Introduction to the analysis of normed linear spaces

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Introduction to the Analysis of Normed Linear Spaces

AUSTRALIAN MATHEMATICAL SOCIETY LECTURE SERIES Editor-in-Chief: Professor J.H. Loxton, School of Mathematics, Physics, Computing and Electronics, Macquarie University, NSW 2109, Australia Editors: Professor C. C. Heyde, School of Mathematical Sciences, Australian National University, Canberra, ACT 0200, Australia Associate Professor W.O. Neumann, Department of Mathematics, University of Melbourne, Parkville, Victoria 3052, Australia Associate Professor C.E.M Pearce, Department of Applied Mathematics, University of Adelaide, SA 5005, Australia

1 2 3 4 5 6 7 8 9 10 11 12 13 14

Introduction to Linear and Convex Programming, N. CAMERON Manifolds and Mechanics, A. JONES, A GRAY & R. HUTTON Introduction to the Analysis of Metric Spaces, J. R. GILES An Introduction to Mathematical Physiology and Biology, J. MAZUMDAR 2-Knots and their Groups, J. HILLMAN The Mathematics of Projectiles in Sport, N. DE MESTRE The Petersen Graph, D. A. HOLTON & J. SHEEHAN Low Rank Representations and Graphs for Sporadic Groups, C. PRAEGER & L. SOICHER Algebraic Groups and Lie Groups, G. LEHRER (ed) Modelling with Differential and Difference Equations, G. FULFORD, P. FORRESTER & A. JONES Geometric Analysis and Lie Theory in Mathematics and Physics, A. L. CAREY & M. K. MURRAY (eds) Foundations of Convex Geometry, W. A. COPPEL Introduction to the Analysis of Normed Linear Spaces, J. R. GILES The Integral: An Easy Approach after Kurzweil and Henstock, L. P. YEE & R. VYBORNY

Australian Mathematical Society Lecture Series. 13

Introduction to the Analysis of Normed Linear Spaces

J. R. Giles

Department of Mathemtics University of Newcastle, Australia

CAMBRIDGE UNIVERSITY PRESS

PUBLISHED BY THE PRESS SYNDICATE OF THE UNIVERSITY OF CAMBRIDGE

The Pitt Building, Trumpington Street, Cambridge, United Kingdom CAMBRIDGE UNIVERSITY PRESS

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www.cup.cam.ac.uk www.cup.org

© J. Giles 2000 This book is in copyright. Subject to statutory exception and to the provisions of relevant collective licensing agreements, no reproduction of any part may take place without the written permission of Cambridge University Press. First published 2000 Printed in the United Kingdom at the University Press, Cambridge

A catalogue record for this book is available from the British Library

ISBN 0 521 65375 4 paperback

for Zeny wife, mother and ytayta

CONTENTS Preface

xi

Chapter I § 1.

NORMED LINEAR SPACE STRUCTURE AND EXAMPLES

Basic properties of normed linear spaces Structure, examples, quotient spaces, continuous linear mappings, isomorphisms, Schauder bases.

§2.

Exercises

19

Classes of example spaces

24

Finite dimensional normed linear spaces, inner product spaces, the :Cp-spaces (1 ~ p ~ oo).

§3.

Exercises

45

Orthonormal sets in inner product spaces

50

Orthonormal bases in Hilbert space. Exercises

Chapter II §4.

SPACES OF CONTINUOUS LINEAR MAPPINGS

Norming mappings and forming duals and operator algebras

64 67 67

Basic structure, dual spaces, real and complex spaces, operator algebras, unital Banach algebras.

§5.

Exercises

86

The shape of the dual

92

Finite dimensional spaces, Hilbert spaces, application: contravariant and covariant vectors, sequence spaces, ('C: [a,b], 11·11 00 ) . Exercises

Chapter III §6.

THE EXISTENCE OF CONTINUOUS LINEAR FUNCTIONALS

The Hahn-Banach Theorem

110

113 113

Proof and corollaries, the dual for subspaces.

§7.

Exercises

120

The natural embedding and reflexivity

122

Examples of reflexive spaces, techniques to prove nonreflexivity, completion of a normed linear space. Exercises

130

Contents

§8.

Subreflexivity

132

The Bishop-Phelps Theorem. Exercises

Chapter IV

§9.

138

THE FUNDAMENTAL MAPPING THEOREMS FOR BANACH SPACES

Baire category theory for metric spaces

139

139

Baire spaces, examples using category techniques, residual subsets of a Baire space.

§10.

Exercises

!53

The Open Mapping and Closed Graph Theorems

!53

Criteria for topological isomorphisms and equivalent norms.

§II.

Exercises

!58

The Uniform Boundedness Theorem

163

The Banach-Steinhaus Theorem, weak boundedness, the du Bois Reymond Theorem.

169

Exercises

Chapter V § 12.

TYPES OF CONTINUOUS LINEAR MAPPINGS

Conjugate mappings

171 171

Structure, topological isomorphisms, applications to establish reflexivity properties.

§13.

Exercises

179

Adjoint operators on Hilbert space

181

Basic properties, self-adjoint operators, numerical range, positive operators, normal and unitary operators.

§ 14.

Exercises

193

Projection operators

196

Projections on Banach spaces, orthogonal projections on Hilbert spaces.

§ 15.

Exercises

204

Compact operators

206

The algebra of compact operators on a Banach space, approximation by finite rank operators in Hilbert space and in Banach space.

214

Exercises

viii

Contents

Chapter VI § 16.

217

SPECTRAL THEORY

The spectrum

217

The spectrum in a unital normed algebra, the Spectral Mapping Theorem for polynomials.

§ 17.

Exercises

221

The spectrum of a continuous linear operator

222

The spectrum in finite dimensional spaces, the spectrum in normed linear spaces, the spectrum of a conjugate operator.

§ 18.

Exercises

227

The spectrum of a compact operator

230

The invariant subspace property of compact operators.

§19.

Exercises

235

The Spectral Theorem for compact normal operators on Hilbert space

237

The Spectral Theorem, application in the solution of Fredholm integral equations.

§20.

Exercises

241

The Spectral Theorem for compact operators on Hilbert space

243

Functional calculus, square roots of positive operators, partial isometries, the Polar Decomposition Theorem.

250

Exercises

Appendix A.l A.2 A.3

252 Zorn's Lemma Numerical equivalence

252 254 256

Hamel basis

Historical notes

258

List of symbols

269

List of spaces

271

Index

272

ix

PREFACE This text is designed as a basic course in functional analysis for senior undergraduate or beginning postgraduate students. For students completing their final undergraduate year, it is aimed at providing some insight into basic abstract analysis which more than ever, is the contextual language of much modern mathematics. For postgraduate students it is aimed at providing a foundation and stimulus for their further research development. It is assumed that the student will be familiar with real analysis and have some

background in linear algebra and complex analysis. It is also assumed that the student will have studied a course in the analysis of metric spaces such as that given in the author's text Introduction to the Analysis of Metric Spaces, Cambridge University Press, 1987.

Reference to this text will be made under the abbreviation AMS § _ . In AMS, most of the example spaces introduced are normed linear spaces and many of the implications of linear structure were explored. For example when closure in metric spaces was discussed it was natural to consider the closure of linear subspaces in normed linear spaces and when continuity was considered it was logical to study the continuity of linear mappings on normed linear spaces. In order to make this text as self-contained as possible, the example spaces are again introduced and the elementary properties of normed linear spaces are treated but in a more sophisticated way. For example, in AMS the properties of finite dimensional normed linear spaces were deduced in a contrived manner from compactness properties of the real numbers, but here the properties are established in a more expeditious manner using topological isomorphism with Euclidean space. Nevertheless, this text may be regarded as a sequel to AMS especially in that we assume all those properties generally associated with the metric topology of the space. In AMS we were concerned with the particular topological structure of normed linear spaces as metric spaces. The cohesive theme in this text is with the structural properties of normed linear spaces in general, associated especially with dual spaces and the algebra of continuous linear operators on normed linear spaces. The course given here depends fundamentally on the Hahn-Banach Theorem which assumes the Axiom of Choice. The author follows the approach that a great deal of the analysis of normed linear spaces can be handled with a knowledge of metric space topology and is usefully done so before the student has studied general topology. In fact the wide applicability of the analysis of normed linear spaces would argue for this approach. He would suggest that from an analysis point of view the most useful application of the analysis of general topological spaces is in the analysis of linear topological spaces which, for a proper understanding, requires a knowledge of normed linear space theory.

Preface However, to follow this approach means that some topics have to be omitted, in particular any discussion of weak topologies, which includes the Banach-Alaoglu Theorem and compactness characterisations of reflexivity. But such material is naturally included in a subsequent course on the analysis of linear topological spaces. Spaces involving the Lebesgue integral are mentioned in the text but the student needs no background knowledge of integration theory for this course and references to the Lebesgue integral can be glossed over without loss of understanding. A glance at the table of contents shows that the text consists mostly of well established material. However, at this stage of his mathematical education the student needs occasionally to be made aware of the great problem areas in the subject and to glimpse the frontiers where recent advances have been made. It is important that the serious student feel that he is studying a subject which is still on the move and one where he may be able to follow where the current ground is being broken. So at several points the student is introduced to material which is more recent and is given a guide to literature where major problems have been tackled. For example, in Section 8 the Bishop--Phelps Theorem is proved; this is a result which is constantly used by researchers in the analysis of Banach spaces. When discussing Schauder bases in Section 7 it is of interest to mention the Basis Problem and its solution by Enflo. When treating the representation of compact operators by finite rank operators in Section 15 it is useful to refer to Enflo's contribution. The Invariant Subspace Problem arises naturally in Spectral Theory and it is convenient to give a proof of Lomonosov's Theorem for compact operators in Section 18. A lecture course using this text can be given in between 25 and 30 lecture hours given an adequate extra tutorial program and provided the students have sufficient background. The material can be tailored to suit such a course length by omitting Section 8 completely and Chapter VI on Spectral Theory. At the end of each section there is a set of exercises which follows the order of presentation of material in the section. At the end of the text there is an appendix which includes the set theory results used at various points of the course. Historical notes are included to show how the subject developed from late nineteenth century beginnings into the analysis of an abstract structure using the axiomatic method. Two recent advances are included to demonstrate the continued vigour of research in the area.

xii

Preface As with AMS, this text has a full index.

Rarely are students told that when a

mathematician uses a textbook he regularly consults its index. So the index in this text gives references to all the significant places where a particular concept is used. There are m~y texts in functional analysis which cover the material presented in this text book in a variety of different ways. The great standard text for many years was Angus E. Taylor, Introduction to functional analysis, John Wiley, 1958 which was updated by Angus E. Taylor and David C. Lay in 1980. This is a great reference but is generally pitched at more advanced students than our text. A text more accessible to students of our course is Erwin Kreysig, Introductory functional analysis with applications, John Wiley, 1978. This text is very popular with engineers and does not rely to any great extent on previous knowledge of general topology. The text which had a formative influence on the author is A.L. Brown and A. Page, Elements offunctional analysis, van Nostrand Reinhold, 1970. This text is pitched at the same level as ours and does not have a general topology or general integration prerequisite. Our text often gives more detail and introduces recent results. An expansive text is George Bachman and Lawrence Narici, Functional Analysis, Academic Press, 1966. But this text covers spectral theory with greater detail and depth. The text George F. Simmons, Introduction to topology and modern analysis, McGraw-Hill, 1963 is very readable, but it is written with more of a general topology background. An excellent text, but one also written from a topological point of view is G.J.O. Jameson, Topology and normed linear spaces, Chapman-Hall, 1974. There are more modem texts such as Walter Rudin, Functional Analysis, McGraw-Hill, 2nd edn, 1991. This presents a far more sophisticated course than ours, but will be found to be a useful reference. There are two modern British texts which should be mentioned. Nicholas Young, An introduction to Hilbert space, Cambridge University Press, 1988 concentrates on Hilbert space with applications in mind. xiii

Preface Bela Bollobas, Linear Analysis, Cambridge University Press, 1990. This contains a very fine lecture course and does refer to new results. It moves somewhat faster than our text and with less detail than many students would find comfortable. I have been influenced by the text J.R. Ringrose, Compact non-self-adjoint operators, van Nostrand Reinhold, 1971 in my approach to spectral theory for compact operators. The author has given lectures on a course such as this to third year honours students at the University of Newcastle for a period of more than ten years. The lecture notes were first produced in duplicated form but have been modified and expanded considerably in the light of experience, to the form presented here. Thanks are due to my colleagues and students in the department for their conversations over many years which have had their effect on the final result. My thanks to Philip Charlton for producing the diagrams which appear in the text. I am particularly indebted to Jan Garnsey who has so patiently and competently typed and retyped the final copy from my handwritten manuscript. I am grateful to Brailey Sims for supplying source material for the historical notes. J.R. Giles The University of Newcastle NSW Australia.

xiv

NORMED LINEAR SPACE STRUCTURE AND EXAMPLES

I.

Everyday Euclidean space which underlies most of our mathematical activity, possesses some remarkable abstract structures which provide a rewarding study in their own right. In particular we could isolate the linear space structure and the metric space structure and the one which is a significantly rich mixture of these two, the normed linear space structure. To study such an abstract structure is profitable for two reasons: the study can be undertaken in a systematic and deductive fashion and there are many diverse mathematical situations which exhibit the same structure to which the developed theory can be applied. Moreover, the study of normed linear spaces has an intriguing interest which is quite distinct from that of metric spaces. Just as the study of Euclidean space gives rise to matrix theory so the linear structure of normed linear spaces enables us to propagate more normed linear spaces from the continuous linear mappings between them and this is the origin of operator theory. § 1. BASIC PROPERTIES OF NORMED LINEAR SPACES

We begin with a review of the defining structure of normed linear spaces, of the fundamental properties of continuous linear mappings and of the notions of basis in normed linear spaces. Our theory is developed from a knowledge of these and we will use them in our discussion of examples and our subsequent construction of associated normed linear spaces in Chapter II. 1.1 Definition. Given a linear space X over [ (or JR.), a mapping 11·11: X

nann for X if it satisfies the following properties: For all x EX, (i)

II x11

~

0,

(ii) II x II = 0 if and only if x = 0, (iii) II Ax II = I A I II x II for all scalar A, and for all x, y EX, (iv) II x+y II

~II

x II +II y II.

--7

JR. is a

2

Normed linear space structure

(The norm is said to assign a "length" to each vector in X.) The pair (X, 11·11) is called a

normed linear space. Different norms can be assigned to the same linear space giving rise to different normed linear spaces. 1.2 Remarks. The norm generates a special metric on the linear space. Given a normed linear space (X, 11·11), a function d: X x X --7 IR defined by d(x,y) =II x-y II is a metric for X and is called the metric generated by the norm. This metric is an invariant metric for X; that is, for any given z EX, for all x,y EX.

d(x+z, y+z) = d(x,y) So given x EX and r > 0, the ball

B(x; r) = { y EX : II y-x II < r} = x + { z EX : II z II < r} = x + B(O; r), is the translate by x of the ball B(O; r). Furthermore, property (iii) implies that given r > 0, the ball centred at the origin with radius r, B(O; r) = r{ y EX : II y II < 1} = r B(O; I), is an r-multiple of the unit ball. Properties (iii) and (iv) tell us that the norm is a convex function on X; that is, for any x,y EX we have II A.x+(I-A)y II~ A II x II+ (I-A) II y II for all 0 ~A~ I. This implies that the ball B(O; I) is a convex set in X; that is, for any x,y

E

B(O; I) we have A.x + (1-A)y

E

B(O; I) for all 0

~A~

I.

But (iii) also tells us that the ball B(O; I) is symmetric; that is, for any x E B(O; I), we have- x E B(O; I).

0

Since a normed linear space has both algebra and analysis structures, the relationship between these two quite different aspects of the space is a matter of special interest. The fruitfulness of the relationship follows from the way the linear space operations are linked to the norm. 1.3 Remark. Given a normed linear space (X, 11·11), from properties (iii) and (iv) we can derive the important norm inequality

In x II -

II y

Ill

~ II x-y II

for all x,y EX

which implies that the norm is a continuous function on X. Properties (iii) and (iv) actually relate addition of vectors and multiplication of a vector by a scalar to the norm and imply that these algebraic operations have a continuity property which we call joint continuity : if x

--7

x0 and y

--7

Yo then x + y

--7

x0 + y0 , and if A --7

"-o then A.x --7 A.ox 0 .

This can be deduced simply from the inequalities: II (x+y)- (x 0+y 0) II

~II

x-x 0 II + II y-y 0 II,

and II AX-A0x0 II ~II A.x-Ax 0 II + II Ax 0-A.ox 0 II ~I A Ill x-x 0 II + I A-Ao Ill x0 II.

0

3

§I. Basic properties

A normed linear space is a generalisation of Euclidean space. 1.4 Example. Euclidean n-space and Unitary n-space. The real (complex) linear space of ordered n-tuples of real numbers IR 0 (of complex numbers ([ 0 ), where for x = (1, 1, A2 , ... , A0 ) we define the norm •n=---llxll2

=

L,IA.kl 2 k=1

in the real case is called Euclidean n-space and is denoted by (IR 0 , 11·11 2) and in the complex case is called Unitary n-space and is denoted by (
(IR 0 , 11·11 00 ), (or ( A.z, ... , An) E IR 0 (or
(i)

(IR 0 , 11·111), (or (
n

llxll 1

= LIAkl. k=1

In both cases it is much simpler than in the Euclidean case to verify that these norms satisfy O the norm properties (i)-(iv). Several example spaces can be considered as different forms of a general function space. 1.6 Examples. For any nonempty set X, the set

n (X)

of bounded real (complex)

functions on X is a linear space under pointwise definition of the linear operations and is a normed linear space with norm defined by II f lloo = sup {I f(x) I : x EX}. (i) When X={ 1, 2, ... , n} then "r.>(X) is the linear space IR 0 (or ([ 0 ) and for x (A 1, A2 , . . . , An) E 1R 0 (or ([ 0 ), II x lloo =max {I Ak I: k E { 1, 2, ... , n)} .

=

4 (ii)

Nonned linear space structure

When X =N the set of natural numbers, then n(N) is the linear space m (or .€ 00 ) of

bounded sequences of scalars and for x ={1.. 1, II

X

/..2 , .•. ,

An, ... } Em,

lloo = sup {I Ak I : k E N } .

(iii) When X = [a,b] the bounded closed interval, then n [a,b] is the linear space of bounded scalar functions On [a,b] and for f E n(a,b], II flloo =sup {I f(x) I: x E[a,b]}.

0

1.7 Definition. A normed linear space which is complete as a metric space with its metric generated by the norm, is called a Banach space. Using the completeness of the scalar field we establish the completeness of the general example given in Example 1.6. This also illustrates the general method used to prove completeness in other examples.

1.8 Theorem. For any nonempty set X, the normed linear space (n (X), 11·11 00 ) is a Banach space.

Proof. Consider a Cauchy sequence {fnl in (n(X), ll·lloo); then given£> 0 there exists a v E N such that II fn-fm lloo < £ sup{ I (fn-fm)(x) I : x EX} < £

that is,

for all m,n > V

;

for all m,n > V

.

for all m,n > V

;

But then for each x EX, I fn(x)- fm(x) I < £

that is, for each x EX, { fn(x)} is a Cauchy sequence of scalars. Since the scalar field is complete, for each x EX we can define a function f on X by f(x) = lim fn(x). We need to show that f E n(X). Since {fn} is Cauchy it is bounded in (n(X), 11·11 00 ) ; that is, there exists a K > 0 such that II fn lloo < K which implies that

I fn(x) I < K

Therefore, I f(x) I :S: K for all

X

for all n EN, for all x EX and all n EN.

EX, and so f E n(X).

Then we need to show that {fnl actually converges to fin (n(X), 11·11 00 ) . We had for each x EX,

I fn(x)- fm(x) I < £ for all m,n > v . Fix n and let m --+ oo; then we have for each x EX, So

I fn(x)- f(x) I :S: £ for all n > v . II fn-f lloo =sup {I fn(x)- f(x) I : x EX} :S: £ for all n > V

that is, {fnl converges to fin (n(X), ll·lloo).

;

0

s

§I. Basic properties

1.9 Remarks. So the normed linear spaces (IR 0 , 11·11 00 ), or (0: 0 , 11·11 00 ), (m, 11·11 00 ) and (~ [a,b], ll·lloo) are Banach spaces. Of course, Euclidean n-space (IR 0 , 11·11 2 ) and Unitary

n-space (0: 0 , 11·112) are both complete, but a proof of this will be deduced from the 0 completeness of (IR 0 , ll·lloo). (or (0: 0 , ll·lloo) by Corollary 2.1.5. We will see, as we develop the fundamental theorems in our theory, that completeness plays a strategically important role. A linear subspace in a normed linear space derives norm structure from the parent space in a natural way. 1.10 Definition. Given a normed linear space (X, 11·11) and a linear subspace Y of X, it is clear that the restriction of the norm 11·11 to Y is also a norm for Y. The restriction is denoted ll·lly and (Y, ll·lly) is a normed linear subspace of (X, 11·11). The following examples are significant linear subspaces of the example spaces so far introduced. 1.11 Examples. In m the linear space of bounded sequences, where for each x {lc 1, !..2 , . . . , A0 , ..• } there exists an Mx > 0 such that I "-n I ~ Mx for all n

=

E

N,

we have the following linear subspaces. c the linear subspace of convergent sequences, where for each x = {!.. 1, !..2, ... , !..0 , • . . } there exists a scalar A such that "-n

-7

A as n

c0 the linear subspace of sequences which converge to zero, .t 2 the linear subspace of sequences x = {!.. 1, !..2 , . . . , "-n• ... } where

1 1 the linear subspace of sequences x

-7

oo,

I. I "-n 12 < oo,

= {!.. 1, !..2 , . . . , "-o .... } where I, I "-n I< oo, and

E0 the linear subspace of sequences with only a finite number of nonzero entries, where for each x

={!..

1,

!.. 2 , . . .

,

"-n· ... }, "-n = 0 for all except a finite number of elements inN.

That 1 2 and 1 1 are linear subspaces will be verified in Example 2.2.11 and 2.3.6. Now all of these are normed linear subspaces of (m, 11·11 00 ). 1.12 Examples.

For a metric space (X, d), consider

~(X,

0

d) the linear space of

bounded functions on (X, d). We have the important linear subspace

~ t; (X,

d) of bounded continuous functions on

(X, d). When (X, d) is compact, then all continuous functions on (X, d) are bounded so ~ l;(X,

d)= l;(X, d) the linear subspace of continuous functions on (X, d).

When X= [a,b], we could also consider the linear subspace 1\.[a,b] of

~[a,b]

which

consists of the Riemann integrable functions on [a,b]. Of course, t; [a,b] is a linear subspace of 1\.[a,b].

6

Nonned linear space structure

We could consider the linear subspace t: I [a,b] of t:[a,b] which consists of those functions with a continuous first derivative on [a,b] and t;oo[a,b] which consists of functions which are infinitely often differentiable on [a,b]. We could also form linear subspaces of ~(X, d) of the form { f E ~(X, d) : f(x 0)

=0} for a given x0 EX. =0}.

So we would have ~ 0 [0, 1] the linear subspace {f E ~ [0, 1] : f(O)

We would also have t: 0 [0, 1] the linear subspace {f E t:[O, 1] : f(O) = 0}. Again all of these are normed linear subs paces of the appropriate bounded function space O with the supremum norm ll·lloo. Often the simplest way to examine a normed linear space for completeness is to use the following metric space link between completeness and closedness, (see AMS §4). 1.13 Proposition. Consider a metric space (X, d) and a subset Y. (i) lf(Y, d

Iy

)is complete then Y is a closed subset of(X, d).

(ii) If (X, d) is complete andY is a closed subset of (X, d) then (Y, d

Iy )

is complete. 1.14 Example. Given a metric space (X, d), the normed linear space (~ t:(X, d), ll·lloo) is complete.

Proof. We show that

~ t:(X,

d) is a closed subset of (~(X, d), 11·11 00 ) and apply

Proposition 1.13. Given a cluster point f of~ t:(X, d) in (~(X, d), 11·11 00 ) we show that f is continuous: Now there exists a sequence {f0 } in ~ t: (X, d) convergent to f ; that is, given

£

>0

there exists a v EN such that whenn>v. Consider x0 EX. Since fv+ 1 is continuous at x0 there exists a

I fv+ 1(x)-fv+l(x 0) I<£

o> 0 such that

when d(x,x 0 ) < 0.

Therefore, I f(x)-f(x 0) I:;:; I f(x)- fv+ 1(x) I+ I fv+ 1(x)-fv+l(x0) I+ I fv+ 1(xo)-f(xo) I :;:; 2 II fv+l-f lloo +I fv+l(x)-fv+l(xo) I

< 3£

when d(x,xo) < o;

that is, f is continuous at x0 .

0

When (X, d) is compact, the Banach space t:(X, d) has a particularly rich structure, (see AMS §9). We often encounter situations where we have a linear space with a real function which has all the norm properties except (ii).

§I. Basic properties

7

1.15 Definition. Given a linear space X over
X~

IR is a

seminorm for X if it satisfies all the norm properties except (ii) and instead of (ii) satisfies (ii)'

p(x) = 0 if x = 0.

The pair (X,p) is called a seminormed linear space. So a seminorm allows the possibility that there exists some x otc 0 for which p(x) = 0. 1.16 Remark. A semi norm p on a linear space X generates a semimetric e on X defined by e(x,y)

= p(x-y).

As with the norm properties noted in Remarks 1.2, property (iii) implies that the set {x EX : p(x) < I } has the properties that, given r > 0 { y EX: p(y-x) < r}

=x + r{ z EX: p(z) < I}

and is convex and symmetric. However, if p is not a norm then ker p is a nontrivial linear subspace and is a subset of {xEX:p(x)
1.17 Definition. Given a linear space X and a proper linear subspace M of X, the quotient space (or factor space) XIM is the linear space of cosets [x] = x + M, with addition and multiplication by a scalar defined by [x] + [y] = (x + M) + (y + M) = (x + y) + M and

A[x]

=A(x+M) = Ax + M =[Ax]

= [x+y] for all [x],[y] E X/M

for all [x] E X/M and scalar A.

A seminormed linear space can be transformed into a normed linear space as a quotient space. 1.18 Definition. Given a seminormed linear

S~{!Ce

(X, p), the associated normed linear

space is the quotient space X/ker p whose elements are the cosets [x] II [x]ll = p(x)

=x+ker p, with norm

for any x E [x].

1.19 Example. Consider ~[a,b] the linear space of Riemann integrable functions on [a,b]. Now the function p( ~[a,b] ~ IR defined by b

p 1(f) = is a seminorm on

~[a,b].

f

I f(t) I dt

8

Normed linear space structure

On the quotient space ~[a,b]/ker p 1 of cosets [f] II [f] 11 1 = p 1(f)

=f+ker p 1, we have the associated norm D

for any f E [f].

When a quotient space is generated from a normed linear space by a proper closed linear subspace then the quotient space has an associated quotient norm.

1.20 Definition. Given a normed linear space (X, 11·11), and a proper closed linear subspace M of (X, 11·11), the quotient norm 11·11 is defined on the quotient space X/M by II [x]ll = d(O, x+M) = inf{ II x+m II: m EM}.

1.21 Remark. It is routine to verify the norm properties (i)-(iv) for the quotient norm. But we note that we need the linear subspace M to be closed to establish that the norm is not just a seminorm; that is, we need to show that d(O, x+M) that II [x]ll 0 implies that [x] 0.

=

=0 implies that x + M = M so

=

D

It is instructive to see how such a quotient space inherits completeness from the parent space.

1.22 Theorem. Given a Banach space (X, 11·11) and a proper closed linear subspace M then the quotient space (XIM, 11·11) is also complete.

Proof. Consider a Cauchy sequence {x 0 +M} in (X/M, 11·11). Then for each kEN there exists a v(k) E N such that I II (x 0 + M)- (xm + M) II < 2 k

for all m,n > v(k).

Consider a subsequence of the form {Xn(k) + M} where for each k EN, n(k) > v(k). I Then II (xn(k) + M)- (xn(k+I) + M) II< 2 k . For each k E N choose xk E Xn(k) + M such that I II xk- xk+I II< 2 k . The sequence {xk} in (X, 11·11) has the property that II xk-Xm II:":: II xk-xk+I II+ ... + II Xm_ 1-xm II I < 2 k-I for all kEN,

form> k

so {xd is a Cauchy sequence in (X, 11·11). But (X, 11·11) is complete so there exists an x E X such that {xk} is convergent to x. Then II (xn(k)+M)- (x+M) II= II (xk+M)- (x+M) II:":: II xk-x II and so {Xn(k)+M} is convergent to x + M in (X/M, 11·11). However, {xn(k)+M} is a convergent subsequence of the original Cauchy sequence {x 0 +M), so {x 0 +M} is also convergent in (X/M, 11·11).

D

9

§I. Basic properties

1.23 Example. Consider 1t [a,b] the linear space of Riemann integrable functions on [a,b] with norm II flloo= sup{ I f(t) I: t E[a,b]}. Now (1\ [a,b ], IHioo) is a Banach space, (see AMS §4). Furthermore, ker PI where b

pi(f) =

f

I f(t) I dt

a

is a closed linear subspace of (wt [a,b], IHioo). On the quotient space

1t [a,b]/ker PI

whose elements are cosets, [f)= f + ker PI• we have the norm II [f) lloo = d(O, f+ker PI)= inf{ II f+g lloo : g E ker PI} . From Theorem 1.22, the quotient space (wt [a,b ]/ker PI, IHioo) is a Banach space.

0

1.24 Continuous linear mappings The algebraic study of linear spaces finds its full development in an examination of the homomorphisms or structure preserving mappings between such spaces. As we might expect, these linear mappings are of similar significance in the development of the analysis of normed linear spaces. However, interest in this case is focused on the continuous linear mappings which preserve the topological and norm structure along with the linear structure. The following characterisation theorem provides an essential tool for discussing these mappings.

1.24.1 Theorem. Given normed linear spaces (X, 11·11) and (Y, 11·11'), a linear mapping T:X~Y

(i)

is continuous if and only if there exists an M > 0 such that II Tx II' ~ M II x II for all x EX,

(ii)

has a continuous inverse on T(X) if and only if there exists an m > 0 such that m II x II ~ II Tx II' for all x E X.

Proof. (i)

If the condition holds then clearly T is continuous at 0 and the linearity ofT implies

that T is continuous on X. Conversely, suppose that the condition does not hold; that is, for each n EN there exists

Xn

EX such that II Txn II' > n II

Xn

II.

k II :~ II II = k~ o as n ~ oo,

Then

II

but

II T
II

;~

that is, T is not continuous at 0.

11 )

II> I for all n EN;

I0

Nonned linear space structure

If there exists an m > 0 such that

(ii)

mllx ll:s;IITxll'

for all x EX.

then ker T = {0} soT is one-to-one and T-1 exists on T(X) and is clearly linear. Writing x = T-1y for y E T(X) we have m II T-1y II :s; II T(T-1y) II'= II y II' II T-1 y II :s; _I_ II y II'

so

m

for ally ET(X);

that is, from (i) T-1 is continuous on T(X). Conversely, if T-1 is a continuous linear mapping on T(X) then, from (i) there exists an M > 0 such that

II T-1y II :s; M II y II

for all y E T(X).

II T-1(Tx) II :s;M II Tx II'

Since y = Tx,

~II x II :s; II Tx II'

and so

for all x EX.

0

1.24.2 Remark. From linearity it follows that a linear mapping T is continuous on a normed linear space (X, 11·11) if and only if T is continuous at any one point of X. It follows that a linear mapping T is either continuous at every point of X or continuous at no

0

point of X. Two of the most commonly occurring linear mappings are as follows.

1.24.3 Examples. Consider the normed linear spaces (t:[O, I], 11·11 00 ) and (t: I [0, I], 11·11 00 ) . Consider the linear mapping I: t; [0, I] ~ t; 1[0, I] defined by

(i)

X

I(f)(x) =

Jf(t) dt

for all x E [0,1].

0

I I(f)(x) I :s; I x Ill f 11 00

Now

for all x E [0,1]

II I( f) 11 00 = max {I I(f)(x) I : x E [0, ll} :s; II f lloo for all f E t: [0, I],

and

so I is continuous.

(ii)

Consider the linear mapping D: t; 1[0, I] ~ t; [0, I] defined by D(f)(x) = f' (x) for all x E [0, 1].

For the sequence {f 0 } in t: I [0, I] where f 0 (x) =*sin mtx we have II f 0 lloo = *

II D(f0 ) lloo;, I for all n EN, soD is not continuous.

~0

but

0

1.24.4 Definitions. Consider normed linear spaces (X, 11·11) and (Y, 11·11') and a linear mapping T: (i)

X~

Y.

Tis said to be a topological isomorphism (or a linear homeomorphism) ifT is also a

homeomorphism; that is, T is linear, continuous, invertible and has a continuous inverse on T(X).

II

§I. Basic properties

(X, 11·11) and (Y, 11·11') are said to be topologically isomorphic if there exists a topological

isomorphism of X onto Y. (ii)

T is said to be an isometric isomorphism if II Tx II'= II x II for all x EX.

(X, II· II) and (Y, 11·11') are said to be isometrically isomorphic if there exists an isometric

isomorphism of X onto Y.

A continuous linear mapping with an algebraic inverse is not necessarily a topological isomorphism. 1.24.5 Example. Consider Example 1.24.3. The continuous linear mapping X

l(f)(x)

= Jf(t) 0

is a one-to-one mapping of "C [0, I] onto D(f)(x) = f'(x)

dt

for all x

E

[0, I]

r:ci [0, I] and has an algebraic inverse for all x E [0,1].

But D is not continuous so I is not a topological isomorphism.

D

Nevertheless, we will see in Section I 0, that there are special circumstances under which we can conclude that a continuous linear mapping with an algebraic inverse is a topological isomorphism. 1.24.6 Remark. Not every isometry of a normed linear space into a normed linear space is linear even if it maps 0 to 0. For example, consider the mapping T of a complex normed linear space (X, 11·11) into itself defined by T(A.x)

=XT(x)

for all x EX and scalar A.

However, it was shown by Stefan Mazur and Stanislaus Ulam in 1932 that an isometry of a real normed linear space into a real normed linear space which maps 0 to 0 is linear; (see Stefan Banach, Theorie des operations lineaires, Chelsea reprint 1932 edition, p.l66, and D Exercise 2.4.13).

We deduce the following characterisation of topological isomorphisms from Theorem 1.24.1. 1.24.7 Corollary. Given normed linear spaces (X, 11·11) and (Y, 11·11'), a linear mapping

T: X ~ Y is a topological isomorphism if and only if there exist m, M > 0 such that m II x II .,:; II Tx II' .,:; M II x II

for all x EX.

12

Normed linear space structure

1.24.8 Remarks. A topological isomorphism of (X, II· II) onto (Y, 11·11') establishes an identity between

(i)

the linear and topological structures of (X, II· II) and (Y, II· II') in that there is a one-to-one correspondence between the points of the spaces which identifies the linear structure and the topological structure. The relation "being topologically isomorphic to" is an equivalence relation on the set of all normed linear spaces. As we study different properties of normed linear spaces there is interest in determining whether the properties are "linear topological invariants"; that is, if all normed linear spaces which are topologically isomorphic also possess the same property. (ii)

Of course an isometric isomorphism is a topological isomorphism and so has the

properties refered to in (i). Clearly an isometric isomorphism is also a metric isometry. An isometric isomorphism of (X, 11·11) onto (Y, 11·11') establishes an identity between the normed linear space ·structures of (X, II· II) and (Y, 11·11').

Such mappings are really

"normed linear space isomorphisms". The relation "being isometrically isomorphic to" is an equivalence relation on the set of all normed linear spaces. We will see in Section 5, how recognition that a normed linear space is isometrically isomorphic to a more familiar normed linear space gives us a firmer grasp of the subject matter. D It is useful to note that completeness is a linear topological invariant for normed linear spaces.

1.24.9 Theorem. Given topologically isomorphic normed linear spaces (X, 11·11) and (Y, 11·11'),

if

(X, 11-11) is complete then (Y, 11·11') is complete.

Proof. Suppose Tis a topological isomorphism of (X, 11·11) onto (Y, 11·11') and that (X, II· II) is complete. Then there exist m, M > 0 such that m II x II .,; II Tx II' .,; M II x II

for all x EX.

Consider a Cauchy sequence { Yn} in (Y, 11·11') and sequence { x 0 yn

=Txn-

m II X0 -xm11.,; II T(x 0 -Xm) II'= II Yn-Ymll' { x0

}

}

in (X, 11·11) where

Then since for all m,n

E

1'\1,

is a Cauchy sequence in (X, 11·11). But (X, 11·11) is complete so there exists an x EX

such that { x 0 } is convergent to x. Writing y

=Tx we have

II y 0 -y II'= II T(x 0 -x) II'.,; M II X0 -x II and so { Yn} is convergent toy

E

Y and we conclude that (Y, 11·11') is complete.

D

Although the following implication seems somewhat trivial it will be of use in our discussion of the spectrum of a continuous linear operator in Section 17.

13

§I. Basic properties

1.24.1 0 Corollary. Consider a topological isomorphism T of a Banach space (X, II· II) into a normed linear space (Y, 11·11'). lfT(X) is dense in (Y, 11·11') then T(X) =Y.

·Proof. From Theorem 1.24.9 we have that (T(X), II·II'T(X)) is complete. So by 0 Proposition 1.13(i), T(X) is closed in (Y, 11·11') and therefore T(X) = Y. We mentioned that a linear space X can be assigned different norms. There is a natural way of partitioning such norms. 1.24.11 Definitions. Consider a linear space X with norms IHI and 11·11'. Norm 11·11 is said to be stronger than norm IHI' and norm 11·11' is said to be weaker than norm 11·11 if the identity mapping from (X, 11·11') into (X, 11·11) is continuous. Norms IHI and 11·11' are said to be equivalent if the identity mapping from (X, II· II') into (X, 11·11) is a topological isomorphism. 1.24.12 Remarks. (i) From Theorem 1.24.1 we see that norm II· II is stronger than norm 11·11' if and only if there exists an M > 0 such that llxii'~MIIxll

for all x EX.

The topology generated by norm IHI is stronger than that generated by norm 11·11'. (ii) From Corollary 1.24.7 we see that norms IHI and 11·11' are equivalent norms on X if and only if there exist m, M > 0 such that m II x II ~ II x II' ~ M II x II

for all x EX.

Equivalent norms generate the same norm topology on X. (iii) From Theorem 1.24.9, we see that if a normed linear space (X, II· II) is complete then it is complete under any equivalent norm 11·11'. Again we will see in Section I0 that there is a significant relation between completeness and equivalent norms. 0 1.24.13 Definition. A linear mapping of a normed linear space (X, II· II) into the scalar field of X is called a linear functional on X. A linear mapping of a normed linear space (X, 11·11) into (X, 11·11) is called a linear operator on X. In drawing attention to the special properties of continuous linear functionals the following algebraic property is of crucial importance. 1.24.14 Definitions. Given a linear space X, a proper linear subspace M is called a linear subspace of codimension one or a hyperplane if for a given x0 EX\ M, every x EX

14

Nonned linear space structure

can be represented in the form x = Ax 0 + y

where A is a scalar andy EM.

It is easily verified that for a linear subspace of codimension one, such a representation holds uniquely. 1.24.15 Lemma. The kernel of a nonzero linear functional f on a linear space X is a

linear subspace of codimension one and a linear subspace M of codimension one is the kernel of a nonzero linear functional on X.

Proof. For x0 EX \ ker f, f(x 0) * 0 and for any x EX, f(x) where y E kerf. x = f(xo) Xo + Y For x0 EX \ M, any x EX can be represented uniquely in the form x = AXo + y

where A is scalar andy EM.

Define a functional f on X by f(x) =A. Then clearly f is linear and M =kerf.

D

1.24.16 Corollary. In a normed linear space (X, 11·11), the kernel of a nonzero linear

functional f is either closed or dense.

Proof. Since f is nonzero, if kerf is closed then kerf is not dense. If kerf is not closed then there exists a cluster point x0 of kerf where x0 EX\ kerf. Now from Lemma 1.24.15, X = sp{ x0 , ker f). But kerf is a linear subspace which contains sp {x0 , ker f). So X = ker f.

D

We deduce the following characterisation of the continuity of linear functionals. 1.24.17 Theorem. Given a normed linear space (X, II· II), a linear functional f on X is

continuous if and only ifker f is closed.

Proof. Iff is continuous then clearly ker f is closed. If f is not continuous then for any r > 0, f is unbounded on B(O; r). But then since f is linear f(B(O; r)) =
D

1.24.18 Remark. This characterisation of continuity does not extend to linear mappings

generally. Consider the identity mapping id of ("C [0, I], 11·11 1) onto ("C [0, 1], 11·11 00) . Here ker id = {0} which is closed but id is not continuous.

D

15

§I. Basic properties

1.25

Normed linear spaces with a basis It is useful to see that normed linear spaces can be classified according to what is

called their density character. 1.25.1 Definition.

A metric space (X, d) is said to be separable if it contains a

countable dense subset. 1.25.2 Examples. (i) Euclidean space (IR 0 , 11·11 2) is separable since it contains the countable set (1) 0 as a dense subset. Similarly, Unitary space (CC 0 , 11·11 2) is separable. (ii) The normed linear space (c 0, ll·lloo) is separable. The linear subspace E0 is dense in (c 0 , ll·lloo) and the subset Eo
D

However, there are some significant nonseparable normed linear spaces. 1.25.3 Example. The normed linear space (m, 11·11 00 ) is not separable. Consider the subset E of sequences consisting of digits 0 and I; this set is uncountable. But also for any x, y

E

E, x of. y, II x-y lloo

= I.

So { B(x;

~): x E E}

is an uncountable family of

disjoint open balls in (m, 11·11 00 ). For any subsetS dense in (m, 11·11 00) we must have that each ball in this family contains at least one point from S. But this implies that S is uncountable.

D

In linear space theory the concept of basis is fundamental. 1.25.4 Definitions. Given a linear space X, a Hamel basis for X is a linearly independent subset of X which spans X. It is clear that a nonempty subset A is a Hamel basis for X if and only if each element x of X has a unique representation as a linear combination of elements of A. 1.25.5 Remarks. It follows from Zorn's Lemma, Appendix A.l.7, that (i)

every linear space X has a Hamel basis and

(ii) for any given linear space X any two Hamel bases are numerically equivalent; D (see Appendix A.3). These remarks enable us to make the following definitions.

16

Nonned linear space structure

1.25.6 Definitions. Given a linear space X, the Hamel dimension of X is the cardinal number of any Hamel basis for X. If X has a finite Hamel basis then X is said to be finite dimensional, otherwise it is said to be infinite dimensional. 1.25.7 Examples. (i)

For the linear space E0, the set {e 1, e2, ... , en, ... } where en

={0, ... , 0, I, 0, ... )

nth place is a Hamel basis for E0 and the Hamel dimension of Eo is countably infinite. For the linear space c0 , the set {e 1, e2, ... , en, ... }given in (i) is not a Hamel basis since its span E0 is a proper linear subspace of c 0 . The set of all sequences (ii)

{ { I, t, t2, ... , tn, ... } : t

E

(0, I)} is a linearly independent subset of c0 and has cardinality

of the continuum. But the cardinality of c0 is that of the continuum. So a Hamel basis for c0 has cardinality of the continuum. 0 1.25.8 Remark. The concepts of Hamel basis and Hamel dimension belong essentially to linear space theory and are not so significant in the analysis of infinite dimensional normed linear spaces because there is no natural relation of a Hamel basis to the topological structure of the space. Hamel basis is mainly of use in drawing parallels between linear 0 space and normed linear space theory and in the construction of counterexamples. The basis concept appropriate in the study of infinite dimensional normed linear spaces is the following. 1.25.9 Definition. Given a normed linear space (X, 11·11), a Schauder basis for X is a sequence {en} such that for each element x EX there exists a unique sequence of scalars n

{"-n} such that x = L "-nen ; that is, II x -

L

n=l

k=l

"-nen II ~ 0 as n ~ ""·

For each n E 1'\1, the coefficient "-n is called the nth coordinate of x with respect to the basis { enl ·

1.25.10 Examples. (i)

In any finite dimensional normed linear space the concepts of Schauder basis and

Hamel basis coincide because both in this case are concerned with finite sequences. (ii) In (E 0 , 11·11 00 ), the sequence {e 1, e2o ... , en, ... } where en

= {0,

... , 0, I, 0, ... }

nth place is both a Schauder basis and a Hamel basis.

17

§ I. Basic properties

(iii) In (c 0, ll·lloo), the same sequence {e 1, e 2, each x = {1.. 1, l.z, ... , A.n •... } ec 0 , n

II x -

L

••• ,

en, ... } is a Schauder basis since for

~ek 11 00 = sup { I ~ I : k > n } --7 0 as n --7 oo

k=l

and so x =

L Anen . n=l But this is not a Hamel basis as the Hamel dimension of c0 is the cardinality of the continuum. 0 Confining our definition of Schauder bases to sequences has the following implications. 1.25.11 Theorem. A normed linear space (X, 11·11) with a Schauder basis {en} is separable.

Proof. We may consider X over the reals and II en II

= I for all n E N.

The set of all

rational linear combinations from {en} is a countable set. Given x =

L

~ek and e > 0 there exists avE N such that

n=l

n

II

L

~ekii<E

for all n ~v.

k=l E

For each k E {1,2, ... , v} choose ak E
v

v

v

v

L akek II~ II x- L A.kek II+ L I ~-ak I< 2e. k=l k=l k=l So the set of rational linear combinations from {en} is dense in (X, 11·11).

Then

II x-

0

1.25.12 Remark. The following famous problem is referred to in Stefan Banach, Theorie des operations lineaires, Chelsea reprint 1932 edition, p.111. The basis problem. Does every separable Banach space have a Schauder basis?

After a great deal of effort by many mathematicians the problem was finally solved in the negative by Per Enflo, Acta Math. 130 (1973), 309-317, where he produced a counterexample. 0 There are certain natural linear functionals associated with a Schauder basis.

18

Normed linear space structure

1.25.13 Definition. Given a normed linear space (X, II· II) with a Schauder basis {en}, for each k EN the linear functional fk on X defined by fk(en) = 8kn is called the kth coordinate functional with respect to the basis {en}. It is not always true that such coordinate functionals are continuous. 1.25.14 Example. Consider the normed linear space (E0 , 11·11 00 ) with Schauder basis { ei, e 2 , . . . , e 0 , basis {ei'· e 2 ',

... }

.•• ,

given in Example 1.25.10(ii). We construct another Schauder

en'· ... } as follows.

Write for all n > I.

and Consider x

={AI> '-2, . . . , A,., 0, ... }.

Then x = ("-I -

±

±

kAk) ei ' + kAkek' and the coordinates of x are uniquely k=2 k=2 determined. So { ei ', e 2 ', ... , en'• ... } is a Schauder basis for (E0 , 11·11 00 ) . But further {en'l is convergent to e 1'. However, for the coordinate functional fi' corresponding to ei' we have fi'( en')= 0 for n ~ 2, but fi'( ei') =I so fi' is not continuous.

0

It is useful to know when such natural functionals are continuous. 1.25.15 Definition.

Given a normed linear space (X, 11·11), a Schauder basis {en} is

said to be monotone if for each x =

n

I.

Akek the sequence {II I. Akek II} is monotone

~I

~I

increasing. In most classical sequence spaces the obvious Schauder bases are monotone. For monotone bases we have the following satisfying property. 1.25.16 Theorem. Given a normed linear space (X, 11·11), with a monotone Schauder

basis {en·), the corresponding coordinate functionals are continuous.

Proof. We may consider II en II= I for all n EN. Then given n EN, n

I fn(x) I =II fn(x) en II= II

L k=l

n-I

Akek-

L k=I

n

Akek II :<;;II

L

n-I

L Akek II k=l for all x EX,

Akek II +II

k=I :<;; 2 II x II

since the basis {en} is monotone; so f 0 is continuous.

0

§ I. Basic properties

19

1.26 EXERCISES I.

Consider the linear space m and the linear subspaces E0,

.t 1, .t 2, c0, c introduced

in Examples !.II. (i)

Order these subspaces by set containment.

(ii)

For which of these subs paces do the following real functions define a norm? (a)

=O·t• A2, ... , An, ... } ,

11·11 1 where for x 00

L

II xll 1 =

I Ak I.

k=l

(b)

11-11 2 where for x

= {A1, A2, ... , An, ... } ,

00

L,lll/.

llxll 2 =

k=l

(c)

ll·llc where for x

={A1, A2, ... , An, ... } ,

I

II x lie= sup{_! I Ak I : n n k=l

E

N}

(iii) Determine whether the following normed linear spaces are closed subspaces of (m, ll·lloo). (a) (Eo, ll·lloo)

(b)

(.l I• ll·lloo)

(c)

(co, ll·lloo)

(iv) Determine whether the following normed linear spaces are complete. (a) (co. ll·lloo) (b) (.ll, ll·lloo) (c) (.ll, ll·llt)

2.

Consider the linear space

~[0,1]

and the linear subspaces 'C:[O,l], 1t [0,1],

t: 1[0,1] and t: 0[0,1] introduced in Examples 1.12. (i)

Order these subspaces by set containment.

(ii)

Determine whether the following normed linear spaces are closed subspaces of ('C:[O,l]. ll·lloo). (a)

'C: 1[0,1]

(b)

'C: 00 [0,1].

(iii) Determine whether the following normed linear spaces are closed subspaces of ('ft [0,1], IHioo). (a)

t: 1[0,1]

(b)

(iv) Determine whether

'ft 0 [0,1].

t: 1[0,1]

is complete with respect to the following

norms. (a) II f lloo =max {I f(t) I: t E [0,11}.

I I

(b)

llf Ill=

I f(t) I dt.

0

(c)

llfll'=llflloo+llf' lloo.

20

3.

Nonned linear space structure

Given M and N linear subspaces of a normed linear space (X, 11·11) where X= M EB N. (i)

Prove that the real function 11·11' on X where

11·11'=11 mll+llnll for every x EX where x = m +nand mE M, n EN, defines a new norm on X. (ii)

Prove that M and N are closed subspaces of (X, 11·11').

(iii) Prove that if (X, 11·11) is complete and M and N are closed subspaces of (X, 11·11) then (X, 11·11') is complete. (iv) Prove that if (X, II· II') is complete then M and N are closed subspaces of (X, 11·11). (v)

4.

If (X, 11·11') is complete is it necessarily true that (X, II· II) is complete?

Consider a nonempty subset Kin a normed linear space (X, 11·11) and the function d(x, K)

=inf {II x-k II : k E K} .

(i)

Prove that d(x, K) is continuous on (X, 11·11).

(ii)

Prove that K is convex if and only if d(x, K) is a convex function on X.

(iii) Given that K is a closed linear subspace of (X, 11·11) prove that d(x, K) is a norm on X/K.

5.

Consider the linear space C [0, I] with norms 11·11 00 and 11·11 1. (i)

Consider the linear functional p0 on C[O,l] where Po (f)= f(O). (a)

(ii)

Prove that Po is continuous on (C [0, 1], ll·lloo).

(b) Show that Po is not continuous on (C[O,I], 11·11 1). Show that ll·lloo and ll·ll1 are not equivalent norms on C[O,I].

(iii) Consider the linear subspace C 0 [0,1]. (a) Prove that C 0 [0, I] is closed in (C[O, 1], 11·11 00 ) . (b) 6.

(i)

Show that C 0 [0,1] is dense in (C[O,l], 11·11 1).

For the linear space m, prove that p: m --7IR where, for x

= p, 1, A. 2 ,

•.. ,

An, ... },

p(x)=L~ 2n is a norm for m. (ii)

Prove that p is weaker than (a) the 11·11 00 -norm form and

(b) the 11·11 1-norm for -t 1• (iii) Show that neither (-t 1, p) nor (m, p) is complete.

§I. Basic properties

7.

21

Consider a linear functional f on a normed linear space (X, 11·11). (i) Prove that f is continuous if and only if the seminorm I f I is continuous. (ii)

Give an example to show that a seminorm p on X is not necessarily continuous if ker p is closed.

(iii) Prove that f is discontinuous if and only if ker I f I is dense.

8.

(i)

Consider the normed linear space (c, 11·11 00 ) and the functional f on c defined for x {A,, A2, ... , An, ... } by

=

=

f(x) =

L

t.

2 Prove that f is linear and continuous and determine ker f. Consider the normed linear spaces (c0 , 11·11 and (1 1, 11·11 1) and the mapping T: c0 ~ 1 1 defined for x 1, /... 2 , . . . , An, ... } by k=l

(ii)

/...

00 )

={/..

Tx={~'. ~2 , (a)

~~

}.

Prove that T is linear and continuous and determine ker T. Determine whether T is a topological isomorphism.

(b) 9.

... ,

Consider the normed linear space (c, ll·lloo) and the functional f on c defined for x 1 , /... 2, ... , An, ... } EC where "-n~ A by f(x) =A. (i) Prove that f is a continuous linear functional on (c, ll·lloo).

={/..

(ii)

Consider the mapping T 1: c

~ co defined by T,x ={A.,-/.., /...2-/..., ... , An-A, ... }. Prove that Tis a continuous linear mapping of (c, 11·11 00 ) onto (c 0 , 11·11 00 ) but is not one-to-one.

(iii) Consider the mapping T2 : c

~

c0 defined by

T 2x = { /..., /... 1-/..., /... 2-/..., ... , An-A, ... }. Prove that Tis a topological isomorphism of (c, ll·lloo) onto (c 0 , ll·lloo). (iv) Given that (c 0 , ll·lloo) is complete deduce that (c, ll·lloo) is complete.

10.

(v)

Define an equivalent norm 11·11 on c such that (c, 11·11) is isomorphically isomorphic to (c 0 , 11·11).

(i)

Consider the functional p 1 defined on (c, ll·lloo) where, for X= {A,, /...2, ... 'An, ... },

p 1 = lim I An I.

n--7=

(a) (b)

Prove that p 1 is a continuous seminorm on (c, ll·lloo). Show that the quotient space (c/ker PI> 11·11) is isometrically isomorphic to the scalar field.

22

Nonned linear space structure

(ii)

Consider the functional p2 defined on (m, 11·11 00) where, for

x

={A

1, A2 , . . . , An, ... }, pz(x) = lim sup I An I. n~

(a)

Prove that p 2 is a continuous semi norm on (m, 11·11 00 ) .

(b)

Show that the quotient space (m/ker p2 , 11·11) is complete and infinite dimensional.

II.

(i)

Prove that the linear spaces t:[a,b] and t:[O,I] are isomorphic under the mapping f ~ f* where f*(t) = f((l-t)a+tb) fortE [0, I].

(ii)

Hence, or otherwise, prove that (a) (t;[a,b], ll·lloo) and (t;[O, 1], ll·lloo) are isometrically isomorphic and (b)

12.

(i)

(t:[a,b], 11·11 1) and (t:[O,I], 11·11 1) are topologically isomorphic.

Prove that the linear spaces, Y = {f E t;(-Jt,Jtj : f(-1t) =f(Jt)}, t:(r) where r is the unit circle { (A,!l) E lR 2 : A2 + 11 2 = I} and t:~(21t)

(ii)

of continuous periodic functions of period 2n,

are isomorphic under the mapping f H f* where f*(COS 8, sin 8) = f(8) for 8 E lR. Hence, or otherwise, prove that (Y, 11·11 00 ) , (t:(r), ll·lloo) and (t:~(27t),

ll·lloo) are isometrically isomorphic.

(iii) Hence, or otherwise, deduce that (a) Y is closed in (t; [-Jt,Jt], 11·11 00 ) and (b) 13.

t:~(21t)

is closed in

(~t:(IR),

11·11

00 ) .

Prove that norms 11·11 and 11·11' are equivalent norms for a linear space X if and only if any one of the following conditions holds. (i)

Convergence of sequences is preserved.

(ii)

Boundedness of sets is preserved.

(iii) Sequences being Cauchy is preserved. 14.

Given a normed linear space (X, 11·11) and a symmetric convex set K with 0 E int K, the gauge p of K is defined by p(x) = inf {A> 0 : x E AK}. (i)

Prove that p is a continuous seminorm on X.

(ii)

Prove that if K is bounded then p is an equivalent norm for X.

§I. Basic properties

15.

(i)

23

Prove that separability is a linear topological invariant for normed linear spaces.

16.

(ii)

Deduce that (a) (c, ll·lloo) is separable, (b) (c, 11-iloo) and (m, 11-iloo) are not topologically isomorphic.

(i)

Prove that the linear subspace Eo is dense in both (c 0, 11·11 00 ) and

(ii)

(11, 11·111). Show that ll·lloo and 11·11 1 are not equivalent norms for E0.

(iii) Deduce that (1 1, 11·11 1) is separable. 17.

Consider a separable Banach space (X, II· II) with a Schauder basis {e0 }. (i)

Prove that 11·11': X -7 lR where for x =

L

A.kek

k=l

n

II x II' = sup { II

=

L

A.kek II : n e N }

k=l

(ii)

18.

is a norm for X and (X, 11·11') is complete. Prove that {e0 } is a monotone basis for (X, 11·11').

Consider the normed linear space ('fl[-1,1], 11-iloo) of polynomials on [-1,1] with norm II p(t) lloo =max {I p(t) I: t e [-1,1] }. (i)

Prove that the sequence {e0 } where e0 = I and e0 = t0 for all n e N, is a Schauder basis for 'P [-I, I] and that the coordinate functionals f0 are given by f0 (p) =

(ii) (iii)

p(n)(O)

n!

Prove that the coordinate functional f0 is continuous but show that for each n eN, f 0 is not continuous. Show that {e0 } is not a Schauder basis for (t:[-1 ,I], 11-iloo) although 'P[-1,1] is dense in ('C:[-1,1], 11-iloo).

Normed linear space structure

24

§2. CLASSES OF EXAMPLE SPACES

As we said previously, a normed linear space is a generalisation of Euclidean space. However, we could conceivably carry out such a generalisation in several different ways. So we study examples of normed linear spaces by classifying them according to a particular mode of generalisation. We begin by examining the class of examples which preserves finite dimensionality and then study the class which is dimension free but preserves the special inner product structure of Euclidean space. 2.1 Finite dimensional normed linear spaces Euclidean n-spaces and Unitary n-spaces have the following fundamental topological properties. 2.1.1 Proposition. Ann-dimensional Euclidean space (Unitary space) (i) is complete and

(ii) has the property that a subset is compact if and only if it is closed and bounded, (the Borel-Lebesgue Theorem), (see AMS §3 and §8). We treat Euclidean n-space as the prototype of all n-dimensional normed linear spaces and the following theorem explains why this is so. 2.1.2 Theorem. All n-dimensional normed linear spaces over IR (or ([) are topologically isomorphic to (IR n, 11·11 2 ), (or (
Proof. Consider an n-dimensional linear space Xn over IR and a basis { e 1,e2 ,

... ,

en} for Xn- Now Xn and IR n are isomorphic under the linear mapping

T: IR n --7 Xn defined by T(/~. 1 ,

A2 ,

... ,

An)= A1e 1 + A2e 2+ ... + Anen.

We show that Tis a topological isomorphism of (Xn, II· II) onto (IR n, 11·11 2 ): Firstly, T: IR n --7 Xn is continuous since n

II T (A 1, A2,

... ,

An) II ;<:;

L

I Ak Ill ek II ;<:;

k=l

n

= M II (A 1, A2 ,

... ,

An) 11 2 where M =

~)ekle. k=l

Secondly, consider the continuous mapping 11·11 o T: IR n-7 IR where 11·11 o T (A 1, A2 , ... , An)= II A1e 1 + A2e 2 + ... + Anen II.

§2. Example spaces

25

The unit sphere S(O; I) being closed and bounded is compact in (Rn, ll·llz), so 11·11 o T attains a minimum m 2 0 at some point (A~, A~, ... , A~) E S(O; I). If m = 0 then

11·11 o T(A.~. A.~ •... , A~)= II T(A.~. A.~, ... , A.~) II= 0, but this is impossible since Tis one-to-one. Therefore m > 0 and we have m :<;;II T(A 1, A. 2 ,

... ,

An) II for all (A. 1, A. 2 ,

... ,

An) ES(O; I)

and so m II (A 1, A. 2 , . . . , An) 11 2 :<;; II T(A. 1, A.2 , •.. , An) II for all (A 1, A. 2 , which implies that T has a continuous inverse on Xn.

... ,

An) E IR n,

D

2.1.3 Remark. Since "being topologically isomorphic to" is an equivalence relation on the set of all normed linear spaces it follows from Theorem 2.1.2 that all n-dimensional normed linear spaces over the same scalar field are topologically isomorphic. D Theorem 2.1.2 has many significant corollaries which spell out particular topological properties for finite dimensional normed linear spaces. The first is a particular application of Theorem 2.1.2. 2.1.4 Corollary. All norms for a given finite dimensional linear space are equivalent. The next follows directly from Proposition 2.l.l(i) and Theorem 1.24.9. 2.1.5 Corollary. Every finite dimensional norrned linear space is complete. We can make the following immediate deduction from this corollary using Proposition 1.13(i). 2.1.6 Corollary. Every finite dimensional subspace of a norrned linear space is closed. Theorem 2.1.2 also enables us to carry over the characterisation of compactness given in the Borel-Lebesgue Theorem to all finite dimensional normed linear spaces. 2.1.7 Corollary. In every finite dimensional norrned linear space a subset is compact if

and only if it is closed and bounded.

Proof. Consider a closed and bounded subset A of a finite dimensional normed linear space (Xn, II· II) over IR. Denote by T the topological isomorphism of (Xn, II· II) onto

(IR n. 11·11 2 ). Since Tis a homeomorphism, T(A) is closed in (IR n, 11·11 2 ) and since Tis a topological isomorphism, T(A) is bounded in (IR n, 11·11 2 ). By the Borel-Lebesgue Theorem, T(A) is compact in (IR n, 11·11 2). But A is the image of T(A) under the continuous D mapping T-1 so it is compact in (Xn, 11·11).

26

Nonned linear space structure

As we pursue our study of normed linear spaces in general, it is important to note that normed linear spaces possessing the characterisation of compactness given in Corollary 2. 1.7 are necessarily finite dimensional. To show this we need the following technical lemma. 2. 1.8 Riesz Lemma. Consider a proper closed linear subspace M of a normed linear space (X, 11·11). For each 0 < 8 < I there exists an x0 EX\ M where II x0 II = I such that d(x 0 , M);::: 8.

Proof. Since M is proper there exists an x 1EX \ M and since M is closed, d(x 1, M) = d > 0. So there exists some x0E M such that II x 1- x0 .

II~

d/8.

Xl-XO

Wnte xl\ = II xi-xO II . Then for all x EM, II x0-x II=

II x1-x0 II

II x 1 - (x 0+ II x 1-x 0 II x) II

and since x0 +II x 1-x0 II x EM we have d II XI) -X II ;::: = 8. d/8

D

2.1. 9 Riesz Theorem. A normed linear space is finite dimensional

if and only if the

closed unit ball is compact.

Proof. Consider a normed linear space (X, II· II) with compact closed unit ball B[O; I]. Now there exists a finite subset { xl' x2 ,

... ,

x0 } in B[O; I] such that

n

U { xk + B(O; i)}

:::2 B[O; 1]. Since M 0

= sp{ x 1, x2, •.. , x0 }

is finite dimensional

k=l

it is, by Corollary 2.1.6, a closed linear subspace of (X, 11·11). Suppose M 0 is proper. Then by Riesz Lemma 2.1.8 there exists an x 314 EX\ M 0 where II x 314 II

= I such that

d(x3/4• M 0 )

;:::

n

But this contradicts

U {xk + B(O; i)}

:::2 B[O; 1].

k=l

So we conclude that X = M 0 and is finite dimensional. Corollary 2.1.7 completes the characterisation.

D

Significantly, Theorem 2.1.2 implies that linear mappings on finite dimensional normed linear spaces are automatically continuous. 2.1.10 Corollary. Every linear mapping of a finite dimensional normed linear space into

a normed linear space is continuous.

~

•

27

§2. Example spaces

Proof. Consider an n-dimensional normed linear space (Xn, 11·11) over lR with a basis { e 1, e 2 , ... , en}. For any linear mapping T of Xn into a normed linear space (Y, 11·11') we have, for x

= A. 1e 1 + A. 2e 2 + ... + A.nen, n

Tx=

L

~ Tek.

k=l n

II Tx II' :::;

So

L

I A.k I II Tek II' :::; K

k=l n

by the Cauchy-5chwarz inequality, where K

=

'L,I1Tek11' 2 . k~l

Since (Xn, 11·11) is topologically isomorphic to (IR n, 11-11 2), there exists an M > 0 such that II (1.. 1, A.2 , •.• , An) 11 2 :::; M II x II for all x EX

II Tx II' :::; KM II x II for all x E X.

so

D

This corollary gives us useful insight into the nature of convergence in any finite dimensional normed linear space.

2.1.11 Corollary. In a finite dimensional normed linear space (Xm, 11·11) with basis { e 1, e 2 , . . . , em}, a sequence { Xn} is convergent to x

if and only if each coordinate

sequence {A.~} is convergent to A.k fork E { 1, 2, ... , m} where x 1 = A.:e, + A.ie 2+ ... + A.~em x 2 = A.ie 1 + ~e 2 + ... + A.~ em

m

Proof. Now

llxn-xll:::;

L

1~-~lllekll

k=l

so coordinatewise convergence implies convergence. Conversely, for each k E { 1, 2, ... , m} the coordinate functional fk: Xm--? lR defined for x

=A. 1e 1 + A.2e 2 + ... + l..mem

fk(x) = A.k is linear, so by Corollary 2.1.1 0, is continuous.

by

D

Again we should note that the property given in Corollary 2.1.1 0 actually characterises finite dimensionality. To prove this we use the fact that every linear space has a Hamel basis, (see Appendix A.3).

28

Normed linear space structure

2.1.12 Theorem. A normed linear space (X, 11·11) is finite dimensional

if and only if every

linear functional on X is continuous.

Proof. Suppose X is infinite dimensional and consider a Hamel basis { e 1J. We may suppose that II eA. II = I for all 'A. Now since X is infinite dimensional { eA,} contains a countably infinite subset {eA.n}. A linear functional is defined on X if we specify its values on a Hamel basis so any linear functional f on X taking values f(eA.) = n for n EN, is not n

continuous on (X, II· II). Corollary 2.1.10 completes the characterisation.

D

In Corollary 1.24.16 we noticed how the continuity of a linear functional could be determined by its kernel being closed. It is worth noting that a similar characterisation does extend to linear mappings with finite dimensional range.

2.1.13 Theorem. A linear mapping T of a normed linear space (X, 11·11) into a finite

dimensional normed linear space (Y, 11·11') is continuous if and only ifker T is closed.

Proof. If T is continuous then ker T is closed. Conversely, suppose that dim T(X) =nand that {e 1,

••• ,

en I is a basis for T(X).

Then for each XE X Tx = f 1(x) e 1+ ... + fn(x) en. Since T is linear, f 1,

••• ,

fn are linear functionals on X. For each k E { 1, 2, ... , n I

consider xk EX such that Txk = ek and Xn

= sp{x 1, ..• , Xnl.

Now X= X 0 EB ker T. Since ker T is a closed linear subspace, d(x, ker T) is a norm for Xn and since Xn is finite dimensional, d(x, ker T) is an equivalent norm for Xn- So there exists a K > 0 such that II x II

~

K d(x, ker T) for all x E X 0 .

Since Xn is finite dimensional fk is continuous on Xn for each k E { 1, 2, ... , n I so there exists an Mk > 0 such that I fk (x) I ~ Mk II x II for all x E X 0

.

Now for z EX, z = x + y for x EX 0 andy Eker T, and I fk (z) I =I fk (x) I ~ Mk II x II

~

K Mk d(x, ker T)

~

K Mk II x+y II= K Mk II z II

so fk is continuous on X. We conclude that Tis continuous on X.

D

The compactness property which, by the Riesz Theorem 2.1.9 characterises finite dimensional normed linear spaces, enables us to establish the existence of points of best approximation in finite dimensional subspaces of a normed linear space.

