Banach Fixed Point Theorems in Generalized Metric Space Endowed with the Hadamard Product

Abstract

In this paper, we prove some Banach fixed point theorems in generalized metric space where the contractive conditions are endowed with the Hadamard product of real symmetric positive definite matrices. Since the condition that a matrix A converges to zero is not needed, this produces stronger results than those of Perov. As an application of our results, we study the existence and uniqueness of the solution for a system of matrix equations.

1. Introduction

The Banach contraction principal was introduced in 1922 by S. Banach [1], and it plays a vital role in mathematics and has many other applications in different branches of science.

The concept of metric space was generalized by Perov in [2], where the set of real numbers was replaced by vector-valued ${\mathbb{R}}^n$. Hence, classical principal contraction mapping was extended for contraction mapping on a vector-valued metric space called the generalized metric space.

In this context, some interesting results have been introduced by many authors (see [3,4,5,6,7,8,9]). The purpose of this work is to present some versions of Banach fixed point theorems in generalized metric space endowed with the Hadamard product. Our concept can be used to solve a system of symmetric positive definite matrix equations. The results using the Hadamard product of a vector-valued metric are strong, since the condition that A is a matrix converging to zero is not needed. In other words, instead of supposing that A is a convergent matrix $M_d\left(\mathbb{R}\right)$, we suppose that A is a matrix in ${\mathbb{R}}^d$ and we apply the Hadamard product to obtain more suitable contraction conditions.

2. Preliminaries

In this section, we recall some notations and auxiliary results that will be used throughout this paper. Let ${\mathbb{R}}^d$ be the set of all $\mathrm{d}\times 1$ real matrices, and if $x,y\in {\mathbb{R}}^d$, $x={(x_1,x_2,\cdots ,x_d)}^t$, $y={(y_1,y_2,\cdots ,y_d)}^t$ by $x\precsim y$ (respectively, $x\prec y$), and we mean $x_i\le y_i$ (respectively, $x_i<y_i$) for all $i=1,2,\cdots ,d$. In addition, ${\mathbb{R}}_{+}^d=\{x\in {\mathbb{R}}^d:x_i\ge 0,\forall i=1,2,\cdots ,d\}$ is the set of positive elements in ${\mathbb{R}}^d$, and we denote $x\succsim \tilde{0}$ if $x_i\ge 0$ for all $i=1,2,\cdots ,d$, where $\tilde{0}$ is the zero element in ${\mathbb{R}}^d$.

By a Hadamard product $x\odot y$ of two $\mathrm{d}\times 1$ symmetric positive definite matrices, we mean the entrywise product. Thus,

$$x\odot y=\left(\begin{array}{c}{x}_1\\ x_2\\ ⋮\\ x_d\end{array}\right)\odot \left(\begin{array}{c}{y}_1\\ y_2\\ ⋮\\ y_d\end{array}\right)=\left(\begin{array}{c}{x}_1.y_1\\ x_2.y_2\\ ⋮\\ x_d.y_d\end{array}\right),$$

and the Hadamard product unit over ${\mathbb{R}}^d$ will be denoted by $I_H^{{\mathbb{R}}^d}=\left(\begin{array}{c}1\\ 1\\ ⋮\\ 1\end{array}\right)$. An element $x={(x_1,x_2,\cdots ,x_d)}^t$ has an inverse if $x_i\ne 0,\forall i=1,2,\cdots ,d$ and is denoted by $x^{-1}=\left(\begin{array}{c}\frac{1}{x_1}\\ \frac{1}{x_2}\\ ⋮\\ \frac{1}{x_d}\end{array}\right)\in {\mathbb{R}}^d$, and x is called invertible if it has an inverse.

Definition 1 

([2]). Let X be a non-empty set. A mapping $\mathrm{d}:X\to X\to {\mathbb{R}}^d$ is called a vector-valued metric on X or the generalized metric space if the following properties are satisfied:

1. 

$\mathrm{d}(x,y)\succsim \tilde{0}$, and $\mathrm{d}(x,y)=\tilde{0}$ if and only if $x=y$;

2. 

$\mathrm{d}(x,y)=\mathrm{d}(y,x)$, for all $x,y\in X$;

3. 

$\mathrm{d}(x,z)\precsim \mathrm{d}(x,y)+\mathrm{d}(y,z)$, for all $x,y,z\in X$

Then, we call $(X,{\mathbb{R}}^d,\mathrm{d})$ the generalized metric space on X.

Example 1. 

Let ${\mathrm{d}}_{\mathbb{R}}:X\times X\to \mathbb{R}$ be a metric on $\mathbb{R}$, and $(X,{\mathrm{d}}_{\mathbb{R}})$ be a metric space on $\mathbb{R}$. Define $\mathrm{d}:X\times X\to {\mathbb{R}}^d$ by

$$\begin{array}{c}\hfill \mathrm{d}(x,y)=\left(\begin{array}{c}{\mathrm{d}}_{\mathbb{R}}(x,y)\\ {\mathrm{d}}_{\mathbb{R}}(x,y)\\ ⋮\\ {\mathrm{d}}_{\mathbb{R}}(x,y)\end{array}\right)\in {\mathbb{R}}^d.\end{array}$$

Then, $(X,{\mathbb{R}}^d,\mathrm{d})$ is the generalized metric space on X.

Definition 2. 

Let $\left\{x_n\right\}$ be a sequence in X, and $x\in X$. If, for every $\widetilde{ϵ}\in {\mathbb{R}}^d$, with $\widetilde{ϵ}\succ \tilde{0}$, and there is $N\in \mathbb{N}$ such that for all $n>N$, $\mathrm{d}(x_n,x)\prec \widetilde{ϵ}$, then $\left\{x_n\right\}$ is called a convergent sequence with respect to ${\mathbb{R}}^d$ and $\left\{x_n\right\}$ converges to x. Moreover, if, for any $\widetilde{ϵ}\succ \tilde{0}$, there exists $N\in \mathbb{N}$ such that for all $n,m>N$, $\mathrm{d}(x_n,x_m)\prec \widetilde{ϵ}$, then $\left\{x_n\right\}$ is called a Cauchy sequence with respect to ${\mathbb{R}}^d$. We say that $(X,{\mathbb{R}}^d,\mathrm{d})$ is a complete generalized metric space if every Cauchy sequence with respect to ${\mathbb{R}}^d$ is convergent.

Example 2. 

