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

Authors: Davood Alimohammadi; Hadis Pazandeh
Journal: Journal of Operators
 (2013)

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## Abstract

LetX 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)*.


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## Body

## 1. Introduction and Preliminaries

Letℝ 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 𝕊𝔽 the set of all λ∈𝔽 with |λ|=1.Let𝔛 be a normed space over 𝔽. We denote by 𝔛* and B𝔛 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(BA*), then φ has an extension to some ψ∈Ext(B𝔛*).We know that if𝔛 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(B𝔛) and Ext(B𝒴).LetX be a compact Hausdorff space. We denote by C𝔽(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∈C𝔽(X). We write C(X) instead as Cℂ(X). For x∈X, we consider the linear functional eC𝔽(X),x on C𝔽(X) defined by eC𝔽(X),x(f)=f(x) (f∈C𝔽(X)), which is called the evaluation functional on C𝔽(X) at x. Clearly, λeC𝔽(X),x∈BC𝔽(X)* for all (x,λ)∈X×𝔽. It is known [2, page 441] that
(1)Ext(BC𝔽(X)*)={λeC𝔽(X),x:(x,λ)∈X×𝕊𝔽}.LetA 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).LetA be linear subspace of C𝔽(X) containing 1X, 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 K𝔽(A)={φ∈A*:∥φ∥=φ(1X)=1}. If x∈X, then eC𝔽(X),x|A∈K𝔽(A) and δx, the point mass measure on X at x, is a representing measure for eC𝔽(X),x|A. We denote by Ch(A,X) the set of all x∈X for which δx is the only representing measure for eC𝔽(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
(2)Γ(C𝔽(X),X)=Ch(C𝔽(X),X)=X.LetA 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.LetX 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.LetX be a topological space. An involution τ on X is called a topological involution on X, if τ is continuous.LetX be a compact Hausdorff space, and let τ be a topological involution on X. We define
(3)C(X,τ)={f∈C(X):f∘τ=f¯}.
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 iX, 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
(4)max{∥f∥X,∥g∥X}≤∥f+ig∥X≤2max{∥f∥X,∥g∥X},
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)=Cℝ(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 Cℝ(X).LetX 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.LetX 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
(5)(Ψf)(x,0)=f(x),(Ψf)(x,1)=f(x)¯(f∈C(X)r;x∈X),
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.LetX 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 1X. 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 Kℝ(A)={φ∈A*:∥φ∥=φ(1X)=1}. If x∈X, then Re(eC(X),x|A)∈Kℝ(A) and the measure mx=(1/2)(δx+δτ(x)) is a real part representing measure for Re(eC(X),x|A). We denote by Ch(A,X,τ) the set of all x∈X for which mx is the only real part representing measure for Re(eC(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 algebraC(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].LetX be a compact Hausdorff space. For each (x,λ)∈X×ℂ, define the map ψC(X),x,λ:C(X)r→ℝ by
(6)ψC(X),x,λ(f)=Re(λf(x)),(f∈C(X)r),
in fact, ψC(X),x,λ=Re(λeC(X),x). Clearly, ψC(X),x,λ∈B(C(X)r)*.Letτ be a topological involution on X. For each (x,λ)∈X×ℂ define the map φC(X,τ),x,λ:C(X,τ)→ℝ by φC(X,τ),x,λ=ψC(X),x,λ|C(X,τ). Kulkarni and Limaye showed [1, Proposition 3] that
(7)Ext(B(C(X)r)*)={ψC(X),x,λ:(x,λ)∈X×Sℂ}.
Applying this result, they obtained the following theorem.Theorem 1 (see [1, Corollary 5]).
LetX be a compact Hausdorff space, and let τ be a topological involution on X. Suppose that
(8)PX,τ={(x,λ)∈X×Sℂ:τ(x)≠x}∪{(x,λ)∈X×Sℝ:τ(x)=x}.
Then
(9)Ext(BC(X,τ)*)={φC(X,τ),x,λ:(x,λ)∈PX,τ}.The classical Banach-Stone theorem states that ifT is a linear isometry from C𝔽(X) onto C𝔽(Y), then there exists a homeomorphism h from Y onto X and a continuous function a∈C𝔽(Y) with a(Y)⊆𝕊𝔽 such that
(10)(Tf)(y)=a(y)f(h(y)),(f∈C𝔽(X),y∈Y).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.
