# Hermitian Operators and Isometries on Banach Algebras of Continuous Maps with Values in Unital CommutativeC⁎-Algebras

Authors: Osamu Hatori
Journal: Journal of Function Spaces
 (2018)

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## Abstract

We study isometries on algebras of the Lipschitz maps and the continuously differentiable maps with the values in a commutative unitalC⁎-algebra. A precise proof of a theorem of Jarosz concerning isometries on spaces of continuous functions is exhibited.


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## Body

## 1. Introduction

In this paper an isometry means a complex-linear isometry. de Leeuw [1] probably initiated the study of isometries on the algebra of Lipschitz functions on the real line. Roy [2] studied isometries on the Banach space Lip(K) of Lipschitz functions on a compact metric space K, equipped with themax normfM=maxf∞,Lf, where L(f) denotes the Lipschitz constant of f. Cambern [3] has considered isometries on spaces of scalar-valued continuously differentiable functions C1([0,1]) with norm given by f=maxx∈[0,1]fx+f′x for f∈C1([0,1]) and determined a representation for the surjective isometries supported by such spaces. Rao and Roy [4] proved that surjective isometries on Lip([0,1]) and C1([0,1]) with respect to the norm f=f∞+f′∞ are of canonical forms in the sense that they are weighted composition operators. They asked whether a surjective isometry on Lip(K) with respect to thesum normfL=f∞+L(f) for f∈Lip(K) is induced by an isometry on K (note that f∞([0,1])+L(f)=f∞([0,1])+f′∞([0,1]) for every f∈Lip([0,1]). The reason is as follows. Let f∈Lip([0,1]). Then f is absolutely continuous. Hence the derivative f′ exists almost everywhere on [0,1], and it is integrable by the theory of the absolutely continuous functions. Furthermore the equality(1)fb-fa=∫abf′tdt,0≤a≤b≤1holds. As L(f)<∞ we see that f′ is essentially bounded. In fact,(2)f′t0=limt→t0ft0-ftt0-t≤Lfassures that f′∞([0,1])≤L(f). By (1) we have(3)fb-fab-a=1b-a∫abf′tdt≤1b-a∫abf′∞0,1dt=f′∞0,1.It follows that L(f)≤f′∞([0,1]). We conclude that L(f)=f′∞([0,1]). Thus f∞([0,1])+L(f)=f∞([0,1])+f′∞([0,1]).) Jarosz [5] and Jarosz and Pathak [6] studied a problem when an isometry on a space of continuous functions is a weighted composition operator. They provided a unified approach for certain function spaces including C1(K), Lip(K), lipα(K), and AC[0,1]. In particular, Jarosz [5, Theorem] proved that aunital isometry between unital semisimple commutative Banach algebras withnatural norms is canonical. By a theorem of Jarosz [5] a surjectiveunital isometry on Lip(K) is an algebra isomorphism when the norm is either the max norm or the sum norm. The situation is very different without assuming the unitality for the isometry with respect to the max norm. There is a simple example of a surjective isometry which is not canonical [7, p.242]. On the other hand, Jarosz and Pathak exhibited in [6, Example 8] that a surjective isometry on Lip(K) with respect to the sum norm is canonical. After the publication of [6] some authors expressed their suspicion about the argument there and the validity of the statement there had not been confirmed until quite recently. Hence the problem on isometries with respect to the sum norm has not been well studied.Jiménez-Vargas and Villegas-Vallecillos in [8] have considered isometries of spaces of Lipschitz maps on a compact metric space taking values in a strictly convex Banach space, equipped with the norm f=maxf∞,Lf; see also [9]. Botelho and Jamison [10] studied isometries on C1([0,1],E) with maxx∈[0,1]fxE+f′xE. See also [11–27]. Refer also to a book of Weaver [28].We propose a unified approach to the study of isometries with respect to the sum norm on Banach algebrasLip(K,C(Y)), lipα(K,C(Y)), and C1(K,C(Y)), where K is a compact metric space, [0,1], or T (T denotes the unit circle on the complex plane), and Y is a compact Hausdorff space. We study isometries without assuming that they preserve unit. As corollaries of a general result we describe isometries on Lip(K,C(Y)), lipα(K,C(Y)), C1([0,1],C(Y)), and C1(T,C(Y)), respectively.The main result in this paper is Theorem14, which gives the form of a surjective isometry U with respect to the sum norm between certain Banach algebras with the values in a commutative unital C∗-algebra. The proof of the necessity of the isometry in Theorem 14 comprises several steps. The crucial part of the proof of Theorem 14 is to prove that U(1)=1⊗h for an h∈C(Y2) with |h|=1 on Y2 (Proposition 15). To prove Proposition 15 we apply Choquet’s theory (cf. [29]) with measure theoretic arguments. A proof of Proposition 15 is completely the same as that of [30, Proposition 9]. Please refer to it. By Proposition 15 we have that U0=(1⊗h¯)U is a surjective isometry fixing the unit. Then by applying a theorem of Jarosz [5] (Theorem 1 in this paper) we see that U0 is also an isometry with respect to the supremum norm. By the Banach-Stone theorem U0 is an algebra isomorphism. Then by applying Lumer’s method (cf. [30]) we see that U0 is a composition operator of type BJ (cf. [31]).Our proofs in this paper make substantial use of the theorem of Jarosz [5, Theorem]. The author believes that it is convenient for the readers to show a precise proof because there need to be some ambitious changes in the original proof by Jarosz.


