Common ﬁxed point of mappings satisfying rational inequalities in ordered complex valued generalized metric spaces

Azam et al. (Numer Funct Anal Optim, 32(3):243–253, 2011) introduced a notion of complex valued metric space and obtained common ﬁxed point result for mappings in such spaces. In this paper, introducing the concept of complex valued generalized metric spaces, some ﬁxed and common ﬁxed point results are obtained. As an application, periodic point property of common ﬁxed point problem for two rational type contractive mappings involved therein is established.

conditions to establish the existence and uniqueness of a solution of an operator equation T x = x. This principal is constructive in nature and is one of the most useful tools in the study of nonlinear equations. There are many generalizations of the Banach's contraction mapping principle in the literature. These generalization were made either by using the contractive condition or by imposing some additional conditions on an ambient space. There have been a number of generalizations of metric spaces such as, rectangular metric spaces, pseudo metric spaces, fuzzy metric spaces, quasi metric spaces, partial metric spaces, probabilistic metric spaces, D-metric spaces and cone metric spaces (see [1,10,15,20,25]). Branciari [8] introduced the notion of a generalized metric space replacing the triangle inequality by a rectangular type inequality. He then extended Banach's contraction principle in such spaces. In 2008, Azam et al. [6] obtained sufficient conditions for existence of unique fixed point of Kannan type mappings defined on generalized metric spaces. Samet [28] and Sarma et al. [29] showed that some propositions in [8] are not true. Moreover, in [29], a rigorous and nice proof of the Banach's contraction principle is presented, by assuming that the generalized metric space is Hausdorff. Common fixed point problem for two maps under several variants of non-commutativity has been studied by many authors. Recently, Azam et al. [7] obtained the generalization of Banach's contraction principal introducing the concept of a complex valued metric space. The existence of fixed points in ordered metric spaces has been initiated in 2004 by Ran and Reurings [22], and further studied by Nieto and Lopez [18]. Several authors have studied the problem of existence and uniqueness of a fixed point for mappings satisfying different contractive conditions in the framework of partially ordered metric spaces (see for example [2][3][4][5]9,12,21,24]).
The aim of this paper is to introduce the concept of a complex valued generalized metric space and to study the fixed and common fixed point results for two mappings satisfying rational inequalities, without exploiting any type of commutativity condition. The results presented in this paper substantially extend and strengthen the results given in [7] and [23].
Consistent with Azam et al. [7] and [23], the following definitions and results will be needed in the sequel.
Let C be the set of complex numbers and let z 1 , z 2 ∈ C. Define a partial order ≤ on C as follows: z 1 ≤ z 2 if and only if Re (z 1 ) ≤ Re(z 2 ), Im(z 1 ) ≤ Im(z 2 ).
In particular, we will write z 1 ≤ z 2 if one of (1), (2) and (3) is satisfied and we will write Some elementary properties of the partial order ≤ on C are the following: (iii) If z 1 ≤ z 2 and r ≥ 0 is a real number, then r z 1 ≤ r z 2 .
for all x, y ∈ X and all distinct u, v ∈ X each one is different from x and y.
Then d is called a complex valued generalized metric on X and (X, d) is called a complex valued generalized metric space.
It is easy to verify that (X, d) is a complex valued generalized metric space when θ ∈ [0, π 2 ]. Note that So d is not a complex valued metric.
The next example gives a method to construct a new complex valued generalized metric space from a family of given generalized metric spaces. Example 1.3 Let {(X n , d n ) : n ∈ K ⊂ N} be a family of disjoint complex valued generalized metric spaces and let X = {X n : n ∈ K }. Define, for all x, y ∈ X , a mapping d : , if x, y ∈ X n , for some n ∈ K 1, if x ∈ X n , y ∈ X m for some m, n ∈ K , m = n.
Clearly, (X, d) is a complex valued generalized metric space. Let X be a complex valued generalized metric space and A ⊆ X. A point x ∈ X is called an interior point of a set A whenever there exists 0 A point x ∈ X is called a limit point of A whenever for every 0 < r ∈ C, B(x, r )∩( A\x) = φ. A subset B ⊆ X is called closed whenever each limit point of B belongs to B.
Let {x n } be a sequence in X and x ∈ X. If for every c ∈ C, with 0 < c there is n 0 ∈ N such that for all n > n 0 , d (x n , x) < c, then x is called the limit of {x n } and we write lim n→∞ x n = x or x n → x as n → ∞. If for every c ∈ C, with 0 < c, there is an n 0 ∈ N such that for all n, m > n 0 , d(x n , x m ) < c, then {x n } is called a Cauchy sequence in X. If every Cauchy sequence is convergent in X , then it is called a complete complex valued generalized metric space.  The following definition is due to Altun([3]).
Definition 1.6 [3] Let (X, ) be a partially ordered set. A pair ( f, g) of self-maps of X is said to be weakly increasing if f x g f x and gx f gx for all x ∈ X . If f = g, then we have f x f 2 x for all x in X and in this case, we say that f is a weakly increasing map.
Example 1.7 Let X = R + be endowed with usual ordering. Let f : X → X be defined by where f and g are two self-maps on X. A common fixed point problem is to find some x in X such that x = f x = gx, and we denote it by C F P( f, g, X ). A nonempty subset W of a partially ordered set X is said to be totally ordered if every two elements of W are comparable.

