bela bollobas-linear analysis_ an introductory course, second edition (1999)

Bela Bollobás

II

an introductory course

SECOND EDITION

CAMBRIDGE MATHEMATICAL TEXTBOOKS

LINEAR ANALYSIS

An Introductory Course

Bela BollobásUniversity of Cambridge

CAMBRIDGEUNIVERSITY PRESS

PUBLISHED BY THE PRESS SYNDICATEOF THE UNIVERSITY OF CAMBRIDGEThe Pitt Building, Trumpington Street, Cambridge CB2 1RP, United Kingdom

CAMBRIDGE UNIVERSITY PRESSThe Edinburgh Building, Cambridge CB2 2RU, UK http://www.cup.cam.ac.uk40 West 20th Street, New York, NY 10011-4211. USA http://www.cup.org10 Stamford Road. Oakleigh, Melbourne 3166, Australia

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

This book is in copyright. Subject to statutory exceptionand to the provisions of relevant collective licensing agreements.no reproduction of any part may take place withoutthe written permission of Cambridge University Press.

First published 1990Second 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 I

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

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 Cambridgeon numerous occasions for the third-year undergraduates readingmathematics. It is intended to be a fairly concise, yet readable and down-to-earth, introduction to functional analysis, with plenty of challengingexercises. In common with many authors, I have tried to write the kind ofbook that I would have liked to have learned from as an undergraduate. Iam convinced that functional analysis is a particularly beautiful andelegant area of mathematics, and I have tried to convey my enthusiasm tothe reader.

In most universities, the courses covering the contents of this book aregiven under the heading of Functional Analysis; the name Linear Analysishas been chosen to emphasize that most of the material in on linear func-tional analysis. Functional Analysis, in its wide sense, includes partialdifferential equations, stochastic theory and non-commutative harmonicanalysis, but its core is the study of normed spaces, together with linearfunctionals and operators on them. That core is the principal topic of thisvolume.

Functional analysis was born around the turn of the century, and withina few years, after an amazing burst of development, it was a well-established major branch of mathematics. The early growth of functionalanalysis was based on 19th century Italian function theory, and was givena great impetus by the birth of Lebesgue's theory of integration. Thesubject provided (and provides) a unifying framework for many areas:Fourier Analysis, Differential Equations, Integral Equations, Approxima-tion Theory, Complex Function Theory, Analytic Number Theory, Meas-ure Theory, Stochastic Theory, and so on.

ix

x Preface

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 andRadon from Germany, Fejér, Haar, von Neumann, Frigyes Riesz andMarcel Riesz from Hungary, Banach, Mazur, Orlicz, Schauder, SierpiElskiand Steinhaus from Poland, Hardy and Littlewood from England, Gel-fand, Krein and Milman from Russia. The abstract theory of normedspaces was developed in the 1920s by Banach and others, and waspresented 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; thismaterial is supplemented with a body of other fundamental results andsome pointers to more recent developments.

The theory presented in this book is best considered as the naturalcontinuation of a sound basic course in general topology. The readerwould benefit from familiarity with measure theory, but he will not be at agreat disadvantage if his knowledge of measure theory is somewhat shakyor even non-existent. However, in order to fully appreciate the power ofthe results, and, even more, the power of the point of view, it is advisableto 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, withemphasis on the many beautiful general results concerning abstract spaces.An important feature of the book is the large collection of exercises, manyof which are testing, and some of which are quite difficult. An exercisewhich is marked with a plus is thought to be particularly difficult. (Need-less to say, the reader may not always agree with this value judgement.)Anyone willing to attempt a fair number of the exercises should obtain athorough grounding in linear analysis.

To help the reader, definitions are occasionally repeated, various basicfacts are recalled, and there are reminders of the notation in severalplaces.

The third-year course in Cambridge contains well over half of the con-tents of this book, but a lecturer wishing to go at a leisurely pace will findenough material for two terms (or semesters). The exercises should cer-tainly provide enough work for two busy terms.

There are many people who deserve my thanks in connection with thisbook. Undergraduates over the years helped to shape the course;numerous misprints were found by many undergraduates, including JohnLongley, Gábor Megyesi, Anthony Quas, Alex Scott and Alan Stacey.

Preface xi

I am grateful to Dr Pete Casazza for his comments on the completedmanuscript. Finally, I am greatly indebted to Dr Imre Leader for havingsuggested many improvements to the presentation.

Cambridge, May 1990 Bela Bollobás

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

B. B.

1. BASIC INEQUALITIES

The arsenal of an analyst is stocked with inequalities. In this chapter wepresent briefly some of the simplest and most useful of these. It is anindication of the size of the subject that, although our aims are verymodest, this chapter is rather long.

Perhaps the most basic inequality in analysis concerns the arithmeticand geometric means; it is sometimes called the AM-GM inequality.The arithmetic mean of a sequence a = (a1,. . . , of n reals is

A(a) =

a is non-negative then the geometric mean is

/n \1/flG(a) = (II a.)

/

where the non-negative nth root is taken.

Theorem 1. The geometric mean of n non-negative reals does notexceed their arithmetic mean: if a = (a1,. .. , then

G(a) A(a). (1)

Equality holds if a1 = =

Proof. This inequality has many simple proofs; the witty proof we shallpresent was given by Augustin-Louis Cauchy in his Cours d'Analyse(1821). (See Exercise I for another proof.) Let us note first that thetheorem holds for n = 2. Indeed,

(a1 —a2)2 = a?—2a1a2+a? 0;

1

2 Chapter 1: Basic inequalities

so

(a1+a2)2 4a1a2

with equality iff a1 = a2.

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

— jj. .am,

m m

— ai+...+am+bi+...+bm2m

If equality holds then, by the induction hypothesis, we have a1 == tZm = b1 = = bm. This implies that the theorem holds when-

ever 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,

with equality if a1 = = aN, in other words if a1 = = 0

In 1906 Jensen obtained some considerable extensions of the AM-GMinequality. These extensions were based on the theory of convex func-tions, founded by Jensen himself.

A subset D of a real vector space is convex if every convex linearcombination of a pair of points of D is in D, i.e. if x,y E D and0 < 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 — 1 points of Dis in D, we find that

and so

= ED.

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

f(tx+(1—t)y) tf(x)+(1—t)f(y) (2)

whenever x,y E D and 0 < t < 1. We call f strictly convex if it is con-vex and, moreover, f(tx+(1—t)y) = tf(x)+(1—t)f(y) and 0< t < Iimply 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 isconvex and it is strictly concave if —f is strictly convex. Clearly, f isconvex 1ff the set {(x,y) E Dxli: y f(x)} is convex.

Furthermore, a function f: D R is convex (concave, ...) iff its res-triction to every interval [a,b} = {ta+(1—t)b:0 t 1) in D is con-vex (concave, ...). Rolle's theorem implies that if f: (a, b) —' R is dif-ferentiable then f is convex if f' is increasing and f is concave if f' isdecreasing. In particular, if f is twice differentiable and f" 0 then f isconvex, while if f" 0 then j is concave. Also, if f" > 0 then f isstrictly convex and if f" < 0 then f is strictly concave.

The following simple result is often called Jensen's theorem; in spiteof its straightforward proof, the result has a great many applications.

Theorem 2. Let f: D — be a concave function. Then

tf(x1) (3)

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

Proof. Let us apply induction on n. As for n = 1 there is nothing toprove and for n = 2 the assertions are immediate from the definitions,let us assume that n 3 and the assertions hold for smaller values of n.


Suppose first that f is concave, and let

E (0,1) with

i = 2,...,n, set t = t1/(1—t1), so that 1 = 1. Then, byapplying the induction hypothesis twice, first for n — 1 and then for 2, wefind that

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 mayassume that not all of x2,. . . ,x,, are equal. But then

(1 —t1)1=2

< (1 t;xI);

so the inequality in (3) is strict. 0

It is very easy to recover the AM-GM inequality from Jensen'stheorem: logx is a strictly concave function from (0, to IL so fora1,... ,

a— loga1 log —,

which is equivalent to (1). In fact, if t1 ,. . . > 0 and t, = 1 then

t,logx1 log tx1, (4)

with equality if x1 = = giving the following extension ofTheorem 1.

Theorem 3. Let a1,.. 0 and ps,... > 0 with P1 = 1.

Then


af' p1a,, (5)

with equality if a1 = =

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

The two sides of (5) can be viewed as two different means of thesequence a1,... , a generalized geometric meanand the right-hand side is a generalized arithmetic mean, with the vari-ous terms and factors taken with different weights. In fact, it is rathernatural to define a further extension of these notions.

Let us fix P = 1: the will play the role ofweights or probabilities. Given a continuous and strictly monotonicfunction (0, — R, the c-mean of a sequence a = (a1,. . . , a,,)

(a1 > 0) is defined as

Mç(a) =

Note that M, need not be rearrangement invariant: for a permutation irthe c-mean of a sequence a1,... ,a,, need not equal the c-mean of thesequence .. , a,,.(.,). Of course, if = = p,, = 1/n then everyc-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 preciselya0.

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 littlesurprising that Jensen's theorem enables us to give an exact answer tothese questions (see Exercise 31).

Theorem 4. Let Pi ,.. . ,p,, > 0 be fixed weights with p, = 1 and letifs: (0, P be continuous and strictly monotone functions, such

that is concave if ç is increasing and convex if is decreasing.Then

Mq,(a)


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

Proof. Suppose that is increasing and is concave. Set b• =and note that, by Jensen's theorem,

=

= =\i=1 I

If - is strictly concave and not all a, are equal then the inequalityabove is strict since not all b• are equal.

The case when is decreasing and is convex is proved analo-gously. 0

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

= = a+x = (XE

ab = (a1b1 abc =

and so on.If p(t) = r (—cx <r < x, r 0) then one usuaLly writes Mr for Mc.

For r > 0 we define the mean M, for all non-negative sequences: ifa = (a1,... (a1 0) then

in \1/r

= p,af= l

Note that if Pi = = p,, = 1/n then M1 is the usual arithmetic meanA, M2 is the quadratic mean and M_1 is the harmonic mean. As animmediate consequence of Theorem 4, we shall see that Mr is a continu-ous monotone increasing function of r.

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


= max a, M0(a) = H af.

Thus M0(a) is the weighted geometric mean of the a1. It is easilychecked that we have M,(a) = for all r r

Theorem 5. Let a = (a1,... be a sequence of positive numbers, notall equal. Then Mr(a) is a continuous and strictly increasing function ofr on the extended real line r

Proof. It is clear that M,(a) is continuous on (—x,O)U(O,x). Toshow 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 increasingfunction of t > 0, and is a concave function, and if r < 0 then t' is

decreasing and (i'S is convex. Hence, by Theorem 4, we have<M5(a).

Let us write A(a) and G(a) for the weighted arithmetic and geometricmeans of a = i.e. set

A(a) = M1(a)=

pa, and G(a) = M0(a) = II af.

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

M,0(a) = lim Mr(a), = urn Mr(a), G(a) = limM,(a).

The proofs of the first two assertions are straightforward. Indeed, let1 m n be such that am = Then for r > 0 we have

Mr(a) = pi/ra.

so am = Since Mr(a) for every r,we have lim,....,. = as required. Also,

= him Mr(a 1)} I = lim Mr(a).

The final assertion, G(a) = lim,....0 requires a little care. Inkeeping with our conventions, for < r < (r 0) let us writear = (ar,. . . Then, clearly,

Mr(a) = A(a')".

Also, it is immediate that


urn (a[ — 1) = = log ar—.oT

r

and so

lim!{A(a')_l} = logG(a). (6)r—.O r

Since

logt (—1

for every t> 0, if r> 0 then

log G(a) = log G(a') log A(a') {A(a') — 1}.

Letting r 0, we see from (6) that the right-hand side tends to log G(a)and so

lim logM,.(a) = urn logA(a') = logG(a),r-4.Q+ r

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 aredue to Holder, Minkowski, Cauchy and Schwarz. Recall that ahermitian form on a complex vector space V is a function p: Vx V Csuch that = Aç(x,z)+A.ç(y,z) and q'(y,x) = for allx,y,z E V and A,p E C. (Thus = Ahermitian form is said to be positive if ç(x, x) is a positive real numberfor alix E V(x 0).

Let ç(•,) be a positive hermitian form on a complex vector space V.Then, given x,y E V, the value

ço(Ax+y,Ax+y) =

is real and non-negative for all A E C. For x 0, setting A =we find that

ço(x,x)p(y,y)

and the same inequality holds, trivially, for x = 0 as weH. This is the


Cauchy—Schwarz inequality. In particular, asn

ç(x,y)1=1

is a positive hermitian form on C'1,

/ \1/2/ \I/2x,p,

1=1 i—I i—I

and so

/ \1/2/ ,,

xy,( I = 1y112

i—i \i=i / \:=1 / \i—1 / \i—1

(7)

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

Theorem 6. Suppose

11p,q>1 and —+—=1.

Then for complex numbers a1,. . . , a,,, b1,. . . , b,, we have

/ \i/,O/ \1/qbk ( I ak I

'I ( I

(8)k—i \k—1 I \k—1 I

with equality if all ak are 0 or = and akbk = e Iakbkl forall k and some t and 0.

Proof. Given non-negative reals a and b, set x1 = a1', x2 =

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

—+—, (9)

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

k—iIakI1'

= k—i= 1.

But then, by (9),


1 1

k=I + q)=_+._=1.

Furthermore, if equality holds then

Iak!° = IbkI" and= k=I

Iakbkl,

implying akbk = e Iakbk .Conversely, it is immediate that under these

conditions we have equality in (8). 0

Note that if Mr denotes the rth mean with weights p, =(I = 1,...,n) and for a = and b = we putab = al = and lbl =then HOlder's inequality states that if p1 + q -' = 1 with p, q > 1, then

M1(labl) Mp(lal)Mq(lbt).

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

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

(p—1)(q—1)= 1, (p—1)qp or (q—1)p=q.

Note that 2 is the only exponent which is its own conjugate. As weremarked earlier, the special case p = q = 2 of Holder's inequality iscalled the Cauchy—Schwarz inequality.

In fact, one calls I and conjugate exponents as well. HOlder's ine-quality is essentially trivial for the pair

M1(labl)

with equality iff there is a 0 such that bk I = I) and akbk =elakbkl whenever ak 0.

The next result, Minkowski's inequality, is also of fundamentalimportance: in chapter 2 we shall use it to define the classical I,, spaces.

Theorem 7. Suppose 1 p < and a1,... are complexnumbers. Then

fn \l/P /n \l/PI

\k=I / \k=1 /


with equality 1ff one of the following holds:(1) allak are 0;

(ii) bk = tak for all k and some t 0;

(iii) p = 1 and, for each k, either = 0 or bk = Ikak for some k 0.

Proof. The assertion is obvious if p = 1 so let us suppose that1 <p not all ak are 0 and not all bk are 0. Let q be the conju-gate of p: p' + q -' = 1. Note that

ak+bkVk=1

Iak+bkV'Iakl + Iak+bkV'Ibk(.

Applying (8), HOlder's inequality, to the two sums on the right-handside with exponents q and p, we find that

1/q up up

IakV)1\k=i \k=1

/ n \h/P / n

= ( Iak+bkV) +(\k=1 /

\k=1 /

we obtain (8). The case of equality follows from that in Holder's ine-quality. 0

Minkowski's inequality is also essentially trivial for p = i.e. for theM,, mean:

with equality if there is an index k such that ak I = a I),IbkI and Iak+bkl = Iakl+IbkI.

The last two theorems are easily carried over from sequences tointegrable functions, either by rewriting the proofs, almost word forword, or by approximating the functions by suitable step functions.Readers unfamiliar with Lebesgue measure will lose nothing if they takef and g to be piecewise continuous functions on [0, 1].


Theorem 8. (HOlder's inequality for functions) Let p and q be conju-gate indices and let f and g be measurable complex-valued functions ona measure space such that fl" and are integrable. Thenfg is integrable and

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

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

(I (Jlfv th)"'+(f IgI" th)11"

Exercises

All analysts spend half their time hunting through theliterature for inequalities which they want to use but can-

not prove.

Harald Bohr

1. Let = fx = (x1)7: x1 n and 0 for every c(i) Show that g(x) = x, is bounded on and attains its

supremum at some point z = EA and x1 = minx, <x2 = maxx1. Set Yi =

Y2 = and = x1 for 3 i n. Show that y =E and g(y) > g(x). Deduce that z = 1 for all i.

(iii) Deduce the AM-GM inequality.2. Show that if c(': (a,b) — (c,d) and cc: (c,d) are convex func-

tions and cc is increasing then (coiIi)(x) = is convex.3. Suppose that f: (a, b) —' (0, is such that log! is convex. Prove

that f is convex.4. Let!: (a,b) (c,b) and cc: be such that and ç1of

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

f(x) = f7(x) for every x E (a,b). Show that f(x) isalso convex.

6. Suppose that f: (0, 1) R is an infinitely differentiable strictlyconvex function. Is it true that f"(x) > 0 for every x E (0, 1)?


7. Let p, q> I be conjugate exponents. By considering the areas ofthe domains

= {(x,y): 0 x a and 0 y

and

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

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

In \1/rMr(a) M,((a1fl) = (

p,a[) (r > 0).\i=I /

If p,q> 1 are conjugate exponents, that is = 1, and(a1 ,.. • , and (b1 ,.. . , 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 functionof r, i.e. is a strictly convex function of r.

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<q<p<r then

/ I, \(r—p)/(r—q)/ ,, \(p—q)/(r—q)

(i=1 /

for all positive reals13. Deduce from the previous exercise that if 0 < q <p < r and

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

b b (r—p)/fr—q) b (p—q)/fr—q)

ía (Lfrth)


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

Mp(a)Mq(b)

for all positive sequences a = (a1,... and b = (b1,.. .Prove also that if p. q, r, s > 0 are such that + q1 + r1 =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) Mr(a) + Mr(b).

16. Let Pp,q(x,y) = (x,y 0). Show that Ppq is concave 1ffp,q I and (p— 1)(q— 1) 1 and that it is convex iffp,q 1 and(p—1)(q—1) 1.

17. Deduce from the result in the previous exercise that M1(ab) andMq(b) are comparable if (p — 1)(q — 1) 1. To be precise, if

p,q >1 and (p—1)(q—I) 1 then

Mp(a)Mq(b)

and if p,q < I and (p—1)(q—1) 1 then

M1(a) Mp(a)Mq(b).

18. = isa decreasing func-tion of x 0 and tends to M1(a) as x Show also that ifr 1 then M,(a + x) — x is an increasing function of x 0 and italso tends to Mi(a) as x —*

19. Prove that if r> 1, a > 0 and a 1 then

aT—1 > r(a—1)

and

a—ir

Deduce that if a, b > 0 and a b then


<ar_br < ra''(a—b)

if r < 0 or r> 1, and that the reverse inequalities hold if0<r<1.

20. Prove Chebyshev's inequality: if r> 0, 0 < a2

and

Mr(0)

unless all the a1 or all the b are equal. Prove also that the me-quality is reversed if a = is monotone increasing and b =

is monotone decreasing. [HINT: Note first that it suffices toprove the result for r = 1. For r = 1 the difference of the twosides is —

21. For ar,... ,a,, > 0 a2) let Irk be the arithmetic mean of theproducts of the form a (1 i1 < <1k n). Show

that the sequence ir0, . . , ii,, is strictly log-concave:

for all k (1 k n—i). [HINT: Apply induction on n. Forn = 2 the assertion is just the AM-GM inequality for two terms.Now let n 3 and denote by ir the appropriate average forOt,... Check that

=

and deduce that for

=

A < B < C < -ire22. Deduce from the inequality in the previous exercise the following

considerable extension of the AM-GM inequality, already knownto Newton: if a1,..., a,, are positive, n 2 and a1 a2 then

> ... > ir,1l/n.

23. Let f: R be a strictly convex function with f(0) 0. Provethat if a1 ,. . . , a,,, 0 and at least two a are non-zero then

f(a,) as).


24. Let f: (0, — (0, be a monotone increasing function such thatf(x) /x2 is monotone decreasing. For a, b > 0 set

f(a,b) = and g(a,b) =

Prove that if a,, b, > 0 then\2 \ \fn

ab,)(

b.))( g(a1, b,))(

b,2\i=1 / \i=i / \i=i / \i=1 / \i=i

25. Prove that if f,g: satisfy (*) for all a,b1 > 0

then they are of the form given in the previous exercise.26. Show that

abi) + b.2)a,2+b12

(± a12)(± b?)

for all real a , b with a? + b? > 0.

27. Let b1 , b2,. .. , be a rearrangement of the positive numbersa1,a2,. . . Prove that

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 satisfy-

ing f(0) = 0. Prove the following inequality due to G. H. Hardy:

{f'(x)}2 dx.

[HINT: Note that

— 1/211(x) \' f(x)f(x)—x

and so

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


30. Show that Theorem 4 characterizes comparable means, i.e. ifM4,(a) for all a = (a1)7 (a, > 0) then is concave if q

is increasing and convex if is decreasing.31. In a paper the authors claimed that if � 0 for k = 1,2,. . ., n then

x," � x1 (2x2 — x1) (3x3 — 2x2). . . (n — 1) i).

Show that this is indeed true if 2x2 � x1, 3x3 � 2x2,. . ., � (n — 1)

x1_ 1' and equality holds if and only ifx1 = ... = Show also that theinequality need not hold if any of the n — 1 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 moy-ennes, Acta Mathematica, 30 (1906), 175—93. Much of this chapter isbased 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 inprint, and although its notation is slightly old-fashioned, it is well worthreading.

Other good books on inequalities are D. S. Mitrinovic, Analytic me-qualities, Springer-Verlag, Berlin and New York, 1970, xii + 400 pp.,and A. W. Marshall and I. 01km, inequalities: Theory of Majorizationand Its Applications, Academic Press, New York, 1979, xx + 569 pp.

2. NORMED SPACES AND BOUNDEDLINEAR OPERATORS

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

As so often happens when starting a new area in mathematics, theratio of theorems to definitions is rather low in this chapter. However,the reader familiar with elementary linear algebra and the rudiments ofthe theory of metric spaces is unlikely to find it heavy going because theconcepts to be introduced here are only slight extensions of various con-cepts arising in those areas. Moreover, the relatively barren patch israther small: as we shall see, even the basic definitions lead to fascinat-ing questions.

A normed space is a pair (V, fl), where V is a vector space overor C and is a function from V to = {r E r O} satisfying

(I) Dxli = 0 iff x = 0;

(ii) liAxD = iA lxii for all x E V and scalar A;(iii) lxii + for all x,y E V.

We call lixii the norm of the vector x: it is the natural generalizationof 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 thesum of the lengths of the other two sides.

In most cases the scalar field may be taken to be either R or C, evenwhen, for the sake of simplicity, we specify one or the other. If wewant to emphasize that the ground field is C, say, then we write 'com-plex 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 identifya normed space X = (V. with its underlying vector space V, andcall the vectors in V the points or vectors of X. Thus x E X means thatx is a point of X, i.e. a vector in V. We also say that is a norm onx.

Every normed space is a metric space and so a topological space, andwe shall often make use of some basic results of general topology.Although this book is aimed at the reader who has encountered metricspaces and topological spaces before, we shall review some of thebasic 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) suchthat (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 dis-tance 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 forv C I' then U,, U,, C i, and (iii) r is closed under finite intersec-tion: if U1 ,..., U1,, r then U1 C r. The elements of the collec-tion r are said to be open (in the topology r). A topological space is apair (X, 'r), where X is a set and r is a topology on X. If it is clear thatthe topology we take is r then we do not mention r explicitly and wecall 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 orthe topology induced by r. In most cases every subset Y is consideredto be endowed with the subspace topology.

Given a topological space X, a set N C X is said to be a neighbour-hood 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 intersectionof a collection of closed sets is closed, every subset A of X is containedin a unique minimal closed set A = {x C X: every neighbourhood of xmeets 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 subsetsof 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 basisfor 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.

20 Chapter 2: Normed spaces and bounded linear operators

A neighbourhood base at a point x0 is a collection v of neighourhoodsof x0 such that every neighbourhood of x0 contains a member of v.

There are numerous ways of constructing new topological spaces fromold ones; let us mention here the possibility of taking products, to bestudied in some detail in Chapter 8. Let (X,o) and (Y,i) be topologicalspaces. The product topology on Xx Y = {(x, y): x E X, y E Y} is thetopology with basis {Ux V: U E o, V r}. Thus a set W C Xx Y isopen if for every (x,y) E W there are open sets U C X and V C Ysuch 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) <r} (x E X, r> 0)

form a basis for a topology. This topology is said to be defined orinduced 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 anymetric.

Given topological spaces (Xj,r1) and (X2,r2), a map f: X1 —+ X2 issaid to be continuous if C for every U E r2, i.e. if the inverseimage of every open set is open.

A bijection f from X1 to X2 such that both f and are continuousis said to be a homeomorphism; furthermore, (X1 ,r1) and (X2,r2) aresaid to be homeomorphic if there is a homeomorphism from X1 to X2.

A sequence in a topological space (X, r) is said to be convergentto a point x0 C X, denoted —p x0 or = x0, if for everyneighbourhood N of x0 there is an n0 such that C N whenevern n0. Writing S for the subspace {n': n = 1,2,.. .}u{0} of R withthe Euclidean topology, we see that x,, = x0 iff the mapf: S —' X, given by f(n = and f(0) = x0 is continuous.

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

If a and i are topologies on a set X and a C r then a is said to beweaker (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) iscontinuous.

The topological spaces occuring in linear analysis are almost alwaysHausdorff 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 pointsx,y E X there are disjoint open sets and U,, such that x E U, andy E Ui,. A topological space (X, r) is compact if every open cover has afinite subcover, i.e. if whenever X = U,,E:f. where each U,, is anopen set, then X U,,EF U,, for some finite subset F of r. A subset Aof a topological space (X, r) is said to be compact if the topology on Ainduced by r is compact. Every closed subset of a compact space iscompact, 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 wehadf(x) <s = sup{f(y): yE K}for every XE K then

U E K: f(x) <r}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) andd(Ax,Ay) = AId(x,y) for all x,y,z X and scalar A. The inducedmetric in turn, defines a topology on X, the norm topology. We shallalways freely consider a normed space as a metric space with theinduced metric and a topological space with the induced topology, andwe shall use the corresponding terminology.

Let X be a normed space. By a subspace of X we mean a linear sub-space 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 isclosed if it is closed in the norm topology. Given a set Z C X, the sub-space 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 containingz.

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 thatd(xn,xm) = —* 0 as min{n,m} then converges tosome point x0 in X (i.e. = —' 0). A completenormed space is called a Banach space. It is easily seen that a subset ofa complete metric space is complete 1ff it is closed; thus a subspace of aBanach space is complete if it is closed.


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

The definition of the norm implies that Br(X0) is closed, is

open, and for r> 0 the interior of is IfltBr(Xü) = Dr(X0). Theclosure of is Dr(Xø) = Br(XO). Furthermore, the boundaryäBr(Xø) of is the sphere Sr(X0).

The definition also implies that if X is a normed space then the mapXx X —p X given by (x, y) '—f x + y is uniformly continuous. Similarly,the map —, X (or CXX X) given by (A,x) Ax is continuous.Note that the norm function . fi from X to is also Continuous.

Let us give a host of examples of normed spaces. Most of these areimportant spaces, whilc some othcrs are presented only to illustrate thedefinitions. The vector spaces we take are vector spaces of sequences orfunctions with pointwise addition and multiplication: if x = and

y = then Ax +/i.y = (Ax, if f = f(t) and g = g(t) are func-tions then = Af(t)

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


Examples 1. (i) The n-dimensional Euclidean space: the vector space isor C" and the norm is

In \1/2

lixil = (\=i /

where x = (x1 ,.. . ,x,,). The former is a real Euclidean space, the latteris a complex one.

(ii) Let S be any set and let be the vector space of all boundedscalar-valued functions on S. For f E let

Ilfil = suplf(s)I.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),

Ilfil = supjf(OI.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 con-tinuous functions on K, with the supremum norm

IlfO = = : x E K}.

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

IIxII=

Ix,

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

= max IxaI

is a norm, the the space it gives is lx".(vi) Let 1 p For x = (x1,. . . ,x,,) 1k" (or C") put

IIXIIp=

kkl.0).

This defines the space l,' (real or complex); the norm is theThe notation is consistent with that in example (v) for li" and also, in anatural way, with that for (see Exercise 6). Note that I"' is exactlythe n-dimensional Euclidean space.


(vii) Let X consist of all Continuous real-valued functions f(t) onthat vanish outside a finite interval, and put

= j If(t)I dt.-

(viii) Let X consist of all continuous complex-valued functions on[0,1] and forfE Xput

/ rl \1/2

11f112 = I jIf(t)12 dt

\0(ix) For 1 p < 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/pIIXIIp

=

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 samenorm As remarked earlier, we have both real and complex formsof these spaces.

(x) For I p the space 1) consists of those Lebesguemeasurable functions on [0, 1] for which

/ \iIP

= (j d:}/

Note that L1(0, 1) is precisely the space of integrable functions.Strictly speaking, a point of 1) is an equivalence class of func-

tions, two functions being equivalent if they agree almost everywhere.Putting it another way, fi and f2 are equivalent (and so are consideredto be identical) if = 0.

(xi) By analogy with the previous examples, the astute reader willguess that L(0, 1) Consists of all essentially bounded Lebesgue measur-able functions on [0, 11, i.e. those functions f for which


= 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

= s 5).

Once again, strictly speaking 1) consists of equivalence classes offunctions bounded on [0, 1].

(xii) Lc,(a,co), etc., are defined similarly tothe spaces in (x) and (xi). Of these, once again, the Hubert spaces

etc., are the most important.(xiii) Let 1) consist of the functions f(t) on (0, 1) having n con-

tinuous and bounded derivatives, with norm

iiiii = sup{± 0< 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

cke, (n = 1,2,...)

with norm1/2

11111 = ( ICkI—n


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

define

+ e2)lixil = inf1iai + lbI + id: a,b,c E and x = ae1+be2+

(xviii) Let V be as in (xvii) and set

114

c(e1 +e2)+d(e1 —e2)andx = ae1+be2+

(xix) Let V be an n-dimensional real vector space and letV1 , L'2,. . . , 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 satisfyconditions (i)—(iii). It is obvious that conditions (i) and (ii) are satisfiedin 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 lastthree examples, the triangle inequality is precisely Minkowski's inequal-ity in one of its many guises. Thus Theorem 1.7 (Minkowski's inequal-ity for sequences) is just the triangle inequality in 1:

ilx+ylip IIxIlp+Ilyllp (1)

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

llflip + (2)

whenever f, g E 1), etc. In turn, the triangle inequality impliesthat the underlying spaces are indeed vector spaces.

Examples (xvii)—(xix) are rather similar, with (xix) being the mostgeneral case; we leave the proof to the reader (Exercise 7). 0

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


function spaces. These spaces have been extensively studied for overeighty years: much is known about them but, in spite of all this atten-tion, many important questions concerning them are waiting to be set-tled. In many ways, the most pleasant and most important of allinfinite-dimensional Banach spaces is '2' the space of square-summablesequences. This space is the canonical example of a Hilberi space.Similarly, of the finite-dimensional normed spaces, the space 1$ is cen-tral: 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, weshall see the others later. -

Let us remark that if X is a normed space then its completion X as ametric space has a natural vector space structure and a natural norm.Thus every normed space is a dense subspace of a Banach space. Weshall expand on this later in this chapter.

Having mentioned the more concise formulations of Minkowski's ine-quality (inequalities (1) and (2)), let us draw attention to the analogousformulations of Holder's inequality. Suppose p and q are conjugateindices, with p 1 and q = permitted. If x = E I,, and

Y = E then

kkYkl iiXiipiiYiiq.k=1

Similarly, 1ff E 1) and g E Lq(O, 1) then fg is integrable and

ilfgiii=

lfgj th lifilpiigiiq.

It is easy to describe the general form of a norm on a vector space Vin terms of its open unit ball. Let X = (V,

fi fi) be a normed space andlet D = {x E V: lxii < 1) be the open unit ball. Clearly

Ii is deter-mined 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

Ax C D andD satisfying (i) is said to be absolutely convex: this property is a

consequence of the triangle inequality. Property (ii) follows from thefact that D is an open ball, while (iii) holds since Dxli < for every xand = 0 1ff y = 0.


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. Thefunction q(x) is the Minkowski functional determined by D. Note thatin conditions (i)—(iii) we do not assume any topology on V.

When studying normed spaces, it is often useful to adopt a geometricpoint 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 sameground field. A linear operator from X to Y is a linear map between theunderlying 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 linearoperators from X to Y is denoted by Y). The image of T isIm 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. Further-more, 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 suchthat

IITxII NIIxII for all x E X.

We shall write Y) for the set of bounded linear operators from Xto Y; = X) is the set of bounded linear operators on X.Clearly Y) is a vector space. The operators in Y)are said to be unbounded. A linear functional on X is a linear operatorfrom X into the scalar field. We write X' for the space of linear func-tionals on X, and X* for the vector space of bounded linear functionalson X. It is often convenient to use the bracket notation for the valueof a functional on an element: for x EX and f€X' we set <f,x> =(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 alinear operator. Then the following conditions are equivalent:

(1) T is continuous (as a map of the topological space X into the topo-logical space Y);

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


Proof. The implication (1) (ii) is trivial.(ii) (iii). Suppose T is continuous at x0. Since Tx0+ B(Y) is a

neighbourhood of T(x0), there is a 6 > 0 such that

x x0+y x0 + ÔB(X) Tx = Tx0 + Ty E Tx0+ B(Y).

Hence & implies IITyII 1 and so IITzD for all z.(iii) (i). Suppose lITxII NIIxII for all x E X. Then if lix—yll <

we have IITx— Tyll <€. 0

Two normed spaces X and Y are said to be isomorphic if there is aLinear map T: X Y which is a topological isomorphism (i.e. abomeomorphism). We call X and Y isometrically isomorphic if there isa linear isometry from X to Y, i.e. if there is a bijective operatorT E Y) such that T' E and IITxII = IIxII for all x X.Two norms, and

11 112' on the same vector space V are said to beequivalent if they induce the same topology on V, i.e. if the formal iden-tity map from X1 = (V, to X2 = (V, 1112) is a topological isomor-phism. As an immediate consequence of Theorem 2, we see that if alinear map is a topological isomorphism then both the map and itsinverse are bounded.

Corollary 3. Let X and Y be normed spaces and let T E Y). ThenT is a topological isomorphism if T E Y) and T' E X).

Two norms, and V on the same vector space V are equivalentif 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 equivalencerelation, and Corollary 3 implies that if a normed space is complete thenit is also complete in every equivalent norm.

The equivalence of norms has an intuitive geometrical interpretationin terms of the (open or closed) unit balls of the normed spaces. Let

II. and fl lb be norms on V, with closed unit balls B1 and B2. Thenand 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 iflxiii ciIxiI2 and iixIi2 dlixiJj for all x E V.

