1 Introduction and preliminaries
Due to the wide application potential, one of the most discussed theorems in nonlinear analysis is the well-known 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. [1‐4, 6‐9]). For the sake of completeness, we recollect the main results of these papers.
Theorem 1.1
([4])
Anzeige
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
\(\mathcal{M}\).
$$ d(Tx, Ty)\leq \alpha \cdot \frac{d(x,Tx) d(y,Ty)}{d(x,y)}+\beta \cdot d(x,y), $$
(1)
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
\(x\in \mathcal{M}\).
$$ d(Tx, Ty)\leq \alpha \cdot d(y,Ty)\frac{1+d(x,Tx)}{1+d(x,y)}+\beta \cdot d(x,y) $$
(2)
Throughout this paper, we shall denote the set of positive numbers and the set of real numbers by \(\mathbb{N}\) and \(\mathbb{R}\), respectively.
Anzeige
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 b-metric space. The notion of the extended b-metric space was introduced by Kamran et al. [5] as an extension of b-metric 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 b-metric if for all \(x,y,z\in \mathcal{M}\) it satisfies
The pair
is called an extended b-metric space.
is called an extended b-metric if for all \(x,y,z\in \mathcal{M}\) it satisfies
\((d_{\theta }1)\)
if and only if \(x=y\);
if and only if \(x=y\);
\((d_{\theta }2)\)
;
;
\((d_{\theta }3)\)
.
.
is called an extended b-metric space.It is clear that an extended b-metric space coincides with the corresponding b-metric 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 b-metric space.
be an extended b-metric 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 \(\lim_{n\rightarrow \infty } x_{n} = x\).
, 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 \(m,n\geq N\).
, for all \(m,n\geq N\).Definition 1.3
([5])
An extended b-metric space
is complete if every Cauchy sequence in \(\mathcal{M}\) is convergent.
is complete if every Cauchy sequence in \(\mathcal{M}\) is convergent.Notice that an extended b-metric need not be continuous.
Lemma 1.1
([5])
Let
be an extended
b-metric space. If
is continuous, then every convergent sequence has a unique limit.
be an extended
b-metric 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 b-metric space. In addition, we assume that
is a continuous functional unless otherwise stated.
represents a complete extended b-metric 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 b-metric 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
\(y\in \mathcal{M}\), \(T^{n}y\rightarrow u\).
such that
(3)
, where
\(x_{n}=T^{n}x_{0}\), \(n \in \mathbb{N}\)
Then
T
has precisely one fixed point
u. Moreover, for each
\(y\in \mathcal{M}\), \(T^{n}y\rightarrow u\).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
\(T^{n}x \rightarrow u\).
(4)
and
(5)
, 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 \(x_{n}=T^{n}x _{0}=Tx_{n-1}\), 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 \(x=x_{n-1}\) and \(y=x_{n}\) in the inequality (4), we derive that
with
Thus,
$$ x_{n}\neq x_{n+1} \quad \text{{for all }} n\in \mathbb{N}_{0}. $$
(6)
(7)
(8)
For refining the inequality above, we shall consider the following cases:
Case (i).
If
, then
, which is a contradiction.
, then
, which is a contradiction.
Case (iii).
Suppose that
; this yields
We shall illustrate that this case is not possible. For this reason, we consider the following subcases:
; this yields
(10)
Case (iii)a.
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
(15)
, we find that
(16)
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
.
(17)
.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 b-metric 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
□
converges to some \(S \in (0,\infty )\). Indeed,
and hence we get the desired result. Thus, we have
(18)
. 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
$$ Tu=T\Bigl( \lim_{n\rightarrow \infty }x_{n}\Bigr)= \lim _{n\rightarrow \infty }T(x_{n})= \lim_{n\rightarrow \infty }x_{n+1}=u. $$
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
\(x_{n}=T^{n}x_{0}\), \(n\in \mathbb{N}\).
(19)
, \(i\in \{1,2,3,4 \}\)
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. □
. Indeed, we have
(20)
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
\(T^{n}x\rightarrow u\).
(21)
Proof
Example 2.1
Let
be a complete extended b-metric 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
be a complete extended b-metric 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,
$$\begin{aligned} \lim_{n\rightarrow \infty }\theta (x_{n}, x_{n+p}) =& \lim_{n\rightarrow \infty }\theta \bigl(T^{n}x, T^{n+p}x \bigr) \\ =& \lim_{n\rightarrow \infty }\theta \biggl(\frac{x}{3^{n}}, \frac{x}{3^{n+p}}\biggr) \\ =& \lim_{n\rightarrow \infty } \biggl(\frac{x}{3^{n}}+ \frac{x}{3^{n+p}}+2 \biggr)< 3=\frac{1}{\alpha +\beta }. \end{aligned}$$
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
\(T^{n}x\rightarrow u\).
(22)
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 \(Tu=u\).
, 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,
(23)
. 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 b-metric space (see Example 3.1 in [10]).
as
$$ \theta (x,y)=\textstyle\begin{cases} \vert x-y \vert ^{3}, &\text{{if }}x\neq y, \\ 1, &\text{{if }}x= y. \end{cases} $$
forms an extended b-metric 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.
$$ Tx=\textstyle\begin{cases} 3, &\text{{if }} x=1, \\ 4, &\text{{otherwise.}} \end{cases} $$
and
for any \(x\in \{3,4,5,\ldots \}\), the inequality (22) becomes
and
. Obviously, we have
and the existence of a fixed point is ensured by Corollary 2.3.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.
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
\(y\in X\), \(T^{n}y\rightarrow u\).
(24)
, 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 \(x=x_{n-1}\) and \(y=x_{n}\) in the inequality (24), we get
Recursively, we derive that
and regarding that
, we derive that
(25)
(26)
, we derive that
(27)
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
(28)
(29)
Accordingly, we have
, that is, \(Tu=u\).
, 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
.
be defined as \(\theta (x,y)=x+y+1\) and
and \(T:\mathcal{M}\rightarrow \mathcal{M}\) be defined by
$$ T\frac{1}{2}=T\frac{1}{4}=\frac{1}{4}\quad \text{{and}}\quad T \frac{1}{8}=\frac{1}{2}. $$
.But
and
which proves that
is an extended b-metric on \(\mathcal{M}\).
$$\begin{aligned}& \theta \biggl(\frac{1}{2},\frac{1}{4} \biggr)= \frac{1}{2}+\frac{1}{4}+1= \frac{7}{4}, \\& \theta \biggl(\frac{1}{2},\frac{1}{8} \biggr)= \frac{1}{2}+\frac{1}{8}+1=\frac{13}{8}, \\& \theta \biggl(\frac{1}{8}, \frac{1}{4} \biggr)= \frac{1}{8}+\frac{1}{4}+1=\frac{11}{8} \end{aligned}$$
is an extended b-metric on \(\mathcal{M}\).We will consider the following cases:
(i)
(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
\(x\in \mathcal{M}\), \(T^{n}x\rightarrow u\).
(30)
3 Conclusion
It is clear that the corresponding results in the setting of both b-metric 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 well-known Banach contraction mapping principle are derived from our results.
Competing interests
The authors declare that they have no competing interests.
Open Access This article is distributed under the terms of the Creative Commons Attribution 4.0 International License (http://creativecommons.org/licenses/by/4.0/), which permits unrestricted use, distribution, and reproduction in any medium, provided you give appropriate credit to the original author(s) and the source, provide a link to the Creative Commons license, and indicate if changes were made.
Publisher’s Note
Springer Nature remains neutral with regard to jurisdictional claims in published maps and institutional affiliations.