EUROPEAN JOURNAL OF PURE AND APPLIED MATHEMATICS Vol. 10, No. 3, 2017, 473-487 ISSN 1307-5543 – www.ejpam.com Published by New York Business Global Common fixed point theorems for three maps in cone pentagonal metric spaces Abba Auwalu1,∗, Evren Hınçal1 1 Department of Mathematics, Near East University, Nicosia-TRNC, Turkey Abstract. In this paper, we prove some common fixed point theorems of three self mappings in non-normal cone pentagonal metric spaces. Our results extend and improve the recent results announced by many authors. 2010 Mathematics Subject Classifications: 47H10, 54H25 Key Words and Phrases: Cone pentagonal metric spaces, Common fixed point, Contraction mapping principle, Weakly compatible maps 1. Introduction The concept of metric space was introduced by Fŕechet [8]. Let (X, d) be a metric space and S : X → X be a mapping. Then S is called Banach contraction if there exists α ∈ [0, 1) such that d(Sx, Sy) ≤ αd(x, y), for all x, y ∈ X. (1) Banach [7] proved that if X is complete, then every Banach contraction mapping has a fixed point. The mapping S is called Kannan contraction if there exists α ∈ [0, 1/2) such that d(Sx, Sy) ≤ α [ d(x, Sx) + d(y, Sy) ] , for all x, y ∈ X. (2) Kannan [14] proved that if X is complete, then every Kannan contraction has a fixed point. He further showed that the conditions (1) and (2) are independent of each other (see, [14, 15]). The study of existence and uniqueness of fixed points of a mapping and common fixed points of two or more mappings has become a subject of great interest. Many authors proved the Banach contraction and Kannan contraction principles in various generalized metric spaces (e.g., see [4, 5, 6, 9, 10, 11, 13, 18]). Long-Guang and Xian [11] introduced the concept of a cone metric space and proved some fixed point theorems for contractive type conditions in cone metric spaces. Later on many ∗Corresponding author. Email addresses: abba.auwalu@neu.edu.tr, abbaauwalu@yahoo.com (A. Auwalu), evren.hincal@neu.edu.tr, evrenhincal@yahoo.co.uk (E. Hınçal) http://www.ejpam.com 473 c© 2017 EJPAM All rights reserved. A. Auwalu, E. Hınçal / Eur. J. Pure Appl. Math, 10 (3) (2017), 473-487 474 authors have (for e.g., [1, 6, 9, 12, 17, 19]) proved some fixed point theorems for different contractive types conditions in cone metric spaces. Recently, Garg and Agarwal [9] introduced the notion of cone pentagonal metric space and proved Banach contraction mapping principle in a normal cone pentagonal metric space setting. Motivated and inspired by the results of [9, 17, 16], it is our purpose in this paper to continue the study of common fixed points of a three self mappings in non-normal cone pentagonal metric space setting. Our results extend and improve the results of [2, 3, 6, 9, 13, 18, 17, 16], and many others. 2. Preliminaries The following definitions and Lemmas, introduced in [1, 3, 6, 9, 11], are needed in the sequel. Definition 1. Let E be a real Banach space and P subset of E. P is called a cone if and only if: (i) P is closed, nonempty, and P 6= {0}; (ii) a, b ∈ R, a, b ≥ 0 and x, y ∈ P =⇒ ax+ by ∈ P ; (iii) x ∈ P and −x ∈ P =⇒ x = 0. Given a cone P ⊆ E, we defined a partial ordering ≤ with respect to P by x ≤ y if and only if y− x ∈ P. We shall write x < y to indicate that x ≤ y but x 6= y, while x� y will stand for y − x ∈ int(P ), where int(P ) denotes the interior of P. A cone P is called normal if there is a number k ≥ 1 such that for all x, y ∈ E, the inequality 0 ≤ x ≤ y =⇒ ‖x‖ ≤ k‖y‖. (3) The least positive number k satisfying (3) is called the normal constant of P. In this paper, we always suppose that E is a real Banach space and P is a cone in E with int(P ) 6= ∅ and ≤ is a partial ordering with respect to P. Definition 2. Let X be a nonempty set. Suppose the mapping ρ : X ×X → E satisfies: (i) 0 < ρ(x, y) for all x, y ∈ X and ρ(x, y) = 0 if and only if x = y; (ii) ρ(x, y) = ρ(y, x) for all x, y ∈ X; (iii) ρ(x, y) ≤ ρ(x, z) + ρ(z, y) for all x, y, z ∈ X. Then ρ is called a cone metric on X, and (X, ρ) is called a cone metric space. The concept of a cone metric space is more general than that of a metric space, because each metric space is a cone metric space where E = R and P = [0,∞) (e.g., see [11]). A. Auwalu, E. Hınçal / Eur. J. Pure Appl. Math, 10 (3) (2017), 473-487 475 Definition 3. Let X be a nonempty set. Suppose the mapping ρ : X ×X → E satisfies: (i) 0 < ρ(x, y) for all x, y ∈ X and ρ(x, y) = 0 if and only if x = y; (ii) ρ(x, y) = ρ(y, x) for all x, y ∈ X; (iii) ρ(x, y) ≤ ρ(x,w) + ρ(w, z) + ρ(z, y) for all x, y, z ∈ X and for all distinct points w, z ∈ X − {x, y} [Rectangular property]. Then ρ is called a cone rectangular metric on X, and (X, ρ) is called a cone rectangular metric space. Remark 1. Every cone metric space is cone rectangular metric space. The converse is not necessarily true (e.g., see [6]). Definition 4. Let X be a non empty set. Suppose the mapping d : X ×X → E satisfies: (i) 0 < d(x, y) for all x, y ∈ X and d(x, y) = 0 if and only if x = y; (ii) d(x, y) = d(y, x) for x, y ∈ X; (iii) d(x, y) ≤ d(x, z) + d(z, w) + d(w, u) + d(u, y) for all x, y, z, w, u ∈ X and for all distinct points z, w, u,∈ X − {x, y} [Pentagonal property]. Then d is called a cone pentagonal metric on X, and (X, d) is called a cone pentagonal metric space. Remark 2. Every cone rectangular metric space and so cone metric space is cone pen- tagonal metric space. The converse is not necessarily true (e.g., see [9]). Let (X, d) be a cone pentagonal metric space. Let {xn} be a sequence in X and x ∈ X. If for every c ∈ E with 0 � c there exist n0 ∈ N and that for all n > n0, d(xn, x) � c, then {xn} is said to be convergent and {xn} converges to x, and x is the limit of {xn}. We denote this by limn→∞ xn = x or xn → x as n → ∞. If for every c ∈ E, with 0 � c there exist n0 ∈ N such that for all n,m > n0, d(xn, xm)� c, then {xn} is called Cauchy sequence in X. If every Cauchy sequence is convergent in X, then X is called a complete cone pentagonal metric space. Definition 5. Let P be a cone defined as above and let Φ be the set of non decreasing continuous functions ϕ : P → P satisfying: (i) 0 < ϕ(t) < t for all t ∈ P \ {0}, (ii) the series ∑ n≥0 ϕ n(t) converge for all t ∈ P \ {0} From (i), we have ϕ(0) = 0, and from (ii), we have limn→0 ϕ n(t) = 0 for all t ∈ P \ {0}. Let T and S be self maps of a nonempty set X. If w = Tx = Sx for some x ∈ X, then x is called a coincidence point of T and S and w is called a point of coincidence of T and S. Also, T and S are said to be weakly compatible if they commute at their coincidence points, that is, Tx = Sx implies that TSx = STx. A. Auwalu, E. Hınçal / Eur. J. Pure Appl. Math, 10 (3) (2017), 473-487 476 Lemma 1. Let T and S be weakly compatible self mappings of nonempty set X. If T and S have a unique point of coincidence w = Tx = Sx, then w is the unique common fixed point of T and S. Lemma 2. Let (X, d) be a cone metric space with cone P not necessary to be