Topics in
Uniform Algebras
and
Banach Function Algebras
by
Taher Ghasemi-Honary
Department of Mathematics,
University for Teacher Education,
49, Roosevelt Avenue,
Teheran,
Iran.
A Thesis submitted for the degree of Ph.D. in the University
of London and for the Diploma of Imperial College.
Department of Mathematics,
Imperial College of Science and Technology,
London, S.W.7.
June, 1976 2
Abstract
In the introductory chapter a brief outline of those parts of commutative Banach algebras and uniform algebras which form a basis for the following chapters is given. In particular the definitions and notations which we shall use throughout the thesis is established.
Since the theory of peak points and boundaries of uniform algebras is well-developed, we give details of the known properties of them in the first chapter.
The main object of this thesis is to study the relation between peak points and the Shilov boundary of Banach function algebras.
In Chapter two the density of peak ooints in the Shilov boundary of a Banach function algebra is discussed and then some examples of Banach function algebras which contain peak sets without peak points are given.
We also give some examples of uniform function spaces with certain properties.
In Chapter three we consider a particular class of Banach function algebras of infinitely differentiable functions on compact plane sets such that the algebras are quasianalytic. This class was introduced by
Dales & Davie [11] who have given an example of a natural Banach function algebra whose set of peak points is countable, and hence is of first category in the Shilov boundary. I have also given such an example on the Swiss cheese for which the set of peak points is uncountable but it is still of first category in the Shilov boundary.
The fact that the maximal ideal space of H is not first countable is proved in Chapter four, and it is used to construct Banach function algebras having certain properties. An example of a uniform algebra on the closed unit disk is also given which is not finitely generated.
Finally Chapter five contains some propositions concerning the question arising from Gamelin [15]. I have given some conditions on the
Banach algebra A with MA = IT such that T = b75- = r(A) or Tc r(A). 3
Acknowledgements
I am indebted to my supervisor, Professor J.G. Clunie for suggesting the topic of this thesis and for his guidance during my studies. He specially helped me in constructing the counter-examples.
I also offer my thanks to Miss Pindelska of the Mathematics
Library, who obtained for me the papers and books which I needed during my studies.
Special thanks are due to the University for Teacher Education in Teheran who supported me financially throughout my postgraduate studies.
I am most grateful to Mrs. M. Robertson for her care and patience in typing the thesis. 4
Contents Page
Chapter one: Introduction 5
1-1 Banach Algebras 5
1-2 Function Algebras 13
1-3 Function Spaces 15
1-4 Uniform Algebras 21
1-5 Peak Points and Boundaries of Uniform Algebras 27
1-6 The Union of Peak Sets 31
Chapter two: Peak Sets and Peak Points of Banach Function Algebras 33
2-1 The existence of the smallest boundary 33
2-2 The density of the peak points in the Shilov boundary 38
2-3 Examples of Banach function algebras having peak sets without peak points 47
2-4 Examples of uniform function spaces 81
Chapter three: Banach Function Algebras on Compact Plane Sets 86
3-1 Definitions and basic properties 86
3-2 Peak points and the Shilov boundary 94
Chapter four: Counter Examples 100
4-1 Preliminaries 100
4-2 Concerning maximal ideal space of H 102
4-3 Examples of uniform algebras having certain properties 106
Chapter five: An Open Question on The Shilov Boundary of Banach Algebras 113
5-1 Preliminaries 113
5-2 Concerning finitely generated uniform algebras 115
5-3 Certain Banach algebras 118
References 121 5
Chapter one
Introduction
The theory of Banach function algebras and in particular uniform algebras draws from three sources: functional analysis, analytic functions of several complex variables and general topology.
The subject of uniform algebras has been receiving an increasing amount of attention in recent years. The theory of peak points and
boundaries for uniform algebras is specially well-developed. But the
theory of Banach function algebras has not been developed too much.
The main object of this thesis is to study the situation which
can occur for Banach function algebras. I shall also study the relation
between peak points and boundaries of uniform function spaces and Banach
function spaces.
In this introductory chapter a brief outline of those parts of
commutative Banach algebras and uniform algebras which form a basis for
the following chapters is given. In particular the definitions and
notations which we shall use throughout the thesis is established.
Since the theory of peak points and boundaries of uniform algebras
is well-developed, we give details of the known properties of them.
Some of the results stated in this chapter are only relevant to the
other parts and will not be referred to in the remainder of the thesis.
References are given in the form [15] or [15; chap. II], which refer
to reference 15. 6
1-1 Banach Algebras
(1.1.1) Let (A, 11• II) be a commutative normed algebra with unit 1 over the complex field C . Since for every f,g e A, Ilfg11 < 11f11 .11g11 we have 11111 > 1. We suppose that 11111 = 1 and it can be shown that this leads to no loss of generality. From now on by a normed algebra we mean a commutative normed algebra over d with unit 1 such that
11111 = 1. If the normed algebra (A, 11'11) is complete under 11.11' it is called a Banach algebra.
Let A be a normed algebra. f e A is invertible if there exists g e A such that f.g = 1. g is called the inverse of f and is denoted -1 1 1 by f or . The family of invertible elements of A is denoted by A .
The spectrum of f e A in A is denoted by aA(f) and is given by
1 crA(f) = {X ee -A c A }.
If A is a Banach algebra, it is well-known that aA(f) # 0. Let A be a normed algebra and A be the norm completion of A so that A is a Banach algebra. Since °A(f) C aA(f) and aA(f) # 0, we have aA(f) # 0. The following theorem is a direct result of aA(f) #cP.
Theorem 1 (Gelfand-Mazur Theorem)
Any normed algebra which is a field is isometrically isomorphic to the complex field C .
Definition. A non-zero complex homomorphism on A is called a character.
Let J be a closed maximal ideal of the normed algebra (A, 11'11).
The quotient algebra A/J, in the usual quotient norm, is also a normed algebra over C with the unit 1 + J. Since 111+g11 > 1 for all g e J we have 111+J11 > 1. As 11111 = 1 we conclude that 111+J11 = 1. 7
Moreover every non-zero element of A/J is invertible and so by the
Gelfand-Mazur Theorem A/J is isometrically isomorphic to C . Since the projection 4):A A/J is a continuous homomorphism with kernel J, then J is the kernel of a continuous character 4) on A; i.e.
J = {f e A: 4)(f) = 01. Conversely, if 4) is a continuous character on
A and M is the kernel of 4), then M¢ is a closed maximal ideal of A.
Consequently the correspondence 4) -* M(1) is a 1-1 correspondence between the continuous characters on A and the closed maximal ideals of A.
Clearly 1¢1I = 4)(1) = 1 for any continuous character 4) on A. It is customary to identify each closed maximal ideal of A with the continuous character that it determines. If A is a Banach algebra, every maximal ideal is closed; equivalently every character on A is continuous.
(1.1.2) Let A be a normed algebra. The maximal ideal space (spectrum or carrier space) of A, denoted by MA, is defined as the set of all closed maximal ideals of A; or equivalently, the set of all continuous characters on A. Clearly MA is a subset of the unit ball of the conjugate space A* of A. We define the topology of MA to be the weak*- * topology that MA inherits from A . In other words, a net {4)a} in MA converges to 4) if and only if ¢a (f) 4)(f) for all f 6 A.
A basis of open neighbourhoods of sbo 6 MA is given by the sets of the form
N(¢0;f1,f2, fri;E) = f4J C MA:14/(fj) 4/0(fi) l < C. 1 where e > 0, n is a positive integer and fl,f2, fn a A. Since A* is a Hausdorff space with the weak*-topology, MA is also Hausdorff. By Alaoglu's theorem, the closed unit ball of A* is weak*-compact and 8 so MA, being a weak -closed subset of the closed unit ball, is also weak -compact. (1.1.3) Let A be a normed algebra. The Gelfan4 transform of f e A is the complex-valued function f on MA defined by t(4) = 4(f). Clearly A = {E:f e Al is a normed algebra of continuous functions on MA under the uniform norm = sup 'NO = sup 14)(f) I. I ri I IM A6 MA 0614A But A may not be uniformly closed in C(MA). Note that IIiIIM = 1 and A If IM I If I I for all f e A. The map T:A 4'A which is defined by A T(f) = 2 is a homomorphism. If A is the norm completion of A then any 4 e MA can be extended to a continuous character tl.) on A such that Ihid I = I14)1 = 1. On the other hand every proper ideal of A is contained in a maximal ideal of A. Therefore f(MA) = QA(f) for all f e A. A direct result of this equality is the following Theorem. Theorem 2. Let A be a Banach algebra. f 6 A is invertible in A if and only if f 0 on MA. The spectral radius of f e A is defined as r(f) = sup fIXI : X e aA(f)1. If A is a Banach algebra we have: inflIfnlin =/1 i Ifn i I n = r(f) = sup{ 12(01 6 MA} = - 4.0D A 9 The radical of a normed algebra A, denoted by rad(A), is defined as the intersection of all maximal ideals of A. If A is a Banach algebra, then rad(A)= {f e A:4(f) = 0 for all 4) c MA f c A:aA(f) = OeMA Definition A is called semi-simple if rad(A) = {0}. If A is a semi-simple normed algebra and we define IIfIl = IIf II for f c A, then A becomes a semi-simple normed algebra under this norm and the mapping T:A + ecomesA b 1-1 and continuous. Therefore A and A are isometrically isomorphic and so we can identify them. Now if to each • e MA we let correspond a IP such that 0) = 0(f), then 11, e M. * With the weak -topology MA is also a compact Hausdorff space and mA. i.e. M MA - A' A and MAA are homeomorphic and so we can identify them too. Moreover A is a Banach algebra if and only if A is a Banach algebra. Theorem 3. If A is a semi-simple Banach algebra under two different norms 11.111 and 11-112, then these two norms are equivalent; i.e. there exist positive constants K1 and K2 such that K1IIfIl < IlfI12 < K2llfll l for all f e A. Proof. Let A l = (A/11.111), A2 = (A.II•112) and define a new norm for A by IlfIl = 'WI]. + lIf112 (f c A). To prove I 1.1 I is complete, let ff lc° be a dauchy sequence in (A, I H I) so that IIf - frail -----4.÷,,,,, 0 n n=1 n 1 n,m and IIf ›. 0. Since A n - fmII 2 n,m l and A2 are Banach algebras there exist f and g in A such that I If - flI t 0 and n 1-r--;; Ilfn - gll ----)"2 n400 °. Therefore 10 Ilfn - M 0 and Ilfn - gll M ----* O. A l n4-00 n4.00 Since A is complete under 11'111 and 11'112, any non-zero complex- valued homomorphism of A is continuous either in 11'111 or in 11'112' A A Thus 1.MA4„, = MA and hence f = g on M„,MA = MA . But Al and A2 are both 2 2 semi-simple, so f = g. Therefore Ilf - fit + 0, i.e. !HI is a complete n n400 norm. Since 11f111 <— 11f11 and 11f112 < 11f1, by the closed graph theorem, 11'111 and 11'11; and 11'112 and 11'11 are equivalent. Therefore (1.111 and 11.112 are equivalent. (1.1.4) The Shilov Boundary Let A be a normed algebra. A subset E of MA is a boundary for A if every function f e A attains its maximum modulus on E. Theorem 4. The intersection of all closed boundaries for A is a boundary for A; it is called the Shilov boundary of A and is denoted by r(A). As a result of ‘i(MA) = aA(f) and the above Theorem we have: Theorem 5. If A is a Banach algebra and f e A. Then boA(f)c= i(r(A)) where baA(f) is the topological boundary of aA(f). (1.1.5) Let A be a normed algebra and F be a subset of A. A sub- algebra B of A is said to be generated by F if it is the smallest closed subalgebra of A which contains F and the unit 1. In other words each element of B is the limit of a sequence of polynomials in the elements of F. F is called the set of generators for B. If A is generated by F and F is finite or countable then A is called, respectively, finitely or countably generated. 11 Let F = X e A:A e Al be a set of generators for A, where, in general, A is uncountable. With X E A we associate a complex plane CA, and letel = 11{Cx:A c Al with the usual product topology. The map A N:MA 4- C which is defined by Tr (CP) = A * is called the canonical mapping of MA into . Considering the weak - topology of MA, it is a liameomorphism, and so Tr (MA) is a compact subset A of CA. If A is countable C is metrizable and so Tr(MA) and hence MA is also metrizable. In particular if A is finitely generated, MA is a compact subset of el where n is the number of generators. If A is A uncountably generated, C is not even first countable but N(MA) and hence MA may be first countable or metrizable [see example (4.2.3)]. Let A be a normed algebra and fl,f2, fn c A. The joint spectrum of f1,f2, f is given by n n a(f ) e C :f - Al,f - An ;belong to a fn) = {(X1, ..., An 1 n proper ideal of 10. If A is a Banach algebra, then crA(fl, fn) = {(21(40, ?n(4)):(!)cM/11. If the Banach algebra A is generated by fl, f , by the above discussion n fn) and so M fn) n Tr(MA) = aA(fl, A =a(f Al' (1.1.6) A subset P of MA is called a peak set of A if there exists f c A such that f = 1 on P and 121 < 1 on MA P. A subset P of MA is called a local peak set of A if there exists f e A and a neighbourhood N of P in MA such that 2 = 1 on P and 121 < 1 on N\ P. 12 Theorem 6 (Rossi's Local Peak Set Theorem) Let A be a Banach algebra. Then every local peak set of A is a peak set of A. Rossi's original result [22] and that given in [16] are stated for uniform algebras only, but by trivial modification we can get the above result [15]. Corollary 1. Let P = F1 U F2 U ...0 Fn be a peak set of A and Fk's be non-empty, pairwise disjoint, closed subsets of MA. Then each F (1 < k < n) is a peak set of A. K In particular if the peak set P has a finite number of components, then each component is also a peak set of A. Corollary 2. If the peak set P of A is finite each point of it is a peak point of A. Corollary 3. If the peak set P of A has no peak point of A then P is infinite and perfect. Theorem 7 (Local Maximum Modulus Principle) Let A be a Banach algebra and N be an open subset of MA. Then for every f e A, we have I 1;1 IN = (r(A)(N)tjbN one of the interesting results of the above Theorem is the following Corollary. Corollary. If MA is homeomorphic to the unit interval I = [0,1] or to the unit circle T = z EC:IzI = 11. Then r(A) = MA. 13 1-2 Function Algebras (1.2.1) Let X be a compact Hausdorff space, and C(X) be the set of all continuous complex valued functions on X. With the usual definition of pointwise addition and multiplication, C(X) is a commutative algebra over the complex field C which contains the constants. Clearly C(X) is a Banach algebra under the uniform norm lifilx = suplf(x)I (f c CVO). x cX A subset F of C(X) separates the points of X if, for any two different points xl and x2 in X, there exists a function f 6 F such ) # f(x ). Since X is a compact Hausdorff space, by Urysohn's that f(x1 2 Lemma, for any xl x2 in X there exists f c C(X) such that f(xl) = 0 and f(x2) = 1. Hence C(X) separates the points of X. (1.2.2) Definitions I. A is a function algebra if it is a subalgebra of C(X) which contains the constants and separates the points of X. If A is normed it is called a normed function algebra. II. If the normed function algebra is complete, it is called a Banach function algebra. III.If the norm of the Banach function algebra (A, 11'11) on X is equivalent to the uniform norm !I'll x, so that A is a uniformly closed subalgebra of C(X), then A is called a uniform algebra. In this case we actually assume that I Ifl I = I If I Ix for every f c A. A trivial example of a uniform algebra is C(X). If A is a function algebra on X then A, the uniform closure of A, is a uniform algebra. But if A is a normed function algebra on X, the norm-completion A of A is a Banach algebra but may not be a Banach function algebra on X. 14 (1.2.3) Let (A, 11'11 ) be a normed function algebra on X such that 11f11x < 11f11 for all f 6 A. For every x e X the map (I)x: A -1'4% , defined by (1)x(f) = f(x) for f e A, is a continuous character on A and so cPx e MA. CPx is called the evaluation homomorphism at x. If the norm of A is not stronger or equal to the uniform norm, cPx may not be continuous. On the other hand if (A,11'11) is a Banach function algebra A. Hence for every f e A we have. on X, 4x is continuous and so (I)x c M Ilf I l x = suplf(x)I = sup Ict)x(f)I < suP1()(f)1 = < 11f11- xeX 4)xeMA cl)eMA A Therefore if f 6 rad(A), f = 0 on MA and so Ilfilx = 0, i.e. f = O. Thus rad(A) = {0}. Consequently any Banach function algebra is semi- simple and hence by Theorem 3, if A is a Banach function algebra under the norms 11'111 and II-112 then these two norms are equivalent. Huaever when 4) is continuous the map F:x (I) is an embedding of X x x into MA and we usually identify X with F(X) = {40x:x e XI in this way. Hence X can be regarded as a (weak*) closed subset of MA. In general F(X) is a proper subset of MA, but there are Banach function algebras A. for which X = F(X) = M For example every (1) e MC(X) is the evaluation homomorphism cpx at some x e X and so X a Mc(x). Definition. Let (A,11-11) be a normed function algebra on X such that - MA; 11f11 X —< 11f11 for every f e A. A is called natural if X = M - i.e. every 4) e MA is given by an evaluation homomorphism (f)x at some x e X. As a consequence of f(MA) = 6A (f) we have the following theorem. Theorem 8. Let A be a natural Banach function algebra on X. If f e A has no zeros on X it is invertible in A. 15 Let (A, 1 1' 11 ) be a semi-simple Banach algebra so that A is a Banach function algebra on MA under the norm IIfII = 11f11- Since A is isometrically isomorphic to A and MA -24 MA, A is natural. Hence any semi-simple Banach algebra A can be identified with a natural Banach function algebra A on its maximal ideal space MA. Although in A this case A and A are isometrically isomorphic but A is always natural whereas A may not be. As a result of Rossi's local peak set theorem we have: Theorem 9. If A is a natural Banach function algebra on X, then every local peak set of A (w.r.t. X) is a peak set of A (w.r.t. X). 