# About Some Families of Nonexpansive Mappings with respect to Renorming

Authors: Juan Rafael Acosta-Portilla; Carlos Alberto Hernández-Linares; Víctor Pérez-García
Journal: Journal of Function Spaces
 (2016)

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## Abstract

We characterize the family of nonexpansive mappings which are invariant under renormings and we also compare the families of nonexpansive mappings under two equivalent norms.


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## Body

## 1. Introduction

A Banach space(X,·) satisfies the fixed point property (FPP) if every nonexpansive mapping T:C→C has a fixed point, where C is a closed convex bounded subset of X. For a long time, it was an open problem if the fixed point property could be equivalent to the reflexivity. It was until 2008, when Lin [1] proved that there exists a nonreflexive Banach space with the FPP. Actually, Lin used a renorming ·L of the space l1 such that (l1,·L) has the FPP. One year later, Domínguez-Benavides proved that every reflexive Banach space can be renormed to have the FPP [2]. After these two articles, the Fixed Point Property and Renorming Theory were clearly connected. However, not all nonreflexive Banach spaces can be renormed to have the FPP as it was proved in [3]; in this work it is proved that l∞ and l1(Γ) where Γ is an uncountable set cannot be renormed to have the FPP. Recently, many works have appeared to be looking for new examples of nonreflexive Banach spaces enjoying the FPP or trying to find some structure on families of equivalent norms with the FPP. In the first sense the works should be mentioned [4–7]. In the second way the works are remarkable [8, 9]. After Lin’s result, the question remains open: does reflexivity implies FPP? In this setting the relevant paper is [10]. The FPP under a renorming also has been studied for other kinds of mappings; for example, see [11, 12].From the papers cited in the previous paragraph the FPP is not an isomorphic property. Moreover, the family of nonexpansive mappings could be changed after a renorming. For this reason, the aim of this paper is to say something about the following question.What happened with the family of nonexpansive mappings with another equivalent norm?In order to do that, we make the following assumptions. Let(X,·) be a normed space and C a nonempty subset of X; then for each ·-Lipschitz function T:C→C we denote by K(T,·) its Lipschitz constant with respect to · and by N(X) the collection of equivalent norms on X. For each ·0 norm on X we define (1)NEC,·0=S:C⟶C∣KS,·0≤1.


## 2. The Family of Nonexpansive Mappings over Every Renorming

In this section we will study some families of nonexpansive mappings over every renorming and we will characterize them. The first approximation is to characterize the setS′(C)=⋂·∈N(X)NE(C,·), when C is a convex, closed, and bounded set in X. It is worthwhile to mention that we prove that at least there are many elements in NE(C,·) as elements in NE([0,1],·). Moreover, we show that their structure is similar.For eachx∈C we call fx:C→C the constant function x and we denote by I:C→C the identity map. We define (2)SC=conv⁡fx∣x∈C∪I.Remark 1.
It is clear that, for each norm· on X, T∈NE(C,·) for all T∈S(C). It is easy to prove that each element T∈S(C) is of the form T=y+αI, for some y∈X and 0≤α≤1.In the rest of this paper we will denote byF the scalar field R or C, associated with the normed space. In order to characterize the set S′(C) we will give the following theorem.Theorem 2.
LetC be a nonempty subset of a normed space X and T:C→C; then the next statements are equivalent.(1)
For each norm· on X one has that T∈NE(C,·).(2)
For each·∈N(X) one has that T∈NE(C,·).(3)
·0∈N(X) exists such that T∈NE(C,·0) and for each x,y∈C, Tx-Ty∈span⁡{x-y} holds.(4)
For eachx,y∈Cα∈F exists with α≤1 such that α(x-y)=Tx-Ty.Proof.
It is straightforward that(1) implies (2).
We are going to prove that(2) implies (3); in order to do this we proceed by contraposition. We may suppose that ·0∈N(X) exist such that T∈NE(C,·0) and x,y∈C such that Tx-Ty∉span⁡{x-y}.
