The Fixed Point Property in 𝑐_𝟎 with an Equivalent Norm

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 𝜔-compact and do not contain asymptotically isometric 𝑐₀—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 𝑙_1∞ and that every infinite-dimensional subspace of 𝑙_1∞ 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,‖⋅‖). A mapping 𝑇∶𝐾→𝐾 is said to be nonexpansive if ‖𝑇𝑥−𝑇𝑦‖≤‖𝑥−𝑦‖,𝑥,𝑦∈𝐾.(1.1) We say that 𝐾 has the fixed point property for nonexpansive mappings (FPP) if every nonexpansive mapping 𝑇∶𝐾→𝐾 has a fixed point, that is, a point 𝑥∈𝐾 such that 𝑇𝑥=𝑥. We say that a Banach space (𝑋,‖⋅‖) has the fixed point property for nonexpansive mappings (FPP) if every nonempty, convex, closed, and bounded subset 𝐾 of (𝑋,‖⋅‖) has the FPP, and we say that the Banach space (𝑋,‖⋅‖) has the weak fixed point property for nonexpansive mappings (𝜔-FPP) if every nonempty, convex and weakly compact subset 𝐾 of (𝑋,‖⋅‖) has the FPP.

In this paper we study the FPP in the Banach space 𝑐₀ with the equivalent norm ‖⋅‖_𝐷 defined by ‖𝑥‖_𝐷=sup_{𝑖,𝑗∈ℕ}||𝑥_𝑖−𝑥_𝑗||,𝑥=𝑥_𝑖∈𝑐₀.(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 𝜔-compact. Specifically, they proved that if 𝐾 is a convex, closed and bounded subset of 𝑐₀, then 𝐾 is 𝜔-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 𝜔-compact contains an asymptotically isometric 𝑐₀-summing basic sequence, 𝑎𝑖𝑠𝑏𝑐₀ sequence for short, that is, a sequence {𝑦_𝑛}_𝑛⊂𝑐₀ such that for all {𝑡_𝑛}_𝑛∈𝑙₁,sup_𝑛∈ℕ1+𝜀_𝑛⁻¹|||||_∞_𝑖=𝑛𝑡_𝑖|||||≤‖‖‖‖_∞_𝑛=1𝑡_𝑛𝑦_𝑛‖‖‖‖≤sup_𝑛∈ℕ1+𝜀_𝑛|||||^∞_𝑖=𝑛𝑡_𝑖|||||,(1.3) for some {𝜀_𝑛}_𝑛⊂ℝ with 0≤𝜀_𝑛+1≤𝜀_𝑛 and lim_𝑛𝜀_𝑛=0. They proved that if a convex, closed and bounded subset 𝐾 of a Banach space contains an 𝑎𝑖𝑠𝑏𝑐₀ 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 𝜔-compact if and only if it has the FPP.

It is easy to see that (𝑐₀,‖⋅‖_𝐷) contains 𝑐₀ isometrically, and then it contains 𝑎𝑖𝑠𝑏𝑐₀ sequences.

First we prove that every infinite-dimensional subspace 𝑌 of (𝑐₀,‖⋅‖_𝐷) has a complemented asymptotically isometric copy of 𝑐₀ and hence by a result proved by Dowling et al. in [4], 𝑌 does not have the FPP. Also, as an immediate consequence we obtain that 𝑌 has an 𝑎𝑖𝑠𝑏𝑐₀ sequence. Nevertheless, we exhibit a nonempty closed, convex and bounded subset of (𝑐₀,‖⋅‖_𝐷), which is not 𝜔-compact and does not contain 𝑎𝑖𝑠𝑏𝑐₀ sequences.

Then for every selection of signs Θ={𝜃_𝑖}, we define the Θ-basis of 𝑐₀ which is equivalent to the summing basis and define the corresponding asymptotically isometric Θ-basic sequence, 𝑎𝑖Θ𝑏𝑐_0𝐷 sequence for short. We prove that if Θ₁≠±Θ₂, then the 𝑎𝑖Θ₁𝑏𝑐_0𝐷 and 𝑎𝑖Θ₂𝑏𝑐_0𝐷 sequences are different in the sense that there exists a nonempty, closed, convex, and bounded subset of (𝑐₀,‖⋅‖_𝐷), which is not 𝜔-compact, contains an 𝑎𝑖Θ₁𝑏𝑐_0𝐷 sequence, and does not contain 𝑎𝑖Θ₂𝑏𝑐_0𝐷 sequences. We also show that the 𝑎𝑖𝑠𝑏𝑐₀ and 𝑎𝑖Θ𝑏𝑐_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 𝑎𝑖Θ𝑏𝑐_0𝐷 sequences.

Next we prove that if a convex and closed subset 𝐾 of a Banach space contains an asymptotically isometric 𝑐_0𝐷-summing basic sequence, that is, an 𝑎𝑖Θ𝑏𝑐_0𝐷 sequence, where Θ is such that 𝜃_𝑖=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 𝑙_1∞. Then, by a result of Dowling et al. in [6], the space 𝑙_1∞=(𝑐₀,‖⋅‖_𝐷)^∗ has “many” 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 𝑙_1∞ 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 {𝑒_𝑛} the canonical basis of 𝑐₀ and by {𝜉_𝑛} the summing basis of 𝑐₀, that is, 𝜉_𝑛=∑^𝑛𝑖=1𝑒_𝑖,𝑛∈ℕ.

García Falset proved in [7] that a Banach space with strongly bimonotone basis and with the weak Banach-Saks property has the 𝜔-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 ‖⋅‖_∞ are equivalent, we get that (𝑐₀,‖⋅‖_𝐷) has the weak Banach-Saks property. Hence we have that (𝑐₀,‖⋅‖_𝐷) has the 𝜔-FPP.

To study the FPP in the space (𝑐₀,‖⋅‖_𝐷) using 𝑎𝑖𝑠𝑏𝑐₀ sequences, we would expect that nonempty, convex, closed and bounded subsets 𝐾 of (𝑐₀,‖⋅‖_𝐷), which are not 𝜔-compact, contain an 𝑎𝑖𝑠𝑏𝑐₀ sequence. This fact is true for some 𝜔-compact sets in (𝑐₀,‖⋅‖_𝐷), since the space 𝑐₀ embeds isometrically in (𝑐₀,‖⋅‖_𝐷). In fact we have the following proposition.

Proposition 1. Let {𝑢_𝑘}_𝑘⊂(𝑐₀,‖⋅‖_𝐷) be a block basis of {𝑒_𝑛} with 𝑢_𝑘=∑^𝑞𝑘_{𝑖=𝑝𝑘}𝑎_𝑖𝑒_𝑖,1≤𝑝₁≤𝑞₁<𝑝₂≤𝑞₂<⋯. If ‖𝑢_𝑘‖_∞=1=𝑎_𝑖𝑘, for some 𝑝_𝑘≤𝑖^𝑘≤𝑞_𝑘, and 𝑦_𝑘=(1/2)(𝑢_2𝑘−𝑢_2𝑘−1), then the space span{𝑦_𝑘} is isometric to (𝑐₀,‖⋅‖_∞).

Proof. Since ‖𝑢_𝑘‖_∞=1=𝑎_𝑖𝑘 for every 𝑘∈ℕ, then |𝑎_𝑗|≤1 for all 𝑗∈ℕ and max_{𝑝2𝑘−1≤𝑖≤𝑞2𝑘−1,𝑝2𝑘≤𝑗≤𝑞2𝑘}||𝑎_𝑖+𝑎_𝑗||=𝑎_{𝑖2𝑘−1}+𝑎_𝑖2𝑘=2.(2.1) Hence, it is straightforward to see that ‖∑^𝑛𝑘=1𝑡_𝑘𝑦_𝑘‖_𝐷=‖∑^𝑛𝑘=1𝑡_𝑘𝑒_𝑘‖_∞.

In the following theorem, we will show, using some results proved by Dowling et al. [4, 8], that every infinite-dimensional subspace 𝑌 of 𝑐_0𝐷 fails to have the FPP.

Theorem 2. Let 𝑌 be an infinite-dimensional subspace of 𝑐_0𝐷. Then 𝑌 has a complemented asymptotically isometric copy of 𝑐₀ and thus 𝑌 does not have the FPP.

