# Convexities and Existence of  the  Farthest Point

Authors: Z. H. Zhang; C. Y. Liu
Journal: Abstract and Applied Analysis
 (2011)

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## Abstract

Five counterexamples are given, which show relations among the
new convexities and some important convexities in Banach space. Under the assumption that Banach spaceX is nearly very convex, we give a sufficient
condition that bounded, weakly closed subset of X has the farthest points. We also give a sufficient condition that the farthest point map is single valued in a
residual subset of X when X is very convex.


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## Body

## 1. Introduction

LetX be a Banach space, and let X* be its dual space. Let us denote by B(X) and S(X) the closed unit ball and the unit sphere of X, respectively. Let x∈S(X),Σ(x)={x*∈S(X*):x*(x)=1}. For any sequence {xn}⊂X, define sep(xn)≡inf{∥xn-xm∥:n≠m}. Let B be a bounded subset of X. We define a real-valued function rB:X→R by (1.1)rB(x)=sup⁡{‖x-y‖:y∈B}
and call rB(x) the farthest distance from x to B. The function rB is convex and Lipschitz-continuous. In fact, |rB(x)-rB(y)|≤∥x-y∥ for all x,y∈X.A pointz∈B is called a farthest point of B if there exists an x∈X such that ∥x-z∥=rB(x).The mappingFB:X→2B defined by FB(x)={z∈B:∥x-z∥=rB(x)} is called the farthest point map of B.The existence of a farthest point ofB is equivalent to the fact that the set (1.2)D={x∈X:‖x-z‖=rB(x)forsomez∈B}
is nonempty. In [1–5], the existence of a farthest point of B is studied. Edelstein [2] showed that if X is uniformly convex space, then the set D defined above is dense in X. Asplund [1] showed that if X is both reflexive and locally uniformly rotund (LUR), then the set D is dense in X. Lau [4] proved that if B is weakly compact subset of X, then D contains a dense Gδ set of X, and if X is reflexive Banach space, then for every bounded, weakly closed subset B in X, D contains a dense Gδ subset of X.We say thatX is strongly convex (resp., very convex/nearly strongly convex/nearly very convex) if any x∈S(X) and {xn}⊂B(X) with x*(xn)→1 as n→∞ for some x*∈Σ(x) imply xn→x as n→∞ (resp., xn→wx as n→∞/{xn} is relatively compact/{xn} is weakly relatively compact).The author [6] proved that strong convexity (resp., very convex/nearly strong convexity/nearly very convex) has important applications in approximation theory. Bandyopadhyay et al. [7] also proposed two generalizations of LUR and weakly locally uniform rotundity (WLUR), which were called almost locally uniform rotundity (ALUR) and weakly almost locally uniform rotundity (WALUR). A Banach space X is said to be ALUR (resp., WALUR) if for any x∈S(X),{xn}⊂B(X) and {xm*}⊂B(X*), the condition limmlimnxm*((xn+x)/2)=1 implies xn→x (reps., xn→wx). Recently, we have proved that ALUR and strong convexity, WALUR and very convex are equivalent, respectively [8]. Sullivan [9] defined very rotund space. A Banach space X is known as very rotund if no x*∈S(X*) is simultaneously a norming element for some x∈S(X) and x**∈S(X**), where x≠x**. The author [10] proved that very rotund space coincides with very convex space. By [6–12], we know that the four new convexities mentioned above have a lot of good properties and applications.It is known that LUR, WLUR, midpoint locally uniform rotundity (MLUR) and weakly midpoint locally uniform rotundity (WMLUR) are four important convexities in the geometric theory of Banach spaces. By [9–12], the relation of the convexities mentioned above is shown in Figure 1 below.Figure 1
The relationship between the convexities.The structure of this paper is as follows. In Section2, We will give five counterexamples, which show the relations among the four new convexities mentioned above and LUR, WLUR, MLUR, WMLUR, and rotund (R).In Section3, we prove that if X is nearly very convex space and if for every x∈X, there exists x0*∈∂rB(x) which attains its norm, then for every bounded, weakly closed subset B of X, the set D defined above contains a Gδ subset of X, which improves the results of Edelstain [2], Asplund [1], and Lau [4]. Finally, we also prove a sufficient condition that the farthest point map FB is single valued in a residual subset of X when X is very convex.


