# Nonvanishing Preservers and Compact Weighted Composition

Authors: Dongyang Chen; Lei Li; Risheng Wang; Ya-Shu Wang
Journal: Abstract and Applied Analysis
 (2013)

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## Abstract

We will give theα-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:LipX→LipY 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.


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## Body

## 1. Introduction

The classical Banach-Stone theorem tells us that, whenX and Y are compact Hausdorff spaces, every linear surjective isometry from C(X) onto C(Y) can be written as a weighted composition operator; that is, it is of the form
(1)(Tf)(y)=J(y)f(φ(y)),
where φ is a homeomorphism from Y onto X and J∈C(Y) with |J(y)|=1 for all y∈Y. 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 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-valuedα-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

An ordered vector spaceE is said to be a vector lattice if max{x,y} exists for any x, y in E. A vector lattice E is said to be a Banach lattice if it is complete under its norm ∥·∥ and satisfies the Riesz law:
(2)|x|≤|y|⟹∥x∥≤∥y∥,
where |x|=max{x,-x}.Let(X,d) be a metric space and E a Banach space; if 0<α≤1, a function f from X to E is said to be α-Lipschitz if
(3)Lα(f)=sup{∥f(x)-f(y)∥dα(x,y):x,y∈X,x≠y}<∞.
The α-Lipschitz function space  Lipα(X,E) is the space of all E-valued α-Lipschitz functions on X. Lipαb(X,E) is the Banach space of all bounded α-Lipschitz functions f:X→E with the α-Lipschitz norm
(4)∥f∥=max{Lα(f),∥f∥∞},
where ∥f∥∞=sup{∥f(x)∥:x∈X}. Furthermore, the little Lipschitz space  lipα(X,E) is then defined to be the closed subspace of Lipα(X,E) of these functions f with the following property: for every ε>0, there exists δ>0 such that ∥f(x1)-f(x2)∥<εdα(x1,x2) whenever d(x1,x2)≤δ. lipαb(X,E) is the subspace of lipα(X,E) consisting of all bounded functions. Notice that when α=1, Lipα(X,E) is just Lipschitz space Lip(X,E). Moreover, if E is a Banach lattice, then Lipα(X,E) is a vector lattice with the usual pointwise order
(5)f≤g⟺f(x)≤g(x),∀x∈X.
However, Lipαb(X,E) is not a Banach lattice since ∥·∥ does not satisfy the Riesz law in general.A mappingφ from Y to X is said to be a α-Lipschitz homeomorphism if it is bijective and φ and φ-1 are both α-Lipschitz. If f is in Lipα(X) and e is a vector in E, denote by f⊗e the function x↦f(x)e in Lipα(X,E). In particular, 1⊗e denotes the constant function x↦e on X. For any function f in Lipα(X,E), the zero set {x∈X:f(x)=0} of f is denoted by z(f) and its cozero set {x∈X:f(x)≠0} is coz(f), and f is said to be nonvanishing if z(f)=∅. An operator T:Lipα(X,E)→Lipα(Y,F) is said to be a nonvanishing preserver if
(6)z(f)=∅⟺z(Tf)=∅,∀f∈Lipα(X,E).T is said to be a Riesz isomorphism if T(f∨g)=Tf∨Tg and T(f∧g)=Tf∧Tg for any f,g∈Lipα(X,E).