Proof. Let {𝜀_𝑘}_𝑘⊂(0,1) be a sequence such that 𝜀_𝑘+1<𝜀_𝑘,𝑘∈ℕ and 𝜀_𝑘→0. As in [9] we construct sequences {𝑛_𝑘}⊂ℕ and {𝑦_𝑘}_𝑘⊂𝑌 such that 𝑛_𝑘<𝑛_𝑘+1,𝑦_𝑘=∑^∞𝑖=𝑛_𝑘𝛼^𝑘𝑖𝑒_𝑖,‖𝑦_𝑘‖_∞=1, and sup_{𝑖≥𝑛𝑘+1}||𝛼^𝑗𝑖||<𝜀^𝑘+24𝑘∀𝑗=1,…,𝑘,andevery𝑘∈ℕ.(2.2) Since ‖𝑦_𝑘‖_∞=1, taking −𝑦_𝑘 instead of 𝑦_𝑘, if necessary, we can suppose that there exists 𝑛_𝑘≤𝑟^𝑘<𝑛_𝑘+1 such that 𝛼^𝑘𝑟_𝑘=1.(2.3)
Define 𝑥_𝑘=(𝑦_2𝑘−1−𝑦_2𝑘)/2.Then, by (2.3) and (2.2), we get that 1−(𝜀_𝑘/2)<‖𝑥_𝑘‖_𝐷<1+(𝜀_𝑘/2) and ^∞_𝑘=1𝑡_𝑘𝑥_𝑘=12^∞_𝑘=1𝑡_𝑘^∞_{𝑖=𝑛2𝑘−1}𝛼^{𝑖2𝑘−1}𝑒_𝑖−^∞_𝑖_=𝑛2𝑘𝛼^𝑖2𝑘𝑒_𝑖=12^∞_𝑘=1𝑡_𝑘^∞_{𝑖=𝑛2𝑘−1}𝛼^{𝑖2𝑘−1}−𝛼^𝑖2𝑘𝑒_𝑖=12^∞_𝑘=1⎛⎜⎜⎝^{𝑛2𝑘+1−1}_𝑖=_{𝑛2𝑘−1}^𝑘_𝑗=1𝑡_𝑗𝛼^{𝑖2𝑗−1}−𝛼^𝑖2𝑗𝑒^𝑖⎞⎟⎟⎠,(2.4) where 𝛼^𝑖2𝑘=0 for 𝑖=𝑛_2𝑘−1,…𝑛_2𝑘−1,𝑘∈ℕ. Then by (2.3) and (2.2), if 𝑘>1, we get ‖‖‖‖^∞_𝑛=1𝑡_𝑛𝑥_𝑛‖‖‖‖_𝐷≥12max^𝑛_2𝑘−1^≤𝑟<𝑛_2𝑘^,𝑛_2𝑘^≤𝑠<𝑛_2𝑘+1|||||_𝑘_𝑗=1≥1𝑡_𝑗𝛼^{𝑟2𝑗−1}−𝛼^𝑟2𝑗−𝛼^{𝑠2𝑗−1}+𝛼^𝑠2𝑗|||||2|||||^𝑘_𝑗=1≥1𝑡_𝑗𝛼^{𝑟2𝑗−12𝑘−1}−𝛼^𝑟2𝑗_2𝑘−1−𝛼^{𝑟2𝑗−1}_2𝑘+𝛼^𝑟2𝑗_2𝑘|||||2||𝑡^𝑘||||𝛼^{𝑟2𝑘−1}_2𝑘−1−𝛼^{𝑟2𝑘−1}_2𝑘+𝛼^𝑟2𝑘_2𝑘||−12^𝑘−1_𝑗=1≥1||𝑡_𝑗||||𝛼^{𝑟2𝑗−1}_2𝑘−1−𝛼^𝑟2𝑗_2𝑘−1−𝛼^{𝑟2𝑗−1}_2𝑘+𝛼^𝑟2𝑗_2𝑘||2||𝑡^𝑘−1||||𝛼^{𝑟2𝑘−1}_2𝑘−1+𝛼^𝑟2𝑘_2𝑘||−||𝛼^{𝑟2𝑘−1}_2𝑘||2^𝑘−1_𝑗=1≥1||𝑡_𝑗||||𝛼^{𝑟2𝑗−1}_2𝑘−1||+||𝛼^𝑟2𝑗_2𝑘−1||+||𝛼^{𝑟2𝑗−1}_2𝑘||+||𝛼^𝑟2𝑗_2𝑘||2||𝑡^𝑘||2−𝜀_𝑘−12^𝑘−1_𝑗=1||𝑡_𝑗||𝜀^𝑘≥||𝑡_𝑘𝑘||1−𝜀^𝑘2−max^{1≤𝑗≤𝑘}||𝑡_𝑗||𝜀^𝑘2.(2.5) On the other hand, if 𝑛_2𝑘−1≤𝑟<𝑛_2𝑘+1,𝑛_2𝑚−1≤𝑠<𝑛_2𝑚+1,𝑘≤𝑚, using (2.2), we get 12|||||^𝑘_𝑗=1𝑡_𝑗𝛼^{𝑟2𝑗−1}−𝛼^𝑟2𝑗−^𝑚_𝑗=1≤1𝑡_𝑗𝛼^{𝑠2𝑗−1}−𝛼^𝑠2𝑗|||||2||𝑡^𝑘+1||||𝛼^{𝑟2𝑘−1}||+||𝛼^𝑟2𝑘||+||𝑡_𝑚||||𝛼^{𝑠2𝑚−1}||+||𝛼^𝑠2𝑚||2^𝑘−1_𝑗=1||𝑡_𝑗||||𝛼^{𝑟2𝑗−1}||+||𝛼^𝑟2𝑗||+^𝑚−1_𝑗=1≤1||𝑡_𝑗||||𝛼^{𝑠2𝑗−1}||+||𝛼^𝑠2𝑗||2||𝑡^𝑘||1+𝜀_𝑘+||𝑡_𝑚||1+𝜀_𝑚+^𝑘−1_𝑗=1||𝑡_𝑗||𝜀^𝑘𝑘−1+^𝑚−1_𝑗=1||𝑡_𝑗||𝜀^𝑚≤1𝑚−12||𝑡^𝑘||1+𝜀_𝑘+||𝑡_𝑚||1+𝜀_𝑚+max_{1≤𝑗<𝑘}||𝑡_𝑗||𝜀_𝑘+max_{1≤𝑗<𝑚}||𝑡_𝑗||𝜀_𝑚≤12max^{1≤𝑗≤𝑘}≤sup_𝑛∈ℕ||𝑡_𝑗||1+𝜀_𝑘+max_{1≤𝑗≤𝑚}||𝑡_𝑗||1+𝜀_𝑚1+𝜀^𝑛max_{1≤𝑗≤𝑛}||𝑡_𝑗||≤sup_𝑛∈ℕ1+𝜀_𝑛||𝑡_𝑛||.(2.6) Then we obtain sup_𝑛∈ℕ||𝑡^𝑛||1−𝜀^𝑛2−max^{1≤𝑗≤𝑛}||𝑡_𝑗||𝜀^𝑛2≤‖‖‖‖^∞_𝑛=1𝑡_𝑛𝑥_𝑛‖‖‖‖_𝐷≤sup_𝑛∈ℕ1+𝜀_𝑛||𝑡_𝑛||.(2.7) Now, define 𝑧_𝑛=𝑥_𝑛/(1+𝜀_𝑛) and 𝑚=(1−𝜀₁)/(1+𝜀₁); then (1−𝜀_𝑛)/(1+𝜀_𝑛)≤‖𝑧_𝑛‖_𝐷 and lim_𝑛‖𝑧_𝑛‖_𝐷=1. On the other hand, 1+𝜀₁𝑚sup_𝑛∈ℕ||𝑡_𝑛||≤=su1p_𝑛∈ℕ−𝜀₁sup_𝑛∈ℕ||𝑡_𝑛||=1−𝜀2¹sup_𝑛∈ℕ||𝑡_𝑛||−𝜀2¹sup_𝑛∈ℕ||𝑡_𝑛||||𝑡^𝑛||1−𝜀^𝑛2−max^{1≤𝑗≤𝑛}||𝑡_𝑗||𝜀^𝑛2.(2.8) Thus 𝑚sup_𝑛∈ℕ||𝑡_𝑛||≤sup_𝑛∈ℕ||𝑡^𝑛||1−𝜀^𝑛2−max^{1≤𝑗≤𝑛}||𝑡_𝑗||𝜀^𝑛2≤‖‖‖‖^∞_𝑛=1𝑡_𝑛𝑧_𝑛‖‖‖‖_𝐷≤sup_𝑛∈ℕ||𝑡_𝑛||.(2.9) Then by Theorem 2 of [8] 𝑌 contains an asymptotically isometric copy of 𝑐₀ and since 𝑌 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], 𝑌 does not have the FPP.

As a consequence of the last theorem, we get that every infinite-dimensional subspace of (𝑐₀,‖⋅‖_𝐷) contains an 𝑎𝑖𝑠𝑏𝑐₀ 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 𝑎𝑖𝑠𝑏c₀ sequences.

Proposition 3. Let {𝜉_𝑛} be the 𝑐₀ summing basis. Then 𝐶=^∞_𝑛=1𝜆_𝑛𝜉_𝑛∶𝜆_𝑛≥0,^∞_𝑛=1𝜆_𝑛=1(2.10) does not have 𝑎𝑖𝑠𝑏𝑐₀ sequences with the norm ‖⋅‖_𝐷.

Proof. Suppose that {𝑦_𝑛} is an 𝑎𝑖𝑠𝑏𝑐₀ sequence in 𝐶 with ‖⋅‖_𝐷 for some sequence {𝜀_𝑛}. Then 𝑦_𝑛=∑^∞𝑖=1𝜆^𝑛𝑖𝜉_𝑖 for some sequence {𝜆^𝑛𝑖} such that 𝜆^𝑛𝑖≥0 and ∑^∞𝑖=1𝜆^𝑛𝑖=1. Fix 0<𝜀<1/4. Passing to a subsequence we can suppose that 𝜀_𝑛+1≤𝜀_𝑛<(1/2)−2𝜀 and 1/(1+𝜀_𝑛)>1−𝜀,𝑛∈ℕ.
Assume first that there exists 𝑀∈ℕ such that for every 𝑛≥𝑀, ∑^{∞𝑖=𝑀+1}𝜆^𝑛𝑖≤(1/2)−𝜀. Let 𝑢_𝑛=∑^𝑀𝑖=1𝜆^𝑛𝑖𝜉_𝑖 and 𝑣_𝑛=∑^{∞𝑖=𝑀+1}𝜆^𝑛𝑖𝜉_𝑖; then 𝑦_𝑛=𝑢_𝑛+𝑣_𝑛. Since {𝑢_𝑛}⊂[𝜉_𝑖]^𝑀𝑖=1 is bounded and dim[𝜉_𝑖]^𝑀𝑖=1=𝑀, passing to another subsequence we can suppose that 𝑢_𝑛→𝑢 for some 𝑢∈𝐶. Then, there exist 𝑛₁,𝑛₂∈ℕ with 𝑀≤𝑛₁<𝑛₂ such that ‖‖𝑢_𝑛1−𝑢_𝑛2‖‖_𝐷<𝜀.(2.11) Since ∑^{∞𝑖=𝑀+1}𝜆^𝑛𝑖≤(1/2)−𝜀,𝑛≥𝑀, we also get ‖‖𝑣_𝑛1−𝑣_𝑛2‖‖_𝐷=max_{𝑀+1≤𝑟≤𝑘<∞}|||||_𝑘_𝑖=𝑟𝜆^𝑛1_𝑖−^𝑘_𝑖=𝑟𝜆^𝑛2_𝑖≤_∞|||||_𝑖=_𝑀+1𝜆^𝑛1_𝑖+^∞_𝑖=_𝑀+1𝜆^𝑛2_𝑖≤1−2𝜀.(2.12) Hence ‖𝑦_𝑛1−𝑦_𝑛2‖_𝐷≤1−𝜀. On the other hand, since {𝑦_𝑛} is an 𝑎𝑖𝑠𝑏𝑐₀ sequence, we have that ‖𝑦_𝑛1−𝑦_𝑛2‖_𝐷≥1/(1+𝜀_𝑛2), which contradicts the fact that 1/(1+𝜀_𝑛2)>1−𝜀.
Assume now that for all 𝑀∈ℕ, there exist 𝑛≥𝑀 such that ∑^{∞𝑖=𝑀+1}𝜆^𝑛𝑖>(1/2)−𝜀. Denote each 𝑦_𝑛 by {𝛼^𝑛𝑖}=∑^∞𝑖=1𝛼^𝑛𝑖𝑒_𝑛, where {𝑒_𝑛} is the canonical basis of 𝑐₀. Then 𝛼^𝑛𝑖=∑^∞𝑗=𝑖𝜆^𝑛𝑗. Since 𝑦₁,𝑦₂∈𝑐₀, there exists 𝑀∈ℕ such that 𝛼^1𝑖,𝛼^2𝑖<𝜀2,𝑖≥𝑀.(2.13) By hypothesis, there exists 𝑛₀∈ℕ such that ∑^{∞𝑖=𝑀+1}𝜆^𝑛0_𝑖>(1/2)−𝜀. Then 32−2𝜀≤‖‖𝑦¹+𝑦₂−𝑦_𝑛0‖‖_𝐷.(2.14) On the other hand, since {𝑦_𝑛} is an 𝑎𝑖𝑠𝑏𝑐₀ sequence, we have that ‖𝑦₁+𝑦₂−𝑦_𝑛0‖_𝐷≤1+𝜀₁, which contradicts the fact that 𝜀₁<(1/2)−2𝜀.

