On coupled generalised Banach and Kannan type contractions

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1 Available online at J. Nonlinear Sci. Appl. 5 (01, Research Article On coupled generalised Banach Kannan type contractions B.S.Choudhury a,, Amaresh Kundu b a Department of Mathematics, Bengal Engineering Science University, Shibpur,Howrah , West Bengal, India b Department of Mathematics, Siliguri Institute of Technology, Sukna, Darjeeling , West Bengal, India. This paper is dedicated to Professor Ljubomir Ćirić Communicated by Professor V. Berinde Abstract In this paper we have proved two theorems in which we have established the existence of coupled fixed point results in partially ordered complete metric spaces for generalised coupled Banach Kannan type mappings. The generalisation has been accomplished by following the line of argument given by Geraghty [Proc. Amer. Math. Soc., 0 (1973, ]. Here the mapping are assumed to satisfy certain contractive type inequalities. We have illustrated our result with two examples. First example is presented to show that our result is a proper generalizations of the corresponding results of Bhaskar et al [Nonlinear Anal. TMA, 65 (7 (006, ]. c 01. All rights reserved. Keywords: Partially ordered set, Contractive-type mapping, Mixed monotone property, Coupled fixed point. 010 MSC: Primary 5H5; Secondary 5H Introduction Preliminaries Fixed point theory in recent has developed rapidly in partially ordered metric spaces; that is, metric spaces endowed with a partial ordering. References [1], [13], [5], [6], [7] are some examples of these works. Fixed point problems have also been considered in generalisation of metric spaces endowed with partial orderings as for example in partially order cone metric spaces [3], in partially ordered G-metric spaces [] in partially ordered probabilistic metric spaces [1]. In their paper Bhaskar Lakshmikantham [16] Corresponding author addresses: binayak1@yahoo.co.in (B.S.Choudhury, kunduamaresh@yahoo.com (Amaresh Kundu Received

2 B.S. Choudhury, A. Kundu, J. Nonlinear Sci. Appl. 5 (01, established a coupled contraction mapping principle in partially ordered metric spaces for mappings having mixed monotone property. An application of their result to differential equations has also been given in the same work. After the publication of this work, several coupled fixed point results have appeared in the literature. The work of Bhaskar et. al; was further generalized to coupled coincidence point theorems in [7] [] under two separate sets of sufficient conditions. Several other coupled fixed coincidence point results were proved in works like those noted in references [3], [6], [11], [17], [9]. Geraghty [15] introduced an extension of the Banach contraction mapping principle in which the contraction constant was replaced by a function having some specified properties. The method applied by Geraghty was utilized to obtain further new fixed point results works like [] [10]. Definition 1.1. [15] Let S is the class of functions β : R + [0, 1 with (i R + = {t R/t > 0}, (ii β(t n 1 implies t n 0. (1.1 With the help of the above class of functions Geraghty [15] had established a generalisation of the Banach contraction principle. Theorem 1.. [15] Let (X, d be a complete metric space let T : X X be a mapping satisfying d(t x, T y β(d(x, y.d(x, y, for x, y X, where β S. Then T has a unique fixed point z X {T n (x} converges to z for each x X. A. Amini-Hari H. Emami, [] has shown that the result which Geraghty had been proved in [15] is also valid in complete partially ordered metric spaces. The following is the result of A. Amini-Hari H. Emami. Theorem 1.3. [] Let (X, be a partially ordered set suppose that there exists a metric d in X such that (X, d is a complete metric space. Let T : X X be a nondecreasing mapping such that d(t x, T y β(d(x, y.d(x, y, for x, y X with x y, where β S. Assume that either T is continuous or X satisfies the following condition: if {x n } is a non decreasing sequence in X such that x n x, then x n x n N. Besides, suppose that for each x, y Xthere exists z X which is comparable to x y. If there exists x 0 X with x 0 T x 0, then T has a unique fixed point. The essential feature of this line of generalisation is that the contraction constant has been replaced by a function belonging to the class S. A further extension of the above result has been done in [10]. Kannan type mappings are a class of contractive mappings which are different from Banach contraction. Like Banach contraction they have unique fixed points in complete metric spaces. However, unlike the Banach condition, there exist discontinuous functions satisfying the definition of Kannan type mappings. Following their appearance in [0], [1], many persons created contractive conditions not requiring continuity of the mappings established fixed points results of such mappings. Today, this line of research has a vast literature. Another reason for the importance of Kannan type mappings is that it characterizes completeness which the Banach contraction principle does not. It has been shown in [30], [3] that the necessary existence of fixed points for Kannan type mappings implies that the corresponding metric space is complete. The same is not true with the Banach contractions. There is an example of an incomplete metric space where every Banach contraction has a fixed point [1]. Kannan type mappings, its generalizations extensions in various spaces have been considered in a large number of works some of which appear in [5], [8], [9], [18], [19], [], [8], [31].

