AAA Abstract and Applied Analysis 1687-0409 1085-3375 Hindawi Publishing Corporation 741050 10.1155/2013/741050 741050 Research Article Nonvanishing Preservers and Compact Weighted Composition Operators between Spaces of Lipschitz Functions Chen Dongyang 1 http://orcid.org/0000-0003-1177-7539 Li Lei 2 Wang Risheng 2 Wang Ya-Shu 3 Camilli Fabio M. 1 School of Mathematical Sciences Xiamen University Xiamen 361005 China xmu.edu.cn 2 School of Mathematical Sciences and LPMC Nankai University Tianjin 300071 China nankai.edu.cn 3 Department of Applied Mathematics Chung Yuan Christian University Chung-Li 32023 Taiwan cycu.edu.tw 2013 30 10 2013 2013 14 06 2013 09 09 2013 2013 Copyright © 2013 Dongyang Chen et al. This is an open access article distributed under the Creative Commons Attribution License, which permits unrestricted use, distribution, and reproduction in any medium, provided the original work is properly cited.

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:LipXLipY is a nonvanishing preserver, then T is a weighted composition operator Tf=h·fφ, where φ:YX 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 X 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 JC(Y) with |J(y)|=1 for all yY. 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  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 . Later, Chen et al. , Ercan and Önal [6, 7], and Miao et al.  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  tackled the Banach-Stone type theorem for lattices of real Lipschitz functions. Later, Jiménez-Vargas and Villegas-Vallecillos  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.  presented a Lipschitz version of the result in , 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·fφ, where φ:YX 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 space E 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|xy, where |x|=max{x,-x}.

Let (X,d) be a metric space and E a Banach space; if 0<α1, 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,yX,xy}<. 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:XE with the α-Lipschitz norm (4)f=max{Lα(f),f}, where f=sup{f(x):xX}. 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)fgf(x)g(x),xX. However, Lipαb(X,E) is not a Banach lattice since · does not satisfy the Riesz law in general.

A mapping φ 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 fe the function xf(x)e in Lipα(X,E). In particular, 1e denotes the constant function xe on X. For any function f in Lipα(X,E), the zero set {xX:f(x)=0} of f is denoted by z(f) and its cozero set {xX: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)=,fLipα(X,E).T is said to be a Riesz isomorphism if T(fg)=TfTg and T(fg)=TfTg for any f,gLipα(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 that 0<α1, 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 (<inline-formula><mml:math xmlns:mml="http://www.w3.org/1998/Math/MathML" id="M121"><mml:mn mathvariant="normal">0</mml:mn><mml:mo><</mml:mo><mml:mi>α</mml:mi><mml:mo><</mml:mo><mml:mn mathvariant="normal">1</mml:mn></mml:math></inline-formula>).

A ( X , E ) = lip α b ( X , E ) , A(Y,F)=lipαb(Y,F).

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.

Suppose that X is a metric space and 0<α1, for any x1x2X, 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: 0f1, f(x1)=0, f(x2)=1 and z(f)={xX:d(x,x2)d(x1,x2)}.

Theorem 1.

Let T: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)),fA(X,E),yY. Here, φ is a homeomorphism from Y onto X and all fiber linear maps Jy:EF are isomorphisms.

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 lip1b(X,E)=lip1(X,E) consisting of all E-valued constant functions defined on X. Let φ be any map from 2 to and T:lip1(,E)lip1(2,E) a linear bijection operator defined by (10)Tf(y)=f(φ(y)),y2. It is obvious that the operator T is a Riesz isomorphism preserving nonvanishing functions with a weighted composition representation, but and 2 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,,fnA(X,E) and n.

Proof of Theorem <xref ref-type="statement" rid="thm3.1">1</xref>.

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 Contexts 1 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 xX and yY, 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 xX, Jφ-1(x):EF 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 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 eE with e=1, we have that (13)infyY(Jy)(e)=infyYT(1e)(y)>0.

Proof.

