Discussion on the fixed point problems with constraint inequalities

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1 Alqahtani et al. Journal of Inequalities and Applications (018) 018:4 R E S E A R C H Open Access Discussion on the fixed point problems with constraint inequalities Badr Alqahtani 1, Rahmatollah Lashkaripour,ErdalKarapınar 3* and Javad Hamzehnejadi * 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 introduce the concept of comparable complete metric spaces and consider some fixed point theorems for mappings in the setting of incomplete metric spaces. We obtain the results of Ansari et al. [J. Fixed Point Theory Appl. 0:6, 018] with weaker conditions. Moreover, we provide some corollaries and examples show that our main result is a generalization of existing results in the literature. Keywords: Comparable metric space; Fixed point; Generalized α-h-φ-contractions; Constraint inequalities 1 Introduction and preliminaries Let Y be a nonempty subset of a metric space (X, d) andt beafunctionthatmapy into itself. A fixed point of the mapping T is an element x Y for which Tx = x. Fixed point theory plays a crucial role in nonlinear functional analysis and many authors have studied this notion. In 19, Banach [7] reported the pioneer metric fixed point result for contraction mappings. Many authors have generalized this significant result in several directions; see e.g. [1 3, 8, 13]. Recently there have been many developments concerning the existence of fixed points for operators defined in a metric space equipped with a partial order. In 016, Jleli and Samet [10] provided sufficient conditions for the existence of a fixed point of T satisfying the two constraint inequalities Ax 1 Bx and Cx Dx, wheret : X X defined on a complete metric space equipped with two partial orders 1 and and A, B, C, D : X X are self-map operators. In the other words, the problem is to investigate the existence a point x X such that Tx = x; Ax 1 Bx; Cx Dx. (1.1) Before presenting the main result obtained in [10], let us recall some basic definitions and remarkable results introduced in [10](seealsoe.g.[4, 5, 9, 15, 16]). The Author(s) 018. This article is distributed under the terms of the Creative Commons Attribution 4.0 International License ( which permits unrestricted use, distribution, and reproduction in any medium, provided you give appropriate credit to the original author(s) and the source, provide a link to the Creative Commons license, and indicate if changes were made.

2 Alqahtani et al. Journal of Inequalities and Applications (018) 018:4 Page of 16 Definition 1.1 Let (X, d) be a metric space. A partial order onx is d-regular if for any two sequences u n } and v n } in X,wehave lim d(u n, u)= lim d(v n, v)=0, u n v n for all n u v. n n Definition 1. Let (X, ) beanorderedset.amappingt : X X is said to be - preserving if x y implies T(x) T(y). Definition 1.3 Let 1 and betwopartialordersonx and operators T, A, B, C, D : X X be given. The operator T is called (A, B, C, D, 1, )-stable if for all x X, Ax 1 Bx CTx DTx. Example 1.4 Let X = R and consider the standard order onx.leta, B, C, D : X X be the operators defined by Ax = x, Bx = x, Cx = exp(x), Dx = exp ( x x + ), Tx = x +1, x R. Then the operator T is (A, B, C, D,, )-stable. Let us denote by the set of all lower semi-continuous functions ψ :[0, ) [0, ) such that ψ 1 0} = 0}. The main theorem presented in [10] is given by the following result. Theorem 1.5 Let (X, d) be a complete metric space endowed with two partial orders 1 and. Let operators T, A, B, C, D : X Xbegiven. Suppose that the following conditions are satisfied: (i) i is d-regular, i =1,; (ii) A, B, C and D are continuous; (iii) there exists x 0 X such that Ax 0 1 Bx 0 ; (iv) T is (A, B, C, D, 1, )-stable; (v) T is (C, D, A, B,, 1 )-stable; (vi) there exists ψ such that (Ax 1 Bx and Cy Dx) d(tx, Ty) d(x, y) ψ ( d(x, y) ). Then the sequence T n x 0 } converges to some x X which is a solution to (1.1). Ansari et al. in [6]provedthatx is the unique solution to (1.1)andremovedthecontinuity of C and D. In our main theorem, we replace the completeness assumption of the space X with weaker conditions. Also we consider a more general condition in assumption (vi). For this purpose, we review the concept of generalized α-h-φ-contraction type mapping and some examples introduced in [14]. Also, we introduce new concepts to remove the completeness assumption of the space X.

