Abstract and Applied AnalysisVolume 2011 (2011), Article ID 574614, 19 pagesdoi:10.1155/2011/574614

Research Article

The Fixed Point Property in with an Equivalent Norm

Berta Gamboa de Buen¹ and Fernando Núñez-Medina²

¹Matemáticas Básicas, Centro de Investigación en Matemáticas (CIMAT), Apartado Postal 402, 36000 Guanajuato, GTO, Mexico²Departamento de Matemáticas Aplicadas, Universidad del Papaloapan (UNPA), 68400 Loma Bonita, OAX, Mexico

Received 7 June 2011; Accepted 27 August 2011

Academic Editor: Elena Litsyn

Copyright © 2011 Berta Gamboa de Buen and Fernando Núñez-Medina. 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.

Abstract.

We study the fixed point property (FPP) in the Banach space with the equivalent norm . The space with this norm has the weak fixed point property. We prove that every infinite-dimensional subspace of contains a complemented asymptotically isometric copy of , and thus does not have the FPP, but there exist nonempty closed convex and bounded subsets of which are not -compact and do not contain asymptotically isometric —summing basis sequences. Then we define a family of sequences which are asymptotically isometric to different bases equivalent to the summing basis in the space and we give some of its properties. We also prove that the dual space of over the reals is the Bynum space and that every infinite-dimensional subspace of does not have the fixed point property.

1. Introduction

We start with some notations and terminologies. Let be a nonempty, convex, closed and bounded subset of a Banach space . A mapping is said to be nonexpansive if We say that has the fixed point property for nonexpansive mappings (FPP) if every nonexpansive mapping has a fixed point, that is, a point such that . We say that a Banach space has the fixed point property for nonexpansive mappings (FPP) if every nonempty, convex, closed, and bounded subset of has the FPP, and we say that the Banach space has the weak fixed point property for nonexpansive mappings (-FPP) if every nonempty, convex and weakly compact subset of has the FPP.

In this paper we study the FPP in the Banach space with the equivalent norm defined by The norm was used by Hagler in [1] to construct a separable Banach space with nonseparable dual such that does not embed in and every normalized weakly null sequence in has a subsequence equivalent to the canonical basis of .

In [2], Dowling et al. gave a characterization of nonempty, convex, closed and bounded subsets of which are not -compact. Specifically, they proved that if is a convex, closed and bounded subset of , then is -compact if and only if every nonempty, convex, closed and convex subset of has the FPP. To do that, the authors showed that every closed, convex and bounded subset of which is not -compact contains an asymptotically isometric -summing basic sequence, sequence for short, that is, a sequence such that for all , for some with and . They proved that if a convex, closed and bounded subset of a Banach space contains an sequence, then there exists a nonempty, convex, closed and bounded subset of without the FPP. The authors used this fact in [3] as a tool to prove that a nonempty, closed, convex and bounded subset of is -compact if and only if it has the FPP.

It is easy to see that contains isometrically, and then it contains sequences.

First we prove that every infinite-dimensional subspace of has a complemented asymptotically isometric copy of and hence by a result proved by Dowling et al. in [4], does not have the FPP. Also, as an immediate consequence we obtain that has an sequence. Nevertheless, we exhibit a nonempty closed, convex and bounded subset of , which is not -compact and does not contain sequences.

Then for every selection of signs , we define the -basis of which is equivalent to the summing basis and define the corresponding asymptotically isometric -basic sequence, sequence for short. We prove that if , then the and sequences are different in the sense that there exists a nonempty, closed, convex, and bounded subset of , which is not -compact, contains an sequence, and does not contain sequences. We also show that the and sequences are different in the last sense for all . Hence, to give a similar result of Theorem 4 of [2] about convex, closed and bounded sets in without the FPP, it is necessary to consider the sequences.

Next we prove that if a convex and closed subset of a Banach space contains an asymptotically isometric -summing basic sequence, that is, an sequence, where is such that for all , then there exists a nonempty, convex, closed and bounded subset of without the FPP.

