A Result on a Pata-Ćirić Type Contraction at a Point

In this manuscript, we define a new contraction mapping, Pata-Ćirić type contraction at a point, that merges distinct contractions defined by Seghal, Pata and Ćirić. We proved that in a complete space, each Pata-Ćirić type contraction at a point possesses a fixed point. We express an example to illustrate the observed result.


Introduction and Preliminaries
In recent decades, the fixed point theory has not only been one of the most interesting research topics in nonlinear functional analysis, but also one of the most dynamic and productive research areas. Although the first published work on this subject was conducted a century ago, the fixed point theory continues to be the center of attention. The reason why the fixed point theory is still the attraction center can be explained as follows: different problems in various disciplines can be turned into fixed point problems. Mostly, solving the fixed problem is easier than the original representation. Note that in the metric fixed point theory, not only the existence (mostly, also uniqueness) of the fixed point (which corresponds to the solution) is guaranteed, but also, it is shown "how to reach" the mentioned fixed point and hence the solution.
Before giving the main results of the manuscript, we shall introduce and fix the notations that we deal with it. Throughout this article, we use the letter S to denote a non-empty set. We presume that d forms a metric over S. Thereafter, the pair (S, d ) denotes metric space. Unless otherwise stated in the particular conditions, the pair (S * , d ) expresses that the corresponding metric space is complete. The expressions R, R + , R + 0 stand for the set of all "real numbers", "positive real numbers", and "non-negative real numbers". In addition, the letters N and N 0 are restricted to indicate the set of all positive integers and all non-negative integers, respectively.
In the following theorem, Bryant [1] proved the analog of Banach's fixed point theorem [2] for not the mapping itself but for its iterated form of the given mapping.
for all w ∈ S.
Inspired from the idea of Bryant [1], we give the following example to illustrate the necessity of the result of Bryant [1] The inequality (1) holds for n = 2. Despite T not being continuous, T0 = 0.

Theorem 2 ([3]).
A continuous self-mapping T on (S * , d ) admits a unique fixed point u ∈ S, if it satisfies the condition: there exists a constant k > 1 such that, for each z ∈ S, there is a positive integer n(z) such that for all w ∈ S.
Immediately after the publication of Seghal's article, Guseman [4] showed that the theorem of continuity was unnecessary, see also [5,6].
Throughout this work, we note as Υ the set of all auxiliary functions ψ : [0, 1] → [0, +∞) which are increasing, continuous at zero, and ψ(0) = 0. Furthermore, for an arbitrary point z 0 in a (S, d ), we consider the function z = d (z, z 0 ), for all z ∈ S, that will be called "the zero of S".

Theorem 4 ([8]).
Suppose that a self-mapping T on (S, d ) is a Pata type contraction at a point. Then, T admits a unique fixed point.

