# Unital Compact Homomorphisms between Extended Analytic Lipschitz Algebras

Authors: Davood Alimohammadi; Maliheh Mayghani
Journal: Abstract and Applied Analysis
 (2011)

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## Abstract

LetX and K be compact plane sets with K⊆X. We define A(X,K)={f∈C(X):f|K∈A(K)}, where A(K)={g∈C(X):g is analytic on int(K)}. For α∈(0,1], we define Lip(X,K,α)={f∈C(X):pα,K(f)=sup{|f(z)-f(w)|/|z-w|α:z,w∈K,z≠w}<∞} and LipA(X,K,α)=A(X,K)∩Lip(X,K,α). It is known that LipA(X,K,α) is a natural Banach function algebra on X under the norm ||f||Lip(X,K,α)=||f||X+pα,K(f), where ||f||X=sup{|f(x)|:x∈X}. These algebras are called extended analytic Lipschitz algebras. In this paper we study unital homomorphisms from natural Banach function subalgebras of LipA(X1,K1,α1) to natural Banach function subalgebras of LipA(X2,K2,α2) and investigate necessary and sufficient conditions for which these homomorphisms are compact. We also determine the spectrum of unital compact endomorphisms of LipA(X,K,α).


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## Body

## 1. Introduction and Preliminaries

We letℂ, 𝔻={z∈ℂ:|z|<1}, 𝔻¯={z∈ℂ:|z|≤1}, 𝔻(λ,r)={z∈ℂ:|z-λ|<r}, and 𝔻(λ,r)¯={z∈ℂ:|z-λ|≤r} denote the field of complex numbers, the open unit disc, the closed unit disc, and the open and closed discs with center at λ and radius r, respectively. We also denote 𝔻(0,r) by 𝔻r.LetA and B be unital commutative semisimple Banach algebras with maximal ideal spaces ℳ(A) and ℳ(B). A homomorphism T:A→B is calledunital if T1A=1B. If T is a unital homomorphism from A into B, then T is continuous and there exists a norm-continuous map φ:ℳ(B)→ℳ(A) such that Tf̂=f̂∘φ for all f∈A, where ĝ is the Gelfand transform g. In fact, φ is equal the adjoint of T*:B*→A* restricted to ℳ(B). Note that T* is a weak*-weak* continuous map from B* into A*. Thus φ is a continuous map from ℳ(B) with the Gelfand topology into ℳ(A) with the Gelfand topology.LetA be a unital commutative semisimple Banach algebra, and let T be an endomorphism of A, a homomorphism from A into A. We denote the spectrum of T by σ(T) and define(1.1)σ(T)={λ∈C:λI-Tis not invertible}.For a compact Hausdorff spaceX, we denote by C(X) the Banach algebra of all continuous complex-valued functions on X.Definition 1.1.
LetX be a compact Hausdorff space. A Banach function algebra on X is a subalgebra A of C(X) which contains 1X, the constant function 1 on X, separates the points of X, and is a unital Banach algebra with an algebra norm ∥·∥. If the norm of a Banach function algebra on X is ∥·∥X, the uniform norm on X, it is called a uniform algebra on X.LetA and B be Banach function algebras on X and Y, respectively. If φ:Y→X is a continuous mapping such that f∘φ∈B for all f∈A and if T:A→B is defined by Tf=f∘φ, then T is a unital homomorphism, which is called the induced homomorphism from A into B by φ. In particular, if Y=X and B=A, then T is called the induced endomorphism of A by the self-map φ of X.LetA be a Banach function algebra on a compact Hausdorff space X. For x∈X, the map ex:A→ℂ, defined by ex(f)=f(x), is an element of ℳ(A) and is called the evaluation homomorphism on A at x. This fact implies that A is semisimple and ∥f∥X≤∥f̂∥ℳ(A) for all f∈A. Note that the map x↦ex:X→ℳ(A) is a continuous one-to-one mapping. If this map is onto, we say that A is natural.Proposition 1.2.
LetX and Y be compact Hausdorff spaces, and let A and B be natural Banach function algebras on X and Y, respectively. Then every unital homomorphism T:A→B is induced by a unique continuous map φ:Y→X. In particular, if X is a compact plane set and the coordinate function Z belongs to A, then φ=TZ and so φ∈B.Proof.
LetT:A→B be a unital homomorphism. Since A and B are unital commutative semisimple Banach algebras, there exists a continuous map ψ:ℳ(B)→ℳ(A) such that Tf̂=f̂∘ψ for all f∈A. The naturality of the Banach function algebra A on X implies that the map JA:X→ℳ(A), defined by JA(x)=ex, is a homeomorphism and so JA-1:ℳ(A)→X is continuous. Since B is a Banach function algebra on Y, the map JB:Y→ℳ(B), defined by JB(y)=ey, is continuous. We now define the map φ:Y→X by φ=JA-1∘ψ∘JB. Clearly, φ is continuous. Let f∈A. Since
(1.2)(Tf)(y)=Tf̂(ey)=(f̂∘ψ)(JB(y))=(f̂∘JA)(φ(y))=f̂(eφ(y))=eφ(y)(f)=f(φ(y))=(f∘φ)(y),
for all y∈Y, we have Tf=f∘φ. Therefore, T is induced by φ.
Now, letX be a compact plane set, and let Z∈A. Then φ=Z∘φ=TZ, and so φ∈B.Corollary 1.3.
