# On Fixed Point Property under Lipschitz and Uniform Embeddings

Authors: Jichao Zhang; Lingxin Bao; Lili Su
Journal: Journal of Function Spaces
 (2018)

---

## Abstract

We first present a generalization ofω⁎-Gâteaux differentiability theorems of Lipschitz mappings from open sets to those closed convex sets admitting nonsupport points and then show that every nonempty bounded closed convex subset of a Banach space has the fixed point property for isometries if it Lipschitz embeds into a super reflexive space. With the application of Baudier-Lancien-Schlumprecht’s theorem, we finally show that every nonempty bounded closed convex subset of a Banach space has the fixed point property for continuous affine mappings if it uniformly embeds into the Tsirelson space T⁎.


---

## Body

## 1. Introduction

The infinite Rademacher theorem [1, 2] states that every Lipschitz mapping from a separable Banach space to a Banach space with the Radon-Nikodým property (in short, RNP) is almost everywhere Gâteaux differentiable, i.e., everywhere Gâteaux differentiable off an Aronszajn null set. It is shown in [3, 4] that Aronszajn null sets (introduced by Aronszajn [1]), Gauss null sets (introduced by Phelps [5]), and cube null sets (introduced by Mankiewicz [6]) coincide. In particular, Aronszajn null sets and Lebesgue null sets coincide in a finite dimensional Banach space [3]. To establish the localized version of the infinite Rademacher theorem, Cheng and Zhang [7] substituted closed convex subsets of a Banach space X for X and proved that every Lipschitz mapping from a closed convex subset C of a Banach space with nonempty nonsupport point set N(C) to a Banach space with the RNP is almost everywhere Gâteaux differentiable.As is well known, dual Banach spaces without separability may fail RNP. The Lipschitz mappings from a separable Banach space to a dual Banach space without separability need not be Gâteaux differentiable anywhere. To circumvent this obstruction, a weaker notion of aω∗-Gâteaux differentiability was introduced (see, for instance, [3, 8]). Heinrich and Mankiewicz [9] proved the infinite ω∗-Rademacher theorem: every Lipschitz mapping from a separable Banach space to the dual space of a separable Banach space is almost everywhere ω∗-Gâteaux differentiable. Therefore, a question naturally arises.Question 1.
The question is whether the infiniteω∗-Rademacher theorem can be localized to those Lipschitz mappings from a closed convex set C admitting nonempty nonsupport point set N(C) to the dual space of a separable Banach space.The aim of this paper focuses on the study of the question above. Based on the ideas of the infiniteω∗-Rademacher theorem, we shall prove the following theorem.Theorem 2.
Suppose thatC is a closed convex set of a separable Banach space X with N(C)≠∅ and that Y is a separable Banach space. Let f:C→Y∗ be a Lipschitz mapping. Then f is ω∗-Gâteaux differentiable off a Aronszajn null set of N(C).As an application of Theorem2, we obtain the following result.Theorem 3.
Every nonempty bounded closed convex subsetC of a Banach space X has the fixed point property for isometries if it Lipschitz embeds into a super reflexive space.With the application of Baudier-Lancien-Schlumprecht’s theorem, we finally get the following result.Theorem 4.
Every nonempty bounded closed convex subsetC of a Banach space X has the fixed point property for continuous affine mappings if it uniformly embeds into the Tsirelson space T∗.The letterX will always be a real Banach space and X∗ its dual. SX denotes the unit sphere of X. For a subset C⊂X, we denote by C¯ the closure of C. For simplicity, we also denote XC=span¯(C) and Cx=∪{λ(C-x):λ>0} for some x∈C.