## 2. Preliminaries

LetY be a compact Hausdorff space. Let E be a real or complex Banach space. The space of all E-valued continuous maps on Y is denoted by C(Y,E). When E=C (resp. R), C(Y,E) is abbreviated by C(Y) (resp. CR(Y)). For a subset S of Y, the supremum norm of F on S is f∞(S)=supx∈SF(x)E for F∈C(Y,E). When no confusion will result we omit the subscript S and write only ·∞. Let K be a compact metric space and 0<α≤1. For F∈C(K,E), put(4)LαF=supx≠yFx-FyEdx,yα.Then Lα is called an α-Lipschitz number of F, or just a Lipschitz number of F. When α=1 we omit the subscript α and write only L(F). The space of all F∈C(K,E) such that Lα(F)<∞ is denoted by Lipα(K,E). When α=1 the subscript is omitted and it is written as Lip(K,E).When0<α<1 the closed subspace (5)lipαK,E=F∈LipαK,E:limx→x0fx0-fxEdx0,xα=0for  everyx0∈Kof Lipα(K,E) is called alittle Lipschitz space. There are a variety of complete norms on Lipα(K,E) and lipα(K,E). In this paper we are mainly concerned with the norm ·L of Lipα(K,E) (resp., lipα(K,E)) which is defined by(6)fL=f∞K+LαF,F∈LipαK,Eresp.lipαK,E.The norm ·M of Lipα(K,E) (resp., lipα(K,E)) is defined by(7)fM=max⁡f∞K,LαF,F∈LipαK,Eresp.lipαK,E.Note that Lip(K,E) (resp., lipα(K,E)) is a Banach space with respect to ·L and ·M, respectively. If E is a Banach algebra, the norm ·L is multiplicative. Hence Lip(K,E) (resp., lipα(K,E)) is a (unital) Banach algebra with respect to the norm ·L if E is a (unital) Banach algebra. The norm ·M fails to be submultiplicative even if E is a Banach algebra. For a metric d(·,·) on K, the Hölder metric is defined by dα for 0<α<1. Lipα((K,d),E) is isometrically isomorphic to Lip((K,dα),E).We are mainly concerned withE=C(Y) in this paper. Then Lipα(K,C(Y)) and lipα(K,C(Y)) are unital semisimple commutative Banach algebras with ·L, when E=CLip(K,C) (resp., lipα(K,C)) is abbreviated to Lip(K) (resp. lipα(K)).LetF∈C(K,C(Y)) for K=[0,1] or T. We say that F is continuously differentiable if there exists G∈C(K,C(Y)) such that(8)limt→t0Ft0-Ftt0-t-Gt0∞Y=0for every t0∈K. We denote F′=G. Put(9)C1K,CY=F∈CK,CY:Fis  continuously  differentiable.Then C1(K,C(Y)) with norm F=F∞+F′∞ is a unital semisimple commutative Banach algebra. If Y is singleton we may suppose that C(Y) is isometrically isomorphic to C and we abbreviate C1(K,C(Y)) by C1(K).By identifyingC(K,C(Y)) with C(K×Y) we may assume that Lip(K,C(Y)) (resp., lipα(K,C(Y))) is a subalgebra of C(K×Y) by the correspondence(10)F∈LipK,CY↔x,y↦Fxy∈CK×Y.Throughout the paper we may suppose that (11)LipX,CY⊂CX×Y,lipαX,CY⊂CX×Y,C1K,CY⊂CK×Y.We say that a subset Q of C(Y) is point separating if Q separates the points of Y. The unit of commutative Banach algebra B is denoted by 1. The maximal ideal space of B is denoted by MB. Suppose that B is a unital point separating subalgebra of C(Y) equipped with a Banach algebra norm. Then B is semisimple because {f∈B:f(x)=0} is a maximal ideal of B for every x∈X and the Jacobson radical of B vanishes.