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 𝑐_0𝐷-summing basic sequence. First we consider the following definition.

Definition 4. Let {𝑥_𝑛} be a bounded basic sequence in a Banach space 𝑋. We say that {𝑥_𝑛} is a convexly closed sequence if the set 𝐶=^∞_𝑛=1𝑡_𝑛𝑥_𝑛∶𝑡_𝑛≥0,^∞_𝑛=1𝑡_𝑛=1(2.15) is closed, that is, if conv{𝑥_𝑛}=𝐶.

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 𝑎𝑖𝑠𝑏𝑐₀ 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 𝑙_𝑝, 1<𝑝<∞, are not convexly closed.

Definition 5. Let {𝑥_𝑛} be a sequence in a Banach space 𝑋. We say that {𝑥_𝑛} is an asymptotically isometric 𝑐_0𝐷-summing basic sequence, 𝑎𝑖𝑠𝑏𝑐_0𝐷 sequence for short, if {𝑥_𝑛} is convexly closed and there exists {𝜀_𝑛}⊂(0,∞) such that 𝜀_𝑛↘0 and sup_{1≤𝑛≤𝑚<∞}1+𝜀_𝑚⁻¹|||||_𝑚_𝑘=𝑛𝑡_𝑘|||||≤‖‖‖‖_∞_𝑛=1𝑡_𝑛𝑥_𝑛‖‖‖‖≤sup_{1≤𝑛≤𝑚<∞}1+𝜀_𝑚|||||^𝑚_𝑘=𝑛𝑡_𝑘|||||,∀𝑡_𝑛∈𝑙₁.(2.16)

Now, we prove that the analogous of the operator defined in [2] is still contractive and then Banach spaces containing 𝑎𝑖𝑠𝑏𝑐_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 {𝜀_𝑛}⊂(0,∞) be a sequence such that 𝜀_𝑛→0 and 𝜀_𝑛<2⁻¹4^−𝑛,𝑛≥2. If 𝐾 contains an 𝑎𝑖𝑠𝑏𝑐_0𝐷 sequence with this {𝜀_𝑛}, then there exists a nonempty, convex and closed subset 𝐶 of 𝐾 and 𝑇∶𝐶→𝐶 affine, nonexpansive, and fixed-point-free. Moreover, 𝑇 is contractive.

Proof. Let {𝑥_𝑛} be an 𝑎𝑖𝑠𝑏𝑐_0𝐷 sequence in 𝐾 with {𝜀_𝑛}⊂(0,∞) such that 𝜀_𝑛<2⁻¹4^−𝑛,𝑛≥2. Set 𝐶=conv𝑥_𝑛=^∞_𝑛=1𝑡_𝑛𝑥_𝑛∶𝑡_𝑛≥0,𝑛∈ℕ𝑦^∞_𝑛=1𝑡_𝑛=1⊂𝐾.(2.17) Thus 𝐶 is nonempty, convex, closed and bounded. Define 𝑇𝑥_𝑛=∑^∞𝑗=1((𝑥_𝑛+𝑗)/2^𝑗),𝑛∈ℕ, and extend 𝑇 linearly to 𝐶, that is, if 𝑥=∑^∞𝑛=1𝑡_𝑛𝑥_𝑛∈𝐶 then define 𝑇(∑^∞𝑛=1𝑡_𝑛𝑥_𝑛)=∑^∞𝑛=1𝑡_𝑛𝑇𝑥_𝑛. It is easy to see that 𝑇(𝐶)⊂𝐶 and that 𝑇 is affine and fixed-point-free, see [2]. We only need to show that 𝑇 is a contractive mapping. Let 𝑥,𝑦∈𝐶, with 𝑥≠𝑦. Then 𝑥=∑^∞𝑛=1𝑡_𝑛𝑥_𝑛 and𝑦=∑^∞𝑛=1𝑠_𝑛𝑥_𝑛,with 𝑡_𝑛,𝑠_𝑛≥0,and ∑^∞𝑛=1𝑡_𝑛=∑^∞𝑛=1𝑠_𝑛=1. Let 𝛽_𝑛=𝑡_𝑛−𝑠_𝑛,𝑛∈ℕ, such that ∑^∞𝑛=1𝛽_𝑛=0. As in [2] we have 𝑇(𝑥)−𝑇(𝑦)=^∞_𝑛=1𝐵_𝑛𝑥_𝑛,(2.18) where 𝐵₁=0 and 𝐵_𝑛=(𝛽₁/2^𝑛−1)+(𝛽₂/2^𝑛−2)+⋯+(𝛽_𝑛−1/2),𝑛≥2. Consequently, ‖𝑇(𝑥)−𝑇(𝑦)‖=‖‖‖‖^∞_𝑛=1𝐵_𝑛𝑥_𝑛‖‖‖‖≤sup_{1≤𝑛≤𝑚<∞}1+𝜀_𝑚|||||^𝑚_𝑘=𝑛𝐵_𝑘|||||.(2.19) Take 𝑛,𝑚∈ℕ with 𝑛≤𝑚. Since ^𝑚_𝑘=𝑛𝐵_𝑘=𝛽¹2^𝑛−1+𝛽²2^𝑛−2+⋯+𝛽^𝑛−1+𝛽₁22^𝑛+𝛽²2^𝑛−1+⋯+𝛽^𝑛−12²+𝛽^𝑛+𝛽2+⋯₁2^𝑚−1+𝛽²2^𝑚−2+⋯+𝛽^𝑛−12^{𝑚−(𝑛−1)}+𝛽^𝑛2^𝑚−𝑛+𝛽^𝑛+12^{𝑚−(𝑛+1)}+⋯+𝛽^𝑚−1=122𝛽^𝑛−1+1+𝛽_𝑛+⋯+𝛽_𝑚−12²+1𝛽_𝑛−2+𝛽_𝑛−1+⋯+𝛽_𝑚−2+⋯2^𝑛−1+1𝛽₁+𝛽₂+⋯+𝛽_{𝑚−(𝑛−1)}+⋯2^𝑚−2𝛽₁+𝛽₂+12^𝑚−1𝛽₁,(2.20) we have 1+𝜀_𝑚|||||^𝑚_𝑘=𝑛𝐵_𝑘|||||≤1+𝜀_𝑚+1+1+2𝜀22𝜀^𝑚−1_𝑚−21+21𝜀_𝑚−1||𝛽_𝑛−1+𝛽_𝑛+⋯+𝛽_𝑚−1||2²11+2𝜀_𝑚−2+1+2𝜀_{𝑚−(𝑛−1)}||𝛽_𝑛−2+𝛽_𝑛−1+⋯+𝛽_𝑚−2||+⋯2^𝑛−111+2𝜀_{𝑚−(𝑛−1)}+1+2𝜀₂||𝛽₁+𝛽₂+⋯+𝛽_{𝑚−(𝑛−1)}||+⋯2^𝑚−211+2𝜀₂||𝛽₁+𝛽₂||+1+2𝜀¹2^𝑚−111+2𝜀₁≤⎛⎜⎜⎝sup_{1≤𝑖≤𝑗≤𝑚}||𝛽₁||1+2𝜀_𝑗⁻¹|||||_𝑗_𝑘=𝑖𝛽_𝑘|||||⎞⎟⎟⎠𝑄_𝑛𝑚,(2.21) where 𝑄_𝑛𝑚=1+𝜀_𝑚1+2𝜀^𝑚−12+1+22²𝜀^𝑚−2+1+2𝜀_{𝑚−(𝑛−1)}+⋯+2^𝑛−1+1+2𝜀^𝑚−𝑛2^𝑛+⋯+1+2𝜀²2^𝑚−2+1+2𝜀¹≤1+12^𝑚−12⋅4^𝑚21+21²+⋯+12^𝑚−1+2⋅41^𝑚−1+⋯+1=1+12^𝑚−1⋅4¹2⋅4^𝑚1−21_𝑚−1+21^2𝑚−1+12^2𝑚−2+⋯+1<1+12^𝑚+14^𝑚1−21_𝑚−1+21^𝑚<1.(2.22) Then we get sup_{1≤𝑛≤𝑚<∞}1+𝜀_𝑚|||||^𝑚_𝑘=𝑛𝐵_𝑘|||||≤sup_{1≤𝑛≤𝑚<∞}1+2𝜀_𝑚⁻¹|||||_𝑚_𝑘=𝑛𝛽_𝑘|||||<sup_{1≤𝑛≤𝑚<∞}1+𝜀_𝑚⁻¹|||||_𝑚_𝑘=𝑛𝛽_𝑘≤‖‖‖‖_∞|||||_𝑛=1𝛽_𝑛𝑥_𝑛‖‖‖‖=‖𝑥−𝑦‖.(2.23) Thus 𝑇 is contractive.