29

§2. Example spaces

2.1.14 Theorem. Given a normed linear space (X, 11·11) and a finite dimensional linear subspace Xm.for any xo EX\ Xm there exists a Yo E Xm such that II x0-y 0 II = d( x0 , Xm)·

=

Proof. Writing d d(x 0 , Xm), consider the closed ball B[x 0 ; 2d]. From Corollary 2.1.6 we have that Xm is closed so B[x 0; 2d] n Xm is closed and bounded in (Xm, ll·llxm). Since Xm is finite dimensional we deduce from Corollary 2.1.7 that that B[x 0 ; 2d] n Xm is compact. Consider a sequence {x 0 I in B[x 0 ; 2d] n Xm such that II x-x 0 II ---7 d as n ---7 oo. Then since B [x 0 ; 2d] n Xm is compact, {x0 I has a subsequence { x 0 k} which is convergent to some Yo

E

xm.

Figure 1. The ball B[x 0 ; 2d] "scoops" a compact set out of Xm. d :0: II x0-y 0 II :0: II x 0-x 0 k II +II x0 k-y 0 II we deduce that II x0-y0 II= d.

Since

---7

d

as k

---7 oo

D

This existence theorem is made somewhat sharper by the following uniqueness theorem. 2.1.15 Definition. A normed linear space (X, 11·11) is said to be rotund if for every x 1' y, II x II= II yll =I we have II x+y II< 2. 2.1.16 Theorem. Consider a rotund normed linear space (X, 11·11) and a linear subspace M./ffor x0 EX\ M there exists a Yo EM such that II x0-y 0 II = d(x 0 , M) then Yo is unique.

Proof. Suppose that there exist Yo and Yo' in M such that II x 0-y 0 II = II x 0-y 0' II = d(x 0 , M). Then d(x 0 , M) =

i

(II x0-y 0 II+ II x 0-y 0' II);::: II x 0 -

~ (y 0 + y0 ')

II ;::: d(x 0 , M).

30

Nonned linear space structure

So

II x0-y 0 II= II x0-y 0 ' II= d(x 0 , M) and II (x 0-y 0 ) + (x 0-y 0 ') II= 2d(x0 , M).

Since (X, 11·11) is rotund we deduce that y0 ' =y0 .

D

2.1.17 Remark. Theorems 2.1.14 and 2.1.16 guarantee the existence and uniqueness of best approximating points in finite dimensional linear subspaces; the actual calculation of the form of such points is another problem altogether. One of the many advantages of inner product space structure is that it enables the calculation of best approximating points in these spaces. D 2.2. Inner product spaces The inner product spaces are undoubtedly the most useful class of example spaces. They result from a generalisation of Euclidean space which is dimension free but which retains an algebraic structure enabling fruitful geometrical insight into the nature of the space. We give our definitions for complex linear spaces but it is to be understood that these include real linear spaces. 2.2.1 Definition.

Given a linear space X over

a::, a mapping(.,.):

X x X --7

a::

is an

inner product on X if it satisfies the following properties: For all x,y,z E X (i) (x+y, z) = (x, z) + (y, z) (ii) (h, y) = A.(x, y) for all A. E a:: (iii) (y, x)

= (x, y)

(iv) (x, x)

~

(v)

(x, x)

0

= 0 if and only if x = 0.

A linear space X with an inner product(.,.) is called an inner product space (or a preHilbert space) and is sometimes denoted formally as a pair (X,(. , .)). 2.2.2 Remarks. Properties (i) and (ii) imply that for any z EX the mapping (. , z): X --7 C is a linear functional on X. From properties (i), (ii) and (iii) it follows that for all x,y,z EX, (x, y+z) = (y+z, x) = (y, x) + (z, x) = (x, y) + (x, z)

and

(x, ~y) = (Af,X) = X(Y,x)

'

= A(x,y) foraiiA.E[.

So for any z E X, the mapping (z, .): X --7 Any mapping from X x X into

a::

a::

is not linear but "conjugate linear".

which satisfies properties (i), (ii) and (iii) is called an

hermitian form on X. An hermitian form which satisfies property (iv) is said to be a positive hermitian form and a positive hermitian form which satisfies property (v) is said to D be a positive definite hermitian form.

3I

§2. Example spaces

From properties (i)-(iv) we derive the following important inequality. 2.2.3 The Cauchy-Schwarz inequality. A positive hermitian form (. , .) on a linear

space X satisfies the property

I (x, y) 12 ~ (x, x) (y, y) for all x,y EX.

Proof.

If (x, x)

= (y, y) = 0 then by (iv) (x- (x, y)y, x- (x, y)y);::: 0

simplifying to -2 I (x, y)

If (y, y)

1'

12 ;:::

0 which implies that (x, y) = 0.

0 then by (iv) (

X -

~ (y, y)

y'

X -

(~ (y, y)

y)

;::>:

0

which simplifies to (x, x)(y, y) -I (x,y) 12 ;::: 0. If (y, y)

=0 but (x, x) 1' 0 we expand (y - ~ (x, x)

x' y -

(~ (x, x)

x);::: 0.

D

The following inequality provides the link with normed linear spaces. 2.2.4 The Minkowski inequality. A positive hermitian form (. , .) on a linear space

X satisfies the property .Yr(x_+_y_,-X+-y-,-)

~

-{(X;YJ + {(Y.Y)

for all x, y EX.

Proof. (x+y, x+y)

= (x, x) + 2 Re(x, y) + (y, y) ~ (x, x) + 2 .,J(x, x)(y, y) + (y, y)

from the Cauchy-Schwarz inequality ,

D 2.2.5. Remarks. (i)

It follows from the Minkowski inequality that a linear space X with a positive

hermitian form (. , .) is a seminormed linear space with seminorm p2 : X -t IR defined by p2(x) = .,J(x,x). A linear space X with an inner product(., .) is a normed linear space with norm 11·11 2 : X -t IR, the norm generated by the inner product, where II x 11 2 = {(X:X). (ii)

In terms of this norm the Cauchy-Schwarz inequality takes the form I (x, y) I ~II x 11 2 II y 11 2 for all x, y EX.

32

Nonned linear space structure

This inequality is important in linking the algebraic inner product to the norm topology and implies that the inner product has a continuity property which we call joint continuity of the inner product: if x ~ x0 andy ~ Yo then (x, y) ~ (x 0 , y0). This can be deduced simply from the following inequalities. I (x, y)- (x 0 , y0) I~ I (x, y-y 0 ) I+ I (x-x 0 , y0)

I~

II x 1111 y-y 0 II+ II x-x 0 1111 y0 II.

0

A linear space X with a positive hermitian form(. , .) is a seminormed linear space. But it can also be transformed in a natural way into an inner product space where the norm generated by the seminorm is the natural norm generated by the seminorm as given in Definition 1.18. 2.2.6 Theorem. Given a linear space X with positive hermitian form(.,.) generating a

seminorm p 2, the quotient space X/ker p 2 is an inner product space with inner product defined by ([x], [yJ) = (x, y) for any x E [x] andy E [y].

Proof. We need to check that the definition of the inner product in X/ker p 2 is well defined. For any x, x' E [x] andy, y' E [y], p 2(x-x') = 0 and p 2(y-y') I (x, y)- (x', y') I~ I (x-x', y) I+ I (x', y-y') I ~

= 0.

So

P2(x-x') P2(Y) +Pix') P2(y'-y) from the Cauchy-Schwarz inequality,

=0.

We need to check that the inner product has property (v). For any x EX, ([x], [x]) So if ([x], [xJ)

= (x, x) = p~(x).

=0 then x E ker P2> but ker p 2 is the zero element in Xlker p 2 .

0

An inner product space has the following simple but very significant norm relation. 2.2.7 The parallelogram law. In an inner product space X, II x+y 11 2+ II x-y 11 2 = 2 ( II x 11 2 +II y 11 2) for all x,y EX;

(that is, the sum of the squares of the lengths of the diagonals of a parallelogram is equal to the sum of the squares of the lengths of the sides).

Proof.

II x+y 11 2 +II x-y 11 2 = (x+y, x+y) + (x-y, x-y) = 2 ( II X 11 2 + II y 11 2)

by straightforward expansion of the inner products.

0

§2. Example spaces

2.2.8 Remark.

33

P. Jordan and J. von Neumann, Math. Ann. 36 (1935), 719-723,

proved that if the norm of a normed linear space satisfies the parallelogram law then the norm is generated by an inner product. This is not very difficult to prove, (see Exercise 2.4.9) but it was one of the first characterisations of inner product spaces among normed linear spaces. (A recent collection of such characterisations was compiled by Dan Amir, 0 Characterisations of inner product spaces, Birkhauser, 1986.) 2.2.9 Definition. An inner product space which is complete as a normed linear space is called a Hilbert space. 2.2.10 Example. Unitary n-space (Euclidean n-space) is an inner product space with inner product(.,.) defined on q:n (or IR.n) for x = (}., 1, A2, ... , An) and y = (J..L 1, J..l2, ... , J..ln) by n

(x, Y) =

L Akiik·

k=l It is quite clear that all the inner product properties (i)-(v) are satisfied. The norm generated by this inner product is the Unitary (Euclidean) norm, n

llxll 2 =

I,IA./.

k=l We deduce from the inner product structure that the Cauchy-Schwarz inequality holds and applying this inequality to (I AJ I, I Azl, ... , I An I) and (I J..L 1 I, I J..L2 1, ... , I J..ln I) E IR.n, (J..L 1, J..L2 , n

we have for (A 1, A2 , ... , An) and n

... , J..ln)

E

q:n that

L I Ak l.ik I = L I Ak I I J..lk I

k=l

k=l

(This derivation of the Cauchy-Schwarz inequality from the inner product structure should be compared with the direct proof, (see AMS § 1). 0 2.2.11 Example. Hilbert sequence space ~ 2 is the linear space whose elements x = { A1, A2, ... , An, ... } are sequences of scalars such that I. I An 12 is convergent, and with coordinatewise definitions of the linear space operations. This is an inner product space with inner product(. , .) defined for x = { Ap A2, ... , An, ... }, I. I An 12 < oo and Y J..li, J..lz, · · · , J..ln, ... }, I. I J..ln 12 < oo, by

={

=

(x, y)

= L Akiik.

k=l Although this is a straightforward generalisation of Example 2.2.10 there are significant points which require checking.

34

Nonned linear space structure

Firstly, we use the Cauchy-Schwarz inequality of Example 2.2.10 to show that an inner product is defined. For any n EN,

~~Ake ~~~.uke < so

I

I

I Ak ilk I < oo and

k=l

00

,

Ak ilk is convergent.

k=l

Secondly, we use the Minkowski inequality of Example 2.2.10 to show that .t 2 is closed under coordinatewise definition of addition. For any n EN, n

IPk +.u/ ~ k=l so

I

I Ak+J..lk 12 < oo and x + y E .t 2 .

