# Some fixed point theorems for rational Geraghty contractive mappings in ordered b-metric spaces

## Abstract

In this paper, new classes of rational Geraghty contractive mappings in the setup of b-metric spaces are introduced. Moreover, the existence of some fixed point for such mappings in ordered b-metric spaces are investigated. Also, some examples are provided to illustrate the results presented herein. Finally, an application of the main result is given.

MSC:47H10, 54H25.

## 1 Introduction

Using different forms of contractive conditions in various generalized metric spaces, there is a large number of extensions of the Banach contraction principle [1]. Some of such generalizations are obtained via rational contractive conditions. Recently, Azam et al. [2] established some fixed point results for a pair of rational contractive mappings in complex valued metric spaces. Also, in [3], Nashine et al. proved some common fixed point theorems for a pair of mappings satisfying certain rational contractions in the framework of complex valued metric spaces. In [4], the authors proved some unique fixed point results for an operator T satisfying certain rational contractive condition in a partially ordered metric space. In fact, their results generalize the main result of Jaggi [5].

Ran and Reurings started the studying of fixed point results on partially ordered sets in [6], where they gave many useful results in matrix equations. Recently, many researchers have focused on different contractive conditions in complete metric spaces endowed with a partial order and obtained many fixed point results in such spaces. For more details on fixed point results in ordered metric spaces we refer the reader to [7, 8] and [9].

Czerwik in [10] introduced the concept of a b-metric space. Since then, several papers dealt with fixed point theory for single-valued and multi-valued operators in b-metric spaces (see, e.g., [1116] and [17, 18]).

Definition 1 Let X be a (nonempty) set and $s\ge 1$ be a given real number. A function $d:X×X\to {\mathbb{R}}^{+}$ is a b-metric if the following conditions are satisfied:

(b1) $d\left(x,y\right)=0$ iff $x=y$,

(b2) $d\left(x,y\right)=d\left(y,x\right)$,

(b3) $d\left(x,z\right)\le s\left[d\left(x,y\right)+d\left(y,z\right)\right]$

for all $x,y,z\in X$.

In this case, the pair $\left(X,d\right)$ is called a b-metric space.

Definition 2 [19]

Let $\left(X,d\right)$ be a b-metric space.

1. (a)

A sequence $\left\{{x}_{n}\right\}$ in X is called b-convergent if and only if there exists $x\in X$ such that $d\left({x}_{n},x\right)\to 0$ as $n\to \mathrm{\infty }$.

2. (b)

$\left\{{x}_{n}\right\}$ in X is said to be b-Cauchy if and only if $d\left({x}_{n},{x}_{m}\right)\to 0$, as $n,m\to \mathrm{\infty }$.

3. (c)

The b-metric space $\left(X,d\right)$ is called b-complete if every b-Cauchy sequence in X is b-convergent.

The following example (corrected from [20]) illustrates that a b-metric need not be a continuous function.

Example 1 Let $X=\mathbb{N}\cup \left\{\mathrm{\infty }\right\}$ and $d:X×X\to \mathbb{R}$ be defined by

Then $d\left(m,p\right)\le \frac{5}{2}\left(d\left(m,n\right)+d\left(n,p\right)\right)$ for all $m,n,p\in X$. Thus, $\left(X,d\right)$ is a b-metric space (with $s=5/2$). Let ${x}_{n}=2n$ for each $n\in \mathbb{N}$. So $d\left(2n,\mathrm{\infty }\right)=\frac{1}{2n}\to 0$ as $n\to \mathrm{\infty }$ that is, ${x}_{n}\to \mathrm{\infty }$, but $d\left({x}_{n},1\right)=2↛5=d\left(\mathrm{\infty },1\right)$ as $n\to \mathrm{\infty }$.

Lemma 1 [21]

Let $\left(X,d\right)$ be a b-metric space with $s\ge 1$, and suppose that $\left\{{x}_{n}\right\}$ and $\left\{{y}_{n}\right\}$ are b-convergent to x and y, respectively. Then

$\frac{1}{{s}^{2}}d\left(x,y\right)\le \underset{n\to \mathrm{\infty }}{lim inf}d\left({x}_{n},{y}_{n}\right)\le \underset{n\to \mathrm{\infty }}{lim sup}d\left({x}_{n},{y}_{n}\right)\le {s}^{2}d\left(x,y\right).$

Moreover, for each $z\in X$, we have

$\frac{1}{s}d\left(x,z\right)\le \underset{n\to \mathrm{\infty }}{lim inf}d\left({x}_{n},z\right)\le \underset{n\to \mathrm{\infty }}{lim sup}d\left({x}_{n},z\right)\le sd\left(x,z\right).$

Let $\mathfrak{S}$ denote the class of all real functions $\beta :\left[0,+\mathrm{\infty }\right)\to \left[0,1\right)$ satisfying the condition

In order to generalize the Banach contraction principle, Geraghty proved the following.

Theorem 1 [22]

Let $\left(X,d\right)$ be a complete metric space, and let $f:X\to X$ be a self-map. Suppose that there exists $\beta \in \mathfrak{S}$ such that

$d\left(fx,fy\right)\le \beta \left(d\left(x,y\right)\right)d\left(x,y\right)$

holds for all $x,y\in X$. Then f has a unique fixed point $z\in X$ and for each $x\in X$ the Picard sequence $\left\{{f}^{n}x\right\}$ converges to z.

Amini-Harandi and Emami [23] generalized the result of Geraghty to the framework of a partially ordered complete metric space as follows.

Theorem 2 Let $\left(X,d,⪯\right)$ be a complete partially ordered metric space. Let $f:X\to X$ be an increasing self-map such that there exists ${x}_{0}\in X$ with ${x}_{0}⪯f{x}_{0}$. Suppose that there exists $\beta \in \mathfrak{S}$ such that

$d\left(fx,fy\right)\le \beta \left(d\left(x,y\right)\right)d\left(x,y\right)$

holds for all $x,y\in X$ with $y⪯x$. Assume that either f is continuous or X is such that if an increasing sequence $\left\{{x}_{n}\right\}$ in X converges to $x\in X$, then ${x}_{n}⪯x$ for all n. Then f has a fixed point in X. Moreover, if for each $x,y\in X$ there exists $z\in X$ comparable with x and y, then the fixed point of f is unique.

In [24], some fixed point theorems for mappings satisfying Geraghty-type contractive conditions are proved in various generalized metric spaces. As in [24], we will consider the class of functions $\beta :\left[0,\mathrm{\infty }\right)\to \left[0,1/s\right)$ such that

Theorem 3 [24]

Let $s>1$, and let $\left(X,D,s\right)$ be a complete metric type space. Suppose that a mapping $f:X\to X$ satisfies the condition

$D\left(fx,fy\right)\le \beta \left(D\left(x,y\right)\right)D\left(x,y\right)$

for all $x,y\in X$ and some $\beta \in \mathcal{F}$. Then f has a unique fixed point $z\in X$, and for each $x\in X$ the Picard sequence $\left\{{f}^{n}x\right\}$ converges to z in $\left(X,D,s\right)$.

Also, by unification of the recent results obtained by Zabihi and Razani [25] we have the following result.

Theorem 4 Let $\left(X,⪯\right)$ be a partially ordered set and suppose that there exists a b-metric d on X such that $\left(X,d\right)$ is a b-complete b-metric space (with parameter $s>1$). Let $f:X\to X$ be an increasing mapping with respect to such that there exists an element ${x}_{0}\in X$ with ${x}_{0}⪯f\left({x}_{0}\right)$. Suppose there exists $\beta \in \mathcal{F}$ such that

$sd\left(fx,fy\right)\le \beta \left(d\left(x,y\right)\right)M\left(x,y\right)+LN\left(x,y\right)$
(1.1)

for all comparable elements $x,y\in X$, where $L\ge 0$,

$M\left(x,y\right)=max\left\{d\left(x,y\right),\frac{d\left(x,fx\right)d\left(y,fy\right)}{1+d\left(fx,fy\right)}\right\}$

and

$N\left(x,y\right)=min\left\{d\left(x,fx\right),d\left(x,fy\right),d\left(y,fx\right),d\left(y,fy\right)\right\}.$

If f is continuous, or, whenever $\left\{{x}_{n}\right\}$ is a nondecreasing sequence in X such that ${x}_{n}\to u\in X$, one has ${x}_{n}⪯u$ for all $n\in \mathbb{N}$, then f has a fixed point. Moreover, the set of fixed points of f is well ordered if and only if f has one and only one fixed point.

The aim of this paper is to present some fixed point theorems for rational Geraghty contractive mappings in partially ordered b-metric spaces. Our results extend some existing results in the literature.

## 2 Main results

Let denotes the class of all functions $\beta :\left[0,\mathrm{\infty }\right)\to \left[0,\frac{1}{s}\right)$ satisfying the following condition:

Definition 3 Let $\left(X,d,⪯\right)$ be a b-metric space. A mapping $f:X\to X$ is called a rational Geraghty contraction of type I if there exists $\beta \in \mathcal{F}$ such that

$d\left(fx,fy\right)\le \beta \left(M\left(x,y\right)\right)M\left(x,y\right)$
(2.1)

for all comparable elements $x,y\in X$, where

$M\left(x,y\right)=max\left\{d\left(x,y\right),\frac{d\left(x,fx\right)d\left(y,fy\right)}{1+d\left(x,y\right)},\frac{d\left(x,fx\right)d\left(y,fy\right)}{1+d\left(fx,fy\right)}\right\}.$

Theorem 5 Let $\left(X,⪯\right)$ be a partially ordered set and suppose there exists a b-metric d on X such that $\left(X,d\right)$ is a b-complete b-metric space (with parameter $s>1$). Let $f:X\to X$ be an increasing mapping with respect to such that there exists an element ${x}_{0}\in X$ with ${x}_{0}⪯f\left({x}_{0}\right)$. Suppose f is a rational Geraghty contraction of type I. If

1. (I)

f is continuous, or,

2. (II)

whenever $\left\{{x}_{n}\right\}$ is a nondecreasing sequence in X such that ${x}_{n}\to u\in X$, one has ${x}_{n}⪯u$ for all $n\in \mathbb{N}$,

then f has a fixed point.

Moreover, the set of fixed points of f is well ordered if and only if f has one and only one fixed point.

Proof Let ${x}_{n}={f}^{n}\left({x}_{0}\right)$ for all $n\ge 0$. Since ${x}_{0}⪯f\left({x}_{0}\right)$ and f is increasing, we obtain by induction that

${x}_{0}⪯f\left({x}_{0}\right)⪯{f}^{2}\left({x}_{0}\right)⪯\cdots ⪯{f}^{n}\left({x}_{0}\right)⪯{f}^{n+1}\left({x}_{0}\right)⪯\cdots .$

We do the proof in the following steps.

