International Journal of Analysis and Applications Volume 17, Number 3 (2019), 342-360 URL: https://doi.org/10.28924/2291-8639 DOI: 10.28924/2291-8639-17-2019-342 PATA-TYPE FIXED POINT RESULTS IN bν(s)-METRIC SPACES FANGYUAN DONG, PEISHENG JI∗ AND XIAOHUI WANG School of Mathematics and statistics, Qingdao University, Shangdong 266071, P. R. China ∗Corresponding author: jipeish 1@sina.com Abstract. The aim of this is to study fixed point theorems in bν (s)-metric spaces under the Pata-type conditions. As consequences, we establish common fixed point results of Pata-type for two maps in bν (s)- metric spaces. 1. Introduction The Banach contraction principle introduced by Banach [6] is one of the most important results in math- ematical analysis. It is the most widely applied fixed point result in many branches of mathematics and generalized in many different directions. Some generalizations of the notion of a metric space have been proposed by some authors, such as, rectangular metric spaces, semi metric spaces, pseudo metric spaces, probabilistic metric spaces, fuzzy metric spaces, quasi metric spaces, quasi semi metric spaces, D metric spaces, and cone metric spaces (see [1, 2, 7, 20, 24, 26, 29–33, 36]). The other direction of investigation is concerned with generalizations of contractive condition (see [3, 10, 11, 18, 19] and others in literature). One of the interesting recent results of this kind was obtained by V. Pata in [23]. Several scholars have already used Pata-type conditions to obtain new fixed point results (see [5, 9, 14–17]). V. Pata obtained the following interesting refinement of the classical Banach Contraction Principle. Received 2018-12-01; accepted 2019-01-30; published 2019-05-01. 2010 Mathematics Subject Classification. 54H25; 47H10. Key words and phrases. Fixed point; bν (s)-metric space; Pata-type contractive mapping; triangular α-admissible mapping. c©2019 Authors retain the copyrights of their papers, and all open access articles are distributed under the terms of the Creative Commons Attribution License. 342 https://doi.org/10.28924/2291-8639 https://doi.org/10.28924/2291-8639-17-2019-342 Int. J. Anal. Appl. 17 (3) (2019) 343 Theorem 1.1. [23] Let (X,d) be a metric space, f : X → X, Let Λ ≥ 0,η ≥ 1 and β ∈ [0,η] be fixed constants and ψ : [0, 1] → [0,∞) be an increasing function, vanishing with continuity at 0. If the inequality d(fx,fy) ≤ (1 −ε)d(x,y) + Λεηψ(ε)[1 + ||x|| + ||y||]β is satisfied for every ε ∈ [0, 1] and all x,y ∈ X, then f has a unique fixed point z ∈ X. Here, ||x|| = d(x,x0) for a chosen point x0 ∈ X. It was also shown by an example that the previous theorem is real generalization of Banach’s result. More results of this kind were subsequently obtained by various authors. b-metric spaces were firstly used by I.A. Bakhtin and S. Czerwik. Definition 1.1. [4,8] Let X be a nonempty set, s ≥ 1 be a given real number. A mapping d : X×X → [0,∞) is called a b-metric with parameter s if for all x,y ∈ X the following holds: (1) d(x,y) = 0 if and only if x = y; (2) d(x,y) = d(y,x) for all x,y ∈ X ; (3) d(x,y ≤ s[d(x,z) + d(z,y)] for all x,y,z ∈ X (b-triangular inequality). Then the pair (X,d) is called a b-metric space. Remark 1.1. In general, b-metric might