1-3 Function Spaces (1.3.1) Definitions Let X be a compact Hausdorff space. I. A is a function space if it is a (linear) subspace of C(X) which contains the constants and separates the points of X. If A is normed as a linear space then it is called a normed function space. II. If A is a complete normed function space it is called a Banach function space. III.If the norm of the Banach function space (A, 11'11) is equivalent to the uniform norm II-IIx so that A is a uniformly closed subspace of C(X), then A is called a uniform function space. In this case we actually assume that 11f11 = 11f11X for every f 6 A. Clearly every function algebra is a function space and every Banach function algebra is a Banach function space. So everything which is true or defined for a function space is also true or defined for a function algebra. 16 If A is a function space, the uniform closure A of A is then a uniform function space. But the norm completion A of the normed function space A is a Banach space and it may not be a Banach function space. Since no multiplication is defined on a function space, the ideals and hence the maximal ideal space are meaningless for function spaces. Therefore we cannot define natural Banach function spaces. Also there is no guarantee that the uniform norm is weaker or equal to the norm of a Banach function space. (1.3.2) Let A be a function space on X and A be the conjugate space ); i.e. A* is the space of all continuous linear functionals of (A,11•11X ) • It is well-known that A* is a Banach space. on (A, 1 1* 1 IX Definition. The state space of A, denoted by K(A), is defined by K(A) = {0 c A*:I 14)11 = c(1) = 1}. We can topologize K(A) with the weak*-topology that it inherits from this topology on A*. By Alaoglu's theorem K(A) is a weak*-compact convex subset of the closed unit hdll in A*. Certainly K(A) is Hausdorff in the weak*-topology. Clearly the evaluation functionals 4x (x c X), defined by Ox(f) = f(x) for f c A, are continuous on (A,11•11x) and 110x1I = Ox(1) = 1. Hence 0x c K(A) and the map F:X K(A), defined by F(x) = Ox for x c X, is a homeomorphism and so X = F(X) C:K(A). Since the set of extreme points of K(A) is contained in F(X), it follows from the Krein-Milman theorem that K(A) = 60(F(X)) * where C (F(X)) is the weak -closed convex hull of F(X). 0 17 (1.3.3) Definitions Let A be a function space on the compact Hausdorff space X. I. A subset S of X is called a boundary of A (w.r.t. X) if every f E A attains its maximum modulus at some point on S; i.e. for every f E A there exists x e S such that I If I IX = If(x) A boundary of A, with respect to X, is denoted by B(A,X). II. A subset M of X is called a minimal boundary of A if M is a boundary of A but there is no other boundary of A which is properly contained in M. If S c:X is a boundary which is contained in any other boundary, then it is called the smallest boundary. If the smallest boundary exists it is unique. III.The Shilov boundary of A (w.r.t. X) is the smallest closed boundary of (w.r.t. X) and it is denoted by r(A,X). IV. The Choquet boundary of A (w.r.t. X) is the set of all x E X for which (1)x is an extreme point of K(A) and it is denoted by Ch(A,X). It is quite well-known that every f e A attains its maximum modulus at some x E X for which (1) is an extreme point of K(A) so that Ch(A,X) x is a boundary of A. Moreover every closed boundary of A contains Ch(A,X) and so Ch(A,X) = r(A,X); i.e. the Shilov boundary exists for function spaces. The existence of the Shilov boundary for function algebras was first proved by Shilov. If X is metrizable, Ch(A,X) is a GS-set in X; i.e. Ch(A,X) is a countable intersection of open sets in X. V. The maximum set Mf of f is defined by Mf = {xcX:If(x)1 = Clearly Mf is closed in X. VI. A subset P of X is called a peak set of A (w.r.t. X) if there exists f E A such that f = 1 on P and Ifl < 1 on XN.P. Clearly every peak set is closed in X and every finite intersection of peak sets of A is also a peak set of A. 18 VII. A closed subset E of X is called a local peak set of A (w.r.t. X) if there exist f E A and a neighbourhood N of E in X such that f= 1 on E and Ifi < 1 on N`.E. VIII. The point x c X is a peak point of A (w.r.t. X) if CO is a peak set of (w.r.t. X). In other words there exists f c A such that f(x) = 1 and If(y)1 < 1 for y 5 x (y c X). The set of peak points of A (w.r.t. X) is denoted by S0(A,X). Clearly S0(A,X) is contained in any boundary of A. In fact S0(A,X) is the intersection of all boundaries of A (w.r.t. X), but it may not be a boundary of A. However, the intersection of all closed boundaries of A (w.r.t. X) is a closed boundary of A and it is indeed r(k,x). e then P = CxcX:f(x) = f(x0)7. is a peak set If f A and x0 Mf 1 of A (w.r.t. X) and the peaking function is g = (1 + f(x0)). S0(A,X) is a boundary of A if and only if any peak set of A contains a peak point of A. A local characterization of the points of r(A,X) is clear: A point x e X is in r(A,X) if and only if for any neighbourhood e Nx of x in X there exists f A such that Mf CNx . IX. A non-empty subset Q of X is called a p-set of A if it is the inter- section of some family of peak sets of A. Clearly any p-set of A is closed in X. A p-set of A is called a minimal set of A if it does not contain any other p-set of A. X. The point q e X is a p-point of A if {q} is a p-set of A. XI. The point x e X is a strong boundary point of A if for any given neighbourhood Nx of x in X, there exists a peak set P of A such that x xcPCN.N . The set of strong boundary points of A (w.r.t. X) is called the strong boundary of A (w.r.t. X) and it is denoted by S(A,X). 19 In fact the notion of strong boundary point and the p-point are the same. It is easily proved that x c X is a p-point of A if and only if x is a strong boundary point of A. In general S(A,X) may not be a boundary of A. But S(A,X) is a boundary of A if and only if every peak set of A contains a p-point of A. If PC= X is a peak set of A and f c A peaks on P we have 00 P = n x e X: If (x) I> 1— n1- 1; n=1 i.e. every peak set of A is a compact Gs-set in X. (1.3.4) We now state some of the properties of function spaces which are known. Theorem 10. Let A be a function space on X and x e X. Suppose there exist constants a,0 with 0 < a < S < 1 such that for every neighbourhood N of x there exists f c A with !WIX < 1 If(x)1 > 0 and 1f1 < a on X•.N. Then x e Ch(A,X). Theorem 11. If A is a function space on X, then S(A,X) CICh(A,X). Theorem 12 (Bishop [4]) Let A be a uniform function space on X and x c X. Suppose there exist constants a,0 with 0 < a < S < 1 such that for every neighbourhood N of x there exists f c A with 11f1lx < 1, f(x)> a and 1f1 < a on X•.N. Then x c S(A,X). The following theorem is usually stated for uniform algebras or Banach function algebras. But, as it is seen by the proof, it is also true for Banach function spaces. 20 Theorem 13. Let (A, be a Banach function space on X. Then any countable intersection of peak sets of A is also a peak set of A. Proof. Let P = fl P and f c A peak on P n C X. Clearly n=1 n n 00 f n f = E n=1 2n1Ifnli is in A and Oa 1 f(x) = E n n=1 2 IIfnII for x e P and 00 1 If E n=1 2nlIfnll for y c X\ P. Hence f c A peaks on P. Using the compactness of X and the above theorem we can easily prove that any p-set of A which is a Ge-set in X is a peak set of A. Since in metrizable spaces any closed set is a Ga-set, any p-set of A is a peak set of A if X is metrizable. Hence there is no distinction between p-sets and peak sets of A when X is metrizable. But this is not true if X is not metrizable; even if X is first countable [see example 2.2.4]. However, if X is first countable any singleton fxl in X is a Ga-set and so any p-point of A is a peak point of A. In other words if A is a Banach function space on a first countable compact Hausdorff space X, then S (A,X) = S(A,X). 0 21 Theorem 14 [20] The metrizable -4-Pace If A is a uniform function space onvX, the peak points of A (w.r.t. X) are dense in the Choquet boundary of A (w.r.t. X); i.e. Ch(A,X)Cgo(A,X). Since Ch(A,X) = r(A,X) we have S0(A,X) = r(A,X); i.e. the set of peak points of A is dense in the Shilov boundary of A. Theorem 15 (Cho [7]) Let A be a function space on X which is generated by a weakly compact convex subset of (C(X), I I•I IX) • Then the peak points of A are dense in the Choquet boundary of A; i.e. Ch(A,X) C: S0(A,X). Hence S0(A,X) = r(A,X) and S(A,X) = r(A,x). As an important Corollary of this Theorem we have: Theorem 16 If A is a Banach function space on the compact metrizable space X, (A,X) = r(A,X). then 0 This theorem has also been proved by H.G. Dales [10] when A is a Banach function algebra on a compact metrizable space X. I have given another proof for this Theorem which is constructive and elementary [see Theorem (2.2.1)]. 1-4 Uniform Algebras (1.4.1) As it was stated before if (A,11•11) is a normed algebra, A is a function algebra on MA. So we can talk about the subsets of MA which are boundaries, p-sets or peak sets of A (w.r.t. MA). But it is customary to write B(A,MA), Ch(A,MA), r(A,MA), So(A,MA) and S(A,MA) instead of B(A,MA), Ch(A,MA ), r(A,MA), s0(AMA) and S(A,MA) respectively. 22 If A is a uniform algebra on X, then 11E U M = IlfilX and so A is also a uniform algebra (on MA). In this case r(A,MA) is in fact a closed subset of X, while X is regarded as a closed subset of MA. (1.4.2) Standard Uniform Algebras Let K be a compact subset of and z = (z n C n l' z2' zn) c C . I. P(K) is the uniform closure of all polynomials in z ,z 1 2, zn on K. II. R(K) is the uniform closure of all rational functions of z1,z2,..., zn on K which have no poles on K. III.H(K) is the uniform closure on K of all functions analytic in some neighbourhood of K. IV. A(K) is the algebra of all continuous functions on K which are analytic on the interior of K. It is easily seen that the above algebras are all uniform algebras and P(K)C:R(K)C:H(K)C:A(K)cl C(K). These algebras have been intensively studied. The main problem is to find out for which K equality holds among either of these inclusions. The swiss cheeses provide examples of compact plane sets K such that K° = int(K) = 4) and P(K) # R(K) A(K) = C(K). A swiss cheese is a compact plane set K obtained from the closed unit disk D by deleting a sequence {D.17 of open disks, such that CO 3 3=1 D. (-1 15 = oo 3 k) and K = EN,(k.) D.) has empty interior. If the radii k j=1 00 r. of D. are sufficiently small so that 2= r. < co, then R(K) A(K). 3 7 j=1 3 [For example see 15, Chap. II]. 23 (1.4.3) Let K be a compact subset of C. Definition. The polynomial convex hull of K, denoted by K, is given by en ={z e :IP(z) I < IIPIIK for all polynomials P on K). The rational convex hull of K, denoted by R-hull (K), is given by R-hull (K) = {z e tn: IR(z)I < IIRIIK for all rational functions with poles off Kl. K is called polynomially convex if K = K, and rationally convex if R-hull (K) = K. A Since R-hull (K) C= K, polynomially convex sets are rationally convex, but the converse is not true. For example, the torus {(z,w) cC2:1z1 = Iwl = 1) is rationally convex but not polynomially_ convex. Clearly k is always closed and polynomially convex; i.e. I = f. One of the well-known Theorems about P(K) and R(K) is the following: = K and MR(K) = R-hull (K). Theorem 17. MP(K) In general the maximal ideal spaces of H(K) and A(K) are not known. But when K is a compact plane set R-hull (K) = K and H(K) = R(K) so that M = M H(K) R(K) = K. Theorem 18. (Aren's Theorem [1,15]) 2 Let U be an open subset of the Riemann sphere S , and let K be a 2 compact subset of S containing U. Suppose the algebra A(K,U) of functions in C(K) which are analytic on U contains a non-constant function. Then A(K,U) is a uniform algebra on K whose maximal ideal space is K. Corollary. Let K be a compact subset of the plane C. since = K. A(K) = A(K,e), by the theorem MA(K) If K C= c n it is clear that P(K) is generated by the coordinate functions zl,z2, zn. Therefore P(K) is finitely generated. It is also true, but not obvious, that R(K) is finitely generated. In fact R(K) is generated by n + 1 functions. But it is not known whether H(K) 24 or A(K) are finitely generated or not. Even when K is a compact plane set we do not know whether A(K) is finitely generated or not. However, H(K) and A(K) are countably generated. If K is polynomially convex, it is easy to show that P(K) = R(K), but the converse is not true in general; i.e. if P(K) = R(K), K may be different from K. Example. Let K = 1(z,w) E C2: lwl < lzl < 1). By Theorem 3.5 [15] we A have P(K) = R(K) = A(K). However K is not rationally convex and in fact , R-hull (K) = K = i(z,w) e e:IziI < 1, IwI < 11 K. As a result of the functional calculus we have the following important approximation theorem: Theorem 19 (Oka-Weil Approximation Theorem) . If K is a compact polynomially (rationally) convex subset of C , then P(K) = H(K) (R(K) = H(K)). Another form of the Oka-Weil approximation theorem, which contains the above theorem, is the following: Theorem 20. Let K be a compact subset of C:n A P(K) = R(K) if and only if MP(K) = MR(K), i.e. K = R-hull (K). R ( ) R(K) = H(K) if and only if M K = M11(K). (1.4.4) Let K be a compact plane set. A K coincides with the union of K and the bounded components of C K =C \K. Hence K is polynomially convex if and only if K is connected. In other words K is polynomially convex if and only if K is A simply connected. Clearly bK, the topological boundary of K, is contained in bK. 25 Theorem 21 (Mergelyan) P(K) = R(K) if and only if K is polynomially convex. Theorem 22 (Runge) R(K) = H(K) for every K C C• Theorem 23 (Vitushkin) R(K) = A(K) if and only if for every bounded open set N in a(N\K) = a(N\Ko) where a is the continuous analytic capacity and K0 = int(K). In particular we have the following useful results: I. If every point of the boundary bK of K belongs to the boundary of some component of KC, then R(K) = A(K). II. If KC has only finitely many components, then R(K) = A(K). In particular if K is polynomially convex, then P(K) = A(K). III.If the diameters of the components of KC are bounded away from zero, then R(K) = A(K) [6 & 15]. Theorem 24 (Hartogs -Rosenthal) If K has zero planar measure then R(K) = C(K). Clearly if the planar measure of K, denoted by m(K), is zero, then K0 = 1), but the converse is not true. For example if K is the swiss cheese so that R(K) A(K) = C(K), although K° = but m(K) > 0. Theorem 25 (Laurentiev) P(K) = C(K) if and only if K is polynomially convex with empty interior. In particular if K is countable, then P(K) = C(K). In fact C(K) is the only uniform algebra on the compact countable set K. 26 Theorem 26 (Ifitushkin) R(K) = C(K) if and only if for every open set N in C y(N\K) = y(N) where y is the analytic capacity. Theorem 27 (Bishop's peak point criterion for rational approximation) R(K) = C(K) if and only if K\So(R(K)) has zero planar measure. If P is a polynomial, by the maximum modulus principle, = I !Pik- If f e P(K) there exists a sequence of polynomials I IPI K {P }w such that IIf 0. Hence n n=1 PnIIK271 IIP n Prtt IIA K = IIPn PmIIK A ti ---4 0 and so {Pn7 converges uniformly on K to an extension 2 n,m-4-00 n=1 of f, satisfying lIfIIK = The isometric isomorphism f 2, which turns out to be the Gelfand transform, allows us to identify P(K) ^ ^ and P(K). Since K is a polynomially convex compact plane set, by the A Mergelyan Theorem, P(K) = P(K) = A(K). In particular if K is the unit circle; i.e. K = T = {z eC : I z I = 1} we have P(T) = P(T) = P(5) = A(5). In fact P(T) consists of the continuous functions on T for which the negative Fourier coefficients vanish; i.e. 2ff n P(T) = e C(T) : f einv f(ei())d8 = 0, n = 1,2, ... 1. 0 The algebra P(T) = A(D) is called the disk algebra. Theorem 28 (Warmer's Maximality Theorem) P(T) is a maximal subalgebra of C(T); i.e. there is no uniform algebra A distinct from P(T) and C(T) satisfying P(T) C A C C(T). Hence R(T) = A(T) = C(T). 27 (1.4.5) We finally state the following useful theorem on finitely generated Banach algebras. Theorem 29. If A is a finitely generated Banach algebra, generated by fl,f2, fn, then cA(fl,f2, fn), the joint spectrum of A a (f"f fn). fl,f2, fn, is polynomially convex and MA - A 1 Corollary. A uniform algebra A on X is finitely generated if and only if A is isometrically isomorphic to P(K), for some compact polynomially convex subset K of C. 1-5 Peak Points and Boundaries of Uniform Algebras (1.5.1) Let X be a compact Hausdorff space. Theorem 30 (Bishop-Deleeuw) Let A be a uniform algebra on X and x c X. The following assertions are equivalent: I. x c Ch(A,X) II. x c S(A,X) III.For any open neighbourhood N of x in X and c > 0 there exists f c A such that !WI < 1, If(x)I > 1 - c and Ifi < c on X\N. X IV. For any Gs-set F containing x, there exists f c A such that If(x)1 = IlfIl x and Mf C F. Corollary 1. If X is metrizable, S0(A,X) = S(A,X) = Ch(A,X) and hence S0(A,X) is a G-set in X. Corollary 2. S(A,X) = Ch(A,X) and fi(k,X) = r(A,X). In particular if X is first countable, S0(A,X) = S(A,X) is a boundary of A (w:r.t. X). In fact S0(A,X) is the smallest boundary of A (w.r.t. X) which is the same as the unique minimal boundary of Bishop [4]. Clearly S0(A,X) = r(A,x) when X is first countable. 28 If X is not first countable the smallest boundary may not exist, (A,X) may not be a boundary of A. The first countability of X i.e. So (A,X) = r(A,X) is also necessary. for S0 Since (C(X),II'll x) is a uniform algebra, by Corollary 2 S(C(X)) = Ch(C(X)). Using Urysohn's Lemma we can easily prove that every point of X is a strong boundary,point of C(X). Hence Ch(C(X)) = r(c(x)) = S(C(X)) = X. Bishop has given the following example in [4] in which the smallest boundary does not exist and S0(A,X) ¢ r(A,x). Example Let X = II IA where IA = [0,1] and A is an uncountable index set. AEA With the usual product topology X is a compact Hausdorff space. Let M = = (xx)AEA e X: xx = 0 for all but countably many A's} N = = (x ) e X: xA A XcA = 1 for all but countably many A's} . It is easy to see that M and N are boundaries of C(X) and Fin N = (1). Hence C(X) has no smallest boundary. In other words So(C(X)) = (1). On the other hand S(C(X)) = r(c(x)) = X and so I0(C(X)) r(c(x)). Consequently none of the points of X are Gs-set, otherwise that point would be a peak point of C(X). Hence X is not first countable. H.G. Dales has even proved in [10] that C(X) has no minimal boundary. (1.5.2) For a long time, it was conjectured that if A is a natural uniform algebra on the compact Hausdorff space X such that So(A,X) = X (or more generally S(A,X) = X), then A = C(X). This conjecture has been shown to be false by Brian Cole [8] who has given an example of a uniform 29 algebra A on a compact metrizable space X such that A C(X), MA = X and So(A,X) = X (see Browder [6]). It still remains open whether there is a natural uniform algebra A on I = [0,1] or T = Cz E C: Izi = 11 different from C(I) or C(T) respectively. D.M. Wells in [24] and D.R. Wilken in [26] have given some conditions on A under which the natural uniform algebra A on I or T is C(I) or C(T) respectively. Note that if A is a natural uniform algebra on I(T) then, by the local maximum modulus principle, I'(A) a i(I'm) :14 T). However, there are natural Banach function algebras on I or T, different from C(I) or C(T) respectively, such that every point of I or T is a peak point of that algebra [see example (3.1.19)] (1.5.3) The existence of the smallest boundary of a uniform algebra on a compact metrizable space was first proved by Bishop in [4] using the following important result: Theorem 31 (Bishop's Lemma [4]) Let A be a uniform algebra on X. Suppose K is a p-set of A and F C K is a p-set of the restriction algebra A I K. Then F is a p-set of A. Corollary. Every minimal set of A is trivial. Since every peak set contains a minimal set, it follows from the above corollary that every peak set contains a p-point. Hence S(A,X) is a boundary. However, the corollary and the above result are not true for Banach function algebras [see Examples (2.3.2) and (2.3.3)]. 