Common fixed point theorems for generalized (φ, ψ)-weak contraction condition in complete metric spaces

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1 Murthy et al. Journal of Inequalities Applications (015) 015:139 DOI /s y R E S E A R C H Open Access Common fixed point theorems for generalized (φ, ψ)-weak contraction condition in complete metric spaces Penumarthy Parvateesam Murthy 1, Kenan Tas * Uma Devi Patel 1 * Correspondence: kenan@cankaya.edu.tr Department of Mathematics Computer Science, Cankaya University, Ankara, 06790, Turkey Full list of author information is available at the end of the article Abstract The intent of this manuscript is to establish some common fixed point theorems in a complete metric space under weak contraction condition for two pairs of discontinuous weak compatible maps. The results proved herein are the generalization of some recent results in literature. We give an example to support our results. MSC: 47H10; 54H5 Keywords: fixed points; (φ, ψ)-weak contraction condition; altering distance function; weak compatible maps; complete metric spaces 1 Introduction Let (X, d) be a metric space. A function T :(X, d) (X, d) is said to satisfy the contraction condition if there exists a real number k (0, 1) for all x, y X such that d(tx, Ty) kd(x, y). (1.1) It is remarkable that a self-map T satisfying condition (1.1)impliesthatT is continuous (see [1]). One can see in the literature on metric fixed point theory that condition (1.1)has been extended generalized by fixed point theorists in many ways for obtaining fixed points, common fixed points, very recently, proximal points in different spaces. The title of this paper implies that the pair of maps in a complete metric space satisfies certain inequality by generalizing the contraction condition of Banach, but the inequality itself does not force the mapping to be continuous. It was an open question for almost four decades after Banach s fixed point theorem [1]. In 1968, Kannan [] answered this question affirmatively by introducing the following inequality: d(tx, Ty) β [ d(x, Tx)+d(y, Ty) ] ( for all x, y X β 0, 1 ). (1.) Rakotch [3] later Boyd Wond [4] generalized the inequality of Banach by introducing a control function asfollows: d(tx, Ty) α(x, y)d(x, y) for all x, y X α :[0, ) [0, 1) (1.3) 015 Murthy et al.; licensee Springer. This article is distributed under the terms of the Creative Commons Attribution 4.0 International License ( which permits unrestricted use, distribution, reproduction in any medium, provided you give appropriate credit to the original author(s) the source, provide a link to the Creative Commons license, indicate if changes were made.

2 Murthyetal. Journal of Inequalities Applications (015) 015:139 Page of 14 d(tx, Ty) φ ( d(x, y) ) for all x, y X φ :[0, ) [0, ), (1.4) it satisfies certain conditions. In a Hilbert space, Alber Guerre-Delabriere [5]introduced a weak contraction condition, which was later extended by Rhoades [6]in a complete metric space, they argued that the result of Alber Guerre-Delabriere [5]still holds in a complete metric space. A mapping T : X X satisfies the following condition: d(tx, Ty) d(x, y) ϕ ( d(x, y) ) (1.5) for all x, y X, ϕ :[0, ) [0, ) is a continuous nondecreasing function such that ϕ(t)=0ifonlyift =0. Remark If ϕ(t)=(1 k)t,wherek (0, 1) in (1.5), then we shall have the condition (1.1)of Banach [1]. So, in view of (1.1), condition (1.5) is a weaker condition, we call this condition a weak contraction condition. Definition 1.Altering distance function [7]) A function ψ :[0, ) [0, ) is called an altering distance function if it satisfies the following conditions: (1) ψ is monotone increasing continuous, () ψ(t)=0if only if t =0. Definition 1. ([8]) A pair of self-mappings A B of a metric space (X, d) issaidtobe weakly compatible if they commute at their coincidence points. In other words, if Ax = Bx for some x X,thenABx = BAx. The first second author of this paper with Choudhury [9] proved a common fixed point theorem in Saks spaces. The main purpose of this note is to establish a few common fixed point theorems by generalizing the results of Murthy et al. [9] for two pairs of discontinuous functions in a complete metric space by using a weaker condition than condition (1.5) called (ϕ, ψ)-weak contraction condition in metric spaces. For example, refer to Rhoades [6], Dutta Choudhury [10], Zhang Song [11], Doric [1], Hosseini [13], Abkar Choudhury [14], Ahmad et al. [15], Hussain et al. [16], etc. Fixed point theorems in a complete metric space Theorem.1 Let (X, d) be a complete metric space, let A, B, ST: X Xbe mappings satisfying ψ ( d(ax, By) ) ψ ( M(x, y) ) φ ( N(x, y) ) (.1) for all x, y X, with x y M(x, y)=max d(sx, Ty), ) ) } d(sx, Ax)+d(Ty, By), d(sx, By)+d(Ty, Ax)