It is clear that Y) and Xt are vector spaces: Y) is a sub-space of Y) and X* is a subspace of R). In fact, they are alsonormed spaces with a natural norm. The operator norm or simply norm


on Y) is given by

11111 = inf{N>O: liTx!i Niixii for alix E X} = sup{iiTxIi: ilxii 1}.

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

11111 = inf{N > 0: f(x)I NIIXII for all xE X} = sup{If(x)l: IIxU 1).

Note that in these definitions the infimum is attained so

IiTxfl II lxii and 11(x) I lifli lixil for all x.

The terminology is justified by the following simple result.

Theorem 4. The function fl defined above, is a norm on Y). IfY is complete then so is Y). In particular, is a completenormed space.

Proof. Only the second assertion needs proof. Let be a Cauchysequence in Y). Then is a Cauchy sequence in Y for everyxE 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 lim

T E Y).Furthermore, given e > 0, there is an n0 such that ii — ii <€ if

n,m Then for x C X and m n0 we have

=

= iihm(Tn—Tm)xIi

=

eiixiI.

Theref ore


€IIxIl (C + TmII)

and so T E Y). Finally, from the same inequality we find that0

If we extend the operator norm to the whole of Y) by putting= sup{I!TxII: lixil 1} = for an unbounded operator, thenY) consists of the operators in Y) having finite norm.

The Banach space is called the dual of X. For T E Y) andg E r define a function rg: X (or C) by

(rg) (x) = g(Tx).

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

=

so that

Irgil IITU and iirii liii.The operator T* is the adjoin: of T. In fact, as we shall see later(Theorem 3.9), ijrii = liii.

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

(rg,x) = (g,Tx).

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

Theorem 5. Let X, V and Z be normed spaces and let T€ (X, F),

S (F', Z). Then ST: X —' Z is bounded linear operator and

1ISTII ilSPlilllI.

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

Proof. Clearly

ll(ST)(x) II = IIS(Tx) ilTxlI ilSllil TII xli. 0


The last result shows that if X is a Banach space then is a unitalBanach algebra: it is an algebra with a unit (identity element) which isalso a Banach space, such that the identity has norm 1 and the norm ofthe 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 1) for all p and

q (1 p,q(iii) Define S E by

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 notan 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). LetD: —p C°" be the differentiation operator: Df = f'. ThenODD = 1.

(v) Let T: be the formal identity map Tf = f. Then11111 = 1.

(vi) In order to define linear functionals on function spaces, let usintroduce the following notation. Given sequences x = (Xk)° andy = let (x,y) = and let

xy E then (x,y) is well defined.Inequality (3), i.e. HOlder's inequality for sequences, can now be res-

tated once again in the following concise form: if p and q are conjugateindices, with 1 E I,,, andy E 'q' thenxy El1 and

(5)

Inequality (5) implies that for y E the function = (x,y) is abounded linear functional on 1,, and hIYIIq• In fact, =and for 1 p < x the correspondence y is a linear isometry whichidentifies with i (see Exercise 1 of the next chapter). The dual of


contains as a rather small subspace (L and are not separable butis); however, is the dual of c0, the closed subspace of L consisting ofsequences 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, forfE

40gW=

fg

Then, by Holder's inequality (4), is a bounded linear functional onand 1140gI1 IIgIIq. In fact, 1I40g11 = IIgIIq' and for 1 the

correspondence g is a linear isometry which identifies Lq(O, 1) With1)*.

(viii) Let C[O, 1] be the space of continuous functions with thesupremum norm: Ilfil = = max{If(t)I: 0 t 1}. Let 0 1

and define CEO, 1] R by = 1(b). Then is a boundedlinear functional of norm 1.

(ix) Let X be the subspace of CEO, 1] consisting of differentiable func-tions With continuous derivative. Then D: X C[O, 1], defined byDf = f', is an unbounded linear operator.

(x) Let V be the vector space of all scalar sequences x = (Xk withfinitely many non-zero terms. Set = (V, where

\l/P

/

and

IIxIL, = max{IxkI: 1 k <

For 1 r,s let Trs: Xr be the formal identity map: TrsX =x. Then is a linear operator for all r and s. If 1 r s thenT,5 is bounded: in fact, = 1, but if r > s then is unbounded(see Exercise 27). 0

As promised earlier in the chapter, let us say a few words about com-pletions. Every metric space has a unique completion, and if the metricis induced by a norm then this completion is a normed space. Beforeexamining the completion of a normed space, let us review briefly thebasic facts about the completion of a metric space. -

A metric space X is said to be the completion of a metric space X if Xis complete and X is a dense subset of X. (A subset A of a topologicalspace T is dense if its closure is the whole of T.) The completion is


unique in the sense that if X is a dense subset of the complete metricspaces Y and Z then there is a unique continuous map q: Y Z whoserestriction to X is the identity, and this map is an isometry onto Z. Thisis 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 limitz Z. Since ç is required to be continuous, we must have = z

and so ç, if it exists, is unique. On the other hand, if we define a mapY 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) = O}.fl—.x m

Define an equivalence relation on [X] by setting —

0, and let X = [X]/— be the collection of equivalenceclasses with respect to —. For 1,9 E X set

d(i,j) =

where E I and E 9. It is immediate that d: XxX [0, is

well defined, i.e. that is independent of the representa-tives chosen. Furthermore, if C I, 9 and i then

d(i, 1) = lim lim , + ,

=

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

Finally, = 0, and if d(i,9) = 0 then so thatx = y. -

The metric space (X,d) is complete. Indeed, let be a Cauchysequence in Xand let 1(k) (k = 1,2,...). Let be such that

<2-k if n,m

Set Xk = It is easily checked that C [X] and that 1(k) tends tothe equivalence class of tbe sequence (Xk

Writing [x] for the equivalence class of the constant sequencex,x,..., we find that the map X—* X, given by x [x], is an isometry.Thus, with a slight abuse of terminology, X can be considered to be asubset of X. Clearly, X is a dense subset of X, since if is a


representative of I E X and d(x,, , <E whenever n, m n0 thend(i,x,,0) = d(i,

If X is not only a metric space but also a normed space then its com-pletion has a natural Banach-space structure.

Theorem 7. For every normed space X there is a Banach space X suchthat X is dense in X. This space X is unique in the sense that if X is adense subspace of a Banach space X then there is a unique continuousmap X X such that the restriction of to Xis the identity; furth-ermore, this map ç is a linear isometry from X to X.

Proof. Considering X as a metric space, with the metric induced by thenorm, i.e. with d(x,y) = lIx—yiI, let X be its completion. Givenx,y X, let x,, x andy,, —' y, where x,,,y,, E X. Then, for scalars Aand the sequence (Ax,, is a Cauchy sequence. Now, if Xis tobe given a normed-space structure such that the norm on X induces themetric on X, then

tim = A limx,,+p limy,, =

Ax +.&y must be defined to be the limit of the Cauchy sequence(Ax,, It is easily checked that with this definition of additionand scalar multiplication X becomes a vector space. Furthermore,lull = d(x,O) = is a norm on this vector space, turning Xinto a Banach space.

The remaining assertions are clear. 0

In fact, Theorem 7 is also an immediate consequence of Theorem 3.10. Thespace X in this result is called the completion of the normed space X. In view ofTheorem 7, if X is a dense subspace of a Banach space Y then Y is oftenregarded as the completion of X. For example, the space C[0, I] of continuousfunctions on [0, 1] is the completion of the space of piecewise linear functionson [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, for1 p < 1) is the completion of the space of polynomials withthe norm

= (L'dx)

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


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

(x,1.€X)n1

is said to the convergent to x E X ifN

x,—xn1

i.e. ifN

urn x— =0.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

convergent. Indeed, let YN = be the Nth partial sum. Wehave to show that is a Cauchy sequence. Given e > 0, choose ann0 such that <€. Then for n0 N < M we have

n—no

M M

Ii = IIXnII <6.nN+1 nN+1

Hence is indeed a Cauchy sequence and so converges to somex E X, so that x =

In an incomplete space X the assertion above always fails for someabsolutely convergent series x,: this is often useful in checkingwhether a space is complete or not.

Before showing this, let us mention a very useful trick when dealingwith Cauchy sequences in metric spaces: we may always assume that weare dealing with a 'fast' Cauchy sequence. To be precise, given a Cau-chy sequence )°, a priori all we know is that there is a sequence

say = 2k, such that d(Xn,Xm) < 1/k if n,m However,we may always assume that our sequence is much 'better' than this:d(Xn,Xm) <2's if m fl, Oi d(xn,xm) <2-2 if m n, or whateversuits us. Indeed, let f(n) > 0 be an arbitary function. Set = f(1) andchoose 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


n, m n2. Continuing in this way, we find a sequence n1 <n2 <such that , x,) <f(k) if I k. Setting Yk = we find that thesubsequence is such that d(yk ,y,) <f(k) whenever k < I. Now ifwe can show that converges to some limit y, then tends tothe same limit y. Indeed, if , ,,) <e for n, m n0 then forn no we have

, y) = urn Yk) urn sup , Yk)k—.oo

since Yk = x,1 and k is sufficiently large. Thus, in provingthat a Cauchy sequence in a metric space is convergent, we mayindeed assume that , say, if m n.

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

Proof. We have already seen that in a Banach space every absolutelyconvergent series is convergent. Suppose then that in our space X everyabsolutely convergent series is convergent. Our aim is to show thatevery Cauchy sequence is convergent.

Let be a Cauchy sequence in X. As we have just seen, we mayassume that d(x1 , = 111n — Xml for n m. Set x0 = 0 andYk = XkXk_1 (k 1). Then = Yk and IJYkfl <2_k41 There-

fore the series Yk is absolutely convergent and has partial sumsx1 , By assumption, is convergent, say to a vector x,

0

Absolutely convergent series in Banach spaces have many propertiesanalogous to those of absolutely convergent series in For example, if

x,, is absolutely convergent to x and n1 , n2,... is a permutation of1,2,... then x,,. is also absolutely convergent to x. However, aswe shall see in a moment, unlike in this property does not character-ize absolutely convergent series in a Banach space.

The use of series often enables one to identify a Banach space with asequence space endowed with a particular norm. This identification ismade 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 <co the space I,, has a basis; the so-called standard or

canonical basis (e1 where e = (0,..., 0, 1,0,...) = (ö,1 , the


ith term is 1 and the others are 0. It is easily seen that if x =(x1 , x2,...) I,, then x = xe1 and this representation is unique(see Exercise 11). In particular, x, e. is convergent (in I,,) if andonly if lxi 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 newspaces from old. We have already seen the simplest way: every (alge-braic) subspace Y of a normed space X is a normed space, with the res-triction of the norm of X to Y as the norm. If X is complete then Y iscomplete if and only if it is closed. As the intersection of a family ofsubspaces is again a subspace, for every set S C X there is a uniquesmallest subspace containing S, called the linear span of S and denotedby tin 5: it is the intersection of all subspaces containing S and also theset of all (finite) linear combinations of elements of S.

linS = fl{W: WisasubspaceofXandSC W}

=s1,. ..,s,, €5; n =

Similarly, the closed linear span of S, denoted lin 5, is the unique smal-lest closed subspace containing S; it is the intersection of all closed sub-spaces 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 asubspace Z, define an equivalence relation — on X by setting x y ifx — y E Z. For x C X, let [xl be the equivalence class of x: putting itanother 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 spaceand Z is a closed subspace of X then we can define a norm fi

lbon X/Z

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 homo-geneity and the triangle inequality are obvious and Ii [0] lb = 0. All thatremains is to show that if [x] 1 = 0 then [x] = 0. To see this, note thatif [xlii = 0 then lix — fi —' 0 for some sequence (zn) C Z. Hencez,, x and so, as Z is closed, x C Z.

We call X/Z, endowed with the quotient normed space of X byZ, and call

11 lbthe quotient norm. Throughout this book, a quotient


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

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

Theorem 9. Let X and Y be normed spaces, T E Y) andZ = Ker T. Let T0: X/Z Y be the linear map induced by T. ThenT0 is a bounded linear operator from the quotient space X/Z to Y, andits norm is precisely 11711.

Proof.

H To[x] II inf{II ill : y x} = 1171111 [xJ

II 11111.

Conversely, let e > 0 and choose an x E X such that lixil = 1 andIITxIl > 11711—c. Then II[x}H 1 and IIT0[x]Il = IIT1V > 11711—c.Hence II > 11111 — 0

In fact, the quotient norm II 110 on X/Z is the minimal norm on X/Zsuch that if T E Y) and Ker T 3 Z, then the operator T0: X/ZY 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, andpy: 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 iseasily seen that Z is a direct sum of its subspaces X and Y if the topol-ogy on Z (identified with Y = {(x, y): x E X, y E Y}) is preciselythe product of the topologies on X and V. Note that, if Z is a directsum of X and V. Z' is a direct sum of X' and V'. and X is isomorphic toX' and Y is isomorphic to Y', then Z is isomorphic to Z'. If Z is adirect sum of X and Y then the projection Px: Z —. X induces an iso-morphism between Z/Y and X.

Conversely, given normed spaces X and V. there are various naturalways of turning Y, the algebraic direct sum of the underlying vec-tor spaces, into a normed space. For example, for 1 p we maytake the norm = Il(lIxII, Thus for 1 p < we take

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


space (X$ Y, is usually denoted by X Y. Considering X and Vas subspaces of X$,, Y, we see that Y is a direct sum of X and V.

Finally, given a family {II II,.: y C fl of norms on a vector space V. if

IIxII =

for every x C V. then flis a norm on V. Note that the analogous

assertion about the infimum of norms does not hold in general (seeExercise 15).

Having got a good many of the basic definitions under our belts, weare ready to examine the concepts in some detail. In the next chapterwe 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)is the sphere Do these statements hold in a general metricspace 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,} (r1, > r > 0).

Does

n1hold? Is there a ball B(x,r) r> 0, contained in B1,?

3. Prove or disprove each of the following four statements. In a com-plete space every nested sequence of closed has anon-empty intersection.

4. Let X = (V,H il) be a normed space and W a subspace of V. Sup-

pose is a norm on W which is equivalent to the restriction ofii II to W. Show that there is a norm II on V that is equivalentto and whose restriction to W is precisely

I I.5. Let and be two norms on a vector space Vand let Wbe a

subspace of V that is Il-dense in V. Suppose that the restrictionsof . II and to W are equivalent. Are . II and

I Inecessarily

equivalent?6. For 1 p let be the on R1, Show that if

I p < r then llXIlr. For which points x do we haveequality?


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

7. Show that the space defined in Examples 1 (xix) is indeed anonned space.

8. Let 1 p,q,r co be such that p1+q1+r1 = 1, withdefined to be 0. Show that for x, y, z 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 ofShow also that 4 (1 p <cc) and c0 are separable Banach spaces(i.e. each contains a countable dense set) while is not separable.

10. Let p, q and r be positive reals satisfying p1+ q' = r1. Showthat for f E 1) and g E Lq(O, 1) the function fg belongs toLr(0,1) and

IIfIIpIl8iIq.

11. Let e, = (O,...,0,1,0,...) = €4, where 1 <x.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.

12. For x = (x1)r E li', the support of x is suppx = 0}. Let1 <cc and let fl'f2,... €4 be non-zero vectors with disjointsupports. Show that X = Iin(f1)T is isometric to 4: in fact, themap X 4 given by f, defines a linear isometry.

13. Let Z be a closed subspace of a normed space X. Show that thequotient norm on the vector space X/Z is the minimal norm suchthat if Y is a normed space, T Y) and Z C Ker T then thenorm 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 —' suchthat p(Ax) = IA Ip(x) and p(x +y) p(x) +p(y) for all vectorsx,y E V and scalar A. [Thus a seminorm p is a norm if p(x) = 0implies x = O.J

Let {p,.: y I) be a family of seminorms on a vector space Vsuch that


0 <p(x) sup{p7(x): y E fl <for every x E V(x 0). Show that p() is a norm on V.

15. Let lixili and 0x112 be norms on a vector space V. Is jixil =min{11x111, 11x112} necessarily a norm?

16. Consider the vector space c0 of complex sequences x = tend-ing to 0. For a sequence E c0, let be the decreasingrearrangement of I )°. Formally, x if

I{k: IxkI >x}j I{k: IXkI

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

Let

IIxII' = : m = 1,2,. .

.

d0 = {x E c0: IIxII' < rx}

Show that is a norm on d0. Is this space complete?17. For x E set

IIxII' =

Show that II II' is a norm on and that 11 is complete in thisnorm. Is IIxII' equivalent to the l1-norm 11x111 =

18. Show that on every infinite-dimensional normed space X there is adiscontinuous (unbounded) linear functional. [By Zorn's lemma Xhas a Hamel basis, i.e. a set {x7: y E fl C X such that everyx E X has a unique representation in the form x = Ax7.

19. Prove that if two norms on the same vector space are notequivalent then at least one of them is discontinuous on the unitsphere in the other norm. Can each norm be discontinuous whenrestricted to the unit sphere of the other?

20. Give two norms on a vector space such that one is complete andthe other is incomplete.

21. Find two inequivalent norms and 1116 on a vector space Vsuch 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 anelement of X. Is the distance of x from Y attained? (Is there apOint Yo E Y such that = inf{IIx—yII: y E Y}?)


23. Let '12'. . . ,f,, be linear functionals on a vector space V. Showthat 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-tionals?

24k. Let X be a Banach space. Suppose )° C X is a sequence suchthat every x X has a unique representation in the form x =

Prove that the set consists of isolated points.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 thethird.

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

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

Let V be a vector space with basis [Thus every x E V(x 0) has a unique expression in the form x =(n1 <n� < <nk and A, 0).] Show that there is no completenorm on V.

31 Let C be a closed convex set in a normed space X such thatC+B(X) D Bl+E(X) for some e > 0. Does it follow that IntC0, i.e. that C contains a ball of positive radius?

Notes

Abstract normed spaces were first defined and investigated by StefanBanach in 1920 in his Ph.D. thesis at the University of Léopol (i.e. thePolish town of Lwôw, now in the Soviet Union). Much of this thesiswas published as an article: Sur les operations dans les ensemblesabstra its et leur applications aux equations intégrales, FundamentaMathematica, 3 (1922), 133—81. A few years later Banach wrote thefirst book wholly devoted to normed spaces and linear operators:Théorie des Operations Linéaires, Warsaw, 1932, vii + 254 pp. Banachgave an elegant account of the work of many mathematicians involvedin the creation of functional analysis, including Frédéric (Frigyes) Riesz,whom he quoted most frequently, Just ahead of himself, Maurice


Fréchet, Alfred Haar, Henri Lebesgue, Stanislaw Mazur, JuliuszSchauder and Hugo Steinhaus. It is perhaps amusing to note that, whenwriting about Banach spaces, Banach used the term 'espace du type(B)'.

This beautiful book of Banach has had a tremendous influence onfunctional analysis; it is well worth reading even today, expecially in itsEnglish translation: Theory of Linear Operators (translated by F. Jel-lett), North-Holland, Amsterdam, 1987, ix + 237 pp. This edition isparticularly 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 andreferences.

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 + 468pp. This volume concentrates on the function-theoretic and measure-theoretic aspects of functional analysis, so it does not have too much incommon with our treatment of linear analysis.

There are a good many monographs on normed spaces and linearoperators, 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 NormedSpaces, Pergamon Press, International Series of Monographs on Pureand Applied Mathematics, vol. 45, Oxford, 1964, xiii + 771 pp., A. N.Kolmogorov and S. V. Fomin, Introductory Real Analysis (translatedand 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, ACourse in Functional Analysis, Graduate Texts in Mathematics, vol. 96,Springer-Verlag, New York, 1985, xiv + 404 pp.

3. LINEAR FUNCTIONALS AND THEHAIIN-BANACH THEOREM

Given a normed space X = (V,II

let us write X' for the algebraicdual of X, i.e. for the vector space V' of linear functionals on V. ThusX*, the space of bounded linear functionals on X, is a subspace of thevector space X'.

We know from the standard theory of vector spaces that every independentset of vectors is contained in a (Hamel) basis (see Exercise 18 of Chapter 2). Inparticular, for every non-zero vector u E V there is a linear functional fE Vwithf(u) 0. Equivalently, V is a large enough to distinguish the elements ofV: for all x, yE V(x y) there is a functionalfE V such thatf(x) Even

more, the dual V' of V is large enough to accommodate V: there is a naturalembedding 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 boundedlinear 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 thatf(x) 0? As X is a normed space, one would like to use X* to obtainsome 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 forevery 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} whileX = (V, fi

fi) is large, say V is infinite-dimensional.The main aim of this chapter is to show that, as far as the questions

above are concerned, Candide and Pangloss were right, tout est pour lemieu.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 presentthe result implying this, namely the Hahn—Banach theorem, we shallpoint out some elementary facts concerning linear functionals.

Let us show that f E X' is bounded if and only if f(B) is not theentire ground field. Let B = B(X) = B(O, 1) be the closed unit ball ofX. If Al 1 then AB = B(O, Al) C B. Hence for f X' we havesuplf(B)I = sup{If(x)l: lixil 1} = where llfll = if f is

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) isthe open unit ball of X then f(D) = {A: IAI < llfll} for all f E X'(f 0). Hence f E X' is bounded 1ff f(B) is not the entire groundfield, as claimed.

it is often useful to think of a (non-zero) linear functional as a hyper-plane in our vector space. An affine hyperplane or simply a hyperplaneH 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 sub-space with dim X/Y = 1. We say that H is a translate of Y. Given anon-zero functional f E X', let K(f) = f'(O) = {x E X: f(x) = 0} bethe 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 linearalgebra.

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

1. Moreover, if f(x0) 0 then every vector x E X has a uniquerepresentation in the form x = y + Ax0, where y K(f) and A is ascalar.

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

linear functionals and hyperplanes not containing 0. 0

Continuous (i.e. bounded) linear functionals are easily characterizedin terms of K(f) or 1(f). Note that Ilfil 111 and only if lf(x)

I< 1 for

all x with 11111 < 1, i.e. if and only if 1(f) is disjoint from the open unitball D(0,1) = {x X: llxll <1).

Chapter 3: Linear functionals and the Hahn—Banach theorem 47

A subset A of a topological space T is nowhere dense if its closure hasempty 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-zerobounded linear functionals and closed hyperplanes not containing0.

Proof. (a) Suppose that f is continuous. Then K(f) and 1(f) areclosed, since they are inverse images of closed sets. If f(x0) 0 thenf(x) f(x + fx0) for 0. Hence K(f) and 1(f) have empty interiors,and so are nowhere dense.

Suppose now that K(f) is not dense in X, say B(x0, r) fl K(f) = 0 forsome x0 E X and r> 0. Then f(B(xo, r)) = f(x0) +rf(B(X)) does notcontain 0, so f(B(X)) is not the entire ground field. Hence, as we haveseen, 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). 0

In fact, B(x0, r) fl K(f) = 0 implies a bound on the norm of f,namely the bound Ilfil f(xo)I/r. Indeed, otherwise If(x)I >

I IJxII/r for some x E X and so

=—

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 functionalscan be extended from subspaces without increasing their norms. Thismeans that all the questions posed at the beginning of the chapter havereassuring answers. Although the proof of the general form of theHahn—Banach theorem uses Zorn's lemma, the essential part of theproof is completely elementary and very useful in itself.

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


and x E X, and is a convex function (as used in Chapter 1), i.e. ifx,y E X, and 0 t 1 then p(tx+(1—Oy) tp(x)+(1—Op(y). Bythe positive homogeneity of p, the second condition is equivalent top(x+y) p(x) +p(y) for all x,y E X, i.e. to the subadditivity of p. Ascustomary for the operations on R = we use the conventionthat = = 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 functionalf E X' is dominated by the convex functional NIIxII if f E X* andliffi N. As usual, a function S —+ R is said to dominate a functioni/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 letfo be a linear functional on a 1-codimensional subspace Y of X. Sup-pose that fo is dominated by p, i.e.

fo(Y) p(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 representa-tion in the form x = y + tz, where y E Y and t E R. The functional fwe are looking for is determined by its value on z, say f(z) = c. Toprove (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 (2) gives an upper bound on c, and for

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

for all y Y. For s > 0 we have fo(Y) —Sc p(y—sz) and so

C>

for all y E Y. The former holds if

Chapter 3: Linear functionals and the Hahn —Banach theorem 49

c 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),

completing the proof. 0

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 isthe 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 alinear functional f X' dominated by p.

If X is a real normed space and Jo E r then to has an extension to afunctional f on X such that 11111 = Ilfoll.

Proof. Set K, = lin{Y,z1,.. . ,z,,}. By Lemma 3 we can define linearfUflCtioflalS to C C 12 C such that and each f,, is dom-inated by p. Define f: X R 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 dominatedby the convex functional p(x) = where N = Ilfoll. Hence thereis an f E Xt extending to and dominated by p. But then f(x)p(x) = NfIxII for all x E X, so that Ilfif N = Iltoll, implying 11111 =IltoIl. 0

The restriction on Y in Theorem 4 is, in fact, unnecessary. As weshall see, this is an easy consequence of Zorn's lemma, the standardweapon of an analyst which ensures the existence of maximal objects.For the sake of completeness, we shall state Zorn's lemma, but beforedoing 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 suchthat (i) a a for every a P. (ii) if a b and b c for a,b,c E P

forsomea,bE Pthena = b.Briefly, is a transitive and reflexive binary relation on P. We call the


pair (P. a partially ordered set; in keeping with our custom concern-ing normed spaces and topological spaces, (P, is often abbreviated toP. A subset C of P is a chain or a totally ordered set if for all a, b E Cwe have a orb a. Anelementm E Pisamaximalelejnentof Pif m a implies that a = m; furthermore, we say that b is an upperbound foralisES. Itcan beshownthattheaxiom of choice is equivalent to the following assertion.

Zorn's lemma. If every chain in a non-empty partially ordered set P hasan 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 cele-brated Hahn—Banach extension theorem.

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

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

Proof. Consider the set = {f,,: y E 1) of all extensions of fo dom-inated 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. Clearlythe relation 'C' is a partial order on is 'less than or equal to' fo if

C fe). If = if,,: y E is a non-empty chain (i.e. - a totallyordered 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, byZorn's lemma, there is a maximal extension. But by Lemma 3 everymaximal 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 canbe guaranteed in complex normed spaces as well. A complex normedspace X can be considered as a real normed space; as such, we denote itby XR. We write for the dual of Xft. It is easily checked that themapping r: r defined by r(f) = Ref (i.e. r(J)(x) = Ref(x) forx E X) is a one-to-one norm-preserving map onto The inverse of ris the map c: —' r defined by c(S) (x) = g(x) — ig(ix). This enablesus to deduce the complex form of the Hahn—Banach extension theorem.


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

Proof. By Theorem 5, we can extend r(f0) to a functional g on XRsatisfying lid = IIr(fo)II = The complex functional f = c(g) E rextends fo and satisfies IlfIl = IIfo II. 0

The Hahn—Banach theorem has many important consequences; wegive some of them here.

Corollary 7. Let X be a normed space, and let x0 X. Then there is afunctional f C such that f(x0) = IIxoII. In particular, Iixoii C 1ffig(xo)P C for all g 5(r).Proof. We may assume that x0 0. Let Y be the 1-dimensional sub-space lin{x0} and define to E VS by f0(Ax0) = A IixoII. Then 1 andits extension f, guaranteed by the& Hahn—Banach theorem, has therequired properties. 0

Corollary 8. Let X be a normed space, and let x0 C X. If f(x0) = 0 forallfErthenx0=0. 0

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

Corollary 7 implies that the map Y) —' given byT r, is an isometry, as remarked after Theorem 2.4, when wedefined the adjoint.

Theorem 9. If X and Y are normed spaces and T C Y) thenr and itrii = 11711.

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

(Tg)(x0) = g(Tx0) = IlTxoII 11711

so that lIrgIl and iirii IIi1I—€.


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" isdefined by v"(f) = f(v) for f C V. Rather trivially, this embedding isan isomorphism if V is finite dimensional. If X is a normed space withdual r, second dual Xt. and x C X, then we write i for the restrictionof x" to X*: I is the linear functional on given by 1(f) = f(x) forf C In other words, with the bracket notation,

(I,f) = (f,x)

for all f C X*. Since Ii(f)I = If(x)I IlfIllIxIl, we have I C (notjust I E (X*)I), and moreover Dxli. The Hahn—Banach theoremimplies 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 = 1and 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 ofX the whole of X**, i.e.

X = then X is said to be reflexive. We know that X* * is completeeven when X is not, so a reflexive space is necessarily complete. How-ever, a Banach space need not be reflexive. For example, 1, is reflexivefor 1 <p and 11 and are not reflexive. Also, as we shall see,every finite-dimensional normed space is reflexive.

Writing C(L) for the Banach space of bounded continuous functionson a topological space L with the uniform norm, it is easily seen thatevery Banach space X is a closed subspace of C(L) for some metricspace 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 linearisometry onto a closed subspace of C(L). We shall see in chapter 8 thatconsiderably more is true: instead of C(L) we may take C(K), the spaceof all continuous functions on a compact Hausdorff space K with theuniform norm.

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

Chapter 3: Linear fwzc:ionals and the Hahn—Banach theorem 53

Given a real vector space X, a function q: R. = is saidto be a concave functional if —q(x) is a convex functional, i.e. if q ispositive 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 tofind 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 neednot 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): puttingx = 0 in (5) we find that f(y) p(y) for all y E Y, and setting x = —ywe see that f(y) —q(—y), i.e. q(—y) f(—y) for all y Y. Ofcourse, 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 thewhole of X.

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

fo(Y) p(x+y) —q(x)

for all y E Y and x X. Then f X' such that

q(x) f(x) p(x)

for every x X.

Proof. The heart of the matter is the analogue of Lemma 3: once wehave managed to extend to a slightly larger (i.e. one dimensionlarger) subspace, the rest follows as before.

Pick a vector z X (z Y) and set Z = lin{Y,z}. Let us show thatfo has an extension f1 to Z satisfying

f1(u)

54 Chapter 3: Linear functionaLs and the Hahn-Banach theorem

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

f1(y+z) =

and

f1(y'—z)—— f0(y')—c

for all y,y' Y and x,x' X. Such a c exists if and only if

—p(x' -I-y' — z) + fo(Y) +p(x +y + z) — q(x)

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

fo(y)±fo(y') =

fo has a suitable extension to Z.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 functionalon a real vector space X such that q is dominated by p: for all x Xwe have q(x) p(x). Then there is a linear functional f E V' such that

q(x) f(x) p(x) for all x X.

Proof. Let Y = (0) C X. Then the trivial linear functional fo on Ysatisfies the condition in Theorem 11, namely

h(O) p(x) — 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 realvector space X. Suppose that for some ccE A and every xE X there is ane(x)> 0 such that + tE A for all t, fri (x). Then A and B can beseparated by a hyperplane, i.e. there is a non-zero linear functionalfE X' anda real number c such thatf(x) c for all x€ A and yE B.


Proof. We may assume that a = 0, i.e., for every xE X there is ani =e(x)>Osuchthat[—€x,exJCA.DefinefunctionspandqonXbysetting, for XE X,

p(x) = inf{t 0: xE tA}; q(x) = sup{t 0: XE tB}.

It is easily checked that p: X R is a convex functional andq: X is a concave functional. Furthermore, as tA fl tB = 0 fort> 0, we have q(x) p(x). Hence, by Corollary 12, there is a non-zerolinear functional f E X' such that q(x) f(x) p(x) for all x E X. Tocomplete the proof, note that if x E A and y B then

f(x) p(x) 1 q(y) f(y).

Hence we may take c = 1. 0

As the last result of this chapter, we shall show that the separationtheorem gives a pleasant description of the closed convex hull of a set ina normed space. The convex hull co S of a set S in a vector space X isthe intersection of all convex subsets of X containing S, so it is theunique 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 inter-section of all closed convex subsets of X containing S, so it is the uniquesmallest closed convex set containing S. As the closure of a convex setis convex, S is the closure of co S.

The following immediate consequence of the separation theoremshows that is the intersection of all closed half-spaces containing S.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.Then = {x E X: f(x) su2f(s) for allf E

SE.)

Proof. Suppose that x0 S. Then B(x0, r) fl S = 0 for somer> 0. Let f E X' (f 0) separate the convex sets B(x0, r) and 5:

f(x) c

for all x E B(x0, r) and y coS. Since the restriction of f to B(xo, r) isbounded above,f€ X* and O,f(x0) > c. 0

56 Chapter 3: Linear functionaLc and the Hahn—Banach theorem

Throughout the book, we shall encounter many applications of theHahn—Banach theorem and its variants. For example, the last resultwill 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 quickproof of the fact that for 1 p the space is complete.]

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 thedual of 1) is Lq(O. 1).

4. Check that for 1 <p < x, the space is reflexive. Check alsothat and c0 are not reflexive.

5. Let X and Y be normed spaces, and set Z = with norm= li(x,y)ll = llxii + lIyil. What is Z?

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

space of all sequences x (x1,x2,...) which are eventually zero:

x,, E and x1 = 0 if n is sufficiently large, with pointwise opera-tions and norm

\1/2

lxii = (= 1

What is X*?

7. Let X be a Banach space. Show that if X*** = X* then X is

reflexive, i.e. = X.8. A linear functional f E 1,. is positive if f(x) 0 when every

x = E 1,. with x, 0 for all n. Show that every positivelinear functional on ic,. is bounded.

9. Show that p0(x) = = lim is a convex functional

Ofl icc. and deduce the existence of a linear functional f:such that

f(x)

for every x = E10. 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.) LetJo C (i.e. let Jo be a bounded real linear functional on Y) such

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 Ythen f0(A1y1 +A2y2) = A1f0(y1) +A2h(y2). Show that fo need nothave an extension to a bounded complex linear functional on thewhole of X, i.e. it need not have an extension to a functionalfEr.

11k. Let X be a Banach space, Y a 1-codimensional subspace of X andX1 and X2 dense subspaces of X. Is X1 nA'2 dense in X? IsX1 fl Y dense in Y? What are the answers if A'1 and X2 have codi-mension 1?

12. Let A' be a normed space, Y a dense subspace of X and Z a closedfinite-codimensional subspace of A'. Is Zfl Y dense in Z?

13. Let V be a subspace of a normed space A'. Show that the closureof Y is

= fl{Kerf: f E Y C Kerf}.

14. Let K be a closed convex set in a real normed space X. Show thatevery boundary point of K has a support functional: for everyx0 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 letfo: A —÷ R. Show that there is a functional f E B(X') such thatf(a) = f0(a) for all a E A if and only if

A(a)f0(a) A(a)aaEF a€F

for every finite subset F of A and for every function A: F R.16. Let V be a subspace of a normed space X and let x 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) such thatV C Ken and f(x) = 1.