not be continuous functions (see example in [4, 8]). Definition 1.2. [7] Let X be a nonempty set. Let d : X × X → [0,∞) be a mapping such that for all x,y ∈ X and distinct points u,v ∈ X, each distinct from x and y: (1) d(x,y) = 0 if and only if x = y; (2) d(x,y) = d(y,x); (3) d(x,y ≤ d(x,u) + d(u,v) + d(v,y) (rectangular inequality). Then d is called a generalized metric and the pair (X,d) is called generalized metric space (or shortly GMS). Remark 1.2. Obviously, each metric space is a generalized metric space, but the converse is not true. Moreover, Sarma et al. [32] and Samet [31] presented examples showing that generalized metric spaces might not be Hausdorff and, again, that generalized metric might be discontinuous. Also, Suzuki showed in [35] that, in general, generalized metric spaces do not have a compatible topology. As a combination of b-metric and generalized metric spaces, b-rectangular metric spaces were introduced and used in [12, 22, 28]. Definition 1.3. [12] Let X be a nonempty set and s ≥ 1 be a fixed real number.. Let d : X ×X → [0,∞) be a mapping such that for all x,y ∈ X and distinct points u,v ∈ X, each distinct from x and y: (1) d(x,y) = 0 if and only if x = y; Int. J. Anal. Appl. 17 (3) (2019) 344 (2) d(x,y) = d(y,x); (3) d(x,y ≤ s[d(x,u) + d(u,v) + d(v,y)] (b-rectangular inequality). Then d is called a b-rectangular metric and the pair (X,d) is called b-rectangular metric space with parameter s. In 2017, Z.D. Mitrovic and S. Radenovic [21] introduced the concept of bµ(s)-metric space as follows. Definition 1.4. [21] Let X be a nonempty set. Let d : X×X → [0,∞) be a mapping and let ν ∈ N, s ≥ 1. Then (X,d) is said to be a bν(s)-metric space if for all x,y ∈ X and for all distinct points u1,u2, · · · ,uν ∈ X, each of them different from x and y, the following hold: (1) d(x,y) = 0 if and only if x = y; (2) d(x,y) = d(y,x); (3) d(x,y ≤ s[d(x,u1) + d(u1,u2) + · · · + d(uν,y)] (bν(s)-metric inequality). They note that: b1(1)-metric space is usual metric space, b1(s)-metric space is b-metric space with coefficient of Bakhtin and Czerwik, b2(1)-metric space is generalized metric space, b2(s)-metric space is rectangular b-metric space with coefficient s of George et al., bν(1)-metric space is ν-generalized metric space of Branciari. Definition 1.5. [21] Let (X,d) be a bν(s)-metric space, {xn} be a sequence in X and x ∈ X. Then (i) The sequence {xn} is said to be convergent in (X,d) and converges to x, if for every ε > 0 there exists n0 ∈ N such that d(xn,x) < ε for all n > n0 and this fact represented by limn→∞xn = x or xn →∞ as n →∞. (ii) The sequence {xn} is said to be Cauchy sequence in (X,d) if for every ε > 0 there exists positive integer N(ε) such that d(xm,xn) < ε for all m,n > N(ε). (iii) (X,d) is said to be a complete bµ(s)-metric space if for every Cauchy sequence in X converges to some x. And they proved the following Theorem: Theorem 1.2. [21] Let (X,d) be a complete bν(s)-metric space and suppose that T : X → X be a self- mapping satisfying: d(Tx,Ty) ≤ λd(x,y) for all x,y ∈ X, where λ ∈ [0, 1). Then T has a unique fixed point. Int. J. Anal. Appl. 17 (3) (2019) 345 Definition 1.6. [34] Let X be a nonempty set, T : X → X and α : X ×X → [0,∞). We say that T is a triangular