30 I have given another proof, in Chapter two, for the existence of the smallest boundary of a uniform algebra on a first countable space. It is rather elementary and constructive. (1.5.4) We now give in brief detail some of the known results on peak points and boundaries of the standard uniform algebras. Let K be a compact plane set. I. r(R(K)) = r(A(K)) = bK. II. Since P(K) = P(R) = R(K), r(P(K)) = r(l(k)) = bk. fi These results are not true any longer if KC ( and n > 1. Example Let K = {(zi,z2, zn) eCn: 1z11 < 1, 1Z2 < 1,"..., Izn 1 < 1}. K is a compact polynomially convex subset of C n. Clearly every point X = (X1,12, An) c Tn = {(zi,z2, zn) e K: 1z11 = 1z21 = = lz = 1} is a peak point of P(K) = R(K) = H(K); the peaking function is n 1 f (z) = f(z zn) = — (1 + z ) u + 37. z ) ... (1 + An n) 1' z2' n 1 2 2 zn 2 1 It is easy to show that Tn is a boundary of P(K) and so S0(P(K)) = r(P(K)) n = T. Clearly T is a proper subset of the topological boundary of K, for bK = {(z ,z ..., z ) e K: lz 1 = 1 or 1z 1 = 1 or ... or lz 1 = 1}. 1 2' n 2 n Theorem 32 (Melnikov's criterion (15]) Let K be a compact plane set, Zo c K, 0 < a < 1, and En be the n+1 n annulus{z EC: a < lz-z < a }. Then: 0 ot, y(E,;\ K) z0 S (R(K)) if and only if E - +0a, where y is the 0 n=1 a analytic capacity. 31 co \ KO) - +o, where a is the Z S (10 ) if and only if E n 0 0 n=1 a continuous analytic capacity. 1-6 The Union of Peak Sets The following Theorem is quite well-known and it is originally due to Glicksberg [for example see 15]. Theorem 33. Let A be a uniform algebra on a compact Hausdorff space X. CO () K be a closed countable union of p-sets K of Suppose K = n n n=1 A (n = 1,2,...). Then K itself is a p-set of A. -set, it follows from the Since a finite union of GS-sets is a GS above theorem that any finite union of peak sets of A is a peak set of A. -set, we cannot But since a countable union of G-sets may not be a GS say that a closed countable union of peak sets of A is a peak set of A. Although it may be true, it does not follow from the above theorem. be peak sets of a uniform algebra A on X. The Let K1 and K2 following direct argument shows that K1 U K2 is a peak set of A. This argument is originally due to Bear. Let f E A peak on K1 and g e A peak on K2. Since -m m(m + 1) + m(m + 1)(m + 2) = 1 + mf + f 2 f3+ (1 - 2! 3! on X, where m is a real number, and the series on the right is uniformly convergent on X, (1 - f)-m e A. Similarly (1 - g)-m e A. Therefore 1/2 1/2 (1 - f) (1 - g) e A. We now define 1/2 1/2 -(1-f) (1-g) h = e • 32 clearly h E A and h = 1 on Ki (.) K2. If x K111 K2 we have If (x) I < 1 and Ig (x) I < 1 and so TT - < arg (1 - f (x) ) < - < arg (I - 2 - g (x) ) < 2 • 1/2 1/2 Hence Re(1 - f(x)) (1 - g(x)) > 0 and so 1/2 1/2 (1-g(x)) h(x) I = e -Re(1-f(x)) < 1 . Therefore h E A peaks on Ki t) K2. However, the above methods fail when A is a Banach function algebra. In [12] Dales has proved that the following restricted form of Theorem 33 holds for Banach function algebras, and he has also shown, by examples, that the restrictions are necessary. Theorem 34. Let A be a natural Banach function algebra on X. Then a finite union of pairwise disjoint p-sets of A is a p-set of A. Corollary. A finite union of pairwise disjoint peak sets of A is a peak set of A. 33 Chapter. two Peak Sets and Peak Points of Banach Function Algebras In this chapter X stands for a compact Hausdorff space and the peak points, peak sets and boundaries are all with respect to X. Otherwise it will be indicated. 2-1 The existence of the smallest boundary The existence of the smallest boundary (Bishop's minimal boundary) for a uniform algebra on a metrizable space X, was first proved by Bishop [4]. In this section I present another proof for the above fact along with some other results. Theorem (2.1.1). Let A be a uniform algebra on X. Then any non-trivial peak set P of A contains a peak set of A which is a proper subset of the given peak set P. Proof. Let p and q be two distinct points of P. Then there exists g e A such that g(p) = 0, g(q) 0 and maxIg(x)1 = 1. Since P is closed there xeP P such that Ig(x )1 = 1. Without loss of generality we can exists x0 0 assume g(x0) = 1. Let fc 1m=1nn be a decreasing sequence of positive numbers and define } so that Ne . Since P is a peak N = {x X:Ig(x)I < 1 + n En n+1 set it is a Grset and so there exists a sequence of open sets {0 n }n=1 suth that 0 0 and P = () 0n . Now we define Gn = N () 0n n+1 —C n n=1 cn so that Igi < 1 + En on Gn, Gn+1 C, Gn and P = AI Gn. 0pt(Sinj Let f E A peak on P and frinfn=1 be a"sequence of positive numbers. For each n (n = 1,2, ... ) there exists a positive integer kn such that 34 k 1gf n1 < n on GC = X\G . We define n n n k 5-- 1 gf n n=1 2n-1 03 so that h e A, Ihl < 2 on P and h(x0) 2. We now prove n=1 2 Ihl < 2 on Pc = X\P. c, then there exists n0 such that for alln>n xE G LetxEP 0 n C G1. Therefore butxeG — n0 C— Gn0 -1 C CO k + + + 1-1 ) + Ih(x)1 < Ig(x) 1(1 a no n-1 1g(x)11f n(x)1 2 n=no+1 2 CO 1 1 < (1 e ) (1 + 21 + + ) + n 2 — n n-1n 0 0-1 n=n +1 2 2 0 Now assume that we took to begin with E = n = 1n . n n 2 Then CO (1 + + + 1 ) + 1 < (1 + 2 -1 2n-1 2n0 2n0 n=n +1 2 o Co 1 1 1 1 1 1 + + + = n-1 n n+1 2n -1 o 0 o 0 212.11-1 2 2 2 2 n=n0+1 1 1 1 4 = 1 + + + 1 2 2n -2 2n -1 2n -1 ' -3- `' o 0 o 2 2 2 Consequently h assumes its maximum modulus within P and so P = fx e P: h(x) = h(x 1 o) = 2} is a peak set of A. Since h(p) = 0, p P 1 and hence P1 is a proper subset of P. 35 Corollary (2.1.2). Let pl,p2,...,pn c P.and the peak set P contains more than n points. Then there is a peak set P1 (1 P which does not contain any of pl,p2,...,pn. Proof. Let q e P and q # pk (1 < k < n). For each k (1 < k < n) there exists gk e A such that gk(pk) = 0 and gk(q) 0. We define g = gig2 ...gn so that g e A, g(pk) = 0 (1 < k < n) and g(q) 0. We now normalise g such that maxIg(x)1 = 1 and g(x0) = 1 for some x0 e P. xeP By the above theorem there exists h e A such that P1 = fx e P: h(x) = h(x0) = 21 is a peak set of A. Since h(pk) = 2g(pk) = 0 (1 < k < n), p P1 (1 < k < n) and so P1 is the desired peak set. Corollary (2.1.3). Each point of a finite peak set of A is a peak point of A. Proof. Let P = fpl,p2,...,pn1 be a peak set of A and consider the point p (1 < m < n). By the above corollary there is a peak set P C P such m 1 that pk P1 for k m (1 < k < n). Hence P1 = {pm} and so pm is a peak point of A. Thus every point of P is a peak point of A. Now we extend Theorem (2.1.1) using the same technique. Theorem(2.1.4). Let A be a uniform algebra on X. Then any non-trivial p-set of A contains a proper p-set of A. Hence any minimal set of A is trivial. Proof. Let Q = () Pa be a non-trivial p-set of A. There exist p, q E Q aeA and f e A such that f(o) = 0, f(q) 0, max If(x)1 = 1 and f(x0) = 1 for xcQ some x c Q. 0 Consider Ql = {x e X: f(x) = 1} and the function g:= 7(1 f). It is easy to see that Ql 0, QN\ Q1 0, Ig1 < 1 on Q, g = 1 on Q1 and IgI < 1 on Q\Qi. 36 1, We now define Nn = fx c X:Ig(x)I < 1 + (n = 1,2,...). Since Qc: r) there exists a peak set P of A such that QC= P-C n Nn. If 1 Nn = x c P, Ig(x)I < 1 + —1 for each n and so Ig(x)I < 1. So we can proceed as Theorem (2.1.1) and find h c A such that h attains its maximum modulus within P. Since h = 2g on P we have h = 2 on andd Ihl = 2Igl < 2 on Q\121, but Ihl < 2 on P\ Q. Hence Q2 = fx c X: h(x) = 21 is a peak set of A. It is easy to see that Qi = Q2/1 Q and p c Q\ Q1. Therefore Qi is a p-set of A and it is also a proper subset of Q. This completes the proof of the Theorem. Corollary (2.1.5). Let pl,p2,...,pn E Q and the p-set Q contains more than n points. Then there is a p-set Q1 such that Qi c Q but it does not contain any of pl,p2,...,pn. Proof. Let q e Q and q pk (1 < k < n). There exist gk c A such that gk(pk) = 0 and gk(q) 0. We define f = gig2...gn so that f(q) # 0 and f(pk) = 0 (1 < k < n). We can then normalise f such that maxlf(x)I = 1 xeQ Q. and f(x0) = 1 for some xo Now consider Qi = {x c X: f(x) = 11 and take g = 2(1 + f). Clearly Ql Q\,Q, 0, g = 1 on Q1, IgI < 1 on Q and IgI < 1 on 0,121. By Theorem (2.1.4) Ql is a p-set of A and it does not contain any of pl,p2,...,pn. Corollary (2.1.6). Any point of a finite p-set of A is a p-point of A. Lemma (2.1.7). Let A be a function space on X. Then every p-set of A contains a minimal set of A. Proof. Let S3= fQa: a c Al be the collection of all p-sets of A which is ordered by inclusion: if Qa, Qa cp, then Qa 4 Qi3 if and only if Qa C1 Q8. Let Q be an arbitrary element of S3 and define 37 S31 {Qa: Qa =QS n Q #(I) for some a c Al. = {Q e (3 : a e A) is a chain Let A be a subset of A such that a 1 in 6,31; i.e. 432 = {Qa e 31:Yaf f:i e A (Qa C: QS or 02i3 C: Qa)), and define F = f) Qa. Since any finite intersection of Qa e '32 is non- aeA empty and X has the finite intersection property, F # 4> and so F e J31• Clearly F is a lower bound for f32 and so any chain in A has a lower bound. Therefore by Zorn's Lemma S31 has a minimal element M which is in fact a minimal set of A. Hence there exist $ e A such that M = Qa n QC:Q. Consequently every p-set of A contains a minimal set of A. In particular every peak set of A contains a minimal set of A. Corollary (2.1.8). Let A be a uniform algebra on X. Then any p-set of A contains a p-point of A. Hence S(A,X), the set of p-points of A, is a boundary of A and so S(A,X) = r(A,X). Proof. It follows from Theorem (2.1.4) that any minimal set of A is a p-point of A. Therefore by the above Lemma any p-set of A contains a p-point of A. Corollary (2.1.9). If X is first countable S0(A,X), the set of peak points of A, is the smallest boundary of A and S0(A,X) = r(A,X). Proof. Since S0(A,X) is the intersection of all boundaries of A, and S0(A,X) = S(A,X), it follows from the above corollary that S0(A,X) is the smallest boundary of A and S0(A,X) = r(A,X). Corollary (2.1.10). Any infinite p-set of A contains infinitely many p-points of A. Proof. Suppose the infinite p-set Q of A contains only a finite number of p-points pl,p2,...,pn of A. By Corollary (2.1.5) there is a p-set 38 Qi(= Q which does not contain any of pl,p2,...,pn. By Corollary (2.1.8) the p-set Q1 contains a p-point of A which is different from any of pl,p2,...,pn. Therefore Q contains at least n 1 p-points of A which is contrary to our assumption. So Q has infinitely many p-points of A. In particular if X is first countable, any infinite peak set contains infinitely many peak points. 2-2 The density of the peak points in the Shilov boundary None of the theorems and corollaries of section (2-1) are true for Banach function algebras. The only one which is true is S0(A,X) = r(A,x) when X is metrizable. This was proved by H.G. Dales in [10]. But in fact, as a result of Theorem 15 MIDIS theorem [7]), it is even true for Banach function spaces on a metrizable space. I now give another proof for the density of peak points in the Shilov boundary which is rather elementary and constructive. Theorem (2.2.1). Let A be a Banach function algebra on a metrizable space X. Then S0(A,X) = r(A,X). Proof. To prove the density of peak points of A in the Shilov boundary of A, when A was a uniform algebra on a metrizable space X, we used the property that any peak set of A contains a peak point of A. But this is not true if A is a Banach function algebra. However, it is sufficient for S0(A,X) = r(A,X) to show that any neighbourhood N0 of a peak set P0 of A contains a peak point of A. is a peak set of A it is also a peak set of A and so it If P0 of A. Thus there exists f E A such that contains a peak point p1 1 = 1 and Ifil < 1 on X\-4,11. Now let N1 be a neighbourhood of pl such that 1711 C No. Take q = max If(x)I, then f31 < 1. Given xsX\N1 39 el > 0 there exists gl E A such that 11g1-f111x < El- Thus for q E X\Ni we have Igi (p1)1 I gl(q)1 > 1 - 2c1 - I. Hence, by taking el small enough such that 1 - 2c1 - 0.1 > 0 we get Ig1(q)1 < Ig1(P)1 for q E X\ Ni; i.e. gl attains its maximum modulus within N1. Let g0 c A peak on Po and take ni > 0. Then we have el + (p1)1 + > 1 + Ig0I nlIgll < 1 + ni( 01) on XVI]. and Igo yl - el). Hence lg0(p1)1 + n1lg1(p1)1 - Ig0I nilgil > n, (1 - 2e1 - al) > 0 on X\N - i.e.Ig0 n Ig I attains its maximum within N . Clearly 1 1 1 icy 1 < 211) such that G = + e l attains its we can find 4)1 <— 1 1 g0 nl g1 (q)1 > n maximum modulus within N1. In fact 11G1 1 - 1G1 1 (1 - 251 - al) > 0 for q c X \Ni. Since G1 A CT the maximum set M of G1 contains a peak point 1 G 1 s _11 p2 of T.Hence there exists f that f (p ) = 1 and p2 c N1 2 2 2 1f21 < 1 on XN\fp21. As before let N2 be a neighbourhood of p2 such that 1712 C Ni and take (32 = max1f2(x)1 so that a < 1. Given e2 > 0 xeX\No there exists g2 E A such that Irg2 - f211x < e2. Thus for q e XN\N2 we have 1g2(p2)1 - Ig2(q)I > 1 - 2E2 - 02. Hence by taking e2 small enough such that 1 - 2e2 - 02 > 0 we can easily show, as before, that . g2 attains its maximum modulus within N2 Given n2 > 0 we have !Gi l n21g2I < I1G111x n2 (E2 + f32) on X\N2. Since 11G111x = 1G1(p2)1 we get 1G1(p2)1 + n21 g2(p2)1 > 11G111x + n2 (1 - e2). Hence 1G1(P2)1 + 1121 g2(p2)1 1G1(q)1 - 7121 g2(q)1 > n2 (1 - 262 - a2) > 0 for q e X\N2. Therefore 1G11 + 1121 g21 attains its maximum within N2. As before we can find (1).1 (0 <. (1)2 < 27) such that G2 = G1 + 112e 2 g2 = 0 02 g0 + nle 1 gl + n2e g2 attains its maximum modulus within N2. In fact 11G211 - 1G2(q)1 > n2(1 - 2E2 - a2) > 0 for q e X \N2. Now we continue in this way to get a sequence of neighbourhoods 41 {M21413.1 such that 17Nn C Nn-1 (n = 1,2,...) and G = + n e + n g0 l g1 + 40 icf) e ng A attains its maximum modulus within N . Moreover for each n n n n (n = 1,2,...) we have I IG II - I G (q) I > (1 - 2c - a ) > 0 for q e X \ N and hence n X n n n n I + nn (1 - 2en - an). I IGn IIX > sup IGn(q) qeX\Nn Take A = IIG II - supIG (q)I and define fa lc° by n n X n n n=1 qEX\Nn S = min(—,A11 A ..' ,A ). Now we define n n 2' n co g = g0 + nne ngn n=1 and choose n such that this series is convergent in the norm of the n Banach function algebra A. To do this let g = Gn + Hn so that CO IIH II < ) 11,11gkII. Suppose n1,n 2' '''' nk-1 has been chosen so that n k=n+1 G ,G ...,G and hence S are determined and define 1 2' k-1 k-1 s k-1 n k 2k .,gll k II < S for k > n we have Since Sk n co 6 k-1 < 6 / 1 6n 2 (n = 1,2,...). IIHnIl x < IIHnll f_ k — n k=n+1 2 k=n+1 2k2 co i(1) n Since S ÷ 0 we have IIH II + 0 and so the series ) n e g is n n n n n44.0 n.+00 n=1 convergent in norm of A. Let p e Nn (n = 1,2,...) such that IGn(p)I = IIGnlIx. Then - and so Ig(P)1— IGn (P)I IH (P)I > IIGn II X - IIHnII X < IIGnII X 41 6 n(q)I + IIHnlIx < IG n Ig(q)I < IGn(q)I + IHn (q)I < sup IG - An +2 — qcX\N n n 6 < I IGn Hx - 2-a < Ig(P) I for q 6 X\Nn. Therefore g attains its maximum modulus within Nn. As n was arbitrary, Mg = {x e X:Ig(x)I = IIgIIX} c_ n • n=1 Nn Since X is compact, it has the finite intersection property, and CO CO (1 N it follows that () N = () N (1). Now if we can as N CNn n n n+1 n=0 n n=0 choose N 's in such a way that they shrink to one point p; i.e. CO n Nn = {p}, then M9 = {p} and so p e N0 is a peak point of A. n=1 If X is metrizable, we can take N to be the sphere S(p ,r ) n n n and choose r small enough such that N C=. N and () N = {p}. This n n+1 n n n=1 completes the proof of the theorem. In the case that X is first countable I do not yet know whether we can find such a sequence of neighbourhoods which shrinks to {Nn }n=1 one point or not. If it is true then the theorem can be extended for first countable space X. To extend the theorem, I tried to find a class of topological spaces which lies between the class of metrizable spaces and the class of first countable spaces. D.R. Creede [9] introduced a class of spaces, called semi-stratifiable spaces, which lies between the class of semi-metric spaces and the class of spaces in which closed sets are G6-sets. To prove the existence of such a sequence of neighbourhoods CNn l , in a n=1 compact Hausdorff semi-stratifiable space X, we need the following Lemma. Lemma (2.2.2). Let E be a closed G 6-set in the compact Hausdorff space X. Then there is a sequence of open sets IG n ln=1c° in X such that -- 42 co G C: (n = E = G and for any open set 0 containing E n+1 Gn n n=1 C=0 for all n > m. there exists an integer m = 1,2,...) such that Gn CO Proof. Let {0n be a sequence of open sets O such that E = f) 0 . }n=1 n n n=1 Since for each n, E and X\On are closed disjoint subsets of the normal space X, by Urysohn's Lemma there exists fn c C(X) such that O < f < 1 on X, f = 0 on X and f = 1 on E. If we define n n n n P = {x c X: f (x) = 1}, it is easy to see that P is a peak set of n n n C(X) and EC= C: (n = 1,2,...) and so E = P ; i.e. E is a peak — Pn — 0n n n=1ri set of C(X). Hence there exists f e C(X) such that f = 1 on E and IfI < 1 on X\E. Now we define G ='{x e X: If(x)1 > 1 - 27 }. Clearly n co n G is open in X, G C: and E = G. n n+1Gn fl n=1 Let 0 be an open set in X which contains E. Since Oc = X\O is closed and f is continuous on X, there exists x0 c Oc such that maxlf(x)I = If(x0)1. As IfI < 1 on Oc, it follows that 6 = If(x0)1 < 1. xc0c 1 If we take m large enough such that 1 —m < 1 - 6 we have 1 - m-> 6 and hence G 0 for all n > m. n—C An important result of the above Lemma is the following: Corollary (2.2.3). Let X be a compact Hausdorff space. If for every