Generalization of Inequalities in Metric Spaces with Applications

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

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Hassan, E. (2023) Generalization of Inequalities in Metric Spaces with Applications. Journal of Applied Mathematics and Physics, 11, 2923-2931. doi: 10.4236/jamp.2023.1110193.

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 $d:X×X\to \left[0,\infty \right]$ 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) $|x-y|\ge 0$ ; $|x-y|=0$ iff $x=y$ ;

ii) $|x-y|=|y-x|$ ;

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

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, ${d}_{1}\left(x,y\right)=|{\xi }_{1}-{\eta }_{1}|+|{\xi }_{2}-{\eta }_{2}|$ . To satisfy the triangle inequality, let $z=\left({\mu }_{1},{\mu }_{2}\right)$ ; then,

$\begin{array}{c}{d}_{1}\left(x,y\right)=|{\xi }_{1}-{\mu }_{1}+{\mu }_{1}-{\eta }_{1}|+|{\xi }_{2}-{\mu }_{2}+{\mu }_{2}-{\eta }_{2}|\\ \le |{\xi }_{1}-{\mu }_{1}|+|{\xi }_{2}-{\mu }_{2}|+|{\mu }_{1}-{\eta }_{1}|+|{\mu }_{2}-{\eta }_{2}|\\ ={d}_{1}\left(x,z\right)+{d}_{1}\left(z,y\right)\end{array}$ (10)

Therefore, $\left({ℝ}^{2},{d}_{1}\right)$ is also a metric space. Let $X:\left\{x=\left({\xi }_{1},{\xi }_{2},\cdots ,{\xi }_{n}\right):{\xi }_{i}\in ℝ\text{\hspace{0.17em}}\text{or}\text{\hspace{0.17em}}C\right\}$ be a set of bounded sequences of real or complex numbers such that $|{\xi }_{j}|\le M$ $\forall j$ , which are all bounded. Then, we also say $|{\xi }_{j}|\le {M}_{x}$ $\forall j$ ; hence, ${L}^{\infty }=\left\{x={\left({\xi }_{i}\right)}_{i=1}^{\infty }:{\mathrm{sup}}_{i}|{\xi }_{i}|<\infty \right\}$ .

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

$\begin{array}{c}d\left(x,y\right)=\underset{i}{\mathrm{sup}}|{\xi }_{i}-{\eta }_{i}|=\underset{i}{\mathrm{sup}}|{\xi }_{i}-{\mu }_{i}+{\mu }_{i}-{\eta }_{i}|\\ \le \underset{i}{\mathrm{sup}}|{\xi }_{i}-{\mu }_{i}|+\underset{i}{\mathrm{sup}}|{\mu }_{i}-{\eta }_{i}|=d\left(x,z\right)+d\left(z,y\right)\end{array}$ (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 $d\left(x,y\right)={\sum }_{i=1}^{\infty }\frac{1}{{2}^{i}}\frac{|{\xi }_{i}-{\mu }_{i}|}{1+|{\xi }_{i}-{\mu }_{i}|}$ , 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, $|a+b|\le |a|+|b|$ , we have $f\left(|a+b|\right)=f\left(|a|+|b|\right)$ , which implies

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

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

$\begin{array}{c}\frac{|{\xi }_{i}-{\eta }_{i}|}{1+|{\xi }_{i}-{\eta }_{i}|}\le \frac{|{\xi }_{i}-{\mu }_{i}|}{1+|{\xi }_{i}-{\mu }_{i}|}+\frac{|{\mu }_{i}-{\eta }_{i}|}{1+|{\mu }_{i}-{\eta }_{i}|}={\sum }_{i=1}^{\infty }\frac{1}{{2}^{i}}\frac{|{\xi }_{i}-{\eta }_{i}|}{1+|{\xi }_{i}-{\eta }_{i}|}\\ \le {\sum }_{i=1}^{\infty }\frac{1}{{2}^{i}}\frac{|{\xi }_{i}-{\mu }_{i}|}{1+|{\xi }_{i}-{\mu }_{i}|}+{\sum }_{i=1}^{\infty }\frac{1}{{2}^{i}}\frac{|{\mu }_{i}-{\eta }_{i}|}{1+|{\mu }_{i}-{\eta }_{i}|}\end{array}$ (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 ${\sum }_{i=1}^{\infty }{|{\xi }_{i}|}^{b}<\infty$ 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

