= (Tx, Tx) —(rx, rx) =

(rTx,x)—(Trx,x)

=

From Theorem 9.2 we know that PT— rr = 0

0

1ff ((rT— rr)x,x) =

for every x.

(b) If T is normal then, by (a), we have Ker T = Ker P. Hence, by Corollary 4,

Chapter 11: Adjoint operators

162

T=

Im

T is hermitian,

= lirlir = Ij(rT)iI, and so

= tI(TT)"Ii implying

0

=

As a consequence of Theorem 8 one can see that Theorem 7 can be

strengthened: if a projection is normal then it is an orthogoani projection (see Exercise 7).

Theorem 9. Let H be a Hubert space and let U E

be such that

Im U = H. Then the following are equivalent:

(a) U is unitary;

(b) U (c) U

is

an isometry: lUxil = lxii for every x E H; the inner product: (Ux, Uy) = (x,y) for all x,y C H.

preserves

Proof. The polarization identity (3) in Chapter 9 implies that U is an isometry 1ff it preserves the inner product. Thus (b) and (c) are equivalent. If U

is

unitary then (Ux,Uy) = (U*Ux,y) = (x,y).

Conversely,

if (Ux,

and so

Therefore U* = U

Uy) =

(x,y) for all x,y U =

U

(U*Ux,y) = (x,y) an isometry, U is invertible.

C H then

0

Our last aim in this chapter is to show that the converse of Theorem 1(v) also holds if X is a Banach space, i.e. 7" is invertible if T is. First we note a simple condition for invertibility. Call T C 24(X, Y) bounded for all x C X and some below if iiTxIl 0.

Theorem 10. Let X be a Banach space, Y a normed space, and let Then T1 iff ImT is dense in Y and T is bounded below.

Proof. The necessity of the two conditions is obvious. Suppose then that the conditions are satisfied. If T is bounded below then it is injective, so T1 C where Z Im T. Since Z is dense in Y, for

Chapter 11: Adjoint operciors

163

every y E Y there is a sequence (Zk)' ifl Z converging to y. Then by is also convergent, say to x. Hence the second condition,

T(Tzk) =

Tx =

hmk._,O Zk

=Y

andsoY=Z. l'his shows that T1

If flTxfl

dxli for all x E X then,

0

1/c.

clearly, 11T'il

Theorem 11. Let X be a Banach space, Y a normed space and let

TE

Y). Then r is invertible 1ff T is.

Proof. We have seen that if T is invertible then so is r. Suppose then that T is invertible. Let us check that the two conditions in Theorem 10 are satisfied.

By Theorem 3 we have (Im T)° = Ker T = (0) and so Im T is dense in Y. To see that T is bounded below, let x C X and let f be a support functional at x, i.e. let f E be such that (x,f) = fixil and 11111 = 1. Then lixil

= (x,r(ry'f)

=

=

Therefore fl Txli

Ii

(Tx,(rY'f) lix

—

completing the proof.

If Im T is not dense in Y, say TX C Z, with Z a closed subspace, then = (Tx,f) = 0 for Z C Kerf for some f C (f 0). Hence (x,

all x E X and so

=

0.

In particular, if r is bounded below then

Im T is dense in Y. This gives us yet another condition for invertibility; let us state it together with Theorems 10 and 11. Theorem 12. Let X be a Banach space, Y a normed space, and let Y). Then the following conditions are equivalent: T

(a) T is invertible;

(b) r is invertible; (c) Im T is dense in Y and T is bounded below;

(d) T and r are both bounded below.

0

The question of invertibility brings us to the study of the spectrum of an operator and the structure of the algebra of bounded linear operators. But that requries a new chapter.

Chapter 11: Adjoint operators

164

Exercises

1. Given a Banach space X and a set K C X, the annihilator of K in

ris

=

(x,f>

=OforallxE K)

and the annihilator of a set L C X* in X (or the preannihilator of L) is = {x

X: (x,f) =

0

for ailfE L}.

= K0 and Show that if K and L are subspaces then = where K° is the polar of K and °L is the prepolar of L. Show also

that = (linK)a = (linK)0

=

and

a,, =

a(ljfl L)

=

a(lin L)

= °(lin L).

2. Give examples showing that for a Banach space X and a subspace

L C r, the sets °L and L° need not be equal under the natural inclusion X C Y). Show that X and Y be normed spaces and T E is the closure of Im T. °(Ker 4. A subspace U of a normed space V is said to be an invariant sub-

space of an operator S E if SU C U, i.e. Su E U for all u U. Let X be a normed space and T E Show that a closed subspace Y of X is an invariant subspace of T 1ff Y° is an invariant subspace of r. 5. Let X and Y be Banach spaces and T E Y). Prove that Im T is closed iff Im T* is closed.

6. Let X be a Banach space. Show that for T E T"/n! converges in norm to an element of exp T. Show also that (exp T) * = exp r and if mutes with T then

the series denoted by

S

(exp S)(exp T) = (exp T)(expS)exp(S+ T).

In the exercises below, H denotes a Hubert space.

com-

Chapter ii: Adjoint operators

7. Let T be a bounded linear operator on a Hubert space H. Show that T has an eigenvector iff r has 1-codimensional closed invariant subspace.

8. Let H be a Hilbert space and P E

a projection: P2 =

P.

Show that the following are equivalent: (a) P is an orthogonal projection; (b) P is hermitian; (c) P is normal;

(d) (Px,x) = IIPxII2 for all x E H. 9. Let U E be a unitary operator. Show that Im(U—I) = Im(U*_!) (I) Ker(U—I) = (ii) For n

and

deduce

that

1 set

= Show that

PMX for every x E H, where M = Ker(U—I).

(One expresses this by saying that 5,, tends to

in the strong operator topology.) 10. Show that if T E is hermitian then exp iT is unitary. 11. The aim of this exercise is to prove the Fuglede—Putnam theorem. Suppose that R,S, T E with R and T normal and RS = ST. (i) Show that

(exp R) S = S(exp T). show that

(ii) By considering exp(R* — R) S exp( T—

IRexpR*)Sexp(_r)II (iii) For! E

IlSil.

and A E C set F(A)

= f(exp(AR*)Sexp(_Ar)).

Show that F(A) is an analytic function and IF(A)I

IlfilliSli for every A C. Apply Liouville's theorem to deduce that F(A) = F(0) = f(S) for every A and hence that

exp(AR*)S = Sexp(Ar).

(iv) Deduce that R*S =

Sr.

166

Chapter 11: AdjoinS operators

Notes

The notion of an adjoint operator was first introduced by S. Banach, Sur les fonctionelles Iinéaires Ii, Studia Math., 1 (1929), 223—39. Our treatment of adjoint operators is standard. The Gelfand—Nalmark theorem was proved in I. M. Gelfand and M. A. Nalmark, On the embedding of normed rings into the ring of operators in Hubert space, Mat. Sbornik N.S., 12(1943), 197—213; for a thorough treatment of the subject see S. Sakai, and Springer- Verlag, New York-Heidel-

berg-Berlin, 1971. For the Fuglede-Putnam theorem in Exercise 11, see chapter 41 in P. R. Halmos, introduction to Hilbert space and the theory of spectral multiplicity, Chelsea, New York, 1951.

12. THE ALGEBRA OF BOUNDED UNEAR OPERATORS

In this chapter we shall consider complex Banach spaces and complex unual Banach algebras, as we shall study the spectra of various elements. Recall that a complex unital Banach algebra is a complex algebra A with an identity e, which is also a Banach space, in which the algebra structure and the norm are connected by lieU = 1 and llabII x* in A such that Ilaillibil for all a,b E A. If there is an involution x = x*+ye, = Ax*, (xy)* = y*x* and llx*xfl = = x, 11x112 then

As we noted earlier, if X is a complex

A is a

is a complex unital Banach algebra, and if H is Banach space then a complex Hilbert space then is a

An element a of a Banach algebra A is

invertible

(in A)

if

ab = ba = e for some b E A; the (unique) element b is the inverse of a, and is denoted by a1. The spectrum of a E A is oA(a) = SpA(a) =

{A

E C: Ae — a is not invertible in A},

and the resolvent set of a is 8A(a)

C\UA(a). A point of ÔA(a) is said to

be a regular point. The function R: ö(a) A given by R(A) = (Ae — a)1 is the resolvent of a. The element ROt) is the resolvent of a at A or, with a slight abuse of terminology, the resolvent of a. The prime example of a Banach algebra we are interested in is the algebra of bounded linear operators on a complex Banach space X; so our algebra elements are operators. In view of this, if T then we define the spectrum and resolvent set of T without any reference to =

{A

E C: Al— T is not invertible}

where I is the identity operator on X, and

p(T) = C\o(T). 167

168

Chapter 12: The algebra of bounded linear operators

If A is a complex unital Banach algebra then A can be considered to the algebra of all bounded linear operators be a subalgebra of acting on the Banach space A, with the element a corresponding to the operator La of left multiplication by a (so that a La, where A is invertible then so La(X) = ax for every x E A). In particular if a with inverse La'. Conversely, if S E is the inverse La E of La, so that X

= (LaS)X = a(Sx)

for every x E A, then with b = Se we have ab = 1 and so a(ba — e) = (ab)a—a = 0. Hence ba—e E KerLa and so ba = e. Thus b is the inverse of a. Also, Ae — a is invertible 1ff A!— La is invertible. Hence crA(a) =

Although the spectrum of an operator T E how T fits into the algebraic structure of

depends only on is of considerable interest to see how the action of T on X affects invertibility. In particular, we may distinguish the points of cr(T) according to the reasons why it

A!— T is not invertible.

What are the obstruction to the invertibility of an operator S E

By the inverse-mapping theorem, S is invertible 1ff KerS = (0) and ImS = X. Thus if S is not invertible then either KerS (0) or ImS X (or both, of course). Of these, the former is, perhaps, the more basic obstruction. Accordingly, let us define the point spectrum of T E =

The elements of

{A

E C: Ker(AI— T)

as

(0)}.

are the eigenvalues of T; for an eigenvalue

the non-zero vectors in Ker(A!— T) are called eigenvectors with eigenvalue A. Furthermore, Ker(AJ— T) is the eigenspace of Tat A. AE

Clearly

C o-(T).

then the two conditions If X is finite-dimensional and S E KerS = (0) and ImS = X coincide. Hence a finite-dimensional space.

However, if X is infinite-dimensional then we may have KerS = (0) and Im S X, so the point spectrum need not be the entire spectrum. More precisely, by Theorem 11.10 , A E o(T) if either Im(A1— T) is not dense in X or A!— T is not bounded below: there is no 0 such that II(AI— T)xII €IIxIl for every x E X. In the former case A is said to belong to the compression spectrum T), and in the latter case, A is

Chapter 12: The algebra of bounded linear operators

169

said to belong to the approximate point spectrum of T, denoted by Uap(T). In other words, =

E C: there is a sequence

{A

C S(X) such that (Al—

Sometimes

Clearly o(T) C

O}.

is called an approximate eigenvector with eigenvalue A. T) and

0(T) =

0ap(T)U(Tcom(T).

the points of the spectrum are classified further: the residual spectrum is 7r(T) (Tcom(T)\C7p(T) and the continuous spectrum is Sometimes

= (7(T)\(Ucom(T)U(Tp(fl). Thus

o(T) = 0p(T)U0c(T)UUr(T), with the sets on the right being pairwise disjoint.

It is immediate from the definition that the approximate point spec-

is a closed set; the point spectrum

trum closed.

need not be

is a non-empty closed subset of the disc We shall show that The latter assertion is an immediate consequence

{A E C: IA I

of the following simple but important result.

Theorem 1. Let TE

<1. Then I

satisfy 1111

o(T) and

Tk,

(I—Ty' = k =0

where

the series on the right is absolutely convergent in the Banach

space

Proof. Note that 11T9 k=O

and so

HT1I'

=

Tk is absolutely convergent.

Hence

Tk=(l_T)+(T_T2)+(T2_T3)+...=l k =0

and, similarly,

T")(I_T) = I.

Chapter 12: The algebra of bounded linear operators

R(A)

= (Al—TY' = A_1(I_f)

171

(3)

= and IIR(A) II

= II (Al — T) — 'II

IA I

— II ill

Note that we do not yet know whether the spectrum a(T) can be empty or not. In proving that it cannot, we shall make use of Banachspace-valued analytic functions, an example of which we have just seen

in(2).

Given a Banach space X and an open set D C C, a function F: D X is said to be analytic if for every z0 E D there is an r=

r(z0)

>

0

such that D(z0, r) = {z E C: I z — =

I

n0

for some a0,a1,... E Xand all z E D(z0,r), with the series in (4) being

absolutely convergent. Thus (2) shows that the resolvent R : p(T) is an analytic function, with values in the Banach space The standard results concerning analytic functions remain valid in this more general setting. For example, as we noted in Exercise 10 of Chapter 11, the analogue of Liouville's theorem holds: a norm-bounded entire function in constant. To spell it out: if F: C X is analytic and M for some M < and all z C then F(z) is constant. Indeed, for f E Xt the function

g(z) = f(F(z)) is a (complex-valued) analytic function and so it is constant: g(z) = g(0) for all z. Hence f E X and so, by the Hahn—Banach theorem, = F(0) for all z. Also, the radius of convergence of (4) is just as in the classical case. Equivalently, the Laurent series

G(z) =

(5) 1/n•

has radius of convergence s = lim Indeed,

if IzI >s then

< IzI for some e >0 and every if n is sufficiently large, <(1

sufficiently large n. Hence implying that (5) is absolutely convergent.

Chapter 12: The algebra of bounded linear operators

172

z <s then there

Conversely, if

is

an infinite sequence n1 <

>

> 1 and so (5) such that But then does not converge. The analytic functions we shall consider will take their values in or, more generally, in a Banach algebra. If D A (i = 1,2) are analytic functions into a Banach algebra A then F1 F2: D1 fl D2 —' A is also an analytic function. Furthermore, if F1(z) =

forz E D(z0,r),

F2(z) =

then where

F1(z)F2(z)

= n'O These observations more than

=

k0

akbfl_k.

suffice to prove the main results of the

chapter. Theorem

5. The spectrum of an operator T E

is not empty.

Proof. Suppose that o-(T) = 0. Since, as we remarked earlier, formula (2) shows that R(A) is an analytic function on the resolvent, which is now the entire plane, R(A) is an entire function. Furthermore, (3) Hence, by Liouville's theorem, shows that IIR(A)II 0 as Al R(A) is constant and, in fact, R(A) = 0 for every A. But this is clearly

0

impossible.

For emphasis, let us put Corollary 4 and Theorems 2 and 5

together.

Theorem 6. For every complex Banach space X and operator TE the spectrum o(T) is a non-empty closed subset of {A E C: IA I p(T) and II71I}. Furthermore, if A

d(A,o'(T)) = then IIR(A)II

:

E r(T)} = d

l/d.

It is interesting to note that Theorem 5 gives an independent proof of the fact that every nXn complex matrix has an eigenvalue. Note also that the analogue of Theorem 5 fails for real spaces, as shown, for example, by the rotation (e1 , e2) '—i (e2, —e1) in R2. The spectrum is

very useful precisely because it allows techniques of complex analysis to be brought into operator theory.