x e X, fX1 is a G6-set in X, then X is first countable. Proof. By the above Lemma there exists a sequence fGn of open sets .r=1n G in X such that rxl = n G and for each open neighbourhood N of x, n=1 G for some n (n = 1,2,...). Thus {G n }n=1c° n CNx is a countable base at x and so every point of X has a countable base; i.e. X is first countable. Note 1. X being compact and Hausdorff in the above Lemma is a necessary condition. Even if X is normal but not compact the corollary may be false. Appert space [23, example 98] is an example of a perfectly normal space which is neither compact nor first countable. Since it is perfectly 43 normal, it is Hausdorff and every closed set is a Gs-set. In particular for every x c X, {x} is a Gs-set. Note 2. Let A be a Banach function algebra on a semi-stratifiable compact Hausdorff space X. Since every closed set in X is a Gs-set, by the Corollary (2.2.3) X is first countable. Hence any peak set of A contains a peak point of T. Thus to prove %(A,X) = r(k,x) what we actually need is to show the existence of the sequence {N }w=1 of open CO nn n f) neighbourhoods N in Theorem (2.2.1) such that Nn contains only n=1 one point. This can be done by Theorem (1.2) from [9]. But unfortunately a semi-stratifiable compact Hausdorff space is metrizable. To show this I refer to a theorem of Bing [3] and some results of Creede [9]. By Corollary (2.2.3) any semi-stratifiable compact Hausdorff space is first countable, Hence by Corollary (1.4) from [9] a compact Hausdorff space is semi-stratifiable if and only if it is semi-metric. Since a Moore space is semi-metric and, hence, semi-stratifiable, by Corollary (4.11) from [9], a compact Hausdorff space is a Moore space if and only if it is semi-stratifiable. But according to a Theorem due to Bing [3] any paracompact Moore space is metrizable. Hence if a compact Hausdorff space is Moore, semi-metric or semi-stratifiable it is metrizable. But there exists a compact Hausdorff space X such that any closed subset of X is a GS-set in X whereas X is not semi-stratifiable. Example 48 in [23] is a compact Hausdorff, perfectly normal and non- metrizable space and hence it is not semi-stratifiable by the above discussion. Note 3. Let A be a Banach function space on a first countable space X. As we know So(A,X) = S(A,X); i.e. every p-point of A is a peak point of A. One may think that, in this case, every p-set of A is also a peak set of A. But this is not true as the following example shows: 44 Example (2.2.4) [23. example 97] and C Let X consist of two concentric circles C1 2 in the plane; let C2 be the outer circle, C1 the inner. We take a subbasis for the topology all singleton sets of C2 and all open intervals on C1 each together with the radial projection of all but its midpoint on C2. It is rather easy to show that X is a first countable compact Hausdorff space which is not metrizable (23]. To prove X is not perfectly normal we show that C1 is a closed subset of X which is not a Gcset in X. Since for every x 6 C2, {x} is open C2 is open and hence C = X\C is closed. Let C be a G -set in X so that C1 = I) On. 1 2 1 (5 n=1 Hence C2 = U On and so On c: C2. Now let F be a closed infinite subset n=1 of C2 and Cz nnlm=1 be a convergent sequence of distinct elements of F such that z z. Since z E FC C2, {z} is an open neighbourhood of z n n.4.°° and so there exists n0 such that z = z for all n > n0; i.e. the sequence n is finite which is contrary to our assumption. Therefore every closed c subset of C2 is finite. In particular 0 is finite (n = 1,2,...). = U 0c, C must be a countable subset of X which is impossible. Since C2 n 2 n=1 Consequently C1 is not a GS-set in X. is a p-set of C(X). For every x c C2,{x} and We now show that C1 X\{x} are both closed and so by the Urysohn's Lemma there exists f C(X) such that f = 1 on X\Cxl and f(x) = O. Thus X\{x} is a peak set of C(X) for every x c C2. Since C1 = n (o,fx1), C1 is a p-set xeC2 of C(X). But according to the above argument C1 is not a Ge-set and so it is not a peak set of C(X). Lemma (2.2.5). Let A be a function algebra on X. Suppose N is a neigh- bourhood of a peak set P of A and xi,x2,...,xn c N. If P contains a x then there exists a peak set P C: N point different from x x n 1 which does not contain any of xi,x2,...,xn. 45 Proof. Let q e P and q xk (1 < k < n). Then for each k there exists f e A such that f (q) = 1 and f (x ) = 0. Since P is a peak set there k k k k exists g e A such that g = 1 on P and Igi < 1 on X\ P. As X\N is closed and Id < 1 on X\ N for large enough m we have lefl < 1 on m ) = 0 (1 < k < n). Thus h X\ N. Hence h = g f e A, h(q) = 1 and h(xk i,x2,...,x . In attains its maximum modulus within N but not in x n other words there exists a peak set pi c. N of A such that xi,x2,...,xn P1. Proposition (2.2.6). Let A be a Banach function algebra on a metrizable space X. Then any neighbourhood of an infinite peak set of A contains infinitely many peak points of A. Proof. Let the neighbourhood N of the infinite peak set P contain only l,p2,...,p of A. By the above Lemma a finite number of peak points p n there exists a peak set P1 C N of A such that pl,p2,...,pn Pl. Since N\fpi,...,p 1 is a neighbourhood of P1 in X by Theorem (2.2.1) it n, contains a peak point pni.1 which is different from pl,p2,...,pn. Hence N contains n + 1 peak points of A which is contrary to our assumption. Since Theorem (2.2.1) is also true for Banach function spaces, the above proposition is also true for Banach function spaces. It is convenient at this point to prove the following useful and elementary result. Proposition (2.2.7). Let A be a function space on X and A be the uniform closure of A on X. Then: (i) r(A,x) =1.(T,x), (ii) ch(A,x) = Ch(T,x). 46 Proof. (i) Let x e r(T,X) and Nx be a neighbourhood of x. By the characteristic property of the Shilov boundary there exists f E A such that Mf C Nx and IlfIIX = 1 so that 0 = max If(y)1 < 1. Since yeX\Nx 1-0 f Tthere exists g e A such that Ilf-g11X < e = and hence 2 12s_ IIgIlx > 1 - e = 111 and Igl < e + 5 = on X\Nx. Therefore on X\Nx; i.e. g attains its maximum modulus only within Igl < IlgIIX Nx and so x e r (A,x) . Since A C A, r (A,x) (= r (T,X) and so r (A,x) = r (ii) Let K(A) = {* e A*: 1145 1I = *(1) = Jibe the state space of (A,11"11x) and K(A) = {* e T*:114)11 = cl)(1) = 11 be the state space of (A, I I Ix). We first prove K(A) = K(A). Let * e K(A) and f E A, then there exists a sequence {f n }n=1 in A such that 11f 0. Since n-f11X ÷ we have 14)(f {$ (f } is convergent n) - 4)(fm )1 < Ilfn -f m X 4.000, ) n,m n n=1 and so we can define ON=lim *(f ). It is easy to show that * e K(A) n+00 n and it is in fact the extension of 4 from A to A. Conversely if * e K(A) we define 4 = 4}A. It is easily seen that the map F: K(A) + K(A) defined by F(*) = * is 1-1 and onto. To prove the continuity of F, let *0 e K(A) and *0 = F(40). Let N, = {* E K(E):14)(f) - *0(f)1 < e: f e A, e > 0} be a neighbourhood O of * in the weak*-topology. For the above e > 0 there exists g e A o We now define N. = {4) E such that Ilg-fIIX < 4 K(A):I*(g) $0(g)1 < 2 ) (PO and let*eF(N.), then I*(g) - *0(g)I < -5- and 90 I*(f) - 1P0(f)1 < - 4)(g)1 + 14)(g) - 4/0(g)I + 14,0(g) - *0(f)1 < I If -g I Ix + 2 + I lg-fl lx < 74' = e. 47 Hence IPc N ; i.e. F(N, ) C=N, . Now if we take an arbitrary 0 0 0 neighbourhood N(tii0;f1,f 2,...,f n; c) = c K(A): (f i ) - 11)0 (f j)1 < c; f i c A (1 of 0 in KIA), as it is an intersection of the form () N (fecan find 'j=1 0 J a neighbourhood NO Vg fg .. .. fg /6) = () N (lg.) such that 0 1 2 n j=10 j F(N N Consequently the map F:K(A) K(A) is continuous. Since (P0 — * 0. K(A) is (weak*) compact and K(A) is (weak*) Hausdorff, F-1 is also continuous and so F is a homeomorphism; i.e. K(A) K(A). Hence ex(K(A)) = ex(K(T)). Since Ch(A,X) = fx a X: 11)x E ex(K(A)} and Ch(A,X) = fx e X: CPx E ex(K(A))} we have Ch(A,X) = Ch(A,X). 2-3 Examples of Banach function algebras having peak sets without peak points What we actually proved in Theorem (2.2.1) is that every neighbourhood of a peak set of A contains a peak point of A. Whereas when A is a uniform algebra we proved every peak set of A contains a peak point of A. So it is natural to ask whether this is also true for Banach function algebras or not. In this section I give some examples of Banach function algebras on compact metrizable spaces in which there are peak sets without any peak point. In other words, the set of peak points of A, So(A,X), is not a boundary of A. H.G. Dales has also given such an example in [10]. But 2 he has constructed a Banach function algebra on a compact subset of C , having a peak set without peak point, whereas I have constructed such a Banach function algebra on a compact subset of C x R = R3. I have also given an example of a Banach function algebra on a compact plane set having a peak set without peak point, but it is not natural. However, it is conjectured that, in a natural Banach function algebra on a 48 compact plane set, every peak set contains a peak point. Example (2.3.1) In this example we construct a non-natural Banach function algebra 3 on a compact subset of C x R = R having a peak set without any peak point. Let X = {(z,t) e C x R = R3: <1+1t1 , It' < 1 . Clearly X 1. Izi 2 is a compact metrizable space in C x R = R3. z-plane 00 m n 2. Let A be the set of all functions f(z,t) = az t such m,n=0 m,n that a E C and la 1 < co. Clearly >•—°D a zm tn is m,n m,n m,n=O m,n=0 m,n absolutely and uniformly convergent on X and so f(z,t) is continuous on X; i.e. A C C(X). Moreover for each fix t0 E [-1,11, f(z,t0) is < 11-1t01 ). analytic on the disk PtP = {(z,t) c X: Izi It is easy to 2 show that A is a function algebra on X. If a m n _ 2:1 m,n z t = 0 with 00 m,n=0 >--- la I < co, then a E — m,n = 0 for all m,n. Hence each f A has a m,n=0 m,n 09 unique series and so we can define IlfIl = ) 7- lam ,n I so that m,n=0 (A,II'll) is a normed function algebra on X. Now we prove A is complete under this norm. 03 (p) n Let f (z,t) = a zrat > m,n be a Cauchy sequence in A so that P m m,n=0 Ilf -f 11 = ) la(P)m,n - m,na(q)1 ÷ 0. Hence there exists M > 0 P q m,n=0 p,c/i-oo 49 such that Ilf II < M for each p (p = 1,2,...). Given c > 0 there P exists k such that for all p,q > k and for every m,n we have l a (p) - a c. Hence for each fixed m,n fam,(P) 1p=1m is a Cauchy I m,n m,qn1 < n sequence of complex numbers and so for every m,n there exists CO (p) a c C such that a a . To prove > is convergent, m,n m,n m,n m,n 204' m,n=0 let N be fixed, then N (p lam,(P) I —< Ilf 11 < M = 1,2,...). m,n=0 n p — Thus when p 4- m we have N < lam,n 1 M. m,n=0 Now let N ° we get CO > la I < M m,n=0 m,n — and so m n f(z,t) = a z t e A. m,n=0 m,n Now we prove Ilf 0. P P4w Given c > 0 there exists k such that for every fixed N and every p,q > k we have Ia(P) - a(q) 1 If -f I 1 < " m,n=0 m,n m,n p q 2 Let q 00, then N Ia(P) - a I < m,n=0 m,n m,n for every p > k. Now let N -4- 00, then CO (p) 1am,n am,n1 = Ilfp-f11 m,n=0 50 for every p > k. Therefore 11'11 is a complete norm for A and so (A,I1-11) is a Banach function algebra on X. 3. A is not natural. In fact MA a .((,71) E C2 :ICI < 1, In' < 1 To prove this let CO m n a z t e A. f(z,t) = m,n m,n=0 Then ) lam,n1 m,n=0 is convergent and so CO l am,n1 0 m=M+1 n=N+1 as M and N co. Hence CO CO M,N m n i IIf - a ztllmri = II > am,n z t I = > a 1 4, - 0. m,n m,n m,n= 0 m=M+1 m=M1-11 M114m n=N+l n=N+1 Therefore A is the norm closure of polynomials in z and t on X and so A = [z,t]; i.e. A is generated by z and t on X. Hence by Theorem 29 2 2 MA = aA(z,t) = {(4(z),Ot)) e C : e MA}. Now let (,1-1) e C such that Id < 1, Inl < 1 and take an arbitrary element CO m n a t f(z,t) = m,n m,n=0 in A. Since CO a I l m, n m,n=0 51 is convergent, 00 m n a m,n C rri,n=0 is also convergent and so we can define a function 4 on A by CO a m nn . 4)(f) = m,n m,n=O It is easy to show that 4 c MA. Since 4)(z) = and 4(t) = p we have ,T1) c aA(z,t) and so {Mn) e C2: ICI < 1, InI < 11 = aA(z,t) a MA. 4. Since for every t0 c (-1,1], f(z,t0) is analytic on the disk P = {(z,t ) c X: 1z1 < lilt } it attains its maximum modulus, o ol to — 2 , on bP = {(z,t ) c X: IzI = 14-1t011 J. Therefore relative to P o t0 t0 2 every f c A attains its maximum modulus only on bX = {(z,t) E X: I = 1-1.12t1 IZI or Iti = 1}. If there is a peak point x of A 0 c bP t0 then any other point of the circle bPt is also a peak point of A. 0 Note that P is a peak set offAt_ t f) or every t E (-1,1]. The t 0 0 2 peaking function is f(z,t) = 1 c A. 4o 1, 5. Consider the peak set Po = {(z,0) c X: Izi < v. We prove the point (7, 0) E Po is not a peak point of A and so by the above discussion P0 does not contain any peak point of A. We first state the following Lemma from the complex function theory. Lemma. Let f be a non-constant continuous complex valued function with continuous complex derivative on the closed disk {z e C: Izi < r}. If max If(z)1 = f(z ) (Iz 1 = r) then z fi(z0) > O. 0 0 0 f (zo) lz kr Suppose on the contrary there exists CO m n f(z,t) = > b z t m,n=O m,n 52 1 in A which peaks at (p 0). Since 00 lb I m,n=0 m,n is convergent the series 03 m n z t m,n=O m,n is uniformly convergent on the cylinder {(z,t) 6 C x R: lzl < 1, It! < 1}. Hence f is extendable to this cylinder and so it is analytic in a 1 neighbourhood of (-f., 0). Therefore we can write f in the form 0 0 am,n(z - 1m2 tn m,n=0 1 1' 0) = 1, which is convergent in Iz 112 < 2- and ItI < 1. Since f(-2 a0 0 = 1 and so 1 1 2 t + a1/0(z - + a1,1t(z - + a2,0(z - + f(z,t) = 1 + a0,1 Hence for t = 0 we have 1 2 (z - + a2/0(z - + f(z,0) = 1 + a1,0(z (z Since 00 m n b z t m,n=0 m,n 1 is convergent in lzl < 1 and It! < 1, f(z,0) is analytic in Izi .....< 7. . 1 By the hypothesis f attains its maximum modulus only at (T, 0) and so 1 1 by the Lemma f(T., 0) > 0. But since f(z,0) = a10 + 2a20 (z - T.) + ... 1 in lz - ld2 <-2 1- we have f'(- 2' 0) = a10 > 0. 53 1 2 t) = 1 +a + (ItI < 1) Now consider f( 0,1 t+a0,2 t obviously f(2, t) is a differentiable function of the real variable 1 t) is a real differentiable function of the t and hence h(t) = Re f(-2' real variable t. Since h(t) attains its maximum modulus only at t = 0 we have h'(0) = O. On the other hand 2 h(t) = 1 + Re a t + Re a0,2 t + 0,1 t + h'(t) = Re a0 1 + 2Re a0,2 Therefore h'(0) = Re a = O. For convenience let a = ib (b real) 0,1 0,1 and a = a > O. Then 1, 1 1 f(z,t) = 1 + ibt + a(z - + a t(z - + 2 11 2 l+t in I z - 2I < Considering the value of f(z,t) at (T, t) for small positive t we have l+t :5__2 f( , t) = 1 + ibt + + 2 2 + a1 ,1 2 a -2- t + ibt + 0(t2). l+t 2 l+t 2 l+t 2 Hence I f (---, t) I -= (Re f (—, t) ) + (Im f ( 2 2 2 2 , t) = 1 + at + 0 (t ) . l+t Since t is positive by choosing t small enough we get t)I > 1 2 1 which is contrary to our hypothesis that f peaks at 2 0). Consequently 1 0) is not a peak point and so the peak set P does not contain any 2' 0 peak point of A. 6. For t 0 we cannot prove, in the same way as before, that 0 1+lIto x0 = ( t ) c X is not a peak point of A. Because in this case --7--' 0 1+It 1+Ito ( oI, t) iis not always in X and so Re f( I, t) may not be maximum 2 2 54 at x = 1+Itdt ) if we assume f peaks at x0. 0 2 0 0 Ito , to) c X (t 0) is In fact every point of the form x0 = (1+--7-- 0 a peak point of A. To prove this we define f on X by (t - t )2 1 f (z,t) = (1 - 0 ) (-2 + zg (t) ) 4 1 1 where g(t) 6 A, g(t and Ig(t)I = g(t) < (1t1 < 1)- 0) = 1 + It0 I1 1 + Itl 1 1+ItrIl Clearly f A and f( g(to) = 1. Since for -, t0) = y+ 2 11 every (z,t) c X, IT+ zg(t)I < 1 we have If(z,t)I < 1 for every (z,t) e XN,{x0}. Hence x0 is a peak point of A. To prove the existence of such a function g(t) we give the following example: (i) For t0 > 0 we define 2t (1 t_ l+t0 to 1 + e g (t) - 2(1 + t0) Clearly g c A, g(t0) - and 0 < g(t) for every t c [-1,1]. To 1 + t0 1 for every t c [-1,1] it is enough to show that prove g(t) f- 1 + I t 2(1 + t ) 0 2 Log 1) Ott (1 - > 0 (1 + 1 0 tO -- for every t e [-1,1]. (ii) For t0 < 0 we define t 2 (1 ) 1-t 0 to 1 + e g (t) - 2(1 - t0) 55 I E A, g(t ) = - It [ and 0 < g(t) for every Clearly g 1 1 t 1 + 0 - 0 0 1 t 6 [-1,1]. To prove g(t) It! for every t c [-1,1] it is < 1 + enough to show that 2(1 - to) 2t 1) —o) 0 Log (1 4. itl 1 - t0 (1 - for every t 6 [-1,1]. 1+10 , t 0}. 7. As a result of 5 and 6 we have S0(A,X) = {(z,t) c X: Izl - 2 Since A is a Banach function algebra on the compact metrizable space X by = 1+ 1 t I 1 Theorem (2.2.1) we have P(A,X) = %(A,x) . {(z,t) c X: Izl 2 J and so r(A,X) bX. Let P(X) be the uniform closure of polynomials in z and t on X. If m n a z t E A then f(z,t) m,n m,n=0 M,N M,N m n t a zmtn11 IIf Ilx < Ilf - ,7- m,n O. m,n=0 am,nZ m,n=0 M,N+00 Hence f e P(X) and so AC. P(X). On the other hand every polynomial in z and t on X belongs to A so that P(X) CT. Therefore T. P(X). } Since P0 = {(z' 0) c X: Izl < 21 is a peak set of A, it is also a peak set of the uniform algebra T. P(X) and so it contains a peak point of P(X) which is certainly on bPo. Hence every point of bP0 is a peak point of P(X). Therefore itl . 0(P(X)) = AX = {(z,t) e X: IzI = 1 + S0(A,X) = S 2 56 Since by the proposition (2.2.7) r(A,X) = r(ii,x) and Ch(A,X) = Ch(A,X) we have r(A,x) = r(P(X)) =A X and Ch(A,X) = Ch(A,X) = .o(DT,X) = AX. Note. This example also shows that S0(A,X) So(T,X) = AX. Example (2.3.2) In this example we construct a natural Banach function algebra on the same compact metrizable space X = f(z,t) 6CxR= R3 : 1z1 < 1 2 10 , Itl <1 — having a peak set without any peak point. The construction of this example is quite similar to the previous one but it is rather more complicated. co 1. Let A be the set of all functions f(z,t) = 7- an(t)zn such n=0— that an(t) and a"(t) exist and are continuous on (-1,1]. Moreover, the series (n+1)2(lan(t)I + + la;(t)I)(1 + Itl)n n=0 co is uniformly convergent on [-1,1]. If f(z,t) = a (t)zn 7- n 6 A then co n=0 a (t)zn 7- n is uniformly convergent on X. Since a (t) and zn are n=0 n continuous on X, f(z,t) is also continuous on X and so AC C(X). It is easy to show that A is a function algebra on X. Now we define the norm 00 of f(z,t) = a (t)z e A as n n=0 57 co 2 1 1+1t1 n I If I I sup (n+1)(lan(t)I + la:1(t)1 + Ian(t) 2 ) . Itl<1 n=0 It is also easy to show that (A,II-II) is a normed function algebra on X. 2. We now show that is a Banach function algebra on X. To do this let f (z,t) = 7- an,p(t)zn be a Cauchy sequence in A so n=0 that Ilf -f II"-÷ 0. Hence for e > 0 there exists k0 such that for P q Pfq' every p,q> k0 and each n (n = 0,1,2,...), and every t c [-1,1] we have lan,p (t) - an,q(t)1 < c, la'n i p(t) - a'n,q WI < e and E. la"n,p (t) - n,qa" WI < Hence for each fixed n, {a n,pp {a' }°°=1 and {a" }t°=1 are le0=1' n,p p n,p p uniformly convergent on (-1,1]. Therefore for each n and te [-1,1] E C such that a" (t) n (t) uniformly on (-1,1]. there exists Cn(t) n,p n*im Since a" (t) is continuous on (-1,1] C (t) is also continuous on (-1,1] nrlo n and so for fixed n (n = 0,1,2,...) t t f a" (T)dT f cn(T)dT -1 n,p p+00 -1 uniformly on [-1,1]. Since a' (t) is convergent on (-1,1] as p m n,p we let b = lim a' (-1) and define n pom n,p t bn (t) = bn + f cn(T )dT 58 Clearly a' (t) b (t) uniformly on (-1,1] and b'(t) = C (t) n,p n n (It' < 1). Since bn(t) is continuous on (-1,1] we have t t -1f an,p (T)dT f b (T)dT n P+o3 -1 uniformly on [-1,1]. As an,p(t) is convergent on [-1,1] as p -.)