α-admissible mapping if (1) α(x,y) ≥ 1 implies α(Tx,Ty) ≥ 1, for x,y ∈ X; (2) α(x,z) ≥ 1, α(z,y) ≥ 1 implies α(x,y) ≥ 1, for all x,y,z ∈ X. Lemma 1.1. [34] Let T is a triangular α-admissible mapping. Assume that there exists x0 ∈ X such that α(x0,Tx0) ≥ 1. Define sequence {xn} by xn = Tnx0. Then α(xm,xn) ≥ 1 for all m,n ∈ N with m < n. The following lemmas will be used for proving our main results. Lemma 1.2. Let (X,d) be a bν(s)-metric space and let {xn} be a sequence in X with distinct elements ( xn 6= xm for n 6= m). Suppose that d(xn,xn+p) tends to 0 as n →∞ for all p = 1, 2, · · · ,ν, and xn → x as n →∞. Then 1 s d(x,y) ≤ lim inf n→∞ d(xn,y) ≤ lim sup n→∞ d(xn,y) ≤ sd(x,y), for all y ∈ X with y 6= x. Proof. Since {xn} be a sequence in X with distinct elements, we can assume that xn is different from x and y for all n ∈ N. By the bν(s)-metric inequality, we have d(x,y) ≤ s[d(x,xn+ν−1) + d(xn+ν−1,xn+ν−2) + · · · + d(xn+1,xn) + d(xn,y)], d(xn,y) ≤ s[d(xn,xn+ν−1) + d(xn+ν−1,xn+ν−2) + · · · + d(xn+1,x) + d(x,y)]. Since d(xn,xn+p) tends to 0 as n → ∞ for all p = 1, 2, · · · ,ν, and xn → x as n → ∞, taking lim infn→∞ on the both sides of the first inequality and taking lim supn→∞ on the both sides of the second inequality, it follows that 1 s d(x,y) ≤ lim inf n→∞ d(xn,y) ≤ lim sup n→∞ d(xn,y) ≤ sd(x,y). Lemma 1.3. Let (X,d) be a bν(s)-metric space and let {xn} be a sequence in X with distinct elements ( xn 6= xm for n 6= m). Suppose that d(xn,xn+p) tends to 0 as n →∞ for all p = 1, 2, · · · ,ν and {xn} is not a Cauchy sequence. Then there exist � > 0 and two sequence {mk} and {nk} of positive integers such that nk > mk + ν, mk ≥ k and � ≤ lim inf k→∞ d(xnk,xmk) ≤ lim sup n→∞ d(xnk,xmk) ≤ s�, � s ≤ lim inf k→∞ d(xnk−1,xmk−1) ≤ lim sup n→∞ d(xnk−1,xmk−1) ≤ s�. Int. J. Anal. Appl. 17 (3) (2019) 346 Proof. Since {xn} is not a Cauchy sequence, there exists � > 0 for which we can choose two subsequences {xmk} and {xnk} of {xn} such that nk is the smallest index for which nk > mk ≥ k and d(xmk,xnk) ≥ �. (1.1) This means that d(xmk,xmk+1) < �,d(xmk,xmk+2) < �, · · · ,d(xmk,xnk−1) < �. (1.2) Since limn→∞d(xn,xn+p) = 0 for all p = 1, 2, · · · ,ν, we can assume that nk > mk + ν. Using (1.1), (1.2) and bν(s)-metric inequality, we have ε ≤ d(xmk,xnk) ≤ s[d(xmk,xnk−ν) + d(xnk−ν,xnk−ν+1) + · · · + d(xnk−2,xnk−1) + d(xnk−1,xnk)] ≤ s[� + d(xnk−ν,xnk−ν+1) + · · · + d(xnk−2,xnk−1) + d(xnk−1,xnk)]. Since limn→∞d(xn,xn+1) = 0, we get � ≤ lim inf k→∞ d(xnk,xmk) ≤ lim sup k→∞ d(xnk,xmk) ≤ s�. Using (1.1) and bν(s)-metric inequality, we have � ≤ d(xmk,xnk) ≤s[d(xmk,xmk−1) + d(xmk−1,xnk−1) + d(xnk−1,xnk−2) + · · · + d(xnk−ν+1,xnk)]. Since limn→∞d(xn,xn+p) = 0 for all p = 1, 2, · · · ,ν, we get � ≤ s lim inf k→∞ d(xnk−1,xmk−1), that is � s ≤ lim inf k→∞ d(xmk−1,xnk−1). Using (1.2) and bν(s)-metric inequality, we have d(xmk−1,xnk−1) ≤ s[d(xmk−1,xmk) + d(xmk,xnk−ν) + d(xnk−ν,xnk−ν+1) + · · · + d(xnk−3,xnk−2) + d(xnk−2,xnk−1)] ≤ s[d(xmk−1,xmk) + � + d(xnk−ν,xnk−ν+1) + · · · + d(xnk−3,xnk−2) + d(xnk−2,xnk−1)]. By taking the upper limit as k →∞ in the above inequality, since limn→∞d(xn,xn+1) = 0, we get lim sup k→∞ d(xmk−1,xnk−1) ≤ s�. Int. J. Anal. Appl. 17 (3) (2019) 347 Thus � s ≤ lim inf k→∞ d(xmk−1,xnk−1) ≤ lim sup k→∞ d(xmk−1,xnk−1) ≤ sε. � Lemma 1.4. Let {an} and {bn} be two sequences of nonnegative numbers. If lim n→∞ bn = 0, lim n→∞ max{an,bn} = a, then limn→∞an = a. 2. Main Results Throughout the paper, F(T) denotes the set of fixed points of the mapping T . For a given bν(s)-metric space (X,d) and a fixed x0 ∈ X, we will denote ||x|| = d(x,x0) for x ∈ X. We denote by Ψ the family of all functions ψ : [0, 1] → [0,∞) which is an increasing function, continuous at 0, with ψ(0) = 0. Theorem 2.1. Let (X,d) be a complete bν(s)-metric space with s ≥ 1, T : X → X and α : X ×X → [0,∞) a given function. Suppose that following conditions are satisfied: (1) T is a triangular α-admissible mapping; (2) there exist Λ ≥ 0,η ≥ 1, β ∈ [0,η] and ψ ∈ Ψ such that for every ε ∈ [0, 1] and for all x,y ∈ X with α(x,y) ≥ 1 and d(Tx,Ty) > 0, sd(Tx,Ty) <(1 −ε)M(x,y) + Λεηψ(ε)[1 + ||x|| + ||y|| + ||Tx|| + ||Ty||]β (2.1) where M(x,y) = max{d(x,y),d(x,Tx),d(y,Ty), d(x,Tx)d(y,Ty) 1 + d(x,y) }; (3) There exists x0 ∈ X such that α(x0,Tx0) ≥ 1; (4) if {xn} is a sequence in X such that α(xn,xn+1) ≥ 1 for all n ∈ N and xn → x as n → ∞, then there exists a subsequence {xn(k)} of {xn} such that α(xn(k),x) ≥ 1 for all k ∈ N. Then T has a fixed point u and {Tnx0} converges to u. Further, if all x,y ∈ F(T), we have α(x,y) ≥ 1, then T has a unique fixed point in X. Int. J. Anal. Appl. 17 (3) (2019) 348 Proof. Let x0 ∈ X satisfies α(x0,Tx0) ≥ 1. We construct the sequence {xn} in X by xn = Txn−1 = Tnx0 for n ∈ N. If xn = xn+1 for some n ∈ N, then xn is a fixed point of T . Consequently, we suppose that xn 6= xn+1 for all n ∈ N. Since T is a triangular α-admissible mapping, by Lemma 1.1, we have α(xn,xm) ≥ 1, for all n,m ∈ N with n < m. (2.2) Step I. We will show that the sequence {d(xn,xn+1)} is decreasing. Indeed, putting ε = 0, x = xn, y = xn+1 in (2.1), we obtain sd(xn,xn+1) = sd(Txn−1,Txn) < M(xn−1,xn), (2.3) where M(xn−1,xn) = max{d(xn−1,xn),d(xn−1,Txn−1),d(xn,Txn), d(xn−1,Txn−1)d(xn,Txn) 1 + d(xn−1,xn) } = max{d(xn−1,xn),d(xn−1,xn),d(xn,xn+1), d(xn−1,xn)d(xn,xn+1) 1 + d(xn−1,xn) } = max{d(xn−1,xn),d(xn,xn+1)}. (2.4) Combining (2.3) and (2.4), we have sd(xn,xn+1) < max{d(xn−1,xn),d(xn,xn+1)}. Hence d(xn,xn+1) < 1 s d(xn−1,xn) (2.5) for all n ∈ N. Thus the sequence {d(xn,xn+1)} is decreasing. Step II. We will prove that xn 6= xm for all n 6= m. Suppose that xn = xm for some n > m, so we have xn+1 = Txn = Txm = xm+1. By (2.5), we have d(xn,xn+1) < d(xn−1,xn) < · · · < d(xm,xm+1) = d(xn,xn+1) a contradiction. Thus xn 6= xm for all n 6= m. Step III. We will show that for p = 1, 2, · · · ,ν, the sequence {d(xn,xn+p)} is bounded. Indeed, since α(xn,xn+p) ≥ 1 and d(xn,xn+p) > 0, putting ε = 0, x = xn, y = xn+p in (2.1), we obtain sd(xn,xn+p) = sd(Txn−1,Txn+p−1) < M(xn−1,xn+p−1), (2.6) Int. J. Anal. Appl. 17 (3) (2019) 349 where M(xn−1,xn+p−1) = max{d(xn−1,xn+p−1),d(xn−1,Txn−1),d(xn+p−1,Txn+p−1), d(xn−1,Txn−1)d(xn+p−1,Txn+p−1) 1 + d(xn−1,xn+p−1) } = max{d(xn−1,xn+p−1),d(xn−1,xn),d(xn+p−1,xn+p), d(xn−1,xn)d(xn+p−1,xn+p) 1 + d(xn−1,xn+p−1) } ≤ max{d(xn−1,xn+p−1),d(xn−1,xn),d(xn−1,xn)2}. Combining (2.6), we have sd(xn,xn+p) < max{d(xn−1,xn+p−1),d(xn−1,xn),d(xn−1,xn)2}. Taking an = d(xn,xn+p) and bn = d(xn,xn+1), since s ≥ 1, we have san < max{an−1,bn−1,b2n−1}. Since sbn < bn−1 ≤ max{an−1,bn−1,b2n−1} and sb2n < b2n−1 ≤ max{an−1,bn−1,b2n−1}, we have max{an,bn,b2n} < 1 s max{an−1,bn−1,b2n−1} (2.7) for all n ∈ N. Thus the sequence {max{an,bn,b2n}}n∈N is decreasing. Thus K = sup{d(xn,xn+p),d(xn,xn+1),d(xn,xn+1)2 : n = 1, 2, · · · ; p = 1, 2, · · · ,ν} < ∞. (2.8) Step IV. We will prove that the sequence cn = d(xn,x0) is bounded. Using (2.5), we deduce the following estimate cn = d(xn,x0) ≤ s[d(x0,xν−1) + d(xν−1,xν−2) + · · · + d(x2,x1) + d(x1,xn+1) + d(xn+1,xn)] ≤ s[cν−1 + (ν − 1)c0] + sd(Tx0,Txn). Therefore, we infer from (2.1) that cn ≤(1 −ε)M(x0,xn) + Λεηψ(ε)[1 + ||xn|| + ||x1|| + ||xn+1||]β + s[cν−1 + (ν − 1)c0], (2.9) Int. J. Anal. Appl. 17 (3) (2019) 350 where M(x0,xn) = max{d(x0,xn),d(x0,Tx0),d(xn,Txn), d(x0,Tx0)d(xn,Txn) 1 + d(x0,xn) } = max{d(x0,xn),d(x0,x1),d(xn,xn+1), d(x0,x1)d(xn,xn+1) 1 + d(x0,xn) } ≤ max{cn,c1,c21}. (2.10) Combining (2.9) and (2.10), as β ≤ η we have cn ≤(1 −ε)[max{cn,c1,c21}] + Λεηψ(ε)[1 + cn + c1 + cn+1] η + s[cν−1 + (ν − 1)c0]. (2.11) Suppose that the sequence cn = d(xn,x0) is not bounded. Then there is a subsequence {cni} satisfying that cni ≥ max{1,c1,c21, 1 + νK} for all i ∈ N and cni →∞. Using (2.8), we have cn+1 = d(xn+1,x0) ≤ s[d(x0,xn) + d(xn,xn+1) + · · · + d(xn+ν−2,xn+ν−1) + d(xn+ν−1,xn+1)] ≤ sd(x0,xn) + sνK = scn + sνK. Thus, for all i ∈ N, (2.11) implies that cni ≤ (1 −ε)cni + Λε ηψ(ε)[1 + (1 + s)cni + sνK] η + s[cν−1 + (ν − 1)c0] ≤ (1 −ε)cni + Λε ηψ(ε)(3scni) η + s[cν−1 + (ν − 1)c0]. Thus we have cni ≤ (1 −ε)cni + aε ηψ(ε)cηni + b for some a,b > 0. Hence εcni ≤ aε ηψ(ε)cηni + b. Now, as in [23], the choice ε = εi = (1 + b)/cni leads to the contradiction 1 ≤ a(1 + b)ηψ(εi) → 0. Hence the sequence cn = d(xn,x0) is bounded. Step V. For p = 1, 2, · · · ,ν, limn→∞d(xn,xn+p) = 0. Int. J. Anal. Appl. 17 (3) (2019) 351 For all ε ∈ (0, 1] and for x = xn,y = xn+p we have sd(xn,xn+p) = sd(Txn−1,Txn+p−1) < (1 −ε)M(xn−1,xn+p−1) + Λεηψ(ε)[1 + ||xn|| + ||xn+p−1|| + ||xn+p||]β ≤ (1 −ε)M(xn−1,xn+p−1) + Bεηψ(ε), B > 0 (2.12) where M(xn−1,xn+p−1) = max{d(xn−1,xn+p−1),d(xn−1,Txn−1),d(xn+p−1,Txn+p−1), d(xn−1,Txn−1)d(xn+p−1,Txn+p−1) 1 + d(xn−1,xn+p−1) } = max{d(xn−1,xn+p−1),d(xn−1,xn),d(xn+p−1,xn+p), d(xn−1,xn)d(xn+p−1,xn+p) 1 + d(xn−1,xn+p−1) } ≤ max{d(xn−1,xn+p−1),d(xn−1,xn),d(xn−1,xn)2}. (2.13) For p = 1, using (2.13) and (2.5), the inequality (2.12) implies that sd(xn,xn+1) = sd(Txn−1,Txn) ≤ (1 −ε)M(xn−1,xn) + Bεηψ(ε) ≤ (1 −ε)d(xn−1,xn) + Bεηψ(ε). (2.14) By (2.5), the sequence {d(xn,xn+1)} is converges. If limn→∞d(xn,xn+1) = d∗ > 0, it follows from (2.14) that d∗ ≤ Bψ(ε), that is d∗ = 0. A contradiction. Thus limn→∞d(xn,xn+1) = 0. In the following, we assume that d(xn,xn+1) < 1 for all n ∈ N. Thus d(xn,xn+1) 2 < d(xn,xn+1) (2.15) for all n ∈ N. Fixed p ≥ 2. (2.12), (2.13) and (2.15) imply that sd(xn,xn+p) ≤ (1 −ε) max{d(xn−1,xn+p−1),d(xn−1,xn)} + Bεηψ(ε), that is, using the notations in step III, an ≤ san ≤ (1 −ε) max{an−1,bn−1} + Bεηψ(ε). (2.16) Int. J. Anal. Appl. 17 (3) (2019) 352 From step III, we see that max{an,bn} is decreasing. Since limn→∞ bn = 0, by Lemma 1.4, we have lim n→∞ an = lim n→∞ max{an,bn} = t. If t > 0, taking the limit as n →∞ on both sides of (2.16), we have t ≤ Bψ(ε) for all ε ∈ (0, 1], that is