Extended Simulation Function via Rational Expressions

In this paper, we introduce some common fixed point theorems for two distinct self-mappings in the setting of metric spaces by using the notion of a simulation function introduced in 2015. The contractivity conditions have not to be verified for all pairs of points of the space because it is endowed with an antecedent conditions. They are also of rational type because the involved terms in the contractivity upper bound are expressed, in some cases, as quotients.


Introduction and Preliminaries
The notion of a simulation function, introduced by Khojasteh, Shukla and Radenović [1] in 2015, has become into one of the most studied concepts in the field of Fixed Point Theory in recent times. To support this affirmation, we refer the reader to the following recent papers: [2][3][4][5] The great success of simulation functions has inspired many other developments (see, e.g., [6][7][8][9][10][11]), even in the fuzzy setting (see, e.g., [12]). In this manuscript, we use the notion of simulation function in order to prove two common fixed point theorems involving two distinct self-mappings. The importance of these results can be evaluated through the great number of possible consequences that it can easily be derived from the main theorems.
Throughout the manuscript, let (X, d) be a metric space. We denote N 0 := N ∪ {0} where N is the set of all positive integers. Further, R represents the real numbers and R + 0 := [0, +∞). Background on Fixed Point theory can be found on [13]. The set of all simulation functions is denoted by Z.
Theorem 1. Let (X, d) be a complete metric space and let T : X → X be a Z-contraction with respect to a certain simulation function η, that is η(d(Tx, Ty), d(x, y)) ≥ 0 for all x, y ∈ X. (1) Then T has a unique fixed point.
We state the following lemma which will be useful for demonstrating our main result.
Proposition 1. ( [7,15]) Let (X, d) be a metric space, let T : X → X be a self-mapping and let {x n } ⊆ X be a Picard sequence of T (that is, x n+1 = Tx n for all n ∈ N). If lim n→∞ d(x n , x n+1 ) = 0, then either {x n } is almost constant or x n = x m for all n, m ∈ N such that n = m.

