Eltigani I. Hassan^1,2
1Department of Mathematics and Statistics, College of Science, Imam Mohammad Ibn Saud Islamic University (IMSIU), Riyadh, KSA.
²Department of Mathematics, College of Applied Science and Industrial, University of Bahri, Bahri, Sudan.
DOI: 10.4236/jamp.2023.1110193   PDF    HTML   XML   250 Downloads   1,070 Views   Citations

Abstract

In this paper, which serves as a continuation of earlier work, we generalize the idea of inequalities in metric spaces and use them to demonstrate that the incomplete metric space can be used to obtain a Banach space.

Keywords

Metric Spaces, Banach Space, Inequalities

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1. Introduction

This paper aims to generalize some inequalities in metric spaces by providing an explanation for the fact that every normed space is a metric space, while the converse is not always true. We also present applications of these concepts in metric spaces, supported by relevant results. The utilization of the parallelogram law, a fundamental property of Hilbert spaces, has enabled several researchers, including Kirk [1] , Reich [2] , Lim [3] , Zalinescu [4] , Poffald and Reich [5] , Prus and Smarzewski [6] , Xu [7] , Gornicki [8] , and Takahashi [9] , to establish equalities and inequalities in metric spaces and successfully solve various problems.

We present an introduction to some of the fundamental properties of a metric space. In essence, a metric space is defined as a non-empty set X such that to each $x,y\in X$ there corresponds a non-negative number called the distance between x and y. The concept of a metric space was initially introduced in 1906 and further developed in 1914. Additionally, a general inequality concerning polygonal inequality that holds true in metric spaces was established in [10] .

A distance on a non-empty set X is defined as a function $d:X\times X\to \left[0,\infty \right]$ if the following properties are satisfied:

(i) $d\left(x,y\right)=0$ iff $x=y$ .

(ii) $d\left(x,y\right)=d\left(y,x\right)$ for all $x,y\in X$ (symmetry).

(iii) $d\left(x,y\right)\le d\left(x,z\right)+d\left(z,y\right)$ for any $x,y,z\in X$ (triangle inequality).

When these properties are met, the pair (X, d) forms a metric space. One of the main goals of this article is to define metric spaces for specific types of spaces, ensuring that all requirements of a metric space are fulfilled.

2. Basic Definitions

We begin by recalling certain fundamental properties of real numbers.

For all $x,y,z\in ℝ$ ,

i) $\left|x-y\right|\ge 0$ ; $\left|x-y\right|=0$ iff $x=y$ ;

ii) $\left|x-y\right|=\left|y-x\right|$ ;

iii) $\left|x-y\right|\ge \left|x-z\right|+\left|z-y\right|$ .

To generalize these properties, let $\left(X,d\right)$ be a metric space and $x={\xi }_1,{\xi }_2,\cdots ,{\xi }_m$ . Then, we have.

$d\left(x,y\right)\le d\left(x,{\xi }_1\right)+d\left({\xi }_1,{\xi }_2\right)+\cdots +d\left({\xi }_m,y\right)$ (1)

$d\left(x,y\right)\le d\left(x,{\xi }_1\right)+d\left({\xi }_1,y\right)$ (2)

$d\left({\xi }_1,y\right)\le d\left({\xi }_1,{\xi }_2\right)+d\left({\xi }_{2,}y\right)$ (3)

$d\left(x,{\xi }_1\right)+d\left({\xi }_1,y\right)\le d\left(x,{\xi }_1\right)+d\left({\xi }_1,{\xi }_2\right)+d\left({\xi }_2,y\right)$ (4)

$d\left(x,y\right)\le d\left(x,{\xi }_1\right)+d\left({\xi }_1,{\xi }_2\right)+d\left({\xi }_2,y\right)$ (5)

$d\left({\xi }_2,y\right)\le d\left({\xi }_2,{\xi }_3\right)+d\left({\xi }_3,y\right)$ (6)

Thus, $d\left(x,y\right)$ satisfies the properties of a metric space.

Definition 1.2. Let $\alpha :X\to X$ . A point x is said to be an α-fixed point of a mapping of $F:X\to X$ if $\alpha \circ x=\alpha \circ F\left(x\right)$ .

Definition 2.2. (α-weakly isotone increasing) Let $\left(X,\le \right)$ be a partially ordered set, $\alpha :X\to X$ , and $F,G$ be two self-mappings of X. The mapping F is said to be G, α-weakly isotone increasing if for all $x\in X$ , we have $\left(\alpha \circ F\right)x\le \left(\alpha \circ G\right)x\le \left(\alpha \circ F\right)\left(\alpha \circ G\right)\left(\alpha \circ F\right)x$ .

