Common fixed point results on an extended b-metric space

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1 Alqahtani et al. Journal of Inequalities Applications 018) 018:158 RESEARCH Open Access Common fixed point results on an extended b-metric space Badr Alqahtani1, Andreea Fulga Erdal Karapınar3* * Correspondence: erdalkarapinar@yahoo.com; erdal.karapinar@atilim.edu.tr 3 Department of Mathematics, Atilim University, Ankara, Turkey Full list of author information is available at the end of the article Abstract In this paper, we investigate the existence of common fixed points of a certain mapping in the frame of an extended b-metric space. The given results cover a number of well-known fixed point theorems in the literature. We state some examples to illustrate our results. MSC: 6T99; 7H10; 5H5 Keywords: Fixed point; Metric space 1 Introduction preliminaries Throughout the manuscript, we denote N0 := N {0}, where N is the positive integers. Further, R represents the real numbers R+0 := [0, ). Following this pioneering result on b-metric, a number of authors have reported several interesting results in this direction see, e.g., [1,, 1, 1, 16, 18] the related references therein). Definition 1.1 Czerwik [11]) Let X be a nonempty set d : X X [0, ) be a function satisfying the following conditions: b1) dx, y) = 0 if only if x = y. b) dx, y) = dy, x) for all x, y X. b3) dx, y) s[dx, z) + dz, y)] for all x, y, z X, where s 1. The function d is called a b-metric the space X, d) is called a b-metric space, in short, bms. The immediate example of b-metric is the following. Example 1.1 Let Y = {x, y, z} X = Y N. Define a mapping d : X X [0, ) such that dx, y) = dy, x) = dx, z) = dz, x) = 1, dy, z) = dz, y) = A, dx, x) = dy, y) = dz, z) = 0, 1 1 dn, m) =, n m The Authors) 018. This article is distributed under the terms of the Creative Commons Attribution.0 International License which permits unrestricted use, distribution, reproduction in any medium, provided you give appropriate credit to the original authors) the source, provide a link to the Creative Commons license, indicate if changes were made.

2 Alqahtani et al. Journal of Inequalities Applications 018) 018:158 Page of 15 where A [, ). Then we find that dx, y) A [ ] dx, z)+dz, y) for x, y, z X. It is evident that X, d)isab-metric space. Notice also that if A >, the stard triangle inequality does not hold X, d) is not a metric space. Remark 1.1 It is clear that for s =1,theb-metric becomes a usual metric. Recently, Kamran [13] introduced a new type of generalized metric space they proved some fixed point theorems on this space. Definition 1. [13]) Let X be a nonempty set θ : X X [1, ). A function d θ : X X [0, ) is called an extended b-metric if, for all x, y, z X,itsatisfies d θ 1) d θ x, y)=0iff x = y; d θ ) d θ x, y)=d θ y, x); d θ 3) d θ x, y) θx, y)[d θ x, z)+d θ z, y)]. The pair X, d θ ) is called an extended b-metricspace,inshortextended-bms. Remark 1. If θx, y)=s for s 1, then we obtain the definition of bms. Example 1. Let X = [0, 1] θ : X X [1, ), θx, y)= x+y+1 x+y.defined θ : X X [0, )as d θ x, y)= 1 xy for x, y 0, 1], x y, d θ x, y)=0 forx, y [0, 1], x = y, d θ x,0)=d θ 0, x)= 1 x for x 0, 1]. Obviously, d θ 1) d θ ) hold. For d θ 3), we distinguish the following cases: i) Let x, y 0, 1].Forz 0, 1],wehave d θ x, y) θx, y) [ d θ x, z)+d θ z, y) ] 1 xy 1+x + y x + y x + y xyz If z =0,then d θ x, y) θx, y) [ d θ x,0)+d θ 0, y) ] 1 xy 1+x + y x + y x + y xy ii) For x 0, 1] y =0,letz 0, 1]. z 1+x+y. 1 1+x+y. d θ x,0) θx,0) [ d θ x, z)+d θ z,0) ] 1 x 1+x x 1+x xz xz 1 + x). In conclusion, for any x, y, z X, d θ x, z) θx, z) [ d θ x, y)+d θ y, z) ]. Hence, X, d θ ) is an extended b-metric space.