- co t and a (t) = a + f b (T)dT. we can define an = lim an,p(-1) n n n 13+'" -1 Hence an,p(t) --* a (t) uniformly on (-1,1] and an(t) = b (t) (It! < 1). 1)4.00 n n Since a" (t) = bl (t) = C (t) and C (t) is continuous on (-1,1], an (t) E A. n n 0,n n 0. Now we show that f(z,t) = 1_ an(t)z c A and Ilf -fll n=0 p+t0 For every finite M and N we have M 2 t t lt (t)1)(+t1 2:1 (n+1) (la (t)1 + la' (01 2 n,p 2 ) < n=N+1 n,p n,p M (n+1)2(la (01 + la' n,q(01 1 + 2 n,q" (t) 1 )(1+2I t l)n + n=N+1 n,q (01 + la' (t)-a' (t)I + (t)-a" (t)l) (71+1)2 (la (t)-a n,p n,q 2 n,p n,q n=N+1 nsp n,q ,l+Itl )n 2 Since Ilf -f II ---> 0, for c > 0 there exists k0 such that for every P q p,q4-00 0 p,q > ko, each t E [-1,1] and every M & N 1 + Itl )n < e (n+1)2(la (t) - an,q(t)1 + ....)( 2 -4- . n=N+1 n,p 59 Since fq c A for a fixed q = k and e > 0 there exists n 0 0 such that for all M & N > n and each t [-1,1] 0 1 + It1 n > (n+1)2(la (t)1 + ....)( ) < . n=N+1 n,q 2 Hence for all M & N > n0 and every p > lc_ and each t e (-1,1] M 2 1 4. ....)(1 +2 ItI n E (n+1) (la (10 s) < — n=N+1 n,p 2 Now let p + co. Then an, --+ a (t), a' (t)-+ an) (t and a" 00-4 an (t) p n n,P n n ,p n for each t e [-1,1] and hence for every M,N > n0 and every t e (-1,1] we have 1...i a"(t) 1) ( 1 + Iti‘n < e (n+1)2(lan(t)1 la;,(t)I < C. n=N+1 2 n 2 ' Therefore 00 (n+1)2(la 1 la"(014. 1 ItI )n n(t'l I I ]ileil(t)1)(1 n=0 2 is uniformly convergent on [-1,1]; i.e. f c A. Now we prove 1 If -fl 1 —4.0. p+00 Since lIf for e > 0 there exists k such that for every P-f q 11 o 10,q+°3 p,q > k0 and every t c [-1,1] and each N we have N (n+1)2(lan,p(t) - a (t)1 + ....)(1 +2 Itl)n < n=0 n,q Let q co, then 2 + It1 (n+1) (la - an(t)1 + ....)(1 )n< --2 n=0 n,p 2 60 Now let N m, then 00 1)(1+12tI)n (n+1)2(lan,p m-an mi la:1,p(t)-al!)(t)I -:-Ia'2 n",p(t)-an"(t) n=0 < — 2 Thus CO -fil = sup >(n+1)2(la Ilfp (t) -an(t)I+ < c Itl<1 n=0 n,p 2 for every p > ko; i.e. Ilfp-fII Consequently (A, II-II) is a 1344' Banach function algebra on X. co nt 3. If f(z,t) = > a (t)zn c A then IIf(z,t) - >-- an mz I , O. n=0 n n=0 N-*co To show this we say for e > 0 there exists n0 such that for every N >n0 and every t e [-1,1] we have lat (01 + .1.1ati ft% < > (n+1)2(Ia (t)1 + n` ' 2 n=N+1 n 2 n' 2 Hence Ilf - >11 an(t)znI1 = an(t)011 = sup Z (n+1)2(lan(t)1+...) n=0 n=N+l Itl<1 1+Itl n for every N > n. ( 2 ) Therefore !If - an (t) n=0 - N4°1° 61 4. In fact A is the norm closure of polynomials in z and t. In other words A = [z,t]; i.e. A is generated by z & t. 03 n e > 0 there exists Proof. Let f(z,t) = 7- an(t)z A. By 3 given n=0 n we have o such that for every N > n0 N IIf - a (t)zn n I1 < E. n=0L Since an(t) is continuous on the closed interval [-1,1], by the Stone- Weierstrass theorem for each fixed n there exists a sequence of polynomials in t which is uniformly convergent to a'x'(t) on (-1,1]. Hence for 3 there exists a polynomial (n0+1) k A (t) = 0 a t nk, m,n k= such that for every t e [-1,1], Ia"(t) - A < . It is m,n (n +1)3 0 easy to show that there is another polynomial Bm+i,m(t) of degree m + 1 (t) = A (t) and such that B141,n m,n 2c 3 for every t c [-1,1]. Ian(t) - B (t)1 < m+1,n (n + 1) 0 (t) of degree m + 2 such that Finally there exists a polynomial Pm+2,n P (t) = B (t) and 1.42,n m+1,n 46 Ian(t) P < for every t E [-1,1]. m+2,n(t)I (n + 1)3 0 k Now let M = m + 2 and P (t) = a t . Then we have M,n L k n k=0 ' 62 n0 0 n0 an(t)zn - IIPm,n(t)znII = IE (an(t) - Pm,n(tznI1 n=0 n=0 n=0 n 0 2 P + lal (t) - (t) = sup (n+1) (lan M,n Bm+1,n I Itkl n=0 11 (1+Itl)n + 1= da"(t) - A (t) Z' m,n I' no n E C - 2 4E 4. 26 + O 6.5E ....< 2E1 (n+1) ( 3 ) -< 1 < 7E. n=0 +1) (n +1) 2(n +1)3 n +1 (n0 3 0 0 n=0 0 Therefore n o o IIf - atm, n mznII = IIf P (t)znII m,n=0 n=0 In n n n 0 0 n 0 n < IIf - a (t)znII + a (t)Z- C 7e = n n 7 PM,n (t)z II < 8E. n=0 n=0 n=0 Thus every f E A is the norm closure of polynomials in z and t on X and so A is generated by the coordinate functions z & t. 5. A is natural; i.e. MA a X. Proof. Since A = [z,t], MA a aA(z,t) = {(0z)4(0) E C2: 4, EMAl. Since X C aA(z,t) it is enough to show that aA(z,t)C: X. To do this let c() e MA, c = 4(z) and n = 4(t). We must show that n e [-1,1] and ICI < 2 Let a,a e R and a 0 and consider the function A(t) - 1 t-(a+if3) 1 in A. Since 4)(A(t)) = ot)._(a+io is finite, n = 4)(t) a + 1 3. As a is arbitrary, fl must be a real number. Now let a e R and la! > 1, then the function -7t.1.c-- -is in A. Since cP (- 1 1 is finite, t-a1-) =(t)-a n-a n a. Hence In! < 1. 63 Now let f(t) be a continuous function on [-1,1] such that f'(t) and f"(t) exist and they are continuous on [-1,1] so that f(t) c A. It is easy to show that for every c > 0 there exists a polynomial P(t) we have in t such that I If-13 I < 2 • Hence for every 4)e MA 1:11 (f)-f(4(t))1 < Iflf)-0(P)1 10(P) -f(0(t))1 < 1If -PII + IP(n)-f(n)1 < 211f-P11 < c. Therefore 0(f) = f(4)(t)) = f(n) where n is real and InI < 1. 1 1+2n1 I To prove Iii we consider two cases. (i) Let n = 0(t) 0. We define a(t) on [-1,1] by I t « 1) a(t) = 21.11.1_ cos -711-- t I ni 2h (-1111 < t < Inl) -t (-1 < t < - In1). It is easy to show that a(n) = I11I, It! < a(t) < 1 for every t c [-1,1 a(t) is continuous on [-1,1], ce(t) & a"(t) exist and they are also continuous on [-1,1]. In fact (iii < t < (Inl < t < 1) It a' (t) = sin (-In' < t < In!) ci(t) t (-In( < t l and Ia (t)I < 1, la"(t)I < 2ini on [-1,1]. 64 Now we define the function g(z,t) on X by CO 1 g(z,t) = > (g(t))n zn n=0 (n+1)6 where g(t) - 1+a(t) . Clearly g(t) is continuous, g'(t) & g"(t) exist and they are continuous on [-1,1]. Moreover 1 < g(t) < 2 on [-1,1] and 4(g(t)) = g(4(t)). Since g i (t) -2 c0(t) 2a"(t)(1+a(t)) + 4002(0 (11.a(t))2 and g"(t) - (1 + a(t))3 we have Igt(t)1 < 2 and le(t)1 < -21+-+ 4 on [-1,1]. To show g(z,t) c A — I we say 6- (n+1)6 (t) li g(t) in-1 1 2Ei 6( + -fIng"(t)gn-1(t) + n=0 (n+1) n(n_1)g,2( )gn-2(0 1 )(1+12t1 )n - 1 Tr 1+Iti n < NT 1 ) 4 (1 + 2n + 2 ( (Tr + 4) + n(n-1)4))1g(t)1 ( 2 n=O (n+1) n 1+It1 n ) < 1 for every t E SinceIg(t)1 ( 2 [-1,1] and there exists n0 such 2 2 that for every n > n 1 + (2 + +71-On + 2n < 3(n+1) we have 0' 111 1 1 3 2 (1+2 t )n < 21 n=n 4 (1 + (2 + 72-On + 2n )1g(t)1n 2 0 (n+1) n=no (n+1) CO 1 As 2 is convergent, A is uniformly convergent on [-1,1] and n=0 (n+1) so g(z,t) c A. Therefore 00 4)(g) = 1 g4(t)n(cP(z))n 'LIn=0 (n+1) 6 65 00 I is convergent. Hence kg(n)! < 1 and so and so 6 (g(11))nel n=0 (n+1) Id < 2 • (ii) Let n = 0. To prove Iiil < 2— it is enough to show that for every < l+c ICI — . For a given c > 0 define > 0, — 2 < < 1) 2c IT - - COS - (-E < t < e) a (t) = 2c -t (-1 < t < -c). Clearly a(t) has all the properties required in case (1). We define g(t) & g(z,t) quite in the same way as (i) and conclude that n n 1 (g(n)) C is convergent and so ICg(n)I < 1. Since (n+1) a(0) = c 2c— < E we get + a(n) 1 + a(0) 1 + c ICI < 2 2 2 I Therefore ICI < . This completes the proof of MA .2.1 X. = {(z,t t 1 6. It is clear that for each t0 c [-1,11, Pt 0) c X: Izi < 1+ I 01,1 0 is a peak set of A and the peaking function is f(z,t) = 1 - (t-t0)2 2 4 n Let f(z,t) = )7 a (t)z A and consider the function n=0 n 00 n f(z,t a (t )z 0 ) n 0 n=0 which is a power series of z. Since lan(to)znI < la (t ) 1 (1+I t01) n o 2 on Pt and the series 0 1+ I t 1 0 n Ia (t )1( n 0 n=0 2 66 is convergent, CO a (t )zn n 0 n=0 is uniformly convergent on Pt and so f(z,t0) is a continuous function 0 on P and analytic on the interior of P t t which is 0 0 1+Itol i(z,t0) E X: IZI < 2 1. Therefore f(z,t0) attains its maximum modulus only on 1+It I bP = {(z,t0) IzI = 0 } t0 2 and so 1+1t1 S (A,X) Cox = {(z,t) c X: 1z1 = o 2 }. 1+It I 140 It is easy to show that if x = ( 2 ° e t ) c bP o o t is a peak O point of A, then any other point of bPt is also a peak point of A. LO 1 7. The point( 0) is not a peak point of A. Proof. Let 1'2 0) = 1'2 0, 0) be a peak point of A and CO Co f(z,t) = f(x,y,t) = a n n Y- n(t)z = > an(t)(x + iy) n=0 n= 1 ' be the peaking function; i.e. f(--- 0) = 1 and If 1 1 } 2 < 1 on X`\{(.-2' 0) . Let (1,m,n) be the unit element of an arbitrary direction through ( 0, 0). Suppose (Z,t) = (x,y,t) = 1'2 0, 0) p(l,m,n) (p > 0) be an arbitrary point on the fixed line (1,m,n) which lies in X so that f(x,y,t) depends only on P and hence we define 1 F(p) = f(-2 + pl,pm,pn). 67 Since f is a continous function of p, F is also a continuous function of p. We now prove F'(p) & F"(p) exist and they are continuous. Since f c A, the series Co 11 1+ I tl n (n+1)2( lan(t) I + (t) + 1)( 2 ) n=0 is uniformly convergent on [-1,1]. Therefore Co n-1 f = Elna (t)(x+iy) , f 2 = n(n-l)a (t)(x+iy)n-2, x n n n=1 x n=2 n ft = ant (t)(x+iy) n=0 co n-1 ff= nal 2 anu(t)(x+iY) 11. xt = (t)(x+iy) t n=0 n=1 and similarly all the other partial derivatives of f, up to the second degree, exist and they are all continuous on X. Hence they are also continuous with respect to x, y and t. Since x = 1 pl, y = pm and t = Pn are continuous with respect to p, the partial derivatives are all continuous with respect to P. Therefore F'(p) & Fu(P) exist and they are continuous. Moreover we have F'(p) = 1 1 12 + pl,pm,pn) + mfy(7T + pl, pm, pn) + nft + pl, pm, pn), F"(p) = fx 12 + f lm + fxtln + f lm + fY 2 2 2 xy yx 2m + fytmn + ftxln + ftymn + f n . t2 2 2 Since 12 + m + n = 1 it follows from the above that IFH(P)1 < If + If 21 + If 21 2(1f If 1 If 1)- x2 1 xy 1 xt yt 68 On the other hand we have If 2 0 1+It1 n-2 < 2 (x"0 y 1 (n+1) lan(1( ) 411f11, x n=0 2 If 2(x,y,t)1 < l ar,,(t),(14.1 ti )n < 211f11 , t n=0 2 CO t1 n-1 Ifxt(x,Y,01 < nla;1(01(1+1 ) < 211fII. n=0 2 Since f = -f , f = if , f = if 2 2 2 yt we get x xy xt If y2 (x,y,t)I < 4I1f11, (x,y,t) = xy(x,y,t)I = If 2 Ifyt (x,y,t)I = Ifxt(x,Y,t)I Therefore IF"(P)1 <411f11 + 411f11 + 211f11 2 (4 11f1 1 +2 11f11+211f11) = 2611f11- Thus for any direction (l,m,n) through (-f, 0, 0) and every p > 0 we have IF"(p)1 < 2611f11. Since F"(p) is continuous F'(p) - F'(0) = f F"(r)dr and so defining 0 a(p) = F' (p) - F'(0) we get la (p) I < f IF" (r) I dr < 26pl Ifl I. 0 Since F' (p) is also continuous, by setting 8(p) = F(p) - F(0) - F' (0)p, we have 8(p) = fa(r)dr and so la(p)1 < f la(r)Idr < 13p211f11 • 69 Hence F(P) = F(0) + F' (0)P + f3(P) = F(0) + F' (0)p + 0(p2) and so 1 f(z,t) = f(x,y,t) = f(- + pl,pm,pn) = F(p) f (- 0, 1 2 12 ' 0, 0) + p(k x(- 12 ' 0 0) + f y(- 12 ' 0, 0)m + ft 0, 0)n) + 0(p ) = 1 ' f (- 0, 0) + (x (-1 ' 0, 0) + yfy 1 2 2 1x2 2 (-12 ' 0, 0) + tft(- 2' 0, 0) + 0(P ). Now consider the function f (z,0) = E a (0)zn. Since f e A n=0 n 0, 0, I n 1 n-1 Ian (0) , (T) & E In an(0) (-& n=0 n=1 are convergent and so .2E1 n-1 an(0) zn and )7- nan(0) z n=0 n=1 1 are uniformly convergent on the closed disk {z e C: lz I < .5. Hence f (z,0) is continuous on { Iz I <1 .0 and it has a continuous complex 4 n-1 1 derivative f' (z,0) = na (0)z on {I z I < 1 n=1 n 1 1 1 Since f peaks at (7-, 0), max If (z,0) I = f 0) = f 0, 0) = 1 1z1<1/2 1 and so by the lemma in example (2,3,1), 0) > 0. Since CO 1 1 f' (-2' 0) = na n(0) (12) n-1 = f x (-2' 0, 0) n=1 1 1 it follows that fx(7, 0, 0) = a > 0 and fy(f, 0, 0) = if x(7,1 0, 0) = ia. Now consider the function co 1 f (7, 0, t) E an (t) (-12-) n . n=0 1 1 Since f peaks at (7, 0, 0), h(t) = Re f (7, 0, t) attains its maximum 1 1 1 modulus only at (7, 0, 0) and so h'( 0, 0) = Re ft(.7, 0, 0) = 0. Hence 70 1 0, 0) = a'(0) (1)n ft 2 n 2 = ib n=0 where b is a real number. Consequently 1 2 f(x,y,t) = f(27, 0, 0) + (x - 2 + y(ia) + t(ib) + 0(p ) or 1 2, 2 112 f(z,t) = 1 + (z - 2-4 a + ibt + o(p ) (p = lz t2 ). l+t a 2 In particular f( 2—, t) = 1 + .T t + ibt + 0(t ) (0 < t < 1), i f(1+t, t.12 and so )1 = 1 + at + 0(t2) (0 < t < 1). Since a > 0 we can choose t small enough such that 1 + at + 0(t2) > 1 l+t t)I > 1 which is contrary to our hypothesis. and so for such t, 2 Therefore 1'2 0) cannot be a peak point of A and hence there is no 1 1, v. o which other peak point on bP0 = {(z,0) e X : Izi = Consequently P is a peak set of A does not contain any peak point of A. It-n 8. For every t0 0 (It0 I < 1) the point a0 = (1+2 I , t0 ) is a peak point of A. Proof. The peaking function is 2 -t ) 0 1 f(z,t) 4 )(2 + 1 + a(t) where (It I < t < 1) t 0 — — 21t01 I 0 1 I) (t) = It 1 cos (-It0 I —< t —< It 0 0 1 -t (-1 < t < -It0 I). 71 As we have already shown a'(t) & a"(t) exist and they are continuous on [-1,1] so that f c A. Since a(t ) = It01, f(141tol t0) = 1 and o 2 ' 1 since 10 < a(t) on [-1,1], I1-1-a(t)1 < —2 --o X. Therefore If(z,t)1 < 1 1 1+ 10 if either Izi < 2 ( 10 < 1) or t to. Now let t = t0 and Id . 1+1t01 1+1t01 Clearly i = 1 only for z = and so 2 +a ft ) 2 1+Itol If(z,t0)1 < 1 if lz = & z . Consequently If(z,t)I < 1 I Ifit°1°2 2 on X\faol; i.e. f peaks at ao. 1+ 10 As a result of 7 and 8, So(A,X) = f(z,t) c X: - , t Cl 9. 2 1+101. and so r(A,x) = S0(A,X) = AX = {(z,t) a X: 1z1 - Since A is the norm closure of polynomials in z and t, we have AC P (X) C- Kand so .27 = P(X). Hence r (P (X)) = NE-dc) = r (A,X) = AX and Ch(A,X) = Ch(K,X) = So(P(X)). But in fact So(P(X)) = AX. To prove this we say Po = f(z,0) e X: Izi < 24 is a peak set of A and so a peak P(X). Hence P contains a peak point of set of the uniform algebra F.= 0 A which is certainly on bP0 = {(z,O) c X: 1z1 = 2} and so any point of bP is a peak point of I= P(X). Therefore S (F,X) = S (P (X)) = AX. 0 0 0 1 10. P = f(z,0) c X: IzI < —4 is a minimal set of A. 0 2 Proof. Let P be a proper subset of Po which is a peak set of A and 1 f a A peak on P. Clearly PCbP0 = {(z,0) s X: = -J}, otherwise f(z,0) which is analytic on {(z,0) a X: 1z1 < 2} andcontinuous on Po 1 attains its maximum modulus at some point of C(z,0) c X: < -f4 which ie is impossible. Since A is invariant under the rotation (z,t) (ze ,t) we can assume, without loss ofgenerality, that (T, 0) c P. Thus f (2' 0) = 1 and If (z,t) I < 1 on X\P. 72 Now we can proceed in the same way as the proof of 7 to show that l+t l+t for small enough t> 0, If(-75--, t)I > 1. But for t > 0 (7F—, t) c xN,p 1+t and so t)1 < 1. Therefore P cannot be a peak set of A and so Po is a minimal (peak) set of A. 1 11. P0 = f(z,0) c X: Izi < .0 is the only peak set of A which has no peak point of A. In other words any other peak set of A has a peak point of A. Proof. Let P be a peak set of A such that P P0. Since Po is a minimal (peak) set of A, P Po = 0. But any peak set of A intersect the Shilov boundary of A, r(A,X) = S (A,X) U bP . Hence P fl S (A,X) 0; i.e. P 0 0 0 contains a peak point of A. 1 toi 12. For any t0 [-1,1], bPt = {(z,t0) c X: Izi = 1+I } is not a to 2 peak set of A. CO Let bP be a peak set of A and f(z,t) = a (t)zn c A peak Proof. t y-- 0 00 n=0 n n on bP . Consider f(zt ) = , - a (t )z which is continuously differentiable t )0 n o co n0 n-1 1+It I (t )z as a complex function on IzI < ° . Therefore fl(z,tz 0 ) = Inan 0 2 n=1 attains its maximum modulus on bP . Since f = 1 on bP , f'(z,t ) = 0 on t t z 0 0 0 bP and so f'(z,t ) = 0 on P . Hence f(z,t ) is constant on Pt whichwhich t0 z 0 t0 0 is contrary to our hypothesis. Consequently bPt is not a Peak set of A. 0 13. For every c(0 < e < 1) and every a(0 < a < 2w) the set 1+1-0 ia F = f( e , c X: c < Iti < 1) -ia 1 is a peak set of A. The peaking function is f(z,t) = ze 1+a(t) 73 (e < t < 1) 2E wt where a(t) = E - -COS - (-e < t < E) W 2E -t (-1 < t < -E). Clearly It1 < a(t), a(E) = e, a'(t), au(t) exist and they are continuous on [-1,1] so that f e A. Moreover f(z,t) = 1+ 1= 1 on F and for -e < t < 6, Iti < a(t) so that If(z,01 < 1 on this part of X. Now consider an arbitrary t such that e < It 1 < 1, then o 0 1 -ia 1 ze-la 1 ) = 1t 1 and on P . Since only for a(t o 1+aze(t )1 — 2 t 2 o 0 0 1+1to1 1+1t01 eta z = we conclude that 2 -ia If(z,t)1 - 11 + ze 1 < 1 2 1+a(t) , 1+1t01 eta, t )1. Consequently If(z,t)1 < 1 on on Pt \ t( 0 0 2 X\F and f(z,t) = 1 on F; i.e. F is a peak set of A. Since F is the disjoint union of the lines 1+1t1 ia 1+Iti ia F - ( e ,t) X: < t < 1) & F2 =4( e ,t) E X: -1 < t < 1 2 2 2 -e,} and since A is a natural Banach function algebra, by the Rossi's local peak set theorem, F1 & F2 are peak sets of A. Let t0 E E-1,11 and p e bPt . If t0 0 there exist E & a such that LO = {p}. Since F and 0 < e < 1, 0 < a < 2Tr and F (1 Pt are peak sets o o of A, p is a peak point of A. Therefore every point of AX\bPo is a peak point of A. In other words z i 1+1t1 S0(A,X) = {(Zit) E X: I 2 t This result was also proved in 8. 74 We now construct a non-natural Banach function algebra on the closed unit disk D with a minimal set which is non-trivial and hence a peak set without any peak points. Example (2.3.3) Let a(r) be defined on [0,1] as follows: Otherwise a(r) is strictly decreasing, continuous and (smoothly) differentiable at r = 2. Wecan consider a as a continuous function on the closed unit disk 5= {z s C: Izi < 11 by setting a(z) = a(Iz!) (z 6 50. Let A be the set of all functions of the form co co f (z) = aam(z)zn with la 12m+n < > m,n > m,n=0 m,n=0 m,n Clearly the above series is uniformly convergent on D and so f e C(D). It is easily seen that A is a function algebra on D. To define a norm for A we first prove the following lemma: Lemma. If CO m a a (z)zn E 0, m,n m,n=0 75 with CO y- 2min la 1 < CO m,n=O m,n then am,n = 0 for all m,n > 0. Proof. Clearly we have 00 CO 03 n n (2_\ ._ a a E ;> a m,naM(z)z m (z))z 0. m,n=0 n=0 m=0 m,n 1 2 ) we have Considering 1 - 2k1 -1 -< lzl < 1 - 1k2 ((k = _ am (1 ) n (? ,n k m )z = O. n=0 m=0 Setting 00 1 An = am,n(K.) , m=0 03 we say > A zn is uniformly convergent in the annulus n=0 n 1 1 < lz! < 1 - 1— and so it is analytic in 1 - 1 < 1z1 < 1 - 1 - 2k-1 - - 2k 2k-1 2k Since 03 A zn = 0 En=0 n in this annulus, by a theorem from analytic function theory, 03 1 )m A = >--- a (- n m=0 m,n k = 0 for all n > O. Now consider the series CO a zm m=0 m,n which is uniformly convergent in D and so 76 CO m A (z) = Y- a n m, n z m=0 is analytic in D for all n > 0. Hence for all n > 0, CO 1 )m A (0) = lim A (z) = lim A 1 = lim > a (-- = 0. n n z+0 k÷00 n k k4° m=0 m,n k Thus for every n > 0 the analytic function CO An (z)= a z y m,n m=0 1 1 vanishes on the set {0,1 ,...