Recently, Basha (2013) addressed a problem that amalgamates approximation and optimization in the setting of a partially ordered set that is endowed with a metric. He assumed that if A and B are nonvoid subsets of a partially ordered set that is equipped with a metric and S is a non-self-mapping from A to B, then the mapping S has an optimal approximate solution, called a best proximity point of the mapping S, to the operator equation Sx=x, when S is a continuous, proximally monotone, ordered proximal contraction. In this note, we are going to obtain his results by omitting ordering, proximal monotonicity, and ordered proximal contraction on S.

1. Introduction

Let S be a non-self-mapping from A to B, where A and B are nonempty subsets of a metric space X. Clearly, the set of fixed points of S can be empty. In this case, one often attempts to find an element x that is in some sense closest to S(x). Best approximation theory and best proximity point analysis are applicable for solving such problems. The well-known best approximation theorem, due to Fan [1], asserts that if A is a nonempty, compact, and convex subset of a normed linear space X and S is a continuous function from A to X, then there exists a point x in A such that the distance of x to S(x) is equal to the distance of S(x) to A. Such a point x is called a best approximation point of S in A. A point x in A is said to be a best proximity point for S, if the distance of x to S(x) is equal to the distance of A to B. The aim of best proximity point theory is to provide sufficient conditions that assure the existence of best proximity points. Investigation of several variants of contractions for the existence of a best proximity point can be found in [1–15]. In most of the papers on the best proximity, the ordering, proximal monotonicity, and ordered proximal contraction on the mapping S play a key role. A natural question arises that it is possible that we can have other ways that may not require the ordering as well as proximal monotonicity and ordered proximal contraction on the mapping S. Very recently, Basha [5] addressed a problem that amalgamates approximation and optimization in the setting of a partially ordered set that is endowed with a metric. He assumed that if A and B are nonvoid subsets of a partially ordered set that is equipped with a metric and S is a non-self-mapping from A to B, then the mapping S has an optimal approximate solution, called a best proximity point of the mapping S, to the operator equation Sx=x, when S is a continuous, proximally monotone, ordered proximal contraction. In this note, we are going to obtain his results by omitting ordering, proximal monotonicity, and ordered proximal contraction on S.

2. Preliminary Results

Let X be a nonempty set and let d be a metric on X. Unless otherwise specified, it is assumed throughout this paper that A and B are nonempty subsets of X. We recollect the following notations and preliminary results:
(1)d(A,B):=inf{d(x,y)∣x∈A,y∈B},A0:={x∈A∣d(x,y)=d(A,B)forsomey∈B},B0:={y∈B∣d(x,y)=d(A,B)forsomex∈A}.

Proposition 1.

Let A and B be two compact subsets of a metric space (X,d). Then both A0 and B0 are nonempty sets.

Proof.

Suppose that both A and B are two compact subsets of a metric space (X,d). Let {(an,bn)}⊆A×B such that d(an,bn)→d(A,B) as n→∞. Since A and B are compact, A×B is also compact. There exists {(ank,bnk)}⊆{(an,bn)} such that
(2)(ank,bnk)⟶(a,b)∈A×Bask⟶∞.
Note that (2) is equivalent to
(3)ank⟶a∈A,bnk⟶b∈Bask⟶∞.
Let us consider
(4)d(A,B)≤d(a,b)≤d(a,ank)+d(ank,bnk)+d(bnk,b)≤d(a,ank)+d(A,B)+d(bnk,b).
By employing (3) and letting k→∞ in (4), we obtain d(a,b)=d(A,B). Hence A0≠⌀ and B0≠⌀. This completes the proof.

Proposition 2.

Let A be a compact and let B be a closed subset of the Euclidian space ℝn with norm ∥·∥. Then both A0 and B0 are nonempty set.

Proof.

Suppose that A is compact and B is closed subset of the Euclidian space (ℝn,∥·∥). Let {(an,bn)}⊆A×B such that ∥an-bn∥→d(A,B) as n→∞. Since A is compact, there exists {ank}⊆{an} such that
(5)ank⟶a∈Aask⟶∞.
Note that
(6)∥bnk∥≤∥bnk-ank∥+∥ank∥≤d(A,B)+∥ank∥,
for all k∈ℕ. This means that {bnk} is bounded. It follows from the Bolzano-Weierstrass theorem and the closeness of B that there exists {bnkj}⊆{bnk} such that
(7)bnkj⟶b∈Basj⟶∞.
Let us consider
(8)d(A,B)≤∥a-b∥≤∥a-ankj∥+∥ankj-bnkj∥+∥bnkj-b∥≤∥a-ankj∥+d(A,B)+∥bnkj-b∥.
By employing (5) and (7) and letting j→∞ in (8), we obtain ∥a-b∥=d(A,B). Hence A0≠⌀ and B0≠⌀. This completes the proof.

Proposition 3.

Let A and B be two nonempty subsets of a metric space (X,d). Then the following are satisfied.

If B is compact and A is closed, then A0 is a closed subset of X.

If A is compact and B is closed, then B0 is a closed subset of X.

