# The Fixed Point Property inc0 with an Equivalent Norm

Authors: Berta Gamboa de Buen; Fernando Núñez-Medina
Journal: Abstract and Applied Analysis
 (2011)

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## Abstract

We study the fixed point property (FPP) in the Banach spacec0 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 l1∞ and that every infinite-dimensional subspace of l1∞ does not have the fixed point property.


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## Body

## 1. Introduction

We start with some notations and terminologies. LetK be a nonempty, convex, closed and bounded subset of a Banach space (X,∥·∥). A mapping T:K→K is said to be nonexpansive if (1.1)‖Tx-Ty‖≤‖x-y‖,x,y∈K.
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 spacec0 with the equivalent norm ∥·∥D defined by (1.2)‖x‖D=sup⁡i,j∈N|xi-xj|,x={xi}∈c0.
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 {yn}n⊂c0 such that for all {tn}n∈l1,(1.3)sup⁡n∈N(1+ɛn)-1|∑i=n∞ti|≤‖∑n=1∞tnyn‖≤sup⁡n∈N(1+ɛn)|∑i=n∞ti|,
for some {ɛn}n⊂ℝ with 0≤ɛn+1≤ɛn and limnɛn=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 subspaceY 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 signsΘ={θi}, we define the Θ-basis of c0 which is equivalent to the summing basis and define the corresponding asymptotically isometric Θ-basic sequence, aiΘbc0D sequence for short. We prove that if Θ1≠±Θ2, then the aiΘ1bc0D and aiΘ2bc0D 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 aiΘ1bc0D sequence, and does not contain aiΘ2bc0D sequences. We also show that the aisbc0 and aiΘbc0D 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 aiΘbc0D sequences.Next we prove that if a convex and closed subsetK of a Banach space contains an asymptotically isometric c0D-summing basic sequence, that is, an aiΘbc0D sequence, where Θ is such that θi=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)* 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 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 {ξn} the summing basis of c0, that is, ξn=∑i=1nei,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{uk}k⊂(c0,∥·∥D) be a block basis of {en} with uk=∑i=pkqkaiei,1≤p1≤q1<p2≤q2<⋯. If ∥uk∥∞=1=aik, for some pk≤ik≤qk, and yk=(1/2)(u2k-u2k-1), then the space span¯{yk} is isometric to (c0,∥·∥∞).Proof.
Since∥uk∥∞=1=aik for every k∈ℕ, then |aj|≤1 for all j∈ℕ and
(2.1)max⁡p2k-1≤i≤q2k-1,p2k≤j≤q2k|ai+aj|=ai2k-1+ai2k=2.
Hence, it is straightforward to see that ∥∑k=1ntkyk∥D=∥∑k=1ntkek∥∞.In the following theorem, we will show, using some results proved by Dowling et al. [4, 8], that every infinite-dimensional subspace Y of c0D fails to have the FPP.Theorem 2.
LetY be an infinite-dimensional subspace of c0D. Then Y has a complemented asymptotically isometric copy of c0 and thus Y does not have the FPP.Proof.
Let{ɛk}k⊂(0,1) be a sequence such that ɛk+1<ɛk,k∈ℕ and ɛk→0. As in [9] we construct sequences {nk}⊂ℕ and {yk}k⊂Y such that nk<nk+1,yk=∑i=nk∞αikei,∥yk∥∞=1, and
(2.2)sup⁡i≥nk+1|αij|<ɛk+24k∀j=1,…,k,andeveryk∈N.
Since ∥yk∥∞=1, taking -yk instead of yk, if necessary, we can suppose that there exists nk≤rk<nk+1 such that
(2.3)αrkk=1.
Definexk=(y2k-1-y2k)/2.Then, by (2.3) and (2.2), we get that 1-(ɛk/2)<∥xk∥D<1+(ɛk/2) and
(2.4)∑k=1∞tkxk=12∑k=1∞tk(∑i=n2k-1∞αi2k-1ei-∑i=n2k∞αi2kei)=12∑k=1∞tk(∑i=n2k-1∞(αi2k-1-αi2k)ei)=12∑k=1∞(∑i=n2k-1n2k+1-1(∑j=1ktj(αi2j-1-αi2j)ei)),
where αi2k=0 for i=n2k-1,…n2k-1,k∈ℕ. Then by (2.3) and (2.2), if k>1, we get
(2.5)‖∑n=1∞tnxn‖D≥12max⁡n2k-1≤r<n2k,n2k≤s<n2k+1|∑j=1ktj(αr2j-1-αr2j-αs2j-1+αs2j)|≥12|∑j=1ktj(αr2k-12j-1-αr2k-12j-αr2k2j-1+αr2k2j)|≥12|tk||αr2k-12k-1-αr2k2k-1+αr2k2k|-12∑j=1k-1|tj||αr2k-12j-1-αr2k-12j-αr2k2j-1+αr2k2j|≥12|tk|(|αr2k-12k-1+αr2k2k|-|αr2k2k-1|)-12∑j=1k-1|tj|(|αr2k-12j-1|+|αr2k-12j|+|αr2k2j-1|+|αr2k2j|)≥12|tk|(2-ɛk)-12∑j=1k-1|tj|ɛkk≥|tk|(1-ɛk2)-max⁡1≤j≤k|tj|ɛk2.