3 Murthyetal. Journal of Inequalities Applications (015) 015:139 Page 3 of 14 N(x, y)=min d(sx, Ty), ) ) } d(sx, Ax)+d(Ty, By), d(sx, By)+d(Ty, Ax), A(X) T(X) B(X) S(X), (.) (A, S) (B, T) areweakcompatiblepairs, (.3) φ :[0, ) [0, ) is such that φ(t)>0, (.4) which is lower semi-continuous for all t >0 φ is discontinuous at t =0with φ(0) = 0, ψ :[0, ) [0, ) is an altering distance function. (.5) Then A, B, S T have a unique common fixed point in X. Proof Let x 0 X be an arbitrary point. Since A(X) T(X) B(X) S(X), then there exists a point x 1 X such that Ax 0 = Tx 1,forx 1 X, thereexistsapointx X such that Bx 1 = Sx. Inductively, we can construct a sequence y n+1 = Ax n = Tx n+1, y n+ = Bx n+1 = Sx n+ for n =0,1,,... We assume, for all n N 0}, y n y n+1. (.6) At first, we shall show that d(y n, y n+1 ) 0asn for all n N 0},whereN is a set of natural numbers. For this, suppose that x = x n, y = x n+1 in (.1), we have ψ ( d(ax n, Bx n+1 ) ) = ψ ( d(y n+1, y n+ ) ) ψ ( M(x n, x n+1 ) ) φ ( N(x n, x n+1 ) ), (.7) where M(x n, x n+1 )=max d(y n, y n+1 ), ( 1 d(yn, y n+1 )+d(y n+1, y n+ ) ), d(yn, y n+ )+d(y n+1, y n+1 ) )} N(x n, x n+1 )=min d(y n, y n+1 ), ( 1 d(yn, y n+1 )+d(y n+1, y n+ ) ), d(yn, y n+ )+d(y n+1, y n+1 ) )}.

4 Murthyetal. Journal of Inequalities Applications (015) 015:139 Page 4 of 14 Then, by the triangular inequality, we have If M(x n, x n+1 ) max d(y n, y n+1 ), ( 1 d(yn, y n+1 )+d(y n+1, y n+ ) ), d(yn, y n+1 )+d(y n+1, y n+ ) )}. d(y n, y n+1 )<d(y n+1, y n+ ), (.8) then we get M(x n, x n+1 ) d(y n+1, y n+ ), (.9) (.7) implies the following: ψ ( d(y n+1, y n+ ) ) ψ ( M(x n, x n+1 ) ) φ ( N(x n, x n+1 ) ) ψ ( M(x n, x n+1 ) ). Using the monotonically increasing property of ψ function, we have d(y n+1, y n+ ) M(x n, x n+1 ). (.10) From (.9)(.10), we have M(x n, x n+1 )=d(y n+1, y n+ ). (.11) Since 0< d(yn+1, y n+ ) d(y n, y n+1 ) d(yn, y n+ ), (.1) we have N(x n, x n+1 )>0,thenfrom(.7), (.11) thepropertyofφ ψ functions, we have ψ ( d(y n+1, y n+ ) ) ψ ( d(y n+1, y n+ ) ) φ ( N(x n, x n+1 ) ) < ψ ( d(y n+1, y n+ ) ), which is a contradiction. Thus we have d(y n+1, y n+ ) d(y n, y n+1 ). (.13) So, we obtain the following: M(x n, x n+1 )=d(y n, y n+1 ), (.14) N(x n, x n+1 )= d(yn, y n+ ) ). (.15)

