Revisiting the Kannan Type Contractions via Interpolation

In the paper we revisited the well-known fixed point theorem of Kannan under the aspect of interpolation. By using the interpolation notion, we propose a new Kannan type contraction to maximize the rate of convergence.


Introduction
After the distinguished fixed point of Banach, one of the pivotal metric fixed point result was reported by Kannan [1,2].A mapping that satisfies Banach contraction inequality is necessarily continuous.In 1968, Kannan [1] introduced a new type of contraction which is an affirmative answer to the natural question below: Whether there is a discontinuous mapping that fulfils certain contractive conditions and posses a fixed point in the frame of complete metric spaces.

Main results
We start our results by the generalization of the definition of Kannan type contraction via interpolation notion, as follows.
Definition 2.1.Let (X, d) be a metric space.We say that the self-mapping T : X → X is an interpolative Kannan type contraction, if there exist a constant λ ∈ [0, 1) and α ∈ (0, 1) such that for all x, y ∈ X with x = T x.
Theorem 2.2.Let (X, d) be a complete metric space and T be an interpolative Kannan type contraction.
Then T has a unique fixed point in X.
Proof.Let x 0 ∈ (X, d).We shall set a constructive sequence {x n } by x n+1 = T n (x 0 ) for all positive integer n.Without loss of generality, we assume that x n = x n+1 for each nonnegative integer n.Indeed, if there exist a nonnegative integer n 0 such that x n 0 = x n 0 +1 = T x n 0 , then, x n 0 forms a fixed point.Thus, we have for each nonnegative integer n.
Taking x = x n and y = x n−1 in (2.1), we derive that which yields that Thus, we deduce that the sequence {d (x n−1 , x n )} is non-increasing and non-negative.As a result, there is a nonnegative constant L such that lim n→∞ d (x n−1 , x n ) = L.We shall indicate that L > 0. Indeed, from (2.3), we derive that Letting n → ∞ in the inequality above, we observe that L = 0.As a next step, we shall show that the sequence {x n } is Cauchy by using a standard arguments based on the triangle inequality.More precisely, we have Letting n → ∞ in the inequality above, we find that the sequence {x n } is Cauchy.Since (X, d) is a complete metric space, there exists x ∈ X such that lim On what follows we shall show that the limit x of the iterative sequence {x n } forms a fixed point for the given self-mapping T. By substituting x = x n and y = x in (2.1), we find that (2.6) Taking n → ∞ in the inequality above, we derive that d(x, T x) = 0 that is, T x = x.