3 Alqahtani et al. Journal of Inequalities and Applications (018) 018:4 Page 3 of 16 Definition 1.6 ([11]) Let T : X X be a mapping and α : X X [0, ) be a function. The mapping T issaid tobe α-admissible if α(x, y) 1 implies α(tx, Ty) 1. An α-admissible mapping T is said to be triangular α-admissible [1]if α(x, y) 1 and α(y, z) 1 imply α(x, z) 1. Lemma 1.7 ([11]) Let T : X Xbeatriangularα-admissible map. Assume that there exists x 0 Xsuchthatα(x 0, Tx 0 ) 1. Define a sequence x n } by x n+1 = Tx n. Then, we have α(x n, x m ) 1 for all m, n N with n < m. Definition 1.8 ([11]) Let (X, d) beametricspaceandα : X X [0, ) be a function. Asequencex n } is said to be α-regular if the following condition is satisfied: If x n } is a sequence in X such that α(x n, x n+1 ) 1 for all n N and x n x X as n, then there exists a subsequence x nk } of x n } such that α(x nk, x) 1 for all k. Recently we introduced a new class of mappings which contain a Geraghty-contraction type mapping and some of its extensions and some of weakly contractive type mappings as a subclass. Definition 1.9 ([14]) Let (X, d) beametricspace. DefineH(X) by the class of all mappings h : X X [0, 1) which satisfies the following condition: lim n h(x n, y n )=1 lim n d(x n, y n )=0, for all sequences x n } and y n } in X such that the sequence d(x n, y n )} is decreasing and convergent. Example 1.10 ([14]) Let h : R R [0, 1), defined by t (i) h 1 (x, y)=,forsomet [0, ). t+x +y (ii) h (x, y)=k,forsomek (0, 1). Then h 1, h H(R). Let F be the class of those functions β :[0, ) [0, 1) satisfying the following condition: β(t n ) 1 implies t n 0. Example 1.11 ([14]) Let (X, d) be a metric space and β F.Defineh 1, h : X X [0, 1), by h 1 (x, y)=β ( d(x, y) ) ; h (x, y)=β ( M a (x, y) ),

4 Alqahtani et al. Journal of Inequalities and Applications (018) 018:4 Page 4 of 16 where β F and for all x, y X M a (x, y)=max d(x, y), d(x, Tx), d(y, Ty), Then, h 1, h H. } d(x, Ty)+d(y, Tx). Definition 1.1 ([14]) Let (X, d) beametricspaceandα : X X R be a function. A mapping T : X X is said to be generalized α-h-φ-contraction if there exist h H(X) and φ such that α(x, y)φ ( d(tx, Ty) ) h(x, y)φ ( M a (x, y) ). One of extensions of the Banach contraction principle that extend, generalize, and improve some existing results, was given by Lashkaripour et al. as follows. Theorem 1.13 ([14]) Let (X, d) be a complete metric space, α : X X R be a function and T : X X be a mapping. Suppose that the following conditions are satisfied: (i) T is a generalized α-h-φ-contraction type mapping; (ii) T is triangular α-admissible; (iii) there exists x 1 X such that α(x 1, Tx 1 ) 1; (iv) T is continuous or for all sequences x n }, y n } X that α(x n, y n ) 0, n N, the following condition is satisfied: lim h(x n, y n )=1 lim d(tx n, Ty n )=0. n n Then T has a fixed point x X, and T n x 1 } converges to x. Next, we introduce the concept of comparable sequences and comparable complete metric spaces. Definition 1.14 Let (X, ) be an ordered space. A sequence x n } is called a comparable sequence, if ( n, k; x n x n+k ) or ( n, k; x n+k x n ). Example 1.15 Let X = R and consider the standard order onx. Then every monotone sequence is comparable sequence. Definition 1.16 Let (X,, d) be an ordered metric space. X is said to be comparable complete if every Cauchy comparable sequence is convergent. It is easy to see that every complete metric space is comparable complete and that the converse is not true. In the next example, X is comparable complete but it is not complete. Example 1.17 Let X = Q.Supposethat x y k, s N 0} : x y =10 k( 1 10 s).

