1. Introduction

In 2007, Huang and Zhang ^[5] introduced the concept of a cone metric space and proved some fixed point theorems in cone metric space. Later on, many authors have generalized and extended the fixed point theorems of Huang and Zhang ^[5]. Fixed point theorems in partially ordered set was studied by Ran and Reurings ^[9], Nieto and Lopez ^[8]. Subsequently, many authors (see, e. g., ^{[1, 2, 6]}) were investigated the fixed point results on ordered metric spaces. Altun and Durmaz ^[4], Altun , Damnjanovic and Djoric ^[3] obtained fixed point theorems in ordered cone metric spaces. Recently, Kadelburg, Pavlovic and Radenovic ^[7] proved some common fixed point theorems in ordered contractions and quasicontractions in ordered cone metric spaces. In this paper, we proved a common fixed point theorem in ordered cone metric spaces without using the continuity. Our result, generalizes the results of ^[7].

The following definitions are in ^[5].

Definition 1.1. ^[5] Let E be a real Banach space and P be a subset of E. The set P is called a cone if and only if:

(a). P is closed, non–empty and P {0};

(b). a, b , a,b , x,y P imply ax+by P;

(c). x P and –x P implies x = 0.

Definition 1.2.^[5] Let P be a cone in a Banach space E, define partial ordering with respect to P by if and only if y-xP. We shall write x y to indicate but x y while x y will stand for y-x int P, where int P denotes the interior of the set P. This cone P is called an order cone.

Definition 1.3.^[5] Let E be a Banach space and PE be an order cone. The order cone P is called normal if there exists L>0 such that for all x, yE,

The least positive number L satisfying the above inequality is called the normal constant of P.

Most of ordered Banach spaces used in applications posses a cone with the normal constant K = 1.

Definition 1.4. ^[5] Let X be a nonempty set of E. Suppose that the map d: X X→ E satisfies:

(d1). 0 d(x, y) for all x, y X and d(x, y) = 0 if and only if x = y;

(d2). d(x, y) = d(y, x) for all x, y X;

(d3). d(x, y) d(x, z) + d(z, y) for all x, y, z X.

Then d is called a cone metric on X and (X, d) is called a cone metric space.

Remark 1.5. ^[7] (1) If u v and v w , then u w.

(2) If 0u c for each c int P, then u = 0.

(3) If a b + c for each c int P, then a b.

(4) If 0 x y and 0 ≤ a , then 0 ax ay .

(5) If 0 x_n y_n , for each n, and , then 0 x y.

(6) If 0 d(x_n, y_n) b_n and b_n → 0, then, d(x_n,x)c where x_n , x are respectively, a sequence and a given point in X.

(7) If E is a real Banach space with a cone P and if a λa where a P and 0 < λ < 1, then a = 0.

(8) If c int P, 0 a_n and a_n → 0, then there exists n₀ such that for all n > n₀ we have a_n c.

2. Main Result

In this section, we prove a common fixed point theorem in an ordered complete cone metric spaces.

Theorem 2.1. Let (X, , d) be an ordered complete cone metric cone space. Let (f, g) be weakly increasing pair of self-maps on X w. r. t. . Suppose that the following conditions hold:

(i) there exists p, q, r, s, t ≥ 0 satisfying p + q + r + s + t < 1 and q = r or s = t, such that

(1)

for all comparable x, yX;

(ii) if a nondecreasing sequence {x_n} converges to xX, then x_n x for all n. Then, f and g have a common fixed point in X.

Proof. Let x₀ X be arbitrary and define a sequence {x_n} by x_2n+1 = fx_2n and x_2n+2 = gx_2n+1 for all nN. Since, (f, g) is weakly increasing , it can be easily shown that the sequence {x_n} is nondecreasing w. r. t. , that is, x₀x₁…x_nx_n+1…. In particular, x_2n and x_2n+1 are comparable, by (1) we have

It follows that

That is,

(2)

Similarly, we obtain

From (1) and (2), by induction, we obtain that

and

Let

In the case q = r,

Now, for n < m we have

Similarly, we obtain

and

Hence, for n < m

where b_n → 0, as n→∞.

By using (8) and (1) of Remark 1.5 and only the assumption that the underlying cone is solid, we conclude that {x_n} is a Cauchy sequence.

Since (X, d) is complete, there exists uX such that x_n →u (as n→∞).

Letting n→+ ∞

(3)

Let c 0 be given. Choose a natural number N₁ such that d(u, gu) c. Then from (3) we get that d(fu, u) c.

Since c is arbitrary, we get that

Noting that → 0 as m→∞, we conclude that as m→∞.

Hence, P is closed, then - d(fu, u) P.

Thus d(fu, u) P (-P). Hence d(fu, u) = 0.

Therefore, fu = u.

And

Letting n→+ ∞

That is, fu = gu.

Now we show that fu = gu = u. By (1), we have

Letting n→+ ∞

Therefore, fu = gu = u and u is a common fixed point of f and g.

Now, we consider the case when condition (ii) is satisfied. For the sequence {x_n} we have x_n → u X(as n→∞) and x_n u(n). By the construction, fx_n → u and gx_n → u(as n→∞).

Let us prove that u is a common fixed point of f and g. Putting x = u and y = x_n in (1)(since they are comparable) we get that

For the first and fourth term of the right hand side we have d(x_n,u)c and d(u, gx_n )c( for cint P arbitrary and n ≥ n₀). For the second term d(u _,f u) ≼ d(u _,x_n) + d(x_{n ,}gx_n) + d( gx_n , fu)(again the first term n the right can be neglected) and for the fifth term d(x_{n ,}f u) ≼ d( x_n , gx_n) + d(gx_{n ,} fu). It follows that

But x_n → u and gx_n → u ⇒ d(x_n, gx_n) c, which means that d(fu, gx_n)<<c, that is, gx_n →fu. It follows that, fu = u and in a symmetric way ( by using that u⊑u), gu =u.

Remark 2.2. If we choose f and g are continuous mappings in the above Theorem 2.1, then we get the Theorem 2.1 of ^[7].

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