Bernard Aupetit
A Primer on Spectral Theory
SpringerVerlag
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Bernard Aupetit
A Primer on Spectral Theory
SpringerVerlag
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Ce livre est dedie a la memoire de mes parents. Marcel Aupetit MarieTlierese 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
these books are direct consequences of results, very often more general, given in chapters III, V and VII. The subharmonicity of the spectrum and the theory of analytic multifunctions also illustrate two striking facts in the history of mathematics. General spectral theory and the abstract theory of subharmonic functions were both invented by F. Riesz (see his Oeuvres computes), but obviously, in the 1920s, he did not realize that the two theories are equivalent in the following sense. Given an analytic family f(A) of operators, the logarithm of the spectral radius of f(A) is subharmonic in
viii
A Printer on Spectral Theory
A, and conversely, given a subharmonic function 0, then by Z. Slodkowski's result mentioned on page 167, there exists an analytic family of operators defined on t, . ach that ¢(A) is equal to the logaritlun of the spectral radius of f (A). Moreover, .n trying to improve some results of F. Hartogs, K. Oka introduced in 1934 the notion analytic multifunction, but obviously, he saw no link with spectral theory, and the notion remained dormant until the 1980s. So, even in mathematics, the words of the great Florentine statesman and historian Francesco Guicciardini are true.
The book is divided into seven chapters and an appendix. In the first chapter, we give a list of basic results in functional analysis without any proofs, except for those of Machado's theorem and Milman's theorem, which are rarely mentioned in :,lie reference books. The results in this chapter will be used throughout the book. The second chapter introduces the reader to some examples of bounded opera, tors on. ]3anach spaces and Hilbert spaces, and in particular gives F. Riesz's theory for compact operators and V.I. Lomonosov's theorem on invariant subspaces.
In the third chapter, we define and give examples of Banach algebras. The fundamental Holomorphic Functional Calculus Theorem is then proved. In §4, the subharmonic variation of the spectrum is given with many important applications, for instance the Spectral Maximum Principle (Theorem 3.4.13), Liouville's Spectral Theorem (Theorem 3.4.14), the Holomorphic Variation of Isolated Spec, tral Values (Theorem 3.4.20), the Subharmonicity of the uth Diameter (Theorem 3.4.24), the Scarcity of Elements with Finite Spectrum (Theorem 3.4.25), and the Identity Principle for operators having spectra with zero as the only limit point.
The fourth chapter introduces the reader to I.M. Gelfand's theory of commutative Banach algebras and to the representation theory of non commutative Banach algebras, with the standard results of N. Jacobson and A. Sinclair. At the end we give a new analytic proof of I. Kaplansky's theorem on locally algebraic operators, and we extend it to locally independent operators.
The fifth chapter, along with §4 of chapter III and with chapter VII, is the most important part of the book. It contains a great number of applications of spectral subharmonicity. The most striking applications are the generalization of B.E. Johnson's theorem on the uniqueness of the norm (Theorem 5.5.1), the Perturbation Theorem by Ineseseential Elements (Theorem 5.7.4), which immediately implies classical results of I.C. Gohberg, A.F. Ruston and B.A. Barnes, and the new subharmonic proof of B.A. Barnes's theorem on the existence of the socle (Theorem 5.7.8).
Preface
The sixth chapter is a classical presentation of the spectral theorem for normal operators on a Hilbert space, with a few applications at the end. In particular it includes the very nice extension of the RussoDye theorem obtained by L.T. Gardner.
This chapter is independent of chapters V and VII, so it can be read just after chapter IV. The seventh chapter is the most difficult one for two reasons. Firstly, it involves difficult mathematics, such as the theory of pseudoconvex sets. Secondly, our treat
ment is superficial. Nevertheless in spite of these drawbacks we have decided to include it in this book in order to induce the reader to learn more on this new subject which has (and certainly will have in the future) very important applications. In particular, the General Pelczynski Conjecture (Theorem 7.2.9) is solved, and applications to the distribution of spectral values in the plane (for instance Theorem 7.3.10) are given.
The appendix is a compendium of all the results needed in the theories of subharmonic functions, of capacity, and of functions of several complex variables. Obviously, no proofs are given because of the strict limit we fixed on the number of pages in our manuscript. They can be found in the standard textboooks mentioned. Each chapter is followed by a list of problems. In some of them extra material is introduced. A few are very difficult to solve; in these cases we suggest that the reader take a look at the given references. We have tried to keep the prerequisites to a strict minimum, and to develop the techniques of spectral subharnnonicity and analytic multifunctions in the smallest number of pages. This is mainly because we believe that it is easier to learn something quickly and deeply in a small book than to learn in a big book. However the reader is assumed to be familiar with the matter which would generally be covered in onesemester courses on algebra, on complex analysis and on functional analysis. The material presented corresponds roughly to two semesters of lectures, supposing that the students are familiar with the basics of subharmonic functions, capacity and functions of several complex variables. Otherwise it would be better to start with an introductory course covering the matter in the appendix.
Firstly we would like to thank Churchill College, Cambridge, and the Scuola Normale Superiore di Pisa, for their hospitality during the academic year 19841985. There we began to write the manuscript of this book. We also thank the Natural Sciences and Engineering Research Council of Canada and the F.C.A.R. Find of the Province of Quebec for their constant financial help which gave us the opportunity to travel and have interesting discussions with many mathe
maticians.
z
A Primer on Spectral Theory
Obviously much thanks are due to the many attentive readers of the different versions of the manuscript, in particular to our friend Jaroslav Zemanek, to our former students Line Baribeau and Frederic Gourdeau, and,to our students Daniel Turcotte, Alain Fournier and Abdelaziz Maouche. Special thanks go to Louise Chamberland, Alain Charbonneau and Alain Fournier for the unpleasant work of typing this manuscript. Quebec February 1990
Bernard Aupetit
CONTENTS
Preface ............. ......................................................... vii
Chapter I. Some Reminders of Functional Analysis ...........................
1
Chapter II. Some Classes of Operators ....................................... 10 §1. FiniteDimensional Operators ....................................... '0
§2. Bounded Linear Operators on a Banach Space ....................... :2 §3. Bounded Linear Operator on a Hilbert Space ........................ 22
Chapter III. Banach Algebras ................................................ 30 §1. Definition and Examples ............................................ 30 §2. Invertible Elements and Spectrum ...................................
35
§3. Holomorphic Functional Calculus .................................... 42 §4. Analytic Properties of the Spectrum .................................
48
Chapter IV. Representation Theory .......................................... 69
§1. Gelfand Theory for Commutative Banach Algebras .................. 69 §2. Representation Theory for NonCommutative Banach Algebras ...... 80
Chapter V. Some Applications of Subharmonicity ............................ 91
§1. Some Elementary Applications ...................................... 91 §2. Spectral Characterizations of Commutative Banach Algebras
........ 92
§3. Spectral Characterizations of the Radical ............................ 95 §4. Spectral Characterizations of FiniteDimensional Banach Algebras ...
96
§5. Automatic Continuity for Banach Algebra Morphisms............... 99
§6. Elements with Finite Spectrum ......................................102
§7. Inessential Elements ................................................ 106 Chapter VI. Representation of C*agebras and the Spectral Theorem ........ 116 §1. Banach Algebras with Involution .................................... 116
ii
A Primer on Spectral Theory
§2. C'algebras ......................................................... 118
§3. The Spectral Theorem .............................................. 129 §4. Applications ........................................................ 135
Chapter VII. An Introduction to Analytic Multifunction ..................... 142
§1. Definitions and General Properties .................................. 143 §2. The OkaNishino Theorem and Its Applications ...................... 151 §3. Distribution of Values of Analytic Multifunctions ....................181 §4. Conclusion ..........................................................
171 Appendix ....................................................................174 §1. Suhharmonic Functions and Capacity ............................... 174 §2. Functions of Several Complex Variables .............................. 180 References ...................................................................185
Author and Subject Index ....................................................186
Chapter I SOME REMINDERS OF FUNCTIONAL ANALYSIS
This short chapter is intended as an aidememoire of some fundamental results in functional analysis that will be used in the rest of this book. Consequently we shall give no proofs as they can be easily found in all the standard textbooks (for instance 17),18)), with two exceptions; the proof of Milman's theorem (Theorem 1.1.12) and the nice and simple proof of Machado's theorem (Theorem 1.1.16) which is a strong extension of the StoneWeierstrass theorem. If X isanonempty open subset of a complete metric space then the union of a countable collection of closed subsets of X with empty interiors also has an empty interior. This book will be concerned with vector spaces over the complex field. THEOREM 1.1.1 (R. BAIRE).
THEOREM 1.1.2 (F. RIESZ). ball is finitedimensional.
Every normed vector space containing a compact
Suppose that Y is a subspace of a normed
THEOREM 1.1.3 (H.HAHNS.BANACH).
vector spare X and f is a linear functional on Y such that 1f(X)1 5 [I=II
,
for x E Y.
Then f extends to a linear functional F on X such that IF(x)I <_ IIxII , for x E X. COROLLARY 1.1.4.
If X is a normed space and a E X, there a fists a linear
functional F on X such that F(a) = )fall and IF(x)I < IIxII, for all x E X. The following corollary is very trsefti in proving that an element of a Banach space can be approximated by elements of a subspace.
A Primer on Spectral Theory
2
COROLLARY 1.1.5.
Suppose that Y is a subspace of a normed space X and a E X.
Then a is not in the closure of Y if and only if there exists a bounded linear functional F on X such that F(a) = 1 and F(x) = 0, for all x in Y. Let X b a Banach space and X' denote its topological dual. Using the weak topology and Theorem 1.1.3 it is possible to get the following result, which will be used in Chapter IV, §1.
Suppose that A and B are two disjoint nonempty convex and compact subsets of X', with respect to the weak `topology. Then there exists a E X such that COROLLARY 1.1.6.
sup{Re f(a): f E A} < inf{Re f(a): f E B}.
Let K be a compact set and C(K) denote the Banach space of continuous functions on K with complex values, equipped with the uniform norm II f II K = sup=EK If(x)J. In that case F. Riesz has identified exactly the topological dual of C(X) in the following manner. He first showed that a bounded linear functional on C(K) can be written asia linear combination of normalized positive linear functionals F, normalized positive meaning that F(f) > 0 for f > 0 in C(K) and F(1) = 1, or equivalently F satisfies F(1) = 1 = IIFII (see Theorem 6.2.16). He then gave a precise representation of normalized positive linear functionals. THEOREM 1.1.7 (F. RIEsz REPRESENTATION THEOREM).
Let K be a compact
set and F be a normalized positive functional on C(K). Then there exist a complete oalgebra M on K containing all Borel sets and a unique probability measure µ on M which represents F in the sense that
F(f) = I f(x)dp(x) for all f E C(K).
Conversely it is obvious that given such a probability measure p on K the mapping f ' fK f(x)dp(x) defines a normalized positive linear functional on C(K). THEOREM 1.1.8 (S.BANACHL.ALAOGLU).
Let X be a Banach space and X' be its topological dual. Then the unit ball U = {F: F E X', IIFII S 1} is weak compact in X'.
Some Reminders of Functional Analysis
3
THEOREM 1.1.9 (M.KREIND.MILMAN). Suppose that X is a locally convex space. If K is a compact convex subset of X then it is the closed convex hull of its extreme points.
In fact it is possible to be more precise. Given K a compact convex subset of X, we denote by ext(K) the closure of the subset of extreme points of K. If p is a probability measure on ext(K), we shall say that x E K is represented by p if
F(x) _ l
ext(K)
F(y) dp(y)
,
for all F E X'.
By the analogue of Corollary 1.1.4 for locally convex linear spaces, x is the unique vector which can be represented by p. THEOREM 1.1.10. Suppose that A is a compact subset of a locally convex space X X. A point z E X is in the closed convex hull of A if and only if there exists a probability measure p on A which represents x. PROOF. If p represents x then F(x) = fA F(y) dp(y), and as X is separated by its bounded linear functionals, x is in the closed convex hull of A. Conversely, by hypothesis, there exists a family in the convex hull of ext(A) which converges to x, or equivalently there exist points xa = L.;=1 )°y' (A° > 0, E AP = 1, y° E ext(A), rr in some directed set) which converge to X. Then x a is represented by the probability measure pa = where by, denotes the Dirac measure concentrated at the point y°. By Theorem 1.1.7 and Theorem 1.1.8 the set of all probability measures on ext(A) may be identified with a weak'compact convex subset of C(ext(A))'. So (pa) contains a subfamily converging to a probability measure p on ext(A). It is then easy to verify that p represents z. 0
We now prove Milman's theorem, which will be used in Chapter IV, §1. LEMMA 1.1.11.
Suppose that X is a locally convex space, that x E X and that K is a nonempty compact convex subset of X. Then z is an extreme point of K if and only if bs is the only probability measure on K which represents z. PROOF. Suppose that z is an extreme point of K and that the measure p represents x. We have to prove that p is supported by {x}. By regularity of p it suffices to prove that p(E) = 0 for each compact set E C K \ {x}. Suppose that p(E) > 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
A Primer on Spectral Theory
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,+(1r)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.BANACHH.STEINHAUS).
sup *EA
+oo.
Then SUPOEA IITO < +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 StoneWeierstrass 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 HahnBanach theorem, KreinMilman theorem, BanachAlaoglu theorem and the Riesz representation theorem (this last proof is given in [10], pp. 57).
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 nonempty subset E of K is said to be Aantisymmetric if whenever h E A and h is real on E then h is constant on E.
Suppose moreover that A is selfadjoint, 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 Aantisymmetric 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 realvalued and in A, we get a contradiction. If f E C(K) and F is a nonempty 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 Aantisymmetric 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 nonempty 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
A Primer on Spectral Theory
6
compact and nonempty. Thus their intersection {z : x E G, If(z) g(x): > d f(K)} is also compact and nonempty, 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 Aantisymmetric. Suppose this is false. There exists h E A which is both real and nonconstant
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
3},
Ez={x:xEE,I
Since E1 and E2 are nonempty proper closed subsets of E, the minimality of E implies that there exist g1, g2 E A such that
IIf  g, IIE, < d f(K)' and Ilf  92IIE, < df(K)
Forn> 1,ieth"=(1h")2"
(1)
Wehave h",k"EA
amd0
(2)
On E1\E2,where 0
()fl. h">12"h">1(3)
OnA \E1, where 2
h")_2
< 1/2"h" < \4)"
\
(4)
So k" converges uniformly to 91 on E1 \ Ez and uniformly to gz on Ez \ E1. Using (1) and (2) we conclude that IIf  k. 11,U < df(K) for n large enough, so d f(E) < d f(K), which is a contradiction. 0 COROLLARY 1.1.17 (M.H.STONEK.WEIERSTRASS). Let K be a compact set and let A be a closed selfadjoint subalgebra of C(K) containing the constant functions and separating the points of K. Then A = C(K). PROOF.
Let f E C(K). By Theorem 1.1.16 there exists a closed Aantisymmetric set E such that d f(E) = d f(K). But we saw previously that E is a singleton {a}, so considering g(x) = f(a)1 we conclude that df(E) = 0. Hence df(K) = 0 which means that f E A, since A is closed. 0
Some Reminders of Functions! Analysis
7
Let K be a compact metric space. We say that F C C(K) is an equicontinuous family if there exist C > 0 and a > 0 such that if (x)  f (y)I <_ Cd(x, y)°, for all x, y E K and f E F, where d denotes the distance on K. Relatively compact subsets of C(K) can be easily characterized by the following result. THEOREM 1.1.18 (C.ARZELAG.ASCOLI).
Let K be a compact metric space.
Then F is relatively compact in C(K) if and only if it satisfies the following two conditions:
(i) F is equicontinuous, (ii) sup fES If(x) I < +00, for each x E K.
A Primer on Spectral Theory
8
EXERCISE I.
Let f be a continuous function on (0,1] with real values.
(i) First suppose that f is differentiable at a point a E 10, 1). Prove that there exists an integer n > I such that If (x)  f(a)l < njx  al, for all x E [0, 11. (ii) Let E _ (f : f E C((0,1]), for which there exists some a in [0, 1], depending
on f, such that If (z)  f(a)I < nix  al, for all x E [0,1]). Prove that E is closed and has no interior point in C([0,1)) for the topology defined by the uniform norm.
(iii) Derive from that the existence of a continuous function on [0,1J which is not differentiable at every point of 10, 1]. EXERCISE 2.
Let X be a nonempty subset of a complete metric space. Let
be a sequence of complex functions defined on X such that the )f) are lower semicontinuous on X and the f,, converge pointwise on X. Prove that there exist a ball B C X and C > 0 such that sup=EB I C (W.F. Osgood's lemma). EXERCISE 3. Let X be a vector space and let A, B be two disjoint convex subsets of X. Using Zorn's lemma prove that there exist two disjoint convex sets Ao, Bo such that A C Ao, B ,C B0 and X = A0 U Bo (S. Kakutani's theorem). EXERCISE 4. Let X be a normed vector space and let A, B be two disjoint convex subsets of X. Suppose that A has interior points. Prove that there exists a bounded linear functional f such that sup=EA f(x) < inf=EB f(x). 
A n x n matrix (a;1) is double stochastic if a;, > 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 KreinMilman 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 eOzp(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 ArzelaAscoli 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. FiniteDimensional Operators Let X be a finitedimensional 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 finitedimensional operators or matrix theory is very wellknown. 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 selfadjoint (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 selfadjoint or normal compact operators on a Hilbert space in §3, and even for arbitrary selfadjoint and normal operators on a Hilbert space in Chapter VI. From the CayleyHamilton 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 twosided 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 i`a 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 nonzero twosided 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 finitedimentional algebras over the complex field have a very simple structure. For instance if A is semisimple (see Chapter III, §1 for the definitions of Jacobson radical and semisimple algebras) we have the very famous: THEOREM 2.1.2 (d.H.M.WEDDERBURNE.ARTIN).
Let A be asemisimple 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, semisimple 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
A Primer on Spectral Theory
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 finitedimensional 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.
Some Classes of Operators
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 (an1) such that X and sequences of complex numbers PROOF.
=al(a1+bj)+...+an1(an_1+bn1)
an+bn
If all the sequences (ap) are bounded, applying the BolzanoWeierstrass 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_lean1+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 infinitedimensional Banach space. Then there exists a bounded linear operator on X having an infinitedimensional range which is a limit of finiterank operators.
PRooF. Let us denote by 3 the linear subspace of finiterank 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
A Primer on Spectral Theory
It is easy to we that $ is a twosided ideal of £(X), the algebra of bounded linear mappings from X into X, and consequently 1 is a closed twosided 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 ecovering 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 finiterank 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 finiterank operators.
Some Classes of 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 finitedimensional 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 twosided 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 finiterank 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. T1 would be compact, and by Theorem 1.1.2 this would imply that X is finitedimensional. 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 finitedimensional.
(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 finitedimensional. 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
A Primer on Spectral Theory
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 finitedimensional.
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 <_ lix  yll the family I f.1 is equicontinuous. But T(B is Because I compact so, by Theorem 1.1.18, there exists a subsequence (f,,,) converging uniformly on T(B). But IIT' fn,  T' f,,; II = sup:ES I(f,,,  fnj)(Tx)l, so the sequence (T' f,,,) converges. By the remark following the definition of compact operators this implies that T` is compact. PROOF.
Suppose now that T' is compact. Let i be the canonical isometry of X into the
bidual X" defined by i(x)(f) = f(z) for x E X, f E X'. Using the usual duality
notations we have (f,i(Tx)) _ (Tz,f) = (x,T'f) = (T*f,i(x)) = (f,T* o i(x)) for all x E X, f E X'. So i o T = T o i. If z E B then i(z) E B", the unit ball of X", so i(T(B)) C T"(B"). By the first part of the theorem, T** is compact so T"(B") has a compact closure. The same is true for i(T(B)). But i being an isometry, T(B) has a compact closure. 0 Wenow give F. Riesz's theory of compact operators.
Let X be a Banach space, let T E L(X) and A E C. Suppose that E, F ere two closed subspaces of X such that E C F, E t F and (T  AI)(F) C E. Then there exists a E F\E with Ilail < 1, such that IITa  Txll > IAI/2, for all LEMMA 2.2.7.
zEE. By hypothesis, there exists b E F\E such that dist(b, E) = a > 0. Let E F. We have flail = 1. For all y E E be such that llb  yll < 2a and let a = I PROOF.
z E E, we have 1
az= IIb
(b  y  zllb  yII) and yII
y + zllb  yII E E,
so 11b  y  zllb  vii II ? a and consequently Ila  zll > 1/2 for all z E E. But for
T(az)=(TAI)(az)+A(a x) and (TAI)(ax) \x EE by hypothesis, so IiTa  Tzil > JAI/2. 0
Some CIus
of Operators
17
THEOREM 2.2.8. Let X be a Banach space and let T E ZC(X). If A 96 0 then T  AI has a closed range and codim R(T  AI) < +oo.
PROOF. By Theorem 2.2.5 (ii), dimN(T  Al) < +oo. So using the argument given before Lemma 2.2.1, we see that there exists a closed subspace M such that
M +N(T  Al) = X and M nN(T  AI) = {0}. Let S be the restriction of T AI to M with values in X. Then S is injective on M because Sx = 0 for x E M implies
z E M fl N(T  AI). Because X = M +N(T  AI) we have R(S) = R(T  Al). We now prove that there exists r > 0 such that r)jzjj < jjSr U
,
for x E M.
(1)
If this inequality is false then there exists a sequence (zn) of elements of M such that Jjxnjj = 1 and limn Sxn = 0. But T is compact, sp there exists a subsequence (x,,,,) such that Txn. converges to some a E X. Then Ax,, converges to a, so. a E M and Sa = lim(ASza,) = 0. Consequently a = 0, which is absurd because IknII = 1. Thus (1) is true. From (1) we conclude that R(S) is closed. For if Sxn converges, then (Sxn) is a Cauchy sequence so, by (1), (xn) is a Cauchy sequence which converges to some b E M and hence limn Sxn = Sb. If R(T  AI) has infinite codimension then there exists a sequence (an) in X such that, for all n, (an) is not in the subspace V._1 generated by 1(T  Al) and a1,... , an1. By the first part Vn is closed. By Lemma 2.2.7 there exists a sequence (b,) such that bn E VV`Vn1, IIbnII < 1 and HHTbn  Tbi'I > IAI/2 > 0, 1 < i < n  1. This implies that the sequence (Tbn) has no converging subsequence which contradicts the fact that T is compact. D
Let X be a Banach space and let T E EC(X). Suppose that A # 0 and define Nk = A(((T  AI)k) and Rk = R((T  AI)k) for k > 1. Then
THEOREM 2.2.9.
(i) the Nk form an increasing sequence of finitedimensional subspaces of X and the Rk form a decreasing sequence of closed subspaces of X having finite codimensions,
(ii) there exists a smallest integer n such that Nk+1 = Nk for k > n, and then Rk+, = Rk for k > n, and we have X =Nn ED Rn,
(iii) N,, and R,, are invariant subspaces of T  AI, the restriction of T  Al to N. has a nth power zero in £(Nn) and the restriction to R. is invertible in £(Rn).
A Primer on Spectral Theory
18
PROOF.
It is obvious that (T  AI)k = (AI)k + Kk where Kk is a compact
operator. So (i) derives immediately from Theorem 2.2.5 (ii) and Theorem 2.2.8. We now prove (ii). Suppose that Nk iA Nk+l for all k > 1. Then we have (T  AI)(Nk+1) C Nk so, by Lemma 2.2.7, there exists a sequence (zk) such that xk E Nk\Nk_1i flxkjj < 1 for k > 1 and IITzk Tx;II ? (AI/2 for i < k. This gives a contradiction to the compactness of T. A similar argument &!.so proves that the sequence (Rk) stabilizes. Let n be the smallest integer such that Nk = Nkt1, for k > n. We now prove that N" fl R, = {0}. If y E N" fl R,, there exists x E X such that y = (T  AI)"r. Moreover (T  AI)"y = 0, and so (T  AI)2"x = 0. Therefore
.z E N2 = N", so y = 0. Because (Rk) stabilizes there exists a smallest m > n such that Rk = Rk..1 for k > m. We have X. C R. and (T  AI)Rm = Rm. We
now prove that Rm=R"so that m=n. If wehave m>n,letzERm_1CR. with z V Rm. Since (T  AI)z E Rm = (T  AI)Rm there exists t E Rm such that
(TAI)t=(TAI)z. Hence ztEN1 CN", butalsoz  tER",soztand 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
Some Classes of Operators
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(TAI) 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 finitedimensional 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 nonzero 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
A Primer on Spectral Theory
20
IIT"II <_ 1/n!. So lim,,_o IIT"II1l" = 0. Consequently SpT = {0} by Theorem 3.2.8 (iii). Nevertheless 0 is not an eigenvalue, because fo f (t) dt = 0 for all x implies
f=0. Given a Banach space X and a linear subspace Y of X, we say that Y is an invariant subspace of the bounded linear operator T defined on X if T(Y) C Y. Obviously {0} and X are invariant subspaces and Y invariant implies Y invariant. So the interesting closed invariant subspaces are the nontrivial ones. Given an operatorT, it is intereitingto characterize all its closed invariant subspaces because in this way we can get more information about the structure and the spectrum of T.
In 1930, 3.vori Neumann proved that every compact operator on a Hilbert space has a soontrivial closed, invariant subspace. This was extended in 1954 by N. Aronszajn and K.T,Smith for compact operators on a Banach space. Using A. Robinson proved in 1966 that the nonstandard analysis, same result is true for * polynomially compact operator T, that is an operator for which there exists a polynomial p such that p(T) is compact. Later on in 1966, P. Halmos obtained the same result using classical analysis. But all the arguments they used are complicated. In 1973, V.I. Lomonosov applied the Schauder Fixed Point Theorem and a nice argument on compactness to obtain a wonderful and simple proof of a theorem extending all the previous ones. H.M. Hilden even succeeded in eliminating Schauder's theorem. We now present their arguments. THEOREM 2.2.11 (V.I. LOMONosov). Let K be a nonzero compact operator on an infinitedimensional Banach space X. Then there exists a nontrivial closed linear subspace of X which is invariant under all bounded linear operators commuting with
K. Suppose that the assertion is false. If A is an eigenvalue of K then we have N(IC  Al) invariant by all the operators commuting with K, so we get a contradiction. Consequeutly, K has no eigenvalues. So 0 is the only spectral value of K. Without loss of generality we may suppose that IIICII = 1. Let so E X be PROOF.
such that IIIC so II > 1, then we have Droll = IIICII Ilxoll > 1. Let S be the closed unit
ball with centre at so. Because lire ii > I we have 0 f S. Moreover for all x E S we have IIKxICxo11 < IIxxoll < 1 so K(S) is included in the closed unit ball with centre at K xo . Because 110  IC xo ll = II Kxo II > 1 we
conclude that 0 0 K(S). Let A be the closed subalgebra in ..(X) of all T such that KT = TK. For all y 76 0, Y = {Ty: T E A} is then an invariant subspace
Some CIRC a of Operators
21
for all T E A and Y # (0), so it is dense in X. Consequently, for all y 96 0 there exists T E A such that II T y  xo II < 1. For T E A, we consider the open set w(T) = {y: IITy  xoll < 1). Because K is a compact operator, the set K(S) is compact. All the w(T) cover X\{0} and 0 0 K(S), so there exists a finite subset I of A such that K(S) is covered by the w(T) with T E a. Consequently if y E K(S) there exists T E 3F such that IITy  xoll < 1. Until now this is the argument of Lomonosov. We now follow Hilden's argument, We have Kxo E K(S) so there exists Tt E a such that IIT1 Kx  xoll < 1, hence Tt Kxo E S, and consequently KT1Kxo E K(S). Thus there exists T2 E a such that IIT2KT1Kxo  xoll < 1, so T2KT1Kxo E S. By induction we can define a sequence (T") in 'such that IIT"KT"t ...T2KTjKxo  xoll < 1.
Let a = max{IITII:T E I}. By induction it is easy to prove that we have T"KT"I T2KTt K = (T" TI )K", so we conclude that: IIT"KT"t ...T2KTTKII < II(aK)"II. But Sp K = (0) implies Sp(aK) _ (0), so lim".,o II(aK)"II = 0 by Theorem 3.2.8
(iii). Consequently lim". II T" contradiction. 0
t
T2 KT1 ICxo 11 = 0, so IIxo II : 1 which is a
Let 7' he a pnlynornially compact operator on an infiniteditnensioual Urunaclt Space X. Then 7' has a nontrivial closed invariant subspace. COROLLARY 2.2.12.
PnooF. Suppose that p(T) = aol + a1T + + a"T" is a compact operator. If p(T) = 0 let x # 0 in X. Then the finitedimensional linear subspace generated by x,Tx,...,T"x is nontrivial and invariant by T. If p(T) # 0 then T commutes with p(T), and we then apply Theorem 2.2.11. 0
Is it true that every bounded linear operator on a Banach space has a nontrivial closed invariant subspace? In 1975, P. Enflo gave a counterexample to that conjecture, but his very long preprint contained many gaps (which can be removed in the opinion of C. Beauzatny). Finally, a complete counterexample was given by C. Read in 1984, and A.M. Davie subsequently produced an even easier one. But the problem is still open for operators on a Hilbert Space. If X is an arbitrary Banach space we have been able to build nontrivial elements in the subalgebra Cf + .LU(X ). Is it possible to exhibit a bounded linear operator which is not in this subalgebrn? Of course, if X is a corcrete Banach space, it is possible. But if X is an arbitr.uy Banach space it is still an open problem to know if CI { £C(X) is different from £(X). The most recent result towards the solution of this problem is given in S. Shelah and J. Steprans, A Banach space on which there are friv oprralor.+, Pror. Amer. Math. Soc. 104 (1988), pp. 101105.