t = 0. A contradiction. Step VI. We will show that {xn} is Cauchy sequence in X. Suppose, to the contrary, that is, {xn} is not a Cauchy sequence. By Step V and Lemma 1.3, there exist δ > 0 and two sequence {mk} and {nk} of positive integers such that nk > mk + ν, mk ≥ k and δ ≤ lim inf k→∞ d(xnk,xmk) ≤ lim sup n→∞ d(xnk,xmk) ≤ sδ, δ s ≤ lim inf k→∞ d(xnk−1,xmk−1) ≤ lim sup n→∞ d(xnk−1,xmk−1) ≤ sδ. (2.17) Now putting x = xmk−1,y = xnk−1 in (2.1), since d(xnk,xmk) > 0 and α(xnk,xmk) ≥ 1, we obtain sδ ≤ sd(xmk,xnk) = sd(Txmk−1,Txnk−1) < (1 −ε)M(xmk−1,xnk−1) + Bεψ(ε), (2.18) where M(xmk−1,xnk−1) = max{d(xmk−1,xnk−1),d(xmk−1,Txmk−1),d(xnk−1,Txnk−1), d(xmk−1,Txmk−1)d(xnk−1,Txnk−1) 1 + d(xmk−1,xnk−1) } = max{d(xmk−1,xnk−1),d(xmk−1,xmk),d(xnk−1,xnk), d(xmk−1,xmk)d(xnk−1,xnk) 1 + d(xmk−1,xnk−1) }. (2.19) Using Step V and (2.19), we have lim sup k→∞ M(xmk−1,xnk−1) ≤ sδ. Taking the limit of supermum as k →∞ in (2.18), sδ ≤ Bψ(ε), that is, δ = 0, a contradiction. Hence {xn} is Cauchy sequence in X. Since (X,d) is complete, there exists u ∈ X such that lim n→∞ d(xn,u) = 0. Int. J. Anal. Appl. 17 (3) (2019) 353 Step VII. We show that u is a fixed point of T. Using Theorem 2.1 (iv), there exists a subsequence {xnk} of {xn} such that α(xnk,u) ≥ 1 for all k ∈ N. Suppose that u 6= Tu, so d(u,Tu) > 0. Since {xn} is a sequence with distinct elements, we can assume that xn 6= Tu for all n ∈ N. Putting x = xnk,y = u in (2.1), we get d(u,Tu) ≤ s[d(u,xnk+ν) + d(xnk+ν−1,xnk+ν−2) + · · · + d(xnk+2,xnk+1) + d(xnk+1,Tu)] < s[d(u,xnk+ν) + d(xnk+ν−1,xnk+ν−2) + · · · + d(xnk+2,xnk+1)]+ (1 −ε)M(xnk,u) + Bεψ(ε), (2.20) where M(xnk,u) = max{d(xnk,u),d(xnk,Txnk),d(u,Tu), d(xnk,Txnk)d(u,Tu) 1 + d(xnk,u) } = max{d(xnk,u),d(xnk,xnk+),d(u,Tu), d(xnk,xnk+1)d(u,Tu) 1 + d(xnk,u) }. (2.21) Using Step V and (2.21), we have lim k→∞ M(xnk,u) = d(u,Tu). Taking the limit as k →∞ in (2.20), using step V, we have d(u,Tu) ≤ (1 −ε)d(u,Tu) + Bεψ(ε), from which we have d(u,Tu) ≤ Bψ(ε), that is d(u,Tu) = 0, a contradiction. Thus u = Tu. Step VIII. Finally, we prove that the fixed point of T is unique. Suppose that u,v are two fixed points of T such that u 6= v. Then by the hypothesis, α(u,v) ≥ 1. Hence, from (2.1) with ε = 0, x = u and y = v we have sd(u,v) = sd(Tu,Tv) < M(u,v) where M(u,v) = max{d(u,v),d(u,Tu),d(v,Tv), d(u,Tu)d(v,Tv) 1 + d(u,v) } = d(u,v). Thus d(u,v) < d(u,v), a contradiction. � Note. In Theorem 2.1, if s > 1, the inequality (2.1) can be replaced by sd(Tx,Ty) ≤(1 −ε)M(x,y)) + Λεηψ(ε)[1 + ||x|| + ||y|| + ||Tx|| + ||Ty||]β. Int. J. Anal. Appl. 17 (3) (2019) 354 Moreover, if ν ≥ 2, we can give another method to prove that {xn} is a Cauchy sequence. In fact, from (2.7) and (2.8), we have max{an,bn,b2n}≤ K sn for all n ∈ N. Using bµ(s)-inequality, for all n,p ∈ N, we have d(xn,xn+pν) ≤ s[d(xn,xn+1) + d(xn+1,xn+2) + · · · + d(xn+ν−1,xn+ν)] + s2[d(xn+ν,xn+ν+1) + d(xn+ν+1,xn+ν+2) + · · · + d(xn+2ν−1,xn+2ν)] · · · + sp[d(xn+(p−1)ν,xn+(p−1)ν+1) + d(xn+(p−1)ν+1,xn+(p−1)ν+2) + · · · + d(xn+pν−1,xn+pν)] ≤ s[ K sn + K sn+1 + · · · + K sn+ν−1 ] + s2[ K sn+ν + K sn+ν+1 + · · · + K sn+2ν−1 ] + · · · + sp[ K sn+(p−1)ν + K sn+(p−1)ν+1 + · · · + K sn+pν−1 ] ≤ sν K sn 1 − 1 sν−1 , d(xn,xn+pν+1) ≤ s[d(xn,xn+1) + d(xn+1,xn+2) + · · · + d(xn+ν−1,xn+ν)] + s2[d(xn+ν,xn+ν+1) + d(xn+ν+1,xn+ν+2) + · · · + d(xn+2ν−1,xn+2ν)] · · · + sp[d(xn+(p−1)ν,xn+(p−1)ν+1) + d(xn+(p−1)ν+1,xn+(p−1)ν+2) + · · · + d(xn+pν−1,xn+pν+1)] ≤ sν K sn 1 − 1 sν−1 , · · · d(xn,xn+pν+ν−1) ≤ s[d(xn,xn+1) + d(xn+1,xn+2) + · · · + d(xn+ν−1,xn+ν)] + s2[d(xn+ν,xn+ν+1) + d(xn+ν+1,xn+ν+2) + · · · + d(xn+2ν−1,xn+2ν)] · · · Int. J. Anal. Appl. 17 (3) (2019) 355 + sp[d(xn+(p−1)ν,xn+(p−1)ν+1) + d(xn+(p−1)ν+1,xn+(p−1)ν+2) + · · · + d(xn+pν−1,xn+pν+ν−1)] ≤ sν K sn 1 − 1 sν−1 , this implies d(xn,xn+m) ≤ sν K sn 1 − 1 sν−1 . Thus {xn} is a Cauchy sequence. Corollary 2.1. Let (X,�,d) be a partially ordered and complete bν(s)-metric space with s ≥ 1. Suppose that following conditions are satisfied: (1) T is a increasing mapping with respect �, that is Tx � Ty if x � y; (2) there exist Λ ≥ 0,L ≥ 0,η ≥ 1, β ∈ [0,η] and ψ ∈ Ψ such that for every ε ∈ [0, 1] and for all x,y ∈ X with x � y and d(Tx,Ty) > 0, sd(Tx,Ty) ≤(1 −ε)M(x,y) + Λεηψ(ε)[1 + ||x|| + ||y|| + ||Tx|| + ||Ty||]β where M(x,y) = max{d(x,y),d(x,Tx),d(y,Ty), d(x,Tx)d(y,Ty) 1 + d(x,y) }; (3) there exists x0 ∈ X such that x0 � Tx0;. (4) xn � x for all n ∈ N whenever {xn} is nondecreasing sequence in X such that xn → x ∈ X. Then T has a fixed point u and {Tnx0} converges to u. Further, if all x,y ∈ F(T), x and y are comparable, then T has a unique fixed point in X. Proof. Define α : X ×X → [0,∞) as α(x,y) =   1, if x � y, 0, otherwise. Clearly, by Theorem 2.1, T has a fixed point. � Int. J. Anal. Appl. 17 (3) (2019) 356 3. Common fixed point results In this section, we prove some common fixed point results for two self-mappings. Following [27], we introduce the notion of f −α-admissible mapping. Definition 3.1. Let X be a non-empty set. And let T,f : X −→ X and α : X×X −→ [0,∞). The mapping T is f −α-admissible if, for all x,y ∈ X such that α(fx,fy) ≥ 1, we have α(Tx,Ty) ≥ 1. Clearly, if f is the identity mapping, then T is α-admissible. Theorem 3.1. Let (X,d) be a complete bν(s)-metric space with s ≥ 1 and α : X ×X → [0,∞) be a given function. Let T,f : X → X be two mappings. Suppose that following conditions are satisfied:: (1) T is an f −α-admissible mapping; (2) there exist Λ ≥ 0,η ≥ 1, β ∈ [0,η] and ψ ∈ Ψ such that for every ε ∈ [0, 1] and for all x,y ∈ X with α(fx,fy) ≥ 1 and d(Tx,Ty) > 0, sd(Tx,Ty) ≤(1 −ε)M(fx,fy) + Λεηψ(ε)[1 + ||fx|| + ||fy|| + ||Tx|| + ||Ty||]β (4.1) where M(fx,fy) = max{d(fx,fy),d(fx,Tx),d(fy,Ty), d(fx,Tx)d(fy,Ty) 1 + d(fx,fy) }; (3) there exists x0 ∈ X such that α(fx0,Tx0) ≥ 1; (4) if {yn} is a sequence in X such that α(yn,yn+1) ≥ 1 for all n ∈ N and yn → y as n → ∞, then there exists a subsequence {yn(k)} of {yn} such that α(yn(k),y) ≥ 1 for all k ∈ N. Then T and f have a point of coincidence in X. Moreover, if T and f are weakly compatible, then T and f have a common fixed point. Further, if all points of coincidence of f and T , we have α(fx,fy) ≥ 1, then T and f have a unique point of coincidence in X.. Before we prove this theorem, we introduce the following lemma. Lemma 3.1. [13] Let X be a non-empty set and let f : X → X be a self-mapping. Then there exists a subset E of X such that fE = fX and f|E is injective. Int. J. Anal. Appl. 17 (3) (2019) 357 Proof. By Lemma 3.1, there exists E ⊆ X such that fE = fX