Some Results on S-Contractions of Type E

In this manuscript, we consider the compositions of simulation functions and E-contraction in the setting of a complete metric space. We investigate the existence and uniqueness of a fixed point for this composite form. We give some illustrative examples and provide an application.

We recollect some basic notions as well as the fundamental definitions to provide a self-contained manuscript.For more details about the tools and notations, we refer e.g., [19,20].We shall use the letters R, R + , N for the reals, nonnegative real numbers, and natural numbers, accordingly.Moreover, we employ the symbols R + 0 = R + ∪ {0} = [0, ∞) and N 0 = N ∪ {0}.
Definition 1 (See [19]).A function σ : R + 0 × R + 0 → R is called simulation function if it checks the following conditions: We will denote by S the family of all simulation functions.It is clear, due to the axiom (s f 2 ), that σ(t, t) < 0 for all t > 0. ( Let Φ be the class of continuous functions φ : [0, ∞) → [0, ∞) which satisfies the condition φ(x) = 0 if, and only if, x = 0.
semi-continuous and such that f (t) < t for all t > 0 and f (0 for all t, s ≥ 0, where g, h : [0, ∞) 2 → (0, ∞) are two continuous functions with respect to each variable such that g(t, s) > h(t, s) for all t, s > 0. (vi) σ 6 (t, s) = s η(s) − t for all t, s ≥ 0, where η : [0, ∞) → [0, 1) is a function with the property lim sup t→r + η(t) < 1 for all r > 0 Let (M, d) be a metric space and σ ∈ S be a simulation function.We say that a function S : M → M is S-contraction with respect to σ [19], if the inequality σ(d(Sp, Sq), d(p, q)) ≥ 0 for all p, q ∈ M, is attained.
Remark 1. From (σ 2 ), we find that d(Sp, Sq) = d(p, q) for all distinct p, q ∈ M, which infers that, whenever S is a S-contraction, S cannot be an isometry.As a result, the fixed point of a S-contraction S (if there exists) is necessarily unique.
Theorem 1 ([19]).For every S-contraction on a complete metric space there exists exactly one fixed point.
In fact, every Picard sequence converges and its limit is the unique fixed point.

Main Results
We state now our main results.For this purpose, we start by defining a new type of S-contraction.
Definition 2. A self-mapping S defined on a complete metric space (M, d) is a S-contraction of type E with respect to σ if there exists σ ∈ S such that σ(d(Sp, Sq), E(p, q)) ≥ 0 for all p, q ∈ M, where, We denote by C E (M) the set of S-contractions of type E with respect to σ defined on M.
We now present the results regarding the existence of a fixed point, since the uniqueness follows from Remark 1.

Theorem 2. There exists a fixed point for every S ∈ C E (M)
Proof.Given an arbitrary p 0 ∈ M we consider the constructive sequence {p n } ⊂ M which is defined by u n+1 = Su n = S n u 0 for all n ∈ N.
We shall assume that p n+1 = p n for all n ∈ N. Indeed, in the opposite case, where p n 0 = p n 0 +1 for some n 0 ∈ N, then Sp n 0 = p n 0 .This completes the proof and we conclude that the given point p n 0 turns to be a fixed point of S.
Consequently, d (p n+1 , p n ) > 0 and from Equation ( 5), it follows, for all n ≥ 1, that In conclusion, for all n = 1, 2, . . ., we have To understand the inequality Equation ( 8), we consider two cases.For the first case, we suppose that d(p n , p n+1 ) ≥ d(p n−1 , p n ).In this case, the inequality Equation ( 8), becomes a contradiction.Thus, the following case occurs Accordingly, we deduce that the sequence {d(p n , p n−1 )} is non-increasing and bounded below by 0. Hence, the sequence {d(p n , p n−1 )} converges to some d * ≥ 0. In the same time, we note that We claim that Suppose, on the contrary that d * > 0.Then, letting t n = d(p n , p n+1 ) and s n = E(p n , p n−1 ) we get from Equation ( 5) and (s This contradiction shows that d * = 0. We will now show that the sequence {p n } is Cauchy.Suppose, on the contrary, that {p n } is not a Cauchy sequence.Then there exists a real positive number ε > 0 and sequences α On the other hand, it is easy to see that and from Equation (11) respectively Equation ( 13) Moreover from Equations ( 6), ( 11) and ( 13), we have Letting This contradiction proves that ε = 0 and thus the sequence {p n } is Cauchy.On the account of the completeness of (M, d) there exists a point p * ∈ M such that We must prove now that p * = Sp * .Arguing by contradiction, we will assume that d(p * , Sp * ) > 0. By the property (s f 2 ), for r ∈ N sufficiently large, we have Considering the sequences which implies together with Equation ( 18) a contradiction.Thus, we have d(p * , Sp * ) = 0, i.e., Sp * = p * .

