On p-Cyclic Orbital MK Contractions in a Partial Metric Space

A cyclic map with a contractive type of condition called p-cyclic orbital M-Kcontraction is introduced in a partial metric space. Sufficient conditions for the existence and uniqueness of fixed points and the best proximity points for these maps in complete partial metric spaces are obtained. Furthermore, a necessary and sufficient condition for the completeness of partial metric spaces is given. The results are illustrated with an example.


Introduction and Preliminaries
An important field of application of fixed point theory exist nowadays in the investigation of the stability of complex continuous-time and discrete-time dynamic systems [1][2][3].Meir-Keeler self-mappings have received important attention in the context of fixed point theory perhaps due to the associated relaxing in the required conditions for the existence of fixed points compared with the usual contractive mappings [4][5][6].A connection between p-cyclic contractive, p-cyclic Kannan and p-cyclic Meir-Keeler contractions was obtained in [7].The notion of orbital contractions introduced in [8] weakens the contraction condition in a different way by assuming that the contractive condition is not satisfied for all pairs (x, y) ∈ X × X.On the other hand, an extension of Banach's fixed point theorem is obtained in [9] by considering a cyclical contractive condition.Some generalizations of cyclic maps that involve Meir-Keeler maps were given in [5,[10][11][12].Another kind of generalization of the Banach contraction principle is by altering the underlying space.Such kinds of generalizations are also obtained in partial metric spaces.To understand partial metric spaces, one may refer to [13][14][15].For generalizations of the Banach contraction theorem, in which the underlying space is a partial metric space, one may refer to [16][17][18][19][20].In this article, we give the conditions for the existence of a unique fixed point and a best proximity point of p-cyclic orbital Meir-Keeler maps in partial metric spaces.
Example 1.For an example of a partial metric space, let us consider I to be the collection of all nonempty, closed and bounded intervals in R.
then it can be shown that ρ is a partial metric over I, and the self-distance of [a, b] is the length b-a.This is related to the real line as follows: |a ), and so, by mapping each a ∈ R to [a, a], we embed the usual metric structure of R into that of the partial metric structure of intervals.
Following [15], we will recall some basic facts and definitions about partial metric spaces.Each partial metric ρ on X induces a T 0 topology τ ρ on X, which has the base of the family of open balls {B ρ (x, ) : x ∈ X, > 0}, where B ρ (x, ) = {y ∈ X : ρ(x, y) < + ρ(x, x)} for all x ∈ X and > 0.
If ρ is a partial metric on X, then the function ρ s : X × X → [0, ∞) given by: is a metric on X.
1.A sequence {x n } in (X, ρ) converges to a point x ∈ X if and only if lim n,m→∞ ρ(x n , x m ) = lim n→∞ ρ(x n , x) = ρ(x, x); 2. A sequence {x n } in (X, ρ) is called a Cauchy sequence if there exists a finite limit lim n,m→∞ ρ(x n , x m ); 3. (X, ρ) is called a complete partial metric space if every Cauchy sequence {x n } converges, with respect to τ ρ , to a point x ∈ X, such that ρ(x, x) = lim n,m→∞ ρ(x n , x m ).

Lemma 1 ([18]
).A sequence {x n } ∞ n=1 is a Cauchy sequence in a partial metric space (X, ρ) if and only if {x n } ∞ n=1 is a Cauchy sequence in the metric space (X, ρ s ).A partial metric space (X, ρ) is complete if and only if the metric space (X, ρ s ) is complete.Moreover, Below, we obtain a new partial metric from the existing metric and partial metric.
Note that every L-function satisfies the condition φ(s) ≤ s for every s ≥ 0. Suzuki generalized Lim's results in [22].The following result of Lim is used in the proof of the main result.Lemma 2 ([22]).Let Y be a nonempty set, and let f , g : Y → [0, ∞).Then, the following are equivalent: 1.For each > 0, there exists a δ > 0, such that f (x) < + δ ⇒ g(x) < .2. There exists an L-function φ (which may be chosen to be a non-decreasing and continuous) such that f (x)

Main Result
For the investigation of the best proximity points, the notion of the uniform convexity of a Banach space plays a crucial role.In [23], Suzuki et al. introduced the notion of the metric space, which satisfies property UC.In a similar way, we generalize this notion in the partial metric space.Definition 4. Let (X, ρ) be a partial metric space and A and B be subsets.The pair (A, B) is said to satisfy the property UC if the following holds: If {x n }, {z n } are sequences in A and {y n } is a sequence in B, such that lim n→∞ ρ(x n , y n ) = dist(A, B) and for any ε > 0, there is n ∈ N, so that ρ(z m , y n ) < dist(A, B) + ε for all n ≥ N, then for any ε > 0, there is N 1 ∈ N, so that ρ(x n , z m ) < ε for m, n ≥ N 1 .Now, let us recall the notion of cyclic maps.Definition 5. Let X be a nonempty set and A 1 , A 2 , ..., A p be nonempty subsets of X.A map T : , for all i = 1, 2, ..., p, where we use the convention A p+1 = A 1 .Definition 6.Let (X, ρ) be a partial metric space and A 1 , A 2 , ..., A p be nonempty subsets of X.A point x ∈ A i is said to be a best proximity point of We introduce the notion of p-cyclic orbital M-Kcontraction as follows: Definition 7. Let (X, ρ) be a partial metric space, A 1 , A 2 , . . ., A p be nonempty subsets of X and T : ∪ p i=1 A i → ∪ p i=1 A i be a p-cyclic map.The map T is called a p-cyclic orbital M-K contraction if for some x ∈ A i , for each > 0, there exists δ > 0, such that the following condition: holds for all n ∈ N ∪ {0} and for all y ∈ A i , where D k ≥ 0, for k = 1, 2, ..., p.
Theorem 1.Let (X, ρ) be a complete partial metric space.Let A 1 , A 2 , ..., A p be nonempty and closed subsets of X.Let T : ∪ ) for all k = 1, 2, . . ., p and (X, ρ) is a partial metric space with property UC, then for every x ∈ A i satisfying (1), 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 .Furthermore, 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).