On the other hand, if n2k-1≤r<n2k+1,n2m-1≤s<n2m+1,k≤m, using (2.2), we get
(2.6)12|∑j=1ktj(αr2j-1-αr2j)-∑j=1mtj(αs2j-1-αs2j)|≤12[|tk|(|αr2k-1|+|αr2k|)+|tm|(|αs2m-1|+|αs2m|)]+12[∑j=1k-1|tj|(|αr2j-1|+|αr2j|)+∑j=1m-1|tj|(|αs2j-1|+|αs2j|)]≤12[|tk|(1+ɛk)+|tm|(1+ɛm)+∑j=1k-1|tj|ɛkk-1+∑j=1m-1|tj|ɛmm-1]≤12[|tk|(1+ɛk)+|tm|(1+ɛm)+max⁡1≤j<k|tj|ɛk+max⁡1≤j<m|tj|ɛm]≤12[max⁡1≤j≤k|tj|(1+ɛk)+max⁡1≤j≤m|tj|(1+ɛm)]≤sup⁡n∈N((1+ɛn)max⁡1≤j≤n|tj|)≤sup⁡n∈N(1+ɛn)|tn|.
Then we obtain
(2.7)sup⁡n∈N(|tn|(1-ɛn2)-max⁡1≤j≤n|tj|ɛn2)≤‖∑n=1∞tnxn‖D≤sup⁡n∈N(1+ɛn)|tn|.
Now, define zn=xn/(1+ɛn) and m=(1-ɛ1)/(1+ɛ1);  then (1-ɛn)/(1+ɛn)≤∥zn∥D and limn∥zn∥D=1. On the other hand,
(2.8)(1+ɛ1)msup⁡n∈N|tn|=(1-ɛ1)sup⁡n∈N|tn|=(1-ɛ12)sup⁡n∈N|tn|-ɛ12sup⁡n∈N|tn|≤sup⁡n∈N(|tn|(1-ɛn2)-max⁡1≤j≤n|tj|ɛn2).
Thus
(2.9)msup⁡n∈N|tn|≤sup⁡n∈N(|tn|(1-ɛn2)-max⁡1≤j≤n|tj|ɛn2)≤‖∑n=1∞tnzn‖D≤sup⁡n∈N|tn|.
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{ξn} be the c0 summing basis. Then
(2.10)C={∑n=1∞λnξn:λn≥0,∑n=1∞λn=1}
does not have aisbc0 sequences with the norm ∥·∥D.Proof.
Suppose that{yn} is an aisbc0 sequence in C with ∥·∥D for some sequence {ɛn}. Then yn=∑i=1∞λinξi for some sequence {λin} such that λin≥0 and ∑i=1∞λin=1. Fix 0<ɛ<1/4. Passing to a subsequence we can suppose that ɛn+1≤ɛn<(1/2)-2ɛ and 1/(1+ɛn)>1-ɛ,n∈ℕ.
Assume first that there existsM∈ℕ such that for every n≥M, ∑i=M+1∞λin≤(1/2)-ɛ. Let un=∑i=1Mλinξi and vn=∑i=M+1∞λinξi; then yn=un+vn. Since {un}⊂[ξi]i=1M is bounded and dim[ξi]i=1M=M, passing to another subsequence we can suppose that un→u for some u∈C. Then, there exist n1,n2∈ℕ with M≤n1<n2 such that
(2.11)‖un1-un2‖D<ɛ.
Since ∑i=M+1∞λin≤(1/2)-ɛ,n≥M, we also get
(2.12)‖vn1-vn2‖D=max⁡M+1≤r≤k<∞|∑i=rkλin1-∑i=rkλin2|≤∑i=M+1∞λin1+∑i=M+1∞λin2≤1-2ɛ.
Hence ∥yn1-yn2∥D≤1-ɛ. On the other hand, since {yn} is an aisbc0 sequence, we have that ∥yn1-yn2∥D≥1/(1+ɛn2), which contradicts the fact that 1/(1+ɛn2)>1-ɛ.
Assume now that for allM∈ℕ, there exist n≥M such that ∑i=M+1∞λin>(1/2)-ɛ. Denote each yn by {αin}=∑i=1∞αinen, where {en} is the canonical basis of c0. Then  αin=∑j=i∞λjn. Since y1,y2∈c0, there exists M∈ℕ such that
(2.13)αi1,αi2<ɛ2,i≥M.
By hypothesis, there exists n0∈ℕ such that ∑i=M+1∞λin0>(1/2)-ɛ. Then
(2.14)32-2ɛ≤‖y1+y2-yn0‖D.
