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Homomorphisms of Certain Banach Function Algebras Proc. Indian Acad. Sci. (Math. Sci.) Vol. 112, No. 2, May 2002, pp. 331–336. © Printed in India Homomorphisms of certain Banach function algebras F BEHROUZI Institute of Mathematics, University for Teacher Education, 599, Taleghani Avenue, Tehran 15614, Iran E-mail: [email protected] MS received 28 May 2001; revised 19 December 2001. n Abstract. In this note, we study homomorphisms with domain D .X/ or Lipα.X; d/ of which ranges are certain Banach function algebras and determine in which cases these homomorphisms are compact. Keywords. Compact homomorphism; semisimple Banach algebra. 0. Introduction Let X be a compact Hausdorff space and C.X/ be the Banach algebra of all continuous complex-valued functions f on X under the uniform norm kf kX = supx∈X |f.x/|.A subalgebra of C.X/ which separates the points of X, contains the constants, and which is a Banach algebra with respect to some algebra norm k·k, is a Banach function algebra on X. If the norm is equivalent to the uniform norm, then it is called uniform algebra. A Banach function algebra A on X is natural if MA, the maximal ideal space of A,isX; equivalently, if every character on A is given by evaluation at a point of X [3]. Let .X; d/ be a compact metric space and let α ∈ .0; 1]. The algebra of all complex- valued functions f on X for which |f.x/−f.y/| p .f / = x;y ∈ X; x 6= y < ∞ α sup dα.x; y/ : is denoted by Lipα.X; d/. These algebras are called Lipschitz algebras of order α and were first studied by Sherbert. The Lipschitz algebras Lipα.X; d/ for α ≤ 1 are natural Banach function algebras on X under the norm kf kα =kfkX+pα.f / [6,7]. For α = 1 we write Lip.X; d/ instead of Lip1.X; d/. Let X be a perfect compact plane set. A complex-valued function f is called differen- tiable on X if at each point z0 ∈ X 0 f.z/−f.z0/ f .z0/ = lim :z∈X; z −→ z 0 z−z0 exists. n The algebra of functions onPX with continuous nth derivatives is denoted by D .X/. n n .r/ n f ∈ D .X/ kf k n = kf kX=r k·k n;D.X/ For , let D r=0 !. Clearly under the norm D is a normed function algebra. The algebra Dn.X/ is not necessarily complete. For this we consider a type of compact plane set. A compact plane set X is connected by rectifiable arcs if any two points of X 331 332 F Behrouzi can be joined by a rectifiable arc lying within X. For such a set, let δ(z1;z2/denote the geodesic distance between z1 and z2, infimum of the lengths of the arcs joining z1 and z2. Clearly, δ defines a metric, the geodesic metric, on X. A compact plane set X is uniformly regular if it is perfect, connected by rectifiable arcs and if the geodesic metric is uniformly equivalent to the Euclidean metric on X. Thus, if X is uniformly regular, there is a constant C such that δ(z1;z2/≤C|z1 −z2| .z1;z2/∈X: (∗) n The algebras .D .X/;k·kDn/; n = 1; 2;::: were introduced by Dales and Davie [2] in 1973. They proved that when X is uniformly regular, these algebras are natural Banach 1 0 function algebras, also for z1;z2 ∈X; f ∈ D .X/; |f.z1/−f.z2/|≤Ckf kX|z1−z2|, where C is the same constant in (∗) [1]. Let A; B be commutative semi-simple Banach algebras and T : A −→ B be a homo- ∗ ∗ morphism. As we know, T , the adjoint of T , maps MB into MA ∪{0}.IfT maps MB into ∗ c ˆ ˆ MA and ' = T |MB , then Tf = f ◦' for every f in A, where f is the Gelfand transform of f . In this case we say T is induced by '. For example, if T is onto or A; B are unital ∗ with identities 1A; 1B respectively and T.1A/=1B, then T maps MB into MA. 1. Homomorphisms with domain Lipα (X, d) Theorem 1. Let B a Banach function algebra on a compact Hausdorff space X such that ˆ kf kMB =kfkX. Then the following statements are equivalent: (i) Inclusion map i : B −→ C.X/ is compact. (ii) Every homomorphism T from B into a uniform algebra A, induced by ', is compact. Proof. (i)H⇒ (ii): Let T : B −→ A be a homomorphism induced by ', where A is Y {f }∞ a uniform algebra on compact Hausdorff space . Assume that n n=1 be a bounded B {f }∞ C.X/ sequence in . Then there exists subsequence nk k=1 such