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1 Introduction

Fixed point theory appeared first in the solution of the certain differential equations, see, e.g., Liouville [15] and Picard [18]. Banach [2] successfully derived the successive approximation method from the proofs of Picard [18], and he initiated the first fixed point theorem: For every contraction on a complete metric space by starting from an arbitrary point one can construct a recursive sequence such that , that is, x^∗ is a fixed point, and it is unique. It should be noted that in this proof the continuity of the mapping is used efficiently, although it is not assumed so. Indeed, the continuity of the operators is a necessary consequence of the “fulfilling contraction” condition. Roughly speaking, “finding the unique fixed point for a given operator” is equivalent to the existence and uniqueness of the solution of the corresponding differential equations. After Banach, a huge number of papers reported to improve, extend, and generalize the metric fixed point theory, which implicitly improved the differential equations theory but not only that. Metric fixed point theory has a wide application potential in almost all quantitative sciences, in particular, theoretical computer science, economics, and engineering.

Besides this improvement in fixed point theory, there are some operators that do not admit a fixed point. In other words, in any point in its domain, we have Accordingly, we could not find a solution for the considered differential equations or some other equations that are fulfilled by the given operator . Roughly speaking, we could not find an exact solution for the given problem. At this handicap, optimization brings an approximate solution via “best proximity point.”

Let be a metric space and A, B be nonempty subsets of it. Suppose, for a mapping , that the corresponding functional equation does not necessarily have a solution. Regarding that d(A, B) is a lower bound for, an approximate solution . to the corresponding functional equation yields the least possible error when , where dist. Here the approximate solution is called a best proximity point of the considered nonself mapping Note that a best proximity point yields the global minimum of the nonlinear programming problem since for all .

As it is emphasized above, the continuity of the given mapping has a crucial role in obtaining the existence and uniqueness of the fixed point. On the other hand, the continuity is a heavy condition for the given mappings. Consequently, the following natural question appears: is it possible to find a fixed point of a given mapping that is not necessarily continuous? An interesting affirmative answer was given by Kirk, Srinivasan, and Veeramani [14] (see also, for example, [7, 9, 16]).

Theorem 1. Suppose that (X, d) is complete metric space and the letters A, B reserved to denote nonempty closed subsets of it. If, for a nonself mapping with there exists such that for all and , then possesses a unique fixed point in A ∩ B.

In Theorem 1, is called cyclic map. In [10], the concept of “cyclic map” was extended as p-cyclic map as follows.

Definition 1. (See [10, Defs. 3.1, 3.2].) be nonempty subsets of a metric space (X, d).

(i) A map . is called a p-cyclic map if for al , where, the map is called cyclic.

(ii) A point is said to be a best proximity point of in Ai if where

For p-cyclic maps, the distances between the adjacent sets play an important role in the existence of a best proximity point. In [17, 21] and [12], the authors investigated the problem of finding a best proximity point for a p-cyclic map in which the distances between the adjacent sets need not be equal.

In [3], the following lemma is proved, and it is used to prove the main results.

Lemma 1. (See [3, Lemma 3].) Let A and B be nonempty closed subsets of a uniformly convex Banach space such that A is convex. be sequences in A and be a sequence in B satisfying:

(i)

(ii) For every there exists such that

Then for every there exists such that for all

In [11], the following notion of cyclic orbital contraction is introduced in which the contraction condition need not be satisfied for all the points.

Definition 2. Let A and B be nonempty subsets of a metric space X. A cyclic map is said to be cyclic orbital contraction if for some, there exists such that

In [13], the following notion of .-cyclic orbital nonexpansive map is introduced.

Definition 3. (See [13, DEf. 6].) Let be nonempty subsets of a metric spaceX. A p-cyclic map is said to be p-cyclic orbital non expansive if for some and for each the following inequality holds:

In [5,13] and [20], the authors investigated the existence of fixed points and best prox- imity points for various types of cyclic orbital contractions. Cyclic orbital contractions can be compared with the notion of “contractive iterate at a point” introduced in [19], later generalized in [4] and [6].

In [1], the following notion of cyclic -contraction is introduced.

Definition 4. (See [1, Def. 1].) Let A and B be nonempty subsets of a metric space be a strictly increasing map. A cyclic map is said to be cyclic φ-contraction if

Proposition 1. (See [8, Prop. 1].) Let be a strictly convex normed linear space. be nonempty and convex subsets of . be a p-cyclic map. Then T has at most one best proximity point in

2 Main results

We introduce a notion called p-cyclic orbital φ-contraction, which is defined as follows

Definition 5. be nonempty subsets of a metric space and be a strictly increasing map such that We say that a p-cyclic map is p-cyclic orbital -contraction if for each and for some, the following inequality holds:

Proposition 2. Every p-cyclic orbital φ-contraction map is p-cyclic orbital none xpansive.