## 3. A Theorem of Jarosz Revisited: Isometries Preserving Unit

Whether an isometry between unital semisimple commutative Banach algebras is of the canonical form depends not only on the algebraic structures of these algebras, but also on the norms in these algebra in most cases. A simple example is a surjective isometry on the Wiener algebra, which need not be canonical. Jarosz [5] defined natural norms and provided a theorem that isometries between a variety of algebras equipped with natural norms are of canonical forms. For the sake of completeness we outline the notations and the terminologies which are due to [5]. The set of all norms on R2 with p(1,0)=1 is denoted by P. For p∈P we put(12)Dp=limt→+0p1,t-1t.Recently Tanabe pointed out by a private communication that D(p) exists and it is finite for every p∈P. (In fact, it is easy to see that p(1,t)-1/t is increasing since p(1,t) is convex. We also see that inft>0p(1,t)-1/t>-∞.) Let X be a compact Hausforff space and A a liner subspace of C(X) which contains constant functions. A seminorm ⦀·⦀ on A is called one-invariant (in the sense of Jarosz) if ⦀f+1⦀=⦀f⦀ for all f∈A. Let p∈P. A norm · on A is called a p-norm if there is a one-invariant seminorm ⦀·⦀ on A such that ·=p(·∞,⦀·⦀). A natural norm is a p-norm for some p∈P.Theorem 1 (Jarosz [5]).
LetX and Y be compact Hausdorff spaces, let A and B be complex-linear subspaces of C(X) and C(Y), respectively, and let p,q∈P. Assume A and B contain constant functions, and let ·A, ·B be a p-norm and q-norm on A and B, respectively. Assume next that there is a linear isometry T from (A,·A) onto (B,·B) with T1=1. Then if D(p)=D(q)=0, or if A and B are regular subspaces of C(X) and C(Y), respectively, then T is an isometry from (A,·∞) onto (B,·∞).In the sequel a unital semisimple commutative Banach algebraA is identified via the Gelfand transforms with a subalgebra of C(MA). A unital semisimple commutative Banach algebra is regular (in the sense of Jarosz [5]). Hence we have by a theorem of Nagasawa [32] (cf. [33]) that the following holds.Corollary 2.
LetA and B be unital semisimple commutative Banach algebras. Assume they have natural norms, respectively. Suppose that T:A→B is a surjective complex-linear isometry with T1=1. Then there exists a homeomorphism φ:MB→MA such that(13)Tfx=f∘φx,f∈A,x∈MB.Proof.
A unital semisimple commutative Banach algebra is regular by Proposition 2 in [5]. Then Theorem 1 ensures that T is a surjective linear isometry from (A,·∞) onto (B,·∞). It is easy to see that T is extended to a surjective linear isometry T~ from the uniform closure A¯ of A onto the uniform closure B¯ of B. Then a theorem of Nagasawa asserts that there exists a homeomorphism φ:MB→MA such that T~(f)(x)=f∘φ(x) (f∈A¯, x∈MB). As T~|A=T we have the conclusion.Corollary 3.
LetKj be a compact metric space for j=1,2. Suppose that T:Lip(K1)→Lip(K2) is a surjective complex-linear isometry with respect to the norm ·L. Assume T1=1. Then there exists a surjective isometry φ:X2→X1 such that (14)Tfx=f∘φx,f∈LipK1,x∈K2.Conversely if T:Lip(K1)→Lip(K2) is of the form as (14), then T is a surjective isometry with respect to both of ·M and ·L such that T1=1.Proof.
As(Lip(Kj),·L) is a unital semisimple commutative Banach algebra with maximal ideal space Kj, Corollary 2 asserts that there is a homeomorphism φ:K2→K1 such that (15)Tfx=f∘φx,f∈LipK1,x∈K2.Then by a routine argument we see that φ is an isometry.
Converse statement is trivial.Without assumingT1=1, we have that T is a weighted composition operator. We exhibit a general result as Theorem 14 (see also [30]).( L i p ( K ) , · M ) need not be a Banach algebra since ·M need not be submultiplicative. On the other hand, ·M is a natural norm in the sense of Jarosz (see [5]) such that limt→+0max1,t-1/t=0. Then by Theorem 1 we have the following.Corollary 4.
LetKj be a compact metric space for j=1,2. Suppose that T:Lip(K1)→Lip(K2) is a surjective complex-linear isometry with respect to the norm ·M. Assume T1=1. Then there exists a surjective isometry φ:X2→X1 such that(16)Tf=f∘φ,f∈LipK1.Conversely if T:Lip(K1)→Lip(K2) is of a similar form as (16), then T is a surjective isometry with respect to both of ·M and ·L such that T1=1.Proof.
As·M is a natural norm, we have by Corollary 2 that there is a homeomorphism φ:K2→K1 such that(17)Tfx=f∘φx,f∈LipK1,x∈K2.Then by a routine argument we see that φ is an isometry.
Converse statement is trivial.Without the assumption thatT1=1 in Corollary 4, one may expect that T is a weighted composition operator. But it is not the case. A simple counterexample is given by Weaver [7, p.242] (see also [28]).As is pointed out in [34] the original proof of Theorem 1 needs a revision in some part and a proof when A and B are algebras of Lipschitz functions is revised [34, Proposition 7]. Although a revised proof for a general case is similar to that of Proposition 7 in [34], we exhibit it here for the sake of completeness of this paper. To prove Theorem 1 we need Lemma 2 in [5] in the same way as the original proof of Jarosz. The following is Lemma 2 in [5].Lemma 5 (Jarosz [5]).
AssumeA is a regular subspace of C(X) with 1∈A and let x0∈Ch(A). Then for any ε>0 and any open neighborhood U of x0, there is an f∈A such that(18)f∞≤1+ε,fx0=1,fx+1≤ε,x∈X∖U,and |Imf(x)|≤ε for all x∈X.Proof.
The proof is essentially due to the original proof of Lemma 2 in [5]. Several minor changes are needed. We itemize them as follows. (i)
Fiveε/2’s between 11 lines and 5 lines from the bottom of page 69 read as ε/3.(ii)
Nextx∈X∖U1 reads as x∈U1 on the bottom of page 69.(iii)
We point out that the term∑j=1k0-1‍(fj(x)-1) which appears on the first line of the first displayed inequalities on page 70 reads 0 if k0=1.(iv)
The term1+ε on the right hand side of the second line of the same inequalities reads as 1+ε/3.(v)
Twoε/2’s on the same line read as ε/3.(vi)