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

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

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

${\sum }_{j=1}^{\infty }|\stackrel{¯}{\xi }\stackrel{¯}{\eta }|\le \frac{1}{p}{\sum }_{j=1}^{\infty }{|{\stackrel{¯}{\xi }}_{j}|}^{p}+\frac{1}{q}{\sum }_{j=1}^{\infty }{|{\stackrel{¯}{\eta }}_{j}|}^{q}\le \frac{1}{p}+\frac{1}{q}=1$ (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

${\stackrel{¯}{\xi }}_{j}=\frac{{\xi }_{j}}{{\left({\sum }_{k=1}^{\infty }{|{\xi }_{k}|}^{p}\right)}^{\frac{1}{p}}}$ and ${\stackrel{¯}{\eta }}_{j}=\frac{{\eta }_{j}}{{\left({\sum }_{m=1}^{\infty }{|{\eta }_{m}|}^{q}\right)}^{\frac{1}{q}}}$ (15)

Clearly, the sequences ${\stackrel{¯}{\xi }}_{j}$ and ${\stackrel{¯}{\eta }}_{j}$ satisfy (13). Hence, using (13), we obtain ${\sum }_{j=1}^{\infty }|{\xi }_{j}{\eta }_{j}|\le {\left({\sum }_{k=1}^{\infty }{|{\xi }_{k}|}^{p}\right)}^{\frac{1}{p}}{\left({\sum }_{m=1}^{\infty }{|{\eta }_{m}|}^{q}\right)}^{\frac{1}{q}}$ , 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,

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

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

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

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

$|{\omega }_{j}|={|{\xi }_{j}+{\eta }_{j}|}^{p}={|{\xi }_{j}+{\eta }_{j}|}^{p-1}\le |{\xi }_{j}|{|{\omega }_{j}|}^{p-1}+|{\eta }_{j}|{|{\omega }_{j}|}^{p-1}$ (17)

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

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

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

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

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

Then, we obtain

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

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

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

Thus,

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

Finally, we conclude that

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

${L}_{\infty }d\left(x,y\right)=\sqrt{{\sum }_{i=1}^{n}{|{\xi }_{i}-{\mu }_{i}|}^{2}}$ (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 ${‖x‖}_{2}={\left[{\sum }_{i=1}^{n}{|{\xi }_{i}|}^{2}\right]}^{\frac{1}{2}}$ ,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 ${‖x‖}_{2}$ is a normed linear space, we prove the properties of a norm:

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

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

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

Proof of completeness: Let $〈{x}_{m}〉$ be a Cauchy sequence in ${ℝ}^{2}$ . Then, for any $\epsilon >0$ , $\exists {n}_{0}\in N$ such that $‖{x}_{m}-{x}_{n}‖<\epsilon$ $\exists m,r\ge {n}_{0}$ 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 ${\left[{\sum }_{i=1}^{n}{|{\xi }_{i}^{\left(m\right)}-{\xi }_{i}^{\left(r\right)}|}^{2}\right]}^{\frac{1}{2}}<\epsilon$ , $\forall m,r\ge {n}_{0}$ .