If both A and B are compact, then A0 and B0 are nonempty and closed.

Proof.

(i) It is trivial in the case of A0=⌀. Suppose that A0≠⌀ and let {an}⊆A0 such that
(9)an⟶a∈Xasn⟶∞.
Note that A0⊆A and A is closed, so we have a∈A. Since an∈A0, there is bn∈B such that d(an,bn)=d(A,B). It follows from the compactness of B that there exists {bnk}⊆{bn} such that
(10)bnk⟶b∈Bask⟶∞.
Now, let {ank}⊆{an} and consider
(11)d(A,B)≤d(a,b)≤d(a,ank)+d(ank,bnk)+d(bnk,b)≤d(a,ank)+d(A,B)+d(bnk,b).
By employing (9) and (10) and letting k→∞ in (11), we obtain d(a,b)=d(A,B). This implies that a∈A0 and, hence, A0 is closed. This completes the proof.

The proof of (ii) is obvious from (i) and also the proof (iii) follows from Proposition 1 and (i) and (ii).

The next result extends Proposition 3.1 of [10] from normed linear spaces to metrizable topological vector spaces.

Proposition 4.

Let X be real topological vector space whose topology is induced by translation invariant metric d with the property
(*)d(tx,x)=(1-t)d(x,0X),∀(0<t<1,x∈X),
where 0X denotes the zero vector of X. Let A and B be two closed subsets of X such that d(A,B)>0. Then
(12)A0⊆bd(A),B0⊆bd(B),
where bd(A) and bd(B) are denoted by the boundary of A and B, respectively.

Proof.

Let a∈A0. Then there exists b∈B such that d(a,b)=d(A,B)>0. It is obvious that A=intA∪bd(A). Let on the contrary a∈intA. Then, there is closed neighborhood of the 0X (the zero vector) and especially positive number ɛ such that a+t(b-a)∈A, for all t∈[0,ɛ]. Let
(13)t0=max{t∈[0,1]:a+t(b-a)∈A}.
It is clear from A∩B=∅ and the closeness of A and B with b∈B that 0<t0<1 and a+t0(b-a)∈A. Hence, it follows from the translation invariant property and (*) that
(14)0<d(a,b)=d(A,B)≤d(a+t0(b-a),b)=d(a+t0(b-a)-a,b-a)=d(t0(b-a),b-a)=(1-t0)d((b-a),0X)=(1-t0)d((b-a+a),0X+a)=(1-t0)d(a,b)<d(a,b),
which is a contradiction. This completes the proof.

The following example shows that there are metrizable topological vector spaces with the properties cited in the previous proposition which are not normable.

Example 5.

Let X be a real vector space and A={Pn}n∈N a countable family of seminorms on X which separates the nonzero points of X from 0X (the zero vector of X). For each y∈X and each index n∈N, define gy,n(x)=Pn(x-y). Let τ be the topology on X generated by the family {g}(y,n)∈X×N. One can see that (X,τ) is a topological vector space (even locally convex space). One can verify the topology τ induced by the translation invariant meter d(x,y)=∑n=1∞2-nmin(Pn(x-y),1). Moreover, for each positive enough small number t, we have d(tx,x)=(1-t)d(x,0X) for each x∈X. However, X is not normable.

3. Main Results

In this section, we provide an existence result for the best proximity point of the mapping S on the metric space X by omitting ordering, proximal monotonicity, and ordered proximal contraction on S.

We begin with an example which shows that it is possible in the finite dimensional Euclidean space that the proximity points set for even a linear mapping (here projection) be empty.

Example 6.

Let X=ℝ2, A={(x,1/x)∣x>0}, and B={(x,0)∣x⩾0}. Define function S:A→B by
(15)S(x,1x)=(x,0),∀(x,1x)∈A.
It is clear that S is continuous (since it is projection). It is not hard to verify that

both A and B are closed subset of X;

d(A,B)=0;

there is no x*∈X such that d(x*,Sx*)=d(A,B) (i.e., there is no best proximity point).

To achieve understanding in Example 6, let us see Figure 1.

Proposition 7.

Let A be a compact subset and let B be a nonvoidsubset of a metric space (X,d). Let S:A→B be continuous with the property that there exists B~∈𝔅 such that B~⊆S(A), where
(16)𝔅={{bn}⊆B∣∃{an}⊆Asuchthatd(an,bn)↓d(A,B)}.
Then, there exists an element x* in A such that
(17)d(x*,Sx*)=d(A,B).

Proof.

Pick {bn}=B~∈𝔅 such that {bn}=B~⊆S(A). Then there exists {an}⊆A such that d(an,bn)↓d(A,B) and bn=San. Since A is compact, there exists {ani}⊆{an} such that
(18)ani⟶x*asi⟶∞.
By using the continuity of S, we can conclude that
(19)d(x*,Sx*)=limi→∞d(ani,Sani)=limi→∞d(ani,bni)=d(A,B).
This completes the proof.