LetX be a compact Hausdorff space, and let A be a natural Banach function algebra on X. Then every unital endomorphism T of A is induced by a unique continuous self-map φ of X. In particular, if X is a compact plane set and A contains the coordinate function Z, then φ=TZ and so φ∈A.Definition 1.4.
LetX be a compact plane set which is connected by rectifiable arcs, and let δ(z,w) be the geodesic metric on X, the infimum of the length of the arcs joining z and w. X is called uniformly regular if there exists a constant C such that, for all z,w∈X, δ(z,w)≤C|z-w|.The following lemma occurs in [1] but it is important and we will be using it in the sequel.Lemma 1.5 (see [1, Lemma 1.5]).
LetH and K be two compact plane sets with H⊆int(K). Then there exists a finite union of uniformly regular sets in int(K) containing H, namely Y, and then a positive constant C such that for every analytic complex-valued function f on int(K) and any z,w∈H,
(1.3)|f(z)-f(w)|≤C|z-w|(‖f‖Y+‖f′‖Y).LetX be a compact plane set. We denote by A(X) the algebra of all continuous complex-valued functions on X which are analytic on int(X), the interior of X, and call   it the analytic uniform algebra on X. It is known that A(X) is a natural uniform algebra on X.LetX and K be compact plane sets such that K⊆X. We define A(X,K)={f∈C(X):f|K∈A(K)}. Clearly, A(X,K)=A(X) if K=X, and A(X,K)=C(X) if int(K) is empty. We know that A(X,K) is a natural uniform algebra on X (see [2]) and call it the extended analytic uniform algebra on X with respect to K.Let(X,d) be a compact metric space. For α∈(0,1], we denote by Lip(X,α) the algebra of all complex-valued functions f for which pα,X(f)=sup{|f(z)-f(w)|/dα(z,w):z,w∈X,z≠w}<∞. For f∈Lip(X,α), we define the α-Lipschitz norm f by ∥f∥Lip(X,α)=∥f∥X+pα,X(f). Then (Lip(X,α),∥·∥Lip(X,α)) is a unital commutative Banach algebra. For α∈(0,1), we denote by lip(X,α) the algebra of all complex-valued functions f on X for which |f(z)-f(w)|/dα(z,w)→0 as d(z,w)→0. Then lip(X,α) is a unital closed subalgebra of Lip(X,α). These algebras are called Lipschitz algebras of order α and were first studied by Sherbert in [3, 4]. We know that the Lipschitz algebras Lip(X,α) and lip(X,α) are natural Banach function algebras on X.Let(X,d) be a compact metric space, and let K be a compact subset of X. For α∈(0,1], we denote by Lip(X,K,α) the algebra of all complex-valued functions f on X for which pα,K(f)=sup{|f(z)-f(w)|/dα(z,w):z,w∈K,z≠w}<∞. In fact, Lip(X,K,α)={f∈C(X):f|K∈Lip(K,α)}. For f∈Lip(X,K,α), we define ∥f∥Lip(X,K,α)=∥f∥X+pα,K(f). Then Lip(X,K,α) under the algebra norm ∥·∥Lip(X,K,α) is a unital commutative Banach algebra. Moreover, Lip(X,α) is a subalgebra of Lip(X,K,α); Lip(X,K,α)=Lip(X,α) if X∖K is finite, and Lip(X,K,α)=C(X) if K is finite. For α∈(0,1), we denote by lip(X,K,α) the algebra of all complex-valued functions f on X for which |f(z)-f(w)|/dα(z,w)→0 as d(z,w)→0 with z,w∈K. In fact, lip(X,K,α)={f∈C(X):f|K∈lip(K,α)}. Clearly, lip(X,K,α) is a closed unital subalgebra of Lip(X,K,α). Moreover, lip(X,α) is a subalgebra of lip(X,K,α); lip(X,K,α)=lip(X,α) if X∖K is finite, and lip(X,K,α)=C(X) if K is finite. The Banach algebras Lip(X,K,α) and lip(X,K,α) are Banach function algebras on X and were first introduced by Honary and Moradi in [5].LetX be a compact plane set. We define LipA(X,α)=Lip(X,α)∩A(X) for α∈(0,1] and lipA(X,α)=Lip(X,α)∩A(X) for α∈(0,1). These algebras are called analytic Lipschitz algebras. We know that analytic Lipschitz algebras LipA(X,α) and lipA(X,α) under the norm ∥·∥Lip(X,α) are natural Banach function algebras on X (see [6]).LetX and K be compact plane sets with K⊆X. We define LipA(X,K,α)=Lip(X,K,α)∩A(X,K) for α∈(0,1] and lipA(X,K,α)=lip(X,K,α)∩A(X,K) for α∈(0,1). Then LipA(X,K,α) and lip(X,K,α) are closed unital subalgebras of Lip(X,K,α) and lip(X,K,α) under the norm ∥·∥Lip(X,K,α), respectively. Moreover, LipA(X,K,α)=LipA(X,α)[lipA(X,K,α)=LipA(X,α)] if K=X, and LipA(X,K,α)=Lip(X,K,α)[lipA(X,K,α)=lip(X,K,α)] if int(K) is empty.The algebrasLipA(X,K,α) and lip(X,K,α) are called extended analytic Lipschitz algebras and were first studied by Honary and Moradi in [5]. They showed that the extended analytic Lipschitz algebras LipA(X,K,α) and lipA(X,K,α) under the norm ∥·∥Lip(X,K,α) are natural Banach function algebras on X [5, Theorem 2.4].Behrouzi and Mahyar in [1] studied endomorphisms of some uniform subalgebras of A(X) and some Banach function subalgebras of LipA(X,α) and investigated some necessary and sufficient conditions for these endomorphisms to be compact, where X is a compact plane set and α∈(0,1].In Section2, we study unital homomorphisms from natural Banach function subalgebras of LipA(X1,K1,α1) to natural Banach function subalgebras of LipA(X2,K2,α2) and investigate necessary and sufficient conditions for which these homomorphisms are compact. In Section 3, we determine the spectrum of unital compact endomorphisms of LipA(X,K,α).


