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Contractions with rational inequalities in the extended bmetric space
Journal of Inequalities and Applications volume 2019, Article number: 220 (2019)
Abstract
In this paper, we prove some fixed point theorems for contractions of rational type in the setting of the extended bmetric spaces. We present some examples to illustrate the validity of our results. Our results improve and generalize a number of fixed point results in the literature.
1 Introduction and preliminaries
Due to the wide application potential, one of the most discussed theorems in nonlinear analysis is the wellknown Banach contraction principle. It has been generalized in several directions, such as, by relaxing the conditions of the abstract spaces, by relaxing the contraction types, and so on. Among them, we shall now mention the interesting papers of Dass and Gupta [3] and Jaggi [4] in which the rational type expressions were considered in the contraction condition (see also, e.g. [1,2,3,4, 6,7,8,9]). For the sake of completeness, we recollect the main results of these papers.
Theorem 1.1
([4])
Let \((\mathcal{M},d)\) be a complete metric space and \(T:\mathcal{M}\rightarrow \mathcal{M}\) be a continuous mapping. If there exist \(\alpha , \beta \in [0,1 )\), with \(\alpha + \beta <1\) such that
for all distinct \(x,y\in \mathcal{M}\), then T possesses a unique fixed point in \(\mathcal{M}\).
Theorem 1.2
([3])
Let \((\mathcal{M},d)\) be a complete metric space and \(T:\mathcal{M}\rightarrow \mathcal{M}\) be a mapping. If there exist \(\alpha , \beta \in [0,1 )\), with \(\alpha +\beta <1\) such that
for all \(x,y\in \mathcal{M}\), then T has a unique fixed point \(u \in X\). Moreover, the sequence \(\{T^{n}x \}\) converges to the fixed point u for all \(x\in \mathcal{M}\).
Throughout this paper, we shall denote the set of positive numbers and the set of real numbers by \(\mathbb{N}\) and \(\mathbb{R}\), respectively.
In this note, we shall reconsider the results of Dass and Gupta [3] and Jaggi [4] in a newly introduced abstract space, known as extended bmetric space. The notion of the extended bmetric space was introduced by Kamran et al. [5] as an extension of bmetric space.
Definition 1.1
([5])
Let \(\mathcal{M}\) be a nonempty set and \(\theta : \mathcal{M}\times \mathcal{M}\rightarrow [1,\infty )\). A function is called an extended bmetric if for all \(x,y,z\in \mathcal{M}\) it satisfies
 \((d_{\theta }1)\) :

if and only if \(x=y\);
 \((d_{\theta }2)\) :

;
 \((d_{\theta }3)\) :

.
The pair is called an extended bmetric space.
It is clear that an extended bmetric space coincides with the corresponding bmetric space, for \(\theta (x,y)=s \geq 1\) where \(s\in \mathbb{R}\) and it turns to be standard metric if \(s=1\).
As expected, the basic topological notions were defined analogously.
Definition 1.2
([5])
Let be an extended bmetric space.

(i)
A sequence \({x_{n}}\) in \(\mathcal{M}\) is said to converge to \(x\in \mathcal{M}\), if for every \(\epsilon >0\) there exists \(N=N(\epsilon )\in \mathbb{N}\) such that , for all \(n\geq N\). In this case, we write \(\lim_{n\rightarrow \infty } x_{n} = x\).

