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):223230, 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 spaces1 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 [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 semimetric spaces, probabilistic metric spaces, Dmetric 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 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 nonincreasing sequence in X such that implies , that is, , or
• if is a nondecreasing 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 selfmapping T on X is called an almost Jaggi contraction if it satisfies the following condition:
for any distinct with , where and with .
Theorem 2Letbe a complete partially ordered metric space. Suppose that a selfmappingTis an almost Jaggi contraction, continuous and nondecreasing. 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
Now
which implies that
By the triangle inequality, for we have
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 nondecreasing. Take . Then, for any with , we have the result. Let us examine in detail. Without loss of generality, we assume that .
holds for any and any with . Thus, all the conditions of Theorem 2 are satisfied.
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 selfmapping T on X is called a Jaggi contraction if it satisfies the following condition:
for any distinct with , where with .
Corollary 5Letbe a complete partially ordered metric space. Suppose that a selfmappingTis a Jaggi contraction, continuous and nondecreasing. Suppose that there existswith. ThenThas a fixed point.
Then is a complete metric space. Let be defined by
Also, iff . Clearly, T is an increasing and continuous selfmapping 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.
that is,
that is,
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 selfmappingTis nondecreasing 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 selfmappingTis nondecreasing and a Jaggi contraction. Suppose there existswith. If the operatoris continuous for some positive integerp, thenThas a unique fixed point.
The following theorem generalizes Theorem 2.
Theorem 9Letbe a complete partially ordered metric space and letTbe a nondecreasing selfmapping defined onX. Suppose that for some positive integerm, selfmappingTsatisfies the following condition:
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 nondecreasing selfmapping defined onX. Suppose that for some positive integerm, the selfmappingTsatisfies the following condition:
for all distinctand for somewith. Suppose there existswith. Ifis continuous, thenThas a unique fixed point.
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 nondecreasing selfmapping defined onX. Suppose that a selfmappingTonXsatisfies the condition
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
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 ,
where . Letting , then the righthand side of the inequality (9) tends to 0. Thus, the sequence is Cauchy.
Since X is complete, there exists a such that
Consider (7)
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 nondecreasing selfmapping defined onX. Suppose that a selfmappingTonXsatisfies the condition
for any pointswith, and for somewith. ThenThas a fixed point.
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.
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 nondecreasing, continuous selfmapping defined onX. Suppose that a selfmappingTsatisfies the following condition:
for allwith, whereandλ, μare nonnegative 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 nondecreasing mapping, we get
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 ,
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:
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 nondecreasing selfmapping defined onX. Suppose that a selfmappingTsatisfies the following condition:
for allwith, whereandλ, μare nonnegative 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 nondecreasing sequence and , then for all . Since T is a nondecreasing 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 nondecreasing 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 .
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 15Theorem 18 , then we obtain Theorem 2.1Theorem 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 nondecreasing selfmapping defined onX. Suppose that a selfmappingTsatisfies the following condition:
for allwith, whereandμis a nonnegative 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 aTorbitally complete partially ordered metric space and letTbe a nondecreasing selfmapping defined onX. Suppose that a selfmappingTsatisfies the following condition:
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, nondecreasing selfmap defined on a complete partially ordered metric space. Suppose thatTsatisfies the following condition:
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 nondecreasing, continuous mapping. Suppose that a selfmappingTsatisfies
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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