On ( ψ , φ ) -Rational Contractions

: In this paper, we examine the notion of ( ψ , φ )-contractions by involving rational forms in the context of complete metric spaces. We note that some well-known ﬁxed point theorems for rational forms can be deduced from our main results. We also consider some examples to indicate the validity of the presented results.


Introduction and Preliminaries
Thousands of results have been published since Banach [1] proved the first fixed point theorem. Some of these results are equivalent to the results published previously, while others were understood to be a sub-result of the previous results. Therefore, recently, publications that collect and consolidate the results in the literature have started to appear.
Very recently, Proinov (2020) [2], to extend and unify many earlier results, proved that the fixed point theorem of Skof (1977) [3], in the setting of metric spaces, covers several existing results, including the attractive results of Wardowski (2012) [4] and Jleli-Samet (2014) [5]. He also proved that the analog of this observation holds true in the context of dislocated metric spaces.
On the other hand, starting from Das-Gupta (1975) [6] and Jaggi (1977) [7], rational expressions were used to prove fixed point theorems. Later, these approaches were modified for Boyd and Wong contractions, φ-contractions, Geraghty contractions, Wardowski contractions, etc. We observe that the concerns of Proinov [2] are valid for fixed point theorems involving rational expression; that is, some published results are equivalent to earlier results or consequences.
In this paper, we prove that the analog of the fixed theorem of Skof [3] with rational expression unifies and extends several fixed point theorems in the literature.
Then, T admits a unique fixed point.
We also recall the main results in which some rational expressions were studied in a contraction condition.
Theorem 2 ([6]). Let (X, d) be a complete metric space and T : X → X be a mapping such that there exist k 1 , k 2 ∈ [0, 1), with k 1 + k 2 < 1 such that: d(T x, T y) ≤ k 1 · d(y, Ty) 1 + d(x, Tx) 1 + d(x, y) + k 2 · d(x, y) (1) for all x, y ∈ X. Then, T has a unique fixed point υ ∈ X, and the sequence {T n x} converges to the fixed point υ for all x ∈ X.
Theorem 3 ([7]). Let (X, d) be a complete metric space and T : X → X be a continuous mapping. If there exist k 1 , k 2 ∈ [0, 1), with k 1 + k 2 < 1 such that: for all distinct x, y ∈ X, then T possesses a unique fixed point in X.
Finally, let us consider the next lemma (which can be found in many papers; see, e.g., [2]), which will be useful in the sequel.

Lemma 1 ([2]
). Let {x n } be a sequence in a metric space (X, d) such that d(x n , x n+1 ) → 0 as n → ∞. If the sequence {x n } is not Cauchy, then there exist e > 0 and the subsequences {q k } and {r k } of positive integers such that: lim

Main Results
Throughout this section, we will consider that φ, ψ : (0, ∞) → R are two functions such that: Definition 1. Let (X, d) be a complete metric space. A mapping T : X → X is a (ψ, φ)-rational contraction of Type 1 if for every distinct x, y ∈ X such that d(T x, T y) > 0, the following inequality: holds, where M 1 is defined by: Theorem 4. Let (X, d) be a complete metric space and T : X → X be a continuous (ψ, φ)-rational contraction of Type 1. Assume that: Then, T admits exactly one fixed point.