5 Alqahtani et al. Journal of Inequalities and Applications (018) 018:4 Page 5 of 16 Clearly, Q with the Euclidean metric is not a complete metric space, but it is comparable complete metric space. If x n } is an arbitrary Cauchy comparable sequence in X, then the sequence is convergent in R.Weprovethatx is a rational number. In the contrary case let x Q c.sincex n } is a comparable sequence, for all m, n N there exist k, s N 0} such that x n x m =10 k( 1 10 s). Suppose that m, then there exists r Q such that x n x =10 r, which is a contradiction. Therefore the space Q with this order is a comparable complete metric space. Note that for all x Q there exists a comparable sequence x n } Q such that lim n x n = x. Definition 1.18 Let (X,, d) be an ordered metric space. A mapping f : X X is comparable continuous in a X if for each comparable sequence a n } in X if a n a, then f (a n ) f (a). Also, f is comparable continuous on X if f is comparable continuous in each a X. Every continuous function is a comparable continuous function, but the converse is not true in general. Example 1.19 Let X = R with the Euclidean metric and usual order. Let f : R R defined by f (x)=[x]. The function f is not a continuous function. Define the relation on R as follows: x y x y. It is easy to see that the function f is a comparable continuous function. Definition 1.0 Let (X, ) beanorderedspaceandt : X X be a mapping. x 0 X is said to be T-comparable if for all n N, x 0 and T n x 0 be comparable and define J T = x 0 X; ( n N : x 0 T n x 0 ) or ( n N; T n x 0 x 0 )}. Example 1.1 Let X = R with the Euclidean metric and usual order. If define T : X X by T(x)=x,thenJ T = R.Alsoif 1 x 0, g(x)= 1 x <0, then J } =. Proposition 1. Let (X, ) be an ordered set and T : X Xbe -preserving. Let x n } be Picard iterative sequence with initial point x 0 J T, i.e. x n = T n (x 0 ). Then x n } is a comparable sequence. Proof Let n, k N and for all k N, x 0 T k x 0 = x k.sincet is -preserving, x 1 = Tx 0 Tx n = x k+1. Inductively for all n N we can prove that x n x n+k.

6 Alqahtani et al. Journal of Inequalities and Applications (018) 018:4 Page 6 of 16 Main result Let be the family of functions φ :[0, ) [0, ) satisfying the following conditions: (1) φ is continuous and non-decreasing; () φ(t)=0if and only if t =0. In the following theorem, which is our first main result, we weaken assumption (ii) and (vi) of Theorem 1.5. Moreover, we remove the completeness assumption of the space in Theorem 1.5. Theorem.1 Let (X, d, ) be a comparable complete metric space(not necessarily complete). Let 1 and be two partial order over X. Also, let operators T, A, B, C, D : X X be given. Suppose that the following conditions are satisfied: (i) i is d-regular, i =1,and T is -preserving and triangular α-admissible; (ii) A, B and T are comparable continuous; (iii) there exists x 0 J T such that Ax 0 1 Bx 0 and α(x 0, Tx 0 ) 1; (iv) T is (A, B, C, D, 1, )-stable; (v) T is (C, D, A, B,, 1 )-stable; (vi) there exist h H(X) and φ such that (Ax 1 Bx and Cy Dy) α(x, y)φ ( d(tx, Ty) ) h(x, y)φ ( M a (x, y) ). Then the sequence T n x 0 converges to some x X which is a solution to (1). Proof From condition (iii), there exists x 0 J T such that Ax 0 1 Bx 0 and α(x 0, Tx 0 ) 1. Define the sequence x n } by x n = Tx n 1, for all n N. Applying Proposition 1., x n } is a comparable sequence. If x n0 = x n0 +1 for some n 0 N, thentx n0 = x n0 +1 = x n0,and hence the proof is completed. Now, let x n x n+1, n =0,1,,...SinceAx 0 1 Bx 0 and T is (A, B, C, D, 1, )-stable, we have Ax 0 1 Bx 0 CTx 0 DTx 0, that is, Cx 1 Dx 1.Hence Ax 0 1 Bx 0 and Cx 1 Dx 1. Continuing this process, by induction, for all n N we get Ax n 1 Bx n and Cx n+1 Dx n+1. (.1) Also, applying Lemma 1.7 for all m, n N with n < m,wehave α(x n, x m ) 1. (.)

