# Coproximinality in the Space of Bounded Functions

Authors: Eyad Abu-Sirhan; Zuhier Altawallbeh
Journal: International Journal of Mathematics and Mathematical Sciences
 (2014)

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## Abstract

As a counterpart to best approximation in Banach spaces, the best coapproximation was introduced by Franchetti and Furi
(1972). In this paper, we shall consider the relation between coproximinality of a nonempty subsetM of a Banach space X and of C(S, M) in C(S, X).


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## Body

## 1. Introduction

LetM be a nonempty subset of a Banach space X. We say that M is proximinal in X if each x∈X there corresponds at least to one point m0∈M such that
(1)∥x-m0∥≤∥x-m∥,∀m∈M.Recently, another kind of approximation fromM has been introduced by Franchetti and Furi [1] who have considered those elements (if any) m0∈M satisfying
(2)∥m-m0∥≤∥x-m∥,∀m∈M,andforanyfixedx∈X.Such an elementm0 is called best coapproximant of x in M. The set of all such elements, satisfying above inequality, is denoted by RM(x). The subset M is called coproximinal in X if RM(x) is nonempty for any x∈X. If RM(x) is singleton for any x∈X, then M is called coChebyshev; see [2–4].It is clear thatRM(x) is convex if M is convex and closed. The kernal of RM is the set defined by
(3)ker⁡RM={x∈X:∥m∥≤∥m-x∥,∀m∈M}.Many results in the theory of best coapproximation have appeared since Franchetti and Furri’s paper, 1972. These results are largely concerned with the question of existence and uniqueness of best coapproximation (see for example [5–8]). Let ℓ∞(S,X) and C(S,X) denote the Banach spaces of all bounded (resp., continuous) functions from a topological space S into a Banach space X and let M be a closed subset of X. It should be remarked that if S is a compact space, then C(S,X) is a subspace of ℓ∞(S,X). In this paper, we discuss the coproximinality of ℓ∞(S,M) and C(S,M) in ℓ∞(S,X) and C(S,X), respectively. For uniqueness and existence of best coapproximation in C(S,X), see [7].Definition 1.
LetM be a coproximinal subset of a Banach space X. A map FM:X→M which associates with each element in X one of its best coapproximation in M (i.e., FM(x)∈RM(x) for all x∈X) is called a coproximity map.Remark 2.
LetM be a coproximinal subset of a Banach space X. We state some basic properties of a proximity map FM.(1)
IfM is coChebyshev, [5], then
(a)
FM(m)=m, for any m∈M.(b)
FM(αx)=αFM(x), for any scalar α and x∈X. (i.e., FM is homogeneous).(c)
FM(m+x)=m+FM(x), for any x∈X and m∈M.(2)
IfM is a subspace of X, then
(a)
∥FM(x)-FM(y)∥≤∥x-m∥+∥y-m∥, for any x,y∈X and m∈M.(b)
∥FM(x)∥≤∥x∥, for all x∈X. (set y=0 and m=0 in (a) and take into account that FM(0)=0).(c)
FM is continuous at x=0.(d)
IfFM is linear, then FM is continuous.Theorem 3 (see [3]).
IfM is coproximinal hyperplane or 1-dimensional subspace of a Banach space X, then M has a continuous coproximity map.Lemma 4 (see [3]).
LetM be a coproximinal subspace of X. Then the following are equivalent.(1)
M has linear coproximity map.(2)
ker⁡RM contains a closed subspace W such that X=W⊕M. Moreover, if (2) holds, then the linear coproximity map for M can be defined by P(m+w)=m, m∈M, and w∈W.Definition 5.
LetX be a Banach space. Consider the following.(a)
A subspaceM of X is called one complemented in X if there is a closed subspace W such that X=W⊕M and the projection P:X→M is contractive.(b)
A linear projectionP is called an M-projection if ∥x∥=max⁡{∥P(x)∥,∥x-P(x)∥} for all x∈X. A closed subspace J of X is called an M-summand if it is the range of an M-projection, [9].(c)
A linear projectionP is called an L-projection if ∥x∥=∥P(x)∥+∥x-P(x)∥ for all x∈X. A closed subspace J of X is called an L-summand if it is the range of an L-projection.
Clearly everyM-summand is one complemented.Theorem 6.
LetM be a subspace of a Banach space X. Then the following are equivalent.(1)
M is one complemented in X.(2)
M is coproximinal in X and has linear coproximity map.Proof.
(
1
)
→
(
2
): Let X=M⊕W, P:X→M a contractive projection, and x∈X. Then x=m1+w1,
(4)∥P(x-m)∥≤∥x-m∥,∀m∈M,∥m-m1∥≤∥x-m∥,∀m∈M.
HenceM is coproximinal in X. Now let w∈W. Then for any m∈M,   ∥P(w-m)∥=∥m∥≤∥w-m∥. Thus W⊂ker⁡RM and M has linear coproximity map by Lemma 4.
(
2
)
→
(
1
): If M has a linear coproximity map, then ker⁡RM contains a closed subspace W such that X=M⊕W. Moreover, P(m+w)=m, m+w∈M⊕W is a linear coproximity map for M and so P is the projection of M along W by Lemma 4. To show that P is contractive, if x∈X, then ∥m-P(x)∥≤∥m-x∥ for all m∈M, and so ∥p(x)∥≤∥x∥ which means that P is contractive.Corollary 7.
IfM is L-summand, then M is coproximinal and has continuous coproximity map.For more information about coproximinal sets, optimal sets, and contractive sets, the reader is referred to [10–18].


