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Abstract and Applied Analysis
VolumeΒ 2011, Article IDΒ 574614, 19 pages
http://dx.doi.org/10.1155/2011/574614
Research Article

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

1MatemΓ‘ticas BΓ‘sicas, Centro de InvestigaciΓ³n en MatemΓ‘ticas (CIMAT), Apartado Postal 402, 36000 Guanajuato, GTO, Mexico
2Departamento de MatemΓ‘ticas Aplicadas, Universidad del Papaloapan (UNPA), 68400 Loma Bonita, OAX, Mexico

Received 7 June 2011; Accepted 27 August 2011

Academic Editor: ElenaΒ Litsyn

Copyright Β© 2011 Berta Gamboa de Buen and Fernando NΓΊΓ±ez-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 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‘›ξ€Ύ,𝑑𝑛,π”–Ξ˜ξ€Έ=βŽ›βŽœβŽœβŽsup𝑛<l,(𝑛,𝑙)βˆˆπ”–Ξ˜ξ€·1+πœ€Ξ˜π‘™βˆ’1ξ€Έβˆ’1|||||π‘™βˆ’1ξ“π‘˜=π‘›π‘‘π‘˜|||||⎞⎟⎟⎠,πΏπœ€ξ€·ξ€½Ξ˜π‘›ξ€Ύ,𝑑𝑛,π”‡Ξ˜ξ€Έ=βŽ›βŽœβŽœβŽsup𝑛<𝑙,(𝑛,𝑙)βˆˆπ”‡Ξ˜ξ€·1+πœ€Ξ˜π‘™βˆ’1ξ€Έβˆ’1|||||π‘™βˆ’1ξ“π‘˜=π‘›π‘‘π‘˜+2βˆžξ“π‘˜=π‘™π‘‘π‘˜|||||⎞⎟⎟⎠,π‘…πœ€ξ€·ξ€½Ξ˜π‘›ξ€Ύ,𝑑𝑛,π”–Ξ˜ξ€Έ=βŽ›βŽœβŽœβŽsup𝑛<𝑙,(𝑛,𝑙)βˆˆπ”–Ξ˜ξ€·1+πœ€Ξ˜π‘™βˆ’1ξ€Έ|||||π‘™βˆ’1ξ“π‘˜=π‘›π‘‘π‘˜|||||⎞⎟⎟⎠,π‘…πœ€ξ€·ξ€½Ξ˜π‘›ξ€Ύ,𝑑𝑛,π”‡Ξ˜ξ€Έ=βŽ›βŽœβŽœβŽsup𝑛<𝑙,(𝑛,𝑙)βˆˆπ”‡Ξ˜ξ€·1+πœ€Ξ˜π‘™βˆ’1ξ€Έ|||||π‘™βˆ’1ξ“π‘˜=π‘›π‘‘π‘˜+2βˆžξ“π‘˜=π‘™π‘‘π‘˜|||||⎞⎟⎟⎠.(2.33)

We are interested in π‘Žπ‘–Ξ˜π‘π‘0𝐷 sequences for which the numbers πœ€Ξ˜π‘› of Definition 10 are small. We are taking {πœ€Ξ˜π‘›}βŠ‚(0,(1/2)).

We know that the set 𝐢 of Proposition 3 does not have π‘Žπ‘–π‘ π‘π‘0 sequences. Now we also prove that 𝐢 does not contain π‘Žπ‘–Ξ˜π‘π‘0𝐷 sequences with the norm ‖⋅‖𝐷 if Ξ˜β‰ Β±Ξ˜0.

Proposition 11. Let Ξ˜β‰ Β±Ξ˜0. The set βˆ‘πΆ={βˆžπ‘›=1π‘‘π‘›πœ‰π‘›βˆΆπ‘‘π‘›β‰₯0andβˆ‘βˆžπ‘›=1𝑑𝑛=1} does not contain π‘Žπ‘–Ξ˜π‘π‘0𝐷 sequences with the norm ‖⋅‖𝐷.