3. Some Concepts to Prove a Metric Space

Let X be a set of ordered pairs of real numbers $\left\{x=\left({\xi }_1,{\xi }_2\right):{\xi }_i\in ℝ\right\}$ ; we define a metric d on ${ℝ}^2$ as

$d\left(x,y\right)=\sqrt{{\left({\xi }_1-{\eta }_1\right)}^2+{\left({\xi }_2-{\eta }_2\right)}^2}$ (7)

where $x=\left({\xi }_1,{\xi }_2\right)$ and $y=\left({\eta }_1,{\eta }_2\right)\in {ℝ}^2$ . Moreover, for Euclidean space ${ℝ}^n$ , ${ℝ}^n:\left\{\left(x={\xi }_1,{\xi }_2,\cdots ,{\xi }_i\right)\right\}$ , ${\xi }_i$ are real. Additionally, for $C^n:z=\left\{\left({\mu }_1,{\mu }_2,\cdots ,{\mu }_i\right)\right\}$ , $z_i$ are complex numbers, and

$d\left(x,y\right)=\sqrt{{\left({\xi }_1-{\eta }_1\right)}^2+{\left({\xi }_2-{\eta }_2\right)}^2+\cdots +{\left({\xi }_n-{\eta }_n\right)}^n}$ on ${ℝ}^n$ (8)

and

$d_1\left(x,y\right)=\sqrt{{\left({\xi }_1-{\eta }_1\right)}^2+{\left({\xi }_2-{\eta }_2\right)}^2+\cdots +{\left({\xi }_n-{\eta }_n\right)}^n}$ on $C^n$ (9)

where $x={\xi }_1,{\xi }_2,\cdots ,{\xi }_n$ , $y={\eta }_1,{\eta }_2,\cdots ,{\eta }_n$ . Then, $d_1\left(x,y\right)=\left|{\xi }_1-{\eta }_1\right|+\left|{\xi }_2-{\eta }_2\right|$ . To satisfy the triangle inequality, let $z=\left({\mu }_1,{\mu }_2\right)$ ; then,

$\begin{array}{c}{d}_1\left(x,y\right)=\left|{\xi }_1-{\mu }_1+{\mu }_1-{\eta }_1\right|+\left|{\xi }_2-{\mu }_2+{\mu }_2-{\eta }_2\right|\\ \le \left|{\xi }_1-{\mu }_1\right|+\left|{\xi }_2-{\mu }_2\right|+\left|{\mu }_1-{\eta }_1\right|+\left|{\mu }_2-{\eta }_2\right|\\ =d_1\left(x,z\right)+d_1\left(z,y\right)\end{array}$ (10)

Therefore, $\left({ℝ}^2,d_1\right)$ is also a metric space. Let $X:\left\{x=\left({\xi }_1,{\xi }_2,\cdots ,{\xi }_n\right):{\xi }_i\in ℝ\text{or}C\right\}$ be a set of bounded sequences of real or complex numbers such that $\left|{\xi }_j\right|\le M$ $\forall j$ , which are all bounded. Then, we also say $\left|{\xi }_j\right|\le M_x$ $\forall j$ ; hence, $L^{\infty }=\left\{x={\left({\xi }_i\right)}_{i=1}^{\infty }:{\sup }_i\left|{\xi }_i\right|<\infty \right\}$ .

We define the distance as $d\left(x,y\right)={\sup }_i\left|{\xi }_i-{\eta }_i\right|$ , where $x={\left({\xi }_i\right)}_{i=1}^{\infty }\in L^{\infty }$ and $y={\left({\eta }_i\right)}_{i=1}^{\infty }\in L^{\infty }$ . Let $z={\left({\mu }_i\right)}_{i=1}^{\infty }\in L^{\infty }$ ,

$\begin{array}{c}d\left(x,y\right)=\underset{i}{\sup }\left|{\xi }_i-{\eta }_i\right|=\underset{i}{\sup }\left|{\xi }_i-{\mu }_i+{\mu }_i-{\eta }_i\right|\\ \le \underset{i}{\sup }\left|{\xi }_i-{\mu }_i\right|+\underset{i}{\sup }\left|{\mu }_i-{\eta }_i\right|=d\left(x,z\right)+d\left(z,y\right)\end{array}$ (11)

Thus, $\left(L^{\infty },d\right)$ is a metric space.

Example 1.3. Suppose that S consists of the set of all bounded and unbounded sequences of complex numbers. Let the metric d be defined as $d\left(x,y\right)={\displaystyle {\sum }_{i=1}^{\infty }\frac{1}{2^i}\frac{\left|{\xi }_i-{\mu }_i\right|}{1+\left|{\xi }_i-{\mu }_i\right|}}$ , which is convergent and finite. To prove $d\left(x,y\right)\le d\left(x,z\right)+d\left(z,y\right)$ , let $z={\mu }_i\in S$ . We consider the function $f\left(t\right)=\frac{t}{1+t}$ , where $t\in ℝ$ . Since $f^{\prime }\left(t\right)=\frac{t}{{\left(1+t\right)}^2}>0$ , the function $f\left(t\right)$ is increasing. Based on the inequality, $\left|a+b\right|\le \left|a\right|+\left|b\right|$ , we have $f\left(\left|a+b\right|\right)=f\left(\left|a\right|+\left|b\right|\right)$ , which implies

$\frac{\left|a+b\right|}{1+\left|a+b\right|}\le \frac{\left|a\right|+\left|b\right|}{1+\left|a\right|+\left|b\right|}\le \frac{\left|a\right|}{1+\left|a\right|}+\frac{\left|b\right|}{1+\left|b\right|}$