On the other hand, since {yn} is an aisbc0 sequence, we have that ∥y1+y2-yn0∥D≤1+ɛ1, which contradicts the fact that ɛ1<(1/2)-2ɛ.In view of the last proposition and motivated by the behavior of thec0 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{xn} be a bounded basic sequence in a Banach space X. We say that {xn} is a convexly closed sequence if the set
(2.15)C={∑n=1∞tnxn:tn≥0,∑n=1∞tn=1}
is closed, that is, if conv¯{xn}=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 thec0 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 lp, 1<p<∞, are not convexly closed.Definition 5.
Let{xn} be a sequence in a Banach space X. We say that {xn} is an asymptotically isometric c0D-summing basic sequence, aisbc0D sequence for short, if {xn} is convexly closed and there exists {ɛn}⊂(0,∞) such that ɛn↘0 and
(2.16)sup⁡1≤n≤m<∞(1+ɛm)-1|∑k=nmtk|≤‖∑n=1∞tnxn‖≤sup⁡1≤n≤m<∞(1+ɛm)|∑k=nmtk|,∀{tn}∈l1.Now, we prove that the analogous of the operator defined in [2] is still contractive and then Banach spaces containing aisbc0D sequences does not have the FPP.Proposition 6.
LetK be a nonempty, convex, closed and bounded subset of a Banach space X. Let {ɛn}⊂(0,∞) be a sequence such that ɛn→0 and ɛn<2-14-n,n≥2. If K contains an aisbc0D sequence with this {ɛn}, 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{xn} be an aisbc0D sequence in K with {ɛn}⊂(0,∞) such that ɛn<2-14-n,n≥2. Set
(2.17)C=conv¯{xn}={∑n=1∞tnxn:tn≥0,n∈Ny∑n=1∞tn=1}⊂K.
Thus C is nonempty, convex, closed and bounded. Define Txn=∑j=1∞((xn+j)/2j),n∈ℕ, and extend T linearly to C, that is, if x=∑n=1∞tnxn∈C then define T(∑n=1∞tnxn)=∑n=1∞tnTxn. 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=1∞tnxn andy=∑n=1∞snxn,with tn,sn≥0,and ∑n=1∞tn=∑n=1∞sn=1. Let βn=tn-sn,n∈ℕ, such that ∑n=1∞βn=0. As in [2] we have
(2.18)T(x)-T(y)=∑n=1∞Bnxn,
where B1=0 and Bn=(β1/2n-1)+(β2/2n-2)+⋯+(βn-1/2),n≥2. Consequently,
(2.19)‖T(x)-T(y)‖=‖∑n=1∞Bnxn‖≤sup⁡1≤n≤m<∞(1+ɛm)|∑k=nmBk|.
Take n,m∈ℕ with n≤m. Since
(2.20)∑k=nmBk=β12n-1+β22n-2+⋯+βn-12+β12n+β22n-1+⋯+βn-122+βn2+⋯+β12m-1+β22m-2+⋯+βn-12m-(n-1)+βn2m-n+βn+12m-(n+1)+⋯+βm-12=12(βn-1+βn+⋯+βm-1)+122(βn-2+βn-1+⋯+βm-2)+⋯+12n-1(β1+β2+⋯+βm-(n-1))+⋯+12m-2(β1+β2)+12m-1(β1),
we have
(2.21)(1+ɛm)|∑k=nmBk|≤(1+ɛm)(1+2ɛm-1211+2ɛm-1|βn-1+βn+⋯+βm-1|+1+2ɛm-22211+2ɛm-2|βn-2+βn-1+⋯+βm-2|+⋯+1+2ɛm-(n-1)2n-111+2ɛm-(n-1)|β1+β2+⋯+βm-(n-1)|+⋯+1+2ɛ22m-211+2ɛ2|β1+β2|+1+2ɛ12m-111+2ɛ1|β1|)≤(sup⁡1≤i≤j≤m(1+2ɛj)-1|∑k=ijβk|)Qnm,
where
(2.22)Qnm=(1+ɛm)(1+2ɛm-12+1+2ɛm-222+⋯++1+2ɛm-(n-1)2n-1+1+2ɛm-n2n+⋯+1+2ɛ22m-2+1+2ɛ12m-1)≤(1+12⋅4m)[(12+122+⋯+12m-1)+(12⋅4m-1+⋯+12m-1⋅41)]=(1+12⋅4m)[(1-12m-1)+(122m-1+122m-2+⋯+12m+1)]<(1+14m)[(1-12m-1)+12m]<1.
Then we get
(2.23)sup⁡1≤n≤m<∞(1+ɛm)|∑k=nmBk|≤sup⁡1≤n≤m<∞(1+2ɛm)-1|∑k=nmβk|<sup⁡1≤n≤m<∞(1+ɛm)-1|∑k=nmβk|≤‖∑n=1∞βnxn‖=‖x-y‖.
Thus T is contractive.Next for any sequence of signs we will define a basis inc0 equivalent to {ξn}, 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={θi}i, that is, θi∈{-1,1},i∈ℕ, let {ζnΘ}n be the sequence defined by (2.24)ζnΘ=∑k=1nθkek,n∈N.