normal. Then for a, c, u, v, w ∈ E, we have (i) If a ≤ ha and h ∈ [0, 1), then a = 0. (ii) If 0 ≤ u� c for each 0� c, then u = 0. (iii) If u ≤ v and v � w, then u� w. Lemma 3. Let (X, d) be a complete cone pentagonal metric space. Let {xn} be a Cauchy sequence in X and suppose that there is natural number N such that: (i) xn 6= xm for all n,m > N ; (ii) xn, x are distinct points in X for all n > N ; (iii) xn, y are distinct points in X for all n > N ; (iv) xn → x and xn → y as n→∞. Then x = y. 3. Main Results In this section, we prove Banach type and Kannan type contraction principles in cone pentagonal metric spaces of a three self mappings. We give some examples to illustrate the results. Theorem 1. Let (X, d) be a cone pentagonal metric space. Suppose the mappings S, f, g : X → X satisfies the contractive condition: d(Sx, fy) ≤ ϕ ( d(gx, gy) ) , (4) for all x, y ∈ X, where ϕ ∈ Φ. Suppose that S(X) ∪ f(X) ⊆ g(X), and g(X) is a complete subspace of X, then the mappings S, f and g have a unique point of coincidence in X. Moreover, if (S, g) and (f, g) are weakly compatible then S, f and g have a unique common fixed point in X. Proof. Let x0 be an arbitrary point in X. Since S(X) ∪ f(X) ⊆ g(X), we can choose x1 ∈ X such that gx1 = Sx0. Also we can choose x2 ∈ X such that gx2 = fx1. Continuing this process, having chosen xn in X, we obtain xn+1 such that gxn+1 = Sxn and gxn+2 = fxn+1, for all n = 0, 1, 2, · · · . A. Auwalu, E. Hınçal / Eur. J. Pure Appl. Math, 10 (3) (2017), 473-487 477 If gxn = gxn+1, then gxn = Sxn = fxn, and xn is a coincidence point of S, f and g. Hence, we assume that xn 6= xn+1 for all n ∈ N. Then, from (4), it follows that d(gxn, gxn+1) = ϕ ( d(Sxn−1, fxn) ) ≤ ϕ ( d(gxn−1, gxn) ) ≤ ϕ2 ( d(gxn−2, gxn−1) ) ... ≤ ϕn ( d(gx0, gx1) ) . (5) In similar way, it again follows that d(gxn, gxn+2) ≤ ϕn ( d(gx0, gx2) ) , (6) d(gxn, gxn+3) ≤ ϕn ( d(gx0, gx3) ) . (7) Similarly, for k = 1, 2, 3, · · · , it further follows that d(gxn, gxn+3k+1) ≤ ϕn ( d(gx0, gx3k+1) ) , (8) d(gxn, gxn+3k+2) ≤ ϕn ( d(gx0, gx3k+2) ) , (9) d(gxn, gxn+3k+3) ≤ ϕn ( d(gx0, gx3k+3) ) . (10) By pentagonal property and (5), we have d(gx0, gx4) ≤ d(gx0, gx1) + d(gx1, gx2) + d(gx2, gx3) + d(gx3, gx4) ≤ d(gx0, gx1) + ϕ ( d(gx0, gx1) ) + ϕ2 ( d(gx0, gx1) ) + ϕ3 ( d(gx0, gx1) ) ≤ 3∑ i=0 ϕi ( d(gx0, gx1) ) , and d(gx0, gx7) ≤ d(gx0, gx1) + d(gx1, gx2) + d(gx2, gx3) + d(gx3, gx4) + d(gx4, gx5) + d(gx5, gx6) + d(gx6, gx7) ≤ 6∑ i=0 ϕi ( d(gx0, gx1) ) . Now, by induction, we obtain for each k = 1, 2, 3, · · · d(gx0, gx3k+1) ≤ 3k∑ i=0 ϕi ( d(gx0, gx1) ) . (11) Also, using (5), (6), and pentagonal property, we have that d(gx0, gx5) ≤ 2∑ i=0 ϕi ( d(gx0, gx1) ) + ϕ3 ( d(gx0, gx2) ) , A. Auwalu, E. Hınçal / Eur. J. Pure Appl. Math, 10 (3) (2017), 473-487 478 and d(gx0, gx8) ≤ 5∑ i=0 ϕi ( d(gx0, gx1) ) + ϕ6 ( d(gx0, gx2) ) . By induction, we obtain for each k = 1, 2, 3, · · · d(gx0, gx3k+2) ≤ 3k−1∑ i=0 ϕi ( d(gx0, gx1) ) + ϕ3k ( d(gx0, gx2) ) . (12) Again, using (5), (7), and pentagonal property, we have that d(gx0, gx6) ≤ 2∑ i=0 ϕi ( d(gx0, gx1) ) + ϕ3 ( d(gx0, gx3) ) , and d(gx0, gx9) ≤ 5∑ i=0 ϕi ( d(gx0, gx1) ) + ϕ6 ( d(gx0, gx3) ) . By induction, we obtain for each k = 1, 2, 3, · · · d(gx0, gx3k+3) ≤ 3k−1∑ i=0 ϕi ( d(gx0, gx1) ) + ϕ3k ( d(gx0, gx3) ) . (13) Using (8) and (11), for k = 1, 2, 3, · · · , we have d(gxn, gxn+3k+1) ≤ ϕn 3k∑ i=0 ϕi ( d(gx0, gx1) ) ≤ ϕn [ 3k∑ i=0 ϕi ( d(gx0, gx1) + d(gx0, gx2) + d(gx0, gx3) )] ≤ ϕn [ ∞∑ i=0 ϕi ( d(gx0, gx1) + d(gx0, gx2) + d(gx0, gx3) )] . (14) Similarly