On the next linen+1/nε/2 reads as ε/3.(vii)
For any1≤k0≤n we infer that (19)1≥1-2k0-1n≥1-2n-1n>-1.Hence we have|f(x)|≤1+ε if x∈U1 by the first displayed inequalities of page 70.(viii)
The inequalityf∞≤ε on the fifth line on page 70 reads as f∞≤1+ε.LetK be a nonempty convex subset of the complex plane and φ∈[0,2π). Put(20)cK,φ=sup⁡a∈R:there  is  ab∈Rwitha+ibeiφ∈K.Note that we may write(21)cK,φ=supReλ:λ∈e-iφK.Let A be a subspace of C(X) for a compact Hausdorff space. For f∈A we put σ(f)=f(X) and σ~(f)=co(f(X)), where co(·) denotes the closed convex-hull. We define the functions(22)cA:A×0,2π→R,cAf,φ=cσ~f,φ,rA:A×R+×0,2π→R+,rAf,t,φ=f+eiφt1∞.Proof of Theorem1.
Letf∈A. First we note that(23)f∞=sup⁡z∈σf=sup⁡z∈σ~fsince σ~(f) is the closed convex-hull of a compact set σ(f)=f(X). We prove the inequalities(24)t+cAf,φ≤rAf,t,φ≤t+cAf,φ2+f∞2,t≥0,φ∈0,2π,which appear on p. 68 in [5]. Put s=cA(f,φ). As σ~(f) is compact, there exists b∈R such that (s+ib)eiφ∈σ~(f). Hence(25)eiφt+s+ib∈eiφt+σ~f=σ~eiφt1+f.As(26)t+s≤t+s≤t+s+ib=eiφt+s+ib≤eiφt1+f∞=rAf,t,φ,we have(27)t+cAf,φ≤rAf,t,φ.Let x∈X. By the definition of s=cA(f,φ), we infer that Re(e-iφf(x))≤s, hence we have Re(t+e-iφf(x))≤t+s for every t≥0. Then (28)Ree-iφfx≤Ret+e-iφfx≤t+s,t≥0.Letting M=maxt+s,Ree-iφfx, we have(29)eiφt+fx2=t+e-iφfx2=Ret+e-iφfx2+Imt+e-iφfx2≤M2+Imt+e-iφfx2≤t+s2+Ree-iφfx2+Ime-iφfx2=t+s2+e-iφfx2.
Asx∈X is arbitrary we have(30)rAf,t,φ=t+e-iφf∞≤t+s2+f∞2=t+cAf,φ2+f∞2.It follows that (24) holds. In the same way we have(31)t+cBg,φ≤rBg,t,φ≤t+cBg,φ2+g∞2,t≥0,φ∈0,2π,for every g∈B. By (24) and (31) we infer that(32)limt→∞rAf,t,φ-t=cAf,φ,limt→∞rBTf,t,φ-t=cBTf,φ.
As⦀·⦀A is 1-invariant we have(33)prAf,t,φ,⦀f⦀A=pf+eiφt1∞,⦀f+eiφt1⦀A.As T is an isometry, T1=1, and ⦀·⦀B is 1-invariant, we have(34)pf+eiφt1∞,⦀f+eiφt1⦀A=qTf+eiφt1∞,⦀Tf+eiφt1⦀B=qTf+eiφt1∞,⦀Tf+eiφt1⦀B=qrBTf,t,φ,⦀Tf⦀B.Thus(35)prAf,t,φ,⦀f⦀A=qrBTf,t,φ,⦀Tf⦀B.It follows that(36)0=limt→∞prAf,t,φ,⦀f⦀A-qrBTf,t,φ,⦀Tf⦀B=limt→∞rAf,t,φp1,⦀f⦀ArAf,t,φ-rAf,t,φ+limt→∞rBTf,t,φ-rBTf,t,φq1,⦀Tf⦀BrBTf,t,φ+limt→∞rAf,t,φ-t-rBTf,t,φ-t=⦀f⦀ADp-⦀Tf⦀BDq+cAf,φ-cBTf,φ.Recall that D(p)=limt→+0(p(1,t)-1)/t and D(q)=limt→+0(q(1,t)-1)/t. It follows that(37)⦀f⦀ADp-⦀Tf⦀BDq=cBTf,φ-cAf,φ,f∈A,φ∈0,2π.
Suppose thatD(p)=D(q)=0. Then we have by (37) that cB(Tf,φ)=cA(f,φ) for every f∈A and φ∈[0,2π). By Lemma 1 in [5] we infer that σ~(Tf)=σ~(f). Thus we have Tf∞=f∞. We have proved that T is an isometry from (A,·∞) onto (B,·∞) if D(p)=D(q)=0.
Suppose thatA and B are regular subspaces of C(X) and C(Y), respectively. Let f∈A. Put(38)Δf=⦀f⦀ADp-⦀Tf⦀BDq.Suppose that Δf≥0. For any r≥0 and any nonempty compact convex subset K⊂C, we have that(39)cK+Kr,φ=cK,φ+rfor all φ∈[0,2π), where K(r)=z∈C:z≤r. Then by (37) we have (40)cσ~Tf,φ=cBTf,φ=cAf,φ+Δf=cσ~f,φ+Δf=cσ~f+KΔf,φfor all φ∈[0,2π). It follows by Lemma 1 in [5] that(41)σ~Tf=σ~f+KΔf,and therefore(42)Tf∞=f∞+Δf.If Δf≤0, then a similar calculation shows that(43)σ~f=σ~Tf+K-Δfand(44)f∞=Tf∞+-Δf.It follows that in any case (Δf≥0, Δf≤0) we obtain(45)Δf=Tf∞-f∞,f∈A.
We will prove that(46)Tf∞-f∞=Δf≥0for all f∈A. Once it is proved, applying the same argument for T-1 instead of T, we see that T-1g∞-g∞≥0 for every g∈B. As T is a bijection, it follows that f∞-Tf∞≥0 for every f∈A. It will follow that Tf∞=f∞ for every f∈A. A proof of (46) is the following. For every ε>0, denote (47)Aε=f∈A:ρσ~f≤ε.The inequality in (46) is deduced by the following assertions which appear in the proof of [5, Theorem]: (1)
T is a continuous mapping from A,·∞ onto B,·∞.(2)
For eachε>0, the set Aε is dense in A,·∞.(3)
For eachε>0 and each f∈Aε, it holds that T(f)∞≥f∞-ε.
Suppose that these assertions are proved. Letf∈A. By (2), for any ε>0, there is a sequence {fn} of functions in Aε such that fn-f∞→0 as n→∞. By (3) we have(48)Tfn∞-fn∞≥-εfor every n. Letting n→∞ we have(49)Tf∞-f∞≥-εby (1). As ε>0 is arbitrary, we have that(50)Tf∞-f∞≥0.
We show proofs of three assertions (1), (2), and (3) above precisely. The proof of (1) is slightly different from the corresponding one in [5, p. 70]. This change is rather ambitious. We also point out that the terms -π/2 and π/2 which appear in the formulae (7) and (8) in [5] seem inappropriate; they read, for example, as 3π/4 and π/4, respectively.
We now proceed to prove the first statement. Aiming for a contradiction, suppose thatT is not continuous from A,·∞ to B,·∞. Let ε be a positive real number less than 1/100. Then there is a function f0∈A such that f0∞≤ε and T(f0)∞=1. Then there exist y0∈Ch(B) such that |T(f0)(y0)|=1 by [29, Proposition 6.3]. Since T is complex-linear we may suppose that T(f0)(y0)=1.
By (41) and (45), we deduce that Δf0=T(f0)∞-f0∞≥1-ε and σ~(T(f0))=σ~(f0)+K(Δf0). As f0∞≤ε and T(f0)∞=1, we infer that σ~(f0)⊂K(ε) and σ~(T(f0)⊂K(1). Let z∈K(1-2ε). Let x∈X. Then |f0(x)|≤ε asserts that w=z-f0(x)∈K(1-ε)⊂K(Δf0). Thus(51)z=f0x+w∈σ~f0+KΔf0=σ~Tf0.Hence K(1-2ε)⊂σ~(Tf0). As T(f0)∞=1 we have(52)K1-2ε⊂σ~Tf0⊂K1.Consider the open neighborhood U0 of y0 in X2 given by(53)U0=y∈X2:Tf0y-1<ε.We infer that U0 is a proper subset of X2 by (52). Then, by [5, Lemma 2], there exists g∈B such that g∞≤1+ε, g(y0)=1, |g(y)+1|<ε for every y∈X2∖U0 and |Img(y)|<ε for all y∈X2. If H denotes the closed rectangle whose vertices are the four points ±(1+ε)±εi, we have(54)σ~g⊂H.Consider now the set(55)L=e3πi/4z:z≤1,Rez≥1-2ε.We claim that T(f0)(X2)∩L≠∅. Suppose that T(f0)(X2)∩L=∅. As T(f0)(X2) is compact, there exists a positive integer n such that T(f0)(X2)∩Ln=∅, where Ln={e3πi/4z:|z|≤1,Rez>1-2ε-1/n}. Then (52) gives T(f0)(X2)⊂K(1)∖Ln. As σ~(T(f0)) is the closed convex-hull of T(f0)(X2), it is contained in the closed convex set K(1)∖Ln. On the other hand, (1-2ε)e3πi/4∈K(1-2ε)⊂σ~(T(f0)) by (52). As (1-2ε)e3πi/4∈L⊂Ln, this contradicts σ~(T(f0))⊂K(1)∖Ln, and this proves our claim. Hence there is y1∈X2 with T(f0)(y1)∈L. As ε≤1/100, it follows that |T(f0)(y1)-1|≥ε and so y1∈X2∖U0. Hence |g(y1)+1|<ε. Thus g(y1)+T(f0)(y1) is in L-1+K(ε). Thus we have(56)1+22-3ε≤cBg+Tf0,3π4.
