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# Fixed point theorems of contractive mappings of integral type

Fixed Point Theory and Applications20132013:300

https://doi.org/10.1186/1687-1812-2013-300

• Received: 4 July 2013
• Accepted: 22 October 2013
• Published:

## Abstract

Three fixed point theorems for three general classes of contractive mappings of integral type in complete metric spaces are proved. Three examples are included.

MSC:54H25.

## Keywords

• contractive mappings of integral type
• fixed point theorems
• complete metric space

## 1 Introduction

Branciari  was the first to study the existence of fixed points for the contractive mapping of integral type. He established a nice integral version of the Banach contraction principle and proved the following fixed point theorem.

Theorem 1.1 Let f be a mapping from a complete metric space $\left(X,d\right)$ into itself satisfying
${\int }_{0}^{d\left(fx,fy\right)}\phi \left(t\right)\phantom{\rule{0.2em}{0ex}}dt\le c{\int }_{0}^{d\left(x,y\right)}\phi \left(t\right)\phantom{\rule{0.2em}{0ex}}dt,\phantom{\rule{1em}{0ex}}\mathrm{\forall }x,y\in X,$

where $c\in \left(0,1\right)$ is a constant and $\phi \in {\mathrm{\Phi }}_{1}$. Then f has a unique fixed point $a\in X$ such that ${lim}_{n\to \mathrm{\infty }}{f}^{n}x=a$ for each $x\in X$.

Afterwards, many authors continued the study of Branciari and obtained many fixed point theorems for several classes of contractive mappings of integral type; see, e.g.,  and the references therein. In particular, in 2011, Liu et al.  extended the result of Branciari  and deduced the following fixed point theorems.

Theorem 1.2 Let f be a mapping from a complete metric space $\left(X,d\right)$ into itself satisfying
${\int }_{0}^{d\left(fx,fy\right)}\phi \left(t\right)\phantom{\rule{0.2em}{0ex}}dt\le \alpha \left(d\left(x,y\right)\right){\int }_{0}^{d\left(x,y\right)}\phi \left(t\right)\phantom{\rule{0.2em}{0ex}}dt,\phantom{\rule{1em}{0ex}}\mathrm{\forall }x,y\in X,$
where $\phi \in {\mathrm{\Phi }}_{1}$ and $\alpha :{\mathbb{R}}^{+}\to \left[0,1\right)$ is a function with
$\underset{s\to t}{lim sup}\alpha \left(s\right)<1,\phantom{\rule{1em}{0ex}}\mathrm{\forall }t>0.$

Then f has a unique fixed point $a\in X$ such that ${lim}_{n\to \mathrm{\infty }}{f}^{n}x=a$ for each $x\in X$.

Theorem 1.3 Let f be a mapping from a complete metric space $\left(X,d\right)$ into itself satisfying
${\int }_{0}^{d\left(fx,fy\right)}\phi \left(t\right)\phantom{\rule{0.2em}{0ex}}dt\le \alpha \left(d\left(x,y\right)\right){\int }_{0}^{d\left(x,fx\right)}\phi \left(t\right)\phantom{\rule{0.2em}{0ex}}dt+\beta \left(d\left(x,y\right)\right){\int }_{0}^{d\left(y,fy\right)}\phi \left(t\right)\phantom{\rule{0.2em}{0ex}}dt,\phantom{\rule{1em}{0ex}}\mathrm{\forall }x,y\in X,$
where $\phi \in {\mathrm{\Phi }}_{1}$ and $\alpha ,\beta :{\mathbb{R}}^{+}\to \left[0,1\right)$ are two functions with
$\alpha \left(t\right)+\beta \left(t\right)<1,\phantom{\rule{1em}{0ex}}\mathrm{\forall }t\in {\mathbb{R}}^{+},\phantom{\rule{2em}{0ex}}\underset{s\to {0}^{+}}{lim sup}\beta \left(s\right)<1,\phantom{\rule{2em}{0ex}}\underset{s\to {t}^{+}}{lim sup}\frac{\alpha \left(s\right)}{1-\beta \left(s\right)}<1,\phantom{\rule{1em}{0ex}}\mathrm{\forall }t>0.$

Then f has a unique fixed point $a\in X$ such that ${lim}_{n\to \mathrm{\infty }}{f}^{n}x=a$ for each $x\in X$.

In 2008, Dutta and Choudhuty  proved the following result.