Finally, we show that the dual space of , over the reals, is the Bynum [5] space . Then, by a result of Dowling et al. in [6], the space has “many” subspaces and contains an asymptotically isometric copy of and does not have the FPP. In fact, we prove that every infinite dimensional subspace of contains an asymptotically isometric copy of and does not have the FPP.

2. The Space

In the sequel, we will denote by the canonical basis of and by the summing basis of , that is, .

García Falset proved in [7] that a Banach space with strongly bimonotone basis and with the weak Banach-Saks property has the -FPP. It is easy to see that the canonical basis of is strongly bimonotone in . On the other hand, since has the weak Banach-Saks property and and are equivalent, we get that has the weak Banach-Saks property. Hence we have that has the -FPP.

To study the FPP in the space using sequences, we would expect that nonempty, convex, closed and bounded subsets of , which are not -compact, contain an sequence. This fact is true for some -compact sets in , since the space embeds isometrically in . In fact we have the following proposition.

Proposition 1. Let be a block basis of with . If , for some , and , then the space is isometric to .

Proof. Since for every , then for all and Hence, it is straightforward to see that .

In the following theorem, we will show, using some results proved by Dowling et al. [4, 8], that every infinite-dimensional subspace of fails to have the FPP.

Theorem 2. Let be an infinite-dimensional subspace of . Then has a complemented asymptotically isometric copy of and thus does not have the FPP.

Proof. Let be a sequence such that and . As in [9] we construct sequences and such that , and Since , taking instead of , if necessary, we can suppose that there exists such that Define Then, by (2.3) and (2.2), we get that and where for . Then by (2.3) and (2.2), if , we get On the other hand, if , using (2.2), we get Then we obtain Now, define and ; then and . On the other hand, Thus Then by Theorem 2 of [8] contains an asymptotically isometric copy of and since does not contain a copy of , by Corollary 11 of [8] it contains a complemented asymptotically isometric copy of . Finally by Proposition 11 of [4], does not have the FPP.

As a consequence of the last theorem, we get that every infinite-dimensional subspace of contains an sequence. Nevertheless, the following result gives an example of a nonempty, convex, closed and bounded subset of which is not weakly compact and without sequences.

Proposition 3. Let be the summing basis. Then does not have sequences with the norm .

Proof. Suppose that is an sequence in with for some sequence . Then for some sequence such that and . Fix . Passing to a subsequence we can suppose that and .Assume first that there exists such that for every , . Let and ; then . Since is bounded and , passing to another subsequence we can suppose that for some . Then, there exist with such that Since , we also get Hence . On the other hand, since is an sequence, we have that , which contradicts the fact that .Assume now that for all , there exist such that . Denote each by , where is the canonical basis of . Then . Since , there exists such that By hypothesis, there exists such that . Then On the other hand, since is an sequence, we have that , which contradicts the fact that .

In view of the last proposition and motivated by the behavior of the summing basic sequence with the norm , we will define the asymptotically isometric -summing basic sequence. First we consider the following definition.

Definition 4. Let be a bounded basic sequence in a Banach space . We say that is a convexly closed sequence if the set is closed, that is, if .

Note that subsequences of convexly closed sequences are again convexly closed and that every basic sequence equivalent to a convexly closed sequence is convexly closed.

It is easy to see that the summing basis, the canonical basis of , and sequences are convexly closed. Moreover, a weakly null basic sequence in a Banach space is not a convexly closed sequence. Hence the canonical basis of and the canonical basis of , , are not convexly closed.

Definition 5. Let be a sequence in a Banach space . We say that is an asymptotically isometric -summing basic sequence, sequence for short, if is convexly closed and there exists such that and

Now, we prove that the analogous of the operator defined in [2] is still contractive and then Banach spaces containing sequences does not have the FPP.

Proposition 6. Let be a nonempty, convex, closed and bounded subset of a Banach space . Let be a sequence such that and . If contains an sequence with this , then there exists a nonempty, convex and closed subset of and affine, nonexpansive, and fixed-point-free. Moreover, is contractive.

