# Perov type T-contractive Mappings on Cone b-Metric Spaces with Generalized
c-Distance

Talat Nazir1, Mujahid Abbas2,3, Sergei Silvestrov4
1Department of Mathematical Sciences, University of South Africa,
Florida 0003, South Africa
<EMAIL_ADDRESS>
2Department of Mathematics, Government College University,
Katchery Road, Lahore 54000, Pakistan
3Department of Mathematics and Applied Mathematics, University of Pretoria,
Hatfield 002, Pretoria, South Africa
<EMAIL_ADDRESS>
4Department of Mathematics and Physics, School of Education, Culture and
Communication,
Mälardalen University, Box 883, 72123 Västerås, Sweden
<EMAIL_ADDRESS>
###### Abstract

Fixed point results of Perov type mapping which satisfy generalized
$T$-contractive conditions in the setup of cone $b$-metric spaces associated
with generalized $c$-distance are proved and illustrated by nontrivial
examples.

##  1 Introduction

Fixed point theory provides certain simple tools to solve various problems
arising in nonlinear analysis. The metric fixed point theory deals with the
structure of sets equipped with some notion of distance functions and
operators satisfying certain contraction conditions.

The concept of a $b$-metric space was described by Bakhtin [4] and Boriceanu
[6] as an extension of a usual metric space. Some useful fixed point results
in the framework of a $b$-metric space were established in [3, 7, 10, 23].

Huang and Zhang [16] extended the metric space structure and defined a cone
metric space by replacing the range of the distance function with a normed
space equipped with an order induced by the cone. Then, they obtained some
fixed point results in a cone metric space (also, see [32]). Later on, the
cone metric space was extended to a cone $b$-metric space in [13, 17].

The notion of $w$-distance on metric spaces was introduced by Kada et al. [22]
for solving non-convex minimization based problems. While, the notion of
$c$-distance on cone metric spaces was defined by Cho et al. [8] that could be
viewed as the cone model of a $w$-distance. The concept of a $wt$-distance was
initiated by Hussain et al. [18] on a $b$-metric space and several fixed point
results were proved using the $wt$-distance on a $b$-metric space (also, see
[15, 24]). Recently, the notion of a generalized $c$-distance in the setup of
cone $b$-metric space was described by Bao et al. [5] and some fixed point
theorems were proved in this new framework.

On the other hand, Perov [25] employed contraction matrices rather than the
contraction constants to extend well known Banach contraction principle. Fixed
point results for Perov type contractions in a cone metric space were
presented in [11, 12]. Abbas et al. [1] obtained common fixed points of
multivalued Perov type contractive mappings the setup of a cone metric space
equipped a directed graph.

In this paper, we prove several fixed point results of Perov type generalized
$T$-contractive mappings defined on cone $b$-metric space equipped with a
generalized $c$-distance. The obtained results unify, extend and generalize
several results in the current literature.

##  2 Preliminaries

In the sequel $\mathbb{N},\mathbb{N}_{0},\mathbb{R}_{+},\mathbb{R}$ will
denote the set of natural numbers, natural numbers with $0$, positive real
numbers and real numbers, respectively.

Let $E$ be a real Banach space. A cone $P$ is the subset of $E$ that satisfies

  1. (i)

$P$ is closed, non-empty and $P\neq\\{\theta\\}$ (where $\theta$ is the zero
element of $E$);

  2. (ii)

$\alpha,\beta\in\mathbb{R}_{+},$ $\alpha,\beta\geq 0$ with $a,b\in P$ implies
$\alpha a+\beta b\in P;$

  3. (iii)

$-P\cap P=\\{\theta\\}$.

With a cone $P$, the partial ordering $\preceq$ on $E$ is defined as:
$u_{1}\preceq u_{2}$ if $u_{2}-u_{1}\in P$ and conversely. Here $u_{1}\prec
u_{2}$ is used to indicate that $u_{1}\preceq u_{2}$ but $u_{1}\neq u_{2}$ and
$u_{1}\ll u_{2}$ means that $u_{2}-u_{1}$ belongs to the interior of $P$
denoted by $intP.$ If $intP\neq\theta,$ then $P$ is known as solid cone.

A cone $P$ is said to be semi monotone or normal if there exists a positive
constant $\kappa>0$ such that $\theta\preceq u\preceq v$ implies that
$\left\|u\right\|\leq\kappa\left\|v\right\|$ for all $u,v\in P$. The smallest
value of $\kappa$ is known as the normal constant of normal cone $P$. The
definition of normal cone $P$ is also equivalent to the following condition
that

$\inf\\{\parallel u_{1}+u_{2}\parallel:u_{1},u_{2}\in P\text{ and}\parallel
u_{1}\parallel=\parallel u_{2}\parallel=1\\}>0.$

If $u=(u_{1},\dots,u_{n})^{T}$ and
$v=(v_{1},\dots,v_{n})^{T}\in\mathbb{R}^{n}$, then $u\preceq v$ means that
$u_{i}\leq v_{i},$ for all $i=1,\dots,n$. In this way, $P$ becomes a normal
cone with $\kappa=1$ defined as
$P=\\{u=(u_{1},\dots,u_{n})^{T}\in\mathbb{R}^{n}:u_{i}\geq 0$ for
$i=1,2,\dots,n\\}$.

######  Definition 2.1 ([13, 17]).

Let $X$ be a nonempty set and $E$ a real Banach space equipped with the
partial ordering induced by the cone $P\subset E$. For a real number $b\geq 1$
and for all $u_{1},u_{2},u_{3}\in X,$ suppose that the mapping $d:X\times
X\rightarrow E$ satisfies the following conditions:

  1. (b1)

$\theta\preceq d(u_{1},u_{2})$ and $d(u_{1},u_{2})=\theta$ if and only if
$u_{1}=u_{2}$;

  2. (b2)

$d(u_{1},u_{2})=d(u_{2},u_{1})$;

  3. (b3)

$d(u_{1},u_{2})\preceq b[d(u_{1},u_{2})+d(u_{2},u_{3})].$

Then $d$ is called a cone $b$-metric on $X$ and the pair $(X,d)$ is known as a
cone $b$-metric space.

Clearly, for $b=1$, the cone $b$-metric space becomes a cone metric space.
Moreover, for $E=\mathbb{R}$ and $P=[0,\infty)$, the cone $b$-metric on $X$
becomes $b$-metric on $X$.

######  Example 2.2.

Let $X=\left[0,1\right],$ $E=X\times X\ $and $P=\\{\left(x,y\right)\in
E:x,y\geq 0\\}=E.$ Define $d:X\times X\rightarrow E$ by
$d(x,y)=\left(\left(\phi\left(x,y\right)\right)^{p},\alpha\left(\phi\left(x,y\right)\right)^{p}\right)$,
where $\alpha\geq 1$ and $p\geq 1\ $and $\phi:X\times
X\rightarrow\mathbb{R}_{+}$ satisfies $\phi\left(x,y\right)=0$ if and only if
$x=y;$ $\phi\left(x,y\right)=\phi\left(y,x\right)\ $and
$\phi\left(x,z\right)\leq\phi\left(x,y\right)+\phi\left(y,z\right).$ Then
$(X,d)$ is a complete cone $b$-metric space with $b\geq 2^{p-1}\geq 1$ along
with $P$ a solid cone.

Note that $d$ trivially satisfies (b1) and (b2). The condition (b3) is
satisfied since $\left(a+b\right)^{p}\leq 2^{p-1}\left(a^{p}+b^{p}\right)$ for
any $a,b\in\mathbb{R}_{+},$ and hence

$\displaystyle d(x,y)$ $\displaystyle=$
$\displaystyle\left(\left(\phi\left(x,y\right)\right)^{p},\alpha\left(\phi\left(x,y\right)\right)^{p}\right)$
$\displaystyle\leq$
$\displaystyle\left(\left(\phi\left(x,z\right)+\phi\left(z,y\right)\right)^{p},\alpha\left(\phi\left(x,z\right)+\phi\left(z,y\right)\right)^{p}\right)$
$\displaystyle\leq$ $\displaystyle
2^{p-1}\left(\left(\phi\left(x,z\right)\right)^{p}+\left(\phi\left(z,y\right)\right)^{p},\alpha\left(\left(\phi\left(x,z\right)\right)^{p}+(\phi\left(z,y\right)\right)^{p}\right)$
$\displaystyle=$ $\displaystyle
2^{p-1}\left(\left(\phi(x,z\right))^{p},\alpha\left(\phi\left(x,z\right)\right)^{p})+(\left(\phi\left(z,y\right)\right)^{p},\left(\phi\left(z,y\right)\right)^{p}\right)$
$\displaystyle\leq$ $\displaystyle
b\left(d\left(x,z\right)+d\left(z,y\right)\right).$

Thus $d$ is a cone $b$-metric on $X$.

######  Definition 2.3.

Let $X$ be a $b$-cone metric space, $a\in E$ where $\theta\ll a$, and
$\\{u_{n}\\}$ a sequence in $X.$ Then $\\{u_{n}\\}$ is called

  1. (i)

Cauchy sequence if there exists an $n_{0}$ in $\mathbb{N}$ such that
$d(u_{n},u_{m})\ll a$ for all $n,m\geq n_{0}$;

  2. (ii)

convergent if there exists an $n_{0}$ in $\mathbb{N}$ and $u\in X$ such that
$d(u_{n},u)\ll a$ for all $n\geq n_{0}$;

  3. (iii)

If every Cauchy sequence in $X$ is convergent, then $X$ is called a complete
cone metric space.