Chapter 12: The algebra of bounded linear operators

173

From Theorem 6 it is a short step to show that the approximate point spectrum is not empty either. Theorem 7. The approximate point spectrum (Tap(T) contains the boun-

dary äo(T) of the spectrum.

Proof. Let A E öu(T). Pick a sequence A1,A2,... E p(T) tending to A. Then, by the second part of Theorem 6, Therefore there —, C X such that y,, 0 and = 1 for every is a sequence =

n. Setting

we find that

= 1 and

II(AI—

Hence

0.

I

)°

is

0

an approximate eigenvector with eigenvalue A.

As an illustration of the concepts and results presented so far, let us examine the spectrum of the right shift operator S on 1,, (1 p defined by S(x1 , x2,...) = (0, x1 , x2,...). Since IISII = 1, the spectrum a(S) is a closed non-empty subset of the closed disc 4 = {A E C: lAP Furthermore, IISxII = IIxII for every x and so IRS—A)xII 1}. By (1—IAI)IIxIJ, implying that aap(S) C 84 = {A E C: IAI = 1}. Theorem 7 we have oap(S) = 84. (Of course, this is easy to check directly.)

How much of the circle 84 belongs to the point spectrum? Suppose that Sx = Ax 0, where A 0. Then 0 = Ax1, x1 = Ax2, x2 = Ax3,

..,

implying that = 0, x2 = 0, x3 = 0 Hence = 0. Therefore the spectrum a(S) is the closed disc 4, the approximate

point spectrum (Tap(S) is the circle 84 and the point spectrum is empty. What is the spectrum of a 'nice' function f of an operator? This is easy to answer when f is a polynomial; an analogous result holds in a much more general case, namely when f is an analytic function on an open set containing the spectrum.

Theorem 9. Let p(t) be a polynomial with complex coefficients. Then for T E the spectrum of p(T) is precisely

p(a(T)) = {p(A): A E r(T)}. Proof. We may assume that the leading coefficient of p(t) p(O) =

0.

Given A0 E C, let p(:)—A0

=

Then

(t—/.Lk).

is

1 and

174

Chapter 12: The algebra of bounded linear operators

p(T) —A0!

=

k=1

This product fails to be invertible if at least one of the factors, say is not invertible, i.e. p-k E o(T). Since the

are the zeros of

p(s)—A0, this happens 1ff p(p-) = A0 for some p- E

The spectral radius of T E

0

is

r(fl =sup{IAI:AE0.(T)}=max{IAI:AEor(T)}. The spectral radius is a simple function of the sequence shown by the following result, Gelfand's spectral-radius formula.

Theorem

S. For T E

we have

r(T) = Proof. By Theorems 6 and 7 we have = r(T")

11Th.

(6)

Hence r(T) On the other

hand, as p(T) 3 {A E C: IAI > r(T)}, relation (3) tells us that the Laurent series

L

n=0

convergent for IA > r(T). Hence, recalling the formula for the radius of convergence, we find that r(T) 0 is

It is easily seen that the spectral radius is an upper semicontinuous function of the operator in the norm topology; in fact,

r(S+7')

r(S)+ lfl

for all S, T E

(see Exercise 8). It is worth recalling that all the results above are true for the spectra of elements of Banach algebras, not only of elements of In the simplest of all Banach algebras, C, every non-zero element is invertible. In fact, C is the only Banach algebra which is a division algebra.

Theorem 10 (Gelfand—Mazur theorem) Let A be a complex unital Banach algebra in which every non-zero element is invertible. Then

A=C.

Chapter 12: The algebra of bounded linear operators

175

Proof. Given a E A, let A E 0(a). Then A —a (= Al—a) is not inverti-

0

bleandsoA—a=0,i.e.a=A.

is invertible 1ff know from Theorem 11.11 that T E is invertible. Hence A!— T is invertible if Al— T is. 1'. E Therefore, recalling that for a Hilbert space H, the dual H is identified A!— T is with H by an anti-isomorphism, we get that for T E invertible if (A!— T)* = A!— r is invertible. Finally, recalling from

We

Theorem 11.8(b) that for a normal operator T we have IIT"II = 1, we have the following result. n

11111"

for

Theorem 11.

(a) For a Banach space X and an operator T E u(T') = 0(T). (b) For a Hilbert space H and an operator T E

we have

we have

a(r) = conju(T) = {A: A C (7(T)}. (c) If T is a normal operator on a Hilbert space then r(T) = II Tfl.

0

Let us introduce another bounded subset of the complex plane associated with a linear operator. Given a Banach space X and an operator TC

the (spatial) nwnerical range of T is

V(T) = {(Tx,f): XE X, f C X, lIxIl = 0111 = f(x) = 1}. With the notation used before Lemma 8.10,

V(T) = {f(Tx): (x,f)

fI(X)}.

Thus to get a point of the numerical range, we take a point x of the unit sphere S(X), a support functional f at x, i.e. a point of the unit sphere

S(Xt) taking value I at x, and evaluate f at Tx. It is clear that the numerical range depends on the shape of the unit ball, not only on the algebra If T is an operator on a Hubert space then V(T) is just the set of values taken by the hermitian form (Tx, x) on the unit sphere: V(T) = {(Tx,x): lixil = 1}. Nevertheless, the numerical range can be easier to handle than the spectrum and is often more informative. It is clear that, just as the spectrum, V(T) is contained in the closed disc of centre 0 and radius 11711. Even more, the closure of V(T) is sandwiched between 0(T) and this disc. But before we show this, we prove that can be only a little bigger than V(T). Theorem 12. For we have TC

a complex Banach space X and an operator

Chapter 12: The algebra of bounded linear operators

176

C V(T),

V(T) C

where V(T) is the closure of V(T).

Proof. The first inclusion is easily seen since if x E S(X), f E = (x, T'f) = and (x,f) = 1 then i E S(X**), (f,i) = I and where, as earlier, i denotes x considered as an element of To see the second inclusion, let C so that there are f C and C S(X**) such that (f,p) = 1 and = Let 0 < e

<

and

KT(x—y),f)i <

S(X) such that

ikTy,f—g>Ii

this implies that V(T) has a point close to

(Ty,g)

1.

<

and

Then, by Theorem 8.11, there exist g C S(X*) and y (y,g) = 1, ix—yli <€ and Of—gil €. Hence

But

<

in B(r*), there is an x C B(X) such that

Since B(X) is

namely

the point

C V(T):

=

lii + the spectrum of T tained in the closure V(T) of the numerical range V(T). Theorem 13. For a complex Banach space X,

Proof.

Suppose that d(A,V(T)) = infflA—,.ti

By Theorem 12 we also have d(A, we have to show that Al— T is

C V(T)} = d >0.

= d. To

prove that o(T)

functional f C S(X) at x, functional with (x,f) = 1. Then (Tx,f) V(T) and so ii(Al—T)xIi

((AI—T)x,f)l = IA—(Tx,f)i

Hence Ii(A1—

for all x C X.

C

invertible.

Given x E S(X). pick a support norm-I

is con-

T)xii

i.e. a

Chapter 12: The algebra of bounded linear operators

177

Similarly, as d(A,v(r)) = d, we have II(AI— r)fII

f E r. Thus both A!— T and (Al— T)* are bounded below. But

0

then, by Theorem 11.12, Al— T is invertible.

Another aspect of the connection between the spectrum and the numerical range given in Theorem 13 is that coo(T), the convex hull of spectrum, is precisely fl ëö V(T), where the intersection is taken over all

numerical ranges V(T) with respect to norms on X equivalent to the given norm. But we shall not give a proof of this result. The rest of the chapter is about a striking application of the spectral-

radius formula to obtain a remarkable theorem related to material beyond the main body of this book. The theorem is Johnson's uniqueness-of-norm theorem, but the beautiful and unexpectedly simple proof we present is due to Ransford. Let us start with a classical inequality concerning complex functions, namely Hadamard's three-circles theorem, stating that if f is analytic in is a convex functhe annulus R1 < Izi < R2 then M, = Thus if f(z) is analytic in the open disc tion of logr for R1
1, then maxlf(z)I max jf(z)I

jzI=R

(7)

IzI=1/R

forallR(1
If(1)12 = Ig(1)l

max If(z)I.

1z11/R

IzI=R

A similar inequality holds for Banach-space-valued analytic functions, with the norm replacing the modulus. In particular, if f(z) is analytic in the open disc Iz I
max IIf(z)II max IIJ(z)II.

JzI=R

1z11/R

for all R (1
=

(8)

E X) and let

178

Chapter 12: The algebra of bounded linear operators

Set

g(z) = p(J(z)) = n =0

Then, by (7) IIf(1)112

= Ig(l)12

maxlg(z)I max Ig(z)l

IzI'=R

IzI=1/R

Ic(f(z))I

= maxlq'(f(z))I max IzkrR

= maxjlf(z)II max If f(z)II,

zIR

1z11/R

proving (8).

In fact, if we have an analytic function with values in a Banach alge-

bra then in inequality (8) we may replace the norm by the spectral radius.

Lemma 13. Let f(z) be an analytic function in the open disc jz I
maxr(f(z)) max r(J(z)).

IzI=R

Proof. From the spectral-radius formula, we know that

is

monotone decreasing to r(J(z)), and r(f(z)) is a continuous function of for zl = R and fz I = hR. Consequently, by Dini's theorem

z

(Theorem 6.5), for every e > 0 there is an n such that max

max{r(J(z))+e} max {r(J(z))i-e}.

max

Izkl/R zl=R Applying Theorem 9 and inequality (8), we find that IzI=R

r(f(l))2 max

max

zI=R

IzI=1/R

IzIR zI=1/R max{r(f(z))+e} max {r(f(z))+€}.

0

As this holds for every e > 0, the result follows.

The radical Rad B of a complex (unital) Banach algebra B is the intersection of all the maximal left ideals of B. The following lemma relates the radical to the spectral radius.

Lemma 15. If b E B is such that r(b'b) =

bE RadB.

0

for all

b' E B then

Chapter 12: The algebra of bounded linear operators

179

Proof Suppose that b Rad B, that is b L for some maximal left ideal L. Then Bb + L is a left ideal properly containing L, and so Bb + L = B. Hence e = b'b + I for some b' E B and 1 L, where e is the identity. But then e—b'b = I E L and so e—b'b is not invertible. Therefore r(b'b)

0

1.

Now we are ready to give Ransford's proof of Johnson's theorem,

which is slightly more than the assertion that if Rad B ={O} then all Banach-algebra norms on B are equivalent. Theorem 16. Let A and B be Banach algebras, with Rad B = {O}. Then every surjective homomorphism 0: A B is automaticaLly continuous. 0 in A and b in B. By the closedProof. Suppose that a,, graph theorem (Theorem 5.8) it suffices to show that b = 0. Since Rad B = {O}, this is the same as showing that b Rad B, and by Lemma 15 this follows if we show that r(b'b) = 0 for all b' E B. Let then b' E B. Pick a,a' E A with 0(a) = b and 0(a') = b'. Set d = 0(c) = b'b. Then —, 0 in A c = a'a, = A by and d, = C — d in B. Define a linear function

=

and set =

=

= d. Note that p,,(l) = c and Since a homomorphism does not increase the spectral radius.

Hence, by Lemma 14, for all n

Letting n —'

1 and R> 1 we have

we find that

IIcII(R'IIdII) and so, letting R —,

we

see that r(d) = r(b'b) =

0,

as desired.

Corollary 17. (Johnson's uniqueness-of-norm theorem) Let B be a complex unital Banach algebra with norm and Rad B = {0}. Then every Banach algebra norm on B is equivalent to Do. 0 lb

Chapter 12: The algebra of bounded linear operators

180

The algebra of bounded linear operators on a Banach space X satisfies the conditions in Corollary 17, so all norms on turning it into a Banach algebra are equivalent. Exercises

In the exercises below, X is a complex Banach space and T E

1. We call T a left (right) divisor of zero if there is an S E such that S 0 and TS = 0 (ST = 0). Show that the point spectrum of T is

=

E C: Al— T is a left divisor of zero)

(A

and the compression spectrum is =

E C: Al — T is a right divisor of zero}.

{A

2. We call T a left (right) topological divisor of zero if there are T1, T2,... E such that 117j1 = 1 and Ti',, —*0 (T,,T— 0). Show that the approximate point spectrum of T is Uap(T) =

(A

E C: Al— T is a left topological divisor of zero).

3. Show that T is a right topological divisor of zero iff E is a left topological divisor of zero. 4. Show that if A E o(T) then Al— T is either a right divisor of zero or a left topological divisor of zero. Deduce that cr(T) = 5.

Let X be reflexive. Prove that if T is not invertible and is neither

a left nor a right divisor of zero then it is both a left topological divisor of zero and a right topological divisor of zero. 6. Show that

C C: Al— r is not surjective}

Oap(T) = and

crap(r) = 7. Suppose that S, T C r(S+ T)

{A

C C: Al— T is not surjective}.

commute: ST =

r(S) +r(T)

and

TS.

r(ST)

Show that

r(S)r(T).

Show also that these inequalities need not hold if S and T are not assumed to commute.

Chapter 12: The algebra of bounded linear operators

8. Show that for S, T E

we have

r(S+T) Show S.

181

also that if r(S) =

0

r(S)+

11111.

then r:

—+

[0,

is continuous at

9. Suppose that r(T) < 1. Show that 11110

=

n0

is a norm on X, equivalent to the original norm 10. Prove that

r(T) = inf{II TU': 11.

II

II'is a norm on X, equivalent to

IlL

Check that the resolvent R(A) of T satisfies the resolvent identity

= (p—A)R(A)R(p) = for all A,p.

p(T).

12. Check that if S, T E

A E p(ST) and A

0, then A E p(TS)

and

= A'+A1T(A—ST)'S. Show also that o(ST) =

Deduce that a(ST) U{0} = o(TS) U{0}.

o(TS) need not hold.

13. Let K be a non-empty compact subset of C. Show that K is the spectrum of a Hubert space operator: there is an operator SE where H is a Hilbert space, such that u(S) = K.

14. Show that if T is a normal operator on a Hilbert space then r(T) =

11711.

15. For w = Show that

E =

define = SUpkIWkI.

of Tare w1,w2,... and o-(T) = 16. For w = (Wk)° define '2

12

by

=

Show also that eigenvalues What is by

l2

= (0,w1x1,w-,x2,...). Express

the spectral radius

and

in terms of the

sequence w. {z E C: lz 1} and H = L2(4). Let T E be the operator of multiplication by Show that o-(T) = 4 and T has no eigenvalue.

17. Let 4 =

I

Chapter 12: The algebra of bounded linear operators

182

18.

Use the Hahn—Banach theorem to show that for T C

we

have

supReV(T) =

there isanx =

sup{c C

X(x

x(c)

such that 11(1 —rc+rT)xIj

lixilfor

0),

allr

is said to be dissipative if sup Re V(T) 19. An operator T C dissipative then Show that if T is IIx—rTxII

for all x C X and r 0. V(T) and 20. Show that if A C

(A!—

0}. 0.