## 2. Coproximinality inℓ∞(S,X) and C(S,X)

LetS be a compact Hausdorff space and X be a Banach space. We denote ℓ∞(S,X) and C(S,X) to be the Banach space of all bounded ( resp., continuous) functions from S into X.Lemma 8 (see [19]).
LetA be a continuous map of a Banach space X into a closed subset Y of X. If S is compact Hausdorff space, then the map defined by Φ(z)=Aoz is continuous from C(S,X) to C(S,Y).Lemma 9.
LetM be a closed subset of a Banach space X. If there is a continuous coproximity map of X onto M, then C(S,M) is coproximinal in C(S,X) and in fact it has a continuous coproximity map.Proof.
Letϕ:X→M be a continuous coproximity map for M. Define
(5)Φ:C(S,X)⟶C(S,M)
by Φ(f)=ϕof, then Φ is continuous by Lemma 8. Let h∈C(S,M), then
(6)∥h(s)-ϕ(f(s))∥≤∥f(s)-h(s)∥
for all s∈S,
(7)∥h-ϕof∥≤∥f-h∥.
Hence(8)∥h-Φ(f)∥≤∥f-h∥
for all h∈C(S,M) and Φ is a continuous coproximity map from C(S,X) onto C(S,M).Lemma 10.
LetM be a closed subset of a Banach space X. If C(S,M) is coproximinal in C(S,X), then M is coproximinal in X. Moreover, if C(S,M) has a continuous coproximity map, then M has a continuous coproximity map.Proof.
Forx∈X, define fx:S→X by fx(s)=x for all s∈S. The map fx∈C(S,X). By assumption, there exists g0∈C(S,M) such that
(9)∥g0-h∥≤∥fx-h∥
for all h∈C(S,M). In particular,
(10)∥g0-fm∥≤∥fx-fm∥
for all m∈M. For any fixed s0∈S,
(11)∥g0(s0)-m∥≤∥x-m∥
for all m∈M and so g0(s0) is a best coapproximation of x. Now let Φ:C(S,X)→C(S,M) be a continuous coproximity map. For any fixed s0∈S, define ϕ:X→M by ϕ(x)=Φfx(s0). Then ϕ is a proximity map. Suppose that xn→x in X. Then Φ(fxn)→Φ(fx). and
(12)∥ϕ(xn)-ϕ(x)∥=∥Φ(fxn)(s0)⟶Φ(fx)(s0)∥≤∥Φ(fxn)-Φ(fx)∥⟶0.
Henceϕ is continuous.Theorem 11.
LetM be a closed subset of a Banach space X. Then the following are equivalent.(1)
M is coproximinal in X and has a continuous coproximity map.(2)
C(S,M) is coproximinal in C(S,X) and has a continuous coproximity map.Proof.
The proof follows from Lemmas9 and 10.Theorem 12.
LetM be a closed subset of a Banach space X and let S be a topological space. Then the following are equivalent.(1)
M is coproximinal in X.(2)
ℓ∞(S,M) is coproximinal in ℓ∞(S,X).Proof.
(
1
)
→
(
2
). Let f∈ℓ∞(S,X). Since M is coproximinal in X, then RM(f(s)) is nonempty for any s∈S. By axiom of choice, we may define a function h:S→M such that h(s)∈RM(f(s)) for all s∈S. Since   ∥h(s)∥≤∥f(s)∥ for all s∈S,h∈ℓ∞(S,M) and h is a best coapproximation of f in ℓ∞(S,X).
(
2
)
→
(
1
). For x∈X, define fx:S→X by fx(s)=x for all s∈S. Since ℓ∞(S,M) is coproximinal in ℓ∞(S,X), the map fx∈ℓ∞(S,X). By assumption, there exists g0∈ℓ∞(S,M) such that
(13)∥g0-h∥≤∥fx-h∥
for all h∈ℓ∞(S,M). In particular,
(14)∥g0-fm∥≤∥fx-fm∥
for all m∈M. For any fixed s0∈S,
(15)∥g0(s0)-m∥≤∥x-m∥
for all m∈M and g0(s0) is a best coapproximation of x.
We recall that.Definition 13.
LetΦ be a set valued mapping, taking points of a topological space S into the family of all subsets of a topological space T. The mapping Φ is said to be lower semicontinuous if, for each open set O in T, the set Φ-={s∈S:Φ(s)∩O≠ϕ} is open.Theorem 14 (see [19, Michael Selection Theorem]).
LetΦ be a lower semicontinuous of a paracompact space S into the family of nonvoid closed convex subsets of a Banach space X. Then Φ has a continuous selection; that is, there exists a continuous map φ:S→X such that φ(s)∈Φ(s) for all s∈S.Definition 15.