Proof. Let be an sequence in with such that . Set Thus is nonempty, convex, closed and bounded. Define , and extend linearly to , that is, if then define . It is easy to see that and that is affine and fixed-point-free, see [2]. We only need to show that is a contractive mapping. Let , with . Then andwith and . Let , such that . As in [2] we have where and . Consequently, Take with . Since we have where Then we get Thus is contractive.

Next for any sequence of signs we will define a basis in equivalent to , the summing basis of , and a sequence asymptotically isometric to it.

Let be the canonical basis of and for any selection of signs , that is, , let be the sequence defined by Since for all , we get that is a basis of equivalent to the summing basis. The sequence is called the -basis of . Let , where . Then the -basis of is the summing basis. If we define , then is nonempty, convex and bounded. Since and are equivalent, we have that is convexly closed in .

The set is not -compact. The following result shows that the set contains neither sequences nor sequences with the norm if .

Proposition 7. For , let be the -basis of considered in . If then the set contains neither sequences nor sequences with the norm .

Proof. Let . Then for some and Suppose that is an sequence (resp. an sequence) with the norm . Let . If there exists such that for all , then for all , Since is an sequence (resp. an sequence) with the norm , then and this is impossible. Now, if , as in the proof of Proposition 3, we obtain a subsequence of with Since is an with the norm , then (resp. ) and making we get that . This contradiction proves the result.

Although the set of the last proposition has neither sequences nor sequences, for some it does not have the FPP.

For , let be the set such that if , then , and if and , then Denote by the cardinality of . If , define , and .

Proposition 8. Let . Then does not have the FPP in the following cases. (1)There exists such that .(2) and .(3)There exists , with , such that for any with we have and also for all or for all .

Proof. Let .(1) If there exists such that , define and by where . The idea is to translate the coefficients of in the block into the last terms of the block . Then it is easy to see that does not have fixed points. To prove that is nonexpansive first observe that if is even the signs of the and with and are the same and are different if is odd. Now let , , and . Then and . Hence are the expressions of and with respect to the canonical basis. Since the coefficients in (2.29) and (2.30) are the same, or the same with opposite signs, with perhaps some repetitions in (2.30), is an isometry.(2) Suppose now that and . In this case, define . Clearly is nonexpansive and fixed-point-free.(3) In this case it is straightforward to see thatthe operator defined by is nonexpansive and does not have fixed points.

Proposition 9. Let . Suppose does not satisfy the hypotheses of the above proposition, and let be a sequence with . Thenthe operator defined by is expansive.

Proof. Since does not satisfy the hypotheses (1) and (2) of the above proposition, there are three possibilities.(I) for every ; then for every there exists such that .(II); then .(III)There exists such that .Let be fixed with and denote . Since does not satisfy the hypotheses (3) of the above proposition, there exist and with such that or there exists with and there exists with .Case 1. For every there exists such that .There are two subcases.Subcase 1.1. There are and with such that .Let and Then and On the other hand, and .Subcase 1.2. For any and with, we have .There are two subsubcases. (1) and (2) .If for every ; then we would have , which implies . Then there is such that . Let . Then .There are two possibilities: (A) there exists such that and (B) for all .(A)Let . We need to consider the following cases.(a)Let and Then and On the other hand, and .(b). Let and . Then and . On the other hand, and .(B) for all . By hypothesis we have that . There are two cases.(a). Then Let and . Then and On the other hand, and .(b). Then Let and . Then and On the other hand, and .In this case there exists such that . If , then .Hence consider the cases: (A) there exists such that and (B) for all and proceed as in the Case (1).Case 2. and .Then with Hence we can proceed as in Subcase 1.2(1)(A) above taking .Case 3. There is such that .Then with Hence we can proceed as in Subcase 1.2(1)(A) above taking .

Next, for every selection of signs , we will define the asymptotically isometric --basic sequences. To this end, let us consider the following notation.

Let

Definition 10. Let be a sequence in a Banach space . We say that is an asymptotically isometric --basic sequence ( sequence for short) if is convexly closed and there exists such that , and holds for all , where

We are interested in sequences for which the numbers of Definition 10 are small. We are taking .

