A Primer on Spectral Theory (Universitext)

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Bernard Aupetit

A Primer on Spectral Theory

Springer-Verlag

llilivhrSittx( Editors (North America): )A1. Ewing, F.W. Gchring, and P.R. 1-lalmos Aupetil: A Primer on Spectral Theory Bergen: Geometry I, 11 (two volumes) Blledinernlansen: Potential Theory Boa,~/Illeecker: Topology and Analysis Chandrasekharan: Classical courier Charlap: 1lichcrbrach Groups and Flat Manifolds Cliff n: Couples Manifolds Without Potential Cohn: A Classical Invilatitn to Algebraic Numbers and Class cicids Curlli: Abstract Linear Algebra Curds: Matrix Groups van Dalen: Logic and Structure Devlin: Fundamentals of Contemporary Set A Formal Background to Mathematics I a/b

Ed*ards: A tkwmal Background to Mathematics 11 a/b Eatery: Stochastic Calculus Futckss/Rokhlln: beginner's Course in Topology in I?pidrmiolotry Fratienthal: Mathematical

Cardiner: A lust Course In Uroup (arding!Tamhfi ir: Algebra for Computer Science ( mlhlllon: Dynahmical Systems on Suiraces t:oitlbilall: and IIuml/Miikh Second Course In Order Ordinary DilYercu...A'tquations 11dtMts/Krlllkas: Lectures on Number Theory Iversoil: Cohomology of Sheaves Kefy/Maltbews: The Non-Euclidean hyperbolic Plane koialydlcln: Introduction to Algebra ICiasnoneMN/hekrovskII: Systems with Hysteresis Lileclting/Rubel: Complex Analysis: A Functional Analysis Approach Marcus: Number Melds McCarthy: Introduction to Arithmetical Functions Meyer: hasenlial Mathematics for Applied Melds Mlnes/IZlehman/Ilallenlnn : A ('curse in ('onsnnclive Algebra

Moise: Intruducltry Problem Course in Analysis and Topolorgy Montesinos: Classical Tesselations and Three Manifolds NlkeUnlShatitevkh: Oeometries and Groups f3lcaendal: Stochastic Differential cquationrs

keel: Notes on Deontelry Rebe`: Elementary Theory of Metric Spaces IEeet IntichaCUIM to Robust and u melon Algebras Nu

Statistical Mcthixls

tcvnrnruuriI f/rr in?h o1

Bernard Aupetit


Springer-Verlag

New York Berlin Heidelberg London Paris Tolcyo Hong Kong Barcelona

I1, Aupd t tiparlmcnl de Mathdmatiqucs el Staliaiique

University Laval

Qubocr Canada OIK 7P4 Editorial Board (North America):

John It. Ewing

P.R. Ilalmos

Ilinomingtnn, IN 47405

F .W. Gehring Department of Mathematics University or Michigan Ann Arldnr, MI 48109

Dcp:rrlmenl or Mathematics Santa Clara University Santa Clara, CA 95053

USA

USA

USA

Department of Mathematics Indiana University

Mathematics Subject Classificalirms: Primary-4611, 46J, 4tiK, 47A; Secondary-3IA, 321'

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01991 Springer-Verlag New York, Inc. All rights reserved. This work may not be translated or copied in whole or in part without the written perml of the publisher (Springer-Verlag New York, Inc., 175 Fifth Avenue, New York, NY I (101(1, USA), except for brief excerpts In connection with reviews or scholarly analysis. Use in connection with any form of Information storage and retrieval, electronic adaptation, computer software, or by similar or dissimilar methodology now known or hereafter developed is forbidden. The use tot Renetat descriptive names. trade manes, trademarks, etc.. in this publication, even if the fnKntcr ate tint especially identified, is not In he taken as a sign than sikht names, as understood by the 'fade Matka and Merchandise Marks Act. may accordingly lie used freely by anyone. Camera-ready copy supplied by the author rising TEX. Printed and bomtd by R.R. Donnelley & Sons, Inc., I larrisonburg, Virginia. hoisted M the United States or America.

-9I165432I ISÙN O-3a . 7im.7 .WMger-Verlag New York Ikriin licidcllwrg ISBN 15411-47MN67 Sprinprr-Veriag Ikriin Ilcidclheig Ncw'Ymk

Ce livre est dedie a la memoire de mes parents. Marcel Aupetit Marie-Tlierese Le Charme

"Le cose passate fanno lute alle future, perche it mondo fu sempre di una medesima sorte, e tutto quello the e e sara 6 stato in altro tempo; e le cose medesime ritornano ma lotto diversi nomi e colon; pero ognutio non le nconosce, ma solo chi a savio e le osserva e considera diligentemente." Francesco Guicciardini

PREFACE

This book grew out of lectures on spectral theory which the author gave at the Scuola Normale Superiore di Pisa in 1985 and at the Universite Laval in 1987. Its aim is to provide a rather quick introduction to the new techniques of subharmonic functions and analytic multifunctions in spectral theory. Of course there are many paths which enter the large forest of spectral theory: we chose to follow those of subharmonicity and several complex variables mainly because they have been discovered only recently and are not yet much frequented. In our book Propri6tds spectrales des algebres de Banach, Berlin, 1979, we made a first incursion, a rather technical one, into these newly discovered areas. Since that time the bushes and the thorns have been cut, so the walk is more agreeable and we can go even further. In order to understand the evolution of spectral theory from its very beginnings,

it is advisable to have a look at the following books: Jean Dieudonne, History of Functional Analysis, Amsterdam, 1981; Antonie F1 0 for such an E. F1om the compactness of E there exists y E E such that p(U n K) > 0 for every neighbourhood U of y. Let U be a closed convex neighbourhood of y such that U f) K C K \ {z}. The set U fl K is compact and convex


4

and 0 < r = µ(U n K) < 1, for otherwise µ(U n K) = 1, which would imply that x E U n K as x is represented by p. Thus we define two new Borel measures on K. by

µ,(B) = p2(B) =

1

r

U(B n u n K)

1 1

r p(B \ (U n K))

for each Borel subset B of K. Let x; be the unique vector which is represented by µ;.

Then p,(UnK) = 1 implies x, E UnK, so x, y6 x. Furthermore p = rp,+(1-r)µ2 implies that x = rx, +(1- r)x2, a contradiction. The converse is easy to prove (see Exercise 1.6). 0

Suppose that K is a compact convex subset of a locally convex space, that L C K and that K is the closed convex hull of L. Then the extreme points of K are contained in the closure of L. THEOREM 1.1.12 (D. MILMAN).

PROOF.

Let x E ext(K). By Theorem 1.1.10 applied to A = E, there exists a

measure p on L which represents x. By Lemma 1.1.11 we have p = 6zt so z E 113

Let X, Y be two Banach spaces and let (Ta),EA be a family of bounded linear mappings from X into Y. Suppose that for each z E X we have TSIEOREM 1.1.13 (S.BANACH-H.STEINHAUS).

sup *EA

+oo.

Then SUPOEA IIT-O < +00.

THEOREM 1.1.14 (OPEN MAPPING THEOREM).

Let X, Y be two Banach spaces and let T be a bounded linear mapping from X onto Y. Then T is open. Moreover if T is injective then T'3 is a bounded linear mapping and there exist a,,0 > 0 such that ojjxjj < u Tah < #11x1j, for all x E X.

The next result is very important. It will be used often to prove continuity of a linear mapping. Let X, Y be two Banach spaces and let T be a linear mapping from X into Y. Let G = {(z, Tz): z E X) C X x Y, denote the graph of T. Then T is bounded if and only if 0 is closed in X x Y. THEOREM 1.1.15 (CLOSED GRAPH THEOREM).

Some Reminders of Functional Analysis

5

Hence in order to prove that T is bounded it suffices to prove the following converges to a E Y then a = 0. property: if (x,,) converges to 0 and If K is a real segment [a, b], K. Weierstrass proved that every f in C([a, b]) may be uniformly approximated by polynomials on [a, b]. This result was generalized by the Stone-Weierstrass theorem which is an essential tool in functional analysis (for instance it will be used in the proof of Theorem 6.2.6). There are many proofs of this theorem, including the original one by M.H. Stone, and also I. Glicksberg's argument. The latter is based on an ingenious idea due to L. de Branges, which is short but requires several tools from functional analysis such as the Hahn-Banach theorem, Krein-Milman theorem, Banach-Alaoglu theorem and the Riesz representation theorem (this last proof is given in [10], pp. 5-7).

We now intend to give a very simple proof due to S. Machado. However, its limitation is that it uses the axiom of choice or equivalently Zorn's lemma. (Obviously, if K is separable, Zorn's lemma is not necessary.)

Let A he a closed subalgebra of C(K) containing the constant 1. A non-empty subset E of K is said to be A-antisymmetric if whenever h E A and h is real on E then h is constant on E.

Suppose moreover that A is self-adjoint, that is h E A implies Te E A, and separates the points of K, that is if x, y E K with x # y there exists h E A such that h(x) # h(y ). Then the only A-antisymmetric sets are singletons. The reason is simple: suppose that two different points x, y are in the antisymmetric set E. Then there exists h E A such that h(x) 34 h(y), so Re h = or Im h = separates x and y. As these two functions are real-valued and in A, we get a contradiction. If f E C(K) and F is a non-empty closed subset of K we introduce the following notation: d f(F) inff Ilf - gIIF If WI IIIIIF = sup

,

=

Obviously we have df(F3) < df(F2) if Fl C F2. Let f E C(K) and let A be a closed subalgebra of C(K) containing the constant 1. Then there exists a closed A-antisymmetric THEOREM 1.1.16 (S. MACHADO).

subset E of K such that d(E) = d f(K). PROOF. Let F be the family of all non-empty closed subsets F of K such that d f(F) = d f(K). Obviously K E F. If C is a subfamily of .F totally ordered by

inclusion then G =11FEC F is also in 7 for the following reasons. Given g E A, we have d1(F) = d,(K) for all F E C, so the sets {x : z F, Jf(x)-g(z)j dj(K)) are


6

compact and non-empty. Thus their intersection {z : x E G, If(z) -g(x): > d f(K)} is also compact and non-empty, which implies that d f(G) = d f(K). By Zorn's lemma there exists a minimal element E in F. We now intend to prove that E is A-antisymmetric. Suppose this is false. There exists h E A which is both real and non-constant

on E. Replacing h by a linear combination of h and 1 we may suppose that minXEE h(x) = 0 and maxXEE h(x) = 1. Let us define

El ={x:zEE,O 0, F,".1 a,, = 1, for all i and E s1 aq = 1, for all j. It is called a permutation matrix if, moreover, EXERCISE 5.

the entries a;1 take only the values 0 and 1. Using the Krein-Milman theorem prove that a double stochastic matrix is a convex combination of permutation matrices. As mentioned in Lemma 1.1.11, if 6, is the only probability measure representing x, prove that x is an extreme point. EXERCISE 6.

EXERCISE 7. Let K3,..., K be n compact sets and K = K1 x ... x K,,. Prove that every f E C(K) can be uniformly approximated on K by finite sums of expressions

fl(x1) X ... X Mx.) where f1 E C(KI),...,f,, E Let K be a locally compact topological space. By Co(K) we denote the algebra of continuous functions on K vanishing at infinity, that is f E Co(K) EXERCISE 8.

Some Reminders of Functional Analysis

9

if and only if {x : x E K, jf(x)I > e) is compact for all e > 0. Extend the StoneWeierstrass theorem to Co(K).

Let f E Co([0,+ool) Prove that it can be uniformly approximated on (0,+oo( by functions of the form e-Ozp(x), where a > 0 and p is a polynomial.

*EXERCISE 9.

Let f E Co((-oo, +oo)). Prove that it can be uniformly approximated on [-co,+oo] by functions of the form e_R='p(z), where a > 0 and p is a

*EXERCISE 10.

polynomial.

Let F be a family of holomorphic functions defined on a domain U of the complex plane. We say that F is a normal family if every sequence of elements of F contains a subsequence which converges uniformly on compact subsets of U. Using Cauchy's inequalities and the Arzela-Ascoli theorem, prove that F is normal

*EXERCISE 11.

if and only if it is uniformly bounded on each compact subset of U (P. Montel's theorem).

Chapter II SOME CLASSES OF OPERATORS

§1. Finite-Dimensional Operators Let X be a finite-dimensional vector space. We know that all norms on X are equivalent and this implies in particular that all linear mappings T from X into X are continuous. Indeed if II ' II is a norm on X and if e1,. .. , e is a basis of X, then we have IITxII S (JAI I +

As IIxIIi = IA1I +

+

maxim IlTe;(I

,

for x = Aiei + .. +

+ $A,,I is a norm on X which is equivalent to II II, we have the

result.

Of course, using matrices, it is possible to build all the linear mappings from X into X. So the algebra .C(X) of all linear mappings from X into X is isomorphic to M (C), the isomorphism depending on the choice of the basis. The theory of finite-dimensional operators or matrix theory is very well-known. There are many classical textbooks on this subject (for instance P. Lancaster and M. Tismenetsky, The Theory of Matrices, Orlando, 1985), so we only intend to give a brief survey of the properties of matrices.

Every matrix can be written as the sum of a diagonal one and a nilpotent one (triangularization of matrices). This decomposition can be even much more precise (Jordan decomposition in diagonal blocks). If a matrix T is self-adjoint (T = `T) or normal (TIT _'TT) then it can be diagonalized. In other words, it means that there exist k < n orthogonal projections P1,. . . , P,,, satisfying p,2 = p,, p.

PA =Ofori9j T=AlP1+...+AkPk.

Some Classes of Operators

11

These projections are the orthogonal projections corresponding to the different eigenspaces of T. This result will be generalized for self-adjoint or normal compact operators on a Hilbert space in §3, and even for arbitrary self-adjoint and normal operators on a Hilbert space in Chapter VI. From the Cayley-Hamilton theorem, the algebra M"(C) is algebraic in the sense

that we have p(T) = 0 for all T E M"(C) where p(a) = det(T - )J). In Chapter III, §3 we shall give a simple and nice analytic proof of this fundamental theorem. has another interesting property: The algebra

The algebra M"(C) is simple, that is, its only two-sided ideals are {0} and M"(C). THEOREM 2.1.1.

Let eii be the matrix having every coefficient equal to 0 except at the intersection of the ià row and the j'b column where it is 1. It is easy to verify that eijek,,, = 0 if j 96 k and eijejm = eim. Let I be a non-zero two-sided ideal of PROOF.

M ,,(C) and a # 0, a E I. Then a = E11=1 aijeij. Suppose that aqp # 0. Then epgaepq = epq E=1 aipeiq = agpepg. Hence ep, E I. Then eij = eipepgegj E 1 Consequently I = M"(C). 0 All finite-dimentional algebras over the complex field have a very simple structure. For instance if A is semi-simple (see Chapter III, §1 for the definitions of Jacobson radical and semi-simple algebras) we have the very famous: THEOREM 2.1.2 (d.H.M.WEDDERBURN-E.ARTIN).

Let A be asemi-simple finite-

dimensional algebra over C. Then there exist integers n1, .... , nk > 1 such that A=M",(C)ED ...ED M",(C)PROOF.

See for instance I. Herstein, Noncommutative Rings, 1968, or Lemma

5.4.1. 0

This result implies, in particular, that a commutative, semi-simple and finitedimensional complex algebra is isomorphic to C", where n denotes its dimension (See Exercise II.1).

If T E

what can be said about the multifunction T '-. SpT =

(A: det(T- Al) = 0)? Using the Implicit Function Theorem it can be shown that it depends continuously on T (this will in fact be a corollary of a much more general result, see Corollary 3.4.5). But is it analytic in some sense? For instance if F is an analytic function from C into M"(C), that is a family of matrices with entries


12

depending analytically on a parameter A, are the elements of Sp F(A) analytic func-

tions of A? This is false in general. Consider for instance F(A) =

0 1

E M2(C) _') 0

for which Sp F(.) = { v 'A-, - %5).

Nevertheless, once more using the Implicit Function Theorem, it is possible to

prove that for A outside of an exceptional set, called the set of branching point, the points of Sp F(A) vary holomorphically. In the previous example there is only one branching point at 0. In fact, this result is a corollary of a much more general result (see Theorem 3.4.25).

Such functions A '-+ Sp F(A) defined on a domain D C C can be identified with multifunctions, which are sometimes called algebroid multifunction. These are functions A +-+ K(A) with A E p defined by

K(A) = {z: z" +a, (.1)z"'1 + ... + an(A) = 0), where

are holomorphic on D.

Such functions were studied for the first time by C. Puiseux around 1850. They are of great importance for algebraic geometry and the theory of Riemann surfaces.

This notion of an algebroid multifunction will be generalized in Chapter VII.

§2. Bounded Linear Operators on a Banach Space It is possible to construct a great number of explicit linear operators on classical Banach spaces like £2, L', I P, 1', co, LOO, C(K) etc. Hilbert spaces are the most interesting examples because of their nice geometrical properties. But what happens in the case of a general Banach space? Certainly it contains all the multiples of the identity operator I, but are there any other linear operators?

It is easy to build operators having finite or cofinite rank. Let X be a Banach space of infinite dimension and let E be a finite-dimensional subspace of X having basis el,... , e" (it is automatically closed in X). Let fl,..., f" be n given bounded linear functionals on X. Then the operator T defined by Tx = fi (z)e1 + ... + fn(z)en is bounded and its range is included in E. Moreover if we suppose that f;(e,) is one if i = j and zero if i 96 j (this is possible by Theorem 1.1.3) then f;(z -Tx) = 0 for i = 1, ... , n and the linear operator I -T has its range equal to Iter f1 n . nKer f", and so it has finite codimension. In fact it is possible to go further.


13

Let a1,...,a,, be n linearly independent vectors in a normed vector space X. Then there exists e > 0 such that if b1i ... , bn E X satisfy

LEMMA 2.2.1.

max;_1,,..,,.11b1{i < e then a1+bl,...,an+b,, are linearly independent.

For n = I it is obvious. Supposing the result to be true for n -1, we shall prove it for n, Suppose that there exist sequences (bit), . . . , (bn) converging to 0 in (an-1) such that X and sequences of complex numbers PROOF.

=al(a1+bj)+...+an-1(an_1+bn-1)

an+bn

If all the sequences (ap) are bounded, applying the Bolzano-Weierstrass theorem, they contain converging subsequences and so a1, ... , an are linearly dependent, which is a contradiction. Without loss of generality suppose for instance that Iimk iai i = +oo. Then we have:

(a1+ci)+O 2

2

(a2+bk)+...+i3n_lean-1+bn_1)=0

with ak = at/ai f o r i = 2, ... , n - 1 and ck = bl - ° Qom. But lim,E ci = 0. So by the induction hypothesis we get a contradiction. The lemma is proved. 0 THEOREM 2.2.2. Let X be an infinite-dimensional Banach space. Then there exists a bounded linear operator on X having an infinite-dimensional range which is a limit of finite-rank operators.

PRooF. Let us denote by 3 the linear subspace of finite-rank operators, by I its closure for the norm topology and by In the set of operators with rank < n. Suppose the theorem false. Then = Un°_0 ,I. But an is closed for the following reasons. Suppose that T E 13n and T V an. Then there exists a sequence (Tk) converging to T with Ti E taln for k = 1, 2,... and there exist n + 1 unit vectors z1i ... , xn+1 E X such that Tx1i... , Txn+1 are linearly independent. If we choose k large enough such that iiT - Till < e for the e obtained in Lemma 2.2.1 applied to the n + 1 vectors a1 = Tx1i...,an+1 = Txn}1, we get a contradiction because the Tkx1 i ... , Tkxn+1 are linearly dependent. So ta. is closed. The linear subspace

is complete so, by Theorem 1.1.1, one of the an, for instance N, contains a ball of T with centre S and radius r > 0. Let T E 3r be arbitrary. Then S and S + AT are in ZiN for A small. Consequently T has rank < 2N. Using the construction we gave at the beginning of §2 we can build operators of arbitrary rank, so we get a contradiction. 0

24


It is easy to we that $ is a two-sided ideal of £(X), the algebra of bounded linear mappings from X into X, and consequently 1 is a closed two-sided ideal of .£(X). If S E I and if B denotes the closed unit ball of X then S(B) is closed and bounded in the range of S, so it is compact in this range and also in all X. Let us show that this property is also true for T E 1. First we need an elementary lemma in topology.

LEMMA 2.2.3. Let F be a closed subset of a complete metric space. Then the following conditions are equivalent:

(i) F is compact, (ii) F has the e-covering property for all e > 0, that is for every e > 0 there exist finitely many balls of radius a covering F. PROOF. It is obvious that (i) implies (ii). So we suppose that (ii) is true. Let U be an open covering of F. To reach a contradiction, suppose that no finite subcollection of U covers F. By hypothesis F is the union of a finite number of closed sets with

diameter < 1. One of these sets, which we call F1, cannot be covered by finitely many elements of U. So we can continue the process with F1 instead of F. We get a decreasing sequence of closed sets F1, F2,. .. such that diarn F < 1/n and no F. can be covered by finitely many elements of U. We choose x E F,,, and is then a Cauchy sequence converging to x E fla 1F,,. But T E U for some U E U. Because diam F. goes to zero and x E F,,, we have F. C U for n sufficiently large which gives a contradiction. 0

Let X be a Banach space with dosed unit ball B. If T E Z(X) is a limit of finite-rank operators then the closure of T(B) is compact in X. THEOREM 2.2.4.

PROOF. By Lemma 2.2.3 we have to prove that T(B) verifies property (ii). Let e > 0 be given. There exists S E a such that SIT - S11 < e/3. Because S(B) is compact there exist x1i ... , X. E B such that S(B) is included in B(xl, I) U U B(x,,, I). Consequently T(B) C B(xl, 3) U U B(xa, 3`} and hence T(B) C

B(xl,E)U.. UB(x,,,e).0 This property suggests the introduction of a new class of operators which are in some sense not far from finite-rank operators.


DEFINITION.

15

Given a Banach space X we say that a linear operator T from X

into X is compact if the closure of the image by T of the closed unit ball is compact in X. in X the This is equivalent to saying that for every bounded sequence sequence contains a converging subsequence. This definition was introduced

by F. Riesz in 1918 in order to generalize the results of I. Redholm on integral equations (see Exercice 11.4). As we shall see in Theorem 2.2.10, compact operators are spectrally very similar to finite-dimensional ones.

Let us denote by £e:(X) the set of compact operators on X. With the argument of Theorem 2.2.4 it is easy to see that £C(X) is a closed two-sided ideal of Z(X) which, of course, contains W. Is it true that ', = U:(X)? Even if this is true for many concrete Banach spaces (we shall give a proof of this approximation of compact operatcrs by finite-rank operators on a Hilbert space in Corollary 2.3.5), it is false in general. This was first proved in 1973 by P. Enflo. Later his argument was greatly simplified by A.M. Davie. Inspite of this, D.C. Kleinecke proved in 1963

that Ir and £C(X) are not far spectrally (see Corollary 5.7.3).

Let T E £(X). By definition the spectrum of T is the set of A E C such that T - AI is not invertible in £(X). It is denoted by Sp T. So A E Sp T if and only if at least one of the following conditions is verified:

(a) the range of T - Al is not all of X, (b) T - Al is not injective. We denote by N(T) the kernel of T and by R(T) its range.

If (b) holds we say that the A is an eigenvalue of T. The corresponding eigenspace N(T - Al) is the set of x such that Tx = Ax. Such an x # 0 is called an eigenvector corresponding to A.

If T E £t(X) and dim X = co then 0 E Sp T: otherwise T would be surjective, thus I = T T. T-1 would be compact, and by Theorem 1.1.2 this would imply that X is finite-dimensional. THEOREM 2.2.5.

Let X be a Banach space and let T E £(C(X). (i) If the range of T is closed, then it is finite-dimensional.

(ii) If A 96 0, then dimN(T -,\I) < +oo. PROOF. (i) If 1Z(T) is closed then it is complete so, by Theorem 1.1.14, T is an open mapping from X onto R(T). This implies that R(T) is locally compact so, by Theorem 1.1.2, 1Z(T) is finite-dimensional. Let Y = N(T - AI). The restriction of T to Y is a compact operator with range Y. So (ii) follows from (i). 0

A


If P is a compact projection, that is P is compact and satisfies p2 = P, then R(P) is closed. This is because if Px converges toy then P2x = Pz converges to Py, so y = Py E IZ(P), and hence by Theorem 2.2.5 (i) R(P) is finite-dimensional.

By definition the adjoint T" E £(X') of an operator T E .C(X) satisfies (Tx, f) = (x,T* f ), for all x in X and all f in the topological dual X'. Let X be a Banach space. Then T is compact if and only if its adjoint T" is compact. THEOREM 2.2.6.

