Proinov type contractions on dislocated b-metric spaces

In this paper, we improve the Proinov theorem by adding certain rational expressions to the definition of the corresponding contractions. After that, we prove fixed point theorems for these modified Proinov contractions in the framework of dislocated b-metric spaces. We show some illustrative examples to indicate the validity of the main results.


Introduction and preliminaries
In the nature of mathematics, there is the purpose of generalizing, expanding, and obtaining the most general forms of existing concepts and results. The concept of metric, which is the most fundamental and solid basis of the analysis study, has been constantly expanded and generalized with this motivation. Examples of the new metrics that have been put forward for this purpose can be counted as quasi-metric, b-metric, partial-metric, symmetric, D-metric, modular metric, fuzzy metric, soft-metric, G-metric, and so on. On the other hand, it was understood that not all of these newly defined metrics provide a new and original structure. For instance, G-metric can be reduced to semi-metric or cone metric to a standard metric. More examples can be given, but here we stop to focus on the main motivation. Two of the new and original generalizations of metric notions are b-metrics [1][2][3][4][5][6][7][8][9][10][11][12][13][14][15][16] and dislocated metrics [17][18][19][20][21]. Very recently, these two notions have emerged under the name of dislocated b-metric [22,23].
Metric fixed point theory is a field of study that needs an abstract metric framework (see, for instance, [24][25][26][27]). Very recently Proinov [28] proved a fixed point theorem that not only unifies but also generalizes a number of well-known results in the framework of a standard metric space. In particular, he proved that Wardowski [29] and Jleli and Samet [30] results are not only equivalent to each other, but also they are a special case of one of the main results of [28].
In this paper, we improve the Proinov type contractions by involving certain rational expression to the corresponding contraction thought by Proinov [28]. After then, we prove fixed point theorems for these modified Proinov contractions in the framework of dislocated b-metrics. We bring forward illustrative examples to show the validity of the main results.
Let S be a nonempty set and N = {1, 2, 3, . . .}. Some examples of rational contractivity conditions are shown in the following results (see also [31]). Theorem 1 ([32]) Let (S, d ) be a complete metric space and Z : S → S be a mapping such that there exist k 1 , k 2 ∈ [0, 1) with k 1 + k 2 < 1 such that for all v, w ∈ S. Then Z has a unique fixed point x ∈ S, and the sequence {Z n v} converges to the fixed point x for all v ∈ S.
for all distinct v, w ∈ S, then Z possesses a unique fixed point in S.
for all u, v, w ∈ S. In this case, the pair (S, d l ) is a dislocated-metric space (shortly d l -MS).
In this case, the tripled (S, b, s) forms a b-metric space (shortly b-MS).
Obviously, for s = 1, we find the notion of metric space.
We mention that, when s = 1, a d b -MS becomes a d l -MS.
exists and tends to be finite.

Proposition 8 ([36])
In a d b -MS the limit of a convergent sequence is unique.
In case every d b -Cauchy sequence is d b -convergent, we say that the space (S, d b , s) is a complete d b -MS. The next lemma will be useful in the sequel.
for every n ∈ N, then the sequence {Z n v 0 } is a d b -Cauchy sequence.
Proof Let v 0 be an arbitrary point in S and the sequence {v n } with We split the proof in two cases, namely s = 1 and s > 1.
1. For s = 1, d b becomes a dislocated metric and by d l 3 ., for n < p, we have Therefore, lim n,p→∞ d b (v n , v p ) = 0, that is, the sequence {Z n v 0 } is Cauchy. 2. For s > 1, we distinguish two sub-cases: and taking into account (4), we get 1), then C n → 0, and we can find l ∈ N such that C n < 1 s . Therefore, by (a), the sequence

Main results
Henceforth, we use the following notations: and, respectively, where c 1 , c 2 , c 3 , c 4 , c 5 are nonnegative real numbers.
, and two continuous mappings Z, U : S → S such that, for every distinct v, w ∈ S with d b (Zv, Uw) > 0, the following inequality holds. Assume that: Proof For an arbitrary (but fixed) point v 0 ∈ S, let {v n } be the sequence defined as follows: for all n ∈ N 0 . First of all, we claim that v n = v n+1 for any n ∈ N 0 . Indeed, if we can find (5), because the functions , belong to , we have Taking (β 1 ) into account, we get where K = c 1 +c 2 s r -c 3 -c 4 < 1, holds due to the first assumption in (β 1 ). In the same way, replacing in (5) v with v 2n-1 and w with v 2n , and keeping in mind (d b 2 ), we have which leads us to Consequently, (7) and (9) show us that for any n ∈ N, where K ∈ (0, 1). By Lemma 10 it follows that {v n } is a Cauchy sequence. Thus, lim n,p→∞ d b (v n , v p ) exists and is finite. Moreover, since the d b -ms is complete, we get that there exists x ∈ S such that lim n→∞ v n = x and Since the mappings Z and U are supposed to be continuous, we have (5) and since , ∈ , we have However, applying (d b 3 ) and taking into account (d b 2 ), Moreover, by (β 2 ) we get which is a contradiction. Therefore, d b (x, y) = 0 and from (d b 1 ) it follows that x = y, that is, the set F S (Z, U) has exactly one element. (S, d b , s) be a complete d b -ms, , ∈ , a number α ∈ [1, ∞), and a continuous mapping Z : S → S such that, for every distinct v, w ∈ S with d b (Zv, Zw) > 0, the following inequality
Proof Let Z = U in Theorem 11. (S, d b , s) be a complete d b -ms, , ∈ , a number α ∈ [1, ∞), and two mappings Z, U : S → S such that, for every distinct v, w ∈ S with d b (Zv, Uw) > 0, the following inequality