Step I: We show that ${lim}_{n\to \mathrm{\infty }}d\left({x}_{n},{x}_{n+1}\right)=0$. Since ${x}_{n}⪯{x}_{n+1}$ for each $n\in \mathbb{N}$, then by (2.1)

$\begin{array}{rl}d\left({x}_{n},{x}_{n+1}\right)& =d\left(f{x}_{n-1},f{x}_{n}\right)\\ \le \beta \left(M\left({x}_{n-1},{x}_{n}\right)\right)M\left({x}_{n-1},{x}_{n}\right),\end{array}$
(2.2)

where

$\begin{array}{rcl}M\left({x}_{n-1},{x}_{n}\right)& =& max\left\{d\left({x}_{n-1},{x}_{n}\right),\frac{d\left({x}_{n-1},f{x}_{n-1}\right)d\left({x}_{n},f{x}_{n}\right)}{1+d\left({x}_{n-1},{x}_{n}\right)},\\ \frac{d\left({x}_{n-1},f{x}_{n-1}\right)d\left({x}_{n},f{x}_{n}\right)}{1+d\left(f{x}_{n-1},f{x}_{n}\right)}\right\}\\ =& max\left\{d\left({x}_{n-1},{x}_{n}\right),\frac{d\left({x}_{n-1},{x}_{n}\right)d\left({x}_{n},{x}_{n+1}\right)}{1+d\left({x}_{n-1},{x}_{n}\right)},\frac{d\left({x}_{n-1},{x}_{n}\right)d\left({x}_{n},{x}_{n+1}\right)}{1+d\left({x}_{n},{x}_{n+1}\right)}\right\}\\ \le & max\left\{d\left({x}_{n-1},{x}_{n}\right),d\left({x}_{n},{x}_{n+1}\right)\right\}.\end{array}$

If $max\left\{d\left({x}_{n-1},{x}_{n}\right),d\left({x}_{n},{x}_{n+1}\right)\right\}=d\left({x}_{n},{x}_{n+1}\right)$, then from (2.2),

$\begin{array}{rl}d\left({x}_{n},{x}_{n+1}\right)& \le \beta \left(M\left({x}_{n},{x}_{n+1}\right)\right)d\left({x}_{n},{x}_{n+1}\right)\\ <\frac{1}{s}d\left({x}_{n},{x}_{n+1}\right)\\
(2.3)

Hence, $max\left\{d\left({x}_{n-1},{x}_{n}\right),d\left({x}_{n},{x}_{n+1}\right)\right\}=d\left({x}_{n-1},{x}_{n}\right)$, so from (2.2),

$d\left({x}_{n},{x}_{n+1}\right)\le \beta \left(M\left({x}_{n-1},{x}_{n}\right)\right)d\left({x}_{n-1},{x}_{n}\right).$
(2.4)

Since $\left\{d\left({x}_{n},{x}_{n+1}\right)\right\}$ is a decreasing sequence, then there exists $r\ge 0$ such that ${lim}_{n\to \mathrm{\infty }}d\left({x}_{n},{x}_{n+1}\right)=r$. We prove $r=0$. Suppose on contrary that $r>0$. Then, letting $n\to \mathrm{\infty }$, from (2.4) we have

$r\le \underset{n\to \mathrm{\infty }}{lim}\beta \left(M\left({x}_{n-1},{x}_{n}\right)\right)r,$

which implies that $\frac{1}{s}\le 1\le {lim}_{n\to \mathrm{\infty }}\beta \left(M\left({x}_{n-1},{x}_{n}\right)\right)$. Now, as $\beta \in \mathcal{F}$ we conclude that $M\left({x}_{n-1},{x}_{n}\right)\to 0$, which yields $r=0$, a contradiction. Hence, $r=0$. That is,

$\underset{n\to \mathrm{\infty }}{lim}d\left({x}_{n-1},{x}_{n}\right)=0.$
(2.5)

Step II: Now, we prove that the sequence $\left\{{x}_{n}\right\}$ is a b-Cauchy sequence. Suppose the contrary, i.e., $\left\{{x}_{n}\right\}$ is not a b-Cauchy sequence. Then there exists $\epsilon >0$ for which we can find two subsequences $\left\{{x}_{{m}_{i}}\right\}$ and $\left\{{x}_{{n}_{i}}\right\}$ of $\left\{{x}_{n}\right\}$ such that ${n}_{i}$ is the smallest index for which

${n}_{i}>{m}_{i}>i\phantom{\rule{1em}{0ex}}\text{and}\phantom{\rule{1em}{0ex}}d\left({x}_{{m}_{i}},{x}_{{n}_{i}}\right)\ge \epsilon .$
(2.6)

This means that

$d\left({x}_{{m}_{i}},{x}_{{n}_{i}-1}\right)<\epsilon .$
(2.7)

From (2.5) and using the triangular inequality, we get

$\epsilon \le d\left({x}_{{m}_{i}},{x}_{{n}_{i}}\right)\le sd\left({x}_{{m}_{i}},{x}_{{m}_{i}+1}\right)+sd\left({x}_{{m}_{i}+1},{x}_{{n}_{i}}\right).$

By taking the upper limit as $i\to \mathrm{\infty }$, we get

$\frac{\epsilon }{s}\le \underset{i\to \mathrm{\infty }}{lim sup}d\left({x}_{{m}_{i}+1},{x}_{{n}_{i}}\right).$
(2.8)

The definition of $M\left(x,y\right)$ and (2.8) imply

$\begin{array}{c}\underset{i\to \mathrm{\infty }}{lim sup}M\left({x}_{{m}_{i}},{x}_{{n}_{i}-1}\right)\hfill \\ \phantom{\rule{1em}{0ex}}=\underset{i\to \mathrm{\infty }}{lim sup}max\left\{d\left({x}_{{m}_{i}},{x}_{{n}_{i}-1}\right),\frac{d\left({x}_{{m}_{i}},f{x}_{{m}_{i}}\right)d\left({x}_{{n}_{i}-1},f{x}_{{n}_{i}-1}\right)}{1+d\left({x}_{{m}_{i}},{x}_{{n}_{i}-1}\right)},\hfill \\ \phantom{\rule{2em}{0ex}}\frac{d\left({x}_{{m}_{i}},f{x}_{{m}_{i}}\right)d\left({x}_{{n}_{i}-1},f{x}_{{n}_{i}-1}\right)}{1+d\left(f{x}_{{m}_{i}},f{x}_{{n}_{i}-1}\right)}\right\}\hfill \\ \phantom{\rule{1em}{0ex}}=\underset{i\to \mathrm{\infty }}{lim sup}max\left\{d\left({x}_{{m}_{i}},{x}_{{n}_{i}-1}\right),\frac{d\left({x}_{{m}_{i}},{x}_{{m}_{i}+1}\right)d\left({x}_{{n}_{i}-1},{x}_{{n}_{i}}\right)}{1+d\left({x}_{{m}_{i}},{x}_{{n}_{i}-1}\right)},\hfill \\ \phantom{\rule{2em}{0ex}}\frac{d\left({x}_{{m}_{i}},{x}_{{m}_{i}+1}\right)d\left({x}_{{n}_{i}-1},{x}_{{n}_{i}}\right)}{1+d\left({x}_{{m}_{i}+1},{x}_{{n}_{i}}\right)}\right\}\hfill \\ \phantom{\rule{1em}{0ex}}\le \epsilon .\hfill \end{array}$

Now, from (2.1) and the above inequalities, we have

$\begin{array}{rl}\frac{\epsilon }{s}& \le \underset{i\to \mathrm{\infty }}{lim sup}d\left({x}_{{m}_{i}+1},{x}_{{n}_{i}}\right)\\ \le \underset{i\to \mathrm{\infty }}{lim sup}\beta \left(M\left({x}_{{m}_{i}},{x}_{{n}_{i}-1}\right)\right)\underset{i\to \mathrm{\infty }}{lim sup}M\left({x}_{{m}_{i}},{x}_{{n}_{i}-1}\right)\\ \le \epsilon \underset{i\to \mathrm{\infty }}{lim sup}\beta \left(M\left({x}_{{m}_{i}},{x}_{{n}_{i}-1}\right)\right),\end{array}$

which implies that $\frac{1}{s}\le {lim sup}_{i\to \mathrm{\infty }}\beta \left(M\left({x}_{{m}_{i}},{x}_{{n}_{i}-1}\right)\right)$. Now, as $\beta \in \mathcal{F}$ we conclude that $M\left({x}_{{m}_{i}},{x}_{{n}_{i}-1}\right)\to 0$, which yields $d\left({x}_{{m}_{i}},{x}_{{n}_{i}-1}\right)\to 0$. Consequently,

$d\left({x}_{{m}_{i}},{x}_{{n}_{i}}\right)\le sd\left({x}_{{m}_{i}},{x}_{{n}_{i}-1}\right)+sd\left({x}_{{n}_{i}-1},{x}_{{n}_{i}}\right)\to 0,$

which is a contradiction to (2.6). Therefore, $\left\{{x}_{n}\right\}$ is a b-Cauchy sequence. b-Completeness of X shows that $\left\{{x}_{n}\right\}$ b-converges to a point $u\in X$.

Step III: u is a fixed point of f.

First, let f be continuous, so we have

$u=\underset{n\to \mathrm{\infty }}{lim}{x}_{n+1}=\underset{n\to \mathrm{\infty }}{lim}f{x}_{n}=fu.$

Now, let (II) holds. Using the assumption on X we have ${x}_{n}⪯u$. Now, we show that $u=fu$. By Lemma 1

$\begin{array}{rl}\frac{1}{s}d\left(u,fu\right)& \le \underset{n\to \mathrm{\infty }}{lim sup}d\left({x}_{n+1},fu\right)\\ \le \underset{n\to \mathrm{\infty }}{lim sup}\beta \left(M\left({x}_{n},u\right)\right)\underset{n\to \mathrm{\infty }}{lim sup}M\left({x}_{n},u\right),\end{array}$

where

$\begin{array}{rl}\underset{n\to \mathrm{\infty }}{lim}M\left({x}_{n},u\right)& =\underset{n\to \mathrm{\infty }}{lim}max\left\{d\left({x}_{n},u\right),\frac{d\left({x}_{n},f{x}_{n}\right)d\left(u,fu\right)}{1+d\left({x}_{n},u\right)},\frac{d\left({x}_{n},f{x}_{n}\right)d\left(u,fu\right)}{1+d\left(f{x}_{n},fu\right)}\right\}\\ =max\left\{0,0\right\}\\ =0.\end{array}$

Therefore, from the above relations, we deduce that $d\left(u,fu\right)=0$, so $u=fu$.

