International Journal of Analysis and Applications
ISSN 2291-8639
Volume 3, Number 2 (2013), 112-118
http://www.etamaths.com

COMMON FIXED POINT THEOREM IN CONE METRIC SPACE

FOR RATIONAL CONTRACTIONS

R. UTHAYAKUMAR AND G. AROCKIA PRABAKAR∗

Abstract. In this paper we prove the common fixed point theorem in cone

metric space for rational expression in normal cone setting. Our results gen-

eralize the main result of Jaggi [10] and Dass, Gupta [11].

1. Introduction

The Banach contraction principle with rational expressions have been expanded
and some fixed and common fixed point theorems have been obtained in [1], [2].
Huang and Zhang [3] initiated cone metric spaces, which is a generalization of
metric spaces, by substituting the real numbers with ordered Banach spaces. They
have considered convergence in cone metric spaces, introduced completeness of cone
metric spaces, and proved a Banach contraction mapping theorem, and some other
fixed point theorems involving contractive type mappings in cone metric spaces
using the normality condition. Later, various authors have proved some common
fixed point theorems with normal and non-normal cones in these spaces [4], [5], [6],
[7], [8]. Quite recently Muhammad arshad et al.[9] have introduced almost Jaggi
and Gupta contraction in Partially ordered metric space to prove the fixed point
theorem. In this paper we prove the common fixed point theorem in cone metric
space for rational expression in normal cone setting. Our results generalize the
main result of Jaggi [10] and Dass, Gupta [11].

1.1. Basic facts and definitions. Let E be a real Banach space and P a 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, x,y ∈ P ⇒ ax + by ∈ P
(iii) P ∩ (−P) = {0}
Given a cone P ⊂ E, we define a partial ordering ≤ with respect to P by x ≤ y if
and only if y −x ∈ P . We shall write x < y indicate that x ≤ y but x 6= y, while
x � y will stand for y −x ∈ intP , intP denotes the interior of P .
The cone P is called normal if there is a number M > 0 such that for all x,y ∈ E,
0 ≤ x ≤ y implies ‖x‖≤ M ‖y‖.
The least positive number satisfying above is called the normal constant of P .
In the following we always suppose E is a Banach space, P is a cone in E with
intP 6= ∅ and ≤ is partial ordering with respect to P .

2010 Mathematics Subject Classification. 54E35, 54E50, 37C25, 54H25.

Key words and phrases. Cone metric space; Rational expression; Fixed point.

c©2013 Authors retain the copyrights of their papers, and all
open access articles are distributed under the terms of the Creative Commons Attribution License.

112



COMMON FIXED POINT THEOREM IN CONE METRIC SPACE 113

Definition 1.1. [3] Let X be a nonempty 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 all x,y ∈ X;
(iii) 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.

Example 1.1. [3] Let E = R2, P = {(x,y) ∈ E|x,y ≥ 0} ⊂ R2, X = R and
d : X ×X → E such that d (x,y) = (|x−y| ,α |x−y|), where α ≥ 0 is a constant.
Then (X,d) is a cone metric space.

Definition 1.2. [3] Let (X,d) be a cone metric space, x ∈ X and {xn} be a
sequence in X. Then
(i) {xn} converges to x whenever for every c ∈ E with 0 � c there is a natural
number N such that d (xn,x) � c for all n ≥ N.
(ii) {xn} is a cauchy sequence whenever for every c ∈ E with 0 � c there is a
natural number N such that d (xn,xm) � c for all n,m ≥ N.

Definition 1.3. [3] Let (X,d) is said to be a complete cone metric space, if every
cauchy sequence is convergent in X.

2. Main Results

Definition 2.1. [9] Let (X,d) be a cone metric space. A self mapping T on X is
called an almost Jaggi contraction if it satisfies the following condition:

(1) d (Tx,Ty) ≤
αd (x,Tx) d (y,Ty)

d (x,y)
+ βd (x,y) + Lmin{d (x,Ty) ,d (y,Tx)}

for all x,y ∈ X, where L ≥ 0 and α,β ∈ [0, 1) with α + β < 1.

Theorem 2.1. Let (X,d) be a complete cone metric space and P a normal cone
with normal constant M. Let T : X → X be an almost Jaggi contraction, for all
x,y ∈ X where L ≥ 0 and α,β ∈ [0, 1) with α + β < 1. Then T has a unique fixed
point in X.