## 2. Some Counterexamples about Convexities

Lemma 2.1 (see [13]).
Let{xα:α∈D} be a net in X if for every ɛ>0, there exists αs∈D such that the tail {xα:α≥αs} has a finite ɛ-net, then {xα} is a relatively compact subset in X.Lemma 2.2.
IfX is nearly strongly convex space, then X has Kadec property, that is, if whenever x∈S(X) and {xα} is a net in S(X) such that xα→wx, then xα→x. Particularly, X has property H, that is, if whenever x∈S(X) and {xn} is a sequence in S(X) such that xn→wx, then xn→x.Proof.
Suppose that net{xα:α∈D}⊂S(X),x∈S(X) such that xα→wx, we will prove that xα→x.Case 1.
If for everyɛ>0, there exists αs∈D such that {xα:α≥αs} has a finite ɛ-net, by Lemma 2.1 and xα→wx, we may obtain that xα→x.Case 2.
If for everyɛ0>0 such that all tails of net {xα:α∈D} have no finite ɛ0-net, we take f∈S(X*) with f in Σ(x). Since xα→wx,f(xα)→f(x)=1. Choosing {βn}⊂D,β1≤β2≤⋯≤βn≤⋯, we know that(2.1)f(xα)>1-1n,
for any α∈D with α≥βn. Take α1∈D such that α1≥β1. For xα1 and β2, we can choose α2∈D,α2≥β2 such that ∥xα2-xα1∥≥ɛ0. Otherwise, {xα:α≥β2}⊂{x∈X:∥x-xα1∥<ɛ0}, that is, the tail {xα:α≥β2} has finite ɛ0-net. This is a contradiction with the assumption. For xα1,xα2, and β3, there exists α3∈D,α3≥β3 such that ∥xα3-xα1∥≥ɛ0 and ∥xα3-xα2∥≥ɛ0. Otherwise, {xα:α≥β3}⊂{x∈X:∥x-xα1∥<ɛ0}∪{x∈X:∥x-xα2∥<ɛ0}, that is, the tail {xα:α≥β3} has finite ɛ0-net. This is a contradiction with the assumption. According to the same method, we may choose a sequence {xαn}⊂{xα:α∈D} such that
(2.2)sep(xαn)=inf⁡{‖xαn-xαm‖:n≠m}≥ɛ0,αn≥βn,n=1,2,….
Hence, we know that {xαn} is not relatively compact.
On the other hand, by (2.1), we have that(2.3)1≥f(xαn)>1-1n.
This shows that f(xαn)→1. Since X is nearly strongly convex, {xαn} is relatively compact which is a contradiction.Example 2.3.
There exists an MLUR space which is not a nearly strongly convex space.
Recall the equivalent norm defined onc0 by Smith [14]. For k∈ℕ, define a mapping Vk:c0→ℝ1,Vk(x)=Vk(ξ1,ξ2,…)=sup{|ξ1-ξi|:i≥k}. Let {αn} be a sequence of positive real numbers, and ∑n=2∞αn2=1. Define two mappings V and T from c0→l2 as follows:(2.4)V(x)=(α2V2(x),α3V3(x),…),T(x)=(ξ1,α2ξ2,α3ξ3,…).
Since {Vk(x)} and {ξk} are both bounded sequences, and ∑n=2∞αn2=1, then we know that {Vk(x)} and T(x) are in l2. For x in c0, let
(2.5)‖x‖G=(‖x‖∞2+‖V(x)‖22+‖T(x)‖22)1/2.
Since T is one to one continuous, linear, and V(x+y)≤V(x)+V(y) for all x,y in c0, the ∥·∥G is a norm, and ∥·∥∞≤∥·∥G≤6∥·∥∞. This shows that ∥·∥G is an equivalent norm on c0.
Smith shows that(c0,∥·∥G) is MLUR. We say that (c0,∥·∥G) has no property H. Indeed, let x=e1,xn=e1+en, then ∥xn∥G→∥x∥G=3 and xn→wx, but xn↛x. By Lemma 2.2, we know that (c0,∥·∥G) is not nearly strongly convex.Example 2.4.
There exists a very convex space which is not nearly strongly convex space.
