# Partial $b_{v}\left(s\right)$ and $b_{v}\left(\theta\right)$ metric spaces
and related fixed point theorems

Ibrahim Karahan  Department of Mathematics, Faculty of Science, Erzurum
Technical University, Erzurum, 25700, Turkey<EMAIL_ADDRESS>and  Irfan Isik  Department of Mathematics, Faculty of Science, Erzurum
Technical University, Erzurum, 25700, Turkey<EMAIL_ADDRESS>
###### Abstract.

In this paper, we introduced two new generalized metric spaces called partial
$b_{v}\left(s\right)$ and $b_{v}\left(\theta\right)$ metric spaces which
extend $b_{v}\left(s\right)$ metric space, $b$-metric space, rectangular
metric space, $v$-generalized metric space, partial metric space, partial
$b$-metric space, partial rectangular $b$-metric space and so on. We proved
some famous theorems such as Banach, Kannan and Reich fixed point theorems in
these spaces. Also, we give definition of partial $v$-generalized metric space
and show that these fixed point theorems are valid in this space. We also give
numerical examples to support our definitions. Our results generalize several
corresponding results in literature.

###### Key words and phrases:

partial $b_{v}\left(s\right)$ metric space, $b_{v}\left(\theta\right)$ metric
space, generalized metric space, Banach fixed point theorem, Reich fixed point
theorem, Kannan fixed point theorems, weakly contractive mappings

###### 1991 Mathematics Subject Classification:

47H10; 54H25; 55M20

##  1\. Introduction and Preliminaries

Metric space was introduced by Maurice Fréchet [1] in 1906. Since a metric
induces topological properties, it has very large application area in
mathematics, especially in fixed point theory. Generalizing of notions is in
the nature of mathematics. So, after the notion of metric space, many
different type generalized metric spaces were introduced by many researchers.
In 1989, Bakhtin introduced the notion of $b$-metric spaces by adding a
multiplier to triangle ineuality. In 1994, Matthews [3] introduced the notion
of partial metric spaces. In this kind of spaces, self-distance of any point
need not to be zero. This space is used in the study of denotational semantics
of dataflow network. In 2000, Branciari [11] introduced rectangular metric
space by adding four points instead of three points in triangle inequality.
These three spaces are the basis of other generalized metric spaces. After all
these spaces, $v$-generalized metric space [11], rectangular $b$-metric spaces
[4], $b_{v}\left(s\right)$ metric space [21], partial $b$-metric space [5] and
partial rectangular $b$-metric space [6] were introduced in recent years.
Below, we give definitions of some generalized metric spaces.

######  Definition 1.

[9] Let $E$ be a nonempty set and $\rho:E\times E\rightarrow[0,\infty)$ a
function. $\left(E,\rho\right)$ is called $b$-metric space if there exists a
real number $s\geq 1$ such that following conditions hold for all $u,w,v\in
E$:

  1. (1)

$\rho(u,w)=0$ iff $u=w$;

  2. (2)

$\rho(u,w)=\rho(w,u)$;

  3. (3)

$\rho(u,w)\leq s[\rho(u,v)+\rho(v,w)].$

Clearly a $b$-metric space with $s=1$ is exactly a usual metric space.

######  Definition 2.

[3] Let $E$ be a nonempty set and $\rho:E\times E\rightarrow[0,\infty)$ a
mapping. $\left(E,\rho\right)$ is called partial metric space if following
conditions hold for all $u,w,v\in E$:

  1. (1)

$u=w$ iff $\rho(u,u)=\rho(u,w)=\rho(w,w)$;

  2. (2)

$\rho(u,u)\leq\rho(u,w);$

  3. (3)

$\rho(u,w)=\rho(w,u);$

  4. (4)

$\rho(u,w)\leq\rho(u,v)+\rho(v,w)-\rho(v,v)$.

It is clear that every metric space is also a partial metric spaces.

######  Definition 3.

[11] Let $E$ be a nonempty set and let $\rho:$ $E\times
E\rightarrow[0,\infty)$ be a mapping. $(E,\rho)$ is called a rectangular
metric space if following conditions hold for all $u,w\in E$ and for all
distinct points $c,d\in E\setminus\left\\{u,w\right\\}$:

  1. (1)

$\rho(u,w)=0$ iff $u=w$;

  2. (2)

$\rho(u,w)=\rho(w,u)$;

  3. (3)

$\rho(u,w)\leq\rho(u,c)+\rho(c,d)+\rho(d,w)$.

######  Definition 4.

[5] Let $E$ be a nonempty set and mapping $\rho:E\times
E\rightarrow[0,\infty)$ a mapping. $\left(E,\rho\right)$ is called partial
$b$-metric space if there exists a real number $s\geq 1$ such that following
conditions hold for all $u,w,v\in E$:

  1. (1)

$u=w$ iff $\rho(u,u)=\rho(u,w)=\rho(w,w)$;

  2. (2)

$\rho(u,u)\leq\rho(u,w);$

  3. (3)

$\rho(u,w)=\rho(w,u);$

  4. (4)

$\rho(u,w)\leq s[\rho(u,v)+\rho(v,w)]-\rho(v,v)$.

######  Remark 1.

[5] It is clear that every partial metric space is a partial $b$-metric space
with coefficient $s=1$ and every $b$-metric space is a partial $b$-metric
space with the same coefficient and zero self-distance. However, the converse
of this fact need not hold.

In 2017, Mitrovic and Radenovic introduced following generalized metric space
which is referred to as $b_{v}(s)$ metric space. Under the suitable
assumptions, this kind of spaces can be reduced to the other spaces.

######  Definition 5.

[21] Let $E$ be a nonempty set, $\rho:$ $E\times E\rightarrow[0,\infty)$ a
mapping and $v\in\mathbb{N}$. Then $(E,\rho)$ is said to be a $b_{v}(s)$
metric space if there exists a real number $s\geq 1$ such that following
conditions hold for all $u,w\in E$ and for all distinct points
$z_{1},z_{2},\ldots,z_{v}\in E\setminus\left\\{u,w\right\\}$:

  1. 1.

$\rho(u,w)=0$ iff $u=w$;

  2. 2.

$\rho(u,w)=\rho(w,u)$;

  3. 3.

$\rho(u,w)\leq
s[\rho(u,z_{1})+\rho(z_{1},z_{2})+\cdots+\rho\left(z_{v},w\right)].$

This metric space can be reduced to $v$-generalized metric space by taking
$s=1$, rectangular metric space by taking $v=2$ and $s=1$, rectangular
$b$-metric space by taking $v=2$, $b$-metric space by taking $v=1$ and usual
metric space by taking $v=s=1.$

##  2\. Main Results

In this part, motivated and inspired by mentioned studies, we introduce
$b_{v}\left(\theta\right)$ (or extended $b_{v}(s)$) metric space and partial
$b_{v}(s)$ metric space. Also we give some fixed point theorems in these
spaces.

First we introduce partial $b_{v}\left(s\right)$ metric space and give some
properties of it.

###  2.1. Partial $b_{v}\left(s\right)$ Metric Spaces

######  Definition 6.

Let $E$ be a nonempty set and $\rho:$ $E\times E\rightarrow[0,\infty)$ be a
mapping and $v\in\mathbb{N}$. Then $(E,\rho)$ is said to be a partial
$b_{v}(s)$ metric space if there exists a real number $s\geq 1$ such that
following conditions hold for all $u,w,z_{1},z_{2},\ldots,z_{v}\in E$:

  1. (1)

$u=w\Leftrightarrow\rho(u,u)=\rho(u,w)=\rho(w,w)$;

  2. (2)

$\rho(u,u)\leq\rho(u,w)$;

  3. (3)

$\rho(u,w)=\rho(w,u)$;

  4. (4)

$\rho(u,w)\leq
s[\rho(u,z_{1})+\rho(z_{1},z_{2})+\ldots+\rho(z_{v-1},z_{v})+\rho(z_{v},y)]-\sum_{i=1}^{v}\rho(z_{i},z_{i})$.

It is easy to see that every $b_{v}(s)$ metric space is a partial $b_{v}(s)$
metric space. However, the converse is not true in general.

######  Remark 2.

In Definition 6;

  1. (1)

if we take $v=2$, then we derive partial rectangular $b$-metric space.

  2. (2)

if we take $v=1,$ then we derive partial $b$-metric space.

  3. (3)

if we take $v=s=1$, then we derive partial metric space.

######  Remark 3.

Let $(E,\rho)$ be a partial $b_{v}(s)$ metric space, if $\rho(u,w)=0$, for
$u,w\in E$, then $u=w$.

###### Proof.

Let $\rho(u,w)=0$ for $u,w\in E$. From the second condition of partial
$b_{v}(s)$ metric space, since $\rho(u,u)\leq\rho(u,w)=0$, we have
$\rho(u,u)=0$. Similarly, we have $\rho(w,w)=0$. So, we get
$\rho(u,w)=\rho(u,u)=\rho(w,w)=0$. It follows from the first condition that
$u=w$.