Proof. Let be a p-cyclic orbital φ-contraction map satisfying (1) for some . Since and φ is a strictly increasing map, we have

Substituting equation (2) in equation (1), we get

This completes the proof.

Proposition 3.be nonempty subsets of a metric space be a strictly increasing map. is a p-cyclic orbital φ-contraction map such that equation (1) holds for some then

Proof. be arbitrary. and for Then is a nonincreasing sequence of non- negative real numbers and bounded below by dist Therefore, converges to (say). This implies that Since is p-cyclic orbital nonexpansive, we have

This implies that

As φ is strictly increasing and for all , we have

From (3) and (4) we get

Hence,

By combining (4) and (5) we have This gives as φ is a strictly increasing map. Hence the proof.

Proposition 4.be nonempty subsets of a metric spacebe a strictly increasing map. is a p-cyclic orbital φ-contraction map satisfying (1) for some then the following hold:

Proof. By using similar argument as in Proposition 3(i)–(iv) can be proved.

The following proposition is useful to prove the main result whose proof follows from Lemma 1, Propositions 3 and 4.

Proposition 5 be nonempty closed convex subsets of a uniformly convex Banach spacebe a strictly increasing map. is a p-cyclic orbital φ-contraction map satisfying (1) for some then the following hold:

Proposition 6.be nonempty subsets of a metric space X. be a strictly increasing map and If is a p-cyclic orbital -contraction map satisfying (1) for some x 2 Ai (1 6 i 6 p), then

Proof. (i) be arbitrary. As is p-cyclic orbital nonexpan- sive, we have

Also,

Thus,

Now by taking limit on both sides and using Proposition 4(i) we get

From the above inequality we get the following chain of inequalities

Hence,

(ii) Let be the metric induced by the norm Now

Thus, and is a best proximity point of in .

Theorem 2. be nonempty closed subsets of a complete metric space be a strictly increasing map and Let be a p-cyclic orbital -contraction map of type one satisfying (1) for some Then there exists a fixed point of , say, such that for any satisfying (1), the sequence converges to . Proof. Let satisfy equation (1). Let us prove that given there exists an such that

by induction on be given. Now

From Proposition 4(i), for we have Hence, there exists an such that

Hence, it is enough to show that

Fix such that (6) holds. Now (7) is true for Assume that (7) is true for some We will prove that (7) is true for in place of Now

Hence, is a Cauchy sequence, and it converges to a limit, say, For in Proposition 4 (i), we get Now

This implies that and therefore, is a fixed point in . Since is p-cyclic, To prove that is unique, suppose such that Now from Proposition 4(i)

Theorem 3. be nonempty closed and convex subsets of a uniformly convex Banach space be the metric induced by the norm. be a strictly increasing map. is a p-cyclic orbital -contraction map of type two, then for every satisfying equation (1), the sequence converges to which is a unique best proximity point of in .

Proof. If for every then has a unique fixed point in by Theorem 2. Let us assume that dist We claim that for every there exists an such that for all

Suppose not. Then there exists an such that for all there existsfor which

By choosing m_k to be the least integer greater than n_k, to satisfy the above inequality, we have

Now by (9), for each ,

By taking limit on both sides of above inequality as and by using Proposition 5(ii) we have

That is,

Now

Now by using p − 1 times p-cyclic orbital nonexpansiveness of T to , we get

Let dist Then the above inequality becomes

By using (11) in (10) we get

That is,

By taking limit on both sides of (12) as and using Proposition 5(i) and (iii) in (12) we have

Since for each we have Hence,

Since by (8) we have Thus, That is, This is a contradiction to the fact that is strictly increasing and Hence the claim.

Now by Proposition 4(i) Combining this with the claim, by Lemma 1 we have the following: for every there exists an such that

Therefore, is a Cauchy sequence in , and it converges to a point By Proposition 6(ii) and Proposition 1, is the unique best proximity point of in .

Example. Consider endowed with the Euclidean metric. be the subsets of defined as follows:

Define as follows:

Define It is easy to see that is a 4-cyclic map and is a strictly increasing map. We note that dist is 4-cyclic orbital -contraction for all points in the set The unique best proximity point of and We see that for all the sequence converges to the unique best proximity point of in the respective set. Further, do not satisfy condition (1), and do not converge to the best proximity point.

References

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Notas de autor

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