LetX and Y be compact Hausdorff spaces. If T is a linear isometry from C𝔽(X) into C𝔽(Y), then there exists a closed subset Y0 of  Y, a continuous map φ from Y0 onto X and a function a∈C𝔽(Y) with ∥a∥Y=1 and a(Y0)⊆𝕊𝔽 such that
(11)(Tf)(y)=a(y)f(h(y)),∀f∈C𝔽(X),∀y∈Y0.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 fromC(X,τ) onto C(Y,η) and certain linear isometries from C(X,τ) into C(Y,η), applying the extreme points in BC(X,τ)* and BC(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 BC(X,τ)* and BC(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 fromC(X,τ) onto C(Y,η). For this purpose, we need the following lemmas.Lemma 3 (see [3, Lemma 5.1.1]).
LetA 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.
LetS 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 thatε>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.
LetT:C(X,τ)→C(Y,η) be a linear isometry from C(X,τ) onto C(Y,η) with T1X=1Y. Then, there exists a homeomorphism h from Y onto X with h∘η=τ∘h on Y such that
(12)(Tf)(y)=f(h(y)),∀f∈C(X,τ),∀y∈Y.Proof.
We claim that(13)Y={y∈Y:T*(φC(Y,η),y,1)∈Ext(BC(X,τ)*)}.
Let y∈Y. By Theorem 1, φC(Y,η),y,1∈Ext(BC(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
(14)T*(Ext(BC(Y,η)*))=Ext(BC(X,τ)*).
Therefore, T*(φC(Y,η),y,1)∈Ext(BC(X,τ)*). Hence, our claim is justified.
We now show that for eachy∈Y there exists a unique x∈X such that
(15)(Tf)(y)=f(x),∀f∈C(X,τ).
Let y∈Y. By (13), we have T*(φC(Y,η),y,1)∈Ext(BC(X,τ)*). Hence, by Theorem 1, there exists (x0,λ0)∈X×Sℂ such that
(16)T*(φC(Y,η),y,1)=φC(X,τ),x0,λ0.
Since 1X∈C(X,τ) and T1X=1Y, by (16), we have
(17)Re(λ0)=Re(λ01X(x0))=φC(X,τ),x0,λ0(1X)=T*(φC(Y,η),y,1)(1X)=φC(Y,η),y,1(T1X)=φC(Y,η),y,1(1Y)=Re(1Y(y)eC(Y),y(1Y))=Re(1Y(y)1Y(y))=Re(1)=1.
Since |λ0|=1 and Re(λ0)=1, we have Im(λ0)=0 and so λ0=1. Thus, by (16), we have
(18)T*(φC(Y,η),y,1)=φC(X,τ),x0,1.
From (18) we deduce that
(19)Re((Tf)(y))=Re(f(x0)),∀f∈C(X,τ).
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
(20)(Tf)(y)=f(x),∀f∈C(X,τ).
This proves the existent of x. To show uniqueness, assume that there exists x′∈X such that
(21)(Tf)(y)=f(x′),∀f∈C(X,τ).
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 maph:Y→X  by  h(y)=x whenever (Tf)(y)=f(x) for all f in C(X,τ). In fact, we have
(22)(Tf)(y)=f(h(y)),∀f∈C(X,τ)∀y∈Y.
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
(23)f((h∘η)(y))=f(h(η(y)))=(Tf)(η(y))=(Tf)(y)¯=f(h(y))¯=f(τ(h(y)))=f((τ∘h)(y)).
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 thath is bijective. Since T-1 is a linear isometry from C(Y,η) onto C(X,τ) with T-11Y=1X, by the above argument, we deduce that there exists a map k:X→Y such that
(24)(T-1g)(x)=g(k(x)),∀g∈C(Y,η),∀x∈X.
To prove the injectivity of h, let y∈Y. For each g∈C(Y,η), by (22) and (24) we have
(25)g((k∘h)(y))=g(k(h(y)))=(T-1g)(h(y))=(T(T-1g))(y)=g(y).
