Abstract

We study the existence and uniqueness of common coupled fixed point of four self-mappings for Geraghty-type contraction using weakly compatible mappings in partially ordered metric spaces with common limit range property (denoted by ), the property of , and so on. It is noted that the continuity of mappings and completeness of spaces can be removed. Our results improve, extend, complement, and generalize several existing results in the literature. Also, some examples are provided to illustrate the usability of our results.

1. Introduction and Preliminaries

In 1922, Banach proposed the contraction mapping theorem which is famously known as the Banach Contraction Principle in metric spaces. It is also noted that Banach Contraction Principle is one of the pivotal results of fixed point theory in metric spaces. From then on, many researchers worked on this. They improved, extended, and generalized this theorem in different ways in the setting of different metric spaces.

In 1982, Sessa first studied common fixed point theorems for weakly commuting pair of mappings [1]. Afterward, in 1986, Jungck [2] weakened weakly commuting mappings to compatible mappings in metric spaces and proved that compatible pair of mappings commute on the sets of coincidence points of the involved mappings. In 1998, Jungck and Rhoades [3] proposed the notion of weak compatibility if they commute at their coincidence points and proved that compatible mappings are weakly compatible but the reverse does not hold. In 2002, Aamri and Moutawakil [4] introduced the property (E.A) of one pair and the common property (E.A) of two pairs and obtained common fixed point theorems in metric spaces. In 2005, Liu et al. [5] used common property (E.A) to obtain the corresponding fixed point theorems. Later, in 2008, Kubiaczyk and Sharma [6] introduced the property (E.A) in PM spaces and got some fixed point theorems. The property (E.A) always requires the completeness (or closedness) of underlying subspaces for the existence of common fixed point. Hence they coined the idea of common limit in the range property (called ) which relaxes the requirement of completeness (or closedness) of the underlying subspace. In 2011, Sintunavarat and Kumam [7] introduced () property and got the fixed point theorem in fuzzy metric spaces. Soon, Imdad et al. [8] introduced () property and obtained some fixed point theorems in Menger spaces. In 2012, Jain et al. [9] extended the concept of CLR property in the coupled case and established the fixed point theorems. In 2013, Karapinar et al. [10] utilized () property in symmetric spaces to obtain some common fixed point theorems. In 2014, Imdad et al. [11] proved some common fixed point theorems for two pairs of non-self-weakly compatible mappings by means of () property in symmetric spaces. In sequel, Roldán-López-De-Hierro and Sintunavarat [12] generalized some results of Jain et al. [9] by using the generalized contractive conditions and the property in fuzzy metric spaces.

Inspired by the above works, this paper utilizes () property to obtain the common fixed point theorems for Geragthy-type contraction in partially ordered metric spaces. For more details about fixed point theorems under Geragthy-type contraction in some generalized metric spaces, see [1318]. At the same time, uniqueness of common fixed point is obtained. Finally, we illustrate some examples to support our results.

To begin with, we give some basic notions with relevant to common fixed point of mappings and .

Definition 1. Let be a partially ordered metric space. An element is called a coupled fixed point of the mapping if

Definition 2. An element is called
(i) a coupled coincidence point of the mapping and if (ii) a common coupled fixed point of the mapping and if (iii) a common fixed point of the mapping and if

Definition 3. Let be a partially ordered metric space and and be two mappings. Then
(i) is said to be weakly compatible if for all , , such that (ii) is said to be compatible if for sequences , in such that

Proposition 4. Let be a partially ordered metric space and and be two mappings. If is compatible, then is weakly compatible.

Proof. Suppose that , and for some , . Consider the constant sequence , and for all . It is obvious that as , and as . Since is compatible, then , and . Thus, , and . So, is weakly compatible.

The following example shows that the pair of weakly compatible mappings is not necessarily compatible.

Example 5. Let be a metric space where . Define and as follows: Letting , , , , , . But Thus, the pair is not compatible.
Also, the coupled coincidence point of and is . It is, namely, that for , , , . Then we get Meanwhile, It implies that is weakly compatible but not compatible.

The above example shows that the notion of weakly compatibility is more general than compatibility.

Definition 6. Let be a partially ordered metric space and be four mappings.
(i) The pair is said to satisfy the property (E.A) if there exists a sequence in such that (ii) Two pairs and are said to satisfy the property (E.A) if there exist two sequences , in such that (iii) The pair is said to have the common limit range property with respect to the mapping (denoted by ) if there exists a sequence in such that (iv) Two pairs and are said to have the common limit range property with respect to mappings and (denoted by ) if there exist two sequences , in such that Let denote the class of all functions which satisfies the following conditions:
(i) for all , ;
(ii) For any two sequences and of nonnegative real numbers

Lemma 7. Let be a partially ordered metric space and be four self-mappings satisfying the following conditions:
(i) [resp., ];
(ii) The pair satisfies the property [resp., the pair satisfies the property];
(iii) is a closed subset of [resp., is a closed subset of ];
(iv) There exists such that, for all , , , ,

Then the pairs and share the property.

