TRIPLE FIXED POINTS IN ORDERED METRIC SPACES


 Georgina McCarthy
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1 Bulletin of Mathematical Analysis and Applications ISSN: , URL: Volume 4 Issue 1 (2012., Pages TRIPLE FIXED POINTS IN ORDERED METRIC SPACES (COMMUNICATED BY SIMEON REICH HASSEN AYDI, ERDAL KARAPINAR Abstract. In this paper, we prove triple fixed point theorems in partially ordered metric spaces depended on another function. The presented results generalize the theorem of Berinde and Borcut [Tripled fixed point theorems for contractive type mappings in partially ordered metric spaces, Nonlinear Anal. 74(15 ( ]. Also, we state some examples showing that our results are effective. 1. Introduction Banach fixed point theorem and its applications are well known. Many authors have extended this theorem, introducing more general contractive conditions, which imply the existence of a fixed point. Existence of fixed points in ordered metric spaces was investigated in 2004 by Ran and Reurings [29], and then by Nieto and López [28]. For some other results in ordered metric spaces, see e.g. [3, 4, 5, 23, 24, 25, 26, 27]. Bhashkar and Lakshmikantham in [11] introduced the concept of a coupled fixed point of a mapping F : X X X and investigated some coupled fixed point theorems in partially ordered complete metric spaces. Later, various results on coupled fixed point have been obtained, see e.g. [1, 6, 7, 8, 12, 17, 18, 19, 20, 30]. On the other hand, Berinde and Borcut [10] introduced the concept of triple fixed point (see also the papers [2, 9, 31]. The following two definitions are from [10]. Definition 1.1. Let (X, be a partially ordered set and F : X X X X. The mapping F is said to has the mixed monotone property if for any x, y, z X x 1, x 2 X, y 1, y 2 X, x 1 x 2 = F (x 1, y, z F (x 2, y, z, y 1 y 2 = F (x, y 1, z F (x, y 2, z, z 1, z 2 X, z 1 z 2 = F (x, y, z 1 F (x, y, z 2, 2000 Mathematics Subject Classification. 46T99, 54H25, 47H10, 54E50. Key words and phrases. Triple fixed point, ordered sets, metric spaces, nonlinear contractions. c 2012 Universiteti i Prishtinës, Prishtinë, Kosovë. Submitted January 29, Accepted March 7,
2 198 HASSEN AYDI, ERDAL KARAPINAR Definition 1.2. Let F : X X X X. An element (x, y, z is called a triple fixed point of F if F (x, y, z = x, F (y, x, y = y and F (z, y, x = z. Also, Berinde and Borcut [10] proved the following result. Theorem 1.1. Let (X,, d be a partially ordered set and suppose there is a metric d on X such that (X, d is a complete metric space. Suppose F : X X X X such that F has the mixed monotone property and there exist j, r, l 0 with j + r + l < 1 such that d(f (x, y, z, F (u, v, w jd(x, u + rd(y, v + ld(z, w, (1 for any x, y, z X for which x u, v y and z w. continuous or X has the following property: (1 if a nondecreasing sequence x n x, then x n x for all n, (2 if a nonincreasing sequence y n y, then y y n for all n. Suppose either F is If there exist x 0, y 0, z 0 X such that x 0 F (x 0, y 0, z 0, y 0 F (y 0, x 0, z 0 and z 0 F (z 0, y 0, x 0, then there exist x, y, z X such that F (x, y, z = x, F (y, x, y = y and F (z, y, x = z, that is, F has a triple fixed point. In this paper we give some triple fixed point theorems for mappings having the mixed monotone property in partially ordered metric spaces depended on another function which are generalization of the main results of Berinde and Borcut [10]. 