Rectangular cone b-metric spaces over Banach algebra and contraction principle

Abstract

Rectangular cone b-metric spaces over a Banach algebra are introduced as a generalization of metric space and many of its generalizations. Some fixed point theorems are proved in this space and proper examples are provided to establish the validity and superiority of our results. An application to solution of linear equations is given which illustrates the proper application of the results in spaces over Banach algebra.

1 Introduction

Liu and Xu in [1] reported the concept of cone metric space over Banach algebra (in short CMS-BA) and proved contraction principles in such space. They replaced the usual real contraction constant with a vector constant and scalar multiplication with vector multiplication in their results and also furnished proper examples to show that their results were different from those in cone metric space and metric space. The concept defined by Liu and Xu [1] was further generalized by Huang and Radenovic [2] by the introduction of a cone b-metric space over a Banach algebra (in short CbMS-BA).

In this paper we have introduced the concept of a rectangular cone b-metric space over a Banach algebra (in short RCbMS-BA) and proved the Banach contraction principle and weak Kannan contraction principle in this space. Simple examples are given illustrating the validity and superiority of our results. We have also given an application of our result to a solution of a system of linear equations.

2 Preliminaries

A linear space $\mathcal{A}$ over ${K} \in\{\mathbb{R},\mathbb {C}\}$ is an algebra if for each ordered pair of elements $x,y \in\mathcal{A}$, a unique product $xy\in\mathcal{A}$ is defined such that for all $x,y,z \in\mathcal{A}$ and scalar α:

• (i) $(xy)z = x(yz)$;

• (iia) $x(y + z) = xy + xz$;

• (iib) $(x + y)z = xz + yz$;

• (iii) $\alpha(xy) = (\alpha x)y = x(\alpha y)$.

A Banach algebra is a Banach space $\mathcal{A}$ over ${K} \in\{ \mathbb{R},\mathbb{C}\}$ such that, for all $x,y \in\mathcal{A,}$ $\|xy\|\leq\|x\|\|y\|$.

For a given cone ${P}\subset\mathcal{A}$ and $x,y \in\mathcal {A}$, we say that $x\preceq y$ if and only if $y-x\in{P}$. Note that ⪯ is a partial order relation defined on $\mathcal{A}$. For more details on the basic concepts of a Banach algebra, solid cone, unit element e, zero element θ, invertible elements in Banach algebra etc. the reader may refer to [1–3].

For basic properties of Banach algebra and spectral radius refer to [1, 4].

In what follows $\mathcal{A}$ will always denote a Banach algebra, P a solid cone in $\mathcal{A}$ and e the unit element of $\mathcal{A}$.

Definition 2.1

[5]

Let P be a solid cone in a Banach space E. A sequence $\{u_{n}\}\subset P$ is said to be a c-sequence if for each $c\gg\theta$ there exists a natural number N such that $u_{n}\ll c$ for all $n>N$.

Remark 2.2

For more on c-sequences see [2, 3, 5, 6].

Lemma 2.3

[5]

Let E be a Banach space.

• (i) If $a,b,c\in E$ and $a\preceq b\ll c$, then $a\ll c$.

• (ii) If $\theta\preceq a\ll c$ for each $c\gg\theta$, then $a=\theta$.

3 Main results

In this section first we introduce the definition of a rectangular cone b-metric space over a Banach algebra (in short RCbMS-BA) and furnish examples to show that this concept is more general than that of CMS-BA and CbMS-BA. We then define convergence and a Cauchy sequence in a RCbMS-BA and then prove fixed point results in this space.

Definition 3.1

Let χ be a nonempty set and $d_{rcb}:\chi\times\chi\rightarrow \mathcal{A}$ be such that for all $x,y,u,v\in\chi$, $x \neq u$, $v\neq y$:

• (RCbM1) $\theta\preceq d_{rcb}(x,y)$ and $d_{rcb}(x,y)=\theta$ if and only if $x=y$;

• (RCbM2) $d_{rcb}(x,y)=d_{rcb}(y,x)$;

• (RCbM3) there exist $s\in{P}$, $e\preceq s$ such that $d_{rcb}(x,y)\preceq s[d_{rcb}(x,u)+d_{rcb}(u,v)+d_{rcb}(v,y)]$.

Then $d_{rcb}$ is called a rectangular cone b-metric on χ and $(\chi,d_{rcb})$ is called a rectangular cone b-metric space over a Banach algebra (in short RCbMS-BA) with coefficient s. If $s=e$ we say that $(\chi,d_{rcb})$ is a rectangular cone metric space over a Banach algebra (in short RCMS-BA).

In the above definition if condition RCbM3 is replaced with

• (CbM3) $d_{rcb}(x,y)\preceq s[d_{rcb}(x,z)+d_{rcb}(z,y)]$ for all $x,y,z\in\chi$,

then $(\chi,d_{rcb})$ is a CbMS-BA as defined in [2].

Note that every CMS-BA is a CbMS-BA and CbMS-BA is a RCbMS-BA but the converse is not necessarily true. Inspired by [1, 2] we furnish the following examples, which will establish our claim.

Example 3.2

Let $\mathcal{A} = \{a = (a_{i,j})_{2\times2} : a_{i,j}\in\mathbb {R}, 1\le i,j \le2\}$, $\|a\| = \sum_{1\le i,j\le2}|a_{i,j}|$, $P = \{a\in\mathcal{A} : a_{i,j}\ge0, 1\le i,j\le2\}$ be a cone in $\mathcal{A}$. Let $\chi=B\cup\mathbb{N}$, where $B=\{\frac {1}{n}: n\in\mathbb{N}\}$. Let $d_{rcb}:\chi\times\chi \rightarrow\mathcal{A}$ be given by

$$d_{rcb}(x,y)=\left \{ \textstyle\begin{array}{l@{\quad}l} \bigl({\scriptsize\begin{matrix}{}0&0\cr0&0\end{matrix}} \bigr) & \mbox{if $x=y$;}\\ \bigl({\scriptsize\begin{matrix}{}2&2\cr3&3\end{matrix}} \bigr) & \mbox{if $x,y\in B$;} \\ \left({\scriptsize\begin{matrix}{}\frac{1}{n^{3}}&\frac{1}{n^{3}}\cr\frac{1}{n^{4}}&\frac{1}{n^{4}}\end{matrix}} \right) & \mbox{if $x = \frac{1}{n}\in B$ and $y\in\{2,3,4,5,6\}$;}\\ \bigl({\scriptsize\begin{matrix}{}1&1\cr1&1\end{matrix}} \bigr) & \mbox{otherwise.} \end{array}\displaystyle \right . $$

Then $(\chi,d_{rcb})$ is a RCbMS-BA over $\mathcal{A}$ with coefficient $s= \bigl ({\scriptsize\begin{matrix}{}3&0\cr0&3\end{matrix}} \bigr )$. But it is not possible to find $s\in{P}$, $e\preceq s$ satisfying condition CbM3 and so $(\chi,d)$ is not a CbMS-BA over a Banach algebra $\mathcal{A}$.

Example 3.3

Let $\chi= [0,2]$ and let $\mathcal{A} = C_{R^{2}}(\chi )$. For $\alpha= (f,g)$ and $\beta= (u,v)$ in $\mathcal{A}$, we define $\alpha.\beta= (f.u,g.v)$ and $\|\alpha\|=\max(\|f\|,\|g\|)$ where $\|f\| = \sup_{x\in\chi}|f(x)|$. Then $\mathcal{A}$ is a Banach algebra with unit $e=(1,1)$, zero element $\theta=(0,0)$and $P=\{ (f,g)\in\mathcal{A}:f(t)\geq0,g(t)\ge0, t\in\chi\}$ a cone in $\mathcal{A}$. Consider $d_{rcb}:\chi\times\chi\rightarrow \mathcal{A}$ given by

$$\left \{ \textstyle\begin{array}{l@{\quad}l} d_{rcb}(x,y)(t)= (0,0) & \mbox{if $x=y$;}\\ d_{rcb}(x,y)(t) = (c+d.t,a+bt^{2}) & \mbox{if $x,y \in B=[0,{1\over 2})$ and}\\ &\mbox{$a,b,c,d$ are some fixed real}\\ &\mbox{numbers;}\\ d_{rcb}(x,y)(t) = (\frac{1}{n^{3}}(c+d.t),\frac{1}{n^{3}}(a+bt^{2})) & \mbox{if $x = \frac{1}{n}(n\ge2)\in B$ and}\\ &\mbox{$y\in\{1,2\}$;}\\ d_{rcb}(x,y)(t) = (| x - y |^{2}(c+d.t),| x - y | ^{2}(a+b.t^{2})) & \mbox{otherwise.} \end{array}\displaystyle \right . $$

Clearly $(\chi,d_{rcb})$ is a RCbMS-BA over $\mathcal{A}$ with $s=(2,2)$. Again it is not possible to find a real number $s\in {P}$, $e \preceq s$ satisfying condition CbM3 and so $(\chi,d_{rcb})$ is not a CbMS-BA over a Banach algebra $\mathcal{A}$.

