Unital Compact Homomorphisms between Extended Analytic Lipschitz Algebras

Abstract

Let 𝑋 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)∶𝑝_𝛼,𝐾(𝑓)=sup{|𝑓(𝑧)−𝑓(𝑤)|/|𝑧−𝑤|^𝛼∶𝑧,𝑤∈𝐾,𝑧≠𝑤}<∞} and Lip_𝐴(𝑋,𝐾,𝛼)=𝐴(𝑋,𝐾)∩Lip(𝑋,𝐾,𝛼). It is known that Lip_𝐴(𝑋,𝐾,𝛼) is a natural Banach function algebra on 𝑋 under the norm ||𝑓||_{Lip(𝑋,𝐾,𝛼)}=||𝑓||_𝑋+𝑝_𝛼,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 Lip_𝐴(𝑋₁,𝐾₁,𝛼₁) to natural Banach function subalgebras of Lip_𝐴(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 Lip_𝐴(𝑋,𝐾,𝛼).

1. Introduction and Preliminaries

We let ℂ, 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 𝔻_𝑟.

Let 𝐴 and 𝐵 be unital commutative semisimple Banach algebras with maximal ideal spaces ℳ(𝐴) and ℳ(𝐵). A homomorphism 𝑇∶𝐴→𝐵 is called unital if 𝑇1_𝐴=1_𝐵. 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 𝑇𝑓=𝑓∘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.

Let 𝐴 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 define𝜎(𝑇)={𝜆∈ℂ∶𝜆𝐼−𝑇isnotinvertible}.(1.1)

For a compact Hausdorff space 𝑋, we denote by 𝐶(𝑋) the Banach algebra of all continuous complex-valued functions on 𝑋.

Definition 1.1. Let 𝑋 be a compact Hausdorff space. A Banach function algebra on 𝑋 is a subalgebra 𝐴 of 𝐶(𝑋) which contains 1_𝑋, 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.

Let 𝐴 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.

Let 𝐴 be a Banach function algebra on a compact Hausdorff space 𝑋. For 𝑥∈𝑋, the map 𝑒_𝑥∶𝐴→ℂ, defined by 𝑒_𝑥(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≤‖𝑓‖_ℳ(𝐴) for all 𝑓∈𝐴. Note that the map 𝑥↦𝑒_𝑥∶𝑋→ℳ(𝐴) is a continuous one-to-one mapping. If this map is onto, we say that 𝐴 is natural.

Proposition 1.2. Let 𝑋 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. Let 𝑇∶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 𝑇𝑓=𝑓∘𝜓 for all 𝑓∈𝐴. The naturality of the Banach function algebra 𝐴 on 𝑋 implies that the map 𝐽_𝐴∶𝑋→ℳ(𝐴), defined by 𝐽_𝐴(𝑥)=𝑒_𝑥, is a homeomorphism and so 𝐽^𝐴−1∶ℳ(𝐴)→𝑋 is continuous. Since 𝐵 is a Banach function algebra on 𝑌, the map 𝐽_𝐵∶𝑌→ℳ(𝐵), defined by 𝐽_𝐵(𝑦)=𝑒_𝑦, is continuous. We now define the map 𝜑∶𝑌→𝑋 by 𝜑=𝐽^𝐴−1∘𝜓∘𝐽_𝐵. Clearly, 𝜑 is continuous. Let 𝑓∈𝐴. Since (𝑇𝑓)(𝑦)==𝑓𝑇𝑓𝑒𝑒_𝑦=𝑓∘𝜓𝐽_𝐵(𝑦)=𝑓∘𝐽_𝐴(𝜑(𝑦))_𝜑(𝑦)=(𝑓∘𝜑)(𝑦)=,𝑒_𝜑(𝑦)(𝑓)=𝑓(𝜑(𝑦))(1.2) for all 𝑦∈𝑌, we have 𝑇𝑓=𝑓∘𝜑. Therefore, 𝑇 is induced by 𝜑.
Now, let 𝑋 be a compact plane set, and let 𝑍∈𝐴. Then 𝜑=𝑍∘𝜑=𝑇𝑍, and so 𝜑∈𝐵.

Corollary 1.3. Let 𝑋 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. Let 𝑋 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]). Let 𝐻 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, ||𝑓(𝑧)−𝑓(𝑤)||≤𝐶|𝑧−𝑤|‖𝑓‖_𝑌+‖‖𝑓^‖‖_𝑌.(1.3)

Let 𝑋 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.