Proof.
Starting with a point x ∈ X, we define the sequence {x n } by: with x n = x n+1 for all n ∈ N (indeed, on the contrary, if there exists j n ∈ N such that x j n = x j n +1 = Tx j n , we get that x j n is a fixed point of T ). Under this consideration, for x = x n−1 and y = x n , we have: and by (4) (since d(Tx n−1 , Tx n ) = d(x n , x n+1 ) > 0), we get: which is equivalent, denoting by ς n = d(x n−1 , x n ), to: (Of course, we can assume that ς n > 0, since on the contrary, we can find l ∈ N such that d(x l−1 , x l ) = ς l = 0. Thus, x l = x l+1 = Tx l and x l is the fixed point of T .) If there exists n ∈ N such that max {ς n , ς n+1 } = ς n+1 , then ψ(ς n+1 ) ≤ φ(ς n+1 ), which contradicts the assumption (f 0 ). Therefore, for all n > 0, we have ς n > ς n+1 , so that the sequence {ς n } is decreasing, and since it is strictly positive, there exists ς ≥ 0 such that lim n→∞ ς n = ς and ς n > ς for all n > 0. Supposing that ς > 0, because M 1 (x n−1 , x n ) = ς n , replacing in (4) and taking into account (f 0 ), we have, It follows that the sequence {ψ(ς n )} is strictly decreasing, and since it is bounded (below) (because ς n > ς and due to the assumption (f 1 )), we can conclude that {ψ(ς n )} is a convergent sequence. Moreover, from the above inequality, the sequence {φ(ς n )} is also convergent as the same limit. Thus, keeping in mind (f 2 ), which is a contradiction. Therefore, ς = 0 and: We claim that {x n } is a Cauchy sequence. Let us suppose by contradiction that the sequence {x n } defined by (6) is not Cauchy. Then, by Lemma 1, there exist e > 0 and two sequences of positive real numbers (q k ) and (r k ) such that: Furthermore, for all k ≥ 1, we have d(x q k +1 , x r k +1 ) > e. Replacing x = x q k +1 and y = x r k +1 in (4) and taking into account (f 0 ), we have: where: Now, by (8) and (9), we have lim k→∞ M 1 (x q k , x r k ) = e, and it follows by (4) that: This contradicts (f 2 ), and then, {x n } is a Cauchy sequence on a complete metric space. Thus, the sequence converges to a point υ ∈ X, that is: and since the mapping T is continuous, we have: Therefore, from the above inequality together with (f 0 ), we get: which is a contradiction. This closes the proof.

Example 1.
Let the set X = [0, 2] and d : X → X be the distance defined as d(x, y) = |x − y| for every x, y ∈ X. Let also T : X → X be a self-mapping with T x = −x 2 +2x+4 8 and two functions ψ, φ : (0, ∞) → R, Since the assumptions (f 1 )-(f 3 ) are satisfied, it remains to check that T is a (ψ, φ)-rational contraction of Type 1. We have: which shows us that T is a (ψ, φ)-rational contraction of Type 1. Furthermore, by Theorem 4, we get that T has a unique fixed point in X, that is x = 0.605551.
Next, we show that the continuity condition of the operator T can be replaced by the assumption of the continuity of only some iterations of T .
then T has a unique fixed point.
Proof. Let {x n } be the sequence defined by (6). By the proof of Theorem 4, we know that this sequence is convergent to some point υ ∈ X, which means that d(x n , υ) = 0. Let c n(j) be a subsequence of {x n }, where n(j) = j · m for all j ∈ N 0 and m > 1 fixed. Moreover, assuming that T 0 is the identity map on x, we have x n(j) = T m x n(j)−m . Then, since T m is continuous, This means that ς is a fixed point of T m . If we assume that υ = T υ, we have for any j = 0, 1, ..., m − 1 that T m−j−1 υ = T m−j υ. By replacing x by T am−j−1 υ and y by T m−j υ, we have: and (4) becomes, Taking into account (f 0 ), it follows that: Now, since the function ψ is nondecreasing, we get: This leads us to: for every k = j, j + 1, ..., m − 1. Taking in the above inequality j = 0 and k = m − 1, we get: This is a contradiction. Consequently, T υ = υ.

Example 2.
Let the set X = [0, 2] be endowed with the usual distance d(x, y) = |x − y| for every x, y ∈ X.
Let the mapping T : It is clear that the mapping T is not continuous and that Theorem 4 cannot be applied. However, we have that T 2 x = 0 for any x ∈ x, so the assumption (f 3 ) holds. Choosing, for example, the functions ψ, φ : (0, ∞) → R, where ψ(s) = e s and φ(s) = s + 3 4 , we have that the assumptions (f 0 ) − (f 2 ) are also satisfied, and we need to check if the inequality (4) holds for all distinct x, y ∈ X with d(T x, T y) > 0.
Of course, since φ(s) = s + 1 is an increasing function, for x ∈ [0, 1] and y ∈ (1, 2], we have: so that all the assumptions of Theorem 5 are satisfied.