5 Murthyetal. Journal of Inequalities Applications (015) 015:139 Page 5 of 14 Now putting (.14)(.15)in(.7), we have ψ ( d(y n+1, y n+ ) ) ψ ( d(y n, y n+1 ) ) φ ( N(x n, x n+1 ) ). (.16) Therefore d(y n, y n+1 ) is a monotonically decreasing sequence of nonnegative real numbers, then there exists a number r >0suchthat lim d(y n, y n+1 )=r >0. (.17) n By virtue of (.6) (.1), we have N(x n, x n+1 )>0.Takingn in (.16) using (.17), we get lim ψ( d(y n+1, y n+ ) ) lim ψ( d(y n, y n+1 ) ) lim φ( N(x n, x n+1 ) ). n n n This implies that ψ(r) ψ(r) lim n φ( N(x n, x n+1 ) ). We observe that the last term on the right-h side of the above inequality is non-zero. We get a contradiction with φ function. Therefore, we have lim n d(y n, y n+1 )=0. Putting x = x n+1 y = x n+ in (.1) arguing as above, we obtain lim d(y n+1, y n+ )=0. n Therefore, for all n N 0},wehave lim d(y n, y n+1 )=0. (.18) n Next, we prove that y n } is a Cauchy sequence. For this, it is enough to show that the subsequence y n } is a Cauchy sequence. To the contrary, suppose that y n } is not a Cauchy sequence, then there exist ɛ > 0 the sequence of natural numbers n(k)} m(k)} such that n(k)>m(k)>k for k N d(y m(k), y n(k) ) ɛ (.19) corresponding to m(k).we can choose n(k) to be the smallest such that (.19)is satisfied. Then we have d(y m(k), y n(k) 1 )<ɛ. (.0) Putting x = x m(k) 1 y = x n(k) 1 in (.1), where for all k N, ψ ( d(y m(k), y n(k) ) ) ψ ( M(x m(k) 1, x n(k) 1 ) ) φ ( N(x m(k) 1, x n(k) 1 ) ), (.1)

6 Murthyetal. Journal of Inequalities Applications (015) 015:139 Page 6 of 14 where M(x m(k) 1, x n(k) 1 )=max d(y m(k) 1, y n(k) 1 ), d(ym(k) 1, y m(k) )+d(y n(k) 1, y n(k) ) ), d(ym(k) 1, y n(k) )+d(y n(k) 1, y m(k) ) )} N(x m(k) 1, x n(k) 1 )=min d(y m(k) 1, y n(k) 1 ), Using the triangle inequality, we have d(ym(k) 1, y m(k) )+d(y n(k) 1, y n(k) ) ), d(ym(k) 1, y n(k) )+d(y n(k) 1, y m(k) ) )}. d(y m(k), y n(k) ) d(y m(k), y n(k) 1 )+d(y n(k) 1, y n(k) ). Taking the limit k,weget lim d(y m(k), y n(k) )=ɛ. (.) k Again for all k,wehave d(y m(k) 1, y n(k) 1 ) d(y m(k), y m(k) 1 )+d(y m(k), y n(k) )+d(y n(k) 1, y n(k) ), d(y m(k), y n(k) ) d(y m(k), y m(k) 1 )+d(y m(k) 1, y n(k) 1 )+d(y n(k) 1, y n(k) ). On letting limit k using (.18)-(.), we get lim d(y m(k) 1, y n(k) 1 )=ɛ. (.3) k Again for all positive integers k,wehave d(y m(k) 1, y n(k) ) d(y m(k) 1, y m(k) )+d(y m(k), y n(k) ), d(y m(k), y n(k) ) d(y m(k), y m(k) 1 )+d(y m(k) 1, y n(k) ). Letting limit k using (.18)-(.3), we get lim d(y m(k) 1, y n(k) )=ɛ. (.4) k Again for all positive integers k,wehave d(y n(k) 1, y m(k) ) d(y n(k) 1, y n(k) )+d(y n(k), y m(k) ), d(y n(k), y m(k) ) d(y n(k), y n(k) 1 )+d(y n(k) 1, y m(k) ).