Let X be a Banach space. Define $\mathrm{d}:X\times X\to {\mathbb{R}}^d$ by

$$\begin{array}{c}\hfill \mathrm{d}(x,y)=\left(\begin{array}{c}\parallel x-y\parallel \\ \parallel x-y\parallel \\ ⋮\\ \parallel x-y\parallel \end{array}\right)\in {\mathbb{R}}^d.\end{array}$$

Since X is complete, $(X,{\mathbb{R}}^d,\mathrm{d})$ is a complete generalized metric space.

Example 3. 

Let $X={\mathbb{R}}^d$ and $\mathrm{d}:{\mathbb{R}}^d\times {\mathbb{R}}^d\to {\mathbb{R}}^d$ defined by

$$\begin{array}{c}\hfill \mathrm{d}(x,y)=\left(\begin{array}{c}|x_1-y_1|\\ |x_2-y_2|\\ ⋮\\ |x_d-y_d|\end{array}\right)\in {\mathbb{R}}^d,\end{array}$$

where $x=\left(\begin{array}{c}{x}_1\\ x_2\\ ⋮\\ x_d\end{array}\right)$, $y=\left(\begin{array}{c}{y}_1\\ y_2\\ ⋮\\ y_d\end{array}\right)\in {\mathbb{R}}^d$. It is clear that $({\mathbb{R}}^d,{\mathbb{R}}^d,\mathrm{d})$ is a generalized metric space.

Theorem 1 

([2,4]). Let $(X,\mathrm{d})$ be a complete generalized metric space and the mapping $f:X\to X$ with the property that there exists a matrix $A\in M_{m,n}\left(\mathbb{R}\right)$ such that $\mathrm{d}\left(f\right(x),g(x\left)\right)\le A\mathrm{d}(x,y)$ for all $x,y\in X$. If A is a matrix convergent towards zero, then

1. 

$\mathrm{Fix}\left(f\right)=\left\{x^{*}\right\}$.

2. 

The sequence of successive approximations ${\left(x_n\right)}_{n\in \mathbb{N}},x_n=f^n\left(x_0\right)$ is convergent and it has the limit $x^{*}$, for all $x_0\in X$.

3. 

One has the following estimation:

$$\begin{array}{c}\hfill \mathrm{d}(x_n,x^{*})\le A^n{(I-A)}^{-1}\mathrm{d}(x_0,x_1)\end{array}$$

(1)

4. 

If $g:X\to X$ satisfies the condition $\mathrm{d}\left(f\right(x),g(x\left)\right)\le \eta$ for all $x\in X$, $\eta \in {\mathbb{R}}^m$ and considering the sequence $y_n=g^n\left(x_0\right)$, one has

$$\begin{array}{c}\hfill \mathrm{d}(y_n,x^{*})\le {(I-A)}^{-1}\eta +A^n{(I-A)}^{-1}\mathrm{d}(x_0,x_1).\end{array}$$

(2)

In the above theorem, to prove the Banach fixed point in the generalized metric space, it is necessary to choose a convergent matrix $A\in M_d\left(\mathbb{R}\right)$.

In the next section, we will not need such a condition because we will apply a contraction condition endowed with the Hadamard product.

3. Main Results

In this section, we will introduce some Banach fixed point theorems endowed with the Hadamard product.

Definition 3. 

Let $(X,{\mathbb{R}}^d,\mathrm{d})$ be a complete generalized metric space. A mapping $T:X\to X$ is said to be contractive and endowed with the Hadamard product if there exists a matrix $A\in {\mathbb{R}}_{+}^d$ with $\tilde{0}\precsim A\prec I_H^{{\mathbb{R}}^d}$ such that

$$\mathrm{d}(Tx,Ty)\precsim A\odot \mathrm{d}(x,y)\mathrm{for}\mathrm{all}x,y\in X$$

(3)

Example 4. 

Let $X=\mathbb{R}$ and $\mathrm{d}:X\times X\to {\mathbb{R}}^d$ be a generalized metric space defined by $\mathrm{d}(x,y)=\left(\begin{array}{c}|x-y|\\ |x-y|\\ ⋮\\ |x-y|\end{array}\right)\in {\mathbb{R}}^d$ and let $A=\left(\begin{array}{c}\frac{1}{2}\\ \frac{1}{2}\\ ⋮\\ \frac{1}{2}\end{array}\right)$ and $T:X\to X$ such that $Tx=\frac{x}{2}$. Then,

$$\mathrm{d}(Tx,Ty)=\frac{1}{2}\left(\begin{array}{c}|x-y|\\ |x-y|\\ ⋮\\ |x-y|\end{array}\right)\in {\mathbb{R}}^d.$$

It is clear that

$$\mathrm{d}(Tx,Ty)\precsim A\odot \mathrm{d}(x,y)=\left(\begin{array}{c}\frac{1}{2}\\ \frac{1}{2}\\ ⋮\\ \frac{1}{2}\end{array}\right)\odot \left(\begin{array}{c}|x-y|\\ |x-y|\\ ⋮\\ |x-y|\end{array}\right)=\left(\begin{array}{c}|x-y|\\ |x-y|\\ ⋮\\ |x-y|\end{array}\right).$$

Therefore, T is a Hadamard contractive mapping with respect to the generalized metric space $(\mathbb{R},{\mathbb{R}}^d,\mathrm{d})$.

Example 5. 

(Numerical example)

Let $X=\{2,4,6\}$ and $\mathrm{d}:X\times X\to {\mathbb{R}}^d$ defined by

$$\begin{array}{c}\hfill \mathrm{d}(x,y)=\left(\begin{array}{c}|x-y|\\ |x-y|\\ ⋮\\ |x-y|\end{array}\right)\in {\mathbb{R}}^d,\end{array}$$

and let $x=2$, $y=4$, $z=6$. Define $T:X\times X$, by $Tx=\frac{x}{3}$. Then,

$$\begin{array}{c}\hfill \mathrm{d}(Tx,Ty)=\left(\begin{array}{c}|Tx-Ty|\\ |Tx-Ty|\\ ⋮\\ |Tx-Ty|\end{array}\right)\in {\mathbb{R}}^d,\end{array}$$

so

$$\begin{array}{c}\hfill \mathrm{d}(Tx,Ty)=\left(\begin{array}{c}\frac{1}{3}|x-y|\\ \frac{1}{3}|x-y|\\ ⋮\\ \frac{1}{3}|x-y|\end{array}\right)=\left(\begin{array}{c}\frac{1}{3}\\ \frac{1}{3}\\ ⋮\\ \frac{1}{3}\end{array}\right)\odot \left(\begin{array}{c}|x-y|\\ |x-y|\\ ⋮\\ |x-y|\end{array}\right)=A\odot \mathrm{d}(x,y),\end{array}$$

where $A=\left(\begin{array}{c}\frac{1}{3}\\ \frac{1}{3}\\ ⋮\\ \frac{1}{3}\end{array}\right)\prec \left(\begin{array}{c}1\\ 1\\ ⋮\\ 1\end{array}\right)=I_H^{{\mathbb{R}}^d}$. Therefore, T is a Hadamard contractive mapping with respect to the generalized metric space $(X,{\mathbb{R}}^d,\mathrm{d})$.