Auxiliary Results
Without loss of generality, let us assume that x ∈ A 1 .The proof of the following lemma follows the technique used in [11].Lemma 3. Let (X, ρ) be a partial metric space.Let A 1 , A 2 , ..., A p be nonempty subsets of X.Let T:∪ p i=1 A i → ∪ p i=1 A i be a p-cyclic orbital M-K contraction map with constants D k equal to zero or D k = dist(A k , A 1+k ).Then, there exists an L-function φ such that for an x ∈ A 1 satisfying (1), the following holds: if ρ(T pn+k−1 x, T k y) > D k , then: for each k = 1, 2, ..., p, for all n ∈ N and for all y ∈ A 1 .
Remark 1. From Lemma 3, it follows that for a p-cyclic orbital M-K contraction map T, the sequence {ρ( Lemma 4. Let (X, ρ) be a partial metric space.Let A 1 , A 2 , ..., A p be nonempty subsets of X.Let T : ∪ p i=1 A i → ∪ p i=1 A i be a p-cyclic orbital M-K contraction map with constants D k equal to zero or D k = dist(A k , A 1+k ).Then, for any x ∈ A 1 satisfying (1), for all y ∈ A 1 and for each k ∈ {0, 1, 2, ..., p − 1}, the sequence {ρ(T pn+k x, T pn+k+1 y)} ∞ n=1 converges to D k+1 .
Corollary 1.Let (X, ρ) be a partial metric space.Let A 1 , A 2 , ..., A p be nonempty subsets of X.Let T:∪ p i=1 A i → ∪ p i=1 A i be a p-cyclic orbital M-K contraction map with constants D k equal to zero.Then, for any x ∈ A 1 , satisfying (1), the following holds: for all n ∈ N, for all y ∈ A 1 and for each k = 0, 1, 2, . . ., p − 1.

Examples
We start with a lemma, which is useful in checking whether a partial metric space is complete.Lemma 6.Let (X, d) be a complete metric space and (X, ρ) be a partial metric space.Let ω : X → [0, +∞) and ρ(x, y) = d(x, y) + max{ω(x), ω(y)}.The partial metric space (X, ρ) is complete if and only if ω satisfies the condition: Proof.(X, ρ) is complete if and only if (X, ρ s ) is complete ( [19]).By the definition of the partial metric ρ, we get that: Sufficiency: Let {x n } ∞ n=1 be a Cauchy sequence in (X, ρ), and ω satisfies the condition in the lemma.From [19], it follows that {x n } ∞ n=1 is a Cauchy sequence in (X, ρ s ).That is, for every ε > 0, there exists N ∈ N, such that the inequality ρ s (x n , x m ) < ε holds for every n, m ≥ N. Thus, from (13), we get 2d(x n , x m ) + |ω(x n ) − ω(x m )| < ε, and therefore, the inequality d(x n , x m ) < ε holds for every n, m ≥ N. Consequently, {x n } ∞ n=1 is a Cauchy sequence in (X, d), hence converging to x.From the assumption that {x n } ∞ n=1 is a Cauchy sequence in (X, ρ), it follows that the limit: (I) Let us assume that lim sup n→∞ ω(x n ) ≤ ω(x).Then, lim n→∞ max{ω(x), ω(x n )} = ω(x).Consequently, we get: (II) Let us assume that lim sup n→∞ ω(x n ) ≥ ω(x).From the assumption of the Lemma, it follows that lim n→∞ max{ω(x), ω(x n )} = ω(x).Consequently, we get: Thus, {x n } ∞ n=1 is a convergent sequence in (X, ρ) in both cases.To prove the necessity condition, let (X, ρ) be a complete partial metric space.We will prove that ω satisfies the conditions of the Lemma.Let us suppose the contrary, i.e., there exists a sequence {x n } ∞ n=1 that is convergent to some point x ∈ X with respect to the metric d, lim sup n→∞ ω(x n ) ≤ M 1 < ∞, and there is ε 0 > 0, so that the inequality |ω( ]), and therefore, it is convergent.Let us denote its limit by z.From (13), it follows that lim n→∞ ρ s (x n , z) = lim n→∞ (d(x n , z) + |ω(x n ) − ω(z)|) = 0. Consequently, lim n→∞ d(x n , z) = 0, and thus, z = x.Therefore, lim n→∞ |ω(x n ) − ω(z)| = 0, which is a contradiction.