#### Abstract

We will give the -Lipschitz version of the Banach-Stone type theorems for lattice-valued -Lipschitz functions on some metric spaces. In particular, when and are bounded metric spaces, if is a nonvanishing preserver, then is a weighted composition operator , where is a Lipschitz homeomorphism. We also characterize the compact weighted composition operators between spaces of Lipschitz functions.

#### 1. Introduction

The classical Banach-Stone theorem tells us that, when and are compact Hausdorff spaces, every linear surjective isometry from onto can be written as a weighted composition operator; that is, it is of the form where is a homeomorphism from onto and with for all . The theorem has many variable extensions concerning isometries, algebra isomorphisms, and disjointness preserving mappings between continuous function spaces; and we refer the surveys [1, 2] for more history about Banach-Stone theorems. Moreover, Kamowitz [3] gave the representation and spectrum of the compact weighted composition operators on the continuous functions.

Cao et al. stated a lattice version of the classical Banach-Stone theorem in [4]. Later, Chen et al. [5], Ercan and Önal [6, 7], and Miao et al. [8] generalized this result. When , are compact Hausdorff spaces and , are Banach lattices, by the main results of [5, 7], we can see that every vector lattice isomorphism from onto preserving the nonvanishing functions must be a weighted composition operator.

Garrido and Jaramillo [9, 10] and Weaver [11] tackled the Banach-Stone type theorem for lattices of real Lipschitz functions. Later, Jiménez-Vargas and Villegas-Vallecillos [12] proved that two little Lipschitz algebras are order isomorphic if and only if the corresponding compact metric spaces are Lipschitz homeomorphic. Recently, Jiménez-Vargas et al. [13] presented a Lipschitz version of the result in [5], in which the underlying spaces should be compact.

Our first goal of this paper is to prove the Banach-Stone type theorem in the setting of lattice-valued -Lipschitz functions. Section 2 is devoted to the preliminaries about vector lattices and -Lipschitz functions. Then we will give the -Lipschitz version of Banach-Stone theorem in Section 3. In particular, when , are bounded metric spaces, if is a nonvanishing preserver, then we will show that is a weighted composition operator , where is a Lipschitz homeomorphism. Our second aim is to give the characterization of compact weighted composition operators on the -Lipschitz functions.

#### 2. Preliminaries

An ordered vector space is said to be a *vector lattice* if exists for any , in . A vector lattice is said to be a *Banach lattice* if it is complete under its norm and satisfies the Riesz law:
where .

Let be a metric space and a Banach space; if , a function from to is said to be *-Lipschitz* if
The *-Lipschitz function space* is the space of all -valued -Lipschitz functions on . is the Banach space of all bounded -Lipschitz functions with the -Lipschitz norm
where . Furthermore, the *little Lipschitz space* is then defined to be the closed subspace of of these functions with the following property: for every , there exists such that whenever . is the subspace of consisting of all bounded functions. Notice that when , is just Lipschitz space . Moreover, if is a Banach lattice, then is a vector lattice with the usual pointwise order
However, is not a Banach lattice since does not satisfy the Riesz law in general.

A mapping from to is said to be a *-Lipschitz homeomorphism* if it is bijective and and are both -Lipschitz. If is in and is a vector in , denote by the function in . In particular, denotes the constant function on . For any function in , the zero set of is denoted by and its cozero set is , and is said to be *nonvanishing* if . An operator is said to be a *nonvanishing preserver* if
is said to be a *Riesz isomorphism* if and for any .

#### 3. Nonvanishing Preservers on Lipschitz Functions

In this section, our results will be valid (with the same proof) for different kinds of spaces. For this reason we first consider several situations to work in. Throughout this section we will assume that , , are metric spaces and , are Banach lattices.

*Context 1. *, .

*Context 2 (). *, .

This means that when we refer to , , , , we assume that all of them are included at the same time in one of the above two contexts.

Suppose that is a metric space and , for any , the function belongs to . Moreover, if , then we can find such that , and the function belongs to . The function defined in (7) or (8) has the property: , , and .

Theorem 1. *Let be a Riesz isomorphism preserving nonvanishing functions. Then carries the form
**
Here, is a homeomorphism from onto and all fiber linear maps are isomorphisms. *

*Remark 2. *When , the previous theorem is not valid for the little Lipschitz space , where is a connected Banach and is a Banach lattice. Note that if is a connected Banach spaces, we have that consisting of all -valued constant functions defined on . Let be any map from to and a linear bijection operator defined by
It is obvious that the operator is a Riesz isomorphism preserving nonvanishing functions with a weighted composition representation, but and are not homeomorphic.

