Abstract.

The aim of this paper is to obtain fixed point results for interpolative Kannan type contraction mappings. The purpose of this paper is two fold: one relates to the fixed point results for multivalued cyclic interpolative Kannan type contractions, and second deals with proving the Perov fixed point result for interpolative Kannan type contraction mappings in the framework of vector valued metric spaces.

1. Introduction

The Banach Contraction Principle [2] holds a significant importance across various mathematical disciplines. Many authors have made generalizations based on different metric structures and types of contractions. For example, Perov [10], [11] extended the result to vector-valued metric spaces whearas Nadler [9] extended it to multivalued mappings. Furthermore, Karapınar [5] proved a fixed point result for interpolative Kannan-type contractions which was later extended by Gaba et al., [4], Errai et al., [3], and Safeer et al. [6]. Moreover, Kirk et al., [7] established the fixed point results for cyclic type contractions.

The aim of this paper is to present (a) some fixed point results for multivalued cyclic interpolative Kannan type contraction mappings and (b) a fixed point results for Perov interpolative Kannan type contractions in the framework of vector-valued metric spaces.

Let $X$ be a nonempty set equipped with a metric $d$ and $CB(X)$ the family of all nonempty closed and bounded subsets of $X.$ For any $A,B\in CB(X),$ define $H:CB(X)\times CB(X)\to R$ as follows:

$$H(A,B)=\max \{\delta (A,B),\delta (B,A)\},$$

where $\delta (A,B)={sup}_{a\in A}d(a,B)$ and $d(a,B)={inf}_{b\in B}d(a,b).$ Note that, $H$ is a metric on $CB(X)$, called the Pompeiu Hausdorff metric induced by $d.$ Moreover, a point $x\in X$ is said to be a fixed point of a multivalued mapping $T:X\to CB(X)$ if $x\in Tx.$ Equivalently, a point $x\in X$ is fixed point if and only if $d(x,Tx)=0.$

Lemma 1.1.

Let $A,B\in CB(X).$ Then for any $q>1$ and $x\in A,$ there exist $y\in B$ such that

$$d(x,y)\le qH(A,B).$$

Definition 1.2 ([8]).

A map $T:X\to CB(X)$ is called multivalued interpolative Kannan type contraction if there exist $\lambda \in [0,1)$ and $\alpha \in (0,1)$ such that

$$H(Tx,Ty)\le \lambda d{(x,Tx)}^{\alpha }d{(y,Ty)}^{1-\alpha },$$

holds for all $x,y\in X\setminus Fix\{T\}.$

Theorem 1.3 ([8]).

If $(X,d)$ is a complete metric space and $T:X\to CB(X)$ is a multivalued interpolative Kannan type contraction. Then $Fix\{T\}\ne \varphi$ provided that $Tx$ is compact for each $x\in X.$

Let us recall the following definition.
Denote $\mathrm{𝟎}:=(0,0,0,\mathrm{\cdots },0)\in {ℝ}^n.$ Define a partial order on ${ℝ}^n$ as follows:
for any $x,y\in {ℝ}^n$, we say $x\le y$ if and only if $x_i\le y_i$ for all $1\le i\le n,$ where $x=(x_1,x_2,x_3,\mathrm{\cdots },x_n)$ and $y=(y_1,y_2,y_3,\mathrm{\cdots }{y}_n).$

Definition 1.4.

Let $X$ be a nonempty set. A mapping $D:X\times X\to {ℝ}^n$ is called the vector valued metric on $X$ if the followings conditions are satisfied:
$(D_1)$ $\mathrm{𝟎}\le D(x,y)$ for all $x,y\in X$ and $D(x,y)=\mathrm{𝟎}$ if and only if $x=y.$
$(D_2)$ $D(x,y)=D(y,x)$ for all $x,y\in X.$
$(D_3)$ $D(x,y)\le D(x,z)+D(z,y),$ for all $x,y,z\in X$
and $(X,D)$ is called a vector valued metric space,

We denote, $M_n({ℝ}^{+})$, ${\theta }_n$, and $I_n$ by the sets of $n\times n$ matrices with nonnegative real elements, the $n\times n$ zero matrix, and the $n\times n$ identity matrix, respectively. For any $A\in M_n(ℝ),$ the spectral radius of $A$, denoted by $\rho (A)$, is defined as

$$\rho (A)=\max \{|\mu |:\mu \text{ is an eigenvalue of }A\}.$$

Moreover, a matrix $A\in M_n(ℝ)$ is said to converge to ${\theta }_n$ if and only if its spectral radius $\rho (A)$ is strictly less than one, that is, (see [12]). Also, a matrix $A\in M_n(ℝ)$ converges to zero if $A^n\to {\theta }_n$ as $n\to \mathrm{\infty }$. Furthermore, if $A$ and $B$ are matrices in $M_n({ℝ}^{+})$ with $A\le B$ (in the component-wise sense), then implies that .