HxII

T)x = 0 then

Hx+(A/—T)yII

Dxli

for all y E X. (Note that if T is dissipative and Tx = 0 then

y\ y 11 x—-+-Ilx—-l r/ r

r\

x—

y

r

for r > 0.) Deduce that if A C aeo V(T) and (Al— T)2z = 0 then (Al—T)z = 0. is said to be hermitian if V(T) C R. (Note 21. An operator T E space; for a Hubert space this definition cointhat X is a Banach

cides with the usual (earlier) definition of a hermitian operator on a Hubert space.) Thus T is hermitian if both iT and —iT are dissipative. Prove that T is hermitian if iiexp(iT)xH

= lxii

and x C X, where

for all r C

S'

expS

=

for S C

22. Let H be

a

Hilbert space and let S C Show that

V(S) C

(Sx,y) and

be such that

(Sx,x)"2(Sy,y)'12

deduce that IISxIi2

for all x,y

C

(Sx,x)iiSii

H. Deduce that if 0

V(S)

then 0 C a(S).

Chapter 12: The algebra of bounded linear operators

183

23. Show that if S is a hermitian operator on a Hubert space then V(S) = coo(S).

(Note that the assertions in the last two exercises are easily deduced from Theorem 11 (c) as well.)

24. Prove that the result in the previous exercise holds for a normal operator S on a Hilbert space. 25. Let M be a proper ideal of a unital Banach algebra B. Show that the closure of M is also a proper ideal. Deduce that every maximal ideal of B is closed.

26. Let M be a maximal ideal of a complex unital Banach algebra B. Show that B/M is also a complex unital Banach algebra.

The aim of the next five exercises

is

to prove the commutative

Gelfand—Nalmark theorem. In these exercises A is a commutative complex unital Banach algebra.

27. Show that if M is a maximal ideal of A then AIM is isometrically isomorphic to C. 28. Let h: A —' C be a non-zero homomorphism. Show that fihil = 1. 29. Let At be the set of maximal ideals of A. By Exercise 27, At may be identified with the set of non-zero homomorphisms h : A —÷ C. By Exercise 28, this is a subset of B(A*). Give At the relative weak* topology (i.e. the topology induced by the weak* topology on A*). Endowed with this topology, we call At the maximal ideal space of A. Prove that At is a compact Hausdorff space. 30. For x E A the Gelfand transform of x is the function 1: At C defined by i(h) = h(x), where h E At is considered as a homomorphism h : A C. Show that the spectrum o(x) is the range of 1. 31. Show that if A is a commutative unital C-algebra with maximal ideal space At then the Gelfand transformation x '—b maps A isometrically and isomorphically onto C(At), the commutative algebra of continuous functions on the compact Hausdorff space

i

At.

32. Let x=

be the algebra of all (doubly infinite) complex sequences such that =

x,j

with convolution product xy = z, where

Chapter 12: The algebra of bounded linear operators

184

Zn = is a commutative Banach algebra. Show also that the maximal ideal space of 11(Z) can be identified with the unit cirC is defined by cle T = {z E C: Izi = 1}, where z:

Show that

z(x) = 33.

Deduce from the result in the previous exercises that if f(t) =

0

x,j where

=

Let H be a Hilbert space and let T E e > 0 there is an invertible operator S IISTS'II
Show

that for every such that

Notes

The original reference to Gelfand's spectral-radius formula is I. M. Gelfand, Normierte Ringe, Mat. Sbornik N. S., 9 (51) (1941), 3—24; this is also one of the references to the Gelfand—Mazur theorem; the other is S. Mazur, Sur les anneaux linéaires, C. R. Acad. Sd. Paris, 207 (1938), 1025—27.

Theorem 12 is from B. Bollobás, An extension to a theorem of Bishop and Phelps, Bull. London Math. Soc., 2 (1970), 181—2, and Theorem 13 is from J. P. Williams, Spectra of products and numerical ranges, J. Math. Anal. and AppI., 17 (1967), 214—20. A good account of numeri-

cal ranges can be found in F. F. Bonsall and J. Duncan, Numerical Ranges II, London Mathematical Society Lecture Note Series, vol. 10, Cambridge University Press, 1973, vii + 179 pp. Johnson's theorem is from B. E. Johnson, The uniqueness of the (compkte) norm topology, Bull. Amer. Math. Soc., 73 (1967), 537—9; its simple proof is from T. J. Ransford, A short proof of Johnson's uniqueness-of-norm theorem, Bull. London Math. Soc., 21 (1989), 487—8.

Chapter 12: The algebra of bounded linear operators

185

The commutative Gelfand—Nalmark theorem is taken from I. M. Gelfand and M. A. Naimark, On the embedding of normed rings into the ring of operators in HUbert space, Mat. Sbornik, 12(1943), 197—213; another classical

reference is M. A. Naimark, Normed Rings, Revised English edition; translated from the Russian by Leo F. Boron, Groningen: Noordhoff, 1964.

13. COMPACT OPERATORS ON BANACH SPACES

For the sake of simplicity we shall assume that all the spaces appearing in this chapter are complex Banach spaces. Our aim is to study a class

of operators closely resembling the operators on finite-dimensional spaces; we shall show that these operators are somewhat similar to the nXn complex matrices.

Vaguely speaking, the operators we shall look at are 'small' in the sense that they map the unit ball into a 'small' set. To be precise, an the image of the unit ball Bx operator T E Y) is compact if under T, is a relatively compact (i.e. totally bounded) subset of Y. Thus T is compact if and only if T is compact if and only if for every bounded sequence C X the sequence has a convergent subsequence. We shall write Y) for the set of compact operators from X into we write for X). Y. Analogously to

compact, i.e. if dimlmT= dimTX< then TE Indeed, set Z = ImT. is a subset of Since Z is finite-dimensional, Bz is compact and so the compact set IIllIBz. We shall denote by Y) the set of (bounded) finite rank operators from X to Y. Examples 1. (i) Every finite rank operator T E

Y) is

(ii) Every bounded linear functional f E X to C.

(iii) Let I be the closed unit interval [0, 1] and let X be the Banach space C(I) of continuous functions with the supremum norm. Let K(x,y) E C(IxI), i.e. let K be a continuous function on the closed unit square Ix I. For f E C(I) define a function Tf E C(I) by 186

Chapter 13: Compact operators on Banach spaces

187

K(x,y)f(y) dy.

(Tf)(x) =

and it is easily seen that, in fact, T E

Then T E

Indeed,

by the Arzelá—Ascoli theorem (Theorem 6.4) we have only to check N is uniformly bounded and equicontinuous. If IK(x,y)I that

for every (x,y) E Ix! then l(Tf)(x)l

and so TBx &

>0

fore

N L If(y)I

dy

NlIfjI,

is uniformly bounded by N. Also,

for E

> 0,

such that if lxi—x21 <8 then IK(xi,y)—K(x2,y)I

iffE

and 1x1—x21

I(Tf)(xi) —

<8

there is a

<e. There-

then IK(xi ,y)

f

—

K(x2,y)j

I dy

f IK(x1,y)—K(x2,y)l dy <e. Thus TBx is indeed equicontinuous and so T is compact, as claimed.

(iv) Let H be

with orthonormal bases we have

and

a Hilbert space

For every operator T IITeaO2= 1=1 1=1

1=1

j=1

1=1

and so

= 1=1

I(Tf1,e1)12 = 1=1 j=1

j=1

lIre,I12 = 1=1

This shows that \1/2 O71IHS =

IITe1II2

(

\i =

1

is independent of the orthonormal basis (e1 An operator T E is said to be a 1-filbert—Schmidt Hubert—Schmidt norm, II T1IHS, is finite. One often writes

operator if its for the

set of Hubert—Schmidt operators on H, and 111112 for indicates, the Hubert—Schmidt norm is the II norm of the sequence formed by the entries of the matrix representation of T. Indeed, set = (Te,,e,> so that T is given by the matrix A = (a11) in the sense that

Chapter 13: Compact operators on Banach spaces

188

xiei) =

aiixi)ei.

Then

\1/2 tI11IHS = 01112 =

j=1

Putting it another way, with a

= we

have a E H for every i, \1/2 II11IHS =

and

(

\i=1

Tx =

/

(x,a1)e1. 1=1

In particular, lIxII2IIaiIl2 = 11x11211

I (x, a.) 2

V

=

and so 11111

II11IHS.

is easily seen that every Hubert—Schmidt operator is compact. Indeed, with the notation as above, put It

A=

xe1 E H: 1x11

=

11a111, i =

< the set A is a compact subset of H (see Exercise 1). Since TBH is a subset of A, it is relatively compact. 0

Since

The class of compact operators is an example of a closed operator ideal. An operator ideal is a function that assigns to every pair X, Y of Banach spaces a subset

Y) of

SE

then STE

and R E

Theorem 2. (a) Y) is a closed subspace of

(b) If TE STE

SE

Y) such that if T E and TR E

Y), Y).

Y).

and

RE

then

and TR E

Proof. (a) Let us show first that Y) is a subspace of Y), i.e. if 5, T E Y) and E E C then be Y). Let

Chapter 13: Compact operators on Banach spaces

189

has a subsequence, say a bounded sequence in X. As S is compact, is convergent. The operator T is also compact, such that and so (xflk) has a subsequence, say (Xmk), such that (TXm*) is convergent. But then ((AS+itLT)Xm&) is also a convergent sequence. Y). SupY) is a closed subset of Now let us show that Y). We have to show that TE pose

TB1 is totally bounded. Given c > 0 let n be such that — <€. is compact, there are x1,.. ,x,, E B1 such that {T,,x1: 1 As m} is an c-net in 1,B1. Thus if x C Bx then there exists an x, such i II

.

<e. Then

that IITx— Tx,II

II(T—

1 i m} is a 3€-net in TB1. (b) Note that a bounded linear operator maps a bounded sequence

into a bounded sequence and a convergent sequence into a convergent

0

sequence.

Since finite rank operators are compact, by Theorem 2(a) every limit of finite rank operators is compact. The problem of whether every compact operator can be obtained in this way was, for many years, one of the best-known problems in functional analysis. After about 40 years the approximation problem was solved in the negative by Per Enflo in 1973, who constructed a separable reflexive Banach space X for which

is the closure of is not the closure of Since whenever X has a (Schauder) basis (see Exercise 7), Enflo's example also showed that not every separable reflexive Banach space has a basis, solving another long-standing question. As so often in mathematics, the counterexample turned out to be the Start of the story: it opened up whole new fields of research on approximation and basis problems. But we cannot go into that in this book.

The operator ideal of compact operators is closed under taking adjoints as well.

Theorem 3. An operator T C

rE

Y)

is compact if and only

if

is compact.

Proof. (i) Suppose first that T C Y) is compact, i.e. the set K = TBx is compact. For a functional f C r let Rf be the restriction of f to K. Clearly Rf E C(K) and, in fact, the map R: Y' —, C(K) is a

bounded linear map, where, as usual, C(K) is taken with the supremum

norm. Let P = RBr. Note that forf C r we have

Chapter 13: Compact operators on Banach spaces

190

fIrfIl = sup{(x, rf>:

x E Bx}

= sup{(Tx,f): x E = sup{(y,f): y E TBx}

= This

sup{(y,f):yEK}

= IIRfII.

shows that rBr is isometric to 'P C C(K), with the isometry

given by

Rf.

Consequently rBr is totally bounded if and only if 'P is. By the Arzelá—Ascoli theorem, 'P is totally bounded if and only if it is uniformly bounded and equicontinuous. Both conditions are easily checked. If f E Br then DrfIl iirii = flTfl, and so 'P is uniformly

bounded. Also, if f E Br and y,y' E K then I(Rf)(y)—(Rf)(y')I = If(y—y')I and

IIy—y'lI,

so J) is equicontinuous.

(ii) Now suppose that r E

map T E

is compact. By part (I), the

is compact, i.e.

is relatively compact.

But under the natural embeddings X C r* and Y C rs we have TBx = T*BX C rBr., and so TBx is also relatively compact. 0

Theorems 2 and 3 state that the compact operators form a closed operator ideal that is also closed under taking adjoints. In particular, for a Banach space X, the set is a closed ideal of the Banach algebra If H is a Hubert space then is a closed ideal of H which is also closed under taking adjoints: it is a closed

The main aim of the chapter is to present the spectral theory of compact operators, due to Frigyes (Friedrich or Frédénc) Riesz. Recall that is the spectrum of an operator T E

o(T) =

{A

E C: T—A1 does not have a bounded inverse},

where I is the identity operator on X. We proved in the previous chapter that for every T E the spectrum of T is a non-empty closed subset of C, contained in the disc {z E C: Izi IITU}. As we shall see, for a compact operator T E the spectrum of T resembles the spectrum of an operator on a finite-dimensional space, i.e. the spectrum of an PZXn matrix. To be precise, if X is infinite-dimensional and T C then r( T) is a countable set whose only accumulation point is 0 and if A C or(T) (A 0) then A is an eigenvalue of T with finitely many linearly independent eigenvectors.

Chapter 13: Compact operators on Banach spaces

191

and a > 0. Then T has only finitely many Theorem 4. Let T E linearly independent eigenvectors with eigenvalues having modulus at least a. Proof. Suppose x1 , x2,... is an infinite sequence of linearly independent

0 and A1 I a for every i. Set By Theorem 4.8(b), there exists a sequence

eigenvectors such that Tx1 = =

. .

C

,x1j.

Xsuch

and note that

E and

=

E

CkXk then

=

xn_1.

= Hence

= 1.

1/a, E X,, and Iz,,fl Indeed, the first two assertions are obvious and if

=

Put

A1x1

if n >

m

then =

IITZnTZrnII

= 1.

does not contain a subseConsequently the bounded sequence such that is convergent, contradicting the compactquence

0

ness of T.

Our next aim is to show that if T is a compact operator and A 0 is not an eigenvalue of T then A o-(T), i.e. Al— T is invertible. Equivalently, cr(T) U {O} = U{O}, i.e. with the possible exception of 0, every point of the spectrum is in the point spectrum. In proving this we may and shall assume that A = 1; so our aim is to prove that S = I— T is invertible. We need two lemmas, both of which are proved in a more general form than necessary.

Lemma 5. Let T E

and set S =

I—

T. Then SX is a closed sub-

space of X. Proof. Set N = KerS; by Theorem 4 we know that N is finitedimensional, say with basis {b1 ,. , bk,, }. Then there is a closed subspace M C X such that X is the direct sum of M and N: X = MEI3N. Indeed, -

if we choose fr,..

.

.

E X* such that f,(b1) = then we may take The projections of x E X into N and M are

Kerf1. f,(x) b• and PM(4 = x —pN(s). Let S0 be the restriction of S to M: S0 = SIM. Then SX = SM = S0M and KerS0 = KerSflM = {0}, and so S0 is injective. Hence

M= PN(4 =

Chapter 13: Compact operators on Banach spaces

192

to prove that SX = S0M is closed, it suffices (and, indeed is necessary) to prove that S0 is bounded below. Suppose that S0 is not bounded below, i.e. 0 for some sequence CM = 1). Since T is compact, has a convergent subsequence, and so we may assume that itself is convergent, say Tx,, —' y. Then

=

=

= 1. But we also have icting Ker S0 = {0}.

