@ Appl. Gen. Topol. 15, no. 1 (2014), 11-24doi:10.4995/agt.2014.2032 c© AGT, UPV, 2014 Quadruple fixed point theorems for nonlinear contractions on partial metric spaces Erdal Karapınar∗,a and Kenan Tas b a Department of Mathematics, Atilim University 06836, İncek, Ankara, Turkey (ekarapinar@atilim.edu.tr, erdalkarapinar@yahoo.com) b Çankaya University, Department of Mathematics and Computer Science, Ankara, Turkey (kenan@cankaya.edu.tr) Abstract The notion of coupled fixed point was introduced by Guo and Laksmikantham [12]. Later Gnana Bhaskar and Lakshmikantham in [11] investigated the coupled fixed points in the setting of partially or- dered set by defining the notion of mixed monotone property. Very re- cently, the concept of tripled fixed point was introduced by Berinde and Borcut [7]. Following this trend, Karapınar[19] defined the quadruple fixed point. In this manuscript, quadruple fixed point is discussed and some new fixed point theorems are obtained on partial metric spaces. 2010 MSC: 47H10; 54H25; 46J10; 46J15. Keywords: Fixed point theorems; Nonlinear contraction; Partial metric space; Partially ordered set; Quadruple Fixed Point. 1. Introduction and Preliminaries The existence of fixed points in partially ordered metric spaces was con- sidered first by Ran and Reurings [37]. After this remarkable paper, several authors have studied such problems (see e.g. [32, 33, 34, 11, 29, 30, 45, 9, 8] ). The notion of coupled fixed point was introduced by Guo and Laksmikantham [12]. After the interesting paper of Gnana Bhaskar and Lakshmikantham [11], many authors focused on coupled fixed point theory and proved several results (see e.g. [29, 30, 45, 9, 8, 18, 17]). ∗Corresponding author Received March 2012 – Accepted September 2012 http://dx.doi.org/10.4995/agt.2014.2032 E. Karapınar and K. Tas We recall the basic definitions and results from which our quadruple fixed point is inspired. The triple (X, d, ≤) is called a partially ordered metric spaces if (X, ≤) is a partially ordered set and (X, d) is a metric space. Further, if (X, d) is a complete metric space, then the triple (X, d, ≤) is called partially ordered complete metric spaces. Definition 1.1 (see [11]). Let (X, ≤) be a partially ordered set and F : X × X → X. We say that F has mixed monotone property if F(x, y) is monotone non-decreasing in x and is monotone non-increasing in y, that is, for any x, y ∈ X, x1 ≤ x2 ⇒ F(x1, y) ≤ F(x2, y), for x1, x2 ∈ X, and y1 ≤ y2 ⇒ F(x, y2) ≤ F(x, y1), for y1, y2 ∈ X. Definition 1.2 (see [11]). An element (x, y) ∈ X × X is said to be a couple fixed point of the mapping F : X × X → X if F(x, y) = x and F(y, x) = y. We endow the product space X × X with the following partial order: (1.1) (u, v) ≤ (x, y) ⇔ u ≤ x, y ≤ v; for all (x, y), (u, v) ∈ X × X. Two results of Bhaskar and