${}^{\ast }$Department of Mathematics, Tafila Technical University, Tafila, Jordan, e-mail: abu-sirhan@ttu.edu.jo.

1 Introduction

Let $Z$ be a Banach space and $Y$ be a closed subspace of $Z$. For a subset $B$ of $Z,$ define

An element $y^{\ast }\in Y$ is said to be a best simultaneous approximant to the subset $B$ if

The theory of best simultaneous approximation has been studied by many authors. Most of these works have dealt with the space of continuous functions with values in a Banach space e.g. [18, 10, 4]. Some recent results for best simultaneous approximation in the Banach space of $P$-Bochner integrable (essentially bounded) functions have been obtained in [5, 6, 9, 16, 19]. We consider here a problem of simultaneous approximation completing the work done in $[6,19]$. In this paper $(\mathrm{\Omega },\mathrm{\Sigma },\mu )$ stands for a complete $\sigma$-finite measure space and $L^{\mathrm{\infty }}(\mu ,X)$ the Banach space of all essentially bounded functions on $(\mathrm{\Omega },\mathrm{\Sigma },\mu )$ with values in a Banach space $X,$ endowed with the usual norm

In [19], it is shown that if $G$ is a reflexive subspace of a Banach space $X$, then $L^{\mathrm{\infty }}(\mu ,G)$ is simultaneously proximinal in $L^{\mathrm{\infty }}(\mu ,X).$ In [6], it is shown that if $G$ is a separable $w^{\ast }$-closed subspace of a dual space $X$, then $L^{\mathrm{\infty }}(\mu ,G)$ is simultaneously proximinal in $L^{\mathrm{\infty }}(\mu ,X).$ The aim of this paper is to show that if $G$ is a closed separable subspace of a Banach space $X$, then $L^{\mathrm{\infty }}(\mu ,G)$ is simultaneously proximinal in $L^{\mathrm{\infty }}(\mu ,X)$ if and only if $G$ is simultaneously proximinal in $X.$

2 Preliminary Results

The following Theorem is a generalization of the distance formula given in [6]. Here ${\chi }_A$ is denoted to the characteristic function of $A.$

Let $f_1,f_2,...,f_m\in L^{\mathrm{\infty }}(\mu ,X)$. Being strongly measurable functions, there exist sequences of simple functions $(f_{(i,n)}{)}_{n=1}^{\mathrm{\infty }},$ $i=1,2,...,m,$such that

for $i=1,2,...,m,$ and for almost all $t$’s. We may write, [9],

$$ where $A(n,j)$ are disjoint and $\stackrel{k\left(n\right)}{\underset{j=1}{\cup }}A(n,j)=\mathrm{\Omega }.$ Define $d_n(\cdot ):\mathrm{\Omega }\to \mathbb{R}$ by

Then

and

for almost all $s$. Thus $d(\{f_i(\cdot ):1\le i\le m\},G)$ is measurable.

Now, for any $h\in L^{\mathrm{\infty }}(\mu ,G),$

Hence

To prove the reverse inequality, let $ϵ>0$ be given and $w_i,i=1,2,...,m,$ be countably valued functions in $L^{\mathrm{\infty }}(\mu ,X)$ such that

We may write $w_i=\sum _{k=1}^{\mathrm{\infty }}{\chi }_{A_k}(\cdot )x_{(i,k)},$ where $A_k$ are disjoint, $\stackrel{\mathrm{\infty }}{\underset{k=1}{\cup }}{A}_k=\mathrm{\Omega },$ and $\mu \left(A_k\right)>0$, for all $k.$ Let $h_k\in G$ be such that

for all $k.$ Let $g=\sum _{k=1}^{\mathrm{\infty }}{\chi }_{A_k}(\cdot )h_k.$ It is clear that $g\in L^{\mathrm{\infty }}(\mu ,G).$ Further

Assume that$g\left(s\right)$ is a best simultaneous approximation of
$f_1\left(s\right),f_2\left(s\right),...,f_m\left(s\right)$ for almost all $s.$ Then

for almost all $s$, and for all $z\in G.$ Then

for almost all $s,$ therefore $g\in L^{\mathrm{\infty }}(\mu ,G).$ By Theorem 2.1,

Therefore $g$ is a best simultaneous approximation for $f_1,f_2,..,f_m$ in
$L^{\mathrm{\infty }}(\mu ,G).$

The condition in Corollary 2.2 is sufficient; $g\left(s\right)$ is a best simultaneous approximation of $f_1\left(s\right),f_2\left(s\right),...,f_m\left(s\right)$ for almost all $s$ in $G,$ implies $g$ is a best simultaneous approximation of $f_1,f_2,...,f_m$ in $L^{\mathrm{\infty }}(\mu ,G).$ For the converse, we need the following easy lemma.