Lemma 1.5 ([1]).

If $A$ is any matrix with spectral radius Then, $(I-A)$ is nonsingular and its inverse is given by

$${(I-A)}^{-1}=I+A+A^2+A^3+\mathrm{\dots }=\sum _{i=0}^{\mathrm{\infty }}{A}^i.$$

The following result generalizes the Banach contraction principle for vector valued metric spaces ( [10], [11]).

Theorem 1.6.

Let $(X,D)$ be a complete vector-valued metric space and $T:X\to X$. If there exists a matrix $A\in M_n(R^{+})$ which converges to ${\theta }_n$ and for all $x,y\in X,$ we have

$$D(Tx,Ty)\le AD(x,y),$$

then $T$ possess a unique fixed point.

1.1. Multivalued cyclic interpolative Kannan type contraction

Let $A$ and $B$ be nonempty subsets of a metric space $(X,d)$. A mapping $T:A\cup B\to A\cup B$ is said to be cyclic if $T(A)\subset B$ and $T(B)\subset A$.
In 2003, Kirk et al. [7] introduced the concept of cyclic contraction and proved the fixed point result for such mappings.
Let $(X,d)$ be a complete metric space and $\{A_1,A_2,\mathrm{\dots },A_p\}$ a collection of nonempty closed subsets of $X$ with ${\cap }_{i=1}^p{A}_i\ne \varphi$ and $Y={\cup }_{i=1}^p{A}_i.$
Suppose that $T:Y\to CB(X)$ is multivalued. For any set $A$ in $Y,$ define $TA=\cup \{Tx:x\in A\}.$

Definition 1.7.

A mapping $T:Y\to CB(X)$ is called multivalued cyclic interpolative Kannan type contraction if
(i) $T{A}_i\subseteq A_{i+1},$ for all $1\le i\le p$ where $A_{p+1}=A_1$ and $T{A}_p\subseteq A_1,$
(ii) For any $(x,y)\in A_i\times A_{i+1},i=1,2,\mathrm{\dots }p$ and $A_{p+1}=A_1$ with $H(Tx,Ty)>0,$ there exist $\lambda \in (0,1)$ such that

$$H(Tx,Ty)\le \lambda d{(x,Tx)}^{\alpha }d{(y,Ty)}^{1-\alpha },$$

holds with $x,y\notin {\cap }_{i=1}^p{A}_i$ and $d(x,Tx)\ne 0,d(y,Ty)\ne 0.$

Theorem 1.8.

Let $X$ be a complete metric space. If $T:Y\to CB(X)$ is a multivalued cyclic interpolative Kannan type contraction. Then $T$ has a fixed point in ${\cap }_{i=1}^p{A}_i.$

Proof 1.9.

Let $x_1\in A_1\subseteq X$ and $x_2\in T{x}_1\subseteq A_2.$ Then, for any $q=\frac{1}{\sqrt{\lambda }}>1,$ it follows from the Lemma 1.1 that there exist $x_3\in T{x}_2\subseteq A_3$ such that

$$d(x_2,x_3)\le qH(T{x}_1,T{x}_2).$$

Also, for $x_3\in T{x}_2\subseteq A_3,$ there exist $x_4\in T{x}_3\subseteq A_4,$ such that

$$d(x_3,x_4)\le qH(T{x}_2,T{x}_3),$$

holds with $q>1.$ Continuing the same way, we obtain that $x_{p-1}\in T{x}_{p-2}\subseteq A_{p-1}.$ For $q>1.$ By Lemma 1.1, there exist $x_p\in T{x}_{p-1}\subseteq A_p$ such that

$$d(x_{p-1},x_p)\le qH(T{x}_{p-2},T{x}_{p-2}).$$

Moreover, for $x_p\in T{x}_{p-1}\subseteq A_p$ there exist $x_{p+1}\in T{x}_p\subseteq A_1,$ such that

$$d(x_p,x_{p+1})\le qH(T{x}_{p-1},T{x}_p).$$

Following arguments similar to those given above, we can construct a sequence $\{x_{mp+i}\}$ in $X$ where $1\le i\le p,m\in N$ with $x_{mp+i}\in A_i.$ Thus for all $m\in N$ and $1\le i\le p,(x_{mp+i})=(x_n)$ is a sequence in ${\cup }_{i=1}^n{A}_i$ such that for any $n\in N$ and $r=\sqrt{\lambda },$ we have