Remark 1.
Every simulation function is a Ψ-simulation function where ψ is taken as the identity function on [0, +∞).
So all simulation functions presented in Example 1 are Ψ-simulation functions. However, the following example shows that there are Ψ-simulation function which are not simulation functions.
Denote by Z Ψ the set of all Ψ-simulation functions. We have just proved that: The first of our main results is the following one.
Theorem 2. Let (X, d) be a complete metric space and let T, S : X → X be two mappings such that, for all x, y ∈ X, 1 2 min{d(x, Tx), d(y, Sy)} ≤ d(x, y) implies η(d(Tx, Sy), m(x, y)) ≥ 0, where η ∈ Z Ψ and, for all x, y ∈ X, If T and S are continuous, then T and S have a unique common fixed point (that is, there is a unique u ∈ X such that Tu = Su = u). In fact, such a point is the unique fixed point of T and the unique fixed point of S.
Proof. For the sake of clarity, we divide the proof into five steps. The first one is necessary in order to guarantee that the consecutive terms of the sequence we will construct can be supposed as distinct.
Step (1): We claim that any fixed point of T or S is a common fixed point of S and T.
Suppose that x ∈ X is a fixed point of T, that is, Tx = x (the same argument is valid if x is a fixed point of S). Reasoning by contradiction, assume that x is not a common fixed of T and S. This means that Sx = x. Therefore, using y = x in (2), we deduce that η(d(Tx, Sx), m(x, x)) ≥ 0. In particular, taking into account that d(x, Sx) > 0 and condition (η 1 ), we deduce that: which is a contradiction. Therefore, we have proved that any fixed point of T (or S) is a common fixed point of S and T, so step 1 is completed. Next, in order to conclude T or S has a fixed point, we define the following sequence. Starting from an arbitrary point x 0 ∈ X, we shall built up a recursive sequence {x n } defined as: x 2n+1 = Tx 2n and x 2n+2 = Sx 2n+1 for each n ∈ N.
If there is some n 0 ∈ N such that x n 0 +1 = x n 0 , then x n 0 is a fixed point of T or S (either Tx n 0 = x n 0 or Sx n 0 = x n 0 ). In this case, step 1 shows that x n 0 is a fixed point of T and S (and it only remains to prove that this point is unique as in step 5). On the contrary case, assume, without loss of generality, that We shall consider the following set that will be useful in later stages of the proof.
Step (2): We claim that lim To prove it, at first we claim that To see this, suppose that k = 2n for some n ∈ N. We have: So from (2) and (η 1 ) we have: Since ψ is strictly increasing, On the other hand, If there is some n 0 ∈ N such that d(x 2n 0 +1 , x 2n 0 +2 ) ≥ d(x 2n 0 , x 2n 0 +1 ), then m(x 2n 0 , x 2n 0 +1 ) = d(x 2n 0 +1 , x 2n 0 +2 ), which contradicts (7). Therefore, for each n ∈ N, Consequently, (6) is proved when k ≥ 0 is an even number. By the same argument, one can check that (6) holds when k is an odd number. Thus, the sequence {d(x n , x n+1 )} n≥1 is non-increasing and bounded from below, so it is convergent. Hence there is a real number γ ≥ 0 such that We claim that γ = 0. To prove the claim, at first suppose that By (2) and definition of A, we observe that For each n ≥ 0 we have Thus (x 2n , x 2n+1 ) ∈ A for each n ≥ 0. Consequently, (9) implies that In order to prove that γ = 0, suppose, by contradiction, that γ > 0. From (8) we have Therefore, from (η 2 ), which contradicts (10). So the claim is proved, that is, Step (3): We claim that {x n } is a Cauchy sequence. Since {d (x n , x n+1 )} → 0, Proposition 1 guarantees that x n = x m for all n, m ∈ N such that n = m (it cannot be almost constant because x n = x n+1 for all n ∈ N). Reasoning by contradiction, suppose that {x 2n } is not a Cauchy sequence. Taking into account Lemma 1, there exist ε 0 > 0 and subsequences {x 2m k } and {x 2n k } of {x n } such that n k is the smallest index for which n k > m k > k and d(x 2m k , x 2n k ) ≥ ε 0 and Therefore, from the definition of m(x, y) we have On the other hand, we claim that for sufficiently large k ∈ N, if n k > m k > k, then Indeed, since n k > m k and {d(x n , x n+1 )} is non-increasing, we have So, the left hand side of inequality (13) is equal to Therefore, we must show that, for sufficiently large k ∈ N, if n k > m k > k, then According to (11), there exists k 1 ∈ N such that for any k > k 1 , There also exists k 2 ∈ N such that for any k > k 2 , Hence, for any k > max{k 1 , k 2 } and n k > m k > k, we have So, one concludes that Thus we obtain that for any k > max{k 1 , k 2 } and n k > m k > k, Then (13) is proved. Therefore, by (2) and definition of A, for sufficiently large k ∈ N, if n k > m k > k, then (x 2n k , x 2m k −1 ) ∈ A . Consequently, for sufficiently large k ∈ N, if n k > m k > k then which contradicts (12). This contradiction proves that {x n } is a Cauchy sequence and, since X is complete, there exists u ∈ X such that {x n } → u as n → ∞.
Step (4): We claim that u is a common fixed point of T and S. Since T and S are continuous, we deduce that Therefore Tu = Su = u, that is, u is a common fixed point of T and S.