For any $a\in\chi$, the open sphere with center a and radius $\lambda\succ \theta$ is given by

$$B_{\lambda}(a) = \bigl\{ b \in\chi: d_{rcb}(a,b) \prec \lambda\bigr\} . $$

Let $\mathcal{U} = \{Y\subseteq\chi: \forall x \in Y, \exists r\succ\theta, \mbox{ such that } B_{r}(x) \subseteq Z\}$. Then $\mathcal{U}$ defines the rectangular b-metric topology for the RCbMS-BA $(\chi,d_{rcb})$.

The definitions of convergent sequence, Cauchy sequence, c-sequence and completeness in RCbMS-BA are along the same lines as for CbMS-BA given in [2] and so we omit these definitions.

Remark 3.4

We refer to Example 3.2 for the following:

• (i) Open balls in RCbMS-BA need not be an open set. For example $B_{\lambda}(\frac{1}{2})$ with $\lambda= \bigl ({\scriptsize\begin{matrix}{}\frac{3}{4}&\frac{3}{4}\cr\frac{3}{4}&\frac{3}{4}\end{matrix}} \bigr )$ is not open because open balls with center 4 are not contained in $B_{\lambda}(\frac{1}{2})$.

• (ii) The limit of a sequence in RCbMS-BA is not unique. For instance $\{\frac{1}{n}\}$ converges to 2 and 3.

• (iii) Every convergent sequence in RCbMS-BA need not be Cauchy. For $d(\frac{1}{n},\frac{1}{n+p})= \bigl ({\scriptsize\begin{matrix}{}2&2\cr3&3\end{matrix}} \bigr ) \nrightarrow\theta$ as $n\rightarrow\infty$, so $\{\frac {1}{n}\}$ is not a Cauchy sequence.

• (iv) A RCbMS-BA need not be Hausdorff, as it is impossible to find $r_{1}, r_{2} \succ0$ such that $B_{r_{1}}(4)\cap B_{r_{2}}(5) = \phi$.

Theorem 3.5

Let $(\chi,d_{rcb})$ be a complete RCbMS-BA over $\mathcal{A}$ with $\theta\preceq s$ and $T:\chi\rightarrow\chi$. If there exist $\lambda\in P$, $r(\lambda)< 1$ such that

$$ d_{rcb}(Tx,Ty)\preceq\lambda d_{rcb}(x,y) $$

(3.1)

for all $x,y\in\chi$, then T has a unique fixed point.

Proof

Let $x_{0}\in\chi$ be arbitrary. Consider the iterative sequence defined by $x_{n+1}=Tx_{n}$ for all $n\geq0$. We divide the proof into three cases.

Case 1: Let $r(\lambda)\in[0,\frac{1}{s})$ ($s>1$). If $x_{n}=x_{n+1}$ then $x_{n}$ is fixed point of T. Moreover, for any $x\in X$ the iterative sequence $\{T^{n}x\}$ ($n\in\mathbb{N}$) converges to the fixed point. So, suppose that $x_{n}\neq x_{n+1}$ for all $n\geq0$. Setting $d_{rcb}(x_{n},x_{n+1})=d_{n}$, it follows from (3.1) that

$$\begin{aligned}& \begin{aligned} d_{rcb}(x_{n},x_{n+1})&=d_{rcb}(Tx_{n-1},Tx_{n}) \\ &\preceq \lambda d(x_{n-1},x_{n}),\end{aligned} \\& d_{n}\preceq\lambda d_{n-1}\prec d_{n-1}, \end{aligned}$$

i.e. the sequence $\{d_{n}\}$ is strictly decreasing and from this it follows that $d_{n} \neq d_{m}$ whenever $n \neq m$. Continuing this process we get

$$ d_{n}\preceq\lambda^{n}d_{0}. $$

(3.2)

Again setting $d^{*}_{n}=d_{rcb}(x_{n},x_{n+2})$ for any $n\in\mathbb {N}$, using (3.1) we get

$$\begin{gathered} d_{rcb}(x_{n},x_{n+2})= d_{rcb}(Tx_{n-1},Tx_{n+1})\preceq\lambda d_{rcb}(x_{n-1},x_{n+1}) \\ d^{*}_{n} \preceq \lambda d^{*}_{n-1}. \end{gathered} $$

Repeating this process we obtain

$$ d_{rcb}(x_{n},x_{n+2})\preceq \lambda^{n}d^{*}_{0}. $$

(3.3)

Since $r(\lambda)<\frac{1}{s}$, we have $r(s\lambda^{2})\le sr(\lambda ^{2})\le sr(\lambda).r(\lambda) < \frac{1}{s} < 1$ and so $e-s\lambda ^{2}$ is invertible and

$$ \bigl(e-s\lambda^{2}\bigr)^{-1}=\sum _{i=0}^{\infty}\bigl(s\lambda^{2} \bigr)^{i}. $$

(3.4)

We consider $d(x_{n},x_{n+p})$ in two cases.

If p is odd, say $2m+1$, then using (3.2) as well as the fact that $d_{n} \neq d_{m}$ whenever $n \neq m$ we obtain

$$\begin{aligned} d_{rcb}(x_{n},x_{n+2m+1})\preceq{}& s \bigl[d_{n} + d_{n+1}+ d_{rcb}(x_{n+2},x_{n+2m+1}) \bigr] \\ \preceq{}&s[d_{n}+d_{n+1}]+s^{2} \bigl[d_{n+2}+d_{n+3} +d_{rcb}(x_{n+4},x_{n+2m+1})\bigr] \\ \preceq{}&s[d_{n}+d_{n+1}]+s^{2}[d_{n+2}+d_{n+3}]+s^{3}[d_{n+4}+d_{n+5}]+ \cdots+s^{m}d_{n+2m} \\ \preceq{}&s\bigl[\lambda^{n}d_{0}+\lambda^{n+1}d_{0} \bigr]+s^{2}\bigl[\lambda ^{n+2}d_{0}+ \lambda^{n+3}d_{0}\bigr]+s^{3}\bigl[ \lambda^{n+4}d_{0}+\lambda ^{n+5}d_{0} \bigr]+\cdots \\ &+s^{m}\lambda^{n+2m}d_{0} \\ \preceq{}&s\lambda^{n}\bigl[e+s\lambda^{2}+\bigl(s \lambda^{2}\bigr)^{2}+\cdots \bigr]d_{0}+s \lambda^{n+1}\bigl[e+s\lambda^{2}+\bigl(s\lambda^{2} \bigr)^{2}+\cdots\bigr]d_{0} \\ ={}&\Biggl[\sum_{i=0}^{\infty}\bigl(s \lambda^{2}\bigr)^{i}\Biggr]s\lambda^{n}d_{0}+ \Biggl[\sum_{i=0}^{\infty}\bigl(s \lambda^{2}\bigr)^{i}\Biggr]s\lambda^{n+1}d_{0} \\ ={}&\bigl(e-\bigl(s\lambda^{2}\bigr)\bigr)^{-1}s \lambda^{n}d_{0}+\bigl(e-\bigl(s\lambda ^{2}\bigr) \bigr)^{-1}s\lambda^{n+1}d_{0} \\ ={}&\lambda^{n}\bigl(e-\bigl(s\lambda^{2}\bigr) \bigr)^{-1}sd_{0}[e+\lambda]. \end{aligned} $$

Since $r(\lambda)< \frac{1}{s}<1$, using Lemma 2.7 of [6], it is easy to see that $\lambda^{n}$ is a c-sequence. Again using Proposition 2.2 of [5], $\lambda^{n}(e-(s\lambda ^{2}))^{-1}sd_{0}[1+\lambda]\rightarrow\theta$ as $n\rightarrow\infty$, and so it follows that, for any $c\in\mathcal{A}$ with $\theta\ll c$, there exists a natural number $N_{1}$ such that, for any $n>N_{1}$, we have

$$ d_{rcb}(x_{n},x_{n+2m+1})\preceq\lambda^{n} \bigl(e-\bigl(s\lambda ^{2}\bigr)\bigr)^{-1}sd_{0}[1+ \lambda]\ll c. $$

(3.5)

