Fixed Point Results on ∆-Symmetric Quasi-Metric Space via Simulation Function with an Application to Ulam Stability

Badr Alqahtani 1, Andreea Fulga 2 and Erdal Karapınar 3,4,* 1 Department of Mathematics, King Saud University, P.O. Box 2455, Riyadh 11451, Saudi Arabia; balqahtani1@ksu.edu.sa 2 Department of Mathematics and Computer Sciences, Universitatea Transilvania Brasov, 500036 Brasov, Romania; afulga@unitbv.ro 3 Department of Mathematics, Atilim University, 06836 Ankara, Turkey 4 Department of Medical Research, China Medical University, Taichung 40402, Taiwan * Correspondence: erdalkarapinar@yahoo.com


Introduction
In the last few decades, one of the hot topics in topology and analysis has been the quasi-metric, which is a natural generalization of the notion of the metric; see, e.g., [1][2][3][4].Roughly speaking, the quasi-metric appears to be obtained by relaxing the symmetric condition from the axioms of the standard metric.Regarding the physical phenomena, the quasi-metric can be more useful to consider in solving real-life problems [5].On the other hand, one can ask an impulsive question about whether there is a positive constant ∆ such that the distance from a point p to q is dominated by ∆ times the distance from q to p.The answer is affirmative, and such spaces are called the ∆-symmetric quasi-metric.These spaces are quite rich and lie between the quasi-metric and metric.
Quasi-metric spaces are very interesting topics for researchers who work in fixed point theory; see, e.g., [6,7] and the references therein.It is an indispensable fact that real-life applications of fixed point theory have a wide range.Indeed, the border of the range is beyond the following question: Do we transfer the real-life problems in the form of f (x) = x or not?Consequently, after the first metric fixed point result of Banach, several authors have reported a number of interesting results in various directions.Here, we mention one of the interesting generalization of the Banach contraction mapping principle that was given by Seghal [8].
Theorem 1. ( [8]) Let (M, d) be a complete metric space and T a continuous self-mapping of M, which satisfies the condition that there exists a real number q, 0 < q < 1 such that for each v ∈ M, there exists a positive integer m(v) such that for each w ∈ M, d(T m(v) v, T m(v) w) ≤ qd(v, w). (1) Then, T has a unique fixed point in M.
This result above was improved by Guseman [9] by removing the continuity condition.Later, K.Iseki [10], J. Matkowski [11], Singh [12] and Ray and Rhoades [13] extended the result of Seghal [8].The result of Kincses and Totik [14] is one of the most improved results in this direction.Theorem 2. ( [14]) Let T be a self-mapping on a metric space (M, d) such that for some q ∈ [0, 1) and for all v, w ∈ M, we can find a positive integer m(v) such that: d(T m(v) v, T m(v) w) ≤ q max d(v, w), d(w, T m(v) v), d(v, T m(v) w) . ( Then, T has a unique fixed point v * . Theorem 3. ( [14]) Let T be a self-map of a metric space (M, d).Assume there exists a nonincreasing function β : (0, ∞) → [0, 1) such that for each v ∈ M, there exists a positive integer m(v) such that for each w ∈ M Then, T has a unique fixed point v * .
The inflation of so many results causes a commotion.Therefore, it is natural to consider combining and unifying the existence results.The notion of the simulation function is one of the successful consequences of this approach.By using the simulation function, it is possible to combine several distinct types of contractions and hence unify a number of existing results in a single theorem.
In this paper, we aim to get not only the most general metric fixed point results in the context of quasi-metric space, but also unify the several existing results in this direction, including the results of Seghal [8], Guseman [9], Kincses and Totik [14].

Preliminaries
We denote the set of non-negative reals by R + 0 .
Definition 1.For M = ∅, a function q : M × M → R + 0 is called quasi-metric if the following assumptions are held: q(s, u) ≤ q(s, t) + q(t, u), for all s, t, u ∈ M.
