Hindawi Publishing Corporation Abstract and Applied Analysis Volume 2013, Article ID 815289, 7 pages http://dx.doi.org/10.1155/2013/815289 Research Article Fixed Point and Common Fixed Point Theorems on Ordered Cone b-Metric Spaces Sahar Mohammad Abusalim and Mohd Salmi Md Noorani School of Mathematical Sciences, Faculty of Science and Technology, Universiti Kebangsaan Malaysia, 43600 Bangi, Selangor Darul Ehsan, Malaysia Correspondence should be addressed to Sahar Mohammad Abusalim; saharabosalem@gmail.com Received 28 February 2013; Accepted 18 April 2013 Academic Editor: Abdelghani Bellouquid Copyright © 2013 S. M. Abusalim and M. S. M. Noorani. This is an open access article distributed under the Creative Commons Attribution License, which permits unrestricted use, distribution, and reproduction in any medium, provided the original work is properly cited. The concept of a cone b-metric space has been introduced recently as a generalization of a b-metric space and a cone metric space in 2011. The aim of this paper is to establish some fixed point and common fixed point theorems on ordered cone b-metric spaces. The proposed theorems expand and generalize several well-known comparable results in the literature to ordered cone b-metric spaces. Some supporting examples are given. 1. Introduction Fixed point theory has attracted many researchers since 1922 with the admired Banach fixed point theorem. This theorem supplies a method for solving a variety of applied dilemma in mathematical sciences and engineering. A large literature on this subject exists, and this is a very active area of research at present. Banach contraction principle has been generalized in dissimilar directions in different spaces by mathematicians over the years; for more details on this and related topics, we refer to [1–6] and references therein. In contemporary time, fixed point theory has evolved speedily in partially ordered cone metric spaces; that is, cone metric spaces equipped with a partial ordering, for some new results in ordered metric spaces see [7]. A coming early result in this bearing was constituted by Altun and Durmaz [8] under the condition of normality for cones. Then, Altun et al. [9] generalized the results of Altun and Durmaz [8] by omitting the assumption of normality condition for cones. Afterward, several authors have studied fixed point and common fixed point problems in ordered cone metric spaces; for more details see [10–17]. In 2011, Hussain and Shah [18] presented cone b-metric spaces as a generalization of b-metric spaces and cone metric spaces; for some new results in b-metric spaces see [19]. They not only constructed some topological properties in such spaces but also ameliorated some current results about KKM mappings in the setting of a cone b-metric space. After some time, many authors have been motivated to demonstrate fixed point theorems as well as common fixed point theorems for two or more mappings on cone b-metric spaces by the incipient work of Hussain and Shah [18] (see [20–23] and