Best proximity points of p-cyclic orbital Meir – Keeler contraction maps

The importance of Mathematics lies in solving equations of the form f(x) = 0. This equation can also be written as f(x) = g(x)−x for some suitable function g. Finding the solution of the equation f(x) = 0 is equivalent to finding the solution of the equation g(x) = x. Theorems, which provide a theory by enhancing the possibilities for the existence of a solution to the given equation g(x) = x, are called fixed point theorems. One such theorem is the famous Banach contraction theorem. It stats that “if (X, d) is a complete metric space and T is a self map on X such that there exists a k, 0 < k < 1, such that d(Tx, Ty) 6 kd(x, y) for all x, y ∈ X , then, for any ξ ∈ X , {Tξ} converges to a unique fixed point. One of the interesting extensions of the classical Banach contraction theorem is Meir– Keeler contraction introduced by Meir and Keeler in [18]. Later, the authors of [16] introduced a class of mappings called cyclic contractive mappings. If (X, d) is a metric space and A1, A2, . . . , Ap (p > 2) are the nonempty


Introduction
The importance of Mathematics lies in solving equations of the form f (x) = 0.This equation can also be written as f (x) = g(x) − x for some suitable function g.Finding the solution of the equation f (x) = 0 is equivalent to finding the solution of the equation g(x) = x.Theorems, which provide a theory by enhancing the possibilities for the existence of a solution to the given equation g(x) = x, are called fixed point theorems.One such theorem is the famous Banach contraction theorem.It stats that "if (X, d) is a complete metric space and T is a self map on X such that there exists a k, 0 < k < 1, such that d(T x, T y) kd(x, y) for all x, y ∈ X, then, for any ξ ∈ X, {T n ξ} converges to a unique fixed point.
One of the interesting extensions of the classical Banach contraction theorem is Meir-Keeler contraction introduced by Meir and Keeler in [18].
Later, the authors of [16] introduced a class of mappings called cyclic contractive mappings.If (X, d) is a metric space and A 1 , A 2 , . . ., A p (p > 2) are the nonempty subsets of X, then a cyclic contraction mapping is defined on the union of A 1 , A 2 , . . ., A p with some contraction type of condition imposed on this map.Some fixed point results were obtained for this map in [16].In [5], the authors introduced a notion of best proximity points as an extension of fixed points in the following manner.
Let A and B be non empty subsets of a metric space, and T : A∪B → A∪B such that T (A) ⊆ B and T (B) ⊆ A. The map T is called a cyclic map.A point x ∈ A ∪ B is said to be a best proximity point if d(x, T x) = dist(A, B), where dist(A, B) = inf{d(x, y) = x ∈ A, y ∈ B}.Hence, best proximity point theorems are direct extensions of fixed point theorems.
In [15], a notion of cyclic orbital contraction map is introduced and defined as follows: Definition 1.Let A and B be non empty subsets of a metric space, and T : A∪B → A∪B such that T (A) ⊆ B and T (B) ⊆ A. If for some x ∈ A, there exists a k x ∈ (0, 1) such that d T 2n x, T y k x d T 2n−1 x, y , n ∈ N, y ∈ A, then T is called a cyclic orbital contraction map.
Also, in [15], the following best proximity theorem is obtained in which the map is of Meir-Keeler type and the underlying space need to be a uniformly convex Banach space.
Theorem 1.Let X be a uniformly convex Banach space.Let A and B be non empty, closed and convex subsets of X. Suppose that T : A ∪ B → A ∪ B satisfy the following conditions: (i) T (A) ⊆ B and T (B) ⊆ A; (ii) For every > 0, there exists a δ > 0 such that for some x ∈ A, Then there exists a best proximity point z ∈ A such that for every x ∈ A satisfying condition (1), the sequence {T 2n x} converges to z ∈ A.
So far, the authors generalized best proximity points of cyclic orbital contractions, which are defined on the union of two sets only.But in [24], the author considered p-cyclic map with Meir-Keeler orbital type in different direction.
In this article, the map which we consider is a p-cyclic map (Definition 2) on which a Meir-Keeler type of contraction is imposed.That is, a notion of p-cyclic orbital Meir-Keeler contraction is introduced.Sufficient conditions are given for the existence of a best proximity point of this map, which is also a unique periodic point of the map in a given set.The main result of this article generalizes the main results of the theorems given in the literature.