Proof. Let {π‘¦π‘˜}βŠ‚πΆ. Then π‘¦π‘˜=βˆ‘βˆžπ‘›=1πœ†π‘˜π‘›πœ‰π‘› for some πœ†π‘˜π‘›β‰₯0with βˆ‘βˆžπ‘›=1πœ†π‘˜π‘›=1.Suppose that {π‘¦π‘˜} is an π‘Žπ‘–Ξ˜π‘π‘0𝐷 with ‖⋅‖𝐷. Since Ξ˜β‰ Β±Ξ˜0, there exist π‘šβˆˆβ„• and {π‘›π‘˜}βŠ‚β„• with 𝑛1<𝑛2<β‹―, such that for all π‘˜βˆˆβ„•, π‘š<π‘›π‘˜ and πœƒπ‘›π‘˜β‰ πœƒπ‘š. Let 𝑑𝑛=0 for π‘›β‰ π‘š,π‘›π‘˜ and π‘‘π‘š=π‘‘π‘›π‘˜=1. Thus ξ‚€1+πœ€Ξ˜π‘›π‘˜βˆ’1ξ‚βˆ’1πœ€3β‰€πΏξ€·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‘›=1𝑑𝑛𝑦𝑛‖‖‖‖𝐷=β€–β€–π‘¦π‘š+π‘¦π‘›π‘˜β€–β€–π·=2.(2.34) Since (2.34) holds for all π‘˜βˆˆβ„•, making π‘˜β†’βˆž in (2.34), we get that 3≀2, which is a contradiction.

Proposition 12. Let Θ1={πœƒ1𝑖}𝑖 and Θ2={πœƒ2𝑖}𝑖 such that Θ1β‰ Β±Ξ˜2 and Θ1,Θ2β‰ Β±Ξ˜0. Let {𝜁Θ1𝑛} be the Θ1-basis of 𝑐0 considered in (𝑐0,‖⋅‖𝐷) and let πΆξ€·Ξ˜1ξ€Έ=ξƒ―βˆžξ“π‘›=1π‘‘π‘›πœΞ˜1π‘›βˆΆπ‘‘π‘›β‰₯0,βˆžξ“π‘›=1𝑑𝑛.=1(2.35) The set 𝐢(Θ1) does not contain π‘Žπ‘–Ξ˜2𝑏𝑐0𝐷 sequences with the norm ‖⋅‖𝐷.

