On coupled fixed point theorems on partially ordered G-metricspaces

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1 Karapınar et al. Journal of Inequalities and Applications 2012, 2012:200 R E S E A R C H Open Access On coupled fixed point theorems on partially ordered G-metricspaces Erdal Karapınar 1, Billûr Kaymakçalan 2* and Kenan Taş 2 * Correspondence: billur@cankaya.edu.tr 2 Department of Mathematics and Computer Science, Çankaya University, Ankara, Turkey Full list of author information is available at the end of the article Abstract In this manuscript, we extend, generalize and enrich some recent coupled fixed point theorems in the framework of partially ordered G-metric spaces in a way that is essentially more natural. MSC: 46N40; 47H10; 54H25; 46T99 Keywords: coupledfixedpoint; coincidencepoint; mixedg-monotone property; ordered set; G-metric space 1 Introduction and preliminaries In [1]Aydi et al. established coupled coincidence and coupled common fixed point results for a mixed g-monotone mapping satisfying nonlinear contractions in partially ordered G-metric spaces. These results generalize those of Choudhury and Maity [2]. Here we generalize, improve, enrich and extend the above mentioned coupled fixed point results of Aydi et al. Throughout this paper, let N denote the set of nonnegative integers, and N be the set of positive integers. Definition 1.1 See [3]) Let X be a non-empty set, and G : X X X R + be a function satisfying the following properties: G1) Gx, y, z)=0if x = y = z, G2) 0<Gx, x, y) for all x, y X with x y, G3) Gx, x, y) Gx, y, z) for all x, y, z X with y z, G4) Gx, y, z) =Gx, z, y) =Gy, z, x) = symmetry in all three variables), G5) Gx, y, z) Gx, a, a)+ga, y, z) for all x, y, z, a X rectangle inequality). Then the function G is called a generalized metric or, more specially, a G-metric on X, and the pair X, G) is called a G-metric space. Every G-metric on X defines a metric d G on X by d G x, y)=gx, y, y)+gy, x, x), for all x, y X. 1.1) Example 1.2 Let X, d) be a metric space. The function G : X X X [0, + ), defined by Gx, y, z)=max { dx, y), dy, z), dz, x) }, 2012 Karapınar et al.; licensee Springer. This is an Open Access article distributed under the terms of the Creative Commons Attribution License which permits unrestricted use, distribution, and reproduction in any medium, provided the original work is properly cited.

2 Karapınar etal. Journal of Inequalities and Applications 2012, 2012:200 Page 2 of 13 or Gx, y, z)=dx, y)+dy, z)+dz, x), for all x, y, z X,isaG-metric on X. Definition 1.3 See [3]) Let X, G) beag-metric space, and let {x n } be a sequence of points of X.Wesaythatx n )isg-convergent to x X if lim n,m + Gx, x n, x m )=0,thatis, for any ε >0,thereexistsN N such that Gx, x n, x m )<ε, for all n, m N. Wecallx the limit of the sequence and write x n x or lim n + x n = x. Proposition 1.4 See [3]) Let X, G) be a G-metric space. The following are equivalent: 1) {x n } is G-convergent to x, 2) Gx n, x n, x) 0 as n +, 3) Gx n, x, x) 0 as n +, 4) Gx n, x m, x) 0 as n, m +. Definition 1.5 See [3]) Let X, G) beag-metric space. A sequence {x n } is called a G- Cauchy sequence if, for any ε >0,thereisN N such that Gx n, x m, x l )<ε for all m, n, l N,thatis,Gx n, x m, x l ) 0asn, m, l +. Proposition 1.6 See [3]) Let X, G) be a G-metric space. Then the following are equivalent: 1) the sequence {x n } is G-Cauchy, 2) for any ε >0,thereexistsN N such that Gx n, x m, x m )<ε,forallm, n N. Proposition 1.7 See [3]) Let X, G) be a G-metric space. A mapping f : X XisGcontinuous at x 0 if and only if it is G-sequentially continuous at x 0, that is, whenever x n ) is G-convergent to x 0,thesequencefx n )) is G-convergent to f x 0 ). Definition 1.8 See [3]) A G-metric space X, G) is called G-complete if every G-Cauchy sequenceis G-convergent in X, G). Definition 1.9 See [2]) Let X, G)beaG-metric space. A mapping F : X X X is said to be continuous if for any two G-convergent sequences {x n } and {y n } converging to x, y respectively, {Fx n, y n )} is G-convergent to Fx, y). Let X, ) be a partially ordered set and X, G) beag-metric space, g : X X be a mapping. A partiallyordered G-metric space, X, G, ), is called g-ordered complete if for each convergent sequence {x n } n=0 X, the following conditions hold: OC 1 ) if {x n } is a non-increasing sequence in X such that x n x implies gx gx n, n N, OC 2 ) if {y n } is a non-decreasing sequence in X such that y n y implies gy gy n, n N. Moreover, a partially ordered G-metric space, X, G, ), is called ordered complete when g is equal to identity mapping in the above conditions OC 1 )andoc 2 ).

