Nonvanishing Preservers and Compact Weighted Composition Operators between Spaces of Lipschitz Functions
Abstract
We will give the α-Lipschitz version of the Banach-Stone type theorems for lattice-valued α-Lipschitz functions on some metric spaces. In particular, when X and Y are bounded metric spaces, if T : Lip(X) → Lip(Y) is a nonvanishing preserver, then T is a weighted composition operator Tf = h · f∘φ, where φ : Y → X is a Lipschitz homeomorphism. We also characterize the compact weighted composition operators between spaces of Lipschitz functions.
1. Introduction
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 X, Y are compact Hausdorff spaces and E, F are Banach lattices, by the main results of [5, 7], we can see that every vector lattice isomorphism T from C(X, E) onto C(Y, F) 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 α-Lipschitz functions. Section 2 is devoted to the preliminaries about vector lattices and α-Lipschitz functions. Then we will give the α-Lipschitz version of Banach-Stone theorem in Section 3. In particular, when X, Y are bounded metric spaces, if T : Lip(X) → Lip(Y) is a nonvanishing preserver, then we will show that T is a weighted composition operator Tf = h · f∘φ, where φ : Y → X is a Lipschitz homeomorphism. Our second aim is to give the characterization of compact weighted composition operators on the α-Lipschitz functions.
2. Preliminaries
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 0 < α ≤ 1, X, Y are metric spaces and E, F are Banach lattices.
Context 1. , .
Context 2 (0 < α < 1). , .
This means that when we refer to X, Y, A(X, E), A(Y, F), we assume that all of them are included at the same time in one of the above two contexts.
Theorem 1. Let T : A(X, E) → A(Y, F) be a Riesz isomorphism preserving nonvanishing functions. Then T carries the form
Remark 2. When α = 1, the previous theorem is not valid for the little Lipschitz space lip1(X, E), where X is a connected Banach and E is a Banach lattice. Note that if X is a connected Banach spaces, we have that consisting of all E-valued constant functions defined on X. Let φ be any map from ℝ2 to ℝ and T : lip1(ℝ, E) → lip1(ℝ2, E) a linear bijection operator defined by
It is easy to prove the following lemma.
Lemma 3. T preserves common zeros, that is,
Proof of Theorem 1. In the above contexts, A(X, E) and A(Y, F) contain constant functions, so Ey = E and Fy = F, where Ey, Fy are defined in [14, Definition 3.8]. Therefore, by [14, Theorem 3.1] we can derive the result.
Proof. We are going to use the Closed Graph Theorem to prove this lemma. Suppose that the sequence of functions {fn} converges to f0 in A(X, E) and {Tfn} converges to g0 in A(Y, F); then for any x ∈ X and y ∈ Y, we have that {fn(x)} converges to f0(x) in E and {(Tfn)(y)} converges to g0(y) in F, respectively. Notice that, for any x ∈ X, Jφ−1(x) : E → F is continuous; then one can derive that
In order to prove that φ is a α-Lipschitz map from Y onto X, we need the following lemma, and some idea of the proof comes from [15, Lemma 5.8].
Lemma 5. For any fixed element e ∈ E with ∥e∥ = 1, we have that
Proof. By Theorem 1 we can also find a map from X to Iso(F, E) (which is the set of all linear isomorphisms from F to E) and a bijection from X onto Y such that
Suppose on the contrary that there exists a sequence {yn} ⊂ Y such that ∥T(1 ⊗ e)(yn)∥ = ∥(Jyn)(e)∥ ≤ 2−2n for all n ∈ ℕ. If {yn} has a limit point y′ in Y, notice that T preserves nonvanishing functions, then we can see that (Jy′)(e) = 0 and hence e = 0. This leads to a contradiction. On the other hand, if there exists a positive scalar τ > 0 such that dα(yn, ym) ≥ τ for any n, m ∈ ℕ with n ≠ m, when we take the norm one element
Theorem 6. Suppose that X, Y are bounded metric spaces in the Contexts 1 and 2; then φ is a α-Lipschitz map from Y onto X.
Proof. We can define the linear map from A(X) to A(Y) by
Assume that f is a positive function in A(X); one can get that, for any y1, y2 ∈ Y,
Suppose that {fn} is a sequence which converges to 0 in A(X) and the sequence converges to g0 in A(Y). For any n ∈ ℕ and y ∈ Y, , and hence we have that {fn(φ(y))} converges to g0(y) for all y ∈ Y. Notice that {fn} converges to 0; one can conclude that {fn(x)} converges to 0 for all x ∈ X, and, since φ is a bijective map from Y onto X, we have that g0(y) = 0 for any y in Y. Therefore, is a closed operator and hence is continuous.