the references therein). In [8], Altun and Durmaz proved the following results under the condition of normality for cones. Theorem 1 (see [8]). Let (π, β) be a partially ordered set, suppose that there exists a cone metric π in π such that the cone metric space (π, π) is complete, and let π be a normal cone with normal constant πΎ. Let π : π → π be a continuous and nondecreasing mapping with respect to β. Suppose that the following three assertions hold: (i) there exists π ∈ [0, 1) such that π(ππ₯, ππ¦) βͺ― ππ(π₯, π¦) for all π₯, π¦ ∈ π with π¦ β π₯; (ii) there exists π₯0 ∈ π such that π₯0 β πx0 . Then π has a fixed point in π. In [9], Altun et al. generalized the above theorem and proved it without normality condition for cones. 2 Abstract and Applied Analysis Theorem 2 (see [9]). Let (π, β) be a partially ordered set and suppose that there exists a cone metric π in π such that the cone metric space (π, π) is complete over a solid cone π. Let π : π → π be a continuous and nondecreasing mapping with respect to β. Suppose that the following two assertions hold: (i) there exist π, π, π ∈ [0, 1) with π + 2π + 2π < 1 such that π (ππ₯, ππ¦) βͺ― ππ (π₯, π¦) + π (π (ππ₯, π₯) + π (ππ¦, π¦)) + π (π (ππ₯, π¦) + π (ππ¦, π₯)) (1) for all π₯, π¦ ∈ π with π¦ β π₯; Then π has a fixed point in π. Theorem 3 (see [9]). Let (π, β) be a partially ordered set and suppose that there exists a cone metric π in π such that the cone metric space (π, π) is complete over a solid cone π. Let π : π → π be a nondecreasing mapping with respect to β. Suppose that the following three assertions hold: (i) there exist π, π, π ∈ [0, 1) with π + 2π + 2π < 1 such that + π (π (ππ₯, π¦) + π (ππ¦, π₯)) (2) + π (π (π¦, ππ₯) + π (π₯, ππ¦)) (4) for all comparative π₯, π¦ ∈ π; (ii) if an increasing sequence {π₯π } converges to π₯ in π, then π₯π β π₯ for all n. In this paper, we prove some fixed point and common fixed point theorems on ordered cone b-metric spaces. Our results extend and generalize several well-known comparable results in the literature to ordered cone b-metric spaces. Throughout this paper, we do not impose the normality condition for the cones, but the only assumption is that the cone π is solid, that is, int π =ΜΈ 0. The following definitions and results shall be needed in the sequel. Let πΈ be a real Banach space and π denotes the zero element in πΈ. A cone π is a subset of πΈ such that (1) π is nonempty closed set and π =ΜΈ {π}; (3) π₯ ∈ π and −π₯ ∈ π imply π₯ = π. (ii) there exists π₯0 ∈ π such that π₯0 β ππ₯0 ; (iii) if an increasing sequence {π₯π } converges to π₯ in π, then π₯π β π₯ for all π. Then π has a fixed point in π. In the same paper, they also presented the following two common fixed point results in ordered cone metric spaces. Theorem 4 (see [9]). Let (π, β) be a partially ordered set and suppose that there exists a cone metric π in π such that the cone metric space (π, π) is complete over a solid cone π. Let π, π : π → π be two weakly increasing mappings with respect to β. Suppose that the following three