Bulletin of Mathematical Analysis and Applications ISSN: 1821-1291, URL: http://www.bmathaa.org Volume 5 Issue 1 (2013), Pages 22-35 FIXED POINT THEOREMS IN π -POLYGONAL CONE METRIC SPACES (COMMUNICATED BY MARK AGRANOVSKY) IVAN GATICA ARAUS, JOSEFA LORENZO RAMIREZ Abstract. In this paper we present a ο¬xed point theorem for mappings deο¬ned in an π -polygonal cone metric space. Some generalizations of ο¬xed point theorems for S-Kannan and S-Chatterjea contractive mappings on cone metric spaces will be also shown. In these results the underlying cone metric space is considered over a Banach space ordered by a normal cone. 1. Introduction and Preliminaries After the extension of Banach Contraction Principle for contractions on vectorvalued metric spaces by Perov in 1964 ([21]), the notion of cone metric space over an ordered Banach space (also called K-metric space) has been used as a natural framework for proving useful ο¬xed point theorems in the theory of diο¬erential equations (see [1], [7], [9], [19], [20] and [22]). For more information on this topic we refer the reader to [23] and the bibliography therein. Recently, from a paper by Long-Guang and Xian [15] many authors have taken up the question of the existence of ο¬xed points for mappings satisfying some contractive conditions in cone metric spaces (see [2], [8], [11], [16] and [18]). Motivated by two recent papers on rectangular cone metric space ([2] and [16]), we unify various results deal with generalizations of the Banach’s Contraction Principle. On the other hand, we present a ο¬xed point result for a S-Hardy-Rogers cone contraction following some ideas from [12]. Among other consequences we obtain an extension of the well known Kannan ο¬xed point theorem appeared in [8]. It is worth pointing out that in the above-mention papers it is usually required the cone to be normal with nonempty interior. Last condition on the cone is very restrictive, for example, the positive cone of the space of functions πΏ1 and the positive cone of the space of sequences βπ (1 ≤ π < ∞) have empty interior. We emphasize that in our approach just a normality assumption is imposed on the cone. Furthermore, we deal with a class of contractive mappings for which the constant of contractiveness is replaced by a positive operator in the spirit of the papers [9] and [23]. 02000 Mathematics Subject Classiο¬cation: 47H10, 47H09. Keywords and phrases. Fixed point, Contraction-type mapping, cone metric space. c 2013 Universiteti i PrishtineΜs, PrishtineΜ, KosoveΜ. β Submitted October 25, 2012. Published January 3, 2013. 22 CONE METRIC SPACE 23 Let (π, β£β£ ⋅ β£β£) be a Banach space. A set πΎ ⊂ π is called a cone if and only if: (1) πΎ is nonempty and πΎ β= {0π }. (2) If πΌ, π½ ∈ πΎ and π, π ∈ β+ 0 , then ππΌ + ππ½ ∈ πΎ. (3) πΎ ∩ (−πΎ) = {0π }. For a given cone πΎ ⊂ π , we can deο¬ne a partial ordering ≤ with respect to πΎ by πΌ ≤ π½ if and only if π½ − πΌ ∈ πΎ. We will refer (π, β£β£ ⋅ β£β£, πΎ) as an ordered Banach space. The following deο¬nitions relate the norm β£β£ ⋅ β£β£ with the cone πΎ: β The cone πΎ is called normal if there exists a number π ≥ 1 such that for all πΌ, π½ ∈ π , 0π ≤ πΌ ≤ π½ implies β£β£πΌβ£β£ ≤ πβ£β£π½β£β£. The least positive number satisfying above is called the normal constant of πΎ. β The cone πΎ is called closed if πΎ is closed with respect to the topology induced by the norm. Deο¬nition 1.1. Let π be a set and ππΎ : π × π :→ πΎ a mapping. We say that ππΎ is a cone metric, if for all π₯, π¦, π§ ∈ π, one has (1) ππΎ (π₯, π¦) = 0π ⇔ π₯ = π¦; (2) ππΎ (π₯, π¦) = ππΎ (π¦, π₯); (3) ππΎ (π₯, π¦) ≤ ππΎ (π₯, π§) + ππΎ (π§, π¦). The pair (π, ππΎ ) is said to be a cone metric space (CMS). Example 1.2. Let π = π = πΏπ [0, 1] such that πΎπ = {π ∈ πΏπ [0, 1] : 0 ≤ π e.c.t}. We deο¬ne ππΎ : πΏπ [0, 1] × πΏπ [0, 1] → πΎπ such that ππΎ (π, π) = β£π − πβ£. It is clear that ππΎ is a cone metric. The proof of