k=l

The norm generated by this inner product is the Hilbert space norm

11 x 112 =

~~~Ake

As a prototype Hilbert space we should show that (.t 2, 11·11 2) is complete.

Proof of completeness. Consider { x0 } a Cauchy sequence in (.t 2 , 11·11 2 ) where

that is, given e > 0 there exists a v E N such that =

II

X 0 -Xm ll2

=

I, IA~ - A~l 2 <

oo

k=l

So for each kEN, I ~ - A.~ I < e

for all m > n > v

that is, {A.~} is a Cauchy sequence in (
for all m > n > v.

35

§2. Example spaces

But (
A.k= lim~ O--)oo

and consider the sequence x = { A. 1, A.2, ... , A. 0 ,

••• } •

We need to show that ~

(i)

X E

l2 ' (that is,

I. I A.k 12
k=l

(ii) { x0 } converges to x; (that is, II x0 -x 11 2 -4 0 as n -4 oo ).

Now we have for any Jl eN that Jl

I. I ~ -

~ 12) < e2

for all m > n > v

k=l

so keeping n fixed and increasing m we have

f I~- A.k 12) < e2

for all n > v.

k=l As this holds for all Jl e N, ~

I. I ~ -

A.k 12) < e2

for all n > v.

k=l

From closure under addition we deduce that x e l 2 and then this statement implies that D { x0 } converges to x. 2.2.12 Examples. Integral inner product spaces. 'C: [a,b] the linear space of all complex (real) continuous functions on the interval (i) [a,b] is an inner product space with inner product(.,.) defined on 'C:[a,b] by b

=f f(t)

(f, g)

g(t) dt.

a

Clearly since f is continuous on the bounded closed interval [a,b], f is bounded and b

b

I

I

a

a

I f(t) g(t) dt I :::;

I f(t) llg(t) I dt:::; II f 11 00 II g 11 00 (b-a) ,

where II f lloo =max {I f(t) I : a:::; t:::; b} so the inner product is defined. The norm generated by this inner product is b

llfll2 =

Jlf(t)l 2 dt. a

However, ('C:[a,b], 11·11 2) is not complete. Consider the real linear space 'C:[0,1] and the sequence {f0 } where 0:::; t:::;.!.

f 0 (t)= 1

2 1 1 1 -
1 =1-(n + 1)(t--) 2 =0

I

1

-+--
36

Nonned linear space structure

which is pointwise convergent to f on [0, I] where f(t) =I

)

I - < t::; 1 2

=0 I

Now

JI fn(t)- fm(t) edt< n~ 1 for all m > n so {fn} is a Cauchy sequence in

(t: [0, 1], 11·11 2 ). But f e 'C [0, I] and there is no continuous function on [0, I] to which {fn} converges with respect to the 11·11 2-norm. An attempt to overcome this problem by enlarging the space raises other difficulties.

(ii)

We could consider "0\[a,b], the linear space of all Riemann integrable functions on the interval [a,b]. In this case b

(f, g)

=f f(t)

g(t) dt

a

is a positive Hermitian form but it is not positive definite. In the real linear space "01. [0, I] the function f defined by f(t) = 0

o::;t
=I I

has

Jf\t) dt = 0. 0

However, from Theorem 2.2.6 we see that the quotient space"O\[a,b] /ker p2 where b

p 2(f) =

Jlf(t)e dt, can be represented in a natural way as an inner product. a

But again the inner product space ("01. [a,b] I ker p 2, 11·11 2) has the same defect we found in (i); that is, it is not complete. Consider the interval [0, I] and the following adaptation of the construction of Cantor's Ternary Set, (see AMS § 10). From [0, 1] remove E 1

=(~,~),the middle open interval of length i .

From [0, I] \ E 1 remove E 2 = (;2 , 372)

U

(~~ , ;~), the union of the middle open intervals

of the remaining two intervals each of length

!· k.

From [0, I]\ (E 1 u E2) remove E 3, the union of the middle open intervals of the remaining four intervals each of length

i ·/

6 .

n-1

In general, from [0, I]\ UEk remove En, the union of the middle open intervals of the I

remaining 2n-l intervals each of length - 1-1- - 1-1 • 2n- 2n+

§2. Example spaces

37 n

Now consider the sequence {fn} of characteristic functions of UEk; that is, I

t ElJEk =0

t E [:.I] I

)

QE.

Since each fn has only a finite number of discontinuities, it is Riemann integrable and

12 I

p22
f n
T

I

and

p~ (fn-fm) =

j (fn(t)- fm(t)) m

=L

2

dt

1

k+T form > n. n+l 2 So { [fn]} is a Cauchy sequence in (wt [0, 1] I ker p 2 , 11·11 2).

-

Now {fn} converges pointwise to f the characteristic function of UEk; that is, I

tEUEk =0

tE

)

[:.I] IQE,

We recall that a characteristic function on [a,b] is Riemann integrable if and only if given E > 0 the set of points of discontinuity of the function can be contained in a finite union of intervals of total length < E. If the set of points of discontinuity off, which is [0, 1] \ U Ek , can be covered by a finite union of intervals of total length

E<

i

I

then the set U Ek contains I

the union of the remaining intervals [0, 1] and so UEk has total length> 1- E > ~. But the I

sum of the length of the intervals in En -+ ~ as n --+ =. Therefore f is not Riemann integrable on [0,1]. Furthermore, there is no Riemann integrable function f on [0,1] such that p 2(fn-f)--+ 0 as n--+ =.We conclude that { [fn]} does not converge in (wt [a,b] I ker p 2 , 11·11 2). (iii) The attempt to overcome this defect with integral spaces has led to some very fruitful advances in analysis, in particular to the theory of the Lebesgue integral. The successful strategy is to determine a class of functions which is larger than the class of Riemann integrable functions and then proceed as in (ii). In this way we hope that we have made the space large enough to contain the limits of the Cauchy sequences. Consider the linear space of complex functions f on [a,b] such that If 12 is Lebesgue integrable with positive Hermitian form(. , .) defined by (f, g)

=t.b) fg dJl.

38

Nonned linear space structure

The inner product space formed as a quotient by ker p2 where

pz(f)

=

f Id

diJ.

[a,b]

is complete and is often referred to as classical Hilbert space and is denoted

(I: 2 [a,b], 11·11 2 ).

D

There is a theoretically elegant way of representing classical Hilbert space

(I: 2[a,b], 11·112) as the Hilbert space which contains ('C [a,b], 11·11 2) as a dense linear subspace. This is helpful conceptually but not so useful in practice. We will discuss this further in Section 7 .12. Perhaps the most important special property of inner product spaces is that they possess a workable notion of orthogonality which is a direct generalisation of orthogonality in Euclidean spaces. This means that in such spaces there is a rich geometrical structure similar to Euclidean space. We formally define an orthogonality relation by the inner product but the usefulness of the notion depends on its geometrical characterisation in terms of the norm. 2.2.13 Definition. Given an inner product space X, for x,y orthogonal toy if (x, y) = 0 and we write x

l_

E

X we say that x is

y. Given subset M of X we say that x is

orthogonal toM if x is orthogonal to every element of M and we write x l_ M.

2.2.14 Remarks. From the defining properties of an inner product we can deduce the following properties about this orthogonality relation. From property (iii) we see that x is orthogonal to y if and only if y is orthogonal to x; that is, the orthogonality relation is symmetric. From property (i) it follows that if x is orthogonal to a subset M then x is orthogonal to sp M and with property (iii) that every element of sp M is orthogonal to x. From property (ii) we deduce that 0 is orthogonal to every x EX. From property (v) we have that 0 is the only element orthogonal to itself. So we deduce from property (v) the result which is so very useful in practice that if x is orthogonal to every element of the space then x = 0; that is, 0 is the only element orthogonal to all the elements of the space. D We now show that orthogonality defined by the inner product is actually identical to a

§2. Example spaces

"least distance" type of orthogonality defined by the norm and important in best approximation problems. 2.2.15 Theorem. In an inner product space X, for x,y eX (x, y) = 0 if and only if II x+A.y II:::: II x II for all scalar A.. Proof. If (x, y) = 0 then II x+A.y 1111 x II;::: I (x+A.y, x) I= Ill x 11 2 + A.(y, x) I= II x 11 2 II x+A.y II ;::: II X II so for all scalar A..

Conversely, if II x+A.y II ;::: II x II for all scalar A., then II x+A.y 11 2 - II x llllx+A.y II ;::: 0 soRe (x, x+A.y) + Re A.(y, x+A.y) -I (x, x+A.y) I;::: 0. Therefore, Re A.(y, x+A.y);::: 0 for all scalar A.. For real A., Re (y, x+A.y) ;::: 0 for A.;::: 0 and :5: 0 for A. :5: 0. But (y, x+A.y) = (y, x) +A. II y 11 2 ~ (y, x) as A.~ 0 so we deduce that Re (y, x) = 0. For A.= ia where a is real, we have Re {a(iy, x+aiy)} ;::: 0 for all real a so by the same argument we have Re (iy, x) = 0, which implies that Im (y, x) = 0 and we conclude that (x, y) = 0. D 2.2.16. Remark. We should explain why the norm condition characterising orthogonality in Theorem 2.2.15 is called "least distance" orthogonality. x+sp{y} X

Figure 2. To illustrate "least distance" orthogonality.

"*

Given x,y eX, x,y 0, consider the two dimensional linear subspace sp{x,y}. Now d(x, sp{y}) = inf{ II x+A.y II : scalar A.} ;::: II x II if and only if x has 0 as a closest point in sp{y}.

39

40

Nonned linear space structure

Equivalently, d(O, x+sp{y})

=inf{

II x+A.y II: scalar A.} 2': II x II

if and only if 0 has x as a closest point in x + sp{ y}.

D

The following result is an interesting consequence of Theorem 2.2.15 which is not otherwise obvious. 2.2.17 Corollary. In an inner product space X,for x,y EX

if and only if

II x+A.y II 2': II x II

for all scalar A.

II y+A.x II 2': II y II

for all scalar A..

2.2.18 Remark. Such symmetry of least distance orthogonality does not hold generally for normed linear spaces other than inner product spaces. In fact for normed linear spaces of dimension greater than or equal to three if this property holds then the norm of the space is generated by an inner product, (see Dan Amir, Characterisations of inner product spaces, Birkhauser, 1986, p. 143). D Given a normed linear space (X, 11·11), a finite dimensional linear subspace Mn and x EX \ Mn, we have seen in Theorem 2.1.14 as a consequence of compactness properties of finite dimensional normed linear spaces, that there exists a closest point to x in Mn. For infinite dimensional linear subspaces generally we do not have such compactness properties and best approximation points do not necessarily exist. However, in Hilbert space the parallelogram law guarantees the existence of best approximation points in all closed linear subspaces. 2.2.19 Theorem. Given any proper closed linear subspace M of a Hilbert space H and

any x E H \ M, there exists a unique y EM such that II x-y II

= d(x, M).

Proof. There exists a sequence {yn} in M such that II x-yn II~ d(x, M). Consider I · I

Yn-x and Ym-x where m *- n. Now

2 (Yn+Ym) EM so 112 (yn+Ym)- x II 2': d(x, M).

By the parallelogram law we have II Yn-Ym 11 2 = 2(11 Yn-x 11 2 +II Ym-x 11 2) -II Yn+Ym-2x 11 2

~ 2(11 Yn-x 11 2 +II Ym-x 11 2) - 4d 2(x, M). It follows that as II Yn-X II~ d(x, M) we have II Yn-Ym II~ 0; that is, {yn} is a Cauchy

sequence in M. But as H is complete and M is a closed linear subspace of H, { Yn} is convergent to some y EM. Now II x-y II 2': d(x, M), but II x-y

II~

II x-yn II+ II Yn- y

II~

d(x, M), so we conclude that

II x-y II = d(x, M). To show that y is unique suppose there is a z E M such that II x-z II = d(x, M). Consider I I x-y and x-z. Now 2 (y+z) EM so II x- 2 (y+z) II 2': d(x, M).

§2. Example spaces

41

Again by the parallelogram law we have II y-z 11 2 = 2(11 x-y 11 2 +II x-z 11 2 ) - II 2x-(y+z) 11 2 5 4d 2(x, M)- 4d 2(x, M) and we conclude that z = y.

=0. D

2.2.20 Remark. Uniqueness could be deduced directly from the rotundity of the inner product space; (see Exercise 2.4.7 and Theorem 2.1.16). 0 Theorem 2.2.19 has the following interpretation in terms of orthogonality. 2.2.21 Corollary. Given any proper closed linear subspace M of a Hilbert space H there exists a nonzero element z E H \ M such that z is orthogonal toM.

Proof. Consider any x E H \ M. From Theorem 2.2.19 we have that there exists a unique y EM such that II x-y II= d(x, M). Write z = x- y. Then for any wE M and scalar A, II z+ Aw II =II x- (y-Aw) II ;:>: d(x, M)

since y-Aw EM

=II zll. From Theorem 2.2.15 we deduce that (z, w) = 0. But this holds for all w EM so z is orthogonal toM.

D

2.3 The ;r; p spaces (I $p$oo) A significant class of normed linear spaces is determined from a generalisation of the Euclidean norm formula and as such includes the inner product spaces as a special case. We begin with a discussion of these norms in finite dimensional spaces.

2.3.1 Example. 1" space (I 5 p 5 oo). p x

The n-dimensionallinear space
= (A 1,A 2 ,

n

II

X

lip =

(I. I Ak iP) lip

I $ p < oo,

k=l

II x II= = max {I Ak I : k E {I, 2, ... , n}} is called an 1" space. The latter is called the p = oo case and is included in the class because p in
=(AJ>A2, ... ,An)

n

max {IAk I: k E {1, 2, ... , n} = lim (LI Ak iP) 11P, (see AMS §I). p~=

0

k=l

The proof that II·IIP obeys the triangle inequality when p = I or oo is straightforward, but when I < p < oo the proof depends on the following inequalities.

42

Nonned linear space structure

2.3.2 Lemma. For positive real numbers a and p

a

pq

P

ap::;;- + p q

Proof. This is actually a special case of the G.M.-A.M. inequality which states that for positive real numbers x and y x"-y I-A.::;; AX + (1-A)y for any given 0
0

p

2.3.3 The Holder inequality. In ["(or lR 0 ), given I < p <

x

=(A], A2, ... , A

0)

oo,

for any

=

and Y (J.LJ.Jl2, ... , Jl0 ),

n

n

n

(I, I Ak P:k I)::;; (LI Ak iP) 1ip (I, I Jlk lq)Iiq k=l

k=l

k=l

where 1_ + 1_ = I. p

q

Proof. For x,y

* 0, put

in the inequality of Lemma 2.3.2 and summing fork

E

{I ,2, ... ,n} we have

n

Ll AkP:k I k=l n

n

k=l

k=l

0

(I, I Ak iP) 1ip (I, I Jlk lq)Iiq

2.3.4 The Minkowski inequality. In Q:" (or lR "),given I < p < oo,for any

=

x = (At,A2, ... ,An) and Y (J.L~oll2• ... ,Jln) n n n ( L I Ak +Jlk iP) lip ) ::;; ( L I Ak iP) lip ) + ( L I Ilk iP) lip k=l k=l k=l Proof.

n

n

n

k=l

k=l

k=l

)

L I Ak +Jlk iP::;; L I Ak+Jlk iP- 1 I Ak I + L I Ak +Jlk w-' I Jlk I.

§2. Example spaces

43

Using the Holder inequality for the sums on the right we have n

n

n

k=l

k=l

k=l

n

n

k=l

k=l

L I A.k +flk iP $ ( L I A.k +flk l(p-l)q) l/q ( L I A.k iP) l/p ) +( L I A.k+flk l(p-l)q) l/q ( L I Ilk iP) l/p n

But (p-1 )q = p. When x + y

"* 0, dividing by ( L I A.k+flk iP)

Itq )

we derive the Minkowski

k=l

inequality.

D

We now generalise to consider such norms in infinite dimensional sequence spaces.

2.3.5 Examples. Sequence .lp spaces (I $ p $ oo) Given I < p < oo, the set .lp is the linear space of sequences of scalars

=

x {A. I' 1.. 2, ... , A0 , . • . } such that L I A.k iP is convergent, under coordinatewise definition of the linear space operations. The linear space .lp with norm II

;

X

liP = (""' I A.k k=l

It

iP) p.

is called an .lp space. The set -loo is the linear space of bounded sequences of scalars x

={A. I' 1.. 2, ... , 1..

0 , ••• }

under coordinatewise definition of the linear space

operations. The linear space -loo with norm II X lloo = sup { I A.k I : k E N } is called an -loo space (or m space). Again this case is included in the class because for any bounded sequence x sup

= {A.p 1..2 , ••.• A.n •... }.

{I~ I: keN}=

lim

Ci I A.k iP) 11P, (see AMS §I).

0

p~k=l

Verification of the particular linear space and norm properties when p = I or oo is straightforward. But when I < p < oo, particular points we need to verify are that .lp is closed under addition and the norm satisfies the triangle inequality. To prove both we establish an extension of the Minkowski inequality.

2.3.6 The Minkowski inequality. Given I < p < oo and any X= { Ap 1..2, ... , A0 , . . . } andy= {flJ,fl2• ... ,fl0 , ... } in .lp then x + y E .lp and

(~I A.k+flk iP) 11P $(~I A.k 1P) 11P +(~I Ilk iP) 11P. k=l

k=l

k=l

Proof. From the finite dimensional Minkowski inequality 2.3.4 we have for any n eN, n

n

k=l

k=l

n

( L I A.k+flk 1P) 11P $ ( L I A.k iP) 11P + ( L I Ilk iP) 11P

$(~I

so L I A.k+flk iP

k=l A.k iP) 11P

+(~I

flk iP) 11P < oo since x,y e .lp, k=l k=l is convergent and x + y e .lp and the general inequality is established. D

44

Nonned linear space structure

Now the normed linear spaces (.tP' ll·llp) where I

~

p

~

oo

are all complete. The

proof of completeness follows closely the pattern of proof given for Hilbert sequence space (.t 2 , 11·11 2 ) in Example 2.2.11 and is left as an exercise for the reader. 2.3.7 Examples. Integral Z'P spaces (I (i)

Given I

~

~

p < oo).

p < oo, 'C [a,b] the linear space of all complex (real) continuous functions

on the interval [a,b] is a normed linear space with norm b

II fliP= (

f I f(t) iP dt)

lip

a

but the counterexample used in Example 2.2.12(i) shows that ('C [a,b], II· lip) is not complete. (ii)

We may attempt to overcome this problem as in Example 2.2.12(ii) and consider

~ [a,b],

given I

the linear space of all Riemann integrable functions on the interval [a,b]. Again, ~

p < oo, it is easy to see that b

Pp(f) = (

j I f(t) iP dt) lip a

is not a norm, but it is a seminorm. To show that the triangle inequality is satisfied in both cases (i) and (ii), we need to establish the Holder and Minkowski inequalities for integrals. For f,g b

JI f(t) I dt, JI g(t) I dt * 0, given I~ p < a

E ~[a,b]

where

b oo

and recalling that ker Pp = ker p 1,

a

(see AMS § 1), we put a

= ---:-b_l_f-'-(t-'-)-1- - and p - -----:-b_l-"g"-(t-"--)_1_ _

(f I f(t) IP dt)

lip

a

cjl g(t) lq dt) 1iq a

in the inequality of Lemma 2.3.2 and integrating over [a,b] we obtain the required Holder inequality. Then following the pattern of proof of the Minkowski inequality 2.3.4 we derive the triangle inequality. We could form the quotient space ( ~ [a,b] I ker Pp , ll·llp) as in Example 1.19, but again Example 2.2.12(ii) shows that this space is not complete. (iii) We successfully overcome this defect as in Example 2.2.12(iii), by the Lebesgue integral. Given I

~

p <

oo,

consider the linear space of all complex (real) measurable

functions f on [a, b] such that If IP is Lebesgue integrable with seminorm

f

I f iP d]J.) lip). [ab] The normed linear space formed as a quotient by ker Pp is complete and is generally Pp(f) = (

referred to as .t;P [a,b] space. For Lebesgue integration of functions on [a,b] we do not confine our attention to bounded functions as we do for Riemann integration. So we should point out that although ~ [a,b] I ker Pp is the same linear space for all I ~ p < oo, the .t:p[a,b] spaces provide 0 different linear spaces for different values of p where I ~ p < oo.

45

§2. Example spaces

Again we note that given I :s; p < oo, (.r;p[a,b], ll·llp) can be represented as the Banach space which contains ('C[a,b], ll·llp) as a dense linear subspace. We will discuss this further in Section 7 .12.

2.4 EXERCISES

I.

Prove that a normed linear space (X, 11·11) is finite dimensional if and only if (i)

its unit sphere S(O; I) is compact,

(ii)

every bounded sequence has a convergent subsequence, or

(iii) there exists a compact neighbourhood of the origin. 2.

(i)

Consider a normed linear space (X, 11·11), and a closed linear subspace M. (a) Given y i" M, prove that M + sp{ y} is closed in (X, 11·11). (b) Given a compact set A in (X, II· II) prove that M +A is closed. (c) Prove that given any finite dimensional linear subspace N the linear subspace M+N is closed in (X, 11·11).

(ii)

Is it true that M+N is closed for closed linear subspaces M and N of a normed linear space (X, 11·11) ? If it is true give a proof, if it is false give a counterexample. (Hint: See Exercise 1.26.3.)

3.

Consider the linear space

~(X)

of bounded functions on a non empty set X.

Prove that ~(X) is finite dimensional if and only if X is finite. (Hint: Consider ~(X) with norm II f lloo =sup{ I f(x) I : x EX}.) 4.

(i)

Consider then x n complex matrix (aij)· Prove that the mapping ~
(., .):
=(J.l.I ,Jlz, ... ,Jln) by

n

(x, y) =

L

aij Ai P:j is an inner product on
i,j=l

and only if (aij) is a positive definite hermitian matrix. (ii)

Consider 1R n with a norm 11·11. Prove that the norm 11·11 is generated by an inner product if and only if the unit sphere is an ellipsoid.

5.

Show that, when I :s; p :s; oo, p

#

2, none of the II·IIP norms on the .tp-spaces is

generated by an inner product. 6.

(i)

Given a finite dimensional normed linear space (Xm, II· II) prove that if for every continuous linear functional f, {f(xn)} is convergent to f(x) then { Xn} is convergent to x.

46

Nonned linear space structure

(ii)

For an inner product space X, prove that if for every y

E

X,

{ (xn, y)} is convergent to (x, y) and {II Xn II} is convergent to II x II then { Xn} is convergent to x. Give an example to show that in an infinite dimensional inner product space, the condition {II Xn II} being convergent to II x II is necessary. 7.

(i)

Prove that in an inner product space, (x,y) =II x 1111 y II

for x,y

if and only if for some real

8.

*0

a > 0, y = ax.

(ii)

Deduce that an inner product space is rotund.

(i)

Prove the following generalisation of the Holder inequality 2.3.3.

Given I < p < oo and

lp + lq

= I and any

x = {A. 1, A. 2, ... , An, ... } E .tp andy= {Jl 1, )l 2, ... , Jln, ... } E .tq then }.:A.k,:lk is absolutely convergent and

I_ I A.k,:lk I ~ (I. I A.k iP) lip (I. I Ilk !P) Iiq. k=l k=l Prove that equality holds for x,y * 0 if and k=l

(ii)

only if for some real

a> 0, y =ax. (iii) Deduce that for I < p < =, the .tp spaces are rotund. 9.

Given a normed linear space (X, 11·11) where the norm satisfies the parallelogram law II x+y 11 2 + II x-y 11 2 = 2(11 x 11 2 +II y 11 2) for all x, y (i)

EX,

when X is a real linear space, the mapping(., .): X x (x, y) =~(II

prove that X~

lR defined by

x+y 11 2 -II x-y 11 2) is an inner product on X and

II x II= {(X,X), (ii)

when X is a complex linear space, the mapping (., .): X x

(x, y)

X~

=~(II x+y

a:

defined by

11 2-11

x-y 11 2)+ ~ i (II x+iy 11 2-II x-iy 11 2)

is an inner product on X and II x II= {(X,X). (This is the famous Jordan-von Neumann characterisation of inner product spaces referred to in Remark 2.2.8. This expression which defines the inner product by the norm is called the polarisation formula.) I 0.

(i)

Prove that in any inner product space, Apollonius' identity holds; that is,

llz-~(x+y)ll=~.,j211 z-x 11 2 +211 z-y 11 2-11 x-y 11 2

.

(This is a formula for the length of a median of a triangle in terms of the lengths of its sides.)

§2. Example spaces

(ii)

47

Prove that in any normed linear space where the norm satisfies Apollonius' identity the norm is generated by an inner product.

I I.

(i)

In any inner product space, prove that x l_ y if and only if

(ii)

Consider an inner product space X.

II x+Ay II =II x-A.y II for all scalar A. (a)

Prove that if x l_ y then II x+y 11 2= II x 11 2+ II y 11 2

(b)

When X is a real linear space, prove that if for x,y eX

II x+y 11 2= II x 11 2+ II y 11 2 then xl_y, When X is a complex linear space, give a counter-example to show that this result does not necessarily hold. (So when X is real, orthogonality is characterised by Pythagoras'

identity.) 12.

(i)

Given any four elements x,y,z,w in a two dimensional inner product space X, prove that II x-z 1111 y-w II :::; II x-y 1111 z-w II + II y-z 1111 x-w II (Hint: Consider elements of the Euclidean plane as complex numbers and use the usual inner product for complex numbers.) (This is Ptolemy's inequality which says that in the Euclidean plane, given any quadrilateral the sum of the products of the lengths of opposite sides is greater than or equal to the product of the lengths of the diagonals.)

(ii)

Under what conditions on x,y,z,w does equality hold?

(iii) Show that the inequality holds for any four elements x,y,z,w in any inner product space X. (A normed linear space X which satisfies Ptolemy's inequality has norm generated by an inner product, see Dan Amir, Characterisations of

inner product spaces, Birkhauser, 1986, p. 54). 13.

A linear operator T on an inner product space X is an inner product isomorphism if (Tx, Ty) = (x, y) for all x,y eX. (i)

Prove that a linear operator Tis an isometric isomorphism if and only if it is an inner product isomorphism.

(ii)

Prove that if X is finite dimensional then an inner product isomorphism Tis onto, but give an example to show that in general an inner product isomorphism T need not be onto.

(iii) Prove that an isometry T from a real inner product space X into a real inner product space Y, where T(O)

=0, is an isometric isomorphism.

48

Nonned linear space structure

(iv) Show that an isometry T from a complex inner product space X into a complex inner product space Y, which has T(O) = 0, is not necessarily an isometric isomorphism. (See Remark 1.24.6). 14.

Consider a Hilbert space H. (i)

Given a closed convex set A in H and x E H \ A, prove that there exists a unique element y E A such that II x-y II = d(x, A).

(ii)

Given a closed convex set A and a compact set Kin H where An K = 0, prove that there exist elements x E A andy E K such that II x-y II = d(A, K).

(iii)

Is it true that, given closed convex sets A and B in H, there exist elements x EX andy E B such that II x-y II= d(A, B)? If it is true give a proof, if false give a counterexample.

15.

Consider 'C[-rc,rc] and the linear subspace M of odd functions on [-rc,rc]; that is f EM if f(t) = -f(-t) for all t E [-rc,rc]. (i)

Prove that the operator F where F(f)(t) =

~ (f(t) + f(-t))

is linear and continuous on ('C[-rc,rc], 11·11 2). (ii)

Deduce that M is a closed linear subspace of ('C[-rc,rc], 11·112).

(iii) Consider the linear subspace N of even functions on [-rc,rc]; that is, fEN if f(t) = f(-t) for all t E [-rc,rc]. Prove that N is a closed linear subspace of ('C [-rc,rc], 11·11 2) and 'C [-rc,rc] = M Ei1 N and M .lN. (iv) Given a normed linear space (X, II· II) with closed linear subspaces M and N such that X= M Ei1 N, is it necessarily true that (X, II· II) is complete? 16.

Given a subset A in a linear space X, a point a E A is called a passing point of A if there exist points x,y E A, x oF- y and 0 < f..< I such that a = h + ( 1-f....)y. A point a E A which is not a passing point of A is called an extreme point of A; that is, an extreme point of A is not in the interior of any line segment in A. (i)

Prove that a one-to-one linear map T from a linear space X into a linear spaceY maps passing points of a subset A of X into passing points of T(A) in Y.

(ii)

Given a subset A of a normed linear space (X, 11·11), prove that the extreme points of A lie in the boundary of A.

(iii) Prove that every normed linear space which is isometrically isomorphic to a (iv)

rotund normed linear space is also rotund. Prove that the closed unit ball in (c0 , 11·11 00 ) has no extreme points.

(v)

Show that (c0 , 11·11 00 ) and (; 1, 11·11 1) are not isometrically isomorphic. (See AMS §4 and §7.)

49

§2. Example spaces

17.

A normed linear space (X, 11·11) is said to be uniformly rotund if given E > 0, for any x,y eX, II x II, II y II:::; I there exists a O(E) > 0 such that II x-y II < E when II

k(x+y) II > I - o.

(i)

Prove that a uniformly rotund normed linear space is rotund.

(ii)

Prove that any inner product space is uniformly rotund

(iii) The following inequalities, called Clarkson's inequalities, hold in .tp spaces for 2:::; p < "", p

p

II x+y II +II x-y II :::; 2 p

p q-1

2( II x II +II y II )

p-1

p

p

(II x II +II y II ) q

:::; II x+y II +II x-y II

q

wherel + l = 1 p q ' and for I < p :::; 2 these inequalities hold in reverse. Use these inequalities to prove that the 1P-spaces where I< p < co are uniformly rotund. (iv) Prove that in any uniformly rotund normed linear space (X, 11·11), any sequence { x 0 } satisfying II x 0 II~ I, II x0 +xm II ~ 2 is a Cauchy sequence. (v)

Prove that if M is a proper closed linear subspace of a uniformly rotund Banach space (X, 11·11), then for any x eM there exists a unique y eM such that II x-y II = d(x, M).

18.

Consider a proper closed linear subspace M of a Hilbert space H. Prove that for every [x] E HIM there exists a unique element

(i)

x0 e [x] such that II x0 II= II [x] II= inf{ll x+m II: me M}. (ii)

19.

Prove that HIM is a Hilbert space with inner product defined by ([x], [y]) = (xo, Yo) .

Consider an incomplete inner product space X with norm 11-11 generated by the inner product, and (X, 11·11) the completion of (X, 11·11); (see AMS §3). Prove that

X is an inner product space with inner product(. , .) defined by ([x], [y])

= lim

n-7~

(X 0 ,

Yn)

where { x0 } and { Yn} are Cauchy sequences in X satisfying { x0 } e [x] and { Yn}

E

[y], and that ([x], [y]) = (x, y) for all x,y eX.

50

Nonned linear space structure

§3. ORTHONORMAL SETS IN INNER PRODUCT SPACES

The notion of orthogonality plays an important role in the development of the theory of inner product spaces. We now investigate the special theory developed around the idea of mutually orthogonal sets of elements. 3.1 Definition. Given an inner product space X, an orthonormal set of elements {eul is a set where (ewe~)=

and

for all a

0

oF-~

for all a.

II eu II= I

It is clear that any orthonormal set is linearly independent. 3.2 Examples. (i) In Hilbert sequence space (.t 2 , 11·11 2), the set {en} where en

=o

{0, ... , 0, I, 0, ... }

fornEN

nth place is an orthonormal set. (ii) In ('C [-1t,1t], 11·11 2 ) the space of real continuous functions on [-1t,1t], with norm llfll 2 =

"

Jf 2 (t)dt, -Tr

the set of functions

I I I I {2rt , -fit cos t, -fit cos 2t, ... , -fit cos nt, ... I

.

I

-fit sm t, -fit

. I . sin 2t, ... , -fit sin nt, ...

is an orthonormal set.

D

Orthonormal sets have great advantages. When we have a linear subspace M of an inner product space X spanned by a sequence {xnl it is often useful to have an orthonormal sequence {en} which also spans M. The following is an explicit process for generating such an orthonormal sequence. 3.3 Gram-Schmidt orthogonalisation. Given a sequence {xnl which is also a linearly independent set in an inner product space X, then there exists an orthonormal sequence {en} such that for each n E N

sp { ek : k E {I ,2, ... , n}} = sp{ xk : k E {I ,2, ... , n} } .

§3. Orthonormal sets

Proof and construction. Given x 1

51

* 0, we define sequences IYnl and {enl as follows: YI

y1 =x1

e 1 = lT"Y;;T

n-1

Yn

=Xn- L

(xn, ek) ek

k=l It is clear from the construction that for any n E N sp{ ek: k

E {

1,2, ... , n}}

*

= sp{ xk: k E { 1,2, ... , n}}

.

For each n EN, Yn 0 since {x 1, ... , xnl is a linearly independent set. computation shows that {en} is an orthonormal set.

Direct

0

3.4 Example. In ('C[-1,1], 11·11 2) the space of real continuous functions on [-1,1] with I

norm II f 11 2 =

Jf 2(t)dt

, the sequence of polynomials

-I

{ 1, t, t2, ... , tn, ... ) is a linearly independent set. The Gram-Schmidt orthogonalisation of this set yields the orthonormal set

(I fit fi(~t2- .!.) fi(~t3- ~t) Y2'Y2 'Y3 2 2 'Y2 2 2 the Legendre polynomials defined in general as p,-Pn(t) where

Pn(t)

=- 1- ~ 2nn! dtn

(t2-l)n

for n

E { 0,

1, 2, ... }.

D

3.5 Definition. Given an inner product space X with orthonormal set {eul, the set {(x, eu)} is called the set of orthogonal coefficients with respect to {eul. 3.6 Example. For any continuous function f on [-7t,1t], the Fourier coefficients 1t

. ~ Jf(t)

1t

dt),

.~

Jf(t) cos nt dt),

1t

.~

Jf(t) sin nt dt),

for all n EN, 27t -1t '11t -1t '11t -1t are orthogonal coefficients with respect to the orthonormal set given in Example 3.2(ii). 'I

D

We show the significance of the orthogonal coefficients in representing an element with respect to a finite orthonormal set.

52

Nonned linear space structure

3.7 Lemma. Consider a finite orthonormal set {e 1, e 2, ••• , enl in an inner product space X. Then {el, ez, ... , enl is a basis for Mn = sp(e 1, e 2, ... , enl and (i) every x E Mn has unique representation in the form n

x = L (x, ek) ek> k=l

n

(ii) for every x EX\ Mn,

L

(x, ek) ek is the unique best approximating point to x in Mw

k=l

Proof. For x = a 1e 1 + a 2e 2 + ... + anen

(i)

for all k E { I , 2, . . . , n}. If x

=0

then ak

= 0 for all

k E (I, 2, ... , n} and so {e 1, e 2 ,

.•. ,

enl is linearly

independent and is a basis for Mw (ii)

Now for each j E {I, 2, ... , n}, n

(x-

L

n

(x, ek) ek, ej)

= (x, ej)- L

k~

n

So x -

L

(x, ek)(ek, ej)

= (x, ej)- (x, ej) = 0.

~I

(x, ek) ek is orthogonal to Mn and by Theore~ 2.2.15,

k=l unique best approximating point to x in Mn.

n

L

(x, ek) ek is the

k=l

D

3.8 Remark. This lemma has immediate application to the problem of finding points of best approximation in finite dimensional linear subspaces discussed in Theorem 2.1.14. Given a finite dimensional linear subspace Xn in an inner product space X and x0 EX \Xn, we can compute the unique point in Xn of best approximation to x0 as follows. Given a basis {x 1, x2, ... , xnl for Xn we generate an orthonormal basis {e 1, e 2, ... , enl for Xn by Gram-Schmidt orthogonalisation. Then from Lemma 3.7, n

L

(x 0 , ek) ek is the unique point in Xn of best approximation to x0 . Of course

k=l n

L

(x 0 , ek) ek is best approximating with respect to the inner product norm and not k=l necessarily with respect to any other norm on X. D In moving from a finite to an arbitrarily large orthonormal set we need the following preliminary property.

3.9 Lemma. Given an orthonormal set {eal in an inner product space X, for any

x E X, the set of nonzero orthogonal coefficients of x is countable; that is, the set S ={a: (x, ea)

'* 0} is countable.

§3. Orthononnal sets

53

Proof. Given £ > 0, consider the set Se = {a: I (x, ea) I>£}. Suppose that { 1, 2, ... , n}

E

Se. Then from Lemma 3.7(ii),

n

II x 11 2 ~

L

I (x, ek) 12 > n e2 .

k=l

< II X 11 2 n- - - 2 -

Therefore

£

which implies that Se is finite. Buts= u {slln : n EN}, so s is countable.

D

We now extend Lemma 3.7(i) for elements in the closed linear space of an orthonormal set in an infinite dimensional inner product space. 3.10 Theorem. Given an orthonormal set {ea} in an inner product space X, consider M = sp {eal· For every x EM, L(X, ea) ea is well defined over all sequential orderings of

the setS

= {a : (x, ea) '* 0} and x =I (x, ea) ea.

Proof. Now x EM can be expressed as an element of sp{ ea} or a cluster point of such elements.

So given£> 0 there exists a finite set {e 1, e2, ... , enl in leal and a set of

scalars {1.. 1, 1..2, ... , An} such that n

L

Ak ek II < £. k=l But then by Lemma 3.7(ii) applied to sp{e 1, e2, ... , en}, II x -

n

II x-

L

n

(x, ek)ek II<:; II x-

k=l Consider M 0

L

Ak ek II.

k=l

=sp { ea : (x, ea) '* 0}.

n

Then

L

(x, ek)ek

E

M 0 and so we deduce that

k=l xEM 0 . From Lemma 3.9, the setS

={a: (x, ea) '* 0} is countable.

ordering {an} of S. Then given £ > 0 there exists an n

E

Consider a sequential

N and a set of scalars

{ll1, fl2, ... , !ln} such that n

II x -

L flkeak II < £. k=l

Again by Lemma 3.7(ii) applied to sp { ea 1, ea 2, ... , ean} n

II x-

n

L (x, eak) eak II .:; II bl

which implies that x-

x-

L flkeak II bl

L (x, eak) eak converges to x. k=l

Since this holds for any sequential

54

Normed linear space structure

ordering {an} of S, we have that I (x, ea) ea is well defined over all sequential orderings of S and we can write

0 To extend Lemma 3.7(ii) to obtain best approximating elements in the closed linear span of an orthonormal set in an infinite dimensional inner product space, we need completeness of the space. 3.11 Theorem . Given an orthonormal set {eal in a Hilbert space H, consider M

=sp {eal.

For every x E H \ M, I(x, ea) ea is well defined over all sequential

orderings of the setS= {a: (x, ea) oF- 0} and I(x, ea) ea is the unique best approximating point to x in M.

Proof.

From Theorem 2.2.19 we have that there exists a y

E

M, the unique best

approximating point to x in M. But then by Theorem 3.10 y =I (y, ea) ea. However, from Theorem 2.2.15, x-y is orthogonal toM, so (x-y, ea) = 0 for all ea; that is,

(y, ea) = (x, ea) for all a.

So we conclude that

0 Theorems 3.10 and 3.11 enable us to make important existence statements about series representations for elements in an inner product space.

3.12 Corollary. Given an orthonormal set {ea}in an inner product space X, x EX has

representation in the form I(x, ea) ea if and only ifx

E

sp {eal· In any case if X is

complete then I(x, ea) ea always defines an element of X. To show that an element has such a series representation we need to have methods to determine whether the element belongs to the closed linear span of an orthonormal set. The following fundamental properties of orthogonal coefficients are useful in this regard. 3.13 Bessel's inequality. (i) Given a finite orthonormal set {e 1, e2, ... , enl in an inner product space X, for

any x EX n

L k=l

I (x, ek) 12 ~ II x 11 2.

§3. Orthonormal sets

55

(ii) Given any orthonormal set {eal in an inner product space X,jor any x EX,

eis well defined over all sequential orderings of the set

L I (x, ea)

S = {a : (x, ea) *- 0} and L I (x, ea) 12 <:; II x 11 2 .

Proof. n (i)

n

L

0 <:; II x-

k=l

n

n

L

=II x 11 2 -

I (x, ek) 12 -

k=l

L

n (e, ek) ek, x -

k=l I (x, ej) 12 +

j=l

n

n

n

L L

L

(x, ej) ej)

j=l (x, ek) (x, e}(ek, ej)

k=l j=l

L I (x, ek) 12 .

=II x 11 2-

(ii)

L

(x, ek) ek 11 2 = (x-

k=l Given x EX, we have from Lemma 3.9 that the set S = {a : (x, ea)

countable, so

L I (x, ea)

12

oF-

0} is

has at most a countable number of nonzero terms. Consider

any sequential ordering {an} of S. Then from (i) we have for any n EN, n

L

But

L

I (x, eak) 12 <:;II x 11 2 so L I (x, ean) 12 <:; II x 11 2 . k=l n=l I (x, ean) 12 is absolutely convergent so the sum L I (x, ean) 12 is independent of the

particular sequential ordering {an} of S; that is, the sum L I (x, ea) 12 is uniquely defined by

L

e. We conclude that,

I (x, ean)

n=l

D 3.14 Definition. Consider an orthonormal set {eal in an inner product space X. We say that x E X satisfies Parseval's formula if Bessel's inequality is an equality; that is, if

L I (x, ea) 12 = II x 11 2 . By this we mean that

L

I (x, ean) 12 = II x 11 2 for some particular sequential ordering {an}

n=l of the setS = {a : (x, ea)

L

that the sum

I

(x, ea)

12

7'-

0}. Again the absolute convergence of L I (x, ean) 12 implies

is uniquely defined by

n=l

L

I (x, ean)

12 .

n=l

We now show that an element belongs to the closed linear span of an orthonormal set if and only if it satisfies Parseval's formula. 3.15 Theorem. Given an orthonormal set {eal in an inner product space X, x E sp{ X a}

if and only if L I (x, ea) 12 =II x 11 2 . Proof. If x

E

sp{ xal then by Theorem 3.10, for any given sequential ordering {an} of

the setS= {a: (x, ea) *- 0}, x=

L (x, ean)ean· n=l

56

Nonned linear space structure

We have then II x 11 2 = (

L (x, eun)eun• L n=l

(x, eum)eum)

m=l

which by the joint continuity of the inner product implies that II

x 11 2 =

L

I (x, eun) 12

n=l

so we can write llxll 2 =II(x,eu)l 2• Conversely, suppose that Parseval's formula holds. Then for a given particular sequential ordering {an} of S, given e > 0 there exists an v

E

N such that

n

II

x 112 =

L

for all n > v.

I (x, euk) 12 < e

k=l

As in the proof of Bessel's inequality 3.13(i) we have n

n

O~llx-L. (x,euk)eukl~~llxi~-L l(x,euk)I 2 k=l

<E

whenn>V.

k=l

0

So we conclude that x E sp {eu}.

We now examine the particular case of the representation of a Riemann integrable function by its Fourier series. 3.16 Example. Consider the real linear space

~ [-1t,1t]

of Riemann integrable functions

on [-1t,1t]. We noted in Example 2.2.12(ii) that Jl

(f, g) =

Jfg(t) dt -Jl

is a positive Hermitian form but is not positive definite and Jl

fret) dt -Jl

is only a seminorm for ~ [-1t,1t]. If we consider the quotient space space defined by ([f], [gl) = (f, g)

~ [-1t,1t]/ker

for any f

E

p2 then we have an inner product on this

[f] and g E [g].

We will proceed by dealing with the linear space

~

[-1t,1t] using our theory for inner

product spaces, but keeping in mind that the elements of ~[-1t,1t] are only representatives of the equivalence classes of the quotient space ~ [-1t,1t]/ ker p2 and our results have to be interpreted accordingly. We show that any Riemann integrable function f on [-1t,1t] can be represented by its Fourier series with respect to the orthonormal set

57

§3. Orthonormal sets

I I I . {{2rt' {itcos nt, {itsm nt : n

w}

E a

;

that is, f(x) =

2~ _[ f(t)dt + ~ ~1

{(_[ f(t)

cos nt dtJ cos nx + (_[ f(t) sin nt dtJ sin nx}

by which we mean that the Fourier series is convergent to fin the inner product norm; that is, the Fourier series is mean square convergent to f. From the theory of integration, it is clear that iff is Riemann integrable on [-1t,1t] then given f.> 0 there exists a step function

f

I f--<1> 12 (t) dt
.

-11

So

~

[-1t,1t] is contained in the mean square closed linear span of the step functions on

[-1t,1t].

We show that an elementary step function <jl(t) = I

defined by

a :::; t :::; a+b where b ;::

0}

otherwise

=0

satisfies Parseval's formula. It will follow from Theorem 3.15 that the linear space of step functions is contained in the mean square closed linear span of the orthonormal set and so ~ [-1t,1t]

is contained in the mean square closed linear span of the orthonormal set. So we

will conclude that any Riemann integrable function f on [-1t,1t] has mean square representation by its Fourier series. Now 1l

p; (<jl) =

f <j>2(t) dt =b. -1t

But

~I I (<jl, en) 12 = 21

1t (_[ <jl(t) dtJ +

=

~ ~J(_[ <jl(t) cos nt dtJ + (_[ <jl(t) sin nt dtJ}

~~ + ~ ~1 {(J:os nt dtJ + (J:in nt dtJ}

= ~~ + ~

I n=l

~2

{(sin n(a+b)- sin na)2 +(cos n(a+b)- cos na)2} 2 1t

using the fact that

L n=l

n2 =

1t2

6 .

cos nb

I ~

n=l

b2 = 21t +

1t

3-

n2 I.

~I

cos nb n2

58

Nonned linear space structure

Now by Bessel's inequality,

I

n=l

l(<J>,en)1 2 ~p~(<j>)

2 nb)>o b -(b 21t + ~-~~cos 3 1t ....... 2 .

so

n

n=l

~ cos nt . .f b . . . 1 S mce ......, -2 -ls um orm y convergent on [-1t,1t] y We1erstrass' M-test, 1t follows that

n

n=l

t2 1t 2 ~ cos nt) . . t- ( 21t + 3--2IS continUOUS on [-1t,1t], (see AMS §3).

n n=l ....

n

But also from the uniform convergence of

I n=l

J (f_ + ~ _ ~ f. {t _

-n

21t

3

1t n=l

cos2 nt it follows, (see AMS §7), that n

~

~ f.

cos nt)} dt=f-(_f_ + t _ sin nt) n2 2 61t 3 1t n=l n3 2 I 2 2 2 = 1t - 3 1t - 3 1t = 0 ·

]n:-n:

Since the integrand is continuous and positive we conclude that b=(b 2 21t

+~-~f. 3

1t n=l

cos nb)· n2

'

that is,

, en) I2 = p2 (<j>). ......, 2 n=l

Of course, in classical Fourier analysis interest is centred on uniform and pointwise convergence and not just on mean square convergence. 0 3.17 Definition. An orthonormal set {ea} in an inner product space X is said to be

maximal in X if it is not a proper subset of any other orthonormal set in X; that is, if it is maximal in the family of orthonormal subsets of X partially ordered by set inclusion. So {eal is maximal if and only if there does not exist an e EX, e all

* 0 such that (eao e)= 0 for

a.

3.18 Examples. (i) In Hilbert sequence space (1 2, 11·11 2 ), consider the orthonormal set { e 1, e2o ... , en, . : .} where en

If e

={A

1,

A2 ,

... ,

= {0, ... , 0, I, 0, ... }

for all n EN.

nth place An, ... } E 1 2 satisfies (en, e)= 0 for all n EN then An= 0 for all

n E N and e = 0. So {en} is maximal.

59

§3. Orthonormal sets

(ii)

In ('C [-n:,n:], 11·11 2) the space of real continuous functions on [-n:,n:], with norm

:

"

II f 112 = Lf 2(t)dt 'the orthonormal set of functions r~t n

E

N} is not maximal

1t

since the function si,n t satisfies 'I n;

Jcos nt sin t dt = 0

for all n e N.

-Jt

D

The following results are similar to those established for Hamel bases in linear spaces. Using the Axiom of Choice in the form of Zorn's Lemma, Appendix A. I, we are able to establish the existence of maximal orthonormal sets. 3.19 Theorem. In any nontrivial inner product space X there exists a maximal

orthonormal set. Further, given an orthonormal set in X there exists a maximal orthonormal set containing it as a subset.

Proof. Given x eX, x

* 0, then {11 ~ 11} is an orthonormal set in X. So X contains an

orthonormal set. Consider any orthonormal set {eu} in X and !J' the family of all orthonormal sets containing {eul as a subset and partially ordered by set inclusion. It is clear that the union of any totally ordered subfamily is an orthonormal set and is an upper bound for the totally ordered subfamily. So by Zorn's Lemma, we deduce that !J' has a maximal orthonormal set. D The following cardinality result for maximal orthonormal sets is important for the definition of the orthogonal dimension of an inner product space. 3.20 Theorem. In any given inner product space X, all maximal orthonormal sets are

numerically equivalent.

Proof. Suppose there exists a finite maximal orthonormal set {e 1, e2, ... , en} and consider Mn

= sp {e 1, e 2, •.. , en}. Suppose that Mn * X, consider x EX \ Mn and

n

L (x, ek) ek . k=l Now by Lemma 3.7(ii), we have that x- x0 write x0 =

* 0 and x- x0 is orthogonal to Mn.

But this contradicts the maximality of {e 1, e2 , dimensional.

... ,

en} so X = Mn and X is finite

As an orthonormal set is linearly independent, {e 1, e2> ... , en} is a basis

for X and so every maximal orthonormal set in X has n elements.

60

Nonned linear space structure

Suppose there exist two infinite maximal orthonormal sets B 1 = {eal and B 2 = {f~) in X. For each a, we have from Lemma 3.9 that the set B 2(ea)

={~ :(f~,ea) ;to 0}

is countable. But also for each

~.

we have that

(f~,

ea) ;to 0 for some a, otherwise, {eal

would not be maximal. Therefore B2 C U B 2(ea)· a So by Proposition, Appendix A.3.4 there exists a one-to-one mapping of B 2 into B 1. By symmetry of argument and the Schroeder-Bernstein Theorem, Appendix A.2.2 we have our result. D This result enables us to make the following definition. 3.21 Definition. The orthogonal dimension of an inner product space X is the cardinal number of a maximal orthonormal set in X. 3.22 Remark. For a finite dimensional inner product space a maximal orthonormal set is also a basis for the space. For an infinite dimensional inner product space, since a maximal orthonormal set is a linearly independent set, the space has infinite Hamel dimension, (see Appendix A.3). However, even for a Hilbert space with countable orthogonal dimension, a Hamel basis has cardinality of the continuum, (see P.R. Halmos, A Hilbert space problem book, D. van Nostrand, 1967, p. 170). 3.23 Definition. An orthonormal set {eal in an inner product space X is said to be an

orthonormal basis for X if for every x EX, we have that x = L.(x, ea) ea; that is, every x EX can be represented in the inner product norm by its orthogonal expansion with respect to {eal· 3.24 Remark. Given an orthonormal set {eal in an inner product space X, it is clear

0

from Corollary 3.12 that {eal is an orthonormal basis for X if and only if X = sp {eal·

3.25 Example. In Example 3.16 we showed that every Riemann integrable function on [-7t,7t]

has mean square representation by its Fourier series. That is, the orthonormal set I I I . {{21[ , {Jtcos nt, {Jtsm nt : n E N

is an orthonormal basis for this space.

w}

D

The following result gives the relation between a maximal orthonormal set and an orthonormal basis.

§3.

Orthonormal sets

61

3.26 Theorem. In an inner product space X, an orthonormal basis is a maximal orthononnal set and if X is complete then a maximal orthonormal set is an orthonormal

basis.

Proof. If an orthonormal set {ea} is not maximal then there exists an e e X, e

*

* 0 such

that (ea, e)= 0 for all a. and then I,(e, ea) ea = 0 so e I,(e, ea) ea. For a complete inner product space X with orthonormal set {ea}, we have from Corollary 3.12 that, for any x eX, the sum I,(x, ea) ea defines an element of X and x - I,(x, ea) ea is orthogonal to e~ for all ~If {ea} is maximal then x - I,(x, ea) ea = 0; that is, x = I,(x, ea) ea. We conclude that {ea} is an orthonormal basis for X.

D

Most of the common inner product spaces are separable and separability is related to countable orthogonal dimension. 3.27 Theorem. An inner product space X is separable

if and only if it has a countable

orthononnal basis.

Proof. Suppose that X is separable. Then there exists a countable set {xnl dense in X. First we generate a linearly independent set IYnl such that sp(ynl = sp(xnl= X.

*

Denote by y 1 the element x1 0 and by y2 the element x first in the sequence {xnl such that y 2 e sp(yJl nl by Yk the element xnk first in the sequence {Xn} such that Yk e sp {y 1, y2 , Then for each k e :f:ll, sp { Yi : i E { 1, 2, ... , k} } = sp { Xj : i

E { 1,

••. ,

Yk-l}.

2, ... , nk} }

and so We now apply Gram-Schmidt orthogonalisation to the sequence IYnl and obtain an orthonormal sequence {en} such that for each n e :f:ll, sp { ek : k e { 1, 2, ... , n} } = sp { Yk : k e { 1, 2, ... , n} } and By Remark 3.24, {enl is an orthonormal basis for X. Conversely, if {enl is a countable orthonormal basis for X then X= sp {enl· Now the set of all rational linear combinations of elements of {en} is countable and is clearly dense in X. So we conclude that X is separable. D

62

Nonned linear space structure

3.28 Remark.

For a separable inner product space X its countable orthonormal--basis

{en} is a Schauder basis. To show this we need only verify that an element x EX has a unique representation in terms of its orthonormal basis. But if x =

I,

Ak ek then since

k=l

{en} is an orthonormal set and the inner product is jointly continuous we have that

O

(x,en)=AnforallnEN.

We notice from Bessel's inequality 3.13(ii) that in any inner product space X, given an orthonormal sequence {en), for any x E X, we have that

L. I (x, en) 12

is convergent.

Completeness is important in specifying those orthogonal sequence expansions which belong to the space. 3.29 Theorem. Consider an orthonormal sequence {en} in a Hilbert space H and a

sequence of scalars {An}. Then this case for x =

L. An en is convergent if and only if L. I An

12

< oo and in

L. An en we have An = (x, en) for all n E N.

Proof. For m > n, m

m

L Akek 11 2 = I,

I Ak 12 k=n k=n soL. Anen is Cauchy if and only if L.l An 12 is Cauchy. But the space is complete so II

L. An en

is convergent if and only if L. I An

12

is convergent. It follows from the fact that

{en} is an orthonormal sequence and the inner product is jointly continuous that

0

An= (x, en) for all n EN.

This result has application for Fourier analysis in classical Hilbert space

;z:; 2 [-1t,1t] of functions f where f 2 is Lebesgue integrable on [-1t,1t], factored over ker p 2 where P2(f) =

f f2 df.l . [-7r,7r]

We deduce the following as a corollary. 3.30 The Riesz-Fischer Theorem. Given sequences of scalars {Un} and {Bn} such that L. I Un

12

< oo and L. I Bn

2

1

< =, then

the series

-~ + -~I, (Un cos nt + B sin nt) '11t '121t converges to some f E ;z:; 2 [-1t,1t] and a, Un and Bn are the Fourier coefficients off with respect to the orthonormal sequence

§3. Orthonormal sets

cos nt {_I_ ...J2rt , '-fit

,

63

sin nt}

'-fit .

We now seek to classify Hilbert spaces on the basis of orthogonal dimension. This generalises a similar classification of linear spaces by Hamel dimension. 3.31 Theorem. Hilbert spaces over the same scalar field and of the same orthogonal

dimension are isometrically isomorphic. Proof. Consider Hilbert spaces H 1 and Hz over the same scalar field with orthonormal bases {ea} and {fa} of the same cardinality. Consider a one-to-one mapping T of {ea} onto {fa}. Given x e H, we have that x = I(e, ea) ea. Consider the extension ofT as a mapping of H 1 and Hz defined by Tx = I(e, ea) T(ea)· We show that such an extension is wen defined. By Lemma 3.9, I(e, ea) has only a countable number of n.onzero terms so I(e, ea) T(ea) has only a countable number of nonzero terms. By Bessel's inequality 3. 13(ii), I I (e, ea) lz < oo so by Theorem 3.29, I(e, ea) T(ea) defines an element in Hz. Now for each x e H, II Tx liz = I I (e, ea) lz = II x liz by Theorem 3.15, so T is norm preserving. But from the linearity and continuity properties of the inner product we deduce that T is linear. So T is an isometric isomorphism of H 1 into Hz. For y e Hz,

I I (y, fa) lzll y liZ< oo

by Theorem 3.15, so x = I(y, fa) T- 1(fa) e H 1 Tx =I(y, fa) fa= y, and which shows that T is onto.

0

This general result has the fo11owing implications. 3.32 Corollary. (i) Every real (complex) n-dimensional inner product space is isometrically isomorphic to (IR 0 , ll·llz) (([ 0 , ll·llz)). (ii) Every real (complex) separable Hilbert space is isometrically isomorphic to real (complex) Hilbert sequence space (.tz, II· liz).

64

Nonned linear space structure

3.33 EXERCISES I.

Consider the linear space 'C[-n:,n:] of real continuous functions on [-n:,n:] with norm 1f

II f 11 2 =

ff 2(t)dt -lf

and the linear subspace M spanned by {sin t, sin 2t}. (i)

For the function f0 on [-n:,n:] where f0 (t) = t, find the unique best approximating function g0 in M.

(ii)

Show that g0 is also best approximating with respect to the norm II f II~= max {I f(t) I : t E [-n:,n:] } but is not unique.

2.

Consider the sequence of polynomials, {I, t, t2, ... , tn, ... } on [-1,1]. Prove that the orthonormal sequence of polynomials which results from GramSchmidt orthogonalisation of this sequence in the space ('C [-1,1], ll·ll2) has the form {eo. el, e2, ... , en, ... }

= sp{e0 , e 1, ... , enl

where sp{ I, t, ... , tn}

for n

E

{0, I, 2, 3, ... } and

Determine the form of the first three Legendre polynomials, P0, P 1 and P 2 .

3.

Consider the linear space 'C [-1,1] of continuous real functions on [-1,1] with norm I

II f 11 2 =

ff 2(t)dt -I

and the linear subspace M spanned by the Legendre polynomials

{if ..fl t, fs (3t2-J)} (i)

.

For the function f0 on M where f0(t) =I t I find the unique best approximating function g0 in M.

(ii)

Is g0 also best approximating with respect to the norm II f II~= max {I f(t) I : t E [-1,1]} and if so is it unique?

65

§3. Orthonormal sets

4.

Consider the linear space t:[-1t,1t) of real continuous functions on [-1t,1t) with norm

f" f

II f 11 2 =

2 (t)dt

.

-1J:

(i)

(ii)

Prove that the linear subspace M of even functions is a closed linear subspace of (t: [-1t,1t], 11·11). I I Prove that the set {. "'=' . '-'cos nt : n E N } is an orthonormal basis for '121t '11t M.

(iii)

For the function f0 on [-1t,1t) where f0 (t) = et find the unique best

(iv)

approximating element g0 in M. Although (t: [-1t,1t], 11·11 2) is not complete verify that the best even function g0 approximation to f0 found by computation in (iii) is actually in t: [-1t,1t).

5.

(i)

The Chebyshev polynomials {Tnl on [-I ,I] are defined as follows: T0 (t)=I,

T 1(t)=t

and

Tn+l(t)=2tTn(t)-Tn-l(t)

forallne:N.

Prove that the best approximation to f0 (t) = t3 from sp( I, t, t 2 } in (t:[-I,I], 11·11~) is tht! function g (t) = t3 - I T (t).

4

0

(ii)

3

Find a, b, c e IR to minimise the integral l

J I t3 -

a - bt - ct2 I dt .

-l

6.

(i)

Given an inner product space X with an orthonormal sequence {en}, prove that for any x EX, lim (x, en) = 0.

(ii)

Consider the linear space t:[-1t,1t) of real continuous functions on

n->~

[-1t,1t) with inner product 1t

(f, g) =

f fg(t) dt. -Jt

Deduce for any f E t:[-1t,1t), 1t

lim

1t

J f(t) sin nt dt = 0

and lim

n~-Jt

J f(t) cos nt dt = 0

n~-Jt

(This is a special case of the Riemann-Lebesgue Lemma). (iii)

Does this result extend to the linear space

~ [-1t,1t)

functions on [-1t,1t) with positive hermitian form 1t

(f, g) =

f fg(t) dt ? -Jt

of Riemann integrable

66

Nonned linear space structure

(iv)

For f0 E 'C [-TC,TC] we are given that 1t

Jf0(t) sin nt dt = 0

for all n eN.

-1t

Prove that f0 is an even function. 7.

In Hilbert sequence space (.t 2, 11·11 2 ) consider the linear subspace E == sp{x 0 , E0 } where x0 == { I,

k, ... , ~ , ... } and the orthonormal set {e2n} where

e 2 n == { 0, ... , 0, I, 0, ... }. Show that the orthogonal series for x, 2n th place L.(x 0 , e2n)e 2n is not convergent in E. Explain why this is so. 8.

Consider the real linear space '4't[-rc,rc] with semi-norm 1f

ff

Pif) =

2 (t)dt

.

-lf

Prove that 'C[-TC,TC] is mean square dense in '4't[-TC,TC] I ker p2.

9.

(i)

Prove that if an inner product space has a countable Hamel basis then it has a countable orthonormal basis.

(ii)

prove that if an inner produce space has a Schauder basis then it has a countable orthonormal basis.

I 0.

Consider an orthonormal sequence {enl in a Hilbert space H. Prove that {en} is maximal if and only if (x, y) =

L

(x, en) (y, en)

for all x, y E H.

k=l

I I.

Consider an inner product space X with an orthonormal set {eo:}. (i)

Prove that if Parseval's relation holds for each x EX then {eo:} is

(ii)

Prove that if X is complete then the converse holds.

maximal.

12.

Consider the linear space of real functions f on IR where for each f the set { x E IR : f(x) oF- 0} is countable and L. I f(x) 12 < oo, with inner product (f, g) =

L. f(x) g(x) .

Prove that this space is a nonseparable Hilbert space.

II.

SPACES OF CONTINUOUS LINEAR MAPPINGS

The richness of linear space structure enables us to construct new families of normed linear spaces from old through the linear mappings between them. Generating and relating spaces in this way is an activity which distinguishes the analysis of normed linear spaces from that of metric spaces. But further it is this process which is the basis of linear operator theory with its important areas of application. §4. NORMING MAPPINGS AND FORMING DUALS AND OPERATOR ALGEBRAS

The natural association of a norm with the linear space of continuous linear mappings between normed linear spaces is the key factor enabling us to consider new normed linear spaces. The newly formed spaces inherit properties from the original spaces. Of particular interest are dual spaces and operator algebras which have a structural richness not available for general normed linear spaces.

4.1 Definitions. (i) Given linear spaces X andY over the same scalar field, the set J::(X,Y) of linear mappings of X into Y is a linear space under pointwise definition of addition and multiplication by a scalar; that is, for T,S E J::(X,Y), T+S: X -7 Y is defined by (T+S)(x) =Tx + Sx aT: X -7 Y is defined by (aT)(x) =aTx. and There is no difficulty in verifying that T +Sand aTE J::(X,Y) and that the linear space properties hold. (ii) Given normed linear spaces (X, 11·11) and (Y, 11·11') over the same scalar field, the set ~(X,Y)

of continuous linear mappings of X into Y is a linear subspace of J::(X,Y).

Closure under the linear operations follows from the closure of continuity under these operations. These spaces are in fact generalisations of the finite dimensional case.

68

§4. Duals and operator algebras

4.2 Example. When X= !Rn (or
Proof. From Corollary 2.1.10 it follows that in this case ;r; (X, Y) =

~(X, Y).

Given a basis { e 1,e 2, ... , en} of X and a basis {f 1,f2, ... , fm} of Y and a linear mapping T: X -7 Y then for each k E { I ,2, ... , n} m

Tek =

L

Ujk fj j=l and [<X_ik] is a uniquely determined m x n matrix of scalars. Since each x EX has unique representation n

x=L.A.kek k=l and T is linear n

Tx =

m

n

L A.kTek = L L (ujk A.k) fj. k=l

j=l k=l

Conversely, given an m x n matrix [ Ujk] of scalars, the mapping T: X -7 Y defined by m

Tx =

n

L L (Ujk A.k) fj j=l k=l

is clearly linear. Moreover, it is easily verified that this one-to-one correspondence between

~(X,Y)

and

Mmxn, the linear space of m x n matrices of corresponding scalars, is also a linear mapping.

D However,

~(X,Y)

derives norm structure from the norms on X andY.

4.3 Theorem. Given normed linear spaces (X, II· II) and (Y, 11·11'),

~(X,Y)

is a normed

linear space with nann II T II = sup { II Tx II' : II x II ::; I }.

Proof. Since T is continuous there exists an M > 0 such that so

II Tx II' ::; M II x II

for all x EX

II Tx II' ::; M

for all II x II ::; I

and therefore the norm 11·11 on ~(X,Y) is well defined. We check those norm properties which do not follow directly from the definition. When II T II = 0 then II Tx II' = 0 for all II x II ::; I and so II Tx II' = 0 for all x EX. From the norm properties of 11·11' we see that Tx = 0 for all x EX and so T = 0. For the triangle inequality we have for S,T E ~(X,Y)

69

Spaces of continuous linear mappings

II S+T II =sup{ II (S+T)x II' : II x II ::; I } ::; sup{ II Sx II'+ II Tx II' : II x II::; I} ::; sup{ II Sx II' : II x II::; I} +sup{ II Tx II' : II x II::; I} =II

s II +liT II.

D

(X, II ·II)

(Y, II· II')

Figure 3. The generation of ;r;(X,Y) and n(X,Y) from (X, 11·11) and (Y, 11·11'). There is an alternative expression for the norm of a continuous linear mapping which is useful. 4.4 Corollary. ForTE n(X,Y), IITII=inf{M:IITxll':s;MIIxllfarall xEX}.

T

T(Bx[O; I])

Bx[O; I]

By[O; M]

Figure 4. Visualisation of the alternative expression for II T II.

70

§4. Duals and operator algebras

Proof. Since II T II= sup{ II Tx II': II x II:<>: I} II T II ~ II Tx II' for all II x II :<>: I then so II T II~ inf{M: II Tx II':<>: M for all II x II:<>: I}. But clearly, inf{M: II Tx II':<>: M for all II x II:<>: I} ~sup{ II Tx II': II x II:<>: I} =II T II.

D

From Corollary 4.4 we have the following important inequality. 4.5 Corollary. ForTE ncx,Y), II Tx II':<>: II T 1111 X II for all X EX. 4.6 Remark.

For a discontinuous linear mapping T from (X, II· II) into (Y, 11·11'), the set

T(B[O; I]) is unbounded and so II T II is not defined. Unlike the inequality in Theorem 1.24.1(i), the inequality in Corollary 4.5 cannot be used to determine continuity. D It is important to see that the normed linear space (n (X,Y),II·II) inherits

completeness from the range space (Y, 11·11'). 4.7 Theorem. Given normed linear spaces (X, 11·11) and (Y, 11·11'),

if (Y, 11·11') is complete

then so also is (n(X,Y), 11·11). Proof. Consider a Cauchy sequence {T n} in (n(X,Y), II· II); that is, given£> 0 there exists a v E N such that 11Tn-Tm11<£ forallm,n>V. So for any given x EX, II Tnx-Tmx II':<>: II Tn-Tm 1111 x II < e II x II for all m,n > v ; that is, {T nX} is a Cauchy sequence in (Y, 11·11'). Since (Y, II· II') is complete there exists an element in Y, which we denote by Tx, such that {T nX} converges to Tx. Consider the mapping T: X -7 Y defined in this way as the pointwise limit of {T n}. We need to show that

(i) T E n(X,Y) and (ii) {Tn} converges toT in (n(X,Y), 11·11).

(i) Now Tis linear: Given x 1, x2 EX and since T n is linear for all n E N, II T(x 2 +x 2 ) - Tx 1-Tx 2 II':<>: II T(x 1+x 2 ) - T nCx 1+x 2 ) II'+ II Tx1-T nxl II'+ II Txz-T nx 2 II'

which is arbitrarily small since {T n} is pointwise convergent toT. Homogeneity is proved by a similar argument. ButT is also continuous: Since {Tn} is a Cauchy sequence in (n(X,Y), 11·11) it is bounded, (see AMS §3); that is, there exists an M > 0 such that II Tn II:<>: M

for all n EN.

71

Spaces of continuous linear mappings

Therefore, given x EX II Tx II' :<>: II Tx-T nx II' + II T nX II' :<>: M II x II since {T n} is pointwise convergent to T. We conclude that T E (ii)

~(X,Y).

We had, for all x EX II T nX-T mX II' < f II x II

for all m,n > v .

So keeping n fixed and increasing m we have for all II x II:<>: I, II Tnx-Tx II':<>:£ for all n > v. II Tn-T II' :<>: £ for all n > v; Therefore, that is, {Tn} converges toT in (~(X,Y), 11·11).

D

We will find the following extension property of continuous linear mappings particularly useful. 4.8 Theorem. Consider a normed linear space (X, 11·11) and a Banach space (Y, 11·11') and T a continuous linear mapping from a dense linear subspace A of (X, II· II) into

(Y, 11·11'). There exists a uniquely determined continuous linear extension T: X

~

Y and

IITII=IITIIAProof. To define T consider x EX \ A and a sequence {~} in A such that {an} converges

to x in (X, 11·11). Then {~} is a Cauchy sequence in A and II T~- T~ II' = II T(an- ~) II':<>: II T 1111 an-~ II' so that {Tan} is a Cauchy sequence in (Y, 11·11'). Since (Y, 11·11') is complete there exists a y E Y such that {Tan} is convergent toy in (Y, 11·11'). We define Tx = y. To show that this definition is independent of the choice of sequence {an} convergent to x we note that for any other sequence{~'} convergent to x we have so

{T~'}

II Tan' - T~ II :<>: II T II II ~·- ~ II converges to the same limit as {Tan} in (Y, 11·11').

Clearly the mapping T: X ~ Y defined in this way is an extension ofT to X. To prove linearity ofT we consider sequences {an} convergent to x 1 and {an'l convergent to x2 in (X, 11·11'). Then II T(x 1+x 2) - Tx 1 - Tx 2 11' :<>:II T(x 1+x 2) - T(an+ an') II'+ II T~- Tx 1 II'+ II Tan'- Tx 2 11'. SoT is additive and homogeneity follows by a similar argument. Since A is a subspace of (X, 11·11) we have IITII>IITIIA. But for sequence {ani convergent toxin (X, II· II) we have IITxll' :IITx-T~II' +IITIIAII~II'

72

§4. Duals and operator algebras

so

IITxll' :s;IITIIAIIxll

and

IITII:s;IITIIA.

Therefore,

IITII=IITIIA.

for all x EX

The uniqueness ofT follows from the fact that A is dense in (X, 11·11).

D

This theorem has the following consequences. 4.9 Corollary. Consider a normed linear space (X, 11·11), a Banach space (Y, 11·11') and a

linear subspace A dense in (X, 11·11). Then

(~(A,Y),

II· II) and (~(X,Y), 11·11) are

isometrically isomorphic. There are special cases of continuous linear mappings which are important to identify and discuss. Such cases are consequences of the presence of linear structure and will be recognised as a development of those treated in a purely linear algebra study.

4. 10 Dual spaces Given normed linear spaces (X, 11·11) and (Y, 11·11') our first special case arises when we take the range spaceY as the scalar field of the domain space X. We then have the normed linear space of continuous linear functionals on (X, 11·11). 4.10.1 Definitions. (i)

Given a linear space X over ([ (or IR), the algebraic dual (or algebraic conjugate)

space is the linear space X:(X,([) (or X:(X,IR)) usually denoted by (ii)

x#.

Given a normed linear space (X, 11·11) over ([(or IR), the dual (or conjugate) space is

the normed linear space (~(X,
for all x EX.

4. 10.2 Remark. Given a normed linear space (X, 11·11), it is obvious that X* is a linear subspace of X#. However, it follows from Theorem 2.1.12 that X*= X# if and only if X is finite dimensional so when X is infinite dimensional X* is always a proper linear D subspace of X#. Since the scalar field is complete, Theorem 4.7 provides the following important property of dual spaces.

Spaces of continuous linear mappings

73

4.10.3 Corollary. Whether a normed linear space (X, II· II) is complete or not, its dual space (X*, 11·11) is always complete. Furthermore, Theorem 4.8 gives us this additional information. 4.10.4 Corollary. Consider a normed linear space (X, 11·11) and a linear subspace A dense in (X, 11·11). Then (A, II·IIA)* is isometrically isomorphic to (X, 11·11)*. We should recall firstly, the precise algebraic situation for the finite dimensional case. 4.10.5 Theorem. For an n-dimensionallinear space Xn over
x:

I e 1, e2, ... , en}, the algebraic dual lf 1, f2, ... , fnl where fork

E

is also an n-dimensionallinear space with basis

I 1,2, ... , n}

fk(e)j

=I =0

when j = k } j if. k

0

Proof. Since each x E Xn has a unique representation in terms of the basis I e1,e2, ... , en}, Xn is isomorphic to a;n (or lR n) under the mapping

x H (A. I> ~ • . . . , A.n) where x

n

= L A.kek .

k=l For each k E I 1, 2, ... , n} consider the linear functional fk on Xn defined for x=

n

L A.kek by fk(x) = A.k.

k=l Consider the set {f 1, f 2, ... , fnl in

x:.

For any linear functional f on Xn, f(x)

n

n

k=l

k=l

= L A.kf(ek) = L f(ek) fk(x); n

L f(ek) fk. k=l # So {f 1, f 2, ... , fnl spans Xn .

that is,

f=

L akfk =0

n

n

n

If

then

n

L ak fk(x) =0 for all x E Xn so for L iikek we have L I ak 12 = 0

k=l k=l which implies that ak = 0 for all k

E

k=l k=l 11,2, ... , n}; that is, {f1, f 2, ... , fnl is linearly

independent. We conclude that {f1, f2, ... , fn} is a basis for

x:.

It is called the basis of

x:

dual to the

0 One of the simplest examples concerns continuous linear functionals generated by an inner product in an inner product space.

74

§4. Duals and operator algebras

4.10.6 Example. In an inner product space X we noted in Remarks 2.2.2 that given z EX, the functional fz defined on X by for all x EX

fz(x) = (x, z) is linear. But the Cauchy-Schwarz inequality gives us that I fz(x) I= I (x, z)

I~

II z 1111 x II

for all x EX

which implies that fz is continuous on X. Now clearly, llfzll=inf {M: lfz(x) I~MIIx II for all xeX} ~II z II. But

fz(11 : 11) = (11 : II , z) = II z II II fz II =sup {I fz(x) I : II x II ~

so

I} =II z II.

D

A continuous linear functional need not attain its norm on the closed unit ball. We will see later that whether it does or not has some significance. 4.10.7 Example. y

={Jli ,Jl2, ... , Jl

Consider the Banach space (c 0, 11·11 00 ). Given 0 , ... }

E 1 1 \ {0}, consider the linear functional fy defined on c 0 by

fy(x) =I A0 iln for x

={A.

1,

A. 2 ,

.•. ,

A0 ,

•••

}

E c0.

Now I fyCx) I ~sup{ I A0 I : n EN }(I I Jln I) ~II

y 11 1 II x lloo

so fy is continuous and II fy

II~

for all x ec 0

II y 11 1•

Now consider the sequence { x0 } where x1

={sgn JlJ, 0 , ...

x2

= { sgn

x0

={sgn JlJ, ... , sgn Jl

JlJ, sgn Jlz, 0, ...

0,

0, ... }

Then x 0 E c 0 and II x 0 11 00 ~ I for all n E N and n

· fy(x 0 ) =

Ll Jlk I -7 II y 11 1

as n -7

""·

k=l

So II fy II = II y 11 1, but there is no element x0 E c0 , II x 0 lloo = I such that fy(x 0 ) =II fy II= II y 11 1.

D

It is instructive to consider the following geometrical interpretation of the norm of a nonzero continuous linear functional on a normed linear space.

This geometrical

interpretation depends on a duality which exists between the linear functionals and certain linear subspaces.

Spaces of continuous linear mappings

75

4.1 0.8 refinition. Given a linear subspace M of codimension 1 and x0 !1' M, the set x0 + M, the translate of M through x0 is called a hyperplane (more strictly an affine

hyperplane) parallel toM. We make the following deduction from Lemma 1.24.15. 4.10.9 Corollary. Every hyperplane M is ofthefonn x 0 + ker f = { x EX : f(x) = f(x 0)}

where f is a nonzero linear functional and x0 EM. Also for any scalar A and nonzero linear functional f the set { x EX : f(x) =A} is a hyperplane.

Proof. We prove the second part. For x 0 !1' kerf we have

f(f~;~)) =A 'Axo {x EX : f(x) = A } = f(xo) + kerf.

so

D

4.1 0.10 Remark. Note that the correspondence between nonzero linear functionals and linear subspaces of codimension one is not one-to-one because ker Af = ker f for a nonzero linear functional f and any A ~ 0. However, there is a one-to-one correspondence between non-zero linear functionals f and hyperplanes of the form Mr = { x EX : f(x) =

1}.

D

The norm of a nonzero continuous linear functional has a geometrical meaning in terms of the distance from the origin of its corresponding hyperplane. 4.10.11 Theorem. Given a normed linear space (X, 11·11), for a nonzero continuous

linear functional f on X, 1 II f II= d(O, Mr)

Proof. Since f is a continuous linear functional, I f(x) I~ II f 1111 x II for all x EX. 1 So for all x

E

Mr,

Therefore

llxii<':ITTTI.

d(O, Mr) <':

1

iTTTI .

But also, since f is continuous, from Theorem 1.24.17 we have that ker f is closed, and so Mr is also closed. As 0 !1' Mr = Mr then d(O, Mr) > 0. Given 0 < £ < d(O, Mr), consider x0 E Mr such that

II x 0 II < d(O, Mr) + £

.

76

§4. Duals and operator algebras

Consider x E X \ ker f. Then f(x)

* 0 and

X

f(x)

Mr and II X II II x0 II< d(O, Mr) + £ ~ 1 f(x) 1 + £. E

I f(x) I< _ll_x_ll_ < II x II II x0 11-£ d(O, Mr)-£ I I f(x) I~ d(O, Mr) II x II for all x EX,

Then Therefore,

I II f II ~ d(O, Mr) .

so We conclude that

II f II

=

I d(O, Mr) .

D

Figure s.· The distance of the hyperplane Mr from 0. The following example illustrates how Theorem 4.10.11 generalises the Euclidean situation. 4.10.12 Example. In Euclidean space (IR 3, 11·11 2) consider the hyperplane M given by

aA. + bJl + cv = I which is { x E lR 3 : f,(x) = (x,z) = I} where x = (A,Jl,V) and z = (a,b,c). 1 which is -11 1 II = Nowd(O,M)= 11 -11 usingExample4.10.6. ,j a2+b2+c2 z z

j

D

Still exploring the link between a nonzero continuous linear functional f and the hyperplane Mr, it is worth noticing how the attainment of the norm off on the unit sphere is related to the existence of closest points in Mr-

77

Spaces of continuous linear mappings

4.10.13 Corollary. Given a nonzero continuous linear functional f on a normed linear space (X, ll·ll),for x0 EX\ ker f II x0 II f(xo)

x0 ) = d(O, Mr) if. and only if. f ( ~ = II f II.

4.10.14 Remark. We pursue further the relation between non-zero continuous linear functionals and their corresponding hyperplanes. Consider a continuous linear functional f on a normed line~r space (X, 11·11), where II f II = 1 and there exists an x0 EX, II x0 II= 1 such that f(xo)

= 1. We say that the hyperplane Mr is a tangent hyperplane to the closed unit

ball at x0 . By this we mean that x0 E Mr and I f(x) I ~ I for all x EX where II x II

~

1. Notice

that d(O, Mr) = 1 = II x0 II.