Setting $a={\xi }_i-{\eta }_i$ and $b={\mu }_i-{\eta }_i$ , we obtain

$\begin{array}{c}\frac{\left|{\xi }_i-{\eta }_i\right|}{1+\left|{\xi }_i-{\eta }_i\right|}\le \frac{\left|{\xi }_i-{\mu }_i\right|}{1+\left|{\xi }_i-{\mu }_i\right|}+\frac{\left|{\mu }_i-{\eta }_i\right|}{1+\left|{\mu }_i-{\eta }_i\right|}={\displaystyle {\sum }_{i=1}^{\infty }\frac{1}{2^i}\frac{\left|{\xi }_i-{\eta }_i\right|}{1+\left|{\xi }_i-{\eta }_i\right|}}\\ \le {\displaystyle {\sum }_{i=1}^{\infty }\frac{1}{2^i}\frac{\left|{\xi }_i-{\mu }_i\right|}{1+\left|{\xi }_i-{\mu }_i\right|}}+{\displaystyle {\sum }_{i=1}^{\infty }\frac{1}{2^i}\frac{\left|{\mu }_i-{\eta }_i\right|}{1+\left|{\mu }_i-{\eta }_i\right|}}\end{array}$ (12)

Because $d\left(x,y\right)\le d\left(x,z\right)+d\left(z,y\right)$ , we conclude that $\left(S,d\right)$ is a metric space.

Example 2.3. Consider the $L^b$ space for $p\ge 1$ , where $L^b:\left\{x=\left({\xi }_1,{\xi }_2,\cdots ,{\xi }_i,\cdots \right)\right\}$ such that ${\displaystyle {\sum }_{i=1}^{\infty }{\left|{\xi }_i\right|}^b}<\infty$ and ${\xi }_i$ are scalars.

Result 1. (Holder’s inequality). Let $x={\xi }_j\in L^b$ and $y={\eta }_j\in L^b$ . Then, the product of these sequences satisfies

${\displaystyle {\sum }_{j=1}^{\infty }\left|{\xi }_j{\eta }_j\right|}\le {\left({\displaystyle {\sum }_{k=1}^{\infty }{\left|{\xi }_k\right|}^p}\right)}^{\frac{1}{p}}{\left({\displaystyle {\sum }_{m=1}^{\infty }{\left|{\eta }_m\right|}^q}\right)}^{\frac{1}{q}}$ (13)

where $p>1$ and $\frac{1}{p}+\frac{1}{q}=1$ .

Proof. Let ${\bar{\xi }}_j$ and ${\bar{\eta }}_j$ be two sequences such that ${\displaystyle {\sum }_{j=1}^{\infty }{\left|{\bar{\xi }}_j\right|}^p}=1$ and ${\displaystyle {\sum }_{j=1}^{\infty }{\left|{\bar{\eta }}_j\right|}^q}=1$ . Taking $\alpha =\left|{\bar{\xi }}_j\right|$ and $\beta =\left|{\bar{\eta }}_j\right|$ as real positive numbers, we use the inequality $\alpha \beta =\frac{{\alpha }^p}{p}+\frac{{\beta }^q}{q}$ to obtain

${\displaystyle {\sum }_{j=1}^{\infty }\left|\bar{\xi }\bar{\eta }\right|}\le \frac{1}{p}{\displaystyle {\sum }_{j=1}^{\infty }{\left|{\bar{\xi }}_j\right|}^p}+\frac{1}{q}{\displaystyle {\sum }_{j=1}^{\infty }{\left|{\bar{\eta }}_j\right|}^q}\le \frac{1}{p}+\frac{1}{q}=1$ (14)

Let us derive Holder’s inequality. Suppose that $x={\xi }_j\in L^b$ and $y={\eta }_j\in L^b$ are non-zero elements with

${\bar{\xi }}_j=\frac{{\xi }_j}{{\left({{\displaystyle \sum }}_{k=1}^{\infty }{\left|{\xi }_k\right|}^p\right)}^{\frac{1}{p}}}$ and ${\bar{\eta }}_j=\frac{{\eta }_j}{{\left({{\displaystyle \sum }}_{m=1}^{\infty }{\left|{\eta }_m\right|}^q\right)}^{\frac{1}{q}}}$ (15)

Clearly, the sequences ${\bar{\xi }}_j$ and ${\bar{\eta }}_j$ satisfy (13). Hence, using (13), we obtain ${\displaystyle {\sum }_{j=1}^{\infty }\left|{\xi }_j{\eta }_j\right|}\le {\left({\displaystyle {\sum }_{k=1}^{\infty }{\left|{\xi }_k\right|}^p}\right)}^{\frac{1}{p}}{\left({\displaystyle {\sum }_{m=1}^{\infty }{\left|{\eta }_m\right|}^q}\right)}^{\frac{1}{q}}$ , which is finite.

Result 2. (Minkowski inequality). Let $x={\xi }_j\in L^b$ , $y={\eta }_j\in L^b$ , and $p\ge 1$ . Then,

${\left({\displaystyle {\sum }_{j=1}^{\infty }{\left|{\xi }_j+{\eta }_j\right|}^p}\right)}^{\frac{1}{p}}\le {\left({\displaystyle {\sum }_{k=1}^{\infty }{\left|{\xi }_k\right|}^p}\right)}^{\frac{1}{p}}+{\left({\displaystyle {\sum }_{m=1}^{\infty }{\left|{\eta }_m\right|}^p}\right)}^{\frac{1}{p}}$

Proof. By setting $p=1$ and $\left|{\xi }_j+{\eta }_j\right|\le \left|{\xi }_j\right|+\left|{\eta }_j\right|$ and applying the triangle inequality, we obtain

${\displaystyle {\sum }_{j=1}^{\infty }\left|{\xi }_j+{\eta }_j\right|}\le {\displaystyle {\sum }_{j=1}^{\infty }\left|{\xi }_j\right|}+{\displaystyle {\sum }_{j=1}^{\infty }\left|{\eta }_j\right|}$ (16)