Lakshmikantham [11] can be unified as follows: Theorem 1.3. Let (X, ≤) be a partially ordered set endowed with a metric d on X such that (X, d) is a complete metric spaces. Let F : X × X → X have the mixed monotone property on X. Assume that there exists a k ∈ [0, 1) with (1.2) d(F(x, y), F(u, v)) ≤ k 2 [d(x, u) + d(y, v)] , for all u ≤ x, y ≤ v. Suppose either F is continuous or X has the following properties: (i) if a non-decreasing sequence {xn} → x, then xn ≤ x, ∀n; (i) if a non-increasing sequence {yn} → y, then y ≤ yn, ∀n. If, in addition, there are x0, y0 ∈ X such that x0 ≤ F(x0, y0) and F(y0, x0) ≤ y0, then, there exists x, y ∈ X such that x = F(x, y) and y = F(y, x). We notice that Theorem 1.3 was extended to class of cone metric spaces in [17]. Inspired by Definition 1.1, Berinde and Borcut [7] introduced the following definition: (1.3) (u, v, w) ≤ (x, y, z) if and only if x ≥ u, y ≤ v, z ≥ w, where (u, v, w), (x, y, z) ∈ X3. Definition 1.4 (see [7]). 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 x1, x2 ∈ X, x1 ≤ x2 =⇒ F(x1, y, z) ≤ F(x2, y, z), c© AGT, UPV, 2014 Appl. Gen. Topol. 15, no. 1 12 Quadruple fixed point theorems y1, y2 ∈ X, y1 ≤ y2 =⇒ F(x, y1, z) ≥ F(x, y2, z), z1, z2 ∈ X, z1 ≤ z2 =⇒ F(x, y, z1) ≤ F(x, y, z2), The following is the main tripled fixed point result of Berinde and Borcut [7]. Theorem 1.5. Let (X, ≤) be partially ordered set and (X, d) be a complete metric space. Let F : X × X × X → X be a continuous mapping having the mixed monotone property on X. Assume that there exist constants a, b, c ∈ [0, 1) such that a + b + c < 1 for which (1.4) d(F(x, y, z), F(u, v, w)) ≤ ad(x, u) + bd(y, v) + cd(z, w) for all x ≥ u, y ≤ v, z ≥ w. If there exist x0, y0, z0 ∈ X such that x0 ≤ F(x0, y0, z0), y0 ≥ F(y0, x0, y0), z0 ≤ F(x0, y0, z0) then there exist x, y, z ∈ X such that F(x, y, z) = x and F(y, z, y) = x and F(z, y, x) = z The notion of metric space was introduced by Maurice René Fréchet [10] in 1906. Pseudometric space, quasimetric space, semimetric space, partial met- ric space are some examples of the generalizations of metric space. In this manuscript, we discuss partial metric space, introduced by Matthews (see e.g. [31]). The concept of the metric space started to apply to computer science around 1970. By using Baire metric, G. Khan [16] modeled a parallel computation. It consists of a set computing via sending unending streams of information by using infinite sequences. Hence, with this paper, reservoir of the theory of metric space started to be used in the branches of computer science, such as, domain theory and semantics. The handicap of this approaches is, in com- puter science, infinite sequence corresponding to unterminated programs. But, in computer science, unterminated program is bad. This un-solicited status solved by Matthews with his suggestion of non-zero self distance in metric con- struction. In the last decade, on partial metric spaces remarkable number of papers were reported (see e.g. [1]-[6],[13]-[15],[24]-[28],[39]-[55]) A mapping p : X × X → [0, ∞) is called partial metric (see e.g.