Step (5): We claim T and S have a unique common fixed point, and it is the unique fixed point of T and the unique fixed point of S.
Suppose that T and S have two distinct common fixed points u, v ∈ X. Therefore Tu = Su = u, Tv = Sv = v and d (u, v) > 0. Therefore Taking into account that the contractivity condition (2) guarantees that η(d(Tu, Sv), m(u, v)) ≥ 0. Therefore, which is a contradiction. As a consequence, T and S have, at most, a unique common fixed point. Furthermore, if x is a fixed point of T or S, step 1 guarantees that x is a common fixed point of T and S, which is unique, so T and S can only have, at most, a unique fixed point. for all x ∈ X.
Clearly (X, d) is complete and T and S are continuous self-mappings on X. Let us show that T and S satisfy the hypotheses of Theorem 2 associated to the function for all t, s ∈ [0, +∞) .
If we take α = 0.9 and ψ (t) = t/2 for all t ∈ [0, +∞), then ψ ∈ Ψ and Item (e 1 ) in Example 2 guarantees that η ∈ Z Ψ . We have to prove that the contractivity condition holds for all x, y ∈ X. Therefore, let x, y ∈ X be two arbitrary points such that and we have to prove that The first case occurs when x = y. In this case, (14) shows that Tx = x or Sx = x. Clearly Tx = x or which means that (15) holds when x = y under the assumption (14). Now assume that x = y. Notice that Therefore, inequality (15) is equivalent to: we will consider the cases x < y and y < x, and the subcases depending on the number that reaches the minimum in (17).
In this subcase, Hence, (14) is equivalent to We show that |3x − 2y| 12 as follows. On the one hand, which holds because x ≤ 8 11 y by (18). On the other hand, which is true because x < y. Joining (19) and (20), we deduce that so the contractivity condition (16) holds.
This subcase is impossible because which contradicts the fact that x < y.
In this subcase, This fact contradicts (14) because: which is impossible since Therefore, this subcase cannot hold.
In this case, Therefore, Since y < x, then which holds because of (21). Hence m(x, y), and the contractivity condition (16) holds.
In any case, we have proved that η(d(Tx, Sy), m(x, y)) ≥ 0 for all x, y ∈ X such that Finally, notice that η is not a simulation function because if t 0 = 2 and s 0 = 1, then As a consequence, Theorem 2 is applicable in order to guarantee that T and S have a common fixed point, which is unique and, in fact, it is the unique fixed point of T and the unique fixed point of S. However, other results using simulation functions are not applicable to this context.
If we pay attention to all details in Example 4, then we observe that, in fact, for all distinct points x, y ∈ X. Hence, the reader can imagine that the term does not play a role in m (x, y) and, indeed, in the proof. Next, we show an example in which this term is a key piece in order to guarantee that T and S have a common fixed point.
Example 5. Let X = [−1, 1] endowed with the usual metric d (x, y) = |x − y| for all x, y ∈ X, and let T, S : X → X given by Tx = 3 10 x and Sx = − 2 5 x for all x ∈ X.
Clearly (X, d) is complete and T and S are continuous self-mappings on X. We are going to show that T and S satisfy all the hypotheses of Theorem 2 associated to Notice that in Example 4 we have proved that η ∈ Z Ψ but η is not a simulation function. Before proving that T and S satisfy the contractivity condition, we want to highlight that condition is false even when x and y are positive and they satisfy the antecedent condition in the contractivity condition. To do that, take x 0 = 0.53 and y 0 = 0.73. Then d (x 0 , y 0 ) = 0.2 and This will prove that the second term in the maximum in , when x = y, will be of great importance. Let x, y ∈ X be such that If x = y, then the previous condition guarantees that Tx = x or Sx = x. This is only possible when x = 0 and, in this case, Then Furthermore, the contractivity condition can be equivalently expressed as: As a consequence, we have to prove that, for all distinct x, y ∈ X, We consider some cases and subcases.
In this case, In any case, condition min 7 |x| 20 , 7 |y| 10 ≤ |x − y| holds. Since In this case, In any case, condition Then we know that: In particular, multiplying by x and y, 5 12 y 2 < xy and 5 12 x y < x 2 .
To prove that this inequality holds, we use (24). Then: Adding the last two inequalities, Then we know that 0 < 6 13 x < y.
Multiplying by x and y, we deduce that 6 13 x 2 < xy and 6 13 xy < y 2 .
In the following result, we replace the continuity of T and S by a distinct contractivity condition. Theorem 3. Let (X, d) be a complete metric space and let T, S : X → X be two mappings such that, for all x, y ∈ X, where η ∈ Z Ψ and r(x, y) = max d(x, y), Then T and S have a unique common fixed point.