This implies that (k∘h)(y)=y since C(Y,η) separates the points of Y. Therefore, k∘h=IY, the identity mapping on Y, and so h is injective.
To prove thath is onto, let x∈X. For each f∈C(X,τ), by (24) and (22) we have
(26)f((h∘k)(x))=f(h(k(x)))=(Tf)(k(x))=(T-1(Tf))(x)=fx.
This implies that (h∘k)(x)=x since C(X,τ) separates the points of X. Therefore k∘h=IX, the identity mapping on X, and so h is onto.
We now check thath:Y→X is continuous. Let y∈Y, and let {yα}α∈J be a net in Y such that limαyα=y in Y. Then we have
(27)limα(Tf)(yα)=(Tf)(y),∀f∈C(X,τ).
By (27) and definition of h, we deduce that
(28)limαf(h(yα))=f(h(y)),∀f∈C(X,τ).
Since C(X,τ) separates the points of X, by (28), we have limαh(yα)=h(y) in X. Therefore, h:Y→X is continuous.
Sinceh: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.
LetT:C(X,τ)→C(Y,η) be an isometry mapping from C(X,τ) onto C(Y,η) with T(0)=0 and T1X=1Y. 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 T1X=1Y, applying Theorem 5, we deduce that there exists a homeomorphism h from Y onto X such that
(29)(Tf)(y)=f(h(y)),∀f∈C(X,τ),∀y∈Y.
It follows that T is a homomorphism. Hence, the proof is complete.We now study the onto case (not necessarily unit preserving).Lemma 7.
Letf∈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
(30)max{∥f+g∥X,∥f-g∥X}≥1+δ.Proof.
Assume that there existsx0∈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,
(31)∥f+g∥X≤1,∥f-g∥X≤1.
Hence, for each δ>0 we have
(32)max{∥f+g∥X,∥f-g∥Y}<1+δ.
This completes the proof.Lemma 8.
LetT:C(X,τ)→C(Y,η) be a linear isometry from C(X,τ) onto C(Y,η), and let a=T1X. Then, |a(y)|=1 for all y∈Y.Proof.
Sincea∈C(Y,η) and ∥a∥Y=∥T1X∥Y=∥1X∥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
(33)max{∥a+g∥Y,∥a-g∥X}≥1+δ.
Letε>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,
(34)max{|1+f(x0)|2,|1-f(x0)|2}≥12[|1+f(x0)|2+|1-f(x0)|2]=1+|f(x0)|2≥1+ε2.
It follows that
(35)max{|1+f(x0)|,|1-f(x0)|}≥1+ε2.
Hence,
(36)max{∥1X+f∥X,∥1X-f∥X}≥1+ε2.
Since ∥a+g∥Y=∥1X+f∥X and ∥a-g∥Y=∥1X-f∥X, we have
(37)max{∥a+g∥Y,∥a-g∥Y}≥1+ε2.
It is enough that we choose δ=1+ε2-1.In the following result, we show that every linear isometry fromC(X,τ) onto C(Y,η) is a weighted composition operator.Theorem 9.
LetT: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
(38)(Tf)(y)=a(y)f(h(y)),∀f∈C(X,τ),∀y∈Y.Proof.
Assume thata=T1X. 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¯)(T1X)=1Y. 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)(1X)=1Y. By Theorem 5, there exists a homeomorphism h from Y onto X with h∘η=τ∘h on Y such that
(39)((a¯·T)(f))(y)=f(h(y)),∀f∈C(X,τ),∀y∈Y.
This implies that
(40)(Tf)(y)=a(y)f(h(y)),∀f∈C(X,τ),∀y∈Y.
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 ofC(X,τ)-spaces of a known Holsztyńskis’s theorem (Theorem 2) for into linear isometries of C𝔽(X)-spaces. We first study unit-preserving linear isometries from C(X,τ) into C(Y,η).Theorem 10.
LetT:C(X,τ)→C(Y,η) be a linear isometry from C(X,τ) into C(Y,η) satisfying the following conditions:(i)
T1X=1Y,(ii)
ify∈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
(41)(Tf)(y)=f(h(y)),∀f∈C(X,τ),∀y∈Y0.Proof.