assertions hold: (i) there exist π, π, π ∈ [0, 1) with π + 2π + 2π < 1 such that + π (π (π¦, ππ₯) + π (π₯, ππ¦)) π (ππ₯, ππ¦) βͺ― ππ (π₯, π¦) + π (π (π₯, ππ₯) + π (π¦, ππ¦)) (2) if π, π are nonnegative real numbers and π₯, π¦ ∈ π, then ππ₯ + ππ¦ ∈ π; for all π₯, π¦ ∈ π with π¦ β π₯; π (ππ₯, ππ¦) βͺ― ππ (π₯, π¦) + π (π (π₯, ππ₯) + π (π¦, ππ¦)) (i) there exist π, π, π ∈ [0, 1) with π + 2π + 2π < 1 such that Then π and π have a common fixed point π₯∗ ∈ π. (ii) there exists π₯0 ∈ π such that π₯0 β ππ₯0 . π (ππ₯, ππ¦) βͺ― ππ (π₯, π¦) + π (π (ππ₯, π₯) + π (ππ¦, π¦)) π, π : π → π be two weakly increasing mappings with respect to β. Suppose that the following three assertions hold: (3) for all comparative π₯, π¦ ∈ π; (ii) π or π is continuous. For any cone π ⊂ πΈ, the partial ordering βͺ― with respect to π is defined by π₯ βͺ― π¦ if and only if π¦ − π₯ ∈ π. The notation of βΊ stands for π₯ βͺ― π¦ but π₯ =ΜΈ π¦. Also, we use π₯ βͺ π¦ to indicate that π¦ − π₯ ∈ int π, where int π denotes the interior of π. A cone π is called normal if there exists the number πΎ such that σ΅© σ΅© π βͺ― π₯ βͺ― π¦ σ³¨⇒ βπ₯β ≤ πΎ σ΅©σ΅©σ΅©π¦σ΅©σ΅©σ΅© , (5) for all π₯, π¦ ∈ πΈ. The least positive number πΎ satisfying the above condition is called the normal constant of π. Definition 6 (see [18]). Let π be a nonempty set and πΈ a real Banach space equipped with the partial ordering βͺ― with respect to the cone π. A vector-valued function π : π × π → πΈ is said to be a cone b-metric function on π with the constant π ≥ 1 if the following conditions are satisfied: (1) π βͺ― π(π₯, π¦) for all π₯, π¦ ∈ π and π(π₯, π¦) = π if and only if π₯ = π¦; (2) π(π₯, π¦) = π(π¦, π₯) for all π₯, π¦ ∈ π; (3) π(π₯, π¦) βͺ― π (π(π₯, π¦) + π(π¦, π§)) for all π₯, π¦, π§ ∈ π. Then π and π have a common fixed point π₯∗ ∈ π. Then pair (π, π) is called a cone b-metric space (or a cone metric type space); we shall use the first mentioned term. Theorem 5 (see [9]). Let (π, β) be a partially ordered set and suppose that there exists a cone metric π in π such that the cone metric space (π, π) is complete over a solid cone π. Let Observe that if π = 1, then the ordinary triangle inequality in a cone metric space is satisfied; however, it does not hold true when π > 1. Thus the class of cone b-metric spaces is Abstract and Applied Analysis 3 effectively larger than that of the ordinary cone metric spaces. That is, every cone metric space is a cone b-metric space, but the converse need not be true. The following examples show the above remarks. Proof. Let π > π ≥ 1. It follows that π (π₯π , π₯π ) βͺ― π π (π₯π , π₯π+1 ) + π 2 π (π₯π+1 , π₯π+2 ) + ⋅ ⋅ ⋅ + π π−π π (π₯π−1 , π₯π ) . 