3 Alqahtani et al. Journal of Inequalities Applications 018) 018:158 Page 3 of 15 Some fundamental concepts, like convergence, Cauchy sequence, completeness in a extended-bms, aredefined asfollows [13]. Definition 1.3 [13]) Let X, d θ ) be an extended-bms. i) A sequence x n in X is said to converge to x X if, for every ɛ >0,thereexists N = Nɛ) N such that d θ x n, x)<ɛ for all n N.Inthiscase,wewrite lim n x n = x. ii) A sequence x n in X is said to be Cauchy if, for every ɛ >0,thereexistsN = Nɛ) N such that d θ x m, x n )<ɛ for all m, n N. Definition 1. An extended-b-metric space X, d θ )iscompleteifeverycauchysequence in X is convergent. Lemma 1.1 Let X, d θ ) be a complete extended-bms. If d θ is continuous, then every convergent sequence has a unique limit. Theorem 1.1 [13]) Let X, d θ ) be an extended-bms such that d θ is a continuous functional. Let T : X Xsatisfy d θ Tx, Ty) kd θ x, y) 1) for all x, y X, where k [0, 1) is such that, for each x 0 X, lim n,m θx n, x m )< 1 k, here x n = T n x 0, n =1,,... Then T has precisely one fixed point u. Moreover, for each y X, T n y u. For our purposes, we need to recall the following definition which is proposed by Popescu [17]. Definition 1.5 Let T : X X α : X X [0, ). We say that T is an α-orbital admissible if, for all x, y X,wehave αx, Tx) 1 α Tx, T x ) 1. ) Definition 1.6 AsetX is regular with respect to mapping α : X X [0, ) if, whenever {x n } is a sequence in X such that αx n, x n+1 ) 1αx n+1, x n ) 1 for all n x n x X as n, then there exists a subsequence {x nk) } of {x n } such that αx nk), x) 1 αx, x nk) ) 1 for all n. S,T) orbital cyclic Definition.1 Suppose that T, S are two self-mappings on a complete extended-bms X, d θ ). Suppose also that there are two functions α, β : X X [0, )suchthat,forany x X, αx, Tx) 1 βtx, STx) 1 βx, Sx) 1 αsx, TSx) 1. 3) Then we say that the pair S, T is an α, β)-orbital-cyclic admissible pair.

4 Alqahtani et al. Journal of Inequalities Applications 018) 018:158 Page of 15 We start with the following lemma which is essential in our main results. Lemma.1 [3]) Let X, d θ ) be an extended b-metric space.if there exists q [0, 1) such that the sequence {x n } for an arbitrary x 0 X satisfies lim n,m θx n, x m )< 1, also q 0<d θ x n, x n+1 ) qd θ x n 1, x n ) ) for any n N, then the sequence {x n } is Cauchy in X. Proof Let {x n } n N be a given sequence. By employing inequality )recursively,wederive that 0<d θ x n, x n+1 ) q n d θ x 0, x 1 ). 5) Since q [0, 1), we find that lim n d θx n, x n+1 )=0. 6) On the other h, by d θ 3), together with triangular inequality, for p 1, we derive that d θ x n, x n+p ) θx n, x n+p ) [d θ x n, x n+1 )+d θ x n+1, x n+p ) ] θx n, x n+p )d θ x n, x n+1 )+θx n, x n+p )d θ x n+1, x n+p ) θx n, x n+p )q n d θ x 0, x 1 )+θx n, x n+p )θx n+1, x n+p ) [ d θ x n+1, x n+ )+d θ x n+, x n+p ) ] θx n, x n+p ) q n d θ x 0, x 1 )+θx n, x n+p )θx n+1, x n+p ) q n+1 d θ x 0, x 1 )+ + θx n, x n+p ) θx n+p 1, x n+p ) q n+p 1 d θ x 0, x 1 ) n+p 1 = d θ x 0, x 1 ) i=1 q i i θx n+j, x n+p ). 7) j=1 Notice the inequality above is dominated by n+p 1 i=1 q i i j=1 θx n+j, x n+p ) n+p 1 i=1 q i i j=1 θx j, x n+p ). On the other h, by employing the ratio test, we conclude that the series i=1 a i, where a i = q i i j=1 θx a j, x n+p ) converges to some S 0, ). Indeed, lim i+1 i a i = lim i qθx i, x i+p ) < 1, hence we get the desired result. Thus, we have S = q i i=1 j=1 i θx j, x n+p ) withthepartialsums n = n q i i=1 j=1 i θx j, x n+p ). Consequently, we observe, for n 1, p 1, that d θ x n, x n+p ) q n d θ x 0, x 1 )[S n+p 1 S n 1 ]. 8) Letting n in 8), we conclude that the constructive sequence {x n } is Cauchy in the extended b-metric space X, d θ ).