Let $g\in L^{\mathrm{\infty }}(\mu ,G),$ then

for all $s\in \mathrm{\Omega }.$

Thus $\underset{1\le i\le m}{sup}{\Vert f_i-g\Vert }_{\mathrm{\infty }}\ge 1.$ Since $g\in L^{\mathrm{\infty }}(\mu ,G)$ was arbitrary, then

To prove the converse inequality. Let $ϵ>0$ be given. Let $f_1^{{}^{\prime }},f_2^{{}^{\prime }},...,f_m^{{}^{\prime }}\in L^{\mathrm{\infty }}(\mu ,X)$ be countably valued functions such that $f_i^{{}^{\prime }}\left(\mathrm{\Omega }\right)\subset$ $f_i\left(\mathrm{\Omega }\right)$,
$i=1,2,\dots ,m,$ and

We may write $f_i^{{}^{\prime }}=\sum _{k=1}^{\mathrm{\infty }}{\chi }_{A_k}(\cdot )x_{(i,k)},$ with the subsets $A_k$ disjoint and measurable, and $x_{(i,k)}\in$ $f_i\left(\mathrm{\Omega }\right).$ For each $k$ take $h_k\in G$ such that

It is clear that $g$ defined by

belongs to $L^{\mathrm{\infty }}(\mu ,G)$ and

Therefore,

Sufficiency of the condition is an immediate consequence of Corollary 2.2. We will show the necessity. Assume that $L^{\mathrm{\infty }}(\mu ,G)$ is simultaneously proximinal in $L^{\mathrm{\infty }}(\mu ,X)$ and take $f_1,f_2,...,f_m$ in $L^{\mathrm{\infty }}(\mu ,X)$. Consider the non-negative measurable function

Take ${\mathrm{\Omega }}_0=\{s\in \mathrm{\Omega }:h\left(s\right)=0\},$ and for each $n=1,2,...,$take
${\mathrm{\Omega }}_n=\{s\in \mathrm{\Omega }:n-1<h\left(s\right)\le n\}.$ Of course, we may forget those ${\mathrm{\Omega }}_n$ which are $\mu -$null sets, so, without loss of generality, we will assume that $\mu \left({\mathrm{\Omega }}_n\right)>0$ for all $n.$ Now for each $n=1,2,...,$ we define $f_{in}:\mathrm{\Omega }\to X$ by

It is clear that $f_{in}$ $\in L^{\mathrm{\infty }}(\mu ,X),$ $i=1,2,...,m,$ and also that

for all $s\in {\mathrm{\Omega }}_n.$ So, it follows from proceeding lemma that

On the other hand, using simultaneous proximinality of $L^{\mathrm{\infty }}(\mu ,G),$ we deduce that there exists $g_n\in L^{\mathrm{\infty }}(\mu ,G)$ such that

Therefore, we have

for almost all $s\in {\mathrm{\Omega }}_n.$ Then,

for almost all $s\in {\mathrm{\Omega }}_n.$ Thus

for almost all $s\in {\mathrm{\Omega }}_n.$ We do notice that, if $s\in {\mathrm{\Omega }}_0,$ then $f_i\left(s\right)=f_j\left(s\right),$ for 1$\le i\le j\le m.$ Hence, it is clear that $g$ defined by

for all $s\in \mathrm{\Omega }$ enjoys the required property.

3 Main Result

The following lemmas will be used to prove our main result.

Let $\mathrm{\Phi }$ be a set-valued mapping, taking each point of a measurable space $\mathrm{\Omega }$ into a subset of a metric space $X.$ We say that $\mathrm{\Phi }$ is weakly measurable if ${\mathrm{\Phi }}^{-1}(O)$ is measurable in $\mathrm{\Omega }$ whenever $O$ is open in $X.$ Here we have put, for any $A$ $\subset$ $X,$

The following theorem is due to Kuratowski [17], it is known as Measurable Selection Theorem.

$\left(2\right)\Rightarrow \left(1\right):$ Let $x_1,x_2,...,x_m$ be any finite number of elements in $X.$ Define $f_i:\mathrm{\Omega }\to X,$ $i=1,2,...,m,$ by

Using simultaneous proximinality of $L^{\mathrm{\infty }}(\mu ,G)$ and Theorem 2.4, we get $g\in L^{\mathrm{\infty }}(\mu ,G)$ such that $g\left(s\right)$ is a best simultaneous approximation of
$f_1\left(s\right),f_2\left(s\right),...,f_n\left(s\right)$ for almost all $s.$ Choose $s_0\in \mathrm{\Omega }$ so that $g\left(s_0\right)$ is a best simultaneous approximation of $x_1,x_2,...,x_m$ in $G.$

$\left(1\right)\Rightarrow \left(2\right):$ Let $f_1,f_2,...,f_m$ be any finite number of elements in $L^{\mathrm{\infty }}(\mu ,X).$ For each $s\in \mathrm{\Omega }$ define

For each $s\in \mathrm{\Omega },$ $\mathrm{\Phi }\left(s\right)$ is closed, bounded, and nonvoid subset of $G.$ We shall show that $\mathrm{\Phi }$ is weakly measurable. Let $O$ be an open set in $X,$ the set

can be also be described as

Since $(\mathrm{\Omega },\mathrm{\Sigma },\mu )$ is complete, $f_i$ is measurable in the classical sense for $i=1,2,...,m,$ by Lemma 3.1. Since subtraction in $X,$ $sup,$ and the norm in $X$ are continuous, then the map

is measurable for any set $A.$ It follows that ${\mathrm{\Phi }}^{-1}(O)$ is measurable. By Theorem 3.3, $\mathrm{\Phi }$ has a measurable selection; i.e., there exists a function  $f:\mathrm{\Omega }\to G$ such that $f(s)\in \varphi (s)$ for each $sϵ\mathrm{\Omega }$ $$and $f$ is measurable in the classical sense. By Lemma 3.2, $f$ is strongly measurable. Hence $f$ is a best simultaneous approximation for $f_1,f_2,...,f_m$ in $L^{\mathrm{\infty }}(\mu ,G)$ by Theorem 2.4.