LetE=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(1X)=1Y implies that 1Y∈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
(42)S*(Ext(BE*))=Ext(BC(X,τ)*).
Now we define
(43)Y0={y∈Y:S*(φC(Y,η),y,1|E)∈Ext(BC(X,τ)*)}.
We first show that Y0 is nonempty. To prove this fact, we show that the following statement, namely (I), holds.
(I) For eachx∈X there exists y0∈Y such that S*(φC(Y,η),y,1|E)=φC(X,τ),x,1.
Letx∈X. Then φC(X,τ),x,1∈Ext(BC(X,τ)*) by Theorem 1. From (42), there exists Λ∈Ext(BE*) such that
(44)S*Λ=φC(X,τ),x,1.
Since E is a linear subspace of C(Y,η), there exists Λ~∈Ext(BC(Y,η)*) such that
(45)Λ~|E=Λ.
By Theorem 1, there exists (y0,γ0)∈Y×𝕊ℂ such that
(46)Λ~=φC(X),y0,γ0.
From (44), (45), and (46), we have
(47)S*(φC(Y,η),y0,γ0|E)=φC(X,τ),x,1.
Since 1X∈C(X,τ) and T1X=1Y, by (47) we have
(48)Re(γ0)=Re(γ01Y(y0))=φC(Y,η),y0,γ0(1Y)=φC(Y,η),y0,γ0(T1X)=φC(Y,η),y0,γ0(S1X)=S*(φC(Y,η),y0,γ0|E)(1X)=φC(X,τ),x,1(1X)=Re(1eC(X),x(1X))=Re(1X(x))=Re(1)=1.
This implies that γ0=1 since |γ0|=1. Hence, by (47) we have
(49)S*(φC(Y,η),y0,1|E)=φC(X,τ),x,1,
and so statement (I) holds.
SinceφC(X,τ),x,1∈Ext(BC(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 eachy∈Y0 there exists a unique point x∈X such that (Tf)(y)=f(x) for all f in C(X,τ).
Lety∈Y0. Then, S*(φC(Y,η),y,1|E)∈Ext(BC(X,τ)*). Hence, by Theorem 1, there exists (x0,λ0)∈X×𝕊ℂ such that
(50)S*(φC(Y,η),y,1|E)=φC(X,τ),x0,λ0.
Since 1X∈C(X,τ) and T1X=1Y, by (50) we have
(51)Re(λ0)=Re(λ0·1X(x0))=φC(X,τ),x0,λ0(1X)=S*(φC(Y,η),y,1|E)(1X)=φC(Y,η),y,1(S1X)=φC(Y,η),y,1(T1X)=φC(Y,η),y,1(1Y)=Re(1eC(Y,η),y(1Y))=Re(eC(Y,η),y(1Y))=Re(1Y(y))=Re(1)=1.
This implies that λ0=1 since |λ0|=1. Thus, by (50) we have
(52)S*(φC(Y,η),y,1|E)=φC(X,τ),x0,1.
From (52) we give
(53)Re((Sf)(y))=Re(f(x0)),∀f∈C(X,τ).
It follows that
(54)Re((Tf)(y))=Re(f(x0)),∀f∈C(X,τ).
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
(55)(Tf)(y)=f(x),∀f∈C(X,τ).
This proves the existent of x. To show uniqueness, assume that there exists x′∈X such that
(56)(Tf)(y)=f(x′),∀f∈C(X,τ).
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 maph: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
(57)(Tf)(y)=f(h(y)),∀f∈C(X,τ)∀y∈Y0.
We prove that h∘η=τ∘h in Y0. Let y∈Y0. For each f∈C(X,τ), by (57) we have
(58)f((h∘η)(y))=f(h(η(y)))=(Tf)(η(y))=(Tf)(y)¯=f(h(y))¯=f(τ(h(y)))=f((τ∘h)(y)).
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 thatY0 is a η-invariant. Let y∈Y0. For each f∈C(X,τ) we have
(59)S*(φC(Y,η),η(y),1|E)(f)=φC(Y,η),η(y),1(Sf)=φC(Y,η),η(y),1(Tf)=Re(1eC(Y),η(y)),1(Tf))=Re((Tf)(η(y)))=Re((Tf)(y)¯)=Re((f(h(y)))¯)=Re(1f(τ(h(y))))=Re(1eC(X),τ(h(y))|C(X,τ)(f))=(Re(1eC(X),τ(h(y))|C(X,τ)))(f)=φC(X,τ),τ(h(y)),1(f).