3 Karapınar etal. Journal of Inequalities and Applications 2012, 2012:200 Page 3 of 13 Definition 1.10 See [4]) An element x, y) X X issaidto beacoupledfixedpoint of the mapping F : X X X if Fx, y) =x and Fy, x) =y. Definition 1.11 See [5]) An element x, y) X X is called a coupled coincidence point of a mapping F : X X X and g : X X if Fx, y) =gx), Fy, x) =gy). Moreover, x, y) X X is called a common coupled coincidence point of F and g if Fx, y) =gx) =x, Fy, x) =gy) =y. Definition 1.12 Let F : X X X and g : X X be mappings. The mappings F and g are said to commute if g Fx, y) ) = F gx), gy) ), for all x, y X. Definition 1.13 See [4]) Let X, ) be a partially ordered set and F : X X X be a mapping. Then F is said to have mixed monotone property if Fx, y) is monotone nondecreasing in x and is monotone non-increasing in y, that is, for any x, y X, x 1 x 2 Fx 1, y) Fx 2, y), for x 1, x 2 X, and y 1 y 2 Fx, y 2 ) Fx, y 1 ), for y 1, y 2 X. Definition 1.14 See [5]) Let X, ) be a partially ordered set and F : X X X and g : X X be two mappings. Then F is said to have mixed g-monotone property if Fx, y) is monotone g-non-decreasing in x and is monotone g-non-increasing in y,that is,for any x, y X, gx 1 ) gx 2 ) Fx 1, y) Fx 2, y), for x 1, x 2 X, 1.2) and gy 1 ) gy 2 ) Fx, y 2 ) Fx, y 1 ), for y 1, y 2 X. 1.3) Let denote the set of functions φ :[0, ) [0, )satisfying a) φ 1 {0})={0}, b) φt)<t for all t >0, c) lim r t + φr)<t for all t >0. Lemma 1.15 See [5]) Let φ.forallt>0, we have lim n φ n t)=0.

4 Karapınar etal. Journal of Inequalities and Applications 2012, 2012:200 Page 4 of 13 Aydi et al. [1] proved the following theorems. Theorem 1.16 Let X, ) be a partially ordered set and G be a G-metric on X such that X, G) is a complete G-metric space. Suppose that there exist φ, F: X X Xand g : X Xsuchthat G Fx, y), Fu, v), Fw, z) ) ) Ggx, gu, gw)+ggy, gv, gz) φ 1.4) 2 for all x, y, u, v, w, z Xwithgw gu gx and gy gv gz. Suppose also that F is continuous and has the mixed g-monotone property, FX X) gx) and g is continuous and commutes with F. If there exist x 0, y 0 Xsuchthatgx 0 Fx 0, y 0 ) and Fy 0, x 0 ) gy 0, then F and g have a coupled coincidence point, that is, there exists x, y) X Xsuchthat gx = Fx, y) and gy = Fy, x). Theorem 1.17 Let X, ) be a partially ordered set and G be a G-metric on X such that X, G, ) is regular. Suppose that there exist φ and mappings F : X X Xand g : X Xsuchthat G Fx, y), Fu, v), Fw, z) ) ) Ggx, gu, gw)+ggy, gv, gz) φ 1.5) 2 for all x, y, u, v, w, z Xwithgw gu gx and gy gv gz. Suppose also that gx), G) is complete, F has the mixed g-monotone property and FX X) gx).ifthereexistx 0, y 0 Xsuchthatgx 0 Fx 0, y 0 ) and Fy 0, x 0 ) gy 0, then F and g have a coupled coincidence point. In this manuscript, we generalize, improve, enrich and extend the above coupled fixed point results. We also state some examples to illustrate our results. This paper can be considered as a continuation of the remarkable works of Berinde [6, 7]. 2 Main results We begin with an example to illustrate the weakness of Theorem 1.16 and Theorem 1.17 above. Example 2.1 Let X = R.DefineG : X X X [0, )by Gx, y, z)= x y + x z + y z for all x, y, z X.ThenX, G)isaG-metric space. Define a map F : X X X by Fx, y)= 1 12 x y and g : X X by gx)= x for all x, y X.Supposex = u = z 2 G Fx, y), Fu, v), Fz, w) ) 1 = G 12 x y, 1 12 u v, 1 12 z + 7 ) 12 w = v y + w y + w v 2.1)