Sy, and so S0y = 0, contrad-

and so

0

Lemma 6. Let S, T E be such that S+ T = I and SX C Y, where Y is a closed proper subspace of A'. Then for every E > 0 there is a point x0 E such that d(Txo, TY) > €. 0 1

Proof. By Theorem 4.8(a), there exists an x0 E Bx such that d(x0, Y)> 1—c. As Tx0 = x0 — Sx0, Sx0 E Y and TI' = (I — S) Y C I', we have

d(x0—Sx0,Y) = d(x0,Y)> 1—c.

0

a compact operator and suppose A

0 is not an

d(Tx0,TY) Theorem

7.

Let T be

eigenvalue of T.

Then

A

o(T).

Proof. By replacing T by T/A, it suffices to prove the result for A = 1. S = I— T and KerS = (0). We have to show that Let then T E S is invertible.

= (n = 0,1,...), so that are closed. Let us By Lemma 5 the subspaces

Let us prove that SX = X. Set = A' D show first that

.

for some n. Indeed, otherwise Y0 3 Y1 3 and all the inclusions are strict. Then, by Lemma 6, one can find ele> ments such that But then, in particular, E if n m, and so has no convergent subse— TYmI! > quence, contradicting the compactness of T. We claim that, in fact, Y0 = Suppose that this is not so. Then Then, there is an m such that Let U E there exists a pOint v such that as Su SYm, I'm + i u—v E KerS, contradicting Su = Sv. But then S(u—v) = 0 and so 0 our assumption. Consequently = I'0, i.e. SX = A', as claimed. The

=

proof is essentially complete. The bounded map S: X —

1—1 map of the Banach space X onto itself and so, by the

mapping theorem, S is

invertible.

X

is a

inverse-

0

Chapter 13: Compact operators on Banach spaces

193

Let us restate the information contained in Theorems 4 and 7 about the spectrum of a compact operator as a single result.

Theorem 8. Let T be a compact operator on an infinite-dimensional Banach space. Then (T) = {O,A1,A2,...}, where the sequence Ai ,A2,... (of non-zero complex numbers) is either finite or tends to 0; furthermore, every A. is an elgenvalue of T, with finite-dimensional

0

eigenspace.

With some more work, we can gel more detailed information about the structure of compact operators. If T is compact and S = 1— T then Im S is a finite-codimensional closed subspace of X; even more, for a suitable n 1, X is the direct sum of KerS" and ImS". We prove this, and a little more, in the following theorem.

Theorem 9. Let X be a Banach space, T E and S = I— T. Set Nk = KerSk and Mk = ImS" (k = 0,1,...), where = Then is an increasing nested sequence of finite-dimensional subspaces and (Mk

is

a decreasing nested sequence of finite-codimensional subspaces.

There is a smallest n 0 such that N,, = Nm for all m n. Furthermore, M,, = Mm for all m n, Xis the direct sum of M,, and N,,, and M,, is an automorphism of M,,. (1— T)", we see that = I— where Tk is compact. Hence, by Lemma 5, Mk is a closed subspace of X. Clearly N0 C N1 C ... and M0 J J ...; furthermore, we know that each

Proof. By expanding 5" =

Nk is finite-dimensional.

As in the proof of Theorem 7, Lemma 6 implies that there is a smallest n such that N,, = N,, + and there is also a smallest m such that a and Mm = Mm' for all Mm = Mm+i. Then N,, = N,,. for all a' Let us turn to the main assertions of the theorem. We prove first that N,, fl M,,. As y E M,,, we have y = S"x for N,, fl M,, = {0}. Let y some x E X. But as y N,,, S"y = 0 and so = 0. Hence x E N2,, = N,,, implying S"x = 0. Thus y = S"x = 0, showing that

N,,flM,, =

{0}.

We claim that for p =

max{n, m} we have X = N,, Indeed, given x E X, we have But = and so there is a such that = Sex. Hence x—y N,, and so vector y Could we have p > a? x = y + (x — y) shows that X = N,, + Clearly not, since then M,, would strictly contain and so we would

Chapter 13: Compact operators on Banach spaces

194

have

#

Thus p =

{0}.

n

and X is the direct sum of

and

is finite-dimensional (see Exercise 4.20). Finally, = = and as

=

=

N1

C

= {0}.

Hence, by the inverse-mapping theorem, the restriction of S to

is

invertible.

0

Putting Theorems 8 and 9 together, we arrive at the crowning achievement of this chapter: a rather precise description of the action of a compact operator. Theorem 10. Let X be an infinite-dimensional Banach space and let T be a compact linear operator on X. Then o(T) = {0,A1,A2,...}, where the sequence A1,A2,... is either finite or tends to 0. For every A = A there is an integer kA I and closed subspaces NA = N(A; T) and = M(A; T) invariant under T, such that NA is finite-dimensional and MA X = The restriction of Al— T to MA is an automorphism of MA, NA

and for

=

A

= Ker(AI—

A=

A

Ker(Al—

we have NA C MM.

Proof. Only the last claim needs justification. The operator T maps MA into itself and NA into itself. Furthermore, T)INA is an autoof the finite-dimensional space NA since if we had (id— T)x = 0 for some x E NA (x 0) then we would have 0 for every n I, contradicting = (AJ—T)'1x = 0. Consequently, for every n 1, and so NA C MM. = NA morphism

0 There

is no doubt that Theorem 10 is a very beautiful theorem. At

first sight it is not only beautiful but very impressive as well: it seems to come close to giving us a very fine decomposition of the space into a direct sum of generalized eigenspaces. Unfortunately, this is rather a mirage: the theorem cannot even guarantee that our compact operator has a non-trivial closed invariant subspace, let alone give a direct-sum decomposition. In fact, non-trivial closed invariant subspaces do exist, as we shall prove in Chapter 16. However, to prove the existence of invariant subspaces we shall need some results to be proved in Chapter 15.

Before we turn to that, in the next chapter we shall show that a

Chapter 13: Compact operators on Banach spaces

195

best possible decomposition can be guaranteed if we deal with a compact normal operator on a Hubert space. Exercises 1.

A=

li,.: lxi

{x = (x,)°

a compact subset of (1 Prove 2. Let K be a closed and bounded subset p 0 there is an n such that K is compact if and only if for every lx1V <€ for every x = E K. that 1) be the vector 3. As in Example I (xiii) of Chapter 2, let space of k times continuously differentiable functions f: (0, 1)

such

that k

11111k = sup >

IfW(()1

1=0

Show

that Xk =

1), II

Ilk) is a Banach space and the formal

identity map i: Xk —p Xk_I (f J) is a compact operator. Y), where X is infinite-dimensional. Show. that the 4. Let T E closure of TS(X) = {Tx: x X, lixil = 1} in Y contains 0. (HINT: 1 C S(X) such that for Consider a sequence n m.) be an orthonormal basis of a Hilbert space H and let 5. Let

Y), where Y is

T

6. Let X be every

a normed space.

Show that

a normed space such that for every finite set

0,

0.

A C X and

X has a decomposition

X=

as a direct sum of two closed subspaces, such that M is

finite-

dimensional,

d(a,M)

for every a

<€

A, and IIPN(X)II

cd(x,M)

for some c > 0 and every x E X, where PN tion

onto N. Show that

is the canonical projec-

is the norm closure of

i.e.

Chapter 13: Compact operators on Banach spaces

196

every compact operator on X is the operator-norm limit of finite rank operators. 7. Let X be a Banach space with a Schauder basis )'. Show that is the closure of be a sequence of non-zero complex numbers tending to 8. Let 0. Show that cr(T) = {0,A1,A2,. .} for some complex operator T .

on some compact Banach space X. Show that if all A, are real then X can be chosen to be a real Banach space.

9. Let X and Y be Banach spaces, let T E JE

Y), and let

Y) be invertible. Show that Im(J— T) is closed in Y and

has finite codimension.

10. Let X be a complex Banach space, and let T E be such that is compact for some n 1. Show that o(T) = {0, A1, A2,. . where the sequence A1 , A2,... is finite or tends to 0, and every A is an elgenvalue of T. What is the relationship between the sub.

spaces N(A; T) and

Ta)?

11. Let X1 , X2,... be Banach spaces and let X = having norm direct sum of these spaces, with x = Ilixill

be the

=

T

Let

T

E E

for every n.

and let 7, —' T in the 12. Let X be a Banach space, (1, C is relatively comoperator-norm topology. Show that is the unit ball of X. Show also that if pact, where B = B(X) A then A C o(T). 13. Let E be the space C[0, 1], endowed with the Euclidean norm

= 11f112

1

(j

If(x)12 dx)

i.e. T let T be as in Example 1 (iii). Show that T C maps the unit ball of the Euclidean space E into a relatively compact set. Show that T C 14. Let H be a Hilbert space and T C 0 whenever x,, converges weakly to 0, i.e. whenever (x,, , x) —' 0 for every x C H. 15. Construct a compact operator T on 1,, (1 p co) such that cr(T) = {0} and 0 is not an eigenvalue of T. and

Chapter 13: Compact operators on Banach spaces

197

16. Let c3 , c2,... be non-negative reals and

{x E 12: x =

C

for every k}.

xkI

Show that if C is a compact subset of '2 then Ck sequences (Ck

is

0.

For what

C compact?

17k. Let K be a compact subset of a normed space. Show that K is contained in the closed absolutely convex hull of a sequence tending to 0: there is a sequence x,, 0 such that K C C, where

C=

: n = 1,2,...}

A1X1

:

1A11

n=

= Notes

Compact operators were first introduced and applied by Hubert, Grundzuge einer ailgemeinen Theorie der linearen Integraigleichungen,

Leipzig, 1912, and F. Riesz, Les systémes d'équations a une infinite d'inconnus, Paris, 1913, and Uber lineare Funktionalgleichungen, Acta Math., 41(1918), 71—98. In presenting the Riesz theory of compact operators we relied on Ch. xi of J. Dieudonné, Foundations of Modern Analysis, Academic Press, New York and London, 1960, xiv + 361 pp. Theorem 3 is due to J. Schauder, Uber lineare vollstetige funktional Operationen, Studia Math., 2 (1930), 185—96. The first solution of the approximation problem was published by P.

Enflo, A counterexample to the approximation problem in Banach spaces, Acta Math., 30 (1973), 309—17; a simplified version of the solution is in A. M. Davie, The approximation problem for Banach spaces, Bull. London Math. Soc., 5 (1973), 261—6.

14. COMPACT NORMAL OPERATORS

In the previous chapter we saw that for every compact operator T on a Banach space X, the space can almost be written as a direct sum of generalized eigenspaces of T. If we assume that X is not merely a Banach space, but a Hubert space, and T is not only compact but compact and normal, then such a decomposition is indeed possible — in fact, there is a decomposition with even better properties. Such a decomposition will be provided by the spectral theorem for compact normal operators: a complete and very simple description of compact normal operators. Thus with the study of a compact normal operator on a Hilbert space we arrive in the promised land: everything fits, everything works out beautifully, there are no blemishes. This is the best of all possible worlds.

We shall give two proofs of the spectral theorem, claiming the existence of the desired decomposition. In the first proof we shall make use of some substantial results from previous chapters, including one of the important results concerning the spectrum of a compact operator. The second proof is self-contained: we shall replace the results of the earlier chapters by easier direct arguments concerning Hilbert spaces and normal operators. To start with, we collect a number of basic facts concerning normal operators in the following lemma. Most of these facts have already been proved, but for the sake of convenience we prove them again.

be a normal operator. Then the following

Lemma 1. Let T E assertions hold. (a) =

for every x E H.

(b) KerT= Kerr. (c)

198

= 11711" for every n

1.

Chapter 14: Compact normal operators (d) r(T) =

199

11711.

then Ker(AI— T) I Ker(j&I— T).

(e) If A

(f) For every A E C, both Ker(AI— T) and (Ker(A1— T))1 are invari-

ant under both T and r. (g) If H is the orthogonal direct sum of the closed subspaces H0 and H1 invariant under T then with T0 = TIH0 and T1 = nH1 we have max{ll Toll,

11111 =

T, is a normal operator on

and

Proof. (a)

As rr =

II T111}

we have

(rTx,x)

=

(TTx,x) =

(b) By part (a), we have Tx = 0

if rx = 0.

llTxIl2

= (Tx,Tx) =

(c) If S E

= rh-i1 (i = 1,2).

with

(rx,rx)

= llrxll2.

is hermitian then

llSxll2

= (Sx,Sx) = (SaSX,x) = (S2x,x)

IlS2llllxll2.

From this it follows that 115211 = 11S112, and by induction = 11S112'". This implies that IISII = IISII" for every n

on m we get

1. As rr

is hermitian, 11Th2 =

= II(Tr)hh =

=

(d) By (c) and the spectral-radius formula (Theorem 12.9),

r(T) =

urn

II

= urn

11Th

=

for a complex Banach (In fact, (c) and (d) are equivalent: if S E for every n 1.) space X then r(S) = IISII if IlShI =

(e) If Tx = Ax and Ty = T) = Ker(jiI— yE A(x,y) = (Tx,y) =

then ry = jiy because by (b) we have Therefore

(x,ry) = (x,1y) =

then (x,y) = 0. (f) As Al— T commutes with T and r, Ker(AI— T) is invariant under

and so if A

both Tand r. Also, let (x,y) = 0 for all yE Ker(AI—T). Then, since Ker(Al—T) is Invariant under T, for y E Ker(Ai— T) we have (Tx,y)

= (x,ry) = 0.

Hence Tx E (Ker(A1— T))'. Similarly,

200

Chapter 14: Compact normal operators

(Px,y) = (x, Ty)

=0

for every y E Ker(Al— T) and so Px = (Ker(AI—

(h) Letx = h0+h1, with li E H, (i = 0,1). Then llx112 = 11h0112+11h1112,

Tx = Th0+Th1 = T01z0+Th1

and IITxll2 = lIToholI2+ llT1h1ll2 + lIT1 11211h1

max{II

112

II T1112}(11h0112+ 1lh1112)

= max{11T0112, 11T1112}11xlI2.

Thus

max{II T111, II T2!I}. The reverse inequality is obvious.

Finally, as H0 and H1 are invariant under T, it follows that are invariant under P. H0 =

H1 =

0

It is worth emphasizing that Lemma I is a collection of elementary and simple facts, except for part (d), which is based on the spectralradius formula. Let us see then the first incarnation of the spectral theorem, claiming the existence of a spectral decomposition for a compact normal operator. Theorem 2. Let T E be a compact normal operator. For an eigenvalue A of T, let HA = Ker(T— Al) be the eigenspace of T belonging to A, and denote by PA the orthogonal projection onto HA. The operator T has countably many non-zero eigenvalues, say A1 , A2 Furthermore, dim HAk for every k, the projections PA are orthogo1, and nal, i.e. "Ak"A, = 0 if k (1) k

where the series is convergent in the norm of

Proof. By Theorem 13.8 and Lemma 1(e), we have to prove only (1). Given >0, choose n 1 such that lAkI <e fork> n. Set

H1

S, = SIH1 (i = 1,2), we have T0

and

= S0 and S1 = 0.

and Therefore, by

Chapter 14: Compact normal operators

201

Lemma 1(g), lIT—SO = max{JIT0—Soll, llT1—S111} = IITill.