2. MAIN RESULTS We start with the following definition (see e.g. [13, 14, 22, 21]. Definition 2.1. Let (X, d be a metric space. A mapping T : X X is said to be ICS if T is injective, continuous and has the property: for every sequence {x n } in X, if {T x n } is convergent then {x n } is also convergent. Let Φ be the set of all functions φ : [0, [0, such that (1 φ is nondecreasing, (2 φ(t < t for all t > 0, (3 lim r t + φ(r < t for all t > 0. From now on, we denote X 3 = X X X. Our first result is given by the following: Theorem 2.1. Let (X, be a partially ordered set and suppose there is a metric d on X such that (X, d is a complete metric space. Suppose T : X X is an ICS mapping and F : X 3 X is such that F has the mixed monotone property. Assume that there exists φ Φ such that ( d(t F (x, y, z, T F (u, v, w φ max{d(t x, T u, d(t y, T v, d(t z, T w} for any x, y, z X for which x u, v y and z w. Suppose either (a F is continuous, or (b X has the following property: (2
3 TRIPLE FIXED POINTS IN ORDERED METRIC SPACES 199 (i if nondecreasing sequence x n x (respectively, z n z, then x n x (respectively, z n z for all n, (ii if nonincreasing sequence y n y, then y n y for all n. If there exist x 0, y 0, z 0 X such that x 0 F (x 0, y 0, z 0, y 0 F (y 0, x 0, y 0 and z 0 F (z 0, y 0, x 0, then there exist x, y, z X such that F (x, y, z = x, F (y, x, y = y and F (z, y, x = z, that is, F has a triple fixed point. Proof. Let x 0, y 0, z 0 X such that x 0 F (x 0, y 0, z 0, y 0 F (y 0, x 0, y 0 and z 0 F (z 0, y 0, x 0. Set x 1 = F (x 0, y 0, z 0, y 1 = F (y 0, x 0, y 0 and z 1 = F (z 0, y 0, x 0. (3 Continuing this process, we can construct sequences {x n }, {y n } and {z n } in X such that x n+1 = F (x n, y n, z n, y n+1 = F (y n, x n, y n and z n+1 = F (z n, y n, x n. (4 Since F has the mixed monotone property, then using a mathematical induction it is easy that Assume for some n N, x n x n+1, y n+1 y n, z n z n+1, for n = 0, 1, 2,... (5 x n = x n+1, y n = y n+1 and z n = z n+1, then, by (4, (x n, y n, z n is a triple fixed point of F. From now on, assume for any n N that at least x n x n+1 or y n y n+1 or z n z n+1. (6 Since T is injective, then by (6, for any n N 0 < max{d(t x n, T x n+1, d(t y n, T y n+1, d(t z n, T z n+1 }. Due to (2 and (4, we have d(t x n, T x n+1 = d(t ( F (x n 1, y n 1, z n 1, T F (x n, y n, z n φ max{d(t x n 1, T x n, d(t y n 1, T y n, d(t z n 1, T z n } and d(t y n+1, T y n = d(t F (y n, x n, y n, T F (y n 1, x n 1, y n 1 φ({d(t y n 1, T y n, d(t x n 1, T x n, d(t y n 1, T y n } = φ(max{d(t y n 1, T y n, d(t x n 1, T x n } φ(max{d(t z n 1, T z n, d(t y n 1, T y n, d(t x n 1, T x n }, d(t z n, T z n+1 = d(t F (z n 1, y n 1, x n 1, T F (z n, y n, x n φ(max{d(t z n 1, T z n, d(t y n 1, T y n, d(t x n 1, T x n }. Having in mind that φ(t < t for all t > 0, so from (7(9 we obtain that 0 < max{d(t x n, T x n+1, d(t y n, T y n+1, d(t z n, T z n+1 } φ(max{d(t z n 1, T z n, d(t y n 1, T y n, d(t x n 1, T x n } < max{d(t z n 1, T z n, d(t y n 1, T y n, d(t x n 1, T x n }. (7 (8 (9 (10
4 200 HASSEN AYDI, ERDAL KARAPINAR It follows that max{d(t x n, T x n+1, d(t y n, T y n+1, d(t z n, T z n+1 } < max{d(t z n 1, T z n, d(t y n 1, T y n, d(t x n 1, T x n }. Thus, {max{d(t x n, T x n+1, d(t y n, T y n+1, d(t z n, T z n+1 }} is a positive decreasing sequence. Hence, there exists r 0 such that lim max{d(t x n, T x n+1, d(t y n, T y n+1, d(t z n, T z n+1 } = r. n + Suppose that r > 0. Letting n + in (10, we obtain that 0 < r lim φ(max{d(t z n 1, T z n, d(t y n 1, T y n, d(t x n 1, T x n } = lim φ(t < r, n + + it is a contradiction. We deduce that t r (11 lim max{d(t x n, T x n+1, d(t y n, T y n+1, d(t z n, T z n+1 } = 0. (12 n + We shall show that {T x n }, {T y n } and {T z n } are Cauchy sequences. Assume the contrary, that is, {T x n }, {T y n } or {T z n } is not a Cauchy sequence, that is, lim d(t