5 Karapınar etal. Journal of Inequalities and Applications 2012, 2012:200 Page 5 of 13 and x Ggx, gu, gz)+ggy, gv, gw)=g 2, u 2, z ) y + G 2 2, v 2, w ) 2 = 1 2[ y v + y w + v w ]. 2.2) It is clear that there is no φ that provides the statement 1.4)ofTheorem1.16. Notice that 0, 0) is the unique common coincidence point of F and g.infact,f0, 0) = g0) = 0. For some coupled fixed point and coupled coincidence point theorems, we refer the reader to [8 34]. We now state our first result which successively guarantees a coupled fixed point. Theorem 2.2 Let X, ) be a partially ordered set and G be a G-metric on X such that X, G) is a complete G-metric space. Suppose that there exist φ, F: X X Xand g : X Xsuchthat [ G Fx, y), Fu, v), Fw, z) ) + G Fy, x), Fv, u), Fz, w) )] φ Ggx, gu, gw)+ggy, gv, gz) ) 2.3) for all x, y, u, v, w, z Xwithgw gu gx and gy gv gz. Suppose also that F is continuous and has the mixed g-monotone property, FX X) gx) and g is continuous and commutes with F. If there exist x 0, y 0 Xsuchthatgx 0 Fx 0, y 0 ) and Fy 0, x 0 ) gy 0, then F and g have a coupled coincidence point, that is, there exists x, y) X Xsuchthat gx = Fx, y) and gy = Fy, x). Proof Given x 0, y 0 X satisfying gx 0 Fx 0, y 0 )andfy 0, x 0 ) gy 0, we shall construct iterative sequences x n )andy n ) in the following way: Since FX X) gx), we can choose x 1, y 1 X such that gx 1 = Fx 0, y 0 )andgy 1 = Fy 0, x 0 ). Analogously, we choose x 2, y 2 X such that gx 2 = Fx 1, y 1 )andgy 2 = Fy 1, x 1 ) due to the same reasoning. Since F has the mixed g-monotone property, we conclude that gx 0 gx 1 gx 2 and gy 2 gy 1 gy 0.By repeating this process, we derive the iterative sequence gx n = Fx n 1, y n 1 ) gx n+1 = Fx n, y n ) and gy n+1 = Fy n, x n ) gy n = Fy n 1, x n 1 ). If for some n 0 we have gx n0 +1, gy n0 +1) =gx n0, gy n0 ), then Fx n0, y n0 )=gx n0 and F = y n0, x n0 )=gy n0,thatis,f and g have a coincidence point. So, we assume that gx n+1, gy n+1 ) gx n, gy n ) for all n N. Thus, we have either gx n+1 = Fx n, y n ) gx n or gy n+1 = Fy n, x n ) gy n. We set t n = Ggx n+1, gx n+1, gx n )+Ggy n+1, gy n+1, gy n ) 2.4)