2 Example 7. Let π = {−1, 0, 1}, πΈ = R , and π = {(π₯, π¦) : π₯ ≥ 0, π¦ ≥ 0}. Define π : π × π → π by π(π₯, π¦) = π(π¦, π₯) for all π₯, π¦ ∈ π, π(π₯, π₯) = π, π₯ ∈ π, and π(−1, 0) = (3, 3), π(−1, 1) = π(0, 1) = (1, 1). Then (π, π) is a complete cone b-metric space but the triangle inequality is not satisfied. Indeed, we have that π(−1, 1) + π(1, 0) = (1, 1) + (1, 1) = (2, 2) βΊ (3, 3) = π(−1, 0). It is not hard to verify that π = 3/2. (7) Now, (6) and π β < 1 imply that π (π₯π , π₯π ) βͺ― π π (π₯π , π₯π+1 ) + π 2 π (π₯π+1 , π₯π+2 ) + ⋅ ⋅ ⋅ + π π−π π (π₯π−1 , π₯π ) Example 8. Let π = R, πΈ = R2 , and π = {(π₯, π¦) ∈ πΈ : π₯ ≥ 0, π¦ ≥ 0}. Define π : π × π → πΈ by π(π₯, π¦) = (|π₯ − π¦|2 , |π₯ − π¦|2 ). Then, it is easy to see that (π, π) is a cone b-metric space with the coefficient π = 2. But it is not a cone metric spaces since the triangle inequality is not satisfied. βͺ― π βπ π (π₯0 , π₯1 ) + π 2 βπ+1 π (π₯0 , π₯1 ) Definition 9 (see [18]). Let (π, π) be a cone b-metric space, {π₯π } a sequence in π and π₯ ∈ π. = π βπ (1 + π β + (π β)2 + ⋅ ⋅ ⋅ + (π β)π−π−1 ) π (π₯0 , π₯1 ) (1) For all π ∈ πΈ with π βͺ π, if there exists a positive integer π such that π(π₯π , π₯) βͺ π for all π > π, then π₯π is said to be convergent and π₯ is the limit of {π₯π }. One denotes this by π₯π → π₯. + ⋅ ⋅ ⋅ + π π−π βπ−1 π (π₯0 , π₯1 ) = (π βπ + π 2 βπ+1 + ⋅ ⋅ ⋅ + π π−π βπ−1 ) π (π₯0 , π₯1 ) βͺ― π βπ π (π₯0 , π₯1 ) σ³¨→ π as π σ³¨→ ∞. 1 − π β (8) (2) For all π ∈ πΈ with π βͺ π, if there exists a positive integer π such that π(π₯π , π₯π ) βͺ π for all π, π > π, then {π₯π } is called a Cauchy sequence in π. According to Lemma 10(2), and for any π ∈ πΈ with π β« π, there exists π0 ∈ N such that for any π > π0 , (π βπ /(1 − π β))π(π₯0 , π₯1 ) βͺ π. Furthermore, from (8) and for any π > π > π0 , Lemma 10(3) shows that (3) A cone metric space (π, π) is called complete if every Cauchy sequence in π is convergent. π (π₯π , π₯π ) βͺ π. The following lemma is useful in our work. Hence, by Definition 9(2) {π₯π } is a Cauchy sequence in π. Lemma 10 (see [24]). (1) If πΈ is a real Banach space with a cone π and π βͺ― ππ where π ∈ π and 0 ≤ π < 1, then π = π. (2) If π ∈ int π, π βͺ― ππ , and ππ → π, then there exists a positive integer π such that ππ βͺ π for all π ≥ π. (3) If π βͺ― π and π βͺ π, then π βͺ π. (4) If π βͺ― π’ βͺ π for each π βͺ π, then π’ = π. Theorem 12. Let (π, β) be a partially ordered set and suppose that there exists a cone b-metric π in π such that the cone b-metric space (π, π) is complete with the coefficient π ≥ 1 relative to a solid cone π. Let π : π → π be a continuous and nondecreasing mapping with respect to β. Suppose that the following three assertions hold: (i) there exist ππ , π = 1, . . . , 5, such that 2π π1 + (π + 1)(π2 + π3 ) + (π 2 + π )(π4 + π5 ) < 2 with ∑5π=1 ππ < 1, 2. Fixed Point Results In this section, we prove some fixed point theorems on ordered cone b-metric space. We begin with a simple but a useful lemma. Lemma 11. Let {π₯π } be a sequence in a cone b-metric space (π, π) with the coefficient π ≥ 1 relative to a solid cone π such that π (π₯π , π₯π+1 ) βͺ― βπ (π₯π−1 , π₯π ) , (9) (6) where β ∈ [0, 1/π ) and π = 1, 2, . . .. Then {π₯π } is a Cauchy sequence in (π, π). π (ππ₯, ππ¦) βͺ― π1 π (π₯, π¦) + π2 π (ππ₯, π₯) + π3 π (ππ¦, π¦) + π4 π (ππ₯, π¦) + π5 π (ππ¦, π₯) (10) for all π₯, π¦ ∈ π with π¦ β π₯; (ii) there exists π₯0 ∈ π such that π₯0 β ππ₯0 . Then π has a fixed