Since ∥∑n=1mtnξn∥∞=∥∑n=1mtnζnΘ∥∞ for all {tn}n=1m⊂𝕂, we get that {ζnΘ} is a basis of c0 equivalent to the c0 summing basis. The sequence {ζnΘ} is called the Θ-basis of c0. Let Θ0={θi}, where θi=1,i∈ℕ. Then the Θ0-basis of c0 is the c0 summing basis. If we define C={∑n=1∞tnζnΘ:tn≥0and∑n=1∞tn=1}, then C is nonempty, convex and bounded. Since ∥·∥∞ and ∥·∥D are equivalent, we have that {ζnΘ} is convexly closed in (c0,∥·∥D).The setC={∑n=1∞tnζnΘ:tn≥0and∑n=1∞tn=1} is not ω-compact. The following result shows that the set C contains neither aisbc0D sequences nor aisbc0 sequences with the norm ∥·∥D if Θ≠±Θ0.Proposition 7.
ForΘ≠±Θ0, let {ζnΘ} be the Θ-basis of c0 considered in (c0,∥·∥D). If
(2.25)C={∑n=1∞tnζnΘ:tn≥0,∑n=1∞tn=1},
then the set C contains neither aisbc0D sequences nor aisbc0 sequences with the norm ∥·∥D.Proof.
Let{yk}⊂C. Then yk=∑n=1∞λnkζnΘ for some λnk≥0and ∑n=1∞λnk=1.Suppose that {yk} is an aisbc0D sequence (resp. an aisbc0 sequence) with the norm ∥·∥D. Let n0=min{n:θn≠θ1}. If there exists 0<ρ<1 such that ∑i=1n0-1λik≤1-ρ for all k≥1, then for all k≥1,
(2.26)‖yk‖D≥∑n=1∞λnk+∑n=n0∞λnk≥1+ρ.
Since {yk} is an aisbc0D sequence (resp. an aisbc0 sequence) with the norm ∥·∥D, then ∥yk∥D≤1+ɛk→1 and this is impossible. Now, if limsupk∑i=1n0-1λik=1, as in the proof of Proposition 3, we obtain a subsequence {yki} of {yk} with ∥yki-yki+1∥D→0.Since {yn} is an aisbc0D with the norm ∥·∥D, then (1+ɛki)-1≤∥yki-yki+1∥D (resp. (1+ɛki+1)-1≤∥yki-yki+1∥D) and making i→∞ we get that 1≤0. This contradiction proves the result.Although the setC of the last proposition has neither aisbc0D sequences nor aisbc0 sequences, for some Θ it does not have the FPP.ForΘ={θi}, let Fn be the set such that if i,j∈Fn, then θi=θj, and if i∈Fn+1 and j∈Fn, then θi≠θj.Denote by rn the cardinality of Fn. If rn<∞, define p0=0,pn-1=minFn-1, and pn=maxFn.Proposition 8.
LetΘ≠±Θ0. Then
(2.27)C={∑n=1∞tnζnΘ:tn≥0,∑n=1∞tn=1}
does not have the FPP in the following cases.(1)
There existsk≥1 such that rn≤rn+k<∞,n∈ℕ.(2)
r1=1 and r2=∞.(3)
There exists{in}, with i1>1, such that for any k,l∈ℕ with ik-1<l<ik we have θl=θik and also θk≠θik for all k≥2  or θk=θik for all k≥2.Proof.
LetΘ≠±Θ0.
(1) If there existsk≥1 such that rn≤rn+k<∞,n∈ℕ, define qn=∑j=n-(k-2)n+1rj,n≥k and T:C→C by
(2.28)T∑n=1∞tnζnΘ=T∑n=1∞∑i=pn-1+1pntiζiΘ=∑n=k∞∑i=wnpn+1ti-qnζiΘ,
where wn=pn+rn+1-rn+1-k+1. The idea is to translate the coefficients of ∑n=1∞tnζnΘ in the block Fn into the last rn terms of the block Fn+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 θi and θj with i∈Fn and j∈Fn+k are the same and are different if k is odd. Now let x=∑n=1∞tnζnΘ, y=∑n=1∞snζnΘ, and x-y=∑n=1∞αnζnΘ. Then αn=tn-sn and ∑n=1∞αn=0. Hence
(2.29)x-y=∑n=1∞∑i=pn-1+2pnθpn(∑n=i∞αn)ei,(2.30)T(x-y)=∑n=k∞(θpn+1∑n=i-qn∞αn)(∑i=pn+1wnei)∑i=wn+1pn+1θpn+1(∑n=i-qn∞αn)ei
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 thatr1=1 and r2=∞. In this case, define T∑n=1∞tnζnΘ=∑n=1∞tnζn+1Θ. 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=1∞tnζnΘ=∑n=1∞tnζinΘ is nonexpansive and does not have fixed points.Proposition 9.
