# Farthest Points and Subdifferential inp-Normed Spaces

Authors: S. Hejazian; A. Niknam; S. Shadkam
Journal: International Journal of Mathematics and Mathematical Sciences
 (2008)

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## Abstract

We study the farthest point mapping in ap-normed space X in virtue of subdifferential
of r(x)=sup{∥x−z∥p:z∈M}, where M is a weakly sequentially compact
subset of X. We show that the set of all points in X which have farthest point in M contains
a dense Gδ subset of X.


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## Body

## 1. Introduction

LetX be a real  linear space. A quasinorm is a
real-valued function on X satisfying the following conditions.
(i)
∥x∥≥0 for all x∈X, and ∥x∥=0 if and only if x=0.(ii)
∥λx∥=|λ|∥x∥ for all λ∈R and all x∈X.(iii)
There is a constantK≥1 such that ∥x+y∥≤K(∥x∥+∥y∥) for all x,y∈X.
The pair (X,∥⋅∥) is called a quasinormed space if ∥⋅∥ is a quasinorm on X. The smallest possible K is called the modules of concavity of ∥⋅∥. By a quasi-Banach space we mean a complete
quasinormed space, that is, a quasinormed space in which every Cauchy sequence
converges in X.This class includes Banach spaces. The most
significant class of quasi-Banach spaces, which are not Banach spaces,
isLp-spaces for 0<p<1 equipped with the Lp-norms ∥⋅∥p. A quasinorm ∥⋅∥ is called a p-norm (0<p<1) if ∥x+y∥p≤∥x∥p+∥y∥p for all x,y∈X. In this case, a quasinormed (quasi-Banach) space is called a p-normed (p-Banach) space. By the Aoki-Rolewicz theorem
[1], each quasinorm
is equivalent to some p-norm. Since it is much easier to work with p-norms than with quasinorms,
henceforth we restrict our attention mainly to p-norms. See [2–4] for more information.If x* is in X*, the dual ofX, and x∈X we write x*(x) as 〈x*,x〉. We also consider quasinorms with K>1. The case where K=1 turns out to be the classical normed spaces,
so we will not discuss it and refer the interested reader to [5–7] for analogue results
concerning normed spaces.In this paper, using some strategies from [5–7], we study the farthest point mapping in a p-normed space X in virtue of subdifferential of r(x)=sup{∥x−z∥p:z∈M}, where M is a weakly sequentially compact subset of X. We show that the set of all points in X which have farthest point in M contains a dense Gδ subset of X.LetX be a p-normed space and let M be a nonempty bounded subset of X.
The mapping QM:X→2M defined by QM(x)={z∈M:∥x−z∥p=supt∈M∥x−t∥p} is called the farthest point map of M.
We call M a remotal (uniquely remotal, resp.) set if for
each x∈X the set QM(x) is not empty (is singleton, resp.) [8–10].