Proof. Since the pair satisfies the property, there exists a sequence in such that In view of conditions (i) and (iii), for , there exists a sequence such that . It follows that Therefore, it suffices to prove that . In fact, by (iv), putting , , we can obtain that Letting , it follows that Thus, .
It yields thatthat is, the pairs and share the property.

Remark 8. It can be pointed that Lemma 7 generalizes Lemma 3.2 in [8], though still from four self-mappings to four self-mappings, but here and are defined in product space . Simultaneously, it is straight forward to notice that Lemma 7 improves Lemma 1 of [11] from symmetric spaces to general partially ordered metric spaces.

2. Main Results

Now, we state and prove our main results.

Theorem 9. Let be a complete partially ordered metric space and be the mappings. Suppose the following assumptions hold:
(i) , ;
(ii) and are continuous;
(iii) and are both compatible;
(iv) There exists such that for all , , , , the inequality (18) holds.
Then , , and have a unique common fixed point in .

Proof. Fix arbitrary , . From condition (i), , there exist , such that Since , there exist , such that Continuing in this way, we can construct four sequences , , , and in such that Step 1. We will show that , in are Cauchy sequences. Putting , in the equality (18), we get Likewise, , in the equality (18); we get: By the mentioned above, we can obtain that Letting , it follows that By the property of , we get Analogously, we can prove that In conclusion, we obtain that Therefore, Now, suppose to the contrary that, , are not Cauchy sequences. Then there exists , for which we can find subsequences , of and , of with such that By view of (35) and triangle inequality, we get Letting , we obtain: Again by means of triangle inequality, we have Thus,Letting , it yields that .
Now, by means of property of , it follows that which implies that which contracts with (37). Thus, , in are Cauchy sequences.
Step 2. We will show that and . Suppose to the contrary that and . Since is complete, there exist , such that Also its subsequences converge as follows: Since and are continuous, , and , , also , and , . Since and are compatible, we have , , and , .
Putting , in the equality (18), we have Letting , we obtain Putting , in the equality (18), we have: Letting , we obtain Combined with inequalities (45) and (47), it is a contraction. Therefore, Step 3. We will show that . Putting and in the equality (18), we get Letting , it yields that Hence It suffices to show that . Putting and in the equality (18), we get It implies that . Thus, . Until now, we show that .
Analogously, we can also obtain that .
Step 4. Set Since and are both compatible, we can get We show that . Suppose to the contrary that . Putting and in the equality (18), we get It implies that ; it is a contradiction! So, we have .
Step 5. We will show that . Putting and in the equality (18), we get It implies that .
In sequel, we show that . Putting and in the equality (18), we get It implies that . Therefore, we get . So, is a unique common fixed point of , , , and in .

Sequentially, continuity of and is removed, and the compatibility of is relaxed to the weak compatibility of ; we propose the following theorem. It is noted that the closedness of or is necessary.

Theorem 10. Let be a complete partially ordered metric space and be the mappings. Suppose the following conditions hold:
(i) , ;
(ii) and are both weakly compatible;
(iii) There exists such that, for all , , , , the inequality (18) holds;
(iv) or is closed in .
Then , , and have a unique common fixed point in .

Proof. Let , . Then followed by Theorem 9, we construct sequences , in . , are two Cauchy sequences; since is complete, , are converge sequences and its subsequences converge as follows: Case 1 ( is closed in ). Since is closed in , there exists , such that Step 1. Putting and in the equality (18), we get Letting , it yields that It follows that . Since is weakly compatible, then ; that is, . Similarly, we can obtain that .
Step 2. If , putting and in the equality (18), we get Letting , it yields that If , putting and in the equality (18), we get Letting , it yields that which is a contradiction! So, we have that It follows that , .
Since , then there exists , such that , .
Step 3. Putting and in the equality (18), we get Letting , it yields that It yields that . Similarly, we can show that . Since is weakly compatible, then ; that is, . Similarly, we can obtain that .
Analogous to Step 2, we can obtain that , . To sum up, we have that Step 4. Putting and in the equality (18), we get Since the property of , . Then, .
Until now, we have proved that . Thus, is a unique common fixed point of , , and .
Case 2 ( is closed in ). This proof of Case 2 is similar as the Case 1 provided above.

In the following, compared to Theorem 10, completeness of is removed. And instead, we assume that () or () has property (E.A).

Theorem 11. Let be a partially ordered metric space and be the mappings. Suppose the following conditions hold:
(i) , ;
(ii) and are both weakly compatible;
(iii) There exists such that for all , , , , the inequality (18) holds;
(iv) is closed in and satisfies the property or is closed in and satisfies the property .
Then , , and have a unique common fixed point in .