## 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 that0<α≤1, X, Y are metric spaces and E, F are Banach lattices.Context 1.
A
(
X
,
E
)
=
Lip
α
b
(
X
,
E
), A(Y,F)=Lipαb(Y,F).Context 2 (0<α<1).
A
(
X
,
E
)
=
lip
α
b
(
X
,
E
), A(Y,F)=lipαb(Y,F).This means that when we refer toX, 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.Suppose thatX is a metric space and 0<α≤1, for any x1≠x2∈X, the function
(7)f(x)=max{0,1-dα(x,x2)dα(x1,x2)}
belongs to Lipαb(X). Moreover, if 0<α<1, then we can find β>0 such that α<β<1, and the function
(8)f(x)=max{0,1-dβ(x,x2)dβ(x1,x2)}
belongs to lipαb(X). The function f defined in (7) or (8) has the property: 0≤f≤1, f(x1)=0, f(x2)=1 and z(f)={x∈X:d(x,x2)≥d(x1,x2)}.Theorem 1.
LetT:A(X,E)→A(Y,F) be a Riesz isomorphism preserving nonvanishing functions. Then T carries the form
(9)(Tf)(y)=(Jy)f(φ(y)),∀f∈A(X,E),y∈Y.
Here, φ is a homeomorphism from Y onto X and all fiber linear maps Jy:E→F are isomorphisms.Remark 2.
Whenα=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 lip1b(X,E)=lip1(X,E) 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
(10)Tf(y)=f(φ(y)),∀y∈ℝ2.
It is obvious that the operator T is a Riesz isomorphism preserving nonvanishing functions with a weighted composition representation, but ℝ and ℝ2 are not homeomorphic.
It is easy to prove the following lemma.Lemma 3.
T preserves common zeros, that is,
(11)⋂i=1nz(fi)≠∅⟺⋂i=1nz(Tfi)≠∅
for any f1,…,fn∈A(X,E) and n∈ℕ.Proof of Theorem1.
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.Lemma 4.
In the Contexts1 and 2, T is automatically continuous.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
(12)(Tfn)(φ-1(x))=(Jφ-1(x))fn(x)⟶(Jφ-1(x))f0(x)=(Tf0)(φ-1(x))
for all x in X. Since φ is a bijection from Y onto X, we get that the sequence {(Tfn)(y)} converges to (Tf0)(y) for all y in Y, and hence g0=Tf0. This means that T is a closed operator from A(X,E) to A(Y,F), and then T is continuous.In order to prove thatφ 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 elemente∈E with ∥e∥=1, we have that
(13)infy∈Y∥(Jy)(e)∥=infy∈Y∥T(1⊗e)(y)∥>0.Proof.
By Theorem1 we can also find a map J~ from X to Iso(F,E) (which is the set of all linear isomorphisms from F to E) and a bijection φ~ from X onto Y such that
(14)(T-1g)(x)=(J~x)g(φ~(x))
for all x∈X and g∈A(Y,F). From the definition of ψ, φ, and φ~, we can see that φ-1=φ~=ψ.
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
(15)bn=T(1⊗e)(yn)∥T(1⊗e)(yn)∥,
then we can derive that
(16)[T-1(T(1⊗e))](φ(yn))=(1⊗e)(φ(yn))=e,[T-1(T(1⊗e))](φ(yn))=(J~φ(yn))T(1⊗e)(yn)
for all n∈ℕ. Therefore, for any n∈ℕ, we know that
(17)∥(J~φ(yn))(bn)∥=∥(J~φ(yn))T(1⊗e)(yn)∥∥T(1⊗e)(yn)∥=1∥T(1⊗e)(yn)∥≥22n.
Moreover, for any n∈ℕ, by the similar manner of (7) and (8) we can define the function ψn(y)∈Ab(Y) such that 0≤ψn≤1, Lα(ψn)≤m for some m>0, ψn(yn)=1 and ψn(y)=0 for all y such that d(y,yn)≥τ/2. When put
(18)h0=∑n=2∞‍ψn⊗bn2n,
we can see that h0 belongs to Ab(Y,F) and h0(yn)=bn/2n for n>1. Then one can conclude that
(19)(T-1h0)(φ(yn))=(J~φ(yn))(h0[φ~(φ(yn))])=(J~φ(yn))(h0(yn))=12n(J~φ(yn))(bn),
and hence
(20)∥(T-1h0)(φ(yn))∥=∥(J~φ(yn))(bn)∥2n>2n,∀n≥2.
