#### 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