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{max1≤i≤m∥xi−y∥:y∈G}. The set G is called simultaneously remotal if, for any (x1,x2,…,xm)∈Xm, there exists g∈G such that ρ(x1,x2,…,xm,G)=max1≤i≤m∥xi−g∥. 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 x∈X, set ρ(x,G)=sup{∥x-y∥:y∈G}. A point g0∈G is called a farthest point of G if there exists x∈X such that ∥x-g0∥=ρ(x,G). For x∈X, the farthest point map FG(x)={g∈G:∥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)={x∈X: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 [1–4]. Remotal sets in vector valued continuous functions was considered in [5]. Related results on Bochner integrable function spaces, Lp(I,X), 1≤p≤∞, are given in [6–8].

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 [9]. Here, we will use the following definition.

Definition 1.1.

Let G be a closed bounded subset of X. A point g∈G is called a simultaneous farthest point of (x1,x2,…,xm)∈Xm if
ρ(x1,x2,…,xm,G)=suph∈Gmax1≤i≤m(‖xi-h‖)=max1≤i≤m(‖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)={g∈G:ρ(x1,x2,…,xm,G)=max1≤i≤m(‖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 f∈L∞(I,X), we set ∥f∥∞=esssup{∥f(s)∥:s∈I}. For G⊂X, we set L∞(I,G)={f∈L∞(I,X):f(s)∈G, almost all s∈I}.

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 f∈L∞(I,X) to a bounded set L∞(I,G). We begin with the following proposition.

Proposition 2.1.

Let f1,f2,…,fm, then max1≤i≤m∥fi∥∞ = esssupmax1≤i≤m(∥fi(t)∥).

Proof.

For 1≤i≤m,
‖fi‖∞=esssup‖fi(t)‖≤esssupmax1≤i≤m(‖fi(t)‖)≤max1≤i≤m(‖fi‖∞).
Hence, max1≤i≤m∥fi∥∞=esssupmax1≤i≤m(∥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→ℝ defined by Φ(t)=ρ(f1(t),f2(t),…,fm(t),G), where f1,f2,…,fm∈L∞(I,X), then Φ∈L∞(I) and
ρ(f1,f2,…,fm,L∞(I,G))=‖Φ‖∞=supg∈L∞(I,G)max1≤i≤m(‖fi-g‖∞).

Proof.