But T1 is a compact normal operator and so, by Theorem 13.8, llT1ll is

precisely the maximum modulus of an eigenvalue of T1. As every

eigenvalue of is an eigenvalue of T, by our choice of n we have 0 Hence (1) does hold. IIT1II Let us state two other versions of the spectral theorem.

Theorem 3. Let T be a compact hermitian operator on an infinitedimensional Hilbert space H. Then one can find a closed subspace of H, a (finite or countably infinite) orthonormal basis of and a sequence of complex numbers v,, 0, such that if x = where E then Tx =

Proof. Let A1,A2,... and HA1,HA2,... be as in Theorem 2. Take a (necessarily finite) orthonormal basis in each and let be the union of these bases. Let H0 be the closed linear span of the orthonorand set v,, = Ak if x,, E HAk. mal sequence 0 Corollary 4. Let T be a compact normal operator on a Hilbert space H. Then H has an orthonormal basis consisting of eigenvalues of T. 0 In fact, compact normal operators are characterized by Theorem 2 (or Theorem 3). Let {x7: y E f) be an orthonormal basis of a Hubert space H, and let T be such that Tx,, = Then T is compact 1ff (2)

for every E > 0 (see Exercise 2).

Our proof of Theorem 2 was based on two substantial results: Theorem 13.8 concerning compact operators on Banach spaces, and the spectral-radius formula. We shall show now how one can prove Theorem 2 without relying on these results. It is a little more convenient to prove Theorem 2 for compact hermuian operators; it is then a simple matter to extend it to normal operators.

Recall that the numerical range V(T) of a Hilbert space operator

TE

is

Chapter 14: Compact normal operators

202

{(Tx,x): x E S(H)} and the numerical radius v(T) is

v(T) = sup{IAI : A E V(T)}.

If T E p.4(H) is hermitian, i.e. r =

T,

then its numerical range is real

since

(Tx,x) = (x,rx) = (x,Tx) = (Tx,x)

every x H and so (Tx,x) is real. In fact, T E is hermitian if its numerical range is real. Also, the spectrum of a hermitian operator is real. We shall not make use of any of the results proved about numerical ranges; the next lemma is proved from first principluses. for

Lemma 5. Let T be a hermitian operator. Then 1111 = v(T). Proof. Set = v(T), so that (Tx,x)I have to show that ill v. Given x S(H), let y E S(H) be (Tx,y) = (x, Ty) = IJTxII and so

for every x

,'

H.

We

(I

IITxIIy.

Then

v, as claimed.

0

6. Let U be a compact hermitian operator on H. Then

U has

ITxH =

(Tx,y) =

=

Hence IITxII Theorem an

such that Tx =

I)y112} = v.

i' for every x E S(H) and

so 11Th

eigenvalue of absolute value

Proof. Set

a = inf (Ux,x) lxii

=1

and

b=

sup

(Ux,x)

11111 = I

that = [a,b]. By Lemma 5, flUfl = max{—a,b}. Replacing U by —U, if necessary, we may assume that hUh = b > 0. We have to show that b is an eigenvalue of U. By the definition of b, there is a sequence C S(H) such that —' b. Since U is a compact operator, by replacing by a so

subsequence,

Then

we may suppose that is convergent, say —' b and = I. As

b because

Chapter 14: Compact normal operators

203

=

=

and —, b,

we have —*0.

Therefore =

Then, on the one hand, Yo = bx0 and, on the Consequently we have Ux0 = bx0. Since 0 (in fact, lixoll = 1), b is indeed an eigenvalue of U.

Put x0 = y0/b.

other hand, IIxofl

1

Ux,, —p Ux0.

Let us now see how Theorem 6 may be used to deduce Theorem 2 for compact hermitian operators. For the sake of variety, we restate Theorem 2 in the following form. Theorem 7. Let H be a Hubert space and let U

be a compact hermitian operator. Then there is a (possibly finite) sequence (Ak) of real numbers and a sequence (Bk) of linear subspaces of H such that (a) Ak .—' 0;

(b) dimHk (c)

(d) if x =

Xk+X, where Xk

Hk and

i E H,' for every k, then

Ux = k

Proof. Let A,., (y E I') be the non-zero elgenvalues of U and let II,, be the eigenspace belonging to A7: H,, = Ker(U—A,,I). We know that H,..1H8 if y 8. and, for every Let us show first that dim H,, 0, there are only finitely many A,, with IA,. I e. Suppose not. Then, by taking an orthonormal basis in each H,, with IA., I e, we find that there is an infinite orthonormal sequence such that Ux,, = where does not contain- a convergent subsequence, €. But then contradicting the compactness of U. I

I

Chapter 14: Compact normal operators

204

This implies that the non-zero eigenvalues may be arranged in a sequence (Ak) such that with = Ker(U—Akl) the conditions (a)—(c) are satisfied. Then, as each H,, is invariant under U, so is the closed linear span M of all the H,, and, consequently, so is M1. Denote by U the restriction of U to M Then U E is also a compact hermitian operator.

As a non-zero eigenvalue of U is also a non-zero eigenvalue of U, it follows from the definition of M and from Theorem 6, that U =

0.

If the sequence (A,,) of non-zero eigenvalues is finite then we are done. Otherwise, let x

=

k1

where Xk E 11k and i E M1.

Xk +

Put

Xk + I

and

= k=I =

Then

and

= k=1

AkXk.

x. As

—.

AkXk

= k=1

H,

0

the continuity of U implies that Ux = y, proving (d).

Before we recover from Theorem 7 the full force of Theorem 2, let us

show that compact hermitian operators are rather like real numbers. An operator T E is said to be positive if it is hermitian and i.e. (Tx,x) 0 for every x E H. Note that if T is any V(T) C (bounded linear) operator on a Hilbert space then rr and are positive (hermitian) operators:

(rTx,x) =

(Tx, Tx)

IITxII2

and

(Trx,x)

= I$Tx112.

Theorem 8. A compact positive operator U on a Hubert space has a unique positive square root V. Every hermitian square root of U is compact.

Proof. Let A 1,A2,... be the non-zero eigenvalues of U, let Hk be the eigenspace belonging to A,, and let M be the closed linear span of the Then = KerU and > 0 for every k. Define V E by Vx = if x E Hk and Vx = 0 if x E M1. Then V is a positive square root of U.

Chapter 14: Compact normal operators

206

,..., be commuting compact nonnal operators on a Hilbert space H. Then H has an orthonormal basis consisting of comTheorem 9. Let T1

mon eigenvectors of all the T1.

C C and k = 1,. , n, the eigenspace Ker(pJ — Tk) is invariant under all the 7. Hence H is the orthogonal direct sum of Proof. For every

. .

the subspaces = ('1

All these spaces are finite-dimensional, with the possible exception of the union of ,o. Taking an orthonormal basis of each .

0

these bases will do.

As our final theorem concerning abstract operators in this chapter, let us note that our results, say Theorem 2 or Theorem 3, give a complete characterization of compact normal operators up to unitary equivalence. Two operators T, T C are said to be unitarily equivalent if for some unitary operator U we have T' = U'TU = U*TU, i.e. if they have the same matrix representation with respect to some orthonormal bases. Let X be the collection of functions n: C\{0} {0, 1,2,. . } whose 1} has no accumulation point (i.e. in C there C\{0}: n(A) support {A is no accumulation point other than 0). In particular, the support is finite or countably infinite. The following result is easily read out of Theorem 2 (see Exercise 14). .

Theorem 10. Let H be an infinite-dimensional complex Hilbert space be the collection of compact normal operators on H. For and let TC

and A C C\{0} set

nr(A) = dim Ker(A1— T).

Then the correspondence T

a surjection furthermore, T and T' are unitarily equivalent if ni.. = nr. defines

—÷

0

We close this chapter by showing how the spectral theorems we have just proved enable us to solve a Fredholm integral equation. Let 1 = [a, b] for some a
(f,g) and norm 111112 =

= Ja

f(t)g(t) dt

Thus the completion of E is L2(0, 1).

Chapter 14: Compact normal operators

E define Uf E E by

Let K(s, t) E C(!x I), and for f (Uf)(s)

207

K(s,

dt.

= Ja

Then K is the kernel of the integral operator U. It is easily checked that

Ue

C(I)) and U maps the unit ball of E into a relatively compact set in C(1). Indeed, this follows from the Arzelá—Ascoli theorem, since for all t then (Uf) (s) — (Uf) (s') if K(s, 1) — K(s', t) I E11f112 by the I

Cauchy—Schwarz inequality.

As the formal identity map C(I) — E, where f f, is continuous, U extends to a compact operator on L2(O, 1); for simplicity, we write U for this extension as well. In fact, (Uf)(s)

= Ja

K(s, t)f(t) dt

forf E L2(O, 1). From now on we suppose also that K(s, t) = K(t,s); in this case U is easily seen to be a hermitian operator. We consider a Fredhoim integral

equation g(s)

= Ja

K(s, t)f(t) dt — Af(s) = ((U—A)f)(s).

For what values of A can we solve this equation, and what can we say

about the solution? As we shall see, this question is ideally suited for the theory we have at our disposal. Let us prove three quick lemmas before giving the answer.

Lemma 11. Let (p,,) be an orthonormal sequence of eigenvectors of U with non-zero eigenvalues (A,,) guaranteed by each of Theorems 2, 3

and 7, and Corollary 4, such that if fJ..p,, for every n then Uf = 0. (The sequences (ç,,) and (A,,) may thus be finite or infinite). Then each A,, is real, N

for every x, where N depends only on K(s, t), and

IK(s,t)I2dsdt. a

Proof. For 0

t

for every m we have

Ja

I put ks(s) = K(s, t). Then, by Bessel's inequality,

Chapter 14: Compact normal operators

208

jb

2

ds.

=

Therefore rb

m

m

dx

= ja

rb

j

Ja

IK(s,t)12

dsdt.

0

a

E L2(0, 1) is an eigenvector of U with eigenvalue

Lemma 12. If

E C(J).

0 then

Proof. By the Cauchy—Schwarz inequality, Uf E C(I) for f E L2(0, 1). Hence = 0 C(l).

Lemma 13. Let g E L2(0, 1), f = Ug and let (ce) be the Fourier coefficients of f with respect to Then the series converges absolutely and uniformly to f(s) in [0, 1]. Proof. Let g = where for =

and

be the decomposition of g in L2(O, 1), every n. As U: L2(0, 1) C(I) is continuous, 0, the uniform convergence follows. Furthermore, +

for every finite set F of indices, \2

/ I

/

\2

/

=

I

\nEF

IanI2)( ( n€F

nEF

/

0

byLemmall.

We are now ready to answer our question about the Fredholm integral equation.

Theorem 14. If A equation

o(U) and g E C(I) then the Fredholm integral Uf—Af = g

with a hermitian kernel K(s, t) = K(:, s) has a unique solution f given by

f(t) = where

1

+

is an orthonormal basis of eigenvectors of U, the series is

Chapter 14: Compact normal operators

209

absolutely and uniformly convergent in [0, 1] and the (as) are the Fourier coefficients of g with respect to ('pa), i.e. CI,

a,

= Ja

dt.

Proof. The unique solution in L2(0, 1) is! = (U—AY'g. Clearly, f has

Fourier coefficients

By Lemma 13, g+Af = Uf

C(1) and

g+Af = where the series is absolutely and uniformly convergent.

0

There are numerous other applications of the elementary spectral theory of compact hermitian operators to integral equations, notably to the Sturm—Liouville equation. However, a proper account of these applications would be much longer than we have space for. In conclusion, let us point out that the spectral theorem for compact

normal operators is only the beginning of the story. The result we proved is a very simple version of a spectral theorem for normal (not necessarily compact) operators. With every normal operator one can associate a so-called spectral measure containing all the information about the operator up to unitary equivalence. Exercises

1. Let H0 be a closed subspace of a Hilbert space H invariant under a normal operator T E necessarily invariant Is H1 = under T? 2. Prove relation (2), i.e. show that if y E f) is an orthonormal is given by Tx,, = basis of a Hilbert space H and T E where E C, then T E 0 there are only finitely many v., of modulus at least e. 3. Let V E be such that V2 is a compact positive operator. Is V necessarily compact? And if V2 = I? 4. Let S. T E be such that IJSxlI = IITxII for all x E H, with S a positive operator. Show that S = (rT)"2 = TI. be a normal operator (i.e. TT* = p1'). Use ele5. Let T mentary linear algebra to show that the matrix of T is diagonal in some orthonormal basis.

Chapter 14: Compact normal operators

210

6.

Let U be the compact operator defined by a hermitian kernel, as in Theorem 14. Show that if U is positive then K(s,s) 0 for every s (0 s 1). Show also that K(s,t)

Akck(s)pk(t)

is the kernel of a positive operator. Deduce that

K(s,t) = where the series is absolutely and uniformly convergent. be an orthonormal basis consisting of eigenvectors of a 7. Let

compact hermitian operator U, with Let = be the multiset : > 0} arranged in a decreasing order, with y1 , the corresponding eigenvectors. Putting it another way: let > 0 be the sequence of nonP2 negative eigenvalues repeated according to their multiplicities. Show that = max{(Ux,x): lixil = 1, x 1y1 for i = 1,...,n—

1}.

Show also that = rninmax{(Ux,x): xE H,,...1, lxii = 1},

where the minimum is over all

(n —

1)-codimensional subspaces

H,,1. Finally, show that = maxmin{(Ux,x): x E H,,, lixil = 1},

where the maximum is over all n-dimensional subspaces F,,. 8. Let U be a positive hermitian operator. Show that llUxlI4

(Ux,x)(U2x, Ux)

for every vector x. Deduce from this that hUll = v(U). 9. Let U E be a positive hermitian operator with Ker U = {0}.

Show that there is a sequence of hermitian operators (U,,)° C such that U,, Ux x and UU,,x — x for every x E H. Can one have U,, U

I as well?

10. Let U E be hermitian. Prove that Im U is a closed subspace of H if U has finite rank.

Chapter 14: Compact nor,nal operators

211

Prove that T is 11. Let T E (a) normal 1ff H has an orthonormal basis consisting of eigenvectors of T; (b) hermitian 1ff it is normal and all its eigenvalues are real; (c) positive iff it is normal and all its eigenvalues are nonnegative reals.

be hermitian. Prove that there are unique positive 12. Let U E such that operators U+, U_ E U=

-

and

U.... =

U....

U.k. = 0.

13. Prove the Fredhoim alternative for hermitian operators: Let U be a compact hermitian operator on a Hilbert space H and consider the following two equations: Ux—x = 0

(2)

Ux—x=x0

(3)

and

where x0 E H. Then either (a) the only solution of (2) is x =

0,

and then (3) has a unique

solution,

or

(b) there are non-zero solutions of (2), and then (3) has a solution 1ff x0 is orthogonal to every solution of (2); furthermore, if (3) has a solution then it has infinitely many solutions: if x is a solution of (3) then x' is also a solution if x — x' is a solution of (2). 14. Give a detailed proof of Theorem 10. In particular, check that the f( is a surjection. map

be normal and, as in Theorem 10, for A E C set I for every A E C n7.(A) = dim Ker(AI— T). Prove that nT(A)

15. Let T E

(including A = 0) if there is a cyclic vector for T, i.e. a vector x0 C H such that lin{x0, Tx0, T2x0.. . } is dense in H. .