${\sum }_{i=1}^{n}{|{\xi }_{i}^{\left(m\right)}-{\xi }_{i}^{\left(r\right)}|}^{2}<{\epsilon }^{2}$ , $\forall m,r\ge {n}_{0}$

${|{\xi }_{i}^{\left(m\right)}-{\xi }_{i}^{\left(r\right)}|}^{2}<{\epsilon }^{2}$ , $\forall m,r\ge {n}_{0}$

$|{\xi }_{i}^{\left(m\right)}-{\xi }_{i}^{\left(r\right)}|<\epsilon$ , $\forall m,r\ge {n}_{0}$

Because $〈{\xi }_{i}^{\left(m\right)}〉$ 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, $‖{x}_{m}-x‖={\left[{\sum }_{i=1}^{n}{|{\xi }_{i}^{\left(m\right)}-{\xi }_{i}|}^{2}\right]}^{\frac{1}{2}}$ , 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 ${L}_{\infty }:\left\{x=\left({\xi }_{i}\right),{\xi }_{i}\in {ℝ}^{n}\text{\hspace{0.17em}}\text{or}\text{\hspace{0.17em}}C:{\mathrm{sup}}_{i}|{\xi }_{i}|<\infty \right\}$ . ${d}_{\infty }d\left(x,y\right)={\mathrm{sup}}_{i}|{\xi }_{i}-{\mu }_{i}|$ , 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 $n\ge N$ , ${d}_{\infty }\left({x}_{m},{x}_{n}\right)<\epsilon$ , $\exists m,r\ge N$ , ${\mathrm{sup}}_{i}|{x}_{m}-{\mu }_{i}^{\left(r\right)}|<\epsilon$ . For each fixed $|{\xi }_{i}^{\left(m\right)}-{\mu }_{i}^{\left(r\right)}|<\epsilon$ , 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 ${\mathrm{sup}}_{i}|{\xi }_{i}^{\left(m\right)}-{\mu }_{i}^{\left(r\right)}|<\epsilon$ for $m,r\ge N$ , each i, and let $r\to \infty$ . Thus, ${d}_{\infty }\left({x}_{m},x\right)={\mathrm{sup}}_{i}|{\xi }_{i}^{\left(m\right)}-{\mu }_{i}^{\left(r\right)}|<\epsilon$ , ${x}_{m}\to x$ because

$|{\xi }_{i}|=|{\xi }_{i}-{\xi }_{i}^{\left(m\right)}|+|{\xi }_{i}^{\left(m\right)}|<\epsilon +{k}_{m}$ (25)

where ${k}_{m}={\mathrm{sup}}_{i}|{\xi }_{i}^{\left(m\right)}|<\infty$ , ${x}_{m}\in {L}_{\infty }$ . $\left({L}_{\infty },{d}_{\infty }\right)$ 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 $d\left(x,y\right)={\sum }_{i=1}^{\infty }\frac{1}{{2}^{i}}\frac{|{\xi }_{i}-{\eta }_{i}|}{1+|{\xi }_{i}-{\eta }_{i}|}$ , 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 $\left(x,0\right)=‖x‖={\sum }_{i=1}^{\infty }\frac{1}{{2}^{i}}\frac{|{x}_{i}|}{1+|{x}_{i}|}$ . In that case, we would have the following:

$‖\alpha x‖={\sum }_{i=1}^{\infty }\frac{1}{{2}^{i}}\frac{|\alpha {x}_{i}|}{1+|\alpha {x}_{i}|}=|\alpha |‖x‖$ (26)

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

Lemma 3.8. Consider ${L}^{p}$ space and $p>1$ : ${L}^{p}:\left\{x={\left({\xi }_{i}\right)}_{i=1}^{\infty },{\xi }_{i}\in ℝ\text{\hspace{0.17em}}\text{or}\text{\hspace{0.17em}}ℂ:{\mathrm{sup}}_{i}{|{\xi }_{i}|}^{p}<\infty \right\}$ . Define ${‖x‖}_{{L}^{p}}={\left({\sum }_{i=1}^{\infty }{|{\xi }_{i}|}^{p}\right)}^{\frac{1}{p}}$ . Therefore, $\left({L}^{p},{‖\text{ }\cdot \text{ }‖}_{p}\right)$ is a normed space, so

