Linear Isometries between Real Banach Algebras of Continuous Complex-Valued Functions

Abstract

Let 𝑋 and 𝑌 be compact Hausdorff spaces, and let 𝜏 and 𝜂 be topological involutions on 𝑋 and 𝑌, respectively. In 1991, Kulkarni and Arundhathi characterized linear isometries from a real uniform function algebra 𝐴 on (𝑋, 𝜏) onto a real uniform function algebra 𝐵 on (𝑌, 𝜂) applying their Choquet boundaries and showed that these mappings are weighted composition operators. In this paper, we characterize all onto linear isometries and certain into linear isometries between 𝐶(𝑋,𝜏) and 𝐶(𝑌,𝜂) applying the extreme points in the unit balls of 𝐶(𝑋,𝜏)^∗ and 𝐶(𝑌,𝜂)^∗.

1. Introduction and Preliminaries

Let ℝ and ℂ denote the field of real and complex numbers, respectively. The symbol 𝔽 denotes a field that can be either ℝ or ℂ. The elements of 𝔽 are called scalars. We also denote by 𝕊_𝔽 the set of all 𝜆∈𝔽 with |𝜆|=1.

Let 𝔛 be a normed space over 𝔽. We denote by 𝔛^∗ and 𝐵_𝔛 the dual space of 𝔛 and the closed unit ball of 𝔛, respectively. For a subset 𝐸 of 𝔛, let Ext(𝐸) denote the set of all extreme points of 𝐸. Kulkarni and Limaye showed [1, Theorem 2] that if 𝐴 is a nonzero linear subspace of 𝔛 and 𝜑∈Ext(𝐵_𝐴^∗), then 𝜑 has an extension to some 𝜓∈Ext(𝐵_𝔛^∗).

We know that if 𝔛 and 𝒴 are normed spaces over 𝔽 and 𝑇∶𝔛→𝒴 is a linear isometry from 𝔛 onto 𝒴 over 𝔽, then 𝑇 is a bijection mapping between Ext(𝐵_𝔛) and Ext(𝐵_𝒴).

Let 𝑋 be a compact Hausdorff space. We denote by 𝐶_𝔽(𝑋) the unital commutative Banach algebra of all continuous functions from 𝑋 into 𝔽, with the uniform norm ‖𝑓‖_𝑋=sup{|𝑓(𝑥)|∶𝑥∈𝑋}, 𝑓∈𝐶_𝔽(𝑋). We write 𝐶(𝑋) instead as 𝐶_ℂ(𝑋). For 𝑥∈𝑋, we consider the linear functional 𝑒_{𝐶𝔽(𝑋),𝑥} on 𝐶_𝔽(𝑋) defined by 𝑒_{𝐶𝔽(𝑋),𝑥}(𝑓)=𝑓(𝑥) (𝑓∈𝐶_𝔽(𝑋)), which is called the evaluation functional on 𝐶_𝔽(𝑋) at 𝑥. Clearly, 𝜆𝑒_{𝐶𝔽(𝑋),𝑥}∈𝐵_{𝐶𝔽(𝑋)}^∗ for all (𝑥,𝜆)∈𝑋×𝔽. It is known [2, page 441] that Ext𝐵_{𝐶𝔽(𝑋)}^∗=𝜆𝑒_{𝐶𝔽(𝑋),𝑥}∶(𝑥,𝜆)∈𝑋×𝕊_𝔽. (1)

Let 𝐴 be a real or complex linear subspace of 𝐶(𝑋). A nonempty subset 𝑆 of 𝑋 is called a boundary for 𝐴 (with respect to 𝑋), if for each 𝑓∈𝐴 the function |𝑓| assumes its maximum on 𝑋 at some 𝑥∈𝑆. We denote by Γ(𝐴,𝑋) the intersection of all closed boundaries for 𝐴. If Γ(𝐴,𝑋) is a boundary for 𝐴, it is called the Shilov boundary for 𝐴 (with respect to 𝑋).

Let 𝐴 be linear subspace of 𝐶_𝔽(𝑋) containing 1_𝑋, the constant function with value 1 on 𝑋. A representing measure for 𝜑∈𝐴^∗ is an 𝔽-valued regular Borel measure 𝜇 on 𝑋 such that 𝜑(𝑓)=∫_𝑋𝑓𝑑𝜇 for all 𝑓∈𝐴. Let 𝐾_𝔽(𝐴)={𝜑∈𝐴^∗∶‖𝜑‖=𝜑(1_𝑋)=1}. If 𝑥∈𝑋, then 𝑒_{𝐶𝔽(𝑋),𝑥}|_𝐴∈𝐾_𝔽(𝐴) and 𝛿_𝑥, the point mass measure on 𝑋 at 𝑥, is a representing measure for 𝑒_{𝐶𝔽(𝑋),𝑥}|_𝐴. We denote by Ch(𝐴,𝑋) the set of all 𝑥∈𝑋 for which 𝛿_𝑥 is the only representing measure for 𝑒_{𝐶𝔽(𝑋),𝑥}|_𝐴. If Ch(𝐴,𝑋) is a boundary for 𝐴, it is called the Choquet boundary for 𝐴 (with respect to 𝑋). We know that Γ𝐶_𝔽(𝑋),𝑋=Ch𝐶_𝔽(𝑋),𝑋=𝑋. (2)

Let 𝐴 be a real linear subspace of 𝐶(𝑋), and let 𝑆 be nonempty subset of 𝑋. We say that 𝐴 is extremely regular at 𝑆 if for every open neighborhood 𝑈 of 𝑆 and for each 𝜀>0 there is a function 𝑓∈𝐴 with ‖𝑓‖_𝑋=1 such that 𝑓(𝑥)=1 for all 𝑥∈𝑆 and |𝑓(𝑦)|<1 for all 𝑦∈𝑋⧵𝑈.

Let 𝑋 be a nonempty set. A self-map 𝜏∶𝑋→𝑋 is called an involution on 𝑋 if 𝜏(𝜏(𝑥))=𝑥 for all 𝑥∈𝑋. A subset 𝑆 of 𝑋 is called 𝜏-invariant if 𝜏(𝑆)⊆𝑆. Clearly, if 𝑆⊆𝑋 is 𝜏-invariant, then 𝜏(𝑆)=𝑆. A 𝜏-invariant measure on 𝐸 is a measure 𝜇 on 𝐸 such that 𝜇∘𝜏=𝜇.

Let 𝑋 be a topological space. An involution 𝜏 on 𝑋 is called a topological involution on 𝑋, if 𝜏 is continuous.

Let 𝑋 be a compact Hausdorff space, and let 𝜏 be a topological involution on 𝑋. We define 𝐶(𝑋,𝜏)=𝑓∈𝐶(𝑋)∶𝑓∘𝜏=𝑓. (3) Then, 𝐶(𝑋,𝜏) is a unital uniformly closed self-adjoint real subalgebra of 𝐶(𝑋) which separates the points of 𝑋 and does not contain 𝑖_𝑋, the constant function with value 𝑖 on 𝑋. Note that Re𝑓∈𝐶(𝑋,𝜏) for all 𝑓∈𝐶(𝑋,𝜏). Moreover, 𝐶(𝑋)=𝐶(𝑋,𝜏)⊕𝑖𝐶(𝑋,𝜏) and max𝑓_𝑋,𝑔_𝑋≤𝑓+𝑖𝑔_𝑋≤2max𝑓_𝑋,𝑔_𝑋, (4) for all 𝑓,𝑔∈𝐶(𝑋,𝜏). In fact, the complex Banach algebra (𝐶(𝑋),‖⋅‖_𝑋) can be regarded as the complexification of the real Banach algebra (𝐶(𝑋,𝜏),‖⋅‖_𝑋). Note that 𝐶(𝑋,𝜏)=𝐶_ℝ(𝑋) if and only if 𝜏 is the identity self-map on 𝑋. Hence, the class of real Banach algebras 𝐶(𝑋,𝜏) is, in fact, larger than the class of real Banach algebras 𝐶_ℝ(𝑋).

Let 𝑋 be a compact Hausdorff space, 𝜏 a topological involution on 𝑋, and 𝐸 a subset of 𝐶(𝑋,𝜏). We say that 𝐸 separates the points of 𝑋/𝜏 if for each 𝑥,𝑦∈𝑋 with 𝑦∉{𝑥,𝜏(𝑥)} there exists a function 𝑓 in 𝐸 such that 𝑓(𝑥)≠𝑓(𝑦). It is known [3, Lemma 1.3.7] that if Re𝐶(𝑋,𝜏)={Re𝑓∶𝑓∈𝐶(𝑋,𝜏)}, then Re𝐶(𝑋,𝜏) separates the points of 𝑋/𝜏.