7 Alqahtani et al. Journal of Inequalities and Applications (018) 018:4 Page 7 of 16 Since x n } is comparable, applying (.1), (.) and (vi), by symmetry, for n =1,,..., we have φ ( d(x n, x n+1 ) ) α(x n 1, x n )φ ( d(x n, x n+1 ) ) = α(x n 1, x n )φ ( d(tx n 1, Tx n ) ) h(x n 1, x n )φ ( M a (x n 1, x n ) ) < φ ( M a (x n 1, x n ) ). (.3) Also, we have M a (x n 1, x n )=max d(x n 1, x n ), d(x n 1, Tx n 1 ), d(x n, Tx n ), d(x } n 1, Tx n )+d(x n, Tx n 1 ) = max d(x n 1, x n ), d(x n 1, x n ), d(x n, x n+1 ), d(x } n 1, x n+1 )+d(x n, x n ) = max d(x n 1, x n ), d(x n, x n+1 ), d(x } n 1, x n+1 ) max d(x n 1, x n ), d(x n, x n+1 ), d(x } n 1, x n )+d(x n, x n+1 ) = max d(x n 1, x n ), d(x n, x n+1 ) }. If M a (x n 1, x n )=d(x n, x n+1 ), applying (.3), we deduce that φ ( d(x n, x n+1 ) ) < φ ( M a (x n 1, x n ) ) = φ ( d(x n, x n+1 ) ), which is a contradiction. Thus, we conclude that M a (x n 1, x n )=d(x n 1, x n ), n N. (.4) Now, from (.3)and(.4), we get φ ( d(x n, x n+1 ) ) < φ ( d(x n 1, x n ) ), n N. The monotony of φ implies that d(x n, x n+1 )<d(x n 1, x n ), n N. We deduce that the sequence d(x n, x n+1 )} is nonnegative and decreasing. Consequently, there exists r 0suchthatlim n d(x n, x n+1 )=r. Weprovethatr =0.Inthecontrary case, suppose that r >0.Thenfrom(.3)and(.4), we have 0< φ(d(x n, x n+1 )) φ(d(x n 1, x n )) h(x n 1, x n ),

8 Alqahtani et al. Journal of Inequalities and Applications (018) 018:4 Page 8 of 16 which implies that lim n h(x n 1, x n )=1.Sinceh H, lim d(x n 1, x n )=0. n This implies that r =0, which is acontradiction. Therefore lim n d(x n, x n+1 )=0. Now, we shall prove that x n } is a Cauchy sequence in comparable complete metric space (X,, d). Suppose, on the contrary, that x n } is not a Cauchy sequence. Thus, there exists ɛ > 0 such that, for all k N,thereexistn k > m k > k such that d(x mk, x nk ) ɛ. Also, choosing m k as small as possible, it may be assumed that d(x mk, x nk 1)<ɛ. Hence for each k N,wehave ɛ d(x mk, x nk ) d(x mk, x nk 1)+d(x nk 1, x nk ) ɛ + d(x nk 1, x nk ). Letting k in the above inequality, we get lim d(x n n k, x mk )=ɛ. Thetriangleinequalityimpliesthat lim n d(x n k +1, x mk )=ɛ, lim d(x n n k, x mk 1)=ɛ, lim d(x n n k +1, x mk +1)=ɛ. (.5) We see that, for all k N,thereexistsi k 0, 1} such that n k m k + i k 1(). Now, applying (.1), for all k >1,wededucethat Ax nk 1 Bx nk and Cx mk i k Dx mk i k, or Ax mk i k 1 Bx mk i k and Cx nk Dx nk. Now, applying (vi), for k N,weconcludethat φ ( d(x nk +1, x mk i k +1) ) α(x nk, x mk i k )φ ( d(x nk +1, x mk i k +1) ) = α(x nk, x mk i k )φ ( d(tx nk, Tx mk i k ) ) h(x nk, x mk i k )φ ( M a (x nk, x mk i k ) ). (.6)

