Abstract

We introduce the notions of the function and function, and then we prove two common fixed point theorems in complete generalized metric spaces under contractive conditions with these two functions. Our results generalize or improve many recent common fixed point results in the literature.

1. Introduction and Preliminaries

In 2000, Branciari [1] introduced the following notion of a generalized metric space where the triangle inequality of a metric space had been replaced by an inequality involving three terms instead of two. Later, many authors worked on this interesting space (e.g., [2–7]).

Let be a generalized metric space. For and , we define that Branciari [1] also claimed that is a basis for a topology on , is continuous in each of the coordinates, and a generalized metric space is a Hausdorff space. We recall some definitions of a generalized metric space as follows.

Definition 1.1 (See [1]). Let be a nonempty set and be a mapping such that for all and for all distinct point each of them different from and , one has the following:(i)if and only if ,(ii),(iii)(rectangular inequality). Then is called a generalized metric space (or shortly ).

We present an example to show that not every generalized metric on a set is a metric on .

Example 1.2. Let with be a constant, and we define by(1), for all ,(2)for all , (3),(4),(5),(6),where is a constant. Then let be a generalized metric space, but it is not a metric space, because

Definition 1.3 (See [1]). Let be a , be a sequence in and . We say that is convergent to if and only if as . We denote by as .

Definition 1.4 (See [1]). Let be a , be a sequence in and . We say that is Cauchy sequence if and only if, for each , there exists such that for all .

Definition 1.5 (See [1]). Let be a . Then is called complete if every Cauchy sequence is convergent in .

In this paper, we also recall the concept of compatible mappings and prove two common fixed point theorems which incorporated the compatible map concept followed. In 1986, Jungck [8] introduced the below concept of compatible mappings.

Definition 1.6 (See [8]). Let be a , and let be two single-valued functions. We say that and are compatible if whenever is a sequence in such that .
In particular, if by taking for all .

Later, many authors studied this subject (compatible mappings), and many results on fixed points and common fixed points are proved (see, e.g., [9–14]).

2. Main Results

In this paper, we first introduce the below concept of the function.

Definition 2.1. We call a function if the function satisfies the following conditions:, .

Lemma 2.2. Let be a function. Then for all , where denotes the th iteration of .

Proof. Let be fixed. If for some , then It follows that for all , and so we get that .
If for all , then we put . Thus, Since is a function, we have that . Therefore, the sequence is strictly decreasing and bounded from below, and so there exists an such that . We claim that . If not, suppose that , then we have that a contradiction. So we obtain that , that is, .

We now state the main common fixed-point theorem for the function in a complete , as follows.

Theorem 2.3. Let be a Hausdorff and complete , and let be a function. Let be four single-valued functions such that for all , Assume that and , and the pairs and are compatible. If or is continuous, then , and have a unique common fixed point in .

Proof. Given that . Define the sequence recursively as follows:
Step 1. We will prove that Using (2.4), we have that for each and so we can conclude that Similarly, we also conclude that Generally, we have that for each By induction, we get that By Lemma 2.2, we obtained that .
We claim that is Cauchy. We claim that the following result holds.Step 2. Claim that, for every , there exists such that if , then .
Suppose that the above statesment is false. Then there exists such that, for any , there are with satisfying that(a) is even and is odd,(b),(c) is the smallest even number such that the condition (b) holds. Taking into account (b) and (c), we have that Letting , we get the following: Letting , we get the following: Using (2.4), (2.13), and (2.15), we have taking , we get that , a contradiction. So is Cauchy. Since is complete, there exists such that . So we have as .Step 3. We will show that is a common fixed point of , , , and .
Assume that is continuous. Then we have as . By the rectangular property, we have Since and are compatible and as , we conclude that as . Taking into account (2.18), (2.19), and (2.20), we have that as . Since for each . Taking and taking into account (2.17), (2.18), (2.19), (2.20), and (2.21), we get that and this is a contradiction unless , that is, .
On the same way, we have that, for each , letting , we obtained the following: and this is a contradiction unless , that is, .
Since , put such that . Then and using (2.4), and this is a contradiction unless , that is, and so .
Since and are compatible and , we have that which implies that . Using (2.4), we also have and this is a contradiction unless , that is, .
From above argument, we get that and so is a common fixed point of , and .Step 4. Finally, to prove the uniqueness of the common fixed point of and , let be another common fixed point of , and . Then using (2.4), we have and this is a contradiction unless , that is, . Hence is the unique common fixed point of , , , and in .

We give the following example to illustrate Theorem 2.3.

Example 2.4. Let with is a constant, and we define by(1)for all ,(2)for all ,(3),(4),(5),(6) and ,where is a constant.
If , , then is a function. We next define , , , by Then all conditions of Theorem 2.3 are satisfied, and is a unique common fixed point of , , , and .

For the case (the identity mapping) and , we are easy to get the below fixed-point theorem.

Theorem 2.5. Let be a Hausdorff and complete , and let be a function. Let be a single-valued function such that for all , Then has a unique fixed point in .

We next introduce the below concept of the function.

Definition 2.6. We call a function if the function satisfies the following conditions: is a strictly increasing, continuous function in each coordinate,() for all ,,,, and .

Example 2.7. Let denote that Then is a function.

We now state the main common fixed point theorem for the function in a complete .

Theorem 2.8. Let be a Hausdorff and complete , and let be a function. Let be four single-valued functions such that for all , Assume that and , and the pairs and are compatible. If or is continuous, then , , , and have a unique common fixed point in .

Proof. Given . Define the sequence recusively as follows:
Step 1. We will prove that Using (2.34) and the definition of the function, we have that for each and so we can conclude that Similarly, we also conclude that Generally, we have that for each Now, for each , if we denote , then is a strictly decreasing sequence. Thus, it must converge to some with . We claim that . If not, suppose that , then Passing to the limit, as , we have that , which is a contradiction. So we get .
We claim that is Cauchy. We claim that the following result holds.Step 2. Claim that, for every , there exists such that if , then .
Suppose that the above statesment is false. Then there exists such that, for any , there are with satisfying that(d) is even and is odd,(e),(f) is the smallest even number such that the condition (e) holds. Taking into account (e) and (f), we have that Letting , we get the following: Letting , we get the following:
Using (2.34), (2.43), and (2.45), we have taking , we get that , a contradiction. So is Cauchy. Since is complete, there exists such that . So we have as .Step 3. We will show that is a common fixed point of , , , and .
Assume that is continuous. Then, we have as . By the rectangular property, we have Since and are compatible and as , we conclude that as . Taking into account (2.48), (2.49), and (2.50), we have that as . Since for each . Letting and taking into account (2.47), (2.48), (2.49), (2.50), and (2.51), we get that and this is a contradiction unless , that is, .
On the same way, we have that, for each , Taking , we obtained that and this is a contradiction unless , that is, .
Since , put such that . Then and using (2.34), and this is a contradiction unless , that is, and so .
Since and are compatible and , we have that which implies that . Using (2.34), we also have and this is a contradiction unless , that is, .
From above argument, we get that and so is a common fixed point of , , , and .Step 4. Finally, to prove the uniqueness of the common fixed point of , , , and , let be another common fixed point of , , , and . Then using (2.34), we have and this is a contradiction unless , that is, . Hence is the unique common fixed point of , , , and in .