SinceTx-Ty∉span⁡{x-y} then Y=span⁡{x-y,Tx-Ty} has dimension 2 and then a projection P from X to Y exists that is ·0-bounded. Since Y is finite and dimensional and the set {x-y,Tx-Ty} is linear and independent, we can construct a norm ·1 in Y such that x-y1<Tx-Ty1; note that Tx-Ty1-x-y1 could be as large as we desire. We consider the function ·2 defined in X by x2=Px1+(I-P)x0; it is easy to check that ·2 is a norm on X. Now we will prove that ·2 is equivalent to ·0. Since ·0 and ·1 are equivalent in Y then 0<l≤u the optimal constants exist such that(3)la0≤a1≤ua0foreacha∈Y.Since P is ·0-bounded, we have for each z∈X(4)z2=Pz1+I-Pz0≤uPz0+I-Pz0≤uP+I-Pz0.On the other hand we have (5)z2=Pz1+I-Pz0≥lPz0+I-Pz0.There are two possibilities, l≥1 or l<1. If l≥1 then (6)z2≥Pz0+I-Pz0≥z0.If l<1 then (7)z2≥lPz0+lI-Pz0=lPz0+I-Pz0≥lz0;thus ·2 is equivalent to ·0.
Now we have(8)x-y2=Px-y1+I-Px-y0=x-y1<Tx-Ty1=PTx-Ty1+I-PTx-Ty0=Tx-Ty2;that is, T∉NE(C,·2).
Now we prove that(3) implies (4). Let x,y∈C; if Tx=Ty the proof is over because α=0 is the required constant; then without loss of generality we may assume that Tx≠Ty; hence x≠y and since Tx-Ty∈span⁡{x-y} then α∈F exist such that α(x-y)=Tx-Ty; thus (9)x-y0≥Tx-Ty0=αx-y0.Therefore |α|≤1.
Now we show that(4) implies (1). Let · be a norm on X and x,y∈C; then α∈F exists with α≤1 such that α(x-y)=Tx-Ty; therefore (10)x-y≥αx-y=Tx-Ty.Thus T∈NE(C,·).Remark 3.
Note thatKT,·2 could be arbitrary large, since in the proof of (2) implying (3) in the previous theorem we can take Tx-Ty1-x-y1 as large as we want.Lemma 4.
LetC be a nonempty subset of a normed space, T:C→C, and x,y,z∈C are distinct such that, for each α∈F, and z≠αx+(1-α)y, αx,y,αx,z,αy,z∈F exist with(11)αx,yx-y=Tx-Ty,αx,zx-z=Tx-Tz,αy,zy-z=Ty-Tz;then αx,y=αx,z=αy,z.Proof.
From (11) it follows that(12)αy,zy-z=αx,zx-z-αx,yx-y.Let spanR⁡{x-y,x-z}={r1(x-y)+r2(x-z)∣r1,r2∈R}. Since the set {x-y,x-z} is linearly independent, then we can define the operator R:span⁡{x-y,x-z}→spanR⁡{x-y,x-z} by Rαx-y+βx-z=Re⁡(α)(x-y)+Re⁡(β)(x-z) for each α,β∈F. It is clear that R is a real linear projection, (13)Rαw=Re⁡αRw,for  eachα∈F,w∈spanR⁡x-y,x-z,and y-z∈spanR⁡{x-y,x-z}. For this and (12) we have(14)Re⁡αy,zy-z=Re⁡αy,zRy-z=Rαy,zy-z=Rαx,zx-z-αx,yx-y=Re⁡αx,zx-z-Re⁡αx,yx-y.We define the linear operator ϕ:spanR⁡{x-y,x-z}→R2 by ϕ(x-y)=(1,0)=e1 and ϕ(x-z)=(0,1)=e2; then (15)Re⁡αx,ye1=ϕRe⁡αx,yx-y,Re⁡αx,ze2=ϕRe⁡αx,zx-z;thus(16)Re⁡αx,ze2-Re⁡αx,ye1=ϕRe⁡αx,zx-z-Re⁡αx,yx-y=ϕRe⁡αy,zy-z=ϕRe⁡αy,zx-z-x-y=Re⁡αy,ze2-e1.We consider the triangles in R2:(17)△0e1e2,△0Re⁡αx,ye1Re⁡αx,ze2.Since (18)Re⁡αx,ye1∈spanR⁡e1,Re⁡αx,ze2∈spanR⁡e2,Re⁡αx,ze2-Re⁡αx,ye1∈span⁡e2-e1,then the triangles are similar; thus (19)Re⁡αx,y=Re⁡αx,z=Re⁡αy,z.In a similar way we prove that Im⁡(αx,y)=Im⁡(αx,z)=Im⁡(αy,z), by considering the operator R(α(x-y)+β(x-z))=Im⁡α(x-y)+Im⁡(β)(x-z).
Thenαx,y=αx,z=αy,z.Theorem 5.
Under assumptions of Theorem2 and if C has at most two elements or for each distinct x,y∈Cz∈C exists such that, for each α∈F, z≠αx+(1-α)y, then the statement, (A)
T=y+αI for some α∈F with α≤1 and y∈X,is equivalent to each statement in that theorem.Proof.
