Abstract
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 and are nonvoid subsets of a partially ordered set that is equipped with a metric and is a non-self-mapping from to , then the mapping has an optimal approximate solution, called a best proximity point of the mapping , to the operator equation , when 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 .
1. Introduction
Let be a non-self-mapping from to , where and are nonempty subsets of a metric space . Clearly, the set of fixed points of can be empty. In this case, one often attempts to find an element that is in some sense closest to . 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 is a nonempty, compact, and convex subset of a normed linear space and is a continuous function from to , then there exists a point in such that the distance of to is equal to the distance of to . Such a point is called a best approximation point of in . A point in is said to be a best proximity point for , if the distance of to is equal to the distance of to . 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 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 . 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 and are nonvoid subsets of a partially ordered set that is equipped with a metric and is a non-self-mapping from to , then the mapping has an optimal approximate solution, called a best proximity point of the mapping , to the operator equation , when 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 .
2. Preliminary Results
Let be a nonempty set and let be a metric on . Unless otherwise specified, it is assumed throughout this paper that and are nonempty subsets of . We recollect the following notations and preliminary results:
Proposition 1. Let and be two compact subsets of a metric space . Then both and are nonempty sets.
Proof. Suppose that both and are two compact subsets of a metric space . Let such that as . Since and are compact, is also compact. There exists such that Note that (2) is equivalent to Let us consider By employing (3) and letting in (4), we obtain . Hence and . This completes the proof.
Proposition 2. Let be a compact and let be a closed subset of the Euclidian space with norm . Then both and are nonempty set.
Proof. Suppose that is compact and is closed subset of the Euclidian space . Let such that as . Since is compact, there exists such that Note that for all . This means that is bounded. It follows from the Bolzano-Weierstrass theorem and the closeness of that there exists such that Let us consider By employing (5) and (7) and letting in (8), we obtain . Hence and . This completes the proof.
Proposition 3. Let and be two nonempty subsets of a metric space . Then the following are satisfied. (i)If is compact and is closed, then is a closed subset of . (ii)If is compact and is closed, then is a closed subset of . (iii)If both and are compact, then and are nonempty and closed.
Proof.
(i) It is trivial in the case of . Suppose that and let such that
Note that and is closed, so we have . Since , there is such that . It follows from the compactness of that there exists such that
Now, let and consider
By employing (9) and (10) and letting in (11), we obtain . This implies that and, hence, 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 be real topological vector space whose topology is induced by translation invariant metric with the property where denotes the zero vector of . Let and be two closed subsets of such that . Then where and are denoted by the boundary of and , respectively.
Proof. Let . Then there exists such that . It is obvious that . Let on the contrary . Then, there is closed neighborhood of the (the zero vector) and especially positive number such that , for all . Let It is clear from and the closeness of and with that and . Hence, it follows from the translation invariant property and that 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 be a real vector space and a countable family of seminorms on which separates the nonzero points of from (the zero vector of ). For each and each index , define . Let be the topology on generated by the family . One can see that is a topological vector space (even locally convex space). One can verify the topology induced by the translation invariant meter . Moreover, for each positive enough small number , we have for each . However, is not normable.
3. Main Results
In this section, we provide an existence result for the best proximity point of the mapping on the metric space by omitting ordering, proximal monotonicity, and ordered proximal contraction on .
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 , , and . Define function by It is clear that is continuous (since it is projection). It is not hard to verify that (i)both and are closed subset of ;(ii);(iii)there is no such that (i.e., there is no best proximity point).
To achieve understanding in Example 6, let us see Figure 1.
Proposition 7. Let be a compact subset and let be a nonvoidsubset of a metric space . Let be continuous with the property that there exists such that , where Then, there exists an element in such that
Proof. Pick such that . Then there exists such that and . Since is compact, there exists such that By using the continuity of , we can conclude that This completes the proof.
The following result establishes an existence result in order to be nonempty best proximity point set for the mapping without assuming any ordering, proximal monotonicity, and ordered proximal contraction on the . 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 and be nonvoidclosed subsets of a complete metric space such that is nonvoidand is totally bounded. Let and satisfy the following conditions: (a) and are continuous;(b) and . Then, there exists an element in such that
Proof. Suppose that . Let and note that . Then, we have So, we have and there exists Equation (22) indicates that . Since , there exists such that . Thus In the next step, since , we obtain . Then, we have and there exists Equation (24) indicates that . Since , there exists such that . Thus Following by this way, we can produce the sequence such that Since and is totally bounded, there exists a subsequence such that is a Cauchy sequence. By using the completeness of , we have Applying the continuity of and , we obtain This completes the proof.
If (the identity mapping), then Theorem 8 reduces to the following corollary.
Corollary 9. Let and be nonvoidclosed subsets of a complete metric space such that is nonvoidand is totally bounded. Let be a continuous function such that . Then, there exists an element in such that
If , then . Then, by Corollary 9, we obtain the following corollary which says that the fixed points set of the mapping is nonempty.
Corollary 10. Let be a nonvoidclosed and totally bounded subset of a complete metric space . Let be continuous. Then, , where denotes the set of all fixed points of .
In the following result, we are going to relax the continuity of the mappings and (see conditions (a) and (c) of Theorem 3.1 in [5]).
Theorem 11. Let be a nonvoidcompact subset and let be a nonvoidsubset of a complete metric space . Let and and define by Suppose that and is lower semicontinuous where is the distance function of the metric space . Then, there exists an element in such that
Proof. By the assumption, we notice that Then, we have By using the lower semicontinuity of , we have that there exists such that This completes the proof.
Corollary 12. Let be a nonvoidcompact subset and let be a nonvoidsubset of a complete metric space . Let and be continuous and surjective. Define by Then, there exists an element in such that
Proof. It is obvious that the continuity and surjectivity of and imply the lower semicontinuity of , where is the distance function of the metric space and , 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.