Proof :
Choose x0 ∈ X. Set x1 = Tx0, xn = Txn−1

d (xn,xn+1) = d (Txn−1,Txn)

≤
[
αd (xn−1,Txn−1) d (xn,Txn)

d (xn−1,xn)
+ βd (xn−1,xn)

+Lmin{d (xn−1,Txn) ,d (xn,Txn−1)}]

≤
[
αd (xn−1,xn) d (xn,xn+1)

d (xn−1,xn)
+ βd (xn−1,xn)

+Lmin{d (xn−1,xn+1) ,d (xn,xn)}]
≤ d (xn,xn+1) + βd (xn−1,xn)

(1 −α) d (xn,xn+1) ≤ βd (xn−1,xn)

d (xn,xn+1) ≤
β

1 −α
d (xn−1,xn)



114 UTHAYAKUMAR AND PRABAKAR

k = β
1−α, α + β < 1 0 < k < 1

and by induction,

d (xn,xn+1) ≤ kd (xn−1,xn)
.

.

≤ knd (x0,x1)

d (xn,xm) ≤ d (xn,xn+1) + d (xn+1,xn+2) + ...... + d (xn+m−1,xm)
≤

(
kn + kn+1 + ..... + kn+m−1

)
d (x0,x1)

≤
kn

1 −k
d (x0,x1)

We get ‖d (xn,xm)‖ ≤ M k
n

1−k ‖d (x0,x1)‖ which implies that d (xn,xm) → 0 as
n → ∞. Hence xn is a cauchy sequence, so by completeness of X this sequence
must be convergent in X.

d (u,Tu) ≤ d (u,xn+1) + d (xn+1,Tu)
≤ d (u,xn+1) + d (Txn,Tu)

≤ d (u,xn+1) +
αd (xn,Txn) d (u,Tu)

d (xn,u)
+ βd (xn,u)

+Lmin{d (xn,Tu) ,d (u,Txn)}

≤ d (u,xn+1) +
αd (xn,xn+1) d (u,u)

d (xn,u)
+ βd (xn,u)

+Lmin{d (xn,u) ,d (u,xn+1)}
≤ d (u,xn+1) + βd (xn,u) + Lmin{d (xn,u) ,d (u,xn+1)}

So using the condition of normality of cone

‖d (u,T (u))‖≤ M (‖d (u,xn+1)‖ + β‖d (xn,u)‖ + Lmin‖d (xn,u) ,d (u,xn+1)‖)

As n → 0 we have ‖d (u,T (u))‖≤ 0. Hence we get u = Tu, u is a fixed point of T.

Definition 2.2. [10] Let (X,d) be a cone metric space. A self mapping T on X is
called Jaggi contraction if it satisfies the following condition:

(2) d (Tx,Ty) ≤
αd (x,Tx) d (y,Ty)

d (x,y)
+ βd (x,y)

for all x,y ∈ X and α,β ∈ [0, 1) with α + β < 1.

Corollary 2.1. Let (X,d) be a complete cone metric space and P a normal cone
with normal constant M. Let T : X → X be a Jaggi contraction

(3) d (Tx,Ty) ≤
αd (x,Tx) d (y,Ty)

d (x,y)
+ βd (x,y)

for all x,y ∈ X and α,β ∈ [0, 1) with α + β < 1. Then T has a unique fixed point
in X.
Proof: Set L = 0 in theorem 2.1.



COMMON FIXED POINT THEOREM IN CONE METRIC SPACE 115

Theorem 2.2. Let (X,d) be a complete cone metric space and P a normal cone
with normal constant M. Suppose the mappings S,T is called an almost Jaggi
contraction if it satisfies the following condition:

(4) d (Sx,Ty) ≤
αd (x,Sx) d (y,Ty)

d (x,y)
+ βd (x,y) + Lmin{d (x,Ty) ,d (y,Sx)}

for all x,y ∈ X where L ≥ 0 and α,β ∈ [0, 1) with α + β < 1. Then each of S,T
has a unique fixed point and these two fixed points coincide.