Lemma 1. 

Let $A,B,C\in {\mathbb{R}}^d$; then,

1. 

$(I_H^{{\mathbb{R}}^d}-A)$ has an inverse ${(I_H^{{\mathbb{R}}^d}-A)}^{-1}=I_H^{{\mathbb{R}}^d}+A+A^{\odot 2}+A^{\odot 3}+\cdots$;

2. 

$A\odot B=B\odot A$;

3. 

If A and B are positive matrices, then $A\odot B$ is a positive matrix;

4. 

$C\odot (A+B)=C\odot A+C\odot B$;

5. 

$C\odot (A\odot B)=(C\odot A)\odot B$;

6. 

If $A\precsim B$, then $A\odot C\precsim B\odot C$.

Proof. 

We use the basic properties of the Hadamard product. □

Next, we will prove the principle of the Banach fixed point type endowed with the Hadamard product [1].

Theorem 2. 

Let $(X,{\mathbb{R}}^d,\mathrm{d})$ be a complete generalized metric space and $T:X\to X$ be a contractive mapping with respect to the Hadamard product, such that

$$\mathrm{d}(Tx,Ty)\precsim A\odot \mathrm{d}(x,y),$$

(4)

with

$$\tilde{0}\precsim A={({\lambda }_1,\cdots ,{\lambda }_d)}^t\prec I_H^{{\mathbb{R}}^d}0\le {\lambda }_i<1\mathrm{for}i=1,2,\cdots ,d.$$

Then, T has a unique fixed point.

Proof. 

Let $x_0$ be an arbitrary element in X. Define a sequence $\left\{x_n\right\}\subset X$ such that $T{x}_n=x_{n+1}$ for all $n=0,1,2,\cdots$.

$$\begin{array}{cc}\hfill \mathrm{d}(x_{n+1},x_n)=\mathrm{d}(T{x}_n,T{x}_{n-1})& \precsim A\odot \mathrm{d}(x_n,x_{n-1})\hfill \\ \hfill & \precsim A^{\odot 2}\odot \mathrm{d}(x_{n-1},x_{n-2})\hfill \\ \hfill & ⋮\hfill \\ \hfill & \precsim A^{\odot n}\odot \mathrm{d}(x_1,x_0).\hfill \end{array}$$

By triangle inequality, we have, for $n>m$,

$$\begin{array}{cc}\hfill \mathrm{d}(x_{n+1},x_m)& \precsim \mathrm{d}(x_{n+1},x_n)+\mathrm{d}(x_n,x_{n-1})+\cdots +\mathrm{d}(x_{m+1},x_m)\hfill \\ \hfill & \precsim (A^{\odot n}+A^{\odot (n-1)}+\cdots +A^{\odot m})\odot \mathrm{d}(x_1,x_0)\hfill \\ \hfill & =\left(\begin{array}{c}{\lambda }_1^n+{\lambda }_1^{n-1}+\cdots +{\lambda }_1^m\\ {\lambda }_2^n+{\lambda }_2^{n-1}+\cdots +{\lambda }_2^m\\ ⋮\\ {\lambda }_d^n+{\lambda }_d^{n-1}+\cdots +{\lambda }_d^m\end{array}\right)\odot \mathrm{d}(x_1,x_0)\hfill \\ \hfill & =\left(\begin{array}{c}{\lambda }_1^m\\ {\lambda }_2^m\\ ⋮\\ {\lambda }_d^m\end{array}\right)\odot \left(\begin{array}{c}1+{\lambda }_1+{\lambda }_1^2+\cdots +{\lambda }_1^{n-m}\\ 1+{\lambda }_2+{\lambda }_2^2+\cdots +{\lambda }_2^{n-m}\\ ⋮\\ 1+{\lambda }_d+{\lambda }_d^2+\cdots +{\lambda }_d^{n-m}\end{array}\right)\odot \mathrm{d}(x_1,x_0).\hfill \end{array}$$

(5)

Using Lemma 1, we obtain

$$\begin{array}{cc}\hfill \mathrm{d}(x_{n+1},x_m)& \precsim \left(\begin{array}{c}{\lambda }_1^m\\ {\lambda }_2^m\\ ⋮\\ {\lambda }_d^m\end{array}\right)\odot \left(\begin{array}{c}\frac{1}{1-{\lambda }_1}\\ \frac{1}{1-{\lambda }_2^2}\\ ⋮\\ \frac{1}{1-{\lambda }_d^m}\end{array}\right)\odot \mathrm{d}(x_1,x_0)\hfill \\ \hfill & =\left(\begin{array}{c}\frac{{\lambda }_1^m}{1-{\lambda }_1}\\ \frac{{\lambda }_2^m}{1-{\lambda }_2^2}\\ ⋮\\ \frac{{\lambda }_n^m}{1-{\lambda }_d^m}\end{array}\right)\odot \mathrm{d}(x_1,x_0)\to \tilde{0}\mathrm{as}n,m\to +\infty .\hfill \end{array}$$

(6)

Thus, $\left\{x_n\right\}$ is a Cauchy sequence in X. Since $(X,{\mathbb{R}}^d,\mathrm{d})$ is a complete metric space, there exists $x\in X$ such that ${lim}_{n\to \infty }{x}_n=x$. Since

$$\begin{array}{cc}\hfill \tilde{0}& \precsim \mathrm{d}(Tx,x)\precsim \mathrm{d}(T{x}_n,Tx)+\mathrm{d}(T{x}_n,x)\hfill \\ \hfill & \precsim A\odot \mathrm{d}(x_n,x)+\mathrm{d}(x_{n+1},x)\to \tilde{0}\mathrm{as}n\to +\infty .\hfill \end{array}$$

(7)

Hence, $\mathrm{d}(Tx,x)=\tilde{0}$. This implies that $Tx=x$, and T has a fixed point.