the following lemma easily follows from the deο¬nition of cone metric. Lemma 1.3. Let (π, ππΎ ) be a cone metric space over an ordered Banach space (π, β£β£ ⋅ β£β£, πΎ) such that πΎ is a normal cone with normal constant π. Then the function π· : π ×π → [0, ∞) deο¬ned by π·(π₯, π¦) = β£β£ππΎ (π₯, π¦)β£β£ satisο¬es the following properties: (1) π·(π₯, π¦) = 0 ⇔ π₯ = π¦; (2) π·(π₯, π¦) = π·(π¦, π₯); (3) π·(π₯, π¦) ≤ π[π·(π₯, π§) + π·(π§, π¦)] for all π₯, π¦, π§ ∈ π. Recently, M.A. Khamsi [17] introduced the concept of a metric type space: for an arbitrary set π, the pair (π, π·) is called a metric type space if π· : π × π → [0, ∞) is a function satisfying properties (1), (2) and (3) in the above lemma. Deο¬ning a topology on this class of spaces, he obtained some metric and topological ο¬xed point theorems (see [18]). However, it is worth pointing out that this deο¬nition of π· corresponds to the concept of quasi-metric and the pair (π, π·) is called a quasimetric space in the literature. If (π, π·) is a quasi-metric space then the topology in π induced by π·, is canonically deο¬ned by means of the theory of uniform structures. The balls π΅(π₯, π) = {π¦ ∈ π : π·(π₯, π¦) < π} for π > 0 form a basis of neighbourhoods of π₯ for the topology induced by the uniformity of π. This is a metric topology since the uniform structure associated to π· has a numerable basis (we refer the reader to Chapter 8 in [10]). Therefore, a sequence {π₯π }π∈β converges to π₯ in π, 24 I. GATICA ARAUS, J. LORENZO RAMIREZ if lim π·(π₯π , π₯) = 0 and π is complete if every Cauchy sequence is convergent in π. π This fact suggests that many ο¬xed point results proved in cone metric spaces do not need full structure because they could follow by using the real valued function π·. However, this approach does not work in order to prove certain ο¬xed point theorems [8]. Example 1.4. Let (Ω, Σ, π) be a positive measure space. For every 1 < π < ∞ consider the Banach space π = πΏπ (Ω). Deο¬ne π· : π × π → [0, +∞) by ∫ π·(π, π) = β£π − πβ£π ππ, π, π ∈ π. Ω Then π· trivially satisο¬es properties (1), (2) in the above Lemma. Now take π, π, β ∈ π. Since π > 1, it is well known that (∫ ) ∫ ∫ π π−1 π π β£π − πβ£ ππ ≤ 2 β£π − ββ£ ππ + β£β − πβ£ ππ , Ω Ω Ω which means π·(π, π) ≤ 2π−1 [π·(π, β) + π·(β, π)]. Therefore π· is a quasi-metric on π. Concerning example 1.4 and the concept of ”quasi-metric space” (which is called in some paper b-metric space), it is worth to notice that several examples of quasimetrics (including example 1.4) and some ο¬xed point results in this framework are given in [3], [4], [5], [6], and the references therein. Following some ideas from [2] next we introduce a class of spaces including cone metric spaces. Deο¬nition 1.5. Let π be a set and ππΎ : π × π :→ πΎ a mapping. We say that ππΎ is a π -polygonal cone metric, if for all π₯, π¦ ∈ π and for all distinct points π§1 , π§2 , . . . , π§π ∈ π, each of them diο¬erent from π₯ and π¦, one has (1) ππΎ (π₯, π¦) = 0π ⇔ π₯ = π¦; (2) ππΎ (π₯, π¦) = ππΎ (π¦, π₯); (3) ππΎ (π₯, π¦) ≤ ππΎ (π₯, π§1 ) + ππΎ (π§1 , π§2 ) + ⋅ ⋅ ⋅ + ππΎ (π§π −1 , π§π ) + ππΎ (π§π , π¦). The pair (π, ππΎ ) is said to be a π -polygonal cone metric space, ππ -CMS for short. If π = 1 the pair (π, ππΎ ) is a cone metric space. The concept of 2-polygonal cone metric space is referred in [2] as cone rectangular metric space. It is clear that a cone metric space is a π -polygonal cone metric space, for all π ≥ 2. √ Example 1.6. Let π = β, (π, β£β£ ⋅ β£β£) = (β2 , β£β£ ⋅ β£β£) with β£β£πΌβ£β£ = πΌ12 + πΌ22 and πΎ = {(πΌ1 , πΌ2 ) : πΌ1 , πΌ2 ≥ 0}. We deο¬ne ππΎ : π × π → πΎ as follow: β§ π₯ = π¦, β¨ (0, 0) : if π₯ ∈ {1, 2} ∧ π¦ ∈ {1, 2} ∧ π₯ β= π¦, ππΎ (π₯, π¦) = (3π, 3) : if β© (π, 1) : if (π₯ ∈ {1, 2}π ∨ π¦ ∈ {1, 2}π ) ∧ π₯ β= π¦. with π ∈ (0, ∞). Then ππΎ is a rectangular cone metric but it is not a cone metric because it lacks the triangular property: ππΎ (1, 