} which has a limit point z = 0 in D. Therefore, using the- same theorem from analytic function theory, for every n > 0, CO m A (z) = a z = 0 n / m,n m=0 on D and so a = 0 for all m,n > 0. This completes the proof of the m,n lemma. Thus each f e A has a unique series of the form 03 m n a a m,n z and so we can define a norm for A by m,n=0 Co 2m+n 11f11 = lam,n1 m,n=0 Quite in the same way as Example (2.3.1), we can show that the above norm is complete and so (A, 11'11) is a Banach function algebra on D. Clearly P = fz c D: IzI < 0 121 is a peak set of A and the peaking function is a e A. Since a(z) = 1 on P 0 for every f e A we have Co f(z) = a zn m,n m,n=0 on P0. As 03 a zn m,n=0 m,n 77 0 } is uniformly convergent on D, f(z) is analytic in 10 = {z e D: Izi < -21 1 and so it attains its maximum modulus (w.r.t. P0) bP = {z e D: I zI = 0 -42 Therefore P°0 does not contain any peak set of A and hence no peak point of A. Now assume PC bP0 is a peak set of A. Without loss of generality 1 we can suppose 2 P. Let 00 m n g(z) = Y- a a (z)z A m,n m,n=0 peak on P so that(2) = 1, g = 1 on P and Ig I < 1 on 15\ P. Then consider the function CO n > a z , h(z) = m,n m,n=0 which is analytic in D and non-constant on the closed disk P0 = {IzJ < 2}. . Since max Ih(z)1 = h(2) = g(-'p = 1, by the lemma in Example (2.3.1) IzI‹ 1/2 we have 03 n-1 h'z(2 = nam ,n• (12) > 0. m=0 n=1 Since 00 00 m n Mi-ri g = > a a z e A, < co m,n lam,n 12 m,n=0 m,n=0 1 = --and a = 1. Hence and so we can expand g in a neighbourhood of z0 2 0 CO ) g(z) = b (a(z) - 1)m 1n )-- m,n - m,n=0 78 1 with b = 1 and so in a neighbourhood of r0= 2we have 0, 0 Co 1n g(r) = b (a(r) - 1)111(r - 2 . m,n=0 m,n Now taking 0(r) = r, we can assume g.as a power series in a and fi; i.e. Co g(r) = g(a(r),0(r)) = bm n(a(r) - 1)111 0(0 - m,n=O 1 1 1 Since a and 13 are differentiable at r0= -2 and a'(-T) = 0, 07(-2) = 1 we have gr (2) = g(1 (4)a (22- ) + (14 = . CO 1 1 m 1 1)n-1 But gi(1,& = 2 nbm,n (a(- 2 ) - 1) (a(-2) — -2 = b m=0 0,1 n=1 1 and so 2 = /301*, Since g = h on Po = I --< 224 and h(r) = am,nrn m,n=0 is uniformly convergent in (0,1], we have Co )7 1 n-1 1 hr' ( 2 = nam,n (. 2) = h (-T) n=1 m=0 and 1 g(r) - g (-i) h(r) - h(-J.-) g'(1) = lim 1 2 1 - lim 1 - h' r(-) 2 . r41/2 r - -2- r41/2- r - y 1 1 Therefore b0,1 = gr' (7) = h' (-2 ) > 0. 79 1 Now Re g(r) = 1 + b0,1(r - + Re bio(a(r) - 1) + terms in higher powers of r - 1 and a(r) - 1. Hence [--. Re g(r)] dr r=1/2 = b0 ,1 > 0. But we assumed g peaks on P and so Re g(-1--) = 1 and IRe g(r)I < Ig(r)1 < 1 (0 < r < 1). Hence Re g(r) has a 1 maximum at r = -2-rand so [,a7. Re g(r)]r=112 = 0 which is incompatible with the preceding result. Consequently P cannot be a peak set of A and so the peak set Po does not contain a proper peak set of A; i.e. Po is a minimal set of A. In particular Po does not contain any peak point of A. Clearly Po has a non-empty interior and so a minimal set may have non- empty interior. To prove A is not natural it is enough to introduce a ¢ E MA such that ¢ ¢z for every z C16. We define the function ¢ : A÷C by CO a 2m+n $(f) = m,n m,n=0 where 0) m a a zn E m,n A. m,n= Since CO la 12m+n< Co, m m,n=OT ,n is well-defined on A and it is easy to show that ¢ c MA. Now let Z 6 D and consider the function a c A. Since a(z) < 1 and ¢(a) = 2 it follows that ¢ ¢ z and so ¢ is not an evaluation homomorphism in MA. Hence M A is not homeomorphic to D; i.e. A is not natural. Finally, note that A = [a,z]; i.e. A is generated by a and z. Hence MA = a (a,z) = f(cp(a)4(z)) c C2 A : ¢ EMA). 80 For each of the previous examples we introduced a peak set without any peak point. But in each case this peak set which has no peak point is connected and minimal. So one might think that in a Banach function algebra on a compact metrizable space, every peak set which has no peak point is minimal and connected. But this is not in fact true, even for a natural Banach function algebra. To show this we use the following example which was constructed by H.G. Dales [10]. Example (2.3.4) Define 0 : R R by 1 (x < 0) (x) = 2 1 - x (x > 0), and K = {(z,w) E C2: Izi < 13(1w1 - 1), ki < 21. Let D(K) be the algebra of all continuous complex-valued functions on K which have continuous second order complex derivatives on K with respect to z and w. Define a norm in D(K) by m+n K. 111E11 = ---T— m.n.1 I la m fn I I m+n<2 az 3117 (D(K),11-11) is a Banach function algebra on K. Now let D0 (K) be the subalgebra of D(K) which is generated by the coordinate functions z and w. (D0 (K) I I I ) is a natural Banach function algebra on K. P = {(z,w) c K: z = 1} = {(1,w) c C2: 1w1 < 1} is a peak set of A which +z has no peak point of A. Clearly the function f(z,w) - 12 peaks on P. P is a minimal set of A and S0 (D0 (K)) = {(z,w) c K: Izi = 0(1w! - 1),. 1 < e K: Izi = 0(1w1 - 1), 1 < 1w1 < 21. r(r)0 (K)) = 0 (p0 (K)) = {(z,w) 81 2 Now let Po = {(1,w)EC2: iwi < OUT(-1,w)EC:11< 11. 2 Clearly P is a peak set of A and the peaking function is f(z,w) 1+z 0 2 It is easy to see that Po does not contain any peak point, but it is neither connected nor minimal. However, in a natural Banach function algebra on a compact metrizable space, if a peak set without peak points is connected, then it might be minimal. 2-4 Examples of uniform function spaces As a result of previous sections, if A is a Banach function algebra on a compact metrizable space X, although go(A,X) = r(A,x), but So(A,X) may not be a boundary of A. On the other hand if A is a uniform algebra on the first countable space X, then So(A,X) is a boundary of A and so -S o(A,X) = r(A,X). As we indicated before (Theorem 16) for a Banach function space A on a metrizable space X, it is still true that -s o(A,x) = r(A,x), but So(A,X) may not be a boundary of A. Now the following question arises that: If A is a uniform function space on a compact metrizable space X, is So(A,X) a boundary of A? The answer is "No" as the following example shows: Example (2.4.1) izi < 1+1t1 1 Let X = {(z,t) eCxR= R3: , itl < 1} I — 2 and A be the Banach function algebra described either in Example (2.3.1) or (2.3.2). Now we define B to be the set of all functions of the form: f(z,t) = a + 6z + yt + 6t2. Clearly B(2 A and it is a function space on X. To prove B is uniformly closed let {f n }n=1 be a sequence in B such that 82 2 f (z, t) =a + 0 z+yt+ (St and fn n n flix 0 where f e C(X). 2 Hence for every (z,t) e X, a + ri z + y t + a t n n n n f(z,t) and so there exist complex numbers a, 0, y and 6 such that a al and 6 as n n n -4- 13. Yn Y n Now we define g(z,t) = a + 0z + yt + dt2. Clearly g e B and Ilfn-glIX = Han-a) + (13n-13)z (Yn-Y)t (6n-6)t211X 21 Ian-al 113 -61 0. nn-YI ISn n+oo Hence g = f and so f which is the uniform limit off n ln=1 is in B. Thus B is a uniform function space on X. Since f(z,t) = 1 - t2 e B peaks on Po = {(z,0) e X: IzI < 24 P is a peak set of B. But as we showed in Examples (2.3.1) and (2.3.2) 0 P 0 does not contain any peak point of A and so it does not contain any peak point of B. Therefore S0 (B,X) is not a boundary of B. In other words there is no smallest boundary of B. Note 1. As we proved in Corollary (2.1.4) in a uniform algebra any minimal set is trivial. But this is not true for uniform function spaces. In the above example Po is a minimal set of A, as it was shown in Example (2.3.2), and so Po is a non-trivial minimal set of B. Another property of the above example is that S (B,X) 0 Ch(B,X), because S0(B,X) is not a boundary of B whereas Ch(B,X) is a boundary of B. Note 2. As we stated in Theorem 33 in a uniform algebra the union of two peak sets is a peak set. But this is not true for a uniform function space. To show this we take B to be the uniform function space described in Example (2.4.1). Let 0 < t < 1 and consider the peak sets 0 83 1+It I o , 1+It I P = f(zt ) c X: < / & P o , t o -t - t(z t ) e X: Izl < 0 2 0 ' 0 2 of B. Let f(z,t) = a + 8z + yt + 6t2 peak on P = Pt 1.) P, so that f = 1 LO on P and Ifl < 1 on X\P. Thus a + 8z + yt0 + 6t0 = 1 for every l+to 2 z(lzi < ) and so R = O. Since we have also a - yt + (St = 1 it -----2- 0 0 2 02 follows that y = 0. Hence f is of the form f(z,t) = 1 + 6(t -t ) 2 2 where 6 C C. Let 6 = d + 162, then 11 + 6 (t -t0 )I < 1 for every t 1 1 2 2 2 2 where It' < 1 and t t0 . Thus for t - t0 > 0, Si < 0 and for 2 2 t - t0 < 0, 6 1 > 0 which is impossible. Therefore such a function f c B which peaks on P does not exist and so P = P UP is not a 0 to0 peak set of B. Note 3. We can also employ the example of Dales [10] which was described in Example (2.3.4) to show that the set of peak points of a uniform function space may not be a boundary. For example let B be the set of all functions of the form f(z,w) = a + 173z + yw. Quite in the same way as in Example (2.4.1) we can show that B is a uniform function space +z on K. Since f(z,w) = 12 E B peaks on P } 1 = {(1,w) e K: IwlI < 1 , P is a peak set of B. But P does not contain any peak point of D 1 1 0 (K) and so it does not contain any peak point of BCD0(K). Hence S0(B,K) is not a boundary of B. It is easy to prove, in the same way as in Note 2, that although P 1 = {(1,w) e K: 1w1 < 1} and P-1 = {(-1,w) e K: Iwl ) are peak sets of B, but P = P 1 P-1 is not a peak set of B. As a result of Theorem (3.3.3) in [21] we have: Proposition (2.4.2). Let A be a Banach function algebra on a compact metrizable space X. If X is infinite then r(A,x) and hence S0(A,X) are also infinite. 84 The above proposition is not true any longer if we drop the algebraic structure of A. To show this we give the following example of a uniform function space A on a compact plane set X such that the set of peak points of A is finite whereas X is infinite. Example (2.4.3) Let X = {(x,y) c C: -1 < x < 1, -1 < y < 1} and A be the set of all functions of the form f(z) = a + az on X where a,a c C. It is easy to show that A is a uniform function space on X. Clearly the maximum set Mz of the coordinate function z e A is z2,z3'z4 }where z = 1+i, z {zl, 1 2 = -1+i, z3 = -1-i and z4 = 1-i. Since the function f(z) = --(11 + c A peaks only at z = z 2 zk k (k = 1,2,3,4), zk is a peak point of A. Now we prove S0(A,X) = {zz2'z3'z4}. Since every f c A is analytic on X° T int(X), f attains its maximum modulus on bX. Suppose on the contrary zo = 1 + iy0 (-1 < y0 < 1) is a peak point of A and f(z) = a + az peaks at z0. Then a all iy0) = 1 and la + azI < 1 for z 0 zo (z c X). So f(z) = 1 - a(l+iyo) + a 1 + a(z - 1 -iyo) and hence f(z ) = f (l+i) = 1 + Si (1-y0 (1-y ) + ia 1 )= 2 0 1(1-y0), f(z ) = f(1-i) = 1 - a (1+y0) = 1 + S2 (1+y0) - ia (1+y ), 4 i 1 0 85 where a = a ia Since lf(z )1 < 1 and If(z4)J 1 2. 1 < 1 we get 11 - a (1 - y0) I < 1 and 11 + a2(1 y0)1 < 1. Since 0 < 1-y0 and 0 < 1 + y0 from the first inequality we conclude S2 > 0 and from the second one we get f32 < 0 which is impossible. Therefore z = 1 + iy (-1 < y < 1) cannot be a peak point of A. 0 0 0 Similarly we can prove z0 = -1 + iy0 (-1 < y0 < 1), z0 = x0 + i and z = x - i (-1 < x < 1) are not peak points of A and so the only peak 0 0 0 points of A are zl, z2, z3 and z4; i.e. So(A,X) = fzi,z2,z3,z41. It is also easy to show that P = {zz21 is not a peak set of A. If it is, then there exists f(z) = a + az e A such that f(zi) = f(z2) = 1 and so a + f3(1 + i) = a + 0(-1+i) = 1. Hence 13 = 0 and a = 1; i.e. f = 1 peaks on P = {zi,z2} which is impossible. 86 Chapter three Banach Function Algebras on Compact Plane Sets In this chapter we consider Banach function algebras of infinitely differentiable functions on perfect, compact plane sets. The subject of this chapter was introduced in the paper [11] of Dales & Davie. In this paper a natural quasianalytic Banach function algebra - is constructed and it is then used to give an example of a natural Banach function algebra such that the set of peak points is countable and so it is of first category in the Shilov boundary. In section (3-2) I use the same Banach function algebra but on the Swiss cheese so that although the set of peak points of it is not countable it is still of first category in the Shilov boundary. I also determine the maximal ideal space of certain Banach function algebras introduced in [11]. Throughout this chapter X will denote a perfect, compact plane set, otherwise it will be indicated. 3-1 Definitions and basic properties Definition (3.1.1). A complex-valued function f on X is called differentiable at z c X if o f(z) - f(z ) 0 ft (z ) = lim 0 z - z÷z 0o 0 (zeX) exists. f is differentiable on X if it is differentiable at every point of X. 87 Definition (3.1.2). The algebra of functions f c C(X) with continuous th k k derivative on X is denoted by D (X). Similarly D (X) is the algebra of functions with derivatives of all orders. Definition (3.1.3). Let {Mk}:=0 be a sequence of positive numbers such that Mk (r = 0,1,2,...,k). (1) M M > (r) r k-r r:(k-r)! lif (k) lix We define D(X,{Mkj) = {f c D (X) < CO } M k=0 k and for f c D(X,{Mic}) ilf(k)l ix 1 1f11 = > M k=0 k k For convenience, we regard D (X) as being an algebra of the type D(X,{Mk}) by setting Mr = r! (r = 0,1,...,k) and -7 5--=1 0 (r = k Hence for f e D (X), k ilfck)ll . 11f11= r!X r=0 Clearly (Dk(X).11'11k) and (D(X,{Mk}).11'11) are normed function algebras on X. We now introduce the type of compact set which we shall consider. A compact set X is rectifiably arcwise connected if any two points of X can be joined by a rectifiable arc lying within X. For such a set, let d(z ,z 1 2) denote the geodesic distance between z1 and z2•, the infimum of the lengths of arcs in X joining z1 and z2. Since X is compact, this infimum is attained. Clearly d defines a metric, the geodesic metric, on X. 88 Definition (3.1.4). A compact plane set is uniformly regular if it is perfect, rectifiably arcwise connected and if the geodesic metric is uniformly equivalent to the Euclidean metric on X. In other words, there exists a constant K such that for every z,w e X d (z,w) < Klz-w l. 1 According to a lemma from [11], D (X) C R(X) where X is uniformly regular and R(X), as usual, is the uniform closure on X of the rational functions with poles off X. Therefore D(X,{Mk}) C Dc°(X) C Dk(X) C Dl (X)C R(X). The following Theorem is also due to Dales & Davie [11]. Theorem (3.1.5). If X is a compact plane set which is a finite union of k uniformly regular sets, then D (X) (k = 1,2,...) and D(X,{Mk}) for co 1 satisfying (1) are Banach function algebras on X with the k k=0 appropriate norm. Definition (3.1.6). Let D be any of the preceding algebras with the appropriate norm. The subalgebra of D which is generated by the coordinate function z is denoted by DP and that one which is generated by the rational functions with poles off X that belong to D is denoted by DR. M = 1 and P = for k = 1,2,.... (we take P = co if M = m). Let Po k kk k k Clearly if Pk is bounded as k 00, none of the preceding algebras is 1 natural for any X. This follows from the fact that, if f (z) - a z-a (a / X), then f / D for a sufficiently close to X. We are, thus, a is unbounded as k co. We include the interested in the case when Pk k algebras D (X) under this lead. 89 The following theorem is due to Dales & Davie [11] and it is quite easy to prove. Theorem (3.1.7) Let X be uniformly regular and Pk + = as k 4 =. Then the Banach function algebra DR = DR(X,{Mk}) is natural; MD LI X. R Proposition (3.1.8) Let X be uniformly regular and Pk = as k =. Then X, {Mk}) = R(X) and hence ii(x,{M ER ( k}) = R(X). In particular Dk(X) = R(X). Proof. Since D1 (X) C R(X) it follows DR C R(X) and so FR C R(X). As Pk = we can easily show that every rational function with poles off X k+= belongs to DR and so the uniform closure on X of the rational functions with poles off X is contained in the uniform closure of DR; i.e. R(X) C 5R. Therefore B R(X). Corollary (3.1.9) r(DR) = R) = r(mx)) = bX, r(D) = NE) = NR 00) = bX, Ch(DR) = Ch (DR)= Ch(R(X)) = SD(R(X)) cbx, Ch(D) = Ch(D) = Ch(R(X)) = So(R(X))CbX. Proposition (3.1.10). If X is uniformly regular then for every sequence {vik})c:=10 satisfying (1), the Banach function algebra DP = Dp(X,fMk1) is uniformly dense in P(X); i.e. 51, = P(X) where P(X) is the uniform closure of polynomials on X. Proof. Let f c D and C > 0 be given. There exists a polynomial in z such that lif-Pilx < Ilf-PII < C where 11.11 is the norm of Dp C: D. Hence f E P(X) and so Dp c; P(X). On the other hand P(X) CI DP. Therefore 5P = P(X). 