Suppose that T is compact. Let (f,) be a sequence in the unit ball of X'. f.(y)I n,letzERm_1CR. with z V Rm. Since (T - AI)z E Rm = (T - AI)Rm there exists t E Rm such that

(T-AI)t=(T-AI)z. Hence z-tEN1 CN", butalsoz - tER",soz--tand this is a contradiction. We now prove that X = N" + R. For all x E X we have (T - AI)"z E R. = Rm and (T - AI)"R" = R", so there exists y E R" such that (T - AI)"y = (T - AI)"z. Thus x - y E N,,, hence the result. We now prove (iii). If x E N,, then (T - AI)"+lx = 0 so (T - AI)x E N. and obviously (T - Al)" = 0 on N,,. If z E R" then x = (T - Al)"y for some y E X, so (T - AI)x E R"+1 = R". By definition, the range of T - AI restricted to R" is R"+1 = R" so this restriction is surjective. This restriction is also injective because

N"f1R"={0}.0 This theorem is particularly interesting when A is an eigenvalue of T. For instance, it immediately implies that for A # 0 the following properties are equivalent:

(i) N(T - Al) = {0}, (ii) 11(T - AI) = X, (iii) A is not in the spectrum of T. In fact the decomposition of X as a direct sum of two closed subspaces E and F which are such that:

(a) dim E < +oo, (b) T(E) C E and the restriction of T - AI on E is nilpotent in £(E), (c) T(F) C F and the restriction of T - AI on F is invertible in £(F), is unique.

Indeed, let x E E, then x= y+ z where y E N", z E R. By hypothesis there exists an integer p such that (T - AI)'x = 0, so (T - AI)'y = -(T - AI)Pz E


19

N" fl R,, = {0} and hence (T - AI)Pz = 0. But the restriction of T - AI on R is invertible, so z = 0. Consequently E C N. A similar argument shows that

N. C E, hence N. = E. If z = y+z E F with y E N,,, z E R. we have

(T - A.I)"x = (T - Al)"z E R so (T - Af)"(F) C R,,. But (T - AI)"(F) = F by (c), so F C R. A similar argument proves that R C F. THEOREM 2.2.10. Let X be a Banach space and let T E £r(X). Then we have the following properties:

(i) the spectrum of T is a compact subset of C having at most 0 as a limit point, (ii) every spectral value A 34 0 is an eigenvalue of T.

PROOF. The compactness of SpT follows from Theorem 3.2.8, which we shall prove later (see also Exercise 11.6). By the remark following Theorem 2.2.9, if A # 0 is a spectral value then JV(T-AI) 0 0. So (ii) is proved. Let Ao E Sp T, Ao # 0 be an eigenvalue and let E(Ao) = N,,, F(Ao) = R. be the corresponding closed subspaces

obtained in Theorem 2.2.9. We have 0 < dim E(Ao) < +oo. Let us denote by T1 (resp. TZ) the restriction of T to E(Ao) (resp. F(Ao)). Then Tl - AOIEta,l is nilpotent on the finite-dimensional subspace E(Ao). So the characteristic polynomial

of Ti is p(z) = (A0 - z)P for some integer p < n. This implies in particular that T1 - AIE(A,) is invertible in £(E(A0)) for A 96 A0. By Theorem 2.2.9 (iii), we have T2 -AOIF(a,) invertible in >r(F(Ao)) so, by Theorem 3.2.3 (which we will prove later), T2 - AIF(a,) is invertible in £(F(Ao)) for !A - AoI < e, with a small enough. These

two results imply that N(T -AI) _ {0} and R(T - Al) = X for 0 < JA - AoI < e, so that T - AI is invertible for 0 < IA - AoI < e. Consequently A0 is isolated in the spectrum T, hence we have (i). 0

If a compact operator has a non-zero spectrum then it has an eigenvalue and it can be decomposed. If it has only 0 as a spectral value then we may have an eigenvalue at 0, but there are also examples without eigenvalue. We now give such an example. EXAMPLE.

Let X = C([0,11) and let T be the Volterra operator defined by

(Tf)(x) = 10" f (t) dt ,

for all f E C((0,11).

We have fITf11 < 11f 11 and j(Tf)(z) = f(x). So if (f") is a sequence of the unit ball of X then (T f,,) is equicontinuous and bounded so, by Theorem 1.1.18, (T f,,) contains a converging subsequence. This shows that T is compact. If 11f 11 < 1 it is obvious that 1T f(z)(< z so by induction fT" f(z)$ < z"/n!. This implies that


20

IIT"II 0.

There exists a no such that k > no implies IAkI S e. So for n > no we have IITn+m - TTII < e. Consequently the sequence converges in L(H) to some operator S. If x E E. then Tnx = Apx for all n > p, so Sx = Ax = Tx. Consequently, S and T coincide on all Ep for p > 0, so on their Hilbertian direct stun H we have S = T. 0 COROLLARY 2.3.5. Let H be a Hilbert space. Then every compact operator on H can be approximated by finite-rank operators.

Let T E i2C(H) and e > 0. Then we also have T* E £E(H), so ReT = (T + T*)/2 and Im T = (T - T*)/2i are self-adjoint compact operators on H. By Theorem 2.2.5 we know that Ek = Pk(H) is finite dimensional. So, by Theorem PROOF.

2.3.4 (ii), there exist two finite-rank operators T1 and Tz such that II Re T-T1 II < e/2 and 11 Im T - T2II < e/2. Then Tl + iT2 has finite rank and IIT - (T1 + iT2)I) < e. 0


27

Theorem 2.3.4 can be reformulated differently. COROLLARY 2.3.6 (FREDHOLM ALTERNATIVE).

Let H be an infinite-dimensional

Hilbert space and let T be a self-adjoint compact operator on H. Denoting by {AkIk>> the set of non-zero eigenvalues of T (of course Ak E R), then we have the following properties:

(i) if A # At, for all k, then the equation Tx - Ax = y has a unique solution in H for all y E H, (ii) if A # 0 and .\ = Ak, for some k, then the equation Tx - Ax = y has a solution in H if and only if y is orthogonal to N(T - Al'); so either this equation has no solution or it has an infinite number of solutions. All this theory can be applied to integral operators T f (x) f k(x, y) f (y) dy with symmetric kernels. In particular it has many applications to the SturmLiouville problem (see J. Dieudonne, Elements d'analyse, Vol 1, Paris, 1963, Chapter 11, §7).


28

EXERCISE 1. Prove that a commutative and semi-simple finite-dimensional complex algebra is isomorphic to C", where n denotes its dimension. What happens if it is a commutative and semi-simple finite-dimensional real algebra?

EXERCISE 2.

Let f be an analytic function from a domain D C C into M,(C).

Denote by ak the ktb-symmetric function of the eigenvalues of a n x n matrix. Prove that A'-a ok(f(A)) is holomorphic on D.

Prove that the commutators [a, b] = ab - ba form a linear subspace of codimension one in M"(C) (Shoda's theorem). Given a linear functional f on

* EXERCISE 3.

M"(C) prove the equivalence of the following properties:

(i) f = a Tr, where a is some complex number and where Tr denotes the trace, (ii) f (ab - ba) = 0, for all a, b E M"(C),

(iii) f (sax-1) = f (a), for all a E M"(C) and z invertible in M"(C), (iv) J f (x)I < Cp(x), where C is some positive constant and where p denotes the

spectral radius. EXERCISE 4.

(i) Suppose that k is continuous on [a, b] x [a, b). For f E C([a, b)) define T f by.

Tf(x) =

Ja

`

k(x,y)f(y)dy

Prove that T is a compact operator on C([a, b)). (ii) Suppose that k E L2([a, b] x (a, 6)). For f E L2([a, b]) define T f as previously. Prove that T is a compact operator on L2([a, b)). EXERCISE 5.

Given a sequence (a") converging to 0, prove that there exists a'

compact operator Ton 12 such that Sp T = {0} U for. [ n E N}. EXERCISE 6. Given T E EE(X) and e > 0 prove directly, without using Theorem 3.2.8, that (A: A E Sp T, [A[ > c} is finite.

EXERCISE 7.

Given T E U (X) and A

76

0 prove that dimN(T - AI) _

codim R(T - AI) = dimH(T* - XI) = codim R(T - XI). EXERCISE 8. Let H be a separable Hilbert space with orthonormal basis el, e3..... We consider T the left weighted shift having weight an = 1/n (resp. S the right


29

weighted shift). Prove that T and S are compact operators on H, that T is nilpotent and that S is not nilpotent. Moreover show that Sp S = {0}.

Let H be a Hilbert space and let I be a closed two-sided ideal of E(H). Prove that I = .NC(H) (Calkin's theorem). EXERCISE 9.

EXERCISE 10.

Let H = L2()0, +oo(). For f E H we define T f by

(Tf)(x) =

i r f(!)dt.

Prove that T E £(H) but that T is not compact. Given a right or left weighted shift T with weight (a.) determine explicitly the spectrum of T. If you do not succeed look at (3).

*EXERCISE 11.

EXERCISE 12. space.

Extend Theorem 2.3.4 for normal compact operators on a Hilbert

Given q, f E C([a, b)) and A E C, the Sturm-Liouville problem consists in finding the solutions of the differential equation

*EXERCISE 13.

y"-gy+Ay=f, with some limit conditions aly(a) + Qly'(a) = 0, a2y(b) + 92y'(b) = 0. Prove that y is a solution of this problem if and only if

y(z) _ -J K(t,z)f(t)dt where K is a convenient continuous function on [a, b] x [a, b]. If you do not succeed look at the book of J. Dieudonne.

Chapter III BANACH ALGEBRAS

.g1. Definition and Examples A complex algebra is a vector space A over the complex field C, with a multiplication satisfying the following properties: x(Yz) = (xY)z, (x + Y)z = xz + yz, x(Y + z) = zy + xz, A(xy) = (Ax)Y = x(AY),

for all x, y,zEAandAEC. If moreover A is a Banach space for a norm II II and satisfies the norm inequality IIxyII 1 and it contains an isometric copy of A (associating to z the diagonal matrix having only z on the diagonal).

I-anach Algebras

33

A much more useful tool is the following. Let I be a dosed two-sided ideal i of A. Then A/I is a Banach space for the norm III.JII = denotes the coset x+I of x. With this norm it is easy to verify that A/I is a Banach algebra, called the quotient algebra of A by the two-aided ideal 1. This comes from

1111+v1115 fix+U+y+vli S llz+ull+lly+vii 51ixy+xv+uy+UVIi < liz+U11 Ily+vii for U, v E 1, so Ii11 + v111 5 illxiil + Iilviil, 1111 vlii 5 iliilli -111vill. This implies

in particular that ilii'111 = IIIiJII 5 liiilil IIIIIII so 1 S Iliilll 1. Then limsup Ilx"II11" < (A( n-»oo

for all (Al > p(x).

(3)

So finally, using (1) and (3), we get: p(z) < lnm mf Ilx' ll'i" < lim sup llx" ll'/" < P(x), "-moo

and the theorem is proved. 0 The formula given by (iii) is called the Beurling-Gelfand formula mainly because it was used intensively by I.M. Gelfand in 1939 and introduced a bit earlier by A. Beurling in harmonic analysis. This formula is very useful. We gave three applications in Chapter 2 (the examples following Theorem 2.2.10, Theorem 2.2.11 and Corollary 2.3.2) but throughout this book we shall encounter a great number of other applications. This formula can be proved using various other methods. If A is a Banach algebra in which every non-zero element is invertible then A is isometrically isomorphic to C. COROLLARY 3.2.9 (I.M.GELFAND-S.MAZUR).

PROOF.

Let x E A. By Theorem 3.2.8 (ii), Spx is non-empty. Let A E Spx.

Then x - Al is not invertible, consequently x = A 1. This implies that Sp x contains

only one point, which we cell a(x). The formula x = a(x)I implies that a is an isomorphism from A onto C. It is an isometry because we have (lxll _- Ia(x)I (I111 = la(x)I. 0


40

If x E A satisfies q(x) = 0 for some polynomial q then by Theorem 3.2.6 the spectrum of x is included in the set of zeros of q. In particular, if p is a projection we have Spp c (0, 1). But if p is a non-trivial projection, that is p 34 0,1, we cannot have Spp = {0) because p(p) = lim IIpnII1/" = lim IIpII1/n = 1, nor can we have Spp = {1}, because p(1 -p) = 1 and Sp(1 -p) _ {0}. So we have Spp = (0, 1) for a non-trivial projection p. REMARK.

COROLLARY 3.2.10. Let A be a Banach algebra. Suppose that x, y E A satisfy xy = yx. Then p(x + y) : p(x) + p(y) and p(xy) < P(x)p(y)

Because (xy)" = x"y" for every integer n > 1, using Theorem 3.2.8 (iii), we conclude that PROOF.

p(xy) = lim II(xy)nlll/n 1 there exists z E A such that z" = x. THEOREM 3.3.6.

PROOF.

By hypothesis 0 is in the unbounded component of C\ Sp x. Hence there

exists a simply connected open set S1 containing Spz and f E H(ft) such that ef(a) = A on Q. Then we apply Theorem 3.3.3 and take y = f(x). For each n > 1 we take z = 01". 0 This implies in particular the non-trivial fact that an invertible n x n matrix is an exponential (obviously the converse is true).

Let A be a Banach algebra. We denote by exp(A) the set of all products of exponentials eXt .. e", where x1, ... , x E A. It is obvious that exp(A) C G(A). But t -+ e`=, - e`s is a continuous function from 10, 11 into G(A) which connects 1 and esl ... es^. So in fact exp(A) is included in the connected component of G(A) containing 1, which is denoted by G1(A) and is called the principal component of G(A).

Banach Algebras

THEOREM 3.3.7 (E.R. LORCH). GI (A). PROOF.

47

If A is a Banaa6 algebra we have exp(A) _

First we prove that exp(A) is open in G(A). Let a E exp(A) and suppose

that (Ix-all < 1/IIa-'((. Then III-a-'xli =

IIa-'(a-x)II < 1 and so p(1-a-'x) < 1.

Consequently Sp(a-i x) is included in the open disk of centre 1 and radius 1. But for Re A > 0 there exists a holomorphic function f such that e/(a) = A. So by Theorem

3.3.3 there exists y = f(a-lz) such that ey = a-lx. Then x = acy E exp(A). We now prove that exp(A) is closed in G(A). If a E exp(A) and a,, = a then (an 1 a) converges to I and so ill - a-'all < 1 for n large. As before, we conclude

that a-'a = es* for some z E A, so a E exp(A). Because exp(A) is closed and open in G(A) and is contained in G,(A), we conclude that exp(A) = G1(A). 0

For x E A we define the exponential spectrum of x, denoted by e(x), by the set of .1 E C such that Al - x V exp(A). Obviously we have Spx C e(x). Let (Sp x)" be the polynomially convex hull of Sp x, that is the union of Sp x with the bounded components of C\ Sp x, and let Ao 0 (Sp x)-. Then by Theorem 3.3.6 there exists y such that Ao1 - x = e', and consequently A0 e(x). In other words, Sp x C e(z) C (Sp x)^. This implies in particular that c(x) is a non-empty compact subset of C because GI(A) is open in A. We now prove a simple and nice result characterizing the spectrum of i, for i E A/I, where I is a closed two-sided ideal of A.

THEOREM 3.3.8 (R. HARTE).

Let T be a continuous morphiem from a Banach algebra A onto a Banach algebra B. Then T(exp(A)) = exp(B). So we have

e(Tx) = n e(x +y) and Sp Tx C n Sp(x + y) C (SpTx)-. yEKer T

yEKer T

In particular, if I is a closed two-sided ideal of A then

Sp i c n Sp(x + y) C (Sp i)-. yEI

It is clear that T(esi ... es^) = Tz, ... eTs-, so the first part is obvious. If A f c(Tx), then Tx- Al = Tu for some is E exp(A). So, taking y = u-x+ Ai, we PROOF.

have x + y - A E exp(A) and Ty = 0, and consequently l,EKer T e(Z + y) C e(Tz). The other inclusion is obvious. Moreover f 'yEKer T SP(z + Y) C n,,,,,, . c(x + y) = e(Tz) C (SpTz)'by the previous remark. The last part follows immediately, caking

B=A/I and Ta=i.0


48

§4. Analytic Properties of the Spectrum Let A be a Banach algebra. The main question in spectral theory is the following: what can be said about the spectrum function x i-+ Sp x when x varies in A? Is that function continuous or analytic in some sense? In order to measure the continuity of the spectrum we introduce a distance on the set of compact subsets of C, called the Eausdorfj distance and defined by A(K1, K2) = max( sup dist(z, Kt), sup dist(z, K2)) zEK2

zEK,

for Kl, K2 compact subsets of C. Let r > 0 and K be a compact subset of C. If K + r denotes {z : dist(z,K) _< r) then obviously K, C K2 + i(Kt,K2) and K2 C Kl + A(K1, K2 ). We shall say that x s-+ Sp z is continuous at a E A if, for every e > 0, there exists b > 0 such that lix -all < b implies A(Sp x, Spa) < E. As usual we shall say that x +-+ Sp x is continuous on E if it is continuous at every point

of E. If for a given e > 0, the number b > 0 is independent of a on E, we say that x s- Sp x is uniformly continuous on E. If E is a cone, that is aE C E for all a > 0, it is equivalent to say that there exists C > 0 such that L(Sp x, Sp y) < CO X - yII for x, y E E. Of course there are examples of algebras where the spectrum behaves nicely.

Let A be a Banach algebra. Suppose that z, y E A commute. Then Spy C Sp z + p(z - y) and consequently we have A(Sp x, Sp y) < p(x - y) < lix - yII Furthermore, if A is commutative then the spectrum function is uniformly continuous on A. THEOREM 3.4.1.

Suppose the inclusion is not true. This means that there exists a E Spy such that dist(a, Sp z) > p(x - y). Therefore by Theorem 3.3.5, we have p((al - x)-')p(z - y) < 1. So, by Corollary 3.2.10, we have p((al - x)-(x - y)) < PROOF.

1.

But al - y = at - x + x - y and at - z is invertible by hypothesis, so

al -y = (al - z)(1 + (al -z)-'(x - y)j is also invertible, which is a contradiction. 0 then the spectrum function is continuous (this derives from the If A = Implicit Function Theorem and also from Corollary 3.4.5, as we shall see below). But in that case it is not uniformly continuous because if we take an

_

n2

n2

1

(n2(n-n2) n-n2

°

n2(n-n2)

1

n-n2-1/n)

Banach Algebras

49

then Ilan - bnll =1/n and A(Spa,,,Spbn) > 1/2, for n large enough.

If A = .X(X) then the spectrum function is also continuous (see Corollary 3.4.5) but not uniformly continuous. In general the spectrum function behaves very badly. Let us consider A = £(H), where H is a Hilbert space. Then the spectrum function is uniformly continuous when restricted to the subspace of self-adjoint operators (see Theorem 6.2.1), but it is not continuous for practically all self-adjoint operators: this difficult result comes from a theorem of B.B. Morrel and J. Morrel.

We only give an easier result due to S. Kakutani showing that the spectrum function can be discontinuous. EXAMPLE. Let (a,,) be the sequence of positive numbers defined by a = e-k if n = 2k(2 + 1). Then in the separable Hilbert space H with orthonormal basis (en) we consider the weighted shift T with weight (a,,). For every k > 1 we define

Tk6£(H)by if n = 2k(21 + 1), for sonie 1, Tken =

otherwise.

It is easy to verify that Tk ++'en = 0, for every n, so the operators Tk are nilpotent, and hence Sp Tk = {0}. We have

(T - Tk)e n

e-ken+, ,ifn=2k(21+1),forsome 0

,

otherwise.

So IIT - Tk1I < e-k, and hence the operators Tk converge to T.

We have Tmen = anon+1 ' ' ' an+m-len+m and consequently we have JIT'"II = supn(anan+l "' an rm-1) By the definition of the sequence (an) we have e-t

alai ... azs -1 =

flexp(-j2t-j-l) i=t

So (alai ...

azs-i)'/(2'-1) >

t=i

eXP(-j/2j+i)).2 Let o =

)

I

j/2i+1. Then

0 < e2a < lim,n...oo IlTmIII/` = p(T). So SpT # {0}. Furthermore, V. Muller has given a more sophisticated example where the spec-

trum function is discontinuous even on the real line. He proved that there exist T, S E .Q(H) such that p(T) > 0 and p(T + AS) = 0 for all rational numbers A in ]0, If (see [I], pp. 36-38). We now give some positive results. The first one has been known for a long time.


30

Let A be a Banach algebra. Then the spectrum function z H Sp x is upper semicontinuous on A, that is for every open set U containing Sp z there exists 6 > 0 such that Iix - Y11 < b implies Spy C U. THEOREM 3.4.2.

Suppose that there exist sequences (yn) and (an) such that x = lime yn, an E Spy fl (C\U). Then iani < iiyeii, so (a,,) is a bounded sequence. By the Bolzano-Weierstrass theorem we may suppose without loss of generality that PROOF.

it converges to a. But a 4 U because C\U is closed, so al - z is invertible. By Theorem 3.2.3, &.1 - y will be invertible for n large, which is a contradiction. U The following theorem is a particular case of an old result of K. Kuratoweki. It says that even if the spectrum function is discontinuous, the set of its points of continuity is rather large. THEOREM 3.4,3. Let A be a Banach algebra. Then the set of points of continuity of z i -s Sp r is a dense G6 -subset of A. PROOF.

It is easy to see that the algebra of real continuous functions on C is

separable. So we suppose that (f,) is a dense sequence of functions in this algebra. We define f,(x) = sup{fe(A) : A E Spx}, for x E A. By Theorem 3.4.2, the f, are upper semicontinuous in the classical sense.

We now prove that x " Spx is continuous at a if and only if the in are continuous at a. Let Ao E Spa be such that f,,(Ao) = fn(a) and let e > 0. Because the function f e is continuous at A0, there exists b > 0 such that IA - A0I < b implies Ifn(A) - fe(Ao)I < e. If the spectrum function is continuous at a, there exists r > 0 such that Iz - al < r implies that there exists A E Sp z such that IA - Aoi < b. Then

fn(z) > f,(A) > fe(Ao) - e. So, with the upper semicontinuity of f,, this implies the continuity of fe. Conversely, suppose that all J. are continuous at a and that x " Sp z is discontinuous at a. Then there exist a sequence (xk) converging to

a, a E Spa, and r > 0 such that B(a, r) fl Sp xk = 0 for all k. There exists f continuous on C such that f(a) = 1 and f(z) = 0 for it - aI > r, so there exists n such that f,(a) > 2/3 and fn(z) < 1/3 for Iz - al > r. Consequently fe(a) > 2/3 and fe(zk) < 1/3 and this is a contradiction because fe(a) = limk fw(zk). Let C be the set of points of continuity of z .-+ Sp z and Ce be the set of points of continuity of f,,. So we have C = fl s1Ce. Using Theorem 1.1.1 it is sufficient to prove that C is a dense G6-subset of A. So let f be an upper semicontinuous function from A into R, let D be its set of points of discontinuity and let (Us)e>1

Banaeh Alsebras

51

be a countable basis of R. In order to prove the theorem we have only to show that

D=UDn n>1

where D is f-'(Un) minus its interior points. If X E D there exist some U. containing f(x) such that f-'(U.) is not a neighbourhood of x, and so x E D. is not Conversely, if x E D,,, then U. is a neighbourhood of f (x) such that f a neighbourhood of z, so x E D. Now D. C 77F .Un) \ (fÌ (U.))° = Of `'(U.), so D has no interior points and is a F,-set. Hence we get the result. 0

The first important results concerning spectral variation are due to J.D. Newburgh. It is amazing to notice that these results are not given in all standard books on Banach algebra theory. They were given publicity for the first time in [1J.

THEOREM 3.4.4 (J.D. NEWBURGH). Let A be a Banach algebra and x E A. Suppose that U, V are two disjoint open sets such that Sp x C UUV and Sp xflU 0.

Then there exists r > 0 such that lix - y11 < r implies spy fl u 9t 0.

By Theorem 3.4.2 there exists d > 0 such that lix - y'! < d implies Spy C U U V. Suppose the theorem to be false. Then there exists a sequence converging to x such that Sp yn C V, for n large enough. Let f be the PROOF.

holomorphic function on U U V defined by 1 on U and 0 on V. By Holomorph+c Functional Calculus we have Iim f(yn) = f(x) and f(yn) = 0 for n large enough. But Sp f (x) = f (Sp x) contains 1, so f (x) 0 0, which gives a contradiction. 0

Suppose that the spectrum of a is totally disconnected. Then x '- Sp z is continuous at a. COROLLARY 3.4.5 (J.D. NEWBURGH).

Let e > 0. Because Spa is totally disconnected it is included in the union U of a finite number of disjoint open sets, intersecting Spa and having diameters less than E. By Theorem 3.4.1, there exists rl > 0 such that fix - all < r, implies Spx C U. Applying Theorem 3.4.4 to U, there exists rs > 0 such that fix - all < PROOF.

r2 implies sup dist(z, Sp x) < E for z E Spa. So JJx - au < min(rl, r2) implies i (Spa,Spx) < e. 0

52


This implies in particular that the spectral function is continuous at all elements having finite or countable spectrum.