## 2. Unital Compact Homomorphisms

We first give a sufficient condition for which a continuous mapφ:X2→X1 induces a unital homomorphism T from a subalgebra B1 of A(X1,K1) into a subalgebra B2 of A(X2,K2).Proposition 2.1.
LetXj and Kj be compact plane sets with int(Kj)≠∅ and Kj⊆Xj, and let Bj be a subalgebra of A(Xj,Kj) which is a natural Banach function algebra on Xj under an algebra norm ∥·∥j, where j∈{1,2}. If φ∈B2 with φ(X2)⊆int(K1), then φ induces a unital homomorphism T:B1→B2. Moreover, if Z∈B1, then φ=TZ.Proof.
The naturality of Banach function algebraB2 on X2 implies that σB2(h)=h(X2), where σA(h) is the spectrum of h∈A in the Banach algebra A. Let f∈B1. Since φ∈B2, φ(X2)⊆int(K1), and f is analytic on int(K1), we conclude that f is analytic on an open neighborhood of σB2(φ). By using the Functional Calculus Theorem [2, Theorem  5.1 in Chapter I], there exists g∈B2 such that ĝ=f∘φ̂ on ℳ(B2). It follows that
(2.1)g(z)=ez(g)=ĝ(ez)=f(φ̂(ez))=f(ez(φ))=f(φ(z))=(f∘φ)(z),
for all z∈X2 and so g=f∘φ. Therefore, f∘φ∈B2. This implies that the map T:B1→B2 defined by Tf=f∘φ is a unital homomorphism from B1 into B2, which is induced by φ. Now let Z∈B1. Then φ=TZ by Proposition 1.2.Corollary 2.2.
LetX and K be compact plane sets with int(K)≠∅ and K⊆X. Let B be a subalgebra of A(X,K) which is a natural Banach function algebra on X under an algebra norm ∥·∥B. If φ∈B with φ(X)⊆int(K), then φ induces a unital endomorphism T of B. Moreover, if Z∈B, then φ=TZ.Proposition 2.3.
Suppose thatαj∈(0,1], zj∈ℂ, 0<rj<Rj, Gj=𝔻(zj,Rj), Ωj=𝔻(zj,rj), Xj=Gj¯, and Kj=Ωj¯, where j∈{1,2}. Then for each ρ∈(r1,R1] there exists a continuous map φρ:X2→X1 with φρ(X2)=𝔻(z1,ρ)¯ such that φρ∈LipA(X2,K2,α2) and φρ does not induce any homomorphism from LipA(X1,K1,α1) to LipA(X2,K2,α2).Proof.
Letρ∈(r1,R1]. We define the map φρ:X2→X1 by
(2.2)φρ(z)={z1+ρ(z-z2)r2|z-z2|≤r2,z1+ρ(z-z1)|z-z2|r2<|z-z2|≤R2.
Clearly, φρ is a continuous mapping, φρ(X2)=𝔻(z1,ρ)¯, and φρ∈LipA(X2,K2,α2). We now define the function fρ:X1→ℂ by
(2.3)fρ(z)={ρ(z-z1)r1|z-z1|≤r1,ρ(z-z1)|z-z1|r1<|z-z1|≤R1.
Then, fρ∈LipA(X1,K1,α1). Since 0<r1r2/ρ<r2 and
(2.4)(fρ∘φρ)(z)={ρ2r1r2(z-z2)|z-z2|≤r1r2ρ,ρ(z-z2)|z-z2|r1r2ρ<|z-z2|≤R2,
we conclude that fρ∘φρ∉LipA(X2,K2,α2). Therefore, φρ does not induce any homomorphism from LipA(X1,K1,α1) to LipA(X2,K2,α2). Hence, the proof is complete.Corollary 2.4.
Suppose thatα∈(0,1], λ∈ℂ, 0<r<R, G=𝔻(λ,R), Ω=𝔻(λ,r), X=G¯, and K=Ω¯. Then for each ρ∈(r,R], there exists a continuous self-map φρ of X with φρ(X)=𝔻(λ,ρ)¯ such that φρ∈LipA(X,K,α) and φρ does not induce any endomorphism of LipA(X,K,α).We now give a sufficient condition for a unital homomorphism from a subalgebraB1 of LipA(X1,K1,α1) into a subalgebra B2 of LipA(X2,K2,α2) to be compact.Theorem 2.5.
Suppose thatαj∈(0,1], Xj and Kj are compact plane sets with int(Kj)≠∅ and Kj⊆Xj, and Bj is a subalgebra of LipA(Xj,Kj,αj) which is a natural Banach function algebra on Xj under the norm ∥·∥Lip(Xj,Kj,αj), where j∈{1,2}. Let φ:X2→X1 be a continuous mapping. If φ is constant or φ∈B2 with φ(X2)⊆int(K1), then φ induces a unital compact homomorphism T:B1→B2.Proof.
Ifφ:X2→X1 is constant, then the map T:B1→B2 defined by Tf=f∘φ is a unital homomorphism from B1 into B2 with dimT(B1)≤1, and so it is compact.