The statement(A) implies that (1) in Theorem 2 is obvious.
Now we show that(4) of Theorem 2 implies (A). If C={x} then Tx=0+1Ix; if C={x,y} with x≠y and α is such that α(x-y)=Tx-Ty then Tz=w+αIz with w=Ty-αy for each z∈C. We suppose that for each distinct x,y∈Cz∈C exists such that, for each α∈F, z≠αx+(1-α)y. For each x,y∈C let αx,y∈F such that (20)αx,yx-y=Tx-Ty.It is clear that αx,y=αy,x. Let x,y∈C with x≠y. It is enough to prove that, for each distinct z1,z2∈C, αz1,z2=αx,y. Let z1,z2∈C; we may suppose without loss of generality that the following cases are exhaustive.(i)
α1,α2∈F exist such that(21)z1=α1x+1-α1y,z2=α2x+1-α2y.(ii)
α1∈F exists such that (22)z1=α1x+1-α1y,and for each α2∈F(23)z2≠α2x+1-α2y.(iii)
For eachα1,α2∈F(24)z1≠α1x+1-α1y,z2≠α2x+1-α2y.We assume (i) and z1≠x; then by hypothesis z∈C exist such that for each α∈F(25)z≠αx+1-αy,z≠αx+1-αz1,z≠αz1+1-αz2;then by Lemma 4(26)αx,y=αx,z=αz,y,αx,z1=αx,z=αz,z1,αz1,z2=αz1,z=αz,z2.Thus αx,y=αx,z=αz,z1=αz1,z2.
We suppose(ii), z1≠x, and z1≠y; then for each α∈F(27)z2≠αx+1-αz1;thus by Lemma 4(28)αx,y=αx,z2=αz2,y,αx,z1=αx,z2=αz2,z1.Then αz1,z2=αx,y.
We suppose(iii) and we may assume that for each α∈F(29)x≠αz1+1-αz2;then by Lemma 4(30)αx,y=αx,z1=αz1,y,αz1,z2=αz1,x=αx,z2.Then αz1,z2=αx,z1=αx,y; hence αz1,z2=αx,y for each distinct z1,z2∈C.Remark 6.
IfC is a nonempty subset of a normed space (X,·), in such a way that it lies inside a one-dimensional affine subspace of X, then D⊂F and a surjective isometry ϕ:(C,·)→(D,·) exist. If,additionally, C is convex, then ϕ can be affine. To prove this, we may suppose that C has at least two points. Let x,y∈C with x≠y and s=x-y; then we define ϕ:(C,·)→(F,·) by ϕ(z)=αzs where αz is the unique element in F such that z=αzx+(1-αz)y; it is not hard to check that ϕ is a surjective isometry from C to D=ϕ(C), and if C is convex, then ϕ is affine.
It is not hard to prove that the setD and the isometry ϕ can be constructed independent of the norm in C; for this it is sufficient to consider the renorming r· on F with r=x-y, ϕ:C→F defined by ϕ(z)=αz, and D=ϕ(C).Theorem 7.
Under assumptions of Theorem2 and if x,y∈C exist such that for each z∈Cα∈F exists with z=αx+(1-α)y, then(1) to(4) in Theorem 2 are equivalent to each of the following statements.(B)
For each norm· on X, D⊂F, f∈NE(F,·), and a surjective isometry ϕ:(C,·)→(D,·) exist with T=ϕ-1fDϕ.(C)
D⊂F, a surjective function ϕ:C→D, and f∈NE(F,·) exist with T=ϕ-1fDϕ, and for each norm · on X, r·∈N(F) exist such that ϕ is a · to r·-isometry.Proof.
First we prove that(B) implies (1) of Theorem 2. Let · be a norm on X; then D⊂F, f∈NE(F,·), and a surjective isometry ϕ:(C,·)→(D,·) exist with T=ϕ-1fDϕ and fD=ϕTϕ-1; it is not hard to prove that T is a ·-Lipschitz function with K(T,·)≤K(f,·)≤1; therefore T∈NE(C,·).
Now we prove that(1) of Theorem 2 implies (B). Let · be a norm on X and x,y∈C such that for each z∈Cαz∈F exist with z=αzx+(1-αz)y; by Remark 6  D⊂F and a surjective isometry ϕ:(C,·)→(D,·) exist; we define g=ϕTϕ-1; thus K(g,·)=K(T,·)≤1 and by Kirszbraun’s Theorem [13, 14], an extension f:(F,·)→(F,·) of g exists such that K(f,·)=K(g,·)≤1.