point π₯∗ ∈ π. Proof. If π₯0 = ππ₯0 , then the proof is finished. Suppose that π₯0 =ΜΈ ππ₯0 . Since π₯0 β ππ₯0 and π is nondecreasing with respect 4 Abstract and Applied Analysis to β, we obtain by induction that π₯0 β ππ₯0 = π₯1 β π1 π₯0 = π₯2 β ⋅ ⋅ ⋅ β ππ−1 π₯0 = π₯π β ππ π₯0 = π₯π+1 β ⋅ ⋅ ⋅ . Then we have, π (π₯π+1 , π₯π ) = π (ππ π₯0 , ππ−1 π₯0 ) = π (π (ππ−1 π₯0 ) , π (ππ−2 π₯0 )) βͺ― π1 π (ππ−1 π₯0 , ππ−2 π₯0 ) + π2 π (ππ−1 π₯0 , ππ−2 π₯0 ) + π3 π (ππ π₯0 , ππ−1 π₯0 ) + π4 π (ππ π₯0 , ππ−2 π₯0 ) + π5 π (ππ−1 π₯0 , ππ−1 π₯0 ) = π1 π (π₯π , π₯π−1 ) + π2 π (π₯π+1 , π₯π ) + π3 π (π₯π , π₯π−1 ) + π4 π (π₯π+1 , π₯π−1 ) + π5 π (π₯π , π₯π ) βͺ― π1 π (π₯π , π₯π−1 ) + π2 π (π₯π+1 , π₯π ) + π3 π (π₯π , π₯π−1 ) (11) Then, one can assert that + (π2 + π π4 ) π (π₯π+1 , π₯π ) . If we use condition (iii) instead of the continuity of π in Theorem 12, we have the following result. Theorem 13. Let (π, β) be a partially ordered set and suppose that there exists a cone b-metric π in π such that the cone bmetric space (π, π) is complete with the coefficient π ≥ 1 relative to a solid cone π. Let π : π → π be a nondecreasing mapping with respect to β. Suppose that the following three assertions hold: (i) there exist ππ , π = 1, . . . , 5, such that 2π π1 + (π + 1)(π2 + π3 ) + (π 2 + π )(π4 + π5 ) < 2 with ∑5π=1 ππ < 1, π (ππ₯, ππ¦) βͺ― π1 π (π₯, π¦) + π2 π (ππ₯, π₯) + π3 π (ππ¦, π¦) + π π4 (π (π₯π+1 , π₯π ) + π (π₯π , π₯π−1 )) . π (π₯π+1 , π₯π ) βͺ― (π1 + π3 + π π4 ) π (π₯π , π₯π−1 ) sequence in π. Since π is complete, there exists π₯∗ ∈ π such that π₯π → π₯∗ . Since π is continuous, then π₯∗ = lim π₯π+1 = lim ππ π₯0 = lim π(ππ−1 π₯0 ) = π(lim ππ−1 π₯0 ) = π(lim π₯π ) = π(π₯∗ ). Therefore, π₯∗ is a fixed point of π. + π4 π (ππ₯, π¦) + π5 π (ππ¦, π₯) for all π₯, π¦ ∈ π with π¦ β π₯; (12) (ii) there exists π₯0 ∈ π such that π₯0 β ππ₯0 ; (iii) if an increasing sequence {π₯π } converges to π₯ in π, then π₯π β π₯ for all π. On the other hand, we have Then π has a fixed point π₯∗ ∈ π. π (π₯π , π₯π+1 ) = π (ππ−1 π₯0 , ππ π₯0 ) = π (π (ππ−2 π₯0 ) , π (ππ−1 π₯0 )) βͺ― π1 π (ππ−2 π₯0 , ππ−1 π₯0 ) + π2 π (ππ−2 π₯0 , ππ−1 π₯0 ) + π3 π (ππ−1 π₯0 , ππ π₯0 ) Proof. As in the Theorem 12, we can construct an increasing sequence {π₯π } and prove that there exists π₯∗ ∈ π such that π₯π → π₯∗ . Now, condition (iii) implies π₯π β π₯∗ for all π. Therefore, we can use condition (i) and so π (ππ₯π , ππ₯∗ ) βͺ― π1 π (π₯π , π₯∗ ) + π2 π (ππ₯π , π₯π ) + π3 π (ππ₯∗ , π₯∗ ) + π4 π (ππ−1 π₯0 , ππ−1 π₯0 ) + π5 π (ππ π₯0 , ππ−2 π₯0 ) + π4 π (ππ₯π , π₯∗ ) + π5 π (ππ₯∗ , π₯π ) . (17) = π1 π (π₯π , π₯π−1 ) + π2 π (π₯π , π₯π−1 ) + π3 π (π₯π+1 , π₯π ) + π4 π (π₯π , π₯π ) + π5 π (π₯π+1 , π₯π−1 ) βͺ― π1 π (π₯π , π₯π−1 ) + π2 π (π₯π , π₯π−1 ) + π3 π (π₯π+1 , π₯π ) + π π5 (π (π₯π+1 , π₯π ) + π (π₯π , π₯π−1 )) . (13) Then, one can assert that π (π₯π+1 , π₯π ) βͺ― (π1 + π2 + π π5 ) π (π₯π , π₯π−1 ) + (π3 + π π5 ) π (π₯π+1 , π₯π ) . 