COUPLED FIXED AND COINCIDENCE POINT THEOREMS FOR GENERALIZED CONTRACTIONS IN METRIC SPACES WITH A PARTIAL ORDER

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1 ITALIAN JOURNAL OF PURE AND APPLIED MATHEMATICS N ( ) 434 COUPLED FIXED AND COINCIDENCE POINT THEOREMS FOR GENERALIZED CONTRACTIONS IN METRIC SPACES WITH A PARTIAL ORDER K. Ravibabu Department of Mathematics G.M.R.I.T, Rajam Srikakulam India konchada 2011@rediffmail.com Ch. Srinivasa Rao Department of Mathematics Mrs. A. V. N. College Visakhapatnam Indiadrcsr41@yahoo.com Ch. Raghavendra Naidu Department of Mathematics Govt. Degree College Palakonda, Srikakulam , India ch.rvnaidu@gmail.com Abstract. In this paper, we establish results on the existence uniqueness of coupled common fixed point theorems coupled coincidence fixed point theorems for such non-linear contraction mappings having a mixed monotone property in partially ordered complete metric spaces with out using continuity.our results generalize extend the results of V. Lakshmikantham L. Ciric [13], Sintunavarat Poom Kumam [16]. Keywords: coupled fixed point, coupled coincidence point, mixed monotone property, partially ordered set. 1. Introduction Recently V. Lakshmikantham L. Ciric [13] generalized the concept of coupled fixed point theorems for non-linear contractions in partially ordered metric spaces. Subsequently Sintunavarat Poom Kumam [16] studied unique coupled fixed point theorem in partially ordered metric spaces. The aim of this paper is to extend the results of T.G. Bhaskar V. Lakshmikantham [5] V. Lakshmikantham L. Ciric [13] Sintunavarat Poom Kumam [16]. Corresponding author

2 COUPLED FIXED AND COINCIDENCE POINT THEOREMS for a mixed monotone non-linear contractive mapping to generalize the notion of a mixed monotone mapping We proved some coupled coincidence coupled common fixed point theorems for a pair of mappings.our results extend the recent fixed point theorems due to V. Lakshmikantham L. Ciric [13], fixed point theorems due to Sintunavarat Poom Kumam [16] include several recent developments. Suppose (X, ) is a partially ordered set. Let F : X X be such that for x, y X, x y F (x) F (y). Then the mapping F is said to be nondecreasing, similarly a non-increasing mapping is defined, Bhaskar Lakshmikantham [5] introduced the following notions of a coupled fixed point theorems. Before going to prove the main result, we need some basic definitions results from the literature. 2. Preliminaries Definition 2.1 (Bhaskar Lakshmikantham [5]). 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 monotone non-increasing in its second argument,that is, for any x, y X, x 1, x 2 X, x 1 x 2 F (x 1, y) F (x 2, y) y 1, y 2 X, y 1 y 2 F (x, y 1 ) F (x, y 2 ). Definition 2.2 (Bhaskar Lakshmikantham [5]). Let X be a nonempty set. 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 Bhaskar Lakshmikantham [5] proved the following two coupled fixed point theorems. Theorem 2.3 (Bhaskar Lakshmikantham [5], Theorem 2.1). 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 continuous mapping having the mixed monotone property on X. Assume there exists a k [0, 1) with (2.3.1) d(f (x, y), F (u, v)) k (d(x, u) + d(y, v)) 2 for each x u y v. 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). Theorem 2.4 (Bhaskar Lakshmikantham [5], Theorem 2.2). 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:

3 436 K. RAVIBABU, CH. SRINIVASA RAO CH. RAGHAVENDRA NAIDU i) if a non decreasing sequence {x n } x, then x n x, for all n N. ii) if a non increasing sequence {y n } y,then y y n, for all n N. Let F : X X X be a continuous mapping having the mixed monotone property on X. Assume that there exists a k [0, 1) with d(f (x, y), F (u, v)) k (d(x, u) + d(y, v)) for all x, y, u, v X, 2 for which x u y v. If there exists 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). Definition 2.5. (V. Lakshmikantham L. Ciric [13]). Let (X, ) be a partially ordered set F : X X X g : X X. The mapping F is said to have the mixed g-monotone property, if F is monotone g-non decreasing in its first argument is monotone g-non increasing in its second argument, that is, for any x, y X x 1, x 2 X, g(x 1 ) g(x 2 ) F (x 1, y) F (x 2, y) y 1, y 2 X, g(y 1 ) g(y 2 ) F (x, y 1 ) F (x, y 2 ). Definition 2.6 (Lakshmikantham Ciric [12]). Let X be a nonempty set. An element (x, y) X X is called a coupled coincidence point of a mapping F : X X X g : X X, if x = g(x) = F (x, y) y = g(y) = F (y, x). Definition 2.7 (Lakshmikantham Ciric [12]). Let X be a nonempty set F : X X X, g : X X. We say F g are commutative, if g(f (x, y)) = F (g(x), g(y)), for all x, y X. Theorem 2.8 (Lakshmikantham Ciric [12], Theorem 2.1). Let (X, ) be a partially ordered set suppose there is a metric d on X, such that (X, d) is a complete metric space. Assume there is a function φ : [0, ) [0, ) with φ(t) < t lim r t + φ(r) < t for each t > 0, also suppose that F : X X X g : X X are such that F has the mixed g-monotone property ( ) d(g(x), g(u)) + d(g(y), g(v)) (2.8.1) d(f (x, y), F (u, v)) φ, 2 for all x, y, u, v X for which g(x) g(u) g(y) g(v). Suppose F (X X) g(x), g is continuous commutes with F also suppose either (a) F is continuous (or) (b) X has the following property. i) if a non decreasing sequence {x n } x, then x n x, for all n N. ii) if a non increasing sequence {y n } y, then y y n, for all n N. If there exist x 0, y 0 X such that g(x 0 ) F (x 0, y 0 ) g(y 0 ) F (y 0, x 0 ), then there exist x, y X such that g(x) = F (x, y) g(y) = F (y, x), that is, F g have a coupled coincidence point.

4 COUPLED FIXED AND COINCIDENCE POINT THEOREMS Theorem 2.9 (Lakshmikantham Ciric [12], Theorem 2.2). In addition to the hypothesis of Theorem 2.8, suppose that F g are commutative for every (x, y), (z, t) X X, there exists a (u, v) X X such that (F (u, v), F (v, u)) is comparable to (F (x, y), F (y, x)) (F (z, t), F (t, z)).then F g have a unique coupled common fixed point, that is, there exists a unique (x, y) X X such that x = g(x) = F (x, y) y = g(y) = F (y, x). In 2013, Sintunavarat Kumam [16] gave an extension of the result of Bhaskar Lakshmikantham [5], Lakshmikantham Ciric [13]. They used this concept to establish the existence of coupled coincidence point coupled common fixed point theorem. Theorem 2.10 (Sintunavarat Kumam [16], Theorem 2.1). Let (X, ) be a partially ordered set suppose there is a metric d on X, such that (X, d) is a complete metric space. Assume there is a function φ : [0, ) [0, ) with φ(t) < t lim r t + φ(r) < t, for each t > 0 also suppose that F : X X X is such that F has the mixed monotone property ( ) d(x, u) + d(y, v) (2.10.1) d(f (x, y), F (u, v)) φ, 2 for all x, y, u, v X for which x u y v. Suppose either (a) F is continuous (or); (b) X has the following property. i) if a non decreasing sequence {x n } x, then x n x for all n N; ii) if a non increasing sequence {y n } y, then y y n for all n N. 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. Theorem 2.11 (Sintunavarat Kumam [16], Theorem 2.2). In addition to the hypothesis of Theorem 2.10, suppose that for every (x, y), (z, t) X X, there exists a (u, v) X X, which is comparable to (x, y) (z, t). Then F has a unique coupled fixed point. Corollary 2.12 (Sintunavarat Kumam [16], Corollary 2.4). Let (X, ) be a partially ordered set suppose there is a metric d on X, such that (X, d) is a complete metric space. Suppose that F : X X X g : X X are such that F has the mixed g-monotone property assume there is a k [0, 1) such that (2.12.1) d(f (x, y), F (u, v)) k (d(g(x), g(u)) + d(g(y), g(v))), 2 for all x, y, u, v X for which g(x) g(u) g(y) g(v).