$d(x_n,x_{n+1})\le$ $qH(T{x}_{n-1},T{x}_n)\le \sqrt{\lambda }d{(x_{n-1},T{x}_{n-1})}^{\alpha }d{(x_n,T{x}_n)}^{1-\alpha }$

$d(x_n,x_{n+1})\le$ $rd{(x_{n-1},x_n)}^{\alpha }d{(x_n,x_{n+1})}^{1-\alpha }$

$d(x_n,x_{n+1})\le$ $r^{1/\alpha }d(x_{n-1},x_n)\le rd(x_{n-1},x_n).$

Thus,

$$d(x_n,x_{n+1})\le rd(x_{n-1},x_n)\le r^{2}d(x_{n-2},x_{n-1})\mathrm{\cdots }\le r^{n-1}d(x_1,x_2).$$

Consequently, for any $n\in N,$ we have

$$d(x_n,x_{n+1})\le r^{n-1}d(x_1,x_2).$$

Hence for any $m,n\in N$ with $m\ge n,$ by triangle inequality we obtain

$d(x_n,x_m)$ $\le d(x_n,x_{n+1})+d(x_{n+1},x_{n+2})+\mathrm{\cdots }+d(x_{m-1},x_m)$

$\le r^{n-1}d(x_1,x_2)+r^{n-2}d(x_1,x_2)+\mathrm{\cdots }+r^{m-2}d(x_1,x_2)$

$=(r^{n-1}+r^{n-2}+\mathrm{\cdots }+r^{m-2})d(x_1,x_2)$

$=({\displaystyle \sum _{i=n-1}^{m-2}}{r}^i)d(x_1,x_2).$

On taking limit as $n,m\to \mathrm{\infty }$ we get $d(x_n,x_m)\to 0.$ Hence $(x_n)$ is a Cauchy sequence in $X.$ As $X$ is complete, it converges to a point $x\in X,$ that is, $x_n\to x.$ It follows from construction of $(x_n)$ that $x_n=x_{mp+i},$ where $n=mp+i,$ with $m\in N$ and $i=1,2,\mathrm{\dots }p.$ Thus $(x_{mp+i})$ is a sequence in $A_i,$ for each fixed $1\le i\le p,$ and all these sequences $(x_{mp+i})\subseteq A_i,1\le i\le p$ are subsequences of $(x_n),$ and converges to $x.$ Since each $A_i$ is closed so for each fixed $1\le i\le p,$ $x_{mp+i}\to x\in A_i.$ Thus $x\in {\cap }_{i=1}^p{A}_i.$ Now, we show that $x$ is the fixed point of $T,$ that is, $d(x,Tx)=0.$ Since for any fixed $1\le i\le p,$, we have

$0\le d(x_{mp+i+1},Tx)\le$ $H(T{x}_{mp+i},Tx)$

$\le$ $\lambda d{(x_{mp+i},T{x}_{mp+i})}^{\alpha }d{(x,Tx)}^{1-\alpha }$

$\le$ $\lambda d{(x_{mp+i},x_{mp+i+1})}^{\alpha }d{(x,Tx)}^{1-\alpha }.$

On taking limit as $m\to \mathrm{\infty },$ we have

$0\le$ $\underset{m\to \mathrm{\infty }}{lim}d(x_{mp+i+1},Tx)\le \underset{n\to \mathrm{\infty }}{lim}\lambda d{(x_{mp+i},x_{mp+i+1})}^{\alpha }d{(x,Tx)}^{\alpha }$

$0\le$ $\underset{m\to \mathrm{\infty }}{lim}d(x_{mp+i+1},Tx)\le 0.$

Hence ${lim}_{m\to \mathrm{\infty }}d(x_{mp+i+1},Tx)=d(x,Tx)=0.$ Thus $x\in Tx.$

Definition 1.10.