A cone $b$-metric space $X$ is said to be complete if every Cauchy sequence
$\\{u_{n}\\}$ in $X$ is convergent in $X$.

If a cone is normal, then the sequence $\\{u_{n}\\}$ converges to a point
$u\in X$ if and only if $d(u_{n},u)\rightarrow 0$ as $n\rightarrow\infty.$

######  Definition 2.4 ([5]).

Let $(X,d)$ be a cone $b$-metric space. For $b\geq 1$, the mapping $q:X\times
X\rightarrow E$ is called a generalized $c$-distance on $X$ if for any
$u_{1},u_{2},u_{3}\in X,$ the following properties are satisfied:

  1. (q1)

$\theta\preceq q(u_{1},u_{2})$;

  2. (q2)

$q(u_{1},u_{3})\preceq b[q(u_{1},u_{2})+q(u_{2},u_{3})]$;

  3. (q3)

If $q(u,v_{n})\preceq e$ for all $n\geq 1$, where $e\in P,$ then
$q(u,v)\preceq be$, where $\\{v_{n}\\}\subseteq$ $X$ be a sequence convergent
to $v\in X$;

  4. (q4)

If $e^{\ast}\in intP$, then an $e\in E$ with $\theta\ll e$ exists such that
for $q(u_{3},u_{1})\ll e$ and $q(u_{3},u_{2})\ll e$ gives $d(u_{1},u_{2})\ll
e^{\ast}$.

For any $b$-metric space $(X,d)$, where $E=\mathbb{R}$ and $P=[0,\infty),$ the
$wt$-distance on a $b$-metric space $X$ is the generalized $c$-distance, but
the converse does not hold in general [18].

Also for $b=1$, the generalized $c$-distance becomes $c$-distance [8]. Note
that, for $b=1,$ $E=\mathbb{R}$ and $P=[0,\infty)$ it becomes the $w$-distance
given in [22]. Moreover, $q(u_{1},u_{2})=\theta$ does not imply $u_{1}=u_{2}$
and $q(u_{1},u_{2})\neq q(u_{2},u_{1})$ for all $u_{1},u_{2}\in X$.

######  Example 2.5.

Let $X=[0,1],$ $E=\mathbb{R}^{2}\ $ and $P=\\{\left(x,y\right)\in E:x,y\geq
0\\}\subseteq E.$ Define $d:X\times X\rightarrow E$ as given in Example 2.2.
Then $(X,d)$ is a complete cone $b$-metric space with $b\geq 2^{p-1}\geq 1$
along with $P$ a solid cone.

Define
$q\left(x,y\right)=\left(\psi\left(y\right),\eta\psi\left(y\right)\right),$
where $\eta\geq 1$ and $\psi:X\rightarrow\mathbb{R}_{+}$ is of the form
$\psi\left(t\right)=at^{p}$ for some $p>1,$ where $a>0.$ Then $q$ is a
generalized $c$-distance on $X.$

Note that $q$ is trivially satisfying (q1) since for all $x,y\in X,$

$q\left(x,y\right)=\left(\psi\left(y\right),\eta\psi\left(y\right)\right)=\left(ay^{p},\eta
ay^{p}\right)\succeq\theta.$

For (q2), for any $x,y,z\in X,$

$\displaystyle q\left(x,z\right)$ $\displaystyle=$
$\displaystyle\left(\psi\left(z\right),\eta\psi\left(z\right)\right)$
$\displaystyle=$ $\displaystyle\left(az^{p},\eta az^{p}\right)$
$\displaystyle\leq$ $\displaystyle b\left(a\left(y^{p}+z^{p}\right),\eta
a\left(y^{p}+z^{p}\right)\right)$ $\displaystyle=$ $\displaystyle
b\left(\left(ay^{p},\eta ay^{p}\right)+\left(az^{p},\eta az^{p}\right)\right)$
$\displaystyle=$ $\displaystyle
b\left(q\left(x,y\right)+q\left(y,z\right)\right).$

For (q3), one needs to show that for any $\left\\{v_{n}\right\\}$ in $X$ that
converges to $v$ in $X,$ if $q(u,v_{n})\preceq e$ for all $n\geq 1$, where
$e=\left(e_{1},e_{2}\right)\in P,$ then $q\left(u,v\right)\preceq be.$ Since
$q\left(u,v_{n}\right)=\left(\psi\left(v_{n}\right),\eta\psi\left(v_{n}\right)\right)=\left(av_{n}^{p},\eta
av_{n}^{p}\right)\preceq e,$ that is $\left(e_{1}-av_{n}^{p},e_{2}-\eta
av_{n}^{p}\right)\in P$ or equivalently, $e_{1}-av_{n}^{p}\geq 0\ $and
$e_{2}-\eta av_{n}^{p}\geq 0,$ taking limit as $n\rightarrow\infty$ yields
$e_{1}-av^{p}\geq 0\ $and $e_{2}-\eta av^{p}\geq 0,$ which gives
$q\left(u,v\right)=\left(\psi\left(v\right),\eta\psi\left(v\right)\right)=\left(av^{p},\eta
av^{p}\right)\preceq e\leq be.$

Finally to show (q4), we need to show that if $e^{\ast}\in intP$, then an
$e\in E$ with $\theta\ll e$ exists such that for $q(u_{3},u_{1})\ll e$ and
$q(u_{3},u_{2})\ll e$ gives $d(u_{1},u_{2})\ll e^{\ast}.$

Condition
$q(u_{3},u_{1})=\left(\psi\left(u_{1}\right),\eta\psi\left(u_{1}\right)\right)=\left(au_{1}^{p},\eta
au_{1}^{p}\right)\ll e=\left(e_{1},e_{2}\right)$ is equivalent to
$e_{1}-\psi\left(u_{1}\right)\geq 0$ and $e_{2}-\eta\psi\left(u_{1}\right)\geq
0$, and the condition
$q(u_{3},u_{2})=\left(\psi\left(u_{2}\right),\eta\psi\left(u_{2}\right)\right)=\left(au_{2}^{p},\eta
au_{2}^{p}\right)\ll e\ $ is equivalent to $e_{1}-\psi\left(u_{2}\right)\geq
0$ and $e_{2}-\eta\psi\left(u_{2}\right)\geq 0.$

Now

$\displaystyle d(u_{1},u_{2})$ $\displaystyle=$
$\displaystyle\left(\left(\phi\left(u_{1},u_{2}\right)\right)^{p},\alpha\left(\phi\left(u_{1},u_{2}\right)\right)^{p}\right)$
$\displaystyle\leq$
$\displaystyle\left(\left(\phi\left(u_{1},u_{3}\right)+\phi\left(u_{3},u_{2}\right)\right)^{p},\alpha\left(\phi\left(u_{1},u_{3}\right)+\phi\left(u_{3},u_{2}\right)\right)^{p}\right)$
$\displaystyle=$ $\displaystyle
2^{p-1}\left(\left(\phi\left(u_{1},u_{3}\right))^{p}+(\phi\left(u_{3},u_{2}\right))^{p},\alpha\left(\left(\phi\left(u_{1},u_{3}\right)\right)^{p}+(\phi\left(u_{3},u_{2}\right)\right)^{p}\right)\right)$
$\displaystyle=$ $\displaystyle
2^{p-1}\left(\left((\phi\left(u_{1},u_{3}\right))^{p},\alpha\left(\left(\phi\left(u_{1},u_{3}\right)\right)^{p})+((\phi\left(u_{3},u_{2}\right))^{p},\alpha(\phi\left(u_{3},u_{2}\right)\right)^{p}\right)\right)$
$\displaystyle\leq$ $\displaystyle
2^{p-1}(\left(k_{1}\psi\left(u_{1}\right),\alpha
k_{1}\psi\left(u_{1}\right)\right)+\left(k_{2}\psi\left(u_{2}\right),\alpha
k_{2}\psi\left(u_{2}\right)\right))$ $\displaystyle=$ $\displaystyle
2^{p-1}(k_{1}\left(e_{1},\frac{\alpha}{\eta}e_{2}\right)+k_{2}\left(e_{1},\frac{\alpha}{\eta}e_{2}\right))\quad\quad\mbox{
{(}$k_{1},k_{2}\geq 0${)}}$ $\displaystyle=$ $\displaystyle
2^{p-1}\left(k_{1}+k_{2}\right)\left(e_{1},\frac{\alpha}{\eta}e_{2}\right)$
$\displaystyle=$ $\displaystyle\left(e_{1}^{\ast},e_{2}^{\ast}\right)$
$\displaystyle=$ $\displaystyle e^{\ast},$

where
$e=\left(e_{1},e_{2}\right)=\left(\frac{1}{2^{p-1}\left(k_{1}+k_{2}\right)}e_{1}^{\ast},\frac{\eta}{2^{p-1}\alpha\left(k_{1}+k_{2}\right)}e_{2}^{\ast}\right).$
Clearly $\theta\ll e,$ and for any $u_{1}$ and $u_{2}$ such that
$q(u_{3},u_{1})=\left(au_{1}^{p},\eta au_{1}^{p}\right)\ll e$ and
$q(u_{3},u_{2})=\left(au_{2}^{p},\eta au_{2}^{p}\right)\ll e$ holds
$d(u_{1},u_{2})\ll e^{\ast}.$

