An admissible Hybrid contraction with an Ulam type stability

In the last three-four decades, there is a blown out in the number of publications in metric fixed point theory. This fact forces researchers to find a way to combine, unify and merge the existing results in a proper way. In this paper, we aim to give an interesting example for this trend. We introduce a new hybrid contraction which not only combine and unify the several existing linear and nonlinear contractions but also extend these results. Let Ψ be the set of functions ψ : [0,∞) → [0,∞) such that


Introduction and preliminaries
In the last three-four decades, there is a blown out in the number of publications in metric fixed point theory. This fact forces researchers to find a way to combine, unify and merge the existing results in a proper way. In this paper, we aim to give an interesting example for this trend. We introduce a new hybrid contraction which not only combine and unify the several existing linear and nonlinear contractions but also extend these results.

Lemma 1.2.
Suppose that for a triangular α-orbital admissible mapping f : X → X there exists x 0 ∈ X such that α(x 0 , f x 0 ) ≥ 1. Then α(xn , xm) ≥ 1, for all n, m ∈ N, where the sequence {xn} is defined by x n+1 = f xn, n ∈ N. Definition 1.3. Let α : X × X → [0, ∞) be a mapping. The set X is called regular with respect to α if for a sequence {xn} in X such that α(xn , x n+1 ) ≥ 1, for all n and xn → x ∈ X as n → ∞ we have α(xn , x ) ≥ 1 for all n.

Main results
We start with a definition of a new notion, namely "admissible hybrid contraction": where q ≥ 0 and λ i ≥ 0, i = 1, 2, 3, 4, 5 such that The concept of "admissible hybrid contraction" is inspired from the notion of "interpolative contractions", see e.g. [4][5][6][7][8][9] The main results of this manuscript is the following theorem: Theorem 2.2. Let (X , d) be a complete metric space and let f be an admissible hybrid contraction, Suppose also that: Then f has a fixed point.
Proof. Starting from an arbitrary point x 0 in X we recursively set-up the sequence {xn}, as xn = f n x 0 for all n ∈ N. Supposing that there exists some m ∈ N such that fνm = x m+1 = xm, we find that xm is a fixed point of f and the proof is finished. So, we can presume from now on that xn ≠ x n−1 for any n ∈ N. Under the assumption (i), f is admissible hybrid contraction, if we substituting in (5) x by x n−1 and y by xn we get Taking into account that f is triangular α−orbital admissible, together with (4) holds and the above inequality becomes Case 1. For the case q > 0 we have If we suppose that d(x n−1 , xn) ≤ d(xn , x n−1 ), since ψ is a nondecreasing function, which is a contradiction. Therefore, for every n ∈ N we have and the inequality (8) yields Let now, m, p ∈ N such that p > m. By the triangle inequality and since d(xm , Since ψ is a c-comparison function the series ∑︀ ∞ j=0 ψ j (d(x 0 , x 1 )) is convergent, so that, denoting by Sn = ∑︀ n j=0 ψ j (d(x 0 , x 1 )) the above inequality becomes: which tells us that {xn} is a Cauchy sequence on a complete metric space, so that, there exists z such that We will prove that this point z is a fixed point of f . If f is continuous, (due to assumption (iii)) In the alternative hypothesis, that f 2 is continuous we have f 2 z = lim n→∞ f 2 xn = z and we want to show that This is a contradiction, so that f z = z.

Case 2.
For the case q = 0 taking x = x n−1 and y = xn we have As in the first case, we have that d( which is a contradiction. Then from (14) we obtain and inductively we get d(xn , x n+1 ) ≤ ψ n (d(x 0 , x 1 )).
By using the same arguments as the case q > 0 we shall easily obtain that {xn} is a Cauchy sequence in a complete metric space and so, there exists z such that limn→∞ xn = z.
We claim that z is a fixed point of f . Under the assumption that f is continuous we have and together with the uniqueness of limit, f z = z. Also, if f 2 is continuous, as in case (1) we have that f z = z and then This contradiction shows us that z = f z.
In all other cases, α(x , y) = 0 and (5) is obviously satisfied. Thus, letting λ 1 = λ 2 = λ 3 = λ 5 = 1 4 , λ 4 = 0 and q = 2 we obtain that f is an admissible hybrid contraction which satisfies the assumptions (i), (ii), (iv) of Theorem 2.2 and then x = 0 is the fixed point of f . Proof. Following the lines in the proof of Theorem 2.2, we already know that for any q ≥ 0, the sequence {xn} is Cauchy, and due to the completeness of the metric space (X , d), there exists a point z such that lim n→∞ d(xn , z) = 0. Since the space X is regular with respect to α, inequality (5) together with the triangular inequality gives us Again, we have to consider two separate cases. For the case p > 0, Since lim  (i) f is triangular α−orbital admissible; (ii) there exists x 0 ∈ X such that α(x 0 , f x 0 ) ≥ 1; (iii) either, f is continuous, or (iv) f 2 is continuous and α(f x , x ) ≥ 1 for any x ∈ Fix f 2 (X ).
If one of the below conditions (c 1 )-(c 3 ) is satisfied, then f has a fixed point z ∈ X , that is, f z = z. where a 1 , a 2 , a 3 , a 4 are non-negative real such that a 1 + a 2 + a 3 + a 4 = 1 and y)), where c 1 , c 2 are non-negative real numbers such that c 1 + c 2 = 1 and We can get a series of corollaries, considering in Corollary 2.4 by assigning ψ ∈ Ψ properly, for example, by taking ψ(t) = kt for any t ≥ 0 with k ∈ [0, 1), and/or α(x , y) = 1 or both. Since it is apparent we skip the details. for any x , y ∈ Fix f (X ) then the fixed point of f is unique.
Proof. Let v ∈ X be another fixed point of f , different from z. By replacing in (5), and taking into account the additional hypotheses, we have which is a contradiction. Thus, z = v, so that f possesses exactly one fixed point. t. Since neither f , nor f 2 are continuous, Theorem 2.2 cannot be applied. On the other hand, is easy to see that f is triangular α−orbital admissible and also the assumptions (2), (3) from Theorem 2.3 are satisfied. Considering q = 0, λ 1 = λ 2 = λ 3 = λ 4 = 1/4 and λ 5 = 0 and taking into account the definition of function α, we remark that the only interesting case is for x = b and y = d . We have in this case: Consequently, the map f has a fixed point, that is x = a.

Ulam type stability
Considered as a type of data dependence, the notion of Ulam stability was started by Ulam [10,11] and deveped by Hyers [12], Rassias [13], etc In this section we investigate the general Ulam type stability in sense of a fixed point problem. Suppose that f : X → X is a self-mapping on a metric space (X , d). The fixed point problem there exists a solution z ∈ X of (20) such that In case that for C > 0, we consider ρ(t) = Ct for all t ≥ 0 then the fixed point equation (20)  Proof. (i) Since from Theorem 2.5 we know that there is an unique z ∈ X such that f z = z, let y * ∈ X such that d(y * , f y * ) ≤ ε, for ε > 0.