[31]) on a nonempty set X if the following conditions are satisfied: (PM1) p(x, y) = p(y, x) (symmetry) (PM2) If p(x, x) = p(x, y) = p(y, y) then x = y (equality) (PM3) p(x, x) ≤ p(x, y) (small self-distances) (PM4) p(x, z) + p(y, y) ≤ p(x, y) + p(y, z) (triangularity) The pair (X, p) is called a partial metric space (PMS). Additionally, a triple (X, p, ≤) is called a partially ordered partial metric space if (X, p) is a partial metric space and (X, ≤) is a partially ordered set. c© AGT, UPV, 2014 Appl. Gen. Topol. 15, no. 1 13 E. Karapınar and K. Tas For a partial metric p on X, the functions dp, dm : X × X → R + given by (1.5) dp(x, y) = 2p(x, y) − p(x, x) − p(y, y) and (1.6) dm(x, y) = max{p(x, y) − p(x, x), p(x, y) − p(y, y)} are (usual) metrics on X. It is clear that dp and dm are equivalent. Moreover, (1.7) lim n→∞ dp(x, xn) = 0 ⇔ p(x, x) = lim n→∞ p(x, xn) = lim n,m→∞ p(xn, xm) Each partial metric p on X generates a T0 topology τp on X with a base of the family of open p-balls {Bp(x, ε) : x ∈ X, ε > 0}, where Bp(x, ε) = {y ∈ X : p(x, y) < p(x, x) + ε} for all x ∈ X and ε > 0. Example 1.6 (see e.g. [31, 24, 3]). Consider X = [0, ∞) with p(x, y) = max{x, y}. Then (X, p) is a partial metric space. It is clear that p is not a (usual) metric. Note that in this case dm(x, y) = |x − y| and dp(x, y) = 1 2 |x − y|. Example 1.7 (see [31]). Let X = {[a, b] : a, b, ∈ R, a ≤ b} and define p([a, b], [c, d]) = max{b, d} − min{a, c}. Then (X, p) is a partial metric spaces. Example 1.8 (see [31]). Let X := [0, 1] ∪ [2, 3] and define p : X × X → [0, ∞) by p(x, y) = { max{x, y} if {x, y} ∩ [2, 3] 6= ∅, |x − y| if {x, y} ⊂ [0, 1]. Then (X, p) is a partial metric space. Example 1.9 (see [31]). Let S be a non-empty set. By Sω, we denote the set of all infinite sequence x = {x0, x1, · · · } over S. For all such sequences x, y ∈ Sω define dS(x, y) = 2 −k, where k is the largest number (possibly ∞) such that xi = yi for each i < k, that is, dS(x, y) = 2 − sup{n|∀i 0. Remark 1.14. Since dp and dm are equivalent, we can take dm instead of dp in the above lemma. Karapınar [19] introduced the concept of quadruple fixed point and proved some quadruple fixed point theorems in partially ordered metric spaces (see also [20]- [23]). The aim of this paper is introduce the concept of quadruple fixed point and prove the related fixed point theorems in the context of partially ordered partial metric spaces. 