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

Author Affiliations

1 Department of Mathematics, International Islamic University, H-10, Islamabad, 44000, Pakistan

2 Department of Mathematics, Atilim University, İncek, Ankara, 06836, Turkey

3 Department of Mathematics, COMSATS Institute of Information Technology, Chack Shahzad, Islamabad, 44000, Pakistan

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Journal of Inequalities and Applications 2013, 2013:248  doi:10.1186/1029-242X-2013-248

 Received: 6 February 2013 Accepted: 2 May 2013 Published: 17 May 2013

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.

### 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 , 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 [1-6]. 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.[10-12] published some new results. Recently, many papers have been reported on partially ordered metric spaces (see, e.g., [9-19] and also [8,20-33]).

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

• if is a non-increasing sequence in X such that implies , that is, , or

• if is a non-decreasing sequence in X such that implies , that is, .

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 be a partially ordered metric space. A self-mapping T on X is called an almost Jaggi contraction if it satisfies the following condition:

(1)

for any distinct with , where and with .

Theorem 2Letbe a complete partially ordered metric space. Suppose that a self-mappingTis an almost Jaggi contraction, continuous and non-decreasing. Suppose there existswith. ThenThas a unique fixed point.

Proof Let and set . If for some , then T has a fixed point. In particular, is a fixed point of T. So, we assume that for all n. Since , then

(2)

Now

which implies that

By the triangle inequality, for we have

(3)

where . Letting in the inequality (3), we get . Thus, the sequence is Cauchy. Since X is complete, there exists a point such that . Furthermore, the continuity of T in X implies that

Therefore, z is a fixed point of T in X. Now, if there exists another point in X such that , then

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

Example 3 Let with the usual metric and usual order ≤. We define an operator as follows:

Then T is continuous and non-decreasing. Take . Then, for any with , we have the result. Let us examine in detail. Without loss of generality, we assume that .

Case 01. If , then

holds for any and any with . Thus, all the conditions of Theorem 2 are satisfied.

Case 02. If , then

holds for any and any with . Hence, all the conditions of Theorem 2 are satisfied.

Case 03. If and , then we can easily evaluate that . Further, we have and . By the help of these observations, we derive that

Notice that is the fixed point of T.

Definition 4 Let be a partially ordered metric space. A self-mapping T on X is called a Jaggi contraction if it satisfies the following condition:

(4)

for any distinct with , where with .

Corollary 5Letbe a complete partially ordered metric space. Suppose that a self-mappingTis a Jaggi contraction, continuous and non-decreasing. Suppose that there existswith. ThenThas a fixed point.

Proof Set in Theorem 2. □

Example 6 Let , be defined by

Then is a complete metric space. Let be defined by

Also, iff . 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 . Then

that is,

• Let . Then

that is,

• Let and . Then

that is,

Then conditions of Corollary 5 hold and T has a fixed point (here, 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 7Letbe a complete partially ordered metric space. Suppose that a self-mappingTis non-decreasing and an almost Jaggi contraction. Suppose there existswith. If the operatoris continuous for some positive integerp, thenThas a unique fixed point.

Proof As in Theorem 2, we define a sequence and conclude that the sequence converges to some point . Thus its subsequence () also converges to z. Also,

Therefore z is a fixed point of . We now show that . Let m be the smallest positive integer such that but (). If , then

which implies that

Regarding (1), we have

Inductively, we get

where . Notice that . Therefore,

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

Corollary 8Letbe a complete partially ordered metric space. Suppose that a self-mappingTis non-decreasing and a Jaggi contraction. Suppose there existswith. If the operatoris continuous for some positive integerp, thenThas a unique fixed point.

Proof Set in Theorem 7. □

The following theorem generalizes Theorem 2.

Theorem 9Letbe a complete partially ordered metric space and letTbe a non-decreasing self-mapping defined onX. Suppose that for some positive integerm, self-mappingTsatisfies the following condition:

(5)

for any distinctwithand for somewithand. Suppose there existswith. Ifis continuous, thenThas a unique fixed point.

Proof Due to Theorem 2, we conclude that has a unique fixed point, say . Consider now

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

Corollary 10Letbe a complete partially ordered metric space and letTbe a non-decreasing self-mapping defined onX. Suppose that for some positive integerm, the self-mappingTsatisfies the following condition:

(6)

for all distinctand for somewith. Suppose there existswith. Ifis continuous, thenThas a unique fixed point.

Proof Set in Theorem 9. □

Now, we give the following example.

Example 11 Let with the usual metric and usual order ≤. We define an operator as follows:

It can be easily seen that T is discontinuous and does not satisfy (1) for any with when , . Now for all . It can be verified that satisfies the conditions of Theorem 9 and 0 is a unique fixed point of .

Theorem 12Letbe a complete partially ordered metric space and letTbe a non-decreasing self-mapping defined onX. Suppose that a self-mappingTonXsatisfies the condition

(7)

for any pointswith, and for somewithand. Suppose there existswith. ThenThas a fixed point.

Proof Define sequences as in Theorem 2. If for some then T has a fixed point. In particular, is a fixed point of T. Therefore, we assume that

(8)

Due to (7), we have

which implies that

Recursively, we obtain that

As in Theorem 2, we prove that is a Cauchy sequence. Indeed, by the triangle inequality, we have for ,

(9)

where . Letting , then the right-hand side of the inequality (9) tends to 0. Thus, the sequence is Cauchy.

