IJMMSInternational Journal of Mathematics and Mathematical Sciences1687-04250161-1712Hindawi Publishing Corporation89059810.1155/2011/890598890598Research ArticleOn Simultaneous Farthest Points in L(I,X)Al-SharifSh.1RawashdehM.2GottwaldSiegfried1Mathematics DepartmentYarmouk UniversityIrbid 21163Jordanyu.edu.jo2Department of Mathematics and StatisticsJordan University of Science and TechnologyIrbid 22110Jordanjust.edu.jo201121072011201120122010230520112011Copyright © 2011 Sh. Al-Sharif and M. Rawashdeh.This is an open access article distributed under the Creative Commons Attribution License, which permits unrestricted use, distribution, and reproduction in any medium, provided the original work is properly cited.

Let X be a Banach space and let G be a closed bounded subset of X. For (x1,x2,,xm)Xm, we set  ρ(x1,x2,,xm,G)=sup{max1imxiy:yG}. The set G is called simultaneously remotal if, for any (x1,x2,,xm)Xm, there exists gG such that  ρ(x1,x2,,xm,G)=max1imxig. In this paper, we show that if G is separable simultaneously remotal in X, then the set of -Bochner integrable functions, L(I,G), is simultaneously remotal in L(I,X). Some other results are presented.

1. Introduction

Let X be a Banach space and G a bounded subset of X. For xX, set ρ(x,G)=sup{x-y:yG}. A point g0G is called a farthest point of G if there exists xX such that x-g0=ρ(x,G). For xX, the farthest point map FG(x)={gG:x-g=ρ(x,G)}, that is, the set of points of G farthest from x. Note that this set may be empty. Let R(G,X)={xX:FG(x)ϕ}. We call a closed bounded set G remotal if R(G,X)=X and densely remotal if R(G,X) is a norm dense in X. The concept of remotal sets in Banach spaces goes back to the sixties. However, almost all the results on remotal sets are concerned with the topological properties of such sets, see . Remotal sets in vector valued continuous functions was considered in . Related results on Bochner integrable function spaces, Lp(I,X), 1p, are given in .

The problem of approximating a set of points x1,x2,,xm simultaneously by a point g (farthest point) in a subset G of X can be done in several ways, see . Here, we will use the following definition.

Definition 1.1.

Let G be a closed bounded subset of X. A point gG is called a simultaneous farthest point of (x1,x2,,xm)Xm if ρ(x1,x2,,xm,G)=suphGmax1im(xi-h)=max1im(xi-g). We call a closed bounded set G of a Banach space X simultaneously remotal if each m-tuple (x1,x2,,xm)Xm admits a farthest point in G and simultaneously densely remotal if the set of points R(G,Xm)={(x1,x2,,xm)Xm:FG(x1,x2,,xm)ϕ}, where FG(x1,x2,,xm)={gG:ρ(x1,x2,,xm,G)=max1im(xi-g)} is norm dense in Xm.

Clearly, if m=1, then simultaneously remotal is precisely remotal.

In this paper we consider the problem of simultaneous farthest point for bounded sets of the form L(I,G) in the Banach space L(I,X), where X is a Banach space.

Throughout this paper, X is a Banach space, G is a closed bounded subset of X and L(I,X), the space of all X-valued essentially bounded functions on the unit interval I. For fL(I,X), we set f=esssup{f(s):sI}. For GX, we set L(I,G)={fL(I,X):f(s)G, almost all sI}.

2. Distance Formula

The farthest distance formula is important in the study of farthest point. In this section, we compute the -farthest distance from an element fL(I,X) to a bounded set L(I,G). We begin with the following proposition.

Proposition 2.1.

Let f1,f2,,fm, then max1imfi = esssupmax1im(fi(t)).

Proof.

For 1im, fi=esssupfi(t)esssupmax1im(fi(t))max1im(fi).   Hence, max1imfi=esssupmax1im(fi(t))

Theorem 2.2.

Let X be a Banach space and let G be a closed bounded subset of X. If a function Φ:I defined by Φ(t)=ρ(f1(t),f2(t),,fm(t),G), where f1,f2,,fmL(I,X), then ΦL(I) and ρ(f1,f2,,fm,L(I,G))=Φ=supgL(I,G)max1im(fi-g).

Proof.

