A New Approach of Some Contractive Mappings on Metric Spaces

: In this paper, we introduce a new contraction-type mapping and provide a ﬁxed-point theorem which generalizes and improves some existing results in the literature. Thus, we prove that the Boyd and Wong theorem (1969) and, more recently, the ﬁxed-point results due to Wardowski (2012), Turinici (2012), Piri and Kumam (2016), Secelean (2016), Proinov (2020), and others are consequences of our main result. An application in integral equations and some illustrative examples are indicated.


Introduction
The Banach contraction principle [1] is a fundamental result in the fixed-point theory. It ensures the existence and uniqueness of fixed points of certain self-maps on metric spaces and provides an iterative method to find the respective fixed points. Therefore, it is a very important and powerful tool in solving the existence problems in pure and applied sciences. More precisely, if T is a self-mapping on a complete metric space (X, d) such that d(Tx, Ty) ≤ cd(x, y), ∀ x, y ∈ X for some c ∈ (0, 1), then there exists a unique x * ∈ X such that Tx * = x * . Moreover, for each x 0 ∈ X, the sequence (T n x 0 ) n converges to x * . In this setting, we say that T is a Banach contraction.
Since then, many researchers generalized and improved the result of Banach by extending the spaces and the operators. Additionally, new areas of application of these results are being discovered.
A function T : X → X is called a Picard operator [2] if it has a unique fixed point x * and, for each x 0 ∈ X, lim n→∞ T n x 0 = x * , where T n is the n-th composition of T. In 1969, Boyd and Wong [3] generalized the Banach contraction principle by replacing the linear condition with a real-valued map called a comparison function. A self-mapping T on a metric space is said to be a ϕ-contraction if d(Tx, Ty) ≤ ϕ d(x, y) , ∀ x, y ∈ X, x = y, where ϕ : (0, ∞) → (0, ∞) is upper semi-continuous from the right mapping and satisfies ϕ(t) < t for every t > 0. They proved that if the metric space is complete, then T is a Picard operator. Later, Wardowski [4] introduced a new type of contractive self-map T on a metric space (X, d), the so-called F-contraction. This is defined by the inequality τ + F(d(Tx, Ty)) ≤ F(d(x, y)), ∀ x, y ∈ X, Tx = Ty, where τ > 0 and F : (0, ∞) → R satisfies the conditions (F 1 )-(F 3 ) defined as follows: (F 1 ) F is strictly increasing, that is, for all α, β ∈ (0, ∞) such that α < β, F(α) < F(β); (F 2 ) for each sequence (α n ) n of positive numbers lim n→∞ α n = 0 if, and only if lim n→∞ F(α n ) = −∞; Wardowski proved that, whenever (X, d) is complete, every F-contraction is a Picard operator. The above result has been extended to new classes of Picard mappings by weakening the conditions (F 1 )-(F 3 ) or by defining new contractive conditions by many authors (see, for example, [5][6][7][8][9][10][11][12][13]).
Very recently, Proinov [6] considered a self-mapping T on a complete metric space satisfying a general contractive-type condition of the form and proved some fixed-point theorems which extend many earlier results in the literature (see some of them in [6]). In this paper, we generalize the fixed-point result given by Proinov ([6], Th. 3.6) by considering general contractive conditions defined by inequality G(d(Tx, Ty)) ≤ H(d(x, y)), for each x, y with Tx = Ty, where G, H : (0, ∞) → R satisfy conditions (C 1 ), (C 2 ) and (C 3 ) defined below.

Results
Let us consider two mappings G, H : (0, ∞) → R satisfying the following conditions: (C 1 ) the set of continuity points of G is dense in (0, ∞); (C 2 ) for every r ≥ t > 0, one has G(r) > H(t); We will denote by G the family of all pairs of functions (G, H) which satisfy conditions The following result is easy to be proved.

Remark 1.
Under hypothesis (C 2 ), condition (C 3 ) is equivalent to Proof. "(C 3 ) ⇒ (C 3 )" Let (t n ) n be a sequence of positive numbers such that t n t > 0.
hence, the series of positive terms ∑ n≥1 (G(t n ) − H(t n )) diverges.
We proceed by contradiction. Let us suppose that there exists t > 0 such that lim inf s t G(s) − H(s) = 0. Then, one can find a sequence (t k ) k , t k > t, such that There is no loss of generality in assuming that (t k ) k is decreasing. By the above, there exists k 1 ∈ N such that Next, one can find k 2 > k 1 such that Inductively, we obtain a subsequence (t k n ) n of (t k ) k such that and so the series ∑ n G(t k n ) − H(t k n ) converges. Moreover, t k n t. This contradicts (C 3 ). Example 1. Let us consider α, β ∈ (0, 1), a > 0, τ > 0 such that αa < a − τ and let us define G, H : R + → R by G(r) = βr, Then, (G, H) ∈ G.
Proof. (C 1 ) Obvious. (C 2 ) Choose r ≥ t > 0. Then Lemma 1. Let F : (0, ∞) → R be a map and (t k ) k a sequence of positive real numbers such that If one of the following conditions holds: Note that the previous lemma gives some classes of functions satisfying property (P). At the same time, there exist functions having property (P), but which do not satisfy any of the conditions of Lemma 1 as it follows from the following example. Example 3. Let (r n ) n be a decreasing sequence of positive numbers converging to 0 and f , g : (0, ∞) → R be two mappings such that lim n→∞ f (r n ) = −∞ and g is bounded from below. Then, the mapping F : (0, ∞) → R given by , t / ∈ {r n , n = 1, 2, . . . } satisfies property (P). If, further, the set of discontinuity points of g is at most countable (in particular if g is monotone on each interval (r n+1 , r n )), then the set of discontinuity points of F is also at most countable.
For the last assertion, if we denote by ∆ F , ∆ g the sets of discontinuities of F and g, respectively, then Note that one can find easily numerous functions satisfying the conditions of Example 3 such as: r n = 1/n, f (r) = −1/r, g(t) = t and so on. Proposition 1. [12,15] Let (x n ) n be a sequence of elements from a metric space (X, d) and ∆ be a subset of (0, ∞) such that (0, ∞) \ ∆ is dense in (0, ∞). If d(x n , x n+1 ) −→ n 0 and (x n ) n is not a Cauchy sequence, then there exist η ∈ (0, ∞) \ ∆ and the sequences of natural numbers (m k ) k , (n k ) k such that Our main result is the following: Theorem 2. Let (X, d) be a complete metric space and G, H be two mappings such that (G, H) ∈ G and one of them satisfies property (P). Let also consider the map T : X → X satisfying the following condition Then T is a Picard operator.
Proof. First of all we remark that, from conditions (C 2 ) and (4), we deduce that T satisfies which implies that T has at most one fixed point.
In order to show the existence of fixed point of T, let x 0 ∈ X be fixed. We define a sequence (x n ) n by x n = Tx n−1 , n ≥ 1, and let us denote d n = d(x n+1 , x n ), n ≥ 0. If there exists n 0 ∈ N such that x n 0 +1 = x n 0 , then x n 0 is a fixed point of T. We next suppose that x n+1 = x n for each n ∈ N. Then, d n > 0 for all n ∈ N and, by (5), the sequence (d n ) n is decreasing. Thus, one can find d ≥ 0 such that d n d. Next, we will prove that d = 0. Indeed, using (4), we get G(d n ) ≤ H(d n−1 ) for all n ≥ 1. From the above, we obtain according to condition (C 3 ) from Remark 1. It follows that lim n→∞ G(d n ) = −∞. At the same time, since d n < d n−1 , we deduce from (C 2 ) that H(d n ) < G(d n−1 ) for all n = 1, 2, . . . , hence, lim n→∞ H(d n ) = −∞. We conclude by hypothesis that lim n→∞ d n = 0. Now, assume that the sequence (x n ) n is not Cauchy, and let ∆ be the set of discontinuities of G. Since (G, H) satisfies (C 1 ), it follows that (0, ∞) \ ∆ is dense in (0, ∞).
Since G is continuous at η, from the last inequality we obtain letting k → ∞ which contradicts (C 3 ). Consequently, (x n ) n is a Cauchy sequence and, X being complete, there exists x * ∈ X such that x n → x * as n → ∞. Finally, condition (5) yields Thus, Tx * = x * .

Example 4.
Let us consider G, H : (0, ∞) → R defined by Then, G is not monotone, satisfies property (P) and (G, H) ∈ G. Furthermore, if X = [0, ∞) is endowed with the standard metric d(x, y) = |x − y| and T : X → X, Tx = x x+1 , satisfies (4), then T is a Picard operator while it does not satisfy the Banach condition.
(C 2 ) Let us consider r ≥ t > 0. Three cases can occur: II. r ≥ t > 1 2 . From the following relations III. r > 1 2 ≥ t. One has r − 1 12 Clearly, G satisfies (P), and it is not monotone. In order to prove the second part of the statement, set x, y ∈ X such that Tx = Ty. Then, x = y, say x < y. The following cases can occur: Then

II.
y−x (1+x)(1+y) > 1 2 . Then G d(Tx, Ty) = ln Therefore, the inequality (4) is fulfilled. Theorem 2 shows that T is a Picard operator (its unique fixed point being x = 0). For the last sentence, let us consider two sequences x n = 1 n , y n = 2 n . Then lim n→∞ d Tx n , Ty n d(x n , y n ) = lim hence, T does not satisfy the Banach condition.
where G, H : (0, ∞) → R are two mappings satisfying the following conditions: Consequently, (C 3 ) is verified. The conclusion now follows from Theorem 2.
Then, the function G • ϕ is upper semi-continuous from the right at a.

Proof. One has
We need to show that lim sup Case II. lim sup t a ϕ(t) = ϕ(a). Then, using (i), (ii),

Remark 2.
If we take in Corollary 1 a nondecreasing and right-continuous function G : (0, ∞) → R and H = G • ϕ, where ϕ is a comparison function, then every self-mapping T on a complete metric space satisfying (1) is a Picard operator.
Proof. By the monotonicity of G, it is obvious that (8) is equivalent to (1). From Lemma 2, it follows that H is upper semi-continuous from the right function. The rest of the conditions from Corollary 1 are clearly verified.
In the next two corollaries, we will highlight that the results given by Secelean and Wardowski [8], Secelean [9], Wardowski [4], and Piri and Kumam [5] can be obtained as particular cases of Theorem 2.

Corollary 2.
Let us consider (X, d) a complete metric space, and T : X → X. We suppose that there exists a nondecreasing function F : (0, ν) → R, ν > diam(X) := sup Then, T is a Picard operator.
(C 1 ) This condition is clearly verified due to the monotonicity of F. (C 2 ) Fix r ≥ t > 0. Then, by the property of ψ, one has (C 3 ) Let us consider a sequence of positive real numbers (t n ) n such that t n t > 0. Then, the sequence F(t n ) n is non-increasing and F(t) ≤ F(t n ) for every n = 1, 2, . . . hence, one can find λ ∈ [F(t), µ) such that F(t n ) λ. Since ψ is right-continuous at λ, one has lim . From Remark 1, we deduce that (C 3 ) is also satisfied.
By Lemma 1, we deduce that G satisfies property (P). Now, the conclusion follows from Theorem 2.
If, in the previous corollary, we take ψ(t) = t − τ for some τ > 0, one obtains an improvement of the results from [4,9], where F satisfies only condition (F 1 ). Corollary 3. Let us consider (X, d) a complete metric space and T : X → X. We suppose that there exist nondecreasing F : (0, ∞) → R and τ > 0, such that Then, T is a Picard operator.
Inspired by [16], we can formulate an improvement of Wardowski's result given in the previous corollary.

Corollary 4.
Let us consider a complete metric space (X, d) and two functions F : (0, ∞) → R, θ : (0, ∞) → (0, ∞). Assume that F has property (P) and the set of its continuity points is dense in (0, ∞). Suppose further that ∑ n θ(t n ) = ∞ for each decreasing sequence of real numbers (t n ) n with a positive limit. If T : X → X is such that then T is a Picard operator.
Since conditions (C 2 ), (C 3 ) are also obviously verified, one can apply Theorem 2. (γ) ϕ d(x, y) + F d(Tx, Ty) ≤ F d(x, y) for all x, y ∈ X with Tx = Ty. Then, T is a Picard operator.
In the following, we will show that one of the main theorems of Proinov can be obtained as a consequence of our results.
Then, T is a Picard operator.
We first note that (i) implies that G satisfies property (P). Next, from (i) and (ii) we deduce that, for some r ≥ t > 0, we have hence, (G, H) satisfies (C 2 ).
Additionally, due to the monotonicity of ψ, we deduce that there exists lim Thus, (G, H) satisfies (C 3 ). The conclusion follows from Theorem 2.
Notice that Corollary 1 can be obtained as a particular case of the previous corollary.

Remark 4.
Example 4 proves that the result of Proinov pointed out in Corollary 7 can be obtained from Theorem 2 without imposing the monotonicity of the function ψ.

Application
Next, we use our main results in order to give an existence and uniqueness result for the solution of a certain integral equation.

Proposition 2.
Let us consider G, H defined in Example 1, and the integral equation under the following conditions: Then, the Equation (9)  The conclusion now follows from Theorem 2 applied to operator T.

Conclusions
We introduce a new type of contractive function on a metric space that generalizes and extends some of the contractions studied in the literature, and we provide a fixed-point theorem that improves many of the known results. Some examples and an application to integral equations are also given. The result we proved can be extended to more general metric spaces.