We claim that(57)σ~g+Tf0⊂coK-1,1∪2+K3ε,where K(z0,r)=z∈C:z-z0≤r. Let y∈X2. Suppose first that |T(f0)(y)-1|<ε. Since g(X2)⊂H by (54), we have(58)Tf0y+gy∈K1,ε+H=H+1+Kε.Suppose next that |T(f0)(y)-1|≥ε. Then y∈X2∖U0 and so |g(y)+1|<ε. Moreover, |T(f0)(y)|≤1. Therefore we have(59)gy+Tf0y∈K1-1+Kε=K-1,1+Kε.It follows from (58) and (59) that(60)g+Tf0X2⊂K-1,1∪H+1+Kε,and hence(61)σ~g+Tf0⊂coK-1,1∪2+K3εas is claimed. Therefore we have(62)cBg+Tf0,π4≤2+3ε.
Putf1=T-1(g). We claim that Δf1≤ε. If Δf1<0, there is nothing to prove. Suppose that Δf1≥0. Then, by (41), we have(63)σ~g=σ~f1+KΔf1.Since σ~(g)⊂H by (54), we have(64)H⊃σ~f1+KΔf1.As H does not include a closed disk with the radius greater than ε, we conclude that Δf1≤ε.
In the following we will consider two cases:0≤Δf1≤ε and Δf1≤0. Suppose first that 0≤Δf1≤ε. Then (64) yields σ~(f1)⊂H. From f0∞≤ε we deduce that σ~(f0)⊂K(ε). Hence we have(65)σf1+f0⊂σf1+σf0⊂σ~f0+σ~f0⊂H+Kε.Since H+K(ε) is convex we have(66)σ~f1+f0⊂H+Kε.From (39) we infer that (67)cAf1+f0,3π4=cσ~f1+f0,3π4≤cH+Kε,3π4=cH,3π4+ε=22+1+2ε.Since T(f1+f0)=g+T(f0), from (56) and (67) we obtain that(68)1-4+2ε≤cBTf1+f0,3π4-cAf1+f0,3π4.By (63) and 1=g(y0), we deduce that 1∈σ~(f1)+K(Δf1). Thus there is z∈σ~(f1) such that |z-1|≤Δf1. It follows that 2/2-Δf1≤cA(f1,π/4); hence we have(69)22-2ε≤cAf1+f0,π4as f0∞≤ε and 0≤Δf1≤ε. We get by (62) and (69) that(70)cBTf1+f0,π4-cAf1+f0,π4≤22+5ε.On the other hand, cB(T(f),φ)-cA(f,φ) is invariant for any φ by (37). From (68) and (70) we deduce that ε≥(2-2)/2(9+2) and this contradicts that ε≤1/100.
For the second case, suppose next thatΔf1≤0. Then, by (43), we have(71)σ~f1=σ~g+K-Δf1,and, by (54), it follows that σ~(f1)⊂H+K(-Δf1). Moreover, σ~(f0)⊂K(ε) since f0∞≤ε. Then(72)σf1+f0⊂σf1+σf0⊂σ~f1+σ~f0⊂H+K-Δf1+Kε.Hence, σ~(f1+f0)⊂H+K(-Δf1)+K(ε). Using (39), we infer that(73)cAf1+f0,3π4≤cH+K-Δf1+Kε,3π4=cH,3π4+-Δf1+ε=22+1+2ε+-Δf1.By (71), we obtain that σ~(f1)⊃g(X2)+K(-Δf1), and, as g(y0)=1, we infer that σ~(f1)⊃1+K(-Δf1). Hence 2/2+(-Δf1)≤cA(f1,π/4), so that(74)22+-Δf1-ε≤cAf1+f0,π4as f0∞≤ε. Since T(f1+f0)=g+T(f0), we obtain by (56) and (73) that(75)1-4+2ε--Δf1≤cBTf1+f0,3π4-cAf1+f0,3π4.We also obtain by (62) and (74) that(76)cBTf1+f0,π4-cAf1+f0,π4≤22+4ε--Δf1.Since cB(T(f),φ)-cA(f,φ) is invariant for any φ by (37), from (75) and (76) we deduce that ε≥(2-2)/2(8+2) and this is impossible since ε≤1/100.
Next we show a proof of the second assertion (2). Letf∈A. We prove that there exists a sequence {fn}⊂A which uniformly converges to f such that ρ(σ~(fn))→0 as n→∞. Without loss of generality we may assume that f∞=1. Then there exists y0∈Ch(A) such that |f(y0)|=f∞ by [29, Proposition 6.3]. We may assume that f(y0)=1. Suppose that n≥4. Put(77)Un=x∈X:fx-1≤1n2and(78)Ω=coK1∪1+1n+K1n2.(In the following we identify R2 and C; that is, we identify (x,y) and x+iy for every x,y∈R.) Since we assume that n≥4 we infer by a simple calculation that(79)1-1n2-δnn,1+1n+δnn×-1n2-δnn,1n2+δnn⊂Ωfor δn with 0<δn≤1/n2. We assume that 0<δn≤1/n2. By [5, Lemma 2] there exists hn∈A such that hn∞≤1+2δn, hn(y0)=1, |hn+1|≤2δn on X∖Un, and |Imhn|≤2δn on X. Put gn=(hn+1)/2. Then gn∈A and(80)gn∞≤1+δn,gny0=1,gn≤δnonX∖Un,Imgn≤δnonX,-δn≤Regn≤1+δnonX.Let x∈Un. Then we have(81)1-1n2≤Refx≤1,-1n2≤Imfx≤1n2and(82)-δnn≤Regnxn≤1n+δnn,-δnn≤Imgnxn≤δnn.Hence(83)1-1n2-δnn≤Refx+gnxn≤1+1n+δnn,-1n2-δnn≤Imfx+gnxn≤1n2+δnn.It follows that we have(84)f+gnnx∈1-1n2-δnn,1+1n+δnn×-1n2-δnn,1n2+δnn⊂Ωfor x∈Un. Suppose that x∈X∖Un. Then(85)fx+gnxn≤1+δnn≤1+1n3<1+1n2,and hence(86)fx+gnxn∈K1+K1n2⊂Ωfor x∈X∖Un. Since 1+1/n=f(y0)+gn(y0)/n, we have by combining (84) and (86) that(87)1+1n∈f+gnnX⊂Ω.As Ω is convex we obtain(88)1+1n∈σ~f+gnn⊂Ω.Recall that for r≥0 and a complex number z0(89)Kz0,r=w∈C:w-z0≤rdenotes the closed disk with center z0 and radius r. We observe that ρ(Ω,1+1/n)=1/n+1/n2. Recall that(90)ρΩ,1+1n=sup⁡r:z∈Ω,1+1n∈Kz,r⊂Ω.Let l1 be the line defined by the equation(91)y=-n2n+1x-1+1n+1n2+1n3,part of which is a part of the boundary of Ω. Let l2 be the line defined by the equation(92)y=-n2n+1x-1-1n2.By some calculation we have that the distance between 1-1/n2 and 1+1/n is 1/n+1/n2 and it coincides with the distance between the point 1-1/n2 and the line l1. Hence we see that(93)1+1n∈K1-1n2,1n+1n2⊂Ω.Thus 1/n+1/n2≤ρ(Ω,1+1/n).
Next we prove thatK(p,r)∩Ωc≠∅ for every p∈Ω with r=|p-(1+1/n)|>1/n+1/n2. It will follow that 1/n+1/n2≥ρ(Ω,1+1/n) and the equality will hold. Let p=(xp,yp)=xp+iyp∈Ω with r=|p-(1+1/n)|>1/n+1/n2. We prove the case where yp≥0. A proof for the case where yp≤0 is the same and we omit it. We divide Ω+=q=xq,yq=xq+iyq∈Ω:yq≥0 into two parts:(94)Ω1=q=xq,yq=xq+iyq∈Ω+:-n2n+1xq-1-1n2≤yq≤-n2n+1x-1+1n+1n2+1n3and(95)Ω2=q=xq,yq=xq+iyq∈Ω+:yq≤-n2n+1xq-1-1n2.Suppose that p∈Ω1 and r=|p-(1+1/n)|>1/n+1/n2. The distance between l1 and l2 is 1/n+1/n2. Hence K(p,r)∩Ωc≠∅. Suppose that p∈Ω2 and r=|p-(1+1/n)|>1/n+1/n2; that is, p≠1-1/n2. Let l′ be the line passing through p which is parallel to l2. Let p′ be the unique point in the intersection of l′ and the x-axis. Then p′=1-1/n2-u for some u≥0. Then the distance between l1 and l′ is 1/n+1/n2+u/(1+1/n), which is equal to the distance between the point p and the line l1. On the other hand(96)p-1+1n≥p′-1+1n=1n+1n2+u.It follows that K(p,r)∩Ωc≠∅. We conclude that if p∈Ω satisfies r=|p-(1+1/n)|>1/n+1/n2, then K(p,r)∩Ωc≠∅. Thus we have(97)ρΩ,1+1n=1n+1n2.Since 1+1/n∈σ~(f+gn/n)⊂Ω, we have(98)ρσ~f+gnn≤ρσ~f+gnn,1+1n≤ρΩ,1+1n=1n+1n2.On the other hand, gn/n∞→0 as n→∞ ensures that f+gn/n-f∞→0 as n→∞. It follows that, for every ε>0, Aε is dense in (A,·∞).
Finally we show a proof of the third assertion (3). As is pointed out in the proof of [5, Theorem], ρ(co(K)+K(c))≥c for any K⊂C and any c≥0. Let ε>0 and f∈Aε. Suppose that Δf≤0. Then by (43) we have σ~(f)=σ~(Tf)+K(-Δf). Hence we have (99)ρσ~f=ρσ~Tf+K-Δf≥-Δf.As ε≥ρ(σ~(f)), we conclude by (45) that(100)Tf∞-f∞=Δf≥-ε.This completes the proof of the theorem.