9 Alqahtani et al. Journal of Inequalities and Applications (018) 018:4 Page 9 of 16 Also, for any k N,wehave M a (x nk, x mk i k )=max d(x nk, x mk i k ), d(x nk, Tx nk ), d(x mk i k, Tx mk i k ), } d(x nk, Tx mk i k )+d(x mk i k, Tx nk ) = max d(x k, x mk i k ), d(x nk, x nk +1), d(x mk i k, x mk i k +1), } d(x nk, x mk i k +1)+d(x mk i k, x nk +1) Since lim k d(x nk, x nk +1)=0, max d(x nk, x mk i k ), d(x nk, x nk +1), d(x mk i k, x mk i k +1), d(x nk, x mk i k )+d(x mk i k, x mk i k +1) + d(x } m k i k, x nk )+d(x nk, x nk +1). lim M a(x nk, x mk i k )= lim d(x n k, x mk i k ). (.7) k k Combining (.6) and(.7) with the continuity of φ,we get lim φ( d(x nk +1, x mk i k +1) ) lim h(x n k, x mk i k ) lim φ( d(x nk, x mk i k ) ). k k k Applying (.5), we deduce that lim h(x n k, x mk i k )=1. k Since h H(X), lim k d(x n k, x mk i k )=0, which is a contradiction. Thus, x n } is Cauchy comparable and so there exists x X such that lim n x n = x.sincet is a comparable continuous function, lim x n+1 = lim Tx n = Tx. n n Therefore Tx = x. (.8) A and B are comparable continuous and x n } is a comparable sequence, therefore lim d( Ax n, Ax ) = lim d(bx n, Bx n )=0. n n Since 1 is d-regular, (.1)impliesthat Ax 1 Bx. (.9)

10 Alqahtani et al. Journal of Inequalities and Applications (018) 018:4 Page 10 of 16 Since T is (A, B, C, D, 1, )-stable, applying (.9), we have CTx DTx. This implies that Cx Dx. (.10) Applying (.8), (.9)and(.10), we deduce that x is a solution of (.1). In the following theorem, we omit the continuity condition of the mapping T in Theorem.1. Theorem. Let (X, d, ) be a comparable complete metric space(not necessarily complete). Let 1 and be two partial order over X. Also, let operators T, A, B, C, D : X X be given. Suppose that the following conditions are satisfied: (i) i is d-regular, i =1,and T is -preserving and triangular α-admissible; (ii) A, B are comparable continuous; (iii) there exists x 0 J T such that Ax 0 1 Bx 0 and α(x 0, Tx 0 ) 0; (iv) the sequence T n x 0 } is α-regular; (v) T is (A, B, C, D, 1, )-stable and (C, D, A, B,, 1 )-stable; (vi) there exist h H(X) and φ such that for all sequences x n }, y n } X where α(x n, y n ) 0, n N, the following conditions are satisfied: lim h(x n, y n )=1 lim d(tx n, Ty n )=0; n n (Ax 1 Bx and Cy Dy) α(x, y)φ ( d(tx, Ty) ) h(x, y)φ ( M a (x, y) ). Then T has a fixed point x X, and T n x 0 } converges to x. Proof From condition (iii), there exists x 0 J T such that Ax 0 1 Bx 0 and α(x 0, Tx 0 ) 1. Define the sequence x n } by x n = Tx n 1, for all n N. Following the proof of Theorem.1, we know that, for n =0,1,..., Ax n 1 Bx n, Cx n+1 Dx n+1 and α(x n, x n+1 ) 1, (.11) and the sequence x n } is convergent to some x X.Also,wehave Ax 1 Bx. (.1) Now, we prove that Tx = x. In the contrary case suppose that Tx x.sincethesequence x n } is α-regular, there exists a subsequence x nk } such that α(x nk, x ) 1 for all k N. Without loss of generality, we assume that α ( x n, x ) 1, n =0,1,,... (.13)