## 2. A Proof of Theorem2

We first recall definitions of support points [5, 7] and Aronszajn null sets [1, 3].Definition 5.
Suppose thatC is a convex set in a Banach space X. A point x∈C is said to be a support point of C if there exists a non-zero functional x∗∈X∗ such that x∗,x=max{x∗,y:y∈C}.
We denote byS(C) the set of all support points of C and by N(C) the set of all nonsupport points of C.LetX be a Banach space. Let λ be the one dimensional Lebesgue measure. For each x∈X∖{0}, set(1)Λx=A⊂X:Ais  a  Borel  set  such  thatλA∩z+Rx=0for  everyz∈X,and for a finite or infinite sequence {xn} of nonzero points in X, set (2)Λxn=A=∪An,An∈Λxnfor  everyn∈N.Definition 6.
Suppose thatA is a Borel set in a separable Banach space X. The set A is called an Aronszajn null set (or simply, a null set) if(3)A∈∩Λxn:xnare  nonzero  vectors  such  that  spanxnis  dense  inX.Note thatN(C) is a Gδ set; hence it is a Borel set (see, for instance, [7, 10]).The following two lemmas are presented in [3, 8].Lemma 7.
LetY be a Banach space with the RNP. Then every Lipschitz function f from an open subset U of Rn into Y is Gâteaux differentiable off a null set of U.Lemma 8.
LetF be an n-dimensional subspace of Banach space X, and let {xk}k=1n be a basis for F. Let λn be the Lebesgue measure on F, and let A be a Borel subset of X such that λn(F∩(A+x))=0 for every x∈X. Then A∈Λ({xk}k=1n}).Definition 9.
Suppose thatC is a closed convex set of a separable Banach space X with N(C)≠∅ and that Y is a separable Banach space. Let f:C→Y∗ be a Lipschitz mapping. Then f is said to be ω∗-Gâteaux differentiable at x∈N(C) if there exists a bounded linear operator Tx:C→Y∗ such that for every h∈Cx and every y∈Y the following limit exists:(4)Txhy=limλ→0⁡fx+λh-fxyλ.In this case Tx is called the ω∗-derivative of f at x and is denote by Df∗(x).Theorem 10.
Suppose thatC is a closed convex set of a separable Banach space X with N(C)≠∅ and that Y is a separable Banach space. Let f:C→Y∗ be a Lipschitz mapping; i.e., there exists K>0 such that f(x)-f(y)≤Kx-y for all x,y∈C. Let {xn} be a sequence which consists of linearly independent vectors in N(C). For every n∈N, set Xn=span{xk:k≤n} and set fn≡f|C∩Xn. Then
(i)fn is ω∗-Gâteaux differentiable off a null set of N(C)∩Xn for every n∈N;
(ii)Dfn∗(x)≤K whenever Dfn∗(x) exists for every n∈N;
(iii) If, in addition, there existsk>0 such that f(x)-f(y)≥kx-y for all x,y∈C, then then Dfn∗(x):Xn→Y∗ is a linear isomorphism for almost every x∈N(C)∩Xn.Proof.