######  Proposition 1.

Let $E$ be a nonempty set such that $d_{1}$ is a partial metric and $d_{2}$ is
a $b_{v}(s)$ metric on $E$. Then $\left(E,\rho\right)$ is a partial $b_{v}(s)$
metric space where $\rho:E\times E\rightarrow[0,\infty)$ is a mapping defined
by $\rho(u,w)=p(u,w)+d(u,w)$ for all $u,w\in E.$

###### Proof.

Let $(E,d_{1})$ be a partial metric space and $(E,d_{2})$ be a $b_{v}(s)$
metric space.Then it is clear that first three conditions of the partial
$b_{v}(s)$ metric space are satisfied for the function $\rho$. Let
$u,w,z_{1},z_{2},\ldots,z_{v}\in E$ be arbitrary points. Then, we have

$\displaystyle\rho(u,w)$ $\displaystyle=$ $\displaystyle
d_{1}(u,w)+d_{2}(u,w)$ $\displaystyle\leq$ $\displaystyle
d_{1}(u,z_{1})+d_{1}(z_{1},z_{2})+\ldots+d_{1}(z_{v},w)-\sum_{i=1}^{v}d_{1}(z_{i},z_{i})$
$\displaystyle+s\left[d_{2}(u,z_{1})+d_{2}(z_{1},z_{2})+\ldots+d_{2}(z_{v},w)\right]$
$\displaystyle\leq$ $\displaystyle
s\left[d_{1}(u,z_{1})+d_{1}(z_{1},z_{2})+\ldots+d_{1}(z_{v},w)-\sum_{i=1}^{v}d_{1}(z_{i},z_{i})\right.$
$\displaystyle\left.+d_{2}(u,z_{1})+d_{2}(z_{1},z_{2})+\ldots+d_{2}(z_{v},w)\right]$
$\displaystyle=$ $\displaystyle
s\left[\rho(u,z_{1})+\rho(z_{1},z_{2})+\ldots+\rho(z_{v},w)-\sum_{i=1}^{v}\rho(z_{i},z_{i})\right]$
$\displaystyle\leq$ $\displaystyle
s\left[\rho(u,z_{1})+\rho(z_{1},z_{2})+\ldots+\rho(z_{v},w)\right]-\sum_{i=1}^{v}\rho(z_{i},z_{i})\text{.}$

So, $(E,\rho)$ is a partial $b_{v}(s)$ metric space.

Now, we give definitions of convergent sequence, Cauchy sequence and complete
partial $b_{v}(s)$ metric space by the following way.

######  Definition 7.

Let $(E,\rho)$ be a partial $b_{v}(s)$ metric space and let
$\left\\{u_{n}\right\\}$ be any sequence in $E$ and $u\in E$. Then:

  1. (1)

The sequence $\left\\{u_{n}\right\\}$ is said to be convergent and converges
to $u$, if $\lim_{n\rightarrow\infty}\rho(u_{n},u)=\rho(u,u)$.

  2. (2)

The sequence $\left\\{u_{n}\right\\}$ is said to be Cauchy sequence in
$(E,\rho)$ if $\lim_{n,m\rightarrow\infty}\rho(u_{n},u_{m})$ exists and is
finite.

  3. (3)

$(E,\rho)$ is said to be a complete partial $b_{v}(s)$ metric space if for
every Cauchy sequence $\left\\{u_{n}\right\\}$ in $E$ there exists $u\in E$
such that

$\lim_{n,m\rightarrow\infty}\rho(u_{n},u_{m})=\lim_{n\rightarrow\infty}\rho(u_{n},u)=\rho(u,u)\text{.}$

Note that the limit of a convergent sequence may not be unique in a partial
$b_{v}(s)$ metric space.

Now we give an analogue of Banach contraction principle. Our proof is very
different from the original proof of Banach contraction principle in usual
metric space.

######  Theorem 1.

Let $(E,\rho)$ be a complete partial $b_{v}\left(s\right)$ metric space and
$S:E\rightarrow E$ be a contraction mapping, i.e., $S$ satisfies

$\rho(Su,Sw)\leq\lambda\rho(u,w)$ (2.1)

for all $u,w\in E,$ where $\lambda\in[0,1)$. Then $S$ has a unique fixed point
$b\in S$ and $\rho(b,b)=0$.

###### Proof.

Let $G=S^{n_{0}}$ and define a sequence $\left\\{u_{n}\right\\}$ by
$Gu_{n}=u_{n+1}$ for all $n\in\mathbb{N}$ and arbitrary point $u_{0}\in E$.
Since $\lambda\in[0,1)$ and $\lim_{n\rightarrow\infty}\lambda^{n}=0$, there
exists a natural number $n_{0}$ such that
$\lambda^{n_{0}}<\frac{\varepsilon}{4s}$ for given $0<\varepsilon<1.$ Then,
for all $u,w\in E$ we get

$\rho(Gu,Gw)=\rho(S^{n_{0}}u,S^{n_{0}}w)\leq\lambda^{n_{0}}\rho(u,w)\text{.}$
(2.2)

So, we have

$\rho(u_{k+1},u_{k})=\rho(Gu_{k},Gu_{k-1})\leq\lambda^{n_{0}}\rho(u_{k},u_{k-1})\leq\lambda^{kn_{0}}\rho(u_{1},u_{0})\rightarrow
0\text{,}$

as $k\rightarrow\infty$. Hence, there exists a $l\in\mathbb{N}$ such that

$\rho(u_{l+1},u_{l})<\frac{\varepsilon}{4s}\text{.}$

Now, let

$B_{\rho}[u_{l},\varepsilon/2]:=\left\\{w\in
E:\rho(u_{l},w)\leq\frac{\varepsilon}{2}+\rho(u_{l},u_{l})\right\\}\text{.}$

We need to prove that $G$ maps the set $B_{\rho}[u_{l},\varepsilon/2]$ into
itself. Since $u_{l}\in B_{\rho}[u_{l},\varepsilon/2]$, it is a nonempty set.
Let $z$ be an arbitrary point in $B_{\rho}[u_{l},\varepsilon/2]$. Then, using
(2.2) we get

$\displaystyle\rho(Gz,u_{l})$ $\displaystyle\leq$ $\displaystyle
s\left[\rho(Gz,Gu_{l})+\rho(Gu_{l}+Gu_{l+1})+\ldots+\rho(Gu_{l+v-2},Gu_{l+v-1})\right.$
$\displaystyle\left.+\rho(Gu_{l+v-1},u_{l})\right]-\sum\limits_{i=0}^{v-1}\rho(Gu_{l+i},Gu_{l+i})$
$\displaystyle\leq$ $\displaystyle
s\left[\rho(Gz,Gu_{l})+\rho(Gu_{l}+Gu_{l+1})+\ldots+\rho(Gu_{l+v-2},Gu_{l+v-1})+\rho(Gu_{l+v-1},u_{l})\right]$
$\displaystyle\leq$ $\displaystyle
s[\lambda^{n_{0}}(\frac{\varepsilon}{2}+\rho(u_{l},u_{l}))+\rho(u_{l+1},u_{l+2})+\ldots+\rho(u_{l+v-1},u_{l+v})+\rho(u_{l+v},u_{l})$
$\displaystyle\leq$ $\displaystyle
s\left\\{\lambda^{n_{0}}(\frac{\varepsilon}{2}+\rho(u_{l},u_{l}))+\rho(u_{l+1},u_{l+2})+\ldots+\rho(u_{l+v-1},u_{l+v})+s[\rho(u_{l},u_{l+1})\right.$
$\displaystyle\left.+\rho(u_{l+1},u_{l+2})+\ldots+\rho(u_{l+v-1},u_{l+v})+\rho(u_{l+v},u_{l+v})]-\sum\limits_{i=1}^{v}\rho(u_{l+i},u_{l+i})\right\\}$
$\displaystyle\leq$ $\displaystyle
s\left\\{\lambda^{n_{0}}(\frac{\varepsilon}{2}+\rho(u_{l},u_{l}))+\left(s+1\right)\rho(u_{l},u_{l+1})+(s+1)\rho(u_{l+1},u_{l+2})+\right.$
$\displaystyle\left.(s+1)\rho(u_{l+2},u_{l+3})+\ldots+(s+1)\rho(u_{l+v-1},u_{l+v})+s\rho(u_{l+v},u_{l+v})\right\\}$
$\displaystyle\leq$ $\displaystyle
s\left\\{\lambda^{n_{0}}(\frac{\varepsilon}{2}+\rho(u_{l},u_{l}))+(s+1)\rho(u_{l},u_{l+1})+(s+1)\rho(u_{l+1},u_{l+2})+\right.$
$\displaystyle\left.(s+1)\rho(u_{l+2},u_{l+3})+\ldots+(s+1)\rho(u_{l+v-1},u_{l+v})+s\lambda^{vn_{0}}\rho(u_{l},u_{l})\right\\}$
$\displaystyle=$
$\displaystyle\rho(u_{l},u_{l})\left[s\lambda^{n_{0}}+s^{2}\lambda^{vn_{0}}\right]+s\lambda^{n_{0}}\frac{\varepsilon}{2}+s^{2}\rho(u_{l},u_{l+1})+$
$\displaystyle
s(s+1)\left[\rho(u_{l+1},u_{l+2})+\ldots+\rho(u_{l+v-1},u_{l+v})\right]$

Since $\lambda^{n_{0}}<\frac{\varepsilon}{4s}$ and
$\rho(u_{l},u_{l+1})\leq\frac{\varepsilon}{4v(s^{2}+s)},$ we have

