Introduction to Operator Theory and Invariant Subspaces

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Author: Bernard Beauzamy

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INTRODUCTION TO OPERATOR THEORY AND INVARIANT SUBSPACES

North-Holland Mathematical Library Hoard of Advisory Editors: M. Artin, H. Bass, J. Eells. W. Feit. P. J. Freyu. F. W. Gehring. H. Halbcrstam , L. V. Horrnandcr, J. H. B. Kcrnpcrrnan , H. A. Lauwcrier, W. A. J. Luxemburg, F. P. Peterson. I. M. Singer and A. C. Zaanen.

VOLUME 42

NORTH-HOLLAND AMSTERDAM ·NEWYORK . OXFORD ·TOKYO

Introduction to OperatorTheory and Invariant Subspaces Bernard BEAUZAMY lnstitut de Co/cui Mathematique U. F.R. de Mathematiques Universite de Paris 7 France

NORTH-HOLLAND AMSTERDAM ·NEWYORK . OXFORD ·TOKYO

e

ELSEVIER SCIENCE PUBLISHERS BV. 19XX

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Library of Congress Cataloging-in-Publicatton Data Bernard, 1949Introduction to operator theory and Invariant subspaces / Bernard Belluzamy. CII. -- (North-Holland uthematlcal lIbrary; v.42) p. Bibliography: p. Includes Index. ISBN 0-444-70521-X (U.S.) 1. Operator theory. 2. InvarIant subspaces. I. TItle. I I. Ser i es. QA329.B44 1966 88-25064 515.7'24--dc19 CIP

Beauza~y.

PRINTED Ij\; THE ;'\jETHERLANDS

Introduction

As a whole, Operator Theory does not exist yet. We may have satisfactory results for some classes of operators, or handle, in a convenient way, some limited situations, but most of the fundamental problems are still unsolved, and it is even not clear that they are correctly formulated. Among classes which have received special attention during a recent past, the best known are the compact operators and the normal ones, for which some powerful tools are available, such as spectral decompositions. But even for these classes, some simple questions are still open. The most fundamental question in Operator Theory is the Invariant Subspace Problem, which is still unsolved in Hilbert spaces. Most attempts, since the beginning of the century, have been made in the positive direction, that is trying to prove that every operator has a non-trivial Invariant Subspace. An obvious remark is that, in order to succeed, one has first to show that, for every operator, there is a point z , the orbit of which (that is : the set {Tn x ; n E IN}) is not dense in the whole space. Indeed, if every orbit is dense, there is no Invariant Subspace. Unfortunately, this second problem is also still open, and, strangely enough, has not received much attention. Moreover, there are stronger versions of it which are also unsolved. The Invariant Subspace Problem does not appear as the last open question in a well-polished theory. It even seems to me, conversely, that the paths which might lead to its solution have not been traced. Fundamental-and extremely natural- questions, such as the behavior of orbits of a given operator, are considered here for the first time. Even the most indulgent reader will not be able to find in the whole Operator Theory (let's call it this way, for simplicity), a result which approaches, by far, the depth of Dvoretzky's Theorem, about sections of convex bodies, in the Geometry of Banach spaces, or of Carleson's Theorem, about convergence of Fourier series, in Harmonic Analysis. Obviously, the whole field is globally

Introduction late, compared to many others in Analysis (not to mention, of course, other branches of mathematics). To this fact, I can see three possible and complementary explanations. First, a basic understanding of Operator Theory relies on a rather deep knowledge of other areas, such as Fourier Analysis or Analytic Functions. Just to explain what are the Invariant Subspaces of the usual shift requires Beurling's theory of the decomposition of H2 functions into inner and outer parts. This theory is not so old: 40 years only. Other tools, connected with boundary values of analytic functions, are fairly recent, and not well understood. The connection with spectral synthesis seems also to come up. In short, we may say that the analytic tools which are needed are not old, and have not made their way yet. The second reason is that the theory has developed only, so far, into a small number of directions, which might not be the right ones. Emphasis has been laid upon a few constructions which have been religiously considered by their adepts, despite their limited range of applicability. The third one is a strong lack of examples. Besides trivial ones, such as shifts of multiplication operators, very few examples of significant operators are known. Indeed, their construction is quite hard, but, precisely for this reason, it should become a kind of moral necessity for the specialits in the field, just to prove that the abstract results they present are not vacuous. A consequence of this lack of examples is that one never knows what is the range of application of the results which exist. A striking case is that of Lomonossov's Theorem, proving the existence of hyperinvariant subspaces for operators commuting with a compact one (see Chapter IV). This theorem was proved in 1974. Only in 1981 was given an example of an operator to which it does not apply. Meanwhile, dozens of papers had been published, dealing with "extensions" of Lomonossov's Theorem, and these extensions did not carry a single example showing they were not vacuous. The present status of Operator Theory may perhaps be compared to what Geometry of Banach Spaces was twenty years ago, before the creation of numerous examples of "pathological" Banach spaces, and the unified theories which followed. When, some years ago, I wrote a book about Geometry of Banach Spaces, the task did not seem too hard: the theory in itself had reached a nice degree of achievement, but no elementary text-book was available to present it.

Introduction

vii

The present situation is exactly opposite. First, as I already stated, the theory does not yet exist, and, second, the books to present it are already quite numerous. So, why should I add one more item to a list which is already too long? To such a good question, I can provide only a bad answer : I won't present things in the same manner. Indeed, we will not try to develop the achievements of the theory, we will show the gaps: this spirit is already clear from what we said. Our goal is to be, at the same time, an Introduction to Operator Theory, and a research book. In the present case, these words should not look incompatible. Let's explain them more in detail. Being an Introduction, the book is intended for students. It requires only a basic knowledge of classical Analysis: measure theory, analytic functions, Hilbert spaces, functional Analysis. The book is self-contained, except for a few technical tools, for which precise references are given. Part I takes things at their beginning: finite-dimensional spaces, general spectral theory. But very soon (Chapter III), new material is presented, leading to new directions for research, and open questions are mentioned. Part II concerns compactness and its applications: not only spectral theory for compact operators (with Invariant Subspaces, Lomonossov's Theorem), but also duality between the space of nuclear operators and the space of all operators on a Hilbert space, a result which is very seldom presented. Part III deals with analytic functions, and contains only few new developments. It seemed necessary to us to include HP spaces, Beurling's theory, etc, because they play a central role in Operator Theory. These results are of course presented in many well-known and well-written books, dealing with Fourier Analysis and Analytic functions. But, quite often, these books contain a general theory which is quite deep and goes far beyond what we need here. This is why we have chosen to present a simplified version, operator-oriented. But on this particular topic, what we present is obviously unsufficient : recent advances on the Invariant Subspace Problem seem to rely upon deep properties of analytic functions. Part IV contains Algebra Techniques: Gelfand's Theory, and applications to Normal Operators. Here again, directions for research are indicated. Part V presents dilations and extensions : Nagy-Foias dilation theory, and the author's work about C1-contractions.

viii

Introduction

Part VI deals with the Invariant Subspace Problem: first, positive results derived from S. Brown's methods. Then, we turn to counter-examples. After P. Enflo's counter- example of an operator with no Invariant Subspaces, further examples were found, by C. Read and the author. C. Read's example was later simplified by himself and then by A.M. Davie j this last version is now quite accessible, and is presented here for the first time. As can be seen from this brief description, a lot of new material is presented, and the level of research is reached, on the Invariant Subspace Problem, both on the positive and the negative directions. Each Chapter ends with some "Complements" which indicate further results, open problems, and give references. To help the reader, we have added exercises and an Index. We also felt free to indicate what are the directions of research which look promising, and, conversely, what are the ones which have, more or less, reached their limits. These opinions are of course strictly personal, but they are motivated by quite a few years of experience ! There are, of course, several aspects of Operator Theory which we do not approach (such as approximation by compact operators) : we had in mind the Invariant Subspace Problem. But we would be glad if the topics presented could attract new people to an area which is quite fascinating and challenging. The notation is rather customary, and is compatible with that of our previous books. A Hilbert space is denoted by H , a Banach space by E or X, an operator by T, and so on. However, in a few occasions, we found the existing terminology improper, confusing, and sometimes misleading. In such cases, we took the liberty to eliminate gothic letters, complicated symbols, in order to be left with the naked ideas. We do not think of the notation as a decoration, but as a mean of communicating the ideas, as simply as possible. A few words about the presentation and the typing. In the bad old days, it was customary to thank our secretaries for their typing, with hypocritical tears in the eyes (see my previous books). Things have changed, and, for once, in the good sense. This book was typeset by the author himself, using Plain TEX. Knuth's learn in its subtleties, but already an elementary knowledge of it allows (as one sees here) an output which is incomparably better than what anyone can produce on a typewriter. Most symbols, commands already

TEX is not easy to

Introduction

ix

exist in 1EX, and require no further work. The only symbol I had to create myself was" f+ ". I am glad to offer to the mathematical community the definition of such an important feature (with no increase in the price of the book) :

\def\notto{\hbox{$ - \rightarrow- \kern - 1.5em\hbox{/ }$ } } So, if you put this line in your "macro file" (list of definitions), whenever you type " \ notto " , you get

f+ .

Mathematical editing had decreased in typographical quality, many publishers presenting books which were not typeset, but only typewritten, and then reproduced photographically. Let's hope that 'lEX will open a new era. The editor I used was PC- Write, which is, in my opinion, the best of all I tried. It allows to redefine the keyboard (so as to produce, for example, greek letters with one keystroke), to store long sequences of words as "macros", and many other things. The book follows two courses I gave at the University of Paris VII, in 1984/85 and 86/87. The final version was mostly written during the fall 87, when I was Visiting Professor at the Ohio State University, Columbus. I found there nice working facilities, including a good library. Computer equipment, however, was quite unsatisfactory. Fortunately, I was invited, for short periods, by other Universities, where I could produce "hard versions" of my work. I acknowledge the warm hospitality of Kent State University, University of Columbia-Missouri, University of Illinois at Urbana-Champaign, which put at my disposition all the modern equipment I needed, such as computers with hard disks, laser printers, and so on. It is a very pleasant duty to mention all people who helped me, in various ways, to prepare the manuscript: Sylvie Guerre, Valerie Martel, Francoise Piquard, Earl Berkson, Gilles Godefroy, Elias Saab, Many others contributed to specific chapters and are mentioned at proper place. My interest in Operator Theory originally comes from conversations with Per Enflo, who had done in this area a pioneering work which was not fully understood, and from which I could benefit. I am glad to acknowledge an influence which is anyway so obvious that I could not hide it, even if I wanted to.

x

Introduction

As G. B. Shaw said, let me now withdraw and lift the curtain. If you find that there is not much behind, please consider at least that I did not paint it with artificial colors.

Paris, december 1987 Bernard Beauzamy

Table of Contents

Introduction Oceano Tex

....

1

PART I : GENERAL THEORY. Chapter I : Operators on Finite-Dimensional Spaces.

5

1. Spectrum of the operator.

5

2. Minimal Polynomial. 3. The analytic functional calculus. 4. Computing the operator norm on a Hilbert space. Exercises on Chapter I. Notes and Comments. Complements on Chapter I

6

17 17

Chapter II : Elementary Spectral Theory.

19

1. The spectrum of an operator.

21

2. The analytic functional calculus. 3. The Invariant Subspace Problem. Exercises on Chapter II. Notes and Comments. Complements on Chapter II.

10 14 15

27 32 35

38 39

Chapter III : The Orbits of a Linear Operator.

41

O. Introduction.

41

1. Basic Facts. A. The image of a baIl by a linear operator. B. Baire Property for operators. C. Rolewicz example of an operator with one hypercyclic point D. The regularity of the sequence (11Tn Il)n~o, 2. Operators having an orbit which tends to infinity. A. Exponential growths and related topics. B. Quasi-exponential growths.

43 43 44 45 47 48 48 53

Table

0/ Contents

C. Polynomial Growths. 3. How often can an orbit come back close to 0 ? 4. Operators with irregular orbits. 5. Hypercyclicity. Exercises on Chapter III. Notes and Comments

57 62 66

71 74

75

Part II : COMPACTNESS AND ITS APPLICATIONS. Chapter IV : Spectral Theory for compact operators.

79

1. Compact operators.

79

2. Spectral Theory for compact operators. 3. Spectral Theory for Normal Compact operators on Hilbert spaces. 4. Spectral decomposition for compact operators on Hilbert spaces. 5. Invariant Subspace for compact operators. 6. Commutation between compact operators and C 1 operators. Exercises on Chapter IV.

81 84 88 89 91 93

Notes and Comments. Complements on Chapter IV

94 95

Chapter V : Topologies on the space of operators.

97

1. Nuclear operators.

97

2. The space .c(H) as a dual space. Exercises on Chapter V.

101 115

Notes and Comments. Complements on Chapter V.

117 117

Part III : BANACH ALGEBRAS TECHNIQUES. Chapter VI : Banach Algebras.

123

1. Spectral Theory.

2. Ideals and Homomorphisms. 3. Commutative Banach Algebras.

123 126 128

4. Functional Calculus.

131

5. C· -algebras.

131

6. An Invariant Subspace Theorem.

134

Exercises on Chapter VI.

137

Notes and Comments.

140

Table 01 Contents Chapter VII : Normal Operators. 1. Algebra generated by a normal operator. 2. Spectral Measures. 3. Integration with respect to a spectral measure. 4. Spectral Representation of a Normal Operator. 5. The spectrum of a multiplication operator. 6. Hyperinvariant Subspaces for Normal Operators. Exercises on Chapter VII. Notes and Comments. Complements on Chapter VII.

xiii

141 141 143 144 145 153 154 156 157 157

Part IV : ANALYTIC FUNCTIONS. Chapter VIII: Banach Spaces of Analytic Functions.

163

1. Harmonic and subharmonic functions.

163

2. Basic facts about Fourier Series. 3. HP Spaces. 4. Jensen's Formula, Jensen's Inequality. 5. Factorization of HP functions : Inner and Outer functions. 6. Factorization of Inner Functions: Blaschke products and. . . 7. The Disk Algebra .A (D) . Exercises on Chapter VIII. Notes and Comments. Complements on Chapter VIII.

166 169 173 177 179 182 187 188 189

Chapter IX : The Multiplication by ei B on H2 (IT) and £2 (IT) .

191

1. The multiplication by ei B in H 2 .

192

2. Szego's Theorem. 3. Multiplication by eie on L 2 (IT ,d6/ 21r ) . 4. Multiplication by x on L 2[0, 1]. Exercises on Chapter IX. Notes and Comments. Complements on Chapter IX.

195 196

201 204

206 206

PART V : DILATIONS and EXTENSIONS. Chapter X : Minimal Dilation of a Contraction.

213

1. Construction of the isometric and unitary dilations.

213

Table of Contents

xiI'

2. Decompositions of the space )I. Exercises on Chapter X. Notes and Comments. Complements on Chapter X.

220 227 228 230

Chapter XI : The HCX) Functional Calculus.

233

1. Construction of the Calculus.

2. The Spectral Theorem for the HCX) functional calculus. 3. The HCX) functional calculus on the algebra generated ... Exercises on Chapter XI. Notes and Comments. Complements on Chapter XI.

233 239 242 247 248 248

Chapter XII : C.-Contractions.

249

1. The extension of a C. -contraction. 2. Representing Functions. 3. Functional Representation of Convergent Series. 4. Connections with Nagy-Foias Dilation Theory. 5. The Algebra .It (n) . 6. Two examples. 7. The spectrum of T and the spectrum of f. 8. Invariant subspaces for the C.-contractions. Exercises on Chapter XII. Notes and Comments. Complements on Chapter XII.

250 251 257 260 261 265 273 277 282 286 286

Part VI : INVARIANT SUBSPACES. Chapter XIII: Positive results.

291

1. When is the HCX) functional calculus isometric? 2. Invariant Subspaces. Exercises on Chapter XIII. Notes and Comments.

291 305

315 315

Chapter XIV : A Counter-Example to the Invariant Subspace Problem. 317 Exercises on Chapter XIV. Notes and Comments

340 344

Index References

347

351

Oceano Tex

Elle con temple I'ecran, tout vert dans les tenebres, Ou les chiffres s'aflichant jettent des lueurs d'algebre. Viendra-t-il, Ie chapitre inscrit dans les memoires ? o touches, que vous savez de lugubres histoires ! Combien de mots, de lignes, se sont evanouis, Dans l'aveugle machine a jamais engloutis ! II

Mais Ie menu s'afliche, rassurant, familier, Tiens, dit-elle en souriant, le voici tout entier.

III Sur la table, entasse; l'article et ses progres, Certain, dernontre presque, et probable a peu pres, Attend, bloc annote, qu'on veuille bien l'imprimer Symbole apres symbole, comme Sisyphe son rocher. Ratures, sens perdu, doute, feuillet manquant, Partout la question triple: -Comment? Ou ? Quand? IV

Mais Ie mot, qu'on Ie sache, est un etre vivant, La main de I'auteur vibre et tremble en le frappant. La pagination a 1'infini s 'echappe, A chaque instant lacune, embfiche, chausse-trappe. Faut-il mettre un backslash ou bien une cedille, Au bout du theorerne et de ses codicilles ? Regardons Ie manuel. Ces choses-la sont rudes, II faut pour les comprendre avoir fait ses etudes !

Oceano Tex

v La tache se termine, le dernier mot s'inserit. Elle appelle TEX. 11 vient, suppute, renifle, eerit, Reclame deux dollars avec une parenthese, Accentue, calcule en toute hypothese Le lieu mobile, obseur, capricieux, changeant, Oil se plait la virgule aux nageoires d'a.rgent. Oh cliquetis des phrases, tohu-bohu, rumeur, De I'univers des mots Ie terrible ecumeur ! VI

TEX termine son travail; apres tant de souffrance, Tant de labeur obtient enfin sa recompense. L'imprimante achevant lentement ses devoirs Eclaire tout a. eoup dans ses jambages noirs Le theoreme orne de tous ses corollaires, Eblouissant Shakespeare et ravissant Euler!

Victor Hugo p.e.c. Bernard Beauzamy

PART I

GENERAL THEORY

Nous imitons, horreur ! la toupie et la boule Dans leur valse et leurs bonds; merne dans nos sommeils La Curiosite nous tourmente et nous roule, Comme un Ange cruel qui fouette des soleils.

This First Part is devoted to a basic study of Operators in general. We start with a Chapter about finite-dimensional spaces, in order to emphasize later the differences with the infinite-dimensional case. We then turn to general Spectral Theory, and finally study the behavior of the iterates of an operator.

Chapter I

Operators on Finite-Dimensional Spaces

1. Spectrum of an operator.

Let E be a finite-dimensional, complex, normed space, and let dim E = d. Let also T be a linear operator from E into itself. First, we observe that T is automatically continuous: indeed, take an algebraic basis (el •...• ed) of E, and write x

=

E:

O:iei, y

=

E:

/3iej. Then:

d

IITx - Tyll

II L)O:j

-

s, )Tei II

~ K max 100i - /3i I ,

I

with K

=

E:

IIT

1

ei ll·

We define the operator norm of T :

IITllop =

sup

IITxll·

(1)

II%II~I

Since the unit ball is compact (see for instance B.B. [1]), and since T is continuous, this supremum is actually attained, and is a maximum. We will usually drop the subscript "op", and we just write

IITII.

We denote by £,(E) the set of linear continuous operators on E, equipped with this norm. When a basis (ell"" ed) is chosen, T can be represented by a matrix

M -- (t·&,J.)&,J .. = I "'0''''' ~ the

i -th column of which is made with the components of T'e, on the basis, that

is Tej = Ei ti,jei' If a vector x is given by its components X = (XI, ... , Xd) on the basis, the vector Tx has M X as components on the same basis. Of course, the matrix depends on the choice of the basis, the operator norm does not.

Chapter I

6

We are not going here to deal with the theory of matrices, which is a huge theory in itself ; we will instead restrict ourselves to the notions related to our future study of operators on infinite dimensional Banach spaces. The characteristic polynomial of T is the determinant of T - AI

CT(A)

=

det (T - AI),

for A E ~,

and this polynomial is also the determinant of the matrix M - AI, if M represents T is a given basis. It is easily seen to be independent from the choice of the basis. This determinant is of the form:

(2)

It has degree d , and therefore has d complex roots, not necessarily distinct. If A is one of them, det (T - AI) = 0, which means that T -- AI is not invertible. Since the space is finite dimensional, this is equivalent to the fact that T - AI is not injective (or not surjective). Consequently, there exists a vector z , x f:- 0, such that (T - AI)x = 0, or Tx = AX. The vector x is called an eigenvector associated with the eigenvalue A. To simplify our notation, we write T - A instead of T - AI. Let now AI, ... , Am be an enumeration of the roots (m ~ d). The set = {AI, ... , Am} is called the spectrum of T. It is a finite subset of the complex plane, consisting of at most d points. It is never empty, but may consist in a single point: the spectrum of the identity is {I}. The spectrum of any projection is {O, I}.

a(T)

Conversely, given any d points Ai, .. " Ad in the plane, it is easy to see that there is an operator T on ~d with a(T) = {AI, ... ,Ad}' Indeed, let (el,'" ,ed) be the canonical basis of «: d, and define Tej = Ajej. Therefore, the spectrum has a simple structure. This does not mean, however, that all questions in finite-dimensional operator theory are necessarily easy. Among the hardest (and, quite often, with no satisfactory solution), let's mention: precise computations of the operator norm, approximation of a given operator by operators in a given class, and so on.

Finite Dimensional Spaces

1

2. Minimal Polynomial. Let p(z) = L~

a/czk be a polynomial with complex coefficients. We de-

fine: N

(1)

p(T) = LakT/c o

We will show that there is a "smallest" polynomial m, such that m(T) = 0. This polynomial will be called the minimal polynomial of T. For this, let (el,.'" ed) be a basis of E. The vectors el, Tel, ... , Td ex cannot be independent: there exists a linear combination: d

L a;I)T j e

l

(2)

= 0,

o

that is, by definition (1), a polynomial 81, d081 ~ d, with 81 (T)el = The same way, we find polynomials 82,... , 8d, with:

8j(T)ej

for all j = 2, ... , d.

0,

=

o.

Put s = 81.82 ... 8d. Then s(T)ej = 0 for all j 8(T)x = 0 for all x in E.

=

1, ... , d, and therefore

We have found a polynomial s, such that 8(T) = O. But 8 has to be reduced: its degree may be d 2 , and we will see later that such a polynomial exists with dO s ~ d. We factor 8(A) = a n~1 (A - Ai)Oi , and we may assume a = 1. If a point Ai is not in the spectrum, the corresponding term T - >'i I is invertible, so can be removed from s . Let

u(>.)

=

a

II

(>. - Ada'.

>'iEu(T)

Now, we look at all x's such that (T - >'dlllX = o. If for all of them we also have (T - >'x) 0 l - I X = 0, we replace QI by QI - 1 in u. We start again with al - 1, and so on, until we cannot proceed further. We then pass to Q2, and so on. More precisely, we define, for i = 1, ... ,m (where m is the number of points in a(T), m ~ d) :

Vi = inf {k E IN ; (T - >..)k x = 0, for all x s.t. (T - Ai)k+ 1 x = O}. Then we put: m

m(>.)

II (A - >'ir', i=1

Chapter I

8

and we have obtained a polynomial m, still satisfying m(T) = O. By construction, all roots Ai of m(A) belong to n(T). Conversely, let AO E n(T), Yo a corresponding eigenvector. Then:

o = m(T)yo

=

m(Ao)Yo,

so m(Ao) = 0, and the roots of m(A) are exactly the points of n(T). The polynomial m(T) is called the minimal polynomial of T, for the following reason :

Proposition 2.1. - The polynomial m divides every polynomial p such that p(T) = O.

Proof. - Let p be such a polynomial. First, as we already observed, p must have the points in n(T) as roots; we eliminate the others, and write: m

II (A -

p(A)

Ai)Qi,

(3)

1=1

We now show that ai 2' Vi, for i = 1, ... , m. Assume on the contrary that, for instance, at < Vt. Then, by the definition of Vt, there is a point Xl such that:

0,

Yt

From (3), we deduce

with q(At)

f-

O. We have TYt

=

AtYt, so :

which contradicts p(T) = 0 and proves our claim.

Corollary 2.2. - If p , q are two polynomials, we have p(T) only if ever)' Ai E n(T) is a root of p - q of order 2' Vi.

= q(T)

Indeed, this says that p - q is divisible by the minimal polynomial.

if and


9

Among all polynomials satisfying p(T) = 0, the most noticeable one is the characteristic polynomial, which we have already defined:

c('\) = det (T - ,\). Theorem 2.3 (Cayley-Hamilton). - The characteristic polynomial satisfies

c(T) = O. Proof. - Elementary linear algebra (see for instance LN. Herstein [1)) allows us to write in a proper basis, the matrix of T in a triangular form:

where '\~, ... ,'\~ are the points of u(T), but this time each of them repeated according to its multiplicity. If (VI,." ,Vd) is this basis, we get: TVI = '\;VI, TV2 =

al,2vI

+ '\~V2'

which means:

(T - '\;)VI = 0, (T - '\~)V2

= al,2vI,

Therefore (T-'\~)VI = 0, (T-'\~)(T-'\~)V2 = 0, ... , (T-'\'l)'" (T-'\d)Vd = O. So the product (T - '\'.) ... (T - '\d) annihilates all the vectors of the basis. Since the ,\i's are the roots of the characteristic polynomial, we get c(T) = O. From Theorem 2.3 and Proposition 2.2 follows obviously that dam ~ d. There are obvious examples in which the minimal polynomial has degree strictly less than the degree of the characteristic polynomial (which is d, the dimension of the space). For example, a projection always satisfies T2 - T = 0, that is

m(,\) =

,\2 - '\.

10

Chapter I

3. The analytic functional calculus. We have already defined p(T), when p is a polynomial. We now extend this definition to a larger class of functions. Let I be a function from ~ into itself. We say that I belongs to the space 7(T) if there exists a neighborhood V of u(T) on which I is analytic (for an elementary theory of analytic functions, we refer the reader to H. Cartan [1]). We recall that I is said to be analytic on a compact set if it is analytic on some neighborhood of this compact set. The neighborhood does not need to be connected, and depends on the function. Let f E 7(T). For every Ai E u(T), we consider the derivatives f(k)(Ad, k < Vi (there are VI + ... + V m = d such derivatives). Let p be a polynomial such that fUr:) (Ail all k < vi. We then put:

= p(k)(Ai),

for all Ai E o(T),

f(T) = p(T) This definition does not depend on the choice of the polynomial p : if q another one with the same properties, then p(T) = q(T), by Corollary 2.2.

IS

We now list some elementary properties of this definition:

Theorem 3.1. - If a) of

+ (3g

I, g E 7(T),

E 7(T), and (01

0,

(3 E ~ ,

+ (3g)(T)

=

o/(T) + fJg(T),

b) f.g E J"(T) , and f.g(T) = I(T)g(T) , c) if f(A) = E~ OkAk, then f(T) = E~ ok T k, d) I(T) = 0

u and only if f(k) (Ai) =

0, for all Ai E u(T) , all k < Vi.

The proof is left to the reader. The first quality of the class l(T) is that it contains functions with values o and 1 only, thus allowing us to build projections which commute with T : For every Ai E u(T), let fi('x) defined by: fi('x) = 1 on some neighborhood of Ai, = 0 on some neighborhood of all other Aj'S, j 1= i. We put E, = e, (T) ; this is an operator, with the following properties:

Proposition 3.2. - For i,j

a)

= 1, ... , m,

we have:

E; = Ei ,

b) EiEj = 0, c) I = E~

for

i 1=

i,

s..

These properties follow immediately from Theorem 3.1.

11


For i = 1, ... , m, we call Xi the image of the operator Ei. Then we get the formula: Indeed, by Proposition 3.2, c), every z can be written as x = L~ EiX, and this decomposition is unique by b). Since E; is in fact defined as a polynomial in T, it clearly commutes with it. Therefore: i = 1, . . . ,m, which means that Xi is invariant by T. Also, we have:

(1) Indeed, E;

=

pi(T), where Pi is a polynomial satisfying: 1~ k

Vj

f- i,

.;)e;(>.)

i= I k=O

One checks immediately that, for i

=

1, ... , m, and k
O be a sequence of functions in :1(T). Then In(T) converges in operator nor~ if and only if the d sequences (/A k ) ('\i))nE IN I for i = 1, ... ,m, k < Vi, converge in ([: . Proof. - 1) Assume that the d complex sequences converge. Then Proposition 3.5 indicates that In(T) is Cauchy, and therefore converges. 2) Assume that the sequence (fn(T))nEIN converges in f.(E). We know that (T - ,\')VI - I E I fo 0, and therefore we may find z such that (T ,\.)Vl- 1 Elx

fo

0. Let y

= Elx,

Yk

= (T

- .Adky, k

This Corollary implies obviously that II R(A) II --+ 00, when d(A) --+ O. Therefore, p(T) is the domain of analyticity of R(.\) : this function cannot be extended analytically on a larger domain.

Chapter II Proposition 1.4. - The spectrum a(T) is non-empty, compact, and contained in the disk {..\ ; 1..\1 < IITII}. Proof. - We set g(..\) = Lk~O Tk /..\k+l. For 1..\1> IITII, this series converges in operator norm, and (..\ - T)g(..\) = I, so g(..\) = R(..\), for 1..\1 > IITII. Assume that the spectrum is empty. Then g(..\) is analytic in the whole plane. But it satisfies Ifg(..\)11 --+ 0, when 1..\1 --+ 00, so, by the maximum modulus principle, we must have g(..\) = 0 for all ..\ E IITIf.

(2)

Ie~O

We put

=

r(T)

max{I..\I; ..\ E a(T)} ,

and r(T) is called the spectral radius of T. The following Proposition gives a way of computing r(T) and extends formula (2) :

Proposition 1.5. - We have: a) r(T) = Iimle---+oo IITk II Ilk

b) if 1..\1 > r(T), R(..\) norm.

< IITII,

= Lk~O

T k / ..\k+ I , this series converging in operator

Proof. - The radius of convergence of the Laurent series Lk~O

Tk / >.k+ I is

I = lim sup IITkllll k. k---+oo

So r(T) = I. All we have to show therefore is that r(T) ~ lim infk---+oo IITIe II Ilk. For this, observe that Tk _..\Ie = (T - ..\)Pk(T), for some polynomial PI.. Therefore, if ..\ E o(T), ..\Ie E a(TIe), for every k ? 0 ; so 1>.11e ~ IITlell, and 1>'1 :S lim infk---+oo IITkllllk. This proves the existence of the limit limk---+oo IITklll/k, and establishes our formula.

Proposition 1.6 (Resolvent Equation). - For ..\,p. E p(T), we have:

R(..\) - R(p.)

= (p. - >.).R(..\).R(p.).

Proof. - We have:

(p. - T)(..\ - T)(R(..\) - R(p.)) from which the result follows.

(p. - ..\) I,

Elementary Spectral Thwry

29

We also recall :

Proposition 1.7. - If

IITIl
0, small enough, the series

c, IAI ~ r(T) + e}. f(T)

=

=

So, if Cr+e

-~

1C

~

f=

2Z1r

=

f=

{

2Z1r

2~"

0

r H

akAk converges uniformly in the is the circle {A E a::; IAI = r(T) + e} , L~

ak Ak R(A)dA

0

ak (

f;ak I:Ti o

Ak R(A)dA

lCr+r

j~O

1

>.k-i-1d>.

c r +.

d) is a consequence of Proposition 1.9. A similar proposition holds for adjoints in a Hilbert space:

Elementary Spectral Theory

Proposition 2.2. - Let T be an operator on a Hilbert space H. For f E l(T), we define:

f(T)· = fd(T·). Proof. - Clearly, fd is analytic on some neighborhood of u(T·) Then from formula (1) follows:

f(Tt

=

Jr

-.1 { f(>.)R(X,T·)d>'

by Proposition 1.9,

Jr

2171"

=

conj(u(T)).

-.1 ( f(>.) R(>., Tr d>' 2171"

=

=

~

J(r

2t7l"

f(X)R(>', T·)d>.

= fU(T·). We will now investigate how the spectrum is transformed by a function of the operator. This question is of course of fundamental importance whenever a functional calculus is defined. In the present case, the result is quite satisfactory, and is described by the next theorem:

Theorem 2.3 (Spectral Mapping Theorem). - If f E l(T),

f(o(T)) = u(f(T)), where f(u(T))

=

{f(>.) ;

x E u(T)}.

Proof. - Let >. E u(T) and define 9 by

f(>.) - f( E)

>.-€

(3)

Then gEl (T), and satisfies :

f(>.) - f( E) Therefore:

(>. - T)g(T)

=

f(>.) - f(T)

If f(>.) - f(T) was invertible (that is fP.) f/: o(J(T))), let A be its inverse. Then A.g(T) is the inverse of >. - T, so >. ~ u(T). This contradicts our original assumption, so f(>') E o(f(T)) , and f(o(T)) C o(f(T)). Conversely, let's assume that IJ tI. f(u(T)). Then the function h(€) = 1/(f(€) - IJ) is in l(T) and satisfies h(€)(f(€) - IJ) = 1. Therefore:

and IJ

tI.

h(T)(f(T) -IJ)

=

I,

u(f(T)). This proves that f(o(T))

:J

o(f(T)).

90

Chapter II

We now study the composition of functions.

Proposition 2.4. - Let I E l(T), g E l(/(T)), and set h h E l(T), and h(T) = g(/(T)) .

=

g

0

I.

Then

Proof. - From the previous theorem, we deduce that h E l(T). Let U be a neighborhood of O'(/(T)) 1 V of O'(T) , as previously, such that g is analytic on U, and I is analytic on V, with I(V) cU. Let's consider the operator:

A(A)

= ~ 217r

If A ~ I(O'(T)), (A - I(~))-l

( R(~, T) d~. Jay A - I(e)

is analytic on O'(T), and A(A) = (A - f)-I(T).

So:

(A - f(T))A(A)

I

~

{

g(A)R(A, f(T))dA

41r

Jau Jay

1

and A

R(A, f(T)) .

=

=

Therefore,

g(f(T)) =

Jau ~ f ( 2 2111"

= =

~

(

2171"

Jay

g(A) R(~,

T) d~dA

A - f(~)

g(f( ~))R(

~, T)d~

=

h(T)

Proposition 2.5. - Let (fn)n~o be a sequence of functions, in l(T), all analytic on the same neighborhood V of O'(T). If (fn)n~o converges to f, uniformly on V, then In(T) converges to I(T) in operator norm. Proof. - Let U be a neighborhood of O'(T) , with In - f uniformly on r, and:

i

Ifn(A) - f(A)I·IIR(A)lldA

Therefore, IIfn(T) - f(T)11

-+

-+

0,

Uc

V, and

when n

r = au.

Then

-+ 00.

o.

Definition. - A point AD E O'(T) is said to be an isolated point of O'(T) if there exists a neighborhood U of A such that Un O'(T) = {AD}. An isolated point Ao of O'(T) is a pole of T if the function A -+ R(A, T) has a pole at Ao. The order of Ao is denoted by IIT(Ao) and is called the order of the pole.

91


Theorem 2.6 (Minimal Equation). - Let f E l(T). Then f(T) = 0 if and only if f is identically 0 on a neighborhood of u(T) , except maybe at a finite n umber of poles AI,' .. , Am , such that for all j = 1, ... , m I f has a zero of order ~ tiT (Ai ) at Ai. Proof. - Assume first that f has the described properties. Let Cj be a small circle around Ai, positively oriented (j = 1, ... , m). Then :

f(T) =

~

f

2111'".

]=1

f f(A)R(A)dA. l., J

Bu t, at Ai

» R (A) has a pole of order tiT (A i)' and f has a zero of order So f(>..)R(A) is analytic on the disk limited by Ci, and f(T) = o. Assume now that f(T) = O. By Theorem 2.3, f(o(T)) = o. Let U be a neighborhood of o(T) on which f is analytic. By a well-known theorem about the zeroes of an analytic function (see H. Cartan [1]), f is identically zero on some neighborhood of any point of o(T) which is not isolated in o(T). Indeed, if A is such a point, if V is a disk around A, with V c U, V contains a sequence of points of o(T) converging to A, and f is identically 0 on V. Let now A E u(T), an isolated point: there is an open disk V c U, with o(T) n V = {A}. We assume that f is not identically 0 on V. We have f(A) = 0 , and so f has a zero of order n at A. So the function gl(e) = (A - e)" / f(e) is analytic on V. Let e be a function identically equal to Ion V and to 0 on a neighborhood of O(T)\{A}. Let g = gl.f Then:

~ tiT (Ai).

(A - T)"e(T) = f(T)g(T)

=

O.

The Laurent series of R( e) on V can be written : +00

R(E) =

L Am(A - elm

(1)

-00

where

where C is a circle around A, positively oriented. So A_ m + 1 = -(A - T)me(T) = 0, for m ~ n. Then (1) implies that A is a pole of order ~ n, and the Theorem is proved.

32

Chapter II

We recall that an operator T' is said to commute with T if TT' = T'T. The set of operators which commute with T is called the commutant of T and denoted by {T}'.

Proposition 2.1. - Let T' be an operator which commutes with T and let f E 1"(T). Then T' commutes with f(T). Proof. - By the definition of f(T), it's enough to show that T' commutes with R('\, T), for ,\ E p(T). But this follows from the two equations: (,\ - T)R('\)T'

T'

(,\ - T) T' R ( x)

T' ,

since (,\ - T) is invertible, ,\ E p(T) . 3. The Invariant Subspace Problem.

Let T be an operator on a Banach space E. We say that a subspace F of E is invariant by T if T F c F. Does there exist a closed linear subspace F, invariant by T, besides the trivial ones: F = E and F = {o} ? (In old mathematical papers, the word "stable" was used instead of "invariant". It seems better, but I take the modern terminology.) This question arises naturally from the theory of eigenvectors in finite dimensional spaces: if x is an eigenvector, of course F = {'\x ; ,\ E ~} is such a subspace. But we have seen an easy example of an operator on 12 (1l) with no eigenvalue. So, some other concept has to be substituted for it, and the concept of non-trivial invariant subspace is the broadest and the most natural one (see, however, the concept of Invariant Convex Subset, introduced in Exercise 10, at the end of this Chapter). A stronger concept is that of hyperinvariant subspace: this is a subspace which is also invariant by all operators which commute with T. IT we fix a point x in E and consider the subspace:

we obtain a closed subspace, invariant by T. Such a subspace will be called elementary invariant subspace. If F% is the whole space, the point x is said to be cydic, it is non-cyclic otherwise. Of course, any Invariant Subspace contains an elementary Invariant Subspace, and, in fact, is just the reunion of all the elementary Invariant Subspace that it contains.


