Interpolative Kannan- Meir-Keeler type contraction

In this short manuscript, we revisit the renowned contraction’s of Meir-Keeler by involving the interpolation theory in the context of complete metric space. We provide a simple example to illustrate the validity of the observed result.


Introduction
There is no need to emphasize it again, as it is very well known how effectively fixed point theory is used in nonlinear functional analysis, topology, and applied mathematics. Instead, it makes more sense to emphasize the effective use of fixed point theory in all branches of qualitative science. In this paper, we combine three exciting ideas and trends in the metric fixed point theory. After the famous fixed point theorem of Banach [1], one of the most significant advances in metric fixed point theory was given by Kannan [3,4]. It was later understood that the Banach contraction and Kannan contractions are independent [1]. On the other hand, in another aspect, Meir-Keeler [6] proposed an interesting contraction inequality that can be called a uniform contraction. Another interesting contraction concept that was recently published is interpolative contraction [5].
This paper aims to revisit the slightly modified version of the Meir-Keeler type contraction that is observed by combining the Kannan type contraction and interpolative contraction. We first recall the basic definitions and results. Definition 1.1. Let (X, d ) be a complete metric space. A mapping T : X → X is said to be a Meir-Keeler contraction on X (on short, MK-contraction), if for every ε > 0, there exists δ > 0 such that for every x, y ∈ X.
[6] On a complete metric space (X, d ), any MK-contraction T : X → X has a unique fixed point.
On a complete metric space (X, d ), any interpolative Kannan-contraction T : X → X has a fixed point.

Main Results
We start this section with the definition of interpolative Kannan-Meir-Keeler type contraction.
Theorem 2.2. On a complete metric space (X, d ), any interpolative KMK-contraction T : X → X has a fixed point.
Proof. Starting with a point x 0 ∈ X, we build the sequence {x m }, by the following rule: for all m ∈ N. Thus, by the assumption (2), we have and then, equivalent, Then, the sequence {d (x m , x m+1 )} is strictly decreasing and since d (x m , x m+1 ) > 0, for every m ∈ N ∪ {0}, it follows that the sequence {d (x m , x m+1 )} tends to a point ω ≥ 0. We claim that ω = 0. Indeed, if we suppose that ω > 0, we can find N ∈ N, such that This is a contradiction, and that's why we get ω = 0.
In order to show that {x m } is a Cauchy sequence, let ε > 0 be fixed and we can consider that δ(ε) can be choose such that δ(ε) < ε. Since lim m→∞ d (x m , x m+1 ) = 0, we can find l ∈ N such that d (x m , x m+1 ) < ε 2 , for m ≥ l, and we claim that , for any p ∈ N. Of course, the above inequality holds for p = 1. Supposing that for some p, (2.3) holds, we will prove it for p + 1. Indeed, using the triangle inequality, together with (2.2) we have Therefore, the sequence {x m } is Cauchy and by the completeness of the space X it follows that there exists We shall show that x * = Tx * . Supposing on the contrary, that x * = Tx * , by (2.2) we have Therefore, d (x * , Tx * ) = 0, that is, x * is a fixed point of the mapping T. We choose γ = 1 2 . Thus, we claim that T satisfies the conditions of Theorem 2.2. Indeed, for ε < 1, with δ = √ 2 − ε,  Consequently, by Theorem 2.2, the mappings T has fixed points; these are C = (2, −1) and all P ∈ X\A.