This implies that
(60)S*(φC(Y,η),η(y),1|E)=φC(X,τ),τ(h(y)),1.
Hence, S*(φC(Y,η),y,1|E)∈Ext(BC(X,τ)*) by Theorem 1. Therefore, η(y)∈Y0 and so Y0 is η-invariant.
Now we show thath is surjective. Let x∈X. According to statement (I), we deduce that there exists a point y0∈Y0 such that
(61)S*(φC(Y,η),y0,1)=φC(X,τ),τ(x),1.
It follows that y0∈Y and we have
(62)Re((Tf)(y0))=Re(f(x)),∀f∈C(X,τ).
Statement (II) implies that there exists a point x′∈X such that
(63)(Tf)(y0)=f(x′),∀f∈C(X,τ).
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
(64)(Tf)(η(y0))=(Tf)(y0)¯=f(x′)¯=f(τ(x′))=f(x),
for all f in C(X,τ), and so x=h(η(y0)). Therefore, h is surjective.
Continually, we show thatY0 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
(65)∥g∥Y=∥Tf∥Y=∥f∥X=|f(x0)|=|(Tf)(y0)|=|g(y0)|.
Therefore, Y0 is a boundary for E.
We now check thath:Y0→X is continuous. Let y∈Y0, and let {yα}α∈J be a net in Y0 such that limαyα=y in Y0. Then, for each f∈C(X,τ) we have limα(Tf)(yα)=(Tf)(y) in X. This implies that for each limαf(h(yα))=f(h(y)) in X. Since C(X,τ) separates the points of X, we have limαh(yα)=h(y) in X. Therefore, h:Y0→X is continuous.
Finally, we show thatY0 is closed in Y. Let y∈Y0¯, the closure of Y0 in Y. Then, there exists a net {yα}α∈J in Y0 such that limαyα=y in Y. Then, for each f∈C(X,τ) we have limα(Tf)(yα)=(Tf)(y), and so limαf(h(yα))=(Tf)(y) in X. Since the net {f(h(yα))}α∈J converges for all f in C(X,τ), we conclude that the net {h(yα)}α∈J converges in X to a point x∈X. Hence, for each f∈C(X,τ) we have limαf(h(yα))=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
(66)S*(φC(Y,η),y,1|E)(f)=(φC(Y,η),y,1|E)(Sf)=(φC(Y,η),y,1)(Tf)=Re(1eC(Y),y(Tf))=Re((Tf)(y))=Re(f(h(y)))=Re(eC(X),h(y)(f))=(Re(1eC(X),h(y)|C(X,τ)))(f)=φC(X,τ),h(y),1(f).
Hence,
(67)S*(φC(Y,η),y,1|E)=φC(X,τ),h(y),1.
This implies that S*(φC(Y,η),y,1|E)∈Ext(BC(X,τ)*), by Theorem 1. Therefore, y∈Y0. So Y0 is closed in Y.Note that ifT: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(1X)=1Y, then conditions (i) and (ii) in Theorem 10 hold.We now study the into case (not necessarily unit-preserving).Theorem 11.
LetT:C(X,τ)→C(Y,η) be a linear isometry from C(X,τ) into C(Y,η) satisfying the following conditions:(i)
|T1X(y)|=1 for all y∈Y,(ii)
ify∈Y and there is a point x∈X such that Re(T1X(y)¯(Tf)(y))=Re(f(x)) for all f in C(X,τ), then (T1X¯)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
(68)(Tf)(y)=a(y)f(h(y)),∀f∈C(X,τ),∀y∈Y0.Proof.
Assume thata=T1X. 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¯)(T1X)=1Y. 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)(1X)=1Y.
Applying (ii), we can easily show that ify∈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, a¯E=(a¯·T)(C(X,τ)). We define S:C(X,τ)→E by Sf=Tf and S1:C(X,τ)→a¯E 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 a¯E. Now, we define
(69)Y0={y∈Y:S*(φC(Y,η),y,a(y)¯|E)∈Ext(BC(X,τ)*)}.