In Theorem 1, there is no question of the distance between the sets.But in this article, we consider a p-cyclic map in which the distance between the adjacent sets play an important role in obtaining a best proximity point.The condition, under which a best proximity point is obtained for cyclic maps, need not be the same for obtaining best proximity point for p-cyclic sets, p 2. Therefore, the main result of this article is not a direct generalization of Theorem 1.

Preliminaries
The following notion of p-cyclic maps was first introduced by Kirk et al. in [16].
Eldred and Veeramani in [5] introduced the concept of best proximity point for a cyclic map defined on union of two sets, which is an approximation of fixed point defined as follows: A fixed point of a p-cyclic map, if it exists, it exists only in the intersection ∩ p i=1 A i .
Lemma 1. (See [23].)Let Y be a non empty set and let f, g : Y → [0, ∞).Then the following are equivalent: (i) For each > 0, there exists a δ > 0 such that f (x) < + δ ⇒ g(x) < .(ii) There exists an L-function φ (which may be chosen non decreasing and continuous Eldred and Veeramani in [5] proved the following lemma, which is an important property of a uniformly convex Banach space.It is used to prove the main results.http://www.mii.lt/NALemma 2. (See [5].)Let X be a uniformly convex Banach space.Let A and B be non empty and closed subsets of X.Let A be convex.Let {x n } and {z n } be sequences in A, and {y n } be a sequence in B such that lim n x n −y n = dist(A, B) and lim n z n −y n = dist(A, B).Then lim n x n − z n = 0. Definition 5. (See [3, p. 42].)We say that the Banach space (X, • ) is strictly convex if x = y whenever x, y ∈ X are such that x = y = 1 and x + y = 2.
The next theorem is stated for Banach spaces in [24], but it holds true for any normed space.We omit the proof because it is similar to that done in [24].The theorems, which give characterization for the strict convexity in Banach spaces [3] and [7], holds true for normed spaces too [8].
Lemma 3. (See [24].)Let A, B be closed subsets of a strictly convex normed space (X, • ) such that dist(A, B) > 0 ,and let A be convex.If x, z ∈ A and y ∈ B be such An excellent overview of the development of the geometry of Banach spaces may be found in [3].Basic concepts about geometry of Banach spaces can be found in two other excellent books [6] and [7].

Main results
We introduce a notion of p-cyclic orbital non expansive map, which is defined as follows: Definition 6.Let (X, d) be a metric space.Let A i , i = 1, . . ., p, be non empty subsets of X.Let T : ∪ p i=1 A i → ∪ p i=1 A i be a p-cyclic map such that for some x ∈ A i (1 i p) and for each k = 1, 2, . . ., p, the following condition is satisfied: Then T is called a p-cyclic orbital non expansive map.
The conditions, for which dist Proof.(a) For any k ∈ {1, 2, . . ., p}, there hold the inequalities Consequently, we get the chain of inequalities Thus, (a) holds true.
(c) To prove that each z k is a periodic point in A i+k , consider, To prove the uniqueness of z 0 as the periodic point, suppose that ξ Since z 0 −T z 0 = dist(A i , A i+1 ) and from the above, ξ−T z 0 = dist(A i , A i+1 ), and since X is strictly convex, it follows that ξ = z 0 .Hence, z 0 is unique.From z k = T k z 0 the uniqueness of each z k as a periodic point follows.
http://www.mii.lt/NAWe would like to point out that all known results about cyclic maps, where the conditions are symmetric as like as (2); see, for example, [13,15,20] etc.The distances between the consecutive sets appear to be equal.It is in contrast to the conditions investigated in [21] and [24], where the conditions, which ensure existence and uniqueness of the best proximity points, hold true for sets with different distances between them.Now we introduce a notion of p-cyclic orbital Meir-Keeler contraction map.
Definition 7. Let (X, d) be a metric space.Let A i , i = 1, . . ., p, be non empty subsets of X.Let T : ∪ p i=1 A i → ∪ p i=1 A i be a p-cyclic map.Then T is called a p-cyclic orbital Meir-Keeler contraction map if for some x ∈ A i (1 i p) and for each k = 0, 1, 2, . . ., (p − 1), the following holds: for every > 0, there exists a δ > 0 such that In the above definition, if we omit the distances between the sets in condition (3), then we get the following condition (4).In this case, a unique fixed point is obtained.
Theorem 2. Let (X, d) be a complete metric space.Let A i , i = 1, . . ., p, be non empty subsets of X.Let T : ∪ p i=1 A i → ∪ p i=1 A i be a p-cyclic map such that for some x ∈ A i and for each k = 1, 2, . . ., p, the following condition is satisfied: for every > 0, there exists a δ > 0 such that Then {T pn x} converges to a limit say, z 0 ∈ A i , which is the unique fixed point of T in ∩ p i=1 A i .