Letφ:X2→X1 be a nonconstant mapping with φ∈B2 and φ(X2)⊆Ω1. Then the map T:B1→B2 defined by Tf=f∘φ is a unital homomorphism from B1 to B2 by Proposition 2.1. To prove the compactness of T, let {fn}n=1∞ be a bounded sequence in B1 with ∥fn∥Lip(X1,K1,α1)≤1 for all n∈ℕ. This implies that {fn|K1}n=1∞ is a bounded sequence in C(K1) which is equicontinuous on (K1,d1α1). By Arzela-Ascoli’s theorem, {fn}n=1∞ has a subsequence {fnj}j=1∞ such that {fnj|K1}j=1∞ is convergent in C(K1). Since fnj|K1∈A(K1) for all j∈ℕ, {fnj|K1}j=1∞ is convergent in A(K1). By Montel's theorem, the sequences {fnj}j=1∞ and {fnj′}j=1∞ are uniformly convergent on the compact subsets of int(K1). Since φ(X2) and K1 are compact sets in the complex plane and φ(X2)⊆int(K1), by using Lemma 1.5, we deduce that there exists a finite union of uniformly regular sets in int(K1) containing φ(X2), namely Y, and then a positive constant C such that for every analytic complex-valued function f on int(K1) and any z,w∈φ(X2)(2.5)|f(z)-f(w)|≤C|z-w|(‖f‖Y+‖f′‖Y).
Therefore, there exists a positive constant C such that
(2.6)|fnj(φ(z))-fnj(φ(w))|≤C|φ(z)-φ(w)|(‖fnj‖Y+‖fnj′‖Y),
for all j∈ℕ and any z,w∈X2. Let j,k∈ℕ. Then, for all z,w∈K2 with φ(z)≠φ(w), we have
(2.7)|((fnj∘φ)-(fnjfnk∘φ))(z)-((fnj∘φ)-(fnjfnk∘φ))(w)||z-w|α2=|(fnj-fnk)(φ(z))-(fnj-fnk)(φ(w))||φ(z)-φ(w)|⋅|φ(z)-φ(w)||z-w|α2≤Cpα2,K2(φ)(‖fnj-fnk‖Y+‖fnj′-fnk′‖Y).
The above inequality is certainly true for all z,w∈K2 with z≠w and φ(z)=φ(w). Therefore,
(2.8)pα2,K2((fnj∘φ)-(fnjfnk∘φ))≤Cpα2,K2(φ)(‖fni-fnj‖Y+‖fni′-fnj′‖Y),
and so
(2.9)‖(fnj∘φ)-(fnjfnk∘φ)‖Lip(X2,K2,α2)≤(1+Cpα2,K2(φ))(‖fnj-fnk‖Y+‖fnj′-fnk′‖Y).
SinceY is a compact subset of int(K1), we deduce that the sequences {fnj}j=1∞ and {fnj′}j=1∞ are convergent uniformly on Y. Therefore, {fnj∘φ}j=1∞ is a Cauchy sequence on Lip(X2,K2,α2), that is {Tfnj}j=1∞ is convergent in Lip(X2,K2,α2). Hence, T is compact.Corollary 2.6.
Suppose thatα∈(0,1], X and K are compact plane sets with int(K)≠∅, and K⊆X. Let B be a subalgebra of LipA(X,K,α) which is a natural Banach function algebra on X with the norm ∥·∥Lip(X2,K2,α2), and let φ be a self-map of X. If φ is constant or φ∈B with φ(X)⊆int(K), then φ induces a unital compact endomorphism of B.Definition 2.7.
(a)
Asector in 𝔻(z0,r) at a point ω∈∂𝔻(z0,r) is the region between two straight lines in 𝔻(z0,r) that meet at ω and are symmetric about the radius to ω.(b)
Iff is a complex-valued function on 𝔻(z0,r) and ω∈∂𝔻(z0,r), then ∠limz→ωf(z)=L means that f(z)→L as z→ω through any sector at ω. When this happens, we say that L is angular (or non-tangential) limit of f at ω.(c)
An analytic mapφ:𝔻(z0,r)→𝔻ρ has an angular derivation at a point ω∈∂𝔻r(z0,r) if for some η∈∂𝔻ρ(2.10)∠lim⁡z→ωη-f(z)ω-z
exists (finitely). We call the limit the angular derivative of φ at ω and denote it by ∠φ′(ω).Lemma 2.8.
Let0<r≤1, and let φ:𝔻(z0,r)→𝔻ρ be an analytic function and ψ:𝔻→𝔻 defined by ψ(z)=(1/ρ)φ(z0+rz). Then φ has angular derivation at ω∈∂𝔻(z0,r) if and only if ψ has angular derivation at (ω-z0)/r∈∂𝔻. Moreover,
(2.11)∠φ′(ω)=rρ∠ψ′(ω-z0r).The following result is a modification of Julia-Caratheodory’s theorem. For further details and proof of Julia-Caratheodory’s theorem, see [7, pages 295–300].Theorem 2.9.