For simplicity, let ${\xi }_j+{\eta }_j={\omega }_j$ ; then,

$\left|{\omega }_j\right|={\left|{\xi }_j+{\eta }_j\right|}^p={\left|{\xi }_j+{\eta }_j\right|}^{p-1}\le \left|{\xi }_j\right|{\left|{\omega }_j\right|}^{p-1}+\left|{\eta }_j\right|{\left|{\omega }_j\right|}^{p-1}$ (17)

By choosing $j=1,2,\cdots ,n$ (any fixed value of n), $x={\xi }_j\in L^b$ and ${\left|{\omega }_j\right|}^{p-1}\in L^q$ because

${\left({\left|{\omega }_j\right|}^{p-1}\right)}^q={\left|{\omega }_j\right|}^{\left(p-1\right)q}$ (18)

Because ${\displaystyle {\sum }_{j=1}^{\infty }{\left|{\omega }_j\right|}^{\left(p-1\right)q}}={\displaystyle {\sum }_{j=1}^{\infty }{\left|{\omega }_j\right|}^p}<\infty$ , we can apply the Holder’s inequality to obtain

${\displaystyle {\sum }_{j=1}^n\left|{\xi }_j\right|{\left|{\omega }_j\right|}^{p-1}}\le {\left({\displaystyle {\sum }_{k=1}^n{\left|{\xi }_k\right|}^p}\right)}^{\frac{1}{p}}{\left({\left({\displaystyle {\sum }_{m=1}^n{\left|{\omega }_m\right|}^{p-1}}\right)}^q\right)}^{\frac{1}{q}}={\left({\displaystyle {\sum }_{k=1}^n{\left|{\xi }_k\right|}^p}\right)}^{\frac{1}{p}}{\left({\displaystyle {\sum }_{m=1}^n{\left|{\omega }_m\right|}^p}\right)}^{\frac{1}{q}}$

${\displaystyle {\sum }_{j=1}^n\left|{\xi }_j\right|{\left|{\omega }_j\right|}^{p-1}}\le {\left({\displaystyle {\sum }_{k=1}^{\infty }{\left|{\xi }_k\right|}^p}\right)}^{\frac{1}{p}}{\left({\displaystyle {\sum }_{m=1}^{\infty }{\left|{\omega }_m\right|}^p}\right)}^{\frac{1}{q}}$ (19)

Then, we obtain

$\begin{array}{c}{\displaystyle {\sum }_{j=1}^n{\left|{\omega }_j\right|}^p}\le {\displaystyle {\sum }_{j=1}^n\left(\left|{\xi }_j\right|+\left|{\eta }_j\right|\right){\left|{\omega }_j\right|}^{p-1}}\\ \le \left({\left({\displaystyle {\sum }_{k=1}^n{\left|{\xi }_k\right|}^p}\right)}^{\frac{1}{p}}+{\left({\displaystyle {\sum }_{k=1}^n{\left|{\eta }_k\right|}^p}\right)}^{\frac{1}{p}}\right){\left({\displaystyle {\sum }_{m=1}^n{\left|{\omega }_m\right|}^p}\right)}^{\frac{1}{q}}\end{array}$ (20)

Taking the limit as $n\to \infty$ , we obtain

${\left({\displaystyle {\sum }_{j=1}^{\infty }{\left|{\omega }_j\right|}^p}\right)}^{1-\frac{1}{q}}\le \left({\left({\displaystyle {\sum }_{k=1}^{\infty }{\left|{\xi }_k\right|}^p}\right)}^{\frac{1}{p}}+{\left({\displaystyle {\sum }_{k=1}^{\infty }{\left|{\eta }_k\right|}^p}\right)}^{\frac{1}{p}}\right)$ (21)

Thus,

${\left({\displaystyle {\sum }_{j=1}^{\infty }{\left|{\omega }_j\right|}^p}\right)}^{\frac{1}{p}}\le \left({\left({\displaystyle {\sum }_{k=1}^{\infty }{\left|{\xi }_k\right|}^p}\right)}^{\frac{1}{p}}+{\left({\displaystyle {\sum }_{k=1}^{\infty }{\left|{\eta }_k\right|}^p}\right)}^{\frac{1}{p}}\right)$ (22)

Finally, we conclude that

${\left({\displaystyle {\sum }_{j=1}^{\infty }{\left|{\xi }_{j+}{\eta }_j\right|}^p}\right)}^{\frac{1}{p}}\le {\left({\displaystyle {\sum }_{k=1}^{\infty }{\left|{\xi }_k\right|}^p}\right)}^{\frac{1}{p}}+{\left({\displaystyle {\sum }_{k=1}^{\infty }{\left|{\eta }_k\right|}^p}\right)}^{\frac{1}{p}}$ (23)

Theorem 3.3. A mapping T of a metric space $\left(X,d\right)$ into a metric space $\left(X,d\right)$ is continuous if and only if the inverse image of any open subset of Y is an open subset of X.

Definition 3.4. (Complete metric space) A continuous metric space $\left(X,d\right)$ is said to be complete if every Cauchy sequence in X converges to an element of X.

Example 3.5. Let $\left({ℝ}^n,d\right)$ be a complete metric space.

