EXISTENCE OF FIXED POINTS IN COMPLETE

5(1) (2010), 91–99

International Journal of Modern Mathematics c

2010 Dixie W Publishing Corporation, U. S. A.

EXISTENCE OF FIXED POINTS IN COMPLETE CONE METRIC SPACES Akbar Azam, Muhammad Arshad and Ismat Beg Received February 9, 2009; Revised April 9, 2009

A.Azam, Department of Mathematics, F.G. Postgraduate College, H-8, Islamabad-44000, Pakistan M.Arshad, Department of Mathematics, International Islamic University, H-10, Islamabad- 44000, Pakistan I. Beg, Center for Advanced Studies in Mathematics, Lahore University of Management Sciences, Lahore-54792, Pakistan Abstract We prove the existence of common fixed points of a pair of self mappings satisfying a Banach type contractive condition in complete cone metric spaces. Our results generalize several well-known results in the literature. Keywords: Fixed point, common fixed point, contractive type mapping, cone metric space. 2000 Mathematics Subject Classification:47H10,54H25, 55M20.

1

Introduction

Recently Guang and Zhang [6] generalized the notion of metric space by replacing the set of real numbers by ordered Banach space and defined cone metric space. They [6] also proved Banach contraction mapping theorem and some other fixed point theorems of contractive type mappings in cone metric spaces. Afterwords, Abbas and Jungck [1], Abbas and Rhoades [2], Azam, Arshad and Beg [4], Arshad, Azam and Vetro [3], Bari and Vetro [5] Ilic and Rakocevic [7], Vetro [10], Raja and Vaezpour [8] further studied coincidence points and common fixed points of contractive type mappings in cone metric spaces. The 91

92

Azam, Arshad and Beg

purpose of this paper is to obtain common fixed points of a pair of self mappings satisfying a generalized contractive condition in a cone metric space.

2

Preliminaries

A subset P of a real Banach space B is called a cone if it has following properties: (i) P is non-empty closed and P 6= {0}; (ii) 0 ≤ a, b ∈ R and x, y ∈ P ⇒ ax + by ∈ P ; (iii) P ∩ (−P ) = {0}. For a given cone P ⊆ B, we can define a partial ordering ≤ on B with respect to P by x ≤ y if and only if y − x ∈ P . We shall write x < y if x ≤ y and x 6= y, while x  y will stands for y − x ∈ intP , where intP denotes the interior of P. The cone P is called normal if there is a number κ ≥ 1 such that for all x, y ∈ B, 0 ≤ x ≤ y ⇒ kxk ≤ κ kyk .

(I)

The least number κ satisfying (I) is called the normal constant of P. For details we refer [6, 9]. In the following we always suppose that B is a real Banach space and P is a cone in B with intP 6= ∅ and 6 is a partial ordering with respect to P. Definition 2.1. Let X be a nonempty set. Suppose that the mapping d : X × X → B, satisfies: 1. 0 ≤ d(x, y), for all x, y ∈ X and d(x, y) = 0 if and only if x = y; 2. d(x, y) = d(y, x), for all x, y ∈ X; 3. 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. Let xn be a sequence in X, and x ∈ X. If for every c ∈ B, with 0  c there is n0 ∈ N such that for all n > n0 , d(xn , x)  c, then {xn } is said to be convergent, {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 ∈ B with 0  c there is n0 ∈ N such that for all n, m > n0 , d(xn , xm )  c, then {xn } is called a Cauchy sequence in X. If every Cauchy sequence is convergent in X, then X is called a complete cone metric space. Let us recall [9] that if P is a normal cone, then xn ∈ X converges to x ∈ X if and only if d(xn , x) → 0 as n → ∞. Furthermore, xn ∈ X is a Cauchy sequence if and only if d(xn , xm ) → 0 as n, m → ∞.

EXISTENCE OF FIXED POINTS

3

93

Fixed Point

Theorem 3.1. Let (X, d) be a complete cone metric space, P be a normal cone with normal constant κ. Suppose the mappings S, T : X → X satisfy: d(Sx, T y) ≤ A d(x, y) + B d(x, Sx) + Cd(y, T y) + D d(x, T y) + E d(y, Sx)

(II)

for all x, y ∈ X where A, B, C, D, E are non negative real numbers with A + B + C + D + E < 1, B = C or D = E. Then S and T have a unique common fixed point. Proof. Let x0 be an arbitrary point in X define x2k+1

=

Sx2k

x2k+2

=

T x2k+1 ,

k = 0, 1, 2, ....

Then, d(x2k+1 , x2k+2 )

= d(Sx2k , T x2k+1 ) ≤ A d(x2k , x2k+1 ) + B d(x2k , Sx2k ) + C d(x2k+1 , T x2k+1 ) +D d(x2k , T x2k+1 ) + E d(x2k+1 , Sx2k ) ≤

[A + B] d(x2k , x2k+1 ) + Cd(x2k+1 , x2k+2 ) + D d(x2k , x2k+2 )

≤

[A + B + D] d(x2k , x2k+1 ) + [C + D] d(x2k+1 , x2k+2 ).