A sequence of sets{An} in a topological space (X,T) is convergent and has the limit A if and only if limsup⁡{An}=A=liminf⁡{An} where limsup⁡{An} is the set of those elements which are limits of points in An for infinitely many n and liminf⁡{An} is the set of those elements which are limits of points in An for cofinitely many (for all but finitely many) n.Lemma 16.
LetS be a compact Hausdorff space, X a Banach space, and G a compact subset of X. For f∈C(S,X), we define πf:S→2G by
(16)πf(s)={g0∈G:∥g0-g∥≤∥g-f(s)∥,∀g∈G}.
Thenπf is lower semicontinuous.Proof.
LetO be an open set in G. We need to prove that
(17)πf-(O)={s∈S:πf(s)∩O≠ϕ}
is open. For this purpose, let {xn} be a sequence in S/πf-(O) and xn→x as n→∞, so xn∉πf-(O) for all n which means that
(18)πf(xn)∩O=ϕ.
Thus πf(xn)⊂G/O for all n, and {πf(xn)} is a sequence of subsets of G. We shall show that πf(xn)→πf(x) as n→∞. First, we show that
(19)limsup⁡πf(xn)⊆πf(x).
Let z∈limsup⁡πf(xn). Then there is a sequence of points {zk} and a subsequence {πf(xnk)} of {πf(xn)} such that zk∈πf(xnk) and zk→z as k→∞. By the definition of πf, we get
(20)∥zk-g∥≤∥f(xnk)-g∥∀g∈G.
Taking k→∞ and nk→∞, and since f is continuous, we have
(21)∥z-g∥≤∥f(x)-g∥∀g∈G,
so z∈πf(x). This proves that limsup⁡πf(xn)⊆πf(x). Now we shall show that πf(x)⊆liminf⁡πf(xn). Let z∉liminf⁡πf(xn). In this case, if zk→z as k→∞ then zk∉πf(xk) for infinitely many k, and, so,
(22)∥zk-gk∥>∥f(xk)-gk∥forsomegk∈G.
We may assume that, without loss of generality, gk→g as k→∞, for some g∈G, since G is compact. Taking k→∞, we have that
(23)∥z-g∥>∥f(x)-g∥,
so z∉πf(x) and we have as a conclusion the follwing result:
(24)liminf⁡πf(xn)⊆limsup⁡πf(xn)⊆πf(x)⊆liminf⁡πf(xn).
Thus, lim⁡πf(xn)=πf(x). Finally, we know that G/O is a closed set and the members of the sequence of sets {πf(xn)} are all subsets of G/O for all n. Now, to complete the proof, we still need to show that
(25)lim⁡πf(xn)⊆GO.
To do so, let z∈limsup⁡πf(xn). By the definition of limsup⁡πf(xn), there is a sequence of points {zk} and a subsequence {πf(xnk)} of the sequence {πf(xn)} such that
(26)zk∈πf(xnk)⊆GO∀k,zk⟶zask⟶∞.
Since G/O is closed, we get z∈G/O. This proves that πf(x)⊆G/O. This means that πf(x)∩O=ϕ, and so x∉πf-(O). Thus πf-(O) is open set and so πf is lower semicontinuous.Theorem 17.
LetM be a compact convex subset of a Banach space X and S be a compact Hausdorff space. Then the following are equivalent.(1)
M is coproximinal in X.(2)
C(S,M) is coproximinal in C(S,X).Proof.
(
2
)
→
(
1
). The proof is similar to that given in Lemma 10.
(
1
)
→
(
2
). Let f∈C(S,X) and πf:S→2M as defined in Lemma 16. Then πf is lower semicontinuous by Lemma 16. Since M is closed, nonvoid, and convex, then πf(s) is closed, nonvoid, and convex. By Michael Selection Theorem, πf has a continuous selection g:S→M. Then g∈C(S,M) and
(27)∥g(s)-h(s)∥≤∥f(s)-h(s)∥,∀s∈S,h∈C(S,M),∥g-h∥≤∥f-h∥,∀h∈C(S,M).
Henceg is a best coapproximation of f in C(S,M).Corollary 18.
LetS be a compact Hausdorff space and let X be a reflexive Banach space. Then B1(X) (the closed unit ball of X) is coproximinal in X if and only if C(S,B1(X)) is coproximinal in C(S,X).Proof.
SinceX is reflexive, then B1(X) is compact by Alaoglu Theorem. As B1(X) is closed and convex subset of X, the result follows from Theorem 17.Problem 19.
IfS is a compact Hausdorff space and X is a Banach space, when C(S,X) is coproximinal in ℓ∞(S,X).



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