It is easy to prove the following lemma.

Lemma 3. * preserves common zeros, that is,
**
for any and .*

*Proof of Theorem 1. *In the above contexts, and contain constant functions, so and , where , are defined in [14, Definition 3.8]. Therefore, by [14, Theorem 3.1] we can derive the result.

Lemma 4. *In the Contexts 1 and 2, is automatically continuous.*

*Proof. *We are going to use the Closed Graph Theorem to prove this lemma. Suppose that the sequence of functions converges to in and converges to in ; then for any and , we have that converges to in and converges to in , respectively. Notice that, for any , is continuous; then one can derive that
for all in . Since is a bijection from onto , we get that the sequence converges to for all in , and hence . This means that is a closed operator from to , and then is continuous.

In order to prove that is a -Lipschitz map from onto , we need the following lemma, and some idea of the proof comes from [15, Lemma 5.8].

Lemma 5. *For any fixed element with , we have that
*

*Proof. *By Theorem 1 we can also find a map from to (which is the set of all linear isomorphisms from to ) and a bijection from onto such that
for all and . From the definition of , , and , we can see that .

Suppose on the contrary that there exists a sequence such that for all . If has a limit point in , notice that preserves nonvanishing functions, then we can see that and hence . This leads to a contradiction. On the other hand, if there exists a positive scalar such that for any with , when we take the norm one element
then we can derive that
for all . Therefore, for any , we know that
Moreover, for any , by the similar manner of (7) and (8) we can define the function such that , for some , and for all such that . When put
we can see that belongs to and for . Then one can conclude that
and hence
This is a contradiction in Contexts 1 and 2 since .

Theorem 6. *Suppose that , are bounded metric spaces in the Contexts 1 and 2; then is a -Lipschitz map from onto .*

*Proof. *We can define the linear map from to by
We have to show that is well defined at first. For any fixed element with , from Lemma 5 we can choose a positive scalar such that for all in , and then it is easy to see that the function which maps to belongs to .

Assume that is a positive function in ; one can get that, for any ,
that is, is a bounded -Lipschitz function. Moreover, in Context 2 we can derive that is also a little Lipschitz function. This means that the function belongs to . Therefore, is a well-defined bijective linear operator from onto , and is also a nonvanishing preserver.

Suppose that is a sequence which converges to in and the sequence converges to in . For any and , , and hence we have that converges to for all . Notice that converges to ; one can conclude that converges to for all , and, since is a bijective map from onto , we have that for any in . Therefore, is a closed operator and hence is continuous.

For any and in , there exists a function such that and and (in fact, has the properties that we need). Here denotes the diameter of . Then we can derive that
Furthermore, we have that
and this means that is a -Lipschitz mapping from onto . Similarly, we can see that is also -Lipschitz, and then is a -Lipschitz homeomorphism.

For the spaces of scalar-valued Lipschitz functions, we give a complete characterization of nonvanishing preservers. But at first we need to recall a special case of [16, Lemma 25].

Lemma 7. *Let , be in Contexts 1 and 2. Suppose that is a linear nonvanishing preserver; then the map given by
**
is a Riesz isomorphism preserving nonvanishing functions. *

*Proof. *For completeness, we will sketch the proof. Observe that is never vanishing. If and , then if and only if if and only if if and only if . In particular, if , then . Let and . Then is a partition of into two open sets.

Suppose that and . Then on and on . Hence . For any , we have that , and and . Then we can derive that
This means that is well defined. Moreover, it is easy to check that is bijective.

From the previous paragraph, if , then . If and , then by the above,
By [17, Lemma 2.3], is biseparating, and hence . It follows that and thus . Therefore, . Thus is a Riesz isomorphism. It is trivial to check that if for any .

Theorem 8. *Suppose that , are bounded metric spaces and is a nonvanishing preserver between the following function spaces: *(i)* and ;*(ii)* and . **Then is a weighted composition operator of the form
**
Here and is a -Lipschitz map.*

*Proof. *By Lemma 7 we have that is a Riesz isomorphism. Then by Theorem 6 we can derive the conclusion.

In Theorem 8, the boundedness of the metric spaces can not be dropped.

*Example 9. *Let be the positive integers with the discrete metric, and we can derive that is not Lipchitz homeomorphic to . By [18, Example ] we can derive that , and then the identity map is a nonvanishing preserver. However, the underlying metric spaces are not Lipschitz homeomorphic.

#### 4. Compact Weighted Composition Operators on Lipschitz Spaces

Suppose that , are metric spaces, , and is a weighted composition operator, that is,
Here and is a -Lipschitz mapping. Put . Recall that is *supercontractive* on if for each there exists such that whenever and . In this section, we will characterize the compact weighted composition operator and consider its spectrum.

