Existence and Uniqueness of Fixed Points of Generalized F-Contraction Mappings

The newest generalization of the Banach contraction through the notions of the generalized F-contraction, simulation function, and admissible function is introduced. The existence and uniqueness of fixed points for a self-mapping on complete metric spaces by the new constructed contraction are investigated. The results of this article can be viewed as an improvement of the main results given in the references.


Introduction and Preliminaries
In 1922, Banach proved the following famous and fundamental result in fixed-point theory [1]. Let (X, d) be a complete metric space. Let T be a contractive mapping on X; that is, there exists q ∈ [0, 1) satisfying d(Tx, Ty) ≤ q.d(x, y), ∀x, y ∈ X. (1) en, there exists a unique fixed point x 0 ∈ X of T. is theorem, which is called the Banach contraction principle that is a forceful tool in nonlinear analysis [9][10][11][12][13][14] and fixed-point theory, is a fascinating subject, with an enormous number of algorithms and applications in various fields of mathematics, see, e.g., [15][16][17][18]. is principle has been generalized in different directions by various researchers. One of them is the following theorem that is presented by Bryant. Theorem 1 (see [2]). If f is a mapping of a complete metric space into itself and if, for some positive integer k, f k is a contraction, then f has a unique fixed point.
It is obvious that f k is continuous but there are examples that show it cannot imply the continuity of f and so eorem 1 is a real extension of the Banach principle.
Theorem 2 (see [19]). Let (X, d) be a complete metric space, q ∈ [0, 1), and T: X ⟶ X be a continuous mapping. If for each x ∈ X there exists a positive integer k � k(x) such that for all y ∈ X, then T has a unique fixed point u ∈ X. Moreover, for any x ∈ X, u � lim n⟶∞ T n x. Several researchers are interested to generalize Banach contraction. Here, we state two of them. Wardowski [8] generalized the Banach contraction as follows.
Definition 2 (see [3,21]). Let ζ: [0, ∞) × [0, ∞) ⟶ R be a mapping, then ζ is called a simulation function if it satisfies the following conditions: s n are sequences in (0, ∞) such that lim n⟶∞ t n � lim n⟶∞ s n > 0 and t n < s n for all n ∈ N, then limsup n⟶∞ ζ t n , s n < 0 (4) We denote the set of all simulation functions by Z.
Theorem 3 (see [4]). Let (X, d) be a complete metric space and T: X ⟶ X a mapping which satisfies the condition: If there exist f ∈ F and τ > 0 such that for each x ∈ X there is a positive integer n(x) such that for all y ∈ X, then T has a unique fixed point z ∈ X and T n (x 0 ) ⟶ z for each x 0 ∈ X, as n ⟶ ∞.
e first aim of this paper is to generalize eorem 2 by introducing a more general contraction type mapping through the notions of the generalized F-contraction, simulation function, and admissible function. en, by the new constructed contraction and suitable conditions, the existence and uniqueness of fixed points are investigated. e following definitions and preliminary results are needed in the next section.
Definition 4 (see [23]). An α-admissible map T is said to have the K-property, while for each sequence x n ⊆ X with α(x n , x n+1 ) ≥ 1 for all n ∈ N 0 , the nonnegative integer numbers, there exists a positive integer number k such that α(Tx n , Tx m ) ≥ 1, for all m > n ≥ k.
Lemma 1 (see [5]). Let F: (0, +∞) ⟶ R be an increasing function and α n be a sequence of positive real numbers. en, the following holds: [24]). Let (X, d) be a metric space and x n be a sequence in X such that lim n⟶∞ d(x n , x n+1 ) � 0. If x n is not a Cauchy sequence, then there exist ε > 0 and two sequences of positive integers n k and m k with

Main Results
In this section, the main achievements of this article are presented. e existence and uniqueness of fixed points of the self-mappings on complete metric spaces satisfying the generalized F-contraction (relation (6) of the following theorem) with suitable assumptions are established by the first theorem. e second theorem can be viewed as a generalized version of Suzuki's theorem given in [21]. Of course it ensures existence of fixed points for self-mappings under suitable hypothesis.

Theorem 4
Let (X, d) be a complete metric space and α: X × X ⟶ (0, +∞) be a symmetric function, where α(x, y) ≥ 1 and T: X ⟶ X be a continuous mapping which satisfies the condition: if there exist F ∈ F, τ > 0, L ≥ 0, and simulation function ζ such that for all x ∈ X there is a positive integer n(x) such that for all y ∈ X and where then T has a unique fixed point.
Proof. We shall built a recursive sequence x k as follows: for the chosen arbitrary point x 0 ∈ X with n 0 � n(x 0 ), we set en, x i 0 turns to be a fixed point of T n i 0 . On the other hand, us, Tx i 0 form a fixed point of T n i 0 . If Tx i 0 � x i 0 , then we conclude that T has a fixed point and that terminate the proof. Suppose, on the contrary, that Hence, However, erefore, So, τ ≤ 0, which is a contradiction. Consequently, we deduce that for all i ∈ N 0 , Journal of Mathematics en, However, and since F is strictly increasing, so, it follows from (14) that So, τ ≤ 0, which is a contradiction. Consequently, Hence, from (14) and (19), we have (20) or In general, one can get Hence, So, from (F 2 ), we have erefore, with notice to (F 3 ), there exists k ∈ (0, 1) such that Now, (22) implies that en, it can be easily seen that So, there exists i 0 ∈ N 0 such that Consequently, if m > n > n 0 , then Since k ∈ (0, 1), the series ∞ j�n 0 1/j (1/k) is convergent. erefore, x i is a Cauchy sequence, and since X is complete, there exists u ∈ X such that x i ⟶ u as i ⟶ ∞.As a next step, we show that u is a fixed point of T n(u) . Indeed, due to the continuity of T, we have and so u is a fixed point of T. For proving the uniqueness of the fixed point, let us consider u and v be two distinct fixed points and n � n(u). So, we have d(u, v) > 0, and hence, we get that d(Tu, Tv) > 0; then, by (6) and (ζ2), (Tu, Tv))).