Finally, suppose that the set of fixed point of f is well ordered. Assume to the contrary that u and v are two fixed points of f such that $u\ne v$. Then by (2.1),

$d\left(u,v\right)=d\left(fu,fv\right)\le \beta \left(M\left(u,v\right)\right)M\left(u,v\right)=\beta \left(d\left(u,v\right)\right)d\left(u,v\right)<\frac{1}{s}d\left(u,v\right),$
(2.9)

because

$M\left(u,v\right)=max\left\{d\left(u,v\right),\frac{d\left(u,u\right)d\left(v,v\right)}{1+d\left(u,v\right)}\right\}=d\left(u,v\right).$

So we get $d\left(u,v\right)<\frac{1}{s}d\left(u,v\right)$, a contradiction. Hence $u=v$, and f has a unique fixed point. Conversely, if f has a unique fixed point, then the set of fixed points of f is a singleton, and so it is well ordered. □

Definition 4 Let $\left(X,d\right)$ be a b-metric space. A mapping $f:X\to X$ is called a rational Geraghty contraction of type II if there exists $\beta \in \mathcal{F}$ such that

$d\left(fx,fy\right)\le \beta \left(M\left(x,y\right)\right)M\left(x,y\right)$
(2.10)

for all comparable elements $x,y\in X$, where

$\begin{array}{rcl}M\left(x,y\right)& =& max\left\{d\left(x,y\right),\frac{d\left(x,fx\right)d\left(x,fy\right)+d\left(y,fy\right)d\left(y,fx\right)}{1+s\left[d\left(x,fx\right)+d\left(y,fy\right)\right]},\\ \frac{d\left(x,fx\right)d\left(x,fy\right)+d\left(y,fy\right)d\left(y,fx\right)}{1+d\left(x,fy\right)+d\left(y,fx\right)}\right\}.\end{array}$

Theorem 6 Let $\left(X,⪯\right)$ be a partially ordered set and suppose that there exists a b-metric d on X such that $\left(X,d\right)$ is a b-complete b-metric space. Let $f:X\to X$ be an increasing mapping with respect to such that there exists an element ${x}_{0}\in X$ with ${x}_{0}⪯f\left({x}_{0}\right)$. Suppose f is a rational Geraghty contractive mapping of type II. If

1. (I)

f is continuous, or,

2. (II)

whenever $\left\{{x}_{n}\right\}$ is a nondecreasing sequence in X such that ${x}_{n}\to u\in X$, one has ${x}_{n}⪯u$ for all $n\in \mathbb{N}$,

then f has a fixed point.

Moreover, the set of fixed points of f is well ordered if and only if f has one and only one fixed point.

Proof Set ${x}_{n}={f}^{n}\left({x}_{0}\right)$. Since ${x}_{0}⪯f\left({x}_{0}\right)$ and f is increasing, we obtain by induction that

${x}_{0}⪯f\left({x}_{0}\right)⪯{f}^{2}\left({x}_{0}\right)⪯\cdots ⪯{f}^{n}\left({x}_{0}\right)⪯{f}^{n+1}\left({x}_{0}\right)⪯\cdots .$

We do the proof in the following steps.

Step I: We show that ${lim}_{n\to \mathrm{\infty }}d\left({x}_{n},{x}_{n+1}\right)=0$. Since ${x}_{n}⪯{x}_{n+1}$ for each $n\in \mathbb{N}$, then by (2.10)

$\begin{array}{rl}d\left({x}_{n},{x}_{n+1}\right)& =d\left(f{x}_{n-1},f{x}_{n}\right)\\ \le \beta \left(M\left({x}_{n-1},{x}_{n}\right)\right)M\left({x}_{n-1},{x}_{n}\right)\\ \le \beta \left(d\left({x}_{n-1},{x}_{n}\right)\right)d\left({x}_{n-1},{x}_{n}\right)\\ <\frac{1}{s}d\left({x}_{n-1},{x}_{n}\right)\\ \le d\left({x}_{n-1},{x}_{n}\right),\end{array}$
(2.11)

because

$\begin{array}{rcl}M\left({x}_{n-1},{x}_{n}\right)& =& max\left\{d\left({x}_{n-1},{x}_{n}\right),\frac{d\left({x}_{n-1},f{x}_{n-1}\right)d\left({x}_{n-1},f{x}_{n}\right)+d\left({x}_{n},f{x}_{n}\right)d\left({x}_{n},f{x}_{n-1}\right)}{1+s\left[d\left({x}_{n-1},f{x}_{n-1}\right)+d\left({x}_{n},f{x}_{n}\right)\right]},\\ \frac{d\left({x}_{n-1},f{x}_{n-1}\right)d\left({x}_{n-1},f{x}_{n}\right)+d\left({x}_{n},f{x}_{n}\right)d\left({x}_{n},f{x}_{n-1}\right)}{1+d\left({x}_{n-1},f{x}_{n}\right)+d\left({x}_{n},f{x}_{n-1}\right)}\right\}\\ =& max\left\{d\left({x}_{n-1},{x}_{n}\right),\frac{d\left({x}_{n-1},{x}_{n}\right)d\left({x}_{n-1},{x}_{n+1}\right)+d\left({x}_{n},{x}_{n+1}\right)d\left({x}_{n},{x}_{n}\right)}{1+s\left[d\left({x}_{n-1},{x}_{n}\right)+d\left({x}_{n},{x}_{n+1}\right)\right]},\\ \frac{d\left({x}_{n-1},{x}_{n}\right)d\left({x}_{n-1},{x}_{n+1}\right)+d\left({x}_{n},{x}_{n+1}\right)d\left({x}_{n},{x}_{n}\right)}{1+d\left({x}_{n-1},{x}_{n+1}\right)+d\left({x}_{n},{x}_{n}\right)}\right\}\\ =& d\left({x}_{n-1},{x}_{n}\right).\end{array}$

Therefore, $\left\{d\left({x}_{n},{x}_{n+1}\right)\right\}$ is decreasing. Then there exists $r\ge 0$ such that ${lim}_{n\to \mathrm{\infty }}d\left({x}_{n},{x}_{n+1}\right)=r$. We will prove that $r=0$. Suppose to the contrary that $r>0$. Then, letting $n\to \mathrm{\infty }$, from (2.11)

$\frac{1}{s}r\le \underset{n\to \mathrm{\infty }}{lim}\beta \left(d\left({x}_{n-1},{x}_{n}\right)\right)r,$

which implies that $d\left({x}_{n-1},{x}_{n}\right)\to 0$. Hence, $r=0$, a contradiction. So,

$\underset{n\to \mathrm{\infty }}{lim}d\left({x}_{n-1},{x}_{n}\right)=0$
(2.12)

holds true.

Step II: Now, we prove that the sequence $\left\{{x}_{n}\right\}$ is a b-Cauchy sequence. Suppose the contrary, i.e., $\left\{{x}_{n}\right\}$ is not a b-Cauchy sequence. Then there exists $\epsilon >0$ for which we can find two subsequences $\left\{{x}_{{m}_{i}}\right\}$ and $\left\{{x}_{{n}_{i}}\right\}$ of $\left\{{x}_{n}\right\}$ such that ${n}_{i}$ is the smallest index for which

${n}_{i}>{m}_{i}>i\phantom{\rule{1em}{0ex}}\text{and}\phantom{\rule{1em}{0ex}}d\left({x}_{{m}_{i}},{x}_{{n}_{i}}\right)\ge \epsilon .$
(2.13)

This means that

$d\left({x}_{{m}_{i}},{x}_{{n}_{i}-1}\right)<\epsilon .$
(2.14)

As in the proof of Theorem 5, we have

$\frac{\epsilon }{s}\le \underset{i\to \mathrm{\infty }}{lim sup}d\left({x}_{{m}_{i}+1},{x}_{{n}_{i}}\right).$
(2.15)

From the definition of $M\left(x,y\right)$ and the above limits,

$\begin{array}{c}\underset{i\to \mathrm{\infty }}{lim sup}M\left({x}_{{m}_{i}},{x}_{{n}_{i}-1}\right)\hfill \\ \phantom{\rule{1em}{0ex}}=\underset{i\to \mathrm{\infty }}{lim sup}max\left\{d\left({x}_{{m}_{i}},{x}_{{n}_{i}-1}\right),\hfill \\ \phantom{\rule{2em}{0ex}}\frac{d\left({x}_{{m}_{i}},f{x}_{{m}_{i}}\right)d\left({x}_{{m}_{i}},f{x}_{{n}_{i}-1}\right)+d\left({x}_{{n}_{i}-1},f{x}_{{n}_{i}-1}\right)d\left({x}_{{n}_{i}-1},f{x}_{{m}_{i}}\right)}{1+s\left[d\left({x}_{{m}_{i}},f{x}_{{m}_{i}}\right)+d\left({x}_{{n}_{i}-1},f{x}_{{n}_{i}-1}\right)\right]},\hfill \\ \phantom{\rule{2em}{0ex}}\frac{d\left({x}_{{m}_{i}},f{x}_{{m}_{i}}\right)d\left({x}_{{m}_{i}},f{x}_{{n}_{i}-1}\right)+d\left({x}_{{n}_{i}-1},f{x}_{{n}_{i}-1}\right)d\left({x}_{{n}_{i}-1},f{x}_{{m}_{i}}\right)}{1+d\left({x}_{{m}_{i}},f{x}_{{n}_{i}-1}\right)+d\left({x}_{{n}_{i}-1},f{x}_{{m}_{i}}\right)}\right\}\hfill \\ \phantom{\rule{1em}{0ex}}=\underset{i\to \mathrm{\infty }}{lim sup}max\left\{d\left({x}_{{m}_{i}},{x}_{{n}_{i}-1}\right),\hfill \\ \phantom{\rule{2em}{0ex}}\frac{d\left({x}_{{m}_{i}},{x}_{{m}_{i}+1}\right)d\left({x}_{{m}_{i}},{x}_{{n}_{i}}\right)+d\left({x}_{{n}_{i}-1},{x}_{{n}_{i}}\right)d\left({x}_{{n}_{i}-1},{x}_{{m}_{i}+1}\right)}{1+s\left[d\left({x}_{{m}_{i}},{x}_{{m}_{i}+1}\right)+d\left({x}_{{n}_{i}-1},{x}_{{n}_{i}}\right)\right]},\hfill \\ \phantom{\rule{2em}{0ex}}\frac{d\left({x}_{{m}_{i}},{x}_{{m}_{i}+1}\right)d\left({x}_{{m}_{i}},{x}_{{n}_{i}}\right)+d\left({x}_{{n}_{i}-1},{x}_{{n}_{i}}\right)d\left({x}_{{n}_{i}-1},{x}_{{m}_{i}+1}\right)}{1+d\left({x}_{{m}_{i}},{x}_{{n}_{i}}\right)+d\left({x}_{{n}_{i}-1},{x}_{{m}_{i}+1}\right)}\right\}\hfill \\ \phantom{\rule{1em}{0ex}}\le \epsilon .\hfill \end{array}$

Now, from (2.10) and the above inequalities, we have

$\begin{array}{rl}\frac{\epsilon }{s}& \le \underset{i\to \mathrm{\infty }}{lim sup}d\left({x}_{{m}_{i}+1},{x}_{{n}_{i}}\right)\le \underset{i\to \mathrm{\infty }}{lim sup}\beta \left(M\left({x}_{{m}_{i}},{x}_{{n}_{i}-1}\right)\right)\underset{i\to \mathrm{\infty }}{lim sup}M\left({x}_{{m}_{i}},{x}_{{n}_{i}-1}\right)\\ \le \epsilon \underset{i\to \mathrm{\infty }}{lim sup}\beta \left(M\left({x}_{{m}_{i}},{x}_{{n}_{i}-1}\right)\right),\end{array}$

which implies that $\frac{1}{s}\le {lim sup}_{i\to \mathrm{\infty }}\beta \left(M\left({x}_{{m}_{i}},{x}_{{n}_{i}-1}\right)\right)$. Now, as $\beta \in \mathcal{F}$ we conclude that $\left\{{x}_{n}\right\}$ is a b-Cauchy sequence. b-Completeness of X shows that $\left\{{x}_{n}\right\}$ b-converges to a point $u\in X$.