99

A similar concept occurs for the Hyperinvariant Subspaces : instead of just taking the iterates of T, we take all operators commuting with T, and define: G % = span {Ax; A commutes with T} Then G z is an elementary hyperinvariant subspace. In the sequel, we will be looking for non-trivial Invariant Subspaces or Hyperinvariant Subspaces. We will often omit the words "non-trivial", and abbreviate the four words in "Invariant Subspace" or "Hyperinvariant Subspace". The Invariant Subspace problem is still open in Hilbert spaces. Its existence is the background for the whole book, and there is not a single chapter which does not contain a part devoted to it. Let's come back to spectra, and justify the claim we made at the end of Section 1. If up(T) is not empty, T has eigenvectors, and so Invariant Subspaces, as we already said. If ur(T) is not empty, the set:

F

=

Imp. - T)

is non-trivial, and is invariant under T. This proves our assertion: if one is looking for Invariant Subspaces, one may assume u(T) = ua(T) \ up(T). One can moreover, in a reflexive space, assume that any sequence of almost eigenvectors converges weakly to 0 (Proposition 1.13). If the spectrum u(T) is not connected, the analytic functional calculus allows us to exhibit Invariant Subspaces :

Theorem 3.1 (F. Riess}, -If u(T) = Ul UU2, where Ul and U2 are disjoint closed sets, then T has Hyperinvariant Subspaces F) and F2 ; moreover:

where TIF denotes the restriction of T to the subspace F.

Proof. - Let U 1 , U 2 be disjoint open sets, containing Ul and U2 respectively. Let el = 1 in U 1 , 0 in U2 , and e2 = 0 in U 1 , 1 in U2 . Then el and e2 are in J"(T), and: edT) + e2(T) = I e~(T)

=

el(T) ,

e~(T)

=

e2(T).

So edT) and e2(T) are projections, which commute with T. Their ranges F) and F 2 are the required subspaces : They are closed, non-trivial, and hyperinvariant, by Proposition 2.7.

Chapter 1I To prove the second statement, we observe that, on F 1 , T = TeI{T). So U(Ttl (T)) = g(u(T)) , where g(A) = Atl (A), by Theorem 2.3. But g(u(T)) = Ul. This proves the Theorem.

Remark. - In the above Theorem, under the assumption that the spectrum is not connected, we exhibit two functions el, e2, such that el (T)e2(T) = 0, and none of the operators e 1 (T) or e2 (T) is o. One should not think that the assumption is, by any means, necessary for the conclusion. Indeed, for the right shift S on l2(71), if we define ad8) = l[o,1I'J(8), a2(8) = 1111',211'J(8) and set el = t ia., e2 = eia~, we have eI{S) =I- 0, e2(S) =I- 0 (as we will see in Chapter VII), and eI{S)e2(S) = o. But in this case, e.(S) and t2(S) are defined by other means than the Riesz's functional calculus (see Chapter VII). So the assumption of non-connectedness is linked with the tools we are using, and not with the conclusion.


95

Exercises on Chapter II. Exercise 1. - Determine itself:

0p,

Or,

/(t) -

0D.

for the operator T from C[O,I) into

g(8) =

fall /(t)dt

Same question if T is considered as an operator on L.[O, 1].

Exercise 2. - Determine itself:

0p,

/(t)

Or,

0D.

for the operator T from C[O,I) into

g(t) = t/(t)

--+

Same question if T is considered as an operator on L 1 [0, 1) .

Exercise 3. - Show that there exists a polynomial P such that P(T) = 0 if and only if the spectrum of T consists in a finite number of eigenvalues. Exercise 4. - Assume that IITII ~ r' < r , and that / is analytic on the closed disk D(O, r). If moreover I/(A) I ~ R for every A with IAI = r, show that:

II/(T)II ~

rR " r-r Exercise 5. - Let T be an operator, C a set contained in p(T). Set d = dist(C,o(T)). Show that there exists a K ;::: 1 such that: n ;::: 0, A E C.

Exercise 6: Upper semi-continuity of the spectrum. - If An is a family of sets, we define: lim sup An

=

nkUn>kAk.

Let cP be the function on £(E) : 4>(T) = o(T), with values into the subsets of Prove that 4> is upper semi- continuous, that is :

(C.

if

t;

--+

T,

Iimsupo(Tn ) C o(T).

[Hint: Take A E lim sup o(Tn ) . Show first that there exists a sequence An" E o(Tn ,, ) such that AnA: --+ A, then use Cor. 1.8.]

Exercise 7 : The spectrum is not continuous. - Let Sk be the weighted shift on 12 (Z) defined by : Slef n

= SlefD

f

if n

n+l

=

t= °

1

k"e n + 1 '

1) Prove that O(Sk) is the unit circle (follow the proof of Proposition 1.13). 2) Prove that the sequence Sk converges to the operator A defined by Af n = f n + l (n:j:. 0), Aeo = 0. 3) Prove that o(A) is the closed unit disk.

Chapter II

96

Exercise 8. - Assume that the spectrum of T does not intersect the open disk D(O,a). Show that:

and that, inside this disk, the resolvent is given by the formula:

L ,XRT-(n+l)

R('x) = -

n~O

Exercise 9. - Let T be an operator such that that the sequence ('L,~ T k ) N is bounded.

IITII

~ 1 and 1 ~

u(T). Show

Exercise 10. - Let T be an operator for which there exists M such that: n2:1

Show that for every

T,

0
0 and a C > 1 with :

Show that: . f -anI·lmln n-oo an+l

1 < -. -

C

b) Let T be an invertible operator on a Banach space E. Assume that there exists C > 1 such that for all x =I- 0 in E, there is a ()(x) > 0 with :

Show that, for every e > 0, there is a n such that :

c) Deduce that:

lle i 8 -

Til > 1 ~

~

.


97

Exercise 12. - Let T be an operator on a reflexive Banach space E, with

IITII = 1

and 1 E u(T).

a) Show that, for any A, JJ. such that 1 + A E p(T), 1 + JJ. E p(T) ,

AR(A

+ 1)(1 - JJ.R(JJ. + 1))

b) Let A Em, 0

< A~

=

AR(JJ. + 1)(1 - AR(A

+ 1))

1. Sho . that:

IIAR(.\ + 1)11 s

1.

c) Fix x E E. Let (An)n~O be a sequence converging to 0, with 0 < An < 1, such that AnR(A n + l)x converges weakly to some point y. Let JJ., 0 < JJ. ~ 1. Let z be the weak limit of some subsequence of IJ.R(JJ. + I)A nR(An + l)x. Using a), show that z = y. Deduce that :

d) Deduce that Ty = y (write T = -((IJ. + 1) - T)

+ (IJ. + 1)1).

e) Let

x

=

{l'

E E; lim JJ.R(IJ.

+ l)l' exists for the norm}

~-+O

Show that, for every A > 0,

IJ.R(IJ. + l)(x + AR(A + l)x)

--+

0,

(use a), exchanging the roles of A and IJ.). Deduce that

x + AR(A + l)x

E

X

Deduce that x + y EX, and then that x EX. f) Deduce that, if 1 is not an eigenvalue of T, for all x E E,

AR(A + l)x

--+

0,

A --+

o.

Exercise 13 : Closed Invariant Convex Subset of the Unit Ball. - We investigate the following question: if T is a non-zero contraction (that is IITII ~ 1) on a Banach space, does there exist a closed convex subset C of the unit ball BE, with C t- {O}, C t- BE, and TC c C ? This problem is weaker than the Invariant Subspace problem, but obviously related. Surprinsinglyenough,

98

Chapter II

it has never been considered. The answer is "yes", and is simple to obtain. We give two proofs. The first one is constructive, and is given now. The second one is posponed until Chapter XIV. a) Solve the problem when T is an isometry, and when r(T) < 1. b) Assume that T is not an isometry, and that r(T) = 1. Take a vector 1, such that:

xo, Ilxoll =

lim IITnxolJ < 1

n-oo

Consider:

Show that C is convex, closed, invariant under T and not reduced to {O}. c) Show that C is not the whole unit ball (consider a ..\, 1..\1 = 1, in the spectrum of T, and take a sequence (Xk)k~O of almost eigenvectors for ..\). Observe that C is also different from any ..\8 E , for ..\ E ct, 1..\1 = 1.

Notes and Comments.

Most results in Section 1 can be found in any book dealing with spectral theory j let's mention Dunford-Schwartz [I], vol. I, and H.R. Dowson [I], among others. Proposition 1.14 was pointed out to us by V. Martel. The analytic functional calculus presented in Section 2 is due to F. Riesz. For both sections, our presentation follows Dunford-Schwartz, with minor changes. For weighted shifts and their spectra, an excellent reference is the survey paper by A. Shields [I]. For exercises on Hilbert spaces, the best source is the book by P. Halmos [21. Exercises 6 and 7 come from there. Exercise 12 follows a nice work of R. Emilion [11. Exercise 13 is due to the author.

99

Elementary Spectral Theory Complements on Chapter II.

There are links between the growth of the sequence liT-nil, n ~ +00 (assuming T to be invertible), and the growth of the resolvent R(A, T), when ,\ approaches the spectrum of T. As an example, we mention the following result, due to A. Atzmon [1] : Proposition. - Let T be an operator on a Banach space E, with u(T) C C. If:

a) There exists c > 0, a,O < a < 1, such that:

then: b)

IIR('\, T) II with f3

a

=

d

=

O(exp( (1 _ 1,\l)p)) ,

1,\1

-4

1-.

= ~ d = 4 PcCt./P. l-Ct. '

Conversely, if b) holds with some constants d > 0, f3 > 0, a) holds with --.f!..- c = 3dCt./P l+P'

Techniques connected with the growth of the resolvent will be used in Part VI, Chapter I, in order to obtain invariant subspaces. Techniques connected with the growth of the backwards iterates will be used in Part V, Chapter III, for the same purpose.

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Chapter III

The Orbits of a Linear Operator

O. Introduction.

The tools most commonly used in Operator Theory are functional calculuses, of which we have already seen an example in the previous Chapter (further ones will be given later). Unfortunately, in order to be applied and produce, for instance, Invariant Subspaces, they require either some special configuration of the spectrum, or the operator to be a contraction (that is IITII ~ 1), or both. Therefore, they can handle only special situations. We introduce now a new concept: the study of the orbits of a linear operator. If y is a point in a Hilbert space H, the orbit of y under the operator T is just the set of the iterates {Tn y ; n E IN}. We do not study each orbit separately. What we do is to find in the space a point y for which the behaviour of the orbit is as good as possible. Indeed, among all points in the space, some may have "regular" orbits, some may have "irregular orbits". By "regular", we mean for instance that the orbit may tend to infinity, or may tend to zero, or may stay inside a ball, or may stay outside a ball. By "irregular" orbits, we mean conversely those which go "up" and "down", that is which satisfy sUPn IITny 11 = +00, inf'., IITnyl1 = O. More irregular orbits are those of hypercyclic points, that is : such an orbit is dense in the whole space. This point of view casts a new light upon several of the deepest questions in Operator Theory j most of them, indeed, can be rephrased as properties of a single orbit. For instance, the Invariant Subspace Problem says: find a point y -# 0, such that the closed linear space spanned by the orbit of y is not the whole space. Indeed, if one is trying to solve the Invariant Subspace Problem in a Hilbert space in the positive direction, that is, to prove that any operator has a nontrivial invariant subspace, a first -and necessary- step in this direction is to prove that for every operator there is a point y, such that its orbit is not

Chapter III dense in the whole space: if already the orbit is dense, so is the linear space it generates! Unfortunately, already this simple question (simple to state) is unsolved in a Hilbert space. In the space iI, conversely, C. Read [5] has constructed an example of an operator such that every point (except of course 0) has a dense orbit. Before that, and using the ideas introduced by P. Enflo to solve the Invariant Subspaces problem in Banach Spaces (see Chapter XIV for a discussion of this question), we could build an operator with a property, called supercyclicity, and very close to hypercyclicity : for every point y, the rays spanned by the orbit (that is {CTn y ; n E IN, C > o} ) are dense in the whole space [9]. The present chapter is mostly concerned by the Hilbert space setting. There, very little is known. The first example of an operator with one hypercyclic point was constructed by S. Rolewicz III (1969). Then, P. Halmos asked the question whether there are operators with a vector space of hypercyclic points [21. This question was solved affirmatively by the author (B. Beauzamy

[121). The study of the regularity of orbits is interesting in itself, independently of the Invariant Subspace Problem. But it is far from being straightforward. Rolewicz's basic example [1] (which is presented here in a slightly improved version) explains why. The operator is very simple (a weighted shift on '2(71)), but there are three different dense sets in the space: - a dense set of hypercyclic vectors: for each of them, the orbit goes everywhere in the space, and thus is extremely irregular, - a dense set of points, the iterates of which tend strongly to 0, - a dense set of points, the iterates of which tend to infinity, exponentially fast. Therefore, to find a point with a regular orbit is not an easy task. But, clearly, if one reduces the operator to a contraction, as Operator Theorists usually do, the problem (and the concept) will completely disappear: all orbits are now restricted to the unit ball, and one looses all information about their behaviour: Did they tend to infinity? Did they oscillate? This explains why the questions we are investigating here (though of a very basic nature) have been overlooked by the specialists in the field. We try here to bring some answers to these fundamental questions, and present positive results. But also, to show the limits of these positive results we present two examples: one of an operator with IITn I --+ +00, but such

Orbits of a Linear Operator

that all orbits come back arbitrarily close to zero, one of an operator such that all orbits oscillate, that is, for all y, sup IIT"YII

= +00, and inf IIT"YII = o.

As we said, the concept we study here is new, and so are the tools we introduce for its study. They refer to a basic understanding of Operator Theory. It's quite surprising that simple results, like Baire Property for Operators (see Section 1, below), though totally elementary, appear here for the first time. These tools are presented in Section 1. The present chapter is organized as follows: - in Section 1, we present Rolewicz's example, together with other basic facts, dealing with the shape of the iterates of a ball under a linear operator. - in Section 2, we give conditions for an operator, in order to have an orbit tending to infinity. These conditions are (of course) upon the growth of the sequence (liT" 11)"~o. For example, we prove that an operator with spectral radius strictly larger than 1 must have a dense set of points, the iterates of which tend to infinity exponentially fast. We prove also that the condition

L I/IIT"II < +00 ensures that

there are orbits tending to infinity. - Section 3 deals with the following question: How often can an orbit come close to 0 ? - Section 4 gives various examples of strange orbits. We show that the condition liT" II ----+ +00 is not sufficient to ensure that there is a point, the orbit of which tends to infinity. We also construct an operator such that for all y, sup IIT"YII = +00, and inf IITR y l1 = o. - Section 5 studies Hypercyclicity. We describe the existence of an hyper-

cyclic point as related to the properties of balls. We give a sufficient condition ensuring that there is a point which is not hypercyclic. 1. Basic Facts.

A. The image of a ball by a linear operator. An important tool for our study will be the consideration of the shape and position of successive iterates (T" B}">o , for a fixed ball B. Indeed, this comes from the following elementary consideration: Our aim is to predict the behaviour of sequences (T"x}">o, and then to find a point x for which this behaviour is as satisfactory as possible. This behaviour, obviously, cannot be determined from the consideration of a finite number of points. On the contrary, if you look at the images of a ball, the operator is totally described by one image of a single ball, no matter how small it is : linearity and continuity

Chapter III make it possible to reconstruct the picture in the whole space. Therefore, it is important to understand well the shape of the images of a ball. Let B be any closed ball in a Banach space. The image T B is a convex set, which is closed if, for instance, the space is reflexive (or the operator T weakly compact). Indeed, T B is then a weakly compact set. This is the case in a Hilbert space. The inverse image T-l B is always a closed convex set. It may be bounded or not. It is bounded if and only if the operator T-l exists as a continuous operator. Let b be the center of B, Tb its image. The image T B is balanced with respect to Tb : Tb + ..\Z is in TB when Tb + z is in TB, for all ..\ E (t, with 1..\1 = 1. The largest distance IITz - Tbll in TB is TIITII, where T is the radius of B. This distance may not be exactly realized, since (even on a Hilbert space) there are operators which do not attain their norm (that is : for which there is a point z, IIxii = 1, with IITxll = sequence of points of norm 1 such that IITxnll -+ SUPn IITb - T(b + rxn)ll.

IITII). But if (x n )n20 is a IITII, this distance d m a x is

The smallest distance IITz - Tbll in TB is T/IIT-lll ; it may be 0 if T- l does not exist as a continuous operator (then we set liT-III = +00). So in this case, the image of a ball has empty interior. Otherwise, it contains a ball. The above considerations apply to T-l B as well. We will always assume T to be injective. So the largest distance inside T-l B , starting from T- l b, is

rIIT-lll, which may be

infinite, and the smallest is

r/IITII.

B. Baire Property for operators. Surprisingly enough, Baire Property is always given for spaces, and not for linear operators, though it seems to be its natural setting.

Proposition l.B.l. - Let T be an operator on a Banach space E, with dense range. Let (On)n2 0 be a countable family of dense open sets. Then the intersection t-o; is dense.

n

Proof. - Let y E E, e > O. Since 0 0 is dense, there exists Uo E 00 with Iluo - yll < £/2. Let B o be an open ball, centered at UQ, with radius TO < £/4, and small enough to be contained in 0 0 • The set TO I is dense; therefore we may find that

IITul - uoll < ro/IITII.

Ul

E 0

We take a ball Bi , centered at

1

nT-I Bo such

Ul,

with radius


< e / 4 2 , and contained in 0 1 n T-l Bo . Continuing inductively, we construct Uk in O» n T-l Bk-l, a ball Bk centered at Uk, contained in Ok n T-l Bk-l, with radius Tk < c/4 k + 1 , such that IITuk - uk-III < rk/IITllk. So we get, for '1

k,j> 0 :

and therefore, for any fixed k , the sequence (Ti uk+i)i~o is a Cauchy sequence in E, which converges to a point Yk E Bk n Ok. Obviously, Tn Yk = Yk-n, for all n, k with n ~ k . So Yo has an infinite chain of backward iterates. Moreover, by construction, Ily - Yoll < e , and Yk E O«, so tro, contains Yo for all k.

Corollary 1.B.2. - If one takes T = Id, one gets the usual Baire property. Corollary 1.B.3. - Let T be an operator with dense range. Then, there is a dense set of points z which have an infinite chain of backward iterates, that is : for all n > 0, there is Yn such that Tn Yn = z , This follows from Proposition 1.B.1 if we take On = E for every n. Corollary 1.B.3 was obtained by J. Esterle [1], who deduced it from a theorem of Mittag-Leffler. It was observed by Chris Lennard that linearity does not play any role here j Proposition 1.B.1. is true for a sequence of complete metric spaces En and a sequence of continuous operators Tn, Tn acting from En into En-I. C. Rolewicz example of an operator with one hypercyclic point. In 1969, S. Rolewicz [1] constructed the first example of an operator on 12 with one hypercyclic point z, that is, such that the iterates T" z are dense in the whole space. We have slightly modified his construction, so to have a class of examples with various supplementary properties. Let 12 (Z) be the space of square summable complex sequences and let (e n )nEll be, as usually, the canonical basis of this space. The support of x is

the set {k, Xk i- O}. The operator A will be a weighted shift on 12(1l), that is, will be defined by : for k E 1l.

Chapter III

Theorem. - If the weights

Wk

satisfy:

n

n-+oo

II

lim

II

lim

Wk

= +00,

Wk

=

and

-n

n-+oo

Wk

> 1 for k > 0,

(1)

< 1 for k < 0,

(2)

a

and

0,

a

°
0, and fix E > O. to O. Put 13k = (a~ - a~+1)l/2, We have: lim

n-+oo

IIT(x +

IITxll 2 +

(1 + E)131 X n) I

(1 + E)2r2(a~

> (1 + E)2r2(a~ and so we can find an index nl such that, if Yl = z

- a~)

~ an ,

+ (1 + E)131x n1 ,

Assume now nl < ... < nk-l have been chosen in such a way that, if

we have:

and

Ilx - Ylil
O is a normalized weakly null sequence in a Banach space E, we have, for all z E E : limsup Ilxn n--++oo

+ zll

max{I/2,ll zll}·

~

Proof. - One distinguishes between the cases Ilzll ~ 1/2 and Ilzll > 1/2. Further details are left to the reader. Since the space is reflexive, for every A E u(T) with IAI weakly null sequence (xn)n~O such that :

TX n

-

AX n

--+

0, when n

~

r, there is a

+00.

For x E E, e > 0, we form

Proper choices of nl < ...

1

0

-2) 0: or'

,

if j

~

k,

from which the result follows as in Theorem 2.A.l ; details are left to the reader.

Chapter III

51!

Let's come back to the Hilbert space setting.

In Theorem 2.A.1, our

assumption concerns the spectral radius : we gave it this way because it is more striking, but it is too strong. Indeed, what we need is just the fact that IITxnl1 ~

T,

and not the fact that TX n

a(T)

=

AXn

-

sup{limsup IITxnl1 ; Ilxnll

~

O. So we define:

= 1,

-+

Xn

0 weakly},

and this quantity (as David Berg pointed out to us ; see Chapter IV, Exercise 10) is just IITlle, norm of T in the Calkin algebra, that is : IITlle

=

inf{IIT

+

KII, K compact}.

Theorem 2.A.7. - Let T be an operator on a Hilbert space, satisfying:

L 1/11Tn l ;

< +00.

n~O

Let (an)n~O be a sequence with L l/a~ < +00 and IITnll e/ a n for every x E H, every e > 0, there exists a point y with: Ily -

xii
(1 - e)IITnlle/an ,

IITnyl1

for all n

~

O.

Proof. - Fix x E Hand e > O. For each k , by the definition of a(Tk), and the equality a(Tk) (X~k»)n~o with:

= IITkll e, there is a normalized weakly null sequence

We form:

for a proper choice of

nl

< ...
O, and each k ~ 0, lITkxnll ~ 1 when n ~ +00. So such a growth requires new techniques, to which we now turn. They hold in any Banach space.

Chapter III

Theorem 2.B.l. - Assume the sequence IITnl1 is increasing and assume there exists a 0

>

0 such that,Eor all n, m ~ 0 :

Then there is a p

>

0 and a dense set of points y such that

The proof of this result is divided into several steps. The first one is a quantitative version of the classical Banach-Steinhaus Theorem :

Proposition 2.B.2. - Let T be an operator on a Banach space. Let (nk}k~o be a strictly increasing sequence of integers, with no = 0, and (ck)k>O a sequence of positive real numbers. In any ball of radius 1 which does not intersect the unit ball, there is a point y such that:

(I) Proof of Proposition 2.B.2. Let B o be any ball of radius 1, which does not intersect the unit ball. Then property (1) is satisfied, for k = 0, by any point in B o . Assume we have found B o => ••. => Bk-l, balls of radius rj = 3 1 - j, with centers Yj, such that (1) holds for all x E Bj (j = 0,1, ... ,k -1).

Lemma 2.B.3. - Let B be a ball with center b and radius r . Let C > o. There exists in B a ball B', with radius r /3, such that for all y E B', we have:

IITyl1

2:

1

3(1 - c)rIlTII·

Proof of Lemma 2.B.3. - Let u, Ilull = 1, be such that IITul1 > (1 - 6/2) IITII. Then:

(1IT(b + 2ru/3}11 + IIT(b - 2ru/3)ID/2 If b' = b ± 2ru/3, we get IITb'll Y = b'+v,

II Tyl1 and clearly B'

c B.

>

IITb'll -

~

~

2r(1 - c/2)II TII/3 .

2r(1 - c/2) IITII/3, and if Ilvll

IITvl1

> r(1 -

c)IITII/3,

0 such that i; :S

We set bn = 1/(6 log Tn)' From the asfor all m, n ~ 1, so the sequence (bn)n~o to O. From these properties follows that 1/ n P , n ~ 1, or

log Tn > n P /6.

(4)

We choose k l / p log k,

for k

>:

1.

Then n"+l -

nk

(k

< [~(k

p

+ 1) lip log(k + 1) -

k l / p log k

+ 1)(I/P)-llog(k + 1)],

Chapter III

56

for k > k o , by an easy computation (here [x] is the largest integer smaller than z }. In order to simplify the notation, we set f (:z:) = log 1"1 z I. So : f(nk+d - k log 3 - f(nk+l - nk)

> f ((k + 1P/P log (k + 1)) - k log 3 - f ( ~ (k + 1) ~

-1

log (k

+ 1))

P

> f ( ~ (k + 1) ~ - 1 log (k + 1)) f ((k + 1)P/4) - k log 3 P

- f(!(k + 1);-llog(k + 1)) P

+ 1)*-llog(k + 1)) 6f((k + 1)~

> f(~(k

P

4

- 1) -

+ 1);-llog(k + 1)]P [(k + l)p/4]P -

> 216[~(k 1

> 26(k + 1)[log(k + 1)]P - k log 3

~

klog3 klog3,

+00 when k

bY(4) ~

+00.

Using Proposition 2.BA, we may now compute mYII,

inf IIT mE[nk,nk+l!

and the estimate in the statement of Theorem 2.B.l follows directly from this computation. This Theorem applies, for instance, to the situation IITn II = en" (0 < a: < 1), since then 10g1"mn = 10g1"mlogTn, or to any case \ITnl\ = ecn" (c > 0, 0 < a: < 1). But is does not apply, conversely, if the growth of the sequence (1"n)n>O is just polynomial. We observe that, in the setting of Theorem 2.B.l, we cannot conclude that the sequence IITny11 is increasing. The first term in the right-hand side of Proposition 2.BA is non-zero if T is invertible. It may allow us to conclude that IITnyl1 ~ +00, when n ~ +00 but requires obviously that I\Tn 11/ IIT-n II ~ +00. Let us illustrate this on an example.

Corollary 2.B.5. - Assume that IITkl1 2 k? for some a: > 0, and also that IIT-k II ~ k ft for some fJ < a:. Then there is a dense set of points y with IITky11 ~ +00. Proof. - By Proposition 2.B.4, we get, if nk < mk < nk+l II T

m kYl1

2

fl 3

1- k nk (nk+l - mk)-ft

2

fl 3

1- k

nk nk~1

,

57

Orbits of a Linear Operator and the result follows if we take nk = 3k~ JOt. C. Polynomial Growths.

The case when the sequence (1ITnll)n~o has only a polynomial growth (for n 2 instance IlT ll -- n ) is not so easy to handle. Two main results can be given. First, in the case when L l/IlTnll < +00, Theorem 2.C.l belows answers our question in a satisfactory way. If we know only that L 1/IIT n Il2 < +00, we have to make an extra assumption, namely that T" is large on a large set. This is studied in Theorem 2.C.5 below. Theorem 2.C.l. - Let T be an operator on a Hilbert space H, such that L l/IIT n ll < +00. Then there is a dense set of points y such that IITnyl1 ~ +00, when n ~ +00. Proof. - Let B be the unit ball, and B' a ball with radius '7 > 0 and center b'. Let M = max{llxll; x E B'}. As before, we put Tn = IITnll. Let (Pn)n~o be an increasing sequence, tending to infinity, such that

'"' L- -e. Tn

(1)

< +00.

n

We will show that there is a ni and a point y in B' with the following property:

(2) This means that for n > nI, Tn y is not in f3nB, or that II T ny11 ~ which implies IITn y 11 ~ +00. Lemma 2.C.2. - For every n ~ 0, for every u E H with IITnul1 convex f3nT-n B is contained in the strip:

B«,

1, the

Proofof Lemma 2.C.2. - This is clear, since if x E PnT-n B, then T":x E f3nB, and I{ Tn x , Tnu) I :S

e.:

For every n ~ 0, we choose n« with 0 < n« < (iJ;n) 2, and Ilxnll = 1, such that:

Xn ,

with

(3)

Chapter III

58

Let Un

= xn/IITnxnll, so IITnunl1 = 1, and : n; = {v j I(Tnv,Tnxn)l::; ,on} {v

=

j

I( v, T·nTnxn)1

::; ,on}.

and it follows from Lemma 2.C.2 that we have:

(4)

,8nT-n BeRn. We will now study the strip R n

.

We have:

(x n, T·nTnx n) = IITnx n l 12 ~

(1 - '1n)2T~

.

We write T·nTn x n = AnXn + A~X~, where Ilx~1I = 1 and x~ Xn. Then An = (T·nTn x n, x n) is real and positive, and

(5)

is orthogonal to

(6) Since we get: IA~12::;

4'1nT~,

::;

IA~I

2~

T~

(7)

•

If vERn, we have :

< ,8nll Tnxnll,

I(V,AnXn)I-I(v,A~X~)1

(8)

and if v E B' , by (7) :

I( u, A~x~)1

< 2M~

T~

•

Therefore, we deduce from (6) and (8) that if vERn

n B' :

(1-'1n)2 T~ I(v,x n )! :S ,8nIITnxnll+2M~T~, and by the choice of n« : ,on

I( V,X n )1

::;

Tn (1 - '1nP

+

2M~

(1 - '1nP

a and a point y in B' which is not in Un>nl R~. Therefore, this point y is not in ,8nT-n B , for all n > nl, and this will prove our claim.

Orbits

0/ a

Linear Operator

59

Choose nl such that ' " /3n 4L--na

and put an = 4/3n/Tn. Assume on the contrary that Un>na R~ covers B'. Since B' is weakly compact, and since the R~ are open weak neighborhoods of 0, a finite number of them already cover B'. So, in order to conclude the proof of Theorem 2.C.I, all we need is the following Proposition, which is due to Th. Bang [1]. The presen t proof was shown to us by P. Enflo :

Proposition 2.C.3. - Let B' be a ball with center b' and radius 77 > 0, and Xl, ... , X n be points with IIXj II = 1. Let also aj > 0 with L~ aj < 77. Then the reunion of the strips j = 1, ... ,no

cannot cover B' .

Proof. - We first need a lemma: Lemma 2.C.4. - Let Zl ... Zn be points in a Hilbert space. Let el ... en be the choice of signs which maximizes 1I b' + L ~ e j Zj II. Then, for every i = 1 ... n , we have: n

I(b' + LejZj,zi)1 2:

Il zil1 2 •

j=1

Proof of Lemma 2.C.4. - We put

Z

= L~

CjZj,

Yi = Lj,ii ejZj. Then:

and so Re (b'+Yi,eiZi) 2: O. But:

(b' + Yi + eiZi, eiZi ) (b' + Yi, cizd + and therefore :

from which the lemma follows.

IIZi 11 2

,

Chapter III

60

We now come back to the proof of Proposition 2.C.3. Let 6 > 0 such that

and ci (i = 1, ... , n) the choice of signs which maximizes n

lib' + (1 + 6) LCjl:tjxjll. I

By Lemma 4, for each i ,

!( b' +

(1 + 6)

L l:tjCjXj, (1 + 6)l:tixi)1

2:: ll:tiI 2 {1 + 6)2 ,

l(b'+(1+6)Ll:tjCjxj,xdl 2:: l:ti(I+6). But y

= b' + (1 + 6) I: l:tjCjXj belongs to B', and for all i = 1 ... n.

This ends the proof of Proposition 2.C.3, and therefore that of Theorem 2.C.1. We now turn to the case when I: I/llTnl1 = +00. Then the discussion is basically upon the size of the set where T" is large. Therefore, in this respect, we introduce the Gelfand numbers cn(T), n > 0 :

x, C

cn(T) = inf{IITlxnll ;

H, codim x; < n}.

Theorem 2.C.5. - Let T be an operator on a Hilbert space H, such that

I: l/ cn(T n)2
O. Set Xo the following properties :

= Yo.

Assume XI, ...

,Xk-I

have been chosen with

a) IIXjl1 = 1, for j = 1, ... ,k - 1, b) xo, xl, ... ,Xk-I are mutually orthogonal, c) if Yj

=

Yo + c:L{ Xl/b" then for j 2"

= 1, ... ,k -

liT 'Yk-III 2:: E: c 2j / 2bj . We now choose

Xk.

1:

Orbits 0/ a Linear Operator

61

We consider the set : Xk

= {x, Re(T+ 2jT2jYk_I,X) =0, for j = 1, ... ,k-l, and (Xl, X)

= 0,

for 1 = 0, ... , k - I}.

This set is an intersection of 2k - 1 hyperplanes, so it has codimension < 2k. By assumption, IIT 2klxk II 2 C2k, so we may find a point Xk, Ilxkll = 1, with:

(2) ( X" Xk) = 0, for I = 0, ... , k - I, II T 2kxkli

(3)

(4)

2 Ck/ 2.

Replacing Xk by -Xk if necessary (which changes nothing to the previous conclusions), we may also assume :

(5) We set Yk = Yk-I + cXk/bk. Since the Xj'S are mutually orthogonal, we get: k

IIYkl1 2 =

Yk-I

+ e L lib; , 1

and if Y = lim Yk = Yo

+ e L~

Xk/bk, this series converges and 00

lIy - Yoll

c(L l/b~

=

)1/2.

(6)

1

Furthermore, if j IIT

2j

< k,

Ykl1 2 = IIT 2i Yk-I + cT 2i Xk/bk 11 2 = IIT2iYk_11l2

2

+ 2Re(T·2iT2iYk_I,cxk/bk) + cllT 2i x k/bkl1 2

IIT2iYk_1112 ,

and finally :

IIT 2j Yk l12 >_

e 2 C22i /4b2i :

(7)

The same way, we get, using (5) :

(8) so c) holds for j = k. We then get from c) : IIT

2i y

II

2 cC2j/ 2bi ,

for all J', so Theorem 2.C.5 is proved.

Chapter III

62

Remark. - The conclusion of Theorem 2.C.1 holds under the more general assumption : 1/C6n{T n )2 < +00,

L

for some 0 > O. To prove it, one replaces the multiples of 2 (which we used), by a convenient arithmetic progression. We don't know if the condition L; 1/IITn Il2 < +00 is sufficient to ensure that there are orbits tending to infinity. We will see in Section 4 that certainly the condition IITnl1 ---+ +00 does not suffice. 3. How often can an orbit come back close to 0 ?

In several situations, the techniques of Section 2 are unsufficient to ensure that there is a point y such that the iterates Tn y tend to infinity. But they are still good enough to conclude that there is a point y the iterates of which never come back into the unit ball, or, in some cases, come back into it very rarely.

Theorem 3.1. - Let T be an operator on a Hilbert space H. We assume that the sequence Tn = IITnl1 is asymptotically increasing, that is : lim inf Tn / Tn n-+oo

Let (xn)n~o

with

Ilxnll

1

> 1.

(1)

= 1 be such that

IITnxnll/Tn

---+ 1.

(2)

If the sequence (Xn)n>O does not tend to zero weakly, there is a point y in H such that: for all n ~ O.

Proof. - We need two Lemmas : Lemma 3.2. - Let (xn)n>O be a sequence in a Hilbert space which is not weakly convergent to O. There exists a a in H, a 0 < 1, and a sequence {nk)k~O of integers, such that: limsup k-+oo

Ilx

n k

-

all < o.

Proof of Lemma 3.2. - There is a sequence (nk)k;::::O and a point a in H, a i= 0, such that Xnk ---+ a weakly. If we put Zk = Xnk - a, Zk ---+ 0 weakly, and:

2. IIzk + al1 2 = k-+oo lim IIzkl12 + Ila11 So IIzkll ---+ (1 - IlaI12)l/2. If 0 = (1 - !llaI12)l/2, for k > k o , we get Ilzkll < o < 1, and so IIxnk - all < o. 1 =

lim

k-+oo

69

Orbits of a Linear Operator Lemma 3.3. - Assume that there is an a

satisfies, for all

i 2: 0

> 0 such that the sequence Tn

:

lim inf Tn/Tn-; n-oo

> a.

(3)

The point a given in Lemma 3.2 satisfies : liminf IITna11 2: a(1 - 6). n-oo

Proof of Lemma 3.3. - We put IITn x n ll IITn"all But, if

i
IITn" xnkll -IIT n" (Xnk - a) II

~

en). Then we have:

(1 - 6)Tn" - ekTn" .

nk

and so

IITi a li 2: ((1 - O)TnA: - ekTn,,)/Tn,,_j . Theorem 3.1 follows from Lemma 3.3, if we take y = xri«, for 1>"1 large enough and 1 large enough. We observe that condition (3) is weaker than condition (1). Theorem 3.1 was stated this way for simplicity. We don't know if any regularity condition is necessary.

A sequence (xn)n>O satisfying (2) will be called a norming sequence for the sequence

(Tn)n~o.

IT the set where the Tn's are large has some specific geometric properties, the answer to our problem is simple :

Theorem 3.4. - Let T be an operator on a Hilbert space, with the following property: there is a point y, Ilyll = 1, and an e > 0 such that if we set: F

= {x; IRe(x,a)1 2: c} ,

and an = sup{IITnxll ; x E F, Ilxll = I}, we have an

---+

+00. Then,

for all

n~O:

e ny11 IIT 2: a n( -2 -

~).nc

Proof. - We choose Xn with Ilxnll = 1, Re(xn,y) ~ s , IITnxn11 > (1- ~)O:n. Then, with x = X n , we get lIey - xII ~ ,,/1 - e 2 and therefore: IleTnyli ~

~

IITnxn11 -IIT n(x n - cy)1l

a n (1 -

-!.n - ~),

Chapter III and so:

as we announced. We now study the case where the sequence (xn)n~O

is weakly null.

Theorem 3.5. - Let T be an operator on a Hilbert space H and assume that there is a norming weakly null sequence (xn)n~O for (Tn)n~o. Then, for any sequence (aJc)k~O' positive with L~ ai = 1, for any e > 0, there is a sequence (nJc)Jc>o such that, in every ball of radius strictly larger than e , we can find a point Y with : for all k ?: 1.

Proof. - Fix Yo in H. Take n 1 large enough so that if n

~

n1 ,

Then:

and so if we set y = Yo ±

ealxn1 ,

we get

Assume now YI, ... , YJc-1 have been defined:

Yj

=

Yo ±

where the sequence j=I, ... ,k-l

Ealx n 1

nl

IITnXnl1

± ... ±

ElXjX n j ,

for j

= 1, ... ,k -

1.

< ... < nk-I is so lacunary that, for 1 = 1, ... ,j

~

(1-

IIT n ; Yj II ?:

~

- ... -

eaj(1 -

~

4~")Tn, - ...

II T n' Y1"II >- ElXl(1 - !4 - ... - !4' - 3!(lX~

~ 4~· 1

~

1,

if n?: nj ,

(b)

)Tn j

(c)

,

+ ... + a~

+1

))Tn.

(d)

65


We now choose nk large enough to have (a) and (b) for i Yk = Yk-l ± gO:kXnk j by a proper choice of the sign, we get: IITn"Ykll ~

1

k.

We set

1

gO:k(1 - 4" - ... - 4 k ) Tn" ,

which is (c). Now, for 1 < k, IITnrykl12

> IITn'Yk_111 2 >

g Ctl

-

21(gO:k Xn",T·

1

1

(1 - 4" - ... - 4')

n'Tn'Yk_I)1

Tn. -

"31

2 ) Ct, - "31 e (2 Ctk-I + ... + 0:'+1

~ 0:, Tnl ,

2

0:,) Tnr

Tn•.