Definition 2.
Let (X, d) be a complete metric space. The mapping T : X → X is said to be a (ψ, φ)-rational contraction of Type 2 if for all x, y ∈ X with d(T x, T y) > 0, the following condition is satisfied: where M 2 is defined by: Theorem 6. Let (X, d) be a complete metric space and T : X → X be a (ψ, φ)-rational contraction of Type 2.
Assume that: (f 1 ) ψ is non-decreasing and lower semi-continuous; (f 4 ) lim sup Then, T admits exactly one fixed point.
Proof. Let {x n } be the sequence defined by (6). Thus, by similar reasoning, we have that ς n = d(x n−1 , x n ) > 0 for every n ∈ N. Therefore, since d(Tx n−1 , Tx n ) > 0, for every n ∈ N, for x = x n−1 and y = x n , we have: Consequently, by (13), we have: which, keeping in mind (f 0 ), is equivalent to: Thus, due to the monotony of the function ψ, ς n+1 < max {ς n , ς n+1 }, so that 0 < ς n+1 < ς n , for each n ∈ N, then there exists ς ≥ 0 such that ς n ς. We claim that ς = 0. If we assume by contradiction that ς > 0, we have: Taking the superior limit in the above inequality and keeping in mind (f 4 ), we get: which is a contradiction. Thus, we have: Now, we claim that {x n } is a Cauchy sequence. Again, arguing by contradiction, by Lemma (1), we have that there exist e > 0 and the sequences of positive real numbers (q k ) and (r k ) such that: Thus, d(x q k +1 , x r k +1 ) = d(Tx q k , Tx r k ) > e > 0 for all k ≥ 1, and from (13), together with (f 0 ) we have: Since ψ is non-decreasing we get d(x q k +1 , x r k +1 ) < M 2 (x q k , x r k ), for each k ≥ 1, where: and taking into account (16) and (17): In this case, letting k → ∞ in (18), we have: which is a contradiction. This shows that {x n } is a Cauchy sequence. By the completeness of the space (x, d), the sequence {x n } converges to a point υ in x, that is: We claim that υ is a fixed point of T . Supposing by contradiction that d(T υ, υ) > 0 and using the same arguments as in the previous theorem, we have that there exists n 0 ∈ N such that d(T υ, x n+1 ) = d(T υ, Tx n ) > 0 for any n ≥ n 0 . Now, by (13) we have: where: On the one hand, from (16) and (20), we get: and then: υ, T υ)).
On the other hand, lim n→∞ d(T υ, Tx n ) = lim n→∞ d(T υ, x n+1 ) = d(T υ, υ). Therefore, taking the inferior limit in (21) when n → ∞ and taking into account the lower semi-continuity of ψ, we have: which is a contradiction. Therefore, we have T υ = υ, and we claim that this is the unique fixed point of T . If we suppose thatυ is also a fixed point of T such that d(T υ, Tυ) = d(υ,υ) > 0 and from (13), we have: with: Thus, by (23), which is a contradiction.
Thus, all the assumptions of Theorem 6 hold, so that T has a unique fixed point.
Proof. Following the lines and using the same notations as in the proof of Theorem 6, by (15), we have that ψ(ς n ) < ψ(max {ς n , ς n+1 }). Since for max {ς n , ς n+1 } = ς n+1 , we get a contradiction. Therefore, we conclude that ς n > ς n+1 . Consequently, on the one hand, we have that there exists a point ς ≥ 0 such that ς n ς. We claim that ς = 0. On the contrary, if we suppose that ς > 0, by (13) together with f 0 , we have: for all n ∈ N. Then, the sequence {ψ(ς n )} is decreasing and also bounded (because (f 1 ) and ς n > ς). Therefore, the sequence {ψ(ς n )} is convergent, and moreover, by the above inequality, the sequence {φ(ς n )} is also convergent to the same limit. Thus, keeping in mind (f 2 ), we have: which is a contradiction, so that, We will show that {x n } is a Cauchy sequence. In order to prove that, arguing by contradiction, by Lemma 1, there exist e > 0 and (q k ), (r k ) two sequences of positive integers such that (3) holds. Since lim k→∞ d(x q k +1 , x r k +1 ) = e+, we have that d(x q k +1 , x r k +1 ) = d(Tx q k , Tx r k ) > 0, and replacing in (13), we get: On the other hand, from the above inequality and (f 2 ), we have: This is a contradiction, so that {x n } is a Cauchy sequence, so it is convergent to some point υ ∈ X (due to the completeness of the metric space (X, d)). If we suppose that d(T υ, υ) > 0, because d(T υ, Tx n ) → d(T υ, υ), we have that there exists n 0 ∈ N such that d(T υ, Tx n ) > 0, for n ≥ n 0 . Then, from (13), 1+d(υ,x n ) and moreover, taking into account (22): υ, T υ)).
Taking the limit as inferior and using (f 5 ), we obtain: This is a contradiction. Therefore, T υ = υ, that is υ is a fixed point of T , and using the same arguments as in Theorem 6, we have that, in fact, this fixed point is unique. Example 4. Let X = [0, ∞) and d be the usual distance on X. Let T : X → X, where T x = 1 2 ln(x 2 + x + 2) and ψ, φ : (0, ∞) → R, where ψ(s) = e s and φ(s) = 1 + s. We check that T is a (ψ, φ)-rational contraction of Type 2. Indeed, if x > y (and it is analogues for the case x < y), then: On the other hand, since: we obtain: Thus, (13) is satisfied, and by Theorem 7, we have that the mapping T has a fixed point.

