Abstract

In this paper, we prove some common fixed point results for two mappings satisfying contraction conditions in complete b-metric spaces. Meanwhile, two examples are presented to support our results.

2010 Mathematics Subject Classification. Primary 47H10.

1. Introduction

In 1922, Banach ¹ proved the Banach contraction principle. Since then, several works have been done about fixed point theory regarding different classes of contractive conditions in some spaces such as: quasi-metric spaces ^{2, 3}, cone metric spaces ^{4, 5}, partially order metric spaces ^{6, 7, 8}, G-metric spaces ⁹.

The concept of metric space was introduced by Czerwik in ¹⁰. After that, several papers have been published on the fixed point theory of various classes of single-valued and multi-valued operators in metric spaces (see ^{2, 11, 12}). Aydi et al. in ¹³ proved common fixed point results for single-valued and multi-valued mappings satisfying a weak contraction in metric spaces. Starting from the results of Berinde ¹⁴, Pacurar ¹⁵ proved the existence and uniqueness of fixed point of contractions on b- metric spaces. Using a contraction condition defined by means of a comparison function, ¹⁶ established results regarding the common fixed points of two mappings. Hussain and Shah in ¹⁷ introduced the notion of a cone b- metric spaces, generalizing both the notions of b- metric spaces and cone metric spaces, they considered topological properties of cone metric spaces and results on KKM mappings in the setting of cone b- metric spaces.

The aim of this paper is to consider and establish some common fixed point results for two mappings satisfying contraction conditions in complete b-metric spaces. Meanwhile, two examples are presented to support our results.

2. Preliminaries

Let and denote the sets of all real numbers and nonnegative numbers respectively. denotes the set of positive integers and . Suppose

{ is upper semicontinuous and nondecreasing in each coordinate variable satisfying condition }

and

{ is upper semicontinuous and nondecreasing in each coordinate variable satisfying condition }.

In order to obtain our main results, we need to introduce some definitions and lemmas.

Definition. Let be a nonempty set and A function is called a metric with constant if

(1) if and only if ;

(2) for all

(3) for all

The pair is called a metric space.

It is obvious a metric space with is a metric space. There are examples of metric spaces which are not metric spaces. (see ¹⁸)

Definition. Let be a sequence in a metric space .

(1) A sequence is called convergent if and only if there is such that when ;

(2) is a Cauchy sequence if and only if when .

As usual, a b-metric space is said to be complete if and only if each Cauchy sequence in this space is convergent.

Lemma 2.1. ¹⁹ Let be nondecreasing and upper semicontinuous. Then for each if and only if

3. Main Results

Now we are ready to prove our main results.

Theorem 3.1. Let be a complete metric space with constant Suppose and are two mappings and one of them is continuous. If there exists such that

(1)

for all , then and have a unique common fixed point

Proof. Let be arbitrary. We define a sequence as follows:

We now suppose that for every n. If not, there exists some such that . If then and from the contraction condition (1) with and , we have

Suppose that It follows from the definition of that

which is a contradiction. Therefore, By the definition of the sequence , it means that That is, is a common fixed point of and B.

If then using the same arguments in the case , it can be shown that is a common fixed point of and B.

From now on, we suppose that for all Now we shall prove that

(2)

We consider two cases:

Case I: From the contraction condition (1) with and , we get

If , by virtue of the definition of , one can obtain

a contradiction. It follows that

Hence,

(3)

Case II: Using the same technique in proving the case I, it can be proved that (2) holds for That is,

(4)

From (3) and (4), we conclude that (2) holds for all

Since for all using Lemma 2.3, we obtain that for all . It follows that

(5)

Now we prove that is a Cauchy sequence. To do this, it is sufficient to show that the subsequence is a Cauchy sequence in . Assume on the contrary that is not a Cauchy sequence. Then there exists for which we can find subsequence and so that is the smallest index for which ,

(6)

and

(7)

Using the triangle inequality in b-metric space and (6), we have

Taking the upper limit as , one can obtain

(8)

Also,

hence,

On the other hand, we get

It follows from (5) and (8) that

Consequently,

(9)

Similarly, we deduce that

(10)

Using the triangle inequality in b-metric space and contraction condition (1), we have

In view of above inequality and (5), (9), (10), one can obtain that

It is a contradiction and it follows that is a Cauchy sequence in . Since is complete, there exists such that

Without loss of generality, we suppose A is continuous. It follows that

This implies that is a fixed point of .

Next, we show that is a fixed point of $B.$ In view of the contraction condition (1), we get that

If suppose that then we have

a contradiction. It follows that That is, is also a fixed point of B.

Assume that is another common fixed point of and , that is, Then

which is a contradiction. It follows that is a unique common fixed point in X. This completes the proof.

If in Theorem 1, then we get that:

Corollary 3.2. Let be a complete metric space with constant and be a continuous mapping. If there exists such that

for all , then has a unique fixed point

Theorem 3.3. Let be a complete metric space with constant Suppose and are two mappings and one of them is continuous. If there exists such that

(11)

for all , then and have a unique common fixed point

Proof. Let be arbitrary. We define a sequence as follows:

We now suppose that for every $n.$ Otherwise, there exists some such that . If from the contraction condition (11) with and , one can obtain

We suppose that By the definition of , we have

a contradiction. Hence, It follows from the definition of the sequence that

That is, is a common fixed point of and B.

Similarly, if we can prove that is a common fixed point of and B.

From now on, we suppose that for all Using the similar argument in the proof of Theorem 3.1, one can deduce that

(12)

It follows from Lemma 2.3 that for all , which implies that

(13)

Next we prove that is a Cauchy sequence. Obviously, it is sufficient to show that the subsequence is a Cauchy sequence in . As in the proof of Theorem 3.1, we obtain that inequalities (9),(10) hold, and

(14)

The triangle inequality in b-metric space and contraction condition (11) ensure that

In light of above inequality and (9), (10), (13) and (14), we have

It is a contradiction. Hence, is a Cauchy sequence in . The completeness of ensures that there exists such that

Without loss of generality, we suppose is continuous. It follows that

That is, is a fixed point of .

Next, we shall prove that is a fixed point of B. By the contraction condition (11), we obtain that

If we suppose that then one can get

which is a contradiction. Hence, we deduce that is also a fixed point of B.

Suppose that and are different common fixed points of and , then we obtain that

a contradiction. Consequently, is a unique common fixed point in X. This completes the proof.

If in Theorem 3, we have the following result.

Corollary 3.4. Let be a complete metric space with constant Suppose be a continuous mapping. If there exists such that

(15)

for all , then has a unique common fixed point

4. Examples

Example 4.1. Let endowed with the metric:

with constant Consider mappings by and Define the mapping by

Clearly, is a complete metric space and is continuous with respect to d. So we verify the contraction condition (1).

By calculus, we have

Therefore, we show that the contraction condition (1) is satisfied. It follows that we can apply Theorem 3.1 and and have a unique common fixed point

Example 4.2. Let endowed with the metric:

with constant Define mappings by

and Consider the mapping by

It is easy to verify that is a complete metric space and is continuous with respect to d. By calculus, we obtain that

That is, the contraction condition (11) holds. Theorem 3.3 ensures that and have a unique common fixed point

Data Availability

No data were used to support this study.

Conflicts of Interest

The authors declare that they have no conflicts of interest regarding the pub-lication of this paper.

Authors Contributions

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

Acknowledgments

The second author would like to acknowledge the grant: Science and Research Project Foundation of Liaoning Province Education Department LQN201902 and the Research Foundation for the Doctoral Program of Shenyang Normal University BS201703, for financial support.

References