Proof. Let {π‘¦π‘˜}βŠ‚πΆ. Then π‘¦π‘˜=βˆ‘βˆžπ‘›=1πœ†π‘˜π‘›πœΞ˜1𝑛 for some πœ†π‘˜π‘›β‰₯0 with βˆ‘βˆžπ‘›=1πœ†π‘˜π‘›=1.Suppose that {π‘¦π‘˜} is an π‘Žπ‘–Ξ˜2𝑏𝑐0𝐷 with the norm ‖⋅‖𝐷.
Suppose first πœƒ11=πœƒ21; since Θ1β‰ Ξ˜2, there exists π‘š>1 such that πœƒ1π‘šβ‰ πœƒ2π‘š.
There are two cases.
Case 1. (1,π‘š)βˆˆπ”–Ξ˜1. In this case (1,π‘š)βˆˆπ”‡Ξ˜2. Let 𝑑𝑛=0 for 𝑛≠1,π‘š and 𝑑1=π‘‘π‘š=1. Thus ξ‚€1+πœ€Ξ˜2π‘šβˆ’1ξ‚βˆ’1πœ€3β‰€πΏξ‚€ξ‚†Ξ˜2𝑛,𝑑𝑛,π”–Ξ˜2ξ‚πœ€βˆ¨πΏξ‚€ξ‚†Ξ˜2𝑛,𝑑𝑛,π”‡Ξ˜2ξ‚β‰€β€–β€–β€–β€–βˆžξ“π‘›=1𝑑𝑛𝑦𝑛‖‖‖‖𝐷=‖‖𝑦1+π‘¦π‘šβ€–β€–π·β‰€2.(2.36) Since πœ€Ξ˜2π‘šβˆ’1<1/2, we get a contradiction.
Case 2. (1,π‘š)βˆˆπ”‡Ξ˜1. In this case (1,π‘š)βˆˆπ”–Ξ˜2. Let 𝑑𝑛=0 for 𝑛≠1,π‘š and 𝑑1=π‘‘π‘š=1. Thus 2β‰€βˆžξ“π‘›=1πœ†1𝑛+βˆžξ“π‘›=1πœ†π‘šπ‘›+βˆžξ“π‘›=π‘šπœ†1𝑛+βˆžξ“π‘›=π‘šπœ†π‘šπ‘›β‰€β€–β€–β€–β€–βˆžξ“π‘›=1𝑑𝑛𝑦𝑛‖‖‖‖𝐷=‖‖𝑦1+π‘¦π‘šβ€–β€–π·πœ€β‰€π‘…ξ‚€ξ‚†Ξ˜2𝑛,𝑑𝑛,π”–Ξ˜2ξ‚πœ€βˆ¨π‘…ξ‚€ξ‚†Ξ˜2𝑛,𝑑𝑛,π”‡Ξ˜2≀1+πœ€Ξ˜2π‘šβˆ’1.(2.37) Since πœ€Ξ˜2π‘šβˆ’1<1/2, we get a contradiction.
Suppose now πœƒ11β‰ πœƒ21; since Θ1β‰ βˆ’Ξ˜2, there exists π‘š such that πœƒ1π‘š=πœƒ2π‘š.
There are two cases.
Case 1. (1,π‘š)βˆˆπ”–Ξ˜1; in this case (1,π‘š)βˆˆπ”‡Ξ˜2. Let 𝑑𝑛=0 for 𝑛≠1,π‘š and 𝑑1=π‘‘π‘š=1. Thus ξ‚€1+πœ€Ξ˜2π‘šβˆ’1ξ‚βˆ’1πœ€3β‰€πΏξ‚€ξ‚†Ξ˜2𝑛,𝑑𝑛,π”–Ξ˜2ξ‚πœ€βˆ¨πΏξ‚€ξ‚†Ξ˜2𝑛,𝑑𝑛,π”‡Ξ˜2ξ‚β‰€β€–β€–β€–β€–βˆžξ“π‘›=1𝑑𝑛𝑦𝑛‖‖‖‖𝐷=‖‖𝑦1+π‘¦π‘šβ€–β€–π·β‰€2.(2.38) Since πœ€Ξ˜2π‘šβˆ’1<1/2, we get a contradiction.
Case 2. (1,π‘š)βˆˆπ”‡Ξ˜1. In this case (1,π‘š)βˆˆπ”–Ξ˜2. Let 𝑑𝑛=0 for 𝑛≠1,π‘š and 𝑑1=π‘‘π‘š=1. Thus 2β‰€βˆžξ“π‘›=1πœ†1𝑛+βˆžξ“π‘›=1πœ†π‘šπ‘›+βˆžξ“π‘›=π‘šπœ†1𝑛+βˆžξ“π‘›=π‘šπœ†π‘šπ‘›β‰€β€–β€–β€–β€–βˆžξ“π‘›=1𝑑𝑛𝑦𝑛‖‖‖‖𝐷=‖‖𝑦1+π‘¦π‘šβ€–β€–π·πœ€β‰€π‘…ξ‚€ξ‚†Ξ˜2𝑛,𝑑𝑛,π”–Ξ˜2ξ‚πœ€βˆ¨π‘…ξ‚€ξ‚†Ξ˜2𝑛,𝑑𝑛,π”‡Ξ˜2≀1+πœ€Ξ˜2π‘šβˆ’1