A Primer on Spectral Theory
22
§3. Bounded Linear Operators on a Hilbert Space There are many examples of bounded linear operators on a Hilbert space. For
instance, if E is a closed linear subspace of H and if El denotes its orthogonal complement then we have H = E ® E. So for every x E H, there is a unique decomposition x = z1 +x2 such that xl E E, x2 E E. The linear mapping P from H onto E defined by Pz = z, is called the orthogonal projection on E. It satisfies 1; itn.obviously I  P is the orthogonal projection on El and P = P2, IIPU. =
ratifies I  P = (I 'P)2 aAJI  PII = I. Let (E,),51 be a family od orthogonal closed subspaces of H such that H is the Hilbertian direct sum of the Ei and let (Ai)iEl be a bounded family of complex numbers. Denoting by Pi the orthogonal projection on Ei, it is easy to verify that for a fixed x E H the family
TFx = E AiPiz iEF
is a Cauchy
r, for e11 finite subsets F of I. Let e > 0. There exists a finite subset F0 of I such that L;EF IIPixII' C e for all finite subsets F of I disjoint from Fo. Consequently, if we take finite subsets Ft and F2 of I such that Fo C F, C F2 we have ITp,x TF's=II' = II EiEF A,PjxII = EiEF,\F, IA,I2IIPixII2 < C2 EiEF,\F, IIPixII2 < C2 e, where sup IA, 5 C. So this implies that the family (Tpx) converges in H. If we denote by T the linear operator defined by
Tx = E AiPix iE1
then this operator is bounded by Theorem 1.1.13. We call it the strong aura of the series EIE! AiPi. It is important to notice that in general such a series converges strongly but not in norm. If the family (A,)iE! 90CI to 0, that is for every e > 0 there exists a finite subset Fo C I such that max.EF IA.I < e, for all finite subsets F C I disjoint from Fo, then the previous series converges in norm. As previously, for an arbitrary x we have
IITp,z  TF,XII2 _ E IAi1'IIPixII' < e' E IIPixII' < e'IIx1I', iEF,\F,
iEF,\F,
so IITp,  Tp, II < e. This implies that (TF) converges in norm in £(H) to some T and we have
T=1: AiPi. iE!
Some Classes of Operators
23
This situation will occur for selfadjoint and normal compact operators on a Hilbert space (see below). Another interesting class of bounded linear operators on a Hilbert space is given by the weighted shifts. Suppose that H is & separable Hilbert space having an orthonormal basis et, e2..... Then for all x = E' I ane. with =t IAn12 < +oo we define the right shift with weight given by a bounded sequence (an) by 00
Tx = F, anAnen+i n=]
It is easy to verify that IITII < maxn Ianl. We can also define the left shift with weight given by a bounded sequence (a,,) by 00
(with the convention that co = 0).
Sx = n=I
It is also easy to verify that IISII 5 max. Ianl. Of course T and S are not invertible because el 4 7Z(T) and el E N(S). Suppose that the weight is one, that is a. = 1 for all n. Then we have ST = I and TS = projection on the orthogonal complement
ofel #I. For more concrete examples of bounded linear operators on a Hilbert space the reader is refered to the classical book by P. Halmos [3].
If T E £(H) then the adjoitct of T denoted by T* satisfies the property (Taly) _ (x IT*y), for all x,y E H.
Then T " T* is an involution on £(H), that is it satisfies the following properties: (a) (T + S)* = T* + S*,
(b) (AT)* = ZT*,
(c) (T*)* = T, (d) (TS)* = S*T*. THEOREM 2.3.1. Let H be a Hilbert space and let T E £(H). Then we have the following properties:
(i) N(T) = R(T)1 and N(T*) = 7Z(T)1, (ii) IITII = IIT*II,
A Primer on Spectral Theory
24
(iii) IIT*T1I = IISII,
(iv) I + T*T is invertible in E(H) and
II(1+T*T)'II < I.
Obviously x E N(T) is equivalent to (Tzly) = (xIT*y) = 0 for all y E H, so x E 1(T*)i. The second relation comes from the first one and (c). We have IITxII2 = (TxITx) = (T*Txlx) < IIT*TIIIIxII < IIT*IIIITII 11x112. Consequently PROOF.
IITI12 < IIT*TII < IIT*111ITI1 This implies that IITII !5 IIT*11. So we have (ii) by (c)
and simultaneously we have property (iii). Moreover
1I(I +T*T)xll2 = ((I +T*T)2xlx) = II2I12 +2IITzII2 +
IIT*TxII2
? IIxJ
.
(1)
So I + T*T is injective. The same inequality implies that its range is closed. If is a Cauchy sequence then is also a Cauchy sequence. By (i) we have R(I +T*T)1 = Af(I + T*T) = {0} so R(I +T*T) = H, hence 1+ T*T is invertible in £(H). Moreover by inequality (1) we have 11(1+T*T)'II'5 1. 0
((I +
Among bounded linear operators three classes of operators are interesting. DEFINITION.
An operator T E £(H) is said to be
(i) aelfadjoint if T = T*,
(ii) normal if TT* = T*T,
(iii) unitary if TT* = T*T =1. Every T E t(H) has a unique decomposition T = R + IS where R, S E £(H) and R = R*, S = S*. These, the real part and imaginary part of T, are given by R = (T + T*)/2, S = (T  T*)/2i. Also it is easy to prove that T is unitary if and only if T is an isometry of H onto itself.
For T E £(H) we denote by p(T) the spectral radius of T, that is p(T) _ max{Ial: A E SpT}.
Let H be a Hilbert space and let S be a selfadjoint operator on H. Then we have the following properties: COROLLARY 2.3.2.
(i) IIS2I1= IISI12 and consequently IISII = p(S),
(ii) Sp.S C R and consequently Sp(T*T) C R+ for all T E L(H). By Theorem 2.3.1 (iii) we have I1S211 = IISII. Consequently by induction we get 1IS2" lI = 11511'". By Theorem 3.2.8, which we shall prove later, we have PROOF.
Some Classes of Operators
25
p(S) = limn.ao IIS°II'I^, so IITII = p(S). Let A = a + i,6 with n,/? E R. Denote
AI  S by S>,. Then SA=S,+i1I. So for all xEMwe have
IISAXII = (SAX Sir) _ (S0XI Saz) + I#I'IIzII2
and consequently IISi=II ? RIIkII
As in the proof of Theorem 2.3.1 we conclude that S.\ = Al  S is invertible for 13 # 0, so Sp S C R. By Theorem 2.3.1 (iv), I+ A2T`T is invertible for all A > 0 (we apply it to AT) so Sp(T*T) cannot contain a negative number. 0
It is easy to prove that the algebra £4(H) of compact operators is stable by the involution T '.4 T*. The argument is very similar to that of Theorem 2.2.6. We now study the particular case of selfadjoint compact operators on a Hilbert space which has several consequences for integral and differential equations. THEOREM 2.3.3. Let H be a Hilbert space and let T E £Q(H) be selfadjoint. Then either IITII or 11T11 is an eigenvalue of T. PROOF.
If T = 0 the result is obvious. So supppose that T # 0. By Corollary
2.3.2 we have SpT C R. Moreover, by Corollary 2.3.2, we have IITII = p(T) so there exists A y6 0, A E SpT such that IAI = 11711. Consequently A = IITII or A = IITII and by Theorem 2.2.10 it is an eigenvalue. 0
We know that selfadjoint n x n matrices can be diagonalized or equivalently can be written as A.P. where the P. are selfadjoint orthogonal projections and the A. are real numbers. This result can be extended to selfadjoint compact operators on a Hilbert space by the following: THEOREM 2.3.4 (SPECTRAL THEOREM FOR SELFADJOINT COMPACT OPERA
Let H be a Hilbert space and let T be a selfadjoint compact operator on H. Let (Ak)k>1 be the discrete set of nonzero eigenvalues TORS ON A HILBERT SPACE).
of T. Also let Eo = N(T) and Ek = N(T  AkI), for k > 1. Then we have the following properties:
(i) for k > 0 the closed subspaces Ek are orthogonal aod their Hilbertian direct sum is H. Moreover if Pk denotes the selfadjoint projection on Ek we have TPk = PkT for all k,
A Primer on Spectral Theory
26
(ii) the series Ek>1 AkPk converges in norm in E(H) and we have
T = E AkPk. k>1
PROOF. By Theorem 2.3.3, the set of nonzero eigenvalues is nonempty. So,by Theorem 2.2.10, it is either finite or a sequence converging to zero. If x E E, and y E Ei with i iA j then we have Tx = six and Ty = Aiy (with the convention that Ao = 0). So we have A,(xly) = (Txly) = Xi(x)y) = Ai(xly),
and consequently (xly) = 0. Let E = 45k>1Ek be the Hilbertian direct sum of the Ek, that is the closure of the algebraic sum of the Ek. Because T(Ek) = Ek and because T is bounded we have T(E) C E. The operator T being selfadjoint we also have T(E1) C El. If T were different from zero on El then its restriction to the closed subspace El would be a selfadjoint compact operator so, by Theorem 2.3.3, it would have a nonzero eigenvalue which would also be a nonzero eigenvalue of T. But this is a contradiction for otherwise we would have El fl Ek t {0) for some k > 1. So EL = Eo. Consequently H = *k>oEk. If SpT is finite it is obvious that Ek>1 Akpk converges. yo suppose that SpT is infinite. Let T. = E1
n+m AkPkXI12
IITn+mx  T%211' = II k=n+1
_
IAkI2IIPkzII' k=n+l
and Ek "+1 IIPkxII2 < IIxI12 because the projections are orthogonal. Let e > 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 finiterank 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 selfadjoint 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 finiterank operators T1 and Tz such that II Re TT1 II < e/2 and 11 Im T  T2II < e/2. Then Tl + iT2 has finite rank and IIT  (T1 + iT2)I) < e. 0
Some Classes of Operators
27
Theorem 2.3.4 can be reformulated differently. COROLLARY 2.3.6 (FREDHOLM ALTERNATIVE).
Let H be an infinitedimensional
Hilbert space and let T be a selfadjoint compact operator on H. Denoting by {AkIk>> the set of nonzero 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).
A Primer on Spectral Theory
28
EXERCISE 1. Prove that a commutative and semisimple finitedimensional complex algebra is isomorphic to C", where n denotes its dimension. What happens if it is a commutative and semisimple finitedimensional real algebra?
EXERCISE 2.
Let f be an analytic function from a domain D C C into M,(C).
Denote by ak the ktbsymmetric 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 (sax1) = 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
Some Classes of Operators
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 twosided 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 SturmLiouville 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 <_ It)I  Ily9I, for all z, y E A, we say that A is a complex Banach algebra.
If A has a unit (which is unique), denoted by 1, we can always suppose that 11111 = 1 because otherwise we can replace 11 II by an equivalent norm
111.111 satisfying
1111111 = 1 and IIIxvI II < IIIxII I IIIYIII for all z, Y E A (see Exercise III.1). If A is a Banach algebra without unit it is always possible to imbed it isometrically in the Banach algebra with unit A = A x C, where the operations and the norm are defined by (z, a) + (y, I9) = (x + y, a + 0), A(z, a) = (Az, Aa), (z, a)  (y,,8) = (zy + Ox + ay,aP), II(z,a)II = IIZII + IaI. In the definition of a Banach algebra, instead of the condition IIxyII < IIx1I . IIYU it is enough to suppose that the left multiplication x + zy and right multiplications z . yx are continuous, because with this hypothesis there exists an equivalent norm satisfying the standard norm
inequality (see Exercise 111.2).
The reader may be surprised by the fact that we consider only Banach algebras over the field of complex numbers and not Banach algebras over the field of real numbers. The reason is simple. Very often the proofs involve the use of analytic tools which are inefficient in the case of the field of real numbers. In many cases, it is possible to extend to real algebras results obtained for complex algebras. However,
Banach Algebras
31
the way to do this is not always obvious. So in this small book we shall restrict ourselves to the complex situation and, except otherwise stated, to the case of an algebra with unit. For the real situation, see the standard textbooks (11], (2] and [6]).
We now give some examples of commutative Banach algebras.
Let K be a compact set. Then C(K), the Banach space of all complex continuous functions on K, with the supremum norm, is a Banach algebra with unit. If K has only n elements then C(K) = C" with coordinatewise multiplication, and of course C is the simplest commutative Banach algebra. If K is a locally compact space, then the Banach space of all complex continuous functions on K which go to zero at infinity, that is continuous functions f such that {x: s E K, If(x)I > e} is compact for all e > 0, with the supremum norm, is a Banach algebra which has no unit if K is not compact. EXAMPLE 1.
EXAMPLE 2.
If K is compact, every closed subalgebra of C(K) is a Banach algebra.
For instance if K is a compact subset of C" the following examples of Banach algebras are interesting: P(K), the algebra of continuous functions on K which are uniform limits of polynomials on K; R(K), the algebra of continuous functions K which are uniform limits of rational functions with poles outside of K; and A(K), the algebra of continuous functions on K which are holomorphic on the interior of K. We have the inclusions P(K) C R(K) C A(K) C C(K), and there are examples in C" (with n > 2) for which all these inclusions are strict. Let G be a locally compact commutative group and let p be its Hear measure. Then L1(G) is a Banach algebra if we define multiplication by convolution EXAMPLE 3.
(f * 9)(x) = IG (xy1)9(y)dp(y)
f
and the norm by the L1norm IlfIi1 = fc If(x)I di (x). For instance if we consider the additive group L1(Z) = 11 then we have
(f * 9)(n) _ > f(n  k)9(k) 4EZ
HftI1= > If(k)I 4EZ
In fact this algebra 11 can be identified with the Wiener algebra W of continuous functions on (0, 21r] having Fourier coefficients a" such that E"EZ Ia" I < too
32
A Primer on Spectral Theory
and f (t) = E"EZ a"e'"t on [0, 21rl, with the operations of pointwise addition and multiplication and the norm IIf II = ikEZ Ianl. The Banach space of integrable functions on [0,1] with multiplication defined by convolution (f *g)(z) = fo f (xt)g(t) di and L'norm IIf II, = fe If (t)I dt is a Banach algebra. EXAMPLE 4.
Let us now give examples of noncommutative Banach algebras.
Let X be a Banach space. Then E(X), the algebra of all bounded linear operators on X, is a Banach algebra with unit for the usual operator norm. EXAMPLE 5.
If X is finite dimensional, with dim X = n, then .C(X) can be identified with M"(C). If dim X > 1 then .C(X) is not commutative. Except for finitedimensional ones, the simplest noncommutative Banach algebra is .C(H) where H is an infinitedimensional Hilbert space. This algebra £(H) has nice properties which we shall study in Chapter VI. Every closed subalgebra of C(X) is also a Banach algebra. For instance, the algebra of compact operators .Ct(X) is a Banach algebra which, by Theorem 1.1.2, has no unit if X is infinitedimensional. Using Exercise 111.3, it is easy to prove that every Banach algebra can be isomorphically represented as a closed subalgebra of £(X) for some Banach space X, but in practice this does not help very much. Starting with some Banach algebras, how can new ones be obtained?
For instance, given A a Banach algebra, it is possible to consider .C(A), the algebra of bounded linear operators on A, and all its closed subalgebras. Given a family (Aj),EI of Banach algebras we can define the Banach algebra product A, as the set of all families (si),e' such that a, E A, and sup;E1 IIziII < +oo with the norm IIail = suPIEI IIziI(. It is easy to verify that H Ai is a Banach algebra which contains an isometric copy of each Ai, (by the mapping z + (zi)IEI where zi = z if i = io and zi = 0 otherwise). If I = {l, ... , n}, obviously this product coincides with Ai x ... x A". fic I
Given a Banach algebra A it is also possible to consider the n x n matriz algebra
M"(A) with the standard operations
(aii)+(be,)=(a,,+bii), (aif).(bit)= (
aikbk7
and the norm II(a,s)II = sup:=1,...,"(II(gi1II + ... + I(ai"II)
M"(A) is not commutative for n > 1 and it contains an isometric copy of A (associating to z the diagonal matrix having only z on the diagonal).
Ianach Algebras
33
A much more useful tool is the following. Let I be a dosed twosided 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 twoaided 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 <_ II1!! = 1, and consequently III111I = 1 if A has a unit.
Let X be a Banach space. Then' and SE(X) are closed twosided ideals of £(X), so we can consider £(X)/' and F(X)/$9:(X). The latter is called EXAMPLE 6.
the Calkin algebra of X. If X is a Hilbert space they coincide, and we then have a Banach algebra with involution because EE(X) is stable by involution. For a given Banach algebra A there is a particular twosided ideal of A called the radical or the Jacobson radical of A which plays a very important role. By convention a left or right ideal in a ring is different from the ring. LEMMA 3.1.1 (W. KRULL).
Let A be a ring with unit 1. Every left (rasp. right) ideal of A is contained in a maximal left (reap. right) ideal. PROOF.
Let Ls be such a left ideal and let a be the family of all left ideals
containing Lo. This family is partially ordered by inclusion. This order is inductive because if E is a chain in ' then it is easy to see that either t f IEC I is a left ideal
or UIEC I = A. But this last case is impossible, because 1 v r for all I E C. By Zorn's lemma there exists a maximal left ideal in a. 0 'LEMMA 3.1.2 (N. JACOBSON).
Let A be an algebra with unit 1 and let z, y E A, A E C, with A qE 0. Then Al my is invertible in A if and only if A1 yz is invertible in A. PROOF.
Suppose that Al  my has an inverse u in A. Then we have (A1 zy)u =
u(A1xy)= 1. So we have: (Al  yx)(yux + 1) = Ayux + Al  y(xy)uz  yz
=Ayux+Aly(Au1)xyx= Al, (yux + 1)(A1  yz) = Ayux + AI  yu(xy)x  yx
= Ayux + A1 y(Au  1)x  yz = Al. Consequently Al  yz is invertible in A. 0
A Primer on Spectral Theory
34
THEOREM 3.1.3.
Let A be a ring with unit 1. Then the following sets are identical:
(i) the intersection of all maximal left ideals of A, (ii) the intersection of all maximal right ideals of A,
(iii) the set of x such that 1  zx is invertible in A, for all z E A, (iv) the set of x such that 1  xz is invertible in A, for all z E A. PROOF. By Lemma 3.1.2, (iii) and (iv) are equivalent. We only prove the equivalence of (i) and (iii), as the equivalence of (ii) and (iv) can be proved by a similar argument. Let x be in the intersection of all maximal left ideals of A. If a = 1  zx is not invertible then Aa is a left ideal of A so, by Lemma 3.1.1, it is contained in
some maximal left ideal Lo. Then xx E Lo and 1  zx E Lo, so 1 E Lo, that is Lo = A which is a contradiction. Conversely, suppose that 1  zx is invertible for all z E A. If x is not in the intersection of all maximal left ideals, it means that there exists a maximal left ideal Lo such that x f Lo. Then Lo + Ax = A and consequently 1  zz E Lo, but this gives a contradiction because it implies that we
have 1ELo.U This twosided ideal of A having properties (i)(iv) is called the radical of A and denoted by Rad A. If Rad A = {0} we say that A is semisimple. Of course simple rings (like are semisimple. There are many examples of semisimple Banach algebras, for instance examples 1, 2, 3. THEOREM 3.1.4. PROOF.
Let
Let X be a Banach space. Then E(X) is semisimple. ,
9t 0 be fixed in X and let It = {T:T E £(X),Tf = 0). It is
obviously a left ideal of .C(X). We prove that it is a maximal left ideal. Suppose that 3 is another left ideal containing 14, with It p< 3. Then .3f = (T f : T E 3} is a linear subspace of X different from {0} which is invariant under all S E £(X).
If 3E # X, let ql E 3f, 171 0 0, and qz 0 3C. Then there exists S E Z(X) such that Sql = qs. But this contradicts the invariance of 3C. So 3f = X, and consequently there exists U E 3 such that UC = f. For an arbitrary T E £(X) we have TU  T E IC, so T E 3 + 11 C 3, and hence E(X) = 3. Consequently RadC(X)c f1 .IE=(o).o Later on we shall give other characterizations of the radical. Even if a ring is not semisimple it is always possible to suppose we are in this situation using the following theorem:
Banach Algebras
35
Let A be a ring with unit 1. Then Al Rad A is semisimple. Moreover if x E A the coset i is invertible in Al Rad A if and only if x is invertible
THEOREM 3.1.5.
in A. PROOF. Let L' be a maximal left ideal of Al Rad A. Then L = {x: x E A, i E L'} is a left ideal of A. It is maximal because if J is a left ideal containing L, then L' C J = {i: x E J}, so that L' = J and thus L = J. Conversely, if L is a maximal left ideal of A, then L is a maximal left ideal of Al Rad A for similar reasons. By property (i) this implies that {x: i E Rad(A/ R,ad A)} is in the intersection of all
maximal left ideals of A, that is the radical of A. Hence Al Rad A is semisimple. It is obvious that if x is invertible in A then i is invertible in A/ Rad A. So suppose the latter property is true. There exists y E A and u, v E Rad A such that xy  I = is,
yx  1 = v. By property (iii), 1 + u and I + v are invertible in A, so we have xy(1 + u)'' = 1 and (1 +v)'yx = 1. Hence x is invertible. 0
§2. Invertible Elements and Spectrum Let A be a Banach algebra. We denote by G(A) the set of invertible elements of A. It is obviously a group containing the unit. We now prove that it is an open subset of A. THEOREM 3.2.1.
Suppose that A is s Banach algebra, x E A and JJxJJ < 1. Then 1  x is invertible and 00
k
kwo
PROOF.
Let JJxJJ = r < I and let s = F,'=u xk. Obviously, for n < m, we have
11s  s.11 < Z'km +1112JJk < Lit,
so
some element a = Fk o xk. Because xs = (1  z)a 1. 0
is a Cauchy sequence converging to
1 we conclude that all  x) =
COROLLARY 3.2.2.
Every left (resp. right, reap. twosided) ideal of A is disjoint from the open unit ball with centre at 1. Consequently every maximal left (resp. right, reap. twosided) ideal of A is closed. In particular Rad A is closed, so Al Rad A is a semisimple Banach algebra. PROOF.
Let L be a maximal left ideal of A. Then L contains no invertible elements, so the intersection of L with the open unit ball centred at 1 is empty by Theorem 3.2.1. Consequently 1 fl B(1, I) = 0 and so Z # A. This implies that L = 1 if L is maximal. By Theorem 3.1.3 (i), Rad A is closed. 0
A Primer on Spectral Theory
36
THEOREM 3.2.3. Suppose that A is a Banacb algebra and that a is invertible.
If fix  all < 1/IIa111, then x is invertible. Moreover the mapping z " x1 is a bomeomorphism from G(A) onto G(A).
PROOF. We have x = a + x  a = all + a'(x  a)). We have IIa'(z  a)II < Ilxall IIa' II < 1 so, by Theorem 3.2.1, l+a'(xa) is invertible, and consequently x is invertible. Moreover x1 = (1 + a'(x  a))'a' _ E4 0(a'(a  z))ka1. Consequently, we have 00
Ilx' a'II 5 IIa'll2llx  allE(Ila'll  IIx  all)k k=o
So x ' x' is continuous, and since it is its own inverse, it is a homeomorphisin. 0 As with operators we can define the spectrum of x, denoted by Sp x, as the set of A E C such that Al  x is not invertible in A. Also we define the spectral radius by p(x) = sup{ JAI : A E Spx).
We have A1 x = A(1 x/A), for A # 0. So Theorem 3.2.1 implies that A1 x is invertible for llx)) < )A), and consequently Spa is a bounded subset of C. In other words, it says that p(x) < Ilxil. Lemma 3.1.2 can be reformulated as saying that Sp(xy) U {0} = Sp(yy) U {0} and consequently p(xy) = p(yz), for all x, y E A. Also Theorem 3.1.3 is equivalent
to saying that the radical of A is the set of z E A such that p(xz) = 0 for all z E A. In particular it implies that the radical is included in the set of quasinilpotent elements, that is the set of elements a such that p(a) = 0. But in general this inclusion is strict (for instance consider M (C), C(X) etc...) and in some cases these two sets are identical (for instance if A is commutative, see the remark following Theorem 4.1.2). Theorem 3.1.5 implies that Spa = Sp i, for the coset i
ofxinA/R.adA. THEOREM 3.2.4.
Let A be a Banach algebra and let z, y E A. Suppose that
xy = 1 and yx # 1. Then Sp x and Spy contain a neighbourhood of 0. PROOF.
By hypothesis x is not invertible. Let p = yx # 1. Then p2 = y(xy)x = p.
Moreover (z  A1)y = xy  Ay = l Ay and y(z  Al) = p  Ay # 1 Ay. If I Ay is invertible, then
(z  A1)Y(1  Ay)1 = 1. Because y and (1  Ay)1 commute we have
y(1Ay)'(xA1)=(1Ay)'y(xA1)=(1Ay)'(pAY)#1
Banach Algebraa
37
and consequently x  Al is not invertible, that is A E Spx. So we have proved that B(0,1/p(y)) C Spx. The argument for Spy is similar. 0 In 1981, S. Berberian and I. Halperin, using a complicated method, proved the following corollary which is in fact an immediate consequence of Theorem 2.2.10 and Theorem 3.2.4. COROLLARY 3.2.5 (S.BERBERIANI.HALPERIN).
Let X be a Banach space and
let T, S E £(X) be such that TS = I and ST # I. Then for every invertible operator U E .Cf,X ), both operators T  U and S  U are not compact. PROOF.
Suppose T  U = K is compact for some U invertible. Then U1T 
I=
U` K = K1 is compact. Taking T1 = U 1 T and S1 = SU we have T1 S1 =
U1(TS)U = I and S1T1 = ST y6 I. So by Theorem 3.2.4, SpT1 contains a disk, but this is absurd by Theorem 2.2.10 since Sp Ti = Sp(I + K1) = 1 + Sp K1 is discrete. 0 THEOREM 3.2.6. Let A be a Banach algebra and let x E A. Then for every nonconstant polynomial p with complex coefficients we have Sp p(z) = p(Sp z).
Let q(z) = C(z  z1) ... (z be a given arbitrary polynomial. Then q(x)=C(xz11)...(xznl). Soq(x)isinvertibleifandonly ifeach ofthexz,l PROOF.
is invertible. Applying this to q(z) = p(z)  A, we get the result. 0
We say that x E A is algebraic of degree n if there exists a polynomial p of degree n such that p(x) = 0, and if q(x) 14 0 for all nonzero polynomials q of degree < n  1. Such a p with leading coefficient 1 is called a minimal polynomial for x (in fact it is unique!). THEOREM 3.2.7.
Given an integer n > 1, the set E. of algebraic elements of A
with degree < n is closed in A. PROOF.
Let (xk) be a sequence of E converging to xo and for each xk let pk be its minimal polynomial. We define a sequence of polynomials qk of degree n by: qk(Z) = Z^de"Pi, pk(Z).
Then the qk can be written: qk(z) = 4 + akz + ... +
ak1z"I
+Z n.
38
A Primer on Spectral Theory
Let Qk...... 8' be the n roots of qk. We want to prove that J8k J < JJxk JI. If Qk = 0, this is obvious. Otherwise $k is a root of pk. If it is not in Sp xk then Xk  611 is invertible, so we have r(xk) = 0, where r(z) _ _' = is a polynomial of degree less K
than the degree of zk, which is absurd. So 0' E Spxk and then ($,'I $ JJxkJJ. But we have
alk
0 k
=001+...+fik
= (1)nQk...#n
So the sequences (ak), ... , (ak') are bounded, because (xk) converges to xs. Using the BolzanoWeierstrass theorem, without loss of generality we may suppose that they respectively converge to ao,... , an,. Then by continuity we have a0 l +
alto + ... + an1xon' + xo. So xo is algebraic of degree < n. 17 THEOREM 3.2.8 (I.M. GELFAND).
Let A be a Banach algebra and x E A. Then
(i) A + (Al  x)' is analytic on C\Spx and goes to 0 at infinity, (ii) Sp x is compact and nonempty, (iii) p(x) = limn.oo IJxi1J'/n.
PROOF. We know that p(x) < JJzJJ. Moreover Spz is closed by Theorem 3.2.3, because Ao 0 Spx and IA  Aol < 1/II(Aol  x)'IJ implies A V Spx. So Spz is compact. Again by Theorem 3.2.3, A + R(A) = (Al  x)' is continuous on the
open set C\Spx. Let A,µ 0 Spx. We have (Al  x)  (µl  x) = (A  µ)l, so multiplying by R(A)R(p) we get R(p)  R(A) = (A  p)R(A)R(µ). Consequently Iimµa JAA R(A)2 by continuity of R. So R is analytic on C\Spx.
Moreover (Al  x)' = j(1  r/A)' for A ¢ Spx. So, for JAI > 2JIxJJ, we have by Theorem 3.2.1 JJ(A1  x)`'JJ s ICI
Hence it goes to zero at infinity. We now prove that Spx is nonempty. Suppose this is false and let f be a bounded linear functional on A. Then by (i), A _4 f((A1  x)') is entire and goes to zero at infinity so, by Liouville's theorem, it is identically zero. This being true for all f E A', by Corollary 1.1.4, we have (Al  x) ' = 0 for all A V Spx, but this is absurd.