and f : E −→ X is one-to-one. Now, define a map h : f(E) −→ f(E) by h(f(x)) = Tx. Since f is one-to-one on E, h is well defined. Note that for all fx,fy ∈ f(E) with α(fx,fy) ≥ 1,d(h(fx),h(fy)) > 0, then (4.1) can be rewrite as sd(h(fx),h(fy)) ≤(1 −ε)M(fx,fy) + Λεηψ(ε)[1 + ||fx|| + ||fy|| + ||h(fx)|| + ||h(fy)||]β, M(fx,fy) = max{d(fx,fy),d(fx,h(fx)),d(fy,h(fy)), d(fx,h(fx))d(fy,h(fy)) 1 + d(fx,fy) }; Thus for all x′,y′ ∈ f(E) with α(x′y′) ≥ 1 and d(hx′,hy′) > 0, we have sd(hx′,hy′) ≤(1 −ε)M(x′,y′) + Λεηψ(ε)[1 + ||x′|| + ||y′|| + ||hx′|| + ||hy′||]β where M(x′,y′) = max{d(x′,y′),d(x′,hx′),d(y′,hy′), d(x′,hx′)d(y′,hy′) 1 + d(x′,y′) }. Since f(E) = f(X) is complete, by using Theorem 2.1, there exists x0 ∈ E such that h(fx0) = fx0. Hence T and f have a point of coincidence in X. It is clear that T and f have a common fixed point whenever T and f are weakly compatible. � In 2017, M. Rangamma and P. M. Reddy [25] established a unique common fixed point theorem for T-contraction of two self mappings on generalized cone b-metric spaces with solid cone. In the following theorem, a unique common fixed point theorem for T-contraction of two self mappings on bν(s)-metric spaces is established. Theorem 3.2. Let (X,d) be a complete bν(s)-metric space and α : X × X → [0,∞) be a given function. Let T,f : X → X be a mappings. Suppose that T is one to one and T(X) is a complete subspace of X, and the following conditions are satisfied: (1) if α(Tx,Ty) ≥ 1 then α(Tfx,Tfy) ≥ 1, and α(x,y) ≥ 1,α(y,z) ≥ 1 implies α(x,z) ≥ 1, x,y,z ∈ X; (2) there exist Λ ≥ 0,η ≥ 1, β ∈ [0,η] and ψ ∈ Ψ such that for every ε ∈ [0, 1] and for all x,y ∈ X with α(Tx,Ty) ≥ 1 and d(Tfx,Tfy) > 0, sd(Tfx,Tfy) ≤(1 −ε)M(Tx,Ty) + Λεηψ(ε)[1 + ||Tx|| + ||Ty|| + ||Tfx|| + ||Tfy||]β (4.3) where M(Tx,Ty) = max{d(Tx,Ty),d(Tx,Tfx),d(Ty,Tfy), d(Tx,Tfx)d(Ty,Tfy) 1 + d(Tx,Ty) }; Int. J. Anal. Appl. 17 (3) (2019) 358 (3) There exists x0 ∈ X such that α(Tx0,Tfx0) ≥ 1; (4) if {xn} is a sequence in X such that α(xn,xn+1) ≥ 1 for all n ∈ N and xn → x as n → ∞, then there exists a subsequence {xn(k)} of {xn} such that α(xn(k),x) ≥ 1 for all k ∈ N. Then f has a fixed point in X. Further, if all x,y ∈ F(f), we have α(Tx,Ty) ≥ 1 then f has a unique fixed point in X. Moreover, if f and T are commuting at the fixed point of f, then f and T have a unique common fixed point in X. Proof. Since T is one to one, the conditions (i) and (ii) can be restated as (i′) if α(Tx,Ty) ≥ 1 then α(TfT−1Tx,TfT−1Ty) ≥ 1, and α(x,y) ≥ 1,α(y,z) ≥ 1 implies α(x,z) ≥ 1, x,y,z ∈ X. (ii′) if α(Tx,Ty) ≥ 1 and d(TfT−1Tx,TfT−1Ty) > 0 implies sd(TfT−1Tx,TfT−1Ty) ≤ (1 −ε)M(Tx,Ty) + Λεηψ(ε)[1 + ||Tx|| + ||Ty|| + ||TfT−1Tx|| + ||TfT−1Ty||]β where M(Tx,Ty) = max{d(Tx,Ty),d(Tx,TfT−1Tx),d(Ty,TfT−1Ty), d(Tx,TfT−1Tx)d(Ty,TfT−1Ty) 1 + d(Tx,Ty) }. Let f′ = TfT−1. Then we have (i′′) f′ is a triangular α-admissible mapping in TX. (ii′′) for all x′,y′ ∈ TX, if α(x′,y′) ≥ 1 and d(f′x,f′y) > 0 implies sd(f′x′,f′y′) ≤(1 −ε)M(x′,y′) + Λεηψ(ε)[1 + ||x′|| + ||y′|| + ||f′x′|| + ||f′y′||]β where M(x′,y′) = max{d(x′,y′),d(x′,f′x′),d(y′,f′y′), d(x′,f′x′)d(y′,f′y) 1 + d(x′,y′) }. Then, by Theorem 2.1, there exist x′ = Tx ∈ TX such that f′Tx = Tx, that is Tfx = Tx. 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