In 1966, A. Brown and R.G. Douglas considered the problem of formulating a maximum principle for the multifunction A +- Sp(f (A)) where f is an analytic function from a domain D of C into a Banach algebra A. In 1968-1970, E. Vesentini solved this question by proving Theorems 3.4.7, 3.4.11 and 3.4.13.

Let D be a domain of C. A function of from D into R U {-oo} is said to be aubharrnonic on D if it is upper semicontinuous on D and satisfies the mean inequality S(ao)

'r

j(Ao+re9)d9

for all closed disks ff(A0i r) included in D. In particular h is harmonic on D if and only if h and -h are subharmonic on D. Subharmonic functions have a great number of very beautiful properties which we cannot give in detail in this small book. So we refer the reader to the appendix for a quick survey, or to a standard textbook ((41, for instance). LEMMA 3.4.6. Let f be an analytic function from a domain D of C into a Banach space X. Then A " log II f(A)II is subharmonic on D.

Obviously this function is continuous. Let $(A0ir) be a closed disk included in D. By Cauchy's theorem we have PROOF.

f(ro) = and consequently Ilf(Ao)II
3,91 < arg z < 92 } and let f be an analytic function from a domain D of C into a Banach algebra A such that Sp f (A) C l U B(0, r), for all A E D. Define

u(A)=max{argz:zESpf(A)nS2}, v(A)=min{argz:zESpf(A)n S2},


54

Then u and -v are subharmonic on D. PROOF. Without loss of generality we may suppose that Sp f (A) fl Q V 6, for all A E D. On n we consider the branch of the logarithm log z = log jzj + i arg z and we define

h(t) = to

i log z

, on Q , on B(0, r)

where a < 01 is a fixed real number. Then h is holomorphic on f2 U B(0, r). By the Holomorphic Functional Calculus we have

Sph(f(A))C {=ilogz:zESpf(A)flfl)U{a). So u(A) = max{Rez : z E Sph(f(A))}. Then we apply Corollary 3.4.9 to ho f. For -v, the proof is similar. 0 If x is in a Banach algebra we define the peripherical spectrum of x to be the set of A E Sp x such that CAI=p(z). THEOREM 3.4.11.

Let f be an analytic function from a domain D of C into a

Banach algebra A. Suppose that there exists As E D such that p(f(A)) < p(f(Ao)), for all A E D. Then the peripherical spectrum of f (A) is constant on D.

PROOF. By the Maximum Principle for subharmonic functions (see Theorem A.1.3) there exists a constant c such that p(f(A)) = c on D. If c = 0 the result is obvious. So suppose c > 0 and that there exist A1, Az E D and z E C such that j z I = c, z E Sp f(Aj), and z Sp f(A2). Let a > 0. Then g(A) = f(A) + azl is analytic and so, by Theorem 3.4.7, A i-+ p(f (A) + azl) is on D. We have Sp g(A) C ff(az, c) C ff(0, (a + 1)c).

Moreover these two disks are tangent at the point (a+ 1)a. Consequently p(g(A3))
0, we may suppose that Sp f (Ao) C 10, 21r[. By upper semicontinuity there exists 6 > 0 such that JA - A0J < 6 implies Spf(A) C )0,2ir[. By Theorem 3.3.3, g(A) = e'1(a) is defined for JA - Aol < 6 and Spg(A)

is included in the unit circle. So by Theorem 3.4.11, Spg(A) = exp(iSpf(A)) is constant for JA - Aol < 6. But z '-+ e" is one-to-one on 10,2r[, so Spf(A) is constant for JA - AoI < 6. Now let E _ (A : A E D,Sp f(A) = Sp f(Ao)). The previous argument shows that E is open and closed in D. So E = D. 0 EXAMPLE. Let x, y E A. Suppose that Sp(x + Ay) C R for allA E C. Then by the previous result we have

Sp(x+Ay)=Spz,

for all AEC.

Dividing by A we get

Sp(px+y)=pSpx, forallp#0, so

p(px + y) = JpJp(x)

for all p # 0.

,

But p +. p(px + y) is subharmonic. So we get p(y) = lim sup p(px + y) = 0 (see Theorem A.1.2). p-..D

yoo

Let f bean analytic function from a domain D of C into a Banach algebra A. Suppose that there exists Ao E D such that Sp f(A) C Sp f(Ao), for all A.E D. Then 8Sp f(Ao) C 8Sp f(A) and Sp f(Ao)- = Sp f(A)', for all A E D. In particular if Sp f(Ao) has no inteTHEOREM 3.4.13 (SPECTRAL MAXIMUM PRINCIPLE).

rior points or if Sp f (A) does not separate the plane for all A E D, then Sp f (A) is constant on D. PROOF.

Suppose that zo E OSpf(Ao) and zo f 8Sp f(A1) for some Al E D.

Of course zo is not an interior point of Sp f (A1) because in that case it would be interior to Sp f (Ao). So zo 0 Sp f (A1), and hence there exists r > 0 such that ff(zo,r) fl Sp f(A1) = 0. Since zo E 8Sp f(Ao) there exists z1 Sp f(Ao) such that Jz1 - zoJ < r/3. Then

dist(z,,Sp f(Ao)) < r/3 and

dist(z,,Sp f(A1)) > 2r/3.


56

But by hypothesis dist(zn,Spf(A)) > dist(z1,Spf(Ao)). So by Corollary 3.4.8 and the Maximum Principle for subharmonic functions we get dist(z1,Sp f(A)) _ dist(zl, Sp f(Ao)) on D. So we have a contradiction at A = A1. Hence 8Sp f(Ao) C0 Sp f (A) for all A E D. Let U(A) be the unbounded component of C\ Sp f (A). Then we have U(Ao) C U(A). Now suppose U(A0) # U(A). Then there exists z E U(A)

such that z E Sp f(Ao) Let z be connected to infinity by an arc I' included in U(A). Let zo be the supremum on r, for the order defined by the parametrization, of the points of Sp f(Ao). Then zo E Sp f(Ao) and it cannot be interior, so zo E 8Sp f(Ao) C 8Sp f(A). But this is a contradiction because U(A) fl 8Sp f(A) = 0. So U(A) = U(Ao) and then Sp f(A)" = Sp f(Ao)" for all A E D. If Sp f(Ao) has no interior points then Sp f(Ao) = aSp f(Ao) C 8Sp f(A) C Sp f(A) C Sp f (A0); if Sp f(A) does not separate the plane then Sp f(A) = Sp f(A)" and so the proof is complete. 0 THEOREM 3.4.14 (LIOUVILLE'S SPECTRAL THEOREM).

Let f be an analytic

function from C into a Banach algebra A. Suppose there exists a bounded set C such that Spf(A)CCforallAEC. Then Spf(A)" is constant on C. The set E = Sp f (A) is compact. Let ro E and e > 0 be such that B(zo, e) fl E = 0. Then, by Corollary 3.4.8, - log dist(zo, Sp f (A)) is subharmonic on C and smaller than - loge. So by Liouville's theorem for subharmonic functions, we conclude that dist(zo, Sp f (A)) is constant on C. Let zl E OE and suppose there exists Al E C such that z1 8Sp f(A1). Then zl f Spf(A3) because Spf(A1) C E and z1 E 8E. There exists r > 0 such that B(zj, r) fl Sp 1(A1) = 0. Let Az be such that PROOF.

B(zl, r/5) f1 Sp f (A,) 34 0

and let xs E B(zl, r/5)\E # 0. We have dist(z2, Sp f(A2)) < 2r/5 and

dist(z2i Sp f(A1)) > 4r/5.

So if we apply the first part to zo = zs we get a contradiction. Consequently 8E C 8Sp f(A) which implies E" C Sp f(A)", but the converse is true, so Sp f(A)' = E Hence Sp f (A)' is constant. 0 REMARK 1.

With that hypothesis it is false in general that Sp f (A) is constant. For instance, on 112(2) with the orthonormal basis we consider the two weighted shifts

at.

0 en+1

,ifn=-1 ,ifn#-1

be

to

,ifn=-1 ,if n#-1.

Ba tach Algebras

57

For A E C we then have

(a +Ab)e n

Aeo

,ifn=-1

en+1

,if n#-1.

By Problem 85 of [3] we can deduce that Spa is the closed unit disk and that' Sp(a + Ab) is the unit circle for A 36 0. So Sp(a + Ab) is bounded but not constant. More easily, we can prove that Spa is included in the closed unit disk, that 0 E Spa, and that Sp(a+Ab) is included in the unit circle for A # 0. It is ouvious that 0 E Sp a

because ae_1 = 0. Fork > 1 we have (a + Ab)ken = en+k for n > 0 or n < -k, and (a + Ab)ken = Aen+k otherwise. So II(a + Ab)k 11 < max(1, CAI), which implies p(a + Ab) < 1 by Theorem 3.2.8. For A y& 0, a + Ab is invertible and its inverse satisfies 1

(a + Ab) -

e _

e_1 ,ifn=0 en-_1

,# if n 0

.

A similar argument shows that p((a+Ab)-1) 1. But Corollary 3.2.10 implies that 1 < p(a + Ab)p((a + Ab)-1) so 1 = p(a + Ab) = p((a + Ab)-1) and hence Sp(a + Ab) is included in the unit circle for A # 0. THEOREM 3.4.15.

Let f be an analytic function from C into a Banach algebra A. Then either Sp f (A)" is constant or UAEC SP f (A)' is dense in C. PROOF.

Suppose there exist zo E C and r > 0 such that B(zo, r) fl Sp f (A)' = 0, for all A E C. Then u(z) = Z is holomorphic on a neighbourhood of Sp f (A)-, for all A E C. It is easy to see that u maps a polynomially convex subset of C\ff(zo, r) onto a polynomially convex set. So, by Theorem 3.3.3, Spu(f(A))- = u(Sp f(A)"). But Spu(f(A))- C B(0,1/r). So, by Theorem 3.4.14, Spu(f(A))" is constant. Because u is injective, we conclude that Sp f(A)" is constant. 0 REMARK 2.

Theorem 3.4.15 will be greatly improved in Chapter VII. We shall see that either Sp f(A)" is constant or C\ UXSC Sp f(A)' is a G6-set having zero capacity. COROLLARY 3.4.16.

Let f be an analytic function from C into a Banach algebra A. Suppose there exists a constant number C such that

mix{) Reu - Rev) : u,v E Sp f(A)} < C


58

for all A E C. Then there exists an entire function h such that

Spf(A)' = h(A) - h(0) +Spf(0)-. In particular this happens if the diameter of Sp f (A) is uniformly bounded on C.

With the notations of Corollary 3.4.9, u(A) and -v(A) are subharmonic. So u(A) - v(A) is subharmonic and bounded, and so is constant. Let a be such PROOF_

that u(A) = v(A) + a for A E C. Then -u = -v - a is subharmonic on C, and consequently u is harmonic on C. Then there exists h entire such that u(A) Reh(A) for all A E C. The function 9(A) = f(A) - h(A)1 is analytic and we have Spg(A) = Sp f(A) - h(A) C (z :Re z < 0). This implies that UAEC Sp g(A) ^ cannot be dense in C. So by Theorem 3.4.15, Sp g(A) - is constant. Then we have

Sp f(A)- = h(A) - h(0)+Sp f(0)". 0 Let f be an analytic function from a domain D of C into a Banach algebra A. Suppose that Sp f(A) = {0,a(A)} for all A E D, where a is a mapping from D into C. Then a is holomorphic on D. THEOREM 3.4.17.

PROOF. By Corollary 3.4.5, a is continuous on D. Let D' be the open subset of D where a(A) 96 0. If D' is empty there is nothing to prove. So suppose D' non-empty. By Radd's extension theorem (see [7], p. 280) it is enough to prove that a is locally

holomorphic on D'. Let Ao E D'. There exist 6 > 0 such that for JA - Aol < 6, we are in the situation of Corollary 3.4.10, in which case u = v. Consequently u, v are harmonic on B(Ao, 6) and there exists k holomorphic on that disk such that u(A) = arga(A) = Imk(A). Taking g(A) = e-k(A) f(A) we have Spg(A) C R for (A -AoI < 6. By Corollary 3.4.12, Spg(A) is constant on B(Ao, 6). So a(A) = Cek(a), for some C, is locally holomorphic. 0

Let f be an analytic function from a domain D of C into a Banach algebra A. Suppose that Sp f(A) = {a(A)) for all A E D, where a is a mapping from D into C. Then a is holomorphic on D. COROLLARY 3.4.18.

PROOF.

The proof is almost identical to the previous one. 0

Banach Algebras

THEOREM 3.4.19.

59

Let f be an analytic function from a domain D of C into a

Banach algebra A. Suppose that Sp f (A) lies on a vertical segment for all A E D. Then there exist a holomorphic function h on D and a fixed ct_)mpact subset K of R such that Sp f(A) = h(A) + iK, for all A E D. PROOF. With the notations of Corollary 3.4.9 we have u(A) = v(A) for A E D. So u, v are harmonic on D. Let us denote by h(A) the element of Sp f (A) with the smallest imaginary part. Fix Ao E D and 6 > 0 such that ff(Ao, 6) C D. On B(Ao, 6) there is a holomorphic function k such that u(A) = Re k(A). Taking

g1(A) = -i(f (A)- k(A)1) on B(Ao, b), we have SP 91(A) C R. So, by Corollary 3.4.12,

Sp gl (A) is constant on B(A0, b). This implies in particular that h is holomorphic on B(Ao, 6). Then h is holomorphic on all D. Once more, arguing with g2(A) _ (A) - h(A)1) on D and applying Corollary 3.4.12, we obtain the result. 0

Let f be an analytic function from D C C into EC(X) and let Ao E D, oo E Sp f (Ao) with ao 54 0. To simplify suppose that ao is an eigenvalue with multiplicity

one, or equivalently that the projection asssociated to N(f (I o) - aGI) has rank one. Then there exist r, b > 0 such that IA - Ao I < 6 implies that Sp f (A) fl B(ao, z )

contains only one eigenvalue a(A). What can be said about this function a? In this particular case it is known that a is holomorphic on B(A0, 6). The classical proof depends strongly on the fact that f(A) E £E(X): see for instance the book by I.C. Gohberg and M.G. Krejn, Introduction a la theorie des operateurs lineaires non auto-adjoints dons an espace hilbertien, Paris, 1971, Chapter 11.

In the next theorem we shall see that this result is true in general. THEOREM 3.4.20 (HOLOMORPIIIC VARIATION OF ISOLATED SPECTRAL VALUES).

Let f be an analytic function from a domain D of C into a Banach algebra A. Suppose there exist A0 E D, ao E Spf(Ao) and r,b > 0 such that IA - AoI < b implies that A E D and that Sp f (A) fl B(ao, r) contains only one point a(A). Then a is holornorphic on a neighbourhood of A. PROOF. If Sp f(Ao) = {ao) then, by Theorem 3.4.2, we may suppose without loss of generality that Sp f(A) _ {a(A)) for IA - AoI < S. Then the result follows from Corollary 3.4.18. Suppose now that Sp f(Ao) is larger than {ao). For the same reasons we may suppose that IA - Ao I < 6 implies Sp f (A) fl OB(ao, r) = 0, and in particular that Sp f (Ao) fl {z: 1z - aol > r) # 0. So by Theorem 3.4.4, without loss of generality, we may suppose that for (A - Aol < 6, Spf(A) is the union of a(A) E B(ao,r) and a non-empty set included in {z: Jr - aoI > r). Let h be the holomorphic function defined by h(z) = z on B(ao,,-) and h(z) = 0 on


60

{z: Iz - ao > r}. By Theorem 3.3.3, h(f (A)) is defined for IA- Ao < 6 and we have

Sp h(f (a)) = (0, a(A)). Then Theorem 3.4.17 implies the result. 0 REMARK 3. If ao is isolated in Sp f (Ao), it is not true in general that Sp f (A) will have an isolated element near ao, for A near A0. Theorem 3.4.4 only says that there will be a small component of Sp f(A) near ao. The previous theorem asserts that it will vary holomorphically if it contains only one point. This theorem will be generalized in Chapter VII (Theorem 7.1.6).

Even if the spectrum funetic is not continuous it has some weak continuity properties, correspo ding to what is called the fine topology in potential theory, that is the smallest topo ogy for which all subharmonic functions are continuous. LEMMA 3.4.21.

Let 01, ... , iOn be upper semicontinuous functions on an open . If #1(Ao) + . + ma(A0) = + O,,(A): A - \o,,\ 0 Ao, A E E), then there exists & sequence

subset D of C. L!t E C D and Ao E D n E

lim sup{01(A) + (Ak) converging to %o such that Ak 96 A0, Ak E E, and 01(Ao) = limt_,o, i= n.

for

PROOF. Let *(A) = 01(A) + + mn(A). By hypothesis there exists a sequence (µk) converging to ;,o, with pk & Ao and pk E E, such that b(Ao) = limk-,,, O(µk). If n = 1 the lemr_ia is obvious. S`-upposing that the lemma is true for n - 1

functions, we prow it for n. If 01(A0) =

then (µk) con-

tains a subsequence, which we denote in the same way for convenience, such that limk-moo 01 (14) _ 01(Ao). Then E 2 ¢i(yk) converges to ¢,(A0) $o, by induction hypothesis, there exists a subsequence (µk) such that lank-»oo 0i(!tk) = 4i(Ao) for i = 2,. .. , n. But it is also true for i = 1, so in this case the proof is complete. If L = q'1(Ao), then there exists a subsequence

(pk,) such that L --

so E12 0.(µk,) converges to A(Ao) - L = + 0a(A0) > E 2 0,(Ao). This is a contradiction because

01(Ao) - L + 42(A0) + + On is upper semicontinuous: 0 02 +

For the definition of non-thin jets at a point see the Appendix. THEOREM 3.4.22 (WEAK LOWER SEMICONTINUITY OF THE BOUNDARY OF THE SPECTRUM). Let .f be an analytic function from a domain D of C into a Banach

61

Banach Algebras

algebra A. Suppose that E C D\{Ao) is non-thin at Ao E D fl P. Then there exists a sequence (Pk) converging to Ao, such that pk E E and

e Sp f(Ao) C asp f(µk) + B(O,1/k), fork > 1. PROOF.

Let e > 0 be given and let B(f r, a/2), ..., B(f,,, a/2) be a finite covering

of 8Sp f(Ao), with fr, ,fn E 8Sp f(Ao). We choose i7i,...,q V Spf(Ao) such that If i - q; j < e/8, for i = 1, ... , n. By Theorem 3.4.2 there exists r > 0 such that D(Ao, r) C D and such that u i(z) = 1 /(z - qi) is holomorphic on a neighbourhood of Sp f(A) for IA - Ao) < r. By Theorem 3.3.3 and 3.4.7, 0i(A) = p(ui(f(A))) > 0 is subharrnonic f o r I A - Ao < r and i = 1, ... , n. Because E is non-thin at A0 we have O;00) = lim sup

{s(A):A -r

Ao, A E E }

.

JJJ

So, by Lemma 3.4.21, there exists a sequence (Ak) converging to Ao such that Ak E E

and mi(Ao) = limk-.,,. COO, for i = 1, ... , n. In particular there exists p(e) E E such that lp(e) - A0J < r and O(µ(e)) > Oi(Ao)/2, for i = 1,...,n. By Theorem 3.3.5, we have dist(r7i, Sp f (p(e))) =

1

1, so 0Spf(A0) C F. If the interior of Sp f(Ao) is empty the conclusion is obvious. So


62

suppose this interior non-empty. It cannot be included in F since F has no interior points and so there exists zo interior to Sp f(Ao) with zo E C\F. But zo can be connected to infinity by an arc 1' included in C\F, which is a contradiction because r must cross e Sp f (Ao) C F. 0

This corollary can be applied to F = H or any Jordan arc.

We now finish this section with two important results. Let A be a Banach algebra and z E A. For an integer n > I we define the n-th spectral diameter of x, denoted 6n(x), by n n+l

6n(x)=max

[J

J

1-Ajl

(I 2. Obviously p is a polynomial of degree n having n roots Al, ... , A". We have 0 = p(A,) = X((A,l - x)") E Sp(A,1 - x)" and consequently A; E Spx and JAI( p(x). We have

p(A) = A" - nX(z)A"-' +

(2)X(x2)A"--s

11

+ ... + (-1)"X(z") = Ho - Ai), i=1


70

and consequently n

n ai = nX(x) (=1

(n)X(.T2). 2

i 2, we conclude that X(x)2 = X(x2) for all x E A. Then we have X((x+y)2) = X(x2+y2+xy+yx) _ (X(.c) + X(y))2 = X(x2) + X(y2) + 2X(x)X(y), and so for all x, y E A.

(1)

(ab - ba)2 + (ab + ba)2 = 2[a(bab) + (bab)a)

(2)

(X(ab - ba))2 +4X(a)2X(b)2 = 4X(a)X(bab)

(3)

X(zy + yx) = 2X(x)X(y)

,

Now the identity

implies

Taking a = x - X(x)1 and 6 = y we have X(a) = 0. Consequently X(ay) = x(ya) and hence X(zy) = X(yx). Finally, from (1) we obtain x(xy) = X(x)X(y). 17 We now intend to show that A has plenty of characters if A is commutative and that these characters play a very important role concerning the representation of the algebra. This important discovery, which has many consequences in spectral theory, in harmonic analysis and in approximation theory, was made by I.M. Gelfand by 1940.

THEOREM 4.1.2 (I.M. GELFAND).

Let A be a commutative Banach algebra. Then

we have the following properties:

(i) X - Ker X defines a bijection from the set of characters of A onto the set of maximal ideals of A,

(ii) for every z E A we have Sp x = {X (x): for all characters X of A). PROOF. (i) If X is a character of A then obviously Ker X is an ideal of A. It is maximal because codim Ker X = 1. Conversely if I is a maximal ideal of A then, by

Representation Theory

71

Corollary 3.2.2, it is closed. Moreover A/I is a commutative algebra with no nontrivial ideals so A/I is a field. By Corollary 3.2.9, there exists an isomorphism 0 from A/I onto C. Taking X as the composition of the canonical morphism A -+ All and 0, X is a character and KerX = I. If Ker X1 = Ker X2, then since X1(1) = X2(1) = I we conclude that X1 = X2, so that the previous mapping is a bijection. (ii) If X is a character then x-X(x)1 is not invertible because (x-X(x)1)y = 1 implies 0 X(y) = 0 = X(1) = 1, a contradiction. Conversely, if x - Al is not invertible then (z - A1)A is an ideal which is, by Lemma 3.1.1, contained in some

maximal ideal I. By (i), there exists a character Xo such that I = KerXo so

Xo((x-Al)1)=Xo(x)-A= 0, and so (ii) is proved. 0 REMARK 1.

In particular, this theorem implies that the set of characters of A, denoted by 9n(A), is not empty. It also implies that in the commutative case, the radical of A coincides with its set of quasi-nilpotent elements. In fact, if we have p(x) = 0 then X(x) = 0 for all x E fJl(A), consequently X(zy) = 0 for all y E A. Hence p(xy) = 0 and, by Theorem 3.1.3 (iii), x E RadA. In some cases it is possible to determine 911(A) explicitly. THEOREM 4.1.3.

Let K be a compact set and A = C(K). Then 9)1(A) can be identified with K. In particular, for every closed ideal I of C(K) there exists a closed subset F of X such that I = { f : f E C(K), f (z) = 0 for all z E F}. PROOF.

For each x E K, f ,-+ f (x) is a character of A, denoted by Xs and called the evaluation at z. Since A separates the points of K then z -4 Xz is an embedding of K into 9)1(A). We now prove that every X E '9l(A) is a X. for some x E K. Suppose this is false. Let X # X. for all x E K. Then for every z E K there exists

f, Asuch that f.(x)00and X(f=)=0. The Vs={y:yEK,fs(y)#0}form

an open covering of K. So there exist x i, ... , zn E K such that the f, = f=; , for

i = 1, ... , n, satisfy X(f;) = 0, for i = 1,...,n, and such that for every x E K there exists an i for which f,(x) 0 0. Let g = flfl + + f f,,. Then X(g) = 0

and g(x) > 0 for all z E K. Consequently g'1 E C(K), which is a contradiction as we would then have X(1) = 0. Hence x - X= is a bijection from K onto 911(A). Let I = { f : f E 1). It is easy to verify that I n I is an ideal of C(K). Let F = {x: x E K, f(z) = 0, for all f E 1 fl I}. Because f(z) = 0 is equivalent to

ff(x)=0wehave F={x:xEK,f(x)=0,forall f EI}. Let J=l=EFKerXs. Then I fl I C I C J. We shall now prove that J C I fl f and the theorem will thus be proved. Let K' be the contraction of K, that is the compact topological space


72

obtained from K by the equivalence relation X

=_yt

(

x=y, ifxgF yEF, ifxEF.