$L_{\infty }d\left(x,y\right)=\sqrt{{\displaystyle {\sum }_{i=1}^n{\left|{\xi }_i-{\mu }_i\right|}^2}}$ (24)

where $x={\xi }_1,{\xi }_2,\cdots ,{\xi }_n$ , $y={\eta }_1,{\eta }_2,\cdots ,{\eta }_n\in {ℝ}^n$ . Now, we are ready to state our main result.

Example 3.6. A Banach space under the norm defined by ${\Vert x\Vert }_2={\left[{\displaystyle {\sum }_{i=1}^n{\left|{\xi }_i\right|}^2}\right]}^{\frac{1}{2}}$ ,where $x={\left({\xi }_i\right)}_{i=1}^n=\left({\xi }_1,{\xi }_2,\cdots ,{\xi }_n\right)\in {ℝ}^n$ , ${\xi }_i\in ℝ$ for all i.

Proof. To prove that ${\Vert x\Vert }_2$ is a normed linear space, we prove the properties of a norm:

(i) $\Vert x\Vert ={\left[{\displaystyle {\sum }_{i=1}^n{\left|{\xi }_i\right|}^2}\right]}^{\frac{1}{2}}\ge 0$ , ${\xi }_i\in ℝ$ for all i. $\Vert x\Vert =0$ implies that ${\left[{\displaystyle {\sum }_{i=1}^n{\left|{\xi }_i\right|}^2}\right]}^{\frac{1}{2}}=0$ ; then, ${\displaystyle {\sum }_{i=1}^n{\left|{\xi }_i\right|}^2}=0$ , ${\left|{\xi }_i\right|}^2=0$ , $\left|{\xi }_i\right|=0$ , and ${\xi }_i=0$ .

(ii) ${\Vert x+y\Vert }^2={\left[{\displaystyle {\sum }_{i=1}^n{\left|{\xi }_i+{\eta }_i\right|}^2}\right]}^{\frac{1}{2}}={\displaystyle {\sum }_{i=1}^n\left|{\xi }_i+{\eta }_i\right|+\left|{\xi }_i+{\eta }_i\right|}$ . $\begin{array}{l}{\Vert x+y\Vert }^2\le {\displaystyle {\sum }_{i=1}^n\left(\left|{\xi }_i\right|+\left|{\eta }_i\right|\right)\left|{\xi }_i+{\eta }_i\right|}={\displaystyle {\sum }_{i=1}^n\left|{\xi }_i\right|\left|{\xi }_i+{\eta }_i\right|}+{\displaystyle {\sum }_{i=1}^n\left|{\eta }_i\right|\left|{\xi }_i+{\eta }_i\right|}\\ ={\displaystyle {\sum }_{i=1}^n\left|{\xi }_i\left({\xi }_i+{\eta }_i\right)\right|}+{\displaystyle {\sum }_{i=1}^n\left|{\eta }_i\left({\xi }_i+{\eta }_i\right)\right|}\le \Vert x\Vert \Vert x+y\Vert +\Vert y\Vert \Vert x+y\Vert \end{array}$ implies that ${\Vert x+y\Vert }^2\le \Vert x+y\Vert \left(\Vert x+y\Vert \right)$ . Then, $\Vert x+y\Vert \le \Vert x\Vert +\Vert y\Vert$ .

(iii) $\Vert \alpha x\Vert ={\left[{\displaystyle {\sum }_{i=1}^n{\left|\alpha {\xi }_i\right|}^2}\right]}^{\frac{1}{2}}={\left[{\displaystyle {\sum }_{i=1}^n{\left|\alpha \right|}^2{\left|{\xi }_i\right|}^2}\right]}^{\frac{1}{2}}=\left|\alpha \right|{\left[{\displaystyle {\sum }_{i=1}^n{\left|{\xi }_i\right|}^2}\right]}^{\frac{1}{2}}=\left|\alpha \right|\Vert x\Vert$ . Hence, it is a normed space.

Proof of completeness: Let $\langle x_m\rangle$ be a Cauchy sequence in ${ℝ}^2$ . Then, for any $\epsilon >0$ , $\exists n_0\in N$ such that $\Vert x_m-x_n\Vert <\epsilon$ $\exists m,r\ge n_0$ and $x_m,x_r\in {ℝ}^2$ .

$x_m=\left({\xi }_1^{\left(m\right)},{\xi }_2^{\left(m\right)},\cdots ,{\xi }_i^{\left(m\right)},\cdots ,{\xi }_n^{\left(m\right)}\right)$ and $x_n=\left({\xi }_1^{\left(r\right)},{\xi }_2^{\left(r\right)},\cdots ,{\xi }_i^{\left(r\right)},\cdots ,{\xi }_n^{\left(r\right)}\right)$ , ${\xi }_i^{\left(m\right)},{\xi }_i^{\left(r\right)}\in ℝ$ , $\forall i$ so ${\left[{\displaystyle {\sum }_{i=1}^n{\left|{\xi }_i^{\left(m\right)}-{\xi }_i^{\left(r\right)}\right|}^2}\right]}^{\frac{1}{2}}<\epsilon$ , $\forall m,r\ge n_0$ .