Let f1,f2,…,fm∈L∞(I,X). Being strongly measurable, there exist m sequences of simple functions (fin), 1≤i≤m such that ∥fin(t)-fi(t)∥→0 as n→∞ for almost all t∈I. 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)|≤max1≤i≤m‖fin(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)=supg∈Gmax1≤k≤m(‖fkn(t)-g‖)=supg∈Gmax1≤k≤m‖∑i=1m(n)χAin(t)(xin(k)-g)‖=∑i=1m(n)χAin(t)supg∈Gmax1≤k≤m‖xin(k)-g‖.
So Φn is a simple function for each n and Limn→∞∥Φn(t)-Φ(t)∥=0 for almost all t∈I. Hence Φ is measurable. Furthermore, for each w∈L∞(I,G),
max1≤i≤m‖fi-w‖∞=max1≤i≤messsup(‖fi(t)-w(t)‖)=esssupmax1≤i≤m(‖fi(t)-w(t)‖),Proposition2.1≤esssupsupg∈Gmax1≤i≤m(‖fi(t)-g‖)=‖ρ(f1(t),f2(t),f3(t),…,fm(t),G)‖∞=‖Φ‖∞.
Thus,
‖Φ‖∞≥ρ(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, 1≤i≤m. We may write γj=∑i=1∞χAixi(j), as in [10]. We may assume ∑i=1∞χAi=1, μ(Ai)>0 for all i. For each i, Choose hi∈G such that
max1≤k≤m‖xi(k)-hi‖>ρ(xi(1),xi(2),…,xi(m),G)-ϵ.
Now, set g∈L∞(I,G) as g(s)=∑i=1∞χAi(s)hi. The inequality
‖γj-g‖∞≤‖γj-fj‖∞+‖fj-g‖∞
implies
max1≤j≤m‖γj-g‖∞≤ϵ+max1≤i≤m‖fj-g‖∞.
Further,
max1≤j≤m‖fj-g‖∞≥max1≤j≤m‖γj-g‖∞-ϵ=max1≤j≤m(esssup(‖γj(t)-g(t)‖))-ϵ=esssupmax1≤j≤m‖∑i=1∞χAi(t)(xi(j)-hi)‖-ϵ,Proposition2.1=esssupmax1≤j≤m∑i=1∞χAi(t)‖xi(j)-hi‖-ϵ=esssup∑i=1∞χAi(t)max1≤j≤m‖xi(j)-hi‖-ϵ≥esssup∑i=1∞χAi(t)ρ(xi(1),xi(2),…,xi(m),G)-2ϵ=esssupρ(γ1(t),γ2(t),…,γm(t),G)-2ϵ.
For 1≤j≤m and a∈G, the inequality
‖fj(t)-a‖≤‖fj(t)-γj(t)‖+‖γj(t)-a‖
implies
ρ(f1(t),f2(t),…,fm(t),G)≤max1≤j≤m‖fj(t)-γj(t)‖+ρ(γ1(t),γ2(t),…,γm(t),G).
Therefore,
max1≤j≤m‖fj-g‖∞≥esssup(ρ(f1(t),f2(t),…,fm(t),G)-max1≤j≤m‖γj(t)-fj(t)‖)-2ϵ≥‖Φ‖∞-esssupmax1≤j≤m‖γj(t)-fj(t)‖-2ϵ≥‖Φ‖∞-3ϵ.
Hence, ρ(f1,f2,…,fm,L∞(I,G))+3ϵ≥∥Φ∥∞. 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,…,fm∈L∞(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, max1≤i≤m(∥fi(t)-g(t)∥)=ρ(f1(t),f2(t),…,fm(t),G) for almost t∈I. Since G is bounded, it follows that g∈L∞(I,G) and
esssupmax1≤i≤m(‖fi(t)-g(t)‖)=esssupρ(f1(t),f2(t),…,fm(t),G).
Theorem 2.2 and Proposition 2.1 implies that
max1≤i≤m(‖fi-g‖∞)=esssupmax1≤i≤m(‖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,…,fm∈L∞(I,X). Then, there exist γ1,γ2,…,γm simple functions such that ∥fi-γi∥∞<ϵ/2, 1≤i≤m. 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 gk∈G such that max1≤i≤m∥xk(i)-yk(i)∥<ϵ/2 and
max1≤i≤m‖yk(i)-gk‖=ρ(yk(1),yk(2),…,yk(m),G)=supg∈Gmax1≤i≤m‖yk(i)-g‖.
Set ϕ=∑k=1nχAkgk and hi=∑k=1nχAkyk(i). Then,
max1≤i≤m‖hi-ϕ‖∞=max1≤i≤messsup‖hi(t)-ϕ(t)‖=esssupmax1≤i≤m‖∑k=1nχAk(t)(yk(i)-gk)‖,Proposition2.1=esssupmax1≤i≤m∑k=1nχAk(t)‖yk(i)-gk‖≥esssupmax1≤i≤m∑k=1nχAk(t)‖yk(i)-g‖
for every g∈G. In particular, for any w∈L∞(I,G), using Proposition 2.1,
max1≤i≤m‖hi-ϕ‖∞≥esssupmax1≤i≤m‖hi(t)-w(t)‖≥max1≤i≤messsup‖hi(t)-w(t)‖=max1≤i≤m‖hi-w‖∞,
Hence, ϕ is a farthest point from the m-tuple (h1,h2,…,hm). But
max1≤i≤m‖hi-γi‖∞=max1≤i≤messsup‖∑k=1nχAk(yk(i)-xk(i))‖=max1≤i≤messsup∑k=1nχAk‖yk(i)-xk(i)‖=esssupmax1≤i≤m∑k=1nχAk‖yk(i)-xk(i)‖,Proposition2.1=esssup∑k=1nχAkmax1≤i≤m‖yk(i)-xk(i)‖≤ϵ2.
Hence,
max1≤i≤m‖fi-hi‖∞≤max1≤i≤m‖fi-γi‖+max1≤i≤m‖γi-hi‖≤ϵ2+ϵ2=ϵ.
This complete the proof of the theorem.

