About Some Families of Nonexpansive Mappings with respect to Renorming

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.

1. Introduction

A Banach space satisfies the fixed point property (FPP) if every nonexpansive mapping has a fixed point, where is a closed convex bounded subset of . 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 of the space such that 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 and where 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 be a normed space and a nonempty subset of ; then for each -Lipschitz function we denote by its Lipschitz constant with respect to and by the collection of equivalent norms on . For each norm on we define

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 set , when is a convex, closed, and bounded set in . It is worthwhile to mention that we prove that at least there are many elements in as elements in . Moreover, we show that their structure is similar.

For each we call the constant function and we denote by the identity map. We define

Remark 1. It is clear that, for each norm on , for all . It is easy to prove that each element is of the form , for some and .

In the rest of this paper we will denote by the scalar field or , associated with the normed space. In order to characterize the set we will give the following theorem.

Theorem 2. Let be a nonempty subset of a normed space and ; then the next statements are equivalent.(1)For each norm on one has that .(2)For each one has that .(3) exists such that and for each , holds.(4)For each exists with such that .

Proof. It is straightforward that implies .
We are going to prove that implies ; in order to do this we proceed by contraposition. We may suppose that exist such that and such that .
Since then has dimension and then a projection from to exists that is -bounded. Since is finite and dimensional and the set is linear and independent, we can construct a norm in such that ; note that could be as large as we desire. We consider the function defined in by ; it is easy to check that is a norm on . Now we will prove that is equivalent to . Since and are equivalent in then the optimal constants exist such thatSince is -bounded, we have for each On the other hand we have There are two possibilities, or . If then If then thus is equivalent to .
Now we have that is, .
Now we prove that implies . Let ; if the proof is over because is the required constant; then without loss of generality we may assume that ; hence and since then exist such that ; thus Therefore .
Now we show that implies . Let be a norm on and ; then exists with such that ; therefore Thus .

Remark 3. Note that could be arbitrary large, since in the proof of implying in the previous theorem we can take as large as we want.

Lemma 4. Let be a nonempty subset of a normed space, , and are distinct such that, for each , and , exist withthen .

Proof. From (11) it follows thatLet . Since the set is linearly independent, then we can define the operator by for each . It is clear that is a real linear projection, and . For this and (12) we haveWe define the linear operator by and ; then thusWe consider the triangles in :Since then the triangles are similar; thus In a similar way we prove that , by considering the operator .
Then .

Theorem 5. Under assumptions of Theorem 2 and if has at most two elements or for each distinct exists such that, for each , , then the statement, (A) for some with and ,is equivalent to each statement in that theorem.

Proof. The statement implies that in Theorem 2 is obvious.
Now we show that of Theorem 2 implies . If then ; if with and is such that then with for each . We suppose that for each distinct exists such that, for each , . For each let such that It is clear that . Let with . It is enough to prove that, for each distinct , . Let ; we may suppose without loss of generality that the following cases are exhaustive.(i) exist such that(ii) exists such that and for each (iii)For each We assume and ; then by hypothesis exist such that for each then by Lemma 4Thus .
We suppose , , and ; then for each thus by Lemma 4Then .
We suppose and we may assume that for each then by Lemma 4Then ; hence for each distinct .

Remark 6. If is a nonempty subset of a normed space , in such a way that it lies inside a one-dimensional affine subspace of , then and a surjective isometry exist. If,additionally, is convex, then can be affine. To prove this, we may suppose that has at least two points. Let with and ; then we define by where is the unique element in such that ; it is not hard to check that is a surjective isometry from to , and if is convex, then is affine.
It is not hard to prove that the set and the isometry can be constructed independent of the norm in ; for this it is sufficient to consider the renorming on with , defined by , and .

Theorem 7. Under assumptions of Theorem 2 and if exist such that for each exists with , then (1) to (4) in Theorem 2 are equivalent to each of the following statements.(B) For each norm on , , , and a surjective isometry exist with .(C), a surjective function , and exist with , and for each norm on , exist such that is a to -isometry.

Proof. First we prove that implies of Theorem 2. Let be a norm on ; then , , and a surjective isometry exist with and ; it is not hard to prove that is a -Lipschitz function with ; therefore .
Now we prove that of Theorem 2 implies . Let be a norm on and such that for each exist with ; by Remark 6   and a surjective isometry exist; we define ; thus and by Kirszbraun’s Theorem [13, 14], an extension of exists such that .
We prove that implies . By Remark 6   and a surjective exist such that, for each norm on , exists so that is an isometry. We define ; by we have that ; thus . Since the Lipschitz constant of a function defined from a subset of to itself is independent of the norm on , then . Hence again by Kirszbraun’s Theorem, an extension exists with .
Now we prove that implies . Let be a norm on ; then exist such that is a to -isometry. We define , , and ; it is clear that is an isometry; thus ; then again by Kirszbraun’s Theorem the conclusion follows.

