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., [11–16] and [17, 18]).

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

(b₁) $d(x,y)=0$ iff $x=y$,

(b₂) $d(x,y)=d(y,x)$,

(b₃) $d(x,z)\le s[d(x,y)+d(y,z)]$

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

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

Definition 2 [19]

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

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

• (b) $\{x_n\}$ in X is said to be b-Cauchy if and only if $d(x_n,x_m)\to 0$, as $n,m\to \mathrm{\infty }$.

• (c) The b-metric space $(X,d)$ 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 $X=\mathbb{N}\cup \{\mathrm{\infty }\}$ and $d:X\times X\to \mathbb{R}$ be defined by

$$d(m,n)=\{\begin{array}{cc}0,\hfill & \text{if }m=n,\hfill \\ |\frac{1}{m}-\frac{1}{n}|,\hfill & \text{if one of }m,n\text{ is even and the other is even or }\mathrm{\infty },\hfill \\ 5,\hfill & \text{if one of }m,n\text{ is odd and the other is odd }(\text{and }m\ne n)\text{ or }\mathrm{\infty },\hfill \\ 2,\hfill & \text{otherwise}.\hfill \end{array}$$

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

Lemma 1 [21]

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

$$\frac{1}{s^2}d(x,y)\le \underset{n\to \mathrm{\infty }}{lim\hspace{0.17em}inf}d(x_n,y_n)\le \underset{n\to \mathrm{\infty }}{lim\hspace{0.17em}sup}d(x_n,y_n)\le s^{2}d(x,y).$$

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

$$\frac{1}{s}d(x,z)\le \underset{n\to \mathrm{\infty }}{lim\hspace{0.17em}inf}d(x_n,z)\le \underset{n\to \mathrm{\infty }}{lim\hspace{0.17em}sup}d(x_n,z)\le sd(x,z).$$

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

$$\beta (t_n)\to 1\text{implies that}{t}_n\to 0,\text{as }n\to \mathrm{\infty }.$$

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

Theorem 1 [22]

Let $(X,d)$ 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

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

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 $\{f^{n}x\}$ 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 $(X,d,⪯)$ 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

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

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 $\{x_n\}$ 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 ℱ of functions $\beta :[0,\mathrm{\infty })\to [0,1/s)$ such that

$$\beta (t_n)\to \frac{1}{s}\text{implies that}{t}_n\to 0,\text{as }n\to \mathrm{\infty }.$$

Theorem 3 [24]

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

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

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 $\{f^{n}x\}$ converges to z in $(X,D,s)$.

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

Theorem 4 Let $(X,⪯)$ be a partially ordered set and suppose that there exists a b-metric d on X such that $(X,d)$ 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 $x_0⪯f(x_0)$. Suppose there exists $\beta \in \mathcal{F}$ such that

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

(1.1)

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

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

and

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

If f is continuous, or, whenever $\{x_n\}$ 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 ℱ denotes the class of all functions $\beta :[0,\mathrm{\infty })\to [0,\frac{1}{s})$ satisfying the following condition:

$$\underset{n\to \mathrm{\infty }}{lim\hspace{0.17em}sup}\beta (t_n)=\frac{1}{s}\text{implies that}{t}_n\to 0,\text{as }n\to \mathrm{\infty }.$$

Definition 3 Let $(X,d,⪯)$ 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

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

(2.1)

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

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

Theorem 5 Let $(X,⪯)$ be a partially ordered set and suppose there exists a b-metric d on X such that $(X,d)$ 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 $x_0⪯f(x_0)$. Suppose f is a rational Geraghty contraction of type I. If

• (I) f is continuous, or,

• (II) whenever $\{x_n\}$ 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 $x_n=f^n(x_0)$ for all $n\ge 0$. Since $x_0⪯f(x_0)$ and f is increasing, we obtain by induction that

$$x_0⪯f(x_0)⪯f^2(x_0)⪯\cdots ⪯f^n(x_0)⪯f^{n+1}(x_0)⪯\cdots .$$

We do the proof in the following steps.