Let 𝑋 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 𝑝_𝛼,𝑋(𝑓)=sup{|𝑓(𝑧)−𝑓(𝑤)|/𝑑^𝛼(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+𝑝_𝛼,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)|/𝑑^𝛼(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 𝑝_𝛼,𝐾(𝑓)=sup{|𝑓(𝑧)−𝑓(𝑤)|/𝑑^𝛼(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+𝑝_𝛼,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)|/𝑑^𝛼(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].

Let 𝑋 be a compact plane set. We define Lip_𝐴(𝑋,𝛼)=Lip(𝑋,𝛼)∩𝐴(𝑋) for 𝛼∈(0,1] and lip_𝐴(𝑋,𝛼)=Lip(𝑋,𝛼)∩𝐴(𝑋) for 𝛼∈(0,1). These algebras are called analytic Lipschitz algebras. We know that analytic Lipschitz algebras Lip_𝐴(𝑋,𝛼) and lip_𝐴(𝑋,𝛼) under the norm ‖⋅‖_{Lip(𝑋,𝛼)} are natural Banach function algebras on 𝑋 (see [6]).

Let 𝑋 and 𝐾 be compact plane sets with 𝐾⊆𝑋. We define Lip_𝐴(𝑋,𝐾,𝛼)=Lip(𝑋,𝐾,𝛼)∩𝐴(𝑋,𝐾) for 𝛼∈(0,1] and lip_𝐴(𝑋,𝐾,𝛼)=lip(𝑋,𝐾,𝛼)∩𝐴(𝑋,𝐾) for 𝛼∈(0,1). Then Lip_𝐴(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, Lip_𝐴(𝑋,𝐾,𝛼)=Lip_𝐴(𝑋,𝛼)[lip_𝐴(𝑋,𝐾,𝛼)=Lip_𝐴(𝑋,𝛼)] if 𝐾=𝑋, and Lip_𝐴(𝑋,𝐾,𝛼)=Lip(𝑋,𝐾,𝛼)[lip_𝐴(𝑋,𝐾,𝛼)=lip(𝑋,𝐾,𝛼)] if int(𝐾) is empty.

The algebras Lip_𝐴(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 Lip_𝐴(𝑋,𝐾,𝛼) and lip_𝐴(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 Lip_𝐴(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 Section 2, we study unital homomorphisms from natural Banach function subalgebras of Lip_𝐴(𝑋₁,𝐾₁,𝛼₁) to natural Banach function subalgebras of Lip_𝐴(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 Lip_𝐴(𝑋,𝐾,𝛼).

2. Unital Compact Homomorphisms

We first give a sufficient condition for which a continuous map 𝜑∶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. Let 𝑋_𝑗 and 𝐾_𝑗 be compact plane sets with int(𝐾_𝑗)≠∅ and 𝐾_𝑗⊆𝑋_𝑗, and let 𝐵_𝑗 be a subalgebra of 𝐴(𝑋_𝑗,𝐾_𝑗) which is a natural Banach function algebra on 𝑋_𝑗 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 algebra 𝐵₂ on 𝑋₂ implies that 𝜎_𝐵2(ℎ)=ℎ(𝑋₂), where 𝜎_𝐴(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 𝜎_𝐵2(𝜑). By using the Functional Calculus Theorem [2, Theorem  5.1 in Chapter I], there exists 𝑔∈𝐵₂ such that 𝑔̂=𝑓∘𝜑 on ℳ(𝐵₂). It follows that 𝑔(𝑧)==𝑓𝑒_𝑧(𝑒𝑔_𝑧)=𝑔̂𝑒_𝑧=𝑓𝜑𝑒_𝑧(𝜑)=𝑓(𝜑(𝑧))=(𝑓∘𝜑)(𝑧),(2.1) 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. Let 𝑋 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 that 𝛼_𝑗∈(0,1], 𝑧_𝑗∈ℂ, 0<𝑟_𝑗<𝑅_𝑗, 𝐺_𝑗=𝔻(𝑧_𝑗,𝑅_𝑗), Ω_𝑗=𝔻(𝑧_𝑗,𝑟_𝑗), 𝑋_𝑗=𝐺_𝑗, and 𝐾_𝑗=Ω_𝑗, where 𝑗∈{1,2}. Then for each 𝜌∈(𝑟₁,𝑅₁] there exists a continuous map 𝜑_𝜌∶𝑋₂→𝑋₁ with 𝜑_𝜌(𝑋₂)=𝔻(𝑧₁,𝜌) such that 𝜑_𝜌∈Lip_𝐴(𝑋₂,𝐾₂,𝛼₂) and 𝜑_𝜌 does not induce any homomorphism from Lip_𝐴(𝑋₁,K₁,𝛼₁) to Lip_𝐴(𝑋₂,𝐾₂,𝛼₂).