Here, the pair (M, q) is called a quasi-metric space.
The quasi-metric notion is a concrete extension of the metric concept.Therefore, as expected, each metric space forms a quasi-metric space, but the converse is not necessarily true.For instance, the well-known functions l, r : R × R → R + 0 , defined by l(s, t) = max{s − t, 0} and r(s, t) = max{t − s, 0}, are quasi-metric, but not a metric.Indeed, d(s, t) := max{l(s, t), r(s, t)} forms a standard Euclidean metric on R.
A sequence {s n } in M converges to s ∈ M if: In a quasi-metric space (M, q), the limit for a convergent sequence is unique.If s n → s, we have for all s ∈ M: lim n→∞ q(s n , t) = q(s, t) and lim n→∞ q(t, s n ) = q(t, s).
Let (M, q) be a quasi-metric space and {s n } be a sequence in M. We say that a sequence {s n } is left-Cauchy if for every > 0, there exists a positive integer N = N( ) such that q(s n , s m ) < for all n ≥ m > N. Analogously, a sequence {s n } is called right-Cauchy if for every > 0, there exists a positive integer N = N( ) such that q(s n , s m ) < for all m ≥ n > N. Furthermore, the sequence {s n } is said to be Cauchy if for every > 0, there exists a positive integer N = N( ) such that q(s n , s m ) < for all m, n > N. It is evident that a sequence {s n } in a quasi-metric space is Cauchy if and only if it is left-Cauchy and right-Cauchy.A quasi-metric space (M, q) is left-complete (respectively, right-complete, complete) if each left-Cauchy sequence (respectively, right-Cauchy sequence, Cauchy sequence) in M is convergent.Definition 2. Suppose that (M, q) is a quasi-metric space, and {s n } ⊂ M and s ∈ M. Let T be a self-mapping.We say that T is: Observe that the simultaneous right and left continuity of a mapping yields the continuity of it.
Definition 3. Suppose that (M, q) is a quasi-metric space.We say that it is ∆-symmetric if there exists a positive real number ∆ > 0 such that: It is clear that if ∆ = 1, then the ∆-symmetric quasi-metric space (M, q) forms a metric space.
Example 1. Suppose that (M, q) is a quasi-metric space, and a function q : M × M → R + 0 is defined as follows: (M, q) is a three-symmetric quasi-metric space, but it is not a metric space.
In the following, we recall the main properties of ∆-symmetric quasi-metric spaces.
Lemma 1. (See, e.g., [15]) Let (M, q) be a ∆-symmetric quasi-metric space and {s n } be a sequence in M and s ∈ M. Then: If {t n } is a sequence in M and q(s n , t n ) → 0, then q(t n , s n ) → 0.
A non-decreasing function ψ : R + We reserve the letter Ψ to denote the family of all c-comparison functions.Note that each c-comparison function forms a comparison function.For more details and examples for comparison and c-comparison functions, see, e.g., [16,17].
Let ϕ : R + 0 → R + 0 be a c-comparison function that satisfies the following condition: We shall use the letter Φ to represent the class of such functions (i.e., the set of all c-comparison functions that satisfies the condition (Ξ)).
In what follows, we recollect the definition of the simulation function.
Definition 4. See, e.g., [18].Suppose that Then, ζ is called a simulation function.Further, the letter S denotes all simulation functions ζ.
A set M is regular with respect to mapping α : The following technical lemma will be used in the proof of the first main result of this paper.

Lemma 2.
Let M be a non-empty set and T : M → M form a triangular α-orbital admissible mapping.Consider the iterative sequence s n = Ts n−1 , n ∈ N. If there exists s 0 ∈ M such that α(s 0 , Ts 0 ) ≥ 1, then for any k ∈ N, we have: and: for all p ∈ {0, 1, 2, ...} .
Proof.Due to the statement of the theorem, there exists a point s 0 ∈ M such that α(s 0 , Ts 0 ) ≥ 1.