${\displaystyle {\sum }_{i=1}^n{\left|{\xi }_i^{\left(m\right)}-{\xi }_i^{\left(r\right)}\right|}^2}<{\epsilon }^2$ , $\forall m,r\ge n_0$

${\left|{\xi }_i^{\left(m\right)}-{\xi }_i^{\left(r\right)}\right|}^2<{\epsilon }^2$ , $\forall m,r\ge n_0$

$\left|{\xi }_i^{\left(m\right)}-{\xi }_i^{\left(r\right)}\right|<\epsilon$ , $\forall m,r\ge n_0$

Because $\langle {\xi }_i^{\left(m\right)}\rangle$ is a Cauchy sequence in $ℝ$ , ${\xi }_i^{\left(m\right)}\to {\xi }_i\in ℝ$ , and $ℝ$ is complete. Let $x={\xi }_1,{\xi }_2,\cdots ,{\xi }_i,\cdots ,{\xi }_n\in ℝ$ , ${\xi }_i\in ℝ$ , $\forall i$ . Now, $\Vert x_m-x\Vert ={\left[{\displaystyle {\sum }_{i=1}^n{\left|{\xi }_i^{\left(m\right)}-{\xi }_i\right|}^2}\right]}^{\frac{1}{2}}$ , where ${\xi }_i^{\left(m\right)}\to {\xi }_i$ , $\forall i$ ; then, ${\xi }_i^{\left(m\right)}-{\xi }_i\to 0$ as $m\to \infty$ , and $x_m-x\to 0$ as $m\to \infty$ implies that $x_m\to x\in {ℝ}^2$ , so ${ℝ}^n$ is a complete space.

Lemma 3.7. Let $\left(L_{\infty },d_{\infty }\right)$ be a complete metric space $L_{\infty }:\left\{x=\left({\xi }_i\right),{\xi }_i\in {ℝ}^n\text{or}C:{\sup }_i\left|{\xi }_i\right|<\infty \right\}$ . $d_{\infty }d\left(x,y\right)={\sup }_i\left|{\xi }_i-{\mu }_i\right|$ , where $x={\left({\xi }_i\right)}_{i=1}^{\infty }$ and $y={\left({\eta }_i\right)}_{i=1}^{\infty }\in L_{\infty }$ .

Claim. We consider $L_{\infty }$ . Given $x_m={\left({\xi }_i^m\right)}_{i=1}^{\infty }$ as a Cauchy sequence in $L_{\infty }$ , for given $\epsilon >0$ , $\exists N\left(\epsilon \right)$ such that for $n\ge N$ , $d_{\infty }\left(x_m,x_n\right)<\epsilon$ , $\exists m,r\ge N$ , ${\sup }_i\left|x_m-{\mu }_i^{\left(r\right)}\right|<\epsilon$ . For each fixed $\left|{\xi }_i^{\left(m\right)}-{\mu }_i^{\left(r\right)}\right|<\epsilon$ , we consider $\left({\xi }_i^{\left(1\right)},{\xi }_i^{\left(2\right)},\cdots \right)$ . $x_i$ behaves as a real or complex Cauchy sequence because $x_m=\left({\eta }_1,{\eta }_2,\cdots ,{\eta }_n,\cdots \right)$ . To show $x\in L_{\infty }$ , we obtain ${\sup }_i\left|{\xi }_i^{\left(m\right)}-{\mu }_i^{\left(r\right)}\right|<\epsilon$ for $m,r\ge N$ , each i, and let $r\to \infty$ . Thus, $d_{\infty }\left(x_m,x\right)={\sup }_i\left|{\xi }_i^{\left(m\right)}-{\mu }_i^{\left(r\right)}\right|<\epsilon$ , $x_m\to x$ because

$\left|{\xi }_i\right|=\left|{\xi }_i-{\xi }_i^{\left(m\right)}\right|+\left|{\xi }_i^{\left(m\right)}\right|<\epsilon +k_m$ (25)

where $k_m={\sup }_i\left|{\xi }_i^{\left(m\right)}\right|<\infty$ , $x_m\in L_{\infty }$ . $\left(L_{\infty },d_{\infty }\right)$ is a complete metric space. Alternative proofs do exist.

Result 3. Every normed space is a metric space, but the converse need not be true in general.

Let S be s set of sequences (bounded or unbounded) of real or complex numbers and define $d\left(x,y\right)={\displaystyle {\sum }_{i=1}^{\infty }\frac{1}{2^i}\frac{\left|{\xi }_i-{\eta }_i\right|}{1+\left|{\xi }_i-{\eta }_i\right|}}$ , where $x={\left({\xi }_i\right)}_{i=1}^{\infty }\in X$ and $y={\left({\eta }_i\right)}_{i=1}^{\infty }\in X$ . Clearly, $\left(S,d\right)$ is a metric space. The question is whether it is a normed space? The answer is no. If it were a normed space, we could define $\left(x,0\right)=\Vert x\Vert ={\displaystyle {\sum }_{i=1}^{\infty }\frac{1}{2^i}\frac{\left|x_i\right|}{1+\left|x_i\right|}}$ . In that case, we would have the following:

$\Vert \alpha x\Vert ={\displaystyle {\sum }_{i=1}^{\infty }\frac{1}{2^i}\frac{\left|\alpha x_i\right|}{1+\left|\alpha x_i\right|}}=\left|\alpha \right|\Vert x\Vert$ (26)

$\Vert \alpha x\Vert \ne \left|\alpha \right|\Vert x\Vert$ which fails to satisfy the norm property. So, $\left(S,\Vert .\Vert \right)$ is not a normed space, but it is a metric space.