2) = (3π, 3) > ππΎ (1, 3) + ππΎ (3, 2) = (2π, 2). CONE METRIC SPACE 25 From the deο¬nition of N-polygonal cone metric we can easily deduce the following lemma. Lemma 1.7. Let (π, ππΎ ) be a ππ -CMS over an ordered Banach space (π, β£β£ ⋅ β£β£, πΎ) such that πΎ is a normal cone with normal constant π. Then the function π· : π × π → [0, ∞) deο¬ned by π·(π₯, π¦) = β£β£ππΎ (π₯, π¦)β£β£ satisο¬es the following properties: (1) π·(π₯, π¦) = 0 ⇔ π₯ = π¦; (2) π·(π₯, π¦) = π·(π¦, π₯); (3) π·(π₯, π¦) ≤ π[π·(π₯, π§1 ) + π·(π§1 , π§2 ) + ⋅ ⋅ ⋅ + π·(π§π −1 , π§π ) + π·(π§π , π¦)] for all distinct points π§1 , π§2 , . . . , π§π ∈ π, each of them diο¬erent from π₯ and π¦. Given a N-polygonal cone metric space (π, ππΎ ), the concept of sequence will be the usual. We next deο¬ne convergence and completeness. Deο¬nition 1.8. Let (π, ππΎ ) be a ππ -CMS over an ordered Banach space (π, β£β£ ⋅ β£β£, πΎ) such that πΎ is a normal cone, π₯ ∈ π and {π₯π }π∈β ⊂ π. We say that {π₯π }π∈β converges to π₯ in π, if lim π·(π₯π , π₯) = 0. We write π₯π → π₯ to denote that π the sequence {π₯π }π∈β is convergent to π₯, that is, π₯π → π₯ ⇔ π·(π₯π , π₯) → 0. In general, for a N-polygonal cone metric space with π ≥ 2, the uniqueness of the limit of a sequence does not hold (see Example 1.6 in [16]). However, the limit is unique for a convergent Cauchy sequence as we show below. Deο¬nition 1.9. Let (π, ππΎ ) be a ππ -CMS over an ordered Banach space (π, β£β£ ⋅ β£β£, πΎ) such that πΎ is a normal cone and {π₯π }π∈β a sequence in π. We say that {π₯π }π∈β is a Cauchy sequence, if for every positive real number π, there exists ˆ ∈ β such that for all π, π ≥ π ˆ , we have that π·(π₯π , π₯π ) < π. π Next Lemma is a extension of Lemma 1.10 presented in [16]. Lemma 1.10. Let (π, ππΎ ) be a ππ -CMS over an ordered Banach space (π, β£β£⋅β£β£, πΎ) such that πΎ is a normal cone with normal constant π and {π₯π }π∈β be a sequence in π. If {π₯π }π∈β is an Cauchy sequence, such that satisο¬es the following conditions: (a): (∃π₯, π¦ ∈ π)(π₯π → π₯ ∧ π₯π → π¦), ˆ ∈ β)(π, π ≥ π ˆ ⇒ π₯π β= π₯π ∧ π₯π β= π₯ ∧ π₯π β= π¦), (b): (∃π then π₯ = π¦. Proof. Suppose {π₯π }π∈β is an Cauchy sequence satisfying conditions (a) and (b). ˆ we have For all π ≥ π π·(π₯, π¦) ≤ π[π·(π₯, π₯π ) + π·(π₯π+1 , π₯π+2 ) + ⋅ ⋅ ⋅ + π·(π₯π+π −1 , π¦)]. Taking limit as π → ∞, we obtain π·(π₯, π¦) = 0 and hence π₯ = π¦. β‘ Deο¬nition 1.11. Let (π, ππΎ ) be a ππ -CMS over an ordered Banach space (π, β£β£ ⋅ β£β£, πΎ) such that πΎ is a normal cone. Then π is called a complete N-polygonal cone metric space, if every Cauchy sequence is convergent in π. Finally, we recall some deο¬nitions and facts about positive operators. Deο¬nition 1.12. Let (π, πΎ) be an ordered vector space and π : π → π an operator. We say that (1) π is positive if π(πΎ) ⊆ πΎ. 26 I. GATICA ARAUS, J. LORENZO RAMIREZ (2) π is monotone non-decreasing if for all πΌ, π½ ∈ π such that π½ − πΌ ∈ πΎ, then π(π½) − π(πΌ) ∈ πΎ. It is clear that every linear positive operator is monotone non-decreasing. Lemma 1.13. Let (π, β£β£⋅β£β£, πΎ) be an ordered Banach space such that πΎ is a normal cone with normal constant π and π : π → π be a linear positive operator. If β£β£ππ (πΌ)β£β£ → 0 for πΌ ∈ πΎ − {0π }, then π(πΌ) − πΌ ∈ / πΎ − {0π }. Proof. Consider πΌ ∈ πΎ−{0π } such that β£β£ππ (πΌ)β£β£ → 0. Suppose that π(πΌ)−πΌ ∈ πΎ, that is πΌ ≤ π(πΌ). Since π is a linear and positive operator we have that πΌ ≤ ππ (πΌ) for all π ∈ β. Hence β£β£πΌβ£β£ ≤ πβ£β£ππ (πΌ)β£β£ for all π ∈ β. Taking limit as π → ∞ we have β£β£πΌβ£β£ = 0, and so πΌ = 0π which is a contradiction. β‘ In the sequel we shall use the following notation: β π΅ + (π ) = {π : π → π /π is a positive bounded linear operator }. β β£β£πβ£β£π = sup{β£β£π(πΌ)β£β£ : πΌ ∈ π ; β£β£πΌβ£β£ ≤ 1}. 