Theorem 13 Let
holds. Assume that: Proof Let v 0 ∈ S be a chosen point and {v n } be the sequence defined by (6) in the proof of Theorem 11. Thus, following the same arguments, we can assume that d b (Zv 2n , Uv 2n+1 ) > 0 and from (13) we get Since by (β 2 ) is nondecreasing, we deduce that which is equivalent to where C = c 1 +c 2 s α -c 3 -c 4 -c 5 < 1 by (β 1 ). Similarly, taking v = v 2n and w = v 2n-1 in (5) and keeping in mind (d b 2 ), we get However, from relations (14), (15), together with Lemma 10, we find that {v n } is a Cauchy sequence in a complete d b -ms. Therefore, there exists x ∈ S such that lim n,p→∞ Without loss of generality, we can suppose that x = v n for any n ∈ N. Supposing that x = Zx, by (5), we have or, taking (β 2 ) into account, However, since On the other hand, and then which contradicts our assumption c 2 ≤ 1. Thus, we get d b (x, Zx) = 0, that is, or, keeping in mind (β 2 ) which is a contradiction. Therefore, As a last step, we claim that x is the unique fixed point of the mappings Z and U. Indeed, if we suppose that there exists another point υ ∈ F S (Z, U) such that x = υ, by (13) we have Since the function is supposed to be nondecreasing, it follows that which is a contradiction. Therefore, the set F S (Z, U) has exactly one element. (S, d b , s) be a complete d b -ms, , ∈ , a number α ∈ [1, ∞), and a mapping Z : S → S such that, for every v, w ∈ S with d b (Zv, Uw) > 0, the following inequality

Theorem 16 Let
Assume that: is nondecreasing. Then the set F S (Z, U) has exactly one element.
Proof Let us take in (20), v = v 2n and w = v 2n+1 , where the sequence {v n } is defined as in Theorem 11. We have Furthermore, taking (β 2 ) and the above relation into account, we get Similarly, taking v = v 2n , respectively w = v 2n-1 , we obtain Now, choosing C = c 1 +sc 2 s α < 1 (by assumption (β 1 )), we have d b (v n , v n+1 ) < Cd b (v n-1 , v n ) for any n ∈ N. Therefore, Lemma 10 leads us to the conclusion that {v n } is a Cauchy sequence. Thus, since the space is complete, there exists x ∈ S such that lim n,p→∞ Supposing that Zx = x, we have Moreover, without loss of generality, we can assume that d b (v n , x) > 0 for any n ∈ N, and then from (20) we get or, by (β 2 ), Returning in (25), we have Letting n → ∞ and keeping in mind (24), we get which is a contradiction. Thus, d b (Zx, x) = 0 and from (d b 1 ) we have x = Zx. Analogously, we have or, by (β 2 ), On the other hand, supposing that d b (x, Ux) > 0, we have Combining the above inequalities and taking limit as n → ∞, we obtain 0 < d b (x, Ux) < 0, which is a contradiction. Therefore, d b (x, Ux) = 0, and then x = Ux. Thus, x is a common fixed point for Z and U, that is, F S (Z, U) = ∅ and it remains to show that the set F S (Z, U) is in fact reduced to a single point. On the contrary, let υ ∈ F S (Z, U) with υ = x. Replaced in (20), we have and, due to (β 2 ), which is a contradiction. Therefore, it follows that x = υ and the set F S (Z, U) has exactly one element. Let d b be the d b -metric on S (with s = 2) given by |v -w| 2 otherwise.
, and a mapping Z : S → S such that, for every distinct v, w ∈ S with d b (Zv, Zw) > 0, the following inequality holds, where Assume that: (β 1 ) c 1 + sc 2 < s α ; (β 2 ) is nondecreasing. Then the set F S (Z) has exactly one element.
Proof Let Z = U in Theorem 16.