A mapping $T:Y\to CB(X)$ is called a multivalued cyclic -interpolative Kannan type contraction if
(i) $T{A}_i\subseteq A_{i+1},$ for all $1\le i\le p$ where $A_{p+1}=A_1$ and $T{A}_p\subseteq A_1.$
(ii) For any $(x,y)\in A_i\times A_{i+1},i=1,2,\mathrm{\dots }p$ and $A_{p+1}=A_1$ with $H(Tx,Ty)>0,$ there exist $\lambda \in (0,1)$ such that

$$H(Tx,Ty)\le \lambda d{(x,Tx)}^{\alpha }d{(y,Ty)}^{\beta },$$

holds with $x,y\notin {\cap }_{i=1}^p{A}_i,x,y\notin Fix(T)$ and $d(x,Tx)\ge 1,d(y,Ty)\ne 0.$

Theorem 1.11.

Let $X$ be a complete metric space. If $T:Y\to CB(X)$ is a multivalued cyclic -interpolative Kannan type contraction. Then $T$ has a fixed point in ${\cap }_{i=1}^p{A}_i.$

Proof 1.12.

Take $x_1\in A_1.$ Following arguments similar to those given in Theorem 1.8, we construct a sequence $\{x_n\}$ in $X$ satisfying:

$$d(x_n,x_{n+1})\le qH(T{x}_n,T{x}_{n-1}),\forall n\in N.$$

From Definition 1.10, we have

$$d(x_n,x_{n+1})\le rH(T{x}_n,T{x}_{n-1})\le \sqrt{\lambda }d{(x_{n-1},T{x}_{n-1})}^{\alpha }d{(x_x,T{x}_n)}^{\beta },$$

that is,

$$d{(x_n,x_{n+1})}^{1-\beta }\le \sqrt{\lambda }d{(x_{n-1},T{x}_{n-1})}^{\alpha },$$

because $x_{n+1}\in T{x}_n.$ Moreover

$$d(x_n,x_{n+1})\le r^{\frac{1}{1-\beta }}d{(x_{n-1},T{x}_{n-1})}^{\frac{\alpha }{1-\beta }}.$$

Indeed, Thus

$$d(x_n,x_{n+1})\le rd(x_{n-1},T{x}_{n-1})\le rd(x_{n-1},x_n).$$

Adopting the similar procedure as in the proof of Theorem 1.8, we can prove that $(x_n)$ is a Cauchy sequence. Furthermore, using the completeness of $X,$ $(x_n)$ converges to fixed point $x\in {\cap }_{i=1}^p{A}_i$ of $T.$

Definition 1.13.

A mapping $T:Y\to CB(X)$ is called a multivalued cyclic $(\lambda ,\alpha +\beta >1)$-interpolative Kannan type contraction if
(i) $T{A}_i\subseteq A_{i+1},$ for all $1\le i\le p$ where $A_{p+1}=A_1$ and $T{A}_p\subseteq A_1.$
(ii) For any $(x,y)\in A_i\times A_{i+1},i=1,2,\mathrm{\dots }p$ and $A_{p+1}=A_1$ with $H(Tx,Ty)>0,$ there exist $\lambda \in (0,1)$ such that

$$H(Tx,Ty)\le \lambda d{(x,Tx)}^{\alpha }d{(y,Ty)}^{\beta },$$

is satisfied for all $x,y\notin {\cap }_{i=1}^p{A}_i,x,y\notin Fix(T)$ and $d(x,Tx)\ne 0,d(y,Ty)\ne 0.$

Theorem 1.14.

Let $X$ be a complete metric space, and $T:Y\to CB(X)$ a multivalued cyclic $(\lambda ,\alpha +\beta >1)$-interpolative Kannan type contraction. If there exists $x\in X$ such that $d(x,Tx)\le 1,$ then $T$ has a fixed point in ${\cap }_{i=1}^p{A}_i.$

Proof 1.15.

