Abstract

Let be a Banach space and let be a closed bounded subset of . For , we set  . The set is called simultaneously remotal if, for any , there exists such that  . In this paper, we show that if is separable simultaneously remotal in , then the set of -Bochner integrable functions, , is simultaneously remotal in . Some other results are presented.

1. Introduction

Let be a Banach space and a bounded subset of . For , set . A point is called a farthest point of if there exists such that . For , the farthest point map , that is, the set of points of farthest from . Note that this set may be empty. Let . We call a closed bounded set remotal if and densely remotal if is a norm dense in . 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, , , are given in [6–8].

The problem of approximating a set of points simultaneously by a point (farthest point) in a subset of can be done in several ways, see [9]. Here, we will use the following definition.

Definition 1.1. Let be a closed bounded subset of . A point is called a simultaneous farthest point of if We call a closed bounded set of a Banach space simultaneously remotal if each -tuple admits a farthest point in and simultaneously densely remotal if the set of points , where is norm dense in .

Clearly, if , then simultaneously remotal is precisely remotal.

In this paper we consider the problem of simultaneous farthest point for bounded sets of the form in the Banach space , where is a Banach space.

Throughout this paper, is a Banach space, is a closed bounded subset of and , the space of all -valued essentially bounded functions on the unit interval . For , we set . For , we set , almost all .

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 to a bounded set . We begin with the following proposition.

Proposition 2.1. Let , then = .

Proof. For , Hence,

Theorem 2.2. Let be a Banach space and let be a closed bounded subset of . If a function defined by , where , then and

Proof. Let . Being strongly measurable, there exist sequences of simple functions , such that as for almost all . We may write . Since is a continuous function of , the inequality implies that Set . Then, So is a simple function for each and for almost all . Hence is measurable. Furthermore, for each , Thus, To prove the reverse inequality. Let be given, since countably valued functions are dense in , there exist countably valued functions in such that , . We may write , as in [10]. We may assume , for all . For each , Choose such that Now, set as . The inequality implies Further, For and , the inequality implies Therefore, Hence, . Since was arbitrary, we have .

Corollary 2.3. Let be a strongly measurable function from to a closed bounded subset of a Banach space , and . If is a simultaneous farthest point of in , then is a simultaneous farthest point of in .

Proof. By assumption, for almost . Since is bounded, it follows that and Theorem 2.2 and Proposition 2.1 implies that and is a simultaneous farthest point of in .

3. Remotal Sets in

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

Theorem 3.1. If is simultaneously densely remotal in , then is simultaneously densely remotal in .

Proof. Let . Then, there exist simple functions such that , . Now, we can write without lost of generality, . Since is simultaneously densely remotal, then there exist an -tuple and such that and Set and . Then, for every . In particular, for any , using Proposition 2.1, Hence, is a farthest point from the -tuple . But Hence, This complete the proof of the theorem.

For a remotal set , the map defined by is a multivalued map in general. Hence for any , the map is a multivalued map from into .

Before proceeding we remind the reader for some facts regarding mult-valued maps. For a Banach space and a measurable space a function is said to be strongly measurable if it is the pointwise limit of a sequence of simple functions almost everywhere. On the other hand is said to measurable in the classical sense if is measurable in for any closed set in , see [10]. A multivalued function is said to measurable in the classical sense if is measurable in for any closed set in , here . A measurable in the classical sense may be extracted from a measurable multivalued function where is a separable Banach space provided that is a closed subset of for each and such that for each , see [11, page 289].

Theorem 3.2. Let be a closed bounded simultaneously remotal in such that is finite dimensional subspace of , then is simultaneously remotal in .

Proof. First, we will prove that is a closed valued map. If is finite set, then it is closed. If is not finite, let , then there exists such that . This implies for every . Taking the limit as , we get for every , and this implies that and is a closed multivalued map.
To prove that is measurable in the classical sense let be any closed subset of . If , then there exists that converges to . Since , then . Choose . Then, has a convergent subsequence being a sequence in a bounded closed subset of a finite dimensional space. But for every . Taking the limit as , we get for every . Hence . Therefore and is closed. Hence, is measurable and if is the vector valued map , , then is a measurable closed multivalued map. By Theorem 6.6.4 in [11, page 289], has a measurable selection say . Further, for every and so Hence, for every and is remotal in .

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 be a separable simultaneously remotal subset of . Then is simultaneously remotal in .

Proof. Let . Using Corollary 2.3, it is sufficient to show that there exists a measurable function defined on such that is a simultaneous farthest point of in . Since are strongly measurable. We may assume that are separable sets in . So there exist a countable partition of such that diam, for each , , for all ; where Consider the partition , where , for . Then diam. For simplicity write as . For each , let be a simultaneous farthest point from in . Define the map from into by is a simultaneous farthest point from . Apply Lemma 3.1 in [6] with and , we get countable partitions in each and therefore countable partition in the whole of in measurable sets and a sequence of subsets such that Repeat the same argument in each with , , and . For each , we get a countable partition of in measurable sets and a sequence of subsets of such that Now, we will use mathematical induction for each ; let be the set of -tuples of natural numbers and . On this consider the partial order defined by if and only if and , . Then by induction for each , we can find a countable partition of of measurable sets and a collection of subsets of such that:(1) and ,(2) and if ,(3)diam for and diam for .
We may assume , for all . For each , let and define from into by . Then, for each and , we have Hence, is a Cauchy sequence in for all . Therefore is a convergent sequence. Let be defined to be the pointwise limit of . Since is strongly measurable for each , we have is strongly measurable. Further for , , and for some , we have For , the inequality implies that Therefore, Taking limit as , we get This completes the proof of the theorem.