7 Murthyetal. Journal of Inequalities Applications (015) 015:139 Page 7 of 14 Letting limit k using (.18)-(.4), we get lim k d(y n(k) 1, y m(k) )=ɛ. (.5) From (.1)-(.5), we get lim k M(x m(k) 1, x n(k) 1 )=ɛ lim N(x m(k) 1, x n(k) 1 )=0. k Letting k in (.1), we get ψ(ɛ) ψ(ɛ) lim k φ( N(x m(k) 1, x n(k) 1 ) ). Using discontinuity of φ at t =0φ(t)>0fort > 0, we observe that the last term on the right-h side of the above inequality is non-zero. Thus we arrive at a contradiction. Hence y n } is a Cauchy sequence. Therefore, the Cauchy sequence y n } is a convergent sequence, it converges to a point z (say) in X. Consequently, the subsequences also converge to z in X: Ax n z, Tx n+1 z, Bx n+1 z Sx n z. Now we shall prove that z is a common fixed point of A, B, S T. Since B(X) S(X), there exists v X such that z = Sv.Letd(z, Av) 0. Putting x = v y = x n+1 in (.1), we get ψ ( d(av, Bx n+1 ) ) ψ ( M(v, x n+1 ) ) φ ( N(v, x n+1 ) ), (.6) where M(v, x n+1 )=max d(sv, Tx n+1 ), ( 1 d(sv, Av)+d(Txn+1, Bx n+1 ) ), d(sv, Bxn+1 )+d(tx n+1 Av) )} N(v, x n+1 )=min d(sv, Tx n+1 ), ( 1 d(sv, Av)+d(Txn+1, Bx n+1 ) ), d(sv, Bxn+1 )+d(tx n+1, Av) )}. Taking n using z = Sv,wehave M(v, z)=max d(sv, z), ) 1 d(sv, Av)+d(z, z), d(sv, z)+d(z, Av) ( )} = ) d(z, Av).

8 Murthyetal. Journal of Inequalities Applications (015) 015:139 Page 8 of 14 Also we have ψ ( d(av, z) ) ψ ( ) ) d(z, Av) lim φ( N(v, x n+1 ) ). n Using discontinuity of φ at t =0φ(t)>0fort > 0, we observe that the last term on the right-h side of the above inequality is non-zero. Therefore we obtain ψ ( d(av, z) ) < ψ ( ) ) d(z, Av). Hence we arrive at a contradiction with the ψ function. Therefore d(z, Av)=0 Av = z Av = z = Sv. Since (A, S) is a weakly compatible pair of maps, so it commutes at their coincidence point v, i.e., ASv = SAv Az = Sz. Now we shall show that Az = Sz = z. For this, putting x = z y = x n+1 in (.1), we get ψ ( d(az, Bx n+1 ) ) ψ ( M(z, x n+1 ) ) φ ( N(z, x n+1 ) ), (.7) where M(z, x n+1 )=max d(sz, Tx n+1 ), ( 1 d(sz, Az)+d(Txn+1, Bx n+1 ) ), d(sz, Bxn+1 )+d(tx n+1, Az) )} N(z, x n+1 )=min d(sz, Tx n+1 ), ( 1 d(sz, Az)+d(Txn+1, Bx n+1 ) ), d(sz, Bxn+1 )+d(tx n+1, Az) )}. Taking n using Az = Sz,weget M(z, z)=d(sz, z). Now (.7)impliesthat ψ ( d(sz, z) ) ψ ( d(sz, z) ) lim n φ( N(z, x n+1 ) ). Using discontinuity of φ at t =0φ(t)>0fort > 0, we observe that the last term on the right-h side of the above inequality is non-zero. Therefore we obtain ψ ( d(sz, z) ) < ψ ( d(sz, z) ), which is contradiction. Therefore d(sz, z) = 0 Sz = z Sz = Az = z. Similarly, we can show that Tz = Bz = z.hencesz = Az = Tz = Bz = z.