We know that the set of Proposition 3 does not have sequences. Now we also prove that does not contain sequences with the norm if .

Proposition 11. Let . The set does not contain sequences with the norm .

Proof. Let . Then for some with Suppose that is an with . Since , there exist and with , such that for all , and . Let for and . Thus Since (2.34) holds for all , making in (2.34), we get that , which is a contradiction.

Proposition 12. Let and such that and . Let be the -basis of considered in and let The set does not contain sequences with the norm .

Proof. Let . Then for some with Suppose that is an with the norm .Suppose first ; since , there exists such that .There are two cases.Case 1. . In this case . Let for and . Thus Since , we get a contradiction.Case 2. . In this case . Let for and . Thus Since , we get a contradiction.Suppose now ; since , there exists such that .There are two cases.Case 1. ; in this case . Let for and . Thus Since , we get a contradiction.Case 2. . In this case . Let for and . Thus Since , we get a contradiction.

Propositions 3, 7, and 11 show that, in contrast with Theorem 4 of the Dowling et al. paper [2] for sequences in , in the space we need an infinite number of sequences (at least and sequences) to have a similar result.

3. The Space

It is known that the dual of the Bynum space is the Bynum space . Below we prove that the dual space of when the scalar field is the set of real numbers is also the Bynum space . Let us suppose then that . First we calculate the extreme points of the unit ball of .

Lemma 13. Let . Then we have

Proof. First note that if then and . Consequently, if with for some , then , for all . Analogously if with for some , then , for all. Let and . Thus .Take and suppose that with . Also suppose that and . Since , there exists such that . Since and , if for some , we have that . Thus . On the other hand, if for some , we also have that , because if we get that , which contradicts that and if then and we also have a contradiction. Therefore, . Hence . Thus . Analogously .Take now . Then there exists such that . Let and . If , define , and . Thus, with and . Therefore, . Suppose now that or . Since and , we have that . Since , there exists such that , which implies that . Consequently, .

Theorem 14. Let . There exists a unique sequence such that and where and .

Proof. Let . Since is a shrinking basis of , there exists a unique sequence such that . As sets and hence . Thus where is the Riesz representation. Consequently, Therefore, . Thus where and .

Corollary 15. is the Bynum space and it has the -FPP.

Remark 16. It is well known that has the fixed point property for left reversible semigroups, that is, whenever is a semigroup such that for any , and is a representation of as nonexpansive mappings on a nonempty -compact convex subset of , there is a common fixed point in for . (see [10–12]). In particular, has the fixed point property. Is this the case for

Next we will see that every infinite-dimensional subspace of contains an asymptotically isometric copy of and then, by a result of Dowling and Lennard [13], it does not have the FPP.

First recall that a Banach space contains an asymptotically isometric copy of if there exists and such that for every and every scalars , In this case we say that is an asymptotically isometric -sequence (-sequence for short).

Observe that if is another sequence in such that for all , where and , then for every and every scalars , and is also an -sequence.

Proposition 17. Let , and let be a strictly increasing sequence in such that If , then is isometrically equivalent to the canonical basis in , that is, for every and every scalars , we have that .

Proof. Let be scalars; then

Theorem 18. Every infinite-dimensional subspace of contains an asymptotically isometric copy of and hence it does not have the FPP.

Proof. Let be an infinite-dimensional subspace of and a sequence in such that , where and Define Changing by , if necessary, we can assume that . If there is a sequence such that , then by Proposition 17, is isometrically equivalent to the canonical basis of . It is straightforward to see that . Then by the above remark, is an -sequence.Suppose that for all and let Then and Now let Suppose that . It is easy to check, using (3.10), that Hence, by Proposition 17, is isometrically equivalent to the canonical basis of .Now, if we define , it is straightforward to see that and by the above remark, is an -sequence.Finally in [13] Dowling and Lennard proved that if a Banach space contains an -sequence, then it does not have the FPP. Hence does not have the FPP.

Acknowledgments

This work was partially funded by Grant SEP CONACYT 102380 and partially supported by Conacyt Scholarship 165342 and CIMAT Scholarship.

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