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

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

(2.2)

where

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

If $max\{d(x_{n-1},x_n),d(x_n,x_{n+1})\}=d(x_n,x_{n+1})$, then from (2.2),

$$\begin{array}{rl}d(x_n,x_{n+1})& \le \beta (M(x_n,x_{n+1}))d(x_n,x_{n+1})\\ <\frac{1}{s}d(x_n,x_{n+1})\\ <d(x_n,x_{n+1}),\end{array}$$

(2.3)

which is a contradiction.

Hence, $max\{d(x_{n-1},x_n),d(x_n,x_{n+1})\}=d(x_{n-1},x_n)$, so from (2.2),

$$d(x_n,x_{n+1})\le \beta (M(x_{n-1},x_n))d(x_{n-1},x_n).$$

(2.4)

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

$$r\le \underset{n\to \mathrm{\infty }}{lim}\beta (M(x_{n-1},x_n))r,$$

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

$$\underset{n\to \mathrm{\infty }}{lim}d(x_{n-1},x_n)=0.$$

(2.5)

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

$$n_i>m_i>i\text{and}d(x_{m_i},x_{n_i})\ge \epsilon .$$

(2.6)

This means that

$$d(x_{m_i},x_{n_i-1})<\epsilon .$$

(2.7)

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

$$\epsilon \le d(x_{m_i},x_{n_i})\le sd(x_{m_i},x_{m_i+1})+sd(x_{m_i+1},x_{n_i}).$$

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

$$\frac{\epsilon }{s}\le \underset{i\to \mathrm{\infty }}{lim\hspace{0.17em}sup}d(x_{m_i+1},x_{n_i}).$$

(2.8)

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

$$\begin{array}{c}\underset{i\to \mathrm{\infty }}{lim\hspace{0.17em}sup}M(x_{m_i},x_{n_i-1})\hfill \\ =\underset{i\to \mathrm{\infty }}{lim\hspace{0.17em}sup}max\{d(x_{m_i},x_{n_i-1}),\frac{d(x_{m_i},f{x}_{m_i})d(x_{n_i-1},f{x}_{n_i-1})}{1+d(x_{m_i},x_{n_i-1})},\hfill \\ \frac{d(x_{m_i},f{x}_{m_i})d(x_{n_i-1},f{x}_{n_i-1})}{1+d(f{x}_{m_i},f{x}_{n_i-1})}\}\hfill \\ =\underset{i\to \mathrm{\infty }}{lim\hspace{0.17em}sup}max\{d(x_{m_i},x_{n_i-1}),\frac{d(x_{m_i},x_{m_i+1})d(x_{n_i-1},x_{n_i})}{1+d(x_{m_i},x_{n_i-1})},\hfill \\ \frac{d(x_{m_i},x_{m_i+1})d(x_{n_i-1},x_{n_i})}{1+d(x_{m_i+1},x_{n_i})}\}\hfill \\ \le \epsilon .\hfill \end{array}$$

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

$$\begin{array}{rl}\frac{\epsilon }{s}& \le \underset{i\to \mathrm{\infty }}{lim\hspace{0.17em}sup}d(x_{m_i+1},x_{n_i})\\ \le \underset{i\to \mathrm{\infty }}{lim\hspace{0.17em}sup}\beta (M(x_{m_i},x_{n_i-1}))\underset{i\to \mathrm{\infty }}{lim\hspace{0.17em}sup}M(x_{m_i},x_{n_i-1})\\ \le \epsilon \underset{i\to \mathrm{\infty }}{lim\hspace{0.17em}sup}\beta (M(x_{m_i},x_{n_i-1})),\end{array}$$

which implies that $\frac{1}{s}\le {lim\hspace{0.17em}sup}_{i\to \mathrm{\infty }}\beta (M(x_{m_i},x_{n_i-1}))$. Now, as $\beta \in \mathcal{F}$ we conclude that $M(x_{m_i},x_{n_i-1})\to 0$, which yields $d(x_{m_i},x_{n_i-1})\to 0$. Consequently,

$$d(x_{m_i},x_{n_i})\le sd(x_{m_i},x_{n_i-1})+sd(x_{n_i-1},x_{n_i})\to 0,$$

which is a contradiction to (2.6). Therefore, $\{x_n\}$ is a b-Cauchy sequence. b-Completeness of X shows that $\{x_n\}$ 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

$$\begin{array}{rl}\frac{1}{s}d(u,fu)& \le \underset{n\to \mathrm{\infty }}{lim\hspace{0.17em}sup}d(x_{n+1},fu)\\ \le \underset{n\to \mathrm{\infty }}{lim\hspace{0.17em}sup}\beta (M(x_n,u))\underset{n\to \mathrm{\infty }}{lim\hspace{0.17em}sup}M(x_n,u),\end{array}$$

where

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

Therefore, from the above relations, we deduce that $d(u,fu)=0$, 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),