90 Corollary (3.1.11) r() ) = NET = Ch (DP) = Ch ) = S0 (P (X) ) = bX . Now to determine the maximal ideal spaces of DI, and D we prove the following general Theorem: Theorem (3.1.12) Let A and B be two normed function algebras on a compact Hausdorff space X such that Ac B, A = B and for every 4) c MB and every f c A, 1*(f)I < !Wile Then MB a M. Proof. Let 4) c M and f c B. There exists a sequence ff r in A such B n n=1 that Ilf -f11 0. Since I*(f n X ) - *(fm)I < Ilf -f II (m,n = 1,2,....) n4c° n n m X and !If n-fm 11X ---÷ 0, it follows that {*(f )1°) is a Cauchy sequence n,m4.0* n n= 1 of complex numbers and so we can define a function * on B by *(f) = lim *(fn). Clearly 4, is linear and multiplicative on B and since n400 4) is non-zero on B, 4, is non-zero on B. Since 1*(fn)I < IlfnlIx (n = 1,2,...) and IlfnIIX---11f11X we have 14/(f)I < 'kit and so n400 X = 4,(1) = 1. Therefore 4, c Now we define the map F: MB ÷ ME:by F(*) = 4,. Clearly F is 1-1 and onto. To prove the continuity of F, let *0 c MB and *0 = F(4)0). Suppose Nil, = {* c B I*(f ) - * (f )1 < c; 1 < k < n, f R. is a 'YO k 0 k k neighbourhood of *0 in Mg. For each k (1 < k < n) there exists gk s A such that 'if k-gk11 X < 4 . Now consider the neighbourhood 1 and let * c F(N4‘ ). Then 'f0 91 hP(fk) 4)0(fk) I lofk) 4) (g0 1 1 * (gk) *0(gk) I + 4 (gk) *0 (fk) ! C E < fk- g k X + l(P(gk) (P0(gk) + I Igk-fki lx` 71-+.Y + T E • Therefore 4 e N, and so F(N )C N ; i.e. F is continuous at 4) . Y 4)0 - 0 Thus F is a homeomorphism and so MB = If A = B in the above theorem, then we get the following important result: Corollary (3.1.13) Let A be a normed function algebra on X such that for every 40 c MA and every f c A, I4)(f)1 < Ilflix- Then MA =- M. particular if A is a natural Banach function algebra on X, then A is also natural (uniform algebra) on X but the converse is not true [see Example (3.1.17)]. Corollary (3.1.14) If X is uniformly regular and Pk 4 co as k 4 co, then M R, where X D is the polynomially convex hull of X. Proof. Since D and D R are Banach function algebras, for every f c D we have IIfII M = lllfn ll l!n = M D DP n As D is natural, 11;11 and so for every 4) 6 M and every R M = 'Id'X D DR f 6 D we have = 14)(f)I < limIlfnll l/n Ilfli'Ix n-*c° Therefore by Corollary (3.1.13), MD = MD . But by proposition (3.1.10) P P P(X) and so MD a MP(X) a X. 92 Corollary (3.1.15) Let A and B be two Banach function algebras on X such that A C B, = ii and I B —< I f IA for every f E A. Then if A is natural B is also natural but the converse is not true [see Example (3.1.17)]. Proof. Since A is a natural Banach function algebra, for every f c A = WWI ' IlfIlx - n+00 A b(0 1 < /n < lim rif Hence for every (I) c MB and f c A, k n111/n IB I n4c0 1I n4001 A . Therefore by Theorem (3.1.12) MB = lifIlX B Since A is natural, by the Corollary (3.1.13) T= B is also natural and so MB ; Mg= 1,4-7; X; i.e. B is natural. Now we can prove the following important result: Corollary"(3.1.16) If X is uniformly regular and Pk 4- 00 as k 0,, then D = D(X,fMk1) k is a natural Banach function algebra on X. In particular D (X) is natural for any finite k (k = 1,2,...). Proof. Clearly DR C D, DR = D= R(X) and for every f c DR,IIfil D 'I fllD R Since DR is natural, by Corollary (3.1.15) D is also natural. Example (3.1.17) ,1 < 14-1t1 Let X= {(z,t) 6 C X R: 1 — 2 , Itt < 1} and consider the Banach function algebras A & B which were constructed in Examples (2.3.1) and (2.3.2) respectively. Since B is natural and T. B = P(X), by Corollary (3.1.13) A is also natural whereas A is not natural. In Corollary (3.1.15) we take B = P(X) and A to be the Banach function algebra in Example (2.3.1). Clearly AT= P(X) and for every 93 f cA. Ilf11 B = I If I IX < 110A- But B = P(X) is natural whereas A is not. Proposition (3.1.18) Let X be uniformly regular and P co as k co. Then k Dp(X,{Mk}) = DR(X,{Mk}) if and only if X is polynomially convex. A Proof. If Dp = DR, MD = MD and so X = X; i.e. X is polynomially convex. Now let X = X and a X. Then z - a # 0 on X and since D is natural z - a is invertible in Dp. Hence every rational function with poles off X is in Dp. Therefore DR C Dp and so Dp = DR. The above proposition may still be true if we replace DR by D. But as far as I know it has not been proved in general. One can easily k k see that D (X) = D (X) if and only if X is polynomially convex. It is now convenient to give an example of a natural Banach function algebra A on the unit interval I = [0,1] such that A # C(I) and S0(A,I) = I. Example (3.1.19) Let X = I = [0,1] and A = Dl (X)= D1(I) = {f c C(I) : f' c C(I)1. As we showed in the previous parts of this section, A is a natural Banach function algebra on I = [0,1] under the norm IIfIIl =IIfIII + I Since I is polynomially convex in C we have D1(I) = DR(I) = 1 D (I) and hence A is generated by the coordinate function x; i.e. A = D1 (I) = [x]. Since A = P (I) = C (I) and A # C(I), A is not a uniform algebra on I. To prove S0(A,I) = I, let x0 c I. (i)If x0 = 0, then f(x) = 1 -xcApeaks atx = O. 0 (ii)If 1/2 < x0 < 1, then f (x) = sin -i--T2r 6 A peaks at x0. 0 (iii)If 0 < x0 < 1/2, we define 94 (0 < x < 2x0 sin 2x2 ) x0 0 f(x) = 12 xnx - -- sin — (2x0 < x < 1) 0 — — Since TTX — cos (0 < X < 2x 2x0 ) 2x0 0 f' (x) = -7 7X cos — (2x 2x x 0 < x < 1) 0 0 is continuous on I, f e A. Clearly f(x0 ) = sin 2 = 1, if(x)I < 1 for x x 0 < x < 2x0 and If(x)1 < 1 for 2x — — 2 0 < x < 1. Therefore f peaks at x0 and hence So(A,I) = I. 3-2 Peak points and the Shilov boundary First I quote some elementary definitions and theorems. Definition (3.2.1) Let X be a topological space. (i) A subset E of X is called nowhere dense (in X) if its closure has empty interior; or alternatively if (30\i) = X. (ii) A subset F of X is called first category in X if it is a countable union of nowhere dense sets in X. (iii)A subset S of X is called second category in X if it is not first category in X. The following Theorem is due to R. Baire [13]. Theorem (3.2.2) CO Let X be a locally compact Hausdorff space. If S = On where n=1 0n is open in X and O n= X (n = 1,2,....), then S = X. Corollary (3.2.3) If X is a locally compact Hausdorff space, it is of second category 95 in itself. Corollary (3.2.44 If X is a locally compact Hausdorff space, any dense Gs-set in X is of second category in X. Proof. Let S = O be a GS r) n -set in X and S = X. Then Sc = X\S = U Oc. n=1 n=1 Since S = X, (X\Tiric) = X\s,‹ = On = X and so Sc is of first category in X. If S is of first category in X, then X = SUSc is also first category in X which is impossible by Corollary (3.2.3). Therefore S is of second category in X. The following Theorem is also due to R. Baire [13]. Theorem (3.2.5) Let X be a locally compact Hausdorff space, then a set of first category in X has empty interior. Proposition (3.2.6) Let A be a uniform algebra on a compact metrizable space X. Then So(A,X), the set of peak points of A (w.r.t. X), is of second category in r(A,X), the Shilov boundary of A (w.r.t. X). Proof. It is well-known that So = s0(A,X) is a GS-set in X and CO s (A,X) = r(A,x). Suppose S = () O where O is open in X. Since 0 0 n n=1 n S C: r = 0 r(A,x), we have CO Co s So = o r = n (0 n r) = ' n=1 n nn=1 G where G n = On (1 r is open in r. Considering r as a compact metrizable space, by Corollary (3.2.4) So is of second category in r. 96 This proposition is not true for Banach function algebras. Dales & Davie [11] have given an example of a natural Banach function algebra A on an uncountable plane set X having only countably many peak points. Since this uncountable plane set X is perfect and r(A,x) = X, then So(A,X) is of first category in F(A,X). Since So(A,X) = NA,x) = X, by Corollary (3.2.4) and the fact that So(A,X) is of first category, So(A,X) is not a Ga-set in X. I have given another example of a Banach function algebra on the Swiss cheese such that the set of peak points is uncountable and it is of first category in the Shilov boundary. To construct this example we need the following definitions and Theorems which one can find in [11]. Definition (3.2.7) co Let A be a subalgebra of D (X) where X is a perfect, compact plane set. A is called a quasianalytic algebra if, for each zo e X the (k) conditions f e A and f (z ) = 0 (k = 0,1,2,...) imply that f = 0 in 0 some proper superset of {z0}. To construct a natural quasianalytic Banach function algebra we impose the following conditions on the sequence -CM lc° which was k k=0 introduced in Definition (3.1.3). SO M k 2 II. M .m I. m k+1 (k = 1,2,...). k=0 k+1 k < Mk-1 Clearly such a sequence exists. For example, take Mo = 1 and M = lc:- log 2 log 3 log (k+1) (k = 1,2,....). Theorem(3.2.8) [11] Let X be uniformly regular and the sequence {Kx}ee satisfy in k=0 (3.1.3), I, II and P m as k cc. Then D (X,{M }) is a natural k R k quasianalytic Banach function algebra on X. 97 Definition (3.2.9) Let X be a compact plane set and zo c bX, where bX is the topological boundary of X. Then z0 is strongly accessible from the complement of X if there exists a closed disk V such that V n x = {2 }. 0 Theorem (3.2.10) [11] Let X be a uniformly regular set, and let A be a natural quasi- analytic Banach function algebra on X containing the coordinate function z. Suppose that zo c bX and Ur) X\,{20} is connected for every U in some basic family of neighbourhoods of 20. Then z0 is a peak point of A if and only if z0 is strongly accessible from the complement of X. Note. If A is a natural quasianalytic Banach function algebra on the uniformly regular set X,which contains the coordinate function z, then A = R(X) and so r(A,X) = r(A,x) = r(R(X)) = bX. Now we are in a position which enables us to give the required example. But for convenience we first state the following lemma which is easily proved. Lemma (3.2.11) Let 21 and z2 be two arbitrary points on a circle. Suppose a(21,22) denotes the minimum of the lengths of the circular arcs joining z1 and z2. Then 6(2l' ) < 2Iz -2 I z2 1 2 ' Example (3.2.12) Let X be the Swiss cheese; that is let X be a compact plane set obtained from the closed unit disk B by deleting a sequence {Dk}71.1 of open disks, such that 50-) DQ = (I)(k / 0 and X = EAU Dk has an k=1 empty interior. For convenience take To = b-D- and Tk = bDk (k = 1,2,...). Clearly X is a perfect, compact plane set. To prove X is uniformly regular, let 21 and z2 be two arbitrary points in X and L be the line 98 segment joining z and z2. Then L intersects a sequence {Dk} 1 n=1 (n) (n) Suppose L intersects D at z of the disks Dk. 1 and z2 kn (n = 1,2,...). We now replace LD by the circular arc on the circle T k n kn ( n) ( ) and z and is of smaller length. If L is which is bounded by z1 2 n the length of this circular arc, then by the Lemma (3.2.11) (n)1 I. < 21i z(n) - z 1 (n = 1,2,...). Therefore if we replace Lil Dk n 1 2 n by the above mentioned circular arc for every n (n = 1,2,...) we get an arc in X joining z1 and z2 so that the length of this arc is less than 21z1-z21. Hence zi and z2 can be connected by a rectifiable arc in X and so X is rectifiably arcwise connected. On the other hand since (S(z1,z2), the geodesic distance between z1 and z2 is the infimum of the lengths of arcs joining z1 and z2, we have ,z ) < 21z -z ,z , E X). (z1 2 1 2 I (z1 2 Hence X is uniformly regular. bX = X. To prove, there exists a basic family Now let z0 ll (XN,{z0}) is connected {Ua}acA of neighbourhoods in X such that Ua 1 for every a 6 A, we take U = S(z0, where S(z0' 1 is a disk of radius n n n 1 — and centre z0 (n = 1,2,...). Clearly {U n }n=1 is a basic family of neighbourhoods of z0. Suppose z1 and z2 are two arbitrary points of U n (X \{z0}). The line segment [z1,z2] intersects infinitely many (k = 1,2,...). As before we replace L /1 D by the circular arc disks Dk k 1 on Tk = bDk which is inside Un = S(zo,Tand hence z1 and z2 can be joined by an arc in Un (tiX. If this arc passes through z0, we can always join z1 and z2 to a third point z3 E Tin() (X \{z0}) to get an arc in Lin n (X \{z0}) joining z1 and z2. Therefore Un (1(X\{zo}) is path-connected and hence it is connected. Now let z E U T . It is quite obvious that we can always find 0 k k=0 99 a disk 0}, and so z V such that V n X = {z 0 is strongly accessible from the complement of X. We now prove the other points of X which are deeper inside X are not strongly accessible from the complement of X. co Let z0 U Tk and V be a closed disk such that Tr\ X = {z0}. k=0 Since X is perfect, it has no isolated point and so z must be on the 0 boundary of V. Since V(1 X = {z0}, either VC: F1, for some k0 0 (k = 1,2,...) or VC. C`,..D. Hence z e T = bD or z c T = bD 0 0 k 0 k0 0 0 which is contrary to our assumption. Therefore such a disk idoes- not exist and so the only points of X = bX which are strongly accessible CO from the complement of X are the points of U T . k k=0 Now consider a natural quasianalytic Banach function algebra A on X such that z 6 A. For example take A = DR (X,{Mk}) where k)k=0 satisfies in the hypothesis of Theorem (3.2.8). Since for every m Z bX = X there exists a basic family fli 1 of neighbourhoods of z 0 6 n n=1 0 in X such that U (1 (XN\fz n 0 1) is connected for any n (n = 1,2,...), by Theorem (3.2.10) the only peak points of A are those which are strongly CO accessible from the complement of X. Therefore S0(A,X) = U T • k=0 Since Tk is a circle in the plane it is closed in X. Let 1 zo e Tk (k = 0,1,2,...) and N = S(z0,-11-)n X be a neighbourhood of z0 z in X. As for any n (n = 1,2,...) N o z contains points which are 0 not on T (k = 0,1,2,...), N k z is not contained in T for any n 0 k (n = 1,2,...). Hence for any k, Tk has no interior point in X and CO so T is nowhere dense in X. Therefore S0(A,X) = U T k k is of first k=0 category in r(A,X) = X. Since A c R(X), S0(A,X) C S0(R(X)). But R(X) is a uniform algebra on X and so by Proposition (3.2.6), So(R(X)) is of second category in moo) = bX = X. Hence S0(A,X) is a proper subset of S0 (R(X)); i.e. R(X) has peak points deeper inside X. 100 Chapter four Counter Examples In this chapter I shall give some examples to answer some of the questions which were raised in chapter one. I shall also prove the main result that M m, the maximal ideal space of H , is not first H countable. 4-1 Preliminaries Definition (4.1.1). Let H(D) be the class of analytic (holomorphic) functions on the open unit disk D = {z c C : 1z1 < 1}. For each f E H(D) and 0< r <1 we define 27 l/p M (r,f) = If(rei8)11) de , op < p < co) 27 0 Mw(r,f) = max If(z)I. 1z1 =r Now for 0 < p < 00 we define HP = {f e H(D) : sup M (r,f) < .1 0 Clearly H is the class of bounded analytic functions on D. For 1 < p < 00 it is quite well-known that HP is a Banach space under the norm lIfil = sup M (r,f) = lim M (r,f) (f c 0). In 0 1 1 f1 1 0. = sup If (z) 1 (f z ED 101 00 The Banach algebra H has been intensively studied in [17]. Let z c D and define 4)z(f) = f(z) for f c H. Clearly cpz c M and it is the evaluation homomorphism at z c D. If we take A = {4)z c M z c D}, it is easy to show that the map n: D A defined by u(z) = z is a homeomorphism and so D = A. Note that the topology of A is the weak*-topology that inherited from M . According to the corona theorem T= M ; i.e. M is the weak*-closure of A a D [see for example 14]. Definition (4.1.2). For 1 < p < co, tP is defined to be the class of all CO p sequences {z in C such that 2: iz I < 00. k}k=1 k t is defined to be k=1 the class of all bounded sequences {z0.17. 1in C. tr) is then a Banach space under the norm 'Ix!' = 1 1))1/P where x = {z )°3=1 c tip (1 < p < co). k=1 kk iis also a Banach space under the norm 11x11 = sup IZk,I where x = Czklk=1 k c Definition (4.1.3). A sequence {zk}7i =1 of complex numbers in the unit disk is said to be uniformly separated if there is 6 > 0 such that for every k (k = 1,2,...) zk z, ] > 6. j=1 1 z.zk j k We now quote the following theorems from [14]. 102 Theorem (4.1.4) If there is a constant M < 1 such that 1 - lzk+11 < M(1 - lzkl) (k = 1,2,...). Then fz lm=1 is uniformly separated. kk Definition (4.1.5). For a given sequence } in D, let T be the {zk k=1 linear operator on HP (0 < p < co) defined by T (f) = {(1 Iz 1 2)1/P f(zk ))k=1 P). k (f e H In particular T.,,(f) = ff(zk)173 ,„1 (f E 1r). In general T need not map H into 2)?. But, however, the following Theorem can be proved [14]. Theorem (4.1.6) (Main interpolation theorem) For 1 < p < co, T (HP) = Qp if and only if fz 1m=1 is uniformly kk separated. Theorem (4.1.7). f e H(D) is continuous in D and absolutely continuous 1 on T = bD if and only if f' e H . co 4-2 Concerning maximal ideal space of H Theorem (4.2.1). M am po, the maximal ideal space of H , is not first countable. Hence H is not countably generated; in other words H is not separable. 103 Proof. Assume on the contrary M . is first countable. Suppose ¢ is H an element of M . for which C = ¢(z) and 1C1 = 1, where z is the H coordinate function. Then ¢ has a countable base {N } n n=1 in M By the corona theorem M . is the weak* closure of A = D, so each Nn H has an element of A. Hence for every n (n = 1,2,...) there exists C e D such that ¢ e N and ¢ ---,s¢ in the weak* topology of M . n n C n+00 . SI n H ).___*¢(f) for every f e H . In particular Therefore ¢ (f) = “Cn n n-400 ¢ (z)---40(z) and so C --+ in the usual topology of the plane. n n+co n na. Now we construct a subsequence { } {CI1}:=1 such that zk 17=1 of 1+Izk1 < iz (k = 1,2,...). --- (1 - lz 1) or equivalently 1 - lzk+1 1 <— 2 k 2 I k+11i Take z = c and consider the disk S(C, 1-k11) Since c -4- c, there 1 1 , 2 n e S(C, ______J1-klI so that 1 +kll < 1c I. Taking z = c we exists C — ' n I. 2 n2 n2 2 2 2 1+1z 1-z,11 have 11 <1z,1. Now consider the disk S(C, 4- ), there exists 2 1 41 2 IC we have e S(C, 1-1z21) so that 1+1z21 < 1. Taking z3 = n3 2 2 n3 1+1z21 < lz 1. 1 3 1 If we continue in this way we get a sequence {z lm 2 k k=1 1 = C and 1+zkl < izk+11 (k = 1,2,...). such that zk n — I k 2 By Theorem (4.1.4) fzkl:=1 is uniformly separated and so by . Theorem (4.1.6) T.(H ) = Rte. Since {zkl:=1 is a subsequence of 1m=1 and f(C ) ---3 ¢(f) for every f e Ham, we have f(z ) 4)(f) n k fc nn n4.0 )}c° is convergent for every f e for every f e H. Thus {f(zkk=1 Since T.(H ) = R , it follows that every sequence in .2, is convergent, which is impossible. Therefore M . is not first countable. H If X is a compact metrizable space, then (C(X), 11'11 X) and hence any uniform algebra on X is countably generated. But this is not true for Banach function algebras as the following example shows: 104 Example (4.2.2). Let K = {z e C : Izi < 2:} and A = HM I K. Clearly if f & f 1-1°' and f IK = f2IK then f = f 1 2 1 1 2 and so there is a 1-1 correspondence between A and He°. Hence for each f e A there exists a unique g e H°5 such that f = g1K. Define IIfIl = suplg(z)1. it zED is easy to show that (A, 11'1I) is a Banach function algebra on K and it is in fact isometrically isomorphic to H . Hence MA -= M H' and since by Theorem (4.2.1) M m is not first countable, MA is not first countable too. Therefore A is not countably generated. We now give an example of a natural Banach function algebra on D = {z c C : 1z1 < 1} which is still not countably generated. Example (4.2.3) Let A = {f f' E Hc*}. Since Hm(= Hi, if f c A then f' c Hi. Therefore by Theorem (4.1.7) every f c A can be extended to a continuous function on D. So we can consider A as a function algebra on D so that A C A(D), where A(D) is the disk algebra. Since z c A, A(D) C 17 and so A = A(D). It is easy to show that A is a Banach function algebra on B under the norm IIfII (f 6 A). To prove MA LI miz, let 4) c MA and f c A. Then there exists M > 0 such that for large enough n -1 Ilfnll = Ilfnll w lInfi fn-1110, < nl IfI I! (11-11f11. + Ilf'11.) <. mnIlf11:-1 and so = lim ilfnill/n < ,11/n nl/nilf111-1/n. Jim p A n4m n4m 105 l/n l/n Since lim M = 1im n = 1 we have n÷m n-4-00 li:P(f)! < I Ifi IM < IlflI co = I If! 1E, - A Therefore by Corollary (3.1.13), MA ; M. But A = A(D) so that MA = Ec hence MA - D; i.e. A is a natural Banach function algebra on D. Since S 0 (A,D) = T = 0, any peak set of A contains a peak point of A. We now prove A is not countably generated. Let a e [0,2n) and define 1+eiaz is ga(z) e 1-e z (z a D). It is easy to show that ga e H and in fact Ilgali m = 1. Moreover for every a,13 E [0,27) and a / P. we have Ilga-0> 1. L . Define fa(z) = f (z a D). 0 Clearly f(4(z) = ga(z) a Hc° and since Ifa(z)I < Ilgall. = 1 for every z e D, it follows that fa e A. Since Ilga-0. > 1 we get > 1 Ilfa-f(3 II = Ilfa-f0c° II ilfa0coi-ft I Hence the open balls Ba = A : Ilf-fa ll < } (a e [0,27)) in A are mutually disjoint. Since the family fBa1ac[0,27) is an uncountable family of open balls in A, A cannot be separable and so it is not countably generated. 