Figure 6. Mr is a tangent hyperplane to the closed unit ball at x0

.

We have been dealing with linear spaces over ([ or IR. Now a linear space X over ([ can also be regarded as a linear space over IR; we will denote this associated linear space by XJR. There is in fact a one-to-one correspondence between x# and (XIR)#. 4.10.15 Theorem. Given a complex linear functional f on a complex linear space X, then Ref is a real linear functional on the associated real linear space XIR.

Given a rea/linear functional fiR on XIR, then a complex linear functional f on X is defined by

Proof. Given f on X it is clear that Ref is real linear functional on XJR. However, for each x EX, f(x)

= Re f(x) + i Im f(x).

78

§4. Duals and operator algebras

But f(ix) = i f(x) soRe f(ix) + i lm f(ix) = Re i f(x) + i Im i f(x) =- lm f(x) + i Re f(x). Equating real parts we have Im f(x) =- Re f(ix) so we have the following decomposition off in terms of Re f, f(x) = Re f(x)- i Re f(ix)

for all x EX.

This suggests that, given fiR on XIR we can define f on X by f(x) = fiR(x)- i fiR(ix). Clearly f is additive and homogeneous for real scalars. We need to check homogeneity for complex scalars. It is sufficient to consider a= i~ where ~is real. Then f(ax) = fiR(ax)- i fiR(iax) =~(fiR (ix)

+ i fiR(x))

= a(fiR(x)- i fiR(ix)) =a f(x). So we conclude that f is a complex linear functional on X.

0

For any given complex normed linear space (X, 11·11) we have an associated real normed linear space (XIR, 11·11). The continuous linear functionals on these spaces are closely related. 4.10.16 Theorem. There is a one-to-one norm preserving correspondence between X* and (XIR)*.

Proof. It is clear from the formula relating f on X to fiR on XIR given in Theorem 4.1 0.15, that f is continuous on (X, 11·11) if and only if fiR is continuous on (XIR, 11·11). We show that this correspondence is norm preserving. Clearly,

I f(x) I~ IRe f(x) I for all x E IR

so

II f II ~ II fIR II.

However, given

f.

> 0 there exists an x0 EX, II x0 II :S: 1 such that

I f( x0 ) I > II f II - f.. Now if f(xo) = eie I f(x 0) I then f(e-i8x 0) =I f(x 0) I= I fiR(e-i8x 0) I so I fiR(e-i8x 0) I> II f II- f.. We conclude that II fiR II= II f II.

0

4.10.17 Remark. We should discuss the relation between hyperplanes in both spaces X and XIR because the order relation associated with IR implies that for AE IR the real hyperplane { x E XIR: fIR (x) = A} separates the space into easily definable half spaces {x EXIR: fiR(x) ~ A.}and {x EXIR: fiR(x) :S: A.}. Now although a complex linear subspace of X is a real linear subspace of XIR, a real linear subspace of XIR is not necessarily a complex linear subspace of X. Further, for a nonzero

79

Spaces of continuous linear mappings

complex linear functional f on X, kerf is a real linear subspace of codimension 2 in XIR. From the formula relating f on X to fiR on XIR given in Theorem 4.10.15 we have that (ker f)IR = ker fiR n ker fiR(i . ) This not only implies that (ker f)IR is contained in ker fIR but also tells us that (ker f)IR is uniquely determined by fiR. Therefore for complex linear functionals on a complex linear space X, there is a one-to-one correspondence between (ker f)IR and ker fiR in XIR although (ker f)IR is of codimension I in ker fIR. Given a complex normed linear space (X, 11·11) and a continuous linear functional f on X,

Mf= {x EXIR: f(x) =I} is contained as a closed real affine hyperplane of the corresponding closed real hyperplane

Mf IR but both are at the same dIstance o

= {x EXIR: fiR(x) = 1},

1 iTTTI

=

I iTTII IR

f rom the ongm. o

o

D

Operator algebras

4.11

Given normed linear spaces (X, II· II) and (Y, 11·11') our other special case arises when we take the range space (Y, 11·11') to be the same as the domain space (X,Ii-11). We then have the normed linear space of continuous linear operators on (X, 11·11). But such operator spaces actually have extra algebraic structure. 4.11.1

Definitions. Given an algebra A over

a:

(or IR ), a norm IHI on A is said to be

an algebra norm if it satisfies the additional norm property: (v) For all x,y E A,

II xy

II~

II x 1111 y II, (the submultiplicative inequality).

The pair (A, II· II) is called a normed algebra. A normed algebra which is complete as a normed linear space is called a complete normed algebra (or a Banach algebra). Different norms can be assigned to the same algebra A giving rise to different normed algebras. Equivalent norms for A as a linear space are not necessarily equivalent algebra norms; that is, they do not necessarily preserve the submultiplicative inequality. If A has a multiplicative identity e and II e II= 1, then (A, 11·11) is said to be a unital normed algebra. 4.11.2 Remark. Property (v) actually links multiplication to the norm and implies that multiplication is jointly continuous: if x ---7 x 0 andy ---7 Yo then xy ---7 x0y0. This can be deduced simply from the inequality II xy-x 0y 0 II~ II x(y-y 0) II+ II (x-x 0)y 0 II

~II

x 1111 y-y 0 II+ II x-x 0 1111 Yo II .

D

80

§4. Duals and operator algebras

4.11.3 Theorem. Given a normed linear space (X, II· II) over [ (or IR), the space of

linear operators, usually denoted by :C(X), and the space of continuous linear operators, usually denoted by n(X), are noncommutative algebras under multiplication defined by composition. Further, (n(X), 11·11) is a unital normed algebra and is complete if(X, 11·11) is complete.

Proof. Given T,S E :C(X) we define TS on X by TS = ToS ToS(x) = T(S(x)) for all x EX.

where

Given T,S E n(X), it is clear that TS is also continuous and so n(X) is a subalgebra of :C (X). Now for T,S E n(X) and all

X

EX

II TS(x) II= II T(S(x)) II ~

II T II II Sx II

since T is continuous

~

II T 1111 S 1111 x II

since S is continuous.

II TS II= sup {II TS(x) II : II xll ~ I }

So

~IITIIIIS

II.

The identity operator I is the multiplicative identity and clearly II I II= I. The completeness of (n(X), II· II) follows from Theorem 4.7.

D

It is worth noting that an operator algebra is a generalisation of the linear space Mn of n x n matrices. 4.11.4 Example. When X= IR n (or a:n) then n(X) is algebra isomorphic to the linear space Mn of n x n matrices with entries from IR (or[).

Proof. Here we are dealing with a special case of Example 4.2. We need only verify that given a basis ( e 1,e2 , ... , en} of X then the linear mapping T ---+ [ aijl of n (X) into Mn, preserves multiplication. Now Mn is a noncommutative algebra under matrix multiplication, n

[aijl

[~ijl = [ L aij ~jk]. j=I

n

Tek =

I, aik ei i=I n

and

S(Tek) =

L

j=I n

so S o T

H

[

L j=I

aij ~jk ] .

~jk

n

n

n

L aij ei = L L

i=I

aij ~jk ei

i=l j=I

D

Spaces of continuous linear mappings

4.11.5 Remark. The operator algebra

(~(X),

81

11·11) can be regarded as the prototype of

noncomrnutative unital normed algebras. We will notice that the operator algebra (~(H), 11·11) where H is a Hilbert space is of particular importance in itself and is a prototype of a special class of noncommutative unital normed algebras. D Many of the properties of the operator algebra

(~(X),

IHI) are derived directly from

its being a unital normed algebra. So we will develop some elementary normed algebra theory. There are considerable advantages in doing so, in that we are made aware of the essential structure from which the properties are derived, (they mostly do not depend on the underlying space (X, 11·11)), and the theory applies in a wider setting than that of operator algebras.

4.11.6 Definition. Given an algebra A with identity e, an element x E A is said to have a multiplicative inverse if there exists an element x-1 E A such that xx-1 = x-lx =e. It is easy to see that if an element x E A has a multiplicative inverse then it is unique. An element in A which has a multiplicative inverse is said to be regular, an element which does not is said to be singular. Clearly the set of regular elements of A form a group under multiplication. Linear algebra considerations provide the following characterisation for operator algebras.

4.11. 7 Theorem. Given a normed linear space (X, 11·11), an element T E I: (X) is regular in J::(X) if and only ifT is one-to-one and onto and an element T E ~(X) is regular in ~(X)

if and only ifT is one-to-one, onto and T-1 is continuous.

This generalises the finite dimensional situation.

4.11.8 Example. A linear operator T on a finite dimensional linear space Xn is regular if and only if T is one-to-one. Given a basis {e 1,e2, ... , en} for Xn and the linear operator Ton Xn has matrix representation [a.ijl with respect to this basis, then Tis regular if and D only if det[ a.ijl '# 0. It is clear that in any algebra with identity, the identity is a regular element. We show that in any unital Banach algebra, elements sufficiently close to the identity are also regular.

4.11.9 Definition. Given an algebra A and x E A, we write x l=x and define inductively the positive integral powers of x by, xn = x(xn-1). Clearly, for all m,n E F:l. xn.xm = xm.xn = xm+n

82

§4. Duals and operator algebras

Given a normed algebra (A, II· II) and xE A, it follows by induction from the submultiplicative inequality that II x" II $11 x II" for all n EN. 4.11.10 Theorem. Given a unital Banach algebra (A, 11·11), ifx

E

A is such that

L x" converges and e-x is regular and (e-xr 1 =e + L x".

lim sup II x" 11 11" < 1 then n--?oo

n=1

Proof. Choose real a such that lim sup II x" 11 11"
Then there exists a v E N such that II x" II< a"

for all n > v.

Therefore the series rx" is absolutely convergent and since (A, 11·11) is complete, rx" is n

convergent; (see AMS §3). Write s 0

=e + L xk and s = e + L xk. k=1

shows that (e-x)s 0 Now s 0 (e-x)s

--7

=s

s and II x" II

= s(e-x) = e;

0

(e-x)

--7

A simple calculation

k=1

= e-x"+ 1.

0 as n

--7 oo

so by the joint continuity of multiplication we have

that is, e-x is regular and (e-x)-1 = e +

L

x".

k=1

D

4.11.11 Corollary. In a unital Banach algebra, any x E A such that II e-x II < I is regular.

Proof. If II e-x II < I then from the submultiplicative inequality we have II (e-x)" 11 11" < I for all n EN and so from Theorem 4.11.1 0, x = e - (e-x) is regular.

S(O; I)

Figure 7. A pictorial representation of the ball B(e; I) of regular elements in any unital Banach algebra.

D

Spaces of continuous linear mappings

83

As an aside, it is of interest to note that the following limit property holds simply as a consequence of the submultiplicative property of the norm. 4.11.12 Proposition. In a normed algebra (A, ll·ll),for any x E A, lim II xn 11 11n always

exists and is equal to inf {II xn 11 11n : n EN } . Proof. Write v

=inf{ II xn 11

11n :

n EN}. Given£> 0 choose mEN such that II xm II 11m< v + £.

For any n E N write n = Pnm + qn where Pn and qn are integers and Pn ~ 0 and

0 ~ qn

~

m- 1. Then by the submultiplicative property, for any n EN

v ~II xn 11 11n =II xmpn xqn 11 11n ~II Xm IIPn/n II X llq"/n

q Since nn --7 0 as n --7 Therefore

oo

< (v + £) mpn/n II x llq 010 • mp we have ~ --7 I as n --7 oo.

lim II x" 11 11" exists and is equal to v.

D

n~~

4.11.13 Remark. Given a unital Banach algebra (A, 11·11) and an element a E A, Theorem 4.11.1 0 shows that e- Ax is regular for scalar A if lim sup II ('Ax)n 11 11" < I; that is, for all n~

scalars A such that I A I< I I lim sup II x" 11 11". n~

For that range of scalars A, 00

(e-'Axr' = e

+I

A"x".

n=l

This series is called the Neumann series for x.

D

We now apply our theory to the solution of an integral equation. 4.11.14 Example. Volterra integral equations Consider the integral equation of the form X

f(x) = g(x) + A where

Jk(x,t) f(t) dt

g is a given continuous function on [a,b], f is the required solution function on [a,b ], k the kernel of the equation is a given continuous function on the triangular region { (x,t) : a~ t ~ x, a~ x ~ b},

and A is a scalar.

84

§4. Duals and operator algebras

The equation has a unique solution function f

E

'C[a,b] for each given g

E

'C[a,b] and

parameter A..

Proof. The Volterra operator K on t: [a,b] defined by X

(K(f))(x) =

Jk(x,t) f(t)

is a continuous linear operator on ('C[a,b], 11·11 00 ) , (see AMS §7). In terms of the Volterra operator, the integral equation can be written (I- A.K)(f) = g. From Remark 4.11.13 we see that I - A.K is regular for all I A. I < I I lim sup II K 0 11 11" n-;oo

and the solution function will be given by f =(I +

~~ A.Kn )cg);

that is, we have a series solution for the Volterra integral equation. In fact, we show that lim sup II K 0 II lin = 0 n-;oo

and so we have such a solution for all scalars A.. Now

IK(f)(x) I ~ (x-a) sup{ I k(x,t) II f(t) I : a~ t :S: x} :S: M II f lloo (x-a) where M =sup{ I k(x,t) I: a~ t ~ x, a~ x ~ b}. We prove by induction that

I K 0 (f)( X) I ~ M 0 II f lloo

(x-~)n n.

for all a :S: x

~

b:

The previous statement is case n = I. Suppose that the statement is true for some n EN. Then I Knt 1(f)(x) I =I K(K"f)(x) I =I

Jk(x,t) (K"f)(t) dt I a

~M

nt]

X

1: f lloo

n.

J(t-a)" dt

= Mnt-111 f II

nt-] (x-a) oo (n+l)!

which concludes the induction. Using this fact we obtain for all n EN II K 0 f II= max {I (K 0 f)(x) I : a~ x :S: b} ::;; M 0 II f lloo so

II Kn II = sup {II Kn II : II f II

~

I } ::;; Mn

(b-~)n . n.

(b-~)n n.

Spaces of continuous linear mappings

Since

(~) 11n ---7 0 n.

85

as n ---7 oo we conclude that lim sup II Kn 11 11n = 0. n-->oo

It follows that, for every g E 'C[a,b] and parameter 'A there exists a unique solution of the

Volterra integral equation given by the Neumann series

0 4.11.15 Remark. In AMS §5 we established the existence of a unique solution using Banach's Fixed Point Theorem and that theorem provided a similar way of finding a solution by iteration. 0 Corollary 4.11.11 tells us that in any unital Banach algebra, B(e; I) is an open ball of regular elements. As we might expect this implies a topological property for the set of regular elements. 4.11.16 Theorem. In any unital Banach algebra (A, 11·11), the set of regular elements is

open.

Proof. Given a regular element x E A andy E A, II e-x-Iy II= II x-I(x-y) II::;; II x-I 1111 x-y II. So from Corollary 4.11.11 we have that x-Iy is regular when II x-y II< 1111 x-I II. But in this casey= xx-Iy is itself regular being as it is, the product of regular elements x and x-Iy. Therefore, {yEA:IIx-yll< 1/llx-111} is a set of regular elements and this implies that the set of regular elements is open.

0

4.11.17 Corollary. In any unital Banach algebra the set of singular elements is closed. Another topological property relating to the regular elements is as follows. 4.11.18 Theorem. On a unital normed algebra (A, 11·11), the mapping x

H

x-1 is a

homeomorphism of the set of regular elements onto itself.

Proof. Since the mapping x

H

x-I is its own inverse mapping it is sufficient to prove

continuity. For regular elements x,y E A, II x-I- y-1 II= II x-I(x-y) y-1 II::;; II x-I 1111 y-1 1111 x-y II. II y-I II ::;; II x-I - y-I II+ II x-I II::;; II x-I 1111 y-I 1111 x-y II+ II x-I II. But So for all regular yEA where II x-y II< I/211 x-I II we have II y-111 < 2 II x-I II and so II x-I- y-1 II< 2 II x-I 11 2 II x-y II, which implies that the mapping x H x-I is continuous at x.

0

86

§4. Duals and operator algebras

4.12 EXERCISES I.

For the following continuous linear mappings T determine II T II. Given k E I 1,2, •.• , n}, the projection Pk: (IR n, 11·11 2) --7IR where (i) (ii) (iii)

2.

Pk0"1, A2, · · · , An) = Ak· T: (IR 3, 11·11 2) --7 (IR 11-11 2) where T (AI,~. A3) = (A3, A1, ~). Do: ('C 1[0,1], 11·11') --t!R where II f II'= II f lloo +II f' lloo and D 0(f) = f'(O).

J.

(iv)

the shiftS: (.t 1, 11·11 1) --7 (.t 1, 11-11 2) where S(IAI, A2o ···,An,··.})= 10, (AI, A2o · .. ,An··.}.

(i)

Given a sequence I a. 1, a.2, ... , a.n, ... } of scalars prove that la.1A 1, a. 2A2o ... , a.n An, ... }E .t 2 for all IA 1, ~ •... ,An, ... }E .t 2 if and only if I a. 1, a.2, ... , Un, ... } is bounded.

(ii)

A diagonal operator Ton (.t 2, 11-11 2) is defined by a bounded sequence of scalars I a. 1, a. 2, ... , a.n , ... } where for X= IA1, A2, ···,An,·· .}E .t2, Tx = la.1A1, U2A2, · · · • UnAn, · · .}.

(a) Prove that Tis continuous and II T II= sup {I Un I : n EN} . (b) Prove that T is a topological isomorphism onto if and only if (iii)

inf { I a.n I : n E N } > 0. Deduce that (m, IHioo) is isometrically isomorphic to a linear subspace of (~(.t 2 ),

3.

Given y

11·11) and that (~(.t 2 ), 11·11) is not separable.

={J.11,J.12, ... , lln• ... }

by fy(x) =

lA.n lin

E

.t 1 consider the functional fy defined on .t 1

for all x ={AI,~ •... , An, ... }

E

.tl.

(i)

Prove that fy is linear and continuous on .t 1 with respect to both the 11-11 2 and 11·11 1 norms.

(ii)

Determine II fy II in both (.t 1, 11·11 2) and (.t 1, 11·11 1) and find whether fy attains its norm on the unit sphere in each case.

4.

Consider the functional F defined on the real linear space 'C[-7t,7t] by 1t

F(f) =

Jf(t) sin t dt. -1t

(i)

Prove that F is linear and determine whether F is continuous on 'C [-7t,7t] with respect to norms ll·ll2, ll·lloo and 11·11 1.

(ii)

Wherever possible for 'C[-7t,7t] with norms 11·11 2, ll·lloo and 11·11 1 determine II F II in the corresponding dual space, and find whether F attains its norm on the unit sphere of the space.

Spaces of continuous linear mappings

5.

(i)

87

Consider a continuous linear mapping T of a Banach space (X, 11·11) into a normed linear space (Y, 11·11') where T(X) is dense in (Y, 11·11') and T-1 exists and is continuous on (T(X), 11·11'). Prove that T maps X onto Y and (Y, 11·11') is complete.

(ii)

Consider Banach spaces (X, 11·11) and (Y, 11·11') and a continuous linear mapping T from a dense linear subspace of (X, 11·11) onto a dense linear subspace of (Y, 11·11'). Given T, the unique continuous linear extension of T mapping (X, 11·11) into (Y, 11·11'), prove that (a) ifT is an isometric isomorphism then Tis an isometric isomorphism onto (Y, 11·11'), (b) if T is a topological isomorphism then T is a topological isomorphism onto (Y, 11·11').

(iii)

Deduce that (a) isometrically isomorphic normed linear spaces have isometrically isomorphic completions, (b) topologically isomorphic normed linear spaces have topologically isomorphic completions.

6.

A real trigonometric polynomial t on IR of degree n is of the form n

L

(ak sin k8 + bk cos k8) where ak, bk E IR fork E {0, 1, ... , n}. k=O Consider the real linear space 'J' n of trigonometric polynomials of degree less

t(S) =

than or equal to n and the linear operator D on 'J' n defined by D(t) = t'(S). (i)

Prove that Dis continuous on ('J' n• 11·11) for any norm 11·11 on 'J' n·

(ii)

For norm 11-11 2 on 'J' n where 11·11 2 is generated by the inner product 1t

(t, s) =

J ts(S)

de

-1t

(iii)

prove that II D II = n. For norm ll·lloo on 'J' n where II t lloo =max {I t(S) I : e prove that II D II

E

[-1t,1t]}

=n.

(This result was first proved by S. Bernstein about 1900. For an elementary proof see R.P. Boas, Math.Mag. 42 (1969), 165-174.) 7.

Consider ("C [0, 1], IHioo) and the set of continuous linear functionals {Px : x E [0, 1] where Px(f) = f(x) for all f E "C [0, 11}. (i) Prove that this set is contained in the unit sphere of ("C[O, 1], IHioo)*. (ii)

Show that ("C [0, 1], ll·lloo)* is not rotund.

88 8.

§4. Duals and operator algebras

(i)

Consider a normed linear space (X, 11·11) with a Schauder basis {en}. Prove that {en} is a monotone basis if and only if II T n II :::; 1 for all n EN where Tn is the finite rank operator defined for each =

x

=I

n

"-nen by T n(x) =

n=l

(ii)

I

A.kek .

k=l

Prove that if a Hilbert space has a countable orthonormal basis then it has a monotone Schauder basis.

9.

(i)

A linear space X has norms II· II and 11·11' where for some M > 0,

(ii)

Deduce that if 11·11 and 11·11' are equivalent norms for X then

II x II :::; M II x II' for all x EX. Prov.e that (X, II· II)* s;;; (X, 11·11')*. (X, 11·11)* = (X, 11·11')* and 11·11* and 11·11'*, norms on X* induced by 11·11 and 11·11', are equivalent norms on X*. (iii)

For t: [-1t,1t). prove that ('C:[-1t,1t), 11·111)* C ('C:[-1t,1t), 11·112)* C ('C:[-1t,1t), 11·11 00 )*.

*

10.

(i)

*

.

Given a linear space X with algebraic dual x#, prove that each x EX generates a linear functional "x on X# defined by "x(f) = f(x) for all f EX#.

(ii)

Hence, or otherwise prove that a linear space X is finite dimensional if

(iii)

and only if its algebraic dual x# is finite dimensional. A normed linear space (X, 11·11) has a finite dimensional dual space X*. Prove that X is finite dimensional.

11.

Prove that (.tJ, 11-iloo)* is isometrically isomorphic (co, 11·11 00 )* and that

(~((.t 1 , 11·11 (c 0 , ll·lloo)), 11·11) is isometrically isomorphic to (~(co, ll·lloo), 11·11). 00 ),

12.

(i)

Given a closed hyperplane M

={x EX: f(x) =A.} in a real normed

linear space (X, 11·11), and a point x0 EX, prove that I A. I

(a)

d(O, M)

= iTTTI ,

(b) d(x 0 , M)

I A.-f(x 0 ) I and II f II

(c) there exists a y0 EM such that II x0-y 0 II= d(x 0 , M) if and only if there (ii)

exists an x EX such that f(x) = II f II II x II. Given closed hyperplanes M 1 {x EX : f(x) = A. 1} and

=

M 2 = {x EX : f(x) = A.2} in (X, 11·11), prove that d(MI,M2) =

I A. 1-A. 2 I II f II

89

Spaces of continuous linear mappings

13.

Given a real normed linear space (X, 11·11), prove that a linear functional

(i)

f is continuous if and only if for any scalar A., the half-space { x EX : f(x) ~ A.} is closed. (ii)

Given an open ball B(x 0 ; r) contained in a closed half-space H

= {x EX: f(x)

~A.}

prove that

inf f(B(x 0 ; r))-A. f(x 0 )-'A d(B(xo; r), H) = II f II = -1-1f-1-1 - r.

14.

Given a nonempty subset A of a real normed linear space (X, 11·11), a hyperplane M

= { x EX : f(x) = A.} is called a hyperplane of support for A if A is contained in

one or other of the half-spaces { x EX : f(x) ~ A.} or {x EX : f(x) ~ A.} and there exists at least one x0 E A such that f(x 0) = A.. (i)

Given a compact set Kin a normed linear space (X, 11·11) and a closed hyperplane M

={x EX: f(x) = 0}, prove that there exists an x0 E K such

that d(x 0 , M) = d(K, M) and that

M 0 = {x EX: f(x) = f(x 0 )} is a hyperplane of support forK at x0 . (ii)

Consider the closed unit ball B[O; 1] in (X, 11·11). Prove that M 1 = {x EX: f(x) = f(x 1)} is a hyperplane of support forB[O; 1] at x 1 EB[O; 1], if and only if I f(x 1) I= II f 1111 x 1 II.

15.

Consider the functional f defined on (c0 , ll·lloo) for x =

by

f(x)

= {A. 1,A. 2, ...

, An, ... }

')...

= I. ~ k=l 2

(i)

Prove that f is linear and continuous and II f II = I, but show that f does not

(ii)

Show that for any x0 10M= {xE X: f(x) = 0} there is no closest point to x0

(iii)

Show that the closed unit ball has no hyperplane of support of the form

attain its norm on the unit sphere. inM. {x EX: f(x) =A.).

16.

Given a normed linear space (X, 11·11) and elements x EX and f EX*, an operator x ® f is defined on X by x ® f(y) = f(y) x (i)

for all y EX.

Prove that x ® f is a continuous linear operator on X and II X® f II= II X 1111 f II.

(ii)

Prove that every continuous linear operator T on X with n-dimensional range T(X) has the form T = x 1 ® f 1 + x2 ® f 2 + ... + xn ® fn where x 1,x 2 , ... , Xn EX and f 1,f2 , ... , fn EX*.

90

§4. Duals and operator algebras

(iii)

Prove that X is finite dimensional if and only if every linear operator on X is continuous.

17.

Given a normed linear space (X, 11·11) with proper closed linear subspace M, consider the quotient space XIM with norm 11·11': XIM ~ R defined by

II [x]ll' = inf {II x+m II: m EM}. (i)

(a) Prove that the quotient mapping 1t:

X~

XIM defined by

1t(x) = x + M is a continuous linear mapping and ll1t II = I. (b) Prove that 1t attains its norm on the closed unit ball of (X, II· II) if and only if for each x E X there exists an m E M such that

II x-m II = d(x, M) . (ii)

The linear subspace M is said to have codimension n in the linear space X if there exists a linearly independent set {e 1,e 2, ... , en} in X\ M such that each x EX can be represented in the form x

= A. 1e 1 + A.2e 2 + ... + Anen + z

for scalars {A. I> ~ •••. , An} and z EM. Prove that if M has codimension n in X then XIM has dimension n. (iii)

A linear functional f on X has M s;;; kerf. (a) Define a linear functional F on XIM by F(x+M) = f(x). Prove that F is continuous on (XIM, 11·11) if and only iff is continuous on (X, 11·11) and II F II = II f II. (b) Prove that if M has finite codimension in X then f is continuous on (X, 11·11).

18.

Given a normed linear space (X, 11·11), prove that 13(X) is a commutative algebra if and only if X is one dimensional. (Hint: If X has more than one dimension use Exercise 16 to construct noncommutative operators with one dimensional range.)

19.

Given a unital normed algebra (A, 11·11) consider the left regular representation map ·a H Ta of A into 13(A) defined by Ta(x) = ax

for all x EA.

Prove that (A, 11·11) is isometrically algebra isomorphic to a subalgebra of (13(A), 11·11) under left regular representation.

Spaces of continuous linear mappings

20.

Consider a unital Banach algebra (A, 11·11). (i)

Prove that if II x II < I then II (e-x)

(ii)

-1

-(e-x)

II~

2

II X 11 I=iTXTf .

Prove that if x E A is regular andy

E

A is such that II yx-1 II < I then

x-y is regular and (x-yr 1 = x- 1 +

f x-\yx- 1)k. k=l

21.

22.

Given a unital Banach algebra A, prove that (i)

if x and xy are regular then y is also regular,

(ii)

if xy and yx are regular then both x and y are regular,

(iii)

if xy

=e and yx =z * e then z2 =z where z * 0, e.

Solve the Volterra integral equations X

(i)

f(x)

= I + A. Jex-t f(t) dt

for 0

~

x

~I,

for 0

~

x ~I.

0 X

(ii)

f(x) = I + A.

J(x-t) f(t) dt 0

91

92

Spaces of continuous linear mappings

§5. THE SHAPE OF THE DUAL

To gain a firmer grasp of the concept of the dual of a normed linear space we need to examine the particular shape of the dual space for several example normed linear spaces.

5.1 Finite dimensional normed linear spaces Given a n-dimensionallinear space Xn over (C (or IR) with basis {e 1, e 2,

... ,

en} we recall from Theorem 4.10.5 that the algebraic dual Xn #is also a linear

space with basis {f1, f 2,

... ,

fnl dual to {e 1, e 2,

•.. ,

enl where

k}

whenj = j ;t k n

Furthermore, since f =

L

f(ek) fk,

k=l

x:

0

is isomorphic to u:n (or IRn) under the mapping

fH (f(e 1), f(e 2),

.•. ,

f(en)).

and every linear functional f on Xn is of the form n

f(x) =

L

n

L

"-k f(ek)

where x =

"-kiik

for some (a. 1, a. 2 ,

A.kek. k=l Being an isomorphism onto [n implies that every linear functional f on Xn has the form k=l n

f(x)

=L

... ,

a.n)

E

u:n

k=l n

where x

=L

A.kek.

k=l

X~ =

x:.

Now given any norm 11·11 on Xn, we have from Theorem 2.1.12 that However, the actual norm on X~ does depend on the norm given on Xn. To

obtain some idea of the form of the actual dual (Xn, 11·11)* we determine the corresponding norm 11·11 on u:n (or IR n) so that (
5.1.1 Example. Consider the normed linear space (
... ,

en} for u:n where ek = (0, ... , 0,1,0, ... , 0). kth place

Now every continuous linear functional f on u:n is of the form n

f(x)

=L k=l

"-k f(ek)

93

§5. The shape of the dual

n

L

I f(x) I ~

So

I A.k II f(ek) I

k=l

~

n

(L

I f( ek) I) max { I A.k I : k

E {

1, 2, ... , n} }

k=l n

L

= (

I f(ek) I) II x lloo

k=l so that

II fll ~ (

n

L

I f(ek) 1).

k=l .

(

0

0

Constder the element x0 = A. 1 , A.2 ,

... ,

A0~J where

o f(eJ A.k = I f(eJ I iff(eJ =0 Then II x0 lloo

~

* 0}

otherwise

1 and n

f(x 0 ) =

L

I f(ek) I

k=l n

II f II =

L

I f(ek) I. k=l This tells us that the isomorphism so

of [

0

* onto

[

0

f H (f(e 1), f(e 2), ... , f(e 0 )) is an isometric isomorphism of ([ 0 , 11·11 00 )* onto ([ 0 , 11·11 1).

0

Consider the normed linear space ([ 0 , ll·llp) where 1 < p < oo with

5.1.2 Example.

standard basis as in Example 5.1.1. Again we have that every continuous linear functional f on [

0

is of the form n

f(x) =

L

A.k f(ek)

k=l So

I f(x) I ~

n

L

I A.k II f(ek) I

k=l

~

(Ik=l I A.k iP) (Ik=l I f(ek) 11P

n

= ( L I f(ek) lq)l/q II x liP k=l n

so that

II f II ~ ( L I f(ek) lq) 11q. k=l

lq)Itq by Holder's inequality where.!+.! = 1

p

q

94

Spaces of continuous linear mappings

=(A.~, A.~, ... , A.~)

Consider the element x 0

A.~

f(ek) I f( ek) lq- 2

if f( ek) "# 0 } .

0 Then

II x0 II =

n

0:

where

otherwise I

I f(ek) lq) 1P since p(q-1) = q,

k=l n

and

f(x 0) =

L

I f(ek) lq::; II f II II x0

liP

k=l n 1 I I llfll ~cr lf(ek)lq)/q since!--=k=l p q

so

n

II f II = ( L I f(ek) lq) 11q. k=l This tells us that the isomorphism Therefore,

fH (f(e 1), f(e 2), ... , f(en)) of ern* onto ern is an isometric isomorphism of cern, ll·llp)* onto (([n, ll·llq).

5.2

D

Hilbert spaces In a Hilbert space, the significant characterisation of orthogonality given in Theorem

2.2.15 and the existence of best approximation points to closed linear subspaces given in Theorem 2.2.19 enables us to establish a satisfactory representation of continuous linear functionals on such spaces. In Example 4.10.6 we noted that the inner product generates continuous linear functionals in a natural way. The following theorem tells us that every continuous linear functional is generated in this manner by the inner product. 5.2.1 The Riesz Representation Theorem for Hilbert Space. For any given continuous linear functional f on a Hilbert space H, there exists a unique z E H

such that f(x) = (x, z)

for all x

E

H.

Proof. Since f is a continuous linear functional, kerf is a closed linear subspace of H. When kerf= H then

f(x) = 0

for all x

E

H

so

f(x) = (x, 0)

for all x

E

H.

When ker f "# H then from Corollary 2.2.21 we have that there exists a nonzero element z 0 E H such that z 0 is orthogonal to ker f. From Lemma 1.24.14 we have that ker f is a linear subspace of codimension one; that is, every x E H can be represented uniquely in the form x = A.z0 + y Now f(x) = A.f(z 0 ). . f(za'J Scaling z 0 we put z = - -2 z0 . II z0 11

where A. is a scalar and y E ker f.

95

§5. The shape of the dual

Then (z0 ,z) = f(z 0 ) and f(x) = Af(z 0) = (A.z0 +y, z) = (x, z)

for all x E H.

To show the uniqueness of z, consider any z' E H such that f(x) = (x, z) = (x, z') for all xE H. Then (x, z-z')

= 0 for all x E H and, as noted in Remarks 2.2.14, this implies that

5.2.2 Remark.

z'

= z. 0

Now every continuous linear functional f on a Hilbert space H is of the

form fz where fz(x) = (x, z)

for all x E H.

We notice from Example 4.1 0.6 that fz(

11~11)= II z II= II fzll

so every continuous linear functional attains its norm on the unit sphere at the unit vector from which it is generated by the inner product. 0 It is of interest to follow the implications of this theorem for finite dimensional Hilbert

spaces, because it helps reveal the rich linear space structure of such spaces which is so useful in developing tensor calculus. 5.2.3 Application. Contravariant and covariant vectors. Given an n-dimensional linear space Xn with basis {el' e 2, ... , en}, we have seen that the dual space Xn * is also an n-dimensionallinear space with basis {f1 .f2, ... , fn} where fi(ej)

=8j

for all i, j E {I, 2, ... , n};

(8i is the Kronecker delta where 8i = I J

J

=0

i =j

*

i j). When Xn is also an inner product space, it follows from the Riesz Representation Theorem

5.2.1 that for each i E {I, 2, ... , n), given fi EX~ there exists an element ei E Xn such that fi(x) = (x, ei)

for all x E Xn .

Since fi(ej) = 8j, it follows that the set (el, e2, ... , en} is also a basis for Xn- Given {e 1, e 2, ... , en} a basis for Xn, the set (el, e2, ... , en} is called the dual basis for Xn. From the definition of this set it is clear that, for each i,j E { I, 2, ... , n}, ei is orthogonal to ej for all i

* j and fi attains its norm at 11 :ii 11 .

Suppose that the inner product on Xn is defined by n

(x, y) =

L

i,j=l

.

.

aij A1 iil

96

Spaces of continuous linear mappings

where x = A. 1e 1 + A. 2e2 + ... + A0 e 0 andy= J.1. 1e 1 + J.1. 2e 2 + ... + J.l. 0 e 0

,

and (aij) is a

positive definite hermitian matrix. Then we can determine the components of each element of the dual basis {el, e2, ... , e 0 } in terms of the original basis {e 1, e 2, ... , e 0 }. Now (ej, ej) = aij· Suppose that ei = aile 1 + ai2e 2 + ... + aine 0 Then fi(ej)

•

= 8ji = (ej, et).

= ail(ej, e 1) + ai2(ej, e 2) + ... + ai"(ej, e

0)

n

= aiiai·J + ai2a2·J + · · • aina-nJ. = £... ~ aik a-k. r

So aik =

k=l cofactor of a·k .. det(a;j) 1 and (aiJ) is the inverse matrix to (aij).

Further, since (ej, ei)

= oJ

we have that (ei, ei)

= aii .

Given any vector x E X0 where x = A. 1e 1 + A.2e 2 + ... + A0 e 0 2 we call (A.\ A. , ... , A.") the contravariant components ofx. But also x = A. 1e 1 + A. 2e 2 + ... + A0 e 0 and we call (A. 1, A.2, ... , A0 ) the covariant components of x. Since we can express the dual basis in terms of the original basis through the matrix (ajj), we can relate the contravariant and covariant components of x in a similar way: n n n x = L t..i ei = L A.i ei = L A.i aikek i=l i=l i,j=l n

so for each k

E

{I, 2, ... , n}, A.k = (x, ek) =

L

A.i aik

i=l n

and of course

A.k = (x, ek) =

L

A.i aik·

0

i=l 5.2.4 Remark. Given a Hilbert space H, the mapping z H fz of H into H* where fz(x) = (x, z)

for all x E H

is, by the Riesz Representation Theorem 5.2.1, one-to-one and onto. But further, this mapping is additive since fz 1+z2(x)

= (x, z 1+z 2) = (x, z 1) + (x, z2)

=(fz 1 + fz)(x)

for all x

E

H.

X E

H.

The mapping is also conjugate homogeneous because faz(X)

= (X, az) = U(X, Z) =

afz(x)

for all

These properties, together with the norm preserving property II fz II = II z II, imply that the mapping is an isometry of H onto H* since

II \ - f'211 =II fz1-'211 =II z 1- z2 11. For a complex Hilbert space this mapping is not linear. But for a real Hilbert space the mapping is linear and is an isometric isomorphism of H onto H*. 0

97

§5. The shape of the dual

Given a Hilbert space H, the mapping z H fz where fz(x) = (x, z)

for all x EX

is a conjugate linear isometry of H onto H*. However, a Hilbert space is self-dual by which we mean that a Hilbert space His actually isometrically isomorphic to its dual H*. We establish this indirectly, using the mapping from the Riesz Representation Theorem 5.2.1 to show that H and H* have the same orthogonal dimension and so by Theorem 3.31 are isometrically isomorphic. 5.2.5. Theorem. A Hilbert space H is isometrically isomorphic to its dual H*.

Proof. Consider a maximal orthonormal set {ea} in Hand the set {fa} in H* where for each a fa(x) = (x, ea)

for all x E H.

Now the mapping ea H fa is one-to-one and onto and II fa II =II ea II = I for all a. But H* is an inner product space with inner product (fx,

fy)

=(y, x)

for all x,y E H,

(fw f~) = (e~, ea) = 0

so

for a*'~

and {fa} is an orthonormal set in H*. If for f E H*, (f, fa)

=0 for all a

then by the Riesz Representation Theorem 5.2.1 there

exists an e E H, II e II = II f II such that f(x) = (x, e) so

(e, ea)

=0

for all x E H for all a.

But {eal is a maximal orthonormal set so e

= 0.

Therefore, f

= 0 and {fal

is a maximal

orthonormal set for H*. We conclude that Hand H* have the same orthogonal dimension and so are isometrically isomorphic. D 5.3 Infinite dimensional sequence spaces For an infinite dimensional normed linear space (X, 11·11) we saw in Remark 4.10.2 that the dual space X* is a proper linear subspace of X#. Its size as well as its norm is determined by the norm II· II on X. We will see how we have to take this into account as we modify the procedure followed with the finite dimensional examples in Section 5.1. 5.3.1 Example. {e 1, e 2 ,

••. ,

Consider the normed linear space (c 0 , 11·11 00 ) with Schauder basis

ek, ... } where ek

Any x = {A. 1,

= {0, ... , 0,1,0, ... }

for each kEN.

kth place can be represented as an infinite series in (c 0 , 11·11 00 ),

A.z, ... , Ak, ... } E c0 x = L A.kek, (see Example 1.25.10(iii)). k=l

98

Spaces of continuous linear mappings

Consider any continuous linear functional f on (c 0 , 11·11 00 ). n

Writing sn = L Akek we have, since f is linear k=l n

L

f(sn) =

Ak f(ek),

k-1

and since f is continuous and II x-sn 11 00

~

0 as n

~

oo, then f(sn)

~

f(x) as n

~

oo ;

that is, every continuous linear functional f on (c0 , 11·11 00 ) can be represented as an infinite series of the form =

f(x) =

L

Ak f(ek) for all x

= {A. 1, A. 2 , ... , A.k•· .. } E c 0

.

k=l We show that the mapping is an isometric isomorphism of (c 0 , 11·11 00 )* onto (1 1, 11·11 1). We begin by showing that {f(e 1), f(e 2),

•.. ,

f(ek), ... } E 1 ~·

Given n EN, consider the element x0 = {A~ , A~ , ... , A~ , ... } E c 0 where if f(ek) of. 0 and I

~ k ~ n }·

otherwise Now II x0 lloo ~ I and n

L

I f(ek) I~ II f 1111 x0 11 00 ~II f II. k=l But this is true for all n EN, so I I f(ek) I< oo; that is, f(x 0 ) =

{f(e 1), f(e 2 ),

... ,

f(ek), ... } E1 1•

But this also implies that

L

I f(ek) I ~II f II.

k=l Now

I f(x) I ~

=

L

I~ II f(ek) I

k=l

~(f

k=l

so that

II f II ~

lf(ek)l)sup{IA.kl:kEN}=(f lf(ek)l)llxlloo k=l

L

I f( ek) I

k=l =

and therefore,

II f II=

L

I f(ek) I.

k=l This tells us that the mapping f H

{f(e 1), f(e 2 ), ... , f(ek), ... }

from (c 0 , ll·lloo)* into (1 1, 11·11 1) is norm preserving. We now show that the mapping is onto. For any {a 1, a 2 , . . . , ak, ... } E 1 1 and x

= {A. 1,

Az, ... , Ak, ... } Ec 0 we have that

the series IA.kak is absolutely convergent. So the linear functional f defined on c 0 by

99

§5. The shape of the dual

f(x)

=

r A.kii

k=l

r I a.k I when II

satisfies I f(x) I :5:

X

llco :5: I' and so is continuous.

k=l

It is clear that the mapping f H { f(e 1), f(e 2), ... , f(ek), ... } it is an isometric isomorphism of (c 0 , ll·llco)* onto (1 1, 11·11 1). that is linear so we conclude Being an isomorphism onto 1 1 implies that every continuous linear functional f on (c0 , ll·llco) has the form for some {a. I' a. 2, ... , a.k, ... } E 1 1 where x

= {A.p A.2, ... , A.k,

... } E c 0 .

0

5.3.2 Remark. We note that since (1p 11·11 1) is isometrically isomorphic to a dual space

0

then it follows from Corollary 4. 10.3 that (1 1, 11·11 1) is complete.

5.3.3 Example. Given I < p < co consider the normed linear space (1p, IHip) with the same sequence {e 1, e 2, ... , ek, ... } in 1p as in Example 5.3.1. For any x {A. 1 , A.2, ... , A.k, ... } E 1p we have that

=

II

X-

r n

A.kek liP=

(

r~

I A.k IP

)I/

p --7

0

as n

--7

co since x E 1P ;

k=n+l

k=l that is,

So the sequence {e 1, e 2, ... ,ek, ... } is also a Schauder basis for (1p, IHip). Consider any continuous linear functional f on (1P, II·IIP). n

Writing s 0

=L

A.kek we have, since f is linear

k=l

n

f(sn) =

r ~ f(ek) k=l

and since f is continuous and II x-s0 liP --7 0 as n --7 co, then f(s 0 )

--7

f(x) as n --7 co;

that is, every continuous linear functional f on (1p, II· lip) can be represented as an infinite series of the form

= L Ak f(ek)

for all X= {AI, A2, ... , Ak, ... } k=l We show that the mapping f H { f(e 1), f(e 2), ... , f(ek), ... } f(x)

is an isometric isomorphism of (1p, ll·llp)* o_nto (1q, ll·llq)

where~ + ~ =

We begin by showing that {f(e 1), f(e 2), ... , f(ek), ... }

E

1q.

E

I.

1p·

Spaces of continuous linear mappings

100

Given n EN, consider the element x0 ={A.~, A.~, ... , A.~, ... } E .tq where

A.~

f(ek) I f(ek) lq- 2

if f(ek) *- 0 and I

0

(fk=l I f(ek) lq )''P and (~,I f(ek) lq )"q ~II f II

Now II x0 liP= so that

~

k

~

n }

otherwise

f(x 0) =

f I f(ek) lq ~II f 1111 x0 liP

k=l

since I

-i

=

~.

But this is true for all n EN, so :L I f(ek) lq < oo; that is, { f(e,), f(e2), ... , f(ek), ... } E .tq· But this also implies that

(k=lf I f(ek) lq )''q ~II fll. Now

I f(x) I~

=

I

I A.k II f(ek) I

k=l

~

(f I

A.k

iP

k=l

)''P (fk=l I f(ek) lq )''q by Holder's inequality

so that

and therefore This tells us that the mapping f H {f(e 1), f(e 2 ),

... ,

f(ek), ... }

from (.tp, ll·llp)* into (.tq, 11·11) is norm preserving. We now show that the mapping is onto. For any {a 1, a 2 , . . . , ak, ... } E .tp we deduce from Holder's inequality that the series :LA.kiik is absolutely convergent. So the linear function f defined on .tp by f(x) =

I

A.kiik

k=l satisfies I f(x) I ~ (

f

I a k lq

k=l

)''q when II x liP~ I, and so is continuous.

It is clear that the mapping fH {f(e 1), f(e 2 ), ... , f(ek), ... } is a linear mapping so we conclude that it is an isometric isomorphism of (.tp, II· lip)* onto (.tq, ll·llq).

101

§5. The shape of the dual

Being an isomorphism onto ..eq implies that every continuous linear functional f on (..eP'II·IIp) has the form

f(x) =

L

A.kak

for some {up a 2 ,

... ,

ak, ... } E ..eq

k=l

5.3.4 Remarks. Since every (..ep, ll·llp) space where 1 < p < oo is isometrically

isomorphic to a dual space (..eq, ll·llq)* where

i~ +

= 1, we deduce from Corollary 4.10.3

that all the (..ep, ll·llp) spaces where I < p < oo are complete. When p = 2 we have classical Hilbert sequence space (..e 2 , 11·11 2) and the mapping f

H

{f(e 1), f(e 2 ),

... ,

f(ek), ... }

is an isometric isomorphism of (..e 2 , 11·11 2 ) onto its own dual (..e 2 , 11·11 2 )*; (see Remark

D

5.2.4).

5.4

The Banach Space (C [a,b], ll·lloo) The form of the continuous linear functionals on the function space (C[a.b], ll·lloo) is

given in another important representation theorem due to F. Riesz. This result is significant in the development of functional analysis because of its generalisations and applications. We will confine ourself to the real Banach space (C[a,b], ll·lloo) of continuous real functions [a,b]. The representation theorem shows that a continuous linear functional on (C[a,b], ll·lloo) can be represented as a Riemann-Stieltjes integral. We develop background theory sufficient to explain the theorem and refer the reader to the text T.M. Apostol, Mathematical Analysis, Addison Wesley, 1957, for a standard treatment of the theory of this integral. 5.4.1 Definitions. Given bounded real functions f and

a on [a,b] and a partition P of

[a,b] where a= t0 < t 1 < ... < t0 = b a Riemann-Stieltjes sum is the real number n

S(P, f, a)=

I

f(~k) (a(tk)- a(tk_ 1))

k=l

where tk-I :s; ~k :s; tk for each k

E

{I, 2, ... , n}.

We say that f is Riemann-Stieltjes integrable with respect to a if there exists a real number I and given£> 0 there exists a partition P£ of [a,b] such that

I S(P, f,

a)- I I < £

for all partitions P, refinements of P£.

In this case I is denoted by b

I=

Jf(t) da(t)

and is called the Riemann-Stieltjes integral off with respect to a.

102

Spaces of continuous linear mappings

5.4.2 Remarks. (i)

We note that if o:(t) = t for all t E [a,b] then S(P,f,o:) = S(P,f) the usual Riemann sum

for f with respect to P. Further, f is Riemann integrable on [a,b] if and only iff is Riemann-Stieltjes integrable with respect to such an o: on [a,b]. (ii)

If f is Riemann-Stieltjes integrable with respect to an o: which has a continuous

derivative o:' on [a,b], then fo:' is Riemann integrable and b

b

f f(t) do:(t) = f f(t) o:'(t) dt ; a

a

(see T.M. Apostol, Mathematical Analysis, p. 197). (iii) Clearly, the following elementary linearity properties hold. (a) If f and g are Riemann-Stieltjes integrable with respect to o: on [a,b] then Af + f..Lg is also Riemann-Stieltjes integrable with respect too: on [a,b] for any A,f..L b

b

E

lR and

b

f (Af+J..Lg)(t) do:(t) =A f f(t) do:(t) + f..L f g(t) do:(t) a

a

a

(b) Iff is Riemann-Stieltjes integrable with respect to both o: and~ on [a,b] then f is Riemann-Stieltjes integrable with respect to b

AO:+f..L~ b

on [a,b] for any A,f..L E lR and b

f f(t) d(AO:+f..L~)(t) =A f f(d) do:(t) + f..L f f(t) d~(t) a

a

a

(See T.M. Apostol, Mathematical Analysis, p. 193). (iv) For our purposes the following existence theorem is important. For any continuous function f and bounded monotone increasing function o: on [a,b], f is Riemann-Stieltjes integrable on [a,b]; (see T.M. Apostol, Mathematical Analysis, p. 211).

0

Among the continuous linear functionals on (t:[a,b], ll·lloo) we draw attention to the positive linear functionals. 5.4.3 Definitions. We define the set t;+[a,b] = {f E t;[a,b]: f(t)::::: 0 for all t E [a,b]} and the partial order relation on t:[a,b] g:::; f if f- g

E

t;+[a,b].

A linear functional F on t; [a,b] is said to be a positivelinear functional if F(f)::::: 0 for all f E t;+[a,b]. It is remarkable that although such positive linear functionals are defined by purely algebraic properties, they are always continuous. 5.4.4 Lemma. Every positive linear functional F on (t:[a,b], 11·11 00 ) is continuous and F(l)=IIFII.

103

§5. The shape of the dual

Proof. Clearly for every f E C[a,b], II f II 1 ± f ~ 0 and so II f II F( I)

± F(f) = F(ll f II 1 ± f) ~ 0.

Since F(l) ~ 0 we have I F(f) I :0::: II f II F( I) and so F is continuous and II F II :0: : F(l). On the other hand since II 1 II= 1 we have F(l) :0::: II F II.

D

The continuous linear functionals on (C [a,b], ll·lloo) can be expressed as the difference of two positive linear functionals. 5.4.5 Lemma. Given a continuous linear functional F on (C [a,b], 11·11 00 ) , there exist

positive linear functionals p+ andF- on C[a,b] such that F =F+- F-. Proof. We define the real functional F+ on t;+[a,b] as follows. For f E t;+[a,b] put F+(f) Since F(O)

= sup{F(g): 0:0::: g :0::: f}.

=0 we have F+(f) ~ 0.

But also since F is a continuous linear functional F+(f) :0::: II F II II f lloo. Clearly F+(Af) = A.F+(f) for all A~ 0. We show that F+ is additive on t;+[a,b]. For f 1h E t;+[a,b], and 0:0::: g 1 :0: : f 1 and 0:0::: g 2 :0::: f 2 we have 0:0::: g 1+g 2 :0::: f 1+f2 and so F+(f 1+f2 ) ~ F(g 1+g 2 ) = F(g 1) + F(g 2 ). Therefore,

F+(f 1+f2)

~

F+(f 1) + F+(f2 ).

Conversely, for 0 :0::: g :0::: f 1+f2 we have 0 :0::: min {f 1, g) :0::: f 1 and 0 :0: : g- min {f1, g) :0::: f2 so that

F+(f 1+f2) = sup{F(g): 0:0::: g :0::: f 1+f2 } :0: : sup{ F(min {f 1,g))}+ sup{F(g-min{f 1,g))} :0::: F+(f 1) + F+(f2 ).

Now F+ can be extended to a linear functional on C[a,b] as follows. For f E C[a,b] we consider f+= lfl 2+ f and f-= lf1 2- f and note that r+ and f-are in t;+[a,b] and f = f+- f-. So we define F+ on C[a,b] by F+(f) =F+(f~- F+(f). We note that -f = f-- f+ so F+(-f) = F+(f ) - F+(f~ =- P(f) and we conclude that F+ is linear. We define the linear function F- on C[a,b] by F-(f) = F+(f) - F(f). Then F- is also continuous. But also since F+ is a positive linear functional F+(f) ~ F(f) for all f E t;+[a,b], so

for all f E t;+[a,b];

104

Spaces of continuous linear mappings

that is, F- is also a positive linear functional.

D

We establish the representation of the continuous linear functionals on ('C[a,b], 11·11 00 ) by first finding a representation of the positive linear functionals on 'C[a,b].

5.4.6 Theorem. For a positive linear functional F on the real Banach space ('C[a,b ],11·11 00) , there exists a bounded monotone increasing function a on [a,b] such that b

F(f) =

Jf(t) d.a(t)

for all f E 'C[a,b].

Proof. We proceed to define the function a on [a,b]. Given t E (a,b) and n E l1:l sufficiently large we define the function
I

I - n(x-t)

t

a

~

x

~

~

x

~

t +

t

1

}

~

+l<x~b

0

n

a

t

+ 1/n

b

Figure 8. The graph of n

0 ~

~

$1,0 (x)

~

I

for all x E [a,b].

Since F is a positive linear functional, the sequence { F($ 1, 0 )} is a decreasing sequence of real numbers bounded below by 0 and so is convergent to a real number which we denote by a(t). Now for a< t 1 < t2
E

l1:l we have 0 ~ $11 ,n(x)

~

$12 ,0 (x)

~

I for all x

E

[a,b] and

therefore and so a(t 1)

~

a(t 2).

We define a(a) = 0 and a(b) = F(l) and then

a is a monotone increasing function on [a,b].

Given a continuous function f on [a,b], then f is uniformly continuous on [a,b], (see AMS §8). So given E > 0 there exists a 8 > 0 such that

105

§5. The shape of the dual

I f(t 1)- f(t 2 ) I< e for t 1,t2 e [a,b] when I t 1-t2 1< 8. Now f is Riemann-Stieltjes integrable with respect to a so there exists a partition Pe of [a,b] such that b

I S(P, f, a) -

Jf(t) da(t) I < e a

for all partitions P, refinements of Pe. Consider a partition P of [a,b], a= t0 < t 1 < ... < tm = b which is a refinement of Pe such that max { I tk-tk-l I : k e { 1, 2, ... , m} } < ~· Choose n e N sufficiently large that

~ < min{l tk-tk-l I: k e {1, 2, ... , m} }. Then only consecutive intervals of [to, tl +

I .... [tm-1' tml n1 ], .... [tk-1' tk + n].

may have points in common. For each k e { 1, 2, . . . , m}. the decreasing sequence of real numbers { F( 1k,n)} converges to a(tk) and so we may choose n EN sufficiently large that a(tk)::;; F(tk,n)::;; a(tk) +

e m

for all k

E {

1, 2, ...• m}.

a

b

We now define the function f0 on [a,b] by m

fo(x) = f(tl) tl,n (x) +

L

f(tk) ( tk,n - tk-1 ,n)(x) .

k=2

A point x e [a,b] belongs to one or two of the intervals from(*). If x belongs to only one interval then either to::;; x < t 1 and f0(x) =f(tJ) 1 or for some k E { 1, 2, ... , m}. tk-l + <X::;; tk and f0(x) = f(tk).

n

Then

I f(x) - f0(x) I < e.

106

Spaces of continuous linear mappings

If x belongs to two intervals then for some k tk ~X< tk +

I

n and so

E { 1,

2, ... , m-1}

fo(X) = f(tk) (tk,n- tk-l•n)Cx) + f(tk+l)(tk+l•n- tk,n)(x).

From the definition of the functions t,n we have f 0(x) = f(tk)(1-n(x-tk)) + f(tk+l) n(x-tk). Since I x-tk I< o and I x-tk+l I< owe have I f(x)-f(tk) I< E and I f(x)-f(tk+l) I< E and I f(x)-f0 (x) I~ I f(x)-f(tk) I (1-n(x-tk)) +I f(x)-f(tk+l) I n(x-tk) <E. We conclude that for all x E [a,b] I f(x)-f0 (x) I < E and so II f-f0 lloo <E.

t3

Figure 10.

+ 1/n

The graphs off and f 0 .

Now since F is a positive linear functional from Lemma 5.4.4 we have I F(f) - F(f0 ) I ~ II F II E. From the definition of f0 we have m

F(f0) = f(t 1) FCt 1,n) +

L

f(tk) (FCtk,n)- FCtk-l•n)) ·

k=2

But since for each k

E {

1, 2, ... , m}, 0

~

F(<j> 1k,n)- a(tk) <Elm, we have

m

I F(f0 ) -

L

f(tk) (a(tk)- a(tk_ 1)) I< 2 E II f II

k=l

using the fact that a(t0) = 0. m

But

L

f(tk) (a(tk)- a(tk_ 1)) is a Riemann-Stieltjes sum S(P, f, a) for f with respect to a k=l where Pis a partition which is a refinement of Pe. Therefore,

§5. The shape of the dual

107

I F(f)- S(P, f, a) I :-:; I F(f)- F(fo) I + I F(fo)- S(P, f, a) I < (II F II + 2 II f II) e and we conclude that b

F(f) =

Jf(t) da(t) .

D

We now extend this result to the representation of continuous linear functionals on the space. To do so we introduce the following ideas.

5.4.7 Definitions. Given a real function a on [a,b] and a partition P of [a,b], a= t0 < t 1 < ... < t 0 = b we write n

V(P, a)=

I.

I a(tk)- a(tk_ 1) I

k=l

The function a is said to be of bounded variation if the set of real numbers {V(P, a): P any partition of [a,b]} is bounded. If a is of bounded variation V(a) = sup{V(P, a): P any partition of [a,bl} is called the total variation of a on [a,b].

5.4.8 Remark. Clearly any bounded monotone function on [a,b] is of bounded variation. But further, functions of bounded variation have the following characterisation by monotone functions. A function a on [a,b] is of bounded variation if and only if a can be expressed as the difference of two monotone increasing functions; (see T.M. Apostol, Mathematical Analysis, p. 168). D We express our general representation theorem in terms of functions of bounded variation.

5.4.9. The Riesz Representation Theorem for ('C [a,b], 11·11 00 ) . For any continuous linear functional F on a real Banach space ('C[a.b], 11·11 00 ) , there exists a function a of bounded variation on [a,b] such that b

F(f) = and

Jf(t) da(t)

V(a) =II F II.

for all f E 'C[a,b],

108

Spaces of continuous linear mappings

Proof. From Lemma 5.4.5 there exist positive linear functionals F+ and F- on t; [a,b] such that F = F+ - F -0 By Theorem 5.406 there exist bounded monotone increasing functions a+ and a- on [a,b] such that b

F+(f) =

b

Jf(t) da+(t)

and F -(f)=

Jf(t) da-(t)

for all f

E

t;[a,b].

b

Then

Jf(t) d(a+- a-)

F(f) = P(f)- F -(f)=

by Remark 5.4o2(iii)(b)o But from Remark 5.408, a= a+- a- is of bounded variation and we have b

J f(t) da(t)

F(f) =

for all f E t; [a,b]o

a

Since F has this representation, given f [a,b], a= t0 < t 1 < o o o< tn

E

t; [a,b] and £ > 0 there exists a partition P of

=b such that n

I F(f)I F(f) I<£+

I.

f(tk) (a(tk)- a(tk_ 1)) I< Eo

k=l

n

So

I.

n

I f(tk) II a(tk)- a(tk_ 1) I::;£+ II f lloo

k=l =£+II fll 00 V(a)o

I.

I a(tk)- a(tk_ 1) I

k=l

It follows that I F(f) I::; V(a) II f lloo

for all f

E

t;[a,b]

II F II::; V(a)o

so

Following the pattern of the proof of Theorem 5.406, given£> 0 consider a partition

P of [a,b], a= t0 < t 1 < 0 0 0 < tm = b such that m

I V(a)-

I.

I a(tk)- a(tk_ 1) I<£ o k=l Since the sequence { F+(tk• n)} converges to a+(tk) and the sequence { F -(tk , n)} converges to a-(tk), the sequence { F(tk,n)} converges to a(tk)o So we choose n EN sufficiently large that I F(tk,n)- a(tk) I< elm for all k

E

{I, 2, 0 0 0, m}o

Consider the continuous function f0* on [a, b] defined by

*

fo(x) = Et,,n(x) +

m

I. Ek(tk,n- tk-I•n)(x)

k=2 where if = -1

*

Now F(f0 ) = e 1F(t 1,n) +

if m

I. Ek(F(tk,n)- F(tk-l•n))o

k=2

§5. The shape of the dual

So we have

*

I F(f0)-

*

I F(f0 ) -

then

m

L £k

109

(a(tk)- a(tk_ 1)) I< 2£ using the fact that a(t0) = 0,

k=l

m

L I (a(tk)- a(tk_ 1)) I< 2£ k=l

and we deduce that I F(f0*)- V(a) I< 3£. Therefore

V(a)-3£
and we conclude that V(a) =II F II.

D

5.4.10 Remarks. (i)

There is an alternative method of proving Theorem 5.4.9 using the Hahn-Banach

Theorem of the next chapter. However, the Hahn-Banach Theorem establishes properties for normed linear spaces in general and its proof depends on the Axiom of Choice. Properties of particular spaces such as that given in Theorem 5.4.9 can be developed directly. We have chosen to present the direct proof even at the cost of a slight increase in technicality. For the alternative method the reader should consult A.L. Brown and A. Page,

Elements of Functional Analysis, Van Nostrand Reinhold, 1970, p. 202. (ii)

In earlier examples where we determine the shape of the dual we find an isometric

isomorph of the dual. Theorem 5.4.9 does not go so far as this. Given a continuous linear function F on (C [a,b], ll·lloo) there is no unique function of bounded variation a on [a,b] such that b

F(f) =

Jf(t) da(t) .

We can add an arbitrary constant to a and we can alter a at its points of discontinuity in (a,b) without altering the value of the integral. A function a of bounded variation on [a,b] is said to be normalised if a( a) = 0 and a is continuous on the right at all points of (a,b ). The set ')1, ~ '\J [a,b] of such normalised functions of bounded variation is a Banach space with norm II a II = V(a). It can be shown that (C [a,b], ll·lloo)* is isometrically isomorphic to

('Jl, ~ '\J [a,b], 11·11). (See A.E. Taylor and D.C. Lay, Introduction to Functional Analysis, Kreiger, 1980, p. 150.) (iii) It is not difficult to extend Theorem 5.4.9 to complex Banach spaces (C [a,b], 11·11 00 ) .

D

II 0

5.5

I.

Spaces of continuous linear mappings

EXERCISES

Prove that the dual space of (
cAf. A.~ •... ' A~)

where

A.~ = ~~~:~~~

if f(ek)

* 0 and k is the first member of {I, 2, ... ,n}

}

where lf(ek)l = max { lf(ek)l : k E {I, 2, ... ,n}} = 0

otherwise

we have f(x 0 ) =II f 1111 xo 11 1.

2.

Consider a Hilbert space with a maximal orthonormal sequence {en}. Prove that for any f E H*, II f II = I, the series LJ(en) en is convergent to some x E H and II x II = I and f(x) = I.

3.

Prove that the dual space of (1 1, 11·11 1) is isometrically isomorphic to (m, 11·11 00 ) by considering the sequence {e 1, e 2,

... ,

ek, ... } in .t 1 where

ek = {0, ... , 0, I, 0, ... } for each k EN, kth place and for every continuous linear functional f on .t 1, and any n E N, an element x0 = {A.~, A.~, ... , A.~} E1 1 where

A.~ = ~~~:~~~

if f(ek)

* 0 and k is the.first member of N

where }·

lf(ek)l =max { lf(ej)l: J E {I, 2, ... ,n}} = 0

otherwise

Deduce that every continuous Iinearfunctional f on (.t 1, 11·11 1) has the form f(x) =

I. A.kak

for some {a 1, a 2 , ... , ak, ... } Em

k=l

where x

4.

(i)

= {A- 1, A. 2, ... , A.k, ... } E .t,.

Using the representation of continuous linear functionals on (c 0 , ll·lloo) and (.t 1, 11·11 1), exhibit in each case a continuous linear functional which

(ii)

does not attain its norm on the closed unit ball. Using the representation of continuous linear functionals on (.tp, ll·llp) where I < p < ""• prove that every continuous linear functional on such a space attains its norm on the closed unit ball.

§5. The shape of the dual

5.

III

Determine the form of the dual of the normed linear spaces (E0, 11·11 00) , (E0 , 11·11 1) and (E0 , 11·11 2). Exhibit in each case a continuous linear functional on the space which does not attain its norm on the closed unit ball.

6.

The real Banach space (c 0 , 11·11 00) is a proper closed linear subspace of the Banach space (c, 11·11 00 ). However, by considering the mapping y H f of . e 1 into c* defined by

L Ilk A.k-1>

f(x) =Ill lim A0 + n--->=

k=l

where x"" {A- 1, A2, ... , A.k, ... } Ec andy"" {f.1 1, ll2• ... , Ilk• ... } E ..e,, or otherwise, prove that the dual spaces (c 0 , ll·lloo)* and (c, 11·11=)* are isometrically isomorphic.

7.

In a Hilbert space H, given z E H consider the continuous linear functional fz defined by fz(x) = (x, z)

for all x E H.

(i)

Prove that the dual space H* is a Hilbert space with inner product

(ii)

Hence, or otherwise, prove that H and H**, the dual of H*, are

(fx, fz) = (z, x). isometrically isomorphic.

8.

(i)

Prove that (E0 ,11·11oo)* is isometrically isomorphic to (c, 11·11=)*.

(ii)

Prove that ('f'[a,b], ll·lloo)* is isometrically isomorphic to (t'>[a,b], 11·11=)* and deduce that ('f'[a,b], ll·lloo)* is a Hilbert space. (Hint: Use Weierstrass' Approximation Theorem, AMS §9.)

9.

(i)

Given a bounded monotone increasing function a on [a,b] prove that the function F on (t'>[a,b ], 11·11 00 ) defined by b

F(f) =

Jf(t) da(t) a

is a positive linear functional and II F II= V(a). (ii)

Given a function a of bounded variation on [a,b] prove that the function F on (t: [a,b], ll·lloo) defined by b

F(f) =

Jf(t) da(t) a

is a continuous linear functional and II F II= V(a).

112

Spaces of continuous linear mappings

(iii)

Given a continuous function a on [a,b], prove that the function F on ('C [a,b], ll·lloo) defined by b

Jf(t) a(t) dt

F(f) =

a

b

is a continuous linear functional and II F II

= JI a(t) I dt. a

(iv)

For the linear functionals F 1 and F2 on ('C[a,b], ll·lloo) where I

F1 =

I

Jt2 f(t) dt and F2 = J(I -2t) f(t) dt 0

0

prove that both F 1 and F 2 are continuous and determine II F 1 II and II F 2 11.

III.

THE EXISTENCE FUNCTIONALS

OF

CONTINUOUS

LINEAR

Given any linear space X, it follows from the existence of a Hamel basis for X and the fact that any linear functional is determined by its values on the Hamel basis, that the algebraic dual X# is generally a "substantial" space. We know, from Remark 4.10.2, that for an infinite dimensional normed linear space (X, 11·11), the dual X* is a proper linear subspace of X#. For the development of a theory of normed linear spaces in general, quite apart from particular examples or classes of examples, it is important to know that given any normed linear space (X, 11·11), its dual X* is also "substantial enough" and by this we mean that we have a dual which generalises sufficiently the properties we are accustomed to associate with the dual of a Euclidean space or indeed, with the duals of the familiar example spaces. We now use the Axiom of Choice in the form of Zorn's Lemma, (see Appendix A. I), to prove the Hahn-Banach Theorem, an existence theorem which is crucial for the development of our general theory. The theorem assures us that for any nontrivial normed linear space there is always an adequate supply of continuous linear functionalii. The immediate application of this result is in the study of the structure of the second dual X** of a normed linear space (X, II· II) and of the relation between the space X and its duals X* and X**. §6. THE HAHN-BANACH THEOREM There are several forms of the Hahn-Banach Theorem, some more general than others. The form sufficient for our purpose asserts the existence of norm preserving extensions of continuous linear functionals from a linear subspace of a normed linear space to the whole space. It is an interesting exercise to generalise this form, (see Exercise 6.13.5), but we will make no use of the generalisation. In fact we mainly use our restricted form of the theorem through its corollaries. The proof of the Hahn-Banach Theorem is an application of Zorn's Lemma but we isolate the computational first step of that proof in the following lemma.

114

The existence of continuous linear functionals

6.1 Lemma. Given a normed linear space (X, 11·11), consider a continuous linear

functional f defined on a proper linear subspace M of X. Given x0 EX\ M, f can be extended to a continuous linear functional f on M0 = sp {x0 ,M} such that II f0 II on M0 is equal to II f II on M.

Proof. We may suppose that II f II = I.

Case 1: X a real linear space Since M is of codimension I in M0 , any x E M0 can be represented uniquely in the form where A. E lR andy EM .. x = A.x 0 + y For fo to be a linear extension off on M we must have fo(Y) = f(y)

for ally EM

f0 (x) = Af0(x0 ) + fo(Y)

and

= Af0 (x 0) + f(y)

and we are free to choose a value for f0(x 0). We show that a value for f0(x 0) can be chosen such that f0 is continuous on M0 and II f0 II= I. Now for f0 continuous on M0 , II f0 II = sup {I f(x) I : II x II ::; I, x E M0 }

~sup{ I f(y) I: II y II::; I, y EM}= II fil =I. But fo is continuous and II f0 II = I if I f0(x) I::; II x II

for all x E M0 .

Now this will happen if the value f0 (x0) satisfies - II A.x 0+y II ::; A.f0(x 0) + f(y)::; II A.x 0+y II; that is, - f (y/A-)- II x0 + y/A. II::; f 0(x0 )::;- f (y/A-) +II x0 + y!A. II

when A."# 0.

(*)

(Notice that when A.< 0, we have I I - f (y/A.) +- II h 0 + y II ::; f0(x 0)::;- f (y/A.) --II A.x 0 + y II A, A, - f (y/A-) -II - (x 0 + y!A.) II ::; f0 (x 0)

::; -

f (y/A.) +II - (x 0 + y/A.) II .)

Now for any tw0 elements y 1, y2 EM, we have But Therefore

f(y 2)- f(yJ) = f(YrYI)::; II YrYI II. II YrYI II::; II x0+y 2 II+ II x0+y 1 II. -f(y 1) -II x0+y 1 II::;- f(y2) +II xo+Y2II.

So

sup{-f(y)-11 x0+y II: y EM} and inf{- f(y) +II x0+y II : y EM}

and

both exist

sup{- f(y) -II x0+y II: y EM} ::; inf{- f(y) +II x0+y II : y EM} .

So it is possible to choose a value for f0 (x 0) between these two and then that value will satisfy inequality (*) which gives II f0 II = I as required.

115

§6. The Hahn-Banach Theorem

Case 2: X a complex linear space We have seen from Theorems 4.10.15 and 4.10.16 that a complex normed linear space (X, 11·11) can be regarded as a real normed linear space (XJR, 11·11) and there is a one-to-one norm preserving correspondence f H fiR of X* onto (XJR)* where f and fiR are related by f(x) =fiR (x)- i fiR (ix)

for x EX.

Now M 0 "' sp { x 0 , M} as a real normed linear space is M 0 JR = sp { x0 , ix 0 , M lR } . From Case I applied twice fiR on MJR can be extended to a continuous linear functional f0 JR on M 0 1R such that II fOJR II = I. But then f0 given by for all x EM 0 extends f on M as a continuous linear functional f0 on M 0 such that II f0 II = I.

0

The Hahn-Banach Theorem uses Zorn's Lemma to carry extensions, like those achieved in Lemma 6.1 for one extra dimension, to the whole space.

6.2 The Hahn-Banach Theorem. For a normed linear space (X, 11·11), consider a continuous linear functional f defined on a proper linear subspace M of X. Then f can be extended as a continuous linear functional fo on X such that II f0 II on X is equal to II f II on M.

Proof. Consider the set ~ of all possible extensions of f as norm preserving continuous linear functionals on linear subspaces containing M, with partial order relation :::; on ~: f 1 :::_ f2 if dom f 1 ~ dom f2 and f 2 is an extension of f 1. From Lemma 6.1, we see that ~ is nonempty. Suppose that lfa} is a totally ordered subset of~. We show that {fa} has an upper bound in ~. For x E U dom fw there exists an a 0 such that x

a

E

dom fa 0 , so we define a

functional fl on U dom fa by

a

f 1(x) =fa (x), 0

and since {fa} is a totally ordered set, f 1 is well defined. Now since {fa} is a totally ordered subset, dom f 1 = U dom fa is a linear subspace of X

a

containing M. But also from the definition we see that f 1 is a linear extension of fa for all a. As such an extension, sup {I f(x) I : II x II :<:; I, x E dom f 1 } ~II f II. If there exists an x E dom fl where II x II:<:; I and fl(x) >II f II then, since there exists an CXo such that x

E

dom fa , we would have fa (x) > II f II which contradicts the defining 0

0

property for elements of~ that II fa II = II f II. So we conclude that II f 1 II = II f II and then 0

f 1 E~.

116

The existence of continuous linear functionals

Since fa::; f 1 for all a, we have that f 1 is an upper bound for {fa} . It now follows from Zorn's Lemma that

~

has a maximal element f0 .

Now dom f0 = X, for otherwise by Lemma 6.1, for x0 EX\ dom f0 we could extend f0 as a norm preserving continuous linear functional on sp { x0, dom f0 } but then fo would not D be maximal in ~. The Hahn-Banach Theorem has the following important corollary which guarantees the existence of certain continuous linear functionals which attain their norm on the closed unit ball, or considered geometrically guarantees the existence of tangent hyperplanes to the closed unit ball. It is this first corollary rather than the theorem itself which we will use in the subsequent development of our analysis. Now the norm of a continuous linear functional f on a normed linear space (X, 11·11) is given by II f II =sup {I f(x) I : II x II::; I}. Given a continuous linear functional f there does not necessarily exist an x EX, II x II = I such that f(x) = II f II. However, the corollary tells us that given an x EX, II x II = I there does exist a continuous linear functional f such that f(x) = II f II. 6.3 Corollary. Given a normed linear space (X, 11·11), for each non-zero x0 EX, there exists a non-zero continuous linear function f0 on X such that f0(x 0) = II f0 1111 x0 II.

Proof. Consider the one dimensional linear subspace M"' sp{x 0 }. Define the functional f onMby f(A.x 0 ) = A. II x0 II

for scalar A..

Then f is a continuous linear functional on M such that f(x 0) =II x0 II and II f II = I. By the Hahn-Banach Theorem 6.2, f can be extended as a continuous linear functional f0 on X such that II f0 II = I. Then f0 on X satisfies f0(x 0) =II f0 1111 x0 II.

D

Let us explore the geometrical interpretation of this result. 6.4 Remark. Corollary 6.3 implies that given x0 EX, II x0 II = I there exists a continuous linear functional f0 on X, II f0 II = I such that f0 (x 0) = I. Geometrically this says that for each x0 EX on the boundary of the closed unit ball B [0; I] there exists a closed tangent hyperplane Mr0 "' { x EX: f0(x) = I} to B[O; I] at x0 where II f0 II= I. Clearly, I

d(O, Mr0) = ffTol1 = I = II x0 II and x0 E Mro· This is the sort of geometrical property we assume quite naturally in Euclidean space.

D

We now show how Corollary 6.3 implies that X* is "substantial" in another sense.

§6. The Hahn-Banach Theorem

117

6.5 Definitions. (i) Given a linear space X, a linear subspace Y of xlf is said to be total on X if for for x EX, f(x) = 0 for all fEY implies that x = 0; or contrapositively, for x "# 0 there exists an fEY such that f(x)

"#

0.

(ii) Given a linear space X, a linear subspace Y of X# is said to separate the points of X if for x,y EX, x "# y there exists an fEY such that f(x) "# f(y).

Clearly, the fact that Y is a linear subspace of X# implies that these properties are equivalent.

6.6 Remark. Given a normed linear space (X, 11·11), Corollary 6.3 implies that the dual X* is total on X or equivalently separates the points of X.

D

The second corollary to the Hahn Banach Theorem implies that in general there are sufficient continuous linear functionals on a normed linear space, not only to separate points, but also to separate points from proper closed linear subspaces. 6.7 Corollary. Given a normed linear space (X, 11·11), for any proper closed linear subspace M and xo EX \ M, there exists a continuous linear functional fo on X such that I fo(M) = 0, f0(x 0) = I and II fo II= d(xo, M)

Proof. Consider M0 = sp{x0 , M). Now any x E M0 has the form x = A.x 0 + y where A. is a scalar and y EM. Define a linear functional f on M0 by f(x) = A.. Then clearly f(M) = 0 and f(x 0) = 1. Since kerf= M which is closed, then f is continuous on M 0. But also, II f II = d(0,1Mr) . However, d(O, Mr)

=inf{ II x II : f(x) = I = f(xo)} = inf{ II x II : f(x-x0) = 0} = inf{ II x0-y II : f(y) = 0} =d(x0, M).

I Sollfll=d(xo, M).

By the Hahn-Banach Theorem there exists a continuous linear functional f 0 on X an extension of f on M0 such that 1

llf0 11=11fll=d(xo, M).

D

6.8 Remark. Corollary 6.7 implies that given a set A in a normed linear space (X, 11·11), a point x0 E spA if and only if for every continuous linear functional f on X where f(A) = 0 we have f(x 0) = 0. If for every continuous linear functional f on X where f(A) = 0 we have f(x 0) = 0 then d(x 0 , spA) = 0, for otherwise by Corollary 6.7 there exists a continuous linear functional f0 on X such that f0(spA) = 0 and f0(x 0) = 1. The converse is immediate. D

118

The existence of continuous linear functionals The Hahn-Banach Theorem can be used to reveal a great deal about the relation

between a normed linear space and its dual. To illustrate this we consider separability of the spaces. The separability of a normed linear space does not necessarily imply the separability of its dual. For example the Banach space (1 1, 11·11 1) is separable but its dual is isometrically isomorphic to (m, 11·11 00 ) which is not separable (see Example 1.25.3 and Exercise 1.26.16). However, Corollary 6.7 enables us to establish the following relation. 6.9 Theorem. A normed linear space (X, 11·11) is separable if its dual (X*, 11·11) is

separable.

Proof. Consider a countable set {fn EX*: n EN} dense in (X*, 11·11). For each n EN there exists an Xn EX, II Xn II ~ I such that I f 0 (xn) I ;::>:~II fn II. Consider M = sp { x 0 : n E N }. Suppose that M *X. Then for x0 EX\ M we have from Corollary 6.7 that there exists an f0 EX* such that f0(M) = 0 and f0(x0) 0. But then

*

I

2 II fn II II f0 II

and

~I

~II

fn(Xn) I =(fn- f0)(xn) I ~II f 0 - f0 II

fn- f0 II + II f 0 II

~

3 II f 0 - f0 II for all n EN.

But this contradicts the density of {fn EX* : n EN} in (X*, 11·11). So we conclude that X= sp { xn: n EN}; that is, (X, 11·11) is separable.

D

The Hahn-Banach Theorem is useful in determining the form of the dual for subspaces and quotient spaces of a normed linear space. 6.10 Definition. For a linear subspace M of a normed linear space (X, II· II) the

annihilator of M is the subset Mj_ = {f EX* : f(x)

=0

for all x EM}.

It is evident that Mj_ is always a closed linear subspace of (X*, II· II); (see Exercise

6.!3.8(i)). 6.11 Theorem. Consider a linear subspace M of a normed linear space (X, 11·11). (i) M* is isometrically isomorphic to X*!Mj_.

(ii) /fM is a closed linear subspace, then (XIM)* is isometrically isomorphic to Mj_.

Proof. (i)

Given f EM* we have from the Hahn-Banach Theorem 6.2 that there exists an

extension f0 EX* and II f0 II

= II f II.

Consider the mapping T: M*

--7

X*/Mj_ defined by T(f)

= fo + Mj_.

§6. The Hahn-Banach Theorem

119

If fb EX* is another extension off then f0 - fb E M_j_ and so this mapping is well defined. Clearly T is linear. Since each element in fo + M_j_ is an extension off,

II f II ::;; inf {II f0+g II : gEM_!_} = II f 0+M_!_ II = II T(f) II. But also since f 0 is a norm preserving extension off,

II f II = II f 0 II ~ inf {II f 0+g II : g EM_!_} = II f 0+M_!_ II = II T(f) II. So T is an isometric isomorphism. But since the restriction of any continuous linear functional on X is a continuous linear functional on M we conclude that T is onto. (ii)

Consider the quotient mapping rc:

X~

X/M defined by rc(x)

= x+M and the

mapping T: (XIM)* ~X* defined by T(h)

= h o rc.

Clearly T is linear. But also

I T(h)(x) I = I h(x+M) I::;; II h 1111 x+M 11·::;; II h 1111 so

II T(h) II::;; II h II

X

II

for all

X

EX

for all h E (XIM)* .

Ifx EM then T(h)(x) = h(x+M) = h(M) = 0 since M is the zero of X/M. Therefore T maps (XIM)* into M_j_. Further for f E M_j_ we can define unambiguously an hE (XIM)* by h(x+M)

= f(x).

Clearly, such an h is a linear functional on XIM and

I h(x+M) I = I f(x) I = I f(x+m) I ::;; II f 1111 x+m II =II f 1111 x+M II

for all mE M

for all x EX .

So h is continuous on XIM and II h II ::;; II f II. But then T maps (X!M)* onto M_j_. Moreover, T is one-to-one because if T(h) = 0 then h(x+M) = 0 for all x EM and so h = 0. So the mapping T can be expressed as T(h)(x) and then

=h(x+M) =f(x)

II h II::;; II f II= II T(h) II