39
Bsnsch Algebras
We now prove (iii). We know that p(x) < llx(I, for all x E A. So, by Lemma 3.2.6 applied to z", we conclude that P(x)" 5 Ilx"II
.
(1)
Let f E A'. Then A'. f((A1  x)') is holomorphic on C\ Sp x by part (i), so in particular if IAI > p(x). For JAI > ilxll, by Theorem 3.2.1, we have (2)
(M)
Consequently this is also true for (Al > p(x) by the Identity Principle. Let A be
fixed such that JAI > p(x). Then for every f E A' we have sup" I L I < +00. By Theorem 1.1.13 applied to A' and to the sequence of T. : A'  C, defined by T"(f) = L4 .I, we conclude that there exists a constant C, depending on A, such that Ilx"ll _< CIA(" for all n > 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 BeurlingGelfand 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 nonzero element is invertible then A is isometrically isomorphic to C. COROLLARY 3.2.9 (I.M.GELFANDS.MAZUR).
PROOF.
Let x E A. By Theorem 3.2.8 (ii), Spx is nonempty. 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
A Primer on Spectral Theory
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 nontrivial 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 nontrivial 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 < noo lim IIx"II1/" noo lim IIynII'/" = P(x)P(y). n+oo
Let a > p(x), p > p(y) and a = x/a, b = y/fl. Then p(a) < 1 and p(b) < 1. So there exists some integer N such that n > N implies max(IIa2" II, 11b 2' II) < 1. Defining yn = omaax" IIakII .
IIb2"1lk,
we have 1/z"
xky2"k
II(x+y)2"II'/2" =
<
n
; (2) akp2"kIIakII
1Y"
.
IIb'"kII
k=0
<(a +p)7p But it is easy to see that the sequence (y,,) is decreasing for n > N, because max
0
Ilak11.Ilb2n{1kll
= Max
(max0:5k<2, IIakII
IIb'"+'*II,2" Max
+l
11akII
IIb'"+'`kll
IN max(IIa'" II, II6'" II). 3/24 So we have p(x + y) = limn ryR/2" < II(x + y)2"11 <_ (a + 0)1im sup,, (a + p) lim supsa, 'r'N = a + ,B, for arbitrary a > p(x), p > p(y). Hence the w
theorem is proved. 0
41
Banach Algebras
Using Gelfand's theory (Chapter IV) it is possible to give an easier proof (see Exercise IV.3).
Suppose that A is a Banach algebra and that B is a closed subalgebra containing
the same unit 1. If x E B, what is the relation between the spectrum of x related to B, denoted by SpBx, and the spectrum of x related to A, denoted by Spwx? Of course, we have Spwx C SpBx, but is it possible to say more? be a sequence of invertible elements of A converging to a noninvertible element. Then lim,, JJx1 IJ = +oo.
THEOREM 3.2.11.
Let
PROOF. Suppose the result is false. Then there exist C > 0 and a subsequence of (x,,), which we denote in the same way, such that Ilz;,1II <_ C. Let x be the limit of the sequence (x,,). We have
and JJx1(z 1 + xn 1(z 
1 for n large enough so, by Theorem 3.2.1, CIJx is invertible, and hence z is invertible. So we get a contradiction. 0
COROLLARY 3.2.12.
Let x E A and let a be in the boundary of Sp x. Then there
exists a sequence (z,,) of elements of A such that
1 and
al) = 0. PROOF.
Because a E OSpx, there exists a sequence
converging to a with a
(z  al)z = (x s_aw
+oo, so
HJ(x 
_
Spz. We take x = (a 
of complex numbers
Then we have (a  a,,)x,,. Consequently +Iaa, 4. By Theorem 3.2.11, 0. The other result is proved similarly. 0 I
IFIT
.
In this situation we say that x  al is a topological divisor of zero. So all boundary points of the spectrum correspond to topological divisors of zero (but the converse is not true in general!).
Let A be a Banach algebra and let B be a closed subalgebra containing the unit 1. We have:
THEOREM 3.2.13.
(i) G(B) is the union of some components of B n G(A), and furthermore the set
OG(B)nBnG(A) is empty,
A Primer on Spectra! Theory
42
(ii) if x E B, then SpBx is the union of SPAX and a (possibly empty) collection of hounded components of C\SPAX, in particular BSpaz C BSPAZ.
(i) It is obvious that G(B) C G(A), so G(B) C B fl G(A). These last two sets are open in B. We prove that B fl G(A) contains no boundary point x of G(B). If such an x = limx", with x" E G(B), exists and is in G(A), by PROOF.
xi on G(A) we conclude that z`i = limzni. In particular continuity of z (11xni11) is bounded so, by Theorem 3.2.11, x is not in the boundary of G(B), a contradiction. Let Il be a component of B n G(A) that intersects G(B), and let U be the complement of G(B). Since BfG(A) contains no boundary points of G(B), it is the union of the two open sets fl n G(B) and Sl (i U. But l is connected, so cl f u is empty. Hence Sl C G(B). So (i) is proved. (ii) Let SPA  C\SpAx and flu = C\SpBx . Obviously RB C RA. A similar argument shows that OSle f f A = 0, so that SZB is the union of some components of PA. Hence (ii) is proved. 0 With the same hypotheses suppose that SpAx does not separate the complex: plane. Then SpBx = SpAx. COROLLARY 3.2.14.
This corollary will be used later when SpAx C R or when SpAx is finite or countable.
§3. Holomorphic Functional Calculus If x is in a Banach algebra A and if p(A) = as+ai A+ .,+&.A" is a polynomial with complex coefficients, we can define without any problem p(x) = aoI + aix + + a"x". Now if r(A) _ 6AI (where p and q are relatively prime) is a rational function with poles outside Spx then, by Theorem 3.2.6, q(x) is invertible, so we can define r(x) by p(x)q(x)'i. Is it possible to extend such definition to a larger class of functions"
Suppose that, f is holomorphic on the disk CAI < R and p(x) < R, so that akAk for JAI < R. This implies in particular that Fk o akJJxkJ) converges, so that f (z) = F,' o akzk converges in A to an element we call f (z). This argument can be applied in particular to define et, the exponential of x.
f (A) = E
U
But if f is only holomorphic on a neighbourhood of Sp z, is it possible to define f (x )? Of course this function cannot be defined by series because there is a problem in glueing the local pieces. But we now show that it can be done using the Cauchy formula for contours. This idea originates from the pioneering work of F. Riesz.
Banach Algebra
43
Let K be a compact subset of C supporting a measure p and let f be a continuous function from K into a Banach algebra A. Then, exactly as is done in the scalar situation, it is possible to define fK f(A) dp(A), and of course we have 0 (fk. f(A) dp(A)) = fK 0(f (A)) dp(A) for all bounded linear functional 0 on A (see for instance [7], pp. 7378).
Let r E A be fixed and let 1 be an arbitrary open set containing Spx. Let r he a smooth oriented contour contained in St\ Sp r and surrounding Spa. The set Sp x may be very complicated, but we may suppose that 0 is a finite 'anion of polygonal closed curves, without multiple points, and such that Spx is contained in the union of the bounded components they limit.
For f E H(Q), the algebra of holomorphic functions on 0, the integral z)_1 is defined and 2*t fr f{A,)(A1  x)'1 dA is welldefined because A . (AI continuous on I'. Moreover this integral is independent of the contour r surrounding Spx for the following reasons. Suppose that there exist two contours F1 i r2 in 12\ Sp x, surrounding Sp x such that
x1 = _
fr,
f(A)(Al 
x)1 dA it
z)1
tai Jr,
f(A)(A1
dA = z2 .
By Theorem 1.1.3 there exists 0 E A' such that +b(z1) 91 0(x2). We have
0(x1) = _ J h(A) dA r,
t
0(x1) =
tai JISh(A) dA
where h(A) = f(A)0((AI  x)"1). But by Theorem 3.2.8 (i), 4 is holomorphic on S2\ Sp x, so by Cauchy's theorem 0(x1) = 0(x2) and we get a contradiction. So we can set f(x) = 2 fr f(A)(A1x)'1 dA for an arbitrary contour 1' having the prey ious properties. But before doing that we must at least verify that this definition coincides with the standard one given for rational functions. LEMMA 3.3.1.
Let x E A, let r be a smooth contour surrounding Sp z and
let r(A) be a rational function having no poles surrounded by F. Then r(z) _ 22*, Jr r(A)(A1  x)1 dA.
PROOF. We first prove this for r(A) = (a  A)"*, with n E Z, and a not surrounded by F. Let
_ 1 r" =. tai r r(A)(AI  x)"1 dA.
A Primer on Spectral Theory
44
When A V Sp x we have the relation (A1 _ x)l = (al  x)1 + (a  A)(al  x)1(Al  x)1. So we have
r"
x)1 fr
 A)" dA + (al 
But the first integral is zero because A ' (a  A)" is holomorphic, so we have
rn+l = (al  x)rn.
(1)
Consequently the formula will be proved for (aA)", n E Z, if we succeed in proving it for n!= 0. We have
f (al  x)I dA = tai , r
r (Al 
x)1 dl
 tai /j1alR (i ± A
+... A
/
dA
fr(Al  x)"1 dA = 1. Thus the first part is proved. Now if r is an arbitrary rational function it can be written as anal a,,l al,kl + ... + an,k+ r(A) = p(A) al  A + ... + (al  A)k, + ... + an  A (an  A)k (2) where p is a polynomial in A and where the al, ... , an are the poles with their respective multiplicities k1,...,kn (decomposition in simple elements). A formal calculation implies that we also have for R > f f xtj. So
r(x) = p(x) + al,l (all  x)1 + ... + al,*,( (all  x)k' + .. . ....+ a,,,l(an l  x)1 + ... + 1  x)k
(3)
So the first part implies the result. 0 COROLLARY 3.3.2 (CAYLEYHAMILTON THEOREM). Let a be a n x n complex matrix and let p(A) = det(a  Al) be its characteristic polynomial. Then p(a) = 0.
Let a, ... , ak be the different eigenvalues of a and let r be the union of k small disjoint circles with centres respectively at al, ... , ark. By Lemma 3.3.1 we PROOF.
have
p(a)
2Ii Jrp(A)(Al 
a)l dA.
But (a A 1) 1 = F.t _Myb(A), where b(A) is a n x n matrix depending analytically on A since its (i, j) entry is the (j,i) cofactor of a  Al, and so is a polynomial in A of degree < n  1. We then have
p(a)=2'i by Cauchy's theorem. 0
45
Banach Algebras
Let A be a Banach algebra and let x E A. Suppose that St is an open set containing Spx and that r is an arbitrary smooth contour included in SZ and surrounding Spx. Then the mapping f .+ f (x) = *, Jr f (A)(A1 x)1 A from H(A) into A has the following properties: THEOREM 3.3.3 (HOLOMORPHIC FUNCTIONAL CALCULUS).
(i) (f1 + f2/)(x) = f1(x) + f2(x), (11) (f1 . f2)(x) = fl (X) . f2(x) = f2(x)  fl (x),
(iii) 1(x) = 1 and I(x) = x (where I(A) = A),
(iv) if
converges to f uniformly on compact subsets of It, then f (x) = lim
(v) Spf(x) = Asp X). (i) is obvious by definition. (iii) follows immediately from Lemma 3.3.1. converges uniformly to f on r and II(A1  x)111 is (iv) is also easy because bounded on this set. We now prove (ii). By Runge's theorem (see (7], p.288) there exist two sequences of rational functions (rk) and (r2), with poles not surrounded PROOF.
by r, converging uniformly respectively to f1 and f2. By Lemma 3.3.1 we have (rkr2)(x) = r'(x) . r2(x). So by (iv), property (ii) is true. We finish by proving (v). If f has no zero on Spx then g = 1/f is holomorphic on a neighbourhood S1 1 of Spx. If necessary we can replace r by a contour r1 c I1 surrounding Spx. We have f (A)g(A) = 1 on r1 and consequently, applying (ii) and (iii) for r1 and Its, we get f (x)g(x) = 1. Thus f (x) is invertible in A. On the other hand, if f (a) = 0 for some a E Spx, then there exists h E H(St) such that f (A) = (a  A)h(A) on It, and consequently f(x) = (al  x)h(z). But al  x is not invertible, so f(x) is not invertible. Therefore fit  f(x) is not invertible if and only if f takes the value f on Spx. 0 We now give several elementary applications of this extremely important theorem. THEOREM 3.3.4. Let A be a Banach algebra. Suppose that x E A has a disconnected spectrum. Let U0, U1 be two disjoint open sets such that Spx C Uo U U1, Sp x fl U0 # 0 and Sp x fl U1 54 0. Then there exists a nontrivial projection p commuting with x, such that
Sp(pz) = (Spx fl Ul) U {0} ,
Sp(x  px) = (Spx fl Uo) U {0}
.
PROOF. We consider the holomorphic function f defined on U = Uo U U, by f(A) = 0 on Uo and f(A) = 1 on U1. We set p = f(x). Because f(A)2 = f(A), we
A Primer on Spectral Theory
46
get p2 = p. By construction p commutes with x, by Theorem 3.3.3 (ii) and (iii). By part (v) of Theorem 3.3.3 we have Spp = (0, 1), sop is nontrivial. Moreover, we have px = f(x)I(x) = (f I)(x). But (f I)(A) = A on U1 and (f I)(A) = 0 on Uo, so by Theorem 3.3.3 (v) we have Sp(px) = (Sp z fl U1) U {0}. For Sp(x  px) the result is obtained similarly. 0
Let A be a Banach algebra. Suppose that x E A and that a V Sp z. Then we have THEOREM 3.3.5.
dist(a, Sp x) PROOF.
1
x)1).
p((al 
Let 11 be an open set containing Spz, but not a. Then f(A) = 1/(a  A)
is holomorphic on 11. So by Theorem 3.3.3 (v) we have
Sp(al  x)1 = {
1
a A
AE
Spz}
.
JJJ
So in particular,
p((al 
z)1)
11
= sup { la
: A E Spx ? Al
=1/inf{jaAl:AESpx}=1/dist(a,Spx).0 Let A be a Banach algebra. Suppose that x E A has a spectrum which does not separate 0 from infinity. Then there exists y E A such that z = ell. In particular, for every integer 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 nontrivial 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 (Ixall < 1/IIa'((. Then IIIa'xli =
IIa'(ax)II < 1 and so p(1a'x) < 1.
Consequently Sp(ai 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(alz) such that ey = alx. 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 nonempty 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 twosided 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 twosided 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 = ux+ 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
A Primer on Spectral Theory
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(nn2) nn2
°
n2(nn2)
1
nn21/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 selfadjoint operators (see Theorem 6.2.1), but it is not continuous for practically all selfadjoint 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 = ek 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
eken+, ,ifn=2k(21+1),forsome 0
,
otherwise.
So IIT  Tk1I < ek, and hence the operators Tk converge to T.
We have Tmen = anon+1 ' ' ' an+mlen+m and consequently we have JIT'"II = supn(anan+l "' an rm1) By the definition of the sequence (an) we have et
alai ... azs 1 =
flexp(j2tjl) i=t
So (alai ...
azsi)'/(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. 3638). We now give some positive results. The first one has been known for a long time.
A Primer on Spectral Theory
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 BolzanoWeierstrass 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 G6subset 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) \ (f`I (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
A Primer on Spectral Theory
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 19681970, 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 <
2a J 0
If
f(A(, +rei*)d8,
IIf(Ao +re1e)II d8.
For every polynomial p, IeP(A) I IIf(A)II = IIeP(a)f(A)II and A  eP(A) f (A) is analytic,
so, by the first part, IeP(')I 11f(A)II is subharmonic. Then the BeckenbachSaks theorem (see Theorem A.1.8) implies that log II f(A)II is subharmonic. 0 THEOREM 3.4.7 (E. VESENTINI). Let f be an analytic function from a domain D of C into a Banach algebra A. Then A " p(f(A)) and A 4 log p(f (A)) are
subharmonic on D.
Bansch Algebras
53
By Theorem 3.2.8 the sequence j log 11f (A)2"11 decreases and converges to log p(f (A)). Because A }+ f (A)2" is analytic, then by Lemma 3.4.6 the function A _ log 11 f (A)2" 11 is subharmonic. So log p(f (A)), being a decreasing limit of subharmonic functions, is subharmonic. Taking the composition of this function with e`, which is convex and increasing, we conclude that p(f(A)) is subharmonic. Cl PROOF.
It is fair to say that B. Schmidt and V. Istraj,escu proved Theorem 3.4.7 practically at the same time, but E. Vesentini was the first to obtain nontrivial applications in spectral theory of this nice and simple theorem. In the rest of this section and mainly in Chapter V, we shall see that Theorem 3.4.7 has a huge number of important applications. COROLLARY 3.4.8.
a Banach algebra A.
Let f be an analytic function from a domain D of C into Suppose that a $ Sp f(A) for all A E D. Then A
1/ dist(a, Sp f (A)) and A +  log dist(a, Sp f (A)) are subharmonic on D. PROOF.
By Theorem 3.3.5 we have dist(a, Sp f (A)) =
ai_ f 0,J) 1
so we apply
Theorem 3.4.7 to the analytic function A 4 (al  f0 COROLLARY 3.4.9.
Let f be an analytic function from a domain D of C into a
Banach algebra A. Define
u(A)=max{Reu:uESpf(A)}, v(A)=min{RevvESpf(A)}.
Then u(A) = log p(ef(')), v(A) = logp(eand u and v are subharmonic on D.
PROOF.
It is obvious that v(A) =  max{Rev:v E Sp(f(A))}, so it is sufficient
to prove that u is subharmonic. We first prove that max{Reu u E Spa) = logp(e=), for z E A. By Theorem 3.3.3 applied to f(z) = e', we have logp(e=) = max(log le"I : u E Spz) = max(Reu : u E Spx). Consequently u(A) =log p(eftal). Then the result comes from Theorem 3.4.7. 0 :
COROLLARY 3.4. 10. Let03,91 < 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},
A Primer on Spectral Theory
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)) <
(a + 1)c because (a + 1)z ¢ Spg(A2) and Spg(A3) C R(az,c). But p(g(A)) < (a + 1)c = p(g(A1)), for every A E C. So by the Maximum Principle, p(g(A)) _ (a + 1)c and this is a contradiction. Consequently {z : z E Sp f(A),Jz{ = p(f(A))} is constant. 0
Let f be an analytic function from a domain D of C into a Banach algebra A. Suppose that Sp f (A) C R for all A E D. Then Sp f (A) is constant on D. COROLLARY 3.4.12.
Banach Algebras
55
PROOF. First we prove that it is locally constant. Let As E D. Replacing f(A) by of (A) + 01, with appropriate a, $ > 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 onetoone 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.
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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 G6set 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
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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' nonempty. 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) = ek(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 autoadjoints 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 nonempty 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
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{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 limkmoo 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 nonthin 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 nonthin 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 nonthin 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
<
2
= 2 dist(t1t, Sp f (Ao)) < 2Jq,  f, I < e/4.
Consequently ff(i,,/4) meets 8 Sp f (µ(e)). But ff(f j, e/2) D $(r7i, a/4), so meets 8 Sp f (p(e)) for all i = 1, ... , n. If f E 8 Sp f (Ao), there exists some f j such that If  f i J< e/2, so there exists some C E 8 Sp f (p(e)) such that if  CCI < e. Consequently 8Sp f(Ao) C 8Sp f(p(e))+B(0,e). 0 In particular, this theorem can be applied if E is a subdomain U of D and A0 E D is a boundary point of U, and also if E is a Jordan are included in D ending at Ao E D (see Theorem A.1.17 and Theorem A.1.18). COROLLARY 3.4.23.
With the hypotheses of Theorem 3.4.22, suppose that there
exists a closed subset F of C such that 8Sp f(A) C F for all A E E. Then asp f(Ao) C F. If moreover F has no interior point and does not separate the plane then Sp f (Ao) C F. PROOF.
By Theorem 3.4.22 we have 8 Sp f(Ao) C F+ B(0,11k) for all k > 1, so 0Spf(A0) C F. If the interior of Sp f(Ao) is empty the conclusion is obvious. So
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suppose this interior nonempty. 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 nth spectral diameter of x, denoted 6n(x), by n n+l
6n(x)=max
[J
J
1Ajl
(I
for n + 1 arbitrary points A1, ... , An+, in Spx. For n = 1 it is the classical diameter of Sp x denoted by 6(x). THEOREM 3.4.24.
Let f be an analytic function from a domain D of C into a
Banach algebra A. Then for arbitrary n > 1 the functions A A " log 6n(f (A)) are subharmonic on D.
6n(f (.1)) and
PROOF. Suppose that n = 1. Let a E A and dal = 1. By Corollary 3.4.9 the length of the projection of Sp z on the line (t5. t E R) is given by log p(e°i) + log p(e°s). Consequently 6(z) = max (log p(e°i) + log p(e
By Theorem 3.4.7, A 64 logp(e°1ta)1+logp(e°fta)) is subharmonic, consequently A , 6(f(A)) satisfies the mean inequality. But by Theorem 3.4.2, it is upper semicontinuous. So A i 6(f (,1)) is subharmonic. We have Jesta)16(f(A)) = 6(ePta> f (A))
for every polynomial p, and .1 ' e't) f (A) is analytic, so by the first part is subharmonic. Then the BeckenbachSaks theorem implies that A H log 6(f (A)) is subharmonic (see Theorem A.1.8). For n > 2, the proof is much more complicated. It uses tensor products of Banach algebras and joint spectrum, and was given for the first time in 1982 by Z. Slodkowski. In Chapter VII we shall indicate how this result can be obtained more geometrically (Theorem 7.1.3 and Theorem 7.1.13). 0 For the definition of,capacity see the Appendix.
Banach Algebras
63
Let f be an analytic function from a domain D of C into a Banach algebra A. Then either the set of A E D such that Sp f (A) is finite is a Bore] set having zero capacity, or there exist an integer n > I and a closed discrete subset E of D such that THEOREM 3.4.25 (SCARCITY OF ELEMENTS WITH FINITE SPECTRUM).
#Spf(A)=nfor AED\Eand #Spf(A)
Let F = {A: # Sp f(A) < +oo} and Fk = {A: # Sp f(A) < k}. Then
A E Fk if and only if log bk(f (A)) = oo. But, by Theorem 3.4.24, log bk(f (A)) is subharmoaic. So, by H. Cartan's theorem (see Theorem A.1.29), either Fk is a G6set having zero capacity for all k > 1, or c(F,,) > 0 for the smallest integer n > I such that log bn(f (A)) _ oo on D, in which case we have Fn = D. The first case implies that F is a Bore] set and that c(F) = 0. The second case implies # Sp f (A) < n, for all A E D. Let E _ { A:,\ E D, # Sp f (A) < n). We shall now prove that E is closed in D. Let (At) be a sequence of elements of E converging to Ao E D. If Ao 0 E then Sp f(Ao) contains n distinct points as,...,an. We choose e > 0 such that all the disks B(ae, e) are disjoint. By Theorem 3.4.4, for a large enough we would have Sp f (A) fl B(a;, e) 36 0 for i = 1,...,n and consequently # Sp f (Ai) = n, which is a contradiction. So E is closed. Let Ao V E. Then Sp f (Ao) = {c , ... , an}. The previous argument, with Theorem 3.4.4, implies in fact that # (Sp f(A)flB(ai, e)) 1 for !A  AoI small enough. Consequently, by Theorem 3.4.20 we have Sp f(A) = {a1(A), ..., an(A)), where the a; are holomorphic on a neighbourhood of A0. For every A E D\E the function '(A) = fy<, 0 such that (A  Al < r, with A 0 E, implies that one of the disks contains at least two points of Spf(A). Consequently j0(A)J < 2eb(f(A})" But b(f(A)) is upper semicontinous at A1i so limaa, O(A) = 0. By the Rado's extension theorem (see [7), p.280), 0 is holomorphic on all D. Consequently its set of zeros, precisely E, is discrete. D We immediately give an application of the scarcity theorem to spectral theory. In the period 19521955, F.V. Atkinson, B.Sz.Nagy and Ju.L. Smul'jan proved independently the following result: let A  f(A) be an analytic function from a domain D C C into the algebra of compact operators on a Banach space and
let z 96 0, then the set of A E D such that z E Sp f (A) is a closed and discrete subset of D. Their argument was essentially based on the fact that the projections
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64
associated to isolated eigenvalues of compact operators have finite rank. If D = C and f(A) = AK for some fixed compact operator K then this result says nothing more than Theorem 2.2.10. For a general f, B. Sz.Nagy believed that this result was deeper than Riesz's theorem. Actually it does not depend on the fact that f (A) is compact but only on the geometry of the graph of the multifunction A Sp f (A), namely that Sp f(A) has at most 0 as a limit point for all A E D. THEOREM 3.4.26.
Let f be an analytic function from a domain D C C into a
Banach algebra A. Suppose that for all A E D the spectrum of f (A) has at most 0 as a limit point. Let z # 0. Then either the set of A E D such that z E Sp f (A) is closed and discrete in D, or z E Sp f (A) for all A E D.
Suppose that z E Sp f(A0) for some Ao E D. We shall prove that A0 is either isolated or interior in the set E = {A: A E D, z E Sp f(A)}. Because z 96 0 there exists an open disk L centred at z and not containing 0 such that PROOF.
LT fl Sp f (Ao) = {z}. By Theorem 3.4.4 and upper semicontinuity of the spectrum, there exists r > 0 such that 1AAoI < r implies i flSp f(A) # 0 and 8Af1Sp f(A) _ 0. Let h be the function defined by
z forzEA {0 forzo&
h(z)_
By Theorem 3.3.3 and the hypothesis, we have # Sp h(f (A)) < +oo and # (0 fl Sp f (A)) = # (Sp h(f (A))\{0}) for IA  AoI < r. So, by Theorem 3.4.25, there exist an integer n > 1, a closed discrete subset F of the disk B(A0, r) and n functions which are holomorphic on B(Ao, r)\F, such that i f1 Sp f(A) _
for A E B(Ao,r)\F.
There exists s such that 0 < a < r and B(Ao, a)flF C {Ao). Then the functions al, ... , an are holomorphic on B(Ao, a), except perhaps at Ao. Moreover, by upper semicontinuity of the spectrum we have lima.ao a;(A) = z for i = 1, 2, ... , n. Therefore the a;'s can be extended holomorphically to the whole disk B(Ao, a). It follows that either &;,(A) = z for some io or there exists t with 0 < t < a such that a;(A) 9& z for all A E B(Ao, t)\ { Ao } and i = 1, ... , n. In the first can A0 is an interior point of E while in the second case A0 is isolated in E. To finish the argument we
consider the set E' of all limit points of E in D. By upper semicontinuity of the spectrum, E is closed, so E' C E. Let p E E'. Since p is not isolated in E it is an interior point of E, hence an interior point of E'. So E' is both closed and open in D. Consequently we have either E' = 0 or E' = D and the proof is complete. 0
Banach Algebras
68
This result can be generalized supposing only that for every A, Sp f(A) is at most countable, with the conclusion that E is either closed and at most countable, or all D. But the proof is much more complicated and uses the theory of analytic multifunction (see Chapter VII).
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66
EXERCISE 1.
Let A be a Banach algebra for a norm II.11 such that 1111134 1. Setting
IIIx111= maXuy11<1 IIxyIJ, prove that IIIxyIII S III=III IIIyIII,
5 If IxNl 5 IIxll and
1111111 = 1, for allx,yEA.
Suppose that A is an algebra which is, at the same time, a Banach space for a norm II ' 11. Suppose moreover that the left multiplications x H xy and right multiplications x . xy are continuous for all y E A. Prove that A is a Banach algebra for a norm I I1 'III equivalent to 11 EXERCISE 2.
11
Given a Banach algebra A prove that there exists a continuous isomorphism of A onto a closed subalgebra of .C(X ), for some suitable Banach space X. EXERCISE 3.
EXERCISE 4.
Pr we that £ (X) + Cf is semisimple.
L,rt H be a [filbert space.
*EXERCISE 5.
£(H)/Ztr(H) is
Prove that the Catkin algebra
Is it still true in the case of a Banach space?
EXERCISE 6. Let p be a projection of a Banach algebra A. Prove that pAp is a closed semisimpl subalgebra of A, with unit p. EXERCISE 7.
Let I be a closed twosided ideal of a Banach algebra A. Prove that
Rad I= f n Rad. t. EXERCISE 8.
Give another proof of Theorem 3.2.8, part (iii).
EXERCISE 9. Given n elements x1,...,x of a Banach algebra A, suppose that x;x, = 0, for i # j. Prove that Sp(x1 + + (Spx1 U . . . USpxn) \ {0}. EXERCISE 10. Let A be a Banach algebra without unit. Denote by e the element (0,1) of A. Prove that
3(p(x)+ Ia1) <_ p(x+ \e) <_ o(x)+ IaI,
for all x E A and .\ E C, where p is the spectral radius in A. EXERCISE 11.
Let x, y be two commuting elements of a Banach algebra A. Prove
that Sp(x + y) C Spx + Spy. Conclude that b(z + y) < 6(x) + 6(y). Give an example of a Banach algebra A and of an element x E A such that x is a topological divisor of zero and 0 V BSpx.
*EXERCISE 12.
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Banach Algebras
Let A be a Banach algebra and let z E A. Suppose that U is an open set containing Sp x, that f is holomorphic on U, that U1 is an open set containing f(Sp x) and that g is holomorphic on U1. Prove that (go f)(x) = 9(f (x)). ExE1tCISE 13.