Main Results
In this section, we prove a common fixed point result for weakly increasing maps on an ordered complex valued generalized metric space.

Theorem 2.1 Let (X, ) be a partially ordered set such that there exists a complete complex valued generalized metric d on X and (S, T ) a pair of weakly increasing self-maps on X.
Suppose that, for every comparable x, y ∈ X we have either

If S or T is continuous or for any nondecreasing sequence x n with x n → z in X we necessarily have x n z for all n ∈ N, then S and T have a common fixed point. Moreover, the set of common fixed points of S and T is totally ordered if and only if S and T have one and only one common fixed point.
Proof First we shall show that if S or T has a fixed point, then it is a common fixed point of S and T . Let u be a fixed point of S. Then from (2.1) with x = y = u, we have for u = T u : Now let x 0 be an arbitrary point of X. If Sx 0 = x 0 , then the proof is finished. Assume that Sx 0 = x 0 . Construct a sequence {x n } in X as follows: Continuing this way, we have If neither S, nor T is continuous, then by given assumption x n u for all n ∈ N. We claim that u is a fixed point of S. If not, then d(u, Su) = z > 0. From (2.1), we obtain and so which on taking limit as n → ∞ gives |z| ≤ a 3 |z| , a contradiction, and so u = Su. Therefore Su = T u = u. Now suppose that set of common fixed points of S and T is totally ordered. We prove that common fixed point of S and T is unique. Assume on contrary that u and v are distinct common fixed points of S and T . By supposition, we can replace x by u and y by v in (2.1;) to obtain which implies that |d(u, v)| ≤ (a 1 + a 2 2 + a 5 ) |d(u, v)| , a contradiction. Hence u = v. Conversely, if S and T have only one common fixed point then the set of common fixed point of S and T being singleton is totally ordered.
Although we studied a common fixed point problem for two mappings to consider a more general result, we could use even one and yet the result would have been new. In Theorem 2.1, take S = T , to obtain the following corollary.

Corollary 2.2
Let (X, ) be a partially ordered set such that there exists a complete complex valued generalized metric d on X and let T be a weakly increasing self-map on X. Suppose that, for every comparable x, y ∈ X, either

If S or T is continuous or for a nondecreasing sequence {x n } with x n → z in X we necessarily have x n z for all n ∈ N. Then S and T have a common fixed point. Moreover, the set of common fixed points of S and T is totally ordered if and only if S and T have one and only one common fixed point.
Proof First we shall show that if S or T has a fixed point, then it is a common fixed point of S and T . Indeed, let u be a fixed point of S. Then from (2.5) with x = y = u, we have for u = T u :

= a[d(u, u)d(u, T u) + d(u, T u)d(u, u)] d(u, T u) + d(u, u)
Hence d(u, T u) = 0 and so u is a common fixed point of S and T . Similarly if u is a fixed point of T , then it is also fixed point of S. Now let x 0 be an arbitrary point of X. If Sx 0 = x 0 , then the proof is finished. Assume that Sx 0 = x 0 . Define a sequence x n in X as follows: Continuing this process we have, x 1 x 2 · · · x n x n+1 · · · . We may assume that d(x 2n , x 2n+1 ) > 0, for every n ∈ N. If not, then x 2n = x 2n+1 for some n and for all those n, x 2n = x 2n+1 = Sx 2n and proof is finished. Now, since x 2n and x 2n+1 are comparable and taking d(x 2n , x 2n+1 ) > 0 for n = 0, 1, 2, 3, . . . , we have Therefore, |d(x n , x m )| ≤ a n 1−a |d(x 0 , x 1 )| , and so |d(x n , x m )| → 0, as m, n → ∞. It follows that x n is a Cauchy sequence in X . Since X is complete, the sequence x n converges to a point u in X . If S or T is continuous, then it is clear that Su = u = T u.
If neither S, nor T is continuous, then by our assumption x n u for all n in N. We claim that u is a fixed point of S. If not, then d(u, Su) = z > 0. Now from (2.5), we have (x n+1 , x n+2 ) .
Taking the limit as n → ∞ gives |z| ≤ 0, a contradiction and so u = Su. Therefore Su = T u = u. Now suppose that set of common fixed points of S and T is totally ordered. We claim that common fixed point of S and T is unique. Assume on the contrary that u and v are distinct common fixed points of S and T . By supposition, we can replace x by u and y by v in (2.5) to obtain Conversely, if S and T have only one common fixed point then the set of common fixed point of S and T being a singleton is totally ordered.
In Theorem 2.3, take S = T , to obtain the following corollary. It is straightforward to check that (X, d) is a complex valued generalized metric space for θ ∈ [0, π 2 ]. Note that