Denoting by a the point of K' corresponding to F, J can be identified with the maximal ideal of functions of C(K') vanishing at a. Identifying in! with its image in C(K') it is easy to verify that (I fl 1) + Cl separates the points of K' and so, by the Stone-Weierstrass theorem, (I n 1) + Cl = C(K'). Hence if f E J then f Al + g with g E I fl I, so Al = f - g E J, and consequently A = 0. So the assertion is proved. 0 In his famous book Tauberian Theorems, N. Wiener proved that a continuous function defined on R, with period 27r, which has an absolutely converging trigono-

metric series (f E W in the terminology of Example 3, Chapter III, §1) and such

that f(x) ,-t 0 for 0 < x < 2x, has also an inverse with an absolutely converging trigonometric series. The original proof is long and complicated. This result was extended by P. Levy for the composition of f with a function h holomorphic on a neighbourhood of the set of values of f. In 1940, I.M. Gelfand surprised the mathematical world by giving both results a very simple proof. We now see his argument.

Let W be the commutative Banach algebra of absolutely converging trigonometric series a_,o ane'"t, with sum, product and norm defined by ll(an)ll =X En _,p lang. Then Wl(W) can be identified with the interval [0, 21rl.

THEOREM 4.1.4.

PROOF. If E [0, 21rl then Xs(f) = n_- aneins where f (t) _ T° ane'nt defines a character of W. We now prove the converse, namely that for every character X E M(W) ther a exists z E [0, 2a) such that X = X.. Applying X to the two functions e" and e-'t we have

IX(e")I a IIe't11 =1 and

IX(e-")I < lle-"11 = 1.

But X(e") . X(e 't) = 1, so IX(e")l = 1. Let x E (0, 2ir) such that X(eie) = e's By continuity of X we obtain immediately that 00

X(f) = E aneins = x:(f) nz-00

SOX =Xs.0

,

for all f E W.


73

COROLLARY 4.1.5 (N. WIENER-P. LEVY). Let f E W and let h be holomorphic on a neighbourhood of the set of values off . Then h o f E W.

By Theorem 4.1.2 (ii) and Theorem 4.1.3 we have Spf = {X(f ): X E YJl(W)} = {X=(f):x E [0,2x]} = f([0,2rr]). By Theorem 3.3.3, we have h o f = PROOF.

h(f)EW.O In particular, if f (x) # 0 for 0 < x < 2x then 1/z is holomorphic on a neighbourhood of the set of values of f, so 1/f E W. We give another application.

Let K he a compact subset of C and let A(K) be the Banach algebra of continuous functions on K which are holomophic on the interior of K. Then 9J2(A(K)) can be identified with K. THEVAEM 4.1.6.

The mapping x

X. is an injective mapping from K into M(A(K)) because the function I(z) = z is in A(K). Conversely we prove that all X E 9J1(A(K)) have the form Xs. We have x = X(I) E K, because otherwise the PROOF.

function defined by 1

z - X(I) is in A(K), so that we get X(9 (I - X(I)1)) = x(l) = 0 which is a contradiction. Since X(I) = x, by continuity and the local series representation of f, we get

X(f)=f(x),so x=x:.0 Let f1, ... , fn E A(K) be such that for all x E K there exists at least one i such that f.(x) # 0. Then there exist 91, ... , g E A(K) such that COROLLARY 4.1.7.

fig, +...+ fn9n=I. PROOF.

Let I be the set of all sums fi 91 +

+ f ngn with 91, ... , gn E A(K).

If I A, then it is contained in some maximal ideal Ker X=, and consequently f, (x) = 0 for i = 1, ... , n which is a contradiction. So I = A, consequently 1. is such a sum. 0 For some commutative Banach algebras A the set of characters Wt(A) may be extremely complicated and very difficult to determine explicitly. This occurs in the next example.

Let U be an open subset of C. We denote by H°°(U) the Banach algebra of functions which are holomorphic and bounded on U. Then the open set U can


74

be embedded in 9l2(A) by x -+ X: (see K. Hoffman, Banach Spaces of Analytic Functions, Englewood Cliffs, 1962). It is a famous problem, called the Corona problem, to determine if U is dense in 911(A) for the Gelfand topology which we define below. This result is true for U simply or finitely connected. The first proof, a cori'iplicated one, was given by L. Carleson. Now there is a rather simple proof of this fact due to T. Wolff (see for instance T.W. Gamelin, Wolff's proof of the Corona Theorem, Israel J. Math. 37 (1980), pp. 113-119). But the problem remains unsolved for a general open set.

Let A be a commutative Banach algebra and let x E A. We define the Gelfand transform of x, denoted i, by the formula DEFINITION.

i(x) = x(x) for x E 911(A). Then i is a function defined on 9J (A) and x'-+ i is a morphism.

The set 91i(A) is included in the unit ball of the topological dual of A. The Gelfand topology on 9J2(A) is by definition the restriction of the weak *-topology on

this set, that is the weakest topology that makes every i continuous. THEOREM 4.1.8 (I.M GELFAND). Let 9J2(A) be the set of characters of a commutative Banach algebra A. We have the following properties:

(i) 932(A) is compact for the Gelfand topology,

(ii) the Gelfand transform is a continuous morphism from A onto a subalgebra A of C(97l(A)) whose kernel is Rad A. It is an isomorphism if and only if A is semi-simple,

(iii) for each x E A the range of i is the spectrum of x and ((i((. = p(x). (iii) is obvious by Theorem 4.1.2(ii). It is also obvious that x " i is a morphism. It is continuous because ((sl(o = p(x) < ((x((. The last part of (ii) comes from (iii). So we now prove (i). By Theorem 1.1.8, it is sufficient to prove that MI(A) is weak *-closed in the unit ball of A'. Let fo E A' be in the weak *-closure of S 1(A). We have only to prove that fo(zy) = fo(x) fo(y), for all x, y E A and fo(1) = 1. We PROOF.

fixx,yEAande>0. LetU={f:fEA',jf(z)-fo(z)(<e,forz=1,x,y,zy}. Then Zr is a neighbourhood of fo for the weak *-topology which contains some X E 1t(A). So we have 11 - fo(1)I = Ix(1) - fo(1)I < E,

fo(xy) - fo(x)fo(y) = fo(xy) - x(xy) + (x(y) - fo(y))x(x) + (x(x) - fo(x))fo(y),


75

and consequently 1 fo(xy) - fo(x)fo(y)l 5 e(1 + 11x11 + Ifo(y)() ,

for every e > 0.

So 931(A) is weak *-closed in the unit ball of A', and hence weak *-compact. 0

COROLLARY 4.1.9. Let T be a morphism from a commutative Banach algebra A into a semi-simple commutative Banach algebra B. Then T is continuous. PROOF.

Suppose lim x = 0 and lim Tx = a. By the Closed Graph Theorem it

is enough to show that a = 0. Let X E 931(B). Then x o T is a character of A, and hence it is continuous. So we have

x(a) = limo x(Txn) = h a(x o T)(x") = 0

,

for every x E 97f(B). So a E Rad B= {0}. 0

COROLLARY 4.1.10.

On a semi-simple commutative Banach algebra all the Banach algebra norms are equivalent. PROOF.

Let 11' 111, 11' 112 be two Banach algebra norms on A. We apply the previous

corollary to the identity mapping from (A, 11 II,) onto (A, 11 112). 0

For the definition of Banach algebras with involution see Chapter VI, §1. COROLLARY 4.1.11.

Every involution on a semi-simple commutative Banach al-

gebra is continuous. PROOF.

Let us consider the new norm 111xu1J = JJx*1J. It follows immediately from

the definition of an involution that III JJJ is submultiplicative and 1111111 = 1. If is a Cauchy sequence for 111 111, then (x*) is a Cauchy sequence for 11 11, and

consequently it converges to a E A. Then lim, 11Jx,n - will = 0. So 111 111 is a Banach algebra norm on A. By Corollary 4.1.10 there exists C > 0 such that JJJxhl1 = JJx`1J < CJJxJJ, for x E A, so the involution is continuous. 0

The three last corollaries will be improved upon in Chapter V.


76

COROLLARY 4.1.12. Let [a, b] be a bounded interval of A.

Then the algebra

CO°([a, b]) has no Banach algebra norm. PROOF.

Suppose that C°°([a, bJ) is a Banach algebra for the norm II ' lI The

identity mapping is a morphism from CO°([a, b)) into C([a, b]), so by Corollary 4.1.9

there exists a > 0 such that 11f 11. < allfll, for all f E C°°([a,b]). Let D: f '- f be the linear mapping from C°°([a, b]) into itself. We now prove that it is continuous. ll fÎI = 0 and lim,o° IIf' - g[l = 0, for some g E C°°([a, b]). Suppose that By the previous inequality we have limI[f^lloo I[fn - gIIo. = 0. So G(t) = f,, g(s) ds is identically zero on [a, b), and consequently g = 0. By the Closed Graph Theorem, D is continuous, hence there exists ,B > 0 such that IIFII :5 011f III for f E C°°([a, b]). For A E C, this second inequality implies that

Taf =,f +

Af' +... + ^ f(^) +... n!

converges in C°°(fa, b]). By Leibniz's formula for the derivativeof products it is easy to verify that Ta is a morphism from C°°([a, b]) into itself. If f E Rad C°°([a, b]) then X=(f) = f (x) = 0 for all a < x < b. So CO°([a, b)) is semi-simple. By Corollary

4.1.9 there exists y > 0 such that IfTafff

ylifII ,

for all f E C°°([a, b]).

The entire function A " Ta f is bounded so, by Liouville's theorem, Ta f = f, and hence f' = 0, for all f'E C°O([a, b]) which is absurd. So the assertion is proved. 0 A commutative Banach algebra is called a function algebra on a compact set K if it is isometrically isomorphic to a closed subalgebra of C(K) which separates the points of K and contains the constants. THEOREM 4.1.13.

A Banach algebra A is a function algebra if and only if IIX2 fl =

IIxII2,fora,11zEA. PROOF. The necessity is obvious. So suppose that 11x211 = 11x112 for all x E A. We first prove that A is commutative. By induction we have IIxz" fl = ]1x1]2 for n > 0.

So, by Theorem 3.2.8 (iii), p(z) = IIzIf. Fixing x, y E A, and using Lemma 3.1.2, for all A E C, we have Ile-As2Ehy1I = p(e'A"xe") =

p(x) So the analytic function A -* e-x"xe''" is bounded in norm. By Liouville's theorem it is constant, and consequently zy = yx for all x, y E A. By Theorem 4.1.8 we have fii[I0° = ffxfl , for the Gelfand transform. This implies in particular that A is closed in C(Wl(A)). But obviously it contains the constants and separates the points of M(A). So the result is proved. 0


77

In Theorems 4.1.4 and 4.1.6 we saw that it is easy to characterize the set of characters of some given commutative algebras. We shall prove that this nice situation always occurs when the algebra is finitely generated.

Given K a compact subset of C" we recall that the polynomial convex hull K of K is defined by k = (z: z E C", Ip(z)I < max Ip(u)I for all polynomials p).

Then we say that K is polynomially convex if K = K. If n = 1 then k is the union of K with its holes. But if n > 2 then k has no topological characterization. Even if K is nicely defined, k may be extremely difficult to determine explicitly. By Exercise IV.5, the set of characters of P(K) can be identified with K. More generally we have: Tur..oREM 4.1.14. Let A be a commutative Banach algebra having n topological generators. Then !32(A) can be identified with a compact polynomially convex subset of C". PROOF. Let x1, ... , xn be topological generators of A. Consequently every element of A is a limit of polynomials in x1, ... , xn. We consider the mapping 0 from WT(A) into C" defined by 0: X -' (X(xi ), ... , X(xn))

We prove that it is a homeomorphism from DA(A) onto K = O(Wt(A)) and that K is polynomially convex in C". By the definition of the Gelfand topology, 0 is continuous. If O(XI) = (X2), then Xn and Xs coincide on the polynomials in z1,... , x,, which are dense in A, so X.t = X2. But 91Z(A) is compact, so 0 is a homeornorphism from 97t(A) onto K. We now prove that K is polynomially convex. Suppose that Ip(z)I < maxEK Ip(u)I for all polynomials p. Then IpWI p(x), so f(x) E coSpx (see Exercise IV.8). 0

The set I of spectral states is convex, contained in the unit ball of A', and weak'-closed. So, by Theorem 1.1.8 it is weak *-compact. By Theorem 1.1.9 it is the closed convex hull of its extreme points. LEMMA 4.1.16.

If f is an extreme spectral state then it is a character of A.

Let 9R denote the set of characters of A. Then cow C ,?i. Suppose there exists fo E 3 such that fo V cow. So, by Corollary 1.1.6, there exists xo E A such PROOF.

that sup {Re f(xo): f E coff) < Re fo(xo). Consequently

Re fo(xo) > max{ReX(xo): X E 9JI} = max{Rez: z E Spxo}

= max{Rez: z E coSpxo) and this gives a contradiction because fo(xo) E coSpxo. So co Y = a. By Theorem 1.1.12 the set of extreme points of I is included in 911. 0 THEOREM 4.1.17 (M. NAGASAWA).

Let A andB be commutative and semi-simple

Banach algebras. Suppose that T is a linear mapping from A onto B such that T1 = 1 and p(Tx) = p(x) for all x E A. Then T is an isomorphism from A onto B. PROOF.

Let f be an extreme spectral state on B and let g = f oT. Then we have

g(1) = 1 and Ig(x)j = if(Tx)I < p(Tx) = p(x) for all x E A. So g is a spectral

79


state on A. Suppose that g = 2.142. where gI, gz are spectral states on A. With the hypotheses, the mapping T is bijective, so for all y E B we have f (Y) =

gi

(T-I Y)

+ g2(T -'Y) 2

As was done previously, it is easy to verify that gl o T-I and 92 o T-1 are spectral states on B. Hence, f being extreme, we have f = gl oT-1 = g2oT'1. Consequently

g = gl = gz and g is an extreme spectral state of A. By Lemma 4.1.16, g is a character on A. Hence f(Txy) = f (Tx) f (Ty) for all x, y E A. This being true for all extreme spectral states f on B, by Theorem 1.1.9 it is true for all spectral states, and in particular for all characters of B. So Tay = TxTy for all x, y E A, because B is semi-simple. 0 LEMMA 4.1.18.

Let K be a compact space. Then f is an extreme point of the

closed unit ball of C(K) if and only if if (z)I = 1, for all z E K. PROOF.

Suppose that If(z)I = 1 for all z E K and suppose that f = "z" where

u,vEC(K)andllull=llvll- 1. By the T.T. West decomposition (see for instance S.R. Caradus, W.E. Pfaffenberger, B. Yood, Catkin Algebras and Algebras of Operators on Banach Spaces, New York, 1974, Theorem 5.3.2, p. 51), we can write R = C + Q with C compact and Q quasi-nilpotent, and so #Spe(T+Q) > 1. 0 PROOF.

§4. Spectral Characterizations of Finite-Dimensional Banach Algebras Let A be a Banach algebra such that A/ Rad A is finite-dimensional. For all x E A the class i is algebraic in A/ Rad A and consequently Sp z is finite. Surprisingly, the converse is true even supposing that the spectrum is finite on a verysmall part of the algebra. This result was used by I. Kaplansky to characterize ring isomorphisms of Banach algebras.

The following lemma is a generalization of the Wedderburn-Artin theorem (Theorem 2.1.2). LEMMA 5.4.1. Let A be a semi-simple Banach algebra. Suppose there exists an integer n > 1 such that for all z E A, x is algebraic of degree < n. Then A is the direct sum of at most n algebras isomorphic to some Mk(C), with k < n. PROOF. Let x be a continuous irreducible representation of A on a Banach space X. If dim X > n, there exist n + 1 linearly independent vectors C1, ... , En+t E X.

So by Theorem 4.2.5 there exists z E A such that x(r)f1 = (1, x(z)f2 26 , ... , (n + 1)f..+l. Consequently 11, 2, ... , n + 1) C Sp lr(x) C Sp z

Applications of Subharmonicity

97

which is a contradiction because x is algebraic of degree n. So k = dim X < n. Consequently, by Exercise IV.13, r(A) Mk(C). Let r,, ... , r,,, be m continuous irrex a,,, (A). ducible representations of A with different kernels and let Bm = rt (A) x For i # j, r,(Kerri) is a two-sided ideal of r,(A) Mk;(C) and consequently

ir,(Ker ri) = ri(A) or ri(Ker ri) _ (0). The same argument with ri(Ker r,) implies that ri(Ker r,) = ri(A) or ri(Ker r,) _ {0}. Finally we obtain A = Kerry+Kerri or Ker r, = Ker ri, but this last case is impossible. By the Chinese Remainder Theorem, the mapping O: x + (r (x) ... , rm(x)) is onto B,,,. Let A,...... ,,, be different complex numbers. There exists x E A such that

rl(x) = All , ... , r,n(x) = Amt. But x is algebraic of degree < n so there exists a polynomial p of degree < n such that p(x) = 0. Then p(Al) = = p(A.. ) = 0 and consequently m < degp < n. If m is the greatest number < n such that we have m continuous irreducible representations rl, . . . , with different kernels, then by Theorem 4.2.1 (ii) we have RadA = Kerry n fl Ker r,,, = {0}. So 0 is an isomorphism of A onto B,,, and the lemma is proved. 0 In a real vector space X we say that a set U is absorbing if there exists a E U such that for all z E X, there exists r > 0 such that a + Ax E U for -r < A < r. For instance, an open sat is absorbing but the converse is not true in general. THEOREM 5.4.2.

Let A be a Banach algebra containing an absorbing set U such that Sp x is finite for all x E U. Then A/ Red A is finite-dimensional. PROOF.

Replacing A by A/ Rad A and U by its image under the canonical mapping from A onto Al RadA, we may suppose without loss of generality that A

Let a E U be such that for all a E A there exists r > 0 such that a + Ax E U for -r < A < r. Considering the analytic function is semi-simple.

A -. a + A(x - a) = f (A), we have Sp f (A) finite for A in some real interval which has a non-zero capacity. So, by Theorem 3.4.25, # Sp(a + A(x - a)) < -I-oo for all A E C. In particular, # Sp z < +oo for all z E A. Let Ak = {x: x E A, # Sp x < k}. By Corollary 3.4.5, Ak is closed. So by Baire's theorem there exists a smallest integer n such that # Sp z < n for x in a ball B(b, a). Applying again the argument at the beginning of this proof, with the absorbing set B(b, a), we con-

clude that # Sp x 0 such that ho + A(h - ho) E U for 0 < A < r. By Theorem 3.4.25 we conclude that # Sp h < +00, for all h E H. Now let z = h + ik E A be arbitrary, with h, k E H. Considering as earlier the analytic function

A h-' h + .1k we have # Sp(h + Ak) < +oo, for A E R. So # Sp(h + Ak) < +oo for all A E C, and in particular for A = i. Then by Theorem 5.4.2, A/ Rad A is finite-dimensional. 0

We now give some small applications which improve a result due to R.E. Edwards. THEOREM 5.4.4.

Let A be a Banach algebra containing a non-empty open set U

of invertible elements such that p(z)p(z-1) = 1, for all z E U. Then A/RadA is isomorphic to C. PROOF.

Let z E U. There exists r > 0 such that JAI < r implies z - Al E U and

so p(z -A 1) p((z - A1)-1) = 1. By Theorem 3.3.5, we conclude that Sp s is included

in a circle centred at A for all IAI < r. Consequently #Spz = 1, for z E U. The proof of Theorem 5.4.2 with n = 1 implies that Al Red A = C. 0 COROLLARY 5.4.5. Let A be a Banach algebra containing a non-empty open set U of invertible elements such that IIz{I Hz-1II = 1, for all z E U. Then A is isomorphic to C. PROOF.

Without toss of generality we may suppose that U is connected. First

we prove that IIxH pz-111 = 1 on G,(A), the connected component of 1 in the

set of invertible elements. Let E _ {z: z E G1(A) such that Hzf I{z-'ll = 1). This is a closed subset of Gl(A) containing 1. We now prove that it is open.

Let a E U and z E E. Then for y E Ua-1 we have: 1 < HzyO . Hy-'z-111 = IIeyaa-' II Ilaa-' y-' z-' II 1 such that # Sp(a, x) < n, for all x E A. LEMMA 5.6.5.

Clearly a E a and a ae't E a for all x E A, and thus etae't - a CLet Ak be the set of x E A such that # Sp(esae`t - a):5 k. We have A = Uk 1Ak PROOF.

and by Corollary 3.4.5, Ak is closed. So by Baire's theorem there exists a smallest integer n such that An contains a ball B(b, r). We fix a E A and consider m(A) =

eb+A(t-b)

ae-k-alt-b)

- a E a.

This function 0 is analytic, and for (AI JJx - bil < r we have # Sp O(A) < n. So, by

Theorem 3.4.25, we have # Sp 4(A) < n for all A E C. This implies that x E A. Hence A = A. Considering ( a - eatae'at A f(A) = { [a, x]

for A 76 0

forA=0

we have # Sp f (A) < n, for A 36 0. Once more by Theorem 3.4.25 we conclude that # SP f (0) = # Sp[a, x] < n. 0 THEOREM 5.6.6.

If for some a E A we have a+,I C a then a is algebraic modulo

the radical of A.

PROOF. By the previous lemma there exists an integer n such that # Sp(a, x) < n, for all x E A. Let x be a continuous irreducible representation of A on a Banach

space X. Let t E X be such that to = t, {1 = x(a) , ... , n+1 = x(a)"+1t are linearly independent in X X. For ao, a 1, ... , a"+l given in X, by Theorem 4.2.5, there exists x E A such that 1r(x) fo = ao, ... , r(x) f"+, = a"+1. So [r(a), x(x)] f, = iE, for

i = 0, ... , n, if we take a1 = x(a)ao, a2 = or(a)al - f 1 , ... , an+t = ir(a)aa - ntn. Hence {Q,. .. , n} C Sp[x(a), r(x)] C Sp[a, x], which gives a contradiction. Consequently {l, ... ,1;n+1 are linearly dependent, and by Theorem 4.2.7R(a) is algebraic

of degree < n + 1. We have a E I and so Spa = {fl , ... , t}. Consequently (x(a) - #11)"+1 x x (x(a) - #,l)"+1 = 0 for every continuous irreducible representation x, and so (a - #11)"+1 x ... X (a - fel)"+1 E RadA. 0

Let T be a bounded linear operator on a Hilbert space H which is not polynomiatly compact. Then there exists U E E(H) such that Sp. U is finite and Spe(T + U) is infinite. COROLLARY 5.6.7.

PROOF. We apply Theorem 5.6.6 to the Calkin algebra £(H)/.CC(H) which is semi-simple. 0

105

Applicatic is of Subharmonicity

Let A be a semi-simple Banach algebra. Suppose that go E A is a non-nilpotent quasi-nilpotent element. Then there exists another quasi-nilpotent element q1 E A such that Sp(qo + ql) is infinite. THEOREM 5.6.8.

PROOF.

Suppose that go -

e=goe_x

E a for all z E A. The same argument as in

the proof of Lemma 5.6.5 implies that # Sp[qo, x) < n for all x E A and some integer n > 1. As in the proof of Theorem 5.6.6 we conclude that qo+' = 0, sc we have a

contradiction. Consequently there exists some x E A such that for q1 = etgoe-`, we have Sp(qo +q1) infinite. 0 If (en)n>o is the standard basis of f2(N), then considering the two operators a, b defined by en+1 aen = 0

if n is odd ,ifnis s even

ben =

if n is odd if n is even

we have a2 = b1 = 0 and (a + b)en = en+1 for n > 0. So a + b is the unilateral shift whose spectrum is the unit circle (see [3), Problem 85). For a general Banach space X, is it possible to build two quasi-nilpotent operators whose sum has infinite spectrum? This problem is difficult because in general X has no topological basis so it is impossible to give an explicit construction. Nevertheless we shall solve the problem using a circuitous method. LEMMA 5.6.9 (S. GRABINER).

Let A be a Banach algebra such that its set of

nilpotent elements contains a linear subspace on which the degree of nilpotency is unbounded. Then A contains a non-nilpotent quasi-nilpotent element which is a limit of nilpotent ones. PROOF. Let M be the set of nilpotent elements of A. We denote by X the closure of the linear subspace contained in M on which the degree of ailpotency is unbounded. Let Ek = {x: x E M fl x, xk = 01. Then m fl x= Uk>lEk and each of the Ek is closed in X. We now show that Ek has no interior point in X. If a is interior to Ek then, for z E X, we have a + A(x - a) E Ek for A small. Hence (a + A(z - a))k = 0 for all A E C, by the Identity Principle. Consequently xk = 0, and the degree of nilpotency is bounded on X, a contradiction.

Now let N be the set of quasi-nilpotent elements of A.

Then N =

fln>1 {x: p(x) < 1/n}, so N is a Go-set by upper-semicontinuity of p. The set M fl X is dense in X so the quasi-nilpotent elements of X form a dense G6-subset of X. By Baire's theorem we conclude that the set of quasi-nilpotent elements of X is not a countable union of closed subsets of X with empty interior, so it is different

from MflX.0


106

THEOREM 5.6.10.