for all x EX

for all hE (X!M)*.

We conclude that

II T(h) II = II h II for all h E (X!M)* soT is an isometric isomorphism of (XIM)* onto M_j_.

D

6.12 Remark. The proofs of Theorem 6.11 can be given using the techniques of conjugate mappings presented in Chapter 5, Section 12. D

120

The existence of continuous linear functionals

6.13 EXERCISES I. A normed linear space (X, 11·11) is said to be smooth if for each x EX, II x II = I, there exists only one continuous linear functional f on X where II f II = I and f(x) = I. (i) Show that (a) (IR 3 , 11·11 1) is not smooth (b) (c 0 , ll·lloo) is not smooth.

and

(ii) Prove that Hilbert space 2.

(i)

i~

smooth.

Prove that a normed linear space (X, 11·11) is rotund if and only if for every f EX* where II f II = f(x) = f(y) = I for x,y EX where II x II =II y II = I we have x =y.

(ii) Prove that (a) if X* is smooth then X is rotund

(b) if X* is rotund then X is smooth.

and 3.

Prove that a normed linear space (X, II· II) is smooth at x EX, II x II l.f

i!ffi.

II x+Ay II- II x II

A.---70

A

.

f

eXIStS Or a

II y

E

= I if and only

X,

and if it is smooth at x then this limit is f(y) where f EX*, II f II= I and f(x) =II x II. 4.

(i)

Given a normed linear space (X, 11·11), prove that for any x EX, II x II

= sup {I f(x) I : f EX*, II f II ~ I } .

(ii) A linear subspace Y of X* is said to be nanning if

II x II

= sup {I f(x) I : fEY, II f II ~ I } .

Prove that such a subspace Y is total on X. (iii) Prove that if a linear subspace Y of X* is dense in X* then Y is norming. But give an example of a linea~ subspace Y of X* which is norming but not dense in X*. 5.

A convex functional <jl on a linear space X is a real functional on X defined by

<jl((l-A)x+Ay) ~ (1-A) <jl(x) + A<jl(y) for all x,y EX and 0 ~ A~ I. Prove the more general form of the Hahn-Banach Theorem. Consider a convex functional <jl on a linear space X, a proper linear subspace M of X and a linear functional f on M such that

Re f(x)

~
(x)

for all x EM.

Then there exists a linear functional f 0 on X an extension off on M, such that

Re f0(x) 6.

~

<jl(x)

for all x EX.

Consider a normed linear space (X, 11·11), a proper closed linear subspace M of X and x0 EX \M.

§6. The Hahn-Banach Theorem (i)

121

Prove that there exists a continuous linear functional f on X such that II f II= I, f(M) = 0 and f(x 0) = d(x 0 , M).

(ii) Prove that there exists a closed hyperplane M0 containing M such that

d(x 0, M 0) = d(x 0, M). 7.

In the Banach space (m, ll·lloo), consider m 0 the smallest closed linear subspace of (m, 11·11 00 ) containing all sequences of the form

P-1, where x (i)

= {1.. 1,

Az-AJ, A3-Az, · · · , An-An-I, · · · l

Az, A3, ... , An, ... } Em.

Show that e = {I, I, I, ... } e m 0 , and prove that there exists a continuous linear functional f* on (m, ll·lloo) such that II f* II= I, f*(e) =I and f*(m 0) = 0.

(ii) The Banach limit of a bounded sequence x

={A1, A.z, ... , An, ... } is

defined by LIM An= f*(x). Prove that

8.

(i)

(a)

LIM An= LIM An+ I·

(b)

LIM An ~ 0 if An ~ 0 for all n E M.

(c)

LIM (aAn+~fln) =a LIM An+~ LIM fln for scalars a,~ andy= {f.L 1, flz, ... , fln, ... } Em.

An ~ lim sup An A.o =lim An·

(d)

lim inf An ~ LIM

(e)

If x E c then LIM

if An E lR for all n E M.

Given a nonempty subset M of a normed linear space (X, 11·11), prove that Mj_ is a closed linear subspace of (X*, 11·11).

(ii) Given a nonempty subset N of (X*, 11·11) we define the set Nj_ = {x EX: f(x) = 0 for all fEN}. Prove that Nj_ is a closed linear subspace of (X, 11·11). (iii) Prove that (Mj_)j_ = M if and only if M is a closed linear subspace of

(X, 11·11). (iv) Prove that N,;;;;; (N j_)j_ but by considering the linear space c0 as a closed linear subspace of (m, ll·lloo) show that j_ (co j_) "' co. (v) Prove that if N is a finite dimensional linear subspace of (X*, 11·11) then N = (N j_)j_. 9.

Given a linear subspace M of a normed linear space (X, II· II), prove that for any fEX*,

d(f,Mj_)=llfiMII

and there exists an f0 E Mj_ such that II f-f0 II

= d(f, M_j_).

122

§7. The natural embedding and reflexivity

§7. THE NATURAL EMBEDDING AND REFLEXIVITY

It is Corollary 6.3 to the Hahn-Banach Theorem which enables us to develop a significant theory of dual spaces and to define the important class of reflexive spaces. 7.1 Definitions. Given a normed linear space (X, 11·11) and its dual X* with norm 11·11 defined by II f II = sup {I f(x) I : II x II ~ I}. the dual space (X*, 11·11) has its own dual (X*)* usually written X** and called the second

dual space (or second conjugate space) of (X, 11·11). We will usually denote elements of X

by x,y,z, elements of X* by f,g,h and elements of X** by F,G,H. The norm of X** is of course defined by II F II = sup {I F(f) I : II f II ~ I}.

The definition of the second dual (X**, 11·11) prompts us to enquire into its relation to the original space (X, 11·11) from which it is generated. For elements x EX and f EX*, we can consider f fixed and x varying over X as we do when we think off as a functional on X. But alternatively, we can consider x fixed and f varying over X*, and when we do this we have x acting as a functional on X*. 7.2 Theorem. Given a normed linear space (X, 11·11) and x EX, the functional~ defined by

~(f)= f(x)

for all f EX*

is a continuous linear functional on X* and~ as an element of(X**, 11·11) satisfies

11~11 =llxll. Proof. Clearly, ~is linear: ~ (f+g) = (f+g)(x) = f(x) + g(x) =~(f)+~ (g) for f, g EX* and

~(at)= uf(x) = u~ (f)

for scalar a and f EX*,

using the fact that in X* addition and multiplication by a scalar are defined pointwise. But also~ is continuous since I ~ (f) I = I f(x) I ~ II x 1111 f II

for all f EX*.

As a continuous linear functional on X* we have that~ EX**. But II~ II = sup {I f(x) I : II f II ~ I} ~II x II. However, by Corollary 6.3 applied to (X, 11·11), for x EX there exists an f EX* such that II f II = I and f(x) = II x II, so that II QII =sup {I f(x) I : II f II ~ I} ~II xll.

Therefore, II ~ II = II x II.

0

The existence of continuous linear functionals

123

Given a nonned linear space (X, II· II) and x eX the identification given in Theorem 7.2 of

Q

as an element of X** suggests that we investigate the mapping

x f--t Q of X into X**. 7.3 Theorem. Given a normed linear space (X, 11·11), the mapping x f--t Qinduced by the

definition of

Qas A

x(f) = f(x)

for all f eX*

is an isometric isomorphism ofX into X**.

Proof. The mapping x f--t Qis linear: 1\ A A AA x+y(f) = f(x+y) = f(x) + f(y) = x(f) + y(f) = (x+y)(f) for all f eX* 1\ A A x+y = x +y. so 1\ A ax(f) = f( ax) = af(x) = ax(f) for all scalar a and f eX* Also 1\ A ax= ax so using the fact that f is linear. From Theorem 7.2 we have that II QII = II x II for all x eX so we conclude that the mapping x f--t Qis an isometric isomorphism.

0

This identification of the original space (X, 11·11) as part of the second dual (X**, 11·11) suggests that we give a name to this mapping. f--t Q of X into X** induced by the definition of Qas Q(f) = f(x) for all f eX*, is called the natural

7.4 Definition. Given a normed linear space (X, 11·11), the mapping x A

embedding of X into X**. We denote by X the image of X in X** under the natural embedding.

Figure 11. The natural embedding x f--t

Qof X into X**.

Now it will be important to determine whether a normed linear space is isometrically isomorphic to its second dual under the natural embedding.

124

§7. The natural embedding and reflexivity

7.5 Definition. A normed linear space (X, 11·11) where the natural embedding x H ~ maps X onto X** is said to be reflexive. 7.6 Remark. Since a dual space is always complete we deduce that a normed linear space (X, 11·11) is isometrically isomorphic to its second dual (X**, 11·11) only if (X, 11·11) is a Banach space. So completeness is a necessary condition for a normed linear space to be 0 reflexive. But it is not a sufficient condition as is shown in the examples below. 7.7 Examples of reflexive spaces We now give some examples of classes of normed linear spaces which are reflexive. 7. 7 .I Finite dimensional normed linear spaces. A finite dimensional linear space is algebraically reflexive. Given an n-dimensionallinear space Xn we have from Theorem

4.1 0.5 that the algebraic dual X~ is n-dimensional and so the second algebraic dual X~# is also n-dimensional. Now the mapping x H ~of Xn into X~ induced by the definition of~ as

~(f)= f(x)

for all f EX~

is linear. But again from Theorem 4.1 0.5 we see that X~ is total on X 0 and this implies A

that if ~(f)= 0 for all f EX~ then x = 0, so the mapping x H ~is one-to-one. But then Xn, A

the image under this mapping, is n-dimensional so X0 = X~#· For an n-dimensional normed linear space (Xn, 11·11) we have

x:

=

X~ so it is clear that (Xn, II· II) is reflexive. 0

7.7.2 Hilbert space. From the Riesz Representation Theorem 5.2.1 we have that for any given continuous linear functional f on a Hilbert space H there exists a unique z E H such that f is of the form f(x) = (x, z)

for all x E H

and we denote this functional by fz. Now the dual space H* is itself a Hilbert space with inner product defined by (fx, fz) = (z, x) for all x,z E H; (see Exercise 5.5.7). Applying the Riesz Representation Theorem 5.2.1 to H* we have for any given continuous linear functional F on H* there exists a unique fx E H* such that F is of the form F(fz) = (fz. fx) = (x, z)

for all fz E H* for all z EH.

But

~(fz) = fz(x) = (x, z)

so

F(f) = ~(f)

that is,

F =~

and we conclude that

H = H**.

for all z E H

for all f E H* ;

A

0

The existence of continuous linear functionals

7.7.3 The (.lp, ll·llp) spaces where I < p <

oo,

125

In Example 5.3.3. we showed that

given I < p < oo the dual space (.lp, II· lip)* is isometrically isomorphic to (lq, ll·llq) where

~ + ~

= I. We showed that every continuous linear functional f on (lp, 11-ilp) can be

represented in the form 00

f(x)=L A.kilk for x={I.. 1,1.. 2 , ... ,A.k, ... }elp k=l and {11,, 112, ... , Ilk· ... } e lq. Now defining for each k eN the continuous linear functional fk on lp by fk(x) = A.k, 00

we can write

f(x) =

L

ilk fk(x)

k=l 00

f = I, ilk fk . k=l For any given F e l ** we have so

p

00

F(f) =

L

ilk F(fk) . k=l However, (.lp, 11-ilp)** is isometrically isomorphic to (lq, 11-ilq)* and so every continuous linear functional F on (lq, 11-ilq) can be represented in the form 00

F(f) =

L

ilk A.k

k=l

for

f= {11 1, 11 2, ... , Ilk> ... } e lq and {1.. 1, 1..2 , ••• , A.k, ... } e lp.

So { F(f 1), F(f2), ••• , F(fk), ... } e lp. But then A

F(f) = x(f) for all f e l F= Q A ** and we conclude that lp = l P .

**

P

where x = { F(f1), F(f2 ),

.•.

F(fk), ... } e lp;

that is,

O

7.8 Remark. It should be noted that it is not sufficient for reflexivity that a Banach space (X, 11·11) be isometrically isomorphic to (X**, 11·11). R.C. James, Proc. Nat. Acad. Sci.

USA 37 ( 1951 ), 174.--177, has given an example of a nonreflexive Banach space with just this property. For reflexivity we must have (X, 11·11) isometrically isomorphic to (X**, 11·11) under the natural embedding. That is why in Examples 7.7.2 and 7.7.3, although it is obvious from Example 5.3.3 that (.lp, ll·llp) is isometrically isomorphic to (lp, ll·llp)** and from Exercise 5.5.7 that a Hilbert space H is isometrically isomorphic to H**, yet for reflexivity we need to establish the isometric isomorphism under the natural embedding and this requires a little more careful computation. 0

126

§7. The natural embedding and reflexivity

7.9 Techniques to prove nonreflexivity To prove nonreflexivity directly we have to show that the natural embedding is into but not onto. Such computation usually involves a knowledge of the form of the dual and the form of the second dual of the space and in other than a few cases this can be quite a complication.

So the nonreflexivity of a normed linear space is usually established by

indirect argument. It is clear that if a normed linear space is not complete then it is not reflexive. But we can sometimes quickly determine non-reflexivity by an appeal to separability. 7.9.1 Theorem. A separable Banach space (X, II· II) with a nonseparable dual

(X*, 11·11) is not reflexive. Proof. It follows from Theorem 6.9 that the second dual (X**, 11·11) cannot be separable. But then (X, 11·11) and (X**, 11·11) cannot be isometrically isomorphic, (see Exercise 1.26.15). D 7.9.2 Example. The Banach space (c 0 , ll·lloo) is separable, (see Example 1.25.2(ii)). But from Example 5.3.1, (c 0, 11·11 00 )* is isometrically isomorphic to (1 1, 11·11 1) and in Exercise 5.5.3 we prove that (1 1, 11·11 1)* is isometrically isomorphic to (m, ll·lloo). So (c 0, 11·11)** is isometrically isomorphic to (m, 11·11 00 ). But (m, ll·lloo) is not separable, (see Example 1.25.3). Therefore we can conclude that (c0 , 11·11 00 ) is not reflexive.

D

A more generally applicable indirect method is a consequence of the following important property of reflexive spaces. 7.9.3 Theorem. On a reflexive normed linear space (X, 11·11) every continuous linear

functional f attains its norm on the closed unit ball of(X, II· II).

Proof. Applying Corollary 6.3 to (X*, 11·11) we have that for any given f EX* there exists a continuous linear functional F on X* such that F(f) = II f II II F II. But we are given that II

X= X**, so F =Q Then

for some x EX.

f(x) = Q(f) =II f 1111 x II.

D

7.9.4 Remark. It follows from Theorem 7.9.3 that if a Banach space has a continuous linear functional which does not attain its norm on the closed unit ball then the space is not reflexive. This provides a technique for proving nonreflexivity which does not even D necessitate a knowledge of the form of the dual space.

127

The existence of continuous linear functionals

7.9.5 Example. Consider the Banach space ('C[-1t,1t], 11·11

00 ) .

We exhibit a continuous

linear functional on the space which does not attain its norm on the closed unit ball, (see Exercise 4. 12.4 ). Consider the linear functional F defined by 1t

F(f)

Jf(t) sin t dt .

=

-1t

Now

_[1 sin t I dt}fll~

IF(f)l$;(

forallfE'C[-1t,1t].

1t

JI sin t I dt = 4. If we choose f

So F is continuous and II F II$;

0

on [-1t,1t] defined by

-1t

$; t < 0 }· Q$;t$;1t

f 0 (t) = -1 I

-7t

1t

Then

F(f0)

= JI sin t I dt = 4 -1t

and since II fo lloo = I we would have II F II = 4 and F attain its norm at f0 . But fo,; 'C[-1t,1t]. However, we can modify f0 to show that II F II= 4.

-1t

Figure 12. A modification of f0 to give a sequence { fn} in 'C [-1t,1t] so that F(fn)

--7

4 as n

--7

oo.

We define a sequence {fn} in 'C[-1t,1t] by I

$;t<--

-Jr

Now II fn lloo = I for all n EN.

n

I

I

= nt

-fi$;t$;fi

=I

-n
I

128

§7. The natural embedding and reflexivity 1t

But F(fn) =

Jfn(t) sin t dt > 4 - ~

forallneN.

-1t

So F(fn) ~ 4 as n ~ oo and therefore II F II = 4. But it is clear that there is no f E 'C [-1t,1t], II f lloo = I such that F(f) = 4. So we conclude that ('C [-1t,1t], IHioo) is not reflexive. D 7.9.6 Remark. It is important to observe that one of the most significant theorems in this theory was proved by R.C. James, Studia Math. 23 (1964), 205-216. By proving the converse of Theorem 7.9.3 he established the following theorem.

The James characterisation of reflexivity. A Banach space is reflexive if and only if every continuous linear functional attains its norm on the closed unit ball of the space. Although this is an extremely useful tool in determining reflexivity, the proof is quite "deep". It is inappropriate to present it in this course and in fact we will not need to make use of it. D Theorem 7.9.3 and the James characterisation of reflexivity point to the significance of the class of reflexive Banach spaces for approximation theory; (see Exercise 7 .13.2). We now show that if a Banach space is reflexive then its dual, or predual is also reflexive. 7.10 Theorem. A Banach space (X, 11·11) is reflexive

if and only if its dual (X*, IHI) is

reflexive. A

Proof. Consider :T EX***. Now :T IX is a continuous linear functional on X, so there exists an f EX* such that A

AA

:T I. (x) = f(x) for all x EX. X

A

If X is reflexive then X = X**. So A

:T (F) = f(F) Now

A

for all F eX**.

A

::r = f and we conclude that x* =X***;

that is, X* is reflexive.

A

Conversely, suppose that X is a proper linear subspace of X**. Since X is a A

Banach space, X is closed so from Corollary 6.7 there exists an :T EX***, :T ~ 0 such A

that :T (X) = 0. A

AA

If X* is reflexive, :T = f for some f EX*, so f (X) = 0 which implies that f(X) = 0; that is, A

f = 0. But this contradicts :T ~ 0, so we conclude that X= X** ; that is, X is reflexive.

D

129

The existence of continuous linear functionals

7.11 Remark. Theorem 7.10 implies that if a Banach space (X, II· II) is not reflexive then none of its duals or preduals is reflexive. D

The completion of a normed linear space

7.12

It is the theoretically powerful Hahn-Banach Theorem which enables us to establish in a painless fashion that with an incomplete normed linear space th:;rc is always a complete normed linear space which contains the original as a dense subspace. 7.12.1 Definition. The completion of an incomplete normed linear space (X, 11·11) is a complete normed linear space (X, II· II) such that (X, 11·11) is isometrically isomorphic to a dense subspace of (X, 11·11). The study of the second dual of a normed linear space enables us to define such a completion. 7.12.2

Theorem. For any incomplete normed linear space (X, 11·11) there exists a

completion

Proof.

ex, 11·11) which is unique up to isometric isomorphisms.

Consider the second dual (X**, 11·11) of (X, 11·11).

By Corollary 4.1 0.3,

A

(X**, 11·11) is complete so it contains X the closure of the natural embedding of X. As a A

closed subspace of a complete space, (X, 11·11) is complete by Proposition 1.13. Since X is 1\

1\

isometrically isomorphic to X under the natural embedding, we have that (X, II· II) is a completion of (X, II· II). The uniqueness up to isometric isomorphisms follows from Theorem 4.8. 7.12.3

D

Corollary. Given an incomplete normed linear space (X, 11·11) with completion

(X, 11·11) then (X, II· II)* is isometrically isomorphic to (X, 11·11)*. 7.12.4 Example. E 0 and .t 1 are dense linear subspaces of (c 0 , ll·lloo). Then (E0 , 11·11 00 ) and (.t 1, 11·11 00 ) have the same completion and (E0 , 11·11 00 )* and (.t 1, 11·11 00 )* and (c 0, 11·11 00 )* are isometrically isomorphic to (.t 1, 11·11 1).

D

When the incomplete space is an inner product space then it is significant that there is a natural extension of the inner product to its completion. 7.12.5 Theorem. For an incomplete inner product space X its completion product space with inner product ( . , . ) defined by

X is an inner

130

§7. The natural embedding and reflexivity

where {Xn} and {yn} are sequences in X convergent on ;: andy in

X.

Proof. Now { (xn, Ynl} is a Cauchy sequence of scalars so lim (xn, Ynl exists. It is not difficult to show that this limit is independent of the choice of sequences {xn} and {y n} in X converging to ;: and

y in X.

Properties (i)-(iv) of the inner product follow from those of

the inner product in X and the limit properties. We check property (v). If(;_,;_)= 0, then lim II xn 11 2 = 0 so (xn} converges to 0 and then ;: = 0. Clearly, the inner product on

Xis an extension of that on X.

D

7 .12.6 Example. (t: [a,b], 11·11 2) is an incomplete inner product space and its completion is a Hilbert space which contains (t: [a,b], 11·11 2 ) as a dense linear subspace. But (t:[a,b], 11·11 2 )* is isometrically isomorphic to (ti[a,b], 11-11 2 )* and as it is a Hilbert space we have by Theorem 5.2.5 that (ti[a,b], 11·11 2 ) is isometrically isomorphic to its dual. So the elements ti[a,b] can be thought of as continuous linear functionals over (t: [a,b], 11·11 2).

D 7.12.7 Remark. Some regard classical Hilbert space (:C 2 [a,b], 11·11 2 ) introduced in Example 2.2.12(iii) as (ti 2 [a,b], 11·11 2) the completion of (t: 2 [a,b], 11·11 2 ). 7.13

EXERCISES

I.

Determine whether the following normed linear spaces are reflexive. (ii) (~ I• 11·11,) (iii) (m, ll·lloo) (i) (Eo, 11·11 00 ) (iv) (c, 11·11 00 )

2.

(v)

(ti[-1t,1t], 11·11 2 )

(vi) (t:~(21t), 11·11 00 )

Consider a reflexive Banach space (X, 11·11). (i)

Given any proper closed linear subspace M and x0 EX \ M, prove that there exists an element Yo EM such that II x0-y 0 II= d(x0, M);

(ii)

(see Exercise 6.13.6). Given any proper closed convex set K and x0 EX\ K, is it true that there exists an element Yo E K such that II x0-y 0 II = d(x 0, K)?

3.

For a reflexive Banach space (X, 11·11) prove that (i)

X is smooth if and only if X* is rotund, and

(ii)

X is rotund if and only if X* is smooth.

D

The existence of continuous linear functionals

4.

131

Consider a linear space X with norms IHI and 11·11'. (i)

Prove that IHI and 11·11' are equivalent if and only if they generate equivalent

(ii)

Prove that a Banach space (X, 11·11) is reflexive if and only if (X, 11·11')

norms 11·11* and 11·11'* on X*; (see Exercise 4.12.9). is reflexive where IHI and 11·11' are equivalent norms on X. 5.

Using the information in Section 5 about the shape of the dual, exhibit on each space (i) (co, ll·lloo), (ii) (.ll, 11·111), (iii) (~[-!,!], ll·lloo). a continuous linear functional which does not attain its norm on the closed unit ball and deduce that none of these spaces is reflexive.

6.

7.

(i)

Using the James characterisation of reflexivity, prove that a uniformly rotund

(ii)

Banach space is reflexive; (see Exercise 2.4.17). Hence, deduce that .lp space (I < p < oo) is reflexive.

Many of the properties of finite dimensional normed linear spaces can be derived from Corollary 6.3 to the Hahn-Banach Theorem without using compactness at all. Using such a method prove that (i)

a finite dimensional normed linear space (Xn, II· II) is reflexive,

(ii)

every linear functional f on Xn is continuous,

(iii) convergence in norm is equivalent to coordinatewise convergence, (that is, Corollary 2.1.11 holds), (iv) the closed unit ball is compact. 8.

Given a linear space X consider the algebraic embedding x 1-7

Q of X

into X##

induced by the definition of ~ as

~(f)= f(x) (i)

Prove that

~

for all f EX#.

the image of X under the algebraic embedding x

1-7

Q is

an

isomorphism of X into X##. (ii)

Prove

that~= X##,

(that is, X is algebraically reflexive), if and only if X is

finite dimensional. (See Theorem 2.1.12).

132

The existence of continuous linear functionals

§8. SUBREFLEXIVITY

In any discussion of approximation theory and reflexivity we notice the significance of continuous linear functionals which attain their norm on the closed unit ball of the space. A key result which gives useful information about the distribution of norm attaining continuous linear functionals in the dual was given by E. Bishop and R.R. Phelps, Bull. Amer. Math. Soc. 67 ( 1961 ), 97-98. They showed that completeness for a normed linear space implies that the set of norm attaining continuous linear functionals is dense in the dual.

This result has been

generalised out of all recognition and in its most general form is called the Ekeland Variational Principle. It has wide application in a remarkable number of areas as the survey article by I. Ekeland, Bull. Amer. Math. Soc. N.S. 1 (1979), 443-474 demonstrates. We confine our attention to the result originally given by Bishop and Phelps and show its usefulness for some geometrical considerations. We begin by developing a geometrical lemma. 8.1 Definition. Given convex sets A and B in a linear space X the convex hull of A and B, denoted by co{A,B} is the set {Aa + (1-A)b: 0 sA::; 1 and a E A, bE B}. Clearly co{A,B} is a convex set in X. -

8.2 Notation. It is convenient to denote by B the closed unit ball of (X, 11·11). 8.3 Lemma. Given a normed linear space (X, 11·11), a nonzero continuous linear functional f

on X and a constant k > 1, consider the sets T = { x EX: x E ker f and I < II x II ::; k} and K = co {B, T}.

Then the functionalll·ll' on X where II x II'= inf{A > 0: x E AK}

is a norm equivalent to the given norm II· II on X.

Figure 13. A pictorial representation of K =co {B, T}.

§8. Subreflexivity

133

Proof. For y E Band x eT we have ay eB and ax ET for all I a I::; I, so a(A.y + (1-A.)x) E K for all 0:5 A. :5 I and I a I :5 I.

"*

Therefore, for x -T- 0, if x E J.LK for some J.1 > 0 then A.x E I A. I J.LK for A. 0 so II A.x II' :51 A. Ill x II'. For x substitute A.x and for A. substitute 1/A. then we have II x II' :S-1- II A.x II' I A. I and we conclude that II A.x II' = I A. Ill x II' for A. -T- 0. But also for x,y -T- 0, if x E A.K andy E J.LK for some A.,J.L > 0 then x/A. and y/J.L E K. Since K is convex,

_l:_ (x!A.) + L

(y/J.L) E K; A.+J.L A.+J.L that is, x+y E K and so x + y E (A.+J.L)K; A.+J.L This implies that II x+y II' :511 x II'+ II y II'. Since ker 11·11' = {0}, 11·11' is a norm for X. But also B ~ K ~ II x II :5 II x II' $

~

II x II

kB which implies that

for all x EX

so II· II' and 11-11 are equivalent norms for X.

D

We develop our theory for Banach spaces over the real numbers and we establish the Bishop-Phelps Theorem for Banach spaces over the real or complex numbers by appealing to Theorem 4.10.16. The main argument of the Bishop-Phelps Theorem is based on the following lemma.

8.4 Lemma. For a real Banach space (X, 11·11) consider f EX*, II f II= I. Given 0 < k :5 I and u E B where f(u) > 0 there exists an x0 E B where f(x 0) > 0 such that k II x0-u II :5 f(x 0)- f(u) and f(y)- f(x 0) < k II y-x0 II for all y E B, y -T- x0 . Proof. We define a sequence {x0 } in B inductively as follows. Choose x 1 = u. For each n EN consider the set

={y E B : f(y)- f(xn) > K II y-xn II}. If sn = 0, write xn+l = xn and if sn "* 0, choose xn+l E sn such that sn

I

f(Xn+l) ~ 2 (f(x 0 ) +Sup{ f(x) : X E S0 } ). For the sequence {x0 } so defined k II X0 -x 0 _J II < f(x 0 ) - f(x 0 _J) so

for all n EN for all m > n.

(i)

(ii)

134

The existence of continuous linear functionals

Now the sequence { f(xn)} is increasing and is bounded above, so it is convergent. But then this implies that the sequence {xn} is Cauchy. Since (X, 11·11) is complete, the sequence {xn} is convergent to some x 0 EB. Since f is continuous at x0, inequality (ii) implies that k II xo-xn II ~ f(xo)- f(xn) and in particular

k II x0-u II

~

for all n E :f::I

f(x 0)- f(u).

Suppose there exists avE B, v oF- x0 such that f(v) Then

~

f(x 0) + K II v-x 0 II.

f(v) > f(x 0) = lim f(xn) n---;~

since f is continuous at x0 . But then v E Sn for all n E :f::I which implies that 2 f(xn+t)- f(xn) ~ sup{f(x) : x E Sn} ~ f(v) and so

f(x 0 )

~

f(v).

But this is a contradiction so f(y)- f(x 0) < k II y-x 0 II for ally E B, y

oF-

x0 .

0

8.5 Theorem. For a real Banach space (X, 11·11), given f EX*, II f II = I and E > 0 there exists

a g EX*, II g II = I and x 0 EX, II x 0 II = I such that g(x 0) = I and I g(x) I ~ E

Proof.

for x E kerf and II x II

~

I.

We need to find agE X*, II g II= I and x0 E K, II x0 II= I such that g(x 0 ) = I

and I g(z) I~ I for all z E T

= { x EX : x E kerf and II x II ~ _!_}. E

Consider K

= co {B,

From Lemma 8.3, K is an equivalent norm ball which contains B. If x0 E B

n bdy K, from Corollary 6.3 we see that there exists agE X*, such that g(x 0) = sup {I g(x) I : x E K} ~ {I g(x) I : x E B}

= II g II ~ g(x0).

Figure 14. K is to one side of the hyperplane {x EX : g(x)

= I}.

T}.

§8. Subreflexivity So to establish the theorem we need to show that B

n bdy K

135

* 0.

Consider u E B but u il bdy K. We may consider f(u) > 0. There exists an a > I such that au E K. Then there exists 0 < 'A < I such that au= 'Ax+ (1-'A)z So

for some x EB and z ET.

x-u = (a-l)u + (1-'A)(x-z) II x-u II :<; (a-1) + (1-'A) II x-z II :<; (I +

and

t) (a-'A) since II x-z II :<; I + t .

But also f(u) < a f(u) = 'A f(x) < f(x). So

f(x)- f(u) =(a-!) f(u) + (1-'A) f(x) >(a-A.) f(u);:: II x-u II f(u; . 1+-

(i)

E

Figure 15. u E B but u il bdy K and au E K for a> I. . f(u) h . In Lemma 8.4, choosmg k = - 1 we ave that there ex1sts an x0 E B where f(x 0) > 0,

1+E

such that

f(u; II x0-u II :<; f(x 0) - f(u) 1+-

(ii)

E

f(y)- f(x 0 ) < f(u; II y-x 0 II 1+-

and

for all y E B and y

* x0 .

(iii)

E

Comparing inequalities (i) and (iii) we see that x0 E bdy K. This completes the proof of the theorem.

0

In order to show the full implications of this theorem we need the following lemma. 8.6 The Parallel Hyperplane Lemma.

For a real nonned linear space (X, 11·11), given 0 < E :<; ~ and f,g EX*, II f II= II g II= I and I g(x) I:<; £for x E kerf and II x II:<; I then either II f-g II:<; 2£ or II f+g II:<; 2£.

Proof. Consider glker f. Now II glker f II :<; E so by the Hahn-Banach Theorem 6.2 there exists an h EX* an extension of glker f to X such that II h II :<;E. Then (g-h)(x) = 0

for x E kerf, so g-h = af for some real a.

136

The existence of continuous linear functionals

Now

I I - I a I I = II g II - II g-h II :<:; II h II :<:; E.

So if a ~ 0 then If a< 0 then

II f-g II= II (1--a)f-h II :<:;I 1--a I+ II h II:<:; 2£. II f+g II= II (l+a)f+h II:<:; ll+a I+ II h II:<:; 2£.

0

Theorem 8.5 and Lemma 8.6 have an immediate corollary, a gloss on Corollary 6.7 to the Hahn-Banach Theorem. 8.7 Corollary. Consider a real Banach space (X, II· II) and any proper closed linear subspace M. Given E > 0 there exists a g EX*, II g II = I and an x0 EX \ M, II x0 II = I such that g(x 0) = I and I g(x) I:<:; £for all x EM and II x II:<:; I and d(x0 , M) ~ 1-2£.

Proof. From the modification of Corollary 6.7 given in Exercise 6.13.6, there exists an f EX*, II f II= I such that f(M) = 0. Now consider kerf. From Theorem 8.5 we have that there exists ag EX*, II g II= I and x0 EX, II x0 II= I such that g(x 0 ) =I and I g(x) I :<:; E for x E kerf and II x II :<:; I so

I g(x) I :<:; E

for x EM and II x II = I.

Again by Exercise 6.13.6 there exists an a such that I a I= I and af(x 0) = d(x 0 , kerf). From Lemma 8.6 we may assume that II f-g II :<:; 2£ so I = g(x 0 ) :<:; I g(x 0) - f(x 0 ) I + I f(x 0 ) I :<:;II f-g II+ d(x 0 , kerf) :<:; 2E + d(x 0 , M). Then d(x 0 , M) ~ I - 2£.

D

8.8 Definition. A normed linear space (X, 11·11) is said to be subreflexive if the set of continuous linear functionals which attain their norm on the closed unit ball is dense in the dual (X*, 11·11). From Theorem 8.5 and Lemma 8.6 we deduce the key theorem of this section. 8.9 The Bishop-Phelps Theorem. Every Banach space is subreflexive. Within a short time of publication of the original Bishop-Phelps Theorem 8.9 it was realised that the proof provided more information than had been claimed. So we present this form which has applications in its own right. 8.10 The improved version of the Bishop-Phelps Theorem. Given a Banach space (X, 11·11) and an f EX*, II f II = I and 0 < E :<:; ~ and u EX, II u II = I such that I f(u)- I I :<:; £2 then there exists agE X*, II g II= I and an x0 EX, II x0 II = I such that g(x 0) =I, II f-g II:<:; 2£ and II x0-u II:<:; 2E.

Proof. We firstly prove the theorem for a real Banach space (X, 11·11). Within the proof of Theorem 8.5 inequality (ii) gives us that

§8. Subreflexivity

137

(I~) f(x 0-u) II x0-u II::;;

f(u)

where f(x 0 ) > 0 and so II f+g II > (f+g)(x 0 ) > I. Lemma 8.6 gives us that II f-g II:::; 2£. But also we see that

But since 0 < £ :::;

kwe have I - £2 ::;; ~ II x0-u II :::; ~ (£+£ 2):::; ~

so

(~ £) = 2£.

The theorem is extended to complex Banach spaces by Theorem 4.10.16 which establishes that there is a norm preserving one-to-one correspondence between the continuous complex linear functionals and the continuous real linear functionals on X. If I f(u)-1 I< £2 then I fIR. (u)-1 I < £2 so by Theorem 8.10 applied to XIR. we have that there exists a g!R. E XIR. *, II g!R. II= I and an x0 EX, II x0 II =I such that g!R.(x 0 ) =I, II (f-g)IR. II= II f!R.-g!R. II::;; 2£ and II x0-u II :::; 2£. Then for gE X* defined as in Theorem 4.10.16 we have II g II = I, g(x 0 ) = I

D

and II f-g II :::; 2£ and II x0-u II :::; 2£.

We now show how the Bishop-Phelps Theorem can be applied. In Remark 7.9.6 we mentioned the James characterisation of reflexivity. The Bishop-Phelps Theorem 8.9 enables us to prove a special case of this result directly. 8.10 Corollary. Consider a Banach space (X, 11·11) where the dual space X*is smooth. If

every continuous linear functional attains its norm on the closed unit ball of the space then X is reflexive.

Proof. By X* being smooth we mean that to each f EX*, II f II = I there exists only one FE X**, II F II= I such that F(f) = I. However, for each f EX*, II f II = I there exists an x EX, II x II = I such that f(x) = I; that is,

Q(f)

= I.

So the set of continuous linear functions on X* which attain their norm on the closed unit ball of A

X* is the set X. A

But by the subreflexivity of (X*, 11·11), the set X is dense in (X**, 11·11). But since (X, 11·11) is A

A

complete, X is closed in (X**, 11·11). So X**= X; that is, X is reflexive.

D

138

The existence of continuous linear functionals

8.11 EXERCISES

I.

Consider a real Banach space (X, 11·11). (i)

Given x EX\ {0} we write D(x)={feX*:IIfll=l and f(x)=llxll}. (a)

Prove that given x EX\ {0} and any y EX and A.> 0

( ) f ( ) < llx+yll - llxll < f X y A. - X+Ay y for any fx eD(x) and fx+A.y eD(x+A.y) and that the inequality is reversed for A.< 0. (b)

Given x EX\ {0} deduce that if, for ally EX and A.* 0 and any selection fX+A.y E D(x+A.y) we have fX+A./Y) converges as A.~ 0 then X is smooth at x.

(ii) Given f eX*, II f II = I a closed slice of the closed unit ball B(X) cut off by f is a

subset { x e B(X) : f(x) ~ I -

o}

for any 0 <

o< I.

Given x eX, II x II = I, prove that for any y eX, II y II = I and 0 <

A.<~ ,

fx+A.yCx) ~ I - 0 for all fx+A.y E D(x+A.y). (iii) Suppose that X has the property that every f eX*, II f II = I cuts off slices of B(X) of arbitrarily small diameter. Prove that (a) (b)

X is rotund, every f eX* attains its norm on B(X), (Hint: Use Cantor's Intersection Theorem AMS §4.)

(c)

X* is smooth,

(d)

X is reflexive.

(iv) Prove that if X is uniformly rotund then X is reflexive.

IV.

THE FUNDAMENTAL MAPPING THEOREMS FOR BANACH SPACES

In addition to the Hahn-Banach Theorem the three mapping theorems, the Open Mapping Theorem, the Closed Graph Theorem and the Uniform Roundedness Theorem are vital for the development of any general theory of Banach spaces. In these theorems we begin to appreciate the importance of the completeness condition. The proofs are based on Baire category arguments which reveal the implications of completeness for the metric topology. So we begin by developing this theory and demonstrate something of its force before applying it to establish the fundamental mapping theorems. §9. BAIRE CATEGORY THEORY FOR METRIC SPACES In complete metric spaces the metric topology has important characteristics and a knowledge of these is indispensable in establishing many significant results in the analysis of normed linear spaces and elsewhere. We recall the following definition from the analysis of metric spaces. 9.1 Definition. its closure A

Given a metric space (X, d), a subset A is said to be dense in (X, d) if

= X.

This means that A is dense in (X, d) if and only if every point of X is either a point of A or a cluster point of A. Equivalently, A is dense in (X, d) if and only if for every x EX and E

> 0, we have B(x;

E)

n A of- 0.

The following concept related to density is used to partition metric spaces into disjoint classes.

140

The fundamental mapping theorems

9.2 Definition. Given a metric space (X, d), a subset A is said to be nowhere dense in XifintA= 0. It is clear that A is nowhere dense in X if and only if C(A) is dense in X; (see Exercise

9.19.1). 9.3 Remark. So a closed set A in (X, d), is nowhere dense if and only if C(A) is dense in X. A closed nowhere dense set A is its own boundary, bdy A= C(A) n A= A. Further, the boundary.of any open or closed set is nowhere dense since, for any open set

-

-

G, bdy G = G n C(G) and C(bdy G) = C(G) u G from which it can be seen that C(bdy G) is dense in X. D 9.4 Examples. (i) In any normed linear space (X, 11·11), (a) (ii)

any finite subset is nowhere dense,

(b) any proper closed linear subspace is nowhere dense. In (IR, I · I), the set of natural numbers N is nowhere dense.

(iii) In [0,1] with the usual metric, Cantor's Ternary Set K is closed and has empty interior so is nowhere dense. (iv) In any discrete metric space (X, d), since every subset is both open and closed, every subset except the null set 0 has nonempty interior so the only nowhere dense subset is 0.0 It is clear that, whether or not a set is nowhere dense depends on the metric and the

space. 9.5 Examples. (i) A proper closed linear subspace Y of a normed linear space (X, 11·11) is not nowhere dense in (Y, ll·lly). The set of natural numbers N as a subset of (IR, I ·I) is a discrete metric space in its relative metric topology and so has no nonempty nowhere dense sets. D

(ii)

9.6 Remark.

Since the concept of nowhere denseness is defined in terms of interior and

closure, concepts which are invariant under equivalent metrics, so this concept is also invariant under equivalent metrics. D Now a finite union of nowhere dense sets is nowhere dense. This leads us to introduce the following definitions.

141

§9. Baire category

9.7 Definitions. Given a metric space (X, d), a subset A is said to be first category (or meagre) in X if A can be represented as the union of a countable number of nowhere dense sets. A subset A which is not of the first category in X is said to be of second category in

X. 9.8 Examples. (i) In any normed linear space (X, 11·11), (a) any countable set is first category, (ii)

(b) any countable union of first category sets is first category. In (IR, I · 1), the set of rational numbers
(iii)

In [0, I] with the usual metric, Cantor's Ternary Set K is uncountable but is first

category because it is nowhere dense. (iv) In [0, I] with the usual metric, consider the set lK constructed from Cantor's Ternary Set K in the following way:

[i, ~].consider the image of K under the mapping f (t) =t (t+l). In the interval [t ,~] ,consider the image of K under mapping f (t) = t (t+ I), in the interval

In the interval

1

21

{~, ~]. consider the image of K under the mapping f22 (t)

t

=

t

(t+4) and in the interval

[~, ~],consider the image K under the mapping f23 (t) = (t+7). Continuing this process at the nth step we put a ~ reduced copy of K into each of the intervals of length ~ 3

3

remaining in [0, 1]. Now each copy of K is nowhere dense in [0, I] and there is a countable number of copies of K in the new set lK , so lK is first category in [0, I] but it is 0 uncountable and dense in [0,1]. 9.9 Definitions. Given a metric space (X, d), a subset A is said to be residual in X if C(A) is first category. (X, d) is said to be a Baire space if every residual set in X is dense

in X. In a Baire space a first category subset has no interior so a residual subset is always dense and second category. A Baire space (X, d) is necessarily second category in itself, for if X were first category then the null set 0 would be dense in X. We develop the important characterisations for Baire spaces which we use in applications.

The fundamental mapping theorems

142

9.10 Theorem. Given a metric space (X, d), the following are equivalent. (i)

(X, d) is a Baire space.

(ii)

For every countable family of dense open sets {Gnl we have nan is dense in X.

(iii) For every countable family of closed sets IFn} such that X= UFn we have U int Fn is dense in X. Proof. (i)~(iii)

Since bdy Fn is nowhere dense, then U bdy Fn is first category in X. But X is a

Baire space so C(U bdy Fn) is dense in X. Now bdy Fn = Fn \ int Fn and since X= UFn we have that C(U bdy Fn)

~

U int Fn and soU int Fn is dense in X.

(iii)~(ii) It is clear that X= non

u (U C(Gn)), the union of a countable family of closed

sets. So int non u (U int C(Gn)) is dense in X. But since Gn is dense in X, int C(Gn) = 0, and then int non is dense in X. But this implies that non is dense in X. (ii)~(i)

Consider a countable family !Enl of nowhere dense sets in X. Then UEn is first

category in X. Since C(En) is open and dense in X so nqEn) is dense in X. But

0

n C(En) s;;; C(UEn) so C(UEn) is dense in X. We are now ready to reveal the special properties of complete metric spaces. 9.11 Baire's Theorem. A complete metric space (X, d) is a Baire space.

Proof.

Consider a countable family of dense open sets {Gnl in X.

any X E X and r > 0, we have non n B(x; r)

We show that for

* 0.

Since G 1 is dense in X there exists an x 1 E G 1 n B(x; r) such that G 1 n B(x; r) contains B[x 1; r 1] where 0 < r 1 < ~ . Since G 2 is dense in X there exists an x2 E G2 n B(x 1; r 1) such that G 1 n G 2 n B(x; r) n B(x 1; r 1) contains B[x 2 ; r 2] where 0 < r2 < ~ . Continuing inductively we have that there exists an Xn E Gn n B(xn-1; rn_ 1) such that G1 n G2 n ... n Gn n B(x; r) n B( Xn-1; rn-J) contains B[xn; rnl where 0 < rn < :{n

.

Now the sequence {B[xn; rnl} is a nested sequence of closed sets whose diameters tend to zero. Since (X, d) is complete, we have by Cantor's Intersection Theorem, (see AMS §4), 0 that there exists ayE nB[xn; rnl and soy E non n B(x; r). Again whether or not a set is first or second category depends on the metric and the space.

143

§9. Baire category

9.12 Example. In any Banach space (X, 11·11), a proper closed linear subspace Y is nowhere dense in X but by Theorem 9.11 is second category in (Y, ll·lly ). D It is clear from Definitions 9.7 that a countable union of first category subsets is first

category and we can use this fact to deduce the category of some subsets using Theorem

9.11. 9.13 Examples. (i) As (IR, I · I) is complete it is a Baire space. The set of rationals

1R so we deduce that the set of irrationals IR \ (ii)

<12

<12

is first category in

is dense and second category in 1R.

As [0, I] with the usual metric is complete it is a Baire space. Now the set lK defined

in Example 9.8(iv) is first category and so its complement is dense and second category

D

~IR.

The following theorem establishes a large class of sets as Baire spaces. 9.14 Theorem. An open subset of a Baire space is a Baire space.

Proof. Consider an open subset G of a metric space (X, d). We show that if A is nowhere dense in G then A is nowhere dense in X. For any subset A -

-

-

of G, the closure of A in G is An G and int (An G)= int An G. If int (An G) = 0 then since A is a subset of G we have int A= 0. Further, if a subset B is dense in X then B n G is dense in G. Therefore, for a Baire space (X, d), any first category subset A of G is first category in X and has C(A) dense in X which implies that C(A) n G is dense in G and we conclude that

I

(G, d a) is a Baire space.

D

9.15 Remarks. The particular properties from Baire Category Theory which are used widely in applications are as follows. (i) A complete metric space (X, d) is second category and so for a countable family of closed sets {F0 } such that X = UF 0 , there exists at least one F 00 which has nonempty interior. (Of course, since (X, d) is a Baire space we know that Uint F 0 is dense in X but this property is not so often needed in applications.) (ii)

A complete metric space (X, d) is a Baire space and so for a countable family of

dense open sets {G0 } we have nGn is dense in X. Baire Category Theory has a major use in providing nonconstructive existence proofs, showing that the desired examples occur as first or second category subsets in some D complete metric space.

144

The fundamental mapping theorems

We demonstrate the techniques of proof using Baire Category Theory and sample something of the results achieved by Baire category methods in the following three examples. These all concern properties of real functions on Baire spaces and their properties are involved in different ways. The first shows a limitation on the set of points of discontinuity of a real function continuous on a dense set, the second shows how a real lower semicontinuous function defined on an open set is continuous on a dense subset of its domain and the third shows how badly a continuous real function can behave from the point of view of its differentiability.

9.16 The set of points of discontinuity of a real function continuous on a dense set We recall the following definitions from real analysis. 9 .16.1 Definitions. Given a real function f on IR, for any bounded interval J we define m(f, J), the oscillation off over J by m(f, J) =sup {I f(x) - f(y) I : x,y

E

J},

and for x0 E IR we define m(f, x0 ), the oscillation off at x0 by m(f, x0 ) = inf { m(f, J) : all such J containing x0 }. 9 .16.2 Theorem. Given a real function f on IR, continuous at the points of a dense set,

the set D of points of discontinuity off is first category. Proof. It is clear that f is continuous at x0 if and only if m(f, x0)

=0.

So for each n EN, we define En

I

= { X E IR : m(f, X) ~ n}

and we show first that En is closed. Consider a cluster point x0 of En. Then for any bounded interval I containing x 0 there exists an x E En and so m( f, I) ~ m( f, x) ~

But then

nI .

m( f, x0) = inf { m( f, I) : all such I containing x0 }

~

i.

So x0 E En and En is closed. Now D

= U {En: n EN}.

If for some En we have int En o1 0 then int D o~ 0 which contradicts the fact that f is continuous on a dense set. Therefore, for each n conclude that D is first category.

E

N , En is nowhere dense and we

0

§9. Baire category

145

9.16.3 Remark. The ruler function f defined on IR by f(x)

~

for rational x =

~ "#

0,

(p,q mutu•lly pdmo) }

0 for irrational x f(O)

is continuous at irrational points and discontinuous at rational points. Now the set of rationals is first category in IR so the ruler function is typical of Theorem 9.16.2. However, Theorem 9.16.2 tells us that it is not possible to have a real function on IR continuous at rational points and discontinuous at irrational points, because the set of irrationals is second category in IR. D 9.17 The continuity of lower semicontinuous functions 9.17.1 Definitions. Given a metric space (X, d), a real function f on X is said to be lower semicontinuous at x0 EX if given E > 0 there exists a > 0 such that f(x) > f(x 0) - e when d(x, x0) <

o o,

that is, if

lim inf f(x)

~

f(x 0 ).

X~XO

A real function f on X is said to be upper semicontinuous at x0 if -f is lower semiE > 0 there exists a > 0

o

continuous at x0; that is, f is upper semicontinuous at x0 if given such that f(x) < f(x 0) + e that is, if

when d(x, x0) <

o,

lim sup f(x) :5 f(x 0 ). X~XO

Of course, f is continuous at x0 if and only if f is both lower semicontinuous and upper semicontinuous at x0 . We have the following global characterisation of lower semicontinuity. 9.17.2 Proposition. Given a real function f on a metric space (X, d), the following are

equivalent. f is lower semicontinuous on X. For every ex. E IR, the set { x EX : f(x) :5 ex.} is closed. (iii) For every ex. E IR, the set { x EX : f(x) >ex.} is open. (i) (ii)

Proof. (i)~(ii) Given

ex. E IR, consider x0 a cluster point of the set { x EX : f(x) :5 ex.}. Since f is

lower semi-continuous at x0 , given E > 0 there exists a

o> 0 such that

f(x) > f(x 0) - e when d(x, x0) < o. Since x0 is a cluster point of { x EX : f(x) :5 ex.} there exists an x ex.~ f(x)

E

B(x 0 ; o) such that

> f(x 0)- e and so f(x 0) :5 ex. which implies that { x EX : f(x) :5 ex.} is closed.

146

The fundamental mapping theorems

(ii)=>(iii) This follows from the set property that for any subset A of 1R, X\ (f-l(A)) = f-I (IR \A). (iii)=>(i) Given x0 EX and£> 0, (iii) implies that { x EX : f(x) > f(x 0 ) - £} is open. But then there exists a o> 0 such that B(x 0 ; o) k: { x EX : f(x) > f(x 0) - £}; that is, f(x) > f(x 0 ) - £ when d(x, x0 ) < which implies that f is lower semicontinuous at x0 .

o 0

How discontinuous can a lower semicontinuous function be?

We use Baire

Category Theory to show that if the lower semicontinuous function is defined on an open subset of a complete metric space then the function is actually continuous on a dense subset of its domain. This is an extremely useful result. 9.17.3 Theorem. A real lower semicontinuous function f on a metric space (X, d) is

continuous at the points of a residual subset of X.

Proof. Consider {U 0 } a countable base for the usual topology on lR and for each n EN the set D 0

= f- 1(0 0 ) \

int f-l(Un) which is nowhere dense in X. Then D = U { D 0

:

n EN}

is first category in X. We show that f is continuous at the points of X \D. Consider x0 EX\ D and k E lR such that f(x 0 ) < k and U0 such that Un k: (-oo, k) and f(x 0 ) we have x0

E

int

f- 1(0 0 )

E

U 0 . Since x0 e' D 0 and f(x 0)

E

Un

.

To show that f is continuous at x0 it is sufficient to show that f(x)

:<==;

k for all x E int f-l(Un)·

Suppose there exists some x 1 E int f-l(U 0 ) with f(x 1) > k. By the lower semicontinuity of fat x 1 there exists an open neighbourhood V of x 1 such that f(x) > k for all x E V. Now

*

V n f-l(U 0 ) 0 and for x2 E V n f-l(U 0 ) we have k < f(x 2) < k which is a contradiction. So we conclude that f is continuous at the points of X \D. 0 9.17.4 Remarks. (i)

It is clear that a similar theorem holds for an upper semicontinuous function f by

applying Theorem 9.17.3 to the function -f. (ii)

Such a result is of significance when (X, d) is a Baire space because then it tells us

that the lower semicontinuous function is continuous at the points of a dense residual 0 subset of X. 9.18 The existence of continuous nowhere differentiable functions Since most of the continuous real functions we manipulate in calculus have only a finite number of points in their domains where they are not differentiable, it is surprising to learn that there exist continuous functions which are nowhere differentiable. The first

§9. Baire category

147

example of such a function was constructed by Karl Weierstrass in the nineteenth century. But Baire Category Theory shows that "most" continuous functions are of this type by showing that the class of functions with at least one derivative is "small". So Baire Category Theory reveals much more about the nature of continuous functions than Weierstrass' example and by a nonconstructive method. 9.18.1 Theorem. In the Banach space ('C [0, 1], 11·11 00 ) the subset 'C 1 of functions which have at least one derivative at a point of[O, I) is first category.

Proof. For each n E N and 0 < h < 1_ we define n

En h = {f E 'C [0, I) : for some t E [0, I - !_), ' n

I -< h If(t+h)-f(t)

} n.

We show first that En,h is closed. Consider a cluster point f0 of En,h in ('C [0, 1], ll·lloo). Then given E > 0 there exists an f E En,h such that II f0-f 11 00 <E. But there exists atE [0, I -

~]

such that

(t) I 2£ If (t+h)-f h ::;; h +n 0

0

::;; n since E > 0 is arbitrary. So f0 EEn,h· But then E n = n {E n, h : all 0 < h < ln

} is also closed.

We show that En is nowhere dense in ('C[O,I], ll·lloo) by showing that that C(En) is dense in ('C[O,I), ll·lloo). Given f E 'C [0, I) and E > 0 choose a polynomial p such that II f-p lloo <E. This is possible because Weierstrass' Approximation Theorem, (AMS, §9), gives us that the polynomials are dense in the continuous functions on [0, 1). Then choose a continuous piecewise linear function q on [0, I) of the following form. q(t) = 2.5mt 1 0::;; t :=;;2.10m )

1

_1_< t

<-l-

2.10m

-

lOrn

_1_
q(t) = q(t ------;u)

!Om

10

where we choose m sufficiently large that 1m < E and 2.5m> n + 2 II p' 11 00 • 2

I Then II q 11 00 = - < E and for all t E [0, I) and h sufficiently small 2m

Iq(t+h~--q(t) I =

2.5m > n + 2 II p' 11 00 •

For the continuous function g =p+q we have II g--q 11 00 :=;;II f-p 11 00 +II q 11 00 < 2£.

148

The fundamental mapping theorems

wm Figure 16. The "sawtooth" function q. But also for all t E [0, I) and h sufficiently small

I g(t+htg(t) I ~ I q(t+h~-q(t) 1-1 p(t+hh- p(t) I > n which implies that g E C(En). We conclude that C(En) is dense in ("C[O,J], 11·11 00 ). We now have that U{En: n EN} is first category in ("C [0, 1], ll·lloo), but "C 1 ~ U {En : n EN}.

D

9.18.2 Remark. Since ("C [0, !], ll·lloo) is a Baire space we conclude that the set of continuous nowhere differentiable functions on [0, I] is a residual set in ("C [0, 1], 11·11 00 ) . Now although Weierstrass' Approximation Theorem implies that "C 00 [0,!] is also a dense subset of ("C [0, I], 11-iloo), it follows from Theorem 9.18.1 that it is a first category subset. The set of nowhere differentiable functions is also dense but is a second category subset.

D 9.19. Residual subsets of a Baire space. 9.19.1 Definitions. Given a metric space (X, d), a subset which can be represented as a countable intersection of open sets is called a G 0 subset and a subset which can be represented as a countable union of closed sets is called an Fcr subset. The families of G0 subsets and Fcr subsets are extensions of the families of open and closed sets. 9.19.2 Remark. It follows from de Morgan's Theorem that in any metric space (X, d), a subset A is a G 0 subset if and only if C(A) is an Fcr subset. So in (IR, I · 1), the set of rationals Ql as a countable set, is an F cr subset and so the set of irrationals is a G 0 subset. In a metric space the open and closed sets have the following special properties.

0

149

§9. Baire category

9.19 .3 Proposition. In a metric space (X, d),

every closed set F is a G 0 subset and (ii) every open set G is an Fcr subset. (i)

Proof. (i)

For each n

E

F\i consider G0 = { B(x;.!.) : x E F}. n

For each n

E

Fli, G 0 is open and F!;;;; G 0 so F!;;;; nG0 •

Consider y

E

nG0 . For each n

E

F\i, there exists an x

or is a cluster point of F. But F is closed soy

E

E

F such that y

E

B(x;.!. ), so y n

E

F

F and we conclude that nG0 !;;;; F.

Therefore F =nG 0 . (ii)

0

This follows by de Morgan's Theorem.

The following theorem gives us further insight into the nature of residual subsets. 9.19.4 Theorem. Consider a metric space (X, d). (i)

A residual subset E contains a G0 subset of X.

(ii) A subset E which contains a dense G0 subset of X is residual.

(iii) When (X, d) is a Baire space, a subset E is residual

if and only ifE contains a dense

G 0 subset of X.

Proof. (i)

Now C(E) is first category and so C(E) =

U E0

where E 0 is nowhere dense for each

1

n E F\i. So E ::! (ii)

n1 C(E

a G 0 subset of X.

0)

If E contains a dense G0 subset of X then E ::1

nG

0

where G0 is dense and open for

1

each n

E

F\i. So C(E) h U C(G0 ) where int C(G 0 ) 1

=0

for each n

E

F\i.

This implies that C(E) is first category and so E is residual. (iii) From (i) we have that if E is residual then E ::1

n C(E

0)

where E 0 is nowhere dense

1

for each n

E

F\i. Then C(E 0 ) is dense for each n E F\i.

Since (X, d) is a Baire space, it follows from Theorem 9.10(ii) that

n C(E

0)

is also dense.

1

0 9.19.5 Remark. The information in Theorem 9.19.4 provides us with an important technique:

!50

The fundamental mapping theorems

In any metric space it is obvious that the intersection of two dense subsets is not necessarily dense; in fact it could be empty. But in a complete metric space, if we have two dense 0 subsets which are also residual then their intersection is also dense and residual.

The following example illustrates how this technique works. 9.19.6 Example. Given a separable Banach space (X, 11·11) and an open subset A of X, consider a real function f on Ax X with the properties that for each y EX, f(x,y) is lower semicontinuous in X and for each y EX, f(x,y) is continuous in y. Then there exists a dense G 0 subset D of A such that f(x,y) is continuous at the points of D for every y EX.

Proof. Now (X, II· II) is a Baire space and by Theorem 9.10, A is a Baire space. Consider {y 0 } a countable dense subset of X. Given n EN, there exists a dense G 0 subset D 0 of A such that f(x,y 0 ) is continuous at the points of Do- By Theorem 9.19.4, D = () D 0 I

is a dense G 0 subset of A and f(x,y 0 ) is continuous at the points of D for all n EN. But given x EX, f(x,y) is continuous in y so we conclude that f(x,y) is continuous at the points of D for every y

E

D

X.

9.20. EXERCISES

I.

Consider a metric space (X, d) and a subset A of X. (i) (ii)

= C(A). Prove that A\ int Ais nowhere dense in X.

Prove that int C(A)

(iii) Prove that the following are equivalent (a)

A is nowhere dense in X,

(b) (c)

C(A) is dense in X, for any open subset G of X there exists an x E G and an r > 0 such that B(x; r)

2.

n A= 0.

Consider a metric space (X, d) and a subset E of X. (i)

Prove that if a subset A of E is nowhere dense in E then it is nowhere dense in X.

(ii) Prove that if a subset A of E is first category in E then it is first category in X. (iii) Show that if a subset A of E is second category in E then it is not necessarily second category in X. (iv) Prove that if a subset A of E is second category in E which is second category in X then A is second category in X.

!51

§9. Baire category

3.

(i)

For a metric space (X, d) with the property that every x EX is a cluster point of X, prove that singleton sets are nowhere dense in X.

(ii) Hence, using Haire's Theorem prove that both lR and Cantor's Ternary Set K are uncountable. 4.

(i)

(a) (b)

Show that the set of rationals
(ii) (a)

Prove that for a metric space (X, d) a real lower semi continuous function f on X has the property that for every open set G in IR, f-1(G) is an Fcr subset of X.

(b) 5.

Give an example to show that the converse is not true in general.

Consider a normed linear space (X, 11·11). Prove that if X is of second category then the unit open ball is of second category and every nonempty open subset of X is second category.

6.

(i)

Using the fact that norms ll·lloo and 11·11 1 are not equivalent on the linear space t;(O,l], deduce that the normed linear space (t;[O,l], 11·11 1) is not second category and so not complete.

(ii) Given a Banach space (X, 11·11), prove that every norm 11·11' on X which is lower semicontinuous with respect to the 11·11-norm is continuous with respect to the 11·11-norm. 7.

For the real function f on lR defined by f(x)

=I =0

x irrational } x rational

prove that there is no sequence {f0 } of continuous real functions pointwise convergent to f. (Hint: Suppose that there exists some such sequence {fn} of continuous real functions pointwise convergent to f. (i)

For each n E F\i, define En ={x

(ii) For each n

E

E

lR: fn(x) ~~}and prove that En is closed.

F\i, define Fn = n {Ek: k ~ n} and prove that Fn is closed. E F\i such that x E Fy.

(iii) Prove that x is irrational if and only if there exists a v (iv) Prove that the set of irrationals is U {Fn : n

E

F\i}.

(v) Prove that (iv) implies that the set of irrationals is first category in lR which it is not.)

152

8.

The fundamental mapping theorems

Prove that on a complete metric space any real function which is the pointwise limit of a sequence of continuous real functions is itself continuous on a residual set. But give an example to show that the pointwise limit may still be discontinuous at the points of a dense set.

9.

Prove that on a complete metric space any real function which is the pointwise limit of a monotone decreasing sequence of real lower semicontinuous functions is itself upper semicontinuous at the points of a residual subset.

I 0.

Consider an infinite dimensional linear space X where X is the union of countably many finite dimensional linear subspaces. (i)

Prove that with any norm 11·11, (X, 11·11) is first category in itself.

(ii) Prove that no infinite dimensional Banach space (X, 11·11) has a countable

Hamel basis. (iii) Show that the linear space E 0 cannot be given a norm 11·11 to make (E 0, II· II) complete. II.

Consider an infinite dimensional Banach space (X, 11·11). Prove that X contains a proper linear subspace of codimension one which is of second category in X and which is not complete.

§I 0. The Open Mapping Theorem

153

§ 10. THE OPEN MAPPING AND CLOSED GRAPH THEOREMS

When we study continuous mappings between metric spaces we are naturally led to examine the continuity of inverse mappings. It is useful to have criteria under which we can tell that a continuous one-to-one onto mapping is a homeomorphism without having to test the inverse mapping for continuity.

Compactness provides a simple criterion in that a

continuous one-to-one mapping from a compact metric space onto a metric space is a homeomorphism, (see AMS §9). However, it is reasonable to ask whether there is some simple criterion which applies to continuous linear mappings between normed linear spaces. We should notice at the outset that not every continuous linear one-to-one onto mapping is a homeomorphism. 10.1 Example. Consider the identity mapping id of ('C [0,1], 11·11 00 ) onto ('C [0, 1], 11·11 1). This mapping is continuous linear one-to-one onto but its inverse is not continuous, (see AMS §7). But we also note that ('C[0,1], ll·lloo) is complete and ('C[0,1], 11·11 1) is not complete so the normed linear spaces cannot be topologically isomorphic, (see Theorem 1.24.9). D The Open Mapping Theorem provides as a corollary an important criterion for a continuous linear mapping between normed linear spaces to be a topological isomorphism. 10.2 Definition. A mapping T from a metric space (X, d) into a metric space (Y, d') is said to be an open mapping if for every open set Gin (X,d), we have that T(G) is an open set in (Y, d'). 10.3 Remark. Not all continuous mappings are open mappings and not all open mappings are continuous. The identity mapping between IR with its usual metric and IR with the discrete metric is a simple example illustrating both these facts. However, a continuous oneto-one onto mapping is a homeomorphism if and only if it is an open mapping, (see AMS § 10). It is from this result that we develop our interest in open mappings. 0 The Open Mapping Theorem gives us conditions under which a linear mapping between normed linear spaces is an open mapping. For convenience we introduce the following notation.

!54

The fundamental mapping theorems

10.4 Notation. Denoting by B the open unit ball in the normed linear space (X, 11·11) and by B' the open unit ball in the normed linear space (Y, 11·11'), we use the following notation. rB rB'

= B(O; r)

=B(O; r)

and x+rB and x+rB'

= B(x; r) in (X, 11·11),

=B(O; r) in (Y, 11·11').

We approach the Open Mapping Theorem through the following lemmas.

10.5 Lemma. For a linear mapping T of a normed linear space (X, 11·11) onto a normed linear space (Y, 11·11') of second category, T(B) is a neighbourhood ofO in (Y, 11·11'). Proof. Since T is onto, Y = U { nT(B) : n EN}. However, (Y, 11·11') is second category so there exists some n0 EN such that n0T(B) has nonempty interior. Any open set in n0T(B) contains points of n0T(B) so there exists a y0 E n0T(B) such that y0 E int n0T(B) . But since translation is a homeomorphism, 0 E int (n 0T(B)- y0 ). Since Yo E n0T(B) there exists an x0 E n0 B such that Yo= Tx 0 and since Tis linear, n0T(B)- Yo= T(n 0B- x0). Now for any x E noB, we have II x-x 0 II < 2n 0 so n0T(B)- Yo s;; 2n 0T(B). Again since translation is a homeomorphism n0T(B)- Yo =-n' 0T"'("'B")---Y-oSo 0 E int (n 0T(B)- y0 ) s;; int 2n 0T(B). Since multiplication by a nonzero scalar is a homeomorphism, we conclude that 0 EintT(B); that is, T(B) is a neighbourhood of 0 in (Y, 11·11').

D

Notice that in this lemma there are very few restrictions on the linear mapping T; it is onto but not necessarily continuous. Lemma 10.5 reveals the surprising consequences of the range space being second category. The next lemma shows that if we assume continuity of the linear mapping from a complete domain space onto a second category range space, we are able to refine the results of Lemma 10.5 further. We frame the statement of the lemma to exhibit the implications of completeness for continuous linear mappings.

10.6 Lemma. For a continuous linear mapping T of a Banach space (X, 11·11) into a normed linear space (Y, 11·11'), if T(B) is a neighbourhood ofO in (Y, 11·11'), then T(B) is also a neighbourhood ofO in (Y, 11·11').

§I 0.

The Open Mapping Theorem

!55

Proof. If 0 E int T(B) then there exists a 0 > 0 such that oB' ~ T(B). For y E oB' we have -0 y E T(B) and so there exists an x 1 E B such that y 1 = Tx 1 and II y-y 1 II < Now

z·

~ B' ~ T(kB) so there exists an x2 Ek B such that y2 = Tx 2 and II (y-y 1)- y 2 II<~ By induction we obtain a sequence {xn} in X such that xnE 2

L, B, Yn = Txn and

0

II y-(yi+Yz + ... +yn) II<-. 2n

n

We write sn

=L xk. k=l

Since II Xn

II<~, we have 2

II Sm-Sn II

~

m

L k=n+l

II Xk II< n1_ 1 2

for all m > n. n

So {snl is a Cauchy sequence in (X, 11·11). But also II sn II~

L II xk II< 2.

k=l Now (X, 11·11) is complete so there exists an x EX such that {snl is convergent to x and II x

II~

2. But n

II Tx-y II'~ II Tx-Tsn II'+ II Tsn-

n

L Yk II'+ II L Yk- y II'

k=l k=l and since Tis continuous and linear we deduce that y =Tx, soy E 3T(B). Then oB'

~ 3T(B) and so~ B' ~ T(B); that is, T(B) is a neighbourhood of 0 in (Y, 11·11'). 0

We are now in a position to state and prove the Open Mapping Theorem. 10.7 The Open Mapping Theorem. A continuous linear mapping T of a Banach space (X, 11·11) onto a Banach space (Y, 11·11') is

an open mapping.

Proof. Consider an open set Gin (X, 11·11) andy E T(G). We show that y E int T(G). Let x E G be such that y = Tx. Since G is open there exists an r > 0 such that x + rB

~G.

Now Haire's Theorem 9.11 gives us that (Y, 11·11') is second category so from Lemmas 10.5 and 10.6 we deduce that there exists a o > 0 such that oB' s;;;; r T(B). Now y + oB' s;;;; y+rT(B) = T(x+rB) s;;;; T(G) soy E int T(G). This implies that T(G) is open in (Y, 11·11').

D

10.8 Remark. We note that Theorem 10.7 could be generalised to have range space (Y, 11·11') a normed linear space of second category. However, we have given the most

common form of the statement of the theorem and it is generally in this context where it has application. D

!56

The fundamental mapping theorems

Most applications of the Open Mapping Theorem are derived from the following corollary which gives a simple criterion for a continuous linear mapping to be a homeomorphism. 10.9 Corollary. A continuous linear one-to-one mapping T of a Banach space (X, 11·11)

onto a Banach space (Y, 11·11') is a topological isomorphism.

Proof. As Tis one-to-one and onto its inverse T-1 exists and is a linear mapping from (Y, II· II') onto (X, 11·11). From the Open Mapping Theorem 10.7, Tis an open mapping so it is also a homeomorphism. 0 Corollary 10.9 has a particularly useful application in determining whether norms are equivalent. 10.10 Corollary. For a linear space X, ifll·ll and 11·11' are norms such that both (X, 11·11) and (X, II· II') are complete and there exists a K > 0 such that

II x II $; K II x II' for all x EX then 11·11 and 11·11' are equivalent norms for X.

Proof. Consider the identity mapping id: (X, 11·11')

-7 (X,

11·11). The inequality

II x II $; K II x II' for all x E X implies that id is continuous. But id is linear one-to-one and onto so by Corollary 10.9, it is a topological isomorphism which implies that 11·11 and 11·11' are equivalent norms for X. 0 The second mapping theorem which is in many cases easier to apply than the Open Mapping Theorem is the Closed Graph Theorem and its proof can be derived directly from the Open Mapping Theorem. 10.11 Definitions. Given metric spaces (X, d) and (Y, d') the product metric dn for

X x Y is defined by dn((x,y), (x',y')) = max { d(x, x'), d'(y, y')} and we call (X x Y, dn) the product space of (X, d) and (Y, d'). It is easy to see that a sequence {(Xn.Yn)} is convergent to (x,y) in (X x Y, dn) if and only if (xnl is convergent to x in (X, d) and IYnl is convergent toy in (Y, d'). Given normed linear spaces (X, 11·11) and (Y, 11·11') over the same scalar field, X x Y is a linear space, the product norm 11-iln for X x Y is defined by II (x,y) lin= max {II x II, II y II'}

§ 10. The Open Mapping Theorem

157

and we call (X x Y, ll·llrr) the product space of (X, 11·11) and (Y, 11·11'). The product norm generates the product metric. Given a mapping T of a set X into a set Y the graph ofT is the subset GT of X x Y defined by GT

={(x,y) : y = Tx, x EX}.

When X andY are linear spaces over the same scalar field then Tis linear if and only if GT is a linear subspace of X x Y. We say that a mapping T of a metric space (X, d) into a metric space (Y, d') has a closed graph if GT is closed in (X x Y, drr). For mappings between metric spaces there is a close relation between continuity and having closed graph. We will explore this relation but first we give a more convenient way of expressing the fact that a mapping has a closed graph. 10.12 Lemma. A mapping T of a metric space (X, d) into a metric space (Y, d') has a closed graph if and only if for every sequence {xnl in X where {xnl is convergent toxin (X, d) and {Txnl is convergent toy in (Y, d'), we have y = Tx.

Proof. Suppose that GT is closed in (X x Y, drr) and that {Xn} is convergent to x in (X, d) and {Txn} is convergent toy in (Y, d'). Then { (xn,TXn)} is convergent to (x,y) in (X x Y, drr) . But since GT is closed, (x,y)

E

GT and therefore y = Tx.

Conversely, if (x,y) is a cluster point of GT in (X x Y, drr) then there exists a sequence { (xn,TXn)} in GT which is convergent to (x,y) in (X x Y, drr)· Then {Xn} is convergent to x in (X, d) and {Txnl is convergent toy in (Y, d').

D

From Lemma I 0.12 it is clear that any continuous mapping between metric spaces always has a closed graph. But further, it is not difficult to prove that a mapping T from a metric space (X, d) into a compact metric space (Y, d') with closed graph is continuous, (see AMS, §8). It is reasonable to ask whether there is a simple criterion for a linear mapping between normed linear spaces with a closed graph, to be continuous. We should note that not every linear mapping between normed linear spaces with closed graph is necessarily continuous. 10.13 Example. The differential operator D from ("CI[O,l], ll·lloo) into ("C[O,I], ll·lloo) defined by D(f) = f' has a closed graph; (this is the classical uniform convergence theorem for differentiation, see AMS §8). However, D is not continuous, We note that ("C [0, I], ll·lloo) is complete but ("C 1[0, I], ll·lloo) is not complete, (see AMS, §4).

D

!58

The fundamental mapping theorems The Closed Graph Theorem provides a criterion for the continuity of linear mappings

between normed linear spaces. Its proof is an elegant application of the Open Mapping Theorem. 10.14 The Closed Graph Theorem. A linear mapping T of a Banach space (X, 11·11) into a Banach space (Y, II· II') with a closed

graph, is continuous. Proof. Consider X renormed with norm 11·11 1 defined by II x 11 1 =II x II+ II Tx II'. Then

II Tx

II'~

II x II+ II Tx II'= II x 11 1 for all x EX soT is a continuous linear mapping from (X, 11·11 1) into (Y, 11·11').

We show that 11·11 1 and II· II are equivalent norms for X. llxll~llxii+IITxll'=llxll 1

Now

forallxEX

so if we show that (X, 11·11 1) is complete then from Corollary I 0.10 to the Open Mapping Theorem we will have that 11·11 and 11·11' are equivalent. We show that (X, 11·11 1) is complete. Consider a Cauchy sequence {xnl in (X, 11·11 1). Then since II Xm-Xn 11, =II Xm-Xn II+ II Txm-TXn II' we have that {xnl is a Cauchy sequence in (X, 11·11) and {Txnl is a Cauchy sequence in (Y, 11·11'). As both (X, 11·11) and (Y, 11·11') are complete there exist an x EX such that {xnl is convergent x in (X, 11·11) and ayE Y such that {Txnl is convergent toy in (Y, 11·11'). ButT has closed graph so y = Tx. Then II Xn-X 11 1 =II Xn-X II+ II Txn-Tx II' so {xnl is convergent toxin (X, 11·11 1), and we conclude that (X, 11·11 1) is complete.

10.15 EXERCISES I.

Consider an open mapping T of a metric space (X, d) into a metric space (Y, d'). (i)

Prove that T maps compact sets in (X, d) to compact sets in (Y, d').

(ii) Show that T does not necessarily map closed sets in (X, d) to closed sets in (Y, d').

2.

(i)

Consider the Banach space ('C [0, 1], ll·lloo) and 'C [0, I] with norm 11·11 1 and norm 11·11 2. Deduce from the Open Mapping Theorem that ('C[O,l], 11·11 1) and ('C[O,l], 11·11 2 ) are not complete.

(ii) A mapping T on 'C [0, I] is defined by t

T f(t) =

Jf(s) 0

ds.

0

159

§10. The Open Mapping Theorem

(a)

Prove that Tis a continuous one-to-one mapping of (t: [0, 1], ll·lloo) onto the linear subspace t:6 [0, I]

(b)

Show that

T- 1 is

= { f e t: 1[0, I]

: f(O) = 0}.

not continuous and explain why this does not contradict

the Open Mapping Theorem. 3.

M and N are closed linear subspaces of a Banach space (X, 11·11) such that X= M EB N. Prove that the norm 11·11' on X defined for z eX where z = x + y, x eM and y eN by

II z II' =II x II + II y II is an equivalent norm for X. 4.

Consider the finite dimensional linear space Xm over lR with basis {e 1, e2 , ... , em} and any norm on Xm. Assuming that (Xm, 11·11) is second category and using the Open Mapping Theorem with the mapping X f-7

(A('

Az'

0

°

0

'

Am)