Let 𝑋 be a compact Hausdorff space. We assume that 𝐶(𝑋)_𝑟 denotes 𝐶(𝑋), regarded as a real normed space by restricting the scalar multiplication to real scalars. Let 𝜏 be the topological involution on 𝑋×{0,1} defined by 𝜏(𝑥,0)=(𝑥,1) (𝑥∈𝑋) and 𝜏(𝑥,1)=(𝑥,0) (𝑥∈𝑋). Then the map Ψ∶𝐶(𝑋)_𝑟→𝐶(𝑋×{0,1},𝜏) defined by Ψ𝑓(𝑥,0)=𝑓(𝑥),Ψ𝑓(𝑥,1)=𝑓(𝑥)𝑓∈𝐶(𝑋)_𝑟;𝑥∈𝑋, (5) is an isometrical isomorphism from 𝐶(𝑋)_𝑟 onto 𝐶(𝑋×{0,1},𝜏). Hence, the class of real Banach algebras 𝐶(𝑌,𝜏) is, in fact, larger than the class of real Banach algebras 𝐶(𝑋)_𝑟.

Let 𝑋 be a compact Hausdorff space, and let 𝜏 be a topological involution on 𝑋. Let 𝐴 be a real subspace of 𝐶(𝑋,𝜏) containing 1_𝑋. A real part representing measure for 𝜑∈𝐴^∗ is a 𝜏-invariant regular real Borel measure 𝜇 on 𝑋 such that 𝜑(𝑓)=∫_𝑋Re𝑓𝑑𝜇 for all 𝑓∈𝐴. Let 𝐾_ℝ(𝐴)={𝜑∈𝐴^∗∶‖𝜑‖=𝜑(1_𝑋)=1}. If 𝑥∈𝑋, then Re(𝑒_{𝐶(𝑋),𝑥|𝐴})∈𝐾_ℝ(𝐴) and the measure 𝑚_𝑥=(1/2)(𝛿_𝑥+𝛿_𝜏(𝑥)) is a real part representing measure for Re(𝑒_{𝐶(𝑋),𝑥}|_𝐴). We denote by Ch(𝐴,𝑋,𝜏) the set of all 𝑥∈𝑋 for which 𝑚_𝑥 is the only real part representing measure for Re(𝑒_{𝐶(𝑋),𝑥}|_𝐴). Clearly, Ch(𝐴,𝑋,𝜏) is a 𝜏-invariant subset of 𝑋. If Ch(𝐴,𝑋,𝜏) is a boundary for 𝐴, it is called the Choquet boundary for 𝐴 with respect to (𝑋,𝜏). We know that Ch(𝐶(𝑋,𝜏),𝑋,𝜏)=𝑋.

The real Banach algebra 𝐶(𝑋,𝜏) and its real linear subspaces were first considered by Kulkarni and Limaye in [4]. For a detailed account of several properties of 𝐶(𝑋,𝜏), we refer to [3].

Let 𝑋 be a compact Hausdorff space. For each (𝑥,𝜆)∈𝑋×ℂ, define the map 𝜓_{𝐶(𝑋),𝑥,𝜆}∶𝐶(𝑋)_𝑟→ℝ by 𝜓_{𝐶(𝑋),𝑥,𝜆}𝑓=Re𝜆𝑓(𝑥),𝑓∈𝐶(𝑋)_𝑟, (6) in fact, 𝜓_{𝐶(𝑋),𝑥,𝜆}=Re(𝜆𝑒_{𝐶(𝑋),𝑥}). Clearly, 𝜓_{𝐶(𝑋),𝑥,𝜆}∈𝐵_{(𝐶(𝑋)𝑟)}^∗.

Let 𝜏 be a topological involution on 𝑋. For each (𝑥,𝜆)∈𝑋×ℂ define the map 𝜑_{𝐶(𝑋,𝜏),𝑥,𝜆}∶𝐶(𝑋,𝜏)→ℝ by 𝜑_{𝐶(𝑋,𝜏),𝑥,𝜆}=𝜓_{𝐶(𝑋),𝑥,𝜆}|_{𝐶(𝑋,𝜏)}. Kulkarni and Limaye showed [1, Proposition 3] that Ext𝐵_{(𝐶(𝑋)𝑟)}^∗=𝜓_{𝐶(𝑋),𝑥,𝜆}∶(𝑥,𝜆)∈𝑋×𝑆_ℂ. (7) Applying this result, they obtained the following theorem.

Theorem 1 (see [1, Corollary 5]). Let 𝑋 be a compact Hausdorff space, and let 𝜏 be a topological involution on 𝑋. Suppose that 𝑃_𝑋,𝜏=(𝑥,𝜆)∈𝑋×𝑆_ℂ(𝑥)∪)∶𝜏≠𝑥(𝑥,𝜆∈𝑋×𝑆_ℝ∶𝜏(𝑥)=𝑥. (8) Then Ext𝐵_{𝐶(𝑋,𝜏)}^∗=𝜑_{𝐶(𝑋,𝜏),𝑥,𝜆}∶(𝑥,𝜆)∈𝑃_𝑋,𝜏. (9)

The classical Banach-Stone theorem states that if 𝑇 is a linear isometry from 𝐶_𝔽(𝑋) onto 𝐶_𝔽(𝑌), then there exists a homeomorphism ℎ from 𝑌 onto 𝑋 and a continuous function 𝑎∈𝐶_𝔽(𝑌) with 𝑎(𝑌)⊆𝕊_𝔽 such that 𝑇𝑓𝑦=𝑎𝑦𝑓ℎ𝑦,𝑓∈𝐶_𝔽(𝑋),𝑦∈𝑌. (10)

This well-known theorem has been generalized in several directions. In 1966, an important generalization was given by Holsztyński [5] by considering into linear isometries as the following.

Theorem 2. Let 𝑋 and 𝑌 be compact Hausdorff spaces. If 𝑇 is a linear isometry from 𝐶_𝔽(𝑋) into 𝐶_𝔽(𝑌), then there exists a closed subset 𝑌₀ of  𝑌, a continuous map 𝜑 from 𝑌₀ onto 𝑋 and a function 𝑎∈𝐶_𝔽(𝑌) with ‖𝑎‖_𝑌=1 and 𝑎(𝑌₀)⊆𝕊_𝔽 such that 𝑇𝑓𝑦=𝑎𝑦𝑓ℎ𝑦,∀𝑓∈𝐶_𝔽(𝑋),∀𝑦∈𝑌₀. (11)

In 1975, another important generalization of Banach-Stone theorem was given by Novinger [6] for certain complex subalgebras of 𝐶(𝑋) and 𝐶(𝑌) considering the concept of their Choquet boundaries. In 1991, Kulkarni and Arundhathi [7] generalized the given results by Novinger for certain real subalgebras of 𝐶(𝑋,𝜏) and 𝐶(𝑌,𝜂) considering the concept of their Choquet boundaries. We can cite some other generalizations of Banach-Stone theorem, for example, the generalization obtained by Cambern [8] for space of vector-valued continuous functions, by Jeang and Wong [9] for spaces of scalar-valued continuous functions vanishing at infinity, by Araujo and Font [10] for certain subspaces of scalar-valued continuous functions considering concepts of their Choquet and Shilov boundaries, and by Jiménez-Vargas and Villegas-Vallecillos [11] for spaces of scalar-valued Lipschitz functions.

Throughout the rest of this paper, 𝑋 and 𝑌 are compact Hausdorff spaces and 𝜏 and 𝜂 are topological involutions on 𝑋 and 𝑌, respectively.

In this paper, we characterize all linear isometries from 𝐶(𝑋,𝜏) onto 𝐶(𝑌,𝜂) and certain linear isometries from 𝐶(𝑋,𝜏) into 𝐶(𝑌,𝜂), applying the extreme points in 𝐵_{𝐶(𝑋,𝜏)}^∗ and 𝐵_{𝐶(𝑌,𝜂)}^∗. Our results in Section 2 are some of the given results by Kulkarni and Arundhathi in [7] that we obtain by applying the extreme points in 𝐵_{𝐶(𝑋,𝜏)}^∗ and 𝐵_{𝐶(𝑌,𝜂)}^∗. The main result in Section 3 is a generalization of the given result by Holsztyński in [5] for certain into isometries.