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

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


f = f(t) E X and a1 ,... , R

set

=

and define

p(f) = n = 1,2,...;

a convex functional and deduce that there is a gen-eralized 'integral' 1(f) such that for f, g E andA,p.,t9 E R we have

(i) 1(Af+ = Al(f) + id(s); (ii) 1(f) 0 if f 0;

(iii) J(J(t + t0)) = I(J(t)); (iv) 1(J( —1)) =

(v) 1(1) = 1 (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 Rsuch that urn SUPt..IOØ

f = f(t) E X and a1,... E R

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) + } = A LIM f(t) + g(t);

(ii) LIM f(t) 0 if lim inff(r) 0;(—C

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

(iv) LIM1 = 1.


19. For x = 4, and n1 < <flk put

1r(x;nl,...,nk) = ±ki—I

and define

p is a convex functional, and deduce the existence of alinear 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 ageneralized limit of the sequence and is usually denoted by

LIM LIM + LIM

LIM LIM

Let rn N be fixed and define = n/rn — What isLIM

20. Show that if in Theorem 13 we drop the condition on A then theassertion 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 reineund angewandte Mathematik, 157 (1927), 214—29; and by S. Banach,Sur les fonctionnelles linéaires II, Studia Math., 1 (1929), 22 3—39. Thecomplex version, namely Theorem 6, was proved by H. F. Bohnenblustand 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 lesoperations 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 andoperators on them, we look at the 'smallest' normed spaces, namely thefinite-dimensional ones. As far as the crude classification of norms isconcerned, these spaces are very simple indeed: any two norms on afinite-dimensional vector space are equivalent. This can be proved inmany different ways; the proof we give here is based on a lemma aboutthe space

Lemma 1. The closed unit ball of is compact.

Proof. We shall show that is sequentially compact. Let (e,)? bethe standard basis of so that

= A1 I.

Given C B = with Xk = we haveI a bounded sequence of complex numbers has a conver-

gent subsequence, by repeatedly selecting subsequences, we can find asubsequence such that converges to some scalar A for everyi (1 i n).

Setting x = A,e1 we find that

lim = 0

and so x E B.

Theorem 2. On a finite-dimensional vector space any two norms areequivalent.

60

Chapter 4: Finite-dimensional normed spaces 61

Proof. Let V be an n-dimensional vector space with basis Letbe the -norm on V given by

=i=I i=1

and letII

be an arbitrary norm on V. It suffices to show thatII

andare equivalent.

Let X1 = (V, III). S1 = S(X1) and let f: S1 be defined byf(x) = lxii. The set is a closed subset of the compact set B(X1) andtherefore it is compact. Furthermore

11(x) —f(y)I ilx—yll —

1=1

lx1—y11

= (max Iieali)Ux—yiiiI

so f is a continuous function on the compact set S1. Hence f attains itsinfimum m and supremum M on Si. Since f(x) = lixIl > 0 for allxE 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 andare 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 dimensionare isomorphic.

Proof. If dim X = dim Y then there is an invertible operatorT C Y). As both T and T1 are bounded, X and Y are iso-morphic. 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 everyequivalent norm. Since, for example, the space is complete, theassertion follows. 0

Corollary 6. In a finite-dimensional space a set is compact 1ff it is closedand bounded. In particular, the closed unit ball and the unit sphere arecompact.

Proof. Recall that is compact, and that a closed subset of a com-pact set is compact. 0

Corollary 7 Every finite-dimensional subspace of a normed space isclosed and complete.

Proof. The assertion is immediate from Corollary 5. 0

In fact, as proved by Frederic Riesz, the compactness of the unit ballcharacterises finite-dimensional normed spaces. We shall prove this bymaking 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.(a) If Y is closed then for every 0 there is a point x S(X) whose

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 dis-tance from Y is 1.

Proof. Let z E X\Y and set Z = lin{Y, z}. Define a linear functionalf: Z R by f(y + Az) = A for y E Y and A E R. Then f is a boundedlinear functional since Kerf = V is a closed subspace of Z. By theHahn—Banach theorem, f has an extension to a bounded linear func-tional 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

F(x—y) F(x)IIx—yIj

Fl = 1—c.


(b) 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

Corollary 9. Let X1 C X2 C be finite-dimensional subspaces of anormed space, with all inclusions proper. Then there are unit vectorsx1,x2,... such that x,, E X,, and = 1 for all n 2.

In particular, an infinite-dimensional normed space contains an infinitesequence of 1-separated unit vectors (i.e. with 1 forn 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 unitball is compact.

Proof. From Corollary 6, all we have to show is that if X is infinite-dimensional then its unit ball B(X) is not compact. To see this, simplytake an infinite sequence x1 , x2,... E B(X) whose existence is

guaranteed by Corollary 9: 11x1 —x1fl 1 for i j. As this sequencehas no convergent subsequence, B(X) is not compact. 0

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

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

Lemma 11. Let x1,... , x, be linearly independent vectors in a realnormed space X of dimension n 2. Then there is a vector E S(X)such that — > 1 for all i (1 i < n).

Proof. We may assume that dim X = n. Let f S(X) be such thatf(x1) = 0 for 1 i < n. (In other words, we require

64 Chapter 4: Finite-dimensional normed spaces

K(f) = lin(x1,.. .

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

K={xES(X):f(x)=Let x K}. Then for 1 i <nwe have —x,) = 1. Since —x1) = —1

the choice of x,, tells us that x,, —x, K, so that —xJ 1. HenceIIx,,—x111 > 1 for every 1(1 i < n). 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 vectorssuch 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 unitvectors. This time we shall rely on the finite-dimensional form of theHahn—Banach theorem. Let us choose vectors x1,x2,... E S(X) andsupport functionals x,x,... as follows. Pick x1 C S(X1) andlet E be a support functional at x1 : = 1. Supposek <n, = n, x1,.. . x,.. .,4 xr(x1) = 1and xr(x1) = 0 for i < j. Then

k k

W fl K(x7)= fl Kerx7

has dimension at least n — k (in fact, precisely n — k) and so we can picka vector Xk+1 Let 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 C S(X) and C such thatlin{x1 ,.. . , x,j = X,, = 1 and = 0 for n <m. In particu-lar, XmII 1 for n m.

The canonical bases of 1,, and = i, where + = 1, areexamples of pairs of sequences of this type. In fact, with =(0,..., 0,1,0,...) 6 and = (0,..., 0,1,0,...) where the l'soccur in the nth places, we find that = = = 1 and

0 for all n m, not only when n <m. A pair of sequencessatisfying these conditions is called a normali.ced biorthogonal system.To be precise, given a normed space X, a biorthogonal system on X


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

Theorem 13. Let X be an n-dimensional real normed space. Then thereis a normalised biorthogonal system (x1 )7, (xr x.

Proof. Let X = [fl) and S = S(X). For u1 ,... , u,, E writev(uj,. . . , for the n-dimensional Euclidean volume of the n-simplexwith vertices 0, u1 ,.. . , Let x1 ,... , x,, E S be such that v(x1,... , x,,)is maximal.

We claim that these vectors will do. What are the functionals (xClearly xr has to be defined by = I and x7(x1) = 0 for j i. Butdoes x7 have norm 1? Indeed it does, since 11x711 > 1 implies thatx7(y1) > I for some y, E S and so

D

The sequence whose existence is guaranteed by Theorem 13 iscalled an A uerbach system.

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

ku1 = and = max1=1

Let X1 and X,. be the normed spaces defined by these norms. ThenTheorem 13 claims precisely that x1,. .. ,x,,, can be chosen in such a waythat

B(X1) C B(X) C B(X,0).

Clearly 8(X1) is the convex hull of the 2n vectors ±x1, ±x2,..., ±x,and is the n-dimensional parallelepiped whose vertices are

(€, E {—1,1}).

Let us return to the opening statement of this chapter: the isomorphicclassification of finite-dimensional normed spaces is trivial, with twospaces being isomorphic if and only if they have the same dimension.Based on this, one could come to the hasty verdict that there is nothingto finite-dimensional normed spaces: they are not worth studying. As it


happens, this would not only be a hasty verdict but it would also beutterly incorrect. There are a great many important and interestingquestions, only the isomorphic classification is not one of them. Allthese questions, many of which are still open, concern the metric pro-perties of the finite-dimensional normed spaces.

Perhaps the most fundamental question is the following: given two n-dimensional normed spaces, how close are they to being isometric? Letus formulate this question precisely.

Let X and Y be isomorphic normed spaces, i.e. let X and Y be suchthat there is a bounded linear operator T C Y) which has abounded 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—Mazurdistance to be

Note that d(X, Y) 1, d(Y, X) = d(X, Y) = d(X*, for any twospaces, and

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

for any three spaces. Thus in some sense it would be more natural tomeasure the 'distance' between X and Y by log d(X, Y).

What does d(X, Y) <d mean in geometric terms? It means preciselythat there is an invertible linear operator T E Y) such that

!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 lineartransformation the unit ball of one of the spaces is sandwiched betweenthe 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 iso-morphic and so d(X, Y) But how large can d(X, Y) be? A com-pactness argument implies immediately that d(X, Y) f(n) for somefunction f: N R. In fact, the following simple consequence ofTheorem 13 gives a bound on d(X, Y).


Corollary 14. Let X = (V,H fi) be an n-dimensional normed space.

Then n.

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

=jA1 I.

Then X1 = (V, lii) is isometric to 1k"; so we have to show thatd(X,X1) n.

We claim that the formal identity J: X3 X is such that 11111 1 andn. Indeed, forx = Ax1 we have

IIJ(x)II=

I IIx,II=

A1 I = lixili

and so 11111 1. (As IIJxiII = lixilli = 1, in fact IIJII = 1.)In order to estimate let (x,' be the other half of the normal-

ised biothogonal system (x1)?, (x7)7; in other words, let 4 be definedby 4(x1) = Given x = Ax1, choose €, E (—1, 1} such that

eA1 = 1A11 for i = 1,... ,n. Set f = r. Then Ilfil n

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, infact, unnecessary, because the result is a trivial consequence of (1), therelation equivalent to Theorem 13. Indeed, B(X,,,) C nB(X1) and soB(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 beingbest possible. In 1948 Fritz John proved an essentially best-possibleupper bound for d(X, Y) by first bounding the distance of ann-dimensional space from rather than from Before giving thisresult, let us think a little about the Banach—Mazur distance from aEuclidean 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, suchthat

SB( Y) c B(X) c dSB( }')

where d = d(X, Y). Now if Y = = (Rn, II11) then B(Y) is theEuclidean unit ball

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

ECB(X)CdE.

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

Theorem 15. (John's theorem) Let X = (R'1, IFII) be a normed spacewith unit ball B = B(X). Then there is a unique ellipsoid D of minimal(Euclidean) volume containing B. Furthermore,

C BC D.

In particular, d(X,lfl n1"2.

Proof. A simple compactness argument shows that the infimum of thevolumes of ellipsoids containing B is attained. Furthermore, every ellip-soid of minimal volume is centred at 0.

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

and1=1 1=1

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


Denoting by the volume of D, we have vol D' = a,, andso a = 1. Let E be the ellipsoid

E = E

Then

BC DflD' CE

and

(volE)2 — 2 — 2a2 — 2a1

(vol D)2 1

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

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

BC P = E lxii

for some c >For a > b > 0 define an ellipsoid Eab by

Ea,b E W': + b2

so that (VOl D)/(vol Ea,b) = ab"'. If x C B then x C DflP and so

+ b2i=2

x? = + b2

a+b2.

Hence B C and vol Ea,b <vol D whenever

+1,2 1 and >

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


are satisfied for some choice of 0 < b <a. This is a trivial matterwhich, unfortunately, looks a little untidy.

Put 0< e < a = and b = 1—e. Then ab"1 > 1.Also,

2_ + +2 2

C C \ C

=

= <1

if e > 0 is sufficiently small. 0

Theorem 15 can be reformulated in terms of inscribed ellipsoids: thereis a unique ellipsoid D of maximal volume contained in B andD C B C n 112D. Furthermore, if B is not a unit ball but only abounded convex body in P'1 then some translate B' of B satisfiesD 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. Thend(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 bygoing through it is, in fact, close to being best possible: in 1981 theRussian mathematician E. D. Gluskin proved that there is a constantc > 0 such that for every a there are n-dimensional normed spaces Xand Y with d(X, Y) cn. At the time this result was extremely surpris-ing; the proof was based on a probabilistic argument which has becomean important tool in the so-called local theory of Banach spaces, thetheory concerning finite-dimensional spaces.

Exercises

1. Let x1 ,. . . , be non-zero elements in a normed space. ForA = (A1)7 C P'1 set

DAlI0 = max{H± : €, C {—1, 1} (i =


Show thatII Ho is a norm on Show also that for A =

andj = E R' we have

IlAllo = = IIiIJomaxIA1I.

2. Let X1 = (V,,1111) (i = 1,...,n) be normed spaces. For 1

p ooandx = E V = set

=

Show that is a norm on V. Show also that any two of thesenorms are equivalent. [The space (V, fi is usually denoted by

it isadirect sum of3. Let be a symmetric norm on R", i.e.

=with 1e11 = O,1,O,...,O)I = I

and let X1,. .. , X,, be normed spaces. Show that

II(xI,x2,... = 11X2112, . .. ,

is a norm on X = . inducing the product topology onthis space. Check that each X1 is a closed subspace of X and thatX 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 n there are unit vectors

u = p.1x1 (i = 1,... ,k)jE U,

such that U1flU,. = 0 for i i' andk

A,u1 c max 1A111=1

for all Aj,...,Ak ER.Let X be a finite-dimensional real normed space, x0 E X andT E Can be dense in X? [HINT: First prove it forcomplex normed spaces.]Let Y be a proper subspace of a finite-dimensional normed spaceX. Can one always find a vector z0 E X (z0 0) such that


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

7. Let Y be a finite codimensional closed subspace of a normed spaceX and let T E Z), where Z is a normed space. Show that ifthe 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

for all A1,... C Prove that

for all A1,... C R.9. Let X and Y be n-dimensional normed spaces. Show that the

Banach—Mazur distance between X and Y is attained, i.e. there isan operator T0 E Y) such that E X) and

d(.X, Y) = inf{IITIIJIT'lI : T C Y)} = 11T01111T0111.

10. Show that if 2 p then

d(1;,!n

11. Let X be a normed space with unit ball B = B(X) and supposethat 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 thatX is finite-dimensional, and deduce Theorem 10.

12. Let Y be a finite-dimensional subspace of an infinite-dimensionalnormed space X. Show that for every 0 there is an x C S(X)such that (1 + €)Itx + y C Y.

13. A sequence in a normed space is said to be a basic sequenceif 0 for every n, and there is a constant K such that

AkxkH AkxkD

for all I m ii and scalars a,.. • , The minimal K satisfying


this is the basis constant of Deduce from the result in theprevious exercise that every infinite-dimensional normed space con-tains a basic sequence.

14. Let be a basic sequence in a Banach space. Show thatis a Schauder basis of its closed linear span, i.e. every x Ehas a unique representation in the form x = Akxk, with theseries convergent in norm.

15. Deduce from Corollary 9 that a Banach space cannot have a count-ably infinite algebraic basis, i.e. if X = where e1 , e2,...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 operatornorm).

17k. Let X be an infinite-dimensional separable Banach space such thatfor every subspace V C X (dim V = n) there is an operator

E B(Y, such that Q JiJ flK. Prove that there is an

operator T E B(X,12) such that K.18. Let B C be a bounded convex body (i.e. IntB 0). Show

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 finite-codimensional subspace then it is a bounded linear functional.

20. Let X1 = (V1, II•IJ) and X2 = (V2, be subspaces of a normedspace X = (V, with X1 finite-dimensional. Show that ifV = then X is the direct sum of X1 and X2, i.e. the pro-jections X = X1 + X2 X1 (i = 1,2) are continuous.

Notes

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

Excellent accounts of much of the excitement concerning finite-dimensional spaces can be found in the following volumes: V. D. Mi!-man and 0. Schechtman, Asymptotic Theory of Finite Dimensional


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

5. THE BAIRE CATEGORY THEOREM ANDTHE CLOSED-GRAPH THEOREM

In this chapter we shall present several of the most fundamental anduseful results in classical functional analysis: the Baire category theorem,the Banach—Steinhaus theorem, the theorem of the condensation ofsingularities, the open-mapping theorem and the closed-graph theorem.All these results are very closely related; indeed, they are practicallyvariants of each other. Nevertheless, it is rather useful to emphasize themany facets of the same phenomenon, because even a subtle variationin the formulation can make quite a difference in an application.

The simplest member of this group of results is due to Baire and con-cerns 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 ifYl) G 0 whenever G is a non-empty open set. We shall give severalequivalent formulations of the Baire category theorem: here is the first.

Theorem 1. Let G1, G2,... be a sequence of dense open subsets of acomplete metric space X. Then G = is dense in X.

Proof. As in a normed space, for x E X and r> 0 we set

Dr(X) = {y X: d(x,y) <r} and = (ye X: d(x,y) r}.

Although need not be Br(X), we have D,.(x) C Br(X) C D,÷E(x) forall 0. Since is a dense open subset of X, for all x E X andr> 0 the open ball D,(X) meets in a non-empty open set so thereare y E X and s > 0 such that C C Ii Dr(X). This is theproperty 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 positivenumbers r1,r2,... as follows. Choose x1 E X and 0 < r1 < 1 such that

75

76 Chapter 5: The Baire category theorem

B,1(x1) C G1flD,(x). Next choose x2 E X and 0 < r2 < 4 such thatBr2(X2) C then choose x3 E X and 0 < r3 < 4 such thatBrj(X3) C etc. By construction, —* 0 and B,1(XI) D

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

x0 E D,(X)flG. 0

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

Theorem 1'. If a complete metric space is the union of countably manyclosed sets then at least one of the closed sets has non-empty interior.

Proof. Let X = where X is a complete metric space and eachF,, is closed. Setting G,, = X\F, we find that (3,, = 0 so that atleast one of the open sets G,, is not dense in X: G,, X. ButIntF,,=X\G,,. 0

Strictly speaking, Theorem 1' says precisely that the set Gin Theorem1 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 conse-quence of Theorem 1'.

A subset Y of a topological space X is said to be nowhere dense in Xif the closure of Y has empty interior: hit Y = 0. Note that a set Y isnowhere dense if and gnly if its closure, Y, is nowhere dense. A subsetZ C X is meagre in X or of the first category in X if it is the union ofcountably many nowhere-dense sets. Clearly, the union of countablymany meagre sets is meagre. A subset of X is of the second category inX 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 0for some n.

Let us use this terminology to give another reformulation of the Bairecategory theorem.

Theorem The complement of a meagre subset of a complete metricspace is dense. In particular, a complete metric space is of the secondcategory.

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

Chapter 5: The Baire category theorem 77

Proof. Suppose Y = 1',, C X with X complete and eachnowhere dense in X. Then 1',, is nowhere dense in X so G,, = is

a dense open set and therefore G = is dense. Clearly

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

meagre. 0

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

The intuitive meaning of a set of the second category is that it is alarge 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' israther 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 thespace, while being of the second category is an intrinsic property of ametric space. For example, given a collection of convex bodies inloosely speaking one may say that 'almost all sets in have property P',meaning that in the natural (Hausdorif) metric on the set of convexbodies having P is of the second category.

It is perhaps worth pointing out that the two notions are only similar butnot 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, andone 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 continuousfunctions. Let fj,f2,...: [0,1] R be continuous functions such thatfor each t E [0, 1] the sequence is bounded. Then there is aninterval [a,bJ C [0,1] (a <b) on which the sequence fl,f2, ... is uni-formly bounded.

Let us see that the Baire category theorem implies the followingextension of this result.

Theorem 2. (Principle of uniform boundedness) Let U be a set of thesecond category in a metric . space X and let be a family of continuousfunctions f: X —' R such that {f(u): f E is bounded for everyii E U. Then the elements of are uniformly bounded in some ballB,(x0) (x0 X; r > 0), i.e. If(x)I n holds for some n and all f Eand x E B,(xo). In particular, the conclusion holds if X is complete and{f(x): f E is bounded for every x X.


Proof. For n 1 let

flf'([—n,n]).I

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

n1so by the definition of a set of the second category, mt 0 for somen.

The second assertion is immediate from Theorem 1". 0

The next result is only a little more than a reformulation of Theorem2 in the setting of linear maps.

Theorem 3. (Banach—Steinhaus theorem) Let X and Y be normedspaces, 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 <

for all u E U. Then JITII N for some N and all T E In particular,the conclusion holds if X is complete and T

T E xby is a ball E X; r > 0) such

that IITxII n for some n and all T E and x E B,.(x0). As in theproof of Theorem 2.2, this implies that E N = fir for all T E

Indeed, let T E and x E B(X). Then 10 + rx and x0 — rx are inBr(XO) and so

IITxO = = N

and therefore 11J N for all T E 0

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

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

urn = for all


Then there is a set U C X of the second category in X such that foru E U we have

IIT,,,,,(u)II = for all n.

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

Hence so is V = U..1 and thus by Theorem 1" the set U = X\V is

of the second category in X. 0

The next major result, the open-mapping theorem, is a consequenceof the Baire category theorem and the important lemma below. Recallthat for a normed space X we write D(X) for the open unit ball andB(X) for the closed unit ball; furthermore, we use the notationD,(X) = rD(X) and = rB(X).

Lemma 5. Suppose X and Y are normed spaces, X is complete,T Y) and T(Dr(X)) 3 D3(Y). Then T(D,(X)) 3

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 thatIIzII < 1—S < 1 and set y = z/(1 —8). We shall show thaty 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 suc-cessively Yi Y such that y,, Yn—i E S"'A and fly,, —ylI <8".Having chosen Yo'••• we can find a suitable y,, sincey E and the set S"'A is dense in

By the definition of A, there exists a sequence C X such thatTx,, = y,, and Putting

x = x,,

we have

Dxli ilx,,il <(1 —Sr' and Tx=

=

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

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


Proof. Set G = T(D(X)). Since T is a linear map, it suffices to showthat G contains a neighbourhood of 0 in Y.

Note that = rG and the closure of rG is rG. SinceY = T(X) = nG, by the Baire category theorem we see thatmt nG 0 for some n and so mt G 0. The set is convex andsymmetric about 0 so mt G 0 implies that G 3 for somer > 0. Indeed, if D(x0, r) C G then D(—xo, r) C G and so the convexityof implies that D(0, r) = DT(Y) C G. But then, by Lemma 5,G = T(D(X)) 3 D,.(Y). 0

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

Theorem 7. (Inverse-mapping theorem) If T is a one-to-one boundedlinear 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 toBy 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 andT 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 mapinto 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 andV endowed with the norm II(x,y)II = Dxli + flyll. This norm induces theproduct topology on Xx V and so, by assumption, 1(T) is a closed sub-set of Z. In fact, Z is easily seen to be complete, so 1(T) is a closedsubspace of the Banach space Z and therefore it is itself a Banachspace. The linear map U: 1(T) —' X, given by U((x,y)) = x, is anorm-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 itmuch easier to prove that an operator is continuous. When is a linearoperator T: X V continuous? If x1 , x2,... E X and —' x0 imply


that —' Tx0. Now if X and Y are Banach spaces, the closed-graphtheorem tells us that in order to prove the continuity of a linear opera-tor T: X —, Y, it suffices to show that if x,, —. and Tx,, then

Tx0, i.e. Tx0 = Yo — a considerable gain!

Exercises

1. Let f: R4 R be a continuous function such that =0 for all x > 0. Show that = 0.

Let f: R R be infinitely differentiable. Suppose for everyx E R there is a such that = 0 for all k Prove thatf is a polynomial.

3. Let K = [0,1], X = and, for n 1, set F,, =

{f E X: 3 t E K such that+ h) — f(t)

n V h with t + h E

Prove that F,, is closed and nowhere dense in CR(K). Deduce thatthe set of continuous nowhere-differentiable real-valued functionson [0,11 is dense in

4. Prove that if a vector space is a Banach space with respect to twonorms then the topologies induced by the norms are eitherequivalent or incomparable (i.e. neither is stronger than the other).

5. Let V be a vector space with algebraic basis e1 , e2,... (sodim V = and every v E V is a unique linear combination of thee,) and let II be a norm on V. Show that is incomplete.

6. LetXbeanormed space andSCX. Showthatif{f(x):xES}isbounded for every linear functional f E X then the set S isbounded: lixil K for some K and all x E S. (Using fancy termi-nology 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 func-tional f E V' which is continuous in one of the norms and discon-tinuous in the other.

8. Let X be a closed subspace of L1(0, 2). Suppose for everyf E L1(0, 1) there is an F C X whose restriction to (0, 1) is f.Show that there is a constant c such that our function F can alwaysbe chosen to satisfy IIF1I clifli.

9. Let 1 p,q and let A = (a11)' be a scalar matrix. Supposefor every x = (x1)° the series is convergent for


every 1, andy = Ely, where y, =Show that the

map A: I,, —p 'q' defined by x y, isa bounded linear map.10. Let [0,1] (n = 1,2,...) be uniformly bounded continu-

ous functions such that

j dx c

for some c > 0. Suppose 0, (n = 1,2,...) and

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

for everyxE [0,1]. Prove that

cnn1

11. For two sequences of scalars x = and y = (y1)° set

(x,y) = x•y,.i=1

Let I <p,q < be conjugate indices and let = E(n = 1,2,...). Show that 0 for every y ifis norm bounded (i.e. K for some K and all n) and

0 for every i. (This is a characterization of sequences intending weakly to 0; cf. Exercise 6.)

12. Let = E (n = 1,2,...). Show that — 0 forevery y c0 iff is bounded and —* 0 for every i. (Simi-larly to Exercise 11, this is a characterization of sequences intending weakly to 0.)

13. Show that for I p < the La-norm

I r' \1/P

11111, = (j If(t) "dt'U

on C[0, ills dominated by the uniform norm Ilfil = IfO) I

and deduce that C[0, 1] is incomplete in the norm14. 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 directsum of the subspaces KerP = and ImP = PX if and only ifP is bounded.

15. Let X and Y be normed spaces and T E Y). Show that thegraph of T is closed 1ff whenever x,, 0 and y then y = 0.


As we shall see in the following exercises, the results of this chapterare very useful in proving the fundamental properties of Schauder bases.Recall from Chapter 2 that a sequence is a Schauder basis or sim-ply a basis of a Banach space X if every x X has a unique representa-tion in the form with the series convergent in norm.

16. Let be a basis of a Banach space X. Define P, by

=

and for x E X set Mxlii = Prove that is a norm onX and X that is complete in this norm as well.

17. Let X, and be as in the previous exercise. Prove thatand [The number is called

the basis constant of cf. Exercise 4.13.]18. Let x1 , x2,... be non-zero vectors in a Banach space X, with

) = X. Prove that is a basis of X if and only if thereis a constant K such that

A,xaO

for all 1 m <n and scalars A1,...19. Define a sequence of functions Xn: [0, 1] [—1, 1] as follows.

Set Xi(t) 1 and for n 2 define k 0 by 2k Set,= and define

1

if <1+1,

( 0 otherwise.

The sequence is called the Haar system. Prove that theHaar system is a basis in 1) for I p

20. Define a sequence of continuous functions ç,1: [0, 1] [0, 1] as

follows. Set 1 and for n 2 put

= Ldu,

where X* is the k th Haar function, defined in the previous exer-cise. The sequence is called the Schauder system. Provethat the Schauder system is a basis in 1) with the uniformnorm. Show also that the basis constant is 1.

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 infinitesequence of linearly independent vectors in a normed space cannotbe 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 clas-sic, cited in Chapter 2. The original paper of R. Baire is Sur Ia conver-gence de variables réelles, Annali Mat. Pura e AppI., 3 (1899), 1—122;

Osgood's theorem is in W. F. Osgood, Non-uniform convergence andthe 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, FundamentaMath., 9 (1927), 51—7.

6. CONTINUOUS FUNCTIONS ON COMPACTSPACES AND THE

STONE-WEIERSTRASS THEOREM

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

11111 = sup = maxlf(x)I.z€K xEK

In fact, as we shall see in Chapter 8, every Banach space is a closedsubspace of C(K) for some compact Hausdorff space K, although thisdoes not really help in the study of a general Banach space.

First we shall show that C(K) is large enough, namely that it containssufficiently many functions; for example, every continuous function on adosed subset of K can be extended to a continuous function on thewhole of K without increasing its norm. This result, which is a specialcase of the Tietze—Urysohn extension theorem, is thus reminiscent ofthe Hahn—Banach theorem: it ensures that the space C(K) is large, justas the Hahn—Banach theorem ensured that the dual space of a normedlinear space was large. It is worth emphasizing that the existence of agood stock of continuous functions on a topological space X cannot betaken for granted, not even when X seems to be very pleasant. Forexample, it can happen that X is a countable Hausdorff space and stillevery continuous function on X is constant (see Exercise 19).

In the second part of the chapter we shall show that C(K) is not toolarge in the sense that it is the norm closure (i.e. the closure in thetopology induced by the norm) of a good many 'small' subspaces offunctions.

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

86 Chapter 6: Continuous functions on compact spaces

A topological space is said to be normal if every pair of disjointclosed sets can be separated by disjoint open sets: if A and B are dis-joint closed sets then there are disjoint open sets U and V such thatA 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 anopen 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 Haus-dorif space. For a E A and b E B, let U(a, b) and V(a, b) be disjointopen neighbourhoods of a and b. Let us fix a point a A. Clearly,B C Uh E B V(a, b) and so, as B is compact, B C V(a, b1) forsome Set

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 aand V(a) contains B.

Now note that U0 A U(a) is an open cover of A and so, as A is

compact, A C U(a,) for some a1,. ..,am E A. Let

U = U U(a1) and V = fl 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 assertionfollows from (a). 0

If A and B arc disjoint closed sets in a metric space X then there is acontinuous 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} Bi xEB

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

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

such that A C and B C f'(l).

Chapter 6: Continuous functions on compact spaces 87

Proof. Let q0 = 0, q1 = I and let q2,q3.... be an enumeration of therationals strictly between 0 and 1. Set LI0 = 0. A0 = A, U1 = )AB andA1 = X. Let us construct a sequence of closed sets A0, A1 ,... and a

sequence of open sets U1 ,.. such that U, C A, and if q <q1 then14,C U1. Suppose we have found A0,...,A,,1 and U0,...,What shall we choose for A,, and U,,? Let be the maximal q1(I < n) which is less than q,, and let q1 be the minimal q1 (1 < n) whichis greater than q,. Then Ak C U, and so, as X is normal, one can finda 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 asuitable continuous function f: for x E X set

f(x) = : x E A,,}.

Clearly 0 f(x) I and f is 0 on A and I on B. All we have to checkis that f is continuous. Suppose then that 0 <f(x0). Thenx0 Ak and so X'\Ak is a neighbourhood of x0; for x X\Ak we havef(x) qk• Similarly, if < q, I then xEU,. So U, is a neighbourhoodof x0 and for x E U, we have f(x) q,. o

We are ready to present one of the main results of this section. ByUrysohn's lemma, if A and B are disjoint closed (possibly empty) sub-sets 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 intothe two-point set 1} then h has an extension to a continuous func-tion g: X—' [—1,1]. The Tietze—Urysohn extension theorem belowclaims that every continuous function 1: C [—1, 1] has such an exten-sion. This result is easily proved by using Urysohn's lemma to provideus with continuous functions on X whose restrictions to C are better andbetter 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 andG01B0 = Then f' = f—G0 IC maps C into [—I, fl — a very goodstart indeed. Next, set

A1 = {x C C: f'(x) — B1 = {x C C: f'(x) 0)

and take in a continuous function H1 0] such that HLIAI =and H21B1 = 0. Then F1 = G0 + is a good approximation of f on C,namely f— FIIC maps C into [— and Continuing in thisway, we get continuous functions F2, F3, ... with 2" and


I — mapping C into Thus F = F, will havethe required properties.

For the sake of completeness, in the proof below we shall present astreamlined 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 Continuousfunction. 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 functionson the whole of X whose restrictions to C are better and better approxi-mations of f. To be precise, set fo =

A0 = {x E C: f0(x) — 4} and 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 con-structed a continuous function

: C—p

define

= {x E C: and = {x E C:

By Urysohn's lemma there is a continuous function

r I 2n 12

such that

= and = i(2)n.

Set

and note that is a continuous map C intoLet f1 ,... and F0. F1 .... be the sequences of functions constructed

in this way. Then for all x X and n 0 and soF(x) = is continuous, being a uniform limit of continuousfunctions, and

IF(x)I 4 = 1ii=0


Furthermore, for x E C and n 0 we have

- k=O=

<(4)fl+1

andsof(x)=F(x). 0

- It may be worth noting that it is trivial to ensure that F(x)I

1: if

F(x) is any continuous extension of f then

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. Thisis a Banach space in the uniform norm Ilfil = sup{If(x) x E K}. Theresults 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 afunction f E C(K) such that f(x) # f(y). Also, if A and B are disjointclosed subsets of K then for some f E C(K) we have f(a) = 0 andf(b)= 1 for all aEA and bEB. Finally, if ACK is closed andf 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 anothersource of functions that can be guaranteed to belong to C(K). Thisleads us to consider relatively compact subsets of C(K), that is subsets inwhich every sequence has a subsequence convergent to some element ofC(K).

Recall that a topological space is compact if every open cover has afinite subcover, it is countably compact if every countable open coverhas a finite subcover and it is sequentially compact if every sequence hasa convergent subsequence. Given a metric space X and N C X, the setN is an €-net if for every x E X there is a pointy EN with d(x,y) <€.The €-nets one tends to consider are finite. A metric space is totallybounded or relatively compact if it contains a finite €-net for everye > 0, or, equivalently, if every sequence has a Cauchy subsequence.For a metric space, the properties of being compact, countably compactand sequentially compact coincide; furthermore, a metric space is com-pact iff it is complete and relatively compact. In particular, a subset ofa complete metric space is relatively compact if its closure is compact; aclosed subset of a complete metric space is compact 1ff it is relativelycompact.


In order to characterize the relatively compact subsets of the completemetric space C(K), let us introduce some more terminology. A setS C C(K) is uniformly bounded if it is bounded in the supremum norm,i.e. if Ilfil N for some N and all f C S. We say that S is equicontinu-ous at a point x C K if for every 0 there is a neighbourhood U of xsuch that if y E and f E S then 11(x) —f(y)I < €. Furthermore, S isequicontinuous if it is equicontinuous at every point.

Theorem 4. (Arzelà—Ascoli theorem) For a compact space K, a setS C C(K) is relatively compact if and only if it is uniformly boundedand equicontinuous.