Take0<r≤1. Let φ:𝔻(z0,r)→𝔻 be a nonconstant analytic function and ω∈∂𝔻(z0,r). Then the following are equivalent:  (i)
liminfz→ω(∥φ∥𝔻r-|φ(z)|)/(r-|z|)=δ<∞,(ii)
∠limz→ω(η-φ(z))/(ω-z) exists for some η∈∂𝔻,(iii)
∠limz→ωφ′(z) exists and ∠limz→ωφ(z)=η∈∂𝔻.The boundary pointη in (ii) and (iii) is the same, and δ>0 in (i). Also the limit of the difference quotients in (ii) coincides with the limit of the derivative in (iii), and both are equal to ωη¯δ.Note that the existence of the angular derivativeφ at ω∈∂𝔻(z0,r), according to Theorem 2.9, is equivalent to liminfz→ω(∥φ∥𝔻(z0,r)-|φ(z)|)/(r-|z-z0|)<∞. In this case the angular derivative of φ at ω is nonzero.Proposition 2.10.
LetX be a compact plane set, and let 𝔻(z0,r)⊆X and K=𝔻(z0,r)¯. Suppose that c∈∂𝔻(z0,r) and φ∈LipA(X,K,1) is a nonconstant function such that |φ(c)|=∥φ∥𝔻(z0,r)¯. Then the angular derivative of φ at c exists and is nonzero.Proof.
LetΓ={z∈𝔻(z0,r):|z-c|/(r-|z-z0|)<2}. For every z∈Γ we have
(2.12)‖φ‖D(z0,r)-|φ(z)|r-|z-z0|=|φ(c)|-|φ(z)|r-|z-z0|≤|z-c|r-|z-z0||φ(z)-φ(c)||z-c|<2p1,K(φ).
Therefore, liminfz→ω(∥φ∥𝔻(z0,r)-|φ(z)|)/(r-|z-z0|)<∞, and, by Theorem 2.9, the proof is complete.Definition 2.11.
(a)
A plane setX at c∈∂X has an internal circular tangent if there exists a disc D in the complex plane such that c∈∂D and D¯∖{c}⊆int(X).(b)
A plane setX is called strongly accessible from the interior if it has an internal circular tangent at each point of its boundary. Such sets include the closed unit disc 𝔻¯ and 𝔻(z0,r)¯∖⋃k=1n𝔻(zk,rk), where closed discs 𝔻(zk,rk)¯ are mutually disjoint in 𝔻(z0,r).(c)
A compact plane setX has peak boundary with respect to B⊆C(X) if for each c∈∂X there exists a nonconstant function h∈B such that ∥h∥X=h(c)=1.Example 2.12.
The closed unit disc𝔻¯ has peak boundary with respect to A(𝔻¯) because, if c∈∂𝔻¯, then the function h:𝔻¯→C defined by h(z)=(1/2)(1+c¯z) belongs to A(𝔻¯) and satisfies ∥h∥𝔻¯=h(c)=1.LetX be a compact plane set. The algebra R(X) consists of all functions in C(X) which can be approximated by rational functions with poles off X. It is known that R(X) is a natural uniform algebra on X.Example 2.13.
LetX be a compact plane set such that ℂ∖X is strongly accessible from the interior. If R(X)⊆B⊆C(X), then X has a peak boundary with respect to B.Proof.
Letz0∈ℂ∖X. Since ℂ∖X is strongly accessible from the interior, for each c∈∂(ℂ∖X), there exists a δ>0 such that |c-z0|=δ and 𝔻(z0,δ)¯⊆int(ℂ∖X). Now, we define the function h:X→ℂ by
(2.13)h(z)=δ2(c¯-z0¯)(z-z0).
Then h∈B, ∥h∥X=h(c)=1.Theorem 2.14.
LetX1 be a compact plane set such that G1=int(X1) is connected, and G1¯=X1. Suppose that X1 has peak boundary with respect to LipA(X1,1). Let Ω1⊆G1 be a bounded connected open set in the complex plane, and let K1=Ω1¯. Let Ω2 be a bounded connected open set in the complex plane, and let K2=Ω2¯ such that K2 is strongly accessible from the interior. Suppose that X2 is a compact plane set such that K2⊆X2. If T:LipA(X1,K1,1)→LipA(X2,K2,1) is a unital compact homomorphism, then T is induced by a continuous mapping φ:X2→X1 such that φ is constant on K2 or φ(K2)⊆G1=int(X1). Moreover, φ=TZ.Proof.
SinceLipA(X1,K1,1) and LipA(X2,K2,1) are, respectively, natural Banach function algebras on X1 and X2, T:LipA(X1,K1,1)→LipA(X2,K2,1) is a unital homomorphism, X1 is a compact plane set, and Z∈LipA(X1,K1,1), we conclude that T is induced by φ=TZ and so φ∈LipA(X2,K2,1) by Proposition 1.2. Suppose that φ is nonconstant on Ω2. Since φ is analytic on Ω2, we deduce that φ(Ω2) is an open subset of X1 and so φ(Ω2)⊆G1. We now show that φ(K2)⊆G1. Suppose that φ(K2)⊈G1. Then there exists c∈∂K2 such that φ(c)∈∂X1. Since X1 has peak boundary with respect to LipA(X1,1), there exists a nonconstant function h∈LipA(X1,1) such that ∥h∥X1=h(φ(c))=1. We now define the sequence {fn}n=1∞ of complex-valued functions on X1 by fn(z)=(1/n)hn(z). Let n∈ℕ. Then
(2.14)‖fn‖X1=1n(‖h‖X1)n=1n,(2.15)p1,K1(fn)=sup⁡⁡{|hn(z)-hn(w)|n|z-w|:z,w∈K1,z≠w}≤sup⁡⁡{|h(z)-h(w)||z-w|:z,w∈K1,z≠w}≤sup⁡⁡{|h(z)-h(w)||z-w|:z,w∈X1,z≠w}≤p1,X1(h).
