IJMMS International Journal of Mathematics and Mathematical Sciences 1687-0425 0161-1712 Hindawi Publishing Corporation 196391 10.1155/2014/196391 196391 Research Article Coproximinality in the Space of Bounded Functions http://orcid.org/0000-0003-1360-5554 Abu-Sirhan Eyad http://orcid.org/0000-0002-0703-3679 Altawallbeh Zuhier Mohapatra Ram N. 1 Department of Mathematics Tafila Technical University, Tafila 6610 Jordan 2014 742014 2014 16 01 2014 25 03 2014 7 4 2014 2014 Copyright © 2014 Eyad Abu-Sirhan and Zuhier Altawallbeh. 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.

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 subset M of a Banach space X and of C(S, M) in C(S, X).

1. Introduction

Let M be a nonempty subset of a Banach space X. We say that M is proximinal in X if each xX there corresponds at least to one point m0M such that (1)x-m0x-m,mM.

Recently, another kind of approximation from M has been introduced by Franchetti and Furi  who have considered those elements (if any) m0M satisfying (2)m-m0x-m,mM,andforanyfixedxX.

Such an element m0 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 xX. If RM(x) is singleton for any xX, then M is called coChebyshev; see .

It is clear that RM(x) is convex if M is convex and closed. The kernal of RM is the set defined by (3)kerRM={xX:mm-x,mM}.

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 ). Let (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 (S,X). In this paper, we discuss the coproximinality of (S,M) and C(S,M) in (S,X) and C(S,X), respectively. For uniqueness and existence of best coapproximation in C(S,X), see .

Definition 1.

Let M be a coproximinal subset of a Banach space X. A map FM:XM which associates with each element in X one of its best coapproximation in M (i.e., FM(x)RM(x) for all xX) is called a coproximity map.

Remark 2.

Let M be a coproximinal subset of a Banach space X. We state some basic properties of a proximity map FM.

If M is coChebyshev, , then

FM(m)=m, for any mM.

FM(αx)=αFM(x), for any scalar α and xX. (i.e., FM is homogeneous).

FM(m+x)=m+FM(x), for any xX and mM.

If M is a subspace of X, then

FM(x)-FM(y)x-m+y-m, for any x,yX and mM.

FM(x)x, for all xX. (set y=0 and m=0 in (a) and take into account that FM(0)=0).

FM is continuous at x=0.

If FM is linear, then FM is continuous.

Theorem 3 (see [<xref ref-type="bibr" rid="B18">3</xref>]).

If M is coproximinal hyperplane or 1-dimensional subspace of a Banach space X, then M has a continuous coproximity map.

Lemma 4 (see [<xref ref-type="bibr" rid="B18">3</xref>]).

Let M be a coproximinal subspace of X. Then the following are equivalent.

M has linear coproximity map.

kerRM contains a closed subspace W such that X=WM. Moreover, if (2) holds, then the linear coproximity map for M can be defined by P(m+w)=m, mM, and wW.

Definition 5.

Let X be a Banach space. Consider the following.

A subspace M of X is called one complemented in X if there is a closed subspace W such that X=WM and the projection P:XM is contractive.

A linear projection P is called an M-projection if x=max{P(x),x-P(x)} for all xX. A closed subspace J of X is called an M-summand if it is the range of an M-projection, .

A linear projection P is called an L-projection if x=P(x)+x-P(x) for all xX. A closed subspace J of X is called an L-summand if it is the range of an L-projection.

Clearly every M-summand is one complemented.

Theorem 6.

Let M be a subspace of a Banach space X. Then the following are equivalent.

M is one complemented in X.

M is coproximinal in X and has linear coproximity map.

Proof.

( 1 ) ( 2 ) : Let X=MW, P:XM a contractive projection, and xX. Then x=m1+w1, (4)P(x-m)x-m,mM,m-m1x-m,mM.

Hence M is coproximinal in X. Now let wW. Then for any mM,   P(w-m)=mw-m. Thus WkerRM and M has linear coproximity map by Lemma 4.

( 2 ) ( 1 ) : If M has a linear coproximity map, then kerRM contains a closed subspace W such that X=MW. Moreover, P(m+w)=m, m+wMW 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 xX, then m-P(x)m-x for all mM, and so p(x)x which means that P is contractive.

Corollary 7.

If M 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 .

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Let S be a compact Hausdorff space and X be a Banach space. We denote (S,X) and C(S,X) to be the Banach space of all bounded ( resp., continuous) functions from S into X.

Lemma 8 (see [<xref ref-type="bibr" rid="B12">19</xref>]).

Let A 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.

Let M 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.

Let ϕ:XM 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 hC(S,M), then (6)h(s)-ϕ(f(s))f(s)-h(s) for all sS, (7)h-ϕoff-h.

Hence (8)h-Φ(f)f-h for all hC(S,M) and Φ is a continuous coproximity map from C(S,X) onto C(S,M).