We prove that(B) implies (C). By Remark 6  D⊂F and a surjective ϕ:C→D exist such that, for each norm · on X, r·∈N(F) exists so that ϕ:(C,·)→(D,r·) is an isometry. We define g=ϕTϕ-1; by (A) we have that T∈NE(C,·); thus g∈NE(D,r·). Since the Lipschitz constant of a function defined from a subset of F to itself is independent of the norm on F, then g∈NE(D,·). Hence again by Kirszbraun’s Theorem, an extension f:(F,·)→(F,·) exists with K(f,·)=K(g,·)≤1.
Now we prove that(C) implies (B). Let · be a norm on X; then r·∈N(F) exist such that ϕ is a · to r·-isometry. We define D′=rD, ψ=rϕ, and g=ψTψ-1; it is clear that ψ:(C,·)→(D′,·) is an isometry; thus K(g,·)=K(T,·)≤1; then again by Kirszbraun’s Theorem the conclusion follows.Corollary 8.
Let(X,·) be a normed space and C a nonempty, convex, closed, and bounded subset of X; then(1)
ifx,y∈C exist such that for each z∈Cα∈F exists with z=αx+(1-α)y, then D⊂F and a surjective isometry ϕ:(C,·)→(D,·) exist in such a way that for each T∈S′(C)f∈NE(D,·) exist with K(f,·)=K(T,·) and T=ϕ-1fϕ;(2)
if for eachx,y∈Cz∈C exists such that, for each α∈F, z≠αx+(1-αy), then S′(C)={T:C→C∣T=y+αI for some y∈X and α≤1}.In generalS′(C) is not equal to S(C); for instance we take C=[-1,1] with the absolute value norm and T:[-1,1]→[-1,1] the operator defined by Tx=-x; then by Remark 1T∉S(C) and by Corollary 8T∈S′(C).Remark 9.
It is important to note that ifC is convex, closed, and bounded, then each element in S′(C) has a fixed point; to check this we consider the following two cases.(1)
IfC lies inside a one-dimensional affine subspace of X, then by Schauder’s Theorem the conclusion is clear.(2)
We suppose that for eachx,y∈Cz∈C exists such that, for each α∈F, z≠αx+(1-α)y. Let T∈S′(C); then by Theorem 5  Tx=y+αx for some α∈F with α≤1 and y∈X; then z∈C exists such that {z,y} is linearly independent; let Y=span⁡{z,y} and C1=Y∩C; it is not hard to prove that T(C1)⊂C1; thus again by Schauder’s Theorem, TC1 has a fixed point.Corollary 10.
Let(X,·) be a normed space and C a nonempty, convex, closed, and bounded subset of X such that for each x,y∈Cz∈C exist with z≠αx+(1-α)y; then for each T∈NE(C,·)∖S′(C), ·1∈N(X) exists with T∉NE(C,·1).
In particular ifT is a nonlinear operator or an operator without fixed points, then a renorming ·1 of X exists in such a way that T is not a ·1-nonexpansive operator and by Remark 3, K(T,·1) is large as one desires.Theorem 11.
Under assumptions of Theorem2, the following statements are equivalent to each statement in that theorem.(D)
For each of the norms· and ·1 on X, K(T,·)=K(T,·1)≤1.(E)
For each·,·1∈N(X), K(T,·)=K(T,·1)≤1.Proof.
It is clear that(D) implies (E); now we prove that (E) implies (3) of Theorem 2. For this we proceed by contraposition; hence we may suppose that ·∈N(X) exist with K(T,·)≤1 and x,y∈C such that Tx-Ty∉span⁡{x-y}; there exists ·1∈NX with KT,·1>1, by using Remark 3 and similar arguments of the proof of (2) implying (3) in Theorem 2.
We prove that(3) of Theorem 2 implies (D). We have the following two cases.(i)
For eachx,y∈Cz∈C exist such that, for each α∈F, z≠αx+(1-α)y.(ii)
x,y∈C exist such that for each z∈Cα∈F exist with z=αx+(1-α)y.We suppose (i); then by Theorem 5, T=y+αI for some y∈X and α≤1; thus for each norm the Lipschitz constant of T is α.
Now we assume(ii); then by (C) in Theorem 7 and the fact that the Lipschitz constant of functions defined from subsets of F to itself is independent of the norm on F, we have K(T,·)=K(T,·1)≤1 for each of the norms · and ·1 on X.


## 3. Comparing Families of Nonexpansive Mappings

Now, we will compare the families of nonexpansive mappings by given conditions in which the families are different. Before that, to prove the two theorems in this section we need to prove two technical lemmas.Definition 12.