Let $x_1\in X$ such that $d(x_1,T{x}_1)\le 1.$ Without any loss of generality we assume that $x_1\in A_1\subseteq X.$ Consider $x_2\in T{x}_1\subseteq A_2$ and for $q=\frac{1}{\sqrt{\lambda }}>1.$ By Lemma 1.1, there exist $x_3\in T{x}_2\subseteq A_3$ such that

$$d(x_2,x_3)\le qH(T{x}_1,T{x}_2).$$

Also by Definition 1.13, we have

$$d(x_2,x_3)\le q\lambda d{(x_1,T{x}_1)}^{\alpha }d{(x_2,T{x}_2)}^{\beta }.$$

As $x_3\in T{x}_2$ , $d(x_2,T{x}_2)\le d(x_2,x_3).$ Thus

$$d(x_2,x_3)\le \sqrt{\lambda }d{(x_1,T{x}_1)}^{\alpha }d{(x_2,x_3)}^{\beta },$$

$$d{(x_2,x_3)}^{1-\beta }\le rd{(x_1,T{x}_1)}^{\alpha },$$

that is,

$$d(x_2,x_3)\le r^{\frac{1}{1-\beta }}d{(x_1,T{x}_1)}^{\frac{\alpha }{1-\beta }}.$$

Since $\frac{\alpha }{1-\beta }>1$ and $d(x_1,T{x}_1)\le 1,$ we have $d{(x_1,T{x}_1)}^{\frac{\alpha }{1-\beta }}\le 1$ and $r^{\frac{1}{1-\beta }}\le r.$ Thus

$$d(x_2,x_3)\le r.$$

Continuing the same way, we can construct a sequence by using Lemma 1.1 and Definition 1.13 such that

$$d(x_n,x_{n+1})\le r^n$$

for all $n\in N.$ Note that,

$$d(x_n,x_m)\le r^n+r^{n+1}+r^{n+2}+\mathrm{\dots },+r^{m-1}=\sum _{i=n}^{m-1}{r}^i.$$

Hence $d(x_n,x_m)\to 0$ as $n,m\to \mathrm{\infty }.$ Thus $(x_n)$ is a Cauchy sequence in $X,$ and completeness of $X$ yields that $(x_n)$ converges to some $x\in X.$ Further, due to the cyclic nature of the mapping, the sequence $(x_n)$ is composed of sub-sequences $x_i\in A_i,mp\le i\le (m+1)p\forall m\in N.$ Thus the sub-sequences $\{x_{mp+i}\}\in A_i\mathrm{\hspace{0.17em}\hspace{0.17em}\hspace{0.17em}1}\le i\le p,m\in N.$ Moreover, the convergence of $(x_n)$ assures that each of the sub-sequence also converges to the same limit $x,$ such that $x_{mp+i}\to x,\forall \mathrm{\hspace{0.17em}\hspace{0.17em}1}\le i\le p.$ Hence $x\in A_i$ for all $1\le i\le p$ because each $A_i$ is closed. Thus $x\in {\cap }_{i=1}^p{A}_i$. Now, we show that $x$ is the fixed point of $T.$ Note that,

$0$ $\le d(x_{n+1},Tx)\le H(T{x}_n,Tx)$

$\le \lambda d{(x_n,T{x}_n)}^{\alpha }d{(x,Tx)}^{\beta }$

$\le \lambda d{(x_n,x_{n+1})}^{\alpha }d{(x,Tx)}^{\beta }$

$\le \lambda r^{n\alpha }d{(x,Tx)}^{\beta }.$

On taking limit as $n\to \mathrm{\infty },$ we obtain that

$$0\le \underset{n\to \mathrm{\infty }}{lim}d(x_{n+1},Tx)\le 0,$$

and

$$\underset{n\to \mathrm{\infty }}{lim}d(x_{n+1},Tx)=d(x,Tx)=0.$$

Thus $x\in Tx,$ that is, $x\in {\cap }_{i=1}^p{A}_i$ is the fixed point of $T.$

Remark 1.16.

In Theorem 1.11, we have which implies that

$$H(Tx,Ty)\le \lambda d{(x,Tx)}^{\alpha }d{(y,Ty)}^{\beta }\le \lambda d{(x,Tx)}^{\alpha }d{(y,Ty)}^{1-\alpha },$$

since $d(y,Ty)\ge 1$. Thus, Theorem 1.11 is satisfied for $\alpha +\beta \le 1.$ However, it does not cover Theorem 1.8 which is more generally applicable without the restriction $d(x,Tx)\ge 1$. In Theorem 1.14 with $\alpha +\beta >1$ implying $\alpha >1-\beta$, we have

$$H(Tx,Ty)\le \lambda d{(x,Tx)}^{\alpha }d{(y,Ty)}^{\beta }\le \lambda d{(x,Tx)}^{1-\beta }d{(y,Ty)}^{\beta },$$

because $d(x,Tx)\le 1$. Thus, Theorem 1.14 is satisfied for $\alpha +\beta \ge 1$ but does not cover Theorem 1.8 which does not restrict with $d(x,Tx)\le 1$.