This is a contradiction in Contexts 1 and 2 since ∥T-1h0∥∞<∞.Theorem 6.
Suppose thatX, 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 mapT~ from A(X) to A(Y) by
(21)(T~f)(y)=f(φ(y)),∀y∈Y.
We have to show that T~ is well defined at first. For any fixed element e∈E with ∥e∥=1, from Lemma 5 we can choose a positive scalar ν such that ∥T(1⊗e)(y)∥≥ν>0 for all y in Y, and then it is easy to see that the function h which maps y to 1/∥T(1⊗e)(y)∥ belongs to A(Y).
Assume thatf is a positive function in A(X); one can get that, for any y1,y2∈Y,
(22)|(T~f)(y1)∥T(1⊗e)(y1)∥-(T~f)(y2)∥T(1⊗e)(y2)∥|=|∥(Jy1)f(φ(y1))e∥-∥(Jy2)f(φ(y2))e∥|≤∥(Jy1)f(φ(y1))e-(Jy2)f(φ(y2))e∥=∥T(f⊗e)(y1)-T(f⊗e)(y2)∥≤Lα(T(f⊗e))dα(y1,y2)≤∥T(f⊗e)∥dα(y1,y2);
that is, (T~f)(y)∥T(1⊗e)(y)∥ is a bounded α-Lipschitz function. Moreover, in Context 2 we can derive that (T~f)(y)∥T(1⊗e)(y)∥ is also a little Lipschitz function. This means that the function (T~f)(y)=(T~f)(y)∥T(1⊗e)(y)∥h(y) belongs to A(Y). Therefore, T~ is a well-defined bijective linear operator from A(X) onto A(Y), and T~ is also a nonvanishing preserver.
Suppose that{fn} is a sequence which converges to 0 in A(X) and the sequence {T~fn} converges to g0 in A(Y). For any n∈ℕ and y∈Y, (T~fn)(y)=fn(φ(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, T~ is a closed operator and hence T~ is continuous.
For anyy1 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
(23)|(T~f0)(y1)-(T~f0)(y2)|≤Lα(T~f0)dα(y1,y2)≤∥T~∥∥f0∥dα(y1,y2).
Furthermore, we have that
(24)d(φ(y1),φ(y2))=|f0(φ(y1))-f0(φ(y2))|=|(T~f0)(y1)-(T~f0)(y2)|≤∥T~∥(D(X)+D(X)1-α)dα(y1,y2),
and this means that φ is a α-Lipschitz mapping from Y onto X. Similarly, we can see that φ-1 is also α-Lipschitz, and then φ is a α-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.
LetA(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
(25)Sf=Tf·T1|T1|
is a Riesz isomorphism preserving nonvanishing functions.Proof.
For completeness, we will sketch the proof. Observe thatT1 is never vanishing. If f∈A(X) and λ∈ℝ, then λ∈rangef if and only if 0∈range(f-λ) if and only if 0∈range(Tf-λT1) if and only if λ∈rangeTf/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 thatf∈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
(26)Sf=Tf·T1|T1|=(T(f+)-T(f-))·T1|T1|=|T(f+)|-|T(f-)|∈A(Y).
This means that S is well defined. Moreover, it is easy to check that S is bijective.
From the previous paragraph, if0≤f∈A(X), then Sf=|Tf|≥0. If f∈A(X) and g=Sf≥0, then by the above,
(27)0≤Sf=Tf·T1X|T1X|=|T(f+)|-|T(f-)|.
By [17, Lemma 2.3], T is biseparating, and hence T(f+)·T(f-)=0. It follows that T(f-)=0 and thus f-=0. Therefore, f≥0. Thus S is a Riesz isomorphism. It is trivial to check that 0∈rangef if 0∈rangeSf for any f∈A(X).Theorem 8.
Suppose thatX, 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).Then T is a weighted composition operator of the form
(28)(Tf)(y)=h(y)f(φ(y)).
Here h=T1 and φ:Y→X is a α-Lipschitz map.Proof.
By Lemma7 we have that T is a Riesz isomorphism. Then by Theorem 6 we can derive the conclusion.In Theorem8, the boundedness of the metric spaces can not be dropped.Example 9.
Letℕ1 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