5 438 K. RAVIBABU, CH. SRINIVASA RAO CH. RAGHAVENDRA NAIDU Suppose F (X X) g(x), g is continuous suppose either (a) F is continuous (or) (b) X has the following property. i) if a non decreasing sequence {x n } x, then x n x for all n N. ii) if a non increasing sequence {y n } y, then y y n for all n N. If there exist x 0, y 0 X such that g(x 0 ) F (x 0, y 0 ) g(y 0 ) F (y 0, x 0 ),then there exist x, y X such that g(x) = F (x, y) g(y) = F (y, x), that is, F g have a coupled coincidence fixed point. 3. Main result In this section, we improve the results in section 2, by replacing the conditions (i) φ(t) < t (ii). lim r t + φ(r) < t by the single condition: φ(t + 0) < t, the average in the argument of φ by maximum. We introduce the class Φ of functions as follows: Φ = {φ/φ : [0, ) [0, ), φ is increasing φ(t + 0) < t t > 0}. We observe that φ Φ φ(t) < t lim r t + φ(r) < t t > 0. Now we prove our main result. Theorem 3.1. Let (X, ) be a partially ordered set suppose there is a metric d on X, such that (X, d) is a complete metric space. Assume there is an increasing function φ : [0, ) [0, ) with φ(t + 0) < t for each t > 0, also suppose that F : X X X is such that F has the mixed monotone property (3.1.1) d(f (x, y), F (u, v)) φ{max(d(x, u), d(y, v))}, for all x, y, u, v X for which x u y v. Suppose either a) F is continuous (or); b) X has the following property. i) if a non decreasing sequence {x n } x, then x n x, for all n N. ii) if a non increasing sequence {y n } y, then y y n, for all n N. 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. Proof. Suppose (3.1.2) u 0 F (u 0, v 0 ) v 0 F (v 0, u 0 ). Define the sequences {u n } {v n } by u 1 = F (u 0, v 0 ) v 1 = F (v 0, u 0 ). In general u n+1 = F (u n, v n ) v n+1 = F (v n, u n ), for n = 0, 1, 2,... From (3.1.2), u 0 F (u 0, v 0 ) = u 1. Therefore u 0 u 1 v 0 F (v 0, u 0 ) = v 1. Therefore v 0 v 1. Now u 2 = F (u 1, v 1 ) F (u 0, v 1 ) F (u 0, v 0 ) = u 1.

6 COUPLED FIXED AND COINCIDENCE POINT THEOREMS Therefore u 2 u 1. And v 2 = F (v 1, u 1 ) F (v 0, u 1 ) F (v 0, u 0 ) = v 1. Therefore v 2 v 1. Similarly u 3 u 2 v 3 v 2. In general u n+1 u n v n+1 v n. Therefore u 0 u 1 u 2... u n u n+1 v 0 v 1 v 2... v n v n+1. Therefore {u n } is {v n } is. First we show that {u n } {v n } are Cauchy sequences. If possible assume either {u n } or {v n } fails to be Cauchy. Then either or Therefore, i.e lim d(u m, u n ) 0 m,n lim d(v m, v n ) 0. m,n max{ lim d(u m, u n ), lim d(v m, v n )} = 0, m,n m,n lim max{ lim d(u m, u n ), lim d(v m, v n )} = 0, m,n m,n m,n i.e, there exist ε > 0, for which we can find sub sequences {m k } {n k } of positive integers with n k > m k > k such that (3.1.3) max{d(u mk, u nk ), d(v mk, v nk )} ε. Further, we choose {n k } to be the smallest positive integer such that n k > m k satisfying (3.1.1). Hence, we have max{d(u mk, u nk ), d(v mk, v nk )} ε (3.1.4) max{d(u mk, u nk 1), d(v mk, v nk 1)} < ε. Now, we prove that I. lim k max{d(u nk, u mk ), d(v nk, v mk )} = ε; II. lim k max{d(u nk 1, u mk 1), d(v nk 1, v mk 1)} = ε; III. lim k max{d(u mk, u nk 1), d(v mk, v nk 1)} = ε. First we prove I: From the triangular inequality we have (3.1.5) d(u nk, u mk ) d(u nk, u nk 1) + d(u nk 1, u mk ) < d(u nk, u nk 1) + ε (3.1.6) d(v nk, v mk ) d(v nk, v nk 1) + d(v nk 1, v mk ) < d(v nk, v nk 1) + ε. From (3.1.3),(3.1.5) (3.1.6) (3.1.7) ε max{d(u nk, u mk ), d(v nk, v mk )} < max{d(u nk, u mk ), d(v nk, v nk 1)}.

7 440 K. RAVIBABU, CH. SRINIVASA RAO CH. RAGHAVENDRA NAIDU On letting k, Therefore Therefore (I) holds. Now we prove II: Therefore Therefore ε lim k max{d(u n k, u mk ), d(v nk, v mk )} ε. lim max{d(u n k, u mk ), d(v nk, v mk )} = ε k d(u nk 1, u mk 1) d(u mk 1, u mk ) + d(u mk, u nk 1) < d(u mk 1, u mk ) + ε ( by (3.1.4)) d(v nk 1, v mk 1) d(v mk 1, v mk ) + d(v mk, v nk 1) < d(v mk 1, v mk ) + ε ( by (3.1.4)). max{d(u nk 1, u mk 1), d(v nk 1, v mk 1)} Max{d(u mk 1, u mk ), d(v mk 1, v mk )} + ε. (3.1.8) lim sup max{d(u nk 1, u mk 1), d(v nk 1, v mk 1)} ε. Now Therefore d(u nk, u mk ) d(u nk, u nk 1) + d(u nk 1, u mk 1) + d(u mk 1, u mk ) d(v nk, v mk ) d(v nk, v nk 1) + d(v nk 1, v mk 1) + d(v mk 1, v mk ). {d(u nk, u mk ), d(v nk, v mk )} max{d(u nk, u nk 1), d(v nk 1, v nk )} Letting k, we get + max{d(v nk 1, v mk 1), d(v mk 1, u mk 1)} + max{d(u mk 1, u mk ), d(v mk 1, v mk )}. 0 lim inf max{d(u nk 1, u mk 1), d(v nk 1, v mk 1)} lim sup max{d(u nk 1, u mk 1), d(v nk 1, v mk 1)} ε. Therefore lim max{d(u nk 1, u mk 1 ), d(v nk 1, v mk 1 )} = ε.