Note that the conditions $q(u_{3},u_{1})=\left(au_{1}^{p},\eta
au_{1}^{p}\right)\ll e$ and $q(u_{3},u_{2})=\left(au_{2}^{p},\eta
au_{2}^{p}\right)\ll e$ are equivalent to

$\displaystyle u_{1}$
$\displaystyle\leq\frac{1}{\sqrt[p]{2^{p-1}a\left(k_{1}+k_{2}\right)}}\sqrt[p]{e_{1}^{\ast}}$
$\displaystyle u_{1}$
$\displaystyle\leq\frac{1}{\sqrt[p]{2^{p-1}a\alpha\left(k_{1}+k_{2}\right)}}\sqrt[p]{e_{2}^{\ast}},$
$\displaystyle u_{2}$
$\displaystyle\leq\frac{\eta}{\sqrt[p]{2^{p-1}a\left(k_{1}+k_{2}\right)}}\sqrt[p]{e_{1}^{\ast}},$
$\displaystyle u_{2}$
$\displaystyle\leq\frac{\eta}{\sqrt[p]{2^{p-1}a\alpha\left(k_{1}+k_{2}\right)}}\sqrt[p]{e_{2}^{\ast}},$

which can also be written as

$\left(u_{1},u_{2}\right)\leq\left(\min\\{\frac{\sqrt[p]{e_{1}^{\ast}}}{\sqrt[p]{2^{p-1}a\left(k_{1}+k_{2}\right)}},\frac{\sqrt[p]{e_{2}^{\ast}}}{\sqrt[p]{2^{p-1}a\alpha\left(k_{1}+k_{2}\right)}}\\},\right.\\\
\left.\min\\{\frac{\eta\sqrt[p]{e_{1}^{\ast}}}{\sqrt[p]{2^{p-1}a\left(k_{1}+k_{2}\right)}},\frac{\eta\sqrt[p]{e_{2}^{\ast}}}{\sqrt[p]{2^{p-1}a\alpha\left(k_{1}+k_{2}\right)}}\\}\right).$

The following useful lemma is needed in the sequel.

######  Lemma 2.6.

Let $q$ be a generalized $c$-distance on $X$ equipped with cone $b$-metric
space $(X,d).$ For sequences $\\{x_{n}\\}$ and $\\{y_{n}\\}$ in $X$, two
sequences $\\{a_{n}\\}$ and $\\{b_{n}\\}$ in $P$ converging to $\theta$ and
for any $u_{1},u_{2},u_{3}\in X$, the following statements hold:

  1. (i)

if for all $n\in\mathbb{N}$, $q(x_{n},u_{1})\preceq a_{n}$ and
$q(x_{n},u_{2})\preceq b_{n}$, then $u_{1}=u_{2}$. In particular, for
$q(u_{3},u_{1})=\theta$ and $q(u_{3},u_{2})=\theta$ implies $u_{1}=u_{2}$;

  2. (ii)

if for all $n\in\mathbb{N}$, $q(x_{n},y_{n})\preceq a_{n}$ and
$q(x_{n},u_{1})\preceq b_{n}$, then $\\{y_{n}\\}$ converges to $u_{1}$;

  3. (iii)

if $q(x_{n},x_{m})\preceq a_{n}$ for all $m,n\in\mathbb{N}$ with $m\geq n,$
then $\\{x_{n}\\}$ is a Cauchy sequence in $X$;

  4. (iv)

if $q(u_{1},x_{n})\preceq a_{n}$ for all $n\in\mathbb{N}$, then $\\{x_{n}\\}$
is a Cauchy sequence in $X$.

Let $M_{n\times n\text{ }}(\mathbb{R}^{+})$ denote the set of all $n\times n$
matrices with non negative elements. It is well known that if $A$ is any
square matrix of order $n$, then $A(P)\subset P$ if and only if $A\in
M_{n,n}(\mathbb{R}^{+})$. A matrix $A\in M_{n,n}(\mathbb{R}^{+})$ is called
convergent to zero if $A^{n}\longrightarrow\Theta$ as
$n\longrightarrow\infty$, where $\Theta$ is the null matrix of size $n.$

Regarding this class of matrices, we have the following classical result in
matrix analysis (see [26], [30] and [31]).

######  Theorem 2.7.

Let $A\in M_{n,n}(\mathbb{R}^{+})$. The following statements are equivalent:

  1. (i)

$A^{n}\rightarrow\Theta$, as $n\rightarrow\infty$;

  2. (ii)

eigenvalues of $A$ lie in the open unit disc, that is, for all $\lambda\in C$
satisfying $\det(A-\lambda I_{n})=0,$ we have $\left|\lambda\right|<1$ ;

  3. (iii)

matrix $I_{n}-A$ is non-singular and
$(I_{n}-A)^{-1}=I_{n}+A+A^{2}+...+A^{m}+...$;

  4. (iv)

matrix $(I_{n}-A)$ is non-singular and $(I_{n}-A)^{-1}$ has non-negative
elements.

Perov [25] obtained the following generalization of Banach contraction
principle for the case when metric $d$ take values in $\mathbb{R}_{+}^{n},$
that is, $d\left(x,y\right)\in\mathbb{R}_{+}^{n}$ and known as generalized
metric space.

######  Theorem 2.8.

Let $(X,d)$ be a complete generalized metric space, $f$ be a self mapping on
$X$ and $A\in M_{n,n}(\mathbb{R}^{+})$ a matrix convergent to zero. If for any
$u,v\in X,$ we have

$d(f(u),f(v))\leq A(d(u,v)),$

then the following statements hold:

  1. 1.

$f$ has a unique fixed point $u^{\ast}\in X.$

  2. 2.

The Picard iterative sequence $u_{n}=f^{n}(u_{0}),$ $n\in\mathbb{N}$ converges
to $u^{\ast}$ for all $u_{0}\in X$.

  3. 3.

$d(u_{n},u^{\ast})\leq A^{n}(I_{n}-A)^{-1}(d(u_{0},u_{1})),$ $n\in\mathbb{N}.$

  4. 4.

if $g:X\rightarrow X$ satisfies the condition $d(f(u),g(u))\leq a$ for all
$u\in X$ and some $a\in\mathbb{R}^{n},$ then for the sequence
$v_{n}=g^{n}(u_{0}),$ $n\in\mathbb{N},$ the inequality

$d(v_{n},u^{\ast})\leq(I_{n}-A)^{-1}(a)+A^{n}(I_{n}-A)^{-1}(d(u_{0},u_{1}))$

is valid for all $n\in\mathbb{N}.$

The role of vector valued norm has much importance in the study of semi-linear
operator systems. For details, we refer to [26]. Let $\mathcal{B}(E)$ be the
set of all bounded linear operators on $E$, and $L(E)$ be the set of all
linear operators on $E$.

Note that $\mathcal{B}(E)$ is a Banach algebra. If $A\in\mathcal{B}(E)$, then

$r(A)=\lim\limits_{n\rightarrow\infty}\left\|A^{n}\right\|^{\frac{1}{n}}=\inf_{n}\left\|A^{n}\right\|^{\frac{1}{n}}$

is the spectral radius of $A$. We write $\mathcal{B}(E)^{-1}$ for the set of
all invertible elements in $\mathcal{B}(E).$ Let us remark that if $r(A)<1$,
then

  1. (i)

series $\sum\limits_{n=0}^{\infty}A^{n}$ is absolutely convergent;

  2. (ii)

the element $I-A$ is invertible in $\mathcal{B}(E)$ and
$\sum\limits_{n=0}^{\infty}A^{n}=(I-A)^{-1}.$

If $A_{1},A_{2}\in\mathcal{B}(E)$ and $A_{1}A_{2}=A_{2}A_{1}$ then
$r(A_{1}A_{2})\leq r(A_{2})r(A_{1}).$ If $A\in\mathcal{B}(E)$ and
$A^{-1}\in\mathcal{B}(E)$ exists, then $r(A^{-1})=1/r(A).$ Further, if
$\parallel A\parallel<1$, then $I-A$ is invertible and

$\left\|(I-A)^{-1}\right\|\leq\frac{1}{1-\left\|A\right\|}.$

Note that $r(A)\leq\left\|A\right\|$.

######  Remark 2.9 ([11]).

Let $X$ be a $b$-cone metric space, $P\subseteq E$ be a cone in $E$ and
$A:E\rightarrow E$ be a linear operator. The following conditions are
equivalent:

  1. a1)

$A$ is increasing, that is, $u\preceq v$ implies that $A(u)\preceq A(v);$

  2. a2)

$A$ is positive, that is, $A(P)\subset P;$

  3. a3)

If $\left\|A\right\|<1$ and $A$ is positive, then inverse of $I-A$ is
positive.

######  Remark 2.10 ([1]).

Let $P\subseteq E$ be a cone in $E.$ Suppose that $A:E\rightarrow E$ is a
bounded linear operator with $A\left(P\right)\subset P$ and $r(A)<1$. If
$u\preceq A(u)$ for $u$ in $P$, then $u=\theta$.