Lemma 3.8. Consider $L^p$ space and $p>1$ : $L^p:\left\{x={\left({\xi }_i\right)}_{i=1}^{\infty },{\xi }_i\in ℝ\text{or}ℂ:{\sup }_i{\left|{\xi }_i\right|}^p<\infty \right\}$ . Define ${\Vert x\Vert }_{L^p}={\left({\displaystyle {\sum }_{i=1}^{\infty }{\left|{\xi }_i\right|}^p}\right)}^{\frac{1}{p}}$ . Therefore, $\left(L^p,{\Vert \cdot \Vert }_p\right)$ is a normed space, so

$d\left(x,y\right)={\left({\displaystyle {\sum }_{i=1}^{\infty }{\left|{\xi }_i-{\eta }_i\right|}^p}\right)}^{\frac{1}{p}}={\Vert x-y\Vert }_{L^p}$ (27)

where $x=\left({\xi }_i\right)\in L^p$ and $y=\left({\eta }_i\right)\in L^p$ . Thus, $\left(L^p,d\right)$ is a complete metric space. Hence, $\left(L^p,{\Vert \cdot \Vert }_p\right)$ is a Banach space.

Theorem 2.9. Let $\left(X,\le \right)$ be a partially ordered set and suppose that there exists a metric d on X such that $\left(X,d\right)$ is a complete metric space. Let $\alpha :X\to X$ and F and G be two self-mappings of $\left(X,d\right)$ such that for comparable $x,y\in X$ ,

$\begin{array}{l}\xi d\left(\alpha \circ F\left(x\right),\alpha \circ G\left(y\right)\right)+\eta d\left(\alpha \circ F\left(x\right),\alpha \left(x\right)\right)+\mu d\left(\alpha \left(y\right),\alpha \circ G\left(y\right)\right)\\ -\min \left\{d\left(\alpha \circ F\left(x\right),\alpha \left(y\right)\right),d\left(\alpha \circ G\left(y\right),\alpha \left(x\right)\right)\right\}\\ \le k\max \left\{d\left(\alpha \left(x\right),\alpha \left(y\right)\right),d\left(\alpha \circ F\left(x\right),\alpha \left(x\right)\right),d\left(\alpha \left(y\right),\alpha \circ G\left(y\right),\frac{1}{2}d\left(\alpha \circ F\left(x\right),\alpha \left(y\right)\right)\right)\right\}\end{array}$

For $\xi ,\eta ,\mu >0$ , $k>0$ , and $\xi >k$ , we assume the following:

(i) F is G,α-weakly increasing and

(ii) X is regular.

Then, F and G have a unique α-fixed point.

Proof. Let $x_0\in X$ . From the sequence $x_n$ with respect to $\alpha$ , we obtain $x_{2n+2}=\alpha \circ F\left(x_{2n+1}\right)=F_{\alpha }\left(x_{2n+1}\right)$ and $x_{2n+1}=\alpha \circ G\left(x_{2n}\right)=G_{\alpha }\left(x_{2n}\right)$ for $n=1,2,\cdots$ . Let $d_n=d\left(\alpha \circ \left(x_n\right)\right)$ , $\left(\alpha \circ \left(x_{n+1}\right)\right)>0$ , $n=1,2,\cdots$ . Because $G_{\alpha }$ is $F_{\alpha }$ -weakly increasing, we have

$\begin{array}{c}{x}_1\le \alpha \circ G\left(x_0\right)\le \alpha \circ F\left(\alpha \circ G\left(x_0\right)\right)=\alpha \circ F\left(x_1\right)=x_2\le \left(\alpha \circ G\right)\left(\alpha \circ F\left(\alpha \circ G\left(x_0\right)\right)\right)\\ =\alpha \circ G\left(\alpha \circ F\left(x_1\right)\right)=\alpha \circ G\left(x_2\right)=x_3\le \alpha \circ G\left(x_1\right)\le \alpha \circ F\left(\alpha \circ G\left(x_2\right)\right)\\ =\alpha \circ F\left(x_3\right)=x_4\le \left(\alpha \circ G\right)\alpha \circ F\left(\alpha \circ G\left(x_2\right)\right)=\left(\alpha \circ G\right)\left(\alpha \circ F\left(x_3\right)\right)=x_5\end{array}$ (28)

By continuing this process, we obtain $x_1\le x_2\le x_3\le \cdots \le x_n\le x_{n+1}\le \cdots$ so $x_{2n}\le x_{2n+1}$ , $\forall n=1,2,\cdots$ . Now, with $x=x_{2n+1}$ and $y=x_{2n}$ , we have

$\begin{array}{l}\left[\xi d\left(\alpha \circ F\left(x_{2n+1}\right)\right),\alpha \circ G\left(x_{2n}\right)+\eta d\left(\alpha \circ F\left(x_{2n+1}\right),x_{2n+2}\right)\right]\\ +\mu d\left(x_{2n},\alpha \circ G\left(x_{2n}\right)\right)-\min \left\{d\left(\alpha \circ F\left(x_{2n+1}\right),x_{2n}\right),d\left(\alpha \circ G\left(x_{2n}\right),x_{2n+1}\right)\right\}\\ \le k\max \left\{d\left(x_{2n+1},x_{2n}\right),d\left(\alpha \circ F\left(x_{2n+1}\right),x_{2n+1}\right),d\left(x_{2n},\alpha \circ G\left(x_{2n}\right),\frac{1}{2}d\left(\alpha \circ F\left(x_{2n+1}\right),x_{2n}\right)\right)\right\}\end{array}$ (29)