Next for any sequence of signs we will define a basis in 𝑐₀ equivalent to {𝜉_𝑛}, the summing basis of 𝑐₀, and a sequence asymptotically isometric to it.

Let {𝑒_𝑛} be the canonical basis of 𝑐₀ and for any selection of signs Θ={𝜃_𝑖}_𝑖, that is, 𝜃_𝑖∈{−1,1},𝑖∈ℕ, let {𝜁^Θ𝑛}_𝑛 be the sequence defined by 𝜁^Θ𝑛=^𝑛_𝑘=1𝜃_𝑘𝑒_𝑘,𝑛∈ℕ.(2.24) Since ‖∑^𝑚𝑛=1𝑡_𝑛𝜉_𝑛‖_∞=‖∑^𝑚𝑛=1𝑡_𝑛𝜁^Θ𝑛‖_∞ for all {𝑡_𝑛}^𝑚𝑛=1⊂𝕂, we get that {𝜁^Θ𝑛} is a basis of 𝑐₀ equivalent to the 𝑐₀ summing basis. The sequence {𝜁^Θ𝑛} is called the Θ-basis of 𝑐₀. Let Θ₀={𝜃_𝑖}, where 𝜃_𝑖=1,𝑖∈ℕ. Then the Θ₀-basis of 𝑐₀ is the 𝑐₀ summing basis. If we define 𝐶={∑^∞𝑛=1𝑡_𝑛𝜁^Θ𝑛∶𝑡_𝑛≥0and∑^∞𝑛=1𝑡_𝑛=1}, then 𝐶 is nonempty, convex and bounded. Since ‖⋅‖_∞ and ‖⋅‖_𝐷 are equivalent, we have that {𝜁^Θ𝑛} is convexly closed in (𝑐₀,‖⋅‖_𝐷).

The set 𝐶={∑^∞𝑛=1𝑡_𝑛𝜁^Θ𝑛∶𝑡_𝑛≥0and∑^∞𝑛=1𝑡_𝑛=1} is not 𝜔-compact. The following result shows that the set 𝐶 contains neither 𝑎𝑖𝑠𝑏𝑐_0𝐷 sequences nor 𝑎𝑖𝑠𝑏𝑐₀ sequences with the norm ‖⋅‖_𝐷 if Θ≠±Θ₀.

Proposition 7. For Θ≠±Θ₀, let {𝜁^Θ𝑛} be the Θ-basis of 𝑐₀ considered in (𝑐₀,‖⋅‖_𝐷). If 𝐶=^∞_𝑛=1𝑡_𝑛𝜁^Θ𝑛∶𝑡_𝑛≥0,^∞_𝑛=1𝑡_𝑛=1,(2.25) then the set 𝐶 contains neither 𝑎𝑖𝑠𝑏𝑐_0𝐷 sequences nor 𝑎𝑖𝑠𝑏𝑐₀ sequences with the norm ‖⋅‖_𝐷.

Proof. Let {𝑦_𝑘}⊂𝐶. Then 𝑦_𝑘=∑^∞𝑛=1𝜆^𝑘𝑛𝜁^Θ𝑛 for some 𝜆^𝑘𝑛≥0and ∑^∞𝑛=1𝜆^𝑘𝑛=1.Suppose that {𝑦_𝑘} is an 𝑎𝑖𝑠𝑏𝑐_0𝐷 sequence (resp. an 𝑎𝑖𝑠𝑏𝑐₀ sequence) with the norm ‖⋅‖_𝐷. Let 𝑛₀=min{𝑛∶𝜃_𝑛≠𝜃₁}. If there exists 0<𝜌<1 such that ∑^{𝑛0−1𝑖=1}𝜆^𝑘𝑖≤1−𝜌 for all 𝑘≥1, then for all 𝑘≥1, ‖‖𝑦_𝑘‖‖_𝐷≥^∞_𝑛=1𝜆^𝑘𝑛+^∞_𝑛_=𝑛0𝜆^𝑘𝑛≥1+𝜌.(2.26) Since {𝑦_𝑘} is an 𝑎𝑖𝑠𝑏𝑐_0𝐷 sequence (resp. an 𝑎𝑖𝑠𝑏𝑐₀ sequence) with the norm ‖⋅‖_𝐷, then ‖𝑦_𝑘‖_𝐷≤1+𝜀_𝑘→1 and this is impossible. Now, if limsup_𝑘∑^{𝑛0−1𝑖=1}𝜆^𝑘𝑖=1, as in the proof of Proposition 3, we obtain a subsequence {𝑦_𝑘𝑖} of {𝑦_𝑘} with ‖𝑦_𝑘𝑖−𝑦_𝑘𝑖+1‖_𝐷→0.Since {𝑦_𝑛} is an 𝑎𝑖𝑠𝑏𝑐_0𝐷 with the norm ‖⋅‖_𝐷, then (1+𝜀_𝑘𝑖)⁻¹≤‖𝑦_𝑘𝑖−𝑦_𝑘𝑖+1‖_𝐷 (resp. (1+𝜀_𝑘𝑖+1)⁻¹≤‖𝑦_𝑘𝑖−𝑦_𝑘𝑖+1‖_𝐷) and making 𝑖→∞ we get that 1≤0. This contradiction proves the result.

Although the set 𝐶 of the last proposition has neither 𝑎𝑖𝑠𝑏𝑐_0𝐷 sequences nor 𝑎𝑖𝑠𝑏𝑐₀ sequences, for some Θ it does not have the FPP.

For Θ={𝜃_𝑖}, let 𝐹_𝑛 be the set such that if 𝑖,𝑗∈𝐹_𝑛, then 𝜃_𝑖=𝜃_𝑗, and if 𝑖∈𝐹_𝑛+1 and 𝑗∈𝐹_𝑛, then 𝜃_𝑖≠𝜃_𝑗.Denote by 𝑟_𝑛 the cardinality of 𝐹_𝑛. If 𝑟_𝑛<∞, define 𝑝₀=0,𝑝_𝑛−1=min𝐹_𝑛−1, and 𝑝_𝑛=max𝐹_𝑛.

Proposition 8. Let Θ≠±Θ₀. Then 𝐶=^∞_𝑛=1𝑡_𝑛𝜁^Θ𝑛∶𝑡_𝑛≥0,^∞_𝑛=1𝑡_𝑛=1(2.27) does not have the FPP in the following cases.(1)There exists 𝑘≥1 such that 𝑟_𝑛≤𝑟_𝑛+𝑘<∞,𝑛∈ℕ.(2)𝑟₁=1 and 𝑟₂=∞.(3)There exists {𝑖_𝑛}, with 𝑖₁>1, such that for any 𝑘,𝑙∈ℕ with 𝑖_𝑘−1<𝑙<𝑖_𝑘 we have 𝜃_𝑙=𝜃_𝑖𝑘 and also 𝜃_𝑘≠𝜃_𝑖𝑘 for all 𝑘≥2 or 𝜃_𝑘=𝜃_𝑖𝑘 for all 𝑘≥2.

Proof. Let Θ≠±Θ₀.
(1) If there exists 𝑘≥1 such that 𝑟_𝑛≤𝑟_𝑛+𝑘<∞,𝑛∈ℕ, define 𝑞_𝑛=∑^{𝑛+1𝑗=𝑛−(𝑘−2)}𝑟_𝑗,𝑛≥𝑘 and 𝑇∶𝐶→𝐶 by 𝑇^∞_𝑛=1𝑡_𝑛𝜁^Θ𝑛=𝑇^∞_𝑝𝑛=1^𝑛_{𝑖=𝑝𝑛−1+1}𝑡_𝑖𝜁^Θ𝑖=^∞_{𝑝𝑛=𝑘}^𝑛+1_{𝑖=𝑤𝑛}𝑡_{𝑖−𝑞𝑛}𝜁^Θ𝑖,(2.28) where 𝑤_𝑛=𝑝_𝑛+𝑟_𝑛+1−𝑟_{𝑛+1−𝑘}+1. The idea is to translate the coefficients of ∑^∞𝑛=1𝑡_𝑛𝜁^Θ𝑛 in the block 𝐹_𝑛 into the last 𝑟_𝑛 terms of the block 𝐹_𝑛+𝑘. Then it is easy to see that 𝑇 does not have fixed points. To prove that 𝑇 is nonexpansive first observe that if 𝑘 is even the signs of the 𝜃_𝑖 and 𝜃_𝑗 with 𝑖∈𝐹_𝑛 and 𝑗∈𝐹_𝑛+𝑘 are the same and are different if 𝑘 is odd. Now let 𝑥=∑^∞𝑛=1𝑡_𝑛𝜁^Θ𝑛, 𝑦=∑^∞𝑛=1𝑠_𝑛𝜁^Θ𝑛, and 𝑥−𝑦=∑^∞𝑛=1𝛼_𝑛𝜁^Θ𝑛. Then 𝛼_𝑛=𝑡_𝑛−𝑠_𝑛 and ∑^∞𝑛=1𝛼_𝑛=0. Hence 𝑥−𝑦=^∞_𝑝𝑛=1^𝑛_{𝑖=𝑝𝑛−1+2}𝜃_𝑝𝑛^∞_𝑛=𝑖𝛼_𝑛𝑒^𝑖,(2.29)𝑇(𝑥−𝑦)=^∞_𝑛=𝑘𝜃^{𝑝𝑛+1∞}_{𝑛=𝑖−𝑞𝑛}𝛼_𝑛⎛⎜⎜⎝^𝑤𝑛_{𝑖=𝑝𝑛+1}𝑒_𝑖⎞⎟⎟⎠^𝑝𝑛+1_𝑖=_𝑤𝑛+1𝜃_𝑝𝑛+1^∞_{𝑛=𝑖−𝑞𝑛}𝛼_𝑛𝑒^𝑖(2.30) are the expressions of 𝑥−𝑦 and 𝑇(𝑥−𝑦) 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 𝑟₁=1 and 𝑟₂=∞. In this case, define 𝑇∑^∞𝑛=1𝑡_𝑛𝜁^Θ𝑛=∑^∞𝑛=1𝑡_𝑛𝜁^Θ𝑛+1. Clearly 𝑇 is nonexpansive and fixed-point-free.
(3) In this case it is straightforward to see thatthe operator 𝑇∶𝐶→𝐶 defined by 𝑇∑^∞𝑛=1𝑡_𝑛𝜁^Θ𝑛=∑^∞𝑛=1𝑡_𝑛𝜁^Θ𝑖_𝑛 is nonexpansive and does not have fixed points.