Recall the equivalent norm on Hilbert space by  Troyanski in Isratescu  [15]. Let X be a Hilbert space, and let {ei}i=0∞ be an orthogonal basis. For any x∈X,x=λ0e0+λ1e1+⋯+λnen+⋯, let(2.6)‖x‖12=max⁡{|λ0|2+|λ2|2+⋯+|λ2n|2+⋯,|λ1|2+|λ3|2+⋯+|λ2n+1|2+⋯}.
It is obvious that ∥·∥1 is an equivalent norm on X. Further, we set
(2.7)‖x‖=(‖x‖12+∑i=1∞12i|λi|2)1/2.
Clearly, this is again a norm on X which is an equivalent original one. Troyanski shows that (X,∥·∥) is R and reflexive [15], but it has no property H. Hence, (X,∥·∥) is very convex, but is not nearly strongly convex by Lemma 2.2.Example 2.5.
There exists a nearly very convex space which is neither very convex space nor nearly strongly convex space.
For anyx∈l2, let(2.8)|‖x‖|=max⁡⁡{12‖x‖2,‖x‖∞}.
Since (1/2)∥x∥2≤|∥x∥|≤∥x∥2,|∥x∥| is an equivalent norm on l2. Since (l2,|∥·∥|) is reflexive, we know that (l2,|∥·∥|) is nearly very convex space. We say that (l2,|∥·∥|) has no property H. Indeed, let x=e1,xn=e1+en, then |∥x∥|=|∥xn∥|=1, but |∥xn-xm∥|=1. Hence, B (l2,|∥·∥|) is not nearly strongly convex by Lemma 2.2. We say that (l2,|∥·∥|) is not very convex. In fact, let e1,2=(1,1,0,…),e1,3=(1,0,1,0,…), then |∥e1,2+e1,3∥|=2. This shows that (l2,|∥·∥|) is not R, and therefore, (l2,|∥·∥|) is not very convex.Example 2.6.
There exists a strongly convex space which is not WLUR space.
LetE=(l2,∥·∥), where x=(a1,a2,…)∈l2,(2.9)‖x‖2={|a1|+(a22+a32+⋯)1/2}2+{(a22)2+⋯+(ann)2+⋯}.
Let X=(Σ⊕E)l2. In [16], it is proved that X is 2R, but is not KUR. Since 2R implies R, reflexive and property H, we get that X is strongly convex space.
X is not WLUR space. Indeed, let {ei}i=1∞ be the natural basis, and x=(e1,e1,0,…),xn=(1/2(e1+en),e1,0,…), then limn→∞∥xn∥=2=∥x∥ and limn→∞∥xn+x∥=8. However,(2.10)d(xn,span⁡{x})=inf⁡{‖xn-y‖:y∈span⁡{x}}≥12.
Choose f∈S(X*) such that f(xn)≥1/2,f(y)=0,forally∈span{x}. It is easily proved that
(2.11)lim⁡n→∞‖x‖x‖+xn‖xn‖‖=12lim⁡n→∞‖x+xn‖=2,
but f((x/∥x∥)+(xn/∥xn∥))≥|f(xn/∥xn∥)|=1/(22). This shows that X is not WLUR space.Example 2.7.
There exists a nearly strongly convex space which is not a strongly convex space.
Forx=(x1,x2) in ℝ2, let ∥x∥=|x1|+|x2|, then (ℝ2,∥·∥) is reflexive and has property H. Hence, it is nearly strongly convex. It is easy to prove that (ℝ2,∥·∥) is not R space. Therefore (ℝ2,∥·∥) is not strongly convex.Remark 2.8 (Smith [17]).
gave three examples(l2,∥·∥W),(l2,∥·∥A),(l1,∥·∥H). He shows that (l2,∥·∥W) is WLUR not MLUR, and (l2,∥·∥A) is MLUR not WLUR, and (l1,∥·∥H) is R not MLUR. It is easily proved that (l1,∥·∥H) is not WMLUR either.By the above five counterexamples and Remark2.8, we know that, except for very convex implied WMLUR, none of the above converse implied relations in the diagram is generally true.