LetX be vectorial space endowed with two norms ·1 and ·2 and C a subset of X. One will say that ·1 is collinear to ·2 in C, if there exists λ>0 such that x1=λx2 for all x∈C. We can omit the subset C if the context is clear.Remark 13.
It is clear that the collinearity between norms is an equivalent relation. LetT:C→C; note that K(T,·1)=K(T,·2) if ·1 is collinear to ·2 in C-C.Lemma 14.
LetX be a normed space. If C is a nontrivial convex subset of X with 0∈C and ·1 and ·2 are two norms over X such that they are not collinear in C, then x,y∈C∖{0} exist such that x1=y1 and y2>x2.Proof.
Since·1 and ·2 are not collinear in C, then it is false that λ>0 exists such that for all z∈C we have z1=λz2; that is, the function f(z)=z1/z2 defined on C∖{0} is not constant; thus x′,y′∈C∖{0} exist such that x′1/x′2>y′1/y′2.
Ifx′1=y′1 then we concluded the proof. Else, we call x the element in {x′,y′} such that x1=min⁡{x′1,y′1}; if x=x′ then we take y=x1/y′1y′, so y1=x1; since C is convex and 0∈C we have that y∈C. Therefore x1=y1 and y1/y2=y′1/y′2; thus x2<y2.
Ifx=y′ the proof is similar.Lemma 15.
LetX be a normed space, C a nonempty subset of X, ·1 and ·2 norms over X; then the following statements are equivalent.(1)
·1 and ·2 are collinear in C-C.(2)
For eachy∈C the norms ·1 and ·2 are collinear in C-y.Proof.
It is clear that(1) implies (2); thus we are going to show that (2) implies (1). We may suppose without loss of generality that C have at least two elements. For each y∈C we define λy>0 as the scalar such that (31)x-y1=λyx-y2∀x∈C.
Lety0∈C; then for all x∈C we have that x-y01=λy0x-y02 and x-y01=λxx-y02; since C has more than one element we have λx=λy0 for all x∈C; that is x-y1=λy0x-y2 for all x,y∈C and then ·1 and ·2 which conclude the proof.Definition 16.
Given normed spacesX1,X2,Y1, and Y2 and for i=1,2, Ci⊂Xi, and Di⊂Yi being convex, one calls that T:C1→C2 and S:D1→D2 are isometrically affine equivalent, in symbols T≃S, if affine surjective isometries ϕi:C1→Di for i=1,2 exist, such that T=ϕ2-1Sϕ1 and S=ϕ2Tϕ1-1.The intuition is that two operators are isometrically affine equivalent if they have essentially the same transformation.(32)The following lemma summarizes some properties of the isometrically affine relation.Lemma 17.
LetT≃S; then one has the following. (1)
T is continuous if and only if S is continuous; moreover T and S have the same modulus of continuity ω(t); hence the affine isometry relation preserves uniform continuity and the Lipschitzian property with the same Lipschitz constant.(2)
T is affine if and only if S is affine.The previous lemma allows us to translate properties of real functions to functions defined between arbitrary Banach spaces; this provides us with elements to ensure the existence of operators with desired properties.Theorem 18.
Let(X,·) be a normed space, C be a nontrivial convex subset of X, and f be a function; then we have the following.(1)
Iff≃g with g:([0,1],·)→([0,1],·) is Lipschitzian and x,y∈C and x≠y, then T:C→C exists as Lipschitzian such that K(T,·)=K(f,·) and T[x,y]≃f.(2)
Iff≃g with g:([0,1],·)→(R,·) is Lipschitzian, then x,y∈C, x≠y, and T:C→C exist as Lipschitzian such that K(T,·)=K(f,·) and T[x,y]≃f.Proof.
We suppose the hypothesis of(1). Since C is convex, then [x,y]⊂C, and since all norms in R are collinear, then the Lipschitz constant k of g is independent of the norm in R.
We consider the operatorϕ:[0,1]→[x,y] defined by ϕr=rx+(1-r)y for each r∈[0,1] and we provide the metric space [0,1] with the norm r0=x-yr for each r∈[0,1]. It is concluded from definition that ([0,1],·0) is ϕ-isometric to the space ([x,y],·). It is known that ([0,1],·0) is a metric space with the binary intersection property and metrically convex; then ([x,y],·) have the same properties.
We callh:[x,y]→[x,y] the operator defined by h=ϕgϕ-1; hence h≃g and g≃f; thus by Lemma 17 we have K(h,·)=k; then by Definition 1.3 and Proposition 1.4 of [15], an extension T:C→[x,y]⊂C of h exists with K(T,·)=K(h,·); then T[x,y]≃f and for Lemma 17T and f have the same Lipschitz constant.
