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
$$ x_{n}\neq x_{n+1} \quad \text{{for all }} n\in \mathbb{N}_{0}. $$
(6)
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:

Case (iii).
Suppose that ; this yields
We shall illustrate that this case is not possible. For this reason, we consider the following subcases:
 Case (iii)_{a}.:

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

Assume that , that is,
Furthermore, from (10), we observe that
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
$$ Tu=T\Bigl( \lim_{n\rightarrow \infty }x_{n}\Bigr)= \lim _{n\rightarrow \infty }T(x_{n})= \lim_{n\rightarrow \infty }x_{n+1}=u. $$
□
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
\(x_{n}=T^{n}x_{0}\), \(n\in \mathbb{N}\).
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
$$\begin{aligned} \lim_{n\rightarrow \infty }\theta (x_{n}, x_{n+p}) =& \lim_{n\rightarrow \infty }\theta \bigl(T^{n}x, T^{n+p}x \bigr) \\ =& \lim_{n\rightarrow \infty }\theta \biggl(\frac{x}{3^{n}}, \frac{x}{3^{n+p}}\biggr) \\ =& \lim_{n\rightarrow \infty } \biggl(\frac{x}{3^{n}}+ \frac{x}{3^{n+p}}+2 \biggr)< 3=\frac{1}{\alpha +\beta }. \end{aligned}$$
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
$$ \theta (x,y)=\textstyle\begin{cases} \vert xy \vert ^{3}, &\text{{if }}x\neq y, \\ 1, &\text{{if }}x= y. \end{cases} $$
Then forms an extended bmetric space (see Example 3.1 in [10]).
Let \(T:\mathcal{M}\rightarrow \mathcal{M}\) be defined by
$$ Tx=\textstyle\begin{cases} 3, &\text{{if }} x=1, \\ 4, &\text{{otherwise.}} \end{cases} $$
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
$$ T\frac{1}{2}=T\frac{1}{4}=\frac{1}{4}\quad \text{{and}}\quad T \frac{1}{8}=\frac{1}{2}. $$
We have \(T^{n}x\rightarrow 0\) for any \(x\in \mathcal{M}\) and .
But
$$\begin{aligned}& \theta \biggl(\frac{1}{2},\frac{1}{4} \biggr)= \frac{1}{2}+\frac{1}{4}+1= \frac{7}{4}, \\& \theta \biggl(\frac{1}{2},\frac{1}{8} \biggr)= \frac{1}{2}+\frac{1}{8}+1=\frac{13}{8}, \\& \theta \biggl(\frac{1}{8}, \frac{1}{4} \biggr)= \frac{1}{8}+\frac{1}{4}+1=\frac{11}{8} \end{aligned}$$
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
\(x\in \mathcal{M}\), \(T^{n}x\rightarrow u\).