for k = 1, 2, 3, · · · , (9) and (12) implies that d(gxn, gxn+3k+2) ≤ ϕn [ 3k−1∑ i=0 ϕi ( d(gx0, gx1) ) + ϕ3k ( d(gx0, gx2) )] ≤ ϕn [ ∞∑ i=0 ϕi ( d(gx0, gx1) + d(gx0, gx2) + d(gx0, gx3) )] . (15) Again, for k = 1, 2, 3, · · · , (10) and (13) implies that d(gxn, gxn+3k+3) ≤ ϕn [ ∞∑ i=0 ϕi ( d(gx0, gx1) + d(gx0, gx2) + d(gx0, gx3) )] . (16) A. Auwalu, E. Hınçal / Eur. J. Pure Appl. Math, 10 (3) (2017), 473-487 479 Thus, by (14), (15), and (16), we have, for each m, d(gxn, gxn+m) ≤ ϕn [ ∞∑ i=0 ϕi ( d(gx0, gx1) + d(gx0, gx2) + d(gx0, gx3) )] . (17) Since ∑∞ i=0 ϕ i ( d(gx0, gx1) + d(gx0, gx2) + d(gx0, gx3) ) converges (by definition 5), where d(gx0, gx1)+d(gx0, gx2)+d(gx0, gx3) ∈ P\{0}, and P is closed, then ∑∞ i=0 ϕ i ( d(gx0, gx1)+ d(gx0, gx2) + d(gx0, gx3) ) ∈ P \ {0}. Hence lim n→∞ ϕn [ ∞∑ i=0 ϕi ( d(gx0, gx1) + d(gx0, gx2) + d(gx0, gx3) )] = 0. Then, for given c� 0, there is a natural number N1 such that ϕn [ ∞∑ i=0 ϕi ( d(gx0, gx1) + d(gx0, gx2) + d(gx0, gx3) )] � c, ∀n ≥ N1. (18) Thus, from (17) and (18), we have d(gxn, gxn+m)� c, for all n ≥ N1. Therefore, {gxn} is a Cauchy sequence in X. Since g(X) is a complete subspace of X, there exists a points u, v ∈ g(X) such that limn→∞ gxn = v = gu. Now, we show that gu = Su. Given c� 0, we choose a natural numbers N2, N3 such that d(v, gxn) � c 4 , ∀n ≥ N2, and d(gxn, gxn+1) � c 4 , ∀n ≥ N3. Since xn 6= xm for n 6= m, by pentagonal property, we have that d(gu, Su) ≤ d(gu, gxn) + d(gxn, gxn+1) + d(gxn+1, gxn+2) + d(gxn+2, Su) = d(v, gxn) + d(gxn, gxn+1) + d(gxn+1, gxn+2) + d(Su, fxn+1) ≤ d(v, gxn) + d(gxn, gxn+1) + d(gxn+1, gxn+2) + ϕ ( d(gu, gxn+1) ) < d(v, gxn) + d(gxn, gxn+1) + d(gxn+1, gxn+2) + d(v, gxn+1) � c 4 + c 4 + c 4 + c 4 = c, for all n ≥ N, where N := max{N2, N3}. Since c is arbitrary, we have d(gu, Su) � c m , ∀m ∈ N. Since c m → 0 as m → ∞, we conclude c m − d(gu, Su) → −d(gu, Su) as m → ∞. Since P is closed, −d(gu, Su) ∈ P. Hence d(gu, Su) ∈ P ∩ −P. By definition of cone we get that d(gu, Su) = 0, and so gu = Su = v. Hence, v is a coincidence point of S and g. Similarly, we can prove that gu = fu = v, which implies that v is a point of coincidence of S, f and g, i.e. gu = fu = Su = v. Next, we show that v is unique. For suppose v′ be another point of coincidence of S, f and g, that is Su′ = fu′ = gu′ = v′, for some u′ ∈ X, then d(v, v′) = d(Su, fu′) ≤ ϕ ( d(gu, gu′) ) = ϕ ( d(v, v′) ) < d(v, v′). A. Auwalu, E. Hınçal / Eur. J. Pure Appl. Math, 10 (3) (2017), 473-487 480 Hence v = v′. Since (S, g) and (f, g) are weakly compatible, by Lemma 1, v is the unique common fixed point of S, f and g. This completes the proof of the theorem. Example 1. Let X = {1, 2, 3, 4, 5}, E = R2 and P = {(x, y) : x, y ≥ 0} is a cone in E. Define d : X ×X → E as follows: d(x, x) = 0, ∀x ∈ X; d(1, 2) = d(2, 1) = (4, 8); d(1, 3) = d(3, 1) = d(3, 4) = d(4, 3) = d(2, 4) = d(4, 2) = (1, 2); d(1, 5) = d(5, 1) = d(2, 5) = d(5, 2) = d(3, 5) = d(5, 3) = d(4, 5) = d(5, 4) = (3, 6). Then (X, d) is a cone pentagonal metric space, but (X, d) is not a cone rectangular metric space because it lacks the rectangular property: (4, 8) = d(1, 2) > d(1, 3) + d(3, 4) + d(4, 2) = (1, 2) + (1, 2) + (1, 2) = (3, 6) as (4, 8)− (3, 6) = (1, 2) ∈ P. Define a mapping S, f and g : X → X as follows: S(x) = 4, ∀x ∈ X. f(x) = { 4, if x 6= 5; 2, if x = 5. g(x) = x, ∀x ∈ X. Clearly S(X)∪ f(X) ⊆ g(X), g(X) is a complete subspace of X. Also, the pairs (S, g) and (f, g) are weakly compatibles. The conditions