Proof.Let x ∈ A i satisfy (4).Define, for each k = 1, 2, . . ., p, the following sets: Then each f k and g k satisfies condition (i) of Lemma 1.Hence, there exists an L-func- From the definition of L-function it follows that for all n ∈ N, y ∈ A i and for each k = 1, 2, . . ., p. From ( 6) and ( 7) it follows that Let s n = d(T pn x, T pn+1 y), where y ∈ A i , n ∈ N. If s n = 0 for some n ∈ N, then from ( 8) it follows that s n → 0 as n → ∞.Assume that s n > 0 for all n ∈ N. Then by ( 6) s n+1 < s n and hence converges to an r 0. If r > 0, then by ( 4) there exists a δ > 0 such that r d T pn x, T pn+1 y < r + δ, n ∈ N.
Then there exists an L-function φ such that which is a contradiction.Hence, r = 0. Therefore, d(T pn x, T pn+1 y) → 0 as n → ∞.
Let k ∈ {1, 2, . . ., p}.Then for k = 0, 1, 2, . . ., (p − 1).Therefore, ∩ p i=1 A i is nonempty.Let us prove that for > 0, there exists an n 0 ∈ N such that d(T pn x, T pm x) < for all n, m n 0 .Let > 0 be given.From (9) it follows that there exists n 0 ∈ N such that the inequality d(T pm+k x, T pm+k−1 x) < δ/p holds for every m n 0 and k ∈ N. Let us prove that d(T pm+1 x, T pn x) < /2 by induction on n.This is true for n = m.Let us assume that this inequality is true for some n n 0 .We need to prove that the inequality holds for n + 1.By the inductive assumption we obtain the inequalities The map T is a p-cyclic orbital Meir-Keeler contraction, and thus, it follows from (10) that d(T pm+1 x, T p(n+1) x) < ε/2.Therefore, from the inequality http://www.mii.lt/NAwe get that d(T pn x, T pm x) < for all n, m n 0 .Hence, {T pn x} is a Cauchy sequence, and thus, it converges to a z ∈ A i .Now using ( 8) and ( 9), we get The notion of L-function, given in Definition 4 and Lemma 1, is used to obtain the following result for a p-cyclic orbital Meir-Keeler contraction map.Lemma 4. Let (X, d) be a metric space.Let A i , i = 1, . . ., p, be non empty subsets of X.Let T : ∪ p i=1 A i → ∪ p i=1 A i be a p-cyclic orbital Meir-Keeler contraction map.Then there exists an L-function φ such that for an x ∈ A i satisfying (3), the following hold: where we use the notation λ p,n,k (x, y) = d(T pn+k x, T k+1 y) − dist(A i+k , A i+k+1 ), and for each k = 1, 2, . . ., p.
Proof.Let x ∈ A i satisfy (3).For each k = 1, 2, . . ., p, define the following sets: be defined as follows: Since T is a p-cyclic orbital Meir-Keeler contraction map, each f k and g k satisfy condition (i) of Lemma 1, and hence, ( 12) and ( 13) hold.
Remark 2. From Lemma 4 it follows that a p-cyclic orbital Meir-Keeler contraction map is p-cyclic orbital non expansive map.
Lemma 5. Let (X, d) be a metric space.Let A i , i = 1, . . ., p, be non empty subsets of X.
Proof.Let s n = d(T pn+k x, T pn+k+1 y) − dist(A i+k , A i+k+1 ).Then s n 0 for all n ∈ N. By Remark 4, s n+1 s n for all n ∈ N. If s n = 0 for some n, then the lemma follows.Suppose s n > 0 for every n ∈ N. Then by Lemma 4 there exists an L-function φ satisfying ( 12) and (13).Since s n+1 s n , {s n } converges to an r 0. Suppose r > 0.Then, for this r > 0, by (3) there exists a δ > 0 such that r s n < r + δ and such that s n+1 < φ(s n ) r.That is, s n+1 < r, which is a contradiction.
Theorem 3. Let X be a uniformly convex Banach space.Let A 1 , A 2 , . . ., A p be non empty, closed and convex subsets of X.Let T : ∪ p i=1 A i → ∪ p i=1 A i be a p-cyclic orbital Meir-Keeler contraction map.Then, for every x ∈ A i satisfying (3), the sequence {T pn x} converges to a unique point z ∈ A i , which is the best proximity point as well as the unique periodic point of T in A i .Also, T k z is a best proximity point of T in A i+k , which is also a unique periodic point of T in A i+k for each k = 1, 2, . . ., (p − 1).