2. Quadruple Fixed Point Theorems Let (X, p, ≤) be a partially ordered partial metric spaces. We consider the following partial order on the product space X4 = X × X × X × X: (2.1) (u, v, r, t) ≤ (x, y, z, w) if and only if x ≥ u, y ≤ v, z ≥ r, t ≤ w where (u, v, r, t), (x, y, z, w) ∈ X4. Regarding this partial order, we state the definition of the following mapping. Definition 2.1. Let (X, ≤) be partially ordered set and F : X4 → X. We say that F has the mixed monotone property if F(x, y, z, w) is monotone non- decreasing in x and z, and it is monotone non-increasing in y and w, that is, for any x, y, z, w ∈ X (2.2) x1, x2 ∈ X, x1 ≤ x2 ⇒ F(x1, y, z, w) ≤ F(x2, y, z, w), y1, y2 ∈ X, y1 ≤ y2 ⇒ F(x, y1, z, w) ≥ F(x, y2, z, w), z1, z2 ∈ X, z1 ≤ z2 ⇒ F(x, y, z1, w) ≤ F(x, y, z2, w), w1, w2 ∈ X, w1 ≤ w2 ⇒ F(x, y, z, w1) ≥ F(x, y, z, w2). c© AGT, UPV, 2014 Appl. Gen. Topol. 15, no. 1 15 E. Karapınar and K. Tas Definition 2.2. An element (x, y, z) ∈ X4 is called a quadruple fixed point of F : X4 → X if (2.3) F(x, y, z, w) = x and F(y, z, w, x) = y and F(z, w, x, y) = z and F(w, x, y, z) = w For a metric space (X, d), the function ρ : X4 → [0, ∞), given by, ρ((x, y, z, w), (u, v, r, t)) := d(x, u) + d(y, v) + d(z, r) + d(w, t) is a metric space on X4, that is, (X4, ρ) is a metric induced by (X, d). The aim of this paper is to prove the following theorem. Theorem 2.3. Let (X, ≤) be partially ordered set and (X, p) be a complete partial metric space. Let F : X4 → X be a mapping having the mixed monotone property on X. Assume that there exists a constant k ∈ [0, 1) such that (2.4) p(F(x, y, z, w), F(u, v, r, t)) ≤ k 4 [p(x, u) + p(y, v) + p(z, r) + p(w, t)] for all x ≥ u, y ≤ v, z ≥ r, w ≤ t. Suppose there exist x0, y0, z0, w0 ∈ X such that x0 ≤ F(x0, y0, z0, w0), y0 ≥ F(y0, z0, w0, x0), z0 ≤ F(z0, w0, x0, y0), w0 ≥ F(w0, x0, y0, z0). Suppose either (a) F is continuous, or (b) X has the following property: (i) if {xn} is a non-decreasing sequence xn → x (respectively, zn → z), then xn ≤ x (respectively, zn ≤ z) for all n, (ii) if {yn} is a non-increasing sequence yn → y(respectively, wn → w), then yn ≥ y (respectively, wn ≥ w) for all n, then there exist x, y, z, w ∈ X such that F(x, y, z, w) = x, F(y, z, w, x) = y, F(z, w, x, y) = z, F(w, x, y, z) = w. Proof. We construct a sequence {(xn, yn, zn, wn)} in the following way: Set x1 = F(x0, y0, z0, w0) ≥ x0, y1 = F(y0, z0, w0, x0) ≤ y0, z1 = F(z0, w0, x0, y0) ≥ z0, w1 = F(w0, x0, y0, z0) ≤ w0, and by the mixed monotone property of F , for n ≥ 1, inductively we get (2.5) xn = F(xn−1, yn−1, zn−1, wn−1) ≥ xn−1 ≥ · · · ≥ x0, yn = F(yn−1, zn−1, wn−1, xn−1) ≤ yn−1 ≤ · · · ≤ y0, zn = F(zn−1, wn−1, xn−1, yn−1) ≥ zn−1 ≥ · · · ≥ z0, wn = F(wn−1, xn−1, yn−1, zn−1) ≤ wn−1 ≤ · · · ≤ w0, c© AGT, UPV, 2014 Appl. Gen. Topol. 