Common best proximity points theorems for H-contractive non-self mappings

Fixed point theory and contractive mappings are popular tools in solving a variety of problems such as control theory, economic theory, nonlinear analysis and global analysis.There are many works on di erent types of contractions to nd a xed point in metric spaces. Improving and extending some kind of those, in this paper, we introduce a new version of H-contradiction for four mappings in a metric space (X,d). Then, we prove the existence and uniqueness of a common best proximity point for four non-self mappings. An example is also given to support our main result. The related xed point theorem are also proved.


Introduction and Preliminaries
Since the rst results of Banach in 1922, various authors have been studying xed points, and, in recent years, best proximity points of mappings in metric spaces. Their discoveries are still being generalized in many directions; see [1] to [10]. In a recent paper, Wardowski [11] presented a new contraction, which called F -contraction and proved a xed point results in complete metric spaces. Then Omidvari et al. [12] proved existence of a unique best proximity point for F -contractive non-self mappings. In this paper, we extend their results by introduce a new version of Wardowski's contraction for four mappings in a complete metric space and estabilish a new common best proximity point theorem. Next, by an example and a xed point result, we support our main results and show some applications of them.
Given two non-empty subsets A and B of a metric space (X, d) , the following notions and notations are used in the sequel .
Denition 1.1. An element u ∈ A is said to be a common best proximity point of the non-self mappings Denition 1.2. The mappings f : A → B and g : A → B are said to be commute proximally if they satisfy the condition that is said to have P-property if and only if for any a 1 , a 2 ∈ A 0

Main Results
We begin our study with following denition Denition 2.1. Let H : R + → R be a mapping satisfying: Then f, g, S and T have unique common best proximity point .
Similarly, a point a 2 ∈ A 0 can be chosen such that g(a 1 ) = S(a 2 ). Continuing this process, we obtain a sequence {a n } ⊆ A 0 such that We wil prove that the sequence {u n } is convergent in A 0 .
Also from (H 3 ) we have ∃k ∈ (0, 1) such that lim n→∞ α k n H(α n ) = 0 On the Other hand, by (5) H(α n ) − H(α 0 ) ≤ −nC Therefor α k n H(α n ) − α k n H(α 0 ) ≤ −nα k n C ≤ 0 Letting n −→ ∞ in the above inequality and using (6) and (7) , we obtain lim n→∞ nα k n = 0 Hence there exists N 1 ∈ N such that nα k n ≤ 1 for all n ≥ N 1 . Therefor for any n ≥ N 1 This means that series ∞ i=1 α i is convergent, then By the triangular inequality and(8) Since{u n } ⊆ A 0 and A 0 is a closed subset of the complete metrice space (X, d), we can nd u ∈ A 0 such that lim n→∞ u n = u . By (1) and because of the fact {f, S} and {g, T } commute proximally, f u 2n−1 = Su 2n and gu 2n = T u 2n+1 . Therefore, the continuity of f, g, S and T and n → ∞ ascertains that f u = gu = T u = Su.
then a is a common best proximity point of the mapping f, g, S and T . Suppose that a = a is another common best proximity point of the mapping f, g, S and T , so that As pair (A, B) has the P-property then from (9) and (10) By theorem 2.1 we also obtain the following common xed point theorem.
Theorem 2.2. Let (X, d) be a complete metric space . Let f, g, S, T : X → X be given continuous mappings and satisfy the H-contractive condition such that S and T commute f and g respectively. Further let f (X) ⊆ T (X), g(X) ⊆ S(X). Then f, g, S and T have unique common xed point.
Proof. We take the same sequence {u n } and u as in the proof of theorem 2.1. Due to the fact that S and T commute f and g respectively we have By continuity of f, g, S, T and n → ∞ we have If f u = gu, since f, g, S, T : X → X satisfy the H-contractive condition and by (11) H(d(f u, gu)) ≤ −C + H(max{d (Su, T (f u, gu)). Therefore C ≤ 0, which is a contraction. Then f u = gu and by (11), f u = gu = Su = T u.
We set a = f u = gu = Su = T u. Because of the fact T commute g we obtain Therefore, H(d(a, ga)) ≤ −C + H(d(a, ga)) and consequently C ≤ 0, that is a contraction by C > 0. Therefore Similarly, we can show that a = f a = Sa.
Hence, by (12) and (13) we deduce that a = f a = ga = Sa = T a. Therefore, a is a common xed point of f, g, S and T . Assume one contrary that, p = f p = gp = Sp = T p and q = f q = gq = Sq = T q but p = q. Consequence H(d(p, q)) ≤ −C + H(d(p, q)), then C ≤ 0, a contradiction. Therefore, f, g, S and T have a unique common xed point.