Now we are going to prove(2). Let f be a function such that g exists as above; then optimal r,s∈R exist such that g([0,1])⊂[r,s]⊂R. There are two possibilities: (33)s-r≤1ors-r>1.If s-r≤1 then we can choose r′≤r≤s≤s′ such that s′-r′=1; thus [r,s]⊂[r′,s′]; we define the operator ϕ:([0,1],·)→([r′,s′],·) by ϕt=tr′+1-ts′; it is clear that ϕ is an isometry; then the operator h=ϕ-1g=ϕ-1gId[0,1] is Lipschitzian and satisfies h≃g≃f which is the hypothesis of (1); thus without loss of generality, we may suppose that s-r>1.
We fix distinctx,y∈C, for each t∈[0,1]; we call at=tr+(1-t)s and define a norm on [r,s] by at-ap0=x-yt-p for each t,p∈[0,1]; we chose a,b∈[r,s] such that a<b and b-a=1, and we define a norm on [0,1] by t1=a-b0t for each t∈[0,1]; we consider the operators ϕ2:([r,s],·0)→([x,y],·) defined by ϕ2(tr+(1-t)s)=tx+(1-t)y and ϕ1:([0,1],·1)→([ϕ2(a),ϕ2(b)],·) defined by ϕ1(t)=tϕ2(a)+(1-t)ϕ2(b) for each t∈[0,1] which are surjective affine isometries.
Thus the operatorh=ϕ2gϕ1-1 satisfies h:[ϕ2(a),ϕ2(b)]→[x,y] and h≃g≃f; then by Lemma 17  h and f have the same Lipschitz constant k and again by Definition 1.3 and Proposition 1.4 of [15], an extension T:C→[x,y]⊂C of h exists with the same Lipschitz constant.Corollary 19.
Let(X,·) be a normed space and C⊂X be a nontrivial convex; then a nonaffine operator always exists in NE(C,·).Proof.
Letf∈Lip⁡([0,1]) be a nonaffine function such that K(f,·)≤1 and distinct x,y∈C; then by Theorem 18, an operator T:C→C exists with K(T,·)=K(f,·)≤1 and T[x,y]≃f; thus by Lemma 17  T is a nonaffine operator.From Remark13 given a nonempty subset C of a normed space X and two collinear norms ·1 and ·2 in C-C, we have that NE(C,·1)=NE(C,·2). Now, we prove the reciprocal of the previous statement.Remark 20.
It is important to note that the collinearity of two norms· and ·1 on C does not necessarily imply that, for each T:C→C, K(T,·)=K(T,·1).
Let·2 and ·1 be, respectively, the euclidean and the one norm on R2; then we define(34)x,y=x,y2,ifsign⁡xy=1x,y1,ifsign⁡xy=-1for each (x,y)∈R2, C={(x,y)∣(x,y)2≤1}, and S the rotation on R2 determined by the matrix cos⁡(π/2)-sin⁡(π/2)sin⁡(π/2)cos⁡(π/2). It is clear that S(C)⊂C, K(S,·2)=1, and K(S,·)>1; since ·2 and · are collinear in the first quadrant, then they are collinear on C+(1,1). Let T=S+(1,1); then T:C+(1,1)→C+(1,1) with ·2 and · collinear on C+(1,1), but K(T,·2)=K(S,·2)≠K(S,·)=K(T,·).Theorem 21.
LetX be a normed space and C a nontrivial convex subset of X. If ·1,·2∈N(X), then the following statements are equivalent.(1)
·1 and ·2 are not collinear in C-C.(2)
NE(C,·1)≠NE(C,·2).(3)
T∈NE(C,·1)∖NE(C,·2) and S∈NE(C,·2)∖NE(C,·1) exist.(4)
Nonaffine mappingsT,S with T∈NE(C,·1)∖NE(C,·2) and S∈NE(C,·2)∖NE(C,·1) exist.Proof.
It is clear that we only need to prove that(1) implies (4). We suppose (1); then by Lemma 15  y∈C exists such that ·1 and ·2 are not collinear in C-y; thus without loss of generality we may assume that 0∈C.
Since·1 and ·2 are not collinear in C and 0∈C, then by Lemma 14  x,y∈C∖{0} exist such that(35)x1=y1,x2<y2.Without loss of generality we may assume that x1=y1=1.