6 Karapınar etal. Journal of Inequalities and Applications 2012, 2012:200 Page 6 of 13 for all n N. Due to the property G2), we have t n >0foralln N. By using inequality 2.3), we obtain Ggx n+1, gx n+1, gx n )+Ggy n+1, gy n+1, gy n ) = G Fx n, y n ), Fx n, y n ), Fgx n 1, gy n 1 ) ) + G Fy n, x n ), Fy n, x n ), Fgy n 1, gx n 1 ) ) φ Ggx n, gx n, gx n 1 )+Ggy n, gy n, gy n 1 ) ). 2.5) Taking 2.4) into account, 2.5) becomes t n φt n 1 ). 2.6) Since φt) <t for all t > 0, it follows that t n is monotone decreasing. Therefore, there is some L 0suchthatlim n + t n = L. Now, we assert that L =0.Suppose,onthecontrary,thatL > 0. Letting n + in 2.6) and using the properties of the map φ,weget L = lim n + t n lim n + φt n 1)<L, which is contradiction. Thus L =0. Hence lim t n = lim Ggx n+1, gx n+1, gx n )+Ggy n, gy n, gy n 1 )=0. 2.7) Next, we prove that gx n )andgy n )arecauchysequencesintheg-metric space X, G). Suppose, on the contrary, that at least one of gx n )andgy n )isnotacauchysequencein X, G). Then there exist ε > 0 and sequencesof naturalnumbers mk)) and lk)) such that for every natural number k, mk) >lk) k and r k = Ggx mk), gx mk), gx lk) )+Ggy mk), gy mk), gy lk) ) ε. 2.8) Now, corresponding to lk), we choose mk) to be the smallest for which 2.8)holds.Hence Ggx mk) 1, gx mk) 1, gx lk) )+Ggy mk) 1, gy mk) 1, gy lk) )<ε. Using the rectangle inequality property G5)), we get ε r k Ggx mk), gx mk), gx mk) 1 )+Ggx mk) 1, gx mk) 1, gx lk) ) + Ggy mk), gy mk), gy mk) 1 )+Ggy mk) 1, gy mk) 1, gy lk) ) = Ggx mk) 1, gx mk) 1, gx lk) )+Ggy mk) 1, gy mk) 1, gy lk) )+t mk) 1 < ε + t mk) ) Letting k + in the above inequality and using 2.7)yields lim r k = ε ) k +

7 Karapınar etal. Journal of Inequalities and Applications 2012, 2012:200 Page 7 of 13 Again, by the rectangle inequality, we have r k = Ggx mk), gx mk), gx lk) )+Ggy mk), gy mk), gy lk) ) Ggx mk), gx mk), gx mk)+1 )+Ggx mk)+1, gx mk)+1, gx lk)+1 ) + Ggx lk)+1, gx lk)+1, gx lk) )+Ggy mk), gy mk), gy mk)+1 ) + Ggy mk)+1, gy mk)+1, gy lk)+1 )+Ggy lk)+1, gy lk)+1, gy lk) ) = t lk) + Ggx mk), gx mk), gx mk)+1 )+Ggy mk), gy mk), gy mk)+1 ) + Ggx mk)+1, gx mk)+1, gx lk)+1 )+Ggy mk)+1, gy mk)+1, gy lk)+1 ). Using the fact that Gx, x, y) 2Gx, y, y) for any x, y X, we obtain from properties G2)- G4) r k t lk) +2Ggx mk), gx mk), gx mk)+1 )+2Ggy mk), gy mk), gy mk)+1 ) + Ggx mk)+1, gx mk)+1, gx lk)+1 )+Ggy mk)+1, gy mk)+1, gy lk)+1 ) = t lk) +2t mk) + Ggx mk)+1, gx mk)+1, gx lk)+1 )+Ggy mk)+1, gy mk)+1, gy lk)+1 ). Next, using inequality 2.3), we have Ggx mk)+1, gx mk)+1, gx lk)+1 )+Ggy mk)+1, gy mk)+1, gy lk)+1 ) = G Fx mk), y mk) ), Fx mk), y mk) ), Fx lk), y lk) ) ) + G Fy lk), x lk) ), Fy mk), x mk) ), Fy mk), x mk) ) ) φ Ggx mk), gx mk), gx lk) )+Ggy mk), gy mk), gy lk) ) ) φr k ). 2.11) Now, using 2.7),2.10), the properties of the function φ, andlettingk + in 2.11), we get ε lim k + φr k)= lim t ε) + φt)<ε, which is a contradiction. Thus, we have proven that gx n )andgy n )arecauchysequences in the G-metric space X, G). Now, since X, G) is complete, there are x, y X such that gx n )andgy n ) are respectively G-convergent to x and y. That is from Proposition 1.4,we have lim Ggx n, gx n, x)= lim Ggx n, x, x)=0, lim Ggy n, gy n, y)= lim Ggy n, y, y)=0. Using the continuity of g, we get from Proposition 1.7 lim G ggx n ), ggx n ), gx ) = lim G ggx n ), gx, gx ) =0, lim G ggy n ), ggy n ), gy ) = lim G ggy n ), gy, gy ) = )