Step III: u is a fixed point of f.

First, let f be continuous, so we have

$u=\underset{n\to \mathrm{\infty }}{lim}{x}_{n+1}=\underset{n\to \mathrm{\infty }}{lim}f{x}_{n}=fu.$

Now, let (II) hold. Using the assumption on X we have ${x}_{n}⪯u$. Now, we show that $u=fu$. By Lemma 1

$\begin{array}{rl}\frac{1}{s}d\left(u,fu\right)& \le \underset{n\to \mathrm{\infty }}{lim sup}d\left({x}_{n+1},fu\right)\\ \le \underset{n\to \mathrm{\infty }}{lim sup}\beta \left(M\left({x}_{n},u\right)\right)\underset{n\to \mathrm{\infty }}{lim sup}M\left({x}_{n},u\right)\\ =0,\end{array}$

because

$\begin{array}{rcl}\underset{n\to \mathrm{\infty }}{lim}M\left({x}_{n},u\right)& =& \underset{n\to \mathrm{\infty }}{lim}max\left\{d\left({x}_{n},u\right),\frac{d\left({x}_{n},f{x}_{n}\right)d\left({x}_{n},fu\right)+d\left(u,fu\right)d\left(u,f{x}_{n}\right)}{1+s\left[d\left({x}_{n},f{x}_{n}\right)+d\left(u,fu\right)\right]},\\ \frac{d\left({x}_{n},f{x}_{n}\right)d\left({x}_{n},fu\right)+d\left(u,fu\right)d\left(u,f{x}_{n}\right)}{1+d\left({x}_{n},fu\right)+d\left({x}_{n},fu\right)}\right\}\\ =& max\left\{0,0\right\}\\ =& 0.\end{array}$

Therefore, $d\left(u,fu\right)=0$, so $u=fu$. □

Definition 5 Let $\left(X,d\right)$ be a b-metric space. A mapping $f:X\to X$ is called a rational Geraghty contraction of type III if there exists $\beta \in \mathcal{F}$ such that

$d\left(fx,fy\right)\le \beta \left(M\left(x,y\right)\right)M\left(x,y\right)$
(2.16)

for all comparable elements $x,y\in X$, where

$\begin{array}{rcl}M\left(x,y\right)& =& max\left\{d\left(x,y\right),\frac{d\left(x,fx\right)d\left(y,fy\right)}{1+s\left[d\left(x,y\right)+d\left(x,fy\right)+d\left(y,fx\right)\right]},\\ \frac{d\left(x,fy\right)d\left(x,y\right)}{1+sd\left(x,fx\right)+{s}^{3}\left[d\left(y,fx\right)+d\left(y,fy\right)\right]}\right\}.\end{array}$

Theorem 7 Let $\left(X,⪯\right)$ be a partially ordered set and suppose that there exists a b-metric d on X such that $\left(X,d\right)$ is a b-complete b-metric space. Let $f:X\to X$ be an increasing mapping with respect to such that there exists an element ${x}_{0}\in X$ with ${x}_{0}⪯f\left({x}_{0}\right)$. Suppose f is a rational Geraghty contractive mapping of type III. If

1. (I)

f is continuous, or,

2. (II)

whenever $\left\{{x}_{n}\right\}$ is a nondecreasing sequence in X such that ${x}_{n}\to u\in X$, one has ${x}_{n}⪯u$ for all $n\in \mathbb{N}$,

then f has a fixed point.

Moreover, the set of fixed points of f is well ordered if and only if f has one and only one fixed point.

Proof Set ${x}_{n}={f}^{n}\left({x}_{0}\right)$.

Step I: We show that ${lim}_{n\to \mathrm{\infty }}d\left({x}_{n},{x}_{n+1}\right)=0$. Since ${x}_{n}⪯{x}_{n+1}$ for each $n\in \mathbb{N}$, then by (2.16)

$\begin{array}{rl}d\left({x}_{n},{x}_{n+1}\right)& =d\left(f{x}_{n-1},f{x}_{n}\right)\\ \le \beta \left(M\left({x}_{n-1},{x}_{n}\right)\right)M\left({x}_{n-1},{x}_{n}\right)\\ \le \beta \left(d\left({x}_{n-1},{x}_{n}\right)\right)d\left({x}_{n-1},{x}_{n}\right)\\ <\frac{1}{s}d\left({x}_{n-1},{x}_{n}\right)\\ \le d\left({x}_{n-1},{x}_{n}\right),\end{array}$
(2.17)

because

$\begin{array}{rcl}M\left({x}_{n-1},{x}_{n}\right)& =& max\left\{d\left({x}_{n-1},{x}_{n}\right),\frac{d\left({x}_{n-1},f{x}_{n-1}\right)d\left({x}_{n},f{x}_{n}\right)}{1+s\left[d\left({x}_{n-1},{x}_{n}\right)+d\left({x}_{n-1},f{x}_{n}\right)+d\left({x}_{n},f{x}_{n-1}\right)\right]},\\ \frac{d\left({x}_{n-1},f{x}_{n}\right)d\left({x}_{n-1},{x}_{n}\right)}{1+sd\left({x}_{n-1},f{x}_{n-1}\right)+{s}^{3}\left[d\left({x}_{n},f{x}_{n-1}\right)+d\left({x}_{n},f{x}_{n}\right)\right]}\right\}\\ =& max\left\{d\left({x}_{n-1},{x}_{n}\right),\frac{d\left({x}_{n-1},{x}_{n}\right)d\left({x}_{n},{x}_{n+1}\right)}{1+s\left[d\left({x}_{n-1},{x}_{n}\right)+d\left({x}_{n-1},{x}_{n+1}\right)+d\left({x}_{n},{x}_{n}\right)\right]},\\ \frac{d\left({x}_{n-1},{x}_{n+1}\right)d\left({x}_{n-1},{x}_{n}\right)}{1+sd\left({x}_{n-1},{x}_{n}\right)+{s}^{3}\left[d\left({x}_{n},{x}_{n}\right)+d\left({x}_{n},{x}_{n+1}\right)\right]}\right\}\\ \le & max\left\{d\left({x}_{n-1},{x}_{n}\right),\frac{d\left({x}_{n-1},{x}_{n}\right)s\left[d\left({x}_{n},{x}_{n-1}\right)+d\left({x}_{n-1},{x}_{n+1}\right)\right]}{s\left[d\left({x}_{n-1},{x}_{n}\right)+d\left({x}_{n-1},{x}_{n+1}\right)+d\left({x}_{n},{x}_{n}\right)\right]}\right\}\\ =& d\left({x}_{n-1},{x}_{n}\right).\end{array}$

Therefore, $\left\{d\left({x}_{n},{x}_{n+1}\right)\right\}$ is decreasing. Similar to what we have done in Theorems 5 and 6, we have

$\underset{n\to \mathrm{\infty }}{lim}d\left({x}_{n-1},{x}_{n}\right)=0.$
(2.18)

Step II: Now, we prove that the sequence $\left\{{x}_{n}\right\}$ is a b-Cauchy sequence. Suppose the contrary, i.e., $\left\{{x}_{n}\right\}$ is not a b-Cauchy sequence. Then there exists $\epsilon >0$ for which we can find two subsequences $\left\{{x}_{{m}_{i}}\right\}$ and $\left\{{x}_{{n}_{i}}\right\}$ of $\left\{{x}_{n}\right\}$ such that ${n}_{i}$ is the smallest index for which

${n}_{i}>{m}_{i}>i\phantom{\rule{1em}{0ex}}\text{and}\phantom{\rule{1em}{0ex}}d\left({x}_{{m}_{i}},{x}_{{n}_{i}}\right)\ge \epsilon .$
(2.19)

This means that

$d\left({x}_{{m}_{i}},{x}_{{n}_{i}-1}\right)<\epsilon .$
(2.20)

From (2.18) and using the triangular inequality, we get

$\epsilon \le d\left({x}_{{m}_{i}},{x}_{{n}_{i}}\right)\le sd\left({x}_{{m}_{i}},{x}_{{m}_{i}+1}\right)+sd\left({x}_{{m}_{i}+1},{x}_{{n}_{i}}\right).$

By taking the upper limit as $i\to \mathrm{\infty }$, we get

$\frac{\epsilon }{s}\le \underset{i\to \mathrm{\infty }}{lim sup}d\left({x}_{{m}_{i}+1},{x}_{{n}_{i}}\right).$
(2.21)

Using the triangular inequality, we have

$d\left({x}_{{m}_{i}},{x}_{{n}_{i}}\right)\le sd\left({x}_{{m}_{i}},{x}_{{n}_{i}-1}\right)+sd\left({x}_{{n}_{i}-1},{x}_{{n}_{i}}\right).$

Taking the upper limit as $i\to \mathrm{\infty }$ in the above inequality and using (2.20) we get

$\underset{i\to \mathrm{\infty }}{lim sup}d\left({x}_{{m}_{i}},{x}_{{n}_{i}}\right)\le \epsilon s.$
(2.22)

Again, using the triangular inequality, we have

$d\left({x}_{{m}_{i}},{x}_{{n}_{i}}\right)\le sd\left({x}_{{m}_{i}},{x}_{{m}_{i}+1}\right)+{s}^{2}d\left({x}_{{m}_{i}+1},{x}_{{n}_{i}-1}\right)+{s}^{2}d\left({x}_{{n}_{i}-1},{x}_{{n}_{i}}\right).$