Proof. Let ψ ∈ Z Ψ be a function associated to η as in Definition 2. First of all, we prove that, under the contractivity condition, any fixed point of T is a common fixed point of S and T (that is, if x ∈ X is such that Tx = x, then Sx = x). To prove it, suppose that x ∈ X is such that Tx = x and, by contradiction, assume that Sx = x. Then Using y = x in (28), since Since d(x, Sx) > 0, this means that which is impossible. This proves the claim. In fact, swaping T and S, we have checked that any fixed point of T or S is a common fixed point of T and S. Next, by starting from an arbitrary point x 0 , we shall built up a recursive sequence {x n } defined as: x 2n+1 = Tx 2n , and x 2n+2 = Sx 2n+1 for each n ∈ N.
If there is n 0 ∈ N such that x n 0 = x n 0 +1 , then x n 0 is a fixed point of T or S, so x n 0 is a common fixed point of T and S. In this case, the existence of some common fixed point of T and S is guaranteed. Through the rest of the proof, we suppose that In particular, We shall consider the following set in that will be useful in later stages of the proof.
We divide the rest of the proof into five steps.
Step (1): We claim that To see this, suppose that k = 2n for some n ∈ N. We have So from (28) and (η 1 ) we have: and as ψ is increasing, then On the other hand, So, if there exists some n 0 ∈ N such that d(x 2n 0 +1 , x 2n 0 +2 ) ≥ d(x 2n 0 , x 2n 0 +1 ), then r(x 2n 0 , x 2n 0 +1 ) = d(x 2n 0 +1 , x 2n 0 +2 ) which contradicts (33). Hence, for each n ∈ N, and also Consequently, (32) is proved when k ≥ 0 is an even number. By the same argument, one can verify that (32) holds when k is an odd number.
Next, reasoning by contradiction, suppose that γ > 0. From (34) we have which contradicts (36). So the claim is proved and we obtain that Step (3): We claim that {x n } is a Cauchy sequence. To show that {x n } is a Cauchy sequence, because of (37), it is enough to show that the subsequence {x 2n } is a Cauchy sequence. On contrary, suppose that {x 2n } is not a Cauchy sequence. By Lemma 1 there exist ε 0 > 0 and subsequences {x 2m k } and {x 2n k } of {x n } such that n k is the smallest index for which n k > m k > k and d(x 2m k , x 2n k ) ≥ ε 0 and Therefore, from the definition of r(x, y) we have On the other hand, we claim that for sufficiently large k ∈ N, if n k > m k > k, then Indeed, since n k > m k and {d(x n , x n+1 )} is non-increasing we have So, the left hand side of inequality (39) is equal to Therefore, we must show that, for sufficiently large k ∈ N, if n k > m k > k, then According to (37), there exists k 1 ∈ N such that for any k > k 1 , There exists k 2 ∈ N such that for any k > k 2 , Hence, for any k > max{k 1 , k 2 } and n k > m k > k, we have So, one concludes that Thus we obtain that for any k > max{k 1 , k 2 } and n k > m k > k, So (39) is proved. Therefore, by (28) and the definition of A, for sufficiently large which contradicts (38). This contradiction demonstrates that {x n } is a Cauchy sequence and, since X is complete, there exists u ∈ X such that {x n } → u as n → ∞.
Step (4): u is a common fixed point of T and S. If u is a fixed point of T or S, we have demonstrated at the beginning of the proof that u is a common fixed point of T and S, and this step is finished. Next, suppose that d (u, Tu) > 0 and d (u, Su) > 0, and we will get a contradiction.
If there are infinitely many positive even integers {2n k } such that x 2n k = u for all k ∈ N, then x 2n k +1 = Tx 2n k = Tu for all k ∈ N, and as {x 2n k +1 } → u, then Tu = u, which is false. The same is true if there are infinitely many positive odd integers {2n k + 1} such that x 2n k +1 = u for all k ∈ N, because x 2n k +2 = Tx 2n k +1 = Su for all k ∈ N. In general, if there is a subsequence {x n k } such that x n k = u for all k ∈ N, then we can guarantee that u is a common fixed point of T and S. On the contrary case, without loss of generality, suppose that d(x n , u) = 0 for each n ≥ 0.
In the same way, one can show that lim n→∞ r(u, x 2n+1 ) = d(Tu, u).
Now, we claim that for each n ≥ 0, at least one of the following inequalities is true: If these inequalities are both false for some n 0 ≥ 0, then we get which is a contradiction, and the claim is proved. So, one can consider the following two subcases. Subcase (4.1): The inequality (42) holds for infinitely many n ≥ 0. In this case, for infinitely many n ≥ 0 we have Therefore (x 2n , u) ∈ A. Thus Consequently, from (35) it is seen that for infinitely many n ≥ 0, η(d(Tx 2n , Su), r(x 2n , u)) ≥ 0.
However, lim k→∞ d(x 2n k +1 , Su) = lim k→∞ r(x 2n k , u) = d(u, Su) > 0, and from (η 2 ) we have lim sup which contradicts (44). Subcase (4.2): The inequality (42) only holds for finitely many n ≥ 0. In this case, there exists n 0 ≥ 0 such that (43) is true for any n ≥ n 0 . Similar to subcase (4.1), one can prove that (43) also leads to a contradiction unless u is a fixed point of T or S.
As a consequence, in any case, u is a common fixed point of T and S.
Step (5): The common fixed point of T and S is unique. Suppose that u and v are two distinct common fixed points of T and S, that is, Tu = Su = u and Tv = Sv = v. Hence d (u, v) > 0 and we deduce from (28) that which is a contradiction. Hence, T and S have a unique common fixed point.