Definition 2.
A self-mapping T , defined on (S, d ), is called Pata-Ćirić type contraction at a point if for every ε ∈ [0, 1] and for each z ∈ S there exists a positive integer n(z) such that the following inequality is fulfilled: and further Continuing in this way, we deduce that for q 1 , q 2 , ..., q k ∈ N.
Let q be an arbitrary but fixed positive integer and k a positive integer, depending on z 0 and q. (We can suppose that q > n 0 and also k > n 0 ). We denote this as Using the triangle inequality, by (5) we have where . Thus, coming back in (10) and taking into account (9) we have Our next purpose is to prove that the sequence { i } is bounded. Taking into account that β ∈ [0, λ]), the above inequality becomes and therefore, We assert that the sequence { i } is bounded. To use the method of Reductio ad Absurdum, we suppose that the sequence { i } is unbounded. Attendantly, we can find a sub-sequence { i n } such that lim n→∞ i n = ∞. For this reason, there is number N 1 ∈ N such that i n > 1 + 3ρ 0 for each n ≥ N 1 . Thus, Since ε n → 0 as n → 0, taking into account the properties of the function ψ, we obtain which is a contradiction. Therefore, our assumption is false, so that the set d (z 0 , Thus, taking into account (7), there exists a positive number K such that for m ≥ m 0 . In order to prove that the sequence {z n } is Cauchy, let k, l ∈ N. Denoting by p 0 = n k+l−1 + n k+l−2 + ... + n k , we have where p 1 in {p 0 , n k−1 , p 0 + n k−1 } is chosen such that Continuing in this way, we found that Dividing by k λ , the previous inequality gives Thereof, the sequence {z n } is Cauchy on (S * , d ). On account of completeness, there is u ∈ S so that lim n→∞ z n = u.
The first goal is to indicate that u = T n(u) u. Otherwise, it turns to T n(u) u = u. Accordingly, we have By the triangle inequality, we have T n(u) u ≤ d (T n(u) u, u) + d (u, z 0 ) = d (T n(u) u, u) + u , and letting q → +∞ in the above inequality we have but this implies that Taking ε → 0, since λ ≥ 1 and using the property of function ψ we have that which is a contradiction. Therefore, d (u, T n(u) u) = 0, that is u is a fixed point of T n(u) .
Suppose that there exists another point v ∈ S such that T n(u) v = v and v = u, by (5) we have Letting ε → 0 and keeping in mind the property of ψ, we have d (u, v ) = 0, that is, the fixed point of T n(u) is unique.
Finally, because T n(u) (Tu) = T (T n(u) u) = Tu taking into account this uniqueness, we have Tu = u.
We are now concerned with the second part of the theorem. Let u be a fix point of T in S.
for i ∈ N 0 and reasoning by contradiction, we will proof that a i ≤ a i−1 for i ≥ 1. Presuming that there is i 0 ∈ N such that a i 0 > a i 0 −1 ≥ 1, since T n(u) u = u, we have, Since, by the triangle inequality, we get Keeping, the continuity of the function ψ in mind, together with the property ψ(0) = 0, we find that a i 0 = 0, letting ε → 0 in the above inequality. It contradicts our assumption a i 0 > a i 0 −1 ≥ 0. We have proved that the sequence {a i } is non-increasing, so that there exists a ≥ 0 such that lim For all i ∈ N we have Thus, By repeating the above arguments, when ε → 0 we get a = 0. Since z, i, j were fixed, we find that lim Therefore, as m → +∞ we have Furthermore, and letting ε → 0, we get δ ≤ 0, which is a contradiction. This means that T n(u) w n → u and since T n(u) u = u, we conclude that T n(u) is continuous at u. It easy to see that the Theorem 3 it cannot be applied, because for z = b, w = c and ε = 0 we have On the other hand, since z a b c e g T 3 z e c a b g choosing λ = β = 1, ψ(ε) = ε 2 and Λ = 4 + 1 1+ε 3 we have: for every z, w ∈ S, so that the mapping T has a unique fixed point.

Application
We will consider in this section the problem of the existence and uniqueness of solutions for a second-order nonlinear integrodifferential equation, as an application of our main results. Let B be a Banach space endowed with the norm · . We denote by E the Banach space of all continuous functions from [0, 1] in B, with the norm (c 1 ) C (0) = I; (c 2 ) C (t)z is strongly continuous in t ∈ R, for any fixed z ∈ B; (c 3 ) C (s + t) + C (s − t) = 2C (s)C (t), for s, t ∈ R.
Let the S (t) : t ∈ R be the associated sine family of that cosine family, defined as follows S(t)z = t 0 C (s)zds, z ∈ B, t ∈ R and two positive constants Θ 1 ≥ 1, Θ 2 such that 1] × B → B be continuous function and the nonlinear integro-differential equation with the initial value where z 0 , z 1 ∈ B are given. Here A : B → B, defined by is the infinitesimal generator of a cosine family {C (t) : t ∈ R}. The function z ∈ S, defined for t ∈ [0, 1] by is a mild solution of the Equation (13) with the initial values (14).
Taking into account the conditions imposed in the hypothesis of the theorem regarding the constant Υ, we can find ε ∈ [0, 1] such that 227 360 (Θ 2 Υ) 2 = ε 2 . We have, then, Therefore, considering ψ(ε) = ε, Λ = 2, λ = β = 1, we can easily see that for z arbitrary in E , there exists n(z) = 2 such that d (T 2 z, T 2 w ) ≤ P (z, w ), for every w ∈ E , that is, the mapping T is a Pata-Ćirić type contraction at a point. Therefore, since all the assumption of Theorem 5 are fulfilled, the problem (13)-(14) admits an unique solution. (Moreover, we mention that in the case n = 1, the presumptions of the Theorem 5 are not satisfied.)