$d\left(x,y\right)={\left({\sum }_{i=1}^{\infty }{|{\xi }_{i}-{\eta }_{i}|}^{p}\right)}^{\frac{1}{p}}={‖x-y‖}_{{L}^{p}}$ (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, $\left({L}^{p},{‖\text{ }\cdot \text{ }‖}_{p}\right)$ 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 $x,y\in X$ ,

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

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

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

or

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

or

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

Letting $H=\mathrm{max}\left\{{d}_{2n},{d}_{2n+1}\right\}$ , 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, $\mathrm{max}\left\{{d}_{2n},{d}_{2n+1},\frac{1}{2}\left({d}_{2n},{d}_{2n+1}\right)\right\}\le H=\mathrm{max}\left\{{d}_{2n},{d}_{2n+1}\right\}$ . From Equation (30), $\left(\xi +\eta \right){d}_{2n+1}+\mu {d}_{2n}\le k\mathrm{max}\left\{{d}_{2n},{d}_{2n+1}\right\}$ 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, $\mathrm{lim}\left(\alpha \circ F\right)\left({x}_{2n+1}\right)={x}_{2n+1}=z$ and $\mathrm{lim}\left(\alpha \circ F\right)\left({x}_{2n+1}\right)=z$ . 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

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

Finally, we arrive at the following conclusion:

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

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.

Conflicts of Interest

The authors declare that the research was conducted in the absence of any commercial or financial relationships that could be construed as a potential conflict of interest.

 [1] Goebel, K. and Kirk, W.A. (1973) A Fixed-Point Theorem for Transformations Whose Iterates Have Uniform Lipschitz Constants. Studia Mathematica, 47, 135-140. https://doi.org/10.4064/sm-47-2-134-140 [2] Reich, S. (1978) An Iterative Procedure for Constructing Zeros of Accretive Sets in Banach Spaces. Nonlinear Analysis, 2, 85-92. https://doi.org/10.1016/0362-546X(78)90044-5 [3] Lim, T.C. (1983) Fixed-Point Theorems for Uniformly Lipschitzian Mappings in Lp-Spaces. Nonlinear Analysis, 7, 555-563. https://doi.org/10.1016/0362-546X(83)90044-5 [4] Poffald, I.E. and Reich, S. (1986) An Incomplete Cauchy Problem. Journal of Mathematical Analysis and Applications, 113, 514-543. https://doi.org/10.1016/0022-247X(86)90323-9 [5] Prus, B. and Smarzewski, R. (1987) Strongly Unique Best Approximation and Centers in Uniformly Convex Spaces. Journal of Mathematical Analysis and Applications, 121, 10-21. https://doi.org/10.1016/0022-247X(87)90234-4 [6] Xu, H.K. (1991) Inequalities in Banach Spaces with Applications. Nonlinear Analysis, 16, 1127-1138. https://doi.org/10.1016/0362-546X(91)90200-K [7] Gornicki, J. (1996) Fixed Points of Involutions. Mathematica Japanica, 43, 151-155. [8] Takahasi, W. (1970) A Convexity in Metric Spaces and Non-Expansive Mapping I. Kodai Mathematical Seminar Reports, 22, 142-149. [9] Ra, A.C.M. and Reurings, M.C.B. (2004) A Fixed-Point Theorem in Partially Ordered Sets and Some Applications to Matrix Equations. Proceedings of the American Mathematical Society, 132, 1435-1443. https://doi.org/10.1090/S0002-9939-03-07220-4 [10] Choudhury, B.S. and Kundu, A. (2010) A Coupled Coincidence Point Result in Partially Ordered Metric Spaces for Compatible Mappings. Nonlinear Analysis TMA, 73, 2524-253. https://doi.org/10.1016/j.na.2010.06.025