(i). Without loss of generality, we assume that0∈N(C). Let {ym} be a dense sequence in SY. Given n∈N, define φm(x)=fn(x)(ym) for every m∈N and x∈C. Then φm are Lipschitz real-valued functions on C. Since {xn} is a sequence which consists of linearly independent vectors in N(C) and Xn=span{xk:k≤n}, we obtain that span(N(C)∩Xn)=Xn, and hence, N(C)∩Xn is nonempty open in Xn (see also p.p, 9 in [11]).
Applying Lemma7 to Xn and N(C)∩Xn, then φm are Gâteaux differentiable off a null set of N(C)∩Xn. Let Dφm(x) denote the derivative of φm at x.
Consider the set(5)W=x∈NC∩Xn:Dφmxexists  for  everym∈N.Then W is not null set in N(C)∩Xn. Choose x∈W; then the following limits exists:(6)Dφmxh=limλ→0⁡fnx+λh-fnxymλ,for every m∈N, h∈Xn. Since f is Lipschitz, fn(x+λh)-fn(x)/λ is a bounded ω∗-Cauchy sequence when λ→0. By Alaoglu theorem, ω∗-limλ→0fn(x+λh)-fn(x)/λ exists in Y∗. Since Dφm(x) is linear for every m∈N, the limit ω∗-limλ→0fn(x+λh)-fn(x)/λ is linear, and hence it is a bounded linear operator from Xn into Y∗ because f is Lipschitz. Therefore, Dfn∗(x) exists for every n∈N and x∈W.
(ii). For everyn∈N, h∈Xn and x∈W, we have(7)Dfn∗xh=supm∈N⁡Dfn∗xhym=supm∈N⁡limλ→0⁡fnx+λh-fnxymλ≤Kh.This completes our assertion (ii).
(iii). We omit the proof because the proof of assertion (iii) is similar to the proof of assertion (c) of Theorem 14.2.18, p. 385 in [8].Proof of Theorem2.
Without loss of generality, we assume that0∈N(C). By Theorem 2.6 in [7], span(C)=span(N(C))=⋃n=1∞nN(C) is a dense subspace of X. Since X is separable, there exists a sequence {xn} of linearly independent vectors in N(C) such that span{xn} is dense in X. For each n∈N, set Xn=span{xk:k≤n}. Let(8)Dn=x∈NC:ω∗-limλ→0⁡fx+λh-fxλexists  for  everyh∈Xn.and let D=∩Dn. Since Dn is a Borel set, so is D. Therefore, for each x∈D,(9)Txh=ω∗-limλ→0⁡fx+λh-fxλ,h∈spanxndefines a bounded linear operator from span{xn} to Y∗.
Sincespan{xn} is dense in X, there exists a unique bounded linear extension T of Tx from span{xn} to X. Therefore, T=Df∗(x) and D is just the set of all ω∗-Gâteaux differentiability points of f in N(C). It remains to show that D is not null.
By Theorem10, (N(C)∖Dn)∩Xn is a null set in N(C)∩Xn. Given z∈X, we define fz by fz(x)=f(x-z),x∈(N(C)+z)∩Xn. Then fz is a Lipschitz mapping on set (N(C)+z)∩Xn, and by Lemma 7, (N(C)∖Dn+z)∩Xn is the set of all non-Gâteaux differentiability points of fz. By Lemma 8, N(C)∖Dn∈Λ({xk:k≤n}). Therefore, N(C)∖D∈Λ({xn}).
Next, for any sequence{xn′} of non-zero vectors in X whose linear span is dense in X, we shall prove N(C)∖D∈Λ({xn′}). If span{xn′}⊂span(C), we just repeat the procedure above. Otherwise, for any x∈{xn′} such that x∉span(C), if (N(C)∖D)∩(z+Rx)≠∅ for every z∈X, then there exists z′∈N(C)∖D such that (N(C)∖D)∩(z+Rx)=(N(C)∖D)∩(z′+Rx)={z′}. Therefore, N(C)∖D∈Λ(x). This implies that N(C)∖D is null set.