2. Fixed Point Theorems for generalized contraction In this section we present an adaptation from classical Banach Contraction Principle for generalized contraction on π -polygonal cone metric spaces. Theorem 2.1. Let (π, ππΎ ) be a complete ππ -CMS over an ordered Banach space (π, β£β£ ⋅ β£β£, πΎ) such that πΎ is a normal cone with normal constant π and π : π → π be a mapping. If for all π₯, π¦ ∈ π we have that (∗) ∑∞ ππΎ (π π₯, π π¦) ≤ π(ππΎ (π₯, π¦)), where π ∈ π΅ (π ) and π=0 β£β£ππ β£β£π < ∞, then π has a unique ο¬xed point π₯∗ in π. Moreover, the iterative sequence π₯π = π π π₯0 converges to π₯∗ for any initial point π₯0 ∈ π and one has the following estimation: + ππΎ (π₯π , π₯∗ ) ≤ ππ (πΌ − π)−1 (ππΎ (π₯0 , π π₯0 )). Proof. Take any point π₯0 ∈ π such that ππΎ (π₯0 , π π₯0 ) ∈ πΎ − {0π } and consider the 2 π sequence {π₯π }∞ π=0 ⊂ π given by π₯1 = π π₯0 , π₯2 = π π₯0 ,..., π₯π = π π₯0 . From the assumption (∗), we have that ππΎ (π₯1 , π₯2 ) = ππΎ (π π₯0 , π 2 π₯0 ) ≤ π(ππΎ (π₯0 , π₯1 )), ππΎ (π₯2 , π₯3 ) = ππΎ (π 2 π₯0 , π 3 π₯0 ) ≤ π(ππΎ (π₯1 , π₯2 )), and ππΎ (π₯π , π₯π+1 ) = ππΎ (π π π₯0 , π π+1 π₯0 ) ≤ π(ππΎ (π₯π−1 , π₯π )), for all π ≥ 1. Since π is monotone non-decreasing it follows that, for any π ∈ β, ππΎ (π₯π , π π₯π ) ≤ ππ (ππΎ (π₯0 , π π₯0 )). Now, we will show that π₯π β= π₯0 for all π ∈ β. Suppose that there exists π ˆ∈β such that π₯πˆ = π₯0 , then ππΎ (π₯0 , π π₯0 ) = ππΎ (π₯πˆ , π π₯πˆ ) ≤ ππˆ (ππΎ (π₯0 , π π₯0 )). π ˆ Therefore, hand, ∑∞ π = πππΎ (π₯0 , π π₯0 ) ∈ πΎ − {0π } and π (π) − π ∈ πΎ. On the other since π=0 β£β£π β£β£π < ∞, we can apply Lemma 1.13 to the operator ππˆ to get a contradiction. We proceed to prove that π has a ο¬xed point by dividing the proof into two cases. CONE METRIC SPACE 27 (Case a): Suppose that π₯π = π₯π for some π, π ∈ β such that 1 ≤ π < π, that is π π π₯0 = π π π₯0 . If we deο¬ne π¦0 := π π π₯0 we have that π¦0 = π π π¦0 where π = π − π > 1. Therefore, we obtain ππΎ (π¦0 , π π¦0 ) = ππΎ (π π π¦0 , π π +1 π¦0 ) ≤ ππ (ππΎ (π¦0 , π π¦0 )). Letting π → ∞, in view of Lemma 1.13 we must have ππΎ (π¦0 , π π¦0 ) = 0π and π¦0 is a ο¬xed point of π . (Case b): Assume that π₯π β= π₯π for all π, π ∈ β such that π β= π. Using π -polygonal property (3) and by deο¬nition of π we have that, ππΎ (π₯0 , π π π₯0 ) ≤ ππΎ (π₯0 , π π₯0 ) + ππΎ (π π₯0 , π 2 π₯0 ) + ⋅ ⋅ ⋅ + ππΎ (π π −2 π₯0 , π π −1 π₯0 ) +ππΎ (π π −1 π₯0 , π π +1 π₯0 ) + ππΎ (π π +1 π₯0 , π π π₯0 ) ≤ (πΌ + π + ⋅ ⋅ ⋅ + ππ −2 )(ππΎ (π₯0 , π π₯0 )) + ππ −1 (ππΎ (π₯0 , π 2 π₯0 )) +ππ (ππΎ (π₯0 , π π₯0 )). Hence π ππΎ (π₯0 , π₯π ) ≤ (πΌ + π + ⋅ ⋅ ⋅ + π ) (π ∑ ) ππΎ (π₯0 , π₯π ) . π=1 Similarly we obtain ππΎ (π₯0 , π₯π +1 ) ≤ (πΌ + π + ⋅ ⋅ ⋅ + ππ )(π (πΎπ(π₯0 , π₯1 )) ) ∑ ≤ (πΌ + π + ⋅ ⋅ ⋅ + ππ ) ππΎ (π₯0 , π₯π ) , π=1 and ππΎ (π₯0 , π₯π +π ) ≤ (πΌ + π + ⋅ ⋅ ⋅ + ππ −1()(ππΎ (π₯0 , π₯1 )) +)ππ (ππΎ (π₯0 , π₯π )) π ∑ ≤ (πΌ + π + ⋅ ⋅ ⋅ + ππ ) ππΎ (π₯0 , π₯π ) , π=1 for any π ∈ {2, . . . , π }. In the same way, for π = 1, ⋅ ⋅ ⋅ , π we have that 2π ππΎ (π₯0 , π₯2π +π ) ≤ (πΌ + π + ⋅ ⋅ ⋅ + π2π −1()ππΎ (π₯0 , π₯1 ) + π (π(π₯0 , π₯π )) ) π ∑ ≤ (πΌ + π + ⋅ ⋅ ⋅ + π2π ) ππΎ (π₯0 , π₯π ) . π=1 Continuing this process, we get for each π ∈ β and π ∈ {1, . . . , π } ππ ππΎ (π₯0 , π₯ππ +π )≤ (πΌ + π + ⋅ ⋅ ⋅ + π ) (π ∑ ) ππΎ (π₯0 , π₯π ) π=1 . It is clear that for any π ∈ β there exist π ∈ β ∪ {0} and π ∈ {1, 2⋅, ⋅, ⋅, π } such that π = ππ + π. From the above we obtain ππ ππΎ (π₯0 , π₯π )≤ (πΌ + π + ⋅ ⋅ ⋅ + π ) (π ∑ π=1 ) ππΎ (π₯0 , π₯π ) . 28 I. GATICA ARAUS, J. LORENZO RAMIREZ In particular π ππΎ (π₯0 , π₯π )≤ (πΌ + π + ⋅ ⋅ ⋅ + π ) (π ∑ ) ππΎ (π₯0 , π₯π ) . π=1 Let π, π ∈ β. From above we have (∗∗) ππΎ (π₯π , π₯π+π ) = ππΎ (π π π₯0 , π π+π π₯0 ) π ≤ ππ (π π₯0()) πΎ (π₯0 , π ) (∑ ) ∑π π π π ≤π π=0 π π=1 ππΎ (π₯0 , π₯π ) . Now, we apply the normality of the cone to get β β( ) ∞ π ∑ ∑ π π β β β£β£ππΎ (π₯π , π₯π+π )β£β£ = π·(π₯π , π₯π+π ) ≤ πβ£β£π β£β£π β£β£π β£β£π π·(π₯0 , π₯π ) . π=0 π=1 Letting π → ∞ we conclude that {π₯π }π∈β is a Cauchy sequence. Since π is complete, there exists π₯∗ ∈ π such that π π π₯0 → π₯∗ . We