15, no. 1 16 Quadruple fixed point theorems Due to (2.4) and (2.5), we have (2.6) p(x1, x2) = p(F(x0, y0, z0, w0), F(x1, y1, z1, w1)) ≤ k 4 [p(x0, x1) + p(y0, y1) + p(z0, z1) + p(w0, w1)] (2.7) p(y1, y2) = p(F(y0, z0, w0, x0), F(y1, z1, w1, x1)) ≤ k 4 [p(y0, y1) + p(z0, z1) + p(w0, w1) + p(x0, x1)] (2.8) p(z1, z2) = p(F(z0, w0, x0, y0), F(z1, w1, x1, y1)) ≤ k 4 [p(z0, z1) + p(w0, w1) + p(x0, x1) + p(y0, y1)] (2.9) p(w1, w2) = p(F(w0, x0, y0, z0), F(w1, x1, y1, z1)) ≤ k 4 [p(w0, w1) + p(x0, x1) + p(y0, y1) + p(z0, z1)] Regarding (2.4) together with (2.6),(2.7),(2.8) we have (2.10) p(x2, x3) = p(F(x1, y1, z1, w1), F(x2, y2, z2, w2)) ≤ k 4 [p(x1, x2) + p(y1, y2) + p(z1, z2) + p(w1, w2)] (2.11) p(y2, y3) = p(F(y1, z1, w1, x1), F(y2, z2, w2, x2)) ≤ k 4 [p(y1, y2) + p(z1, z2) + p(w1, w2) + p(x1, x2)] (2.12) p(z2, z3) = p(F(z1, w1, x1, y1), F(z2, w2, x2, y2)) ≤ k 4 [p(z1, z2) + p(w1, w2) + p(x1, x2) + p(y1, y2)] (2.13) p(w2, w3) = p(F(w1, x1, y2, z1), F(w2, x2, y2, z2)) ≤ k 4 [p(w1, w2) + p(x1, x2) + p(y1, y2) + p(z1, z2)] Recursively we have (2.14) p(xn+1, xn+2) = p(F(xn, yn, zn, wn), F(xn+1, yn+1, zn+1, wn+1)) ≤ k 4 [p(xn, xn+1) + p(yn, yn+1) + p(zn, zn+1) + p(wn, wn+1)] (2.15) p(yn+1, yn+2) = p(F(yn, zn, wn, xn), F(yn+1, zn+1, wn+1), xn+1) ≤ k 4 [p(yn, yn+1) + p(zn, zn+1) + p(wn, wn+1) + p(xn, xn+1)] (2.16) p(zn+1, zn+2) = p(F(zn, wn, xn, yn), F(zn+1, wn+1, xn+1, yn+1)) ≤ k 4 [p(zn, zn+1) + p(wn, wn+1) + p(xn, xn+1) + p(yn, yn+1)] (2.17) p(wn+1, wn+2) = p(F(wn, xn, yn, zn), F(wn+1, xn+1, yn+1, zn+1)) ≤ k 4 [p(wn, wn+1) + p(xn, xn+1) + p(yn, yn+1) + p(zn, zn+1)] c© AGT, UPV, 2014 Appl. Gen. Topol. 15, no. 1 17 E. Karapınar and K. Tas For simplicity, we can use the matrix notation as follow. Set M =     1 4 1 4 1 4 1 4 1 4 1 4 1 4 1 4 1 4 1 4 1 4 1 4 1 4 1 4 1 4 1 4     , Dn =     p(xn+1, xn) p(yn+1, yn) p(zn+1, zn) p(wn+1, wn)     and R = ( 1 4 1 4 1 4 1 4 ) . Notice that (2.18) RM = R and Mn = M for all n ∈ N. So we have, (2.19) D1 ≤ kD0, (2.20) D2 ≤ kMD1 ≤ k 2 M 2 D0 = k 2 MD0, and, inductively (2.21) Dn ≤ kMDn−1 ≤ k n MD0. (2.22) p(xn+1, xn+2) ≤ kRDn     p(xn, xn+1) p(yn, yn+1) p(zn, zn+1) p(wn, wn+1)     Hence, by (2.18),(2.4) and (2.5), we have (2.23) p(xn+1, xn+2) = p(F(xn, yn, zn, wn), F(xn+1, yn+1, zn+1, wn+1)) ≤ k 4 [p(xn, xn+1) + p(yn, yn+1) + p(zn, zn+1) + p(wn, wn+1)] ≤ kRDn ≤ k n+1RMD0 ≤ k n+1RD0. We shall show the sequences {xn} are Cauchy easily by using (2.14)-(2.21). Without loss of generality, we may assume that m > n. By using (2.14)-(2.21) together with triangle inequality, we obtain that (2.24) p(xm, xn) ≤ p(xm, xm−1) + p(xm−1, xm−2) + · · · + p(xn+1, xn) ≤ km−1RD0 + · · · + k nRD0 ≤ kn(1 + · · · + km−n−1)RD0 ≤ kn 1 1−k RD0 Letting n → ∞ in (2.24) and recalling that k ∈ [0, 1), we get that lim n→∞ p(xn, xm) = 0. By definition, dp(xn, xm) = 2p(xn, xm) − p(xn, xn) − p(xm, xm) ≤ 2p(xn, xm). Thus, we have (2.25) lim n→∞ dp(xn, xm) = 0. c© AGT, UPV, 2014 Appl. Gen. Topol. 