Based on the definition of the iterative sequence {s n } ⊂ M and taking into account that T is α-orbital admissible, we get: Recursively, we derive that: On the other side, using the condition (TO) of Definition 5, we deduce that: Recursively, for all p ∈ {1, 2, ...}, we have: Definition 6.Let α : M × M → R + 0 be a function.We say that a self-mapping T : M → M satisfies the condition (U) if:

Main Results
We state our first main results.
Theorem 4. Let (M, q) be a complete ∆-symmetric quasi-metric space, T be a continuous self-mapping and α : M × M → R + 0 , ζ ∈ S, ϕ ∈ Ψ. Suppose that for every x ∈ M, there is a positive integer n = n(s) such that the inequality: is fulfilled for all t ∈ M.Moreover, assume that: (i) T forms a triangular α-orbital admissible; (ii) there exists s 0 ∈ M with the property α(s 0 , Ts 0 ) ≥ 1 and α(Ts 0 , s 0 ) ≥ 1; Then, T has a fixed point u ∈ M.
Proof.Fix s ∈ M. By assumption, there exist n = n(s) such that for all t ∈ M: Regarding the condition (ζ1), we find: which yields that: Now, we shall define r 1 (s) and r 2 (s) as follows: for any s ∈ M.
Due to (Σ), Now, we claim that there exist c ∈ (0, ∞), with c > l such that: Observe that the assumption ( 18) holds for m = 0. On the contrary, we suppose that there is a positive integer k so that: Regarding the triangle inequality, together with the inequalities ( 16) and ( 12), we have that: Taking, the assumption (19) into account and keeping ϕ ∈ φ in mind, we get: This is a contradiction, since we already supposed that a k ≥ c, and we have a k − ϕ(a k ) > l due to (Ξ).Hence, the set q(s, T m•n(s)+p ) : m ∈ N is bounded.Due to fact that p ∈ {0, 1, ..., n(s) − 1}, we conclude that the set {q(s, T m s) : m ∈ N} is bounded, and so, r 1 (s) = sup {q(s, T m s) : m ∈ N} < +∞.
On the other hand, the space (M, q) is the ∆-symmetric, so, for all s ∈ M and m ∈ N, we have: From here, we conclude that the set {q(T m s, s) : m ∈ N} is also bounded and r 2 (s) < +∞.Therefore, for all s ∈ M and all m 1 , m 2 ∈ N, Now, let s 0 ∈ M be an arbitrary point.If Ts 0 = s 0 , then s 0 forms a fixed point of T, which completes the proof.Hence, we assume that Ts 0 = s 0 .Starting from s 0 , we construct inductively a sequence {s k }, by: where n k = n k (s).Since: by induction, we deduce that: which means that {s k } k ≥ 0 is a subsequence of the orbit T k s 0 : k ∈ N .Furthermore, we observe that: s k+i = T n 0 +n(1)...+n k +n k+1 +...+n k+i−1 s 0 = T n k +n k+1 +...+n k+i−1 (T n 0 +n 1 +...+n k−1 s 0 ) = T n k +n k+1 +...+n k+i−1 s k . ( We shall prove that {s n } is right-Cauchy and left-Cauchy in (M, q).First of all, notice that from (20), for any k ∈ N and replacing s = s k−1 and t = T n k s k−1 in ( 16), we get: Hence, since ϕ is increasing, we obtain that: By using the triangular inequality and ( 24), for all i ≥ 1, we get: q(x k , x k+l ) ≤ q(s k , s k+1 ) + q(s k+1 , s k+2 )... + q(s k+l−2 , s k+l−1 ) + q(s k+l−1 , s k+l ) However, assuming that Ts 0 = s 0 , we have r 1 (s 0 ) ≥ q(s 0 , T n k s 0 ) > 0, and by (Σ), the series Therefore, for an arbitrary ε > 0, there which ensures that {x k } is a right-Cauchy sequence.On account of Lemma 1, we derive that {x k } is a left-Cauchy sequence in (M, q).Therefore, it is Cauchy in the complete quasi-metric space (M, q).It yields that there exists u ∈ M such that: For the rest of the proof, we consider that {s k } is the sequence defined above and u is the limit of this sequence.