8 COUPLED FIXED AND COINCIDENCE POINT THEOREMS Therefore (II) holds. Now d(u nk, u mk ) d(u nk, u nk 1) + d(u nk 1, u mk ) d(u nk, u nk 1) + d(u nk 1, u mk 1) + d(u mk 1, u mk ) d(v nk, v mk ) d(v nk, v nk 1) + d(v nk 1, v mk ) d(v nk, v nk 1) + d(v nk 1, v mk 1) + d(v mk 1, v mk ). Therefore max{d(u nk, u mk ), d(v nk, v mk )} max{d(u nk, u nk 1), d(v nk, v nk 1)} + max{d(u nk 1, u mk ), d(v nk 1, v mk )} max{d(u nk, u nk 1), d(v nk, v nk 1)} + max{d(u nk 1, u mk 1), d(v nk 1, v mk 1)} + max{d(u mk 1, u mk ), d(v mk 1, v mk )}. On letting k, from (I), (II), we get (III), since ε 0 + lim max{d(u nk 1, u mk ), d(v nk 1, v mk )} ε. Now we have ε lim inf max{d(u nk 1, u mk 1), d(v nk 1, v mk 1)} lim sup max{d(u nk 1, u mk 1), d(v nk 1, v mk 1)} = ε. Since u nk 1 u mk 1 v nk 1 v mk 1. From (3.1.1),we get (3.1.9) d(u nk, u mk ) = d(f (u nk 1, v nk 1), F (u mk 1, v mk 1)) φ{max{d(u nk 1, u mk 1), d(v nk 1, v mk 1)}} similarly (3.1.10) d(v nk, v mk ) = d(f (v nk 1, u mk 1), F (v mk 1, u nk 1)) φ{max{d(v nk 1, v mk 1), d(u mk 1, u nk 1)}}. From (3.1.9) (3.1.10),we have (3.1.11) ε s k φ(p k ) < p k. Where s k = max{d(u nk, u mk ), d(v nk, v mk )}. Therefore s k ε. p k = max{d(u nk 1, u mk 1), d(v nk 1, v mk 1)}. On letting k, from (3.1.11), we have ε lim φ(p k ) ε. Therefore ε = lim φ(p k ) < ε, by (I), a Contradiction. Hence {u n } {v n } are Cauchy sequences. Suppose u n u v n v. Suppose (a) holds. Then F is continuous, hence u n+1 = F (u n, v n ) F (u, v). So u = F (u, v), since u n+1 u. Similarly v n+1 = F (v n, u n ) F (v, u). So

9 442 K. RAVIBABU, CH. SRINIVASA RAO CH. RAGHAVENDRA NAIDU v = F (v, u), since v n+1 v. Therefore (u, v) is a coupled fixed point of F. Now suppose (b) holds. Then u n u v n v for all n. d(u n+1, F (u, v)) = d(f (u n, v n ), F (u, v)) φ{max (d(u n, u), d(v n, v))} max{d(u n, u), d(v n, v)} 0 as n i.e, d(u n+1, F (u, v)) 0 as n. i.e, u n+1 F (u, v). So u = F (u, v). Similarly v = F (v, u). Therefore (u, v) is a coupled fixed point of F. Lemma 3.2. Under the hypothesis of Theorem 3.1, suppose (x, y) is a coupled fixed point of F (u, v) is comparable to (x, y). Write u 0 = u v 0 = v construct the sequences {u n } {v n } by u n+1 = F (u n, v n ) v n+1 = F (v n, u n ) for n = 0, 1, 2, 3,... Then {u n } {v n } are Cauchy sequences {u n } x {v n } y. Proof. Case (i): Suppose (u, v) (x, y), so that (3.2.1) u x v y. Write (3.2.2) u 0 = u v 0 = v, construct the sequences {u n } {v n } by (3.2.3) u n+1 = F (u n, v n ) v n+1 = F (v n, u n ). From (3.2.1) (3.2.2) u 0 x v 0 y for every n. Now we have to show that u n x v n y for every n. From (3.2.3),u 1 = F (u 0, v 0 ) F (x, v 0 ) F (x, y) = x. Therefore u 1 x. And v 1 = F (v 0, u 0 ) F (y, u 0 ) F (y, x) = y. Therefore v 1 y. Similarly u 2 x v 2 y. Hence by induction u n x v n y for every n. As in theorem 3.1, we can show that {u n } {v n } are Cauchy sequences. Suppose {u n } a {v n } b. Now (3.2.4) d(u n+1, x) = d(f (u n, v n ), F (x, y)) φ{max (d(u n, x), d(v n, y))} < max{d(u n, x), d(v n, y)} ( since u n x v n y) (3.2.5) d(v n+1, y) = d(f (v n, u n ), F (y, x)) φ{max (d(v n, y), d(u n, x))} < max{d(v n, y), d(u n, x)} ( since u n x v n y).