15, no. 1 18 Quadruple fixed point theorems Since (X, p) is a complete partial metric space, then by Lemma 1.11, (X, dp) is a complete metric space. Thus, {xn} converges in (X, dp), say x. Again by 1.11, we have (2.26) p(x, x) = lim n→∞ p(xn, xm) = lim n→∞ p(xn, x) = 0. Analogously, one can prove that {yn}, {zn} and {wn} are Cauchy sequences. Since (X, dp) is complete metric space, there exists x, y, z, w ∈ X such that (2.27) p(y, y) = lim n→∞ p(yn, ym) = lim n→∞ p(yn, y) = 0, p(z, z) = lim n→∞ p(zn, zm) = lim n→∞ p(zn, z) = 0, p(w, w) = lim n→∞ p(wn, wm) = lim n→∞ p(wn, w) = 0. Suppose now the assumption (a) holds. Then by (2.26) and (2.27), we have (2.28) x = lim n→∞ xn = lim n→∞ F(xn−1, yn−1, zn−1, wn−1) = F( lim n→∞ xn−1, lim n→∞ yn−1, lim n→∞ zn−1, lim n→∞ wn−1) = F(x, y, z, w) Analogously, we also observe that (2.29) y = lim n→∞ yn = lim n→∞ F(xn−1, wn−1, zn−1, yn−1) = F(x, w, z, y) z = lim n→∞ zn = lim n→∞ F(zn−1, yn−1, xn−1, wn−1) = F(z, y, x, w) w = lim n→∞ wn = lim n→∞ F(zn−1, wn−1, xn−1, yn−1) = F(z, w, x, y) Thus, we have F(x, y, z, w) = x, F(x, w, z, y) = y, F(z, y, x, w) = z, F(z, w, x, y) = w. Suppose now the assumption (b) holds. Since {xn}, {zn} are non-decreasing and xn → x, zn → z and also {yn}, {wn} are non-increasing and yn → y, wn → w, then by assumption (b) we have xn ≥ x, yn ≤ y, zn ≥ z, wn ≤ w for all n. Due to (2.26) and (2.27), we have (2.30) p(F(x, y, z, w), F(x, y, z, w)) ≤ k 4 [p(x, x) + p(y, y) + p(z, z) + p(w, w)] = 0. Consider now, (2.31) p(xn, F(x, y, z, w)) = p(F(xn−1, yn−1, zn−1, wn−1), F(x, y, z, w)) ≤ k 4 [p(xn−1, x) + p(yn−1, y) + p(zn−1, z) + p(wn−1, w)] Letting n → ∞ in (2.31), by Lemma 1.12 we get (2.32) p(x, F(x, y, z, w)) ≤ k 4 [p(x, x) + p(y, y) + p(z, z) + p(w, w)] c© AGT, UPV, 2014 Appl. Gen. Topol. 15, no. 1 19 E. Karapınar and K. Tas Regarding (2.26) and (2.27), we conclude that p(x, F(x, y, z, w)) = 0. Hence, by (2.26),(2.30),(2.32) and definiton (2.33) dp(x, F(x, y, z, w)) = 2p(x, F(x, y, z, w))−p(F(x, y, z, w), F(x, y, z, w))−p(x, x) = 0. Thus, we have x = F(x, y, z, w). analogously we we get F(y, z, w, x) = y, F(z, w, x, y) = z, F(w, x, y, z) = w. Thus, we proved that F has a quadruple fixed point. � 3. Uniqueness of Quadruple Fixed Point In this section we shall prove the uniqueness of quadruple fixe point. For a product X4 of a partial ordered set (X, ≤) we define a partial ordering in the following way: For all (x, y, z, t), (u, v, r, t) ∈ X4 (3.1) (x, y, z, w) ≤ (u, v, r, t) ⇔ x ≤ u, y ≥ v, z ≤ r, w ≥ r. We say that (x, y, z, w) is equal (u, v, r, t) if and only if x = u, y = v, z = r and w = t. Theorem 3.1. In addition to hypothesis of Theorem 2.3, suppose that for all (x, y, z, t), (u, v, r, t) ∈ X ×X ×X ×X, there exists (a, b, c, d) ∈ X ×X ×X ×X that is comparable to (x, y, z, t) and (u, v, r, t), then F has a unique quadruple fixed point. Proof. The set of quadruple fixed point of F is not empty