Let f1,f2,,fmL(I,X). Being strongly measurable, there exist m sequences of simple functions (fin), 1im such that fin(t)-fi(t)0 as n for almost all tI. We may write fjn=i=1m(n)χA(i,n)xin(j). Since ρ(x1,x2,,xm,G) is a continuous function of (x1,x2,,xm)Xm, the inequality |ρ(f1n(t),f2n(t),,fmn(t),G)-ρ(f1(t),f2(t),,fm(t),G)|max1imfin(t)-fi(t)0 implies that |ρ(f1n(t),f2n(t),,fmn(t),G)-ρ(f1(t),f2(t),,fm(t),G)|0. Set Φn(t)=ρ(f1n(t),f2n(t),,fmn(t),G). Then, Φn(t)=supgGmax1km(fkn(t)-g)=supgGmax1kmi=1m(n)χAin(t)(xin(k)-g)=i=1m(n)χAin(t)supgGmax1kmxin(k)-g. So Φn is a simple function for each n and LimnΦn(t)-Φ(t)=0 for almost all tI. Hence Φ is measurable. Furthermore, for each wL(I,G), max1imfi-w=max1imesssup(fi(t)-w(t))=esssupmax1im(fi(t)-w(t)),Proposition  2.1esssupsupgGmax1im(fi(t)-g)=ρ(f1(t),f2(t),f3(t),,fm(t),G)=Φ. Thus, Φρ(f1,f2,,fm,L(I,G)). To prove the reverse inequality. Let ϵ>0 be given, since countably valued functions are dense in L(I,X), there exist countably valued functions γ1,γ2,,γm in L(I,X) such that fi-γi<ϵ, 1im. We may write γj=i=1χAixi(j), as in . We may assume i=1χAi=1, μ(Ai)>0 for all i. For each i, Choose hiG such that max1kmxi(k)-hi>ρ(xi(1),xi(2),,xi(m),G)-ϵ. Now, set gL(I,G) as g(s)=i=1χAi(s)hi. The inequality γj-gγj-fj+fj-g implies max1jmγj-gϵ+max1imfj-g. Further, max1jmfj-gmax1jmγj-g-ϵ=max1jm(esssup(γj(t)-g(t)))-ϵ=esssupmax1jmi=1χAi(t)(xi(j)-hi)-ϵ,Proposition  2.1=esssupmax1jmi=1χAi(t)xi(j)-hi-ϵ=esssupi=1χAi(t)max1jmxi(j)-hi-ϵesssupi=1χAi(t)ρ(xi(1),xi(2),,xi(m),G)-2ϵ=esssupρ(γ1(t),γ2(t),,γm(t),G)-2ϵ. For 1jm and aG, the inequality fj(t)-afj(t)-γj(t)+γj(t)-a implies ρ(f1(t),f2(t),,fm(t),G)max1jmfj(t)-γj(t)+ρ(γ1(t),γ2(t),,γm(t),G). Therefore, max1jmfj-gesssup(ρ(f1(t),f2(t),,fm(t),G)-max1jmγj(t)-fj(t))-2ϵΦ-esssupmax1jmγj(t)-fj(t)-2ϵΦ-3ϵ. Hence, ρ(f1,f2,,fm,L(I,G))+3ϵΦ. Since ϵ was arbitrary, we have Φ=ρ(f1,f2,,fm,L(I,G)).

Corollary 2.3.

Let g be a strongly measurable function from I to a closed bounded subset G of a Banach space X, and f1,f2,,fmL(I,X). If g(t) is a simultaneous farthest point of f1(t),f2(t),,fm(t) in G, then g is a simultaneous farthest point of f1,f2,,fm in L(I,G).

Proof.

By assumption, max1im(fi(t)-g(t))=ρ(f1(t),f2(t),,fm(t),G) for almost tI. Since G is bounded, it follows that gL(I,G) and esssupmax1im(fi(t)-g(t))=esssupρ(f1(t),f2(t),,fm(t),G). Theorem 2.2 and Proposition 2.1 implies that max1im(fi-g)=esssupmax1im(fi(t)-g(t))=ρ(f1,f2,,fm,L(I,G)) and g is a simultaneous farthest point of f1,f2,,fm in L(I,G).

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In this section, we raise the question: if G is a simultaneously remotal set in X, is L(I,G) simultaneously remotal in L(I,X)? We get a positive answer to this question in the case that G is a separable simultaneously remotal subset of X or in the case that Span G is finite dimensional subspace of X. We begin with the following theorem.

Theorem 3.1.

If G is simultaneously densely remotal in X, then L(I,G) is simultaneously densely remotal in L(I,X).