Let X be a Banach space of infinite dimension. Then there

exist two quasi-nilpotent and compact operators TT,T2 on X such that Sp(Ti + T2) is infinite.

PROOF. We prove that A = £(!(X)+CI satisfies the hypothesis of Lemma 5.6.9. Let (Xk)k>o be a sequence of finite-dimensional linear subspaces of X such that

Xo = {0} and Xk is strictly included in Xk+i, for k > 0. Let F,, = {T:T E £(X),T(X) C X,, and T(Xk) C Xk_1 for k = 1,...,n} and let F = U,,>>F,. Then F is a linear subspace of the set of nilpotent elements of A and contains elements with degree of nilpotency as large as we want as X is infinite-dimensional. Moreover, by Exercise III.4, A is semi-simple. So, using Theorem 5.6.8, the theorem is proved. 0

§7. Inessential Elements There are many results in spectral theory concerning the relation between the spectrum of an operator and its essential spectrum, that is, the spectrum of the class of this operator in the quotient algebra obtained from the closed two-sided ideal of compact operators. These include the theorems of B.A. Barnes, I.C. Gohberg, D.C. Kleinecke and A.F. Ruston which are given below. In this section we show that the hypothesis that the elements of the closed twosided ideal are compact is irrelevant. The essential assertion is that these elements have. a spectrum which is either finite or a sequence converging to zero. With this point of view many results in spectral theory can be extended and greatly simplified. We only present a selected list of new dishes obtained by this "nouvelle cuisine". The main ingredient in these arguments is Theorem 3.4.26.

Let I be a two-sided ideal (not necessarily closed) of a Banach algebra A. We

say that I is inessential if, for every x E I, the spectrum of x has at most 0 as a limit point. For instance in £(X) the set I and the set £C(X) of compact operators are two-sided inessential ideals. Given a two-sided ideal I of A we denote by kh(I) the intersection of all kernels of continuous irreducible representations it of A such

that I C Ker R. It is easy to see that I C 7 C kh(I), and that kh(I) is the inverse image of the radical of A/1.

Let x be in A and a be isolated in the spectrum of z. We define the projection associated to x and a by

p=

1 f (All -x)-ida 2Iri

107


where r is a contour around a, separating a from the remaining spectrum of x. In fact p does not depend on the contour r, as long as r separates a from the rest of the spectrum. Thus we can suppose that 1' is a small circle with centre at or.

Let I he a two-sided ideal of A and let x E kh(I). Suppose that 0 is isolated in the spectrum of x. Then the projection associated to x and a is in I. LEMMA 5.7.1.

a

PROOF.

Let r be a circle centred at a, separating a from 0 and from the rest of

the spectrum. For A E F we have

(A1 -

x)-i

+

=

So we have

P=

1

dA

x

x(A1 -

x)-r

l

2ai r A + tai r A (Al - x}-1 dA.

The first term is zero and the second term is in kh(I), sop E kh(I). Let p denote the coset of p in A/I. Then p E Rad(A/7) and so p(li) = 0, where p denotes the spectral radius. But p is also a projection, consequently p = 0, and hence p E T. Moreover pIp is a closed subalgebra of A, hence a Banach algebra with identity p, in which pIp is a dense two-sided ideal, and so pIp = p7p. Then p = p3 E pip = pip C I. Q

The argument shows that I and kh(I) have the same set of projections. The following result is an improvement of a classical result of D.C. Kleinecke. THEOREM 5.7.2.

Let I and J he two two-sided inessential ideals of A having the same set of projections. Denoting by x + I (resp. T + J) the coset of x in All (resp.

A/J), then x + I is invertible in All if and only if x + J is invertible in A/J. If moreover I and J are closed, then Sp(x + I) = Sp(x + J), for all x E A.

Suppose that x +J is invertible in A/J but that x + I is not invertible in All. Without loss of generality we may suppose that x + I is not right invertible. Then there exists y E A such that a = xy - 1 E J. If 1 + a is invertible then xy(1 +a)-' = 1, and sox +1 is right invertible. Consequently --1 E Spa. Because J is inessential, -1 is isolated in Spa. By Lemma 5.7.1, the corresponding projection p is in J, so by hypothesis it is also in I. By the Holomorphic Functional Calculus it is easy to see that -1 f Sp(a - ap), so that xy - ap = 1 -I- a - ap has an inverse z in A. Consequently (x + I)(yz + I) = I + I which is a contradiction. By a symmetric argument we get that x + I is invertible if and only if x + J is invertible. Replacing x by x - Al we get the last part of the theorem. 0 PROOF.


108

In particular this result can be applied with 3r, I and .CC(X), all of which have the same set of projections. So we get the following:

Let X be a Banach space and let T be a bounded linear operator on X. Then we have Sp(T+',) = Sp(T+ZC(X )) = Sp.(T). COROLLARY 5.7.3 t, D.C.KLEINECKE).

We shall see below that if I is a two-sided inessential ideal then 7 and kh(I) are also inessential. Thus Theorem 5.7.2 can be used in that case. Let I be a fixed inessential two-sided ideal of A. For x in A, we define D(x) in the following way: A

A V D(x) b

Spx

or A is an isolated spectral value of z with

the corr esponding projection in I. It is easy to verify that D(x) is compact and that Sp z \ D(z) is discrete, and hence finite or countable. It is also obvious that D(x - Al) = D(z) - A for every A E C. The next result is a strong improvement of a theorem obtained previously by

I.C. Gohberg for A = .C(X) and I = U(X) (am for instance L.C. Gohberg and M.G. Krejn, Introduction s In thforie des ophuteurs liniairea non dana un espace hilbertien, Chapter 1, Theorem 5.1 and Lemma 5.2).

Let I be a two aided inessential ideal of a Banach algebra A. Fbr z E A and y E I we have the THEOREM 5.7.4 (PERTURBATION BY INESSENTIAL ELEMENTS).

following properties:

(i) if G is a connected component of C \ D(z) intersecting C \ Sp(z + y) then it is a component of C \ D(z + y), (ii) the unbounded connected components of C \ D(x) and C \ D(z + y) coincide, in particular D(z) and D(z + y) have the same external boundaries, (iii) if i denotes the coset of z in A/7 then we have Sp i C D(x) and D(z)" = (Sp i)^, where - denotes the polynomially convex hull of the set. PROOF.

(i)

Let G' = G \ Spx and let A E G'. Then G' is a domain such that

G \ G' is discrete. We have

Al - (z +y) = (Al - x)(1 - (Al - x)-'y).

(1)

1o9


Let f (A) _ (A1- x)-ly, which is analytic on G', and has values in I. By hypothesis and Theorem 3.4.26, we have either 1 E Sp f (A) for all A E G', or {A: A E G',1 E Sp f(A)} closed and discrete in G. Suppose we are in the first situation. Because G \ Sp(x + y) is a non-empty open set, G' \ Sp(x + y) is non-empty. Equation (1) implies G' C Sp(x + y), so we get a contradiction. Hence for all A E G' we must have Al - (z + y) invertible except on a closed discrete subset, and consequently Al - (x + y) is invertible for A E G except on a discrete subset. Let a be such a point of the discrete subset of G, and suppose that a E Sp x. Then there exists a small circle r, with centre at a, isolating a from the rest of the spectrum of z and from the rest of the discrete subset of G. If A E r, then Al - (x + y) and Al - x are invertible. Moreover we have

(Al - (x + y))-' = (Al -

x)-i + (Al - (z + y))-1 y(Al -

x)-'.

(2)

The last term of the second member is in I, and because a E Spz \ D(z) we have 1

tai

r(Al-z)-'dAEI. r

Then by (2) and Lemma 5.7.1 we have (z) fr(Al - (z + y))-' dA E I and consequently a V D(x + y). If a 4 Sp x, the same argument works except that (1 ) fr(A1 - x)-' dA = 0. Then G C C \ D(z + y). Let H be the connected component of C \ D(z + y) containing G. If H fl (C \ Sp x) is empty, then G C H C Sp z

and G C C \ D(x). Thus G C Spz \ D(z) which is absurd because this last set is discrete. So H intersects C \ Sp x and we may apply the previous argument to H to conclude that H C C \ D(x), and hence that H = G. (ii) This property follows immediately from property (i) if we notice that the intersection of the unbounded components of C \ D(z) and C \ D(x + y) contains

the set of z such that Jz) > max(I'xlI, JJx + y1l). (iii) Let a 34 0, a E Spx \ D(x), and let p be its associated projection, which is in .1. We have

Spi =. Sp(x - xp) C Sp(z - xp). By the Holomorphic Functional Calculus, a V Sp(z - zp), and thus a f Spi. Hence Sp i C D(x) U {0). Taking A V Sp i, we have

Spi-A=Sp(x--Al)CD(z-A1)=D(x)-A, and so Spi C D(x). In particular, we have (Spi)'C D(x)^. Denoting by 19,, the external boundary of a set, we have 8,D(z) = 8eD(x + y) and thus 8eD(z) C Sp(z + y), for every y E I. But by upper semicontinuity of the spectrum this inclusion is also true for y E 7. By Theorem 3.3.8, we have 0,D(z) C (Spi), hence D(x)^= ((9.D(x))'C (Spi) (Spi)"and the proof is complete. 0


110

Let H be a Hilbert space. Taking X E C(H) and y E .C(H), it is false

REMARK.

in general that D(x) = D(z + y). By inessential perturbations, some holes may appear. For instance on H = 1P(Z), taking the two weighted shifts

,ifn=-1 aen =

,ifn-r llx - y!;. By Theorem 3.3.5, we have dist(A, Sp z) = 1/p((Jl1-x)'1). But (Al-x)'' is normal so, by (iv), we have dist(A, Sp z) = 1/(((A1-x)'1((. Consequently Il(a1-x)Ì(x-y)(l < Il(A1 - x)-'II II(x - Y)II < 1, so P(x)P(x*,

Al - y= (Al-x)11+(Al-x)-'(x-y)} is invertible, which is a contradiction. Interchanging x and y we conclude that A(Spz,Spy) < lIx _. yII. 13


iso REMARK.

Property (iv) of Theorem 6.2.1 implies the uniqueness of the C'-algebra

norm.

Let A be a commutative C'-algebra and let x E A. For every character X we have X(z*) = X(x). COROLLARY 6.2.2.

PROOF.

Let x = At + ik, with h, k self-adjoint. If X is a character then X(h) E Sp It

and X(k) E Spk, so by property (iii) of Theorem 6.2.1, X(h) and x(k) are real numbers. Consequently X(x*) = X(x). 0

In the situation of a C'-algebra, Lemma 6.1.3 can be improved.

Let A be a C'-algebra and let B be a closed subalgebra of A stable by involution, oowtaining the unit 1. Then for every x E B we have COROLLARY 6.2.3.

SPAT = SPBZ. PROOF. O b v i o u s l y we have SPA X C SPB Z' So suppose that x E B and x is in-

vertible :n A. Then we must show that r is invertible in B. Now x' and xx' are also invertible in A. By Theoreaz 6.2.1 (iii), we have Sp;4(zx') C R \ {0}. Consequently C \ SpA(xx') is connected. By Corollary 3.2.14, SpA(rx') = SPB(xx*), so zz` is

invertible in B. Hence z'i = z'(xx')'n E B. 0 COROLLARY 6.2.4 (V.PTAX-J.ZEMANEK).

Let M = (aid) be a normal n x n

matrix and let r be the square root of E,_2 $asl an eigenvalne A of M sut that fail - Al < r.

E%2 jail 1z. Then there exists

PROOF. Let P be the projection having zero entries, except on the first line and the first column where each entry is 1, and let Q = I-P. Define N = PMP+QMQ. We have IIM - Nil = r so, by property (v) of Theorem 6.2.1, we have L(Sp M, Sp N) < r. But a1I E Sp Ni, so we get the result. 0

Let x, y be two ekmenls of a Banach algebra with involution A. Supposing that ry = yz, is it true ths* x'y = yz`? In general the answer is negative. But if A is a C'-algebra and if z is normal then it is true. This result can even be slightly generalized by the following result. THEOREM 6.2.5 (B.FUGLEDE-C.R.Pt'TNAM-MI.ROSENBLUM).

Let x, y, z be AJe-

inents of a C'-algebra A. Assume that r, y are t:nrmaJ and that rz = zy. Then

r'z = zy'.

Representation Theory and Spectral Theorem

121

From the hypothesis we conclude that x"z = zy" for all n > 1. So easz = zeA1' or z = e'asze'r, for all A E C. Consequently en`s*ze-av* = eas*'1:zelr'a9'* because x and y are normal. But by continuity of the involution, the two elements are unitary. So JIu1 JJ = JIuz!! = 1. Considering u1 = eJ:*'Is and u2 = ea:*ze-'r* we have IIf(A)II < IIUIzu211 < pxJJ. So by the analytic function f(A) = Liouville's theorem, f(A) is constant and equal to z. Hence ea=*z = zea' for all A E C and this implies x*z = zy*. 0 PROOF.

The next theorem is one of the most important results in spectral theory. It is in fact the key to the proof of the spectral theorem for normal operators on a Hilbert space. Surprisingly it says that the onl' commutative C*-algebras (with unit) are the algebras C(K). Let A be a commutative C*algebra and let 9931 be its set of characters. Then A is isometrically isomorphic to THEOREM 6.2.6 (I.M.GELFAND-M.A.NAIMARK).

By Corollary 6.2.2 we have (x*)- = (x")'. Then the Gelfand transform x -+ x- is a *-morphism from A into C(Wl). For any x E A we have 14xJJ = p(x) = JIllloo by Theorem 6.2.1 (iv) and Theorem 4.1.8. So x -+ x is an isometry. In particular it is injective. Because the Gelfand transform is an isometry, A is PROOF.

complete for 11 11., so it is closed in C(9931). It is also a subalgebra of C(Wl), stable

by conjugation, containing the constants (because 1 = 1) and separating the points of an. So by the Stone-Weierstrass theorem we conclude that A = C(993l). 0 If A is a C*-algebra without unit then 9931 is locally compact, but not compact.

In that case A is isometrically isomorphic to Cs(99f1), the algebra of continuous functions on '931 which tend to zero at infinity (see Exercise VI.3). Given an element x of a Banach algebra and f holomorphic on a neighbourhood of the spectrum of x we have seen in Theorem 3.3.3 that it is possible to define f(z) such that Sp f(z) = f(Sp x). If x is normal in a C*-algebra then it is possible to extend this holomorphic functional calculus to a continuous one. Even more, in §3 we shall extend it to bounded Borel functions.

Let A be a C*-algebra and let x be normal in A. For every function f continuous on Spx it is possible to define a normal element f(x) in A such that we have the following properties: THEOREM 6.2.7 (CONTINUOUS FUNCTIONAL. CALCULUS).

(1) the mapping f - f (x) is an algebra morphism from C(Sp.r) into A such that 1(z) = 1, 1(x) = z (where 1(A) = A) and f(x)* = f(x),

122


(ii) Sp f(x) = f(Spx), so in particular Ilf(x)II = sup{)f(A)l : A E Spx),

(iii) if (f,) converges uniformly to f in C(Spx) then ff(x) converges to f(x). PROOF. Let B be the closed and commutative subalgebra of A generated by 1, x, x*. It is stable by involution, so by Theorem 6.2.6, it is isometrically isomorphic

to C(91t), where 91? denotes its set of characters. We now show that 9Jt can be identified with SpA z. Let 4; be the mapping from 9)2 into Spgx defined by

By Theorem 4.1.2, it is surjective. If x1(z) = Xz(x), then by Corollary 6.2.2 X, (x*) = X2(x*), so by continuity Xi and X2 coincide on all B, that is X, = X2 By definition of the Gelfand topology, the mapping 4s is a continuous bijection from 9Jt onto SpBx. By Corollary 6.2.3, Spgx = SPBX. Since Mt is compact it is homeomorphic to Spgx. Let I/ denote the isometric isomorphism from C(Spx) onto B. For f E C(Sp x) we define

f(x) ='y(f) Property (i) is obvious. By Corollary 6.2.3 we have SPA 1(x) = Sp8f(x) _ f(SpBx) = f(SpAX). Consequently lif(x)H = p(f(x)) = sup{If(A)I : A E Spx}. Property (iii) follows from f(x)I) = sup{Ifn(A) - f(A)I : A E Spx}. 0 If m is a normal n x n matrix then it is well-known that it is diagonalizable and that the eigenspaces associated to different eigenvalues are orthogonal. In other words, if {A11... , Ak} are the different eigenvalues of m, we have m = F k, Aipi,

where the pi are self-adjoint orthogonal projections, that is they verify p? = pi,

p; = pi, for all 1 < i < k, pipj = 0 for i

and p, +

+ pk = I. In fact this

result follows from Theorem 6.2.7. COROLLARY 6,2.8. Let A be a C*-algebra and let x be a normal element of A ,having a finite spectrum {A,, ... , Ak}. Then there exist self ac{joint orthogonal projections pa,...,p, in the commutative closed subalgebra generated by 1, z, x* such that pi + +pk = 1 and x = E t Aipi

PROOF. On Sp x = { A,, ... , Ak } we define the k functions X, , ... , Xk by

X(A) =

1

ifA=Ai

0

if A 16 Ai

Obviously we have X;=Xi,X,_T7,XiXj=0fori96 E k=1 A;Xi = I, where I(A) = A, on Spx. Then we apply Theorem 6.2.7 to the pi = X,(x). 0


123

An element x of a C*-algebra A is said to be positive, denoted z > 0, if it is self-adjoint and if its spectrum contains only positive real numbers. For instance if h is self-adjoint then by Theorem 6 ? 1 (iii) we have Sp h2 C {A : A E R}, so

h2>0. The positive elements of a C*-algebra play an important role which will be illustrated by several of the following theorems. THEOREM 6.2.9.

Let A be a C*-algebra and let h be a self-adjoint element of A.

Then there exists a unique decompositon h = u - v such that u >- 0, v > 0 and

uv=vu=0. PROOF.

By Theorem 6.2.1 we have Sp h C R. On the real line th, function f (t) = t

can be written f = f+ - f- where f+(t) = max(t,0), f- (t) = iaax(-t,0) and

we

have f+ f - = 0. Then we apply Theorem 6.2.7 to h and the three functions f, f +, f to prove the existence of u and v. If we have another decomposition h = r - s with r > 0, s > 0 and rs = sr = 0, then hr = rh and he = sh, so r and s commute with all the elements in the C*-algebra generated by 1 and h. This implies that u, v, r, s commute in pairs. Let C be a commutative C*-algebra containing 1, u, v, r, a. Applying the Gelfand transform to C we lave u - v = r - s, u6 = Pi = 0 and u, 6, r", e > 0. This is only possible if u = r, 6 = e, so u = r and

v=s.0

THEOREM 6.2.10 (I.M.GELFAND-M.A.NAIMARK). Let A be a C*-algebra and

letz>0inA. Then there exists aunique yEAsuch that y'=zand y>0. Moreover y commutes with all the elements that commute with x.

PROOF. We have Sp z C 10, +oo[. We apply Theorem 6.2.7 to f(A) = A'12. Then f (z) is self-adjoint and satisfies f (z)2 = x by property (i). Moreover Sp f (x) C

(0, +oo[ so y = f (z) > 0. If ax = za, then by induction ap(x) = p(x)a for all

polynomials p, so a commutes with all elements of the subalgebra generated by 1 and x, in particular with y.. We now prove the uniqueness of y. Suppose that

z2 = z with z > 0. Then zz = z-' = zz so, by the previous argument, y and z

commute. Let C be the closed subalgebra generated by 1, y, z. It is a commutative C*-algebra containing x. Applying the Gelfand transform to C we conclude that

y=zsoy=z.0

If x > 0 this unique element y is called the square root of z and is denoted by xhl2.


124

THEOREM 6.2.11 (M.FUKAMIYA-I.KAPLANSKY).

Let A be a C*-algebra. Then

we have the following properties:

(i) ifu,vEA with u>Oandv>0then u+v> (ii) if x E A then xx* > 0, (iii) if x E A then I + xx* is invertible. PROOF.

Suppose u > 0. and v > 0. Then u + v is self-adjoint so, by Theorem

6.2.1 (iii), we have Sp(u + v) C R. Let a = [Hull, 13 = 11vil, so that Sp u C [0, a] and $p v C [0, /3]. Cc,nsequently we have

Sp(al - u) C [0, a] and Sp(fl1 - v) C [0, (#].

So we get 1+al - till = p(al - u) < a and 1131 - v[] _. p(,61 - v) < Q. Hence 11(a + S)1 - (u + v)II < a +,6. But we know that Sp(u + v) C R so, by the previous inequality, Sp(u + v) is contained in the closed disk of radius a + 0 with centre at a+Q and hence u+v > 0. We now prove (ii). Let y = xx* which is self-adjoint and let B be chosen .ts in Lemma 6.1.3 applied to y. By Theorem 6.2.6, B is isometrically isomorphic to C(9R) and of course we have SPA Y = Spg y = {X(y) : x E Wt). Then y is a real function on 9JI. We have to show that 0 on 91t. Since B = C(97t) there exists z E B such that

z=jyj-yOil 0J1.

(1)

Because z is real, by Theorem 6.2.7, z = z*. Let u = zx = h + ik with h, k self-adjoint. Then uu* = zxx*z* = zyz = z2y

(2)

u*u = 2h2 + 2k2 - uu* = 2h2 + 2k2 - z2y.

(3)

But h, k are self-adjoint so their spectra are real, consequently h2 > 0 and k2 > 0. Because y is real we have (z2y)' < 0 on R. Since z2y E B it follows that -z2y > 0. So (3) and (1) imply that u*u > 0. By Lemma 3.1.2, we have Sp(uu*) C Sp(u*u) U {0), so uu' > 0. This implies that (z2y)" > 0 on 9J1. With the previous argument we conclude that (z2y)' = 0 on 9J2, and this is possible only if (yj by (1). So y > 0. Property (iii) is obvious because -1 Sp(xx'). 0 Theorem 6.2.11 implies in particular that (x.r*)'/2 is well-defined for all x E A.


125

COROLLARY 6.2.12 (POLAR DECOMPOSITION OF AN INVERTIBLE ELEMENT).

Let A be a C*-algebra and let x be invertible in A. Then there exists a unique decomposition x = hu with h > 0 and u unitary. Moreover we have h = (zx*)1/2.

If x is invertible, so is xx*, hence Sp(xx*) C)0,+oo( Consequently, by Theorem 6.2.7 (ii), (xx*)1/2 is invertible. Let u = (xi*)-l/sx. It is also invertible and uu* = (xx*)J1 /2xx*(xx*)-112 = 1. So u-1 = u' and u is unitary. If x = kv with k > 0 and v unitary then xx* = kvv*k = k2, so by Theorem 6.2.10, we have k = (zx*)112 and v = u. D PROOF.

We now give a very beautiful application of this last result which implies a classical result due to B. Russo and H.A. Dye. THEOREM 6.2.13 (L.T.GARDNER). Let A be a C*-algebra. Denote by U the set of unitary elements of A. If f z[ < 1 there exist an integer n > 2 and n elements ul,... , un in U such that x = n(ul +--- + u,,).

PROOF. We fix u e U and take hall < 1. First we prove that there exist ul, us E U We have y= z(l+u-la)andj(u-'a`l < IHalD2 criP(U,)

(1)

.

c=1

Since each P(U,) is self-adjoint we have '(s)' = F(s). Let Vj,... , V. be another similar partition and let i be a simple function such that t = fli on Vi. Then V-1 fi(s)fi(t) = >2 rr,f9,P(U;)P(V5) = Z-1

n V,) = fi(st).

(2)

An analogous argument shows that fi is linear. If x, y E H, (1) leads to

(4'(s)zly) _ >2 a+(P(U;)xl y) _ E asP=,y(U;) =

JK

4 dP=,tr

So we get

III (s)xII2 = (fi(s)`fi(s)llz) = ($(ss)xlz)

=1

1.912

dPr,z .

K

Consequently

Ih I(s)zl 5 llsll.llzll,

(3)

by property (c) of the resolution of the identity. On the other hand, let i be chosen

such that jail = Ilsll. and let x be in the range of P(U;). Then

fi(s)x = a'P(U,)x = air

(4)


132

because the projections have orthogonal ranges. It follows from (3) and (4) that t141(s)II = 11311.