Proof. (a) Suppose that S is a totally bounded subset of C(K). Letx E K and 0. The set S contains a finite c-net: there is a finite set{f1 f,,} CS such that if fE S then hf—fill <c. Then, trivially,11111 <c + and so S is uniformly bounded. Furthermore, asf, is continuous, for every x C K there is a neighbourhood U1 of x suchthat Ifi(x)—fi(y)I <c for yE U,. Then U = fl'1 U is a neighbour-hood of x showing that S is equicontinuous at x. Indeed, for fE S let I

(1 i n) be such that 111—1,11 <e. Then for y E U we have

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 be an open neighbourhood of x such thatIf(x)—f(y)I <€ if fE S and yE Then K = LJEK let

be 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 afinite-dimensional vector space are equivalent, we could have chosenany norm on C".) A bounded set in 1 is totally bounded because abounded closed set in 1 is compact; therefore R is totally bounded sothere are functions ,. . . , f,,,, E S such that the set

1 i m}

is 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

all j (I n). Let now xE K be an arbitrary point. Then xEfor 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


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

Theorem 5. Let K be a compact topological space and letCR(K). Suppose that for every x C K the sequence

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

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

0 <€.

Note that if n and y C U,, then

0 fo(Y) Jn(Y) fo(y) <€.

Since is certainly a cover of K and K is compact, K is coveredby a finite number of these sets: K = for some pointsIl,... ,Xk. The last inequality implies that if n n0 = max{nR1,. . .

then IJo(y)fn(y)I <e for ally, and so 0

It is worth noting that all three conditions are needed in Theorem 5,namely that K is compact, the sequence of continuous functionsis pointwise increasing and the pointwise limit, Jo is a continuous func-tion (see Exercise 13).

Now let us turn to one of the most fundamental results in functionalanalysis, 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 sub-spaces, more precisely, it is the uniform closure of a good many subalge-bras. Recall that a topological space is said to be locally compact ifevery point has a compact neighbourhood. For a locally compact Haus-dorif 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, isthe closure of the set {x: f(x) 0).


Note that if L is compact then C(L) is just the set of all continuousfunctions; similarly. is the set of all continuous functions withcompact support and C0(L) is the set of all continuous function f on Lfor which {x E L: If(x)l e} is compact for every e > 0.

If f and g are functions on the same space, define Af. f+g and fg bypointwise operations. With these operations, C(L) is a commutativealgebra. This algebra has a unit, the identically I function. Bydefinition, C C0(L) C C(L) and CC(L) and C0(L) are subalgebrasof C(L). We know that the vector space C(L) is a Banach space withthe uniform (supremum) norm:

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

Theorem 6. Let L be a locally compact Hausdorff space. The functionis a norm on C(L) and the space C(L) is complete in this norm;

C0(L) is a closed subspace of C(L) and is a dense subspace ofC0(L). Furthermore, forf,g E C(L) one has

Ilfil (1)

and I. the identically I function, has norm 1.

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

some x1 ,. . . E K(,. Setting K1 = we find that K1 is acompact set such that K0 C tnt K1.

By Urysohn's lemma (Theorem 2) there is a continuous functionh: K1 [0, 1] that is I on and 0 on K1\lnt K1. Extend h to a func-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 thenorm of e is I then it is called a unital normed algebra. If the norm iscomplete then the algebra is a Banach algebra. We saw in Chapter 2that the set of bounded linear operators on a normed space X, isa unital normed algebra and if X is complete then is a unital


Banach algebra. The trivial part of Theorem 6 shows that C(L) is acommutative unital Banach algebra, C0(L) is a commutative Banachalgebra and is a commutative normed algebra.

The spaces CR(L), and are defined analogously; theyconsist of the appropriate continuous real-valued functions. Thesespaces are not only real normed algebras but they are also lattices underthe natural lattice operations fvg and fAg.

Given real-valued functions f and g on a set S, define new functionsfvg and fAg on S by setting, for x E S,

(fvg)(x) = max{f(x),g(x)} and (fAg)(x) = min{f(x),g(x)}.

We call fvg the join off and g and fAg the meet off and g. The joinand the meet are the lattice operations.

Of course, the join of two functions is just their maximum and themeet is just their minimum. It is clear that CR(L), andare all closed under the lattice operations.

For a set A of bounded functions on a set S. the uniform closure A ofA is the closure of A in the uniform topology on the space ofbounded functions on S with the uniform norm (Example 2.1 (ii)). Thusa function f on S belongs to A if for every 0 there exists a g E Asuch that If(x) <e for all x E S.

In this chapter we shall study the uniformly closed subalgebras ofC(L), C0(L), CR(L) and so we are particularly interested in uni-form approximations by elements from various subsets of these algebras.As the next result shows, the lattice operations are very useful when welook for such approximations.

Theorem 7. Let K be a compact Hausdorif space and let A be a subsetof which is closed under the lattice operations. Then A, the uni-form closure of A, is precisely the set of those continuous (real-valued)functions on K that can be arbitrarily approximated at every pair ofpoints by functions from A.

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

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

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


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

<€ and <E•

Put = {z E K: <zf(z) +€}. Then is an open setcontaining x and y. Fixing x, we find that K = and so afinite number of the sets also cover K, say K = By

assumption, the minimum of the corresponding functions• fr,,,,,,

namely = also belongs to A. Clearly

>f(x)—E and <f(y)+€ for allyE K.

Let = {y E K: f(y) —e}. Then is open and K =

Kso that, as K is compact, K = for some

x1, • . ,X,.•,E K. Let =L1vL2v" be the maximum of the

corresponding functions. By assumption, E A, and

f(x)—€<f((x)<f(x)+e forallxEK.

As there is such an fE for every 0, f E A. 0

Lemma 8. A uniformly closed subalgebra A of gb(S), the space ofbounded real-valued functions on a set S is also closed under the latticeoperations.

Proof. Since (fvg)(x) = max{f(x),g(x)} =andfAg = f+g—fvg, it suffices to show that A iff E A. Furth-ermore, as for n 1 we have nfl = nlfl, andf A holds if nf E A,it suffices to show that if f A and 11111 = supx€s If(x) I < 1 thenflEA.

Let then f E A, flffl < 1 and e > 0. Let Ck(t—3)k be the Tay-

lor series for about z = 3. As this series converges uniformlyon [0, 1], there is a polynomial

P(:)k=O

Ck(t—3)k

such that

for all 0 t 1.

Hence

IP(s2) —(s2+€2)"2J < for all —1 s 1


and so

IP(s2) — Is II < 2E for alt —1 s 1.

Set Q(s) = P(s2) — P(O). Then Q is a real polynomial with constantterm 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€ foralixEKand so

IIQ(f) —fillshowing that IfI can be approximated by the function Q(f) E A within4€. Since A is uniformly closed, Ill E A. 0

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 iffor 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 spaceand let A be an algebra of real-valued continuous functions on K thatseparates the points of K. Then A, the uniform closure of A, is eitherCR(K) or the algebra of all continuous real-valued functions vanishing ata single point x. E K.

Proof. (a) Suppose first that there is no point E K such thatf E A. It is easily checked that then for all x y

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

x and y by elements of A. In fact, we can do rather better than onlyapproximate: since (J(x), f(y)) and (f2(x) , f2(y)) are linearly indepen-dent vectors in fl2, some linear combination of them is (g(x),g(y)). Sofor 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. Wehave to show that if g E CR(K) and g

the constant functions to A, i.e. let B C CR(K) be thealgebra 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

C E R, such that <€. As = = Owe have PcI <eand so PIf —gil <2€. Hence g E A. 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 compacuficazionof L is the topological space on = where and a set Scis open iff(a) SflL is open in Land (b) E S then LIS is compact. It is easilychecked that is a compact Hausdorif space inducing the original topologyon L. Set

= {f E = O}.

Then C,?(L) is a closed subalgebra of CR(L). Furthermore, if L is alocally compact Hausdorif space then f fl L is an isometric isomor-phism between and This is why the elements ofare 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 offunctions 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 andg(y) 0. Note that by Theorem 9 if a subalgebra A of CR(K) separatesthe points of K strongly then A = CR(K).

Theorem 9'. Let L be a locally compact Hausdorif space and let A be asubalgebra of strongly separating the points of L. Then A isdense in i.e. A = 0

It is obvious that in Theorem 9' we cannot replace by CR(L)because itself is a uniformly closed subalgebra of CR(L). Furth-ermore, we cannot replace by C0(L) either. This can be seenfrom the following example. Let 4 = {z E C:

I1} be the closed

unit disc in C and let A consist of those continuous functions on 4 thatare analytic in the interior of 4. Then A is a closed subalgebra ofC0(4) = C(4), it strongly separates the points but does not contain thefunctionf(z) = C0(4).

The example above is, in fact, not an ad hoc example but one that goesto the heart of the matter: what we lack is complex conjugation. Thisleads us to a form of the Stone—Weierstrass theorem for complex func-tions, which is an easy consequence of the second variant, Theorem 9'.

Theorem 10. (Stone—Weierstrass theorem for complex functions) Let Lbe a locally compact Hausdorif space. Suppose A is a complex


subalgebra of C0(L) which is closed under complex conjugation andstrongly separates the points of L. Then A = C0(L).

Proof. Let be the set of real-valued functions in A. Then AR is a realsubalgebra of A and it is also a subalgebra of As AR contains(f+ 1)/2 and (f — f) /2i for every f A, it satisfies the conditions ofTheorem 9'. Therefore, C0R C A and so

1C0R(L) cA and C0(L) = cA. 0

Let us note some immediate consequences of the Stone—Weierstrass

theorem; the first is the original theorem due to Weierstrass.

Corollary 11. Every continuous real-valued function on a boundedclosed subset of IR?* can be uniformly approximated by polynomials. 0

Corollary 12. Every continuous complex-valued function on the circleT = {z E C: IzI = 1} can be uniformly approximated by trigonometricpolynomials ok = -it

Corollary 13. Let K and L be compact Hausdorff spaces and endowKxL with the product topology. Then every continuous functionf: 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 i n). 0

The Stone—Weierstrass theorem is very useful in showing that various

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

consisting of continuous functions is dense. Since every continuous real-valued function on [0,11 can be uniformly approximated by polynomi-als, and the uniform norm dominates the La-norm, the space of polyno-mials is also dense in 1].

Exercises

1. Show that a metric space is compact 1ff it is countably compact ifit is sequentially compact.

2. Show that a metric space is compact if and only if it is totallybounded and complete.

3. Show that a subset of a complete metric space is totally bounded ifand only if its closure is compact.


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

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

6. Let CR(X) be the normed space of bounded continuous real-valued functions on a topological space X, with the uniform(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 arenon-empty disjoint closed subsets of X and a, b E R then there is afunction f E V such that f(X) C [a, b], fI A = a and fIB = b.Prove that V is dense in Ca(X).

8. Let K be a compact metric space and S C C(K). Show that S isequicontinuous if for every e > 0 there is a 8 > 0 such thatIf(x) — f(y)

I<e whenever d(x, y) <8. Is this true for every

metric space K?9. Prove the following complex version of the Tietze—Urysohn exten-

sion theorem. Let A be a closed subset of a normal topologicalspace X and let f: A —* C be a continuous function such thatIlfilA = SUPXEA 1(x) I = 1. Show that there is a continuous func-tion F: C such that flFflx = = 1.

10. Let L be a locally compact Hausdorff space. Show that a setS C is totally bounded if it is bounded, equicontinuous andfor 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.

11. Let B, = {zE C : Izl D = {zE C : < 1) and let F be aset of functions analytic on D such that If(z)l M, if IzI = r < 1.Show that for r < 1 the set FIB, {IIB,: f E F} is a relativelycompact subset of C(B,). Deduce that every sequence C Fcontains a subsequence which is uniformly convergent onevery 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 uni-formly bounded analytic functions. Prove that there is a subse-quence which is uniformly convergent on every compactsubset of U.

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


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 foreach point (x0,y0) E G, at least one solution of

y'(x) = f(x,y)

passes through (x0,y0). [HINT: Let V be a closed neighbourhoodof (x0,y0) and set K = (x,y) 1'). Choose a <x0 < b such that

f(x,y): x [a,b], x YYo <K1 C U.x—xo

The aim is to show that our differential equation has a solutionthrough (x0,y0), defined on [a, b]. Find such a solution by consid-ering piecewise linear approximations, say with division pointsx0±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 canyou say about K?

17k. We say that e > 0 is a Lebesgue number of a cover U,.EF U,. of ametric space if, for every point x, the ball B(x, €) is contained insome U,.. Show that a metric space (X, d) is compact 1ff for everymetric equivalent to d every open cover has a (positive) Lebesguenumber.

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 uniformmetric

d(f,g) = sup d(f(x),g(x)) = maxd(J(x),g(x)).xEK xEK

Show that a subset S of C(K,L) is totally bounded if and only if it isequicontinuous (i.e. for all x E K and E > 0 there is a S > 0 suchthat if d(x,y) <8 then d(f(x),f(y)) <€ for every f E S) and theset {f(x): f E S} is a totally bounded subset of L for every x K.

Construct a countable Hausdorif space on which every continuousfunction is constant. The first such example was given by Urysohnin 1925.

20. Let K be a compact Haudsorif space. What are the maximalideals of CR(K)? And the maximal closed subalgebras?


21. Let K be a compact Hausdorif space. For a continuous surjectionof K onto a compact Hausdorff space K', let

be given by ç(f)(x) = (fco)(x) = Show thatis a closed unital subalgebra of CR(K) and every closed unitalsubalgebra of CR(K) can be obtained in this way. Give a similardescription of the closed subalgebras of CR(K).

22. Let X be a normal topological space and let U1,. . . be opensets covering X. Prove that there are continuous functionsfe,. : X— [0,11 such that fk(x) = 1 for every x and

= 0 for x Uk (k = 1,. . . ,n). (The functions f1,. . . , j, are

said to form a partition of unity subordinate to the cover

Notes

The relevant results of Urysohn and Tietze are in P. Urysohn, (..Yber dieMdchtigkeit der zusammenhangenden Mengen, Math. Ann., 94 (1925),262—95 and H. Tietze, Uber Funkrionen, die auf einer abgeschlossenenMenge stetig sind, J. für die reine und angewandte Mathematik, 145(1915), 9—14. Heinrich Tietze proved a special case of the extensiontheorem, while Paul Urysohn (who was Russian but tended to publish inGerman) proved Theorems 2 and 3 in the generality we stated them.Various special cases of the extension theorem were proved by a goodmany people, including Lebesgue, Brouwer, de Ia Vallée Poussin, Bohrand Hausdorif. For the Arzelà—Ascoli theorem see C. Arzelã, Sulkserie di funzioni (paiie prima), Memorie Accad. Sci. Bologna, 8 (1900),131—86. Of course, there are several excellent books on general topol-ogy; J. L. Kelley, General Topology, Van Nostrand, Princeton, NewJersey, 1963, xiv and 298 pp., is particularly recommended.

The original version of Theorem 9 is in K. T. Weierstrass, Uber dieanalytische Darsteilbarkeit sogenannier willkürlicher Funktionen reelerArgumente, S.-B. Deutsch Akad. Wiss. Berlin, KI. Math. Phys. Tech.,1885, 633—39 and 789—805. The Stone—Weierstrass theorem itself is inM. H. Stone, Applications of the theory of Boolean rings to generaltopology, Trans. Amer. Math. Soc., 41(1937), 375—81; an exposition ofthe field can be found in M. H. Stone, The generalized Weierstrassapproximation 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.Given a set Sand a functionf: S, a point x ES satisfying 1(x) = xis said to be a fixed point of 1. There is a large and very importantbody of 'fixed-point theorems' in analysis, that is, results claiming thatevery function satisfying certain conditions has a fixed point. Theoremsof this kind often enable us to solve equations satisfying rather weakconditions. The aim of this section is to present the most fundamentalof these fixed-point theorems and some of its great many applications.In Chapter 15 we shall return to the topic to prove some more sophisti-cated results.

A map f: X —' X of a metric space into itself is said to be a contrac-tion if

d(J(x),f(y)) kd(x,y) (1)

for some k < 1 and all x,y E X. One also calls f a contraction withconstant k. It is immediate from the definition that every contractionmap is continuous; in fact, it is uniformly continuous. Although weshall not make much use of this terminology, a function f betweenmetric 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 isa contraction map if it satisfies a Lipschitz condition with constant lessthan 1.

The result below is usually referred to as Banach's fixed-point theoremor the contraction-mapping theorem.

Theorem 1. Let f: X —+ X be a contraction of a (non-empty) completemetric 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 apoint 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. (2)

Indeed, (2) is trivially true for n = 1; assuming that it holds for n, wehave

=

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 pointx C X. Since f is continuous, the sequence converges to f(x).But = converges to x and so f(x) = x. The uniquenessof the fixed point is even simpler: if f(x) = x and f(y) = y thend(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-

ful 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 x0and setting = for n 1. Let us note the following slightextension of Theorem 1.

Theorem 2. Let X be a complete metric space and let f: X X be suchthat f" is a contraction map for some n 1. Then f has a unique fixed

Proof. We know from Theorem 1 that has a unique fixed point x,say. Since

f'1(f(x)) = f(f(x)) = f(x),

f(x) is also a fixed point of f and so f(x) = x. As a fixed point of f isalso a fixed point of f's, the map f does have a unique fixed point. 0

The very simple results above and their proofs have a large number ofapplications. In the rest of this chapter we shall be concerned with two

Chapter 7: The contraction-mapping theorem

directions: first we shall study maps between Banach spaces and then weshall present some applications to integral equations.

The proof of Theorem 1, sometimes called the method of successiveapproximations, is very important in the study of maps between Banachspaces. Here is a standard example concerning functions of two van-ables which are Lipschitz in their second variable. Recall that for anormed space X we write Dr(X) = {x E X: lixil <r} for the open ballof centre 0 and radius r.

Theorem 3. Let X and Y be Banach spaces and let

F: Dr(X))<Ds(Y) 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 < 1.

Then there is a unique map y: DS(Y) such that

y(x) = F(x,y(x)) (5)

for every x E D,(X), and this map y is continuous.

Proof. For x E D,(X), define yo(x) = 0, = F(x,yo(x)) = F(x,0)and d = 11y1(x) — y0(x) II = IIyi(x) Note that, by (4), d <s(i — k).

Furthermore, if n 1 and E DS(Y) then set

=

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

=


In turn, this implies that (7) holds for n:

ILvo(x)II+ i—O

Uy1+i(x)—y1(x)II

< 5,

and so, in particular, EAs F is a continuous function, each D,(X) D5(Y) is continu-

ous. Inequalities (6) and (7) imply that the sequence is uniformlyconvergent 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 = IIF(x,y(x)) — F(x,yo(x))II —yo(x)

and so y(x) = y0(x). 0

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) X be a con-traction with constant k < 1 such that

< 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 off to U is a homeomorphism of U onto D,(X).

Proof. Define F: D,(X) X X by

F(x,y) = x—e(y).

Then F satisfies the conditions of Theorem 3. Indeed, (3) holds byassumption; furthermore, if x E thefl

IIF(x,0)D = 11xe(0)II < r+r = s(1—k),

and so (4) also holds. Hence, by Theorem 3, there is a unique function

Chapter 7: The contraction-mapping theorem 105

g: D,.(X) 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 haveto check that U is an open neighbourhood of 0 and f: U —+ D,(X) is ahomeomorphism. Clearly,f is a one-to-one map of U onto D,(X). Since g isunique, the continuity off implies that U =7 '(D,(X)) is open. As g is alsocontinuous,fis indeed a homeomorphism. Finally,f(0) = e (0) E D,.(X) and soOEMU. 0

From Theorem 4 it is a small step to the inverse-function theorem forBanach 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

whenever 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 differenti-

able at x0 E U if there is a linear operator T E Y) such that

f(x0+h) = f(x0)+Th+o(h).

The operator T is the derivative of f at x0: in notation, f'(x0) = T.Thus f is differentiable at x0 with derivative T E Y) 1ff for every

0 there is a 8 > 0 such that

Ilf(xo+h)—f(xo)--ThH eIIhII

whenever lihIl <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 follow-ing simple analogue of the mean-value theorem.

Lemma 5. Let X and Y be Banach spaces, U C X an open convex setand f: U Y a differentiable function, with


Ilf'(x)II 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 sequencesC U such that x0 Yo' and for a 0 either = and

+ = + or + = + y,,) and + = and

k' k and all a 0. But then there is a z0 E U such thatz0, — z0 and

IIXn YnII = lix,, zoil + iIzo—ynli,

as the segments [x,, , y4,,] are 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 suitablecircumstances, a function f is a homeomorphism in some neighbour-hood.

Theorem 6. Let X and Y be Banach spaces, let U be an open neigh-bourhood of a point x0 E X and let f: U Y be such that f'(x) existsand is continuous in U. Suppose

Then f is a homeomorphism of an open neighbourhood U0 of x0 ontoan open neighbourhood V0 of f(x0), is continuously differentiableon V0 and


(f_1)'(y) =

fory E V0.

Proof. We may assume that x0 = 0 and f(xo) = 0. Furthermore, settingT0 = f'(O) and replacing f by X X, we may assume thatY = 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,

Ik(x)—e(y)II 411x—yII

for all x,y E J)5(X). Hence, by Theorem 4, f(x) = x+e(x) is a homeo-morphism of some open neighbourhood U0 of 0 onto V0 = D514(X).

The rest is plain sailing. For Yo E V0 set x0 = f'(y0) E andT = f'(x0). Furthermore, for y0+k E V0 set x0+h = f'(y0+k). Ask —* 0, we have h 0, and so k = Th+o(h) and

h =

Consequently,

f'(y0+k) = f'(f(xo+h)) = x0+h =

proving that = T' = The continuity offollows from the fact that the map T T1 is continuous. 0

The contraction-mapping theorem has numerous applications to dif-ferential and integral equations. Here we shall confine ourselves tothree 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 thereis 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 operatorL: CEO,!] C[0,1], defined by

Cs

(Lx)(s) = I K(s,t)x(t) dt


is called the Volterra integral operator with kernel K(s, t). In fact, wecould assume that K(s, t) is defined only on the closed triangle0 t s 1, since for s < t we do not use K(s, t). Write K1(s, t) =K(s,t) and for n 1 set

t)=

K(s, u)Kju,t)du.

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 out-side the closed triangle 0 t s 1. It is easily shown by inductionon n that is precisely the Volterra integral operator with kernelIndeed, if this is true for n then

+ 1x) (s)

=f K(s, u) t)x(t) dt du

= f J dtdu00

Cu jS

= J Jdux(t) dt0t

'-U

= Jdt.

0

The function K(s, t) is continuous on the closed unit square and so it isbounded, say

IK(s,t)I M

for all 0 s, t 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


Hence if n is sufficiently large, say n 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 claimsthat 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 fixedpoint. 0

The next application concerns a non-linear variant of the Volterraintegral operator.

Theorem 8. Let K(s, t) and w(s, t) be continuous real functions on theunit 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 functiony(s) on [0, 1] such that

y(s) = v(s)+ J K(s, 0 w(t, y(t)) dt.

Proof. Define L: C[0, 1] 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.


To show this, we shall turn to trickery. For a > 0, we introduce anew 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-IhEx — EyhL

= J JK(s, t){w(t, x(t)) — w(t, y(t))} cit ds

0 0

i_i i_s

MNJ J

—y(t) I dt ds00i.1 i_I

= MNJ J

e —y(r) cit dtOt

= MNJ dt

MN

This shows that for a > MN the map

L: Xi., —.

is 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. EIn 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) inTheorem 8.

Theorem 9. Let K(s, 1, u) be a continuous function on 0 s, I 1,u 0, such that

IK(x,:,ui) —K(s,I,u2)I —u2J,

where N(s, z) is a continuous function satisfying

J0


for every 0 s 1. Then for every v E CEO, 1] there is a unique func-tion y E CEO, 1] such that

y(s) = u(s)+ 10 K(s,t,y(t)) di.

Proof. Define L: CEO, 1] —' C[O, 1] by

(Lx)(s)=

K(s,t,x(t)) di,

so that a solution of our integral equation is precisely a fixed point ofthe map T: CEO,!] — C[O, 1], given by Tx = v + Lx. Forx,y C[O,1] we have

(Lx — Ly)(s)= 1 {K(s, t,x(t)) — K(s, t,y(t))}

10 N(s, t) Ix(t) —y(t) I d:

kflx—yfl.

Hence IILx—LyII implying that L is a contraction and there-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 otherfixed-point theorems. But for the time being we return to the study offormed spaces.

Exercises

1. Show that a contraction of an incomplete metric space into itselfneed not have a fixed point.

2. Let A and B be disjoint subsets of a metric space, with A compactand 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 andB are closed.

3. Show that if f is a map of a complete metric space X into itselfsuch that d(f(x),f(y)) <d(x,y) for all x,y X (x y) then f

Chapter 7: The contraction-mapping theorem

need not have a fixed point. Show however that if X is compactthen 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 ife d(x,y) <8 then d(f(x),f(y)) <e. Prove that f has a uniquefixed point.

5. Let f be a map of a compact metric space X into itself. Supposethat for every 0 there is a & = 8(€) > 0 such that ifd(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

that if , + 1) —* 0 as n then the sequence con-verges to a fixed point of f.

6. Let f be a map of a complete metric space X into itself such that

d(J(x),f(y))

for all x,y E X, where is a monotone increasing

function such that = 0 for every t> 0. Deduce fromthe result in the previous exercise that f has a unique fixed pointa, and = a for every x E X.

7. Let f be a continuous map of a (non-empty) compact Hausdorffspace into itself. Show that there is a (non-empty) compact subsetsuch that f(A) = A.

8. Show that on the space c0 the norm function x =lixil = is 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

i—I

is not differentiable at any point.

In the next five exercises, X, Y and Z are Banach spaces.

10. Let f,g: X be

differentiable at x0 and

a an an

open set C Y. Suppose that f is differentiable at x0 E U and(J'(x0))' E Show that f1: V—p U is differentiable at


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 Zcontinuous maps with f(U) C V. Suppose that f is differentiableat a point x0 E U and g is differentiable at Yo = f(x0) E V. Showthat 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,..., letU — Y be differentiable at every point of U. Suppose that

every x0 U has a neighbourhood N(x0) on which Hfh(x) II is

uniformly convergent. Show that if is convergent forsome x E U then it is convergent for all x U, the map

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) —f(y) —f'(z)(x—y)II IIx—yII sup If(s) —f'(z)II,.cE(x,y]

where, as usual, [x, y] is the closed segment {tr + (1 — t) y: 01).

Notes

The contraction-mapping theorem is from S. Banach, Sur les operationsdans les ensembles abstraits et leur applications aux equations integroJes,Fundamenta Mathematica, 3 (1922), 133—81. Of course, it can be foundin most books on general topology and basic functional analysis. A richsource 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 aspace, it is often useful to consider topologies other than the normtopology we have introduced so far. Two of the most important ofthese topologies are the weak topology and the weak-star topology.Before we define these topologies and prove Alaoglu's theorem, thefundamental result concerning the weak-star topology, we shall reviewsome additional concepts in point set topology and present a proof ofTychonov's theorem.

Given a topological space (X, i-), a collection a of subsets of X is saidto be a sub-basis for i if a C r and every member of r is a union offinite intersections of sets from a. Equivalently, a is a sub-basis for r ifa C and for all U E 1 and x E U there are sets S1,. . . ,S,, such thatx E S. C U. More formally, a is a sub-basis for T If

= U (1 5: C (1)J

where is the set of all finite subsets of a. In most cases we mayomit the second term on the right, X}, especially if the intersectionof an empty collection of subsets of X is taken to be X. Not only dosub-bases tend to be more economical than bases, but they also enableus to use an arbitrary family of sets to define a topology. Indeed, ifa C is any collection of subsets of X then the collection i definedby (1) is a topology on X.

The possibility of using an arbitrary set system to define a topologyenables us to use a set of functions into topological spaces to define atopologyonaset. LetXbeasetandforeachyEl'letf,, beamapof X into a topological space (X,, , r,,). Then there is a unique weakest

114

Chapter 8: Weak topologies and duality 115

topology r on X such that every f,,: (X, r) (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 everyx E U there are ,. . . , y, E F and U1 E ..., U, E i,,, such that

x E fl C U.

The topology i defined in this way is called the weak topology generatedby and is denoted by o(X, where = y E I).

A map! from a topological space (X',T') into (X,o(X, is continu-ous 1ff f,.of: (X',r') —' is continuous for every y E I'. l'his so-called universal property characterizes the weak topology.

The product topology is one of the prime examples of a weak topol-ogy. Let be a topological space for each y F and letX = X1, be the Cartesian product of the sets X., (y E F). ThusX is the collection of all the functions x: F UYEr X,, such thatx(y) E X.1,. One usually writes x7 for x(y) and sets x = also,x,, is the 'y-component of x. Let p.,,: X —' X,, be the projection ontoX,,; thus p,,(x) = x,,. The product topology on X is the weak topologygenerated by the projections p.,. (y F). To spell it out: a set U C X isopen in the product topology if for every point x = (X,,),,EF there areindices y,,. . . , y,, and open sets U,,,..., U,. E such that

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

(1 = {y = (y,,) E X: y.,1 (i = 1,. ..,n)} C U.

Thus to guarantee that a point y = (y,.) belongs to a fixed neighbour-hood 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,the weak topology a(X, X generated by thebounded linear functionals on X. Thus U C X is open in the weaktopology (w-open) 1ff for every x E U there are bounded linear func-tionals and positive reals such that

{y X: (i = 1,...,n)} C U.

By replacing byf1/€1, may take every to be 1.

116 Chapter 8: Weak topologies and duality

The weak-star topology cr(X*, X) on the dual r of a normed space Xis 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 (seeTheorem 3.10). In more detail: a set G C X is open in the weak-startopology if for every g E G there are points x1,. .. , x, E Xand positive reals el,... , e,,, such that

{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 isprecisely 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 tobe appropriate to mention Tukey's lemma, which is equivalent to Zorn'slemma (and so to the axiom of choice). A system of subsets of a setS is said to be of finite character if a subset A of S belongs to if andonly if every finite subset of A belongs to

Lemma 1. (Tukey's lemma) Let be a set system of finite characterand let F E Then has a maximal element containing F.

Proof. Let = {A C F C A}. Order the elements of by inclu-sion: 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 allfinite subsets of D belong to ?F, and F C D. Clearly, D C is anupper bound for Hence, by Zorn's lemma, has a maximal ele-ment. 0

A system of subsets of a set has the finite-intersection property ifF, 0 for all F1,. . . , F,, C It is easily seen that a topological

space X is compact if the intersection of all subsets of a system ofclosed sets which has the finite-intersection property is non-empty.Indeed, = {X\F: F E is a collection of open sets. Since no finitecollection of sets in covers X, the entire system also fails to coverX, so there is a point x C X which belongs to no member of Hencex C F.

A trivial reformulation of this characterization of compactness is thefollowing; a topological space is compact if whenever is a system of

Chapter 8: Weak topologies and duality

subsets with the finite-intersection property then flAE.A A 0. This isthe characterization we shall use in the proof below.

Theorem 2. (Tychonov's theorem) The product of a collection of com-pact sets is compact.

Proof. Let {X,,: y 1) be a collection of compact spaces and letX X,, be endowed with the product topology. We have toshow that X is compact, i.e. that if is a system of subsets of X withthe finite-intersection property then A

Let be the collection of all set systems on X with the finite-intersection property. Then is of finite character and so, by Tukey'slemma, there is a maximal set system with the finite-intersection pro-perty containing Since

fl fl A,

we may assume that our set system is itself is maximal for the finite-intersection property.

The maximality of implies that the intersection of any two sets in .s4belongs to as does the intersection of finitely many sets in .si2. Conse-quently, if B C X meets every set in .sii then B E In particular, ifA and A C B C X then BE

So far we have made no use of the fact that X is a product space, letalone a product of compact spaces. Let us do so now. For each y E I',the system {p,,(A): A of subsets of X,, has the finite-intersectionproperty. 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 setmeets every element of Hence and so

fl E

whenever ,. . . , y,,} C 1' and U,, is a neighbourhood of x7.Let x = (x,J,,Er X. We claim that x E A. Indeed, for

every neighbourhood U of x there are indices y1 ,... , -y,, E I and neigh-bourhoods U,, of x (i . . , n) such that

ri C U.


Consequently, (3) implies that U E s4, i.e. U intersects every element ofHence every neighbourhood of x intersects every A and so

x E A for every A E This proves our claim and completes the proofof our theorem. 0

From here it is a small step to Alaoglu's theorem, the fundamentalresult concerning the weak-star topology; in view of the fact that boththe 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 anormed space X is compact in the weak-star topology.

Proof. = R: IIxII}ifXisarealspaceand= E C: IIxII} if Xis a complex space. Furthermore, let D

be the product 11xEX endowed with the product topology. Sincea 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 topologyimply that is a continuous map of B into D, and it is clear that ç isone-to-one. Furthermore, again by the definitions of the product topol-ogy and the weak-star topology, is a continuous map from c(B*)onto 8*. Thus B* is homeomorphic to the subset of D. There-fore, to complete the proof, it suffices to show that c(B*) is a closedsubset of the compact space D.

Let then = D be in the closure of Define a scalarfunction f on X by setting f(x) = for x C X. We claim that f is alinear functional, i.e. f E X'. To see this, let x,y C X and let a andbe scalars. Since C for every n C N there is a continuouslinear functional E B such that

1(x) + + < (4)

As + p9y) = + inequality (4) implies that

f(ax = af(x)

i.e.f C X'. Since If(x)I = Qxfl for every x X, f is not only a


linear functional, but it is also continuous and 11111 1, i.e. f E B.Finally, by the definition of f, we have = C completingthe proof. 0

As an immediate consequence of Alaoglu's theorem, we can see thata subspace of the Banach space of continuous functions on a compactHausdorif 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 Kis endowed with the weak-star topology. In particular, everynormed space is isometrically isomorphic to a subspace of C(K) forsome compact Hausdorff space K.

Proof. Let X and K be as in the theorem; thus K is a compact Haus-dorff space, namely endowed with the weak-star topology. Forx E X let f, be the restriction of i E (defined by i(x*) = x(x) for

E to K = B(r). By the definition of the weak-star topology,f E C(K); indeed, the weak-star topology on r is precisely the weak-est topology in which every function i is continuous.

The map x - of X into C(K) is clearly linear. Furthermore, bythe Hahn—Banach theorem,

HfxII = sup{lfx(x*)I: K}

= : f C = lixil,

and so the map X C(K) is a linear isometry. 0

It is customary to write w-topology for the weak topology and w-topology for the weak-star topology. Similarly, one may speak of w-open sets, w-neighbourhoods, and so on, all taken in theappropriate spaces. Thus a w-open subset of X is a o(X, X *)..opensubset, and the of a subset of X* is the closure of the set inthe As before, we consider X as a subspace of

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 thesection we shall consider only real spaces.

The weak topology is weaker than the norm topology, so everyw-closed set is also norm closed. The following result claims that forconvex sets the converse is also true.


Theorem 5. A convex set C of a normed space X is closed if it is w-closed.

Proof. Suppose that C is closed and x0 C. Then d = d(x0, C) > 0.