Proposition 9. Let Θ≠±Θ₀. Suppose Θ does not satisfy the hypotheses of the above proposition, and let {𝑖_𝑛} be a sequence with 𝑖₁>1. Thenthe operator 𝑇∶𝐶→𝐶 defined by 𝑇∑^∞𝑛=1𝑡_𝑛𝜁^Θ𝑛=∑^∞𝑛=1𝑡_𝑛𝜁^Θ𝑖_𝑛 is expansive.

Proof. Since Θ does not satisfy the hypotheses (1) and (2) of the above proposition, there are three possibilities.(I)𝑟_𝑛<∞ for every 𝑛≥2; then for every 𝑘 there exists 𝑛 such that 𝑟_𝑛+𝑘<𝑟_𝑛.(II)𝑟₂=∞; then 𝑟₁>1.(III)There exists 𝑘>2 such that 𝑟_𝑘=∞.
Let {𝑖_𝑛} be fixed with 𝑖₁>1 and denote 𝑖₀=0. Since Θ does not satisfy the hypotheses (3) of the above proposition, there exist 𝑘 and 𝑙 with 𝑖_𝑘−1<𝑙<𝑖_𝑘 such that 𝜃_𝑙≠𝜃_𝑖𝑘 or there exists 𝑘₁≥2 with 𝜃_𝑘1=𝜃_𝑖𝑘1 and there exists 𝑘₂≥2 with 𝜃_𝑘2≠𝜃_𝑖𝑘2.
Case 1. For every 𝑘 there exists 𝑛 such that 𝑟_𝑛+𝑘<𝑟_𝑛.
There are two subcases.
Subcase 1.1. There are 𝑘 and 𝑙 with 𝑖_𝑘−1<𝑙<𝑖_𝑘 such that 𝜃_𝑙≠𝜃_𝑖𝑘.
Let 𝑥=(1/8)𝜁^Θ1+(3/8)𝜁^Θ𝑘−1+(1/2)𝜁^Θ𝑘 and 𝑦=(1/16)𝜁^Θ1+(3/16)𝜁^Θ𝑘−1+(3/4)𝜁^Θ𝑘.Then 𝑥−𝑦=(1/16)𝜁^Θ1+(3/16)𝜁^Θ𝑘−1−(1/4)𝜁^Θ𝑘=−(1/16)∑^{𝑘−1𝑖=2}𝜃_𝑖𝑒_𝑖−(1/4)𝜃_𝑘𝑒_𝑘 and ‖𝑥−𝑦‖_𝐷≤5/16.On the other hand, 𝑇𝑥−𝑇𝑦=(1/16)𝜁^Θ𝑖₁+(3/16)𝜁^Θ𝑖_𝑘−1−(1/4)𝜁^Θ𝑖_𝑘=−(1/16)∑^𝑖𝑘−1_𝑗=𝑖1+1𝜃_𝑗𝑒_𝑗−(1/4)∑^𝑖𝑘_{𝑗=𝑖𝑘−1+1}𝜃_𝑗𝑒_𝑗 and ‖𝑇𝑥−𝑇𝑦‖_𝐷=1/2.
Subcase 1.2. For any 𝑘∈ℕ and 𝑙 with𝑖_𝑘−1<𝑙<𝑖_𝑘, we have 𝜃_𝑙=𝜃_𝑖𝑘.
There are two subsubcases. (1) 𝜃₁=𝜃_𝑖1 and (2) 𝜃₁≠𝜃_𝑖1.
(1)𝜃₁=𝜃_𝑖1
If 𝜃_𝑘=𝜃_𝑖𝑘 for every 𝑘; then we would have 𝐹₁=ℕ, which implies Θ=±Θ₀. Then there is 𝑘 such that 𝜃_𝑘≠𝜃_𝑖𝑘. Let 𝑠=min{𝑙∶𝜃_𝑙≠𝜃_𝑖𝑙}. Then 𝑠>1.
There are two possibilities: (A) there exists 𝑟>𝑠 such that 𝜃_𝑟=𝜃_𝑖𝑟 and (B) 𝜃_𝑘≠𝜃_𝑖𝑘 for all 𝑘≥𝑠.(A)Let 𝑘+1=min{𝑟>𝑠∶𝜃_𝑟=𝜃_𝑖𝑟}. We need to consider the following cases.(a)𝜃_𝑘=𝜃_𝑘+1.Let 𝑥=(1/2)𝜁^Θ𝑘−1+(1/2)𝜁^Θ𝑘+1 and 𝑦=(3/4)𝜁^Θ𝑘−1+(1/4)𝜁^Θ𝑘+1.Then 𝑥−𝑦=−(1/4)𝜁^Θ𝑘−1+(1/4)𝜁^Θ𝑘+1=𝜃_𝑘+1((1/4)𝑒_𝑘+(1/4)𝑒_𝑘+1) and ‖𝑥−𝑦‖_𝐷=1/4.On the other hand, 𝑇𝑥−𝑇𝑦=−(1/4)𝜁^Θ𝑖_𝑘−1+(1/4)𝜁^Θ𝑖_𝑘+1=(1/4)∑^𝑖𝑘_{𝑗=𝑖𝑘−1+1}𝜃_𝑗𝑒_𝑗+(1/4)∑^𝑖𝑘+1_{𝑗=𝑖𝑘+1}𝜃_𝑗𝑒_𝑗=−(1/4)𝜃_𝑘+1∑^𝑖k_{𝑗=𝑖𝑘−1+1}𝑒_𝑗+(1/4)𝜃_𝑘+1∑^𝑖𝑘+1_{𝑗=𝑖𝑘+1}𝑒_𝑗 and ‖𝑇𝑥−𝑇𝑦‖_𝐷=1/2.(b)𝜃_𝑘≠𝜃_𝑘+1. Let 𝑥=(1/2)𝜁^Θ𝑘−1+(1/2)𝜁^Θ𝑘+1 and 𝑦=(3/4)𝜁^Θ𝑘+(1/4)𝜁^Θ𝑘+1. Then 𝑥−𝑦=(1/2)𝜁^Θ𝑘−1−(3/4)𝜁^Θ𝑘+(1/4)𝜁^Θ𝑘+1=−(1/2)𝜃_𝑘𝑒_𝑘+(1/4)𝜃_𝑘+1𝑒_𝑘+1=𝜃_𝑘+1((1/2)𝑒_𝑘+(1/4)𝑒_𝑘+1) and ‖𝑥−𝑦‖_𝐷=1/2. On the other hand, 𝑇𝑥−𝑇𝑦=(1/2)𝜁^Θ𝑖_𝑘−1−(3/4)𝜁^Θ𝑖_𝑘+(1/4)𝜁^Θ𝑖_𝑘+1=−(1/2)∑^𝑖𝑘_{𝑗=𝑖𝑘−1+1}𝜃_𝑗𝑒_𝑗+(1/4)∑^𝑖𝑘+1_{𝑗=𝑖𝑘+1}𝜃_𝑗𝑒_𝑗=−(1/2)𝜃_𝑘∑^𝑖𝑘_{𝑗=𝑖𝑘−1+1}𝑒_𝑗+(1/4)𝜃_𝑘∑^𝑖𝑘+1_{𝑗=𝑖𝑘+1}𝑒_𝑗 and ‖𝑇𝑥−𝑇𝑦‖_𝐷=3/4.(B)𝜃_𝑘≠𝜃_𝑖𝑘 for all 𝑘≥𝑠. By hypothesis we have that 𝑠>2. There are two cases.(a)𝜃_𝑠−1=𝜃_𝑠. Then 𝜃_𝑖𝑠−1≠𝜃_𝑖𝑠.Let 𝑥=(1/4)𝜁^Θ𝑠−2+(1/2)𝜁^Θ𝑠−1+(1/4)𝜁^Θ𝑠 and 𝑦=(1/4)𝜁^Θ𝑠−1+(3/4)𝜁^Θ𝑠. Then 𝑥−𝑦=(1/4)𝜁^Θ𝑠−2+(1/4)𝜁^Θ𝑠−1−(1/2)𝜁^Θ𝑠=−𝜃_𝑠((1/4)𝑒_𝑠−1+(1/2)𝑒_𝑠) and ‖𝑥−𝑦‖_𝐷=1/2.On the other hand, 𝑇𝑥−𝑇𝑦=(1/4)𝜁^Θ𝑖_𝑠−2+(1/4)𝜁^Θ𝑖_𝑠−1−(1/2)𝜁^Θ𝑖_𝑠=−(1/4)𝜃_𝑠∑^𝑖𝑠−1_{𝑗=𝑖𝑠−2+1}𝑒_𝑗+(1/2)𝜃_𝑠∑^𝑖𝑠_{𝑗=𝑖𝑠−1+1}𝑒_𝑗 and ‖𝑇𝑥−𝑇𝑦‖_𝐷=3/4.(b)𝜃_𝑠−1≠𝜃_𝑠. Then 𝜃_𝑖𝑠−1=𝜃_𝑖𝑠.Let 𝑥=(1/4)𝜁^Θ𝑠−2+(1/4)𝜁^Θ𝑠−1+(1/2)𝜁^Θ𝑠 and 𝑦=(3/4)𝜁^Θ𝑠−1+(1/4)𝜁^Θ𝑠. Then 𝑥−𝑦=(1/4)𝜁^Θ𝑠−2−(1/2)𝜁^Θ𝑠−1+(1/4)𝜁^Θ𝑠=−(1/4)𝜃_𝑠−1𝑒_𝑠−1+(1/4)𝜃_𝑠𝑒_𝑠=𝜃_𝑠((1/4)𝑒_𝑠−1+(1/4)𝑒_𝑠) and ‖𝑥−𝑦‖_𝐷=1/4.On the other hand, 𝑇𝑥−𝑇𝑦=(1/4)𝜁^Θ𝑖_𝑠−2−(1/2)𝜁^Θ𝑖_𝑠−1+(1/4)𝜁^Θ𝑖_𝑠=−(1/4)𝜃_𝑖𝑠−1∑^𝑖𝑠−1_{𝑗=𝑖𝑠−2+1}𝑒_𝑗+(1/4)𝜃_𝑖𝑠∑^𝑖𝑠_{𝑗=𝑖𝑠−1+1}𝑒_𝑗=𝜃_𝑠−1(−(1/4)∑^𝑖𝑠−1_{𝑗=𝑖𝑠−2+1}𝑒_𝑗+(1/4)∑^𝑖𝑠_{𝑗=𝑖𝑠−1+1}𝑒_𝑗) and ‖𝑇𝑥−𝑇𝑦‖_𝐷=1/2.
(2)𝜃₁≠𝜃_𝑖1
In this case there exists 𝑘 such that 𝜃_𝑘=𝜃_𝑖𝑘. If 𝑠=min{𝑙∶𝜃_𝑙=𝜃_𝑖𝑙}, then 𝑠>1.
Hence consider the cases: (A) there exists 𝑟>𝑠 such that 𝜃_𝑟≠𝜃_𝑖𝑟 and (B) 𝜃_𝑘=𝜃_𝑖𝑘 for all 𝑘≥𝑠 and proceed as in the Case (1).
Case 2. 𝑟₂=∞ and 𝑟₁>1.
Then 𝜃_𝑝1≠𝜃_𝑖𝑝1 with 1<𝑝₁.Hence we can proceed as in Subcase 1.2(1)(A) above taking 𝑘=𝑝₁.
Case 3. There is 𝑠>1 such that 𝑟_𝑠+1=∞.
Then 𝜃_𝑝𝑠≠𝜃_𝑖𝑝𝑠 with 1<𝑝_𝑠.Hence we can proceed as in Subcase 1.2(1)(A) above taking 𝑘=𝑝_𝑠.