11 Alqahtani et al. Journal of Inequalities and Applications (018) 018:4 Page 11 of 16 Applying (.11), (.13), for n =0,1,...,weget φ ( d ( x n+1, Tx )) = φ ( d ( Tx n, Tx )) α ( x n, x ) φ ( d ( Tx n, Tx )) h ( x n, x ) φ ( ( M a xn, x )). (.14) Also, we have ( M a xn, x ) = max d ( x n, x ), d(x n, Tx n ), d ( x, Tx ), d(x n, Tx )+d(x, Tx n ) = max d ( x n, x ), d(x n, x n+1 ), d ( x, Tx ), d(x n, Tx )+d(x, x n+1 ) Since lim n d(x n, x )=0,lim n M a (x n, x )=d(x, Tx ). Applying (.14)andthecontinuity of φ,wegetlim n h(x n, x )=1,andso } }. d ( x, Tx ) = lim n d( Tx n, Tx ) =0. This is a contradiction. Therefore Tx = x.sincet is (A, B, C, D, 1, )-stable, applying (.1), we have CTx DTx. Therefore, Cx Dx.Thisimpliesthatx is a solution of (1.1). Forthe uniqueness of the solution of (1.1) we will consider the following condition. (H1) For all x, y Fix(T),thereexistsz X such that α(x, z) 1 and α(z, y) 1. Theorem.3 Adding condition (H1) to the hypotheses of Theorem.1 (resp. Theorem.), we see that x is the unique fixed point of T. Proof Let y X be another solution of (1.1), that is, Ty = y, Ay 1 By, Cy Dy. (.15) we show that x = y.inthecontrarycase,letx y.thereexistsz X such that α ( x, z ) 1 and α ( z, y ) 1. Since T is triangular α-admissible, we have α(x, y ) 1. Now, Ax 1 Bx and Cy Dy, which implies that φ ( d ( x, y )) = φ ( d ( Tx, Ty )) α ( x, y ) φ ( d ( Tx, Ty )) h ( x, y ) φ ( ( M a x, y )) < φ ( M a ( x, y )). (.16)

12 Alqahtani et al. Journal of Inequalities and Applications (018) 018:4 Page 1 of 16 On the other hand, we have ( M a x, y ) = max d ( x, y ), d ( x, Tx ), d ( y, Ty ), d(x, Ty )+d(y, Tx } ) = d ( x, y ). (.17) Applying (.16) and(.17), we have φ(d(x, y )) < φ(d(x, y )), which is a contradiction. This implies that x = y, and so the fixed point of T is unique. Example.4 Let X = [, 3) and define relation onr as follows: x y [x]=[y] and x y. The space X with the Euclidean metric is not a complete metric space, but it is comparable complete metric space. We take 1 = =.LetT : X X be the mapping defined by T(x)= 1 ( x [x] ), x X. For all x, y X such that x y, wehavetx Ty. Therefore T is -preserving. consider the mappings A, B, C, D : X X defined by D(x)= 4x +1, x x 0, 5 x 1, 4 A(x)= B(x)= x + x <0, 1 x <1, 4 x x 1, 4x +1 x 0, C(x)= D(x)= 0 x <1, x x<0. Obviously, i isd-regular, i =1,.Moreover,A and B are comparable continuous mappings. If for some x X, wehaveax Bx, thenx [0, 1 4 ] [1, 5 ] which implies that 4 Tx [0, 1 ]. Therefore 8 C(Tx)=0 4Tx +1=DTx. Thus T is (A, B, C, D, 1, )-stable. If for some x X, wehavecx Dx then x [0, 1 4 ], which implies that Tx [0, 1 ]. Therefore 8 ATx = Tx 1 4 = BTx. Thus T is (C, D, A, B,, 1 )-stable. Define h : X X [0, 1) and α : X X R as follows: 1 [x]=[y], α(x, y)= and h(x, y)= 1 0 otherwise,.