So our assumption is satisfied at step k. Let Y = lim Yk IITnr yll ~

2(

g Ctk

j

we get from (d) :

for I = 1,2, ...

as we announced. We observe that there is quite a substantial difference between the statement of Theorem 3.4 and Proposition I.B.6. Indeed, in the latter, the sequence (nkh>o was arbitrary, and in the former it is not (more precisely, it has to satisfy a certain lacunarity condition : there is a function f such that any sequence satisfying n1l:+1 > f(nk) will suit). Corollary 3.6. - There is a dense set of points Y such that

L II T nYIl / lI T n ll =

+00.

n~O

Corollary 3.7. - Assume that the sequence Tn is increasing. Then, if nk-I < j ::; nk : .

g

liT' yll 2: "3 O:k . Proof.-Indeed,IIT;YII > iO:kTn,,/Tn,,-; ~

iCtk.

Therefore, in any ball of size e , we can find points Y the iterates of which will approach 0 arbitrarily slowly. In other words, if for a point x we set

which is the first time the iterates of x enter the ball of radius 1/ k , centered at 0, then: for any ball B(g) of radius e not containing O.

Chapter III

66

This last statement can be improved:

Theorem 3.8. - Let T be an operator on a Hilbert space H, with r(T) = 1. Assume moreover that Cnu(T) contains a point A which is not an eigenvalue. Then, for any decreasing sequence (a")n~O, positive and tending to 0, for any e > 0, for any Xo, Yo in H, there is a point y with:

Ily - yoll
k.

IITkxk11

4. Operators with irregular orbits.

All the operators we have met so far, satisfying IITnl1 --+ +00, have the property that for some y, IITny11 --+ +00. This is the case even for Rolewicz's example (see Section 1). We present here, first, an example of an operator with no such vector. It will be a shift on the space 12(IN) ; as usually, (en)n>O is the canonical basis. A. An operator satisfying arbitrarily close to O.

= \l'log n, such that each orbit comes back

be an increasing sequence of integers, satisfying, for all k 2': 0 :

Let (nk)k~o

Put Bk

liT" II

= ink + 1, ... , nk+d. ui,

]

=

We define Tej

{(lOg n) 1/2" , if j E B n 0 ,

= wjej, with: ,

n = 1, 2, ... ;

otherwise .

Wj'S are decreasing from B n to B n + 1 , Tk attains its norm on en .. +l, and it is exactly y'iOik. We now prove that for all y, inf" IIT"YII = o.

Then, since the

Orbits of a Linear Operator Assume conversely that for some D > 0, some

67

y with lIyll = 1, (1)

for all n. a) We first consider the case when y = Lt~ by the first point in each Bk. Then: Ti

akenk+1 , that is, y is supported

if i > k; if i ~ k.

{O, (y'logk)i/kenk+i+l'

_

enk+l so

II T i y ll 2 =

L lakI 2(1og k )i /

k

(2)

k?i

Set in = en log n, for n ~ 1. We first show that if k 2: in+l (log k)i n / k ~

(3)

2

First, the function x ---+ (log x)A/x (A > 1) is decreasing for x 2: e , and so the maximal value of k ---+ (log k)i n / k for k 2: in+l is for k = in+l . Next, elementary computations show that : ((n

+

l)1og(n

+ 1))

en Jogn-(n+ I) log(n+ I)

---+

1, n

---+

+00 ,

so (3) follows. Assume (1) holds. Then:

L

lakl 2(log k)in/ k > D,

for all n.

k?in

But:

L

lak/ 2(log k)in/ k

< 2

k?in+1

and thus for n

~

L

lakl 2

---+

0, n

---+

+00

k?in+l

no , in+l

L

lakI 2(logk)in / k > D/2

k=in

But (log k)in/ k

< login

= n log n, so in+l

nlogn

L

lakl 2 2: D/2, for n 2: no

(4)

k=in

Now, if we put Sn = :L{::/n lakl 2 , we have (4), since :L l/n log n = +00.

"Li

oo

Sn

= 1,

and this contradicts

68

Chapter III b) General case. We write Y = L:k~

1

Yk , where Yk is supported by Bk

L a,Ti e,

Ti Yk

IEBIc

and 0, { (""log k)i/kei+l,

L

IITi Yk l12

if j + I 2: k; if i + I < k.

ja,1211Tie,112

IEBIc

For a fixed j, the largest quantity among the IITi e11l 2 , I E Bk, is (""log k)ijk . So: IITi yk l12 ~ L(Vlogk)i/kIlYkI12 k~i

and we are back to the computation of a).

Remark. - Banach - Steinhaus Theorem says that there is a point Y which satisfies sUPn IITny11 = +00. Let's see where is such a point here. Put:

ki

= n2i~+1'

Y

=

Lekj/j!+€ i~O

This series converges in 12 , and :

Remark. - As it is presented, the operator is not injective. We can obtain an injective one, with the same properties, the following way: instead of wi = 0, we put wi = 1/2, and we repeat it a number of times sufficiently large; that is, we increase the lacunarity of the sequence (ni )i~o. Remark. - We don't know what exact rate of growth on the IITn II is necessary in order to get the present conclusion. For a weighted shift, the estimates IlTn ll = n€ does not allow the same result: there is always a point y the orbit of which goes to infinity. This point is of the following type: y = L: n -Cl e n lc , for a > 1/2 conveniently chosen. So there is a large gap between the estimates of Section 2 and the present example.

Orbits

0/ a

Linear Operator

69

This operator, as well as all the ones we have met so far, has a point which orbit tends to O. So in view of these examples, one might ask the following question : Given an operator T on a Banach space, does one of the following hold: - either there is a point x such that T" X always stays in the unit ball, - or there is a point z , such that Tn x never comes into the unit ball ? The answer to this question is also negative, as the following example shows: B. An operator such that, for every Y

i- 0,

sUPnllTnYl1

=

00, infnllTnYl1

Here again, the operator will be a weighted shift on 12(IN) , for

i

~

= O.

Tej = wjej,

O.

Let (Pn)n>O, (qn)n>O be two sequences of integers, with :

o = Po < qo < We put B n

PI

= {Pn, .. ' ,qn

< 2pI < ql < ... < v« < 2pn < qn < ...

- I}, B~ = {qn,'" ,Pn+1 - I}

The weights will be constant inside each B n and B~. They are 1/2 in each B~ (n ~ 0), and will be strictly larger than 1 in each B«, They are greater in B n than in B n + I . Set Pn =

IljEB.. Wj, P~

PnP~

= IljEB~

=

Wj.

We require: for n

1/2,

~

n ,

(1)

0

for n

~

1

(2)

21'..

II

Wj

< 1 + 1/2 n

,

for n

~

(3)

1

1'..

So we get, by (2) : IITp..+l- l ep,.11

=

n l k: -. +00, n - 00

and therefore, for all k , SUPR IITRekl1 = +00. Consequently, for every Y i- 0 : sup IITRYII = +00. n

We now show that inf R IITnyl1 = o. We write y = Yk has support only between Pk and Pk+1 - 1, that is : 1'''+1- 1

Yk

L

j=p"

Qjej

Eci

oo

u», where each

Chapter III

70

PA:+1- 1

L

TP" Yk =

a,W, ••• Wl+ p,. el+ p ,. + 1

PA:

To estimate the norm of this vector, we will distinguish several cases : First case: n ~ k. Then the product equal to Pn + 1 < Pk. So, by (3),

WI ••• Wl+ Pn

contains a number of terms

Second case : n > k. Set n = k + j, with j > o. Then, since we know that both products wPA: ••• Wl-l and Wl+ p,.+1 ••• w PH j + 1 - 1 are ~ 1/2, we get

n such that T" Xo E B, Tn' Xo E B'. Therefore, Tn' - n B intersects B'. 2) Assume that P(T) holds. The space being separable, let (Bn)n~o be an enumeration of the open balls, with rational radii, centered on a dense countable family. Let B be any open ball. By assumption, there is an index nl such that T':"» B 1 n B is non-empty. Since it is an open set, it must contain some open ball U1 of radius < 1/2. Now, there is an index n2 such that T-n'J B 2 n U 1 is non-empty, so it contains

Chapter III

72

an open ball U2 of radius < 1/4, and so on. The intersection nnUn contains a single point x, and Tn/r. x E B" for all k, so x is hypercyclic.

Corollary 5.3. - Assume T to be invertible. Then T has an hypercyclic point if and only if T- 1 has an hypercyclic point. Proof. - Property P (T) is obviously equivalent to property P (T-l ) . Corollary 5.4. - Assume that T is invertible and has one hypercyclic point. Then there is a dense G6 of points y which are hypercyclic both for T and T-l, that is: {Tn y j n ~ o} is dense; {T-n y j n ~ o} is also dense. This situation occurs, for instance, for the modified Rolewicz's example, we presented in Section 1. In this case (if we choose the weights correctly: say wi = 2 for j > 0, wi = 1/2 for i < 0), we get a situation which is 1 • Both have spectral radius 2, so there are totally "symmetric" for T and points y such that both IITn y li and IIT-nYIi grow up exponentially fast (by Theorem 2.8), and there are points y' such that both {Tn y' ; n ~ o} and {T-n y' j n ~ O} are dense in the whole space.

r-

So we see that the iterates of a given ball may intersect any other given ball. But if they don't too often, there is in the first a point, the iterates of which never enter the second:

Theorem 5.5. - Let T be an operator on a Banach space E and (n,,),,>o be an increasing sequence of integers such that: inf "~o

II Tn/r.1I

=

6

> 0

3"

Assume there exists two balls B(b, r), B'(b', r'), with B not containing 0, r' = 6, such that : rnBnB' = 0 except maybe for n E {n"

j

k

~ O}.

Then there is in B a point y such that

Tn y never belongs to the ball B" with center b', radius r' /2. Proof. - We put B o = B. Assume that we have built balls B" C ... C Bi C B, with radius = 1/3", such that:

r"

Ti B m n B"

= 0, if j < n m +l ,

m

= 0, ... ,k.

(1)

We now look at Tnlc+ 1 B" : this set may intersect B". We will construct B,,+ 1.

Orbits Let e < 1/4, and let

0/ a Linear Operator

Uk+l

,with

Uk+l =

79

1, such that:

Therefore one of the two points Tnk+ 1 Vk+l or Tnk+l Wk+l is at distance from b' at least (1 - e)~rkIITnlt+l II. Say it's the first one. If !lull < rk+l = 1/3k+ 1 , we obtain:

IITnlt+l (Vk+l + u)

- b'll

~

IITnk+1 Vk+l -

~

((1 - e) 3 k2+ 1

>

3 k:

1

b'll -IIT n k+ 1 11/3k+ 1 -

3 k1+1 )IIT

(1 - 2e) IITnlt+J II

and Bk+l is the ball of center Vk+l, radius we get that Tn y never comes into E".

rk+l'

nlt+l

II

~ s/2

Let y E

nkBk.

From (1)

Unfortunately, we don't know of any concrete situations where Theorem 5.5 can be applied.

Chapter III

Exercises on Chapter III

Exercise 1. - Let T be an operator on a Hilbert space H. Assume that it satisfies Ln?:o 1/IITn112 < +00 I and that there exists an orthonormal sequence (xn)n?:O of points in H, such that: n 2: 0. Prove that there is a point y such that IITn y ll -

00.

Exercise 2. - Let T be an operator on a Hilbert space H. We assume that there exists a ball B(O, r) and a sequence (xn)n>O I with IIx nll = 1, such that: a) Ln?:o 1/IIT nxnll < +00 b) for every point z in B(O, r), every n 2: 0, every j = 1, ... , n - 1, I

Prove that there is a point y such that IITny11 -

00.

Exercise 3. - State and prove Proposition 2.B.2 for a family Tn of operators, instead of the iterates of a single one. Exercise 4. (This exercise uses some material from Chapter XIII) - Let T be and Xo in a contraction on a Hilbert space, such that u(T) :J C. Let e > H I with IIxoll = 1. Let (In)n?:o be any family of polynomials such that:

° I

L

1/11 1n(T)1l 2
e

(L 1/11 1n(T)11 2)1/ 2 , n?:O

there is a point y such that, for all n 2: 1,

a) Show that, for every n 2: 0, there exists a weakly null sequence (X~)k, such that:

75


(Choose An, with 1.1, Chapter XIII.)

IAnl =

1, such that In(A n)

= IIInll oo , and use Proposition

b) Put "t« = Illn(T)II, and consider:

xoll < e},

Cn = {y; Illn(T)y Take TJ > 0, e < p < (1 + TJ)e, and consider: p 1 Yo + - X'. 11

Show that for and set:

i

1

large enough, this point is not in C1. Fix such ai,

i

=

ii ,

-P x·111

Yl = Yo +

11

c) Consider : Yl

Show that for

i

+ 12 -P

2

X'.

1

large enough, this point is neither in C1 nor in C2 • Set: p 2 Y2

=

Yl

+-

12

Xh .

d) Repeat the argument, and take y = lim Yn' Show that Y is in none of the Cn 's. Show moreover that the sequence in may be chosen so that :

Ily - Yol1 2
0, n 2: o} was dense; (B. Beauzamy [9]), and an operator with all vectors hypercyclic was recently constructed by C. Read (C. Read [5]) on II and related spaces (see Chapter XIV for a discussion of these matters). The present chapter brings several restrictions upon the existence of such an operator on a Hilbert space, saying that the iterates IITn II cannot increase too quickly: of course, a point with IITnyl1 ~ +00 cannot be hypercyclic.

76

Chapter III

These restrictions can be compared with an old result of J. Wermer [1] if u(T) contains more than a point and if LnE1llog IIT nll/(l + n 2 ) < +00, then T has non-trivial invariant subspaces, and therefore some non-hypercyclic points. Wermer's condition says, roughly speaking, that IIT n l1 and liT-nil don't increase too fast. However, there is a large gap between Wermer's result (besides the fact that it also uses T-l), and the results presented here, so we certainly cannot conclude that there is no operator on a Hilbert space with all orbits dense. On the contrary, an operator which is rather close to having this property has recently been constructed by the author (B. Beauzamy [12]) : it has an hypercyclic vector Xo, and for every polynomial p with complex coefficients, p(T)xo is also hypercyclic.

PART II

COMPACTNESS AND ITS APPLICATIONS

Singuliere fortune oii le but se deplace, Et, rr'etant nulle part, peut etre n'imparte au ! OU l'Homme, dont jamais I'esperance n'est lasse, Pour trouver Ie repos court toujours comme un fou !

In this Second Part, we leave the general frame in order to concentrate on compactness. First, compact operators for themselves, then, on a Hilbert space, the space of nuclear operators, which appear to be the predual of the space £(H). This duality is studied in detail, since it allows us to describe interesting topologies on £(H).

Chapter IV

Spectral Theory for compact operators

1. Compact operators.

Let T be an operator on a Banach space E. We say that T is compact if the image T(BE) of the closed unit ball of E is relatively compact in E : that is T(BE) is compact in E, for the norm. This is obviously equivalent to the following statement: for every sequence (Xn)n~O, bounded in E, we can find a subsequence (x~)n~O such that (Tx~)n>o converges. Let's observe that if E is reflexive (and, in particular, for a Hilbert space), T(BE) is closed. Indeed, BE is weakly compact, T is weakly continuous, so T(BE) is weakly compact, and therefore closed in norm (see B.B. [1], pp. 37 and 49). So we see that on a reflexive space, T is compact if and only if T(BE) is compact. If E is infinite - dimensional, a compact operator cannot be surjective. Indeed, let's assume that T is surjective. Then it's an open mapping (B.B. [1], p. 13), which means that it transforms an open set into an open set. SO T(B E) is open, and relatively compact. By F. Riesz theorem (see B.B. [1], p. 17), this implies that E has finite dimension. As a special case, we see that the identity is never compact when the dimension of E is infinite. This argument can be immediately extended to the following:

Proposition 1.1. - Let T be a compact operator, with closed range. Then this range is finite-dimensional. The image of a compact operator may be dense in the whole space, and therefore be infinite-dimensional. This is the case, for instance, of the operator on 12 (IN) defined by : K(xn)n~O

= (xn/(n +

l))n~o

(The fact that this operator is compact is a consequence of b) in Proposition 1.2, below.)

80

Chapter IV But conversely, any finite rank operator is compact.

Proposition 1.2. - a) If T is compact, and U I , U2 are any operators, U 1TU2 is compact. is a sequence of compact operators, converging in norm to b) H (Tn)n~o an operator T, T is also compact.

Proof. - a) is obvious. For b), we use a diagonal procedure: let (xkh~o be a bounded sequence in E. First, we extract a subsequence xii) such that T1xi l ) converges in E, then, from xii) , a subsequence xi2 ) such that T2xi2 ) converges in E, and so on. Put Yj = x;il, the "diagonal" subsequence. Then (TkYj),. converges for all k's, and so (Ty,.),. converges, which proves that T is compact. Another proof is as follows: given e > 0, we choose n such that IITTnll < «n, and then we cover Tn(BE) by a finite number of balls of radius c/2. The balls with same centers and radius e will cover T(BE).

If K (E) is the vector space of all compact operators from E into itself, Proposition 1.2 can be summarized as follows: K (E) is a closed two-sided ideal of .c(E). Remark 1.3. - On a Hilbert space, every compact operator is the limit of a sequence of finite rank operators.

Indeed, let T be compact, e > 0, and (Xl,." I x n ) be points such that the reunion of the balls B(Xil c) covers T(BH) . Let F = span {Xl, ... ,x n } , P the orthogonal projection onto F. Then, for x E BH,

IITx - PTxl1 = so

liT -

inf IITx -

yEF

yll < C,

PTII ~ c.

Proposition 1.4. - The operator T is compact (from E into itself) if and only if rr is compact (from E* into itself). Proof. - a) Assume T to be compact. Let in the unit ball of E*. Then:

Ifn(Y) - fn(z)1 ::;

Ily - zll,

(In)n~O

be a sequence of elements

for all Y, z EE,

and so this sequence is equicontinuous on BE. Let's consider it as a sequence of functions on the compact K = T(BE) : it is bounded and equicontinuous.

Spectral Theory for Compact Operators

By Arzela-Ascoli theorem, a subsequence some function g. This can be written :

(f~)

81

converges uniformly on K, to

.- g(Tx),

f~(Tx)

uniformly for x E BE, or :

converges in E*, and "T is compact. so (tT f~)n>o b) Conversely, assume that -r is compact. By a), »r is compact from EU into itself. So ttTBE•• is relatively compact in EU. But ttTIE = T, and the norm induced by EU on E is the norm of E (see B.B. [11, p.25). So T( BE) is relatively compact in E* * , and therefore in E. 2. Spectral Theory for compact operators. We now study the operator A - T, T compact.

Theorem 2.1. - Let T be compact and A i= O. Then: a) For every n 2: 1, K er (A - T)n is finite-dimensional,

b) For every n 2: 1, the image of (A - T)n is closed,

c) There is a n 2: 1 such that, for all k 2: 0, 1m (A - T)n = 1m (A - Tt+ k . Proof. - a) We write:

(A - T)n = AnI - An-IT + ... + (_l)nT n = AnI

+A

where A is compact. Let Fn = Ker(A - T)n. On Fn , we get I

= -lA

An

so the identity of F n is compact, and F n is finite-dimensional. b) We first consider the case n = 1. By a), there is a closed subspace M 1 in E, such that E = M 1 EEl F 1 , and

Im(A - T) We need a lemma :

82

Chapter IV

Lemma 2.2. - Let A =j:. 0, M 1 a closed subspace of E, with M 1 n Ker (A T) = {O}. Then the restriction of (A - T) to M 1 is an isomorphism, and (A - T)M1 is closed. Proof of Lemma 2 2. ~ Let's consider the restriction of (A - T) to M 1 , denoted by (A - T)IM 1 • This operator is injective, let's show that the inverse in is continuous. If this was not the case, we could find a sequence (xn)n~O M 1 , with Ilxnll = 1, and (A - T)x n --. O. But (xn)n~O has a subsequence x~ such that Tx~ converges to some point y, and AX~ - y --. 0, so x~ converges to u/ A, with y/ A =j:. 0, and Ay = Ty : a contradiction. So (A - T)IM 1 is an isomorphism from M 1 onto its image (A - T)M 1 , which is therefore closed. The Lemma is proved. This proves b) for n = 1 j for n 2: 1, we repeat the argument by induction, considering T on 1m (oX - T)n-l .

3) Let's assume that, on the contrary, the sequence of subspaces

is strictly decreasing. We need a small lemma :

Lemma 2.3. - Let G be a proper subspace of E Ilzll = 1 , such that liz - yll 2: 1/2, for all y E G. Proof of Lemma 2.3. - Let g E E*, Ilzll = 1 , such that g(z) 2: 1/2.

Ilgll =

j

there is a point z E E,

1 , 9 = 0 on G, and choose z,

We return to the proof of 3). For every n 2: 1, take Zn E G n, Ilznll = 1, such that IlZn - ylI ~ 1/2, for all y E G n+1 . So Ilzn ~ zmll ~ 1/2, for n =j:. m. We have also (oX - T)G n = G n + 1 • For k > 0, we set : 1

Zn,k = Zn - Zn+k - >:(A - T)(zn - Zn+k).

We have Zn - Zn+k E G n , so (A- T)(zn ~ Zn+k) E G n + 1 , and Zn+k E G n+ 1 . So Ilzn,kll 2: 1/2. But AZn,k = TZ n - TZn+k' and, though the sequence (zn)n~O is bounded, no subsequence of (Tzn)n>o may be convergent. This contradicts the compactness of T and proves our claim.

Theorem 2.4 (Spectral Decomposition of Compact Operators). - The spectrum of a compact operator is either finite, or consists of a sequence tending to O. The point 0 is always in u(T).

89


Every A E u(T), with A f. 0, is an eigenvalue of T, the corresponding subspace, F>., is finite-dimensional.

Proof. - We first show that every A =1= 0 in the spectrum is an eigenvalue. To see this, it's enough to show that if A - T is injective, it is also surjective. By Theorem 2.1, c), there is an n > 0 such that 1m (,\ - T)n = 1m (A T)n+l. Let Y E E. We can find x such that:

which means

(A - T)n(y - (A - T)x)

= 0,

and this implies Y = (,\ - T)x, since A - T is injective. Now, a single argument will show at once that o(T) is either finite or countable (and, in the latter case, made of a sequence converging to 0), and that F>. (A f. 0) is finite-dimensional. All we have to show is that, for all 6 > 0, there is only a finite number of independent vectors which are eigenvectors for some eigenvalue A, IAI 2: 6. Assume the converse. For some 6 > 0, we may find a sequence (An)n>O of eigenvalues, fJ < IAn I :S IITII, and, for each n , an eigenvector x n , the family (xn)n~O being independent (that is : any finite subfamily is independent). Let En be the (closed) subspace spanned by Xl, ••• , x n. Since the Xi'S are eigenvectors, TEn C En, (>t.n - T)E n C E n- l . Since E n- l is strictly contained in En, by Lemma 2.3 we can find a point

Yn E En IIYnl1 = 1 , llYn -

xii

2: 1/2, for all x E E n- l 1

1

zn,m = Yn - An TYn + Am TYm

.

For n

E E n- l

> m, put:

,

and so

IIT(Yn) - T(Ym)1I = llYn - Zn,mll 2: 1/2. An Am But IIYn/Anll < 1/6, and the bounded sequence Yn/An has no subsequence such that TYn/ An converges: this contradiction proves the Theorem.

Remarks.

-r

1. We have seen (Chapter II) that T and have the same spectrum, and that if one is compact, so is the other. Therefore, if A -# 0 is an eigenvalue for T, it is also an eigenvalue for and conversely.

rr,

(This is not true in general for non-compact operators: T may have eigenvalues, and tT none.)

8.1

Chapter IV

2. We have seen that 0 E u(T). But 0 does not need to be an eigenvalue, which means that T may be injective, though compact. This is the case of the operator K described at the beginning of this chapter. More generally, for a sequence (an)n~O -+ 0, let's consider the operator:

This operator is compact, and all the an's are eigenvalues. If all an's are the operator is injective.

i-

0,

3. A compact operator may have no eigenvalue, and, in this case, its spectrum reduces to {O}. As an example of this situation, let's consider the weighted shift on h(IN) :

n E IN, with W n -+ O. This operator is compact, since it is the compose of en -+ wne n, and of the usual shift. Since all the elements of the image of T" have their first k coordinates equal to 0, T has no eigenvalue .\ i- O. It is injective if all wn's are i- O. The spectral radius of T is 0, since there is no eigenvalue.

Definition. - An operator T with u(T) = {O} is called quasi-nilpotent. For a compact, quasi-nilpotent operator, Theorem 2.4 gives no information at all, and therefore should not be considered as a satisfactory description. We will come back on this later, but first, for a special class of compact operators on Hilbert spaces, we will obtain more precise information. 3. Spectral Theory for Normal Compact operators on Hilbert spaces. In this Section, the space is a (complex, infinite-dimensional) Hilbert space H. An operator T on it is said to be normal if it commutes with its adjoint:

We will study normal operators in detail later (Chapter VII). However, we need here a simple Proposition:

Proposition 3.1. - Let T be a normal operator. Then:

a) Ker T

= Ker T" ,

85


b) If F>. (T) is the set of eigenvectors for T, corresponding to the eigenvalue A, then F>.(T) = FX(T*). For A f u , F>.(T) and F#l(T) are orthogonal.

Proof. - a) T and T* commute, so K er T is invariant by T*, and the restrictions of T and T* are still adjoints. But inside K er T, T = 0, so T* = o. This proves that K er T c K er T*. Since T* * = T, the reverse inclusion follows. b) We apply the a) to A - T, X- T*. Now, if A f u , for x E F>.(T), y E F#l(T) , we have: A(X, y) = (Tx, y) = (x, T* y) = Il(X, y),

so (x, y) =

o.

Theorem 3.2 (Spectral Decomposition for Normal Compact Operators). - Let T be a normal compact operator on a Hilbert space H. Then: a) There exists at least one eigenvalue A, with

IAI =

[IT[I,

b) The space H can be decomposed into: H = ffi>'El1 p(T)F>. .

For the associated decomposition, x = (X>.hEl1 p (T ) , we have: Tx = (..\X>.hEO"p(T) , T*x = (XX>.hEO"p(T) .

Conversely, any operator defined by such a formula is normal compact.

Proof. - We may assume that IITII = 1. Let (Xn)n~O Ilxnll = 1 , IITxnl1 ---. IITII. Then:

lIT*TII

~

(T*Tx n , x n )

---.

1 =

be a sequence with

lIT11 2,

which implies the formula:

(1) Set

U

= T*T. Then

U

is self-adjoint (that is:

Iluxn- xnl12 = IIuxnl1 2So 1 belongs to ua(u), and, since But, for n ~ 1,

so: lim n

IITnll l / n =

2(ux n,x n )

Ilull =

*=

U),

and:

+ IIx nl1 2 -.

0, n ---.

1 , this implies r(u) = 1.

lim II(T*T)nlll/2n n

U

=

~

and r(T) = 1. We have therefore obtained the following:

1,

00.

86

Chapter IV

Lemma 3.3. ~ For any normal operator, r(T) =

IITII.

Now, since moreover T is compact, by Theorem 2.4, there must be an eigenvalue A, IAI = r(T) , and a) is proved. For b), we observe that each F>.. is invariant by T and T+. Let's put

the sum being direct by Proposition 3.1. Then F is invariant by T and T+. Let Flo be its orthogonal: it is also invariant. If Flo was not reduced to {O}, the restriction of T to it would have an eigenvalue, which would contradict the definition of F. If x = (X>..hEC7 p ( T ) , we clearly get Tx be an operator of the form :

= (Ax>..hEC7 p ( T ) . Conversely, let T

where A is a sequence converging to O. One checks immediately that T+ can be written: and so T is normal. Let's check it is compact. Let e > O. We decompose A into Al U A2, where:

and we set: UI

((x>..))

(AX>")AEA 1 ,

U2((X>..))

(AxAhEA~.

Then T = UI + U2, U2 has finite rank since A2 is finite, and each subspace corresponding to A E A2 has finite dimension. Moreover, I!ulil < e. So T is the limit, in operator norm, of a sequence of operators with finite rank. Therefore, T is compact (Proposition 1.2, b)). So, for any normal compact operator T on a Hilbert space, we can write a spectral decomposition :

Tx =

L An(z, en)en,

(2)

n~O

where An are the eigenvalues of T (not necessarily distinct), and (en)n~o Hilbertian basis, each en being an eigenvector for the corresponding An.

a

87


This will allow us to write a spectral decomposition also for compact, nonnormal operators on Hilbert spaces. But first, we need a few tools, which will also be used in the next chapter. Let T be a compact operator on H, and set, as before, u a compact, self-adjoint operator, which is positive, namely:

(ux, x)

= T* T. This is

\Ix E H.

~ 0 ,

Lemma 3.4. - A positive, compact, self-adjoint operator on H has a square root: there is an operator v on H which is also positive, compact, self-adjoint and satisfies:

II vxl1 2

(ux, x) ,

Ilvll=

\Ix E H,

M·

Proof. - We write the spectral decomposition of u given by Theorem 3.2 :

ux = LAn(x,en)en, n~O

where An are the eigenvalues of u , and (en)n~O a Hilbertian basis, each en being an eigenvector for the corresponding An. Since (ue n, en) = An, the An's are real positive. Set:

vx =

L

A(x,en)e n,

n~O

then obviously this operator is compact, positive, self-adjoint and satisfies v 2 u. Moreover :

IIvxl1 2 =

L[A(x,e n )]2 n~O

L Anl(x , en)1

2

n~O

and

(ux, x) =

(L An(x, en)en, L (x, em)e m) n~O

=

L

m~O

2

An l(x , en )1

n~O

Finally,

Ilvll = sUPn{VX:}, by

Lemma 3.3 again, and so

Ilvll = M.

=

Chapter IV

88

To simplify our notation, we write ITI instead of y'T*T. Let M be a closed subspace of H. An operator V is called a partial isometry from M onto V (M) if I[V xl[ = lIxll, for x EM, and V x = 0 for x E M.L. This definition implies that V(M) is closed.

Lemma 3.5. -Let T be an operator on a Hilbert space H. There is a partial isometry V from (K er T).L to 1m T, such that:

=

T

VITI,

ITI = V*T.

Proof. - For S E £(H), one has, for all x E H, IlSxll 2 = (S* Sx, x), and thus: KerS*S = KerS, 1mS*S = (KerS*S).L = (KerS).L

Applying this to S

= 1mS*.

= T, and then to S = ITI, we get : Ker T

=

KerT*T = Ker ITI,

1m T* = 1m T*T = 1m ITI. But, for x E H :

Thus the application V defined on Im T by VITlx = Tx extends to an isom-----.L etry of Irn ITI. We take also V = 0 on 1m ITI ,so we get a partial isometry, with VITI = T. From this, we deduce T* = ITIV*, T*T = ITIV*T, ITI 2 = ITIV*T, so V*T = ITI on (Ker ITI).L = (Ker T).L. Since V*T = ITI also on K er T, the second formula is established. 4. Spectral decomposition for compact operators on Hilbert spaces.

Theorem 4.1. - Let T be a compact operator on a Hilbert space H. Then, for all x in H, we can write a decomposition: Tx

=

L An(x,xn)Yn, n~O

where (An)n~o is the sequence of eigenvalues of y'T*T, (xn)n~O an orthonormal sequence of corresponding eigenvectors, and (Yn)n~O an orthonormal sequence of eigenvectors for TT* .

Proof. - By Theorem 3.2, ITlx

L An(x, xn)X n, n~O


89

where (An)n~o is the sequence of eigenvalues of ITI, (xn)n~O an orthonormal sequence of corresponding eigenvectors. By Lemma 3.5, T = VITI, so

Tx

=

VITlx

=

L

An(X,Xn)VX n .

n~O

Set Yn = Vx n. Then the sequence (Yn)n~O is orthonormal in H, since V is an isometry on 1m ITI. The Yn's are eigenvectors for TT'" , since:

TT '" Yn

=

TT '" V Xn

=

T IT I Xn

==

V IT 12 Xn

=

2 AnV Xn

=

2 AnYn ,

and this proves the Theorem. It's quite interesting to compare spectral decompositions obtained for normal compact operators (Theorem 3.1) and for compact ones (Theorems 2.4 and 4.1). The first one may be regarded as satisfactory: the space is decomposed into parts, on which the operator is represented by a multiplication. The second one is not: despite the "similarity" (at first glance) between Theorem 3.2 and Theorem 4.1, the latter does not provide any information which can be easily used. It even does not ensure the existence of Invariant Subspaces. We now turn to this question. 5. Invariant Subspaces for Compact Operators.

We come back to the case of a compact operator T on a Banach space E.

Theorem 5.1 (Lomonossov). - Every non-zero compact operator has Hyperinvariant Subspaces. Proof. - We may of course assume (as we already said in Chapter II, 1), that T has no eigenvalue. Take a point Xl such that TXI i= a and take Xo = AXI, for A large enough, to get : inf{IITxll ; x E B(xo, I)} > O. We just write B for B(xo, 1). Let A = {T}' be the algebra of all operators which commute with T. We will show that there is a point Yo E E, Yo i= 0, such that :

IIT'yo - xoll 2: 1,

'tiT' EA.

(1)

This implies the Theorem: let F = span{T'yo ; T' E A}. Then F is an Hyperinvariant Subspace for T, and, since it does not intersect the interior of B, it cannot be the whole space.

Chapter IV

90

To prove (1), assume conversely that:

vv i- 0, 3T' E A, with IIT'y -

xoll
o in .c(X), such that, for all x EX, all eE X·, Anx --+ 0 and tAn e--+ 0, the following holds : For every e > 0, there is a N such that:

Then:

Theorem. - Let X be a Banach space satisfying (P), and let T be a noncompact operator on X. Let Tn be a sequence of compact operators such that: T;» --+ Tx, for all x E X for all

Then there exists a sequence Lan = 1 and : where K

of positive real numbers such that

(an)n~o

liT - Kil

e E X·

=

IlTlle'

= LanTn.

Moreover (though this result is not explicitly stated), the proof shows that, for every '1 > 0, the operator K may be chosen such that IITK - KTII < '1. The class of Banach spaces satisfying (P) has not received any satisfactory description, but it contains at least the I p spaces, 1 < P < 00. Conversely, it was proved by Benyamini - Lin [1] that in L p (1 < P < 00 1 P ::f- 2), there exists an operator without best compact approximant.

Chapter V

Topologies on the space of Operators

So far, we have seen only one topology on £(E), namely the topology of the norm. The purpose of this chapter is to introduce several others, which will be used later. We restrict ourselves to the Hilbert space setting; the extension to the Banach case is discussed in the "Complements" at the end of this chapter. An essential tool for this study is the following fact: the space £(H) is a dual space, namely the dual of the space of nuclear operators, which we now introduce. 1. Nuclear operators. Let H be a Hilbert space. An operator N on H will be called nuclear if in H, with Ilznll of complex numbers, with

there exist two sequences, (Zn)n~O, for all n, and a sequence (A.n)n~o such that, for all x in H,

(fn)n~o,

= 1, Ilfnll = 1

Ln IAnl

. its spectral decomposition: N(x>.) = Ax>., x>. E F>.. We know that all A's are real. Put:

H 2 = ED>.. ,

and define N 1 by N 1 = NIH 1 on HI, = 0 on H 2, and N 2 = -NIH 2 on H2, = 0 on HI. Then N 1 , N 2 are nuclear, self-adjoint, positive, and N = N 1 -N2.

Topologies on the Space of Operators

101

2. The space £(H) as a dual space. We are going to prove the following result :

Theorem 2.1. - The space .c(H) can be identified with the dual of the space J/(H) of nuclear operators, in the duality: (T, N) -+ tr (NT),

where T is in .c(H), N in JI(H).

For the proof, we require several lemmas :

Lemma 2.2. - If N is a nuclear operator,

IINIIJI

= sup{ltr (UN)I

j

U E .c(H),

II UII

:S I}.

Proof of Lemma 2.2. - First,

Itr(UN)1 S Next, taking U

= V·

sup

IIUNIIJI S IIUII·IINII.AI·

given by Proposition 1.1 b), we get:

Itr (UN)[

~

Itr (V· N)I =

tr INI

= IINII.AI'

11U1I~1

by Proposition 1.1, c), and this proves Lemma 2.2. We now introduce the weak topology on .c(H). For general facts about weak topologies on Banach spaces, we refer the reader to B.B. [1], Part I, Chapter 2. The weak topology on .c(H) is defined by the family of semi-norms Pz,y, x,y in H :

Pz,y(T) = I(Tx, y)l· An elementary neighborhood of 0 is of the form:

where Xl, ..• , X n , YI, ... , Yn are points in Hand e > O. A net (T a ) converges to T weakly if and only if for all z , Y in H,

(TaX, y)

-+

(Tx, y).

Chapter V

10e

Lemma 2.3. - The closed unit ball of £'(H) is compact for the weak topology. Proof of Lemma 2.3. - Let B = BH be the closed unit ball of H. It is compact for the weak topology on H (Alaoglu's Theorem, see B.B. [I], p. 40). Therefore the product BB is compact for the product topology (Tychonoff"s Theorem). For T in £'(H) , with IITII = 1, we consider the application:

A : T

--+

(TX)ZE8 E BB .

Obviously, A is injective from B£ (H) into B B , and is an homeomorphism if B£(H) is endowed with the weak topology. Let's show that the image of A is closed. Let (TQ ) be a net in B£(H), such thatA(T --+ t/J E BB, with t/J = (t/Jz)zEB. Clearly, t/J is linear. Write t/J(x) instead of t/Jz, and define T by TO = 0, Tx = IIxll.t/J(x/llxll) if z =1= o. Then AT = t/J. So ImA is closed, therefore compact, and B£(H) is compact for the weak topology. We write now £, for £,(H). Let £...... be the set of linear functionals on L, which are continuous for the weak topology. We define, for all x, y in H : Q )

Wz,y (T)

=

(Tx, y) .

Due to the definition of neighborhoods (1), an element only if it can be represented as a finite sum:

e is in .c", if and

n

E=

LWZi,Yi ,

(2)

1

for some points x}, ... , x n , Yl, ... ,Yn in H. We also consider £. , which is just the dual of L; equipped with the dual norm: II/II =

sup If(T) I ,

(3)

IITlI~l

if / E

e•.

The inclusion L . . . c L" holds. Indeed, if z , Y E H, we have:

IWz,y(T)1

= I(Tx,Y)1 < II x ll· IIYII·IIT II,

which proves that Wz,y E L" , and that

We can consider the closure of £'" in L" for the norm (3) ; we get a Banach space £ •. We will show that £ can be identified to the dual of £., and that L; can be identified with JI(H).