## 4. Hermitian Operators on a Banach Algebras of Continuous Maps Whose Values Are in a Uniform Algebras

LetX and Y be compact Hausdorff spaces. Let B be a unital subalgebra of C(X) which separates the points of X. Throughout this section we assume B is a Banach algebra with the norm ·B and A is a uniform algebra on Y. Recall that a uniform algebra on Y is a uniformly closed subalgebra of C(Y) which contains constants and separates the points of Y. For functions f∈C(X) and g∈C(Y), let f⊗g∈C(X×Y) be the function defined by f⊗g(x,y)=f(x)g(y) for (x,y)∈X×Y, and for a subspace EX of C(X) and a subspace EY of C(Y), put(101)EX⊗EY=∑j=1nfj⊗gj:n∈N,fj∈EX,gj∈EY,and(102)1⊗EY=1⊗g:g∈EY.Throughout the section B~ is a unital subalgebra of C(X×Y) with a Banach algebra norm ·B~. We assume that B⊗A⊂B~. Note that B~ separates the points of X×Y since A separates the points of Y and B separates the points of X. We assume that there exists a compact Hausdorff space M and a complex-linear map D:B~→C(M) such that kerD=1⊗A. We assume that FB~=F∞(X×Y)+DF∞M for every F∈B~. Hence D is continuous. Defining(103)⦀F⦀=DF∞M,F∈B~,⦀·⦀ is aone-invariant seminorm in the sense of Jarosz; ⦀·⦀ is a seminorm on B~ such that ⦀F+1⦀=⦀F⦀ for every F∈B~. Hence the norm ·B~ is anatural norm (see [5, p.67]) Note that B~ is a regular subspace of C(X×Y) in the sense of Jarosz [5, Proposition 2].Lumer’s seminal paper [35] opened up a useful method of finding isometries which is often referred to as Lumer’s method. It involves the notion of Hermitian operators and the fact that UHU-1 must be Hermitian if H is Hermitian and U is a surjective isometry.Definition 6.
LetA be a unital Banach algebra. We say that e∈A is a Hermitian element if(104)exp⁡iteA=1for every t∈R. The set of all Hermitian elements of A is denoted by H(A).IfA is a unital C∗-algebra, then H(A) is the set of all self-adjoint elements of A. Hence H(Mn(C)) is the set of all Hermitian matrices, and H(C(Y))=CR(Y).Definition 7.
LetE be a complex Banach space. The Banach algebra of all bounded operators on E is denoted by B(E). We say that T∈B(E) is a Hermitian operator if T∈H(B(E)).Note that a Hermitian element of a unital Banach algebra and a Hermitian operator are usually defined in terms of numerical range or semi-inner product. Here we define them by an equivalent form (see [36]). By the definition of a Hermitian operator we have the following.Proposition 8.
LetEj be a complex Banach space for j=1,2. Suppose that V:E1→E2 is a surjective isometry and H:E1→E1 is a Hermitian operator. Then VHV-1:E2→E2 is a Hermitian operator.Proposition 9.
An elementF∈B~ is Hermitian if and only if there exists f∈A∩CR(Y) such that F=1⊗f.Proof.
Suppose thatF∈B~ is a Hermitian element. Then(105)exp⁡itF∞X×Y+DexpitF∞M=exp⁡itFB~=1for every t∈R. Suppose that there exists a point (x,y)∈X×Y with ImF(x,y)≠0, where Im denotes the imaginary part of a complex number. Suppose that ImF(x,y)>0. Then(106)exp-iF∞X×Y≥exp-iFx,y=expImFx,y>1.Suppose that ImF(x,y)<0. Then(107)exp⁡iF∞X×Y≥expiFx,y=exp-ImFx,y>1.In any case we have there exists t∈R such that(108)exp⁡itFB~≥expitF∞X×Y>1,which contradicts our assumption. We have that(109)F∈CRX×Y.Thus for every (s,t)∈X×Y and t∈R, expitFs,t=1. Hence exp(itF)∞(X×Y)=1 for every t∈R. By (105) we have D(exp(itF)∞(M)=0, which ensures that D(exp(itF))=0 for every t∈R. Thus exp⁡itF∈1⊗A for every t∈R. We have (110)exp⁡itF-1t-iF=∑n=2∞iFntn-2n!t,and hence for every t∈R with |t|≤1 we have(111)exp⁡itF-1t-iFB~≤∑n=2∞FB~ntn-2n!t≤∑n=2∞FB~nn!t≤exp⁡FB~t.It follows that(112)expitF-1t-iF∞X×Y≤exp⁡itF-1t-iFB~→0as t→0. Since exp(itF)-1/t∈1⊗A, for each t∈R there exists gt∈A such that(113)expitF-1t=1⊗gt.By (112) we have(114)1⊗g1/n-iF∞X×Y→0as n→∞. We have that {1⊗g1/n} is a Cauchy sequence in C(X×Y); thus we infer that{g1/n} is a Cauchy sequence in C(Y). Since A is uniformly closed as it is a uniform algebra, there exists g∈A such that(115)g1/n-g∞Y→0,and hence(116)1⊗g1/n-1⊗g∞X×Y→0as n→∞. It follows by (114) that iF=1⊗g; thus(117)F=1⊗-ig∈1⊗A.By (109) we see that -ig∈CR(Y); thus we have f=-ig∈A∩CR(Y) and F=1⊗f.
Suppose conversely thatB~∋F=1⊗f for f∈A∩CR(Y). We infer that F∈CR(X×Y) and |expitFx,y=1 for every t∈R and (x,y)∈X×Y. Hence exp(itF)∞(X×Y)=1 for every t∈R. Since (118)expitF=expit1⊗f=1⊗expitf∈1⊗A,we have D(exp(itF))=0. It follows that(119)expitFB~=exp⁡itF∞X×Y+DexpitF∞M=1for every t∈R. We conclude that F is a Hermitian element in B~.Note thatf∈A is Hermitian if and only if f∈A∩CR(Y) by [37, Proposition 5]. Hence Proposition 9 asserts that F is a Hermitian element in B~ if and only if F=1⊗f for a Hermitian element f in A.Proposition 10.
Suppose thatU:B~→B~ is a surjective unital isometry. Then U is an algebra isomorphism.Proof.
As we have already mentioned,B~ is a regular subspace (in the sense of Jarosz) with a natural norm. Then by Theorem 1U is also an isometry with respect to the supremum norm on X×Y. Then U is uniquely extended to a surjective isometry, with respect to the supremum norm, U~, from the uniform closure B~¯ onto itself. Since B~¯ is a uniform algebra, a theorem of Nagasawa [32] asserts that U~ is an algebra isomorphism since U~1=1. Thus U is an algebra isomorphism from B~ onto itself.Theorem 11.
A bounded operatorT:B~→B~ is a Hermitian operator if and only if T(1) is a Hermitian element in B~ and T=MT(1), the multiplication operator by T(1).Proof.
By Proposition10, every surjective unital isometry on B~ is multiplicative. Then by [37, Theorem 4], we have the conclusion.