$\displaystyle\rho(Gz,u_{l})$ $\displaystyle\leq$
$\displaystyle\rho(u_{l},u_{l})\left[s\frac{\varepsilon}{4s}+s^{2}\frac{\varepsilon^{v}}{(4s)^{v}}\right]+s\frac{\varepsilon}{4s}\frac{\varepsilon}{2}+$
$\displaystyle(s^{2}+s)\left[\rho(u_{l},u_{l+1})+\rho(u_{l+1},u_{l+2})+\ldots+\rho(u_{l+v-1},u_{l+v})\right]$
$\displaystyle\leq$
$\displaystyle\rho(u_{l},u_{l})+\frac{\varepsilon}{4}+(s^{2}+s)v\frac{\varepsilon}{4v(s^{2}+s)}$
$\displaystyle=$
$\displaystyle\frac{\varepsilon}{2}+\rho(u_{l},u_{l})\text{.}$

So, $Gz\in B_{\rho}[u_{l},\varepsilon/2]$. Therefore, $G$ maps
$B_{\rho}[u_{l},\varepsilon/2]$ into itself. Since $u_{l}\in
B_{\rho}[u_{l},\varepsilon/2]$ and $Gu_{l}\in B_{\rho}[u_{l},\varepsilon/2]$,
we obtain that $G^{n}u_{l}\in B_{\rho}[u_{l},\varepsilon/2]$ for all
$n\in\mathbb{N}$, that is, $u_{m}\in B_{\rho}[u_{l},\varepsilon/2]$ for all
$m\geq l$. On the other hand, from definition of partial $b_{v}\left(s\right)$
metric space, since
$\rho(u_{l},u_{l})\leq\rho(u_{l},u_{l+1})<\frac{\varepsilon}{4v(s^{2}+s)}<\frac{\varepsilon}{2}$,
we have

$\rho(u_{n},u_{m})<\frac{\varepsilon}{2}+\rho(u_{l},u_{l})<\varepsilon$

for all $n,m>l$. This means that the sequence $\left\\{u_{n}\right\\}$ is a
Cauchy sequence. Completeness of $E$ implies that there exists $b\in E$ such
that

$\lim_{n\rightarrow\infty}\rho(u_{n},b)=\lim_{n,m\rightarrow\infty}\rho(u_{n},u_{m})=\rho(b,b)=0\text{.}$
(2.3)

Now, we need to show that, $b$ is a fixed point of $S$. For any
$n\in\mathbb{N}$ we get

$\displaystyle\rho(b,Sb)$ $\displaystyle\leq$ $\displaystyle
s\left[\rho(b,u_{n+1})+\rho(u_{n+1},u_{n+2})+\ldots+\rho(u_{n+v-1},u_{n+v})+\right.$
$\displaystyle\left.\rho(u_{n+v},Sb)\right]-\sum\limits_{i=1}^{v}\rho(u_{n+i},u_{n+i})$
$\displaystyle\leq$ $\displaystyle
s\left[\rho(b,u_{n+1})+\rho(u_{n+1},u_{n+2})+\ldots+\rho(u_{n+v-1},u_{n+v})+\rho(u_{n+v},Sb)\right]$
$\displaystyle\leq$ $\displaystyle
s\left[\rho(b,u_{n+1})+\rho(u_{n+1},u_{n+2})+\ldots+\rho(u_{n+v-1},u_{n+v})+\rho(Su_{n+v-1},Sb)\right]$
$\displaystyle\leq$ $\displaystyle
s\left[\rho(b,u_{n+1})+\rho(u_{n+1},u_{n+2})+\ldots+\rho(u_{n+v-1},u_{n+v})+\lambda\rho(u_{n+v-1},b)\right].$

So, it follows from (2.3) that $\rho(b,Sb)=0$. So, $b$ is a fixed point of
$S$.

Now, we show that $S$ has a unique fixed point. Let $a,b\in E$ be two distinct
fixed points of $S$, that is, $Sa=a$, $Sb=b$. Then, contractivity of mapping
$S$ implies that

$\rho(a,b)=\rho(Sa,Sb)\leq\lambda\rho(a,b)<\rho(a,b)\text{,}$

which is a contradiction. So, it folllows that $\rho(a,b)=0$, that is, $a=b$.
Moreover, for a fixed point $a$, let assume that $\rho(a,a)>0$. Then we get
$\rho(a,a)=\rho(Sa,Sa)\leq\lambda\rho(a,a)<\rho(a,a)$ which is a
contradiction. So, we have $\rho(a,a)=0$.

Now, we prove an analogue of Kannan fixed point theorem.

######  Theorem 2.

Let $(E,\rho)$ be a complete partial $b_{v}\left(s\right)$ metric space and
$S:E\rightarrow E$ a mapping satisfying the following condition:

$\rho(Su,Sy)\leq\lambda\left[\rho(u,Su)+\rho(w,Sw)\right]$ (2.4)

for all $u,w\in E$, where $\lambda\in[0,\frac{1}{2})$,
$\lambda\neq\frac{1}{s}$. Then $S$ has a unique fixed point $b\in E$ and
$\rho(b,b)=0$.

###### Proof.

.First we show the existence of fixed points of $S$. Let define a sequence
$\left\\{u_{n}\right\\}$ by $u_{n}=S^{n}u_{0}$ for all $n\in\mathbb{N}$ and an
arbitrary point $u_{0}\in E$ and $\sigma_{n}=\rho(u_{n},u_{n+1})$. If
$\sigma_{n}=0$, then for at least one $n,$ $u_{n}$ is a fixed point of $S$.
So, let assume that $\sigma_{n}>0$ for all $n\geq 0$. Since $S$ is a Kannan
mapping, it follows from (2.4) that

$\displaystyle\sigma_{n}$ $\displaystyle=$
$\displaystyle\rho(u_{n},u_{n+1})=\rho(Su_{n-1},Su_{n})$ $\displaystyle\leq$
$\displaystyle\lambda\left[\rho(u_{n-1},Su_{n-1})+\rho(u_{n},Su_{n})\right]$
$\displaystyle=$
$\displaystyle\lambda\left[\rho(u_{n-1},u_{n})+\rho(u_{n},u_{n+1})\right]$
$\displaystyle=$
$\displaystyle\lambda\left[\sigma_{n-1}+\sigma_{n}\right]\text{.}$

Therefore, we get $\sigma_{n}\leq\frac{\lambda}{1-\lambda}\sigma_{n-1}$. On
repeating this process we obtain

$\sigma_{n}\leq\left(\frac{\lambda}{1-\lambda}\right)^{n}\sigma_{0}\text{.}$

From hypothesis, since $\lambda\in[0,\frac{1}{2})$, we have

$\lim_{n\rightarrow\infty}\sigma_{n}=\lim_{n\rightarrow\infty}\rho(u_{n},u_{n+1})=0.$
(2.5)

So, for every $\varepsilon>0$, there exists a natural number $n_{0}$ such that
$\sigma_{n}<\varepsilon/2$ and $\sigma_{m}<\varepsilon/2$ for all $n,m\geq
n_{0}$. From (LABEL:(i)), we have

$\displaystyle\rho(u_{n},u_{m})$ $\displaystyle=$
$\displaystyle\rho(Su_{n-1},Su_{m-1})$ $\displaystyle\leq$
$\displaystyle\lambda\left[\rho(u_{n-1},Su_{n-1})+\rho(u_{m-1},Su_{m-1})\right]$
$\displaystyle=$
$\displaystyle\lambda\left[\rho(u_{n-1},u_{n})+\rho(u_{m-1},u_{m})\right]$
$\displaystyle=$ $\displaystyle\lambda\left[\sigma_{n-1}+\sigma_{m-1}\right]$
$\displaystyle<$
$\displaystyle\frac{\varepsilon}{2}+\frac{\varepsilon}{2}=\varepsilon$

for $n,m>n_{0}$. Hence, $\left\\{u_{n}\right\\}$ is Cauchy sequence in $E$ and
$\lim_{n,m\rightarrow\infty}\rho(u_{n},u_{m})=0$. It follows from the
completeness of $E$ that there exists $b\in E$ such that

$\lim_{n\rightarrow\infty}\rho(u_{n},b)=\lim_{n,m\rightarrow\infty}\rho(u_{n},u_{m})=\rho(b,b)=0\text{.}$

Now,we show that $b$ is a fixed point of $S$. From definition of Kannan
mappings and partial $b_{v}\left(s\right)$ metric space, we have

$\displaystyle\rho(b,Sb)$ $\displaystyle\leq$ $\displaystyle
s\left[\rho(b,u_{n+1})+\rho(u_{n+1},u_{n+2})+\ldots+\rho(u_{n+v-1},u_{n+v})+\right.$
$\displaystyle\left.\rho(u_{n+v},Sb)\right]-\sum\limits_{i=1}^{v}\rho(u_{n+i},u_{n+i})$
$\displaystyle\leq$ $\displaystyle
s\left[\rho(b,u_{n+1})+\rho(u_{n+1},u_{n+2})+\ldots+\rho(u_{n+v-1},u_{n+v})+\rho(u_{n+v},Sb)\right]$
$\displaystyle\leq$ $\displaystyle
s\left[\rho(b,u_{n+1})+\rho(u_{n+1},u_{n+2})+\ldots+\rho(u_{n+v-1},u_{n+v})+\rho(Su_{n+v-1},Sb)\right]$
$\displaystyle\leq$ $\displaystyle
s\left[\rho(b,u_{n+1})+\rho(u_{n+1},u_{n+2})+\ldots+\rho(u_{n+v-1},u_{n+v})+\right.$
$\displaystyle\left.\lambda\left\\{\rho(u_{n+v-1},Su_{n+v-1})+\rho(b,Sb)\right\\}\right].$

So, it follows from the last inequality that

$\displaystyle\rho(b,Sb)$ $\displaystyle\leq$
$\displaystyle\frac{s}{(1-s\lambda)}\left[\rho(b,u_{n+1})+\rho(u_{n+1},u_{n+2})\right.$
$\displaystyle\left.+...+\rho(u_{n+v-1},u_{n+v})+\lambda\rho(u_{n+v-1},Su_{n+v-1})\right]\text{.}$

Since $\lambda\neq\frac{1}{s}$ and $\left\\{u_{n}\right\\}$ is a Cauchy and
convergent sequence, we have $\rho(b,Sb)=0$, so $Sb=b$. It means that $b$ is a
fixed point of $S$.