## 3. Fixed Point Property for Isometries

In this section we applyω∗-Gâteaux differentiability of Lipschitz mappings to the fixed point theory. As a result, we obtain Theorem 3.LetC be a nonempty bounded closed convex subset of a Banach space X. Recall that a mapping T:C→C is said to be isometry if (10)Tx-Ty=x-y,whenever x,y∈C. We say that C has the fixed point property for isometries if every isometry T:C→C has a fixed point.Definition 11.
Suppose thatC is a subset of a Banach space X and that Y is a Banach space. We say that C Lipschitz embeds into Y∗ provided that there is a mapping f:C→Y∗ and constants k,K>0 such that for all x,y∈C, (11)kx-y≤fx-fy≤Kx-y.In this case, f is said to be Lipschitz embedding and the smallest possible constant K is called the Lipschitz constant of the mapping, in short, Lip(f)=K.Proposition 12.
Suppose thatC is a closed convex separable set in a Banach space X. Then N(C)≠∅ in XC.Proof.
We may assume that0∈C. Since C is separable, XC is a separable space. It follows immediately from Proposition 1 in [10] that N(C)≠∅.Theorem 13.
Suppose thatC is a separable closed convex set of a Banach space X, and that Y is a separable Banach space. Then C Lipschitz embeds into Y∗ if and only if XC is linearly isomorphic to a subspace of Y∗.Proof.
Sufficiency. It is clearly trivial.
Necessity. From Proposition 12, N(C)≠∅ in XC. We may assume that 0∈N(C). It follows from Theorem 2.6 in [7] that C0=⋃n=1∞‍nC=span(C) is a dense subspace of XC. Note that XC is separable, thus there exists a sequence {xn} of linearly independent vectors in N(C) such that span{xn} is dense in XC. Let f:C→Y∗ be a Lipschitz embedding. For each n∈N, set Xn=span{xk:k≤n}. Consider the set of all points x∈N(C) for which there is a bounded linear operator Tn:Xn→Y∗ such that(12)limλ→0⁡fx+λh-fxyλ=Tnhy,h∈Xn,y∈Y,and(13)kh≤Tnh,h∈XnFrom the proof of Theorem 10, the conclusion follows using argument similar to the proof of Theorem 2.Proposition 14.
Suppose thatC is a separable closed convex set of a Banach space X, and that Y is a dual Banach space. Then C Lipschitz embeds into Y if and only if XC is linearly isomorphic to a subspace of Y.Proof.
It suffices to show the necessity. LetY=Z∗ and let f:C→Z∗ be a Lipschitz embedding. Since C is a separable, Xf(C) is a separable subspace of Z∗. By Corollary 14.2.22 in [8], we can find a separable Banach space W⊂Z and linearly isometric embeddings i1:Xf(C)→W∗ and i2:W∗→Z∗. Hence, i1f:C→W∗ is a Lipschitz embedding. By Theorem 13, XC is linearly isomorphic to a subspace of W∗. Note that W∗ is linearly isometric isomorphic to a subspace of Z∗. Therefore, XC is linearly isomorphic to a subspace of Y.Proof of Theorem3.
LetC be a nonempty bounded closed convex subset of a Banach space X and let C be Lipschitz embedded into a super reflexive space Y. We still assume 0∈C. We can also assume that C is separable because the fixed point property for isometries of a bounded closed convex set in a Banach space is separably determined. Superreflexivity of Y entails Y=Y∗∗ and hence Y∗∗ is also a super reflexive space. By Proposition 14, XC is linearly isomorphic to a closed subspace of Y∗∗. This implies XC is a super reflexive space. Therefore, C has the fixed point property for isometries by Maurey’s fixed point theorem [12].Remark 15.
The converse Theorem3 does not hold. In fact, for a nonempty uniformly convexifiable set C of Banach space X, C has the fixed point property for isometries (See [13, Theorem 4.2]). However, Beauzamy [14] proved that there exists a uniformly convexifiable set in a Banach space which can not linearly embedded into a super reflexive space.