2. Onto Linear Isometries

We first determine unit-preserving linear isometries from 𝐶(𝑋,𝜏) onto 𝐶(𝑌,𝜂). For this purpose, we need the following lemmas.

Lemma 3 (see [3, Lemma 5.1.1]). Let 𝐴 and 𝐵 be real linear subspaces of  𝐶(𝑋,𝜏) and 𝐶(𝑌,𝜂), respectively. Suppose that 𝑥∈𝑋, 𝑦∈𝑌, 𝐴 is extremely regular at {𝑥,𝜏(𝑥)} and 𝐵 is extremely regular at {𝑦,𝜂(𝑦)}. Let 𝑇∶𝐴→𝐵 be a linear isometry from 𝐴 onto 𝐵 and Re((𝑇𝑓)(𝑦))=Re(𝑓(𝑥)) for all 𝑓∈𝐴. Then, (𝑇𝑓)(𝑦)=𝑓(𝑥) for all 𝑓∈𝐴 or (𝑇𝑓)(𝑦)=𝑓(𝜏(𝑥)) for all 𝑓∈𝐴.

Lemma 4. Let 𝑆 be a nonempty 𝜏-invariant closed subset of 𝑋. Then, 𝐶(𝑋,𝜏) is extremely regular at 𝑆. In particular, 𝐶(𝑋,𝜏) is extremely regular at {𝑥,𝜏(𝑥)} for all 𝑥 in 𝑋.

Proof. Assume that 𝜀>0 and 𝑈 is an open neighborhood of 𝑆 in 𝑋. By Urysohn’s lemma, there exists a function 𝑔∈𝐶(𝑋) with 0≤𝑔(𝑥)≤1 for all 𝑥∈𝑋, 𝑔(𝑥)=1 for all 𝑥∈𝑆, and 𝑔(𝑥)=0 for all 𝑥∈𝑋⧵𝑈. Define the function 𝑓∶𝑋→ℂ by 𝑓=(𝑔∘𝜏)𝑔. Then, 𝑓∈𝐶(𝑋,𝜏), ‖𝑓‖_𝑋=1, 𝑓(𝑥)=1 for all 𝑥∈𝑆 and |𝑓(𝑥)|<𝜀 for all 𝑥∈𝑋⧵𝑈. Hence, 𝐶(𝑋,𝜏) is extremely regular at 𝑆.

Theorem 5. Let 𝑇∶𝐶(𝑋,𝜏)→𝐶(𝑌,𝜂) be a linear isometry from 𝐶(𝑋,𝜏) onto 𝐶(𝑌,𝜂) with 𝑇1_𝑋=1_𝑌. Then, there exists a homeomorphism ℎ from 𝑌 onto 𝑋 with ℎ∘𝜂=𝜏∘ℎ on 𝑌 such that 𝑇𝑓𝑦=𝑓ℎ𝑦,∀𝑓∈𝐶(𝑋,𝜏),∀𝑦∈𝑌. (12)

Proof. We claim that 𝑌=𝑦∈𝑌∶𝑇^∗𝜑_{𝐶(𝑌,𝜂),𝑦,1}∈Ext𝐵_{𝐶(𝑋,𝜏)}^∗. (13) Let 𝑦∈𝑌. By Theorem 1, 𝜑_{𝐶(𝑌,𝜂),𝑦,1}∈Ext(𝐵_{𝐶(𝑌,𝜂)}^∗). Since 𝑇 is a linear isometry from 𝐶(𝑋,𝜏) onto 𝐶(𝑌,𝜂), we conclude that 𝑇^∗ is a linear isometry from 𝐶(𝑌,𝜂)^∗ onto 𝐶(𝑋,𝜏)^∗ and so 𝑇^∗Ext𝐵_{𝐶(𝑌,𝜂)}^∗=Ext𝐵_{𝐶(𝑋,𝜏)}^∗. (14) Therefore, 𝑇^∗(𝜑_{𝐶(𝑌,𝜂),𝑦,1})∈Ext(𝐵_{𝐶(𝑋,𝜏)}^∗). Hence, our claim is justified.
We now show that for each 𝑦∈𝑌 there exists a unique 𝑥∈𝑋 such that 𝑇𝑓𝑦=𝑓(𝑥),∀𝑓∈𝐶(𝑋,𝜏). (15) Let 𝑦∈𝑌. By (13), we have 𝑇^∗(𝜑_{𝐶(𝑌,𝜂),𝑦,1})∈Ext(𝐵_{𝐶(𝑋,𝜏)}^∗). Hence, by Theorem 1, there exists (𝑥₀,𝜆₀)∈𝑋×𝑆_ℂ such that 𝑇^∗𝜑_{𝐶(𝑌,𝜂),𝑦,1}=𝜑_{𝐶(𝑋,𝜏),𝑥0,𝜆0}. (16) Since 1_𝑋∈𝐶(𝑋,𝜏) and 𝑇1_𝑋=1_𝑌, by (16), we have Re𝜆₀=Re𝜆₀1_𝑋𝑥₀=𝜑_{𝐶(𝑋,𝜏),𝑥0,𝜆0}1_𝑋=𝑇^∗𝜑_{𝐶(𝑌,𝜂),𝑦,1}1_𝑋=𝜑_{𝐶(𝑌,𝜂),𝑦,1}𝑇1_𝑋=𝜑_{𝐶(𝑌,𝜂),𝑦,1}1_𝑌=Re1_𝑌𝑦𝑒_{𝐶(𝑌),𝑦}1_𝑌=Re1_𝑌𝑦1_𝑌𝑦=Re(1)=1. (17) Since |𝜆₀|=1 and Re(𝜆₀)=1, we have Im(𝜆₀)=0 and so 𝜆₀=1. Thus, by (16), we have 𝑇^∗𝜑_{𝐶(𝑌,𝜂),𝑦,1}=𝜑_{𝐶(𝑋,𝜏),𝑥0,1}. (18) From (18) we deduce that Re𝑇𝑓𝑦=Re𝑓𝑥₀,∀𝑓∈𝐶(𝑋,𝜏). (19) Since 𝑇 is a linear isometry from 𝐶(𝑋,𝜏) onto 𝐶(𝑌,𝜂) and by Lemma 4, 𝐶(𝑋,𝜏) is extremely regular at {𝑥₀,𝜏(𝑥₀)}, and 𝐶(𝑌,𝜂) is extremely regular at {𝑦,𝜂(𝑦)}, we conclude that (𝑇𝑓)(𝑦)=𝑓(𝑥₀) for all 𝑓 in 𝐶(𝑋,𝜏) or (𝑇𝑓)(𝑦)=𝑓(𝜏(𝑥₀)) for all 𝑓∈𝐶(𝑋,𝜏) by Lemma 3. We assume that 𝑥=𝑥₀ whenever (𝑇𝑓)(𝑦)=𝑓(𝑥₀) for all 𝑓 in 𝐶(𝑋,𝜏) and 𝑥=𝜏(𝑥₀) whenever (𝑇𝑓)(𝑦)=𝑓(𝜏(𝑥₀)) for all 𝑓 in 𝐶(𝑋,𝜏). Then, 𝑥∈𝑋 and we have 𝑇𝑓𝑦=𝑓(𝑥),∀𝑓∈𝐶(𝑋,𝜏). (20) This proves the existent of 𝑥. To show uniqueness, assume that there exists 𝑥^′∈𝑋 such that 𝑇𝑓𝑦=𝑓𝑥^′,∀𝑓∈𝐶(𝑋,𝜏). (21) Then, 𝑓(𝑥)=𝑓(𝑥^′) for all 𝑓 in 𝐶(𝑋,𝜏). This implies that 𝑥=𝑥^′ since 𝐶(𝑋,𝜏) separates the points of 𝑋.
Now we define the map ℎ∶𝑌→𝑋  by  ℎ(𝑦)=𝑥 whenever (𝑇𝑓)(𝑦)=𝑓(𝑥) for all 𝑓 in 𝐶(𝑋,𝜏). In fact, we have 𝑇𝑓𝑦=𝑓ℎ𝑦,∀𝑓∈𝐶(𝑋,𝜏)∀𝑦∈𝑌. (22) Now we prove that ℎ∘𝜂=𝜏∘ℎ on 𝑌. Let 𝑦∈𝑌. Then, 𝜂(𝑦)∈𝑌. From (22), for each 𝑓∈𝐶(𝑋,𝜏) we have 𝑓𝑦ℎ𝜂𝑦=𝜂𝑦=ℎ∘𝜂=𝑓𝑇𝑓𝑇𝑓𝑦=𝑓ℎ𝑦=𝑓𝜏ℎ𝑦=𝑓(𝜏∘ℎ)𝑦. (23) This implies that (ℎ∘𝜂)(𝑦)=(𝜏∘ℎ)(𝑦) since 𝐶(𝑋,𝜏) separates the points of 𝑋. Therefore, ℎ∘𝜂=𝜏∘ℎ on 𝑌.
Continually, we prove that ℎ is bijective. Since 𝑇⁻¹ is a linear isometry from 𝐶(𝑌,𝜂) onto 𝐶(𝑋,𝜏) with 𝑇⁻¹1_𝑌=1_𝑋, by the above argument, we deduce that there exists a map 𝑘∶𝑋→𝑌 such that 𝑇⁻¹𝑔(𝑥)=𝑔(𝑘(𝑥)),∀𝑔∈𝐶𝑌,𝜂,∀𝑥∈𝑋. (24) To prove the injectivity of ℎ, let 𝑦∈𝑌. For each 𝑔∈𝐶(𝑌,𝜂), by (22) and (24) we have 𝑔(𝑘∘ℎ)𝑦=𝑔𝑘ℎ𝑦=𝑇⁻¹𝑔ℎ𝑦=𝑇𝑇⁻¹𝑔𝑦=𝑔𝑦. (25) This implies that (𝑘∘ℎ)(𝑦)=𝑦 since 𝐶(𝑌,𝜂) separates the points of 𝑌. Therefore, 𝑘∘ℎ=𝐼_𝑌, the identity mapping on 𝑌, and so ℎ is injective.
To prove that ℎ is onto, let 𝑥∈𝑋. For each 𝑓∈𝐶(𝑋,𝜏), by (24) and (22) we have 𝑓𝑥(ℎ(𝑘(𝑥=(𝑘(𝑥=𝑇((ℎ∘𝑘)())=𝑓)))𝑇𝑓))₋₁𝑇𝑓(𝑥)=𝑓𝑥. (26) This implies that (ℎ∘𝑘)(𝑥)=𝑥 since 𝐶(𝑋,𝜏) separates the points of 𝑋. Therefore 𝑘∘ℎ=𝐼_𝑋, the identity mapping on 𝑋, and so ℎ is onto.
We now check that ℎ∶𝑌→𝑋 is continuous. Let 𝑦∈𝑌, and let {𝑦_𝛼}_𝛼∈𝐽 be a net in 𝑌 such that lim_𝛼𝑦_𝛼=𝑦 in 𝑌. Then we have lim_𝛼𝑇𝑓𝑦_𝛼=𝑇𝑓𝑦,∀𝑓∈𝐶(𝑋,𝜏). (27) By (27) and definition of ℎ, we deduce that lim_𝛼𝑓ℎ𝑦_𝛼=𝑓ℎ𝑦,∀𝑓∈𝐶(𝑋,𝜏). (28) Since 𝐶(𝑋,𝜏) separates the points of 𝑋, by (28), we have lim_𝛼ℎ(𝑦_𝛼)=ℎ(𝑦) in 𝑋. Therefore, ℎ∶𝑌→𝑋 is continuous.
Since ℎ∶𝑌→𝑋 is a bijective continuous mapping, 𝑌 is a compact space, and 𝑋 is a Hausdorff space, we conclude that ℎ⁻¹∶𝑋→𝑌 is continuous, and so ℎ is a homeomorphism. Hence, the proof is complete.