By Theorem 1 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 xX and gA(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(1e)(yn)=(Jyn)(e)2-2n for all 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 with nm, when we take the norm one element (15)bn=T(1e)(yn)T(1e)(yn), then we can derive that (16)[T-1(T(1e))](φ(yn))=(1e)(φ(yn))=e,[T-1(T(1e))](φ(yn))=(J~φ(yn))T(1e)(yn) for all n. Therefore, for any n, we know that (17)(J~φ(yn))(bn)=(J~φ(yn))T(1e)(yn)T(1e)(yn)=1T(1e)(yn)22n. Moreover, for any n, by the similar manner of (7) and (8) we can define the function ψn(y)Ab(Y) such that 0ψn1, 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=2ψnbn2n, 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,n2. This is a contradiction in Contexts 1 and 2 since T-1h0<.

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 T~ from A(X) to A(Y) by (21)(T~f)(y)=f(φ(y)),yY. We have to show that T~ is well defined at first. For any fixed element eE with e=1, from Lemma 5 we can choose a positive scalar ν such that T(1e)(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(1e)(y) belongs to A(Y).

Assume that f is a positive function in A(X); one can get that, for any y1,y2Y, (22)|(T~f)(y1)T(1e)(y1)-(T~f)(y2)T(1e)(y2)|=|(Jy1)f(φ(y1))e-(Jy2)f(φ(y2))e|(Jy1)f(φ(y1))e-(Jy2)f(φ(y2))e=T(fe)(y1)-T(fe)(y2)Lα(T(fe))dα(y1,y2)T(fe)dα(y1,y2); that is, (T~f)(y)T(1e)(y) is a bounded α-Lipschitz function. Moreover, in Context 2 we can derive that (T~f)(y)T(1e)(y) is also a little Lipschitz function. This means that the function (T~f)(y)=(T~f)(y)T(1e)(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 and yY, (T~fn)(y)=fn(φ(y)), and hence we have that {fn(φ(y))} converges to g0(y) for all yY. Notice that {fn} converges to 0; one can conclude that {fn(x)} converges to 0 for all xX, 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 any y1 and y2 in Y, there exists a function f0A(X) such that f0D(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~f0dα(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.

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 (25)Sf=Tf·T1|T1| is a Riesz isomorphism preserving nonvanishing functions.

Proof.

For completeness, we will sketch the proof. Observe that T1 is never vanishing. If fA(X) and λ, then λrangef if and only if 0range(f-λ) if and only if 0range(Tf-λT1) if and only if λrangeTf/T1. In particular, if f0, then Tf/T10. Let Y+={yY:(T1)(y)>0} and Y-={yY:(T1)(y)<0}. Then Y+Y- is a partition of Y into two open sets.

Suppose that fA(X) and f0. Then Tf0 on Y+ and Tf0 on Y-. Hence Tf·T1/|T1|=|Tf|A(Y). For any fA(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, if 0fA(X), then Sf=|Tf|0. If fA(X) and g=Sf0, then by the above, (27)0Sf=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, f0. Thus S is a Riesz isomorphism. It is trivial to check that 0rangef if 0rangeSf for any fA(X).

Theorem 8.

Suppose that X, Y are bounded metric spaces and T is a nonvanishing preserver between the following function spaces:

0<α1 and T:Lipα(X)Lipα(Y);

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:YX is a α-Lipschitz map.

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 1 be the positive integers with the discrete metric, and we can derive that 1 is not Lipchitz homeomorphic to . By [18, Example 1.6.4] we can derive that Lipb()=Lipb(1)=, and then the identity map I:Lipb()Lipb(1) is a nonvanishing preserver. However, the underlying metric spaces are not Lipschitz homeomorphic.

4. Compact Weighted Composition Operators on Lipschitz Spaces

Suppose that X, Y are metric spaces, 0<α1, and T:Lipαb(X)Lipαb(Y) is a weighted composition operator, that is, (29)(Tf)(y)=h(y)f(φ(y)),yY,fLipαb(X). Here h=T1 and φ:YX is a α-Lipschitz mapping. Put Y0={yY:h(y)=0}. Recall that φ:YX is supercontractive on YY if for each ε>0 there exists δ>0 such that d(φ(y1),φ(y2))<εd(y1,y2) whenever y1,y2Y 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 that T is compact. For any y0YY0, 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 yU0. 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 fn1/n and |fn(x1)-fn(x2)|dα(x1,x2) for any x1,x2X. This implies that {fn} is a bounded sequence in Lipαb(X). If T is compact, then there exists a subsequence {fnk} such that Tfnkg0Lipαb(Y). Since fn0 uniformly, for any yY, we have that (32)|(Tfnk)(y)|=|h(y)fnk(φ(y))|h|fnk(φ(y))|0, and then g0=0. This means that Tfnk0 in Lipαb(Y).