If p is even say 2m, using (3.2) and (3.3) as well as the fact that $d_{n} \neq d_{m}$ whenever $n \neq m$ we obtain

$$\begin{aligned} d_{rcb}(x_{n},x_{n+2m})\preceq{}& s \bigl[d_{rcb}(x_{n},x_{n+1})+d_{rcb}(x_{n+1},x_{n+2})+d_{rcb}(x_{n+2},x_{n+2m}) \bigr] \\ &\preceq s[d_{n}+d_{n+1}] \\ &+s^{2} \bigl[d_{rcb}(x_{n+2},x_{n+3})+d_{rcb}(x_{n+3},x_{n+4})+d_{rcb}(x_{n+4},x_{n+2m})\bigr] \\ \preceq{}&s[d_{n}+d_{n+1}]+s^{2}[d_{n+2}+d_{n+3}]+s^{3}[d_{n+4}+d_{n+5}]+ \cdots \\ & +s^{m-1}[d_{2m-4}+d_{2m-3}]+s^{m-1}d_{rcb}(x_{n+2m-2},x_{n+2m}) \\ \preceq{}&s\bigl[\lambda^{n}d_{0}+\lambda^{n+1}d_{0} \bigr]+s^{2}\bigl[\lambda ^{n+2}d_{0}+ \lambda^{n+3}d_{0}\bigr]+s^{3}\bigl[ \lambda^{n+4}d_{0}+\lambda ^{n+5}d_{0} \bigr]+\cdots \\ &+s^{m-1}\bigl[\lambda^{2m-4}d_{0}+\lambda ^{2m-3}d_{0}\bigr]+s^{m-1}\lambda^{n+2m-2}d^{*}_{0} \\ \preceq{}&s\lambda^{n}\bigl[e+s\lambda^{2}+s^{2} \lambda^{4}+\cdots \bigr]d_{0}+s\lambda^{n+1} \bigl[e+s\lambda^{2}+s^{2}\lambda^{4}+\cdots \bigr]d_{0} \\ &+s^{m-1}\lambda^{n+2m-2}d^{*}_{0} \\ \preceq{}&\Biggl[\sum_{i=0}^{\infty}\bigl(s \lambda^{2}\bigr)^{i}\Biggr]s\lambda ^{n}d_{0}+ \Biggl[\sum_{i=0}^{\infty}\bigl(s \lambda^{2}\bigr)^{i}\Biggr]s\lambda ^{n+1}d_{0}+s^{m-1} \lambda^{n+2m-2}d^{*}_{0} \\ \preceq{}&\bigl(e-\bigl(s\lambda^{2}\bigr)\bigr)^{-1}s \lambda^{n}d_{0}+\bigl(e-\bigl(s\lambda ^{2}\bigr) \bigr)^{-1}s\lambda^{n++}d_{0}+s^{m-1} \lambda^{n+2m-2}d^{*}_{0} \\ \preceq{}&\bigl(e-\bigl(s\lambda^{2}\bigr)\bigr)^{-1}s \lambda^{n}d_{0}[e+\lambda ]+s^{m-1} \lambda^{n+2m-2}d^{*}_{0} \\ \preceq{}&\bigl(e-\bigl(s\lambda^{2}\bigr)\bigr)^{-1}s \lambda^{n}d_{0}[e+\lambda ]+s^{m-1} \lambda^{2m-2}\lambda^{n}d^{*}_{0} \\ \preceq{}&\bigl(e-\bigl(s\lambda^{2}\bigr)\bigr)^{-1}s \lambda^{n}d_{0}[e+\lambda ]+s^{2m-1} \lambda^{2m-2}\lambda^{n}d^{*}_{0} \\ \preceq{}&\bigl(e-\bigl(s\lambda^{2}\bigr)\bigr)^{-1}s \lambda^{n}d_{0}[e+\lambda]+\sum _{i=0}^{\infty}s^{i+1}\lambda^{i} \lambda^{n}d^{*}_{0} \\ ={}& \bigl(e-\bigl(s\lambda^{2}\bigr)\bigr)^{-1}sd_{0}[e+ \lambda]\lambda^{n} \\ &+\biggl(\frac {e}{s}-\lambda\biggr)^{-1}d^{*}_{0} \lambda^{n} \quad \biggl(\mbox{since } r(\lambda) < \frac{1}{s}, \mbox{by Lemma 1.6 of [6]}\biggr). \end{aligned}$$

Note that $r(\lambda)< \frac{1}{s}<1$ and so using Lemma 2.7 of [6], it is easy to see that $\lambda ^{n}$ is a c-sequence. Again using Proposition 2.2 of [5], $(e-(s\lambda^{2}))^{-1}sd_{0}[e+\lambda ]\lambda^{n}+(\frac{e}{s}-\lambda)^{-1}d^{*}_{0}\lambda^{n}\rightarrow \theta$ as $n\rightarrow\infty$ and so it follows that, for any $c\in\mathcal{A}$ with $\theta\ll c$, there exists a natural number $N_{2}$ such that, for any $n>N_{2}$, we have

$$ d_{rcb}(x_{n},x_{n+2m+1})\preceq\bigl(e-\bigl(s \lambda^{2}\bigr)\bigr)^{-1}sd_{0}[e+\lambda ] \lambda^{n}+\biggl(\frac{e}{s}-\lambda\biggr)^{-1}d^{*}_{0} \lambda^{n}\ll c . $$

(3.6)

Let $N_{0} = \operatorname{Max}\{N_{1}, N_{2}\}$. Then for all $n \ge N_{0}$ we have

$$ d_{rcb}(x_{n},x_{n+p})\ll c . $$

(3.7)

Thus $\{x_{n}\}$ is a Cauchy sequence and since $(\chi,d_{rcb})$ is complete, we can find $u\in\chi$ such that

$$ \lim_{n\rightarrow\infty}x_{n}=u. $$

(3.8)

Since $d_{n} \neq d_{m}$ whenever $n \neq m$ there exists $k\in\mathbb {N}$ such that $d_{rcb}(u,Tu) \neq\{d_{k}, d_{k+1},\ldots \}$. Then for any $n > k$

$$\begin{aligned} d_{rcb}(u,Tu)\preceq{}& s\bigl[d_{rcb}(u,x_{n})+d_{rcb}(x_{n},x_{n+1})+d_{rcb}(x_{n+1},Tu) \bigr] \\ ={}&s\bigl[d_{rcb}(u,x_{n})+d_{n}+d_{rcb}(Tx_{n},Tu) \bigr] \\ \preceq{}&s\bigl[d_{rcb}(u,x_{n})+d_{n}+\lambda d_{rcb}(x_{n},u)\bigr] \\ \preceq{}& s\bigl[(e+\lambda)d_{rcb}(x_{n},u) + \lambda^{n}d_{0}\bigr]\rightarrow \theta\quad\mbox{as } n \rightarrow\infty, \end{aligned} $$

i.e. $Tu=u$. Now if $Tv=v$ and $d_{rcb}(u,v)\neq\theta$ then using (3.1) one can easily deduce that $d_{rcb}(u,v)=0$, and so the fixed point is unique.

Case 2: Let $r(\lambda)\in[\frac{1}{s},1)$ ($s>1$). In this case, we have $r(\lambda)^{n}\rightarrow0$ as $n\rightarrow \infty$, and so there exists $n_{0}\in N$ such that $r(\lambda )^{n_{0}}<\frac{1}{s}$. Note that $r(\lambda^{n_{0}}) \le r(\lambda)^{n_{0}}<\frac{1}{s}$. Also by (3.1),

$$\begin{aligned} d_{rcb}\bigl(T^{n_{0}}x,T^{n_{0}}y \bigr)={}&d_{rcb}\bigl(T\bigl(T^{n_{0}-1}x\bigr),T \bigl(T^{n_{0}-1}y\bigr) \\ \preceq{}&\lambda d_{rcb}\bigl(T^{n_{0}-1}x\bigr),T^{n_{0}-1}y \bigr) \\ ={}&\lambda d_{rcb}\bigl(T\bigl(T^{n_{0}-2}x\bigr),T \bigl(T^{n_{0}-2}y\bigr) \\ \preceq{}&\lambda^{2}d_{rcb}\bigl(T^{n_{0}-2}x \bigr),T^{n_{0}-2}y\bigr) \\ &\cdots \\ \preceq{}&\lambda^{n_{o}} d_{rcb}(x,y). \end{aligned} $$

Thus by case 1, $T^{n_{0}}$ has a unique fixed point $u^{*}\in X$. Now we have

$$ T^{n_{0}}\bigl(Tu^{*}\bigr)= T^{n_{0}+1}\bigl(u^{*}\bigr)=T \bigl(T^{n_{0}}u^{*}\bigr)=Tu^{*}, $$

(3.9)

i.e. $Tu^{*}$ is also a fixed point of $T^{n_{0}}$. Hence, by the uniqueness of the fixed point of $T^{n_{0}}$ we get $Tu^{*}= u^{*}$. Now suppose $Tu=u$ and $Tv=v$. Then $T^{n_{0}}u=T^{n_{0}-1}(Tu)=T^{n_{0}-1}u=\cdots=Tu=u$ and $T^{n_{0}}v=T^{n_{0}-1}(Tv)=T^{n_{0}-1}v=\cdots=Tv=v$. By the uniqueness of the fixed point of $T^{n_{0}}$ we get $u=v$.