of Theorem 1 holds for all x, y ∈ X, where ϕ(t) = 1 3 t, and 4 is the unique common fixed point of the mappings S, f and g. Corollary 1. Let (X, d) be a cone pentagonal metric space. Suppose the mappings S, f, g : X → X satisfies the contractive condition: d(Sx, fy) ≤ λd(gx, gy), for all x, y ∈ X, where λ ∈ [0, 1). Suppose that S(X) ∪ f(X) ⊆ g(X), and g(X) is a complete subspace of X, then the mappings S, f and g have a unique point of coincidence in X. Moreover, if (S, g) and (f, g) are weakly compatible then S, f and g have a unique common fixed point in X. Proof. Define ϕ : P → P by ϕ(t) = λt. Then it is clear that ϕ satisfies the conditions in definition 5. Hence the results follows from Theorem 1. A. Auwalu, E. Hınçal / Eur. J. Pure Appl. Math, 10 (3) (2017), 473-487 481 Corollary 2. (see [4]) Let (X, d) be a cone pentagonal metric space. Suppose the mappings S, g : X → X satisfies the contractive condition: d(Sx, Sy) ≤ ϕ ( d(gx, gy) ) , for all x, y ∈ X, where ϕ ∈ Φ. Suppose that S(X) ⊆ g(X), and g(X) or S(X) is a complete subspace of X, then the mappings S and g have a unique point of coincidence in X. Moreover, if S and g are weakly compatible then S and g have a unique common fixed point in X. Proof. Putting f = S in Theorem 1. This completes the proof. Corollary 3. Let (X, d) be a cone pentagonal metric space. Suppose the mappings S, g : X → X satisfies the contractive condition: d(Sx, Sy) ≤ λd(gx, gy), for all x, y ∈ X, where λ ∈ [0, 1). Suppose that S(X) ⊆ g(X), and g(X) or S(X) is a complete subspace of X, then the mappings S and g have a unique point of coincidence in X. Moreover, if S and g are weakly compatible then S and g have a unique common fixed point in X. Proof. Putting f = S in Theorem 1. The results follows from Corollary 1. Corollary 4. (see [16]) Let (X, d) be a cone rectangular metric space. Suppose the map- pings S, f, g : X → X satisfies the contractive condition: d(Sx, fy) ≤ λd(gx, gy), for all x, y ∈ X, where λ ∈ [0, 1). Suppose that S(X) ∪ f(X) ⊆ g(X), and g(X) is a complete subspace of X, then the mappings S, f and g have a unique point of coincidence in X. Moreover, if (S, g) and (f, g) are weakly compatible then S, f and g have a unique common fixed point in X. Proof. This follows from the Remark 2 and Theorem 1. Corollary 5. (see [17]) Let (X, d) be a cone rectangular metric space. Suppose the map- pings S, g : X → X satisfies the contractive condition: d(Sx, Sy) ≤ ϕ ( d(gx, gy) ) , for all x, y ∈ X, where ϕ ∈ Φ. Suppose that S(X) ⊆ g(X), and g(X) or S(X) is a complete subspace of X, then the mappings S and g have a unique point of coincidence in X. Moreover, if S and g are weakly compatible then S and g have a unique common fixed point in X. A. Auwalu, E. Hınçal / Eur. J. Pure Appl. Math, 10 (3) (2017), 473-487 482 Proof. This follows from the Remark 2 and Corollary 2. Corollary 6. (see [2]) Let (X, d) be a cone pentagonal metric space. Suppose the mapping S : X → X satisfy the following: d(Sx, Sy) ≤ ϕ ( d(x, y) ) , for all x, y ∈ X, where ϕ ∈ Φ. Then S has a unique fixed point in X. Proof. Putting g = I in Corollary 2, where I is the identity mapping. This completes the proof. Corollary 7. (see [17]) Let (X, d) be a cone rectangular metric space. Suppose the map- ping S : X → X satisfy the following: d(Sx, Sy) ≤ ϕ ( d(x, y) ) , for all x, y ∈ X, where ϕ ∈ Φ. Then S has a unique fixed point in X. Proof. This follows from the Remark 2 and Putting g = I in Corollary 2. Corollary 8. (see [9]) Let (X, d) be a cone pentagonal metric space and P be a normal cone with normal constant k. Suppose the mapping S : X → X satisfies the contractive condition: d(Sx, Sy) ≤ λd(x, y), for all x, y ∈ X, where λ ∈ [0, 1). Then S has a unique fixed point in X. Proof. Putting g = I in Corollary 3, where I is the identity mapping. This completes the proof. Corollary 9. (see [6]) Let (X, d) be a cone rectangular metric space and P be a normal cone with normal constant k. Suppose the mapping S : X → X satisfies: d(Sx, Sy) ≤ λd(x, y), for all x, y ∈ X, where λ ∈ [0, 1). Then S has a unique fixed point in X. Proof. Putting g = I in Corollary 3 and Remark 2, the results follows. Theorem 2. Let (X, d) be a cone pentagonal metric space. Suppose the mappings S, f, g : X → X satisfies the contractive condition: d(Sx, fy) ≤ λ [ d(gx, Sx) + d(gy, fy) ] , (19) for all x, y ∈ X, where λ ∈ [0, 1/2). Suppose that S(X) ∪ f(X) ⊆ g(X), and g(X) is a complete subspace of X, then the mappings S, f and g have a unique point of coincidence in X. Moreover, if (S, g) and (f, g) are weakly compatible then S, f and g have a unique common fixed point in X. A. Auwalu, E. Hınçal / Eur. J. Pure Appl. Math, 10 (3) (2017), 473-487 483 Proof. Let x0 be an arbitrary point in X. Define, like in Theorem 1, a sequence {gxn} in X such that gxn+1 = Sxn and gxn+2 = fxn+1, for all n = 0, 1, 2, · · · . We assume that xn 6= xn+1, for all n ∈ N. Then, from (19), it follows that d(gxn, gxn+1) = d(Sxn−1, fxn) ≤ λ ( d(gxn−1, Sxn−1) + d(gxn, fxn) ) ≤ λ ( d(gxn−1, gxn) + d(gxn, gxn+1) ) . So that, d(gxn, gxn+1) ≤ λ 1− λ d(gxn−1, gxn) ≤ rd(gxn−1, gxn), where r = λ 1− λ ∈ [0, 1) ≤ r2d(gxn−2, gxn−1) ... ≤ rn ( d(gx0, gx1) ) . (20) In similar way, it again follows that d(gxn, gxn+2) ≤ rn ( d(gx0, gx2) ) , (21) and d(gxn, gxn+3) ≤ rn ( d(gx0, gx3) ) . (22) Similarly, for k = 1, 2, 3, · · · , It further follows that d(gxn, gxn+3k+1) ≤ rn ( d(gx0, gx3k+1) ) , (23) d(gxn, gxn+3k+2) ≤ rn ( d(gx0, gx3k+2) ) , (24) d(gxn, gxn+3k+3) ≤ rn ( d(gx0, gx3k+3) ) . (25) Using the same argument in the proof of Theorem 1, we can show that {gxn} is a Cauchy sequence in X. Since g(X) is a complete subspace of X, there exists a points u, v ∈ g(X) such that limn→∞ gxn = v = gu. Now, we show that gu = Su. Given c � 0, we choose a natural numbers M1,M2,M3 such that d(v, gxn) � c(1−λ) 3 , ∀n ≥ M1, d(gxn, gxn+1) � c(1−λ) 3 , ∀n ≥ M2 and d(gxn+1, gxn+2)� c(1−λ) 3(1+λ) , ∀n ≥M3. Since xn 6= xm for n 6= m, by pentagonal property, we have that d(gu, Su) ≤ d(gu, gxn) + d(gxn, gxn+1) + d(gxn+1, gxn+2) + d(gxn+2, Su) A. Auwalu, E. Hınçal / Eur. J. Pure Appl. Math, 10 (3) (2017), 473-487 484 ≤ d(v, gxn) + d(gxn, gxn+1) + d(gxn+1, gxn+2) + d(fxn+1, Su) ≤ d(v, gxn) + d(gxn, gxn+1) + d(gxn+1, gxn+2) + λ ( d(gu, Su) + d(gxn+1, fxn+1) ) < d(v, gxn) + d(gxn, gxn+1) + d(gxn+1, gxn+2) + λ ( d(gu, Su) + d(gxn+1, gxn+2) ) d(gu, Su) ≤ 1 1− λ ( d(v, gxn) + d(gxn, gxn+1) + (1 + λ)d(gxn+1, gxn+2) ) � c 3 + c 3 + c 3 = c, for all n ≥M, where M := max{M1,M2,M3}. Since c is arbitrary, we have d(gu, Su) � c m , ∀m ∈ N. Since c m → 0 as m → ∞, we conclude c m − d(gu, Su) → −d(gu, Su) as m → ∞. Since P is closed, −d(gu, Su) ∈ P. Hence d(gu, Su) ∈ P ∩ −P. By definition of cone we get that d(gu, Su) = 0, and so gu = Su = v. Hence, v is a point of coincidence of S and g. Similarly, we can prove that gu = fu = v, which implies that v is a point of coincidence of S, f and g, i.e. gu = fu = Su = v. Next, we show that v is unique. For suppose v′ be another