3 B.S. Choudhury, A. Kundu, J. Nonlinear Sci. Appl. 5 (01, Definition 1.. [0, 1] A mapping T : X X, where (X, d is a metric space, is called a Kannan type mapping if there exists 0 < λ < 1 such that, for all x, y X, the following inequality holds: d(t x, T y λ [d(x, T x + d(y, T y]. (1. Let (X, be a partially ordered set F : X X. The mapping F is said to be non-decreasing if for all x 1, x X, x 1 x implies F (x 1 F (x non-increasing if for all x 1 x implies F (x 1 F (x. Definition 1.5. [16] Let (X, be a partially ordered set F : X X X. The mapping F is said to have the mixed monotone property if F is monotone non-decreasing in its first argument is monotone non-increasing in its second argument; that is, for any x, y X, x 1, x X, x 1 x = F (x 1, y F (x, y y 1, y X, y 1 y = F (x, y 1 F (x, y. Definition 1.6. [16] An element (x, y X X, is called a coupled fixed point of the mapping F : X X X if F (x, y = x F (y, x = y. The following coupled contraction mapping theorem was established by Bhaskar et al [16]. Theorem 1.7. [16] Let (X, be a partially ordered set suppose there is a metric d on X such that (X, d is a complete metric space. Assume that X has the following property: 1. if a non-decreasing sequence {x n } x, then x n x, for all n,. if a non-increasing sequence {y n } y, then y y n, for all n. Let F : X X X be a mapping having the mixed monotone property on X. Assume that there exists a k [0, 1 such that for x, y, u, v X with x u, y v the following inequality holds d(f (x, y, F (u, v k [d(x, u + d(y, v]. (1.3 If there exists x 0, y 0 X such that x 0 F (x 0, y 0 y 0 F (x 0, y 0, then there exist x, y X such that x = F (x, y y = F (y, x. In the present work, following the idea of Geraghty [15], we established two coupled coincidence point theorems to generalize Banach Kannan type contractions in partially ordered metric spaces. Two illustrative examples are given. One of our theorems extends the work of Bhaskar et. al [16].. Main Results Theorem.1. Let (X, be a partially ordered set suppose there is a metric d on X such that (X, d is a complete metric space. Let F : X X X be a mapping such that F has the mixed monotone property satisfies d(x, u + d(y, v d(x, u + d(y, v d(f (x, y, F (u, v β( (, (.1 for all x, y, u, v X with x u y v β S. Also suppose that (a F is continuous or (b X has the following properties: i if a non-decreasing sequence {x n } x, then x n x, for all n 0, (. ii if a non-increasing sequence {y n } y, then y n y, for all n 0. (.3 If there are x 0, y 0 X such that x 0 F (x 0, y 0 y 0 F (y 0, x 0, then there exist x, y X such that x = F (x, y y = F (y, x, that is, F has a coupled fixed point in X.