Thus, we have: lim Keeping ( 26) and ( 29) in mind together with the uniqueness of a limit, we conclude that u = Tu, that is u is a fixed point of T.
In the following theorem, we remove the assumption of the continuity of the mapping T.
Theorem 5. Let (M, q) be a complete ∆-symmetric quasi-metric space, T be a self-mapping and α : M × M → R + 0 , ζ ∈ S, ϕ ∈ Ψ. Suppose that for every x ∈ M, there is a positive integer n = n(s) such that the inequality: is fulfilled for all t ∈ M. Suppose also that: (i) T is a triangular α-orbital admissible; (ii) there exists s 0 ∈ M such that α(s 0 , Ts 0 ) ≥ 1 and α(Ts Then, T has a fixed point u ∈ M. Furthermore: (a) for each x ∈ M, lim m→∞ T m s = u; (b) the mapping T n(u) is continuous at u.

Proof.
As in Theorem 4, we can construct an iterative sequence {s n } that converges to a point u ∈ M, which means that: We claim, under the assumption that M is α-regular, that u is a fixed point of T n (u), that is T n (u)u = u.First of all, replacing in (30) s = s k−1 and t = T n(u) s k−1 , we have for n k−1 = n(s k−1 ): or, keeping in mind (ζ 1 ), Taking into account that ϕ is monotone increasing, we can write the chain of inequalities: Note that ϕ ∈ Ψ is a c-comparison function, and hence, it satisfies (Ω) condition.Thus, we get that: Furthermore, from (35) and Lemma 1, we have: By contradiction, we assume that T n(u) u = u, and let ε 1 = q(u, T n(u) u) > 0 and ε 2 = q(T n(u) u, u) > 0. Without loss of generality, we suppose that ε 1 ≤ ε 2 .
Again, since ϕ ∈ Ψ, (Γ) holds, which means that: From (31) and respectively (36), there exists k 0 ∈ N such that for any k ≥ k 0 , there hold: If M is regular with respect to α, then there exists a subsequence {s k } of {s n } such that α(u, s k ) ≥ 1 for all k and using the triangle inequality, we obtain: which is a contradiction.As a consequence, T n(u) u = u.
Finally, we prove that the statement (b) holds.For this purpose, we will show that q(u, T n(u) t k ) ≤ q(u, t k ) where {t k } is an arbitrary sequence in M such that: Assume, by contradiction, that there exists some k ∈ N such that: Replacing in (30), we have: Since ζ ∈ S and using the α-regularity of M, we have: which is a contradiction.As a consequence, lim k→∞ q(u, T n(u) t k ) = 0. Since q(T n(u) t k , u) ≤ Mq(u, T n(u) t k ) for all k ∈ N, we get that lim k→∞ q(T n(u) t k , u) = 0, and therefore: which means that T n(u) is continuous at s = u.
We notice that to guarantee the uniqueness, we need to add an additional condition.
Theorem 6. Adding the condition (U): If u is a fixed point of T n(u) for any t ∈ M, α(u, t) ≥ 1 to the statement of Theorem 4, respectively 5, we obtain the uniqueness of the fixed point.
Proof.By Theorem 4, we know that T n(u) has at least one fixed point.We suppose that there exists Then, due the condition (U): which is a contradiction.In this case, it is obvious that Tu = T(T n(u) u) = T n(u) (Tu) which shows that Tu is a fixed point of T, and due to its uniqueness, we conclude that Tu = u.