Corollary 2. eorem 3 is contained in eorem 4 by taking
e following example shows that if the mapping T satisfies the condition of eorem 4, it cannot guarantee in general the continuity of the mapping T. Example 1. Let X � R denote the real numbers with the usual metric d. Define function T: X ⟶ X by en, T discontinues at each point of X, and T 2 � 1. If α is an arbitrary element of [0, 1), then ∀x ∈ X, ∃n x � 2; ∀y ∈ X: d T n x x, T n x y � 0 ≤ α d(x, y).

(40)
Now, it is obvious that the function ζ(t, s) � αs − t of condition (6) of eorem 4 on [0, ∞) × [0, ∞) is a simulation function and T satisfies following condition: but T discontinues at each point of X. Moreover, T satisfies all the assumptions of eorem 4, when L � 0 and the unique fixed point of T is x � 1 and Picard's iteration of T; that is, if y ∈ X is an arbitrary point of X, then T n (y) is convergent to the fixed point.

Theorem 5.
Let (X, d) be a complete metric space and α: X × X ⟶ (0, +∞) a symmetric function, where α(x, y) ≥ 1. Assume that T: X ⟶ X is a mapping in which there exist F ∈ G, τ > 0, and the simulation function ζ such that for all x, y ∈ X with T n(x) x ≠ T n(x) y, where n(x) is a positive integer and 1/2d(x, where m(x, y) is defined as in eorem 4, satisfying the following conditions: n ⟶ ∞ and α(x n , x n+1 ) ≥ 1 for all n ∈ N 0 , then α(x n , x) ≥ 1 for all n ∈ N 0 , and (iv) T has the K-property, then T has a fixed point in X.
Proof. Let x 0 ∈ X be an arbitrary point. e recursive sequence x k is inductively constructed as follows: n 0 � n(x 0 ), and we set x 1 � T n 0 x 0 and inductively get x i+1 � T n i x i with n i � n(x i ).
We assert that x i ≠ x i+1 for all i ∈ N 0 . Suppose, on the contrary, that there exists i 0 ∈ N 0 such that en, x i 0 turns to be a fixed point of T n i 0 . On the other hand, us, Tx i 0 form a fixed point of T n i 0 . If Tx i 0 � x i 0 , then we conclude that T has a fixed point and that terminate the proof. Suppose, on the contrary, that Tx i 0 ≠ x i 0 and hence d(T n i 0 (Tx i 0 ), T n i 0 (x i 0 )) > 0. en, by (42), we have Hence,

Journal of Mathematics
However, erefore, So, τ ≤ 0, which is a contradiction. Consequently, we deduce that, for all i ∈ N 0 , x i ≠ x i+1 . en, d(x i+1 , x i ) > 0, and so Now, by (42), Hence, However, If d( and since F is strictly increasing, so, it follows from (50) that Hence, τ ≤ 0, which is a contradiction. erefore, Hence, from (50) and (55), we have Consequently, Hence, So, from (G 2 ), we have Now, we claim that x i is a Cauchy sequence. If it is not true, then by Lemma 2, there exist ε 0 > 0 and two sequences of positive integers n k and m k with n k > m k > k such that d(x m k , x n k ) > ϵ 0 , d(x m k , x n k − 1 ) < ε 0 , and On the other hand, since lim k⟶∞ d(x n k , x m k − 1 ) � ε 0 > 0 and lim k⟶∞ d(x n k , x n k +1 ) � 0, with considering a subsequence if it is needed, one can suppose that there exists k 1 ∈ N such that for any k > k 1 and n k > m k > k, So, it is obvious that, for all k > k 1 and n k > m k > k, Also, using the K-property, there exists k 2 ∈ N such that If k ≥ max k 1 , k 2 , then it follows from (65) that Letting n ⟶ ∞, the continuity of F through (L1) and (62) implies which is contradicted by τ > 0. Consequently, x i is a Cauchy sequence in the complete metric space X. Hence, there exists u ∈ X such that x i ⟶ u, as n ⟶ ∞. To complete the proof, we show that u is a fixed point of T. We first claim, for all n ≥ 0, that In fact, if we assume that, for some i 0 ≥ 0, both of them are false, then Hence, (55) implies which is a contradiction and the claim is proved. Now, let us begin with the first part of (68); that is, suppose that and on the contrary, assume that Tu ≠ u. Without loss of generality, one can imagine that Tx i ≠ Tu, for all i ∈ N 0 (because if x i+1 � Tx i � Tu for infinite values of i, then uniqueness of the limit concludes that Tu � u). en, from (45) and (iii), we get And since F is continuous on (0, +∞), and d(u, Tu) > 0, as i ⟶ ∞, we get τ + F(d(u, Tu)) ≤ F lim i⟶∞ m x i , u . However, which is contradicted, as τ > 0. So, d(u, Tu) � 0, i.e., Tu � u.
Finally, if we assume that the second part of (68) is true, i.e., then by using the same manner, we can prove that d(u, Tu) � 0, i.e., Tu � u. Suppose that u and v are two fixed points of T. If u ≠ v, then d(Tu, Tv) > 0. Furthermore, α(u, v) ≥ 1, because u, v ∈ Fix(T). It is also clear that 1/2d(u, Tu) � 0 < d (u, v).

Data Availability
e data used to support the findings of this study are included within the article.

Conflicts of Interest
e authors declare that they have no conflicts of interest.