(5)

Now suppose f E Lm(P). There exists a sequence of simple measurable functions sk that converges to f in the norm of LOD(P). By (5), the corresponding operators -t(3k) form a Cauchy sequence in £(H), so they converge in norm to an operator

that we call O(f). It is easy to see that 4'(f) does not depend on the particular choice of the sk. Then (5) is also true for f E LOO(P). Properties (ii) and (iii), being true for simple functions, also hold with f E LOO(P). Finally if S commutes with every P(U), it commutes with Z(s) whenever s is simple, and consequently, by continuity, with every member of O(L°°(P)). 0 By convention we set

4'(f)=JKfdP. Let A be a closed, commutative selfacUoint subalgebra of £(H) which contains the identity and let K be the set of THEOREM 6.3.2 (SPECTRAL THEOREM).

characters of A. Then we have the following properties:

(i) there exists a unique resolution of the identity P on the Borel subsets of K such that

T=I TdP x

for every T E A, where T is the Gelfand transform of T,

(ii) P(U) 34 0 for every non-empty open subset U of K,

(iii) S E £(H) commutes with every T E A if and only if S commutes with every projection P(U). PROOF. By Theorem 6.2.6, T -+ T is an isometric *-isomorphism from A onto C(K). Let z,y E H be given. Then T -+ (Ts $y) is a bounded linear functional on C(K), of norm less than or equal than JJrJJ JJyll. By Theorem 1.1.7, there exists a unique regular complex measure ps,# on K such that

(TxIy) = I T dps,,,, for T E A. K

(6)

When t is real, T is self-adjoint, hence ps,,, = p,,,, . By (6) and the uniqueness of p,,, we conclude that ps,,,(U) is, for every Borel set U C K, a sesquilinear functional. Since JJp.,,JJ < JJ=JJ JJy1J1 it follows that

r JK'

(7)


133

is a bounded sesquilinear functional on H, for every bounded Borel function f on K. Consequently there exists an operator 4'(f) E .C(H) such that

(`Z'(f)xly) = fK f dux,y, for z,y E H.

(8)

By (6) we have 'Y(T) = T, so T is an extension of the mapping f -+ T from C(K) onto A. If f is real then (4'(f )x ly) is the complex conjugate of (*(f)VJz), so T(f) is self-adjoint.

We now prove that '(f g) ='P(f)'I'(g), for bounded Borel functions f and g. If S, T E A then (ST )^ = S"T . So we have

IK

ST dp,,V = (STzly) =

SdµTx,9 .

(9)

JK

Since A = C(K) we have T doss = uT,,,. Consequently formula. (9) remains true if we replace S by f . Then

Ix fTdp ,r = IK f dµTz,r = (W(f)Tzly) = (Tx z) ='KTdIA x,:

(10)

where z = I(f )'y. Once again we can replace T by g in this formula so the assertion is proved.

We now define the resolution of the identity P. If U is a Borel subset of K we set

P(U) = $(xu)

(11)

It is easy to verify that this is a resolution of the identity. Moreover (P(U)xly) _ µ.,,(U) so, by (6), we have

T=

J TdP.

(12)

K

We now prove the uniqueness of P. The regularity of the complex Borel measures P1,,, shows that each Ps,5, is uniquely determined by (6). This follows from the uniqueness assertion in Theorem 1.1.7. Since (P(U)zly) = P1,11(U), each projection P(U) is also uniquely determined by (6). So the proof of (i) is complete. Suppose

now that P(U) = 0, for U open. If T E A and t has its support in U then, by (12), ve have T = 0, and hence t = 0. By Urysohn's lemma, U must be empty. So (ii)

134


is proved. We choose S E £(H), x, y E H and we put z = S"y. For any TEA and any Borel subset i1 of K we have r

(STxIy) = (Txlz) = KTdP,,, , (TSxly) =

JK

!Ws=,y ,

(13) (14)

(SP(U)xly) = (P(U)xlz) = P=,=(U),

(15)

(P(U)Sx1y) = Ps:,y(U)

(16)

If ST = TS for every T E A, the measures in (13) and (14) are equal, so that SP(U) = P(U)S, by (15) and (16). The same argument proves the converse. This completes the proof. 0 The reader must keep in mind that, in general, the projections P(U) are outside the algebra A except, of course, if A = £(H). It is a classical result in matrix theory that a commuting family of self-adjoint n x n matrices can be simultaneously diagonalized. It is easy to see that this result is a corollary of Theorem 6.3.2. A much more important theorem is the following. THEOREM 6.3.3 (SPECTRAL THEOREM FOR NORMAL OPERATORS).

Let T E

£(H) be a normal operator. Then there exists a unique resolution of the identity P on the Borel subsets of Sp T which satisfies

T=J

AdP(A).

(17)

Sp T

F1urthermore every projection P(U) commutes with every S E £(H) which commutes with T. Let A be the smallest closed subalgebra of Z(H) generated by 1, T, T*. It is a commutative self-adjoint subalgebra of £(H). By Corollary 6.2.3, SPAT is equal to Sp T, the spectrum of T corresponding to £(H). So, by Theorem 6.2.6, A is isometrically isomdrphic to C(SpT). The existence of P follows from Theorem 6.3.2. We now prove the uniqueness. If P exists and satisfies (17), by Theorem PROOF.

6.3.1, we have

P(T,T`) =

JSp T

P(A, a) dP(A)

(18)

for an arbitrary polynomial in two variables with complex coefficients. By the Stone-Weierstrass theorem, these polynomials are dense in C(SpT). Consequently the projections P(U) are uniquely determined by the integrals (18), hence by T. If TS = ST then, by Theorem 6.2.5, we have T*S = ST*, so S commutes with all the elements of A. Consequently S commutes with the projections P(U). D


135

We shall refer to this P as the spectral decomposition of T. The previous arguments can be summarized in order to extend the Continuous Functional Calculus to bounded Borel functions. THEOREM 6.3.4 (SYMBOLIC CALCULUS FOR NORMAL OPERATORS).

Let T E

£(H) be a normal operator. If f is a bounded Borel function on SpT we set

f(T) = /

spT

f dP,

where P is the unique resolution of the identity associated to T by Theorem 6.3.3. The mapping f - f (T) has the following properties:

(i) it is an algebra morphism from the algebra of all bounded Borel functions on

SpT into .(ht) such that 1(T) = 1, I(T) = T (where I(A) = .1 on SpT) and f(T)* = f(T), (!i) IIf(T)II < sup{I f(A)I : A E SpT}, the equality being true if f E C(SpT),

(iii) if f converges uniformly to f, then f(T) (iv) if S E J2(H) and TS = ST, then f(T)S = Sf(T) for every bounded Bore] function f ,

(v) T is the limit in norm of linear combinations of projections P(U). If the normal operators T sit in some closed self-adjoint subalgebra A of C(H) we know, by Theorem 6.2.7, that for f continuous, the f (T) are normal operators in A. But if f is a bounded Borel function we can only say, by Theorem 6.3.4, that the f(T) are normal operators in the bicommutant A" of A, that is the set of operators commuting with all the operators commuting with A. In general A" is much larger than A.

§4. Applications It is not true that every T E L(H) has a polar decomposition (See.Exercise VI.18). But we extend Corollary 6.2.12 to arbitrary normal operators. Let T E E(H) be normal. Then it has a polar decomposition T = PU where P is positive, U is unitary, and P and U commute with each other and with T. THEOREM 6.4.1.

PROOF.

Let p and u be the two bounded Borel functions on SpT defined by

p(.1) = JAI, u(A) _

if A # 0 and u(0) = 1. Using Theorem 6.3.4 we put P = p(T)


136

and U = u(T). By construction P is self-adjoint. The relations AA = 1A12, uii = 1 and A = p(A)u(A) imply that p2 = TT*, consequently P = (TT`)112, UU` = I and T = PU. So the result is proved. 0 THEOREM 6.4.2.

Let U E £(H) be unitary. There exists Q E £(H), self-adjoint,

such that U = e'Q Since U is unitary, its spectrum lies on the unit circle. Consequently there exists a real bounded Borel function f on Sp U that satisfies exp(i f (A)) = A, for A E Sp U. We apply Theorem 6.3.4 and take Q = f (U). Obviously Q is self-adjoint and verifies U = e'Q. 0 PROOF.

The group of all invertible operators in £(H) is connected and every invertible operator is the product of two exponentials. COROLLARY 6.4.3.

PROOF.

Let T be invertible in £(H). By Corollary 6.2.12, we have T = PU,

where P = (TT * )112 is invertible and U is unitary. Because SpP C ]0, +oo(, the logarithm is a continuous real function on Sp P, and so by Theorem 6.2.7 we have P = es, for some self-adjoint S E £(H). By Theorem 6.4.2, we have U = e'Q for some self-adjoint Q E £(H). So T = ese'Q. The continuous function t -+ else"Q, with 0 < t < 1, defines a continuous arc from I to T. So the group of invertible elements is connected. 0

It is natural to ask whether every invertible T E .£(H) is an exponential or, equivalently, if the product of two exponentials is an exponential. This is true for finite-dimensional algebras (see Remark following Theorem 3.3.6), but false in general (see (81, pp.317-319).

A related problem is to know if the set of exponentials is open in general. A negative answer follows from J.B. Conway and B.B. Morrel, Roots and Logarithms of Bounded Operators on Hilbert Space, Journal of Functional Analysis 70 (1987), pp.171-193, where they prove that T is in the interior of the set of exponentials if and only if 0 is in the unbounded component of Sp T. LEMMA 6.4.4.

Let A be a

and let I be a left ideal of A (reap. right

ideal). Denote by A the set of finite subsets of I, ordered by inclusion. There exists family (Uo),,EA such that

(i) U. E I, ua > 0, )IuQII < 1,

(ii) lima xua = x, for each x E 1 (resp. lima uax = x).

137


PROOF.

For o = {x,.. . , x, } E A we put and ua = va (va

V. = x1*xl + ... +

1

'

Because the real function t -+ t (t + ?) -1 takes its values between 0 and 1, fort > 0, we have ua>0and 0 0, consequently

(xi(ua - 1))*(xi(ua - 1)) < E (xi(ua - 1))*(xi(ua - 1)) < 4n i=1 So we have 11xi(ua - 1) 112 < -L. Consequently we have lima zua = x, for each z E 1, and this is also true for the elements of I. 0 Let A be a C*-algebra and Jet I be a closed two-sided ideal of A. Then I is stable by involution and A/I is a C*-algebra with the standard involution and the quotient norm. THEOREM 6.4.5.

PRooF. let

Let (ua)OEA be a family having the properties given in Lemma 6.4.4 and We have lluax* x*II = Ilzua - XI(.

-

But uaz* E I, and consequently x` E I. To prove that All is a C*-algebra it is sufficient to prove that III=1112 - (III'xIII We first prove that So lima uax* = x

IIIi(II = lima 11x - xual(. If Y E I we have lima yua = y, so

limsup Ilx-xuall = limsup IIz-zua+y-yua(( = limsup I((x+y)(1-ua)11 < l(x+y(I, a

a

a

because III - uall 5 1. Consequently

IIIxIII > lim sup IIz - zua11 > li minf ((z - zua(l > in f I(x + !III = (1Ix(II


138

So the assertion is proved. Now for every z E I we have 11Ix1112 = liom Ilx - xuall2 = 1)m Il (x - xua)'(x - xua)II

x'xua - uox'x + uox'xuall

= lim a

= 1)m 11x'x + z - zuo - x+xuo - uax`Z - uaz + uaZU0 + uax*xuall

= lim 11(1 - ua)(x*x + z)(1 - ua)ll < llx'x +

Z11.

a

Hence III=1112 < III="x111.17

Consequently the Calkin algebra £(H)/U_(H) is a C'-algebra. We finish with a small result related to Markov covariances. THEOREM 6.4.6.

Let a be in a C'-algebra A. Then lie-'all < 1, for all t > 0, if

and only if a+a'>Oand Re Spa>0. PROOF.

First we prove that the two conditions are necessary. Suppose t > 0. By

Exercise 111.15, we have n

But

1, and consequently p(e-t(°+°'))

= lle-t(a+` )11 < 1 .

The spectrum of a+a' being real, we conclude that a+a' > 0. Moreover II a-t° 11 < 1. So by the Holomorphic Functional Calculus, Re Spa > 0. We now prove that the two concitions are sufficient. We have t

1-

f Je-°*(a. + a)e-a ds > 0

(1)

by hypothesis. S.) IIe-t4112 = IIe-We-t°11 < 1 and then I'(t) = 11e-,all < 1, for all t > 0. This implies, in particular, that 1'(t) is decreasing on (0, +oo). In order to prove the result we have to prove that r cannot be equal to 1 on some interval

(0, uJ, u > 0. Suppose r(t) = 1 for 0 < t < u. By Theorem 6.2.17 there exists an extreme state o on A such that 1. Because of continuity and positivity of the extreme state, we conclude that o(e-11*(a + a)e-'°) = 0 for 0 < s < u, so o(e-We-94) = 1 for 0 < t < u. Applying the Identity Principle to


the entire function Hence

139

we conclude that o(e-" 'e-'a) = 1 for all t > 0.

1 = o(e-eo'e-re)

1,

so file -all = 1, for all t > 0. Then by Theorem 3.2.8 we get P(e-ca) = lim lie

n-oo

'1I" = 1,

a contradiction. The proof is now complete. 0

140

EXERCISE' 1.


Prove that A(&) and C(K) are not C*-algebras.

Consider M.(C) with the involution m* defined by the conjugate of the transpose Kim, for m E M (C). Determine explicitly the C*-algebra norm on EXERCISE 2.

Prove that a commutative C*-algebra without unit is isometrically isomorphic to Co(lt). EXERCISE 3.

Let A be a C*-algebra without unit. Prove that there exists a unique C'-algebra norm on the Banach algebra with unit A = A X C. EXERCISE 4.

EXERCISE 5.

Prove that a C*-algebra is semi-simple.

EXERCISE 6. Let A, B be two C*-algebras and let F be a *-morphism from A onto B. Prove that 4L is isometric if and only if 0 is injective.

Let A be a finite-dimensional C*-algebra and let x E A be such that jJ2lj < 1. Prove that there exist two unitary elements u1,u2 such that x

*EXERCISE 7.

EXERCISE 8. Let r be the unit circle and let u be the identity function u(z) = z in C(I'). Prove that u is unitary but cannot be written as e'h, for some continuous real function h defined on r. **EXERCISE 9.

Prove that in Theorem 6.2.13 and Corollary 6.2.14, U can be replaced

by E. First use Theorem 6.2.13, then use the fact that r2,.

x = 2xi f

(Al + a)(1 + Ax*)-' d)

0

for x normal, UUxll < 1, JAI = 1, and that (Al + x)(1 + is unitary with Sp(A1 + z)(1 + Az*)'' not meeting the half-line {-a) - a > 0}. If you do not succeed look at [1], pp. 109-113 where you will find some suggestions. Let A be a C*-algebra for a norm 11 11 and let I ( be another Banach algebra norm on A such that je's j = 1, for all self-adjoint elements h. Prove that the two norms coincide. This result is improved upon in the following difficult exercise. EXERCISE 10.

***EXERCISE 11.

Let A be a Banach algebra with involution such that 0e'"1+ = 1, for

all self-adjoint elements h. Prove that A is a C*-algebra for this norm. If you do not succeed look at [1[, pp. 122-123.


EXERCISE 12.

141

Let A be a separable C'-algebra. Prove that there exists an iso-

metric *-morphism from A into £(I2).

Let A be a C'-algebra. Suppose that there exists on A another x-* such that IIzx#li = 11x112, for all x E A. Prove that x' = x#, involution x for all x E A. Conclude that two C'-algebras are isomorphic if and only if they are EXERCISE 13.

*- isomorphic.

Let A be a C'-algebra and let x E A. We define V(x) = {s(x) : z E 6), where 6 is the set of states. Prove that V(x) is a compact and convex set containing the spectrum of x. If x is normal prove that V(x) = coSpx. Give an EXERCISE 14.

example showing that in general V(x) # co Sp x.

Let T be a quasi-nilpotent bounded operator on a Hilbert space H. Suppose that TT= is compact. Prove that T is compact. EXERCISE 15.

EXERCISE 16.

Prove that a normal operator T E E(H) is compact if and only if

it satisfies the following two conditions:

(i) Sp T has at most 0 as a limit point,

(ii) if A # 0 then dim 9(T - AI) is finite. EXERCISE 17.

Let T E E(H) be normal and compact. Prove that T has an

eigenvalue A with IAI = IITII and that f(T) is compact if f E C(SpT) and f(0) = 0. To prove this last property use Mergelyan's theorem (see [7], p. 423). EXERCISE 18.

On 12 prove that the right shift, with weight of. = 1, has no polar

decomposition.

Let A be a C'-algebra and let B be a Banach algebra with involution. Suppose that 4' is an injective s-morphism from A into B. Prove that EXERCISE 19.

114'(x)11 ? NxII, for all x E A. This result can be improved by the following exercise which is a generalization of Exercise IV.11.

Let A be a C`-algebra and let B be a Banach algebra. Suppose that 0 is an injective morphism from A into B. Prove that there exists C > 0 such that 114'(x)11 > CIIxfl, for all x E A. This result is due to S.B. Cleveland. Using an idea of A. Rodriguez Palacios it can be proved with the help of Theorem 5.5.1.

***EXERCISE 20.

Chapter VII AN INTRODUCTION TO ANALYTIC MULTIFUNCTIONS

As we explained in the Preface this chapter will be a quick incursion, as the crow flies, into the important new field of analytic multifunctions. A complete treatment would need a full book. In particular it would be necessary to recall the great number of results we need in the theory of functions of several complex variables. Of course we are not able to do this, in such a limited number of pages, so we shall suppose that the reader is familiar with all these prerequisites, which are contained for instance in [5,9] or in many other books on the field, and which are summarized in the Appendix, §2. In the last ten years the use of subharmonic functions in spectral theory and in the theory of uniform algebras has given a lot of interesting new results (see Chapters III and V for spectral theory; Exercises VII.8-9-10 and [101 for the theory of uniform algebras).

In the 1980s the important analytic tool of analytic multifunctions'came back to life. The origin of this concept goes back to some work by K. Oka related to singularity sets in C2, and in some sense, has its roots in the work of Charles Puiseux, a student of Augustin Cauchy, who obtained, in relation with algebraic geometry, several interesting results about algebroid functions. Of course K. Oka never gave applications to functional analysis.

This concept was resuscitated in 1980 for the following reasons. In 1977 the author introduced the following conjecture: if f is an analytic function from a domain D of C into a Banach algebra and if the set of ,\ for which Sp f (A) is at most countable has a positive capacity, then Sp f (A) is at most countable for all A in D. He also explained how this conjecture would solve another one (a generalization of a conjecture due to A. Pelczynski concerning C'-algebras), which can be expressed in the following way: let A be a Banach algebra with involution, and suppose that

Analytic Multifunctions

143

Sp h is at most countable for all h = h*, then Sp x is at most countable for all x in A, and A has a "particular" algebraic structure. Another reason for this revival is the discovery of the strange parallelism between Theorem 3.4.25 and the famous E. Bishop's theorem on analytic structure ([10], Chapter 11). In 1977, John Wermer and the author were able to extend Bishop's theorem using the subharmonic technique exploited in the proof of Theorem 3.4.25 (see Capacity and Uniform Algebras, Journal of Functional Analysis 28 (1978), pp. 386-400).

So, behind the resemblance of spectral theory and theory of fibres on a uniform algebra, new objects to discover were hidden: the analytic multifunction.

§1. Definitions and General Properties The convenient definition was discovered by Z. Slodkowski (Analytic Set- Valued

Functions and Spectra, Mathematische Annalen 256 (1981), pp. 363-386). In fact, in his paper, Z. Slodkowski gave two definitions, the first one being valid only for analytic multifunction from C into C, the second one for analytic multifunction from C" into Cm, and he proved their equivalence for m = n = I. We shall use the second definition as it leads to interesting results very quickly.

The majority of results and ideas mentioned in this section are due to B. Aupetit, T.J. Ransford, Z. Slodkowski, H. Yamaguchi and A. Zralbi. DEFINITION.

Let K be a mapping from an open subset U of C" into tae set of

non-empty compact subsets of C'". We shall say that K is an analytic multifunction on U if it satisfies the following properties:

(i) K is upper semicontinuous on U, (ii) for every relatively compact open subset V of U and every function 0 plurisubharmonic on a neighbourhood of the restriction of the graph of K to V, the function y5(A) = max{qS(z): z E K(A)}

is plurisubharmonic on V.

A function is plurisubharmonic if and only if it is locally plurisubharmonic, and consequently a multifunction is analytic if and only if it is locally analytic. In (ii) we may suppose that 0 is a C°°-strictly plurisubharmc is function since ¢ can be approximated by such functions.


144

EXAMPLES.

(1) If h is holomorphic on D C C", with values in C', then A '- {h(A)} is an analytic multifunction on D. (2) Let Ko, K1 be two compact subsets of C". For A E D C C, the multifunction K(A) = AKo + KI is analytic on D. (3) Let f be an analytic function from D C C into M"(C). Then A i-. Sp f (A) is an analytic multifunction on D (see Exercise VII.2). Much more important results will be given later. An interesting class of multifunction is the following.

An upper semicontinuous multifunction K from D C C" into Cm is said to have holomorphic selections if, for each Ao E D and each zo E 8K(Ao), there exists h holomorphic on a neigbourhood U of Ao, with values in Ct,'such that DEFINITION.

zo=h(A0)andh(A)EK(A),forAEU. Such multifunction having holomorphic selections are continuous analytic multifunction (see Exercise VII.3). Examples (1) and (2) are in this class, but not (3) globally on D because of the branching points. The following theorem has a very technical proof which will not be given here. We refer the reader to the papers of B. Aupetit, B.Aupetit & A.Zraibi, Z. Slodkowski, and T.J. Ransford. THEOREM 7.1.1.

The following properties hold.

(i) If (K,) is a sequence of analytic multifunction defined on D C C" with values

inC'",such that K,1(A)c K,(A) for each A ED, then K=fly ,K, is an analytic multifunction from D into C'.

(ii) If K1i ... , K, are analytic multifunction from D C C" into Cm then K = KI U ... U K, is an analytic multifunction from D into C'".

(iii) If K is an analytic multifunction from D C C" into C' and if L is an analytic multifunction from G C C' into Ck, where G is an open set containing all the K(A) for A E D, then L o K defined by (L o K)(A) _ {L(z): z E K(A)}

is an analytic multifunctiun from D into Ck. (14-) If K1,.. . , K. are analytic multifuuctions from D C C" into Cm, then K = K1 x x K. is an analytic multifunction from D into Cmp. -

(v) Let K be an upper semicontinuous multifunction from D C C" into C'. Then K is an analytic multifunction on D if and only if t K(at + b) is an analytic rnulrifunction ou {t: t E C, at + b E D), for every a, b E C".

Analytic Multifunetions

145

Let K be an analytic multifunction from L C C" into CTM and suppose that L is an upper . semi continuous multifunction from D into Cm such that OL(A) C K(A) C L(A), for each A E D. Then L is an analytic multifunction. In particular K" is an analytic multifunction. THEOREM 7.1.2.

PRooF. Let Dl be open in D and let 0 be plurisubharmonic on a neighbourhood

of the graph of L restricted to D1. If A E Di then we have, by the Maximum Principle, O(A) = max{4(A, z): z E L(A)} = max{o(A, z): z E 8L(A))

max{4(A,z):z E K(A)} = 01(A) < max{4(A,z):z E L(A)). So r('(A) _ 01 (A), and 01 is subharmonic by hypothesis. Obviously K" satisfies the two inclusions, and it is easy to verify that it is upper semicontinuous. D

Let K be an analytic multifunction from D C C into C. Then log p(K(A)), A - log b"(K(A)) for n > 1, and A .-- logc(K(A)) are subharA monic on D (where c denotes the capacity). THEOREM 7.1.3.

PROOF.

For the first function, this is immediate from the definition using the

plurisubharmonic function ¢(A, z) = log Izj. For the second one, let O(Ai zii ... i zn+1) = n(n + 1)

u

1 0 be such that

fw-wo' a}. Then Ea = U07,so which is a contradiction. Thus E, = 0, E., = E,lal, and consequently A E which means that 7(k) < 7 for all A E D, and so D7+'K(A) = D7K(A). 0


157

The classical Cantor-Bendixson theorem says that every closed subset of C is the disjoint union of a perfect set and an at most countable set. It can be generalized in the following form: COROLLARY 7.2.6. Let K be an analytic multifunction from a domain D C C into C. Then for each A E D, K(A) is the disjoint union of two sets L(A), M(A) such that:

(i) either L(A) = 0, for all A E D or L is an analytic multifunction from D into C such that DL(A) = L(A), for all A E D, (ii) M(A) is at most countable for all A E D. PROOF.

Let y < w1 as in the statement of Theorem 7.2.5 and 14-t L(A) = D"K(A),

M(A) = K(A) \ L(A). By Theorem 7.2.5 and the definition of y, part (i) is true. For every A E D, D°K(A) \ D°-1 K(A) consists of isolated points, so it is at most countable. Then L(A) is a countable union of at most countable sets, hence it is at most countable. 0 Let K be an analytic multifunction from D C C into C and let F be a closed subset of D having non-zero capacity. Suppose that A E F implies K(A) at most countable. Then there exists A0 E F such that DK(Ao) # K(Ao). THEOREM 7.2.7 (B.AUPETIT-J. ZEMANEK).