For any y1 and y2 in Y, there exists a function f0 ∈ A(X) such that ∥f0∥ ≤ D(X) + D(X) 1−α and f0(φ(y1)) = d(φ(y1), φ(y2)) and f0(φ(y2)) = 0 (in fact, f0(x) = d(x, φ(y2)) has the properties that we need). Here D(X) denotes the diameter of X. Then we can derive that
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 A(X), A(Y) be in Contexts 1 and 2. Suppose that T : A(X) → A(Y) is a linear nonvanishing preserver; then the map S : A(X) → A(Y) given by
Proof. For completeness, we will sketch the proof. Observe that T1 is never vanishing. If f ∈ A(X) and λ ∈ ℝ, then λ ∈ range f if and only if 0 ∈ range(f − λ) if and only if 0 ∈ range(Tf − λT1) if and only if λ ∈ range Tf/T1. In particular, if f ≥ 0, then Tf/T1 ≥ 0. Let Y+ = {y ∈ Y : (T1)(y) > 0} and Y− = {y ∈ Y : (T1)(y) < 0}. Then Y+ ∪ Y− is a partition of Y into two open sets.
Suppose that f ∈ A(X) and f ≥ 0. Then Tf ≥ 0 on Y+ and Tf ≤ 0 on Y−. Hence Tf · T1/|T1| = |Tf | ∈ A(Y). For any f ∈ A(X), we have that f+, f− ∈ A(X), and |T(f+)| = T(f+) · T1/|T1| and |T(f−)| = T(f−) · T1/|T1|. Then we can derive that
From the previous paragraph, if 0 ≤ f ∈ A(X), then Sf = |Tf | ≥ 0. If f ∈ A(X) and g = Sf ≥ 0, then by the above,
Theorem 8. Suppose that X, Y are bounded metric spaces and T is a nonvanishing preserver between the following function spaces:
- (i)
0 < α ≤ 1 and T : Lipα(X) → Lipα(Y);
- (ii)
0 < α < 1 and T : Lipα(X) → Lipα(Y).
Proof. By Lemma 7 we have that T 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 ℕ1 be the positive integers with the discrete metric, and we can derive that ℕ1 is not Lipchitz homeomorphic to ℕ. By [18, Example 1.6.4] we can derive that Lipb(ℕ) = Lipb(ℕ1) = ℓ∞, and then the identity map I : Lipb(ℕ) → Lipb(ℕ1) is a nonvanishing preserver. However, the underlying metric spaces are not Lipschitz homeomorphic.
4. Compact Weighted Composition Operators on Lipschitz Spaces
Theorem 10. Suppose that T is compact. For any y0 ∈ Y∖Y0, there is an open neighborhood U0 of y0 such that φ is supercontractive on U0 and φ(U0) is totally bounded.
Proof. Since h(y0) ≠ 0, we can find an open neighborhood U0 of y0 such that |h(y)|≥|h(y0)|/2 > 0 for all y ∈ U0. Suppose on the contrary that there exist {xn}, {yn} ⊂ U0 such that d(xn, yn) → 0 and
Let
On the other hand, for any n ∈ ℕ, by the Mean Value Theorem we have that
On the other hand, suppose on the contrary that φ(U0) is not totally bounded, then there exist a constant τ > 0 and zn = φ(un) ∈ φ(U0) such that dα(zn, zm) > τ whenever n ≠ m. Let
Theorem 11. Suppose that φ is supercontractive on Y∖Y0 and φ(Y∖Y0) is totally bounded; then the weighted composition operator defined by (29) is compact.
Proof. Let be a bounded sequence, that is, ∥fn∥ ≤ M for some M > 0. Since φ(Y∖Y0) is totally bounded, there exists a subsequence of {fn}, which is also denoted by {fn}, such that {fn} is convergent uniformly in φ(Y∖Y0). Denote the limit by f0(x) for all x ∈ φ(Y∖Y0). It is easy to verify that f0 is a bounded Lipschitz function in φ(Y∖Y0). By the similar argument of [18, Theorem 1.5.6] we can extend f0 to be a bounded Lipschitz function in , which is also denoted by f0. It suffices to show that {Tfn} converges to Tf0 in .
Since T is a weighted composition operator, it is easy to see that {Tfn} converges to Tf0 uniformly on Y. Let ε > 0 be given. Since φ is supercontractive on Y∖Y0, there exists δ > 0 such that
We will show that Lα(Tfn − Tf0) → 0 by dividing into four cases as the following arguments. For any y1, y2 ∈ Y with y1 ≠ y2.