EXEactsE 14. Let A be a Banach algebra and let x E A. Suppose that f (x) = 0, whore f is holomorphic on a neighbourhood of Sp x. Prove that a is algebraic. EXERCISF' 15.
Let A be a Banach algebra and let r.,y E A. Prove that er4Y = Is it true in general that c=+r = e:. fr?
EXERCISE 16. Let T be a n x n matrix and let f be holomorphic in a neighbourhood of the spectrum of T. Suppose that a1, ... , a are n distinct numbers in the domain of definition of f Prove that f(T)f(a')xfTaft
i= 1
EXERCISF 17.
3#I
ai
 aj
Let A be a commutative Banach algebra. Prove that the group of
invertible elements of A i3 either connected or has an infinite number of components. What happens in the noncommutative case? (See V. Paulsen's result).
Extend Corollary 3.4.12, supposing that Spf(A) C r for all A E D, where t in an analytic arc of the complex plane. EXERCISE 18.
EXERCISE 19.
Let f be an analytic function from an open set D C C into a Banach
algebra A. Suppose that E is a closed disk included in D and that Sp f(A) C R, for all A E 8L . Prove that Sp f(A) C R and is constant on T. Let u be an element of a Banach algebra A which is a projection modulo the radical. Prove that there exists a projection in A wich is equal to is EXERCISE 20.
modulo the radical. EXERCISE 21. Let a, b be two elements of a Banach algebra A. Suppose that Sp(a + Ab) contains only one element, denoted by a(a + ab), for all A E C. Prove
that a(u + Ab) = a(a) + Aa(b), for all A E C. As a corollary prove that if A is semisimple and Sp x contains one element for all z E A then A is isomorphic to C. *EXERCISE 22.
algebra A.
Let f be an analytic function from a domain D C C into a Banach
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A Primer on Spectral Theory
(i) Suppose that for every A E D the convex hull of the spectrum of f (A) is a linear segment [p(A)a(A), a(A)] where 0 < p(A) < 1. Prove that a is holomorphic on
D and that p is contant on D.
(ii) Suppose that for every A E D the convex hull of the spectrum of f(A) is [0, a(A)]. Prove that a is holomorphic on D.
Chapter IV REPRESENTATION THEORY
§1. Gelfand Theory for Commutative Banach Algebras Let A be a Banach algebra. A linear functional X on A is called a character of A if it is multiplicative and not identical to 0 on A. This last condition is equivalent to saying that x(l) = 1 because X(x) = x(x)x(l). If x is a character of A it is easy to verify that X(x) E Sp(x), for all x E A, because (x  X(z)1)y = y(x  X(x)1) =
1 leads to an absurdity. Consequently IX(x)j < p(z) S lizil, so a character is continuous and of norm one.
Many algebras have no characters, for instance M"(C) for n > 1, £(H), etc. (see Exercise W.1).
In 19671968, using analytic tools, A. Gleason, J.P. Kahane and W. Zelazko gave a nice characterization of characters. We now give a very elementary proof of this result which is due to M. Roitman and Y. Sternfeld.
Let A be a Banach algebra. Then X E A' is a character of A if and only if(z) E Spa, for aliz E A. THEOREM 4.1.1.
PROOF.
Suppose that X(x) E Spa, for all x E A. In particular X(1) E Sp1, so
X(1) = 1. Let p(A) = X((,\1 z)") for n > 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
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70
and consequently n
n ai = nX(x) (=1
(n)X(.T2). 2
i
However, n
a
\ f
n
n
=>A+2
A;aj
A + n(n  1)X(x2).
So n2IX(x)2 X(x2)I < np(x)2+nlx(x2)I. This being true for all n > 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 xX(x)1 is not invertible because (xX(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((xAl)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 quasinilpotent 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
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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 StoneWeierstrass 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) nz00
SOX =Xs.0
,
for all f E W.
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COROLLARY 4.1.5 (N. WIENERP. 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
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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. 113119). 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 semisimple,
(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),
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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 semisimple 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 semisimple 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 semisimple 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.
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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 f^II = 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 semisimple. 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 IleAs2Ehy1I = p(e'A"xe") =
p(x) So the analytic function A * ex"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
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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 <_
max Ip(x(xi),...,X(xn))I = xE A)
xE
()
Wp(xl,...,x"M
P(P(xn,...,x")) < IIp(xn,...,xn)D Hence Xo(p(xn,... , x,,)) = p(z) defines a character on the algebra of polynomials in x1,. .. , x,, which can be extended continuously to A. Since z = (Xo(xj ), ... , Xo(x.))
we have zEK.0 The original proof of Nagasawa's theorem, stated below, uses Choquet's bound
ary and is rather complicated. We now give a simplified proof based on ideas of G. Niestegge.
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Let A be a commutative Banach algebra. We say that a linear functional f on A is a spectral state if it satisfies f(1) = 1 and If(x)I < p(x) for all x E A. The last condition implies that If (x)j < Ijx fl, so f is continuous. Obviously a character is a spectral state. For every subset E of a vector space, its convex hull is denoted by co E.
LEMMA 4.1.15. A linear functional f is a spectral state on A if and only if f (X) E coSpx for all x E A. PROOF.
If f(x) E coSpx for all x E A then f(1) E coSpl = {1}, so f(1) = 1,
and moreover )f(x)) < max{lz):z E coSpx} = p(x). Conversely, for A
coSpx we have if (x)  AI = If (x  A1)I < p(x  Al), which means that f (z) is in the disk centred at A with radius p(x  Al). But co Sp x is the intersection of all such disks B(A, p(x  Al)) for I AI > 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 semisimple
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
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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
(TI Y)
+ g2(T 'Y) 2
As was done previously, it is easy to verify that gl o TI and 92 o T1 are spectral states on B. Hence, f being extreme, we have f = gl oT1 = 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 semisimple. 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. Taking u3=u/f, v1=v/f, we have 1z where IIuIII = IIvIII < 1. Let z E K be arbitrary. Then we have 2 = ul(z) + v1(z) and lu1(z)I < 1, Ivl(z)I < 1. Consequently ul(z) = vl(z) = 1, so ul = vi = 1, and hence f is extreme. Conversely suppose that IIf II = 1 and If (z)I < 1 at some point of K. Then, by continuity, we have the same inequality on a closed set E 96 K having a nonempty interior. Because K is normal, there exists h E C(K) with support included in E and such that sup:EE Ih(z)I < 1  sup=EE If(z)I. Now we have
f=(f+h)2(fh) andIIf+hII=IIfhp= sup I(f±h)(z)I=1, sEK\E
and consequently f is not extrelnal. 0 COROLLARY 4.1.19 (S. BANACIIM. STONE).
Let K1, K2 be two compact sets. Suppose that T is a linear isometry from C(K1) onto C(K2). Then there exist u E C(K2) such that Iu(x)I = 1 for all x E K2 and a homeomorphism h from K2 onto K1 such that
(Tf)(x) = u(x)f(h(x)) for aJlzEK2 and f EC(K1). PROOF.
Let u = T1. Then IIu1I = 1. Suppose that u
with II f1I, IIg1I < 1.
Then 1 = T `/+2 and IIT'f II, IIT1yII S 1. By Lemma 4.1.18, 1 = TI If =
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T1g, so u = f = g, and hence u is extremal. Consequently lu(x)i = 1 for all x E K2. Let S be the linear mapping defined by Sf(x) _ s We have 11Sfil = II Tf 11 = Ilf 11 and S1 = 1. Because the norm coincides with the spectral radius on C(K) we conclude, by Theorem 4.1.17, that S is an isomorphism from C(K1) on
C(K2). Consequently f ' Sf(x) is a character, and so has the form f " f(h(x)) for some point h(x) E K1. It is now easy to verify that h is bijective and continuous and so is a homeomorphism. 0
REMARK 2.
In the hypothesis of Corollary 4.1.19 it is sufficient to suppose that
TO = 0, T(if) = iTf and IITf  Tgli = Ilf  gil, for all f, g E C(K1). Then the linearity of T results from a classical theorem due to S. Mazur and S. Ulam (see S. Banach, Theorie des operations lineaires, New York, 1955, pp. 166168).
§2. Representation Theory for NonCommutative Banach Algebras Let A be a Banach algebra. We say that a is a representation of A on a complex vector space X of dimension larger or equal to one if a is a nontrivial morphism from A into the algebra of linear operators on X. If a linear subspace Y of X satisfies
a(x)Y C Y for all x E A, we say that Y is invariant under 7r(z). A representation it is said to be irreducible if the only linear subspaces of X invariant under a(x) are {0} and X. This representation is said to be bounded if X is a Banach space and if s(z) is a bounded linear operator on X for all x E A. Moreover it is said to be continuous if it is bounded and if there exists a constant C > 0 such that IIir(x)iI 5 Cllxll for all x E A.
For instance if A is commutative then qvery character X is a continuous irreducible representation on X = C. If A = .C(X) for some Banach space X then a(x) = x is a continuous irreducible representation on X. Let L be a maximal left ideal of A (which exists by Lemma 3.1.1). It is closed by Corollary 3.2.2, so X = AIL is a Banach space for the norm IIIaiil = i rEL Ila+ull. There is a natural representation sr of A on X defined by a(x)a = za.
This continuous representation is called the left regular representation associated to L. It is irreducible because L is a maximal left ideal. The kernel of the representation is (L : A) = {z: x E A, zA C L}.
Reps. sentstion Theory
THEOREM 4.2.1.
81
We have the following properties:
(i) for every irreducible representation w of A there exists a maximal left ideal L such that Ker w = (L : A), consequently Ker w is a closed twosided ideal of A,
(ii) the radical of A is the intersection of the kernels of all continuous irreducible representations of A, (iii) for every r in A the spectrum of x is the union of all the spectra of the a(z) in the corresponding algebras w(A) for all continuous irreducible representations w.
PROOF.
and let C
(i) Let 7r be an irreducible representation on a complex vector space X x E A} contains and is invariant 0 be in X. Then F =
under r, so F = X. Let L = {x: x E A, w(x)1 = 0). Because F = X we have L
A, so L is a left ideal of A. Let J be a proper left ideal of A containing L.
Then either J = L or {w(z)e: x E J} is different from {0} and is invariant under w.
z E J} = X. Consequently there exists e E J such that w(e) = C, so xe  x E L for all x E A. Then for any x E A we have z = (x  ze) +ze E L+ J C J, so that A = J, which is a contradiction. Consequently L is a maximal left ideal. So
Obviously Kerw c (L : A). If X E (L : A) then xA C L, so w(z)w(y)C = 0 for every y E A. Consequently w(z)X = 0, hence w(z) = 0, that is z E Kerw. (ii) If x E Rad A then xA C Rad A C L for all maximal left ideals L. Consequently z E (L : A) for all such maximal left ideals L. By part (i), x is in the
intersection of the kernels of all continuous irreducible representations of A. Conversely if x is in this intersection then x E (L : A) for all maximal left ideals L.
Hence x=zlELforallsuch L,sothat xERadA. (iii) It is obvious that C SPAS. Conversely let A E SPAS. We prove that there exists a continuous irreducible representation w of A on a Banach space X such that A E Sp4(X)w(z) C Replacing z by Al  x, we may
suppose that z is not invertible in A. Suppose first that x has no left inverse in A. Then there exists a maximal left ideal L containing x. Considering the left regular representation associated to L, we have w(z)1 = x = 0, so w(x) is not invertible in L`(A/L). Suppose now that z has a left inverse t in A. Of course
it does not have a right inverse. Let u = 1  sl. Thus uz = z  x(lz) = 0. If u E Rad A then zl = I  u is invertible, which implies that z has a right inverse and hence a contradiction. Consequently u RadA. By part (ii), there exists a continuous irreducible representation w on a Banach space X such that w(u) 0 0. Since w(u)w(x) = 0 this implies that w(z) is not invertible in E(X). 0
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REMARK 1. In fact the notions of bounded irreducible representations and continuous irreducible representations are the same (see [2), Theorem 7, p. 128). REMARK 2.
If A is a commutative algebra then maximal left ideals coincide with
maximal ideals and (L : A) = L. By Corollary 3.2.9, AIL
C. So in this case
irreducible representations coincide with characters. Conversely, if all continuous irreducible representations of a Banach algebra are characters, then by Theorem 4.2.1 (ii), Al Rad A is commutative. So representation theory on Banach spaces with dimension greater than one is only interesting for noncommutative Banach algebras. In some sense Theorem 4.2.1 is a generalization of Theorem 4.1.2.
Let A be a Banach algebra and let x be a continuous irreducible representation of A on a Banach space X. Then C = {T:T E THF:oREM 4.2.2 (1. SCHUR).
11(X),T7r(r) = x(x)T, for all x E A) is isomorphic to C. PROOF.
It is obvious that C is a closed subalgebra of E(X) containing the identity.
Let T # 0 be in C. Then Tx(x)(X) = w(x)T(X) C T(X) for every x E A, so T(X) is invariant under r, and consequently T(X) = X. Also KerT is invariant under x, but Ker T = X is impossible, so T is invertible as a linear operator. By the Open Mapping Theorem, T is invertible in .C(X) and its inverse satisfies Trl x(x) = 7(x)T1 for all x E A, so T1 E C. By Corollary 3.2.9, C is isomorphic to C. 0 We now prove a very important theorem due to N. Jacobson. In fact this result can be extended to the general situation of non. commutative rings (see I. Herstein, Noncommuiative Rings, 1968). LEMMA 4.2.3. Let w be a continuous irreducible representation of A on a Banach space X. If El, f2 are linearly independent in X there exists a E A such that w(a)II = 0 and a(a)42 910.
Suppose that w(x)41 = 0 implies x(x)42 = 0. Let Li = {x: x E A, x(x)ei = 0), for i = 1, 2. They are both maximal left ideals and L1 C L2, PROOF.
so L1 = L2 = L. The linear mappings T; from AIL into X, defined for i = 1, 2 by Ti(a) = x(a)41, are bounded and bijective. Let D = T2Ti 1, which is a bounded linear operator on X. Let E E X and suppose that 4 = r(b)41. Then we have x(a)D4 = w(a)T2T, 1({) = w(a)T2(b) = w(a)w(b)42 = w(ab)6,
Dx(a)4 = T2Ti lx(ab)ty1 = T2(ab) = x(ab)42i
Representation Theory
83
so D,(a) = n(a)D for all a E A. By Theorem 4.2.2 there exists A 34 0 such that D = Al. Then T2 = AT1. So taking a = 1 we get S2 =
which gives a
contradiction. 0 Let wr be a continuous irreducible representation of A on a Banach
LEMMA 4.2.4.
are linearly independent in X there exists a E A such that space X. If zr(a)Ci = 0 for 1 < i < n  1 and ir(a)* 0. PROOF.
This result is true for n = 2 by the previous lemma. We proceed by
induction. Suppose n > 2 and that the property is true for n1 linearly independent
vectors in X. There exists al E A such that r(al)ti = 0 for 1 < i < n  2 and 1r(at)t and xr(aj)t,, 0. If 7r(a1)G_1 = 0, we have finished. If are linearly independent then, by Lemma 4.2.3, there exists a2 E A such that n(a2)7r(a1)G_1 = 0 and ir(a2)x(a1)t,, A 0. So we take a = a2a1. We now suppose f are A # 0. The vectors S1,. .. ,1n2, that liucarly independent so there exists a3 E A such that a(a3)t;i = 0 for 1 < i < n  2 and 1r(a3)(AG_1  fin) # 0. If a(a3)Sn_1 = 0, we have finished. So suppose
a(a3Xn_1 # 0. If
and ir(a3) are linearly independent there exists
a4 E A such that a(a1)1r(a3)en1 = 0 and 1r(a4)7r(a3)C. 34 0. Then we take a = a4a3. So suppose once again that we have;11r(a3)Sn_1 = ir(a3){n. By hypothesis, A # p. Because a(a3)Sn_1 0 0 there exists as E A such that x(as)a(a3)r;n_1 = ir(at)e _tt1. Taking a =tt a1  as/a3 we have tr(a)ct; =`c 0 for I < i < n1 and r(a)C. = fi(al)SnAfi(a1)4n1 px(a$a3)tn1 = (A  A)n(a1)fn1 54 0.13
Let x be a continuous irreducible representation of A on a Banach space X. If 41,. .., n are linearly independent in X and if r11, ... , qn are in X there exists a E A such that 7r(a)£i = rji, THEOREM 4.2.5 (JACOBSON DENSITY THEOREM).
for i.=1,...,n. PROOF.
By Lemma 4.2.4 there exists b4 E A such that 0.
0 if i t k and
Then there exists ck E A such that 1(ck)x(bk)G = r!k We take
a=c1b1 If a is an irreducible representation of A over a vector space X of dimension n, then ir(A) is isomorphic to (see Exercise IV.13). The next corollary is very useful.
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84
With the hypotheses of Theorem 4.2.5, we supare linearly independent. Then there exists a invertible pose further that in A such that w(a)(; = qi for i = 1,... , n. COROLLARY 4.2.6 (A. SINCLAIR).
PROOF.
Let F be the finitedimensional linear subspace of A generated by Given the hypothesis, there exists a linear mapping T from
(i, .
F onto F, which is invertible in .C(F), such that T(; = q;, for i = 1,... , n. The algebra Z(F) is isomorphic to a matrix algebra Mk(C) for some k < 2n, so it is a Banach algebra. Because Spe(F)T is finite, by Theorem 3.3.6, there exists R E £(F) By theorem 4.2.5, such that T = eR. Let B be a basis of F containing there exists a E A such that w(a)( = R( for all ( E B. Consequently w(a) and R coincide on F. In particular, w(a)k and Rk coincide on F for all integer k > 1. So, by continuity of w, we get
w(eo)(j =e R(;=T(;=q;,
foci=l,...,n
and e° is invertible in A. Q
To finish this section we give a result which was proved by I. Kaplansky in his book Infinite Abelian Groups, Ann Arbor, 1969, using the theory of finitely generated modules. Let T be a linear operator on a complex vector space X. We say that T is locally algebraic if for every 4 E X there exists a nontrivial polynomial p such that p(T)4 = 0. The standard result of 1. Kaplansky states that boundedly locally algebraic (the degree of p is bounded independently of () implies algebraic (for another proof see the book by H. Radjavi and P. Rosenthal, Invariant Subspaces, Berlin, 1973). This important result has many consequences. We present an analytic proof of the result which is very interesting because it implies a surprising generalization.
Let X be a complex vector space and let T be a linear operator from X into X. Suppose that there exists an integer n > 1 such that (, T4,. .. , T"( are linearly dependent for all ( E X. Then T is algebraic of degree less than or equal to n. THEOREM 4.2.7 (I. KAPLANSKY).
Suppose that n is the smallest integer having this property. Hence there exists (o E X such that (o, T(o, ... , T"'(u are linearly independent but (o, T(o, ... , T"(o are not. Then there exists a monic polynomial po of degree n such that pa(T)to = 0 and if p is another monic polynomial of degree n PROOF.
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Repreentation Theory
p = po Let q E X be an arbitrary fixed vecsuch that tor. We now prove that po(T)q = 0. Let F be the linear subspace generated by fo, Tfo,
,
T"fo, ri, Ti?,. .. , T"q. Then dim F <_ 2n. For A E C we set
fo(A) = fo + aq E F
f, (A) =Tfo(A) E F
fn1(A) =T"'fo(A) E F g(A) = T"fo(A) E F.
Because fo(0), ... , f"_1(0) are linearly independent in F there exist n linear functionals on F, denoted by o, ... , 0"1, such that Oi(fj(0)) = 6ii
,
for 0 < i, j < n  1.
(1)
We define
Mfo(a))
4o(f1(A)) 01(f1(A))
0n1(f0(.X))
O"1(f1(A))
00(fo(A))
... ...
Oo(f"10))
01(f"10))
... mn1(f"10))
which is a polynomial of degree < n, satisfying A(O) = 1. Let E be the finite set of its zeros. From the hypothesis we conclude that for A gf E there exist CIO (A), ... , a"._1(A) E C such that
g(A) = ao(A)fo(A) + ... + a"1(A)fn1(A),
(2)
so we have
00(g(A)) = ao(A)Oo(fo(A)) + ... +a"1(A)Oo(fn1(A)) (3)
a"1(A)4"1(f"1(a))
"1(9(a)) =
By Cramer's formulae the a, coincide on C\E with rational functions. Relation (2) can be written:
(pa(T)fo(A)=0, for A0Ewith ...  a0(A)1, Sl pa(T) = T" a"1(A)T'"' 
(4)
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Let us denote by #1(A),   , #"(A) the roots of the polynomial p,,. We have 
(T01(A)1)...(Tfl"(A)1)fo(A) =0, forAfE,
(5)
and obviously (T32(A)1) ... (Tpn(A)1 fo(A) 96 0 for X 4 E, by the definition of E. So (5) implies that 81(A) is in the spectrum of T. A similar argument implies that Q2(A), ... , Q"(A) are also in the spectrum of T. Consequently IPi(A)I < JIT1H, for i = 1, ... , n and A E, where IITh is the operator norm corresponding to a given norm on the invariant subspace F. So the symmetric functions ao(A), . , a"_1(A) are also bounded on C\E. Because the a; coincide with rational functions on C\E, we conclude from Liouville's theorem 'that there are constant numbers 7o... , 7n1 E C
such that cq(A) = 7t, for A 0 E. Let p(z) = z"  7"_1 z"1     70 Then 
p(T)fo(A) = 0 on C\E, but also on C, by continuity in A. In particular 0, for an arbitrary q E X. Hence po(T) = 0, so p= T is algebraic of degree < n. 0 COROLLARY 4.2.8 Let X be a complex Banach space and let T be a bounded linear operator from X into X. Suppose that for every t; in X there exists an
integer n (depending on f) such that e, T f, ... , T"C are linearly dependent. Then T is algebraic. PROOF.
Fork >I let Xk =
E X and ,Tt,...,Tkt are linearly dependent).
Then Xk is the set off for which there exists a monic polynomial p with degree < k, such that p(T)e = 0. By continuity of T and Lemma 2.2.1, the sets Xk are closed.
By hypothesis X = Uk 1Xk. Consequently, by Theorem 1.1.1, there exist a ball B(to, r) and a smallest integer m such that t: E X,", for r. Let rl E X be a fixed arbitrary vector. Then to + trl E B(to, r) if 1tl  flq) < r. With this hypothesis there exists a monic polynomial p of degree < m such that p(T)(to + tq) = 0. Moreover there exists a monic polynomial po of degree < m such that po(T)£o = 0. Hence po(T)p(T)rt = 0. So q E X2,". We then apply Theorem 4.2.7 with n = 2m
to get the result. 0 The proof of the next theorem is a slight modification of the argument used in the proof of Theorem 4.2.7.
Let X and Y be two complex vector spaces and let T1t...,T" be linear operators from X into Y. Suppose that for every t E X the vectors Tl 4, . , T. t; are linearly dependent. Then there exist a 1, ... , a" E C not all zero such that S = a1T1 + + a"T" has finite rank < n  1. Moreover, if X = Y and THEOREM 4.2.9.
the Ti commute, then S2 = 0.
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Representation Theory
PROOF.
If for all ( E X, the vectors T1f,...,Tn1( are linearly dependent, it is
So suppose that there exists (o E X such that T10,. . , Tnl fo are linearly independent and T10,. .. , T, o are not. Then there exist as,...,an_1 E C such that enough to prove the result with Ti , .
. , T,_1.
(6)
(Tn + anITn_1 + ... a1T1)(o = 0.
Let 9 E X be an arbitrary fixed vector and let F be the linear subspace of Y generated by T: (o, , Tn(o, Ti q, . , Tn? . Then dim F s 2(n  1). For A E C we set fo(A) = Co + Aq
f1(A)=T1fo(A)EF (7)
fn1(A) = Tntfo(A) E F 9(A) = Tnfo(A) E F.
Because fl(0},..., fn_1(0) are linearly independent in F there exist n  1 linear functionals on F, denoted by 01, ... , On1 , such that
O;(fi(0))= 5,,,
for 1si,j sn1.
(8)
We define
,&(,k)_
01 (f) (A))
01(f2(A))
.
Y'1(fnI(A))
02(f1(A))
42(f2(A))
...
02(fn1(A))
10n1(f1(A))
0n1(f2(A))
which is a polynomial of degree $ n  1, satisfying A(0) = 1, and
4'1(f1(A))
...
01(90))
...
01 (fn1(A))
02(fl(A))
...
42(9(A))
...
02(fn1(A))
0n1(f1(A))
0n1(9(A))
... 0n1(fn1(A ))
T
i 1 h column
which is also a polynomial of degrees n  1 , satisfying A,(0) = o;, by (6) and (8). If E denotes the set of zeros of A then, arguing as in tht proof of Theorem 4.2.7, we conclude that
(A(A)TT  An1(A)T.1  ...  A1(A)T1)fo(A) = 0
(9)
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on C\E, but, by continuity in A, this relation is true on all C. Let a = 1 and let 61,. .. , #,, be the coefficients of A respectively in 1(A),.. . ,
& _1(A), 0(.1). (which does not depend on q!), R = 91T1 +Setting S = a1 T1 +(which depends on q!) and looking at the coefficients of degree 0 and I in A, from (9) we obtain: Seo = 0 Sq+No=O. Consequently Sq is in the linear subspace generated by T1 CO, , T.1{o So S has finite rank < n  1. If moreover the T; commute, then S and R commute, so
SIq = SNo = RS{o = 0. Hence S' = 0. 0 REMARK 3. Let P and Q be two different projections, having the same range of dimension 1, defined on a complex vector space X. For every C E X the vectors PC and QC are dependent, and obviously there are linear combinations of P and Q having rank one. But aP. +,3Q 0 for every a, Q E C. So in general it is impossible to have S = 0 in Theorem 4.2.9.
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Representation Theory
*EXERCISE I.
Prove that M"(C) and E(H), where H is a Hilbert space, have no
characters. EXERCISE 2. Let A be a commutative Banach algebra and let x E A. Prove that there exists a character X such that p(x) = IX(x)I. EXERCISE 3.
Prove Corollary 3.2.10 and Theorem 3.4.1 using Theorem 4.1.2.
Let K1i K2 be two compact spaces. Suppose that there exists a linear isometry from C(KI) onto C(K2 ). Prove that KI and K2 are homeomorphic. EXERCISE 4.
Let K be a compact subset of C". Prove that M(P(K)) is homeomorphic to K. Suppose now that every continuous function on K can be uniformly approximated on K by polynomials. Conclude that K = K. EXERCISE 5.
***EXERCISE 6.
Give an example of a curve r in c" (n > 2) such that I' # r. (If
you do not succeed look at (10]). For such an example conclude that c(r) # P(F). EXERCISE 7. Let S be a completely regular Hausdorff space and let Cb(S) be the algebra of bounded continuous complexvalued functions on S. Prove that Cb(S) is a Banach algebra for the norm IIf1I = sups if Prove that Cb(S) is isomorphic to C(X), for some compact set X such that S is homeomorphic to a dense subset of X, and every bounded continuous complexvalued function on S extends continuously to X (X is called the StoneOech cornpactification of S).
Let K be a compact and convex subset of the complex plane. Prove that K is the intersection of all closed disks containing K. EXERCISE 8.
EXERCISE 9. Let DI and D2 be two domains of the complex plane. Characterize explicitly all the linear isometries from A(D1) onto A(D2).
EXERCISE 10. Let A be a commutative Banach algebra. A boundary for A is a closed subset E of 27l(A) such that supxEE IX(f)I = SUPxEWA) IX(f)I, for every f E A. Prove that the intersection of all boundaries of A is a boundary of A (called the Ji1ov boundary of A). Given a compact subset K of C determine explicitly the Silov boundaries of C(K) and A(K).
Let K be a compact set and let I be a submultiplicative norm on C(K) (which is not complete). Prove that Ilf Ii. <_ If I, for all f E C(K). This important result is due to I. Kaplansky. If you do not succeed look at the book by
**EXERCISE 11.
A. Sinclair mentioned at the end of Chapter V, §5.
I
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A Primer on Spectral Theory
Try to prove the following result of H.Rossi. Let .4 be a commutative Banach algebra and fix Xo E 97l(A)\S, where S denotes the Nov boundary of A. Let U be a neighbourhood of Xe with U CTi(A)\S. Then for all f E A we
***EXERCISE 12.
have
IXo(f)I <_ sup IX(f)I. XEMU
If you do not succeed please read 1101, Chapter 9.
EXERCISE 13. U t A be a Banach algebra and let it bean irreducible representation of A on a linear vector space of dimension n. Prove that 7r(A) is isomorphic to
Is it. possible to extend Theorem 4.2.9 to a countable family of linear operators from X into Y?
*EXERCISE 14.
Chapter V SOME APPLICATIONS OF SUBHATIMONICITY
The powerful technique of subharmonic functions which we introduced in Chapter III, §4, has a great number of applications in spectral theory.
§1. Some Elementary Applications We start with a simple application which improves Theorem 3.4.2 and the Gersgorin theorem for matrices.
Let a be an element of a Banach algebra and let U be a bounded open set containing Spa. Then sup, 8u p((a  Al)1(x  a)) < 1 implies Sp x CU.
THEOREM 5.1.1.
If A E C\ZT, then aAl is invertible, so we define f(A) = (aA1)1(xa). This function is analytic on C \ ii and goes to 0 at infinity. By the Maximum Principle for subharmonic functions on unbounded open sets (see the remark following Corollary A.1.4), we conclude from the hypothesis that p(f(A)) < 1 on C \ U. So 1 + f(,\) is invertible for A ¢ U. Hence (a ,\1)(1 + f (A)) = x  Al is invertible for A V U. Consequently Spa C U. 0 PROOF.