Letα0∈(x2/y2,1); then z=α0y satisfies x2<z2<y2; we define the operator f:[0,x]→[0,y] by (36)fαx=αy,if0≤α≤α0z,ifα0≤α≤1.Since (37)f1-α00+α0x=z=α0y≠α0α0y=1-α00+α0fx=1-α0f0+α0fx,then f is nonaffine.
We affirm thatK(f,·1)=1; in fact, let α,β∈[0,1]; then by symmetry we only have the following three cases.(i)
0≤α,β≤α0; then (38)fαx-fβx1=αy-βy1=α-βy1=α-βx1=αx-βx1.(ii)
α0≤α,β≤1, then f(αx)-f(βx)1=z-z1=0≤αx-βx1.(iii)
0≤α<α0<β≤1; then (39)fαx-fβx1=αy-z1=αy-α0y1=α-α0y1=α-α0x1<α-βx1=αx-βy1.Now we prove that K(f,·2)>1; for this we have (40)f0-fx2=0-z2=z2>x2=0-x2.Since x1=y1=1, then the sets [0,x] and [0,y] are affine ·1-isometric; then by Remark 6, they are affine isometric to ([0,1],·) and g:([0,1],·)→([0,1],·) exists with f≃g; thus by Theorem 18 an extension T:C→[0,y]⊂C of f exists, with the same ·1-Lipschitz constant that means K(T,·1)=1; thus T∈NE(C,·1), but T is ·2-Lipschitz and K(T,·2)≥K(f,·2)>1. Then T∉NE(C,·2).
In a similar way we prove the existence ofS∈NE(C,·2)∖NE(C,·1), by exchanging the order of the norms and applying Lemma 14.Remark 22.
We notice that given a convex subsetC that lies inside a one-dimensional affine subspace of X, then for each pair of norms ·1 and ·2 in X not necessarily equivalent, we have that NE(C,·1)=NE(C,·2); this is because C is affine isometric to a closed convex subset of F for a norm, and all norms in F are collinear; then by Theorem 21 the conclusion follows.Corollary 23.
LetX be a normed space and C⊂X such that x,y∈C exist so that for each z∈Cα∈F exists with z=αx+(1-α)y; then, for each norm · on X, S′(C)=NE(C,·).Remark 24.
From the last theorem we infer that it is not possible to compare the families of nonexpansive mappings under noncollinear renormings with respect to the contention relation because, under noncollinear renorming, nonexpansive mappings always win and lose.Corollary 25.
LetX be a normed space, and C⊂X such that for each x,y∈Cz∈C exists with z≠αx+(1-α)y for each α∈F; then for each norm · on X, NE(C,·)≠S′(C).Corollary 26.
The nonexpansive operators constructed in the proof of Lemma15 do not belong to S′(C).


## 4. Examples

In the present section we support the theoretical results of the previous ones by providing some examples.In general, given a Banach spaceX and a convex subset C of X, it is hard to construct examples of nontrivial nonexpansive operators from C to C; by trivial we understand the elements in the sets S(C),S′(C) or more general affine mappings from C to itself. In this sense, when a new Banach space is studied to know some of its geometric properties, for instance, the FPP, then it is natural to ask for the existence of some nontrivial nonexpansive operators defined in that space. An example of last situation is the renorming ·L of l1 studied by Lin [1], for which at the moment in the literature only few examples of nonexpansive mappings exist.Example 27.
Let(l1,·1) be the Banach space of absolutely summable series of real numbers endowed whit the norm (xn)1=∑n=0∞xn. It is well known that (l1,·|1) does not have the FPP; the classical example is the right-shift operator R defined from C to C, where (41)C=conv⁡en∣n∈N¯·1,Rx1,x2,…=0,x1,….For each k∈N let νk be the seminorm in l1 defined by νk((xn))=∑n=k∞xn; thus the ·L norm in l1 is defined by (42)xnL=supk⁡γkνkxnwhereγk=8k1+8k.
Now we construct some examples of nontrivial·L-nonexpansive operators defined from C to itself such that they are not ·1-nonexpansive and the opposite.
Let(43)g1=e1,g2=1-γ1γ2e1+γ1γ2e2,g3=1-γ1γ2e1+γ1γ2-γ1γ3e2+γ1γ3e3;then g11=g21=g31=1 and g1,g2,g3∈C.
As in Lemma15, we consider the set C-g2 and define (44)x′=g1-g2=γ1γ2e1-e2,y′=g3-g2=γ1γ3e3-e2;then x′1=2(γ1/γ2) and y′1=2(γ1/γ3); since 2γ1≥γ2,2γ2≥γ3 and γ1≠γ2 we have (45)x′L=γ12γ1γ2=γ1x′1,y′L=max⁡γ12γ1γ3,γ22γ1γ3,γ1=2γ1γ2γ3=γ2y′1;hence ·1 and ·L are not collinear in C-g2; in fact, by Lemma 15, they are not collinear in C-C.