Corollary 6. Let 𝑇∶𝐶(𝑋,𝜏)→𝐶(𝑌,𝜂) be an isometry mapping from 𝐶(𝑋,𝜏) onto 𝐶(𝑌,𝜂) with 𝑇(0)=0 and 𝑇1_𝑋=1_𝑌. Then 𝑇 is an isomorphism.

Proof. Since (𝐶(𝑋,𝜏),‖⋅‖_𝑋) and (𝐶(𝑌,𝜂),‖⋅‖_𝑌) are real Banach algebras and 𝑇 is an isometry from 𝐶(𝑋,𝜏) onto 𝐶(𝑌,𝜂) with 𝑇(0)=0, we conclude that 𝑇 is a linear isometry by Mazur-Ulam theorem (see [12]). Since 𝑇1_𝑋=1_𝑌, applying Theorem 5, we deduce that there exists a homeomorphism ℎ from 𝑌 onto 𝑋 such that 𝑇𝑓𝑦=𝑓ℎ𝑦,∀𝑓∈𝐶(𝑋,𝜏),∀𝑦∈𝑌. (29) It follows that 𝑇 is a homomorphism. Hence, the proof is complete.

We now study the onto case (not necessarily unit preserving).

Lemma 7. Let 𝑓∈𝐶(𝑋,𝜏) with ‖𝑓‖_𝑋=1. Then, |𝑓(𝑥)|=1 for all 𝑥∈𝑋, if for every 𝜀>0 there exists 𝛿>0 such that 𝑔∈𝐶(𝑥,𝜏) with ‖𝑔‖_𝑋≥𝜀 implies that max𝑓+𝑔_𝑋,𝑓−𝑔_𝑋≥1+𝛿. (30)

Proof. Assume that there exists 𝑥₀∈𝑋 such that |𝑓(𝑥₀)|≠1. Then, |𝑓(𝑥₀)|<1 since ‖𝑓‖_𝑋=1. Let 𝜀=(1/3)(1−|𝑓(𝑥₀)|). Then, 0<𝜀<1/3 and |𝑓(𝑥₀)|<1−2𝜀. Take 𝑉={𝑥∈𝑋∶|𝑓(𝑥)|<1−𝜀}. Since 𝑓∈𝐶(𝑋,𝜏), we deduce that 𝑉 is a 𝜏-invariant open subset of 𝑋 and {𝑥₀,𝜏(𝑥₀)}⊆𝑉. By Urysohn’s lemma, there exists 𝑔₁∈𝐶(𝑋) such that 0≤𝑔₁(𝑥)≤1 for all 𝑥∈𝑋, 𝑔₁(𝑥₀)=𝑔₁(𝜏(𝑥₀))=1, and 𝑔₁(𝑥)=0 for all 𝑥∈𝑋⧵𝑉. Let 𝑔=𝜀𝑔₁(𝑔₁∘𝜏). Then, 𝑔∈𝐶(𝑋,𝜏), 0≤𝑔(𝑥)≤𝜀 for all 𝑥∈𝑋, 𝑔(𝑥₀)=𝑔(𝜏(𝑥₀))=𝜀, and 𝑔(𝑥)=0 for all 𝑥∈𝑋⧵𝑉. Therefore, 𝑓+𝑔_𝑋≤1,𝑓−𝑔_𝑋≤1. (31) Hence, for each 𝛿>0 we have max𝑓+𝑔_𝑋,𝑓−𝑔_𝑌<1+𝛿. (32) This completes the proof.

Lemma 8. Let 𝑇∶𝐶(𝑋,𝜏)→𝐶(𝑌,𝜂) be a linear isometry from 𝐶(𝑋,𝜏) onto 𝐶(𝑌,𝜂), and let 𝑎=𝑇1_𝑋. Then, |𝑎(𝑦)|=1 for all 𝑦∈𝑌.

Proof. Since 𝑎∈𝐶(𝑌,𝜂) and ‖𝑎‖_𝑌=‖𝑇1_𝑋‖_𝑌=‖1_𝑋‖_𝑋=1, by Lemma 7, it is sufficient to show that for each 𝜀>0 there exists 𝛿>0 such that 𝑔∈𝐶(𝑌,𝜂) with ‖𝑔‖_𝑌≥𝜀 implies that max𝑎+𝑔_𝑌,𝑎−𝑔_𝑋≥1+𝛿. (33)
Let 𝜀>0, and let 𝑔∈𝐶(𝑌,𝜂) with ‖𝑔‖_𝑌≥𝜀. The surjectivity of 𝑇 implies that there exists 𝑓∈𝐶(𝑋,𝜏) such that 𝑇𝑓=𝑔. Since 𝑇 is an isometry, we have ‖𝑓‖_𝑋=‖𝑔‖_𝑌≥𝜀. This implies that there exists 𝑥₀∈𝑋 such that |𝑓(𝑥₀)|≥𝜀. Therefore, max1+𝑓𝑥₀²,1−𝑓𝑥₀²≥121+𝑓𝑥₀²+1−𝑓𝑥₀²=1+𝑓𝑥₀²≥1+𝜀². (34) It follows that max1+𝑓𝑥₀,1−𝑓𝑥₀≥√1+𝜀². (35) Hence, max1_𝑋+𝑓_𝑋,1_𝑋−𝑓_𝑋≥√1+𝜀². (36) Since ‖𝑎+𝑔‖_𝑌=‖1_𝑋+𝑓‖_𝑋 and ‖𝑎−𝑔‖_𝑌=‖1_𝑋−𝑓‖_𝑋, we have max𝑎+𝑔_𝑌,𝑎−𝑔_𝑌≥√1+𝜀². (37) It is enough that we choose 𝛿=√1+𝜀²−1.