of Xm into lR m where x = A1e 1 + A2e 2 + ... + A.mem, prove that (Xm, 11·11) is topologically isomorphic to (IRm, 11·11 2). 5.

(i)

Consider a linear mapping T of a normed linear space (X, 11·11) into a normed linear space (Y, 11·11'). (a) Prove that if T has closed graph then T has closed kernel. (b) But show that a linear mapping with a closed kernel does not necessarily have a closed graph; (see AMS §7).

(ii) (a) Prove that a linear functional on a normed linear space is continuous if it

has a closed graph. (b) Show that a linear mapping of a normed linear space into a finite dimensional normed linear space with closed graph is not necessarily continuous. 6.

For the mapping T of (t: 1[0,1], ll·lloo) into (t:[O,I], ll·lloo) defined by T(f)

=f' + f

prove that (i)

T has closed graph,

(ii) T is not continuous, (iii) (t: 1[0, I], ll·lloo) is not complete.

160

7.

The fundamental mapping theorems

A linear space X is a Banach space with respect to both norms 11·11 and II· II' and has the property that if a sequence {Xn} in X is convergent with respect to both norms 11·11 and II· II' then the limit point is unique. Prove that norms 11·11 and 11·11' are equivalent.

8.

(i)

For a linear mapping T of a Banach space (X, 11·11) into a Banach space (X, 11·11'), prove that iff o T EX* for all fEY* then T is continuous.

(ii) For linear operators T and S on a Hilbert space H we have (Tx, y)

= (x, Sy) for all x,y E H.

Prove that both T and S are continuous. (iii) For a linear operator T on a Hilbert space H, prove that the functional
=(Tx, x)

is continuous if and only if T is continuous. 9.

(i)

Show that Lemma 10.6 can be generalised to the following statement. For a linear mapping T from a Banach space (X, 11·11) into a normed linear space (Y, II· II') with closed graph, ifT(B) is a neighbourhood ofO in (Y, 11·11') then T(B) is also a neighbourhood ofO in (Y, 11·11').

(ii) Hence, show that the Open Mapping Theorem can be generalised to the

following statement. A linear mapping T from a Banach space (X, 11·11) onto a Banach space (Y, 11·11')

with a closed graph is an open mapping. (iii) Further, prove the following statement.

A linear one-to-one mapping T from a Banach space (X, II· II) onto a Banach space (Y, 11·11') with a closed graph is a homeomorphism. I 0.

(i)

Consider a normed linear space (X, 11·11), a proper closed linear subspace N and the quotient space (X/N, 11·11). Prove that the quotient mapping n : X -7 X/N defined by

n(x) = x + N.

is a continuous, linear and open mapping. (ii) Given a continuous mapping T of a Banach space (X, 11·11) onto a Banach space (Y, 11·11'), prove that the mapping T of the quotient space (XIker T, 11·11) onto (Y, 11·11') defined by T(x+ker T) is a topological isomorphism.

=Tx

§10. The Open Mapping Theorem

161

(iii) Show that the Open Mapping Theorem can be generalised to the following statement

A continuous linear mapping T of a Banach space (X, 11·11) into a normed linear space (Y, 11·11') where T(X) is second category in (Y, 11·11'), is an open onto mapping and (Y, II· II') is complete. II.

We saw how the Closed Graph Theorem was deduced from the Open Mapping Theorem. However, show that the Open Mapping Theorem can be deduced from the Closed Graph Theorem. (Hint: Consider a continuous linear mapping T of a Banach space (X, 11·11) onto a Banach space (Y, 11·11'). As in Exercise 10, consider the mapping

-

X

T: ker T

---7

-

Y defined by T(k+ker T) = Tx.

Show that T is one-to-one, onto and has a closed graph and therefore T has a closed graph. Apply the Closed Graph Theorem to give that T is continuous and deduce that Tis an open mapping.)

12.

(i)

Consider a complex Banach algebra A. A linear functional <jl on A is said to be

multiplicative if also <jl(xy) = <jl(x) <jl(y) for all x,y EA. Prove that every multiplicative linear functional <jl on A is continuous and II <jl II:-:;; I. (Hint: Suppose on the contrary, that there exists an x 0 E A, II x0 II < I and <jl(x 0) = I. Consider Yo=

L x~

and show that x0 + x0y0 =Yo a contradiction to

k=l

the contrary hypothesis.) (ii) A commutative Banach algebra is said to be semisimple if the set of multiplicative linear functions is total on A. Prove that every algebra homomorphism from a complex Banach algebra A into a semisimple complex commutative Banach algebra B is continuous. (iii) Consider a semisimple complex commutative algebra A. Prove that all norms under which A is complete are equivalent.

162

13.

The fundamental mapping theorems

Consider a separable Banach space (X, 11·11) with a Schauder basis {en}. Consider X with norm 11·11': X ~ IR where for x

=L A.kek, k=l n

II x II'= sup{ II L A.kek II : n EN} k=l

(i)

Prove that 11·11 and 11·11' are equivalent norms for X.

(ii) Hence, deduce that for any separable Banach space with a Schauder basis, the

coordinate functionals are continuous. (Hint: See Exercise 1.26.17 and Theorem 1.25.16)

163

§II. The Uniform Boundedness Theorem

§II. THE UNIFORM BOUNDEDNESS THEOREM

Our motivation for the study of uniform convergence is to determine the properties of the limit mapping of a pointwise convergent sequence of scalar mappings on a metric space. The three classical theorems on uniform convergence of sequences of real functions on an interval of the real line show that uniform convergence conditions are sufficient to guarantee the "good behaviour" of the limit function. In general, there is particular interest in the continuity of the limit mapping of a pointwise convergent sequence of continuous mappings. In Dini's Theorem (AMS §9) the compactness of the domain space ensures that a pointwise convergent monotone sequence of continuous mappings with a continuous limit mapping is uniformly convergent. As an extension of this type of investigation it is reasonable to ask, for a pointwise convergent sequence of continuous linear mappings between normed linear spaces, whether the limit mapping is continuous and whether convergence must necessarily be stronger than pointwise convergence. We will derive a useful result in this line as a consequence of a more general question concerning boundedness of a set of continuous linear mappings.

11.1 Definitions. Consider a set ~ of continuous linear mappings from a normed linear space (X, 11·11) into a normed linear space (Y, 11·11'). We say that the set

~

is pointwise bounded if for each x EX the set {Tx: T E

~)

is a

bounded set in (Y, 11·11'). We say that the

set~

is uniformly bounded

if~

is a bounded set in

(~(X,Y),

11·11) the

normed linear space of continuous linear mappings from X into Y.

11.2 Remark. If the set

~ is uniformly bounded then there exists an M > 0 such that II T II S: M for all T E ~ •

Therefore, for any x EX, II Tx II' S: II T II II x II S: M II x II so

~

for all T

E ~.

is pointwise bounded. However, the converse is not true in general.

11.3 Example. Consider the set of continuous linear functionals {fn: n EN) on (E 0 , 11·11 00 ) defined, for x {1.. 1, l..z, . .. , An, ... ,}, by

=

fn(x) =nAnNow for any given x0 EX there exists an n0 EN such that An= 0 for all n > n0, so I f0 (Xo) IS: no II x0 lloo for all n EN;

D

164

The fundamental mapping theorems

that is, {fn: n EN} is pointwise bounded. Nevertheless, II fn II = n for every n E N, so {fn: n EN} is not uniformly bounded. But notice that the normed linear space

(E 0 , 11·11

00 )

D

is not complete.

On the other hand, for a set of continuous linear mappings on a Banach space, pointwise boundedness does imply uniform boundedness. This is the content of the Uniform Roundedness Theorem. This theorem, like the Open Mapping Theorem uses Baire category arguments and in particular Haire's Theorem 9.11 that a complete normed linear space is second category. 11.4 The Uniform Roundedness Theorem.

If a set 'J' of continuous linear mappings from a Banach space (X, 11·11) into a normed linear space (Y, 11·11') is pointwise bounded then it is uniformly bounded.

Proof. For each n EN, write Fn = { x EX : II Tx II' :<: : n for all T E 'J'}. 'J' is continuous, T-'(B'[O; nl) is closed and so Fn is closed. But also since 'J' is pointwise bounded, for each x EX there exists an n EN such that

Since each T x

E

E

Fn; that is, X= U {Fn: n EN}. Now (X, 11·11) is complete, so by Haire's Theorem 9.11

there exists some n 0 EN such that Fn 0 has nonempty interior. Therefore there exists some x 0 EX and r > 0 such that x 0 + rB

~

Fn 0 , so

II T(x 0+rB) II':<::: n0 Now since Tis linear, T(rB)

which implies that

II Tx II' :<: : 2n 0

for all x E rB

II Tx II' :<: : 2 n°

for all x E B and all T

r

2n 0 Therefore, II T II :<: : -r- for all T 11.5 Remark.

for all T E 'J'.

= T(x 0+rB)- Tx0 , so

E

E

'J'; that is, 'J' is uniformly bounded.

'J'.

D

We note that, as with the Open Mapping Theorem, the Uniform

Roundedness Theorem could be generalised to have domain space (X, 11·11) a normed linear space of second category. However, the form given in Theorem 11.4 is generally sufficient for our purposes. D One of the most important consequences of this theorem concerns the problem we mentioned in introducing the Uniform Roundedness Theorem, the continuity of the limit of a pointwise convergent sequence of continuous linear mappings.

§II. The Uniform Boundedness Theorem

165

11.6 Definition. We say that a sequence {T n} of continuous linear mappings of a normed linear space (X, 11·11) into a normed linear space (Y, II· II') is pointwise convergent to a mapping T if for each x EX, { T n(x)} is convergent to Tx in (Y, 11·11'). Of course the pointwise limit of a sequence of continuous linear mappings is itself linear. But it need not be continuous. 11.7 Example. Consider the normed linear space (.t 1 , ll·lloo) and the sequence {fn} of linear functionals on

.t I defined,

for x = {A 1,

A:z, ... , An, ... }, by

n

I. Ak . k=I Now I fn(x) I~ n II x lloo for all x EX, so for each n EN, fn is continuous. fn(X) =

But {fn} is pointwise convergent to the linear functional f on f(x) =

.t I where

I. Ak k=I

I. I Ak I for each n EN. k=n+I However, for Xn {I, I, ... , I, 0, ... }, II Xn lloo =I but f(xn) = n so f is not nth place continuous on (.t 1, ll·lloo)· since x E .t 1 and I fn(x)- f(x) I ~

=

But notice again that the normed linear space (.t 1, ll·lloo) is not complete.

D

11.8 Remark. We should notice that the pointwise convergence of a sequence of continuous linear mappings {T n} to a continuous linear mapping T does not necessarily imply that the set {Tn: n EN} is uniformly bounded. In Example 11.3 we have a sequence {fnl which is not uniformly bounded. But for any given x0 E E0 , x0 = {A~, A~, ... , A~ ... } there exists an n0 EN such that A~= 0 for all n > n0 . Then fn(x 0 ) = 0 for all n > n0 , so {fnl is pointwise convergent to the zero functional.

0

However, for a sequence of continuous linear mappings on a Banach space we have the following significant corollary to the Uniform Roundedness Theorem. 11.9 The Banach-Steinhaus Theorem.

If {T n} is a sequence of continuous linear mappings of a Banach space

(X, 11·11) into a

normed linear space (Y, 11·11'), pointwise convergent to a mapping T, then T is a continuous linear mapping of (X, 11·11) into (Y, 11·11').

166

The fundamental mapping theorems

Proof. Since {Tnl is pointwise convergent then the set {Tn: n eN} is pointwise bounded. From the Uniform Boundedness Theorem 11.4, {T n : n e N} is uniformly bounded; that is, there exists an M > 0 such that

II T n II ~ M

for all n e N.

II T 0 (x) II' ~ M II x II

Therefore, But for each x eX,

II Tx II'

~

for all n e N and x eX.

II Tx-Tn(x) II'+ II Tn(x) II' ,

so as {Tn} is pointwise convergent to T we conclude that

II Tx II'

~

M II x II

for all x eX;

that is, T is continuous.

0

However, the Banach-Steinhaus Theorem does not imply that pointwise convergence of a sequence of continuous linear mappings on a Banach space is any stronger than pointwise convergence; it does not provide a generalisation of Dini's Theorem. 11.10 Example. Consider the Banach space (c0 , 11·11 00 ) and the sequence of continuous

linear functionals {fnl on c 0 where for x =

p. 1, ~ •.•• , A.n •... },

fn(X) =An. Now {fn} is pointwise convergent to the zero functional. From the Uniform Boundedness Theorem 11.4, the set {fn : n e N} is uniformly bounded. But for en= {0, ... , 0, 1, 0, ... }, nth place

II fn II ~ f 0 (en) = 1 so {fnl is not convergent to the zero functional under the dual norm on (c0 , 11-iloo).

0

Another important application of the Uniform Boundedness Theorem is in characterising the boundedness of a set in a normed linear space. 11.11 Definitions. A nonempty set A in a normed linear space (X, 11·11) is said to be weakly bounded if f(A) is a bounded set of scalars for each f eX*. 1\

A nonempty set B in the dual (X*, 11·11) is said to be weak * bounded if x(B) is a bounded set of scalars for each x e X. There are occasions when we deduce the boundedness of a set from its weak boundedness or weak* boundedness, which is sometimes easier to test. 11.12. Theorem. (i) A nonempty set A in a normed linear space (X, 11·11) is bounded if and only if it is weakly

bounded. (ii) lf(X, 11·11) is complete, a nonempty set Bin the dual (X*, 11·11) is bounded it is weak

* bounded.

if and only if

167

§II. The Uniform Boundedness Theorem

Proof. For any x EX and f EX*, I f(x) I

~

II f 1111 x II.

So if A is bounded in (X, 11·11) then it is weakly bounded and if B is bounded in (X*, 11·11) then it is weak * bounded. Conversely, A

(i)

f(A)

= {f(x) : x E A} = {x(f) : x E A} is bounded for each f EX*; that is, the set

{ ~ : x E A} is pointwise bounded on (X, 11·11). But (X*,II·II) is always complete so from the Uniform Roundedness Theorem, the set A

{ x: x E A} is uniformly bounded on (X*, 11·11), which implies that A is bounded in (X, 11·11). A

(ii)

x(B)

={f(x) : x E B} is bounded for each x EX; that is the set {f: fEB} is pointwise

bounded on (X, 11·11). But here we assume that (X, 11·11) is complete so from the Uniform Roundedness Theorem the set {f: fEB} is uniformly bounded on (X, 11·11), which implies that B is bounded in

0

(X*, 11·11). 11.13

Remark. In Example 11.3 we have a set in the dual of an incomplete normed linear

space which is weak * bounded but not bounded. So the completeness condition in Theorem 0 I 1.12(ii) is significant. Another application of the Uniform Roundedness Theorem concerns bilinear mappings. 11.14 Definitions. Given linear spaces X, Y and Z over the same scalar field, a mapping

p of X x Y into Z is said to be a bilinear mapping if for given x EX, the associated mapping p,: Y ~ Z defined by p,(y) = P(x,y) and for given y E Y, the associated mapping Py: Y ~ Z defined by Py(Y) = P(x,y) are both linear. If (X, 11·11), (Y, II· II') and (Z, II· II") are normed linear spaces and for given x EX, Px is

continuous and for given y E Y, Py is continuous then p is said to be separately continuous. If p is continuous on X x Y with the product norm then p is said to be jointly continuous. If a bilinear mapping is jointly continuous then clearly it is separately continuous. The Uniform Roundedness Theorem enables us to establish a converse result. ll.IS Theorem. Consider normed linear spaces (X, 11·11), (Y, 11·11') and (Z, 11·11") and a

bilinear mapping then

p of X x Y into Z which is separately continuous.

p is jointly continuous.

If (X, 11·11) is complete

168

The fundamental mapping theorems

Proof. For each y E Y, II y II = I consider the associated continuous linear mapping

~Y

of

(X, 11·11) into (Z, 11·11"). Then there exists a Ky > 0 such that II ~y(x) II" :;::; Ky II x II But then the set { ~Y : y E Y, II y II

for all x EX.

= I } is pointwise bounded on (X, 11·11).

Since (X, 11·11) is

complete, by the Uniform Boundedness Theorem the set is uniformly bounded on (X, 11·11); that is, there exists a K > 0 such that II ~y(x) II" :;::; Kll x II for all x EX and y E Y, II y II= I. But since ~is bilinear this implies that II ~(x,y) II" :;::; Kll x 1111 y II for all x EX and y E Y

0

and this inequality gives us that~ is jointly continuous. We now present an interesting application which exhibits the power of the Uniform Boundedness Theorem. Although we have seen, (from Example 3.16) that every f E 'C [-7t,7t] has mean square representation by its Fourier series, it was shown by duBois Reymond in 1876 that the Fourier series for such a function may actually fail to converge pointwise to the function. He actually constructed a continuous function whose Fourier series is divergent at 0. The proof given here is an existence proof and it does also give us

more information about the pointwise convergence of a Fourier series, (see Exercise 11.17 .5). 11.16 The du Bois Reymond Theorem.

There exists an f

E

'C[-1t,1t] whose Fourier series is divergent at 0.

Proof. Given f E 'C [-7t,7t] we write

a

n

°

I,

Sn (t; f)= 2 +

(ak cos kt

+ ~k sin kt)

k=l

the nth partial sum of the Fourier series for f, where a 0 , ak and ~k are the Fourier coefficients for f defined with respect to the orthonormal set I I I . {{21[ '{1[ cos nt , {1t sm nt : n E

w}

N

.

We show that { Sn(O; f)} is divergent for some f E 'C[-1t,1t]. Suppose that { Sn(O; f)} converges for every f E 'C[-1t,1t]. Now for any f E 'C[-1t,1t].

sn (0; f)=

a

20 +

n

I,

ak.

k=l

From the formula for the Fourier coefficients we have 1t

Sn (0 ; f)=~

f f(t) Dn(t) dt

-lt

I

where

Dn(t) =

n

2 + I,

cos kt.

k=l

But

sin (n+t)t . t fortE(-7t,7t)\{O}. 2 s1n 2

I69

§II. The Uniform Boundedness Theorem

Notice that cjl 0 (f) = S0 (0; f) is a linear functional on 'C[-1t,1t] and 1t

I cjl 0 (f)

I:;::;~

II flloo

f I D (t) I dt. 0

-lt

So
II
~

f I D (t) I dt. 0

-lt

~

n In I sin (n+±)t I d J21 sin(2n+ I)t I dt But J I DnCt) I dt = 2 I I D (t) I dt ;:: 2 ------"-- t = 2 1t

0

-lt

0

t

0

0 (k+l)lt

(k+l)lt

i

n I

;:: 2

2n+l

J

I sin(2~+I)t I

k=O~

n-I

dt ;:: 2 k~

2n+l

(~~~ ~1t

11

2n+I

sin(2n+ I )t I dt

2n+l

4 n-I I =ni.k+l k=O which is divergent and this contradicts the boundedness of {II
11.17

EXERCISES

I.

Use the Uniform Boundedness Theorem II.4 to show that the normed linear spaces (E 0 , 11·11 00 ) and (1 1, ll·lloo) are not complete.

2.

(i)

For a normed linear space (X, 11·11), {f0 } is a bounded sequence in (X*, 11·11). Prove that the set {x EX : lim f0 (x) exists} is a closed linear subspace of

(X, 11·11). (ii) For a Banach space (X, 11·11), {f0 } is a pointwise bounded sequence in (X*, 11·11). (a) Prove that Y

={x EX : lim f (x) exists} is a closed linear subspace 0

(X, 11·11). (b) Prove that if Y is second category in (X, 11·11) then Y =X. (c) Prove that the set {x EX: lim f 0 (x) = 0} is a closed linear subspace of (X, 11·11).

170

3.

The fundamental mapping theorems

(i)

For a sequence {an} of complex numbers, prove that if the series LanAn converges for all

X

= {AI, /...z, : . . , An, ... } E c then the series Lan is

absolutely convergent. (ii) For I < p < oo and a sequence {an} of complex numbers, prove that if the Series LanAn converges for all then { al' a 2 ,

4.

... ,

an, ... }

E

X=

lp

Az, ... , An, ... } E lp I I where p + q = I. {A,,

Consider the normed linear space 'fl[O,I] with norm I

II p

f p(t)

11 1 =

I

dt

0

I

Show that

~(p, q) = Jpq(t) dt is a bilinear functional on

'fl [0, I]

x 'fl [0, I] which is

0

separately continuous but not jointly continuous.

5.

(i)

Consider (X, IHI) a Banach space and (Y, 11·11') a normed linear space and

~

a

subset of -r, (X, Y). Suppose that for some x0 EX the set {Tx 0 : T E ~ } is unbounded in (Y, 11·11'). Prove that the set { x EX : {Tx : T E ~} is bounded} is first category in (X, 11·11). (ii) For any given f

a

E

t; [-1t,1t] consider the partial sum of the Fourier series off n

°

I, (ak cos kt + ~k sin kt) k=l (a) Prove that for each t E [-1t,1t], the set

Sn (t; f)= 2 +

{f

E

t;[-1t,1t]: the sequence { Sn(t; f)} is bounded},

is first category in (t; [-1t,1t], 11·11 00 ) . (b) Prove that there exists a continuous function on [-1t,1t] whose Fourier series is divergent at each point of a dense subset of [-1t,1t]. 6.

Consider an algebra A which is also a Banach space (A, 11·11) where multiplication is a continuous function of x for fixed y and is a continuous function of y for fixed x. (i)

Prove that there exists a k > 0 such that II xy

II~

k II x 1111 y II for all x,y EA.

(ii) Prove that there is an equivalent norm for A under which A is a Banach algebra. (iii) Suppose that A has an identity e. Using the left regular representation of A in -r,(A) defined by

a H Ta

where Tax= ax for all x E A

prove that 11·11' where II a II'= II Ta II is an equivalent norm for A under which A is a unital Banach algebra.

V.

TYPES OF CONTINUOUS LINEAR MAPPINGS

A continuous linear mapping between normed linear spaces generates in a natural way, a continuous linear mapping between the duals of those spaces called its conjugate mapping. In a Hilbert space, because of the close relation between the space and its dual, a continuous linear operator has associated with it and its conjugate another continuous linear operator on the space, called its adjoint. Projection operators on Banach and Hilbert spaces are an important set of operators especially for the decomposition of the space into simpler component subspaces. Compact operators on Banach and Hilbert spaces are natural generalisations of finite rank operators and many of their properties are derived from those of finite rank operators. §12. CONJUGATE MAPPINGS The process of forming conjugate mappings is a natural extension of the duality between Banach spaces and their duals. Although the technical development of the idea may at first seem contrived, nevertheless it provides a remarkably powerful technique in application. 12.1 Definition. Given a continuous linear mapping T of a normed linear space (X, 11·11) into a normed linear space (Y, 11·11'), the mapping T' of (Y, 11·11')* into (X, 11·11)* called the

conjugate mapping ofT is defined by T'(g) =goT; (T' g)(x) = g(Tx) that is,

for all x EX.

12.2 Remark. Since Tis a continuous linear mapping of (X, IHI) into (Y, 11·11') and g is a continuous linear functional on (Y, 11·11'), the composite mapping go T is a continuous linear 0 functional on (X, 11·11); that is, T' as defined does map Y* into X*. We now determine the particular properties of the conjugate mapping. 12.3 Theorem. Given a continuous linear mapping T of a normed linear space (X, 11·11) into a normed linear space (Y, 11·11'),

172

(i)

Types of continuous linear mappings

the conjugate mapping T' of (Y*, 11·11') into (X*, 11·11) is continuous and linear and

liT' II=IITII, (ii) the mapping T H T' is an isometric isomorphism of-r,(X,Y) into -r,(Y*, X*).

Proof. (i)

Since composition is distributive over addition and is homogeneous over multiplication

by a scalar, we have that T'(g 1+g 2) = (g 1+g 2) oT = g 1oT + g2oT = T'(g 1)+ T'(g 2) for g 1, g2 E Y*. and

T'(ag) =ago T = aT'(g)

for a scalar and g

E

Y*,

so T' is linear. Since T is a continuous linear mapping we have that II T' g II= II goT II :0:: II T 1111 g II

for all g E Y*

so T' is continuous and II T' II :0:: II T II . However, given £ > 0 there exists an x0 EX, II x0 II :0:: I such that II Tx 0 II' > II T II - e . By Corollary 6.3 of the Hahn-Banach Theorem, for Tx0

E

Y there exists a g0 E Y*,

II g0 II = I such that g0(Tx 0) = II Tx 0 II. Now II T' II;:: II T 'g0 II;:: I T'g0(x 0 ) I =I g0(Tx 0) I =II Tx 0 II'> II T II-£. Therefore, II T' II = II T II.

Figure 17

T': Y*

~X*

generated from T:

by T' g(x) = g(Tx)

for all x EX.

X~

Y.

§ 12.

(ii)

Conjugate mappings

173

Again, since composition is distributive over addition and is homogeneous over

multiplication by a scalar, we have (T 1+T 2)'(g) = go(T 1+T 2 ) = goT 1 + goT 2 = T 1'(g) + T2'(g) = (T 1' + T 2')(g) and

for all g E Y*

(aT)'(g) = go aT= a(g o T) = aT' (g)

so the mapping T

H

for a scalar and all g E Y*,

T' is linear.

However, since II T' II= II T II, the mapping T H T' is an isometric isomorphism of into ~(Y*,X*).

~(X,Y)

0

12.4 Remark. It is useful to examine the behaviour of conjugates under composition. (i) Given a continuous linear mapping T from (X, 11·11) into (Y, 11·11') and a continuous linear mappingS from (Y, 11·11') into (Z, 11·11"), then SoT:X~Z

and

(SoT)': Z*

Then

So (ii)

(So T)'(h)

~X*.

= h o (SoT)

for h E Z*

=T'(hoS)

wherehoSEY*

= T' o S'(h).

for all hE Z*

(SoT)' =T'oS'. For the special case where Tis a continuous linear operator on (X, 11·11) then we saw in

Section 4.11 that

~(X)

is a normed algebra with identity. Then

~(X*)

is also a normed

algebra with identity. So in this case we need the composition properties of conjugates given in (i). But also for the identity operator I on (X, 11·11), I'(f) = f o I = f for all f EX so I' is the identity operator on (X*, 11·11). We notice that, because products are reversed under the mapping T H T' of ~(X) into ~(X*), the mapping is not an algebra isomorphism, but it does preserve identities.

0

12.5 Example. Consider finite dimensional linear spaces Xn and Ym and T a linear mapping of Xn and Y m· Consider a basis ( e 1, e2, dual basis ( g 1, g 2,

... ,

•.. ,

en} for Xn and a basis for Ym with

* (see Theorem 4.10.5). gm} for Ym;

Now T has matrix representation T = [ajk] j E (I, 2, . . . , m} } k

E

(I, 2, . . . , n}

with respect to the bases for Xn and Y m· For x

=A. 1e 1 + A.2e2 + ... + "-nen

and g

=J..I. 1g 1 + J..I. 2g2 + ... + Jlmgm

n

we have

Tx = l:ajk "-k j=l m

and

g(Tx) =

n

l: l:

j=l k=l

ai "-k j:i. J

J

Types of continuous linear mappings

174

n

and therefore

T'g

=I aij ili

·

j=l

We conclude that T' has matrix representation T' = [akjl j

E

{I, 2, . . . , m}

}

kE{l,2, . . . ,n} the transpose of the matrix representation forT.

D

12.6 Example. Consider the continuous linear operator Ton (c 0 , 11·11 00 ) defined for

x

={A

1,

A.2 ,

... ,

An, ... } by Tx = { A. 1,

~ , ••• , ~ , • • • } •

We note that any continuous linear functional f on (c 0 , ll·lloo) is of the form f(x) =

L

A.n iln where { jl 1, jlz, ... , !ln, ... } E ~ 1; (see Example 5.3.1).

So T'f(x) = f(Tx) =

L:

generated by {Ill, ~2

A. nn lln· Therefore T'f is the continuous linear functional on c 0

, ... ,

llnn , ... }

E

~ 1.

D

The following example seems trivial but we will see that it is quite useful. 12.7 Example. Consider any normed linear space (X, 11·11) with proper linear subspace A. The inclusion mapping in: A

~

X is defined by

in( a) =a Now in':

X*~

for all a EA.

A* is defined by (in'f)(a) = f(in(a)) = f( a)

in'(f) = f

so

for all a E A

IA-

Moreover from the Hahn-Banach Theorem 6.2 we have that every continuous linear functional f on A can be extended as a continuous linear functional f on X so { f A : f EX*} = A*. That is, in' maps X* onto A*.

I

D

We apply this result in Theorem 12.12. A number of applications of conjugate mappings concern one-to-one and onto relations. There is a duality between these concepts as the following result shows. 12.8 Theorem. Consider a continuous linear mapping T from a normed linear space (X, 11·11) into a normed linear space (Y, 11·11'). Then T' is one-to-one if and only ifT has

dense range.

175

§12. Conjugate mappings

Proof. If T'g = 0 for some g eY* we have T'g(x) = 0 for all x eX. So g(Tx) = 0 for all x EX; that is, g(T(X)) = 0. If T(X) = Y then, since g is continuous, g = 0; that is, T' is one-to-one. Conversely, suppose T(X) "# Y. Then by Corollary 6.7 to the Hahn-Banach Theorem there exists a nonzero g0 E Y* such that g0(T(X)) = 0. Then g0(Tx) = 0 for all x EX which implies that T' g0 = 0. But then T' is not one-to-one.

0

12.9 Definition. Given a continuous linear mapping T from a normed linear space (X, 11·11) into a normed linear space (Y, 11·11'), the conjugate mapping T ': Y* --7 X* has its own conjugate mapping T": X**

--7

Y** called the second conjugate ofT.

12.10 Remark. It is natural to ask about the relation between T: X conjugate T": X**

--7

--7

Y and its second

Y**. If we regard X as embedded in X** andY as embedded in Y**

under the natural embedding, then T" is an extension ofT from X to X**. Consider T" 15( • Then for all x EX A

A

(T" x)(g)

= x(T'g) = g(Tx)

1\ = Tx(g)

for all g E Y*.

1\

A

T"(x) = Tx .

So

A

A A

A

A

Therefore, if we define a mapping T: X

--7

Y by

1\

T(x) = Tx A

we see that T" is an extension ofT to X**.

0

It is of interest to investigate duality relations between a continuous linear mapping and its conjugate when one of these mappings is a topological isomorphism. 12.11 Theorem. Consider a continuous linear mapping T of a normed linear space (X, 11·11) into a normed linear space (Y, 11·11').

1fT is a topological isomorphism of X onto Y then T' is a topological isomorphism of Y* onto X* an.d(T')- 1 = (T- 1)'. (ii) If (X, 11·11) is complete and T' is a topological isomorphism ofY* onto X* then Tis a

(i)

topological isomorphism of X onto Y.

Proof. (i)

If T is a topological isomorphism of X onto Y

then ToT- 1 =I on Y and T- 1 oT =I on X. Using the properties of conjugates under composition, Remark 12.4, we have (ToT- 1)' = (T- 1)' oT' =I' on Y = I on Y* and

(T- 1 oT)' = T' o(T- 1)' = I' on X= I on X*.

176

Types of continuous linear mappings

So we conclude from algebraic considerations that T' has an inverse mapping (TT 1 = (T- 1)' from Y* onto X* and since T- 1 is continuous then from Theorem 12.3(i), (T- 1)' = (T ')- 1 is continuous. (ii)

If T' is a topological isomorphism of Y* onto X*, then from (i) it follows that T" is a

topological isomorphism of X** onto Y**. Therefore there exist m, M > 0 such that m II F II ~ II T "F II' ~ M II F II for all FE X**. A

So restricting T" to X we have from Remark 12.1 0, that m II x II

II Tx II' ~ M II x II

~

for all x E X

which implies that Tis a topological isomorphism of X into Y. We show that T(X) = Y. Since (X, 11·11) is complete then T(X) is complete and so is closed in (Y, 11·11'). ButT' is one-to-one so from Theorem 12.8, T(X) = Y.

0

We now demonstrate how conjugate mapping techniques can produce unexpected powerful results. Some of the most interesting applications are to do with establishing reflexivity properties for Banach spaces. 12.12 Theorem. Given a reflexive Banach space (X, 11·11), a normed linear space (Y, 11·11')

topologically isomorphic to (X, 11·11) is also reflexive.

Proof. Given that T is a topological isomorphism of (X, IHI) onto (Y, 11·11') we have by Theorem 12.ll(i) that T' is a topological isomorphism of Y* onto X* and T" is a topological isomorphism of X** onto Y**. A

From Remark 12.10 we see that T" is an extension ofT which is a topological isomorphism "

1\

1\

1\

of X onto Y. However, since (X, 11·11) is reflexive, X** = X and soT" = T and we conclude A

0

that Y** = Y; that is, (Y, II· II') is reflexive.

12.13 Theorem. Given a reflexive Banach space (X, II· II), every closed linear subspace A

is reflexive.

Proof. Consider the inclusion mapping in: A Now

in': X*

and

in": A**

-7

-7

X.

A* A

-7

X**= X

since X is reflexive.

For any G E A** and f EX* we have that A

in" G(f) = x(f). A

so

G(in'(f)) = x(f).

But from Example 12.7 we have that

in'(f)

= f IA A

so

G(fiA)= x(f) forallfEX*

(*)

Since A is a proper closed linear subspace of X we have from Corollary 6.7 to the HahnBanach Theorem that for any x EX\ A there exists an f EX* such that f(A) = 0 and f(x) 0.

*

§ 12. Conjugate mappings

177

A

So in ( *) we have that x satisfies x e A and A

A

in" maps G to a in X. A

A

A

in" maps a to a in X.

But

Since in' is onto, we have from Theorem 12.8 that in" is one-to-one. A

A

Therefore, G =a and we conclude that A**= A; that is, A is reflexive.

0

Now let us return to the definition of a conjugate mapping. For noncontinuous linear mappings we have difficulty in defining a conjugate as in Definition 12.1. However, in some cases the pursuit of a generalisation is of value. 12.14 Definition. Given a linear mapping T of a normed linear space (X, 11·11) into a normed linear space (Y, 11·11') which is not necessarily continuous. ForgeY*, goT is a linear functional on (X, 11·11) which may not be continuous. For the set D(T') = {g eY*: goT is continuous on X} we define the mapping T' of D(T') into X* called the conjugate mapping ofT by that is,

T'(g) = goT; T' (g)(x) = g(Tx)

for all x eX.

Of course if T is continuous on X then this definition is that given in Definition 12.1. The usefulness of this generalisation depends on the size of D(T'). We show that for linear mappings with closed graph this set is quite substantial. 12.15 Theorem. For a linear mapping T of a normed linear space (X, II· II) into a normed linear space (Y, 11·11') with closed graph, D(T') is total in Y*.

Proof. We need to show that given a Yo e Y, Yo'* 0, there exists a g0 e D(T') such that g0(y0) 0. Since Tis linear, (0, y0) ii!: GT. Since GT is a closed linear subspace in (X x X, 11·111t) we have from Corollary 6.7 to the Hahn-Banach Theorem that there exists a nonzero continuous linear functional h on (X x X, 11·111t) such that

'*

h(O,y 0)

'* 0

and h(x,Tx) = 0

Define g0 e Y* by

for all x eX.

g0(y) = h(O,y) so that g0(y 0)

'* 0.

f(x) = h(x,O) 0 = h(x,Tx) = f(x) + g0(Tx)

for all x eX.

If we define f e X* by then So

g 0 o T =- f eX*, which implies that g0 eD(T'). It is instructive to see how the size of D(T') is related to the continuity ofT.

0

178

Types of continuous linear mappings

12.16 Theorem. For the linear mapping T of a normed linear space (X, 11·11) into a normed

linear space (Y, 11·11'), D(T') = Y if and only ifT is continuous.

Proof. If Tis continuous then clearly D(T') = Y*. Conversely, if D(T') = Y* then sup{ I g(Tx) I :II x II::;; I} =II goT II for each g E Y*. That is, the set { II Tx II : II x II::;; I } is weakly bounded so by Theorem 11.12(i) from the Uniform Roundedness Theorem, the set { II Tx II : II x II::;; I} is bounded which implies that T is continuous. D No matter what the size of D(T'), the conjugate T' has regular properties of its own. 12.17 Theorem. For a linear mapping T of a normed linear space (X, 11·11) into a normed linear space (Y, 11·11'), the conjugate mapping T' has a closed graph in (Y* x X*, ll·lln)·

Proof. Consider a sequence {gn} in D(T ') which is convergent to g

E

Y* and where the

sequence {T'gn} is convergent to some f EX* Then for x EX, {gn(Tx)} is convergent to g(Tx) but gn(Tx)

=T' gn(x) ~ f(x) as n ~

oo

for all x EX.

Therefore, g(Tx) = f(x) for all x EX. Since f is continuous we have that g E D(T') and T' g = f so T' has closed graph.

D

From Theorem 12.17 we can deduce a further property. 12.18 Corollary. For a linear mapping T of a normed linear space (X, 11·11) into a Banach

space (Y, 11·11'), D(T') is closed.

Proof. Consider g

E

D(T). For any sequence {gn} in D(T') convergent tog we have

II T'gm-T'gn II::;; II T' 1111 gm-gn II

for all m,n EN

so {T'gnl is a Cauchy sequence in (Y, 11·11'). But (Y, 11·11') is complete so there exists a y E Y such that {T'gnl is convergent toy. ButT' has closed graph soy= T' g and therefore D D(T') is closed.

§ 12. Conjugate mappings

179

12.19. EXERCISES

1.

A continuous linear mapping T of a nonned linear space (X, 11·11) into a nonned linear space (Y, 11·11') is said to be nonn increasing if II Tx II' 2': II x II for all x EX. Prove that when Tis norm increasing and (X, 11·11) is complete then Tis onto if and only if T' is one-to-one.

2.

Consider a continuous linear mapping T of a normed linear space (X, 11·11) into a normed linear space (Y, 11·11'). Prove that Tis an isometric isomorphism of (X, 11·11) onto (Y, II· II') if and only if T' is an isometric isomorphism of (Y*, 11·11') onto (X*, 11·11).

3.

Given a normed linear space (X, 11·11) and a linear subspace A dense in X, prove that A* is isometrically isomorphic to X* under the conjugate of the inclusion mapping in: A~ X where in( a) = a; (see Theorem 4.8).

4.

Given Banach spaces (X, 11·11) and (Y, 11·11'), prove that if (X, 11·11) is reflexive and there exists a continuous linear mapping T of (X, II· II) onto (Y, 11·11') then (Y, II· II') is reflexive.

5.

Given a Banach space (X, 11·11) with dual space X* and second dual X**, denote the natural embedding of X into X** by Q and the natural tmbedding of X* into X*** by Q 1 ; that is, and (i)

(Qx)(f) = f(x)

for all f EX*

(Q 1f)(F) = F(f)

for all FE X**.

Prove that Q'Q 1 is the identity mapping on X*.

(ii) Hence prove that if X is reflexive then X* is reflexive. 6.

Consider a closed linear subspace M of a nonned linear space (X, 11·11) and the inclusion mapping in: M

~X

where in(m) = m.

Consider the conjugate mapping in': X* H M* and prove that (i)

ker in'= M.l and

(ii) M* is isometrically isomorphic to X*JM.L. 7.

Consider a closed linear subspace M of a normed linear space (X, 11·11) and the quotient space (X/M, 11·11). Prove that if (X, II· II) is reflexive then (X/M, II· II) is reflexive. (See Theorem 6.ll(ii).)

180 8.

Types of continuous linear mappings

For a closed linear subspace M of a normed linear space (X, 11·11) we write M_j__j_ = {FE X** : F(f)

=0

for all f E M_j_}.

Prove that (i)

Mil is isometrically isomo~hic toM**,

(ii) if (X, 11·11) is reflexive then M

=Mil,

(iii) if XIM is reflexive then for every FE X** there exists an x EX such that F(f) = ~(f) for all f E M_j_ and if also M is reflexive then (X, II· II) is reflexive. 9.

For Banach spaces (X, II· II) and (Y, 11·11') prove that the mapping T

H

T' of

n(X, Y) into n(X*, Y*) is onto if and only if (Y, 11·11') is reflexive.

I 0.

Consider normed linear spaces (X, 11·11) and (Y, 11·11') and a linear mapping T of X into Y. Prove that T is continuous if and only if for every sequence { Xn} in X where f(xn) --7 0 as n --7

oo

for every f EX*, we have g(Txn) --7 0 as n --7 00 for

every g E Y*.

II.

Using the Uniform Roundedness Theorem 11.4 prove the following special case of the Closed Graph Theorem 10.14. lfT is a linear mapping of a Banach space (X, 11·11) into a reflexive Banach space (Y, 11·11') and T has a closed graph, then Tis continuous.

§ 13. Adjoint operators

18 1

§ 13. ADJOINT OPERATORS ON HILBERT SPACE

Given a continuous linear operator T on a Hilbert space H, we are able to associate with T another continuous linear operator on H derived from its conjugate T' on H* and the mapping relating H and H* given in the Riesz Representation Theorem 5.2.1 and discussed in Remark 5.2.4. 13.1 Notation. We recall that the mapping z fz(x)

~

fz of H into H* defined by

=(x, z)

for all x E H

is one-to-one and onto and is conjugate linear and norm-preserving. We denote this mapping by cjl. Now cjl is an isometry onto so is invertible and cj>-1 is also an isometry onto. 13.2 Definition. for a continuous linear operator Ton a Hilbert space H we define the adjoint operator T* on H by T*

= cj>- 1 o T' o cjl.

T'

Figure 18. T* = cj>-1 oT' ocjl. 13.3 Remark. Since cjl, cj>-1 and T' are additive so too is cj>-1 o T' o T. Since cjl and cj>-1 are both conjugate homogeneous and Tis homogeneous, then cj>-1 o T' o cjl is homogeneous. Therefore T*

= cj>-1 o T' o cjl is linear.

Both cj> and cj>-1 as isometries are continuous and T' is continuous so cj>- 1o T' o cjl is continuous. We conclude that T*

=cj>- 1o T' o cjl is a continuous linear operator on H.

D

182

Types of continuous linear mappings

The relation between a continuous linear operator T and its adjoint T* can be expressed more directly through the inner product on H. 13.4 Theorem. Given a continuous linear operator Ton a Hilbert space H, (Tx, z)

=(x, T*z)

for all x,z E H.

Moreover, this statement defines the adjoint operator T* on H. Proof. Given z E H, we have from the definition of the conjugate T' on H*,

= fz(Tx) = (Tx, z)

T' fz(x)

for all x E H

and from the definition ofT*, T' fz(x) = fT*z(x) = (x, T*z)

So

=(x, T*z)

(Tx, z)

for all x E H.

for all x,z E H.

If S is an operator on H such that for all x,z E H

(Tx, z) = (x, Sz)

=(x, Sz)

then

(x, T*z)

So given z E H,

(x, T*z-Sz) = 0

Therefore,

T*z =Sz S = T*.

that is,

for all x,z E H. for all x E H.

for all z EH;

0

13.5 Remark. We showed that T* is a continuous linear operator on H but this could be established directly using the statement in Theorem 13.4 as a definition. For any z 1, z 2 EH and all x EH (x, T*(z 1+z 2)) = (Tx, z 1+z 2) = (Tx, z 1) + (Tx, z 2)

=(x, T* z 1) + (x, T* z 2) =(x, T* z 1+T* z 2) T* (z 1+z 2) = T* z 1+ T* z 2.

so Similarly, we show that

T*(ax) = aT*x

for scalar a and x E H.

So T* is linear. But also for all x E H,

II T*x 112 = (T*x, T*x) = (TT*x, x) :<:;;

II TT*x II II x II

:<:;;

II T 1111 T*x 1111 x II

since T is continuous. Therefore,

II T*x II ::;; II T 1111 x II

for all x E H

and we conclude that T* is continuous and liT* II:<:;IITII.

0

The taking of adjoints for continuous linear operators on Hilbert space generalises the following matrix operation in the finite dimensional case. 13.6 Example. Consider a linear operator Ton a finite dimensional inner product space Hn. We examine the form of the adjoint operator T* on Hn-

§ 13. Adjoint operators

183

With respect to an orthonormal basis {e 1, e 2, ... , e 0 } for H 0 , the linear operator T has matrix representation T= [ajkl,

For x

=A. 1e 1 + A.2e 2 + ... + A e 0

j,k E {1,2, ... , n}. and z

0

=J.1. 1e 1 + 112e 2 + ... + J.l. e 0

0,

n

we have

Tx =I. a.k A.k k=l J n

n

Iii

(Tx, z) = I. ajk A.k so (x, T*z) = I. A.k ajk I! j,k=l j,k=l J Therefore, T* has matrix representation

and

T*=(iik)•

j,kE{l,2, ... ,n},

the conjugate transpose of the matrix representation ofT.

We now examine properties of the operator T

0 T* on n(H). These properties

H

could be deduced from the definition using the conjugate mapping, but we use the direct method depending on the definition of the adjoint by the inner product given in Theorem 13.4.

Given a continuous linear operator Ton a Hilbert space H, the

13.7 Theorem.

mapping T H T* on n(H) has the properties (i) (T 1+T 2)* = T 1*+T/ (ii)

(aT)* =

aT*

for scalar a

(iii) (T1 T 2)* = T /TI * (iv) T** = T (v)

II T* II= II T II

(vi) II T*T II= II T 112. Proof. The proofs of properties (i)-(iv) all follow the same pattern. We prove property (iv). Given z EH, (x, T**z) = (T*x, z) = (z, T*x) = (Tz, x) = (x, Tz) for all x E H. So

T**z =Tz

that is,

T** = T.

(v)

for all z EH;

We have already shown in Remark 13.5 that II T*

From (iv),

II T II =II T**

II~

II~

II T II.

II T* II and so II T* II =II T II.

(vi) From Theorem 4.11.3 we have that n(H) is a normed algebra and so II T*T II~ II T* 1111 T II= II T 112 But also

~

Therefore, and we conclude that

by (v).

II Tx 112 = (Tx, Tx) = (T*Tx, x) II T*Tx II II x II ~ II T*T 1111 x 112

II T 112 ~ II T*T II II T*T II= II T 112.

for all x E H.

0

184

Types of continuous linear mappings

The adjoint mapping on a Hilbert space H is an element of the operator algebra ~(H).

This suggests that this operation can be generalised to normed algebras.

13.8 Definitions. Given an algebra A over the complex numbers
(x+y)* = x* + y*

(ii)

(ax)*= ax*

for scalar a

(iii) (xy )* = y*x * (iv) x** = x, is called an involution on A. An algebra A with an involution is called a *-algebra. With a normed algebra (A, II· II) we need the involution to be related to the norm. A Banach *-algebra (A, II· II) where the involution and the norm are related by (vi) II x*x II = II x 112

for all x E A

is called a B*algebra . 13.9 Remarks. (i)

The submultiplicative property of the norm (Definition 4.11.1) enables us to deduce

from (vi) and (iv) of Definitions 3.8 that (v)

llx*ll=llxll.

(ii)

This abstract structure, a B* algebra is a generalisation of ~(H) for a Hilbert space

H. It can be seen from Theorem 13.7 that

~(H)

is the prototype of B* algebras.

(iii) A major inquiry in the development of Banach algebra theory was to determine whether, given a B* algebra A, there exists a Hilbert space H so that A can be represented as a subalgebra of ~(H); (by "represented" we mean that there exists an isometric algebra isomorphism which preserves the * operation). The problem was first solved in the affirmative by I.M. Gelfand and M.A. Naimark, Math. Sbornik 12 (1943), 411504.

0

13.10 Self-adjoint operators For a complex Hilbert space H, the involution on properties of

~(H)

reveals important structural

~(H).

13.10.1 Definition. A continuous linear operator T on a Hilbert space H is said to be self-adjoint if T* = T.

13.10.2 Example. In Example 13.6 we had a linear operator Ton a finite dimensional inner product space Hn with orthonormal basis {e 1, e2, ... , en} with matrix representation j,k

E

{1,2, ... , n}.

§ 13. Adjoint operators

185

It is clear that T is self-adjoint if and only if ajk = akj

j, k

E

{I, 2, ... , n}.

So the self-adjoint operators are those with hermitian symmetric matrix representation. We know that in matrix theory this set of matrices is of crucial significance.

0

!3.10.3 Remarks. (i)

The set of self-adjoint operators is always nontrivial; it always contains the zero

operator and the identity operator but for any continuous linear operator Ton H we have from properties (i), (iii) and (iv) of Theorem 13.7 that T + T*, TT* and T*T are all selfadjoint operators on H. Properties (i) and (ii) of Theorem 13.7 show that the set of selfadjoint operators is a real linear subspace of ~(H). (ii) The set of self-adjoint operators on His also closed in

~(H):

For a sequence {T nl of self-adjoint operators convergent to a continuous linear operator Tin (~(H), 11·11) we have II T-T* II

:<=:;

II T-T n II + II Tn-T n* II + II T n*-T* II = II T-T n II + II (Tn-T)* II by property (i) of Theorem 13.7 by property (v) of Theorem 13.7

and we conclude that T* = T. (iii) But further, given a continuous linear operator Ton H, using properties (i), (ii) and (v) of Theorem 13.7 we see that the operators T+T* T-T* - 2and 2 1 are self-adjoint. But then (!+T*) . (!-T*) T=~-2- +1~21 that is, every continuous linear operator T on H can be expressed in terms of self-adjoint operators in this way and it is clear that such a representation is unique. In this respect, the set of self-adjoint operators in ~(H) acts like the real numbers the complex numbers [.

lR in

0

In discussing properties of a continuous linear operator on a Hilbert space over the complex numbers, the following set in the complex plane gives a picture which determines many properties of the operator. 13.10.4 Definitions.

For a linear operator T on an inner product space X, the

numerical range ofT is the set of complex numbers W(T) ={ (Tx, x) : II x II = I }. When this set is bounded, the numerical radius ofT is the real number w(T) =sup{ I (Tx, x) I: II x II= I}.

186

Types of continuous linear mappings

13.10.5 Remarks. (i)

It is clear that if T is continuous then I (Tx, x) I :<=:; II T II

for all II x II

=I

so W(T) is a bounded set in the complex plane contained in B[O; II T II]. Further (ii)

w(T) :<=:;II T II.

In a finite dimensional inner product space Hn with orthonormal basis

{e 1, e 2,

..• ,

en} and linear operator T with matrix representation T = [ajkl

we have, for x

j,k

E

{1,2, ... , n}

= A. 1e 1 + A.2e 2 + ... + "-nen that n

(Tx, x)

=L

ajk Ak xj

j,k=l

so W(T) is a quadratic form associated with the matrix. In the earliest work done on Hilbert space, interest was centred on such a quadratic form. It then came to be realised that the significance of the object lies in its abstract formulation. (iii) The Toeplitz-Hausdorff Theorem states that the numerical range of a linear operator on a Hilbert space is always a convex subset of the complex plane. (See a proof in F.F. Bonsall and J. Duncan, Numerical Ranges II, Cambridge University Press (1973), pp 5-6.) However, we will not need to use this fact.

Figure 19.

The numerical range of a continuous linear operator T, a subset of the complex plane

We will see later in Section 17 that the concept of numerical range of a linear operator has been usefully generalised to normed linear spaces. 0

§ 13. Adjoint operators

187

Self-adjoint operators can be neatly characterised in terms of their numerical range. But to establish this characterisation we first develop the following numerical range properties. 13.10.6 Lemma. For a linear operator Ton a complex inner product space X, 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) for all x,y EX, the general polarisation formula, and so (i) T = 0 if and only ifW(T) = {0}, (ii) when T is continuous, ~II T II~ w(T) ~II T II.

Proof. The polarisation formula follows by straightforward calculation. (i) IfW(T) = (0} then from the general polarisation formula (Tx, y) = 0 for all x,y EX from which it follows that T = 0. The converse is obvious. (ii)

From the polarisation formula 41 (Tx, y) I ~ w(T) (II x+y 112 +II x-y 112 +II x+iy 112 +II x-iy 112)

and by the parallelogram law ~

I (Tx, y) I So

w(T) (II x 112 +II y 112).

I (Tx, y) I ~ 2 w(T) for all II x II, II y II

~

I.

But

II T II= sup{ II Tx II: II x II~ I} =sup{ (Tx,

So

II T II

1 i~ll) : II x II ~ I}

~sup {I (Tx, y) I : II x II, II y II ~ I } . ~2

w(T)

~ II T II ~ w(T) ~ II T II.

and we have

0

13.10.7 Remark. It should be noted that although this lemma holds for complex inner product spaces it does not hold in general for real inner product spaces. Consider Euclidean space (IR 2 , 11·11 2) and linear operator T with matrix representation

0 -1

T= [ I

J

0 .

For any x = (/c 1, A2) we have (Tx, x) = (A 1 A2) [ But clearly T

* 0.

~ ~ J( ~: ) = (A -

2

-A 1) (

~:

)

=0

0

13.10.8 Theorem. A continuous linear operator T on a complex Hilbert space H is self-adjoint if and only ifW(T) is real.

188

Types of continuous linear mappings

Proof. If T is self-adjoint then, for any x E H (Tx, x)

= (x, Tx) = (Tx, x)

and so W(T) is real. Conversely, if W(T) is real then, for any x E H (Tx, x)

=

(Tx, x)

= (x, T*x) = (T*x, x)

and so ((T*-T)x, x) = 0. This implies that W(T*-T) = {0} and from Lemma 13.10.6(i) we have T*

=T.

0

Perhaps the most important property of the numerical range is that it helps us to locate the elements in

~(H)

which have continuous inverses.

13.10.9 Theorem. Given a continuous linear operator Ton a Hilbert space H,for any

A"' W(T) the continuous linear operator T- AI has a continuous inverse on H. Proof. For A"' W(T) we write d;; d ('A, W(T)) > 0. For x E H, II x II

= I, II (T-'AI) x II ~ I ((T-'AI)x, x) I =I (Tx, x)- A I ~ d

(*)

II (T-'AI)x II ~ d II x II for all x E H, so which implies that T-'AI is a topological isomorphism.

Now (T-'AI)(H) is a closed linear subspace of H. Suppose that (T-'AI)(H) is a proper subspace of H. Then by Corollary 2.2.21 there exists a z E H, II z II = I such that z is orthogonal to (T-'AI)(H), so (( T-'AI)z, z) = 0 which contradicts(*). So T-'AI is a topological isomorphism of H onto H.

D

Using the characterisation for self-adjoint operators given in Theorem 13.10.8 we can make the following deduction. 13.10.10 Corollary. Given a self-adjoint operator Ton a complex Hilbert space H,for any complex number A;; a+ i~ where~* 0, the continuous linear operator T-'AI has a continuous inverse on H. Theorem 13.10.8 also enables us to define other sets of self-adjoint operators using the order relation of the real numbers. 13.10.11 Definition. A self-adjoint operator T on a Hilbert space H is said to be a positive operator if W(T) ~ 0; that is, Tis a positive operator if A~ 0 for all A E W(T). 13.10.12 Remarks. (i)

The set of positive operators is always nontrivial; it contains the zero and identity

operators and for any continuous linear operator T on H we have that T*T and TT* are

§ 13. Adjoint operators

189

positive operators since (T*Tx, x) = (x, T*Tx) = (Tx, Tx) =II Tx 112:?: 0 for all x E H. (ii)

The set of positive operators is a positive cone in B (H); that is, for a positive

operator T and a :?: 0 we have that aT is also a positive operator and for positive operators T 1 and T 2 we have that T 1+T 2 is also a positive operator. This positive cone is also closed in B (H): For a sequence {T n} of positive operators convergent to a selfadjoint operator T in (B(H), 11·11) we have I (T 0 x, x)- (Tx, x)

I~

II T 0 -T 1111 x 112 for all x EH

so (Tx,x):?: 0 for all x E H. (iii) The positive cone of positive operators induces a partial order relation on the set of self-adjoint operators. For self-adjoint operators T 1 and T 2 we say that T 1 ~T 2 if T 2-T 1 :?: 0. From Lemma 13.10.6(i) it follows that if T 1 ~ T2 and T 2 ~ T 1 then T 1 = T 2.

0

As in Corollary 13.10.10 we have the following property for positive operators. 13.10.13 Corollary. Given a positive operator Ton a complex Hilbert space H, for any complex number A= a+ ip where a< 0 or p 0, the continuous linear operator

*

T-'AI has a continuous inverse on H. The following particular case is important. 13.10.14 Corollary. For any continuous linear operator Ton a Hilbert space H, the

continuous linear operators I+T*T and I+TT* are topological isomorphisms ofH onto H. 13.10.15 Remark. In the attempt to establish the Gelfand-Naimark Representation Theorem forB* algebras mentioned in Remarks 13.9(iii), a key step was to use the B* algebra structure to prove that e+x*x and e+xx* are regular for every element x of a unital B* algebra. We notice that the proof of Corollary 13.10.14 uses the underlying Hilbert space structure on which the operators act. The Gelfand-Naimark Theorem had to achieve this result without the assumption of any such underlying structure.

0

13.11 Normal and unitary operators For a complex Hilbert space, further structural properties of B(H) are revealed by drawing attention to other special sets of continuous linear operators on H. 13.11.1 Definitions. A continuous linear operator Ton a Hilbert space H is said to be a normal operator if T*T = TT*, and is said to be a unitary operator if T*T = TT* = I. We will first explore the properties of normal operators.

190

Types of continuous linear mappings

13.11.2 Remarks. (i)

Clearly self-adjoint operators are normal so the set of normal operators contains the

closed real linear subspace of self-adjoint operators in n(H). But also given any normal operator Ton Hand any scalar A., then A.T is also normal. Further, the set of normal operators is closed in n (H): For a sequence {T n} of normal operators convergent to a continuous linear operator T in (n(H), 11·11) we have II T*T-TT* II :::; II T*T-T*T0 II + II T*T0 -T 0 *T0 II + II T 0 *T0 - TnTn* II +II TnT n*-TnT* II+ II TnT*-IT* II :::; 2 II T* 1111 T-T n II + 2 II (T-T n)* 1111 T n II and so T*T = TT*.

(ii)

Given a complex finite dimensional inner product space Hn with an orthonormal

basis and a linear operator Ton Hn we saw in Example 13.6 that the matrix representation ofT* is the conjugate transpose of the matrix representation ofT. The linear operator T 0 commutes with its adjoint T* if and only if the matrix representations do the same.

Normal operators have the following characterisation. 13.11.3 Theorem. A continuous linear operator T on a complex Hilbert space H is

normal if and only if II Tx II = II T*x II for all x E H. Proof. Now II Tx 112 = (Tx, Tx) = (x, T*Tx) and II T*x 112 = (T*x, T*x) = (x, TT*x) so II Tx II = II T*x II for all x E H if and only if (x, (T*T-TT*)x) = 0 for all x E H; that is, by Lemma 13.10.6(i) if and only if T*T = TT*.

0

The limit developed in Proposition 4.11.12 is very significant for normal operators. 13.11.4 Theorem. A normal operator Ton a complex Hilbert space H has the

properties (i) (ii)

II T211 =II T 11 2 and lim IIT0 11 11"=11TII. n-7~

Proof. (i)

From Theorem 13.11.3 we have that II T2x II = II T*Tx II

so

for all x EH

II T2 II = II T*T II.

But from Theorem 13.7(vi) we have that II T*T II= II T 11 2. Therefore (ii)

II T211 =II T 112.

Since powers ofT are also normal operators we have that II T 2" II = II T 11 2" for all n E N .

§ 13. Adjoint operators

191

We have from Proposition 4.11.12 that lim II T 0 11 11" exists. So lim II T 0 11 11" = lim II T 2" 11 112" =II T II.

D

We now discuss the properties of unitary operators. 13.11.5 Remark. (i)

The algebraic formulation of the definition tells us that T is unitary if and only if T

has a continuous inverse and T -I = T*. (ii)

Such operators generalise rotation operators in Euclidean space. For example, in (IR 2 , 11·11 2) the rotation operator Te which rotates by an angle e about the origin with matrix representation, Te= [

cos e

-sin e ]

sin e *

T

has

e

=[

cos e

cos e

sine ]

-sine

cos e

and clearly T8* = T8-1 . Equivalently, in (
0 Te z = eie z T 8*z=e-i8=T() 1 z

and

forallzE
(iii) We noted in Remarks 13.10.3(iii) that the set of self-adjoint operators acts like the real numbers in the complex numbers. We see now that the set of unitary operators acts like the set of complex numbers with modulus I. D There are useful equivalent formulations for unitary operators. 13.11.6 Theorem. For a continuous linear operator T on a Hilbert space H the following conditions are equivalent. (i) (ii)

T*T =I, (Tx, Ty) = (x, y) for all x,y E H,

(iii) II Tx II = II x II

for all x EH.

Proof. (i) ~ (ii)

IfT*T =I then (Tx, Ty) = (x, T*Ty) = (x, y)

(ii) so

~(iii)

for all x,y E H.

If (Tx, Ty) = (x, y) for all x,y E H then (Tx, Tx) = (x, x)

IITxll =II xll

for all x E H.

192

Types of continuous linear mappings

(iii) ==> (i) If II Tx II = II x II

for all x E H

then (x, x) =II x 112 = II Tx 112 = (Tx, Tx) = (x, T*Tx) ((T*T-I)x,x)=O forallxEH. so By Lemma 13.10.6(i), T*T =I.

D

13.11.7 Remarks. (i) Property (iii) says that T is an isometric isomorphism of H into H and (ii) says that T is an inner product isomorphism. So (ii) {::} (iii) says that a continuous linear operator Ton His an inner product isomorphism if and only if Tis an isometric isomorphism; (see Exercise 2.4.13). (ii)

If H is finite dimensional then since T is one-to-one it is also onto and this is

enough to give us that Tis unitary. However, in general there exist continuous linear operators which satisfy the conditions of Theorem 13.11.6 but are not onto and so are not unitary. For example, the shift operator Ton (.t 2, 11·11 2) defined by

.T(P'-t.

Az, ···,An,··.})= {0, A1, Az, ···,An,··.}

is an isometric isomorphism of (.t 2, 11·11 2) onto a proper closed linear subspace of (.t 2 , 11·11 2 ).

D

For a characterisation of unitary operators in general we need the following result. 13.11.8 Theorem. A continuous linear operator T on a Hilbert space H is unitary if

and only ifT is an isometric isomorphism ofH onto H.

Proof.

Taking into account the algebraic formulation of the definition of a unitary

operator we need only prove that if T is an isometric isomorphism of H onto H then TT* =I. Now T-1 exists on Hand from Theorem 13.11.6 we have T-1 = (T*T) T-1 = T*(TT-1) = T*. So TT* =TT-l =I and we conclude that Tis unitary.

D

Along the same lines as Corollary 13.10.10 for self-adjoint operators and Corollary 13.10.13 for positive operators we have the following result for unitary operators. 13.11.9 Theorem. Given a unitary operator Ton a complex Hilbert space H, for all A E ([,I A I* I, the operator T-AI has a continuous inverse on H.

Proof. Since both T and T* are unitary and, from Theorem 13.11.3 are isometric isomorphisms, II T II= II T* II= I. So from Theorem 13.10.9 we see that both T-Al and T*-AI have continuous inverses on H for I A I > I. Also both T and T* have continuous inverses on H. Now consider 0 < I A I < I. Then T*- ~I has a continuous inverse on H. But T-AI = - A(T* - ~ I)T so T-AI has a continuous inverse-~ T*(T* - ~ 1)- 1 on H.

0

§13. Adjoint operators

193

!3.12. EXERCISES

1.

Consider a Hilbert space H. (i)

Prove that Tis a topological isomorphism of H onto H if and only if T* is a topological isomorphism of H onto H.

(ii) Prove that if T is a topological isomorphism of H onto H then (T -I)* = (T*)-1 . 2.

(i)

In 2-dimensional Unitary space (([ 2, 11·11 2) consider the operator T with two distinct eigenvalues a 1 and a 2 with corresponding eigenvectors x 1 and x2 with II x 1 11 2 =II x2 11 2 = I. Prove that W(T) is a closed elliptical disc (with interior) with foci at a 1 and a 2 and major axis ~

Ia

-a2I

1

1-(x I' x 2 )

(ii) In 3-dimensional Unitary space

. . l(x 1, x 2)11a 1-a21 an d mmor axis . ~ l-(x 1, x 2) (([ 3 ,

11·11 2) consider the operator T with matrix

representation

Prove that W(T) is an equilateral triangle (with interior) whose vertices are the three cube roots of 'A. 3.

Consider a linear operator T on a complex inner product space X. Using the general polarisation formula, or otherwise, prove that T is continuous if and only if W(T) is a bounded set in the complex plane. Does this characterisation for the continuity ofT hold for real inner product spaces? Given a complex Hilbert space H and n(H) the Banach algebra of continuous linear operators on H prove that the numerical radius is an equivalent linear space norm but show that it is not an equivalent algebra norm for n(H).

4.

Given a continuous linear operator Ton a Hilbert space H where there exists an x0

E

H, II x0 II = I such that II T II = (Tx 0 , x0).

Prove that x0 is an eigenvector ofT with eigenvalue II T II. 5.

Consider a self-adjoint operator Ton a Hilbert space H. Prove that II T II =sup { (Tx, x) : II x II = I }. (Hint: Consider the expression 411 Tx 112 ( T(ax+

~),ax+~)-( T(ax- ~),ax-~) where a= II Tx 11 112.)

194

6.

Types of continuous linear mappings

(i)

Consider the positive operator T on a Hilbert space H. (a) Prove that ( . , . ) defined by ( x, y) = (Tx, y)

for all x,y EH

is a positive hermitian form on H. (b) Prove that I (Tx, y) ~ (Tx, x)(Ty, y) for all x,y E H, and deduce that II Tx 11 2 ~ (Tx, x) II T II for all x E H.

e

(ii) Prove that for any sequence {T nl of positive operators on H, II T nX II n 7.

(i)

--7

=for each x E H if and only if (T nX, x)

--7

0 as n

--7

--7

0 as

=for each x E H.

For a sequence of continuous linear operators {T n} on a Hilbert space H we are given that { (T nX, y)} is convergent to (Tx, y) for all x,y E H. (a) Prove that T so defined is a continuous linear operator on H; (see also Exercise 10.15.8(i)). (b) Prove that if the operators T n are self-adjoint then so too is T.

(ii) Consider a sequence of self-adjoint operators {T nl on H such that T n ~ T n+ 1 for all n EN, where { (T nX, x)} is convergent for each x E Hand where there exists an K > 0 such that I (T nX, x) I ~ K for all II x II= I. Prove that (a) there exists a continuous linear operator T on H such that {T nX} is convergent to Tx for each x E H and (b) Tis a self-adjoint operator on H. (Hint: Use inequality Exercise !3.12.6(i)(b) to show that II TmX-Tnx 114 ~ ((Tmx, x)-(Tnx, x)) K II x 11 2 for all m >nand x EH.) 8.

A continuous linear operator T on a Hilbert space H is called a contraction if II T II (i)

~

Prove that T is a contraction if and only if I- T*T ~ 0.

(ii) Given a continuous linear operatorS which is invertible on H, prove that TS-1 is a contraction if and only if T*T 9.

~

S*S.

Given a continuous linear operator T on a complex Hilbert space write H = l (T + T*) and K = l (T- T*). 2

(i)

2

Prove that Tis a normal operator if and only if HK =KH.

(ii) Given that T is normal, prove that T has a continuous inverse H if and only if H2+K2 has a continuous inverse on Hand in this case T- 1 = T*(H2+K2)- 1. (iii) Prove that T is unitary if and only if H2+K2 = I. I 0.

Consider a normal operator T on a complex Hilbert space H. (i)

Prove that (a) ker T2 = ker T and (b) T2(H) = T(H).

(ii) Prove that T(H) is dense in H if and only if T is one-to-one.

I.

§ 13. Adjoint operators

11.

195

Consider a continuous linear operator T on a complex Banach space X. (i)

Prove that V(T) k: () {A.e ([:I A.-a I :5: II T-al II}. ae([

(ii) Prove that . II l+aT 11-1 sup Re V(T) = I1m a--)0+

for

a

ae IR.

(iii) Prove that V(T) k: IR if and only if

lim a--)o

12.

II l+iaT 11-1

a

0

for

ae JR.

Consider a self-adjoint operator Ton a complex Hilbert space H. (i)

Prove that the continuous linear operators T +il and T-il have continuous inverses on H.

(ii) The Cayley transform ofT is a continuous linear operator U on H defined by U = (T-ii)(T+il)-1. Prove that U is a unitary operator. (iii) Prove that T can be recovered from the Cayley transform U to give

T = i (I+U)(I-U)-1. (Note: The Cayley transform generalises the Mobius transform for complex numbers

z-i w= z+i which maps the real numbers into the unit circle. To recover we have z =i (1+w)

1-w

and (iii) gives the analogous result for 'B(H).) 13.

Given a complex Hilbert space H, a subalgebra A of 'B(H) is said to be self-adjoint if it contains the identity operator and the adjoint of each of its elements. If A is also a closed subalgebra it is called a C* algebra. Given an arbitrary subset M of 'B(H), the commutant of M denoted by Me is the set of all continuous linear operators on H which commute with every operator in M. Prove that (i)

Me is a unital Banach subalgebra of 'B(H),

(ii) if M is self-adjoint then Me is a C* algebra.

196

Types of continuous linear mappings

§ 14. PROJECTION OPERATORS

We now examine the properties of the special set of projection operators on a Banach space and on a Hilbert space. Such operators play a significant role in the decomposition of a space into component subspaces. 14.1 Definition. A linear operator P on a linear space X is called a projection operator (an

algebraic projection) if it is idempotent; that is, p2 = P. For linear spaces there is an important relation between projection operators and direct sum decomposition of the linear space, expressed in the following result. 14.2 Lemma.

Consider a linear space X.

(i)

/fP is a projection on X then X= P(X) $ ker P.

(ii)

lfX = M $ N where M and N are linear subspaces of X then the linear operator P on X defined by

Px = m where x = m+n, m EM and n EN is a projection operator on X.

Proof. (i) For any x EX, we have x = Px + (I-P)x. Now Px E P(X) and P(I-P)x = (P-P2)x = 0, so (I-P)x E ker P. Therefore, X= P(X) + ker P. For Px E P(X) n ker P we have Px = P(Px) = 0 since Px E ker P, so Px = 0. We conclude that X= P(X) EB ker P. (ii)

For X = M EB N with operator P defined on X by Px = m

where x = m+n, m EM and n EN

the unique decomposition implied by the direct sum gives us that P is linear. But also P2x = m, so p2 = P; that is, P is a projection operator. 14.3 Corollary.

D

On a linear space X, a linear operator Pis a projection operator if and

only ifi-P is a projection operator.

Proof. (I-P)2 = I - 2P + p2 = I- P if and only if p2 = P.

D

In a normed linear space the natural projection operators are continuous and the natural decomposition of the space is by closed linear subspaces.

197

§ 14. Projection operators

!4.4 Theorem. Given a normed linear space (X, 11·11), ifP is a continuous projection operator on X,

(i)

then X= P(X) EB ker P where P(X) and ker Pare closed linear subspaces of X. (ii) Given a Banach space (X, II· II), if X= M EB N where M and N are closed linear

subspaces of X then the linear operator P defined on X by Px = m where x = m+n, m EM and n EN , is a continuous projection on X.

Proof. (i)

Since P is continuous, ker P is closed. But also I-P is continuous so P(X) = ker(I-P)

is also closed. (ii)

Renorm X by II x II'= II m II +II n II .

Since (X, 11·11) is a Banach space and M and N are closed linear subspaces then (X, 11·11') is also a Banach space; (see Exercise 1.26.3(iii)). II Px II =II m II :<;;II m II+ II n II =II x II' for all x EX,

Since

then P is a continuous linear mapping of (X, 11·11') into (X, 11·11). However,

II x II= II m II + II n II= II x II' for all x EX

so by Corollary I 0.10 to the Open Mapping Theorem we have that 11·11' and 11·11 are equivalent norms for X. Therefore, P is a continuous projection operator on (X, 11·11).

0

14.5 Remark. Given a linear space X and a linear subspace M of X there always exists at least one linear subspace N complementary toM in X; that is, such that X= M EB N. Given a normed linear space (X, 11·11) and a closed linear subspace M of X, if there exists a closed linear subspace N of X such that X = M EB N we say that M is complemented in X. In terms of projection operators, Theorem 14.4 implies that a closed linear subspace M is complemented in X if and only if there exists a continuous projection operator P from X onto M. F.J. Murray, Trans. Amer. Math. Soc. 41, (1937), 138-152 has given an example of a closed linear subspace M in (lp, ll·llp) where I < p < =, p

* 2, for which there does not

exist any closed linear subspace complementary toM; that is, where M is not the range of any continuous projection operator. 0 In Hilbert space we have an orthogonality relation which gives a naturally preferred way of forming direct sum decompositions of the space and the preferred continuous projection operators on the space. We first investigate orthogonality of subspaces in Hilbert space. 14.6 Definition. Given sets A and B in an inner product space X, we say that A is

orthogonal to B if every element of A is orthogonal to every element of B; that is, (x,y)=O forallxEAandyEB.

198

Types of continuous linear mappings

For a set A, the orthogonal complement of A, denoted by A_!_, is defined by A_!_={xEX:(x,y)=O forallyEA}. 14.7 Lemma. For any set A in an inner product space X, the orthogonal complement A j_ is

a closed linear subspace of X.

Proof. Now Aj_ =

n{

ker fy :yEA} where, given y E A, fy is the continuous linear

functional on X defined by fy(x) = (x, y) for all x EX; (see Example 4.10.6). So ker fy is a closed linear subspace of X for each y E A and therefore

n{

ker fy : y E A} is a

closed linear subspace of X.

D

The following property is worth noting. 14.8 Corollary. For any set A in an inner product space X, (spA)_!_= A_!_.

Proof. Clearly (sp Al ~ Aj_. If y .1 A then y .1 spA. If x is a cluster point of sp A then there exists a sequence {X0 } in sp A such that {xn} is convergent to x. For y EX, I (y, x)- (y, xn) I;<:; II y II II x-xn II. So if y E Aj_ then (y, x0 ) = 0 for all n EN and (y, x) = 0; that is, y E (sp A)j_.

D

The following theorem justifies the term "orthogonal complement". 14.9 Theorem. Given any closed linear subspace M of a Hilbert space H, then H = M EB M_!_.

Proof. Clearly M n M_!_ = {0} so M + M_!_ = M EB M_!_. We show firstly that M EB Mj_ is closed. If x is a cluster point of M EB Mj_ then there exists a sequence {Xn} in M EB Mj_ such that {x 0 } is convergent to x in H. For each n EN we write Xn = u 0 + v 0

where un EM and v 0 E Mj_.

Since ( u 0 , vm) = 0 for all m, n EN II Xn- Xm 112 =II Un- Urn 112 +II

V0-

vm 112.

So {u 0 } is a Cauchy sequence in M and {vn} is a Cauchy sequence in Mj_ . Since both M and Mj_ are closed and H is complete there exist u EM and v E Mj_ such that {un} is convergent to u and {v 0 } is convergent to v.

199

§ 14. Projection operators

II

But

X 0-

(u+v) II ::; II

U0 -

u II +II v0 - vII

so x = u+v E M EB M.l which implies that M EB M.l is closed. Suppose that M EB M.l is a proper closed linear subspace of H. Then from Corollary 2.2.21, there exists a z 1' 0 such that z is orthogonal to M EB M.l. But then z is orthogonal to M so z E M.l, and z is orthogonal to M.l. Therefore, (z, z) = 0 which contradicts z conclude that H = M EB M.l.

* 0.

We

0

This result gives us another useful elementary property. 14.10 Corollary. Given any closed linear subspace M of a Hilbert space H, Mil= M. Proof. Now H = M EB M.l and H = M.l EB Mil. But M ~ M.l.l_ Consider x E H where

x = m + n, m EM and n E M.l = y + n', y E M.l.l and n' E M.l.

Then y- m = n- n' E M.l. But y -mE Mil soy = m and we conclude that Mil= M.

0

We use orthogonal complements to show the close relation between the kernel and range of a continuous linear operator and its adjoint. 14.11 Theorem. For a continuous linear operator Ton a Hilbert space H (i)

T(H).l = ker T*,

(ii)

T(H) = (ker T*).l.

Proof. (i) Now y E T(H).l if and only if (Tx, y) = 0 for all x E H, that is, (x, T*y) = 0 for all x E H which is T*y = 0, that is, if and only if y E ker T*. But T(H).l = T(H).l by Corollary 14.8. (ii)

It follows from Corollary 14.1 0, that T(H) = T(H)il = (ker

T*l from (i).

0

14.12 Remark. Theorem 14.9 implies that in Hilbert space, direct sum decompositions by closed linear subspaces exist; that is, given a closed linear subspace M of a Hilbert space H, there exists a closed linear subspace complementary to M, in particular M.l. So from Theorem 14.4(ii) we see that given any closed linear subspace M of a Hilbert space H there exists a continuous projection operator on H with range M, in particular the projection operator P with kernel M.l. O

200

Types of continuous linear mappings

So in Hilbert space we are naturally directed to examine those continuous projection operators whose range and kernel are orthogonal subspaces. 14.13 Definition. On an inner product space X, a projection operator P where P(X) is orthogonal to ker P is called an orthogonal projection. Such operators have the following characterisation. 14.14 Theorem. On an inner product space X, a projection operator P on X is an

orthogonal projection if and only if (Px, y) = (x, Py)

for all x,y EX.

Proof. If P(X) and ker P are orthogonal then for any x = m + n and y m, m' E P(X) and n, n' E ker P,

= m'+ n' where