Proof.Let > 0 be given.Since T is a p-cyclic orbital Meir-Keeler contraction map, there exists an x ∈ A i and a δ > 0 satisfying (3).Without loss of generality, let δ < .By Lemma 5, Hence, by Lemma 2, T p(n+1)+1 x − T pn+1 x → 0 as n → ∞.Therefore, it is possible to choose an n 1 ∈ N such that and, by Lemma 5, Fix n n 1 .We show that by the method of induction.It is obvious that condition ( 16) is true for m = n.Assume that condition ( 16) is true for an m > n.To prove this condition for m + 1, consider Now http://www.mii.lt/NAUsing ( 14) in (17), we obtain Hence, ( 16) holds for (m + 1) in place m.Consider ( 16) and (15).By Lemma 5 the following holds: for m > n n 1 , Hence, {T pn x} is a Cauchy sequence and converges to a z ∈ A i .By Proposition 1 z is a best proximity point of T in A i , and z is a unique periodic point of T in A i .Let ξ ∈ A i satisfy (3).Then by what we have proved, {T pn ξ} converges to an η ∈ A i such that η − T η = dist(A i , A i+1 ) and T p η = η.But z is the unique periodic point of T in A i .Hence, η = z.By proposition 1 T k z is a best proximity point of T in A i+k for each k = 1, 2, . . ., p.
It is obvious that if condition (3) is satisfied for all x ∈ A i , then the obtained best proximity point is unique.Theorem 3 is a generalization of Theorem 1, and the following theorem proved in [13].
Theorem 4. (See [13].)Let X be a uniformly convex Banach space, and let A 1 , A 2 , . . ., A p (p 2) be non empty, closed and convex subsets of X.Let T be a p -cyclic map such that for every x ∈ A i and y ∈ A i+1 , the following condition is satisfied: for every > 0, there exists δ > 0 such that d(x, y) < dist(A i , A i+1 ) + + δ =⇒ d(T x, T y) < dist(A i+1 , A i+2 ) + .(18) Then, for any x ∈ A i , the sequence {T pn x} converges to a unique z ∈ A i , which is a best proximity point of T in A i .Moreover, this point is a unique periodic point of T in A i .Further, T k z = z k is a best proximity point of T in A i+k for each k = 1, 2, . . ., p.
From Theorem 4 we observe that a best proximity point is obtained if condition (18) is satisfied for all x ∈ A i and y ∈ A i+1 and for all i = 1, 2, . . ., p. From Theorem 3 we observe that a best proximity point of T is obtained even if condition (3) is satisfied for an x ∈ A i , for all y ∈ A i and for some i, 1 i p.
Remark 3. From Proposition 1 and Theorem 3 we observe that if X is a uniformly convex Banach space and A i , i = 1, . . ., p, are closed convex subsets of X and if T :

Examples and applications
We will illustrate the above results with some examples, and we will give an application for integral operators.We will define a map T , and we will prove that T satisfies condition (3) for k = 1.The proofs that the map T satisfies conditions (3) for k = 0, 2, 3, . . ., p − 1 can be done a similar fashion.
We will show with the first example the difference between p-cyclic Meir-Keeler and p-cyclic orbital (Meir-Keeler) contraction maps.
, and let us denote the sets We define the function f : R → R + by f (t) = sign(t)(λ|t| + (1 − λ)α), and we define a map T : 3, 4. The distance between the consecutive sets is equal to 2α.
We will show that T is a p-cyclic orbital Meir-Keeler contraction map.Let us choose x = (α, α) ∈ A 1 .Let ε > 0 be arbitrary chosen.Let us put Consequently, T is 4-cyclic orbital Meir-Keeler contraction map.It is easy to observe that x is a best proximity point of T in A 1 .
If α = 0, then we get that T satisfies (4), and by Theorem 2 there is a unique fixed point z 0 ∈ ∩ 4 i=1 A i .It can be observed in a similar fashion that T is not 4-cyclic Meir-Keeler map and is not a p-cyclic orbital contraction.http://www.mii.lt/NAWe will present an example in infinite dimensional Banach space, where the map T is defined as an integral operator.We will need the fact that the inequality holds for every ε ∈ (0, +∞).First, we will show that g (ε) > 0 for every ε ∈ (0, +∞).Indeed, using the inequalities 2(1 + 2ε)
Let us recall that L 2 [−1, 1] is the space of all classes of measurable functions f such that