4 B.S. Choudhury, A. Kundu, J. Nonlinear Sci. Appl. 5 (01, Proof. By the condition of the theorem there exist x 0, y 0 X such that x 0 F (x 0, y 0 y 0 F (y 0, x 0. We define x 1, y 1 X as x 1 = F (x 0, y 0 y 1 = F (y 0, x 0. Then x 1 x 0 y 1 y 0. I n the same way, using the mixed monotone property of F we define x = F (x 1, y 1 y = F (y 1, x 1. Then x = F (x 1, y 1 F (x 0, y 1 F (x 0, y 0 = x 1 y = F (y 1, x 1 F (y 1, x 0 F (y 0, x 0 = y 1. Continuing the above procedure we have two sequences {x n } {y n } in X such that Due to the mixed monotone property of F, we have, x n+1 = F (x n, y n y n+1 = F (y n, x n for all n 0. (. x 0 F (x 0, y 0 = x 1 F (x 1, y 1 = x... x n = F (x n 1, y n 1 x n+1 = F (x n, y n... (.5 y 0 F (y 0, x 0 = y 1 F (y 1, x 1 = y... y n = F (y n 1, x n 1 y n+1 = F (y n, x n.... (.6 From (.1, (., (.5 (.6, for all n 1, it follows that d(x n, x n+1 = d(f (x n 1, y n 1, F (x n, y n Let, for all n 0, β( d(x n 1, x n + d(y n 1, y n ( d(x n 1, x n + d(y n 1, y n d(y n+1, y n = d(f (y n 1, x n 1, F (y n, x n β( d(y n 1, y n + d(x n 1, x n ( d(y n 1, y n + d(x n 1, x n. a n = d(x n, x n+1, b n = d(y n, y n+1 δ n+1 = d(x n, x n+1 + d(y n, y n+1. (.7 Then from the above two inequalities, for all n 1, δ n+1 = d(x n+1, x n + d(y n+1, y n β( d(y n 1, y n + d(x n 1, x n (d(y n 1, y n + d(x n 1, x n β( d(y n 1, y n + d(x n 1, x n δ n (.8 δ n. Therefore the sequence {δ n } is a monotone decreasing sequence of non-negative real numbers. Hence there exists δ 0 such that lim n δ n = δ. Assume δ > 0. (.9 Then from (.8 we have δ n+1 β( d(y n 1, y n + d(x n, x n 1 < 1. δ n Letting n in the above inequality, using (.9, we get By virtue of (1.1, this implies that lim β(d(y n 1, y n + d(x n, x n 1 = 1. n lim n {d(x n 1, x n + d(y n 1, y n } = lim n δ n = 0

5 B.S. Choudhury, A. Kundu, J. Nonlinear Sci. Appl. 5 (01, which is a contraction with (.9. Hence lim a n = lim d(x n 1, x n = 0 lim b n = lim d(y n 1, y n = 0. (.10 n n n n Next we show that {x n } {y n } are Cauchy sequences. If possible, let at least one of {x n } {y n } be not a Cauchy sequence. Then there exists ɛ > 0 sequences of natural numbers {m(k} {l(k} for which m(k > l(k k, such that for all k 1, either d(x l(k, x m(k ε or d(y l(k, y m(k ε. Then for all k 1, d k = d(x l(k, x m(k + d(y l(k, y m(k ε. (.11 Now corresponding to l(k we can choose m(k to be the smallest positive integer for which (.11 holds. Then, for all k 1, d(x l(k, x m(k 1 + d(y l(k, y m(k 1 < ε. (.1 Further from (.7, (.11 (.1, for all k 1, we have ε d k = d(x l(k, x m(k + d(y l(k, y m(k d(x l(k, x m(k 1 + d(x m(k 1, x m(k + d(y l(k, y m(k 1 + d(y m(k 1, y m(k = d(x l(k, x m(k 1 + d(y l(k, y m(k 1 + a m(k 1 + b m(k 1 < ε + a m(k 1 + b m(k 1. Taking the limit as k, using (.10 we have From (.1, (., (.5, (.6, for all k 1, we obtain lim d k = ε. (.13 k d(x l(k+1, x m(k+1 = d(f (x l(k, y l(k, F (x m(k, y m(k β( d(x l(k, x m(k + d(y l(k, y m(k Also from (.1, (., (.5, (.6, for all k 0, we have d(y l(k+1, y m(k+1 = d(f (y l(k, x l(k, F (y m(k, x m(k ( d(x l(k, x m(k + d(y l(k, y m(k. (.1 β( d(y l(k, y m(k + d(x l(k, x m(k ( d(y l(k, y m(k + d(x l(k, x m(k. (.15 Again, for all k 1, we have d k = d(x l(k, x m(k + d(y l(k, y m(k d(x l(k, x l(k+1 + d(x l(k+1, x m(k+1 + d(x m(k+1, x m(k +d(y l(k, y l(k+1 + d(y l(k+1, y m(k+1 + d(y m(k+1, y m(k d(x l(k+1, x m(k+1 + d(y l(k+1, y m(k+1 + d(x l(k, x l(k+1 + d(x m(k+1, x m(k +d(y l(k, y l(k+1 + d(y m(k+1, y m(k d(x l(k+1, x m(k+1 + d(y l(k+1, y m(k+1 d(x l(k+1, x l(k + d(x l(k, x m(k + d(x m(k, x m(k+1 +d(y l(k+1, y l(k + d(y l(k, y m(k + d(y m(k, y m(k+1 d k + d(x l(k+1, x l(k + d(x m(k+1, x m(k + d(y l(k+1, y l(k +d(y m(k, y m(k+1.