Examples
Example 2. Let the set M = 0, 5  3 ∪ {2} and d(p, q) = |p − q|, d : M → M. Suppose that σ : [0, ∞) × [0, ∞) → R is defined as σ(s, t) = s 2 − t and hence σ ∈ S. Define a map S : M → M as follows Notice that for p = 2 and q = 5 3 , we have and for these values there is no k Thus, the mapping S is not a contraction.
On the other hand, it is S-contraction of type E. For the proof of our claim, we need to consider two distinct cases: Case 1. q = 2, p < 1.Then, we find that we have that As a result, we have and also We deduce that S is a S-contraction of type E. Further, all conditions of Theorem 2 are fulfilled and p = 1 is a fixed point of S. Finally, we mention that the uniqueness of the fixed point follows from the Remark 1.
One can easily get that Moreover, we have Firstly, we claim that S is not a contraction.Indeed, for p = 4 and q = 5, we could not find a real constant k 2 ∈ [0, 1) such that d(T4, T5) = 2 ≤ k 2 d(4, 5) is satisfied.So, S is not a contraction.Now, we shall show that S is a S-contraction of type E. For this purpose, we examine all possible cases: As a last case, for p = 4, q = 5, we derive σ (d(S4, S5), E(4, 5)) = σ(2, 4) = 4  2 − 2 = 0. Evidently, we conclude that S ∈ C E (M).
In addition, all conditions of Theorem 2 are attained and p = 3 is a fixed point of S. As in the above example, the uniqueness results from the Remark 1.

Consequences and Application
In this section we give one corollary and we consider an application of the main result in which the solution for an integral equation can be described.
Proof.It is enough to take σ(t, s) = σ 1 (t, s) in Example 1 and apply Theorem 2.
It is clear that by choosing simulation function σ from Example 1 and applying Theorem 2, we get further corollaries, as we got Corollary 1. So, we skip this list of corollaries by using the analogy.
Let M = C(I, R) be the set of all continuous functions on I = [0, 1] equipped with a metric d(x, y) = x − y = sup{|x(s) − y(s)| : s ∈ I}, for all x, y ∈ M. Then (M, d) forms a complete metric space.We investigate the integral equation where the functions η : I × R → R and ξ : I → R are continuous and K : I × I → R + 0 is a function such that K(s, •) ∈ L 1 (I) for all s ∈ [0, 1].We handle the map S : M → M which is defined by Theorem 3. The Equation ( 21) has a unique solution in M, if the following conditions are fulfilled: (a1) there exists µ ∈ Φ with µ(s) < s for all s > 0 satisfying for all u ∈ I and for all x 1 , x 2 ∈ M. (a2) followed by the inequality assumed Proof.Note that any fixed point of Equation ( 21) is a solution of the integral Equation (21).On account of (a1) and (a2), we find that where, E(x 1 , x 2 ) = x 1 − x 2 + x 1 − Sx 1 − x 2 − Sx 2 .Hence, we derive that Therefore we have This implies that all the conditions of Corollary 1 and hence Theorem 2 are satisfied.Thus, the operator S has a unique fixed point which is the solution of the integral Equation ( 21) in M. Hence, we conclude that the integral Equation (23) has exactly one solution in C(I, R).