Next, for every selection of signs Θ≠±Θ₀, we will define the asymptotically isometric 𝑐_0𝐷-Θ-basic sequences. To this end, let us consider the following notation.

Let 𝔖𝔇_ΘΘ=(𝑛,𝑚)∶𝜃_𝑛=𝜃_𝑚,=(𝑛,𝑚)∶𝜃_𝑛≠𝜃_𝑚.(2.31)

Definition 10. Let {𝑥_𝑛} be a sequence in a Banach space 𝑋. We say that {𝑥_𝑛} is an asymptotically isometric 𝑐_0𝐷-Θ-basic sequence (𝑎𝑖Θ𝑏𝑐_0𝐷 sequence for short) if {𝑥_𝑛} is convexly closed and there exists {𝜀^Θ𝑛}⊂(0,(1/2)) such that 𝜀^Θ𝑛↘0, and 𝐿𝜀^Θ𝑛,𝑡_𝑛,𝔖_Θ∨𝐿𝜀^Θ𝑛,𝑡_𝑛,𝔇_Θ≤‖‖‖‖^∞_𝑛=1𝑡_𝑛𝑥_𝑛≤𝑅𝜀_Θ𝑛‖‖‖‖,𝑡_𝑛,𝔖_Θ∨𝑅𝜀^Θ𝑛,𝑡_𝑛,𝔇_Θ(2.32) holds for all {𝑡_𝑛}∈𝑙₁, where 𝐿𝜀^Θ𝑛,𝑡_𝑛,𝔖_Θ=⎛⎜⎜⎝sup_{𝑛<l,(𝑛,𝑙)∈𝔖Θ}1+𝜀^Θ𝑙−1⁻¹|||||_𝑙−1_𝑘=𝑛𝑡_𝑘𝐿𝜀_Θ𝑛|||||⎞⎟⎟⎠,,𝑡_𝑛,𝔇_Θ=⎛⎜⎜⎝sup_{𝑛<𝑙,(𝑛,𝑙)∈𝔇Θ}1+𝜀^Θ𝑙−1⁻¹|||||_𝑙−1_𝑘=𝑛𝑡_𝑘+2^∞_𝑘=𝑙𝑡_𝑘𝑅𝜀_Θ𝑛|||||⎞⎟⎟⎠,,𝑡_𝑛,𝔖_Θ=⎛⎜⎜⎝sup_{𝑛<𝑙,(𝑛,𝑙)∈𝔖Θ}1+𝜀^Θ𝑙−1|||||^𝑙−1_𝑘=𝑛𝑡_𝑘𝑅𝜀_Θ𝑛|||||⎞⎟⎟⎠,,𝑡_𝑛,𝔇_Θ=⎛⎜⎜⎝sup_{𝑛<𝑙,(𝑛,𝑙)∈𝔇Θ}1+𝜀^Θ𝑙−1|||||^𝑙−1_𝑘=𝑛𝑡_𝑘+2^∞_𝑘=𝑙𝑡_𝑘|||||⎞⎟⎟⎠.(2.33)

We are interested in 𝑎𝑖Θ𝑏𝑐_0𝐷 sequences for which the numbers 𝜀^Θ𝑛 of Definition 10 are small. We are taking {𝜀^Θ𝑛}⊂(0,(1/2)).

We know that the set 𝐶 of Proposition 3 does not have 𝑎𝑖𝑠𝑏𝑐₀ sequences. Now we also prove that 𝐶 does not contain 𝑎𝑖Θ𝑏𝑐_0𝐷 sequences with the norm ‖⋅‖_𝐷 if Θ≠±Θ₀.

Proposition 11. Let Θ≠±Θ₀. The set 𝐶={∑^∞𝑛=1𝑡_𝑛𝜉_𝑛∶𝑡_𝑛≥0and∑^∞𝑛=1𝑡_𝑛=1} does not contain 𝑎𝑖Θ𝑏𝑐_0𝐷 sequences with the norm ‖⋅‖_𝐷.

Proof. Let {𝑦_𝑘}⊂𝐶. Then 𝑦_𝑘=∑^∞𝑛=1𝜆^𝑘𝑛𝜉_𝑛 for some 𝜆^𝑘𝑛≥0with ∑^∞𝑛=1𝜆^𝑘𝑛=1.Suppose that {𝑦_𝑘} is an 𝑎𝑖Θ𝑏𝑐_0𝐷 with ‖⋅‖_𝐷. Since Θ≠±Θ₀, there exist 𝑚∈ℕ and {𝑛_𝑘}⊂ℕ with 𝑛₁<𝑛₂<⋯, such that for all 𝑘∈ℕ, 𝑚<𝑛_𝑘 and 𝜃_𝑛𝑘≠𝜃_𝑚. Let 𝑡_𝑛=0 for 𝑛≠𝑚,𝑛_𝑘 and 𝑡_𝑚=𝑡_𝑛𝑘=1. Thus 1+𝜀^Θ𝑛_𝑘⁻¹⁻¹3≤≤𝐿‖‖‖‖_∞𝜀^Θ𝑛,𝑡_𝑛,𝔖_Θ∨𝐿𝜀^Θ𝑛,𝑡_𝑛,𝔇_Θ_𝑛=1𝑡_𝑛𝑦_𝑛‖‖‖‖_𝐷=‖‖𝑦_𝑚+𝑦_𝑛𝑘‖‖_𝐷=2.(2.34) Since (2.34) holds for all 𝑘∈ℕ, making 𝑘→∞ in (2.34), we get that 3≤2, which is a contradiction.

Proposition 12. Let Θ₁={𝜃^1𝑖}_𝑖 and Θ₂={𝜃^2𝑖}_𝑖 such that Θ₁≠±Θ₂ and Θ₁,Θ₂≠±Θ₀. Let {𝜁^Θ1_𝑛} be the Θ₁-basis of 𝑐₀ considered in (𝑐₀,‖⋅‖_𝐷) and let 𝐶Θ₁=^∞_𝑛=1𝑡_𝑛𝜁^Θ1_𝑛∶𝑡_𝑛≥0,^∞_𝑛=1𝑡_𝑛=1.(2.35) The set 𝐶(Θ₁) does not contain 𝑎𝑖Θ₂𝑏𝑐_0𝐷 sequences with the norm ‖⋅‖_𝐷.