## 2. Main Results

LetX be a p-normed space and let M be a bounded subset of X.
For each x∈X,
we define the subdifferential of a function f at x by
(2.1)∂f(x)={x*∈X*:sgn〈x*,y−x〉|〈x*,y−x〉|p+f(x)≤f(y)∀y∈X}.This set may be empty even if we
consider X to be a Banach space [7, Example 3.8]. In a p-normed space, it may happen that ∂r(x)≠∅,
although we should note that a p-normed space may have a trivial dual as well
as it might have a nontrivial dual, see [11, Chapter 3], for some examples. To see the nonemptiness,
suppose that X is a p-normed space, x∈X,
and M={x}.
Thus, r(x)=0 and obviously 0∈∂r(x).
Also for each x*∈X* with ∥x*∥≤1,
we have(2.2)|〈x*,y−x〉|p≤∥x*∥p∥y−x∥p≤∥y−x∥p≤r(y)=r(y)−r(x)(y∈X).It follows that sgn〈x*,y−x〉|〈x*,y−x〉|p+r(x)≤r(y)(y∈X),
and so ∂r(x)≠∅.
Throughout the rest, we assume ∂r(x)≠∅ when we deal with this set. For an arbitrary
nonempty bounded subset M of X,
finding the set of all x for which ∂r(x)≠∅ remains an open question.Lemma 2.1.
LetX be a p-Banach space and let M be a bounded subset in X.
Then for each x∈X,
each element of ∂r(x) has norm less than or equal to 1 and hence ∂r(x) is w*-compact.Proof.
Letx∈X and x*∈∂r(x).
We have sgn〈x*,y−x〉|〈x*,y−x〉|p+r(x)≤r(y)(y∈X). By definition of r(x) we have |r(y)−r(x)|≤∥x−y∥p for all y∈X [10, 12].
Hencesgn〈x*,y−x〉|〈x*,y−x〉|p≤∥x−y∥p and therefore ∥x*∥≤1.Note that−r(x)≤−∥x−y∥p≤−sgn〈x*,x−y〉|〈x*,x−y〉|p=sgn〈x*,y−x〉|〈x*,y−x〉|p,
thus infz∈Msgn〈x*,z−x〉|〈x*,z−x〉|p≥−r(x).Now we have the following proposition which is
interesting on its own right.Proposition 2.2.
LetX be a p-Banach space and let M be a bounded subset of X.
Then the set F={x∈X:infz∈Msgn〈x*,z−x〉|〈x*,z−x〉|p>−r(x) for some x*∈∂r(x)} is of the first category in X.Proof.
Let(2.3)Fn:={x∈X:infz∈Msgn〈x*,z−x〉|〈x*,z−x〉|p≥−r(x)+1nfor somex*∈∂r(x)}. Then F=⋃n=1∞Fn. We will show that for each n,
(i)
Fn is a norm closed subset of X;(ii)
Fn has empty interior. To see (i), let {xm} be a sequence in Fn which converges to an element x in X.
For each m,
choose xm*∈∂r(xm) such that(2.4)infz∈Msgn〈xm*,z−xm〉|〈xm*,z−xm〉|p≥−r(xm)+1n.By Lemma 2.1∥xm*∥≤1 for all m.
Without loss of generality, we assume that {xm*} converges weak* to x*. For every y∈X,
we have(2.5)|〈xm*,y−xm〉−〈x*,y−x〉|≤|〈xm*,y−xm〉−〈xm*,y−x〉|+|〈xm*,y−x〉−〈x*,y−x〉|≤∥xm−x∥+|〈xm*−x*,y−x〉|.This shows that {〈xm*,y−xm〉}m=1∞ converges to 〈x*,y−x〉. Since xm*∈∂r(xm),(2.6)sgn〈xm*,y−xm〉|〈xm*,y−xm〉|p+r(xm)≤r(y)(y∈X),or equivalently(2.7)〈xm*,y−xm〉|〈xm*,y−xm〉|p−1+r(xm)≤r(y)(y∈X).It follows that(2.8)〈x*,y−x〉|〈x*,y−x〉|p−1+r(x)≤r(y)(y∈X)and hence(2.9)sgn〈x*,y−x〉|〈x*,y−x〉|p+r(x)≤r(y)(y∈X).This shows that x*∈∂r(x). It follows from (2.4) that(2.10)sgn〈xm*,z−xm〉|〈xm*,z−xm〉|p≥−r(xm)+1n(z∈M).We use the fact that sgn〈xm*,y−xm〉|〈xm*,y−xm〉|p=〈xm*,y−xm〉|〈xm*,y−xm〉|p−1 once more to obtain the inequality(2.11)sgn〈x*,z−x〉|〈x*,z−x〉|p≥−r(x)+1n(z∈M).Therefore x∈Fn.