To prove the uniqueness part, let y be another fixed point. Now,

$$\begin{array}{c}\hfill \tilde{0}\precsim \mathrm{d}(x,y)=\mathrm{d}(Tx,Ty)\precsim A\odot \mathrm{d}(x,y)\end{array}$$

and we have

$$\begin{array}{c}\hfill \tilde{0}\precsim \mathrm{d}(x,y)\odot (I_H^{{\mathbb{R}}^d}-A)\precsim \tilde{0}\end{array}$$

(8)

since $A\prec I_H^{{\mathbb{R}}^d}$, which results in a contradiction unless $\mathrm{d}(x,y)=\tilde{0}$, and so $x=y$. Then, the fixed point is unique, and this proves the theorem. □

In the above theorem, the condition that $m>0$ is not necessary. This is much stronger than the results introduced by Perov [2].

Theorem 3 

(Kannan type [10]). Let $(X,{\mathbb{R}}^d,\mathrm{d})$ be a complete generalized metric space. Suppose that $T:X\to X$ satisfies the contractive condition

$$\begin{array}{c}\hfill \mathrm{d}(Tx,Ty)\precsim A\odot \left(\mathrm{d}(Tx,x)+\mathrm{d}(Ty,y)\right)\end{array}$$

(9)

for all $x,y\in X$, where $\tilde{0}\precsim A\prec I_H^{{\mathbb{R}}^d}/2$, with $A={({\lambda }_1,{\lambda }_2,\cdots ,{\lambda }_d)}^t$, $0\le {\lambda }_i<\frac{1}{2}$, $i=1,2,3,\cdots ,d$. Then, T has a unique fixed point.

Proof. 

Choose $x_0\in X$. Set $x_1=T{x}_0,x_2=T{x}_1,\cdots ,x_{n+1}=T{x}_n$. We have

$$\begin{array}{cc}\hfill \mathrm{d}(x_{n+1},x_n)=\mathrm{d}(T{x}_n,T{x}_{n-1})& \precsim A\odot \left(\mathrm{d}(T{x}_n,x_n)+\mathrm{d}(T{x}_{n-1},x_{n-1})\right)\hfill \\ \hfill & =A\odot \left(\mathrm{d}(x_{n+1},x_n)+\mathrm{d}(x_n,x_{n-1})\right).\hfill \end{array}$$

Thus,

$$\begin{array}{cc}\hfill \mathrm{d}(x_{n+1},x_n)& \precsim \left(\frac{A}{I_H^{{\mathbb{R}}^d}-A}\right)\odot \mathrm{d}(x_n,x_{n-1})\hfill \\ \hfill & =K\odot \mathrm{d}(x_n,x_{n-1})\hfill \\ \hfill & \precsim \cdots \precsim K^{\odot n}\odot \mathrm{d}(x_1,x_0),\hfill \end{array}$$

where $K=\frac{A}{I_H^{{\mathbb{R}}^d}-A}={\left(\frac{{\lambda }_1}{1-{\lambda }_1},\frac{{\lambda }_2}{1-{\lambda }_2},\cdots ,\frac{{\lambda }_d}{1-{\lambda }_d}\right)}^t$.

For $n>m$,

$$\begin{array}{cc}\hfill \mathrm{d}(x_n,x_m)& \precsim \mathrm{d}(x_n,x_{n-1})+\mathrm{d}(x_{n-1},x_{n-2})+\cdots +\mathrm{d}(x_{m+1},x_m)\hfill \\ \hfill & \precsim \left(K^{\odot (n-1)}+K^{\odot (n-2)}+\cdots +K^{\odot m}\right)\odot \mathrm{d}(x_1,x_0)\hfill \\ \hfill & =\left(\begin{array}{c}{\left(\frac{{\lambda }_1}{1-{\lambda }_1}\right)}^{n-1}+{\left(\frac{{\lambda }_1}{1-{\lambda }_1}\right)}^{n-2}+\cdots +{\left(\frac{{\lambda }_d}{1-{\lambda }_d}\right)}^m\\ {\left(\frac{{\lambda }_2}{1-{\lambda }_2}\right)}^{n-1}+{\left(\frac{{\lambda }_2}{1-{\lambda }_2}\right)}^{n-2}+\cdots +{\left(\frac{{\lambda }_d}{1-{\lambda }_d}\right)}^m\\ ⋮\\ {\left(\frac{{\lambda }_d}{1-{\lambda }_d}\right)}^{n-1}+{\left(\frac{{\lambda }_d}{1-{\lambda }_d}\right)}^{n-2}+\cdots +{\left(\frac{{\lambda }_d}{1-{\lambda }_d}\right)}^m\end{array}\right)\odot \mathrm{d}(x_1,x_0)\to 0\mathrm{as}n,m\to +\infty .\hfill \end{array}$$

Hence, $\left\{x_n\right\}$ is a Cauchy sequence. By the completeness of X, there is $x\in X$ such that $x_n\to x$ as $n\to +\infty$. Since

$$\begin{array}{cc}\hfill \tilde{0}\precsim \mathrm{d}(Tx,x)& \precsim \mathrm{d}(T{x}_n,Tx)+\mathrm{d}(T{x}_n,x)\hfill \\ \hfill & \precsim A\odot [\mathrm{d}(T{x}_n,x_n)+\mathrm{d}(Tx,x)]+\mathrm{d}(x_{n+1},x)\hfill \end{array}$$

$$\begin{array}{cc}\hfill \tilde{0}\precsim \mathrm{d}(Tx,x)\precsim & \left(\frac{I_H^{{\mathbb{R}}^d}}{I_H^{{\mathbb{R}}^d}-A}\right)\odot [A\odot \mathrm{d}(T{x}_n,x_n)+\mathrm{d}(x_{n+1},x)]\hfill \\ \hfill & =\left(\begin{array}{c}\frac{1}{1-{\lambda }_1}\\ \frac{1}{1-{\lambda }_2}\\ ⋮\\ \frac{1}{1-{\lambda }_d}\end{array}\right)\odot \left[\left(\begin{array}{c}{\lambda }_1\\ {\lambda }_2\\ ⋮\\ {\lambda }_d\end{array}\right)\odot (\mathrm{d}(x_{n+1},x_n)+\mathrm{d}(x_{n+1},x)\right]\hfill \\ \hfill & =\left(\begin{array}{c}\frac{{\lambda }_1}{1-{\lambda }_1}\\ \frac{{\lambda }_2}{1-{\lambda }_2}\\ ⋮\\ \frac{{\lambda }_d}{1-{\lambda }_d}\end{array}\right)\odot \left[\mathrm{d}(x_{n+1},x)+\mathrm{d}(x,x_n)\right]+\mathrm{d}(x_{n+1},x).\hfill \end{array}$$

(10)

Thus $\mathrm{d}(Tx,x)=\tilde{0}$ as $n\to +\infty$; therefore, $Tx=x$.