Set D = {x E X: d(x0, x) <d}. By the separation theorem (Theorem3.13) there is a bounded linear functional f E X* such that

s = sup(x*,x) inf(x*,x).xEC xED

Then

U = {x E X: (x*,x) > s}

is a w-neighbourhood of x0 and Un C = 0. This shows that C is alsow-closed. As we remarked earlier, the converse is trivial. fl

Theorem 6. is the w*.closure of 8(X) in

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 and8(X) C B(r*), we have B C B(X**). Furthermore, as theof a convex set is convex, B is a convex set which is closed in the normtopology of X

Suppose that, contrary to the assertion of the theorem, BThen there is a point E B(X**)\B so that, by the separationtheorem (Theorem 3.13) there is a bounded linear functional 4 * * on

* such that

(4**,b) 1 <

for all b E B. Let 4 be the restriction of xr* to the subspace X ofr*. Then

(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 concerningreflexivity. Note that the w-topology on a normed space X is preciselythe restriction to X of the on

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


compact if it is closed, B(X) is a subset of ifB(X) = B(X**). But this means precisely that B(X) is w-compact 1ffx=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. SinceB(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 isprecisely o-(Y, Hence B(Y) is w-compact and so, by Theorem 7, Yis reflexive. 0

Theorem 9. Let X be a Banach space. The following assertions areequivalent:

(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.

Proof. The implication (i) (ii) is trivial (and so is (1) (iii)). Sup-pose that (ii) holds. Since is it is also w-compact.Hence, by Theorem 7, is reflexive. Thus (ii) (iii).

Finally, suppose that (iii) holds. As X is a Banach space, the ballB(X) is norm closed in Therefore, by Theorem 5, B(X) is w-closed, i.e. Since = 1*, this means thatB(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 beingcontinuous 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.

Equivalently, there is an 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 closedunit sphere S(X).) It is a rather deep theorem of R. C. James that thisproperty characterizes reflexive spaces: a Banach space X is reflexive ifevery bounded linear functional attains its norm on B(X).


Our next aim is to show that every Banach space comes rather closeto having this property: the functionals which attain their suprema onthe unit ball are dense in the dual space. (This property used to becalled subreflexivity. As the result claims that every Banach space issubreflexive, the term has gone out of use.) In fact, we shall provesomewhat 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 byPx and onto X. by pr, we have = 5(X). Furthermore,px.(fI(X)) is precisely the set of functionals in S(X*) which attain theirnorms 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.

We shall show that not only is dense in 5(r), which is pre-cisely the subreflexivity of X. but also every point (x, f) of 118(X) isclose to some point (y,g) of 11(X):

lix—yli + Ill—gil <q(8), (5)

where 0 as 6 —* 0.

The proof of this theorem is based on the following lemma, whoseproof is left to the reader (see Exercise 13).

Lemma 10. Let X be a real normed space and let f,g E be suchthat

whenever x E B(X) and f(x) = 0, where 0 <€ < Then either

hf—gil 2e or iif±ghi 2e. 0

Theorem 11. (Bishop—Phelps—Bollobás theorem) Let X be a Banachspace and let x0 5(X) and Ia E be such that

ifo(xo)11 <8 =

where 0 < 1. Then there exist x1 5(X) and E such thatf1(x1) = 1, hhfo—fiii e and ilxo—xiII


Proof. In the notation introduced before Lemma 10, the theorem claimsthat for every (xo,fo) E 118(X) there is an (x1 , E 11(X) such thatIIfofiII and I)xo—xiII <€; in particular, (5) holds with =

say

As linear functionals are determined by their real parts and the mapf '- Ref is an isometry, we may assume that Xis a real space.

Let

k

and define a partial order on B = B(X) by setting x y if

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 maxi-mal 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 <t < 1,

Yi E B and z1 E C such that

x1 = sy1+(t—s)z1.

Since fo(xo) > 0 and x y implies that f0(x) fo(y), we see that

f0(x0) = sfo(yi) <fo(Yi)

and so

f0(y1—x1) = (1—s)f0(y1) > >0. (6)

Furthermore,


y1—x1 = (l—s)y1—(t—s)z1,

and so

< (7)

Inequalities (6) and (7) give

I 2\ 1+2/€Oyi—xiII (1—s)(1+—J =E/

showing that x1 Yi• But, as x1 is maximal, we have x1 = contrad-icting (6), say. This proves that Int D, as claimed.

By the separation theorem (Theorem 3.13) there is a linear functional

f1 E such thatf1(x1) = 1 andf1(x) 1 for allx ED. Then 1

1 and f1(x) for all x E BflHence, by Lemma 10,

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 tonumerical ranges. Our last aim in this chapter is to prove anothermajor result of functional analysis, the Krein—Milman theorem.Although in its proper generality this result has nothing to do with dualspaces and weak topologies, we present it here since it has some naturalapplications concerning dual spaces.

Our proof of the Krein—Milman theorem is based on anotherequivalent form of the axiom of choice, namely on the well-orderingprinciple, 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 asmallest 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.

Proof. A subset B of an ordered set (A, is said to be an initial seg-ment if a E A, b C B and a b imply that a E B.

LetS be a set and let P = y El) be the set of all wellordered sets such that is a subset of S. For (5,,, and (Se, in


Pset

if 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,,, y be a chain. Set 5' = S,, and for

if T S' T then t' E 5,, forsome y E I', and the smallest element of r n 5,, in (5,,, E,,) is also asmallest element of 1' in (5, E').

Since every chain has an upper bound, by Zorn's lemma fromChapter 3, (P, has a maximal element (Se, To complete theproof, all we have to show is that S0 = S. Suppose that this is not so,say s1 S\S0. Extend the order to an order E1 on = S0U{s1} bysetting s s1 for all s E S0. Then (Si, is a well ordered set and(Se, <(S1, contradicting the maximality of (S0, 0

Let K be a convex subset of a vector space. A point in K is said tobe an extreme point of K if it is not in the interior of any line segmententirely in K. Thus a E K is an extreme point of K if wheneverb,c K, 0< t < 1 and a = tb+(1—t)c, we have a = b = c. The setof extreme points of K is denoted by Ext K.

Recall from chapter 3 that the convex hull of a subset S of a vectorspace Vis

coS=

;x1 : x E 5, 0 (i = 1,... ,,z), = 1 (n =

If S is in a normed space then S, the closed convex hull of S, is theclosure of coS; by Theorem 3.14, is the intersection of all closedhalf-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, weshall be satisfied with a somewhat artificial form of this result. Given avCctor space V with dual V', for F C V' and S C V the F-closed convexhull 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

126 Chapter 8: Weak topologiesand duality

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 topologyon a vector space V and let F C V'. Suppose that F separates thepoints of V(i.e. iff(v) = 0 for alIf C F then v = 0) and eachf C Fisr-continuous. Let K be a non-empty r-compact F-convex subset of V.Then ExtK 0 and K = COFEXtK.

Proof. Let y C I') be a well ordering of F. Then there is a func-tion s: R, y such that for all y C F we have

s7=max{f.,(x):xCK,f8(x)=s8for6<y}. (8)

This is immediate from the fact that F (i.e. F) is well ordered and if s5is determined for all 6 E F preceding y F then

Kfl fl8<7

is a non-empty compact set on which the continuous function 1., attainsits supremum, 57•

The existence of the function s implies that there is a point a G Ksuch that f7(a) = s,, for all y f. This point a is an extreme point ofK. 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 thepoints of V. we have a = b = c. Thus ExtK / 0.

Set K' = COFEXt K. Then K' C K since K is F-convex. To completethe proof, we have to show that, in fact, K' K. Suppose that this isnot so. Then there is a functional Jo C F such that

maxf0(x) <maxf0(x). (9)x€K xEK

Choosing a well ordering of F in which to is the first element, the pointa produced by the procedure above is such that h(a) =But a K', contradicting (9). 0

Corollary 14. Let K be a compact convex subset of a Banach space.0

Corollary 15. Let X be a normed space. Then =

Proof. By Alaoglu's theorem, is Furthermore, it isX-convex since if C X and Jo then f0(x0) > 1 for somex0 C B(X) and f(x0) 1 for all x B(X). 0


The last result implies that has 'many" extreme points:sufficiently many to guarantee a 'large" closed convex hull, namely

Occasionally this fact is sufficient to ensure that a given space isnot a dual space. For example, it is easily seen that Ext B(c0) = 0; soc0 is not a dual space. Indeed, let x = E B(c0). Then IXkI <for some k and so, with = = for n k, Yk = Xk + and

= we have y = z = E B(c0) and x = 4(y+z), sothat x Ext B(c0). Thus Ext B(c0) is indeed empty.

The Krein—Milman theorem is one of the fundamental results of func-tional analysis, leading to deep theorems in operator theory and abstractharmonic analysis, like the Gelfand—Naimark theorem concerningirreducible *..representations of C-algebras, and the Gelfand—Ralkovtheorem about unitary representations of locally compact groups. Butthese results will not be presented in this book. We shall return to com-pact convex sets in Chapters 13—16, when we study compact operatorsand 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 normedspace X then the weak topology on Y is precisely the topologyinduced 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 apoint 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 thenX/Y is reflexive.

4. Show that

d(x,y) =

defines a metric on and the restriction of this metric toinduces the weak-star topology on

5. Let X be an infinite-dimensional normed space. Show that the w-closure of S(X) = {x E X: lixil = 1} is B(X) = {x E X: lixil 1}.


6. Let be a sequence in 1,, (1 <p weakly converging tox0 (thus x0 in the w-topology, i.e. ,x') (xo,x*)for every E = lq). Prove that if —' lixoil thenIiXn'XoII ' 0.

7. Show that a sequence C l, (1 <p < converges weakly to 01ff is bounded (i.e. (Ix,,II K for some K and all n) and

= 0 for every i, where = ,

8. Let = C (n = 1,2,...). Show that 0for every y C c0 1ff is bounded and = 0 for every i.

9. Formulate and prove assertions analogous to the previous twoexercises for the spaces 1) and Lq(O, 1).

10. Let X be an infinite-dimensional normed space. Show that thenorm function lxii (x E X) is not weakly continuous at any pointof X.

11. Let be the standard basis of 12 and let

A = {em+men: 1 m <n}.

Prove 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 forevery x C X there is a point of M nearest to x 1ff f attains itsnorm on B(X).

13. Prove Lemma 10: if X is a real normed space and f,g E aresuch that (Kerf) fl B(X) C g — ' [ — €, €} for some 0 < then

Ill—gil 2€ or i)f+giI 2€. [HINT: If f ±g then there area,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 ofB(X) into M parallel to N. Let i-fl' be the norm on M with unitball D, and note that with fo = fJM and g0 = giM we haveilafo+Pgoii' = iiaf+flgfl for all E R. This shows that itsuffices to prove the lemma in the case when X = M, i.e. when Xis 2-dimensional. Use a simple geometric argument to completethe proof.]

14. Show that the assertion of Lemma 10 is best possible: for0 < 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 maximalelement x1 x0 in the proof of Theorem 11.

16. Check that Theorem 11 can be strengthened a little: if

ifo(xo) — fl < where 0 <e < then we may demand


and IIxo—xtII <€+€2. (This variant of the result isessentially best possible.)

17. Let K C R" be a compact convex set. Show that if n = 2 thenExt 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.

Show that the closed convex set has no extreme points.21. Let I be an infinite-dimensional Banach space whose unit ball has

only finitely many extreme points. Show that X is not a dualspace, i.e. there is no normed space Ysuch that X = r.

22. Show that CR[0, 1] is not a dual space.

Notes

Tychonov's theorem is from A. Tychonoff, Uber em Funktionenraum,Mathematische Annalen, 111 (1935), 762—66; somewhat earlier, anincomplete version of the result appeared in A. Tychonoff, (Jber die topo-logische Erweilerung von Räumen, Mathematische Annalen, 102 (1929),544—61. (The modern transliteration of the name 'Tychonov' is closer tothe Russian original than the one used in German before the second worldwar. Similarly, the old-fashioned usage of Markoff, Tschebischeff, Egor-off, etc., is due to German influence.) A proof of the thoerem can befound 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 twoyears 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 offunctionals attaining their norms on S(X) is dense in is from E. Bishopand 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. LondonMath. Soc., 2(1970), 181—2. Our presentation followed that given in F. F.Bonsall and J. Duncan, Numerical Ranges I!, London Math. Soc. LectureNote 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 nexttwo chapters we shall look at the 'nicest' examples of these spaces,namely Euclidean spaces and Hilbert spaces. These spaces are thenatural generalizations of the n-dimensional Euclidean space

A hermitian form on a complex vector space V is a map f: Vx V C

such that for all x,y,z V and C we have

(I) =

(ii) f(y,x) = f(x,y) for all x,y V.

Similarly, a hermitian form on a real vector space V is a mapf: Vx V R satisfying (i) and (ii). In that case (ii) says simply that f issymmetric, so a real hermitian form is just a symmetric bilinear form. Iff is a hermitian form on a complex vector space V then g = Ref is areal 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 onecomplex 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. Notethat f(x,x) is real for all x C V and f(Ax,Ax) = IAI2f(x,x) for all x C Vand 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 iff(x,y) = 0; orthogonahty is sometimes expressed by writing x .Ly. Avector orthogonal to itself is said to be isotropic. We call f degenerate ifsome non-zero vector is orthogonal to the entire space; otherwise theform 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 allx,y E Vwe have

If(x,y)12 f(x,x)f(y,y) (1)

and

f(x+y,x+y)"2 f(x,x)"2+f(y,y)'12. (2)

Purthermore, f is non-degenerate 1ff 0 is the only isotropic vector withrespect to f.

Proof. In proving (1), we may replace x by Ax if IA I = 1 so we mayassume 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 x is isotropic then (1) implies that x is orthogonal. to everyvector in V. 0

Inequality (1) is, of course, just the Cauchy—Schwarz inequality; ine-

quality (2) is a variant of Minkowski's inequality for p = 2.

In view of Theorem 1, a positive hermitian form is non-degenerate ifit is a positive-definite hermitian form, i.e. a hermitian form f such thatifx Othenf(x,x) >0.

An inner-product space is a vector space V together with a positivedefinite hermitian form on V. This positive-definite hermitian form issaid to be the inner product or scalar product on V and we shall write it

(.,.) (and, occasionally, as (, )). Thus (,) is such that for allx,y, z E V and for all scalars A and we have

(i) = A(x,z)+p.(y,z);(ii) (y,x) = (x,y);(iii) (x,x) 0, with equality if x = 0.

More often than not, it will not matter whether our inner-productspace is a complex or a real inner-product space. As the complex casetends 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 vectorspace V then lxii = (x,x)'12 is a norm on V. A normed space is said tobe a Euclidean space or a pre-Hilbert space if its norm can be derivedfrom an inner product. A complete Euclidean space is called a Hubertspace. Clearly every subspace of a Euclidean space is a Euclidean spaceand every closed subspace of a Hilbert space is a Hubert space.

The Cauchy—Schwarz inequality states that l(x,y)l lIxIlIlyll; in par-ticular, the inner product is jointly continuous in the induced norm.The inner product defining a norm can easily be recovered from thenorm. 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)

in the real case. Therefore in a Euclidean space we may, and oftenshall, use the inner product defining the norm. For this reason, we usethe 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 exten-sion. If T is a linear operator (not necessarily bounded) on a complexEuclidean 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 besuch 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. 0

It is worth pointing Out that Theorem 2 cannot be extended to realEuclidean spaces (see Exercise 2).

In a Euclidean space, the theorem of Pythagoras holds and so doesthe parallelogram law.

Theorem 3. Let E be a Euclidean space. If x1,...,x,, are pairwiseorthogonal vectors then


2

xl =

Furthermore, if x,y E then

IIx+y112+ = 211x112+211y112.

Proof. Both (5), the Pythagorean theorem, and (6), the parallelogramlaw, are immediate upon expanding the sides as sums of products. Tospell it out,

, 2 /,, a \ a

= x.) = (x1 ,x1) + (x1 ,x,)\i=1 I 1=1 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 (seeExercise 3). This shows, in particular, that a normed space is Euclideaniff all its two-dimensional subspaces are Euclidean. Furthermore, acomplex normed space is Euclidean iff it is Euclidean when consideredas a real normed space.

The last assertion is easily justified without the above characterizationof Euclidean spaces. Indeed, if E is an Euclidean space then Re(x.y) isa real inner product on the underlying real vector space of E and itdefines the same norm on E. Conversely, if V is a complex vector spaceand (.,•) is a real inner product on V (i.e. is an inner product onVft) such that the induced norm satisfies IIAxII = A IIIxH for all x E Vand 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 andsatisfying 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 aseparable infinite-dimensional Hilbert space.

(iii) Let E be the vector space of all eventually zero sequences ofcomplex numbers (i.e. x = (x,)° belongs to E if x = 0 whenever i is

134 Chapter 9: Euclidean spaces and Hubert spaces

sufficiently large), with inner product (x,y) = Then E is adense subspace of 12; it is an incomplete Euclidean space.

(iv) It is immediate that

(f,g)= J

f(t)g(t) dta

is an inner product on the vector space C([a, b]); the norm defined isthe 12-norm

/ \1/2

11f112 = (j If(t)12d1

and not the uniform norm. This is an incomplete Euclidean space (seeExercise 14).

(v) Also,

(f,g)= Ja

is 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 If(OI2dt\0

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

x = (x1)? and y = in V define

f(x,y) = a,1x191.1,1 = 1

Then I is a hermitian form if = a9 for all i, j. Also, every hermitianform on V can be obtained in this way. 0

Theorem 5. The completion of a Euclidean space is a Hilbert space.

Proof. The assertion is that the inner product on a Eucideanspace E can be extended to the completion E of E. Let (x,,) andbe Cauchy sequences in E and let i and 9 be their equivalence classesin E. Define


=

Since

i(Xn,Yn)(Xm,Ym)i i(Xnm,Yn)l+l(Xm,Ynym)I

+ flXmil —ymIl,

the limit exists. It is easily seen to depend only on I and j and not onthe particular representatives (y,,). Furthermore, (, •)is an innerproduct on E, extending E and defining precisely the norm on the com-pletion E. 0

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, andx2 F2. If F is a subspace of E then F and F' are orthogonal sub-spaces and Ffl F' = {O} since if x E Ffl F' then x I x and so x =0. Ifx F and y E F' then lix +yii2 = lIxii2+ iiyl12, so the projectionsx+y x and x+y '-+ y are both bounded. Thus F+ F' is the orthogo-nal direct swn of F and F'. In general, F+ F' need not be the entirespace, not even when F is closed (see Exercise 6). However, when F iscomplete, 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 aunique representation in the form

x=x,+x2 (x1€F,x2EF').Furthermore, if y E F and y x1 then

ilx—yil > lix—xili = ilx2il.

Proof. 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.


Let then E F be such that

< d2+!

By identity (6), the parallelogram law,

—ymIl2 = 211Xyn112+211Xym112 II2XYn —ymII2

22—+—,n m

and so is a Cauchy sequence. Since F is complete, y,, x1 forsome x1 E F. Then IIx—xiII = d.

Let us show that x2 = x — x1 is orthogonal to F. Suppose this is notso. Then F has an element, say y, such that (x2,y) 0. Replacing yby z = (x2 , y)y we have (x2 , z) E R and (x2, z) > 0. But then forsufficiently 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 0 then, as (x1—y)J.x2, we have

IIx—y112 = flx1 = lxi —y112+ lIx2ll2 > 1lx2112 = d2. 0

In view of Theorem 6, if F is a complete subspace of a Euclidean spaceE then we call F' the orthogonal complement of F in E. The mapPF: E = F+ F' —. E defined by x = x1 +x2 x1 is called the orthogonalprojection 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) = (PFX,PFY) = (x,PFy).

If F (0) then = 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 theclosed 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 -+ Cdefined 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

11111 ll.xoll. Furthermore, If(xo) I HxoII2 and so 11111 Theexistence of an orthogonal decomposition implies that all bounded linearfunctionals on H can be obtained in this way.

Theorem 9. (Riesz representation theorem) Let f be a bounded linearfunctional on a Hubert space H. Then there is a unique vector x0 E Hsuch that f(x) = (x,x0) for all x H. Furthermore, 11111 = lIxoll.

Proof. We may assume that f 0. Then M = Kerf is a closed one-codimensional 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 inthe form

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) = 0 for allx E H, then

IIxo—x6112 = = 0.

The assertion litV = lixoll was shown immediately before the theorem.0

Corollary 10. Let H be a Hilbert space. For y E H let E bedefined by = (x,y)(x E H). Then the map H H* defined byy is an isometric anti-isomorphism between H and H, i.e. ify f and z g then Ay+pg Af+1g. If H is a real Hubert spacethen the map is an isometric isomorphism between H and H. 0


The last result enables us to identify a Hilbert space with its dual; thisidentification is always taken for granted. It is important to rememberthat the identification is an anti-isomorphism since this will make theadjoint of a Hubert-space operator seem a little different from theadjoint of a Banach-space operator.

Exercises

1. Let f be a hermitian form on a real vector space U. LetV = 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).

Show that I is a hermitian form on the complex vector space Vand Il U = f. Show also that f is positive 1ff f is, and that f isdegenerate 1ff f is.

2. Give an operator T E such that (Tx,x) = 0 for all x EandflTIJ=1.

3. Show that a normed space E is a Euclidean space if and only if theparallelogram law holds in E, i.e.

lix +y112 + —yfl2 = 211x112 + 211y112.

4. Let E be a Euclidean space. Show that for x1,.. . ,x,, E Ewe have

=11 i1

where the summation is over all 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-tion: = p and II P11 = 1. Show that P is the orthogonal projec-tion 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.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 IIyIlIIz—xli + IIzIIIIx—yU.

10. 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 productinducing the norm?

11. Let F be a closed subspace of a Hilbert space H. Show that thequotient space H/F is also a Hilbert space.

12. Let I' be an arbitrary set and let 12(f) be the vector space of allfunctions f: C such that >.,EI' If(v)I2 In particular, iff E 12(fl then {y E 1': 0} is countable. Show that 12(F) is aHilbert space with then norm

/

11111 = I 11(7)12

13. Let {H7: y E f) be a family of Hilbert spaces. Let H be the vec-tor space of functions f: F UYEI such that f(y) E and

yEr

Show that

/ \112

IIfIl = I 111(7)112

is a norm on H and with this norm H is a Eucidean space. Is Hnecessarily a Hubert space?

14. Turn CEO, 1] into a Euclidean space by setting

= Lf(t)g(t) dt.

Show that this space is incomplete.

140 Chapter 9: Euclidean and Hubert spaces

15. f: [0,1] Ft such thatf' E L2(0,1) and f(0) = 0. Prove that Visa real Hubert space with inner product

= Lf'(t)g'(:) dr.

16. A real matrix A is called orthogonal if (Ax,Ay) = (x,y) forall x,y E Prove that A is orthogonal 1ff it maps orthogonalvectors 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 andImP±KerP.

18. Let A be a non-empty subset of a Hubert space H and letT 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 sub-space and T is the orthogonal projection onto A: T =

19. Show that the unit ball of '2 contains an infinite set A such thatIIx—yII > for all x,y E A (x y). Show also that A must con-sist of norm-i vectors.

Notes

The concept of an abstract Hubert space was introduced by J. von Neu-mann, Eigenwerz Theorie Hermitescher funktional Operatoren, Math.Ann., 102 (1930), 49—131. Earlier special realizations of a Hubert spacehad been considered by several people. In particular, from 1904 to 1910David Hilbert published some fundamental papers on integral equationswhich led him to consider some 'Hilbert spaces' of functions. It seemsthat the name 'Hubert space' was first used by F. Riesz in his book, Lessystèmes d'equanons a une infinite d'inconnus, Paris, 1913, for what weknow as 12: "Considérons l'espace Hilbertien; nous y entendonsl'ensemble des systémes (Xk) tels que 12 converge."

JO. ORTHONORMAL SYSTEMS

In this chapter we continue the study of Euclidean spaces and Hubertspaces. As we shall see, every separable Euclidean space contains theexact analogue of the canonical basis of = (Ce', This means thatwe can use 'coordinates', which in this context we call Fouriercoefficients, to identify the points of the space, these coefficients behav-ing very much like the coordinates in or R". If our space is in fact aHubert space then the space is thus naturally identified with 12, telling usthat all separable Hilbert spaces are isometric.

Let E be a Euclidean space and let S C E be a set of vectors. If E isthe closed linear span of S, i.e. linS = E, then we call S fundamental ortotal. We call S orthogonal if it consists of non-zero pairwise orthogonalelements. An orthogonal set of unit vectors is said to be an orthonormalset. An orthogonal set is complete if it is a maximal orthogonal set. Acomplete orthonormal set in a Hilbert space is said to be an orthonormalbasis. We shall see, amonst other results, that every orthonormal basis isa Schauder basis of the best kind, with basis constant 1 (see Exercises16—18 of Chapter 5), and every Hilbert space contains an orthonormalbasis.

Occasionally, we call a set S of vectors a system of vectors, especiallywhen s is not written in the form of a sequence.

Theorem 1. A fundamental orthogonal set in a Euclidean space is com-plete. 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. SetM = linS. Then Mi = (0) so M = (Mi)i = H. Thus S is a funda-mental system. 0

141

142 Chapter 10: Orthonormal systems

The Gram—Schmidt orthogonalization process enables us to replace asequence of linearly independent vectors by an orthonormal sequence.

Theorem 2. Let x1 , x2,... be linearly independent vectors in a Euclideanspace E. Then there exists an orthonormal sequence Yi such thatlin{x1,. . .,x,j = lin{y1,. . . ,yj for every n.

Proof. Forn = 0,1,... set

= and =

where isthe orthogonal projection onto so that (x — I M,,for every x E E. Then 0 since M, and EAlso, Set = Then the sequencehas the required properties. Indeed, by the construction, M,, =lin{y1,. . ,yj and so the y are linearly independent. Furthermore,

and 0

In fact, it is easy to define explicitly the orthogonal projectionappearing in the proof of Theorem 2; namely, set = 0 and for n 1

define

PMX=

The Gram—Schmidt orthogonahzation process enables us to show thatevery separable space (i.e. every space containing a countable dense set)

contains a fundamental orthonormal sequence. Also, by Zorn's lemmaevery Eucidean space contains a complete orthonormal system.

Theorem 3.(a) Every separable Euclidean space contains a fundamental orthonor-

mal sequence.(b) In a Euclidean space, every orthonormal system is contained in a

complete orthonormal system. In a Hilbert space, every orthonor-mal system is contained in an orthonormal basis.

Proof. (a) Let be a dense sequence in a Euclidean space E. Dis-card if it is in Lin{x1,. . . and apply Theorem 2 to the sequenceobtained.

(b) Let S0 be an orthonormal system in a Euclidean space E. Set

= {S: S is an orthonormal system in E}.

Then I is partially ordered by inclusion and every totally ordered subsetI' of I has an upper bound in I, namely S. Hence, by Zorn's

Chapter 10: Orthonormal systems 143

lemma, .Z has a maximal element Thus is a complete orthonor-mal system containing So. If E is a Hubert space, then the system S1 isa fundamental system. 0

4. (1) Let T be the unit circle in the complex plane:T {z E C: Iz I = 1}. Equivalently, T is the real line modulo 2ir:

T = {e1': t E R} = 0 t 2ir}.

Let E be the Euclidean space CW with inner product

i j2w

(f,g)=y-J f(t)g7Jjdt,IT0

where we have used the second convention: f and g are continuousfunctions from [O,2ir] to C, with f(0) = f(21T) and g(0) = g(2ir). Thefunctions = e*hhl (n = 0, ±1, ±2,...) form an orthonormal system inE. By the Stone—Weierstrass theorem, the linear span of these func-tions is dense in C(T) endowed with the uniform norm. Hence

n = 0, ±1,.. .} is a fundamental, and so complete, system in E.The completion of E is L2(1J.

(ii) Let E be the Euclidean space C[— 1, 1] with inner product

=&.

The functions 1,:, t2,... are linearly independent in E and, by theStone—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 nthLegendre polynomial

=

where D is the differentiation operator (see Exercise 1).(iii) 1,sint,cos:,sin2,cos2t,... is an orthogonal sequence in the Hil-

bert space 1.2(0, it). Since the subspace of continuous functions is densein L2(0, it), by the Stone—Weierstrass theorem this sequence is a funda-mental orthogonal sequence.

(iv) For n = 0,1,... define the nth Rademacher function r, E 1.2(0, 1)

by

— 1 if is even

— —1 if is odd.

144 Chapter 10: Orthonormai systems

Equivalently, = sign sin 2"irt. (Note that we view as an elementof L2(0, 1); so it is, strictly speaking, an equivalence class of functionsdiffering on sets of measure 0. Thus it is also customary to define

t < t

an it is easilyseen to be incomplete.

(v) Let a < b and let ç E C(a,b) be a positive function.For f and g in the space b) of continuous functions with compactsupport, set

rb(f,g)

= Jdt.

a

Then is an inner product on b); denote its completion byThe space L2(ç) consists of all measurable functions f(t) on

[a, bJ such that

dt <

Now let a = 0, b = and = et. Then the functions 1,t,t2,form a fundamental system in L2(ç) and the polynomials obtained fromthem by the orthogonahzation process are the Laguerre polynomials upto constant factors, i.e. multiples of (n = 0, 1,...) (see Exer-cise 5).

(vi) Consider the space constructed in Example (v) butwith a = b = and ç(t) = e'. Orthogonalizing the sequence1, t, t2,... we obtain multiples of the Hermite polynomials, i.e. multiples

= 0,1,...) (see Exercise 6). 0

If )7 is an orthonormal basis in an n-dimensional Euclidean spaceE, then every vector x E E is a linear combination of the Qk:X = Furthermore Ck = Also, if

=and

= k=1dkpk,

then

(x,y)= k—i

Ckdk and 11x112

= k—ickI2.

In other words, the correspondence x identifies E with


The main reason why an orthonormal basis (cok)r in a Hubert space His 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 thereare unique coefficients (ck)T such that x = ckpk. This sequence(ck)r satisfies 11x112 =

12 and every sequence satisfying

I12 arises as a sequence of coefficients. Thus an orthonor-

mal basis enables us to identify H with 12, just as any n-dimensionalEuclidean space can be identified with

Theorem 5. Let be an orthonormal sequence in a Hubert spaceH. Then, for a scalar sequence c = (Ck)°, the series is con-

vergent iff C C (i.e. 2 If the series is convergent then

leo= 1ICI12 = (

k=1

Proof. Set x,, = Ck(pk. Then, for 1 n m, by the Pythagoreantheorem, we have

lIXn_Xm112k=n+1

Hence is convergent if lCk 2 is convergent. Relation (1)holds since, again by the Pythagorean theorem,

k=1ICkH

A slightly different formulation of Theorem 5 goes as follows. Letbe 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 forevery k. This is usually called the Riesz—Fischer theorem. It looks par-ticularly easy (it is particularly easy) because the space is assumed to becomplete. The original form of the theorem concerned L2[O, 11, wherethe completeness, which is far from trivial, had to be proved.

Let be an orthonormal sequence in a Hilbert space H. Forx C H, set Ck = (k = 1,2,...). We call c1,c2,... the Fouriercoefficients of x with respect to the series

k=1

is the Fourier series of x with respect to

146 Chapter 10: Orthononnal systems

It is easily seen that if is an orthonormal basis then every vectoris the sum of its Fourier series and so, in particular, it is determined byits Fourier coefficients. In fact, we have the following slightly more gen-eral result.

Theorem 6. Let be an orthonormal sequence in a Hilbert space Hand let M = lin(cok)r. Then, for all x,y E H, we have

k1= PM(x);

k=1Rx,ck)12 = IIPM(x)112

(iii)k=1

= (PM(x),PM(y)) = (x,PM(y)) = (PM(x),y).

Also, suppose c = C 12, i.e. ICkI Then there is aunique vector u C M with Fourier coefficients Ck = (k =1,2,...), namely u = Ck(pk. Furthermore, v C H has Fouriercoefficients if u = PMV.

Proof. Set

= k=1

Then, for 1 k n, we have = 0 and so x =x a 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); further-more, (ii) follows from (i) and (1).


To see (iii), set

=

Then = lim and so

(PMx,y) = (x,PMy) = (PMX,PMY)

= =

= timIc=1

=

The proof of the second part is equally easy. By Theorem 5, the seriesis convergent to some vector u M. Then

= c1ço1, Qk) = Ckfri

for every k, and so u does have Fourier coefficients c1 , c2 Also, ifu = PMV then

(V,ck) = (v,PMpk) = (PMV,ck) = (u,ck)

and so u and v have the same Fourier coefficients. Finally, if v E Hand = Ck for every k then = 0 for every k and sov—u is orthogonal to = M. Therefore u = PMV, as claimed. 0

The following useful corollary is amply contained in the result above.

Corollary 7.(a) (Bessel's inequality) If is an orthonormal sequence in a

Euclidean space E and x E E then

k=1

(b) (Parseval's identities) If is a complete orthonormal sequencein a Hubert space H and x, y E H then

k—i= 11x112 and

k—I= (x,y). 0

148 Chapter JO: Orthonormal syste,ns

Parseval's identities imply that every infinite-dimensional Hubertspace 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) and IThen the map H given by x I is a linear isometry of H onto

Corollary 9. Every n-dimensional Euclidean space is linearly isometricto and every separable infinite-dimensional Hubert space is linearlyisometric to '2 0

Having seen that all separable infinite-dimensional Hilbert spaces areisometric to 12, what can we say about other Hilbert spaces? As inExercise 11 of Chapter 9, given a set F, we denote by 12(fl the vectorspace of complex-valued functions f on F with countable support andsuch that Then

11111 = ( lf(y)12

is a norm on '2(i) and with this norm 12(F) is a Hilbert space. Thusis precisely '2 and 12({I , 2,.. . , n}) is

Theorem 10. Every Hubert space is isometrically isomorphic to a space12(fl.

Proof. Let H be a Hubert space. Then H contains a complete ortho-normal system, say {q.,: y E f'}. Then, by Bessel's inequality, for everyx 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


Setting i(y) = the map H 12(fl given by x i is a linearisometry of H onto l2(fl. 0

The fact that every separable (infinite-dimensional) Hubert space isisometrically isomorphic to '2 is very important when we are studyingthe abstract properties of Hubert spaces. Nevertheless, in applicationsHilbert spaces often appear as spaces of functions, and then we areinterested in the connections between the Hubert space structure andthe properties of the functions.

One of the most important Hilbert function spaces is the com-pletion of C(T), the space of continuous functions on 'the circle T,defined in Example 4(i). In the standard orthonormal basis ii =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)

for every f E with denoting the Hubert space norm, i.e. thenorm 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 theStone—Weierstrass theorem, which tell us that f can be uniformlyapproximated by a trigonometric polynomial. (Of course, we used pre-cisely the Stone—Weierstrass theorem to show that {eml: =0,±1,±2,...} is a complete system.) In particular, for every >0there 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. Sincethe projection operator has norm 1, we have

IISnffII = <2€.