## 4. Fixed Point Property for Continuous Affine Mappings

In this section, with the application of Baudier-Lancien-Schlumprecht’s theorem, we obtain Theorem4.LetC be a nonempty bounded closed convex subset of a Banach space X. Recall that a mapping T:C→C is said to be affine if (14)Tλx+1-λy=λTx+1-λTy,whenever x,y∈C and λ∈[0,1]. We say that C has the fixed point property for continuous affine mappings if every continuous affine mapping T:C→C has a fixed point.LetX and Y be two Banach spaces and C be a subset of X, and let f:C→Y be a mapping. Set (15)ρft=supfx-fyY:x-yX≥t,and(16)ωft=sup⁡fx-fyY:x-yX≤t.We say that f is uniformly continuous if limt→0ωf(t)=0 and ρf(t)>0 for all t>0. C is said to be uniformly embedded into Y provided that there is a mapping f:C→Y such that f is injective and both f and f-1 are uniformly continuous.Proposition 16.
Suppose thatC is a nonempty bounded closed convex subset of a Banach space X. If C affinely uniformly embeds into a reflexive space Y, then C has the fixed point property for continuous affine mappings.Proof.
Suppose thatC affinely uniformly embeds into a reflexive space Y. Then there exist an affinely uniformly continuous mapping T:C→Y. We may assume that 0∈C. Defining a mapping T1(x)=T(x)-T(0) for all x∈span¯(C) is a linearly uniformly continuous mapping from span¯(C) to Y. Thus, span¯(C) is a reflexive space. It follows from Schauder-Tychonoff theorem [15] that C has the fixed point property for continuous affine mappings.Remark 17.
The affinity in Proposition16 is necessary. Mazur [16] proved that the unit ball BL2 of L2 is uniformly homomorphic onto the unit ball BL1 of L1. It is easy to see that BL2 has the fixed point property for continuous affine mappings by Schauder-Tychonoff theorem [15] and BL1 does not have the fixed point property for continuous affine mappings by Theorems 3.2 in [17]. On the other hand, the Mazur theorem [16] also implies that weak compactness is usually not preserved under uniform embeddings.Recall the construction of the Tsirelson spacesT∗ and T originally designed by Tsirelson [18]. Let E,F⊂N and n∈N. We denote E<F if max(E)<min(F) and n≤E if n≤min(E). Here we set max(∅)=0 and min(∅)=∞. We say that a sequence {En}j=1n⊂N is admissible if n≤E1<E2<⋯<En. Let {ej} be the canonical basis of c00. For every x=∑j=1∞‍αjej∈c00, we put x0=xc0=maxj∈N|αj| and then define inductively for k=1,2,⋯(17)xk=max⁡xk-1,12max⁡∑j=1nEjxk-1:Enj=1nis  admissible.Put(18)x=limk→∞xk=maxk∈N⁡xk.Then ∥·∥ a norm on c00 and T is defined to be the completion of c00 with respect to the norm ∥·∥. We denote the dual of T by T∗ which is nowadays usually referred to as T∗. For more detail, we refer the reader to [19].For any infinite subsetM of N, let [M]ω={A⊂M:Aisinfinite}. For each k∈N, let [M]k={A⊂M:|A|=k}, where |A| denotes the cardinality of the set A. Elements of [N]k will always be listed in an increasing order; i.e., for every m¯={m1,m2,…,mk}∈[M]k, we assume that m1<m2<⋯<mk. Recall that the Hamming metric is defined by (19)dHm¯,n¯=i∈1,2,…,min⁡m¯,n¯:mi≠mi+max⁡m¯,n¯-min⁡m¯,n¯,where m¯,n¯∈[N]k. Note that the metric dH, can be seen as the graph metric on the Hamming graph over a countable alphabet, denoted Hkω(N) or simply Hkω, where two vertices are adjacent if they differ in exactly one coordinate. Let h:N×N→N be a bijective. Then the map ϕk:Hkω→T∗ defined by ϕk(n¯)=1/2∑i=1k‍yh(i,ni) is a Lipschitz mapping with Lip(ϕk)=1 (see, for instant, [20]).Proof of Theorem4.
By Schauder-Tychonoff theorem [15], it suffices to show that C is weakly compact. Suppose, to the contrary, that C is not weakly compact. Then, by James’ theorem [21], there exist δ>0 and a sequences {xn}⊂C such that for all k∈N,(20)dcox1,x2,…,xk,coxk+1,xk+2,⋯≥δ.Choose n¯∈[N]2k. Then(21)1k∑i=1kxni-1k∑i=k+12kxni≥δ,which implies that(22)∑i=1kxni-∑i=k+12kxni≥kδ.For each n¯∈Hkω, define a map ϕk:Hkω→C by ϕk(n¯)=1/k∑i=1k‍xni. Then ϕk is a Lipschitz mapping with Lip(ϕk)=2/k.
Letf be a uniform embedding from C into T∗. Then Lip(f∘ϕk)≤ωf(2/k). By Theorem 4.4 in [20], for all k≥1, there exists M′∈[N]ω such that(23)f∘ϕkm¯-f∘ϕkn¯≤5Lip⁡f∘ϕk,for all m¯,n¯∈[M′]k. This implies(24)f∘ϕkm¯-f∘ϕkn¯≤5ωf2k.In particular, for all k≥1, choose m¯,n¯∈[M′]k such that m1<m1<⋯<mk<n1<n2⋯<nk. We obtain(25)0<ρfδ≤5ωf2k.This is a contradiction when k is sufficiently large and which completes our proof.



---

*Source: 4758546-2018-10-21.xml*