Now we show the uniqueness of fixed point. But first, we need to show that if
$b\in E$ is a fixed point of $S$, then $\rho(b,b)=0$. Let assume to the
contrary that $\rho(b,b)>0$. Then, from (2.4) we have

$\rho(b,b)=\rho(Sb,Sb)\leq\lambda\left[\rho(b,Sb)+\rho(b,Sb)\right]=2\lambda\rho(b,b)<\rho(b,b)\text{,}$

which is a contradiction. So, assumption is wrong, namely, $\rho(b,b)=0$. Now,
we can show that $S$ has a unique fixed point. Suppose $a,b\in E$ be two
distinct fixed points of $S$. Then we have $\rho(b,b)=\rho(a,a)=0$, and it
follows from (2.4) that

$\displaystyle\rho(b,a)$ $\displaystyle=$
$\displaystyle\rho(Sb,Sa)\leq\lambda\left[\rho(b,Sb)+\rho(a,Sa)\right]$
$\displaystyle=$ $\displaystyle\lambda\left[\rho(b,b)+\rho(a,a)\right]=0$

Therefore, we have $\rho(b,a)=0$ and so $b=a$. Thus $S$ has a unique fixed
point. This completes the proof.

######  Theorem 3.

Let $(E,\rho)$ be a complete partial $b_{v}\left(s\right)$ metric space and
$S:E\rightarrow E$ a mapping satisfying:

$\rho(Su,Sw)\leq\lambda\max\left\\{\rho(u,w),\rho(u,Su),\rho(w,Sw)\right\\}$
(2.6)

for all $u,w\in E$ and $\lambda\in\left[0,\frac{1}{s}\right)$. Then, $S$ has a
unique fixed point $b\in E$ and $\rho(b,b)=0$.

###### Proof.

We begin with the existence of fixed points of $S$. Let $u_{0}\in E$ be an
arbitrary initial point and let $\left\\{u_{n}\right\\}$ be a sequence defined
by $u_{n+1}=Su_{n}$ for all $n$. If $u_{n}=u_{n+1}$ for at least one natural
number $n$, then it is clear that this point is a fixed point of $S$. So, let
assume that $u_{n+1}\neq u_{n}$ for all $n$. Now, it follows from (2.6) that

$\displaystyle\rho(u_{n+1},u_{n})$ $\displaystyle=$
$\displaystyle\rho(Su_{n},Su_{n-1})$ $\displaystyle\leq$
$\displaystyle\lambda\max\left\\{\rho(u_{n},u_{n-1}),\rho(u_{n},Su_{n}),\rho(u_{n-1},Su_{n-1})\right\\}$
$\displaystyle=$
$\displaystyle\lambda\max\left\\{\rho(u_{n},u_{n-1}),\rho(u_{n},u_{n+1}),\rho(u_{n-1},u_{n})\right\\}$
$\displaystyle=$
$\displaystyle\lambda\max\left\\{\rho(u_{n},u_{n-1}),\rho(u_{n},u_{n+1})\right\\}\text{.}$

Set $L=\max\left\\{\rho(u_{n},u_{n-1}),\rho(u_{n},u_{n+1})\right\\}$. There
exist two cases. If $L=\rho(u_{n},u_{n+1})$, then we get
$\rho(u_{n+1},u_{n})\leq\lambda\rho(u_{n+1},u_{n})<\rho(u_{n+1},u_{n})$ which
is a contradiction. So, we must have $L=\rho(u_{n},u_{n-1})$ and then we have

$\rho(u_{n+1},u_{n})\leq\lambda\rho(u_{n},u_{n-1})\text{.}$

On repeating this process, we obtain

$\rho(u_{n+1},u_{n})\leq\lambda^{n}\rho(u_{1},u_{0})$ (2.7)

for all $n$. On the other hand, since $\lambda^{n}\rightarrow 0$ for
$n\rightarrow\infty$, there exists a natural number $n_{0}$ such that
$0<\lambda^{n_{0}}s<1$. For $m,n\in\mathbb{N}$ with $m>n$, by using inequality
(2.7), we obtain

$\displaystyle\rho(u_{n},u_{m})$ $\displaystyle\leq$ $\displaystyle
s\left[\rho(u_{n},u_{n+1})+\rho(u_{n+1},u_{n+2})+\ldots+\rho(u_{n+v-3},u_{n+v-2})\right.$
$\displaystyle\left.+\rho(u_{n+v-2},u_{n+n_{0}})+\rho(u_{n+n_{0}},u_{m+n_{0}})+\rho(u_{m+n_{0}},u_{m})\right]$
$\displaystyle-\sum\limits_{i=1}^{v-2}\rho(u_{n+i},u_{n+i})-\rho(u_{n+n_{0}},u_{n+n_{0}})-\rho(u_{m+n_{0}},u_{m+n_{0}})$
$\displaystyle\leq$ $\displaystyle
s\left[\rho(u_{n},u_{n+1})+\rho(u_{n+1},u_{n+2})+\ldots+\rho(u_{n+v-3},u_{n+v-2})\right.$
$\displaystyle\left.+\rho(u_{n+v-2},u_{n+n_{0}})+\rho(u_{n+n_{0}},u_{m+n_{0}})+\rho(u_{m+n_{0}},u_{m})\right]$
$\displaystyle\leq$ $\displaystyle
s\left(\lambda^{n}+\lambda^{n+1}+\cdots+\lambda^{n+v-3}\right)\rho(u_{0},u_{1})$
$\displaystyle+s\lambda^{n}\rho(u_{v-2},u_{n_{0}})+s\lambda^{n_{0}}\rho(u_{n},u_{m})+s\lambda^{m}\rho(u_{n_{0}},u_{0}).$

So, we get

$\displaystyle\left(1-s\lambda^{n_{0}}\right)\rho(u_{n},u_{m})$
$\displaystyle\leq$ $\displaystyle
s\left(\lambda^{n}+\lambda^{n+1}+\cdots+\lambda^{n+v-3}\right)\rho(u_{0},u_{1})$
$\displaystyle+s\lambda^{n}\rho(u_{v-2},u_{n_{0}})+s\lambda^{m}\rho(u_{n_{0}},u_{0}).$

By taking limit from both side, we have

$\lim_{n,m\rightarrow\infty}\rho(u_{n},u_{m})=0$

Therefore, $\left\\{u_{n}\right\\}$ is a Cauchy sequence in $E$. By
completeness of $E,$ there exists $b\in E$ such that

$\lim_{n\rightarrow\infty}\rho(u_{n},b)=\lim_{n,m\rightarrow\infty}\rho(u_{n},u_{m})=\rho(b,b)=0\text{.}$
(2.8)

Now, we show that $b$ is a fixed point of $S$. From definition of partial
$b_{v}\left(s\right)$ metric space and inequality (2.6), we have

$\displaystyle\rho(b,Sb)$ $\displaystyle\leq$ $\displaystyle
s\left[\rho(b,u_{n+1})+\rho(u_{n+1},u_{n+2})+\ldots+\rho(u_{n+v-1},u_{n+v})+\right.$
$\displaystyle\left.\rho(u_{n+v},Sb)\right]-\sum\limits_{i=1}^{v}\rho(u_{n+i},u_{n+i})$
$\displaystyle\leq$ $\displaystyle
s\left[\rho(b,u_{n+1})+\rho(u_{n+1},u_{n+2})+\ldots+\rho(u_{n+v-1},u_{n+v})+\rho(u_{n+v},Sb)\right]$
$\displaystyle\leq$ $\displaystyle
s\left[\rho(b,u_{n+1})+\rho(u_{n+1},u_{n+2})+\ldots+\rho(u_{n+v-1},u_{n+v})+\rho(Su_{n+v-1},Sb)\right]$
$\displaystyle\leq$ $\displaystyle
s\left[\rho(b,u_{n+1})+\rho(u_{n+1},u_{n+2})+\ldots+\rho(u_{n+v-1},u_{n+v})+\right.$
$\displaystyle\left.\lambda\max\left\\{\rho(u_{n+v-1},b),\rho(u_{n+v-1},u_{n+v}),\rho(b,Sb)\right\\}\right].$

Set
$F=\max\left\\{\rho(u_{n+v-1},b),\rho(u_{n+v-1},u_{n+v}),\rho(b,Sb)\right\\}$.
There exists three cases:

1\. If $F=\rho(u_{n+v-1},b)$, then we get

$\rho(b,Sb)\leq
s\left[\rho(b,u_{n+1})+\rho(u_{n+1},u_{n+2})+\ldots+\rho(u_{n+v-1},u_{n+v})+\lambda\rho(u_{n+v-1},b)\right].$

So, it follows from (2.8) that $\rho(b,Sb)=0$.