The very easy relation (2) leaves open the question whether theFourier series of f E L2(T) tends to f pointwise in some sense. Forexample, it may not be unreasonable to expect that if f is continuousthen tends to f(t) for every t. Sadly, this is not true, as was

150 Chapter 10: Orthonormai systems

first shown by du Bois Reymond in 1876. Nevertheless, Fej6r proved in1900 that there is a simple way to recover a continuous function fromthe partial sums of its Fourier series. To be precise, Fej& showed thatthe Fourier series of a continuous function if Cesdro swnmable and thesum is the function itself: if f is continuous then

Sd,n+1 k=O

i.e. the average of the partial sums, tends uniformly to f. Fej6r'stheorem was of tremendous importance: it launched the modern theoryof Fourier series.

Aithougli need not converge to f(t), the set of points at whichit fails to tend to f(t) cannot be too large. It was asked by du Bois Rey-mond whether the partial sums Sn! of a function f C(T) tend to falmost everywhere, and later Lusin conjectured that the answer is in theaffirmative for every f L2(T). For many years this was one of the

most famous conjectures in analysis; finally, it was proved by Carlesonin 1966. The very intricate proof is one of the greatest triumphs of hardanalysis; not surprisingly, it is far beyond the scope of this book.

Exercises

1. Show that the Legendre polynomials

D ngr,2 i — n 1— n ,

RI.' 1J /

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 from1,t,t2,... by the Gram—Schmidt orthogonalization process, count-ing P0(z) = 1 the 0th term, P1(t) = : the first, P2(t) = 4(3g2_ 1) thesecond, etc.

2. Show that = 1 and = (—1)".3. Deduce from the previous exercises that

—(2n— = 0

for all n 2.

4. Show that D((:2— 1)Ph(t)) is orthogonal to for k <n, anddeduce that

= 0

for every n 1.

Chapter 10: Orthonormal systems

5. Check that the Laguerre polynomials etDlt(e_hthl) (n = 0, 1,...) areindeed orthogonal polynomials in where ç': [0, oo) —' R is

e', as in Example 4(v).6. the analogous assertion about the Hermite polynomials

et and weight function = : —' R, as inExample 4 (vi).

7. Let L2(ç) be as defined in Example 4(v), with a positive weightfunction (a, b) R. Suppose that 1,t, t2,... L2(q). Let

be the polynomials obtained by orthogonalizing thesequence 1, t, t2 (We do no: normalize the sequence; so p,,O)is the unique monic polynomial of degree n orthogonal to everyfork<n.)

(1) Show that

=

(ii) Show that for n 2 we have

= (t — )p,, — — _2(1)

where

— (tPn—i,Pn—i) — IIPn—i1122 and

IIPn _iIl lIP,, —211

8. Let x1,. . . ,x, be unit vectors in a Euclidean space and let

. ,y,, be the sequence obtained from it by Gram—Schmidtorthogonalization, with

=Show that 1, with equality only if y =

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 orthog-onal.

10. Let x1,. . . be linearly independent vectors in a Euclideanspace, with N = ("p). Show that there are orthonormal vectorsYi ,.. • , y,, such that y• = A1x1, where A1 , A2,. . . , A,, are dis-

joint subsets of {1,2,...,N}.


11. Define the Gram determinant of a sequence x1 , ...... , in a Hil-bert space as

i.e. as the determinant of the n xn matrix whose entry in the ithrow and jth column is the inner product (x1,x1). Show thatG(x1 ,.. . , 0, with equality iff the set {x1,.. . x,j is linearlydependent.

Show also that if {x1 ,. . . , x,,} is a linearly independent set span-fling a subspace then

=

\ G(x1,x2,. . .

12. Prove the following weak form of the Riemann—Lebesgue lemma:if f E C(T) then

.1:dt 0 —*

13. The aim of this exercise is to prove Fe/Er's theorem. For f E L2(T)set

= (Skf)(t).

(i) Show that

t) f(x) — x)

where Ku(s), the FejEr kernel, is defined as

Ku(s)=

rs

(ii) Prove that if 0 <s < 2ir then

IKu(s) = —Ifl+1\ sInks

and

= = n+1.


(iii) Deduce from (i) and (ii) that Ku(s) 0 for all s E T,Ku(s) 0 uniformly for 0 < s 2ir—S < 2ir and

1— I2ir

(iv) Deduce Fejér's theorem: if f C(T) then

—* f(t)

uniformly in t as n —p(v) Show also that if f E and f is continuous at then

5

14k. Let H be a Hubert space and let S1 = y f) and S2 ={y.,: y E fl be such that (Xa , = (where = 0 if aand ôaa = 1). 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 wehave hardly scraped the surface of Fourier analysis proper, the reader isencouraged to consult a book on the topic; T. W. Körner, FourierAnalysis, Cambridge University Press, 1988, xii + 591 pp., is particu-larly recommended.

The original Riesz—Fischer theorem was proved in F. Riesz, Sur lessystèmes 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 indepen-dent discovery, precisely described by Fischer in the introduction of hispaper: "Le 11 mars, M. Riesz a présenté a l'Académie une Note sur lessystèmes orthogonaux de fonctions (Comptes Rendus, 18 mars 1907).J'étais arrivé au mime résultat et je l'ai démontré dans une conferencefait a Ia Société mathématique a Brunn, deja le 5 mars. Ainsi monindépendance est évidente, mais la priorité de Ia publication revient aM. Riesz."

Carleson's theorem, mentioned at the end of the chapter, was provedin L. Carleson, Convergence and growth of partial sums of Fourierseries, Acta Math., 116 (1966), 135—57.

Exercise 9 is from J. Hadamard, REsolution d'une question relative auxdEterminants, Bulletin Sd. Math. (2), 17 (1893), 240—348, and Exercise


13 is from L. Fejér, Sur les fonctions bornées et inlegrables, ComptesRendus, 131 (1900), 984—7 and Investigations of Fourier series (in Hun-garian), Mat. és Fiz. Lapok, 11(1902), 49—68; see also Untersuchungenuber Fouriersche Reihen, Math. Annalen, 58 (1904), 51—69.

11. ADJOINT OPERATORS

In the next four sections we give a brief account of the theory ofbounded linear operators on Banach spaces. Our aim is to presentseveral general concepts and prove some of the fundamental results.We are mainly interested in the spectral theory, to be treated in threesections, but before we embark on that, in this chapter we study thebasic properties of the adjoint of an operator.

Throughout this chapter we shall use the product notation for thevalue of a linear functional on an element: (x,f) = (f,x) = f(x) for x ina vector space V and f in V', the dual of V. In particular, if X is anormed space and X* is the dual of X, i.e. the Banach space of allbounded linear functionals on X, then (,) is the bilinear form onXxr 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 her-initian 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 theadjoin: or dual r of an operator T E Y) is the unique mapr: r X such that

(x,Tg) = (Tx,g) (1)

155

156 Chapter 11: Adjoint operators

for all x E X and g E r. Indeed, for g E r, we have a function Pgon X defined by (Pg)(x) = (Tx,g); it is easily seen that this function islinear and, in fact, Pg E Y*. In turn, the second dual of T is theunique 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 Y).Then(a) P E and IIP1I = 11711;

(b) for scalars A1 and A2, we have (A1 T1+A2T2)* = A17? +A21r;(c) with the natural inclusions X C * and Y C r * we have

T,i.e. P*x Txforallx€X;(d) if Z is a normed space and S E then (ST)* = PS*;(e) if T is invertible, i.e. T' E then P is also invertible and

(T*)1 =

Proof. Part (a) is Theorem 3.9, part (b) is immediate from the definitionof the adjoint, and (c) follows from Theorem 3.10, giving the naturalembeddings X—* X** and Y—* To see (d), note that for x E Xand h we have

(x,PS*h) = (Tx,S*h) = (STx,h).

Finally, (e) follows since T' T = 1x. where is the identity operatoron X and so 'r = = Similarly, Iy• =Hence (Pr' = 0

Let now H and K be Hubert spaces, with their inner products writtenas (, ). We know from the Riesz representation theorem in Chapter 8that a Hilbert space is naturally isometric with its dual but the isometryis an anti-isomorphism. Thus x E H can be considered as a functionalon H: it acts on H as multiplication on the right: (,x). Then forTE we have P E

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 functionf: 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 is immedi-

ate that 1' is a linear map from K to H. Since

l.f(x) I II TxIl 11711 Dxli

we have

Dull = liryli = 11111 IITllIlylI

and 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 EThen r E IT'll = 11711 and, for all scalars A1 and A2 we have

(3)

(T1T2)* = T17* (4)

and

r*T (5)

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, this is also clear from Theorem 1(b) and the fact that the

natural isometry between a Hubert space and its dual is an anti-

isomorphism.Finally, let T E Then

117112 = sup IITxlI2 = sup (Tx, Tx)11x01 IjxH=1

= sup i(rTx,x)I

IIrlll IlrlIlIllI = 0


Let us emphasize again that in (3) we have to take the conjugates ofthe coefficients, unlike in Theorem 1(b), which is the analogous state-ment for normed spaces. This is because in (3) we take 1? as a mapfrom 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 anti-isomorphism. Theorem 2 is of special interest in the case H = K. Aninvolution on a complex Banach algebra is a map x such that (3),(4) and (5) hold:

(A1x1+A2x2) (xix2)* = x.

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 r. Hence everynorm-closed subalgebra of which is closed under taking adjoints(i.e. every closed .subalgebra of is also a C*.algebra. What isremarkable is that the converse of this statement is also true: everyalgebra is isomorphic to a closed *..subalgebra of This is thecelebrated Gelfand—Naimark theorem; although it is beyond the scopeof this book, we shall present some exercises concerning it at the end ofthe next chapter. Qiven a normed space X with dual X, for a subspaceK C X define the annihilator of K inr as

K°= {fcr:(x,f) = OforallxE K}.

Similarly, for a subspace L C r, the annihilator of L in X (or thepreannihilator 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) I1 for all f E }.

Of course, if K and L are subspaces, as we have chosen them, or atleast if they are unions of one-dimensional subspaces, then the twodefinitions 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 Y). Then

Ker T = °(Im T) and Ker r = (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) 0 for all x E X}

= {g Y*: (Tx,g) = 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 inclusionX C However, if X if reflexive and so X is identified withthen for every set L C X* we have °L = L°. If H is a Hubert spacethen not only is H a reflexive space but also the dual H is identifiedwith H. With this identification, L° = = LL for every set L C H.Hence Theorem 3 has the following immediate consequence.

Corollary 4. Let H and K be Hubert spaces and let T EThen

Ker T = (Im T)1 and Ker T = (Im T)

It is worth noting that Im T = °(Ker does not hold in generalsince Im T need not be closed. However, if Im T is closed then we dohave Im T = °(Ker T) (see Exercise 3).

Let H be a Hubert space. An operator T is called hermizianor seif-adjoint if r = T. Thus T is self-adjoint if

(Tx,y) = (x, Ty) for all x,y E H.

Clearly an operator T E is self-adjoint if (x,y) = (Tx,y) is a her-mitian form on H. If S and T are commuting seif-adjoint operators thenST is also hermitian since

(STx,y) = (Tx,Sy) = (x,TSy) = (x,STy).


In particular, if T is seif-adjoint then so is T" for every n 1. Notealso that if T is seif-adjoint then by Theorem 2,

II = II 7'' lii = 117112,II = II = II

etc. Therefore IIT2kII = IITII2*. Also, if 1 n 2" then

11T2k1I = IITnT2*_hhII 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 the operator rr is a positive self-adjointoperator. Indeed,

(rT)' = = rT and (rTx,x) = (Tx, Tx) = IITxlI2 0.

Replacing T by r, we see that is also a positive self-adjoint opera-tor.

Theorem 5. Let H be a complex Hilbert space. Then every operatorT E has a representation in the form T = + iT2, where T1 andT2 are hermitian, and this representation is unique.

Proof. Set

T1 = and T2 =

Then T1 and T2 are hermitian and T = T1 + IT2. The uniqueness followsfrom the fact that if T1 and T2 are hermitian and T1 + IT2 = 0 then

T1+iT2 = (T1+iT2)* = T1—iT2

andsoT2=OandT1=0. 0

Examples 6. (I) Let T be the right translation on '2' i.e. letT((x1,x2,...)) = ((0,x1,x2,...)). Then r is the left translation:

= ((x2,x3,...)). Clearly 11711 = !lrII = 1, is theidentity I but I:

TT*((xi,x2,x3,...)) = ((0,x2,x3,...)).

(ii) For C[O, 1} define TQ: L2(O, 1) L2(0, 1) as multiplicationby

(Tçf)(t) = ço(t)f(t) (0 t 1).

Chapter 11: Adjoint operators 161

Then

and IIT,Il = I 1}.

Clearly Tç is a positive hermitian operator 1ff is a non-negative real-valued function.

(iii) Let M be a closed subspace of a Hubert space and let be theorthogonal projection onto M. Then is a positive self-adjoint opera-tor and (as every projection) satisfies = 0

It is easily seen that the properties in Example 6(111) characterise theorthogonal projections in a Hubert space.

Theorem 7. Let H be a Hilbert space and let P E be a self-adjoint projection: P2 = P = P. Then M = ImP is closed and P is theorthogonal projection of H onto M: 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 directsum of KerP and ImP. o

In addition to hermitian operators and orthogonal projections, let usIntroduce two other important classes of operators. Given a Hilbertspace H, an operator T E is said to be normal if 7T = r T, andunitary if T is invertible and its inverse is r. Note that every hermitianoperator is normal, and so is every unitary operator. In the followingresults 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.(b) If Tis normal then = for everyn land

KerT = Kerr = (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 1ff ((rT— rr)x,x) =0 for every x.

(b) If T is normal then, by (a), we have Ker T = Ker P. Hence, byCorollary 4,


Im T =

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 bestrengthened: if a projection is normal then it is an orthogoani projec-tion (see Exercise 7).

Theorem 9. Let H be a Hubert space and let U E be such thatIm U = H. Then the following are equivalent:

(a) U is unitary;(b) U is an isometry: lUxil = lxii for every x E H;(c) U preserves the inner product: (Ux, Uy) = (x,y) for all x,y C H.

Proof. The polarization identity (3) in Chapter 9 implies that U is anisometry 1ff it preserves the inner product. Thus (b) and (c) are equiva-lent.

If U is unitary then

(Ux,Uy) = (U*Ux,y) = (x,y).

Conversely, if (Ux, Uy) = (x,y) for all x,y C H then (U*Ux,y) = (x,y)and so U = U an isometry, U is invertible.Therefore U* = U 0

Our last aim in this chapter is to show that the converse of Theorem1(v) also holds if X is a Banach space, i.e. 7" is invertible if T is. Firstwe note a simple condition for invertibility. Call T C 24(X, Y) boundedbelow if iiTxIl for all x C X and some 0.

Theorem 10. Let X be a Banach space, Y a normed space, and letThen T1 iff ImT is dense in Y and T is

bounded below.

Proof. The necessity of the two conditions is obvious. Suppose thenthat the conditions are satisfied. If T is bounded below then it is injec-tive, 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 bythe second condition, is also convergent, say to x. Hence

Tx = T(Tzk) = hmk._,O Zk = Y

andsoY=Z.l'his shows that T1 If flTxfl dxli for all x E X then,

clearly, 11T'il 1/c. 0

Theorem 11. Let X be a Banach space, Y a normed space and letT E Y). Then r is invertible 1ff T is.

Proof. We have seen that if T is invertible then so is r. Suppose thenthat T is invertible. Let us check that the two conditions in Theorem10 are satisfied.

By Theorem 3 we have (Im T)° = Ker T = (0) and so Im T is densein Y. To see that T is bounded below, let x C X and let f be a supportfunctional at x, i.e. let f E be such that (x,f) = fixil and 11111 = 1.Then

lixil = = (x,r(ry'f)= (Tx,(rY'f)

Thereforefl Txli Ii — lix completing the proof.

If Im T is not dense in Y, say TX C Z, with Z a closed subspace, thenZ C Kerf for some f C (f 0). Hence (x, = (Tx,f) = 0 forall x E X and so = 0. In particular, if r is bounded below thenIm 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 letT Y). Then the following conditions are equivalent:

(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 ofan operator and the structure of the algebra of bounded linear opera-tors. But that requries a new chapter.


Exercises

1. Given a Banach space X and a set K C X, the annihilator of K inris

= (x,f> =OforallxE K)

and the annihilator of a set L C X* in X (or the preannihilator ofL) is

= {x X: (x,f) = 0 for ailfE L}.

Show that if K and L are subspaces then = K0 and =where K° is the polar of K and °L is the prepolar of L. Show alsothat

= = (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 subspaceL C r, the sets °L and L° need not be equal under the naturalinclusion X C

X and Y be normed spaces and T E Y). Show that°(Ker is the closure of Im T.

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 allu U. Let X be a normed space and T E Show that aclosed subspace Y of X is an invariant subspace of T 1ff Y° is aninvariant subspace of r.

5. Let X and Y be Banach spaces and T E Y). Prove that Im Tis closed iff Im T* is closed.

6. Let X be a Banach space. Show that for T E the seriesT"/n! converges in norm to an element of denoted by

exp T. Show also that (exp T) * = exp r and if S com-mutes with T then

(exp S)(exp T) = (exp T)(expS)exp(S+ T).

In the exercises below, H denotes a Hubert space.

Chapter ii: Adjoint operators

7. Let T be a bounded linear operator on a Hubert space H. Showthat T has an eigenvector iff r has 1-codimensional closed invari-ant 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.(I) Show that Im(U—I) = Im(U*_!) and deduce that

Ker(U—I) =(ii) For n 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 strongoperator 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).

(ii) By considering exp(R* — R) S exp( T— show that

IRexpR*)Sexp(_r)II IlSil.

(iii) For! E 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 forevery 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. Ourtreatment of adjoint operators is standard. The Gelfand—Nalmarktheorem was proved in I. M. Gelfand and M. A. Nalmark, On theembedding 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, seechapter 41 in P. R. Halmos, introduction to Hilbert space and the theory ofspectral multiplicity, Chelsea, New York, 1951.

12. THE ALGEBRA OF BOUNDEDUNEAR OPERATORS

In this chapter we shall consider complex Banach spaces and complexunual Banach algebras, as we shall study the spectra of various ele-ments. Recall that a complex unital Banach algebra is a complex alge-bra A with an identity e, which is also a Banach space, in which thealgebra structure and the norm are connected by lieU = 1 and llabIIIlaillibil for all a,b E A. If there is an involution x x* in A such that

= x, = x*+ye, = Ax*, (xy)* = y*x* and llx*xfl =

11x112 then A is a As we noted earlier, if X is a complexBanach space then is a complex unital Banach algebra, and if H isa complex Hilbert space then is a

An element a of a Banach algebra A is invertible (in A) ifab = ba = e for some b E A; the (unique) element b is the inverse ofa, 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 tobe 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 atA or, with a slight abuse of terminology, the resolvent of a.

The prime example of a Banach algebra we are interested in is thealgebra of bounded linear operators on a complex Banach spaceX; so our algebra elements are operators. In view of this, if Tthen we define the spectrum and resolvent set of T without any referenceto

= {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 tobe a subalgebra of the algebra of all bounded linear operatorsacting on the Banach space A, with the element a corresponding to theoperator La of left multiplication by a (so that a La, whereLa(X) = ax for every x E A). In particular if a A is invertible then so

La E with inverse La'. Conversely, if S E is the inverseof 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 theinverse of a. Also, Ae — a is invertible 1ff A!— La is invertible. Hence

crA(a) =

Although the spectrum of an operator T E depends only onhow T fits into the algebraic structure of it is of considerableinterest to see how the action of T on X affects invertibility. In particu-lar, we may distinguish the points of cr(T) according to the reasons whyA!— T is not invertible.

What are the obstruction to the invertibility of an operator S EBy the inverse-mapping theorem, S is invertible 1ff KerS = (0) andImS = X. Thus if S is not invertible then either KerS (0) orImS X (or both, of course). Of these, the former is, perhaps, themore basic obstruction.

Accordingly, let us define the point spectrum of T E as

= {A E C: Ker(AI— T) (0)}.

The elements of are the eigenvalues of T; for an eigenvalueA E the non-zero vectors in Ker(A!— T) are called eigenvectorswith eigenvalue A. Furthermore, Ker(AJ— T) is the eigenspace of Tat A.Clearly C o-(T).

If X is finite-dimensional and S E then the two conditionsKerS = (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 notdense in X or A!— T is not bounded below: there is no 0 such thatII(AI— T)xII €IIxIl for every x E X. In the former case A is said tobelong 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 byUap(T). In other words,

= {A E C: there is a sequence C S(X)

such that (Al— O}.

Sometimes is called an approximate eigenvector with eigenvalue A.Clearly o(T) C T) and

0(T) = 0ap(T)U(Tcom(T).

Sometimes the points of the spectrum are classified further: the resi-dual spectrum is 7r(T) (Tcom(T)\C7p(T) and the continuous spectrum is

= (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-

trum is a closed set; the point spectrum need not beclosed.

We shall show that is a non-empty closed subset of the disc{A E C: IA I The latter assertion is an immediate consequenceof the following simple but important result.

Theorem 1. Let TE satisfy 1111 <1. Then I o(T) and

(I—Ty' = Tk,k =0

where the series on the right is absolutely convergent in the Banachspace

Proof. Note that

k=O11T9 HT1I' =

and so Tk is absolutely convergent.

Hence

Tk=(l_T)+(T_T2)+(T2_T3)+...=lk =0

and, similarly,

T")(I_T) = I.


R(A) = (Al—TY' = A_1(I_f)=

(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 beempty or not. In proving that it cannot, we shall make use of Banach-space-valued analytic functions, an example of which we have just seenin(2).

Given a Banach space X and an open set D C C, a functionF: D X is said to be analytic if for every z0 E D there is anr = r(z0) > 0 such that D(z0, r) = {z E C: I z —

I<r} C D and

=n0

for some a0,a1,... E Xand all z E D(z0,r), with the series in (4) beingabsolutely convergent. Thus (2) shows that the resolvent R : p(T)

is an analytic function, with values in the Banach spaceThe standard results concerning analytic functions remain valid in this

more general setting. For example, as we noted in Exercise 10 ofChapter 11, the analogue of Liouville's theorem holds: a norm-boundedentire 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 theHahn—Banach theorem, = F(0) for all z.

Also, the radius of convergence of (4) is just as inthe classical case. Equivalently, the Laurent series

G(z) = (5)

has radius of convergence s = lim 1/n•

Indeed, if IzI >s then < IzI for some e >0 and everysufficiently large n. Hence <(1 if n is sufficiently large,implying that (5) is absolutely convergent.


Conversely, if z <s then there is an infinite sequence n1 <such that > But then > 1 and so (5)

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) areanalytic functions into a Banach algebra A then F1 F2: D1 fl D2 —' A is

also an analytic function. Furthermore, if

F1(z) = F2(z) = forz E D(z0,r),

then

F1(z)F2(z)

= n'Owhere

= k0akbfl_k.

These observations more than suffice to prove the main results of thechapter.

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 isnow the entire plane, R(A) is an entire function. Furthermore, (3)shows that IIR(A)II 0 as Al Hence, by Liouville's theorem,R(A) is constant and, in fact, R(A) = 0 for every A. But this is clearlyimpossible. 0

For emphasis, let us put Corollary 4 and Theorems 2 and 5 together.

Theorem 6. For every complex Banach space X and operatorT E the spectrum o(T) is a non-empty closed subset of{A E C: IA I II71I}. Furthermore, if A p(T) and

d(A,o'(T)) = : E r(T)} = d

then IIR(A)II l/d.

It is interesting to note that Theorem 5 gives an independent proof ofthe fact that every nXn complex matrix has an eigenvalue. Note alsothat the analogue of Theorem 5 fails for real spaces, as shown, forexample, by the rotation (e1 , e2) '—i (e2, —e1) in R2. The spectrum isvery useful precisely because it allows techniques of complex analysis tobe brought into operator theory.


From Theorem 6 it is a short step to show that the approximate pointspectrum 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 thereis a sequence C X such that y,, 0 and = 1 for everyn. Setting = we find that = 1 and

II(AI— I0.

Hence )° is an approximate eigenvector with eigenvalue A. 0

As an illustration of the concepts and results presented so far, let usexamine the spectrum of the right shift operator S on 1,, (1 pdefined by S(x1 , x2,...) = (0, x1 , x2,...). Since IISII = 1, the spectruma(S) is a closed non-empty subset of the closed disc 4 = {A E C: lAP1}. Furthermore, IISxII = IIxII for every x and so IRS—A)xII

(1—IAI)IIxIJ, implying that aap(S) C 84 = {A E C: IAI = 1}. ByTheorem 7 we have oap(S) = 84. (Of course, this is easy to checkdirectly.)

How much of the circle 84 belongs to the point spectrum? Supposethat 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 anopen set containing the spectrum.

Theorem 9. Let p(t) be a polynomial with complex coefficients. Thenfor 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) is 1 andp(O) = 0. Given A0 E C, let

p(:)—A0=

(t—/.Lk).

Then


p(T) —A0!= k=1

This product fails to be invertible if at least one of the factors, sayis 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 0

The spectral radius of T E is

r(fl =sup{IAI:AE0.(T)}=max{IAI:AEor(T)}.

The spectral radius is a simple function of the sequenceshown 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

Ln=0

is convergent for IA > r(T). Hence, recalling the formula for theradius of convergence, we find that r(T) 0

It is easily seen that the spectral radius is an upper semicontinuousfunction of the operator in the norm topology; in fact,

r(S+7') r(S)+ lflfor all S, T E (see Exercise 8).

It is worth recalling that all the results above are true for the spectra ofelements of Banach algebras, not only of elements of In the sim-plest of all Banach algebras, C, every non-zero element is invertible. Infact, C is the only Banach algebra which is a division algebra.

Theorem 10 (Gelfand—Mazur theorem) Let A be a complex unitalBanach algebra in which every non-zero element is invertible. ThenA=C.


Proof. Given a E A, let A E 0(a). Then A —a (= Al—a) is not inverti-bleandsoA—a=0,i.e.a=A. 0

We know from Theorem 11.11 that T E is invertible 1ff

1'. E is invertible. Hence A!— T is invertible if Al— T is.Therefore, recalling that for a Hilbert space H, the dual H is identifiedwith H by an anti-isomorphism, we get that for T E A!— T isinvertible if (A!— T)* = A!— r is invertible. Finally, recalling fromTheorem 11.8(b) that for a normal operator T we have IIT"II = 11111" forn 1, we have the following result.

Theorem 11.(a) For a Banach space X and an operator T E we have

u(T') = 0(T).(b) For a Hilbert space H and an operator T E 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 associ-ated with a linear operator. Given a Banach space X and an operatorT C 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 unitsphere S(X), a support functional f at x, i.e. a point of the unit sphereS(Xt) taking value I at x, and evaluate f at Tx. It is clear that thenumerical range depends on the shape of the unit ball, not only on thealgebra If T is an operator on a Hubert space then V(T) is justthe 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 beeasier to handle than the spectrum and is often more informative. It isclear that, just as the spectrum, V(T) is contained in the closed disc ofcentre 0 and radius 11711. Even more, the closure of V(T) is sandwichedbetween 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 a complex Banach space X and an operatorT C we have


V(T) C C V(T),

where V(T) is the closure of V(T).

Proof. The first inclusion is easily seen since if x E S(X), f Eand (x,f) = 1 then i E S(X**), (f,i) = I and = (x, T'f) =

where, as earlier, i denotes x considered as an element ofTo 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 < 1.

Since B(X) is in B(r*), there is an x C B(X) such that

< and <

Then, by Theorem 8.11, there exist g C S(X*) and y S(X) such that(y,g) = 1, ix—yli <€ and Of—gil €. Hence

KT(x—y),f)i < and ikTy,f—g>Ii

But this implies that V(T) has a point close to namely the point(Ty,g) C V(T):

=

lii +

Theorem 13. For a complex Banach space X, the spectrum of T is con-

tained in the closure V(T) of the numerical range V(T).

Proof. Suppose that

d(A,V(T)) = infflA—,.ti C V(T)} = d >0.

By Theorem 12 we also have d(A, = d. To prove that o(T) Cwe have to show that Al— T is invertible.

Given x E S(X). pick a support functional f C S(X) at x, i.e. a

norm-I 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— T)xii

for all x C X.


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. Butthen, by Theorem 11.12, Al— T is invertible. 0

Another aspect of the connection between the spectrum and thenumerical range given in Theorem 13 is that coo(T), the convex hull ofspectrum, is precisely fl ëö V(T), where the intersection is taken over allnumerical ranges V(T) with respect to norms on X equivalent to thegiven 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 materialbeyond the main body of this book. The theorem is Johnson'suniqueness-of-norm theorem, but the beautiful and unexpectedly simpleproof 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 inthe annulus R1 < Izi < R2 then M, = is a convex func-tion of logr for R1 <r < Thus if f(z) is analytic in the open discI

<R0, where R0> 1, then

maxlf(z)I max jf(z)I (7)jzI=R IzI=1/R

forallR(1 <R< R0).To see (7), note simply that g(z) = f(z)f(1/z) is analytic in (an open

set containing) the annulus hR Iz R and so attains its maximumthere on the boundary. In particular,

If(1)12 = Ig(1)l maxig(z)i max If(z)I.IzI=R 1z11/R

A similar inequality holds for Banach-space-valued analytic functions,with the norm replacing the modulus. In particular, if f(z) is analytic inthe open disc Iz I <R0, with values in a Banach space X then

111(1)112 max IIf(z)II max IIJ(z)II. (8)JzI=R 1z11/R

for all R (1 <R < Re). Indeed, let f(z) = E X) and letE S(X*) be a support functional at f(1) =

= =


Set

g(z) = p(J(z)) =n =0

Then, by (7)

IIf(1)112 = Ig(l)12 maxlg(z)I max Ig(z)lIzI'=R IzI=1/R

= maxlq'(f(z))I max Ic(f(z))IIzkrR

= 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 spectralradius.

Lemma 13. Let f(z) be an analytic function in the open disc jz I <R0,with values in a Banach algebra A. Then

r(f(l))2 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 ofz for zl = R and fz I = hR. Consequently, by Dini's theorem(Theorem 6.5), for every e > 0 there is an n such that

max max max{r(J(z))+e} max {r(J(z))i-e}.IzI=R Izkl/R zl=R

Applying Theorem 9 and inequality (8), we find that

r(f(l))2

max maxIzIR zI=1/R

max{r(f(z))+e} max {r(f(z))+€}.zI=R IzI=1/R

As this holds for every e > 0, the result follows. 0

The radical Rad B of a complex (unital) Banach algebra B is the inter-section of all the maximal left ideals of B. The following lemma relatesthe radical to the spectral radius.

Lemma 15. If b E B is such that r(b'b) = 0 for all b' E B thenbE RadB.


Proof Suppose that b Rad B, that is b L for some maximal leftideal L. Then Bb + L is a left ideal properly containing L, and soBb + L = B. Hence e = b'b + I for some b' E B and 1 L, where e isthe identity. But then e—b'b = I E L and so e—b'b is not invertible.Therefore r(b'b) 1. 0

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 allBanach-algebra norms on B are equivalent.

Theorem 16. Let A and B be Banach algebras, with Rad B = {O}. Thenevery surjective homomorphism 0: A B is automaticaLly continuous.

Proof. Suppose that a,, 0 in A and b in B. By the closed-graph theorem (Theorem 5.8) it suffices to show that b = 0. SinceRad B = {O}, this is the same as showing that b Rad B, and byLemma 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'. Setc = a'a, = d = 0(c) = b'b. Then —, 0 in A

and d, = — d in B. Define a linear function C A by

=

and set

= =

Note that p,,(l) = c and = d.Since a homomorphism does not increase the spectral radius.

Hence, by Lemma 14, for all n 1 and R> 1 we have

Letting n —' 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 com-plex unital Banach algebra with norm

lband Rad B = {0}. Then

every Banach algebra norm on B is equivalent to Do. 0


The algebra of bounded linear operators on a Banach space Xsatisfies the conditions in Corollary 17, so all norms on turning itinto 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 suchthat S 0 and TS = 0 (ST = 0). Show that the point spectrum ofT is

= (A E C: Al— T is a left divisor of zero)

and the compression spectrum is

= {A E C: Al — T is a right divisor of zero}.

2. We call T a left (right) topological divisor of zero if there areT1, 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 neithera left nor a right divisor of zero then it is both a left topologicaldivisor of zero and a right topological divisor of zero.

6. Show that

Oap(T) = C C: Al— r is not surjective}

and

crap(r) = {A C C: Al— T is not surjective}.

7. Suppose that S, T C commute: ST = TS. Show that

r(S+ T) r(S) +r(T) and r(ST) r(S)r(T).

Show also that these inequalities need not hold if S and T are notassumed to commute.


8. Show that for S, T E we have

r(S+T) r(S)+ 11111.

Show also that if r(S) = 0 then r: —+ [0, is continuous atS.

9. Suppose that r(T) < 1. Show that

11110= n0

is a norm on X, equivalent to the original norm10. Prove that

r(T) = inf{II TU': II II'is a norm on X, equivalent to IlL

11. 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.

Deduce that a(ST) U{0} = o(TS) U{0}. Show also that o(ST) =o(TS) need not hold.

13. Let K be a non-empty compact subset of C. Show that K is thespectrum of a Hubert space operator: there is an operatorS E where H is a Hilbert space, such that u(S) = K.

14. Show that if T is a normal operator on a Hilbert space thenr(T) = 11711.

15. For w = E define 12 by =Show that = = SUpkIWkI. Show also that eigenvaluesof Tare w1,w2,... and o-(T) = What is

16. For w = (Wk)° define '2 l2 by

= (0,w1x1,w-,x2,...).

Express and the spectral radius in terms of thesequence w.

17. Let 4 = {z E C: lzI

1} and H = L2(4). Let T E be theoperator of multiplication by Show that o-(T) = 4 and T has noeigenvalue.


18. Use the Hahn—Banach theorem to show that for T C we

have

supReV(T) = sup{c C there isanx = x(c) X(x 0),

such that 11(1 —rc+rT)xIj lixilfor allr 0}.

19. An operator T C is said to be dissipative if sup Re V(T) 0.

Show that if T is dissipative then

IIx—rTxII HxII

for all x C X and r 0.

20. Show that if A C V(T) and (A!— 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 11 y\ yx—-+-Ilx—-l x—r r\ r/ r

for r > 0.) Deduce that if A C aeo V(T) and (Al— T)2z = 0 then(Al—T)z = 0.

21. An operator T E is said to be hermitian if V(T) C R. (Notethat X is a Banach space; for a Hubert space this definition coin-cides with the usual (earlier) definition of a hermitian operator ona Hubert space.) Thus T is hermitian if both iT and —iT are dis-sipative. Prove that T is hermitian if

iiexp(iT)xH = lxii

for all r C and x C X, where

S'expS

=

for S C22. Let H be a Hilbert space and let S C be such that

V(S) C Show that

(Sx,y) (Sx,x)"2(Sy,y)'12

and deduce that

IISxIi2 (Sx,x)iiSii

for all x,y C H. Deduce that if 0 V(S) then 0 C a(S).