Notes

There are many good accounts of applications of the spectral theorem for

compact hermitian opertors to differential and integral equations. We followed i. Dieudonné, Foundations of Modern Analysts, Academic Press, New York and London, 1960, xiv + 361 pp. Here are some of the

212

Chapter 14: Compact normal operators

other good books to consult for the Sturm—Liouville problem, Green's functions, the use of the Fredholm alternative, etc: D. H. Griffel, Applied Functional Analysis, Ellis Horwood, Chichester, 1985, 390 pp., I. J. Maddox, Elements of Functional Analysis, 2nd edn., Cambridge University Press, 1988, xii + 242 pp., and N. Young, An Introduction to Hubert Space, Cambridge University Press, 1988, vi + 239 pp.

15. FiXED-POINT THEOREMS

In Chapter 7 we proved the doyen of fixed-point theorems, the contraction-mapping theorem. In this chapter we shall prove some considerably more complicated results: Brouwer's fixed-point theorem and some of its consequences. It is customary to deduce Brouwer's theorem from some standard results in algebraic topology, but we shall present a self-contained combinatorial proof. Before we can get down to work, we have to plough through some definitions.

A flat (or an affine subspace) of a vector space V is a set of the form F = x+ W, where W is subspace of V. If W is k-dimensional then we call F a k-flat. As the intersection of a set of flats is either empty or a flat, for every set S C V there is a minimal flat F containing 5, called the flat spanned by S. Clearly

F

A,x1:

x. E S,

A=

1, n

=

=

be points in a vector space. We say that these points Let x0 , x1 ,. . are in general position if the minimal flat containing them is kdimensional, i.e. if the vectors x1 — x0, x3 — x0,. , X,, — x0 span a kdimensional subspace. Equivalently, they are in general position if = 0 whenever = 0 and = 0 or, = = = in other words, if the points are distinct and {x1 —x0,x2—x0,. . ,x1, —x0} is a linearly independent set of vectors. For 0 k n, let x0,x1,. . . ,Xk be k+ 1 points in R" in general position. The k-simplex o = (x0,x1,. ,Xk) with vertices x0,x1,. . ,Xk is the . .

.

. .

.

following subset of R": 213

Chapter 15: Fixed-point theorems

214

k

k

p.,

=

1,

p.,> 0 for all i

1=0

The skeleton of a is the set {x0 , x1 ,. .. , x,j and the dimension of a is k. Usually we write 0k for a simplex of dimension k and call it a ksimplex. A 0-simplex is called a vertex. A simplex a1 is a face of a simplex a2 if the skeleton of crj is a subset of the skeleton of a2. Note that the closure of the simplex a = (x0, x1,... , in R" is

5=

,Xk] k

k

=

p.

1=0

p., =

x1:

0

1, p.,

i=0

C {0,1,...,n}},

=

i.e. the closure of a is precisely the union of all faces of a, including itself. Also, 5 is precisely x1,. . , X,,, }, the convex hull of the vertices, and a is the interior of this convex hull in the k-flat spanned by .

the vertices. A finite set K of disjoint simplices in

is called a simplicial complex

if every face of every simplex of K is also a simplex of K. We also call K a simplicial decomposition of the set 1K I = U {u: E K}, the body of K. If K is a simplicial complex and a, r E K then the closed simplices 5 and are either disjoint or meet in a closed face of both. We are ready to prove the combinatorial basis of Brouwer's theorem.

Lemma 1. (Sperner's lemma) Let K be a simplicial decomposition of a closed n-simplex 5 = [x0, x1,. .. , Let S be the set of vertices of K and let y: S — {0, 1,... ,n} be an (n+ 1)-colouring of S such that the colours of the vertices contained in a face [x¼, x.R,. . , x•] of a- belong to Call an n-simplex a" multicoloured if the vertices of o" {i0, , i,.}. are coloured with distinct colours. Then the number of multicoloured n-simplices of K is odd. .

Proof. Let us apply induction on n. For n =

0

the assertion is trivial;

so assume that n 1 and the result holds for n — 1. Call an (n — 1)-face of K marked if its vertices are coloured with

0, 1,... , n —1, with each colour appearing once. For an n-simplex a-" E K, denote by m(o-") the number of marked (n — 1)-faces of a-". Note that a multicoloured n-simplex has precisely one marked (n — 1)-

face, and an n-simplex, which is not multicoloured, has either no

Chapter 15: Fixed-point theorems

215

marked face or two marked faces. Therefore the theorem claims that

m(K) =

(1) (1"EK

is

odd.

Now let us look at the sum in (1) in another way. What is the contriE K to m(K)? Jf is not marked, bution of an (n— 1)-simplex the contribution is 0. In particular, if a-" is in a closed (n — 1)-face of = [x0,x1,... then the contribution of is 0. a other than E K is in and is marked then the contribution of tf o" is a face of exactly one n-simplex of K. Furthermore, if u"—1 is in a, i.e. in the interior of the original n-simplex, then

contributes 1 to m(u") if a"1 is a face of a": as there are two such n-simplices cr", the total contribution of to m(K) is 2. Hence, modulo 2, m(K) is congruent to the number of marked (n— 1)-simplices in ö0. By the induction hypothesis, this number is odd. Therefore so is m(K), completing the proof. 0 Given points XO,Xi,.. , of R" in general position, for every point x of the k-dimensional affine plane through x0,x1,. .. there are unique reals , A2,. , A, such that . .

Ac

x=x0+

A.(x1—x0). i=1

such that x Hence, there are unique reals p.o, i,... , = and = 1. These p., (i = 0,1,... ,k) are called the barycentric coordinates of x with respect to (xo,x1 ,... , Xk). Also, if p., = 1

then

p.,x,

E

Furthermore, the closed half-space of

contain-

ing 1k and bounded by the (k—1)-flat spanned by X0,X1,...,Xk_1 is characterized by 0. The barycentric coordinates can be used to define a very useful simplicial decomposition. Given a simplicial complex K, the barycentric subdivision sd K of K is the simplicial decomposition of 1K I obtained as follows. For a simplex a = (x0,x1,. . ,Xk) K set .

k

c,.

thus

plices

=

1

Lxi;

is the barycentre of a-. The complex sd K consists of all simsuch that a proper face of a-i +1 ce,,. . , .

(i=0,l,...,k—1).

Chapter 15: Fixed-point theorems

216

To define the r-times iterated barycentnc subdivision of K, set 1. Thussd1K= sdK. sd°K = Kandsd'K= The mesh of K, written mesh K, is the maximal diameter of a simplex of K. Equivalently, it is the maximal length of a 1-simplex of K. Note that if = (x0,x1,. . ,x,) (i = 0,1,... ,k) are faces of a k-simplex = (x0, x1,. . , and r = , then the diameter of = r is less than k/(k + 1) times the diameter of if. Therefore, if K is any simplicial complex then for every 0 there is an r such that mesh sdrK < €. Let Y be a subset of a topological space X, and let a = {A,,: y E 1'} .

. .

.

be a collection of subsets of X. We call a a covering of Y if Y C UEJ. A,,. Furthermore, a is a closed covering if each A,,

is

closed, and it is an open covering if each A,, is open. In what follows, the underlying topological space X is always Sperner's lemma has the following important consequence.

be a closed covering of a closed n.

Corollary 2. Let {A0,A1,.. . simplex a = [x0, x1,. .

.

, x,,

}

such that each closed face [x¼, x11,.

A.. Then

a is contained in

A.

a

. .

,

x.] of

A is

Proof. As we may replace A by

compact. The compactness of the sets A0, A1 ,. , implies that it suffices to show that for every e > 0 there are points a E A. (i = . .

0,1,...,n) such that Ia,—a11 < e if i

j.

Let K be a triangulation of & such that every simplex of K has diameter less than €; as we have seen, for K we may take an iterated barycentnc subdivision of a. Given a vertex x of K contained in a face . ,x1) of u, we know that x E U.,0 As,. Set y(x) = min{i1 : x E Aj. The colouring y of the vertex set of K satisfies the conditions of Lemma 1 and so K has a multicoloured n-simplex Let then

0.

.

y(a1) = i

But then a1 E A, as required.

(i = 0,1,...,n).

0

From here it is a short step to one of the most fundamental fixedpoint theorems, namely Brouwer's fixed-point theorem. A closed n-cell is a topological space homeomorphic to a closed n-simplex. Theorem 3. (Brouwer's fixed-point theorem) Every continuous mapping of a closed n-cell into itself has a fixed point.

Chapter 15: Fixed-point theorems

217

Proof. We may assume that our n-cell is exactly a closed simplex 0" —p 0" is a continuous map, 0" = [xo,xt,... , x,, 1. Suppose that sending a point =

=

1=0

to

=

(IL;

=

i).

1=0

For each i, let

A= . ,A,,} is a closed covering of 0". If a point belongs to a closed face [xc, x•1,. .. ,x.] of 0" then = 0 x= = 1. Since for i {i0, p4 = 1, there is } and so and so x E Consequently, an index j such that p.4'

Then {A0,A1 ,.

. .

,x,] C U_0 A,,

showing that the conditions of Corollary 2 are satisfied. Thus there is a pointx in all the A; such an xis a fixed point of ç. 0

The following lemma enables us to apply Theorem 3 to a rather pleasant class of spaces, namely the compact convex subsets of finitedimensional spaces, i.e. the bounded closed convex subsets of finitedimensional spaces.

Lemma 4. Let K be a non-empty compact convex subset of a finitedimensional normed space. Then K is an n-cell for some n. Proof. We may assume that K contains at least two points (and hence it contains a segment) since otherwise there is nothing to prove. We may also suppose that K is in a real normed space and hence that K is a compact convex subset of = (IR", II•II) for some n. Further-

more, by replacing R" by the flat spanned by K and translating it, if necessary, we may assume that 0 E mt K. Finally, let 5 be an n-simplex containing 0 in its interior, and define a 5 as follows: for x E R" define homeomorphism K

n(x) = ,zK(X) = inf{t:t> 0, x C tK}, and

Chapter 15: Fixed-point theorems

218

m(x) = mo(x) = inf{t: and

t> 0, x E t&},

for x E K set

ifx=0,

0

1

n(x) j—x m(x)

0

.

Corollary 5. Let K be a non-empty compact convex subset of a finite-

dimensional normed space. Then every continuous map f: K—' K has a fixed point. Proof. This is immediate from Theorem 3 and Lemma 4.

0

Our next aim is to prove an extension of Corollary 5 implying, in particular, that the corollary is true without the restriction that the normed space is finite-dimensional. This is based on the possibility of approximating a compact convex subset of a normed space by compact convex subsets of finite-dimensional subspaces. Unfortunately, the simple lemma we require needs a fair amount of preparation.

Let S = {x1 ,.. • , x,, } be a finite subset of a normed space X. For e > 0 let N(S, e) be the union of the open balls of radius centred at

xI,... k

N(S,€) = U D(x1,€). 1=1

For x E N(S,e) define A(x) = max{0,€—IIx—x1Ij} (i = 1,... ,k) and set A(x) A.(x). !f x E N(S,€) then x belongs to at least one open = ball D(x1,€), and for that index i we have A•(x) > 0. Hence A(x) > 0 for every x N(S, e). Define the Schauder projection : N(S, e) —' co{x1,. . . by

=L

A(x)

Here k

co{x1,. ..,Xk} =

k

Ax1: A

0,

A1 = 1

i=1

is the convex hull of the points x1 , -

, xp: the intersection of all convex sets containing all the points x — 1,... , xk. This convex hull is, in fact, compact, since it is the continuous image of a closed (k — 1)-simplex in is the standard basis of R" Indeed, if = say, then the . -

II -

.

Chapter 15: Fixed-point theorems ek] is a bounded closed subset of Furthermore, ç: co{x1 xk}, given by

closed simplex 5 =

it is compact.

219

[e1

k

k

k

0 and

where A,

A,x1,

A-e,

and so

A- =

1.

1=1

is

a continuous map.

is a continuous map from

Lemma 6. The Schauder projection N(S,E) to co{x1,. .. ,x,,} and IS,E(x)—xII < E

for all x E N(S,e).

Proof. Only (2) needs any justification. If x E N(S, e) then k

A,(x)

=

A,(x)

=

i=I But if A.(x) >0 then 11x1—xll <.e and so A,(x)

k's.f(x)xlI

<E.

A,(xi >0

Here then is the promised extension of Corollary 5, point theorem.

Schauder's

fixed-

Theorem 7. Let A be a (non-empty) closed convex subset of a normed space X and let f: A A be a Continuous map such that K = f(A) is compact. Then f has a fixed point. Proof. Let n

1. As K is

compact, there is a finite set

=

such

C K

that

KC Set tion

{xi,..,x*)

= co{x1 ,. Of

Corollary

5,

fore, by (2),

. .

D(xaJ) = and

denote by

the

Schauder projection

1/n) —, K,,. We have K,, C A, so by Lemma 6 the restncto K,, is a continuous map of K,, into itself. Hence, by there is a point x,, E K, such that p,,(f(x,,)) = x,. There-

Chapter 15: Fixed-point theorems

220

<

(3)

belongs to the compact set K, the sequence has a — cc, where x as k x E K. But convergent subsequence, say cc, and so 1(x) = x. then, by (3), x as k D As each

As a beautiful application of Brouwer's theorem, we prove Perron's theorem concerning eigenvalues of positive matrices.

Theorem 8. A matrix whose entries are all positive has a positive eigenvalue with an eigenvector whose coordinates are all positive.

be an nXn matrix with a11 > 0 for all i and j. Let be the standard basis in R". The closed (n — 1)-simplex ö = [e1 ,.. , e,, j is a 'face' of the unit sphere

Proof. Let A =

(a11)

.

S(lr) =

C

=

IIxIIi

1].

I

=

=

given by x The continuous map a Ax/IIAxII1 has a fixed point 0 x (x1)?. Clearly, Ax = Ax for some A > 0 and x1 > 0 for all i.

From Theorem 7 it is a short step to a version of the Markov— Kakutani fixed-point theorem. An affine map of a vector space V into itself is a map of the form where S: V—f V is a linear map. Equivalently, T: V V is an affine map if Aix1)

A1T(x1)

=

whenever x

V. A

0 and

A

=

1.

Theorem 9. Let K be an non-empty compact convex subset of a normed space X and let be a commuting family of continuous affine maps on

X such that T(K) C K for all T C

Then

some x0 E K is a fixed

point of all the maps T E Proof. For T C

let

be the set of fixed points of T in K: KT = {x C K: Tx = x}.