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

(2.9)

because

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

So we get $d(u,v)<\frac{1}{s}d(u,v)$, 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 $(X,d)$ 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

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

(2.10)

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

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

Theorem 6 Let $(X,⪯)$ be a partially ordered set and suppose that there exists a b-metric d on X such that $(X,d)$ 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 $x_0⪯f(x_0)$. Suppose f is a rational Geraghty contractive mapping of type II. If

• (I) f is continuous, or,

• (II) whenever $\{x_n\}$ 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 $x_n=f^n(x_0)$. Since $x_0⪯f(x_0)$ and f is increasing, we obtain by induction that

$$x_0⪯f(x_0)⪯f^2(x_0)⪯\cdots ⪯f^n(x_0)⪯f^{n+1}(x_0)⪯\cdots .$$

We do the proof in the following steps.

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

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

(2.11)

because

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

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

$$\frac{1}{s}r\le \underset{n\to \mathrm{\infty }}{lim}\beta (d(x_{n-1},x_n))r,$$

which implies that $d(x_{n-1},x_n)\to 0$. Hence, $r=0$, a contradiction. So,

$$\underset{n\to \mathrm{\infty }}{lim}d(x_{n-1},x_n)=0$$

(2.12)

holds true.

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

$$n_i>m_i>i\text{and}d(x_{m_i},x_{n_i})\ge \epsilon .$$

(2.13)

This means that

$$d(x_{m_i},x_{n_i-1})<\epsilon .$$

(2.14)

As in the proof of Theorem 5, we have

$$\frac{\epsilon }{s}\le \underset{i\to \mathrm{\infty }}{lim\hspace{0.17em}sup}d(x_{m_i+1},x_{n_i}).$$

(2.15)

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

$$\begin{array}{c}\underset{i\to \mathrm{\infty }}{lim\hspace{0.17em}sup}M(x_{m_i},x_{n_i-1})\hfill \\ =\underset{i\to \mathrm{\infty }}{lim\hspace{0.17em}sup}max\{d(x_{m_i},x_{n_i-1}),\hfill \\ \frac{d(x_{m_i},f{x}_{m_i})d(x_{m_i},f{x}_{n_i-1})+d(x_{n_i-1},f{x}_{n_i-1})d(x_{n_i-1},f{x}_{m_i})}{1+s[d(x_{m_i},f{x}_{m_i})+d(x_{n_i-1},f{x}_{n_i-1})]},\hfill \\ \frac{d(x_{m_i},f{x}_{m_i})d(x_{m_i},f{x}_{n_i-1})+d(x_{n_i-1},f{x}_{n_i-1})d(x_{n_i-1},f{x}_{m_i})}{1+d(x_{m_i},f{x}_{n_i-1})+d(x_{n_i-1},f{x}_{m_i})}\}\hfill \\ =\underset{i\to \mathrm{\infty }}{lim\hspace{0.17em}sup}max\{d(x_{m_i},x_{n_i-1}),\hfill \\ \frac{d(x_{m_i},x_{m_i+1})d(x_{m_i},x_{n_i})+d(x_{n_i-1},x_{n_i})d(x_{n_i-1},x_{m_i+1})}{1+s[d(x_{m_i},x_{m_i+1})+d(x_{n_i-1},x_{n_i})]},\hfill \\ \frac{d(x_{m_i},x_{m_i+1})d(x_{m_i},x_{n_i})+d(x_{n_i-1},x_{n_i})d(x_{n_i-1},x_{m_i+1})}{1+d(x_{m_i},x_{n_i})+d(x_{n_i-1},x_{m_i+1})}\}\hfill \\ \le \epsilon .\hfill \end{array}$$