10 COUPLED FIXED AND COINCIDENCE POINT THEOREMS Write A n = max{d(u n, x), d(v n, y)}. Then by (3.2.4) (3.2.5) (3.2.6) A n+1 φ(a n ) < A n. Write lim n A n = α. Then by (3.2.6), lim φ(a n ) = α. But by hypothesis lim φ(a n ) = φ(α + 0) < α, if α > 0, which is a contradiction. Therefore α = 0. Therefore lim n A n = 0. Therefore (3.2.7) max{d(u n+1, x), d(v n+1, y)} 0 as n. Therefore d(u n+1, x) = 0 d(v n+1, y) = 0 i.e,d(a, x) = 0 since u n+1 a. Therefore a = x d(b, y) = 0 since v n+1 b. Therefore b = y. Therefore {u n } x {v n } y. Case (ii): Suppose (u, v) (x, y). Then u x v y. We assume that (u 0, v 0 ) = (u, v) (x, y). Now u 1 = F (u 0, v 0 ) F (x, v 0 ) F (x, y) = x. Therefore u 1 x v 1 = F (v 0, u 0 ) F (y, u 0 ) F (y, x) = y. Therefore v 1 y. Similarly u 2 = F (u 1, v 1 ) F (x, v 1 ) F (x, y) = x. Therefore u 2 x v 2 = F (v 1, u 1 ) F (y, u 1 ) F (y, x) = y. Therefore v 2 y. Thus by induction follows that u n x v n y for every n. As in Theorem (3.1), we can show that {u n } {v n } are Cauchy sequences. Suppose {u n } a {v n } b. Now d(u n+1, x) = d(f (u n, v n ), F (x, y)) φ{max (d(u n, x), d(v n, y))} < max{d(u n, x), d(v n, y)} (since u n x v n y) d(v n+1, y) = d(f (v n, u n ), F (y, x)) φ{max (d(v n, y), d(u n, x))} < max{d(v n, y), d(u n, x)} (since u n x v n y). Therefore max{d(u n+1, x), d(v n+1, y)} 0 as n (as in (3.2.7)) i.e, d(u n+1, x) = 0 d(v n+1, y) = 0 i.e, d(a, x) = 0 since u n+1 a. Therefore a = x, d(b, y) = 0 since v n+1 b. Therefore b = y, {u n } x {v n } y. Theorem 3.3. In addition to the hypothesis of Theorem 3.1, suppose that for every (x, y), (z, t) X X, there exists a (u, v) X X, which is comparable to (x, y) (z, t). Then F has a unique coupled fixed point. Proof. Suppose (x, y) (z, t) are coupled fixed points of F. Suppose (u, v) is comparable with (x, y) (z, t). Let {u n } {v n } be as in Lemma 3.2, since (u, v) is comparable with (x, y), {u n } x {v n } y. Similarly, since (u, v) is comparable with (z, t), again by Lemma 3.2, {u n } z {v n } t. Therefore x = z y = t. Therefore F has a unique coupled fixed point.

