Abstract

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

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 of Lipschitz functions on a compact metric space , equipped with the max norm , where denotes the Lipschitz constant of . Cambern [3] has considered isometries on spaces of scalar-valued continuously differentiable functions with norm given by for and determined a representation for the surjective isometries supported by such spaces. Rao and Roy [4] proved that surjective isometries on and with respect to the norm are of canonical forms in the sense that they are weighted composition operators. They asked whether a surjective isometry on with respect to the sum norm for is induced by an isometry on (note that for every . The reason is as follows. Let . Then is absolutely continuous. Hence the derivative exists almost everywhere on , and it is integrable by the theory of the absolutely continuous functions. Furthermore the equalityholds. As we see that is essentially bounded. In fact,assures that . By (1) we haveIt follows that . We conclude that . Thus .) 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 , , , and . In particular, Jarosz [5, Theorem] proved that a unital isometry between unital semisimple commutative Banach algebras with natural norms is canonical. By a theorem of Jarosz [5] a surjective unital isometry on 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 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 ; see also [9]. Botelho and Jamison [10] studied isometries on with . 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 algebras , , and , where is a compact metric space, , or ( denotes the unit circle on the complex plane), and is a compact Hausdorff space. We study isometries without assuming that they preserve unit. As corollaries of a general result we describe isometries on , , , and , respectively.

The main result in this paper is Theorem 14, which gives the form of a surjective isometry with respect to the sum norm between certain Banach algebras with the values in a commutative unital -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 for an with on (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 is a surjective isometry fixing the unit. Then by applying a theorem of Jarosz [5] (Theorem 1 in this paper) we see that is also an isometry with respect to the supremum norm. By the Banach-Stone theorem is an algebra isomorphism. Then by applying Lumer’s method (cf. [30]) we see that 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

Let be a compact Hausdorff space. Let be a real or complex Banach space. The space of all -valued continuous maps on is denoted by . When (resp. ), is abbreviated by (resp. ). For a subset of , the supremum norm of on is for . When no confusion will result we omit the subscript and write only . Let be a compact metric space and . For , putThen is called an -Lipschitz number of , or just a Lipschitz number of . When we omit the subscript and write only . The space of all such that is denoted by . When the subscript is omitted and it is written as .

When the closed subspace of is called a little Lipschitz space. There are a variety of complete norms on and . In this paper we are mainly concerned with the norm of (resp., ) which is defined byThe norm of (resp., ) is defined byNote that (resp., ) is a Banach space with respect to and , respectively. If is a Banach algebra, the norm is multiplicative. Hence (resp., ) is a (unital) Banach algebra with respect to the norm if is a (unital) Banach algebra. The norm fails to be submultiplicative even if is a Banach algebra. For a metric on , the Hölder metric is defined by for . is isometrically isomorphic to .

We are mainly concerned with in this paper. Then and are unital semisimple commutative Banach algebras with , when (resp., ) is abbreviated to (resp. ).

Let for or . We say that is continuously differentiable if there exists such thatfor every . We denote . PutThen with norm is a unital semisimple commutative Banach algebra. If is singleton we may suppose that is isometrically isomorphic to and we abbreviate by .

By identifying with we may assume that (resp., ) is a subalgebra of by the correspondenceThroughout the paper we may suppose that We say that a subset of is point separating if separates the points of . The unit of commutative Banach algebra is denoted by . The maximal ideal space of is denoted by . Suppose that is a unital point separating subalgebra of equipped with a Banach algebra norm. Then is semisimple because is a maximal ideal of for every and the Jacobson radical of 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 with is denoted by . For we putRecently Tanabe pointed out by a private communication that exists and it is finite for every . (In fact, it is easy to see that is increasing since is convex. We also see that .) Let be a compact Hausforff space and a liner subspace of which contains constant functions. A seminorm on is called one-invariant (in the sense of Jarosz) if for all . Let . A norm on is called a -norm if there is a one-invariant seminorm on such that . A natural norm is a -norm for some .