Example 2. Let M = [0, 4], q : M × M → R + 0 , defined by: It easy to see that (M, q) is a complete two-symmetric quasi-metric space.Let T : M → M, We shall prove that T satisfies the conditions of Theorem 5.If s ∈ [0, 1], then Ts, T 2 s ∈ [0, 1] and Hence, T is an α-orbital mapping.For α(s, t) ≥ 1, we consider the following cases: Therefore, T is a triangular α-orbital admissible mapping, so the assumption (i) of Theorem 5 is satisfied.
Further, Condition (ii) is satisfied, since α(0, T0) = α(0, 0) = 2.Moreover, if {s n } is a sequence in M such that α(s n , s n+1 ) ≥ 1 for all n ∈ N and lim n→∞ s n = s ∈ M, then s = 0 and α(0, s n )geq1 for all n.Hence, M is α-regular.Due to the manner in which we defined the function α, the following cases are the interesting ones, letting, for example ζ(t, s) = s 2 − t and ϕ(t) = t 3 .
we have s > t, T n s = s 2 n+1 , T n t = t 2 n and T n s ≥ T n t.Hence, For s = 1 2 and t = 3 2 , we have Therefore, for any s ∈ M, there exists n(s) ∈ N, for example n(s) = 4, such that for every t ∈ M, all assumptions of Theorem 4 are satisfied.Then, T has a (unique) fixed point u = 0. On the other hand, q(T 1 2 , T Hence T is not a contraction.Moreover, we cannot find the functions ζ ∈ S, ϕ ∈ φ such that: This shows that Theorem 5 is indeed a generalization of known results. Theorem 7. Let (M, q) be a complete ∆-symmetric quasi-metric space, a self-mapping T and a map α : M × M → R + 0 .Suppose that there exist ζ ∈ S, ϕ ∈ φ such that for every s ∈ M, there is a positive integer n = n(s) such that for all t ∈ M: for each s, t ∈ M, where: S(s, t) = max q(s, t), q(s, T n(s) s), q(s, T n(s) t) .
Then, T has a fixed point u ∈ M. Furthermore, for each s ∈ M, lim m→∞ T m s = u.
Proof.We remark firstly that if s ∈ M, then there exist n = n(s) such that for all y ∈ M: or, with (ζ1 Furthermore, The functions r 1 (s) and r 2 (s), defined by (17), are finite, which is observed by following step by step the lines in the proof of Theorem 4. By substituting s by T n(s) and y by T (k−1)n(x)+p s in (45) and taking (Γ) into account, we have: S(s, T (k−1)n(s)+p s) = max q(s, T (k−1)n(s)+p s), q(s, T n(s) s), q(s, T n(s) (T (k−1)n(s)+p s)) = max q(s, T (k−1)n(s)+p s), q(s, T n(s) s), q(s, T kn(s)+p s) Denoting by a k = q(s, T kn(s)+p s) for k ∈ N, from the triangle inequality, we have: a k = q(s, T kn(s)+p s) ≤ q(s, T n(s) s) + q(T n(s) s, T kn(s)+p s) = q(s, T n(s) x) + q(T n(s) s, T n(s) (T (k−1)n(s)+p s)).
Following the reasoning in the previous theorem, notice that we can find c ∈ (0, ∞), with c > l = max {a 0 , q(s, T n (s)s)} ≥ a 0 such that t − ϕ(t) > l for all t ∈ R + 0 .We shall prove that a k < c for all k ∈ N.For this, we assume, on the contrary, namely, that there exists a positive integer i such that a i−1 < c ≤ a i .We have then: or a i − ϕ(a i ) ≤ l, which is a contradiction.Therefore, the set {q(s, T n s) : n ∈ N} is bounded and r 1 (s) < ∞.Since the space (M, q) is ∆-symmetric, we have for all s ∈ M and all k ∈ N that: q(T n s, s) ≤ Mq(s, T n s) ≤ Mr 1 (s), which shows that the set {q(T n s, s) : n ∈ N} is also bounded.Hence, r 2 (s) < ∞.Let s 0 ∈ M be arbitrary.We will show now that the sequence {s i } constructed inductively by: is a Cauchy sequence.By this construction of sequence {s i }, it follows that for any i ∈ N and l ≥ 1, If we denote by s 0 = m i+l−1 + ... + m i+1 + m i , using the condition (42) or, equivalently, (45), for s = s i−1 and t = T p 0 t i−1 , we have: S(s i−1 , T p 0 s i−1 ) = max q(s i−1 , T p 0 s i−1 ), T(s i−1 , T m i−1 s i−1 ), q(s i−1 , T p 0 +m i−1 s i−1 )) .