It implies that [1 − C − D]d(x2k+1 , x2k+2 ) ≤ [A + B + D] d(x2k , x2k+1 ). That is, 

 A+B+D d(x2k+1 , x2k+2 ) 6 d(x2k , x2k+1 ). 1−C −D Similarly, d(x2k+2 , x2k+3 )

= ≤

d(Sx2k+2 , T x2k+1 )   A+C +E d(x2k+1 , x2k+2 ). 1−B−E

Now by induction, we obtain for each k = 0, 1, 2, ...   A+B+D d(x2k , x2k+1 ) d(x2k+1 , x2k+2 ) 6 1−C −D    A+B+D A+C +E 6 d(x2k−1 , x2k ) 1−C −D 1−B−E     A+B+D A+C +E A+B+D 6 d(x2k−2 , x2k−1 ) 1−C −D 1−B−E 1−C −D    k  A+C +E A+B+D A+B+D d(x0 , x1 ), ≤ ... 6 1−C −D 1−B−E 1−C −D

94

Azam, Arshad and Beg

and 

 A+C +E d(x2k+2 , x2k+3 ) ≤ d(x2k+1 , x2k+2 ) 1−B−E    k+1 A+C +E A+B+D ≤ ... ≤ d(x0 , x1 ). 1−B−E 1−C −D Let



A+B+D F = 1−C −D





 A+C +E ,G= . 1−B−E

In case B = C       A+B+D A+B+E A+B+D A+B+E FG = = 0 such that p < n < m and      F (F G)p (F G)p d(xm , xn ) ≤ Max (1 + G) , (1 + F ) d(x0 , x1 ). 1 − FG 1 − FG

EXISTENCE OF FIXED POINTS

95

Since, P is a normal cone with normal constant κ, therefore,       F (F G)p (F G)p kd(xn , xm )k ≤ κ Max (1 + G) , (1 + F ) kd(x0 , x1 )k . 1 − FG 1 − FG Thus,    (F G)p F (F G)p , (1 + F ) → 0 as p → ∞. Max (1 + G) 1 − FG 1 − FG 



Therefore, d(xn , xm ) → 0 as n, m → ∞. Hence { xn } is a Cauchy sequence. Since X is complete, there exists u ∈ X such that xn → u. Now, d(u, Su) ≤

d(u, x2n ) + d(x2n , Su)

≤

d(u, x2n ) + d(T x2n−1 , Su)

≤

d(u, x2n ) + A d(u, x2n−1 ) + B d(u, Su) + Cd(x2n−1 , T x2n−1 ) +D d(u, T x2n−1 ) + E d(x2n−1 , Su)

≤ d(u, x2n ) + A d(u, x2n−1 ) + B d(u, Su) + Cd(x2n−1 , x2n ) +D d(u, x2n ) + E d(x2n−1 , u) + E d(u, Su)] ≤ (1 + D) d(u, x2n ) + (A + E) d(u, x2n−1 ) + Cd(x2n−1 , x2n ) +(B + E) d(u, Su). It further implies that d(u, Su) − (B + E) d(u, Su) ≤ (1 + D) d(u, x2n ) + (A + E) d(u, x2n−1 ) +C d(x2n−1 , x2n ). That is,    1+D A+E d(u, x2n ) + d(u, x2n−1 ) 1−B−E 1−B−E   C + d(x2n−1 , x2n ). 1−B−E 

d(u, Su) ≤

Hence,    1+D A+E kd(u, Su)k ≤ κ kd(u, x2n )k + κ kd(u, x2n−1 )k 1−B−E 1−B−E   C +κ kd(x2n−1 , x2n )k . 1−B−E 

Letting n → ∞ , we have kd(u, Su)k = 0. It implies that d(u, Su) = 0 and hence u = Su. Similarly, by using d(u, T u) ≤ d(u, x2n+1 ) + d(x2n+1 , T u),

96

Azam, Arshad and Beg

we can show that u = T u. It implies that u is a common fixed point of S, T. Next we show that S and T have unique common fixed point. For this assume that, there exists another point u∗ in X such that u∗ = Su∗ = T u∗ . Now, d(u, u∗ )

= d(Su, T u∗ ) ≤ Ad(u, u∗ ) + Bd(u, Su) + Cd(u∗ , T u∗ ) + D d(u, T u∗ ) + Ed(u∗ , Su) ≤ Ad(u, u∗ ) + Bd(u, u) + Cd(u∗ , u∗ ) + D d(u, u∗ ) + Ed(u, u∗ ) ≤ (A + D + E)d(u, u∗ ).