Now we make the construction of Lemma14 in this particular case. It is clear that(46)x′1x′L=1γ1>1γ2=y′1y′L.Since γ3≥γ2, then y′1≤x′1; thus we define x=y′ and y=(y′1/x′1)x′; then by (46) (47)x1=y1,yL=y′1x′1x′L<y′L=xL.We get α0∈(yL/xL,1); thus z=α0x∈[0,x] and yL<zL<xL. We consider the function f:[0,y]→[0,x] defined by (48)fαy=αx,if0≤α≤α0,z,ifα0≤α≤1;then by the same argument of the proof of Theorem 21, the existence of a nonaffine mapping T∈NE(C,·1)∖NE(C,·L) follows.
Now we prove the existence of a nonaffine mappingS∈NE(C,·L)∖NE(C,·1). From (46) it follows that(49)x′Lx′1=γ1<γ2=y′Ly′1.Since γ1/γ2≤γ2/γ3, then x′L≤y′L; hence we define x=x′ and y=(x′L/y′L)y′; thus by (49) (50)xL=yL,y1=x′Ly′Ly′1<x′1=x1.Let λ0∈(y1/x1,1); hence w=λ0x satisfies y1<w1<x1; we define the function g:[0,y]→[0,x],(51)gλy=λx,if0≤λ≤λ0,w,ifλ0≤λ≤1;then again by the same argument of the proof of Theorem 21, we ensure the existence of a nonaffine mapping S∈NE(C,·L)∖NE(C,·1).Remark 28.
Letγ=(γk) be a nondecreasing sequence in (0,1) such that γk→1; then by [7], the renorming (52)xnγ=supk⁡γkνkxnof (l1,·1) has the FPP. Note that the previous example is valid for each renorming ·γ in which(i)
γ1≠γ2,(ii)
2γ1≥γ2 and 2γ2≥γ3,(iii)
γ1/γ2≤γ2/γ3.Example 29.
Now we characterize the family of nonexpansive mappingsT:C→C over any renorming, where C is as in Example 27; that is, we characterize the set S′(C). The next calculations are effortless if we consider the equality (53)C=xn∈l1∣xn≥0,∑xn=1.It is clear that C does not lie inside a one-dimensional affine subspace of l1; then by Corollary 8, for each T∈S′(C), (yn)∈l1 and λ≤1 exist such that T(xn)=(yn)+λ(xn) for each (xn)∈C.
We affirm that each entry of(yn) is nonnegative and 0≤λ≤1; for this, if yk<0 for some k∈N then the k-entry of Tek+1 is yk; thus Tek∉C; hence for each k∈N we have yk≥0. If -1≤λ<0, let k∈N such that 0≤yk<-λ; then the k-entry of Tek is yk+λ<0; hence 0≤λ≤1.
Let(xn)∈C; then (54)1=Txn1=yn+λxn=∑yn+λxn=∑yn+∑λxn=∑yn+λ;thus 0≤λ≤1 and (yn)∈(1-λ)C, which agrees with notation of Remark 1  S′(C)⊂{T=(yn)+λI∣0≤λ≤1and(yn)∈(1-λ)C}; the proof of the other contention is a direct calculation; thus (55)S′C=T=yn+λI∣0≤λ≤1,yn∈1-λC.In this case S(C)=S′(C); for this, let 0≤λ≤1 and (yn)∈(1-λ)C; thus (1-λ)-1(yn)∈C; then T=(1-λ)[(1-λ)-1(yn)]+λI∈S(C).Example 30.
Let(X,·) be a normed space and B its unit ball. We characterize the set S′(B). By Corollary 8, for each T∈S′(B), y∈X and λ≤1 exist with Tx=y+λx for each x∈B; we affirm that λ≤1 and y∈(1-λ)B; in fact, let λ≤1, y∈(1-λ)B, and x∈B; then (56)Tx=y+λx≤y+λx≤1-λ+λ=1;hence, with notation of Remark 1, {T=y+λI∣λ≤1andy∈(1-λ)B}⊂S′(B). Now we prove the other contention. Let λ≤1 and y∉(1-λ)B; then y>1-λ; we define x=λ-1(λ/y)y; thus x∈B; since(57)y+λx=y+λyy=1+λyy=y+λ>1-λ+λ=1,then y+λx∉B; hence (58)S′B=T=y+λI∣λ≤1,y∈1-λB.



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