Theorem 1 (Jarosz [5]). Let and be compact Hausdorff spaces, let and be complex-linear subspaces of and , respectively, and let . Assume and contain constant functions, and let , be a -norm and -norm on and , respectively. Assume next that there is a linear isometry from onto with . Then if , or if and are regular subspaces of and , respectively, then is an isometry from onto .

In the sequel a unital semisimple commutative Banach algebra is identified via the Gelfand transforms with a subalgebra of . 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. Let and be unital semisimple commutative Banach algebras. Assume they have natural norms, respectively. Suppose that is a surjective complex-linear isometry with . Then there exists a homeomorphism such that

Proof. A unital semisimple commutative Banach algebra is regular by Proposition 2 in [5]. Then Theorem 1 ensures that is a surjective linear isometry from onto . It is easy to see that is extended to a surjective linear isometry from the uniform closure of onto the uniform closure of . Then a theorem of Nagasawa asserts that there exists a homeomorphism such that (, ). As we have the conclusion.

Corollary 3. Let be a compact metric space for . Suppose that is a surjective complex-linear isometry with respect to the norm . Assume . Then there exists a surjective isometry such that Conversely if is of the form as (14), then is a surjective isometry with respect to both of and such that .

Proof. As is a unital semisimple commutative Banach algebra with maximal ideal space , Corollary 2 asserts that there is a homeomorphism such that Then by a routine argument we see that is an isometry.
Converse statement is trivial.

Without assuming , we have that is a weighted composition operator. We exhibit a general result as Theorem 14 (see also [30]).

need not be a Banach algebra since need not be submultiplicative. On the other hand, is a natural norm in the sense of Jarosz (see [5]) such that . Then by Theorem 1 we have the following.

Corollary 4. Let be a compact metric space for . Suppose that is a surjective complex-linear isometry with respect to the norm . Assume . Then there exists a surjective isometry such thatConversely if is of a similar form as (16), then is a surjective isometry with respect to both of and such that .

Proof. As is a natural norm, we have by Corollary 2 that there is a homeomorphism such thatThen by a routine argument we see that is an isometry.
Converse statement is trivial.

Without the assumption that in Corollary 4, one may expect that 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 and 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]). Assume is a regular subspace of with and let . Then for any and any open neighborhood of , there is an such thatand for all .

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)Five ’s between 11 lines and 5 lines from the bottom of page 69 read as .(ii)Next reads as on the bottom of page 69.(iii)We point out that the term which appears on the first line of the first displayed inequalities on page 70 reads if .(iv)The term on the right hand side of the second line of the same inequalities reads as .(v)Two ’s on the same line read as .(vi)On the next line reads as .(vii)For any we infer that  Hence we have if by the first displayed inequalities of page 70.(viii)The inequality on the fifth line on page 70 reads as .

Let be a nonempty convex subset of the complex plane and . PutNote that we may writeLet be a subspace of for a compact Hausdorff space. For we put and , where denotes the closed convex-hull. We define the functions