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

$$\begin{array}{rl}\frac{\epsilon }{s}& \le \underset{i\to \mathrm{\infty }}{lim\hspace{0.17em}sup}d(x_{m_i+1},x_{n_i})\le \underset{i\to \mathrm{\infty }}{lim\hspace{0.17em}sup}\beta (M(x_{m_i},x_{n_i-1}))\underset{i\to \mathrm{\infty }}{lim\hspace{0.17em}sup}M(x_{m_i},x_{n_i-1})\\ \le \epsilon \underset{i\to \mathrm{\infty }}{lim\hspace{0.17em}sup}\beta (M(x_{m_i},x_{n_i-1})),\end{array}$$

which implies that $\frac{1}{s}\le {lim\hspace{0.17em}sup}_{i\to \mathrm{\infty }}\beta (M(x_{m_i},x_{n_i-1}))$. Now, as $\beta \in \mathcal{F}$ we conclude that $\{x_n\}$ is a b-Cauchy sequence. b-Completeness of X shows that $\{x_n\}$ 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

$$\begin{array}{rl}\frac{1}{s}d(u,fu)& \le \underset{n\to \mathrm{\infty }}{lim\hspace{0.17em}sup}d(x_{n+1},fu)\\ \le \underset{n\to \mathrm{\infty }}{lim\hspace{0.17em}sup}\beta (M(x_n,u))\underset{n\to \mathrm{\infty }}{lim\hspace{0.17em}sup}M(x_n,u)\\ =0,\end{array}$$

because

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

Therefore, $d(u,fu)=0$, so $u=fu$. □

Definition 5 Let $(X,d)$ 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

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

(2.16)

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

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

Theorem 7 Let $(X,⪯)$ be a partially ordered set and suppose that there exists a b-metric d on X such that $(X,d)$ 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 $x_0⪯f(x_0)$. Suppose f is a rational Geraghty contractive mapping of type III. If

• (I) f is continuous, or,

• (II) whenever $\{x_n\}$ 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 $x_n=f^n(x_0)$.

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

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

(2.17)

because

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

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

$$\underset{n\to \mathrm{\infty }}{lim}d(x_{n-1},x_n)=0.$$

(2.18)

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

$$n_i>m_i>i\text{and}d(x_{m_i},x_{n_i})\ge \epsilon .$$

(2.19)

This means that

$$d(x_{m_i},x_{n_i-1})<\epsilon .$$

(2.20)

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

$$\epsilon \le d(x_{m_i},x_{n_i})\le sd(x_{m_i},x_{m_i+1})+sd(x_{m_i+1},x_{n_i}).$$

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

$$\frac{\epsilon }{s}\le \underset{i\to \mathrm{\infty }}{lim\hspace{0.17em}sup}d(x_{m_i+1},x_{n_i}).$$

(2.21)

Using the triangular inequality, we have

$$d(x_{m_i},x_{n_i})\le sd(x_{m_i},x_{n_i-1})+sd(x_{n_i-1},x_{n_i}).$$

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

$$\underset{i\to \mathrm{\infty }}{lim\hspace{0.17em}sup}d(x_{m_i},x_{n_i})\le \epsilon s.$$

(2.22)

Again, using the triangular inequality, we have

$$d(x_{m_i},x_{n_i})\le sd(x_{m_i},x_{m_i+1})+s^{2}d(x_{m_i+1},x_{n_i-1})+s^{2}d(x_{n_i-1},x_{n_i}).$$

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

$$\underset{i\to \mathrm{\infty }}{lim\hspace{0.17em}sup}d(x_{m_i+1},x_{n_i-1})\ge \frac{\epsilon }{s^2}.$$

(2.23)