= (m, m'+ n') = (m, m') = (m + n, m') = (m, m') . Conversely, if x E P(X) andy E ker P then Px = x and Py = 0 so that (x, y) = (Px, y) = (x, Py) = 0 (Px, y)

and

(x, Py)

and so P(X) and ker P are orthogonal.

D

It is interesting to note that orthogonal projections are automatically continuous. 14.15 Theorem. An orthogonal projection P on a inner product space X is continuous and

ifP

* 0 then II P II = I.

Proof. For x

= m + n where m E P(X) and n E ker P we have Px = m and (Px, n) = 0.

So

II x 112 =II Px 112 + II n 112

and

II Px II

~

II x II

for all x EX;

that is, P is continuous. But also II P II ~ I. Since p2 Therefore, II P II = I.

= P then II P II ~II P 112 and so II P II ~ I

if P

* 0. D

14.16 Corollary. On a Hilbert space H, an orthogonal projection Pis a positive operator.

Proof. Clearly Theorem 14.14 implies that Pis self-adjoint. Then for all x E H (Px, x) = (P2x, x) = (Px, Px) = II Px 11 2 ~ 0.

D

We have the following important characterisation of the kernel of an orthogonal projection on a Hilbert space.

§ 14. Projection operators

201

14.17 Theorem. /fP is an orthogonal projection on a Hilbert space H, then ker P = P(H)l. and P(H) = ker pl._

Proof. From Theorem 14.4(i), P(H) is closed. From Corollary 14.16, Pis self-adjoint. So our result follows from Theorem 14. II (i) and Corollary 14.10. 0 We now investigate the structure of the set of orthogonal projections on a Hilbert space H as a subset of 'r.>(H). We do this because we will ultimately show that certain classes of operators on a Hilbert space can be represented by orthogonal projections. 14.18 Theorem. Consider P and Q orthogonal projections on an inner product space X. Then PQ is an orthogonal projection if and only ifPQ = QP and in this case PQ(X) = P(X)

n Q(X).

Proof. If PQ is an orthogonal projection on X then from Theorem 14.14, (PQx, y) = (x, PQy) = (Px, Qy) = (QPx, y) for all x,y EX, soPQ=QP. Conversely, if PQ = QP then (PQ)2 = PQPQ = P2Q2 = PQ. But also, (PQx, y) = (Qx, Py) = (x, QPy) = (x, PQy) for all x,y EX, so we conclude from Theorem 14.14 that PQ is an orthogonal projection. If x e P(X) n Q(X) then Px = x and Qx = x so PQx = x and P(X) n Q(X) b:: PQ(X). If x e PQ(X) then x = PQx = P(Qx) so x E P(X). But also x e QP(X) so x e Q(X). Then PQ(X) b:: P(X) n Q(X). Therefore,

PQ(X) = P(X) n Q(X).

0

14.19 Definition. Orthogonal projections P and Q on an inner product space X are said to

be orthogonal to each other if PQ = 0. The following theorem gives some explanation of this property. 14.20 Theorem. Consider orthogonal projections P and Q on a Hilbert space H. Then P and Q are orthogonal to each other if and only ifP(H) is orthogonal to Q(H).

Proof. If P(H) is orthogonal to Q(H) then Q(H) b:: P(H)l. so from Theorem 14.14 Q(H) b:: ker P and so PQ(H) = {0}; that is, PQ = 0. Conversely, if PQ = 0 then for all x e Q(H) we have Px = P(Qx) = 0. So Q(H) b:: ker P = P(H)l. which implies that P(H) is orthogonal to Q(H).

0

202

Types of continuous linear mappings

14.21 Theorem. Consider P and Q orthogonal projections on a Hilbert space H. Then P+Q is an orthogonal projection on H if and only ifP and Q are orthogonal to each other and

then (P+Q)(H) = P(H) + Q(H).

Proof. If P and Q are orthogonal projections which are orthogonal to each other, then from Theorem 14.20, PQ = 0 = QP. Then (P+Q)* = P* + Q* = P + Q but also (P+Q)2 = p2 + PQ + QP + Q2 = P + Q. So from Theorem 14.14, P+Q is an orthogonal projection. Conversely, if P+Q is an orthogonal projection, for x E P(H) we have Px = x so II x 112 =II Px 112 :<;;II Px 112 +II Qx 112 = (Px, x) + (Qx, x) = ((P+Q)x, x) = II (P+Q)x 1111 x II :<;; II x 112 since II P+Q II :<;; I. Therefore, II Qx II = 0 and so x E ker Q and P(H) C ker Q = Q(H).l by Theorem 14.17, and P(H) is orthogonal to Q(H). But this argument also shows that for x E P(H) we have x E ker Q and so x E (P+Q)(H). Therefore P(H) .s;: (P+Q)(H). The argument is symmetric with respect toP and Q so Q(H) .s;: (P+Q)(H). Therefore, P(H) + Q(H) .s;: (P+Q)(H). However, for any x E (P+Q)(H) we have x = (P+Q)x and we conclude that

E

P(H) + Q(H) so

(P+Q)(H) .s;: P(H) + Q(H) (P+Q)(H) = P(H) + Q(H).

0

Projection operators play a role in simplifying the action of a linear operator by studying its restriction to special component subspaces. 14.22 Definitions. Given a linear operator T on a linear space X we say that a linear subspace M of X is invariant under T if T(M) .s;: M. Given linear subs paces M and N of X such that X = M EB N we say that a linear operator T on X is reduced by M and N if both M and N are invariant under T. 14.23 Remarks. For a linear operator Ton a linear space X with an invariant subspace M, we can regard the restriction ofT to M as a linear operator on M. When a linear operator T on a linear space X is reduced by linear subs paces M and N we can study T on X by T restricted to M and T restricted to N and such restrictions may be considerably simpler than the original T. For example, when X is n-dimensional then we will have our greatest simplification if there exists a set of n one-dimensional subspaces which reduce the linear operator T. This is equivalent to X having a basis {e 1, e2, ... , e0 } each element of which is an eigenvector ofT. 0 We can characterise the reduction of a linear operator in terms of associated projection operators.

203

§14. Projection operators

14.24 Theorem. Consider a linear space X and linear subspaces M and N where X= M EB Nand denote by P and Q the projection operators associated with M and N. Then a linear operator T on X is reduced by M and N if and only if PTx =TPx and QTx =TQx for all x EX.

Proof: Write x = m + n where mE M and n EN. Suppose that T is reduced by M and N. Then PTx

= PTm + PTn = Tm = TPx since Tm EM = P(X),

and similarly QTx =TQx since Tn EN =Q(X). Conversely, Tm = TPx = PTx EM and Tn = TQx

= QTx EN.

D

Of special interest is the case of closed invariant subspaces for continuous linear operators on a Hilbert space.

14.25 Theorem. Consider a continuous linear operator Ton a Hilbert space H. A closed linear subspace M ofH is invariant under T if and only ifMl. is invariant under T*.

Proof: Given that T(M) h M then for all x EM, (Tx, y) = 0 for ally E MJ.. Then (x, T*y) = 0 so T*y E MJ..

D

This result has the following implication.

14.26 Corollary. Given a continuous linear operator Ton a Hilbert space H, if the closed linear subspace M ofH is invariant under T then T is reduced by M and Ml..

14.27 Remarks. The following famous problem has been the object of considerable research over many years. The invariant subspace problem. Does a continuous linear operator Ton a Banach space X have a nontrivial invariant closed linear subspace? Every continuous linear operator Ton a Banach space X has trivial invariant subspaces {0} and X. On a finite dimensional complex linear space every linear operator has a nonzero eigenvector so the problem really concerns infinite dimensional Banach spaces. The general problem was solved in the negative by Per Enflo, Acta Math. 158 (1987), 213-313 and independently by C.J. Read, Bull. London Math. Soc. 16 (1984), 337-401, who in Bull. London Math. Soc. 17 (1985), 305-317 gave a counterexample in the Banach space (.t 1, 11·11 1). The solution is still not known for continuous linear operators in Hilbert space. However, for certain classes of linear operators on a complex Banach spac~ it is known that the solution is positive and we will have more to say about this in Section 18. D

204

Types of continuous linear mappings

14.28 EXERCISES

I.

(i)

Consider a finite dimensional subspace Mn of an inner product space X. Prove that there exists an orthogonal projection of X onto Mn.

(ii) Consider a finite dimensional subspace Mn of a normed linear space (X, 11·11). Prove that there exists a continuous projection of X onto Mn. 2.

(i)

Given a proper closed linear subspace M of a Hilbert space H, consider the mapping P of H into M defined for each

XE

H by

II x-Px II = d(x, M). Prove that P is linear, continuous and an orthogonal projection operator on H. (ii) Given a proper closed linear subspace M of a rotund reflexive Banach space X, consider the mapping P of X into M defined for each

XE

X by

II x-Px II = d(x, M). Prove that Pis linear, continuous and a projection operator on X with II P II 3.

Consider a Banach space (X, II· II) with linear subspaces M 1, M 2,

... ,

= I.

Mn such that

X= M 1 EB M 2 EB ... EB Mn with corresponding projection operations P 1, P2, M 1, M 2 ,

... ,

.•. ,

Pn from X onto

Mn.

Prove that the subspaces M 1, M 2,

.•. ,

Mn are closed if and only if the projections

P 1, Pz, ... , Pn are continuous. 4.

Consider a closed linear subspace M of a Banach space (X, 11·11) such that XIM has finite dimension. Prove that every projection of X onto M is continuous.

5.

Consider a Banach space (X, 11·11) with closed linearly independent subs paces M and N. Prove that M EB N is closed if and only if there exists ad> 0 such that II m-n II

6.

~

d for all m E M and n E N and II m II

= II n II = I.

Consider a normed linear space (X, 11·11) with a closed linear subpace M. Prove that M is complemented in X if (i)

M is finite dimensional, or

(ii) XIM is finite dimensional. 7.

Consider a continuous projection operator P on a Hilbert space H. Prove that (i)

if II P II

~

I then P is an orthogonal projection

(ii) if Pis a normal operator then Pis an orthogonal projection.

205

§ 14. Projection operators

8.

Consider a continuous projection operator P on a normed linear space (X, 11·11). (i)

Prove that the conjugate operator P' on X* is also a continuous projection operator.

(ii) Prove that P'(X*) = (ker P)j_ = { fE X* : f(ker P) = 0} and ker P' = P(X)_l_ = { fE X* : f(ker P) = 0}. (iii) Prove that P'(X*) is finite dimensional if and only if P(X) is finite dimensional and dim P'(X*) = dim P(X).

9.

Consider a continuous linear operator Ton a Banach space (X, II· II). Prove that if T is reduced by the closed subspaces M and N then the conjugate operator T' is reduced by the subspaces M_l_

10.

(i)

={fE X* : f(M) = 0} , N_l_ ={fE X* : f(N) = 0} .

Consider a linear operator Ton a linear space X reduced by linear subspaces M and N. Prove that

;J and Tl ~/exist,

(a)

T-1 exists if and only ifTI

(b)

ifT-1 exists then T-1 is reduced by M and Nand T-Il M =TI-l and T-Il N =TI-l M N.

(ii) Consider a continuous linear operator Ton a Banach space (X, 11·11) reduced by closed linear subspaces M and N. Prove that T-1 exists and is continuous if and only if Tl

II.

;J

and Tl ~ 1 both exist and are continuous.

Consider a normal operator Ton a Hilbert space H and M a closed linear subspace of H where M and Mj_ are invariant under T. Prove that (i)

II T II= max {II TIM II, II TIM_j_ II},

(ii) TIM is a normal operator on M, (iii) TIM* = T*IM'

12.

Consider a sequence {P n} of continuous projection operators on a Banach space X, pointwise convergent to an operator P on X. (i)

Prove that Pis a continuous projection operator on X.

(ii) Prove that if X is a Hilbert space and {P n} is a sequence of orthogonal projection operators then P is also an orthogonal projection operator on X. 13.

Consider a sequence {P n} of orthogonal projection operators on a Hilbert space H where Pj Pk = 0 for j, kEN, j

oF-

k.

Prove that the operator P on H defined by P =

L

P n is an orthogonal projection on

n=l Hand P(H) is the closure of the span of

U P 0 (H). n=1

206

Types of continuous linear mappings

§ 15. COMPACT OPERATORS

When we generalise the study of linear mappings on finite dimensional to infinite dimensional normed linear spaces it is reasonable that we select for special attention those continuous linear mappings which retain some of the special properties of those on finite dimensional spaces.

An important set of such linear mappings is the set of compact

mappings. 15.1 Definition. A linear mapping T of a normed linear space (X, 11·11) into a normed linear

space (Y, 11·11') is said to be compact if for each bounded sequence {xnl in (X, 11·11) the sequence {Txnl has a subsequence convergent in (Y, 11·11'). The reason for the term "compact" mapping is evident from the following characterisation. 15.2 Theorem. A linear mapping T of a normed linear space (X, 11·11) into a normed linear

space (Y, II· II') is compact if and only if for each bounded set Bin (X, 11·11), the set T(B) is compact in (Y, 11·11').

Proof. Suppose that Tis compact and B is a bounded set in X. Consider a sequence {y 0 } in ---

T(B). For each n E N, choose Xn

E

I

B such that II y n- Txn II < ;;- .

Now the sequence {Txn} has a subsequence {Txnk} convergent to say y

E

T(B) and then

{Ynk} is also convergent toy. So T(B) is compact. Conversely, for any bounded sequence {Xn} in X, the set { Txn : n E N } is compact 0 so {Txn} has a convergent subsequence and therefore T is compact. The following elementary lemma is useful 15.3 Lemma. A linear mapping T of a normed space (X, 11·11) into a normed linear space

(Y, 11·11') is compact if and only if for each sequence {xnl in the closed unit ball of(X, 11·11) the sequence {Txnl has a subsequence convergent in (Y, 11·11'). Proof. Consider a sequence {xn} in (X, 11·11) where there exists an r > 0 such that II Xn II :-::; r x x xn for all n EN. Then II __n II:-::; I for all n EN. If {T( __n)} has a subsequence {T(__l<_)} r r r convergent toy in (Y, 11·11') then subsequence { T(x 0 k)} is convergent tory in (Y, 11·11'). We conclude that T is compact. The converse is obvious.

0

207

§ 15. Compact operators

Compact mappings are automatically continuous. 15.4 Theorem. A compact mapping T of a normed linear space (X, 11·11) into a normed

linear space (Y, 11·11') is continuous.

Proof. If T is not continuous then for each n E N there exists an x0 EX such that II Tx 0 II> n II x0 II. But then the sequence

{~}is bounded but the sequence {T(~)} llx 11 llx 11 0

cannot have a convergent subsequence so T is not compact.

0

0

15.5 Remark. Not all continuous linear mappings are compact. Consider the identity operator on any normed linear space and the closed unit ball of the space. If the identity operator is compact then the closed unit ball is compact and by Riesz Theorem 2.1.9, this implies that the space is finite dimensional. So the identity operator on any infinite dimensional normed linear space is continuous but not compact. 0 The compactness property generalises the finite dimensional situation. 15.6 Theorem. A linear mapping T of a finite dimensional normed linear space (X 0 , 11·11)

into a normed linear space (Y, 11·11') is compact.

Proof. Since T is automatically continuous by Corollary 2.1.1 0, for a bounded subset B of (X 0 , 11·11), T(B) is bounded in (Y, 11·11'). But T(B) is a bounded subset of the finite dimensional subspace (T(X 0 ), 11-ii'T<Xnl). Therefore, T(B) is compact by Corollary

2.1.7.

0

However, Theorem 15.6 serves to motivate the following idea. 15.7 Definition. A linear mapping T of a normed linear space (X, 11·11) into a normed linear space (Y, 11·11') is said to be of finite rank if T(X) is finite dimensional. 15.8 Remarks. (i)

Not all finite rank mappings are continuous. Consider a discontinuous linear

functional f on an infinite dimensional normed linear space (X, II· II) and the associated one dimensional linear operator Ton X defined for a given z EX by Tx = f(x) z. Then T is finite rank but not continuous. (ii) However, a continuous finite rank mapping is always compact. The proof follows as in Theorem 15.6. 0

208

Types of continuous linear mappings

It is interesting to see how the compactness criterion actually gives us information about the range of a compact mapping. 15.9 Theorem. Consider a compact mapping T of a normed linear space (X, II· II) into a normed linear space (Y, 11·11'). (i)

T(X) is separable.

(ii) /fT(X) is of second category in itself then Tis offinite rank.

Proof. (i)

For the closed unit ball B of (X, 11·11), T(X)

= U nT(B) .. n=l

But T(B) is compact so it is separable. Then U nT(B)is also separable and we conclude n=l

that T(X) is separable. (ii) If also (T(X), II·II'T(X)) is second category then by Lemma 10.5, for the closed unit ball B of (X, 11·11), T(B) is also a neighbourhood of the origin in (T(X), ll·ll'nx)). Since (T(X), II·II'T(X)) has a compact neighbourhood of the origin we conclude from Riesz Theorem 2.1.9 that T(X) is finite dimensional. One of the main applications of compact operators is in the theory of integral equations. It is with an eye to this setting that we consider an example of a compact operator on the Banach space (t:[a,b], ll·lloo). For a discussion of compactness in this space we need the following concept. 15.10 Definition. A subset A of (t:[a,b], ll·lloo) is said to be equicontinuous on [a,b] if given

£

> 0 and x0 E [a,b] there exists a 8(e,x0) > 0 such that for all f E A,

If(x)- f(x 0) I < £

for all x E [a,b] where I x-x 0 I< 8.

We need to use the following characterisation of compactness in (t: (X), 11·11 00 ) , where (X, d) is a compact metric space, (see AMS §9). 15.11

The Ascoli-Arzela Theorem.

Given a compact metric space (X, d), a subset A of (t: (X) 11·11 00 ) has

Acompact if and

only if A is bounded and equicontinuous. 15.12 Example. Consider the Fredholm operator K defined on t:[a,b] by b

(Kf)(x)

= Jk(x,t)

f(t) dt

a

with kernel k a continuous complex function on the square region (x,t) : a.,::; x .,::; b, a.,::; t.,::; b}. Now K is a continuous linear operator on S (t:[a,b], 11·11 00 ) , (see AMS §7), but further K is a compact operator.

={

209

§ 15. Compact operators

Proof. We use the characterisation given in Theorem 15.2. Consider a bounded set B in ('C[a,b], ll·lloo). Then there exists an M > 0 such that II f lloo <;; M

for all fEB.

We need to show that K(B) is compact. By the Ascoli-Arzela Theorem 15.11 we need only show that K(B) is bounded and equicontinuous. Since K is a continuous linear operator it follows that K(B) is bounded. Now k is uniformly continuous on the square region S. So given £ > 0 there exists a 8 > 0 such that

Ik(x,t)- k(x 2,t) I < £

for all t E [a,b] and x 1,xz E [a,b] and I x 1-x 2 1 < o.

Therefore, for any fEB b

I (Kf)(x 1) - (Kf)(x 2) I <;;

JI k(x 1,t)- k(x 2,t) II f(t) I dt a

<;; (b-a) II flloo sup {I k(x 1,t) -k(x 2 ,t) I: a<;; t <;; b}

< £(b-a)M

for I x 1-x 2 I< 8 and all fEB;

O

that is, K(B) is equicontinuous.

We now investigate the structure of the set of compact mappings among the continuous linear mappings. 15.13 Notation. Given normed linear spaces (X, 11·11) and (Y, 11·11'), we denote by

%(X,Y) the set of compact mappings of X into Y and for the case where Y

= X,

by

% (X) the set of compact operators on X. 15.14 Theorem. Given nonned linear spaces (X, 11·11) and (Y, 11·11'), then 'K(X,Y) is (i)

(ii)

a linear subspace of~(X,Y) and a closed linear subspace of~(X,Y) if(Y, 11·11') is complete.

Proof. (i) Given compact mappings SandT consider {xnl a bounded sequence in (X, 11·11). Then there exists a subsequence { Xnk} such that {Sxnk} is convergent and a further subsequence { Xnk,t } such that { Txnk.t} is convergent. Therefore {(aS+ ~T)xnk } is convergent for any scalars a and~. So 'K(X, Y) is a linear .t

subspace of (ii)

~(X,

.

Y).

Consider a sequence {T n} of compact mappings convergent to a continuous linear

mapping Tin (~(X,Y), 11·11) and a bounded sequence {xnl in (X, 11·11). Since T 1 is compact there exists a subsequence {xn,I} of {xnl such that {T 1xn,I} is convergent in (Y, 11·11'). Since T 2 is compact there exists a subsequence {xn,zl of {xn,I} such that {T 2 xn,z} is convergent in (Y, 11·11'). Continuing this process we consider the diagonal sequence

210

Types of continuous linear mappings

{xn,nl = {x 1, 1 , x 2 , 2 ,

..• ,

Xn,n• ... }.

This is a subsequence of each subsequence {Xn,l }. {Xn, 2 }.

... , { Xn,k}

, ... by

construction. Given £ > 0 there exists a v EN such that II Tn-T II<£

for all n

~v.

So II Txm,m- Txn,n II ~ II Txm,m-T vXm,m II + II T vXm,m- T vXn,n II + II T vXn,n- Txn,n II ~II T-Tv II II xm,m II+ II Tvxm,m- Tvxn,n II+ II T-Tv II II xn,n II.

But {Tvxn,nl is convergent so we conclude that {Txn,nl is Cauchy in (Y, 11·11'). Since

(Y, 11·11'), is complete then {Tx 0 ,0 } is convergent; that is, Tis compact. So we conclude that 'K(X,Y) is closed in (n(X,Y), 11·11).

D

Theorem 15.14(ii) provides a useful method for proving that a linear operator is compact. 15.15 Example. Given Hilbert sequence space (.t 2 , 11·11 2 ), the diagonal operator T defined for x

= {A!o

~ •...

, A0 ,

...}

by Tx

= {a 1A. 1, a 2 A. 2 , . . . , U A 0

0 , ••• }

where

U0 -7 0 as n --7 ""• is compact.

Proof. For each n EN, consider the finite rank operator F 0 where F 0 x = {a 1 A. 1, a 2 A. 2 , ..• , U 0 A0 , 0, ... }. Clearly for each n EN, Fn is linear and continuous. Since Un--7 0 as n --7 ""• given£> 0 there exists avE N such that I an I<£ when n > v. Then II(T-Fn)xll 2 =

~

I.

2

2

lakiiA.kl <e

k=n+l IIT-Fnll<£ foralln>v;

So

,------:::---~

I.

k=n+l

2

IA.kl.

that is, Tis the limit of continuous finite rank operators Fn which are compact. By Theorem 15.14(ii) we conclude that Tis compact.

D

The following structural property of 'K(X) is of interest. 15.16 Theorem. Given a compact operator T and a continuous linear operatorS on a normed linear space (X, 11·11), then STand TS are both compact operators on (X, 11·11).

Proof. Consider a bounded sequence {x0 } in (X, 11·11). Then since Sis continuous the sequence {Sx 0

}

is also bounded. Since T is compact the sequence { T(Sxn)} has a

convergent subsequence; that is, TS is a compact operator. But also the sequence { Txn} has a convergent subsequence { Tx 0 k} and since S is continuous the sequence { S(Txnk)} is also convergent; that is, ST is a compact operator. D

§ 15. Compact operators

211

This result has a consequence for the continuity of inverses.

!5.17 Corollary. Given a compact operator Ton an infinite dimensional normed linear space (X, 11·11), if T -I exists on (X, 11·11) then T -I is not continuous.

Proof. If T- 1 is continuous on X then by Theorem 15.16 we have that I= T- 1T is also compact. But from Remark 15.5 we see that I is not compact so T- 1 is not continuous. 0 We now show how the compactness property is inherited by conjugate mappings.

15.18 Schauder's Theorem. A compact mapping T of a normed linear space (X, II· II) into a normed linear space (Y, II· II') has compact conjugate mapping T'of (Y*, 11·11') into (X*, 11·11).

Proof. Since T is compact, for the closed unit ball B in (X, 11·11), T(B) is compact in (Y, 11·11'). Consider {f0 } a sequence in the closed unit ball of (Y*,II·II'). In ('C(T(B), ll·lloo) the set A= {f 0 l_: n EN} is bounded and since T(B)

I f0 (y)- f 0 (y 1) I<;; II y-y 1 II' for all n EN then A is also equicontinuous. Therefore by the Ascoli-Arzela Theorem 15.11, A is compact in ('C(T(B), 11·11 00 ) . So {f0 } has a subsequence {fnk} where

{f0 kl-: n EN} is convergetn in ('C(T(B), 11·11 T(B)

00 ) .

But

liT' f 0 k-T' f0 j II= sup {I (f0 k-fn) (Tx) I: x EB }. So {T' f 0 k } is a Cauchy sequence in (X*, 11·11). But (X*, 11·11) is complete so {T' f 0 k } is convergent in (X*, 11·11). Then by Lemma 15.3, T' is compact.

0

For Hilbert space, the following theorem reveals the special structure of compact operators as limits of finite rank operators.

15.19 Theorem. Given a continuous linear operator T on a Hilbert space H, the following conditions are equivalent. (i)

(ii)

Tis compact. For any orthonormal set {ea} in Hand f.> 0 the set

{a:

I (Tew ea) I ~f.} is finite.

(iii) There exists a sequence {F0 } of continuous finite rank operators on H such that

II T-F0 ll-7 0 as n

--7

oo.

212

Types of continuous linear mappings

Proof. (i) =} (ii) ForT compact suppose that (ii) does not hold. Then for some orthonormal set {ea} and some r > 0, the set

{a : I (Tea, ea) I ~ r}

is infinite. So it has a countably

infinite subset which gives rise to an orthonormal sequence {en} such that I (Ten, en) I ~ r for all n E M . Since Tis compact there is a subsequence {enk} such that {Tenk} converges to an x E H.

Discarding a finite number of terms of this sequence we may suppose that

II Te 0 k- x II < ~

for all k E N.

I (Tenk' enk)- (x, e0 k) I= I (Te 0 k- x, e0 k) I~ II Te 0 k- x II < ~

Then

for all kEN.

and so

But this contradicts Bessel's inequality, Theorem 3.13. (ii) =}(iii) Given n EN, consider the family I

I (Teao ea) I > ~ for all

::r of all orthonormal sets {eal in H where

a.

::r partially ordered by set inclusion. Now the union of any totally ordered subfamily of orthonormal sets from ::r is itself a member of ::r and is an upper bound for this subfamily. It follows by Zorn's Lemma that ::r has a maximal member {e~}. As a member of ::r, {e~} is finite.

By (ii) each such orthonormal set is finite. Consider

Consider the finite dimensional linear subspace M = sp {e~}. Then

I (Tx, x) I <.!. n

for otherwise,

::r contains {e~} u

for all x E M.l and II x II = I

{x} which contradicts the maximality of

{e~}.

Consider the projection P n of H onto M and write x = (1-P0 )z where z E H. Then

I (T(I-P 0 )z, (I-P0 )z) I<.!.;

that is,

I ((1-P0 ) T(I-P0 )z, z) I <.!.

n

for all z E Hand II z II ~ I.

n

Therefore, by Lemma 13.10.6(ii), II (1-P0 )T(I-P0 ) II < ~ . n

The continuous linear operator Fn = PnT + TPn- PnT P n has finite rank and II T-F0 II < ~. n (iii) =} (i) Since continuous finite rank operators are compact, the result follows from

0

Theorem 15.14(ii).

For separable Banach spaces with a Schauder basis we have a comparable result. But to establish this we need an important properties of Schauder bases. 15.20 Lemma. Consider a separable Banach space (X, 11·11) with a Schauder basis {e0 }. There exists a K ~ I such that for any x

= L "-k ek, k=l

n

II

L k=l

"-k ek II~ K II

L k=l

"-k ek II

for all n EN.

213

§ 15. Compact operators

Proof. Consider the Banach space X with norm n

II x II' = sup {II I A.k ek II : n E N }. k=l

(see Exercise 1.26.17).

n

II x II = II I A.k ek II <;; sup {II I A.k ek II : n E N } = II x II' k=l k=l By Corollary I 0.10 to the Open Mapping Theorem, norms 11·11 and 11·11' are equivalent, so

Now

there exists a K > 0 such that n

~

sup {II I Ak ek II : n EN} <;; K II I A.k ek II for all x = I A.k ek. k=l k=l k=l Obviously, K ~ I.

D

15.21 Lemma. For every compact subset A in a separable Banach space (X, 11·11) with a Schauder basis {e 0 },

lim sup II x-P0 x II= 0 n-+>o xEA

where for x = I Ak ek we define for each n E N, the projection operator P0 on X by k=l n

P0 x=IA.kek· k=l Proof. Since A is compact, given £ > 0 there exists an £-net {y 1, y2,

••• ,

Ym) in X

such that n

A<;;; U B(yk; E) ; (see AMS §8). k=l For x E A there exists some Yko such that II x-yko II<£. But also there exists a v

E

N such that, for all k E { I, 2, ... , m) II Yk-PnYk II<£

for all n > v.

Therefore, II x-P0 x II<;; II x-Yko II+ II yk 0-P0 yko II+ II P0 Yk0- P0 x II <;;II x-Yko II (1+11 P0 II)+ II yk 0-PnYko II <E(l+IIP0 II)+E

foralln>v.

From Lemma 15.20 we have that n

so Then

II P0 X II = II I A.k ek II <;; K II I Ak ek II = K II x II k=l k=l II P0 II <;; K for all n E N. sup II x-P 0 x II< (2+K)E

for all x EX,

for all n > v.

D

XEA

15.22 Theorem. Consider a continuous linear operator Ton a separable Banach space (X, 11·11) with a Schauder basis {e 0 ). Then Tis compact if and only if there exists a sequence {F0 } of continuous finite rank operators on X such that II T-F0

II~

0 as n ~"".

214

Types of continuous linear mappings

Proof. For each n E N consider the natural projection P n on X defined for each x=

I

n

Akekby Pn(x)=

k=l

I

AkekwhereiiPnii::;K.

k=l

Suppose that T is compact. Then for the closed unit ball B of (X, II· II), T(B) is compact. Therefore by Lemma 15.21, given£> 0 there exists avE N such that II Tx- PnTx II:::;; (2+K)£ for all x EB and n > v. IIT-PnT11<(2+K)£ foralln>v.

So

But· P nT is a continuous finite rank operator on (X, 11·11). The converse follows again from Theorem 15.14(ii). 15.23 Remark. Theorems 15.19 and 15.21 state that on a Hilbert space or on a separable Banach space with Schauder basis every compact operator can be represented as the limit of continuous finite rank operators. In 1932, Stefan Banach asked whether such a representation holds for compact linear operators on any Banach space. However, Per Enflo, Acta Math. 130 (1973), 309-317, constructed a separable reflexive Banach space having a compact operator which is not the limit of continuous finite rank operators. So it was deduced that the space constructed by Enflo does not have a Schauder basis. This provided a negative solution to the Basis Problem. D 15.24 EXERCISES I.

(i)

Consider a compact mapping T of a Banach space (X, 11·11) into a Banach space (Y, 11·11'). Prove that (a) T(X) is closed in (Y, 11·11') if and only if T(X) is finite dimensional,

(b) every closed linear subspace of T(X) is finite dimensional. (ii) Prove that on a Banach space (X, II· II) every compact projection operator P has P(X) finite dimensional. (iii) Prove that on a Hilbert space H every compact self-adjoint operator T has T(X) finite dimensional. 2.

(i)

Given a bounded sequence of complex numbers (an} prove that the diagonal operator T on (m, 11·11 00 ) defined for x = { A1, A2, Tx = { a 1A1, a 2A2 , n -7

••. ,

•.. ,

An, ... } by

an An, ... } is compact if and only if an

-7

0 as

oo.

(ii) Consider a continuous linear operatorS on (m, 11-11 00 ) and an operator P on (~ (m), 11·11 00 ) defined for x

where ( e 1,

= { A1, A2, ... , An, ...

} Em by

P(S)(x) = { S(e 1) A1, S(e 2 ) A2 , ... , S(en) An, ... } e 2, ... , en, ... } is the usual basis form.

215

§ 15. Compact operators

(a) (b)

Prove that Pis a continuous projection operator on (~(m), 11·11 00 ) . Prove that if Sis a compact operator on (m, 11-11 00 ) then P(S) is a compact operator on (m, 11-11 00 ) .

3.

Consider continuous linear operators S and Ton a Banach space (X, 11·11). We know that ST =I does not necessarily imply that TS

=I.

However, if T is a

compact operator prove that (i)

S(l- T)

=I if and only if (1- T)S =I, and

(ii) if S has inverse I-T then I- (1-T)- 1 is compact.

4.

Consider a compact mapping T of a normed linear space (X, 11·11) into a normed linear space (Y, 11·11'). Prove that if a sequence {xnl in X has the property that { f(xn)} -7 0 as n -7 oo for all f eX* then II Txn II -7 0 as n -7

5.

(i)

oo,

Consider a compact mapping T of an infinite dimensional Banach space (X, 11-11) into a Banach space (Y, 11·11'). Prove that { Tx : II x II = I } contains 0

inY. (ii) Consider a separable Hilbert space H with an orthonormal basis {en!. (a) Prove that if Tis a compact mapping of H into a normed linear space then Ten -7 0 as n -7 oo. (b) Prove that if T is a continuous linear mapping of H into a Banach space where L II Ten 11 2 < oo then T is compact. 6.

Prove the converse of Schauder's Theorem 15.18 A continuous linear mapping T from a Banach space (X, 11-11) into a Banach space (Y, 11·11') is compact if its conjugate mapping T' from (Y*, 11·11') into (X*, II· II') is compact.

7.

Consider continuous linear operators T and S on a Banach space (X, 11·11) where T has a continuous inverse on X and T - S is compact. Prove that (i)

S(X) is closed and has finite codimension,

(ii) either S has a continuous inverse on X or is not one-to-one.

8.

Consider a continuous linear operator T on a complex Hilbert space H. (i) Prove that T is compact if and only if T*T is compact. (ii) Suppose that lim II TR II l/n =0. Prove that T is compact if and only if T T* is compact. (Hint: Use Theorem 15.19(ii).)

216 9.

Types of continuous linear mappings

(i)

Consider an incomplete normed linear space (X, 11·11) and a compact operator T on (X, 11·11). Prove that the unique continuous linear extension Ton the completion (X, 11·11) is also compact.

(ii) Consider the Fredholm operator K defined on "C[O, I] by I

(Kf)(x) =

Jk(x,t) f(t) dt 0

with kernel k a continuous complex function on the square region S = { (x, t) : 0:::;; x :::;; I, 0:::;; t:::;; I}. Prove that' (a) K is a compact operator on ("C[O, I]. 11·11), (b) I 0.

K is a compact operator on the Hilbert space ( J:; [0,

I]. 11·11 2).

Consider a compact operator Ton a Banach space (X, 11·11) and M a closed invariant subspace ofT. Prove that the mapping x + M H Tx + M is a compact operator on the quotient space (X/M, 11·11).

VI.

SPECTRAL THEORY

In Section 4 we noted that, given a normed linear space the algebra of continuous linear operators on the space is a noncommutative unital normed algebra and as such, interest lies in studying the set of regular elements in this algebra. Spectral theory develops this line of investigation by determining, for a continuous linear operator T on a complex normed linear space X, properties of the set of scalars {A E
§ 16. THE SPECTRUM

As many of the properties of an operator algebra can be derived directly from its being a unital normed algebra, it is more straightforward to begin by developing our theory for elements of a unital normed algebra. The spectrum of an element of a unital normed algebra is a useful representation in the complex plane of the singular elements associated with any given element of the algebra. 16.1 Definitions. Given an element x in a complex unital normed algebra (A, 11·11), the spectrum of x is the set of complex numbers cr(x) {A E (C : x- Ae is singular},

=

and the resolvent set of xis the complement of cr(x) p(x) = {A E

(C :

x- A.e is regular}.

For any element of a complex unital normed algebra it is important to locate its spectrum as a subset of the complex plane and the following properties give us our first general information in this regard. To avoid unnecessary complications we will assume completeness; that is, mostly we will work in a complex unital Banach algebra, and in particular we will consider the algebra of operators on a complex Banach space.

218

Spectral theory

16.2 Theorem. For any element x in a complex unital Banach algebra (A, 11·11), cr(x) is compact.

Proof. Consider the mapping from II x II then II x/A II < 1 and from Corollary 4.11.11 we deduce that e - x/A is regular. Then x- Ae is regular so we conclude that cr(x) >;;;; {A E
D

Because of the spectral property given in Theorem 16.2, it is useful to define the following constant.

16.3 Definition. Given an element x in a complex unital Banach algebra (A, 11·11), the spectral radius of x is defined as v(x) =sup {I A I: A Ecr(x) }. 16.4 Remark. The spectral radius is the radius of the smallest closed disc centred at 0 and containing the spectrum. From Theorem 16.2 we have cr(x) ::; { A : I A I ::; II x II} andso

v(x)::;llxll

0

forallxEA.

We now establish deeper properties for the spectrum. We show that in a complex unital Banach algebra the spectrum is nonempty and we have a formula for the spectral radius in terms of the norm although the spectrum is a purely algebraic notion. To do so we need to use complex analysis of analytic functions.

16.5 Definition. Given an element x in a complex unital normed algebra (A, 11·11), the resolvent operator R maps p(x) into A and is defined by R(A) = (x- Aer 1• 16.6 Lemma. Given an element x in a complex unital normed algebra (A, 11·11), (i)

R(A)- R(IJ.) = (A-IJ.) R(A) R(~J.)for all A, 11 E p(x) and II R(A) II~ 0 as A~

(ii) for any f E A*, foR is analytic on p(x). (iii) When A is complete, for all I A I> v(x),

foR(A)=-.!. (f(e)

A

+I

n=l

f(xn))·

A0

oo,

and

219

§ 16. The spectrum

Proof. (i)

Since x-1 - y - 1 = x - 1 (y - x) y - 1 for all regular elements x, y E A, we have R(A.)- R(J.L) = (A.-J.L) R(A.) R(J.L)

so R is continuous on p(x). But also from Theorem 4.11.18, (e- x/A.r 1 ~ e as

II R(A.) II = II (x-A.e)-1 II = - 1-11 (e- xtA.r 1 II I A. I

so

(ii)

For any f E A* and A., J1. E p(x),

A.~

oo

~o

as

A.~ oo.

A.* J1.

f o R(A.) - f o R(J.L) = f(R(A.) R(J.L)) A.-J.L and so

lim f o R(A.)- f o R(J.L) = f(R2(J.L)) A.~Jl A.-J.L

and then foR is analytic on p(x). (iii) For I A. I > II x II we have from Theorem 4.11.10 that R(A.)

= (x-A.e)-1 = - -1 ( e + A

Loo

-'S._ ") •

n=1 A."

For any f EA* we have foR(A)=

_l

(f(e)

+

A But f o R is analytic on p(x)

;;;?. { AE

L

n=1

f(x"))· A."

v(x)} so this is the Laurent series expansion

D

for f o R on { AE v(x)}.

Lemma 16.6 enables us to establish the following spectral properties. 16.7 Theorem. Given an element x in a complex unital normed algebra (A, 11·11), (i)

cr(x)

* 0 and

(ii) when A is complete v(x)= lim llx"lllin =inf{llx"ll 110 :nEN}. n~~

Proof. (i)

Suppose that cr(x) = 0, then p(x) = (C and for any f E A*, f o R is an entire function.

But also since f E A*, I f o R(A.) I ~II f 1111 R(A.) II so from Lemma 16.6(i), f o R(A.)

~

0 as A.

~ oo

By Liouville's Theorem we deduce that f o R(A.) = 0 for all A. E
* 0.

220 (ii)

Spectral theory

From Lemma 16.6(iii) we have that for any f E A* and any real a> v(x), the series

L

f(xn) is convergent and so the sequence {f(xn)} is bounded. But this holds for all n=l an an f E A*, so by the Theorem 11.12(i), we conclude that there exists a K > 0 such that

II xn II< K

for all n EN .

an

Then II xn II lin< a K 11 n for all n EN and so lim sup II xn 11 11 n

lim sup II xn 11lin ~ v(x). n-+x>

Now for v(x) > 0 there exists a A E a: such that I A I < v(x) where x- Ae is singular. If also lim sup II xn II lin< I A I n-+x>

then by Theorem 4.11.1 0, x - Ae is regular. So we conclude that v(x) = lim sup II xn 11 11n. n-+x>

From Proposition 4.11.12 we have v(x)= lim 11xn11 11n =inf{llxnll 11n:nEN}.

D

n~~

16.8 Remark. We note that we invoked the completeness condition when we used series arguments like those of Theorem 4.11.1 0 and Corollary 4.11.11. Completeness guarantees the boundedness of the spectrum and a meaningful definition of spectral radius. D We now present a useful algebraic property of the spectrum. 16.9 The Spectral Mapping Theorem for polynomials. Consider a complex unital normed algebra (A, II· II) and a polynomial p with complex

coefficients. For every x E A, cr(p(x)) = p(cr(x)) = {p(A): A E<J(x) }.

Proof. Clearly p(x) EA. For any A E a: consider the polynomial p(t) - A. Now if p is of degree n then by the Fundamental Theorem of Algebra, p(t)- A has n roots, A1, "-2,

... , An

and we can write n

for some a E a:. i=l If A E cr(p(x)) then there exists some i0 E { 1, 2, ... , n} such that x- Ai 0e is singular, that p(x)- Ae =a

IT (x-Aie)

is Aio E cr(x). Then A= p(Ai 0 ) E p(cr(x)) and so cr(p(x)) (;;;; p(cr(x)). Conversely, if A E p(cr(x)) then there exists i0 E { 1, 2, ... , n} such that Aio E cr(x) and A= p(Ai 0 ).

However, if Ae cr(p(x)) then every factor x- Aie, of p(x)- Ae is

regular so Ai ecr(x) for all i E { 1, 2, ... , n}. So we conclude that p(cr(x)) (;;;; cr(p(x)).

D

221

§ 16. The spectrum

!6.1 0 EXERCISES

I.

An element z of a complex unital Banach algebra (A, 11·11) is called a topological divisor of zero if there exists a sequence {z0 } in A where II z0 II = 1 for all n EN

and either z0 z --7 0 or z:z, (i)

--7

0 as n --7 ""· Prove that

every divisor of zero is a topological divisor of zero,

(ii) the set of topological divisors of zero is closed in A, (iii) every topological divisor of zero is a singular element, (iv) the boundary of the set of singular elements is contained in the set of topological divisors of zero. 2.

Consider a complex unital Banach algebra A and a complex unital Banach subalgebra A'. Prove that for any element x E A', (i)

<J A'(x) ~

crA(x) and

(ii) acr A(x) ~ acr A'(x). 3.

Consider a complex unital Banach algebra (A, 11·11). Prove that each of the following conditions implies that (A, 11·11) is isometrically isomorphic to (i)

a:.

A is a division algebra; that is, every nonzero element of A has an inverse.

(ii) Zero is the only topological divisor of zero. (iii) There exists an M > 0 such that II xy II ~ M II x 1111 y II for all x, y EA. (iv) For every regular element x E A, II x- 1 II=~ . 4.

Consider a complex unital normed algebra A. Prove that (i) if x E A is regular then x- 1 - e/A = (x- 1/A)(Ae-x) and deduce that cr(x -1) = { IIA : AE cr(x)}, (ii) if we are given that e- xy is regular for x, y E A then e- yx is regular where (e-yxr 1 = e + y(e-xyr 1 x and deduce that for any x, yEA cr(xy) \ {0} = cr(yx) \ {0}.

5.

Consider the complex unital commutative Banach algebra (t:[O, 1], II· II"") and f E t; [0, I]. Prove that (i)

f is either regular or is a topological divisor of zero,

(ii) A E cr(f) if and only if XE crcf), (iii) f = 6.

f

if and only if cr(f) ~ IR.

Consider a complex unital Banach algebra A. Prove that for any given x E A the spectral radius v(x) has the following properties (i)

v('Ax) = I A I v(x),

(ii) v(xy) = v(yx), (iii) if xy = yx then v(xy)

~

v(x) v(y) and v(x+y)

~

v(x) + v(y).

222

Spectral theory

§17. THE SPECTRUM OF A CONTINUOUS LINEAR OPERATOR

We now consider the particular case of the algebra of continuous linear operators on a normed linear space. Although Section 16 gives us general structural properties of these as elements of the unital normed algebra of operators, here it is important to determine whether a particular continuous linear operator has an inverse on the space and whether the inverse is also continuous. 17. 1 The spectrum in linear spaces We begin with purely algebraic considerations in ;r; (X), the algebra of linear operators on a linear space X. It is clear that a linear operator T on a linear space X has an inverse T -I on X if and only if T is one-to-one and onto. So we have the following definitions. 17.1.1 Definitions. Given a linear operator Ton a linear space X, an eigenvalue ofT is a scalar A where there exists an xi' 0 such that Tx =Ax. Given an eigenvalue A ofT, the

elements x EX such that Tx = Ax are called eigenvectors ofT and the linear space { x EX : Tx = Ax} is called the eigenspace associated with the eigenvalue A. The set of eigenvalues ofT is called the point spectrum ofT and is denoted by Pcr(T). Clearly A E Pcr(T) if and only if T- AI is not one-to-one. There is a fundamental relation between the eigenvalues and their associated eigenspaces. 17.1.2 Theorem. Given a linear operator Ton a linear space X with distinct eigenvalues At, A2, ... , A0 , any set { x 1, x2, ... , x0 } of corresponding eigenvectors is linearly

independent. Proof. We use proof by induction. Now {xtl is linearly independent. Suppose that {x 1, x2, ... , xk} is linearly independent and consider a 1x 1 + a 2x2 + ... + akxk + ak+lxk+l = 0. Then T(a 1x 1 + a 2x 2 + ... + ak+ 1xk+l) = a 1Tx 1 + a 2Tx 2 + ... + ak+ 1Txk+l = atAtxl + a2A2x2 + ... + ak+tAk+lxk+l = 0.

Then al(At-Ak+l)xl + aiA2-Ak+l)x2 + ... + ak(Ak-Ak+l)xk = 0 and since { x 1, x2, ... , xk} is linearly independent at(At-Ak+l) = a2(ArAk+l) = · · · = ak(Ak-Ak+l) = 0. Since the eigenvalues are distinct, we have a 1 = a 2 = ... = ak = 0. Since xk+l i' 0 we conclude that a 1 = a 2 = ... = ak and so { x 1, x2, ... , xk+l} is linearly independent. The following is an elementary consequence.

= ak+l = 0 D

223

§ 17. The spectrum of a continuous linear operator

17.1.3 Corollary. Given a linear operator Ton a linear space X, for eigenspaces M 1 and M 2 associated with eigenvalues A. 1 and A.z, M 1 n M 2 = {01. So we have some insight into the structure of the point spectrum of a linear operatQr on a finite dimensional linear space. 17.1.4 Corollary. For a linear operator Ton an n-dimensionallinear space Xn, the point spectrum Pcr(T) contains no more than n elements. Proof. Xn contains the direct sum of its eigenspaces so T cannot have more than n eigenvalues. D For finite dimensional linear spaces, the failure of a linear operator to have an inverse can be reduced to its failing to be one-to-one. 17 .1.5 Theorem. A linear operator T on an n-dimensionallinear space Xn is one-to-one

if and only if it is onto. Proof. Consider { e 1, e2 , ••. , en} a basis for Xn. Suppose that T is one-to-one and consider the set { Te 1, Te 2 ,

I

••• ,

I

Ten}. If for scalars

akek=Osinc~

f

akek)=Oso k=l k=l Tis one-to-one. But { e 1, e 2 , •.. , en} is a basis for Xn so a 1 = a 2 = ... =an = 0. Then {Te 1, Te2 , ... , Ten} is linearly independent and so is a basis for Xn. We conclude {a 1,a2 ,

•••

,an} we have

akTek=OthenT(

k=l

that T is onto. Conversely, if T is onto then for each ek E Xn where k

E { I,

2, ... , n I there exists n

an xk

E

Xn such that ek = Txk. Then for such a set { x 1, x2 ,

..• ,

xn}, if

I. akxk = 0 then k=l

n

n

I. ak Txk = 0 and I. akek = 0 so a 1 = a 2 = ... =an = 0 since { e 1, e2, ••• , en} is k=l k=l basis for Xn. Then {x 1, x2 , •.. , Xn} is linearly independent and so is a basis for Xn·

a

n

IfTx = 0 then x =

I. ~k xk for scalars ~k where k E {I, 2, ... , nl. k=l

n

0=

So

n

I. ~k Txk = I. ~kek

k=l k=l and then ~ 1 = ~ 2 = ... = ~n = 0 since { e 1, e2 , . . . , en} is a basis for Xn. Therefore D x = 0 and we conclude that T is one-to-one. 17.2 The spectrum in normed linear spaces Given a continuous linear operator Ton a normed linear space (X, 11·11), Tis a regular element of ~(X) if and only if T -I exists and is continuous on X. When X is finite dimensional we have seen in Theorem 17 .1.5 that T -I exists on X if and only if T is

224

Spectral theory

one-to-one and the continuity ofT and T -I is automatic by Corollary 2.1.1 0. When X is infinite dimensional the situation is a little more complicated. In general, a continuous linear operator Ton a Banach space (X, 11·11) is not invertible in (i)

~(X)

if and only if

T is not one-to-one, or

(ii) T is one-to-one but T(X) is not dense in (X, 11·11), or (iii) Tis one-to-one and T(X) is dense in (X, 11·11) but T- 1 is not continuous on T(X). So we are led to the following decomposition of the spectrum ofT. 17 .2.1 Definitions. The spectrum of a continuous linear operator T on a complex Banach space (X, 11·11) ,

cr(T)

={A

E

a: : AI- Tis singular in ~(X)}

can be separated into three disjoint component sets: the point spectrum consists of the eigenvalues ofT Pcr(T) {A E a: : AI - T is not one-to-one},

=

the residual spectrum is the set Rcr(T) = {A E

a: : AI - T is one-to-one but (AI-T)(X) is not dense}

and the continuous spectrum is the set Ccr(T) = {A E a: : AI - T is one-to-one and (AI-T)(X) is dense but (AI-Tr 1 is not continuous on (AI-T)(X)}. So

cr(T) = Pcr(T) u Rcr(T) u Ccr(T).

17.2.2 Remarks.

(i) Theorem 17 .1.5 shows that for a continuous linear operator T on a complex finite dimensional normed linear space, cr(T) = Pcr(T). (ii) We recall from Corollary 1.24.10 that for a continuous linear operator Ton a Banach space (X, 11·11), if AI- T is one-to-one and (AI- T)(X) is dense and (AI- Tr 1 is continuous on (AI- T)(X) then AI- Tis onto and so AI- Tis regular in ~(X). 0

I 7. 3 The spectrum and numerical range It is important to be able to determine the spectrum of a continuous linear operator T on a complex Banach space (X, 11·11), but it is not always a simple set to specify. From Section 16 we saw that cr(T) is a compact set contained in the closed disc of radius II T II. In Hilbert space H the closure of the numerical range ofT is a smaller set contained in the closed disc of radius II T II which also contains the spectrum. In Theorem 13.10.9 we showed that for any A f1' W (T), T - AI is regular in

~(H).

This result has the following

expression in terms of the spectrum ofT. 17.3.1 Theorem. For a continuous linear operator Ton a complex Hilbert space,

cr(T) !:;;; W (T).

§ 17. The spectrum of a continuous linear operator

225

Theorem 13.10.8 then has an immediate implication for the spectrum of self-adjoint operators. 17.3.2 Corollary. For a self-adjoint operator Ton a complex Hilbert space, cr(T) ~[-II T II, II T II]. For the spectrum of positive operators we have the following result. 17.3.3 Corollary. For a positive operator Ton a complex Hilbert space, cr(T) ~ [0, II T II]. For the spectrum of normal operators, Theorems 13.11.4 and 16.2 have the following implication. 17 .3.4 Corollary. For a normal operator Ton a complex Hilbert space, v(T) = w(T) = II T II

and there exists a A E cr(T) such that I A I = II T II. For the spectrum of unitary operators, numerical range considerations do not give us further insight. Theorem 13.11.9 has the following expression in terms of the spectrum. 17.3.5 Corollary. For a unitary operator Ton a complex Hilbert space cr(T) ~ { A E ([ : I A I = I }. In the 1960s there was a successful generalisation of the concept of numerical range of a continuous linear operator on a normed linear space. This development was largely due to F.F. Bonsall and an exposition of the basic theory can be found in F.F. Bonsall and J. Duncan, Numerical Ranges I and II, Cambridge University Press (1971) and (1973).

17.3.6 Definition. Given a complex normed linear space (X, 11·11), for each x EX, II x II = I, consider the set D(x) = { f EX* : II f II = I, f(x) = I}. Corollary 6.3 to the HahnBanach Theorem guarantees that for each x EX, II x II = 1, we have that D(x) is nonempty. For a continuous linear operator T on X we call the set of complex numbers V(T) = {f(Tx): x EX, II x II= I, f ED(x)} the numerical range ofT. Clearly, when X is an inner product space, V(T) = W(T). The result which gave the stamp of success to this work is the generalisation of Theorem 17 .3. I. 17.3.7. Theorem. For a continuous linear operator Ton a complex Banach space

(X, 11·11), cr(T)

~

V(T).

226

Spectral theory

Proof. For A f1' V(T) we write d

=d('A, V(T)) > 0.

For x EX, II x II = I and f E D(x),

II (T-'AI)x II~ I f((T-'AI)x) I~ d

II (T-'AI)x II ~ d II x II

so

for all x E X

which implies that T -'AI is a topological isomorphism. Now (T-'AI)(X) is a closed linear subspace X. Suppose that (T-'AI)(X) is a proper subspace. Then by Corollary 8.7 there exists an x0 f1' (T-'AI)(X), II x0 II= 1 and an f0 E D(x 0) such that d I f0(x) I~ - - for all x E (T-'AI)(X) and II x II~ I. 211T-'AIII d But then 2 ~ I f0( (T-'AI)x 0 ) I = I f0(Tx 0) - 'A I ~ d('A, V (T)) ~ d, a contradiction. So we conclude that T -AI is onto and A f1' cr(T).

D

17.4 The spectrum of a conjugate operator In Theorem 12.11 we studied the relation between the regularity of a continuous linear operator on a normed linear space and its conjugate on the dual space. This result has the following implications for the spectrum. 17 .4.1 Theorem. For a continuous linear operator Ton a complex normed linear space (X, 11·11) with conjugate operator T' on (X*, 11·11), (i)

cr(T' ) \;;; cr(T) and

(ii)

if (X, 11·11) is complete then cr(T') = cr(T). We are particularly interested in the relation on Hilbert space where for A scalar, T-'AI

is regular if and only if T* -'AI is regular. 17.4.2 Corollary. For a continuous linear operator Ton a complex Hilbert space H with

adjoint T* on H,

cr(T*) = {I: 'A E cr(T)}.

The point spectrum of a normal operator on Hilbert space has particularly satisfying properties which suggest a decomposition of the space. 17.4.3 Theorem. Consider a normal operator Ton a complex Hilbert space H.

-

-

if and only if 'A is an eigenvalue ofT*, and A and A have the

(i)

A is an eigenvalue ofT

(ii)

If A and 11 are distinct eigenvalues forT then the corresponding eigenspaces Mt.. and

same eigenspace: M11 are orthogonal.

Proof. (i)

Since T -'AI is also normal, by Theorem 13.11.3 we have that II (T-'AI)x II = II (T*-AI)x II

and our result follows.

for all x E H,

§17. The spectrum of a continuous linear operator

(ii)

227

For x E MA. andy EMil we have A.(y, x) = (y, A.x) = (y, T*x)

by (i)

= (Ty, x) = (!J.y, x) = !J.(y, x). Since A."* 1J. then (y, x) = 0.

D

17.5 EXERCISES

1.

(i)

Given a continuous linear operator Ton a complex Banach space with conjugate operator T', prove that (a)

(ii)

R cr(T) k: Pcr(T' ) ,

u Rcr(T'),

(b)

Pcr(T) k: Pcr(T')

(c)

Ccr(T') k: Ccr(T).

Given a continuous linear operator T on a complex Hilbert space with adjoint T*, prove that

2.

(i)

(a)

A. E Rcr(T) if and only if A. E Pcr(T*),

(b)

if A E Pcr(T) then A E Pcr(T*)

(c)

A. E Ccr(T*) if and only if A. E Ccr(T).

u Rcr(T*),

Consider the right shift operator SIR on (1 2 , 11·11 2) defined for X=

{A. 1, A. 2, ... , An, ... } by SIR (x) = { 0, A.1, A.2, ... , A.n, ... } .

Prove that

(ii)

(a)

Pcr(SIR) = 0,

(b)

Rcr(SIR)={A.EU::IA.I
(c)

Ccr(SIR)={A.EU::IA.I=l}.

The left shift operator Sn.. on (1 2, 11·11 2) is defined for X=

{A. 1, A2, ... , An, ... } by S n.. (X) = { A2, A3, ... , An, . . . } .

Prove that (a)

3.

* = Sn.., SIR

(b)

Pcr(Sn..) = Rcr(SIR),

(c)

Rcr(Sn..) = Pcr(SIR),

(d)

Ccr(Sn..) = Ccr(SIR)·

Consider the diagonal operator Ton (1 2 , 11·11 2) defined by the bounded sequence of scalars { a 1, a2> ... , an, ... } where for X=

{A. 1 , A. 2, ... , An, ... } E 1 2 , Tx = { a 1 A. 1, a 2 A. 2 ,

(i)

... ,

Prove that (a)

Pcr(T) = {an : n EN},

(b)

cr(T) = {an : n E N } .

anA.n, ... } .

228

Spectral theory

(ii)

Show that every nonempty compact subset of the complex plane is the spectrum of some diagonal operator; (See Exercise 4.12.2.)

4.

Consider the complex Banach space ('C[O,l], 11·11 00 ). (i)

For the integral operator I defined by X

Jf(t) dt

I(f)(x) =

0

prove that

(ii)

(a)

(b)

(OJ =Rcr(l).

For the integral operator I on the subspace 'C 0 [0, I] determine cr(T).

(iii) For the multiplication operator M defined by M(f)(x) = x f(x) determine Pcr(M), Rcr(M) and Ccr(M). 5.

Given a continuous linear operator Ton a complex Banach space (X, II· II), 'A E
£

> 0 there exists an

x EX, II x II = I such that II (T-'AI)(x) II < £. The approximate spectrum ofT, denoted APcr(T) is the set of all approximate eigenvalues ofT. (i) Prove that APcr(T) is closed. (ii)

Prove that 'A E APcr(T) if and only if 'AI-T does not have a continuous inverse on ('AI-T)(X).

(iii) Prove that Pcr(T) u Ccr(T) <;;; APcr(T) <;;; cr(T). (iv) ForT' the conjugate operator, prove that (a) APcr(T) = {'A E
7.

APcr(T') = {'A E
Consider a normal operator T on a complex Hilbert space H. Prove that (i)

Rcr(T) = 0,

(ii)

if H is separable then Pcr(T) is countable.

Consider a continuous linear operator Ton a complex Banach space (X, 11·11) and a closed linear subspace M of X. Prove that (i) Pcr(T) :::2. Pcr(TIM), (ii)

Rcr(T) <;;; Rcr(TIM),

(iii) Ccr(TIM) <;;; Ccr(T) u Pcr(T), (iv) p(T) <;;; p(TIM) u Rcr(TIM).

229

§ 17. The spectrum of a continuous linear operator

8.

Consider a continuous linear operator T on a complex Banach space X. (i)

The operator T is said to be dissipative if sup Re V(T) II (1-rT)x II

such an operator (ii)

~

~

0. Prove that for

II x II for all x EX and r ~ 0.

The operator Tis said to be hennitian if V(T)

~

IR. Prove that T is

hermitian if and only if both iT and- iT are dissipative. (iii) Prove that for an hermitian operator T, II (I± irT)x II 9.

~

II x II

for all x EX and r> 0.

Consider a continuous linear operator Ton a complex Banach space X. (i) Prove that if A E V(T) and I A I= II T II and X is rotund then A is an eigenvalue ofT. (ii)

Given that T is a dissipative operator, prove that II T 2x II ~ r (II rx+Tx II- r II x II) for all x EX and r ~ 0.

a

(iii) Given that Tis a continuous linear operator, prove that if A E co V(T) and (AI-T) 2x

= 0 for some x oF- 0 then (AI-T)x = 0.

Deduce that if T is an hermitian operator and AE cr(T) and (AI-T) 2 x =0 for some x oF- 0 then (AI-T)x

= 0.

230

Spectral theory

§18. THE SPECTRUM OF A COMPACT OPERATOR

Compact operators have properties which give their spectrum a very simple structure. The key to revealing this structure is the following property: Given a compact operator T on a normed linear space, the linear operator I - T is one-to-one if and only if I-T is onto. We develop the proof of this property through stages. 18.1 Theorem. Given a compact operator Ton a normed linear space (X, 11·11),

if

I-T is one-to-one then I-T has a continuous inverse on (1-T)(X). Proof. Suppose that I - T does not have a continuous inverse. Then for each n E N there exists x0 E X such that II (I-T)x 0 II< l11 x0 II. n

Now { ~~~~II} is a bounded sequence and since T is compact there is a subsequence { 11:::11 } Xnk such that {T( 11 xnk11 )} is convergent to some y E Y. But Xnk I II (1-T) (- 11 ) II<11 Xnk nk Xnk so { llxnkll} is also convergent toy. Then Ty = y. However, II y II= I and then I-T is not one-to-one.

D

18.2 Lemma. Given a compact operator Ton a normed linear space (X, 11·11),

if I-T is

one-to-one on X then (1-T)(X) is a closed linear subspace of(X, 11·11). Proof. Consider a cluster pointy of (1-T)(X). Then there exists a sequence { Yn} in (I-T)(X) \ {y} such that {y n} is convergent to y. Write yn = (I-T)(x 0 ). Then from Theorem 18.1 there exists an m > 0 such that mllxii<;;II(I-T)xll and

forallxEX

mllx 0 11<;;11(1-T)x0 ll=lly 0 11

forallnEN.

Since { Yn} is convergent, { Yn} is bounded and so { x is bounded. Since Tis compact, {x 0 } has a subsequence {x 0 k} such that {Tx 0 k} is convergent. As x0 = Yn + Tx 0 , then 0 }

{x 0 k} is also convergent to say, x. Then Ynk = (I-T)(x 0 k) is convergent toy= (1-T)(x); this tells us that y E (1-T)(X) and (I-T)(X) is closed.

D

18.3 Remark. If we had assumed that the normed linear space (X, II· II) is complete then

it would have followed directly from Theorem 18.1 and Theorem 1.24.9 that (1-T)(X) is closed in (X, 11·11). D 18.4 Lemma. Given a compact operator Ton a normed linear space (X, ll·ll),for each

n E N, (I-T)" = I- S 0 where S 0 is a compact operator on X.

§ 18. The spectrum of a compact operator

231

By Theorem 15.16, Sn is compact.

D

We are now in a position to prove the first part of our key property.

18.5 Theorem. Given a compact operator Ton a normed linear space (X, 11·11), if I-T is one-to-one then !-T is onto.

Proof. We write, ltn

=(1-T)n (X) for each n EN.

Then clearly

X :;;21\1 :;;21\z:;;;! ... :;;21\n:;;;! .... Suppose that ltn

ltn+l for each n EN. By

"#

Lemmas 18.2 and 18.4, ltn+I is closed in ltn, so by Riesz Lemma 2.1.8 for each n EN there exists an Xn E ltn, II Xn II= I such that d(xn, ltn+l) ~ ~. Now for n > m we have II Txm-Txn II = II Xm- (I-T)xm-Xn + (I-T)xn II

~

2I

since (1-T)xm + Xn- (1-T)xn E ltm+I· But then although { Xn} is bounded, {Txn} cannot have a convergent subsequence and this contradicts T being compact. So there exists some J.1. EN such that 1\~ = 1\~+I and then ltn+I = ltn for all n ~ J.l.. Since I - T is one-to-one it follows that (I-T)n is one-to-one for all n EN. Given x EX, (1-T)~x = (1-T) 2 ~y for some y EX so (1-T)~(x- (1-T)~y) = 0. Since (I-T)~ is one-to-one, x = (1-T)~y. That is, 1\~ =X. But this implies that (1-T)(X) = X.

D

Theorems 18.1 and 18.5 have the following implications for the spectrum of a compact operator.

18.6 Theorem. Consider a compact operator T on a complex normed linear space (X, 11·11). (i)

0 E cr(T). ~(X)

(ii) For A"# 0, T- AI is regular in

if and only ifT- AI is one-to-one.

Proof. (i) follows from Corollary 15.17. (ii) follows from T- AI=- A(l- Tf").) and TJA is a compact operator.

The following structural property is immediate.

18.7 Corollary. For a compact operator Ton a complex normed linear space (X, 11·11), cr(T)

= {0} u

Pcr(T).

D

232

Spectral theory

A description of this spectral structure can now be given. 18.8 Theorem. Given a compact operator Ton a complex normed linear space (X, 11·11), Pcr(T) is countable and has 0 as its only possible cluster point.

Proof. We prove that for any given£> 0, the set P E = { A. is finite. Then Pcr(T) \ {0}

E

Pcr(T) : I A. I 2: £}

=U

P 110 and so Pcr(T) is countable. n=l Suppose on the contrary, that there exists an r > 0 such that Pr is infinite. Then there exists a sequence { "-n} in Pr where "-n

* "-ro for all n * m.

Consider a sequence { x0 } of eigenvectors where for each n E N, Tx 0 = "-nxnFrom Theorem 17 .1.2, { x0 : n E N } is a linearly independent set. Consider M 0 = { x 1, X2> . . . , X0 }. Then for each n EN, since M 0 is finite dimensional it is closed and from the linear independence of the set { x 0 : n EN}, M 0_1 ~ M"' From Riesz Lemma 2.1. 8 we have for each n

k.

that d(y 0, M 0_1) 2:

E

N, there exists a y n E M 0, II y n II = I such

Now form< n, II Ty 0 - Tym II= II "-nYn- C"-nYn- Ty 0 + Tym) II. n

But any x E M 0 has the form x =

I

akxk.

k=l n

So (T-A-0 1)(x)

=I

ak(A.k-A.n)xk and (T-A-01)(M 0 )s;;; M 0 _ 1•

k=l

Therefore, (A-0 1-T)(y 0 ) + Ty m E Mn-1 II Ty 0

and so

-

1"-ol r Tym 112:2 2:2.

But then, although { yn} is a bounded sequence, {Ty n} cannot have a convergent subsequence and this contradicts T being compact. So we conclude that cr(T) is countable.

D

18.9 Remark. Quite independently of our general theory we have shown that the spectrum of a compact operator on a complex normed linear space is nonempty and compact. D We examine the relation between the regularity of a compact operator and its conjugate.

We begin with a result similar to Theorem 12.11 but without using

completeness. 18.10 Theorem. Consider a compact operator T on a normed linear space (X, 11·11). Then

I-T is a topological isomorphism on X if and only if I-T' is a topological isomorphism on X*.

233

§ 18. The spectrum of a compact operator

Proof. We need only show the result comparable to Theorem 12.11 (ii). If I-T' is a topological isomorphism on X* then by Theorem 12.8 its conjugate I-T" is one-to-one on X** so I-T is one-to-one on X. By Theorems 18.1 and 18.5, I-T is a 0 topological isomorphism on X. As in Theorem 17.4.1 we can relate the spectra of a compact operator and its conjugate. The result now holds without completeness. 18.11 Corollary. For a compact operator Ton a complex nonned linear space (X, 11·11)

with conjugate operatorT',

cr(T) = cr(T' ).

The converse of Theorem 18.5 is established using conjugate operators. 18.12 Theorem. Given a compact operator Ton a normed linear space (X, 11·11),

if

I - T is onto then I-T is one-to-one.

Proof. Since I-T is onto, by Theorem 12.8, its conjugate I-T' is one-to-one on X*. Since Tis compact, by Schauder's Theorem 15.18, its conjugate T' is also compact. It follows from Theorems 18.1 and 18.5 that I-T' is a topological isomorphism on X* and so our result follows from Theorem 18.10. 0 We now determine the special properties of the eigenspaces corresponding to the eigenvalues of a compact operator. 18.13 Theorem. For a compact operator T on a normed linear space (X, 11·11), ker(l-T) is finite dimensional.

Proof. Write B

={x

E

ker(l-T) : II x II ~ I} which is the closed unit ball in

(ker(I-T), IHiker(I-T))· Since ker(I-T) = {x EX: Tx = x}, then T(B) =B. Since Tis compact and ker(I-T) is closed in (X, 11·11) then T(B) is closed in (X, 11·11) and so is compact. Then B is compact and so by Riesz Theorem 2.1.9, ker(I-T) is finite dimensional.

0

18.14 Corollary. Given a compact operator Ton a complex normed linear space (X, 11·11),/or an eigenvalue A# 0, the eigenspace corresponding to A is finite dimensional.

Proof. The eigenspace { x EX : (AI-T)(x) = 0} = ker (I- Tf'A.).

0

18.15 The invariant subspace property of compact operators It was mentioned in Remarks 14.27 that for certain classes of continuous linear

operators on a Banach space there is a positive solution to the Invariant Subspace Problem.

234

Spectral theory

In fact N. Aronszajn and K. Smith, Ann. Math. 60 (1954), 345-350, proved that every compact operator on an infinite dimensional complex Banach space has a nontrivial closed invariant subspace. Their result follows with a simpler proof from a more general result proved by V.I. Lomonosov, Funk. Anal. i Prilozen 7 (1973), 55-56. We present a proof based on a modification of Lomonosov's argument. 18.15.1 Lomonosov's Theorem.

Every nonzero compact operator Ton a complex Banach space (X, 11·11) has a nontrivial closed invariant subspace. Proof. We writer= { S E ~(X) : ST = TS} which is a closed subalgebra of ~(X). For

each y EX, y "# 0 we write r(y) = {Sy; S Er}. Since r is closed under composition, S(r(y)) ~ r(y) for all SE r so if r(y) "#X for some y EX then r(y) is a suitable nontrivial closed invariant subspace forT. Suppose that r(y) =X for ally EX. Choose x0 EX such that Tx 0 "# 0. Then x 0 "# 0 and by the continuity ofT there exists an open ball B centred at x0 such that

II xll

~

2I II x0 II

and II Tx

II~

2I II Tx 0 II

for all x E B.

Given y E T(B), since r(y) =X then x0 E r(y). Since B is an open neighbourhood of x0 there exists an open neighbourhood W of y and someS E r such that S(W)

~B.

Since Tis a compact operator, T(B) is compact. Now 0

e T(B).

W 1, W 2 ,

..• ,

But since T(B) is compact there exists a finite open cover W 0 of T(B) such that Sk(Wk)

~

B for some Sk Er and k E {I, 2, ... , n}.

Now Txo ET(B) so Txo lies in some wk] and XI= skjTxo EB. Then Tx 1 = TSk 1Tx 0 ET(B) so TSk 1Tx 0 lies in some Wk 2 and x 2 = Sk 2TSk 1Tx 0 E B. Continuing, we obtain a sequence { x 0 } where Xn = sknT ... skjTxo E B. Putting k =max {II Sk 1 II, II Sk 2 II, ... , II Skn II} > 0 we have

k· II x

0

II :S; II x 0 II :S; k" II T" 1111 x0 II

This gives us v(T) = lim II T" 11 11" =

for all n EN.

k > 0.

n---7~

Since T is a compact operator it has an eigenvalue A"# 0. But then ker (AI-T) is a nontrivial closed invariant subspace of X under T.

The proof actually establishes more than the theorem statement. 18.15.2 Corollary. Given a nonzero compact operator Ton a complex Banach space (X, 11·11) the nontrivial closed invariant subspace under Tis also invariant under every

continuous linear operatorS which commutes with T.

235

§ 18. The spectrum of a compact operator

Proof. If r(y) ot- X for some y EX, y ot- 0, then r(y) is invariant under S. If r(y) = X for all y EX, y ot- 0 then T has an eigenvalue A. ot- 0. So, if x E ker(A.I-T), T(Sx) = S(Tx) = S('A.x) = A.S(x) then Sx E ker(A.I-T) which implies that ker(A.I-T) is invariant under S.

0

18.16 Remarks. (i)

A quasinilpotent operator T on a complex Banach space X is a continuous linear

operator whose spectrum cr(T) = {0}. C.J. Read J. London Math. Soc. (2), 56 (1997), 595-606, has given an example of a quasinilpotent operator which does not have an invariant subspace. (ii)

The theory of compact operators on a normed linear space is structurally very rich.

We have only given sufficient of the theory to develop the spectral properties we need for the Spectral Theorem in Section 19. The possible further development of the theory can be glimpsed in the proof of Theorem 18.5. The illuminating Riesz-Schauder Theory of compact operators encompasses the study of the ranges and kernels of iterations of the mapping I-T where Tis a compact operator. We leave some of these developments to be explored in Exercise 18.17 .6. A fuller account of this theory can be found in A.L. Brown and A. Page, Elements of Functional Analysis, van Nostrand Reinhold, 1970, pp.248-

0

255. 18.17 EXERCISES I.

Prove that for a continuous finite rank operator Ton a Banach space (X, 11·11), the spectrum cr(T) is a finite set consisting only of eigenvalues.

2.

Consider a compact operator Ton a Banach space (X, 11·11). (i)

(ii)

Prove that, for any A. ot- 0, (T-A.I)(X) is a closed linear subspace of (X, 11·11). Prove that for any regular operatorS on (X, 11·11), (S-T)(X) is closed in (X, 11·11) and has finite codimension.

3.

The right shift operatorS on (lz, 11·11 2) is defined by S({A.I, /...2, ... , 'An, ... })= { 0, /...1, /...2, ... , 'An, ... }. (i)

(ii)

Prove that (a)

'AI- Sis one-to-one for all A. E a:,

(b)

cr(S) ={A. E
The operator Ton (1 2, 11·11 2) is defined by T({A.t, /...2, ···'An,···})= Prove that (a)

TS is a compact operator,

(b)

Pcr(TS) = 0 .

c·;'

~2 , · · · 'n~I

'· · ·} ·

236

4.

Spectral theory

The operator Ton (1 2 , 11·11 2) is defined by T{A, 1, Az, A. 3 ,

... ,

An, ... } = {0, A. 1, ~2

,

"i, ... ,~n, ... }.

Prove that

5.

(i)

Tis compact,

(ii)

cr(T) = Rcr(T) = {0}.

Consider a sequence {T n} of compact operators on a Banach space (X, 11·11) and an operator Ton (X, 11·11) such that Tn ~ T as n ~ if AnE cr(T n) for all n EN and An ~A as n ~

6.

oo

oo

in (~(X), 11·11). Prove that

then A E cr(T).

Consider a compact operator Ton a normed linear space (X, 11·11). In Theorem 18.5 it was actually proved, writing ~n = (1-T)n(X) for each n EN, that there exists a J.L E N such that ~n::::) ~n+l #

for all n E {I, 2, ... , J.L-I} and

~n = ~n+l for all n ~

(i)

Writing 'Jtn

J.L.

= ker(I-T)n for each n EN, prove that there exists a

v E N such that 'Jl,n

C #

'Jtn+I for all n E {I, 2, ... , v-1} and

'Jl,n = 'Jtn+I for all n ~ v. (ii)

Prove that (b)

X= ~u EB 'Jt 11 , ~ 11 and ')1, 11 are invariant subspaces ofT, and deduce that

(c)

v = J.L.

(a)

(iii) Prove that, as for a linear operator on a finite dimensional linear space, codim (1-T)(X) =dim ker(I-T). 7.

Consider an incomplete normed linear space (X, 11·11) and a compact linear operator Ton (X, 11·11) with a unique continuous linear extension Ton (X, 11·11). Writing

~n

= (1-T)n(X) and

-Din = (I-T)nc)() for each n E N prove that if J.L is the least natural number such that ~n = ~n+l for all n ~ Jl

then it is the least natural number such that ~n = ~n+l for all n ~

J.L.

§ 19.

§19.

The spectral theorem for compact normal operators

237

THE SPECTRAL THEOREM FOR COMPACT NORMAL OPERATORS ON HILBERT SPACE.

We have seen in Theorem 17.4.3 that for a normal operator on a complex Hilbert space, the eigenspaces associated with distinct eigenvalues are orthogonal. It is this property which enables us to construct a spectral theorem for such operators on a finite dimensional inner product space. 19.1 The Finite Dimensional Spectral Theorem. For a normal operator Ton a complex finite dimensional inner product space Hn there

exists an orthonormal basis for Hn consisting of eigenvectors ofT; that is, T can be represented as a diagonal matrix with respect to some orthonormal basis for Hn. We have seen in Section 18 that compact operators have a simple spectral structure. We show in this Section that we can establish a spectral theorem for compact normal operators on Hilbert space which includes the finite dimensional case. Consider a compact normal operator T on a complex Hilbert space H. From Theorem 18.8 the set of eigenvalues of T is countable. Consider the eigenvalues in sequence P-n} where I An I ~ I An+ 1 I for all n E N. By Corollary 18.14, for each n E N, Nn

=ker(An I-T) is

finite dimensional and is the eigenspace of An· Since we are in an

inner product space, for each n EN we can choose an orthonormal basis for Nn, say {un 1, Unz, ... , Unkn}. Since Tis normal, by Theorem 17.4.3 we have for n 7' m that Nn is orthogonal to Nm. So the sequence of eigenvectors { ull, u,z, ... , ulkl' uz,, Uzz, ... 'Uzkz' ... , un,, Unz, ... , Unkn, ... is an orthonormal set in X. We denote by F the closed linear span of this set. We now relate the range and kernel ofT to the subspace F and its orthogonal complement F.i. 19.2 Theorem. For a compact normal operator Ton a complex Hilbert space H, T(F) ~ F and F.l = ker T.

Proof. Since F is the closed linear span of eigenvectors it is clear that T(F)

~F.

Since T is normal, it follows from Theorem 17 .4.3 that the eigenvectors of T are eigenvectors ofT* so T*(F) ~F. Therefore, for X E F andy E F.i we have (x, Ty) = (T*x, y) = 0 so T( F.i) ~ F.i. Similarly, (x, T*y) = (Tx, y) = 0 so T*( F.i) ~ F.i. So F reduces both T and T* and TIF.l is a normal operator on Hilbert space F.i. Suppose that TIF.l

7' 0.

Then by Corollary 17.3.4 there exists a A E cr(Tip.l) such that

I A I= II TIF.lll. But TIF.l is also compact on F.i so by Theorem 18.8, A is an eigenvalue

238

Spectral theory

of Tip_!_• but then A is an eigenvalue ofT. So A= An for some n EN. But Nn s;;; F and this contradicts F

n F_j_ = {0}. Therefore Tip_!_= 0 which implies that F_j_ s;;; ker T.

*

*

If F_j_ ker T then by Theorem 2.2.19 and Corollary 14.8 there exists a z 0 and z E ker T and z E F. But this contradicts T being one-to-one on F. So we conclude that

~=~~

0

We are now in a position to present the spectral theorem for compact normal operators on Hilbert space. 19.3 The Spectral Theorem. For the compact normal operator Ton the complex Hilbert space H,for each x E H kn

oo

Tx =

I, I,

An(x, Unk)unk·

n=l k=k 1

Proof. For x

E

H we can write x = y + z where y E F and z E F_j_. Then from Theorem

19 .I we have that Tx = Ty. Since F is the closed linear span of the orthonormal set { ull• ul2• · · ·, Utk!' Uzl• Uzz, · · ·, Uzkz• · · ·, Unl• Un2• · · ·, Unkn• · · · then by Theorem 3.10 we have that kn

I,

Tx =

I,

(Tx, Unk)unk

n=l k=k 1 kn

I,

I,

(x, T* Unk)unk

n=l k=k 1 kn

I,

I,

0

An(x, Unk)unk by Theorem 17.4.3(i).

n=l k=k 1 From this theorem we can derive an extension of Corollary 17.3.4. 19.4 Corollary. Given a compact normal operatorT on a complex Hilbert space H, foreachn EN I An I= v(Tip~_ 1 )

where Fn = sp {ull• ul2• ... , UJkj• Uzl• Un, ... , uzkz• ... , Unl• Un2• ... , Unkn}.

Proof. For x

kp

_]_

E

I, I,

F n-l , Tx

Ap(x, upk) upk·

p=n k=k 1 kp So

I,

(Tx, x) =

I,

AP I (x, upk) 12.

p=n k=k 1 Since { ull, ul2• · · · • ulk1• Uzl• Uzz, · · · • Uzkz• · · · • unl• Un2• · · · • Unkn• · · · orthonormal set, from Bessel's inequality 3.13 k

I,

~

n=l k=k 1

Ap I (x, unk) 12 :"':: II x 11 2.

is an

§ 19. The spectral theorem for compact normal operators

239

Also since I "-m I $: I An I for all m ;:: n, kp

I (Tx, x) I $:

L L

I Ap II (x, upk) 12

p=n k=k 1 kp $: I An I

L L

I (x, upk) 12

p=n k=k 1 $; I