Proof. Let {𝑦_𝑘}⊂𝐶. Then 𝑦_𝑘=∑^∞𝑛=1𝜆^𝑘𝑛𝜁^Θ1_𝑛 for some 𝜆^𝑘𝑛≥0 with ∑^∞𝑛=1𝜆^𝑘𝑛=1.Suppose that {𝑦_𝑘} is an 𝑎𝑖Θ₂𝑏𝑐_0𝐷 with the norm ‖⋅‖_𝐷.
Suppose first 𝜃¹¹=𝜃²¹; since Θ₁≠Θ₂, there exists 𝑚>1 such that 𝜃^1𝑚≠𝜃^2𝑚.
There are two cases.
Case 1. (1,𝑚)∈𝔖_Θ1. In this case (1,𝑚)∈𝔇_Θ2. Let 𝑡_𝑛=0 for 𝑛≠1,𝑚 and 𝑡₁=𝑡_𝑚=1. Thus 1+𝜀^Θ2𝑚−1⁻¹3≤≤𝐿‖‖‖‖_∞𝜀^Θ2_𝑛,𝑡^𝑛,𝔖_Θ2∨𝐿𝜀^Θ2𝑛,𝑡^𝑛,𝔇_Θ2_𝑛=1𝑡_𝑛𝑦_𝑛‖‖‖‖_𝐷=‖‖𝑦₁+𝑦_𝑚‖‖_𝐷≤2.(2.36) Since 𝜀^Θ2_𝑚−1<1/2, we get a contradiction.
Case 2. (1,𝑚)∈𝔇_Θ1. In this case (1,𝑚)∈𝔖_Θ2. Let 𝑡_𝑛=0 for 𝑛≠1,𝑚 and 𝑡₁=𝑡_𝑚=1. Thus 2≤^∞_𝑛=1𝜆^1𝑛+^∞_𝑛=1𝜆^𝑚𝑛+^∞_𝑛_=𝑚𝜆^1𝑛+^∞_𝑛_=𝑚𝜆^𝑚𝑛≤‖‖‖‖^∞_𝑛=1𝑡_𝑛𝑦_𝑛‖‖‖‖_𝐷=‖‖𝑦₁+𝑦_𝑚‖‖_𝐷≤𝑅𝜀^Θ2_𝑛,𝑡^𝑛,𝔖_Θ2∨𝑅𝜀^Θ2𝑛,𝑡^𝑛,𝔇_Θ2≤1+𝜀^Θ2𝑚−1.(2.37) Since 𝜀^Θ2_𝑚−1<1/2, we get a contradiction.
Suppose now 𝜃¹¹≠𝜃²¹; since Θ₁≠−Θ₂, there exists 𝑚 such that 𝜃^1𝑚=𝜃^2𝑚.
There are two cases.
Case 1. (1,𝑚)∈𝔖_Θ1; in this case (1,𝑚)∈𝔇_Θ2. Let 𝑡_𝑛=0 for 𝑛≠1,𝑚 and 𝑡₁=𝑡_𝑚=1. Thus 1+𝜀^Θ2𝑚−1⁻¹3≤≤𝐿‖‖‖‖_∞𝜀^Θ2_𝑛,𝑡^𝑛,𝔖_Θ2∨𝐿𝜀^Θ2𝑛,𝑡^𝑛,𝔇_Θ2_𝑛=1𝑡_𝑛𝑦_𝑛‖‖‖‖_𝐷=‖‖𝑦₁+𝑦_𝑚‖‖_𝐷≤2.(2.38) Since 𝜀^Θ2_𝑚−1<1/2, we get a contradiction.
Case 2. (1,𝑚)∈𝔇_Θ1. In this case (1,𝑚)∈𝔖_Θ2. Let 𝑡_𝑛=0 for 𝑛≠1,𝑚 and 𝑡₁=𝑡_𝑚=1. Thus 2≤^∞_𝑛=1𝜆^1𝑛+^∞_𝑛=1𝜆^𝑚𝑛+^∞_𝑛_=𝑚𝜆^1𝑛+^∞_𝑛_=𝑚𝜆^𝑚𝑛≤‖‖‖‖^∞_𝑛=1𝑡_𝑛𝑦_𝑛‖‖‖‖_𝐷=‖‖𝑦₁+𝑦_𝑚‖‖_𝐷≤𝑅𝜀^Θ2_𝑛,𝑡^𝑛,𝔖_Θ2∨𝑅𝜀^Θ2𝑛,𝑡^𝑛,𝔇_Θ2≤1+𝜀^Θ2𝑚−1.(2.39) Since 𝜀^Θ2_𝑚−1<1/2, we get a contradiction.

Propositions 3, 7, and 11 show that, in contrast with Theorem 4 of the Dowling et al. paper [2] for 𝑎𝑖𝑠𝑏𝑐₀ sequences in 𝑐₀, in the space (𝑐₀,‖⋅‖_𝐷) we need an infinite number of sequences (at least 𝑎𝑖𝑠𝑏𝑐₀ and 𝑎𝑖Θ𝑏𝑐_0𝐷 sequences) to have a similar result.

3. The Space (𝑐₀,‖⋅‖_𝐷)^∗

It is known that the dual of the Bynum space 𝑐₀₁ is the Bynum space 𝑙_1∞. Below we prove that the dual space of (𝑐₀,‖⋅‖_𝐷) when the scalar field is the set of real numbers is also the Bynum space 𝑙_1∞. Let us suppose then that 𝕂=ℝ. First we calculate the extreme points of the unit ball of (𝑐₀,‖⋅‖_𝐷).

Lemma 13. Let 𝑋=(𝑐₀,‖⋅‖_𝐷). Then we have ℰ𝐵_𝑋=𝑥_𝑛∈𝑆_𝑋∶𝑥_𝑛∈{1,0},𝑛∈ℕ∪𝑥_𝑛∈𝑆_𝑋∶𝑥_𝑛∈{−1,0},𝑛∈ℕ.(3.1)

Proof. First note that if {𝑥_𝑛}∈𝑆_𝑋 then |𝑥_𝑛−𝑥_𝑚|≤1,𝑛,𝑚∈ℕ and |𝑥_𝑛|≤1,𝑛∈ℕ. Consequently, if {𝑥_𝑛}∈𝑆_𝑋 with 𝑥_𝑛0=1 for some 𝑛₀∈ℕ, then 0≤𝑥_𝑛≤1, for all 𝑛∈ℕ. Analogously if {𝑥_𝑛}∈𝑆_𝑋 with 𝑥_𝑛0=−1 for some 𝑛₀∈ℕ, then −1≤𝑥_𝑛≤0, for all𝑛∈ℕ. Let 𝐴={{𝑥_𝑛}∈𝑆_𝑋∶𝑥_𝑛∈{1,0},𝑛∈ℕ} and 𝐵={{𝑥_𝑛}∈𝑆_𝑋∶𝑥_𝑛∈{−1,0},𝑛∈ℕ}. Thus 𝐴,𝐵⊂𝑆_𝑋.
Take 𝑥={𝑥_𝑛}∈𝐴 and suppose that 𝑥=(𝑦+𝑧)/2 with 𝑦,𝑧∈𝑆_𝑋. Also suppose that 𝑦={𝑦_𝑛} and 𝑧={𝑧_𝑛}. Since 𝑥∈𝐴, there exists 𝑛₀∈ℕ such that 𝑥_𝑛0=1. Since 𝑥_𝑛=(𝑦_𝑛+𝑧_𝑛)/2 and 𝑦_𝑛,𝑧_𝑛≤1, if 𝑥_𝑛=1 for some 𝑛∈ℕ, we have that 𝑥_𝑛=𝑦_𝑛=𝑧_𝑛. Thus 𝑥_𝑛0=𝑦_𝑛0=𝑧_𝑛0=1. On the other hand, if 𝑥_𝑛=0 for some 𝑛∈ℕ, we also have that 𝑥_𝑛=𝑦_𝑛=𝑧_𝑛, because if 𝑦_𝑛<0 we get that |𝑦_𝑛−𝑦_𝑛0|>1, which contradicts that |𝑦_𝑛−𝑦_𝑚|≤1,𝑛,𝑚∈ℕ and if 𝑦_𝑛>0 then 𝑧_𝑛<0 and we also have a contradiction. Therefore, 𝑥=𝑦=𝑧. Hence 𝑥∈ℰ(𝐵_𝑋). Thus 𝐴⊂ℰ(𝐵_𝑋). Analogously 𝐵⊂ℰ(𝐵_𝑋).
Take now 𝑥={𝑥_𝑛}∈𝑆_𝑋⧵{𝐴∪𝐵}. Then there exists 𝑛₀∈ℕ such that 0<|𝑥_𝑛0|<1. Let 𝑎=inf_𝑛𝑥_𝑛 and 𝑏=sup_𝑛𝑥_𝑛. If 𝑥_𝑛0∈(𝑎,𝑏), define 𝑐=min(|𝑥_𝑛0−𝑎|,|𝑥_𝑛0−𝑏|),𝑦_𝑛=𝑧_𝑛=𝑥_𝑛,𝑛≠𝑛₀,𝑦_𝑛0=𝑥_𝑛0−𝑐, and 𝑧_𝑛0=𝑥_𝑛0+𝑐. Thus, 𝑥_𝑛=(𝑦+𝑧)/2 with 𝑦,𝑧∈𝑆_𝑋 and 𝑥≠𝑦,𝑥≠𝑧. Therefore, 𝑥∈𝑆_𝑋⧵ℰ(𝐵_𝑋). Suppose now that 𝑥_𝑛0=𝑎 or 𝑥_𝑛0=𝑏. Since 0<|𝑥_𝑛0|<1 and sup_{𝑛,𝑚∈ℕ}|𝑥_𝑛−𝑥_𝑚|=1, we have that 0∈(𝑎,𝑏). Since 𝑥_𝑛→0, there exists 𝑛₁>𝑛₀ such that 𝑥_𝑛1∈(𝑎,𝑏), which implies that 𝑥∈𝑆_𝑋⧵ℰ(𝐵_𝑋). Consequently, ℰ(𝐵_𝑋)⊂𝐴∪𝐵.

Theorem 14. Let 𝑓∈(𝑐₀,‖⋅‖_𝐷)^∗. There exists a unique sequence {𝑐_𝑛}∈𝑙₁ such that 𝑓=∑^∞𝑛=1𝑐_𝑛𝑒^∗𝑛 and ‖𝑓‖_𝐷=max^∞_𝑛=1𝑐^+𝑛,^∞_𝑛=1𝑐^−𝑛,(3.2) where 𝑐^+𝑛=max(𝑐_𝑛,0) and 𝑐^−𝑛=−min(𝑐_𝑛,0).