8 Karapınar etal. Journal of Inequalities and Applications 2012, 2012:200 Page 8 of 13 Since gx n+1 = Fx n, y n )andgy n+1 = Fy n, x n ), employing the commutativity of F and g yields ggx n+1 )=g Fx n, y n ) ) = Fgx n, gy n ), ggy n+1 )=g Fy n, x n ) ) = Fgy n, gx n ). 2.13) Now, we shall show that Fx, y) =gx and Fy, x) =gy. The mapping F is continuous, and since the sequences gx n )andgy n ) are respectively G-convergent to x and y, using Definition 1.9, thesequencefgx n, gy n )) is G-convergent to Fx, y). Therefore, from 2.13), ggx n+1 )) is G-convergent to Fx, y). By uniqueness of the limit and using 2.12), we have Fx, y)=gx. Similarly, we can show that Fy, x)=gy.hence, x, y) is a coupled coincidence point of F and g. This completesthe proof. The following example illustrates that Theorem 2.2 is an extension of Theorem Example 2.3 Let us reconsider Example 2.1.DefineamapF : X X X by Fx, y)= 1 12 x y and g : X X by gx)= x for all x, y X.ThenFX X)=[0, )=gx)=x.weobserve 2 that G Fx, y), Fu, v), Fz, w) ) + G Fy, x), Fv, u), Fw, z) ) 1 = G 12 x y, 1 12 u v, 1 12 z + 7 ) 12 w 1 + G 12 y x, 1 12 v u, 1 12 w + 7 ) 12 z = v y + w y + w v u x + z x + z u v y + w y + w v u x + z x + z u = 8 ) v y + w y + w v ) u x + z x + z u 12 and 1 Ggx, gu, gz)+ggy, gv, gw)=g 2 x, 1 2 u, 1 ) 1 2 z + G 2 y, 1 2 v, 1 ) 2 w = 1 [ ) x u + x z + u z 2 + y v + y w + v w )]. 2.14)

9 Karapınar etal. Journal of Inequalities and Applications 2012, 2012:200 Page 9 of 13 Then, the statement 2.3)ofTheorem2.2 is satisfied for φt)= 2 t and 0, 0) is the desired 3 coupled coincidence point. In the next theorem, we omit the continuity hypothesis of F. Theorem 2.4 Let X, ) be a partially ordered set and G be a G-metric on X such that X, G, ) is g-ordered complete. Suppose that there exist φ and mappings F : X X Xandg: X Xsuchthat [ ) )] G Fx, y), Fu, v), Fw, z) + G Fy, x), Fv, u), Fz, w) φ Ggx, gu, gw)+ggy, gv, gz) ) 2.15) for all x, y, u, v, w, z Xwithgw gu gx and gy gv gz. Suppose also that gx), G) is complete, F has the mixed g-monotone property and FX X) gx).ifthereexistx 0, y 0 Xsuchthatgx 0 Fx 0, y 0 ) and Fy 0, x 0 ) gy 0, then F and g have a coupled coincidence point. Proof Proceeding exactly as in Theorem 2.2, wehavethatgx n )andgy n )arecauchysequences in the complete G-metric space gx), G). Then, there exist x, y X such that gx n gx and gy n gy. Sincegx n )isnon-decreasingandgy n ) is non-increasing, using the regularity of X, G, ), we have gx n gx and gy gy n for all n 0. If gx n = gx and gy n = gy for some n 0, then gx = gx n gx n+1 gx = gx n and gy gy n+1 gy n = gy,which implies that gx n = gx n+1 = Fx n, y n )andgy n = gy n+1 = Fy n, x n ), that is, x n, y n )isacoupled coincidence point of F and g. Then, we suppose that gx n, gy n ) gx, gy) for all n 0. Using the rectangle inequality, 2.15) and property φt)<t for all t >0,weget G Fx, y), gx, gx ) + G Fy, x), gy, gy ) G Fx, y), gx n+1, gx n+1 ) + Ggxn+1, gx, gx) + G Fy, x), gy n+1, gy n+1 ) + Ggyn+1, gy, gy) = G Fx, y), Fx n, y n ), Fx n, y n ) ) + Ggx n+1, gx, gx) + G Fy, x), Fy n, x n ), Fy n, x n ) ) + Ggy n+1, gy, gy) φ Ggx, gx n, gx n )+Ggy, gy n, gy n ) ) + Ggx n+1, gx, gx)+ggy n+1, gy, gy) < Ggx, gx n, gx n )+Ggy, gy n, gy n ) + Ggx n+1, gx, gx)+ggy n+1, gy, gy). Letting n + in the above inequality, we obtain G Fx, y), gx, gx ) + G Fy, x), gy, gy ) =0, which implies that gx = Fx, y) andgy = Fy, x). Thus we proved that x, y) isacoupled coincidence point of F and g.