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

$$\begin{array}{c}\underset{i\to \mathrm{\infty }}{lim\hspace{0.17em}sup}M(x_{m_i},x_{n_i-1})\hfill \\ =\underset{i\to \mathrm{\infty }}{lim\hspace{0.17em}sup}max\{d(x_{m_i},x_{n_i-1}),\frac{d(x_{m_i},f{x}_{m_i})d(x_{n_i-1},f{x}_{n_i-1})}{1+s[d(x_{m_i},x_{n_i-1})+d(x_{m_i},f{x}_{n_i-1})+d(x_{n_i-1},f{x}_{m_i})]},\hfill \\ \frac{d(x_{m_i},f{x}_{n_i-1})d(x_{m_i},x_{n_i-1})}{1+sd(x_{m_i},f{x}_{m_i})+s^3[d(x_{n_i-1},f{x}_{m_i})+d(x_{n_i-1},f{x}_{n_i-1})]}\}\hfill \\ =\underset{i\to \mathrm{\infty }}{lim\hspace{0.17em}sup}max\{d(x_{m_i},x_{n_i-1}),\frac{d(x_{m_i},x_{m_i+1})d(x_{n_i-1},x_{n_i})}{1+s[d(x_{m_i},x_{n_i-1})+d(x_{m_i},x_{n_i})+d(x_{n_i-1},x_{m_i+1})]},\hfill \\ \frac{d(x_{m_i},x_{n_i})d(x_{m_i},x_{n_i-1})}{1+sd(x_{m_i},x_{m_i+1})+s^3[d(x_{n_i-1},x_{m_i+1})+d(x_{n_i-1},x_{n_i})]}\}\hfill \\ \le \epsilon .\hfill \end{array}$$

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

$$\begin{array}{rl}\frac{\epsilon }{s}& \le \underset{i\to \mathrm{\infty }}{lim\hspace{0.17em}sup}d(x_{m_i+1},x_{n_i})\\ \le \underset{i\to \mathrm{\infty }}{lim\hspace{0.17em}sup}\beta (M(x_{m_i},x_{n_i-1}))\underset{i\to \mathrm{\infty }}{lim\hspace{0.17em}sup}M(x_{m_i},x_{n_i-1})\\ \le \epsilon \underset{i\to \mathrm{\infty }}{lim\hspace{0.17em}sup}\beta (M(x_{m_i},x_{n_i-1})),\end{array}$$

which implies that $\frac{1}{s}\le {lim\hspace{0.17em}sup}_{i\to \mathrm{\infty }}\beta (M(x_{m_i},x_{n_i-1}))$. Now, as $\beta \in \mathcal{F}$ we conclude that $\{x_n\}$ is a b-Cauchy sequence. b-Completeness of X shows that $\{x_n\}$ 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

$$\begin{array}{rl}\frac{1}{s}d(u,fu)& \le \underset{n\to \mathrm{\infty }}{lim\hspace{0.17em}sup}d(x_{n+1},fu)\\ \le \underset{n\to \mathrm{\infty }}{lim\hspace{0.17em}sup}\beta (M(x_n,u))\underset{n\to \mathrm{\infty }}{lim\hspace{0.17em}sup}M(x_n,u),\end{array}$$

where

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

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

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

Corollary 1 Let $(X,⪯)$ be a partially ordered set and suppose that there exists a b-metric d on X such that $(X,d)$ 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 $x_0⪯f(x_0)$. Suppose that

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

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

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

or

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

or

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

If f is continuous, or, for any nondecreasing sequence $\{x_n\}$ 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 $(X,⪯)$ be a partially ordered set and suppose that there exists a b-metric d on X such that $(X,d)$ 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 $x_0⪯f(x_0)$. Suppose

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

or

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

or

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

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 $\{x_n\}$ 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 $(X,⪯,d)$ 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 $x_0⪯f^m(x_0)$ and

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

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

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

or

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

or

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

for some positive integer m.

If $f^m$ is continuous, or, for any nondecreasing sequence $\{x_n\}$ 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 $\psi :[0,\mathrm{\infty })\to [0,\mathrm{\infty })$ such that

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

for all $t>0$.

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

• (a) $\psi (t)<t$ for all $t>0$;

• (b) $\psi (0)=0$.

As an example ${\psi }_1(t)=kt$, for all $t\ge 0$, where $k\in [0,1)$, and ${\psi }_2(t)=ln(t+1)$, for all $t\ge 0$, are in Ψ.

Theorem 8 Let $(X,⪯)$ be a partially ordered set and suppose that there exists a b-metric d on X such that $(X,d)$ 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 $x_0⪯f(x_0)$. Suppose that

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

(2.24)

where

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

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 $x_0⪯f(x_0)$ and f is increasing, we obtain by induction that

$$x_0⪯f(x_0)⪯f^2(x_0)⪯\cdots ⪯f^n(x_0)⪯f^{n+1}(x_0)⪯\cdots .$$

Putting $x_n=f^n(x_0)$, 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 $d(x_n,x_{n+1})>0$, for all $n\in \mathbb{N}$.