Corollary 8. Let be a normed space and a nonempty, convex, closed, and bounded subset of ; then(1)if exist such that for each exists with , then and a surjective isometry exist in such a way that for each exist with and ;(2)if for each exists such that, for each , , then for some and .

In general is not equal to ; for instance we take with the absolute value norm and the operator defined by ; then by Remark 1 and by Corollary 8.

Remark 9. It is important to note that if is convex, closed, and bounded, then each element in has a fixed point; to check this we consider the following two cases.(1)If lies inside a one-dimensional affine subspace of , then by Schauder’s Theorem the conclusion is clear.(2)We suppose that for each exists such that, for each , . Let ; then by Theorem 5   for some with and ; then exists such that is linearly independent; let and ; it is not hard to prove that ; thus again by Schauder’s Theorem, has a fixed point.

Corollary 10. Let be a normed space and a nonempty, convex, closed, and bounded subset of such that for each exist with ; then for each , exists with .
In particular if is a nonlinear operator or an operator without fixed points, then a renorming of exists in such a way that is not a -nonexpansive operator and by Remark 3, is large as one desires.

Theorem 11. Under assumptions of Theorem 2, the following statements are equivalent to each statement in that theorem.(D)For each of the norms and on , .(E)For each , .

Proof. It is clear that implies ; now we prove that implies of Theorem 2. For this we proceed by contraposition; hence we may suppose that exist with and such that ; there exists with , by using Remark 3 and similar arguments of the proof of implying in Theorem 2.
We prove that of Theorem 2 implies . We have the following two cases.(i)For each exist such that, for each , .(ii) exist such that for each exist with .We suppose ; then by Theorem 5, for some and ; thus for each norm the Lipschitz constant of is .
Now we assume ; then by in Theorem 7 and the fact that the Lipschitz constant of functions defined from subsets of to itself is independent of the norm on , we have for each of the norms and on .

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. Let be vectorial space endowed with two norms and and a subset of . One will say that is collinear to in , if there exists such that for all . We can omit the subset if the context is clear.

Remark 13. It is clear that the collinearity between norms is an equivalent relation. Let ; note that if is collinear to in .

Lemma 14. Let be a normed space. If is a nontrivial convex subset of with and and are two norms over such that they are not collinear in , then exist such that and .

Proof. Since and are not collinear in , then it is false that exists such that for all we have ; that is, the function defined on is not constant; thus exist such that .
If then we concluded the proof. Else, we call the element in such that ; if then we take , so ; since is convex and we have that . Therefore and ; thus .
If the proof is similar.

Lemma 15. Let be a normed space, a nonempty subset of , and norms over ; then the following statements are equivalent.(1) and are collinear in .(2)For each the norms and are collinear in .

Proof. It is clear that implies ; thus we are going to show that implies . We may suppose without loss of generality that have at least two elements. For each we define as the scalar such that Let ; then for all we have that and ; since has more than one element we have for all ; that is for all and then and which conclude the proof.

Definition 16. Given normed spaces , and and for , , and being convex, one calls that and are isometrically affine equivalent, in symbols , if affine surjective isometries for exist, such that and .

The intuition is that two operators are isometrically affine equivalent if they have essentially the same transformation. The following lemma summarizes some properties of the isometrically affine relation.

Lemma 17. Let ; then one has the following. (1) is continuous if and only if is continuous; moreover and have the same modulus of continuity ; hence the affine isometry relation preserves uniform continuity and the Lipschitzian property with the same Lipschitz constant.(2) is affine if and only if 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 be a normed space, be a nontrivial convex subset of , and be a function; then we have the following.(1)If with is Lipschitzian and and , then exists as Lipschitzian such that and .(2)If with is Lipschitzian, then , , and exist as Lipschitzian such that and .

Proof. We suppose the hypothesis of . Since is convex, then , and since all norms in are collinear, then the Lipschitz constant of is independent of the norm in .
We consider the operator defined by for each and we provide the metric space with the norm for each . It is concluded from definition that is -isometric to the space . It is known that is a metric space with the binary intersection property and metrically convex; then have the same properties.
We call the operator defined by ; hence and ; thus by Lemma 17 we have ; then by Definition and Proposition of [15], an extension of exists with ; then and for Lemma 17 and have the same Lipschitz constant.
Now we are going to prove . Let be a function such that exists as above; then optimal exist such that . There are two possibilities: If then we can choose such that ; thus ; we define the operator by ; it is clear that is an isometry; then the operator is Lipschitzian and satisfies which is the hypothesis of ; thus without loss of generality, we may suppose that .
We fix distinct , for each ; we call and define a norm on by for each ; we chose such that and , and we define a norm on by for each ; we consider the operators defined by and defined by for each which are surjective affine isometries.
Thus the operator satisfies and ; then by Lemma 17   and have the same Lipschitz constant and again by Definition and Proposition of [15], an extension of exists with the same Lipschitz constant.