9 Murthyetal. Journal of Inequalities Applications (015) 015:139 Page 9 of 14 Now we shall show that z is the unique common fixed point of A, B, S T. Let z 1 be also a fixed point of A, B, S T. Putting x = z y = z 1 in (.1), we have ψ ( d(z, z 1 ) ) ψ ( d(z, z 1 ) ) φ ( d(z, z 1 ) ), (.8) a contradiction. Hence d(z, z 1 )=0 z = z 1.HenceA, B, S T have a unique common fixed point in X. When we take S = T = I identity map, we get the following theorem. Theorem. Let (X, d) be a complete metric space. Let A, B : X X be two self-mappings which satisfy the following inequality: ψ ( d(ax, By) ) ψ ( M(x, y) ) φ ( N(x, y) ), (.9) where x, y X, x y, M(x, y)=max d(x, y), ) ) } d(x, Ax)+d(y, By), d(x, By)+d(y, Ax) N(x, y)=min d(x, y), ) ) } d(x, Ax)+d(y, By), d(x, By)+d(y, Ax) ; (1) φ :[0, ) [0, ) is such that φ(t)>0which is lower semi-continuous for all t >0, φ is discontinuous at t =0with φ(0) = 0, () ψ :[0, ) [0, ) is an altering distance function. Then there exists a unique fixed point of A, B in X. When we take ψ(t)=t in Theorem.1 Theorem., we get the following corollaries. Corollary.3 Let (X, d) be a complete metric space. Let A, B, ST: X Xbeselfmappings which satisfy the following inequality: d(ax, By) M(x, y) φ ( N(x, y) ), (.30) where x, y X, x y M(x, y)=max d(sx, Ty), 1 N(x, y)=min d(sx, Ty), 1 ( ) ) } d(sx, Ax)+d(Ty, By), d(sx, By)+d(Ty, Ax), ( ) ) } d(sx, Ax)+d(Ty, By), d(sx, By)+d(Ty, Ax) ; (1) A(X) T(X) B(X) S(X), () (A, S) (B, T) are weak compatible pairs, (3) φ :[0, ) [0, ) is such that φ(t)>0which is lower semi-continuous for all t >0, φ is discontinuous at t =0with φ(0) = 0. Then A, B, S T have a unique common fixed point in X.

10 Murthyetal.Journal of Inequalities Applications (015) 015:139 Page 10 of 14 Corollary.4 Let (X, d) be a complete metric space. Let A, B : X X be two self-mappings which satisfy the following inequality: d(ax, By) M(x, y) φ ( N(x, y) ), (.31) where x, y X, x y, M(x, y)=max d(x, y), 1 N(x, y)=min d(x, y), 1 ( ) ) } d(x, Ax)+d(y, By), d(x, By)+d(y, Ax) ( ) ) } d(x, Ax)+d(y, By), d(x, By)+d(y, Ax) ; (.3) φ :[0, ) [0, ) is such that φ(t)>0which is lower semi-continuous for all t >0, φ is discontinuous at t =0with φ(0) = 0. Then there exists a unique fixed point of A, BinX. Now, we are ready to establish the following theorem in which we are going to replace with N(x, y)=min d(sx, Ty), ) ) } d(sx, Ax)+d(Ty, By), d(sx, By)+d(Ty, Ax) N(x, y)=min d(sx, Ty), ( 1 ) } d(sx, By)+d(Ty, Ax). Theorem.5 Let (X, d) be a complete metric space. Let A, B, ST: X Xbeselfmappings which satisfy the following inequality: ψ ( d(ax, By) ) ψ ( M(x, y) ) φ ( N(x, y) ) (.33) for all x, y Xwithx y M(x, y)=max d(sx, Ty), ) ) } d(sx, Ax)+d(Ty, By), d(sx, By)+d(Ty, Ax) N(x, y)=min d(sx, Ty), ( 1 ) } d(sx, By)+d(Ty, Ax) ; (1) A(X) T(X) B(X) S(X), () (A, S) (B, T) are weak compatible pairs, (3) φ :[0, ) [0, ) is a lower semi-continuous function with φ(t)>0for all t (0, ) φ(0) = 0, (4) ψ :[0, ) [0, ) is an altering distance function which in addition is strictly monotone increasing. Then A, B, S T have a unique common fixed point in X.

11 Murthyetal.Journal of Inequalities Applications (015) 015:139 Page 11 of 14 One can easily prove this theorem by using the technique given in the proof of Theorem.1. When we take S = T = I identity map, we get the following theorem. Theorem.6 Let (X, d) be a complete metric space. Let A B : X X be self-mappings which satisfy the following inequality: ψ ( d(ax, By) ) ψ ( M(x, y) ) φ ( N(x, y) ) (.34) for all x, y Xwithx y, M(x, y)=max d(x, y), ) ) } d(x, Ax)+d(Ty, By), d(x, By)+d(y, Ax) N(x, y)=min d(x, y), ( 1 ) } d(x, By)+d(y, Ax) ; (1) φ :[0, ) [0, ) is a lower semi-continuous function with φ(t)>0for all t (0, ) φ(0) = 0, () ψ :[0, ) [0, ) is an altering distance function which in addition is strictly monotone increasing. Then there exists a unique fixed point of A, B in X. When we take ψ(t)=t in Theorem.5 Theorem.6, we get the following corollaries. Corollary.7 Let (X, d) be a complete metric space. Let A, B, ST: X Xbeselfmappings which satisfy the following inequality: d(ax, By) M(x, y) φ ( N(x, y) ), (.35) where x, y X, x y M(x, y)=max d(sx, Ty), ) ) } d(sx, Ax)+d(Ty, By), d(sx, By)+d(Ty, Ax), N(x, y)=min d(sx, Ty), ( 1 ) } d(sx, By)+d(Ty, Ax) ; (1) A(X) T(X) B(X) S(X), () (A, S) (B, T) are weak compatible pairs, (3) φ :[0, ) [0, ) is a lower semi-continuous function with φ(t)>0for all t (0, ) φ(0) = 0. Then A, B, S T have a unique common fixed point in X. Corollary.8 Let (X, d) be a complete metric space. Let A, B : X X be two self-mappings which satisfy the following inequality: d(ax, By) M(x, y) φ ( N(x, y) ) (.36)