Proof. Let 𝜌∈(𝑟₁,𝑅₁]. We define the map 𝜑_𝜌∶𝑋₂→𝑋₁ by 𝜑_𝜌(𝑧)=⎧⎪⎪⎨⎪⎪⎩𝑧¹+𝜌𝑧−𝑧²𝑟₂𝑧₁||𝑧−𝑧₂||≤𝑟₂,+𝜌𝑧−𝑧¹||𝑧−𝑧₂||𝑟²<||𝑧−𝑧₂||≤𝑅₂.(2.2) Clearly, 𝜑_𝜌 is a continuous mapping, 𝜑_𝜌(𝑋₂)=𝔻(𝑧₁,𝜌), and 𝜑_𝜌∈Lip_𝐴(𝑋₂,𝐾₂,𝛼₂). We now define the function 𝑓_𝜌∶𝑋₁→ℂ by 𝑓_𝜌(𝑧)=⎧⎪⎪⎨⎪⎪⎩𝜌𝑧−𝑧¹𝑟₁𝜌𝑧−𝑧₁||𝑧−𝑧₁||≤𝑟₁,||𝑧−𝑧₁||𝑟¹<||𝑧−𝑧₁||≤𝑅₁.(2.3) Then, 𝑓_𝜌∈Lip_𝐴(𝑋₁,𝐾₁,𝛼₁). Since 0<𝑟₁𝑟₂/𝜌<𝑟₂ and 𝑓_𝜌∘𝜑_𝜌(𝑧)=⎧⎪⎪⎨⎪⎪⎩𝜌²𝑟₁𝑟₂𝑧−𝑧₂||𝑧−𝑧₂||≤𝑟¹𝑟₂𝜌𝑧−𝑧₂𝜌,||𝑧−𝑧₂||𝑟¹𝑟₂𝜌<||𝑧−𝑧²||≤𝑅₂,(2.4) we conclude that 𝑓_𝜌∘𝜑_𝜌∉Lip_𝐴(𝑋₂,𝐾₂,𝛼₂). Therefore, 𝜑_𝜌 does not induce any homomorphism from Lip_𝐴(𝑋₁,𝐾₁,𝛼₁) to Lip_𝐴(𝑋₂,𝐾₂,𝛼₂). Hence, the proof is complete.

Corollary 2.4. Suppose that 𝛼∈(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 𝜑_𝜌 of 𝑋 with 𝜑_𝜌(𝑋)=𝔻(𝜆,𝜌) such that 𝜑_𝜌∈Lip_𝐴(𝑋,𝐾,𝛼) and 𝜑_𝜌 does not induce any endomorphism of Lip_𝐴(𝑋,𝐾,𝛼).

We now give a sufficient condition for a unital homomorphism from a subalgebra 𝐵₁ of Lip_𝐴(𝑋₁,𝐾₁,𝛼₁) into a subalgebra 𝐵₂ of Lip_𝐴(𝑋₂,𝐾₂,𝛼₂) to be compact.