Case 3: $s=1$. The proof follows from case 1. □

Remark 3.6

In an open problem in [7] the authors have asked whether it is possible to increase the range of λ in Theorem 2.1 of [7] from $(0,{1\over s})$ to $(0,1)$. Since every rectangular b-metric space is a RCbMS-BA, Theorem 3.5 gives a positive answer to the question posed by the authors.

Definition 3.7

Let $(\chi,d_{rcb})$ be a RCbMS-BA, $\theta\preceq s$ and $T:\chi\rightarrow\chi$. Then T is called a weak Kannan contraction iff there exist $L,\lambda\in P$ such that $0\le r(\lambda )< \frac{1}{s+1}$, and

$$ d_{rcb}(Tu,Tv)\preceq\lambda\bigl[d_{rcb}(u,Tu)+d_{rcb}(v,Tv) \bigr] + L.\alpha (u,v)\quad \forall u,v\in\chi $$

(3.10)

and $\alpha(u,v) = d_{rcb}(u,Tv)$ or $d_{rcb}(v,Tu)$.

Theorem 3.8

Let $(\chi,d_{rcb})$ be a complete RCbMS-BA with $\theta\preceq s$, and $T:\chi\rightarrow\chi$ be a mapping. If T is a weak Kannan contraction mapping then T has a fixed point. Further if

$$ L< 1 \quad\textit{or} \quad d_{rcb}(Tx,Ty)\preceq L^{*}. \bigl[d_{rcb}(x,Tx)+d(y,Ty)\bigr] $$

(3.11)

for some $L^{*}\in P$, then the fixed point is unique.

Proof

Let $x_{0}\in\chi$ be arbitrary. Consider the iterative sequence defined by $x_{n+1}=Tx_{n}$ for all $n\geq0$. Let $d_{rcb}(x_{n},x_{n+1})=d_{n}$ and suppose $\alpha(x,y) = d_{rcb}(x,Ty)$. It follows from (3.10) that

$$\begin{gathered} d_{rcb}(x_{n},x_{n+1})=d_{rcb}(Tx_{n-1},Tx_{n}), \\ d_{n}\preceq \lambda[d_{n-1}+d_{n}]. \end{gathered} $$

If $\alpha(x,y) = d_{rcb}(x,Ty)$

$$\begin{gathered} d_{rcb}(x_{n},x_{n+1})=d_{rcb}(Tx_{n-1},Tx_{n}), \\ d_{n}\preceq \lambda[d_{n-1}+d_{n}]. \end{gathered} $$

Thus in both cases

$$\begin{gathered} d_{n}\preceq\lambda[d_{n-1}+d_{n}], \\ d_{n}\preceq(e-\lambda)^{-1}\lambda d_{n-1}=\beta d_{n-1}, \end{gathered} $$

where $\beta=(e-\lambda)^{-1}\lambda$. Repeating this process we obtain

$$ d_{n}\preceq\beta^{n}d_{0}. $$

(3.12)

Also, for $\alpha(x,y) = d_{rcb}(x,Ty)$

$$\begin{aligned} d_{rcb}(x_{n},x_{n+2})={}& d_{rcb}(Tx_{n-1},Tx_{n+1})=d_{rcb}(Tx_{n+1},Tx_{n-1}) \\ \preceq{}&\lambda[d_{n-1}+d_{n+1}] + L. d_{n} \\ \preceq{}&\lambda[d_{n-1}+d_{n+1}] + L.d_{n}; \end{aligned} $$

for $\alpha(x,y) = d_{rcb}(y,Tx)$

$$\begin{aligned} d_{rcb}(x_{n},x_{n+2})={}& d_{rcb}(Tx_{n-1},Tx_{n+1}) \\ \preceq{}&\lambda[d_{n-1}+d_{n+1}] + L. d_{n} \\ \preceq{}&\lambda[d_{n-1}+d_{n+1}] + L.d_{n}. \end{aligned} $$

Thus we have

$$\begin{aligned} d_{rcb}(x_{n},x_{n+2})\preceq{}& \lambda\bigl[\beta^{n-1}d_{0}+\beta ^{n+1}d_{0} \bigr] + L\bigl[\beta^{n}d_{0}\bigr] \\ \preceq{}&\lambda\beta^{n-1}\bigl[e+\beta^{2} \bigr]d_{0} + L.\beta^{n-1}\beta d_{0} \\ \preceq{}&\eta\beta^{n-1}, \end{aligned} $$

where $\eta=(\lambda(e+\beta^{2}) + L\beta)d_{0} \in P$. Thus we have

$$ d_{rcb}(x_{n},x_{n+2})\preceq\eta \beta^{n-1}. $$

(3.13)

Note that $r(\lambda)< 1$ and so $(e-\lambda)^{-1}$ is invertible and $(e-\lambda)^{-1}= \sum_{i=0}^{\infty}\lambda^{i}$. Therefore $r(\beta )=r((e-\lambda)^{-1}\lambda) = (\sum_{i=1}^{\infty}\lambda^{i}) \le \sum_{i=1}^{\infty}r(\lambda)^{i} = \frac{r(\lambda)}{1-r(\lambda)}$ (as $r(\lambda)< 1$). Thus we have $r(\beta)< \frac{1}{s}$. Therefore

$$r\bigl(s\beta^{2}\bigr)=sr\bigl(\beta^{2} \bigr) \leq sr(\beta)r(\beta) =\frac{1}{s} < 1 $$

so $e-s\beta^{2}$ is invertible and

$$ \bigl(e-s\beta^{2}\bigr)^{-1}=\sum _{i=0}^{\infty}\bigl(s\beta^{2} \bigr)^{i}. $$

(3.14)

We will analyze $d_{rcb}(x_{n},x_{n+p})$ as follows: For some odd p say $2m+1$

$$\begin{aligned} d_{rcb}(x_{n},x_{n+2m+1})\preceq{}& s \bigl[d_{n}+d_{n+1}+d_{rcb}(x_{n+2},x_{n+2m+1}) \bigr] \\ \preceq{}&s[d_{n}+d_{n+1}]+s^{2}\bigl[d_{n+2}+d_{n+3} +d_{rcb}(x_{n+4},x_{n+2m+1})\bigr] \\ \preceq{}&s[d_{n}+d_{n+1}]+s^{2}[d_{n+2}+d_{n+3}]+s^{3}[d_{n+4}+d_{n+5}]+ \cdots +s^{m}d_{n+2m} \\ \preceq{}&s\bigl[\beta^{n}d_{0}+\beta^{n+1}d_{0} \bigr]+s^{2}\bigl[\beta^{n+2}d_{0}+\beta ^{n+3}d_{0}\bigr]+s^{3}\bigl[\beta^{n+4}d_{0}+ \beta^{n+5}d_{0}\bigr]\hspace{-0.5pt}+\cdots \\ &+s^{m}\beta^{n+2m}d_{0} \\ \preceq{}&s\beta^{n}\bigl[1+s\beta^{2}+s^{2} \beta^{4}+\cdots\bigr]d_{0}+s\beta ^{n+1}\bigl[1+s \beta^{2}+s^{2}\beta^{4}+\cdots\bigr]d_{0} \\ ={}&\Biggl[\sum_{i=0}^{\infty}\bigl(s \beta^{2}\bigr)^{i}\Biggr]s\beta^{n}d_{0}+ \Biggl[\sum_{i=0}^{\infty}\bigl(s \beta^{2}\bigr)^{i}\Biggr]s\beta^{n+1}d_{0} \\ ={}&\bigl(e-\bigl(s\beta^{2}\bigr)\bigr)^{-1}s \beta^{n}d_{0}+\bigl(e-\bigl(s\beta^{2}\bigr) \bigr)^{-1}s\beta^{n+1}d_{0}. \end{aligned}$$