5 Alqahtani et al. Journal of Inequalities Applications 018) 018:158 Page 5 of 15 Theorem.1 Let T, S be two self-mappings on a complete extended-bms X, d θ ) such that the pair T, Sformsanα, β)-orbital-cyclic admissible pair. Suppose that i) for each x 0 X, lim n,m θx n, x m )< 1 k k, where x n = Sx n 1 x n+1 = Tx n for each n N; ii) there exists x 0 X such that αx 0, Tx 0 ) 1; iii) either S T arecontinuous, or iii*) if x n is a sequence in X such that x n u, then αu, Tu) 1 βu, Su) 1. Moreover, if for all x, y Xk [0, 1 ) αx, Tx)βy, Sy)d θ Tx, Sy) k [ d θ x, Tx)+d θ y, Sy) ], 9) then the pair of the mappings T, S possesses a common fixed point u, that is, Tu = u = Su. Proof By assumption ii), there exists a point x 0 X such that αx 0, Tx 0 ) 1. Take x 1 = Tx 0 x = Sx 1. By induction, we construct a sequence {x n } such that x n = Sx n 1 x n+1 = Tx n n N. 10) We have αx 0, x 1 ) 1, since S, T)isanα, β-orbital-cyclic admissible pair, we get αx 0, x 1 ) 1 βtx 0, STx 0 )=βx 1, x ) 1 βx 1, x ) 1 αsx 1, TSx 1 )=αx, x 3 ) 1. Applying again 3), αx, x 3 ) 1 βtx, STx )=βx 3, x ) 1 βx 3, x ) 1 αsx 3, TSx 3 )=αx, x 5 ) 1. Recursively, we obtain αx n, x n+1 ) 1 for all n N, 11) βx n+1, x n+ ) 1 for all n N. 1) Without loss of generality, we assume that x n x n+1 for each n N 0. Indeed, if x n0 = x n0 +1 for some n 0 N 0,thenu = x n0 forms a common fixed point for S T, which finalizes the proof. More precisely, to see that u is the common fixed point of S T, weshall examine the following two cases. First, we assume that n 0 is even, that is, n 0 =k. Inthis