The following result establishes an existence result in order to be nonempty best proximity point set for the mapping S without assuming any ordering, proximal monotonicity, and ordered proximal contraction on the S. It is worth noting that it is only an existence result without applying any iteration method (see Theorem 3.1 of [5]).

Theorem 8.

Let A and B be nonvoidclosed subsets of a complete metric space (X,d) such that A0 is nonvoidand A is totally bounded. Let S:A→B and g:A→A satisfy the following conditions:

S and g are continuous;

S(A0)⊆B0 and A0⊆g(A0).

Then, there exists an element x* in A0 such that
(20)d(gx*,Sx*)=d(A,B). Proof.

Suppose that A0≠⌀. Let x0∈A0 and note that S(A0)⊆B0. Then, we have
(21)Sx0∈S(A0)⊆B0={y∈B∣d(x,y)=d(A,B)forsomex∈A}.
So, we have Sx0∈B and there exists
(22)x0′∈Asuchthatd(x0′,Sx0)=d(A,B).
Equation (22) indicates that x0′∈A0. Since A0⊆g(A0), there exists x1∈A0 such that x0′=gx1. Thus
(23)d(gx1,Sx0)=d(A,B).
In the next step, since x1∈A0, we obtain Sx1∈S(A0)⊆B0. Then, we have Sx1∈B and there exists
(24)x1′∈Asuchthatd(x1′,Sx1)=d(A,B).
Equation (24) indicates that x1′∈A0. Since A0⊆g(A0), there exists x2∈A0 such that x1′=gx2. Thus
(25)d(gx2,Sx1)=d(A,B).
Following by this way, we can produce the sequence {xn}⊆A0 such that
(26)d(gxn+1,Sxn)=d(A,B),∀n∈ℕ.
Since A0⊆A and A is totally bounded, there exists a subsequence {xni}⊆{xn} such that {xni} is a Cauchy sequence. By using the completeness of X, we have
(27)xni⟶x*asi⟶∞.
Applying the continuity of S and g, we obtain
(28)d(gx*,Sx*)=limi→∞d(gxni+1,Sxni)=d(A,B).
This completes the proof.

If g=I (the identity mapping), then Theorem 8 reduces to the following corollary.

Corollary 9.

Let A and B be nonvoidclosed subsets of a complete metric space (X,d) such that A0 is nonvoidand A is totally bounded. Let S:A→B be a continuous function such that S(A0)⊆B0. Then, there exists an element x* in A0 such that
(29)d(x*,Sx*)=d(A,B).

If A=B, then A=A0=B0=B. Then, by Corollary 9, we obtain the following corollary which says that the fixed points set of the mapping S is nonempty.

Corollary 10.

Let A be a nonvoidclosed and totally bounded subset of a complete metric space (X,d). Let S:A→A be continuous. Then, F(S)≠⌀, where F(S) denotes the set of all fixed points of S.

In the following result, we are going to relax the continuity of the mappings S and g (see conditions (a) and (c) of Theorem 3.1 in [5]).

Theorem 11.

Let A be a nonvoidcompact subset and let B be a nonvoidsubset of a complete metric space (X,d). Let S:A→B and g:A→A and define g⊗S:A→A×B by
(30)(g⊗S)(x)=(gx,Sx),∀x∈A.
Suppose that (g⊗S)(A)=A×B and d∘(g⊗S) is lower semicontinuous where d is the distance function of the metric space (X,d). Then, there exists an element x* in A such that
(31)d(gx*,Sx*)=d(A,B).

Proof.

By the assumption, we notice that
(32){(a,b)∣(a,b)∈A×B}=A×B=(g⊗S)(A)={(g⊗S)(a~)∣a~∈A}.
Then, we have
(33)d(A,B)=inf{d(a,b)∣(a,b)∈A×B}=inf{d((g⊗S)(a~))∣a~∈A}=inf{(d∘(g⊗S))(a~)∣a~∈A}.
By using the lower semicontinuity of (d∘(g⊗S)), we have that there exists x*∈A such that
(34)d(gx*,Sx*)=(d∘(g⊗S))(x*)=min{(d∘(g⊗S))(a~)∣a~∈A}=inf{(d∘(g⊗S))(a~)∣a~∈A}=d(A,B).
This completes the proof.

Corollary 12.

Let A be a nonvoidcompact subset and let B be a nonvoidsubset of a complete metric space (X,d). Let S:A→B and g:A→A be continuous and surjective. Define g⊗S:A→A×B by
(35)(g⊗S)(x)=(gx,Sx),∀x∈A.
Then, there exists an element x* in A such that
(36)d(gx*,Sx*)=d(A,B).

Proof.

It is obvious that the continuity and surjectivity of S and g imply the lower semicontinuity of d∘(g⊗S), where d is the distance function of the metric space (X,d) and (g⊗S)(A)=A×B, respectively. Applying Theorem 11, we have the desired result.

Conflict of Interests

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

Acknowledgments

The authors would like to thank the two anonymous referees for their valuable comments which were helpful in improving the paper. Moreover, the third author would like to thank the National Research Council of Thailand, Grant R2557B051 for financial support.

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