EXISTENCE OF BEST PROXIMITY POINTS SATISFYING TWO CONSTRAINT INEQUALITIES

. In this paper, we prove the existence of best proximity point and coupled best proximity point on metric spaces with partial order for weak proximal contraction mappings such that these critical points satisfy some constraint inequalities.


Introduction and preliminaries
Most of the mathematical problems in ordinary differential equation, partial differential equation, game theory, operation research etc. which appear in the nature of applicability, will have the solution which satisfies the objective in which some constraints are binded together with the problem. So, it is quite natural to find out the fixed point of an operator satisfying the constraints which are coupled with the objective.
Over a century, after the famous Banach contraction principle, many fixed point results were given by mathematicians proving the existence of fixed point for weak contraction mappings on metric spaces [6], [2], [3], [4], [5], [9]. After all these developments in the literature of fixed point theory, Lakshmikantham and Ciric [11] in 2009 introduced the concept of coupled fixed point which has wide range of applicability in partial differential equations.
Later, Choudhury and Maity [4] introduced the cyclic coupled fixed point and gave the existence of strong coupled fixed point. All these mappings were generalized by various authors [6], [8], [10], [1] by the concept called proximity point which are also known to be generalized fixed point. Our results will also be hammering away at establishing the existence of proximity points on metric space with partial order.
Let (X, d) be a metric space endowed with two partial orders 1 and 2 and P, Q, R, S, T : X → X be five self-operators. Recently, Samet and Jleli [7] have contemplated on the existence of a point x ∈ X which satisfies the following. (1) Definition 1.1. Let be a partial order on complete metric space X. Then, is d-regular if {a n }, {b n } are sequences in X, we have lim n→∞ d(a n , a) = lim n→∞ d(b n , b) = 0 with a n b n for all n ∈ N, then a b, where (a, b) ∈ X × X. Definition 1.2. Let X be a nonempty set endowed with two partial orders 2 and 1 . Let T, P, Q, R, S : X → X be given operators. We say that the operator T is (P, Q, R, S, 2 , 1 )-stable, if the following condition is satisfied for x ∈ X: Let (X, d) be a complete metric space endowed with two partial orders 1 and 2 . Let P, Q, R, S, T : X → X be given operators. Suppose that the following conditions are satisfied: If P x 1 Qx, Ry 2 Sy implies d(T x, T y) ≤ d(x, y) − θ(d(x, y)), where θ ∈ Θ. Then, there exists a point x ∈ X satisfying (1).
They also raised the question of the existence of best proximity point together with constraint inequalities.
In the sequel, let (X, d) be a complete metric space with two nonempty closed subsets A and B endowed with partial orders 1 on A and 2 on B. Let us consider the operators P, Q which are self maps on A 0 and R, S be self maps on B 0 , where A 0 = {x ∈ A : d(x, y) = d(A, B) for some y ∈ B}, Let T be an operator from A 0 to B 0 . In this paper, our intend is to give the existence of best proximity points and coupled best proximity points satisfying two constraint inequalities. One of our result also generalizes a result in [4].

Existence of best proximity point satisfying two constraint inequalities
In this section, our aim is to evince that there exists a element x ∈ A such that x is a proximity point of an operator T and also satisfies two constraint inequalities.
Definition 2.1. An operator T : A → B is said to be (T, P, Q, R, S, 1 , 2 )-stable if for x ∈ A 0 , P x 1 Qx implies R(T x) 2 S(T x), and T is said to be (T, P, Q, R, S, 1 , 2 ) P -stable if for y ∈ B 0 , (1) ϕ is continuous and non-decreasing, Theorem 2.4. Let A and B be two nonempty closed subsets of a complete metric space (X, d) endowed with partial orders 1 on A 0 and 2 on B 0 . Let P, Q be self-maps on A 0 and R, S be self-maps on T is (T, P, Q, R, S, 1 , 2 )-stable and (T, P, Q, R, S, 1 , 2 ) P -stable. Then there exists a point x ∈ A 0 satisfying the followings: Hence, again using (iii) we get Similarly, we can construct sequences {x n } and {T x n } in A 0 and B 0 , respectively satisfying P x n 1 Qx n and R(T x n ) 2 S(T x n ).
Since T is a C-proximal ϕ-contraction, we have Therefore, d(x n , x n+1 ) is decreasing and lim n→∞ d(x n , x n+1 ) = 0 by the definition of ϕ.
We claim that {x n } is a Cauchy sequence. Given that > 0, there exists We prove that {x n } is a Cauchy sequence by the method of induction. Fix m ≥ N ( ) and assume that d(x m , x i ) < , for all i = m + 1, . . . , n. Therefore, Thus {x n } is a Cauchy sequence and hence converges to some element x ∈ A 0 . Let .
Since P and Q are continuous maps and 1 is d-regular, we obtain P x 1 Qx. Now, using conditions (i) and (ii), we get R(T x) 2 S(T x). Thus x is an proximity point of T and satisfies the constraint inequalities. Corollary 1. Let (X, d) be a complete metric space with two partial orders 1 and 2 to itself. Let T, Q, R, S are self-maps on X satisfying that (i) there exists ϕ ∈ Φ such that P x 1 Qx, Ry 2 Sy implies d(T x, T y) ≤ ϕ(d(x, y)), Then there exists a point x ∈ X such that T x = x.
Proof. The proof is the same as Theorem 2.4 with A = B = X.

Best proximity points for cyclic coupled mapping
In this section, we give the existence of proximity points for coupled maps of cyclic type with respect to A and B. In the next section, we prove the same with proximity point also satisfying two constraints. The result of this section generalizes the result of [4].

Definition 3.1. Let A and B be two nonempty subsets of a metric space (X, d).
A mapping F : X × X → X is called coupled proximal mapping on A and B if F : A × B → B and F : B × A → A satisfy the inequality d (F (y 1 , x 1 for some k ∈ (0, 1 2 ), where x 1 , x 2 , u ∈ A 0 , y 1 , y 2 ∈ B 0 and d(u, F (x 2 , y 2 )) = d (A, B). Definition 3.2. Let A and B be two closed subsets of a metric space (X, d). Then A and B is said to satisfy the P -property if for x 1 , x 2 ∈ A 0 and y 1 , y 2 ∈ B 0 satisfies and so, d(x n , u n+1 ) ≤ c d(x n , u n−1 ) where c = k 1−k . Now, by using the above inequalities, we have the following: where M = max{d(x 0 , u 1 ), d(u 0 , x 1 )}. Define a sequence {z n } by z 2n = u n , n > 0 z 2n−1 = x n , n > 0.
{z n } is a Cauchy sequence. Therefore, {x n } and {y n } are Cauchy sequences which converge to x ∈ A and y ∈ B, respectively. Then d(x, y) = d(A, B) by the continuity of d and P -property.
Therefore, d(x, u ) = 0 as n → ∞ which implies d(x, F (x, y)) = d (A, B). Similarly, we can prove that d(y, F (y, x)) = d(A, B) which concludes that (x, y) is the strong coupled proximal point of F .

Existence of coupled best proximity point satisfying two constraint inequalities
This section apart from the previous section shows that two points (x 0 , y 0 ) satisfying the inequalities are enough for the existence of coupled best proximity instead of an arbitrary pair (x, y). Let (X, d) be a complete metric space with A and B as mentioned above. Define F : X × X → X be a coupled mapping with respect to A and B. Let P, Q, R, S are the same as pre-defined. Our intend is to find (x, y) ∈ A × B such that Definition 4.1. Let A and B be two nonempty subsets of a metric space (X, d). A mapping F : X × X → X is called C-coupled proximal mapping on A and B if F : A × B → B and F : B × A → A satisfy the following: where x 1 , x 2 ∈ A 0 , y 1 , y 2 ∈ B 0 and d(u, F (x 2 , y 2 )) = d(A, B) for some k ∈ (0, 1 2 ). Definition 4.2. Let A and B be two subsets of X with partial orders 1 and 2 . Let F : X × X → X be a mapping satisfying that Then F is said to be (F, P, Q, R, S, 1 , 2 ) coupled stable if x ∈ A and y ∈ B, then R(F (x, y)) 2 S(F (x, y)) and P (F (y, x)) 1 Q(F (y, x), whenever, P x 1 Qx and Ry 2 Sy. Theorem 4.3. Let (X, d) be a complete metric space and A, B be two nonempty closed subsets of X. Let P, Q be self-mappings on A and R, S be self-mappings on B. Let F : X × X → X be C-coupled proximal mapping with respect to A and B. Suppose that (i) { i , i = 1, 2} is d-regular on A 0 and B 0 , (ii) P, Q, R, S are continuous, (iii) there exist x 0 ∈ A and y 0 ∈ B such that P x 0 1 Qx 0 and Ry 0 1 Sy 0 , (iv) F is (F, P, Q, R, S, 1 , 2 ) coupled stable, (v) A and B satisfy the P -property.
Then there exists a point (x, y) ∈ A × B which satisfies (A, B), where y = F (x, y), P x 1 Qx, Ry 2 Sy.
Proof. Let x 0 ∈ A, y 0 ∈ B such that P x 0 1 Qx 0 and Ry 0 2 Sy 0 . Let {x n } and {y n } be two sequences defined by F (x n , y n ) = y n+1 , F (y n , x n ) = x n+1 and u n be defined as d(u n , y n ) = d (A, B). Since F is (F, P, Q, R, S, 1 , 2 ) coupled stable, each of {x n } and {y n } satisfies P x n 1 Qx n and Ry n 2 Sy n . where M = max{d(x 0 , u 1 ), d(u 0 , x 1 )}. Define a sequence {z n } by z 2n = u n , n > 0 z 2n−1 = x n , n > 0.
By the following fact that Thus, as n → ∞, d(x, u ) = 0, we conclude that d(x, F (x, y)) ≤ d (A, B), i.e., d(x, F (x, y)) = d(A, B). Since P, Q are continuous maps and 1 is d-regular, we have P x 1 Qx. From the conditions (i) and (ii), we get that Ry 2 Sy. Thus (x, y) satisfies (2).