Topologies on the Space oj Operators

109

The first part follows from :

Lemma 2.4. - £(H) can be identified to the dual of the Banach space £ •. Proof of Lemma 2.4. - We first consider £ ..... as equipped with the weak topology 0(£"", £), topology of pointwise convergence on elements of L, Then £ is the dual of £ ..... , 0(£ ..... , £). Indeed, this is a special case of a general statement in Functional Analysis: if E and F are topological vector spaces in duality, the dual of F equipped with o(F, E) is E (see B.B. [1], pp. 24-27, or Bourbaki [1]). Here, E = £(H), equipped with the weak topology, and its dual is £"". So £ is the dual of £ ..... equipped with 0(£"" £), and the unit ball of £ is compact for this topology. The identity mapping:

(8.c,0(£, £"'))

(8.c, weak topology)

~

is continuous, since the linear functionals T ---t (Tx, y) (for z , y III H) are weakly continuous. Therefore, the inverse mapping is also continuous, and both topologies are identical, which proves our claim. Now, the dual of £'" endowed with 0(£"", £) is the same as the dual of L; endowed with 0(£., £), or the dual of L; endowed with the norm. This proves the Lemma.

Lemma 2.5. - Let (Xi)i~O,

L

II xil1

be two sequences with:

(ydi~O 2

is an isometry. Let's now show that it is surjective. All we have to show is that the image of ~ is dense in L; : the image of an isometry is always closed. Also, since this image is a vector subspace, it is dense in £. for the norm defined by formula (5) if and only if it is dense for the weak topology o(£.,£(H)) (B.B. [1], p. 34). Assume on the contrary that 4.>()'/) is not dense in L; for 0(£., £(H)). Since (again) it is a vector subspace, it must be contained in a closed hyperplane: there exists U i- 0 in £(H) such that, for every N in )I, ¢N(U) = o. We show that this is impossible. Since U i- 0, lUI i- O. Let Zl, ... ,Zn, with Ilzill = 1, mutually orthogonal, such that IU I i- 0 on F = span {Zl' ... , zn}, and let P be the orthogonal projection onto F. Then IUIP i- 0 on F, and \UIP = 0 on FJ... If x E H , we have:

(Px, x) and so P is a positive operator. Let N = PV·, where V· is given by Proposition 1.1, b), and satisfies lUI = V·U. We have:

¢N(U) = tr (NU) = tr (PV·U) = tr (PIUI) = tr (IUIP)

i-

0,

since IUIP is a positive non-zero operator. Since P is nuclear (as any finiterank operator), we get a contradiction, which proves our Lemma. We now finish the proof of Theorem 2.1 : 4.> is a surjective isometry, so we can use it to identify £(H) with the dual of ),/ (H), since it was the dual of L«, by Lemma 2.4. This proves our Theorem. From this Theorem follows that T

T

--+

sup{ltr(NT)1

j

--+

II T II t:(H)

N E ),/(H),

and

IINIIN:S

I}

Chapter V

106

are two equivalent norms on £(H). But we can show now that they are equal. Indeed, put E = JI(H), with its norm, E· the dual space, which is £(H), endowed with the norm IITII = sup{ltr (NT)I i N E J/(H), IINII.w ~ I}, and E;, the space £(H) with the usual operator norm. The dual norm of Ei is IINII(E;). = sup{ltr (NT)! i T E £(H), 11TIIop ~ I}, and this norm induces a norm on E by the same formula, the dual of which is Ei. But by Lemma 2.2, this induced norm is exactly the nuclear norm; therefore Ei = E+, and we obtain the formula: IITllop = sup{ltr (NT)j ; N E JI(H), IINII.w

< I}.

This may look like a complicated proof for a formula which is obvious anyway. Indeed, let's give a direct proof of it. One one hand, if IINII.w

< 1,

jtr (NT)l

O. We can find x E H, Ilxll = 1, with IITxl1 > (1 - e) IITII, and then y, with Ilyll = 1, such that (Tx, y) = IITxll. Consider now the rank one operator x 0 y. It is nuclear, and has nuclear norm at most 1. Furthermore, To (x 0 y) = Tx 0 y , so by formula (8), Section1 : tr (T

0

(x 0 y)) = tr (Tx 0 y)

(Tx, y) > (1 - c) IITII,

and our formula is proved. From now on, we make the (isometric) identification corresponding to and we just say that £(H) "is" the dual of JI(H).

~,

Since £(H) is a dual, it has a weak-* topology, namely a(£(H), J/(H)), which we now describe. We call ultraweak topology on £(H) the topology defined by the family of semi-norms :

(6) where (x,k2:0, (yil,~o

are two sequences of points in H, satisfying:

(7)

107

Topologies on the Space of Operators

Theorem 2.7. - The topology G(£(H), N(H)) coincides with the ultraweak topology.

e

Proof. - 1) Let be a linear functional on £(H), continuous with respect to the semi-norms (6). Then, there are sequences (Xdi~O, (Yi)i~O' satisfying (7), such that:

I€(T)I ~ P(Xi),(Yi)(T) , for T E £(H). By Lemma 2.5, € belongs to L«, which means that € is continuous for G(£(H), N(H)). 2) Conversely, we wish now to show that the elements of

£. are continuous

for the ultraweak topology. We start with elements of £"".

Lemma 2.8. - IE € E £"", there exists n eI, ... ,en, e~, ... , e~

~

1, and two orthonormal sequences

in H, a sequence of positive real numbers AI, ... , An,

such that

n

€

L AiWei.< .

=

I

Proof of Lemma 2.8. - We know by (2) that

€ can

be written:

p

LW

€ =

x i • yi

,

I

for some finite sequences xl, ... , X p i YI, ... , Yp in H. Consider two families (Xi)i~q, (y;k~q, such that the applications: p

It : Y ~ L(Y, Yi)Xi 1 q

h :

Y ~

L(y,y;·)x~. I

e

are equal. Let's put = 2:1 Wx'_.y'.. J J If u., v are in H, with the rank 1 mapping u. ® v, we get :

€' (u ® v) =

L (u ® v (xj), y;. ) i

L (xi, v) (u, y;.) i

= (v, L (u, yj)xi) i = (v,h(u)) = (v, It (u)) =

€(u®v),

108

Chapter V

by a similar argument. So € = €' on all rank one operators, and therefore on all finite rank operators. But if P is the orthogonal projection onto

obviously €(TP)

= €(T) , €,(TP)

= €,(T) , so €(T) = €,(T) for every T E

£(H). Now, we consider the operator: p

U

Y

-+

L(Y,Yi)Xi I

and write U = V lUI (Proposition 1.1). lUI is positive, self-adjoint, has finite rank, so there is an orthonormal basis el, ... ,en, made of eigenvectors, associated to the eigenvalues AI, ... , An, with Ai ~ o. Therefore:

i = 1, .. . ,n, n

Uy =

L

Ai(Y, ei)e~

,

I

with e~ = Vei. The sequence (e~) on ImlUI), and

is orthonormal (because V is an isometry n

L Ai(y,ei)e~

,

1

so

€ = L~=l

AiW e i .io IIxill2 < e/2, and

V = {T E .c(H) ; Then, if T E V

n B(H),

IITxill
O is an element of 12(H). Let on 12(H) defined by T((zi)i~o) = (TZi)i~O. Then:

14>(T) I ~

l'

(2: IlziI12)l/2
can be identified to a linear functional on it , continuous for the norm: therefore, there exists an element y in it , such that:

(Tx, jj) H '

4>(1')

00

4>(T)

=

L(Txi,Yi), o

Chapter V

114

which proves c). The fact that d) and e) are equivalent follows from Proposition 2.9, the fact that f) and g) are equivalent follows from Proposition 2.10. Clearly, a) implies d) which implies f). To show that f) implies b), we observe that if the restriction 4>ISM is ultra-strongly continuous, condition (1) above holds for T E BM, and therefore for all T EM: so 4> is ultra-strongly continuous on M.

Proposition 2.12. - Let M be an ultraweakly closed subspace of £(H) and 1. M its pre-annihilator (or pre-orthogonal), that is :

1.M

=

{N E J/(H)

j

tr (NT)

= 0,

for all T EM}.

Then the dual of the quotient J/(H)j1. M is M, and the weak topology u(M, JI j 1. M) is the topology induced on M by u(.c, J/).

Proof. - This is a general result in Functional Analysis: since M is closed in = (1. M)1. , and this result can be found in B.B. [1], Proposition

0(£, JI), M 3, p. 43.

In Part VI, we will apply the results of this chapter to the subalgebra of .c(H) generated by a single operator.

Topologies on the Space

0/ Operators

115

Exerclaes on Chapter V.

Exereise 1. - Let (en)n>O be the canonical basis of '2(1N), and Un be the operator defined by :

= (x, en)eo .

U« (x) a) Prove that

U~

is given by : U~

(x) = (x, eo)en ,

that the sequence (Un)n~O tends to 0 ultra-strongly, that the sequence U~ does not tend to 0 strongly, and that Un does not tend to 0 in norm. b) Deduce from a) that the norm topology is strictly stronger than the ultra-strong topology. c) Deduce from a) that the mapping T -+- T· is not continuous for the strong topology, nor for the ultra-strong topology. d) Show that the sequence U~ tends to 0 weakly. Deduce that the strong topology is strictly stronger than the ultraweak topology (and than the weak topology).

Exercise 2. - Let W be a neighborhood of 0 in £(H), for the ultra-strong topology. Show that there is a point z such that : sup

IITzll = +00.

TEW

(Hint: assume the converse, apply Banach-Steinhaus theorem and use exercise 1, b), to get a contradiction.)

Exercise 3. - Let (en)n~O be the canonical basis of 12(IN) , and Pn be the orthogonal projection onto F n = {Aen ; A E et}. Let Tm,n = Pm + mPn. a) Let (Xi)i>O be a sequence of points in H = '2(1N), with L IIxil12 < 00, and c > o. Show that one can choose m and then n large enough so that :

L IlT

2

m ,n X i 11

< e.

i

Deduce that the set .M = {Tm,n; m, n E IN} has 0 as an accumulation point, for the ultra-strong topology. (nkh~o be two increasing sequences of integers. Show b) Let (mk)k~O, that there exist x, y, such that (T mA: ,nA: x, y) does not tend to O. Deduce that no sequence (TrnA: ,nA:)k in .M converges weakly to o. c) Deduce from a), b) that £(H) is not metrizable for any of the weak, ultraweak, strong, ultra-strong topologies.

116

Chapter V

Exercise 4. - a) Let N be a finite rank operator, P the projection onto ImN. Show that: sup Itr (UPN)! = IINIIJI . 11U1I~1

Deduce that:

= IINIIJI .

Itr (KN)I

sup K compact, IIKII$1

b) Deduce the same formula for any N nuclear. c) Show that the dual of K (H) can be identified to )I (H), by means of the duality (K, N) ~ tr (K N).

Exercise 5. - Let Pn be the orthogonal projection, in l2(IN), on {Afn ; A E ~} . Show that, for any bounded sequence of scalars (an)n~o, the series L anPn converges for the strong operator topology, to an operator U, and that:

IIUII =

sup lanl· n

Deduce that t.(H) , equipped with its norm, contains a subspace isometric to loo(IN) (see B.B. [1], p. 107 for the terminology). Deduce that t. (H) is not a separable Banach space.

Exercise 6. -a) Let Bt. the unit ball of t.(H). Show that the application (U,T) ~ UT is continuous from Bt. x t.(H) into t.(H) , when they are both equipped with the strong topology. b) Deduce that the composition of operators is separately continuous for the strong topology. and (Tk)k~o of operators, which c) Deduce that for all sequences (UJc)k~O are strongly convergent to U and T, the sequence (UkTk) converges strongly to UT. Exercise 7. - We now show that on the product t. (H) x t. (H) , the composition of operators is not continuous for the strong topologies. Let z , Y be two points in H, with norm 1, and:

v

= {T E t.(H) ; I(Tx,Y)1 ::; I}.

Let also xl, ... , x n be a finite sequence of points in H, and

W = {T E t.(H) ;

IITxil1 ::;

We will find U and T in W such that TU

~

1, i

= 1, ... ,n}.

V : this will prove the assertion.

Topologies on the Space 01 Operators

117

a) Using exercise 2, show that there is a point z E H, that: sup

IITzl1

=

IIzl1 =

1 , such

+00.

TEW

b) Construct U E W , such that U x = AZ for some ..\ > c) Choose R E W such that IIRUxl! > 1. d) Construct an operator U', IIU'I! ~ 1, such that:

o.

(U'RUx, y) ~ 1.

e) Put T = U' R and conclude.

Exercise 8. - We equip £(H) with the weak topology. Show that the mappings T -+ T·, T -+ UT, T -+ TU (U fixed) are continuous.

Notes and Comments :

We have followed the presentation of J. Dixmier [1], with a few notational changes. We are greatly indebted to Francoise Piquard and Gilles Godefroy for their help in writing the Banach space version, which now follows.

Complements on Chapter V :

A. The space £(E) as a dual space. Let E be a Banch space, with dual E·. We consider the rank-one operators: for x in E, z" in E· , and z in E. On the space l(E·), we put the topology of point-wise convergence on the rank-one operators:

Ta

-+

T if and only if «(Ta

-

T)x·, x)

-+

o.

118

Chapter V

Then the unit ball B.c(E-) is compact for this topology. So .c(E*) is the dual of some space Z, which we now describe. First, we observe that by a computation analogous to the one we did for the proof of Theorem 2.1,

IITII =

sup{tr (T

0

z"

x ® x*) ; x E E,

E E*,

lIxll.llx*11 = I}.

Hence, by Hahn-Banach Theorem,

Bz = conv {x ®

z"

i x E E,

z"

E E*,

Ilxll.llx*11 =

I},

where the closure is taken for the norm of Z. Formally, this norm is given from the duality with £(E*), but it can be expressed more explicitly. The set of linear combinations of rank-one operators x ® z" is called the algebraic tensor product of E and E* , and denoted by E ® E*. On this space, the norm is by definition the gauge functional of the unit ball, that is the gauge of:

conv {x ®

z" ; x E E, z"

E E*,

Ilxll.llx*11

= I},

that is : Since the algebraic tensor product E ® E* is dense in Z, every u in Z can be written: 00

U

=

LU

n '

1

where, for all n, Un is in E ® E* and satisfies U can be written:

lIunll

(1)

u

with:

= I/2 n. This shows that

00

L

Ilxnll·l x~1I

:O, for all n, such that : XZ n

-

0,

of elements of A, with

z~ x -

0,

n - + 00.

For instance, I(t) = t is a divisor of 0 in C([O, 1]).

Ilznll

= Ilz~

II =

1

Banach Algebras

Proposition 1.4. - Every point in the boundary elements in A is a divisor of O.

aG

the group of invertible

Remark: Such a point is not itself in G, since G is open. Proof. - There exists a sequence X;I X

tt.

G, since

Xn

E G and x 1

:s: XZ

The same way,

ZnX ----t

:s:

Ilx~lx-I11

o. Set

which implies 1/llx~III-Jo

in G, with X n -Jo z , We know that G. So, by Proposition 1.1,

(xn)n~O

tt.

Zn

Ilx~11111x-xnll,

= x;I/llx~lll.

n = (x - xn)zn + 1/llx~111

Then

Ilznll = 1, and

-Jo o.

0, and the Proposition is proved.

Let now B be a closed subalgebra of A, containing 1. For x E B, we distinguish between: UB (x), spectrum of x considered as an element of B, and 0' A (z}, spectrum of x considered as an element of A. The comparison is as follows:

Theorem 1.5. - Let B be a closed subalgebra of A, containing 1. For every x E B, 0' A(x) e UB (x), and the boundary aUB (x) is contained in auA(x). Proof. - The first assertion is clear. For the second, let A E aUB (x). Then A- x is in aG B , boundary of the invertible elements in B. So A- x is a divisor of 0 in B, therefore in A. This implies that A- x is not invertible in A, and so A E UA(x). But A E aUB (x), and so there exists a sequence (An)n~O, An -----+ A, An E PB(x). Then An EPA(X), so A E auA(x) . We observe, however, that UA(X) and UB(X) need not be the same. As an example, consider the right shift S on 12(7L), and let 8 be the closed su balgebra of (H) generated by S, that is the closure in (H) of finite sums Lk~O Clk Sk. Trivial computations show that S-1 is not in this algebra, so o E UB (S). From Theorem 1.5 follows that UB (S) = D, though 0' C(H) (S) =

.c

.c

c.

Corollary 1.6. - IE PA(X) is connected, UA(X) = UB(X). Indeed, Us (x)\u A(x) is open, by Theorem 1.5 (and equal to PA (x)nUB (x)), and: which is the reunion of two disjoint open sets. An application of this Corollary is the following : if we have an operator T on H, and if we know that u(T) elR, then, in order to determine u(T) , we may work in any subalgebra of £(H) containing T and I.

Chapter VI

11!6

2. Ideals and Homomorphisms. Let 1 be an ideal in A, the quotient space Allis an algebra, where the multiplication is defined by multiplication of elements in each class. If x is the class of z , we also recall that the quotient norm is : IIzll

=

inf{llxli j x EX}.

The quotient mapping q, defined by :

q(x) = ±,

x E A,

from A into All, is an homomorphism. One checks easily that if 1 is closed, the quotient AlI is complete, therefore a Banach algebra, and that it has a unit, as soon as A does. An ideal 1 is said to be proper if 1 ¥= A. A proper ideal is said to be maximal if, for every proper ideal 1', the inclusion 1 c l' implies 1 = l' (in other words, there is no proper ideal which is strictly bigger). A proper ideal cannot contain any invertible element: otherwise, it would contain 1, and therefore the whole algebra.

Proposition 2.1. - Every maximal ideal is closed. contained in some maximal ideal.

Every proper ideal is

Proof. - Let 1 be a maximal ideal. The closure I is also an ideal, containing 1. But, since 1 does not contain any invertible element, it does not intersect the open ball {x E A jill - z II < I}, and neither does I. This proves that I is proper, and maximality implies 1 = I. Let now 1 be a proper ideal. We consider the family of all proper ideals which contain 1, ordered by inclusion. This set is inductive: if (Ii) is a subfamily which is totally ordered, the reunion uili is a proper ideal (since, again, none of them intersects the ball {z E A j 111 - z] < I}), which is a majorant for 1. Therefore, by Zorn's axiom (see for instance B.B. [1], p. 20), there are maximal elements: they are maximal ideals containing 1.

Example. - In the algebra C ([0,1]), the set of functions which vanish on a prescribed set A C [0, 1] is a closed ideal:

FA

= {f

E C([O, 1]) ; f(t)

= 0,

Vt E A}.

A maximal ideal is obtained when A is reduced to a single point: F{a}

=

{f E C([O, 1])

j

f(a) = O}.

127

Banach Algebras

An homomorphism mapping satisfying :

t/J

between two Banach Algebras A and B is a linear

t/J(xy) = t/J(x)t/J(y) ,

for all x, yEA

The multiplicativity property provides automatic continuity:

Proposition 2.2. - Every homomorphism t/J from A into G:: is continuous, and if t/J :j:. 0, then t/J(1) = 1, Iit/JII = 1 , and t/J(x) E u(x) , for all x EA. Proof. - a) The kernel K er t/J is an ideal in A. Since it is an hyperplane, this ideal is maximal, therefore closed, by Proposition 2.1. b) Let x, Ilxli = 1 . Then:

so It/J(x) I ::; 11t/J111/n. Letting n ~ +00, we get It/J(x) I ::; 1, so IIt/JII < 1. But if 1/; :j:. 0, there is a point x such that t/J(x) :j:. 0. Then t/J(x) = t/J(l).t/J(x), and 1/;(1) = 1. This proves also that IIt/JII = 1. c) For x E A, t/J(x) - x E K er 1/;, which is a proper ideal, and therefore contains no invertible elements. So t/J(x) E u(x). A non-zero homomorphism from A into

(V is

called a character. The spec-

trum of A, denoted by X(A), is the set of all characters on A. It is of course a

subset of A*, the dual of A. We put on A* the topology u(A*, A), that is the topology of pointwise convergence on elements of A. The theory which follows should remind the reader about similar results in Banach spaces: one shows that the unit ball of E* is weakly compact, and one considers the space E as a closed subspace of E**, identifying each x to the functional on E* : € ~ €(x) (see B.B. [11, Part I, chap. II, Sections 2, 3).

Proposition 2.3. - As a subset of the unit ball of A*, endowed with the topology u(A*, A), the spectrum X(A) is compact. Proof. - All we have to show is of course that x(A) is closed for u(A*, A). Let € E BA- , be an accumulation point of x(A) for u(A*, A). We know that, for all z , y in A, all e > 0, the set : VZ,Y;l:

= {" E A* ; I€(x) - ,,(x)1

< e , I€(y) - ,,(y) I < s , I€(xy) - " (xy) I < s}

is a neighborhood of € for u(A*, A). Since € is an accumulation point of X(A), this neighborhood contains a character f/J. But since:

€(xy) - €(x)E(Y) = E(xy) - fjJ(xy) + t/J(x)(fjJ(y) - €(y))

+ (t/J(x)

- €(x))€(y),

Chapter VI

128

this implies :

+ ItP(x) I + I€(y) I) ~ e(1 + IIxll + Ilyll)· E: > 0, we get €(xy) = €(x)€(y). The same way,

I€(xy) - €(x)€(y)I

~ e(1

Since this is true for every obtain €(I) = 1. So € is in X(A), and our Proposition is proved. For x in A, we define its Gelfand's Transform

x from A into C(x(A)). The correspondence x

--+

II x llC(x(A»

:

{tP --+ tP(x)},

--+

x is called the Gelfand's morphism. .If = {x; x

We have:

x by

we

We put:

E A}.

sup{lx(tP)1 i tP E x(A)} sup{ltP(x)1 ; 1/J E x(A)}

< r(x)

by Proposition 2.2

< Ilxll and this last formula proves that the application x

--+

x is continuous, from A

into C(X(A))So far, we have not exhibited any non-trivial ideals or characters, and in some cases there are none (see exercises). But further information can be obtained for the case of a commutative Banach algebra, to which we now turn. Though the whole algebra f,(H) is not commutative, commutative Banach algebras come naturally when one looks at the algebra generated by a single operator and identity, or by I, T and T'" in the case of a normal operator. 3. Commutative Banach Algebras.

Theorem 3.1. - Let A be a commutative Banach algebra. For every x in A,

u(x) = {x(tP) ; tP E x(A)}, and therefore

r(x) = sup{x(tP)

i

t/J

E x(A)}

=

II xllc(x(A»-

Proof. - IT A E u(x), x' = A - x is contained in a proper ideal, namely:

Ax' = {ux';

u E

A}.

This is an ideal, since A is commutative, and it does not contain I, otherwise x' would be invertible. This ideal is contained in a maximal ideal M. We now show that M is the kernel of some character.

Banach Algebras

1£9

Lemma 3.2. - An ideal is maximal if and only if it is the kernel of some character. Proof. - We have already seen that the kernel of a character was a maximal ideal. Conversely, let .M be a maximal ideal. Then the quotient AIM is a Banach algebra; let q be the quotient map. If some proper ideal I was contained in AIM, q-l(I) would be a proper ideal, containing M, with q-l(I) i- M. This is impossible, since M is maximal. So AI M contains no non-trivial ideal. Thus, every element is invertible (if an element x is not invertible, Ax' is a proper ideal), and by Theorem 1.3, AI.M is isometric to ~. Let 4> be the isometry from AI.M onto 0 q is a character on A with K er t/J = M.

The Radical of A, denoted by Rad(A), is the intersection of all maximal ideals contained in A. The algebra is said to be semi-simple if Rad(A) = {O}. By Theorem 3.1, we have: Corollary 3.3. - The following statements are equivalent:

a) x E Rad(A) , b) z E n,pEX(A){Y ; t/J(Y)

=

O},

c) X(t/J) = 0, for all t/J E X ( A),

d) r(x) = 0, e) lim IIx nll l / n

= o.

Theorem 3.4. - Let A be a commutative Banach algebra. morphism x ~ x has the following properties :

a) i(t/J)

= 1,

The Gelfand's

for all t/J E x(A) ,

b) A contains the constant functions, and separates points in A : that is, for all x and Y in A, there is a t/J in A, with t/1(x) i- t/J(y) .

c)

x is invertible in

d)

IIxllc(x(A» =

C(x(A)) if and only if x is invertible in A.

lim IlxnIl 1/ n,

e) Gelfand's morphism is bijective from A onto semi-simple.

A if and

only if A is

Proof. - a) follows from Proposition 2.2, since for every character, t/1(I) = 1.

b) We have A(t/J) = t/1(oX) = oX, and therefore A contains the constant functions. Moreover, if for every z E A, x(t/Jd = X(t/12), t/Jdx) = tP2(X) , and so

t/Jl

=

t/J2'

Chapter VI

190

c) % is invertible in C(x(A)) if and only if %(1/1) 1= 0, for all 1/1 in X(A). This is equivalent to 0 fI. 0' (x) , by Theorem 3.1, or to x invertible in A. d) This computation was made during the proof of Proposition 2.2, b). e) The mapping x --+ is injective if and only if Rad(A) = {O}.

x

We now apply this theory to the subalgebra of £(H) generated by a single operator. Let T be an operator on a Hilbert space H. Let's consider M, the largest commutative subset of £(H) containing T. The existence of M is ensured by Zorn's axiom. Of course, M contains I ,T, T2 ... , and R(A, T) for A E p(T). Moreover, M is an algebra: if AI, A 2 are in M, so are Al + A 2 , A IA 2 , and M is closed in operator norm. IT an operator A is in M, it commutes with every M in M. IT A is invertible in £(H), its inverse also commutes with every M in M, which proves that A-I is also in M. This shows that for every A in M, O'(A) is the same in £(H) and in M. If T is a normal operator, .M also contains T+ .

We may apply Gelfand's Theory to .M, which is a commutative Banach algebra, and we obtain:

Proposition 3.5. - We have:

O'(T) = {T(1/1)

j

1/1 E x(M)}.

In other words, for every A E X(.M), there is a character 'r/J of the algebra

.M such that : 1/1(T)

=

A.

The proof is an obvious consequence of Theorem 3.1 and of the fact that:

O'M(T) = O'(T). This Proposition will be used later (Chapter XIII) in order to show that some functions of an operator are not O. Indeed, if f(T) E .M,

1/1(f(T)) = f(t/J(T)) = f(A), and if we can choose A such that f(A) =I- 0, we obtain that f(T) is non-zero.

Banach Algebras

191

4. Functional Calculus. Let A be a commutative Banach algebra with unit. Let x E A, and 1 (x) be the set of functions which are analytic on o(x). For f E 1(x) , we define:

f(x) =

~ f 2,,1'"

l-

f(A)(A - x)-ldA,

where r is a curve around o(x) contained in the set of analyticity of f, and positively oriented. The properties of this functional calculus are the same as for the functional calculus described in Part I, Chap. II, 2 ; one gets, for instance:

f(o(x)) = o(f(x)). 5. C*-algebras. A mapping x --+ x" from A into itself, is an involution if, for every x, y in A, every A in «:,

(x*)* = x,

(x + y)* = z"

+ v' ,

(AX)* = Xx*, (xy)*

=

y*x·.

A C· -algebra is an algebra equipped with an involution, such that:

(1) We observe that 1· = 1.

Examples. - £(H) and K (H) are non-commutative C· -algebras, the involution being T --+ T* , the adjoint of T. - H X is a compact topological space, the space C(X) of continuous functions on X is a commutative C· -algebra, the involution being f --+ 7. In a C* -algebra, an element x is said to be normal if xx* = x" x, self-

adjoint if z"

= x.

Proposition 5.1. - Let A be a commutative C· -algebra. Then:

a) IIx*1I

=

Ilxll, for all

x E A,

b) x* is invertible if and only if x is,

d) If x is normal,

IIxl12 = Ilx 211,

e) u(x*) = conj(o(x)) ,

r(x) =

Ilxll,

Chapter VI

lSI!

where conj(n(x)) denotes {X; A E n(x)}.

Proof. - a) We have, by (1) :

and therefore

IIxll ::; Ilx*II, and the converse inequality follows,

replacing x by

z" . b) If x is invertible, xx- 1 = 1, so (x-1)*x* = l.

IIx 211 2 = II(x 2)*x211 = II(x*px211 II(x*x)*(x*x)11 = IIx*x11 = IIxll 4 , and so c) IT x is normal,

2

for n = 2 k , k = 1,2, ... , which implies Ilxnll 1/ n = Ilxll, and r(x) = d) and e) have the same proof as in Part I, Chapter II.

Ilxll.

Thus, by c), we have a new proof of Theorem 3.2, Chapter IV : a normal compact operator has an eigenvalue A with IAI = IITII. Indeed, we see here that for a normal operator, r(T) = IITII, so there is a point A E n(T) , with IAI = 11TH. But for a compact operator, such a A must be an eigenvalue, by Theorem 2.4, Chapter IV. H A, Bare C* -algebras, an homomorphism 4J is a " E X(AT,T" )

----+

p(4J(T), 4J(T+ ))},

with 4>(T+) = 4>(T) . By Proposition VI.5.5, the spectrum of T is the same in £(H) and AT,T" . By Theorem VI.5.4, t is an homeomorphism from X(AT,T·) onto u(T). Using this homeomorphism, we can identify these two sets, namely replace :

by:

{A E o(T)

----+

p(A,I)},

and we know that this last set is dense in C(o(T)). Considering G on the whole algebra AT,T- (and not only on polynomials), we get an *-isometry : G : AT,T.

----+

C(o(T)).

(1)

Chapter VII The image of T is the function {¢ ---+ ¢(T)} , which has been replaced, in our identification, by the function {-X ---+ -X}, -X E u(T). Let ~ be the inverse mapping of G, from C(u(T)) onto AT,T. : to each I E C(u(T)) we associate an operator I(T) in the algebra AT,T· . Of course, if I = p is a polynomial, we get the usual p(T). Now, if I is analytic on u(T), we write:

I(T) =

~

( f(-X)(-X - T)-ld-X. 2~1r Jr By Proposition VI.5.5, we know that if -X E p(T), (-X - T)-l E AT,T•. Therefore, the whole integral is in AT,T•. Finally, since ~ = c:», we get:

ITT)

=

I

I,

E C(u(T)),

and

(f(T)r = J(T*), by the *-properties of G. This new functional calculus is much more general than the one we defined in Chapter II : f needs only to be continuous on u(T), instead of being analytic on a neighborhood of u(T). We immediately give an application of this new functional calculus : Proposition 1.2. - A positive self-adjoint operator T has a square root: there is a positive self-adjoint operator U satisfying U 2 = T. Proof. - We know that since T is self-adjoint, u(T) cIR (Proposition V.5.3). Since T is positive, a(T) cIR+ (since every -X E u(T) is in aa(T) , there is a normalized sequence (xn)n~O such that TX n - -Xx n ---+ 0, so (Tx n, x n) ---+ -X, and the numbers (Txn,x n) are positive). We consider the function f : -X ---+ -X 1/4 on u(T) : it is continuous on IR+. Set Y = I(T). Then y* = f(T)* = J(T*) = f(T), since f is real and T self-adjoint. So Y is self-adjoint. Set U = v>. Then U is self-adjoint, and positive, since:

(UX,X) Finally, U2

= y 4 = T.

=

(y 2x , x )

=

Indeed, set g(-X)

(Yx,Yx)

= -X 4 •

= IIYxI1 2 •

Then go f(-X)

= -X

on u(T), so

g(Y) = T. The formulas proved in Chapter II, Section 2 (Proposition I, 2, Theorem 3, Proposition 4) are still valid here. But there is no need to discuss them now, since we are going to extend this new functional calculus further.

Normal Operators

2. Spectral Measures. We begin with some general considerations about vector valued integration ; in the next Section, we will apply them to the case of normal operators. Let Ii. be a locally compact space. A spectral measure E is a function defined on the Borel subsets of Ii. , with values in £(H) satisfying: a) for every Borel subset 6 b) E(0)

C

ti., E(6) is a self-adjoint projection in H,

= 0, E(ti.) = I,

c) for all z , y in E, if we define E z ,1I by :

E Z'1I(6) = (E(6)x, y), we obtain a regular measure on ti., equipped with the Borel a- algebra. (We recall that a measure JJ. is said to be regular if for all Borel subsets:

J.L(A)

inf{JJ.(O) ; 0 open containing A} sup{JJ.(K) ; Kcompact contained inA}.)

d) IT 61 , 62 are two Borel subsets of Ii. , with 61 n 62

= 0, then:

e)

This last property implies :

E(6) = E(6)2

E(6t E(6).

So:

E z,z(6)

=

(E(6)x,x)

=

IIE(6)xfl2,

and for every x in H, Ez,z is therefore a positive measure on Ii. , of total mass:

Chapter VII 3. Integration with respect to a spectral measure. We now develop an integration theory with respect to a spectral measure. By many respects, it is analogous to the usual integration theory. We first consider step-junctions : n

f =

L

Ci 16. ,

i:::::l

where Cl, ••• , Cn are complex numbers, 61 , ••• ,6,., Borel subsets of ~, and 16 is the characteristic function of 6 : it has value 1 in 6, 0 outside. Then, by definition, the integral of f with respect to E will be the operator:

/ IdE

= /

tCil"dE

=

tCiElii.

i:::::l

i:::::l

For z , y in H, we get : n

((/ fdE)x, y)

L ci(E(6i)X, y) I::::: 1

,.,

L Ci Ez,y(6i) 1:::::1

/ fdEz,y. Therefore,

1((/ fdE)x,x)1 And thus, sup IIzll~1

1((/ fdE)x,x)1 ~

sup If(t)l· t

But:

and so sup

1((/ IfldE)x, y) I < sup If(t) I, t

IIzll~I,lIyll~l

which means that:

II/ IfldEl1 ~ s~p If(t)l·

(2)

Normal Operators Applying (1) to

I

instead of

III, we get also:

11/ IdEIl

~ s~p

(3)

I/(t)l·

We call 800 (.6.) the algebra of bounded Borel functions on .6., equipped with the norm

11/1100 = Step functions are dense in BOO(~). of fIdE to functions I E Boo (~),

sup tEA

I/(t)l·

Therefore, we may extend the definition by posing :

/ IdE =

li~

IndE,

/

where (In)n~o is a sequence of step-functions, converging to I in BOO(~). This definition does not depend on the sequence (/n)n~o, This way, we get (2) and (3) for IE BOO(.6.) , and

((/ldE)X,y)

=

IdEx,y,

/

I

E

BOO(~),

x,y E

H.

4. Spectral Representation of a Normal Operator. We now use spectral measures to extend the functional calculus exhibited in Section 1. We first come back to the general setting of a commutative C·algebra, and then we apply the result to the algebra generated by a single normal operator.

Theorem 4.1. - Let if be a commutative C· -algebra contained in £(H), and let ~ = X(A). There exists a unique spectral measure E on ~ such that:

/A iue ,

U =

'rf U E A.

Moreover: a) The *-isometry

fI

--+

fA ius, from

an *-homomorphism :

I

--+

i

onto A can be extended as

C(~)

IdE,

from BOO(.6.) into £(H), with:

II

i

IdEl1

~ 11/1100,

I

E

BOO(~)

Chapter VII b) For every Borel subset fJ c

~,

every U in A, U and E(fJ) commute.

c) An operator V in £(H) commutes with all U in A if and only i6t commutes with all the projections E(fJ) , fJ E ~. Proof. - We know (Theorem 5.2, Chap. VI) that the application U ~ U is an *-isometry from C(~) onto A. Therefore, for all x, y in H, the inequality :

I(Ux,y)1
"dE.

u(T)

With this measure, we define, for I E BOO (u(T)) ,

I(T)

f.

=

IdE.

u(T)

This application extends the functional celculus, previously defined for I E l(T) , and then for I E C(u(T)). We now list some properties of this functional calculus:

Theorem 4.3. - Let T be a normal operator. We have:

a) If 11,12 E BOO(u(T)) , a,f3 E

~,

(all + f3h)(T) = ah(T) + f3h(T),

150

Chapter VII

b) (/(T))*

!(T*), for IE BOO (u(T)) ,

II/(T)II < 11/1100' for I E Boo(u(T))' d) 11/(T)xI1 2 = JU(T) 1J1 2 dE z ,z , x E H, IE Boo(u(T)) ,

c)

e) If U commutes with T and T* , it commutes with all operators I(T),

I E Boo(u(T)) , f) If In, I E 8 OO (u (T )), and if In converges to I uniformly on u(T)' then In(T) -+ I(T), in operator norm. If In(x) converges to I(x), Vx E u(T) , then In(T) ---+ I(T) for the strong operator topology. Proof. - a) to e) are just rephrasing of Theorem 4.2, f) follows from c) and

d). Remark. - One can show that if U commutes with T, it commutes with T· (Fuglede's Theorem, see Douglas [1] and Exercise 7). The spectral measure E is called a resolution of Identity. Indeed,

I = E(o(T)) =

fIdE. JU(T)

We have seen that the mapping I ---+ I(T) transformed characteristic functions into projections. Since step functions are dense in 8 OO (u (T )), we see that every normal operator is the limit (in operator norm), of a sequence of linear combinations of projections.

L2 •

The simplest example of normal operator is a multiplication operator on Fix a function 4> in L oo and consider the mapping: M~:

Then M; = Mj"

and

M~

I

---+

¢f.

is normal. We will show that, conversely, every

normal operator can be represented this way. We say that a point z E H is cyclic for (T, T·) if the set {U z; U E is dense in H. This means also that the set of finite combinations is dense in H.

L

AT,T*}

ak,ITkT· 1z

Theorem 4.4. - Let T be a normal operator. We assume that there is a point z E H which is cyclic for (T, T·). Then, there is a positive regular Borel measure J.L on u(T), and a unitary transformation A from H onto L 2 (u (T ), J,L ), such that for every g E L 2 (o(T ), J.L),

A E o(T).

Normal Operators

151

If 1 E BOO(u(T))' and if I(T) is the previously defined operator, then:

where MJ is the multiplication by

1

on L 2 (u(T ),Il ) .

Proof. - Let E be the spectral measure given by Theorem 4.2. Put Il Then, for 1 E BOO(u(T)),

(/(T)z, z) = Let H o = {/(T)z ;

Ho

=:)

1 E 8°O(u(T))}.

1

u(T)

=

Ez,z'

Idll.