Note that $r(\beta)< \frac{1}{s}<1$ and using Lemma 2.7 of [6], $\beta^{n}$ is a c-sequence. Again using Proposition 2.2 of [5], $(e-(s\beta^{2}))^{-1}s\beta ^{n}d_{0}+(e-(s\beta^{2}))^{-1}s\beta^{n+1}d_{0} \rightarrow\theta$ as $n\rightarrow\infty$. It follows that, for any $c\in A$ with $\theta\ll c$, there exists $N_{1}\in\mathbb{N}$ such that, for any $n>N_{1}$, we have

$$ d_{rcb}(x_{n},x_{n+2m+1})\preceq\bigl(e-\bigl(s \beta^{2}\bigr)\bigr)^{-1}s\beta ^{n}d_{0}+ \bigl(e-\bigl(s\beta^{2}\bigr)\bigr)^{-1}s\beta^{n+1}d_{0} \ll c. $$

(3.15)

For some even p, say 2m,

$$\begin{aligned} d_{rcb}(x_{n},x_{n+2m})\preceq{}& s \bigl[d_{n}+d_{n+1}+d_{rcb}(x_{n+2},x_{n+2m}) \bigr] \\ \preceq{}&s[d_{n}+d_{n+1}]+s^{2} \bigl[d_{n+2}+d_{n+3} +d_{rcb}(x_{n+4},x_{n+2m})\bigr] \\ \preceq{}&s[d_{n}+d_{n+1}]+s^{2}[d_{n+2}+d_{n+3}]+s^{3}[d_{n+4}+d_{n+5}]+ \cdots \\ & +s^{m-1}[d_{2m-4}+d_{2m-3}]+s^{m-1}d_{rcb}(x_{n+2m-2},x_{n+2m}) \\ \preceq{}&s\bigl[\beta^{n}d_{0}+\beta^{n+1}d_{0} \bigr]+s^{2}\bigl[\beta ^{n+2}d_{0}+ \beta^{n+3}d_{0}\bigr]+s^{3}\bigl[ \beta^{n+4}d_{0}+\beta ^{n+5}d_{0}\bigr]+ \cdots \\ &+s^{m-1}\bigl[\beta^{2m-4}d_{0}+ \beta^{2m-3}d_{0}\bigr]+s^{m-1}\eta \beta^{n+2m-3}d_{0} \\ \preceq{}&s\beta^{n}\bigl[1+s\beta^{2}+s^{2} \beta^{4}+\cdots\bigr]d_{0}+s\beta ^{n+1}\bigl[1+s \beta^{2}+s^{2}\beta^{4}+\cdots\bigr]d_{0} \\ &+s^{2m-1}\eta\beta^{2m-2}\beta^{n-1}d_{0} \\ ={}&\Biggl[\sum_{i=0}^{\infty}\bigl(s \beta^{2}\bigr)^{i}\Biggr]s\beta^{n}d_{0}+ \Biggl[\sum_{i=0}^{\infty}\bigl(s \beta^{2}\bigr)^{i}\Biggr]s\beta^{n+1}d_{0} +\eta\Biggl[\sum_{i=0}^{\infty}s^{i+1} \beta^{i}\Biggr]\beta^{n-1}d_{0} \\ ={}&\bigl(e-\bigl(s\beta^{2}\bigr)\bigr)^{-1}s\beta \beta^{n-1}d_{0}+\bigl(e-\bigl(s\beta ^{2}\bigr) \bigr)^{-1}s\beta^{2}\beta^{n-1}d_{0} \\ &+\eta\biggl(\frac{e}{s}-\beta \biggr)^{-1}\beta^{n-1}d_{0} \quad\biggl(\mbox{since } r(\beta)< \frac{1}{s}\biggr). \end{aligned} $$

Note that $r(\beta)< \frac{1}{s}<1$ and using Lemma 2.7 of [6], $\beta^{n-1}$ is a c-sequence. Again using Proposition 2.2 of [5] $(e-(s\beta^{2}))^{-1}s\beta\beta ^{n-1}d_{0}+(e-(s\beta^{2}))^{-1}s\beta^{2}\beta^{n-1}d_{0}+\eta(\frac {e}{s}-\beta)^{-1}\beta^{n-1}d_{0} \rightarrow\theta$ as $n\rightarrow\infty$. It follows that, for any $c\in A$ with $\theta \ll c$, there exists $N_{2} \in\mathbb{N}$ such that, for any $n>N_{2}$,

$$ \begin{aligned}[b] d_{rcb}(x_{n},x_{n+2m})\preceq{}& \bigl(e-\bigl(s\beta^{2}\bigr)\bigr)^{-1}s\beta\beta ^{n-1}d_{0}+\bigl(e-\bigl(s\beta^{2}\bigr) \bigr)^{-1}s\beta^{2}\beta^{n-1}d_{0} \\ &+\eta\biggl(\frac {e}{s}-\beta\biggr)^{-1}\beta^{n-1}d_{0} \\ \ll{}& c. \end{aligned} $$

(3.16)

Let $N_{0} = \operatorname{Max}\{N_{1}, N_{2}\}$. Then for all $n \ge N_{0}$ we have

$$ d(x_{n},x_{n+p})\ll c. $$

(3.17)

Thus $\{x_{n}\}$ is a Cauchy sequence and by completeness of $(\chi ,d_{rcb})$ there exists $u\in\chi$ such that

$$ \lim_{n\rightarrow\infty}x_{n}=u. $$

(3.18)

Since $d_{n} \neq d_{m}$ whenever $n \neq m$ there exists $k\in\mathbb {N}$ such that $d(u,Tu) \neq\{d_{k}, d_{k+1}, \ldots \}$. Then for any $n > k$

$$\begin{aligned}& \begin{aligned}d_{rcb}(u,Tu)&\preceq s\bigl[d_{rcb}(u,x_{n})+d_{rcb}(x_{n},x_{n+1})+d_{rcb}(x_{n+1},Tu) \bigr] \\ &=s\bigl[d_{rcb}(u,x_{n})+d_{n}+d_{rcb}(Tx_{n},Tu) \bigr] \\ &\preceq s\bigl[d_{rcb}(u,x_{n})+d_{n}+\lambda \bigl\{ d_{rcb}(x_{n},Tx_{n})+d_{rcb}(u,Tu) \bigr\} + L.\alpha(x_{n},u)\bigr] \\ &=s\bigl[d_{rcb}(u,x_{n})+d_{n}+\lambda\bigl\{ d_{rcb}(x_{n},x_{n+1})+d_{rcb}(u,Tu)\bigr\} \bigr] + L. d_{rcb}(u,x_{n+1}) \\ &\preceq s\bigl[d_{rnb}(u,x_{n})+d_{n}+\lambda\bigl\{ d_{rnb}(x_{n},x_{n+1})+d_{rcb}(u,Tu)\bigr\} \bigr] + L.d_{rnb}(u,x_{n+1}),\end{aligned} \\& (e-s\lambda)d_{rcb}(u,Tu)\preceq s\bigl[d_{rnb}(u,x_{n})+(e+ \lambda )\beta^{n}d_{0}\bigr] + L.d_{rnb}(u,x_{n+1}) . \end{aligned}$$

Note that $r(s\lambda) \le sr(\lambda) < \frac{1}{3} < 1$ and so $e-s\lambda$ is invertible. Also, $r(\beta)< \frac{1}{s}<1$. Hence using Lemma 2.7 of [6], $\beta^{n}$ is a c-sequence and use of Proposition 2.2 of [5] gives $s(e-\lambda s)^{-1}[d_{rcb}(u,x_{n})+(e+\lambda)\beta^{n}d_{0}] + L.d_{rnb}(u,x_{n+1})\rightarrow\theta$ as $n\rightarrow\infty$. It follows that, for $c\in A$ and $\theta\ll c$, there exists $N_{3} \in \mathbb{N}$, such that, for any $n>N_{3}$,

$$ d_{rcb}(u,Tu)\preceq s(e-\lambda s)^{-1}\bigl[d_{rcb}(u,x_{n})+(1+ \lambda )\beta^{n}d_{0} + L.d_{rcb}(u,x_{n+1})\bigr] \ll c, $$

(3.19)

i.e. $Tu=u$. Uniqueness follows easily from (3.11). □

Theorem 3.9

Let $(\chi,d_{rcb})$ be a complete RCbMS-BA with $s\geq1$ and $T:\chi\rightarrow\chi$ be a mapping. If there exists $\lambda\in P$ such that $0\le r(\lambda)< \frac {1}{s+1}$, and

$$ d_{rcb}(Tx,Ty)\preceq\lambda\bigl[d_{rcb}(x,Tx)+d_{rcb}(y,Ty) \bigr], $$

(3.20)

for all $x,y\in\chi$ then T has a unique fixed point.

Proof

Note that (3.20) implies (3.10) and (3.11). Hence the result follows from Theorem 3.8. □

Corollary 3.10

Theorem 3.2 of [5] and Theorem 2.2 of [1].