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 deducedfrom 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 maxi-mal 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 commutativeGelfand—Nalmark theorem. In these exercises A is a commutative com-plex unital Banach algebra.

27. Show that if M is a maximal ideal of A then AIM is isometricallyisomorphic 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 maybe identified with the set of non-zero homomorphisms h : A —÷ C.By Exercise 28, this is a subset of B(A*). Give At the relativeweak* topology (i.e. the topology induced by the weak* topologyon A*). Endowed with this topology, we call At the maximal idealspace 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 Cdefined by i(h) = h(x), where h E At is considered as a homomor-phism 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 maximalideal space At then the Gelfand transformation x '—b i maps Aisometrically and isomorphically onto C(At), the commutativealgebra of continuous functions on the compact Hausdorff spaceAt.

32. Let be the algebra of all (doubly infinite) complex sequencesx = such that

=x,j

with convolution product xy = z, where


Zn =

Show that is a commutative Banach algebra. Show also thatthe maximal ideal space of 11(Z) can be identified with the unit cir-cle T = {z E C: Izi = 1}, where z: C is defined by

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 Show that for everye > 0 there is an invertible operator S such thatIISTS'II <r(T)+e.

Notes

The original reference to Gelfand's spectral-radius formula is I. M. Gel-fand, Normierte Ringe, Mat. Sbornik N. S., 9 (51) (1941), 3—24; this isalso one of the references to the Gelfand—Mazur theorem; the other isS. 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 Bishopand Phelps, Bull. London Math. Soc., 2 (1970), 181—2, and Theorem 13is 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, NumericalRanges 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'suniqueness-of-norm theorem, Bull. London Math. Soc., 21 (1989),487—8.


The commutative Gelfand—Nalmark theorem is taken from I. M. Gel-fand and M. A. Naimark, On the embedding of normed rings into the ring ofoperators in HUbert space, Mat. Sbornik, 12(1943), 197—213; another classicalreference is M. A. Naimark, Normed Rings, Revised English edition; trans-lated 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 appearingin this chapter are complex Banach spaces. Our aim is to study a classof operators closely resembling the operators on finite-dimensionalspaces; we shall show that these operators are somewhat similar to thenXn complex matrices.

Vaguely speaking, the operators we shall look at are 'small' in thesense that they map the unit ball into a 'small' set. To be precise, anoperator T E Y) is compact if the image of the unit ball Bxunder T, is a relatively compact (i.e. totally bounded) subset of Y.Thus T is compact if and only if T iscompact if and only if for every bounded sequence C X thesequence has a convergent subsequence.

We shall write Y) for the set of compact operators from X intoY. Analogously to we write for X).

Examples 1. (i) Every finite rank operator T E Y) is compact, i.e.if dimlmT= dimTX< then TE Indeed, set Z = ImT.Since Z is finite-dimensional, Bz is compact and so is a subset ofthe compact set IIllIBz.

We shall denote by Y) the set of (bounded) finite rank opera-tors from X to Y.

(ii) Every bounded linear functional f EX to C.

(iii) Let I be the closed unit interval [0, 1] and let X be the Banachspace C(I) of continuous functions with the supremum norm. LetK(x,y) E C(IxI), i.e. let K be a continuous function on the closed unitsquare Ix I. For f E C(I) define a function Tf E C(I) by

186

Chapter 13: Compact operators on Banach spaces 187

(Tf)(x)=

K(x,y)f(y) dy.

Then T E and it is easily seen that, in fact, T E Indeed,by the Arzelá—Ascoli theorem (Theorem 6.4) we have only to check

that is uniformly bounded and equicontinuous. If IK(x,y)I Nfor every (x,y) E Ix! then

l(Tf)(x)l N L If(y)I dy NlIfjI,

and so TBx is uniformly bounded by N. Also, for E > 0, there is a

& >0 such that if lxi—x21 <8 then IK(xi,y)—K(x2,y)I <e. There-fore iffE and 1x1—x21 <8 then

I(Tf)(xi) —

fIK(xi ,y) — 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 a Hilbert space with orthonormal bases and

For every operator T we have

IITeaO2=1=1 1=1 1=1 1=1 j=1

and so

= I(Tf1,e1)12 = lIre,I12 =1=1 1=1 j=1 j=1 1=1

This shows that\1/2

O71IHS = (IITe1II2

\i = 1

is independent of the orthonormal basis (e1An operator T E is said to be a 1-filbert—Schmidt operator if its

Hubert—Schmidt norm, II T1IHS, is finite. One often writes for theset of Hubert—Schmidt operators on H, and 111112 for

II indicates, the Hubert—Schmidt norm is thenorm of the sequence formed by the entries of the matrix representationof T. Indeed, set = (Te,,e,> so that T is given by the matrixA = (a11) in the sense that

188 Chapter 13: Compact operators on Banach spaces

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 Tx = (x,a1)e1.

\i=1 / 1=1

In particular,

V

=I (x, a.) 2 lIxII2IIaiIl2 = 11x11211

and so 11111 II11IHS.It is easily seen that every Hubert—Schmidt operator is compact.

Indeed, with the notation as above, put

A ==

xe1 E H: 1x11 11a111, i =

Since < the set A is a compact subset of H (see Exercise1). Since TBH is a subset of A, it is relatively compact. 0

The class of compact operators is an example of a closed operatorideal. An operator ideal is a function that assigns to every pair X, Yof Banach spaces a subset Y) of Y) such that if T E Y),SE and R E then STE and TR E Y).

Theorem 2.(a) Y) is a closed subspace of Y).(b) If TE SE and RE then

STE 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 C then E Y). Let be


a bounded sequence in X. As S is compact, has a subsequence, saysuch that is convergent. The operator T is also compact,

and so (xflk) has a subsequence, say (Xmk), such that (TXm*) is conver-gent. But then ((AS+itLT)Xm&) is also a convergent sequence.

Now let us show that Y) is a closed subset of Y). Sup-pose T E Y). We have to show thatTB1 is totally bounded. Given c > 0 let n be such that II — <€.As is compact, there are x1,.. . ,x,, E B1 such that {T,,x1: 1i m} is an c-net in 1,B1. Thus if x C Bx then there exists an x, suchthat <e. Then

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 convergentsequence. 0

Since finite rank operators are compact, by Theorem 2(a) every limitof finite rank operators is compact. The problem of whether every com-pact operator can be obtained in this way was, for many years, one ofthe best-known problems in functional analysis. After about 40 yearsthe approximation problem was solved in the negative by Per Enflo in1973, who constructed a separable reflexive Banach space X for which

is not the closure of Since is the closure ofwhenever X has a (Schauder) basis (see Exercise 7), Enflo's

example also showed that not every separable reflexive Banach spacehas a basis, solving another long-standing question. As so often inmathematics, the counterexample turned out to be the Start of the story:it opened up whole new fields of research on approximation and basisproblems. But we cannot go into that in this book.

The operator ideal of compact operators is closed under takingadjoints as well.

Theorem 3. An operator T C Y) is compact if and only ifr E is compact.

Proof. (i) Suppose first that T C Y) is compact, i.e. the setK = TBx is compact. For a functional f C r let Rf be the restrictionof f to K. Clearly Rf E C(K) and, in fact, the map R: Y' —, C(K) is abounded linear map, where, as usual, C(K) is taken with the supremumnorm. Let P = RBr. Note that forf C r we have


fIrfIl = sup{(x, rf>: x E Bx}

= sup{(Tx,f): x E

= sup{(y,f): y E TBx}

= sup{(y,f):yEK} = IIRfII.

This shows that rBr is isometric to 'P C C(K), with the isometrygiven by Rf.

Consequently rBr is totally bounded if and only if 'P is. By theArzelá—Ascoli theorem, 'P is totally bounded if and only if it is uniformlybounded and equicontinuous. Both conditions are easily checked.

If f E Br then DrfIl iirii = flTfl, and so 'P is uniformlybounded. Also, if f E Br and y,y' E K then

I(Rf)(y)—(Rf)(y')I = If(y—y')I IIy—y'lI,

and so J) is equicontinuous.(ii) Now suppose that r E is compact. By part (I), the

map T E is compact, i.e. is relatively compact.But under the natural embeddings X C r* and Y C rs we haveTBx = T*BX C rBr., and so TBx is also relatively compact.0

Theorems 2 and 3 state that the compact operators form a closedoperator ideal that is also closed under taking adjoints. In particular,for a Banach space X, the set is a closed ideal of the Banachalgebra If H is a Hubert space then is a closed ideal of Hwhich is also closed under taking adjoints: it is a closed

The main aim of the chapter is to present the spectral theory of com-pact operators, due to Frigyes (Friedrich or Frédénc) Riesz. Recall thatthe spectrum of an operator T E is

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 previouschapter that for every T E the spectrum of T is a non-emptyclosed subset of C, contained in the disc {z E C: Izi IITU}. As weshall see, for a compact operator T E the spectrum of T resem-bles the spectrum of an operator on a finite-dimensional space, i.e. thespectrum of an PZXn matrix. To be precise, if X is infinite-dimensionaland T C then r( T) is a countable set whose only accumulationpoint is 0 and if A C or(T) (A 0) then A is an eigenvalue of T withfinitely many linearly independent eigenvectors.


Theorem 4. Let T E and a > 0. Then T has only finitely manylinearly independent eigenvectors with eigenvalues having modulus atleast a.

Proof. Suppose x1 , x2,... is an infinite sequence of linearly independenteigenvectors such that Tx1 = A1x1 0 and A1 I a for every i. Set

= . . ,x1j. By Theorem 4.8(b), there exists a sequenceC Xsuch E and = = 1.

Put = and note that Iz,,fl 1/a, E X,, andE Indeed, the first two assertions are obvious and if

= CkXk then

= xn_1.

Hence if n > m then

IITZnTZrnII = = 1.

Consequently the bounded sequence does not contain a subse-quence such that is convergent, contradicting the compact-ness of T. 0

Our next aim is to show that if T is a compact operator and A 0 isnot an eigenvalue of T then A o-(T), i.e. Al— T is invertible. Equiva-lently, 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 wemay and shall assume that A = 1; so our aim is to prove that S = I— Tis invertible. We need two lemmas, both of which are proved in a moregeneral 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 finite-dimensional, say with basis {b1 ,. - . , bk,, }. Then there is a closed subspaceM 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 takeM = Kerf1. The projections of x E X into N and M arePN(4 = 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


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 somesequence C M = 1). Since T is compact, has a con-vergent subsequence, and so we may assume that itself is conver-gent, say Tx,, —' y. Then

= =

and so = 1. But we also have Sy, and so S0y = 0, contrad-icting Ker S0 = {0}. 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 apoint x0 E such that d(Txo, TY) > 1 €. 0

Proof. By Theorem 4.8(a), there exists an x0 E Bx such thatd(x0, Y)> 1—c. As Tx0 = x0 — Sx0, Sx0 E Y and TI' = (I — S) Y C I',we have

d(Tx0,TY) d(x0—Sx0,Y) = d(x0,Y)> 1—c. 0

Theorem 7. Let T be a compact operator and suppose A 0 is not an

eigenvalue of T. Then A o(T).

Proof. By replacing T by T/A, it suffices to prove the result for A = 1.Let then T E S = I— T and KerS = (0). We have to show thatS is invertible.

Let us prove that SX = X. Set = (n = 0,1,...), so that

= A' D . By Lemma 5 the subspaces are closed. Let usshow first that = for some n. Indeed, otherwise Y0 3 Y1 3and all the inclusions are strict. Then, by Lemma 6, one can find ele-ments E such that > But then, in particular,

— TYmI! > if n m, and so has no convergent subse-quence, contradicting the compactness of T.

We claim that, in fact, Y0 = Suppose that this is not so. Thenthere is an m such that Let U E Then,as Su I'm + i SYm, there exists a pOint v such that

Su = Sv. But then S(u—v) = 0 and so 0 u—v E KerS, contradicting

our assumption. Consequently = I'0, i.e. SX = A', as claimed.The proof is essentially complete. The bounded map S: X — X is a

1—1 map of the Banach space X onto itself and so, by the inverse-mapping theorem, S is invertible. 0


Let us restate the information contained in Theorems 4 and 7 aboutthe spectrum of a compact operator as a single result.

Theorem 8. Let T be a compact operator on an infinite-dimensionalBanach space. Then (T) = {O,A1,A2,...}, where the sequenceAi ,A2,... (of non-zero complex numbers) is either finite or tends to 0;furthermore, every A. is an elgenvalue of T, with finite-dimensionaleigenspace. 0

With some more work, we can gel more detailed information aboutthe structure of compact operators. If T is compact and S = 1— T thenIm S is a finite-codimensional closed subspace of X; even more, for asuitable 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. SetNk = KerSk and Mk = ImS" (k = 0,1,...), where = Thenis 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. Further-more, M,, = Mm for all m n, Xis the direct sum of M,, and N,,, and

M,, is an automorphism of M,,.

Proof. By expanding 5" = (1— T)", we see that = I— where Tk iscompact. Hence, by Lemma 5, Mk is a closed subspace of X. ClearlyN0 C N1 C ... and M0 J J ...; furthermore, we know that eachNk is finite-dimensional.

As in the proof of Theorem 7, Lemma 6 implies that there is a smal-lest n such that N,, = N,, + and there is also a smallest m such thatMm = Mm+i. Then N,, = N,,. for all a' a and Mm = Mm' for all

Let us turn to the main assertions of the theorem. We prove first thatN,, fl M,, = {0}. Let y N,, fl M,,. As y E M,,, we have y = S"x forsome x E X. But as y N,,, S"y = 0 and so = 0. Hencex E N2,, = N,,, implying S"x = 0. Thus y = S"x = 0, showing thatN,,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 avector y such that = Sex. Hence x—y N,, and sox = y + (x — y) shows that X = N,, + Could we have p > a?Clearly not, since then M,, would strictly contain and so we would


have # {0}. Thus p = n and X is the direct sum of andas is finite-dimensional (see Exercise 4.20).

Finally, = = and

= = 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 crowningachievement of this chapter: a rather precise description of the action ofa compact operator.

Theorem 10. Let X be an infinite-dimensional Banach space and let Tbe a compact linear operator on X. Then o(T) = {0,A1,A2,...}, wherethe 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) andMA = M(A; T) invariant under T, such that NA is finite-dimensional andX =

The restriction of Al— T to MA is an automorphism of MA,

NA = Ker(AI— Ker(Al—

and for = A A = A we have NA C MM.

Proof. Only the last claim needs justification. The operator T mapsMA into itself and NA into itself. Furthermore, T)INA is an auto-morphism of the finite-dimensional space NA since if we had(id— T)x = 0 for some x E NA (x 0) then we would have(AJ—T)'1x = 0 for every n I, contradicting =0. Consequently, = NA for every n 1, and so NA C MM.

0

There is no doubt that Theorem 10 is a very beautiful theorem. Atfirst sight it is not only beautiful but very impressive as well: it seems tocome close to giving us a very fine decomposition of the space into adirect sum of generalized eigenspaces. Unfortunately, this is rather amirage: the theorem cannot even guarantee that our compact operatorhas a non-trivial closed invariant subspace, let alone give a direct-sumdecomposition. In fact, non-trivial closed invariant subspaces do exist,as we shall prove in Chapter 16. However, to prove the existence ofinvariant subspaces we shall need some results to be proved in Chapter15. Before we turn to that, in the next chapter we shall show that a


best possible decomposition can be guaranteed if we deal with a com-pact normal operator on a Hubert space.

Exercises

1.

A = {x = (x,)° li,.: lxi

a compact subset of2. Let K be a closed and bounded subset (1 p Prove

that K is compact if and only if for every 0 there is an n suchthat lx1V <€ for every x = E K.

3. As in Example I (xiii) of Chapter 2, let 1) be the vector

space of k times continuously differentiable functions f: (0, 1)

such thatk

11111k = sup > IfW(()11=0

Show that Xk = 1), II Ilk) is a Banach space and the formalidentity map i: Xk —p Xk_I (f J) is a compact operator.

4. Let T E Y), where X is infinite-dimensional. Show. that theclosure of TS(X) = {Tx: x X, lixil = 1} in Y contains 0. (HINT:Consider a sequence C S(X) such that 1 for

n m.)5. Let be an orthonormal basis of a Hilbert space H and let

T Y), where Y is a normed space. Show that 0.

6. Let X be a normed space such that for every finite set A C X andevery 0, 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 is the canonical projec-

tion onto N. Show that is the norm closure of i.e.


every compact operator on X is the operator-norm limit of finiterank operators.

7. Let X be a Banach space with a Schauder basis )'. Show thatis the closure of

8. Let be a sequence of non-zero complex numbers tending to0. Show that cr(T) = {0,A1,A2,. . .} for some complex operator Ton some compact Banach space X. Show that if all A, are realthen X can be chosen to be a real Banach space.

9. Let X and Y be Banach spaces, let T E Y), and letJ E Y) be invertible. Show that Im(J— T) is closed in Y andhas finite codimension.

10. Let X be a complex Banach space, and let T E be such thatis 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 isan 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 = be thedirect sum of these spaces, with x = having norm

Ilixill=

Let TE T

E for every n.12. Let X be a Banach space, (1, C and let 7, —' T in the

operator-norm topology. Show that is relatively com-pact, where B = B(X) is the unit ball of X. Show also that if

A then A C o(T).13. Let E be the space C[0, 1], endowed with the Euclidean norm

11f112

=

(j1

If(x)12 dx)

and let T be as in Example 1 (iii). Show that T C i.e. Tmaps the unit ball of the Euclidean space E into a relatively com-pact set.

14. Let H be a Hilbert space and T C Show that T C0 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.


16. Let c3 , c2,... be non-negative reals and

C {x E 12: x = xkI for every k}.

Show that if C is a compact subset of '2 then Ck 0. For whatsequences (Ck is C compact?

17k. Let K be a compact subset of a normed space. Show that K iscontained in the closed absolutely convex hull of a sequence tend-ing 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 infinited'inconnus, Paris, 1913, and Uber lineare Funktionalgleichungen, ActaMath., 41(1918), 71—98. In presenting the Riesz theory of compactoperators we relied on Ch. xi of J. Dieudonné, Foundations of ModernAnalysis, Academic Press, New York and London, 1960, xiv + 361 pp.

Theorem 3 is due to J. Schauder, Uber lineare vollstetige funktionalOperationen, 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 Banachspaces, Acta Math., 30 (1973), 309—17; a simplified version of the solu-tion 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 aBanach space X, the space can almost be written as a direct sum of gen-eralized eigenspaces of T. If we assume that X is not merely a Banachspace, but a Hubert space, and T is not only compact but compact andnormal, then such a decomposition is indeed possible — in fact, there is adecomposition with even better properties. Such a decomposition willbe provided by the spectral theorem for compact normal operators: acomplete and very simple description of compact normal operators.Thus with the study of a compact normal operator on a Hilbert spacewe arrive in the promised land: everything fits, everything works outbeautifully, there are no blemishes. This is the best of all possibleworlds.

We shall give two proofs of the spectral theorem, claiming theexistence of the desired decomposition. In the first proof we shall makeuse of some substantial results from previous chapters, including one ofthe important results concerning the spectrum of a compact operator.The second proof is self-contained: we shall replace the results of theearlier chapters by easier direct arguments concerning Hilbert spacesand normal operators.

To start with, we collect a number of basic facts concerning normaloperators in the following lemma. Most of these facts have alreadybeen proved, but for the sake of convenience we prove them again.

Lemma 1. Let T E be a normal operator. Then the followingassertions hold.

(a) = for every x E H.(b) KerT= Kerr.(c) = 11711" for every n 1.

198

Chapter 14: Compact normal operators 199

(d) r(T) = 11711.

(e) If A then Ker(AI— T) I Ker(j&I— T).(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 wehave

11111 = max{ll Toll, II T111}

and T, is a normal operator on with = rh-i1 (i = 1,2).

Proof. (a) As rr = we have

llTxIl2 = (Tx,Tx) = (rTx,x) = (TTx,x) = (rx,rx) = llrxll2.

(b) By part (a), we have Tx = 0 if rx = 0.

(c) If S E is hermitian then

llSxll2 = (Sx,Sx) = (SaSX,x) = (S2x,x) IlS2llllxll2.

From this it follows that 115211 = 11S112, and by induction on m we get= 11S112'". This implies that IISII = IISII" for every n 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 =

(In fact, (c) and (d) are equivalent: if S E for a complex Banachspace X then r(S) = IISII if IlShI = for every n 1.)

(e) If Tx = Ax and Ty = then ry = jiy because by (b) we havey E T) = Ker(jiI— Therefore

A(x,y) = (Tx,y) = (x,ry) = (x,1y) =

and so if A then (x,y) = 0.

(f) As Al— T commutes with T and r, Ker(AI— T) is invariant underboth Tand r.

Also, let (x,y) = 0 for all yE Ker(AI—T). Then, since Ker(Al—T) isInvariant 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 112

max{II 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 H1 =

H0 = are invariant under P. 0

It is worth emphasizing that Lemma I is a collection of elementaryand simple facts, except for part (d), which is based on the spectral-radius formula.

Let us see then the first incarnation of the spectral theorem, claimingthe existence of a spectral decomposition for a compact normal operator.

Theorem 2. Let T E be a compact normal operator. For aneigenvalue A of T, let HA = Ker(T— Al) be the eigenspace of T belong-ing to A, and denote by PA the orthogonal projection onto HA. Theoperator T has countably many non-zero eigenvalues, say A1 , A2

Furthermore, dim HAk for every k, the projections PA are orthogo-nal, i.e. "Ak"A, = 0 if k 1, and

(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 and andS, = SIH1 (i = 1,2), we have T0 = S0 and S1 = 0. Therefore, by


Lemma 1(g),

lIT—SO = max{JIT0—Soll, llT1—S111} = IITill.

But T1 is a compact normal operator and so, by Theorem 13.8, llT1llis precisely the maximum modulus of an eigenvalue of T1. As everyeigenvalue of is an eigenvalue of T, by our choice of n we haveIIT1II Hence (1) does hold. 0

Let us state two other versions of the spectral theorem.

Theorem 3. Let T be a compact hermitian operator on an infinite-dimensional Hilbert space H. Then one can find a closed subspaceof H, a (finite or countably infinite) orthonormal basis of and asequence 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 theunion of these bases. Let H0 be the closed linear span of the orthonor-mal sequence and set v,, = Ak if x,, E HAk. 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 (orTheorem 3). Let {x7: y E f) be an orthonormal basis of a Hubertspace H, and let T be such that Tx,, = Then T is com-pact 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 thespectral-radius formula. We shall show now how one can proveTheorem 2 without relying on these results. It is a little more con-venient to prove Theorem 2 for compact hermuian operators; it is then asimple matter to extend it to normal operators.

Recall that the numerical range V(T) of a Hilbert space operatorTE is


{(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 realsince

(Tx,x) = (x,rx) = (x,Tx) = (Tx,x)

for every x H and so (Tx,x) is real. In fact, T E is hermitianif its numerical range is real. Also, the spectrum of a hermitian opera-tor is real. We shall not make use of any of the results proved aboutnumerical ranges; the next lemma is proved from first principluses.

Lemma 5. Let T be a hermitian operator. Then 1111 = v(T).

Proof. Set ,' = v(T), so that (Tx,x)I for every x H. Wehave to show that

(I ill v.Given x S(H), let y E S(H) be such that Tx = IITxIIy. Then

(Tx,y) = (x, Ty) = IJTxII and so

ITxH = (Tx,y) =

= I)y112} = v.

Hence IITxII i' for every x E S(H) and so 11Th v, as claimed. 0

Theorem 6. Let U be a compact hermitian operator on H. Then U hasan eigenvalue of absolute value

Proof. Set

a = inf (Ux,x) and b = sup (Ux,x)lxii = 1 11111 = I

so that = [a,b]. By Lemma 5, flUfl = max{—a,b}. Replacing Uby —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

subsequence, we may suppose that is convergent, sayThen b because —' b and = I. As


=

=

and

—, b,

we have

—*0.

Therefore

=

Put x0 = y0/b. Then, on the one hand, Yo = bx0 and, on theother hand, Ux,, —p Ux0. Consequently we have Ux0 = bx0. Since

IIxofl 1 (in fact, lixoll = 1), b is indeed an eigenvalue of U. 0

Let us now see how Theorem 6 may be used to deduce Theorem 2 forcompact hermitian operators. For the sake of variety, we restateTheorem 2 in the following form.

Theorem 7. Let H be a Hubert space and let U be a compacthermitian operator. Then there is a (possibly finite) sequence (Ak) ofreal 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,, bethe eigenspace belonging to A7: H,, = Ker(U—A,,I). We know thatH,..1H8 if y 8.

Let us show first that dim H,, and, for every 0, there are onlyfinitely many A,, with IA,. I e. Suppose not. Then, by taking anorthonormal basis in each H,, with IA., I e, we find that there is aninfinite orthonormal sequence such that Ux,, = where

I I€. But then does not contain- a convergent subsequence,

contradicting the compactness of U.


This implies that the non-zero eigenvalues may be arranged in asequence (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 Mof all the H,, and, consequently, so is M1. Denote by U the restrictionof 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, itfollows from the definition of M and from Theorem 6, that U = 0.

If the sequence (A,,) of non-zero eigenvalues is finite then we aredone. Otherwise, let

x= k1

Xk + where Xk E 11k and i E M1.

Put

= k=IXk + I and

= k=1AkXk.

Then = and —. x. As

= k=1AkXk H,

the continuity of U implies that Ux = y, proving (d). 0

Before we recover from Theorem 7 the full force of Theorem 2, let usshow that compact hermitian operators are rather like real numbers.An operator T E is said to be positive if it is hermitian andV(T) C i.e. (Tx,x) 0 for every x E H. Note that if T is any(bounded linear) operator on a Hilbert space then rr and arepositive (hermitian) operators:

(rTx,x) = (Tx, Tx) IITxII2 and (Trx,x) = I$Tx112.

Theorem 8. A compact positive operator U on a Hubert space has aunique positive square root V. Every hermitian square root of U iscompact.

Proof. Let A 1,A2,... be the non-zero eigenvalues of U, let Hk be theeigenspace belonging to A,, and let M be the closed linear span of the

Then = KerU and > 0 for every k. Define V E byVx = if x E Hk and Vx = 0 if x E M1. Then V is a positivesquare root of U.


Theorem 9. Let T1 ,..., be commuting compact nonnal operators ona Hilbert space H. Then H has an orthonormal basis consisting of com-mon eigenvectors of all the T1.

Proof. For every C C and k = 1,. . . , n, the eigenspace Ker(pJ — Tk)is invariant under all the 7. Hence H is the orthogonal direct sum ofthe subspaces

= ('1

All these spaces are finite-dimensional, with the possible exception of. ,o. Taking an orthonormal basis of each the union of

these bases will do. 0

As our final theorem concerning abstract operators in this chapter, letus note that our results, say Theorem 2 or Theorem 3, give a completecharacterization of compact normal operators up to unitary equivalence.Two operators T, T C are said to be unitarily equivalent if for someunitary operator U we have T' = U'TU = U*TU, i.e. if they have thesame matrix representation with respect to some orthonormal bases.

Let X be the collection of functions n: C\{0} {0, 1,2,. . . } whosesupport {A C\{0}: n(A) 1} has no accumulation point (i.e. in C thereis no accumulation point other than 0). In particular, the support isfinite or countably infinite. The following result is easily read out ofTheorem 2 (see Exercise 14).

Theorem 10. Let H be an infinite-dimensional complex Hilbert spaceand let be the collection of compact normal operators on H. ForT C and A C C\{0} set

nr(A) = dim Ker(A1— T).

Then the correspondence T defines a surjection —÷

furthermore, T and T' are unitarily equivalent if ni.. = nr. 0

We close this chapter by showing how the spectral theorems we havejust proved enable us to solve a Fredholm integral equation.

Let 1 = [a, b] for some a <b, and write E for the Euclidean spaceC(1) endowed with the inner product

(f,g)= J

f(t)g(t) dta

and norm 111112 = Thus the completion of E is L2(0, 1).


Let K(s, t) E C(!x I), and for f E define Uf E E by

(Uf)(s)= J

K(s, dt.a

Then K is the kernel of the integral operator U. It is easily checked thatU e C(I)) and U maps the unit ball of E into a relatively compactset in C(1). Indeed, this follows from the Arzelá—Ascoli theorem, sinceif

IK(s, 1) — K(s', t) I for all t then (Uf) (s) — (Uf) (s') E11f112 by the

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 forthis extension as well. In fact,

(Uf)(s)= J

K(s, t)f(t) dta

forf E L2(O, 1).

From now on we suppose also that K(s, t) = K(t,s); in this case U iseasily seen to be a hermitian operator. We consider a Fredhoim integralequation

g(s)= J

K(s, t)f(t) dt — Af(s) = ((U—A)f)(s).a

For what values of A can we solve this equation, and what can we sayabout the solution? As we shall see, this question is ideally suited forthe 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 Uwith non-zero eigenvalues (A,,) guaranteed by each of Theorems 2, 3and 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 eachA,, is real,

N

for every x, where N depends only on K(s, t), and

JIK(s,t)I2dsdt.

a a

Proof. For 0 t I put ks(s) = K(s, t). Then, by Bessel's inequality,for every m we have


2

=

jbds.

Thereforem m rb rb

= jdx

J j IK(s,t)12 dsdt. 0a a a

Lemma 12. If E L2(0, 1) is an eigenvector of U with eigenvalue0 then E C(J).

Proof. By the Cauchy—Schwarz inequality, Uf E C(I) for f E L2(0, 1).Hence = C(l). 0

Lemma 13. Let g E L2(0, 1), f = Ug and let (ce) be the Fouriercoefficients of f with respect to Then the series con-

verges absolutely and uniformly to f(s) in [0, 1].

Proof. Let g = + be the decomposition of g in L2(O, 1),where for every n. As U: L2(0, 1) C(I) is continuous,

= and 0, the uniform convergence follows. Furthermore,for every finite set F of indices,

/ \2 / \2

I = I/ \nEF

(IanI2)(

n€F nEF /

byLemmall. 0

We are now ready to answer our question about the Fredholmintegral equation.

Theorem 14. If A o(U) and g E C(I) then the Fredholm integralequation

Uf—Af = g

with a hermitian kernel K(s, t) = K(:, s) has a unique solution f given by

1f(t) = +

where is an orthonormal basis of eigenvectors of U, the series is


absolutely and uniformly convergent in [0, 1] and the (as) are theFourier coefficients of g with respect to ('pa), i.e.

CI,

a,= J

dt.a

Proof. The unique solution in L2(0, 1) is! = (U—AY'g. Clearly, f hasFourier 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 spectraltheory of compact hermitian operators to integral equations, notably tothe Sturm—Liouville equation. However, a proper account of theseapplications would be much longer than we have space for.

In conclusion, let us point out that the spectral theorem for compactnormal operators is only the beginning of the story. The result weproved is a very simple version of a spectral theorem for normal (notnecessarily compact) operators. With every normal operator one canassociate a so-called spectral measure containing all the informationabout the operator up to unitary equivalence.

Exercises

1. Let H0 be a closed subspace of a Hilbert space H invariant undera normal operator T E Is H1 = necessarily invariantunder T?

2. Prove relation (2), i.e. show that if y E f) is an orthonormalbasis of a Hilbert space H and T E is given by Tx,, =where E C, then T E 0 there are onlyfinitely 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.5. Let T be a normal operator (i.e. TT* = p1'). Use ele-

mentary linear algebra to show that the matrix of T is diagonal insome orthonormal basis.


6. Let U be the compact operator defined by a hermitian kernel, as inTheorem 14. Show that if U is positive then K(s,s) 0 for everys (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.7. Let be an orthonormal basis consisting of eigenvectors of a

compact hermitian operator U, with = Letbe the multiset : > 0} arranged in a decreasing

order, with y1 , the corresponding eigenvectors. Putting itanother way: let P2 > 0 be the sequence of non-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 subspacesH,,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,,)° Csuch 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 sub-

space of H if U has finite rank.

Chapter 14: Compact nor,nal operators 211

11. Let T E Prove that T is(a) normal 1ff H has an orthonormal basis consisting of eigenvec-

tors 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 non-

negative reals.12. Let U E be hermitian. Prove that there are unique positive

operators U+, U_ E such that

U = - and U.... = U.... U.k. = 0.

13. Prove the Fredhoim alternative for hermitian operators: Let U be acompact hermitian operator on a Hilbert space H and consider thefollowing two equations:

Ux—x = 0 (2)

and

Ux—x=x0 (3)

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 solu-

tion 1ff x0 is orthogonal to every solution of (2); furthermore,if (3) has a solution then it has infinitely many solutions: if xis a solution of (3) then x' is also a solution if x — x' is a solu-tion of (2).

14. Give a detailed proof of Theorem 10. In particular, check that themap f( is a surjection.

15. Let T E be normal and, as in Theorem 10, for A E C setn7.(A) = dim Ker(AI— T). Prove that nT(A) I for every A E C(including A = 0) if there is a cyclic vector for T, i.e. a vectorx0 C H such that lin{x0, Tx0, T2x0.. . . } is dense in H.

Notes

There are many good accounts of applications of the spectral theorem forcompact hermitian opertors to differential and integral equations. Wefollowed i. Dieudonné, Foundations of Modern Analysts, AcademicPress, New York and London, 1960, xiv + 361 pp. Here are some of the


other good books to consult for the Sturm—Liouville problem, Green'sfunctions, the use of the Fredholm alternative, etc: D. H. Griffel, AppliedFunctional Analysis, Ellis Horwood, Chichester, 1985, 390 pp., I. J. Mad-dox, Elements of Functional Analysis, 2nd edn., Cambridge UniversityPress, 1988, xii + 242 pp., and N. Young, An Introduction to HubertSpace, Cambridge University Press, 1988, vi + 239 pp.

15. FiXED-POINT THEOREMS

In Chapter 7 we proved the doyen of fixed-point theorems, thecontraction-mapping theorem. In this chapter we shall prove some con-siderably more complicated results: Brouwer's fixed-point theorem andsome of its consequences. It is customary to deduce Brouwer's theoremfrom some standard results in algebraic topology, but we shall present aself-contained combinatorial proof.

Before we can get down to work, we have to plough through somedefinitions.

A flat (or an affine subspace) of a vector space V is a set of the formF = x+ W, where W is subspace of V. If W is k-dimensional then wecall F a k-flat. As the intersection of a set of flats is either empty or aflat, for every set S C V there is a minimal flat F containing 5, calledthe flat spanned by S. Clearly

F=

A,x1: x. E S, A = 1, n =

Let x0 , x1 ,. . be points in a vector space. We say that these pointsare in general position if the minimal flat containing them is k-dimensional, i.e. if the vectors x1 — x0, x3 — x0,. . . , X,, — x0 span a k-dimensional 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 posi-

tion. The k-simplex o = (x0,x1,. . . ,Xk) with vertices x0,x1,. . . ,Xk is thefollowing subset of R":

213

214 Chapter 15: Fixed-point theorems

k k

p., = 1, p.,> 0 for all i1=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 k-simplex. A 0-simplex is called a vertex.