Thus
(2.16)‖fn‖Lip(X1,K1,1)≤1n+p1,X1(h)≤1+p1,X1(h),
by (2.14) and (2.15). This implies that {fn}n=1∞ is a bounded sequence in LipA(X1,K1,1). The compactness of homomorphism T implies that there exists a subsequence {fnj}j=1∞ of {fn}n=1∞ and a function g in LipA(X2,K2,1) such that
(2.17)lim⁡j→∞‖Tfnj-g‖Lip(X2,K2,1)=0.
This implies that
(2.18)lim⁡j→∞‖Tfnj-g‖X2=0.
On the other hand, we have ∥Tfnj∥X2≤1/nj for all j∈ℕ by (2.14). Hence,
(2.19)lim⁡j→∞‖Tfnj‖X2=0.
By (2.18) and (2.19), g=0. Therefore, by (2.17) we have
(2.20)lim⁡j→∞‖Tfnj‖Lip(X2,K2,1)=0.
This implies that
(2.21)lim⁡j→∞p1,K2(fnj∘φ)=0.
Assume that ε>0. By (2.21), there exists a natural number N such that for each j∈ℕ with j≥N(2.22)sup⁡⁡{|(fnj∘φ)(z)-(fnj∘φ)(w)||z-w|:z,w∈K2,z≠w}<ɛ.
In particular,
(2.23)sup⁡⁡{|((h∘φ)(z))nN-((h∘φ)(w))nN|nN|z-w|:z,w∈K2,z≠w}<ε.
This implies that
(2.24)1nNsup⁡⁡{|((h∘φ)(z))nN-((h∘φ)(c))nN||z-c|:z∈K2,z≠c}<ε.
Since c∈∂K2 and K2 is strongly accessible from the interior, there exists an open disc D=𝔻(z0,r) such that c∈∂D and D¯∖{c}⊆int(K2). Since φ is analytic on int(D¯)⊆int(K2) and h is analytic on φ(D¯)⊆int(X1), we deduce that h∘φ is analytic on int(D¯). On the other hand, we can easily show that
(2.25)p1,D¯(h∘φ)≤p1,X1(h)p1,K2(φ)<∞.
Therefore, h∘φ∈LipA(X2,D¯,1). Since ∥h∥X1=h(φ(c))=1, we conclude that
(2.26)(h∘φ)(c)=‖h∘φ‖D¯=1.
We claim that h∘φ is constant on D. If h∘φ is nonconstant on D, then, by Proposition 2.10, ∠(h∘φ)′(c) exists and is nonzero and since (h∘φ)nN(c)=1, (h∘φ)nN(c)∈∂D. If Γ is a sector in D at c∈∂D, then
(2.27)1nN∠lim⁡z→cz∈Γ|(h∘φ)nN(z)-(h∘φ)nN(c)z-c|≤ε
by (2.24). Thus
(2.28)1nN∠((h∘φ)nN)′(c)≤ε.
But
(2.29)∠((h∘φ)nN)′(c)=nN(h∘φ)nN-1(c)⋅∠(h∘φ)′(c).
Hence, by (2.28) we have
(2.30)∠(h∘φ)′(c)=1nN∠((h∘φ)nN)′(c)≤ε.
Since ɛ is assumed to be a positive number, we conclude that ∠(h∘φ)′(c)=0, contradicting to ∠(h∘φ)′(c)≠0. Hence, our claim is justified. Since φ is nonconstant on K2, φ is a nonconstant analytic function on connected open set D. This implies that φ(D) is a connected open set in the complex plane. This implies that h is constant on connected open set G1. The continuity of h on X1=G1¯ follows that h is constant on G1¯=X1. This contradicting to h is nonconstant on X1. Therefore, φ(K2)⊆G1.Corollary 2.15.
LetX be a compact plane set such that G=int(X) is connected and G¯=X. Let Ω⊆G be a bounded connected open set in the complex plane, and let K=Ω¯. Suppose that K is strongly accessible from the interior and X has peak boundary with respect to LipA(X,1). If T is a unital compact endomorphism of LipA(X,K,1), then T is induced by a continuous self-map φ of X such that φ is constant on K or φ(K)⊆G=int(X). Moreover, φ=TZ.Lemma 2.16.
LetG and Ω be bounded connected open sets in the complex plane with Ω⊆G, and let X=G¯ and K=Ω¯. Then for each c∈G∖K there exists a function fc∈LipA(X,K,1) such that fc is not analytic at c.Proof.
Letc∈G∖K. Then there is a positive number r such that
(2.31){z∈C:|z-c|≤r}⊆G∖K.
We now define the function fc:X→ℂ by
(2.32)fc(z)={z-cz∈X,|z-c|≥r,(1+r)(z-c)1+|z-c|z∈X,|z-c|<r.
It is easily seen that fc∈LipA(X,K,1) and fc is not analytic at c.Definition 2.17.
LetX and K be compact plane sets such that K⊆X. We say that K has peakK-boundary with respect to B⊆A(X,K) if for each c∈∂K there is a function h∈B such that h is nonconstant on K and ∥h∥X=h(c)=1.Example 2.18.
Letr∈(0,1] and K=𝔻r¯. Suppose that LipA(𝔻¯,K,1)⊆B⊆A(𝔻¯,K). Then K has peak K-boundary with respect to B.Proof.
We first assume thatr=1. If for each c∈∂𝔻¯ the function h:𝔻¯→ℂ defined by h(z)=(1/2)(1+c¯z), then h∈B, h is nonconstant on K=𝔻¯ and h(c)=1=∥h∥𝔻¯.