Then:

{/(T)z; 1 E C(u(T))} = {Uz; U E AT,T·},

and since z is cyclic for (T,T·), H o is dense in H. We define A o , from H o into L 2 (u(T ),Il ) , by :

Ao(J(T)z) = 1 , If I(T)z

= g(T)z, then: o

=

11(1 -

g)(T)zI12 =

1

u(T)

If - gl2dEz ,z ,

which shows that A o is well-defined. It is an isometry, which can be extended into an isometry A from H into L 2 (u(T ),Il )' The image is the closure of BOO(u(T)) in L2 (u(T), Ill, which is L 2 (u(T ),Il) : the operator A is surjective. If g E B00 (u(T)) ,

Aof(T)g(T)z

Ao(Jg)(T)z

fg ,

which shows that:

and:

and our Theorem is proved. If there is no vector cyclic for (T, T·), one may split the space H into pieces in which there are such vectors, and one gets:

lSf

Chapter VII

Theorem 4.5. - Let T be a normal operator. There is a locally compact metric space X, with a Borel a -algebra T, a regular positive measure IL on T, a bounded continuous function h from X into (C, a unitary transformation A, from H onto L 2(X, T ,IL), such that:

We won't use this result in the sequel, and we refer the reader to DunfordSchwartz [I} for its proof. We now turn to the study of a simple example:

Example. - The operator M of multiplication by ei 8 on L 2 (TI, dO/ 21r ) (we just write M instead of M e i8).

Proposition 4.6. - The spectral measure E oE M is defined by :

where MIll is the multiplication by 16, characteristic function of 6, Borel subset oE TI, in the space L 2(TI,dO/21r).

Proof. - First, this operator is the same as the right shift on 12(71) , if we identify a function in L 2 (TI, dO /211") with its Fourier series. Therefore, its spectrum is the unit circle (see Chapter II). Now, let 9 E L oo (TI , dO/ 21r ) . The operator g(M) is the operator of multiplication by g, on L 2 (TI, dO /21r). Therefore: Mg = and if we take 9

= 16,

!

gdE,

we get the Proposition.

The scalar spectral measures E f,f are given by :

and more generally, for 9 E Boo (IT),

which gives : E

f,f

=

1112

dO . 21r

The invariant subspaces of this operator will be studied in Chapter IX.

Normal Operators

159

Remark 4.7. - An operator T satisfying, for every x E H :

IITxl1 = liT· xII , is normal.

Proof. - We write:

IIT*x1l2 = (x,TT*x) ,

and, with B, = T*T, B 2 = TT* , we get:

But B l

-

B 2 is self-adjoint, and:

IIB l

-

B 2 11 =

sup

112:11=1

1{(Bl

-

B 2 )x,x)1

0.

This remark applies to a surjective isometry, for instance the right shift on

l2(7.1). But it does not apply to an isometry which is not surjective: for instance, the right shift on l2(IN) is not a normal operator. 5. The spectrum of a multiplication operator. Let q, be a function in L oo ([0, 1], dt), and let M rfJ be the operator of multiplication by q, on L 2([0, I],dt). As we already said, this is a normal operator. We now describe its spectrum. Assume first q, to be continuous. Then A - MrfJ is invertible if and only if A - q, is an invertible function in L oo ([0,1], dt) , which means that A - r:/> does not take the value 0. So we get immediately that o(Mr/Jo) = q,([0, 1]). If q, is in Loo([O,I],dt) but not necessarily continuous, things are not so easy. Indeed, q, is only defined a.e., though l1(MrfJ) is a well-defined set. So, we introduce a new notion, the essential range of r:/>, denoted by RrfJ :

RrfJ

=

{A E e

j

W, neighborhood of A, p(q,-l (V)) > O},

where P is the Lebesgue measure on [0,1]. Then we get:

154

Chapter VII

Theorem 5.1. - u(M¢)

=

R¢.

Proof. - 1) Let A E R¢. Let Vn be the disk, centered at A, with radius lin, and let Un = ¢-I(Vn). Choose functions In in L 2([0,1]), with II In II = 1, and In = on U~ : this is possible since P(U n ) > 0. Then:

°

II(A - ¢)/nl ~

= / IA - ¢(t)121In(t)1 2dt =

1 IA -

¢(t) 121In(t)1 2dt

Un

~ supesstEunlA - ¢(t) 12 < I/n 2

,

by the choice of Un. So A - ¢ is invertible.

2) Conversely, let A f/; R¢. There exists a neighborhood V of A which satisfies P(¢-I(V)) = 0. Therefore, there exists an e > such that IIA¢lloo ~ e. So 11 (A - ¢) is in L oo , and is the required inverse for A - M¢.

°

6. Hyperinvariant Subspaces for Normal Operators. Theorem 4.2 gives obviously that every normal operator has Hyperinvariant Subspaces. Indeed, u(T) is not reduced to a single point: if u(T) = {a}, C(u(T)) is (C, so AT,T. is (C, which implies H = (C , contradicting our original assumption according to which H was infinite-dimensional (another argument is : if u(T) = {a}, r(T - a) = 0, so liT - all = 0). Therefore, we can find two disjoint Borel subsets 61 and 62 , with:

Put PI = E(c5d, P2 = E(c52) . Then PI, P2 are two projections, and F I = K er PI is the required Hyperinvariant Subspace. It is hyper invariant by Theorem 4.3, e). We observe, however, that the restriction of a normal operator to an Invariant Subspace is not a normal operator in general (we have already met the easy example of the right shift on 12(IN), restriction to 12(IN) of the right shift on 12(1L)). The reason is that the adjoint of TIF is not T*IF in general, unless F is itself invariant by T* : indeed, this adjoint is P(T*IF), where P is the orthogonal projection onto F. Such an operator: restriction of a normal operator to one of its invariant subspaces, is called subnormal. The question of the existence of Invariant Subspaces for subnormal operators remained open for a long time. Though one

Normal Operators

155

may think that the techniques developed in this chapter give strong information on the operator, they are unsufficient to solve this problem, which was settled only recently (1978) by Scott Brown [1], who gave a positive answer. His approach does not rely at all on Spectral Measures j it will be presented (in a slighly different context), in Chapter XIII. The question of existence of Hyperinvariant Subspaces for subnormal operators remains open. Another class, which is also an extension of normality, has received a positive solution to the Invariant Subspace problem. We say that an operator T is hyponormal if the commutant T·T - TT· is a positive operator. This is of course the case of any normal or subnormal operator. Very recently (1987), S. Brown [2] has shown that every hyponormal operator has Invariant Subspaces.

156

Chapter VII

Exercises on Chapter VII.

Exercise 1. - Let T be a normal operator on H, and E be its spectral measure. Show that A is an eigenvalue of T if and only if E ({A}) -I- O. If .\ is an eigenvalue, show that E({.\}) is the orthogonal projection onto Ker (A-T). Exercise 2. - Let T be normal, and put ITI = VT*T. Let V be such that T = VITI (cf. Chapter IV, Lemma 3.5). Show that there exists ¢, Borel function on o(T) , such that V = ¢(T). Deduce that V and ITI commute. Exercise 3. - Define T, from L 2 l1R) into itself, by : T f(t)

= f(t + 1).

Show that T is normal and find its spectral decomposition.

Exercise 4. - Let A be a self-adjoint operator. Show that: for all t EIR.

Exercise 5. - Show that for a normal operator, o(T) = oa(T). Exercise 6. Let T be a normal operator. Show that T is unitary (resp. selfadjoint) (reap. positive) if and only if o(T) C C (resp. o(T) C IR), (resp. o(T) cIR+ ). Exercise 7. - (Fuglede's Theorem). Let T be a normal operator, and A commuting with T. a) We put U = (T + T*)/2, V = (T - T*)/2i. Show that U and V are self-adjoint and that, for any A E ~ ,

b) Deduce that:

c) Show that AT* - AT can be written as iW>., where W>. is self-adjoint. Show that:

Normal Operators

157

Ded uce that :

d) Consider the mapping:

A

-+

eAT- Ae- AT- .

Show that this is a bounded entire function, with values in £(H). Deduce from Liouville's Theorem that it is constant, and conclude that: eAT- A = Ae>.T-.

e) Deduce that T+ A

= AT+

(Fuglede's Theorem).

Notes and Comments.

Our presentation can be found in many sources. Further information can be found in Dunford-Schwartz [1], J. Conway [1], and many others. A detailed study of essential ranges and multiplication operators, as well as generalizations (Toeplitz operators) can be found in the book of R. G. Douglas [1]. For vector valued integration, the canonical reference is Diestel-Uhl [11. Exercises 1 to 4 come from the book by J. Conway [11. Exercise 7 (a short proof of Fuglede's Theorem) is due to M. Rosenblum and is reproduced from H.R. Dowson [11.

Complements on Chapter VII.

As it is written in the above pages, the previous theory of spectral decompositions for normal operators does not make sense in a Banach space, since T and T+ do not operate on the same space, and so we cannot wonder whether they commute or not. However, spectral decompositions do make sense, and it is a legitimate question to ask in which Banach spaces and for what operators they hold.

158

Chapter VII

A nice theory in this direction has been built recently by E. Berkson, T .A. Gillespie, P. Muhly [1], [2], for invertible operators with bounded iterates :

After equivalent renorming, such an operator becomes a surjective isometry. It's quite natural to investigate first surjective isometries, since they are normal in Hilbert spaces and make sense in Banach spaces. Moreover, one knows (Proposition 1.12, Chapter II), that the spectrum of an isometry is always contained in the unit circle. Let X be a Banach space. A spectral family of projections is an application

t E IR

--+

E (t) E

.c (X) ,

where E(t) is a projection from X into itself, satisfying the following properties :

a) SUPtaR IIE(t) II < 00, b) E(s)E(t)

= E(t)E(s) = E(s), if s :S t.

c) For each to in IR, E(t) is right continuous at t = to, for the strong operator topology, and has a left limit for this topology. This left limit is denoted by E(t d) E(t)

--+

o)' I, when

t --+ +00, and

E(t)

--+

0, when t

--+

-00.

In the sequel, the above family will be concentrated on [0,211"1, which means that: for t :S 0, E(t) = 0 ,

E(t)

=I ,

for t

~

211".

An operator U is said to be trigonometrically well bounded if it can be written:

u =

fG>

eilJdE(s),

J[O,2'1f]

where E(.) is a spectral family of projections concentrated on [0,211"]. Such an operator admits a BV -functional calculus, where BV is the algebra of functions with bounded variation on the Torus. If we denote by L~ a summation with n t- 0, their theorem says:

159

Normal Operators

Theorem (E. Berkson, T.A. Gillespie, P. Muhly). - Let X be a reflexive Banach space, U an operator on it with bounded iterates. The following properties are equivalent:

a) U is trigonometrically well bounded, b) sup II N,t

N

1

L' -en

.

m t

U"1I
0, we set :

B~

(

1It- 81>6

B~

=

P:(9 - t)(g(t) - tf(ei8)) dt , 211"

P:(9 - t)(g(t) - tf(ei8» dt .

(

1I

t - 8 15: 6

IT

(We recall

Irl > 6 > 0, a

direct computation shows that:

IP r' (r ) I ~

°

B~,

2r(1 - r (1 _ 2r cos 6

when r

and this quantity tends to For

211"

~

2 )

+ r 2 )2 ,

1- . Therefore,

B~

0, when r

~

we write :

B~

=

-

(

11'15: 0

dr P:(T)(g(O - r) - (9 - r)f(e i8)) 2 . 11"

1

But P:(T) is an odd function, since P; is even. Therefore,

B~

=

(O P:(T)(g(O _ r) _ (8 _ T)f(ei8)) dr

-

10

211"

-

f

o

»-211"dr

"0

P:(T)(g(9 - r) - (8 - r)f(e'

-0

(O P:(r)(g(9 + r) _ g(8 _ r) _ 2r f(ei8)) dr

10

211"

= (O P'(T)T(g(8 + T) - g(8 - T) _ f(e i8)) dT .

10

r

2r

211"

~

1- .

169

A nalytic Functions But :

g(O + r) - g(O - r) = 2r

f.8+t 8-t

-+

1 ~.

-+

f(e i 8 ) ,

2~

f E LI(II,dB/21r). Therefore, for every

when T -+ 0, for almost every 0, since e > 0, if 6 is small enough, we have:

and then we let r

f(e i t ) dt

This proves our Proposition.

The convolution with the Poisson Kernel can be extended to measures : Proposition 2.3. - Let JL be a finite Baire measure on the unit circle, and define:

f(re i . ) Then

=

i:

f is harmonic in the open disk dJLr =

P,(O - t)d!,(t) .

and the measures :

f(re l

'8

dO

) -

2~

converge to JL in the weak-* topology of measures, when r

-+ 1- .

The proof of Harmonicity is as in Lemma 1.5 above, and if 9 is a continuous function, one checks immediately (using the fact that the Poisson Kernel is an approximate unit) that:

when r

-+ 1- .

3. HP Spaces. For p, 1 ~ P ~ 00, we denote by HP (II) the space of (equivalence classes) of functions f(e i 8 ) in Lp(II,dO/2~), whose Fourier coefficients cn(J) are 0 for n < 0. In other words,

(1) For f(e i8 ) E HP, we define:

f(z) =

L

cnz n

,

zED.

(2)

n~O

Since (Cn)n>O D.

-+

0, this series defines an analytic function inside the open disk

Chapter VIII

170

The reader will observe that we keep the same notation (/) for the function defined on C and its analytic continuation inside the disk. For p, 1 < p :S

00,

We now set, for 1 :S p
, '¢ are in H?",

so 4>'¢ E HOO . As a Corollary to Proposition 3.1 above, we give the following:

Theorem 3.'1 (F. and M. Riesz). - H JI. is a complex measure on Il , and if :

i:

.in"dIL(8) = 0,

for n = 1,2, ... ,

then JL is absolutely continuous with respect to Lebesgue measure: there is a function f in HI such that:

f dO . 2~

Proof. - For zED, we define :

j(z)

= /'11" _'II"

dJL(t).

1 - ze- l t

00

Pr(O - t) =

LTlnlein(fJ-t) , -00

Analytic Functions we have also :

f(z) =

r:

179

Pr(9 - t)dl'(t) .

(I)

Hence:

So

I

is in HI. But then we have: -·8 I(re' ) =

fir

dt . 211"

Pr(S - t)/(t) -

-1("

(2)

and the conclusion now follows if we compare (I) and (2) and apply Proposition

2.3. 4. Jensen's Formula. Jensen's Inequality.

The following Proposition gives a useful example of an harmonic function: Proposition 4.1. - Let n be a simply connected domain, and assume that I is analytic in n and does not vanish in this region. Then there is a function F analytic in n, such that eF = I. As a corollary, log I/(z)1 is harmonic in n. Proof. - Fix Zo in

n and put: F(z)

=

l I(~)d~

,

where r is any path from Zo to z, contained in n. Then F is analytic in n, and F'(z) = I(z), z E o. Since I does not vanish in n, I' /I is analytic in n, and we can apply this to f' / I instead of I : we get a function F, such that F' = 1'/1. We can add a constant to F, so eF(zo) = I(zo). Put h = le- F : then h' = 0, and h(zo) = I, so h is identically equal to I, and I = eF • Take now 9 = ~F. Then 9 is harmonic, III = eg , and log III = g. Proposition 4.2. - J~1r

log 11

-

ei81dS /21r =

o.

Proof. - Let n = {z ; ~z < I}. Since n is simply connected, and 1 - z f. 0 in n, there is a F analytic in n, such that eF(z) = 1- z, and F(O) = O. Since ~(I - z) > 0 in n, we have, for zEn,

UlF(z) = log 11 For small 6 > 0, let

r

- zl ,

I~F(z)1

< 11"/2.

be the path :

f(t) =

6~ t

~

211" - 6,

Chapter VIII and 1 the circular arc whose center is at 1, and passes from ei 5 to e- i 5 within the unit disk. Then :

27f -6 log 11 - et81_ . dO =

1 a

21r

1

ri-.-

I.

r

2Z1r

dz z

F(z)-

=

1 !R-.-

1

2Z1r,..,

dz

F(z)-, Z

by Cauchy's Theorem, and since F(O) = O. But the last integral is smaller than Co log 1/0, where C is a constant, since the length of 1 is smaller than 1rO. Letting 0 -+ 0, this gives the result.

Theorem 4.3 (Jensen's Formula). - Let n = D(O,R) , 1 be analytic in n, with 1(0) -I- O. Let 0 < r < R, and al,"" aN be the zeros of 1 in D(O, r), listed according to their multiplicity. Then:

IJ(O)!

ITN -) r I

n=l an

=

exp

17f

~

log I/(re i 8)1- . 21r

-7("

Proof. - We order the roots so that aIt ... , am are in D(O, r), lam+ll ···laNI = r . Put m

g(z)

=

I( ) z

2

_

N

II r(an-z) r - anz II

n=l

n=m+l

(an-z)'

so g(z) is analytic on some disk D(O, r + E:), E: > 0, and has no zero in this disk. Therefore, by Proposition 4.1, log Ig(z)I is harmonic in this disk, which implies: dO log Ig(rei 8 ) I- = log Ig(O) I .

1-7f 7("

But g(O)

1(0) n~

21r

I:nl . To obtain the formula, we observe finally that:

Indeed, each term r2 -

r(a n

iinz - z) ,

has modulus 1, and for n > m, each term

satisfies:

by Lemma 4.2.

n~m,

Analytic Functions

175

Corollary 4.4 (Jensen's Inequality). - If / E HI and /(0) =I 0,

"1 _,,"

dO log 1/(eSo)l- ~ 21r

log 1/(0)1.

Proof. - We cannot take immediately r = 1 in the previous theorem, since is not analytic in a larger disk D(O, R), with R > 1. For any r, 0 < r < 1, by Jensen's Formula:

"1 1"

_,,"

Fix e >

dO log I/(reSo)l- ~

I

log 1/(0)1.

21r

o. A fortiori,

_,,"

dO log(l/(reSO)1 + c) > log 1/(0)1. 21r

When r - 1-,

a.e. and in L I

•

Thus,

"1

_,,"

dO (log(l/(eSo)1 + c) ~

log 1/(0)1.

21r

Now, when c - 0, the sequence log(l/(eSO)1

+ c)

is decreasing, and thus:

Corollary 4.5. - For every function / in HI, not identically 0, the function log 1/( fSO) I is integrable. Proof. - First, we observe that:

and so all we have to show is that:

"1

_,,"

dO log I/(eSo) I > 21r

-00.

IT 1(0) =I 0, this follows from Corollary 4.4. IT 1(0) zng(z), with g(O) =I 0, and apply the Theorem to g.

=

0, we write I(z)

=

Chapter VIII

176

We observe that, as stated here, Jensen's inequality is a discontinuous estimate, in the following sense : assume that we have a sequence of functions (In)n~o, with In(O) --+ 0, and In --+ I (say: uniformly on II). Then, for each n, we apply Jensen's inequality to In, thus getting estimates which become worse and worse, when n --+ 00, since log I/n(O)1 --+ -00, though at the limit we apply Jensen's inequality to g = 1/zlc. This phenomenon is partly due to the fact that Jensen's inequality takes into account only the first coefficient of the Taylor expansion of I. A stronger version, taking into account a prescribed number k of coefficients in this Taylor expansion has been obtained by P. Enflo and the author (see the "Complements", at the end of this Chapter, in which the general problem of continuity of the functional :

will also be studied).

Corollary 4.6. - If a function I in HI is not identically 0, it cannot vanish identically on a set of positive measure. Indeed, since log III is integrable, the measure of the set where it takes the value -00 must be O. From Jensen's inequality, we deduce the following one, which can be interpreted as a similar inequality, involving the measure Pr(t - O)dO /21r instead of the measure dO /21r :

Proposition 4.7. - For any function

I

'll"

_'II"

I E HI, for 0 ~ r < 1, t E II,

"0 dO log I/(e' )IPr(t - 0) ~ 21r

. log I/(re't)l.

Proof. - Fix zo, Zo = rei! , and apply Jensen's inequality to the function:

4>(z) = I( z +

:0 ).

1 + zoz

We get:

I

'll"

_'II"

log 14>(f

dO

io)l-

21r

~ log 14>(0)1 = log If(zo) I·

Let A be the left-hand side of the above equation. Observing that (f i O+z)/(l +

A nalytic Functions

177

ze i 8 ) has modulus 1, we get, after a change of variables :

A

=

! !

i8(

~

2)

I I/( i8)1 e 1 - ,. _~ og e -zoe 2i8 + (1 + ,.2)e i8 ~ i8 1 _,.2 dB log I/(e )1 .812 _~ 11 - Zoe' 21r

L _~

Zo

dO 21r

log I/(ei 8 )IPr (t - 0) -dO , 21r

as we announced. 5. Factorization of HP functions: Inner and Outer functions.

We recall our terminology: l(e i 8 ) is the value of the function on the unit circle, obtained as a radial limit a.e., and I(z) is the value inside the unit disk. An inner function is a function m in H?", satisfying Im(ei8)1 = 1 a.e., and thus Im(z)1 ~ 1 for every zED, since the numbers Moo(m,r) are increasing with v (see Section 1). For instance, m(e i 8 ) = fin8 (n ~ 0) is an inner function. We will see other examples later, with Blaschke products and singular functions. An outer function F is a function in HI such that:

F(z) = where

IC

IC

exp

!

~ _~

eit + -.-t-

z dt k(t) - , e' - z 21r

(1)

is a complex number of modulus 1, and k(t) a real integrable function.

The meaning of formula (1) is not clear at this point. The following Proposition clarifies the links between k and F :

Proposition 5.1. - For a function F in HI,

k(O) = log IF(e i 8 )I , Proof. - For z

a.e.

= re i8 , we decompose:

where Pv , Qr are real: P; is the Poisson Kernel and Qr the conjugate Kernel:

Q (0) _ r

-

2,. sin 8 1 - 2r cos 9 + ,.2

Chapter VIII

178

Therefore, (1) can be written, for z

F(z) =

= rei 8 ,

exp(P,. * k + iQ,.

It

* k).

So we get:

=

IF(z) 1

exp(P,. * k) .

(2)

Since k E £1, P,. * k -----. k a.e., when r -----. 1- (see Section 2). Since also F(re i 8 ) -----. F(e i 8 ) (Proposition 3.1), our Proposition is proved. We observe that by formula (1), F cannot vanish inside the open unit disk.

Proposition 5.2. ~ Let F be a function in H 1 , not identically O. The following are equivalent: a) F is an outer function,

b) 1£ 1 is a function in H1 such that I/(e i8)1= IF(ei8)1 a.e., then:

IF(z)l

I/(z) I,

~

c) log IF(O)I

\jz E D,

I '" '" + 1

=

_'"

Proof. - 1) a) implies b). Let F be outer, and IF(ei8)\ on II. Then by Proposition 5.1,

F(z)

=

It

exp

dO

log IF(ei8)1- . 27r

1

in H1 such that I/(e i8)1=

ei8 z . dO -'-8- log I/(e'8)1 - .

_'" e' - z

27r

By formula (2) above, with k replaced by log 1/(ei8)1, log IF(re i8)1 =

f'" P,.(O - t) log 1/(eit)1 dt .

1-",

27r

Now, by Proposition 4.7, the right-hand side is larger than log I/(re i8)1. This proves our claim. 2) b) implies a). We consider the outer function: G(z) =

It

exp

I

'"

eit + z

-'-t-log

_'" e' - z

. dt IF(ed)l- . 27r

Since F is integrable, G is in HI, and:

lG(ei8)1

=

IF(ei8)1,

c.e.

Therefore, lG(z)1 ;::: IF(z) 1 in D, by b). Using again the fact that G cannot vanish in D, we find that F / G is analytic in D, has modulus 1 on C, and modulus ~ 1 in D : this is possible only if F /G is constant, and this constant must be of modulus 1. This proves that F is an outer function.

A nalytic Functions

179

3) a) implies c) is obvious on the definition of an outer function. 4) c) implies a). We define G as above. Then ]FI/IGI ~ 1 in D, = 1 on C, and IF(O) I/IG(O) I = 1. Therefore, F /G is constant. We now turn to a first decomposition of functions in HI

Proposition 5.3. - Let f be a function in HI, non identically O. Then f can be factored into m.F', where m is inner and F an outer function. This factorization is unique, up to a constant of modulus 1. Proof. - We put :

F(z)

=

exp

"1

eit + z "t dt eit _ z log If(e' )1 211" •

-,,"

This is an outer function. Put now:

f(z) m(z) = F(z) , one obtains a function analytic in the open disk, and Im(eit)1 = 1, by Proposition 5.1. Therefore, m is an inner function. If f = mlFl is another decomposition, IFI = IFll, so IF(z)1 = IFt{z) I in D by Proposition 5.2, and F = KFl , for some KE ~ ,with IKI = 1. 6. Factorization of Inner Functions: Blaschke products. Singular Functions.

In this paragraph, we study inner functions and get for them further factorizations.

Theorem 6.1. - Let

(an)n~o

be a sequence of complex numbers, with

n

Then the infinite product:

converges uniformly on each disk Izi ~ r < 1. Each an is a zero of B(z), with multiplicity equal to the number of times it is repeated in the sequence, and B(z) has no other zero in D. Finally, IB(z)1 < 1 in D, and IB(ei 8)1 = 1 a.e.

Proof. - Put ; bn

=

Ian I an - z an 1 - iinz

,

Bn

B'n

Chapter VI/I

180 Fix r < 1. For

Izl < r,

which shows that B(z) converges uniformly in the disk of radius r, and that B(z) is analytic in D(O, r}. Furthermore, each an is a zero of B with correct multiplicity, and B(z) =I- 0 otherwise. That IB(z) I < 1 if Izi < 1 is obvious, since it is true for partial products. The radial limit B(e i8) therefore exists a.e., and IB(e i 8 )1 ~ 1. It remains to show that IB(ei8)1 = 1 a.e. But for any function f in the monotonicity of M, (r, J) (Section 3, formula (4» gives :

n»,

We apply this to B~ = B/B n, and we observe that Ibk(e i8)1 lB n (ei8)1= 1 a.e., thus:

1 a.e., so

So:

I

""

IB~(reiO)1

-11"

dfJ 21r

~

111" IB~(eio)1 _,,"

dO = 21r

I""

IB(eiO)1 dO .

-11"

21r

But Bn(z) --+ B(z) uniformly on {lzl = r}, so B~ --+ 1 uniformly on this circle. This implies that f~,,"IB( ei8)IdfJ /21r ~ 1, and since IB(ei8)I ~ 1, we get i8 IB(e )1= 1 a.e. A function of the form :

with Ln(1 - Ian I) < 00, is called a Blaschke product. M is a positive integer, the set of an's may be infinite, finite or empty.

Proposition 6.2. - Let m be an inner function. Then the zeros an of m inside the open disk D satisfy Ln(1 - Ian I) < 00. Proof. Dividing if necessary by a power of z, we may assume m(O) =I- O. Let laol ~ latl ~ ... be the enumeration of the zeros of m inside the unit disk. Fix an integer N and choose r, with IaN I ~ r < 1. By Jensen's Formula,

" I

-,,"

log Im(re

iO) I -~

21r

~ log Im(O) I +

N

L log -anrII ' 0

Analytic Functions

181

which implies : N

2:o IOg -anr s 1

-log]m(O)I·

1

Letting r

-+

1- , we get : N 1 Llog-,-, < C, o an

with C = - log Im(O) I. But for 0 < x < 1, log l~:t

2: z . So we obtain:

and the series converges.

Theorem 6.3. - Every inner function m can be factored as m = B.G, where B is a Blaschke product and G is an inner function with no zero in the open disk. Proof. - We may assume that m has infinitely many zeros an, otherwise this is trivial. Dividing by zM if necessary, we may also assume that m(O) =f O. Put, as before, bn (z )

= -lanl

an -_ z , B(z ) an 1 - anz

= n°O bn () z

.

0

We know that this product converges, by Proposition 6.2 and Theorem 6.1. Put B n = n~bk, 9n = mlBn. For fixed n, e , IBn(z)1 > 1 ~ e, if ]zl is sufficiently close to 1. Therefore,

s~p Ign(re

i 8)1

s

1

1_ c '

and by monotonicity, this holds for all r < 1. So, letting e -+ 0, we get:

But gn -+ G has no zeros.

= m] B,

uniformly on each circle

Izl = r ,

So G is in H'" and

Chapter VIII

182

11], Duren [1]) that the function G, inner

It can be shown (cf. Hoffman

with no zeros inside the disk, can be written under the form :

G(z) = exp

I

'"

+z

ei8 -.-8-

_'" e' - z

dJl(6) ,

where Jl is a positive measure, singular with respect to Lebesgue measure. Such a function is called a singular inner function. The simplest example (obtained with the Dirac measure at 6 = 0) is the function:

z+l G(z) = exp - - . z-l 7. The Disk Algebra A(D). The Disk Algebra A(D) is the space of functions which are analytic inside the open disk D and continuous on the closed disk D : therefore, it is C(TI) n BOO , equipped with the norm :

IIflloo =

sup 8En

1/(ei 8 )f =

sup

1/(z)1 .

zED

This is of course a separable Banach space: the polynomials P+ are dense for the norm. This Banach space has an interesting property, connected with the extension of functions defined on a compact of 0 measure. This property will be used, in an Operator Theory context, in the next Part.

Theorem 7.1 (Fatou). - Let K be a compact in TI, with 0 Lebesgue measure. There exists a function 4> in A(D), such that:

Remark. - By virtue of Corollary 4.6, K must be of measure 0 for such a function to exist.

Proof. - Since K is compact, I1\K is the reunion of a countable family (In)n~o of open intervals, pairwise disjoint. Let en = P(In). For every n, let Yn be a positive function, of class Cion In, satisfying Yn < lie, Yn = 0 at the end-points of In, and:

A nalytic Functions

189

We put y = 0 on K, Y = Yn on In. The function y has the following properties : - 0 ~

- K

Y ~

lie,

= {O j y(O) = O},

- y is of class Cion IT\K and continuous on IT,

- log y is integrable. Put now w = log y. Then w satisfies : a) w = -00 on K, and if dist(O, K) --+ 0, w(8) --+ -00, b) w ~ -Ion IT, c) w is continuous on IT and of class Cion IT\K, d) w is integrable. We now set:

h(z) = =

I

'l " _'II"

eiB

+z

eiB _

dO

z w(O) 271"

(Pr * w)(O) + i(Qr * w)(O) ,

if z = re i O , where P; is the Poisson Kernel and Qr the conjugate Kernel (see Section 2). Since P; ~ 0, from b) follows that rRh :os: -1 in D. Set ur(O) = (Pr * w)(O) ; then U r --+ w, when r --+ 1- , uniformly on IT, since w is continuous. We now show that, since w is Cion IT\K, the functions v r (8) = (Qr * w)(O) converge uniformly on every closed interval contained in IT\K. This will imply that h(ei O) is continuous on IT\K.

Lemma 7.2. - If w is of class Cion a closed interval 10 - 00 1 ~ a, on this interval, V r converges uniformly to :

v (0) = _

Ill"

W

(0 + t) - w (0 - t) dt 2tan(t/2)

_'II"

271"

Proof. - Take 00 = O. The function

ePO(t)

=

w(O + t) - w(O - t) 2 tan(tI2)

is integrable on 1= {t ; 0 ~ t ~ min(O + a, a - O)}, and:

vr(O) - v(O)

=

=

I'll" cPB(t)(l- 2rsinttan(t/2)) dt -ll" 1 - 2r cos t + r 271"

,I .

_,..

2

r) 2

eP B(t)

(1 1 _ 2r cos t

+ r2

dt 21r .

Chapter VIII

184

And one checks for the Kernel gr(t)

(1 - r)2

=

1 - 2rcost + r 2

the following properties: 0 < gr (t) < 1, and gr (t)

--+

0, r

--+

1- , except for

t = O. The Lemma follows by a computation similar to the one made for the

Poisson Kernel (Proposition 3.1). We now come back to the proof of the Theorem. We set 9 = 1/h. Then, since h does not vanish in D, 9 is analytic in D, continuous in D, by Lemma 7.2, and the zeros of 9 on C are exactly the points of K. This proves our Theorem. We observe that ~g < 0 on D\K.

Corollary '1.3. - Let K a compact contained in the unit circle, with 0 Lebesgue measure. There is a function ¢ in A(D), with ¢(z) = 1 for every z E K, and I¢(z) I < 1 if z E D\K. Proof. - We take ¢ = erJ , where 9 is given by the previous theorem. Theorem 7.4 (Rudin, Carleson). - Let K a compact contained in the unit circle, with 0 Lebesgue measure, and let / be a continuous function on K, with complex values. There exists a function F in A(D), such that the restriction of F to K is f, and such that : IIFlloo = sup 1/(z)l· zEK

Proof. - Let ¢ be the function given by Corollary 7.3. For every h in A(D), we have:

Write ¢nh

= h + v«, where v«

is in A(D) and

v« = 0

on K. We get:

inf{llh + -rP1100 ; -rP E A(D), tP = 0 on K} inf II h n

~

+ tPlloo

inf II¢nhll n

lim II¢nhll n

sup Ih(z)l. zEK

But conversely, for every -rP in A(D), with .,p = 0 on K : Ilh + t/Jlloo

=

sup Ih(z) + tP(z) I > sup Ih(z) + ,p(z)[ zEC

zEK

sup Ih(z)l, zEK

Analytic Functions

185

and so : inf{llh +

1/11100 ; 1/1 E A(D), 1/1 = 0 on K} = sup Ih(z)l.

(1)

zEK

Call AK the set of restrictions to K of functions in A(D), equipped with the norm induced by C(K), and call S the subspace of A(D) consisting of functions 1/1, 1/1 = 0 on K. Finally, let R be the restriction operator, from A(D) into C(K). Formula (1) means exactly that for every h in A(D), IlhIIA(D)/s

=

(2)

IIRhllC(K) .

This means that R is an isometry from the Banach space A(D)/ S, into C(K), and therefore has closed range in C(K). The restriction operator from A(D) into C(K) has obviously the same range. We now show that this range is dense in C(K) : the restriction operator from A(D) into C(K)will therefore be surjective. Assume not. Since the range is a vector space, it is contained in a closed hyperplane: there exists a measure J.I. on K such that :

VI E A(D).

/ R(f)dJ.l. = 0,

By Hahn-Banach Theorem, J.I. may be extended to a measure on C(II) , also denoted by It. Therefore :

/ 1 lK In particular taking

1 = ei n 8 , / e

in8

dJ.l.

= 0,

n ~ 0,

VI E A(D).

we find:

lK dlt = 0,

Vn

~

O.

By the F. and M. Riesz Theorem (Theorem 3.7), this implies that l KdJ.l. is absolutely continuous with respect to Lebesgue measure. But since P(K) = 0, this implies It = 0, and the image of R is dense. So, R, from A(D)/S onto C(K) is a surjective isometry. This does not quite finish the proof of our theorem: given I in C(K), we want to find F in

A(D) with IIFlloo

= 11/1100'

We may of course assume that 11/1100 = 1. The fact that R is a surjective isometry implies that there are functions ¢ E A(D) and 1/1 E S, with:

¢

= 1 on

K,

14>1 < 1 on D\K,

(1)

186

Chapter VIII

,p Put 9

= 0 on K,

= + ,p. We have Igi ~ 0

There is an Cl such that:

1

II + ,plloo ~

(2)

2.

1 on K, Igl ~ 2 on C. Let 0

= {z E

1

be the set:

1

C ; Ig(z)! < 1 + 4"}'

> 0 such that II < 1 - Cion Oi, so we can find an rnl in IN on O~ .

On 0

1,

we get :

For k = 1,2, ... , let: 1

Ok

= {zEC; Ig(z)1 < 1+ 4k } '

Then the Ok'S are decreasing and contain K. Assume ml been chosen so that : !jm1gl + ... 2

+ _1_ mk- 1g l < k 1I 2 -

1-

~k 2

< ...
mk-lgl

1

< (2 + ... + < 1-

Using the fact that there is an ck a rnk such that:

+

1

4k- 1 )

1 2k •

> 0 such that I I < 1 - Ck on Ok we find

~1mlgl+"'+2~I4>mkgl and on

1

2 k - 1)(1

I

1) of the function:

f d,k (t)

2d

=

t. log -(t---l-)-((-:-~- -:~-)

Ie-+-I---l)

(4)

Let C(d, k) be the best constant satisfying (3) ; the precise value of C(d, k) is unknown. However, it was shown in [13] that, for d = 1/2,

C(I/2,k)

-2k log 2 ,

~

(5)

and that, asymptotically, when k --. +00,

C(d, k)

~

-2k.

(6)

The precise value of C(d, k) has been computed by A. K. Rigler, S. Y. Trimble, R. S. Varga [1] for the class of Hurwitz polynomials: these polynomials have real positive coefficients, and their roots are either real negative, or pairwise conjugate, with negative real part. For this class, the best constant they find (in the case d = 1/2) is -2k log 2, the upper bound in (5), and it is obtained for the polynomial (z + 1)2k /2 21e • The problem is also solved for other values of d and the extremal polynomials are given; we refer the reader to the above mentioned paper for an exposition of the complete results.

Chapter IX.

The Multiplication by ei 9 on H2(ll) and L 2(ll ) .

We consider in this Chapter the operator M of multiplication by ei 9 : f ---+ ei 9 f . Clearly, this operator is an isometry, both on H2 and on £2. The difference is that it is surjective in £2, and not in H 2 • IT we identify each function in L 2 with the sequence of Fourier coefficients:

/ =

L

C1c

ei k9

---+

(c1ch E z

,

1cEZ the operator M becomes the right shift on , 2 (1l ) : indeed, (c1chEZ is replaced by (ck-dkEZ when / is replaced by ei 9 f. The same holds on H2

/

L

C1c

ei 1c9

---+

(ckh E IN '

1cEIN and: S : (co, Cl, C2, ..• )

---+

(0, co, Cl, C2,' •• ),

is the right shift on ' 2(IN). In both cases, we use the words "right shift" to indicate that this operator pushes to the right. Some people use "bilateral shift" to designate the shift on 12(1l). We find this terminology quite confusing: it is not the operator which is "bilateral", it is the space !

The operators M or S have obvious invariant subspaces. In ' 2 (IN), the sets:

are clearly closed and invariant for S. The definition of Fn also makes sense in 12(1l). However, if we look at M on £2 [Il}, we see immediately another type of Invariant Subspace: Fix a

1ge

Chapter IX

measurable subset A elI, with 0 < peA) < 1 (P(A) is the Lebesgue measure of A), and consider :

This is a closed subspace of L2(II,d6/21r}, invariant by M, not equal to {a} since peA) > 0, and not equal to the whole space since peA) < 1. Let's call Invariant Subspaces of the form F n "of type 1"', and of the form G A "of type II" (we will give precise definitions later). We already see that the Invariant Subspaces depend on the frame: in H2 and L 2 the situation does not look the same : only type I seems to exist in H2, and both types exist in L 2 • We now investigate this question more in detail. 1. The multiplication by ei O in H2 .

Theorem 1.1. - Let F be a closed subspace of H2, F M. There exists an inner function m such that :

t-

{O}, invariant by

(1) Conversely, any subspace of this form is closed and invariant for M.