10 Karapınar etal. Journal of Inequalities and Applications 2012, 2012:200 Page 10 of 13 Corollary 2.5 Let X, ) be a partially ordered set and G be a G-metric on X such that X, G) is a complete G-metric space. Suppose that there exist k [0, 1),F: X X Xand g : X Xsuchthat [ ) )] G Fx, y), Fu, v), Fw, z) + G Fy, x), Fv, u), Fz, w) k [ Ggx, gu, gw)+ggy, gv, gz) ] for all x, y, u, v, w, z Xwithgw gu gx and gy gv gz. Suppose also that F is continuous, has the mixed g-monotone property, FX X) gx) and g is continuous and commutes with F. If there exist x 0, y 0 Xsuchthatgx 0 Fx 0, y 0 ) and Fy 0, x 0 ) gy 0, then F and g have a coupled coincidence point. Proof Taking φt)=kt with k [0, 1) in Theorem 2.4, we obtain Corollary 2.5. Corollary 2.6 Let X, ) be a partially ordered set and G be a G-metric on X such that X, G, ) is g-ordered complete. Suppose that there exist k [0, 1), F: X X Xand g : X Xsuchthat [ ) )] G Fx, y), Fu, v), Fw, z) + G Fy, x), Fv, u), Fz, w) k [ Ggx, gu, gw)+ggy, gv, gz) ] for all x, y, u, v, w, z Xwithgw gu gx and gy gv gz. Suppose also that gx), G) is complete, F has the mixed g-monotone property, FX X) gx).ifthereexistx 0, y 0 X such that gx 0 Fx 0, y 0 ) and Fy 0, x 0 ) gy 0, then F and g have a coupled coincidence point. Proof Taking φt)=kt with k [0, 1) in Theorem 2.4, we obtain Corollary 2.6 Remark 2.7 Taking g = l x the identity mapping) in Corollary 2.5, weobtain[2, Theorem 3.1]. Taking g = I x in Corollary 2.6,weobtain[2,Theorem3.2]. Now we shall prove the existence and uniqueness theorem of a coupled common fixed point. If X, ) is a partially ordered set, we endow the product set X X with the partial order defined by x, y) u, v) x u, v y. Theorem 2.8 In addition to the hypothesis of Theorem 2.2, supposethatforallx, y), x, y ) X X),thereexistsu, v) X XsuchthatFx, y), Fu, v)) is comparable with Fx, y), Fy, x)) and Fx, y ), Fy, x )). Suppose also that φ is a non-decreasing function. Then F and g have a unique coupled common fixed point, that is, there exists a unique x, y) X Xsuchthat x = gx = Fx, y) and y = gy = Fy, x). Proof From Theorem 2.2, the set of coupled coincidences is non-empty. We shall show that if x, y)andx, y ) are coupled coincidence points, that is, if gx = Fx, y), gy)=fy, x),