2\. If $F=\rho(u_{n+v-1},u_{n+v})$, then we get

$\rho(b,Sb)\leq
s\left[\rho(b,u_{n+1})+\rho(u_{n+1},u_{n+2})+\ldots+\left(1+\lambda\right)\rho(u_{n+v-1},u_{n+v})\right].$

Again by using (2.8), we obtain that $\rho(b,Sb)=0$.

3\. If $F=\rho(b,Sb)$ then we get

$\left(1-s\lambda\right)\rho(b,Sb)\leq
s\left[\rho(b,u_{n+1})+\rho(u_{n+1},u_{n+2})+\ldots+\rho(u_{n+v-1},u_{n+v})\right].$

Since $\lambda\in\left[0,\frac{1}{s}\right)$, we obtain that $\rho(b,Sb)=0$,
that is, $Sb=b$. Thus, $b$ is a fixed poit of $S$.

Now we show the uniqueness of fixed point of $S$. Suppose on the contrary that
$a$ and $b$ are two distinct fixed points of $S$ and $\rho(a,b)>0$. It follows
from (2.6) that

$\displaystyle\rho(a,b)$ $\displaystyle=$
$\displaystyle\rho(Sa,Sb)\leq\lambda\max\left\\{\rho(a,b),\rho(a,Sa),\rho(b,Sb)\right\\}$
$\displaystyle=$
$\displaystyle\lambda\max\left\\{\rho(a,b),\rho(a,a),\rho(b,b)\right\\}$
$\displaystyle=$ $\displaystyle\lambda\rho(a,b)<\rho(a,b)\text{,}$

which is a cotradiction. Therefore, we must have $\rho(a,b)=0$ and so $a=b$.
Hence, $S$ has a unique fixed point.

In definition 6, if we take $s=1$, then we derive following definition of
partial $v$-generalized metric space.

######  Definition 8.

Let $E$ be a nonempty set and $\rho:$ $E\times E\rightarrow[0,\infty)$ be a
mapping and $v\in\mathbb{N}$. Then $(E,\rho)$ is said to be a partial
$v$-generalized metric space if following conditions hold for all
$u,w,z_{1},z_{2},\ldots,z_{v}\in E$:

  1. (1)

$u=w\Leftrightarrow\rho(u,u)=\rho(u,w)=\rho(w,w)$;

  2. (2)

$\rho(u,u)\leq\rho(u,w)$;

  3. (3)

$\rho(u,w)=\rho(w,u)$;

  4. (4)

$\rho(u,w)\leq\rho(u,z_{1})+\rho(z_{1},z_{2})+\ldots+\rho(z_{v-1},z_{v})+\rho(z_{v},y)-\sum_{i=1}^{v}\rho(z_{i},z_{i})$.

In Theorems 1,2 and 3, if take $s=1$, then we derive following fixed point
theorems in partial $v$-generalized metric space.

######  Corollary 1.

Let $(E,\rho)$ be a complete partial $v$-generalized metric space and
$S:E\rightarrow E$ be a contraction mapping, i.e., $S$ satisfies

$\rho(Su,Sw)\leq\lambda\rho(u,w)$

for all $u,w\in E,$ where $\lambda\in[0,1)$. Then $S$ has a unique fixed point
$b\in S$ and $\rho(b,b)=0$.

######  Corollary 2.

Let $(E,\rho)$ be a complete partial $v$-generalized metric space and
$S:E\rightarrow E$ a mapping satisfying the following condition:

$\rho(Su,Sy)\leq\lambda\left[\rho(u,Su)+\rho(w,Sw)\right]$

for all $u,w\in E$, where $\lambda\in[0,\frac{1}{2})$. Then $S$ has a unique
fixed point $b\in E$ and $\rho(b,b)=0$.

######  Corollary 3.

Let $(E,\rho)$ be a complete partial $v$-generalized metric space and
$S:E\rightarrow E$ a mapping satisfying:

$\rho(Su,Sw)\leq\lambda\max\left\\{\rho(u,w),\rho(u,Su),\rho(w,Sw)\right\\}$

for all $u,w\in E$ and $\lambda\in\left[0,1\right)$. Then, $S$ has a unique
fixed point $b\in E$ and $\rho(b,b)=0$.

###  2.2. $b_{v}\left(\theta\right)$ Metric Spaces

In 2017, Kamran et al. introduced following generalized metric space which
they call extended $b$-metric space.

######  Definition 9.

[7] Let $E$ be a nonempty set and let $\theta:E\times E\rightarrow[1,\infty)$
be a function. A function $\rho_{\theta}:E\times E\rightarrow[0,\infty)$ is
called an extended $b$-metric if for all $u,v,w\in E$ it satisfies:

  1. (1)

$\rho_{\theta}(u,w)=0$ iff $u=w$;

  2. (2)

$\rho_{\theta}(u,w)=\rho(w,u)$;

  3. (3)

$\rho_{\theta}(u,w)\leq\theta\left(u,w\right)[\rho_{\theta}(u,v)+\rho_{\theta}(v,w)].$

The pair $\left(E,\rho_{\theta}\right)$ is called an extended $b$-metric
space.

It is clear that if $\theta\left(u,w\right)=s$ for all $u,w\in E,$ then we
obtain $b$-metric space.

From this point of view, we introduce following generalized metric space
called as $b_{v}(\theta)$ (or extended $b_{v}(s)$ ) metric space.

######  Definition 10.

Let $E$ be a nonempty set, $\theta:E\times E\rightarrow[1,\infty)$ a function
and $v\in\mathbb{N}$. Then $\rho_{\theta}:E\times E\rightarrow[0,\infty)$ is
called $b_{v}(\theta)$ metric if for all $u,z_{1},z_{2},...,z_{v},w\in E$,
each of them different from each other, it satisfies

  1. (1)

$\rho_{\theta}(u,w)=0$ iff $u=w$;

  2. (2)

$\rho_{\theta}(u,w)=\rho_{\theta}(w,u)$;

  3. (3)

$\rho_{\theta}(u,w)\leq\theta\left(u,w\right)[\rho_{\theta}(u,z_{1})+\rho_{\theta}(z_{1},z_{2})+\cdots+\rho_{\theta}\left(z_{v},w\right)].$

The pair $\left(E,\rho_{\theta}\right)$ is called $b_{v}(\theta)\ $metric
space.

######  Remark 4.

It is clear that if for all $u,w\in E$

  1. (1)

$\theta\left(u,w\right)=s$, then we obtain $b_{v}(s)$ metric space,

  2. (2)

$v=1$, then we obtain extended $b$-metric space,

  3. (3)

$\theta\left(u,w\right)=s$ and $v=1$, then we obtain $b$-metric space,

  4. (4)

$\theta\left(u,w\right)=s$ and $v=2$, then we obtain rectangular $b$-metric
space,

  5. (5)

$\theta\left(u,w\right)=1$ and $v=2$, then we obtain rectangular metric space,

  6. (6)

$\theta\left(u,w\right)=1$, then we obtain $v$-generalized metric space,

  7. (7)

$\theta\left(u,w\right)=1$ and $v=1$, then we obtain usual metric space.

######  Example 1.

Let $E=\mathbb{N}$. Define mappings
$\theta:\mathbb{N}\times\mathbb{N}\rightarrow[1,\infty)$ and
$\rho_{\theta}:\mathbb{N}\times\mathbb{N}\rightarrow[0,\infty)$ by
$\theta\left(u,w\right)=3+u+w$ and

$\rho_{\theta}\left(u,w\right)=\left\\{\begin{array}[]{ll}6,&\text{if
}u,w\in\left\\{1,2\right\\}\text{ and }u\neq w\\\ 1,&\text{if }u\text{ or
}w\notin\left\\{1,2\right\\}\text{ and }u\neq w\\\ 0,&\text{if
}u=w\end{array}\right.$

for all $u,w\in\mathbb{N}$. Then, it is easy to see that
$\left(E,\rho_{\theta}\right)$ is a $b_{v}\left(\theta\right)$ metric space
with $v=5$.

Definitions of Cauchy sequence, convergence and completeness can be easily
extended to the case of $b_{v}\left(\theta\right)$ metric space by the
following way.

######  Definition 11.

Let $\left(E,\rho_{\theta}\right)$ be a $b_{v}\left(\theta\right)$ metric
space, $\left\\{u_{n}\right\\}$ a sequence in $E$ and $u\in E$. Then,

  1. a)

$\left\\{u_{n}\right\\}$ is said to converge to $u$ in
$\left(E,\rho_{\theta}\right)$ if for every $\varepsilon>0$, there exists
$n_{0}\in\mathbb{N}$ such that $\rho_{\theta}\left(u_{n},u\right)<\varepsilon$
for all $n\geq n_{0}$ and this convergence is denoted by $u_{n}\rightarrow u$.

  2. b)

$\left\\{u_{n}\right\\}$ is said to be Cauchy sequence in
$\left(E,\rho_{\theta}\right)$ if for every $\varepsilon>0$, there exists
$n_{0}\in\mathbb{N}$ such that
$\rho_{\theta}\left(u_{n},u_{n+p}\right)<\varepsilon$ for all $n\geq n_{0}$
and $p>0.$

  3. c)

$\left(E,\rho_{\theta}\right)$ is said to be complete if every Cauchy sequence
in $E$ is convergent in $E$.

Now, we are in the position to prove fixed point theorems in $b_{v}(\theta)$
metric spaces. But first, we prove following lemmas which we need in the proof
of main theorems.