11 444 K. RAVIBABU, CH. SRINIVASA RAO CH. RAGHAVENDRA NAIDU Theorem 3.4. 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 there is an increasing function φ : [0, ) [0, ) with φ(t + 0) < t for each t > 0, also suppose that F : X X X g : X X are such that F has the mixed g-monotone property (3.4.1) d(f (x, y), F (u, v)) φ{max(d(g(x), g(u)), d(g(y), g(v)))}, for all x, y, u, v X for which g(x) g(u) g(y) g(v). Suppose that F (X X) g(x), g is continuous commutes with F also Suppose either a) F is continuous (or); b) X has the following property. i) if a non decreasing sequence {x n } x, then x n x for all n N. ii) if a non increasing sequence {y n } y, then y y n for all n N. If there exist x 0, y 0 X such that g(x 0 ) F (x 0, y 0 ) g(y 0 ) F (y 0, x 0 ), then there exist x, y X such that g(x) = F (x, y) g(y) = F (y, x), that is, F g have a coupled coincidence point. Proof. Suppose that x 0, y 0 X such that (3.4.2) g(x 0 ) F (x 0, y 0 ) g(y 0 ) F (y 0, x 0 ). Since (3.4.3) F (X X) g(x) we can choose x 1, y 1 X such that g(x 1 ) = F (x 0, y 0 ) g(y 1 ) = F (y 0, x 0 ). Again from (3.4.3), we can choose x 2, y 2 X such that g(x 2 ) = F (x 1, y 1 ) g(y 2 ) = F (y 1, x 1 ). Continuing this process, inductively we construct the sequences {g(x n )} {g(y n )} in X such that (3.4.4) g(x n+1 ) = F (x n, y n ) g(y n+1 ) = F (y n, x n ) for n = 0, 1, 2... We show that (3.4.5) g(x n ) g(x n+1 ) for n = 0, 1, 2,... (3.4.6) g(y n ) g(y n+1 ) for n = 0, 1, 2,... From (3.4.2), g(x 0 ) F (x 0, y 0 ) g(y 0 ) F (y 0, x 0 ). Therefore from (3.4.3), g(x 1 ) = F (x 0, y 0 ) g(y 1 ) = F (y 0, x 0 ). Therefore g(x 0 ) g(x 1 ) g(y 0 ) g(y 1 ). Thus (3.4.5) (3.4.6) hold for n = 0. Suppose that (3.4.5) (3.4.6) hold for some n 0. Then, since g(x n ) g(x n+1 ) g(y n ) g(y n+1 ), as F has the mixed g-monotone property, from (3.4.3), (3.4.7) g(x n+1 )=F (x n, y n ) F (x n+1, y n ), F (y n+1, x n ) F (y n, x n )=g(y n+1 ).

12 COUPLED FIXED AND COINCIDENCE POINT THEOREMS Again from (3.4.3), (3.4.8) g(x n+2 ) = F (x n+1, y n+1 ) F (x n+1, y n ), F (y n+1, x n ) F (y n+1, x n+1 ) = g(y n+2 ). From (3.4.7) (3.4.8), we get g(x n+1 ) g(x n+2 ) g(y n+1 ) g(y n+2 ). Thus by mathematical induction we conclude that (3.4.5) (3.4.6) hold for all n 0. Therefore (3.4.9) g(x 0 ) g(x 1 ) g(x 2 )...g(x n ) g(x n+1 )... (3.4.10) g(y 0 ) g(y 1 ) g(y 2 )...g(y n ) g(y n+1 )... Therefore {g(x n )}is {g(y n )}is. Denote δ n = max{d((g(x n ), g(x n+1 )), d((g(y n ), g(y n+1 ))}. We show that (3.4.11) δ n φ(δ n 1 ). Since g(x n 1 ) g(x n ) g(y n 1 ) g(y n ) from (3.4.3) (3.4.1), we have (3.4.12) d((g(x n ), g(x n+1 )) = d((f (x n 1, y n 1 ), (F (x n, y n )) φ{max((d(g(x n 1 ), g(x n )), (d(g(y n 1 ), g(y n )))} = φ(δ n 1 ). Similarly (3.4.13) d((g(y n ), g(y n+1 )) = d((f (y n 1, x n 1 ), (F (y n, x n )) φ{max((d(g(y n 1 ), g(y n )), (d(g(x n 1 ), g(x n )))} = φ(δ n 1 ). From (3.4.12) (3.4.13) we obtain (3.4.11). From (3.4.11),since φ(t) < t for t > 0, it follows that sequence {δ n } is monotone decreasing. Therefore there is some δ 0 such that lim n δ n = δ. We show that δ = 0. Suppose to the contrary that δ > 0. Then taking the limit as n on both sides of (3.4.11), we have δ = lim n δ n lim n φ(δ n 1 ) = φ(δ + 0) < δ, a contradiction. Thus δ = 0, (3.4.14) lim n {max((d(g(x n), g(x n+1 )), d(g(y n ), g(y n+1 )))} = 0. Now we prove that {g(x n )} {g(y n )} are Cauchy sequences. Suppose to the contrary that at least one of {g(x n )} or {g(y n )} is not a Cauchy sequence. Then there exist an ϵ > 0, two sub sequences of integers {l k } {m k }, m k > l k k with (3.4.15) r k = max(d(g(x lk ), g(x mk )), d(g(y lk ), g(y mk )) > ϵ for k = 1, 2, 3...