Definition 3.
Let (X, d) be a complete metric space. The mapping T : X → X is said to be a (ψ, φ)-rational contraction of Type 3 if for all x, y ∈ X, when max {d(x, T y), d(y, T x)} = 0, then d(T x, T y) > 0, and the following condition is satisfied: Theorem 8. Let (X, d) be a complete metric space and T : X → X be a (ψ, φ)-rational contraction of Type 3. The mapping T admits exactly one fixed point provided that: (f 1 ) ψ is non-decreasing and lim sup s→s 0 + φ(s) < ψ(s 0 +), for any s 0 > 0.
Proof. Let {x n } be the sequence defined by (6). Thus, by similar reasoning, we have that x n = d(x n−1 , x n ) > 0 for every n ∈ N. Therefore, since d(T x n−1 , T x n ) > 0, for every n ∈ N, for x = x n−1 and y = x n , by (25), we have: which, keeping in mind (f 0 ), gives us: Thus, from (f 1 ), 0 < d(x n , x n+1 ) < d(x n−1 , x n ) for each n ∈ N, so the sequence (d(x n , x n+1 )) is convergent to some ς ≥ 0. We claim that ς = 0. In the case that ς > 0, from (25), Taking the limit as superior in the above inequality and keeping in mind (f 1 ), we get: This is a contradiction, and then, we have: Now, we claim that {x n } is a Cauchy sequence. Again, arguing by contradiction, by Lemma (1), we have that there exist e > 0 and the sequences of positive real numbers (q k ) and (r k ) such that: Thus, it follows that d(x q k +1 , x r k +1 ) = d(Tx q k , Tx r k ) > e > 0 for all k ≥ 1, and from (25), together with (f 0 ), we have: Since ψ is non-decreasing, we get: for each k ≥ 1.
Taking into account (27) and (28): In this case, we get e = 0, which shows us that {x n } is a Cauchy sequence, and by the completeness of the space (X, d), (x n ) converges to a point υ in x, that is: We claim that υ is a fixed point of T . Supposing by contradiction that d(T υ, υ) > 0 and using the same arguments as in the previous theorem, we have that there exists n 0 ∈ N such that d(T υ, x n+1 ) = d(T υ, Tx n ) > 0 for any n ≥ n 0 . Now, by (25), we have: Now, from (f 1 ), we have: and letting n → ∞, we get 0 < lim n→∞ d(T υ, Tx n ) < 0, which is a contradiction. Therefore, we have T υ = υ. Finally, we claim that this is the unique fixed point of T . If we suppose thatυ is also a fixed point of T such that d(T υ, Tυ) = d(υ,υ) > 0 and from (25) Thus, by (f 1 ), which is a contradiction.
We can state many corollaries from our main results. For example, choosing ψ(s) = s and φ(s) = β(s)s in Theorem 4, we have: Corollary 1. Let (X, d) be a complete metric space and β : (0, ∞) → (0, 1) be a function such that lim sup s→s 0 + β(s) < 1 for every s 0 > 0. A continuous mapping T : X → X has a unique fixed point provided that: d(T x, T y) ≤ β(M 1 (x, y))M 1 (x, y), for all x, y ∈ X with d(T x, T y) > 0.
If in Theorem 7, we take φ(s) = κψ(s), we get the following corollary.

Conclusions
In this paper, we were interested in finding some conditions on the functions ψ and φ that guarantee that T has a unique fixed point in terms of rational expression. Our main results offered improvements to known results by applying weaker conditions on the self-map of a complete metric space. Here we mentioned just one corollary for each type of (ψ, φ)-rational contraction by choosing different functions ψ and φ, but it is clear that many similar consequences can be listed, consequences that actually represent independent results.
Funding: This research received no external funding.