Taking the upper limit as $i\to \mathrm{\infty }$ in the above inequality and using (2.20) we get

$\underset{i\to \mathrm{\infty }}{lim sup}d\left({x}_{{m}_{i}+1},{x}_{{n}_{i}-1}\right)\ge \frac{\epsilon }{{s}^{2}}.$
(2.23)

From the definition of $M\left(x,y\right)$ and the above limits,

$\begin{array}{c}\underset{i\to \mathrm{\infty }}{lim sup}M\left({x}_{{m}_{i}},{x}_{{n}_{i}-1}\right)\hfill \\ \phantom{\rule{1em}{0ex}}=\underset{i\to \mathrm{\infty }}{lim sup}max\left\{d\left({x}_{{m}_{i}},{x}_{{n}_{i}-1}\right),\frac{d\left({x}_{{m}_{i}},f{x}_{{m}_{i}}\right)d\left({x}_{{n}_{i}-1},f{x}_{{n}_{i}-1}\right)}{1+s\left[d\left({x}_{{m}_{i}},{x}_{{n}_{i}-1}\right)+d\left({x}_{{m}_{i}},f{x}_{{n}_{i}-1}\right)+d\left({x}_{{n}_{i}-1},f{x}_{{m}_{i}}\right)\right]},\hfill \\ \phantom{\rule{2em}{0ex}}\frac{d\left({x}_{{m}_{i}},f{x}_{{n}_{i}-1}\right)d\left({x}_{{m}_{i}},{x}_{{n}_{i}-1}\right)}{1+sd\left({x}_{{m}_{i}},f{x}_{{m}_{i}}\right)+{s}^{3}\left[d\left({x}_{{n}_{i}-1},f{x}_{{m}_{i}}\right)+d\left({x}_{{n}_{i}-1},f{x}_{{n}_{i}-1}\right)\right]}\right\}\hfill \\ \phantom{\rule{1em}{0ex}}=\underset{i\to \mathrm{\infty }}{lim sup}max\left\{d\left({x}_{{m}_{i}},{x}_{{n}_{i}-1}\right),\frac{d\left({x}_{{m}_{i}},{x}_{{m}_{i}+1}\right)d\left({x}_{{n}_{i}-1},{x}_{{n}_{i}}\right)}{1+s\left[d\left({x}_{{m}_{i}},{x}_{{n}_{i}-1}\right)+d\left({x}_{{m}_{i}},{x}_{{n}_{i}}\right)+d\left({x}_{{n}_{i}-1},{x}_{{m}_{i}+1}\right)\right]},\hfill \\ \phantom{\rule{2em}{0ex}}\frac{d\left({x}_{{m}_{i}},{x}_{{n}_{i}}\right)d\left({x}_{{m}_{i}},{x}_{{n}_{i}-1}\right)}{1+sd\left({x}_{{m}_{i}},{x}_{{m}_{i}+1}\right)+{s}^{3}\left[d\left({x}_{{n}_{i}-1},{x}_{{m}_{i}+1}\right)+d\left({x}_{{n}_{i}-1},{x}_{{n}_{i}}\right)\right]}\right\}\hfill \\ \phantom{\rule{1em}{0ex}}\le \epsilon .\hfill \end{array}$

Now, from (2.16) and the above inequalities, we have

$\begin{array}{rl}\frac{\epsilon }{s}& \le \underset{i\to \mathrm{\infty }}{lim sup}d\left({x}_{{m}_{i}+1},{x}_{{n}_{i}}\right)\\ \le \underset{i\to \mathrm{\infty }}{lim sup}\beta \left(M\left({x}_{{m}_{i}},{x}_{{n}_{i}-1}\right)\right)\underset{i\to \mathrm{\infty }}{lim sup}M\left({x}_{{m}_{i}},{x}_{{n}_{i}-1}\right)\\ \le \epsilon \underset{i\to \mathrm{\infty }}{lim sup}\beta \left(M\left({x}_{{m}_{i}},{x}_{{n}_{i}-1}\right)\right),\end{array}$

which implies that $\frac{1}{s}\le {lim sup}_{i\to \mathrm{\infty }}\beta \left(M\left({x}_{{m}_{i}},{x}_{{n}_{i}-1}\right)\right)$. Now, as $\beta \in \mathcal{F}$ we conclude that $\left\{{x}_{n}\right\}$ is a b-Cauchy sequence. b-Completeness of X shows that $\left\{{x}_{n}\right\}$ b-converges to a point $u\in X$.

Step III: u is a fixed point of f.

When f is continuous, the proof is straightforward.

Now, let (II) hold. By Lemma 1

$\begin{array}{rl}\frac{1}{s}d\left(u,fu\right)& \le \underset{n\to \mathrm{\infty }}{lim sup}d\left({x}_{n+1},fu\right)\\ \le \underset{n\to \mathrm{\infty }}{lim sup}\beta \left(M\left({x}_{n},u\right)\right)\underset{n\to \mathrm{\infty }}{lim sup}M\left({x}_{n},u\right),\end{array}$

where

$\begin{array}{rcl}\underset{n\to \mathrm{\infty }}{lim}M\left({x}_{n},u\right)& =& \underset{n\to \mathrm{\infty }}{lim}max\left\{d\left({x}_{n},u\right),\frac{d\left({x}_{n},f{x}_{n}\right)d\left(u,fu\right)}{1+s\left[d\left({x}_{n},u\right)+d\left({x}_{n},fu\right)+d\left(u,f{x}_{n}\right)\right]},\\ \frac{d\left({x}_{n},fu\right)d\left({x}_{n},u\right)}{1+sd\left({x}_{n},f{x}_{n}\right)+{s}^{3}\left[d\left(u,fu\right)+d\left(u,f{x}_{n}\right)\right]}\right\}\\ =& max\left\{0,0\right\}\\ =& 0.\end{array}$

Therefore, from the above relations, we deduce that $d\left(u,fu\right)=0$, so $u=fu$. □

If in the above theorems we take $\beta \left(t\right)=r$, where $0\le r<\frac{1}{s}$, then we have the following corollary.

Corollary 1 Let $\left(X,⪯\right)$ be a partially ordered set and suppose that there exists a b-metric d on X such that $\left(X,d\right)$ is a b-complete b-metric space, and let $f:X\to X$ be an increasing mapping with respect to such that there exists an element ${x}_{0}\in X$ with ${x}_{0}⪯f\left({x}_{0}\right)$. Suppose that

$d\left(fx,fy\right)\le rM\left(x,y\right)$

for all comparable elements $x,y\in X$, where

$M\left(x,y\right)=max\left\{d\left(x,y\right),\frac{d\left(x,fx\right)d\left(y,fy\right)}{1+d\left(x,y\right)},\frac{d\left(x,fx\right)d\left(y,fy\right)}{1+d\left(fx,fy\right)}\right\}$

or

$\begin{array}{rcl}M\left(x,y\right)& =& max\left\{d\left(x,y\right),\frac{d\left(x,fx\right)d\left(x,fy\right)+d\left(y,fy\right)d\left(y,fx\right)}{1+s\left[d\left(x,fx\right)+d\left(y,fy\right)\right]},\\ \frac{d\left(x,fx\right)d\left(x,fy\right)+d\left(y,fy\right)d\left(y,fx\right)}{1+d\left(x,fy\right)+d\left(y,fx\right)}\right\},\end{array}$

or

$\begin{array}{rcl}M\left(x,y\right)& =& max\left\{d\left(x,y\right),\frac{d\left(x,fx\right)d\left(y,fy\right)}{1+s\left[d\left(x,y\right)+d\left(x,fy\right)+d\left(y,fx\right)\right]},\\ \frac{d\left(x,fy\right)d\left(x,y\right)}{1+sd\left(x,fx\right)+{s}^{3}\left[d\left(y,fx\right)+d\left(y,fy\right)\right]}\right\}.\end{array}$

If f is continuous, or, for any nondecreasing sequence $\left\{{x}_{n}\right\}$ in X such that ${x}_{n}\to u\in X$ one has ${x}_{n}⪯u$ for all $n\in N$, then f has a fixed point.

Corollary 2 Let $\left(X,⪯\right)$ be a partially ordered set and suppose that there exists a b-metric d on X such that $\left(X,d\right)$ is a b-complete b-metric space, and let $f:X\to X$ be an increasing mapping with respect to such that there exists an element ${x}_{0}\in X$ with ${x}_{0}⪯f\left({x}_{0}\right)$. Suppose

$d\left(fx,fy\right)\le ad\left(x,y\right)+b\frac{d\left(x,fx\right)d\left(y,fy\right)}{1+d\left(x,y\right)}+c\frac{d\left(x,fx\right)d\left(y,fy\right)}{1+d\left(fx,fy\right)}$

or

$\begin{array}{rcl}d\left(fx,fy\right)& \le & ad\left(x,y\right)+b\frac{d\left(x,fx\right)d\left(x,fy\right)+d\left(y,fy\right)d\left(y,fx\right)}{1+s\left[d\left(x,fx\right)+d\left(y,fy\right)\right]}\\ +c\frac{d\left(x,fx\right)d\left(x,fy\right)+d\left(y,fy\right)d\left(y,fx\right)}{1+d\left(x,fy\right)+d\left(y,fx\right)},\end{array}$

or

$\begin{array}{rcl}d\left(fx,fy\right)& \le & ad\left(x,y\right)+b\frac{d\left(x,fx\right)d\left(y,fy\right)}{1+s\left[d\left(x,y\right)+d\left(x,fy\right)+d\left(y,fx\right)\right]}\\ +c\frac{d\left(x,fy\right)d\left(x,y\right)}{1+sd\left(x,fx\right)+{s}^{3}\left[d\left(y,fx\right)+d\left(y,fy\right)\right]}\end{array}$

for all comparable elements $x,y\in X$, where $a,b,c\ge 0$ and $0\le a+b+c<\frac{1}{s}$.

If f is continuous, or, for any nondecreasing sequence $\left\{{x}_{n}\right\}$ in X such that ${x}_{n}\to u\in X$ one has ${x}_{n}⪯u$ for all $n\in \mathbb{N}$, then f has a fixed point.

