#### Abstract

Using the concept of common (*E.A*) property, we prove a common fixed point theorem for three pairs of weakly compatible self-maps satisfying a new contractive condition in the framework of a generalized metric space. Our results do not rely on any commuting or continuity condition of mappings. An example is provided to support our result. The results obtained in this paper differ from the recent relative results in the literature.

#### 1. Introduction

The study of fixed points and common fixed points of mappings satisfying certain contractive conditions has been at the center of rigorous research activity. In 2006, Mustafa and Sims [1] introduced the concept of generalized metric spaces or simply -metric spaces as a generalization of the notion of a metric space. Based on the notion of generalized metric spaces, Mustafa et al. [2–5], Obiedat and Mustafa [6], Aydi et al. [7, 8], Gajié and Stojakovié [9], and Zhou and Gu [10] obtained some fixed point results for mappings satisfying different contractive conditions. Shatanawi [11] obtained some fixed point results for -maps in -metric spaces. Chugh et al. [12] obtained some fixed point results for maps satisfying property in -metric spaces. Study of common fixed point problems in -metric spaces was initiated by Abbas and Rhoades [13]. Subsequently, many authors obtained many common fixed point theorems for the mappings satisfying different contractive conditions (see [14–32] for more details). Recently, Abbas et al. [33] and Mustafa et al. [34] obtained some common fixed point results for a pair of mappings satisfying (*E.A*) property under certain generalized strict contractive conditions in -metric spaces. Long et al. [35] obtained some common coincidence and common fixed points results of two pairs of mappings when only one pair satisfies (*E.A*) property in the framework of a generalized metric space.

The aim of this paper is to study common fixed point of three pairs of mappings for which only two pairs need to satisfy common (*E.A*) property in the framework of -metric spaces. Our results do not rely on any commuting or continuity condition of mappings.

#### 2. Definitions and Preliminary Results

In this section, we present the necessary definitions and results in -metric spaces.

*Definition 1 (see [1]). *Let be a nonempty set, and let be a function satisfying the following axioms:(G1) if ; (G2), for all with ; (G3), for all , and with ; (G4) (symmetry in all three variables); (G5) for all , and , (rectangle inequality).Then the function is called a generalized metric, or more specifically a -metric on and the pair are called a -metric space.

It is known that the function on -metric space is jointly continuous in all three of its variables, and if and only if (see [1] for more details and the reference therein).

*Definition 2 (see [1]). *Let be a -metric space, and let be a sequence of points in ; a point in is said to be the limit of the sequence if , and one says that sequence is -convergent to .

Thus, if in a -metric space , then for any there exists (throughout this paper we mean by the set of all natural numbers) such that , for all .

Proposition 3 (see [1]). *Let be a -metric space; then the followings are equivalent:*(1) is -*convergent to *.(2).(3).(4).

*Definition 4 (see [1]). *Let be a -metric space. A sequence is called -Cauchy sequence if, for each , there exists such that for all , that is, if as .

*Definition 5 (see [1]). *A -metric space is said to be -complete (or a complete -metric space) if every -Cauchy sequence in is -convergent in .

Proposition 6 (see [1]). *Let be a -metric space. Then the following are equivalent.*(1)*The sequence *-*Cauchy*.(2)*For every *, *there exists ** such that *, *for all *.

Proposition 7 (see [1]). *Let be a -metric space. Then the function is jointly continuous in all three of its variables.*

*Definition 8 (see [36]). *Let and be self-maps of a set . If for some in , then is called a coincidence point of and , and is called point of coincidence of and .

*Definition 9 (see [36]). *Two self-mappings and on are said to be weakly compatible if they commute at coincidence points.

*Definition 10 (see [33]). *Let be a -metric space. Self-maps and on are said to satisfy the -(*E.A*) property if there exists a sequence in such that and are -convergent to some .

*Definition 11. *Let be a -metric space and , and four self-maps on . The pairs and are said to satisfy common property if there exist two sequences and in such that , for some .

*Definition 12 (see [17]). *Self-mappings and of a -metric space are said to be compatible if and , whenever is a sequence in such that , for some .