Proof. - Obviously, a subspace F given by (1) is invariant for M. It is also closed, because the multiplication by an inner function m is an isometry from H2 into itself (since Im(eiO)1 = 1 a.e.) and the image of an isometry is always closed. Let now F be a closed subspace, invariant by M. If all functions f(z), EM, vanish at 0, we may write, for some k > 0, F = zk Fa, where Fa is invariant, and contains a function f which does not vanish at o. This way, we need only to consider the case when F contains such a function, say fa. We denote by 9 the orthogonal projection of the constant function 1 onto F. This means :

f

1

=

9

+1-

g,

where 9 E F, 1 - 9 E FJ.... The function 1 is not in FJ..., since:

Therefore, 9

f:. o.

We will show that

Igl

is constant on II.

Multiplication by ei8

Igl 2 nt-O, But

199

takes only real values. Taking conjugates, we get, for all n E 1l,

!

'8

ll'"

e,n

'8 2 dO Ig(e')1 -

21r

-ll'"

= 0,

which implies that Igi is constant : there is a A E

Ig(e i 8)1

=

such that :

~

A a.e.

We deduce that Ig(z)1 ~ A in D, and g/A is an inner function. Let's show that F = g.H2. First, we observe that, since 9 E F and F is invariant, the functions g, e i8 g, e 2i8 g, . . . , are a 11·In F ,an d so are th e fi rnit e sums '" L..,..k~O a/ce ik8 g, tha t is the products p.g by polynomials. Let now f be in H 2 , and (Jn) n~O be a sequence of polynomials in P+ , converging to f in H2 (for instance the partial sums of the Fourier series). Then, since 9 is in H?", ot« --+ gf in H2. Therefore, gf E F and gH2 C F. Assume this inclusion to be strict. Then, there exists a function f in F, f .1 9 H2. In particular, for n 2:: 0,

! L

ll'"

. dO e- 1n8fg - = 0.

(2)

21r

-ll'"

But 1 - 9 .1 ein 8 f , since 1 - 9 .1 F :::> ein8F. So, for n e i n 8 (1 -

g)f -dO

21r

-ll'"

Therefore,

! So:

l '" ein8f -dO -ll'"

21r

=

=

2:: 0,

0.

!ll'" ein8f-gdO- . 21r

-ll'"

l '" e,.n 8f g -dO ! -ll'" 2,..-

=

0,

and by (2) we get that this equality holds for every n E ~.

°

°

This proves that fg = 0. Since Igi = A t- a.e., we must have f = a.e., and F = gH2. As we already said, we get an inner function if we replace 9 by m

= g/A.

Chapter IX

19-4

Corollary 1.2 (Benrling's Theorem) - Let f be in H2. The vector space spanned by functions ei n 8f , n ~ 0, is dense in H2 if and only if f is an outer function. Proof. - We put

Ff

=

{ f, ei8f , e2i9f , ... } span

Clearly, Ff is invariant by M. a) Assume Ff = H2. We write the decomposition:

f = m.F, into inner and outer parts. Then :

which implies that m is constant, and that f is outer. b) Assume f to be outer. Since Ff is a closed invariant subspace for M, by Theorem 1.1, there is an inner function m such that :

This implies in particular that there is a function h in H2 such that :

f We write h

= hi.ho, where

=

m.h.

hi is inner, and h o is outer. Then:

and mh, is inner. Since f itself is outer, the uniqueness of the decomposition into inner and outer parts implies that mk; is constant, which in turn implies that both m and hi are constant. Therefore, F] = H 2 • We observe that the above proof gives in fact the more precise result:

Corollary 1.3. - If f is a function in H2, and f = /i.fo is its decomposition into inner and outer parts, we have:

Multiplication by ei 8

195

2. Szego's Theorem. We now derive Szego's Theorem from Beurling's Theory. There are, however, direct proofs (see Hoffman [1]). Let A0 be the subspace of A(D) consisting of the functions / which satisfy /(0) = O.

Theorem 2.1 [Ssegd]. - Let h be a real positive integrable function on II. Then: dO (1) inf 11 - g(e18 )12 h(O) -dO = exp log h(O) - . gEAo

j lf

.

jlf

271"

-If

211"

-If

Proof. - Assume first that log h is integrable, that is J~lf log h dO/211" > Put: 1 ei 8 + z . dO G(z) = exp -'-8log h(eI 8 ) - • l

-00.

jlf

2

-If

e

-

Z

21T'

Then G is an outer function,

and G 2(0)

= exp

I.

-If

log h(O) -dO . 27r

If 9 E Ao, we have:

But the Oth Fourier coefficient of (1- g)G is G (0) (since / (0) = 0). Therefore:

This proves one inequality in (1). Conversely, since G is outer, by Beurling's Theorem, there are polynomials Pn(z) such that PnG converges to the constant function G(O) in H2. Then Pn(O)G(O) --+ G(O) =j:. 0, and Pn(O) --+ 1. We can assume Pn(O) = 1. Thus we can write Pn = 1 - gn, and gn is in Ao . But then:

which proves the converse inequality.

196

Chapter IX

Now, if we assume f~ff logh(e i 8 )d8/ 21r = - 00, we observe that for every e > 0, log(h + e) is integrable, and by the preceding computations, ~

inf

I I

gEAo

= exp

ff

-ff

ff

log(h(8)

~

d8

+ E:) - •

-ff

When e

d8

.

11 - g(e ' 8 )12(h(8) + e) -211" 211"

0, this last expression tends to 0, and we get :

Corollary 2.2. - Let / be in L 2 (TI, d8/211"). Then

jf and only jf:

I

ff

log 1/(ei8)1

-ff

d8

=

-

2'1['

-00.

This follows immediately from Szegd's Theorem, applied to h =

1/1 2 •

3. Multiplication by ei B on L 2 (TI, d8/ 21r ) . Let now J.L be a positive Borel measure on TI. The operator M : / ~ eie/ is a surjective isometry from L 2 (TI, dJ.L) onto itself. As we already said at the beginning of this Chapter, in the case where J.L is the Lebesgue measure, M can be identified to the right shift on 12(1L). We now describe its Invariant Subspaces.

Proposition 3.1. - A closed subspace F contained in L 2 (#-, ) satisfies MF = F if and only if there exists a measurable subset A C TI, such that: F = {/ E L 2

j

/(6)

0 c.e. on A C }

=

Proof. - Set L 2(A) = {/ E L 2 j /(8) = 0 a.e. on A C } . Clearly, L 2(A) is closed, and ML 2(A) =L 2(A). Conversely, let F be a closed subspace of L 2 (#-, ), and assume ei BF = F. Then also e- i BF = F, and, more generally, einB F = F, \In E 1L. We first study the case when F is generated by a single function [; For / E L 2 (#-, ), we set: Ff = span {e in8/

j

n E

Z},

Multiplication by ei B

197

where the closure is taken in L 2 (IL) . By Stone- Weierstrasa Theorem, every function gEe (II) is the uniform limit of trigonometric polynomials, that is of finite sums L~K ateik8. Therefore, for every function 9 E C(II), 9 I E Ff. Let h be a function in Fj. Then:

In

gIlt

for all 9 E C(II),

= 0,

and {h E L 1 (IL). So Ih = 0 a.e., and Ih = O. Let B = {8 j 1(8) = O}. Then all functions in Ff are 0 a.e. on B. Conversely, if tP = 0 a.e. on B, and h is orthogonal to F, tPh = 0, by what we just proved. Therefore, tP E F.L.L = F, and Ff = L2(A) , with A = BC. The Proposition is therefore proved for Fj . We now turn to the general case: F satisfies M F = F. Let (In)n>o be a dense sequence of functions in F. For each n, let :

An = {O

j

fn(O) f O}.

Then:

with A

= UA n .

Proposition 3.2. - A closed subspace F

C

L 2(II, IL) satisfies :

if and only if there exists a Borel function tP such that:

(1) Proof. - a) If ¢ satisfies (1), ei B¢ H 2 C H2, so F is invariant by M, and:

b) Conversely, assume that F satisfies nn~oein8 F = {O}. If G c F is a closed subspace, we denote by F e G the orthogonal complement of G in F. So we put:

Chapter IX

198

Then, for n

~

0,

and therefore : '\J7n~Oe

F

F

in8L C

In.

e (EBn~oein8L)

,

= {O},

= nn~oein8F

which means :

=

F

in8L

In.

'\J7n~Oe

=

Let 4> be any function in L, with 114>112

4> ..1 Therefore,

e

in 8

•

1. We get, for n ~ 1

4>.

[1r 14>1 2

e i n 8 dJ,t

=

O.

ff

Taking conjugates, we obtain the same equality, this time for all n E Z, n =j:. O. This implies that the measure 14>1 2 dJ,t has all Fourier coefficients equal to 0, except the Oth, which is 1. So we get:

(2) The space L has dimension 1 : assume that there is a function 4>' in L with 4>' ..1 4>, in L 2 ( ,.,.) . Since 4> ..L e i n 8 4>' , for n > 0, we get:

[~~ t/>~'e-'n8dll

=

> 0,

0,

n

0 ,

n>

(3)

and since 4>' ..1 e i n 84> , for n > 0, we get :

[~~ t/>'~e-'nodll

o.

Taking conjugates,

[~~ t/>~'' n8dll

=

0,

n > O.

(4)

Using finally the fact that 4> ..1 4>' , we obtain from (3) and (4) :

!_~~ t/>~'

e'nodll

=

0,

n E :1:,

which means that 4>4>',.,. = 0, or 4>'14>1 2 ,.,. = O. But 14>12d,.,. = ~:, so 4>' = 0 a.e. L, each element f in F can be written : Since F = EBn~oein8

f

=

La

ne

in8

4> ,

n~O

with

L lanl 2 < 00.

So F

= 4>.H 2 ,

and our Proposition is proved.


199

This Proposition can be strenghtened :

Proposition 3.3. - If M F c F and M F ¢ such that:

Proof. - As previously, we set L

=

G

i- F, there exists a Borel function

= F e ei 8 F.

Put:

L EB ei 8 L EB e2i 8 L EB ••• ,

and L o = FeG. The operator M is surjective on L o : if we take a function h in L o such that h ...L ML o , Mh EMF, so h ...L Mh, and thus h E L. Therefore, h E L o n L implies h = O. From Proposition 3.1, we deduce that there is a Borel subset A C II such that L o = L 2(A). We now take ¢ E L, and we repeat on G the proof of Proposition 3.2 : we find J¢1 2 dJ,t = dO/21r. But ei n 8 ¢ ...L 1A.¢, since Lo...L ei n 8 L , for n> 0, and 1A4> E L o = L 2(A). This means:

[ffff e

in8

j-fff

e

o,

1A 1¢1 2 dJ,t

in8

dO 1A = 0,

n> O.

n> O.

21r

Taking conjugates, we get that, for every n E 7l, n

I.-ff

ei n 8 1 -dO = 0 A

2

1r

=f. 0,

'

which means that 1A is constant. This is possible only if A has measure 0, and we are back to the previous case.

Proposition 3.4. - Let I E L 2(II,dO/21r). The functions ei n 8 1, n > 0, generate L 2 if and only if we have simultaneously:

a) P{O ; 1(0) = o) = 0, b) f~ff

log 1/(0)IdO /211" =

Proof. - As before, we set :

-00.

Chapter IX

200

1) Assume first F, = L2. Then, by Proposition 3.1, a) holds. We now show that: inf jW" 11 _ (ale itl + ... + akeik8)121f12 dO = O. (1) kjah···,ak

27r

-Ir

Lemma 3.5. - Let I E L 2 , such that P{9 ; 1(9) functions ei n 8 I, n ~ 0, generate L 2 if and only if :

=

O}

= o.

Then the

- { ei8,e 2i8 , . .. } , 1 E span

in the space L 2(II, 1/1 2 d9/ 27r) .

Proof of Lemma 3.5. - Assume first that the functions ein8I, n ~ 0, generate L 2 . For every E > 0, we can find a finite sequence of scalars ao, ... , ak , such that: W" "8 ~ " "8 2 d9 [e-' I - L" aje" II -W" 27r j=o

j

and so,

which means that : - { ei8,e 2i8, ... } , span 1 E -

in the space L 2(II, 1/1 2d9/ 27r) . Conversely, one checks immediately that, if this holds, also e- i 8 , e- 2i8, ... , belong to span{ei8,e2i8, ...}, in the space L 2 (II,1/ 12d9/27r). Therefore, in this space: span {e in8 ; n ~ I}

=

span {e in8 ; n E 1l}.

This equality implies that in the space L 2 (II, dO /27r), span {e in8I

;

n ~ I} = span {e in8I

I

By Proposition 3.1, this last space is L 2 , since Lemma and therefore (1).

;

n E :IL}.

satisfies a). This proves the

By Szegd's Theorem (Section 2), we have: inf gEA o

L -W"

and this means that f~1r

11 -

gl21/1 2-d9 = 27r

log I/ld9/27r =

exp

L -W"

log 1/1 2 -dO ,

27r

-00.

2) Conversely, let's assume that a) and b) hold. Once again, by Szegd's Theorem, (I) is satisfied, and by Lemma 3.5, we get F] = L2.

eoi

Multiplication by ei fJ 4. Multiplication by x on L 2[O,I].

We now turn to the study of the multiplication by x on the space L 2[O,I]. Obviously, if A is a measurable subset of [0,1], and if we set, as before,

we obtain a closed invariant subspace of Mr. in L2[0,1]. It is non-trivial if and only if < P(A) < 1 (of course, P is now the Lebesgue measure on 10,1]).

°

Conversely, we have:

Proposition 4.1. - Any closed subspace of L 2[0, 1], invariant for Mr., is of the form L 2(A), for some measurable set A C [0,1]. Proof. - Let F c 10,IJ with Mr.F C F. We may of course assume that F is not the whole space. Take a function f E F, f 1= 0. First, we consider:

Since Ff is not the whole space, there is a function 9 in L 2[0, 1], 9 =f. 0, with 9 1.. Ff. Therefore,

i

1

x· f(x)g(x)dx

=

0,

k

~

o.

This implies that for every polynomial p,

f

p(x)f(x)g(x)dx

O.

Since polynomials are dense in L 2 [0, I], we get:

° a,e. = ° a.e. on

f(x)g(x) =

°

AC , < P(A) < 1, and Set A = {x ; I(x) 1= O}. Then f Ff C L 2(A). If this inclusion was strict, we could find a function h in L 2(A), h 1.. Ff. 1 Then, as before, we would have f0 pfhdx = 0, for every polynomial p, which implies I.h = o. So I would be 0 on the set where h is =f. 0, which is a subset of A of positive measure. This contradicts the definition of A, and proves our assertion for F]: For F itself, we do as in the proof of Proposition 3.1 : we take a dense sequence (/n)n:;:::o in F, and we obtain F = L 2(A), where A = uA n , each An being the set {In 1= O}.

eoe

Chapter IX

Corollary 4.2. - For every function

where A = {x ; I(x)

1

in L 2[0, 1],

i- O}.

Corollary 4.3. - For every polynomial p,

Therefore, the operator M z on L 2[0, 1] has an interesting property, in terms of cyclic vectors. There is a cyclic vector 10 (namely the polynomial 1) such that for every polynomial p, p(Mz)/o is also cyclic. Let's understand what this means for the functional calculuses and the Invariant Subspace problem. We have seen (Proposition 4.1) that the non-cyclic elements were the functions 1 with non-trivial support: P({/(x) = O}) > 0. We may wonder: starting with a cyclic element, like 10 = 1, how can we get to a non-cyclic one, 1, using a functional calculus ? This question makes sense in general (that is : not only for this operator), and is precisely at the origin of the attempts to use functional calculuses to solve the Invariant Subspace Problem: given an operator T, we start with a cyclic point xo, and we try to find a function 1 such that I(T) is well-defined and I(T)xo is non-cyclic. Such attempts will be described in the next chapters. Here, the answer is fairly easy to give: if f is a polynomial, I(Mz ) 1 = f will still be cyclic, and the same remains true if f is in the Disk Algebra or in H?? . The HOO functional calculus has not been introduced yet (see Chapter XI), so let's accept temporarily the following fact (which is easy to admit) : for 1 E H?", the operator I(Mz ) is the multiplication by f(x) on [0, 1[. Of course, since f is analytic on a neighborhood of the segment [0,11, we get that 1 still satisfies P({/(x) = O}) = 0. If we want to get to a non-cyclic function, we have to take 1 in L oo , vanishing on a set of positive measure on [0,1]. In the present case, f(Mz ) makes sense when f E L oo ' because the operator M z is normal, and u(Mz ) = 10,11 (see Chapter VII, Section 5), and f is bounded on u(Mz ) . But such a functional calculus, admitting functions which are not analytic, just bounded on u(T), does not exist for other classes than normal operators. This is why, in my opinion, the idea of solving the Invariant Subspace Problem by means of a functional calculus does not look very promising.


e09

We have studied in detail three operators: multiplication by ei 8 on H2, and on L 2 , and multiplication by x on L 2[0,1]. We observe that, quite interestingly, the types of Invariant Subspaces encountered are quite different. The first one has Invariant Subspaces of type m.H2, the second one has Invariant Subspaces of type m.H2 and L 2(A), and the third one, of type L 2(A) only. All separable Hilbert spaces are of course isometric, and these three operators look "more or less" the same. But their Invariant Subspaces are quite different conceptually. This indicate that the type of Invariant Subspaces (and therefore the means which are used to find them) must heavily depend on the underlying space and on the specific representation of the operator. This also casts some doubts upon the possible success of methods like Banach Algebra methods, which do not use the structure of the underlying Hilbert space. They may be quite suitable for some classes of operators, under some specific assumptions (we have met such a situation in Chapter VI, Section 6), but I don't see how they could handle the whole problem. All this explains why the solution of the Invariant Subspace Problem is not easy: we have to find general means of exhibiting Invariant Subspaces of various natures. One can imagine that a given method may produce Invariant Subspaces of one type, for some class of operators, but it does not look simple to find a method which would work in all cases. Moreover, the three examples we have studied are very trivial ones : they are just multiplication operators. But if things look already bad at such an early stage, the future does not look very promising !

Chapter IX

Exercises on Chapter IX

Exercise 1. - Let H be a Hilbert space, 12 (IN, H) the space

Let S be the right shift on 12 (IN, H) , and let G be a closed invariant subspace of S in this space. Show that there is a Hilbert space HI and a function F, defined on D, with values in £(H1,H), such that: a) F is analytic on D and if zED, IIF(z)11 ~ 1, b) F(e i 8 ) is defined a.e. and is an isometry, c) the subspace G is of the form F.l2(IN, HI).

Exercise 2. - Let / be in H?", Show that 1, I, /2, ... form an orthonormal basis of H2 if and only if I(z) = Az, for some A, IAI = 1. Exercise 3. - Let F be the subspace of H2 : F

=

{I E H 2 i there is N(J) such that TI n ~ N(J) , /(1 - l/n 2 )

= a}.

Determine the closure of F in H2.

Exercise 4. - Show that the functions {e i n 8 I ; n ~ o) span L 2(Il) if and only if the functions {ei n 8lin ~ o} span L 2(Il). Exercise 5. - Let m be a non-constant inner function, M m the operator of multiplication by m on H2. a) Show that M m is a c.n.u, isometry (see definition of c.n.u. in the next chapter). b) Show that M m is unitarily equivalent to the multiplication by ei 8 if and only if there exist A, IA\ = 1, and a, [o] < 1, such that: m () z =

A z-a 1- ii:z '

zE D.

Exercise 6. - Let 4>, t/J be in L oo , with 14>1 = t/JI 1 a.e. 4>.H2 = t/J.H2 if and only if 4> = At/J, for some A, IAI = 1.

Show that

Multiplication by ei 8 Exercise 7.

~

I

Let

205

be an outer function. For A > 0, set :

UA

=

min(A, -log lJ(e i 8 )I) .

(1)

1) Let IA be the outer function such that log IIAI = UA and IA(O)/(O) > O. Prove that IIA I ~ eA , and that UA (ei B) + log I/(e i 8 ) I s O.

2) Show that IluA+logl/llll--+O,when A--+ +oo,andthat IA(O)I(O)--+ 1.

3) Choose a sequence (An)n>o, increasing fast enough, so that:

L (1 -

IAJO)I(O))
1 such that II An II < 1, and that T is a contraction. We want to show that the sequence IITkll tends to 0, exponentially fast. 1) Show that I - An is invertible, and that its inverse B is a positive operator. 2) Set: C

=

.!:.((n-l)I+(n-2)T+ ... +T n- 1 ) n

Prove that I - An = (I - T)C. 3) Set D = BC. Show that D is positive, and that it is the inverse of

I-T. 4) Prove that for all m, D(I - Tm+l) = 1+ T

D

~

+ ... + T m,

and that

D(I - Tm+l). (the statement D

~

D' means that D - D' is positive.)

5) Deduce that D + DT + .,. + DTm -:; D2. 6) Show that the sequence DTm is decreasing, and deduce that:

mT m -:; mlrl?" -:; D + DT + ... + DT m . 7) Deduce that a {) < 1 such that :

IlTm ll -:; IID2 11/m, for

all

m. k

Deduce that the sequence

IITmll

~

Fix mo, show that there is

o.

tends to 0 exponentially fast.

8) Give an example showing that the result is false if the positivity assumption is removed.

Notes and Comments:

All results presented in this Chapter can be found in Nagy-Foias [1]. We have simplified the presentation and the notations. Exercises 1 to 6 also comes

Minimal Dilation

eeg

from Nagy-Foia., [11. Exercise 7 is a presentation of a nice result of A. Brunei and L. Sucheston [11.

Chapter X

290

Complements on Chapter X :

1. Von Neumann Inequality. We have met (Proposition 1.7) Von Neumann's Inequality:

IIp(T)11

~

max Ip(z)1

(1)

Izl~l

for any contraction T on a Hilbert space and any polynomial p. Two questions can be asked: if (1) holds for every contraction on a space E, must E be (isometric to ) a Hilbert space? and does (1) characterize contractions? For the first question, the answer is yes. The next theorem is due to C. Foiaa [11:

Theorem. - 1£ (l) holds for every contraction on a Banach space E, E is isometric to a Hilbert space. Proof. - Take y in E, lIyll = 1, and Xo in E·, with the rank-one operator T'x = xQ(x)y. Then IITII = 1. Fix consider:

IAI =

IIxoll = 1. Consider a, with lal < 1 and

1.

Then IBa(A) I = 1 (see Chapter VIII, Section 6). By (1), for all x E E,

Replacing x by (1 ~ aT)x, we get:

II (T ~ a)xll < II (1 -

aT)xll,

that is :

The point z being fixed, we choose Then:

for all

z, y E E, a E r (U) x exists in )I. Since 4> E Hoo, it is bounded on the unit circle, and 4>(U) can be defined by means of the integral (see Chapter VII, Theorem 4.2) :

¢(U) = /.

u(U)

where E is the spectral measure of U. satisfies:

¢dE , By Theorem 4.3, Chapter VII, it

/ 1¢(e i 8 )

-

4>r(e i 8 )12 dE x, x(8)

/ 14>(e i 8 )

-

¢r(ei 8 )12 /%(8) :: '

since the measures Ez,z are absolutely continuous with respect to Lebesgue measure (see Chapter X, Theorem 2.5). Of course, the function /z is in £1. Furthermore, we know that 4>r -. 4> a.e, (Chapter VIII, Section 2), and that i8

sup l¢r(e )1 r,8

1100

(Chapter VIII, Section 1). From Lebesgue's Theorem, we deduce that, in )I,

¢r(U)x - 4>(U)x -. 0, and that:

114>(U)11

~

114>1100'

(5)

We observe that the convergence obtained is a strong convergence, not a convergence in operator norm. Since T = PU, we get:

(6) and this implies that 4>r(T)x converges, when r -. 1- , to a limit in H. We denote this limit by 4>(T)x, and from (5) and (6) we get:

¢(T)x = P¢(U)x, and

II¢(T) II < 114>1100, as we announced.

HOO Functional Calculus

1195

We now determine the adjoint of the operator ¢'(T). If ¢' E H oo, we denote by ¢'d the function defined by (see Chapter II, Proposition 2.2) :

A ED. This function is in H?", and if ¢'(eiB) = Ek~O akeiU , the Taylor series of ¢'d is : ¢'d(ei 8 ) = (ik eik8 .

L

k~O

Proposition 1.2. - The adjoint of ¢'(T) is given by :

Proof. - First, quite obviously, if T is c.n.u., so is T·. For r < 1, one checks immediately that :

When r ---+ 1- , the right-hand side converges, at every point z , to ¢'d(T'")x, and the left-hand side, to ¢'(T)'" z . This proves our Proposition. The mapping ¢' phism:

---+

¢'(T) , from H'" into £ (H), is an algebra homomor-

(4)?/J)(T) = ¢(T).p(T) ,

t/>,?/J E H oo

•

It defines a functional calculus. Obviously, if t/> is a polynomial, ¢'(T) coincides with the usual definition.

If 4>

= Lk~O

akeik8,with L lak 1
(T) = Lk~O

akT k .

Finally, if T is normal, we also get:

¢'(T)

=!.

¢'dE ,

O'(T)

where E is the spectral measure of T. In short, the Hoo functional calculus coincides with the previously known ones when both make sense. We now study the continuity properties of the H'" functional calculus.

Theorem 1.3. - Let T be a c.n.u. contraction. Then:

a) If ¢'n

---+

¢' in H?", then ¢'n(T)

b) If sup., II4>nlloo < 00 and ¢'n strongly and ultrastrongly.

---+ ---+

¢'(T) in £(H). ¢' a.e. on C, then t/>n(T)

---+

t/>(T),

296

Chapter XI

c) If . But fz is real valued, so, by Chapter VIII, Proposition 3.2, must be constant, that is fz = C, a.e. We deduce from the fact that G was in HJ that C . F is in HJ, meaning that C> F(O) = O. Since an outer function never vanishes inside the unit disk, C = 0, and fz = 0 a.e. But:

and so z =

o.

Let's show finally that F(T) has dense range. The operator T* is c.n.u. and F" is also outer (check on the definition) and in Boo. So F"(T*) = F(T)* is injective, and F(T) has dense range. Conversely, we have: Proposition 1.6. - For every function u which is not outer, there exists a completely non-unitary contraction T on a Hilbert space, such that u(T) = O. Proof. - We decompose u = m.F, with m inner, non constant, and we build T such that m(T) = O. We know (cf. Chapter VIII) that m.H2 is strictly contained in H2. Set H = H2 e m.H2 : this will be the required Hilbert space. Let M be the multiplication by z, on H2. One checks easily that the adjoint M* is given by : M*f

1

- (f(z) - f(O)) , z

HOD Functional Calculus

£99

which implies that M?" f ----t 0 when n ----t 00, for every function f in H2. The space m.H2 is invariant by M, so H is invariant by M*. We put:

We obtain a contraction T on H, with T':" = M*nI H n ----t 00. This implies that M and Tare c.n.u.

----t

0, strongly, when

By definition, if P is the projection of H2 onto H T

for n

~

P(MIH),

1. Therefore, for every w in HOD,

w(T) = P(w(M)IH)'

But we know that w(M) is the operator of multiplication by w on H2. In particular, for f in H, we have:

=

Pm(M)f

But m.] is in m.H 2 , so P(m.J)

= 0, and

m(T)f

=

P(m.J).

this implies m(T)

= O.

2. The Spectral Theorem for the HOD functional calculus. For the analytic functional calculus, we met (Chapter II, Theorem 2.3), the theorem : f(u(T)) = a(f(T)),

where

(1)

f is analytic on a(T).

We are now dealing with contractions, so a(T) c D. But f is not analytic on a neighborhood of D, only inside D. So formula (1) does not make perfect sense: if u(T) n C is not empty, how should f(a(T) n C) be understood? The function I is only defined a.e. on C, by means of a radial limit. So the left-hand side of (1) is not well defined, though the right-hand side is. For points inside the open disk, there is no difficulty:

Chapter XI

Proposition 2.1. - Let Then:

f

be a function in Hoo, and T a c.n. u. contraction.

f(o(T) n D)

o(f(T)).

C

(2)

Proof. - Set, for -X ED:

g(JL) = {(f(JL) - f(>..))/(I-£ - -X) /'(-X)

if JL E D\{A} if JL = -X

Then g is in HCXJ. Moreover,

g(T)(T - A) = (T - A)g(T) = f(T) - /(A). So, if A E o(T), A - T is not invertible, and /(T) - f(A) is not invertible either. This proves our claim. If / is in A(D), formula (2) can be extended to :

f(o(T))

(3)

o(/(T)).

C

More precisely, we have :

Theorem 2.2. - Let T be a c.n.u, contraction with spectral radius 1, and let A E o(T), with [AI = 1. Assume that / E H'" is continuous at A, that is :

f(A) =

lim

z-A,lzl e no ,

a contradiction. Let e = inf n distE.(/o,nBL). Let: e

x;

2BE. + nBL ,

=

n 2:: O.

This is a compact set, for u(E"', E), which does not contain 10, by (3). So there is a point Yn in E, llYn II = 1, and a real Pn > 0, such that:

(4)

10(Yn) > Pn e

1 + nl(Yn)1 < Pn , 9 E BE·, IE BL, n 2:: o. 2g(Yn) From (5), taking f = 0, 9 such that g(Yn) = 1, we deduce:

e; and from (4),

IPnl < 1.

(5)

(6)

2:: ~,

From (5), we also deduce:

I!(Yn) I

s

(7)

Pn .

n Extracting a subsequence from the Yn's if necessary (we keep the same notation), we can assume that there is a I such that: and I 2:: e/2 .

(8)

We consider: This is a vector subspace of Co, by (7). In the space (c) of convergent sequences, condition (8) implies that the distance of the sequence (/0 (Yn))n to G is 2:: e/2. The dual of (c) is 11, so there is a sequence of complex numbers (Cn ) such that:

LICnl < 00, n

LCnl(Yn) = 0,

f

E

BL ,

n

L Cnlo(Yn) =1. n

Take z =

Ln CnYn

: this is the required point, and our Lemma is proved.

2.. 6

Chapter XI

Here, since \If is an isometry, L n Bl(H) is the image of the ball of Hoo, which is u(L oo , Ld compact. Therefore, B,f(H) is u(£(H), J/ (H)) compact, and therefore closed. By Lemma 3.2, L itself is closed for u(!(H), J/(H)). This proves that L = AT, and that \If is surjective. But then, 'If is an injective and continuous map from the compact BHao, equipped with u(L oo,L 1 ) , onto &:AT' equipped with u(AT,J/T). Therefore, 'Ill in an homeomorphism between these two spaces, equipped with their weak topologies, and our Theorem is proved.

H'" Functional Calculus

Exercises on Chapter XI.

Exercise 1. - Let T be a c.n.u. contraction on H t and let h in Boo be a non-zero function. Show that for every x in H such that h(T)x = 0, we have Tn x --+ 0 when n --+ 00. Exercise 2. - Let B a be the inner function:

Ba(z) =

z-a 1- az

--~,

Construct an operator T such that Ba(T)

lal < 1. = O.

Exercise 3. - (Mean ergodic Theorem). Let T be a contraction on a Hilbert space. Prove that the Cesaaro averages ~ L~-l Tk converge strongly in H, when n --+ 00. Exercise 4. - Let T be an invertible isometry on a Banach space E. Let

A(II) be the algebra of absolutely convergent Fourier series: / is in A(II) if / = L:~oo ci ei j 8 , with L:~oo ICil < 00. Show that f(T) can be defined by means of a normally convergent series, and that the usual properties of a functional calculus hold.

Exercise 5. - Let 8 be the right shift on 12(Z) and (Pn)n~O

be a sequence

of polynomials. a) Show that Pn(8)

--+

0 in £(12 ) if and only if Pn --+ 0 in L oo(I1).

b) Show that Pn(8) --+ 0 strongly if and only if (Pn)n>O is bounded in L oo (Il) and Pn --+ 0 in L 2 (II) . c) Show that Pn (8) --+ 0 weakly if and only if (Pn)n~O is bounded in L oo (II) and, for every j 2:: 0, the j th coefficient of Pn --+ 0 when n --+ 00.

Exercise 6. - Let 8 be the right shift on 12(Z), To every element of the algebra generated by 8 t that is to every finite sum Lk>O akSk, we associate the polynomial Lk~O

akz k•

-

a) Show that the closure of As for the operator norm can be identified to the algebra A(D) . b) Show that the closure of As for the strong topology can be identified to Hoo. c) Show that the closure of As for the weak topology can also be identified to BOO.

Chapter XI

Exercise 7. - Let S be the right shift on 12 (1L). To every element of the algebra generated by S and S-1 t that is to every finite sum E:=-M alcS k , we associate the trigonometric polynomial E~=-M alczlc. a) Show that the closure of the algebra AS,S-l generated by Sand S-I, for the operator norm, can be identified to C(IT) . b) Show that the closure of to L CXJ •

AS,S-l

c) Show that the closure of identified to L CXJ •

for the strong topology can be identified

AS,S-l

for the weak topology can also be

Notes and Comments: Section 1 comes from Nagy-Foias [1]. In Section [1] ; our proof follows J. Esterle [2]. 2, Theorem 2.2 is due to Foi~-Mlak

Complements on Chapter XI :

On the algebra generated by the right shift S, the weak and ultraweak topologies coincide. This property holds for a large class of operators, as described by Bercovici-Foiac-Pearcy [1]. A first example of an operator for which this property does not hold was built by D. Westwood [1]. Another was built by G. Cassier [11. Cassier's example has some interesting properties. Let T be the operator. Then the algebra AT is equal to the commutant {T}/, and also to the bicommutant {T}". Moreover, there is a nuclear operator A with the property that no finite rank operator B satisfies : tr (AU)

=

tr

(BU)

t

for all BEAT .

Chapter XII

C1-Contractions

Let E be a Banach space, and T a contraction on E. We say that T is a C 1-contraction if : For every x =j:. 0 in E, T" 0, when n ---l' +00. This is obviously the case for isometries. Here are two other examples:

r

Example 1. - On l2(lL), the weighted right shift : n E 'IL, is a contraction if

Iwnl

~ 1,

for all n E Z. It is a C 1- contraction if:

n

II

IWkl

f+ 0,

n

---l'

+00,

o

and Wk =I 0, for every k E 7L. For instance, for n ~ 0 gives a C 1-contraction.

Wn

= 1/4 for n

< 0 and

Wn

= 1

Example 2. - On the Hardy space J[2, the multiplication operator MtjJ, by a function 4> in Hoo, is a contraction as soon as 114>1100 ~ 1. It is a C 1-contraction when: P{O ; 14>(e i 8)1 = 1} > O. Indeed, put A

= {O; 14>( ei8) 1= 1}. 11 M ;

1112

Then, for

1 =I 0,

(/11" l4>n(ei8)/(eiO) 12 -11" ~ I/(eiO)1 2 dO)I/2

=

(!

A

dO) 1/2 2~

2~

> 0, since we know that a non-zero function in H 2 cannot vanish identically on a set of positive measure. In both examples above, the operator is c.n.u. except if n E 7L in the first case, or if 4> is identically 1, in the second.

W

These examples will be studied in great detail in Section 6.

n

= 1, for all

Ghapter XII

250

1. The extension of a Gt-contraction.

e,

In order to study T, CI-contraction on E, we will build a new space and extend T as an isometry on it. However, e here contains E as a dense subspace: this is quite different from the situation we encountered when we constructed the unitary dilation. The connections between both constructions will be studied in Section 3. We define a new norm on E, denoted by ~.~,

by the formula:

This is a norm because T is a GI-contraction. Obviously, for x E E,

(1) and

~xl

s Ilxll·

(2)

We denote by E the completion of E for the norm [. ~. The operator T can be extended, by formula (1), into an isometry from into itself, denoted by

e

T. From formula (2) follows that the canonical embedding from E into con tinuous and has norm 1. If E is a Hilbert space, so is

e, with the scalar product (for

e is

z , y E E),

(3) The limit on the right-hand side exists, since (Tm x, Tmy) is the sum of four quantities, decreasing with m : IITm(x ± y)lI, IITm(x ± iy)ll· One easily sees under what conditions the norms 11.11 and ~.~ lent:

are equiva-

Proposition 1.1. - The norms 11.11 and ~.~ are equivalent if and only if there exists on E a norm 11.111, equivalent to 11.11, for which T is an isometry. Proof. -a) Assume first that ~.~ is equivalent to 11.11 Then just take 11·111 since for ~.~, T is an isometry. b) Assume that there exists a norm 11.111, with :

ell·III ~

11·11 ~ GII·III .

= ~.~,

fSl

C I -coniraetions Then, for all x in E,

cllxlll >

C

C Ilxll·

So: ~x~

which implies that ~.~

~

c C IIxll,

and 11.11 are equivalent.

Therefore, in general (that is : except if T is an isometry for some equivalent norm on E), and E do not coincide. E is not a closed subspace of but a dense subspace. We now study the question: when is T surjective ? on

e,

e

e

t

Proposition 1.2. is a surjective isometry on of T is dense in E for the norm [. ~ .

e if and only if the image t is dense e. is a dense subspace of e,

Proof. - 1) Assume first 1m T to be dense in E for ~.~. Then Irn and since it is closed (T being an isometry), we get 1m f = in

e,

2) If f is a surjective isometry, T(E) = 1m T which means that Im T is dense in E for ~.~.

Corollary 1.3. - If T has dense range in E (for the original norm of E), is a surjective isometry. Indeed, a fortiori, this image is dense for

f

~.~.

2. Representing Functions.

In this Section and the next ones, we assume E to be a Hilbert space, and T to be a Cl-contraction, with dense range. The assumption "T has dense range" is a natural one when one is looking for Invariant Subspaces or Hyperinvariant Subspaces, since the closure of the image is hyperinvariant by T.

Being a surjective isometry, f is therefore normal. We denote by J-Lx,1I the scalar spectral measures (see Chapter VII), for z , y E So we may write, for i , g in BOO(II),

e.

(1) As a special case, this holds for

f,

g in

A(D).

Chapter XII

Let's compute the Fourier coefficients of JJz,,,. For k E 'Il, we have:

(2) Put simply CA:(x,y) = [T-kx,y]. If x, Y E l ,we can write:

CA:(x,y) = lim (Tm-A: x, Tmy). n-oo

(3)

We now show:

Proposition 2.1. - If T is c.n.u., the measures JJz,y, for x , y E l, are absolutely continuous with respect to Lebesgue measure. Proof. - We could deduce this from the results of Chapter X and the connections with the m.u.d. which we will study in Section 3. However, we prefer to give a direct proof. First, we observe that :

CA:(x,y)

=

1

4(CA:(x + y,x + Y)-Ck(X - y,x - y)+ + iCk(X + iy, x + iy) - iCk(X - iy, x - iy)X-4)

So we need only to establish the result for the measures J.Lz,z, which we just denote by JJz. We have to do it for every x E l , but we start with the case

xEE (x#O). Assume that, for some x E E, J.Lz is not absolutely continuous with respect to Lebesgue measure. Then there is a compact K c TI, with P(K) = 0, and

J.Lz(K) >

o.