13 446 K. RAVIBABU, CH. SRINIVASA RAO CH. RAGHAVENDRA NAIDU We may also assume (3.4.16) max (d(g(x lk ), g(x mk )), d(g(y lk ), g(y mk )) ϵ by choosing m k to be the smallest number exceeding l k for which(3.4.15) holds. From (3.4.15), (3.4.16) by the triangular inequality ϵ < max{(d(g(x lk ), g(x mk 1)) + d(g(x mk 1), g(x mk ))), Taking the limit as k we get by (3.4.14) (3.4.17) lim k r k = ϵ + (d(g(y lk ), g(y mk 1)) + d(g(y mk 1), g(y mk ))}. Since from (3.4.3) (3.4.1), g(x lk ) g(x mk ) g(y lk ) g(y mk ) we have (3.4.18) (d(g(x lk+1 ), g(x mk+1 )) = d(f (x lk, y lk ), F (x mk, y mk ) φ{max((d(g(x lk ), g(x mk )), d(g(y lk ), g(y mk ))} = φ(r k ). Similarly (3.4.19) (d(g(y lk+1 ), g(y mk+1 )) = d(f (y lk, x lk ), F (y mk, x mk ) φ{max((d(g(y lk ), g(y mk )), d(g(x lk ), g(x mk ))} = φ(r k ). From (3.4.18) (3.4.19), ϵ < r k+1 φ(r k ). Taking k, using (3.4.14) (3.4.17) we get ϵ lim k (φ(r k)) = φ(ε + 0) < ϵ, a contradiction. Thus our supposition is wrong. Therefore {g(x n )} {g(y n )} are Cauchy sequences. Since X is complete,there exist x, y Xsuch that (3.4.20) lim n g(x n) = x From (3.4.20) continuity of g, (3.4.21) lim n g(g(x n)) = g(x) Since F g commute, from (3.4.21) lim g(y n) = y. n lim g(g(y n)) = g(y). n (3.4.22) g(g(x n+1 )) = g(f (x n, y n )) = F (g((x n ), g(y n )) (3.4.23) g(g(y n+1 )) = g(f (y n, x n )) = F (g((y n ), g(x n )).

14 COUPLED FIXED AND COINCIDENCE POINT THEOREMS We show that g(x) = F (x, y) g(y) = F (y, x). Suppose (a) holds. Taking the limit as n in (3.4.22) (3.4.23) by (3.4.20),(3.4.21) continuity of F, we get g(x) = lim n g(g(x n+1)) = lim n F (g((x n), g(y n )) = F ( lim g((x n), lim g(y n)) n n = F (x, y) g(y) = lim n g(g(y n+1)) = lim n F (g((y n), g(x n )) = F ( lim g((y n), lim g(x n)) n n = F (y, x). Therefore g(x) = F (x, y) g(y) = F (y, x). Suppose (b) holds. Since {g(x n )} is non decreasing g(x n ) x {g(y n )} is non increasing g(y n ) y from hypothesis we have g(x n ) x g(y n ) y, for all n. Then by the triangular inequality (3.4.22),(3.4.23) (3.4.1),we get d(g(x), F (x, y))) d(g(x), g(g(x n+1 ))) + d(g(g(x n+1 )), F (x, y)) = {d(g(x), g(g(x n+1 ))) + d(f (g(x n ), g(y n ))), F (x, y))} {d(g(x), g(g(x n+1 )))+ φ(max(d(g(g(x n ), g(x))), d(g(g(y n ), g(y))))} 0 as n. Therefore d(g(x), F (x, y))) 0. Therefore g(x) = F (x, y). Similarly g(y) = F (y, x). Therefore F g have a coupled coincidence point. Lemma 3.5. Suppose (x, y) is coupled coincidence point of F g (g(u), g(v)) is comparable with (g(x), g(y)).write u 0 = u v 0 = v construct the sequences {g(u n )} {g(v n )} by g(u n+1 ) = F (u n, v n ) g(v n+1 ) = F (v n, u n ) for n = 0, 1, 2,... Then g(u n ) g(x) g(v n ) g(y). Proof. Case (i): (g(u), g(v)) (g(x), g(y)). Then g(u) g(x) g(v) g(y). Write u 0 = u v 0 = v. Then (g(u 0 ), g(v 0 )) = (g(u), g(v)) (g(x), g(y)). Therefore g(u 0 ) g(x) g(v 0 ) g(y). Choose u 1,v 1 such that g(u 1 ) = F (u 0, v 0 ) g(v 1 ) = F (v 0, u 0 ). Therefore g(u 1 ) = F (u 0, v 0 ) F (x, v 0 ) F (x, y) = g(x). Therefore g(u 1 ) g(x) g(v 1 ) = F (v 0, u 0 ) F (y, u 0 ) F (y, x) = g(y). Therefore g(v 1 ) g(y). In general (g(u n ), g(v n )) (g(x), g(y). Now d(g(u n ), g(x)) = d(f (u n 1, v n 1 ), F (x, y)) φ{max(d(g(u n 1 ), g(x)), d(g(v n 1 ), g(y)))}