Proof.

Let f1,f2,,fmL(I,X). Then, there exist γ1,γ2,,γm simple functions such that fi-γi<ϵ/2, 1im. Now, we can write without lost of generality, γi=k=1nχAkxk(i). Since G is simultaneously densely remotal, then there exist an m-tuple (yk(1),yk(2),,yk(m)) and gkG such that max1imxk(i)-yk(i)<ϵ/2 and max1imyk(i)-gk=ρ(yk(1),yk(2),,yk(m),G)=supgGmax1imyk(i)-g. Set ϕ=k=1nχAkgk and hi=k=1nχAkyk(i). Then, max1imhi-ϕ=max1imesssuphi(t)-ϕ(t)=esssupmax1imk=1nχAk(t)(yk(i)-gk),Proposition  2.1=esssupmax1imk=1nχAk(t)yk(i)-gkesssupmax1imk=1nχAk(t)yk(i)-g for every gG. In particular, for any wL(I,G), using Proposition 2.1, max1imhi-ϕesssupmax1imhi(t)-w(t)max1imesssuphi(t)-w(t)=max1imhi-w, Hence, ϕ is a farthest point from the m-tuple (h1,h2,,hm). But max1imhi-γi=max1imesssupk=1nχAk(yk(i)-xk(i))=max1imesssupk=1nχAkyk(i)-xk(i)=esssupmax1imk=1nχAkyk(i)-xk(i),Proposition  2.1=esssupk=1nχAkmax1imyk(i)-xk(i)ϵ2. Hence, max1imfi-himax1imfi-γi+max1imγi-hiϵ2+ϵ2=ϵ. This complete the proof of the theorem.

For a remotal set GX, the map T:Xm2G defined by T(x1,x2,,xm)=F(x1,x2,,xm,G)  ={gG:ρ(x1,x2,,xm,G)=max1imxi-g} is a multivalued map in general. Hence for any fL(I,X), the map Tf is a multivalued map from I into G.

Before proceeding we remind the reader for some facts regarding mult-valued maps. For a Banach space X and a measurable space I a function f:IX is said to be strongly measurable if it is the pointwise limit of a sequence of simple functions almost everywhere. On the other hand f is said to measurable in the classical sense if f-1(K) is measurable in I for any closed set K in X, see . A multivalued function T:IX is said to measurable in the classical sense if T-1(K) is measurable in I for any closed set K in X, here T-1(K)={tI:T(t)Kϕ}. A measurable in the classical sense g may be extracted from a measurable multivalued function T:IX where X is a separable Banach space provided that T(t) is a closed subset of X for each tI and such that g(t)T(t) for each tI, see [11, page 289].

Theorem 3.2.

Let G be a closed bounded simultaneously remotal in X such that span(G) is finite dimensional subspace of X, then L(I,G) is simultaneously remotal in L(I,X).

Proof.

First, we will prove that T is a closed valued map. If T(x1,x2,,xm) is finite set, then it is closed. If T(x1,x2,,xn) is not finite, let yT(x1,x2,,xm)¯, then there exists ynT(x1,x2,,xm) such that yny. This implies max1imxi-gmax1imxi-yn, for every gG. Taking the limit as n, we get maxxi-gmaxxi-y for every gG, and this implies that yT(x1,x2,,xm) and T is a closed multivalued map.

To prove that T is measurable in the classical sense let B be any closed subset of G. If (x1,x2,,xm)T-1(B)¯, then there exists (x1(n),x2(n),,xm(n))T-1(B) that converges to (x1,x2,,xm). Since (x1(n),x2(n),,xm(n))T-1(B), then T(x1(n),x2(n),,xm(n))Bϕ. Choose ynT(x1(n),x2(n),,xm(n))B. Then, yn has a convergent subsequence ynky being a sequence in a bounded closed subset of a finite dimensional space. But max1imxi(nk)-ynkmax1imxi(nk)-g for every gG. Taking the limit as nk, we get max1imxi-ymax1imxi-g for every gG. Hence yT(x1,x2,,xm)B. Therefore (x1,x2,,xm)T-1(B) and T-1(B) is closed. Hence, T is measurable and if γ is the vector valued map γ:IXm, γ(t)=(f1(t),f2(t),,fm(t)), then Tγ is a measurable closed multivalued map. By Theorem 6.6.4 in [11, page 289], Tγ has a measurable selection say g. Further, max1imfi(t)-g(t)max1imfi(t)-h(t), for every hL(I,G) and so esssupmax1imfi(t)-g(t)=max1imesssupfi(t)-g(t)max1imesssupfi(t)-h(t). Hence, max1imfi-gmax1imfi-h, for every hL(I,G) and L(I,G) is remotal in L(I,X).