Corollary 2.4 Let (X, ) be a partially ordered set such that there exists a complete complex valued generalized metric d on X and T be weakly increasing self-map on X. Suppose that, for every comparable x, y ∈ X, either d(T x, T y) ≤ a[d(x, T x)d(x, T y) + d(y, T y)d(y, T x)] d(x, T y) + d(y, T x)
is not a complex valued metric space. We define S, T : X → X by Sx = 1 f or x ∈ X and T x = 1, for x ∈ {1, 2, 3} 2, for x = 4.
Note that Sx T Sx and T x ST x for all x ∈ X. Now for a = 3 5 and b = 1 5 , we consider the following cases: (i) If x ∈ X and y ∈ X \4, then we have Sx = T x = 1 and so d(Sx, T y) = 0 and (2.5) is satisfied obviously. (ii) When x = 1and y = 4, then Sx = 1, T y = 2 and (iii) For x = 2 and y = 4, Sx = 1, T y = 2 and

Periodic point results
A fixed point p of T is also a fixed point of T n for every n ∈ N. However, the converse is false. For example, consider, X = [0, 1], and define T by T x = 1 − x. Then T has a unique fixed point 1 2 and every even iterate of T is the identity map, which has every point of [0, 1] as a fixed point. On the other hand, if X = [0, π], T x = cos x, then every iterate of T has the same fixed point as T (cf. [1,11,16]).
If a map T satisfies F(T ) = F(T n ) for each n ∈ N, where F (T ) is the set of fixed point of T , then it is said to have property P [16]. The set O(x, ∞) = x, T x, T 2 x, ... is called the orbit of x. Put λ = (a 3 +a 5 ) 1−a 4 . Obviously 0 ≤ λ < 1 and we have d(u, T u) = d(T u, T n u) ≤ λd(T n−1 u, T n u) ≤ λ 2 d(T n−2 u, T n−1 u) ≤ · · · ≤ λ n d(u, T u).

Thus we have d(u, T u) = d(T u, T n u)
≤ ad(T n−1 u, T n u) ≤ a 2 d(T n−2 u, T n−1 u) ≤ · · · ≤ a n d(u, T u).

BA-cone metric spaces
According to recent paper of Osturk and Basaris [19], it follows that complex valued metric space introduced in [7] and [23] is in fact BA-cone metric space, that is., cone metric space over a solid cone in commutative division Banach algebra B. Recall that a division algebra is an algebra with identity e, in which every nonzero element has an inverse called a unit, that is., x is an unit if and only if there exists an inverse y such that x y = yx = e. Then we write y = x −1 and observe that x −1 is unique when it exists (see [17] and [26]).*** and f : [0, 1] → R, f (t) = e t . If 2b > a 2 then B is not Banach algebra, that is., (X, d) is not a BA-cone metric space.
It is worth to notice that all results from [7,14,23,[30][31][32] and [33] are true and in the context of BA-cone metric spaces over normal solid and monotone cone. For example, Theorem 4. from [7] in this new context has the following form. for all x, y ∈ X, where λ, μ are nonnegative real with λ + μ < 1. Then S, T have a unique common fixed point.
The proof for uniqueness is as in [7].
Further, Theorem 2.1 from [23] in the context of BA-cone metric spaces has the following form. The proof is similarly to the previous one and therefore we omit it. for all x, y ∈ X, where λ, μ, γ are nonnegative reals with λ + μ + γ < 1. Then S, T have a unique common fixed point. Open Access This article is distributed under the terms of the Creative Commons Attribution License which permits any use, distribution, and reproduction in any medium, provided the original author(s) and the source are credited.