x m, T x n 0, or lim d(t y m, T y n 0, n,m + n,m + or lim d(t z m, T z n 0. This means that there exists ε > 0 for which we can n,m + find subsequences of integers (m k and (n k with n k > m k > k such that max{d(t x mk, T x nk, d(t y mk, T y nk, d(t z mk, T z nk } ε. (13 Further, corresponding to m k we can choose n k in such a way that it is the smallest integer with n k > m k and satisfying (13. Then max{d(t x mk, T x nk 1, d(t y mk, T y nk 1, d(t z mk, T z nk 1} < ε. (14 By triangular inequality and (14, we have Thus, by (12 we obtain Similarly, we have d(t x mk, T x nk d(t x mk, T x nk 1 + d(t x nk 1, T x nk < ɛ + d(t x nk 1, T x nk. lim d(t x m k, T x nk lim d(t x m k, T x nk 1 ε. (15 lim d(t y m k, T y nk lim d(t y m k, T y nk 1 ε. (16 lim d(t z m k, T z nk lim d(t z m k, T z nk 1 ε. (17 Again by (14, we have d(t x mk, T x nk d(t x mk, T x mk 1 + d(t x mk 1, T x nk 1 + d(t x nk 1, T x nk d(t x mk, T x mk 1 + d(t x mk 1, T x mk +d(t x mk, T x nk 1 + d(t x nk 1, T x nk < d(t x mk, T x mk 1 + d(t x mk 1, T x mk + ε + d(t x nk 1, T x nk. Letting k + and using (12, we get lim d(t x m k, T x nk lim d(t x m k 1, T x nk 1 ε. (18
5 TRIPLE FIXED POINTS IN ORDERED METRIC SPACES 201 lim d(t y m k, T y nk lim d(t y m k 1, T y nk 1 ε. (19 lim d(t z m k, T z nk lim d(t z m k 1, T z nk 1 ε. (20 Using (13 and (18(20, we have lim max{d(t x m k, T x nk, d(t y mk, T y nk, d(t z mk, T z nk } k + = lim max{d(t x m k 1, T x nk 1, d(t y mk 1, T y nk 1, d(t z mk 1, T z nk 1} k + = ε. Now, using inequality (2 we obtain d(t x mk, T x nk = d(t ( F (x mk 1, y mk 1, z mk 1, T F (x nk 1, y nk 1, z nk 1 φ (21 max{d(t x mk 1, T x nk 1, d(t y mk 1, T y nk 1, d(t z mk 1, T z nk 1} d(t y mk, T y nk (22 = d(t ( F (y mk 1, x mk 1, y mk 1, T F (y nk 1, x nk 1, y nk 1 φ max{d(t y mk 1, T y nk 1, d(t x mk 1, T x nk 1} and d(t z mk, T z nk (23 = d(t ( F (z mk 1, y mk 1, x mk 1, T F (z nk 1, y nk 1, x nk 1 φ max{d(t x mk 1, T x nk 1, d(t y mk 1, T y nk 1, d(t z mk 1, T z nk 1}. We deduce from (22(24 that max{d(t x mk, T x nk, d(t y mk, T y nk, d(t z mk, T z nk } φ(max{d(t x mk 1, T x nk 1, d(t y mk 1, T y nk 1, d(t z mk 1, T z nk 1}. Letting k + in (25 and having in mind (21, we get that 0 < ε lim φ(t < ε, t ε + it is a contradiction. Thus, {T x n }, {T y n } and {T z n } are Cauchy sequences in (X, d. Since X is a complete metric space, {T x n }, {T y n } and {T z n } are convergent sequences. Since T is an ICS mapping, there exist x, y, z X such that lim x n = x, n + Since T is continuous, we have lim T x n = T x, n + (24 (25 lim y n = y, and lim z n = z. (26 n + n + lim T y n = T y, and lim T z n = T z. (27 n + n + Suppose now the assumption (a holds, that is, F is continuous. By (4, (26 and (27 we obtain x = lim x n+1 = lim F (x n, y n, z n = F ( lim x n, lim y n, lim z n = F (x, y, z, n + n + n + n + n + y = lim y n+1 = lim F (y n, x n, y n = F ( lim y n, lim x n, lim y n = F (y, x, y, n + n + n + n + n +
6 202 HASSEN AYDI, ERDAL KARAPINAR and z = lim z n+1 = lim F (z n, y n, x n = F ( lim z n, lim y n, lim x n = F (z, y, x. n + n + n + n + n + We have proved that F has a triple fixed point. Suppose now the assumption (b holds. Since {x n }, {z n } are nondecreasing with x n x, z n z and also {y n } is nonincreasing with y n y, then by assumption (b we have x n x, y n y and z n z, for all n. Consider now d(t x, T F (x, y, z d(t x, T x n+1 + d(t x n+1, T F (x, y, z = d(t x, T x n+1 + d(t F (x n, y n, z n, T F (x, y, z d(t x, T x n+1 + φ(max{d(t x n, T x, d(t y n, T y, d(t z n, T z}. (28 Taking n and using (27, the righthand side of (28 tends to 0, so we get that d(t x, T F (x, y, z = 0. Thus, T x = T F (x, y, z and