Now, if y is another fixed point of T, then

$$\tilde{0}\precsim \mathrm{d}(x,y)\precsim \mathrm{d}(Tx,Ty)\precsim A\odot \left(\mathrm{d}(Tx,x)+\mathrm{d}(Ty,y)\right)=\tilde{0}.$$

Hence, $x=y$. Therefore, the fixed point is unique. □

Theorem 4 

(Chatterjea type [11]). Let $(X,{\mathbb{R}}^d,\mathrm{d})$ be a complete generalized metric space and $T:X\to X$ be a mapping that satisfies the contractive condition

$$\begin{array}{c}\hfill \mathrm{d}(Tx,Ty)\precsim A\odot \left(\mathrm{d}(Tx,y)+\mathrm{d}(Ty,x)\right)\end{array}$$

(11)

for all $x,y\in X$, where $\tilde{0}\precsim A\prec I_H^{{\mathbb{R}}^d}/2$, with $A={({\lambda }_1,{\lambda }_2,\cdots ,{\lambda }_d)}^t$, $0\le {\lambda }_i<\frac{1}{2}$, $i=1,2,3,\cdots ,d$. Then, T has a unique fixed point.

Proof. 

Choose $x_0\in X$. Set $x_{n+1}=T{x}_n$, for $n=0,1,2,\cdots$. We have

$$\begin{array}{cc}\hfill \mathrm{d}(x_{n+1},x_n)=\mathrm{d}(T{x}_n,T{x}_{n-1})& \precsim A\odot \left(\mathrm{d}(T{x}_n,x_{n-1})+\mathrm{d}(T{x}_{n-1},x_n)\right)\hfill \\ \hfill & \precsim A\odot \left(\mathrm{d}(x_{n+1},x_{n-1})\right)\hfill \\ \hfill & \precsim A\odot \left(\mathrm{d}(x_{n+1},x_n)+\mathrm{d}(x_n,x_{n-1})\right).\hfill \end{array}$$

Thus,

$$\begin{array}{cc}\hfill \mathrm{d}(x_{n+1},x_n)& \precsim \left(\frac{A}{I_H^{{\mathbb{R}}^d}-A}\right)\odot \mathrm{d}(x_n,x_{n-1})\hfill \\ \hfill & =K\odot \mathrm{d}(x_n,x_{n-1})\hfill \\ \hfill & \precsim \cdots \precsim K^{\odot n}\odot \mathrm{d}(x_1,x_0),\hfill \end{array}$$

where $K=\frac{A}{I_H^{{\mathbb{R}}^d}-A}={\left(\frac{{\lambda }_1}{1-{\lambda }_1},\frac{{\lambda }_2}{1-{\lambda }_2},\cdots ,\frac{{\lambda }_d}{1-{\lambda }_d}\right)}^t$.

For $n>m$,

$$\begin{array}{cc}\hfill \mathrm{d}(x_n,x_m)& \precsim \mathrm{d}(x_n,x_{n-1})+\mathrm{d}(x_{n-1},x_{n-2})+\cdots +\mathrm{d}(x_{m+1},x_m)\hfill \\ \hfill & \precsim \left(K^{\odot (n-1)}+K^{\odot (n-2)}+\cdots +K^{\odot m}\right)\odot \mathrm{d}(x_1,x_0).\hfill \end{array}$$

Following the proof of Theorem 3,

$$\mathrm{d}(x_n,x_m)\to \tilde{0}\mathrm{as}n,m\to +\infty .$$

Hence, $\left\{x_n\right\}$ is a Cauchy sequence. Since X is complete, it implies that there is an element $x\in X$ such that $x_n\to x$ as $n\to +\infty$. Since

$$\begin{array}{cc}\hfill \mathrm{d}(Tx,x)& \precsim \mathrm{d}(Tx,T{x}_n)+\mathrm{d}(T{x}_n,x)\hfill \\ \hfill & \precsim A\odot \left(\mathrm{d}(Tx,x_n)+\mathrm{d}(T{x}_n,x)\right)+\mathrm{d}(T{x}_n,x)\hfill \\ \hfill & \precsim A\odot \left(\mathrm{d}(Tx,x_n)+\mathrm{d}(x_{n+1},x)\right)+\mathrm{d}(x_{n+1},x)\hfill \\ \hfill & \precsim A\odot \left(\mathrm{d}(Tx,x)+\mathrm{d}(x_n,x)+\mathrm{d}(x_{n+1},x)\right)+\mathrm{d}(x_{n+1},x)\hfill \\ \hfill & \precsim A\odot \mathrm{d}(Tx,x)\mathrm{as}n\to +\infty .\hfill \end{array}$$

This gives a contradiction, so we must have $\mathrm{d}(Tx,x)=\tilde{0}$. This implies that $Tx=x$. Thus, x is a fixed point of T.

Now, suppose that $y\ne x$ is another fixed point of T, since

$$\tilde{0}\precsim \mathrm{d}(x,y)\precsim \mathrm{d}(Tx,Ty)\precsim A\odot \left(\mathrm{d}(Tx,y)+\mathrm{d}(Ty,x)\right)=2A\odot \mathrm{d}(x,y).$$

Therefore, $\mathrm{d}(x,y)=\tilde{0}$, $x=y$. Thus, the fixed point is unique. □

Theorem 5. 

Let $(X,{\mathbb{R}}^d,\mathrm{d})$ be a complete generalized metric space endowed with the Hadamard product, and let $T:X\to X$ be a mapping such that, for each $x,y\in X$, we have

$$\begin{array}{c}\hfill \mathrm{d}(Tx,Ty)\precsim A\odot \mathrm{d}(x,y)+B\odot \mathrm{d}(y,Tx)\end{array}$$

(12)

where $A={\left(a_i\right)}_{1\le i\le d}\in {\mathbb{R}}_{+}^d$, $B={\left(b_i\right)}_{1\le i\le d}\in {\mathbb{R}}_{+}^d$, with $\tilde{0}\precsim A+2B\prec I_H^{{\mathbb{R}}^d}$. Then, T has a unique fixed point.

Proof. 