shall now show that π π₯∗ = π₯∗ . Without any loss of generality, we can assume π π π₯0 β= π₯∗ and π π π₯0 β= π π₯∗ for all π ∈ β. Therefore, since π π π₯0 β= π π π₯0 for all π, π ∈ β such that π β= π, we obtain ππΎ (π₯∗ , π π₯∗ ) ≤ ππΎ (π₯∗ , π₯π+1 ) + ππΎ (π₯π+1 , π₯π+2 ) + ⋅ ⋅ ⋅ + ππΎ (π₯π+π −1 , π₯π+π ) +ππΎ (π₯π+π , π π₯∗ ) ≤ (ππ+1 + ππ+2 + ⋅ ⋅ ⋅ + ππ+π −1 )(ππΎ (π₯0 , π₯1 ))+ ππΎ (π₯∗ , π₯π+1 ) + π(ππΎ (π₯π+π −1 , π₯∗ )) = (ππ β π )(ππΎ (π₯0 , π₯1 )) + ππΎ (π₯∗ , π₯π+1 ) + π(ππΎ (π₯π+π −1 , π₯∗ )), where π = (π + π2 + ⋅ ⋅ ⋅ + ππ −1 ). Thus, π·(π₯∗ , π π₯∗ ) ≤ πβ£β£(ππ β π )(ππΎ (π₯0 , π₯1 )) + ππΎ (π₯∗ , π₯π+1 ) + π(ππΎ (π₯π+π −1 , π₯∗ ))β£β£ ≤ π(β£β£ππ β£β£π β£β£π β£β£π π·(π₯0 , π₯1 ) + π·(π₯∗ , π₯π+1 ) + β£β£πβ£β£π π·(π₯π+π −1 , π₯∗ )). It is clear that π·(π₯∗ , π₯π ) → 0 and β£β£ππ β£β£π → 0, therefore π·(π₯∗ , π π₯∗ ) = 0. Hence π π₯∗ = π₯∗ . In addition, letting π → ∞ in (∗∗) we get β( β ) ∞ π ∑ ∑ ∗ πβ πβ ππΎ (π₯π , π₯ ) ≤ π π ππΎ (π₯0 , π₯π ) = ππ (πΌ − π)−1 (ππΎ (π₯0 , π π₯0 )). π=0 π=1 Uniqueness: Suppose that π₯∗ is not a unique ο¬xed point of π . That is, there exists π¦ ∗ ∈ π such that π₯∗ β= π¦ ∗ and π π¦ ∗ = π¦ ∗ . Since π₯∗ β= π¦ ∗ then ππΎ (π₯∗ , π¦ ∗ ) ∈ πΎ − {0π } and ππΎ (π₯∗ , π¦ ∗ ) = ππΎ (π π₯∗ , π π¦ ∗ ) ≤ π(ππΎ (π₯∗ , π¦ ∗ )). Therefore there exists π = ππΎ (π₯∗ , π¦ ∗ ) ∈ πΎ − {0π } such that π(π) − π ∈ πΎ which is a contradiction by Lemma 1.13. We conclude that π₯∗ = π¦ ∗ . β‘ CONE METRIC SPACE 29 It is easy to prove that lim (β£β£ππ β£β£π )1/π = inf{(β£β£ππ β£β£π )1/π : π = 1, 2, ...}. Thus π→∞ ∑∞ the series π=0 β£β£ππ β£β£π converges if and only if lim (β£β£ππ β£β£π )1/π < 1. As a conseπ→∞ quence we can formulate the following corollaries. Corollary 2.2. Let (π, ππΎ ) be a complete ππ -CMS over an ordered Banach space (π, β£β£ ⋅ β£β£, πΎ) such that πΎ is a normal cone with normal constant π and π : π → π be a mapping. If for all π₯, π¦ ∈ π we have that ππΎ (π π₯, π π¦) ≤ π(ππΎ (π₯, π¦)), where π ∈ π΅ (π ) and lim (β£β£ππ β£β£π )1/π < 1, then π has a unique ο¬xed point in π. + π→∞ Corollary 2.3. Let (π, ππΎ ) be a complete ππ -CMS over an ordered Banach space (π, β£β£ ⋅ β£β£, πΎ) such that πΎ is a normal cone with normal constant π and π : π → π be a mapping. If for all π₯, π¦ ∈ π we have that ππΎ (π π₯, π π¦) ≤ π(ππΎ (π₯, π¦)), + where π ∈ π΅ (π ) and β£β£πβ£β£π < 1, then π has a unique ο¬xed point in π. Since a rectangular metric space is a 2-polygonal cone metric space, we obtain the following consequence. Corollary 2.4. ([2]) Let (π, ππΎ ) be a complete rectangular cone metric space over an ordered Banach space (π, β£β£ ⋅ β£β£, πΎ) such that πΎ is a normal cone with normal constant π ≥ 1 and π : π → π be a mapping. If for all π₯, π¦ ∈ π we have that ππΎ (π π₯, π π¦) ≤ π ⋅ ππΎ (π₯, π¦), where π ∈ [0, 1[ is a real constant. Then π has a unique ο¬xed point in π. 3. Fixed point theorems of S-Kannan and S-Chatterjea cone contractive mappings We begin this section with a more general deο¬nition of π-Hardy-Rogers contraction than that from [12], in order to obtain some news ο¬xed point results on cone metric spaces. Deο¬nition 3.1. Let (π, ππΎ ) be a CMS over an ordered Banach space (π, β£β£ ⋅ β£β£, πΎ) and π, π : π → π two mappings. We say that π ∑ is a S-Hardy-Rogers cone 5 contraction, if there exists {ππ }5π=1 ⊂ π΅ + (π ) with π=1 β£β£ππ β£β£π < 1 such that for all π₯, π¦ ∈ π we have ππΎ (ππ π₯, ππ π¦) ≤ π1 (ππΎ (ππ₯, ππ¦)) + π2 (ππΎ (ππ₯, ππ π₯)) + π3 (ππΎ (ππ¦, ππ π¦)) +π4 (ππΎ (ππ₯, ππ π¦)) + π5 (ππΎ (ππ¦, ππ π₯)). In particular, if (1) π2 ≡ π3 ≡ π4 ≡ π5 ≡ Θ, then π is a S-Banach cone contraction; (2) π4 ≡ π5 ≡ Θ, then π is a S-Reich cone contraction; (3) π1 ≡ π4 ≡ π5 ≡ Θ, then π is S-Kannan cone contraction; (4) π1 ≡ π2 ≡ π3 ≡ Θ, then π is S-Chatterjea cone contraction, where Θ is the null operator. If we put in the previous deο¬nition π = πΌπ, (π, ππΎ ) a metric space, and ππ (πΌ) = ∑5 ππ πΌ where ππ > 0 such that π=1 ππ < 1, we have that π is a Hardy-Rogers contraction (see [14]). 