Now we give the definitions of following Perov type $T$-Hardy Rogers
contractions in the setup of $b$-cone metric spaces via generalized
$c$-distance.

######  Definition 2.11.

1.11 Let $(X,d)$ be a $b$-cone metric space, $q$ a $c$-distance on $X.$
Suppose that $f,T:X\rightarrow X$ are two self mappings. A mapping $f$ is said
to be:

  1. (I)

a Perov $THR_{1}-$contraction if there exists a linear bounded operator
$A:E\rightarrow E$ with $r(bA)<1$ and $A\left(P\right)\subset P$ such that for
all $x_{1},x_{2}\in X$,

$q(Tfx_{1},Tfx_{2})\leq A(U_{T}(x_{1},x_{2})),$ (1.1)

holds, where

$U_{T}(x_{1},x_{2})\in\\{q(Tx_{1},Tx_{2}),q(Tx_{1},Tfx_{1}),q(Tx_{2},Tfx_{2})\\}.$

  2. (II)

a Perov $THR_{2}-$contraction if there exist linear bounded operators
$A_{1},A_{2},A_{3}:E\rightarrow E$ with $A_{k}\left(P\right)\subset P$ for
$k=1,2,3$, $(I-A_{3})^{-1}\in\mathcal{B}(E)$ and
$r[b(I-A_{3})^{-1}(A_{1}+A_{2})]<1$ such that for all $x_{1},x_{2}\in X,$

$q(Tfx_{1},Tfx_{2})\leq
A_{1}\left(q(Tx_{1},Tx_{2})\right)+A_{2}\left(q(Tx_{1},Tfx_{1})\right)+A_{3}\left(q(Tx_{2},Tfx_{2})\right),$
(1.2)

holds.

######  Definition 2.12.

Let $(X,d)$ be a $b$-cone metric space and $P$ a solid cone. A mapping
$T:X\rightarrow X$ is said to be

  1. a)

continuous if, for all $\\{u_{n}\\}$ in $X$ such that
$\lim\limits_{n\rightarrow\infty}u_{n}=u^{\ast}$ implies that
$\lim\limits_{n\rightarrow\infty}Tu_{n}=Tu^{\ast}.$

  2. b)

sequentially convergent if, for every sequence $\\{u_{n}\\}$ in $X$ such that
$\\{Tu_{n}\\}$ is convergent implies that $\\{u_{n}\\}$ is also convergent.

##  3 Fixed Points Results

In this section, we obtain several fixed point results for $T$-Hardy Rogers
and Perov type contractive mappings via $c$-distances in the setup of cone
$b$-metric spaces.

We start with the following result.

######  Theorem 3.1.

Let $(X,d)$ be a complete cone $b$-metric space, $P$ a solid cone and $q$ a
generalized $c$-distance on $X.$ Suppose that $T:X\rightarrow X$ is a one to
one, continuous and sequentially convergent. If the mapping $f:X\rightarrow X$
is a Perov $THR_{1}-$contraction, then $f$ has a unique fixed point
$u^{\ast}\in X.$ And for any $x_{0}\in X,$ the iterative sequence
$\\{f^{n}x_{0}\\}$ converges to the fixed point of $f$. Moreover
$q(Tu^{\ast},Tu^{\ast})=\theta$ provided that $u^{\ast}=fu^{\ast}$.

###### Proof.

Let $x_{0}$ be any element in $X.$ Set $x_{1}=fx_{0},$
$x_{2}=fx_{1}=f^{2}x_{0},..._{\cdot}$ Define a sequence $\\{x_{n}\\}$ in $X$
by $x_{n+1}=f^{n}x_{0}=fx_{n}.$ As the mapping $f$ is a Perov
$THR_{1}-$contraction, we have

$q(Tx_{n},Tx_{n+1})=q(Tfx_{n-1},Tfx_{n}))\leq A(U_{T}(x_{n-1},x_{n})),$ (3.1)

where

$\displaystyle U_{T}(x_{n-1},x_{n})$ $\displaystyle\in$
$\displaystyle\\{q(Tx_{n-1},Tx_{n}),q(Tx_{n-1},Tfx_{n-1}),q(Tx_{n},Tfx_{n})\\}$
$\displaystyle=$ $\displaystyle\\{q(Tx_{n-1},Tx_{n}),q(Tx_{n},Tx_{n+1})\\}.$

If $U_{T}(x_{n-1},x_{n})=q(Tx_{n},Tx_{n+1}),$ then $q(Tx_{n},Tx_{n+1})\leq
A(q(Tx_{n},Tx_{n+1}))$, which by the Remark 2.10 implies that
$q(Tx_{n},Tx_{n+1})=\theta.$ Thus, for all $n\in\mathbb{N},$

$q(Tx_{n},Tx_{n+1})\leq A(q(Tx_{n-1},Tx_{n})).$

Continuing this way, we get $q(Tx_{n},Tx_{n+1})\leq A^{n}(q(Tx_{0},Tx_{1})).$
For any $m,n\in\mathbb{N}$ with $m>n,$ it follows from $(q_{2})$ that

$\displaystyle q(Tx_{n},Tx_{m})$ $\displaystyle\leq
b[q(Tx_{n},Tx_{n+1})+q(Tx_{n+1},Tx_{m})]$ $\displaystyle\leq
bq(Tx_{n},Tx_{n+1})+b[bq(Tx_{n+1},Tx_{n+2})+q(Tx_{n+2},Tx_{m})]$
$\displaystyle\leq\cdot\cdot\cdot$ $\displaystyle\leq
bq(Tx_{n},Tx_{n+1})+b^{2}q(Tx_{n+1},Tx_{n+2})+\cdots+b^{m-n}q(Tx_{m-1},Tx_{m})$
$\displaystyle\leq(bA^{n}+b^{2}A^{n+1}\cdots+b^{m-n}A^{m-1})(q(Tx_{0},Tx_{1}))$
$\displaystyle\leq bA^{n}\left(I-bA\right)^{-1}(q(Tx_{0},Tx_{1})).$ (3.2)

Let $c\gg 0.$ Choose $\delta>0$ such that $c+N_{\delta}(\theta)\subseteq P\,,$
where $N_{\delta}(\theta)=\\{x\in E:\left\|x\right\|<\delta\\}.$ Also, choose
$N_{1}\in\mathbb{N}$ such that $bA^{n}(I-bA)^{-1}(q(Tx_{0},Tx_{1}))\in
N_{\delta}(\theta)$ for all $n>N_{1}$. Thus for all $m>n>N_{1},$ we have

$q(Tx_{n},Tx_{m})\preceq bA^{n}(I-bA)^{-1}(d(Tx_{0},Tx_{1}))\ll c$

which implies that $\\{Tx_{n}\\}$ is a Cauchy sequence in $X$. Since $X$ is
complete, there exists an element point $v\in X$ such that $Tx_{n}\rightarrow
v$ as $n\rightarrow\infty$.

Since $T$ is subsequently convergent, $\\{x_{n}\\}$ has a convergent
subsequence. So, there is $x^{\ast}\in X$ and subsequence $\\{x_{n_{i}}\\}$
such that $x_{n_{i}}\rightarrow x^{\ast}$ as $i\rightarrow\infty$. As $T$ is
continuous, we obtain
$\lim\limits_{n_{i}\rightarrow\infty}Tx_{n_{i}}=Tx^{\ast}$. The uniqueness of
the limit implies that $Tx^{\ast}=v$. By (q${}_{\text{3}}$), we obtain

$q(Tx_{n},Tx^{\ast})\leq bA^{n}(I-bA)^{-1}(q(Tx_{0},Tx_{1})).$ (3.3)

Also

$\begin{array}[]{cc}q(Tx_{n},Tfx^{\ast})&=q(Tfx_{n-1},Tfx^{\ast})\\\ &\leq
A(U_{T}(x_{n-1},x^{\ast})),\end{array}$ (3.4)

where

$\displaystyle U_{T}(x_{n-1},x^{\ast})$ $\displaystyle\in$
$\displaystyle\\{q(Tx_{n-1},Tx^{\ast}),q(Tx_{n-1},Tfx_{n-1}),q(Tx^{\ast},Tfx^{\ast})\\}$
$\displaystyle=$
$\displaystyle\\{q(Tx_{n-1},Tx^{\ast}),q(Tx_{n-1},Tx_{n}),q(Tx^{\ast},Tfx^{\ast})\\}.$

Now, if $U_{T}(x_{n-1},x^{\ast})=q(Tx_{n-1},Tx^{\ast}),$ then we obtain

$\displaystyle q(Tx_{n},Tfx^{\ast})$ $\displaystyle\leq$ $\displaystyle
A\left(q(Tx_{n-1},Tx^{\ast})\right)$ $\displaystyle\leq$ $\displaystyle
bA^{n-1}(I-bA)^{-1}(q(Tx_{0},Tx_{1})).$

Whenever $U_{T}(x_{n-1},x^{\ast})=q(Tx_{n-1},Tx_{n}),$ then we have

$\displaystyle q(Tx_{n},Tfx^{\ast})$ $\displaystyle\leq$ $\displaystyle
A\left(q(Tx_{n-1},Tx_{n})\right)$ $\displaystyle\leq$ $\displaystyle
bA^{n}(I-bA)^{-1}(q(Tx_{0},Tx_{1})).$