Let us now have a positive integer p 1 ∈ {p 0 , m i−1 , p 0 + m i−1 } such that S(s i−1 , T p 0 s i−1 ) = q(s i−1 , T p 1 s i−1 ).Replace in (50): q(s i , T p 0 s i ) ≤ ϕ(q(s i−1 , T p 1 s i−1 )). (51) Hence, ϕ is monotone increasing, and we obtain that: for p 0 , p 1 , ..., p i ∈ N.However, ϕ is a c-comparison function, and from (Ω), there exist ε > 0 such that, ϕ i (r 1 (s 0 )) < 0 for any i ∈ N. Hence, q(s i , s i+l ) < ε for arbitrary i ∈ N and l ≥ 1.We conclude that {l i } is a right-Cauchy sequence on (M, q).Since (M, q) is supposed to be ∆-symmetric, then, from Lemma 1, it is a Cauchy sequence.As the space is complete, there exists u ∈ M such that s n → u, which means that: If we use the hypothesis that the map T is continuous, we obtain: We want to show now that T n(u) u = u, that is u is a fixed point of T n(u) , under the assumption that M is α-regular.First, we show that: lim i→∞ q(s i , T n(u) s i ) = 0 (54) and: lim i→∞ q(T n(u) s i , s i ) = 0 (55) By proceeding as above, from (45), we get: where r i ∈ {n(u), m i−1 , n(u) + m i−1 } is an appropriate index such that S(s i−1 , T n(u) s i−1 ) = q(s i−1 , T r 1 s i−1 ).Using the same arguments as previously given and taking into account that ϕ is monotone increasing, we obtain that: q(s i , T n(u) s i ) ≤ ϕ i (q(s 0 , T r i s 0 ) → 0.
Theorem 8.Besides all assumptions of Theorem 7, suppose that the condition (U) is satisfied.Then, T has a unique fixed point.
Proof.The existence of a fixed point is observed by Theorem 7.For the proof of the uniqueness, we use the method of reductio ad absurdum.Let u, v ∈ M be two fixed points of T n(u) with u = v.Then, we have from hypotheses (U) that α(u, v) ≥ 1 and: which is a contradiction.Hence, u is the unique fixed point for T n(u) .Then, Tu = T(T n(u) u) = T n(u) (Tu) implies that Tu = u, which shows that u is the unique fixed point of T.

Conclusions
In this paper, we obtain fixed point theorems in quasi-metric spaces by using simulations functions.Notice that the class of simulation functions is quite rich; see, e.g., [6,7,[18][19][20]22]. Accordingly, for each simulation function, we find not only the existing results in the literature, but also some new results; in particular, by letting simulation function ζ(t, s) = qs − t, q ∈ (0, 1) in Theorem 4.
On the other hand, we have used an auxiliary function α in simulation functions that unifies the results in standard (quasi-)metric and results in a partially ordered (quasi-)metric and also the frame of the cyclic mappings.As was shown in several papers (see, e.g., [26,28,35]), by taking the auxiliary function α in a proper way, we shall get analogous results in a partially ordered structure and in the setting of cyclic maps.Regarding these aspects and several possible combinations of them, one can get a long list of corollaries of the results of this paper.Regarding the length of the paper, we avoid listing these consequences explicitly.We underline also that we give a simple form of Ulam stable results.It is easily improved in some other directions.