By Theorem 7, KT 0 and, as T is a continuous affine map, KT is a compact convex subset of K. if S C then S maps KT into itself since if Tx = x then T(Sx) = S(Tx) = Sx and so Sx KT. Consequently, if for some T1 C and S C

Chapter 15: Fixed-point theorems

221

is a compact convex set mapped into itself by S. Hence, then by Theorem 7,

K5nfl Kz #0. This implies that the family of sets {KT: T E intersection property. As each belongs to every Kr, i.e. Tx0 =

x0

has the finite-

is compact, there is a point x0 which for every T E 0

One should remark that it is easy to prove Theorem 9 without relying on Theorem 7. Indeed, for T and n 1 the afflne map

maps K into itself and = TE n l} is a commuting family of affine maps of K into itself. From this it follows that the system has the finite-intersection property. of compact sets {S(K): S Hence there is a point x0 such that x0 E TE and

This point x0 is a fixed point of every T E K then

Indeed, if

=

x0 for

T(x0)—x0= Since

is

a bounded sequence, we have T(x0) = x0. Exercises

1. Let X be a Banach space and let f: B(X) X be a contraction from the closed unit ball into X (i.e. d(f(x),f(y)) kd(x,y) for all x,y E B(X) and some k < 1). By considering the map g(x) = {x + f(x)}, or otherwise, prove that if f(S(X)) C B(X) then f has a fixed point. 2. Deduce the following assertion from Corollary 2. Let {A0,A1,.. be a closed covering of a closed simplex = [x0 , x1 ,... , such that, for each 1 (0 I n) the set A is disjoint from the closed (n — 1)-face not containing x1 (i.e.

'opposite' the vertex x). Then

A

0.

Chapter 15: Fixed-point theorems

222

3. Use the result in the previous exercise to prove that if ö = [x0,x1 ,... ,x,] is a closed simplex and f: a 5 is a continuous map

such that for

f(f) C ?,

every closed (k — 1)-face

of 5 we have

then f is a surjection.

4. Prove that Brouwer's fixed-point theorem is equivalent to each of

the following three assertions, where B" = B(11") and S"' = (In fact, we could take B" = B(X) and S"' = S(X) for any n-dimensional real normed space X.)

(i) S"' is not contractible in itself, i.e. there is no continuous map S" 'x [0, 1] —p such that for some x0 E we have x xE (ii) There is no retraction from B" onto i.e. there is no continuous mapf: B" —p such thatf(x) = x for all x S"'. (iii) Whenever f: B" R" is a continuous map without a fixed point then there is a point x E such that x = Af(x) for some 1

O
The

closed faces

of

C,,

R":

1x11

1 for every i}.

arc

and

—1}

,n). For each i (i = 1,...,n) let A be a closed subset of C,, separating and F1, i.e. = U7 U U1, where U1 are disjoint open subsets of with F$ C and (i

F,

= 1,...

C

Prove that fl"1 A.

0.

8. Let q' be a continuous one-to-one map from a compact Hausdorif space K into a Hausdorif space. Show that ç is a homeomorphism between K and ç(K). 9. Prove that every compact metric space is homeomorphic to a closed subset of the Hubert cube: =

12:

2' for every i}.

Chapter 15: Fixed-point theorems

223

Prove that every continuous map of the Hilbert cube into itself has a fixed point. (Check that the map f: B(12) —' B(12) given by S is the right shift f(x) = and e1 = (1,0,0,...).) 11. Show that a continuous map of B(12) into itself need not have a

10.

fixed point.

12. Let C be a closed subset of a compact metric space K, and let f be a continuous map of C into a normed space X. Use Schauder projections and the Tietze—Urysohn extension theorem (Theorem 6.3) to prove that f has a continuous extension F: K X.

The aim of the next three exercises is to make it easy for the reader to prove another beautiful fixed-point theorem.

13. Let a, b, c and x be points

in

a Hilbert space such that

b = 4(a+c) and

0< r

IIx—aD

lix—cil

IIx—bII

r+€

2r.

Deduce from the parallelogram law that 46,

IIa—cH

= 2re+E2, and so ha—cu

where

14. Let C be a subset of B(12), and let f: C C be a non-expansive map, i.e. let f be such that d(f(x),f(y)) d(x,y) for all x,y E C. Suppose x1,x2 and a = +x2) E C, and E fOr i = 1,2. Set c = f(a) and b = (a + c). Assuming that lxi —bfl

11x2—bhl,

lixi—bhi

hixi —all

check that

and hlxi—cil

lIxi—alH-€.

Deduce from the result in the previous exercise that hla—f(a)!l 15.

Let C be a non-empty closed convex subset of

f: C

C

be a non-expansive map. Set

B(12),

and let

Chapter 15: Fixed-point theorems

224

= {x E C: IIf(x) —xli

1/n},

show that 0 for all n. Put and note that the monotone increas= inf{lIxlI: x E ing sequence converges to some d 1. Use the result of the previous exercise to show that and

diamF,, = sup{IIx—ylI : x,y E f,j

0

as n

Conclude from this that f has a fixed point. 16. Let K(s, t) be continuous for 0 s, r 1, and let f(t, u) be continuous and bounded for 0 t 1 and —00 < u <00. Suppose that I

K(s,

t) I

M and lf(t, u)

N for all s, t and u (0

I

s, t

1).

Define an operator T: C[O, 1] —* C[O, 1] by (Tu)(s) =

J

K(s, t)f(t, u(t)) di.

Check that T maps CEO, 1] into the closed ball of radius MN, say C = {u E C[O,1]: hull MN}. Apply the Arzelà—Ascoli theorem to show that TC is a relatively compact subset of C[O, 1].

Make use of the Schauder fixed-point theorem to prove that the Hammerstein

equation

u(s)

=

K(s, t)f(t, u(t)) dt

has a continuous solution. Notes

This chapter is based on the book of J. Dugundji and A. Granas, Fixed Point Theory, vol. I, Polish Scientific Publishers, Warsaw, 1982, 209 pp., which is a rich compendium of beautiful results. Another interesting book on the topic is D. R. Smart, Fixed Point Theorems, Cambridge University Press, 1974.

The more usual approach to fixed-point

theorems is via homology, homotopy and degrees of maps; this can be

found in most books on algebraic topology. The case n = 3 of Brouwer's fixed-point theorem (Theorem 3) was proved in L. E. J. Brouwer, On continuous one-to-one transformations of surfaces into themselves, Proc. Kon. Ned. Ak. V. Wet. Ser. A, 11(1909), 788—98; the first proof of the full result was given by J. Hadamard, Sur quelques

Chapter 15: Fixed-point

225

applications de l'indice de Kronecker; Appendix in J. Tannery, Introduction a la ThEorie des Fonctions d'une Variable, vol. II, 2me éd., 1910;

Brouwer himself proved the general case in 1912. Sperner's lemma is from E. Sperner, Neuer Beweis für die lnvarianz der Dimensionzahl und des Gebietes, Abh. Math. Scm. Hamb. Univ., 6 (1928), 265—72, and Theorem 7 is from 3. Schauder, Der Fixpunktsatz in Funktionalräumen, Studia Math., 2 (1930), 171—80. The original version of Theorem 9 is in A. A. Markoff, Quelques théorEmes sur les ensembles abéliens, C. R. Acad. Sci. URSS (N.S), 1 (1936), 311—3.

16. INVARIANT SUBSPACES

Given a complex Banach space X, which operators T E have nontrivial closed invariant subspaces? This question, the so-called invariantsubspace problem, is the topic of this brief last chapter. Until fairly recently, it was not known whether there was any operator T without a non-trivial (closed) invariant subspace, and it is still not known whether there is such an operator on a (complex) Hubert space.

Much of the effort concerning the invariant-subspace problem has gone into proving positive results, i.e. results claiming the existence of invariant subspaces for operators satisfying certain conditions. Our main aim in this chapter is to present the most beautiful of these positive results, Lomonosov's theorem, whose proof is surprisingly simple. As we remarked earlier, the Riesz theory of compact operators on Banach spaces culminated in a very pleasing theorem, Theorem 13.8, which nevertheless, did not even guarantee the existence of a single nontrivial invariant subspace. This deficiency was put right, with plenty to spare, in Chapter 14, but only for a compact normal operator on a Hilbert space. Now we return to the general case to prove Lomonosov's

theorem, which claims considerably more than that every compact operator has a non-trivial invariant subspace. Before we present this result, we need some definitions and a basic result about compact convex sets.

As in Chapters 13 and 14, all spaces considered in this chapter are complex spaces. Furthermore, as every linear operator on a finitedimensional complex vector space has an eigenvector, we shall consider only infinite-dimensional spaces.

Given a Banach space X and an operator T E call a subspace Y C X in variant under T or T-in variant if V is closed and TY C V. We 226

Chapter 16: Invariant subspaces

227

also say that Y is an invariant subspace of T. Clearly, Y = {O} and Y = X are T-invariant subspaces for every T, so we are interested only in other invariant subspaces, the so-called non-trivial invariant subspaces.

We call a subspace Y a hyperin variant subspace for T if V is an Sinvariant subspace for every S E 2LftX) commuting with T. Since T commutes with itself, every hyperinvariant subspace is an invariant subspace.

Note that if T and S commute then Ker T is S-invariant since if x E Ker T then TS(x) = ST(x) = S(O) = 0 and so S(x) E Ker T. Hence if T has no non-trivial hyperinvariant subspaces then either T is a multiple of the identity or it has no eigenvalue, i.e. = 0. In particuand o(T) {0} then, by Theorem 8 of Chapter 13. lar, if T E the operator T has a non-trivial hyperinvariant subspace. What is the easiest way of constructing a T-invariant subspace? Pick a vector x E X (x 0) and set =

Tx.

Then TY0(x) C Y(x) a

space then Y(x) = X for all x torforTifY(x) =X.

T has no non-trivial invariant sub0. A vector x is said to be a cyclic vec-

if T has no non-trivial invariant subspace then every non-zero vector is a cyclic vector for T. The converse of this is also trivially true: if every non-zero vector is a cyclic vector for T then T has no non-trivial invariant subspace since if V is a non-trivial invariant subspace then no Thus

vector in Y is a cyclic vector for T.

Unfortunately, but not surprisingly, this shallow argument is no advance on the invariant-subspace problem; nevertheless, it tells us that we have to concentrate on cyclic vectors. In the proof of Lomonosov's theorem, we need a basic result concerning closed convex hulls of compact sets.

Theorem 1. (Mazur's theorem) The closed convex hull of a compact set in a Banach space is compact. Proof. K =

e>

0

Let A be a compact subset of a Banach space X

and let

We have to show that K is totally bounded, i.e. for every it contains a finite €-net. Since A is compact, it contains a finite

Chapter 16: Invariant subspaces

228

se-net, say {x1 ,. .. ,x,,}. Thus {x1 ,. . ,x,j C A and for every x E A The set P = co{x1 ,. .. ,x,j is also there is an x1 such that IIx—x111 < .

compact, and so it contains a finite the set = {x

{Yi

.

.

Furthermore,

E X: d(x,P) <

is convex and contains A, and therefore contains coA. But then with M = {x E X: flx—y111 < we have coA C

for some i}

C M and so {y1 ,..., Ym} is an c-net in K =

Here then is the main result of the Chapter. Theorem 2. (Lomonosov's first theorem) Let T be a non-trivial compact operator on an infinite-dimensional complex Banach space X. Then T has a non-trivial hyperinvariant subspace Proof. We may assume that

= 1. Set

Tfl

SA = {T}'

ST = TS}

{S E

and pick a point x0 E X such that lixojI > 11Th. Set B0 = note that 0 B0 and () E TB0. Suppose first that there is a point Yo E X (Yo 0) such that IIT'yo—xohI

1) and

1

for every T' E SA. Then we are done since T' E SA}

V=

is a non-trivial hyperinvariant subspace of X. Suppose then that there is no Yo 0 satisfying (1). Then for every

y E X (y

0) there is an operator T' E SA such that H

Since

7)'

X11fl < 1.

TB0 is a compact set not containing 0, there are operators

SA such that for every yE TB0 there is a

(1

i

a)

satisfying

<

Now we define a map jection. For y E TB0 and 1

i

1.

X reminiscent of the Schauder pron set

Chapter 16: Invariant subspaces

229

A1(y) = max{O, 1— 117y11}

and

A(y)

A1(y).

=

Relation (2) implies that A(y) >

0

for every y E

Therefore

may define = This map

—

X is continuous and so

B0. Consequently, by Mazur's theorem, K vex subset of B0. Hence

ëÔ

is a compact subset of is a compact con-

K

is a continuous map of a compact convex set into itself and so, by Schauder's theorem (Theorem 15.7), it has a fixed point z0 E K: A1(Tz0)

T,Tz0 = z0.

Set

A(Tz0)

Then S E SA, Sz0 =

z0

0, and so Y = KerU—S)

{0}

is a T-invariant subspace. As S is compact, Y is fInite-dimensional. But then TIY is an operator on a complex finite-dimensional space and so it has an eigenvalue A. However then Ker(AI-- T) is a non-trivial hyperin0 variant subspace for T.

A slight variation in the proof shows that an even larger class of operators have hyperinvariant subspaces. commutes Theorem 3. (Lomonosov's second theorem) If T E with a non-zero compact operator and is not a multiple of the identity then it has a hyperinvariant subsp&e.

Chapter 16: Invariant subspaces

231

= — If, on the other hand, p(T) = 0, then ak T", and so 0. 'x} is a T-invariant subspace for every x 1(x) = lin{x, Tx,..., —

0

In spite of the simplicity of its proof, Lomonosov's second theorem is a very powerful result. At the moment, it is not clear how large a class of operators Corollary 4 applies to; in fact, for a while it was not clear that which is not covered by Corollary 4. there is any operator T E The invariant-subspace problem for Banach spaces was solved, in the negative, only fairly recently: Per Enflo and Charles Read constructed complex Banach spaces and bounded linear operators on them which do not have non-trivial invariant subspaces. The original proofs were formidably difficult and the spaces seemed to be rather peculiar spaces. Later, Charles Read gave an easily accessible proof, and showed that his construction works, in fact, on In view of these great results, the invariant-subspace problem for Hilbert spaces has become a very major problem in functional analysis. In fact, it is not impossible that the answer is in the affirmative even on reflexive spaces, i.e. that every bounded linear operator on an infinitedimensional reflexive complex Banach space has a non-trivial invariant subspace. Exercises

1. Let X be a non-separable Banach space. Show that every has a non-trivial invariant subspace.

T

2. Show that the following result can be read out of the proof of Theorem 2. Let SA be a subalgebra of

whose elements do

not have a non-trivial common invariant subspace. Then if TC and T 0 then there is an operator A E SA such that Ker(I—AT)

3. Let T1 ,...,

T,, C be commuting operators. Show that they have a non-trivial common invariant subspace. 4. Deduce from Theorem 1 and Exercise 5.5 the following extension of

Theorem 4.10. If the unit ball of a Banach space Xis a-compact then X is finite-dimensional. 5+ + . Solve

the invariant-subspace problem for Hubert spaces.

232

Chapter 16: Invariant subspaces Notes

Mazur's theorem is from S. Mazur, Uber die kleinste konvexe Menge, die

eine gegebene kompakie Menge enthäl:, Studia Math., 2 (1930), 7—9. Theorems 2 and 3 are from V. I. Lomonosov, On invariant subspaces of families of operators, commuting with a compact operator (in Russian), Funk. Analiz i ego Prilozh, 7 (1973), 55—6; to be precise, Theorem 3 is given as a remark added in proof. Corollary 5 is from N. Aronszajn and K. Smith, Invariant subspaces of completely continuous operators, Ann. Math., 60 (1954), 345—50.