In the following steps, we will complete the proof.

Step I: We will prove that

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

Using condition (2.24), we obtain

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

because

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

If $max\{d(x_{n-1},x_n),d(x_n,x_{n+1})\}=d(x_n,x_{n+1})$, then

$$\begin{array}{rcl}d(x_n,x_{n+1})& \le & sd(x_n,x_{n+1})=sd(f{x}_{n-1},x_n)\\ \le & \psi (M(x_{n-1},x_n))<M(x_{n-1},x_n)\le d(x_n,x_{n+1}),\end{array}$$

(2.25)

which is a contradiction. Hence, $max\{d(x_{n-1},x_n),d(x_n,x_{n+1})\}=d(x_{n-1},x_n)$, so from (2.25),

$$\begin{array}{rcl}d(x_n,x_{n+1})& \le & sd(x_n,x_{n+1})=sd(f{x}_{n-1},x_n)\\ \le & \psi (M(x_{n-1},x_n))<M(x_{n-1},x_n)\le d(x_{n-1},x_n).\end{array}$$

(2.26)

Hence,

$$d(x_n,x_{n+1})\le sd(x_n,x_{n+1})\le \psi (d(x_{n-1},x_n)).$$

By induction,

$$\begin{array}{rl}d(x_{n+1},x_n)& \le \psi (d(x_n,x_{n-1}))\le {\psi }^2(d(x_{n-1},x_{n-2}))\\ \le \cdots \le {\psi }^n(d(x_1,x_0)).\end{array}$$

(2.27)

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

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

(2.28)

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

$$n_i>m_i>i\text{and}d(x_{m_i},x_{n_i})\ge \epsilon .$$

(2.29)

This means that

$$d(x_{m_i},x_{n_i-1})<\epsilon .$$

(2.30)

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

$$\epsilon \le d(x_{m_i},x_{n_i})\le sd(x_{m_i},x_{m_i+1})+sd(x_{m_i+1},x_{n_i}).$$

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

$$\frac{\epsilon }{s}\le \underset{i\to \mathrm{\infty }}{lim\hspace{0.17em}sup}d(x_{m_i+1},x_{n_i}).$$

(2.31)

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

$$\begin{array}{c}\underset{i\to \mathrm{\infty }}{lim\hspace{0.17em}sup}M(x_{m_i},x_{n_i-1})\hfill \\ =\underset{i\to \mathrm{\infty }}{lim\hspace{0.17em}sup}max\{d(x_{m_i},x_{n_i-1}),\frac{d(x_{m_i},f{x}_{m_i})d(x_{n_i-1},f{x}_{n_i-1})}{1+d(x_{m_i},x_{n_i-1})},\hfill \\ \frac{d(x_{m_i},f{x}_{m_i})d(x_{n_i-1},f{x}_{n_i-1})}{1+d(f{x}_{m_i},f{x}_{n_i-1})}\}\hfill \\ =\underset{i\to \mathrm{\infty }}{lim\hspace{0.17em}sup}max\{d(x_{m_i},x_{n_i-1}),\frac{d(x_{m_i},x_{m_i+1})d(x_{n_i-1},x_{n_i})}{1+d(x_{m_i},x_{n_i-1})},\hfill \\ \frac{d(x_{m_i},x_{m_i+1})d(x_{n_i-1},x_{n_i})}{1+d(x_{m_i+1},x_{n_i})}\}\hfill \\ \le \epsilon .\hfill \end{array}$$

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

$$\begin{array}{rl}\epsilon & =s\cdot \frac{\epsilon }{s}\le s\underset{i\to \mathrm{\infty }}{lim\hspace{0.17em}sup}d(x_{m_i+1},x_{n_i})\\ \le \underset{i\to \mathrm{\infty }}{lim\hspace{0.17em}sup}\psi (M(x_{m_i},x_{n_i-1}))\\ \le \psi (\epsilon )<\epsilon ,\end{array}$$

which is a contradiction. Consequently, $\{x_n\}$ is a b-Cauchy sequence. b-Completeness of X shows that $\{x_n\}$ 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 $\{x_n\}$ 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 $\{x_n\}$ 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