So Fn is a closed subset of  X.
To see (ii), suppose that someFk has nonempty interior. Then, there exists an
open ball U in X of radius λ(2r(y0))1/p for some λ>0 and center at y0 such that U⊆Fk.
Let α=1/λp,β=1/(λ+1)p,
and ϵ≤((1+λ)p−1)/k(α+β+1)((1+λ)p)min{r1/p(y0),1} and choose z0∈M such that(2.12)r(y0)−ϵ<∥y0−z0∥p≤r(y0).Let x0=y0+λ(y0−z0) then(2.13)x0−z0=(1+λ)(y0−z0).Choose x1∈U⊆Fk such that(2.14)∥x1−x0∥p=ϵ.Then there exists x1*∈∂r(x1) such that(2.15)infz∈Msgn〈x1*,z−x1〉|〈x1*,z−x1〉|p≥−r(x1)+1k.We will show
that(2.16)sgn〈x1*,y0−x1〉|〈x1*,y0−x1〉|p+r(x1)>r(y0).This will contradict the fact
that x1* is a subdifferential of r at x1 and the proof would be completed. To achieve a
contradiction, we will consider four cases as
follows.
(i)
sgn〈x1*,z0−x1〉<0 and sgn〈x1*,y0−x1〉>0.(ii)
sgn〈x1*,z0−x1〉>0 and sgn〈x1*,y0−x1〉>0.(iii)
sgn〈x1*,z0−x1〉>0 and sgn〈x1*,y0−x1〉<0.(iv)
sgn〈x1*,z0−x1〉<0 and sgn〈x1*,y0−x1〉<0.
We investigate case (i) in detail. The other cases can
be studied similarly. First of all note that(2.17)z0−x1=z0−y0+y0−x1=y0−x0λ+y0−x1=(1+1λ)(y0−x1)+1λ(x1−x0).Now, we have(2.18)r(y0)−r(x1)<∥y0−z0∥p+ϵ−r(x1)=1|1+λ|p∥x0−z0∥p+ϵ−r(x1)(by (2.13))≤1(1+λ)p[∥x0−x1∥p+∥x1−z0∥p]+ϵ−r(x1)=(1−1(1+λ)p)(−r(x1))+(1(1+λ)p+1)ϵ(bythedefinationofr(⋅))≤(1−1(1+λ)p)[sgn〈x1*,z0−x1〉|〈x1*,z0−x1〉|p−1k]+(1(1+λ)p+1)ϵ≤(1−1(1+λ)p)[−|〈x1*,z0−x1〉|p−1k]+(1(1+λ)p+1)ϵ≤(1−1(1+λ)p)[−|〈x1*,λ+1λ(y0−x1)+x1−x0λ〉|p−1k]+(1(1+λ)p+1)ϵ(by(2.17))≤(1−1(1+λ)p)[−|〈x1*,λ+1λ(y0−x1)〉|p+|〈x1*,x1−x0λ〉|p−1k]+(1(1+λ)p+1)ϵ(by the factthat−|a+b|p≤−|a|p+|b|p)≤(1−1(1+λ)p)(λ+1λ)p[−〈x1*,y0−x1〉|p]+ϵλp(1−1(1+λ)p)+(1−1(1+λ)p)(−1k)+(1(1+λ)p+1)ϵ≤(1−1(1+λ)p)[−|〈x1*,y0−x1〉|p]≤sgn〈x1*,y0−x1〉|〈x1*,y0−x1〉|p.We
recall that a setG⊂X is said to be weakly sequentially compact if
each sequence of elements of G contains a subsequence converging weakly to
some element x∈G.Theorem 2.3.
LetM be a weakly sequentially compact subset in a p-Banach space X.
Then the set {x∈X:∥x−z∥p=r(x) for some z∈M} contains a dense Gδ-set in X.
In particular, the set of farthest points of S is nonempty.Proof.
LetF and Fn be defined as in Proposition 2.2
and let D(M)=X∖F.
Then(2.19)D(M)=X∖⋃n∈NFn=⋂n∈N(X∖Fn),where each X∖Fn is an open dense subset of X.
Hence D(M) is a dense Gδ-set in X.
For each x∈D(M) and x*∈∂r(x),
we have(2.20)infz∈Msgn〈x*,z−x〉|〈x*,z−x〉|p=−r(x).By the weak compactness of M,
there exists a point z∈M with sgn〈x*,z0−x〉|〈x*,z0−x〉|p=−r(x). Hence(2.21)r(x)≥∥x−z0∥p≥sgn〈x*,x−z0〉|〈x*,x−z0〉|p=−sgn〈x*,z0−x〉|〈x*,z0−x〉|p=r(x).This shows that D(M)⊆{x:∥x−z∥p=r(x) for some z∈M}.



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