Proof

Note that, for $k\in P$ with $r(k) < {1\over 2}$,

$$d_{rcb}(Tx,Ty) \preceq k\bigl\{ d_{rcb}(x,Ty) + d_{rcb}(y,Tx)\bigr\} $$

implies

$$\begin{aligned} d_{rcb}(Tx,Ty) &\preceq k\bigl\{ d_{rcb}(x,Tx) + d_{rcb}(Tx,y) + d_{rcb}(y,Ty) + d_{rcb}(y,Tx)\bigr\} \\ &\preceq k\bigl\{ d_{rcb}(x,Tx) + d_{rcb}(y,Ty)\bigr\} + 2kd_{rcb}(y,Tx), \end{aligned} $$

where $r(k) < {1\over 2}$ and $r(2k) < 1$. Thus T satisfies conditions (3.10) and (3.11) of Theorem 3.8, with $s=1$. Since every CMS-BA is a RCbMS-BA with $s=1$, the proof follows from Theorem 3.8. □

Corollary 3.11

Theorem 3.3 of [5] and Theorem 2.3 of [1].

Proof

Since every CMS-BA is a RCbMS-BA with $s=1$, the proof follows from Theorem 3.9. □

Example 3.12

Let $\mathcal{A} = \{a = (a_{i,j})_{2\times2} : a_{i,j}\in\mathbb {R}, 1\le i,j \le2\}$, $\|a\| = \max_{i}\sum_{j=1}^{2}|a_{i,j}|$, $P = \{ a\in\mathcal{A} : a_{i,j}\ge0, 1\le i,j\le2\}$ be a cone in $\mathcal{A}$. Let χ = $A \cup B$, where $A = [0,\frac {1}{2}]$ and $B = [1,2]$. Let $d_{rcb}:\chi\times\chi \rightarrow\mathcal{A}$ be given by

$$\left \{ \textstyle\begin{array}{l@{\quad}l} d_{rcb}(0,{1\over 2}) = d_{rcb}({1\over 3},{1\over 4})=d_{rcb}({1\over 5},{1\over 6}) = \bigl({\scriptsize\begin{matrix}{}0.6&0.6\cr0.6&0.6\end{matrix}} \bigr) ;&\\ \quad d_{rcb}(0,{1\over 3})=d_{rcb}({1\over 2},{1\over 5})=d_{rcb}({1\over 4},{1\over 5}) = \bigl({\scriptsize\begin{matrix}{}0.2&0.2\cr0.2&0.2\end{matrix}} \bigr) ;&\\ \quad d_{rcb}(0, {1\over 4})=d_{rcb}({1\over 2},{1\over 3})=d_{rcb}({1\over 4},{1\over 6}) = \bigl({\scriptsize\begin{matrix}{}0.4&0.4\cr0.4&0.4\end{matrix}} \bigr) ;&\\ \quad d_{rcb}(0,{1\over 5})=d_{rcb}({1\over 2},{1\over 6})=d_{rcb}({1\over 3},{1\over 6}) = \bigl({\scriptsize\begin{matrix}{}0.5&0.5\cr0.5&0.5\end{matrix}} \bigr) ;&\\ \quad d_{rcb}(0,{1\over 6})=d_{rcb}({1\over 2},{1\over 4})=d_{rcb}({1\over 3},{1\over 5}) = \bigl({\scriptsize\begin{matrix}{}0.3&0.3\cr0.3&0.3\end{matrix}} \bigr) ;&\\ \quad d_{rcb}(u,v) = \bigl({\scriptsize\begin{matrix}{}0&0\cr0&0\end{matrix}} \bigr) &\mbox{if $u=v$;}\\ d_{rcb}(u,v)= \bigl({\scriptsize\begin{matrix}{}1&1\cr1&1\end{matrix}} \bigr) &\mbox{if $u,v \in A - \{0,{1\over 2},{1\over 3},{1\over 4},{1\over 5},{1\over 6}\}$;}\\ d_{rcb}(u,v) = \left({\scriptsize\begin{matrix}{}\frac{1}{2n}&\frac{1}{2n}\cr\frac{1}{2n}&\frac{1}{2n}\end{matrix}} \right) &\mbox{if $u = \frac{1}{n}\ (n\ge2)\ \in A$ and $v\in\{1,2\}$;}\\ d_{rcb}(u,v) = \bigl({\scriptsize\begin{matrix}{}| u - v |^{2}&| u - v |^{2}\cr| u - v|^{2}&| u - v |^{2}\end{matrix}} \bigr) &\mbox{otherwise.} \end{array}\displaystyle \right . $$

Then $(\chi,d_{rcb})$ is a RCbMS-BA over $\mathcal{A}$ with $s= \bigl ({\scriptsize\begin{matrix}{}2&0\cr0&2\end{matrix}} \bigr )$. However, for $u={1\over n}$ and $v={1\over m}$ it is impossible to find $s\in{P}$, $e\preceq s$ such that $d_{rcb}({1\over n}, {1\over m}) \preceq s(d_{rcb}({1\over n}, 2) + d_{rcb}(2, {1\over m}))$ for all ${n, m \in\mathbb{N}}$, and so $(\chi,d_{rcb})$ is not a CbMS-BA over $\mathcal{A}$. Also $(\chi,d_{rcb})$ is not a CMS-BA over $\mathcal{A}$ as $d_{rcb}({1\over 3},{1\over 4}) = \bigl ({\scriptsize\begin{matrix}{}0.6&0.6\cr0.6&0.6\end{matrix}} \bigr ) \succ d_{rcb}({1\over 3},{1\over 5})+d({1\over 5},{1\over 4}) = \bigl ({\scriptsize\begin{matrix}{}0.5&0.5\cr0.5&0.5\end{matrix}} \bigr )$. Define T by

$$Tu=\left\{ \textstyle\begin{array}{l@{\quad}l} {9\over 20}, &u \in B\cup\{{1\over 6}\}; \\ {1\over 2}-u, &u \in D=\{{1\over n} : n\geq3,n\neq6\}; \\ {1\over 4}, &u\in A-\{D\cup{1\over 6}\}. \end{array}\displaystyle \right. $$

Then T satisfies condition (3.10) For $\alpha(x,y) \neq\theta$ then it is enough if we take L sufficiently large. If $\alpha(u,v) = \theta$, we proceed as follows.

Case (i): $u \in B\cup\{{1\over 6}\}$, $v\in D$, $d_{rcb}(u,Tv)=d_{rcb}(u,{1\over 2}-v)$; $d_{rcb}(v,Tu)=d_{rcb}(v,{9\over 20})$; $\alpha(u,v)=\theta$ iff $u+v={1\over 2}$ or $v={9\over 20}$. Since $v\in D$, $v={9\over 20}$. $u+v={1\over 2}$ only at $u={1\over 6}$ and $v={1\over 3}$. Then $d_{rcb}(Tu,Tv)=d_{rcb}({9\over 20},{1\over 6})= \bigl ({\scriptsize\begin{matrix}{}0.080278&0.080278\cr0.080278&0.080278\end{matrix}} \bigr )$; $d_{rcb}(u,Tu)=d_{rcb}({1\over 6},{9\over 20})= \bigl ({\scriptsize\begin{matrix}{}0.080278&0.080278\cr0.080278&0.080278\end{matrix}} \bigr )$; $d_{rcb}(v,Tv)=d_{rcb}({1\over 3},{1\over 6})= \bigl ({\scriptsize\begin{matrix}{}0.5&0.5\cr0.5&0.5\end{matrix}} \bigr )$. Clearly we can find $\lambda= \bigl ({\scriptsize\begin{matrix}{}k&0\cr0&k\end{matrix}} \bigr )$ with $k\in(0,{1\over 3})$ satisfying (3.10).

Case (ii): $u \in B\cup\{{1\over 6}\}$, $v\in A-\{D\cup{1\over 6}\}$, $d_{rcb}(u,Tv)=d_{rcb}(u,{1\over 4})$; $d_{rcb}(v,Tu)=d_{rcb}(v,{9\over 20})$; $\alpha(u,v)=0$ only at $u={1\over 4}$ or $v={9\over 20}$. But $u\neq {1\over 4}$. Let $v={9\over 20}$ and $u \in B\cup\{{1\over 6}\}$. Then $d_{rcb}(Tu,Tv)=d_{rcb}({9\over 20},{1\over 4})= \bigl ({\scriptsize\begin{matrix}{}0.04&0.04\cr0.04&0.04\end{matrix}} \bigr )$; $d_{rcb}(v,Tv)=d_{rcb}({9\over 20},{1\over 4})= \bigl ({\scriptsize\begin{matrix}{}0.04&0.04\cr0.04&0.04\end{matrix}} \bigr )$; $d_{rcb}(u,Tu)=d_{rcb}(x,{9\over 20})= \bigl ({\scriptsize\begin{matrix}{}\vert x-{9\over 20}\vert^{2}&\vert x-{9\over 20}\vert^{2}\cr\vert x-{9\over 20}\vert^{2}&\vert x-{9\over 20}\vert^{2}\end{matrix}} \bigr )$; when $u={1\over 6}$, $d_{rcb}(u,Tu)= \bigl ({\scriptsize\begin{matrix}{}0.0802777&0.0802777\cr0.0802777&0.0802777\end{matrix}} \bigr )$. Clearly there exist $k\in(0,{1\over 3})$ such that $\lambda= \bigl ({\scriptsize\begin{matrix}{}k&0\cr0&k\end{matrix}} \bigr )$ satisfying (3.10).