By Fatou's Theorem (see Chapter VIII, Section 7), there is a function h in A(D), with h = 1 on K, and Ihl < 1 on D \ K. By the HOO functional calculus, h(T) can be defined. We consider the sequence (hn(T)x)n~o. This is a bounded sequence, since :

So there exists a subsequence h n " (T)x which converges weakly in E to a point z in E. Since the canonical injection from E into l is continuous, it is also weakly continuous, and this sequence also converges weakly to z in l .

259

C 1 -eoniractions For any g in A(D), we have, by (1) : lim [h nk (T)x,g(T)x]

[z,g(T)x]

k--+oo

=

lim k--+oo

111" hnk(eiB)g(eiB)d#lz (8) -II"

IK g(ei8)d#lz(8) . In particular, {z, xl = IJ,z(K) > 0, and so z 1= o. Using now the Rudin-Carleson Theorem (Chapter VIII, Section 7), we take a function ¢, with ¢(e iB) = e- i8 , for 8 E K, and 114>1100 ~ 1. We are going to show that T¢(T)z have:

= z.

Indeed, for every g in A(D), we

[(T4>(T) - I)z,g(T)xl = lim [h nk(T)(T4>(T) - I)x,g(T)x] k--+oo

=

lim k--+oo

=

111" (ei8¢(ei8) -

1) h nk (ei8) g(e i8) dlJ,z(tJ)

-II"

IK (eiB4>(eiB) -

1) g(ei8) dlJ,z (6)

= 0, by the definition of ¢. So we obtain, for every k

~

0 :

[(T¢(T) - I)z, (T¢(T) - I)hnkx] and, letting k

0,

-+ 00,

II (T¢(T)

- I)zll

0,

as we announced. Set V = 4>(T). Then V is a contraction on E, which commutes with T. For every finite sequence of complex numbers, (ak)k>o, we have:

(5) and thus:

IIT(L akTkz)11 k~O

This proves that T is an isometry on

Furthermore, T is surjective on Fz , since ¢(T)z E Fz , and z = T¢(T)z. This contradicts the fact that T is c.n.u., and proves our claim, when x E E.

Chapter XII

Since the measure "'z, x E E, is absolutely continuous with respect to Lebesgue measure, the series LkEZ ck(x)e i k8 is the Fourier series of an integrable function, which we denote by Az (B), and this function is the density of J.'z with respect to Lebesgue measure. So formula (1) can be written:

- - =I

[f(T)x, g(T)x]

dB , i 8)A f(e'°8)g(e z(B) -

(6)

271"

for x E E, f, 9 in Boo [Il] . We deduce from (4) :

(7) The functions Az are real valued, and more precisely positive, but the functions >'z,y are not real valued in general. To finish the proof of Proposition 2.1, we now take z E e. Since E is dense in we can find a sequence (xn.)n~O of points in E with X n --+ Z in We have, for every 9 in Loo(ll) :

e,

e.

In particular, for every Borel subset A ell,

If P(A) = 0, then J.'zJA) = 0, so "'z(A) = 0. This proves that J.'z is also absolutely continuous with respect to Lebesgue measure. Its density, denoted by >'%(8), has the Ck(Z) for Fourier coefficients, and:

Az(B)

"-J

L ck(z)e i k8 . kEZ

Our Proposition is proved; formula (6) extends without modifications to the case where z , y E e. From now on, we assume T to be c.n.u. We now investigate the transformation formulae for the functions >'z,II' x, YE

e.

Proposition 2.2. - Let f, 9 be in L oo (ll , d8/ 271" ) , and z , y E x' = f(T)x, y' = g(T)y. Then:

e.

We set

ess

C 1 -eontractions Proof. - We compute the kth Fourier coefficient of

" f = =

e

-ik8

dB 27r

Az'~'-

-W'

Az',~'

•

[1'-kx',y'j = !1'- k/(1')x, g(1')yJ

=

L:' -W'

e- 1k 8 / g A

z,1I

and this is the k th Fourier coefficient of

_dO11'" 2

by formula (1),

/g Az ,1I.

We observe that, since l' is a surjective isometry, 00(1') c C (Chapter II, Proposition 1.13). We have: Corollary 2.3. - All functions AZ ' 1I' x, Y E l , are 0 a.e. if eie

t. 00(1').

Proof. - Let / E Loo(II) , with /(e i 8 ) = 1 on 00(1'), and 0 on the complement. We know that /(1') = I (Chapter VII, Theorem 4.2). Applying Proposition 2.2 with 9 = /, x' = z , y' = y, we get :

which proves the claim. Proposition 2.4. - The spectrum of l' has positive measure, and :

00(1')

C

oo(T) n C.

Proof. - The first part follows obviously from the previous Corollary : for every x E l,

~X12

=

I

Az dfJ

211'"

imply that if x =F 0, Az is not identically o. This shows that 00(1') has positive measure. For the second part, we already observed that 00(1') C C. Let now A E ~ , A t. oo(T). Then A- T has an inverse, R(A, T). Since R(A, T) and T commute, we get, for m E IN, z E E,

Therefore,

Chapter XII

256

This shows that R(A, T) is continuous on E for the norm extended to into an operator which satisfies :

e,

0.1.

So it can be

OR(A, TH < IIR(A, T) II, and:

(A -

T)Rp., T)

R(A,T)(A - T)

I,

which shows that A i 0(1').

Remark 2.5. - During the above proof, we made the following observation: if an operator A from E into E commutes with T, it can be extended into an operator .A, from into satisfying :

e

e,

OA~

~

IIAII·

We also observe that the previous Proposition implies quite obviously that for any c.n.u. Cl-contraction, the set o(T) n C has positive measure. The function Az will be called "representing function" of the operator T at the point x. We now study several applications of these functions. The first one is an Invariant Subspaces Theorem, obtained by means of an estimate of the distance between a point and the space generated by its iterates.

Proposition 2.6.

~

eXP!IOg A%(O) dO 21r

Let x E E. We have:

=

inf {disfE(Tmx,span {T m+ 1x,T m+ 2x, .. .})} 2. (a) m~O

When T is invertible, this last quantity is equal to : inf {distE(Tmx,span{Tm-Ix,Tm-2x, ... })}2,

(b)

m~O

and both infima may be taken for mE'll.

Proof. - For every polynomial p(T) = Lk~O inf IITm(x - p(T)x) 11 2 m~O

= Ox =

akTk, we can write:

p(T)x0 2

j'" 11 _~

p(e i 8 )12 Az(6) d6 . 21r

If now we take the infimum of both sides upon all polynomials, we get a), by Szegd's Theorem (Chapter IX, Theorem 2.1). If T is invertible, we consider the polynomials q(T) = Lk$O bkTk, and we obtain b). The fact that the infimum may be taken for mE'll comes from the following observation: since T is a contraction, the infimum is actually a limit, when m ~ 00.

C t -contractions

e57

Corollary 2.7. - If T is invertible, log Az is integrable if and only if:

and this occurs if and only jf :

By means of the functions Az , we obtain a kind of formalization (or generalization), valid for the Cl-contractions, of the description of the Invariant Subspaces of the right shift on 12(Z) (see Chapter IX). Indeed, we see:

Theorem 2.8. - Let T be a c.n.u. C l -contraction, and let Az be the representing functions. If for some x E E, log Az is integrable, the operator T has Invariant Subspaces. This condition is obviously satisfied for the right shift, with, for example, x = el, and Az = 1. The Invariant Subspaces we obtain correspond to the Invariant Subspaces of type F n (or, more generally, of m.H2) we saw in Chapter IX, Section 1. The condition

is the abstract generalization of their definition (by "abstract" , we mean "valid for a larger class of operators"). However, it should be pointed out that the tools used to obtain this generalization are the same, namely Szego's Theorem, and no new concept has been introduced. What we proved in fact is that Szego's Theorem has a wider range of application than we originally mentioned. We will see later the corresponding generalization for the second type of Invariant Subspaces of the usual shift: the type L 2(A)' and we will come back to the meaning of both results. We now turn to a second application of the representing functions : the construction of a functional calculus, for convergent series. 3. Functional Representation of Convergent Series. In this Section, we assume T to be invertible and c.n.u. A functional calculus, as we met several times in the previous Chapters,

allows us to give a meaning to the operator f(T) , when f is taken in some class of functions. For instance, if we have a Fourier series, EkEZ akeik8, in

eS8

Chapter XII

some cases we can give a meaning to the operator E akTk. In this Section, we consider the inverse problem: a convergent series E akTkxo being given, what can we say about the properties of the "function" defined (formally) by the Fourier series EkEZ ake ik8 ? The following Proposition answers this question:

Proposition 3.1. - Let Xo E E, with Xo =1= O. Let

(1)

L akTkxo kEZ be a convergent series in E, with sum z . The Fourier series

L ake kEZ

ik8

(2)

defines a function 4J in the space £2 [Il, ..\zo dB /21r) , in the following sense: the partial sums E~N ake ik8 converge in this space, when M, N - 00. This function 4J satisfies moreover:

(3)

a.e.

Proof. - First, we observe that akTkxo - 0, when Ikl- 00. Since IITkxoll ~ 1 for k < 0 and lITkxol1 -f+ 0 when k --+ +00, we get in both cases that ak - 0 when Ikl - 00. The Fourier series EkEZ a"e ik8 therefore defines a distribution on Il , and more precisely a pseudo-fundion. We refer the reader to J.P. Kahane 11] for this concept, which we are not going to use here. For m, N E IN, we set : M

XM,N

LakTkxo, -N M

4JM,N(B) = Since the series (1) converges, XM,N therefore in t . But: M

~XM,N-XM/,N/~

= ~Lakfkxo-N ~

=

1

ik8 L a"e . -N -+ s:

in E, when M, N

--+ 00,

and

M'

Lakfkxo~2 -N' 2

~

_~ I4JM,N(B) - 4JM',N' (B) 1 ..\ zo (B) 21r '

which shows that (4JM,N) is a Cauchy sequence in L2(..\ z odB /27r), and therefore proves our first claim.

eS9

C 1 -eoniractions

1 E L oo (II) , we have

Furthermore, for every function

~/(f)x~2

=

lim

~f(f)XM,N

:

~2

M,N~oo

lim

M,N~oo

I""

_)("

1/(0)12 14>M,N(0) 12 .\%0 (0) dO 211"

(1/(8)1'1(8)1' x•• (8) :: ' and this proves our Proposition. We observe that we don't need the convergence of the series (1) to be in E, but only in C, which is strictly weaker. We even don't need this convergence to hold in the usual sense: if the series is Abel summable (and Abel summabiiity is weaker than convergence of partial sums), the same result holds :

Proposition 3.2. - Let o < r < 1, the series :

XQ

E l, with

Xo

=F O. Assume that for every r,

are convergent, and Jet's denote their sums by X,.. If, when r ----4c 1-, x,. ----4c x in l, then the Fourier series (2) defines a function in L 2 (.\%0 dO /211"), in the following way : if we set, for 0 < r < 1,

4>,.(0)

=

L akrlkleik8 , kEZ,

there exists a function 4> in L 2 (.\:1:0 dO /211") such that :

when r

----4c

1-.

from which the first claim follows. The second is established as in the previous Proposition, replacing 4>M,N by 4>,..

Chapter XII

e60

4. Connections with Nagy-Foi as Dilation Theory. From T, c.n.u. contraction on H, we can build the minimal unitary dilation (J/, U) , as we explained in Chapter X. The connections between f on and ()I, U) are as follows :

e

Proposition 4.1. - For z , y E E, the numbers Ck(X, y) are the Fourier coefficients of the scalar spectral measure of the operator R., *-residual part of (J/, U). Proof. - It relies on a way of building the m.u.d. slightly different from the one presented in Chapter X. We only give the outlines and refer to M. Rome [11 and B. Beauzamy, M. Rome [11 for details. Let N be the closed subspace of 11 (1l, H) generated by sequences of the form (... ,0, -Ty,y, ...) ,

yE H,

and let ZH be the quotient of 11(Z,H) by N. If (YkhE 71 is in 11(1l,H), and fj is its class in Z H ,

The space Z H is a Hilbert space, for the scalar product :

H can be identified to a closed subspace of 1.(1l, H) by means of

u(x) =

(oo.,O,~,o,oo.),

where the Oth coordinate is underlined. Let u be the quotient operator, and 8 be the right shift on It (71, H), S the quotient operator. Then:

uT Set D

= (I

=

- T·TP/2, D = ImD, K l

s «.

= 12(1l, D),

K

= K;

ffi zH, and put:

4>(y) = (oo.,0,Dy,DTy,DT 2Y,oo.)ffiU(Y). Then 4> is an isometry. Define V on K by V

= 8- 1 on

Ks , S on Z H. Then:

261

Cl -eoniractions

for all Yl, Y2 E H , all n E IN. V is a unitary dilation of T, and one checks it is minimal. To identify the *-residual part of the dilation, we put: L = (V

0

4> - 4> 0 T)( H) ,

M = span{V n L; n E 7L}, and we know that R. = K eM. Here, M restriction R. = V Ifl.. is here S on Z H .

= K 1,

and so R.

=

Z

H. The

Finally, one obtains: Proposition 4.2. - If T has dense range, the injection extends to an isometry from onto Z H. Moreover,

e

tioT

ti

from H into Z H

SoU.

=

5. The Algebra A (11) . In order to continue our study of the C1-contractions, we need a few properties of the algebra of functions with absolutely sumrnable Fourier series. We denote by A(11) the algebra of functions on 11, the Fourier series of which is absolutely convergent:

L

1(8)

ak eikB,

kEZ

1I/11A

=

L

lakl,

kEZ

and A(11) is an algebra for the usual pointwise multiplication of functions. The reader is advised to carefully distinguish between A(I1) and A(D), the disk algebra, studied in Chapter VIII. Due to the definition of the norm, A(11) is isometric to the Banach space by the correspondence I -4 (akhEz. We refer the reader to J.P. Kahane [1] for a detailed study of A(I1).

11C~),

Functions in A(11) are obviously continuous, since they are uniforn limits of the partial sums of the Fourier series. We say that a function I in A(11) has strict support if its support (which is the closure of the set {8 ; 1(8) 1= O}) is a strict subset of 11 (this terminology is not extremely good, but it is better, anyway, than the one we used in some papers, with "compact" support. Since 11 itself is compact, every support has to be compact !).

e6e

Chapter XII

Therefore, a function with strict support will vanish identically on some interval of II. There is a restriction on the rate of decrease of the Fourier coefficients (when k -+ ±oo), in order that the function f may have strict support. These coefficients cannot decrease too quickly, when Ikj -+ 00. For instance, if lakI ~ 1/2 k , for k ~ 0, the function

will be analytic in the corona {1/2 < Izl < I}, continuous and therefore integrable on the unit circle. We have not seen, for the corona, any theory like the one developed in Chapter VIn for HI, but the same result holds: such a function cannot vanish on a set of positive measure of C without being identically zero. The way of proving this is as follows: we cover the corona by two simply connected sets DI and D 2 , the first one has in its boundary {Izl = 1 j I arg zl ~ 1r /2} and {lzl = 1/2 j I arg zl ~ 'If /2}, the second one {je] = 1 j [arg zl ~ 1r/2} and {izi = 1/2 j [arg zl >-: 'If/2}. We then make a conformal mapping to transform D 1 into the unit disk. This way, we see that the function f cannot vanish on a set of positive measure of the half-circle {izi = 1 ; Iarg zl ~ 'If/2}, and we repeat the argument for the second half. We now consider the sub-algebras of A(II) of the following form:

Ap = {f E A(II) ; f =

L

akeik8,

kEZ

L

laklpk < oo},

kEZ

where P = (Pk) satisfies: Pn ~

1

j

Pm+n ~

Pm·Pn

for all m,n E IN.

(1)

Such algebras are called Beurling Algebras. For them, a necessary and sufficient condition is known, concerning the weights (Pk), in order that Ap contains functions with strict support. If this is the case, the algebra is said to be non-Quasi-analytic (the converse, quasianalytic, refers to the above property of analytic functions: they cannot vanish on a set of positive measure without being identically zero). This condition comes from Paley-Wiener theorem, concerning the Fourier transform of functions with compact support. It can be expressed the following way:

C 1 -eontractions

fJ69

Theorem 5.1. - Let Ap be a Beurling Algebra in A(JI). It contains functions with strict support if and only if : logPn L l+n nE'll

O

Xkl

2

+

L

IXkI 2 ) 1/ 2 >

O.

k$O

This property obviously implies that there is no subspace where S is unitary, so S is c.n.u. Let's now determine what are Sand t . If x =

L

xne n, easy computations show that, for k ~ 0, Sk X

L

=

Xn-kWn-kWn-k+I·· .Wn-Ien ,

nE'Il

and if Y = LYnfn

E nEZ

Xn-m-kWn-m-k··· Wn-lYn-mWn-m •.. Wn-l

e66

Chapter XlI

If we set n = n' + m :

(s m+kX , smy) =

~ Xn'-kWn'-k··· L..J

Wn'+m-lYn'Wn, ... Wn'+m-l

nEZ If k ~ 0, when m --. ~ Xn'-kYn' L..J n'~k

+00,

+

(sm+k x, smy) has the limit:

~ L..J

Xn'-kYn'

/4 k - n ' +

~ Xn'-kYn' - /4 L..J

k - 2n '

n'

(3)

-00.

Applying Corollary 2.7, we see that, for every x

t= 0,

which means that no point belongs to the closed linear span of its iterates. The operator has no cyclic vector : for every x t= 0,

T-1x

f/.

span{x,Tx, ...},

and therefore has a lot of Invariant Subspaces, B. A Multiplication Operator on H2(II). Let K be a simply connected subset of the closed unit disk, with nonempty interior, not containing 0, and such that the boundary r of K contains an interval of the unit disk. Let 4> be a conformal map from D onto K O • Let's assume furthermore that r is regular enough, so that cP extends to an homeomorphism from D onto K (this assumption is for simplicity i it is not essential). The function cP is in H?", and, more precisely, in A(D). It satisfies of course 114>1100 = 1. Let's call :

and I'

= 4>(1).

We consider the operator M~ of multiplication by cP, from H2 into itself. This is obviously a contraction. For any function f E H2, not identically 0,

which shows that

11M; /11 f+

°:

M~

Since K does not contain 0, M~ The function 1/4> is also in A(D) .

is a C 1 - contraction. is invertible, and its inverse is

Ml/~.

268

Chapter XII

For some 0: > 0, we can find an interval /2 where 1¢(ei8)1 < 1 - 0:. Therefore, for n 2 0,

I Minl ~

=

{1",n(~

/k·)12 ::

..

2 (_I_)2n /, I/(ei8)12 dO 1 - 0:

> (_I_)2n 0 -

where 0 = 0(/)

=

fl~

1-0:

211"

1~

'

1/1 2 dO/ 211" > O. So we get:

Lemma 6.B.l. - For every

I in H 2 , there are

6 > 0, C > 1, such that:

for all n 2

o.

This proves that for every l . the increase of the sequence IIMin III is exponential. Here again, the operator must be c.n.u. We know moreover (cf. Chapter VII) that the spectrum of M¢ is K. This operator has some very interesting properties which cast some light on the range of application of theories like unitary dilations. In order to develop them, we need two results about approximation of continuous functions by polynomials: Runge's Theorem and Mergelya s Theorem (the second is stronger than the first). The reader will find these two results in W. Rudin [I], p. 271 and p. 390.

Runge's Theorem. - Let K be a compact in the plane, with connected complement. Let f be a function analytic on a neighborhood of K. Then there is a sequence of polynomials converging to f uniformly on K. Mergelyan's Theorem. - Let K be a compact in the plane, with connected complement. Let f be a function continuous on K, analytic in KO. Then there is a sequence of polynomials converging to f uniformly on K. Proposition 6.B.2. - For every

f

f in H2,

E span{MtPI,M~f,

...}.

Proof. - Let VI be a neighborhood of K, V2 a neighborhood of 0, with VI n V2 = 0, and let g = 1 on VI, 0 on V 2 • This function is analytic on

269

C 1 -coniractions

a neighborhood of K U {O}, and, by Runge's Theorem, can be uniformly approximated by polynomials : for every e > 0, we can find a polynomial p(z) with: 11 - p(z)1 < E:/2, zE K. Ip(O)1 < ~/2, Replacing p by p - p(O) , we obtain:

11 -

Ip(O)1 = 0, We write p(z)

= Lk~l

p( z) I < e ,

zEK.

(1)

akz k. So we get, for every 8 ElI,

11 -

L ak 0 such that l4Jn(z) - AI > 6, for all z E D, and all n ~ 0, and the inverse of A - T is the multiplication operator, on En, by 1/(A - 4Jn). Therefore,

u(T) n C We now describe

T and

=

I'

U

J' .

! .

Lemma 7.4. - The space ! is

t Proof. - For F = (In)n~o,

~F~2

=

=

(~L.

(I.) ) •

we have:

liT: Inll~

~~ooL n

by Lebesgue Theorem

from which the Lemma follows.

e77

C 1 -eontractions

f is the multiplication by tPn in each L2 (In)' Precisely, if F E (n~ L 2(In )h t then TF = (tPnfn)n. Finally, we see that 0"(1') = I'. Indeed, a) O"(T) c r . HAt. lit with IAI = 1, and if ei S E In, then IA-4>n(e i S) / ~ = dist(A,I'), so: The operator

F

{)

= (fn)n~O,

and A b)

l'

is surjective. Since it is clearly injective, our claim follows.

aCT)

:J I'. Indeed, this is the case, since

aCT)

contains each O"(Mq,n) =

I'. This way, we see that 0"(1') n C contains I' and J' , though 0"(1') is reduced to I'. We observe that we are allowed to take as lithe complement arc of J' , though they have the same end-points. We need just to avoid to take these end-points among the (a~). For the operator T obtained this way, we will get

aCT) :J C, aCT) = I'. The sets that are spectra of C1-contractions have been determined by L. Kerchy [1]. 8. An Invariant Subspace Theorem for the C,-contractions. We are going to prove the following result, due to the author [31 :

Theorem 8.1. - Let E be a Banach space, T a C 1 , invertible contraction which is not a multiple of the identity. We assume that there exists a point x¥-O such that :

(1) Then T has Hyperinvariant Subspaces.

Remark. - The invertibility assumption is not essential, and condition (1) may be replaced by an assumption on a chain of approximate inverses of a single point. See [3].

Proof. - We may assume that T has no eigenvalue. Indeed, if A is an eigenvalue, the space K ere A - T) is closed, invariant under every operator which commutes with T, and is not the whole space since T is not a multiple of the identity.

Chapter XlI

278

The point x being given by (I), we put Ok = IIT-I k lx ll ~ for k E 1l. This sequence obvioulsy satisfies the assumptions of Lemma 5.2. Therefore, for every arc r = [8o~81] ~ there exists a function tP~ with support in r ~ with, if

¢

=

L

lej 10j
0 k=O

for all t E TI.

Set u = 2:.:=0 TkOtP(t). All these functions are in A(IT), and u does not vanish I/u is also in A(TI). on TI, so, by Wiener-Levy Theorem (Theorem 5.2)~ The operator u(T) is well-defined, by a norm-convergent series, since u E A(TI) (recall that T is a surjective isometry). Moreover, it is invertible, and its inverse is (l/u)(T). Therefore, it is injective, and in particular, u(T)x i= O. Thus, for some k ~ 0 ::; k

s N,

This construction can be realized for every arc r. Let's take successively r N = [0, 27r / N] ~ for N = 1, 2, ... , and let's denote by tPN,kN the function obtained (or one of them) by the previous construction: this is a translate of 4> N ~ which satisfies : The supports of ¢N,k N have a width which tends to O. Let a = eiOo be an accumulation point of the middles of these supports. By construction, this means that, for every V, neighborhood of a, there is a function ¢N,kN' for N large enough, the support of which is entirely contained in V. We now consider the function 10(8) = ei 8 - ei 80 • It vanishes at 8 = 80 . By Wiener-Ditkin's Theorem (Theorem 5.3)~ there exists a sequence of functions (fn)n"2:0 in A(ll), convergent to 10 in A(ll), and each In is identically 0 on some neighborhood of 80 •

C t -eontractions

Since [«

--+

£19

fo in A(TI) , we get: IIfn(T) - fo(T)

II

--+

o.

(3)

But:

fo(T)x since

~Tx - ei8ox~

=

lim IITm(Tx - ei 8o x)1 1= m~oo

lim

IITmYII,

m~oo

with y = Tx - ei 90x, and y =I 0 since T has no eigenvector. Since moreover T is C t , we see that limm~oo IITmy11 > o. By (3), we may find no such that jno(T)x =I o. Put simply f = fno' for no chosen this way. The function j is identically 0 on some neighborhood Vo of (Jo. Choose No large enough so that <jJ = <jJNo,kNo has support in Vo. The functions f and <jJ are therefore disjointly supported. They both are in A (II) , satisfy f(f)x =I 0, <jJ(T)x =I 0, and <jJ moreover satisfies (2). Since the functions are disjointly supported, we have j(T)<jJ(f) = o. Set z = <jJ(f)x. Then we know that z =I 0, j(f)z = 0 (4).

Lemma 8.2.

~

The point z is in E.

Proof. - We consider the series EkEZ ckTkx. It converges in E, by (2). Let z' be its sum : then z' E E. The series converges also in <jJ(f)x = z, so z' = z is in E. This proves the Lemma. We now consider :

F = {y E E

j

j(f)y = o}

e, and its sum is

En Kerj(T).

This is a closed subspace of E. Indeed, if Yn --+ y, Yn E F and Y E E, then Yn --+ Y in and f(T)Yn = 0, so f(f)y = o. This subspace is invariant by T, and, more generally, by any operator U which commutes with T. Indeed, as we already observed, U can be extended into an operator U, which commutes with i , therefore with j(f). So :

e,

f(f)uy

=

j(f)Uy

=

Uf(f)y

= 0,

and U y is in E. The subspace F is not reduced to {O}, since it contains z, by (4) and Lemma 8.2. It is not the whole space, since f(f)x =I o. This proves our theorem.

Chapter XII

280

We see that we don't need the invertibility of T. All what is needed is that, for some point x , there is a sequence (xn)n>O of points in E, with:

T" X n

L

n~O

=

x,

logllxnll < 1 + n2

n ~ 0,

(1)

00.

(2)

We obtain:

Corollary 8.3 (Colojoara - Foiag}, - If E is reflexive and if both T and T* are C 1 -contractions, then T has Hyperinvariant Subspaces. Proof. - We are going to show that there are points z with a bounded sequence X n satisfying (1) and (2). Take € in E*, with II€II = 1. Then Ilr*n€11 o. Put 6 = lim IIT*n€ll. Take Yn E E, llYn II = 1, such that T*n €(Yn) = IIT*nell. Then €(Tn yn) = IIT*nell ~ 6. So the sequence Tn Yn does not tend to 0 weakly. Some subsequence Tnle Ynle has a weak limit x =j:. O. Fix m =j:. O. The sequence (Tn le -mYnle)k is bounded, has a weak accumulation point X m . But then TmTnle -m Ynle has Tm x m as accumulation point, so Tm x m = z , Since Ilxm II = 1 , our Corollary is proved.

r

In the course of this Chapter, we have met two Invariant Subspaces Theorems for the C 1 - contractions. The last one (Theorem 8.1) applies when the backward iterates do not grow too fast, and, if one applies it to the usual right shift on 12(1l) , gives subspaces of the type L 2(A) = {f j / = 0 on A C } . The first one (Theorem 2.8) shows that if log >':Z: is integrable, the point x does not belong to the closed linear span of its iterates, which is formally analogous ; ak = to the Invariant Subspaces, for the right shift, of type Fn = {(ak)k~o o for k < n} (corresponding in fact to m.H 2 ) . So we have found for C 1 contractions a formal description of both types of Invariant Subspaces of the right shift on 12(1l). Unfortunately, the conjonction of both types does not provide an answer to the question : does every C 1 - contraction have Invariant Subspaces ? Indeed, Theorem 8.1 does not apply if the backward iterates of each point grow too fast (and moreover we proved in [31 that the condition given on backward iterates was best possible to obtain this type of Invariant Subspaces). In order to be able to apply Theorem 2.8, we need log >':Z: to be integrable for some x . But the functions >':Z: are supported by a(T) (Corollary 2.3) and therefore the statement requires a(T) = C. But the example 6.B has backward iterates exponentially growing, for every point, and a(T) is a proper

C 1 -contractions

281

interval of C. So none of the results apply to it. As we said, this example has Invariant Subspaces for a trivial reason : it is just a function of the usual shift: M; = 4> (Mei' ). But how to recognize in general that an operator is a function of a given one? Moreover, we now observe a clear weakness in the above results: the assumptions are not invariant if T is replaced by a function f(T) , though the conclusion is. For instance, T may have slowly increasing backward iterates, and f(T) quickly increasing ones. So these results should by no means be considered as satisfactory.

282

Chapter XII

Exercises on Chapter XII.

Exercise 1. - Let T be a C1-contraction, and U be an operator such that: Unx - Tnx

--+

0,

n

+00

--+

for all x in E. Show that U = T .

Exercise 2. - Let T be a C1-contraction on a Hilbert space. Assume that T· is also a C1-contraction. Show that there exists a unitary operator U, and operators 8 1, 8 2 , continuous, injective, with dense range, such that:

Exercise 3. - Let T be a C 1-contraction on a Hilbert space E. Let defined as in Section 1.

i , e be

1) Show that the formula (Ax, y) = [z, yj,

e into

for x, y in E defines an operator A from adjoint, and has dense range F. Show that: F =

{x E E

; sup

E which is injective, self-

IIT·- n xll < oo}.

n~O

2) Can you give a meaning to these properties if E is only a Banach space ? 3) Determine the operator A in the special case of the operator Mq, on H 2 which was studied in Section 6.B. 4) We come back to the case where E is a Hilbert space. For x E F (defined in 1)), let LkE71 akT·-kx a series, convergent in E, with sum z. If z = 0, what can be said about the Fourier series LkE7[, akeik8 ?

Exercise 4. - Let A E u(1'). We want to show that there exists a sequence (xn)n~O of points in E, with Ilxnll --+ 1, ~Xn~ --+ 1, II(A - T)xnll --+ 0, ~(A

-

T)xn~

--+

O.

a) Show that there exists a sequence (Zn)n~O

~ Zn~ = 1,

~ (A - f) Zn ~

of points in E such that: --+

o.

C 1 -contractions

£89

b) Show that there exists a sequence (kn)n~o with k~ ~ k n for all n, every sequence (k~)

of integers, such that, for

c) Show that there exists a sequence (hn)n>o of integers, such that, for every sequence (h~) with h~ ~ h n for all n,

d) Put properties.

Un

Show that

max(kn,k~).

TU" Zn has the required

Xn

Exercise 5. - Let T be a C.-contraction on a Hilbert space E. We assume that T has a cyclic vector Xo ; let /0 be the representing function for this point, and let Jl.o = /0 dO /21r . Let m

1. We assume that Xo is cyclic for T'":

~

1) Prove that there exists a set Am C II, with Jl.o(A m) = 1, and a sequence (Pn)n~O

of polynomials, with: n

2) Let Bk = Am

--+ 00,

n [2k1rJm,2(k + 1}1r/m[, B ,k

0 E Am .

0 ~ k

< m, and

= {2k1r 0- ; 0 E Bk } m

Show that the sets B k are pairwise disjoint; deduce that P(A m } ~ l/m, where P is the Lebesgue measure.

3) Show that this is impossible for m large enough. Deduce that there exists a N such that TN has no cyclic vector. 4) Is the integer N universal? In other words, is there a N which fits all C 1 -contractions?

5) Find N in the case of the operator M fJ , Section 5.B. Exercise 6. of 8 2 ?

~

Let 8 be the right shift on 12(7L). What are the cyclic vectors

Chapter XII

Exercise 7. - Assumme T to be invertible, and satisfying : " logiiTRII LJ -1----=..:+-n-2~ nE?L

< 00.

Put Pn = IITnll, n E ?L, and let Ap be the corresponding Beurling algebra. We want to show that for every lEAp, l(a(T)) c a(I(T)) . a) Show that a(T) c C. b) let I = Lj aje ij fJ E Ap , and put, for N ~ 0, PN Let .\ = ei80 E a(T). Define s» by :

PN(e i8) _ PN(e i8o)

=

= L:!N aje ij fJ.

(ei8 _ ei8o)qN(ei8).

Show that qN E Ap , and that:

(e i8 _ ei8o)qN(ei8) ~

l(e i8) - l(e i8o) ,

when N

--+ 00.

c) Deduce that: (.\ - T)qN(T) ~

f(.\) - I(T)

in £(E). Deduce that f(.\) E a(I(T)).

Exercise 8. - Let

M~

be the multiplication operator studied in Section 6.

a) Show that the closure in operator norm of the algebra AM. can be identified to the space A(K) of functions which are continuous on K, analytic in KO. b) Show that the closure of this algebra for the strong operator topology or for the weak topology can be identified to BOO : every operator in these algebras is of the form :

f

--+

t/J 4> I,

for some function t/J E BOO .

Exercise 9. - Let T be a CI-contraction on a reflexive Banach space E. We assume that there exists a bounded sequence (Xj)j;::-o such that Tj Xj --h 0 weakly. a sequence of a) Show that there exists a sequence of integers (nk)J.:~O' positive numbers (o.j), with, for all k ~ 0, L::~ll (}:j = I, and a point z in E, with z i= 0, such that: --+

z,

in norm, when k

--+

+00.

eS5

C 1 -coniraciions

b) Deduce that there is a bounded sequence (Yk)k~O that: in norm, when k --+

of points in E, such

+00.

c) Using Theorem 8.1 and Corollary 8.3, show that T has Hyperinvariant Subspaces.

Exercise 10. - Let T be a C 1 -cont ract ion on a Banach space E. We assume that there exists a point Xo in E, and E: > 0, fJ > 0, such that :

We want to show that if T has no eigenvector, and if a series converges in E :

L "IkT-kxo

=

0,

k~O

then all "Ik'S are O. a) Show that there exists a C

> 0 such that: for all k

~

O.

b) Let g = ECkeik8 E A(JI). For m ~ 0, we set:

L

tPm(g) =

ckT lc+ m,

lc~-m

Let 4>(8) = Ek~O"lke-ik8.

Show that tPm(4))XO

--+

0, when m

--+

+00.

c) Let f(z) = Ek~O "Ikzk. Deduce from a) that this function is analytic in the disk D(O,l + g). Assume that f is not identically O. Show that it can be factored : L

f(z) =

II (z -

zl)h(z),

1=1

where ZI ••• ZL are the zeros of f on the unit circle, repeated according to their multiplicity, and h a function with no zero on the unit circle. d) Deduce that 1/h(e- i8) is in the algebra A(JI). Put hi = 1/ h(e- i8),

B

L = Il,=l (ea'8 - Zl).

e) Show that if u, v are in A(JI), and if tPm(u)XO then tPm (uv )xo --+ O. f) Deduce that tPm(4)' hi) eigenvec tor.

--+

--+

O. Show that if tPm(B)

0 when m --+

--+

+00,

0, then T has an

286

Chapter XII

Notes and Comments.

Results of Sections 1, 2 are due to M. Rome and the author [I] j the proof of Proposition 2.1 was communicated to us by A. Atzmon. Section 3 is due to M. Rome [I]. Section 4 is due to M. Rome and the author [I]. Section 5 : For further results about non-quasi-analytic classes, see Y. Domar [1], Beurling - Malliavin [1], J.P. Kahane [1], Y. Katznelson [1]. Lemma 5.2 is in A. Atzmon [2]. Section 6 is due to the author [5], [8]. Section 7, Proposition 7.1 is due to M. Rome and the author [I]. The rest of the Section is due to the author- [8]. Section 8 is due to the author [3]. Exercise 4 is due to the author [7]. Exercise 5 comes from Nagy - Foias [2]. Exercise 7 is due to A. Atzmon [1]. Exercise 9 is due to the author, and so is exercise 10 [2\.

Complements on Chapter XII.

1) The Theorem of Section 8 was extended by A.Atzmon [4], who gave results containing both this theorem and Wermer's Theorem mentioned in Chapter III. One of them is as follows :

Theorem. - Let T be an operator on a Banach space E. Assume there exist sequences (x n )nE1l in E and (Yn)nE1l in E·, such that, for all n :

Assume moreover that there exists a Beurling sequence (Pn) such that: for all n E 1l,

(a)

and that there is an integer k such that:

n

----+

±oo.

(b)

C l -eontraetions

e87

Then either T is a multiple of identity or it has Hyperinvariant Subspaces.

A similar result holds if the roles of X n and lin are reversed, but it's unknown if we can omit condition b), that is, assume only : log IIT"xoll L 1+ k kEZ 2

for some point

Xo

< 00,

in E.

2) There are links between the cyclic vectors for the usual shift on co(Z) and the Invariant Subspaces for the Cl-contractions (see B. Beauzamy [4]). Let H be a Hilbert space and T be an invertible Cl-contraction on H. Let A be the operator defined by the formula (see exercise 3) :

Let

f = L"EZ Ckt~ik8

E

A(II). As in exercise 10 above, for m ~ 0, we set:

tPm (I)

=

L

clcTIc+ m •

Ic~-m

Then:

Proposition. - At least one of the following holds:

a) There exists a function f in A(II), not identically 0, which satisfies tPm(J)x --+ 0, when m --+ +00, for all x E H (we say that T satisfies an asymptotic functional equation).

b) For every z , Y in H, the sequence ((ATlcx,Y)hEZ is in co(Z) and is cyclic in this space for the right shift. c) T has Invariant Subspaces.

The case a) is excluded if the set of the zeros of the function f is a set of harmonic synthesis of type Z A+ (the reader is referred to [4) for the definitions).


PART VI

INVARIANT SUBSPACES

Tel Ie vieux vagabond, pietinant dans Ia boue, Reve, Ie nez en l'air, de brillants paradis j Son oeil ensorcele decouvre une Capoue Partout oil la chandelle illumine un taudis. Charles Baudelaire ("Le Voyage").

During the course of the previous chapters, we have encountered many results about the Invariant Subspace problem, for various classes of operators. We will now study this problem specifically, and, during this Sixth Part, give two recent results: a positive one, for contractions on a Hilbert space, and a negative one: an example of an operator on the space ll, with no Invariant Subspaces.

Chapter XIII

Positive results

In this Chapter, we consider T, c.n.u. contraction on a Hilbert space. We have seen in Chapter XI how the HOO functional calculus was defined. In a first Section, we are going to investigate the following question: When is this functional calculus isometric ? 1. When is the HOO functional calculus isometric?

To say that the functional calculus is isometric means that:

114>(T) II = 114>1100

,¢ E

u»,

To say that it is isomorphic means that there exists a constant 0 such that:

First, we study this problem for ¢ E A(D), the disk algebra.