######  Lemma 1.

Let $\left(E,\rho_{\theta}\right)$ be a $b_{v}(\theta)$ metric space,
$S:E\rightarrow E$ a mapping and $\left\\{u_{n}\right\\}$ a sequence in $E$
defined by $u_{n+1}=Su_{n}=S^{n}u_{0}$ such that $u_{n}\neq u_{n+1}$. Suppose
that $c\in[0,1)$ such that

$\rho_{\theta}\left(u_{n+1},u_{n}\right)\leq
c\rho_{\theta}\left(u_{n},u_{n-1}\right)$

for all $n\in\mathbb{N}$. Then $u_{n}\neq u_{m}$ for all distinct
$n,m\in\mathbb{N}$.

###### Proof.

Since the proof is very similar with the proof of Lemma 1.11 of [21], we omit
it.

######  Lemma 2.

Let $\left(E,\rho_{\theta}\right)$ be a $b_{v}(\theta)$ metric space with a
bounded function $\theta$ and $\left\\{u_{n}\right\\}$ a sequence in $E$
defined by $u_{n+1}=Su_{n}=S^{n}u_{0}$ such that $u_{n}\neq u_{m}$ for all
$n,m\in\mathbb{N}$. Assume that there exist $c\in[0,1)$ and
$k_{1},k_{2}\in\mathbb{R}^{+}\cup\left\\{0\right\\}$ such that

$\rho_{\theta}\left(u_{m},u_{n}\right)\leq
c\rho_{\theta}\left(u_{m-1},u_{n-1}\right)+k_{1}c^{m}+k_{2}c^{m}$ (2.9)

for all $n,m\in\mathbb{N}$. Then $\left\\{u_{n}\right\\}$ is a Cauchy sequence
in $E$.

###### Proof.

It is easy to see that $\left\\{u_{n}\right\\}$ is Cauchy if $c=0$. So, we
should assume that $c\neq 0$. Since function $\theta\left(u,w\right)$ is
bounded, there exists a number $n_{0}\in\mathbb{N}$ such that

$0<c^{n_{0}}\theta\left(u,w\right)<1$ (2.10)

for all $u,w\in E$. From hypothesis of lemma, we can write

$\displaystyle\rho_{\theta}\left(u_{n+1},u_{n}\right)$ $\displaystyle\leq$
$\displaystyle
c\rho_{\theta}\left(u_{n},u_{n-1}\right)+k_{1}c^{n+1}+k_{2}c^{n}$
$\displaystyle\leq$ $\displaystyle
c\left(c\rho_{\theta}\left(u_{n-1},u_{n-2}\right)+k_{1}c^{n}+k_{2}c^{n-1}\right)+k_{1}c^{n+1}+k_{2}c^{n}$
$\displaystyle=$ $\displaystyle
c^{2}\rho_{\theta}\left(u_{n-1},u_{n-2}\right)+2\left(k_{1}c^{n+1}+k_{2}c^{n}\right)$
$\displaystyle\vdots$ $\displaystyle\leq$ $\displaystyle
c^{n}\rho_{\theta}\left(u_{1},u_{0}\right)+n\left(k_{1}c^{n+1}+k_{2}c^{n}\right).$

Similarly, for all $k\geq 1$, we can write

$\rho_{\theta}\left(u_{m+k},u_{n+k}\right)\leq
c^{k}\rho_{\theta}\left(u_{m},u_{n}\right)+k\left(k_{1}c^{m+k}+k_{2}c^{n+k}\right).$

If $v\geq 2$, then from the definition of $b_{v}(\theta)$ metric space, we get

$\displaystyle\rho_{\theta}\left(u_{n},u_{m}\right)$ $\displaystyle\leq$
$\displaystyle\theta\left(u_{n},u_{m}\right)\left[\rho_{\theta}\left(u_{n},u_{n+1}\right)+\rho_{\theta}\left(u_{n+1},u_{n+2}\right)\right.$
$\displaystyle+\cdots+\rho_{\theta}\left(u_{n+v-3},u_{n+v-2}\right)+\rho_{\theta}\left(u_{n+v-2},u_{n+n_{0}}\right)$
$\displaystyle\left.+\rho_{\theta}\left(u_{n+n_{0}},u_{m+n_{0}}\right)+\rho_{\theta}\left(u_{m+n_{0}},u_{m}\right)\right].$

Then, we have

$\displaystyle\rho_{\theta}\left(u_{n},u_{m}\right)$ $\displaystyle\leq$
$\displaystyle\theta\left(u_{n},u_{m}\right)\left[\left(c^{n}+c^{n+1}+\cdots+c^{n+v-3}\right)\rho_{\theta}\left(u_{0},u_{1}\right)\right.$
$\displaystyle+\left(k_{1}c+k_{2}\right)\left(nc^{n}+\left(n+1\right)c^{n+1}+\cdots+\left(n+v-3\right)c^{n+v-3}\right)$
$\displaystyle+c^{n}\rho_{\theta}\left(u_{v-2},u_{n_{0}}\right)+nc^{n}\left(k_{1}c^{v-2}+k_{2}c^{n_{0}}\right)$
$\displaystyle+c^{n_{0}}\rho_{\theta}\left(u_{n},u_{m}\right)+n_{0}c^{n_{0}}\left(k_{1}c^{n}+k_{2}c^{m}\right)$
$\displaystyle+\left.c^{m}\rho_{\theta}\left(u_{n_{0}},u_{0}\right)+mc^{m}\left(k_{1}c^{n_{0}}+k_{2}\right)\right].$

So, we obtain

$\displaystyle\rho_{\theta}\left(u_{n},u_{m}\right)\left(1-c^{n_{0}}\theta\left(u_{n},u_{m}\right)\right)$
$\displaystyle\leq$
$\displaystyle\theta\left(u_{n},u_{m}\right)\left[\left(c^{n}+c^{n+1}+\cdots+c^{n+v-3}\right)\rho_{\theta}\left(u_{0},u_{1}\right)\right.$
$\displaystyle+\left(k_{1}c+k_{2}\right)\left(nc^{n}+\left(n+1\right)c^{n+1}+\cdots+\left(n+v-3\right)c^{n+v-3}\right)$
$\displaystyle+c^{n}\rho_{\theta}\left(u_{v-2},u_{n_{0}}\right)+nc^{n}\left(k_{1}c^{v-2}+k_{2}c^{n_{0}}\right)+n_{0}c^{n_{0}}\left(k_{1}c^{n}+k_{2}c^{m}\right)$
$\displaystyle+\left.c^{m}\rho_{\theta}\left(u_{n_{0}},u_{0}\right)+mc^{m}\left(k_{1}c^{n_{0}}+k_{2}\right)\right].$

Since $\lim_{n\rightarrow\infty}nc^{n}=0$ and
$1-c^{n_{0}}\theta\left(u_{n},u_{m}\right)>0$, using (2.9), we have
$\rho_{\theta}\left(u_{n},u_{m}\right)\rightarrow 0$ as
$n,m\rightarrow\infty$. This means that $\left\\{u_{n}\right\\}$ is a Cauchy
sequence. Since $b_{v}\left(s\right)$ metric space is a
$b_{2v}\left(s^{2}\right)$ metric space, if $v=1$, then
$\left\\{u_{n}\right\\}$ is Cauchy.

Now we can give Banach fixed point theorem in complete
$b_{v}\left(\theta\right)$ metric space.

######  Theorem 4.

Let $\left(E,\rho_{\theta}\right)$ be a complete $b_{v}(\theta)$ metric space
with a bounded function $\theta$ and $S:E\rightarrow E$ a contraction mapping,
i.e., there exists a constant $c\in\left[0,1\right)$ such that

$\rho_{\theta}\left(Su,Sw\right)\leq c\rho_{\theta}\left(u,w\right)$ (2.11)

for all $u,w\in E$. Then $S$ has a unique fixed point.

###### Proof.

Let $u_{0}\in E$ be an arbitrary initial point and let
$\left\\{u_{n}\right\\}$ be a sequence defined by
$u_{n+1}=Su_{n}=S^{n+1}u_{0}$ and $u_{n}\neq u_{n+1}$ for all $n\geq 0$. It
follows from Lemma 1 that $u_{n}\neq u_{m}$ for all $n,m\in\mathbb{N}$. Since
$S$ is a contraction mapping, we can write

$\rho_{\theta}\left(u_{n},u_{m}\right)=\rho_{\theta}\left(Su_{n-1},Su_{m-1}\right)\leq
c\rho_{\theta}\left(u_{n-1},u_{m-1}\right).$

From Lemma 2, we have $\left\\{u_{n}\right\\}$ is a Cauchy sequence. So, it
follows from completeness of $E$ that there exists an element $u\in E$ such
that $u_{n}\rightarrow u$. Now, we show that $u\in FixS$, i.e., $u=Su$.