2π1 + π2 + π3 + π π4 + π π5 π (π₯π , π₯π−1 ) 2 − (π2 + π3 + π π4 + π π5 ) Taking π → ∞, we have π(π₯∗ , ππ₯∗ ) βͺ― (π3 + π5 )π(π₯∗ , ππ₯∗ ) π(π₯∗ , ππ₯∗ ). Since (π3 + π5 ) < 1, Lemma 10(1) shows that π(π₯∗ , ππ₯∗ ) = π; that is, π₯∗ = ππ₯∗ . Therefore π₯∗ is a fixed point of π. 3. Common Fixed Point Results Now, we give two common fixed point theorems on ordered cone b-metric spaces. We need the following definition. (14) Adding (12) and (14), we get π (π₯π+1 , π₯π ) βͺ― (16) (15) = ππ (π₯π , π₯π−1 ) , where π = (2π1 + π2 + π3 + π π4 + π π5 )/(2 − (π2 + π3 + π π4 + π π5 )) < 1/π . According to Lemma 11, we have {π₯π } is a Cauchy Definition 14 (see [9]). Let (π, β) be a partially ordered set. Two mappings π, π : π → π are said to be weakly increasing if ππ₯ β πππ₯ and ππ₯ β πππ₯ hold for all π₯ ∈ π. Theorem 15. Let (π, β) be a partially ordered set and suppose that there exists a cone b-metric π in π such that the cone bmetric space (π, π) is complete with the coefficient π ≥ 1 relative to a solid cone π. Let π, π : π → π be two weakly increasing mappings with respect to β. Suppose that the following three assertions hold: Abstract and Applied Analysis 5 (i) there exist ππ , π = 1, . . . , 5, such that 2π π1 + (π + 1)(π2 + π3 ) + (π 2 + π )(π4 + π5 ) < 2 with ∑5π=1 ππ < 1, π (ππ₯, ππ¦) βͺ― π1 π (π₯, π¦) + π2 π (π₯, ππ₯) + π3 π (π¦, ππ¦) βͺ― π1 π (π₯2π+1 , π₯2π ) + π2 π (π₯2π+1 , π₯2π+2 ) + π3 π (π₯2π , π₯2π+1 ) + π4 π (π₯2π , π₯2π+2 ) + π5 π (π₯2π+1 , π₯2π+1 ) (18) + π4 π (π¦, ππ₯) + π5 π (π₯, ππ¦) βͺ― (π1 + π3 ) π (π₯2π , π₯2π+1 ) + π2 π (π₯2π+1 , π₯2π+2 ) + π π4 (π (π₯2π , π₯2π+1 ) + π (π₯2π+1 , π₯2π+2 )) for all comparative π₯, π¦ ∈ π; = (π1 + π3 + π π4 ) π (π₯2π , π₯2π+1 ) (ii) π or π is continuous. + (π2 + π π4 ) π (π₯2π+1 , π₯2π+2 ) . Then π and π have a common fixed point π₯∗ ∈ π. (21) Proof. Let π₯0 be an arbitrary point of π and define a sequence {π₯π } in π as follows: π₯2π+1 = ππ₯2π and π₯2π+2 = ππ₯2π+1 for all π > 0. Note that, since π and π are weakly increasing, we have π₯1 = ππ₯0 β πππ₯0 = ππ₯1 = π₯2 and π₯2 = ππ₯1 β πππ₯1 = ππ₯2 = π₯3 , and continuing this process we have π₯1 β π₯2 β ⋅ ⋅ ⋅ β π₯π β π₯π+1 β ⋅ ⋅ ⋅ . That is, the sequence {π₯π } is nondecreasing. Now, since π₯2π and π₯2π+1 are comparative, we can use the inequality (18), and then we have Hence, (1 − (π2 + π π4 )) π (π₯2π+2 , π₯2π+1 ) (22) βͺ― (π1 + π3 + π π4 ) π (π₯2π , π₯2π+1 ) . Adding inequalities (20) and (22), we get π (π₯2π+1 , π₯2π+2 ) βͺ― π (π₯2π+1 , π₯2π+2 ) = π (ππ₯2π , ππ₯2π+1 ) (2π1 + π2 + π3 + π π4 + π π5 ) π (π₯2π , π₯2π+1 ) 2 − (π2 + π3 + π π4 + π π5 ) = ππ (π₯2π , π₯2π+1 ) , (23) βͺ― π1 π (π₯2π , π₯2π+1 ) + π2 π (π₯2π , ππ₯2π ) where π = (2π1 +π2 +π3 +π π4 +π π5 )/(2−(π2 +π3 +π π4 +π π5 )) < 1/π . Similarly, it can be shown that + π3 π (π₯2π+1 , ππ₯2π+1 ) + π4 π (π₯2π+1 , ππ₯2π ) + π5 π (π₯2π , ππ₯2π+1 ) βͺ― π1 π (π₯2π , π₯2π+1 ) + π2 π (π₯2π , π₯2π+1 ) π (π₯2π+3 , π₯2π+2 ) βͺ― ππ (π₯2π+ , π₯2π+2 ) . (24) π (π₯π+1 , π₯π+2 ) βͺ― ππ (π₯π , π₯π+1 ) . (25) Therefore, + π3 π (π₯2π+1 , π₯2π+2 ) = (π1 + π2 + π π5 ) π (π₯2π , π₯2π+1 ) According to Lemma 11, we have {π₯π } is a Cauchy sequence in π. Since π is complete, there exists π₯∗ ∈ π such that π₯π → π₯∗ . Suppose that π is continuous. Then, π₯∗ = lim π₯π+1 = lim ππ π₯0 = lim π(ππ−1 π₯0 ) = π(lim ππ−1 π₯0 ) = π(lim π₯π ) = π(π₯∗ ). Therefore, π₯∗ is a fixed point of π. Now, we need to show that π₯∗ is a fixed point of π. Since π₯∗ β π₯∗ , we can use the inequality (18) for π₯ = π¦ = π₯∗ . Then we have + (π3 + π π5 ) π (π₯2π+1 , π₯2π+2 ) . π (ππ₯∗ , ππ₯∗ ) βͺ― π1 π (π₯∗ , π₯∗ ) + π2 π (π₯∗ , ππ₯∗ ) + π3 π (π₯∗ , ππ₯∗ ) + π4 π (π₯2π+1 , π₯2π+1 ) + π5 π (π₯2π , π₯2π+2 ) βͺ― (π1 + π2 ) π (π₯2π , π₯2π+1 ) + π3 π (π₯2π+1 , π₯2π+2 ) + π π5 (π (π₯2π , π₯2π+1 ) + π (π₯2π+1 , π₯2π+2 )) (19) + π4 π (π₯∗ , ππ₯∗ ) + π5 π (π₯∗ , ππ₯∗ ) = π1 π (π₯∗ , π₯∗ ) + π2 π (π₯∗ , π₯∗ ) + π3 π (π₯∗ , ππ₯∗ ) Hence, (1 − (π3 + π π5 )) π (π₯2π+1 , π₯2π+2 ) βͺ― (π1 + π2 + π π5 ) π (π₯2π , π₯2π+1 ) . + π4 π (π₯∗ , π₯∗ ) + π5 π (π₯∗ , ππ₯∗ ) (20) = π3 π (π₯∗ , ππ₯∗ ) + π5 π (π₯∗ , ππ₯∗ ) = (π3 + π5 ) π (π₯∗ , ππ₯∗ ) . On the other hand and by symmetry we have π (π₯2π+2 , π₯2π+1 ) = π (ππ₯2π+1 , ππ₯2π ) βͺ― π1 π (π₯2π+1 , π₯2π ) + π2 π (π₯2π+1 , ππ₯2π+1 ) + π3 π (π₯2π , ππ₯2π ) + π4 π (π₯2π , ππ₯2π+1 ) + π5 π (π₯2π+1 , ππ₯2π ) (26) Hence, π (π₯∗ , ππ₯∗ ) βͺ― (π3 + π5 ) π (π₯∗ , ππ₯∗ ) . (27) ∗ ∗ Since (π3 + π5 ) < 1, Lemma 10(1) shows that π(π₯ , ππ₯ ) = π; that is, π₯∗ = ππ₯∗ . Therefore π₯∗ is a fixed point of π. Therefore, π and π have a common fixed point. The proof is similar when π is a continuous mapping. 6 Abstract and Applied Analysis Theorem 16. Let (π, β) be a partially ordered set and suppose that there exists a cone b-metric π in π such that the cone bmetric space (π, π) is complete with the coefficient π ≥ 1 relative to a solid cone π. Let π, π : π → π be two weakly increasing mappings with respect to β. Suppose that the following three assertions hold: (i) there exist ππ , π = 1, . . . , 5 such that 2π π1 + (π + 1)(π2 + π3 ) + (π 2 + π )(π4 + π5 ) < 2 with ∑5π=1 ππ < 1, Since π₯π → π₯∗ and ππ₯π → π₯∗ , then by Definition 9(1) and for π β« π there exists π0 ∈ N such that for all π > π0 , π(π₯π , π₯∗ ) βͺ π(1 − (π2 + π4 ))/2(π1 + π3 + π4 ), and π(ππ₯π , π₯∗ ) βͺ π(1 − (π2 + π4 ))/2(π2 + π3 + π4 + π5 ). Then we have π (ππ₯π , ππ₯∗ ) = π (ππ₯∗ , ππ₯π ) βͺ― π (ππ₯, ππ¦) βͺ― π1 π (π₯, π¦) + π2 π (π₯, ππ₯) + + π3 π (π¦, ππ¦) + π4 π (π¦, ππ₯) + π5 π (π₯, ππ¦) , (28) βͺ for all comparative π₯, π¦ ∈ π; (ii) if an increasing sequence {π₯π } converges to π₯ in π, then π₯π β π₯ for all π. Then π and π have a common fixed point π₯∗ ∈ π. π (ππ₯∗ , ππ₯π ) βͺ― π1 π (π₯∗ , π₯π ) + π2 π (π₯∗ , ππ₯∗ ) + π3 π (π₯π , ππ₯π ) π2 + π3 + π4 + π5 π (ππ₯π , π₯∗ ) 1 − (π2 + π4 ) π1 + π3 + π4 π (1 − (π2 + π4 )) 1 − (π2 + π4 ) 2 (π1 + π3 + π4 ) + = Proof. As in Theorem 15, we can construct an increasing sequence {π₯π } and prove that there exists π₯∗ ∈ π such that π₯π → π₯∗ , also; by the construction of π₯π , ππ₯π → π₯∗ . Now, condition (iii) implies π₯π β π₯∗ for all π. Putting π₯ = π₯∗ and π¦ = π₯π in (28), we get π2 + π3 + π4 + π5 π (1 − (π2 + π4 )) 1 − (π2 + π4 ) 2 (π2 + π3 + π4 + π5 ) π π + 2 2 = π. (31) Now