Choose $x_0\in X$. Define a sequence $\left\{x_n\right\}\subseteq X$ by $x_{n+1}=T{x}_n$, for $n\in \mathbb{N}$. We have

$$\begin{array}{cc}\hfill \mathrm{d}(x_{n+1},x_n)=\mathrm{d}(T{x}_n,T{x}_{n-1})& \precsim A\odot \mathrm{d}(x_n,x_{n-1})+B\odot \mathrm{d}(x_{n-1},T{x}_n)\hfill \\ \hfill & \precsim A\odot \mathrm{d}(x_n,x_{n-1})+B\odot \mathrm{d}(x_{n+1},x_{n-1})\hfill \\ \hfill & \precsim A\odot \mathrm{d}(x_n,x_{n-1})+B\odot \left(\mathrm{d}(x_{n+1},x_n)+\mathrm{d}(x_n,x_{n-1})\right).\hfill \end{array}$$

Thus,

$$(I_H^{{\mathbb{R}}^d}-B)\odot \mathrm{d}(x_{n+1},x_n)\precsim (A+B)\odot \mathrm{d}(x_n,x_{n-1}).$$

This implies that

$$\begin{array}{cc}\hfill \mathrm{d}(x_{n+1},x_n)& \precsim \left(\frac{A+B}{I_H^{{\mathbb{R}}^d}-B}\right)\odot \mathrm{d}(x_n,x_{n-1}),\hfill \end{array}$$

and using $\frac{A+B}{I_H^{{\mathbb{R}}^d}-B}=K={\left(k_i\right)}_{1\le i\le d}\in {\mathbb{R}}_{+}^d$, we obtain

$$\begin{array}{cc}\hfill \mathrm{d}(x_{n+1},x_n)& \precsim K\odot \mathrm{d}(x_n,x_{n-1}),\hfill \end{array}$$

so

$$\begin{array}{c}\hfill \mathrm{d}(x_{n+1},x_n)\precsim K^{\odot n}\odot \mathrm{d}(x_1,x_0),\end{array}$$

For $n>m$,

$$\begin{array}{cc}\hfill \mathrm{d}(x_n,x_m)& \precsim \mathrm{d}(x_n,x_{n-1})+\mathrm{d}(x_{n-1},x_{n-2})+\in +\mathrm{d}(x_{m+1},x_m)\hfill \\ \hfill & \precsim \left(K^{\odot (n-1)}+K^{\odot (n-2)}+\in +K^{\odot m}\right)\odot \mathrm{d}(x_1,x_0)\hfill \\ \hfill & \precsim \left(\frac{K^{\odot m}}{1-K}\right)\odot \mathrm{d}(x_1,x_0).\hfill \end{array}$$

(13)

Using the same argument as in Theorem 2, we obtain

$$\mathrm{d}(x_n,x_m)\precsim \left(\begin{array}{c}\frac{k_1^m}{1-k_1}\\ \frac{k_2^m}{1-k_2}\\ ⋮\\ \frac{k_d^m}{1-k_d}\end{array}\right)\odot \mathrm{d}(x_1,x_0)\to \tilde{0}\mathrm{as}m\to +\infty .$$

Thus, $\left\{x_n\right\}$ is a Cauchy sequence in X. By using the completeness of the metric space, there exists $x\in X$ such that $x_n\to x$ as $n\to +\infty$. By using triangle inequality, we have

$$\begin{array}{cc}\hfill \mathrm{d}(Tx,x)& \precsim \mathrm{d}(T{x}_n,Tx)+\mathrm{d}(T{x}_n,x)\hfill \\ \hfill & \precsim A\odot \mathrm{d}(x_n,x)+B\odot \mathrm{d}(x,T{x}_n)+\mathrm{d}(T{x}_n,x).\hfill \end{array}$$

$$\tilde{0}\precsim \mathrm{d}(Tx,x)\precsim A\odot \mathrm{d}(x_n,x)+B\odot \mathrm{d}(x,x_{n+1})+\mathrm{d}(x_{n+1},x)\to \tilde{0}\mathrm{as}n\to +\infty$$

and this gives $\mathrm{d}(Tx,x)=\tilde{0}$ and so $Tx=x$ and T has a fixed point $x\in X$.

Now, if $y\ne x$ is another fixed point of T,

$$\begin{array}{cc}\hfill \tilde{0}\precsim \mathrm{d}(x,y)=\mathrm{d}(Tx,Ty)& \precsim A\odot \mathrm{d}(x,y)+N\odot \mathrm{d}(y,Tx)\hfill \\ \hfill & =A\odot \mathrm{d}(x,y)+B\odot \mathrm{d}(x,y)=(A+B)\odot \mathrm{d}(x,y)\hfill \\ \hfill & \precsim \left(I_H^{{\mathbb{R}}^d}-(A+B)\right)\odot \mathrm{d}(x,y)\precsim \tilde{0}.\hfill \end{array}$$

This results in a contradiction, and we obtain $\mathrm{d}(x,y)=\tilde{0}$. So $x=y$. Thus, T has a unique fixed point. □

Theorem 6. 

Let $(X,{\mathbb{R}}^d,\mathrm{d})$ be a complete generalized metric space and let the self-mapping $T:X\to X$ satisfy

$$\begin{array}{c}\hfill \mathrm{d}(Tx,Ty)\precsim A\odot \mathrm{d}(x,y)+B\odot \left(\mathrm{d}(x,Ty)+\mathrm{d}(y,Tx)\right)\end{array}$$

(14)

for all $x,y\in X$ where $A,B\in {\mathbb{R}}_{+}^d$, with $\tilde{0}⪯A+2B\prec I_H^{{\mathbb{R}}^d}$. Then, T has a unique fixed point.

Proof. 

For $x_0\in X$ and $n\ge 0$, define $x_{n+1}=T{x}_n$, for $n\in \mathbb{N}$. Then,

$$\begin{array}{cc}\hfill \mathrm{d}(x_{n+1},x_n)=\mathrm{d}(T{x}_n,T{x}_{n-1})& \precsim A\odot \mathrm{d}(x_n,x_{n-1})+B\odot \left(\mathrm{d}(x_n,T{x}_{n-1})+\mathrm{d}(x_{n-1},T{x}_n)\right)\hfill \\ \hfill & =A\odot \mathrm{d}(x_n,x_{n-1})+B\odot \left(\mathrm{d}(x_n,x_n)+\mathrm{d}(x_{n-1},x_{n+1})\right)\hfill \\ \hfill & =A\odot \mathrm{d}(x_n,x_{n-1})+B\odot \mathrm{d}(x_{n+1},x_{n-1})\hfill \\ \hfill & \precsim A\odot \mathrm{d}(x_n,x_{n-1})+B\odot \left(\mathrm{d}(x_{n+1},x_n)+\mathrm{d}(x_n,x_{n-1})\right)\hfill \end{array}$$

Thus,

$$(I_H^{{\mathbb{R}}^d}-B)\odot \mathrm{d}(x_{n+1},x_n)\precsim (A+B)\odot \mathrm{d}(x_n,x_{n-1}).$$