since T is injective, we get that x = F (x, y, z. Analogously, we find that F (y, x, y = y and F (z, y, x = z. Thus, we proved that F has a triple fixed point. Theorem 2.1. This completes the proof of Corollary 2.1. Let (X, be a partially ordered set and suppose there is a metric d on X such that (X, d is a complete metric space. Suppose T : X X is an ICS mapping and F : X 3 X is such that F has the mixed monotone property. Assume that there exists φ Φ such that d(t F (x, y, z, T F (u, v, w φ ( d(t x, T u + d(t y, T v + d(t z, T w 3 for any x, y, z X for which x u, v y and z w. Suppose either (a F is continuous, or (b X has the following property: (i if nondecreasing sequence x n x (respectively, z n z, then x n x (respectively, z n z for all n, (ii if nonincreasing sequence y n y, then y n y for all n. If there exist x 0, y 0, z 0 X such that x 0 F (x 0, y 0, z 0, y 0 F (y 0, x 0, y 0 and z 0 F (z 0, y 0, x 0, then there exist x, y, z X such that F (x, y, z = x, F (y, x, y = y and F (z, y, x = z, that is, F has a triple fixed point. Proof. It suffices to remark that d(t x, T u + d(t y, T v + d(t z, T w max{d(t x, T u, d(t y, T v, d(t z, T w}. 3 Then, we apply Theorem 2.1 because that φ is nondecreasing. Corollary 2.2. Let (X, be a partially ordered set and suppose there is a metric d on X such that (X, d is a complete metric space. Suppose T : X X is an
7 TRIPLE FIXED POINTS IN ORDERED METRIC SPACES 203 ICS mapping and F : X 3 X is such that F has the mixed monotone property. Assume that there exists k [0, 1 such that d(t F (x, y, z, T F (u, v, w k max{d(t x, T u, d(t y, T v, d(t z, T w} for any x, y, z X for which x u, v y and z w. Suppose either (a F is continuous, or (b X has the following property: (i if nondecreasing sequence x n x (respectively, z n z, then x n x (respectively, z n z for all n, (ii if nonincreasing sequence y n y, then y n y for all n. If there exist x 0, y 0, z 0 X such that x 0 F (x 0, y 0, z 0, y 0 F (y 0, x 0, y 0 and z 0 F (z 0, y 0, x 0, then there exist x, y, z X such that F (x, y, z = x, F (y, x, y = y and F (z, y, x = z, that is, F has a triple fixed point. Proof. It follows by taking φ(t = kt in Theorem 2.1. Corollary 2.3. Let (X, be a partially ordered set and suppose there is a metric d on X such that (X, d is a complete metric space. Suppose T : X X is an ICS mapping and F : X 3 X is such that F has the mixed monotone property. Assume that there exists k [0, 1 such that d(t F (x, y, z, T F (u, v, w k (d(t x, T u + d(t y, T v + d(t z, T w (29 3 for any x, y, z, u, v, w X for which x u, v y and z w. Suppose either (a F is continuous, or (b X has the following property: (i if nondecreasing sequence x n x (respectively, z n z, then x n x (respectively, z n z for all n, (ii if nonincreasing sequence y n y, then y n y for all n. If there exist x 0, y 0, z 0 X such that x 0 F (x 0, y 0, z 0, y 0 F (y 0, x 0, y 0 and z 0 F (z 0, y 0, x 0, then there exist x, y, z X such that F (x, y, z = x, F (y, x, y = y and F (z, y, x = z, that is, F has a triple fixed point. Proof. It suffices to take φ(t = kt in Corollary 2.1. Remark 1. Taking T = Id X, the identity on X, in Corollary 2.3, we get Theorem 1.1 of Berinde and Borcut (with j = l = r = k 3. Now, we shall prove the existence and uniqueness of a triple fixed point. For a product X 3 of a partially ordered set (X,, we define a partial ordering in the following way: For all (x, y, z, (u, v, r X 3 (x, y, z (u, v, r x u, y v and z r. (30 We say that (x, y, z and (u, v, w are comparable if (x, y, z (u, v, r or (u, v, r (x, y, z. Also, we say that (x, y, z is equal to (u, v, r if and only if x = u, y = v and z = r.