Theorem 2.5. Suppose that 𝛼_𝑗∈(0,1], 𝑋_𝑗 and 𝐾_𝑗 are compact plane sets with int(𝐾_𝑗)≠∅ and 𝐾_𝑗⊆𝑋_𝑗, and 𝐵_𝑗 is a subalgebra of Lip_𝐴(𝑋_𝑗,𝐾_𝑗,𝛼_𝑗) which is a natural Banach function algebra on 𝑋_𝑗 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. If 𝜑∶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.
Let 𝜑∶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 {𝑓_𝑛}^∞𝑛=1 be a bounded sequence in 𝐵₁ with ‖𝑓_𝑛‖_{Lip(𝑋1,𝐾1,𝛼1)}≤1 for all 𝑛∈ℕ. This implies that {𝑓_𝑛|_𝐾1}^∞𝑛=1 is a bounded sequence in 𝐶(𝐾₁) which is equicontinuous on (𝐾₁,𝑑^𝛼1₁). By Arzela-Ascoli’s theorem, {𝑓_𝑛}^∞𝑛=1 has a subsequence {𝑓_𝑛𝑗}^∞𝑗=1 such that {𝑓_𝑛𝑗|_𝐾1}^∞𝑗=1 is convergent in 𝐶(𝐾₁). Since 𝑓_𝑛𝑗|_𝐾1∈𝐴(𝐾₁) for all 𝑗∈ℕ, {𝑓_𝑛𝑗|_𝐾1}^∞𝑗=1 is convergent in 𝐴(𝐾₁). By Montel's theorem, the sequences {𝑓_𝑛𝑗}^∞𝑗=1 and {𝑓^𝑛_𝑗}^∞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)||𝑓(𝑧)−𝑓(𝑤)||≤𝐶|𝑧−𝑤|‖𝑓‖_𝑌+‖‖𝑓^‖‖_𝑌.(2.5) Therefore, there exists a positive constant 𝐶 such that |||𝑓_𝑛𝑗(𝜑(𝑧))−𝑓_𝑛𝑗(𝜑(𝑤))|||≤𝐶||𝜑(𝑧)−𝜑(𝑤)||‖‖𝑓_𝑛𝑗‖‖_𝑌+‖‖𝑓^𝑛_𝑗‖‖_𝑌,(2.6) 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 |||𝑓_𝑛𝑗∘𝜑−𝑓_𝑛𝑘∘𝜑(𝑧)−𝑓_𝑛𝑗∘𝜑−𝑓_𝑛𝑘∘𝜑(𝑤)||||𝑧−𝑤|^𝛼2=|||𝑓^𝑛𝑗−𝑓_𝑛𝑘(𝜑(𝑧))−𝑓^𝑛_𝑗−𝑓_𝑛𝑘(𝜑(𝑤))|||||𝜑(𝑧)−𝜑(𝑤)||⋅||𝜑(|𝑧𝑧)−−𝑤𝜑|(^𝛼2𝑤)||≤𝐶𝑝_𝛼2,𝐾2(𝜑)‖‖𝑓_𝑛𝑗−𝑓_𝑛𝑘‖‖_𝑌+‖‖𝑓^𝑛_𝑗−𝑓^𝑛_𝑘‖‖_𝑌.(2.7) The above inequality is certainly true for all 𝑧,𝑤∈𝐾₂ with 𝑧≠𝑤 and 𝜑(𝑧)=𝜑(𝑤). Therefore, 𝑝_𝛼2,𝐾2𝑓^𝑛_𝑗∘𝜑−𝑓_𝑛𝑘∘𝜑≤𝐶𝑝_𝛼2,𝐾2(𝜑)‖‖𝑓_𝑛𝑖−𝑓_𝑛𝑗‖‖_𝑌+‖‖𝑓^𝑛_𝑖−𝑓^𝑛_𝑗‖‖_𝑌,(2.8) and so ‖‖𝑓_𝑛𝑗∘𝜑−𝑓_𝑛𝑘∘𝜑‖‖_{Lip(𝑋2,𝐾2,𝛼2)}≤1+𝐶𝑝_𝛼2,𝐾2(𝜑)‖‖𝑓_𝑛𝑗−𝑓_𝑛𝑘‖‖_𝑌+‖‖𝑓^𝑛_𝑗−𝑓^𝑛_𝑘‖‖_𝑌.(2.9)
Since 𝑌 is a compact subset of int(𝐾₁), we deduce that the sequences {𝑓_𝑛𝑗}^∞𝑗=1 and {𝑓^𝑛_𝑗}^∞𝑗=1 are convergent uniformly on 𝑌. Therefore, {𝑓_𝑛𝑗∘𝜑}^∞𝑗=1 is a Cauchy sequence on Lip(𝑋₂,𝐾₂,𝛼₂), that is {𝑇𝑓_𝑛𝑗}^∞𝑗=1 is convergent in Lip(𝑋₂,𝐾₂,𝛼₂). Hence, 𝑇 is compact.

Corollary 2.6. Suppose that 𝛼∈(0,1], 𝑋 and 𝐾 are compact plane sets with int(𝐾)≠∅, and 𝐾⊆𝑋. Let 𝐵 be a subalgebra of Lip_𝐴(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)A sector 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)If 𝑓 is a complex-valued function on 𝔻(𝑧₀,𝑟) and 𝜔∈𝜕𝔻(𝑧₀,𝑟), then ∠lim_𝑧→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 map 𝜑∶𝔻(𝑧₀,𝑟)→𝔻_𝜌 has an angular derivation at a point 𝜔∈𝜕𝔻_𝑟(𝑧₀,𝑟) if for some 𝜂∈𝜕𝔻_𝜌∠lim_𝑧→𝜔𝜂−𝑓(𝑧)𝜔−𝑧(2.10) exists (finitely). We call the limit the angular derivative of 𝜑 at 𝜔 and denote it by ∠𝜑^(𝜔).