This implies that

$$\begin{array}{cc}\hfill \mathrm{d}(x_{n+1},x_n)& \precsim \left(\frac{A+B}{I_H^{{\mathbb{R}}^d}-B}\right)\odot \mathrm{d}(x_n,x_{n-1}),\hfill \end{array}$$

and thus

$$\begin{array}{cc}\hfill \mathrm{d}(x_{n+1},x_n)& \precsim K\odot \mathrm{d}(x_n,x_{n-1}),\hfill \end{array}$$

where $K=\frac{A+B}{I_H^{{\mathbb{R}}^d}-B}$, we obtain for all $n\ge 0$, so

$$\begin{array}{c}\hfill \mathrm{d}(x_{n+1},x_n)\precsim K^{\odot n}\odot \mathrm{d}(x_1,x_0).\end{array}$$

Now, for $n>m$, we have

$$\begin{array}{cc}\hfill \mathrm{d}(x_n,x_m)& \precsim \mathrm{d}(x_n,x_{n-1})+\mathrm{d}(x_{n-1},x_{n-2})+\cdots +\mathrm{d}(x_{m+1},x_m)\hfill \\ \hfill & \precsim \left(K^{\odot (n-1)}+K^{\odot (n-2)}+\cdots +K^{\odot m}\right)\odot \mathrm{d}(x_1,x_0)\hfill \\ \hfill & \precsim \left(\frac{K^{\odot m}}{1-K}\right)\odot \mathrm{d}(x_1,x_0)\to \tilde{0}\mathrm{as}n,m\to +\infty .\hfill \end{array}$$

(15)

Therefore, $\left\{x_n\right\}$ is a Cauchy sequence in $(X,{\mathbb{R}}^d,\mathrm{d})$. Taking into account that X is a complete metric space, there must exist $x\in X$ such that $x_n\to x$ as $n\to +\infty$. Thus,

$$\begin{array}{cc}\hfill \mathrm{d}(Tx,x)& \precsim \mathrm{d}(Tx,T{x}_n)+\mathrm{d}(T{x}_n,x)\hfill \\ \hfill & \precsim A\odot \mathrm{d}(x,x_n)+B\odot \left(\mathrm{d}(T{x}_n,x)+\mathrm{d}(Tx,x_n)\right)+\mathrm{d}(T{x}_n,x)\hfill \\ \hfill & \precsim A\odot \mathrm{d}(x,x_n)+B\odot \left(\mathrm{d}(x_{n+1},x)+\mathrm{d}(Tx,x)\right)+\mathrm{d}(x_{n+1},x)\hfill \end{array}$$

as $n\to +\infty$, and we obtain

$$\mathrm{d}(Tx,x)\precsim B\odot \mathrm{d}(Tx,x),$$

since $B\prec I_H^{{\mathbb{R}}^d}$; therefore, there is a contradiction, and we obtain $\mathrm{d}(Tx,x)=\tilde{0}$ and $Tx=x$. Hence, x is a fixed point of T.

Now, we show that T has a unique fixed point. For this, assume that there exists another fixed point $y\in X$ such that $Ty=y$.

$$\begin{array}{cc}\hfill \mathrm{d}(x,y)=\mathrm{d}(Tx,Ty)& \precsim A\odot \mathrm{d}(x,y)+B\odot \left(\mathrm{d}(x,Ty)+\mathrm{d}(y,Tx)\right)\hfill \\ \hfill & =A\odot \mathrm{d}(x,y)+B\odot \left(\mathrm{d}(x,y)+\mathrm{d}(x,y)\right)=A\odot \mathrm{d}(x,y)+2B\odot \mathrm{d}(x,y)\hfill \\ \hfill & \precsim (A+2B)\odot \mathrm{d}(x,y).\hfill \end{array}$$

Since $(A+2B)\prec I_H^{{\mathbb{R}}^d}$ and $\mathrm{d}(x,y)\succsim \tilde{0}$, we must therefore have $\mathrm{d}(x,y)=\tilde{0}$. Thus, $x=y$. This completes the proof. □

4. Application

As an application of our results, we study the existence and uniqueness of the solution for a system of matrix equations.

Theorem 7. 

Suppose that $X={\mathbb{R}}^d$, and $\mathrm{d}:X\times X\to {\mathbb{R}}^d$ defined by

$$\mathrm{d}(a,b)=\left(\begin{array}{c}|a_1-b_1|\\ |a_2-b_2|\\ ⋮\\ |a_d-b_d|\end{array}\right)\in {\mathbb{R}}^d,$$

where $a={\left(a_i\right)}_{1\le i\le d}\in {\mathbb{R}}^d$, and $b={\left(b_i\right)}_{1\le i\le d}\in {\mathbb{R}}^d$. Then, $(X,{\mathbb{R}}^d,\mathrm{d})$ is the generalized metric space. Let $c=\left(c_i\right)\in {\mathbb{R}}^d$ satisfying ${\sum }_{i=1}^d\left|c_i\right|<1$, and $q=\left(q_i\right)\in {\mathbb{R}}_{+}^d$. Then, the equation of matrices

$$a-\sum _{i=1}^d{c}_i\odot a=q$$

has a unique solution.

Proof. 

Define $T:{\mathbb{R}}^d\to {\mathbb{R}}^d$ by

$$\begin{array}{c}\hfill T\left(a\right)=T\left(\left(a_i\right)\right)=\sum _{i=1}^d{c}_i\odot a_i+q_i.\end{array}$$

(16)

Thus, we have

$$\begin{array}{cc}\hfill \mathrm{d}(Ta,Tb)=\mathrm{d}(T\left(a_i\right),T\left(b_i\right))& =\left(\begin{array}{c}|T\left(a_1\right)-T\left(b_1\right)|\\ |T\left(a_2\right)-T\left(b_2\right)|\\ ⋮\\ |T\left(a_d\right)-T\left(b_d\right)|\end{array}\right)\hfill \end{array}$$

by applying (16)

$$\begin{array}{cc}\hfill \mathrm{d}(Ta,Tb)=\mathrm{d}(T\left(a_i\right),T\left(b_i\right))& =\left(\begin{array}{c}|(\sum _{i=1}^d{c}_i\odot a_1+q_1)-(\sum _{i=1}^d{c}_i\odot b_1+q_1)|\\ |(\sum _{i=1}^d{c}_i\odot a_2+q_2)-(\sum _{i=1}^d{c}_i\odot b_2+q_2)|\\ ⋮\\ |(\sum _{i=1}^d{c}_i\odot a_d+q_d)-(\sum _{i=1}^d{c}_i\odot b_d+q_d)|\end{array}\right)\hfill \\ \hfill & =\left(\begin{array}{c}|\sum _{i=1}^d{c}_i\odot (a_1-b_1)|\\ |\sum _{i=1}^d{c}_i\odot (a_2-b_2)|\\ ⋮\\ |\sum _{i=1}^d{c}_i\odot (a_d-b_d)|\end{array}\right)\hfill \\ \hfill & \precsim \sum _{i=1}^d\left|c_i\right|\odot I_H^{{\mathbb{R}}^d}\odot \left(\begin{array}{c}|a_1-b_1|\\ |a_2-b_2|\\ ⋮\\ |a_d-b_d|\end{array}\right)=I_H^{{\mathbb{R}}^d}\odot \mathrm{d}(a,b).\hfill \end{array}$$

This satisfies all conditions in Theorem 2, where $\tilde{0}\precsim A\prec I_H^{{\mathbb{R}}^d}$. □

In the following, we will give another application of the existence and uniqueness of integral equations to support our results.