Since X is complete, there exists a such that

(10)

Consider (7)

(11)

Letting in (11), we get

which is possible only if . Thus, .

Now, we show that z is the unique fixed point of T. Assume, on the contrary, that the operator T has another fixed point . Keeping (7) in mind, we obtain that

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

Corollary 13Letbe a complete partially ordered metric space and letTbe a non-decreasing self-mapping defined onX. Suppose that a self-mappingTonXsatisfies the condition

(12)

for any pointswith, and for somewith. ThenThas a fixed point.

Proof Set in Theorem 12. □

Example 14 Let , be defined by

Then is a complete metric space. Let be defined by

Also, iff . Suppose that and such that . 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 . Then

That is,

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

### 3 Further results

Theorem 15Letbe a complete partially ordered metric space and letTbe a non-decreasing, continuous self-mapping defined onX. Suppose that a self-mappingTsatisfies the following condition:

(13)

for allwith, whereandλ, μare non-negative reals such that. If there existswith, thenThas a fixed point.

Proof By assumption, there exists with . If , then the proof is finished. So, we suppose that . Since T is a non-decreasing mapping, we get

(14)

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

Case 1. If , then using the contractive condition (13), we get

Hence, we derive that

where . Moreover, by the triangular inequality, we have, for ,

and this proves that as .

So, is a Cauchy sequence and, since X is a complete metric space, there exists such that . Further, the continuity of T implies

Thus z is a fixed point.

Case 2. If , then . This implies that , a contradiction. Thus there exists a fixed point z of T. □

Example 16 Let with the usual metric and usual order ≤. We define an operator in the following way:

(15)

It is clear that T is continuous on . Now, for and any such that . Without loss of generality, we assume that . So, we have

for all . Also, there exists such that

is satisfied. This shows that conditions of Theorem 15 hold and T has a fixed point .

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

Theorem 17Letbe a complete partially ordered metric space and letTbe a non-decreasing self-mapping defined onX. Suppose that a self-mappingTsatisfies the following condition:

(16)

for allwith, whereandλ, μare non-negative reals with. And also suppose thatXhas the (OC) property. If there existswith, thenThas a fixed point.

Proof We only have to check that . As is a non-decreasing sequence and , then for all . Since T is a non-decreasing mapping, then for all or, equivalently, for all . Moreover, as and , we get . Suppose that . Using a similar argument as that in the proof of Theorem 15 for , we obtain that is a non-decreasing sequence and for certain . Again, using (OC), we have that . Moreover, from , we get for and for because for as and are comparable and distinct for .

Case 1. If , then applying the contractive condition (16), we get

Making in the above inequality, we obtain

As , , thus . Particularly, and consequently, which is a contradiction. Hence, we conclude that .

Case 2. If , then . Taking the limit as , we get . Then , which implies that , a contradiction. Thus . □

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

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

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

Proof We distinguish two cases.

Case 1. If y and z are comparable and . Now we have two subcases that are as follows:

(i) If , then using the contractive condition, we have

As , so by the last inequality, we have a contradiction. Thus .

(ii) If , then , a contradiction. Thus .

Case 2. If y and z are not comparable, then by a given condition there exists comparable to y and z. Monotonicity implies that is comparable to and for  .

If there exists such that , then as y is a fixed point, the sequence is constant, and consequently . On the other hand, if for . Now we have two subcases as follows:

(i) If , then using the contractive condition, we obtain, for ,

This implies that

By induction we get

Taking limit as in the above inequality, we get

as . Using a similar argument, we can prove that

Now, the uniqueness of the limit gives that .

(ii) If , then . Then

Using a similar argument, we can prove that

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

Remark 19 If in Theorem 15-Theorem 18 , 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 in the theorems of Section 3.

Theorem 20Letbe a complete partially ordered metric space and letTbe a non-decreasing self-mapping defined onX. Suppose that a self-mappingTsatisfies the following condition:

(17)

for allwith, whereandμis a non-negative real with. Suppose also that eitherTis continuous orXsatisfies the condition (OC). If there existswith, thenThas a fixed point.

If 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 21Letbe aT-orbitally complete partially ordered metric space and letTbe a non-decreasing self-mapping defined onX. Suppose that a self-mappingTsatisfies the following condition:

(18)

for all distinctwithand forwith, where. If there existswiththenThas at least one fixed point.

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

Corollary 22LetTbe a continuous, non-decreasing self-map defined on a complete partially ordered metric space. Suppose thatTsatisfies the following condition:

(19)

for any distinctwith, wherewith. Suppose there existswith. ThenThas a fixed point.

The following result is the consequence of Theorem 12.

Corollary 23Letbe a partially ordered metric space. Letbe a non-decreasing, continuous mapping. Suppose that a self-mappingTsatisfies

(20)

for anywith, and for somewithand. Suppose that there existswith. ThenThas a fixed point.

### Competing interests

The authors declare that they have no competing interests.

### Authors’ contributions

All authors contributed equally and significantly in writing this article. All authors read and approved the final manuscript.

### Acknowledgements

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

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