point of coincidence, that is gu′ = fu′ = Su′ = v′, for some u′ ∈ X, then d(v, v′) = d(Su, fu′) ≤ λ ( d(gu, Su) + d(gu′, fu′) ) ≤ λ ( d(v, v) + d(v′, v′) ) . Hence v = v′. Since (S, g) and (f, g) are weakly compatible, by Lemma 1, v is the unique common fixed point of S, f and g. This completes the proof of the theorem. Corollary 10. (see [5]) Let (X, d) be a cone pentagonal metric space. Suppose the map- pings S, g : X → X satisfies the contractive condition: d(Sx, Sy) ≤ λ [ d(gx, Sx) + d(gy, Sy) ] , for all x, y ∈ X, where λ ∈ [0, 1/2). Suppose that S(X) ⊆ g(X), and S(X) or g(X) is a complete subspace of X, then the mappings S and g have a unique point of coincidence in X. Moreover, if S and g are weakly compatible then S and g have a unique common fixed point in X. Proof. Putting f = S in Theorem 2. This completes the proof. Corollary 11. (see [16]) Let (X, d) be a cone rectangular metric space. Suppose the mappings S, f, g : X → X satisfies the contractive condition: d(Sx, fy) ≤ λ [ d(gx, Sx) + d(gy, fy) ] , for all x, y ∈ X, where λ ∈ [0, 1/2). Suppose that S(X) ∪ f(X) ⊆ g(X), and g(X) is a complete subspace of X, then the mappings S, f and g have a unique point of coincidence in X. Moreover, if (S, g) and (f, g) are weakly compatible then S, f and g have a unique common fixed point in X. Proof. This follows from the Remark 2 and Theorem 2. A. Auwalu, E. Hınçal / Eur. J. Pure Appl. Math, 10 (3) (2017), 473-487 485 Corollary 12. (see [3]) Let (X, d) be a complete cone pentagonal metric space and P be a normal cone with normal constant k. Suppose the mapping S : X → X satisfies the contractive condition: d(Sx, Sy) ≤ λ [ d(x, Sx) + d(y, Sy) ] , (26) for all x, y ∈ X, where λ ∈ [0, 1/2). Then (i) S has a unique fixed point in X. (ii) For any x ∈ X, the iterative sequence {Snx} converges to the fixed point. Proof. Putting g = I in Corollary 10. This completes the proof. Corollary 13. (see [18]) Let (X, d) be a cone rectangular metric space. Suppose the mappings S, g : X → X satisfies the contractive condition: d(Sx, Sy) ≤ λ [ d(gx, Sx) + d(gy, Sy) ] , for all x, y ∈ X, where λ ∈ [0, 1/2). Suppose that S(X) ⊆ g(X), and S(X) or g(X) is a complete subspace of X, then the mappings S and g have a unique point of coincidence in X. Moreover, if S and g are weakly compatible then S and g have a unique common fixed point in X. Proof. This follows from the Remark 2 and Corollary 10. Corollary 14. (see [13]) Let (X, d) be a complete cone rectangular metric space and P be a normal cone with normal constant k. Suppose the mapping S : X → X satisfies the contractive condition: d(Sx, Sy) ≤ λ [ d(x, Sx) + d(y, Sy) ] , (27) for all x, y ∈ X, where λ ∈ [0, 1/2). Then (i) S has a unique fixed point in X. (ii) For any x ∈ X, the iterative sequence {Snx} converges to the fixed point. Proof. Putting g = I in Corollary 10 and Remark 2. This completes the proof. Example 2. Let X = {1, 2, 3, 4, 5}, E = R2 and P = {(x, y) : x, y ≥ 0} is a cone in E. Define d : X ×X → E as follows: d(x, x) = 0,∀x ∈ X; d(1, 2) = d(2, 1) = (4, 8); d(1, 3) = d(3, 1) = d(3, 4) = d(4, 3) = d(2, 4) = d(4, 2) = (1, 2); d(1, 5) = d(5, 1) = d(2, 5) = d(5, 2) = d(3, 5) = d(5, 3) = d(4, 5) = d(5, 4) = (3, 6). REFERENCES 486 Then (X, d) is a cone pentagonal metric space, but (X, d) is not a cone rectangular metric space because it lacks the rectangular property: (4, 8) = d(1, 2) > d(1, 3) + d(3, 4) + d(4, 2) = (1, 2) + (1, 2) + (1, 2) = (3, 6) as (4, 8)− (3, 6) = (1, 2) ∈ P. Define a