LetΘ≠±Θ0. Suppose Θ does not satisfy the hypotheses of the above proposition, and let {in} be a sequence with i1>1. Thenthe operator T:C→C defined by T∑n=1∞tnζnΘ=∑n=1∞tnζinΘ is expansive.Proof.
SinceΘ does not satisfy the hypotheses (1) and (2) of the above proposition, there are three possibilities.(I)
rn<∞ for every n≥2; then for every k there exists n such that rn+k<rn.(II)
r2=∞; then r1>1.(III)
There existsk>2 such that rk=∞.
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 k and l with ik-1<l<ik such that θl≠θik or there exists k1≥2 with θk1=θik1 and there exists k2≥2  with θk2≠θik2.
Case 1.
For everyk there exists n such that rn+k<rn.
There are two subcases.
Subcase 1.1.
There arek and l with ik-1<l<ik such that θl≠θik.
Letx=(1/8)ζ1Θ+(3/8)ζk-1Θ+(1/2)ζkΘ and y=(1/16)ζ1Θ+(3/16)ζk-1Θ+(3/4)ζkΘ.Then x-y=(1/16)ζ1Θ+(3/16)ζk-1Θ-(1/4)ζkΘ=-(1/16)∑i=2k-1θiei-(1/4)θkek and ∥x-y∥D≤5/16.On the other hand, Tx-Ty=(1/16)ζi1Θ+(3/16)ζik-1Θ-(1/4)ζikΘ=-(1/16)∑j=i1+1ik-1θjej-(1/4)∑j=ik-1+1ikθjej and ∥Tx-Ty∥D=1/2.
Subcase 1.2.
For anyk∈ℕ and l withik-1<l<ik, we have θl=θik.
There are two subsubcases. (1)θ1=θi1 and (2) θ1≠θi1.
(1)θ1=θi1
Ifθk=θik for every k; then we would have F1=ℕ, which implies Θ=±Θ0. Then there is k such that  θk≠θik. Let s=min{l:θl≠θil}. Then s>1.
There are two possibilities: (A) there existsr>s such that θr=θir and (B) θk≠θik for all k≥s.(A)
Letk+1=min{r>s:θr=θir}. We need to consider the following cases.
(a)
θk=θk+1.Letx=(1/2)ζk-1Θ+(1/2)ζk+1Θ and y=(3/4)ζk-1Θ+(1/4)ζk+1Θ.Then x-y=-(1/4)ζk-1Θ+(1/4)ζk+1Θ=θk+1((1/4)ek+(1/4)ek+1) and ∥x-y∥D=1/4.On the other hand, Tx-Ty=-(1/4)ζik-1Θ+(1/4)ζik+1Θ=(1/4)∑j=ik-1+1ikθjej+(1/4)∑j=ik+1ik+1θjej=-(1/4)θk+1∑j=ik-1+1ikej+(1/4)θk+1∑j=ik+1ik+1ej and ∥Tx-Ty∥D=1/2.(b)
θk≠θk+1.Letx=(1/2)ζk-1Θ+(1/2)ζk+1Θ and y=(3/4)ζkΘ+(1/4)ζk+1Θ. Then x-y=(1/2)ζk-1Θ-(3/4)ζkΘ+(1/4)ζk+1Θ=-(1/2)θkek+(1/4)θk+1ek+1=θk+1((1/2)ek+(1/4)ek+1) and ∥x-y∥D=1/2. On the other hand, Tx-Ty=(1/2)ζik-1Θ-(3/4)ζikΘ+(1/4)ζik+1Θ=-(1/2)∑j=ik-1+1ikθjej+(1/4)∑j=ik+1ik+1θjej=-(1/2)θk∑j=ik-1+1ikej+(1/4)θk∑j=ik+1ik+1ej and ∥Tx-Ty∥D=3/4.(B)
θk≠θik for all k≥s. By hypothesis we have that s>2. There are two cases.
(a)
θs-1=θs. Then θis-1≠θis.Let x=(1/4)ζs-2Θ+(1/2)ζs-1Θ+(1/4)ζsΘ and y=(1/4)ζs-1Θ+(3/4)ζsΘ. Then x-y=(1/4)ζs-2Θ+(1/4)ζs-1Θ-(1/2)ζsΘ=-θs((1/4)es-1+(1/2)es) and ∥x-y∥D=1/2.On the other hand, Tx-Ty=(1/4)ζis-2Θ+(1/4)ζis-1Θ-(1/2)ζisΘ=-(1/4)θs∑j=is-2+1is-1ej+(1/2)θs∑j=is-1+1isej and ∥Tx-Ty∥D=3/4.(b)
θs-1≠θs. Then θis-1=θis.Let x=(1/4)ζs-2Θ+(1/4)ζs-1Θ+(1/2)ζsΘ and y=(3/4)ζs-1Θ+(1/4)ζsΘ. Then x-y=(1/4)ζs-2Θ-(1/2)ζs-1Θ+(1/4)ζsΘ=-(1/4)θs-1es-1+(1/4)θses=θs((1/4)es-1+(1/4)es) and ∥x-y∥D=1/4.On the other hand, Tx-Ty=(1/4)ζis-2Θ-(1/2)ζis-1Θ+(1/4)ζisΘ=-(1/4)θis-1∑j=is-2+1is-1ej+(1/4)θis∑j=is-1+1isej=θs-1(-(1/4)∑j=is-2+1is-1ej+(1/4)∑j=is-1+1isej) and ∥Tx-Ty∥D=1/2.