An I

II X 11 2•

But clearly, (Tunk• Unk) =An for all k E {k,, k2, ... , knl and the result follows from Corollary 17.3.4.

D

We now show how the Spectral Theorem 19.3 can be used to give an expression for inverses related to a compact normal operator on a Hilbert space. 19.5 Theorem. Given the compact normal operator Ton the complex Hilbert space H, for any A !I' o(T) and all x E H (AI-T)- 1x = l x + l

A

A

Proof. For x E H write y = (AI-T)- 1x. Then from the Spectral Theorem 19.3, kn X= (AI-T)y = Ay- L L An(y, U0 k)Unk· n=l k=k 1 kn A I oo y--x= L L _n (y, Unk)unk· So A n=l k=k 1 A But formE Nand 1 E {k,, k2, ... , km}, (y,um,t)-i (x,um,t)=(y-ix,um,£) kn

=

A

L L ,n

n=l k=k 1

(y, unk)(unk• urn.£)

11.

Am =T(y, Um,t) and so

I (y, Um,t) = - - (x, Um,t). A-Am

D

Consequently,

19.6 Example. The solution of Fredholm integral equations. In Section 15 we considered the Fredholm operator K defined on t: [a, b] by b

(Kf)(x)

= Jk(x,t)

f(t) dt

a

with kernel k a continuous complex function on a square region S

= { (x,t) : a$: x $: b, a$: t $: b}.

We showed that K is a compact operator on (t:[a,b], 11-11 00 ) .

240

Spectral theory

It is not difficult to show that K is a compact operator on the incomplete inner product space ('C [a,b], 11·11 2 ), (see Exercise 15.23.9(ii)(a)). Further the unique continuous linear extension K on the Hilbert space completion (:C 2 [a,b], 11·11 2 ) is also a compact operator, (see Exercise 15.23.9(ii)(b)). Consider x0

E

[a, b] and f

E

:C 2 [a,b].

b

fI k(x,t) - k(x ,t) II f(t) I dt

~

I (Kf)(x)- (Kf)(x 0) I

Now

0

a

b

(fI k(x,t)- k(x ,t) edt)

~

0

112

11 f 11 2 .

a

Since k is continuous then Kf E 'C [a,b]. Therefore, if A is an eigenvalue of K with eigenvector f so that Kf =Af, then we can conclude that f E 'C[a,b].

-*

-

Now K the adjoint of K on :C 2 [a, b] is defined by b

(K*f)(x)

=( Jk(t,x) f(t) dt a

b

since

b

b

J ( Jk(x,t) f(t) a

a

b

J( J

dt) g(x) dx =

a

.,-k.,--(t-,x.,-)-g.,-(x-,-) dx) f(t) dt for all f, g

If the kernel k has the property that I k(x,t) I =I k(t,x) I for all (x, t) b

then II Kf II = (

fI k(x,t) 1 I f(t) edt) 2

a

E

'C[a,b].

a

E

S

b

112 =

(

f I k(t,x) 1 I f(t) 1 dt) 2

2

112

a

and so we would conclude from Theorem 13.11.3 that K is a normal operator. Applying the Spectral Theorem 19.3 to the compact normal operator K on ( :C 2 [a,b], 11·11 2) we have the following decompositon for the generating operator K on ('C[a,b], 11·11 2), kn K(f) = L L A0 (f, fnk)fnk n=l k=k 1 where {f01 , ... , fnkn} is an orthonormal basis for the eigenspace associated with the eigenvalue A0 for each n E F::l. But further we can apply Theorem 19.5 to solve the Fredholm integral equation b

g(x) = Af(x)-

Jk(x,t) f(t) dt a

where g E 'C[a, b] and I k(x,t) I= I k(t,x) I for all (x,t)

E

S. This equation has the form

g = (AI-K)(g) and when A e cr(K) = cr(K), (see Exercise 4.12.5) then it has a unique solution f E :C 2 [a,b] where I 1 1 oo kn A b -f = (AI-K)- g =- g +- L L _ n ( Jg(t) f 0 k(t) dt) fnk· A A n=I k=k 1 A-A 0 a

0

§ 19. The spectral theorem for compact normal operators

241

19.7 EXERCISES

I.

Consider a Hilbert space H with orthonormal sequence {e 1, e 2, ... , en, ... } A.z, ... , A.n, ... } where

and a sequence of nonzero complex numbers { A. 1, I A. 1 I ~ I 1..2 I ~ ... ~ I A.n I ~ ... and An ~ 0 as n ~ =. (i)

Prove that for any x e H, ll..n (x, en)en defines an element of H.

(ii)

Prove that the operator T on H defined by Tx = ll..n(x, en)en is linear and continuous and that II T II = I A. 1 I.

(iii) Prove that T is a compact operator on H. (Hint: Use Theorem 15.19.) (iv) Prove that the adjoint T* has the form T*x (v)

=I~n (x, en)en

and deduce that T is normal. For A.* An· 0 for all n eN, prove that A.I-T is invertible and derive an expression for (A.l-T)- 1x in terms of the orthonormal sequence { e 1, e 2 ,

.•• ,

en, ... } and the sequence of complex numbers

{ A.,, A.z, . . . , An, ... } . 2.

Consider a compact normal operator Ton a Hilbert space H. For each eigenvalue A. denote by Pt.. the orthogonal projection onto ker(T-A.I). Prove that the projection operators are orthogonal, that is, Pt..i Pt..k =0 for j, k e N, j

* k,

00

and

T=

L

A.nPA.n

n=l

where { 1.. 1, 1..2 , 3.

••• ,

An, ... } are the eigenvalues of T.

Prove that for every positive compact operator T on a complex Hilbert space H there exists a unique positive operatorS on H such that S2 = T and S is also compact.

4.

Given a compact normal operator Ton a complex Hilbert space H, prove that (i)

W(T) =co Pcr(T),

(ii)

T is self-adjoint if all its eigenvalues are real,

(iii) T is positive if all its eigenvalues are non-negative. 5.

Consider a compact normal operator T on a complex Hilbert space H. Prove that if T has only a finite number of eigenvalues and is one-to-one then H is finite dimensional.

242

6.

Spectral theory

Consider the real function k defined on the square region S

={(x, t): 0 ~ x ~ 1, 0 ~ t ~ 1} by

k(x,t) = (1-x)t

0

~

x

= (1-t)x

0

~

t

~ ~

t } x

and the operator K on (t:[a,b], 11·11 2) defined by I

(Kf)(x) =

Jk(x,t) f(t) dt. 0

(i)

Prove that K is a compact operator and that its extension K on ( l:; 2 [a,b], 11·11 2 ) is self-adjoint.

(ii)

Show that the eigenvalues ofT are

(n~) 2

with corresponding eigenvectors

sin n1t for n EN. (iii) Given a continuous function g on [0, 1] solve the Fredholm integral equation I

g(x) = f(x)-

Jk(x,t) f(t) dt. 0

243

§20. The spectral theorem for compact operators

§20.

THE SPECTRAL THEOREM FOR COMPACT OPERATORS ON HILBERT SPACE

To extend the Spectral Theorem to all compact operators we need to develop other technical properties of self-adjoint and positive operators on Hilbert space. We study them via their mapping into a more amenable space. 20.1

Functional Calculus

Given a self-adjoint operator Ton a complex Hilbert space H we have from Theorem 16.2 and Corollary 17 .3.2 that the spectrum cr(T) is a compact subset of the real numbers. The set 'C(cr(T)) of complex valued continuous functions on cr(T) is a Banach space with respect to the supremum norm II f lloo =sup{ I f(A) I : A E cr(T)}. But also 'C(cr(T)) with pointwise definition of multiplication and pointwise definition of involution by taking conjugates, is a unital commutative B* algebra. On cr(T), a polynomial p has the form

p(A) = ao +alA.+ ... + anA." where al, alo ... 'an

E ([

we can define a polynomial operator pin '0'3(H) by p(T) = a0I + a 1T + ... +anT". We denote by 'fl( cr(T)) the linear subspace of polynomials on cr(T). The formal association of polynomials in 'C(cr(T)) with polynomial operators in '0'3(H) has considerable structure. Theorem. Given a self-adjoint operator Ton a complex Hilbert space H, the mapping p(A) 1--t p(T)from 'fl(cr(T)) into '0'3(H) is an algebra* homomorphism; that is, (p 1 + p2)(T) = p 1(T) +PiT) 20.1.1

p(A.T) = A.p(T) p 1piT) = p 1(T) o p 2(T)

Further,

p(T) = p(T)*. II p 11 00 =II p(T) II for all p E 'fl(cr(T))

so the mapping is an isometric isomorphism. Proof. The * property is the only algebraic property that needs checking. Since Tis self-adjoint, A E cr(T) is real, so we have for p(A) = aa +alA+ ... + ~A.", that p(A.) = a 0 + a 1A. + ... +anA.". Now

p(T)* = (aoi + a 1T + ... +anT")*

which by Theorem 13.7 gives p(T)* = a 0I + a 1T + ... +anT" and

p(T)* = p(T).

Clearly p(T) is a normal operator, so by Corollary 17 .3.4, II p(T) II = v(p(T)) =sup{ I A. I: A. E cr(p(T))} = sup {I p(A) I : A. E cr(T)} by the Spectral Mapping Theorem I 6.9

= II p lloo•

0

244

Spectral theory

By the Stone-Weierstrass Theorem (AMS §9), ~(cr(T)) is dense in (t:(cr(T)), 11·11 00 ) . By Theorem 4.8 there exists a unique continuous linear extension from t:(cr(T)) into

~(H)

which is an isometric isomorphism. Further, the continuity of the algebraic and * operations on t:(cr(T)) guarantee that the isometric isomorphism is also an algebra* isomorphism. 20.1.2 Definition. Given a self-adjoint operator T on a complex Hilbert space H, the mapping f H f(T) of t; ( cr(T)) into ~(H) discussed in Theorem 20.1.1 is called a functional calculus forT. 20.2 Square roots of positive operators We have seen in Section 13.10 that the set of self-adjoint operators in the set of continuous linear operators on a complex Hilbert space has a role analogous to the set of real numbers in the set of complex numbers and the set of positive operators has a role analogous to the set of positive real numbers. Just as there always exists a unique positive square root of a positive real number so we have similar behaviour exhibited for positive operators. 20.2.1 Definition. Given T a positive operator on a complex Hilbert space H, we say that the self-adjoint operator S on H is a square root ofT if S2 = T and when S is a positive operator we call S a positive square root ofT. To show that every positive operator has a unique positive square root we need the following elementary properties from the functional calculus. 20.2.2 Lemma. Consider a self-adjoint operator T on a complex Hilbert space H and f E t:(cr(T)). (i)

/ff(A) is real for all A E cr(T) then f(T) is a self-adjoint operator.

(ii)

/ff(A) is non-negative real for all A E cr(T) then f(T) is a positive operator.

Proof. (i) Given£> 0, by the Stone-Weierstrass Theorem there exists apE ~(cr(T)) such that II f-p II<£.

If f(A) is real for all A E cr(T) we can choose p such that p(A) is real for all A E cr(T). Then p(T) is self-adjoint and

II f(T)-p(T) II < £.

But by Remark 13.10.12(ii), the set of self-adjoint operators in

~(H)

is closed, so f(T) is

self-adjoint. (ii) If f(A) ~ 0 for all A E cr(T) then there exists a real valued function gEt: (cr(T)) where g(A) ~ 0 for all A E cr(T) and g2

= f.

Now by (i), g(T) is self-adjoint. Since f(T) have (f(T)x, x) =II g(T)x 11 2 for all x E H and so f(T) is a positive operator.

= (g(T)) 2 we 0

§ 20. The spectral theorem for compact operators

245

20.2.3 Theorem. Every positive operator T on a complex Hilbert space H has a unique positive square root S.

Proof. By Corollary 17.3.3, cr(T) consists of non-negative real numbers, so the real valued function f on cr(T) defined by f(A.) = 1.. 112 is continuous on cr(T) and (f(A.)) 2 =A. So the continuous linear operatorS = f(T) has the property that S 2 = T. From Lemma 20.2.2(ii) we conclude that S is a positive operator. We now show that Sis unique. Given£> 0 there exists apE 'C(cr(T)) such that II f-p 11 00 < £. Then II S-p(T) II<£. Suppose that V is a positive operator such that vz = T. Then by the Spectral Mapping Theorem 16.8, cr(T) = { J.L2 : J.l E cr(V)}. For s: cr(V) -t cr(T) defined by s(J.l) = J.L 2 we have q E 'C(cr(V)) where q = p o sand since cr(V) consists of non-negative real numbers f o s = id on cr(V). II V - p(T) II = II V - p o s (V) II = II id- q lloo

But

:::; II f- p 1100 II s 1100 <£II X 1100. We conclude that V =Sand that Sis unique.

0

20.2.4 Notation. Given a positive operator T on a complex Hilbert space H we denote by T 112 the unique positive square root ofT.

Theorem 20.2.3 has many interesting consequences. 20.2.5 Corollary. For a positive operator Ton a complex Hilbert space H, ifO E W(T) then

0 E Pcr(T). 1

1

Proof. II T tz x II= (T tz x, T Itz x) = (Tx, x). So if (Tx, x) = 0 for some x E H then T 112 x = 0. Then Tx = T 112T 112 x = 0.

0

20.2.6 Corollary. Given a continuous linear operator Ton a complex Hilbert space H,for

the positive operator T*T, II (T*T) 112 x II = II Tx II for all x E H.

Furthermore, (T*T)l/2 is the only positive operator on H with this property.

Proof. II (T*T) 112 x

11 2

= ((T*T) 112 x, (T*T) 112 x) = (T*Tx, x) =II Tx 11 2 for all x E H.

Suppose that S is a positive operator on H such that II Sx II = II Tx II (S 2x,

11 2

x) = II Sx = II Tx Then So by Lemma 13.10.6(i), S 2 = T*T.

11 2

for all x E H. = (T*Tx, x) for all x E H.

0

246

Spectral theory

For a compact positive operator, using the Spectral Theorem 19.3, we have an explicit form for its positive square root. 20.2.7 Theorem. For a compact positive operator T on a complex Hilbert space H,

T 112 is a compact operator.

Proof. Using Theorem 19.3 and its notation we have that, for each x E H, oo

Tx

kn

=I I

An(x, Unk) Unk· n=l k=k 1 From Corollary 17.3.3 it follows that the eigenvalues { A1, A2, ... , An, ... } are positive real numbers. We define the operatorS on H by

IISxll~~ llxll

Then

and so S is continuous. Again clearly S is positive and S 2 = T. So by Theorem 20.2.3, S = T 112 . For each mE F::l, we define the finite rank operator Sm by m kn _ ,-::Sm(x) = I I ·\/An (x, Unk) Unk· n=l k=k 1 Then

(T112- Sm)(x) =

I

kn

I

_ ,-::\1 An (x, Unk) Unk

n=m+l k=k 1

and

,,2 II T - Sm II~ -yA-m+ I

~

--t 0 as m --too.

So by Theorem 15.19(iii), T 112 is also compact.

0

20.2.8 Corollary. For a compact operator Ton a complex Hilbert space H, the linear

*

operator (T T) 20.3

,,2 zs. also compact.

Partial lsometries In Section 13.10 we saw that any continuous linear operator on a Hilbert space can be

expressed as the sum of self-adjoint operators. There is another class of continuous linear operators on a Hilbert space which is particularly useful for the decomposition of any continuous linear operator.

§20. The spectral theorem for compact operators

247

20.3.1 Definition. A continuous linear operator V on a Hilbert space H is called a partial isometry if there exists a closed subspace M of H such that II Vx II= II x II and

Vx = 0

for all x EM

for all x E M.l.

We call M the initial space of V and N = V(M) the final space of V. From Theorem 14.9,

M EEl M.l = H soN= V(H). Clearly, every isometric isomorphism and every orthogonal projection is a partial isometry. Partial isometries have the following elementary properties. 20.3.2 Theorem. Consider a partial isometry V on a complex Hilbert space H with initial and final spaces M and N. Denoting by P 1 and P2 the orthogonal projections ofH onto M and N,

then (i) V*V =PI (ii) VV* = P2 and (iii) V* is a partial isometry with initial and final spaces N and M.

Proof. (i) Now V = VP 1 so for x EH, (V*Vx, x) = (Vx, Vx) = (VP 1x, VP 1x) =(P 1x,P 1x) sinceiiVxll=llxll forallxEM =(P 1x,x). By Lemma 13.10.6(i) we have that V*V = P 1•

(ii) Again V = P 2V so V* = V*P2 . For x EH, P2 x = Vy for some y EH. Then VV*x = VV*P 2x = VV*Vy = VP,y = Vy = P2x and so VV* = P2 . (iii) For x EN we have II V*x 11 2 = (VV*x, x) = (P 2x, x) =II x 11 2 and for x E N.l we have P2 x = 0 so V*x = V*P 2x = 0. Therefore, V* is a partial isometry with initial space N.

0

For the decomposition we need the following elementary properties. 20.3.3 Lemma. For a continuous linear operator Ton a complex Hilbert space H, (i) ker T*T = ker T and (ii) T*T(H) = T*(H) .

Proof. (i) This follows from II Tx 11 2 = (T*Tx, x) for all x EH, and Corollary 13.10.6. (ii) From Theorem 14.ll(ii), T*T(H) = (ker T*T).l = (ker T).l from (i) = T*(H)

from Theorem 14.ll(ii)

0

248

Spectral theory

20.3.4 The Polar Decomposition Theorem. Consider a continuous linear operator T on a complex Hilbert space Hand write M = T*(H) and N = T(H). There exists a partial isometry V with initial space M andfmal space Nand a positive operatorS on H such that

T = VS and S = V*T and such V and S are uniquely determined.

Proof. Consider S = (T*T) 112 . By Corollary 20.2.6, II Sx II= II Tx II

for all x eH.

By Lemma 20.3.3(ii), S(H) = S 2(H) = T*T(H) = T*(H) . So there exists an isometric isomorphism Sx f-7 Tx mapping from S(H) a dense subspace of M onto T(H) a dense subspace of N. Now from Theorem 4.8 this mapping has a unique isometric isomorphic extension W from M onto N such that WSx =Tx for all x e H. The continuous linear operator V on H defined by Vy = Wy for y eM

=0 has the property that

II Vy II = II y II

so

for y eM.L

II VSx II = II Tx II = II Sx II

for all x e H

for all y eM.

Then V is a partial isometry with initial space M and final space N. Further, Tx =VSx for all x e H. Now by Theorem 20.3.2(i), V*V is the orthogonal projection of H onto M and M = S(H), so we have Sx = V*VSx = V*Tx for all x e H. To prove uniqueness, suppose that there exists a partial isometry V 1 and a positive operator S 1 such that T =V 1S 1 and S 1 =V * 1 T. Then by Theorem 20.3.2(ii), V 1V~ is the orthogonal projection onto N and N have

= T(H), so we

* * 1 = S 1 S 1 = T*V 1V 1T = T*T.

S2

But the positive square root of T*T is unique so S 1 =S. For each x e H, VSx = Tx = V 1Sx. That is, Vy = V 1y for ally e S(H). But as V and V 1 are continuous on M so Vy = V 1y

for ally eM= S(H).

Also from Theorem 20.3.2(i), V *1 V 1 is the orthogonal projection onto M and by Lemma 20.3.3(i) and Theorem 14.14, ker V 1 = ker V~V 1 = M.L so

Vx = V 1x for all x eM.L.

0

§20. The spectral theorem for compact operators

249

20.4 The Spectral Theorem. We now apply the Polar Decomposition Theorem 20.3.4 to extend the Spectral Theorem 19.3 for compact normal operators to compact operators in general. Consider a compact operator Ton a complex Hilbert space H, then by Corollary 20.10,

s := (T*T) 112 is also a compact positive operator. So we have a sequence p.. 1, A.2, ... , An, ... } of positive eigenvalues of S where A.n ~ A.n+l for all n EN and an orthonormal sequence {u 11 , u 12 ,

... ,

ulk]' u 21 , u 22 ,

.•. ,

u2k 2, ... , unl• un 2, ... , Unkn• ...

where for each n EN, { unl, un2• ... , Unkn• ... } is an orthonormal basis for Nn the eigenspace associated with A.n. 20.4.1 Theorem. Given a compact operator Ton a complex Hilbert space H, there exists an

orthonormal sequence {v 11 , v 12 , ... , vlkl' v 21 , v 22 ,

... ,

v 2k2, ... , vnl• vn 2, ... , Vnkn• ... }

such that oo

Tx =

kn

L, L,

A.n(x, Unk)vnk·

n=l k=k 1 oo kn

Proof. By Theorem 19.3, Sx =

L L

A.n(x, unk)unk·

n=l k=k 1 But by the Polar Decomposition Theorem 20.3.4 oo kn Tx =VSx = L L A.n(x, unk)vnk n=l k=k 1 where Vunk = Vnk for all n EN and all k E {k 1, k 2, ... , kn}· Now Vis a partial isometry with initial space S(H). But also by Theorem 20.3.2(i), V*V is an orthogonal projection onto S(H). Now the orthogonal sequence {u 11 , u 12 , . . . , u 1kl' u 21 , u 22 , . . . , u2k 2, ... , unl• un 2, ... , Unkn ... }

=Unk for all n EN and all k E {k 1, k 2, ... , kn}. (vnkm• Vnkp) = (Vunkm• Vunkp) = (V*Vunkm• Unkp) = (unkm• Unkp)

lies in S(H), so V*Vunk Therefore,

so {v 11 , v 12 , ... , v 1k 1, v 21 , v 22 ,

... ,

v2k2, ... , vnl• vn 2, ... , Vnkn• ...

orthogonal sequence and we have our result.

is an

D

20.5 Remark. The Spectral Theorem extends to normal operators on Hilbert space, but to develop the general case would require more background than we allow for here. A gentle account of this development can be found in G. Bachman and L. Narici, Functional Analysis, Academic Press, 1966, or a more sophisticated account can be found in W. Rudin, Functional Analysis, McGraw Hill, 2nd edn. 1991. D

250

Spectral theory

20.5 EXERCISES I.

Consider a self-adjoint operator Ton a complex Hilbert space H. (i) Prove that if cr(T) ~ [0, oo) then T is a positive operator on H. (ii)

Generalise this result in (i) to prove that W(T) = co cr(T).

(Hint: Use the functional calculus.) 2.

Consider a continuous linear operator Ton a complex Hilbert space H. Prove that (i)

if S is a self-adjoint operator on H and for all x e H II Sx II = II Tx II then S = (T*T) 112•

(ii) 3.

if T is a positive operator then II T II = II T 112 11 2.

Consider a positive operator Ton a complex Hilbert space H. Prove that

I~ (Tx, x) 112 (Ty, y) 112 for all x, y eH, (ii) 11Txii~IITII 112 (Tx,x) 112 forallxeH (i)

I (Tx, y)

and deduce that (Tx, x) = 0 if and only if Tx = 0. 4.

Consider a self-adjoint operator Ton a complex Hilbert space H. Prove that there exist positive operators T+ and T- so that T = T+- T-and T+T- = T-T+ = 0. (Hint:ConsiderT+=k (.fi2+T) and T-=t (.fi2-T).)

5.

6.

7.

(i)

Consider a positive operator Ton a complex Hilbert space H. Determine

(ii)

cr(T 112) in terms of elements of cr(T). Consider positive operators T 1 and T 2 on H where T 1T2 = T2T 1. 112

ltz..,lt2 12 .

=T1

(a)

Provethat(T 1T 2)

(b)

Determine cr((T 1Ti 12) in terms of elements of cr(T 1) and cr(T2).

Consider a continuous linear operator T on a complex Hilbert space H. Prove that i~ometry

if and only if T*T is a projection,

(i)

T is a partial

(ii)

Tis a partial isometry if and only if T = TT*T.

Consider { lln} a decreasing sequence of positive real numbers which is finite or convergent to 0 and {en} and {un} orthonormal sequences in a Hilbert space H. (i)

Prove that the operator T defined on H by Tx = L!ln (x, en)un

(ii)

is a compact operator on H. Prove that T*y = LJ.Ln (y, un)en.

(iii)

Prove that (T*T) 112x = L!ln (x, en)en.

§20. The spectral theorem for compact operators

8.

(i)

251

Given a positive operator Ton a complex Hilbert space H, prove that if Tis invertible on H then so is T 112.

(ii)

Given a continuous linear operator T on a complex Hilbert space H, prove that if T is invertible then it has a unique polar decomposition T = US where U is a unitary operator and S is a positive operator on H.

APPENDIX A.l

Zorn's Lemma We confine ourselves to developing sufficient set theory to explain the statement of

Zorn's Lemma and to enable us to apply it. A.l.l Definitions. Given a nonempty set X, a partial order relation :::; on X is a relation defined by the properties: (i) x < x for every x EX (ii)

X

:S y

and y :S

X~ X

(iii) x :::; y andy:::; z

~

(reflexivity) =y

x :::; z

(antisymrnetry) (transitivity).

The set X together with the partial order relation :::;, sometimes denoted (X, :::;), is called a partially ordered set. It is clear that any subset of a partially ordered set is a partially ordered set in its own right. Elements x, y, EX are said to be comparable with respect to the partial order relation:::; if x :::; y or y :::; x. There may be pairs of elements in X which are not comparable. If every pair of elements is comparable with respect to the partial order relation :::; then the relation is said to be a total order relation and (X, :S) is called a totally ordered set, (or a chain). Every subset of a totally ordered set is a totally ordered set in its own right. A partially ordered set has totally ordered subsets.

A.l.2 Examples. (i)

Consider lR the set of real numbers.

The relation "less than or equal to" denoted by :s; is a total order relation on JR. (ii) Consider X a nonempty set and ~ the family of all subsets of X. The relation "set inclusion" denoted by s;;; is a partial order relation on~. Not all subsets are comparable; for example, for A, B s;;; X, A, B, have A

t

B and B

t

*0

and A n B = 0 we

A.

An example of a totally ordered subset is an increasing sequence of sets; that is, {An} where A I s;;; Az s;;; .•. s;;; An s;;; . . . . (iii) Consider F:i the set of natural numbers. The relation "divides" denoted by I is a partial order relation on F:i. Again not all elements are comparable; for example, for 3 and 5, 3

f5

and 5

f 3.

The set

{zn : n E F:i } is a totally ordered subset. (iv) Consider 'C[O, I] the linear space of continuous real mappings on [0, 1]. The relation "is dominated by" denoted by :s; defined by f :s; g when f(x) :s; g(x) for all x E [0, I] is a partial order relation on 'C [0, I]. Not all elements of 'C[O, I] are comparable; for example, f and g where f(x) = x2, g(x) = I - x2. The set {fn : fn(x) = I - xn, n E F:i} is a totally ordered subset.

0

253

Appendix

A. 1.3 Definitions. Given a partially ordered set set (X, ::;) we say that an element x0 EX is

maxima/if Xo ::: X for X EX

:=}

X = x0 .

So a maximal element is not 'less than or equal to' any other element of X. This does not mean that it is the 'greatest' element in X, for it is not necessarily comparable with every element of

X. A maximal element does not necessarily always exist, and if it does it is not necessarily a unique element of X. Similarly, we say that an element Yo EX is minimal if x ::; Yo

for x EX

:=}

x = y0 .

A.l.4 Examples. (i) In (~, ~), ~ has a single maximal element X. (ii)

In (N, 1), N does not have a maximal element.

In the subset A= {I, 2, ... , 10) we see that 6, 7, 8, 9, 10 are maximal elements of A.

D A.l.5 Definitions. Given a partially ordered set (X, ::;), for subset A

~X

we say that x0E X

is an upper bound for A if a<x

for all a E A.

Similarly, we say that Yo EX is a lower bound for A if y0 ::; a for all a E A. An upper or lower bound for A is not necessarily a member of A but is comparable with every member of A. We say that x0 EX is a least upper bound or a supremum of A if x0 is an upper bound for A and x0 ::; x for all upper bounds x of A, and y0 EX is a greatest lower bound or an infimum of A if Yo is a lower bound for A and y ::; Yo for all lower bounds y of A. A subset A of a partially ordered set (X, ::;) need not have an upper bound and if it does it does not necessarily have a supremum. However, if A has a supremum then it is a unique element of X. A similar statement could be made for lower bounds and infimum. In view of this we denote the supremum of the set A by sup A and the infimum by inf A. A.l.6 Examples. (i) In(~,~), consider a subfamily CJ.

~~-

Now X is an upper bound for CJ. and 0 is a lower bound for CJ.. sup CJ. is the union of all sets in CJ. and inf CJ. is the intersection of all sets in CJ..

254 (ii)

Appendix

In (N, I) consider a subset A "' {4, 6, 8).

An upper bound for A is any n EN divisible by 4, 6 and 8; for example, 24, 48, ... sup A is the lowest common multiple of the set A, so sup A = 24. A lower bound for A is any n EN which divides into 4, 6 and 8; for example, I, 2. inf A is the greatest common divisor of the set A so inf A= 2. (iii) In (IR, ~), consider the subset
E

{I, ... , n)}.

supA=v{fk:ke{l, ... ,n)} and inf A =

11 { fk

:k

E { I,

... , n) } ;

For the set B = { fn : fn(x) = 1-x", n EN} we see that B has f0 as an upper bound where f0(x) = I for x e [0, 1], and in fact f0 = sup B.

0

We are now in a position to state Zorn's Lemma which asserts the existence of certain elements in certain partially ordered sets. A.l.7 Zorn's Lemma.

In a nonempty partially ordered set where every totally ordered subset has an upper bound, there exists at least one maximal element. It can be shown that Zorn's Lemma is equivalent to the Axiom of Choice which is a set theory axiom independent of and additional to those more generally accepted axioms of set theory. Among mathematical logicians there is considerable discussion about the Axiom of Choice. Among analysts, the constructive analysts are very critical of its use. However, classical functional analysis does rely at significant points on the Axiom of Choice in the fonn of Zorn's Lemma. A.2

Numerical equivalence

A.2.1 Definitions. Given a family of all subsets of some universal set we say that a subset X is numerically equivalent to a subset Y if there exists a one-to-one mapping of X onto Y. Clearly, numerical equivalence is an equivalence relation on the family of all subsets which partitions subsets into equivalence classes. We call the equivalence classes cardinal numbers. For a universal set which contains the set of real numbers, given n EN, the equivalence class which contains the subset {I, 2, ... , n) is called the cardinal number n. Any set which belongs to this class is said to be finite and have n elements. A set is said to be infinite if it is not finite and is said to be countably infinite if it is numerically equivalent to the set of natural numbers N. A set is said to be countable if it is either finite or countably infinite and is said to be uncountable if it is not countable.

Appendix

255

The following theorem is important in establishing numerical equivalence of sets. A.2.2 The Schroeder-Bernstein Theorem.

For nonempty sets X andY if there exists a one-to-one mapping f of X into Yanda one-toone mapping g ofY into X, then there exists a one-to-one mapping h of X onto Y.

Proof. We divide X into three subsets as follows. For x EX, if x E g(Y) call g-1(x) E Y the first ancestor of x, if g-1(x) E f(X) call f-1(g-1(x)) EX the second ancestor of x, if f-1(g-1(x)) E g(Y) call g-1( f-1(g-1(x))) E Y the third ancestor of x. y

X

X

y

X

X

g-1 (f-l(g-l(x)) f-1(g-1(x))

E f(X)

E g(Y)

Figure 20. Tracing ancestors. Continuing this process we see that there are three possibilities: (i)

x has an infinite number of ancestors, we define the set

(ii)

x has an even number of ancestors; that is, the last ancestor of x is a member of X; we

Xi

= { x EX : x has an infinite number of ancestors}.

define the set Xe

={x EX : x has an even number of ancestors}.

(iii) x has an odd number of ancestors; that is, the last ancestor of x is a member of Y; we define the set X0 x EX : x has an odd number of ancestors}.

={

Now the sets Xi, Xe and X 0 are mutually disjoint and their union is X. We form a division of Y into sets Yi, Ye and Y0 in the same way. We have that f maps Xi onto Yi and Xe onto Y0 and g -1 maps X 0 onto Ye• so we define the mapping h: X

-7

Y by h(x) = f(x) =

XE XiuXe }

g-1(x) XE X 0

and h is a one-to-one mapping of X onto Y.

D

256

Appendix

A.3 Hamel basis Important theory for infinite dimensional linear spaces is constructed on the assumption of Zorn's Lemma. A.3.1 Definition. A Hamel basis for a linear space X is a linearly independent set which spans X. It is clear that a nonempty subset A of X is a Hamel basis for X if and only if each element x EX can be expressed uniquely as a linear combination of elements of A. A.3.2 Examples (i)

In lR n the set { e 1, e 2, ek

... ,

en} where

={0, ... , 0, I, 0, ... , 0}

for k

E { I,

... , n}

kth place is a Hamel basis for lR n. (ii)

In c 0 the set {en} where en

={0, ... , 0, I, 0, ... } for n

E

F:i

nth place is a linearly independent set but it is not a Hamel basis for c 0 since sp{ en : n E F:i} = E 0 . The set is a Hamel basis for E 0 .

D

The proof of the following existence result is an application of Zorn's Lemma. A.3.3 Theorem. Every nontrivial linear space X has a Hamel basis.

Proof. Consider the family

::r of all linearly independent sets in X, partially ordered by set

inclusion. For any x EX, x "' 0, the set {x} is a linearly independent set so Consider a totally ordered subfamily in

::r is nonempty.

::r.

It is clear that the union of sets in this subfamily is a linearly independent set and is an upper bound for the totally ordered subfamily. By Zorn's Lemma Now for any x

::r has a maximal linearly independent set A.

e A it follows that Au {x} is a linearly independent set in X.

So there exist scalars {A., A. 1,

••• ,

"-n} and a subset { e 1,

.•• ,

en} of A such that

n

Ax +

L

A.kek = 0. Now A."' 0 since { e 1,

••• ,

en} is a linearly independent set, so

k=l

I n x =-- L A.kek. A. k=l We conclude that A spans X and therefore A is a Hamel basis for X.

We have a significant cardinality result for Hamel bases of linear spaces.

0

257

Appendix

We need the following cardinality property, which is itself dependent on Zorn's Lemma. A.3.4 Proposition. Given an infinite set A, the set U { N 3 : a E A} where N 3 is countable for each a E A, has the same cardinality as A. A.3.5 Theorem. In any given nontrivial linear space X, any two Hamel bases are

numerically equivalent.

Proof. The case when X is finite dimensional we take as a well known linear algebra result. Consider the case when X is infinite dimensional. Suppose that there exist two Hamel bases, B 1 ea} and B2 f13} in X. Each eaE B 1 is a linear combination of elements from B2 •

={

={

For each eaE B, write B 2 (ea) as the finite subset of B 2 which produces ea by linear combination. Every member of B 2 is a member of at least one of the sets B2 (ea); for otherwise, if there exists an f13 E B 2 such that f13 e B 2(ea) for all a., then f13 is a linear combination of elements from B 1 , say { eak: k E {I, ... , n}} and each eak is a linear combination of elements from B2 , so f13 is a linear combination of elements from B2 and then B 2 is not linearly independent. Therefore, B 2 ~ U B2(ea)·

a.

So by Proposition A.3.4. there exists a one-to-one mapping of B 2 into B 1. By symmetry of argument and the Schroeder-Bernstein Theorem A.2.2 we have our result. D So we are led to the following definition. A.3.6 Definition. The Hamel dimension of a linear space X is the cardinal number of a Hamel basis for X. This definition is consistent with the definition of dimension for finite dimensional linear spaces and is simply an extension of this notion for infinite dimensional linear spaces.

HISTORICAL NOTES

§I. Introduction A persistent theme of mathematics in the nineteenth century was the pursuit of rig our in establishing foundations, in definition of terms and in the presentation of proofs. In analysis, the great proponents of rigour were Augustin-Louis Cauchy early in the century and Karl Weierstrass towards the end. The axiomatising of the real number system by Julius Dekekind and Georg Cantor had a determining effect and became the foundation on which analysis depends. The outstanding development of mathematics in the twentieth century has been the study of structure and the promotion of the axiomatic method in exploring the implications of that structure. It is worthwhile reflecting on the principal stages which led up to this development to appreciate the unifying effect of the study of structure and the simplifying power that an axiomatic analysis of structure has brought to a variety of applications. The significance of structure first became apparent from an algebraic point of view. Until the middle of the nineteenth century mathematicians had been dealing with well determined mathematical objects: numbers, points, curves, functions, . . . . But it came to be realised that the algebraic manipulation of these different objects had a remarkable similarity. The essence of these manipulations did not lie in the nature of the objects but in the rules for handling them which were often the same for different types of objects. The precise formulation of this perception had to wait for the development of set-theoretic concepts and language and it was only towards the end of the century that an abstract group defined on an arbitrary underlying set was to become an area of serious study. By the tum of the century no similar development had occurred in analysis. Extensions of the ideas of limits and continuity which had been formulated were always relative to special objects such as functions or curves. The possibility of defining such notions on an arbitrary set by a generalised notion of distance was first put forward by Maurice Frechet in his doctoral thesis of 1906. On the simple structure of a metric space it is possible to extend most of the arguments concerning neighbourhoods, limits and continuity which are familiar in Euclidean space. Until the middle of the nineteenth century Euclid's geometry was considered a beautiful logical and orderly description of the natural spatial world. In the first half of the nineteenth century the advances in projective geometry and the geometry of many dimensions had started to change this view. The solution of the problem of the Parallel Axiom of Euclid radically changed the outlook of mathematicians. The axioms were no longer self-evident truths but independent hypotheses although they may have been suggested by some real

259

Historical notes

world situation. The importance of Euclid's achievement is that it is a prototype of the axiomatic method; its usefulness is that it is a convenient model of our spatial world locally.

David Hilbert's refinement of Euclid's axioms at the tum of the century, extracting from them any remaining intuitive ideas, had considerable influence in promoting the axiomatic method. He stressed that the axioms concerned undefined objects called, point, line and plane and the relations between these objects and that they were independent of physical reality. All of Euclid's geometry was then deduced consistently from Hilbert's axioms by the axiomatic method. The outcome of the pursuit of rigour in the nineteenth century has been the analysis of structures by the axiomatic method in the twentieth century. Jean Dieudonne declares: "there can be no rigorous proof except in the context of an axiomatic theory in which objects and basic relations have been specified and the axioms by which they are connected have been exhaustively listed". t §2. The development of function spaces It was Weierstrass who defined the concept of uniform convergence in 1841 and had made its importance generally appreciated. Weierstrass' Approximation Theorem of 1885, showing that polynomials are uniformly dense in the continuous functions, (AMS §9), gave further impetus to the idea of a junction space, where the functions are treated as points in a space of functions. The Ascoli-Arzela Theorem of 1883 which characterises compactness of a set of continuous functions, (AMS §9), was a landmark in establishing the focus on function spaces. Many of the problems in mathematical physics studied by mathematicians in the nineteenth century resolved themselves into finding the solution of an integral equation. For example, Fourier studying heat flow in 1811 considered the equation f(x) =

J;

cos(xt) y(t) dt.

Abel's investigation of the tautochrome problem in 1822 led to the equation f(x)

= fx .~

Jo

'I

x-t

dt, f(O)

= 0.

In both cases the aim is to find the function y given the function f. Early in the century individual problems were considered in isolation and solved by particular methods. But towards the end of the century mathematicians had begun the quest for a general theory for the solution of integral equations. In 1895 Jean-Marie LeRoux and in 1896 Vita Volterra published the first existence and uniqueness theorems for general classes of integral equations. Volterra studied equations of the form y(x)

= f(x) + "-J; K(x,t)

y(t) dt.

t Jean Dieudonne, Mathematics- the Music of Reason, translated by H. G. and J.C. Dales, Springer Verlag, Berlin 1992, p 237.

260

Historical notes

The function K(x,t) is the kernel of the equation and is symmetric if K(x, t) = K(t, x). Given functions f and K the aim is to find the function y. Abel's equation is related to this form. A decisive contribution to the theory of integral equations was made by Ivar Fredholm in 1900 and published in full in 1903. He studied a class of integral equations more general than those of Volterra and of the form y(x)

=f(x) +"-I: K(x,t)

y(t) dt.

Fourier's heat flow equation is related to this form. The papers of Volterra and particularly Fredholm devising a general theory gave impetus to the consideration of integral operators acting on a function space. Hilbert was immediately attracted to Fredholm's work. He published six papers on integral equations and by the fourth of these papers, published in 1906, he had begun spectral analysis of compact operators on function spaces. He had laid the foundation for what came to be called spectral theory on Hilbert space. For the Fredholm integral equation with f and symmetric K continuous he showed that

I:

the eigenvalues for the homogeneous equation y(x)

= A.

K(x,t) y(t) dt

are real and can be ordered. He then defined eigenfunctions corresponding to eigenvalues and showed that they form an orthogonal family which give solutions to the homogeneous equation.

He defined Fourier coefficients for a function with respect to normalised

eigenfunctions and showed that any function f of the form f(x)

=I:

K(x,t) g(t) dt

for some continuous function g, can be expressed as a Fourier series in the eigenfunctions of the homogeneous equation. He then reworked his theory based on a complete orthogonal system of functions and showed that what is essential for the spectral theory of an integral operator is that it is a compact operator not that it has a representation in terms of integrals. It was Hilbert's student Erhard Schmidt who in his doctoral thesis of 1905 introduced a simplified approach to Hilbert's theory of integral equations.

In his paper of 1908, he

developed a geometric approach to Hilbert space theory and set the pattern for the modern theory of abstract Hilbert space. Hilbert, in his work on integral equations, had considered a function as represented by its Fourier coefficients in an expansion with respect to an orthonormal sequence of functions. These coefficients are sequences { x 0 } where

'Lx~ <

oo,

Schmidt considered these sequences as co-ordinates of a point in an infinite

dimensional space, .trspace. In his analysis of .trspace which included complex sequences, Schmidt introduced the notation for an inner product and norm. He used the Cauchy-Schwarz and Bessel inequalities and proved that J. 2-space is complete, a vital property for the development of the general theory. He had in his thesis used the Gram-Schmidt orthogonalisation process and he now developed the notion of orthogonal projection.

261

Historical notes

Of crucial importance was the theory of Lebesgue integration which was developed by Henri Lebesgue in his doctoral thesis presented in 1902. Building on the work of Emile Borel on measure, Lebesgue constructed a new general and powerful theory of integration. This theory was much more satisfactory than Riemann's and had the advantage that under mild conditions the limit of an integral is the integral of the limit. It was then possible to consider function spaces which are complete metric spaces. In 1907, Frederic Riesz making use of Lebesgue's new theory of integration and simultaneously with Ernst Fischer, proved the Riesz-Fischer Theorem which implied that the space :C: 2 [a,b] of all Lebesgue square integrable functions on [a,b] with mean square norm is complete and that it is isometrically isomorphic to .t 2-space. Following the publication of this theorem both Schmidt and Frechet remarked that the space J:: 2 [a,b] has a geometry analogous to .lz-space. In the same year Frechet and Riesz independently obtained the representation theorem for continuous linear functionals on J:: 2 [a,b] which was later generalised by Riesz for abstract Hilbert space. In 1909, Riesz obtained a representation theorem for continuous linear functionals on "C[a,b] with the supremum norm by Stieltjes integrals. In 1910 Riesz generalised the study of J:: 2 [a,b]-space by considering the J::p[a,b] spaces where I < p < oo consisting of functions whose pth power is Lebesgue integrable on [a,b] with the p-norm. He used Holder and Minkowski inequalities and established the duality between spaces :C:p[a,b] and :C:q[a,b] where

i

+~

= I.

He extended the Riesz-

Fischer Theorem for these spaces and proved that these spaces are complete. He obtained the representation theorem for continuous linear functionals for these spaces. This was a major step in generalising from Hilbert function space to more general function spaces. Riesz's study of integral equations in I: p[a,b]-space where I < p <

oo

laid the

foundation for operator theory on abstract normed linear spaces. He considered integral equations which generate linear operators on J::p[a,b]-space and he defined continuity and the norm of continuous linear operators on such spaces. He also introduced the notion of the conjugate of an operator: For a continuous linear operator Ton J::p[a,b] where I < p < oo, given g E :C:q[a,b] where

.!.+.!. p q

=I

'

J:

T(f(x)) g(t) dx

defines a continuous linear functional on J::p[a,b]. By the Riesz Representation Theorem there exists a function hE :C:q[a,b] where

J:

i~ +

T(f(x)) g(x) dx =

= I, unique in the Lebesgue sense, such that

J:

f(x) h(x) dx.

The operator T' conjugate toT is defined on :C:q[a,b] by T'(g) = h

262

Historical notes

and he showed that T' is also continuous linear and II T' II =II T II. He considered inverses of operators and related them to inverses of their conjugates. The work of Riesz on compact operators, which he published in 1918, was extraordinarily powerful. He gave a definition for compactness of an operator which greatly improved on that of Hilbert and he developed it for more general Banach spaces. Although his theory was developed for the function space t: [a, b) he regarded it to be of general applicability. He used the term norm for the supremum norm in t: [a,b]. He essentially derived the spectral decomposition for compact operators on Banach spaces. His work was revised and extended in 1930 by Juliusz Schauder and for this reason is called the RieszSchauder Theory of compact operators.

Mathematicians in the eighteenth and nineteenth century were also concerned with problems in the calculus of variations. Here the aim is to find a function y which satisfies certain constraints and minimises or maximises an integral of the form J(y)

J:

=

F(x,y ,y') dt.

For example, in 1696 John Bernoulli proposed the brachistochrone problem, that is, to find the curve joining two points down which a particle will slide in minimum time. This led to finding the function y which minimises the integral J(y)

=lab

I+ [y' (x)f dx. a- y(x)

Volterra interpreted the general calculus of variations problem as one of minimising the functional Jon the function space "C[a,b]. Jacques Hadamard in 1903 took up the study of functionals in relation to the calculus of variations. The term functional is due to him. He began the study of linear functionals on a function space. The term functional analysis was first used by Paul Levy in his book published in 1922 when the main interest was on functionals acting on a function space. §3. The axiomatic analysis of normed linear spaces In his doctoral thesis of 1920, the Polish mathematician Stefan Banach defined an abstract complete normed linear space with axiomatic structure as is now commonly in use.

Independently and at the same time Hans Hahn published the same set of axioms. We have seen that the norm had been defined for particular function spaces and in 1921 Eduard Helly explored a similar structure for sequence spaces. Banach had considered normed linear spaces over the real numbers but in 1923 Norbert Wiener pointed out that the theory could be extended with wider applications using normed linear spaces over the complex numbers. After the first World War, mathematics in Poland experienced a particularly influential period of growth. The story of Banach's recruitment into the mathematical academic

Historical notes

263

community is of interest because he came from a poor background and had to begin earning his own living at the age of fifteen. In 1916, Professor Hugo Steinhaus walking in Cracow Park overheard two students discussing the new theory of Lebesgue measure and integration; they were Stefan Banach and Otto Nikodym. Steinhaus formed a mathematical club which began to meet in his rooms. Steinhaus claimed that Banach was his "greatest mathematical discovery". Considering the achievement of Frechet in defining abstract metric spaces and the similarities in treatment of the different function spaces it seems odd that it took so long to introduce an axiomatic analysis of normed linear spaces. Nevertheless, there was a gradual understanding that many diverse problem areas shared this discernable common structure which was amenable to axiomatic formulation and solution as a consequence of the axiomatic method. The stunning first success of this abstract analysis was Banach's general formulation of his fixed point theorem for contraction mappings which appeared in his doctoral thesis, (AMS §5). It was Helly who brought a geometric aspect to the study. Exploring geometrical notions in Euclidean space he showed that a closed symmetric convex body with 0 as an interior point, as discussed by Herman Minkowski, could be considered as a closed unit ball under an appropriate norm. This enabled a useful visualisation of problems. The geometry of normed linear spaces is a significant metric generalisation of Euclidean geometry. Normed linear spaces form a subclass of metric spaces. But there had been an earlier generalisation to a larger class of spaces. In his book of 1914, Felix Hausdorff defined a general topological space based on the idea of neighbourhood rather than distance and he was able to develop a rich analysis of limits and continuity. His second book, published in 1927, where he concentrated on metric topology, was particularly influential for the Banach school. The development of general topology led to the significant study of weak topologies on normed linear spaces which was used to great effect in Banach's book of 1932. But this aspect of the analysis of normed linear spaces follows naturally as a particular application of the more general analysis of locally convex spaces. What makes the study of normed linear spaces particularly fruitful is the linkage between the algebraic and topological structures, the axiomatic prescription of a continuity condition associated with the basic algebraic operations of addition and multiplication by a scalar. It seems that Frechet was the first to make this observation explicitly. This led naturally to the definition of a linear topological space which is both a linear space and a topological space where there is a similar continuity condition associated with the algebraic operations, in particular where neighbourhoods of a point are translations of scalar multiples of neighbourhoods of the origin.

264

Historical notes

The convexity of the balls in a normed linear space is a significant property and a linear topological space with a neighbourhood base of convex subsets was first given general definition as a locally convex space by John von Neumann in 1935. Many more example spaces were available to be investigated by this expansion of the theory. Progress in the analysis of abstract normed linear spaces depends crucially on the

Hahn-Banach Theorem. Minkowski working geometrically had established what amounted to the Hahn-Banach theorem for finite dimensional normed linear spaces, that is, to each point in the boundary of the closed unit ball there exists at least one hyperplane of support containing the point. He established the dual property with the space of all continuous linear functionals under the dual norm. Helly generalised these ideas for certain sequence spaces which are separable normed linear spaces. However, he confronted the problem that his infinite dimensional spaces might not be self-dual. He gave the first example of a non-reflexive Banach space by exhibiting a continuous linear functional which does not attain its norm on the closed unit ball of the space. In 1927 Hahn, examining the work of Helly, proved the extension form of the theorem for normed linear spaces without any assumption of separability. For a separable space induction can be used, but he introduced the method of transfinite induction for the first time into the study of normed linear spaces. He finally defined the dual of a normed linear space with dual norm and considered the natural embedding of the space into its second dual and also the notion of reflexivity. He had, with this paper, begun the study of duality theory for general normed linear spaces. In 1929 Banach, unaware of Hahn's work, published the same theorem but in a more general form and more obviously as an extension theorem. Given a linear functional on a subspace dominated by a convex functional, his argument guaranteed the existence of a dominated extension of the linear functional to the whole space. Banach's form of the theorem really belongs in the theory of linear spaces quite apart from any topology. It was crucial in the development of the theory of locally convex spaces. Although the general proof was devised using transfinite induction, since the 1940s the proof is generally given using Zorn's Lemma. The generalisation of the theorem to normed linear spaces over the complex numbers was given by Bohnenblust and Sobczyk. The Hahn-Banach Theorem in its more general form implies many fundamental existence properties on a normed linear space X over the real numbers: A continuous convex function

on an open convex subset A is said to be subdifferentiable at

x0 E A if there exists a continuous linear functional f on X such that f(x-x 0)

~

<j>(x)- <j>(x 0)

for all x EA.

The Hahn-Banach Theorem implies that

Historical notes

265

A closed convex set K is said to have a support point x0 E K if there exists a continuous linear functional f on X such that f(x 0 ) = sup f(K). The Hahn-Banach Theorem implies that if K has non-empty interior then every point in the boundary of K is a support point of K. Two nonempty convex sets K 1 and K2 are said to be separated by a hyperplane if there exists a continuous linear functional f on X and a E IR such that K 1 ~{xEX:f(x)~a}

and K 2 ~{xEX:f(x)~a}. The Hahn-Banach Theorem implies that if int K 1 "- 0 and K2 n int K 1 = 0 then K 1 and K2 can be separated by a closed hyperplane. The development of set theory at the end of the nineteenth century had a profound influence. In 1899, Rene Baire proved that in Euclidean space a countable intersection of dense open sets is dense, a result which could be extended for any complete metric space. Baire's work enabled the adoption of non-constructive proofs as a technique in analysis. In 1922 Hahn proved the Uniform Boundedness Theorem using the method of "the gliding hump", but in 1927 Banach and Steinhaus used Baire Category Theory to give a much simpler proof of the theorem. In 1929 Banach using Baire Category Theory again proved both the Open Mapping Theorem and the Closed Graph Theorem and included these in his book of 1932. These results provided convincing evidence of the great power of the axiomatic study of abstract normed linear spaces. The journal Fundamenta Mathematicae specialising in point-set topology was first published in 1920 and soon gained international status. Under the impetus of the analysts of Lvov, the journal Studia Mathematica was founded in 1929 specialising in functional analysis. In 1932 Banach published his book, Theorie des operations lineaires, containing a comprehensive account of the state of the theory of normed linear spaces up to that time. The book had great appeal and became a landmark foundational text. Mathematicians began to see the power of the methods and to apply them. In many areas of research normed linear space terminology began to be used. Soon the theory of Banach spaces became a vital study for intending research mathematicians. The development of the analysis of normed linear spaces was greatly influenced in the 1930s by the group of mathematicians associated with Stefan Banach at Lvov. The friends became known as the Scottish Cafe group because they habitually gathered at the coffee house of an afternoon and worked into the evening setting and solving problems related to their latest research. Eventually in 1935 a large notebook, 'The Scottish Book' was bought

266

Historical notes

to enter problems with rewards and solutions. By the time of the German invasion of Eastern Poland in 1941 the book had entries for 193 problems. The book was hidden during the war and in 1956 was translated and printed. Some of the problems are still unsolved and some have remarkable tales relating to their solution. Problem 153, posed by Stanislaus Mazur in November 1936 was shown by Alexander Grothendieck in 1955 to be an equivalent to the Approximation Problem and is related to the Basis Problem. These problems were solved in the negative by Per Enflo in 1972 and Mazur, a surviving member of the original Scottish Cafe group, presented him with the designated prize, a live goose. With hindsight it seems incredible that the identification of Hilbert function space and Hilbert sequence space did not lead sooner to an axiomatic definition of an inner product space. However, this eventuated only after the axiomatic development of metric spaces by Frechet and normed linear spaces by Banach and the axiomatic method had achieved some degree of acceptance. A stimulus to the study of Hilbert spaces came in 1923 from the newly emerging quantum theory when it was realised that spectral theory begun by Hilbert could be used as a mathematical instrument to study quantum mechanics. In 1929, John von Neumann presented an axiomatic approach to Hilbert space and operators on that space. His aim was to develop spectral theory for classes of operators on this abstract space. His axioms are for a separable space. Since all separable Hilbert spaces are isometrically isomorphic to 1 2-space, this might explain why the abstract generalisation took so long to formulate. In three seminal papers von Neumann developed his spectral theory of operators intrinsically from the axioms defining Hilbert space. The notions of a densely defined operator and of an operator with closed graph were particularly useful for quantum theory. He elucidated a complete spectral theory for normal operators. He formulated the spectral analysis of an operator and determined spectral properties for different types of operators. In short, von Neumann essentially established spectral theory for operators on Hilbert space as it stands today. Von Neumann began the serious study of the algebra of operators on a separable Hilbert space in his second paper on spectral theory. He introduced the weak topology on the algebra of operators and, in collaboration with Francis Murray in 1935, he began to investigate what have been called von Neumann algebras. Again it is surprising that it was not until 1941 that the study of abstract normed

algebras was begun by lzrail Gelfand. As with the structure specified for a normed linear space there is an axiomatic prescription of a continuity condition associated with multiplication. He extended spectral theory to elements of normed algebras and, in collaboration with Mark Naimark, began to study algebras with an involution and B*-algebras. The famous Gelfand-Naimark Theorem of 1943 establishes that every unital B*-algebra is isometrically* isomorphic to a subalgebra of the algebra of operators on some Hilbert space and this paved the way for a new interpretation of Hilbert's spectral theory.

267

Historical notes

§4. Two later significant advances (i)

Variational principles

The Bishop-Phelps Theorem published by Errett Bishop and Robert Phelps in I 961 has been generalised out of all recognition to a variety of forms with remarkably wide applications especially in the theory of optimisation. Bishop and Phelps themselves generalised their result to show that for a closed bounded convex set K in a Banach space the support points of K are dense in the boundary of K and the continuous linear functionals supporting K are dense in the dual. A generalisation was given by Arne Bri'lndsted and Tyrrell Rockafellar in I 965 concerning the subdifferentiability of proper lower semicontinuous convex functions on a Banach space. The outstanding generalisation called Ekeland's Variational Principle proved by Ivar Ekeland in 1964 concerns the points of minimality of real valued proper lower semicontinuous functions on a complete metric space. The original proof of the Bishop-Phelps Theorem used Zorn's Lemma but subsequently a proof was given using only the completeness of the space. In fact Francis Sullivan in I 98 I showed that the Ekeland property actually characterises completeness of the metric space. The applications of the Ekeland Variational Principle were surveyed by Ivar Ekeland in 1979. The Principle can provide a proof to various fixed point theorems, it is of considerable use in non-convex optimisation situations and it has application in differentiability theory on Banach spaces. In I 987, Jonathan Borwein and David Preiss produced what they called a Smooth Variational Principle which gives an improved result when the norm of the Banach space has a differentiable norm. A real valued function \jl on an open subset A of a normed linear space X is said to be Gateaux differentiable at x E A if lim \ji(X+Ay)-\jl(x)

A.-tO

A.

exists and is continuous and linear in y for all y EX, and is Frechet differentiable at x if also this limit is approached uniformly for all y E S(X). They showed that if a Banach space X has a Gateaux (Frechet) differentiable norm at each point of S(X) then every lower semicontinuous function \jl on X is Gateaux (Frechet) subdifferentiable at the points of a dense subset of its domain. If \jl is convex then it is Gateaux (Frechet) differentiable at such points. A real valued function

\jl

on an open subset A of a normed linear space X is said to be

locally Lipschitz and x E A if there exists K > 0 and 8 > 0 such that

I \jl(y)- \jl(z) I :o; K II y-z II

for ally, z E B(x; 8).

In I 990, David Preiss proved the technically deep result that a locally Lipschitz function on an open subset of an Asplund space is Frechet differentiable at the points of a dense subset of its domain.

268

Historical notes

(ii) Asplund spaces A continuous convex function on an open convex subset of a Euclidean space is Frechet differentiable at the points of a dense Gli subset of its domain. In 1933, Mazur showed that a comparable Gateaux differentiability property holds for every continuous convex function on an open convex subset of a separable Banach space. The examination of such differentiability properties for continuous convex functions was taken up again by Edgar Asplund in 1968 and interest in the topic was renewed. In particular, he showed that Mazur's result could be strengthened to Frechet differentiability for a Banach space with separable dual. Nowadays, a Banach space which possesses the Euclidean space property for Frechet differentiability is called an Asplund space. The study of this class of spaces gained significance because of the duality between Asplund spaces and spaces with the Radon-Nikodym property. For some time there had been investigation into the possible extension of the classical Radon-Nikodym Theorem to vector valued measures. In 1967 Mark Rieffel discovered that the extension was possible for vectors from a Banach space which exhibited a special geometrical property. Following a major contribution by Hugh Maynard, in 1973 a geometrical characterisation of the RadonNikodym property was achieved by Bill Davis and Robert Phelps and independently by Bob Huff. That geometrical property had been linked to the differentiability property defining an Asplund space by Isaac Namioka and Robert Phelps in 1975 and in the same year Charles Stegall proved that a Banach space is an Asplund space if and only if its dual has the RadonNikodym property. The Preiss result that a locally Lipschitz function on an open subset of an Asplund space is densely Frechet differentiable shows that such spaces are particularly significant for optimisation theory.

Principal references: Garrett Birkhoff and Erwin Kreysig, The establishment offunctional analysis, His tori a Mathematica II, (1984 ), 258-321.

Jean Dieudonne, History offunctional analysis, North Holland, Amsterdam. Mathematics Studies 49, 1981.

Robert Phelps, Convex Functions, monotone operators and differentiability. Springer Verlag, Berlin.Lecture

Notes in Mathematics 1364, 2nd edn, 1993.

LIST OF SYMBOLS (page where first introduced) APcr(T) ~(X,Y) ~(X)

Ccr(T) ~(cr(T))

D(x)

cln

228 67 80 224 243 138 156

fiR

77

p+

103 103 148 148 14 176 209 209 67 80 75 68 80 118 180 121 109 236 218 144 144 222 243 224 236 218 217 227 227

p-

Fa Gil id in

%(X,Y) %(X) l:;(X,Y) l:;(X)

Mr Mmxn Mn M..l Mil

N.L 'JI.~V[a,b]

'JI.n v(x)

ro(f, J) ro(f, x 0 ) Pcr(T) P(cr(T)) Rcr(T) n,n n,(A.) p(x)

Sn. SIR

270 S(P, f, a) O"(T) O"(X)

T' T*

101 224 217 171 181

X

71 107 107 225 185 185 122 123

X

129

T V(a) V(P, a) V(T) W(T) w(T) II

X II

x#

72

x*

72

XIR x®f

77

89

LIST OF SPACES (page where first introduced)

bounded function spaces

~(X) ~(X,

4 d)

~'C(X,

continuous function spaces

5 d)

5

~ 0 [0, I]

6

'C[a,b]

5

'C 1[a,b]

6

-c=[a,b]

19 22

'C[r]

finite dimensional spaces

'C 'f [21t] 'C 0 [0, I]

22

'C( cr(T))

224

!Rn a;n

polynomial spaces sequence spaces

6

3 3

_tn ,(I $;p$;oo)

41

~[a,b]

J:: 2[a,b]

5 38

J::p[a,b], (1 $; p $; oo)

44

'f[a,b]

15

p

integral spaces

5

~[a,b]

c

5

co

5

Eo

5

,tl

5

,t2

5

_tp,(J $;p$;oo)

m

43 4

INDEX

adjoint operator 181-184 affine hyperplane 75-77 algebra homomorphism 161 algebra norm 79 algebraic conjugate 72 algebraic dual 72 algebraic embedding 131 algebraically reflexive 124, 131 Amir 33, 40, 47 analytic 219 ancestor 255 annihilator 118, 121 Apollonius' identity 46 Apostol 101 approximate eigenvalue 228 approximate spectrum 228 Aronszajn and Smith 234 Ascoli-Arzela Theorem 208 associated normed linear space 7 Axiom of Choice 113, 254 Bachman and Narici 249 Baire category 140-152 Baire space 141 Baire's Theorem 142 Banach II Banach algebra 79-170 Banach limit 121 Banach space 4 Banach-Steinhaus Theorem 165 basis 15 Basis Problem 17, 214 Bessel's inequality 54 best approximation 29, 52, 64 bilinear mappings 167 Bisho{r-Phelps Theorem 136 Boas 87 Bonsall and Duncan 186, 225 Borel-Lebesgue Theorem 24

Index

bounded variation 107 Brown and Page 109, 235 B* algebra 184 Cantor's Ternary Set 36, 140 cardinal number 254 Cauchy-Schwartz inequality 24, 31, 74 Cauchy sequence 4 Cayley transform 195 chain 252 Chebyshev polynomials 65 Clarkson's inequalities 49 classical Hilbert space 38, 62 closed graph 157, 178 Closed Graph Theorem 158 codimension one 13, 75, 152 codimension n 90 commutant 195 compact 24, 206 compact normal operator 237-242 compact operator 206-216, 230--236 comparable 252 complemented 197 complete 4 complete normed algebra 79 completion 49, 87, 129-130 complex linear functional 77 complex linear space 77 conjugate mapping 171-180,226 conjugate space 72 continuous linear mapping 9 continuous spectrum 224 contraction 194 contravariant vector 95 convex functional 2, 120 convex set 2, 132 convex hull 132 coordinate functional 18, 23, 162 coset 7

273

274

Index

countable 232, 254 countable orthonormal basis 61 countably infinite 254 covariant vector 95 C* algebra 195 dense 13, 71, 87, 140 diagonal operator 86, 210, 227 dimension 16 Dini's Theorem 163 discontinuity 144 dissipative operator 229 division algebra 221 dual space 72, 92-112 du Bois Reymond Theorem 168 eigenspace 222 eigenvalue 222 eigenvector 222 Ekeland 132 Enflo 17,203,214 equicontinuous 208 equivalent norms 13, 132, 156, 162 Euclidean space 3, 24-30, 33, 63, 76, 191 even function 48 extension 71 extreme point 48 factor space 7, 118 final space 247 finite dimensional space 16,24-30,73, 80--81, 88,92-94, 124, 131, 173, 182,204, 207,223,241 Finite Dimensional Spectral Theorem 237 finite rank operator 207 first category 141 Fourier coefficients 62 Fourier series 56-58, 168 Fredholm integral equation 239,242 Fredholm operator 208, 216 F0 set 148

Index

functional 13 functional calculus 243 gauge 22 G 0 set 148 Gelfand and Naimark 184 Gelfand-Naimark Theorem 189 general polarisation formula 187 GM-AM inequality 42 Gram-Schmidt orthogonalisation 50, 52, 64 greatest lower bound 253 Hahn-Banach Theorem 115 Halmos 60 Hamel basis 15, 152, 256 Hamel dimension 61, 27,257 hermitian form 30 hermitian operator 229 Hilbert sequence space 33 Hilbert space 33, 94-97, 124 Holder inequality 47 hyperplane 13, 75 hyperplane of support 89 inclusion mapping 174 infimum 253 infinite 254 initial space 247 inner product isomorphism 4 7, 192 inner product space 30--45, 74 integral inner product spaces 35 invariant metric 2 invariant subspace 202 Invariant Subspace Problem 203 invariant subspace property for compact operators 233 involution 184 isometric isomorphism II, 63, 87, 123 isometry 11 isomorphism 12, 87

275

276

Index

James 125, 128 James characterisation of reflexivity 128, 131, 137 joint continuity 2, 32, 79, 167 Jordan and von Neumann 33 kernel 14 Laurent series 219 least upper bound 253 left regular representation 90, 170 left shift operator 227 Legendre polynomials 51, 64 linear functional 13 linear homeomorphism 10 linear operator 13 linear topological invariants 12 Liouville's Theorem 219 Lomonosov's Theorem 234 lower bound 253 lower semicontinuous 145 matrices 68, 80, 182, 184 maximal element 116, 253 maximal orthonormal set 58 Mazur and Ulam II meagre 141 mean square 57 metric generated by the norm 2 Minkowski inequality 31, 43 minimal element 253 Moebius transform 195 monotone 18, 23, 88, 107 multiplicative 161 multiplicative inverse 81 Murray 197 natural embedding 122-125, 175 Neumann series 83 nonreflexive 126 norm I norm generated by the inner product 31

Index

norm increasing 179 normal operator 189 normalised I 09 normed algebra 79, 184 normed linear space 2 normed linear subspace 5 norming 120 nowhere dense 140 nowhere differentiable functions 146 numerical equivalence 59, 254 numerical radius 185, 225 numerical range 185,224-225 odd function 48 open mapping 153 Open Mapping Theorem 155 operator 13 operator algebra 41, 79-86 orthogonal 38, 41, 197 orthogonal coefficients 51 orthogonal complement 198 orthogonal dimension 60 orthogonal projection 200 orthonormal basis 60 orthonormal set 50, 53 oscillation 144 parallel 75 Parallel Hyperplane Lemma 135 parallelogram law 32 Parseval's formula 55 partial isometry 246 partial order relation 252 partially ordered set 252 passing point 48 point spectrum 224 pointwise bounded 163 pointwise convergent 165

277

278

Index

pointwise definition 3, 67 Polar Decomposition Theorem 248 polarisation formula 46 positive cone 189 positive definite hermitian 30 positive linear functional 102 positive operator 188, 200 positive square root 244 pre-Hilbert space 30 product metric !56 product norm !56 product space !56 projection 86 projection operator 196-205 Ptolemy's inequality 47 Pythagoras' identity 47 quasinilpotent operator 235 quotient space 7, 32, 49, 118, 179,216 quotient mapping 90 Read 203,235 real hyperplane 78 real linear functional 77 reduced 202 reflexive 124, 176, 179 regular 81, 85 represented 184 residual 141, 148-150 residual spectrum 224 resolvent operator 218 resolvent set 217 Riemann-Lebesgue Lemma 65 Riemann-Stieltjes integral I 02 Riesz Lemma 26 Riesz Representation Theorem for Hilbert space 94 Riesz Representation Theorem for (t:[a,b], 11·11=) 107 Riesz Theorem 26 Riesz-Fischer Theorem 62

Index

Riesz-Schauder theory 235 right shift operator 227, 235 rotund 29, 87, 120, 204 Rudin 249 ruler function 145 sawtooth function 147 Schauder basis 16-18, 23, 66, 88, 162, 213 Schauder's Theorem 211 Schroeder-Bernstein Theorem 60, 255 second category 141 second conjugate 122 second conjugate mapping 175 second dual 122 self-adjoint 184, 195 semimetric 7 seminorm 7 semisimple 161 separable 15, 17, 61, 62, 118,208 separate 117 separately continuous 167 sequence spaces 97-101 sequential ordering 53 singular 81 slice 138 smooth 120 Spectral Mapping Theorem for Polynomials 220 spectral radius 218 Spectral Theorem for Compact Operators 249 Spectral Theorem for Compact Normal Operators 238 spectrum 217-236 square roots of a positive operator 244 Stone-Weierstrass Theorem 244 stronger norm 13 submultiplicative inequality 79 subreflexive 132-138 supremum 253 symmetric 2

279

280

Index

tangent hyperplane 77, 116 Taylor and Lay 109 topological divisor of zero 221 topological isomorphism II, 87, 175 total 117, 177 total order relation 252 total variation 107 totally ordered set 252 translate 75 uncountable 254 Uniform Boundedness Theorem 164 uniformly bounded 163 uniformly rotund 49, 131 unital normed algebra 79 unitary operator 189 Unitary space 3, 24-30, 33, 63 upper bound 253 upper semicontinuous 145 Volterra integral equations 83, 91 weaker norm 13 weakly bounded 166 weak * bounded 166 Weierstrass' Approximation Theorem 15, 147 Zorn's Lemma 59, 115, 252-254

Th1s text Is a basic course m functional analysts for senior undergraduate and beginning postgraduate students. It alms at providing some Insight Into bask: abstract analysts which IS now the contextual language of much modern mathematics. Although It IS assumed that the student will have famlharl r? with elementary real and complex analysis and linear algebra and have studied a course m the analysis of metric spaces, a knowledge of mtegration theory or general topology is not requ1red . The theme of this text concerns structural properties of normed linear spaces m general. especially associated with dual spaces and contmuous linear oper· ators on normed linear spaces. Butt he implications of the •:enera ll beor~· ar e Illustrated w•th a great variety of example spaces.

Edltor·la-<:bief Professor j.H . Loxton, Deputy V1ce ·Chancellor (Academic ~ MacquarJe University, NSW 2109, Australia Editors

Professor C.C. Heyde, School of Mathematical Sc1ences, Australian National University, Canberra, ACT 0200, Australia Associate Professor W.D Neumann, Department of Mathematic s. University of Melbourne, Parkville, Victoria 3052. Australia Associate Professor C.E.M. Pearce, Department of Applied MathematK>, University of Adelaide, Adelaide, SA 5005. Australia

The Australian Mathematical Society Lecture Ser1es Is intended to operate at the frontiers of mathematics itself and of Its teachmg, and therefore will contain both research monographs and textbooks suitable for graduate or undergraduate students.

CAMBRIDGE U NIV I>Jl. ~l

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9 780521 653756

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