Proof :
Let x1 ∈ S (xo) and x2 = T (x1) such that x2n+1 = S (x2n), x2n+2 = T (x2n+1)

d (x2n+1,x2n+2) = d (Sx2n,Tx2n+1)

≤
[
αd (x2n,Sx2n) d (x2n+1,Tx2n+1)

d (x2n,x2n+1)
+ βd (x2n,x2n+1)

+Lmin{d (x2n,Tx2n+1) ,d (x2n+1,Sx2n)}]

≤
[
αd (x2n,x2n+1) d (x2n+1,x2n+2)

d (x2n,x2n+1)
+ βd (x2n,x2n+1)

+Lmin{d (x2n,x2n+2) ,d (x2n+1,x2n+1)}]
d (x2n+1,x2n+2) ≤ αd (x2n+1,x2n+2) + βd (x2n,x2n+1)

(1 −α) d (x2n+1,x2n+2) ≤ βd (x2n,x2n+1)

d (x2n+1,x2n+2) ≤
β

1 −α
d (x2n,x2n+1)

d (x2n+1,x2n+2) ≤ kd (x2n,x2n+1)(5)

where k = β
1−α, α + β < 1

d (x2n+3,x2n+2) = d (S (x2n+2) ,T (x2n+1))

≤
[
αd (x2n+2,Sx2n+2) d (x2n+1,Tx2n+1)

d (x2n+2,x2n+1)
+ βd (x2n+2,x2n+1)(6)

+Lmin{d (x2n+2,Tx2n+1) ,d (x2n+1,Sx2n+2)}]

≤
[
αd (x2n+2,x2n+3) d (x2n+1,x2n+2)

d (x2n+2,x2n+1)
+ βd (x2n+2,x2n+1)

+Lmin{d (x2n+2,x2n+2) ,d (x2n+1,x2n+2)}]
d (x2n+3,x2n+2) ≤ αd (x2n+2,x2n+3) + βd (x2n+2,x2n+1)

(1 −α) d (x2n+3,x2n+2) ≤ βd (x2n+2,x2n+1)

d (x2n+3,x2n+2) ≤
β

1 −α
d (x2n+2,x2n+1)

d (x2n+3,x2n+2) ≤ kd (x2n+2,x2n+1)

k = β
1−α, α + β < 1

Add equation (6) and (7) we get

∞∑
n=1

d (xn,xn+1) ≤
∞∑
n=1

knd (x0,x1)(7)

=
k

1 −k
d (x0,x1)



116 UTHAYAKUMAR AND PRABAKAR

We get ‖d (xn,xn+1)‖≤ M k1−k ‖d (x0,x1)‖ which implies that d (xn,xn+1) → 0 as
n → ∞. Hence {xn} is a cauchy sequence, so by completeness of X this sequence
must be convergent in X. We shall prove that u is a common fixed point of S and
T .

d (u,Tu) ≤ d (u,x2n+1) + d (x2n+1,Tu)
≤ d (u,x2n+1) + d (Sx2n,Tu)

≤ d (u,x2n+1) +
[
αd (x2n,Sx2n) d (u,Tu)

d (x2n,u)
+ βd (x2n,u)

+Lmin{d (x2n,Tu) ,d (u,Sx2n)}]

≤ d (u,x2n+1) +
[
αd (x2n,x2n+1) d (u,u)

d (x2n,u)
+ βd (x2n,u)

+Lmin{d (x2n,u) ,d (u,x2n+1)}]
≤ d (u,x2n+1) + βd (x2n,u) + Lmin{d (x2n,u) ,d (u,x2n+1)}

So using the condition of normality of cone

‖d (u,T (u))‖≤ M (‖d (u,x2n+1)‖ + β‖d (x2n,u)‖ + Lmin‖d (x2n,u) ,d (u,x2n+1)‖)

As n → 0 we have ‖d (u,T (u))‖≤ 0. Hence we get u = Tu, u is a fixed point of T .
Similarly

d (u,S (u)) ≤ d (u,x2n+2) + d (x2n+2,Su)
≤ d (u,x2n+2) + d (Su,Tx2n+1)

≤ d (u,x2n+2) +
[
αd (u,Su) d (x2n+1,Tx2n+1)

d (u,x2n+1)
+ βd (u,x2n+1)

+Lmin{d (u,Tx2n+1) ,d (x2n+1,Su)}]

≤ d (u,x2n+2) +
[
αd (u,u) d (x2n+1,x2n+2)

d (u,x2n+1)
+ βd (u,x2n+1)

+Lmin{d (u,x2n+2) ,d (x2n+1,u)}]
≤ d (u,x2n+2) + βd (u,x2n+1) + Lmin{d (u,x2n+2) ,d (x2n+1,u)}

So using the condition normality of cone

‖d (u,S (u))‖≤ M (‖d (u,x2n+1)‖ + β‖d (u,x2n+1)‖ + Lmin‖d (u,x2n+2) ,d (x2n+1,u)‖)

As n → 0 we have ‖d (u,S (u))‖≤ 0. Hence we get u = Su, u is a fixed point of S.

Definition 2.3. [9] Let (X,d) be a cone metric space. A self mapping T on X is
called Dass and Gupta contraction if it satisfies the following condition:
(8)

d (Tx,Ty) ≤
αd (y,Ty) [1 + d (x,Tx)]

1 + d (x,y)
+βd (x,y)+Lmin{d (x,Tx) ,d (x,Ty) ,d (y,Tx)}

for all x,y ∈ X, where L ≥ 0 and α,β ∈ [0, 1) with α + β < 1.

Theorem 2.3. Let (X,d) be a complete cone metric space and P a normal cone
with normal constant M. Let T : X → X be a Dass and Gupta contraction, for all
x,y ∈ X where L ≥ 0 and α,β ∈ [0, 1) with α + β < 1. Then T has a unique fixed
point in X.



COMMON FIXED POINT THEOREM IN CONE METRIC SPACE 117

Proof :
Choose x0 ∈ X. Set x1 = Tx0, xn = Txn−1

d (xn,xn+1) = d (Txn−1,Txn)

≤
[
αd (xn,Txn) [1 + d (xn−1,Txn−1)]

1 + d (xn−1,xn)
+ βd (xn−1,xn)

+Lmin{d (xn−1,Txn−1) ,d (xn−1,Txn) ,d (xn,Txn−1)}]

≤
[
αd (xn,xn+1) [1 + d (xn−1,xn)]

1 + d (xn−1,xn)
+ βd (xn−1,xn)

+Lmin{d (xn−1,xn) ,d (xn−1,xn+1) ,d (xn,xn)}]
(1 −α) d (xn,xn+1) ≤ βd (xn−1,xn)

d (xn,xn+1) ≤
β

1 −α
d (xn−1,xn)

k = β
1−α, α + β < 1 0 < k < 1

and by induction,

d (xn,xn+1) ≤ kd (xn−1,xn)
.

.

≤ knd (x0,x1)
d (xn,xm) ≤ d (xn,xn+1) + d (xn+1,xn+2) + ...... + d (xn+m−1,xm)

≤
(
kn + kn+1 + ..... + kn+m−1

)
d (x0,x1)

≤
kn

1 −k
d (x0,x1)

We get ‖d (xn,xm)‖ ≤ M k
n

1−k ‖d (x0,x1)‖ which implies that d (xn,xm) → 0 as
n → ∞. Hence xn is a cauchy sequence, so by completeness of X this sequence
must be convergent in X.

d (u,T (u)) ≤ d (u,xn+1) + d (xn+1,Tu)
≤ d (u,xn+1) + d (Txn,Tu)

≤ d (u,xn+1) +
αd (u,Tu) [1 + d (xn,Txn)]

1 + d (xn,u)
+ βd (xn,u)

+Lmin{d (xn,Txn) ,d (xn,Tu) ,d (u,Txn)}

≤ d (u,xn+1) +
αd (u,u) [1 + d (xn,xn+1)]

1 + d (xn,u)
+ βd (xn,u)

+Lmin{d (xn,xn+1) d (xn,u) ,d (u,xn+1)}
≤ d (u,xn+1) + βd (xn,u) + Lmin{d (xn,xn+1) ,d (xn,u) ,d (u,xn+1)}

So using the condition normality of cone

‖d (u,T (u))‖≤ M (‖d (u,xn+1)‖ + β‖d (xn,u)‖ + Lmin‖d (xn,xn+1) ,d (xn,u) ,d (u,xn+1)‖)
As n → 0 we have ‖d (u,T (u))‖≤ 0. Hence we get u = Tu, u is a fixed point of T .
Corollary 2.2. Let (X,d) be a complete cone metric space and P a normal cone
with normal constant M. Let T : X → X a Dass, Gupta rationl contraction

(9) d (Tx,Ty) ≤
αd (y,Ty) [1 + d (x,Tx)]

1 + d (x,y)
+ βd (x,y)



118 UTHAYAKUMAR AND PRABAKAR

for all x,y ∈ X and α,β ∈ [0, 1) with α + β < 1. Then T has a unique fixed point
in X.
Proof: Set L = 0 in theorem 2.4.

Acknowledegement

This research work has been supported by University Grants Commission (UGC
- SAP II) New Delhi, India.

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Department of Mathematics, The Gandhigram Rural Institute - Deemed University,
Gandhigram - 624 302, Dindigul, Tamil Nadu, India

∗Corresponding author