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

(2.32)

where

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

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

$$\underset{n\to \mathrm{\infty }}{lim\hspace{0.17em}sup}M(u,x_{n-1})=0.$$

(2.33)

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

$$\begin{array}{rl}\frac{1}{s}d(fu,u)& \le \underset{n\to \mathrm{\infty }}{lim\hspace{0.17em}sup}d(fu,x_n)\\ \le \underset{n\to \mathrm{\infty }}{lim\hspace{0.17em}sup}\psi (M(u,x_{n-1}))\\ =0.\end{array}$$

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

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

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

or

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

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

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

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, $f(x_0)=3$, so $x_0⪯f{x}_0$. Define first the b-metric d on X by $d(0,1)=6$, $d(0,3)=9$, $d(1,3)=\frac{1}{2}$, and $d(x,x)=0$. Then $(X,d)$ is a b-complete b-metric space with $s=\frac{18}{13}$. Let $\beta \in \mathcal{F}$ is given by

$$\beta (t)=\frac{13}{18}{e}^{\frac{-t}{9}},t\ge 0$$

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

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

This is because

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

Also,

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

because

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

Also,

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

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

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

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

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

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

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

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

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

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

$$x⪯y\text{iff}x(t)\le y(t)\text{for all }t\in [0,T].$$

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

Consider the first-order periodic boundary value problem

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

(3.1)

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

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

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

(3.2)

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

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

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

(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

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

Consider

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

where $t\in I$.

Using the variation of parameters formula, we get

$$x(t)=x(0)e^{-\lambda t}+{\int }_0^t{e}^{-\lambda (t-s)}\delta (s)ds,$$

(3.4)

which yields

$$x(T)=x(0)e^{-\lambda T}+{\int }_0^T{e}^{-\lambda (T-s)}\delta (s)ds.$$

Since $x(0)=x(T)$, we get

$$x(0)[1-e^{-\lambda T}]=e^{-\lambda T}{\int }_0^T{e}^{\lambda (s)}\delta (s)ds$$

or

$$x(0)=\frac{1}{e^{\lambda T}-1}{\int }_0^T{e}^{\lambda s}\delta (s)ds.$$

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

$$x(t)={\int }_0^{T}G(t,s)\delta (s)ds,$$

where

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

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

$$Sx(t)={\int }_0^{T}G(t,s)F(s,x(s))ds.$$

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

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

$$\begin{array}{rl}{2}^{p-1}|Sx(t)-Sy(t)|& =2^{p-1}|{\int }_0^{T}G(t,s)F(s,x(s))ds-{\int }_0^{T}G(t,s)F(s,y(s))ds|\\ \le 2^{p-1}{\int }_0^T|G(t,s)|\left[\right|F(s,x(s))-F(s,y(s))\left|\right]ds\\ \le 2^{p-1}{\int }_0^T|G(t,s)|\frac{\lambda }{2^{p-1}}\sqrt[p]{ln\left(\right|x(t)-y(t){|}^p+1)}ds\\ \le \lambda \sqrt[p]{ln(d(x,y)+1)}[{\int }_0^t\frac{e^{\lambda (T+s-t)}}{e^{\lambda T}-1}ds+{\int }_t^T\frac{e^{\lambda (s-t)}}{e^{\lambda T}-1}ds]\\ =\lambda \sqrt[p]{ln(d(x,y)+1)}\left[\frac{1}{\lambda (e^{\lambda T}-1)}(e^{\lambda (T+s-t)}{|}_0^t+e^{\lambda (s-t)}{|}_t^T)\right]\\ =\lambda \sqrt[p]{ln(d(x,y)+1)}\left[\frac{1}{\lambda (e^{\lambda T}-1)}(e^{\lambda T}-e^{\lambda (T-t)}+e^{\lambda (T-t)}-1)\right]\\ =\sqrt[p]{ln(d(x,y)+1)}\\ \le \sqrt[p]{ln(M(x,y)+1)},\end{array}$$

or, equivalently,

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

which shows that

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

where

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

or

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

or

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

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

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

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

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

(3.5)

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