Proof of Theorem 1. Let . First we note thatsince is the closed convex-hull of a compact set . We prove the inequalitieswhich appear on p. 68 in [5]. Put . As is compact, there exists such that . HenceAswe haveLet . By the definition of , we infer that , hence we have for every . Then Letting , we haveAs is arbitrary we haveIt follows that (24) holds. In the same way we havefor every . By (24) and (31) we infer thatAs is -invariant we haveAs is an isometry, , and is -invariant, we haveThusIt follows thatRecall that and . It follows thatSuppose that . Then we have by (37) that for every and . By Lemma 1 in [5] we infer that . Thus we have . We have proved that is an isometry from onto if .
Suppose that and are regular subspaces of and , respectively. Let . PutSuppose that . For any and any nonempty compact convex subset , we have thatfor all , where . Then by (37) we have for all . It follows by Lemma 1 in [5] thatand thereforeIf , then a similar calculation shows thatandIt follows that in any case (, ) we obtainWe will prove thatfor all . Once it is proved, applying the same argument for instead of , we see that for every . As is a bijection, it follows that for every . It will follow that for every . A proof of (46) is the following. For every , denote The inequality in (46) is deduced by the following assertions which appear in the proof of [5, Theorem]: (1) is a continuous mapping from onto .(2)For each , the set is dense in .(3)For each and each , it holds that . Suppose that these assertions are proved. Let . By (2), for any , there is a sequence of functions in such that as . By (3) we havefor every . Letting we haveby (1). As is arbitrary, we have thatWe 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 and which appear in the formulae and in [5] seem inappropriate; they read, for example, as and , respectively.
We now proceed to prove the first statement. Aiming for a contradiction, suppose that is not continuous from to . Let be a positive real number less than . Then there is a function such that and . Then there exist such that by [29, Proposition 6.3]. Since is complex-linear we may suppose that .
By (41) and (45), we deduce that and . As and , we infer that and . Let . Let . Then asserts that . ThusHence . As we haveConsider the open neighborhood of in given byWe infer that is a proper subset of by (52). Then, by [5, Lemma 2], there exists such that , , for every and for all . If denotes the closed rectangle whose vertices are the four points , we haveConsider now the setWe claim that . Suppose that . As is compact, there exists a positive integer such that , where . Then (52) gives . As is the closed convex-hull of , it is contained in the closed convex set . On the other hand, by (52). As , this contradicts , and this proves our claim. Hence there is with . As , it follows that and so . Hence . Thus is in . Thus we haveWe claim thatwhere . Let . Suppose first that . Since by (54), we haveSuppose next that . Then and so . Moreover, . Therefore we haveIt follows from (58) and (59) thatand henceas is claimed. Therefore we havePut . We claim that . If , there is nothing to prove. Suppose that . Then, by (41), we haveSince by (54), we haveAs does not include a closed disk with the radius greater than , we conclude that .
In the following we will consider two cases: and . Suppose first that . Then (64) yields . From we deduce that . Hence we haveSince is convex we haveFrom (39) we infer that Since , from (56) and (67) we obtain thatBy (63) and , we deduce that . Thus there is such that . It follows that ; hence we haveas and . We get by (62) and (69) thatOn the other hand, is invariant for any by (37). From (68) and (70) we deduce that and this contradicts that .
For the second case, suppose next that . Then, by (43), we haveand, by (54), it follows that . Moreover, since . ThenHence, . Using (39), we infer thatBy (71), we obtain that , and, as , we infer that . Hence , so thatas . Since , we obtain by (56) and (73) thatWe also obtain by (62) and (74) thatSince is invariant for any by (37), from (75) and (76) we deduce that and this is impossible since .
Next we show a proof of the second assertion (2). Let . We prove that there exists a sequence which uniformly converges to such that as . Without loss of generality we may assume that . Then there exists such that by [29, Proposition 6.3]. We may assume that . Suppose that . Putand(In the following we identify and ; that is, we identify and for every .) Since we assume that we infer by a simple calculation thatfor with . We assume that . By [5, Lemma 2] there exists such that , , on , and on . Put . Then andLet . Then we haveandHenceIt follows that we havefor . Suppose that . Thenand hencefor . Since , we have by combining (84) and (86) thatAs is convex we obtainRecall that for and a complex number denotes the closed disk with center and radius . We observe that . Recall thatLet be the line defined by the equationpart of which is a part of the boundary of . Let be the line defined by the equationBy some calculation we have that the distance between and is and it coincides with the distance between the point and the line . Hence we see thatThus .
Next we prove that for every with . It will follow that and the equality will hold. Let with . We prove the case where . A proof for the case where is the same and we omit it. We divide into two parts:andSuppose that and . The distance between and is . Hence . Suppose that and ; that is, . Let be the line passing through which is parallel to . Let be the unique point in the intersection of and the -axis. Then for some . Then the distance between and is , which is equal to the distance between the point and the line . On the other handIt follows that . We conclude that if satisfies , then . Thus we haveSince , we haveOn the other hand, as ensures that as . It follows that, for every , is dense in .
Finally we show a proof of the third assertion (3). As is pointed out in the proof of [5, Theorem], for any and any . Let and . Suppose that . Then by (43) we have . Hence we have As , we conclude by (45) thatThis completes the proof of the theorem.