Finally if $U_{T}(x_{n-1},x^{\ast})=q(Tx^{\ast},Tfx^{\ast}),$ then we get that

$\displaystyle q(Tx_{n},Tfx^{\ast})$ $\displaystyle\leq$ $\displaystyle
A\left(q(Tx^{\ast},Tfx^{\ast})\right)$ $\displaystyle\leq$ $\displaystyle
bA\left(q(Tx^{\ast},Tx_{n})\right)+bA\left(q(Tx_{n},Tfx^{\ast})\right),$

that is,

$\displaystyle q(Tx_{n},Tfx^{\ast})$ $\displaystyle\leq$ $\displaystyle
bA\left(I-bA\right)^{-1}\left(q(Tx^{\ast},Tx_{n})\right)$ $\displaystyle\leq$
$\displaystyle bA^{n}\left(I-bA\right)^{-1}\left(q(Tx_{0},Tx_{1})\right).$

Then by Lemma 2.6 (i), (3.3) and (3.4), we have $Tx^{\ast}=Tfx^{\ast}$. As the
map $T$ is one to one, so $x^{\ast}=fx^{\ast},$ that is, $x^{\ast}$ is the
fixed point of $f.$

Now by taking $x^{\ast}=fx^{\ast},$ we have

$q(Tx^{\ast},Tx^{\ast})=q(Tfx^{\ast},Tfx^{\ast})\leq
A(U_{T}(x^{\ast},x^{\ast})),$ (3.5)

where

$\displaystyle U_{T}(x^{\ast},x^{\ast})$ $\displaystyle\in$
$\displaystyle\\{q(Tx^{\ast},Tx^{\ast}),q(Tx^{\ast},Tfx^{\ast}),q(Tx^{\ast},Tfx^{\ast})\\}$
$\displaystyle=$
$\displaystyle\\{q(Tx^{\ast},Tx^{\ast}),q(Tx^{\ast},Tx^{\ast}),q(Tx^{\ast},Tx^{\ast})\\}$
$\displaystyle=$ $\displaystyle\\{q(Tx^{\ast},Tx^{\ast})\\}.$

Thus, $q(Tx^{\ast},Tx^{\ast})\leq A(q(Tx^{\ast},Tx^{\ast})).$ By Remark 2.10,
we get $q(Tx^{\ast},Tx^{\ast})=\theta.$

To prove the uniqueness of fixed point, suppose that there exists another
point $y^{\ast}$ in $X$ such that $y^{\ast}=fy^{\ast}$. Then we have

$q(Tx^{\ast},Ty^{\ast})=q(Tfx^{\ast},Tfy^{\ast})\leq
A(U_{T}(x^{\ast},y^{\ast})),$ (3.6)

where

$\displaystyle U_{T}(x^{\ast},y^{\ast})$ $\displaystyle\in$
$\displaystyle\\{q(Tx^{\ast},Ty^{\ast}),q(Tx^{\ast},Tfx^{\ast}),q(Ty^{\ast},Tfy^{\ast})\\}$
$\displaystyle=$
$\displaystyle\\{q(Tx^{\ast},Ty^{\ast}),q(Tx^{\ast},Tx^{\ast}),q(Ty^{\ast},Ty^{\ast})\\}$
$\displaystyle=$ $\displaystyle\\{q(Tx^{\ast},Ty^{\ast}),\theta\\}.$

Now if $U_{T}(x^{\ast},y^{\ast})=q(Tx^{\ast},Ty^{\ast})$, then we have

$q(Tx^{\ast},Ty^{\ast})\leq A(q(Tx^{\ast},Ty^{\ast})),$

which by the Remark 2.10 implies $q(Tx^{\ast},Ty^{\ast})=\theta$. Thus, by
Lemma 2.6 (i), we obtain $Tx^{\ast}=Ty^{\ast}.$

In case $U_{T}(x^{\ast},y^{\ast})=\theta$, we have $q(Tx^{\ast},Ty^{\ast})\leq
A(\theta)=\theta,$ that is, $q(Tx^{\ast},Ty^{\ast})=\theta$. Again, by the
Lemma 2.6 (i), it follows that $Tx^{\ast}=Ty^{\ast}.$ Since $T$ is one to one,
so $x^{\ast}=y^{\ast}$. Thus $f$ has a unique fixed point. ∎

The following non trivial example shows the validity of Theorem 3.1.

######  Example 3.2.

Let $X=[0,1],$ $E=\mathbb{R}^{2}\ $and $P=\\{\left(x,y\right)\in E:x,y\geq
0\\}\subseteq E.$ Define $d:X\times X\rightarrow E$ as given in Example 2.2.
Then $(X,d)$ is a complete cone $b$-metric space with $b\geq 2^{p-1}\geq 1$
along with $P$ a solid cone.

Define
$q\left(x,y\right)=\left(\psi\left(y\right),\eta\psi\left(y\right)\right),$
where $\eta\geq 1$ and $\psi:X\rightarrow\mathbb{R}_{+}$ is of the form
$\psi\left(t\right)=at^{p}$ for some $p>1,$ where $a\in\mathbb{R}_{+}.$ Then
$q$ is a generalized $c$-distance on $X$ as shown in Example 2.5.

Define a selfmap $T:X\rightarrow X$ by $T\left(x\right)=\lambda x$ for some
$\lambda\in(0,1].$ Clearly, $T$ is a one to one, continuous and sequentially
convergent map.

Define $f:X\rightarrow X$ by

$f(x)=\left\\{\begin{array}[]{ll}\beta x^{2}\text{\ }&\text{if }x\neq
1\text{,}\\\ &\\\ \beta x&\text{if }x=1,\end{array}\right.$

where $\beta\in\left(0,\gamma\right).$ A linear bounded operator
$A:E\rightarrow E$ is taken as $A=\left[\begin{array}[]{cc}a_{11}&a_{12}\\\
a_{21}&a_{22}\end{array}\right]$ with $a_{ij}\geq 0$ for
$i,j\in\left\\{1,2\right\\}$. We take $\beta^{p}\leq a_{11}+\eta a_{12}$ and
$\eta\beta^{p}\leq a_{21}+\eta a_{22}.$

Set
$\left\|\mathbf{x}\right\|=\max\\{\left|x_{1}\right|,\left|x_{2}\right|\\}$
for $\mathbf{x}=\left[\begin{array}[]{c}x_{1}\\\ x_{2}\end{array}\right],$
$x_{i}\in\mathbb{R},$ $i=1,2,$ and

$\left\|A\mathbf{x}\right\|=\max\\{\left|a_{11}x_{1}+a_{12}x_{2}\right|,\left|a_{21}x_{1}+a_{22}x_{2}\right|\\}.$

Then,

$\displaystyle\left\|A\right\|$
$\displaystyle=\sup\limits_{\left\|\mathbf{x}\right\|=1}\left\|A\mathbf{x}\right\|=\sup\limits_{\left\|\mathbf{x}\right\|=1}\left[\max\\{\left|a_{11}x_{1}+a_{12}x_{2}\right|,\left|a_{21}x_{1}+a_{22}x_{2}\right|\\}\right]$
$\displaystyle=\max\\{\left|a_{11}+a_{12}\right|,\left|a_{21}+a_{22}\right|\\}.$

Now we are to show that

$q(Tf\left(x_{1}\right),Tf\left(x_{2}\right))\leq A(U_{T}(x_{1},x_{2})),$

holds, where

$U_{T}(x_{1},x_{2})\in\\{q(T\left(x_{1}\right),T\left(x_{2}\right)),q(T\left(x_{1}\right),Tf\left(x_{1}\right)),q(T\left(x_{2}\right),Tf\left(x_{2}\right))\\}.$

Now for all $x,y\in X$ with $y\neq 1$, we obtain that

$\displaystyle q(Tf\left(x\right),Tf\left(y\right))$
$\displaystyle=\left[\begin{array}[]{c}\psi\left(Tf\left(y\right)\right)\\\
\eta\psi\left(Tf\left(y\right)\right)\end{array}\right]^{T}=\left[\begin{array}[]{c}a\left(Tf\left(y\right)\right)^{p}\\\
\eta a\left(Tf\left(y\right)\right)^{p}\end{array}\right]^{T}$
$\displaystyle=\left[\begin{array}[]{c}a\left(\lambda\beta y^{2}\right)^{p}\\\
\eta a\left(\lambda\beta
y^{2}\right)^{p}\end{array}\right]^{T}\leq\left[\begin{array}[]{c}(a_{11}+\eta
a_{12})\left[a\left(\lambda y\right)^{p}\right]\\\ (a_{21}+\eta
a_{22})\left[a\left(\lambda y\right)^{p}\right]\end{array}\right]^{T}$
$\displaystyle=\left[\begin{array}[]{cc}a_{11}&a_{12}\\\
a_{21}&a_{22}\end{array}\right]\left[\begin{array}[]{c}a\left(\lambda
y\right)^{p}\\\ \eta a\left(\lambda y\right)^{p}\end{array}\right]^{T}$
$\displaystyle=\left[\begin{array}[]{cc}a_{11}&a_{12}\\\
a_{21}&a_{22}\end{array}\right]\left[\begin{array}[]{c}\psi\left(T\left(y\right)\right)\\\
\eta\psi\left(T\left(y\right)\right)\end{array}\right]^{T}=A(q(T\left(x\right),T\left(y\right))),$

where

$q\left(T\left(x\right),T\left(y\right)\right)\in
U_{T}\left(x,y\right)=\\{q\left(T\left(x\right),T\left(y\right)\right),q(T\left(x\right),Tf\left(x\right)),q(T\left(y\right),Tf\left(y\right))\\}.$