On the other hand, for any 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 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 nm. Let (34)fn(x)=1-e-dα(x,zn),xX; then it is easy to see that fn(zn)=0 and fn1. Moreover, for any nm, 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 that φ is supercontractive on YY0 and φ(YY0) is totally bounded; then the weighted composition operator defined by (29) is compact.

Proof.

Let {fn}Lipαb(X) be a bounded sequence, that is, fnM for some M>0. Since φ(YY0) is totally bounded, there exists a subsequence of {fn}, which is also denoted by {fn}, such that {fn} is convergent uniformly in φ(YY0). Denote the limit by f0(x) for all xφ(YY0). It is easy to verify that f0 is a bounded Lipschitz function in φ(YY0). 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).

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 YY0, there exists δ>0 such that (36)d(φ(y1),φ(y2))d(y1,y2)<ε whenever y1,y2YY0 and 0<d(y1,y2)<δ.

We will show that Lα(Tfn-Tf0)0 by dividing into four cases as the following arguments. For any y1,y2Y with y1y2.

Case  1. If y1,y2Y0, we have that (Tfn)(yi)=(Tf0)(yi)=0 for i=1,2.

Case  2. If y1,y2YY0 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,y2YY0 and d(y1,y2)>δ, we have that (39)|T(fn-f0)(y1)-T(fn-f0)(y2)|dα(y1,y2)2Tfn-Tf0δα.

Case  4. If y1YY0 and y2Y0, 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 that Lα(Tfn-Tf0)0 and then TfnTf0. 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 T:Lipαb(X)Lipαb(Y) is a nonzero weighted composition operator of the form (29).

If T is compact, then, for any y0YY0, there is an open neighborhood U0 of y0 such that φ is supercontractive on U0 and φ(U0) is totally bounded.

If φ is supercontractive on YY0 and φ(YY0) is totally bounded, then T is compact.

Also here, the result of  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:

0<α1 and T:Lipαb(X)Lipαb(Y);

0<α<1 and T:Lipαb(X)Lipαb(Y).

Then T is compact if and only if φ is supercontractive on YY0.

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 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 have X=Y. Define φ0(x)=x and φn(x)=φ(φn-1(x)) for all xX by induction. A point x0X is said to be the fixed point of φ of order 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 XX0 and φ(XX0) 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 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 0σ(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 gLipb(X) such that g(x0)=1. There is no fLipb(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 λσ(T).

When n2, let δ:=min{d(φi(x0),φj(x0)):0ijn-1}, and define (43)g=1ni=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 λσ(T).

On the other hand, for each fLipb(X) with λf=Tf, for some λ{0}𝒮, we will prove that f=0. This implies that λσ(T) and completes the proof.

From the assumption λf=Tf=h·fφ, for all xX and n, we derive that (44)λnf(x)=h(x)h(φ(x))h(φn-1(x))f(φn(x)).

Given any zX, let ={φn(z):n{0}} and 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 (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 (XX0 and is finite). Let ={z,φ(z),,φn0(z)}. Then there exists 0kn0 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 (XX0, is infinite and 𝒩 is infinite). Notice that {φn(z):n𝒩}(XX0)φ(XX0). Since φ(XX0) is totally bounded, we can derive that {φn(z):n𝒩} 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|2hd(z1,z2)  (φ(z1),a)when  0<d(z1,z2)<δ1. Choose N such that d(φn(z),x-)<δ1 for all nN, and we have, for any n, that (49)|λnf(φN(z))|=|h(φN(z))h(φN+1(z))ii=h(φN+n-1(z))f(φn+N(z))|hn|f(φn+N(z))|=hn|f(φn+N(z))-f(x-)|hnL(f)d(φn+N(z),x-)hnL(f)|λ|2hd(φn+N-1(z),x-)hnL(f)(|λ|2h)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 (XX0, is infinite and 𝒩 is finite). We can choose N0 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)|hN0δ0-N0(δ0λ)nf, for all n>N0. This implies that f(z)=0 as δ0<|λ|.

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.

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