6 Alqahtani et al. Journal of Inequalities Applications 018) 018:158 Page 6 of 15 case, we have x k = x k+1 = Tx k,thatis,x k is a fixed point of T. Now we shall prove that x k = x k+1 = Tx k = Sx k+1. Suppose on the contrary that d θ Tx k, Sx k+1 )>0.Byletting x = x k y = x k+1 in 9) keeping in mind 11)1), we get that 0<d θ x k+1, x k+ )=d θ Tx k, Sx k+1 ) αx k, Tx k )βx k+1, Sx k+1 )d θ Tx k, Sx k+1 ) k [ d θ x k, Tx k )+d θ x k+1, Sx k+1 ) ] kd θ x k+1, Sx k+1 )=kd θ x k+1, x k+ ), a contradiction. Hence, we conclude that d θ Tx k, Sx k+1 )=0x k = x k+1 = Tx k = Sx k+1,thatis,x k = x k+1 = u is a common fixed point of T S. Analogously, one can derive the same conclusion for the case n 0 is odd, that is, n 0 =k 1. Thus, throughout the proof, we suppose that x n x n+1 for all n N 0. 13) In what follows, we shall prove that the sequence {x n } is Cauchy. For this purpose, it is sufficient to examine the following two cases. Case a): Let x = x n y = x n+1. Then, by inequality 9)using11), 1), we get 0<d θ x n+1, x n+ ) = d θ Tx n, Sx n+1 ) αx n, Tx n )βx n+1, Sx n+1 )d θ Tx n, Sx n+1 ) k [ d θ x n, Tx n )+d θ x n+1, Sx n+1 ) ] = k [ d θ x n, x n+1 )+d θ x n+1, x n+ ) ], 1) from here d θ x n+1, x n+ ) qd θ x n, x n+1 ) 15) for each n N 0,whereq = k 1 k <1withk [0, 1 ). Case b): Let x = x n y = x n 1. Then, by inequality 9)using11), 1), we get 0<d θ x n+1, x n ) = d θ Tx n, Sx n 1 ) αx n, Tx n )βx n 1, Sx n 1 )d θ Tx n, Sx n 1 ) k [ d θ x n, Tx n )+d θ x n 1, Sx n 1 ) ] = k [ d θ x n, x n+1 )+d θ x n 1, x n ) ] 16) d θ x n, x n+1 ) qd θ x n 1, x n ) 17) for each n N 0,whereq = k 1 k <1withk [0, 1 ). Combining 15)17), we canconcludethat d θ x m, x m+1 ) qd θ x m 1, x m ) 18)

7 Alqahtani et al. Journal of Inequalities Applications 018) 018:158 Page 7 of 15 for all m N. From Lemma.1, taking into account i), lim n,m θx n, x m )< 1 k = 1 k q,we obtain that {x m } is a Cauchy sequence. By completeness of X, d θ ), there is some point u X such that lim x n = u. 19) n Naturally, we also have x n u x n+1 u. 0) Due to the continuity of the mappings T S,weget u = lim n x n+1 = lim n Tx n = T lim n x n = Tu u = lim n x n+1 = lim n Sx n = S lim n x n = Su. Let us consider now the alternative hypothesis iii*). Taking x = u y = x n+1 in 9) taking into account 1), we get d θ Tu, x n+ )=d θ Tu, Tx n+1 ) αu, Tu)βx n+1, Sx n+1 )d θ Tu, Sx n+1 ) k [ d θ u, Tu)+d θ x n+1, x n+ ) ]. 1) Letting n,weobtain d θ Tu, u)= lim d [ θtu, x n+ ) k lim dθ u, Tu)+d θ x n+1, x n+ ) ] n n = kd θ u, Tu)<d θ u, Tu), ) which implies d θ Tu, u)=0.hence,wegetthattu = u. Analogously, regarding 11) ), we observe that d θ x n+1, Su)=d θ Tx n, Su) αx n, Tx n )βu, Su)d θ Tx n, Su) k [ d θ x n, Tx n )+d θ u, Su) ]. 3) Now, letting n in the inequality above, we derive that d θ u, Su)= lim n d θx n+1, Su) k lim n [ d θ x n, Tx n )+d θ u, Su) ] = kd θ u, Su)<d θ u, Su). ) Hence, we find that Su = u. Accordingly, we conclude that T S have a common fixed point u.

8 Alqahtani et al. Journal of Inequalities Applications 018) 018:158 Page 8 of 15 Example.1 Let X = [0, 1] d θ : X X [0, )definedby d θ x, y)= 1 xy for x, y 0, 1], x y, d θ x, y)=0 forx, y [0, 1], x = y, d θ x,0)=d θ 0, x)= 1 x for x 0, 1] when 1+x+y if x 0, 1], x+y θx, y)= 1 ifx = y =0. Then X, d θ ) is an extended-bms see Example 1.). Let T : X X, S : X X,definedas 1 ifx = 1, Tx)= 1 if x = 1, x+1 otherwise, respectively 1 ifx { 1 Sx)=, 1 }, x otherwise, two functions α, β : X X [0, )definedby 1 ifx, y) {1, 1), 1 αx, y)=,1),1, 1 )}, 0 otherwise, 1 ifx, y) {1, 1), 1 βx, y)=,1),1,1)}, 0 otherwise. We show that the pair T, S forms an α, β)-orbital-cyclic admissible pair. Indeed, for x =1, α1, T1) = α1, 1) 1 βt1, ST1) = β1, 1) 1 β1, S1) = β1, 1) 1 αs1, TS1) = α1, 1) 1. For x = 1 : 1 α, T 1 ) ) 1 = α,1 1 β T 1, ST 1 ) = β1, 1) 1