6 B.S. Choudhury, A. Kundu, J. Nonlinear Sci. Appl. 5 (01, Taking the limit as k in the above two inequalities, using (.10 (.13, we have lim {d(x l(k+1, x m(k+1 + d(y l(k+1, y m(k+1 } = ε. (.16 k Adding (.1, (.15, using the notation of (.11, we have, d(x l(k+1, x m(k+1 + d(y l(k+1, y m(k+1.β( d k (d k d k (Since β S. Letting k in the above inequality, using (.13 (.16 we obtain ε lim k β( d k.ε ε, which implies lim k β( d k = 1. Since β S, it follows that lim k d k = lim k {d(x l(k, x m(k + d(y l(k, y m(k } = 0, which, by the virtue of (.13, contradicts the fact that ε > 0. Therefore, {x n } {y n } are Cauchy sequences in X hence they are convergent in the complete metric space (X, d. Let x n x as n (.17 y n y as n. (.18 Next we prove that x = F (x, y y = F (y, x. Let condition (a of the theorem.1 hold, that is, F is continuous. From (., (.17 (.18 we have respectively x = lim x n+1 = F ( lim x n, lim y n = F (x, y n n n y = lim y n+1 = F ( lim y n, lim x n = F (y, x. n n n Let condition (b of the theorem.1 hold. From (.5, (.6, (.17 (.18 we have that {x n } is non-decreasing such that x n x {y n } is non-increasing such that y n y as n. Then by (. (.3 we have, for all n 0, Then by (.1, using (.19, for all n 0, we have x n x y n y. (.19 d(f (x, y, x n+1 = d(f (x, y, F (x n, y n β( d(x, x n + d(y, y n ( d(x, x n + d(y, y n ( d(x, x n + d(y, y n (Since β S. Taking n in the above inequality, using (.17, we have d(f (x, y, x = 0, that is, x = F (x, y. Similarly, we have y = F (y, x. Thus we have proved that F has a coupled fixed point in X. This completes the proof of the theorem. Our next theorem is a Kannan type coupled fixed point result. Theorem.. Let (X, be a partially ordered set suppose there is a metric d on X such that (X, d is a complete metric space. Let F : X X X be a mapping such that F has the mixed monotone property satisfies d(f (x, y, F (u, v β(m(x, y, u, v.(m(x, y, u, v, (.0 d(x, F (x, y + d(y, F (y, x + d(u, F (u, v + d(v, F (v, u where, M(x, y, u, v = for all x, y, u, v X with x u y v β S,. Also suppose that (a F is continuous or (b X has the properties noted in (. (.3. If there exist x 0, y 0 X such that x 0 F (x 0, y 0 y 0 F (y 0, x 0, then there exist x, y X such that x = F (x, y y = F (y, x, that is, F has a coupled fixed point in X.