Lemma 10.

Let M 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.

For xX, define fx:SX by fx(s)=x for all sS. The map fxC(S,X). By assumption, there exists g0C(S,M) such that (9)g0-hfx-h for all hC(S,M). In particular, (10)g0-fmfx-fm for all mM. For any fixed s0S, (11)g0(s0)-mx-m for all mM 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 s0S, define ϕ:XM by ϕ(x)=Φfx(s0). Then ϕ is a proximity map. Suppose that xnx in X. Then Φ(fxn)Φ(fx). and (12)ϕ(xn)-ϕ(x)=Φ(fxn)(s0)Φ(fx)(s0)Φ(fxn)-Φ(fx)0.

Hence ϕ is continuous.

Theorem 11.

Let M be a closed subset of a Banach space X. Then the following are equivalent.

M is coproximinal in X and has a continuous coproximity map.

C(S,M) is coproximinal in C(S,X) and has a continuous coproximity map.

Proof.

The proof follows from Lemmas 9 and 10.

Theorem 12.

Let M be a closed subset of a Banach space X and let S be a topological space. Then the following are equivalent.

M is coproximinal in X.

(S,M) is coproximinal in (S,X).

Proof.

( 1 ) ( 2 ) . Let f(S,X). Since M is coproximinal in X, then RM(f(s)) is nonempty for any sS. By axiom of choice, we may define a function h:SM such that h(s)RM(f(s)) for all sS. Since   h(s)f(s) for all sS,h(S,M) and h is a best coapproximation of f in (S,X).

( 2 ) ( 1 ) . For xX, define fx:SX by fx(s)=x for all sS. Since (S,M) is coproximinal in (S,X), the map fx(S,X). By assumption, there exists g0(S,M) such that (13)g0-hfx-h for all h(S,M). In particular, (14)g0-fmfx-fm for all mM. For any fixed s0S, (15)g0(s0)-mx-m for all mM and g0(s0) is a best coapproximation of x.

We recall that.

Definition 13.

Let Φ 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-={sS:Φ(s)Oϕ} is open.

Theorem 14 (see [<xref ref-type="bibr" rid="B12">19</xref>, Michael Selection Theorem]).

Let Φ 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:SX such that φ(s)Φ(s) for all sS.

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.

Let S be a compact Hausdorff space, X a Banach space, and G a compact subset of X. For fC(S,X), we define πf:S2G by (16)πf(s)={g0G:g0-gg-f(s),gG}.

Then πf is lower semicontinuous.

Proof.

Let O be an open set in G. We need to prove that (17)πf-(O)={sS:πf(s)Oϕ} is open. For this purpose, let {xn} be a sequence in S/πf-(O) and xnx as n, so xnπf-(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)limsupπf(xn)πf(x). Let zlimsupπf(xn). Then there is a sequence of points {zk} and a subsequence {πf(xnk)} of {πf(xn)} such that zkπf(xnk) and zkz as k. By the definition of πf, we get (20)zk-gf(xnk)-ggG. Taking k and nk, and since f is continuous, we have (21)z-gf(x)-ggG, so zπf(x). This proves that limsupπf(xn)πf(x). Now we shall show that πf(x)liminfπf(xn). Let zliminfπf(xn). In this case, if zkz as k then zkπf(xk) for infinitely many k, and, so, (22)zk-gk>f(xk)-gkforsomegkG. We may assume that, without loss of generality, gkg as k, for some gG, since G is compact. Taking k, we have that (23)z-g>f(x)-g, so zπf(x) and we have as a conclusion the follwing result: (24)liminfπf(xn)limsupπf(xn)πf(x)liminfπf(xn). Thus, limπf(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)limπf(xn)GO. To do so, let zlimsupπf(xn). By the definition of limsupπf(xn), there is a sequence of points {zk} and a subsequence {πf(xnk)} of the sequence {πf(xn)} such that (26)zkπf(xnk)GOk,zkzask. Since G/O is closed, we get zG/O. This proves that πf(x)G/O. This means that πf(x)O=ϕ, and so xπf-(O). Thus πf-(O) is open set and so πf is lower semicontinuous.

Theorem 17.

Let M be a compact convex subset of a Banach space X and S be a compact Hausdorff space. Then the following are equivalent.

M is coproximinal in X.

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 fC(S,X) and πf:S2M 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:SM. Then gC(S,M) and (27)g(s)-h(s)f(s)-h(s),sS,hC(S,M),g-hf-h,hC(S,M).

Hence g is a best coapproximation of f in C(S,M).

Corollary 18.

Let S 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.

Since X 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.

If S is a compact Hausdorff space and X is a Banach space, when C(S,X) is coproximinal in (S,X).

Conflict of Interests

The authors declare that there is no conflict of interests regarding the publication of this paper.

Acknowledgment

Finally the author would like to thank the referees for their valuable advice.

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