For a remotal set G⊆X, the map T:Xm→2G defined by T(x1,x2,…,xm)=F(x1,x2,…,xm,G)={g∈G:ρ(x1,x2,…,xm,G)=max1≤i≤m‖xi-g‖}
is a multivalued map in general. Hence for any f∈L∞(I,X), the map T∘f 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:I→X 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 [10]. A multivalued function T:I→X 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)={t∈I:T(t)∩K≠ϕ}. A measurable in the classical sense g may be extracted from a measurable multivalued function T:I→X where X is a separable Banach space provided that T(t) is a closed subset of X for each t∈I and such that g(t)∈T(t) for each t∈I, 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 y∈T(x1,x2,…,xm)¯, then there exists yn∈T(x1,x2,…,xm) such that yn→y. This implies
max1≤i≤m‖xi-g‖≤max1≤i≤m‖xi-yn‖,
for every g∈G. Taking the limit as n→∞, we get max∥xi-g∥≤max∥xi-y∥ for every g∈G, and this implies that y∈T(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))∩B≠ϕ. Choose yn∈T(x1(n),x2(n),…,xm(n))∩B. Then, yn has a convergent subsequence ynk→y being a sequence in a bounded closed subset of a finite dimensional space. But max1≤i≤m‖xi(nk)-ynk‖≥max1≤i≤m‖xi(nk)-g‖
for every g∈G. Taking the limit as nk→∞, we get
max1≤i≤m‖xi-y‖≥max1≤i≤m‖xi-g‖
for every g∈G. Hence y∈T(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:I→Xm, γ(t)=(f1(t),f2(t),…,fm(t)), then T∘γ is a measurable closed multivalued map. By Theorem 6.6.4 in [11, page 289], T∘γ has a measurable selection say g. Further,
max1≤i≤m‖fi(t)-g(t)‖≥max1≤i≤m‖fi(t)-h(t)‖,
for every h∈L∞(I,G) and so
esssupmax1≤i≤m‖fi(t)-g(t)‖=max1≤i≤messsup‖fi(t)-g(t)‖≥max1≤i≤messsup‖fi(t)-h(t)‖.
Hence,
max1≤i≤m‖fi-g‖∞≥max1≤i≤m‖fi-h‖∞,
for every h∈L∞(I,G) and L∞(I,G) is remotal in L∞(I,X).

Finally, following the steps in the prove of Theorem 3.6 in [6], 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=1∞a countable partition of I such that diam(fi(Ini))<1/2, for each i, 1≤i≤m, for all n; where
diam(S)=sup{‖x-y‖:x,y∈S}.
Consider the partition Ik1k2⋯km=⋂i=1mIki, where (Ik1,Ik2,…,Ikm)⊆∏i=1m{Ini}, for 1≤n<∞. Then diam(fi(Ik1k2⋯km))<1/2. For simplicity write Ik1k2⋯km as {In}n=1∞. For each t∈I, 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 [6] 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
An⊆En,μ*(An)=μ(En),diam(g0(An))<12anddiam(fi(En))<12,1≤i≤m.
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):1≤k<∞} of En in measurable sets and a sequence {A(n,k):1≤k<∞} 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:1≤n<∞}. On this Δ consider the partial order defined by (m1,m2,…,mi)<(n1,n2,…,nj) if and only if i≤j and mk=nk, k=1,2,…,i. Then by induction for each n, we can find a countable partition {Eα:α∈Δn} of I of measurable sets and a collection {Aα:α∈Δn} of subsets of I such that:

Aα⊆Eα and μ*(Aα)=μ(Eα),

Eβ⊆Eα and Aβ⊆Aα if α≤β,

diam(fi(Eα))<1/2n for i=1,2,…,m and diam(g0(Aα))<1/2n for α∈Δn.

We may assume Aα≠ϕ, for all α. For each α∈Δ, let tα∈Aα and define gn from I into G by gn(t)=∑α∈ΔnχEα(t)g0(tα). Then, for each t∈I and n≤m, we have ‖gn(t)-gm(t)‖=‖∑α∈ΔnχEα(t)g0(tα)-∑β∈ΔmχEβ(t)g0(tβ)‖≤‖∑β∈ΔmχEβ(t)(g0(tα)-g0(tβ))‖≤∑β∈Δm‖(g0(tα)-g0(tβ))‖χEβ≤12n.
Hence, (gn(t)) is a Cauchy sequence in G for all t∈I. Therefore (gn(t)) is a convergent sequence. Let g:I→G 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 t∈I, 1≤n<∞, and t∈Eα for some α∈Δn, we have
max1≤i≤m‖fi(t)-gn(t)‖=max1≤i≤m‖fi(t)-g0(tα)‖≥max1≤i≤m|‖fi(tα)-g0(tα)‖-‖fi(t)-fi(tα)‖|≥max1≤i≤m|‖fi(tα)-g0(tα)‖-12n|≥ρ(f1(tα),f2(tα),…,fm(tα),G)-12n.
For 1≤i≤m, the inequality
‖fi(t)-a‖≤‖fi(t)-fi(tα)‖+‖fi(tα)-a‖≤12n+‖fi(tα)-a‖
implies that
ρ(f1(t),f2(t),…,fm(t),G)≤12n+ρ(f1(tα),f2(tα),…,fm(tα),G).
Therefore,
max1≤i≤m‖fi(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)=max1≤i≤m‖fi(t)-g(t)‖.
This completes the proof of the theorem.

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