Lemma 2.8. Let 0<𝑟≤1, and let 𝜑∶𝔻(𝑧₀,𝑟)→𝔻_𝜌 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, ∠𝜑^(𝜔)=𝑟𝜌∠𝜓^𝜔−𝑟𝑧⁰.(2.11)

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. Take 0<𝑟≤1. Let 𝜑∶𝔻(𝑧₀,𝑟)→𝔻 be a nonconstant analytic function and 𝜔∈𝜕𝔻(𝑧₀,𝑟). Then the following are equivalent: (i)liminf_𝑧→𝜔(‖𝜑‖_𝔻𝑟−|𝜑(𝑧)|)/(𝑟−|𝑧|)=𝛿<∞, (ii)∠lim_𝑧→𝜔(𝜂−𝜑(𝑧))/(𝜔−𝑧) exists for some 𝜂∈𝜕𝔻, (iii)∠lim_𝑧→𝜔𝜑^(𝑧) exists and ∠lim_𝑧→𝜔𝜑(𝑧)=𝜂∈𝜕𝔻.

The boundary point 𝜂 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 𝜔𝜂𝛿.

Note that the existence of the angular derivative 𝜑 at 𝜔∈𝜕𝔻(𝑧₀,𝑟), according to Theorem 2.9, is equivalent to liminf_𝑧→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. Let 𝑋 be a compact plane set, and let 𝔻(𝑧₀,𝑟)⊆𝑋 and 𝐾=𝔻(𝑧₀,𝑟). Suppose that 𝑐∈𝜕𝔻(𝑧₀,𝑟) and 𝜑∈Lip_𝐴(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. Let Γ={𝑧∈𝔻(𝑧₀,𝑟)∶|𝑧−𝑐|/(𝑟−|𝑧−𝑧₀|)<2}. For every 𝑧∈Γ we have ‖𝜑‖_{𝔻(𝑧0,𝑟)}−||𝜑(𝑧)||𝑟−||𝑧−𝑧₀||=||𝜑𝑟(−𝑐)||||𝑧−−||𝜑𝑧(₀𝑧||)||≤𝑟−|𝑧||𝑧−−𝑐|𝑧₀||||𝜑(𝑧|𝑧)−−𝜑𝑐|(𝑐)||<2𝑝^1,𝐾(𝜑).(2.12) Therefore, liminf_𝑧→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 set 𝑋 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 set 𝑋 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)⧵⋃^𝑛𝑘=1𝔻(𝑧_𝑘,𝑟_𝑘), where closed discs 𝔻(𝑧_𝑘,𝑟_𝑘) are mutually disjoint in 𝔻(𝑧₀,𝑟). (c)A compact plane set 𝑋 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 disc 𝔻 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+𝑐𝑧) belongs to 𝐴(𝔻) and satisfies ‖ℎ‖_𝔻=ℎ(𝑐)=1.

Let 𝑋 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. Let 𝑋 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. Let 𝑧₀∈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 ℎ(𝑧)=𝛿²𝑐−𝑧₀𝑧−𝑧₀.(2.13) Then ℎ∈𝐵, ‖ℎ‖_𝑋=ℎ(𝑐)=1.