Let x = (a,1) be a n x n matrix and let a be the diagonal matrix with all, ... , ann on the diagonal. Suppose that fore > 0, the disks B(al 1, e), ... , B(an n, e) have disjoint or identical boundaries (this last case occurs if a,, = ajj for i # j). Then liz  all < e implies Sp x C B(a1 1, e) U . U B(an n, e). COROLLARY 5.1.2 (S.A.GER§GORIN).
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92
PRooF. W e h a v e p((a  A1)1(x  a)) S ]](a  A1)' [[ ]]x  all. But the diagonal
matrix (a  Al)' has the diagonal coefficients 1/(all  A),, 1/(a..  A), and Al)_11] = maxi 1/laii  Al = 1/e when A lies on the boundary of hence ]E(a U = B(al 1, e) U
U B(a,,,,, e). We then apply Theorem 5.1.1. 0
.
THEOREM 5.1.3 (D.C.KLEINECKEF.V.SIROKOV). Let a, b be in a Banach alge
bra. Suppose that a(ab  ba) = (ab  ba)a. Then p(ab  ba) = 0. PROOF.
Let [x, y] be the commutator ay  yx of z and y. Then
eAabeAa
=
= b + A[a, b], for all A E C. But p ++ p(pb + [a, b]) is subharmonic on C, so by Theorem A.1.2 we have b+ A11a,JN + 2= 1[a, b]] +
p([p, b]) = iim sup p(pb + [a, b]) = 0. 0 N ...0
Poo
Let a, b be in a Banach algebra. Suppose that (ab ba)a = 0, or that a(ab  ba) = 0, and that 0 is in the boundary of some component of C \ Sp a. Then p(ab  ba) = 0. THEOREM 5.1.4.
PROOF. Suppose for instance that (ab  ba)a = 0, the other case being studied similarly. Then we have (Al  a)b = (b  z(ab  ba))(A1  a) for all A 0 0. Hence (Al  a)b(Al  a)1 = b  X(ab  ba), for all A f Spa. Let U be the component of C \ Spa such that 0 E &U. Then ]A]p(b) = p(Ab (ab  ba))
for A E U.
Because A ++ p(Ab  (ab  ba)) is subharmonic, we conclude from Theorem A.1.2 that p(ab  ba) = lim sup p(Ab  (ab  ba)) = 0. 0 AEU
A0
In particular, if T is a compact operator on an infinitedimensional Banach space, then (TS  ST )T = 0 implies p(TS  ST) = 0.
§2. Spectral Characterizations of Commutative Banach Algebras If A is commutative we know from Corollary 3.2.10 and Theorem 3.4.1 that the spectral radius is subadditive, submultiplicative and uniformly continuous on A. By Theorem 3.1.5, the same result is also true supposing A/ Rad A to be commutative. Surprisingly, the converse is true.
First we strengthen a lemma due to C. Le Page.
Given a Banach algebra A, then by definition 2(A), the centre rnodulo the radical of A, is the set of a E A such that ax  za E R.ad A for all z E A.
Applications of Subhsrmonicity
THEOREM 5.2.1.
93
Let a E A be such that # Sp(ax  xa) = 1 for all x E A. Then
a E Z(A). PROOF.
Let r be a continuous irreducible representation of A on a Banach space
X. Suppose there exists { E X such that C, q = r(a)e and r(a)q are linearly independent. Then, by Theorem 4.2.5, there exists z E A such that
r(x)! = 0
r(x) i = e r(x)r(a)17 = 17
Then ir(ax  xa)f = r(x)q = e and r(ax  xa)q = q + q = 0. So we have {0, 1) C Sp (ax  za) C Sp(ax  xa) which is a contradiction. Consequently for t: E X, the vectors a;, r(a)f and r(a)2e are linearly dependent. By Theorem 4.2.7, r(a) is algebraic of degree _< 2. Without loss of generality we may suppose that r(a)2 = yl, for some y E C. As a matter of fact if we have r(a)2 = or(a) +,61, then taking a' = a  1 we have Sp(a'x  xa') = Sp(ax  xa), for all x E A, and r(a')2 = (fi + a2/4)1. If r(a) is not algebraic of degree 1, there exists i; E X such that F and q = r(a)C are linearly independent. By Theorem 4.2.5 there exists x E A such that
r x(x)f = . Then r(ax  xa)% = q  (t; + q) _ t: and r(ax  xa)q = r(a)(e + q)  r(x)ry£ _ q +7y  7{ = q. Consequently {1,1} C Spr(az  xa) C Sp(ax  xa) which gives a contradiction. Hence a(a) is algebraic of degree 1, that is r(a) = al for some a E C. Then r(ax  xa) = 0 for all continuous irreducible representations r. So by Theorem 4.2.1 (ii), we have ax  za E Rad A, for all x E A. 0 THEOREM 5.2.2.
Let a E A. Then the following properties are equivalent:
(i) a E Z(A), (ii) there exists M > 0 such that p(a + x) < M(1 + p(x)), for every x E A,
(iii) there exists N > 0 such that p((a  a1)'x) < Np((a  .11)1)p(x) for every
xEAandA¢Spa. Changing A for A/ Rad A if necessary, we may suppose without loss of generality that A is semisimple. Then (i) implies (ii) and (iii), by Corollary 3.2.10, PROOF.
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94
taking M > max(1, p(a)) and N > 1. We now prove that (ii) implies (i). Let u E A be fixed and A E C. We define an analytic function from C into A by a  ea"aea` A
AA) _
[a, u]
,
for A&0
,
for A = 0.
(1 + For A # 0 we have p(f(A)) < i (1 + p(a)). So the subharmonic function A '+ p(f (A)) goes to zero at infinity. Hence it is identically 0. Consequently p(au  ua) = 0, for all u E A. By Theorem 5.2.1, a E Z(A). If we look at the proof of Theorem 3.4.1 we conclude that (iii) implies Spy C Spa+Np(ya), for every y E A. In particular we have p(a + x) < p(a) + Np(x) < M(1 + p(x)) for M > max(p(a), N). So (iii) implies (ii). 13 COROLLARY 5.2.3.
Let A be a Banach algebra. Then the following properties are
equivalent:
(i) Al Rad A is commutative,
(ii) p is subadditive on A, that is there exists M > 0 such that p(x+y) < M(p(x)+ p(y)), for all x,y E A,
(iii) p is submultiplicative on A, that is there exists N > 0 such that p(xy) < Np(x)p(y), for all x, y E A,
(iv) p is uniformly continuous on A, which implies that there exists C > 0 such that lp(x)  p(y)l < CJIx  y1l, for all x,y E A. We now give another characterization of Z(A) due to S. Grabiner, along with a new proof. We recall that 6 denotes the diameter. THEOREM 5.2.4.
We have a E Z(A) if and only if sup:EA
6(ezae=
 a) < +oo.
Suppose that a and b commute. Then we have 6(a + b) < 2p(a + b) < 2(p(a) + p(b)) by Corollary 3.2.10. In fact it is possible to prove that 6(a + b) < 6(a) + 6(b), see Exercise I11.11. Suppose a E Z(A). Then 6(e=ae=  a) < 2(p(a) + p(exae=)) = 4p(a), using the remark just before Theorem 3.2.4. We now prove PROOF.
the converse. Let u E A be fixed and let A E C. Then A r, 6(1(A)) is subharmonic on C by Theorem 3.4.24, f (A) being defined as in the proof of Theorem 5.2.2. But it goes to zero at infinity. So it is identically zero. Consequently 6([a,u]) = 0, or equivalently # Sp(au  ua) = 1, for all u E A. Hence we have a E Z(A) by Theorem 5.2.1.13
Applications of Subharmonicity
95
Q3. Spectral Characterizations of the Radical THEOREM 5.3.1 (J.ZEMANEK).
Let A be a Banach algebra. Then the following
properties are equivalent:
(i) a is in the Jacobson radical of A,
(ii) Sp(a + x) = Spx, for all x E A, (iii) p(a + x) = 0, for all quasinilpotent elements x in A, (iv) p(a + x) = 0, for all quasinilpotent elements s in a neighbourhood of 0 in A,
(v) there exists C > 0 such that p(x) < CIIx  all, for all x in a neighbourhood of a in A. PROOF.
It is easy to see that (i) implies (ii) and (v) (with C = 1). It is obvious
that (ii) implies (iii), which implies (iv). We now prove that (iii) implies (i). Taking x = 0 we have p(a) = 0, and so p(,=,,'=) = 0, for all z E A. Consequently
p(a  erae'=) = 0, for all x E A. By Theorem 5.2.4, we have a E Z(A). For every continuous irreducible representation w we have x(a) = al, for some a E C, and moreover p(a(a)) = 0. Hence x(a) = 0. By Theorem 4.2.1 (ii), a E RadA. We now prove that (iv) implies (iii). Suppose that p(a + x) = 0 for all quasinilpotent elements z such that JJgJJ < r. Let q 4 0 be an arbitrary quasinilpotent element. Then p(a + Aq) = 0 for JAI JJgJJ < r. But A logp(a + Aq) is subharmonic and is oo on the disk {A: JAI < r/JJgJJ}, so by H. Cartan's theorem it is oo everywhere. Hence p(a + q) = 0. To finish we prove that (v) implies (iv). Let q be a quasinilpotent element of A. We have p(Aa + q) = IAIp(a + q/A) < CJIgII, for JAI large enough. So the subharmonic function A F p(Aa + q) is bounded on C, and hence constant. Consequently p(a + q) = p(q) = 0. 0 Property (v) implies that if a quasinilpotent element a is not in the radical of A then the spectral radius is continuous but not lipechitzian at a. THEOREM 5.3.2.
Let A be a semisimple Banach algebra and let a E A. 'There exists a E C such that a = al if and only if # Sp(a + q) = 1 for all quasinilpotent elements q of A. PROOF.
If a = al then Sp(a + q) = {a} for all q quasinilpotent. We prove
the converse. Let q be a fixed quasinilpotent element. For A E C there exists h(A) E C such that Sp(a + Aq) _ {h(A)}. By CoroLbry 3.4.18, h is entire. But lim supjaj_,, I h a , < p(q) = 0 because E Sp( +q) for A # 0. So, by Liouville's
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theorem for entire functions, h(A) is constant. Consequently Sp(a+ Aq)  Spa, for A E C. But Spa = {a} implies a = al + qo for some quasinilpotent element qo. So Sp(qo + q) = {0} for all quasinilpotent elements q. By Theorem 5.3.1 (iii),
qo ERadA={0} andsoa=al.0 Given a bounded linear operator T on a Hilbert space H, we define the essential ,spectrum of T, denoted Sp. T, as the spectrum of the coset of T in the Calkin algebra .C(H)/Ct*(II). By Exercise 111.5, we know that this Calkin algebra is semisimple.
Suppose that abounded linear operator Ton a Hilbert space is not the sum of a scalar and a compact operator. Then there exists a quasinilpotent operator Q such that T + Q has more than one point in its essential spectrum. COROLLARY 5.3.3.
By Theorem 5.3.2 applied to the Calkin algebra, there exists R E Z(H) such that p(R) = 0 and # Sp,(T +R) > 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 quasinilpotent, and so #Spe(T+Q) > 1. 0 PROOF.
§4. Spectral Characterizations of FiniteDimensional Banach Algebras Let A be a Banach algebra such that A/ Rad A is finitedimensional. 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 WedderburnArtin theorem (Theorem 2.1.2). LEMMA 5.4.1. Let A be a semisimple 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
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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 twosided 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 finitedimensional. 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 semisimple.
A . a + A(x  a) = f (A), we have Sp f (A) finite for A in some real interval which has a nonzero capacity. So, by Theorem 3.4.25, # Sp(a + A(x  a)) < Ioo 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 < t, for aU z E A. Let a be a continuous irreducible represeptation of A. The argument at the beginning of the proof of Lemma 5.4.1 implies that dim X < n. So let z E A be arbitrary with Spa = m < n. Suppose that al, ...,at E Spr(z), with 1 < m. By the CayleyHamikou
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theorem we have (7r(x)  at 1)" x
.
x (sr(x)  all)" = 0 and consequently
x (lr(x)  a,l)" = 0. This being true for all such repre
(ir(x)  all)" x
x (x  am1)" = 0 because A is semisimple. Hence x is algebraic of degree < n2. We now use Lemma 5.4.1 to finish the proof. 0
sentations s, we have (x  a, 1)" x
Let A be a Banach algebra with involution. Suppose that the real vector subspace H of selfadjoint elements contains an absorbing subset U such that Sp h is finite for all h E U. Then Al Rad A is finitedimensional. COROLLARY 5.4.3.
PROOF.
As U is an absorbing set, there exists ho E U which satisfies the following:
for h E H given, there exists r > 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 finitedimensional. 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 nonempty open set U
of invertible elements such that p(z)p(z1) = 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 nonempty open set U of invertible elements such that IIz{I Hz1II = 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 pz111 = 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 Ua1 we have: 1 < HzyO . Hy'z111 = IIeyaa' II Ilaa' y' z' II <_ II=II 11ya1{ Ha' {{
Ila{I
Ila,y111.
hs1
M = 1. Moreover
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xUa1 is a connected set containing z, so zy e E for y in the neighbourhood Ua`1 of the unit. Consequently E = G1(A). Suppose that z E Rad A, z j4 0. Then x, = 1 t= E G1(A) fort > 0. Hence IIxg1I (Ixi 1II = 1. When t goes to +oo, x1 goes to x, hence lim IIx 1II = 1/IIxII. By Lemma 3.2.11, x is invertible which is absurd. So Rad A = {0}. Moreover we have 1 < p(z)p(x1) < 11.T11  11x'11 on G, (A). So by
Theorem 5.4.4, A is isomorphic to C. 0
§5. Automatic Continuity for Banach Algebra Morphisms With Corollary 4.1.10 we saw that on a commutative semisimple Banach algebra, all the Banach algebra norms are equivalent. In the 1950s, I. Kaplansky conjectured that the same result is true for noncommutative semisimple Banach algebras. This problem was solved only in 1967 by B.E. Johnson. His proof, which is not so easy, uses mainly representation theory (see 12), pp. 128131 or (1), pp. 161163). Using subharmonic functions we now give a very simple proof of an extension of this result.
Let A and B be two Banach algebras and let T be a linear mapping from A into B. We define the separating apace of T by
6(T) = {a: a E B, 3(zn) in A,n00 lim x = 0 and noo lim Tx. = a). It is a closed linear subspace of B and, by the Closed Graph Theorem, T is continuous if and only if 6(T) = {0}. THEOREM 5.5.1.
Let A and B be two Banach algebras. Suppose that T is a linear
mapping from A into B such that p(Tx) < p(z) for every z E A. Then a E 6(T) implies p(Tz) < p(a +Tx), for all z E A. In particular 6(T) flT(A) is included in the set of quasinilpotent elements of B.
Let a E 6(T) and (zn) be such that limn zn = 0 and limn...,, Txn = a. Let z E A and A E C be arbitrary. Then limn_.,,(Az + z) = z and p(T(Axn + x)) = p(ATz + Tz) < p(Azn + z) by hypothesis. So PROOF.
lim sup p(ATzn + Tz) < lim sup p(Azn + z) < p(z ), R00
by upper sernicontinuity of p on A. We set 0,(A) = p(ATz + Tx), which is subharmonic. Consequently
`(A) = limsup#.(A) 5 AX)
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satisfies the mean inequality on C, but in general is not upper semicontinuous. We set O(A) = him sup 0(p) MA
to be its upper regularization, which is subharmonic on C. We have #(A) S ((A) < p(x), for all A E C. So by Liouville's theorem for subharmonic functions, t& is constant. So p(Tx) _ 4(0) < ti(0) = t(,(A) for all A E C. By upper semicontinuity of p on B we have O(A) < p(Aa + Tx)
and consequently %6(A) < lim sup p(pa + Tx) < p(Aa + Tx). 'V _A
So we conclude that p(Tx) < p(Aa + Tx) for all A E C, and in particular for A = 1. If a E 6(T) fl T(A) then a = Tu for some u E A. Taking x = u, we get p(a) = 0, so the result. 0 THEOREM 5.5.2.
Suppose that we have the hypotheses of Theorem 5.5.1 with B semisimple, and moreover that T is onto. Then T is continuous. PROOF.
Let a E 6(T) with a = Tu. Taking x = y  u, we get p(Ty  a) < p(Ty)
for all y E A. So p(a + q) = 0 for all quasinilpotent elements q of B. By Theorem 5.3.1 (iii) we have a E Rad B = (0). 0 COROLLARY 5.5.3 (B.E. JOHNSON).
Let A and B be two Banach algebras, with B semisimple. Suppose that T is a morphism from A onto B. Then T is continuous. PROOF. If T is a morphism we obviously have SpTx C Spa, so p(Tx) < p(x) for all x E A. We then apply Theorem 5.5.2.0 REMARK. In the proof of the corollary it is not necessary to use Theorem 5.3.1 (iii) because in this case 6(T) is it closed twosided ideal of B contained in the set of quasinilpotent elements of B, and hence 6(T) C Rad B = (0}.
From Corollary 5.5.3 we obtain immediately the equivalence of Banach algebra norms and the continuity of involution on a semisimple Banach algebra (see the proofs of Corollary 4.1.10 and Corollary 4.1.11).
This last result on continuity of involution is very important for it implies that whenever we have to study a spectral problem on a Banach algebra with involution,
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we may suppose that the involution is continuous (we transfer the involution to A/ Rad A which is semisimple, and this does not modify the spectrum). In the theory of automatic continuity, the following problem has been known for a long time and remains unsolved: if T is a morphism from a Banach algebra A into a semisimple Banach algebra B, having a dense range in B, is it true that T is continuous? The only partial solution we know is the following:
Let T be a morphism from a Banach algebra A into a semisimple Banach algebra B. Suppose that T(A) is dense and has at most countable THEOREM 5.5.4.
codimension in B. Then T is continuous and onto.PROOF . Because T(A) is dense in B it is easy to verify that 6(T) is a twosided closed ideal of B. Let a E 6(T). Then a°O  1 E 6(T) for all a E C. The cosets corresponding to these elements in the quotient linear space B/T(A) are linearly dependent because T(A) has at most countable codimension in B. Hence there exist 0:1, .On) 1, ... , different from 0 such that
u = 8i(e°ta  1) + ... + #.W"  1) E T(A). So u E 6(T) n T(A) and hence, by Theorem 5.5.1, p(u) = 0. By the Holomorphic Functional Calculus the spectrum of a is included in the set of zeros of the function
f(z) = th(e°,:  1) + ... + pn(e°"`  1), which is not identically zero. So the spectrum of a is finite. By Corollary 3.4.5 the spectrum function is continuous at a. But there exists a sequence (zn) such
that limn...,, x = 0 and llmn Tzn = a, and so p(a) = lim,, p(Txn) < lira"_.,, 0. Consequently every element of 6(T) is quasinilpotent. So 6(T) C Rad B = {0}. From a classical theorem by T. Kato, we conclude that T is onto. o For more details on these problems concerning automatic continuity, see the little book by A. Sinclair, Automatic Continuity of Linear Operators, Cambridge, 1976.
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§6. Elements with Finite Spectrum THEOREM 5.6.1.
Let f be an analytic function from a domain D C C into a
Banach algebra. Suppose that for every A E D the element f(A) is algebraic. Then there exist an integer n > 1 and n holomorphic functions on D, denoted by a1i ... , an, such that a1(A)f(A)n1 +
f(A)n +
... + an(A)1 = 0,
for allAED. Because f(A) is algebraic its spectrum is finite. By Theorem 3.4.25 there exist an integer k, a closed discrete subset E of D and k functions hl,..., hk, locally PROOF.
holomorphic on D \ E, such that Spf(A) = {h1(A),...,hk(A)} for A E D \ E, and
#Spf(A)
(h1(A) + ... + hk(A)) h:(A)hj(A)
7s(A) _ 1
7k(A) =(1)kh1(A) ... hk(A)
These functions are locally holomorphic on D \ E. We prove that 71,.. . , 7k can be extended continuously on D. Let Ao E E and r > 0 be such that 'ff(Ao, r)nE = {Ao). Then Sp f (Ao) = {$i,.. . , fit} for some f < k, and Sp f (A) = {h1(A),... , hk(A)} for
0 < 1A  AoI < r. Without loss of generality we may suppose that 0 0 Sp f(Ao) (we change f(,\) for f(,\) + al with some a E C). We then choose t disjoint open disks Al i ... , At, centred respectively at Q1, ... , /e and not containing 0. There exists s < r such that JA  AoI < s implies Sp f(A) C Al U ... U At. Applying the Holomorphic FLnctionalCalculus and Theorem 3.4.25, we conclude that there exist
integers nii ... , nt > I such that n1 +
+ ne = k and #(Sp f(,\) n A;) = n; for
0 < IA  AoI < a. Then 71 can be extended continuously at A0 by
71(),o) = (nit, + ... + nePe) For the other symmetric functions the continuous extension is similar. Now from Morera's theorem, the k functions 71, ... , 7k are holomorphic on D. Once more by the Holomorphic Function Calculus, g(A) = f(A)k + 71(A) f (A)k1 + +7k(A)1 is quasinilioSawt for all A E D. But for each A fixed, 9(A) is algebraic, so 9(A)
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is nilpotent. Let Dm = (A:A E D,g(A)m = 0), which is closed in D. Then By Baire's theorem and the Identity Principle for analytic functions D = U I we conclude that there exists some integer m0 > 1 such that g(a)m" = 0 on D. Expanding this expression, we get the result. 0
Let X be a Banach space and let f be an analytic function from a domain D C C into £(X). Suppose that for every A E D the element f (A) is polynomially compact. Then there exist n holomorphic functions on D, denoted by a), ... , a,,, such that COROLLARY 5.6.2.
f(A)' +
+0,.(,\)l E CE(X),
for all A E d.
PROOF. We apply Theorem 5.6.1 to £(X)/U(X). 0 For a given Banach algebra A we denote by a the set of elements of A wich have
finite spectrum. This set may be extremely complicated. It contains in particular the set of quasinilpotent elements and the set of projections. In the rest of this section we investigate some properties of J. THEOREM 5.6.3.
Suppose that a + Ab E a for all,\ E C. Then we have
(aa1)(b81)' E I for all a E C and 0 E C \ Sp b. Let 6 V Spb. We have a + A(b  01) E 3F for all A E C. But a E Sp(a + \(b  01)) is equivalent to (a  al)(b  /31)' + Al noninvertible and consequently to A E Sp((a  al)(b  /31)'). We now apply Theorem 3.4.26 to PROOF.
the analytic function f (A) = a + A(b  01). Suppose that {A: a E Sp f (A)} is closed
and discrete. By the previous equivalence we conclude that Sp((a  al)(b  #I)') is compact and discrete, hence finite. If we have a E Sp f (A) for all A E C then E Sp(z + b  #1) for A 96 0, so 0 E limsupl),Io, Sp(z + b  /31) C Sp(b  $1), which gives a contradiction. Hence the theorem is proved. 0 COROLLAR', 5.6.4. PROOF.
If for some a E A we have a + a c 3 then as C a.
Let b E 3F. Adding a constant if necessary, we may suppose that b is
invertible, and so b' E a. By hypothesis a + Ab' E a for all A E C. By the previous theorem with a = /3 = 0, we get ab = a(b')' E . 0
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If for some a E A we have a + a C 1 then there exists an integer n > 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(tb)
aekaltb)
 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 semisimple. 0
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Applicatic is of Subharmonicity
Let A be a semisimple Banach algebra. Suppose that go E A is a nonnilpotent quasinilpotent element. Then there exists another quasinilpotent 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 quasinilpotent 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 nonnilpotent quasinilpotent 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 quasinilpotent elements of A.
Then N =
fln>1 {x: p(x) < 1/n}, so N is a Goset by uppersemicontinuity of p. The set M fl X is dense in X so the quasinilpotent elements of X form a dense G6subset of X. By Baire's theorem we conclude that the set of quasinilpotent elements of X is not a countable union of closed subsets of X with empty interior, so it is different
from MflX.0
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THEOREM 5.6.10.
Let X be a Banach space of infinite dimension. Then there
exist two quasinilpotent 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 finitedimensional 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 infinitedimensional. Moreover, by Exercise III.4, A is semisimple. 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 twosided 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 twosided 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 twosided inessential ideals. Given a twosided 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
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Applications of Subharmonicity
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 twosided 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 twosided 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 twosided 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.
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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 twosided 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 twosided 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)
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Applications of Subharmonicity
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 nonempty open set, G' \ Sp(x + y) is nonempty. 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(Alz)'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
SpiA=Sp(xAl)CD(zA1)=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
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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 =
,ifnr<1'
=10to ,ifn=1 ,ifn#1
we have b of rank one, and so in £2(H), and we have D(a) = Spa = {z: IzI < 1}, D(a + b) = Sp(a + b) = {z: IzI = 1 }. In 1954, using a rather complicated argument, A.F. Ruston proved that if T E
£(X) has an essential spectral radius equal to zero, then the spectrum of T is either finite or a sequence converging to zero, and the projections associated to the nonzero spectral values are in W. By Corollary 5.7.3, the condition that p(T) = 0
in the Calkin algebra £(X)/Z4 (X) is equivalent to saying that for every E > 0
there exists an integer N such that for every n > N there exists T E I with IIT"  T"II < e". A.F. Ruston called such an operator asymptotically quasicompact. This result derives immediately from the following:
Let I be a twosided inessential ideal of a Banach algebra A. Let z E A and suppose that p(i) = 0, where i denotes the coset of x in A/i. Then the spectrum of x has at most 0 as a limit point and, for every nonzero spectral value of x, the associated projection is in I. COROLLARY 5.7.5.
If p(i) = 0, by Theorem 5.7.4 (iii) we have D(x)° = {0}, so D(x) = 0. Hence we get the result. 0 PROOF.
Let I be a twosided inessential ideal of a Banach algebra A. Then kh(I) is inessential, so in particular 7 is inessential.
COROLLARY 5.7.6.
PROOF.
If x E kh(I) then i E Rad(A/7), so p(i) = 0. We then apply the previous
corollary. Q
If a Banach algebra A has minimal left ideals (resp. minimal right ideals) then by definition its eocle, denoted by b.w(A), is the , .i of the minimal left ideals (it
is also equal to the sum of minimal right ideas, so it is a twosided ideal). The reader will find more information on the socle in )I], pp. 7887. Using a rather complicated method, B.A. Barnes proved that every element of kh(soc(A)) has at most 0 as a limit point in its spectrum. This proof was simplified by J.C. Alexander and M.R. Smyth. In fact this result derives from Corollary 5.7.6.
Applications of SuWharmonicity
COROLLARY 5.7.7.
111
In a Banach algebra with minimal left (or right) ideals,
kh(soc(A)) is an inessential ideal.
PROOF. By Corollary 5.7.6 it is sufficient to prove that soc(A) is inessential. But for every x E soc(A), the spectrum of x is finite. For instance this can be seen using the fact that the algebra xAx is finite dimensional, so x is algebraic (see Exercise V.11). 0 If dim A < +oo then the socle of A is nonzero because A = soc(A). Conversely if A is semisimple and A = soc(A) then, by Theorem 5.4.2, A is finitedimensional. If X is a Banach space and A = E(X) then the socle is nonzero because it contains all finiterank operators. It would be interesting to have more examples of Banach algebras with nonzero socle.
The next result was proved by B.A. Barnes (see On the existence of minimal ideals in a Banach algebra, Trans. AMS 133 (1968), pp. 511517). He first obtained the commutative case using a deep result called the Silov Idempotent Theorem (see [10[, Chapter 8). In the proof given below we eliminate this argument by using a subharmonic one.
Let A be a semisimple Banach algebra such that the spectrum of every element of A is at most countable. Then soc(A) 9t {0}.
THEOREM 5.7.8 (B.A. BARNES).
Suppose that soc(A) _ (0). If for all x E A we have # Sp x = 1 then, by Theorem 5.4.4 (or Exercise 111.21), A ^_ C, so we get a contradiction. Hence there exists xo E A such that Sp xo contains at least two isolated points ao, a,. We choose PROOF.
two disjoint open disks Do, Dl, of radius < 1, respectively centred at ao, al, and such that 'Do n Sp xo = {ao } and D, n Sp xo = {a, }. Translating,xo if necessary, we may suppose that 0 0 Sp xo. By Theorem 3.4.2, there exists ro < 1 such that l[x  xolf < ro implies Don Spx # 0, D,nSpx 94 0 and Spxn(ODouhD,u{0}) = 0. For i = 0, 1, let pi be the projection associated to xo and a,. We have pixo = xopi, POPI = PiPo = 0 and Sppixo = 10, a,). Consider the subalgebras Ai = piApi. They are semisimple Banach algebras with identity pi (see Exercise 111.6). We have SPA; PixPi C SPA
Consequently the spectrum is at most countable on these subalgebras. By hypothesis, xp is invertible, so the previous inclusions prove that SPA,pixopi = {a,}. Let r < ro/(11po[[2 + [[p,1[2) be such that [[x  zo[[ < r implies SPA, pixopi C D,, for i = 0, 1. Consider the two sets Gi = {x: 11x  xol[ < t', # SPA; PixPi > 1}.