11 Karapınar etal. Journal of Inequalities and Applications 2012, 2012:200 Page 11 of 13 gx = Fx, y )andgy = Fy, x ), then gx = gx and gy = gy. 2.16) By assumption, there exists u, v) X X such that Fu, v), Fv, u)) is comparable with Fx, y), Fy, x)) and Fx, y ), Fy, x )). Without loss of generality, we can assume that Fx, y), Fy, x) ) Fu, v), Fv, u) ) and F x, y ), F y, x )) Fu, v), Fv, u) ). Put u 0 = u, v 0 = v and choose u 1, v 1 X such that gu 1 = Fu 0, v 0 )andgv 1 = Fv 0, u 0 ). Then, similarly as in the proof of Theorem 2.2, we can inductively define sequences gu n )and gv n )inx by gu n+1 = Fu n, v n )andgv n+1 = Fv n, u n ). Further, set x 0 = x, y 0 = y, x 0 = x, y 0 = y and, in the same way, define the sequences gx n ), gy n ), gx n )andgy n ). Since Fx, y), Fy, x) ) =gx1, gy 1 )=gx, gy) Fu, v), Fv, u) ) =gu 1, gv 1 ), then gx gu 1 and gv 1 gy. UsingthatF is a mixed g-monotone mapping, one can show easily that gx gu n and gv n gy for all n 1. Thus from 2.15), we get Ggu n+1, gx, gx)+ggy, gy, gv n+1 )=G Fu n, v n ), Fx, y), Fx, y) ) + G Fy, x), Fy, x), Fv n, u n ) ) φ Ggu n, gx, gx)+ggv n, gy, gy) ). Without loss of generality, we can suppose that gu n, gv n ) gx, gy) for all n 1. Since φ is non-decreasing, from the previous inequality, we get Ggu n+1, gx, gx)+ggv n+1, gy, gy) φ n Ggu 1, gx, gx)+ggv 1, gy, gy) ) for each n 1. Letting n + in the above inequality and using Lemma 1.15,weobtain lim Ggu n+1, gx, gx)=0 and lim Ggv n+1, gy, gy)= ) Analogously, we derive that lim G gu n+1, gx, gx ) =0 and lim G gv n+1, gy, gy ) = ) Hence, from 2.17), 2.18) and the uniqueness of the limit, we get gx = gx and gy = gy. Hence the equalities in 2.16) aresatisfied.sincegx = Fx, y) andgy = Fy, x), by commutativity of F and g,wehave ggx)=g Fx, y) ) = Fgx, gy) and ggy)=g Fy, x) ) = Fgy, gx). 2.19)

12 Karapınar etal. Journal of Inequalities and Applications 2012, 2012:200 Page 12 of 13 Denote gx = z and gy = w,thenby2.19), we get gz = Fz, w) and gw = Fw, z). 2.20) Thus, z, w) is a coincidence point. Then, from 2.16)withx = z and y = w,wehavegx = gz and gy = gw,thatis, gz = z and gw = w. 2.21) From 2.20), 2.21), we get z = gz = Fz, w) and w = gw = Fw, z). Then, z, w) is a coupled common fixed point of F and g. To prove the uniqueness, assume that p, q) is another coupled common fixed point. Then by 2.16), we have p = gp = gz = z and q = gq = gw = w. Theorem 2.9 Let X, ) be a partially ordered set and G be a G-metric on X such that X, G) is a complete G-metric space and X, G, ) is regular. Suppose that there exist φ and F : X X X having the mixed monotone property such that G Fx, y), Fu, v), Fw, z) ) + G Fy, x), Fv, u), Fz, w) ) φ Gx, u, w)+gy, v, z) ), for all x, y, u, v, w, z Xwithw u xandy v z. If there exist x 0, y 0 Xsuchthat x 0 Fx 0, y 0 ) and Fy 0, x 0 ) y 0, then F has a coupled fixed point. Furthermore, if y 0 x 0, then x = y, that is, x = Fx, x). Proof Following the proof of Theorem 2.4 with g = I x,wehaveonlytoshowthatx = Fx, x). Since y 0 x 0,wegety y n y 1 y 0 x 0 x 1 x n x. Thus, we have y x.supposethatgx, x, y) > 0. Using inequality 2.15), we have Gx, x, y)+gx, y, y) =G Fx, y), Fx, y), Fy, x) ) + G Fx, y), Fy, x), Fy, x) ) φ Gx, x, y)+gy, y, x) ) < Gx, x, y)+gx, y, y), a contradiction. Thus, Gx, x, y)=0andx = y = Fx, x). Competing interests The authors declare that they have no competing interests. Authors contributions All authors have contributed equally. All authors read and approved the final manuscript. Author details 1 Department of Mathematics, Atılım University, İncek, Ankara 06836, Turkey. 2 Department of Mathematics and Computer Science, Çankaya University, Ankara, Turkey. Received: 2 July 2012 Accepted: 23 August 2012 Published: 7 September 2012

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