9 Alqahtani et al. Journal of Inequalities Applications 018) 018:158 Page 9 of 15 1 β, S 1 ) ) 1 = β,1 1 α S 1 ), TS1 = α1, 1) 1. For x = 1 : 1 α, T 1 ) 1 = α, 1 ) 1 β T 1, ST 1 ) ) 1 = β,1 1 1 β, S 1 ) ) 1 = β,1 1 α S 1 ), TS1 = α1, 1) 1. We have thus proved that T is α orbital admissible, sure, because α 1, T 1 ) 1, assumption ii) is satisfied. If x 0 { 1, 1,1},thenx n = T n x 0 =1,so lim n,m θx n, x m )= 3 <3=1 k k, where we choose k = 1 < 1.Otherwise,foreachx 0 X { 1, 1,1}, wehavex n 1 = n k=1 1 )n + x 0 n, x n = x n 1 lim n x n =1.So, lim θx n, x m )= 3 n,m <3=1 k k. Hence, i) is also verified. We have 1 d θ 1, T1) = 0, d θ, T 1 ) 1 =, d θ, T 1 ) =8, 1 d θ 1, S1) = 0, d θ, S 1 ) 1 =, d θ, S 1 ) = d θ T1, S1) = 0, d θ T1, S 1 ) =0, d θ T1, S 1 ) =0, d θ T 1 ), S1 =0, d θ T 1, S 1 ) =0, d θ T 1, S 1 ) =0, d θ T 1 ), S1 =, d θ T 1, S 1 ) =, d θ T 1, S 1 ) =. Because in the other cases αx, y)=0βx, y) = 0, it is enough to investigate the following situations: Case a): For x {1, 1 } y {1, 1, 1 }, d θ Tx, Sy)=0 so inequality 9)issatisfied.

10 Alqahtani et al. Journal of Inequalities Applications 018) 018:158 Page 10 of 15 Case b): Let x = 1, y =1.Then =d θ T 1 ) 1, S1 = α, 1 )β1, 1)d θ T 1 ), S1 1 [ 1 d θ, T 1 ) ] + d θ 1, S1) = 1 [8 + 0] = 8. Case c): Let x = 1, y = 1.Then =d θ T 1, S 1 ) 1 = α 1, T 1 [ 1 d θ, T 1 ) 1 + d θ, S 1 ) 1 β )d θ T 1, S 1 ), S 1 )] = 1 [8 + ] = 10. Case d): Let x = 1, y = 1.Then 1 =α 1 ) 1 β, T 1 [ 1 d θ, T 1 )d θ T 1, S 1 ), S 1 ) + d θ 1, S 1 ] = 1 [8 + ] = 1. Therefore, all the conditions of Theorem. are satisfied T has a unique fixed point, x =1..1 α, β)-orbital-cyclic Definition. Let X be a nonempty set, T : X X, α, β : X X [0, ). We say that T is an α, β)-orbital-cyclic admissible mapping if αx, Tx) 1 impliesβ Tx, T x ) 1 βx, Tx) 1 impliesα Tx, T x ) 1 5) for all x X. Corollary.1 Let T be a self-mapping on a complete extended-bms X, d θ ) such that the mapping T forms an α, β)-orbital-cyclicadmissiblemapping. Suppose that i) for each x 0 X, lim n,m θx n, x m )< 1 k k, where x n = T n x 0, n =1,,...; ii) there exists x 0 X such that αx 0, Tx 0 ) 1 βx 0, Tx 0 ) 1; iii) either T iscontinuous, or iii*) if x n is a sequence in X such that x n u, then αu, Tu)) 1 βu, Tu) 1. Moreover, if for all x, y Xk [0, 1 ) αx, Tx)βy, Ty)d θ Tx, Ty) k [ d θ x, Tx)+d θ y, Ty) ], 6) then the pair of the mappings T possesses a fixed point u, that is, Tu = u. Proof It is sufficient to take S = T in Theorem.1.