$\displaystyle\rho_{\theta}\left(u,Su\right)$ $\displaystyle\leq$
$\displaystyle\theta\left(u,Su\right)\left[\rho_{\theta}\left(u,u_{n+1}\right)+\rho_{\theta}\left(u_{n+1},u_{n+2}\right)\right.$
$\displaystyle\left.+\cdots+\rho_{\theta}\left(u_{n+v-1},u_{n+v}\right)+\rho_{\theta}\left(u_{n+v},Su\right)\right]$
$\displaystyle=$
$\displaystyle\theta\left(u,Su\right)\left[\rho_{\theta}\left(u,u_{n+1}\right)+\rho_{\theta}\left(u_{n+1},u_{n+2}\right)\right.$
$\displaystyle\left.+\cdots+\rho_{\theta}\left(u_{n+v-1},u_{n+v}\right)+\rho_{\theta}\left(Su_{n+v-1},Su\right)\right]$
$\displaystyle\leq$
$\displaystyle\theta\left(u,Su\right)\left[\rho_{\theta}\left(u,u_{n+1}\right)+\rho_{\theta}\left(u_{n+1},u_{n+2}\right)\right.$
$\displaystyle\left.+\cdots+\rho_{\theta}\left(u_{n+v-1},u_{n+v}\right)+c\rho_{\theta}\left(u_{n+v-1},u\right)\right].$

Since $\theta$ is a bounded function and $\left\\{u_{n}\right\\}$ is Cauchy
with $u_{n}\rightarrow u$, we have $\rho_{\theta}\left(u,Su\right)=0$. This
means that $u\in FixS$. Next, we need to show that $u$ is a unique fixed
point. Let assume to the contrary that there exists another fixed point $w$.
Since

$\rho_{\theta}\left(u,w\right)=\rho_{\theta}\left(Su,Sw\right)\leq
c\rho_{\theta}\left(u,w\right)<\rho_{\theta}\left(u,w\right),$

we get $u=w$ that is $u$ is the unique fixed point of $S$.

######  Remark 5.

In Theorem 4,

  1. (1)

if we take the constant $v=1$ and the function $\theta\left(u,w\right)=1$ for
all $u,w\in E$, then we derive classical Banach fixed point theorem in usual
metric spaces.

  2. (2)

if we take $\theta\left(u,w\right)=s$ for all $u,w\in E$ where $s\geq 1$, then
we derive Theorem 2.1 of [21] in $b_{v}\left(s\right)$ metric spaces.

  3. (3)

if $v=1$ and $\theta\left(u,w\right)=s$ for all $u,w\in E$, then we derive
Theorem 2.1 of [29] in $b$-metric spaces.

  4. (4)

if $v=2$ and $\theta\left(u,w\right)=s$ for all $u,w\in E$, then we derive
Theorem 2.1 of [30] and so main theorem of [4] in rectangular $b$-metric
spaces.

  5. (5)

if $\theta\left(u,w\right)=1$ for all $u,w\in E$, then we derive main result
of Branciari [11] in $v$-generalized metric spaces.

In literature, there exist various type of contraction mappings. Weakly
contractive mapping is one of this type of contractions which generalize usual
contractions. A mapping $S:E\rightarrow E$ is called weakly contractive if
there exists a continuous and nondecreasing function $\psi\left(t\right)$
defined from $\mathbb{R}^{+}\cup\left\\{0\right\\}$ onto itself such that
$\psi\left(0\right)=0,\psi\left(t\right)\rightarrow\infty$ as
$t\rightarrow\infty$ and for all $u,w\in E$

$\rho_{\theta}\left(Su,Sw\right)\leq\rho_{\theta}\left(u,w\right)-\psi\left(\rho_{\theta}\left(u,w\right)\right).$
(2.12)

Now, we generalize Banach fixed point theorem for weakly contractive mappings
in $b_{v}\left(\theta\right)$ metric space.

######  Theorem 5.

Let $E$ be a complete $b_{v}\left(\theta\right)$ metric space and $S$ a weakly
contractive mapping on $E$. Then $S$ has a unique fixed point.

###### Proof.

Let $u_{0}\in E$ be an arbitrary initial point. Define sequence
$\left\\{u_{n}\right\\}$ by $u_{1}=Su_{0}$,
$u_{2}=Su_{1}=S^{2}u_{0},\ldots,u_{n+1}=Su_{n}=S^{n}u_{0}$. If $u_{n}=u_{n+1}$
for all $n\in\mathbb{N}$ where $\mathbb{N}$ is the set of positive integer,
then proof is trivial. So, let assume that $u_{n}\neq u_{n+1}$ for all $n$.
Moreover, the case that $u_{n}\neq u_{m}$ for all different $n$ and $m$ can be
easily proved. From (2.12), we can write

$\displaystyle\rho_{\theta}\left(u_{n+1},u_{n+p+1}\right)$ $\displaystyle=$
$\displaystyle\rho_{\theta}\left(Su_{n},Su_{n+p}\right)$ $\displaystyle\leq$
$\displaystyle\rho_{\theta}\left(u_{n},u_{n+p}\right)-\psi\left(\rho_{\theta}\left(u_{n},u_{n+p}\right)\right)$

for all $n,p\in\mathbb{N}$. Let
$\alpha_{n}=\rho_{\theta}\left(u_{n},u_{n+p}\right)$. Since $\psi$ is
nondecreasing, we have

$\alpha_{n+1}\leq\alpha_{n}-\psi\left(\alpha_{n}\right)\leq\alpha_{n}.$ (2.13)

Thus, the sequence $\left\\{\alpha_{n}\right\\}$ has a limit $\alpha\geq 0$.
Now we should show that $\alpha=0$. Assume to the contrary that $\alpha>0$.
Using (2.13), we have

$\psi\left(\alpha_{n}\right)\geq\psi\left(\alpha\right)>0.$

So, we get

$\alpha_{n+1}\leq\alpha_{n}-\psi\left(\alpha\right).$

Hence, we obtain $\alpha_{N+m}\leq\alpha_{m}-N\psi\left(\alpha\right)$ which
is a contradiction for large enough $N$. This proves that $\alpha=0$. This
means that $\left\\{u_{n}\right\\}$ is Cauchy. Completeness of $E$ implies
that there exists a point $u\in E$ such that $u_{n}\rightarrow u$. Now, we
show that $u$ is a fixed point of $S$. Using (2.12) and definition of
$\rho_{\theta}$, we get

$\displaystyle\rho_{\theta}\left(u,Su\right)$ $\displaystyle\leq$
$\displaystyle\theta\left(u,Su\right)\left[\rho_{\theta}\left(u,u_{n+1}\right)+\rho_{\theta}\left(u_{n+1},u_{n+2}\right)\right.$
$\displaystyle\left.+\ldots+\rho_{\theta}\left(u_{n+v-1},u_{n+v}\right)+\rho_{\theta}\left(u_{n+v},Su\right)\right]$
$\displaystyle=$
$\displaystyle\theta\left(u,Su\right)\left[\rho_{\theta}\left(u,u_{n+1}\right)+\rho_{\theta}\left(u_{n+1},u_{n+2}\right)\right.$
$\displaystyle\left.+\ldots+\rho_{\theta}\left(u_{n+v-1},u_{n+v}\right)+\rho_{\theta}\left(Su_{n+v-1},Su\right)\right]$
$\displaystyle\leq$
$\displaystyle\theta\left(u,Su\right)\left[\rho_{\theta}\left(u,u_{n+1}\right)+\rho_{\theta}\left(u_{n+1},u_{n+2}\right)+\ldots+\right.$
$\displaystyle\left.\rho_{\theta}\left(u_{n+v-1},u_{n+v}\right)+\rho_{\theta}\left(u_{n+v-1},u\right)-\psi\left(\rho_{\theta}\left(u_{n+v-1},u\right)\right)\right].$

Since $\rho_{\theta}\left(u_{n},u_{n+p}\right)\rightarrow 0$ and
$u_{n}\rightarrow u$ as $n\rightarrow\infty$ and $\psi\left(0\right)=0$, we
have $u$ is a fixed point of $S$.

To prove the uniqueness of fixed point, we can assume that there exist one
more fixed point $w$. Since $S$ is a weakly contractive mapping, we have

$\rho_{\theta}\left(u,w\right)=\rho_{\theta}\left(Su,Sw\right)\leq\rho_{\theta}\left(u,w\right)-\psi\left(\rho_{\theta}\left(u,w\right)\right)<\rho_{\theta}\left(u,w\right).$

So $u=w$ This finishes the proof.

######  Remark 6.

In Theorem 5,

1 if we take the constant $v=1$, the function $\theta\left(u,w\right)=1$ for
all $u,w\in E$ and $\psi(t)=ct$, then we derive classical Banach fixed point
theorem.

2.if we take $\psi(t)=ct$ and $\theta\left(u,w\right)=s$ where
$s\in\left[1,\infty\right)$, then we derive Theorem 2.1 of [21]

3.if $v=1,$ $\theta\left(u,w\right)=s$ and $\psi(t)=ct$, then we derive
Theorem 2.1 of [29].

4.if $v=2$, $\theta\left(u,w\right)=s$ and $\psi(t)=ct$, then we derive
Theorem 2.1 of [30] and so main theorem of [4].

5.if $v=1$ and $\theta\left(u,w\right)=s$, then we derive main theorem of
[28].

Now, we give Reich fixed point theorem.

######  Theorem 6.

Let $\left(E,\rho_{\theta}\right)$ be a complete $b_{v}(\theta)$ metric space
with a bounded function $\theta$ and $S:E\rightarrow E$ a mapping satisfying:

$\rho_{\theta}\left(Su,Sw\right)\leq\alpha\rho_{\theta}\left(u,w\right)+\beta\rho_{\theta}\left(u,Su\right)+\gamma\rho_{\theta}\left(w,Sw\right)$
(2.14)

for all $u,w\in E$ where $\alpha,\beta,\gamma$ are nonnegative constants with
$\alpha+\beta+\gamma<1$ and $\Gamma_{1}<\frac{1}{\Gamma_{2}}$ where
$\Gamma_{1}=\min\left\\{\beta,\gamma\right\\}$ and
$\Gamma_{2}=\max\left\\{\theta\left(u,Su\right),\theta\left(Su,u\right)\right\\}$.
Then $S$ has a unique fixed point. Moreover, sequence $\left\\{u_{n}\right\\}$
defined by $u_{n}=Su_{n-1}$ converges strongly to the unique fixed point of
$S$.