12 Murthyetal.Journal of Inequalities Applications (015) 015:139 Page 1 of 14 for all x, y Xwithx y, M(x, y)=max d(sx, Ty), ) ) } d(sx, Ax)+d(Ty, By), d(sx, By)+d(Ty, Ax) N(x, y)=min d(sx, Ty), ( 1 ) } d(sx, By)+d(Ty, Ax) ; φ :[0, ) [0, ) is a lower semi-continuous function with φ(t)>0for all t (0, ) φ(0) = 0. Then there exists a unique fixed point of A, B in X. Example.9 Let X = [0, ] be endowed with the Euclidean metric d(x, y)= x y,let A, B, S T X be defined by A(x)= B(x)= 0 ifx =0, x 5 +1 otherwise, S(x)= 0 ifx =0, x 5 +1 otherwise, T(x)= 0 ifx =0, 3x 5 +1 otherwise, 0 ifx =0, 4x 5 +1 otherwise, where x, y X. A(X)=[0, 7 5 ], S(X)=[0,11 5 ], B(X)=[0,9 5 ], T(X)=[0,13 ]. Here A(X) T(X) 5 B(X) S(X), (A, S) (B, T) are weakly compatible maps at x =0butnotcompatible maps. Take ψ(t)=t φ(t)= t if t >0, 0 ift =0. Now we have to check the inequality of Theorem.1 for the following cases. Case 1: If x =0y =0: ψ ( d(ax, By) ) = ψ ( Ax By ) =0, ψ ( M(x, y) ) =0 φ ( N(x, y) ) =0, hence ψ(d(ax, By)) =ψ(m(x, y)) φ(n(x, y)). Case : If x =0y 0: ψ ( d(ax, By) ) = ψ ( Ax By ) = ψ ( y 5 ) = y 5 [ M(x, y)=max Sx Ty, ) 1 Sx Ax + Ty By, [ )] 4y = max 5, y, 1 [ N(x, y)=min Sx Ty, 1 ( y 5 ( Sx By + Ty Ax ) ] + 4y 5 = 4y 5, ( ) ) ] Sx Ax + Ty By, Sx By + Ty Ax

13 Murthyetal.Journal of Inequalities Applications (015) 015:139 Page 13 of 14 [ 4y = min 5, y, )] y 5 + 4y 5 = y, ψ ( M(x, y) ) ( N(x, y) ) ( ) ( ) 4y = ψ y 5 φ = 4y 5 ) y y 5 = ψ( d(ax, By) ). Case 3: If x 0y =0: ψ ( d(ax, By) ) = ψ ( Ax By ) = ψ ( Ax ) ( ) x = ψ = x [ M(x, y)=max Sx Ty, ) 1 Sx Ax + Ty By, [ )] 3x = max 5, x, 1 [ N(x, y)=min Sx Ty, 1 [ 3x ( 3x 5 ( Sx By + Ty Ax ) ] + x = 3x 5, ( ) ) ] Sx Ax + Ty By, Sx By + Ty Ax = min 5, x, )] 3x 5 + x = x, ψ ( M(x, y) ) φ ( N(x, y) ) ( ) ( ) 3x = ψ x 5 φ = 3x 5 ) x x = ψ( d(ax, By) ). Case 4: If x 0y 0: ψ ( d(ax, By) ) = ψ ( Ax By ) ( ) x = ψ 5 +1 y 5 1 = x 5 y 5 [ M(x, y)=max Sx Ty, ) 1 Sx Ax + Ty By, ) ( Sx By + Ty Ax ) ] [ 3x = max 5 4y ( 5, x + y, 3x 5 y 5 + 4y 5 x )], 3x 5 4y x if 0 < y < 5 5, M(x, y)= 3x 5 4y 5,( x 5 + y 5 ), 1 3x ( 5 y 5 + 4y 5 x x ) ify = 5 5, ( x 5 + y 5 ), 1 3x ( 5 y 5 + 4y 5 x x ) if 5 5 < y x, [ N(x, y)=min Sx Ty, 1 ( ) ) ] Sx Ax + Ty By, Sx By + Ty Ax