11 Alqahtani et al. Journal of Inequalities Applications 018) 018:158 Page 11 of 15 Example. Let X = [0, ] define d θ : X X [0, )θ : X X [1, )by x y) if x, y [1, ], d θ x, y)= x y otherwise, respectively x + y +1 ifx, y [1, ], θx, y)= 1 otherwise. Let the self-map T : X X be defined by x if x [0, 1) 8 Tx)= x +3x ifx [1, ]. Define also α, β : X X [0, )by ifx, y [0, 1], αx, y)= 0 otherwise, 1 ifx, y [0, 1], βx, y)= ifx =,y =0, 0 otherwise. We show that T is α, β)-orbital-cyclic admissible. Let x, y X such that αx, Tx) 1 βx, Tx) 1. Then x, y [0, 1). On the other h, if x [0, 1), then Tx 1T x 1. It follows that αtx, T x) 1βTx, T x) 1. Thus, the assertion holds. For x =0,we have T0 =0α0, T0) 1, respectively, β0, T0) 1, so assumption ii) is satisfied. Let now {x n } be a sequence in X such that x n x. Then{x n } [0, 1] x [0, 1]. This implies that αx, Tx) 1. For x 0 [0, 1), we get T n x 0 = x 0 8 n lim n,m θt n x 0, T m x 0 )=1.Ifx 0 [1, ], Tx 0 1,lim n,m θt n x 0, T m x 0 ) = 1. So, assumption i) is satisfied for k = 1 3.Wehavethe following cases: a): For x, y [0, 1), we get d θ x, y)= x y, d θ Tx, Ty)= 1 x y, 8 d θ x, Tx)= 7 8 x, d θy, Ty)= 7 8 y. Replaced in 6)we get αx, Tx)βy, Ty)d θ Tx, Ty) 1 3 [d θ x, Tx)+d θ y, Ty) ]

12 Alqahtani et al. Journal of Inequalities Applications 018) 018:158 Page 1 of 15 or x y 1 [ 7x y ] = 8 7x +7y, which is true for any x, y [0, 1). b): For x =1y =,weknowthatα1, T1) = α1, 0) 1βT1, T 1) = β0, 0) 1, β, T) = β, 0) 1αT, T ) = α0, 0) 1. But in this case 6) isobvious, because d θ T1, T) = 0. c): For x [0, 1) y =,6)becomes αx, Tx)β, T)d θ Tx, T) 1 [ dθ x, Tx)+d θ, T) ] 3 or x = x 8 1 [ ] 7x = 7x +16. d): Forallothercases, αx, Tx) =0orβx, Tx) = 0, for this reason inequality 6) holds. Therefore, all the conditions of Corollary.1 are satisfied T has a fixed point, x =0. Corollary. Let T be a self-mapping on a complete extended-bms X, d θ ) such that T is an α-orbital admissible mapping. Suppose that i) for each x 0 X, lim n,m θx n, x m )< 1 k k, where x n = T n x 0, n =1,,...; ii) there exists x 0 X such that αx 0, Tx 0 ) 1; iii) either T iscontinuous, or iii*) if x n is a sequence in X such that x n u, then αu, Tu)) 1. Moreover, if for all x, y Xk [0, 1 ) αx, Tx)αy, Ty)d θ Tx, Ty) k [ d θ x, Tx)+d θ y, Ty) ], 7) then the pair of the mappings T possesses a fixed point u, that is, Tu = u. Proof It is sufficient to take βx, y)=αx, y) in Corollary.1. Example.3 Let X = [0, ] be endowed with extended b-metric d θ : X X [0, ), defined by d θ x, y)=x y),whereθ : X X [1, ), θx, y)=x + y +1.LetT : X X such that x+1 if x [0, 1], 3 Tx = x if x 1, ]. Define also α : X X 0, )as 1 ifx, y) {[0, 1 αx, y)= ] [0, 1 ]} {[ 1,1] [ 1,1]}, 0 otherwise.