(2)θ1≠θi1
In this case there existsk such that  θk=θik. If s=min{l:θl=θil}, then s>1.
Hence consider the cases: (A) there existsr>s such that θr≠θir and (B) θk=θik for all k≥s and proceed as in the Case (1).
Case 2.
r2=∞ and r1>1.
Thenθp1≠θip1 with 1<p1.Hence we can proceed as in Subcase 1.2(1)(A) above taking k=p1.
Case 3.
There iss>1 such that rs+1=∞.
Thenθps≠θips with 1<ps.Hence we can proceed as in Subcase 1.2(1)(A) above taking k=ps.Next, for every selection of signsΘ≠±Θ0, we will define the asymptotically isometric c0D-Θ-basic sequences. To this end, let us consider the following notation.Let(2.31)SΘ={(n,m):θn=θm},DΘ={(n,m):θn≠θm}.Definition 10.
Let{xn} be a sequence in a Banach space X. We say that {xn} is an asymptotically isometric c0D-Θ-basic sequence (aiΘbc0D sequence for short) if {xn} is convexly closed and there exists {ɛnΘ}⊂(0,(1/2)) such that ɛnΘ↘0, and
(2.32)L({ɛnΘ},{tn},SΘ)∨L({ɛnΘ},{tn},DΘ)≤‖∑n=1∞tnxn‖≤R({ɛnΘ},{tn},SΘ)∨R({ɛnΘ},{tn},DΘ)
holds for all {tn}∈l1, where
(2.33)L({ɛnΘ},{tn},SΘ)=(sup⁡n<l,(n,l)∈SΘ(1+ɛl-1Θ)-1|∑k=nl-1tk|),L({ɛnΘ},{tn},DΘ)=(sup⁡n<l,(n,l)∈DΘ(1+ɛl-1Θ)-1|∑k=nl-1tk+2∑k=l∞tk|),R({ɛnΘ},{tn},SΘ)=(sup⁡n<l,(n,l)∈SΘ(1+ɛl-1Θ)|∑k=nl-1tk|),R({ɛnΘ},{tn},DΘ)=(sup⁡n<l,(n,l)∈DΘ(1+ɛl-1Θ)|∑k=nl-1tk+2∑k=l∞tk|).We are interested inaiΘbc0D sequences for which the numbers ɛnΘ of Definition 10 are small. We are taking {ɛnΘ}⊂(0,(1/2)).We know that the setC of Proposition 3 does not have aisbc0 sequences. Now we also prove that C does not contain aiΘbc0D sequences with the norm ∥·∥D if Θ≠±Θ0.Proposition 11.
LetΘ≠±Θ0. The set C={∑n=1∞tnξn:tn≥0and∑n=1∞tn=1} does not contain aiΘbc0D sequences with the norm ∥·∥D.Proof.
Let{yk}⊂C. Then yk=∑n=1∞λnkξn for some λnk≥0with ∑n=1∞λnk=1.Suppose that {yk} is an aiΘbc0D with ∥·∥D. Since Θ≠±Θ0, there exist m∈ℕ and {nk}⊂ℕ with n1<n2<⋯, such that for all k∈ℕ, m<nk and θnk≠θm. Let tn=0 for n≠m,nk and tm=tnk=1. Thus
(2.34)(1+ɛnk-1Θ)-13≤L({ɛnΘ},{tn},SΘ)∨L({ɛnΘ},{tn},DΘ)≤‖∑n=1∞tnyn‖D=‖ym+ynk‖D=2.
Since (2.34) holds for all k∈ℕ, making k→∞ in (2.34), we get that 3≤2, which is a contradiction.Proposition 12.
LetΘ1={θi1}i and Θ2={θi2}i such that Θ1≠±Θ2 and Θ1,Θ2≠±Θ0. Let {ζnΘ1} be the Θ1-basis of c0 considered in (c0,∥·∥D) and let
(2.35)C(Θ1)={∑n=1∞tnζnΘ1:tn≥0,∑n=1∞tn=1}.
The set C(Θ1) does not contain aiΘ2bc0D sequences with the norm ∥·∥D.Proof.
Let{yk}⊂C. Then yk=∑n=1∞λnkζnΘ1 for some λnk≥0 with ∑n=1∞λnk=1.Suppose that {yk} is an aiΘ2bc0D with the norm ∥·∥D.
Suppose firstθ11=θ12; since Θ1≠Θ2, there exists m>1 such that θm1≠θm2.
There are two cases.
Case 1.