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112
These two sets are open by Theorem 3.4.4. Suppose that B(zo, r) \ Go contains an open set U. Then we have # SPA, y = 1, for all y in the open subset poUpo of Ao. By Exercise V.4 we have poApo = Cpo, so po is in the socle of A by Exercise V.10, which is absurd because po 96 0. Consequently Go is dense in B(xo, r) and the same is true for G. In particular Go n G1 36 0. Let Y E Go n G1. Then SPAS p;yp; contains at least two isolated points ai o, ai 1 E Di (i = 0,1). We set x1 = POOPo + P1 YP1 + (xo Poxo  P1 xl )
By Exercise 111.9 we have
SPx1 : (SPPoypo U SPP1yP1 U SP(xo Poxo P1x1)) \ {0}
D {aoo,aol) alo,al l}. We then continue the process with the projections pi i associated to x1 and the a; i (i, j = 0,1). Let D be the set of finite diadic sequences, that is of finite sequences of Os and Is. If s, s' E D, we say that s < s' if s' is obtained from s by adding some Os or Is (for instance 0 1 < 0101). Then by induction on n it is possible to build a sequence (x,) of elements of A, a sequence of radii r" < 1/2", and for each s E D a disk D having the following properties: (1)
Hiz,,  x"+1II < 1/2" and ff(x",r,,) C B(xn1,rn1),
(ii) if s < s' then 1, C a,,, otherwise 35, and V,, are disjoint, (iii) a, E Sp x" n D,, for all s of length n + 1, e
(iv) diarn(D,) < 1/2"1 if .9 has length n.
By condition (a), the sequence (z") converges to some element z E A. Now if o is an infinite diadic sequence, we can define a(a) to be the unique element of the intersection of all the D,., where o" denotes the finite diadic sequence obtained from a by taking only its first n elements. By condition (c), we have Sp x"nD,,, & 0, and so by upper semicontinuity of the spectrum we have a(a) E Sp x. Moreover if
a # a', then for some n, D,,, and B,s are disjoint, and consequently a(a) 36 a(a'). Because the set of a is uncountable, Sp x is uncountable, so we get a contradiction. Hence soc(A) # (0). Thus we get a contradiction and the proof is complete. 0 Is it possible to give the precise algebraic structure of Banach algebras for which the spectrum of every element is at most countable, at least in the separable case? The answer is yes, by Theorem 5.7.9.
In the following, a E S1 will mean that a is an ordinal of the first or second class (see W. Sierpii ski, Cardinal and Ordinal Numbers, Warsaw, 1965).
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113
Let A be an arbitrary Banach algebra. We take Ao = Al Rad A and inductively A onto we define A. = An, A,, ... , An.... are denoted by Oo, ¢t, ... , 0.... and their kernels by to = Rad A, In), I1 = kh(aoc(A)), ..., 1n = Ker 0n, ... We then define A,,,, with I,,, = and 0,,,, where m is the first infinite ordinal. For every a E A it is possible to define A. and V+a, by transfinite induction, in the following way:
 if a is not a limit ordinal, A = Aa1/kh(soc(A,_1)) and 0a =
o Wo,
where 'Ir.1 is the canonical morphism from Aa_1 onto A
 if a is a limit ordinal we take I, = kh(U5<,I5), A, = A/I, and ¢, the corresponding canonical morphism.
By definition we shall say that Aa is the aCatkin algebra associated to A. It is easy to verify that it is semisimple. Let A be a separable Banach algebra such that the spectrum of every element of A is at most countable. Then there exist ao E 0 and an ordinal
THEOREM 5.7.9.
composition sequence (l, )o
1 < codim I..+1 < +oo, Iae+, = A and 1a+1 /Ia is a modular annihilator algebra for a < ao. PROOF.
The I. are closed in the separable space A, and the family (la),EA is
increasing for inclusion. A classical result of topology (see C. Kuratowaki, Topologie I, Varsovie, 1958, p. 146) implies that this family stabilizes at some 9 E R. If Ap #
{0}, then because SPA, ¢5(x) C Sp x, the algebra Ap satisfies the hypotheses of Theorem 5.7.8. Hence soc(Ap) # {0} which implies that Ip}1 # Ip, a contradiction. Consequently Ip = A. Then from the definition of Ip and from Corollary 3.2.2, ft
is not a limit ordinal. We take ao = jl  1. Then by Exercise V.12, A,, is finitedimensional. Moreover Ia+1 /Ia can be identified with kh(soc(A,)) so, by Corollary 5.7.5 and Corollary 5.7.7, every element of Ia+1 /I, has its spectrum with at most 0 as a limit point. By B.A. Barnes's characterization of modular annihilator algebras and the fact that Radkh(soc(A,)) = kh(aoc(A0)) fl Had A. _ {0} (Exercise 111.7), we obtain that 1a+1 /l, is modular annihilator. 0
We do not give the definition of a modular annihilator algebra, because it is rather technical. We refer the reader to [6).
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EXERCISE 1. Let a, b be two n x n matrices such that a(ab  ba) = 0. Prove that ab  ba is nilpotent. EXERCISE 2. Let A be a Banach algebra such that 6(x + y) < b(x) + 6(y), for all x, y E A. Prove that A is commutative modulo its radical.
Let A be a real Banach algebra. Suppose that p is subadditive or subrnultiplicative or uniformly continuous on A. What can be said about A?
*EXERCISE 3.
EXERCISE 4. Let A be a semisimple Banach algebra. Let U be a nonempty open subset of A such that Sp x contains one point for all x in U. Prove that A is isomorphic to C. EXERCISE 5.
Extend Theorem 5.3.1 and Theorem 5.4.2 to a real Banach algebra.
Let A be a semisimple Banach algebra such that for every x E A there exists y E A satisfying x = xsy.
*EXERCISE 6.
(i) Prove that A is commutative. (ii) Prove that A is finitedimensional.
(iii) Conclude that A is isomorphic to C", for some integer n. *EXERCISE 7.
Prove that a Noetherian Banach algebra is finitedimensional.
Let A be a Banach algebra and let (x: x E A,# Sp x < n) Suppose that a + " C 3,,, for some a E A. Prove that a is algebraic of degree less than or equal to n. EXERCISE 8.
EXERCISE 9.
Let X be a Banach space of infinite dimension. Prove that there
exist two quasinilpotent and compact operators TI,T2 on X such that Sp(TIT2) is infinite. EXERCISE 10.
Let A be a Banach algebra. Prove that for every minimal left ideal
(resp. right ideal) I of A there exists a projection p E A such that I = Ap (esp. pA) and pAp = Cp. Such a p is called a minimal projection. Is the converse true?
Let a E soc(A). Prove that aAa is finitedimensional. If you do not succeed look at (1J, p.81.
*EXERCISE 11.
Let A be a semisimple Banach algebra such that A = kh(soc(A)). Prove that A is finitedimensional. EXERCISE 12.
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115
EXERCISE 13. Let I be a twosided ideal of a Banach algebra A. Prove that kh(kh(I)) = kh(I). **EXERCISE 14.
Extend Theorem 5.7.8 to a real Banach algebra.
Extend Theorem 5.7.8 to a Banach algebra with involution, supposing that the spectrum of every selfadjoint element is at most countable.
***EXERCISE 15.
Chapter VI REPRESENTATION THEORY FOR C* ALGEBRAS
AND THE SPECTRAL THEOREM
§1. Banach Algebras with Involution Among Banach algebras there are very interesting ones called Banach algebras with involution. A mapping z . x* from a Banach algebra A into itself is called an involution on A if it satisfies the following properties for all z, y e A and A E C:
(i) (z + y)* = x* + y*
,
(ii) (Az)* = Ax*,
(iii) (xy)* = y*z*
,
(iv) (z*)* = X. There are many examples of such Banach algebras with involution. Among the commutative ones there are C(K), the algebra of continuous functions on a compact set K, with the involution f + f ; the disk algebra A(A), with the involution defined
by f*(z) = f(i); and the group algebra L'(G) for a locally compact commutative group G (for instance R), with the involution f *(z) = f (x) (supposing that the operation on G is denoted by +). For noncommutative Banach algebras with involution we have £(H), the algebra of bounded operators on a filbert space H, with the standard involution; all closed subalgebras of £(H) which are invariant by this involution, and so in particular £C(H), the algebra of compact operators on H; and also L1(G), the group algebra of a locally compact noncommutative group G, with the involution f *(z) = m(z1) f (z' ), m being the modular function.
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Let A be a Banach algebra with involution. Then any z E A such that z = z* is called selfadjoint, any x E A such that xz* = z*z is called normal, and any x E A such that zx* = x*x = 1 is called unitary. THEOREM 6.1.1.
Let A be a Banach algebra with involution and let z E A. Then
(i) x + x*, i(x  z*), xx* and x*x are selfadjoint,
(ii) x has a unique representation x = h + ik with h, k selfadjoint given by h }1(x + x*), k = ;(z  x*), (iii) x is normal if and only if in the decomposition x = h+ik, with h, k selfadjoint, we have hk = kh,
(iv) the unit 1 is selfadjoint and x is invertible if and only if x* is invertible, in which case (z*)' = (z')*, (v) Spz* _ (a : A E Spx}, in particular p(x*) = p(x), (vi) z E Rad A if and only if x* E Rad A, in particular A/ Rad A has an involution (i)* = (x*)*. PROOF. Statement (1) is obvious. In (ii) we only prove the uniqueness. If x = u + iv, with u, v selfadjoint, then u  h = i(k  v). Taking the conjugate we get u  h = i(k  v), so k = v and u = h. If x is normal it is obvious that h and k commute. Conversely if h and k commute then zx* = (h+ ikXh  ik) = hz + k3 = (h  ik)(h + ik) = x*x. Since 1* = 1.1*, (i) implies that 1 isselfadjoint, and the remaining part of (iv) comes from (3) applied to x and x1. If we apply (iv) then x  Al is invertible if and only if z*  a1 is invertible, so we get (v). If z E RadA, then by Theorem 3.1.3, p(xy) = 0 for all y E A. Thus, by (v), p(y*z*) = 0 for all y E A, which implies by (4) that p(zx*) = 0 for all z E A. Once again by Theorem 3.1.3, we conclude that x * E Rad A. If i denotes the coset of x in Al Rad A then (i)* = (x*)' is welldefined by (v) and it is easy to verify that it is an involution on A/RadA. 0
As a consequence of Theorem 5.5.2 we have the following important fact. THEOREM 6.1.2 (B.E.JoHNSON). Let A a semisimple Banach algebra with involution. Then the involution is continuous on A.
LEMMA 6.1.3 (P.CIVINB.YooD). Let A be a Baneb algebra with involution and z be normal in A. Then there exists a closed and commutative subalgebra B, containing z, stable by involution, such that Spgz = Sp4x.
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PROOF.
Let C be the family of subsets E of A such that a, b E E U E* implies
ab = ba. The family C contains {x} and is inductive for the order defined by inclusion. So, by Zorn's lemma, there exists a maximal set B E C containing x.
Because B is maximal we have B U B* = B so B = B*. A similar argument with and B  B shows that B is a commutative subalgebra of A. We now prove B + B, that B is closed. If a, b E ff U 77* then a = lima,,, b = limb,,, with a,,, b E B U B*, but a,, b,, = b,, a, and so ab = be. Obviously we have SPA X C Spa x. Conversely if A E Spa X and A f Sp,1 x, then x  Al is invertible in A with (x  Al)1 f B, but B U {(x  Al)1 } E C. So we get a contradiction. Hence SPA X = Spa x. 0 THEOREM 6.1.4.
Let A be a Banach algebra with involution. Suppose that x E A
is a selfadjoint element such that its spectrum contains no real number A < 0. Then there exists a selfadjoint element y E A such that x = y2.
PROOF. By Lemma 6.1.3 there exists a closed and commutative subalgebra B of A, stable by involution, such that SPA X = Spa x, so if we succeed in proving the theorem for B, the proof will be finished. In other words, without loss of generality we may suppose that A is commutative. Let D be the complement inC of the set of real numbers A < 0. There exists f holomorphic on D such that f2(z) = z and
f(1) = 1. Since Spx C D we may define y = f(x) which satisfies y2 = x by the Holomorphic Functional Calculus. We now prove that y = y*. Since D is simply connected, by Runge's theorem (see (7), Chapter 13) there .exists a sequence (p,,) of polynomials converging to f uniformly on each compact subset of D. Let q be the polynomials with real coefficients defined by 2gn(z) = p,(z)+ pn(z)
Since f(i) = f (z), the polynomials q, converge to f, uniformly on each compact subset of D. Let y,, = Then y,, = y*. By Theorem 3.3.3 (iv), y = limna, y,,. If the involution is continuous then the proof is complete. Suppose not: then, by Theorem 6.1.2 applied to A/ Rad A, we conclude that y = y*, so y  y* E Rad A. By Theorem 6.1.1 we have y = h + 1k, with h, k selfadjoint, so k E Rad A. The algebra A being supposed commutative, we have x = h2  k2 + 2ihk so hk = 0. But 0 Spx and k2 E Rad A implies h2 invertible, so h is invertible. Hence we have
k=h'1(hk)=0.D §2. C*algebras If f E C(K), then ((f f ((. = ((f1120. If A is a closed subalgebra of £(H), stable by involution, for some Hilbert space H, then by Theorem 2.3.1 (iii) we
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have )jxx*)j = Ijxj)2, for all x E A. These examples suggest the introduction of a particular class of Banach algebras with involution called C*algebras. By definition they are those A such that Ilxx*ll = IIzII2, for all x E A. Of course there are examples of Banach algebras with involution which are not C*algebras. For instance, A(L) with the involution f *(z) =7(i) and L'(R) with the involution f *(x) = f (x) (see Exercise VI.1). THEOREM 6.2.1.
Let A be a C* algebra. Then we have the following properties:
(i) (Ix((= ljx*lj, for all x E A. so the involution is continuous,
(ii) if h is selfadjoint then e'h is unitary, (iii) if h is selfadjoint then Sp h c R, (iv)
llxlj2 for all x E A, in particular for all x normal we have p(x) = jjxjj,
(v) if x, y are normal then L (Sp x, Spy) < 11z  y(l, consequently the spectrum function is uniformly continuous on the set of normal elements. PROOF. We have 11x112 = jlxx*II !5 Ilxll 1Ix*ll so Ilx(( < llx*jj. Consequently llx*lI
Il(x*)*Il = jjxll, hence (i) is proved. This implies in particular that the involution is where continuous. If h = h* we have ei" = 1 + z + jr + so (eik)* = + AT. Consequently p,s(ih)* =
This implies (ii). We now prove (iii). Suppose that a + if E Sph, with ( # 0. Taking k = j(h  al) we have k = k* and i E Sp k, consequently 1 + ik is not invertible in A. Given any A E R we now have (A + 1)1  (Al  ik) not invertible in A so, in particular, IA + 11:5 jIA1  ikll. Consequently (A + 1)2 < I(al  ikl12 = II(Al  ik)(Al  ik)*JI = jjA2 + k2115 A2 + Ilkll2. So we get 2A < 1! k211 _ 1, for all A E R, and this gives a contradiction. Hence (iii) is proved. Suppose that h = h*. Then we have j1h2)) = jlhjj2. By induction it is easy to prove that 110" 11 = 11h112".
Consequently, using Theorem 3.2.8, we get p(h) = limnoo ((h2"((1/2 Taking h = xx* we get p(xx*) = llxx*Il = Ijx1j2. If x and z* commute, by Corollary 3.2.10, we have 11x112 = P(xx*) S = P(x)2 < IIx$12. So P(x) = Ijxjj. To finish we prove (v). Suppose that there exists A E Spy such that dist(A, Sp x) > llx  y!;. By Theorem 3.3.5, we have dist(A, Sp z) = 1/p((Jl1x)'1). But (Alx)'' is normal so, by (iv), we have dist(A, Sp z) = 1/(((A1x)'1((. Consequently Il(a1x)`I(xy)(l < Il(A1  x)'II II(x  Y)II < 1, so P(x)P(x*,
Al  y= (Alx)11+(Alx)'(xy)} is invertible, which is a contradiction. Interchanging x and y we conclude that A(Spz,Spy) < lIx _. yII. 13
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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 selfadjoint. 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.PTAXJ.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 = IP. 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.FUGLEDEC.R.Pt'TNAMMI.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'.
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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*zeav* = 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.GELFANDM.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 StoneWeierstrass 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),
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(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 wellknown 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 selfadjoint 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
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An element x of a C*algebra A is said to be positive, denoted z > 0, if it is selfadjoint and if its spectrum contains only positive real numbers. For instance if h is selfadjoint 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 selfadjoint 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.GELFANDM.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 selfadjoint 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.
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THEOREM 6.2.11 (M.FUKAMIYAI.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 selfadjoint 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 selfadjoint 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=jyjyOil 0J1.
(1)
Because z is real, by Theorem 6.2.7, z = z*. Let u = zx = h + ik with h, k selfadjoint. Then uu* = zxx*z* = zyz = z2y
(2)
u*u = 2h2 + 2k2  uu* = 2h2 + 2k2  z2y.
(3)
But h, k are selfadjoint 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 welldefined for all x E A.
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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 u1 = 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+ula)andj(u'a`l < IHalD<1. such that = So y is invertible and H`yll < 1. By Corollary 6.2.12 we have y = (yy*)1/sv, with v unitary. By the Holomorphic Functional Calculus w = (yy4)1/2 + i(1  yy*)1 /2 is welldefined, and moreover ww* = w*w = 1 and (yy*)1 /2 = "' w* . The product of two unitary elements being unitary, the first assumption is proved. We now consider the sequence defined by a + Ilk yo = u,
yk+1 =
2
Using the first part, by induction, it is easy to prove that there exist 2k unitary elements u1,... , uz. such that yk =
(ul
+... +u241)
Suppose now that lIzIl < 1. We set a = x +
(z  u) where k is the smallest
integer such that Ilall < 1. Then we have z = yk. So the result is proved. D COROLLARY 6.2.14 (B.RussoH.A.DYE). included in the convex hull of U.
In a C*algebra the open unit ball is
If A is finitedimensional then it is possible to prove that IIxiI < 1 impliest x = 1(ul + uz }, for some ul, u1 E U (see Exercise VI.9).
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If h is selfadjoint it is obvious that e'h is unitary. But in general the converse is not true, even for a commutative C*algebra (see Exercise VI.8). If E denotes the set of es', with h selfadjoint, it is possible to improve Theorem 6.2.13 and Corollary 6.2.14, replacing U by E (see Exercise VI.9). With Theorem 6.2.6 we have explicitly characterized commutative C*algebras. Is it possible to do the same for noncommutative ones? We finish this section with this problem. But in order to solve it we need some preliminaries.
Let A be a C*algebra. We say that a linear functional f on A is positive on A if f(xx*) > 0, for all x E A. If A is commutative, so isomorphic to C('9)t), positive linear functionals correspond to positive measures on '.112. If A = £(H), for some Hilbert space H, then f(T) = (Tf Il;), for some f E H, is obviously positive. With Theorem 6.2.17 we shall see that there are many positive linear functionals on A. THEOREM 6.2.15.
Let A be a C*algebra and let f be a positive functional on A. Then we have the following properties:
(1) f(x*) = f(x), for all x E A, (ii) If(zy*)I2 _< f(xz*)f(YY*), for all z,y E A,
(iii) If(z)12 < f(1)f(xz*) < f(1)2p(zz*), in particular If(z)I < f(1)IIxII, for all z E A, and this implies that IIfII = f(1). PROOF.
Let x, y E A and A E C. Since f ((z + Ay)(x + Ay)*) > 0 we have
f(zx*) + Af(yx*) + a f(xy*) + AAf(YY*) > 0,
for all A E C. Taking A = 1 and A = i we conclude that f(yx*) = /(y*) So with y = 1 we get (i). If f(zy*) = 0 then (ii) is obvious. If f(zy*) # 0, take A = tf(zy*)/If(zy*)I with t E R. Then we have f(xz*)+2tif(zy*)I+t'f(yy*) > 0 for all t E R. So we get (ii). Taking y = 1 in (ii) we obtain If (x)I s < f (1) f (xx*). If t > p(zx*) then by Theorem 6.1.4 there exists a selfadjoint element h such that t  xx* = h2, so tf(1)  f(zx*) > 0. This being true for all t > p(zx*), we have f(xx*) < f(1)p(zx*). The rest of (iii) comes from Theorem 6.2.1 (iv). 0 In fact positive linear functionals can be characterized very easily. THEOREM 6.2.16. Let A be a C*algebra. A linear functional f on A is positive if and only if it is bounded and II f 11 = f (l).
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127
PROOF. The first part is proved in Theorem 6.2.15 (iii). Suppose that Ilf11= f(1). We want to prove that f is positive. First we show that f (h) E R if h is selfadjoint. Let f(h) = a + i/3, with a,,8 E R. We set u = t1  ih, for t E R. Then we have ilull2 = Iiuu'II = 1(t2 + h211 5
We also have l f (u
)12
= It  is + 1312 = t2 + a2 +
IIuI12 < If
(U)12
t2
/22
+ Ilhll2 .
+ 2(3t. Hence
 2/31 + Ilh112 < IIu112  2(3t + 11h112 .
So 2/3t < Ilh112, for all t E R, and hence /3 = 0. We must now prove that f(h) > 0 for h = xx*. Without loss of generality we may assume that IIaII < 1 and f(1) = 1. Then Sp h C 10,1] and this implies IllhII = p(1h) < 1. Also f(h) = 1 f(1h) > 1  Ill  hit > 0. This completes the proof. 0
We say that f is a state on A if f is a positive linear functional, and moreover
f (1) = 1. By the previous theorem, f is a state if and only if it is a bounded linear functional satisfying Ilf11 = f(1) = 1. If A = C(99), states correspond to probability measures on M. Let 6 denote the set of states of A. It is easy to see that it is a closed convex subset of the unit ball of the topological dual A' of A. So by Theorem 1.1.8 it is weakly compact. By Theorem 1.1.9, we conclude that it is the weak closure of the set of extreme points of 6. First we show that 6 is large, and this implies the existence of extreme states. THEOREM 6.2.17. Let A be a C*algebra and let B be a C*subalgebra of A containing the unit. Suppose that a is a state on B. Then there exists a state s on A extending a. In particular if x E A there exists an extreme state so on A such
that so(x*x) = IInII2.
PROOF. On B the linear functional or satisfies IIaII = a(1) = 1. By the HahnBanach theorem there exists s E A' such that IIaII = Hall and s extends a on all A. So, by Theorem 6.2.16,.s is a state. If z E A, let B be the C*subalgebra generated by 1 and x*x. By Theorem 4.1.2 and compactness of the spectrum there exists a character X on B such that X(z*z) = p(z'z) = 11z112. But X is a state on B so,
by the first part, there exists a state s on A such that s(x*x) = IIz112. Let r be the set of states s on A such that s(z'x) = pnII2. It is a convex weakly compact subset of the unit ball of A' so, by Theorem 1.1.9, it contains an extreme element so. Suppose that so = U 4U with 31, 32 E G. Then we have 11x112 = $o(x*r) = 2(st(x*z) + s2(x*z)) < 1lz*zll = IIXII2
so necessarily we have s1(x*x) = s2(z*z) = IIx112 that is to say..91,32 E E, so 30 = 91 = 92, because 30 is extreme in r. Hence so is extreme in G. 0
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For commutative C'algebras, that is C(931), extreme states correspond to probability measures concentrated at a point, so they are the characters of A. Let A be a C'algebra and let $ be a state of A. Then N. = {x : x E A, s(x*x) = 0) is a closed left ideal of A and there exists an inner product on A/N, defined by LEMMA 6.2.18.
(x+N,Iy+N,)=s(y*x), for allx,yE A, such that A/N, is a preHilbertian space for that inner product. PROOF.
If x E N, then, by Theorem 6.2.15 (ii), we have Is(zx)I2 < s(x*x)s(zz*)
and so s(zx) = 0, for all z E A. In particular s((yx)'(yx)) = s((x*y*y)x) = 0, hence yx E N for all y E A. Then N, is a left ideal of A. It is closed because s is continuous. Because N. is a left ideal the inner product is welldefined on A/N,. The rest of the proof is left to the reader as an exercise. 0 COROLLARY 6.2.19.
Let A be a C'algebra. Then there exist a Hilbert space H
and a morphism T from A into £(H), such that T(1) = I, T(z*) = T(x)' and IIT(x)II _< IIxII, for all z E A. PROOF.
Let s be a state of A and define T on A, with values in £(A/N,), by
T(x)(y+N,)=xy+N,. It is easy to verify that T is a morphism from A into £(A/N,) and obviously we have
T(1) = I. We have (T(x)(y+N.)IT(x)(y+N,)) = (xy+N,Ixy+N,) = a(y*x*xy). But for y E A fixed, if we define t(x) = s(y'xy), it is a positive linear functional on A so, by Theorem 6.2.15 (iii), we have t(x'x) < t(1)IIx112 = s(y'y)IIz112 = (y + N,Iy + N.)IIx112. Hence IIT(x)II < IIx1I, for all z E A. This inequality enables us to extend the operators T(z) to bounded linear operators on the completion H
of A/N for the norm defined by the inner product. 0 Let A be a C'algebra. Then there exist a Hilbert space H and an isometric *morphism from A onto a dosed selfac(joint subaigebra of £(H). THEOREM 6.2.20 (I.M.GELFANDM.A.NAIMARK).
PROOF. Let s be a state of A. We define H,, the completion of A/N,, and T the *morphism from A into £(H,), as in Corollary 6.2.19. Let H be the Hilbert space
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Representation Theory and Spectral Theorem
which is the direct sum of the (H,),Ee. Let a E A. For any x = (x,),Ee E H we have IIT.(a)I1211x.I
IIT.(a)(x.)II' <<
2 < Ilall'
JIx,JJ' < +oo,
E6 so we can define an element T(a)(x) in H such that (T(a)(x)), = T,(a)(x,). We .E6
.E6
have obtained a mapping T(a) from A into H. It is obvious that T(a) is linear, and from the above inequalities it is bounded, so T(a) E .C(H) with JJT(a)IJ < JJaJJ. Because the T. are *morphisms, we verify easily that T is also a *morphism from A into .C(H). By Theorem 6.2.17 there exists some a E lri such that s(a*a) = JJaJJ' so JJT.(a)112 > (T,(a)(1 + N.)JT.(a)(1 + N.))  a(a*a) = 11a112. Consequently IIT(a)JJ = JJaJJ. Therefore T is an isometry. Consequently T(A) is a closed self
adjoint subalgebra of £(H). Obviously T(1) = I. 0 The Hilbert space H obtained in the previous theorem is extremely large in general. But if A is separable we may suppose that H = 12 (see Exercise VI.12). Theorem 6.2.20 is also true if we suppose A without unit (see Exercise VIA). Theorem 6.2.20 is very important because it says that a C*algebra can always be considered as a closed selfadjoint subalgebra of .C(H) for some suitable Hilbert space H.
53. The Spectral Theorem Throughout this section we suppose that the C*algebra under consideration is £(H) or a closed selfadjoint subalgebra of E(H) containing the identity, for a suitable Hilbert space H. Given a normal operator T we intend to extend Theorem 6.2.7 for all bounded Borel functions on SpT. First we need some definitions and preliminary results.
Let Wt be the oalgebra of all Borel subsets of a compact set K. We define a resolution of the identity to be a mapping P : 91t " £(H) with the following properties:
(i) P(O) = 0, P(K) = I, (ii) each P(U) is a selfadjoint projection, for U E'n11,
(iii) P(U n V) = P(U)P(V), for U, V E Wt, (iv) if u n v = 41, then P(U U V) = P(U) + P(V ), for U, V E Wt, (v) for every z, y E H the set function Ps,, defined by PP,,,(U) = (P(U)x!y), is a complex regular Borel measure on WI.
A Primer on Spectral Theory
130
To get a very simple example we take H = L2([0,1]), fit the aalgebra of Bore] subsets of (0, 1), and we define P by
P(U)f for U E 971, f E L2([b,1]).
It is easy to verify that a resolution of the identity P has the following properties:
(a) any two of the projections P(U) commute with each other, (b) if U and V are disjoint the associated projections are orthogonal, (c) if x E H then P=,s is a positive measure of total variation JJxJJ2, (d) by (iv), P is finitely additive, and if U is the countable union of disjoint U then norm of the series F,1 P(U) the. projections P(U.) is one, so that in general P is not countably additive,
(e),.by (v), for a fixed x E U, the function U  P(U)x is a countably additive Hvalued measure on 9r,
(f) if U., E '971 and
0, for n = 1, 3.... , and if U = U'1 U. then
P(U) 0. Given an essentially bounded Borel function on K we now intend to give meaning to the integral
fdP. 1K
Let P be a resolution of the identity and let f be a complex Borel function on K. Using the fact that there exists a countable base for the topology of the complex plane and using property (f) we conclude that there exists a largest open set V in C such that P(f '(V)) = 0. By definition the essential range of f is the complement of V. We say that f is essentially bounded if its essential range is bounded, hence compact. In that case we denote by Ilf J[. the largest value of JAI, for A in the essential range. We denote by L°°(P) the Banach algebra of essentially bounded complex Borel functions on K, with norm JJ [[a,, and the convention that two functions f and g coincide if JJ f  gLJ = 0. THEOREM 6.3.1.