Let (U, be a countable base of C. We introduc , the set F° = F \ c(FnU.) = 0). By Appendix A.1.21, c(F°) > 0. Sup)ose that DK(A) _ K(A) for A E Fc. Then, by Theorem 7.1.7, K(A) is finite for all A E D, and so DK(A) = 0 which gives a contradiction. Hence there exists Al E F` such that K(A1) is infinite. But K(A1) is countable and so is not perfect., and consequently it contains an isolated point zo. The same argument with K(A1) \ {zo), which is compact and countably infinite, shows that K(A1) contains at least two isolated points zo, z1. For i = 0,1 we choose open disks Di with centres at z, having disjoint closures and such that K(A1) n 3, = {z;). We then choose 0 < r < 1/2 such that ff(A, i r) C D and such that JA - Al I < r implies A'(A) n Mi = 0, for i = 0,1. Because DK(A1) = K(A1), each z, is isolated in K(A1) but is not a good isolated point. Applying Theorems 7.1.5 and 7.1.7, we conclude that the two sets PROOF.

E; = JA: JA - Al I < r, K(A) n A, finite) have capacity zero. But F` n B(A1, r) does not have capacity zero, by definition of F, and E0 U E1 has capacity zero, so there exists A2 E F` n B(A1, r) such that K(A2) n A, is infinite, for i 0, 1. As before we can find four distinct isolated points in K(A, ), say zoo, zo 1 in 0(, and z1 o, z11 in A1.

A Primer on Spectra! Theory

158

So take four open -iisks Ai i centred respectively at zi having disjoint closures such that Do o U Do 1 C Do,

A10UO11 COI, K(.\2) n [Ti j = {z11}.

By induction we can construct a sequence A" E F` such that

(1) IAn+1 - A"I < 2-", n = 1,2,- - -,

(ii) K(A,,) contains at least 2" distinct isolated points zi,,,.i. where ik takes the values 0,1,

is the centre of an open disk A,,..... i,,, with the property that all

(iii) each

these 2" disks have disjoint closures and Ai,,...,C Now (A") is a Cauchy sequence in the closed set F, so converges to some A0 E F.

To obtain a contradiction we show that K(Ao) is uncountable. Let I = (i1, i2, ...) be an arbitrary sequence of Os and Is. Since (zi, , zi, 61 zi, i, is , ...) is bounded, it contains a subsequence converging to zi which is in K(A0) because K is upper semicontinuous. If f 3L J then there exists a smallest integer k such that ik #.7k, so we have

zI E

is

zJ E

and the two disks are disjoint by construction, so zi # zJ. But the set of sequences I is uncountable, so K(Ao) is infinite and uncountable. This is a contradiction. 0 THEOREM 7.2.8 (SCARCITY THEOREM FOR COUNTABLE ANALYTIC MULTIFUNCTIONS). Let I( be an analytic multifunction from a domain D C C into C. Then

either the set of A, for which K(A) is at most countable, has capacity zero, or K(A) is at most countable for all A E D. In the latter situation there exists -1 < w1 such that DIK(A) = 0, for all A E D. PROOF.

Suppose that the set of A for which.K(A) is at most countable does not

have capacity zero. Then there exists a compact set F C D such that c(F) > 0, and such that K(A) is at most countable for A E F. By Corollary 7.2.6, K(A) is the disjoint union of L(A) and M(A), with either L(A) = 0 for A E D, or L analytic multivalued with DL(A) = L(A) for all A E D. In the latter situation L(A)

is at most countable on F so, by Theorem 7.2.7, there exists A0 E F such that DL(Ao) 96 L(Ao), which is a contradiction. So L(A) = D7K(A) = 0, for all A E D. Moreover K(A) = M(A) is at most countable on D. 0


159

REMARK. This result is the best possible. Let F be a compact set having capacity zero. By Evans's theorem (Theorem A.1.24), there exists u subharmonic on C, such that F = (A: A E C, u(A) = -co). We define the multifunction K by

K(A) = {z:z E C,lzl < e'(a)}. It is an analytic multifunction defined on C which satisfies K(A) = {0} on F and which is uncountable on C \ F. We now give the solution to the General Pelczyiiski Conjecture (first mentioned

in ill, p.86), the problem which was, in fact, the main motivation behind the introduction of analytic multifunctions.

Let A be a Banach'algebra containing an absorbing set U such that Sp.r is at most countable for all x E U. Then the spectrum of every element of A is at most countable. If moreover A is separable, then it satisfies the properties THEOREM 7.2.9.

of Theorem 5.7.9. PROOF. The argument is similar to that at the beginning of the proof of Theorem 5.4.2, except that we use Theorem 7.2.8 instead of Theorem 3.4.25. 0

Let A be a Banach algebra with involution. Suppose that the real vector subspace H of self-adjoint elements contains an absorbing subset U such that Sp h is at most countable for all h E H. Then every element of A has an at most countable spectrum. COROLLARY 7.2.10.

PROOF. The argument is a slight modification of that used in the proof of.Corollary 5.4.3, replacing Theorem 3.4.25 by Theorem 7.2.8. 0 We now give an application of the Oka-Nishino theorem to the Identity Principle.

It is easy to see that Theorem 3.4.26 can be paraphrased, with an almost identical proof, in the following manner. THEOREM 7.2.11.

Let K be an analytic multifunction from a domain D of C into

C. Suppose that for all A E D the set K(A) has at most 0 as a limit point. Let z 34 0 be given. Then either Z = {A: A E D, z E K(A)) is a dosed discrete subset of

D or it is ailD. The same argument even proves the following.


i6o

Let K be an analytic multifunction from D into C and let z E C be fixed. Then every point of the set COROLLARY 7.2.12.

Z={A:AED,zEK(A)\DK(A)} is either isolated or interior. If K is a countable analytic multifunction, the analogue of Theorem 7.2.1 cannot be true. For instance, let Ko = {l/n: n = 1,2,...}U{0} and let K(.\) = A+ K0, which is an analytic multifunction on C. Then Z = {A: 1 E K(.\)} is neither discrete nor C. Nevertheless we have the following.

Let K be an at most countable analytic multifunction from a domain D of C into C and let z E C be fixed. Then the THEOREM 7.2.13 (B.AUPETIT-J.ZEMANEK).

set

Z={.1:.ED,zEK(.\)} is either at most countable or it is all D.

By Theorem 7.2.8, there exists a smallest ordinal -y < wi such that D'YK(A) = 0 for all \ E D. Then Z = Uo 2.

An interesting problem would be to extend Theorem 7.2.13 to a larger class. For instance, suppose that K is an analytic multifunction from a domain D of C

Analytic Multifunction

261

into C such that K(A) has capacity zero, for all A E D (such an example is given by Sp f (A), where the operators f (.X) are quasi-algebraic in the sense of P. Halmos: see (2), pp. 251-253). Is it true that Z is a closed set of capacity zero? This problem is related to A. Sadullaev's result implying that-the graph of K is a polar subset of C2.

But is it completely polar? In a recent preprint entitled Pseudoconcave seta and algebraic lemniscates (in Russian), A. Saduliaev solves this last problem positively. If his arguments are correct then the first question else has a positive answer, but the proof is very difficult.

§3. Distribution of Values of Analytic Multifunctions The famous theorem of Picard asserts that a non-constant entire function takes

all the values of the complex plane except perhaps one point. But what happens for the union of all the spectral values of f (A) if f is an analytic function from C into M,, (C)? This problem was partly studied by E. Borel, G. Valiron and G. Remoundos, but their arguments are not always very convincing (even H. Cartan gave some insights on the general situation, but with a false conclusion on the number of exceptional points). In the first part of this section we shall describe the work of A. Zraibi on the solution of the previous problem with the help of Nevanlinna theory.

In the second part we intend to show the intimate connection between such analytic multifunctions and pseudoconvex open subsets of C2. This connection reduces many problems on analytic multifunction - and hence many spectral problems --- to purely geometrical problems on pseudoconvex sets. This geometrical idea gives a very simple proof of the generalization of Picard's theorem to arbitrary analytic multifunctions.

Let F be meromorphic for f zj < R:5 +oo, and let 0 < r < R. We define

m(r, F)

N(r, F) =

2a

f

log+ +F(reè)l d8,

jr n(t) - n(0)

dt + n(0) log r,

where n(t) denotes the number of poles, with their multiplicity, in the disk B(0, t), and

T(r, F) = m(r, F) + N(r, F). R. Nevanlinna proved the following inequality:


162

If F is merornorphic for IzI < R 0 be such that B(a,, e) fl B(ai, e) = 0 for 1 # j. Then there exist a > 0 and integers n1, ... , np such that #(K(A) fl B(a;, e)) = n1 for 0 < JA - Aol < a and i = 1, ... , p. LEMMA 7.3.3.

By Theorem 7.1.7, there exist an integer n and a closed discrete subset F F the lemma is obvious. So we suppose that Ao E F and let Y7 > 0 be such that B(A0, r)) n F = {ao } PROOF.

of D such that #K(A) = n on D \ F and #F(A) < n on F. If Ao

and B(Ao,ri) C D. Then K(ao) _ (01,...,a,} with p < n, and let e > 0 be such that B(a,, e) fl B(ai, e) 3k 0 for i i4 j. By the Localization Principle (Theorem 7.1.5), the multifunctions A -. K(.1) fl B(a,, e) are analytic multifunction on a disk B(Ao, a) with a < 77. So there exist integers ni and closed discrete subsets F, of B(A0, a) such that #(K(A)f1B(a,, e)) = ni on B(Ao, a)\F;, #(K(A)flB(a;, e)) < n;

on F,, and nl + - + n, = n. By upper semicontinuity we may suppose that K(A) is included in U°-1B(a;,e) for A E B(Ao,a), so we have #K(A) = n for A E B(.o,a) \ {,\o}. 0 The integer n; is called the spectral multiplicity of a;.

164


Let F(A,u) = u" + A1(A)u"-' + + A, (A) be defined on C2, where the A,(A) are non-constant entire functions. Then F has at most 2n - 1 exceptional values in the sense of Picard --- that is, for every u E C there exists .1 E C such that F(A, u) = 0, except perhaps for at most 2n - I values of u. LEMMA 7.3.4.

Let uo be an exceptional value. Then F(A, uo) is entire and has no zeros. Consequently either F(A, uo) is constant or F(A, uo) = c""), for some non-constant entire function h. PROOF.

First we suppose that F(A, u) has n distinct exceptional values u1,. .. , u,, such that F(A,ui) = ki, for all A E C. The identities

u' +A,(A)u;' f

u" +AI(A)un'"'

=kn

define a Cramer system for the unknowns A1(A) because the determinant is a Vandermonde determinant n-1

uI

u2-t

n-2

ul

tie

u1 un-2 n n

uI ...

.

U2

Un

1 1

30. 1

Consequently Ai(A) is constant for all i = ],_^ and this is a contradiction. Hence there exist at most n -I exceptional values tiI, ... , un+t for which F(A, ui) is constant. Now we suppose that vo, v 1 , . .. , vn are n + 1 distinct exceptional values such

that F(A, vi) = eh,(), where h, is a non-constant entire function. We have the following identities:

Aa(A)vu -' + ... + An(A) = ek-(A) -- vo

A, (A)v'-t+...+An(A)=e1" ' -vn. Let vn 0 U"I

vn-t

0

Vo

1

Un-' I

v,

1

U"

tart

#0.

Q

` I U"n

Analytic Multifunction

165

We have q = En 0(-1)'v; qi, where vn-I

vn-2 0

0

qj =

...

vo

tin-1 n-1

00

vi+ I

vn-2

Ivn-1

n

...

vn

is also a Vandermonde determinant. So we have

q=

k'(1)

E(-1)'q;

- ` E Ak(A)vn-k}

i=0

k=1 n

n

In

= E(-1)'q,ek'(X) __ t( -1)'q, i=0

i=1

( k=I

Ak(A)v; -k)

i=0

because the last term in the previous equation is eh0(a) - vn ek1(a)

-

0

,n1 1

I ek.(a) - v

vn-1 0

,n-1 1

v0

1

v1

1

=0. vn

1I

Because q; # 0, the determinant q can be written E 1 eM1), and the functions e9' are linearly independent. So by Lemma 7.3.2, we get a contradiction. Hence F has at most n exceptional values of this type, and so 2n - 1 exceptional values. 0 Let K be a non-constant analytic multifunction on C. Suppose that K( 1) is finite on a set E having a non-zero capacity. Then there exists a smallest integer n such that #K(A) < n for all A E C and C \ UAECK(A) has at THEOREM 7.3.5.

most 2n-l points. The first part comes from theorem 7.1.7. So outside F we have K(A) _ {aI(A),...,an(A)}, where the a; are. locally holomorphic. Let PROOF.

n

F(A, ii) = 11(ai(A) - u) = u" + A, (,\)u'-' + ... + An(A) for A i=1

F.

166

A PrinMr on Spectral Theory

This function can be extended analytically to all C2 by Lemma 7.3.3, counting each ai(A) with its multiplicity if A E F. The Ai(A) are well-defined in all C, and they are entire because they can be expressed as symmetric functions of the ai (in fact we use Rado's Extension Theorem). Moreover they are not all constant since K is not constant. So u is not in UXCCK(A) if and only if u is exceptional for F. We then apply Lemma 7.3.4. 0

This result is the best possible because, given arbitrary 2n - 1 distinct points, it is possible to construct a finite analytic nuiltifunctioe on C avoiding these points. Let a , ,- .. , a2,'- 1 be given distinct points, and consider the following analytic

function from C into M,,(C) defined by

/ C, ea + al f(A) _

\

...

(-1)n+lCnea

-C2eA

C3ea

I

as

0

0

0

1

a3

0

0 0

0

0

0

1

an

(-1)nCn-le'

We have [n-I

det(f(A) - z) _ (al - z)...(a. - z) + eA

Ci(ai+l

z)...(a. - z) + CfI

.

Let n-1

P(z) _

C1(ai+1 - z)...(an - z) + Cn.

i- i By induction it is possible to choose the constants C1,. .. , Cn such that we have

P(z) = (an+1 - z)

(a2.-l - z), and consequently

det(f(A) - z) = (a1 - z)...(a. - z) + el(an+1 - z)...(a2n-l - z).

Then the analytic multifunction A - Sp f (A) _ {z: det(f (A) - z) exactly the 2n - 1 points a1i ... , a2,1

0} avoids

We shall now be interested in improving Picard's Theorem when the analytic multifunction takes an infinite number of values. Theorem 3.4.14 and Theorem 3.4.15 can be paraphrased to obtain the following two results. THEOREM 7.3.6.

Let K bean analytic multifunction on C and let E = UAECK(A). Then the boundary of E is included in the boundary of K(A), for all A. In particular,

if K is bounded then K(A)' is constant.

167

Analytic Multifunction.

Let K be an analytic multifunction on C Then either K(A) - is constant or UxrCK(a)^ is dense in C. THEOREM 7.3.7.

A. Zraibi and the author obtained the following generalization of P` card's theorem to analytic multifunctions: if K is an analytic multifunction on C, then either

K'A)' is constant or the complement of the union of the sets K(A)- is a G6-set having zero capacity. The original proof uses Frostman's theorem and is rather complicated. We now intend to give an easy and more geometric proof. As explained in §1, Z. Siodkowski noticed that if K is an analytic multifunction then the complement of its graph is a pseudoconvex open subset of C2. Moreover, in an important paper (Analytic set-valued functions and spectra, Math. Ann. 256 (1981), pp 363--386), he discovered the striking fact that the theory of analytic inultifuncLions with values in C, the theory of fibres for uniform algebras, and spectral theory are locally equivalent. More precisely, if K is an analytic multifunction on an open subset U of C, with values in C, then for every relatively compact subdomain V of U there exists an analytic function from V into L(12) such that K(A) = Sp f(A),

for all A E V, and there exist a uniform algebra A on a compact set K and two elements f, g E A such that V C C \ f (K) and K(A) = g(f -1(A)) for all A E V. The following lemma will show that it is always possible to associate a lot of analytic multifunction to a pseudoconvex open subset of C2. So, in fact, the four theories of pseudoconvex open subsets of C2, analytic multifunction with values in C, spectra of analytic families of operators and fibres associated to uniform algebras are equivalent in the sense that any result in one of these theories will give new information in the other theories. This is one of the reasons why we believe that a deeper knowledge of the geometry of pseudoconvex open subsets of C2 will have a great impact on spectral theory.

Let f2 be a non-empty pseudoconvex open subset of C2 and let (.\o, a) E D. Suppose that D is the open set of A E C such that (A, a) E Q. Then the multifunction K defined on D by LEMMA 7.3.8.

K(A) = I z

a

} U {a}

is analytic.

It is obvious that K(A) is non-empty for A E D. The set K(A) is closed because if un E K(A), limun = u, then either u = a (so u E K(A)), or u 0 a and tim 1/(u-a) But (A,a+l/(un-a)) fl implies (A,a+l/(u- a)) 4 ft, so u E K(A). Moreover K(A) is a compact subset of C because if un E K(A) with PROOF.


168 Jim

+00, it follows from u = 1/(z - a), with (A, zn) 0 St, that Jim z = a,

so (,\,a)

St, which is a contradiction.

Let us now show that St1 = {(.\, z): A E D, z E C, z 4 K(A)} is pseudoconvex.

Because aEK(A)for all AED,we have (A,a)g01 for A E D, so St1={(A,z):AED,zEC\{a},(A, +a)EIt

z-a

//

11

u-'(12) n [D x (C \ {a)))

where u(z) = a + 1/(z - a) is analytic on D x (C \ {a}). Thus f21 is pseudoconvex because u-1(l) and D x (C \ {a}) are pseudoconvex. D THEOREM 7.3.9.

Let f) be a pseudoconvex open subset of C' and let U be a

domain of C such hat U x {0} C St. Then we have the following properties:

(i) either the set )f A E U such that U x {0} C ft is a G6-set of capacity zero, or

UxCCc, (ii) either the set of A E U such that (A) x C C St, except for a finite number of points, is a G; -set of capacity zero, or (U x C) fl ft is the complement of an analytic variety.

This is o )vious by applying Lemma 7.3.8 with a= 0, and Theorem 7.1.3, if we remark that j A) x C C St is equivalent to K(A) = {0} and that {A) x C C St, except for a finite dumber of points, is equivalent to K(A) being finite. 0 PROOF.

We are now able to give a generalization of Picard's theorem to analytic multifunctions. THEOREM 7.3.10 (PICARD THEOREM FOR ANALYTIC MULTIFUNCTIONS). Let

K be an analytic r iultifunction on C. If U is a component of C \ K(Ao ), for some Ao E C, then either U is a component of C \ K(A), for all A E C, or U \ UAECK(A) is a G6-set of capacity zero. In particular if we consider the analytic multifunction K then either Ki A)" is constant or C \ UAECK(A)" is a G6-set of capacity zero. Moreover, if K" is not constant and is not algebroid, then the set F of z for which {A: z E K(A)"} is tnite, is a G6-set of capacity zero. PROOF. We may :;uppose that Ao = 0. Then St = {(A, z): z V K(A)} is pseudoconvex, so the same is true for St' = {(z, A): (A, z) E St). Moreover U x {0} C ff. So we get the first part by applying Theorem 7.3.9 (1). Considering K(A)", the only

Analytic Multifunction,

169

component U is the unbounded one, so by Theorem 7.3.6 and Theorem 7.3.9 (i) we get the result. We now prove the lest part. Let u = C \ IIµECK(µ)" Because the intersection of C \ K(A)' and C \ K(µ)" is always non-empty, U is connected. Moreover F C U. So by Theorem 7.3.9 (ii), we conclude that either F is a G6-set having zero capacity or F = U. In the last situation, using the argument given in the proof of Theorem 7.3.4, we conclude that K' is algebroid. 0 Is this result the best one? Given a compact set C of capacity zero, is it possible to construct an analytic multifunction K on C such that C \ UAECK(A)" = C? Is it even possible to do this for K(A) = Sp f(A), where f is an analytic function from C into the algebra of compact operators on some Banach space? A. Zraibi obtained the following particular cases:

- If C is a subset of C not containing 0 and having at most 0 as a limit point, then there exists an analytic multifunction K such that C \ UAECK(A)" = C (see Exercise VII.12).

- If C is a compact subset of C of capacity zero then there exists an analytic multifunction K such that C \ UAECK(A) = C (but the problem is that K(A) has holes and the sets K(A)- cover all the plane!). It is interesting to note that Theorem 7.3.9 gives a new proof of Tsyji's theorem concerning the distribution of values of entire functions of two complex variables (M. Tsuji, Potential Theory in Modern Function Theory, Second Edition, New York, 1975). The original proof is complicated and uses conformal mapping.

THEOREM 7.3.11 (M. Tsuji). Let G(A, p) be an entire function on C2 which is not of the form G(A, p) = eH(a,P), with H entire on C2. Then there exists a G6-set E having zero capacity such that for p ¢ E there exists A in C satisfying G(A, µ) = 0. Moreover if G is not algebroid - that is there are no entire functions ar, ... , an such that G(A, C) = a"(µ)A" + + ar(µ)A + ao(p) - then there exists a G6-set F having zero capacity such that for µ F there exist an infinite number of A satisfying G(A, µ) = 0. PROOF.

Let U = {µ: G(0, p) 0 0}. If U = 0 then G(0,µ) = 0, for all µ, so

G(A, µ) = A'kH(A, µ), for some integer k > 1 and some entire function H for which H(0, µ) # 0. So we can reduce the general situation to the case where G(0, µ) 96 0. In this case, C\U is closed and discrete, so U is a domain. Let fl = {(A, µ): G(A, p) # 01. This is a pseudoconvex open subset of C2 because it is the complement of an analytic variety. Moreover {0) x U C 0. So, by Theorem 7.3.9, either the set of µ such that C x {µ} C fl is a G6-set of capacity zero, or C x { p) C S2 for all µ. Suppose


170

that we are in this last situation. Then G(A,p) 96 0 for A E C and A E U. In other words, G(A, p) = 0 implies p E C \ U. But for p fixed the set {A: G(A, p) = 0} is either discrete or G(A, p) __ 0 as a function of A. The first case implies that the zeros of G are isolated but, by Hartogs's result, this is impossible for an entire function of two variables. If now p V U implies G(A, p) = 0, we can suppose for instance that

0 0 U, and so G(A, 0) = 0. If we write G(A, p) = En°_o a"(p)A" we conclude that a"(0) = 0, so p divides a"(p) for all n. Hence there exists a greatest integer k such that G(A, p) = pkK(A, p) with K entire and K(A, 0) # 0. Then K(A, p) has isolated zeros on the line C x {0}, wich is a contradiction. So the first part of the theorem is proved. The proof of the last part is very similar to the proof of Theorem 7.3.10. 0

Given K an analytic multifunction on C and 0 < a < 1, it is easy to verify that A '-+ aK(A) + (1 - a)K(A) is analytic. This implies that A .-* coK(A) = Uo 0, lim,,._..,° r = 0 and 1

4(a) R} that is continuously extendable to the boundary and such that

w(z)=z+ao+ zl +zz +

,

near infinity.

Then c(K) = R. COROLLARY A.1.26. K K is a closed disk of radius R-then c(K) = R.


180

COROLLARY A.1.27.

If I is a dosed segment of length L then c(I) = L14.

We now see that compact sets having zero capacity are very small in some sense.

Let h be an increasing function on [0,11 such that h(0) = 0. Given a bounded subset E of C we consider all the coverings of E by a finite or countable number of squares having sides parallel to the coordinate axes and length of the sides less than e. We put H`(E) = inf F, h(d,) < +oo, for all the coverings E C U90 1=1 C; having the previous properties, di denoting the length of the sides of C',. If e decreases to zero then H'(E) increases. By definition,

H(E) = bin H`(E) < +oo. 010

This quantity is the h-Hausdorff measure of E. If h(t) = t°, with a > 0, then the corresponding quantity Ha(E) is the a-Hausdorff measure of E. If a = 2 it is exactly the outer Lebesque planar measure. If a = 1 it is the outer linear measure. Obviously, because E is bounded, we have H2(E) < +oo. It is not difficult to prove that H,,(E) < +oo implies Hs(E) = 0 for 8 > a. THEOREM A.1.28.

If K is a compact subset of C having zero capacity then it is totally disconnected and H0(K) = 0 for all a > 0. We say that a subset of the complex plane is locally of capacity zero if all its bounded subsets have capacity zero. THEOREM A.1.29 (H. CABTAH).

Let 0 be subharmonic on a domain D of C and not identically -oo. Then {z: z E D, O(z) = -co} is a Go-set which is locally of capacity zero.