due to Theorem 2.3. Assume, now, (x, y, z, t) and (u, v, r, t) are the quadruple fixed point of F , that is, F(x, y, z, w) = x, F(u, v, r, t) = u, F(y, z, w, x) = y, F(v, r, t, u) = v, F(z, w, x, y) = z, F(r, t, u, v) = r, F(w, x, y, z) = w, F(t, u, v, r) = t, We shall show that (x, y, z, w) and (u, v, r, t) are equal. By assumption, there exists (a, b, c, d) ∈ X×X×X×X that is comparable to (x, y, z, t) and (u, v, r, t). Define sequences {an}, {bn}, {cn} and {dn} such that a = a0, b = b0, c = c0, d = d0 and (3.2) an = F(an−1, bn−1, zn−1, dn−1), bn = F(bn−1, cn−1, dn−1, an−1), cn = F(cn−1, dn−1, an−1, bn−1), dn = F(dn−1, an−1, bn−1, cn−1). c© AGT, UPV, 2014 Appl. Gen. Topol. 15, no. 1 20 Quadruple fixed point theorems for all n. Since (x, y, z, w) is comparable with (a, b, c, d), we may assume that (x, y, z, w) ≥ (a, b, c, d) = (a0, b0, c0, d0). Recursively, we get that (3.3) (x, y, z, w) ≥ (an, bn, cn, dn) for all n. By (3.3) and (2.4), we have (3.4) p(x, an+1) = p(F(x, y, z, w), F(an, bn, cn, dn)) ≤ k 4 [p(x, an) + p(y, bn) + p(z, cn) + p(w, dn)] (3.5) p(bn+1, y) = p(F(bn, cn, dn, an), F(y, z, w, x)) ≤ k 4 [p(bn, y) + p(cn, z) + p(dn, w) + p(an, x)] (3.6) p(z, cn+1) = p(F(z, w, x, y), F(cn, dn, an, bn)) ≤ k 4 [p(z, cn) + p(w, dn) + p(x, an) + p(y, bn)] (3.7) p(dn+1, w) = p(F(cn, dn, an, bn), F((w, x, y, z))) ≤ k 4 [p(dn, w) + p(an, x) + p(bn, y) + p(cn, z)] Set γn = p(x, an) + p(y, bn) + p(z, cn) + p(w, dn). Then, due to (3.7)-(3.7), we have (3.8) γn+1 ≤ kγn ≤ k nγ0, for all n. � Since 0 ≤ k < 1, the sequence {γn} is decreasing and bounded below. Thus, there exists γ ≥ 0 such that lim n→∞ γn = γ. Now, we shall show that γ = 0. Letting n → ∞ in (3.8), and having mind 0 ≤ k < 1, we obtain that γ ≤ 0. Therefore, γ = 0. That is, lim n→∞ γn = 0. Consequently, we have (3.9) limn→∞ p(x, an) = 0, limn→∞ p(y, bn) = 0, limn→∞ p(z, cn) = 0, limn→∞ p(w, dn) = 0. Analogously, we show that (3.10) limn→∞ p(u, an) = 0, limn→∞ p(v, bn) = 0, limn→∞ p(r, cn) = 0, limn→∞ p(s, dn) = 0. Combining (3.9) and (3.10) yield ,by uniqueness of the limit, that (x, y, z, w) and (u, v, r, t) are equal. Now, in the following example neither the continuity of the mapping F is satisfied nor the conditions (a) and (b) given in Theorem 2.3 hold, but we still obtain a quadruple fixed point result. c© AGT, UPV, 2014 Appl. Gen. Topol. 15, no. 1 21 E. Karapınar and K. Tas Example 3.2. Let X = [0, ∞), and p(x, y) = max{x, y} be a partial metric. Let ”≤” be the usual order on positive half-line. Notice that dp(x, y) = |x − y| becomes the corresponding metric. It is clear that (X, p) is a complete partial metric space. Now define F : X4 → X as F(x, y, z, w) = { x−y+z−w 8 , if x + z ≥ y + w, 0, otherwise . Then it is easy to see that F has the mixed monotone property. 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