Theorem 2.14. Let 𝑋₁ be a compact plane set such that 𝐺₁=int(𝑋₁) is connected, and 𝐺₁=𝑋₁. Suppose that 𝑋₁ has peak boundary with respect to Lip_𝐴(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∶Lip_𝐴(𝑋₁,𝐾₁,1)→Lip_𝐴(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. Since Lip_𝐴(𝑋₁,𝐾₁,1) and Lip_𝐴(𝑋₂,𝐾₂,1) are, respectively, natural Banach function algebras on 𝑋₁ and 𝑋₂, 𝑇∶Lip_𝐴(𝑋₁,𝐾₁,1)→Lip_𝐴(𝑋₂,𝐾₂,1) is a unital homomorphism, 𝑋₁ is a compact plane set, and 𝑍∈Lip_𝐴(𝑋₁,𝐾₁,1), we conclude that 𝑇 is induced by 𝜑=𝑇𝑍 and so 𝜑∈Lip_𝐴(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 Lip_𝐴(𝑋₁,1), there exists a nonconstant function ℎ∈Lip_𝐴(𝑋₁,1) such that ‖ℎ‖_𝑋1=ℎ(𝜑(𝑐))=1. We now define the sequence {𝑓_𝑛}^∞𝑛=1 of complex-valued functions on 𝑋₁ by 𝑓_𝑛(𝑧)=(1/𝑛)ℎ^𝑛(𝑧). Let 𝑛∈ℕ. Then ‖‖𝑓_𝑛‖‖_𝑋1=1𝑛‖ℎ‖^𝑋1^𝑛=1𝑝𝑛,(2.14)_1,𝐾1𝑓_𝑛=sup||ℎ^𝑛(𝑧)−ℎ^𝑛(𝑤)||𝑛|𝑧−𝑤|∶𝑧,𝑤∈𝐾¹≤sup||ℎ(𝑧)−ℎ(𝑤)||,𝑧≠𝑤|𝑧−𝑤|∶𝑧,𝑤∈𝐾¹≤sup||ℎ(𝑧)−ℎ(𝑤)||,𝑧≠𝑤|𝑧−𝑤|∶𝑧,𝑤∈𝑋¹≤𝑝_1,𝑋1,𝑧≠𝑤(ℎ).(2.15) Thus ‖‖𝑓_𝑛‖‖_{Lip(𝑋1,𝐾1,1)}≤1𝑛+𝑝^1,𝑋1(ℎ)≤1+𝑝_1,𝑋1(ℎ),(2.16) by (2.14) and (2.15). This implies that {𝑓_𝑛}^∞𝑛=1 is a bounded sequence in Lip_𝐴(𝑋₁,𝐾₁,1). The compactness of homomorphism 𝑇 implies that there exists a subsequence {𝑓_𝑛𝑗}^∞𝑗=1 of {𝑓_𝑛}^∞𝑛=1 and a function 𝑔 in Lip_𝐴(𝑋₂,𝐾₂,1) such that lim_𝑗→∞‖‖𝑇𝑓_𝑛𝑗−𝑔‖‖_{Lip(𝑋2,𝐾2,1)}=0.(2.17) This implies that lim_𝑗→∞‖‖𝑇𝑓_𝑛𝑗−𝑔‖‖_𝑋2=0.(2.18) On the other hand, we have ‖𝑇𝑓_𝑛𝑗‖_𝑋2≤1/𝑛_𝑗 for all 𝑗∈ℕ by (2.14). Hence, lim_𝑗→∞‖‖𝑇𝑓_𝑛𝑗‖‖_𝑋2=0.(2.19) By (2.18) and (2.19), 𝑔=0. Therefore, by (2.17) we have lim_𝑗→∞‖‖𝑇𝑓_𝑛𝑗‖‖_{Lip(𝑋2,𝐾2,1)}=0.(2.20) This implies that lim_𝑗→∞𝑝_1,𝐾2𝑓^𝑛_𝑗∘𝜑=0.(2.21) Assume that 𝜀>0. By (2.21), there exists a natural number 𝑁 such that for each 𝑗∈ℕ with 𝑗≥𝑁sup⎧⎪⎨⎪⎩|||𝑓^𝑛𝑗∘𝜑(𝑧)−𝑓_𝑛𝑗∘𝜑(𝑤)||||𝑧−𝑤|∶𝑧,𝑤∈𝐾²,𝑧≠𝑤⎫⎪⎬⎪⎭<𝜀.(2.22) In particular, sup||((ℎ∘𝜑)(𝑧))^𝑛𝑁−((ℎ∘𝜑)(𝑤))^𝑛𝑁||𝑛_𝑁|𝑧−𝑤|∶𝑧,𝑤∈𝐾²,𝑧≠𝑤<𝜀.(2.23) This implies that 1𝑛_𝑁sup||((ℎ∘𝜑)(𝑧))^𝑛𝑁−((ℎ∘𝜑)(𝑐))^𝑛𝑁|||𝑧−𝑐|∶𝑧∈𝐾²,𝑧≠𝑐<𝜀.(2.24) 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 𝑝_1,𝐷(ℎ∘𝜑)≤𝑝_1,𝑋1(ℎ)𝑝_1,𝐾2(𝜑)<∞.(2.25) Therefore, ℎ∘𝜑∈Lip_𝐴(𝑋₂,𝐷,1). Since ‖ℎ‖_𝑋1=ℎ(𝜑(𝑐))=1, we conclude that (ℎ∘𝜑)(𝑐)=‖ℎ∘𝜑‖_𝐷=1.(2.26) 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 1𝑛_𝑁∠lim_{𝑧→𝑐𝑧∈Γ}||||(ℎ∘𝜑)^𝑛_𝑁(𝑧)−(ℎ∘𝜑)^𝑛𝑁(𝑐)𝑧−𝑐||||≤𝜀(2.27) by (2.24). Thus 1𝑛_𝑁∠(ℎ∘𝜑)^𝑛𝑁^(𝑐)≤𝜀.(2.28) But ∠(ℎ∘𝜑)^𝑛𝑁^(𝑐)=𝑛_𝑁(ℎ∘𝜑)^𝑛𝑁−1(𝑐)⋅∠(ℎ∘𝜑)^(𝑐).(2.29) Hence, by (2.28) we have ∠(ℎ∘𝜑)^(𝑐)=1𝑛_𝑁∠(ℎ∘𝜑)^𝑛𝑁^(𝑐)≤𝜀.(2.30) Since 𝜀 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. Let 𝑋 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 Lip_𝐴(𝑋,1). If 𝑇 is a unital compact endomorphism of Lip_𝐴(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. Let 𝐺 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 𝑓_𝑐∈Lip_𝐴(𝑋,𝐾,1) such that 𝑓_𝑐 is not analytic at 𝑐.