Proposition 1.1. - Let T be a c.n.u. contraction on a Hilbert space. The functional calculus is isometric on A(D) if and only if the spectrum of T contains the unit circle. Proof. - We know that the inequality :

114>(T)11

1100

always holds, by Chapter XI, Proposition 1.1. So we study the converse inequality. a) Let's assume that u(T) contains C. Then every point A of C is in ao(T), so in ol1(T). Therefore, there is a sequence (xn)n~O, IIxnll = 1, such that TX n - AXn --+ O. Take now ¢ in A(D) with II¢II 00 , and A E C such that

I¢(A)1 = 1. Then, with the above sequence

(Xn)n~O,

n--+oo,

Chapter XIII

which implies :

114>(T)X n II

--+

14>(A) I =

1,

and so :

114>(T) II

=

1.

b) Assume conversely that u(T) does not contain some point >'0 E C. Then some arc r around Ao is in p(T). We may of course assume (for simplicity) that Ao = 1. So I - T is invertible, and so is I - rT , for r close enough to 1. Moreover, there exist 0 < ro < 1 and C such that : r

~

ro .

This means: r

~

ro .

Consider the function :

which is in A(D). Then

l14>r(T)1I

II4>r 1100 =

~

C, though:

1

1_ r

--+

+00,

which proves that the functional calculus is not isometric (and, more precisely, is not isomorphic). We now turn to the study of the same question for the HOO functional calculus. The Theorem we now prove is due to C. Apostol [11. The answer provided is not as satisfactory as in Proposition 1.1, but it says that, for a contraction the spectrum of which contains C, we can always assume the functional calculus to be isometric if we are looking for Hyperinvariant Subspaces. Theorem 1.2 (C. Apostol). - Let T be a contraction on a Hilbert space, such that u(T) contains the unit circle. If the H'" functional calculus is not isometric, the operator has Hyperinvariant Subspaces. Proof. - First, we observe that, by Chapter XI, Proposition 1.2, this functional calculus is isometric for T if and only if it is for T*. Furthermore, we know that if both T and T* are Cl-contractions, they both have Hyperinvariant Subspaces, by Chapter XII, Corollary 8.3. Therefore, we can assume that one of them is not a CI-contraction. Next, we observe that if F is an Hyperinvariant

e99

Positive Results

Subspace for T, Flo is an Hyperinvariant Subspace for T·. So we can assume that T, rather than T·, is not Ct. This means that:

= {z ; T" X

F

-.

0, n -. +oo}

is not {o}. But this set is a closed hyperinvariant subspace for T, so it must be the whole space, otherwise our problem is solved. In other words, we may assume: Tnx -. 0, n -. +00 , for all x E H. A contraction satisfying this property will be called a Co-contraction. It is convenient to represent such an operator by means of a left shift on the space 12(IN,H) (in short: 12(H)). We recall that:

n

Proposition 1.3. - Let H be a Hilbert space, and T be a C o - contraction on it. Let S be the right shift on 12(H) :

There exists a subspace X of 12(H), invariant by the adjoint S· of S, and a surjective isometry V from H onto X, such that: T = V-I (S·lx)V.

Proof. - As we already did in Chapter X, we introduce: D

= JI-T*T,

which is positive, self-adjoint, and satisfies (see Chapter X, Proposition 1.3) :

from which follows clearly: 00

xEH.

We set:

Vx

(Dx,DTx,DT 2x, ...),

(1)

Chapter XIII which is, by (1), an isometry from H into '2(H). It satisfies :

therefore:

(2)

on H.

From this formula follows that the image of V is invariant under S* in 12(H). Set X = 1m V. This is a closed subspace of 12(H) and V is a surjective isometry from H onto X. Moreover, (2) indicates that:

(3) and our Proposition is proved. To simplify the notation, we write A

= S*'x.

We observe that, instead of 12(IN, H), we might have taken l2 (IN, F) , with F = ImD. The Proposition can be rephrased as follows: any Co-contraction on a Hilbert space is unitarily equivalent to the restriction of a left shift to one of its invariant subspaces. But the reader should not expect us to deduce from this result that any Co-contraction has Invariant Subspaces (this is unknown). The shift involved here is not the usual left shift on l2 (IN) : it is a shift on a product space l2 (IN, H) (some people say: a. shift of infinite multiplicity, but we will avoid this terminology). There is no deep statement behind Proposition 1.3 (and we have seen how easy the proof was), but this representation of the operator will be convenient for what follows. For any A,

IAI

IIPPAII

1,

- or oX E p(A), and then: inf{IIP - A)-lxll ; x EX, IIxll = I}

1

and by Lemma 1.5, IIPPAI!2 ~ 1 -IIU(A)11 2 a2 (1 -IAI)2 ~ 1 - a2

,

by Lemma 1.4, 3), and the result follows. A subset G of the open disk D is said to be dominating (for the HOO

functions) if, for any ¢ in H?", sup{I¢(z)1 ; z E G} = lI¢lIoo. This is obviously the case for any corona {a < Izl < I}, with 0 < a < 1, since we know that the numbers

Moo(¢,r)

=

sup{I¢(z)1 ; Izl = r}

are increasing with r (Chapter VIII, Section 3).

e98

Chapter XIII

A set G is dominating if and only if almost every point in C is the nontangential limit of a sequence of points in G.

Proposition 1.7. - Let T be a Co -contraction, such that u(T) contains the unit circle. If, for some a, 0 < a < 1, the set Za is not dominating, the operator T has Hyperinvariant Subspaces. Proof. - If Za is not dominating, some subset C1 of the unit circle, of stricly positive Lebesgue measure, has the following property: for every 8 in C1, ts is not in Za if t < 1 is close enough to 1 : this means indeed that 8 is not the radial limit of points in Za. More formally, for every 1, ts i Za. Set

8

Eel, there exists

R, = {t8;

E: a

E: a

> 0 such that, if

ea

.. We have obtained:

Lemma 1.8. - For the operator S· (or, more generally, for any contraction for which the functional calculus is isometric), for every .\ ED, there exists an element C'A in Jls-, such that:

4>(.\), [or every function ¢ E H?" . The set of the C'A, .\ Era is large enough to generate the unit ball of

Jls- : Lemma 1.9. -If

ra

is dominating,

Proof. - For every ¢ E HOO , we have:

11¢(S*)11

=

II¢lIoo sup 14>(.\)1 'AED

sup 14>(.\) I 'AEr ..

The set aeonv { C>. j .\ Era} is therefore a convex su bset of BMs _ which norms all elements of As-. The equality follows from Hahn-Banach Theorem.

Lemma 1.10. - Let NEB Ms -

Then:

.

dist Ms - (N, aeonv(Bx

Proof. - Let e > 0 be given. C>'I' .. ' ,C'A ... , such that:

18)

Bx))

< a.

By Lemma 1.9, we can find al, ... , am,

m

liN -

EaiC>'iIlMs 1

0, there is a N in B)I,. such that:

Using the assumption, we can find n ~ 1, aiJ ... , an in (S.)) > 114>1100 -

lail

~ 1,

E.

= tr (4)(S*)x.: 18> Y.:)

tr ((x.: 18> Yi) 4>(S*»

= (4)(S*)x.:, Y.:)

= (4)(A)x.:, Yi) , since the points x.:, So we obtain :

Yi

are in X.

114>1100 -

1 e < 1_ a

n

L

lailll4>(A)II·ll xiH·IIYill

1

< 1 ~ a 114>(A)1I , and the Lemma follows. We can now prove the Theorem. As we already said, we can restrict ourselves to the case where T is a Co-contraction. Assume that T has no Hyperinvariant Subspaces. By Proposition 1.7, Za is dominating for every a, o < a < 1, and so is r a' By Proposition 1.13 and Lemma 1.14, for every 4> E Hoo,

114>(A)1I = 114>(T) II

~

(1 -

and so:

114>(T)II and the Theorem is proved.

=

114>1100,

a)II4>lloo ,

Positive Results

905

2. Invariant Subspaces. We are now going to prove:

Theorem 2.1. - Let T be a Co-contraction on a Hilbert space H. If for every a, 0 < a < 1, the set Z~ of points ..\ E D such that there exists a normalized weakly null sequence (zi")),,~o with:

is dominating, T has Invariant Subspaces.

Proof. - We use the notations of Section 1. The Theorem will be proved if we can find two points z , y in H such that: Co

=

x®

v,

(1)

Indeed, for every h E HOC) ,

h(O) = (h(T)x, y). Taking first h = 1, we see that (z, y) = 1, so x and y are non-zero. Furthermore, taking h(z) = zk, k ~ 1, we obtain (Tkx,y) = 0, so span {Tx, T 2x, ... } is orthogonal to y and is the required Invariant Subspace. To obtain x and y, it is clearly enough to use an approximation procedure:

Lemma 2.2. - If there exist two sequences x" ® y"

--+

Ilx" - x"+lll < IIY" - y,,+Ii1
O - and (Yn)n>O. We start with Xo = Yo = o. Assume that we have built the sequences Xo, ... , Xn-l, Yo, ... ,Yn-l, such that: IIXk - xk-ll! ~ I/2 k

IIYk - Yk-lll ~ I/2 lIXk @Yk

-

k

,

,

k

COllJlT ~ I/4 +

1

k = 1,

,n - 1

(5)

k = 1,

,n-I

(6)

k = 0, ... ,n - 1.

,

We wish to build Xn, Yn satisfying (5), (6), (7) for k We know that :

We want to build

Xn

= Xn-l + Un,

Yn

= n.

= Yn-l + V n , such that:

Ilunl! < I/2

n

(9)

llvn II < I/2 n II(xn -

l

+ un)

We write e instead of 1/4

n

@ ,

(10)

n 1 (Yn-l + v n ) - ColIJlT < 1/4 +

and u, v instead of Un,

Lemma 2.3. - For every a, 0

(7)

.

(11)

Vn .

< a < 1,

Proof. - This is Lemma 1.9 above, with S· .

instead of

Z~

r a, and

T instead of

Let now fJ > o. For every a > 0, by Lemma 2.3 applied to e BJl T , we can find a finite sum L~ aiC>.. , with Ai E Z~ for all i, such that: N

II LaiC>.. - (Xn-l ® Yn~l

- CO}lIJiT < fJ

(12)

1

and: N

Llail < c.

(I3)

1

We now have to compare the norms of such quantities in )IT and )Is·. We recall that V is the embedding of H into 12(H) :

Vx

=

(Dx,DTx,DT 2x, .. .).

Positive Resu.lts

L

Lemma 2.4. - For any finite sum

II x ® y -

L aiCAJ..vT

aiG Ai, every points x, y in H, IlVx

=

907

e Vy -

L aiGAJ..v

S ••

Proof of Lemma 2.4. - For every function h in Hoo, tr (h(T)(x ® y -

L aiCAJ)

= (h(T)x, y) -

L aih(.Ai) L aih(Ai)

= (Vh(T)x, VY)h(H) -

= (h(S*)Vx, VY)'2(H)

= tr

-

L aih(Ai)

(h(S*)(Vx®Vy- LaiGAJ),

IIhll oo = 1, we

and if we take the supremum of both sides upon h in H'" with obtain the Lemma.

,AN are now given by (12) and (13):

The numbers N, a1, ... ,aN,A1 they depend upon a, 11. For i = 1 PiP: = -ai. We set:

, N, we choose

Pi

and

Pi

such that

N

L Piztd

U

(14)

1 V

=

N

~fJ~z(v;)

LJ

J

A;

(15)

,

1

and we will show that if a, 11 are small enough, and if III < ... < liN are conveniently chosen, the points u and v satisfy our requirements (9), (10), (11). To simplify our notation, we put:

A

=

(X n-1

+ u) ® (Yn-1 + v).

We want to compute IIA - Coll..vT ; according to Lemma 2.4, we can compute instead IIA - Goll.N s • . We write an expansion: N

A - Go

= X n-1 ® Yn-1

Co +

-

L PiP: (ztd ® zl7 ») i

(a)

i=l N

+ x ® LfJj zi;;)

(b)

i=1 N

+ LfJi zti )

®Y

(e)

i=1

+ LJ ~ p.fJ-~

I

J

z(vd

x,

e z(v;) A;'

it:i

and we will take care of each line separately.

(d)

Chapter XIII

908

zi

v

) is "close" (by definition) to being an eigenvector for T (or First, each V ) , for S·). So it is convenient to consider a point in 12(H) which is really an eigenvector for S·, for the eigenvalue A :

zi

v

Lemma 2.5. - Let A E D, 0 < a < 1, and that:

Then, there exists

Furthermore, if v~

IlzA11 =

E H, with

II (A - T)ZAII < a(1 - lAD· K er (A - S*) c 12(H) such that ileA I

eA E

VzAl1 < v'2a.

ZA

zi

v

)

~ 0 weakly in H, so does

ei

v

)

1, such

= 1 and

IleA -

in '2(H), when

00.

Proof of Lemma 2.5. - We observe that :

(16) Indeed,

II (S·

- A)V ZA I = IIVTz A

-

AZA II
. the orthogonal projection from 12(H) onto Ker (A - S·), we deduce from Lemma 2.6 : IIp>.Vz -

Vzll = IIU·zll
.Vz>.

IIP>.Vz>.1I This proves Lemma 2.5. In the terms (a), (b), (c), (d) appearing in the decomposition of A - Co, d instead of d , and then the we consider the corresponding terms with differences.

et

Lemma 2.7. - The term (a) has a norm in Proof of Lemma 2.7. - In

)Is·,

zt

)IT

at most

fJ +

V2ea.

we write: N

(a) = VXn-l ® VYn-l - Co -

L aiei:» ® ei:d i=l

N

+ '" (e(lId L....J 0:'' > 'i

® e(II.-) - V Z(lIo} ® V z(lId) . Ai

i=l

Let (a') be the first line, (a") the second. By Lemma 1.11,

Ai

x,

Chapter XIII

910

So (a') can be rewritten : N

(a')

=

%"-1

~ Y"-l - Co -

L

aiC).i ,

i=l

which implies by (12) that II (a') liNT < ,.,. For (a"), we write simply : N

L laililetil e (etd - V ztd ) IIN

II (a") IINs • ~

s•

i=l N

+

L lailll(ei:d -

Vzi~d)

~ VztdllN s •

i=l

~ 2V2ea, by the same computation as in the proof of Lemma 2.2. We now turn to the term (b) :

Lemma 2.8. - There is an infinite subset IN 1 of IN , such that if VI < ... < VN are in IN 1, the term (b) has norm in )IT at most n + 4,.,y'€ (1 + ,.,). Proofof Lemma 2.8. - For every V1 < ... < with Ilh v1,...,vNlloo ~ 1, such that:

VN,

there is a function hV1, ...,VN ,

N

IIx ~

L /3jzl;;) liNT

N

(h Vl ,...,VN (T)x,

=

,=1

L /3jzl;;)) .

,=1

We can find an infinite subset of IN, IN'l' such that, if IN~ , the limit of h V 1 , ••• ,VN exists weakly in H?", when Let h o be this limit. We have:

VI VI

<
- el:

i

)

11 2 ) 1/ 2

·IIVYII,

i=1

if the sequence

VI

< ... < VN is chosen in IN2

C IN~,

and therefore :

To finish, we have to compute the norm of (d). The computations are similar to the previous ones. Lemma 2.10. - There is a subset IN3 c IN2 such that if VI < ... < in IN3, the term (d) has norm in )IT at most 71 + (1 + tJ) 2 e y'2 a .

Proof of Lemma 2.10. - We write, in

(d)

)ls* :

VN

are

Positive Results and call (d') the first line, (d") the second. For (d'), we find h in HOD, with lIhll oo

II (d') II .vs • =

919

= 1, such that:

L fJiPih(.xi){ei~o),

V z1;;»),

ioFi

and we find a subset IN~

of IN2 such that if

VI

< ... < VN are in IN;,

if i,j= 1, ... ,N,with ifj. We get:

L IPillfJil/4i+i

II(d')II.vs • s

< " ,

i,j

and for (d") N

N

II(d")II.vs • s I LfJi(Vzto) - ei';il)ll·lI LfJ'jVzi;i)1I i=1

i=1

~ (1 +,,) if

VI

N

N

i=1

i=1

(L IPi1 211 Vzt a> - e1:;) 11 2)1/ 2 (1 +,,) (L IPiI 2) 1/ 2 ,

< ... < VN belong to a subset INa of

and our Lemma is proved. We now choose We observe also that:

IN~.

a

So we get:

and" small enough to get (11).

N

Ilull s

(1 +

")(L IPiI 2)1/ 2

l/a(I-I).I). Set finally z = x/llxll.

IlzlI

= 1, and :

II(). - T)zll

lIyll < = ~

a (1 - 1).1)·

(2)

Assume that the spectrum of T contains the unit circle. By Section 1, if we are looking for Invariant Subspaces, we may assume that for every a, the set of ). 's satisfying (1) is dominating. This ensures that there exists a z satisfying (2). But Theorem 2.1 requires not only a point z, but a weakly null sequence. So we cannot conclude immediately from these arguments that any operator with spectrum containing the unit circle has Invariant Subspaces. However, this is true, and was proved by S. Brown - B. Chevreau - C. Pearcy [2] by some refinements of the techniques presented here. The question of the existence of Hyperinvariant Subspaces for such operators is still open.

Positive Results

915

Exercises on Chapter XIII.

Exercise 1. - Let T be a Co-contraction on H and let U be its minimal unitary dilation. Show that, for every x, y in H,

Exercise 2. - Let that:

z).

as in Lemma 2.5, and

f).

IIPsn f ). 112 >

lim

given by this Lemma. Prove 1 - 2a 2

•

).Ez~,I).I-l-

Exercise 3. - Prove that for any x in H, the function :

is weakly sequentially continuous from H into )Is· (that is : if a sequence (Yn)n~O converges weakly in H to a point y, the sequence x ® Yn converges weakly in )Is. to x ® y). Exercise 4. - Show that in

)Is. :

v x ® Vy = x ® Py ,

x,yE H.

Exercise 5. - Assume that (xn)n>O is a sequence of points in H such that 0 in )IT, for all y in H. Show that V X n ® Z ~ 0, for every z in

Xn ® y ~

l2(H) . Exercise 6. - Let T be a contraction with dominating spectrum. Let A E o(T) n D, and let (xn)n~O be a weakly null sequence such that (A - T)x n ~ o. Show that X n ® y ~ 0, in )IT, for any y E H.

Notes and Comments:

The basic ideas developed in this Chapter come from Scott Brown's work [1], [2], [3]. Theorem 1.2 is due to C. Apostol [II. Corollary 2.11 appears in S. Brown - B. Chevreau - C. Pearcy [1], and the result mentioned in the Remark in S. Brown - B. Chevreau - C. Pearcy [2].

916

Chapter XIII

Our terminology differs slightly from the one adopted by some authors : they call "Co-contraction" a contraction T such that there is a function / in HOO with f(T) = o. In Nagy-Foiaa's terminology, an operator satisfying Tn x ~ 0, when n ~ 00 is called Co. (Coo if the adjoint has the same property, CO,I if the adjoint is a CI-contraction). Here, we don't use the adjoint at all, so we could simplify the terminology. Moreover, the use of Co to indicate that the operator is annihilated by an analytic function does not sound quite natural. The most natural question suggested by this work is : does any operator with r(T) = IITII have Invariant Subspaces? This question was stated long ago. We will also mention it at the end of next Chapter, since the results presented there also support this conjecture.

Chapter XIV

A Counter-Example to the Invariant Subspace Problem in Banach Spaces

The major progress in Modern Operator Theory is obviously P. Enflo's counter-example to the Invariant Subspace problem in Banach spaces (P. Enflo [1]), which opened new lines of thought. The very idea of building an example of an operator enjoying given properties was new in this field, and no previous attempt had ever been made to solve the Invariant Subspace problem in the negative direction. The ideas introduced in Enflo [1] had many applications, including to the positive direction on Hilbert spaces. Two other examples followed the lines initiated by P. Enflo: C. Read

[1] and the present author (B. Beauzamy [9]), the latter enjoying a stronger property, called "supercyclicity". C. Read's example was later strenghtened and simplified by himself (C. Read [2], [3]), so as to produce an operator on II with no Invariant Subspaces (See the "Notes and Comments" at the end of the Chapter for historical comments). C. Read's latest example was then simplified by A.M. Davie [unpublished], who communicated it to us. Though this construction is not the strongest (it does not possess any property like supercyclicity or hypercyclicity), it is by far the simplest, and we present it here. It produces a very elementary example of an operator on II with no Invariant Subspace. In the "Complements" at the end of this Chapter, we refer to more sophisticated examples and consider the problem in Hilbert spaces. Before turning to the construction of the example, we start with a few basic observations. Let T be an operator on a Banach space E. 1) To say that the operator T has no Invariant Subspaces is obviously equivalent to the fact that every non-zero point is cyclic. Indeed, if x is non-cyclic,

is a non-trivial Invariant Subspace.

Chapter XIV

918

Conversely, if T has a non-trivial Invariant Subspace, F, taking x E F, we see that Fr. C F, so x is not cyclic. are linearly 2) Take a cyclic point Xo for T. Then the iterates (Tkxoh:~o independent (since the space is infinite dimensional). Consider a finite sum Lk>O akTkxo. To such a sum, associate the polynomial p(z) = Lk>O akzk (thi; is well defined, by what we just said). On the space P+ of polynomials, pu t the norm : IIpll = II ak z k II·

L

k~O

Then the completion of P+ under this norm is a Banach space, which is obviously isometric to E, since the finite sums Lk~O akTlexO were dense. We keep for it the notation E. Moreover, to the sum :

we associate the polynomial :

L k~O

a/czk+l

=

z(L a/czle) . k~O

In other terms, T becomes the multiplication by z on the space P+, equipped with the norm 11.11. This can be expressed as follows:

Lemma. - Every operator with a cyclic vector can be represented as the multiplication by z on the completion of the polynomials in z , under some norm.

There is obviously nothing deep in this Lemma : it is a mere rephrasing of the original situation. But we understand that there is no restriction of generality if we try to find T as the multiplication by z on a space of polynomials. Moreover, it allows us to simplify the notations: if Xo is a cyclic point, and if p, I are polynomials, instead of p(T)xo we just write p , and instead of l(T)p(T)xo, we write l.p. The norm coming from the operator norm is written as lI'llop, that is : II l 1i op = IIl(T) II· 3) Under this representation, the point Xo becomes the polynomial 1, which is cyclic in E for the multiplication by z (by definition). To say that any point x in E is cyclic means that:

A Counter-Example

919

To obtain this property, since 1 itself is cyclic, it is enough to ensure that 1 E span {x, zx, z2x, .. .}.

This can be written: For every e > 0, there exists a polynomial I such that :

(1)

III x-III < c.

(In the original space, and with the original notation, this property would be written: 1I/(T)x - xoll < c.)

4) In order to produce an operator with no Invariant Subspace, we have to satisfy (1) for every non-zero x, and not just for polynomials. Indeed, satisfying it just for polynomials would ensure them to be cyclic : we would have a dense set of cyclic vectors. But this is by no means uncommon: many operators do have a dense set of cyclic vectors, such as the right shift of 12(71) (see Chapter IX, Exercise 8), or even of hypercyclic vectors (see Chapter III), and they still have Invariant Subspaces. H all polynomials are cyclic, take x E E, and a sequence (Pn)n~O of polynomials converging to x in E. Then, for every e > 0, there is a polynomial In such that:

II/nPn -

111 < c.

If we have

suPll/nllop
0, there exists a polynomial I such that : II/p

-111
0, there is a sequence of polynomials (Pn)n>O, Pn --+ x, with bounded operator norms.

Chapter XIV

S20

5) There is certainly no uniqueness in the choice of the I's satisfying (2). Indeed, if by some II we can move p close to some XI, that is

then XI close to some X2 by 12 , ••• , Xn-I close to X n by In, and finally X n close to 1 by 'n+l, then the product In+l.ln ... 1t will move p close to 1, within C if CI, ••• Cn+1 were chosen small enough. Moreover, this product is likely to have considerable operator-norm. Fix now C > O. Say that I "acts on p" if it satisfies (2) and if moreover it has small operator norm, that is for instance : 1I'l1op ~

inf{II"lIop i I' satisfies (2)} + 1.

In what set around p can we expect the operator norms of the I's acting on this set to remain uniformly bounded? The best would of course be the whole unit sphere. However, this is not possible: this would contradict the fact that the spectrum of T is non-empty (see Exercise 3 below). So we cannot expect to find a constant C such that 1111lop ~ C, for all l's acting on the unit sphere. Therefore, we have to bound the norm 1I.lIop "locally", that is near each point z , with a bound depending on x. The type of the set, containing z , on which Illliop is bounded will determine the structure of the construction. It is a neighborhood of x in P. Enfto [11, B. Beauzamy [91, but on the present construction, it is computed using the coordinates on some finite-dimensional subspaces. So far, we have given only statements which were either trivial or too vague, but were required to explain on what lines an example can be found. We now turn to the construction itself. Theorem. - There exists an operator on

II

with no Invariant Subspace.

Proof. - We first recall or introduce some notation.

n

~

Let, as before, P+ be the set of polynomials with complex coefficients. For 0, let Pn be the set of polynomials with degree ~ n. If p

= E;~o

a;z; , we put: Ipl

=

Lla;l. ;~o

This is the 11 norm, which we denoted by Ipll. For simplicity, we drop the subscript 1.

SHl

A Counter-Example Moreover, for n

~

0, we put: n

E ajz j

=

Pn(p)

;=0 and val(p)

= min{j

; aj -1O}. This is the valuation of p.

Lemma 1. - Let n ~ 0, E:, s, M > o. There exists K > 0 such that, for every m, 0 ~ m ~ n, every polynomial 9 E P«, satisfying Igi ~ M and IPm(g)1 ~ s, there exists q E P«, with:

and

Iql

~ K.

Remark. - The reader will observe that the assumptions on 9 are of "concentration at low degrees", as we discussed at the end of Chapter VIII.

= 0, ... , n, we set

Proof. - For m

Om =

{p E

Pn ;

:

Ipl

~

M and IPm(p) I ~

()}.

This is a compact subset of Pn • For 9 E Om, we write the Euclidean division of zm by g, according to increasing powers of z : zm

with val(r)

=

gq

+r

,

> n. So we get: zm - r ,

gq

Therefore, there exists an open neighborhood U of 9 in Pn , such that, if g'E U,

We cover Om by U lJ • • • , UK.,. , corresponding to polynomials glJ ... ,gK... , and, for k = 1, ... ,Km , if 9 E Uk, we have

Finally, we take:

K and our Lemma is proved.

max max m

k

Iqkl ,

se»

Chapter XIV

Let now (an)n~o and (bn)n~o with ao = bo = I, and :

,

1.

of P+ as follows:

We define a basis (/n)n~o

10 = I, -If n 2: 2 and 1

~

r

~

n - I, or n = r

if ran ~ k ~ ran +

= I,

Vn-r-I ,

he

if (r - I)a n + V n- r < k < ran,

Ik

-If n 2: 2 and 1

~

r

~

(a)

Zk-I1.. ),

21(r- t )11.. -kllb..-

1

zk

,

(h)

,

(c)

n - I,

if r(a n + bn) < k ~ (n - I)a n + rbn , if (n - I)a n + (r - I)b n < k < r(a n + hn) , We also define formula (d) with n

an_r(Zk -

fA:

he = lie

zk - bn zle-bn

= 21(r- i >b..-kl/nl1.. zA: , (d)

by formula (c) with n = 1 when k = al < k < al + hi-

al

+ bl

,

and by

= 1 when

For fixed n 2: I, we have defined fie for V n-l < k ~ V n . Indeed, the sequence (b), (a), (b), (a), _.. allows us to go from V n-l to (n - I)a n : V n-l -+

an

-+

an +V n-2

-+

2a n

-+

2a n +V n-3

-+ ... -+

(n-I)a n

-+

(n-I)a n

•

Then the sequence (d), (c), (d), (c), ... allows us to go from (n - I)a n to (n - I)(an + bn ) = V n :

(n - I)a n -+

-+

an + bn

(n - l)a n + 2hn

-+

-+

(n - l)a n + bn

...

-+

-+

2(a n + hn)

(n - I)(a n + bn)

-+

(n - I)(a n + hn) .

In the first case, to ensure that the numbers involved are in the right order, we

need:

(r - I)a n + V n- r < ran,

A Oounter-Ezampl« which means V n -

r

(n - I)a n - if r < n - 2 :

+ (r + I)b n ,

r

~

n - 2),

= 1. k + 8 < (r + 2)(a n + bn ) , we have:

Ilzk+all = 2(k+a-(r+!)b n)/nan ::; 2(2(n-l)a n

- if r

~

-

!bn)/nan .

n- 2 :

IIz k + all = 2(k+B- !an+I )/b n 2(b n+2(n-l)a n -!an+ al/b n .

~

and the same way, if r < n - I,

and if r

=n -

I, Ilzk+a-b n

II

= 2(k+a-b n- !an+d/bn ::; 2(2(n-l)a n-!a n+d/ bn ,

and we get IITa /kll ~ I in all cases, choosing bn

,

an+l large enough.

A Counter-Example

997

2) We now consider the case n > m. - If k

+s

~

(n - 1)an + rb n

,

- If k + s > (n - 1)an + rb n , - if r < n - 1, we have k + s < (r

IIz k+ 6 11

+ 1)(an + bn ) ,

= 2«k+6)-(r+!)b.. )/na.. 2«n-l)a.. +2b m

~

-

!b.. )/na n

,

~ l/b~

and the same estimate holds for IIz k +6 - b.. II. - if r = n - 1, we have the same estimate for

Ilz k+611 = 2(k+6-!a ~

2(v.. +bm-

n

Ilzk +6- bn II, and

+ 1 )/b ..

i a .. +d/ b..

and we get:

IIT

6

/kll ~ l.

d) (n - 1)an + (r - 1)bn < k < r(a n + bn ) . Then:

T 6/ k =

2((r-i )b.. -k)/na.. Zk+6.

The hypothesis k > (m-1)a m again implies n 2:: m, and we consider separately the cases n = m and n > m.

1) n = m. Since s > bn , k + s > (n - 1)an + rb n • We distinguish between four different subcases :

a) k + s > (n - 1)(an + bn ) •

II Z k + 611 = 2(k+6- !a..+d/bn ~ 2(nb.. -!a .. +d/ b.. ,

IlT

6

/kll

~ 2bn/an+(nbn-!an+d/bn ~

if

an+1

is large enough.

1,

Chapter XIV

998

{l) k + 8 < (r + 1) (an + bn ) , r < n - 1. Then:

since

8 -

bn ~ (n - 1) an .

'"Y) (r + 1)(a n + bn) ~ k + Then, by (2) :

~ (n - 1)a n

8

+ (r +

1)b n, with r

rb n - nan ,

~ 2(r-!)b..-k)/na..

< -

2(na ..-!b..)/na.. br+ 2

~

1,

n

if bn is large enough.

6) k + s > (n - l)a n + (r + l)b n with r < n - 2. Since 8 < 2b n, we have k + 8 < (r + 2)(a n + bn) I and so : I

IIz k+"11 = 2(k+.-(r+!)b..)/na.. ~

if bn is large enough.

2(k+(n-l)a..-(r+!)b..)/na.. ,

A Counter-Example

999

2) n> m. - IT k

+8
0, every family (Yk)Ie~O of points in E \ {O}, dense in E, there exists a family (B:")m~o of open balls, all centered on points Y;, such that uBm = E \ {O}, a family (Cm)m~O of positive reals, such that if Y; E B k, there is a polynomial ';,k with:

Exercise 3. ~ Let Xo be a cyclic point for T. Let oX E O'a(T), and (Yn)n>O, llYn II = 1, be a sequence of corresponding almost eigenvectors, that is (see Chapter II) : n -+ 00. Take: e

0, there is a polynomial lz such that:

then: sup z,lIzll=1

II/ z (T )lI op = +00.

Exercise 4. - Dependence upon e of the Invariant Subspace property. To say that T has no Invariant Subspaces means: For every e > 0, for every x, y, with IIxli polynomial I such that :

II/(T)X -

yll
0, every z , there exists a polynomial I such that:

II/(T)(T - I)x -

xii
0, x E E. Take I given by a). Show that:

~ II/(T)(T _ I)x _ xii + 11/(T)1111 x - Tn x II.

II x + ... + Tn-Ix II n

n

(Hint: introduce the term I(T)(T - 1) :t+T:t+.~±Tn-l:t

.)

c) Deduce that, for n large enough, 11

x

+ ... + Tn-Ix n

il

< 2e.

Chapter XIV

d) Deduce that if T is a contraction on a Banach space, such that there are points x, Xo and ex < 1 such that, for all n, IITAx - xoll < ex, then T has Invariant Subspaces. (This is clear from Exercise 9, Chapter XII, when E is reflexive, but does not follow otherwise.)

Exercise 7. - Closed invariant convex subset of the unit ball (see also Chapter II). Let T be a non-zero contraction on a Banach space. We want to show that there is a closed convex subset C of the unit ball, with C =j:. 8, C =j:. 0, such that TC C C. a) Show that if T has no convex invariant set, for any x E B, with

= 8. z, y in E

Ilxll =

1,

conv{x,Tx,T 2x, ... }

b) Deduce that for every E: > 0, every there exists a polynomial I = L:j~O aj zj , with that: 1I (T )x - yll < g.

aj

with ~ 0 and

IIxii = lIyll = 1, L: aj = 1, such

'

c) Use Exercise 3 above to obtain a contradiction.

Notes and Comments:

Before the Banach spaces examples were found, A. Atzmon [2] produced an example of an operator on a nuclear Frechet space with no Invariant Subspaces. This example relies on properties of some spaces of analytic functions, and is quite interesting. There is no link, however, between his approach and the one which was used later in the Banach space case. P. Enflo's paper [1] was submitted to Acta Mathematica in february 1981. The paper was not well-written, and was quite complicated and not easy to understand, so refereeing took some time, as well as the minor changes which were made, so the paper was definitely accepted in 85, and appeared in 87. Using the ideas of P. Enflo [1), but simplifying and strengthening them, the present author also produced a Banach space example B. Beauzamy [9). This example had a stronger property, called 8upercyclicity : for every x =j:. 0, the set {CTn x ; C > 0 , n E IN} was dense in the whole space.

A Counter-Example This example was presented at the Functional Analysis Seminar, University of Paris VI - Paris VII, in february 84. It appeared in Integral Equations and Operator Theory in june 85. At the same time, C. Read also produced a counter-example (C. Read [1]). His paper benefitted of quick refereeing and "jumped over" the waiting line for publication in the Bulletin 01 the London Mathematical SocietU, so it appeared in july 84. Similar facilities were also offered, by the editors of this Journal, to the present author, who declined. To give such precision in uncommon, and would be of no interest, if C. Read had not, several times, unelegantly and unsuccessfully, tried to claim priority towards the solution of the problem. This behavior might be condemned with stronger words, but we remember we are presently writing for posterity. We have insisted upon several ideas introduced by P. Enflo, and which were used in all later counter-examples. But we would not like to give the impression that the other counter-examples are trivial modifications of his. C. Read's example on II (C. Read [3]), presented here in a simplified version is a major achievement in the theory (I am glad to point this out, despite the previous paragraph), and his latest result of an operator in 11 with all orbits dense (all points hypercyclic) is the "state of the art" in the Banach space setting. In the frame of Hilbert spaces, things are not so advanced. The operator which is "closest" to a counter-example" is the one built by the present author [12] : it has one hypercyclic point %0, and for every polynomial p with complex coefficients, p(T)xo is also hypercyclic. Therefore, the operator has a vector space of hypercyclic points (thus solving a question raised by P. Halmos [2]), but it may still have points which are not cyclic at all, thus having Invariant Subspaces. In my opinion, though being the "closest", this operator is still quite "far" from having no Invariant Subspaces, so far, even, that I have no idea whether such a counter-example exists or not. IT one compares the positive results of the previous chapter and the negative ones of this one, one sees the very large gap in between, and one understands better, perhaps, the opinion stated in the introduction about the state of development of the theory. A lot is still to be done, on a problem which is rather challenging, if not rewarding.


INDEX

absolutely convex hull algebraic tensor product almost eigenvectors approximate unit approximation property

118 24 165 119

Beurling Algebras Beurling sequence Blaschke product

262 264 180

Calkin Algebra Cayley Transform Cesaro averages of the partial sums character (in an algebra) characteristic polynomial commutative algebra commuting with T (operator) compact completely non-unitary operator concentration d at degree k contraction convolution operators cyclic for (T, T*) cyclic point C* -algebra generated by a normal operator Co-contraction

95 236 167 127 6 123 32 79 213

dilation of an operator divisor of 0 dominating

212 124 297

eigenvalue eigenvector elementary hyperinvariant subspace elementary invariant subspace essential norm

6 6 33 33 94

299

190

91 208 150

32 141 293

Index

9.48

essential range essential spectrum extension into a unitary operator

153 95 214

finite support (vector with)

26

Gelfand's Transform

128

harmonic function hypercyclic point hyperinvariant subspace hyponormal operator

163 71 32 155

ideal inner function Invariant Subspace Invariant Convex subset involution isolated point of the spectrum isometric dilation isometry

123 177 32 32 131 30 215 213

Jensen's functional

189

Mahler's measure maximal ideal minimal isometric dilation minimal unitary dilation minimal polynomial multiplicity of an isometry multipliers

189 126 216 216 7 227 208

non-quasi-analytic algebra non-tangential convergence normal element in an algebra normal operator norming sequence nuclear norm nuclear operator nuclear operators on a Banach space

262 171 131 84 63 97 97 119

ou ter function

177

Index

partial isometry partial sums of a Fourier series Poisson Kernel pole of an operator positive operator on H positive operator on L p projection of an operator projective completed tensor product proper ideal pseudo-function pure isometry

88

167 165 30 86

228 215 118

126 258 227

quasi-nilpotent operator quasi-triangular

84 95

radial convergence radical (of an algebra) reducing subspace regular measure regularly closed subspace residual part of a dilation resolution of Identity resolvent of an operator resolvent set of an operator resolvent set (in an algebra) right shift operator

171 129 213 143 244 222 150 21 21 124 20

Schwarz map self-adjoint element in an algebra semi-simple algebra similar to a contraction (operator) singular inner function spectral family of projections spectral measure spectral radius (in an algebra) spectral radius of T spectrum of an operator spectrum (in an algebra) spectrum (of an algebra) spectrum in a finite dimensional space

120 131 129 231 182 158 143 124 22 21 124 127 6

Index

950

*-homomorphism *-residual part of a dilation step-functions strict support (function with) strong topologies sub-algebra subharmonic function subnormal operator supercyclicity support of a finite sequence

132 222 144 261 110 123 163 154 344 45

trace of an operator trigonometrically well bounded operator

98 158

ultrastrong topology ultraweak topology ultraweakly closed algebra unit in an algebra unitary dilation unitary operator unitary algebra

110 106 242 123 211 209 123

valuation of a polynomial

321

weak topology weighted shift

101 84

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