15 448 K. RAVIBABU, CH. SRINIVASA RAO CH. RAGHAVENDRA NAIDU d(g(v n ), g(y)) = d(f (v n 1, u n 1 ), F (y, x)) φ{max(d(g(v n 1 ), g(y)), d(g(u n 1 ), g(x)))}. Let r k = max{d(g(u n ), g(x)), d(g(v n ), g(y))} φ(r k 1 ) < r k 1. Therefore {r k } is a decreasing sequence. Suppose r k α. Hence r k φ(r k 1 ) < r k 1. Letting k,we get lim k φ(r k ) = α. But by hypothesis lim φ(r k ) = φ(α + 0) < α, if α > 0. Therefore α = 0. Therefore lim k r k = 0. Therefore max{d(g(u n ), g(x)), d(g(v n ), g(y))} 0 as n. Therefore d(g(u n ), g(x)) 0 as n d(g(v n ), g(y)) 0 as n. Therefore g(u n ) g(x) g(v n ) g(y). Case(ii): Suppose (g(u), g(v)) (g(x), g(y)). In this case the proof is similar to case (i). Theorem 3.6. In addition to the hypothesis of Theorem 3.4, suppose (x, y), (z, t) X X there exist u, v X X such that (g(u), g(v)) is comparable with (g(x), g(y)) (g(z), g(t)).then F g have unique coupled coincidence point,in the sense that g(x) = g(z) g(y) = g(t). Proof. Suppose (x, y)(z, t) are coupled coincidence points of F g. Suppose (g(u), g(v)) is comparable with (g(x), g(y)) (g(z), g(t)). Let {u n } {v n } be in Lemma 3.5,g(u n ) g(x) g(v n ) g(y). Similarly, since (g(u), g(v)) is comparable with (g(z), g(t)), again by Lemma 3.5, g(u n ) g(z) g(v n ) g(t). Therefore g(x) = g(z) g(y) = g(t). Now F (x, y) = g(x) = g(z) = F (z, t) F (y, x) = g(y) = g(t) = F (t, z). Note: In view of the comment made at the beginning of this section, Theorem 2.3, 2.4, 2.8, follow as corollaries to Theorems 3.1, 3.3, of this section. Corollary 3.7. Let (X, ) be a partially ordered set suppose there is a metric d on X, such that (X, d) is a complete metric space. Suppose that F : X X X g : X X are such that F has the mixed g-monotone property assume there is a k [0, 1) such that d(f (x, y), F (u, v)) k{max(d(g(x), g(u)), d(g(y), g(v)))}, for all x, y, u, v X for which g(x) g(u) g(y) g(v). Suppose that F (X X) g(x), g is continuous also Suppose either (a) F is Continuous (or); (b) X has the following property. i) if a non decreasing sequence {x n } x,then x n x for all n N. ii) if a non increasing sequence {y n } y,then y y n for all n N. If there exist x 0, y 0 X such that g(x 0 ) F (x 0, y 0 ) g(y 0 ) F (y 0, x 0 ), then there exist x, y X such that g(x) = F (x, y) g(y) = F (y, x), that is, F g have a coincidence point.

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17 450 K. RAVIBABU, CH. SRINIVASA RAO CH. RAGHAVENDRA NAIDU [12] R. Kannan, Some results on fixed points, Bull. Calcutta Math. Soc., 60 (1968), [13] V. Lakshmikantham, L. Ciric, Coupled fixed point theorems for nonlinear contractions in partially ordered metric spaces, Nonlinear Anal., 70 (2009), [14] K.P.R. Sastry, Ch. Srinivasarao, N. Apparao S.S.A. Sastry, A Coupled Fixed point theorem for Geraghty contractions in partially ordered metric spaces, ISSSN : , Vol. 4, Issue 3 (version 1), 2014, [15] K.P.R. Sastry, Ch. Srinivasarao, N. Apparao, S.S.A. Sastry, A Fixed point theorem of strict genralized type weakly contractive maps in orbitally complete metric spaces when the control function is not necessarily continuous, ISSSN: , Vol. 18, Issue 1 (version 18), 2013, [16] W. Sintunavarat, P. Kumam, Coupled fixed point results for Non linear integral equations, Journal of Egyptian Mathematical Society, 21 (2013), [17] W. Sintunavarat, Y.J. Cho, P. Kumam, Coupled fixed point theorems for contraction mapping induced by cone ball-metric in partially ordered spaces, Fixed Point Theory Appl., (2012), 2012:128. Accepted: