Open Access

Some unique fixed point theorems for rational contractions in partially ordered metric spaces

Journal of Inequalities and Applications20132013:248

https://doi.org/10.1186/1029-242X-2013-248

Received: 6 February 2013

Accepted: 2 May 2013

Published: 17 May 2013

Abstract

In this paper, we prove some unique fixed point results for an operator T satisfying certain rational contraction condition in a partially ordered metric space. Our results generalize the main result of Jaggi (Indian J. Pure Appl. Math. 8(2):223-230, 1977). We give several examples to show that our results are proper generalization of the existing one.

MSC:47H10, 54H25, 46J10, 46J15.

Keywords

fixed point rational contractions partially ordered metric spaces

1 Introduction

Fixed point theory is one of the famous and traditional theories in mathematics and has a broad set of applications. In this theory, contraction is one of the main tools to prove the existence and uniqueness of a fixed point. Banach’s contraction principle, which gives an answer on the existence and uniqueness of a solution of an operator equation T x = x , is the most widely used fixed point theorem in all of analysis. This principle is constructive in nature and is one of the most useful tools in the study of nonlinear equations. There are many generalizations of Banach’s contraction mapping principle in the literature [16]. These generalizations 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, quasi semi-metric spaces, probabilistic metric spaces, D-metric spaces and cone metric spaces

The basic topological properties of ordered sets were discussed by Wolk [7] and Monjardet [8]. The existence of fixed points in partially ordered metric spaces was considered by Ran and Reurings [9]. After this paper, Nieto et al. [1012] published some new results. Recently, many papers have been reported on partially ordered metric spaces (see, e.g., [919] and also [8, 2033]).

The triple ( X , d , ) is called partially ordered metric spaces (POMS) if ( X , ) is a partially ordered set and ( X , d ) is a metric space. Further, if ( X , d ) is a complete metric space, the triple ( X , d , ) is called partially ordered complete metric spaces (POCMS). Throughout the manuscript, we assume that X . A partially ordered metric space ( X , d , ) is called ordered complete (OC) if for each convergent sequence { x n } n = 0 X , the following condition holds: either

  • if { x n } is a non-increasing sequence in X such that x n x implies x x n n N , that is, x = inf { x n } , or

  • if { x n } is a non-decreasing sequence in X such that x n x implies x n x n N , that is, x = sup { x n } .

In this manuscript, we prove that an operator T satisfying certain rational contraction condition has a fixed point in a partially ordered metric space. Our results generalize the main result of Jaggi [34].

2 Main results

We start this section with the following definition.

Definition 1 Let ( X , d , ) be a partially ordered metric space. A self-mapping T on X is called an almost Jaggi contraction if it satisfies the following condition:
d ( T x , T y ) α d ( x , T x ) d ( y , T y ) d ( x , y ) + β d ( x , y ) + L min { d ( x , T y ) , d ( y , T x ) }
(1)

for any distinct x , y X with x y , where L 0 and α , β [ 0 , 1 ) with α + β < 1 .

Theorem 2 Let ( X , d , ) be a complete partially ordered metric space. Suppose that a self-mapping T is an almost Jaggi contraction, continuous and non-decreasing. Suppose there exists x 0 X with x 0 T x 0 . Then T has a unique fixed point.

Proof Let x 0 X and set x n + 1 = T x n . If x n 0 = x n 0 + 1 for some n 0 N , then T has a fixed point. In particular, x n 0 is a fixed point of T. So, we assume that x n x n + 1 for all n. Since x 0 T x 0 , then
x 0 x 1 x n x n + 1 .
(2)
Now
d ( x n + 1 , x n ) = d ( T x n , T x n 1 ) α d ( x n 1 , T x n 1 ) d ( x n , T x n ) d ( x n , x n 1 ) + β d ( x n , x n 1 ) + L min { d ( x n , x n ) , d ( x n 1 , x n + 1 ) } ,
which implies that
d ( x n + 1 , x n ) ( β 1 α ) d ( x n , x n 1 ) ( β 1 α ) n d ( x 1 , x 0 ) .
By the triangle inequality, for m n we have
d ( x n , x m ) d ( x n , x n + 1 ) + d ( x n + 1 , x n + 2 ) + + d ( x m 1 , x m ) ( k n + k n + 1 + + k m 1 ) d ( x 0 , T x 0 ) . k n 1 k d ( x 0 , T x 0 ) ,
(3)
where k = β 1 α . Letting n in the inequality (3), we get d ( x n , x m ) = 0 . Thus, the sequence { x n } is Cauchy. Since X is complete, there exists a point z X such that x n z . Furthermore, the continuity of T in X implies that
T z = T ( lim n x n ) = lim n T x n = lim n x n + 1 = z .
Therefore, z is a fixed point of T in X. Now, if there exists another point w z in X such that T w = w , then
d ( w , z ) = d ( T w , T z ) α d ( w , T w ) d ( z , T z ) d ( w , z ) + β d ( w , z ) = β d ( w , z ) < d ( w , z ) ,

a contradiction. Hence u is a unique fixed point of T in X. □

Example 3 Let X = [ 0 , 1 ] with the usual metric and usual order ≤. We define an operator T : X X as follows:
T x = { x 10 if  x [ 0 , 1 2 ] , x 5 1 20 if  x ( 1 2 , 1 ] .

Then T is continuous and non-decreasing. Take β = 1 4 . Then, for any α [ 0 , 1 ) with α + β < 1 , we have the result. Let us examine in detail. Without loss of generality, we assume that y x .

Case 01. If x , y [ 0 , 1 2 ] , then
d ( T x , T y ) = 1 10 | x y | 1 4 | x y | = 1 4 d ( x , y ) α d ( x , T x ) d ( y , T y ) d ( x , y ) + 1 4 d ( x , y ) + L min { d ( x , T y ) , d ( y , T x ) }

holds for any L 1 and any α [ 0 , 1 ) with α + β < 1 . Thus, all the conditions of Theorem 2 are satisfied.

Case 02. If x , y ( 1 2 , 1 ] , then
d ( T x , T y ) = 1 5 | x y | 1 4 | x y | = 1 4 d ( x , y ) α d ( x , T x ) d ( y , T y ) d ( x , y ) + 1 4 d ( x , y ) + L min { d ( x , T y ) , d ( y , T x ) }

holds for any L 0 and any α [ 0 , 1 ) with α + β < 1 . Hence, all the conditions of Theorem 2 are satisfied.

Case 03. If x ( 1 2 , 1 ] and y [ 0 , 1 2 ] , then we can easily evaluate that 1 20 | 2 x 1 | 1 20 . Further, we have 9 20 d ( x , T y ) = | x y 10 | 1 and 3 20 d ( y , T x ) = | x 5 1 20 y | 9 20 . By the help of these observations, we derive that
d ( T x , T y ) = | x 5 1 20 y 10 | = | x 10 y 10 + x 10 1 20 | 1 10 | x y | + 1 20 | 2 x 1 | α d ( x , T x ) d ( y , T y ) d ( x , y ) + 1 4 d ( x , y ) + L min { d ( x , T y ) , d ( y , T x ) } .

Notice that 0 X is the fixed point of T.

Definition 4 Let ( X , d , ) be a partially ordered metric space. A self-mapping T on X is called a Jaggi contraction if it satisfies the following condition:
d ( T x , T y ) α d ( x , T x ) d ( y , T y ) d ( x , y ) + β d ( x , y )
(4)

for any distinct x , y X with x y , where α , β [ 0 , 1 ) with α + β < 1 .

Corollary 5 Let ( X , d , ) be a complete partially ordered metric space. Suppose that a self-mapping T is a Jaggi contraction, continuous and non-decreasing. Suppose that there exists x 0 X with x 0 T x 0 . Then T has a fixed point.

Proof Set L = 0 in Theorem 2. □

Example 6 Let X = [ 0 , ) , d : X × X R + be defined by
d ( x , y ) = { max { x , y } if  x y , 0 if  x = y .
Then ( X , d ) is a complete metric space. Let T : X X be defined by
T x = { x 8 ( 1 + x ) if  0 x 2 , x 12 if  2 < x .

Also, x y iff x y . Clearly, T is an increasing and continuous self-mapping on X. We shall prove that conditions of Corollary 5 hold and T has a fixed point.

Proof For the proof of this example, we have the following cases.

  • Let 0 x < y 2 . Then
    d ( T x , T y ) = max { T x , T y } = y 8 ( 1 + y ) 1 4 ( x + y ) = 1 4 ( x y y + y ) = 1 4 [ max { x , x 8 ( 1 + x ) } max { y , y 8 ( 1 + y ) } max { x , y } + max { x , y } ] = 1 4 ( d ( x , T x ) d ( y , T y ) d ( x , y ) + d ( x , y ) ) ,
that is,
d ( T x , T y ) 1 4 d ( x , T x ) d ( y , T y ) d ( x , y ) + 1 4 d ( x , y ) .
  • Let 2 < x < y . Then
    d ( T x , T y ) = max { T x , T y } = y 12 1 4 ( x + y ) = 1 4 ( x y y + y ) = 1 4 [ max { x , x 12 } max { y , y 12 } max { x , y } + max { x , y } ] = 1 4 ( d ( x , T x ) d ( y , T y ) d ( x , y ) + d ( x , y ) ) ,
that is,
d ( T x , T y ) 1 4 d ( x , T x ) d ( y , T y ) d ( x , y ) + 1 4 d ( x , y ) .
  • Let 0 x 2 and 2 < y . Then
    d ( T x , T y ) = max { T x , T y } = y 12 1 4 ( x + y ) = 1 4 ( x y y + y ) = 1 4 [ max { x , x 8 ( 1 + x ) } max { y , y 12 } max { x , y } + max { x , y } ] = 1 4 ( d ( x , T x ) d ( y , T y ) d ( x , y ) + d ( x , y ) ) ,
that is,
d ( T x , T y ) 1 4 d ( x , T x ) d ( y , T y ) d ( x , y ) + 1 4 d ( x , y ) .

Then conditions of Corollary 5 hold and T has a fixed point (here, x = 0 is a fixed point of T). □

In the next theorem, we establish the existence of a unique fixed point of a map T by assuming only the continuity of some iteration of T.

Theorem 7 Let ( X , d , ) be a complete partially ordered metric space. Suppose that a self-mapping T is non-decreasing and an almost Jaggi contraction. Suppose there exists x 0 X with x 0 T x 0 . If the operator T p is continuous for some positive integer p, then T has a unique fixed point.

Proof As in Theorem 2, we define a sequence x n and conclude that the sequence x n converges to some point z X . Thus its subsequence x n k ( n k = k p ) also converges to z. Also,
T p z = T p ( lim k x n k ) = lim k x n k + 1 = z .
Therefore z is a fixed point of T p . We now show that T z = z . Let m be the smallest positive integer such that T m z = z but T q z ( q = 1 , 2 , , m 1 ). If m > 1 , then
d ( T z , z ) = d ( T z , T m z ) α d ( z , T z ) d ( T m 1 z , T m z ) d ( z , T m 1 z ) + β d ( z , T m 1 z ) + L min { d ( z , T m z ) , d ( T m 1 z , T z ) }
which implies that
d ( T z , z ) β 1 α d ( z , T m 1 z ) .
Regarding (1), we have
d ( z , T m 1 z ) = d ( T m z , T m 1 z ) α d ( T m 1 z , T m z ) d ( T m 2 z , T m 1 z ) d ( T m 2 z , T m 1 z ) + β d ( T m 1 z , T m 2 z ) .
Inductively, we get
d ( z , T m 1 z ) = d ( T m z , T m 1 z ) k d ( T m 1 z , T m 2 z ) k m 1 d ( T z , z ) ,
where k = β 1 α . Notice that k < 1 . Therefore,
d ( T z , z ) k m d ( T z , z ) d ( T z , z ) ,

a contradiction. Hence T z = z . The uniqueness of z follows as in Theorem 2. □

Corollary 8 Let ( X , d , ) be a complete partially ordered metric space. Suppose that a self-mapping T is non-decreasing and a Jaggi contraction. Suppose there exists x 0 X with x 0 T x 0 . If the operator T p is continuous for some positive integer p, then T has a unique fixed point.

Proof Set L = 0 in Theorem 7. □

The following theorem generalizes Theorem 2.

Theorem 9 Let ( X , d , ) be a complete partially ordered metric space and let T be a non-decreasing self-mapping defined on X. Suppose that for some positive integer m, self-mapping T satisfies the following condition:
d ( T m x , T m y ) α d ( x , T m x ) d ( y , T m y ) d ( x , y ) + β d ( x , y ) + L min { d ( x , T y ) , d ( y , T x ) }
(5)

for any distinct x , y X with x y and for some α , β [ 0 , 1 ) with α + β < 1 and L 0 . Suppose there exists x 0 X with x 0 T m x 0 . If T m is continuous, then T has a unique fixed point.

Proof Due to Theorem 2, we conclude that T m has a unique fixed point, say z X . Consider now
T z = T ( T m z ) = T m ( T z ) .

Thus, Tz is also a fixed point of T m . But, by Theorem 2, we know that T m has a unique fixed point z. It follows that z = T z . Hence, z is the unique fixed point of T. □

Corollary 10 Let ( X , d , ) be a complete partially ordered metric space and let T be a non-decreasing self-mapping defined on X. Suppose that for some positive integer m, the self-mapping T satisfies the following condition:
d ( T m x , T m y ) α d ( x , T m x ) d ( y , T m y ) d ( x , y ) + β d ( x , y )
(6)

for all distinct x , y X and for some α , β [ 0 , 1 ) with α + β < 1 . Suppose there exists x 0 X with x 0 T m x 0 . If T m is continuous, then T has a unique fixed point.

Proof Set L = 0 in Theorem 9. □

Now, we give the following example.

Example 11 Let X = [ 0 , 1 ] with the usual metric and usual order ≤. We define an operator T : X X as follows:
T x = { 0 , x [ 0 , 1 3 ] , 1 3 , x ( 1 3 , 1 ] .

It can be easily seen that T is discontinuous and does not satisfy (1) for any α , β [ 0 , 1 ) with α + β 1 when x = 1 3 , y = 1 . Now T 2 ( x ) = 0 for all x [ 0 , 1 ] . It can be verified that T 2 satisfies the conditions of Theorem 9 and 0 is a unique fixed point of T 2 .

Theorem 12 Let ( X , d , ) be a complete partially ordered metric space and let T be a non-decreasing self-mapping defined on X. Suppose that a self-mapping T on X satisfies the condition
d ( T x , T y ) α d ( T y , y ) [ 1 + d ( x , T x ) ] 1 + d ( x , y ) + β d ( x , y ) + L min { d ( x , T x ) , d ( x , T y ) , d ( y , T x ) }
(7)

for any points x , y X with x y , and for some α , β [ 0 , 1 ) with α + β < 1 and L 0 . Suppose there exists x 0 X with x 0 T x 0 . Then T has a fixed point.

Proof Define sequences x n as in Theorem 2. If x n 0 = x n 0 + 1 for some n 0 N then T has a fixed point. In particular, x n 0 is a fixed point of T. Therefore, we assume that
x n x n + 1 for all  n N .
(8)
Due to (7), we have
d ( x n , x n + 1 ) = d ( T x n 1 , T x n ) α d ( x n , T x n ) [ 1 + d ( x n 1 , T x n 1 ) ] 1 + d ( x n 1 , x n ) + β d ( x n 1 , x n ) + L min { d ( T x n 1 , x n 1 ) , d ( x n 1 , T x n ) , d ( x n , T x n 1 ) } = α d ( x n + 1 , x n ) [ 1 + d ( x n , x n 1 ) ] 1 + d ( x n 1 , x n ) + β d ( x n 1 , x n ) , + L min { d ( x n 1 , x n ) , d ( x n 1 , x n + 1 ) , d ( x n , x n ) } ,
which implies that
d ( x n + 1 , x n ) ( β 1 α ) d ( x n , x n 1 ) .
Recursively, we obtain that
d ( x n + 1 , x n ) ( β 1 α ) n d ( x 1 , x 0 ) .
As in Theorem 2, we prove that { x n } is a Cauchy sequence. Indeed, by the triangle inequality, we have for m n ,
d ( x n , x m ) d ( x n , x n + 1 ) + d ( x n + 1 , y n + 2 ) + + d ( x m 1 , x m ) ( k n + k n + 1 + + k m 1 ) d ( x 0 , T x 0 ) k n 1 k d ( x 0 , x 1 ) ,
(9)

where k = β 1 α < 1 . Letting n , then the right-hand side of the inequality (9) tends to 0. Thus, the sequence { x n } is Cauchy.

Since X is complete, there exists a z X such that
lim n x n = lim n T x n = z .
(10)
Consider (7)
d ( T z , x n + 1 ) = d ( T z , T x n ) α d ( z , T z ) [ 1 + d ( x n , T x n ) ] 1 + d ( z , x n ) + β d ( z , T x n ) + L min { d ( z , T z ) , d ( z , T x n ) , d ( x n , T z ) } .
(11)
Letting n in (11), we get
d ( T z , z ) α d ( z , T z ) [ 1 + d ( z , z ) ] 1 + d ( z , z ) + β d ( z , z ) = α d ( T z , z ) ,

which is possible only if d ( T z , z ) = 0 . Thus, T z = z .

Now, we show that z is the unique fixed point of T. Assume, on the contrary, that the operator T has another fixed point u z . Keeping (7) in mind, we obtain that
d ( z , u ) = d ( T z , T u ) α d ( T u , u ) [ 1 + d ( z , T z ) ] 1 + d ( z , u ) + β d ( z , u ) + L min { d ( z , T z ) , d ( z , T u ) , d ( u , T z ) } = α d ( u , u ) . [ 1 + d ( z , z ) ] d ( z , u ) + β d ( z , u ) = β d ( z , u ) ,

a contradiction. Hence z is a unique fixed point of T in X. □

Corollary 13 Let ( X , d , ) be a complete partially ordered metric space and let T be a non-decreasing self-mapping defined on X. Suppose that a self-mapping T on X satisfies the condition
d ( T x , T y ) α d ( T y , y ) [ 1 + d ( x , T x ) ] 1 + d ( x , y ) + β d ( x , y )
(12)

for any points x , y X with x y , and for some α , β [ 0 , 1 ) with α + β < 1 . Then T has a fixed point.

Proof Set L = 0 in Theorem 12. □

Example 14 Let X = [ 0 , 1 ] , d : X × X R + be defined by
d ( x , y ) = | x y | .
Then ( X , d ) is a complete metric space. Let T : X X be defined by
T x = 1 6 x 3 .

Also, x y iff x y . Suppose that β = 1 3 and α [ 0 , 1 ) such that α + β < 1 . Clearly, T is an injective, continuous and sequentially convergent mapping on X. We shall prove that conditions of Corollary 8 hold and T has a fixed point.

Proof For the proof of this example, we have the following cases.

Let x , y [ 0 , 1 ] . Then
d ( T x , T y ) = 1 6 | x 3 y 3 | = 1 6 | ( x y ) ( x 2 + x y + y 2 ) | 1 3 | x y | = 1 3 d ( x , y ) .
That is,
d ( T x , T y ) α d ( y , T y ) [ 1 + d ( x , T x ) ] 1 + d ( x , y ) + β d ( x , y ) .

Hence, conditions of Corollary 8 hold and T has a fixed point (here x = 0 is a fixed point of T). □

3 Further results

Theorem 15 Let ( X , d , ) be a complete partially ordered metric space and let T be a non-decreasing, continuous self-mapping defined on X. Suppose that a self-mapping T satisfies the following condition:
d ( T x , T y ) { λ d ( x , y ) + μ d ( x , T x ) d ( x , T y ) + d ( y , T x ) d ( y , T y ) d ( y , T x ) + d ( x , T y ) if A 0 , 0 if A = 0 ,
(13)

for all x , y X with y x , where A = d ( y , T x ) + d ( x , T y ) and λ, μ are non-negative reals such that λ + μ < 1 . If there exists x 0 X with x 0 T x 0 , then T has a fixed point.

Proof By assumption, there exists x 0 X with x 0 T x 0 . If x 0 = T x 0 , then the proof is finished. So, we suppose that x 0 T x 0 . Since T is a non-decreasing mapping, we get
x 0 T x 0 T 2 x 0 T n x 0 T n + 1 x 0
(14)

by iteration. Put x n + 1 = T x n . If there exists n 0 N such that x n 0 = x n 0 + 1 , then from x n 0 = x n 0 + 1 = T x n 0 , we get x n 0 is a fixed point, and the proof is finished. Suppose that x n x n + 1 for n N . Since the points x n and x n 1 are comparable for all n N due to (14), we have the following two cases.

Case 1. If A = d ( x n 1 , T x n ) + d ( x n , T x n 1 ) 0 , then using the contractive condition (13), we get
d ( x n + 1 , x n ) = d ( T x n , T x n 1 ) λ d ( x n , x n 1 ) + μ d ( x n , T x n ) d ( x n , T x n 1 ) + d ( x n 1 , T x n ) d ( x n 1 , T x n 1 ) d ( x n 1 , T x n ) + d ( x n , T x n 1 ) λ d ( x n , x n 1 ) + μ d ( x n , x n + 1 ) d ( x n , x n ) + d ( x n 1 , x n + 1 ) d ( x n 1 , x n ) d ( x n 1 , x n + 1 ) + d ( x n , x n ) λ d ( x n , x n 1 ) + μ d ( x n 1 , x n + 1 ) d ( x n 1 , x n ) d ( x n 1 , x n + 1 ) ( λ + μ ) d ( x n , x n 1 ) .
Hence, we derive that
d ( x n + 1 , x n ) h n d ( x 1 , x 0 ) ,
where h = ( λ + μ ) < 1 . Moreover, by the triangular inequality, we have, for m n ,
d ( x m , x n ) d ( x m , x m 1 ) + d ( x m 1 , x m 2 ) + + d ( x n + 1 , x n ) ( h m 1 + h m 2 + + h n ) d ( x 1 , x 0 ) h n 1 h d ( x 1 , x 0 ) ,

and this proves that d ( x m , x n ) 0 as m , n .

So, { x n } is a Cauchy sequence and, since X is a complete metric space, there exists z X such that lim n x n = z . Further, the continuity of T implies
T z = T ( lim n x n ) = lim n T x n = lim n x n + 1 = z .

Thus z is a fixed point.

Case 2. If A = d ( x n 1 , T x n ) + d ( x n , T x n 1 ) = 0 , then d ( x n + 1 , x n ) = 0 . This implies that x n = x n + 1 , a contradiction. Thus there exists a fixed point z of T. □

Example 16 Let X = [ 0 , 1 ] with the usual metric and usual order ≤. We define an operator T : X X in the following way:
T x = 2 x + 3 4 ( x 2 + x + 5 4 ) .
(15)
It is clear that T is continuous on [ 0 , 1 ] . Now, for λ = 16 25 and any μ [ 0 , 1 ) such that λ + μ < 1 . Without loss of generality, we assume that x y . So, we have
d ( T x , T y ) = 1 4 | 2 x + 3 x 2 + x + 5 4 2 y + 3 y 2 + y + 5 4 | = | 2 x y ( y x ) + 3 ( y x ) ( x + y ) + 3 ( y x ) 5 2 ( y x ) 4 ( x 2 + x + 5 4 ) ( y 2 + y + 5 4 ) | = | 2 x y + 3 ( x + y ) + 1 2 4 ( x 2 + x + 5 4 ) ( y 2 + y + 5 4 ) | | x y | 16 25 | y x | = 16 25 d ( x , y )
for all x , y X . Also, there exists x 0 = 0 X such that
x 0 = 0 T x 0

is satisfied. This shows that conditions of Theorem 15 hold and T has a fixed point 1 2 [ 0 , 1 ] .

We may remove the continuity criteria on T in Theorem 15 as follows.

Theorem 17 Let ( X , d , ) be a complete partially ordered metric space and let T be a non-decreasing self-mapping defined on X. Suppose that a self-mapping T satisfies the following condition:
d ( T x , T y ) { λ d ( x , y ) + μ d ( x , T x ) d ( x , T y ) + d ( y , T x ) d ( y , T y ) d ( y , T x ) + d ( x , T y ) if A 0 , 0 if A = 0 ,
(16)

for all x , y X with y x , where A = d ( x , T x ) + d ( y , T y ) and λ, μ are non-negative reals with λ + μ < 1 . And also suppose that X has the (OC) property. If there exists x 0 X with x 0 T x 0 , then T has a fixed point.

Proof We only have to check that z = T z . As { x n } X is a non-decreasing sequence and x n z X , then z = sup { x n } for all n N . Since T is a non-decreasing mapping, then T x n T z for all n N or, equivalently, x n + 1 T z for all n N . Moreover, as x 0 x 1 T z and z = sup { x n } , we get z T z . Suppose that z T z . Using a similar argument as that in the proof of Theorem 15 for x 0 T x 0 , we obtain that { T n z } is a non-decreasing sequence and lim n T n z = y for certain y X . Again, using (OC), we have that y = sup { T n z } . Moreover, from x 0 z , we get x n = T n x 0 T n z for n 1 and x n T n z for n 1 because x n z T z T n z for n 1 as x n and T n z are comparable and distinct for n 1 .

Case 1. If d ( T n z , T x n ) + d ( x n , T n + 1 z ) 0 , then applying the contractive condition (16), we get
d ( x n + 1 , T n + 1 z ) = d ( T x n , T ( T n z ) ) λ d ( x n , y ) + μ d ( x n , T x n ) d ( x n , T n z ) + d ( T n z , T x n ) d ( T n z , T n + 1 z ) d ( T n z , T x n ) + d ( x n , T n + 1 z ) λ d ( x n , T n z ) + μ d ( x n , x n + 1 ) d ( x n , T n z ) + d ( T n z , x n + 1 ) d ( T n z , T n + 1 z ) d ( T n z , x n + 1 ) + d ( x n , T n + 1 z ) .
Making n in the above inequality, we obtain
d ( z , y ) λ d ( z , y ) .

As λ < 1 , d ( x , z ) = 0 , thus z = y . Particularly, z = y = sup { T n z } and consequently, T z z which is a contradiction. Hence, we conclude that T z = z .

Case 2. If A = d ( T n z , T x n ) + d ( x n , T n + 1 z ) = 0 , then d ( x n + 1 , T n + 1 z ) = 0 . Taking the limit as n , we get d ( z , y ) = 0 . Then z = y = sup { T n z } , which implies that T z z , a contradiction. Thus T z = z . □

Now we prove the sufficient condition for the uniqueness of the fixed point in Theorem 15 and Theorem 17, that is,

U: for any y , z X , there exists x X which is comparable to y and z.

Theorem 18 Adding the above mentioned condition to the hypothesis of Theorem  15 (or Theorem  17), one obtains the uniqueness of the fixed point of T.

Proof We distinguish two cases.

Case 1. If y and z are comparable and y z . Now we have two subcases that are as follows:
  1. (i)
    If d ( z , T y ) + d ( y , T z ) 0 , then using the contractive condition, we have
    d ( y , z ) = d ( T y , T z ) λ d ( y , z ) + μ d ( y , T y ) d ( y , T z ) + d ( z , T y ) d ( z , T z ) d ( z , T y ) + d ( y , T z ) λ d ( y , z ) + μ d ( y , y ) d ( y , z ) + d ( z , y ) d ( z , z ) d ( z , y ) + d ( y , z ) = λ d ( y , z ) .
     
As λ < 1 , so by the last inequality, we have a contradiction. Thus y = z .
  1. (ii)

    If d ( z , T y ) + d ( y , T z ) = 0 , then d ( y , z ) = 0 , a contradiction. Thus y = z .

     

Case 2. If y and z are not comparable, then by a given condition there exists x X comparable to y and z. Monotonicity implies that T n x is comparable to T n y = y and T n z = z for n = 0 , 1 , 2 ,  .

If there exists n 0 1 such that T n 0 x = y , then as y is a fixed point, the sequence { T n x : n n 0 } is constant, and consequently lim n T n x = y . On the other hand, if T n x y for n 1 . Now we have two subcases as follows:
  1. (i)
    If d ( T n 1 y , T n x ) + d ( T n 1 x , T n y ) 0 , then using the contractive condition, we obtain, for n 2 ,
    d ( T n x , y ) = d ( T n x , T n y ) λ d ( T n 1 x , T n 1 y ) + μ d ( T n 1 x , T n x ) d ( T n 1 x , T n y ) + d ( T n 1 y , T n x ) d ( T n 1 y , T n y ) d ( T n 1 y , T n x ) + d ( T n 1 x , T n y ) λ d ( T n 1 x , y ) + μ d ( T n 1 x , T n x ) d ( T n 1 x , y ) + d ( T y , T n x ) d ( y , y ) d ( y , T n x ) + d ( T n 1 x , y ) λ d ( T n 1 x , y ) + μ d ( T n 1 x , T n x ) d ( T n 1 x , y ) d ( y , T n x ) + d ( T n 1 x , y ) .
     
This implies that
d ( T n x , y ) λ d ( T n 1 x , y ) + μ d ( T n 1 x , y ) = λ d ( T n 1 x , T n 1 y ) + μ d ( T n 1 x , T n 1 y ) λ ( λ d ( T n 2 x , y ) + μ d ( T n 2 x , y ) ) + μ ( λ d ( T n 2 x , y ) + μ d ( T n 2 x , y ) ) = λ 2 d ( T n 2 x , y ) + 2 λ μ d ( T n 2 x , y ) + μ 2 d ( T n 2 x , y ) = ( λ + μ ) 2 d ( x , y ) .
By induction we get
d ( T n x , y ) ( λ + μ ) n d ( x , y ) .
Taking limit as n in the above inequality, we get
lim n T n x = y
as λ + μ < 1 . Using a similar argument, we can prove that
lim n T n x = z .
Now, the uniqueness of the limit gives that y = z .
  1. (ii)
    If d ( T n 1 y , T n x ) + d ( T n 1 x , T n y ) = 0 , then d ( T n x , y ) = 0 . Then
    lim n T n x = y .
     
Using a similar argument, we can prove that
lim n T n x = z .

Now, the uniqueness of the limit gives that y = z . This completes the proof. □

Remark 19 If in Theorem 15-Theorem 18 μ = 0 , then we obtain Theorem 2.1-Theorem 2.3 of [10].

We get the following fixed point theorem in partially ordered metric spaces if we take λ = 0 in the theorems of Section 3.

Theorem 20 Let ( X , d , ) be a complete partially ordered metric space and let T be a non-decreasing self-mapping defined on X. Suppose that a self-mapping T satisfies the following condition:
d ( T x , T y ) { μ d ( x , T x ) d ( x , T y ) + d ( y , T x ) d ( y , T y ) d ( y , T x ) + d ( x , T y ) if A 0 , 0 if A = 0 ,
(17)

for all x , y X with y x , where A = d ( y , T x ) + d ( x , T y ) and μ is a non-negative real with 0 μ < 1 . Suppose also that either T is continuous or X satisfies the condition (OC). If there exists x 0 X with x 0 T x 0 , then T has a fixed point.

If ( X , ) satisfies the condition used in Theorem 18, then the uniqueness of a fixed point can be proved.

4 Applications

In this section we state some applications of the main results. The first result is the consequence of Theorem 2.

Corollary 21 Let ( X , d , ) be a T-orbitally complete partially ordered metric space and let T be a non-decreasing self-mapping defined on X. Suppose that a self-mapping T satisfies the following condition:
0 d ( T x , T y ) d s α 0 d ( x , T x ) d ( y , T y ) d ( x , y ) d s + β 0 d ( x , y ) d s + L 0 min { d ( x , T x ) , d ( y , T x ) } d s
(18)

for all distinct x , y X with x y and for α , β [ 0 , 1 ) with α + β < 1 , where L 0 . If there exists x 0 X with x 0 T x 0 then T has at least one fixed point.

Similarly, the following result is the consequence of Corollary 5.

Corollary 22 Let T be a continuous, non-decreasing self-map defined on a complete partially ordered metric space ( X , d , ) . Suppose that T satisfies the following condition:
0 d ( T x , T y ) d s α 0 d ( x , T x ) d ( y , T y ) d ( x , y ) d s + β 0 d ( x , y ) d s
(19)

for any distinct x , y X with x y , where α , β [ 0 , 1 ) with α + β < 1 . Suppose there exists x 0 X with x 0 T x 0 . Then T has a fixed point.

The following result is the consequence of Theorem 12.

Corollary 23 Let ( X , d , ) be a partially ordered metric space. Let T : X X be a non-decreasing, continuous mapping. Suppose that a self-mapping T satisfies
0 d ( T x , T y ) d s α 0 α d ( T y , y ) [ 1 + d ( x , T x ) ] 1 + d ( x , y ) d s + β 0 d ( x , y ) d s + L 0 min { d ( x , T x ) , d ( x , T y ) , d ( y , T x ) } d s
(20)

for any x , y X with x y , and for some α , β [ 0 , 1 ) with α + β < 1 and L 0 . Suppose that there exists x 0 X with x 0 T x 0 . Then T has a fixed point.

Declarations

Acknowledgements

The authors express their gratitude to the anonymous referees for constructive and useful remarks, comments and suggestions.

Authors’ Affiliations

(1)
Department of Mathematics, International Islamic University
(2)
Department of Mathematics, Atilim University
(3)
Department of Mathematics, COMSATS Institute of Information Technology

References

  1. Edelstein M: On fixed points and periodic points under contraction mappings. J. Lond. Math. Soc. 1962, 37: 74–79.MathSciNetView ArticleGoogle Scholar
  2. Hardy GC, Rogers T: A generalization of fixed point theorem of S. Reich. Can. Math. Bull. 1973, 16: 201–206. 10.4153/CMB-1973-036-0MathSciNetView ArticleGoogle Scholar
  3. Kannan R: Some results on fixed points - II. Am. Math. Mon. 1969, 76: 71–76.View ArticleGoogle Scholar
  4. Reich S: Some remarks concerning contraction mappings. Can. Math. Bull. 1971, 14: 121–124. 10.4153/CMB-1971-024-9View ArticleGoogle Scholar
  5. Smart DR: Fixed Point Theorems. Cambridge University Press, Cambridge; 1974.Google Scholar
  6. Wong CS: Common fixed points of two mappings. Pac. J. Math. 1973, 48: 299–312. 10.2140/pjm.1973.48.299View ArticleGoogle Scholar
  7. Wolk ES: Continuous convergence in partially ordered sets. Gen. Topol. Appl. 1975, 5: 221–234. 10.1016/0016-660X(75)90022-7MathSciNetView ArticleGoogle Scholar
  8. Monjardet B: Metrics on partially ordered sets - a survey. Discrete Math. 1981, 35: 173–184. 10.1016/0012-365X(81)90206-5MathSciNetView ArticleGoogle Scholar
  9. Ran ACM, Reurings MCB: A fixed point theorem in partially ordered sets and some application to matrix equations. Proc. Am. Math. Soc. 2004, 132: 1435–1443. 10.1090/S0002-9939-03-07220-4MathSciNetView ArticleGoogle Scholar
  10. Nieto JJ, Lopez RR: Contractive mapping theorems in partially ordered sets and applications to ordinary differential equations. Order 2005, 22: 223–239. 10.1007/s11083-005-9018-5MathSciNetView ArticleGoogle Scholar
  11. Nieto JJ, Lopez RR: Existence and uniqueness of fixed point in partially ordered sets and applications to ordinary differential equation. Acta Math. Sin. Engl. Ser. 2007, 23(12):2205–2212. 10.1007/s10114-005-0769-0MathSciNetView ArticleGoogle Scholar
  12. Nieto JJ, Pouso L, Rodríguez-López R: Fixed point theorems in ordered abstract spaces. Proc. Am. Math. Soc. 2007, 135: 2505–2517. 10.1090/S0002-9939-07-08729-1View ArticleGoogle Scholar
  13. Agarwal RP, El-Gebeily MA, O’Regan D: Generalized contractions in partially ordered metric spaces. Appl. Anal. 2008, 87: 1–8. 10.1080/00036810701714164MathSciNetView ArticleGoogle Scholar
  14. Bhaskar TG, Lakshmikantham V: Fixed point theory in partially ordered metric spaces and applications. Nonlinear Anal., Theory Methods Appl. 2006, 65: 1379–1393. 10.1016/j.na.2005.10.017MathSciNetView ArticleGoogle Scholar
  15. Choudhury BS, Kundu A: A coupled coincidence point result in partially ordered metric spaces for compatible mappings. Nonlinear Anal., Theory Methods Appl. 2010, 73: 2524–2531. 10.1016/j.na.2010.06.025MathSciNetView ArticleGoogle Scholar
  16. Hong S: Fixed points of multivalued operators in ordered metric spaces with applications. Nonlinear Anal., Theory Methods Appl. 2010, 72: 3929–3942. 10.1016/j.na.2010.01.013View ArticleGoogle Scholar
  17. Lakshmikantham V, Ćirić LB: Couple fixed point theorems for nonlinear contractions in partially ordered metric spaces. Nonlinear Anal., Theory Methods Appl. 2009, 70: 4341–4349. 10.1016/j.na.2008.09.020View ArticleGoogle Scholar
  18. Ozturk M, Basarir M: On some common fixed point theorems with rational expressions on cone metric spaces over a Banach algebra. Hacet. J. Math. Stat. 2012, 41(2):211–222.MathSciNetGoogle Scholar
  19. Rouzkard F, Imdad M: Some common fixed point theorems on complex valued metric spaces. Comput. Math. Appl. 2012. doi:10.1016/j.camwa.2012.02.063Google Scholar
  20. Ahmad J, Arshad M, Vetro C: On a theorem of Khan in a generalized metric space. Int. J. Anal. 2013., 2013: Article ID 852727Google Scholar
  21. Altun I, Damjanovic B, Djoric D: Fixed point and common fixed point theorems on ordered cone metric spaces. Appl. Math. Lett. 2010, 23: 310–316. 10.1016/j.aml.2009.09.016MathSciNetView ArticleGoogle Scholar
  22. Amini-Harandi A, Emami H: A fixed point theorem for contraction type maps in partially ordered metric spaces and application to ordinary differential equations. Nonlinear Anal., Theory Methods Appl. 2010, 72: 2238–2242. 10.1016/j.na.2009.10.023MathSciNetView ArticleGoogle Scholar
  23. Arshad M, Azam A, Vetro P: Some common fixed results in cone metric spaces. Fixed Point Theory Appl. 2009., 2009: Article ID 493965Google Scholar
  24. Arshad M, Ahmad J, Karapınar E: Some common fixed point results in rectangular metric spaces. Int. J. Anal. 2013., 2013: Article ID 307234Google Scholar
  25. Aydi H, Karapınar E, Shatanawi W:Coupled fixed point results for ( ψ , φ ) -weakly contractive condition in ordered partial metric spaces. Comput. Math. Appl. 2011, 62(12):4449–4460. 10.1016/j.camwa.2011.10.021MathSciNetView ArticleGoogle Scholar
  26. Azam A, Fisher B, Khan M: Common fixed point theorems in complex valued metric spaces. Numer. Funct. Anal. Optim. 2011, 32(3):243–253. 10.1080/01630563.2011.533046MathSciNetView ArticleGoogle Scholar
  27. Beg I, Butt AR: Fixed point for set-valued mappings satisfying an implicit relation in partially ordered metric spaces. Nonlinear Anal. 2009, 71: 3699–3704. 10.1016/j.na.2009.02.027MathSciNetView ArticleGoogle Scholar
  28. Dricia Z, McRaeb FA, Devi JV: Fixed-point theorems in partially ordered metric spaces for operators with PPF dependence. Nonlinear Anal., Theory Methods Appl. 2007, 67: 641–647. 10.1016/j.na.2006.06.022View ArticleGoogle Scholar
  29. Karapınar E: Couple fixed point on cone metric spaces. Gazi Univ. J. Sci. 2011, 24(1):51–58.Google Scholar
  30. Karapınar E, Luong NV: Quadruple fixed point theorems for nonlinear contractions. Comput. Math. Appl. 2012, 64(6):1839–1848. 10.1016/j.camwa.2012.02.061MathSciNetView ArticleGoogle Scholar
  31. Luong NV, Thuan NX: Coupled fixed points in partially ordered metric spaces and application. Nonlinear Anal., Theory Methods Appl. 2011, 74: 983–992. 10.1016/j.na.2010.09.055MathSciNetView ArticleGoogle Scholar
  32. Samet B: Coupled fixed point theorems for a generalized Meir-Keeler contraction in partially ordered metric spaces. Nonlinear Anal. 2010, 74(12):4508–4517.MathSciNetView ArticleGoogle Scholar
  33. Zhang X: Fixed point theorems of multivalued monotone mappings in ordered metric spaces. Appl. Math. Lett. 2010, 23: 235–240. 10.1016/j.aml.2009.06.011MathSciNetView ArticleGoogle Scholar
  34. Jaggi DS: Some unique fixed point theorems. Indian J. Pure Appl. Math. 1977, 8(2):223–230.MathSciNetGoogle Scholar

Copyright

© Arshad et al.; licensee Springer 2013

This article is published under license to BioMed Central Ltd. This is an Open Access article distributed under the terms of the Creative Commons Attribution License (http://creativecommons.org/licenses/by/2.0), which permits unrestricted use, distribution, and reproduction in any medium, provided the original work is properly cited.