Interpolative Reich – Rus – Ćirić Type Contractions on Partial Metric Spaces

By giving a counter-example, we point out a gap in the paper by Karapinar (Adv. Theory Nonlinear Anal. Its Appl. 2018, 2, 85–87) where the given fixed point may be not unique and we present the corrected version. We also state the celebrated fixed point theorem of Reich–Rus–Ćirić in the framework of complete partial metric spaces, by taking the interpolation theory into account. Some examples are provided where the main result in papers by Reich (Can. Math. Bull. 1971, 14, 121–124; Boll. Unione Mat. Ital. 1972, 4, 26–42 and Boll. Unione Mat. Ital. 1971, 4, 1–11.) is not applicable.


Introduction and Preliminaries
In 1922, Banach laid the foundations of metric fixed point theory by proposing his prominent fixed point result.To put a finer point on it, Banach observed that if a self-mapping T, defined on complete metric space (X, ρ), fulfills the contraction inequality, i.e., there exists a constant k ∈ [0, 1) such that ρ(Tξ, Tη) ≤ kρ(ξ, η) for all ξ, η ∈ X, then it possesses a unique fixed point in X.In 1968, Kannan [1] proposed a new fixed point result.He considered the following contraction type: ρ (Tξ, Tη) ≤ λ [ρ(ξ, Tξ) + ρ(η, Tη)] for all ξ, η ∈ X, where λ ∈ [0, 1  2 ).Very recently, in [2], the acclaimed theorem of Kannan was revisited by taking the interpolation theory into account.For a metric space (X, ρ), the self-mapping T : X → X is said to be an interpolative Kannan type contraction, if there are constants λ ∈ [0, 1) and α ∈ (0, 1) such that for all ξ, η ∈ X with ξ = Tξ.The main result in [2] via an interpolative Kannan type contraction is Theorem 1 ([2]).In the framework of a complete metric space (X, ρ), if a mapping T : X → X forms an interpolative Kannan type contraction, then it possesses a unique fixed point in X.
Note that when the inequality (2) holds for all ξ, η ∈ X, and if T possesses a fixed point (say v ∈ X), then Tξ = v for each ξ ∈ X, that is, T is a constant mapping, which is the trivial case, so the fixed point of T is unique.The appropriate condition on ξ and η in (2) should be ξ, η ∈ X\Fix(T), where Fix(T) = {u ∈ X, Tu = u}.In this case, the author [2] ensured the existence of a unique fixed point.There is a gap, that is, such fixed point is not necessarily unique.The following example illustrates our concern.
As a correction of Theorem 1, we should state Theorem 2. Let (X, ρ) be a complete metric space.A self-mapping T : X → X possesses a fixed point in X, if there exist constants λ ∈ [0, 1) and α ∈ (0, 1) such that The following theorem was proved by Reich, Rus and Ćirić [3][4][5][6][7] independently to combine and improve both Banach and Kannan fixed point theorems.Theorem 3. In the framework of a complete metric space (X, ρ), if T : X → X forms a Reich-Rus-Ćirić contraction mapping, i.e., for all ξ, η ∈ X, where λ ∈ 0, 1 3 , then T possesses a unique fixed point.
Notice that several variations of Reich contractions (3) can be stated.We may state the following: In this paper, we shall investigate the validity of the interpolation approach for Reich contractions in the context of partial metric spaces that was introduced by Matthews [8].
In this case, (X, p) is said to be a partial metric space.
Definition 2. In the framework of a partial metric space (X, p), we say that p(ξ n , ξ m ) exists and is finite; For what follows, we shall recall the following lemma that can be derived easily (see [8]).
Lemma 1.Let p be a partial metric on a non-empty set X and ρ p be the corresponding standard metric space on the same set X.
(a) A sequence {ξ n } is fundamental in the framework of a partial metric (X, p) if and only if it is a fundamental sequence in the setting of the corresponding standard metric space (X, ρ p ).(b) A partial metric space (X, p) is complete if and only if the corresponding standard metric space (X, ρ p ) is complete.Moreover, (c) If ξ n → ζ as n → ∞ in a partial metric space (X, p) with p(ζ, ζ) = 0, then we have In this paper, we initiate the notion of interpolative Reich-Rus-Ćirić type contractions on partial metric spaces.We also present two examples illustrating our approach.

Main Results
We start this section by introducing the notion of interpolative Reich-Rus-Ćirić type contractions.Definition 3. In the framework of a partial metric space (X, p), a mapping T : X → X is called an interpolative Reich-Rus-Ćirić type contraction, if there are constants λ ∈ [0, 1) and α, β ∈ (0, 1) such that p (Tξ, for all ξ, η ∈ X\Fix(T), Theorem 4. In the framework of a partial metric space (X, p), if T : X → X is an interpolative Reich-Rus-Ćirić type contraction, then T has a fixed point in X.
Proof.We take an arbitrary point ξ 0 ∈ (X, p) and build an iterative sequence {ξ n } by ξ n = T n (ξ 0 ) for each positive integer n.If there exists n 0 such that ξ n 0 = ξ n 0 +1 , then ξ n 0 is a fixed point of T. The proof is completed.Henceforwards, assume that ξ n = ξ n+1 for each n ≥ 0. By substituting the values ξ = ξ n and η = ξ n−1 in ( 7), we find that By a calculation, we derive from the inequality (8).We conclude that {p (ξ n−1 , ξ n )} is a non-increasing sequence with non-negative terms.Thus, there is a nonnegative constant such that lim n→∞ p (ξ n−1 , ξ n ) = .Note that ≥ 0. Indeed, from (9), we deduce that Regarding λ < 1, and by taking n → ∞ in the inequality (10), we deduce that = 0.For what follows, we shall prove that {ξ n } is a fundamental (Cauchy) sequence by employing standard tools.More precisely, starting with the triangle inequality, we shall get the following estimation: Letting n → ∞ in the inequality (11), we ascertain that {ξ n } is a fundamental sequence.Hence, lim n,m→∞ p(ξ n , ξ m ) = 0, that is, {ξ n } is a fundamental sequence in (X, p).By Lemma 1, {ξ n } is also Cauchy in (X, ρ p ).More particularly, since (X, p) is complete, (X, ρ p ) is also complete.Hence, there exists ξ ∈ X such that which implies that lim As a next step, we make evident that the limit ξ of the iterative sequence {ξ n } is a fixed point of the given mapping T. Assume that ξ = Tξ, so p(ξ, Tξ) > 0. Recall that x n = Tx n for each n ≥ 0. By letting ξ = ξ n and η = ξ in (7), we determine that p (ξ n+1 , Tξ) =p (Tξ n , Tξ) Letting n → ∞ in the inequality (14), we find out p(ξ, Tξ) = 0, so ξ = Tξ, which is a contradiction.Thus, Tξ = ξ.
The following examples illustrate Theorem 4.
The following is an immediate consequence of our main result.