13 Alqahtani et al. Journal of Inequalities and Applications (018) 018:4 Page 13 of 16 If Ax Bx, Cy Dy and α(x, y)=1,thenx, y [0, 1 ]. Therefore 4 α(x, y)d(tx, Ty) = 1 x y = h(x, y)d(x, y) h(x, y)m a (x, y). Let φ(t)=t, t 0. Therefore (Ax 1 Bx and Cy Dx) α(x, y)φ ( d(tx, Ty) ) h(x, y)φ ( M a (x, y) ). The hypotheses of Theorem.1 are satisfied. Therefore (1.1) hastheuniquesolution x =0. Note that the mappings A, B and T are not continuous and (X, d) is not a complete metric space. 3 Consequences Now, we consider some special cases, where in our result we deduce several well-known fixed point theorems of the existing literature. Corollary 3.1 ([6]) Let (X, d) be a complete metric space endowed with two partial orders 1 and. Let T, A, B, C, D : X X be given operators. Suppose that the following conditions are satisfied: (i) i is d-regular, i =1,; (ii) A and B are continuous; (iii) there exists x 0 X such that Ax 0 1 Bx 0 ; iv T is (A, B, C, D, 1, )-stable; (v) T is (C, D, A, B,, 1 )-stable; (vi) there exists ψ such that Ax 1 Bx, Cy Dy d(tx, Ty) d(x, y) ψ ( d(x, y) ). (3.1) Then the sequence T n x 0 } converges to some x X, which is a solution to (1.1). Moreover, x istheuniquesolutionto(1.1). Proof Define h : X X [0, 1), by d(x,y) ψ(d(x,y) if x y, d(x,y) h(x, y)= 0 ifx = y. (3.) Let x n }, y n } X be such that sequence d(x n, y n )} is decreasing and lim n d(x n, y n )=r. Suppose that lim n h(x n, y n )=1.Weshowthatlim n d(x n, y n ) = 0. In the contrary case, let lim n d(x n, y n )=r >0.Sinceψ is lower semi-continuous, d(x n, y n ) ψ(d(x n, y n ) lim sup h(x n, y n )=lim sup = r ψ(r) =1, n n d(x n, y n ) r

14 Alqahtani et al. Journal of Inequalities and Applications (018) 018:4 Page 14 of 16 which implies that ψ(r)=0,and so r =0.This is a contradiction.therefore lim d(x n, y n )=0. n This implies that h H(X). Let, for some x, y X, Ax 1 Bx, Cy Dy. Then applying (3.1)and(3.)weconcludethat d(tx, Ty) h(x, y)d(x, y) h(x, y)m a (x, y). Also for all x, y X define α(x, y) = 1. The hypotheses of Theorem.1 are satisfied. Hence there exists a unique x X such that x istheuniquesolutionto(1.1). In Theorem.1, by setting 1 =, C = B and D = A, we get the following corollary. Corollary 3. Let (X,, d) be a comparable complete metric space(not necessarily complete) with partial order 1. Also, let operators T, A, B : X Xbegiven. Suppose that the following conditions are satisfied: (i) 1 is d-regular and T is -preserving and triangular α-admissible; (ii) A, B and T are comparable continuous; (iii) there exists x 0 J T such that Ax 0 1 Bx 0 and α(x 0, Tx 0 ) 1; (iv) for all x X, we have Ax 1 Bx BTx 1 ATx; (v) for all x X, we have Bx 1 Ax ATx 1 BTx; (vi) there exist h H(X) and φ such that (Ax 1 Bx and By 1 Ay) α(x, y)φ ( d(tx, Ty) ) h(x, y)φ ( M a (x, y) ). Then the sequence T n x 0 } converges to some x XsatisfyingTx = x and Ax = Bx. By setting A = D = I x and C = B we have the following common fixed point theorem. Corollary 3.3 Let (X,, d) be a comparable complete metric space(not necessarily complete) with partial order 1. Also, let operators T, A, B : X Xbegiven. Suppose that the following conditions are satisfied: (i) 1 is d-regular and T is -preserving and triangular α-admissible; (ii) B and T are comparable continuous; (iii) there exists x 0 J T such that x 0 1 Bx 0 and α(x 0, Tx 0 ) 1; (iv) for all x X, we have x 1 Bx BTx 1 Tx;