It is easy to see that
(70)Y0={y∈Y:S1*(φC(Y,η),y,1|a¯E)∈Ext(BC(X,τ)*)}.
By given arguments in the proof of Theorem 10, Y0 is a η-invariant closed boundary for a¯E, and there exists a continuous function h from Y0 onto X with h∘η=τ∘h such that
(71)((a¯·T)(f))(y)=f(h(y)),∀f∈C(X,τ),∀y∈Y0.
We now show that Y0 is a boundary for E. Let g∈E. Then, a¯g∈a¯E. Hence, there exists a point y0∈Y0 such that
(72)∥a¯g∥Y=|(a¯g)(y0)|.
Since |a(y)|=1 for all y∈Y, we conclude that
(73)∥a¯g∥Y=∥g∥Y.
From (72) and (73), we have
(74)∥g∥Y=|a¯g(y0)|=|a¯(y0)g(y0)|=|a(y0)¯g(y0)|=|a(y0)¯||g(y0)|=|a(y0)||g(y0)|=|g(y0)|.
Therefore, Y0 is a boundary for E.
Finally, we show that(75)(Tf)(y)=a(y)f(h(y)),∀f∈C(X,τ),∀y∈Y0.
Let f∈C(X,τ) and y∈Y0. From (73), we have
(76)((a¯·T)(f))(y)=f(h(y)).
Applying (76), we deduce that
(77)(Tf)(y)=a(y)a(y0)¯(Tf)(y)=a(y)((a¯)(Tf))(y)=a(y)((a¯·T)f)(y)=a(y)f(h(y)).
Hence, the proof is complete.Note that ifT:C(X,τ)→C(Y,η) is a linear isometry from C(X,τ) onto C(Y,η), then |T1x(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.
LetT:C(X,τ)→C(Y,η) be a linear isometry from C(X,τ) into C(Y,η) satisfying the following conditions:(i)
|T1X(y)|=1 for all y∈Y,(ii)
ify∈Y and there is a point x∈X such that Re(T1X(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
(78)(Tf)(y)=b(y)f(k(y)),∀f∈C(X,τ),∀y∈Y1,
then Y1⊆Y0, b=a|Y1 and k=h|Y1.Proof.
Lety∈Y1. Since 1X∈C(X,τ), taking f=1X in the above expression of Tf, we have b(y)=(T1X)(y)=a(y). Thus, b=a|Y1.
LetE=T(C(X,τ)) and S:C(X,τ)→E defined by Sf=Tf. Then,
(79)Y0={y∈Y:S*(φC(Y,η),y,a(y)¯|E)∈Ext(BC(X,τ)*)}.
Let y∈Y1. Then, the above expression of Tf reads as follows:
(80)(Tf)(y)=b(y)f(k(y)),∀f∈C(X,τ).
From (80), for each f∈C(X,τ) we have
(81)S*(φC(Y,η),y,a(y)¯|E)(f)=(φC(Y,η),y,a(y)¯|E)(Tf)=Re(a(y)¯eC(Y),y(Tf))=Re(a(y)¯(Tf)(y))=Re(a(y)¯a(y)f(k(y)))=Re(|a(y)|2f(k(y)))=Re(f(k(y)))=(Re(1eC(X),k(y)))(f)=φC(X,τ),k(y),1(f).
This implies that
(82)S*(φC(Y,η),y,a(y)¯|E)=φC(X,τ),k(y),1.
From (82), we deduce that S*(φC(Y,η),y,a(y)¯|E) is an extreme point of ExtBC(X,τ)* by Theorem 1. Therefore, y∈Y0. So Y1⊆Y0.
We now show thatk=h|Y1. Let y∈Y1. By (80), we have
(83)(Tf)(y)=a(y)f(k(y)),∀f∈C(X,τ).
Since Y1⊆Y0, we deduce that y∈Y0, and so by Theorem 11 we have
(84)(Tf)(y)=a(y)f(h(y)),∀f∈C(X,τ).
Since |a(y)|=1, by (83) and (84) we have
(85)f(k(y))=f(h(y)),∀f∈C(X,τ).
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.



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*Source: 506129-2013-10-24.xml*