(ii)
A sequence \({x_{n}}\) in \(\mathcal{M}\) is said to be Cauchy if for every \(\epsilon >0\) there exists \(N=N(\epsilon )\in \mathbb{N}\) such that , for all \(m,n\geq N\).
Definition 1.3
([5])
An extended bmetric space is complete if every Cauchy sequence in \(\mathcal{M}\) is convergent.
Notice that an extended bmetric need not be continuous.
Lemma 1.1
([5])
Let be an extended bmetric space. If is continuous, then every convergent sequence has a unique limit.
For the sake of simplicity, throughout the paper, we assume that represents a complete extended bmetric space. In addition, we assume that is a continuous functional unless otherwise stated.
In what follows, we recollect the main results of Kamran et al. [5] which is an analog of the Banach contraction principle in the context of extended bmetric space.
Theorem 1.3
([5])
Let \(T:\mathcal{M}\rightarrow \mathcal{M}\) be mapping. If there exists such that
for all \(x,y\in \mathcal{M} \), where for each \(x_{0}\in \mathcal{M}\), , where \(x_{n}=T^{n}x_{0}\), \(n \in \mathbb{N}\) Then T has precisely one fixed point u. Moreover, for each \(y\in \mathcal{M}\), \(T^{n}y\rightarrow u\).
The main purpose of this paper is to extend the results of Kamran et al. [5] for the wellknown fixed point results, including the interesting theorems of Dass and Gupta [3] and Jaggi [4] in the frame of an extended bmetric space.
2 Main results
Theorem 2.1
Let \(T:\mathcal{M}\rightarrow \mathcal{M}\) be a continuous mapping such that, for all distinct \(x,y\in \mathcal{M}\),
where and
Suppose also that, for each \(x_{0}\in \mathcal{M}\), , where \(x_{n}=T^{n}x_{0}\), \(n\in \mathbb{N}\). Then T has a fixed point u. Moreover, for each \(x\in \mathcal{M}\), we have \(T^{n}x \rightarrow u\).
Proof
By presumptions, for given \(x_{0}\in \mathcal{M}\) we construct the sequence \(\{x_{n} \}\) in \(\mathcal{M}\) as \(x_{n}=T^{n}x _{0}=Tx_{n1}\), for \(n\in \mathbb{N}\). If \(x_{n_{0}}=x_{n_{0}+1}=T x _{n_{0}}\) for some \(n_{0}\in \mathbb{N}_{0}:=\mathbb{N}\cup \{0\}\), then \(x^{\ast }=x_{n_{0}}\) forms a fixed point for T which completes the proof. Consequently, throughout the proof, we assume that
By taking \(x=x_{n1}\) and \(y=x_{n}\) in the inequality (4), we derive that
with
Thus,
For refining the inequality above, we shall consider the following cases:

Case (i).
If , then , which is a contradiction.

Case (ii).
If , then the inequality (4) turns into the inequality below:
(9) 
Case (iii).
Suppose that ; this yields
(10)We shall illustrate that this case is not possible. For this reason, we consider the following subcases:
 Case (iii)_{a}.:

Suppose that , that is,
(11)On the other hand, from (10), we have
(12)By a simple calculation, we derive, from the inequality above, that
which contradicts the assumption (11).
 Case (iii)_{b}.:

Assume that , that is,
(13)Furthermore, from (10), we observe that
(14)A simple evaluation implies, from the inequality above, that
which contradicts the assumption (13). Hence, Case (iii) does not occur.
Consequently, we can state that the inequality (9) holds for all these three cases and by applying it recursively we obtain
Since , we find that
On the other hand, by \((d_{\theta }3)\), together with the triangular inequality, for \(p\geq 1\), we derive that
Notice the inequality above is dominated by .
On the other hand, by employing the ratio test, we conclude that the series converges to some \(S \in (0,\infty )\). Indeed, and hence we get the desired result. Thus, we have
Consequently, we observe, for \(n\leq 1\), \(p\leq 1\), that
Letting \(n\rightarrow \infty \) in (18), we conclude that the constructed sequence \(\{x_{n}\}\) is Cauchy in the extended bmetric space . Regarding the assumption of the completeness, we conclude that there exists \(u\in \mathcal{M}\) such that \(x_{n}\rightarrow u\) as \(n\rightarrow \infty \). Due to the continuity of T, we shall show that the limit point u is a fixed point of T. Indeed, we have
□
Corollary 2.1
A continuous mapping \(T:\mathcal{M}\rightarrow \mathcal{M}\) has a fixed point provided that, for all distinct \(x,y\in \mathcal{M}\),
where , \(i\in \{1,2,3,4 \}\) with and for each \(x_{0}\in \mathcal{M}\), , where \(x_{n}=T^{n}x_{0}\), \(n\in \mathbb{N}\).
Proof
The proof follows from Theorem 2.1 by letting . Indeed, we have
Regarding the analogy, we skip the details. □
Corollary 2.2
Let \(T:\mathcal{M}\rightarrow \mathcal{M}\) be a continuous mapping. Suppose that, for all distinct \(x,y\in \mathcal{M}\), we have the inequality
where \(\alpha , \beta \in [0,1 )\), \(\alpha +\beta <1\). Suppose also that, for each \(x_{0}\in \mathcal{M}\), \(\lim_{n,m\rightarrow \infty } \theta (x_{n}, x_{m})<\frac{1}{\alpha + \beta }\), where \(x_{n}=T^{n}x_{0}\), \(n\in \mathbb{N}\). Then T has a unique fixed point u. Moreover, for each \(x\in \mathcal{M}\), we have \(T^{n}x\rightarrow u\).
Proof
The proof follows from Corollary 2.1 by letting , where and , . □
Example 2.1
Let be a complete extended bmetric space, where \(\mathcal{M}= [0,\infty )\) and , and \(\theta : \mathcal{M}\times \mathcal{M}\rightarrow [1,\infty )\) is defined as \(\theta (x,y)=x+y+2\). Let \(T:\mathcal{M}\rightarrow \mathcal{M}\) be defined by \(Tx=\frac{x}{3}\). Obviously,
and, choosing \(\alpha =\frac{2}{9}\) and \(\beta =\frac{1}{9}\), we have
By routine calculation, we obtain
Therefore, all conditions of Corollary 2.2 are satisfied. Thus, T has a fixed point.
Corollary 2.3
Let \(T:\mathcal{M}\rightarrow \mathcal{M}\) be a mapping such that, for all \(x,y\in \mathcal{M}\) we have
where \(\alpha , \beta \in [0,1 )\), \(\alpha +\beta <1\). Suppose also that, for each \(x_{0}\in \mathcal{M}\), \(\lim_{n,m\rightarrow \infty } \theta (x_{n}, x_{m})<\frac{1}{\alpha + \beta }\), where \(x_{n}=T^{n}x_{0}\), \(n\in \mathbb{N}\). Then T has a unique fixed point u. Moreover, for each \(x\in \mathcal{M}\), we have \(T^{n}x\rightarrow u\).
Proof
Letting , where and , in Corollary 2.1, and following the steps of the proof of Theorem 2.1, we know that there exists \(u\in \mathcal{M}\) such that \(T^{n}x\rightarrow u\). We must prove that this point is the unique fixed point of T. Indeed,
Letting \(n\rightarrow \infty \) in the above inequality we get . Hence \(Tu=u\).
In order to show the uniqueness, suppose that there exists \(v\in \mathcal{M}\) such that \(Tu=u\neq v=Tv\). Then
which is a contradiction. Thus, we have completed the proof. □
Example 2.2
Let \(\mathcal{M}= \{1,2,3,4,\ldots \}\) and define as
where \(\theta :\mathcal{M}\times \mathcal{M}\rightarrow [1,\infty )\) is a function defined by
Then forms an extended bmetric space (see Example 3.1 in [10]).
Let \(T:\mathcal{M}\rightarrow \mathcal{M}\) be defined by
Let also \(\alpha =\frac{1}{16}\) and \(\beta =\frac{1}{8}\). Of course, since and for any \(x\in \{3,4,5,\ldots \}\), the inequality (22) becomes
For the case \(x=2\) and \(y=1\), we have and . Obviously, we have
For all other cases, and the existence of a fixed point is ensured by Corollary 2.3.
Corollary 2.4
Theorem 1.3 is concluded from Corollary 2.2 and hence Theorem 2.1.
Proof
It is sufficient to take \(\alpha =0\). □
2.1 Removing the necessity of the continuity of the functional
In the following theorem, we relax the condition by removing the continuity of the functional in the following setting.
Theorem 2.2
Let \(T:\mathcal{M}\rightarrow \mathcal{M}\) be a mapping that satisfies the inequality
for all \(x,y\in \mathcal{M}\), where , be such that, for each \(x_{0}\in \mathcal{M}\), , where \(x_{n}=T^{n}x_{0}\), \(n\in \mathbb{N}\). Then T has a unique fixed point u. Moreover, for each \(y\in X\), \(T^{n}y\rightarrow u\).
Proof
By taking \(x=x_{n1}\) and \(y=x_{n}\) in the inequality (24), we get
Recursively, we derive that
and regarding that , we derive that
On the other hand, by following the same lines in the previous theorem, we conclude that \(\{x_{n}\}\) is a Cauchy sequence. Since \(\mathcal{M}\) is complete, there exists \(u\in \mathcal{M}\) such that the sequence \(\{x_{n}=T^{n}x_{0}\}\) converges to u, that is,
As a next step, we shall prove that u is a fixed point of T. By using (24) and the triangle inequality, we have
Letting \(n\rightarrow \infty \), and taking (27) and (28) into account, we get
Accordingly, we have , that is, \(Tu=u\).
Lastly, we shall indicate that this fixed point is unique. Suppose, on the contrary, that it is not unique. Thus, there exists another fixed point v of T that is distinct from u. Now, by using (24) and ,
which shows that \(u=v\). Therefore, T has a unique fixed point. □
Example 2.3
Let \(\mathcal{M}= \{\frac{1}{2}, \frac{1}{4}, \frac{1}{8} \}\) and the functions \(\theta :\mathcal{M}\times \mathcal{M}\rightarrow [1, \infty )\), be defined as \(\theta (x,y)=x+y+1\) and
Let and \(T:\mathcal{M}\rightarrow \mathcal{M}\) be defined by
We have \(T^{n}x\rightarrow 0\) for any \(x\in \mathcal{M}\) and .
But
and
which proves that is an extended bmetric on \(\mathcal{M}\).
We will consider the following cases:

(i)
For \(x=\frac{1}{2}\), \(y=\frac{1}{4}\), we have , so the condition from Theorem 2.2 is satisfied.

(ii)
For \(x=\frac{1}{2}\), \(y=\frac{1}{8}\)

(iii)
For \(x=\frac{1}{4}\), \(y=\frac{1}{8}\)
Therefore all conditions of Theorem 2.2 are satisfied. Hence T has a unique fixed point, \(x=\frac{1}{4}\).
Corollary 2.5
Let \(T:\mathcal{M}\rightarrow \mathcal{M}\) be a mapping that satisfies
for all \(x,y\in \mathcal{M}\), where \(\alpha , \beta \in [0,1 )\), \(\alpha +\beta <1\) are such that, for each \(x_{0}\in \mathcal{M}\), \(\lim_{n,m\rightarrow \infty } \theta (x_{n}, x_{m})<\frac{1}{\alpha + \beta }\), where \(x_{n}=T^{n}x_{0}\), \(n\in \mathbb{N}\). Then T has a unique fixed point u. Moreover, for each \(x\in \mathcal{M}\), \(T^{n}x\rightarrow u\).
3 Conclusion
It is clear that the corresponding results in the setting of both bmetric space and standard metric space can be included in our results, by letting \(\theta (x,y)=s\geq 1\) and \(\theta (x,y)= 1\), respectively, in the related places. In particular, the main results of Dass and Gupta [3], Jaggi [4] and also the wellknown Banach contraction mapping principle are derived from our results.
Change history
29 August 2019
The authors would like to express their sincere appreciation to the Deanship of Scientific Research at King Saud University for funding this group No. RG1437017. The corrected funding of the original article [1] is listed below.
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Alqahtani, B., Fulga, A., Karapınar, E. et al. Contractions with rational inequalities in the extended bmetric space. J Inequal Appl 2019, 220 (2019). https://doi.org/10.1186/s1366001921766
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DOI: https://doi.org/10.1186/s1366001921766