1.2. Perov intertopaltive Kannan type contraction

Definition 1.17.

Let $(X,D)$ be a complete vector valued metric space with $D:X\times X\to {ℝ}^n$. A self mapping $T:X\to X$ is said to be Perov interpolative Kannan type contraction if there exist a matrix $A\in M_n({ℝ}^{+})$ which converges to zero or having spectral radius and $\alpha \in (0,1)$ such that

$$D(Tx,Ty)\le AD{(x,Tx)}^{\alpha }D{(y,Ty)}^{1-\alpha },$$

is satisfied for all $x,y\in X$ with $x,y\notin Fix(T).$

Theorem 1.18.

Let $T:X\to X$ be Perov interpolative Kannan type contraction. Then $T$ has a fixed point.

Proof 1.19.

Let $x_0\in X,$ we construct a sequence $(x_n)$ as follows:
$x_{n+1}=T(x_n)=T^n(x_0)$ for all positive integers $n.$ Without any loss of generality, we assume that $x_n\ne x_{n+1}$ for each nonnegative integer $n.$ Thus, we have

$$D(x_{n+1},x_{n+2})=D(T{x}_n,T{x}_{n+1})\le AD{(x_n,T{x}_n)}^{\alpha }D{(x_{n+1},T{x}_{n+1})}^{1-\alpha }.$$

This yields

$$D{(x_{n+1},x_{n+2})}^{\alpha }\le AD{(x_n,T{x}_n)}^{\alpha }$$

$$D(x_{n+1},x_{n+2})\le A^{1/\alpha }D(x_n,x_{n+1}).$$

Take $A^{1/\alpha }=B.$ Hence

$$D(x_{n+1},x_{n+2})\le BD(x_n,x_{n+1}).$$ (1)

In a similar fashion, we can write

$$D(x_n,x_{n+1})=D(T{x}_{n-1},T{x}_n)\le AD{(x_{n-1},T{x}_{n-1})}^{\alpha }D{(x_n,T{x}_n)}^{1-\alpha }$$

$$D(x_n,x_{n+1})\le A^{1/\alpha }D(x_{n-1},x_n)$$

$$D(x_n,x_{n+1})\le BD(x_{n-1},x_n).$$ (2)

By combining (1) and (2), we obtain

$$D(x_n,x_{n+1})\le B^{2}D(x_{n-1},x_n).$$ (3)

Following the arguments similar to those given above, we get the following relation for all $n\in N$

$$D(x_n,x_{n+1})\le B^{n}D(x_0,x_1).$$ (4)

Note that,

$D(x_m,x_{m+p})$ $\le D(x_m,x_{m+1})+D(x_{m+1},x_{m+2})+\mathrm{\cdots },D(x_{m+p-1},x_{m+p})$

$\le (B^m+B^{m+1}+B^{m+2}+,\mathrm{\cdots },B^{m+p-1})D(x_0,x_1)$

$\le (B^m+B^{m+1}+B^{m+2}+\mathrm{\cdots })D(x_0,x_1)$

$=B^m(I+B+B^2+B^3+\mathrm{\cdots })D(x_0,x_1).$

Since $B$ is convergent to zero, matrix $(I-B)$ is non-singular and

$${(I-B)}^{-1}=I+B+B^2+\mathrm{\cdots }.$$

Therefore,

$$D(x_m,x_{m+p})\le B^m{(I-B)}^{-1}D(x_0,x_1),$$

As $B^m\to {\theta }_n$ on taking limit as $m\to \mathrm{\infty }.$ So $D(x_m,x_{m+p})\to \mathrm{𝟎}$ as $m\to \mathrm{\infty }.$ Hence, the sequence $(x_m)$ is a fundamental (Cauchy), and using the completeness of the space $(X,D),$ there exist $z\in X$ such that $D(x_n,z)\to \mathrm{𝟎}$ as $n\to \mathrm{\infty },$ that is,

$$\underset{n\to \mathrm{\infty }}{lim}D(x_n,z)=\mathrm{𝟎}.$$ (5)

We now show that $z$ is a fixed point of $T.$ Note that,

$\mathrm{𝟎}$ $\le D(z,Tz)=\underset{n\to \mathrm{\infty }}{lim}D(x_n,Tz)$

$=\underset{n\to \mathrm{\infty }}{lim}D(T{x}_{n-1},Tz)$

$\le A\underset{n\to \mathrm{\infty }}{lim}D{(x_{m-1},T{x}_{m-1})}^{\alpha }D{(z,Tz)}^{1-\alpha }=\mathrm{𝟎}.$

Hence

$$D(z,Tz)=\mathrm{𝟎}⟹z=Tz.$$

Example 1.20.