Corollary 3 Let $\left(X,⪯,d\right)$ be an ordered b-complete b-metric space, and let $f:X\to X$ be an increasing mapping with respect to such that there exists an element ${x}_{0}\in X$ with ${x}_{0}⪯{f}^{m}\left({x}_{0}\right)$ and

$d\left({f}^{m}x,{f}^{m}y\right)\le \beta \left(M\left(x,y\right)\right)M\left(x,y\right)$

for all comparable elements $x,y\in X$, where

$M\left(x,y\right)=max\left\{d\left(x,y\right),\frac{d\left(x,{f}^{m}x\right)d\left(y,{f}^{m}y\right)}{1+d\left(x,y\right)},\frac{d\left(x,{f}^{m}x\right)d\left(y,{f}^{m}y\right)}{1+d\left({f}^{m}x,{f}^{m}y\right)}\right\}$

or

$\begin{array}{rcl}M\left(x,y\right)& =& max\left\{d\left(x,y\right),\frac{d\left(x,{f}^{m}x\right)d\left(x,{f}^{m}y\right)+d\left(y,{f}^{m}y\right)d\left(y,{f}^{m}x\right)}{1+s\left[d\left(x,{f}^{m}x\right)+d\left(y,{f}^{m}y\right)\right]},\\ \frac{d\left(x,{f}^{m}x\right)d\left(x,{f}^{m}y\right)+d\left(y,{f}^{m}y\right)d\left(y,{f}^{m}x\right)}{1+d\left(x,{f}^{m}y\right)+d\left(y,{f}^{m}x\right)}\right\},\end{array}$

or

$\begin{array}{rcl}M\left(x,y\right)& =& max\left\{d\left(x,y\right),\frac{d\left(x,{f}^{m}x\right)d\left(y,{f}^{m}y\right)}{1+s\left[d\left(x,y\right)+d\left(x,{f}^{m}y\right)+d\left(y,{f}^{m}x\right)\right]},\\ \frac{d\left(x,{f}^{m}y\right)d\left(x,y\right)}{1+sd\left(x,{f}^{m}x\right)+{s}^{3}\left[d\left(y,{f}^{m}x\right)+d\left(y,{f}^{m}y\right)\right]}\right\}\end{array}$

for some positive integer m.

If ${f}^{m}$ is continuous, or, for any nondecreasing sequence $\left\{{x}_{n}\right\}$ in X such that ${x}_{n}\to u\in X$ one has ${x}_{n}⪯u$ for all $n\in \mathbb{N}$, then f has a fixed point.

Let Ψ be the family of all nondecreasing functions $\psi :\left[0,\mathrm{\infty }\right)\to \left[0,\mathrm{\infty }\right)$ such that

$\underset{n\to \mathrm{\infty }}{lim}{\psi }^{n}\left(t\right)=0$

for all $t>0$.

Lemma 2 If $\psi \in \Psi$, then the following are satisfied.

1. (a)

$\psi \left(t\right) for all $t>0$;

2. (b)

$\psi \left(0\right)=0$.

As an example ${\psi }_{1}\left(t\right)=kt$, for all $t\ge 0$, where $k\in \left[0,1\right)$, and ${\psi }_{2}\left(t\right)=ln\left(t+1\right)$, for all $t\ge 0$, are in Ψ.

Theorem 8 Let $\left(X,⪯\right)$ be a partially ordered set and suppose that there exists a b-metric d on X such that $\left(X,d\right)$ is a b-complete b-metric space, and let $f:X\to X$ be an increasing mapping with respect to such that there exists an element ${x}_{0}\in X$ with ${x}_{0}⪯f\left({x}_{0}\right)$. Suppose that

$sd\left(fx,fy\right)\le \psi \left(M\left(x,y\right)\right),$
(2.24)

where

$M\left(x,y\right)=max\left\{d\left(x,y\right),\frac{d\left(x,fx\right)d\left(y,fy\right)}{1+d\left(x,y\right)},\frac{d\left(x,fx\right)d\left(y,fy\right)}{1+d\left(fx,fy\right)}\right\}$

for all comparable elements $x,y\in X$. If f is continuous, then f has a fixed point. Moreover, the set of fixed points of f is well ordered if and only if f has one and only one fixed point.

Proof Since ${x}_{0}⪯f\left({x}_{0}\right)$ and f is increasing, we obtain by induction that

${x}_{0}⪯f\left({x}_{0}\right)⪯{f}^{2}\left({x}_{0}\right)⪯\cdots ⪯{f}^{n}\left({x}_{0}\right)⪯{f}^{n+1}\left({x}_{0}\right)⪯\cdots .$

Putting ${x}_{n}={f}^{n}\left({x}_{0}\right)$, we have

${x}_{0}⪯{x}_{1}⪯{x}_{2}⪯\cdots ⪯{x}_{n}⪯{x}_{n+1}⪯\cdots .$

If there exists ${n}_{0}\in \mathbb{N}$ such that ${x}_{{n}_{0}}={x}_{{n}_{0}+1}$ then ${x}_{{n}_{0}}=f{x}_{{n}_{0}}$ and so we have nothing to prove. Hence, we assume that $d\left({x}_{n},{x}_{n+1}\right)>0$, for all $n\in \mathbb{N}$.

In the following steps, we will complete the proof.

Step I: We will prove that

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

Using condition (2.24), we obtain

$d\left({x}_{n+1},{x}_{n}\right)\le sd\left({x}_{n+1},{x}_{n}\right)=sd\left(f{x}_{n},f{x}_{n-1}\right)\le \psi \left(M\left({x}_{n},{x}_{n-1}\right)\right),$

because

$\begin{array}{rcl}M\left({x}_{n-1},{x}_{n}\right)& =& max\left\{d\left({x}_{n-1},{x}_{n}\right),\frac{d\left({x}_{n-1},f{x}_{n-1}\right)d\left({x}_{n},f{x}_{n}\right)}{1+d\left({x}_{n-1},{x}_{n}\right)},\\ \frac{d\left({x}_{n-1},f{x}_{n-1}\right)d\left({x}_{n},f{x}_{n}\right)}{1+d\left(f{x}_{n-1},f{x}_{n}\right)}\right\}\\ =& max\left\{d\left({x}_{n-1},{x}_{n}\right),\frac{d\left({x}_{n-1},{x}_{n}\right)d\left({x}_{n},{x}_{n+1}\right)}{1+d\left({x}_{n-1},{x}_{n}\right)},\frac{d\left({x}_{n-1},{x}_{n}\right)d\left({x}_{n},{x}_{n+1}\right)}{1+d\left({x}_{n},{x}_{n+1}\right)}\right\}\\ \le & max\left\{d\left({x}_{n-1},{x}_{n}\right),d\left({x}_{n},{x}_{n+1}\right)\right\}.\end{array}$

If $max\left\{d\left({x}_{n-1},{x}_{n}\right),d\left({x}_{n},{x}_{n+1}\right)\right\}=d\left({x}_{n},{x}_{n+1}\right)$, then

$\begin{array}{rcl}d\left({x}_{n},{x}_{n+1}\right)& \le & sd\left({x}_{n},{x}_{n+1}\right)=sd\left(f{x}_{n-1},{x}_{n}\right)\\ \le & \psi \left(M\left({x}_{n-1},{x}_{n}\right)\right)
(2.25)

which is a contradiction. Hence, $max\left\{d\left({x}_{n-1},{x}_{n}\right),d\left({x}_{n},{x}_{n+1}\right)\right\}=d\left({x}_{n-1},{x}_{n}\right)$, so from (2.25),

$\begin{array}{rcl}d\left({x}_{n},{x}_{n+1}\right)& \le & sd\left({x}_{n},{x}_{n+1}\right)=sd\left(f{x}_{n-1},{x}_{n}\right)\\ \le & \psi \left(M\left({x}_{n-1},{x}_{n}\right)\right)
(2.26)

Hence,

$d\left({x}_{n},{x}_{n+1}\right)\le sd\left({x}_{n},{x}_{n+1}\right)\le \psi \left(d\left({x}_{n-1},{x}_{n}\right)\right).$

By induction,

$\begin{array}{rl}d\left({x}_{n+1},{x}_{n}\right)& \le \psi \left(d\left({x}_{n},{x}_{n-1}\right)\right)\le {\psi }^{2}\left(d\left({x}_{n-1},{x}_{n-2}\right)\right)\\ \le \cdots \le {\psi }^{n}\left(d\left({x}_{1},{x}_{0}\right)\right).\end{array}$
(2.27)

As $\psi \in \Psi$, we conclude that

$\underset{n\to \mathrm{\infty }}{lim}d\left({x}_{n},{x}_{n+1}\right)=0.$
(2.28)

Step II: Now, we prove that the sequence $\left\{{x}_{n}\right\}$ is a b-Cauchy sequence. Suppose the contrary, i.e., $\left\{{x}_{n}\right\}$ is not a b-Cauchy sequence. Then there exists $\epsilon >0$ for which we can find two subsequences $\left\{{x}_{{m}_{i}}\right\}$ and $\left\{{x}_{{n}_{i}}\right\}$ of $\left\{{x}_{n}\right\}$ such that ${n}_{i}$ is the smallest index for which

${n}_{i}>{m}_{i}>i\phantom{\rule{1em}{0ex}}\text{and}\phantom{\rule{1em}{0ex}}d\left({x}_{{m}_{i}},{x}_{{n}_{i}}\right)\ge \epsilon .$
(2.29)

This means that

$d\left({x}_{{m}_{i}},{x}_{{n}_{i}-1}\right)<\epsilon .$
(2.30)

From (2.29) and using the triangular inequality, we get

$\epsilon \le d\left({x}_{{m}_{i}},{x}_{{n}_{i}}\right)\le sd\left({x}_{{m}_{i}},{x}_{{m}_{i}+1}\right)+sd\left({x}_{{m}_{i}+1},{x}_{{n}_{i}}\right).$

Taking the upper limit as $i\to \mathrm{\infty }$, we get

$\frac{\epsilon }{s}\le \underset{i\to \mathrm{\infty }}{lim sup}d\left({x}_{{m}_{i}+1},{x}_{{n}_{i}}\right).$
(2.31)

From the definition of $M\left(x,y\right)$ and the above limits,

$\begin{array}{c}\underset{i\to \mathrm{\infty }}{lim sup}M\left({x}_{{m}_{i}},{x}_{{n}_{i}-1}\right)\hfill \\ \phantom{\rule{1em}{0ex}}=\underset{i\to \mathrm{\infty }}{lim sup}max\left\{d\left({x}_{{m}_{i}},{x}_{{n}_{i}-1}\right),\frac{d\left({x}_{{m}_{i}},f{x}_{{m}_{i}}\right)d\left({x}_{{n}_{i}-1},f{x}_{{n}_{i}-1}\right)}{1+d\left({x}_{{m}_{i}},{x}_{{n}_{i}-1}\right)},\hfill \\ \phantom{\rule{2em}{0ex}}\frac{d\left({x}_{{m}_{i}},f{x}_{{m}_{i}}\right)d\left({x}_{{n}_{i}-1},f{x}_{{n}_{i}-1}\right)}{1+d\left(f{x}_{{m}_{i}},f{x}_{{n}_{i}-1}\right)}\right\}\hfill \\ \phantom{\rule{1em}{0ex}}=\underset{i\to \mathrm{\infty }}{lim sup}max\left\{d\left({x}_{{m}_{i}},{x}_{{n}_{i}-1}\right),\frac{d\left({x}_{{m}_{i}},{x}_{{m}_{i}+1}\right)d\left({x}_{{n}_{i}-1},{x}_{{n}_{i}}\right)}{1+d\left({x}_{{m}_{i}},{x}_{{n}_{i}-1}\right)},\hfill \\ \phantom{\rule{2em}{0ex}}\frac{d\left({x}_{{m}_{i}},{x}_{{m}_{i}+1}\right)d\left({x}_{{n}_{i}-1},{x}_{{n}_{i}}\right)}{1+d\left({x}_{{m}_{i}+1},{x}_{{n}_{i}}\right)}\right\}\hfill \\ \phantom{\rule{1em}{0ex}}\le \epsilon .\hfill \end{array}$