Case 1. If y1, y2 ∈ Y0, we have that (Tfn)(yi) = (Tf0)(yi) = 0 for i = 1,2.
Case 2. If y1, y2 ∈ Y∖Y0 and 0 < d(y1, y2) < δ, we have that
Case 3. If y1, y2 ∈ Y∖Y0 and d(y1, y2) > δ, we have that
Case 4. If y1 ∈ Y∖Y0 and y2 ∈ Y0, we have that h(y2) = 0 and then
Hence we derive that Lα(Tfn − Tf0) → 0 and then Tfn → Tf0. This means that T 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 α ∈ (0,1). Suppose that is a nonzero weighted composition operator of the form (29).
- (1)
If T is compact, then, for any y0 ∈ Y∖Y0, there is an open neighborhood U0 of y0 such that φ is supercontractive on U0 and φ(U0) is totally bounded.
- (2)
If φ is supercontractive on Y∖Y0 and φ(Y∖Y0) is totally bounded, then T is compact.
Also here, the result of [19] also refers to the case where T is a composition operator.
Corollary 13. Suppose that X, Y are compact metric spaces, and T is a weighted composition operator of the form (29) between the following function spaces:
- (i)
0 < α ≤ 1 and ;
- (ii)
0 < α < 1 and .
When T is a composition operator, that is, h = T1 = 1 in the form (29), then Y0 = ∅ and we can establish the following results in [20, Theorem 1.1].
Corollary 14. Suppose that X, Y are metric spaces and is a composition operator; then T is compact if and only if φ is supercontractive and φ(Y) is totally bounded.
In the following part of this section we have X = Y. Define φ0(x) = x and φn(x) = φ(φn−1(x)) for all x ∈ X by induction. A point x0 ∈ X is said to be the fixed point of φ of order n, n ∈ ℕ, if φn(x0) = x0 and φi(x0) ≠ x0 for any i = 0,1, …, n − 1.
Theorem 15. Let X be a complete metric space and T : Lipb(X) → Lipb(X) a weighted composition operator of form (29) satisfying: φ is supercontractive on X∖X0 and φ(X∖X0) is totally bounded. Then we can derive that σ(T) = {0} ∪ 𝒮, where
Proof. Suppose that x0 is a fixed point of φ of order n. If h(φk(x0)) = 0 for some k, we can see that T is not surjective and hence 0 ∈ σ(T).
Assume that h(φk(x0)) ≠ 0 for any k = 0,1, 2, …, n − 1 and λn = h(x0) ⋯ h(φn−1(x0)).
When n = 1, we have that λ = h(x0) and φ(x0) = x0. There exists g ∈ Lipb(X) such that g(x0) = 1. There is no f ∈ Lipb(X) such that (λ − T)f = g. Indeed, if such f exists, we can derive that
When n ≥ 2, let δ : = min{d(φi(x0), φj(x0)) : 0 ≤ i ≠ j ≤ n − 1}, and define
On the other hand, for each f ∈ Lipb(X) with λf = Tf, for some λ ∉ {0} ∪ 𝒮, we will prove that f = 0. This implies that λ ∉ σ(T) and completes the proof.
From the assumption λf = Tf = h · f∘φ, for all x ∈ X and n ∈ ℕ, we derive that
Given any z ∈ X, let ℱ = {φn(z) : n ∈ ℕ ∪ {0}} and 𝒩 = {n ∈ ℕ:|h(φn(z))| ≥ δ0}; here δ0 is any fixed number with 0 < δ0<|λ|. We provide that f(z) = 0, which implies that f = 0, by dividing into the following cases.
Case I (ℱ∩X0 ≠ ∅). If there exists i0 such that , by (44) we can see that
Case II (ℱ ⊂ X∖X0 and ℱ is finite). Let . Then there exists 0 ≤ k ≤ n0 such that . This means that φk(z) is a fixed point of φ of order n0 − k + 1. By (44), we have that
Case III (ℱ ⊂ X∖X0, ℱ is infinite and 𝒩 is infinite). Notice that {φn(z) : n ∈ 𝒩}⊂(X∖X0)∩φ(X∖X0). Since φ(X∖X0) is totally bounded, we can derive that {φn(z) : n ∈ 𝒩} converges to a point . Moreover, since φ is supercontractive. Then we have that and . By (44) we can see that . Since φ is supercontractive, there exists δ1 > 0 such that
Case IV (ℱ ⊂ X∖X0, ℱ is infinite and 𝒩 is finite). We can choose N0 ∈ ℕ such that |h(φn(z))| < δ0 for n > N0. From (44), we have that
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.