Case (iii): $u\in D$, $v \in A-\{D\cup{1\over 6}\}$, $d_{rcb}(u,Tv)=d_{rcb}(u,{1\over 4})$; $d_{rcb}(v,Tu)=d_{rcb}(v,{1\over 2}-u)$; $\alpha(u,v)=0$ at $u={1\over 4}$ and $u+v={1\over 2}$. At $u={1\over 4}$, $d_{rcb}(Tu,Tv)=({1\over 4},{1\over 4})= 0$. Hence condition (3.10) is satisfied. At $u={1\over n}$ and $v={1\over 2}-{1\over n}$, $d_{rcb}(Tu,Tv)=d_{rcb}({1\over 2}-{1\over n},{1\over 4})= \bigl ({\scriptsize\begin{matrix}{}\vert{1\over 4}-{1\over n}\vert^{2}&\vert{1\over 4}-{1\over n}\vert^{2}\cr\vert{1\over 4}-{1\over n}\vert^{2}&\vert{1\over 4}-{1\over n}\vert^{2}\end{matrix}} \bigr )$; $d_{rcb}(u,Tu)=d_{rcb}({1\over n},{1\over 2}-{1\over n})= \bigl ({\scriptsize\begin{matrix}{}\vert{2\over n}-{1\over 2}\vert^{2}&\vert{2\over n}-{1\over 2}\vert^{2}\cr\vert{2\over n}-{1\over 2}\vert^{2}&\vert{2\over n}-{1\over 2}\vert^{2}\end{matrix}} \bigr )$; $d_{rcb}(v,Tv)=d_{rcb}({1\over 2}-{1\over n},{1\over 4})= \bigl ({\scriptsize\begin{matrix}{}\vert{1\over 4}-{1\over n}\vert^{2}&\vert{1\over 4}-{1\over n}\vert^{2}\cr\vert{1\over 4}-{1\over n}\vert^{2}&\vert{1\over 4}-{1\over n}\vert^{2}\end{matrix}} \bigr )$; Note that $d_{rcb}(u,Tu)+d_{rcb}(v,Tv) = \vert{2\over n}-{1\over 2}\vert^{2} + \vert{1\over 4}-{1\over n}\vert^{2} = 5\vert{1\over 4}-{1\over n}\vert^{2} = 5d_{rcb}(Tu,Tv)$. Hence (3.10) is satisfied with $\lambda= \bigl ({\scriptsize\begin{matrix}{}{1\over 4}&0\cr0&{1\over 4}\end{matrix}} \bigr )$.

Other cases follow similarly. Indeed condition (3.10) is satisfied. Note that $d_{rcb}(u,Tu) + d_{rcb}(v,Tv) \neq\theta$ for any $u,v\in\chi$ and so T satisfies (3.11) for sufficiently large $L^{*}$. Theorem 3.8 is thus applicable and $\operatorname{Fix}(T) = \{{1\over 4}\}$. However, condition (3.20) is not satisfied at $u = {1\over 6}$ and $v = {1\over 4}$ as $d_{rcb}(Tu,Tv) = d_{rcb}({9\over 20},{1\over 4}) = \bigl ({\scriptsize\begin{matrix}{}0.04&0.04\cr0.04&0.04\end{matrix}} \bigr ) \succ{1\over 3}[d_{rcb}(u,Tu) + d_{rcb}(v,Tv)]={1\over 3}[d_{rcb}({1\over 6},{9\over 20})+d_{rcb}({1\over 4},{1\over 4})]= \bigl ({\scriptsize\begin{matrix}{}0.02676&0.02676\cr0.02676&0.02676\end{matrix}} \bigr )$. Hence Theorem 3.9 is not applicable.

Example 3.13

Let $\chi=[0,{1\over 2}]$ and $d_{rcb}(x,y)=|x-y|$. Let $Tx = x^{2}$ for all $x,y\in\chi$. Then Theorem 3.8 is applicable on T and $\operatorname{Fix}(T) = \{0\}$. However, Corollary 3.11 is not applicable on T. If we take $X=[0,1]$ then T satisfies (3.10) but neither $L<1$ nor T satisfy (3.11). 0 and 1 are two fixed points of T.

Now, we will apply Theorem 3.5 to study the existence and uniqueness of solutions of a system of linear equations.

Consider the following system of linear equations:

$$ \left \{ \textstyle\begin{array}{l} a_{11}x_{1} + a_{12}x_{2} + \cdots + a_{1n}x_{n} = b_{1},\\ a_{21}x_{1} + a_{22}x_{2} + \cdots + a_{2n}x_{n} = b_{2},\\ \vdots\\ a_{n1}x_{1} + a_{n2}x_{2} + \cdots + a_{nn}x_{n} = b_{n}, \end{array}\displaystyle \right . $$

(3.21)

with $a_{ij}, b_{i}\ (i = 1 \cdots n, j=1\cdots n)\ \in\mathbb{C}$.

Theorem 3.14

If $\sum_{i=1, i\neq j}^{n}|a_{ij}| + |1-a_{jj}| < 1$, then (3.21) has a unique solution.

Proof

Consider the Banach algebra $\mathcal{A} = \{a = (a_{ij})_{n\times n} : a_{ij}\in\mathbb{C}, 1\le i,j \le n\}$, with e being the identity matrix of order n, multiplication defined as ordinary matrix multiplication and $\|a\| = \sum_{i=1}^{n} |a_{ij}|$. Let $P = \{a\in\mathcal{A} : a_{i,j}\ge0, 1\le i,j\le2\}$ be a cone in $\mathcal{A}$. Let $\chi= \mathbb{C^{n}}$. Define $d_{rcb}: \chi\times\chi\rightarrow\mathcal{A}$ by

$$d_{rcb}(x,y)=\left \{ \textstyle\begin{array}{l@{\quad}l} {1\over 2^{\alpha}} \left({\scriptsize\begin{matrix}{}x_{1}-y_{1}&x_{1}-y_{1} & \ldots &x_{1}-y_{1}\cr x_{2}-y_{2}&x_{2}-y_{2}& \ldots &x_{2}-y_{2}\cr \vdots&\vdots&&\vdots\cr x_{n}-y_{n}&x_{n}-y_{n}& \ldots &x_{n}-y_{n}\end{matrix}} \right) & \mbox{if $x,y \in X$,}\\ &\mbox{$x-y \in[{-1\over 2},{1\over 2}] \times [{-1\over 2},{1\over 2}]\times \cdots \times[{-1\over 2},{1\over 2}]$,} \\ &\mbox{$\alpha$ is the largest integer such that}\\ &\mbox{$\alpha\in\{2,3,4,5,6,7,8,9,10\}$ and}\\ &\mbox{$\max_{i}|x_{i}-y_{i}|< {1\over \alpha}$;}\\ \left({\scriptsize\begin{matrix}{}{1\over 10^{p}}|x_{1}-y_{1}|&{1\over 10^{p}}|x_{1}-y_{1}| & \ldots &{1\over 10^{p}}|x_{1}-y_{1}|\cr{1\over 10^{p}}|x_{2}-y_{2}|&{1\over 10^{p}}|x_{2}-y_{2}|& \ldots &{1\over 10^{p}}|x_{2}-y_{2}|\cr \vdots&\vdots&&\vdots\cr{1\over 10^{p}}|x_{n}-y_{n}|&{1\over 10^{p}}|x_{n}-y_{n}|& \ldots &{1\over 10^{p}}|x_{n}-y_{n}|\end{matrix}} \right) & \mbox{if $x=(x_{1},x_{2},\ldots,x_{n})\in X $:}\\ &\mbox{$ x_{1}+x_{2}+\cdots+x_{n} = p \in\mathbb {N}$ and}\\ & \mbox{$y = (y_{1},y_{2},\ldots,y_{n})\in\mathbb{N}\times \mathbb{N}\times \cdots \times\mathbb{N}$;}\\ \left({\scriptsize\begin{matrix}{}x_{1}-y_{1}&x_{1}-y_{1} & \ldots &x_{1}-y_{1}\cr x_{2}-y_{2}&x_{2}-y_{2}& \ldots &x_{2}-y_{2}\cr \vdots&\vdots&&\vdots\cr x_{n}-y_{n}&x_{n}-y_{n}& \ldots &x_{n}-y_{n}\end{matrix}} \right) & \mbox{otherwise.} \end{array}\displaystyle \right . $$

Then $(\chi,d_{rcb})$ is a RCbMS-BA over $\mathcal{A}$ with $s=10$. Let $T : \chi\rightarrow X$ be defined by

$$Tx = (I-A)x + B \quad\mbox{for all } x \in\chi, $$

where $A_{n\times n}$, $\chi_{{\it s}_{n\times1}}$ and $B_{n\times1}$ are the coefficient matrices of (3.21).