7 B.S. Choudhury, A. Kundu, J. Nonlinear Sci. Appl. 5 (01, Proof. Following the proof of the theorem (.1 we have two sequences {x n } {y n } given by (. satisfying (.5 (.6. Then {x n } is an increasing sequence {y n } is a decreasing sequence in X. Now by (.1, (., (.5 (.6 the fact that β S, we have for all n 1, d(x n+1, x n = d(f (x n, y n, F (x n 1, y n 1 Let, for all n 1, β(m(x n, y n, x n 1, y n 1 (M(x n, y n, x n 1, y n 1 d(y n+1, y n = d(f (y n, x n, F (y n 1, x n 1 β(m(y n, x n, y n 1, x n 1 (M(y n, x n, y n 1, x n 1 a n = d(x n+1, x n, b n = d(y n+1, y n δ n+1 = d(x n+1, x n + d(y n+1, y n. (.1 Then adding the above two inequalities using (.1, for all n 1, we have δ n+1 = d(x n+1, x n + d(y n+1, y n.β(m(x n, y n, x n 1, y n 1 (M(x n, y n, x n 1, y n 1 where, M(x n, y n, x n 1, y n 1 = d(x n, x n+1 + d(y n, y n+1 + d(x n 1, x n + d(y n 1, y n = δ n + δ n+1 Therefore, for all n 1, it follows that (by (.1. δ n+1 β(m(x n, y n, x n 1, y n 1 ( δ n + δ n+1 (. δ n + δ n+1, since β S, that is, δ n+1 δ n. Therefore the sequence {δ n } is a monotone decreasing sequence of non-negative real numbers. Hence there exists δ 0 such that lim n δ n = δ. Assume δ > 0. (.3. Then by (. we have δ n+1 ( δ β(m(x n + δ n, y n, x n 1, y n 1 < 1. n+1 Letting n in the above inequality, using (.3, we get Due to (1.1 the above limit in (. implies that lim β(m(x n, y n, x n 1, y n 1 = 1, (. n lim {d(x n, x n+1 + d(y n, y n+1 + d(x n 1, x n + d(y n 1, y n } = lim {δ n+1 + δ n } = 0. n n But this contradicts (.3. Hence lim a n = lim d(x n+1, x n = 0 lim b n = lim d(y n+1, y n = 0. (.5 n n n n Next we show that {x n } {y n } are Cauchy sequences. If possible, let at least one of {x n } {y n } be not a Cauchy sequence. Then there exists ɛ > 0 sequences of natural numbers {m(k} {l(k} for which m(k > l(k k,

8 B.S. Choudhury, A. Kundu, J. Nonlinear Sci. Appl. 5 (01, such that for all k 1, either d(x l(k, x m(k ε or d(y l(k, y m(k ε. Then for all k 1, d k = d(x l(k, x m(k + d(y l(k, y m(k ε. (.6 Now corresponding to l(k we can choose m(k to be the smallest positive integer for which (.6 holds. Then, d(x l(k, x m(k 1 + d(y l(k, y m(k 1 < ε. (.7 Further from (.1, (.6 (.7, for all k 0, we have ε d k = d(x l(k, x m(k + d(y l(k, y m(k d(x l(k, x m(k 1 + d(x m(k 1, x m(k + d(y l(k, y m(k 1 + d(y m(k 1, y m(k = d(x l(k, x m(k 1 + d(y l(k, y m(k 1 + a m(k 1 + b m(k 1 < ε + a m(k 1 + b m(k 1. Taking the limit as k, using (.5 we have From (., (.5, (.6 (.0, for all k 0, we obtain lim d k = ε. (.8 k d(x l(k, x m(k = d(f (x l(k 1, y l(k 1, F (x m(k 1, y m(k 1 β(m(x l(k 1, y l(k 1, x m(k 1, y m(k 1 (M(x l(k 1, y l(k 1, x m(k 1, y m(k 1. (.9 Also by (., (.5, (.6 (.0, for all k 0, we have d(y l(k, y m(k = d(f (y l(k 1, x l(k 1, F (y m(k 1, x m(k 1 β(m(y l(k 1, x l(k 1, y m(k 1, x m(k 1 (M(y l(k 1, x l(k 1, y m(k 1, x m(k 1. (.30 Adding (.9 (.30, we have, d k = d(x l(k, x m(k + d(y l(k, y m(k Further, by (.5,.β(M(x l(k 1, y l(k 1, x m(k 1, y m(k 1 (M(x l(k 1, y l(k 1, x m(k 1, y m(k 1. (.31 lim M(x l(k 1, y l(k 1, x m(k 1, y m(k 1 k d(x l(k 1, x l(k + d(y l(k 1, y l(k + d(x m(k 1, x m(k + d(y m(k 1, y m(k = lim = 0. (.3 k Taking k in (.31, using (.8 (.3, we obtain ε = 0, which is a contradiction. Therefore, {x n } {y n } are Cauchy sequences in X hence they are convergent in the complete metric space (X, d. Let x n x as n (.33 y n y as n. (.3 Next we prove that x = F (x, y y = F (y, x. Let condition (a of the theorem. hold, that is, F is continuous. From (., (.30 (.31 we have respectively x = lim x n+1 = F ( lim x n, lim y n = F (x, y, n n n y = lim n y n+1 = F ( lim n y n, lim n x n = F (y, x.