§2. Functions of Several Complex Variables The list of results given below is the strict minimum we need. So it is a bit skeletal. If the reader needs more substance he must read [5,9,101 or any of the following books: L. Bers, Intvvdaetion to Several Complex Variables, New York, 1964; M. Fields, Several Complex Variables and Complex Manifolds,.4 and II, Cambridge,

1982; B. Fuks. Special Caters in the Theory of Analytic Function., of Several Complex Variables, Providence, 1965; it. Grauert and K. Fritzsche, Several Complex Variables, New York, 1976; R. Gunning and H. RASsi, Analytic Functions of Several Complex Variables, Englewood Cliffs, 1965; S.G. Krantz, Function Theory of Several Complex Variables, New York, 1982; P. Lelong, Fonctions analytiques et fonctions entiires (n variables), Montreal, 1968; and R. Narasimhan cited in §1.

Appendix

181

Let D be an open subset of C" and suppose that f is separately holomorphic on D. Then f is holomorphic on D. THEOREM A.2.1 (F. HARTOGS).

We know that there exist open subsets D of C" (n > 2) such that each holomorphic function on D can be extended holomorphically on a greater open :.et. So we are now interested in open sets which do not have this property.

We say that an open subset D of C" (n > 2) is an open set of holomorphy if D is non-empty and if there exists a function f holomorphic on D which cannot be extended holomorphically on a greater open set: that is, for each a E 8D there exists a neighbourhood V such that for any connected neighbourhood U of a, included in V, there is no holomorphic function F which coincides with f on a non-empty open subset of U fl D.

If K is a compact subset of D we define the H(D)-hull of K by the following: KD = {x: x E D, If (x)I < if (u)I, for every f E H(D)}. It is obvious that K C KD. Considering f(z) = exp(z ), where (zlt) is the scalar product of z with a vector e E C", we conclude that KD is included in the convex hull of K. It is also obvious that KD is closed in D, but in general it is not compact in D.

Let D be a non-empty open subset of C". Then the following properties are equivalent: THEOREM A.2.2 (H.CARTAN-P.THULLEN).

(i) D is an open set of holomorphy,

(ii) if K is an arbitrary compact subset of D then kD is also a compact subset of D.

COROLLARY A.2.3.

If D is a non-empty convex open subset of C" then D is an

open set of holomorphy.

In particular open balls are open sets of holomorphy. THEOREM A.2.4.

Let (D; );EI be a family of open sets of holomorphy in C". Then the interior of f1;EID, is also an open set of holomorphy in C".

Let D and D' be open sets of holomorphy respectively in C" and C" and let u be a holomorphic map from D into C'", that is all its components u1, ... , u," are holomorphic on D. Then THEOREM A.2.5.

is an open set of holomorphy in C.


182

Let D bean open set ofholomorphyin C" and let fr,..., fk E H(D). Then D1 = {z: z E D, f fe(z) I < 1,i = 1,..., k} is an open set ofholomorphy COROLLARY A.2.6. in

C"

It is bell-known that an open subset U of R" is convex if and only if - log dist(z, 8U) is convex for z E U. This suggests the following definition.

We say that an open subset D of C" is pseudoconvez if it is non-empty and if '-log dist(z, OD) is plurisubharmonic on D. Let (Dk) be an increasing sequence of pseudoconvex open sets of C". Then their union is pseudoconvex in C". THEOREM A.2.7.

Let (D;);Er be a family of pseudoconvex open subsets of C". Then the interior of f1;EIDi is pseudoconvex in C". THEOREM A.2.8.

If K is a compact subset of the open set D C C", we define the P(D)-hull of K, denoted KD, by KD = {z: z E D, 4,(z) < m ax 4,(u), for all 0 plurisubharmonic on D}. uEK

It is obvious that K C KD C KD.

Let D be a non-empty open subset of C". Then the following properties are equivalent: THEOREM A.2.9.

(i) D is pseudoconvex,

(ii) there exists 0 continuous and plurisubharmonic on D such that the sets D, {z: z E D, 4,(z) < c} are relatively compact in D for all real numbers c, (iii) the same property without requiring the continuity of 0,

(iv) if K is compact in D then KD is also a compact subset of D. THEOREM A.2.1O.

Let D be an open set of holomorphy in C". Then D is pseu-

doconvex in C".

An open subset D of C" is pseudoconvex if and only if there exists a plurisubharmonic function 4, 'on D that goes to +oo when z goes to the boundary of D. THEOREM A.2.11.

Appendix

183

Let D1 C C"' , ... , Dk C C"' be pseudoconvex open sets and nk. Then Di x . . x Dk is pseudoconvex in C".

THEOREM A.2.12.

let n = nl +

Let D be a pseudoconvex open subset of C" and u be a biholomorphic mapping from D onto D' C C". Then D' is pseudoconvex. THEOREM A.2.13.

The next result is a generalization of Theorem A.2.11 which is used in Chapter VII.

Let D be a non-empty open subset of C". A real function 0 defined on D is called a vertical function for D if, for any a E C such that {(a,z'):z' E C"-1} intersects D, we have

a

lim 4'(a, z') = +oo whenever (a, z'') is a boundary point of D.

Let D be a non-empty open subset of C". Then D is pseudoconvex if and only if there exists a vertical function 0 for D w iich is plurisubharmonic on D. THEOREM A.2.14.

It is a classical result in the theory of convex sets (perha is not well-known) that an open subset D of R" is convex if and only if, for all a in the boundary of D, there exists an open neighbourhood V of a such that V n ZI is convex. This is a very striking property because, at first sight, the definition o' convexity is not a local one. For pseudoconvex open sets there is a similar result. Let D be a non-empty open subset of C". Then D is pseudoconvex if and only if, for every boundary point of D, there ex -sts an open neighbourhood V of that point such that V n D is pseudoconvex. THEOREM A.2.15.

The next result, along with Theorem A.2.17, is much used in Chapter VII. Let D be apseudoconvex open subset of C' . Then there exists 0 E C°O(D) such that THEOREM A.2.16.

(i) 0 is strictly piurisubharmonic on D, (ii) for all real numbers c, the sets D, = {z: z E D, 4'(z) < c} an relatively compact

inDandD=tJ 1D,t. By Theorem A.2.10 we know that open sets of holomorphy c re pseudoconvex in C". The converse problem, that all pseudoconvex open subsets if C" are open sets


284

of holomorphy, wa , settled by E. Levi in 1911. It was solved, for n = 2, by K. Oka in 1942, and for n > 2 by K. Oka, F. Norguet and H.J. Bremermann in 1953-1954. By Theorem P .2.16 and the Behnke-Stein theorem, which asserts that the union of an increasing se luence of open sets of holomorphy is an open set of holomorphy, the problem is to r. rove that D, = {z: z E D, O(z) < c) is an open set of holomorphy

for ¢ E C'(D) which is strictly plurisubharmonic on D. In fact it is necessary to solve Cousin's problem. For example K. Oka proved the following lemma. Let D be a bounded doi gain in C" and 01,0 E R such that a < Q, and suppose that

Dl = {z: z E D, Re z2 > a} and D2 = {z: z E D, Re x1 < 6} are domains of holomorphy. Ther. D is a domain of holomorphy.

For n = 2, the case of interest to us, the proof of Levi's problem is not too difficult (see for instance the book by B. F1iks, pp. 194-215). For n > 2, many proofs have been given since 1955, using Oka's original method or functional analysis or partial differential equations (Ehrenpreis, Grauert, Narasimhan, Andreotti-Grauert, J.J. Kohn etc.).

Let D be an open subset of C" and let a E 8D. We say that D has a variety of support at a if there exist a neighbourhood U of a and f holornorphic on U such that {z: z E U, f(::) = 0} n 15 = {a}. It is well-known that convex subsets of C" have hyperplanes of support at each boundary point. So this suggests the following general problems which are unsolved until today.

(a) Given a pseudoconvex subset D of C", at which boundary points does D have varieties of support?

(b) For which classes of rather smooth pseudoconvex subsets of C" is it possible to have varieties of support through all boundary points?

We give only a partial solution to (b) which is due to K. Oka. The original proof is rathe complicated, particularly when grad O(A0, zo) = 0. There is a more elementary solution due to A. Zraibi. Using Narasimhan's lemma (see the book by S.G. Krantz, p.111) or the Levi-Krzoska theorem (see [9), pp. 157-162) it is even possible to extend this for D C C". but, in fact, we do not need this general result. THEOREM A.2.17.

Let D be an open subset of C2 and let 0 be a Cs-strictly

plurisubharmonic function on D such that Do = {(A, x): A, x e C, 4(A, z) < 0) is relatively compact in D. Then for all (Ao, zo) E 8Do, either there exist r > 0 and h holomorphic on lr(Ao,r) such that h(A0) = zo and (A,h(A)) V Do for IA - Aol < r,

or there exist s > 0 and k holomorphic on B(zo, a) such that k(zo) = A0 and

(k(z),z)V Dfor z - zoo<s.

REFERENCES

1. Aupetit, Bernard: Proprietis spectrales des algebres de Banach. Lecture Notes in Mathematics 735. Springer-Verlag, Berlin-Heidelberg-New York, 1979.

2. Bonsall, Frank F. and Duncan, John: Complete Normed Algebras. Ergebnisse der Mathematik and ihrer Grenzgebiete 80. Springer-Verlag, New YorkHeidelberg-Berlin, 1973. 3. Halmos, Paul R.: A Hilbert Space Problem Book. D. Van Nostrand, Princeton, 1967.

4. Hayman, W.K. and Kennedy, P.B.: Subharmonic Functions. Volume I. London Mathematical Society Monographs 9. Academic Press, London-New York-San Iansisco, 1976.

5. Hormander, Lars: An Introduction to Complex Analysis in Several Variables. North-Holland Mathematical Library 7. North-Holland Publishing Co., Amsterdam-London, 1973. 6. Rickart, Charles E.: General Theory of Banach Algebras. The University Series in Higher Mathematics. D. Van Nostrand, Princeton, 1960.

7. Rudin, Walter: Real and Complex Analysis. Second Edition. McGraw-Hill Series in Higher Mathematics. McGraw-Hill Book Co., New York 1974. 8. Rudin, Walter: Functional Analysis. McGraw-Hill Series in Higher Mathematics. McGraw-Hill Book Co., New York, 1973. 9. Vladimirov, V.S.: Methods of the Theory of Functions of Many Complex Variables. M.I.T. Press, Cambridge, Mass., 1966.

10. Wermer, John: Banach Algebras and Several Complex Variables. Second Edition. Graduate Texts in Mathematics 35. Springer-Verlag, New YorkHeidelberg-Berlin, 1976.

AUTHOR AND SUBJECT INDEX

Absorbing set 97 Adjoint of an operator 23 Alaoglu, L. 2,5 Banach-Alaoglu theorem 2,5 Alexander, H. 171 Alexander, J.C. 110 Alfsen, E.M. 171 Algebra

a-Calkin 113 Banach 30 Calkln 33,96 complex 30 n x n matrix 32 quotient 33 semi-simple 34 Wiener 31 Algebraic element 37 Algebroid multifunction 12 Analytic multifunction 143 Antisymmetric set 5 Aronazajn, N. 20 Artin, E. 11,96 Wedderburn-Artin theorem 11,96 Arzela, C. 7,9 Arzelk-Ascoli theorem 7,9 Ascoli, G. 7,9 Arzela-Ascoli theorem 7,9 Asymptotically quasi-compact operator 110

Atkinson, F.V. 63 Aupetit, B. 143,144,150,151,156,157, 160,167,171,185

Automatic continuity 99 Baire, R. 1,97,103,104,105 Baire's theorem 1,97,103,104,105 Banach, S. 1,2,4,5,79,80,127

Banach algebra 30 Banach-Alaoglu theorem 2,5 Banach-Steinhaus theorem 4 Banach-Stone theorem 79 Hahn-Banach theorem 1,5,127 Baribeau, L. 171 Barnes, B.A. viii,106,110,111,113 Basener, R. Basener's theorem on analytic structure 173 Beauzamy, C. 21 Beckenbach, E.F. 52,62,176 Beckenbach-Sake theorem 52,62,176 Behnke-Stein theorem 184 Benslimane M. 171 Berberian, S. 37 Berndtsson, B. 171 Bernstein, A.R. 20 Bers, L. 180

Author and Subject Index

Beurling, A. 39 Beurling-Gelfand formula 39 Bishop, E. 143,151,172

Bishop's theorem on analytic structure 143,151,172 Bonsall, F.F. 185 Borel, E. 161,162 Branges, L. de 5

Branching point 12 Brelot, M. 174,177 Bremermann, H.J. 184 Brown, A. 52

Calculus Holomorphic functional calculus 45, 51,54,101,102,107,109,118,125,138 Continuous functional calculus 121 Calkin, J.W. 33,96,158 Calkin algebra 33,96 a-Calkin algebra 113 Cantor, G. 155,157 Cantor-Bendixson theorem 157 Capacity 62,177,178 inner capacity 178 outer capacity 178 Capacitable set 178 Caradus, S. R. 96 Carleson, L. 74 Cartan, H. 63,95,161,176,180,181

H. Cartan's theorem 63,95,176,180 Cartan-Thullen theorem 181 Cauchy, A. 9,43,44,52,142 Cauchy's inequalities 9 Cauchy's theorem 43,44,52 Cayley, A. 11,44,97 Cayley-Hamilton theorem 11,44,97 Cech, E. 89 Stone-Cech compactification 89 Centre rnodulo the radical 92

Character 69 Choquet, G. 77,178 Choquet's boundary 77 Chinese remainder theorem 97 Civin, P. 117 Cleveland, S.H. 141 Closed graph theorem 4,75,76,99

Continuity of spectrum 48 Conway, J.B. 136 Corona problem 74,171 Cousin's problem 184 C`-algebra 118 Davie, A.M. 15,21 Derived set 155 a-derived set 155 Diameter (n-th) 62 Dieudonne J. vii, 27,29 Double stochastic matrix 8 Douglas R.G. 52 Duncan, J. 185 Dye, H.A. ix,125

Edwards, R.E. 98 Eigenspace 15 Eigenvalue 15 Eigenvector 15 Enflo, P. 15,21 Equicontinuous family 7 Equilibrium value 178 Essential range of a function 130

Essential spectrum 96 Essentially bounded function 130 Evaluation at a point 71 Evans, G.C. 159,179 Evans's theorem 159,179 Exponential spectrum 47 Extension point for a pseudoconvex open set 151

187

188


Fekete, M. 179 Fields, M. 180 Fine topology 60 Fredholm, I. 15,27 Fredholm alternative 27 Fritzsche, K. 180 Fuglede, B. 120 Fukamiya, M. 123 Fuks, B. 180,184 Function algebra 76

Gamelin, T.W. 74 Gardner, L.T. ix,125 Gelfand, I.M. viii,38,39,41,69,70,72,74, 121,123,128,171 Gelfand theory viii,41,69 Gelfand transform 74 Gelgand topology 74 Gelfand-Mazur theorem 39 Gelfand-Naimark theorem for commutative algebras 121 Gelfand-Naimark theorem for noncommutative algebras 128 Gelfand-Naimark theorem for Jordan-Banach algebras 171 Beurling-Gelfand formula 39 Gerigorin, S.A. 91 Gleason, A. 69 Glicksberg, I. 5 Gohberg, I.C. viii,59,106,108 Good selection 147 Good isolated point 151 Grabiner, S. 94,105 Grauert, H. 180 Gunning, It. 180

Hahn, H. 1,5,127 Hahn-Banach theorem 1,5,127 Halmos, P. 20,23,161,185

Halperin, I. 37 Hamilton, W.R. 11,44,97 Cayley-Hamilton theorem 11,44,97 Harte, It. 47 Hartogs, F. viii,145,146,149,181 Hausdorff, F. 48,180 Hausdorff distance 48 Hausdorff measure 180 Hayman, W.K. 185 Herstein, I. 11,84 Hilden,H.M. 20,21 Hoffman, K. 74 Holomorphic functional calculus viii, 45,51,54,101,102,107,109,118,125, 138

Holomorphic selections 144 Holomorphic variation of isolated spectral values viii,59 Holomorphic variation of isolated points 147 Hormander, L. 185

Identity principle for analytic functions 103,105,138,159 Implicit function theorem 11,12,48,172 Inessential element 106 Inessential ideal 106 Invariant subspace 20 Involution

in £(H) 23 in a Banach algebra 75,76 Istriiraescu, V. 53

Jacobson, N. viii,33,82,83 Jacobson radical 34 Johnson, B.E. viii,99,100,117 Jordan, P. 171 Jordan algebra 171 Jordan-Banach algebra 171


Kahane, J.-P. 69 Kaidi, A. 171 Kakutani, S. 8,49

Kakutani's theorem 8 Kaplansky,I. viii,84,89,96,99,123 Kato, T. vii,101 Kennedy, P.B. 185 King-Lai Hiong 162 Kleinecke D.C. 15,92,106,107,108 Krantz,S.G. 180,184 Krein, M. 3,5,8 Krein-Milman theorem 3,5,8 Krejn, M. 59,108 Krull, W. 33 Kuratowski, K. 50,113,156

Lancaster, P. 10 Lelong, P. 180 Le Page, C. 92 Levi, E. 184 Levi-Krzoska theorem 184 Levy, P. 72,73 Liouville, J. 27,38,56,76,86,95,100,121, 176

Liouville's theorem 38,76,86,95,121 Liouville's theorem for subharmonic functions 56,100,176 Liouville's spectral theorem viii,56 Sturm-Liouville problem 27 Local characterization of subharmonic functions 176 Localization principle 146,163 Locally algebraic operators 84 boundedly 84 Locally of capacity zero 180 Logarithmic potential 177 Lomonosov, V.I. viii,20 Lorch, E. R. 47

189

Machado, S. viii,1,5 Matrix double stochastic 8 normal 10 permutation 8 self-adjoint 10 Maximum principle for subbarmonic functions 54,56,91,145,148,175 Maximum principle for spectrum 55 Mazur, S. 39,80 Gelfand-Mazur theorem 39 Mazur-Ulam theorem 80 Mergelyan's theorem 141 Milman, D. viii,1,3,4,5,8 Milman's theorem viii,1,3,4 Krein-Milman theorem 3,5,8 Minimal polynomial 37 projection 114 Modular annihilator algebra 113 Monna, A.F. vii Montel, P. 9,176 Montel's theorem 9 Montel-Rado theorem 176 Morera's theorem 102 Morrel, B.B. 49,136 Morrel, J. 49 Multifunction algebroid 12 analytic 143 Muller, V. 49

Nagasawa, N. 77,,78 Naimark, M.A. 121,123,128,171 Gelfand-Naimark theorem for commutative algebras 121 Gelfand-Naimark theorem for noncommutative algebras 128

190


Gelfand-Naimak theorem for Jordan-Banach algebras 171 Narasimhan, R. 145,174,180,184 Narasimhan's lemma 184 Neumann, J. von 20,171 Nevanliuna, R. 161,162 Newburgh, J.D. 51 Niestegge, G. 77 Nishino, T. 151,153,155 Oka-Nishino theorem for pseudoconvex sets 153 Oka-Nishino theorem for analytic multifunctions 151,155,159 Non-thin set 60,177 Norguet, F. 184 Normal element 117 family 9 matrix 10 operator 24 Oka, K. viii,142,151,153,155,159,177, 184

theorem for pseudoconvex sets 153 Oka-Nishino theorem for analytic multifunctions 151,155,159 Oka-Rothstein theorem 61,177 Open mapping theorem 4,84 Open set of holomorphy 181 Operator adjoint of an 23 compact 15 normal 24 real part, imaginary part 24 self-adjoint 24

unitary 24 Volterra 19 Ordinal of the first or second class 112 Orthogonal projection 22 Osgood, W.F. 8 Osgood's lemma 8

Paulsen, V. 67 Pelczynski, A. ix,142,159 General Pelczynski conjecture ix, 159

Permutation matrix 8 Perturbation by inessential elements viii,108

Pfaffenberger, W.E. 96 Picard, E. 161,163,166,167,168 Picard's theorem 161,166 Picard's theorem for matrices 165 Picard's theorem for analytic multifunctions 167,168 Polar decomposition 125,135 Polynomial convex hull 77 Polynomially convex set 77 Positive element 123 functional 126 Principal component of the group of invertible elements 46 Privaloff, I. 174 Product of algebras 32 Projection orthogonal 22 minimal 114 Pseudoconvex open set 182 Ptak, V. 120 Puiseux, C. 12,142 Putnam, C.R. 120

spectral radius of an 24

spectrum of an 15

Quasi-algebraic element 161


Quasi-nilpotent element 36 Quotient algebra 33

Radical 34 Jacobson radical 34

Spectral characterizations 95 Radius (spectral) 24,36 Radjavi, H. 84 Rad6, T, 58,63,166,174,176 Rado 's extension theorem 58,63, 166

Montel-Rado theorem 176 Ransford, T.J. 143,144,151,155,170,171 Read, C. 21 Rellich, F. vii Remoundos, G. 161,163 Representation 80 bounded 80 continuous 80 irreducible 80 left regular 80 Resolution of the identity 129 Rickart, C.E. 185 Riesz, F. vii,viii,1,2,5,15,16,42,64,177 Riesz's characterization of finitedimensional Banach spaces 1 Riesz's representation theorem 2,5, 177

Riesz's theorem for compact operators 64 Riesz's theorem for subharmonic functions 177 Robinson, A. 20 Rodriguez Palacios, A. 141,171 Roitman, M. 69 Rosenblum, M. 120 Rosenthal, P. 84 Rossi, H. 90,180

191

Rudin, W. 146,185 Rudin's characterization of holomorphic functions 146 Runge's theorem 45,118 Russo, B. ix,125 Ruston, A.F. viii,106,110

Sadullaev, A. 161 Saks, S. 52,62,176 Beckenback-Sales theorem 52,62,176 Scarcity of elements with finite spectrum viii,63 Scarcity of elements with finite values 147

Scarcity theorem for countable analytic multifunction 158 Schauder fixed point theorem 20 Schmidt, D. 53 Schur, I. 82 Selection holomorphic 144 good 147 Self-adjoint

matrix 10 operator 24 element 117 Semi-simple algebra 34 Separating space 99 Shelah, S. 21 Shestakov, I.P. 171 Shift weighted 23 left weighted 23

right weighted 23 Shirhsov, A.I. 171

Shoda's theorem 28 Shultz, F.W. 171 Sierpinski, W. 112

192

A Primer on Spectral Theory-

Silov, G.E. 89,111,150

tilov boundary 89,150

Silov idempotent theorem 111 Sinclair, A. viii,84,89,101 Sirokov, D.C. 92 Slin'ko, A.M. 171 Slodkowski, A. viii,62,143,144,147,149, 150,151,167,171

Smith, K.T. 20 §mul'jan, Ja.L. 63 Smyth,M.R. 110 Socle 110

Spectral decomposition 135 diameter (n-th) 62 interpolation 171 maximum viii,77 multiplicity 163 radius 24,36 state 78 Spectral theorem for self-adjoint compact operators 25

general 132 for normal operators 134 Spectrum essential 96 exponential 47 of an element 36

Stone-tech compactification 89 Stone-Weierstrass theorem 1,5,6,9 72,121,134 Stormer, E. 171 Strong sum 22 Sturm-Liouville problem 27,29 Subharmonic function 52 Szego, G. 179 Szokefalvi-Nagy, B. vii,63,64

Tchebycheff polynomial 179 Thin set 177 Thullen, P. 181 Cartan-Thullen theorem 181 Tismenetsky, M. 10 Topological divisor of zero 41

dual 2 Tsuji, M. 169,174 Tsuji's theorem 169

Ulam, S. 80 Uniform continuity of spectrum 48 Upper regularization of a function 175 Urysohn's lemma 133 Unitary element 117 operator 24

of an operator 15 peripherical 54 Square root of a positive element 123 State 127 Steinhaus, H. 4 Banach-Steinhaus theorem 4 Steprans, J. 21

Vesentini, E. 52 Vesentini's theorem 52 Vladimirov, V.S. 185 Volterra operator 19

Sternfeld, Y. 69 Stone, M.H. 1,5,6,9,72,79,89,121,134 Banach-Stone theorem 79

Weak lower semicontinuity of the boundary of the spectrum 60

Valiron, G. 161 Variety of support 184 Vertical function 183


Wedderburn, J.H.M. 11,96 Wedderburn-Artin theorem 11,96 Weierstrass, K. 1,5,6,9,72,121,134 Stone-Weierstrass theorem 1,5,6,9, 72,121,134

Wermer, J. 143,151,171,185 T.T. West decomposition 96 Wiener, N. 31,72,73 Wigner, E. 171 Williams, J.P. 170 Wolff, T. 74

193

Yamaguchi, H. 14; Yood, B. 96,117 Youngson, M.A. 17

Zelazko, W. 69 Zemanek, J. 95,120 151,156,157,160,171 Zhevlakov, K. A. 171 Zorn's lemma 5,6,8,33,118 Zraibi, A. 143,144,761,167,169,171,184

A Primer on Spectral Theory (Universitext)

A Primer on Spectral Theory (Universitext)

A Primer on Spectral Theory (Universitext)