8 204 HASSEN AYDI, ERDAL KARAPINAR Theorem 2.2. In addition to hypothesis of Theorem 2.1, suppose that that for all (x, y, z, (u, v, r X 3, there exists (a, b, c X X X such that (F (a, b, c, F (b, a, b, F (c, b, a is comparable to (F (x, y, z, F (y, x, y, F (z, y, x and (F (u, v, r, F (v, u, v, F (r, v, u. Then, F has a unique triple fixed point (x, y, z. Proof. The set of triple fixed points of F is not empty due to Theorem 2.1. Assume, now, (x, y, z and (u, v, r are two triple fixed points of F, that is, F (x, y, z = x, F (u, v, r = u, F (y, x, y = y, F (v, u, v = v, F (z, y, x = z, F (r, v, u = r. We shall show that (x, y, z and (u, v, r are equal. By assumption, there exists (a, b, c X 3 such that (F (a, b, c, F (b, a, b, F (c, b, a is comparable to (F (x, y, z, F (y, x, y, F (z, y, x and (F (u, v, r, F (v, u, v, F (r, v, u. Define sequences {a n }, {b n } and {c n } such that a 0 = a, b 0 = b, c 0 = c, and for any n 1 a n = F (a n 1, b n 1, c n 1, b n = F (b n 1, a n 1, b n 1, c n = F (c n 1, b n 1, a n 1, for all n. Further, set x 0 = x, y 0 = y, z 0 = z and u 0 = u, v 0 = v, r 0 = r, and on the same way define the sequences {x n }, {y n }, {z n } and {u n }, {v n }, {r n }. Then, it is easy that x n = F (x, y, z, y n = F (y, x, y,, z n = F (z, y, x, u n = F (u, v, r, v n = F (v, u, v, r n = F (r, v, u, for all n 1. Since (F (x, y, z, F (y, x, y, F (z, y, x = (x 1, y 1, z 1 = (x, y, z is comparable to (F (a, b, c, F (b, a, b, F (c, b, a = (a 1, b 1, c 1, then it is easy to show (x, y, z (a 1, b 1, c 1. Recursively, we get that By (33 and (2, we have and (31 (32 (x, y, z (a n, b n, c n for all n. (33 d(t x, T a n+1 = d(t F (x, y, z, T F (a n, b n, c n φ(max{d(t x, T a n, d(t y, T b n, d(t z, T c n } d(t b n+1, T y = d(t F (b n, a n, b n, T F (y, x, y φ(max{d(t a n, T x, d(t b n, T y} φ(max{d(t b n, T y, d(t a n, T x, d(t c n, T z}, d(t z, T c n+1 = d(t F (z, y, x, T F (c n, b n, a n φ(max{d(t z, T c n, d(t y, T b n, d(t x, T a n }; It follows from (34(36 that max{d(t z, T c n+1, d(t y, T b n+1, d(t x, T a n+1 } φ(max{d(t z, T c n, d(t y, T b n, d(t x, T a n }. Therefore, for each n 1, max{d(t z, T c n, d(t y, T b n, d(t x, T a n } φ n (max{d(t z, T c 0, d(t y, T b 0, d(t x, T a 0 }. (37 (34 (35 (36
9 TRIPLE FIXED POINTS IN ORDERED METRIC SPACES 205 It is known that φ(t < t and lim φ(r < t imply lim r t + φn (t = 0 for each t > 0. Thus, from (37, lim max{d(t z, T c n, d(t y, T b n, d(t x, T a n } = 0. This yields that lim d(t x, T a n = 0, lim d(t z, T c n = 0. Analogously, we show that lim d(t u, T a n = 0, lim d(t r, T c n = 0. lim d(t y, T b n = 0, lim d(t v, T b n = 0, Combining (38 and (39 yields that (T x, T y, T z and (T u, T v, T r are equal. The fact that T is injective gives us x = u, y = v and z = w. Now we state some examples showing that our results are effective. Example 2.1. Let X = [ 1 2, 64] with the metric d(x, y = x y, for all x, y X and the usual ordering. Clearly, (X, d is a complete metric space. Let T : X X and F : X 3 X be defined by ( xz 1 6 T x = ln(x + 1 and F (x, y, z = 8, x, y, z X. y It is clear that T is an ICS mapping, F has the mixed monotone property and continuous. Set k = 1 2. Taking x, y, z, u, v, w X for which x u, v y and z w, we have d(t F (x, y, z, T F (u, v, w = 1 (ln x + ln z 2 ln y (ln u + ln w 2 ln v ln x ln u ln y ln v + 1 ln z ln w ( ln x ln u + ln y ln v + ln z ln w (38 (39 = k (d(t x, T u + d(t y, T v + d(t z, T w, 3 which is the contractive condition (29. Moreover, taking x 0 = 