Suppose thatX, Y are metric spaces, 0<α≤1, and T:Lipαb(X)→Lipαb(Y) is a weighted composition operator, that is,
(29)(Tf)(y)=h(y)f(φ(y)),∀y∈Y,f∈Lipαb(X).
Here h=T1 and φ:Y→X is a α-Lipschitz mapping. Put Y0={y∈Y:h(y)=0}. Recall that φ:Y→X is supercontractive on Y′⊂Y if for each ε>0 there exists δ>0 such that d(φ(y1),φ(y2))<εd(y1,y2) whenever y1,y2∈Y′ and 0<d(y1,y2)<δ. In this section, we will characterize the compact weighted composition operator T and consider its spectrum.Theorem 10.
Suppose thatT 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.
Sinceh(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
(30)d(φ(xn),φ(yn))d(xn,yn)>ε0
for some ε0>0. Without loss of generality we can assume that dα2(xn,yn)<1/n.
Let(31)fn(x)=1-e-ndα(x,φ(yn))n;
we can derive that ∥fn∥∞≤1/n and |fn(x1)-fn(x2)|≤dα(x1,x2) for any x1,x2∈X. This implies that {fn} is a bounded sequence in Lipαb(X). If T is compact, then there exists a subsequence {fnk} such that Tfnk→g0∈Lipαb(Y). Since fn→0 uniformly, for any y∈Y, we have that
(32)|(Tfnk)(y)|=|h(y)fnk(φ(y))|≤∥h∥∞|fnk(φ(y))|⟶0,
and then g0=0. This means that Tfnk→0 in Lipαb(Y).
On the other hand, for anyn∈ℕ, by the Mean Value Theorem we have that
(33)|(Tfn)(xn)-(Tfn)(yn)|dα(xn,yn)=|h(xn)fn(φ(xn))|dα(xn,yn)≥|h(y0)|21-e-ndα(φ(xn),φ(yn))ndα(xn,yn)=|h(y0)|2e-nξndα(φ(xn),φ(yn))dα(xn,yn)≥|h(y0)|2e-Lαα(φ)ε0α.
Here 0<ξn<dα(φ(xn),φ(yn))≤Lαα(φ)dα2(xn,yn)<Lαα(φ)/n. Therefore, we can derive that Lα(Tfn)↛0, and this is a contradiction.
On the other hand, suppose on the contrary thatφ(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
(34)fn(x)=1-e-dα(x,zn),∀x∈X;
then it is easy to see that fn(zn)=0 and ∥fn∥≤1. Moreover, for any n≠m, we can derive that
(35)∥Tfn-Tfm∥∞≥|h(un)fm(zn)|≥|h(y0)|2(1-e-dα(zn,zm))≥|h(y0)|2(1-e-τ).
Therefore, {Tfn} has no Cauchy subsequence, and hence T is not compact. This leads to a contradiction.Theorem 11.
Suppose thatφ is supercontractive on Y∖Y0 and φ(Y∖Y0) is totally bounded; then the weighted composition operator defined by (29) is compact.Proof.
Let{fn}⊂Lipαb(X) 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 Lipαb(X), which is also denoted by f0. It suffices to show that {Tfn} converges to Tf0 in Lipαb(Y).
SinceT 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
(36)d(φ(y1),φ(y2))d(y1,y2)<ε
whenever y1,y2∈Y∖Y0 and 0<d(y1,y2)<δ.
We will show thatLα(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
(37)|T(fn-f0)(y1)-T(fn-f0)(y2)|≤|[h(y1)-h(y2)](fn-f0)(φ(y1))|+|h(y2)[(fn-f0)(φ(y1))-(fn-f0)(φ(y2))]|≤Lα(h)dα(y1,y2)|(fn-f0)(φ(y1))|+∥h∥∞(Lα(fn)+Lα(f0))dα(φ(y1),φ(y2)).
Moreover, by (36) we can derive that
(38)dα(φ(y1),φ(y2))=dα(φ(y1),φ(y2))dα(y1,y2)dα(y1,y2)≤εαdα(y1,y2).
Case  3. If y1,y2∈Y∖Y0 and d(y1,y2)>δ, we have that
(39)|T(fn-f0)(y1)-T(fn-f0)(y2)|dα(y1,y2)≤2∥Tfn-Tf0∥∞δα.
Case  4. If y1∈Y∖Y0 and y2∈Y0, we have that h(y2)=0 and then
(40)|T(fn-f0)(y1)-T(fn-f0)(y2)|=|h(y1)(fn-f0)(φ(y1))|=|h(y1)-h(y2)|·|(fn-f0)(φ(y1))|≤Lα(h)dα(y1,y2)|(fn-f0)(φ(y1))|.
Hence we derive thatLα(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.