30 I. GATICA ARAUS, J. LORENZO RAMIREZ Example 3.2. Let π = πΆ([0, 1]), πΎ = {π ∈ π : π ≥ 0}, π = β and ππΎ : π × π → πΎ deο¬ned by ππΎ (π₯, π¦) = β£π₯ − π¦β£ππ‘ , where ππ‘ ∈ π . Then (π, ππΎ ) is a cone 1 metric space. We consider the functions π, π : π → π deο¬ned by ππ₯ = 2π₯ +2 and π π₯ = 3π₯. Obviously, T is not a cone contraction but it is a S-Banach cone contraction from the following: ππΎ (ππ π₯, ππ π¦) = β£ππ π₯ − ππ π¦β£ππ‘ = β£ 1 1 1 1 − β£ππ‘ = β£ππ₯ − ππ¦β£ππ‘ = ππΎ (ππ₯, ππ¦). 6π₯ 6π¦ 3 3 Example 3.3. Let π = πΆ([0, 1]), πΎ = {π ∈ π : π ≥ 0}, π = β and ππΎ : π ×π → πΎ deο¬ned by ππΎ (π₯, π¦) = β£π₯ − π¦β£ππ‘ , where ππ‘ ∈ π . Then (π, ππΎ ) is a cone metric space. We consider the functions π, π : π → π deο¬ned by ππ₯ = π₯2 and π π₯ = π₯2 . Then 2 π₯ π¦ 2 π‘ π‘ ππΎ (ππ π₯, ππ π¦) = β£ππ π₯ − ππ π¦β£π = − π 4 4 ≤ 1 [β£ππ₯ − ππ π₯β£ + β£ππ¦ − ππ π¦β£] ππ‘ 3 1 [ππΎ (ππ₯, ππ π₯) + ππΎ (ππ¦, ππ π¦)]. 3 Therefore T is a S-Kannan cone contraction but, it is easy to see that it is not a Kannan contraction. Moreover, it is not diο¬cult to show that T is besides a S-Chatterjea cone contraction. ≤ Deο¬nition 3.4. Let (π, ππΎ ) be a CMS over an ordered Banach space (π, β£β£ ⋅ β£β£, πΎ) and π : π → π a mapping. We say that: (1) π is a sequentially continuous mapping, if for all sequence {π₯π }π∈β in π, such that π₯π → π₯∗ then ππ₯π → ππ₯∗ ; (2) π is a sequentially convergent mapping, if for all sequence {π₯π }π∈β in π, such that ππ₯π → π₯∗ then there exists π¦ ∗ ∈ π such that π₯π → π¦ ∗ . Our main result requires the following lemma. Lemma 3.5. Let (π, ππΎ ) be a CMS over an ordered Banach space (π, β£β£⋅β£β£, πΎ) such that πΎ is a normal cone, {π₯π }π∈β a sequence in π and π₯∗ ∈ π. If ππΎ (π₯π , π₯∗ ) → 0, then for all π¦ ∈ π ππΎ (π₯π , π¦) → ππΎ (π₯∗ , π¦). Proof. Suppose that ππΎ (π₯π , π₯∗ ) → 0 and let π¦ ∈ π. By triangular inequality for ππΎ we have −ππΎ (π₯π , π₯∗ ) ≤ ππΎ (π₯π , π¦) − ππΎ (π₯∗ , π¦) ≤ ππΎ (π₯π , π₯∗ ), ∀π ∈ β. Since πΎ is a normal cone, we may apply Theorem 1.1.1 of [13] to obtain β£β£ππΎ (π₯π , π¦) − ππΎ (π₯∗ , π¦)β£β£ → 0, that is, ππΎ (π₯π , π¦) → ππΎ (π₯∗ , π¦). β‘ Theorem 3.6. Let (π, ππΎ ) be a complete CMS over an ordered Banach space (π, β£β£⋅β£β£, πΎ) such that πΎ is a normal cone with normal constant π and π, π : π → π two mappings, such that π is a S-Hardy-Rogers cone contraction. Then (1) For all π₯0 ∈ π, {ππ π π₯0 }π∈β is a Cauchy sequence. (2) There exists π₯∗ ∈ π such that ππ π π₯0 → π₯∗ . CONE METRIC SPACE 31 (3) If π is one to one, sequentially continuous and sequentially convergent, then there exists a unique π¦ ∗ such that π π¦ ∗ = π¦ ∗ and for every π₯0 ∈ π, the sequence {π π π₯0 }π∈β converges to π¦ ∗ . Proof. Let π₯0 ∈ π. We consider the sequence {π₯π }π∈β ⊂ π such that π₯1 = π π₯0 , π₯2 = π 2 π₯0 ,..., π₯π = π π π₯0 and ππΎ (π₯0 , π π₯0 ) ∈ πΎ −{0π }. It is clear that π₯π+1 = π π₯π for all π ∈ β. Since π is a S-Hardy-Rogers cone contraction, we have the following ππΎ (ππ₯1 , ππ₯2 ) = ππΎ (ππ π₯0 , ππ π₯1 ) ≤ π1 (ππΎ (ππ₯0 , ππ₯1 )) + π2 (ππΎ (ππ₯0 , ππ π₯0 )) + π3 (ππΎ (ππ₯1 , ππ π₯1 )) +π4 (ππΎ (ππ₯0 , ππ π₯1 )) + π5 (ππΎ (ππ₯1 , ππ π₯0 )) ≤ π1 (ππΎ (ππ₯0 , ππ₯1 )) + π2 (ππΎ (ππ₯0 , ππ₯1 )) + π3 (ππΎ (ππ₯1 , ππ₯2 )) +π4 (ππΎ (ππ₯0 , ππ₯1 )) + π4 (ππΎ (ππ₯1 , ππ₯2 )), because π4 ∈ π΅ + (π ) and π5 (ππΎ (ππ₯1 , ππ₯1 )) = 0π . Therefore [πΌ − π3 − π4 ](ππΎ (ππ₯1 , ππ₯2 )) ≤ [π1 + π2 + π4 ](ππΎ (ππ₯0 , ππ₯1 )). Also ππΎ (ππ₯2 , ππ₯1 ) = ππΎ (ππ π₯1 , ππ π₯0 ) ≤ π1 (ππΎ (ππ₯1 , ππ₯0 )) + π2 (ππΎ (ππ₯1 , ππ π₯1 )) + π3 (ππΎ (ππ₯0 , ππ π₯0 )) +π4 (ππΎ (ππ₯1 , ππ π₯0 )) + π5 (ππΎ (ππ₯0 , ππ π₯1 )) ≤ π1 (ππΎ (ππ₯1 , ππ₯0 )) + π2 (ππΎ (ππ₯1 , ππ₯2 )) + π3 (ππΎ (ππ₯0 , ππ₯1 )) +π5 (ππΎ (ππ₯0 , ππ₯1 )) + π5 (ππΎ (ππ₯1 , ππ₯2 )), that is [πΌ − π2 − π5 ](ππΎ (ππ₯2 , ππ₯1 )) ≤ [π1 + π3 + π5 ](ππΎ (ππ₯1 , ππ₯0 )). Thus [2πΌ −π2 −π3 −π4 −π5 ](ππΎ (ππ₯1 , ππ₯2 )) ≤ [2π1 +π2 +π3 +π4 +π5 ](ππΎ (ππ₯0 , ππ₯1 )). If we put π := [2πΌ − π2 − π3 − π4 − π5 ]−1 β [2π1 + π2 + π3 + π4 + π5 ], then ππΎ (ππ₯1 , ππ₯2 ) ≤ π (ππΎ (ππ₯0 , ππ₯1 )), which implies that π·(ππ₯1 , ππ₯2 ) ≤ πβ£β£π β£β£π π·(ππ₯0 , ππ₯1 ). On the other hand, ππΎ (ππ₯2 , ππ₯3 ) = ππΎ (ππ π₯1 , ππ π₯2 ) ≤ π1 (ππΎ (ππ₯1 , ππ₯2 )) + π2 (ππΎ (ππ₯1 , ππ π₯1 )) + π3 (ππΎ (ππ₯2 , ππ π₯2 )) +π4 (ππΎ (ππ₯1 , ππ π₯2 )) + π5 (ππΎ (ππ₯2 , ππ π₯1 )) ≤ π1 (ππΎ (ππ₯1 , ππ₯2 )) + π2 (ππΎ (ππ₯1 , ππ₯2 )) + π3 (ππΎ (ππ₯2 , ππ₯3 )) +π4 (ππΎ (ππ₯1 , ππ₯2 )) + π4 (ππΎ (ππ₯2 , ππ₯3 )). Hence [πΌ − π3 − π4 ](ππΎ (ππ₯2 , ππ₯3 )) ≤ [π1 + π2 + π4 ](ππΎ (ππ₯1 , ππ₯2 )). Also we have 32 I. GATICA ARAUS, J. LORENZO RAMIREZ ππΎ (ππ₯3 , ππ₯2 ) = ππΎ (ππ π₯2 , ππ π₯1 ) ≤ π1 (ππΎ (ππ₯2 , ππ₯1 )) + π2 (ππΎ (ππ₯2 , ππ π₯2 )) + π3 (ππΎ (ππ₯1 , ππ π₯1 )) +π4 (ππΎ (ππ₯2 , ππ π₯1 )) + π5 (ππΎ (ππ₯1 , ππ π₯2 )) ≤ π1 (ππΎ (ππ₯2 , ππ₯1 )) + π2 (ππΎ (ππ₯2 , ππ₯3 )) + π3 (ππΎ (ππ₯1 , ππ₯2 )) +π5 (ππΎ (ππ₯1 , ππ₯2 )) + π5 (ππΎ (ππ₯2 , ππ₯3 )). By symmetry of ππΎ , we have [πΌ − π2 − π5 ](ππΎ (ππ₯3 , ππ₯2 )) ≤ [π1 + π3 + π5 ](ππΎ (ππ₯2 , ππ₯1 )). Thus ππΎ (ππ₯2 , ππ₯3 ) ≤ π (ππΎ (ππ₯1 , ππ₯2 )) ≤ π 2 (ππΎ (ππ₯0 , ππ₯1 )), Following the above reasoning we obtain ππΎ (ππ₯π , ππ₯π+1 ) ≤ π π ππΎ (ππ₯0 , ππ₯1 ), for all π ∈ β. Given π ∈ π΅ + (π ) notice that if β£β£π β£β£π < 1, then there exists [πΌ − π ]−1 and 1 . Also, if π ∈ π΅(π ) such that there exists π−1 , then for β£β£[πΌ − π ]−1 β£β£π ≤ 1 − β£β£π β£β£π π−1 β£β£π−1 β£β£π all π ∈ β+ , [ππ]−1 = and β£β£[ππ]−1 β£β£π = . Take in to account these π π facts we proceed as follows. π2 + π3 + π4 + π5 and π½ := 2[π1 +π ]. Thus π = [2[πΌ−π ]]−1 β Denote by π := 2 ∑5 [π½]. It is clear that β£β£π β£β£π < 1, because π=1 β£β£ππ β£β£π < 1. Therefore there exists 1 [πΌ−π ]−1 , in consequence there exists [2[πΌ−π ]]−1 and β£β£[2[πΌ − π ]]−1 β£β£π ≤ . 2 − 2β£β£π β£β£π Hence β£β£π β£β£π ≤ β£β£[2[πΌ − π ]]−1 β£β£π β£β£π½β£β£π ≤ β£β£π½β£β£π 2 − 2β£β£π β£β£π = β£β£2π1 + π2 + π3 + π4 + π5 β£β£π 2 − β£β£π2 + π3 + π4 + π5 β£β£π ≤ 2β£β£π1 β£β£π + β£β£π2 β£β£π + β£β£π3 β£β£π + β£β£π4 β£β£π + β£β£π5 β£β£π < 1, 2 − β£β£π2 β£β£π − β£β£π3 β£β£π − β£β£π4 β£β£π − β£β£π5 β£β£π ∑5 because π=1 β£β£ππ β£β£π < 1. We next prove that {ππ₯π }π∈β is a Cauchy sequence. For each π ≥ 1 we have ππΎ (ππ₯π , ππ₯π+π ) ≤ ππΎ (ππ₯π , ππ₯π+1 ) + ππΎ (ππ₯π+1 , ππ₯π+2 ) + ⋅ ⋅ ⋅ ⋅ ⋅ ⋅ + ππΎ (ππ₯π+π−1 , ππ₯π+π )] ≤ (π π + π π+1 + ⋅ ⋅ ⋅ + π π+π−1 )ππΎ (ππ₯0 , ππ₯1 ). Hence CONE METRIC SPACE 33 π·(ππ₯π , ππ₯π+π ) ≤ π[(β£β£π β£β£ππ + β£β£π β£β£π+1 + ⋅ ⋅ ⋅ + β£β£π β£β£ππ+π−1 )π·(ππ₯0 , ππ₯1 )] π π ≤ πβ£β£π β£β£π [(1 + β£β£π β£β£π + ⋅ ⋅ ⋅ + β£β£π β£β£π−1 + ⋅ ⋅ ⋅ )π·(ππ₯0 , ππ₯1 )] π πβ£β£π β£β£ππ = π·(ππ₯0 , ππ₯1 ). 1 − β£β£π β£β£π Therefore, π·(ππ₯π , ππ₯π+π ) → 0 as π → ∞, hence {ππ₯π }π∈β is a Cauchy sequence. Since π is complete, there exists π₯∗ ∈ π such that ππ₯π → π₯∗ . Proof of (3) Suppose that π is one to one, sequentially continuous and sequentially convergent mapping. Since ππ₯π → π₯∗ , there exists π¦ ∗ ∈ π such that π₯π → π¦ ∗ and it must be the case that ππ₯π → ππ¦ ∗ . Therefore ππ¦ ∗ = π₯∗ . We ο¬rst show that π π¦∗ = π¦∗ . For each π ≥ 1 ππΎ (ππ π¦ ∗ , ππ¦ ∗ ) ≤ ππΎ (ππ π¦ ∗ , ππ₯π ) + ππΎ (ππ₯π , ππ¦ ∗ ). We observe that ππΎ (ππ π¦ ∗ , ππ₯π ) = ππΎ (ππ π¦ ∗ , ππ π₯π−1 ) ≤ π1 (ππΎ (ππ¦ ∗ , ππ₯π−1 )) + π2 (ππΎ (ππ¦ ∗ , ππ π¦ ∗ )) +π3 (ππΎ (ππ₯π−1 , ππ₯π )) + π4 (ππΎ (ππ¦ ∗ , ππ₯π )) +π5 (ππΎ (ππ₯π−1 , ππ π¦ ∗ )). Which implies [πΌ − π2 − π5 ](ππΎ (ππ π¦ ∗ , ππ¦ ∗ )) ≤ ππΎ (ππ₯π , ππ¦ ∗ ) + π1 (ππΎ (ππ¦ ∗ , ππ₯π−1 )) +π3 (ππΎ (ππ₯π−1 , ππ₯π )) + π4 (ππΎ (ππ¦ ∗ , ππ₯π )) +π5 (ππΎ (ππ₯π−1 , ππ π¦ ∗ ) − ππΎ (ππ π¦ ∗ , ππ¦ ∗ )). Hence ππΎ (ππ π¦ ∗ , ππ¦ ∗ ) ≤ [πΌ − π2 − π5 ]−1 (ππΎ (ππ₯π , ππ¦ ∗ ) + π1 (ππΎ (ππ¦ ∗ , ππ₯π−1 )) +π3 (ππΎ (ππ₯π−1 , ππ₯π )) + π4 (ππΎ (ππ¦ ∗ , ππ₯π )) +π5 (ππΎ (ππ₯π−1 , ππ π¦ ∗ ) − ππΎ (ππ π¦ ∗ , ππ¦ ∗ ))). β£β£ππΎ (ππ π¦ ∗ , ππ¦ ∗ )β£β£ = π·(ππ π¦ ∗ , ππ¦ ∗ ) ≤ πβ£β£[πΌ − π2 − π5 ]−1 β£β£β£β£ππΎ (ππ₯π , ππ¦ ∗ ) + π1 (ππΎ (ππ¦ ∗ , ππ₯π−1 )) +π3 (ππΎ (ππ₯π−1 , ππ₯π )) + π4 (ππΎ (ππ¦ ∗ , ππ₯π )) +π5 (ππΎ (ππ₯π−1 , ππ π¦ ∗ ) − ππΎ (ππ π¦ ∗ , ππ¦ ∗ ))β£β£ ≤ πβ£β£[πΌ − π2 − π5 ]−1 β£β£[π·(ππ₯π , ππ¦ ∗ ) + β£β£π1 β£β£π π·(ππ₯π−1 , ππ¦ ∗ ) +β£β£π3 β£β£π π·(ππ₯π−1 , ππ₯π ) + β£β£π4 β£β£π π·(ππ₯π , ππ¦ ∗ ) +β£β£π5 β£β£π β£β£ππΎ (ππ₯π−1 , ππ π¦ ∗ ) − ππΎ (ππ¦ ∗ , ππ π¦ ∗ )β£β£]. Denote by π := π·(ππ π¦ ∗ , ππ¦ ∗ ), ππ := π·(ππ₯π , ππ¦ ∗ ), ππ := πβ£β£π β£β£π π·(ππ₯0 , ππ₯1 ) and ππ = β£β£ππΎ (ππ₯π , ππ π¦ ∗ ) − ππΎ (ππ¦ ∗ , ππ π¦ ∗ )β£β£. Thus π ≤ πβ£β£[πΌ − π2 − π5 ]−1 β£β£[ππ + β£β£π1 β£β£π ππ−1 + β£β£π3 β£β£π ππ−1 + β£β£π4 β£β£π ππ + β£β£π5 β£β£π ππ−1 ]. It is clear that ππ , ππ → 0 as π → ∞. On the other hand, by Lemma 3.5 we have ππ → 0 as π → ∞. Thus, π·(ππ π¦ ∗ , ππ¦ ∗ ) = 0. This fact implies ππ π¦ ∗ = ππ¦ ∗ , and from the injectivity of π it follows that π π¦ ∗ = π¦ ∗ . We shall show that π¦ ∗ is the unique ο¬xed point of π . Suppose that there exists another π₯∗ ∈ π such that π π₯∗ = π₯∗ and π₯∗ β= π¦ ∗ . Hence, from the injectivity of 34 I. GATICA ARAUS, J. LORENZO RAMIREZ π we get ππΎ (ππ₯∗ , ππ¦ ∗ ) ∈ πΎ − {0π }. Applying that π is a S-Hardy-Rogers cone contraction, we have that ππΎ (ππ₯∗ , ππ¦ ∗ ) ≤ = ππΎ (ππ π₯∗ , ππ π¦ ∗ ) π1 (ππΎ (ππ₯∗ , ππ¦ ∗ )) + π2 (ππΎ (ππ₯∗ , ππ π₯∗ )) + π3 (ππΎ (ππ¦ ∗ , ππ π¦ ∗ )) +π4 (ππΎ (ππ₯∗ , ππ π¦ ∗ )) + π5 (ππΎ (ππ¦ ∗ , ππ π₯∗ )). = [π1 + π4 + π5 ](ππΎ (ππ₯∗ , ππ¦ ∗ )), which is a contradiction by Lemma 1.13. Therefore π¦ ∗ is the unique ο¬xed point of π and the proof is complete. β‘ Taking ππ₯ = π₯ in the above theorem, the conditions π2 ≡ π3 ≡ π4 ≡ π5 ≡ Θ yields Banach’s ο¬xed point theorem, while π1 ≡ π4 ≡ π5 ≡ Θ yields Kannan’s ο¬xed point theorem and π1 ≡ π2 ≡ π3 ≡ Θ yields Chatterjea ο¬xed point theorem in the context of cone metric spaces. Corollary 3.7. Let (π, ππΎ ) be a complete CMS over an ordered Banach space (π, β£β£ ⋅ β£β£, πΎ) such that πΎ is a normal cone and π : π → π be a mapping. If for all π₯, π¦ ∈ π we have that ππΎ (π π₯, π π¦) ≤ π(ππΎ (π₯, π¦)), where π ∈ π΅ + (π ) and β£β£πβ£β£π < 1, then π has a unique ο¬xed point in π. Corollary 3.8. ([8]) Let (π, ππΎ ) be a complete CMS over an ordered Banach space (π, β£β£ ⋅ β£β£, πΎ) such that πΎ is a normal cone. If π is a self-mapping on π such that ππΎ (π π₯, π π¦) ≤ π(ππΎ (π₯, π π₯)) + π (ππΎ (π¦, π π¦)), for all π₯, π¦ ∈ π, where π, π ∈ π΅ + (π ) and β£β£πβ£β£π + β£β£π β£β£π < 1, then π has a unique ο¬xed point in π. Corollary 3.9. Let (π, ππΎ ) be a complete CMS over an ordered Banach space (π, β£β£ ⋅ β£β£, πΎ) such that πΎ is a normal cone. If π is a self-mapping on π such that ππΎ (π π₯, π π¦) ≤ π(ππΎ (π₯, π π¦)) + π (ππΎ (π¦, π π₯)), for all π₯, π¦ ∈ π, where π, π ∈ π΅ + (π ) and β£β£πβ£β£π + β£β£π β£β£π < 1, then π has a unique ο¬xed point in π. Acknowledgments. This research is partially supported by the Plan Andaluz de InvestigacioΜn de la Junta de AndalucΔ±Μa FQM-127 and by MEC Grant MTM200910696-C02-01. References [1] R. P. Agarwal, Contraction and approximate contraction with an application to multi-point boundary value problems, J. Comput. Appl. 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Perov, On the Cauchy problem for a system of ordinary diο¬erential equations, Pviblizhen. Met. Reshen. Diο¬er. Uvavn., vol. 2, (1964), pp. 115134. [22] Ioan A. Rus, Approximation of common ο¬xed point in a generalized metric space, Anal. NumeΜr. TheΜor. Approx., No. 8 (1979), pp. 236-241. [23] P.P. Zabrejko, K-Metric and K-Normed linear spaces: survey, Collect. Math. vol. 48, 4-6 (1997), pp. 825-859. Ivan Gatica Araus Departamento de MatemaΜtica, Universidad AndreΜs Bello, Republica 252, Santiago, Chile E-mail address: ivan.gatica@unab.cl Josefa Lorenzo Ramirez Departamento de AnaΜlisis MatemaΜtico, Universidad de Sevilla, St. Tarfia s/n, Sevilla, Espana E-mail address: ploren@us.es