In the following result, we show that every linear isometry from 𝐶(𝑋,𝜏) onto 𝐶(𝑌,𝜂) is a weighted composition operator.

Theorem 9. Let 𝑇∶𝐶(𝑋,𝜏)→𝐶(𝑌,𝜂) be a linear isometry from 𝐶(𝑋,𝜏) onto 𝐶(𝑌,𝜂). Then, there exist a function 𝑎∈𝐶(𝑌,𝜂) with |𝑎(𝑦)|=1 for all 𝑦∈𝑌 and a homeomorphism ℎ from 𝑌 onto 𝑋 with ℎ∘𝜂=𝜏∘ℎ on 𝑌 such that 𝑇𝑓𝑦=𝑎𝑦𝑓ℎ𝑦,∀𝑓∈𝐶(𝑋,𝜏),∀𝑦∈𝑌. (38)

Proof. Assume that 𝑎=𝑇1_𝑋. Then 𝑎∈𝐶(𝑌,𝜂). By Lemma 8, we have |𝑎(𝑦)|=1 for all 𝑦∈𝑌. Clearly, (𝑎)(𝑇𝑓)∈𝐶(𝑌,𝜂) for all 𝑓∈𝐶(𝑋,𝜏) and (𝑎)(𝑇1_𝑋)=1_𝑌. We define the map 𝑎⋅𝑇∶𝐶(𝑋,𝜏)→𝐶(𝑌,𝜂) by (𝑎⋅𝑇)(𝑓)=(𝑎)(𝑇𝑓). Then, 𝑎⋅𝑇 is a linear isometry from 𝐶(𝑋,𝜏) onto 𝐶(𝑌,𝜂), and (𝑎⋅𝑇)(1_𝑋)=1_𝑌. By Theorem 5, there exists a homeomorphism ℎ from 𝑌 onto 𝑋 with ℎ∘𝜂=𝜏∘ℎ on 𝑌 such that (𝑎⋅𝑇)𝑓𝑦=𝑓ℎ𝑦,∀𝑓∈𝐶(𝑋,𝜏),∀𝑦∈𝑌. (39) This implies that 𝑇𝑓𝑦=𝑎𝑦𝑓ℎ𝑦,∀𝑓∈𝐶(𝑋,𝜏),∀𝑦∈𝑌. (40) Hence, the proof is complete.

3. Into Linear Isometries

We formulate our main result in this section which is a version for into linear isometries of 𝐶(𝑋,𝜏)-spaces of a known Holsztyńskis’s theorem (Theorem 2) for into linear isometries of 𝐶_𝔽(𝑋)-spaces. We first study unit-preserving linear isometries from 𝐶(𝑋,𝜏) into 𝐶(𝑌,𝜂).

Theorem 10. Let 𝑇∶𝐶(𝑋,𝜏)→𝐶(𝑌,𝜂) be a linear isometry from 𝐶(𝑋,𝜏) into 𝐶(𝑌,𝜂) satisfying the following conditions:(i)𝑇1_𝑋=1_𝑌,(ii)if 𝑦∈𝑌 and there is a point 𝑥∈𝑋 such that Re((𝑇𝑓)(𝑦))=Re(𝑓(𝑥)) for all 𝑓∈𝐶(𝑋,𝜏), then 𝑇(𝐶(𝑋,𝜏)) is extremely regular at {𝑦,𝜏(𝑦)}. Then, there exist a 𝜂-invariant closed boundary 𝑌₀ for 𝑇(𝐶(𝑋,𝜏)) and a continuous map ℎ from 𝑌₀ onto 𝑋 with ℎ∘𝜂=𝜏∘ℎ on 𝑌₀ such that 𝑇𝑓𝑦=𝑓ℎ𝑦,∀𝑓∈𝐶(𝑋,𝜏),∀𝑦∈𝑌₀. (41)