Proof. Let 𝑐∈𝐺⧵𝐾. Then there is a positive number 𝑟 such that {𝑧∈ℂ∶|𝑧−𝑐|≤𝑟}⊆𝐺⧵𝐾.(2.31) We now define the function 𝑓_𝑐∶𝑋→ℂ by 𝑓_𝑐(𝑧)=⎧⎪⎨⎪⎩𝑧(1−+𝑐𝑟)(𝑧−𝑐)𝑧∈𝑋,|𝑧−𝑐|≥𝑟,1+|𝑧−𝑐|𝑧∈𝑋,|𝑧−𝑐|<𝑟.(2.32) It is easily seen that 𝑓_𝑐∈Lip_𝐴(𝑋,𝐾,1) and 𝑓_𝑐 is not analytic at 𝑐.

Definition 2.17. Let 𝑋 and 𝐾 be compact plane sets such that 𝐾⊆𝑋. We say that 𝐾 has peak 𝐾-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. Let 𝑟∈(0,1] and 𝐾=𝔻_𝑟. Suppose that Lip_𝐴(𝔻,𝐾,1)⊆𝐵⊆𝐴(𝔻,𝐾). Then 𝐾 has peak 𝐾-boundary with respect to 𝐵.

Proof. We first assume that 𝑟=1. If for each 𝑐∈𝜕𝔻 the function ℎ∶𝔻→ℂ defined by ℎ(𝑧)=(1/2)(1+𝑐𝑧), then ℎ∈𝐵, ℎ is nonconstant on 𝐾=𝔻 and ℎ(𝑐)=1=‖ℎ‖_𝔻.
We now assume that 0<𝑟<1. For each 𝑐∈𝜕𝐾, set 𝑧₀=(1+𝑟)𝑐/𝑟. Then 𝑧₀∈ℂ⧵𝔻. Define the function ℎ∶𝔻→ℂ by ℎ(𝑧)=⎧⎪⎪⎨⎪⎪⎩−𝑟𝑐𝑧−𝑧₀𝑧∈−𝑟||𝑧−𝑧₀𝔻,||𝑧−𝑧₀||≥1,||𝑐𝑧−𝑧₀𝑧∈𝔻,||𝑧−𝑧₀||<1.(2.33) It is easily seen that ℎ∈Lip_𝐴(𝔻,𝐾,1) and ‖ℎ‖_𝔻=1=ℎ(𝑐).

Lemma 2.19. Let Ω 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. Let 𝑋₁ 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 Lip_𝐴(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∶Lip_𝐴(𝑋₁,𝐾₁,1)→Lip_𝐴(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. Since Lip_𝐴(𝑋₁,𝐾₁,1) and Lip_𝐴(𝑋₂,𝐾₂,1) are, respectively, natural Banach function algebras on 𝑋₁ and 𝑋₂, 𝑇∶Lip_𝐴(𝑋₁,𝐾₁,1)→Lip_𝐴(𝑋₂,𝐾₂,1) is a unital homomorphism, 𝑋₁ is a compact plane set, and 𝑍∈Lip_𝐴(𝑋₁,𝐾₁,1), we conclude that 𝑇 is induced by 𝜑=𝑇𝑍 and so 𝜑∈Lip_𝐴(𝑋₂,𝐾₂,1) by Proposition 1.2.
To prove 𝜑(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∈Lip_𝐴(𝑋₁,𝐾₁,1) such that 𝑓 is not analytic at 𝜑(𝜆). But 𝑓∘𝜑=𝑇𝑓∈Lip_𝐴(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 Lip_𝐴(𝑋₁,𝐾₁,1), there exists a function ℎ∈Lip_𝐴(𝑋₁,𝐾₁,1) such that ℎ is not constant on 𝐾₁ and ‖ℎ‖_𝑋1=ℎ(𝜑(𝑐))=1.(2.34) 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. Let 𝑋 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 Lip_𝐴(𝑋,𝐾,1) and 𝐾 is strongly accessible from the interior. If 𝑇 is a unital compact endomorphism of Lip_𝐴(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 of Lip_𝐴(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. Let 𝑋 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=0𝜑_𝑛(𝑋) is finite, and if 𝑋 is connected, ⋂^∞𝑛=0𝜑_𝑛(𝑋) is singleton where 𝜑_𝑛 is the 𝑛th iterate of 𝜑, that is, 𝜑₀(𝑥)=𝑥 and 𝜑_𝑛(𝑥)=𝜑(𝜑_𝑛−1(𝑥)). If ⋂^∞𝑛=0𝜑_𝑛(𝑋)={𝑥₀}, then 𝑥₀ is a fixed point for 𝜑. In fact, if 𝐹=⋂^∞𝑛=0𝜑_𝑛(𝑋), then 𝜑(𝐹)=𝐹.