Finally, following the steps in the prove of Theorem 3.6 in , we proof the following main result in the paper.

Theorem 3.3.

Let G be a separable simultaneously remotal subset of X. Then L(I,G) is simultaneously remotal in L(I,X).

Proof.

Let (f1,f2,,fm)(L(I,X))m. Using Corollary 2.3, it is sufficient to show that there exists a measurable function g defined on I such that g(t) is a simultaneous farthest point of (f1(t),f2(t),,fm(t)) in G. Since f1,f2,,fm are strongly measurable. We may assume that f1(I),f2(I),,fm(I) are separable sets in X. So there exist {Ini}n=1a countable partition of I such that diam(fi(Ini))<1/2, for each i, 1im, for all n; where diam(S)=sup{x-y:x,yS}. Consider the partition Ik1k2km=i=1mIki, where (Ik1,Ik2,,Ikm)i=1m{Ini}, for 1n<. Then diam(fi(Ik1k2km))<1/2. For simplicity write Ik1k2km as {In}n=1. For each tI, let g0(t) be a simultaneous farthest point from (f1(t),f2(t),,fm(t)) in G. Define the map g0 from I into G by g0(t) is a simultaneous farthest point from (f1(t),f2(t),,fm(t)). Apply Lemma 3.1 in  with ϵ=1/2 and I=In=A, we get countable partitions in each In and therefore countable partition in the whole of I in measurable sets {En}n=1 and a sequence of subsets {An}n=1 such that AnEn,μ*(An)=μ(En),diam(g0(An))<12anddiam(fi(En))<12,1im. Repeat the same argument in each En with ϵ=1/22, I=En, and A=An. For each n, we get a countable partition {E(n,k):1k<} of En in measurable sets and a sequence {A(n,k):1k<} of subsets of I such that A(n,k)E(n,k)An,μ*(A(n,k))=μ(E(n,k)),diam(g0(A(n,k)))<122anddiam(fi(E(n,k)))<122,i=1,2,,m. Now, we will use mathematical induction for each n; let Δn be the set of n-tuples of natural numbers and Δ={Δn:1n<}. On this Δ consider the partial order defined by (m1,m2,,mi)<(n1,n2,,nj) if and only if ij and mk=nk, k=1,2,,i. Then by induction for each n, we can find a countable partition {Eα:αΔn} of I of measurable sets and a collection {Aα:αΔn} of subsets of I such that:

AαEα and μ*(Aα)=μ(Eα),

EβEα and AβAα if αβ,

diam(fi(Eα))<1/2n for i=1,2,,m and diam(g0(Aα))<1/2n for αΔn.

We may assume Aαϕ, for all α. For each αΔ, let tαAα and define gn from I into G by gn(t)=αΔnχEα(t)g0(tα). Then, for each tI and nm, we have gn(t)-gm(t)=αΔnχEα(t)g0(tα)-βΔmχEβ(t)g0(tβ)βΔmχEβ(t)(g0(tα)-g0(tβ))βΔm(g0(tα)-g0(tβ))χ  Eβ12n. Hence, (gn(t)) is a Cauchy sequence in G for all tI. Therefore (gn(t)) is a convergent sequence. Let g:IG be defined to be the pointwise limit of (gn). Since gn is strongly measurable for each n, we have g is strongly measurable. Further for tI, 1n<, and tEα for some αΔn, we have max1imfi(t)-gn(t)=max1imfi(t)-g0(tα)max1im|fi(tα)-g0(tα)-fi(t)-fi(tα)|max1im|fi(tα)-g0(tα)-12n|ρ(f1(tα),f2(tα),,fm(tα),G)-12n. For 1im, the inequality fi(t)-afi(t)-fi(tα)+fi(tα)-a12n+fi(tα)-a implies that ρ(f1(t),f2(t),,fm(t),G)12n+ρ(f1(tα),f2(tα),,fm(tα),G). Therefore, max1imfi(t)-gn(t)ρ(f1(t),f2(t),,fm(t),G)-12n-1. Taking limit as n, we get ρ(f1(t),f2(t),,fm(t),G)=max1imfi(t)-g(t). This completes the proof of the theorem.

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