Proof. Let 𝐸=𝑇(𝐶(𝑋,𝜏)). Since 𝑇∶𝐶(𝑋,𝜏)→𝐶(𝑌,𝜂) is a linear isometry, we deduce that 𝐸 is a uniformly closed linear subspace of 𝐶(𝑌,𝜂). Moreover, 𝑇(1_𝑋)=1_𝑌 implies that 1_𝑌∈𝐸. We define the map 𝑆∶𝐶(𝑋,𝜏)→𝐸 by 𝑆𝑓=𝑇𝑓. Then, 𝑆 is a linear isometry from the real Banach space (𝐶(𝑋,𝜏),‖⋅‖_𝑋) onto the real Banach space (𝐸,‖⋅‖_𝑌). Therefore, 𝑆^∗ is a linear isometry from 𝐸^∗ onto 𝐶(𝑋,𝜏)^∗ and so 𝑆^∗Ext𝐵_𝐸^∗=Ext𝐵_{𝐶(𝑋,𝜏)}^∗. (42) Now we define 𝑌₀=𝑦∈𝑌∶𝑆^∗𝜑_{𝐶(𝑌,𝜂),𝑦,1𝐸}∈Ext𝐵_{𝐶(𝑋,𝜏)}^∗. (43) We first show that 𝑌₀ is nonempty. To prove this fact, we show that the following statement, namely (I), holds.
(I) For each 𝑥∈𝑋 there exists 𝑦₀∈𝑌 such that 𝑆^∗(𝜑_{𝐶(𝑌,𝜂),𝑦,1}|_𝐸)=𝜑_{𝐶(𝑋,𝜏),𝑥,1}.
Let 𝑥∈𝑋. Then 𝜑_{𝐶(𝑋,𝜏),𝑥,1}∈Ext(𝐵_{𝐶(𝑋,𝜏)}^∗) by Theorem 1. From (42), there exists Λ∈Ext(𝐵_𝐸^∗) such that 𝑆^∗Λ=𝜑_{𝐶(𝑋,𝜏),𝑥,1}. (44) Since 𝐸 is a linear subspace of 𝐶(𝑌,𝜂), there exists Λ∈Ext(𝐵_{𝐶(𝑌,𝜂)}^∗) such that Λ|_𝐸=Λ. (45) By Theorem 1, there exists (𝑦₀,𝛾₀)∈𝑌×𝕊_ℂ such that Λ=𝜑_{𝐶(𝑋),𝑦0,𝛾0}. (46) From (44), (45), and (46), we have 𝑆^∗𝜑_{𝐶(𝑌,𝜂),𝑦0,𝛾0𝐸}=𝜑_{𝐶(𝑋,𝜏),𝑥,1}. (47) Since 1_𝑋∈𝐶(𝑋,𝜏) and 𝑇1_𝑋=1_𝑌, by (47) we have Re𝛾₀=Re𝛾₀1_𝑌𝑦₀=𝜑_{𝐶(𝑌,𝜂),𝑦0,𝛾0}1_𝑌=𝜑_{𝐶(𝑌,𝜂),𝑦0,𝛾0}𝑇1_𝑋=𝜑_{𝐶(𝑌,𝜂),𝑦0,𝛾0}𝑆1_𝑋=𝑆^∗𝜑_{𝐶(𝑌,𝜂),𝑦0,𝛾0𝐸}1_𝑋=𝜑_{𝐶(𝑋,𝜏),𝑥,1}1_𝑋=Re1𝑒_{𝐶(𝑋),𝑥}1_𝑋=Re1_𝑋(𝑥)=Re(1)=1. (48) This implies that 𝛾₀=1 since |𝛾₀|=1. Hence, by (47) we have 𝑆^∗𝜑_{𝐶(𝑌,𝜂),𝑦0,1𝐸}=𝜑_{𝐶(𝑋,𝜏),𝑥,1}, (49) and so statement (I) holds.
Since 𝜑_{𝐶(𝑋,𝜏),𝑥,1}∈Ext(𝐵_{𝐶(𝑋,𝜏)}^∗) for all 𝑥 in 𝑋, statement (I) implies that 𝑌₀ is nonempty.
We next show that the following statement, namely, (II), holds.
(II) For each 𝑦∈𝑌₀ there exists a unique point 𝑥∈𝑋 such that (𝑇𝑓)(𝑦)=𝑓(𝑥) for all 𝑓 in 𝐶(𝑋,𝜏).
Let 𝑦∈𝑌₀. Then, 𝑆^∗(𝜑_{𝐶(𝑌,𝜂),𝑦,1}|_𝐸)∈Ext(𝐵_{𝐶(𝑋,𝜏)}^∗). Hence, by Theorem 1, there exists (𝑥₀,𝜆₀)∈𝑋×𝕊_ℂ such that 𝑆^∗𝜑_{𝐶(𝑌,𝜂),𝑦,1𝐸}=𝜑_{𝐶(𝑋,𝜏),𝑥0,𝜆0}. (50) Since 1_𝑋∈𝐶(𝑋,𝜏) and 𝑇1_𝑋=1_𝑌, by (50) we have Re𝜆₀=Re𝜆₀⋅1_𝑋𝑥₀=𝜑_{𝐶(𝑋,𝜏),𝑥0,𝜆0}1_𝑋=𝑆^∗𝜑_{𝐶(𝑌,𝜂),𝑦,1𝐸}1_𝑋=𝜑_{𝐶(𝑌,𝜂),𝑦,1}𝑆1_𝑋=𝜑_{𝐶(𝑌,𝜂),𝑦,1}𝑇1_𝑋=𝜑_{𝐶(𝑌,𝜂),𝑦,1}1_𝑌=Re1𝑒_{𝐶(𝑌,𝜂),𝑦}1_𝑌=Re𝑒_{𝐶(𝑌,𝜂),𝑦}1_𝑌=Re1_𝑌𝑦=Re(1)=1. (51) This implies that 𝜆₀=1 since |𝜆₀|=1. Thus, by (50) we have 𝑆^∗𝜑_{𝐶(𝑌,𝜂),𝑦,1𝐸}=𝜑_{𝐶(𝑋,𝜏),𝑥0,1}. (52) From (52) we give Re𝑆𝑓𝑦=Re𝑓𝑥₀,∀𝑓∈𝐶(𝑋,𝜏). (53) It follows that Re𝑇𝑓𝑦=Re𝑓𝑥₀,∀𝑓∈𝐶(𝑋,𝜏). (54)
According to (54) and condition (ii), 𝐸 is extremely regular at {𝑦,𝜂(𝑦)}. On the other hand, 𝐶(𝑋,𝜏) is extremely regular at {𝑥₀,𝜏(𝑥₀)} by Lemma 4. Hence, by Lemma 3, we deduce that (𝑆𝑓)(𝑦)=𝑓(𝑥₀) for all 𝑓 in 𝐶(𝑋,𝜏) or (𝑆𝑓)(𝑦)=𝑓(𝜏(𝑥₀)) for all 𝑓 in 𝐶(𝑋,𝜏). We assume that 𝑥=𝑥₀ whenever (𝑆𝑓)(𝑦)=𝑓(𝑥₀) for all 𝑓 in 𝐶(𝑋,𝜏) and 𝑥=𝜏(𝑥₀) whenever (𝑆𝑓)(𝑦)=𝑓(𝜏(𝑥₀)) for all 𝑓 in 𝐶(𝑋,𝜏). Then, 𝑥∈𝑋 and we have 𝑇𝑓𝑦=𝑓(𝑥),∀𝑓∈𝐶(𝑋,𝜏). (55) This proves the existent of 𝑥. To show uniqueness, assume that there exists 𝑥^′∈𝑋 such that 𝑇𝑓𝑦=𝑓𝑥^′,∀𝑓∈𝐶(𝑋,𝜏). (56) From (55) and (56) we have 𝑓(𝑥)=𝑓(𝑥^′) for all 𝑓 in 𝐶(𝑋,𝜏). This implies that 𝑥=𝑥^′ since 𝐶(𝑋,𝜏) separates the points of 𝑋. Thus, statement (II) holds.
Now we define the map ℎ∶𝑌₀→𝑋 by ℎ(𝑦)=𝑥 whenever (𝑇𝑓)(𝑦)=𝑓(𝑥) for all 𝑓 in 𝐶(𝑋,𝜏). The statement (II) implies that ℎ is well-defined. By definition of ℎ, we have 𝑇𝑓𝑦=𝑓ℎ𝑦,∀𝑓∈𝐶(𝑋,𝜏)∀𝑦∈𝑌₀. (57) We prove that ℎ∘𝜂=𝜏∘ℎ in 𝑌₀. Let 𝑦∈𝑌₀. For each 𝑓∈𝐶(𝑋,𝜏), by (57) we have 𝑓𝑦ℎ𝜂𝑦=𝜂𝑦=ℎ∘𝜂=𝑓𝑇𝑓𝑇𝑓𝑦=𝑓ℎ𝑦=𝑓𝜏ℎ𝑦=𝑓(𝜏∘ℎ)𝑦. (58) This implies that (ℎ∘𝜂)(𝑦)=(𝜏∘ℎ)(𝑦) since 𝐶(𝑋,𝜏) separates the points of 𝑋. Hence, ℎ∘𝜂=𝜏∘ℎ on 𝑌₀.
Continually, we show that 𝑌₀ is a 𝜂-invariant. Let 𝑦∈𝑌₀. For each 𝑓∈𝐶(𝑋,𝜏) we have 𝑆^∗𝜑_{𝐶(𝑌,𝜂),𝜂(𝑦),1𝐸}𝑓=𝜑_{𝐶(𝑌,𝜂),𝜂(𝑦),1}𝑆𝑓=𝜑_{𝐶(𝑌,𝜂),𝜂(𝑦),1}𝑇𝑓=Re1𝑒_{𝐶(𝑌),𝜂(𝑦)),1}𝜂𝑦𝑇𝑓=Re𝑇𝑓=Re𝑇𝑓𝑦=Re𝑓ℎ𝑦=Re1𝑓𝜏ℎ𝑦=Re1𝑒_{𝐶(𝑋),𝜏(ℎ(𝑦))𝐶(𝑋,𝜏)}𝑓=Re1𝑒_{𝐶(𝑋),𝜏(ℎ(𝑦))𝐶(𝑋,𝜏)}𝑓=𝜑_{𝐶(𝑋,𝜏),𝜏(ℎ(𝑦)),1}𝑓. (59) This implies that 𝑆^∗𝜑_{𝐶(𝑌,𝜂),𝜂(𝑦),1𝐸}=𝜑_{𝐶(𝑋,𝜏),𝜏(ℎ(𝑦)),1}. (60) Hence, 𝑆^∗(𝜑_{𝐶(𝑌,𝜂),𝑦,1}|_𝐸)∈Ext(𝐵_{𝐶(𝑋,𝜏)}^∗) by Theorem 1. Therefore, 𝜂(𝑦)∈𝑌₀ and so 𝑌₀ is 𝜂-invariant.
Now we show that ℎ is surjective. Let 𝑥∈𝑋. According to statement (I), we deduce that there exists a point 𝑦₀∈𝑌₀ such that 𝑆^∗𝜑_{𝐶(𝑌,𝜂),𝑦0,1}=𝜑_{𝐶(𝑋,𝜏),𝜏(𝑥),1}. (61) It follows that 𝑦₀∈𝑌 and we have Re𝑇𝑓𝑦₀=Re𝑓(𝑥),∀𝑓∈𝐶(𝑋,𝜏). (62) Statement (II) implies that there exists a point 𝑥^′∈𝑋 such that 𝑇𝑓𝑦₀=𝑓𝑥^′,∀𝑓∈𝐶(𝑋,𝜏). (63) From (62) and (63), we have Re(𝑓(𝑥))=Re(𝑓(𝑥^′)) for all 𝑓 in 𝐶(𝑋,𝜏). This implies that 𝑥^′∈{𝑥,𝜏(𝑥)} since Re𝐶(𝑋,𝜏) separates the points of 𝑋/𝜏. If 𝑥^′=𝑥, then by (63) and definition of ℎ we have 𝑥=ℎ(𝑦₀). If 𝑥^′=𝜏(𝑥), then by (63) and definition of ℎ we have 𝑇𝑓𝜂𝑦₀=𝑇𝑓𝑦₀=𝑓𝑥^′=𝑓𝜏𝑥^′=𝑓(𝑥), (64) for all 𝑓 in 𝐶(𝑋,𝜏), and so 𝑥=ℎ(𝜂(𝑦₀)). Therefore, ℎ is surjective.
Continually, we show that 𝑌₀ is a boundary for 𝐸. Let 𝑔∈𝐸. Then, there exists a function 𝑓∈𝐶(𝑋,𝜏) such that 𝑇𝑓=𝑔. Since 𝑓 is a continuous complex-valued function on the compact space 𝑋, there exists a point 𝑥₀∈𝑋 such that ‖𝑓‖_𝑋=|𝑓(𝑥₀)|. The surjectivity of ℎ∶𝑌₀→𝑋 implies that there exists a point 𝑦₀∈𝑌₀ such that ℎ(𝑦₀)=𝑥₀. Hence, (𝑇𝑓)(𝑦₀)=𝑓(𝑥₀), and so 𝑔_𝑌=𝑇𝑓_𝑌=𝑓_𝑋=𝑓𝑥₀=𝑇𝑓𝑦₀=𝑔𝑦₀. (65) Therefore, 𝑌₀ is a boundary for 𝐸.
We now check that ℎ∶𝑌₀→𝑋 is continuous. Let 𝑦∈𝑌₀, and let {𝑦_𝛼}_𝛼∈𝐽 be a net in 𝑌₀ such that lim_𝛼𝑦_𝛼=𝑦 in 𝑌₀. Then, for each 𝑓∈𝐶(𝑋,𝜏) we have lim_𝛼(𝑇𝑓)(𝑦_𝛼)=(𝑇𝑓)(𝑦) in 𝑋. This implies that for each lim_𝛼𝑓(ℎ(𝑦_𝛼))=𝑓(ℎ(𝑦)) in 𝑋. Since 𝐶(𝑋,𝜏) separates the points of 𝑋, we have lim_𝛼ℎ(𝑦_𝛼)=ℎ(𝑦) in 𝑋. Therefore, ℎ∶𝑌₀→𝑋 is continuous.
Finally, we show that 𝑌₀ is closed in 𝑌. Let 𝑦∈𝑌₀, the closure of 𝑌₀ in 𝑌. Then, there exists a net {𝑦_𝛼}_𝛼∈𝐽 in 𝑌₀ such that lim_𝛼𝑦_𝛼=𝑦 in 𝑌. Then, for each 𝑓∈𝐶(𝑋,𝜏) we have lim_𝛼(𝑇𝑓)(𝑦_𝛼)=(𝑇𝑓)(𝑦), and so lim_𝛼𝑓(ℎ(𝑦_𝛼))=(𝑇𝑓)(𝑦) in 𝑋. Since the net {𝑓(ℎ(𝑦_𝛼))}_𝛼∈𝐽 converges for all 𝑓 in 𝐶(𝑋,𝜏), we conclude that the net {ℎ(𝑦_𝛼)}_𝛼∈𝐽 converges in 𝑋 to a point 𝑥∈𝑋. Hence, for each 𝑓∈𝐶(𝑋,𝜏) we have lim_𝛼𝑓(ℎ(𝑦_𝛼))=𝑓(𝑥). Therefore, (𝑇𝑓)(𝑦)=𝑓(ℎ(𝑦)) for all 𝑓 in 𝐶(𝑋,𝜏). This implies that for each 𝑓∈𝐶(𝑋,𝜏) we have 𝑆^∗𝜑_{𝐶(𝑌,𝜂),𝑦,1𝐸}𝑓=𝜑_{𝐶(𝑌,𝜂),𝑦,1𝐸}=𝜑𝑆𝑓_{𝐶(𝑌,𝜂),𝑦,1}𝑇𝑓=Re1𝑒_{𝐶(𝑌),𝑦}𝑦𝑓ℎ𝑦𝑒𝑇𝑓=Re𝑇𝑓=Re=Re_{𝐶(𝑋),ℎ(𝑦)}𝑓=Re1𝑒_{𝐶(𝑋),ℎ(𝑦)𝐶(𝑋,𝜏)}𝑓=𝜑_{𝐶(𝑋,𝜏),ℎ(𝑦),1}𝑓. (66) Hence, 𝑆^∗𝜑_{𝐶(𝑌,𝜂),𝑦,1𝐸}=𝜑_{𝐶(𝑋,𝜏),ℎ(𝑦),1}. (67) This implies that 𝑆^∗(𝜑_{𝐶(𝑌,𝜂),𝑦,1}|_𝐸)∈Ext(𝐵_{𝐶(𝑋,𝜏)}^∗), by Theorem 1. Therefore, 𝑦∈𝑌₀. So 𝑌₀ is closed in 𝑌.