## 3. Convexities and Existence of the Farthest Point

Before proceeding to this part, let’s recall that the subdifferential of convex functionf on Banach space X is defined by (3.1)∂f(x)={x*∈X*:x*(y-x)≤f(y)-f(x),∀y∈X}.x→∂f(x) is called subdifferential mapping.Remark 3.1.
It was shown in [4] that if B is a bounded closed subset in Banach space X, then for any x∈X and x*∈∂rB(x), we have that ∥x*∥≤1 and thus
(3.2)sup⁡{x*(x-z):z∈B}≤‖x*‖⋅rB(x)≤rB(x).
Hence, for any x∈X and x*∈∂rB(x), we have that
(3.3)inf⁡z∈Bx*(z-x)≥-rB(x).Lemma 3.2 (Lau [4]).
LetX be a Banach space and B a bounded subset in X, then the set
(3.4)F={x∈X:inf⁡z∈Bx*(z-x)>-rB(x)forsomex*∈∂rB(x)}
is a first category in X.Theorem 3.3.
LetX be a nearly very convex Banach space and B a bounded, weakly closed subset of X. Further, for any x∈X, if there exists an x0*∈∂rB(x) which attains its norm, then
(3.5)D={x∈X:‖x-z‖=rB(x)forsomez∈B}
contains a dense Gδ set of X. In particular, the set of farthest points of B is nonempty.Proof.
DefineF as in Lemma 3.2 and
(3.6)Fn={x∈B:inf⁡z∈Bx*(z-x)≥-rB(x)+1nforsomex*∈∂rB(x)}.
Let Q=X∖F, then
(3.7)Q=X∖⋃n=1∞Fn=⋂n=1∞(X∖Fn),
where each X∖Fn is an open, dense subset in B. Hence, Q is a dense Gδ set in X.
Now, we proveQ⊂D as follows. For any x∈Q, take x0*∈∂rB(x) such that x0* attains its norm. Since infz∈Bx0*(z-x)≥-rB(x), by the definition of Q, we have that infz∈Bx0*(z-x)≤-rB(x). Take sequence {zn}⊂B such that(3.8)lim⁡n→∞x0*(zn-x)=-rB(x).
Clearly, (zn-x)/(-rB(x))∈B(X). Given that X is nearly very convex and that B is weakly closed, there exist z0∈B and subsequence {znk}⊂{zn} such that (znk-x)/(-rB(x))→w(z0-x)/(-rB(x)) as k→∞, that is, znk→wz0 as k→∞. Hence, x0*(z0-x)=-rB(x). It follows that
(3.9)rB(x)≥‖x-z0‖≥|x0*(z0-x)|=rB(x).
Thus, ∥z0-x∥=rB(x). This shows that Q⊂D.Corollary 3.4 (Lau [4]).
IfX is a reflexive Banach space. Then for every bounded, weakly closed subset B in X, the set
(3.10)D={x∈X:‖x-z‖=rB(x)forsomez∈B}
contains a dense Gδ set of X, and hence, the set of farthest points of B is nonempty.Corollary 3.5 (Asplund [1]).
IfX is a reflexive LUR Banach space, then Corollary 3.4 holds for every bounded closed subset B in X.Theorem 3.6.
LetX be a very convex Banach space and B a bounded, weakly closed subset of X. For any x∈X, if there exists an x0*∈∂rB(x) which attains its norm, then the farthest point map FB is single valued in a residual subset of X.Proof.
By Theorem3.3,
(3.11)Q=X∖F={x∈X:sup⁡z∈Bx*(x-z)=rB(x)foranyx*∈∂rB(x)}
is a dense Gδ subset of X, where F is defined as in Lemma 3.2.
Now we prove thatFB(x) is single valued for all x∈Q.
IfFB is not single valued on Q, then there are x0∈Q and z1,z2∈B with z1≠z2 such that ∥x0-zi∥=rB(x0),i=1,2. By Hahn-Banach theorem, we have xi*∈S(X*),i=1,2 such that(3.12)xi*(x0-zi)=‖x0-zi‖,i=1,2.
For any z′∈B,y∈X, we have that
(3.13)xi*(y-x0)=xi*(y-z′)-xi*(x0-z′)≤sup⁡z∈Bsup⁡x*∈B(X*)x*(y-z)-xi*(x0-z′)=rB(y)-xi*(x0-z′).
Thus,
(3.14)sup⁡z∈Bxi*(x0-z)≤rB(y)-xi*(y-x0).
Since x0∈Q,rB(x0)≤rB(y)-xi*(y-x0). This shows that xi*∈∂rB(x0). Let x0*=(1/2)x1*+(1/2)x2*, then x0*∈∂rB(x0) due to the convexity of ∂rB(x0), and supz∈Bx0*(x0-z)=rB(x0). Take a sequence {zn}⊂B such that
(3.15)lim⁡n→∞x0*(x0-zn)=rB(x0).
It follows that limn→∞xi*(x0-zn)=rB(x0),i=1,2. By (3.12), xi*∈Σ((x0-zi)/(rB(x0))),i=1,2. Because X is very convex, (x0-zn)/(rB(x0))→w(x0-zi)/(rB(x0)),i=1,2 as n→∞. According to uniqueness of weak limit point, we have that z1=z2, which is a contradiction.Remark 3.7.
For closed-convex subsetB and bounded closed, relatively weakly compact K in X, Ni and Li [18] proved that the set of all points in B such that the farthest problem max{x,K} is well posed is a dense Gδ subset in B provided that B is both strictly convex and Kadec with respect to K. This shows that the farthest point map FB is single valued in a residual subset of X. By Example 2.4 in this paper, there exists a Banach space where assumptions in Theorem 3.6 are satisfied, but its unit ball B(X) is not Kadec. Let B(X)=B=K, then we know that conditions of Theorem 3.6 are different from conditions of the result by Ni and Li. Hence, the result by Ni and Li does not imply Theorem 3.6.



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