Let P be a resolution of the identity. Then the formula
('W)zI y) = L f dPs,r , for x, y E H ,
Representation Theory and Spectral Theorem
132
defines an isometric isomorphism fi of the Banach algebra L(P) onto a closed commutative selfadjoint subalgebra of £(H) which contains the identity. This isomorphism has the following properties:
(i) 0(1) is the identity of L(H), (ii) fi(f)e = fi(f), for f E LOO(P), (iii) Ijfi(f )xl12 = J. If l2 dP=,., for x E H and f E LOO(P),
(iv) S E .C(H) commutes with every P(U) if and only if S commutes with every
fi(f). PR.UUF.
and let
Property (i) is obvious. Let U11. .. , U be a partition of K, with U, E'Jf, be it simple function such that: s = ai on U;. We define fi(s) E .C(H) by
fi(s) _ >2 criP(U,)
(1)
.
c=1
Since each P(U,) is selfadjoint 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 V1 fi(s)fi(t) = >2 rr,f9,P(U;)P(V5) = Z1
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)
A Primer on Spectral Theory
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 nonempty 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 selfadjoint, 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)
Representation Theory and Spectral Theorem
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 selfadjoint.
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
A Primer on Spectral Theory
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 selfadjoint 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 selfadjoint 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 StoneWeierstrass 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
Representation Theory and Spectral Theorem
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 selfadjoint 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)
A Primer on Spectral Theory
136
and U = u(T). By construction P is selfadjoint. 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), selfadjoint,
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 selfadjoint 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 selfadjoint S E £(H). By Theorem 6.4.2, we have U = e'Q for some selfadjoint 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 finitedimensional algebras (see Remark following Theorem 3.3.6), but false in general (see (81, pp.317319).
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.171193, 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
Representation Theory and Spectral Theorem
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
1))*(xi(ua  1)) _ (ua  1)va(ua  1) = nz va1
\
i=1
The real function t  t(t +
van 1)
n}2 is less than or equal to a, for t > 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 twosided 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 Ilxxuall = limsup IIzzua+yyua(( = limsup I((x+y)(1ua)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
A Primer on Spectral Theory
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(et(°+°'))
= llet(a+` )11 < 1 .
The spectrum of a+a' being real, we conclude that a+a' > 0. Moreover II at° 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)ea ds > 0
(1)
by hypothesis. S.) IIet4112 = IIeWet°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(e11*(a + a)e'°) = 0 for 0 < s < u, so o(eWe94) = 1 for 0 < t < u. Applying the Identity Principle to
Representation Theory and Spectral Theorem
the entire function Hence
139
we conclude that o(e" 'e'a) = 1 for all t > 0.
1 = o(eeo'ere)
1,
so file all = 1, for all t > 0. Then by Theorem 3.2.8 we get P(eca) = lim lie
noo
'1I" = 1,
a contradiction. The proof is now complete. 0
140
EXERCISE' 1.
A Primer on Spectral Theory
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 semisimple.
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 finitedimensional 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 halfline {a)  a > 0}. If you do not succeed look at [1], pp. 109113 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 selfadjoint 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 selfadjoint elements h. Prove that A is a C*algebra for this norm. If you do not succeed look at [1[, pp. 122123.
Representation Theory and Spectral Theorem
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 quasinilpotent 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 smorphism 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.8910 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. 386400).
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. 363386). 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
nonempty 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.
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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".
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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
log !zi
z, ,
which is plurisubharmonic on C"+2. By Theorem 7.1.1 (iv), we know that
A  K(A) x ... x K(A), n + 1 times, is hit analytic multifunction. So, by definition,
log6"(K(A)) =
z1 E K(A),...,zn+1 E K(A))
is plurisubharmonic. By Theorem A.1.22, we know that c(K(\)) is the decreasing limit of 6"(K(A)), when n goes to infinity. So, by Theorem A.1.1 (vi) for subharlnonic functions, we get the last result. 0 Let D be an open subset of C", h be holomorphic on D and 11 = {(A, z): A E
D, z # h(A)) be the complement of the graph of h. Considering the function y(, a) = 1/(z  h(A)), which is holomorphic on f2, it is obvious that 11 is an open s(t (+f holomorphy and so it is pseudoconvex. F. Hartogs proved the converse, namely
that if It is a locally bounded function from D into C such that S1 is an open set of holomorphy, then h is holomorphic on D. The original proof uses subharmonic A gi rncnts (see R. Narasiuthan, Several Complex Variables, p. 56). With Theorem 7.1.: i), H:Lrtogs's result can be reduced to the following generalization.
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Let h be a function from D C C" into C". Then h is holomorphic on D if and only if A + {h(A)} is an analytic multifunction on D. THEOREM 7.1.4.
It is obvious that h holomorphic implies that A  {h(A)} is an analytic multifunctibn. Suppose now that A " {h(A)} is an analytic multifunction on D. PROOF.
By Theorem 7.3.1 (v) and F. Hartogs's theorem on functions which are holomorphic in each variable (15], p.28 or Theorem A.2.1), it is sufficient to prove the theorem
for m = n = 1. Because the result is purely local, we suppose that we are on a included in D and we prove that h is holomorphic on A. Let A be the set of f E C(a) such that there exists an analytic multifunction K defined on a neighbourhood of a with K(A) = {f(A)), for all A E T By Theorem 7.1.1, closed disk
it follows that A is a subelgebra of C(s) containing the polynomials. By the Maximum Principle we have, for f E A, max{If(A)I: A E !3} = max{,f(A)J: A E 80}.
So, by W. Rudin's characterization of holomorphic functions ([71, p. 280), all the elements of A, and in particular h, are holomorphic on 0. 0
A great number of results obtained in Chapter III for Sp f(A), where f is an analytic function from a domain D C C into a Banach algebra, can be extended with very similar arguments to analytic multifunctions (see Exercise V11.6). The three following theorems are very important.
Let D be a domain of C" and let K be an analytic multifunction from D into C'". Suppose that there exist two disjoint open sets U, V in C" such that K(A) C U U V, for every A E D. Then either K(A) fl U = O, for every A E D, or K(A) fl U 910, for every A E D. In the latter case L(A) = K(A) fl U defines an analytic multifunction on D. THEOREM 7.1.5 (LOCALIZATION PRINCIPLE).
PROOF. The two open sets D x U and D x V are disjoint so mo, defined to be 1 on the former one and 0 on the latter, is plurisubharmonic on a neighbourhood of the graph of K. Let QA) = max{4'o(A, z): z E K(A)}. Then 00 is plurisubharmonic
on D. Also t'0(A)=0if K(A)f1U=0,and Oo(A)=1if K(A)f1U#0. Then +Go is either identically zero or identically one on D, so we obtain the first part of the theorem. Now let DI be an open subset of D and 0 be plurisubhermonic on a neighbourhood of the graph of L restricted to D1. Without loss of generality we may suppose that 0 is defined on an open subset of D x U. So we may extend 4 to a function 01 plurisubharmonic on a neighbourhood of the graph of K such that 4'I(A) = 4'(A) on the open subset of D x U and 4'1(A) = oo on D x V. Then 14(A) = max{O(A, z): z E L(A)) = max{4'1(A, z): z E K(A)). So 0 is plurisubharmonic on D1, and hence L is analytic multivalued. 0
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Let D be a domain of C" and K be an analytic multifunction from D into C'". Suppose that U is an open subset of C' such that #(K(A) fl U) = I and K(A) n 8U = 0, for A E D. Then there exists h holomorphic on D, with values in C', such that THEOREM 7.1.6 (HOLOMORPHIC VARIATION OF ISOLATED POINTS).
K(A) n U = {h(A)}.
PROOF. We apply Theorem 7.1.5 and Theorem 7.1.4. 0 THEOREM 7.1.7 (SCARCITY OF ELEMENTS WITH FINITE VALUES). Let D be a
domain of C" and K be an analytic multifunction from D into C'". Then either {A: A E D, #K(A) < oo} is pluripolar in D or there exist an integer N and a closed analytic subvariety F of D such that #K(A) = N on D \ F and #K(A) < N on F. Moreover, in this last situation, for each Ao E D \ F there exist N functions hzi... , hN with values in C', holomorphic on a neigbourhood of Ao, such that K(A) = {hI(A),..., hN(A)} on this neighbourhood. PROOF. The proof is very similar, except for some technical points, to the proof of Theorem 3.4.25. We leave it to the reader as an exercise. 0
In order to prove Z. Slodkowski's result we need two theorems, the first one being an improvement of the result proved in Exercise VII.1, and the second one resulting from the classical theorem on the exhaustion of pseudoconvex open sets by regular strictly pseudoconvex open sets (Theorem A.2.16). The following definition extends the notion of holomorphic selections. For instance, example (3) has no holomorphic selections at the branching points but has good selections everywhere, where a good selection is defined as follows. DEFINITION. An upper semicontinuous multifunction K from D C C" into C is said to have good selections if, for each Ao E D and each zo E 8K(Ao), either there exist r > 0 and h holomorphic for JA  Aol < r, with complex values, such that h(Ao) = zo and h(A) E K(A) for IA  Aol < r, or there exist s > 0 and k holomorphic for Ezzoo < a, with values in D, such that k(zo) = Ao and z E K(k(z))
for $zzo{
THEOREM 7.1.8.
Let K be an upper semicontinuous multiftnction from D C C into C, having good selections. Then K is a continuous analytic multifunction on D.
PROOF. We leave the proof of continuity to the reader. Let i be plurisubharmonic on a neighbourhood of the graph of K on D, and let +6(A) = max{q(A, z): z E K(A)}.
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Then /G is upper 3emicontinuous. Let tK be a closed disk included in D and let p
be a polynomial such that O(A) _< Rep(A) on 8A. In order to show that 0 is subharmonic we have to prove that the previous inequality is true on A. Suppose that there exists A E A such that ,5(A) > Rep(A), and let AO in A be such that +(,(Ao)  Rep(Ao) = sup{b(A)  Rep(A): A E 2K}. Then E = {A: A E N, O(Ao) Rep(Ao) = >/i(A)  Rep(A)} is compact in A, so we can suppose that dist(E, BA) = dist(Ao, OA) = 2E. We have O(A0) _ O(Ao, zo) for some zo E 8K(Ao). Because K has good selectionis, in the first situation there exist r such that 0 < r < e and h holomorphic for I 1  Aol < r such that h(Ao) = zo and h(A) E K(A). Then u(A) = O(A, h(A))Rep(.%)e(Ao)+Rep(Ao) is subharmonic for IAA01 < r, and moreover u(Ao) = 0 and u(A) < 0 for IA  AoI < r. By the Maximum Principle, u(A) = 0 for JA  Ao I < r, )ut this contradicts the fact that dist(E, 8A) = dist(Ao, 80). In
the second situat on there exist s > 0 and k holomorphic for Iz  zol < s such that k(B(zo,s)) c_ B(Ao,a), k(zo) = Ao and z E K(k(z)) for Iz  zol < a. Then v(z) = O(k(z), z)  Rep(k(z))  A(Ao) + Rep(Ao) is subharmonic for 1z  zoI < r,
satisfies v(zo) = 3, and v(s) < 0 for Iz  zol < a. By the Maximum Principle, v(z) = 0 for Ez  sot < s. This is a contradiction, as k is open, then k(B(zo,s)) contains a disk B AO, a) with a > 0 and so on B(Ao, a) there exists some k(z) with
v(z) < 0. 0 Let K be an upper semicontinuous multifunction from D C C into C such that the complement of its graph is pseudoconvex in C2. Then there exist an increasing, sequence of relatively compact open subsets D. of D, exhausting D, and a sequen ce of analytic multifunctions, respectively defined on D,,, having good selections, and such that C for each A E D and K(A) = for each A E D. In particular K is an analytic multifunction on D. THEOREM 7.1.9.
Let D = {A: A E D, dist(A, 8D) > 1/n). Then Dn is compact in D and D = Un 1D, . The set C,, = UAE5.K(A) is compact, so we can construct an increasing sequence of open disks U. such that U. D C,,. Let n = {(A, z): A E D, z V K(A)). Because it is pseudoconvex, there exists 0 E C°°(Sl), 0 strictly PROOF.
plurisubharmonic on i2, such that ilk = {(A, z): (A, z) E Sl, O(A, z) < k) is relatively
compact and satisfies Sl = Uk 1 ik. So we can construct inductively an increasing sequence such that f1k, D D1 X 8U1 , ... , (D X for all integers n. Then we define on D by {z: z E U,,, (A, z' ilk ). This set K,,(A) is compact and nonempty because K(A) C K,,(A) for all A E D,,. By Theorem A.2.17, K,, has good selections so, by Theorem 7.1.8, K. is analytic multivalued on D,,. Let A E D,, and z E K,,+1(A), so
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0 k. + , , that z E and (A, z) 0 Stk.+, f l&., hence z E Kn(A). Because which gives a contradiction. So z E U and (A, z) S U00 _zf24,, = fl, it is obvious that for A E D. we have fl,n>nKn(A) = K(A), and hence K = limn . Kn. By Theorem 7.1.1 (i), K is analytic rtiultivalued on D. 0
Let K be an upper semicontinuous multifunction from D C C into C. Then the following are equivalent: THEOREM 7.1.10 (Z.SLODKOwsKI).
(i) K is an analytic multifunction, (ii)  log dist(z, K(A)) is plurisubharmonic on fl, the complement of its graph, (iii) Il is pseudoconvex in C2. PROOF. By Exercise VII.5, (i) implies (ii). If (ii) is verified then  log dist(z, K(A)) is a plurisubharmonic vertical function for fl, so S2 is pseudoconvex (see Theorem A.2.14). If (iii) is verified then by Theorem 7.1.9, K is analytic multivalued. 0 The following result is in fact equivalent to a classical result due to F. Hartogs. It gives interesting examples of discontinuous analytic multifunctions. THEOREM 7.1.11.
Let D be open in C and ¢ be defined on D with values in
R U (oo). Then the multifunction K, defined by K(A) = {z: IzI < e0'0} on D, is an analytic multifunction on D it and only if ¢ is subharmonic on D. PROOF.
If K is an analytic multifunction on D then, by Theorem 7.1.3,
logp(!i(A)) = O(A) is subharnlonic on D. Conversely if 0 is subliarmonic, then 0 is upper semi continuous, so K is upper semicontinuous on D. By Theorem 7.1.10 we have to show that S2={(A,z):AE D,IzI >(m(a)}
is pseudoconvex. We know that {(A, z): A E D, ° 54 0} is pseudoconvex. The function (A, .) .. 4zIe is plurisubharmonic on D x C, which is pseudoconvex, and so the set {(A, z): A E D, IzIeOtal < 1 } is pseudoconvex. Then f2' = {(A, z): A E D, z 34 0, Iz It 0(A) < 1) is pseudoconvex, being the intersection of two pseudoconvex sets. But A' = A, z' = 1/z defines a biholomorphic transformation from S2' onto SZ. So fl is pseudoconvex. 0 COROLLARY 7.1.12 (F. HARrocs). Let D be open in C and 0 be a positive function on D such that {(A, z): A E D, I z I < /i(A)} is pseudoconvex in C2. Then  log v(A) is suhharmonic on D.
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Taking O(A) _
00
_ n1
log IA  1/nI + log n na
,which is welldefined and aubhar
monic on C, w e have 0(1/n) _ oo f o r n = 1, 2, ..., and 0(0) = 0. The corresponding analytic multivalued function K satisfies K(1/n) = {0} and K(0) = {z: Izl < 1), so it is discontinuous at 0. By condensation of the singularities it is even possible to construct an analytic multifunction K discontinuous on a dense set.
We now finish this section with the most important theorems: these are the motivation behind the theory of analytic multifunction. THEOREM 7.1.13 (B.AUPETITZ.SLODKOwsKI).
Let f be an analytic function
from an open set D C C into a Banach algebra. Then A s+ Sp f(A) is an analytic multifunction on D.
PROOF. By Theorem 3.4.2, A i' Sp f(A) is upper semicontinuous on D. The function (A,z) + (zl is analytic on it, so q$(A,z) = (f(zl f(A))_l11
f(A))1
log dist(A, OD) is plurisubharmonic on S2. We now show that lien #(A, z) = +oo when (A, z) goes to (Ao, zo) E 852. Suppose that there exist M > 0 and two sequences (An),
(zn) such that lim A. = Ao, lim zn = zo, (An, zn) E f2 and #(A., zn) < M. Because dist(A.,BD) > e we conclude that Ao E D, so zoIL  f(A0) is not invertible. But we have
zo1  f (A0) = z01  Zn1 + z, 1  f (An) + f(An)  f(A0)
_ (Zn1  f(An))I1 + (zn1  f(A.))lu(An,ze)l If A,, goes to A0 and zw goes to zo, then u(A,,zn) goes to 0 and we have
II(zni f(AR))'u(An,zn)II 5 (M+log
1,
for n large enough. Thus zoI  f (Ao) is invertible, which is a contradiction. By Theorem A.2.11, f2 is pseudoconvex, so A r+ Sp f (A) is analytic multivalued on D, by Theorem 7.1.10.0 We recall that the gilov boundary is defined in Exercice IV.10. THEOREM 7.1.14 (Z.SLODKOwsKI).
Let A be a commutative Banach algebra. Denote by 932 its set of characters and by S its Silov boundary. Let f,g be two elements of A and let W be a component of f (tV?) \ f (S). Then A  g(f 1(A)) _ (X(g): x(f) = A} is an analytic multifunction on W. The proof is too complicated to be given here. It can be found in several papers by Z. Slodkowski.
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A weak form of this theorem, due to J. Wermer, will be found in Exercise VII.8.
Theorem 7.1.14 and Theorem 7.1.7 imply immediately a strong generalization of E. Bishop's theorem on analytic structure (110), Chapter 11). Finally, why are Theorems 7.1.10, 7.1.13, 7.1.14 of interest? At first glance it seems very strange to start with some spectral problems or some questions relating to uniform algebras and then to consider the rather difficult theory of pseudoconvex open sets. What do we gain by doing this? The main reason is that many problems
in spectral theory and in the theory of uniform algebras are reduced to purely geometrical problems concerning pseudoconvex open subsets of C2. Theorem 7.1.6 is a good illustration of this. In the next two sections we shall discover new ones.
§2. The OkaNishino Theorem and Its Applications We now intend to give the proof of the very important OkaNishino theorem. The original proof due to T. Nishino, contains many obscure points. The following presentation is largely inspired by the ideas of B. Aupetit, T. Nishino, T.J. Ransford, Z. Slodkowski and J. Zemanek. Very often it will be rather superficial since the arguments involved are extremely technical.
Let K be an analytic multifunction from D C C into C. We say that zo E K(Ao) is a good isolated point of K(Ao) if there exist r,s > 0 such that DEFINITION.
B(zo,s) fl K(Ao) = {zo}, and B(zo,s) fl K(A) is finite for IA  AoI < r. This definition implies that zo is isolated in K(Ao). But conversely an isolated point of K(Ao) is not necessarily a good isolated point. To see this, consider the following example of the analytic multifunction defined for A E C by
K(A) = {0} U
2 < jzi < jAj)
.
Of course 0 is isolated in K(0), but it is not a.good isolated point. In fact this definition of a good isolated point is purely geometrical. It means that in a neighbourhood of (A0,zo), the graph of K is an analytic variety. This notion is closely related to that of extension points for pseudoconvex sets which we now define. DEFINITION. Let 12 be a pseudoconvex open subset of C2. We say that a point a E C2 \ S2 is an extension point for S2 if there exist an open neighbourhood U of a and a nonzero holomorphic function f on U, with values in C, such that
U\S2=(z:zEU,f(z)=0). We define 5l' to be the union of Sl and its extension points. Clearly SY is open and contains fI.
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THEOREM 7.2.1. Let K be an analytic multifunction from D CC into C. If zo is a good isolated point of K(Ao) then (Ao, zo) is an extension point forSl = {(A, z): A E D, z V K(A)). Conversely, if zo E 8K(Ao) and (Ao, zo) is an extension point for 1, then zo is a good isolated point of K(Ao). PROOF. Suppose first that zo is a good isolated point of K(Ao). By Theorem 7.1.5 and Theorem 7.1.7 applied to A ., B(zo,s) fl K(A) for JA  Aol < r, we conclude that there exist a smallest integer n and a closed discrete set E in B(Ao, r) such
that #(B(zo,s) fl K(A)) = n for A E B(Ao,r) \ E and #(B(zo,a) fl K(A)) < n for A E E. Taking a smaller r if necessary, we may suppose that E fl B(Ao,r) C where hl,.. , h. are holomorphic {Ao}. So B(zo,s)fK(A) _ for 0 < JA  Aol < r. The symmetric functions in hi(A),...,h,,(A) define global holomorphic functions for 0 < JA  Aol < r. The singularity at A0 is removable for these symmetric functions because they can be extended continuously, using a similar argument to that in the proof of theorem 3.4.5. So f(A, z) = rj ,(zh,(A)), which is expressed only with the symmetric functions of the h,(A), is holomorphic on B(A0, r) x B(zo, s). But z E K(A), for IAA01
Suppose now that zo E 8K(Ao) and that (Ao, zo) is an extension point for R. Let f and U be as in the definition of an extension point. If h(z) = f(A0, z) = 0, locally around zo, then zo is interior to which is a contradiction. So the zeros of h are isolated. Let s > 0 be such that $(zo,s) fl K(Ao) = (zo) and let r > 0 be such that IA  AoI < r implies OB(zo, s) fl K(A) = A. We can also suppose that r is small enough for the zeros of f (A, z) to be isolated for all A fixed satisfying JA  AoI < r. Consequently I A  AoI < r implies that B(zo, s) fl K(A) is finite. 0
Let M,N be open disks in C and U = NxM. Let f be holomorphic on U, with values in C, not identically zero on U. Let Z be the set of (z, rl) E C2 such that LEMMA 7.2.2.
(i) D(z,q)={w:wEM,z+rtwENJ 76 0, (ii) g(w) = f (z +"qw, w) is identically zero on D(z, q).
Then Z is at most countable. Suppose Z to be infinite and uncountable. Then there exists (zo, rio) E Z such that each of its neighbourhoods contains an infinite and uncountable subset of Z. Let D = D(zo,go), which is open in C. Since f is not identically zero on U, the set W = {w:w E M, g(() = f((, w) _ 0 on N} PROOF.
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is closed and discrete. So D \ W is open and nonempty. Let wo E D \ W and r > 0 be such that
fwwo'
(z  zol < r, (qgoI
Let f) be a pseudoconvex open subset of
C2 . Then f2' is also pseudoconvex.
SKETCH of PROOF. Suppose that R' is not pseudoconvex. Then we may choose r > 1 and F: C X B(0, r) F+ C2 such that:
(i) F is a biholomorphic mapping onto an an open subset of C2, (ii) Iz) 5 1 and IwI = I implies F(z,w) E fl',
(iii) IzI < I and lwj < I implies F(z,w) E f1',
(iv) F(1,0) ¢ 12',
(v) if F = (FI,F2) then the two functions f;: B(0, r) ++ C, defined by f;(w) Fi(l,w), are nonconstant.
_
Let A = F''(1). Then we have A pseudoconvex in C x B(O,r). Also, since F is biholomorphic, we have A'= F'(n'). Thus
(zI<1and IwI5limply (z,w)EA', (zISIand (wI=1imply (z,w)EA', and (1,0) ¢ A'.
We consider the setR={w:wEC,Iw)
there exists 0 < s < 1 such that s <'wI < 1 implies (1, w) E A'. Since R is discrete it must be at most countable, so we may chooses such that, further, Jw) = s implies
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(1, w) E A. Then {w: Iwi < s, (1, w) A) is relatively compact in B(0, s). There exists t > 0 such that the multifunction K, defined on B(1, t) by K(z) = {w: w E C, IwI < s, (z, w) f A},
is analytic. Suppose that Iz  11 < t, Izl < 1 and w E 8K(z). Then (z,w) E A' \ A, so it is an extension point for A, and consequently an extension point for the intersection of A and B(1, t) x B(0, s). By Theorem 7.2.1, to is a good isolated point for K(z). But this holds for all w E OK(z), so K(z) is finite for Iz  II < t and IzJ < 1. By Theorem 7.1.7, K(z) is finite for all z satisfying Iz  11 < t. In particular, 0 E K(1) is a good isolated point so, by Theorem 7.2.1, (1,0) E A'. But this is a contradiction.
Second caae: R is not discrete. We may choose a countable set of pairs (f;,U,), with the U, open polydisks Ni x M; and the f; holomorphic and not identically zero
on U;, such that (z,w) is an extension point for A if and only if f:(z,w) = 0 for some i. There exists b with 0 < 6 < I such that
IzI<1+b,IwI=1 implies (z, w)EA'. By Lemma 7.2.2, there exists q E C with I,l < b such that the sets {w: IwI < r, f;(z +r1w,w) = 01
(1)
are all discrete, for any z E C.
Let J(z, w) = (z + r1w, w) and 0 = J' (A). Because J is a biholomorphic mapping, a is pseudoconvex and A' = J'(A'). Also since Ir1I < b, IzI < 1 and I w I = 1 imply (z,w) E 0'. If we set
to = inf(IzI: (z, w) ¢ 0, for some Iwl < 1), then 0 < to < 1. Let zo, wo be such that IzoI = to, Iwol 5 1 and (zo, wo) 0 0'. Then Iwol < 1. So we have the following situation
0 is pseudoconvex in C x B(0, r),
)zl < to and Iw) < 1 imply (z,w) E 8',
Iz)
0',
(zo, too) 0 8' where IzoI = to, IwoI < 1,
and by (1), {w : IwI < r, (zo, w) E 0'\8} is discrete. Now using an argument similar to that of the first case, with (0, 1) replaced by (zo, too), we get a contradiction. 13
Analytic Multifunction
155
For more details on this proof, look at T.J. Ransford, Analytic Multivalued Functions, Doctoral Thesis, Cambridge, 1984.
By definition, we denote by DK(A) the set of points of K(A) which are not good isolated points. It is easy to prove that DK(A) is compact and satisfies K(A)' C DK(A) C K(A), where K(A)' denotes the set of limit points of K(A). By transfinite induction we can define D°K(A) for all ordinal numbers a by
D°K(A) = D(D°''K(A)), if a is not a limit ordinal, D°K(a) = fl,6<°D8K(A), if a is a limit ordintd, with the convention that D°K(A) = K(A). G. Cantor introduced the notion of an aderived set of a closed set C defined by
C(a) = (C(°1))', if a is not a limit ordinal,
C(O) = l9
By transfinite induction it is easy to prove that K(A)( ordinal numbers a.
C D°K(A), for all
THEOREM 7.2.4 (K.OJAT.NisHINo). Let K be an analytic multifunction from a domain D C C into C and let a be an ordinal number. Then either D°K(A) 34 0
for all A E D and D°K: A " D°K(A) is an analytic multifunction on D, or Do K(,\) = 0 for all X E D. In the latter case let y be the smallest ordinal such that D IK(A) = 0, for all A E D. Then ry is not a limit ordinal and there exist an integer
n and a closed discrete subset F of D such that #D"'K(A) = n, for A E D \ F,
and #Df'K(A)
to 0', it is not difficult to see that either L(A) = 0 for all A E D or L(A) 96 0 for all A E D, in which case L is an analytic multifunction on D. It is easy to see that 8DK(A) C 8K(A) so, by Theorem 7.2.1, we have 8DK(A) C L(A) C DK(A), for
allAED. IfL(A),40forallAED,then DK(A)76 0,forall A E D, and it is easy to see that DK is upper semicontinuous. Hence by Theorem 7.1.2, DK is an analytic multifunction on D. If L(A) = 0 for all A E D, then ODK(A) = 0 and
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A Primer on Spectral Theory
consequently DK(A) _ = 0, for all A E D. This says that all points of K(A) are good isolated points of K(A ). Hence K(A) is finite for all A E D, and we apply Theorem 7.1.7 to obtain the conclusion.
Suppose now that a is an arbitrary ordinal number. In order to use transfinite induction, we suppose that the previous properties have been proved for DAK with fi < a. If there exists 8 < or such that DOK(A) = 0 for all A E D, it is obvious that
D°K(A) = 0 for all : E D. So suppose that DOK(A) # 0, for all 0 < a and all A E D. By hypothesis all the DAK are analytic multifunctions on D. If a is not a limit ordinal, by the t rst part of the proof D°K(A) = D(D` K(A)) is either nonempty and D°K is an analytic multifunction on D, or D°K(A) = 0 for all A E D. In this latter case, D''K(A) is finite for all A E D and we conclude as previously. If a is a limit ordina, then {fi: Q < a} is a countable wellordered set such that 0 <,B' < a implies D 3'K(A) C DAK(A). Thus ns<°D,6K(A) is nonempty and, by Theorem 7.1.1(i), D°K is analytic multivalued on D. Using transfinite induction, we then get the result. 0
In fact we shall see in the next theorem that it is not necessary to consider D°K(A) for all a > wl, where wi denotes the first uncountable ordinal number, because D°K(A) stabilizes after some 7 < w1. This was proved by B. Aupetit and J. Zemanek for K(A) at most countable, but the proof is the same in the general case.
Let K be an analytic multifunction from D C C into C. Then there exists 7 < w, such that D7K(A) = D°K(A) for all a such that 7 < a < w, and for allAED.
THEOREM 7.2.5.
By definition of D°K(A) and transfinite induction, it suffices to prove that there exists y < wl such that D'K(A) = D7+'K(A) for all A E D. First we fix A E D. By Kuratowski's theorem (see Theorem 5.7.9), there exists a smallest ordinal 7(A) < w, su.h that D°K(A) = D7(a)K(A) for 7(A) < a < wi. Let E° = {A:A E D,7(A) > a}. Then Ea = U0<0
Now if D, K(Ao) 96 D0+'K(A0), then D8K(Ao) contains a good isolated point zo. This implies that DQK(A) contains good isolated points for A in a neighbourhood
of A0. Consequently E. is open. Then F. = D \ E. is closed, and a < fi implies F. C Fs. So, by Kuratowski's theorem, there exists 7 < w, such that y < a < wl
implies E°=E,. IlutA0E,(a)for all AED. If AEE,then 7(A)>7,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
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157
The classical CantorBendixson 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 14t 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 nonzero 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.AUPETITJ. 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
Analytic Multifunctions
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 selfadjoint 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 OkaNishino 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.