Proof. Let 𝑓∈(𝑐₀,‖⋅‖_𝐷)^∗. Since {𝑒_𝑛} is a shrinking basis of (𝑐₀,‖⋅‖_𝐷), there exists a unique sequence {𝑐_𝑛}⊂𝕂 such that 𝑓=∑^∞𝑛=1𝑐_𝑛𝑒^∗𝑛. As sets (𝑐₀,‖⋅‖_𝐷)^∗=(𝑐₀)^∗ and hence 𝑓∈(𝑐₀)^∗. Thus 𝑓=𝑅{𝑎_𝑛} where 𝑅∶𝑙₁→𝑐^∗0 is the Riesz representation. Consequently, 𝑐_𝑛=𝑓𝑒_𝑛=𝑅𝑎_𝑛𝑒_𝑛=𝑎_𝑛.(3.3) Therefore, {𝑐_𝑛}={𝑎_𝑛}∈𝑙₁. Thus ‖𝑓‖_𝐷==ssuupp_{𝑥∈𝐵𝑋}⎧⎪⎨⎪⎩||𝑓|||||(𝑥_𝑛∈𝐹)||=sup_{𝐵𝑥∈ℰ𝑋}||𝑓(𝑥)||𝑐_𝑛|||||∶𝐹⊂ℕ,𝐹finite=max_∞⎫⎪⎬⎪⎭_𝑛=1𝑐^+𝑛,^∞_𝑛=1𝑐^−𝑛,(3.4) where 𝑐^+𝑛=max(𝑐_𝑛,0) and 𝑐^−𝑛=−min(𝑐_𝑛,0).

Corollary 15. (𝑐₀,‖⋅‖_𝐷)^∗ is the Bynum space 𝑙_1∞ and it has the 𝜔-FPP.

Remark 16. It is well known that 𝑙₁(𝑐₀)^∗ has the 𝜔^∗ fixed point property for left reversible semigroups, that is, whenever 𝑆 is a semigroup such that 𝑎𝑆∩𝑏𝑆≠∅ for any 𝑎,𝑏∈𝑆, and 𝒮={𝑇_𝑠∶𝑠∈𝑆} is a representation of 𝑆 as nonexpansive mappings on a nonempty 𝜔^∗-compact convex subset 𝐾 of 𝑙₁, there is a common fixed point in 𝐾 for 𝒮. (see [10–12]). In particular, 𝑙₁ has the 𝜔^∗ fixed point property. Is this the case for (𝑐₀,‖⋅‖_𝐷)^∗?

Next we will see that every infinite-dimensional subspace of 𝑙_1∞ contains an asymptotically isometric copy of 𝑙₁ and then, by a result of Dowling and Lennard [13], it does not have the FPP.

First recall that a Banach space (𝑋,‖⋅‖) contains an asymptotically isometric copy of 𝑙₁ if there exists {𝑥_𝑛}_𝑛⊂𝑋 and {𝜀_𝑛}⊂(0,1),𝜀_𝑛→0 such that for every 𝑘∈ℕ and every scalars 𝑏₁,…,𝑏_𝑘,^𝑘_𝑖=11−𝜀_𝑖||𝑏_𝑖||≤‖‖‖‖^𝑘_𝑖=1𝑏_𝑖𝑥_𝑖‖‖‖‖≤^𝑘_𝑖=11+𝜀_𝑖||𝑏_𝑖||.(3.5) In this case we say that {𝑥_𝑛}_𝑛 is an asymptotically isometric 𝑙₁-sequence (𝑎𝑖𝑙₁-sequence for short).

Observe that if {𝑦_𝑛}_𝑛 is another sequence in 𝑋 such that ‖𝑦_𝑛−𝑥_𝑛‖<𝛿_𝑛 for all 𝑛, where {𝜀_𝑛+𝛿_𝑛}⊂(0,1) and 𝛿_𝑛→0, then for every 𝑘 and every scalars 𝑏₁,…,𝑏_𝑘,^𝑘_𝑖=11−𝜀_𝑖−𝛿_𝑖||𝑏_𝑖||≤‖‖‖‖^𝑘_𝑖=1𝑏_𝑖𝑦_𝑖‖‖‖‖≤^𝑘_𝑖=11+𝜀_𝑖+𝛿_𝑖||𝑏_𝑖||(3.6) and {𝑦_𝑛} is also an 𝑎𝑖𝑙₁-sequence.

Proposition 17. Let {𝑢_𝑖}_𝑖⊂𝑙_1∞, and let {𝑛_𝑖} be a strictly increasing sequence in ℕ such that 𝑢_𝑖=∑^𝑛𝑖+1_{𝑗=𝑛𝑖+1}𝑎^𝑖𝑗𝑒_𝑗.If ∑^𝑛𝑖+1_{𝑗=𝑛𝑖+1}(𝑎^𝑖𝑗)⁺=∑^𝑛𝑖+1_{𝑗=𝑛𝑖+1}(𝑎^𝑖𝑗)⁻, then {𝑢_𝑖}_𝑖 is isometrically equivalent to the canonical basis in 𝑙₁, that is, for every 𝑘∈ℕ and every scalars 𝑏₁,…,𝑏_𝑘, we have that ‖∑^𝑘𝑖=1𝑏_𝑖𝑢_𝑖‖=∑^𝑘𝑖=1|𝑏_𝑖|.

Proof. Let 𝑏₁,…,𝑏_𝑘 be scalars; then ^𝑘_𝑖=1||𝑏_𝑖||≤^𝑘_𝑖=1𝑏^{+𝑖𝑛𝑖+1}_𝑗=_𝑛𝑖+1𝑎^𝑖𝑗⁺+^𝑘_𝑖=1𝑏^{−𝑖𝑛𝑖+1}_𝑗=_𝑛𝑖+1𝑎^𝑖𝑗⁻=^𝑘_𝑛𝑖=1^𝑖+1_{𝑗=𝑛𝑖+1}𝑏^𝑖𝑎^𝑖𝑗⁺≤‖‖‖‖^𝑘_𝑖=1𝑏_𝑖𝑢_𝑖‖‖‖‖_1∞≤^𝑘_𝑖=1||𝑏_𝑖||.(3.7)

Theorem 18. Every infinite-dimensional subspace of 𝑙_1∞ contains an asymptotically isometric copy of 𝑙₁ and hence it does not have the FPP.

Proof. Let 𝑌 be an infinite-dimensional subspace of 𝑙_1∞,{𝜀_𝑛}⊂(0,(1/2)),𝜀_𝑛↘0 and {𝑥_𝑛} a sequence in 𝑆_𝑌 such that 𝑥_𝑖=∑^∞𝑗=𝑚_𝑖+1𝑎^𝑖𝑗𝑒_𝑗, where 0=𝑚₀<𝑚₁<⋯ and ∑^∞𝑗=𝑚_𝑖+1+1|𝑎^𝑖𝑗|<𝜀_𝑖/8.Define 𝑤_𝑖=^𝑚𝑖+1_𝑗=_𝑚𝑖+1𝑐_+𝑖𝑎^𝑖𝑗𝑒_𝑗,=1‖‖𝑤_𝑖‖‖_𝑚1∞^𝑖+1_{𝑗=𝑚𝑖+1}𝑎^𝑖𝑗⁺𝑐≤1,_−𝑖=1‖‖𝑤_𝑖‖‖_𝑚1∞^𝑖+1_{𝑗=𝑚𝑖+1}𝑎^𝑖𝑗⁻≤1.(3.8) Changing 𝑤_𝑖 by −𝑤_𝑖, if necessary, we can assume that 𝑐^+𝑖=1,𝑛∈ℕ. If there is a sequence {𝑘_𝑖} such that 𝑐^−𝑘_𝑖=1, then by Proposition 17, {𝑤_𝑘𝑖/‖𝑤_𝑘𝑖‖_1∞} is isometrically equivalent to the canonical basis of 𝑙₁. It is straightforward to see that ‖𝑥_𝑘𝑖−(𝑤_𝑘𝑖/‖𝑤_𝑘𝑖‖_1∞)‖_1∞<(1/4)𝜀_𝑘𝑖. Then by the above remark, {𝑥_𝑘𝑖} is an 𝑎𝑖𝑙₁-sequence.
Suppose that 𝑐^−𝑖≠1 for all 𝑖 and let 𝛼_𝑖=1−𝑐^−2𝑖1−𝑐^−2𝑖𝑐^−2𝑖−1,𝛽_𝑖=1−𝑐^−2𝑖−11−𝑐^−2𝑖−1𝑐^−2𝑖.(3.9) Then 0≤𝛼_𝑖<1,0≤𝛽_𝑖<1 and 𝛼_𝑖𝑐^+2𝑖−1+𝛽_𝑖𝑐^−2𝑖=𝛼_𝑖𝑐^−2𝑖−1+𝛽_𝑖𝑐^+2𝑖=1.(3.10)
Now let 𝑣_𝑖=𝛼_𝑖𝑤_2𝑖−1‖‖𝑤_2𝑖−1‖‖_1∞−𝛽_𝑖𝑤_2𝑖‖‖𝑤_2𝑖‖‖_1∞.(3.11) Suppose that 𝑣_𝑖=∑^𝑚2𝑖+1_{𝑗=𝑚2𝑖−1+1}𝑏^𝑖𝑗𝑒_𝑗. It is easy to check, using (3.10), that ^𝑚2𝑖+¹_{𝑗=𝑚2𝑖−1+1}𝑏^𝑖𝑗⁺=^𝑚2𝑖+1_𝑗=𝑚_2𝑖−1+1𝑏^𝑖𝑗⁻=1.(3.12) Hence, by Proposition 17, {𝑣_𝑖} is isometrically equivalent to the canonical basis of 𝑙₁.
Now, if we define 𝑦_𝑛=𝛼_𝑛𝑥_2𝑛−1−𝛽_𝑛𝑥_2𝑛∈𝑌, it is straightforward to see that ‖𝑦_𝑛−𝑣_𝑛‖_1∞<𝜀_𝑛 and by the above remark, {𝑦_𝑛} is an 𝑎𝑖𝑙₁-sequence.
Finally in [13] Dowling and Lennard proved that if a Banach space contains an 𝑎𝑖𝑙₁-sequence, then it does not have the FPP. Hence 𝑌 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.