Note that if 𝑇∶𝐶(𝑋,𝜏)→𝐶(𝑌,𝜂) is a linear isometry from 𝐶(𝑋,𝜏) onto 𝐶(𝑌,𝜂), then 𝑇(𝐶(𝑋,𝜏))=𝐶(𝑌,𝜂) and so 𝑇(𝐶(𝑋,𝜏)) is extremely regular at {𝑦,𝜂(𝑦)} for all 𝑦 in 𝑌. Therefore, if 𝑇∶𝐶(𝑋,𝜏)→𝐶(𝑌,𝜂) is a linear isometry from 𝐶(𝑋,𝜏) on 𝐶(𝑌,𝜂) with 𝑇(1_𝑋)=1_𝑌, then conditions (i) and (ii) in Theorem 10 hold.

We now study the into case (not necessarily unit-preserving).

Theorem 11. Let 𝑇∶𝐶(𝑋,𝜏)→𝐶(𝑌,𝜂) be a linear isometry from 𝐶(𝑋,𝜏) into 𝐶(𝑌,𝜂) satisfying the following conditions:(i)|𝑇1_𝑋(𝑦)|=1 for all 𝑦∈𝑌, (ii)if 𝑦∈𝑌 and there is a point 𝑥∈𝑋 such that Re(𝑇1_𝑋(𝑦)(𝑇𝑓)(𝑦))=Re(𝑓(𝑥)) for all 𝑓 in 𝐶(𝑋,𝜏), then (𝑇1_𝑋)𝑇(𝐶(𝑋,𝜏)) is extremely regular at {𝑦,𝜏(𝑦)}. Then, there exist a 𝜂-invariant closed boundary 𝑌₀ for 𝑇(𝐶(𝑋,𝜏)), a function 𝑎∈𝐶(𝑌,𝜂) with |𝑎(𝑦)|=1 for all 𝑦∈𝑌, and a continuous map ℎ from 𝑌₀ onto 𝑋 with ℎ∘𝜂=𝜏∘ℎ on 𝑌₀ such that 𝑇𝑓𝑦=𝑎𝑦𝑓ℎ𝑦,∀𝑓∈𝐶(𝑋,𝜏),∀𝑦∈𝑌₀. (68)