###### Proof.

Let $\left\\{u_{n}\right\\}$ be a sequence defined by
$u_{n+1}=Su_{n}=S^{n+1}u_{0}$ where $u_{0}\in E$ is an arbitrary initial
point. If $u_{n}=u_{n+1}$ for all $n\in\mathbb{N}$, it is easy to see that
$u_{0}$ is a fixed point of $S$. Now, we assume that $u_{n}\neq u_{n+1}$ for
all $n$. From (2.14) and definition of $\left\\{u_{n}\right\\}$, we have

$\displaystyle\rho_{\theta}\left(u_{n+1},u_{n}\right)$ $\displaystyle=$
$\displaystyle\rho_{\theta}\left(Su_{n},Su_{n-1}\right)$ $\displaystyle\leq$
$\displaystyle\alpha\rho_{\theta}\left(u_{n},u_{n-1}\right)+\beta\rho_{\theta}\left(u_{n},Su_{n}\right)+\gamma\rho_{\theta}\left(u_{n-1},Su_{n-1}\right)$
$\displaystyle=$
$\displaystyle\alpha\rho_{\theta}\left(u_{n},u_{n-1}\right)+\beta\rho_{\theta}\left(u_{n},u_{n+1}\right)+\gamma\rho_{\theta}\left(u_{n-1},u_{n}\right).$

Then, we get

$\displaystyle\rho_{\theta}\left(u_{n+1},u_{n}\right)$ $\displaystyle\leq$
$\displaystyle\frac{\alpha+\gamma}{1-\beta}\rho_{\theta}\left(u_{n},u_{n-1}\right)$
$\displaystyle\leq$
$\displaystyle\left(\frac{\alpha+\gamma}{1-\beta}\right)^{n}\rho_{\theta}\left(u_{1},u_{0}\right).$

Since $\alpha+\beta+\gamma<1$, then it is clear that
$0\leq\frac{\alpha+\gamma}{1-\beta}<1$. So, we obtain

$\lim_{n\rightarrow\infty}\rho_{\theta}\left(u_{n+1},u_{n}\right)=0.$ (2.15)

Also, since we assume that $u_{n}\neq u_{n+1}$ for all $n$ and
$\rho_{\theta}\left(u_{n+1},u_{n}\right)\leq\frac{\alpha+\gamma}{1-\beta}\rho_{\theta}\left(u_{n},u_{n-1}\right)$,
then it follows from Lemma 1 that $u_{n}\neq u_{m}$ for all
$n,m\in\mathbb{N}$. So, we have

$\displaystyle\rho_{\theta}\left(u_{n},u_{m}\right)$ $\displaystyle=$
$\displaystyle\rho_{\theta}\left(Su_{n-1},Su_{m-1}\right)$ $\displaystyle\leq$
$\displaystyle\alpha\rho_{\theta}\left(u_{n-1},u_{m-1}\right)+\beta\rho_{\theta}\left(u_{n-1},Su_{n-1}\right)+\gamma\beta\rho_{\theta}\left(u_{m-1},Su_{m-1}\right)$
$\displaystyle=$
$\displaystyle\alpha\rho_{\theta}\left(u_{n-1},u_{m-1}\right)+\beta\rho_{\theta}\left(u_{n-1},u_{n}\right)+\gamma\beta\rho_{\theta}\left(u_{m-1},u_{m}\right)$
$\displaystyle\leq$
$\displaystyle\alpha\rho_{\theta}\left(u_{n-1},u_{m-1}\right)+\left(\beta\left(\frac{\alpha+\gamma}{1-\beta}\right)^{n-1}+\gamma\left(\frac{\alpha+\gamma}{1-\beta}\right)^{m-1}\right)\rho_{\theta}\left(u_{1},u_{0}\right)\text{.}$

It follows from Lemma 2 that $\left\\{u_{n}\right\\}$ is a Cauchy sequence.
So, from the completeness of $E$, we obtain that there exists a point $u\in E$
such that $u_{n}\rightarrow u$. Now, we show that $u$ is a fixed point of $S$,
i.e., $\rho_{\theta}\left(u,Su\right)=0$. Since

$\displaystyle\rho_{\theta}\left(u,Su\right)$ $\displaystyle\leq$
$\displaystyle\theta\left(u,Su\right)\left[\rho_{\theta}\left(u,u_{n+1}\right)+\rho_{\theta}\left(u_{n+1},u_{n+2}\right)+\cdots+\rho_{\theta}\left(u_{n+v-1},u_{n+v}\right)+\rho_{\theta}\left(u_{n+v},Su\right)\right]$
$\displaystyle\leq$
$\displaystyle\theta\left(u,Su\right)\left[\rho_{\theta}\left(u,u_{n+1}\right)+\rho_{\theta}\left(u_{n+1},u_{n+2}\right)+\cdots+\rho_{\theta}\left(u_{n+v-1},u_{n+v}\right)+\rho_{\theta}\left(Su_{n+v-1},Su\right)\right]$
$\displaystyle\leq$
$\displaystyle\theta\left(u,Su\right)\left[\rho_{\theta}\left(u,u_{n+1}\right)+\rho_{\theta}\left(u_{n+1},u_{n+2}\right)+\cdots+\rho_{\theta}\left(u_{n+v-1},u_{n+v}\right)\right.$
$\displaystyle\left.+\alpha\rho_{\theta}\left(u_{n+v-1},u\right)+\beta\rho_{\theta}\left(u_{n+v-1},u_{n+v}\right)+\gamma\rho_{\theta}\left(u,Su\right)\right],$

we have

$\displaystyle\left(1-\gamma\theta\left(u,Su\right)\right)\rho_{\theta}\left(u,Su\right)$
$\displaystyle\leq$
$\displaystyle\theta\left(u,Su\right)\left[\rho_{\theta}\left(u,u_{n+1}\right)+\rho_{\theta}\left(u_{n+1},u_{n+2}\right)+\cdots\right.$
$\displaystyle+\left.\rho_{\theta}\left(u_{n+v-1},u_{n+v}\right)+\alpha\rho_{\theta}\left(u_{n+v-1},u\right)+\beta\rho_{\theta}\left(u_{n+v-1},u_{n+v}\right)\right].$

Since $\Gamma_{1}<\frac{1}{\Gamma_{2}}$, we get
$\left(1-\gamma\theta\left(u,Su\right)\right)\in\left[0,1\right)$. So, it
follows from (2.15) and convergence of $\left\\{u_{n}\right\\}$ that
$\rho_{\theta}\left(u,Su\right)=0$. This means that $u$ is a fixed point of
$S.$ Now, we need to show that $u$ is a unique fixed point. Let assume that
there exist another fixed point $v$. Then, we have

$\displaystyle\rho_{\theta}\left(u,v\right)$ $\displaystyle=$
$\displaystyle\rho_{\theta}\left(Su,Sv\right)\leq\alpha\rho_{\theta}\left(u,v\right)+\beta\rho_{\theta}\left(u,Su\right)+\delta\rho_{\theta}\left(v,Sv\right)$
$\displaystyle=$ $\displaystyle\alpha\rho_{\theta}\left(u,v\right).$

Since $\alpha<1$, we obtain that $\rho_{\theta}\left(u,v\right)=0,$ i.e., $u$
is the unique fixed point of $S$.

######  Remark 7.

In Theorem 6, if we take $\theta\left(u,w\right)=s$ for all $u,w\in E$ where
$s\geq 1$, then we derive Theorem 2.4 of [21].

In Reich fixed point theorem, if we get $\alpha=0,$ then we obtain following
generalized Kannan fixed point theorem in $b_{v}\left(\theta\right)$ metric
spaces.

######  Theorem 7.

Let $E$ be a complete $b_{v}\left(\theta\right)$ metric space and $S$ a
mapping on $E$ satisfying:

$\rho_{\theta}\left(Su,Sw\right)\leq\beta\rho_{\theta}\left(u,Su\right)+\gamma\rho_{\theta}\left(w,Sw\right)$

for all $u,w\in E$ where $\beta$ and $\gamma$ are nonnegative constants with
$\beta+\gamma<1$ and $\Gamma_{1}<\frac{1}{\Gamma_{2}}$ where
$\Gamma_{1}=\min\left\\{\beta,\gamma\right\\}$ and
$\Gamma_{2}=\max\left\\{\theta\left(u,Su\right),\theta\left(Su,u\right)\right\\}$.
Then $S$ has a unique fixed point.

######  Remark 8.

In Theorem 7,

  1. (1)

if $v=1$ and $\theta\left(u,w\right)=1$ for all $u,w\in E$ where $s\geq 1$,
then we obtain Kannan fixed point theorem [8] in complete usual metric spaces.

  2. (2)

if $v=2$ and $\theta\left(u,w\right)=s$ for all $u,w\in E$ where $s\geq 1$,
then we derive Theorem 2.4 of [4].

  3. (3)

if $v=2$ and $\theta\left(u,w\right)=1$ for all $u,w\in E$ where $s\geq 1$,
then we obtain main theorem of [31] without the assumption of orbitally
completeness of the space and the main theorem of [27].

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