mapping S, f and g : X → X as follows: S(x) = 4, ∀x ∈ X. f(x) = { 4, if x 6= 5; 2, if x = 5. g(x) =  3, if x = 1; 1, if x = 2; 2, if x = 3; 4, if x = 4; 5, if x = 5. Clearly S(X)∪ f(X) ⊆ g(X), g(X) is a complete subspace of X. Also, the pairs (S, g) and (f, g) are weakly compatibles. The conditions of Theorem 2 holds for all x, y ∈ X, where λ = 1 3 , and 4 is the unique common fixed point of the mappings S, f and g. Acknowledgements This research project was supported by the Center of Excellence, Near East University, Nicosia-TRNC, Mersin 10, Turkey. References [1] M Abbas and G Jungck. Common fixed point results for non commuting mappings without continuity in cone metric spaces. Journal of Mathematical Analysis and Applications, 341(1):416–420, 2008. [2] A Auwalu. Banach fixed point theorem in a cone pentagonal metric spaces. Journal of Advanced Studies in Topology, 7(2):60–67, 2016. [3] A Auwalu. Kannan fixed point theorem in a cone pentagonal metric spaces. Journal of Mathematics and Computational Sciences, 6(4):515–526, 2016. [4] A Auwalu and E Hınçal. Common fixed points of two maps in cone pentagonal metric spaces. Global Journal of Pure and Applied Mathematics, 12(3):2423–2435, 2016. REFERENCES 487 [5] A Auwalu and E Hınçal. Kannan - type fixed point theorem in cone pentagonal metric spaces. International Jounal of Pure and Applied Mathematics, 108(1):29–38, 2016. [6] A Azam, M M Arshad, and I Beg. Banach contraction principle on cone rectangular metric spaces. Applicable Analysis and Discrete Mathematics, 3(2):236–241, 2009. [7] S Banach. Sur les opérations dans les ensembles abstraits et leur application aux équations intégrales. Fundamenta Mathematicae, 3:133–181, 1922. [8] M Fŕechet. Sur quelques points du calcul fonctionnel. Rendiconti del Circolo Matem- atico di Palermo, 22:1–74, 1906. [9] M Garg and S Agarwal. Banach contraction principle on cone pentagonal metric space. Journal of Advanced Studies in Topology, 3(1):12–18, 2012. [10] R George, S Janković, K Reshma, and S Shukla. Rectangular b-metric space and contraction principles. Journal of Nonlinear Scienceand Applications, 8(6):1005–1013, 2015. [11] L Huang and X Zhang. Cone metric spaces and fixed point theorems of contractive mappings. Journal of Mathematical Analysis and Applications, 332(2):1468–1476, 2007. [12] D Ilić and V Rakoćević. Common fixed points for maps on cone metric space. Journal of Mathematical Analysis and Applications, 341(2):876–882, 2008. [13] M Jleli and B Samet. The kannans fixed point theorem in a cone rectangular metric space. Journal of Nonlinear Sciences and Applications, 2(3):161–167, 2009. [14] R Kannan. Some results on fixed points. Bulletin of Calcutta Mathematical Society, 60:71–76, 1968. [15] R Kannan. Some results on fixed points ii. American Mathematics Monthly, 76:405– 408, 1969. [16] S Patil and J Salunke. Fixed point theorems for expansion mappings in cone rectan- gular metric spaces. General Mathematics Notes, 29(1):30–39, 2015. [17] R Rashwan and S Saleh. Some fixed point theorems in cone rectangular metric spaces. Mathematica Aeterna, 2(6):573–587, 2012. [18] M Reddy and M Rangamma. A common fixed point theorem for two self maps in a cone rectangular metric space. Bulletin of Mathematics and Statistics Research, 3(1):47–53, 2015. [19] S Rezapour and R Hamlbarani. Some notes on the paper cone metric spaces and fixed point theorems of contractive mappings. Journal of Mathematical Analysis and Applications, 345(2):719–724, 2008.