1 = z 0 and y 0 = 64, we have x 0 F (x 0, y 0, z 0, y 0 F (y 0, x 0, y 0 and z 0 F (z 0, y 0, x 0. Therefore, all the conditions of Corollary 2.3 hold and (8, 8, 8 is the unique triple fixed point of F, since also the hypotheses of Theorem 2.2 hold. On the other hand, we can not apply Theorem 1.1 to this example because the condition (1 does not hold (for j = l = r = k 3 where k is arbitrary in [0, 1. Indeed, for x = z = 1 2, y = v = 64 and u = w = 1, (1 becomes d(f (x, y, z, F (u, v, w =8 ( k 3 (d(x, u + d(y, v + d(z, w = k 3,
10 206 HASSEN AYDI, ERDAL KARAPINAR that is, k 12(1 ( , 309 > 1, which is a contradiction because of k < 1. We conclude that our results generalize the result of Berinde and Borcut given by Theorem 1.1. Example 2.2. Let X = R with d(x, y = x y and natural ordering. Let T : X X and F : X 3 X be defined by T x = x 12 and F (x, y, z = 2 9 (x y + z. It is obvious that T is an ICS mapping, F has the mixed monotone property and continuous. Set φ(t = 2t 3 Φ. Clearly, all conditions of Theorem 2.1 are satisfied and (0, 0, 0 is the desired triple fixed point. Finally, following Example 2.9 in [21], we give a simple example which shows that if T is not an ICS mapping then the conclusion of Theorem 2.1 fails. Example 2.3. Let X = R with the usual metric and the usual ordering. F : X 3 X be defined by F (x, y, z = 2x y + 1, for all, x, y, z X. then F has the mixed monotone property and F is continuous. Also, there exist x 0 = 1, y 0 = 0 and z 0 = 1 such that x 0 F (x 0, y 0, z 0, y 0 F (y 0, x 0, y 0 and z 0 F (z 0, y 0, x 0. Let T : X X be defined by T (x = 1 for all x X, then T is not an ICS mapping. It is obvious that the condition (2 holds for any φ Φ. However, F has no triple fixed point. References [1] M. Abbas, M. A. Khan and S. Radenović, Common coupled fixed point theorem in cone metric space for w compatible mappings, Appl. Math. Comput. 217 ( [2] M. Abbas, H. Aydi and E. Karapınar, Tripled fixed points of multivalued nonlinear contraction mappings in partially ordered metric spaces, Abstract and Applied Analysis, Volume 2011, Article ID , 12 pages. [3] R.P. Agarwal, M.A. ElGebeily and D. O Regan, Generalized contractions in partially ordered metric spaces, Appl. Anal. 87 ( [4] I. Altun and H. Simsek, Some fixed point theorems on ordered metric spaces and application, Fixed Point Theory Appl (2010. Article ID [5] H. Aydi, Coincidence and common fixed point results for contraction type maps in partially ordered metric spaces, Int. Journal of Math. Analysis, Volume 5 (13 ( [6] H. Aydi, Some coupled fixed point results on partial metric spaces, International Journal of Mathematics and Mathematical Sciences, Volume 2011, Article ID , 11 pages. [7] H. Aydi, B. Samet and C. Vetro, Coupled fixed point results in cone metric spaces for wcompatible mappings, Fixed Point Theory Appl. 2011:27, (2011. [8] H. Aydi, M. Abbas and M. Postolache, Coupled coincidence points for hybrid pair of mappings via mixed monotone property, Journal of Advanced Mathematical Studies, 5 (1 ( [9] H. Aydi, E. Karapınar, M. Postolache, Tripled coincidence point theorems for weak ϕ contractions in partially ordered metric spaces, Fixed Point Theory Appl. (in press. [10] V. Berinde and M. Borcut, tripled fixed point theorems for contractive type mappings in partially ordered metric spaces, Nonlinear Anal. 74(15 ( [11] T.G. Bhaskar and V. Lakshmikantham, Fixed point theory in partially ordered metric spaces and applications, Nonlinear Anal. 65 ( [12] Binayak S. Choudhury, N. Metiya and A. Kundu, Coupled coincidence point theorems in ordered metric spaces, Ann. Univ. Ferrara. 57 ( [13] K.P. Chi, On a fixed point theorem for certain class of maps satisfying a contractive condition depended on an another function, Lobachevskii J. Math. 30 (4 ( Let