Consequences
In this section, we illustrate the applicability of the previous theorems by showing they cover a lot of distinct cases which permit us to deduce several results under different hypotheses and contractivity conditions. For instance, the following corollary is an immediate consequence of Theorem 2 by removing the restriction 1 2 min{d(x, Tx), d(y, Sy)} ≤ d(x, y).

Corollary 1.
Let (X, d) be a complete metric space and T and S be two continuous self-maps on X such that, for all x, y ∈ X, where η ∈ Z Ψ and Then T and S have a unique common fixed point.

Corollary 2.
Let (X, d) be a complete metric space and let T be continuous self-map on X. If there exists η ∈ Z Ψ such that, for all x, y ∈ X, then T has a unique fixed point.
Proof. Choose S = T in Theorem 2 and we get the proof.
The following corollary is a consequence of Corollary 2. , if x = y, then T has a unique fixed point.

Corollary 4.
Let (X, d) be a complete metric space and let T and S be two continuous self-maps on X such that, for every x, y ∈ X, m(x, y) So (η 2 ) holds and we can apply Theorem 2 to complete the proof.

Remark 2.
In some of the following results we will consider two functions ψ, φ : [0, +∞) → [0, +∞) such that: (i) ψ is a continuous non-decreasing function and ψ(t) = 0 if, and only if, t = 0. (ii) φ is lower semi-continuous with φ(t) = 0 if, and only if, t = 0. Corollary 5. Let (X, d) be a complete metric space and let T and S be two continuous self-maps on X such that, for every x, y ∈ X, where ψ and φ are given as in Remark 2 and Then T and S have a unique common fixed point.
Proof. Since φ is lower semi-continuous, if lim n→∞ s n = > 0, So one can apply Corollary 4.

Corollary 6.
Let (X, d) be a complete metric space and let T and S be two continuous self-maps on X such that, for every x, y ∈ X, where ψ ∈ Ψ, m(x, y) is defined as in Corollary 5 and ρ : [0, +∞) → [0, 1) is a function such that ρ(t) = 0 if, and only if, t = 0, and lim sup t→s ρ(t) < 1 for each s > 0. Then T and S have a unique common fixed point.
Corollary 7. Let (X, d) be a complete metric space and let T and S be two continuous maps on X such that, for every x, y ∈ X, Therefore (η 2 ) holds and one can apply Theorem 2 to complete the proof. So, by applying the same argument as in Corollary 8, one can prove the following consequence.
Corollary 9. Let (X, d) be a complete metric space and T and S be two continuous self-maps on X such that for every x, y ∈ X, m(x, y))), where ψ ∈ Ψ, m(x, y) is defined as in Corollary 5 and φ : [0, +∞) → [0, +∞) is an upper semi-continuous function such that φ(t) < t for each t > 0, and φ(t) = 0 if, and only if, t = 0. Then T and S have a unique common fixed point.