Now, from (2.24) and the above inequalities, we have

$\begin{array}{rl}\epsilon & =s\cdot \frac{\epsilon }{s}\le s\underset{i\to \mathrm{\infty }}{lim sup}d\left({x}_{{m}_{i}+1},{x}_{{n}_{i}}\right)\\ \le \underset{i\to \mathrm{\infty }}{lim sup}\psi \left(M\left({x}_{{m}_{i}},{x}_{{n}_{i}-1}\right)\right)\\ \le \psi \left(\epsilon \right)<\epsilon ,\end{array}$

which is a contradiction. Consequently, $\left\{{x}_{n}\right\}$ is a b-Cauchy sequence. b-Completeness of X shows that $\left\{{x}_{n}\right\}$ b-converges to a point $u\in X$.

Step III: Now we show that u is a fixed point of f,

$u=\underset{n\to \mathrm{\infty }}{lim}{x}_{n+1}=\underset{n\to \mathrm{\infty }}{lim}f{x}_{n}=fu,$

as f is continuous. □

Theorem 9 Under the same hypotheses as Theorem  8, without the continuity assumption of f, assume that whenever $\left\{{x}_{n}\right\}$ is a nondecreasing sequence in X such that ${x}_{n}\to u\in X$, ${x}_{n}⪯u$ for all $n\in \mathbb{N}$. Then f has a fixed point.

Proof By repeating the proof of Theorem 8, we construct an increasing sequence $\left\{{x}_{n}\right\}$ in X such that ${x}_{n}\to u\in X$. Using the assumption on X we have ${x}_{n}⪯u$. Now we show that $u=fu$. By (2.24) we have

$d\left(fu,{x}_{n}\right)=d\left(fu,f{x}_{n-1}\right)\le \psi \left(M\left(u,{x}_{n-1}\right)\right),$
(2.32)

where

$\begin{array}{rl}M\left(u,{x}_{n-1}\right)& =max\left\{d\left(u,{x}_{n-1}\right),\frac{d\left(u,fu\right)d\left({x}_{n-1},f{x}_{n-1}\right)}{1+d\left(fu,f{x}_{n-1}\right)},\frac{d\left(u,fu\right)d\left({x}_{n-1},f{x}_{n-1}\right)}{1+d\left(u,{x}_{n-1}\right)}\right\}\\ =max\left\{d\left(u,{x}_{n-1}\right),\frac{d\left(u,fu\right)d\left({x}_{n-1},{x}_{n}\right)}{1+d\left(fu,{x}_{n}\right)},\frac{d\left(u,fu\right)d\left({x}_{n-1},{x}_{n}\right)}{1+d\left(u,{x}_{n-1}\right)}\right\}.\end{array}$

Letting $n\to \mathrm{\infty }$,

$\underset{n\to \mathrm{\infty }}{lim sup}M\left(u,{x}_{n-1}\right)=0.$
(2.33)

Again, taking the upper limit as $n\to \mathrm{\infty }$ in (2.32) and using Lemma 1 and (2.33),

$\begin{array}{rl}\frac{1}{s}d\left(fu,u\right)& \le \underset{n\to \mathrm{\infty }}{lim sup}d\left(fu,{x}_{n}\right)\\ \le \underset{n\to \mathrm{\infty }}{lim sup}\psi \left(M\left(u,{x}_{n-1}\right)\right)\\ =0.\end{array}$

So we get $d\left(fu,u\right)=0$, i.e., $fu=u$. □

Remark 1 In Theorems 8 and 9, we can replace $M\left(x,y\right)$ by the following:

$\begin{array}{rcl}M\left(x,y\right)& =& max\left\{d\left(x,y\right),\frac{d\left(x,fx\right)d\left(x,fy\right)+d\left(y,fy\right)d\left(y,fx\right)}{1+s\left[d\left(x,fx\right)+d\left(y,fy\right)\right]},\\ \frac{d\left(x,fx\right)d\left(x,fy\right)+d\left(y,fy\right)d\left(y,fx\right)}{1+d\left(x,fy\right)+d\left(y,fx\right)}\right\}\end{array}$

or

$\begin{array}{rcl}M\left(x,y\right)& =& max\left\{d\left(x,y\right),\frac{d\left(x,fx\right)d\left(y,fy\right)}{1+s\left[d\left(x,y\right)+d\left(x,fy\right)+d\left(y,fx\right)\right]},\\ \frac{d\left(x,fy\right)d\left(x,y\right)}{1+sd\left(x,fx\right)+{s}^{3}\left[d\left(y,fx\right)+d\left(y,fy\right)\right]}\right\}.\end{array}$

Example 2 Let $X=\left\{0,1,3\right\}$ and define the partial order on X by

$⪯:=\left\{\left(0,0\right),\left(1,1\right),\left(3,3\right),\left(0,3\right),\left(3,1\right),\left(0,1\right)\right\}.$

Consider the function $f:X\to X$ given as

$\mathbf{f}=\left(\begin{array}{ccc}0& 1& 3\\ 3& 1& 1\end{array}\right),$

which is increasing with respect to . Let ${x}_{0}=0$. Hence, $f\left({x}_{0}\right)=3$, so ${x}_{0}⪯f{x}_{0}$. Define first the b-metric d on X by $d\left(0,1\right)=6$, $d\left(0,3\right)=9$, $d\left(1,3\right)=\frac{1}{2}$, and $d\left(x,x\right)=0$. Then $\left(X,d\right)$ is a b-complete b-metric space with $s=\frac{18}{13}$. Let $\beta \in \mathcal{F}$ is given by

$\beta \left(t\right)=\frac{13}{18}{e}^{\frac{-t}{9}},\phantom{\rule{1em}{0ex}}t\ge 0$

and $\beta \left(0\right)\in \left[0,\frac{13}{18}\right)$. Then

$d\left(f0,f3\right)=d\left(3,1\right)=\frac{1}{2}\le \beta \left(M\left(0,3\right)\right)M\left(0,3\right)=9\beta \left(9\right).$

This is because

$\begin{array}{rl}M\left(0,3\right)& =max\left\{d\left(0,3\right),\frac{d\left(0,f0\right)d\left(3,f3\right)}{1+d\left(f0,f3\right)},\frac{d\left(0,f0\right)d\left(3,f3\right)}{1+d\left(0,3\right)}\right\}\\ =max\left\{d\left(0,3\right),\frac{d\left(0,3\right)d\left(3,1\right)}{1+d\left(3,1\right)},\frac{d\left(0,3\right)d\left(3,1\right)}{1+d\left(0,3\right)}\right\}=9.\end{array}$

Also,

$d\left(f0,f1\right)=d\left(3,1\right)=\frac{1}{2}\le \beta \left(M\left(0,1\right)\right)M\left(0,1\right)=6\beta \left(6\right),$

because

$\begin{array}{rl}M\left(0,1\right)& =max\left\{d\left(0,1\right),\frac{d\left(0,f0\right)d\left(1,f1\right)}{1+d\left(f0,f1\right)},\frac{d\left(0,f0\right)d\left(1,f1\right)}{1+d\left(0,1\right)}\right\}\\ =max\left\{d\left(0,1\right),\frac{d\left(0,3\right)d\left(1,1\right)}{1+d\left(3,1\right)},\frac{d\left(0,3\right)d\left(1,1\right)}{1+d\left(0,1\right)}\right\}=6.\end{array}$

Also,

$d\left(f1,f3\right)=d\left(1,1\right)=0\le \beta \left(M\left(1,3\right)\right)M\left(1,3\right).$

Hence, f satisfies all the assumptions of Theorem 5 and thus it has a fixed point (which is $u=1$).

Example 3 Let $X=\left[0,1\right]$ be equipped with the usual order and b-complete b-metric given by $d\left(x,y\right)=|x-y{|}^{2}$ with $s=2$. Consider the mapping $f:X\to X$ defined by $f\left(x\right)=\frac{1}{16}{x}^{2}{e}^{-{x}^{2}}$ and the function β given by $\beta \left(t\right)=\frac{1}{4}$. It is easy to see that f is an increasing function and $0\le f\left(0\right)=0$. For all comparable elements $x,y\in X$, by the mean value theorem, we have

$\begin{array}{rl}d\left(fx,fy\right)& =|\frac{1}{16}{x}^{2}{e}^{-{x}^{2}}-\frac{1}{16}{y}^{2}{e}^{-{y}^{2}}{|}^{2}\\ \le \frac{1}{8}|{x}^{2}{e}^{-{x}^{2}}-{y}^{2}{e}^{-{y}^{2}}{|}^{2}\\ \le \frac{1}{8}|x-y{|}^{2}\le \frac{1}{4}d\left(x,y\right)=\beta \left(d\left(x,y\right)\right)d\left(x,y\right)\\ \le \beta \left(M\left(x,y\right)\right)M\left(x,y\right).\end{array}$

So, from Theorem 5, f has a fixed point.

Example 4 Let $X=\left[0,1\right]$ be equipped with the usual order and b-complete b-metric d be given by $d\left(x,y\right)=|x-y{|}^{2}$ with $s=2$. Consider the mapping $f:X\to X$ defined by $f\left(x\right)=\frac{1}{4}ln\left({x}^{2}+1\right)$ and the function $\psi \in \Psi$ given by $\psi \left(t\right)=\frac{1}{4}t$, $t\ge 0$. It is easy to see that f is increasing and $0\le f\left(0\right)=0$. For all comparable elements $x,y\in X$, using the mean value problem, we have

$\begin{array}{rl}d\left(fx,fy\right)& =|\frac{1}{4}ln\left({x}^{2}+1\right)-\frac{1}{4}ln\left({y}^{2}+1\right){|}^{2}\\ \le \frac{1}{4}|x-y{|}^{2}\\ =\frac{1}{4}d\left(x,y\right)=\psi \left(d\left(x,y\right)\right)\le \psi \left(M\left(x,y\right)\right),\end{array}$

so, using Theorem 8, f has a fixed point.