Let $X=\mathcal{C}([0,1],{\mathbb{R}}^d)$ be the algebra of a continuous function on [0, 1] with pointwise addition and multiplication, with the norm ${\parallel f\parallel }_{\infty }=sup\left\{\right|f\left(t\right)|,t\in [0,1]\}$. Then, $\mathcal{C}([0,1],{\mathbb{R}}^d)$ with the given norm is a complete real Banach space.

Define $\mathrm{d}:\mathcal{C}([0,1],{\mathbb{R}}^d)\times \mathcal{C}([0,1],{\mathbb{R}}^d)\to {\mathbb{R}}^d$ as

$$\begin{array}{c}\hfill \mathrm{d}(f,g)=\left(\begin{array}{c}sup\left\{\right|f\left(t\right)-g\left(t\right)|,t\in [0,1\left]\right\}\\ sup\left\{\right|f\left(t\right)-g\left(t\right)|,t\in [0,1\left]\right\}\\ ⋮\\ sup\left\{\right|f\left(t\right)-g\left(t\right)|,t\in [0,1\left]\right\}\end{array}\right)\in {\mathbb{R}}^d.\end{array}$$

(17)

It is clear that $\left(\mathcal{C}([0,1],{\mathbb{R}}^d),{\mathbb{R}}^d,\mathrm{d}\right)$ is a generalized metric space.

Theorem 8. 

Consider the integral equation

$$\begin{array}{c}\hfill f\left(t\right)={\int }_0^{1}k(t,s,f\left(s\right))\mathrm{d}s+h\left(t\right),wheret\in [0,1]andh\left(t\right)\in \mathcal{C}([0,1],{\mathbb{R}}^d).\end{array}$$

(18)

Suppose that

1. 

$k\left(t,s,.\right):[0,1]\times [0,1]\times {\mathbb{R}}^d\to {\mathbb{R}}^d$;

2. 

There exists a continuous real-valued function $\phi :[0,1]\times [0,1]\to \mathbb{R}$ and $\lambda \in (0,1)$ such that

$$\begin{array}{c}\hfill |k(t,s,f\left(t\right))-k(t,s,g\left(t\right))|\le \lambda |\phi (t,s)\left(f\left(s\right)-g\left(s\right)\right)|\mathrm{for}t,s\in [0,1],f,g\in \mathcal{C}([0,1],{\mathbb{R}}^d).\end{array}$$

(19)

3. 

${sup}_{t\in [0,1]}{\int }_0^1\left|\phi (t,s)\right|\mathrm{d}s\le 1$; then, the integral Equation (18) has a unique solution on $\mathcal{C}([0,1],{\mathbb{R}}^d)$.

Proof. 

Let $T:\mathcal{C}([0,1],{\mathbb{R}}^d)\to \mathcal{C}([0,1],{\mathbb{R}}^d)$ be

$$Tf\left(t\right)={\int }_0^{1}k(t,s,f\left(s\right))\mathrm{d}s+h\left(t\right),t\in [0,1].$$

Set $A=\lambda I_H^{{\mathbb{R}}^d}=\left(\begin{array}{c}\lambda \\ \lambda \\ ⋮\\ \lambda \end{array}\right)\in {\mathbb{R}}^d$, where $I_H^{{\mathbb{R}}^d}=\left(\begin{array}{c}1\\ 1\\ ⋮\\ 1\end{array}\right)\in {\mathbb{R}}^d$ and $\lambda \in (0,1)$. Now, we will use the definition of the generalized metric space as in Equation (17):

$$\begin{array}{cc}\hfill \mathrm{d}(Tf,Tg)& =\left(\begin{array}{c}sup\left\{\right|Tf\left(t\right)-Tg\left(t\right)|,t\in [0,1\left]\right\}\\ sup\left\{\right|Tf\left(t\right)-Tg\left(t\right)|,t\in [0,1\left]\right\}\\ ⋮\\ sup\left\{\right|Tf\left(t\right)-Tg\left(t\right)|,t\in [0,1\left]\right\}\end{array}\right)\in {\mathbb{R}}^d\hfill \\ \hfill & =\underset{t\in [0,1]}{sup}\left\{\right|Tf\left(t\right)-Tg\left(t\right)\left|\right\}·I_H^{{\mathbb{R}}^d}\hfill \\ \hfill & =\underset{t\in [0,1]}{sup}\{{\int }_0^1\left|k\right(t,s,f\left(s\right)-k(t,s,g\left(s\right)|\mathrm{d}s\})·I_H^{{\mathbb{R}}^d}\hfill \\ \hfill & \precsim \underset{t\in [0,1]}{sup}\{{\int }_0^1\lambda |\phi (t,s)\left(f\left(s\right)-g\left(s\right)\right)\left|\mathrm{d}s\right\})·I_H^{{\mathbb{R}}^d}\hfill \\ \hfill & \precsim \lambda I_H^{{\mathbb{R}}^d}\odot \underset{t\in [0,1]}{sup}\left\{\right|f\left(s\right)-g\left(s\right)\left|\right\}{I}_H^{{\mathbb{R}}^d}\hfill \\ \hfill & =A\odot \underset{s\in [0,1]}{sup}\left\{\right|f\left(s\right)-g\left(s\right)\left|\right\}{I}_H^{{\mathbb{R}}^d}\mathrm{as}{I}_H^{{\mathbb{R}}^d}\odot I_H^{{\mathbb{R}}^d}=I_H^{{\mathbb{R}}^d}\hfill \\ \hfill & =A\odot \mathrm{d}(f,g),\hfill \end{array}$$

(20)

$$\mathrm{since}A\prec I_H^{{\mathbb{R}}^d}.$$

(21)

Thus, T has a unique fixed point on $\mathcal{C}([0,1],{\mathbb{R}}^d)$, with the generalized metric endowed with the Hadamard product. □