Definition 3.
Let (X, d) be a complete metric space, let T : X → X be a continuous mapping and let η ∈ Z. Then T is called a generalized Z-contraction with respect to η if the following condition is satisfied: , if x = y.

Theorem 4.
Every generalized Z-contraction on a complete metric space has a unique fixed point.

Proof.
It is an obvious consequence of Corollary 1 by letting S = T.
Theorem 5. Let (X, d) be a complete metric space and let T : X → X be continuous mapping such that, for all x, y ∈ X, where ψ and φ are given as in Remark 2 and Then T has a unique fixed point.
Proof. It follows from Corollary 5 when S = T.
Theorem 6. Let (X, d) be a complete metric space and let T and S be two continuous self-maps on X such that, for every x, y ∈ X, , if x = y.
Then T and S have a unique common fixed point.
The following example shows that Theorem 2 is a genuine generalization of the Corollary 1.
The following corollary is a consequence of Theorem 3.
Then T and S have a unique common fixed point.
Corollary 11. Let (X, d) be a complete metric space and let T be self-map on X. If there exists η ∈ Z Ψ such that, for all x, y ∈ X, then T has a unique fixed point.
Proof. Putting S = T in Theorem 3 we get the proof.
The following corollary is an application of Corollary 11.

Corollary 12.
Let (X, d) be a complete metric space and let T be self-map on X. If there exists η ∈ Z Ψ such that, for all x, y ∈ X, η(d(Tx, Ty), r T (x, y)) ≥ 0, where r T (x, y) is defined as in Corollary 11, then T has a unique fixed point.
Corollary 13. Let (X, d) be a complete metric space and let T and S be two self-maps on X such that, for every x, y ∈ X, for all t, s ≥ 0. In the proof of Corollary 4 we showed that η is a Ψ-simulation function. Hence, we can apply Theorem 3 to complete the proof.

Corollary 14.
Let (X, d) be a complete metric space and let T and S be two self-maps on X such that, for every x, y ∈ X, Then T and S have a unique common fixed point.
Corollary 16. Let (X, d) be a complete metric space and let T and S be two maps on X such that, for every x, y ∈ X, Proof. Take η(t, s) = ψ(s) − φ(s) − ψ(t) for all t, s ≥ 0. Then similar to the proof of Corollary 13, one can see that η is a Ψ-simulation function. So, by applying Theorem 3, the proof is completed.
Corollary 17. Let (X, d) be a complete metric space and let T and S be two self-maps on X such that, for every x, y ∈ X, Proof. Define η(t, s) = φ(ψ(s)) − ψ(t) for all t, s ≥ 0. In the proof of Corollary 8 we proved that η is a Ψ-simulation function. Then we can apply Theorem 3 to complete the proof.
Applying the same argument as in Corollary 17, one can prove the following result.

Definition 4.
Let (X, d) be a complete metric space, let T : X → X be a mapping and let η ∈ Z. Then T is called a generalized Z-contraction with respect to η if the following condition is satisfied: η(d(Tx, Ty), r T (x, y)) ≥ 0, for all x, y ∈ X, where r T (x, y) = max d(x, y), [1 + d(x, Tx)] d(y, Ty) 1 + d(x, y) .

Theorem 7.
Every generalized Z-contraction on a complete metric space has a unique fixed point.
Proof. It follows from Corollary 11. Theorem 8. Let (X, d) be a complete metric space and let T : X → X be a continuous map such that, for all x, y ∈ X, 1 2 d(x, Tx) ≤ d(x, y) implies that: ψ(d(Tx, Ty)) ≤ ψ(r T (x, y)) − φ(r T (x, y)) where r T (x, y) = max d(x, y), Then T has a unique fixed point.
Then T and S have a unique common fixed point.
The following example shows that Theorem 3 is a genuine generalization of the Corollary 10.