Theorem 10. *Suppose that is compact. For any , there is an open neighborhood of such that is supercontractive on and is totally bounded.*

*Proof. *Since , we can find an open neighborhood of such that for all . Suppose on the contrary that there exist such that and
for some . Without loss of generality we can assume that .

Let
we can derive that and for any . This implies that is a bounded sequence in . If is compact, then there exists a subsequence such that . Since uniformly, for any , we have that
and then . This means that in .

On the other hand, for any , by the Mean Value Theorem we have that
Here . Therefore, we can derive that , and this is a contradiction.

On the other hand, suppose on the contrary that is not totally bounded, then there exist a constant and such that whenever . Let
then it is easy to see that and . Moreover, for any , we can derive that
Therefore, has no Cauchy subsequence, and hence is not compact. This leads to a contradiction.

Theorem 11. *Suppose that is supercontractive on and is totally bounded; then the weighted composition operator defined by (29) is compact. *

*Proof. *Let be a bounded sequence, that is, for some . Since is totally bounded, there exists a subsequence of , which is also denoted by , such that is convergent uniformly in . Denote the limit by for all . It is easy to verify that is a bounded Lipschitz function in . By the similar argument of [18, Theorem ] we can extend to be a bounded Lipschitz function in , which is also denoted by . It suffices to show that converges to in .

Since is a weighted composition operator, it is easy to see that converges to uniformly on . Let be given. Since is supercontractive on , there exists such that
whenever and .

We will show that by dividing into four cases as the following arguments. For any with .*Case 1*. If , we have that for .

*Case 2*. If and , we have that
Moreover, by (36) we can derive that
*Case 3*. If and , we have that
*Case 4*. If and , we have that and then

Hence we derive that and then . This means that is a compact operator.

By the similar argument, one can conclude the following results for the scalar-valued little Lipschitz function spaces.

Theorem 12. *Let . Suppose that is a nonzero weighted composition operator of the form (29).*(1)*If is compact, then, for any , there is an open neighborhood of such that is supercontractive on and is totally bounded.*(2)*If is supercontractive on and is totally bounded, then is compact. *

Also here, the result of [19] also refers to the case where is a composition operator.

Corollary 13. *Suppose that , are compact metric spaces, and is a weighted composition operator of the form (29) between the following function spaces: *(i)* and ;*(ii)* and .**Then is compact if and only if is supercontractive on . *

When is a composition operator, that is, in the form (29), then and we can establish the following results in [20, Theorem 1.1].

Corollary 14. *Suppose that , are metric spaces and is a composition operator; then is compact if and only if is supercontractive and is totally bounded.*

In the following part of this section we have . Define and for all by induction. A point is said to be the fixed point of of order , , if and for any .

Theorem 15. *Let be a complete metric space and a weighted composition operator of form (29) satisfying: is supercontractive on and is totally bounded. Then we can derive that , where
*

*Proof. *Suppose that is a fixed point of of order . If for some , we can see that is not surjective and hence .

Assume that for any and .

When , we have that and . There exists such that . There is no such that . Indeed, if such exists, we can derive that
and this is impossible. This means that .

When , let , and define
Here . Then, similar to the argument of [3, Proposition 3], we can derive that .

On the other hand, for each with , for some , we will prove that . This implies that and completes the proof.

From the assumption , for all and , we derive that

Given any , let and ; here is any fixed number with . We provide that , which implies that , by dividing into the following cases.*Case I* (). If there exists such that , by (44) we can see that
This implies that .*Case II* ( and is finite). Let . Then there exists such that . This means that is a fixed point of of order . By (44), we have that
and since . Once again, by (44), we derive that
and .*Case III* (, is infinite and is infinite). Notice that . Since is totally bounded, we can derive that converges to a point . Moreover, since is supercontractive. Then we have that and . By (44) we can see that . Since is supercontractive, there exists such that
Choose such that for all , and we have, for any , that
That is,
Since is arbitrary, we can derive that , and then since (44).*Case IV* (, is infinite and is finite). We can choose such that for . From (44), we have that
and then
for all . This implies that as .

#### Acknowledgments

The authors would like to express their thanks to the referees for several helpful comments which improved the presentation of this paper. Research of the second author was partially supported by NSF of China (11301285, 11371201). The fourth author was supported by Department of Applied Mathematics and the Research Group for Nonlinear Analysis and Optimization, post-doctoral fellowship at the National Sun Yat-sen University when this work was started, and this research is supported in part by Taiwan NSC grant 102-2115-M-033-006.