15 Alqahtani et al. Journal of Inequalities and Applications (018) 018:4 Page 15 of 16 (v) for all x X, we have Bx 1 x Tx 1 BTx; (vi) there exist h H(X) and φ such that (x 1 Bx and By 1 y) α(x, y)φ ( d(tx, Ty) ) h(x, y)φ ( M a (x, y) ). Then the sequence T n x 0 } converges to some x XsatisfyingTx = x and Bx = x. 4 Conclusions In this note, we replace the completeness assumption of the space X with a weaker condition by introducing the concept of comparable complete metric spaces. So, we address a fixed point in the setting of incomplete metric spaces by using the constraint inequalities. Acknowledgements The authors thanks to anonymous referees for their remarkable comments, suggestion and ideas that helps to improve this paper. The first and third authors extend their appreciation to Distinguished Scientist Fellowship Program (DSFP) at King Saud University (Saudi Arabia). Funding We declare that funding is not applicable for our paper. Competing interests The authors declare that they have no competing interests. Authors contributions All authors contributed equally and significantly in writing this article. All authors read and approved the final manuscript. Author details 1 Department of Mathematics, King Saud University, Riyadh, Saudi Arabia. Department of Mathematics, University of Sistan and Baluchestan, Zahedan, Iran. 3 Department of Mathematics, Atilim University, Ankara, Turkey. Publisher s Note Springer Nature remains neutral with regard to jurisdictional claims in published maps and institutional affiliations. Received: 4 July 018 Accepted: 16 August 018 References 1. Aksoy, U., Karapinar, E., Erhan, Ý.M.: Fixed points of generalized alpha-admissible contractions on b-metric spaces with an application to boundary value problems. J. Nonlinear Convex Anal. 17(6), (016). Almezel, S., Chen, C.-M., Karapinar, E., Rakocevíc, V.: Fixed point results for various α-admissible contractive mappings on metric-like spaces. Abstr. Appl. Anal. 014,Article ID (014) 3. Alsulami, H., Gulyaz, S., Karapinar, E., Erhan, I.M.: Fixed point theorems for a class of alpha-admissible contractions and applications to boundary value problem. Abstr. Appl. Anal. 014, Article ID (014) 4. Altun, I., Al Arifi, N., Jleli, M., Lashin, A., Samet, B.: A fixed point theorem for JS-contraction type mappings with applications to polynomial approximations. Filomat 31,15 (017) 5. Ansari, A.H., Jacob, G.K., Samet, B.: An optimization problem under partial order constraints on a metric space. J. Fixed Point Theory Appl. 0(1), 6 (018) 6. Ansari, A.H., Kumam, P., Samet, B.: A fixed point problem with constraint inequalities via an implicit contraction. J. Fixed Point Theory Appl. 017, (017) 7. Banach, S.: Sur les opérations dans les ensembles abstraits et leur application aux équations intégrales. Fundam. Math. 3, (19) 8. Hammache, K., Karapinar, E., Ould-Hammouda, A.: On admissible weak contractions in b-metric-like space. J. Math. Anal. 8(3), (017) 9. Jleli, M., Karapinar, E., Samet, B.: On the approximation of solutions to a fixed point problem with inequality constraints in a Banach space partially ordered by a cone. In: Advances in Nonlinear Analysis via the Concept of Measure of Noncompactness, pp Springer, Singapore (017) 10. Jleli, M., Samet, B.: A fixed point problem under two constraint inequalities. Fixed Point Theory Appl. 016, 18 (016) Karapinar, E.: A discussion on α-ψ-geraghty contraction type mappings. Filomat 8(4), (014).

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