We now assume that0<r<1. For each c∈∂K, set z0=(1+r)c/r. Then z0∈ℂ∖𝔻¯. Define the function h:𝔻¯→ℂ by
(2.33)h(z)={-rc(z-z0)z∈D¯,|z-z0|≥1,-r|z-z0|c¯(z-z0)z∈D¯,|z-z0|<1.
It is easily seen that h∈LipA(𝔻¯,K,1) and ∥h∥𝔻¯=1=h(c).Lemma 2.19.
LetΩ be a connected open set in the complex plane, and let φ be a one-to-one analytic function on Ω. If f is a continuous complex-valued function on φ(Ω) and f∘φ is analytic on Ω, then f is an analytic function on φ(Ω).Proof.
By [8, Theorem 7.5 and Corollary 7.6 in Chapter IV], we deduce that φ(Ω) is a connected open set in the complex plane, φ′(z)≠0 for all z∈Ω, and φ-1:φ(Ω)→Ω is an analytic function on φ(Ω). Since f=f∘φ∘φ-1 on φ(Ω), we conclude that f is analytic on φ(Ω).Theorem 2.20.
LetX1 be a compact plane set such that G1=int(X1) is connected and G1¯=X1. Suppose that K1 is a compact subset of X1 such that Ω1=int(K1) is connected, K1=Ω1¯, and K1 has peak K1-boundary with respect to LipA(X1,K1,1). Let Ω2 be a bounded connected open set in the complex plane, and let K2=Ω2¯ such that K2 is strongly accessible from the interior. Suppose that X2 is a compact plane set such that K2⊆X2. If T:LipA(X1,K1,1)→LipA(X2,K2,1) is a unital compact homomorphism and TZ is one-to-one on Ω2, then T is induced by a continuous mapping φ:X2→X1 such that φ=TZ and φ(K2)⊆Ω1=int(K1).Proof.
SinceLipA(X1,K1,1) and LipA(X2,K2,1) are, respectively, natural Banach function algebras on X1 and X2, T:LipA(X1,K1,1)→LipA(X2,K2,1) is a unital homomorphism, X1 is a compact plane set, and Z∈LipA(X1,K1,1), we conclude that T is induced by φ=TZ and so φ∈LipA(X2,K2,1) by Proposition 1.2.
To proveφ(K2)⊆Ω1, we first show that φ(Ω2)⊆K1. Since φ is a one-to-one analytic mapping on Ω2, we conclude that φ(Ω2) is an open set in the complex plane. This follows that φ(Ω2)⊆int(X1)=G1 since φ(Ω2)⊆X1. Suppose that φ(Ω2)⊈K1. Then there exists λ∈Ω2 such that φ(λ)∈G1∖K1. By Lemma 2.16, there exists a function f∈LipA(X1,K1,1) such that f is not analytic at φ(λ). But f∘φ=Tf∈LipA(X2,K2,1), so that f∘φ is analytic on Ω2. Since f is continuous on φ(Ω2) and φ is a one-to-one analytic function on Ω2, we conclude that f is analytic on φ(Ω2) by Lemma 2.19. This contradicts to the fact f is not analytic at φ(λ)∈φ(Ω2). Therefore, φ(Ω2)⊆K1 so φ(Ω2)⊆int(K1)=Ω1 since φ(Ω2) is an open set in the complex plane. Since φ is continuous on K2, φ(Ω2)⊆Ω1, K2=Ω2¯, and K1=Ω1¯, we can easily show that φ(K2)⊆K1. We now show that φ(K2)⊆Ω1. Suppose that φ(K2)⊈Ω1. Then there exists c∈∂K2 such that φ(c)∈∂K1. Since K1 has peak K1-boundary with respect to LipA(X1,K1,1), there exists a function h∈LipA(X1,K1,1) such that h is not constant on K1 and
(2.34)‖h‖X1=h(φ(c))=1.
Applying the similar argument used in the proof of Theorem 2.14, we can prove that h is constant on K1. This contradiction shows that φ(K2)⊆Ω1.Corollary 2.21.
LetX be a compact plane set such that G=int(X) is connected and G¯=X. Let K be a compact subset of X such that Ω=int(K) is connected and K=Ω¯. Suppose that K has peak K-boundary with respect to LipA(X,K,1) and K is strongly accessible from the interior. If T is a unital compact endomorphism of LipA(X,K,1) and TZ is a one-to-one mapping on Ω, then T is induced by a continuous self-map φ of X such that φ=TZ and φ(K)⊆Ω=int(K).


## 3. Spectrum of Unital Compact Endomorphisms

In this section we determine the spectrum of a unital compact endomorphism of a subalgebra ofLipA(X,K,α) which is a natural Banach function algebra with the norm ∥·∥Lip(X,K,α).The following result is a modification of [9, Theorem 1.7] for unital compact endomorphisms of natural Banach function algebras.Theorem 3.1.
LetX be a compact Hausdorff space and B a natural Banach function algebra on X. If T is a unital compact endomorphism of B induced by a self-map φ:X→X, then ⋂n=0∞φn(X) is finite, and if X is connected, ⋂n=0∞φn(X) is singleton where φn is the nth iterate of φ, that is, φ0(x)=x and φn(x)=φ(φn-1(x)). If ⋂n=0∞φn(X)={x0}, then x0 is a fixed point for φ. In fact, if F=⋂n=0∞φn(X), then φ(F)=F.Theorem 3.2.