A Primer on Spectral Theory
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.AUPETITJ.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<°<..Z°, where Z. = {A:.\ E D,z E PROOF.
D°K(.X) \ D°+1K(.\)}. By Corollary 7.2.12, applied to the analytic multifunction D°K, we conclude that D°K has only isolated or interior points. Because the set of ordinals less than y is countable, Z is the disjoint union of an open set and of an at most countable set. If the interior of Z is empty then Z is at most countable and we have finished. If not, we shall show that Z = D. First we note that Z is closed in D by upper semicontinuity of K, and so the boundary of Z in D is at most countable. Let F be the closure of the interior of Z in D. It is enough to prove that
F = D. To do this we have to prove that F is open. Let a E F and let r > 0 be such that ff(a, r) C D. There exists b in the interior of E such that Ia  bI < r. The set of halflines r, with origin at b and such that r n B(a, r) contains a boundary point of Z, is at most countable. So the interior of Z is dense in B(a, r) and hence F J B(a, r). 13 Obviously Theorem 7.2.13 is not true if K is a general analytic multifunction. For example if we take K(.\) _ {z: IzJ < 1} U {z: IzI < dal},
then we have 2 E K(A) if and only if JAI > 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 quasialgebraic in the sense of P. Halmos: see (2), pp. 251253). Is it true that Z is a closed set of capacity zero? This problem is related to A. Sadullaev's result implying thatthe 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 nonconstant 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(re`e)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:
A Primer on Spectral Theory
162
If F is merornorphic for IzI < R < +oo, and if 00 co = F(0) # oo. then for p < r < R we have LEMMA 7.3.1.
rn (p,
I
+16. < 4 loge' T (r, F) + 3 log+ 1 + 4 log+ r + 2 log+ 1 + 4 log+ log+ I) Ico$ F rp
Using the relation
_
F
yr
and the previous lemma, it
is
possible to prove that there exist constant numbers ak, a'' , bk, bk, b'k, ck depending only on k such that /
M [ p,
(k)
F
< ak + ak log+ log+ J F(0) + bk 1og+ I
+bk log
a + bk log r +
r p
+ ck log+ T(r, F).
See KingLai Hiong, Extension dun thiorime de M.R. Nevenlinna, Paris, 1957. The following lemma is a weak form of a theorem due to E. Bore!. LEMMA 7.3.2. Let q1, ... , On be n linearly independent entire functions such that 41 + 0,2 + .. + Yin = 1. Then at least one of the ¢, has a zero.
SKE'rcii of PROOF. F o r a l l k =1, 2, ... , n 1, we have mlk) + + 0") = 0. The n functions 0; being linearly independent, the Wronskian D is not identically zero, A. where 01
02
...
0.
...
D
011
612
o(n!)
o(nI)
01n
We also introduce 1
I
1
s
i
II
02
#1
1
...
...
ti
ion
Let D, be the minor determinant corresponding to the ith element in the first :isle of D, and let A, be the corresponding minor determinant in A. We have 01 = 02 Dm., ' es' ,, = n, . Consequently m(r, 01) = m(r, A/A1) < N(r, A)  N(r, l/A) + m(r, A) + m(r, A1) + 0(1).
163
Analytic Multifunction
If all the Oi are not vanishing, since 0 = DI 02 N(r,1/0) = N(r, D)  N(r, 11D), and so
we get that N(r, 0) 
T(r, 01) = m(r, 01) < N(r, D)  N(r, l/D) + m(r, Al) + m(r, E) + 0(1). If T(r) denotes the greatest of the T(r, cb;), for i = 1, ... , n, then applying the previous inequalities we get
T(r) < O(logT(r)+logr), except perhaps on some segments having a finite total length. So liminfr... ogrr < +oo, which gives a contradiction, because this condition implies that the O, are polynomials. 0 Now we intend to generalize Picard's theorem to finite analytic multifunction by exploiting an idea of G. Remoundos (Eztension aux fonctiona aigebrofdea du theoreme de M. Picard et ses generalisations, Paris, 1938). First we introduce the notion of spectral multiplicity.
Let K be an analytic multifunction defined on an open subset D of C such that K(A) is finite for all A in D. Let K(Ao) = feel,. .., ap } and e > 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;.
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A Primer on Spectral Theory
Let F(A,u) = u" + A1(A)u"' + + A, (A) be defined on C2, where the A,(A) are nonconstant 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 nonconstant 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 n1
uI
u2t
n2
ul
tie
u1 un2 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 nonconstant 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
vnt
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 vnI
vn2 0
0
qj =
...
vo
tin1 n1
00
vi+ I
vn2
Ivn1
n
...
vn
is also a Vandermonde determinant. So we have
q=
k'(1)
E(1)'q;
 ` E Ak(A)vnk}
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
vn1 0
,n1 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 nonconstant analytic multifunction on C. Suppose that K( 1) is finite on a set E having a nonzero 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 2nl 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.
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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 welldefined 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)nCnle'
We have [nI
det(f(A)  z) _ (al  z)...(a.  z) + eA
Ci(ai+l
z)...(a.  z) + CfI
.
Let n1
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)...(a2nl  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.
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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 G6set 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 setvalued functions and spectra, Math. Ann. 256 (1981), pp 363386), 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 nonempty 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 nonempty 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/(ua) But (A,a+l/(una)) 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.
A Primer on Spectral Theory
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
za
//
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 u1(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 G6set 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 G6set of capacity zero. In particular if we consider the analytic multifunction K then either Ki A)" is constant or C \ UAECK(A)" is a G6set 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 G6set 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 nonempty, U is connected. Moreover F C U. So by Theorem 7.3.9 (ii), we conclude that either F is a G6set 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 G6set 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 G6set 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 G6set of capacity zero, or C x { p) C S2 for all µ. Suppose
A Primer on Spectral Theory
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<,,
Let K be a nonconstant convex analytic multifunction on C. Suppose that f1aECli(1) 0. Then UAECK(A) = C. THEOREM 7.3.12.
Let a E (ECK(A) and z E C. The halfline with origin at a containing z has nonzero capacity, so by Theorem 7.3.10 there exists 6 E K(Ao) such that z is on the segment (a, 61. But (a, 61 C K(ao), so z E K(ao). 0 PROOF.
As a corollary we immediately obtain the following result of J.P. Williams. COROLLARY 7.3.13. Let a,b be two noncommuting elements of a Banach algebra A. We define W (x) _ (f (z): f E A', ((f U = f (l) = 1) to be the numerical range of X. Then U)ECW(eabaeab) = C.
PROOF. The multifunction 1 14 W(e.0aeab) is trivially analytic because it has entire selections. Morevover, it is not constant because ab # ba, and Spa C fAECW(eabae'ab).So we get the result by Theorem 7.3.12. 0 In fact these lasi. results are particular cases of the following result, which was obtained by T.J. Rar.sford with the help of covering spaces and lifts of multifunctions.
Let K be an analytic multifunction on C and suppose that K(A) is connected for all \ E C. Then either K(A)" is constant or the union of all K(A)" covers all the plane except perhaps one point. THEOREM 7.3.14.
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171
For the proof, see the thesis of T.J. Ransford. This result implies in particular that for a convex analytic multifunction defined on the complex plane, only one of the following possibilities occurs:
(i) if UAECK(A) avoids two points of C, then K(A) is constant on C, (ii) if U,ECK(A) avoids one point a E C then K(A) has the form K(A) = a +
eti(a)Ko, where h is an entire function and Ke is a fixed compact convex set, (iii) UAECK(A) = C.
§4. Conclusion We hope that this quick introduction to analytic multifunction has inspired the reader to learn more on this new field. Further material can be found in the publications of H. Alexander & J. Wermer, B. Aupetit, B. Aupetit & J. Zemanek, B. Aupetit & A. ZraIbi, T.J. Ransford, Z. Slodkowski, and J. Wermer. In these, many applications to spectral theory and to the theory of uniform algebras are given.
Other applications of great interest have been given to the theory of spectral interpolation (T.J. Ransford) and, very surprisingly, to the Corona Problem mentioned in Chapter IV, §1 (see B. Berndtsson and T.J. Ransford, Analytic Multifunctions, the sequations and a proof of the Corona Problem, Pacific J. Math. 1986; Z. Slodkowski, An analytic setvalued selection and its applications to the corona theorem, Trans. Amer. Math. Soc. 1986; and Z. Slodkowski, On bounded analytic functions in finitely connected domains, Trans. Amer. Math. Soc. 1987). In order to generalize the quantum mechanical formalism, P. Jordan, J. von Neumann and E. Wigner introduced, in 1934, the notion of what is now called a Jordan algebra. This theory has been intensively studied since then (see for instance the recent book by K.A. Zhevlakov, A.M. Slin'ko, I.P. Shestakov, A.I. Shirshov, Rings That Are Nearly Associative, New York, 1982). Several mathematicians stud
ied, with some success, the theory of complete normed Jordan algebras, that is the JordanBanach algebras (see for instance the important paper of E.M. Alfsen, F.W. Shultz and E. Stormer, A GelfandNeumark theorem for Jordan algebras, Adv.
in Math. 1978). The absence of associativity in JordanBanach algebras implies great difficulties for calculations. But in the last five years the use of subharmonic methods and of the theory of analytic multifunctions has given a great number of important results in this nonassociative theory. Obviously we shall not give any details, but the reader will find these new methods and results in the publications of B. Aupetit, B. Aupetit & L. Baribeau, B. Aupetit & M.A. Youngson, B. Aupetit & A. Zraibi, M. Benslimane & A. Kaidi & A. Rodriguez Palacios.
172
A Primer on Spectral Theory
EXERCISE 1. Let K bean upper semicontinuous multifunction from D CC into C, having holomorphic selections. Prove that K is a continuous analytic multifunction on D.
EXERCISE 2. Let f be an analytic function from D C C into M"(C). Prove that A " Sp f (A) is an analytic multifunction on D (Let D' be the set of A E D such that det(z  f (A)) has distinct roots. Using the Implicit Function Theorem prove the result on D' and then, by continuity, extend it to D).
Let (K") be a sequence of analytic multifunctions defined on D C C" with values in C. Suppose that for each A E D there exists a nonempty compact subset K(A) of C' such that EXERCISE 3.
lim A(K,,(A), K(.1)) = 0,
n.00
uniformly on each compact subset of D. Prove that K is an analytic multifunction on D, which is continuous if the K,, are continuous.
Let K1, K2 be two analytic multifunctions from D C C" into C'". Prove that K1 + K2 is an analytic multifunction. Conclude that the convex hull of an analytic multifunction is also an analytic multifunction. EXERCISE 4.
EXERCISE 5.
Let K be an analytic multifunction from D C C" into C and let
f) = {(A, z): A E D, z 0 K(\)} be the complement of its graph. Prove that (A, z)  log dist(z, K(A)) is plurisubharmonic on Q. EXERCISE 6.
Extend Theorems
to the situation of analytic multi
functions. EXERCISE 7. Let K be an analytic multifunction from D C C into C. Suppose that, for some Ao E D, K(Ao) is totally disconnected. Prove that K is continuous at A0.
With the notation of Theorem 7.1.14, prove that the function A log p(g(f (A))) is subharmonic on W.
*EXERCISE 8.
**EXERCISE 9.
Try to prove Theorem 7.1.14.
*EXERCISE 10.
Using Theorem 7.1.14 and Theorem 7.1.7, give a new proof of E. Bishop's theorem on analytic structure (1101, Theorem 11.2). Prove that it is
even possible to replace the hypothesis " G of positive planar measure" by the weaker hypothesis " G of positive capacity".
Analytic Multifunction
173
Using Theorem 7.1.14 and Theorem 7.2.8, give a new proof of R. Basener's theorem on analytic structure ((10], Theorem 20.3). Prove that it is even sufficient to suppose that the fibres are at most countable on a subset G of W
*EXERCISE 11.
having positive capacity. *EXERCISE 12.
Let C be a sul,sct of the complex plane not containing 0 and having
at most 0 as a limit point. Prove that there exists to Pnalytic multifunction K defined on C with values in C, such that K(A) has at most 0 as a limit point for each A E C, and satisfying C \ UACCK(A) = C.
Let K be an analytic multifunction from a domain D of C into C. Suppose that K(A) is convex and never contains 0 for A E D. Denote by a(A) the angle under which K(A) is seen from 0. Prove that a is subharmonic on D. If moreover D = C, conclude that K(A) has the form
*EXERCISE 13.
K(A) = eb(a)Ko where h is an entire function and KO is a fixed compact and convex set not containing 0.
*EXERCISE 14. C.
Let K be an analytic multifunction from an open set D of C into
(i) Suppose first that K(A) is a segment [0, a(A)], for each A E D. Prove that a is holomorphic on D.
(ii) Suppose now that K is continuous on D and that K(A) is a segment for each A E D. Prove that a + P and a# are holomorphic on D. Conclude that there exists an analytic function f from D into M2(C) such that K(A) = coSp f(A), for every A E D.
Let K be an analytic multifunction from an open set D of C into C. Suppose that K(A) has at most zero as a limit point, for each A E D. Does
**EXERCISE 15.
there exist an analytic function f from D into the algebra of compact operator,, 12, such that K(A) = Sp f(A), for each A E D?
APPENDIX
This appendix is essentially a list of results without proofs. The reader intending to learn more about the deep results given in these two sections will be obliged to refer to the selection of books mentioned below.
§1. Subharmonic Functions and Capacity Unfortunately, the most important results on subharmonic functions and ca
pacity are scattered in many books, for instance in [4,5,9) and in the following books: M. Brelot, Elements de is theorie classique du potentiel, Paris, 1965; R. Narasimhan, Several Complex Variables, Chicago, 1971; I. Privaloff, Subharmonic Functions (in Russian), Moscow, 1937; T. Rado, Subharmonic Functions, New York, 1949; M. Tsuji, Potential Theory in Modern Function Theory, New York 1975. For a brief survey, look at the appendix of [1]. THEOREM A.1.1.
Let D be an open subset of the complex plane.
(i) If 01 and 02 are subharmonic on D then 01 + 42 is subharmonic on D. (ii) If d is subharmonic on D and if a is a positive number then a m is subharmonic on D. (iii) If 01 and 02 are subharmonic on D then max(O1, 02) is subbarmonic or. D.
(iv) If 0 is subharmonic on D and if f is a real, convex and increasing function on R, then f o m is subharmonic on D (by convention f(oo) = lim f(x) when x goes to oo).
(v) If
is a sequence of subharmonic functions on D converging uniformly to on each compact subset of D, then 0 is subharmonic on D.
(vi) If
is a decreasing sequence of subharmonic functions on D then 0 = lim is subharmonic on D.
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175
(vii) If is locally integrable on D, and satisfies the mean inequality, and if O*(z) = limsup{O(u):u + z} > 4(z), called the upper regularization of 0, satisfies 0*(z) < +oo for all z in D, then 0* is subharmonic on D_
(viii) Let K be a compact set and p be a positive measure on K. Suppose that (&t)tEK is a family of subharmonic functions on D such that (t,z) ' &t(z) is measurable on K x 813(a, r) for all $(a, r) C D, and such that t :+ Ot(z) is itintegrable for all z E D. Then O(z) = fK Ot(z)dp(t) is subharmonic on D. For (b subharmonic on D, a E D and r > 0 such that $(a, r) C D, we introduce the following notation: f2l, O(a
N
+ reie)dO ,
2n
THEOREM A.1.2. have
M(a, r, 40) = max d(a + rfie). 0<9<2a
If 0 is subharmonic on an open set D, then for every a in D we
O(a) _ limsup'(z) = lira A'(a,r,q) = lim r 4 Z;da
r 0 r>0
r_0 r>O
THEOREM A.1.3 (MAXIMUM PRINCIPLE FOR SURHARMONIC FUNCTIONS).
Let ¢ be a subharmonic function on a domain D. Suppose that there exists a in D such that O(z) < tb(a) for all z in D. Then O (z) = ¢(a) for all z in D. COROLLARY A.1.4. Let D be a bounded domain and 4 be subharmonic on D. Suppose that there exists M such that for every f E 8D we have lim sup ; F '(z) <
M. Then we have 4(z) < M on D or 0 is constant on D.
ZED
If D is unbounded, the same result is true if we suppose further that lim sup,O" ¢(z) < M. ZED
THEOREM A.1.5.
Let 0 be upper semicontinuous on an open subset D of C. Then 0 is subhartonic on D if and only if, for every closed disk included in D and for every polynomial p, the inequality O(z) _< Rep(z) on the boundary of the disk implies the same inequality on all the disk. COROLLARY A.1.6.
Let 01,02 be two positive functions such that log O1 and
log ¢2 are subharmonic on an open set D. Then log(#1 +
) is subharmonic on D.
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176
COROLLARY A.1.7.
Let. D be an open subset of C. Then  log dist(A, 8D) is
subharmo»ic on D. Let be positive on an open set D. Then log k is subharmonic on D if and only if z  leA410(z) is subharmonic THEOR.Ehi A.1.8 (E.F.PECKENBACHS.SAKS).
on D fo: eiery polynomial p. THEOREM A.1.9 (LOCAL CHARACTERIZATION OF SUBHARMONIC FUNCTIONS).
Suppose t..at W is upper sernicontinuous on an open set D and suppose that for each a ir2 D there exists a sequence (re) satisfying r > 0, lim,,._..,° r = 0 and 1
4(a) <2r
z,r
0(a fol
for all n = 1, 2.... Then 0 is su bharmonic on D. In particular 4 is subharmonic on D if and only if it is locally subharrnonic. COROLLARY A.1.10. Ho is of class C2 on an open set D then 0 is subharmonic
on DIf and only i{L4'ispositive on D.
If = 0, then ¢ is
THEOREM A.1.11 (LIOUVILLE's THEOREM FOR SUBHARMONIC FUNCTIONS).
is subharmonic on the complex plane and if lim infr_°O 'x log y constant. THEOREM A.1.12 (T. RAD6).
Let 0 be a subharmonic function on an open set D. There exists an increasing sequence of open sets D. which exhaust D and are relatively compact in D, and a decreasing sequence of functions ¢ subharmonic on D such that 0. E C°°(DA) and 4'(z) = 0,,(z) on D. Let h be holomorphic on an open set D and 0 be subharmonic on a neighbourhood of h(D). Then 0 o h is subharmonic on D. THEOREM A.1.13.
Using Theorem A.1.12 it is possible to obtain the following improvement of Theorem A.1.8. THEOREM A.1.14 (P.MONTELT.RADO). Let 0 be positive on an open set D. Then log 4' is subharmonic on D if and only if z F le°`14'(z) is subharmonic on D, for all complex numbers a.
The following theorem, which is an important step in the proof of H. Cartan's theorem, uses Theorem 1.1.7 in its proof.
177
Appendix
Suppose that 4) is subharmonic and not identical to oo on a domain D. Then there exists a unique Borel positive measure o on D such that for every compact subset K of D and every z in D, we have THEOREM A.1.15 (F. RIES2).
log Iz  e l dp(f) + h(z),
O(z) =
where h is harmonic on the interior of K.
We now study the notions of thin and nonthin sets, first introduced by M. Brelot.
We say that a subset E of C is nonthin at a if a E V and if for all 0 subharmonic on a neighbourhood of a we have ¢(a) = lim sup 4)(t).
za z#a
zEE
For instance, by Theorem A.1.2, C \ {a} is nonthin at a. If a E "r \ E, then E is thin at a, if there exists 0 subharmonic in a neighbourhood of a such that 4(a) > lim sup,. O(z). spa
CEE
It is easy to verify that E nonthin at a and E C F implies F nonthin at a. THEOREM A.1.16 (M. BREI.oT).
Let K be a compact subset of an open set D in C. Suppose that there exists 0 subharmonic on D \ K, such that 4)(z) < 0 on D \ K and lim ta 4)(z) = 0 for some a E OK. Then K is nomthin at a. zED\K
THEOREM A.1.17 (K.OKAW.ROTHSTEIN).
If 1' is an arc in C then it is nonthin
at each of its points. THEOREM A.1.18.
Let U be a domain in C. Then U is nonthin at each of its
boundary points. We now introduce the notion of capacity for Borel subsets of the complex plane,
which is in some sense a measure of their size. First we begin with compact sets. For K compact and p a positive measure supp9rted by K, the logarithmic potential is defined by
u(z) =
flog lzeIdp().
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178
Then u is subharmonic on the complex plane and harmonic outside of K. We suppose now that p is a probability measure concentrated on K. For z E K we have log Cz  tJ < log 6(K), where 6(K) denotes the diameter of K. Consequently,
I(p) =
JK
u(z) dp = ff log Iz  f I dp(C)dp(z) KxK
exists and is bounded above by log b(K). We now introduce the equilibrium value V(K) of K, given by
oo < V(K) = sup I(p) < log 6(K) v
for all the probability measures p concentrated on K. The capacity of K is c(K) _ e"(K). Of course 0 < c(K) < b(K). This quantity is difficult to calculate but we shall give other geometrical definitions latter. It is obvious that c(K,) < c(K2) if
K,CK2. Suppose that K is a compact subset of C such that V(K) > oo, that is c(K) > 0. Then there exists a unique probability measure p supported
THEOREM A.1.19.
by K such that I(p) = V(K). We now extend the definition of capacity to any subset of C. If E C C we define the inner capacity of E by
c`(E) = sup c(K) < +oo KCE
for all compact subsets K of E, and we define the outer capacity of E by
c+(E) = tijec(U) < +oo for all open sets U containing E. The following properties are easy to prove:
(i) for every E C C we have c(E) < c+(E),
(ii) if E C F then c(E) < c(F) and c+(E) < c+(F), (iii) if E' is the image of E by the transformation z + az + Q then c(E') _ laic(E) and c+(E') = talc+(E), (iv) if U is open then c+(U) = c(U). We say that a set E is capacitable if c(E) = c+(E). In that case we denote by c(E) this common value. Of course open sets are capacitable. In the next theorem we obtain that compact sets are capacitable. A very deep result of G. Choquet tells us that bounded Borel sets, more generally bounded analytic sets, are capacitable.
179
Appendix
THEOREM A.1.20. Let K be a compact subset of C and (Un) be a sequence
of bounded open sets such that 'U,,+, C U. and K = n, 1U.. Then c'(K) = limn...,, c(Un), and consequently K is capacitable. THEOREM A.1.21. If (En) is a sequence of Borel sets with inner capacity zero
then their union is a Borel set of inner capacity zero. We denote by Yn the set of polynomials of the farm p(z) _ 'z'8 + alz'l + + an. Also, we denote by tn(K) the quantity infpeV.. max:EK (p(z)I. A compactness argument in the finitedimensional vector space T. shows that tn(K) maXzEK Ipn(z)I for some pn E Yn, which is in fact unique for K given and is called the nth Tchebycheff polynomial of K. THEOREM A.1.22 (M.FEKETEG.SzEGO). have
For any compact subset K of C we
c(K) = n.oo lim bn(K) = lira (tn(K))l1". n=oo COROLLARY A.1.23.
For any compact subset K of C we have c(K) = c(OeK),
where O. K denotes the outer boundary of K. THEOREM A.1.24 (G.C. EVANS). Let K be a compact subset of C having zero capacity. There exists a probability measure p, supported by K, such that, for
the corresponding logarithmic potential u(z) = f K log Iz  f I dd(b), we have K = {z: u(z) = oo}. THEOREM A.1.25.
Let K be a compact and connected subset of C. We denote by D the unbounded component of C \ K. Suppose that there exists a conformal mapping w from D onto {z: IzI > 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 Rthen c(K) = R.
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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 hHausdorff measure of E. If h(t) = t°, with a > 0, then the corresponding quantity Ha(E) is the aHausdorff 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 Goset 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 nonempty 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 nonempty 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 nonempty open subset of C". Then the following properties are equivalent: THEOREM A.2.2 (H.CARTANP.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 nonempty 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.
A Primer on Spectral Theory
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 bellknown 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 nonempty 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 nonempty 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 nonempty 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 nonempty 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 wellknown) 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 nonempty 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
A Primer on Spectral Theory
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 19531954. By Theorem P .2.16 and the BehnkeStein 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. 194215). 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, AndreottiGrauert, 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 wellknown 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 LeviKrzoska theorem (see [9), pp. 157162) 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 Csstrictly
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. SpringerVerlag, BerlinHeidelbergNew York, 1979.
2. Bonsall, Frank F. and Duncan, John: Complete Normed Algebras. Ergebnisse der Mathematik and ihrer Grenzgebiete 80. SpringerVerlag, New YorkHeidelbergBerlin, 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, LondonNew YorkSan Iansisco, 1976.
5. Hormander, Lars: An Introduction to Complex Analysis in Several Variables. NorthHolland Mathematical Library 7. NorthHolland Publishing Co., AmsterdamLondon, 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. McGrawHill Series in Higher Mathematics. McGrawHill Book Co., New York 1974. 8. Rudin, Walter: Functional Analysis. McGrawHill Series in Higher Mathematics. McGrawHill 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. SpringerVerlag, New YorkHeidelbergBerlin, 1976.
AUTHOR AND SUBJECT INDEX
Absorbing set 97 Adjoint of an operator 23 Alaoglu, L. 2,5 BanachAlaoglu theorem 2,5 Alexander, H. 171 Alexander, J.C. 110 Alfsen, E.M. 171 Algebra
aCalkin 113 Banach 30 Calkln 33,96 complex 30 n x n matrix 32 quotient 33 semisimple 34 Wiener 31 Algebraic element 37 Algebroid multifunction 12 Analytic multifunction 143 Antisymmetric set 5 Aronazajn, N. 20 Artin, E. 11,96 WedderburnArtin theorem 11,96 Arzela, C. 7,9 ArzelkAscoli theorem 7,9 Ascoli, G. 7,9 ArzelaAscoli theorem 7,9 Asymptotically quasicompact 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 BanachAlaoglu theorem 2,5 BanachSteinhaus theorem 4 BanachStone theorem 79 HahnBanach 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 BeckenbachSake theorem 52,62,176 BehnkeStein 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 BeurlingGelfand 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 aCalkin algebra 113 Cantor, G. 155,157 CantorBendixson 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 CartanThullen theorem 181 Cauchy, A. 9,43,44,52,142 Cauchy's inequalities 9 Cauchy's theorem 43,44,52 Cayley, A. 11,44,97 CayleyHamilton theorem 11,44,97 Cech, E. 89 StoneCech 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 aderived set 155 Diameter (nth) 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
A Primer on Spectral Theory
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 GelfandMazur theorem 39 GelfandNaimark theorem for commutative algebras 121 GelfandNaimark theorem for noncommutative algebras 128 GelfandNaimark theorem for JordanBanach algebras 171 BeurlingGelfand 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 HahnBanach theorem 1,5,127 Halmos, P. 20,23,161,185
Halperin, I. 37 Hamilton, W.R. 11,44,97 CayleyHamilton 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 JordanBanach algebra 171
Author and Subject Index
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 KingLai Hiong 162 Kleinecke D.C. 15,92,106,107,108 Krantz,S.G. 180,184 Krein, M. 3,5,8 KreinMilman 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 LeviKrzoska 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 SturmLiouville 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 selfadjoint 10 Maximum principle for subbarmonic functions 54,56,91,145,148,175 Maximum principle for spectrum 55 Mazur, S. 39,80 GelfandMazur theorem 39 MazurUlam theorem 80 Mergelyan's theorem 141 Milman, D. viii,1,3,4,5,8 Milman's theorem viii,1,3,4 KreinMilman 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 MontelRado 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 GelfandNaimark theorem for commutative algebras 121 GelfandNaimark theorem for noncommutative algebras 128
190
A Primer on Spectral Theory
GelfandNaimak theorem for JordanBanach 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 OkaNishino theorem for pseudoconvex sets 153 OkaNishino theorem for analytic multifunctions 151,155,159 Nonthin 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 OkaNishino theorem for analytic multifunctions 151,155,159 OkaRothstein 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 selfadjoint 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
Quasialgebraic element 161
Author and Subject Index
Quasinilpotent 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
MontelRado 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 BeckenbackSales 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 Selfadjoint
matrix 10 operator 24 element 117 Semisimple 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 (nth) 62 interpolation 171 maximum viii,77 multiplicity 163 radius 24,36 state 78 Spectral theorem for selfadjoint compact operators 25
general 132 for normal operators 134 Spectrum essential 96 exponential 47 of an element 36
Stonetech compactification 89 StoneWeierstrass theorem 1,5,6,9 72,121,134 Stormer, E. 171 Strong sum 22 SturmLiouville problem 27,29 Subharmonic function 52 Szego, G. 179 SzokefalviNagy, B. vii,63,64
Tchebycheff polynomial 179 Thin set 177 Thullen, P. 181 CartanThullen 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 BanachSteinhaus 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 BanachStone theorem 79
Weak lower semicontinuity of the boundary of the spectrum 60
Valiron, G. 161 Variety of support 184 Vertical function 183
Author and Subject Index
Wedderburn, J.H.M. 11,96 WedderburnArtin theorem 11,96 Weierstrass, K. 1,5,6,9,72,121,134 StoneWeierstrass 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