The invariant-subspace problem for Banach spaces was solved in P. Enflo, On the invariant subspace problem in Banach spaces, Acta Math., 158 (1987), 213—313, and C. J. Read, A solution to the invariant Sub. space Problem, Bull. London Math. Soc., 16 (1984), 337—401. A simplified and stronger version of Enflo's solution can be found in

B. Beauzamy, Un opérazeur sans sous-espace invariant non-trivial: simplification de l'example de P. Enflo, Integral Equations and Operator Theory, 8 (1985), 314—84.

Read's result concerning is in A solution to the Invariant Subspace Bull. London Math. Soc., 17 (1985), 305—17. Problem on the space An interesting account of the results concerning the invariantsubspace problem can be found in B. Beauzamy, Introduction to Opera-

tor Theory and In variant Subspaces, North Holland, Amsterdam, 1988, xiv + 358 pp.

INDEX OF NOTATION

B(X), closed unit ball, 22

B(xo, r), closed ball of radius r and centre x0, 22 Br(XO), dosed ball of radius r and centre x0, 22 space of bounded linear operators on X, 28 Y), space of bounded linear operators, 28 Y), the space of compact operators, 186 Y), the space of finite rank operators, 186 the space of Hubert—Schmidt operators, 181

ci,, the barycentre of the simplex a, 215

Eo 5, the closed convex hull of 5, 55 coS, the convex hull of S, 55

C-algebra, 167 C(K), space of continuous functions on a compact Hausdorff space K, 23

C(L), space of bounded continuous functions on L, 23 space of continuous functions with compact support, 91 space of continuous realvalued functions with compact support, 93

C0(L), space of continuous functions vanishing at infinity, 91 CR(L), space of bounded continuous real-valued functions, 93 space of continuous realvalued functions vanishing at infinity, 93

d(X, Y), Banach-Mazur distance between X and Y, (p6 D(x,r), open ball, 20 D(x0, r), open ball of radius r and centre x0, 22 D,(x0), open ball of radius r and centre x0, 22 ÔA(a), the resolvent set of a in the algebra A, lffl 4, closed unit disc in the complex plane, 96

fvg, the join of f and g, 93 fAg, the meet of f andg, 93 I IS, restriction of a Ito S, 25 gb(S), space of bounded functions on 5,23 Im T, image of T, 28 k-simplex, 213 annihilator of K, 164

K°, polar of K, 158

233

Index of notation

234

weak topology generated by

11 -norm, 23 23

o-(X.

finS, linear span, 38 tinS, closed linear span, 38 linZ, linear span, 21 one-point compactification of L,

y(X, X), weak topology on a normed

8L, preannihilator of L, 164 °L, prepolar of L, 158 Y), space of linear operators, 28 mesh K, the mesh of the simplicial complex K, 216 orthogonal projection onto F. 136

p-mean, 5 Rademacher function, 143 r(T), the spectral radius of T, 174 Rad fit, the radical of the Banach algebra B, 178 p(T), the resolvent set of the operator T. 168

sd K, the barycentric subdivision of K, 235

of vectors orthogonal to 135 S(X), unit sphere, 22 S(x0, r), sphere of radius r and centre S

set

22

sphere of radius r and centre 22

set of finite subsets of 0, LL4 cr(T). the spectrum of the operator T, 167

115

space, 1.15

a(X*, X), weak-star topology, 1.1.6 the spectrum of a in the algebra A, 167 0ap(T), the approximate point spectrum of the operator T, 169 the continuous spectrum of the operator T, 169 Ocom(T), the compression spectrum of the operator T, 1.68 T), the point spectrum of the operator T, 1.68 T), the residual spectrum of the operator T, 1.69

T-invariant, 226 adjoint of T, 31 Hilbert-Schmidt norm, 182 Hilbert—Schmidt norm, 187

v(T), the numerical radius of T, 202 V(T), the numerical range of T, 21)1 x set of vectors orthogonal to x, 135 of x, 23 lixil,,,

X, completion of X, 35 X,, space of linear functionals on X, 28,45

X, dual of X. 31 r, space of bounded linear functionals on X,

INDEX OF TERMS

absolute value of an operator, 205

Banach's fixed-point theorem, 101

absolutely convergent series, 3.6 absolutely convex set, 27 adjoint of an operator, 155 adjoint operator, 31 affine hyperplane, 46 affine map, 220 affine subspace, 213 Alaoglu's theorem, 118 algebra, commutative, 92 algebraic dual of a normed space, 45 AM-GM inequality, 1 analytic, 111 annihilator of a set, 164 annihilator of a subspace, 158 approximate eigenvector, 169 approximate point spectrum, 169 approximation problem, 189 arithmetic mean, 6 weighted, 7 Arzelà—Ascoli theorem, 90 Auerbach system, 65

Banach—Mazur distance, (16 Banach—Steinhaus theorem, 78

Banach algebra, 92 unital, 32 Banach limit, 59 Banach space, 21

barycentre, 215 barycentric coordinates, 215 barycentric subdivision, 215 basic sequence, 72 basis, 19. 83 canonical, 32 Hamel, 42 Schauder, 83 standard, 32 basis constant, 83 Bernstein and Robinson, theorem of, 230

Bessel's inequality, 1.41 biorthogonal system, 64 normalised, 64 Bishop—Pheips—Bollobas theorem, 122 body, 214 bounded below, 162 bounded linear operator, 28 bracket notation, 28 Brouwer's fixed-point theorem, 216

canonical basis, 37 Carleson's theorem, 1511 235

Index of terms

236

Cauchy sequence, 21

Cauchy—Schwarz inequality, Cesàro summable, 150 chain, 50

131

classical function spaces, 22 classical sequence spaces, 26 closed, 19 closed n-cell, 216 closed convex hull, 55 closed covering, 216 closed-graph theorem, 80 closed linear span, 38 closure, 19 coarser topology, 20 commutative algebra, 92 compact, 89 countably, 89 relatively, 89

sequentially. 89 compact operator, 186 compact space, 21 complete, 21 complete orthogonal set of vectors, 141

completely regular topological space, 98

completion of a metric space, 33 completion of a normed space, 35 complex unital Banach algebra, 162 compression spectrum, 168 concave function, 3 concave functional, 53 conjugate, 111

constant k contraction with, 101

Lipschitz condition with, 101 continuous maps, 20 continuous spectrum, 169 contractible in itself, 222 contraction, 101

contraction with constant k, 101 contraction-mapping theorem, 101 convergent series, 36 convex function, 3 convex functional, 42 convex hull, 55 convex subset, 2 countably compact, 89 covering, 216 closed, 216 open, 216 cyclic vector, 227 degenerate form, 130 dense subset, 33 dense topological space, 75 derivative, 105 differentiable map, 105 dimension, 214 direct sum of subspaces, 39 dissipative operator, 182 distance, 19 divisor of zero, 180 topological, 180 dominate, 48 dual of an operator, 155 dual space, 31 eigenspace, 168 eigenvalue, 168 eigenvector, 168 approximate, 169 Enflo's theorem, 231 equicontinuous, 90 equicontinuous at a point, 90 equivalence class of functions, 25 equivalent norms, 29 essential supremum, 25 Euclidean space, 132 extension of a linear functional, 42 extreme point, 125

Index of terms face, 214 marked, 214

Fejér's theorem, 150 finer topology, 20 finite character, 116 finite-intersection property, 116 finite rank operator, 186 fixed point, 101 flat, 213 form

degenerate, 130 hermitian, 130 non-degenerate, 130 positive, 130 symmetric, 130 Fourier coefficent, 145, 149 Fourier series, 145 Fredholm alternative for hermitian operators, 211 Fredhoim integral equation, 207 function concave, 3 convex, 3 strictly concave, 3 strictly convex, 3 fundamental set of vectors, 141

Hadamard's inequality, 1.51 Hahn—Banach extension theorem, 50 complex form, 50 Hamel basis, 42 Hammerstein equation, 224 harmonic mean, 6 Hausdorff topology, 21 Hermite polynomials, 1.44 hermitian form, 130 hermitian form, positive, 8 hermitian operator, 159 on a normed space, 182 132 Hilbert space, Hilbert-Schmidt norm, 187 homeomorphic spaces, 20 homeomorphism, 20 hyperinvariant subspace, 227 hyperplane, 46 Holder's inequality, 9 for functions, 12 image, 28 in general position, 2.11 induced topology, 19 inequality AM-GM, I Cauchy—Schwarz, 9

GeLfand transform, 1.83

Gelfand's spectral-radius formula, 124 Gelfand—Mazur theorem, 174 Gelfand—Nalmark theorem, 158 generalized limit, 59 geometric mean, 1 weighted, 7 Gluskin's theorem, 20 Gram determinant, 152 Gram—Schmidt orthogonalization process, 142 Haar system, 83

237

HOlder's, 9 Minkowski's, 10 initial segment, 124 inner product, 131 inner-product space, 131

mt. interior, 22 invariant subspace, 227 invariant-subspace problem, 226 invariant under, 226 inverse, 1.62

inverse-mapping theorem, 80 invertible, 167 involution, 162

Index of terms

238

isometrically isomorphic spaces, 29 isometry, 162 isomorphic spaces, 29 isotropic vectors, 130

Jensen's theorem, 3 John's theorem, 68 Johnson's uniqueness-of-norm theorem, 179 join, 93 kernel, 28, 108 kernel of an integral operator, 207 Krein—Milman theorem, 126 Laguerre polynomials, 144 lattice operations, 93 Laurent series, 121 Lebesgue number, 99 left shift, 32 Legendre polynomial, 143 linear functional, 28 linear operator, 28 bounded, 28 unbounded, 28 linear span, 38 Lipschitz condition with constant k, 101

local theory of Banach spaces, 20 locally compact space, 91 Lomonosov's first theorem, 228 Lomonosov's second theorem, 229 marked face, 214 Markov—Kakutani fixed-point theorem, 220 maximal element, 50 maximal ideal space, 183 Mazur's theorem, 227 meagre set, 76

mean arithmetic, 6 harmonic, 6 quadratic, 6 meet, 93 mesh, 216 method of successive approximations, 103

metric, 19 metric space, 19 completion of a, 33 Minkowski functional, 28 Minkowski's inequality, 131 for functions, 12 modulus of an operator, 205 multicoloured simplex, 214

n-dimensional Euclidean space, 21 neighbourhood, 19 neighbourhood base, 20 non-degenerate form, 130 non-expansive map, 223 non-trivial, 227 norm, 19 Hilbert-Schmidt, 181 operator, 29 of a functional, 30 smooth, 51 supremum, 92 92 uniform, norm topology, 21 normal operator, 161 normahsed biorthogonal system, 64 normed algebra, 92 unital, 92 normed space, 18 completion of a, 35 nowhere dense set, 76 nowhere dense subset, 41 numerical radius, 202

Index of terms numerical range, 201 spatial, 175

239

quadratic mean, 6 quotient norm, 38 quotient normed space, 38

one-point compactification, 96

open covering. 216 open-mapping theorem, 29 operator ideal, 188 operator norm, 29 order, 49 partial, 49 ordered set, 124 orthogonal complement, 136 orthogonal direct sum, 135 orthogonal matrix, 140 orthogonal set of vectors, 141 orthogonal subspaces, 135 orthogonal vectors, 130 parallelogram law, 133 Parseval's identities, 147 partial order, 49 partial sum, 149 partially ordered set, 50 partition of unity, 100 Perron's theorem, 220 point, 19 point spectrum, 168 polar of a set, 158 polarization identities, 132 positive form, 130 positive hermitian form, B pre-Hitbert space, 132 preannibilator of a set, 164 of a subspace, 158 prepolar of a set, 158 principle of uniform boundedness, 77 probability, 5 product topology, 20, 115 Pythagorean theorem, 133

Rademacher function, 143 radical, 128 Read's theorem, 231 regular point, 167 relatively compact, 89 residual spectrum, 169 resolvent, 167 resolvent identity, 181 resolvent set, 162 restriction of a function. 25 Riemann—Lebesgue lemma, 152 Riesz representation theorem, 137 Riesz—Fischer theorem, 145 right shift, 32 scalar product, 131 Schauder basis, 3L 83 Schauder projection, 218 Schauder system, 83 Schauder's fixed-point theorem, 219 second dual, 156 self-adjoint operator, 159 seminorm, 41 separable, 22 separate, 89 separates the points strongly, 96 separation theorem, 54 sequentially compact, 89 set of the first category, 76 set of the second category, 76 set system of finite character, 116 simplex, 214 multicoloured, 214 simplicial complex, 214 simplicial decomposition, 214 skeleton, 214

240

Index of terms

smooth norm, 51 spatial numerical range, 1.25 spectral decomposition, 2(X) spectral measure, spectral radius, 114 spectrum, 1.67

approximate point, 169 compression, continuous, 169 point, residual. 1.69

Spemer's lemma, 214 standard basis, 31 Stone—Weierstrass theorem, 95 for complex functions, 96 strictly concave function, 3 strictly convex function, 3 strong operator topology, 165 stronger topology, 20 Sturm—Liouville equation, 209 sub-basis for a topology, L1.4 subadditive functional, 48 subreflexive space, 122 subspace, 21 subspace topology, 19 sum of a series, 36 superadditive functional, 5.3 support functional, 51, 115 support plane, 51 92 supremum norm, symmetric form, 130 system set of vectors, 141 Tietze—Urysohn extension theorem, 88 topological divisor of zero, 180 topological space, 19 topology, 19 coarser, 20 finer, 20 Hausdorif, 21 induced, 19

norm, 21 product, 20, 115 strong operator, 165 stronger, 20 subspace, 19 weak, 115 weaker, 20 total set totally,

of

vectors, 141

89

totally ordered set, 50 translate of a subspace, 46 triangle inequality, 18 Tukey's lemma, 116 Tychonov's theorem, 117 unbounded linear operator, 28 uniform closure, 93 uniform norm, 92 uniformly bounded, 90 unital Banach algebra, 32 unital normed algebra, 92 unitarily equivalent operators, unitary operator, 161 universal property, 115 upper bound, 50 Urysohn's lemma, 86 vanishing at vector,

96

19

vertex, 214 Volterra integral operator, 108 weak topology, 115 weak-star topology a(X', X), 116 weaker topology, 20 weakly bounded, 81 weight, 5 weighted arithmetic mean, 7 weighted geometric mean, 7 well-ordered set, 124

Zorn's lemma, 50

Now revised and up-dated, this brisk introduction to functional analysis is intended for advanced undergraduate students, typically final year, who have had some background in real analysis. The author's aim is not just to cover the

standard material in a standard way, but to present results of applications in

contemporary mathematics and to show the relevance of functional analysis to other areas. Unusual topics covered include the geometry of finite-

dimensional spaces, invariant subspaces, fixed-point theorems, and the Bishop—Phelps theorem. An outstand-

ing feature is the large number of exercises, some straightforward, some challenging, none uninteresting.

Bela Bollobás is an active mathematician who works on combinatorics and functional analysis. He has published Graph Theory and Combinatorics, both

textbooks, and two research monographs, Extremal Graph Theory and Random Graphs.

CAMBRIDGE UNIVERSITY PRESS ISBN

0-521-65577-3

MIII

9 780521 655774