Suppose thatX is a compact plane set with int(X)≠∅, Ω is a connected open set in the complex plane with Ω⊆int(X), and K=Ω¯. Let B be a subalgebra of A(X,K) containing the coordinate function Z which is a natural Banach function algebra on X with an algebra norm ∥·∥B. Let T be a unital compact endomorphism of B induced by a self-map φ of X. If φ(X)⊆int(K) and z0 is a fixed point of  φ, then
(3.1)σ(T)={0,1}∪{(φ′(z0))n:n∈N}.Proof.
Clearly 0 and also1∈σ(T) since T(1X)=1X. If φ is constant then the proof is complete. Let λ∈σ(T)∖{0,1}. The compactness of T implies that there exists f∈B∖{0} such that Tf=f∘φ=λf. Since φ(z0)=z0∈int(K), f(z0)=0. We claim that f(j)(z0)≠0 for some j∈ℕ. If f(n)(z0)=0 for all n∈ℕ, then f=0 on an open disc with center z0 and so on Ω. By maximum modules principle, it follows that f=0 on X since φ(X)⊆Ω, λ∈ℂ∖{0}, and λf(z)=f(φ(z)) for all z∈X. This contradicts to f≠0. Hence, our claim is justified. Let m=min{n∈ℕ:f(n)(z0)≠0}. Then f(k)(z0)=0 for all k∈{0,…,n-1} and f(m)(z0)≠0. By m times differentiation of f∘φ=λf, we have (φ′(z0))mf(m)(φ(z0))=λf(m)(z0), and therefore λ=(f′(z0))m. Then σ(T)∖{0,1}⊆{(φ′(z0))n:n∈ℕ}.
Conversely, first we show that, ifλ∈σ(T) with |λ|=1, then λ=1. Let λ∈σ(T) and |λ|=1. The compactness of T implies that there exists g∈B∖{0} such that g∘φ=λg. It follows that |g∘φ|=|g|. Since φ(K)⊆int(K)=Ω and g is analytic on the connected open set Ω, we conclude that g is constant on Ω by maximum modules principle. Since φ(X)⊆Ω, g∘φ=λg, and λ∈ℂ∖{0}, we deduce that g is constant on X. Applying again g∘φ=λg implies that λ=1 since g∈B∖{0}.
We now claim thatφ′(z0)∈σ(T). If φ′(z0)∉σ(T), then there exists a nonzero linear operator S:B→B such that
(3.2)(T-φ′(z0)I)S=I.
Since Z-z01X∈B, h=S(Z-z01X)∈B and so
(3.3)h∘φ-φ′(z0)h=Z-z01X,
by (3.2). By differentiation at z0, we have
(3.4)0=h′(φ(z0))φ′(z0)-φ′(z0)h′(z0)=1,
this is a contradiction. Hence, our claim is justified.
We now show that(φ′(z0))n∈σ(T) for all n∈ℕ. If φ′(z0)=0 or |φ′(z0)|=1, the proof is complete. Suppose that φ′(z0)≠0 and |φ′(z0)|≠1. If (φ′(z0))j∉σ(T) for some j∈ℕ with j>1, then there exists a nonzero linear operator Sj:B→B such that
(3.5)(T-(φ′(z0))jI)Sj=I.
Since (Z-z01X)j∈B, hj=Sj(Z-z01X)j∈B and so
(3.6)hj∘φ-(φ′(z0))jhj=(Z-z01X)j,
by (3.5). By j-1 times differentiation at z0, we have
(3.7)hj(z0)=hj′(z0)=⋯=hj(j-1)(z0)=0,
and by j times differentiation at z0, we have
(3.8)0=(φ′(z0))jhj(j)(φ(z0))-(φ′(z0))jhj(j)(z0)=j!,
this is a contradiction. Thus, (φ′(z0))n∈σ(T) for all n∈ℕ. This completes the proof.Corollary 3.3.
LetB and T satisfy the conditions of Theorem 3.2, and let B be a natural Banach function algebra with the norm ∥·∥α,K. If F is a finite set such that φ(F)=F, then there exist z0∈F and m∈ℕ such that
(3.9){λm:λ∈σ(T)}={0,1}∪{(φm′(z0))n:n∈N}.Proof.
SinceF is a finite set and φ(F)=F, there exist z0∈F and m∈ℕ such that φm(z0)=z0. Since φ(X)⊆int(K), so z0∈int(K). If φ is constant, then the proof is complete. When φ is nonconstant, we define T̃:B→B by T̃f=f∘φm. Then T̃ is a compact endomorphism of B induced by φm by Corollary 2.6 and φm(z0)=z0. Therefore,
(3.10)σ(T̃)={0,1}∪{(φm′(z0))n:n∈N}
by Theorem 3.2. Since Tf=f∘φ and T̃f=f∘φm for all f∈B, we have T̃=Tm. By Spectral Mapping Theorem, σ(Tm)={λm:λ∈σ(T)}. Therefore,
(3.11){λm:λ∈σ(T)}={0,1}∪{(φm′(z0))n:n∈N}.Corollary 3.4.
Suppose thatX is a compact plane set with int(X)≠∅, Ω is a connected open set in the complex plane with Ω⊆int(X), and K=Ω¯. Take α∈(0,1]. Let φ be a self-map of X with φ(X)⊆Ω such that φ∈LipA(X,K,α), and let φ(z0)=z0 for some z0∈Ω. If T is a unital endomorphism of LipA(X,K,α) induced by φ, then T is compact and
(3.12)σ(T)={0,1}∪{(φ′(z0))n:n∈N}.



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*Source: 146758-2011-10-05.xml*