If $y=1$, then we have

$\displaystyle q(Tf\left(x\right),Tf\left(y\right))$
$\displaystyle=\left[\begin{array}[]{c}\psi\left(Tf\left(y\right)\right)\\\
\eta\psi\left(Tf\left(y\right)\right)\end{array}\right]^{T}=\left[\begin{array}[]{c}a\left(Tf\left(y\right)\right)^{p}\\\
\eta a\left(Tf\left(y\right)\right)^{p}\end{array}\right]^{T}$
$\displaystyle=\left[\begin{array}[]{c}a\left(\lambda\beta\right)^{p}\\\ \eta
a\left(\lambda\beta\right)^{p}\end{array}\right]^{T}\leq\left[\begin{array}[]{c}(a_{11}+\eta
a_{12})\left[a\left(\lambda\right)^{p}\right]\\\ (a_{21}+\eta
a_{22})\left[a\left(\lambda\right)^{p}\right]\end{array}\right]^{T}$
$\displaystyle=\left[\begin{array}[]{cc}a_{11}&a_{12}\\\
a_{21}&a_{22}\end{array}\right]\left[\begin{array}[]{c}a\left(\lambda\right)^{p}\\\
\eta
a\left(\lambda\right)^{p}\end{array}\right]^{T}=\left[\begin{array}[]{cc}a_{11}&a_{12}\\\
a_{21}&a_{22}\end{array}\right]\left[\begin{array}[]{c}\psi\left(T\left(y\right)\right)\\\
\eta\psi\left(T\left(y\right)\right)\end{array}\right]^{T}$
$\displaystyle=A(q(T\left(x\right),T\left(y\right))),$

where

$q\left(T\left(x\right),T\left(y\right)\right)\in
U_{T}\left(x,y\right)=\\{q\left(T\left(x\right),T\left(y\right)\right),q(T\left(x\right),Tf\left(x\right)),q(T\left(y\right),Tf\left(y\right))\\}.$

So, the mapping $f:X\rightarrow X$ is a Perov $THR_{1}-$contraction. All
conditions of Theorem3.1 are satisfied. Moreover, $u^{\ast}=0$ is a fixed
point of $f$ and $q(u^{\ast},u^{\ast})=0$.

The following result is the direct consequence of Theorem 3.1.

######  Corollary 3.3.

Let $(X,d)$ be a complete $b$-cone metric space, $q$ a generalized
$c$-distance on $X$ and $P$ a solid cone. If there exists a linear bounded
operator $A:E\rightarrow E$ with $r(bA)<1$ and $A\left(P\right)\subset P$ such
that an inequalities

$q(fx_{1},fx_{2})\leq A(U(x_{1},x_{2}))$ (3.7)

hold for all $x_{1},x_{2}\in X$, where
$U(x_{1},x_{2})\in\\{q(x_{1},x_{2}),q(x_{1},fx_{1}),q(x_{2},fx_{2})\\}.$ Then
$f$ has a unique fixed point $u^{\ast}\in X.$ Moreover, for any $x_{0}\in X,$
iterative sequence $\\{f^{n}x_{0}\\}$ converges to the fixed point
$u^{\ast}\in X$ and $q(u^{\ast},u^{\ast})=\theta$.

###### Proof.

Taking $T$ as an identity map in the Theorem 3.1, the result follows. ∎

######  Example 3.4.

Let $X=[0,1],$ $E=X\times X\ $and $P=\\{\left(x,y\right)\in E:x,y\geq 0\\}=E.$
Define $d:X\times X\rightarrow E$ by
$d(x,y)=\left(|x-y|^{p},\alpha|x-y|^{p}\right)$, where $\alpha\geq 1$ and
$p>1.\ $Then $(X,d)$ is a cone $b$-metric space. Let $q:X\times X\rightarrow
E$ be given by

$q(x,y)=\left(y^{p},\alpha y^{p}\right).$

Then $q$ is a generalized $c$-distance. We fix $p=2.$ Define $f:X\rightarrow
X$ by

$f(x)=\left\\{\begin{array}[]{ll}\dfrac{x^{2}}{10}\text{\ }&\text{if }x\neq
1\text{,}\\\ &\\\ \dfrac{x}{10}&\text{if }x=1.\end{array}\right.$

Define a linear bounded operator $A:E\rightarrow E$ by
$A=\left[\begin{array}[]{cc}\frac{2}{10}&\frac{1}{10}\\\
\frac{3}{10}&\frac{1}{10}\end{array}\right]$. Set
$\left\|\mathbf{x}\right\|=\max\\{\left|x_{1}\right|,\left|x_{2}\right|\\}$
for $\mathbf{x}=\left[\begin{array}[]{c}x_{1}\\\ x_{2}\end{array}\right],$
$x_{i}\in\mathbb{R},$ $i=1,2.$ For an arbitrary $\mathbf{x}\in E,$ we have

$\displaystyle\left\|A\mathbf{x}\right\|$ $\displaystyle=$
$\displaystyle\max\left\\{\left|\frac{2}{10}x_{1}+\frac{1}{10}x_{2}\right|,\left|\frac{3}{10}x_{1}+\frac{1}{10}x_{2}\right|\right\\}$
$\displaystyle\leq$
$\displaystyle\max\left\\{\frac{3}{10}\max\left\\{\left|x_{1}\right|,\left|x_{2}\right|\right\\},\frac{4}{10}\left|x_{2}\right|\right\\}=\frac{3}{10}\left\|\mathbf{x}\right\|.$

Thus, $\left\|A\right\|\leq\frac{3}{10}.$ If
$\mathbf{x}=\left[\begin{array}[]{c}1\\\ 1\end{array}\right],$ then
$\left\|\mathbf{x}\right\|=1$ and

$\left\|A\mathbf{x}\right\|=\max\\{\left|\frac{2}{10}+\frac{1}{10}\right|,\left|\frac{3}{10}+\frac{1}{10}\right|\\}=\max\\{\frac{3}{10},\frac{4}{10}\\}=\frac{3}{10}.$

Hence $\left\|A\right\|=\frac{3}{10}=0.3.$ Clearly
$\left\|bA\right\|=\left\|2A\right\|=0.6<1$ and $A\left(P\right)\subset P.$

Now for all $x,y\in X$ with $y\neq 1$, we obtain that

$\displaystyle q(fx,fy)$
$\displaystyle=\left[\begin{array}[]{c}\frac{y^{4}}{100}\\\ \frac{\alpha
y^{4}}{100}\end{array}\right]^{T}\leq\left[\begin{array}[]{c}(\frac{2}{10}+\frac{\alpha}{10})y^{2}\\\
(\frac{3}{10}+\frac{\alpha}{10})y^{2}\end{array}\right]^{T}$
$\displaystyle=\left[\begin{array}[]{cc}\frac{2}{10}&\frac{1}{10}\\\
\frac{3}{10}&\frac{1}{10}\end{array}\right]\left[\begin{array}[]{c}y^{2}\\\
\alpha y^{2}\end{array}\right]^{T}=A(q(x,y)),$

where $q\left(x,y\right)\in
U\left(x,y\right)=\\{q\left(x,y\right),q(x,fx),q(y,fy)\\}.$ If $y=1$, then

$\displaystyle q(fx,fy)$
$\displaystyle=\left[\begin{array}[]{c}\frac{y^{2}}{100}\\\ \frac{\alpha
y^{2}}{100}\end{array}\right]^{T}\leq\left[\begin{array}[]{c}(\frac{2}{10}+\frac{\alpha}{10})y^{2}\\\
(\frac{3}{10}+\frac{\alpha}{10})y^{2}\end{array}\right]^{T}$
$\displaystyle=\left[\begin{array}[]{cc}\frac{2}{10}&\frac{1}{10}\\\
\frac{3}{10}&\frac{1}{10}\end{array}\right]\left[\begin{array}[]{c}y^{2}\\\
\alpha y^{2}\end{array}\right]^{T}=A(q(x,y)),$

where

$q\left(x,y\right)\in
U\left(x,y\right)=\\{q\left(x,y\right),q(x,fx),q(y,fy)\\}.$

Hence, all the conditions of Corollary 3.3 are satisfied. Moreover,
$u^{\ast}=0$ is a fixed point of $f$ and $q(u^{\ast},u^{\ast})=0$.

The following result is the direct consequence of Theorem 3.1 which extends
and generalizes [14, Theorem 1].

######  Corollary 3.5.

Let $(X,d)$ be a complete cone metric space, $q$ a generalized $c$-distance on
$X$ and $P$ a solid cone. Suppose that $T:X\rightarrow X$ is a one to one,
continuous function and sequentially convergent and $f:X\rightarrow X$ . If
there exists a linear bounded operator $A:E\rightarrow E$ with $r(bA)<1$ and
$A\left(P\right)\subset P$ such that

$q(Tfx_{1},Tfx_{2})\leq A(q(x_{1},x_{2})),$ (3.8)

hold for all $x_{1},x_{2}\in X$, then $f$ has a unique fixed point
$u^{\ast}\in X.$ Moreover, for any $x_{0}\in X,$ the iterative sequence
$\\{f^{n}x_{0}\\}$ converges to the fixed point $u^{\ast}\in X$ and
$q(Tu^{\ast},Tu^{\ast})=\theta$.