again, according to Definition 9(1) it follows that ππ₯π → ππ₯∗ . It follows that ππ₯∗ = π₯∗ . In a similar way and using that π₯∗ β π₯∗ , we can prove that ππ₯∗ = π₯∗ . Therefore, π and π have a common fixed point. Now, we present two examples to illustrate our results. In the first example (the case of a normal cone), the conditions of Theorem 12 are fulfilled, but Theorem 2 of Altun et al. [9, Theorem 12] cannot be applied. In the second example (the case of a nonnormal cone), the conditions of Theorem 12 are fulfilled, but Theorem 3 of Altun et al. [9, Theorem 13] cannot be applied. + π4 π (π₯π , ππ₯∗ ) + π5 π (π₯∗ , ππ₯π ) = π1 π (π₯π , π₯∗ ) + π2 π (ππ₯∗ , π₯∗ ) + π3 π (π₯π , ππ₯π ) + π4 π (π₯π , ππ₯∗ ) + π5 π (ππ₯π , π₯∗ ) βͺ― π1 π (π₯π , π₯∗ ) + π2 (π (ππ₯∗ , ππ₯π ) + π (ππ₯π , π₯∗ )) π1 + π3 + π4 π (π₯π , π₯∗ ) 1 − (π2 + π4 ) (29) + π3 (π (π₯π , π₯∗ ) + π (π₯∗ , ππ₯π )) + π4 (π (π₯π , π₯∗ ) + π (π₯∗ , ππ₯π ) Example 17. Let π = [0, 1] endowed with the standard order and πΈ = R2 and let π = {(π₯, π¦) : π₯, π¦ ≥ 0}. Define π : π × π → πΈ as in Example 8. Define π : π → π by π(π₯) = π₯2 /6. Then π is a continuous and nondecreasing mapping with respect to β. Then we have π (ππ₯, ππ¦) = π ( +π (ππ₯π , ππ₯∗ )) σ΅¨σ΅¨ 2 σ΅¨2 σ΅¨ σ΅¨2 π¦2 σ΅¨σ΅¨ σ΅¨σ΅¨ π₯2 π¦2 σ΅¨σ΅¨ σ΅¨π₯ = (σ΅¨σ΅¨σ΅¨σ΅¨ − σ΅¨σ΅¨σ΅¨σ΅¨ , σ΅¨σ΅¨σ΅¨σ΅¨ − σ΅¨σ΅¨σ΅¨σ΅¨ ) 6 σ΅¨σ΅¨ σ΅¨σ΅¨ 6 6 σ΅¨σ΅¨ σ΅¨σ΅¨ 6 + π5 π (ππ₯π , π₯∗ ) = (π1 + π3 + π4 ) π (π₯π , π₯∗ ) + (π2 + π3 + π4 + π5 ) π (ππ₯π , π₯∗ ) + (π2 + π4 ) π (ππ₯∗ , ππ₯π ) . = 1 σ΅¨σ΅¨ σ΅¨2 σ΅¨ σ΅¨2 σ΅¨2 σ΅¨ σ΅¨π₯ + π¦σ΅¨σ΅¨σ΅¨ (σ΅¨σ΅¨σ΅¨π₯ − π¦σ΅¨σ΅¨σ΅¨ , σ΅¨σ΅¨σ΅¨π₯ − π¦σ΅¨σ΅¨σ΅¨ ) 36 σ΅¨ βͺ― 4 σ΅¨σ΅¨ σ΅¨2 σ΅¨2 σ΅¨ (σ΅¨π₯ − π¦σ΅¨σ΅¨σ΅¨ , σ΅¨σ΅¨σ΅¨π₯ − π¦σ΅¨σ΅¨σ΅¨ ) 36 σ΅¨ βͺ― 4 π (π₯, π¦) , 36 Hence, π (ππ₯∗ , ππ₯π ) βͺ― π1 + π3 + π4 π (π₯π , π₯∗ ) 1 − (π2 + π4 ) + π2 + π3 + π4 + π5 π (ππ₯π , π₯∗ ) . 1 − (π2 + π4 ) π₯ 2 π¦2 , ) 6 6 (30) (32) where π1 = 4/36, π2 = π3 = π4 = π5 = 0. It is clear that the conditions of Theorem 12 are satisfied. Therefore, π has a fixed point π₯ = 0. Abstract and Applied Analysis 7 Example 18. Let π = [0, ∞) endowed with the standard order 1 [0, 1] with βπ’β = βπ’β∞ + βπ’σΈ β∞ , π’ ∈ πΈ and let and πΈ = πΆR π = {π’ ∈ πΈ : π’(π‘) ≥ 0 on [0, 1]}. It is well known that this cone is solid, but it is not normal. Define a cone metric π : π× π → πΈ by π(π₯, π¦)(π‘) = |π₯ − π¦|2 ππ‘ . Then (π, π) is a complete cone b-metric space with the coefficient π = 2. Let us define π : π → π by π(π₯) = π₯/2. Then π is a continuous and nondecreasing mapping with respect to β. Then we have π is an increasing mapping; also we have σ΅¨σ΅¨ 1 1 σ΅¨σ΅¨2 π (ππ₯, ππ¦) (π‘) = σ΅¨σ΅¨σ΅¨σ΅¨ π₯ − π¦σ΅¨σ΅¨σ΅¨σ΅¨ ππ‘ 2 σ΅¨ σ΅¨2 1σ΅¨ σ΅¨2 = σ΅¨σ΅¨σ΅¨π₯ − π¦σ΅¨σ΅¨σ΅¨ ππ‘ 4 1σ΅¨ σ΅¨2 βͺ― σ΅¨σ΅¨σ΅¨π₯ − π¦σ΅¨σ΅¨σ΅¨ ππ‘ + 4 1 σ΅¨σ΅¨σ΅¨ π₯ σ΅¨σ΅¨σ΅¨2 π‘ σ΅¨σ΅¨ σ΅¨σ΅¨ π 5 σ΅¨σ΅¨ 2 σ΅¨σ΅¨ (33) 1 1 βͺ― π (π₯, π¦) (π‘) + π (ππ₯, π₯) (π‘) , 4 5 where π1 = 1/4, π2 = 1/5, π3 = π4 = π5 = 0. It is clear that the conditions of Theorem 12 are satisfied. 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