Let $X=[0,\mathrm{\infty })$, and $D:X\times X\to R^2$ a metric defined as:

Suppose that $T:X\to X$ is defined as follows:

Take, Clearly, $\rho (A)=1/2.$ We now discuss the following cases.
Case I: If $x,y\in [0,4],$ then

$$D(Tx,Ty)=\left(\begin{array}{c}d(Tx,Ty)\\ d(Tx,Ty)\end{array}\right)=\left(\begin{array}{c}d(1,1)\\ d(1,1)\end{array}\right)=\left(\begin{array}{c}0\\ 0\end{array}\right).$$

Also

$AD{(x,Tx)}^{1/2}D{(y,Ty)}^{1/2}$

$=\left(\begin{array}{c}3/4{(x+1)}^{1/2}{(y+1)}^{1/2}\\ 1/8{(x+1)}^{1/2}{(y+1)}^{1/2}\end{array}\right)\ge \left(\begin{array}{c}0\\ 0\end{array}\right).$

Case II: If $x\in [0,4],$ and $y\in (4,\mathrm{\infty }),$ then

$$D(Tx,Ty)=\left(\begin{array}{c}d(Tx,Ty)\\ d(Tx,Ty)\end{array}\right)=\left(\begin{array}{c}d(1,e^{-2y})\\ d(1,e^{-2y})\end{array}\right)=\left(\begin{array}{c}1+e^{-2y}\\ 1+e^{-2y}\end{array}\right)\le \left(\begin{array}{c}1+e^{-8}\\ 1+e^{-8}\end{array}\right).$$

Also

$AD{(x,Tx)}^{1/2}D{(y,Ty)}^{1/2}$

$=\left(\begin{array}{c}(3/4){(x+1)}^{1/2}{(y+e^{-2y})}^{1/2}\\ (1/8){(x+1)}^{1/2}{(y+e^{-2y})}^{1/2}\end{array}\right)$

$\ge \left(\begin{array}{c}(3/4){(4+e^{-8})}^{1/2}\\ (1/8){(4+e^{-8})}^{1/2}\end{array}\right)\ge \left(\begin{array}{c}1+e^{-8}\\ 1+e^{-8}\end{array}\right).$

Similarly, it holds for the case $x\in [0,4]$ and $y\in (4,\mathrm{\infty })$.
Case III: If $x,y\in (4,\mathrm{\infty }),$ then

$$D(Tx,Ty)=\left(\begin{array}{c}d(Tx,Ty)\\ d(Tx,Ty)\end{array}\right)=\left(\begin{array}{c}d(e^{-2x},e^{-2y})\\ d(e^{-2x},e^{-2y})\end{array}\right)=\left(\begin{array}{c}{e}^{-2x}+e^{-2y}\\ e^{-2x}+e^{-2y}\end{array}\right)\le \left(\begin{array}{c}2e^{-8}\\ 2e^{-8}\end{array}\right).$$

Also

$AD{(x,Tx)}^{1/2}D{(y,Ty)}^{1/2}$

$=\left(\begin{array}{c}(3/4){(x+e^{-2x})}^{1/2}{(y+e^{-2y})}^{1/2}\\ (1/8){(x+e^{-2x})}^{1/2}{(y+e^{-2y})}^{1/2}\end{array}\right)$

$\ge \left(\begin{array}{c}(3/4)(4+e^{-8})\\ (1/8)(4+e^{-8})\end{array}\right)\ge \left(\begin{array}{c}2e^{-8}\\ 2e^{-8}\end{array}\right).$

Thus, in all the cases the required interpolative condition holds. Moreover, $x=1$ is a fixed point of $T.$

Conclusion

If we set $A_i=X$ for all $1\le i\le p$ and consider the collection of compact subsets, then  the Theorem 1 of [8] becomes a special case of Theorem 1.8. Furthermore, if we take $n=1$ in Theorem 1.18, then the result of [5] becomes a special case of our Theorem .