## 3 Application

In this section, we present an application where Theorem 8 can be applied. This application is inspired by [9] (also, see [26] and [27]).

Let $X=C\left(\left[0,T\right]\right)$ be the set of all real continuous functions on $\left[0,T\right]$. We first endow X with the b-metric

$d\left(u,v\right)=\underset{t\in \left[0,T\right]}{max}{\left(|u\left(t\right)-v\left(t\right)|\right)}^{p}$

for all $u,v\in X$ where $p>1$. Clearly, $\left(X,d\right)$ is a complete b-metric space with parameter $s={2}^{p-1}$. Secondly, $C\left(\left[0,T\right]\right)$ can also be equipped with a partial order given by

Moreover, as in [9] it is proved that $\left(C\left(\left[0,T\right]\right),⪯\right)$ is regular, that is, whenever $\left\{{x}_{n}\right\}$ in X is an increasing sequence such that ${x}_{n}\to x$ as $n\to \mathrm{\infty }$, we have ${x}_{n}⪯x$ for all $n\in \mathbb{N}\cup \left\{0\right\}$.

Consider the first-order periodic boundary value problem

$\left\{\begin{array}{l}{x}^{\prime }\left(t\right)=f\left(t,x\left(t\right)\right),\\ x\left(0\right)=x\left(T\right),\end{array}$
(3.1)

where $t\in I=\left[0,T\right]$ with $T>0$ and $f:\left[0,T\right]×\mathbb{R}\to \mathbb{R}$ is a continuous function.

A lower solution for (3.1) is a function $\alpha \in {C}^{1}\left[0,T\right]$ such that

$\left\{\begin{array}{l}{\alpha }^{\prime }\left(t\right)\le f\left(t,\alpha \left(t\right)\right),\\ \alpha \left(0\right)\le \alpha \left(T\right),\end{array}$
(3.2)

where $t\in I=\left[0,T\right]$.

Assume that there exists $\lambda >0$ such that for all $x,y\in X$ we have

$|f\left(t,x\left(t\right)\right)+\lambda x\left(t\right)-f\left(t,y\left(t\right)\right)-\lambda y\left(t\right)|\le \frac{\lambda }{{2}^{p-1}}\sqrt[p]{ln\left(|x\left(t\right)-y\left(t\right){|}^{p}+1\right)}.$
(3.3)

Then the existence of a lower solution for (3.1) provides the existence of an unique solution of (3.1).

Problem (3.1) can be rewritten as

$\left\{\begin{array}{l}{x}^{\prime }\left(t\right)+\lambda x\left(t\right)=f\left(t,x\left(t\right)\right)+\lambda x\left(t\right),\\ x\left(0\right)=x\left(T\right).\end{array}$

Consider

$\left\{\begin{array}{l}{x}^{\prime }\left(t\right)+\lambda x\left(t\right)=\delta \left(t\right)=F\left(t,x\left(t\right)\right),\\ x\left(0\right)=x\left(T\right),\end{array}$

where $t\in I$.

Using the variation of parameters formula, we get

$x\left(t\right)=x\left(0\right){e}^{-\lambda t}+{\int }_{0}^{t}{e}^{-\lambda \left(t-s\right)}\delta \left(s\right)\phantom{\rule{0.2em}{0ex}}ds,$
(3.4)

which yields

$x\left(T\right)=x\left(0\right){e}^{-\lambda T}+{\int }_{0}^{T}{e}^{-\lambda \left(T-s\right)}\delta \left(s\right)\phantom{\rule{0.2em}{0ex}}ds.$

Since $x\left(0\right)=x\left(T\right)$, we get

$x\left(0\right)\left[1-{e}^{-\lambda T}\right]={e}^{-\lambda T}{\int }_{0}^{T}{e}^{\lambda \left(s\right)}\delta \left(s\right)\phantom{\rule{0.2em}{0ex}}ds$

or

$x\left(0\right)=\frac{1}{{e}^{\lambda T}-1}{\int }_{0}^{T}{e}^{\lambda s}\delta \left(s\right)\phantom{\rule{0.2em}{0ex}}ds.$

Substituting the value of $x\left(0\right)$ in (3.4) we arrive at

$x\left(t\right)={\int }_{0}^{T}G\left(t,s\right)\delta \left(s\right)\phantom{\rule{0.2em}{0ex}}ds,$

where

$G\left(t,s\right)=\left\{\begin{array}{ll}\frac{{e}^{\lambda \left(T+s-t\right)}}{{e}^{\lambda T}-1},& 0\le s\le t\le T,\\ \frac{{e}^{\lambda \left(s-t\right)}}{{e}^{\lambda T}-1},& 0\le t\le s\le T.\end{array}$

Now define the operator $S:C\left[0,T\right]\to C\left[0,T\right]$ by

$Sx\left(t\right)={\int }_{0}^{T}G\left(t,s\right)F\left(s,x\left(s\right)\right)\phantom{\rule{0.2em}{0ex}}ds.$

The mapping S is nondecreasing [26]. Note that if $u\in C\left[0,T\right]$ is a fixed point of S then $u\in {C}^{1}\left[0,T\right]$ is a solution of (3.1).

Let $x,y\in X$. Then we have

$\begin{array}{rl}{2}^{p-1}|Sx\left(t\right)-Sy\left(t\right)|& ={2}^{p-1}|{\int }_{0}^{T}G\left(t,s\right)F\left(s,x\left(s\right)\right)\phantom{\rule{0.2em}{0ex}}ds-{\int }_{0}^{T}G\left(t,s\right)F\left(s,y\left(s\right)\right)\phantom{\rule{0.2em}{0ex}}ds|\\ \le {2}^{p-1}{\int }_{0}^{T}|G\left(t,s\right)|\left[|F\left(s,x\left(s\right)\right)-F\left(s,y\left(s\right)\right)|\right]\phantom{\rule{0.2em}{0ex}}ds\\ \le {2}^{p-1}{\int }_{0}^{T}|G\left(t,s\right)|\frac{\lambda }{{2}^{p-1}}\sqrt[p]{ln\left(|x\left(t\right)-y\left(t\right){|}^{p}+1\right)}\phantom{\rule{0.2em}{0ex}}ds\\ \le \lambda \sqrt[p]{ln\left(d\left(x,y\right)+1\right)}\left[{\int }_{0}^{t}\frac{{e}^{\lambda \left(T+s-t\right)}}{{e}^{\lambda T}-1}\phantom{\rule{0.2em}{0ex}}ds+{\int }_{t}^{T}\frac{{e}^{\lambda \left(s-t\right)}}{{e}^{\lambda T}-1}\phantom{\rule{0.2em}{0ex}}ds\right]\\ =\lambda \sqrt[p]{ln\left(d\left(x,y\right)+1\right)}\left[\frac{1}{\lambda \left({e}^{\lambda T}-1\right)}\left({e}^{\lambda \left(T+s-t\right)}{|}_{0}^{t}+{e}^{\lambda \left(s-t\right)}{|}_{t}^{T}\right)\right]\\ =\lambda \sqrt[p]{ln\left(d\left(x,y\right)+1\right)}\left[\frac{1}{\lambda \left({e}^{\lambda T}-1\right)}\left({e}^{\lambda T}-{e}^{\lambda \left(T-t\right)}+{e}^{\lambda \left(T-t\right)}-1\right)\right]\\ =\sqrt[p]{ln\left(d\left(x,y\right)+1\right)}\\ \le \sqrt[p]{ln\left(M\left(x,y\right)+1\right)},\end{array}$

or, equivalently,

${2}^{p-1}{\left(|Sx\left(t\right)-Sy\left(t\right)|\right)}^{p}\le ln\left(M\left(x,y\right)+1\right),$

which shows that

${2}^{p-1}d\left(Sx,Sy\right)\le ln\left(M\left(x,y\right)+1\right),$

where

$M\left(x,y\right)=max\left\{d\left(x,y\right),\frac{d\left(x,Sx\right)d\left(y,Sy\right)}{1+d\left(x,y\right)},\frac{d\left(x,Sx\right)d\left(y,Sy\right)}{1+d\left(Sx,Sy\right)}\right\}$

or

$\begin{array}{rcl}M\left(x,y\right)& =& max\left\{d\left(x,y\right),\frac{d\left(x,Sx\right)d\left(x,Sy\right)+d\left(y,Sy\right)d\left(y,Sx\right)}{1+{2}^{p-1}\left[d\left(x,Sx\right)+d\left(y,Sy\right)\right]},\\ \frac{d\left(x,Sx\right)d\left(x,Sy\right)+d\left(y,Sy\right)d\left(y,Sx\right)}{1+d\left(x,Sy\right)+d\left(y,Sx\right)}\right\},\end{array}$

or

$\begin{array}{rcl}M\left(x,y\right)& =& max\left\{d\left(x,y\right),\frac{d\left(x,Sx\right)d\left(y,Sy\right)}{1+{2}^{p-1}\left[d\left(x,y\right)+d\left(x,Sy\right)+d\left(y,Sx\right)\right]},\\ \frac{d\left(x,Sy\right)d\left(x,y\right)}{1+{2}^{p-1}d\left(x,Sx\right)+{2}^{3p-3}\left[d\left(y,Sx\right)+d\left(y,Sy\right)\right]}\right\}.\end{array}$

Finally, let α be a lower solution for (3.1). In [26] it was shown that $\alpha ⪯S\left(\alpha \right)$.

Hence, the hypotheses of Theorem 8 are satisfied with $\psi \left(t\right)=ln\left(t+1\right)$. Therefore, there exists a fixed point $\stackrel{ˆ}{x}\in C\left[0,T\right]$ such that $S\stackrel{ˆ}{x}=\stackrel{ˆ}{x}$.

Remark 2 In the above theorem, we can replace (3.3) by the following inequality:

$|f\left(t,x\left(t\right)\right)+\lambda x\left(t\right)-f\left(t,y\left(t\right)\right)-\lambda y\left(t\right)|\le \frac{\lambda }{{2}^{\frac{{p}^{2}-1}{p}}}\sqrt[p]{{e}^{-M\left(x,y\right)}M\left(x,y\right)}$
(3.5)

for all $x\ne y\in X$.

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## Acknowledgements

The authors are grateful to the referees for valuable remarks that helped them to improve the exposition in the paper.

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Correspondence to Abdolrahman Razani.

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All authors contributed equally and significantly in writing this paper. All authors read and approved the final manuscript.

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Shahkoohi, R.J., Razani, A. Some fixed point theorems for rational Geraghty contractive mappings in ordered b-metric spaces. J Inequal Appl 2014, 373 (2014). https://doi.org/10.1186/1029-242X-2014-373