9 B.S. Choudhury, A. Kundu, J. Nonlinear Sci. Appl. 5 (01, Let condition (b of the theorem. hold. Using (.5, (.6, (.33 (.3 we have that {x n } is non-decreasing such that x n x {y n } is non-increasing such that y n y as n. Then by (. (.3 we have, for all n 0, x n x y n y. (.35 By the virtue of (.35, since β S, from (.0, for all n 0, we have d(f (x, y, x n+1 = d(f (x, y, F (x n, y n β( d(x, F (x, y + d(y, F (y, x + d(x, x n+1 + d(y, y n+1 ( d(x, F (x, y + d(y, F (y, x + d(x, x n+1 + d(y, y n+1 < ( d(x, F (x, y + d(y, F (y, x + d(x, x n+1 + d(y, y n+1. (.36 Similarly, due to (.35 the fact that β S, from (.0, for all n 0, we have d(f (y, x, y n+1 = d(f (y, x, F (y n, x n β( d(x, F (x, y + d(y, F (y, x + d(x, x n+1 + d(y, y n+1 ( d(x, F (x, y + d(y, F (y, x + d(x, x n+1 + d(y, y n+1 < ( d(x, F (x, y + d(y, F (y, x + d(x, x n+1 + d(y, y n+1. (.37 Adding (.36 (.37, for all n 0, we obtain d(f (x, y, x n+1 + d(f (y, x, y n+1 <.( d(x, F (x, y + d(y, F (y, x + d(x, x n+1 + d(y, y n+1 Taking n, using (.5, (.33 (.3, we have d(f (x, y, x + d(f (y, x, y ( d(f (x, y, x + d(f (y, x, y, which implies that d(f (x, y, x + d(f (y, x, y = 0, that is, x = F (x, y y = F (y, x. Thus (x, y is a coupled fixed point of F in X. This completes the proof of the theorem.. Remark.3. Since inequality (.0 bears the same idea of Kannan s inquality described in (1., we recognised (.0 as a coupled Kannan type inequality. 3. Example In this section we have two examples which illustrated the results of theorems.1. respectively. Example 3.1. Let X = [0, 1]. Then (X, is a partially ordered set with x y whenever x y. Let d(x, y = x y for x, y [0, 1]. Then (X,d is a complete metric space. Let F : X X X be defined{ as 1 F (x, y = [(x y 1 (x y ], if x, y [0, 1], x y, 0, if x < y. Let β : [0, [0, 1 be defined as

10 B.S. Choudhury, A. Kundu, J. Nonlinear Sci. Appl. 5 (01, β(t = { 1 t, if t 1, α < 1, if t > 1. For the two points, x 0 = 0 y 0 = c > 0 in X we have x 0 = 0 = F (x 0, y 0 y 0 = c > c [1 c ] = F (y 0, x 0. Let x u y v (equivalently u x y v. The inequality (.1 is trivially satisfied except in the following two cases. Case-1 x y u v (x y u v. Then d(f (x, y, F (u, v = d( 1 [(x y 1 (x y ], 1 [(u v 1 (u v ] = 1 (u v 1 (u v 1 (x y + 1 (x y = 1 [(u x + (y v] 1 [{(u v + (x y}{(u v (x y}] 1 [(u x + (y v] 1 [{(u v (x y} ] = 1 [(u x + (y v] 1 [{(u x + (y v} ]. Therefore, d(f (x, y, F (u, v 1 [(u x + (y v] 1 [(u x + (y v] [ ] [ ] (u x + (y v (u x + (y v = ( [ ] [ ] (u x + (y v (u x + (y v = 1 d(x, u + d(y, v d(x, u + d(y, v = β( (. Case- x < y u v (y x u v. We have, d(f (x, y, F (u, v = d(0, 1 (u v 1 (u v = 1 (u v 1 (u v = 1 (u v + x x 1 (u v + x x < 1 (u x v + y 1 (u v + y x (since y > x = 1 [(u x + (y v] 1 [(u x + (y v]. Then, d(f (x, y, F (u, v < 1 [(u x + (y v] 1 [ (u x + (y v = d(x, u + d(y, v = β( ( ] [ (u x + (y v d(x, u + d(y, v. [(u x + (y v] [ ] [ ] (u x + (y v (u x + (y v ] = ( 1 Thus in both of the above two cases inequality (.1 is satisfied. Also the functions F β satisfy all the conditions required by them in theorem.1. Then, by an application of theorem.1, F has a coupled fixed point. Here (0, 0 is a coupled fixed point of F in X. Example 3.. Let X = [0, 1]. Then X with the usual order be a partially ordered set. Let d be the usual metric on X. Then (X, d is a complete metric space. We define F : X X X as { 1 F (x, y = 16, if x 1 y < 1, 0, otherwise.