13 Alqahtani et al. Journal of Inequalities Applications 018) 018:158 Page 13 of 15 We prove that Corollary. can be applied to T for k = 1,butTheorem1.1 cannot be applied to T. We show that T is an α-orbital admissible mapping. If x, y [0, 1 ], then Tx 1 T x 1. Thus, αx, Tx) 1impliesαTx, T x) 1. Similarly, we get that αx, Tx) 1 implies αtx, T x) 1 for all x, y [ 1,1], so T is α-orbital admissible. In reason of the above arguments, α0, T0) = α0, 1 ) 1. Thus, assertion ii) holds. 3 Note that, for each x 0 X, T n x 0 = n k=1 1 3 )n + x 0 3 n lim n T n x 0 = 1.Hence, lim θ T n x 0 ), T m x 0 ) ) = 1 1 k +1=<3= n,m k. So assumption i) is satisfied, because α 1, T 1 )=α 1, 1 ) 1, assumption iii*) is also satisfied. Let x, y [0, 1 ], or x, y [ 1,1].Wehave dtx, Ty)=αx, Tx)αy, Ty)dTx, Ty)= x y) 9 k [d θ x, Tx)+d θ y, Ty) ] = 1 [ x 1) + 9 ] y 1) 9 = x +y x y Replaced in inequality 8), we get x xy + y 9 or, equivalently, x +y x y +, 36 8xy x y + 0 x 1)y 1) 0. Hence, inequality 8) is satisfied. In other cases, inequality 8) is obviously satisfied, because αx, y) = 0. Therefore, all conditions of Corollary. are satisfied T has a unique fixed point, x = 1. Let x =y =3.Then d θ T, T3) = d θ 9, 16) = 5 > k = k d θ, 3) for any k <1.. Uniqueness Notice that in this section we investigate the existence of common) fixed points of certain operators. For the uniqueness of a fixed point of the observed results, we will consider the following hypothesis. H) For all x, y CFixT),wehaveαx, Tx) 1 βy, Sy) 1. Here, CFixT) denotes the set of common fixed points of T S. Theorem. Adding condition H) to the hypotheses of Theorem.1, we obtain that u is theuniquefixedpointoft.

14 Alqahtani et al. Journal of Inequalities Applications 018) 018:158 Page 1 of 15 Proof Suppose, on the contrary, that v is another fixed point of T.FromH),thereexists v X such that αu, Tu) 1 βv, Sv) 1. 8) Since T satisfies 9), weget that d θ Tu, Sv) αu, Tu)βv, Sv)d θ Tu, Sv) k [ d θ u, Tu)+d θ v, Sv) ], which yields that d θ u, v) 0. Since the inequality above is possible only if d θ u, v)=0,thatis,u = v. This is a contradiction. Thus we proved that u is the uniquefixed point of T. Notice also that instead of hypothesis H), one can suggest different conditions, see, e.g., [15]. 3 Conclusions It is clear that one can list several consequences from our results. By letting θx, y) =s, constant, with 1 s < k 1 in Theorem.1 analogously, in Corollary.1 Corollary.), k we get corresponding fixed point results in the setting of stard b-metric space. On the other h, regarding the techniques used in [15], one can derive another set of corollaries, by choosing the admissible mapping in a proper way. In this way, for example, several existing fixed point results in the literature in the setting of partially ordered metric spaces can be derived. Furthermore, the analogs of fixed point results for cyclic contractions can be found. Acknowledgements The first third authors extend their appreciation to Distinguished Scientist Fellowship Program DSFP) at King Saud University Saudi Arabia). The authors thank anonymous referees for their remarkable comments, suggestion, ideas that helped to improve this paper. Funding Not applicable. Competing interests The authors declare that they have no competing interests. Authors contributions All authors contributed equally significantly in writing this article. All authors read approved the final manuscript. Author details 1 Department of Mathematics, King Saud University, Riyadh, Saudi Arabia. Department of Mathematics Computer Sciences, Universitatea Transilvania Brasov, Brasov, Romania. 3 Department of Mathematics, Atilim University, Ankara, Turkey. Publisher s Note Springer Nature remains neutral with regard to jurisdictional claims in published maps institutional affiliations. Received: 10 April 018 Accepted: 30 May 018

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