106 4-3 Examples of uniform algebras having certain properties If X is a non-metrizable compact Hausdorff space, we cannot construct a countably generated Banach function algebra on X. Otherwise MA and hence X would be metrizable. But if X is metrizable, then any uniform algebra on X is countably generated. Now a question arises at this point; whether any uniform algebra on a compact metrizable, or on a compact plane set, is finitely generated or not. As we know P(X) and R(X) are finitely generated on any compact subset X of Cn. But we do not know whether H(X) or A(X) are also finitely generated or not. We now give an example of a uniform algebra on a compact plane set which is countably generated but not finitely generated. Example (4.3.1) For M > 0 we define the functions F M and GM on the complex plane C by: r (0 < r < M) 2 i(e+2Tr(1— 2-2 M e )) ie F (z) = F (re ) - 22 (M < r < 2M) M e i8 0 (2M < r) , r (0 < r < M) 2 i(e+7(1- e -r GM(z) = G (rere M ) 22 i6 (M < r < 2M) M e (2M < r). Clearly FM and GM are both continuous on C. 107 Now we define for n = 2,3,.... fn (z)= F (z - -13-.7), gn(z) = G 1 1 71 (z 4n(n+1) 4n(n+1) and 1 1 D = {z e C : n IZ - nl— 2n(n+1) 1. It is easily seen that fn and gn are continuous functions on C and D D = (1)(m # n). Assuming that z - 1 = pe n m i0 define for n = 2,3,... 1 1 ) I 4(n+1)( P) (0 < p < 4n(n+1) — — 4n(n+1) f (z) = 1 1 1 (P- ) + 2(p )e ( p < 4n(n+1) 4n(n+1) 4n(n+1) 2n (n+1)) and complete the definition of fl on C by setting CO f (z) = z (z c C\ U D ) . 1 n n=2 It is also easy to show that fl is continuous on C. We now show that the functions fl, f2, g2, f3, g3, ... separate the points of D. co 1 (i) Let z1, z2 s (C \U D ) (U {7.11-}) and zi z2. Then fi(zi) fl (z2) n=2 n n=2 since on this set f (z) = z. 1 l (ii) Let z c D n■\ f l for some n > 2, and z2 1n l n = — or z2 lies outside Dn so that z z . Then f (z ) f (z ). 1 2 n 1 n 2 108 (iii) Let zl, z2 e DnViiil for n > 2 and zi z2. Then either fn(z1) fn(z2) or gn(z1) gn(z2). We now define A to be the uniform algebra on E. {z c c : 1z1 < 1} generated by fly f2, g2, f3, g3, • Let hi,h2,...,hN be non-constant functions in A. Clearly each h k (1 < k < N) is the uniform limit on D of a sequence of polynomials in f ,f ,g ,f 1 2 2 3 ,g3 ,... • Let 1 y = zec: lz - - 1 n 4n(n+1) (n > 2) and note that on yn the only functions from fl,f2,g2,f3,g3,... which do not vanish are f and g . Hence on y , each hk is a uniform limit n n n of polynomials in f and g (or rather their restriction to y ). But n n n for z c y (n > 2) n fn(z) = gn(z) 1 ie z - 1 ' n 4n(n+l) e and hence if f is a sum of terms each a product in h - h - hN(0) 1 1 of degree at least 2, then f(z)dz = 0. Yn Because on yn, f has the behaviour of a function with a single pole at z = 1of residue 0. Now any polynomial P in h ,h ,...,h can be written as the sum of 1 2 N such a function f and a function of the form ao + al (hl-h1(0))+ + aN(hm-hm(0)). 109 Thus by setting A = (h (z) h (0))dz k,n k k (k = 1,2,...,N; n > 2) Yn we have N P(z)dz = II a A (n > 2). k k,n k=1 Yn Consider the linear equations N+2 E A = 0 (k = 1,2,...,N) n k,n n=2 in the unknowns E E ...,E 2' 3' N+2. Since there are more unknowns than equations these equations have a non-trivial solution; i.e. one in which not all ofE2, 3 ,...,EN+2 are 0. Assume now that 2' denotes such a solution and define the one chain y by N+2 Y = E EnY • n=2 n Then for k = 1,2,...,N, N+2 N+2 J (hk(z) - hk(0))dz = T n f (hk(z) - hk(0))dz = o n=2 yn ;E:n=2 cnXk, = Hence, in the earlier notations, N+2 N+2 N N N+2 P(z)dz = P(z)dz = En A = a E A = 0, n j E ak k,n 2= k E n=2 y n=2 k=1 n k,n n k=1 n=2 110 and so f f(z)dz = 0 for every f lying in the uniform closure of the algebra generated by h1,h2,...,hm. Now, since (t,2,_3"—/EN+2) is a non-trivial solution of the equations N+2 = 0 (k = 1,2,...,N) nXk,n n=2 there exists m (2 < m < N+2) such that 74 0 and so N+2 f f (z)dz = E f f (z)dz = 0. m 7- n m y n=2 y n ,h2,...,hu Hence the above algebra generated by h1 does not contain fm and so this algebra must be different from A. As this is the case for any N (N = 1,2,....) it follows that A is not finitely generated. Consequently A is a uniform algebra on the closed unit disk D which is countably generated but not finitely generated. According to Theorem 8 (Chapter one) if A is a natural Banach function algebra on X, then any non-zero function f e A is invertible in A. Now a question may arise here that, if A is a Banach function algebra on X such that any non-zero element of A is invertible in A, then is A natural? The answer is 'No' as the following example shows: Example (4.3.2) 2 1 Let K = {(z,w) C : Iwl Izi < < 1}. Applying Theorem (3.5) from [15] we can show that K = R-hull(K) = {(z,w) e C2 : lzl < 1, lwl < 1} and = K and K K, P(K) is not natural. P(K) = R(K) = A(K). Since MP(K) Now let f e P(K) such that f # 0 on K. Since K is compact and e C(K), attains its maximum modulus somewhere on K. So there exists 6 > 0 such that 6 < Ifl on K. As f e P(K), for every e1 (0 < e1 < 6) 111 there exists a polynomial P in z and w such that Ilf-PII < e1. Hence K 0 < 8 e < If(z)1 - e < IP(z)1 for z e 1 1 K and so I 1 1 = If(z) - P(z)1 < El 1 (z E k). f(z) P(z) f(z) . P(z)I 6(6-E1) Given e > 0 we can choose e small enough such that 1 < e , and 1 8(8e ) 1 hence E. 11'f1 - 1-P 11 K < 1 1 Since P(K) = R(K), e P(K) and so —E P(K); i.e. f is invertible in P(K). As we see although every non-zero f in P(K) is invertible in P(K) but P(K) is not natural. Open question Let A be a natural uniform algebra on a compact Hausdorff space X, and let K be a closed subset of X. Suppose that x belongs to the interior of K (w.r.t. X) and that x is a peak point of A K. Is x a peak point of A? Proposition (4.3.3) The answer is 'Yes' if A = R(X) or A = A(X) for any compact plane set X. In particular if X is polynomially convex, the answer is 'Yes' also for A = P(X). Proof. Clearly R(X) and A(X) are natural uniform algebras on X C C. Suppose x belongs to the interior of K (w.r.t. X) and that it is a peak point of R(X) 1K or A(X) 1K. Since R(X) I KC R(K) and A(X) IK C A(K) it follows that x e bK. As x is an interior point of K X) there exists a disk S(x,e) such that S(x,e)() XCkand so x EbX. 112 Now let 1 1 E = z e C : < lz-x l < n 11+1 n } (n = 1,2,...). 2 2 Choosing k large enough, we have En C S(x,e) for every n > k. Thus E n xC S(x,e) f X C k and so EK = E \ X (n > k). Since n n (S (x, E ) x)° = s bc, f X° C K° it follows that E () X x° c K°, n C S (x , e) and so En\ = E \ X° (n > k) where K° and X° are the interiors of K and X respectively (w.r.t. C). By Melnikov's criterion (Theorem 32), if x is a peak point of R(X) TKCR(K) or A(X) j K C A(K), then n 2n y(E\ K) = +co or E 2 a(E \ K°) = +co respectively, n=1 n=1 where y is the analytic capacity and a is the continuous analytic capacity. Since En\ K = En:\ X and En\ = En\ X° (n > k) we have n 2ny(E \ X) = +co or xi 2 a(En\ X°) = respectively. n n=1 n=1 Therefore by the sufficiency of Melnikov's criterion, x is a peak point of R(X) or A(X) respectively and this completes the proof of the proposition. 113 Chapter five An Open Question on the Shilov Boundary of Banach Algebras As we know the disk algebra A(D) is a natural uniform algebra on D so that its Shilov boundary is the topological boundary of D; I' (A(D)) = T = b5= {z e C : Izi = 11. But there are natural uniform algebras on D such that T is a proper subset of their Shilov boundaries [see Example (5.1.1)). The question which is now arising is that: If A is a natural uniform algebra on D, does r(A,D) contain the topological boundary T of D? Does r(A,D) even meet T? Even Gamelin has raised the following more general question in [15): If A is a Banach algebra such that MA = D, then does r(A,D) contain T or does r(A,D) meet T? In this chapter we shall deal with the above questions, and we will try to find conditions on A such that the Shilov boundary of A is exactly T or contains T. 5-1 Preliminaries We first give an example of a natural uniform algebra A on D such that T is a proper subset of r(A,D). Example (5.1.1). Let U = S(O,p) = {z e C : Izi Take A = A(D,U) to be the uniform algebra of continuous functions on D which are analytic in U. According to Aren's theorem (Theorem 18), M = D. For r (p < r < 1) we define f by: A - r 114 z (Iz1 < r) fr r (z) rz i z i2 (r < Izi < 1), which is analytic in IzI < r and continuous on D. Hence f c A = A(D,U). It is easy to show that f r r attains its maximum modulus only on Id = r. In fact every point of IzI = r is a peak point of A. Since r is arbitrary every point of the annulus D\ U = {z e C p < Izi < 1} is a peak point of A and so So(A,E) = r(A,5) = 5\u. As we see T = 365 is a proper subset of r(A,E). We now state the following lemma which is quite well-known and one can find it in [13]. Lemma (5.1.2). Let M and N be subsets of Rn, bM and bN be the topological o o boundaries of M and N and M & N be the interiors of M and N respectively. If M = N then bM = bN and 140 = N0. Corollary (5.1.3). Let K be a compact plane set such that K o Then bK = T and K L' D. Hence by the Jordan curve theorem K is simply connected; i.e. K = K. The following proposition is also needed later on: Proposition (5.1.4). Let the Banach algebras A and B be isomorphic and r (A) = r(B). (A = B). Then MA :4 MB Proof. Let F : A+Bbe the isomorphism between A and B. Clearly F is -1 linear, multiplicative, 1-1, onto, continuous and F is also continuous. -I Define 7: MA -4- MB by r(c) = ¢ o F for (I) E MA. It is easy to show -1 that tp = (1) o F is a non-zero homomorphism on B. To prove 11) is -1 continuous on B, let g E B, then by the continuity of F there exists K such that 115 1 4) (g) 1 = 14)(F-1(g))1 < 11F-1(g) 1 1 A < KlIg l I B and hence 4) e MB. We can also show, in the usual way, that 7T is a homeomorphism and so MA MB. To prove7MAH=r0E0,lettP0=71.VETIMIMarldtbe a -1 90 neighbourhood of 4) in MB. Since w is continuous on MB'B, 1 N¢ - ) is a neighbourhood of 4)10 and so there exists f c A such 90 90 that f attains its maximum modulus only within N ¢ ; i.e. 0 !RUM = 11;1114430. Taking g = F(f) we have A 11;11 = maxiflg)1 = maxly5(F-1(g))1 = max14)(f)1 = 11 11 B tpeMB (j)cM M A 41614A A and liglim = max14)(g)1 = max14)(F-1(g))1 = max I (p(f) I = I rf 1 IiNNct , 0 0 "0 "0 TO Therefore ; i.e. 11;11m = 11g11N g attains its maximum modulus only A within N, and so 40 c r(B). Similarly we can prove every 4) e r(B) is "0 in Tr(r(A)) and hence Tr(r(A)) r(B) ; i.e. r(A) r(B). 5-2 Concerning finitely generated uniform algebras Let A = [f1,f2,...,fn] be a finitely generated uniform algebra on the compact Hausdorff space X. If K = {(fi(x),...,fn(x)) E Cn : x e X), then it is easy to show that A is isometrically isomorphic to P(K). and r(A,X) = r(P(K)). Therefore by Proposition (5.1.4), MA =- MP(K) a k Clearly the map 7: X ÷ K, defined by 7(x) = (fi(x),...,fn (x)) (x e X), is a homeomorphism and so X K. Therefore MA = X if and only if k = K. Similarly we can say, A = C(X) if and only if P(K) = C(K). 116 We now consider the following particular cases: I. Let X = I = [0,1]. Then K = J = f(fi(x),...,fn(x)) e Cn : x E I/ is a Jordan arc in Cn (n > 1). Thus the uniform algebra A = A on I is natural if and only if J = J. Similarly A = C(I) if and only if P(J) = C(J). Note that r(A,I) ; r(P(J)). A So far there is no example of a Jordan arc J such that J = J but P(J) C(J). Even it is conjectured that for every polynomially convex Jordan arc J, P(J) = C(J). J. Wermer has given an example of a Jordan arc J in C3 [25] such A that J is not rationally convex. Hence J J and so P(J) C(J). 2 Rudin has also given an example of a Jordan arc in C such that J is not A rationally convex. Therefore J J and P(J) # C(J). A If we take K = J where J is the Wermer's or Rudin's Jordan arc, then although K is polynomially convex, but P(K) C(K), otherwise P(J) = C(J) which is impossible. II. Let X = T = {z e C : Izi = 1}. Then K = J ={(fi(z),...,fn(z)) c Cn: Z E T} is a Jordan curve (simple closed arc) in Cn. Hence A = is natural on T if and only if J = J. Also A = C(T) if and only if P(J) = C(J). Note r(A,T) ; r(P(J)). In particular if A = [f]; i.e. if A is singly generated, then J is a Jordan curve in the complex plane A and so by the Jordan curve theorem J J and P(J) C(J). Therefore a singly generated uniform algebra on T cannot be natural and so A = [f] C(T). This means that C(T) is not singly generated, in fact C(T) = [z,Z] by the Stone-Weierstrass theorem. Now let J be a Jordan arc (Jordan curve) in Cn. There exist continuous functions fl,f2,...,fn on I(T) which separate the points of I(T) so that 7(x) = (fl(x),...,fn(x)) for x e I (x e T) is a homeomorphism between I(T) and J. Hence A = [f1,...,fn] is a uniform algebra on I(T) 117 and so A is isometrically isomorphic to P(J). Consequently MA a J and r(A) z r(P(J)). Since A = [fl,...,f n] is a Banach algebra, by the local maximum modulus principle (Theorem 7), if MA ; I then r(A,I) .1-I (if MA ; T then r(A,T) = T). As I = J (T = J) we get the following result from the above discussion: Proposition (5.2.1). If J is a polynomially convex Jordan arc (Jordan curve) in Cn then r(P(J)) = J. III. Let X = D = {z E C : IZI < 1}. The uniform algebra A = (f n on D is natural if and only if K = K where K = f(f (z),..., f (z)) 6 Cn : 1 n z E D}. If Tr: D -± K is defined by Tr (z) = (f1(z) fn (z) ). Then it is a homeomorphism and so Tr(T) = J is a Jordan curve which is contained in K. Since D ; K then K has empty interior (w.r.t. Cn) if n 12. Clearly r (A,15) if and only if J C r (p (K)) , T r (A,TD) (I) if and only if 17 n r (l() ) Since for n > 2, r(P(K)) is not known, we cannot conclude from the above discussion whether T C F(A,E), T rl r(A,F) (1) or not. If A = [f], then K = ff(z)6C:zeEI is a compact plane set and hence r(P(K)) = bK. Since K a D by Corollary (5.1.3) K = K and bK = T. Therefore MA a mp(K) I K I D and NA,B) r(P(K)) a bK. Hence A =. [f] is a natural uniform algebra on Esuch that r(A,D) = T. In the next section we try to extend this result for any singly generated Banach algebra or a nanach algebra having an element f such that f is 1-1 on MA. 118 5-3 Certain Banach algebras As we stated in (1.1.4), f(MA) = aA(f) and baA(f) C:i'(rtA)), where A is a Banach algebra and f e A. Proposition (5.3.1). If A = [f] is a singly generated Banach algebra, then bcA(f) = i'(r(A)). Proof. Let z = i(¢ ) E f(r(A)) but 2 ba (f). Since a (0 is a o 0 0 A A compact plane set, there exists e > 0 such that N = S(zo,e) CaA(f). zO -1 (f) ) is Since A = [f], f:M aA is a homeomorphism and so N = f z A 410 o a neighbourhood of (150 in MA. As cf)0 e r(A) there exists g 6 A such that O g attains its maximum modulus only within NC and hence Ilallm = IMIN00 -A ^-1 A Since f is continuous on K = aA(f) and g is continuous on A MA =f (aA(f)), it follows thath=gaf e C(K). To prove h e P(K), let e > 0 given. There exists a polynomial P such that IIP(f)-gli < E. Hence Ilh-PIIK = max,I h(z)-P(z)I = max1h(f(0))-P(f(0)1 = maxIg(0)-P(f)(0)1 zeK (1)cMA cf)cMA = Iri-P()111/ < II -P(f)II < c. A Therefore h e P(K). On the other hand 111111 K = max1h(z)I = max1h(f(4)))1 = maxIg(OI = IrglI , A MA zeK $EM (PeMA = max1h(2)1 = max1h(f(WI = maxI;MI = II;IIN zO zeN BEN OEN 0 0 0 0 119 Since Ws li = ti A ll1 from the above we get 111111 11911 1191,N., = 111111N ; mA K z IC) 0 i.e. h attains its maximum modulus within zN C:— Aa (f) = K. But 0 h c P(K) cannot attain its maximum modulus in the interior of K. So we get contradiction. Therefore z c ba o A(f) and so i'(r (A)) = baA(f). Corollary (5.3.2). Let A = [f] be a singly generated Banach algebra such that MA K where K is a compact plane set. Then r(A) = bK. Proof. Since f is 1-1, MA ; aA(f) and so ; K.K. Since WO is a compact plane set by Corollary (5.1.3) baA(f) = bK. Therefore by Proposition (5.3.1) NA) ; baA(f) = bK. Proposition (5.3.3) Let A be a Banach algebra having an element f such that f is 1-1 on MA. If MA = X where X is a compact plane set, then bX C r(A). Proof. As we stated before f(MA) = aA (f) and baA(f)C -- f(r(A)). Since f is 1-1 on MA, MA ,14 aA(f) and so aA(f) E-1 X. Therefore by Corollary (5.1.3) baA(f) :1-=.1 by. Hence baA(f) = NbX) C 'i(r(A)) and so bX (= r(A). Corollary (5.3.4). Let A be a natural Banach function algebra on the compact plane set K. If A contains a 1-1 function, then bKC r(A,K). Proposition (5.3.5). There exists a natural, finitely generated, uniform algebra on D which does not contain any 1-1 function. Proof. Let A = {f c A(5) : f'(0) = 0). By Cesaro means we can easily 2 3 show that A = [z ,z ]. Since z2 and z3 together separate the points of D, A is a uniform algebra on D which is generated by z2 and z3. Clearly T = S (A,T) = r(A,D). To prove the naturality of A let 4) c M and f c A. o A c° 2 3 Then there exists a sequence of polynomials {P } in z and z such n n=1 that IIP (z2,z3) - fl O. Since 1¢(z I 1 it follows n 3) I < IIz3I D 120 that 4)(z3) e T. We define w = (4)(z3))1/3 such that w2 = (4(z3))213 2 2 3 ) and hence 4)(f) = lim P (g ,w ) (w e 50. Since P 2,z3) 4(z n(z f(z) n- n n-)-co on B. we have f(w) = 4(f); i.e. 4) E MA is the evaluation homomorphism of some w e D. Thus M z D. A Now let f e A be a 1-1 function. Since f is analytic in D it is univalent and so f'(z) 0 for every z e D. But since f c A, f'(0) = 0 and so we get contradiction. Therefore A does not contain a 1-1 function. Proposition (5.3.6). There exist a singly generated Banach algebra B of continuous functions on D such that MB L." D but B does not contain any 1-1 function. 2 Proof. Take B = [z ] which is a uniformly closed subalgebra of A = [z2,z3]. Since z2 is not 1-1, B does not separate the points of D and so it is not a uniform algebra. But it is certainly a Banach algebra. Since A = [z2,z3] does not contain any 1-1 function, B also has not any 1-1 function. We now prove MB = D. Take f(z) = z2. Since B = [f], f is a homeomorphism and so MB ; uB(f). Since 14)(f)1 < 11f11D— 11z211D— = I we have 4)(f) = ((1)) E for every 4) e MB. Hence uB(f) CIET. Now let v...T e D and h e B, there exists a sequence of polynomials {P lc° of f = z2 such that n n=1 IIP -hi' 0. Since w112 E n m*00 h() = lim P (w) = h . n n4-03 ,Therefore we can define 4(h) = h()A7). It is easy to show that 4) E MB. Since 4(f) = f(47) = w, aB(f) = D. Hence MB Z T. Note that by Corollary (5.3.2), r(B,D) = T. In fact So(B,B) = T. 121 References 1. Arens, R. 'The maximal ideals of certain function algebras' P.J.M. 8(1958), 641-648. 2. Bauer, H. 'Siloischer Rand and Dirichletsches problem' Ann. Inst. Fourier, 11(1961), 89-136. 3. Bing, R.H. 'Metrization of topological spaces' Canad. J. Math. 3(1951), 175-186. 4. Bishop, E. 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