11 TRIPLE FIXED POINTS IN ORDERED METRIC SPACES 207 [14] K.P. Chi and H.T. Thuy, A fixed point theorem in 2metric spaces for a class of maps that satisfy a contractive condition dependent on an another function, Lobachevskii J. Math. 31(4 ( [15] B.S. Choudhury and A. Kundu, A coupled coincidence point result in partially ordered metric spaces for compatible mappings, Nonlinear Anal. 73 ( [16] J. Harjani and K. Sadarangani, Fixed point theorems for weakly contractive mappings in partially ordered sets, Nonlinear Anal. 71 ( [17] E. Karapınar, Couple Fixed Point on Cone Metric Spaces, Gazi University Journal of Science 24 (1 ( [18] E. Karapınar, Coupled fixed point theorems for nonlinear contractions in cone metric spaces, Comput. Math. Appl. 59 (12 ( [19] V. Lakshmikantham and Lj.B. Ćirić, Coupled fixed point theorems for nonlinear contractions in partially ordered metric spaces, Nonlinear Anal. 70 ( [20] N.V. Luong and N.X. Thuan, Coupled fixed points in partially ordered metric spaces and application, Nonlinear Anal. 74 ( [21] N.V. Luong, N.X. Thuan and T.T. Hai, Coupled fixed point theorems in partially ordered metric spaces depended on another function, Bulletin of Mathematical Analysis and Applications, Volume 3 Issue 3(2011, Pages [22] S. Moradi and M. Omid, A fixedpoint theorem for integral type inequality depending on another function, Int. J. Math. Anal. (Ruse 4, no , (2010. [23] H.K. Nashine and I. Altun, Fixed point theorems for generalized weakly contractive condition in ordered metric spaces, Fixed Point Theory Appl., Volume 2011, Article ID , 20 pages. [24] H.K. Nashine and B. Samet, Fixed point results for mappings satisfying (ψ, φweakly contractive condition in partially ordered metric spaces, Nonlinear Anal. 74 ( [25] H.K. Nashine, B. Samet and J. K. Kim, Fixed point results for contractions involving generalized altering distances in ordered metric spaces, Fixed point Theory Appl. 2011,2011:5. [26] H.K. Nashine, B. Samet and C. Vetro, Monotone generalized nonlinear contractions and fixed point theorems in ordered metric spaces, Math. Comput. Modelling 54 ( [27] H.K. Nashine and H. Aydi, Common fixed point theorems for four mappings through generalized altering distances in ordered metric spaces, Annali dell Universit di Ferrara, doi: /s y, (2012. [28] J.J. Nieto and R. R. López, Contractive mapping theorems in partially ordered sets and applications to ordinary differential equations, Order 22 (2005, [29] A.C.M. Ran and M.C.B. Reurings, A fixed point theorem in partially ordered sets and some application to matrix equations, Proc. Amer. Math. Soc. 132 ( [30] B. Samet, Coupled fixed point theorems for a generalized MeirKeeler contraction in partially ordered metric spaces, Nonlinear Anal. 7 (12 ( [31] B. Samet and C. Vetro, Coupled fixed point, finvariant set and fixed point of Norder, Ann. Funct. Anal. 1 (2 ( (H. Aydi Université de Sousse. Institut Supérieur d Informatique et des Technologies de Communication de Hammam Sousse. Route GP14011, H. Sousse, Tunisie address: (E. Karapınar Department of Mathematics, Atilim University 06836, İncek, Ankara, Turkey address:
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