Abstract and Applied Analysis

Abstract and Applied Analysis / 2011 / Article

Research Article | Open Access

Volume 2011 |Article ID 574614 | https://doi.org/10.1155/2011/574614

Berta Gamboa de Buen, Fernando NΓΊΓ±ez-Medina, "The Fixed Point Property in with an Equivalent Norm", Abstract and Applied Analysis, vol. 2011, Article ID 574614, 19 pages, 2011. https://doi.org/10.1155/2011/574614

The Fixed Point Property in π‘πŸŽ with an Equivalent Norm

Academic Editor: Elena Litsyn
Received07 Jun 2011
Accepted27 Aug 2011
Published29 Oct 2011

Abstract

We study the fixed point property (FPP) in the Banach space 𝑐0 with the equivalent norm ‖⋅‖𝐷. The space 𝑐0 with this norm has the weak fixed point property. We prove that every infinite-dimensional subspace of (𝑐0,‖⋅‖𝐷) contains a complemented asymptotically isometric copy of 𝑐0, and thus does not have the FPP, but there exist nonempty closed convex and bounded subsets of (𝑐0,‖⋅‖𝐷) which are not πœ”-compact and do not contain asymptotically isometric 𝑐0β€”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 (𝑐0,‖⋅‖𝐷), and we give some of its properties. We also prove that the dual space of (𝑐0,‖⋅‖𝐷) over the reals is the Bynum space 𝑙1∞ and that every infinite-dimensional subspace of 𝑙1∞ does not have the fixed point property.

1. Introduction

We start with some notations and terminologies. Let 𝐾 be a nonempty, convex, closed and bounded subset of a Banach space (𝑋,β€–β‹…β€–). A mapping π‘‡βˆΆπΎβ†’πΎ is said to be nonexpansive if ‖𝑇π‘₯βˆ’π‘‡π‘¦β€–β‰€β€–π‘₯βˆ’π‘¦β€–,π‘₯,π‘¦βˆˆπΎ.(1.1) We say that 𝐾 has the fixed point property for nonexpansive mappings (FPP) if every nonexpansive mapping π‘‡βˆΆπΎβ†’πΎ has a fixed point, that is, a point π‘₯∈𝐾 such that 𝑇π‘₯=π‘₯. We say that a Banach space (𝑋,β€–β‹…β€–) has the fixed point property for nonexpansive mappings (FPP) if every nonempty, convex, closed, and bounded subset 𝐾 of (𝑋,β€–β‹…β€–) has the FPP, and we say that the Banach space (𝑋,β€–β‹…β€–) has the weak fixed point property for nonexpansive mappings (πœ”-FPP) if every nonempty, convex and weakly compact subset 𝐾 of (𝑋,β€–β‹…β€–) has the FPP.

In this paper we study the FPP in the Banach space 𝑐0 with the equivalent norm ‖⋅‖𝐷 defined by β€–π‘₯‖𝐷=sup𝑖,π‘—βˆˆβ„•||π‘₯π‘–βˆ’π‘₯𝑗||ξ€½π‘₯,π‘₯=π‘–ξ€Ύβˆˆπ‘0.(1.2) The norm ‖⋅‖𝐷 was used by Hagler in [1] to construct a separable Banach space 𝑋 with nonseparable dual such that 𝑙1 does not embed in 𝑋 and every normalized weakly null sequence in 𝑋 has a subsequence equivalent to the canonical basis of 𝑐0.

In [2], Dowling et al. gave a characterization of nonempty, convex, closed and bounded subsets of 𝑐0 which are not πœ”-compact. Specifically, they proved that if 𝐾 is a convex, closed and bounded subset of 𝑐0, then 𝐾 is πœ”-compact if and only if every nonempty, convex, closed and convex subset of 𝐾 has the FPP. To do that, the authors showed that every closed, convex and bounded subset of 𝑐0 which is not πœ”-compact contains an asymptotically isometric 𝑐0-summing basic sequence, π‘Žπ‘–π‘ π‘π‘0 sequence for short, that is, a sequence {𝑦𝑛}π‘›βŠ‚π‘0 such that for all {𝑑𝑛}π‘›βˆˆπ‘™1,supπ‘›βˆˆβ„•ξ€·1+πœ€π‘›ξ€Έβˆ’1|||||βˆžξ“π‘–=𝑛𝑑𝑖|||||β‰€β€–β€–β€–β€–βˆžξ“π‘›=1𝑑𝑛𝑦𝑛‖‖‖‖≀supπ‘›βˆˆβ„•ξ€·1+πœ€π‘›ξ€Έ|||||βˆžξ“π‘–=𝑛𝑑𝑖|||||,(1.3) for some {πœ€π‘›}π‘›βŠ‚β„ with 0β‰€πœ€π‘›+1β‰€πœ€π‘› and limπ‘›πœ€π‘›=0. They proved that if a convex, closed and bounded subset 𝐾 of a Banach space contains an π‘Žπ‘–π‘ π‘π‘0 sequence, then there exists a nonempty, convex, closed and bounded subset of 𝐾 without the FPP. The authors used this fact in [3] as a tool to prove that a nonempty, closed, convex and bounded subset of 𝑐0 is πœ”-compact if and only if it has the FPP.

It is easy to see that (𝑐0,‖⋅‖𝐷) contains 𝑐0 isometrically, and then it contains π‘Žπ‘–π‘ π‘π‘0 sequences.

First we prove that every infinite-dimensional subspace π‘Œ of (𝑐0,‖⋅‖𝐷) has a complemented asymptotically isometric copy of 𝑐0 and hence by a result proved by Dowling et al. in [4], π‘Œ does not have the FPP. Also, as an immediate consequence we obtain that π‘Œ has an π‘Žπ‘–π‘ π‘π‘0 sequence. Nevertheless, we exhibit a nonempty closed, convex and bounded subset of (𝑐0,‖⋅‖𝐷), which is not πœ”-compact and does not contain π‘Žπ‘–π‘ π‘π‘0 sequences.

Then for every selection of signs Θ={πœƒπ‘–}, we define the Θ-basis of 𝑐0 which is equivalent to the summing basis and define the corresponding asymptotically isometric Θ-basic sequence, π‘Žπ‘–Ξ˜π‘π‘0𝐷 sequence for short. We prove that if Θ1β‰ Β±Ξ˜2, then the π‘Žπ‘–Ξ˜1𝑏𝑐0𝐷 and π‘Žπ‘–Ξ˜2𝑏𝑐0𝐷 sequences are different in the sense that there exists a nonempty, closed, convex, and bounded subset of (𝑐0,‖⋅‖𝐷), which is not πœ”-compact, contains an π‘Žπ‘–Ξ˜1𝑏𝑐0𝐷 sequence, and does not contain π‘Žπ‘–Ξ˜2𝑏𝑐0𝐷 sequences. We also show that the π‘Žπ‘–π‘ π‘π‘0 and π‘Žπ‘–Ξ˜π‘π‘0𝐷 sequences are different in the last sense for all Θ. Hence, to give a similar result of Theorem 4 of [2] about convex, closed and bounded sets in (𝑐0,‖⋅‖𝐷) without the FPP, it is necessary to consider the π‘Žπ‘–Ξ˜π‘π‘0𝐷 sequences.

Next we prove that if a convex and closed subset 𝐾 of a Banach space contains an asymptotically isometric 𝑐0𝐷-summing basic sequence, that is, an π‘Žπ‘–Ξ˜π‘π‘0𝐷 sequence, where Θ is such that πœƒπ‘–=1 for all 𝑖, then there exists a nonempty, convex, closed and bounded subset of 𝐾 without the FPP.

Finally, we show that the dual space of (𝑐0,‖⋅‖𝐷), over the reals, is the Bynum [5] space 𝑙1∞. Then, by a result of Dowling et al. in [6], the space 𝑙1∞=(𝑐0,‖⋅‖𝐷)βˆ— has β€œmany” subspaces and contains an asymptotically isometric copy of 𝑙1 and does not have the FPP. In fact, we prove that every infinite dimensional subspace of 𝑙1∞ contains an asymptotically isometric copy of 𝑙1 and does not have the FPP.

2. The Space (𝑐0,‖⋅‖𝐷)

In the sequel, we will denote by {𝑒𝑛} the canonical basis of 𝑐0 and by {πœ‰π‘›} the summing basis of 𝑐0, that is, πœ‰π‘›=βˆ‘π‘›π‘–=1𝑒𝑖,π‘›βˆˆβ„•.

GarcΓ­a Falset proved in [7] that a Banach space with strongly bimonotone basis and with the weak Banach-Saks property has the πœ”-FPP. It is easy to see that the canonical basis of 𝑐0 is strongly bimonotone in (𝑐0,‖⋅‖𝐷). On the other hand, since 𝑐0 has the weak Banach-Saks property and ‖⋅‖𝐷 and β€–β‹…β€–βˆž are equivalent, we get that (𝑐0,‖⋅‖𝐷) has the weak Banach-Saks property. Hence we have that (𝑐0,‖⋅‖𝐷) has the πœ”-FPP.

To study the FPP in the space (𝑐0,‖⋅‖𝐷) using π‘Žπ‘–π‘ π‘π‘0 sequences, we would expect that nonempty, convex, closed and bounded subsets 𝐾 of (𝑐0,‖⋅‖𝐷), which are not πœ”-compact, contain an π‘Žπ‘–π‘ π‘π‘0 sequence. This fact is true for some πœ”-compact sets in (𝑐0,‖⋅‖𝐷), since the space 𝑐0 embeds isometrically in (𝑐0,‖⋅‖𝐷). In fact we have the following proposition.

Proposition 1. Let {π‘’π‘˜}π‘˜βŠ‚(𝑐0,‖⋅‖𝐷) be a block basis of {𝑒𝑛} with π‘’π‘˜=βˆ‘π‘žπ‘˜π‘–=π‘π‘˜π‘Žπ‘–π‘’π‘–,1≀𝑝1β‰€π‘ž1<𝑝2β‰€π‘ž2<β‹―. If β€–π‘’π‘˜β€–βˆž=1=π‘Žπ‘–π‘˜, for some π‘π‘˜β‰€π‘–π‘˜β‰€π‘žπ‘˜, and π‘¦π‘˜=(1/2)(𝑒2π‘˜βˆ’π‘’2π‘˜βˆ’1), then the space span{π‘¦π‘˜} is isometric to (𝑐0,β€–β‹…β€–βˆž).

Proof. Since β€–π‘’π‘˜β€–βˆž=1=π‘Žπ‘–π‘˜ for every π‘˜βˆˆβ„•, then |π‘Žπ‘—|≀1 for all π‘—βˆˆβ„• and max𝑝2π‘˜βˆ’1β‰€π‘–β‰€π‘ž2π‘˜βˆ’1,𝑝2π‘˜β‰€π‘—β‰€π‘ž2π‘˜||π‘Žπ‘–+π‘Žπ‘—||=π‘Žπ‘–2π‘˜βˆ’1+π‘Žπ‘–2π‘˜=2.(2.1) Hence, it is straightforward to see that β€–βˆ‘π‘›π‘˜=1π‘‘π‘˜π‘¦π‘˜β€–π·βˆ‘=β€–π‘›π‘˜=1π‘‘π‘˜π‘’π‘˜β€–βˆž.

In the following theorem, we will show, using some results proved by Dowling et al. [4, 8], that every infinite-dimensional subspace π‘Œ of 𝑐0𝐷 fails to have the FPP.

Theorem 2. Let π‘Œ be an infinite-dimensional subspace of 𝑐0𝐷. Then π‘Œ has a complemented asymptotically isometric copy of 𝑐0 and thus π‘Œ does not have the FPP.

Proof. Let {πœ€π‘˜}π‘˜βŠ‚(0,1) be a sequence such that πœ€π‘˜+1<πœ€π‘˜,π‘˜βˆˆβ„• and πœ€π‘˜β†’0. As in [9] we construct sequences {π‘›π‘˜}βŠ‚β„• and {π‘¦π‘˜}π‘˜βŠ‚π‘Œ such that π‘›π‘˜<π‘›π‘˜+1,π‘¦π‘˜=βˆ‘βˆžπ‘–=π‘›π‘˜π›Όπ‘˜π‘–π‘’π‘–,β€–π‘¦π‘˜β€–βˆž=1, and sup𝑖β‰₯π‘›π‘˜+1||𝛼𝑗𝑖||<πœ€π‘˜+24π‘˜βˆ€π‘—=1,…,π‘˜,andeveryπ‘˜βˆˆβ„•.(2.2) Since β€–π‘¦π‘˜β€–βˆž=1, taking βˆ’π‘¦π‘˜ instead of π‘¦π‘˜, if necessary, we can suppose that there exists π‘›π‘˜β‰€π‘Ÿπ‘˜<π‘›π‘˜+1 such that π›Όπ‘˜π‘Ÿπ‘˜=1.(2.3)
Define π‘₯π‘˜=(𝑦2π‘˜βˆ’1βˆ’π‘¦2π‘˜)/2.Then, by (2.3) and (2.2), we get that 1βˆ’(πœ€π‘˜/2)<β€–π‘₯π‘˜β€–π·<1+(πœ€π‘˜/2) and βˆžξ“π‘˜=1π‘‘π‘˜π‘₯π‘˜=12βˆžξ“π‘˜=1π‘‘π‘˜ξƒ©βˆžξ“π‘–=𝑛2π‘˜βˆ’1𝛼𝑖2π‘˜βˆ’1π‘’π‘–βˆ’βˆžξ“π‘–=𝑛2π‘˜π›Όπ‘–2π‘˜π‘’π‘–ξƒͺ=12βˆžξ“π‘˜=1π‘‘π‘˜ξƒ©βˆžξ“π‘–=𝑛2π‘˜βˆ’1𝛼𝑖2π‘˜βˆ’1βˆ’π›Όπ‘–2π‘˜ξ€Έπ‘’π‘–ξƒͺ=12βˆžξ“π‘˜=1βŽ›βŽœβŽœβŽπ‘›2π‘˜+1βˆ’1𝑖=𝑛2π‘˜βˆ’1ξƒ©π‘˜ξ“π‘—=1𝑑𝑗𝛼𝑖2π‘—βˆ’1βˆ’π›Όπ‘–2𝑗𝑒𝑖ξƒͺ⎞⎟⎟⎠,(2.4) where 𝛼𝑖2π‘˜=0 for 𝑖=𝑛2π‘˜βˆ’1,…𝑛2π‘˜βˆ’1,π‘˜βˆˆβ„•. Then by (2.3) and (2.2), if π‘˜>1, we get β€–β€–β€–β€–βˆžξ“π‘›=1𝑑𝑛π‘₯𝑛‖‖‖‖𝐷β‰₯12max𝑛2π‘˜βˆ’1β‰€π‘Ÿ<𝑛2π‘˜,𝑛2π‘˜β‰€π‘ <𝑛2π‘˜+1|||||π‘˜ξ“π‘—=1π‘‘π‘—ξ‚€π›Όπ‘Ÿ2π‘—βˆ’1βˆ’π›Όπ‘Ÿ2π‘—βˆ’π›Όπ‘ 2π‘—βˆ’1+𝛼𝑠2𝑗|||||β‰₯12|||||π‘˜ξ“π‘—=1π‘‘π‘—ξ‚€π›Όπ‘Ÿ2π‘—βˆ’12π‘˜βˆ’1βˆ’π›Όπ‘Ÿ2𝑗2π‘˜βˆ’1βˆ’π›Όπ‘Ÿ2π‘—βˆ’12π‘˜+π›Όπ‘Ÿ2𝑗2π‘˜ξ‚|||||β‰₯12||π‘‘π‘˜||||π›Όπ‘Ÿ2π‘˜βˆ’12π‘˜βˆ’1βˆ’π›Όπ‘Ÿ2π‘˜βˆ’12π‘˜+π›Όπ‘Ÿ2π‘˜2π‘˜||βˆ’12π‘˜βˆ’1𝑗=1||𝑑𝑗||||π›Όπ‘Ÿ2π‘—βˆ’12π‘˜βˆ’1βˆ’π›Όπ‘Ÿ2𝑗2π‘˜βˆ’1βˆ’π›Όπ‘Ÿ2π‘—βˆ’12π‘˜+π›Όπ‘Ÿ2𝑗2π‘˜||β‰₯12||π‘‘π‘˜||ξ€·||π›Όπ‘Ÿ2π‘˜βˆ’12π‘˜βˆ’1+π›Όπ‘Ÿ2π‘˜2π‘˜||βˆ’||π›Όπ‘Ÿ2π‘˜βˆ’12π‘˜||ξ€Έβˆ’12π‘˜βˆ’1𝑗=1||𝑑𝑗||ξ‚€||π›Όπ‘Ÿ2π‘—βˆ’12π‘˜βˆ’1||+||π›Όπ‘Ÿ2𝑗2π‘˜βˆ’1||+||π›Όπ‘Ÿ2π‘—βˆ’12π‘˜||+||π›Όπ‘Ÿ2𝑗2π‘˜||β‰₯12||π‘‘π‘˜||ξ€·2βˆ’πœ€π‘˜ξ€Έβˆ’12π‘˜βˆ’1𝑗=1||𝑑𝑗||πœ€π‘˜π‘˜β‰₯||π‘‘π‘˜||ξ‚€πœ€1βˆ’π‘˜2ξ‚βˆ’max1β‰€π‘—β‰€π‘˜||𝑑𝑗||πœ€π‘˜2.(2.5) On the other hand, if 𝑛2π‘˜βˆ’1β‰€π‘Ÿ<𝑛2π‘˜+1,𝑛2π‘šβˆ’1≀𝑠<𝑛2π‘š+1,π‘˜β‰€π‘š, using (2.2), we get 12|||||π‘˜ξ“π‘—=1π‘‘π‘—ξ‚€π›Όπ‘Ÿ2π‘—βˆ’1βˆ’π›Όπ‘Ÿ2π‘—ξ‚βˆ’π‘šξ“π‘—=1𝑑𝑗𝛼𝑠2π‘—βˆ’1βˆ’π›Όπ‘ 2𝑗|||||≀12ξ€Ί||π‘‘π‘˜||ξ€·||π›Όπ‘Ÿ2π‘˜βˆ’1||+||π›Όπ‘Ÿ2π‘˜||ξ€Έ+||π‘‘π‘š||ξ€·||𝛼𝑠2π‘šβˆ’1||+||𝛼𝑠2π‘š||+1ξ€Έξ€»2ξƒ¬π‘˜βˆ’1𝑗=1||𝑑𝑗||ξ‚€||π›Όπ‘Ÿ2π‘—βˆ’1||+||π›Όπ‘Ÿ2𝑗||+π‘šβˆ’1𝑗=1||𝑑𝑗||ξ‚€||𝛼𝑠2π‘—βˆ’1||+||𝛼𝑠2𝑗||≀12||π‘‘π‘˜||ξ€·1+πœ€π‘˜ξ€Έ+||π‘‘π‘š||ξ€·1+πœ€π‘šξ€Έ+π‘˜βˆ’1𝑗=1||𝑑𝑗||πœ€π‘˜+π‘˜βˆ’1π‘šβˆ’1𝑗=1||𝑑𝑗||πœ€π‘šξƒ­β‰€1π‘šβˆ’12ξ‚Έ||π‘‘π‘˜||ξ€·1+πœ€π‘˜ξ€Έ+||π‘‘π‘š||ξ€·1+πœ€π‘šξ€Έ+max1≀𝑗<π‘˜||𝑑𝑗||πœ€π‘˜+max1≀𝑗<π‘š||𝑑𝑗||πœ€π‘šξ‚Ήβ‰€12ξ‚Έmax1β‰€π‘—β‰€π‘˜||𝑑𝑗||ξ€·1+πœ€π‘˜ξ€Έ+max1β‰€π‘—β‰€π‘š||𝑑𝑗||ξ€·1+πœ€π‘šξ€Έξ‚Ήβ‰€supπ‘›βˆˆβ„•ξ‚΅ξ€·1+πœ€π‘›ξ€Έmax1≀𝑗≀𝑛||𝑑𝑗||≀supπ‘›βˆˆβ„•ξ€·1+πœ€π‘›ξ€Έ||𝑑𝑛||.(2.6) Then we obtain supπ‘›βˆˆβ„•ξ‚΅||𝑑𝑛||ξ‚€πœ€1βˆ’π‘›2ξ‚βˆ’max1≀𝑗≀𝑛||𝑑𝑗||πœ€π‘›2ξ‚Άβ‰€β€–β€–β€–β€–βˆžξ“π‘›=1𝑑𝑛π‘₯𝑛‖‖‖‖𝐷≀supπ‘›βˆˆβ„•ξ€·1+πœ€π‘›ξ€Έ||𝑑𝑛||.(2.7) Now, define 𝑧𝑛=π‘₯𝑛/(1+πœ€π‘›) and π‘š=(1βˆ’πœ€1)/(1+πœ€1); then (1βˆ’πœ€π‘›)/(1+πœ€π‘›)≀‖𝑧𝑛‖𝐷 and lim𝑛‖𝑧𝑛‖𝐷=1. On the other hand, ξ€·1+πœ€1ξ€Έπ‘šsupπ‘›βˆˆβ„•||𝑑𝑛||=ξ€·1βˆ’πœ€1ξ€Έsupπ‘›βˆˆβ„•||𝑑𝑛||=ξ‚€πœ€1βˆ’12supπ‘›βˆˆβ„•||𝑑𝑛||βˆ’πœ€12supπ‘›βˆˆβ„•||𝑑𝑛||≀supπ‘›βˆˆβ„•ξ‚΅||𝑑𝑛||ξ‚€πœ€1βˆ’π‘›2ξ‚βˆ’max1≀𝑗≀𝑛||𝑑𝑗||πœ€π‘›2ξ‚Ά.(2.8) Thus π‘šsupπ‘›βˆˆβ„•||𝑑𝑛||≀supπ‘›βˆˆβ„•ξ‚΅||𝑑𝑛||ξ‚€πœ€1βˆ’π‘›2ξ‚βˆ’max1≀𝑗≀𝑛||𝑑𝑗||πœ€π‘›2ξ‚Άβ‰€β€–β€–β€–β€–βˆžξ“π‘›=1𝑑𝑛𝑧𝑛‖‖‖‖𝐷≀supπ‘›βˆˆβ„•||𝑑𝑛||.(2.9) Then by Theorem 2 of [8] π‘Œ contains an asymptotically isometric copy of 𝑐0 and since π‘Œ does not contain a copy of 𝑙1, by Corollary 11 of [8] it contains a complemented asymptotically isometric copy of 𝑐0. Finally by Proposition 11 of [4], π‘Œ does not have the FPP.

As a consequence of the last theorem, we get that every infinite-dimensional subspace of (𝑐0,‖⋅‖𝐷) contains an π‘Žπ‘–π‘ π‘π‘0 sequence. Nevertheless, the following result gives an example of a nonempty, convex, closed and bounded subset of (𝑐0,‖⋅‖𝐷) which is not weakly compact and without π‘Žπ‘–π‘ π‘c0 sequences.

Proposition 3. Let {πœ‰π‘›} be the 𝑐0 summing basis. Then 𝐢=βˆžξ“π‘›=1πœ†π‘›πœ‰π‘›βˆΆπœ†π‘›β‰₯0,βˆžξ“π‘›=1πœ†π‘›ξƒ°=1(2.10) does not have π‘Žπ‘–π‘ π‘π‘0 sequences with the norm ‖⋅‖𝐷.

Proof. Suppose that {𝑦𝑛} is an π‘Žπ‘–π‘ π‘π‘0 sequence in 𝐢 with ‖⋅‖𝐷 for some sequence {πœ€π‘›}. Then 𝑦𝑛=βˆ‘βˆžπ‘–=1πœ†π‘›π‘–πœ‰π‘– for some sequence {πœ†π‘›π‘–} such that πœ†π‘›π‘–β‰₯0 and βˆ‘βˆžπ‘–=1πœ†π‘›π‘–=1. Fix 0<πœ€<1/4. Passing to a subsequence we can suppose that πœ€π‘›+1β‰€πœ€π‘›<(1/2)βˆ’2πœ€ and 1/(1+πœ€π‘›)>1βˆ’πœ€,π‘›βˆˆβ„•.
Assume first that there exists π‘€βˆˆβ„• such that for every 𝑛β‰₯𝑀, βˆ‘βˆžπ‘–=𝑀+1πœ†π‘›π‘–β‰€(1/2)βˆ’πœ€. Let 𝑒𝑛=βˆ‘π‘€π‘–=1πœ†π‘›π‘–πœ‰π‘– and 𝑣𝑛=βˆ‘βˆžπ‘–=𝑀+1πœ†π‘›π‘–πœ‰π‘–; then 𝑦𝑛=𝑒𝑛+𝑣𝑛. Since {𝑒𝑛}βŠ‚[πœ‰π‘–]𝑀𝑖=1 is bounded and dim[πœ‰π‘–]𝑀𝑖=1=𝑀, passing to another subsequence we can suppose that 𝑒𝑛→𝑒 for some π‘’βˆˆπΆ. Then, there exist 𝑛1,𝑛2βˆˆβ„• with 𝑀≀𝑛1<𝑛2 such that ‖‖𝑒𝑛1βˆ’π‘’π‘›2‖‖𝐷<πœ€.(2.11) Since βˆ‘βˆžπ‘–=𝑀+1πœ†π‘›π‘–β‰€(1/2)βˆ’πœ€,𝑛β‰₯𝑀, we also get ‖‖𝑣𝑛1βˆ’π‘£π‘›2‖‖𝐷=max𝑀+1β‰€π‘Ÿβ‰€π‘˜<∞|||||π‘˜ξ“π‘–=π‘Ÿπœ†π‘›1π‘–βˆ’π‘˜ξ“π‘–=π‘Ÿπœ†π‘›2𝑖|||||β‰€βˆžξ“π‘–=𝑀+1πœ†π‘›1𝑖+βˆžξ“π‘–=𝑀+1πœ†π‘›2𝑖≀1βˆ’2πœ€.(2.12) Hence ‖𝑦𝑛1βˆ’π‘¦π‘›2‖𝐷≀1βˆ’πœ€. On the other hand, since {𝑦𝑛} is an π‘Žπ‘–π‘ π‘π‘0 sequence, we have that ‖𝑦𝑛1βˆ’π‘¦π‘›2‖𝐷β‰₯1/(1+πœ€π‘›2), which contradicts the fact that 1/(1+πœ€π‘›2)>1βˆ’πœ€.
Assume now that for all π‘€βˆˆβ„•, there exist 𝑛β‰₯𝑀 such that βˆ‘βˆžπ‘–=𝑀+1πœ†π‘›π‘–>(1/2)βˆ’πœ€. Denote each 𝑦𝑛 by {π›Όπ‘›π‘–βˆ‘}=βˆžπ‘–=1𝛼𝑛𝑖𝑒𝑛, where {𝑒𝑛} is the canonical basis of 𝑐0. Then 𝛼𝑛𝑖=βˆ‘βˆžπ‘—=π‘–πœ†π‘›π‘—. Since 𝑦1,𝑦2βˆˆπ‘0, there exists π‘€βˆˆβ„• such that 𝛼1𝑖,𝛼2𝑖<πœ€2,𝑖β‰₯𝑀.(2.13) By hypothesis, there exists 𝑛0βˆˆβ„• such that βˆ‘βˆžπ‘–=𝑀+1πœ†π‘›0𝑖>(1/2)βˆ’πœ€. Then 32β€–β€–π‘¦βˆ’2πœ€β‰€1+𝑦2βˆ’π‘¦π‘›0‖‖𝐷.(2.14) On the other hand, since {𝑦𝑛} is an π‘Žπ‘–π‘ π‘π‘0 sequence, we have that ‖𝑦1+𝑦2βˆ’π‘¦π‘›0‖𝐷≀1+πœ€1, which contradicts the fact that πœ€1<(1/2)βˆ’2πœ€.

In view of the last proposition and motivated by the behavior of the 𝑐0 summing basic sequence with the norm ‖⋅‖𝐷, we will define the asymptotically isometric 𝑐0𝐷-summing basic sequence. First we consider the following definition.

Definition 4. Let {π‘₯𝑛} be a bounded basic sequence in a Banach space 𝑋. We say that {π‘₯𝑛} is a convexly closed sequence if the set 𝐢=βˆžξ“π‘›=1𝑑𝑛π‘₯π‘›βˆΆπ‘‘π‘›β‰₯0,βˆžξ“π‘›=1𝑑𝑛=1(2.15) is closed, that is, if conv{π‘₯𝑛}=𝐢.

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 𝑐0 summing basis, the canonical basis of 𝑙1, and π‘Žπ‘–π‘ π‘π‘0 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 𝑐0 and the canonical basis of 𝑙𝑝, 1<𝑝<∞, are not convexly closed.

Definition 5. Let {π‘₯𝑛} be a sequence in a Banach space 𝑋. We say that {π‘₯𝑛} is an asymptotically isometric 𝑐0𝐷-summing basic sequence, π‘Žπ‘–π‘ π‘π‘0𝐷 sequence for short, if {π‘₯𝑛} is convexly closed and there exists {πœ€π‘›}βŠ‚(0,∞) such that πœ€π‘›β†˜0 and sup1β‰€π‘›β‰€π‘š<βˆžξ€·1+πœ€π‘šξ€Έβˆ’1|||||π‘šξ“π‘˜=π‘›π‘‘π‘˜|||||β‰€β€–β€–β€–β€–βˆžξ“π‘›=1𝑑𝑛π‘₯𝑛‖‖‖‖≀sup1β‰€π‘›β‰€π‘š<βˆžξ€·1+πœ€π‘šξ€Έ|||||π‘šξ“π‘˜=π‘›π‘‘π‘˜|||||𝑑,βˆ€π‘›ξ€Ύβˆˆπ‘™1.(2.16)

Now, we prove that the analogous of the operator defined in [2] is still contractive and then Banach spaces containing π‘Žπ‘–π‘ π‘π‘0𝐷 sequences does not have the FPP.

Proposition 6. Let 𝐾 be a nonempty, convex, closed and bounded subset of a Banach space 𝑋. Let {πœ€π‘›}βŠ‚(0,∞) be a sequence such that πœ€π‘›β†’0 and πœ€π‘›<2βˆ’14βˆ’π‘›,𝑛β‰₯2. If 𝐾 contains an π‘Žπ‘–π‘ π‘π‘0𝐷 sequence with this {πœ€π‘›}, then there exists a nonempty, convex and closed subset 𝐢 of 𝐾 and π‘‡βˆΆπΆβ†’πΆ affine, nonexpansive, and fixed-point-free. Moreover, 𝑇 is contractive.

Proof. Let {π‘₯𝑛} be an π‘Žπ‘–π‘ π‘π‘0𝐷 sequence in 𝐾 with {πœ€π‘›}βŠ‚(0,∞) such that πœ€π‘›<2βˆ’14βˆ’π‘›,𝑛β‰₯2. Set 𝐢=convξ€½π‘₯𝑛=ξƒ―βˆžξ“π‘›=1𝑑𝑛π‘₯π‘›βˆΆπ‘‘π‘›β‰₯0,π‘›βˆˆβ„•π‘¦βˆžξ“π‘›=1𝑑𝑛=1βŠ‚πΎ.(2.17) Thus 𝐢 is nonempty, convex, closed and bounded. Define 𝑇π‘₯𝑛=βˆ‘βˆžπ‘—=1((π‘₯𝑛+𝑗)/2𝑗),π‘›βˆˆβ„•, and extend 𝑇 linearly to 𝐢, that is, if βˆ‘π‘₯=βˆžπ‘›=1𝑑𝑛π‘₯π‘›βˆˆπΆ then define βˆ‘π‘‡(βˆžπ‘›=1𝑑𝑛π‘₯π‘›βˆ‘)=βˆžπ‘›=1𝑑𝑛𝑇π‘₯𝑛. It is easy to see that 𝑇(𝐢)βŠ‚πΆ and that 𝑇 is affine and fixed-point-free, see [2]. We only need to show that 𝑇 is a contractive mapping. Let π‘₯,π‘¦βˆˆπΆ, with π‘₯≠𝑦. Then βˆ‘π‘₯=βˆžπ‘›=1𝑑𝑛π‘₯𝑛 andβˆ‘π‘¦=βˆžπ‘›=1𝑠𝑛π‘₯𝑛,with 𝑑𝑛,𝑠𝑛β‰₯0,and βˆ‘βˆžπ‘›=1𝑑𝑛=βˆ‘βˆžπ‘›=1𝑠𝑛=1. Let 𝛽𝑛=π‘‘π‘›βˆ’π‘ π‘›,π‘›βˆˆβ„•, such that βˆ‘βˆžπ‘›=1𝛽𝑛=0. As in [2] we have 𝑇(π‘₯)βˆ’π‘‡(𝑦)=βˆžξ“π‘›=1𝐡𝑛π‘₯𝑛,(2.18) where 𝐡1=0 and 𝐡𝑛=(𝛽1/2π‘›βˆ’1)+(𝛽2/2π‘›βˆ’2)+β‹―+(π›½π‘›βˆ’1/2),𝑛β‰₯2. Consequently, ‖‖‖‖‖𝑇(π‘₯)βˆ’π‘‡(𝑦)β€–=βˆžξ“π‘›=1𝐡𝑛π‘₯𝑛‖‖‖‖≀sup1β‰€π‘›β‰€π‘š<βˆžξ€·1+πœ€π‘šξ€Έ|||||π‘šξ“π‘˜=π‘›π΅π‘˜|||||.(2.19) Take 𝑛,π‘šβˆˆβ„• with π‘›β‰€π‘š. Since π‘šξ“π‘˜=π‘›π΅π‘˜=𝛽12π‘›βˆ’1+𝛽22π‘›βˆ’2𝛽+β‹―+π‘›βˆ’12+𝛽12𝑛+𝛽22π‘›βˆ’1𝛽+β‹―+π‘›βˆ’122+𝛽𝑛2+𝛽+β‹―12π‘šβˆ’1+𝛽22π‘šβˆ’2𝛽+β‹―+π‘›βˆ’12π‘šβˆ’(π‘›βˆ’1)+𝛽𝑛2π‘šβˆ’π‘›+𝛽𝑛+12π‘šβˆ’(𝑛+1)𝛽+β‹―+π‘šβˆ’12=12ξ€·π›½π‘›βˆ’1+𝛽𝑛+β‹―+π›½π‘šβˆ’1ξ€Έ+122ξ€·π›½π‘›βˆ’2+π›½π‘›βˆ’1+β‹―+π›½π‘šβˆ’2ξ€Έ+1+β‹―2π‘›βˆ’1𝛽1+𝛽2+β‹―+π›½π‘šβˆ’(π‘›βˆ’1)ξ€Έ+1+β‹―2π‘šβˆ’2𝛽1+𝛽2ξ€Έ+12π‘šβˆ’1𝛽1ξ€Έ,(2.20) we have ξ€·1+πœ€π‘šξ€Έ|||||π‘šξ“π‘˜=π‘›π΅π‘˜|||||≀1+πœ€π‘šξ€Έξ‚΅1+2πœ€π‘šβˆ’1211+2πœ€π‘šβˆ’1||π›½π‘›βˆ’1+𝛽𝑛+β‹―+π›½π‘šβˆ’1||+1+2πœ€π‘šβˆ’22211+2πœ€π‘šβˆ’2||π›½π‘›βˆ’2+π›½π‘›βˆ’1+β‹―+π›½π‘šβˆ’2||++β‹―1+2πœ€π‘šβˆ’(π‘›βˆ’1)2π‘›βˆ’111+2πœ€π‘šβˆ’(π‘›βˆ’1)||𝛽1+𝛽2+β‹―+π›½π‘šβˆ’(π‘›βˆ’1)||++β‹―1+2πœ€22π‘šβˆ’211+2πœ€2||𝛽1+𝛽2||+1+2πœ€12π‘šβˆ’111+2πœ€1||𝛽1||ξ‚Άβ‰€βŽ›βŽœβŽœβŽsup1β‰€π‘–β‰€π‘—β‰€π‘šξ€·1+2πœ€π‘—ξ€Έβˆ’1|||||π‘—ξ“π‘˜=π‘–π›½π‘˜|||||βŽžβŽŸβŽŸβŽ π‘„π‘›π‘š,(2.21) where π‘„π‘›π‘š=ξ€·1+πœ€π‘šξ€Έξ‚΅1+2πœ€π‘šβˆ’12+1+2πœ€π‘šβˆ’222++β‹―+1+2πœ€π‘šβˆ’(π‘›βˆ’1)2π‘›βˆ’1+1+2πœ€π‘šβˆ’π‘›2𝑛+β‹―+1+2πœ€22π‘šβˆ’2+1+2πœ€12π‘šβˆ’1≀11+2β‹…4π‘šξ‚1ξ‚Έξ‚΅2+1221+β‹―+2π‘šβˆ’1ξ‚Ά+ξ‚΅12β‹…4π‘šβˆ’11+β‹―+2π‘šβˆ’1β‹…41=ξ‚€1ξ‚Άξ‚Ή1+2β‹…4π‘šξ‚1ξ‚Έξ‚΅1βˆ’2π‘šβˆ’1ξ‚Ά+ξ‚΅122π‘šβˆ’1+122π‘šβˆ’21+β‹―+2π‘š+1<ξ‚€1ξ‚Άξ‚Ή1+4π‘šξ‚1ξ‚Έξ‚΅1βˆ’2π‘šβˆ’1ξ‚Ά+12π‘šξ‚Ή<1.(2.22) Then we get sup1β‰€π‘›β‰€π‘š<βˆžξ€·1+πœ€π‘šξ€Έ|||||π‘šξ“π‘˜=π‘›π΅π‘˜|||||≀sup1β‰€π‘›β‰€π‘š<βˆžξ€·1+2πœ€π‘šξ€Έβˆ’1|||||π‘šξ“π‘˜=π‘›π›½π‘˜|||||<sup1β‰€π‘›β‰€π‘š<βˆžξ€·1+πœ€π‘šξ€Έβˆ’1|||||π‘šξ“π‘˜=π‘›π›½π‘˜|||||β‰€β€–β€–β€–β€–βˆžξ“π‘›=1𝛽𝑛π‘₯𝑛‖‖‖‖=β€–π‘₯βˆ’π‘¦β€–.(2.23) Thus 𝑇 is contractive.

Next for any sequence of signs we will define a basis in 𝑐0 equivalent to {πœ‰π‘›}, the summing basis of 𝑐0, and a sequence asymptotically isometric to it.

Let {𝑒𝑛} be the canonical basis of 𝑐0 and for any selection of signs Θ={πœƒπ‘–}𝑖, that is, πœƒπ‘–βˆˆ{βˆ’1,1},π‘–βˆˆβ„•, let {πœΞ˜π‘›}𝑛 be the sequence defined by πœΞ˜π‘›=π‘›ξ“π‘˜=1πœƒπ‘˜π‘’π‘˜,π‘›βˆˆβ„•.(2.24) Since β€–βˆ‘π‘šπ‘›=1π‘‘π‘›πœ‰π‘›β€–βˆž=β€–βˆ‘π‘šπ‘›=1π‘‘π‘›πœΞ˜π‘›β€–βˆž for all {𝑑𝑛}π‘šπ‘›=1βŠ‚π•‚, we get that {πœΞ˜π‘›} is a basis of 𝑐0 equivalent to the 𝑐0 summing basis. The sequence {πœΞ˜π‘›} is called the Θ-basis of 𝑐0. Let Θ0={πœƒπ‘–}, where πœƒπ‘–=1,π‘–βˆˆβ„•. Then the Θ0-basis of 𝑐0 is the 𝑐0 summing basis. If we define βˆ‘πΆ={βˆžπ‘›=1π‘‘π‘›πœΞ˜π‘›βˆΆπ‘‘π‘›β‰₯0andβˆ‘βˆžπ‘›=1𝑑𝑛=1}, then 𝐢 is nonempty, convex and bounded. Since β€–β‹…β€–βˆž and ‖⋅‖𝐷 are equivalent, we have that {πœΞ˜π‘›} is convexly closed in (𝑐0,‖⋅‖𝐷).

The set βˆ‘πΆ={βˆžπ‘›=1π‘‘π‘›πœΞ˜π‘›βˆΆπ‘‘π‘›β‰₯0andβˆ‘βˆžπ‘›=1𝑑𝑛=1} is not πœ”-compact. The following result shows that the set 𝐢 contains neither π‘Žπ‘–π‘ π‘π‘0𝐷 sequences nor π‘Žπ‘–π‘ π‘π‘0 sequences with the norm ‖⋅‖𝐷 if Ξ˜β‰ Β±Ξ˜0.

Proposition 7. For Ξ˜β‰ Β±Ξ˜0, let {πœΞ˜π‘›} be the Θ-basis of 𝑐0 considered in (𝑐0,‖⋅‖𝐷). If 𝐢=βˆžξ“π‘›=1π‘‘π‘›πœΞ˜π‘›βˆΆπ‘‘π‘›β‰₯0,βˆžξ“π‘›=1𝑑𝑛,=1(2.25) then the set 𝐢 contains neither π‘Žπ‘–π‘ π‘π‘0𝐷 sequences nor π‘Žπ‘–π‘ π‘π‘0 sequences with the norm ‖⋅‖𝐷.

Proof. Let {π‘¦π‘˜}βŠ‚πΆ. Then π‘¦π‘˜=βˆ‘βˆžπ‘›=1πœ†π‘˜π‘›πœΞ˜π‘› for some πœ†π‘˜π‘›β‰₯0and βˆ‘βˆžπ‘›=1πœ†π‘˜π‘›=1.Suppose that {π‘¦π‘˜} is an π‘Žπ‘–π‘ π‘π‘0𝐷 sequence (resp. an π‘Žπ‘–π‘ π‘π‘0 sequence) with the norm ‖⋅‖𝐷. Let 𝑛0=min{π‘›βˆΆπœƒπ‘›β‰ πœƒ1}. If there exists 0<𝜌<1 such that βˆ‘π‘›0βˆ’1𝑖=1πœ†π‘˜π‘–β‰€1βˆ’πœŒ for all π‘˜β‰₯1, then for all π‘˜β‰₯1, β€–β€–π‘¦π‘˜β€–β€–π·β‰₯βˆžξ“π‘›=1πœ†π‘˜π‘›+βˆžξ“π‘›=𝑛0πœ†π‘˜π‘›β‰₯1+𝜌.(2.26) Since {π‘¦π‘˜} is an π‘Žπ‘–π‘ π‘π‘0𝐷 sequence (resp. an π‘Žπ‘–π‘ π‘π‘0 sequence) with the norm ‖⋅‖𝐷, then β€–π‘¦π‘˜β€–π·β‰€1+πœ€π‘˜β†’1 and this is impossible. Now, if limsupπ‘˜βˆ‘π‘›0βˆ’1𝑖=1πœ†π‘˜π‘–=1, as in the proof of Proposition 3, we obtain a subsequence {π‘¦π‘˜π‘–} of {π‘¦π‘˜} with β€–π‘¦π‘˜π‘–βˆ’π‘¦π‘˜π‘–+1‖𝐷→0.Since {𝑦𝑛} is an π‘Žπ‘–π‘ π‘π‘0𝐷 with the norm ‖⋅‖𝐷, then (1+πœ€π‘˜π‘–)βˆ’1β‰€β€–π‘¦π‘˜π‘–βˆ’π‘¦π‘˜π‘–+1‖𝐷 (resp. (1+πœ€π‘˜π‘–+1)βˆ’1β‰€β€–π‘¦π‘˜π‘–βˆ’π‘¦π‘˜π‘–+1‖𝐷) and making π‘–β†’βˆž we get that 1≀0. This contradiction proves the result.

Although the set 𝐢 of the last proposition has neither π‘Žπ‘–π‘ π‘π‘0𝐷 sequences nor π‘Žπ‘–π‘ π‘π‘0 sequences, for some Θ it does not have the FPP.

For Θ={πœƒπ‘–}, let 𝐹𝑛 be the set such that if 𝑖,π‘—βˆˆπΉπ‘›, then πœƒπ‘–=πœƒπ‘—, and if π‘–βˆˆπΉπ‘›+1 and π‘—βˆˆπΉπ‘›, then πœƒπ‘–β‰ πœƒπ‘—.Denote by π‘Ÿπ‘› the cardinality of 𝐹𝑛. If π‘Ÿπ‘›<∞, define 𝑝0=0,π‘π‘›βˆ’1=minπΉπ‘›βˆ’1, and 𝑝𝑛=max𝐹𝑛.

Proposition 8. Let Ξ˜β‰ Β±Ξ˜0. Then 𝐢=βˆžξ“π‘›=1π‘‘π‘›πœΞ˜π‘›βˆΆπ‘‘π‘›β‰₯0,βˆžξ“π‘›=1𝑑𝑛=1(2.27) does not have the FPP in the following cases.(1)There exists π‘˜β‰₯1 such that π‘Ÿπ‘›β‰€π‘Ÿπ‘›+π‘˜<∞,π‘›βˆˆβ„•.(2)π‘Ÿ1=1 and π‘Ÿ2=∞.(3)There exists {𝑖𝑛}, with 𝑖1>1, such that for any π‘˜,π‘™βˆˆβ„• with π‘–π‘˜βˆ’1<𝑙<π‘–π‘˜ we have πœƒπ‘™=πœƒπ‘–π‘˜ and also πœƒπ‘˜β‰ πœƒπ‘–π‘˜ for all π‘˜β‰₯2 or πœƒπ‘˜=πœƒπ‘–π‘˜ for all π‘˜β‰₯2.

Proof. Let Ξ˜β‰ Β±Ξ˜0.
(1) If there exists π‘˜β‰₯1 such that π‘Ÿπ‘›β‰€π‘Ÿπ‘›+π‘˜<∞,π‘›βˆˆβ„•, define π‘žπ‘›=βˆ‘π‘›+1𝑗=π‘›βˆ’(π‘˜βˆ’2)π‘Ÿπ‘—,𝑛β‰₯π‘˜ and π‘‡βˆΆπΆβ†’πΆ by π‘‡βˆžξ“π‘›=1π‘‘π‘›πœΞ˜π‘›=π‘‡βˆžξ“π‘π‘›=1𝑛𝑖=π‘π‘›βˆ’1+1π‘‘π‘–πœΞ˜π‘–=βˆžξ“π‘π‘›=π‘˜π‘›+1𝑖=π‘€π‘›π‘‘π‘–βˆ’π‘žπ‘›πœΞ˜π‘–,(2.28) where 𝑀𝑛=𝑝𝑛+π‘Ÿπ‘›+1βˆ’π‘Ÿπ‘›+1βˆ’π‘˜+1. The idea is to translate the coefficients of βˆ‘βˆžπ‘›=1π‘‘π‘›πœΞ˜π‘› in the block 𝐹𝑛 into the last π‘Ÿπ‘› terms of the block 𝐹𝑛+π‘˜. Then it is easy to see that 𝑇 does not have fixed points. To prove that 𝑇 is nonexpansive first observe that if π‘˜ is even the signs of the πœƒπ‘– and πœƒπ‘— with π‘–βˆˆπΉπ‘› and π‘—βˆˆπΉπ‘›+π‘˜ are the same and are different if π‘˜ is odd. Now let βˆ‘π‘₯=βˆžπ‘›=1π‘‘π‘›πœΞ˜π‘›, βˆ‘π‘¦=βˆžπ‘›=1π‘ π‘›πœΞ˜π‘›, and βˆ‘π‘₯βˆ’π‘¦=βˆžπ‘›=1π›Όπ‘›πœΞ˜π‘›. Then 𝛼𝑛=π‘‘π‘›βˆ’π‘ π‘› and βˆ‘βˆžπ‘›=1𝛼𝑛=0. Hence π‘₯βˆ’π‘¦=βˆžξ“π‘π‘›=1𝑛𝑖=π‘π‘›βˆ’1+2πœƒπ‘π‘›ξƒ©βˆžξ“π‘›=𝑖𝛼𝑛ξƒͺ𝑒𝑖,(2.29)𝑇(π‘₯βˆ’π‘¦)=βˆžξ“π‘›=π‘˜ξƒ©πœƒπ‘π‘›+1βˆžξ“π‘›=π‘–βˆ’π‘žπ‘›π›Όπ‘›ξƒͺβŽ›βŽœβŽœβŽπ‘€π‘›ξ“π‘–=𝑝𝑛+1π‘’π‘–βŽžβŽŸβŽŸβŽ π‘π‘›+1𝑖=𝑀𝑛+1πœƒπ‘π‘›+1ξƒ©βˆžξ“π‘›=π‘–βˆ’π‘žπ‘›π›Όπ‘›ξƒͺ𝑒𝑖(2.30) are the expressions of π‘₯βˆ’π‘¦ and 𝑇(π‘₯βˆ’π‘¦) 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), 𝑇 is an isometry.
(2) Suppose now that π‘Ÿ1=1 and π‘Ÿ2=∞. In this case, define π‘‡βˆ‘βˆžπ‘›=1π‘‘π‘›πœΞ˜π‘›=βˆ‘βˆžπ‘›=1π‘‘π‘›πœΞ˜π‘›+1. Clearly 𝑇 is nonexpansive and fixed-point-free.
(3) In this case it is straightforward to see thatthe operator π‘‡βˆΆπΆβ†’πΆ defined by π‘‡βˆ‘βˆžπ‘›=1π‘‘π‘›πœΞ˜π‘›=βˆ‘βˆžπ‘›=1π‘‘π‘›πœΞ˜π‘–π‘› is nonexpansive and does not have fixed points.

Proposition 9. Let Ξ˜β‰ Β±Ξ˜0. Suppose Θ does not satisfy the hypotheses of the above proposition, and let {𝑖𝑛} be a sequence with 𝑖1>1. Thenthe operator π‘‡βˆΆπΆβ†’πΆ defined by π‘‡βˆ‘βˆžπ‘›=1π‘‘π‘›πœΞ˜π‘›=βˆ‘βˆžπ‘›=1π‘‘π‘›πœΞ˜π‘–π‘› is expansive.

Proof. Since Θ does not satisfy the hypotheses (1) and (2) of the above proposition, there are three possibilities.(I)π‘Ÿπ‘›<∞ for every 𝑛β‰₯2; then for every π‘˜ there exists 𝑛 such that π‘Ÿπ‘›+π‘˜<π‘Ÿπ‘›.(II)π‘Ÿ2=∞; then π‘Ÿ1>1.(III)There exists π‘˜>2 such that π‘Ÿπ‘˜=∞.
Let {𝑖𝑛} be fixed with 𝑖1>1 and denote 𝑖0=0. Since Θ does not satisfy the hypotheses (3) of the above proposition, there exist π‘˜ and 𝑙 with π‘–π‘˜βˆ’1<𝑙<π‘–π‘˜ such that πœƒπ‘™β‰ πœƒπ‘–π‘˜ or there exists π‘˜1β‰₯2 with πœƒπ‘˜1=πœƒπ‘–π‘˜1 and there exists π‘˜2β‰₯2 with πœƒπ‘˜2β‰ πœƒπ‘–π‘˜2.
Case 1. For every π‘˜ there exists 𝑛 such that π‘Ÿπ‘›+π‘˜<π‘Ÿπ‘›.
There are two subcases.
Subcase 1.1. There are π‘˜ and 𝑙 with π‘–π‘˜βˆ’1<𝑙<π‘–π‘˜ such that πœƒπ‘™β‰ πœƒπ‘–π‘˜.
Let π‘₯=(1/8)𝜁Θ1+(3/8)πœΞ˜π‘˜βˆ’1+(1/2)πœΞ˜π‘˜ and 𝑦=(1/16)𝜁Θ1+(3/16)πœΞ˜π‘˜βˆ’1+(3/4)πœΞ˜π‘˜.Then π‘₯βˆ’π‘¦=(1/16)𝜁Θ1+(3/16)πœΞ˜π‘˜βˆ’1βˆ’(1/4)πœΞ˜π‘˜βˆ‘=βˆ’(1/16)π‘˜βˆ’1𝑖=2πœƒπ‘–π‘’π‘–βˆ’(1/4)πœƒπ‘˜π‘’π‘˜ and β€–π‘₯βˆ’π‘¦β€–π·β‰€5/16.On the other hand, 𝑇π‘₯βˆ’π‘‡π‘¦=(1/16)πœΞ˜π‘–1+(3/16)πœΞ˜π‘–π‘˜βˆ’1βˆ’(1/4)πœΞ˜π‘–π‘˜βˆ‘=βˆ’(1/16)π‘–π‘˜βˆ’1𝑗=𝑖1+1πœƒπ‘—π‘’π‘—βˆ‘βˆ’(1/4)π‘–π‘˜π‘—=π‘–π‘˜βˆ’1+1πœƒπ‘—π‘’π‘— and ‖𝑇π‘₯βˆ’π‘‡π‘¦β€–π·=1/2.

Subcase 1.2. For any π‘˜βˆˆβ„• and 𝑙 withπ‘–π‘˜βˆ’1<𝑙<π‘–π‘˜, we have πœƒπ‘™=πœƒπ‘–π‘˜.
There are two subsubcases. (1) πœƒ1=πœƒπ‘–1 and (2) πœƒ1β‰ πœƒπ‘–1.
(1)πœƒ1=πœƒπ‘–1
If πœƒπ‘˜=πœƒπ‘–π‘˜ for every π‘˜; then we would have 𝐹1=β„•, which implies Θ=±Θ0. Then there is π‘˜ such that πœƒπ‘˜β‰ πœƒπ‘–π‘˜. Let 𝑠=min{π‘™βˆΆπœƒπ‘™β‰ πœƒπ‘–π‘™}. Then 𝑠>1.
There are two possibilities: (A) there exists π‘Ÿ>𝑠 such that πœƒπ‘Ÿ=πœƒπ‘–π‘Ÿ and (B) πœƒπ‘˜β‰ πœƒπ‘–π‘˜ for all π‘˜β‰₯𝑠.(A)Let π‘˜+1=min{π‘Ÿ>π‘ βˆΆπœƒπ‘Ÿ=πœƒπ‘–π‘Ÿ}. We need to consider the following cases.(a)πœƒπ‘˜=πœƒπ‘˜+1.Let π‘₯=(1/2)πœΞ˜π‘˜βˆ’1+(1/2)πœΞ˜π‘˜+1 and 𝑦=(3/4)πœΞ˜π‘˜βˆ’1+(1/4)πœΞ˜π‘˜+1.Then π‘₯βˆ’π‘¦=βˆ’(1/4)πœΞ˜π‘˜βˆ’1+(1/4)πœΞ˜π‘˜+1=πœƒπ‘˜+1((1/4)π‘’π‘˜+(1/4)π‘’π‘˜+1) and β€–π‘₯βˆ’π‘¦β€–π·=1/4.On the other hand, 𝑇π‘₯βˆ’π‘‡π‘¦=βˆ’(1/4)πœΞ˜π‘–π‘˜βˆ’1+(1/4)πœΞ˜π‘–π‘˜+1βˆ‘=(1/4)π‘–π‘˜π‘—=π‘–π‘˜βˆ’1+1πœƒπ‘—π‘’π‘—βˆ‘+(1/4)π‘–π‘˜+1𝑗=π‘–π‘˜+1πœƒπ‘—π‘’π‘—=βˆ’(1/4)πœƒπ‘˜+1βˆ‘π‘–k𝑗=π‘–π‘˜βˆ’1+1𝑒𝑗+(1/4)πœƒπ‘˜+1βˆ‘π‘–π‘˜+1𝑗=π‘–π‘˜+1𝑒𝑗 and ‖𝑇π‘₯βˆ’π‘‡π‘¦β€–π·=1/2.(b)πœƒπ‘˜β‰ πœƒπ‘˜+1. Let π‘₯=(1/2)πœΞ˜π‘˜βˆ’1+(1/2)πœΞ˜π‘˜+1 and 𝑦=(3/4)πœΞ˜π‘˜+(1/4)πœΞ˜π‘˜+1. Then π‘₯βˆ’π‘¦=(1/2)πœΞ˜π‘˜βˆ’1βˆ’(3/4)πœΞ˜π‘˜+(1/4)πœΞ˜π‘˜+1=βˆ’(1/2)πœƒπ‘˜π‘’π‘˜+(1/4)πœƒπ‘˜+1π‘’π‘˜+1=πœƒπ‘˜+1((1/2)π‘’π‘˜+(1/4)π‘’π‘˜+1) and β€–π‘₯βˆ’π‘¦β€–π·=1/2. On the other hand, 𝑇π‘₯βˆ’π‘‡π‘¦=(1/2)πœΞ˜π‘–π‘˜βˆ’1βˆ’(3/4)πœΞ˜π‘–π‘˜+(1/4)πœΞ˜π‘–π‘˜+1βˆ‘=βˆ’(1/2)π‘–π‘˜π‘—=π‘–π‘˜βˆ’1+1πœƒπ‘—π‘’π‘—βˆ‘+(1/4)π‘–π‘˜+1𝑗=π‘–π‘˜+1πœƒπ‘—π‘’π‘—=βˆ’(1/2)πœƒπ‘˜βˆ‘π‘–π‘˜π‘—=π‘–π‘˜βˆ’1+1𝑒𝑗+(1/4)πœƒπ‘˜βˆ‘π‘–π‘˜+1𝑗=π‘–π‘˜+1𝑒𝑗 and ‖𝑇π‘₯βˆ’π‘‡π‘¦β€–π·=3/4.(B)πœƒπ‘˜β‰ πœƒπ‘–π‘˜ for all π‘˜β‰₯𝑠. By hypothesis we have that 𝑠>2. There are two cases.(a)πœƒπ‘ βˆ’1=πœƒπ‘ . Then πœƒπ‘–π‘ βˆ’1β‰ πœƒπ‘–π‘ .Let π‘₯=(1/4)πœΞ˜π‘ βˆ’2+(1/2)πœΞ˜π‘ βˆ’1+(1/4)πœΞ˜π‘  and 𝑦=(1/4)πœΞ˜π‘ βˆ’1+(3/4)πœΞ˜π‘ . Then π‘₯βˆ’π‘¦=(1/4)πœΞ˜π‘ βˆ’2+(1/4)πœΞ˜π‘ βˆ’1βˆ’(1/2)πœΞ˜π‘ =βˆ’πœƒπ‘ ((1/4)π‘’π‘ βˆ’1+(1/2)𝑒𝑠) and β€–π‘₯βˆ’π‘¦β€–π·=1/2.On the other hand, 𝑇π‘₯βˆ’π‘‡π‘¦=(1/4)πœΞ˜π‘–π‘ βˆ’2+(1/4)πœΞ˜π‘–π‘ βˆ’1βˆ’(1/2)πœΞ˜π‘–π‘ =βˆ’(1/4)πœƒπ‘ βˆ‘π‘–π‘ βˆ’1𝑗=π‘–π‘ βˆ’2+1𝑒𝑗+(1/2)πœƒπ‘ βˆ‘π‘–π‘ π‘—=π‘–π‘ βˆ’1+1𝑒𝑗 and ‖𝑇π‘₯βˆ’π‘‡π‘¦β€–π·=3/4.(b)πœƒπ‘ βˆ’1β‰ πœƒπ‘ . Then πœƒπ‘–π‘ βˆ’1=πœƒπ‘–π‘ .Let π‘₯=(1/4)πœΞ˜π‘ βˆ’2+(1/4)πœΞ˜π‘ βˆ’1+(1/2)πœΞ˜π‘  and 𝑦=(3/4)πœΞ˜π‘ βˆ’1+(1/4)πœΞ˜π‘ . Then π‘₯βˆ’π‘¦=(1/4)πœΞ˜π‘ βˆ’2βˆ’(1/2)πœΞ˜π‘ βˆ’1+(1/4)πœΞ˜π‘ =βˆ’(1/4)πœƒπ‘ βˆ’1π‘’π‘ βˆ’1+(1/4)πœƒπ‘ π‘’π‘ =πœƒπ‘ ((1/4)π‘’π‘ βˆ’1+(1/4)𝑒𝑠) and β€–π‘₯βˆ’π‘¦β€–π·=1/4.On the other hand, 𝑇π‘₯βˆ’π‘‡π‘¦=(1/4)πœΞ˜π‘–π‘ βˆ’2βˆ’(1/2)πœΞ˜π‘–π‘ βˆ’1+(1/4)πœΞ˜π‘–π‘ =βˆ’(1/4)πœƒπ‘–π‘ βˆ’1βˆ‘π‘–π‘ βˆ’1𝑗=π‘–π‘ βˆ’2+1𝑒𝑗+(1/4)πœƒπ‘–π‘ βˆ‘π‘–π‘ π‘—=π‘–π‘ βˆ’1+1𝑒𝑗=πœƒπ‘ βˆ’1βˆ‘(βˆ’(1/4)π‘–π‘ βˆ’1𝑗=π‘–π‘ βˆ’2+1π‘’π‘—βˆ‘+(1/4)𝑖𝑠𝑗=π‘–π‘ βˆ’1+1𝑒𝑗) and ‖𝑇π‘₯βˆ’π‘‡π‘¦β€–π·=1/2.

(2)πœƒ1β‰ πœƒπ‘–1
In this case there exists π‘˜ such that πœƒπ‘˜=πœƒπ‘–π‘˜. If 𝑠=min{π‘™βˆΆπœƒπ‘™=πœƒπ‘–π‘™}, then 𝑠>1.
Hence consider the cases: (A) there exists π‘Ÿ>𝑠 such that πœƒπ‘Ÿβ‰ πœƒπ‘–π‘Ÿ and (B) πœƒπ‘˜=πœƒπ‘–π‘˜ for all π‘˜β‰₯𝑠 and proceed as in the Case (1).

Case 2. π‘Ÿ2=∞ and π‘Ÿ1>1.
Then πœƒπ‘1β‰ πœƒπ‘–π‘1 with 1<𝑝1.Hence we can proceed as in Subcase 1.2(1)(A) above taking π‘˜=𝑝1.

Case 3. There is 𝑠>1 such that π‘Ÿπ‘ +1=∞.
Then πœƒπ‘π‘ β‰ πœƒπ‘–π‘π‘  with 1<𝑝𝑠.Hence we can proceed as in Subcase 1.2(1)(A) above taking π‘˜=𝑝𝑠.

Next, for every selection of signs Ξ˜β‰ Β±Ξ˜0, we will define the asymptotically isometric 𝑐0𝐷-Θ-basic sequences. To this end, let us consider the following notation.

Let π”–Ξ˜=ξ€½(𝑛,π‘š)βˆΆπœƒπ‘›=πœƒπ‘šξ€Ύ,π”‡Ξ˜=ξ€½(𝑛,π‘š)βˆΆπœƒπ‘›β‰ πœƒπ‘šξ€Ύ.(2.31)

Definition 10. Let {π‘₯𝑛} be a sequence in a Banach space 𝑋. We say that {π‘₯𝑛} is an asymptotically isometric 𝑐0𝐷-Θ-basic sequence (π‘Žπ‘–Ξ˜π‘π‘0𝐷 sequence for short) if {π‘₯𝑛} is convexly closed and there exists {πœ€Ξ˜π‘›}βŠ‚(0,(1/2)) such that πœ€Ξ˜π‘›β†˜0, and πΏπœ€ξ€·ξ€½Ξ˜π‘›ξ€Ύ,𝑑𝑛,π”–Ξ˜ξ€Έπœ€βˆ¨πΏξ€·ξ€½Ξ˜π‘›ξ€Ύ,𝑑𝑛,π”‡Ξ˜ξ€Έβ‰€β€–β€–β€–β€–βˆžξ“π‘›=1𝑑𝑛π‘₯π‘›β€–β€–β€–β€–πœ€β‰€π‘…ξ€·ξ€½Ξ˜π‘›ξ€Ύ,𝑑𝑛,π”–Ξ˜ξ€Έπœ€βˆ¨π‘…ξ€·ξ€½Ξ˜π‘›ξ€Ύ,𝑑𝑛,π”‡Ξ˜ξ€Έ(2.32) holds for all {𝑑𝑛}βˆˆπ‘™1, where πΏπœ€ξ€·ξ€½Ξ˜π‘›