## 5. Banach Algebras ofC(Y)-Valued Maps

Suppose thatX is a compact Hausdorff space. Suppose that B is a unital point separating subalgebra of C(Y) equipped with a Banach algebra norm. Then B is semisimple because {f∈B:f(x)=0} is a maximal ideal of B for every x∈X and the Jacobson radical of B vanishes. The inequality f∞≤fB for every f∈B is well known. We say that B isnatural if the map e:Y→MB defined by y↦ϕy, where ϕy(f)=f(y) for every f∈B, is bijective. We say that B isself-adjoint if B is natural and conjugate-closed in the sense that f∈B implies that f¯∈B for every f∈B, where ·¯ denotes the complex conjugation on Y.Definition 12.
LetX and Y be compact Hausdorff spaces. Suppose that B is a unital point separating subalgebra of C(X) equipped with a Banach algebra norm ·B. Suppose that B is self-adjoint. Suppose that B~ is a unital point separating subalgebra of C(X×Y) such that B⊗C(Y)⊂B~ equipped with a Banach algebra norm ·B~. Suppose that B~ is self-adjoint. We say that B~ is anaturalC(Y)-valuezation ofB if there exists a compact Hausdorff space M and a complex-linear map D:B~→C(M) such that kerD=1⊗C(Y) and D(CR(X×Y)∩B~)⊂CR(M) which satisfies(120)FB~=F∞X×Y+DF∞M,F∈B~.The term “a naturalC(Y)-valuezation of B” comes fromthe natural norm defined by Jarosz [5]. In fact the norm ·B~ is a natural norm in the sense of Jarosz [5].Note that(X,C(Y),B,B~) need not be an admissible quadruple defined by Nikou and O’Farrell [38] (cf. [31]) since we do not assume that {F(·,y):F∈B~,y∈Y}⊂B, which is a requirement for the admissible quadruple. On the other hand if (X,C(Y),B,B~) is an admissible quadruple of type L defined in [30], then B~ is a natural C(Y)-valuezation of B due to Definition 12.Example 13.
LetB=C1([0,1]) and B~=Lip([0,1],C(Y)) for Y=p, a singleton. Then Lip([0,1]) is algebraically isomorphic to Lip([0,1],C(Y)). Suppose that M is the maximal ideal space of L∞([0,1]) and D:B~→C(M) is defined by f↦Γ(f′), where Γ denotes the Gelfand transform in L∞([0,1]). Then B~ is a natural C(Y)-valuezation of B. The Banach algebra Lip([0,1]) with the norm f∞([0,1])+f′∞([0,1]) is isometrically isomorphic to B~.LetY be a compact Hausdorff space. Note that a closed subalgebra B~ of Lip((K,dα),C(Y)) which appears in Example 12 in [30] is an example of a natural C(Y)-valuezation of B. The Banach algebras C1([0,1],C(Y)) and C1(T,C(Y))) which appear in Examples 16 and 17 in [30], respectively, are also examples of natural C(Y)-valuezations of C1([0,1]).