Proof. Assume that 𝑎=𝑇1_𝑋. Then, 𝑎∈𝐶(𝑌,𝜂), and by (i) we have |𝑎(𝑦)|=1 for all 𝑦∈𝑌. Clearly, (𝑎)(𝑇𝑓)∈𝐶(𝑌,𝜂) for all 𝐶(𝑌,𝜂), and (𝑎)(𝑇1_𝑋)=1_𝑌. We define the map 𝑎⋅𝑇∶𝐶(𝑋,𝜏)→𝐶(𝑌,𝜂) by (𝑎⋅𝑇)(𝑓)=(𝑎)(𝑇𝑓). It is easy to see that 𝑎⋅𝑇 is a linear isometry from 𝐶(𝑋,𝜏) into 𝐶(𝑌,𝜂) such that (𝑎⋅𝑇)(1_𝑋)=1_𝑌.
Applying (ii), we can easily show that if 𝑦∈𝑌 and there exists a point 𝑥∈𝑋 such that Re(((𝑎⋅𝑇)𝑓)(𝑦))=Re(𝑓(𝑥)) for all 𝑓 in 𝐶(𝑋,𝜏), then (𝑎⋅𝑇)(𝐶(𝑋,𝜏)) is extremely regular at {𝑦,𝜂(𝑦)}. Let 𝐸=𝑇(𝐶(𝑋,𝜏)). Then, 𝑎𝐸=(𝑎⋅𝑇)(𝐶(𝑋,𝜏)). We define 𝑆∶𝐶(𝑋,𝜏)→𝐸 by 𝑆𝑓=𝑇𝑓 and 𝑆₁∶𝐶(𝑋,𝜏)→𝑎𝐸 by 𝑆₁𝑓=(𝑎⋅𝑇)𝑓. Clearly, 𝑆 is a linear isometry from 𝐶(𝑋,𝜏) onto 𝐸, and 𝑆₁ is a linear isometry from 𝐶(𝑋,𝜏) onto 𝑎𝐸. Now, we define 𝑌₀=𝑦∈𝑌∶𝑆^∗𝜑_{𝐶(𝑌,𝜂),𝑦,𝑎(𝑦)𝐸}∈Ext𝐵_{𝐶(𝑋,𝜏)}^∗. (69) It is easy to see that 𝑌₀=𝑦∈𝑌∶𝑆₁^∗𝜑_{𝐶(𝑌,𝜂),𝑦,1𝑎𝐸}∈Ext𝐵_{𝐶(𝑋,𝜏)}^∗. (70) By given arguments in the proof of Theorem 10, 𝑌₀ is a 𝜂-invariant closed boundary for 𝑎𝐸, and there exists a continuous function ℎ from 𝑌₀ onto 𝑋 with ℎ∘𝜂=𝜏∘ℎ such that (𝑎⋅𝑇)𝑓𝑦=𝑓ℎ𝑦,∀𝑓∈𝐶(𝑋,𝜏),∀𝑦∈𝑌₀. (71) We now show that 𝑌₀ is a boundary for 𝐸. Let 𝑔∈𝐸. Then, 𝑎𝑔∈𝑎𝐸. Hence, there exists a point 𝑦₀∈𝑌₀ such that 𝑎𝑔_𝑌=𝑎𝑔𝑦₀. (72) Since |𝑎(𝑦)|=1 for all 𝑦∈𝑌, we conclude that 𝑎𝑔_𝑌=𝑔_𝑌. (73) From (72) and (73), we have 𝑔_𝑌=𝑎𝑔𝑦₀=𝑎𝑦₀𝑔𝑦₀=𝑎𝑦₀𝑔𝑦₀=𝑎𝑦₀𝑔𝑦₀=𝑎𝑦₀𝑔𝑦₀=𝑔𝑦₀. (74) Therefore, 𝑌₀ is a boundary for 𝐸.
Finally, we show that 𝑇𝑓𝑦=𝑎𝑦𝑓ℎ𝑦,∀𝑓∈𝐶(𝑋,𝜏),∀𝑦∈𝑌₀. (75) Let 𝑓∈𝐶(𝑋,𝜏) and 𝑦∈𝑌₀. From (73), we have (𝑎⋅𝑇)𝑓𝑦=𝑓ℎ𝑦. (76) Applying (76), we deduce that 𝑇𝑓𝑦=𝑎𝑦𝑎𝑦₀𝑇𝑓𝑦=𝑎𝑦(𝑎)𝑦𝑦𝑇𝑓=𝑎(𝑎⋅𝑇)𝑓𝑦=𝑎𝑦𝑓ℎ𝑦. (77) Hence, the proof is complete.

Note that if 𝑇∶𝐶(𝑋,𝜏)→𝐶(𝑌,𝜂) is a linear isometry from 𝐶(𝑋,𝜏) onto 𝐶(𝑌,𝜂), then |𝑇1_𝑥(𝑦)|=1 for all 𝑦 in 𝑌 by Lemma 8 and 𝑇(𝐶(𝑋,𝜏)) is extremely regular at {𝑦,𝜂(𝑦)} for all 𝑦 in 𝑌 by Lemma 4, and so the conditions (i) and (ii) in Theorem 11 hold.

The following result shows that the triple {𝑌₀,𝑎,𝜂} associated to isometry 𝑇 in Theorem 11 possesses a universal property.

Corollary 12. Let 𝑇∶𝐶(𝑋,𝜏)→𝐶(𝑌,𝜂) be a linear isometry from 𝐶(𝑋,𝜏) into 𝐶(𝑌,𝜂) satisfying the following conditions:(i)|𝑇1_𝑋(𝑦)|=1 for all 𝑦∈𝑌, (ii)if 𝑦∈𝑌 and there is a point 𝑥∈𝑋 such that Re(𝑇1_𝑋(𝑦)(𝑇𝑓)(𝑦))=Re(𝑓(𝑥)) for all 𝑓 in 𝐶(𝑋,𝜏), then 𝑇(𝐶(𝑋,𝜏)) is extremely regular at {𝑦,𝜏(𝑦)}. Let 𝑌₀ and ℎ be as in Theorem 11. If 𝑌₁ is a nonempty 𝜂-invariant subset (not necessarily closed) of  𝑌 and 𝑏∈𝐶(𝑌,𝜂) with |𝑏(𝑦)|=1 for all 𝑦∈𝑌₁ and 𝑘∶𝑌₁→𝑋 is a continuous map (not necessarily onto) with ℎ∘𝜂=𝜏∘𝑘 on 𝑌₁ such that 𝑇𝑓𝑦=𝑏𝑦𝑓𝑘𝑦,∀𝑓∈𝐶(𝑋,𝜏),∀𝑦∈𝑌₁, (78) then 𝑌₁⊆𝑌₀, 𝑏=𝑎|_𝑌1 and 𝑘=ℎ|_𝑌1.

Proof. Let 𝑦∈𝑌₁. Since 1_𝑋∈𝐶(𝑋,𝜏), taking 𝑓=1_𝑋 in the above expression of 𝑇𝑓, we have 𝑏(𝑦)=(𝑇1_𝑋)(𝑦)=𝑎(𝑦). Thus, 𝑏=𝑎|_𝑌1.
Let 𝐸=𝑇(𝐶(𝑋,𝜏)) and 𝑆∶𝐶(𝑋,𝜏)→𝐸 defined by 𝑆𝑓=𝑇𝑓. Then, 𝑌₀=𝑦∈𝑌∶𝑆^∗𝜑_{𝐶(𝑌,𝜂),𝑦,𝑎(𝑦)𝐸}∈Ext𝐵_{𝐶(𝑋,𝜏)}^∗. (79) Let 𝑦∈𝑌₁. Then, the above expression of 𝑇𝑓 reads as follows: 𝑇𝑓𝑦=𝑏𝑦𝑓𝑘𝑦,∀𝑓∈𝐶(𝑋,𝜏). (80) From (80), for each 𝑓∈𝐶(𝑋,𝜏) we have 𝑆^∗𝜑_{𝐶(𝑌,𝜂),𝑦,𝑎(𝑦)𝐸}𝑓=𝜑_{𝐶(𝑌,𝜂),𝑦,𝑎(𝑦)𝐸}𝑇𝑓=Re𝑎𝑦𝑒_{𝐶(𝑌),𝑦}𝑇𝑓=Re𝑎𝑦𝑇𝑓𝑦=Re𝑎𝑦𝑎𝑦𝑓𝑘𝑦=Re𝑎𝑦²𝑓𝑘𝑦𝑓𝑘𝑦==ReRe1𝑒_{𝐶(𝑋),𝑘(𝑦)}𝑓=𝜑_{𝐶(𝑋,𝜏),𝑘(𝑦),1}𝑓. (81) This implies that 𝑆^∗𝜑_{𝐶(𝑌,𝜂),𝑦,𝑎(𝑦)𝐸}=𝜑_{𝐶(𝑋,𝜏),𝑘(𝑦),1}. (82) From (82), we deduce that 𝑆^∗(𝜑_{𝐶(𝑌,𝜂),𝑦,𝑎(𝑦)}|_𝐸) is an extreme point of Ext𝐵_{𝐶(𝑋,𝜏)}^∗ by Theorem 1. Therefore, 𝑦∈𝑌₀. So 𝑌₁⊆𝑌₀.
We now show that 𝑘=ℎ|_𝑌1. Let 𝑦∈𝑌₁. By (80), we have 𝑇𝑓𝑦=𝑎𝑦𝑓𝑘𝑦,∀𝑓∈𝐶(𝑋,𝜏). (83) Since 𝑌₁⊆𝑌₀, we deduce that 𝑦∈𝑌₀, and so by Theorem 11 we have 𝑇𝑓𝑦=𝑎𝑦𝑓ℎ𝑦,∀𝑓∈𝐶(𝑋,𝜏). (84) Since |𝑎(𝑦)|=1, by (83) and (84) we have 𝑓𝑘𝑦=𝑓ℎ𝑦,∀𝑓∈𝐶(𝑋,𝜏). (85) This implies that 𝑘(𝑦)=ℎ(𝑦) since 𝐶(𝑋,𝜏) separates the points of 𝑋. Thus, 𝑘=ℎ|_𝑌1. Hence, the proof is complete.

Acknowledgment

The authors would like to thank the referee for some valuable and useful comments.