14 Murthyetal.Journal of Inequalities Applications (015) 015:139 Page 14 of 14 [ 3x = min 5 4y ( ) 5, x + y, 3x 5 y 5 + 4y 5 x )], x 5 + y 5 if 0 < y x 5, x N(x, y)= 5 + y 5, 1 3x ( 5 y 5 + 4y 5 x 5 ) if x 5 < y < x 5, 3x 5 4y 5, x 5 + y 5, 1 3x ( 5 y 5 + 4y 5 x x ) ify = 5 5, 3x 5 4y x if 5 5 < y x, ψ ( d(ax, By) ) = x 5 y 5 ψ( M(x, y) ) φ ( N(x, y) ). So the inequalities hold in each of the cases. Hence all the conditions of Theorem.1 hold A, B, S T havetheuniquecommonfixedpointatx =0inX. 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 Pure Applied Mathematics, Guru Ghasidas Vishwavidyalaya, Bilaspur, Chhatisgarh , India. Department of Mathematics Computer Science, Cankaya University, Ankara, 06790, Turkey. Acknowledgements The first author (PP Murthy) wishes to thank University Grants Commission, New Delhi, India for the MRP (File number 4-3/013 (SR)). Received: 1 October 014 Accepted: 30 March 015 References 1. Banach, S: Sur les opérations dans les ensembles abstraits et leur application aux équations intégrales. Fundam. Math. 3, ). Kannan, R: Some results on fixed points. Bull. Calcutta Math. Soc. 6, (1968) 3. Rakotch, A: A note on contraction mappings. Proc. Am. Math. Soc. 13, (196) 4. Boyd, DW, Wong, TSW: On nonlinear contractions. Proc. Am. Math. Soc. 0, (1969) 5. Alber, YI, Guerre-Delabriere, S: Principles of weakly contractive maps in Hilbert spaces. In: Gohberg, I, Lyubich, Y (eds.) New Results in Operator Theory Its Applications, vol. 98, pp. 7-. Birkhäuser, Basel (1997) 6. Rhoades, BE: Some theorems on weakly contractive maps. Nonlinear Anal. 47, (001) 7. Khan, MS, Swalesh, M, Sessa, S: Fixed points theorems by altering distances between the points. Bull. Aust. Math. Soc. 30, 1-9 (1984) 8. Jungck, G, Rhoades, BE: Fixed points for set valued functions without continuity. Indian J. Pure Appl. Math. 9, 7-38 (1998) 9. Murthy, PP, Tas, K, Choudhury, BS: Weak contraction mappings in Saks spaces. Fasc. Math. 48, (01) 10. Dutta, PN, Choudhury, BS: A generalization of contraction principle in metric spaces. Fixed Point Theory Appl. 008, Article ID (008) 11. Zhang, Q, Song, Y: Fixed point theory for generalized ψ-weak contractions. Appl. Math. Lett., (009) 1. Doric, D: Common fixed point for generalized (, ϕ)-weak contractions. Appl. Math. Lett., (009) 13. Hosseini, VR: Commonfixed for generalized (ϕ-ψ)-weak contractions mappings condition of integral type. Int. J. Math. Anal. 4(31), (010) 14. Abkar, A, Choudhury, BS: Fixed point result in partially ordered metric spaces using weak contractive inequalities. Facta Univ., Ser. Math. Inform. 7(1), 1-101) 15. Ahmad, J, Arshad, M, Vetro, P: Coupled coincidence point results for (, ϕ)-contractive mappings in partially ordered metric spaces. Georgian Math. J. 1(), (014) 16. Hussain, N, Arshad, M, Shoaib, A, Fahimuddin: Common fixed point results for α-ψ-contractions on a metric space endowed with graph. J. Inequal. Appl. 014, 136 (014)