that it is convergent in . {Tf }∞ A It is sufficient to show that nk k=1 is a Cauchy sequence in .Wehave kTf − Tf k =kTfc − Tfc k nk nj Y nk nj MA ˆ ˆ =kfnk ◦'−fnj ◦'kMA ˆ ˆ ≤kfnk −fnjkMB =kfnk −fnjkX: So kTfnk − Tfnj kY −→ 0ask;j −→ ∞ , that is {Tfnk } is a Cauchy sequence in A. (ii)H⇒ (i): is trivial. Theorem 2. Let .X; d/ be a compact metric space. Then every homomorphism from Lipα.X; d/ into uniform algebra A is compact. Proof. As we know, Lipα.X; d/ is natural Banach function algebra. So by Theorem 1, it is sufficient to show that every bounded sequence in Lipα.X; d/ has a subsequence which C.X/ {f }∞ Lip .X; d/ is convergent in . For this, let n n=0 be a bounded sequence in α . Then {f }∞ X n n=1 is uniformly bounded and it also is equicontinuous on . So by the Arzela–Ascoli Theorem, {fn} contains a uniformly convergent subsequence. Homomorphisms of certain Banach function algebras 333 Sherbert in [6] proved that if .X1;d1/ and .X2;d2/ are compact metric spaces then T : Lip.X1;d1/ −→ Lip.X2;d2/ is a homomorphism if and only if Tf = f ◦ '.f ∈ Lip.X1;d1// where ' : X2 −→ X 1 satisfies d1.'.x/; '.y// ≤ Md2.x; y/ for some constant M>0, all x;y ∈ X2. Kamowitz and Scheinberg in [4] have also proved Theorem 3. Let .X1;d1/ and .X2;d2/ be compact metric spaces and let T : Lip .X1;d1/ −→ Lip.X2;d2/be a homomorphism induced by map ' : X2 −→ X 1 , that is Tf = f ◦ '. Then T is compact if and only if ' is supercontraction, i.e. d1.'.x/; '.y// lim = 0 as d2.x; y/ −→ 0 : d2.x; y/ Using this theorem we have Theorem 4. Let X be a uniformly regular set and let .Y; d/ be a compact metric space. If T : Lip.Y; d/ −→ D 1 .X/ is a compact homomorphism, induced by ' : X −→ Y , then ' is a constant map. Proof. Since X is uniformly regular, there exists a constant C>0 such that for every 1 0 1 f ∈ D .X/; z1;z2 ∈X; |f.z1/−f.z2/|≤Ckf kX|z2−z1|. Thus D .X/ ⊆ Lip.X; |·|/. 1 In fact, the norm k·kD1 of D .X/ and Lipschitz norm of Lip.X; |·|/are equivalent on D1.X/, and D1.X/ is a closed subalgebra of Lip.X; |·|/. Thus if T : Lip.Y; d/ −→ D 1 .X/ is compact, then T : Lip.Y; d/ −→ Lip.X; |·|/is also compact. So by Theorem 3 we have d.'.z1/; '.z2// lim = 0as|z1−z2|−→0: |z1 − z2| Let z0 be a fixed point in X. We define the map F : Y −→ R by F.y/ = d.y;'.z0//. Since F ∈ Lip.Y; d/; F ◦ ' ∈ D1.X/. Moreover, for every z; w ∈ X we have F ◦ '.w/ − F ◦ '.z/ d.'.w/; '.z// ≤ : w − z |w − z| So .F ◦ '/0.z/ = 0. Since X is connected, F ◦ ' is constant. Hence for every z ∈ X; F ◦ '.z/ = F ◦ '.z0/ = 0. By the definition of F;'.z/ = '.z0/ for all z ∈ X, and proof is complete. 2. Homomorphisms with domain Dn.X/ Kamowitz and Scheinberg in [5] have proved that Theorem 5. Let A be a uniform algebra. Then every homomorphism from C1.[0; 1]/ into A is compact. The next result is the generalization of this theorem. Theorem 6. Let X be a uniformly regular set and let T be a homomorphism from Dn.X/ into a uniform algebra A. Then T is compact. 334 F Behrouzi Proof. Since X is uniformly regular there exists a constant C such that for every f ∈ 1 0 ∞ D .X/ and z1;z2 ∈ X; |f.z1/−f.z2/|≤CkfkX|z1−z2|. Let {fk}k= be a sequence n ∞ 1 D .X/ kf k n ≤ k = ; ;::: {f } in such that k D 1 for 1 2 . Thus k k=1 is uniformly bounded {f }∞ and equicontinuous. By Arzela–Ascoli Theorem there exists subsequence kj j=1 which is convergent in C.X/. By Theorem 1 every homomorphism from Dn.X/ into uniform algebra A is compact. Remark 7. Let X be a uniformly regular set and let T be a homomorphism from Dn.X/ into a uniform algebra A induced by '. Clearly, ' ∈ Aˆ and if A is a natural uniform n algebra, ' ∈ A. Conversely, assume ' ∈ A and '.Mˆ A/ ⊆ int.X/, since every f ∈ D .X/ is analytic on int.X/, by Functional Calculus Theorem, f ◦ ' ∈ A. Therefore the map S : Dn.X/ −→ A defined by Sf = f ◦ ' is a homomorphism. We shall require the following lemma which is contained in ([1] Chapter I of Part Six). Julia Lemma. Let U¯ be the closed unit disc in C and let ' be a continuously differentiable ¯ ¯ 0 function on U.If|'.c/|=k'kU¯ for some c ∈ U, then either ' is constant or ' .c/ 6= 0.
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