Then the system of linear equation (3.21) is equivalent to $x = Tx$. We will show that T satisfies (3.1). Let $x,y \in\chi$.

Case 1. $x-y \in[{-1\over 2},{1\over 2}]\times[{-1\over 2},{1\over 2}]\times \cdots \times[{-1\over 2},{1\over 2}]$ and α is the largest integer such that $\max_{i}\{|x_{i}-y_{i}| : i = 1,2,\ldots, n\} < {1\over \alpha }$, $\alpha\in\{2,3,4,5,6,7,8,9,10\}$. Then $Tx = (\gamma_{1},\gamma_{2}, \ldots , \gamma_{n})$; $\gamma_{i} = \sum_{j=1, j \neq i}^{n}a_{ij}x_{j}+(1-a_{ii})x_{i}$, $Ty = (\eta_{1},\eta_{2}, \ldots , \eta_{n})$; $\eta_{i} = \sum_{j=1, j \neq i}^{n}a_{ij}x_{j}+(1-a_{ii})x_{i}$ and $Tx-Ty = (\lambda_{1},\lambda_{2}, \ldots , \lambda_{n})$ where $\lambda_{i} = \sum_{j=1, j \neq i}^{n}a_{ij}(x_{j}-y_{j})+(1-a_{ii})(x_{i}-y_{i})$ and $|\lambda_{i}| = \gamma_{i}-\eta_{i} = \sum_{j=1, j \neq i}^{n}|a_{ij}||x_{j}-y_{j}|+|1-a_{ii}||x_{i}-y_{i}| < {1\over \beta}$, $\beta\ge \alpha$. Thus we have

$$\begin{aligned} d_{rcb}(Tx,Ty)={}& {1\over 2^{\beta}} \begin{pmatrix} \lambda_{1}&\lambda_{1} & \ldots &\lambda_{1}\cr\lambda _{2}&\lambda_{2}& \ldots &\lambda_{2}\cr\vdots&\vdots&&\vdots\cr\lambda_{n}&\lambda _{n}& \ldots &\lambda_{n} \end{pmatrix} \\ ={}& {1\over 2^{\beta}} \begin{pmatrix} (1-a_{11})&a_{12} & \ldots &a_{1n}\cr a_{21}&(1-a_{22})& \ldots &a_{2n}\cr\vdots&\vdots&&\vdots\cr a_{n1}&a_{n2}& \ldots &(1-a_{nn}) \end{pmatrix} \\ &\times \begin{pmatrix} x_{1}-y_{1}&x_{1}-y_{1} & \ldots &x_{1}-y_{1}\cr x_{2}-y_{2}&x_{2}-y_{2}& \ldots &x_{2}-y_{2}\cr\vdots&\vdots&&\vdots\cr x_{n}-y_{n}&x_{n}-y_{n}& \ldots &x_{n}-y_{n} \end{pmatrix} \\ \le{}& \begin{pmatrix} (1-a_{11})&a_{12} & \ldots &a_{1n}\cr a_{21}&(1-a_{22})& \ldots &a_{2n}\cr\vdots&\vdots&&\vdots\cr a_{n1}&a_{n2}& \ldots &(1-a_{nn}) \end{pmatrix} \\ &\times{1\over 2^{\alpha}} \begin{pmatrix} x_{1}-y_{1}&x_{1}-y_{1} & \ldots &x_{1}-y_{1}\cr x_{2}-y_{2}&x_{2}-y_{2}& \ldots &x_{2}-y_{2}\cr\vdots&\vdots&&\vdots\cr x_{n}-y_{n}&x_{n}-y_{n}& \ldots &x_{n}-y_{n} \end{pmatrix} . \end{aligned}$$

Case 2. $x-y \notin[{-1\over 2},{1\over 2}]\times[{-1\over 2},{1\over 2}]\times \cdots \times[{-1\over 2},{1\over 2}]$. Then we have

$$\begin{aligned} d_{rcb}(Tx, Ty) ={}& \begin{pmatrix} (1-a_{11})&a_{12} & \ldots &a_{1n}\cr a_{21}&(1-a_{22})& \ldots &a_{2n}\cr\vdots&\vdots&&\vdots\cr a_{n1}&a_{n2}& \ldots &(1-a_{nn}) \end{pmatrix} \\ &\times \begin{pmatrix} x_{1}-y_{1}&x_{1}-y_{1} & \ldots &x_{1}-y_{1}\cr x_{2}-y_{2}&x_{2}-y_{2}& \ldots &x_{2}-y_{2}\cr\vdots&\vdots&&\vdots\cr x_{n}-y_{n}&x_{n}-y_{n}& \ldots &x_{n}-y_{n} \end{pmatrix} . \end{aligned} $$

Thus in both cases we have $d(Tx, Ty) \le\gamma.d(x,y)$, where

$$\gamma= \begin{pmatrix} (1-a_{11})&a_{12} & \ldots &a_{1n}\cr a_{21}&(1-a_{22})& \ldots &a_{2n}\cr\vdots&\vdots&&\vdots\cr a_{n1}&a_{n2}& \ldots &(1-a_{nn}) \end{pmatrix} $$

and $r(\gamma) \le\|\gamma\| = \sum_{i=1, i\neq j}^{n}|a_{ij}| + |1-a_{jj}| < 1$. Thus T satisfies (3.1) and so by Theorem 3.5 the system of linear equations (3.21) has a unique solution. □

Theorem 3.15

If $\sum_{j=1, j\neq i}^{n}|a_{ij}| + |1-a_{ii}| < 1$, then the conclusion of Theorem 3.14 still holds.

Proof

Let $\mathcal{A}$ and P be as in the proof of Theorem 3.14 and $\|a\| = \sum_{j=1}^{n} |a_{ij}|$. Let $\chi= \mathbb {C^{n}}$. Let $d_{rcb}:\chi\times\chi\rightarrow\mathcal{A}$ be given by

$$d_{rcb}(x,y)=\left \{ \textstyle\begin{array}{l@{\quad}l} \left ({\scriptsize\begin{matrix}{}|x_{1}-y_{1}|&|x_{1}-y_{1}| & \ldots &|x_{1}-y_{1}|\cr |x_{2}-y_{2}|&|x_{2}-y_{2}|& \ldots &|x_{2}-y_{2}|\cr \vdots&\vdots&&\vdots\cr| x_{n}-y_{n}|&|x_{n}-y_{n}|& \ldots &|x_{n}-y_{n}|\end{matrix}} \right) &\mbox{$\forall x,y \in X$.} \end{array}\displaystyle \right . $$

Then $(\chi,d_{rcb})$ is a RCbMS-BA over $\mathcal{A}$ with $s=1$. Define the self map T of χ by

$$Tx = (I-A)x + B \quad\mbox{for all } x \in X, $$

where $A_{n\times n}$, $X_{n\times1}$ and $B_{n\times1}$ are the coefficient matrices of (3.21).

Then the system of linear equations (3.13) is the problem $x = Tx$. We will show that T satisfies (3.1). Let $x,y \in\chi$. Then

$$d_{rcb}(Tx, Ty) = \begin{pmatrix} (1-a_{11})&a_{12} & \ldots &a_{1n}\cr a_{21}&(1-a_{22})& \ldots &a_{2n}\cr\vdots&\vdots&&\vdots\cr a_{n1}&a_{n2}& \ldots &(1-a_{nn}) \end{pmatrix} d_{rcb}(x,y) = \gamma.d_{rcb}(x,y) $$

with $\gamma= I - A$ and $r(\gamma) \le\|\gamma\| = \sum_{i=1, i\neq j}^{n}|a_{ij}| + |1-a_{jj}| < 1$. Thus T satisfies (3.1) and so by Theorem 3.5 the system of linear equations (3.21) has a unique solution. □