## 6. Isometries on NaturalC(Y)-Valuezations

The main theorem in this paper is the following.Theorem 14.
Suppose thatBj~ is a natural C(Yj)-valuezation of Bj⊂C(Xj) for j=1,2. We assume that(121)1⊗hFBj~=FBj~for every F∈Bj~ and h∈C(Yj) with |h|=1 on Yj for j=1,2. Suppose that U:B1~→B2~ is a surjective complex-linear isometry. Then there exists h∈C(Y2) such that |h|=1 on Y2, a continuous map φ:X2×Y2→X1 such that φ(·,y):X2→X1 is a homeomorphism for each y∈Y2, and a homeomorphism τ:Y2→Y1 which satisfies(122)UFx,y=hyFφx,y,τy,x,y∈X2×Y2for every F∈B1~.In short a surjective isometry betweenC(Y)-valuezations is a weighted composition operator of a specific form: the homeomorphism X2×Y2→X1×Y1,(x,y)↦(φ(x,y),τ(y)) has the second coordinate that depends only on the second variable y∈Y2. A composition operator induced by such a homeomorphism is said to be of type BJ in [31, 37] after the study of Botelho and Jamison [39].Quite recently the author of this paper and Oi [30, Theorem 8] proved a similar result of Theorem 14 for admissible quadruples of type L. To prove it we apply Proposition 3.2 and the following comments in [31]. Instead of this we prove Theorem 14 by Lumer’s method, with which a proof is simpler than that in [30].In the following in this section we assume thatBj~ is a natural C(Yj)-valuezation of B⊂C(Xj) for j=1,2. We assume that(123)1⊗hFBj~=FBj~for every F∈Bj~ and h∈C(Yj) with |h|=1 on Yj for j=1,2. Suppose that U:B1~→B2~ is a surjective complex-linear isometry. A crucial part of a proof of Theorem 14 is to prove Proposition 15.Proposition 15.
Suppose thatX2 is not a singleton. There exists h∈C(Y2) with |h|=1 on Y2 such that U(1B~1)=1B2⊗h.A similar result for admissible quadruples of type L is proved in [30, Proposition 9]. If we assumed that(124)F·,y:F∈Bj~,y∈Yj⊂Bj,then Bj~ were an admissible quadruple of type L. Although Bj~ in this paper need not be an admissible quadruple of type L, a proof of Proposition 15 is completely the same as that in [30, Proposition 9] since we do not make use of the condition (124) in the proof of [30, Proposition 9]. The condition (124) is needed in [30] when we apply Proposition 3.2 and the following comments in [31].


## 7. Proof of Theorem14: An Application of Lumer’s Method

Proof of Theorem14.
A proof for the case whereX1=x1 and X2=x2 are singletons is the same as the proof of Theorem 8 in [30].
Suppose thatX2 is not a singleton. By Proposition 15 there exists h∈CR(Y2) with |h|=1 on Y2 such that U(1)=1⊗h. Letting U0:B1~→B2~ by U0(F)=(1⊗h¯)U(F), F∈B1~, we see by the hypothesis (1⊗h)FB2~=FB2~ for every F∈B2~ that U0 is a surjective unital isometry from B1~ onto B2~. Then Corollary 2 asserts that U0 is an algebra isomorphism. Let f∈CR(Y1). By Proposition 9, 1⊗f is a Hermitian element in B1~. Then by Theorem 11, M1⊗f is a Hermitian operator on B1~. By Proposition 8  U0M1⊗fU0-1 is a Hermitian operator on B2~. Then by Theorem 11 there exists S(f)∈CR(Y2) such that U0M1⊗fU0-1=M1⊗S(f). Hence an operator S:CR(Y1)→CR(Y2) is defined. Since U0 is an algebra isomorphism, it is easy to see that S is a real algebra isomorphism from CR(Y1) onto CR(Y2). Then Sˇ:C(Y1)→C(Y2) defined by Sˇ(f)=S(Ref)+iS(Imf) for f∈C(Y1) gives a complex algebra isomorphism. Gelfand theory asserts that there is a homeomorphism τ:Y2→Y1 such that Sˇ(f)=f∘τ, f∈C(Y1). It follows that(125)U0M1⊗fU0-1=M1⊗f∘τ,f∈CY1.Since U0-1(1)=1 we have(126)U01⊗f=1⊗f∘τ,f∈CY1.Define Φ:B1→B2~ by Φ(a)=U0(a⊗1), a∈B1. Since U0 is an algebra isomorphism, the map Φ is a unital homomorphism. Since the maximal ideal space of B1 is X1 and the maximal ideal space of B2~ is X2×Y2, there is a continuous map φ:X2×Y2→X1 such that (127)Φax,y=aφx,y,a∈B1,x,y∈X2×Y2.It follows by (126) and (127) that(128)U0a⊗fx,y=U0a⊗11⊗fx,y=U0a⊗1x,yU01⊗fx,y=aφx,yfτy=a⊗fφx,y,τy,x,y∈X2×Y2for every a∈B1 and f∈C(Y1). Thus(129)U0∑aj⊗fjx,y=∑aj⊗fjφx,y,τy,x,y∈X2×Y2for every ∑‍(aj⊗fj)∈B1⊗C(Y1). By the Stone-Weierstrass theorem B1⊗C(Y1) is uniformly dense in C(X1×Y1); hence any element in B1~ is uniformly approximated by B1⊗C(Y1). As U0 is also an isometry with respect to the uniform norm, we see that(130)U0Fx,y=Fφx,y,τy,x,y∈X1×Y1for every F∈B1~ and(131)UFx,y=1⊗hx,yFφx,y,τy=hyFφx,y,τy,x,y∈X1×Y1.As U0 is an algebra isomorphism, the map X2×Y2→X1×Y1 defined by (x,y)↦(φ(x,y),τ(y)) gives a homeomorphism. Therefore, for every y∈Y2, the map(132)φ·,y:X2→X1is a homeomorphism.
Suppose thatX1 is not a singleton. By the same way as in the last part of the proof of Theorem 8 in [30] we have that X2 is not a singleton. Then we have the conclusion by the previous argument.


## 8. Application of Theorem14

We exhibit applications of Theorem14.Corollary 16 ([4, Theorem 3.3]).
Suppose thatU:Lip([0,1])→Lip([0,1]) is a surjective isometry with respect to the norm defined by f∞([0,1])+f′∞([0,1]) for f∈Lip([0,1]). Then U(1) is a constant function of unit modulus such that(133)Ufx=U1fx,f∈Lip0,1,x∈0,1,or(134)Ufx=U1f1-x,f∈Lip0,1,x∈0,1.The converse statement also holds.Proof.
By Example13 we may suppose that Lip([0,1]) is a Banach algebra of C(Y)-valuezation. Applying Theorem 14 we have that U(1)=1⊗h for h∈C(Y) with |h|=1. Since our Y is a singleton, U(1) is a constant function of unit modulus. We also see that the corresponding continuous map φ:[0,1]×Y→[0,1] can be considered as a homeomorphism from [0,1] onto [0,1]; therefore we have that (135)Ufx=U1fφx,f∈Lip0,1,x∈0,1.The rest is a routine argument to prove that φ is an isometry; hence φ(x)=x, x∈[0,1] or φ(x)=1-x, x∈[0,1].
The converse statement is trivial.Corollaries 14, 15, 18, and 19 in [30, Section 6] follow here with a similar proof.



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