11 B.S. Choudhury, A. Kundu, J. Nonlinear Sci. Appl. 5 (01, Let β(t = { 1 t, if t 1, 8 α < 1, if t > 1. Then all the properties required by F β in the theorem. are satisfied. The inequality (.0 is trivially satisfied except in the following two cases. Case-1: x [ 1, 1], y [0, 1. u, v [ 1, 1] d(f (x, y, F (u, v = 1 16 d(x, F (x, y + d(y, F (y, x + d(u, F (u, v + d(v, F (v, u Now = x y + u + v Then the inequality (.0 becomes 1 16 (x y + u + v (x y + u + v. 3 Since 3 x + y + u + v 7, the right h side of the above inequality is bounded below by 0.35, Hence (.0 is satisfied. Case-: x [ 1, 1], y [0, 1 u, v [0, 1 ] d(f (x, y, F (u, v = y + u + v 1 Therefore, 16 (x 1 Since 1 x + y + u + v 5 (x y + u + v. 3 1, therefore the value of the right h side is greater than 16, which shows that (.0 is satisfied. This shows that theorem. is applicable to this example Here (0, 0 is a coupled fixed point of F in X. Remark: If, in particular, we consider the function β(t = k, 0 < k < 1, then the inequality (.1 reduces to inequality (1.3, thus the result of Bhaskar et al; is a special case of theorem.1. Then result of T. Gnana Bhaskar V. Lakshmikantham in [16] is not applicable to example 3.1. Further inequality (1.3 is not satisfied for the choice of x = 0, y = 0, u = c v = 0. This shows that the result of theorem.1 is an actual improvement over the corresponding results of Bhaskar et. al [16]. Acknowledgements The work is partially supported by Council of Scientific Industrial Research, India(Project No. 5(0168/ 09/EMR-II. The first author gratefully acknowledges the support. References [1] R. P. Agarwal, M. A. El-Gebeily D. O Regan, Generalized contractions in partially ordered metric spaces, Appl. Anal., 87(1 (008, [] A. Amini-Hari H. Emami, A fixed point theorem for contraction type maps in partially ordered metric spaces application to ordinary differential equations, Nonlinear Anal. TMA, 7(5 (010, , 1, 1.3 [3] H. Aydi, Some coupled fixed point results on partial metric spaces, Int. J. Math. Math. Sci., 001, (011, Article ID 67091, 11 pages, doi: /011/ [] B. S. Choudhury P. Maity, Coupled fixed point results in generalized metric spaces, Math. Comput. Modelling, 5 (011, [5] B. S. Choudhury K.Das, Fixed points of generalized Kannan type mappings in generalized Menger spaces, Commun. Korean Math. Soc., (009, [6] B. S. Choudhury, N. Metiya A. Kundu, Coupled coincidence point theorems in ordered metric Spaces, Ann. Univ. Ferrara, 57 (011, [7] B. S. Choudhury A. Kundu, A coupled coincidence point result in partially ordered metric spaces for compatible mappings, Nonlinear Anal. TMA, 73 (010,

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