*Definition 13 (see [16]). *A pair of self-mappings of a -metric space is said to be weakly commuting if

*Definition 14 (see [16]). *A pair of self-mappings of a -metric space is said to be -weakly commuting, if there exists some positive real number such that

#### 3. Main Results

In this section, we obtain some unique common fixed point results for six mappings satisfying certain generalized contractive conditions in the framework of a generalized metric space. We start with the following result.

Theorem 15. *Let be a -metric space. Suppose mappings satisfying the following conditions:
**
for all , where is the function satisfying , for all . If one of the following conditions is satisfied, then the pairs , , and have a common point of coincidence in :*(i)*the subspace is closed in , , , and two pairs of and satisfy common property;*(ii)*the subspace is closed in , , , and two pairs of and satisfy common property;*(iii)*the subspace is closed in , , , and two pairs of and satisfy common property.**Moreover, if the pairs , , and are weakly compatible, then , , , , , and have a unique common fixed point in .*

*Proof. *First, we suppose that the subspace is closed in , , , and two pairs of and satisfy common property. Then by Definition 11 we know that, there exist two sequences and in such that
for some .

Since , there exists a sequence in such that . Hence . Next, we will show . In fact, if , then from condition (3), we can get
On letting and based on the property of , we can obtain
It is contradiction; so .

Since is a closed subspace of and , there exists a in such that . We claim that . Suppose not; then by using (3) we obtain
Taking on the two sides of the above inequality and using the property of , we can get
Which is contradiction, and so . Hence, is the coincidence point of pair .

By the condition and , there exists a point in such that . Now, we claim that . In fact, if , then from (3) we have
Letting on the two sides of the above inequality and using the property of , we can obtain
It is contradiction. Hence , and so is the coincidence point of pair .

Since and , there exists a point in such that . We claim that . If not, from (3) and the property of , we have
It is contradiction. Hence , and so is the coincidence point of pair .

Therefore, in all the above cases, we obtain . Now, weakly compatibility of the pairs , , and gives that , , and .

Next, we show that . In fact, if , then from (3) we have
which is contradiction; hence , and so . Similarly, it can be shown that and ; so we get , which means that is a common fixed point of , , , , , and .

Next, we will show the common fixed point of , , , , *,* and is unique. Actually, suppose that ; is another common fixed point of , , , , , and ; then by condition (3) we have
It is a contradiction, unless ; that is, mappings , and have a unique common fixed point.

Finally, if condition (ii) or (iii) holds, then the argument is similar to that above; so we delete it.

This completes the proof of Theorem 15.

Now, we give an example to support Theorem 15.

*Example 16. *Let be a -metric space with
We define mappings , and on by
Note that , , , , , and are discontinuous mappings. Clearly, the subspace is closed in , , , , and the pairs , , and are weakly compatible. Also, the pairs and satisfy common property; indeed, and for each are the required sequences. The control function is defined by .

First, we let
To prove (3), let us discuss the following cases.

*Case 1. *For , then we have , and hence (3) is obviously satisfied.

*Case 2. *For , then we have
Therefore,

*Case 3. *For , , then we have
Next we divide the study in two subcases.(a)If , then we have
(b)If , then we get
Hence, we can get

*Case 4. *For , , then we have
Next we divide the study in two subcases.(a) If , then we obtain
Thus we have
(b) If , then we can get
So, we have

*Case 5. *For , , then we have
Next we divide the study in two subcases.(a) If , then we can get
Thus we have
(b) If , then we obtain
Hence we get

*Case 6. *For , , and , then we have
Next we divide the study in four subcases.(a) If and , then we have
Thus we have
(b) If and , then we have
Therefore, we can get
(c) If and , then we get
Therefore, we can get
(d) If and , then we have
Thus, we can get

*Case 7. *For , , and , then we have
Next we divide the study in four subcases.(a) If and , then we have
Thus, we can get
(b) If and , then we have
Hence, we obtain
(c)If and , then we get
Therefore, we have
(d) If and , then we have
Thus, we obtain

*Case 8. *For , , and , then we have
Next we divide the study in four subcases.(a) If and , then we get
Thus, we obtain
(b) If and , then we have
Thus, we obtain
(c) If and , then we obtain
Thus, we obtain
(d) If and , then we have