Corollary 19. Let be a normed space and be a nontrivial convex; then a nonaffine operator always exists in .

Proof. Let be a nonaffine function such that and distinct ; then by Theorem 18, an operator exists with and ; thus by Lemma 17   is a nonaffine operator.

From Remark 13 given a nonempty subset of a normed space and two collinear norms and in , we have that . Now, we prove the reciprocal of the previous statement.

Remark 20. It is important to note that the collinearity of two norms and on does not necessarily imply that, for each , .
Let and be, respectively, the euclidean and the one norm on ; then we definefor each , , and the rotation on determined by the matrix . It is clear that , , and ; since and are collinear in the first quadrant, then they are collinear on . Let ; then with and collinear on , but .

Theorem 21. Let be a normed space and a nontrivial convex subset of . If , then the following statements are equivalent.(1) and are not collinear in .(2).(3) and exist.(4)Nonaffine mappings with and exist.

Proof. It is clear that we only need to prove that implies . We suppose ; then by Lemma 15   exists such that and are not collinear in ; thus without loss of generality we may assume that .
Since and are not collinear in and , then by Lemma 14   exist such thatWithout loss of generality we may assume that .
Let ; then satisfies ; we define the operator by Since then is nonaffine.
We affirm that ; in fact, let ; then by symmetry we only have the following three cases.(i); then (ii), then .(iii); then Now we prove that ; for this we have Since , then the sets and are affine -isometric; then by Remark 6, they are affine isometric to and exists with ; thus by Theorem 18 an extension of exists, with the same -Lipschitz constant that means ; thus , but is -Lipschitz and . Then .
In a similar way we prove the existence of , by exchanging the order of the norms and applying Lemma 14.

Remark 22. We notice that given a convex subset that lies inside a one-dimensional affine subspace of , then for each pair of norms and in not necessarily equivalent, we have that ; this is because is affine isometric to a closed convex subset of for a norm, and all norms in are collinear; then by Theorem 21 the conclusion follows.

Corollary 23. Let be a normed space and such that exist so that for each exists with ; then, for each norm on , .

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. Let be a normed space, and such that for each exists with for each ; then for each norm on , .

Corollary 26. The nonexpansive operators constructed in the proof of Lemma 15 do not belong to .

4. Examples

In the present section we support the theoretical results of the previous ones by providing some examples.

In general, given a Banach space and a convex subset of , it is hard to construct examples of nontrivial nonexpansive operators from to ; by trivial we understand the elements in the sets or more general affine mappings from 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 of studied by Lin [1], for which at the moment in the literature only few examples of nonexpansive mappings exist.

Example 27. Let be the Banach space of absolutely summable series of real numbers endowed whit the norm . It is well known that does not have the FPP; the classical example is the right-shift operator defined from to , where For each let be the seminorm in defined by ; thus the norm in is defined by Now we construct some examples of nontrivial -nonexpansive operators defined from to itself such that they are not -nonexpansive and the opposite.
Let then and .
As in Lemma 15, we consider the set and define then and ; since and we have hence and are not collinear in ; in fact, by Lemma 15, they are not collinear in .
Now we make the construction of Lemma 14 in this particular case. It is clear thatSince , then ; thus we define and ; then by (46) We get ; thus and . We consider the function defined by then by the same argument of the proof of Theorem 21, the existence of a nonaffine mapping follows.
Now we prove the existence of a nonaffine mapping . From (46) it follows thatSince , then ; hence we define and ; thus by (49) Let ; hence satisfies ; we define the function ,then again by the same argument of the proof of Theorem 21, we ensure the existence of a nonaffine mapping .

Remark 28. Let be a nondecreasing sequence in such that ; then by [7], the renorming of has the FPP. Note that the previous example is valid for each renorming in which(i),(ii) and ,(iii).

Example 29. Now we characterize the family of nonexpansive mappings over any renorming, where is as in Example 27; that is, we characterize the set . The next calculations are effortless if we consider the equality It is clear that does not lie inside a one-dimensional affine subspace of ; then by Corollary 8, for each , and exist such that for each .
We affirm that each entry of is nonnegative and ; for this, if for some then the -entry of is ; thus ; hence for each we have . If , let such that ; then the -entry of is ; hence .
Let ; then thus and , which agrees with notation of Remark 1  ; the proof of the other contention is a direct calculation; thus In this case ; for this, let and ; thus ; then .

Example 30. Let be a normed space and its unit ball. We characterize the set . By Corollary 8, for each , and exist with for each ; we affirm that and ; in fact, let , , and ; then hence, with notation of Remark 1, . Now we prove the other contention. Let and ; then ; we define ; thus ; sincethen ; hence

Competing Interests

The authors declare that there is no conflict of interests regarding the publication of this paper.

Acknowledgments

All authors were partially supported by CONACyT Project 243722. The first author was fully supported by CONACyT national grant.