######  Theorem 3.6.

Let $(X,d)$ be a complete cone $b$-metric space, $q$ a generalized
$c$-distance on $X\ $and $P$ a solid cone. Suppose that $T:X\rightarrow X$ is
a one to one, continuous function and sequentially convergent. If the mapping
$f:X\rightarrow X$ is a Perov $THR_{2}-$contraction, then $f$ has a unique
fixed point $u^{\ast}\in X.$ Also, for any $x_{0}\in X,$ the iterative
sequence $\\{f^{n}x_{0}\\}$ converges to the fixed point of $f$. Moreover
$q(Tu^{\ast},Tu^{\ast})=\theta$ provided that $u^{\ast}=fu^{\ast}.$

###### Proof.

Let $x_{0}$ be any point in $X.$ Define a sequence $\\{x_{n}\\}$ in $X$ by
$x_{1}=fx_{0},$ $x_{2}=fx_{1}=f^{2}x_{0},$ $...,$
$x_{n+1}=fx_{n-1}=f^{n}x_{0}.$ Then for $n\geq 0,$ we have

$\displaystyle q(Tx_{n},Tx_{n+1})$ $\displaystyle=$ $\displaystyle
q(Tfx_{n-1},Tfx_{n})$ $\displaystyle\leq$ $\displaystyle
A_{1}(q(Tx_{n-1},Tx_{n}))+A_{2}(q(Tx_{n-1},Tfx_{n-1}))$
$\displaystyle+A_{3}(q(Tx_{n},Tfx_{n}))$ $\displaystyle=$ $\displaystyle
A_{1}(q(Tx_{n-1},Tx_{n}))+A_{2}(q(Tx_{n-1},Tx_{n}))$
$\displaystyle+A_{3}(q(Tx_{n},Tx_{n+1}))$ $\displaystyle=$
$\displaystyle(A_{1}+A_{2})(q(Tx_{n-1},Tx_{n}))+A_{3}(q(Tx_{n},Tx_{n+1})),$

which implies that

$q(Tx_{n},Tx_{n+1})\leq(A_{1}+A_{2})(I-A_{3})^{-1}(q(Tx_{n-1},Tx_{n})).$

Thus, for all $n\in\mathbb{N\cup\\{}0\\},$

$q(Tx_{n},Tx_{n+1})\leq A(q(Tx_{n-1},Tx_{n})),$

where $A=(A_{1}+A_{2})(I-A_{3})^{-1}$ with
$(b||A_{1}||+||A_{2}||)(I-||A_{3}||)^{-1}<1.$ Continuing this way, we have for
$n\in\mathbb{N\cup\\{}0\\},$

$q(Tx_{n},Tx_{n+1})\leq A^{n}(q(Tx_{0},Tx_{1})).$

For $m,n\in\mathbb{N}\mathbb{\cup\\{}0\\}$ with $m>n,$ it follows from (q2)
that

$\displaystyle q(Tx_{n},Tx_{m})$ $\displaystyle\leq
b[q(Tx_{n},Tx_{n+1})+q(Tx_{n+1},Tx_{m})]$ $\displaystyle\leq
bq(Tx_{n},Tx_{n+1})+b[bq(Tx_{n+1},Tx_{n+2})+q(Tx_{n+2},Tx_{m})]$
$\displaystyle\leq\cdot\cdot\cdot$ $\displaystyle\leq
bq(Tx_{n},Tx_{n+1})+b^{2}q(Tx_{n+1},Tx_{n+2})+\cdots+b^{m-n}q(Tx_{m-1},Tx_{m})$
$\displaystyle\leq(bA^{n}+b^{2}A^{n+1}\cdots+b^{m-n}A^{m-1})(q(Tx_{0},Tx_{1}))$
$\displaystyle\leq bA^{n}\left(I-bA\right)^{-1}(q(Tx_{0},Tx_{1})).$

Let $c\gg 0.$ Choose $\delta>0$ such that $c+N_{\delta}(\theta)\subseteq P\,,$
where $N_{\delta}(\theta)=\\{x\in E:\left\|x\right\|<\delta\\}.$ Also, choose
$N_{1}\in\mathbb{N}$ such that $bA^{n}(I-bA)^{-1}(q(Tx_{0},Tx_{1}))\in
N_{\delta}(\theta)\ $ for all $n>N_{1}$. Thus for all $m>n>N_{1},$ we have

$q(Tx_{n},Tx_{m})\preceq bA^{n}(I-bA)^{-1}(d(Tx_{0},Tx_{1}))\ll c$

which implies that $\\{Tx_{n}\\}$ is a Cauchy sequence in $X$. Since $X$ is
complete, there exists an element $v\in X$ such that $Tx_{n}\rightarrow v$ as
$n\rightarrow\infty$.

Since $T$ is subsequently convergent, $\\{x_{n}\\}$has a convergent
subsequence. So, there are $x^{\ast}\in X$ and a subsequence $\\{x_{n_{i}}\\}$
such that $x_{n_{i}}\rightarrow x^{\ast}$ as $i\rightarrow\infty$. As $T$ is
continuous, we obtain that
$\lim\limits_{n_{i}\rightarrow\infty}Tx_{n_{i}}=Tx^{\ast}$. The uniqueness of
the limit implies that $Tx^{\ast}=v$. By (q3), we get that

$q(Tx_{n},Tx^{\ast})\leq bA^{n}(I-bA)^{-1}(q(Tx_{0},Tx_{1})),$ (3.9)

where $A=A_{1}+A_{2}+A_{3}$ with
$b\left(\left\|A_{1}\right\|+\left\|A_{2}\right\|+\left\|A_{3}\right\|\right)<1.$
Also,

$\begin{array}[]{l}q(Tx_{n},Tfx^{\ast})=q(Tfx_{n-1},Tfx^{\ast})\\\ \quad\leq
A_{1}\left(q(Tx_{n-1},Tx^{\ast})\right)+A_{2}\left(q(Tx_{n-1},Tfx_{n-1})\right)+A_{3}\left(q(Tx^{\ast},Tfx^{\ast})\right)\\\
\quad=A_{1}\left(q(Tx_{n-1},Tx^{\ast})\right)+A_{2}\left(q(Tx_{n-1},Tx_{n})\right)+A_{3}\left(q(Tx^{\ast},Tfx^{\ast})\right)\\\
\quad\leq
A^{n-1}(I-A)^{-1}(q(Tx_{0},Tx_{1}))+A^{n}(I-A)^{-1}(q(Tx_{0},Tx_{1}))\\\
\quad\quad\quad\quad\quad\quad\quad+\left(I-A\right)^{-1}A^{n}\left(q(Tx_{0},Tx_{1})\right),\end{array}$
(3.10)

where $A=A_{1}+A_{2}+A_{3}$ with
$b\left(\left\|A_{1}\right\|+\left\|A_{2}\right\|+\left\|A_{3}\right\|\right)<1.$
By Lemma 2.6 (i), and (3.9) and (3.10), we get that $Tx^{\ast}=Tfx^{\ast}$. As
the map $T$ is one to one, we obtain that $x^{\ast}=fx^{\ast}.$ Now by taking
$x^{\ast}=fx^{\ast},$ we have

$\displaystyle q(Tx^{\ast},Tx^{\ast})$ $\displaystyle=$ $\displaystyle
q(Tfx^{\ast},Tfx^{\ast})$ $\displaystyle\leq$ $\displaystyle
A_{1}\left(q(Tx^{\ast},Tx^{\ast})\right)+A_{2}\left(q(Tx^{\ast},Tfx^{\ast})\right)+A_{3}\left(q(Tx^{\ast},Tfx^{\ast})\right)$
$\displaystyle=$ $\displaystyle
A_{1}\left(q(Tx^{\ast},Tx^{\ast})\right)+A_{2}\left(q(Tx^{\ast},Tx^{\ast})\right)+A_{3}\left(q(Tx^{\ast},Tx^{\ast})\right)$
$\displaystyle=$
$\displaystyle\left(A_{1}+A_{2}+A_{3}\right)\left(q(Tx^{\ast},Tx^{\ast})\right)$
$\displaystyle=$ $\displaystyle A(q(Tx^{\ast},Tx^{\ast})),$

where $A=A_{1}+A_{2}+A_{3}$ with
$b\left(\left\|A_{1}\right\|+\left\|A_{2}\right\|+\left\|A_{3}\right\|\right)<1.$
By Remark 2.10, we obtain that $q(Tx^{\ast},Tx^{\ast})=\theta.$

To prove the uniqueness of fixed point, suppose that there exists another
point $y^{\ast}$ in $X$ such that $y^{\ast}=fy^{\ast}$. Then,

$\displaystyle q(Tx^{\ast},Ty^{\ast})$ $\displaystyle=$ $\displaystyle
q(Tfx^{\ast},Tfy^{\ast})$ $\displaystyle\leq$ $\displaystyle
A_{1}(q(Tx^{\ast},Ty^{\ast}))+A_{2}(q(Tx^{\ast},Tfx^{\ast}))+A_{3}(q(Ty^{\ast},Tfy^{\ast}))$
$\displaystyle=$ $\displaystyle
A_{1}(q(Tx^{\ast},Ty^{\ast}))+A_{2}(q(Tx^{\ast},Tx^{\ast}))+A_{3}(q(Ty^{\ast},Ty^{\ast}))$
$\displaystyle=$ $\displaystyle
A_{1}(q(Tx^{\ast},Ty^{\ast}))+A_{2}(\theta)+A_{3}(\theta)$ $\displaystyle=$
$\displaystyle A_{1}(q(Tx^{\ast},Ty^{\ast})).$

As $\left\|A_{1}\right\|<1$ and $A_{1}\left(P\right)\subset A_{1},\,$ by the
Remark 2.10, we obtain $q(Tx^{\ast},Ty^{\ast})=\theta$. Also, by Lemma 2.6(i),
we get $Tx^{\ast}=Ty^{\ast}.$

Since $T$ is one to one, so $x^{\ast}=y^{\ast}$. Thus $f$ has a unique fixed
point. ∎

The following result is the direct consequence of Theorem 3.6 which extends
and generalizes [14, Corollary 1].