A simplex a1 is a face of a simplex a2 if the skeleton of crj is a subsetof the skeleton of a2.

Note that the closure of the simplex a = (x0, x1,... , in R" is

5 = ,Xk]

k k= p. x1: p., = 1, p., 0

1=0 i=0

= C {0,1,...,n}},

i.e. the closure of a is precisely the union of all faces of a, includingitself. Also, 5 is precisely x1,. . . , X,,, }, the convex hull of the ver-tices, and a is the interior of this convex hull in the k-flat spanned bythe vertices.

A finite set K of disjoint simplices in is called a simplicial complexif every face of every simplex of K is also a simplex of K. We also callK a simplicial decomposition of the set 1K I = U {u: E K}, the bodyof K. If K is a simplicial complex and a, r E K then the closed sim-plices 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 aclosed n-simplex 5 = [x0, x1,. .. , Let S be the set of vertices of Kand let y: S — {0, 1,... ,n} be an (n+ 1)-colouring of S such that thecolours of the vertices contained in a face [x¼, x.R,. . . , x•] of a- belong to{i0, , i,.}. Call an n-simplex a" multicoloured if the vertices of o"are coloured with distinct colours. Then the number of multicolouredn-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 with0, 1,... , n —1, with each colour appearing once. For an n-simplexa-" 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 contri-

bution of an (n— 1)-simplex E K to m(K)? Jf is not marked,the contribution is 0. In particular, if a-" is in a closed (n — 1)-face ofa other than = [x0,x1,... then the contribution of is 0.tf E K is in and is marked then the contribution of

o" is a face of exactly one n-simplex of K. Furthermore, ifu"—1 is in a, i.e. in the interior of the original n-simplex, thencontributes 1 to m(u") if a"1 is a face of a": as there are two suchn-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)-simplicesin ö0. By the induction hypothesis, this number is odd. Therefore so ism(K), completing the proof. 0

Given points XO,Xi,.. , of R" in general position, for every point xof the k-dimensional affine plane through x0,x1,. .. there areunique reals , A2,. . . , A, such that

Ac

x=x0+ A.(x1—x0).i=1

Hence, there are unique reals p.o, i,... , such that x =and = 1. These p., (i = 0,1,... ,k) are called the barycentriccoordinates 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 ischaracterized by 0.

The barycentric coordinates can be used to define a very useful simpli-cial decomposition. Given a simplicial complex K, the barycentric sub-division sd K of K is the simplicial decomposition of 1K I obtained as fol-lows. For a simplex a = (x0,x1,. . . ,Xk) K set

k1

c,. = Lxi;

thus is the barycentre of a-. The complex sd K consists of all sim-plices ce,,. . . , such that a proper face of a-i +1(i=0,l,...,k—1).


To define the r-times iterated barycentnc subdivision of K, setsd°K = Kandsd'K= 1. Thussd1K= sdK.

The mesh of K, written mesh K, is the maximal diameter of a simplexof K. Equivalently, it is the maximal length of a 1-simplex of K. Notethat if = (x0,x1,. . . ,x,) (i = 0,1,... ,k) are faces of a k-simplex

= = (x0, x1,. . . , and r = . . , then the diameter ofr is less than k/(k + 1) times the diameter of if. Therefore, if K is anysimplicial complex then for every 0 there is an r such thatmesh 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 ifY 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 hasthe following important consequence.

Corollary 2. Let {A0,A1,.. . be a closed covering of a closed n.simplex a = [x0, x1,. . . , x,, } such that each closed face [x¼, x11,. . . , x.] ofa is contained in A.. Then A. aProof. As we may replace A by A iscompact. The compactness of the sets A0, A1 ,. . . , implies that itsuffices 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 then 0. Let K be a triangulation of & such that every simplexof K has diameter less than €; as we have seen, for K we may take aniterated barycentnc subdivision of a. Given a vertex x of K containedin 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

y(a1) = i (i = 0,1,...,n).

But then a1 E A, as required. 0

From here it is a short step to one of the most fundamental fixed-point theorems, namely Brouwer's fixed-point theorem. A closed n-cellis a topological space homeomorphic to a closed n-simplex.

Theorem 3. (Brouwer's fixed-point theorem) Every continuous mappingof a closed n-cell into itself has a fixed point.


Proof. We may assume that our n-cell is exactly a closed simplex0" = [xo,xt,... , x,, 1. Suppose that 0" —p 0" is a continuous map,sending a point

= 1=0=

to

=(IL;

1=0= i).

For each i, let

A =

Then {A0,A1 ,. . ,A,,} is a closed covering of 0". If a point

x = belongs to a closed face [xc, x•1,. .. ,x.] of 0" then = 0

for i {i0, } and so = 1. Since p4 = 1, there isan index j such that p.4' and so x E Consequently,

. . ,x,] C U_0 A,,

showing that the conditions of Corollary 2 are satisfied. Thus there is apointx in all the A; such an xis a fixed point of ç. 0

The following lemma enables us to apply Theorem 3 to a ratherpleasant class of spaces, namely the compact convex subsets of finite-dimensional spaces, i.e. the bounded closed convex subsets of finite-dimensional spaces.

Lemma 4. Let K be a non-empty compact convex subset of a finite-dimensional normed space. Then K is an n-cell for some n.

Proof. We may assume that K contains at least two points (and hence itcontains a segment) since otherwise there is nothing to prove.

We may also suppose that K is in a real normed space and hence thatK 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, ifnecessary, we may assume that 0 E mt K.

Finally, let 5 be an n-simplex containing 0 in its interior, and define ahomeomorphism K 5 as follows: for x E R" define

n(x) = ,zK(X) = inf{t:t> 0, x C tK},

and


m(x) = mo(x) = inf{t: t> 0, x E t&},

and for x E K set

10 ifx=0,

n(x) . 0j—xm(x)

Corollary 5. Let K be a non-empty compact convex subset of a finite-dimensional normed space. Then every continuous map f: K—' K hasa 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 par-ticular, that the corollary is true without the restriction that the normedspace is finite-dimensional. This is based on the possibility of approxi-mating a compact convex subset of a normed space by compact convexsubsets of finite-dimensional subspaces. Unfortunately, the simplelemma we require needs a fair amount of preparation.

Let S = {x1 ,.. • , x,, } be a finite subset of a normed space X. Fore > 0 let N(S, e) be the union of the open balls of radius centred atxI,...

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 setA(x) = A.(x). !f x E N(S,€) then x belongs to at least one openball D(x1,€), and for that index i we have A•(x) > 0. Hence A(x) > 0for every x N(S, e). Define the Schauder projection : N(S, e) —'

co{x1,. . . by

A(x)= L

Herek k

co{x1,. ..,Xk} = Ax1: A 0, A1 = 1i=1

is the convex hull of the points x1 , - . - , xp: the intersection of all convexsets 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 inR" Indeed, if is the standard basis of =

II - say, then the


closed simplex 5 = [e1. ek] is a bounded closed subset of and soit is compact. Furthermore, ç: co{x1 xk}, given by

k k k

A-e, A,x1, where A, 0 and A- = 1.1=1

is a continuous map.

Lemma 6. The Schauder projection is a continuous map fromN(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) thenk

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, Schauder's fixed-point theorem.

Theorem 7. Let A be a (non-empty) closed convex subset of a normedspace X and let f: A A be a Continuous map such that K = f(A) iscompact. Then f has a fixed point.

Proof. Let n 1. As K is compact, there is a finite set

= {xi,..,x*) C K

such that

K C D(xaJ) =

Set = co{x1 ,. . . and denote by the Schauder projection1/n) —, K,,. We have K,, C A, so by Lemma 6 the restnc-

tion Of to K,, is a continuous map of K,, into itself. Hence, byCorollary 5, there is a point x,, E K, such that p,,(f(x,,)) = x,. There-

fore, by (2),


< (3)

As each belongs to the compact set K, the sequence has aconvergent subsequence, say — x as k cc, where x E K. Butthen, by (3), x as k cc, and so 1(x) = x. D

As a beautiful application of Brouwer's theorem, we prove Perron'stheorem concerning eigenvalues of positive matrices.

Theorem 8. A matrix whose entries are all positive has a positive eigen-value with an eigenvector whose coordinates are all positive.

Proof. Let A = (a11) be an nXn matrix with a11 > 0 for all i and j. Letbe the standard basis in R". The closed (n — 1)-simplex ö =

[e1 ,.. . , e,, j is a 'face' of the unit sphere

S(lr) = = C IIxIIi=

I

=1].

The continuous map a given by x Ax/IIAxII1 has a fixed pointx (x1)?. Clearly, Ax = Ax for some A > 0 and x1 > 0 for all i. 0

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 intoitself is a map of the form where S: V—f V is a linearmap. 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 normedspace X and let be a commuting family of continuous affine maps onX such that T(K) C K for all T C Then some x0 E K is a fixedpoint 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 acompact convex subset of K. if S C then S maps KT into itself sinceif Tx = x then T(Sx) = S(Tx) = Sx and so Sx KT. Consequently, iffor some T1 C and S C


then is a compact convex set mapped into itself by S. Hence,by Theorem 7,

K5nfl Kz #0.

This implies that the family of sets {KT: T E has the finite-intersection property. As each is compact, there is a point x0 whichbelongs to every Kr, i.e. Tx0 = x0 for every T E 0

One should remark that it is easy to prove Theorem 9 without relyingon Theorem 7. Indeed, for T and n 1 the afflne map

maps K into itself and = T E n l} is a commuting fam-ily of affine maps of K into itself. From this it follows that the systemof compact sets {S(K): S has the finite-intersection property.Hence there is a point x0 such that x0 E T E and

This point x0 is a fixed point of every T E Indeed, if =x0 for K then

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 contractionfrom the closed unit ball into X (i.e. d(f(x),f(y)) kd(x,y) for allx,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 hasa 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.


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 continuousmap such that for every closed (k — 1)-face of 5 we havef(f) C ?, then f is a surjection.

4. Prove that Brouwer's fixed-point theorem is equivalent to each ofthe following three assertions, where B" = B(11") and S"' =

(In fact, we could take B" = B(X) and S"' = S(X) forany n-dimensional real normed space X.)

(i) S"' is not contractible in itself, i.e. there is no continuousmap S" 'x [0, 1] —p 1 such that for some x0 E wehave x x E

(ii) There is no retraction from B" onto i.e. there is nocontinuous mapf: B" —p such thatf(x) = x for all x S"'.

(iii) Whenever f: B" R" is a continuous map without a fixedpoint then there is a point x E such that x = Af(x) for someO<A<1.

5. Let C be a closed convex subset of a Hubert space H. Show thatfor every x E H there is a unique point of C nearest to x, i.e.there is a unique point E C such that

d(x,q,(x)) = inf{d(x,y): y E C}.

Show also that the function H C is continuous.6. Combine Theorem 3 with the assertion in the previous exercise to

deduce Corollary 5.7. Let C,, = be the n-dimensional cube:

= {x = (x1)? R": 1x11 1 for every i}.

The closed faces of C,, arc

and —1}

(i = 1,... ,n). For each i (i = 1,...,n) let A be a closed subsetof C,, separating and F1, i.e. = U7 U U1, where

U1 are disjoint open subsets of with F$ C andF, C Prove that fl"1 A. 0.

8. Let q' be a continuous one-to-one map from a compact Hausdorifspace K into a Hausdorif space. Show that ç is a homeomorphismbetween K and ç(K).

9. Prove that every compact metric space is homeomorphic to aclosed subset of the Hubert cube:

= 12: 2' for every i}.


10. Prove that every continuous map of the Hilbert cube into itselfhas a fixed point. (Check that the map f: B(12) —' B(12) given byf(x) = S is the right shiftand e1 = (1,0,0,...).)

11. Show that a continuous map of B(12) into itself need not have afixed point.

12. Let C be a closed subset of a compact metric space K, and let f bea continuous map of C into a normed space X. Use Schauder pro-jections 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 readerto prove another beautiful fixed-point theorem.

13. Let a, b, c and x be points in a Hilbert space such thatb = 4(a+c) and

0 < r IIx—aD IIx—bII lix—cil r+€ 2r.

Deduce from the parallelogram law that

IIa—cH 46,

where = 2re+E2, and so ha—cu14. 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,

check that

lixi—bhi hixi —all

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 B(12), and letf: C C be a non-expansive map. Set


= {x E C: IIf(x) —xli 1/n},

and show that 0 for all n.Put = inf{lIxlI: x E and note that the monotone increas-

ing sequence converges to some d 1. Use the result of theprevious exercise to show that

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 continu-

ous and bounded for 0 t 1 and —00 < u <00. Suppose thatIK(s, t) I M and lf(t, u)

IN for all s, t and u (0 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, sayC = {u E C[O,1]: hull MN}.

Apply the Arzelà—Ascoli theorem to show that TC is a relativelycompact subset of C[O, 1].

Make use of the Schauder fixed-point theorem to prove that theHammerstein 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, FixedPoint Theory, vol. I, Polish Scientific Publishers, Warsaw, 1982, 209 pp.,which is a rich compendium of beautiful results. Another interestingbook on the topic is D. R. Smart, Fixed Point Theorems, CambridgeUniversity Press, 1974. The more usual approach to fixed-pointtheorems is via homology, homotopy and degrees of maps; this can befound in most books on algebraic topology. The case n = 3 ofBrouwer's fixed-point theorem (Theorem 3) was proved in L. E. J.Brouwer, On continuous one-to-one transformations of surfaces intothemselves, 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, Introduc-tion a la ThEorie des Fonctions d'une Variable, vol. II, 2me éd., 1910;Brouwer himself proved the general case in 1912. Sperner's lemma isfrom E. Sperner, Neuer Beweis für die lnvarianz der Dimensionzahl unddes Gebietes, Abh. Math. Scm. Hamb. Univ., 6 (1928), 265—72, andTheorem 7 is from 3. Schauder, Der Fixpunktsatz in Funktionalräumen,Studia Math., 2 (1930), 171—80. The original version of Theorem 9 is inA. 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 non-trivial closed invariant subspaces? This question, the so-called invariant-subspace problem, is the topic of this brief last chapter. Until fairlyrecently, it was not known whether there was any operator T without anon-trivial (closed) invariant subspace, and it is still not known whetherthere is such an operator on a (complex) Hubert space.

Much of the effort concerning the invariant-subspace problem hasgone into proving positive results, i.e. results claiming the existence ofinvariant subspaces for operators satisfying certain conditions. Ourmain aim in this chapter is to present the most beautiful of these posi-tive results, Lomonosov's theorem, whose proof is surprisingly simple.

As we remarked earlier, the Riesz theory of compact operators onBanach spaces culminated in a very pleasing theorem, Theorem 13.8,which nevertheless, did not even guarantee the existence of a single non-trivial invariant subspace. This deficiency was put right, with plenty tospare, in Chapter 14, but only for a compact normal operator on a Hil-bert space. Now we return to the general case to prove Lomonosov'stheorem, which claims considerably more than that every compactoperator has a non-trivial invariant subspace. Before we present thisresult, we need some definitions and a basic result about compact con-vex sets.

As in Chapters 13 and 14, all spaces considered in this chapter arecomplex spaces. Furthermore, as every linear operator on a finite-dimensional complex vector space has an eigenvector, we shall consideronly infinite-dimensional spaces.

Given a Banach space X and an operator T E call a subspaceY 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} andY = X are T-invariant subspaces for every T, so we are interested onlyin other invariant subspaces, the so-called non-trivial invariant sub-spaces.

We call a subspace Y a hyperin variant subspace for T if V is an S-invariant subspace for every S E 2LftX) commuting with T. Since Tcommutes with itself, every hyperinvariant subspace is an invariant sub-space.

Note that if T and S commute then Ker T is S-invariant since ifx E Ker T then TS(x) = ST(x) = S(O) = 0 and so S(x) E Ker T. Henceif T has no non-trivial hyperinvariant subspaces then either T is a multi-ple of the identity or it has no eigenvalue, i.e. = 0. In particu-lar, if T E and o(T) {0} then, by Theorem 8 of Chapter 13.the operator T has a non-trivial hyperinvariant subspace.

What is the easiest way of constructing a T-invariant subspace? Picka vector x E X (x 0) and set

= Tx.

Then TY0(x) C

Y(x)

a T has no non-trivial invariant sub-space then Y(x) = X for all x 0. A vector x is said to be a cyclic vec-torforTifY(x) =X.Thus 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-trivialinvariant subspace since if V is a non-trivial invariant subspace then novector in Y is a cyclic vector for T.

Unfortunately, but not surprisingly, this shallow argument is noadvance on the invariant-subspace problem; nevertheless, it tells us thatwe have to concentrate on cyclic vectors.

In the proof of Lomonosov's theorem, we need a basic result concern-ing closed convex hulls of compact sets.

Theorem 1. (Mazur's theorem) The closed convex hull of a compact setin a Banach space is compact.

Proof. Let A be a compact subset of a Banach space X and letK = We have to show that K is totally bounded, i.e. for everye > 0 it contains a finite €-net. Since A is compact, it contains a finite

228 Chapter 16: Invariant subspaces

se-net, say {x1 ,. .. ,x,,}. Thus {x1 ,. . . ,x,j C A and for every x E Athere is an x1 such that IIx—x111 < The set P = co{x1 ,. .. ,x,j is alsocompact, and so it contains a finite {Yi . . Furthermore,the set

= {x E X: d(x,P) <

is convex and contains A, and therefore contains coA. But then with

M = {x E X: flx—y111 < for some i}

we have coA C 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 compactoperator on an infinite-dimensional complex Banach space X. Then Thas a non-trivial hyperinvariant subspace

Proof. We may assume that Tfl = 1. Set

SA = {T}' {S E ST = TS}

and pick a point x0 E X such that lixojI > 11Th. Set B0 = 1) andnote that 0 B0 and () E TB0.

Suppose first that there is a point Yo E X (Yo 0) such that

IIT'yo—xohI 1

for every T' E SA. Then we are done since

V = T' E SA}

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

H7)' X11fl < 1.

Since TB0 is a compact set not containing 0, there are operatorsSA such that for every yE TB0 there is a (1 i a)

satisfying

< 1.

Now we define a map X reminiscent of the Schauder pro-jection. For y E TB0 and 1 i n set


A1(y) = max{O, 1— 117y11}

and

A(y)=

A1(y).

Relation (2) implies that A(y) > 0 for every y E Thereforemay define

=

This map — X is continuous and so is a compact subset ofB0. Consequently, by Mazur's theorem, K ëÔ is a compact con-vex subset of B0. Hence

K

is a continuous map of a compact convex set into itself and so, bySchauder'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. Butthen TIY is an operator on a complex finite-dimensional space and so ithas an eigenvalue A. However then Ker(AI-- T) is a non-trivial hyperin-variant subspace for T. 0

A slight variation in the proof shows that an even larger class ofoperators have hyperinvariant subspaces.

Theorem 3. (Lomonosov's second theorem) If T E commuteswith a non-zero compact operator and is not a multiple of the identitythen it has a hyperinvariant subsp&e.


If, on the other hand, p(T) = 0, then =—

ak T", and so1(x) = lin{x, Tx,..., — 'x} is a T-invariant subspace for every x 0.

0

In spite of the simplicity of its proof, Lomonosov's second theorem is avery powerful result. At the moment, it is not clear how large a class ofoperators Corollary 4 applies to; in fact, for a while it was not clear thatthere is any operator T E which is not covered by Corollary 4.

The invariant-subspace problem for Banach spaces was solved, in thenegative, only fairly recently: Per Enflo and Charles Read constructedcomplex Banach spaces and bounded linear operators on them which donot have non-trivial invariant subspaces. The original proofs were for-midably difficult and the spaces seemed to be rather peculiar spaces.Later, Charles Read gave an easily accessible proof, and showed that hisconstruction works, in fact, on

In view of these great results, the invariant-subspace problem for Hil-bert spaces has become a very major problem in functional analysis. Infact, it is not impossible that the answer is in the affirmative even onreflexive spaces, i.e. that every bounded linear operator on an infinite-dimensional reflexive complex Banach space has a non-trivial invariantsubspace.

Exercises

1. Let X be a non-separable Banach space. Show that everyT has a non-trivial invariant subspace.

2. Show that the following result can be read out of the proof ofTheorem 2. Let SA be a subalgebra of whose elements donot have a non-trivial common invariant subspace. Then ifT C and T 0 then there is an operator A E SA suchthat Ker(I—AT)

3. Let T1 ,..., T,, C be commuting operators. Show that theyhave a non-trivial common invariant subspace.

4. Deduce from Theorem 1 and Exercise 5.5 the following extension ofTheorem 4.10. If the unit ball of a Banach space Xis a-compact then Xis 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, dieeine gegebene kompakie Menge enthäl:, Studia Math., 2 (1930), 7—9.Theorems 2 and 3 are from V. I. Lomonosov, On invariant subspaces offamilies of operators, commuting with a compact operator (in Russian),Funk. Analiz i ego Prilozh, 7 (1973), 55—6; to be precise, Theorem 3 isgiven as a remark added in proof. Corollary 5 is from N. Aronszajnand 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 inB. Beauzamy, Un opérazeur sans sous-espace invariant non-trivial:simplification de l'example de P. Enflo, Integral Equations and OperatorTheory, 8 (1985), 314—84.

Read's result concerning is in A solution to the Invariant SubspaceProblem on the space Bull. London Math. Soc., 17 (1985), 305—17.

An interesting account of the results concerning the invariant-subspace 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, 22B(xo, r), closed ball of radius r and

centre x0, 22Br(XO), dosed ball of radius r and cen-

tre x0, 22space of bounded linear opera-

tors on X, 28Y), space of bounded linear

operators, 28Y), the space of compact opera-

tors, 186Y), the space of finite rank

operators, 186the space of Hubert—Schmidt

operators, 181

ci,, the barycentre of the simplex a,215

Eo 5, the closed convex hull of 5, 55coS, the convex hull of S, 55C-algebra, 167C(K), space of continuous functions

on a compact Hausdorff space K,23

C(L), space of bounded continuousfunctions on L, 23

space of continuous functionswith compact support, 91

space of continuous real-valued functions with compactsupport, 93

C0(L), space of continuous functionsvanishing at infinity, 91

CR(L), space of bounded continuousreal-valued functions, 93

space of continuous real-valued functions vanishing atinfinity, 93

d(X, Y), Banach-Mazur distancebetween X and Y, (p6

D(x,r), open ball, 20D(x0, r), open ball of radius r and cen-

tre x0, 22D,(x0), open ball of radius r and cen-

tre x0, 22ÔA(a), the resolvent set of a in the

algebra A, lffl4, closed unit disc in the complex

plane, 96

fvg, the join of f and g, 93fAg, the meet of f andg, 93I IS, restriction of a Ito S, 25gb(S), space of bounded functions on

5,23

Im T, image of T, 28

k-simplex, 213annihilator of K, 164

K°, polar of K, 158

233

234 Index of notation

11 -norm, 2323

finS, linear span, 38tinS, closed linear span, 38linZ, linear span, 21

one-point compactification of L,

8L, preannihilator of L, 164°L, prepolar of L, 158

Y), space of linear operators, 28

mesh K, the mesh of the simplicialcomplex K, 216

orthogonal projection onto F. 136p-mean, 5

Rademacher function, 143r(T), the spectral radius of T, 174Rad fit, the radical of the Banach alge-

bra B, 178p(T), the resolvent set of the operator

T. 168

sd K, the barycentric subdivision of K,235

S set of vectors orthogonal to 135S(X), unit sphere, 22S(x0, r), sphere of radius r and centre

22sphere of radius r and centre22

set of finite subsets of 0, LL4cr(T). the spectrum of the operator T,

167

o-(X. weak topology generated by115

y(X, X), weak topology on a normedspace, 1.15

a(X*, X), weak-star topology, 1.1.6the spectrum of a in the alge-

bra A, 1670ap(T), the approximate point spec-

trum of the operator T, 169the continuous spectrum of the

operator T, 169Ocom(T), the compression spectrum of

the operator T, 1.68T), the point spectrum of the

operator T, 1.68T), the residual spectrum of the

operator T, 1.69

T-invariant, 226adjoint of T, 31

Hilbert-Schmidt norm, 182Hilbert—Schmidt norm, 187

v(T), the numerical radius of T, 202V(T), the numerical range of T, 21)1

x set of vectors orthogonal to x, 135lixil,,, of x, 23X, completion of X, 35X,, space of linear functionals on X,

28,45X, dual of X. 31r, space of bounded linear function-

als on X,

INDEX OF TERMS

absolute value of an operator, 205

absolutely convergent series, 3.6

absolutely convex set, 27adjoint of an operator, 155adjoint operator, 31affine hyperplane, 46affine map, 220affine subspace, 213Alaoglu's theorem, 118algebra, commutative, 92algebraic dual of a normed space, 45AM-GM inequality, 1analytic, 111annihilator of a set, 164

annihilator of a subspace, 158

approximate eigenvector, 169approximate point spectrum, 169approximation problem, 189arithmetic mean, 6

weighted, 7Arzelà—Ascoli theorem, 90Auerbach system, 65

Banach algebra, 92unital, 32

Banach limit, 59Banach space, 21

Banach's fixed-point theorem, 101Banach—Mazur distance, (16Banach—Steinhaus theorem, 78barycentre, 215barycentric coordinates, 215barycentric subdivision, 215basic sequence, 72basis, 19. 83

canonical, 32Hamel, 42Schauder, 83

standard, 32basis constant, 83

Bernstein and Robinson, theorem of,230

Bessel's inequality, 1.41biorthogonal system, 64

normalised, 64Bishop—Pheips—Bollobas theorem, 122body, 214bounded below, 162bounded linear operator, 28bracket notation, 28Brouwer's fixed-point theorem, 216

canonical basis, 37Carleson's theorem, 1511

235

236 Index of terms

Cauchy sequence, 21

Cauchy—Schwarz inequality, 131

Cesàro summable, 150

chain, 50

classical function spaces, 22classical sequence spaces, 26closed, 19closed n-cell, 216closed convex hull, 55closed covering, 216closed-graph theorem, 80closed linear span, 38closure, 19coarser topology, 20commutative algebra, 92compact, 89

countably, 89relatively, 89

sequentially. 89compact operator, 186compact space, 21complete, 21complete orthogonal set of vectors,

141

completely regular topological space,98

completion of a metric space, 33completion of a normed space, 35complex unital Banach algebra, 162compression spectrum, 168concave function, 3concave functional, 53conjugate, 111

constant kcontraction with, 101

Lipschitz condition with, 101continuous maps, 20continuous spectrum, 169contractible in itself, 222contraction, 101

contraction with constant k, 101

contraction-mapping theorem, 101convergent series, 36convex function, 3convex functional, 42convex hull, 55convex subset, 2countably compact, 89covering, 216

closed, 216open, 216

cyclic vector, 227

degenerate form, 130dense subset, 33dense topological space, 75derivative, 105differentiable map, 105dimension, 214direct sum of subspaces, 39dissipative operator, 182distance, 19divisor of zero, 180

topological, 180dominate, 48dual of an operator, 155dual space, 31

eigenspace, 168eigenvalue, 168eigenvector, 168

approximate, 169Enflo's theorem, 231

equicontinuous, 90equicontinuous at a point, 90equivalence class of functions, 25

equivalent norms, 29essential supremum, 25Euclidean space, 132

extension of a linear functional, 42extreme point, 125

Index of terms 237

face, 214marked, 214

Fejér's theorem, 150finer topology, 20finite character, 116finite-intersection property, 116finite rank operator, 186fixed point, 101flat, 213form

degenerate, 130hermitian, 130non-degenerate, 130positive, 130symmetric, 130

Fourier coefficent, 145, 149Fourier series, 145Fredholm alternative for hermitian

operators, 211Fredhoim integral equation, 207function

concave, 3convex, 3strictly concave, 3strictly convex, 3

fundamental set of vectors, 141

GeLfand transform, 1.83

Gelfand's spectral-radius formula, 124Gelfand—Mazur theorem, 174Gelfand—Nalmark theorem, 158generalized limit, 59geometric mean, 1

weighted, 7Gluskin's theorem, 20Gram determinant, 152Gram—Schmidt orthogonalization pro-

cess, 142

Haar system, 83

Hadamard's inequality, 1.51Hahn—Banach extension theorem, 50

complex form, 50Hamel basis, 42Hammerstein equation, 224harmonic mean, 6Hausdorff topology, 21Hermite polynomials, 1.44hermitian form, 130

hermitian form, positive, 8hermitian operator, 159

on a normed space, 182Hilbert space, 132

Hilbert-Schmidt norm, 187homeomorphic spaces, 20homeomorphism, 20hyperinvariant subspace, 227hyperplane, 46Holder's inequality, 9

for functions, 12

image, 28in general position, 2.11induced topology, 19inequality

AM-GM, ICauchy—Schwarz, 9

HOlder's, 9Minkowski's, 10

initial segment, 124inner product, 131inner-product space, 131mt. interior, 22invariant subspace, 227

invariant-subspace problem, 226invariant under, 226inverse, 1.62

inverse-mapping theorem, 80invertible, 167involution, 162

238 Index of terms

isometrically isomorphic spaces, 29isometry, 162isomorphic spaces, 29isotropic vectors, 130

Jensen's theorem, 3John's theorem, 68Johnson's uniqueness-of-norm

theorem, 179join, 93

kernel, 28, 108kernel of an integral operator, 207Krein—Milman theorem, 126

Laguerre polynomials, 144lattice operations, 93Laurent series, 121Lebesgue number, 99left shift, 32Legendre polynomial, 143linear functional, 28linear operator, 28

bounded, 28unbounded, 28

linear span, 38

Lipschitz condition with constant k,101

local theory of Banach spaces, 20locally compact space, 91Lomonosov's first theorem, 228Lomonosov's second theorem, 229

marked face, 214Markov—Kakutani fixed-point

theorem, 220maximal element, 50maximal ideal space, 183Mazur's theorem, 227meagre set, 76

meanarithmetic, 6harmonic, 6quadratic, 6

meet, 93mesh, 216method of successive approximations,

103

metric, 19metric space, 19

completion of a, 33Minkowski functional, 28Minkowski's inequality, 131

for functions, 12modulus of an operator, 205multicoloured simplex, 214

n-dimensional Euclidean space, 21neighbourhood, 19neighbourhood base, 20non-degenerate form, 130non-expansive map, 223non-trivial, 227norm, 19

Hilbert-Schmidt, 181operator, 29of a functional, 30smooth, 51supremum, 92

uniform, 92

norm topology, 21normal operator, 161normahsed biorthogonal system, 64normed algebra, 92

unital, 92normed space, 18

completion of a, 35nowhere dense set, 76nowhere dense subset, 41numerical radius, 202

Index of terms 239

numerical range, 201

spatial, 175

one-point compactification, 96open covering. 216open-mapping theorem, 29operator ideal, 188operator norm, 29order, 49

partial, 49ordered set, 124orthogonal complement, 136orthogonal direct sum, 135orthogonal matrix, 140orthogonal set of vectors, 141orthogonal subspaces, 135orthogonal vectors, 130

parallelogram law, 133Parseval's identities, 147partial order, 49partial sum, 149partially ordered set, 50partition of unity, 100Perron's theorem, 220point, 19point spectrum, 168polar of a set, 158polarization identities, 132positive form, 130positive hermitian form, Bpre-Hitbert space, 132preannibilator

of a set, 164of a subspace, 158

prepolar of a set, 158principle of uniform boundedness, 77probability, 5product topology, 20, 115Pythagorean theorem, 133

quadratic mean, 6quotient norm, 38quotient normed space, 38

Rademacher function, 143radical, 128Read's theorem, 231regular point, 167relatively compact, 89residual spectrum, 169resolvent, 167resolvent identity, 181resolvent set, 162restriction of a function. 25Riemann—Lebesgue lemma, 152Riesz representation theorem, 137Riesz—Fischer theorem, 145right shift, 32

scalar product, 131Schauder basis, 3L 83Schauder projection, 218Schauder system, 83Schauder's fixed-point theorem, 219second dual, 156self-adjoint operator, 159seminorm, 41separable, 22separate, 89separates the points strongly, 96separation theorem, 54sequentially compact, 89set of the first category, 76set of the second category, 76set system of finite character, 116simplex, 214

multicoloured, 214simplicial complex, 214simplicial decomposition, 214skeleton, 214

240 Index of terms

smooth norm, 51spatial numerical range, 1.25spectral decomposition, 2(X)spectral measure,spectral radius, 114spectrum, 1.67

approximate point, 169compression,

continuous, 169point,residual. 1.69

Spemer's lemma, 214standard basis, 31Stone—Weierstrass theorem, 95

for complex functions, 96strictly concave function, 3strictly convex function, 3strong operator topology, 165stronger topology, 20Sturm—Liouville equation, 209sub-basis for a topology, L1.4subadditive functional, 48subreflexive space, 122subspace, 21subspace topology, 19sum of a series, 36superadditive functional, 5.3support functional, 51, 115support plane, 51supremum norm, 92

symmetric form, 130system set of vectors, 141

Tietze—Urysohn extension theorem, 88topological divisor of zero, 180topological space, 19topology, 19

coarser, 20finer, 20

Hausdorif, 21induced, 19

norm, 21product, 20, 115strong operator, 165stronger, 20subspace, 19weak, 115

weaker, 20

total set of vectors, 141

totally, 89

totally ordered set, 50translate of a subspace, 46triangle inequality, 18Tukey's lemma, 116Tychonov's theorem, 117

unbounded linear operator, 28uniform closure, 93uniform norm, 92

uniformly bounded, 90unital Banach algebra, 32unital normed algebra, 92unitarily equivalent operators,unitary operator, 161universal property, 115upper bound, 50Urysohn's lemma, 86

vanishing at 96

vector, 19

vertex, 214Volterra integral operator, 108

weak topology, 115weak-star topology a(X', X), 116weaker topology, 20weakly bounded, 81weight, 5weighted arithmetic mean, 7weighted geometric mean, 7well-ordered set, 124

Zorn's lemma, 50

Now revised and up-dated, this briskintroduction to functional analysis isintended for advanced undergraduatestudents, typically final year, who havehad some background in real analysis.The author's aim is not just to cover thestandard material in a standard way,but to present results of applications incontemporary mathematics and toshow the relevance of functionalanalysis to other areas. Unusual topicscovered include the geometry of finite-dimensional spaces, invariant sub-spaces, fixed-point theorems, and theBishop—Phelps theorem. An outstand-ing feature is the large number ofexercises, some straightforward, somechallenging, none uninteresting.

Bela Bollobás is an active mathemati-cian who works on combinatorics andfunctional analysis. He has publishedGraph Theory and Combinatorics, bothtextbooks, and two research mono-graphs, Extremal Graph Theory andRandom Graphs.

CAMBRIDGEUNIVERSITY PRESS

ISBN 0-521-65577-3

MIII9 780521 655774

bela bollobas-linear analysis_ an introductory course, second edition (1999)

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