(1,m)∈𝔖Θ1. In this case (1,m)∈𝔇Θ2. Let tn=0  for n≠1,m and t1=tm=1. Thus
(2.36)(1+ɛm-1Θ2)-13≤L({ɛnΘ2},{tn},SΘ2)∨L({ɛnΘ2},{tn},DΘ2)≤‖∑n=1∞tnyn‖D=‖y1+ym‖D≤2.
Since ɛm-1Θ2<1/2, we get a contradiction.
Case 2.
(1,m)∈𝔇Θ1. In this case (1,m)∈𝔖Θ2. Let tn=0  for n≠1,m and t1=tm=1. Thus
(2.37)2≤∑n=1∞λn1+∑n=1∞λnm+∑n=m∞λn1+∑n=m∞λnm≤‖∑n=1∞tnyn‖D=‖y1+ym‖D≤R({ɛnΘ2},{tn},SΘ2)∨R({ɛnΘ2},{tn},DΘ2)≤(1+ɛm-1Θ2).
Since ɛm-1Θ2<1/2, we get a contradiction.
Suppose nowθ11≠θ12; since Θ1≠-Θ2, there exists m such that θm1=θm2.
There are two cases.
Case 1.
(1,m)∈𝔖Θ1;  in this case (1,m)∈𝔇Θ2. Let tn=0  for n≠1,m and t1=tm=1. Thus
(2.38)(1+ɛm-1Θ2)-13≤L({ɛnΘ2},{tn},SΘ2)∨L({ɛnΘ2},{tn},DΘ2)≤‖∑n=1∞tnyn‖D=‖y1+ym‖D≤2.
Since ɛm-1Θ2<1/2, we get a contradiction.
Case 2.
(1,m)∈𝔇Θ1. In this case (1,m)∈𝔖Θ2. Let tn=0  for n≠1,m and t1=tm=1. Thus
(2.39)2≤∑n=1∞λn1+∑n=1∞λnm+∑n=m∞λn1+∑n=m∞λnm≤‖∑n=1∞tnyn‖D=‖y1+ym‖D≤R({ɛnΘ2},{tn},SΘ2)∨R({ɛnΘ2},{tn},DΘ2)≤(1+ɛm-1Θ2).
Since ɛm-1Θ2<1/2, we get a contradiction.Propositions3, 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 aiΘbc0D sequences) to have a similar result.


## 3. The Space(c0,∥·∥D)*

It is known that the dual of the Bynum spacec01 is the Bynum space l1∞. 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 l1∞. Let us suppose then that 𝕂=ℝ. First we calculate the extreme points of the unit ball of (c0,∥·∥D).Lemma 13.
LetX=(c0,∥·∥D). Then we have
(3.1)E(BX)={{xn}∈SX:xn∈{1,0},n∈N}∪{{xn}∈SX:xn∈{-1,0},n∈N}.Proof.
First note that if{xn}∈SX then |xn-xm|≤1,n,m∈ℕ and |xn|≤1,n∈ℕ. Consequently, if {xn}∈SX with xn0=1 for some n0∈ℕ, then 0≤xn≤1, for all n∈ℕ. Analogously if {xn}∈SX with xn0=-1 for some n0∈ℕ, then -1≤xn≤0, for alln∈ℕ. Let A={{xn}∈SX:xn∈{1,0},n∈ℕ} and B={{xn}∈SX:xn∈{-1,0},n∈ℕ}. Thus A,B⊂SX.
Takex={xn}∈A and suppose that x=(y+z)/2 with y,z∈SX. Also suppose that y={yn} and z={zn}. Since x∈A, there exists n0∈ℕ such that xn0=1. Since xn=(yn+zn)/2 and yn,zn≤1, if xn=1 for some n∈ℕ, we have that xn=yn=zn. Thus xn0=yn0=zn0=1. On the other hand, if xn=0 for some n∈ℕ, we also have that xn=yn=zn, because if yn<0 we get that |yn-yn0|>1, which contradicts that |yn-ym|≤1,n,m∈ℕ and if yn>0 then zn<0 and we also have a contradiction. Therefore, x=y=z. Hence x∈ℰ(BX). Thus A⊂ℰ(BX). Analogously B⊂ℰ(BX).
Take nowx={xn}∈SX∖{A∪B}. Then there exists n0∈ℕ such that 0<|xn0|<1. Let a=infnxn and b=supnxn. If xn0∈(a,b), define c=min(|xn0-a|,|xn0-b|),yn=zn=xn,n≠n0,yn0=xn0-c, and zn0=xn0+c. Thus, xn=(y+z)/2 with y,z∈SX and x≠y,x≠z. Therefore, x∈SX∖ℰ(BX). Suppose now that xn0=a or xn0=b. Since 0<|xn0|<1 and supn,m∈ℕ|xn-xm|=1, we have that 0∈(a,b). Since xn→0, there exists n1>n0 such that xn1∈(a,b), which implies that x∈SX∖ℰ(BX). Consequently, ℰ(BX)⊂A∪B.Theorem 14.