It implies that u = u∗ , which completes the proof of the theorem. Corollary 3.1 ( [2]). Let (X, d) be a complete cone metric space, P be a normal cone with normal constant κ. Suppose the mappings S, T : X → X satisfy: d(Sx, T y) ≤ α d(x, y) + β [d(x, Sx) + d(y, T y)] + γ [d(x, T y) + d(y, Sx)] for all x, y ∈ X, where, α, β, γ are non negative real numbers with α+2 β +2γ < 1. Then S and T have a unique common fixed point.. Example 3.1. Let X = {1, 2, 3} , B = R2 and P = {(x, y) ∈ B | x, y ≥ 0} ⊂ R2 . Define d : X × X → R2 as follows:  (0, 0) if x = y     ( 5 , 5) if x 6= y and x, y ∈ X − {2} 7 d(x, y) =  (1, 7) if x 6= y and x, y ∈ X − {3}    4 ( 7 , 4) if x 6= y and x, y ∈ X − {1} . Define the mappings S, T : X → X as follows: S (x) = 1 for each x ∈ X ( T (x) =

1 3

if x 6= 2 if x = 2.

Note that 5 d(S(3), T (2)) = ( , 5). 7 Now,

EXISTENCE OF FIXED POINTS

97

α d(3, 2) + β [d(3, S(3)) + d(2, T (2))] + γ [d(3, T (2)) + d(2, S(3))] 4 = α( , 4) + β [d(3, 1) + d(2, 3)] + γ [d(3, 3) + d(2, 1)] 7   4 5 4 = α( , 4) + β ( , 5) + ( , 4) + γ [0 + (1, 7) ] 7 7 7   4α + 9β + 7γ = , 4α + 9β + 7γ 7   5α + 10β + 10γ < , 5α + 10β + 10γ 7   5 (α + 2β + 2γ) , 5 (α + 2β + 2γ) < 7 5 < ( , 5) = d(S(3), T (2) as α + 2β + 2γ < 1. 7 It follows that the mappings S and T do not satisfy the conditions of Corollary 3.2, whereas, for 5 A = B = C = D = 0, E = , 7 all the conditions of Theorems 3.1 are satisfied. Corollary 3.2. Let (X, d) be a complete cone metric space, P be a normal cone with normal constant κ. Suppose the mapping T : X → X satisfies: d(T x, T y) ≤ α d(x, y) + β [d(x, T x) + d(y, T y)] + γ [d(x, T y) + d(y, T x)] for all x, y ∈ X where α, β, γ are non negative real numbers with α + 2 β + 2γ < 1. Then T has a unique fixed point. Proof. Set S = T, in Corollary 3.2. Corollary 3.3 ( [2]). Let (X, d) be a complete cone metric space, P be a normal cone with normal constant κ. Suppose that the mapping T : X → X satisfies:. d(T x, T y) ≤ A d(x, y) + B d(x, T x) + Cd(y, T y) + D d(x, T y) + E d(y, T x) (III) for all x, y ∈ X, where, A, B, C, D, E are non negative real numbers with A + B + C + D + E < 1. Then T has a unique fixed point. Proof. Set S = T, in Theorem 3.1. Corollary 3.4 ( [2]). Let (X, d) be a complete cone metric space, P be a normal cone with normal constant κ. Suppose the mapping T : X → X satisfies: d(T x, T y) ≤ α d(x, y) + β [d(x, T x) + d(y, T y)] for all x, y ∈ X, where, α, β ≥ 0 with α + 2 β < 1. Then T has a unique fixed point.

98

Azam, Arshad and Beg

Corollary 3.5 ( [2, 6]). Let (X, d) be a complete cone metric space, P be a normal cone with normal constant κ. Suppose the mapping T : X → X satisfies: d(T x, T y) ≤ α d(x, y) for all x, y ∈ X where 0 ≤ α < 1. Then T has a unique fixed point. Corollary 3.6. Let (X, d) be a complete cone metric space, P be a normal cone with normal constant κ. Suppose the mapping T : X → X satisfies: d(T x, T y) ≤ β [d(x, T x) + d(y, T y)] for all x, y ∈ X where 0 6 β < 21 .Then T has a unique fixed point. Corollary 3.7. Let (X, d) be a complete cone metric space, P be a normal cone with normal constant κ. Suppose the mapping T : X → X satisfies: d(T x, T y) ≤ γ [d(x, T y) + d(y, T x)] for all x, y ∈ X, where, 0 6 γ < 21 .Then T has a unique fixed point. Example 3.2. Let X = {1, 2, 3} , B = R2 and P = {(x, y) ∈ B | x, y ≥ 0} ⊂ R2 . Define d : X × X → R2 as follows:  (0, 0)     (4, 2) 7 7 d(x, y) = 1  (1, )    1 21 (2, 4)

if x = y if x 6= y and x, y ∈ X − {2} if x 6= y and x, y ∈ X − {3} if x 6= y and x, y ∈ X − {1}.

Define a mapping T : X → X as follows: ( T (x) =

3 1

if x 6= 2 if x = 2.

Note that 4 2 d(T (3), T (2) = d(3, 1) = ( , ). 7 7 Now,

99

EXISTENCE OF FIXED POINTS

α d(3, 2) + β [d(3, T (3)) + d(2, T (2))] + γ [d(3, T (2)) + d(2, T (3))] = = = = < <