######  Corollary 3.7.

Let $(X,d)$ be a complete cone $b$-metric space, $q$ a $c$-distance on $X$ and
$P$ a solid cone. If there exists linear bounded operators
$A_{1},A_{2},A_{3}:E\rightarrow E$ with $A_{k}\left(P\right)\subset P$ for
$k=1,2,3$, $(I-A_{3})^{-1}\in\mathcal{B}(E)$ and
$r[b(I-A_{3})^{-1}(A_{1}+A_{2})]<1$ such that

$q(fx_{1},fx_{2})\leq
A_{1}\left(q(x_{1},x_{2})\right)+A_{2}\left(q(x_{1},fx_{1})\right)+A_{3}\left(q(x_{2},fx_{2})\right)$
(3.11)

holds for all $x_{1},x_{2}\in X$. Then $f$ has a unique fixed point
$u^{\ast}\in X.$ Also, for any $x_{0}\in X,$ iterative sequence
$\\{f^{n}x_{0}\\}$ converges to the fixed point $u^{\ast}\in X$. Moreover
$q(u^{\ast},u^{\ast})=\theta$.

###### Proof.

Taking $T$ as an identity map in Theorem 3.6, the result follows. ∎

######  Example 3.8.

Let $E=\mathbb{R}^{2},$ $X=[0,1]$ and $P=\\{(x,y)\in E:x,y\geq 0\\}.$ Let the
the cone $b$-metric $d:X\times X\rightarrow E$ be given by
$d(x,y)=(|x-y|^{p},\alpha|x-y|^{p})$ where $\alpha\geq 1$. Define $q:X\times
X\rightarrow E$ by $q(x,y)=(y^{p},\alpha y^{p}).$ Define $f:X\rightarrow X$ by

$f(x)=\left\\{\begin{array}[]{cc}\frac{x^{2}}{4}&\text{if }x\neq 1,\\\ &\\\
\frac{1}{4}&\text{if }x=1.\end{array}\right.$

Define the bounded linear operators $A_{1},A_{2},A_{3}:E\rightarrow E$ by
$A_{1}=\left[\begin{array}[]{cc}\frac{1}{5}&\frac{1}{6}\\\
0&\frac{3}{13}\end{array}\right],$
$A_{2}=\left[\begin{array}[]{cc}\frac{1}{7}&0\\\
0&\frac{1}{8}\end{array}\right],$
$A_{3}=\left[\begin{array}[]{cc}\frac{1}{6}&0\\\
0&\frac{2}{9}\end{array}\right]$. Set
$\left\|\mathbf{x}\right\|=\max\\{\left|x_{1}\right|,\left|x_{2}\right|\\}$
for $\mathbf{x}=\left[\begin{array}[]{c}x_{1}\\\ x_{2}\end{array}\right],$
$x_{i}\in\mathbb{R},$ $i=1,2.$ For an arbitrary $\mathbf{x}\in E,$ we have

$\displaystyle\left\|A_{1}\mathbf{x}\right\|$ $\displaystyle=$
$\displaystyle\max\left\\{\left|\frac{1}{5}x_{1}+\frac{1}{6}x_{2}\right|,\left|0x_{1}+\frac{3}{13}x_{2}\right|\right\\}$
$\displaystyle\leq$
$\displaystyle\max\left\\{\frac{11}{30}\max\left\\{\left|x_{1}\right|,\left|x_{2}\right|\right\\},\frac{3}{13}\left|x_{2}\right|\right\\}=\frac{11}{13}\left\|\mathbf{x}\right\|.$

Thus, $\left\|A_{1}\right\|\leq\frac{11}{13}.$ If
$\mathbf{x}=\left[\begin{array}[]{c}1\\\ 1\end{array}\right],$ then
$\left\|\mathbf{x}\right\|=1,$ and

$\left\|A_{1}\mathbf{x}\right\|=\max\\{\left|\frac{1}{5}+\frac{1}{6}\right|,\left|0+\frac{3}{13}\right|\\}=\max\\{\frac{11}{30},\frac{3}{13}\\}=\frac{11}{30}.$

Hence $\left\|A_{1}\right\|=\frac{11}{30}.$ Similarly,
$\left\|A_{2}\right\|=\frac{1}{7}$ and $\left\|A_{3}\right\|=\frac{2}{9}.$
Thus,

$\left\|A_{1}\right\|+\left\|A_{2}\right\|+\left\|A_{3}\right\|=0.7317<1.$

Then, for all $x,y\in X$ and $y\neq 1,$ we have

$\displaystyle q(fx,fy)$ $\displaystyle=$ $\displaystyle
q(\frac{x^{2}}{4},\frac{y^{2}}{4})=\left[\begin{array}[]{c}\frac{y^{2p}}{16}\\\
\frac{\alpha
y^{2p}}{16}\end{array}\right]^{T}\leq\left[\begin{array}[]{c}\frac{1}{5}y^{2}+\frac{1}{6}\alpha
y^{2}+\frac{1}{112}x^{4}+\frac{1}{96}y^{4}\\\ \frac{3}{13}\alpha
y^{2}+\frac{1}{128}\alpha x^{4}+\frac{1}{72}\alpha
y^{4}\end{array}\right]^{T}$ $\displaystyle=$
$\displaystyle\left[\begin{array}[]{cc}\frac{1}{5}&\frac{1}{6}\\\
0&\frac{3}{13}\end{array}\right]\left[\begin{array}[]{c}y^{2}\\\ \alpha
y^{2}\end{array}\right]+\left[\begin{array}[]{cc}\frac{1}{7}&0\\\
0&\frac{1}{8}\end{array}\right]\left[\begin{array}[]{c}\frac{x^{4}}{16}\\\
\frac{\alpha
x^{4}}{16}\end{array}\right]+\left[\begin{array}[]{cc}\frac{1}{6}&0\\\
0&\frac{2}{9}\end{array}\right]\left[\begin{array}[]{c}\frac{y^{4}}{16}\\\
\frac{\alpha y^{4}}{16}\end{array}\right]$ $\displaystyle\leq$ $\displaystyle
A_{1}(q(x,y))+A_{2}(q(x,fx))+A_{3}(q(y,fy)).$

Also for $x,y\in X$ with $y=1,$ we obtain

$\displaystyle q(fx,fy)$ $\displaystyle=$ $\displaystyle
q(\frac{1}{4},\frac{1}{4})=\left[\begin{array}[]{c}\frac{1}{16}\\\
\frac{\alpha}{16}\end{array}\right]^{T}\leq\left[\begin{array}[]{c}\frac{1}{5}+\frac{1}{6}\alpha+\frac{1}{112}x^{4}+\frac{1}{96}\\\
\frac{3\alpha}{13}+\frac{\alpha}{128}x^{4}+\frac{1}{72}\alpha\end{array}\right]^{T}$
$\displaystyle=$
$\displaystyle\left[\begin{array}[]{cc}\frac{1}{5}&\frac{1}{6}\\\
0&\frac{3}{13}\end{array}\right]\left[\begin{array}[]{c}1\\\
\alpha\end{array}\right]+\left[\begin{array}[]{cc}\frac{1}{7}&0\\\
0&\frac{1}{8}\end{array}\right]\left[\begin{array}[]{c}\frac{x^{4}}{16}\\\
\frac{\alpha
x^{4}}{16}\end{array}\right]+\left[\begin{array}[]{cc}\frac{1}{6}&0\\\
0&\frac{2}{9}\end{array}\right]\left[\begin{array}[]{c}\frac{1}{16}\\\
\frac{\alpha}{16}\end{array}\right]$ $\displaystyle\leq$ $\displaystyle
A_{1}(q(x,y))+A_{2}(q(x,fx))+A_{3}(q(y,fy)).$

Thus mapping $f$ satisfies the condition (3.11). Hence all the conditions of
Corollary 3.7 are satisfied. Moreover, $u^{\ast}=0$ is a fixed point of $f$,
and $q(u^{\ast},u^{\ast})=0.$

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