or

$\begin{array}{l}[\xi d\left(x_{2n+2},x_{2n+1}\right)+\eta d\left(x_{2n+2},x_{2n+1}\right)+\mu d\left(x_{2n},x_{2n+1}\right)\\ -\min \left\{d\left(x_{2n},x_{2n+2}\right),d\left(x_{2n+1},x_{2n+1}\right)\right\}]\\ \le k\max \left\{d\left(x_{2n+1},x_{2n}\right),d\left(x_{2n+2},x_{2n+1}\right),d\left(x_{2n},x_{2n+1}\right),\frac{1}{2}d\left(x_{2n+1},x_{2n}\right)\right\}\end{array}$ (30)

or

$\xi d_{2n+1}+\eta d_{2n+1}+\mu d_{2n}-\min \left\{d_{2n},d_{2n+1},0\right\}\le k\max \left\{d_{2n},d_{2n+1},\frac{1}{2}\left(d_{2n},d_{2n+1}\right)\right\}$ (31)

Letting $H=\max \left\{d_{2n},d_{2n+1}\right\}$ , we have $d_{2n}\le H$ , and $d_{2n+1}\le H$ implies that $\frac{1}{2}\left(d_{2n},d_{2n+1}\right)\le H$ . Therefore, $\max \left\{d_{2n},d_{2n+1},\frac{1}{2}\left(d_{2n},d_{2n+1}\right)\right\}\le H=\max \left\{d_{2n},d_{2n+1}\right\}$ . From Equation (30), $\left(\xi +\eta \right)d_{2n+1}+\mu d_{2n}\le k\max \left\{d_{2n},d_{2n+1}\right\}$ if $d_{2n}\le d_{2n+1}$ . Then, $\left(\xi +\eta \right)d_{2n+1}+\mu d_{2n}\le k{d}_{2n+1}$ , so $\left(\xi +\eta -k\right)d_{2n+1}\le -\mu d_{2n}$ , $d_{2n+1}\le g{d}_{2n}$ , where $g=\frac{-\mu }{\xi +\eta -f}\le 1$ , and if $d_{2n+1}\le d_{2n}$ , then $\left(\xi +\eta \right)d_{2n+1}+\mu d_{2n}\le g{d}_{2n}$ implies that $d_{2n+1}\le \frac{g-\mu }{\xi +\eta }{d}_{2n}$ and $d_{2n+1}\le g{d}_{2n}$ , where $g=\frac{f-\mu }{\xi +\eta }{d}_{2n}<1$ . Therefore, $d_{2n+1}\le g{d}_{2n}\le g^2{d}_{2n-1}\le \cdots \le g^{2n+1}{d}_0\to 0$ as $n\to \infty$ . Thus, $\left\{x_n\right\}$ is a Cauchy sequence in X, X is complete, and there exists a point $z\in X$ such that $\left\{x_{2n}\right\}$ converges to z. Hence, $\lim \left(\alpha \circ F\right)\left(x_{2n+1}\right)=x_{2n+1}=z$ and $\lim \left(\alpha \circ F\right)\left(x_{2n+1}\right)=z$ . Because $\left\{x_{2n}\right\}$ is a nondecreasing sequence, if X is regular, it follows that $x_{2n}\le z$ , $\forall n$ . Now, if we put $x=x_{2n+1}$ and $y=z$ , we obtain

$\begin{array}{l}[\xi d\left(\alpha \circ F\right)x_{2n+1},\left(\left(\alpha \circ G\right),z\right)+\eta d\left(x_{2n+1},\left(\alpha \circ F\right)x_{2n+1}\right)+\mu d\left(z,\left(\alpha \circ G\right)z\right)\\ -\min \left\{d\left(\left(\alpha \circ G\right)\left(x_{2n+1}\right),z\right),d\left(\left(\alpha \circ G\right)z,x_{2n+1}\right)\right\}]\\ \le k\max \left\{d\left(x_{2n+1},z\right),d\left(\left(\alpha \circ F\right)\left(x_{2n+1}\right),\left(x_{2n+1}\right),d\left(z,\left(\alpha \circ G\right)z\right),\frac{1}{2}d\left(\left(\alpha \circ F\right)\left(x_{2n+1}\right),z\right)\right)\right\}.\end{array}$

Finally, we arrive at the following conclusion:

$\begin{array}{l}\xi d\left(z,\alpha \circ F\right)\left(z\right)+\eta d\left(z,z\right)+\mu d\left(z,\alpha \circ G\left(z\right)\right)-\min \left\{d\left(z,z\right),d\left(z,\alpha \circ G\left(u\right)\right)\right\}\\ \le k\max \left\{d\left(z,z\right),d\left(z,z\right),d\left(z,\alpha \circ G\left(z\right),\frac{1}{2}d\left(z,z\right)\right)\right\}.\end{array}$

Alternatively, we can simplify it as $\left(\xi +\mu -g\right)d\left(z,\alpha \circ G\left(z\right)\right)\le 0$ or $\alpha \circ G\left(z\right)=z$ , given that $\xi >1+g$ . Thus, z is a fixed point of G. Additionally, using similar reasoning with $x=z$ , $y=x_{2n}$ , we obtain $\alpha \circ z=\alpha \circ F\left(z\right)$ . Hence, z is a common α-fixed point of F and G.

Acknowledgements

The authors express their gratitude to the Deanship of Scientific Research at Imam Mohammad Ibn Saud Islamic University, which funded their work via Research Group no. RG-21-09-51.