Theorem 3.2. Suppose that 𝑋 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 𝜎(𝑇)={0,1}∪𝜑^𝑧₀^𝑛∶𝑛∈ℕ.(3.1)

Proof. Clearly 0 and also 1∈𝜎(𝑇) since 𝑇(1_𝑋)=1_𝑋. 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, if 𝜆∈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 that 𝜑^(𝑧₀)∈𝜎(𝑇). If 𝜑^(𝑧₀)∉𝜎(𝑇), then there exists a nonzero linear operator 𝑆∶𝐵→𝐵 such that 𝑇−𝜑^𝑧₀𝐼𝑆=𝐼.(3.2) Since 𝑍−𝑧₀1_𝑋∈𝐵, ℎ=𝑆(𝑍−𝑧₀1_𝑋)∈𝐵 and so ℎ∘𝜑−𝜑^𝑧₀ℎ=𝑍−𝑧₀1_𝑋,(3.3) by (3.2). By differentiation at 𝑧₀, we have 0=ℎ^𝜑𝑧₀𝜑^𝑧₀−𝜑^𝑧₀ℎ^𝑧₀=1,(3.4) 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 𝑆_𝑗∶𝐵→𝐵 such that 𝑇−𝜑^𝑧₀^𝑗𝐼𝑆_𝑗=𝐼.(3.5) Since (𝑍−𝑧₀1_𝑋)^𝑗∈𝐵, ℎ_𝑗=𝑆_𝑗(𝑍−𝑧₀1_𝑋)^𝑗∈𝐵 and so ℎ_𝑗∘𝜑−𝜑^𝑧₀^𝑗ℎ_𝑗=𝑍−𝑧₀1_𝑋^𝑗,(3.6) by (3.5). By 𝑗−1 times differentiation at 𝑧₀, we have ℎ_𝑗𝑧₀=ℎ^𝑗𝑧₀=⋯=ℎ^{𝑗(𝑗−1)}𝑧₀=0,(3.7) and by 𝑗 times differentiation at 𝑧₀, we have 0=𝜑^𝑧₀^𝑗ℎ^𝑗(𝑗)𝜑𝑧₀−𝜑^𝑧₀^𝑗ℎ^𝑗(𝑗)𝑧₀=𝑗!,(3.8) this is a contradiction. Thus, (𝜑^(𝑧₀))^𝑛∈𝜎(𝑇) for all 𝑛∈ℕ. This completes the proof.

Corollary 3.3. Let 𝐵 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 {𝜆^𝑚∶𝜆∈𝜎(𝑇)}={0,1}∪𝜑^𝑚𝑧₀^𝑛∶𝑛∈ℕ.(3.9)

Proof. Since 𝐹 is a finite set and 𝜑(𝐹)=𝐹, there exist 𝑧₀∈𝐹 and 𝑚∈ℕ such that 𝜑_𝑚(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 𝑇∶𝐵→𝐵 by 𝑇𝑓=𝑓∘𝜑_𝑚. Then 𝑇 is a compact endomorphism of 𝐵 induced by 𝜑_𝑚 by Corollary 2.6 and 𝜑_𝑚(𝑧₀)=𝑧₀. Therefore, 𝜎𝑇={0,1}∪𝜑^𝑚𝑧₀^𝑛∶𝑛∈ℕ(3.10) by Theorem 3.2. Since 𝑇𝑓=𝑓∘𝜑 and 𝑇𝑓=𝑓∘𝜑_𝑚 for all 𝑓∈𝐵, we have 𝑇=𝑇^𝑚. By Spectral Mapping Theorem, 𝜎(𝑇^𝑚)={𝜆^𝑚∶𝜆∈𝜎(𝑇)}. Therefore, {𝜆^𝑚∶𝜆∈𝜎(𝑇)}={0,1}∪𝜑^𝑚𝑧₀^𝑛∶𝑛∈ℕ.(3.11)

Corollary 3.4. Suppose that 𝑋 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∈Lip_𝐴(𝑋,𝐾,𝛼), and let 𝜑(𝑧₀)=𝑧₀ for some 𝑧₀∈Ω. If 𝑇 is a unital endomorphism of Lip_𝐴(𝑋,𝐾,𝛼) induced by 𝜑, then 𝑇 is compact and 𝜎(𝑇)={0,1}∪𝜑^𝑧₀^𝑛∶𝑛∈ℕ.(3.12)