Proof.
Case 1. is closed in and satisfies the property .
Step 1. Since satisfies the property , then there exists a sequence in such that Since , there exists a sequence in such that . Hence . Let us show that . In fact, by means of the equality (18), putting , , we get Letting , it follows that It yields that . Since is closed in , then for some . Thus, .
Step 2. Putting , in the equality (18), we get Letting , it yields that . Thus, . It follows that . Since is weakly compatible, then ; that is, .
Step 3. Since , then there exists such that . We claim that . Putting , in the equality (18), we get It yields that . Thus, . Since is weakly compatible, then ; that is, .
Step 4. Putting , in the equality (18), we get Since the property of , it yields that . Thus, .
Step 5. Putting , in the equality (18), we getSince the property of , it yields that . Thus, . Hence, we obtain that . So, is a common fixed point of , , and .
Step 6 (uniqueness). Suppose that there is another common fixed point of , , and . It is obviously that . Putting , in the equality (18), we get Since the property of , it yields that . Thus, . So, is unique common fixed point of , , and .
Case 2. is closed in and satisfies the property .
The proof can be followed by the line of process of Case 1. To avoid repetition, we omit it.

The next theorem is that the property (E.A) of or can be replaced by the property of and . At the same time, the containment relationships of (or ) and (or ) are removed. As we all know, under the property, it is not necessary to assume the completeness of the spaces, which is an important advantage compared with most of the theorems in fixed point theory.

Theorem 12. Let be a partially ordered metric space and be the mappings. Suppose the following conditions hold:
(i) and share the property;
(ii) and are both weakly compatible;
(ii) There exists such that for all , , , , the inequality (18) holds.
Then , , and have a unique common fixed point in .

Proof. Since the pairs and share the property, there exist two sequences , in such that Since , there exists a point such that . Putting and in the equality (18), it yields that Letting , it follows that Thus, . Hence, , which shows is a coincidence point of the pair .
As , there exists a point such that ; putting , in the equality (18), we have Letting , it follows that Thus, . Hence, , which shows is a coincidence point of the pair .
Since the pair is weakly compatible, and by the previous proof, ; then ; it yields that . And since the pair is weakly compatible, and by the previous proof, ; then ; it yields that . Letting in the inequality (18), we obtain Since , then ; it follows that Thus,  . Therefore, .
Sequentially, letting , in the inequality (18), we obtain Since , then ; it follows that Thus, . By the mentioned above, we have that . Hence, , , and have a common fixed point .
Uniqueness. Assume that is another common fixed point of , , and . It follows that . Letting , in the equality (18), we obtain Since , then ; it follows that and we obtain that . Thus, , , and have a unique common fixed point .

If we utilize Lemma 7, then we have the following.

Corollary 13. Let be a complete partially ordered metric space and be the mappings. Suppose the following conditions hold:
(i) All the conditions in Lemma 7 hold;
(ii) and are both weakly compatible.
Then , , and have a unique common fixed point in .

Proof. By Lemma 7, we can get that and share the property. Then by use of Theorem 12, we can obtain the conclusions.

Instead of the property of and in Theorem 12, we utilize the common property (E.A) to obtain fixed point theorems. It is noticed that it should require the closedness of and .

Theorem 14. Let be a partially ordered metric space and be the mappings. Suppose that the inequality (18) of Lemma 7 and the following hypotheses hold:
(i) The pairs and have the common property (E.A);
(ii) and are closed subsets in ;
(iii) Both the pairs and are weakly compatible.
Then , , and have a unique common fixed point.

Proof. If the pairs and have the common property (EA), then there exist two sequences and in such that Since is closed, then for some . And is closed; then for some . The rest of the proof can run on the lines of Theorem 12.

Corollary 15. The result of Theorem 14 holds if condition (ii) is replaced by condition (ii′):
(ii′) and where denoted the closure of , denoted the closure of .

Corollary 16. The result of Theorem 14 holds if condition (ii) is replaced by condition (ii′′):
(ii′′) and are closed subsets in and and .

In order to show that the common property of two pairs and can be deduced from containment of and property of , we propose the following theorem.

Theorem 17. Let be a partially ordered metric space and be the mappings. Suppose that the inequality (18) of Lemma 7 and the following hypotheses hold:
(i) ;
(ii) is closed in and satisfies the property .
Then and share the common property .

Proof. Since satisfies the property , there exists a sequence such thatSince , for each , there exists a corresponding such that . Therefore, we have that It suffices to show that .
Letting and in the inequality (18), we obtain Letting , we obtain that It follows that .
Thus, we have that Thus, and share the common property .

Remark 18. Theorem 17 shows that our common property of two pairs and is weaker than containment of and property of . It is, namely, that Theorem 17 is indeed a generalization of Theorem 3.4 in [14].

Conflicts of Interest

The authors declare that they have no conflicts of interest.

Authors’ Contributions

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

Acknowledgments

Xiao-lan Liu was partially supported by National Natural Science Foundation of China (Grant no. 61573010), Science Research Fund of Science and Technology Department of Sichuan Province (no. 2017JY0125), Scientific Research Fund of Sichuan Provincial Education Department (no. 16ZA0256), and Scientific Research Project of Sichuan University of Science & Engineering (no. 2014RC01, no. 2014RC03, and no. 2017RCL54). Mi Zhou was partially supported by Hainan Provincial Natural Science Foundation of China (Grant no. 118MS081) and Scientific Research Fund of Hainan Province Education Department (Grant no. Hnjg2016ZD-20).