Letα∈(0,1). Suppose that T:Lipαb(X)→Lipαb(Y) is a nonzero weighted composition operator of the form (29).(1)
IfT 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)
Ifφ 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 thatX, 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 T:Lipαb(X)→Lipαb(Y);(ii)
0<α<1 and T:Lipαb(X)→Lipαb(Y).Then T is compact if and only if φ is supercontractive on Y∖Y0.WhenT 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 thatX, Y are metric spaces and T:Lipαb(X)→Lipαb(Y) 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 haveX=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.
LetX 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
(41)𝒮={λ:λn=h(x0)h(φ(x0))⋯h(φn-1(x0)),hhhx0isafixedpointofφofordern}.Proof.
Suppose thatx0 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 thath(φk(x0))≠0 for any k=0,1,2,…,n-1 and λn=h(x0)⋯h(φn-1(x0)).
Whenn=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
(42)0=λf(x0)-h(x0)f(x0)=λf(x0)-h(x0)f(φ(x0))=g(x0)=1,
and this is impossible. This means that λ∈σ(T).
Whenn≥2, let δ:=min{d(φi(x0),φj(x0)):0≤i≠j≤n-1}, and define
(43)g=1n∑i=0n-1‍1λn-i-1h(x0)⋯h(φi-1(x0))×max{0,1-d(x,φi(x0))δ}.
Here h(φ-1(x0)):=1. Then, similar to the argument of [3, Proposition 3], we can derive that λ∈σ(T).
On the other hand, for eachf∈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 assumptionλf=Tf=h·f∘φ, for all x∈X and n∈ℕ, we derive that
(44)λnf(x)=h(x)h(φ(x))⋯h(φn-1(x))f(φn(x)).
Given anyz∈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 h(φi0(z))=0, by (44) we can see that
(45)λi0+1f(z)=h(z)h(φ(z))⋯h(φi0(z))f(φi0+1(z))=0.
This implies that f(z)=0.
Case II (ℱ⊂X∖X0 and ℱ is finite). Let ℱ={z,φ(z),…,φn0(z)}. Then there exists 0≤k≤n0 such that φn0+1(z)=φk(z). This means that φk(z) is a fixed point of φ of order n0-k+1. By (44), we have that
(46)λn0-k+1f(φk(z))=h(φk(z))h(φk+1(z))⋯h(φn0(z))f(φn0+1(z)),
and f(φk(z))=0 since λ∉{0}∪𝒮. Once again, by (44), we derive that
(47)λkf(z)=h(z)h(φ(z))⋯h(φk-1(z))f(φk(z))=0,
and f(z)=0.
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 x-∈X. Moreover, φn(z)→x- since φ is supercontractive. Then we have that |h(x-)|≥δ0 and φ(x-)=x-. By (44) we can see that f(x-)=0. Since φ is supercontractive, there exists δ1>0 such that
(48)d(φ(z1),φ(z2))<|λ|2∥h∥∞d(z1,z2)(φ(z1),a)when0<d(z1,z2)<δ1.
Choose N∈ℕ such that d(φn(z),x-)<δ1 for all n≥N, and we have, for any n∈ℕ, that
(49)|λnf(φN(z))|=|h(φN(z))h(φN+1(z))⋯ii=h(φN+n-1(z))f(φn+N(z))|≤∥h∥∞n|f(φn+N(z))|=∥h∥∞n|f(φn+N(z))-f(x-)|≤∥h∥∞nL(f)d(φn+N(z),x-)≤∥h∥∞nL(f)|λ|2∥h∥∞d(φn+N-1(z),x-)≤⋯≤∥h∥∞nL(f)(|λ|2∥h∥∞)nd(φN(z),x-)=L(f)(|λ|2)nd(φN(z),x-).
That is,
(50)|f(φN(z))|≤12nL(f)d(φN(z),x-).
Since n is arbitrary, we can derive that f(φN(z))=0, and then f(z)=0 since (44).
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
(51)λnf(z)=h(z)h(φ(z))⋯h(φn-1(z))f(φn(z)),
and then
(52)|f(z)|≤∥h∥∞N0δ0-N0(δ0λ)n∥f∥∞,
for all n>N0. This implies that f(z)=0 as δ0<|λ|.



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*Source: 741050-2013-10-30.xml*