Letf∈(c0,∥·∥D)*. There exists a unique sequence {cn}∈l1 such that f=∑n=1∞cnen* and
(3.2)‖f‖D=max⁡(∑n=1∞cn+,∑n=1∞cn-),
where cn+=max(cn,0) and cn-=-min(cn,0).Proof.
Letf∈(c0,∥·∥D)*. Since {en} is a shrinking basis of (c0,∥·∥D), there exists a unique sequence {cn}⊂𝕂 such that f=∑n=1∞cnen*. As sets (c0,∥·∥D)*=(c0)* and hence f∈(c0)*. Thus f=R{an} where R:l1→c0* is the Riesz representation. Consequently,
(3.3)cn=f(en)=R{an}(en)=an.
Therefore, {cn}={an}∈l1. Thus
(3.4)‖f‖D=sup⁡x∈BX|f(x)|=sup⁡x∈E(BX)|f(x)|=sup⁡{|∑n∈Fcn|:F⊂N,Ffinite}=max⁡(∑n=1∞cn+,∑n=1∞cn-),
where cn+=max(cn,0) and cn-=-min(cn,0).Corollary 15.
(c0,∥·∥D)* is the Bynum space l1∞ and it has the ω-FPP.Remark 16.
It is well known thatl1(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={Ts: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 ofl1∞ 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 {xn}n⊂X and {ɛn}⊂(0,1),ɛn→0 such that for every k∈ℕ and every scalars b1,…,bk,(3.5)∑i=1k(1-ɛi)|bi|≤‖∑i=1kbixi‖≤∑i=1k(1+ɛi)|bi|.
In this case we say that {xn}n is an asymptotically isometric l1-sequence (ail1-sequence for short).Observe that if{yn}n is another sequence in X such that ∥yn-xn∥<δn for all n, where {ɛn+δn}⊂(0,1) and δn→0, then for every k and every scalars b1,…,bk,(3.6)∑i=1k(1-ɛi-δi)|bi|≤‖∑i=1kbiyi‖≤∑i=1k(1+ɛi+δi)|bi|
and {yn} is also an ail1-sequence.Proposition 17.
Let{ui}i⊂l1∞, and let {ni} be a strictly increasing sequence in ℕ such that ui=∑j=ni+1ni+1ajiej.If ∑j=ni+1ni+1(aji)+=∑j=ni+1ni+1(aji)-, then {ui}i is isometrically equivalent to the canonical basis in l1, that is, for every k∈ℕ and every scalars b1,…,bk, we have that ∥∑i=1kbiui∥=∑i=1k|bi|.Proof.
Letb1,…,bk be scalars; then
(3.7)∑i=1k|bi|≤∑i=1kbi+∑j=ni+1ni+1(aji)++∑i=1kbi-∑j=ni+1ni+1(aji)-=∑i=1k∑j=ni+1ni+1(biaji)+≤‖∑i=1kbiui‖1∞≤∑i=1k|bi|.Theorem 18.
Every infinite-dimensional subspace ofl1∞ contains an asymptotically isometric copy of l1 and hence it does not have the FPP.Proof.
LetY be an infinite-dimensional subspace of l1∞,{ɛn}⊂(0,(1/2)),ɛn↘0 and {xn} a sequence in SY such that xi=∑j=mi+1∞ajiej, where 0=m0<m1<⋯ and ∑j=mi+1+1∞|aji|<ɛi/8.Define
(3.8)wi=∑j=mi+1mi+1ajiej,ci+=1‖wi‖1∞∑j=mi+1mi+1(aji)+≤1,ci-=1‖wi‖1∞∑j=mi+1mi+1(aji)-≤1.
Changing wi by -wi, if necessary, we can assume that ci+=1,n∈ℕ. If there is a sequence {ki} such that cki-=1, then by Proposition 17, {wki/∥wki∥1∞} is isometrically equivalent to the canonical basis of l1. It is straightforward to see that ∥xki-(wki/∥wki∥1∞)∥1∞<(1/4)ɛki. Then by the above remark, {xki} is an ail1-sequence.
Suppose thatci-≠1 for all i and let
(3.9)αi=1-c2i-1-c2i-c2i-1-,βi=1-c2i-1-1-c2i-1-c2i-.
Then 0≤αi<1,0≤βi<1 and
(3.10)αic2i-1++βic2i-=αic2i-1-+βic2i+=1.
Now let(3.11)vi=αiw2i-1‖w2i-1‖1∞-βiw2i‖w2i‖1∞.
Suppose that vi=∑j=m2i-1+1m2i+1bjiej. It is easy to check, using (3.10), that
(3.12)∑j=m2i-1+1m2i+1(bji)+=∑j=m2i-1+1m2i+1(bji)-=1.
Hence, by Proposition 17, {vi} is isometrically equivalent to the canonical basis of l1.
Now, if we defineyn=αnx2n-1-βnx2n∈Y, it is straightforward to see that ∥yn-vn∥1∞<ɛn and by the above remark, {yn} 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.



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*Source: 574614-2011-10-29.xml*