Theorem 1.4 Let f be a mapping from a complete metric space $\left(X,d\right)$ into itself satisfying
$\psi \left(d\left(fx,fy\right)\right)\le \psi \left(d\left(x,y\right)\right)-\phi \left(d\left(x,y\right)\right),\phantom{\rule{1em}{0ex}}\mathrm{\forall }x,y\in X,$

where $\psi ,\phi :{\mathbb{R}}^{+}\to {\mathbb{R}}^{+}$ are both continuous and monotone nondecreasing functions with $\psi \left(t\right)=\phi \left(t\right)=0$ if and only if $t=0$. Then f has a unique fixed point $a\in X$ such that ${lim}_{n\to \mathrm{\infty }}{f}^{n}x=a$ for each $x\in X$.

However, to the best of our knowledge, no one studied the following contractive mappings of integral type:
$\psi \left({\int }_{0}^{d\left(fx,fy\right)}\phi \left(t\right)\phantom{\rule{0.2em}{0ex}}dt\right)\le \psi \left({\int }_{0}^{d\left(x,y\right)}\phi \left(t\right)\phantom{\rule{0.2em}{0ex}}dt\right)-\varphi \left({\int }_{0}^{d\left(x,y\right)}\phi \left(t\right)\phantom{\rule{0.2em}{0ex}}dt\right),\phantom{\rule{1em}{0ex}}\mathrm{\forall }x,y\in X,$
(1.1)
where $\left(\phi ,\varphi ,\psi \right)\in {\mathrm{\Phi }}_{1}×{\mathrm{\Phi }}_{2}×{\mathrm{\Phi }}_{3}$;
$\psi \left({\int }_{0}^{d\left(fx,fy\right)}\phi \left(t\right)\phantom{\rule{0.2em}{0ex}}dt\right)\le \alpha \left(d\left(x,y\right)\right)\psi \left({\int }_{0}^{d\left(x,y\right)}\phi \left(t\right)\phantom{\rule{0.2em}{0ex}}dt\right),\phantom{\rule{1em}{0ex}}\mathrm{\forall }x,y\in X,$
(1.2)
where $\left(\phi ,\psi ,\alpha \right)\in {\mathrm{\Phi }}_{1}×{\mathrm{\Phi }}_{3}×{\mathrm{\Phi }}_{5}$;
$\begin{array}{rl}\psi \left({\int }_{0}^{d\left(fx,fy\right)}\phi \left(t\right)\phantom{\rule{0.2em}{0ex}}dt\right)\le & \alpha \left(d\left(x,y\right)\right)\varphi \left({\int }_{0}^{d\left(x,fx\right)}\phi \left(t\right)\phantom{\rule{0.2em}{0ex}}dt\right)\\ +\beta \left(d\left(x,y\right)\right)\psi \left({\int }_{0}^{d\left(y,fy\right)}\phi \left(t\right)\phantom{\rule{0.2em}{0ex}}dt\right),\phantom{\rule{1em}{0ex}}\mathrm{\forall }x,y\in X,\end{array}$
(1.3)

where $\left(\phi ,\psi ,\varphi \right)\in {\mathrm{\Phi }}_{1}×{\mathrm{\Phi }}_{3}×{\mathrm{\Phi }}_{4}$ and $\left(\alpha ,\beta \right)\in {\mathrm{\Phi }}_{6}$.

It is clear that the above contractive mappings of integral type include these mappings in Theorems 1.1-1.4 as special cases. The purpose of this paper is to investigate the existence of fixed points for contractive mappings (1.1)-(1.3) of integral type. Under certain conditions, we prove the existence, uniqueness and iterative approximations of fixed points for contractive mappings (1.1)-(1.3) of integral type in complete metric spaces. Three examples with uncountably many points are constructed.

## 2 Preliminaries

Throughout this paper, we assume that ${\mathbb{R}}^{+}=\left[0,+\mathrm{\infty }\right)$, ${\mathbb{N}}_{0}=\mathbb{N}\cup \left\{0\right\}$, denotes the set of all positive integers, $\left(X,d\right)$ is a metric space, $f:X\to X$ is a self-mapping and
${d}_{n}=d\left({f}^{n}x,{f}^{n+1}x\right),\phantom{\rule{1em}{0ex}}\mathrm{\forall }\left(n,x\right)\in {\mathbb{N}}_{0}×X,$

${\mathrm{\Phi }}_{1}$ = {$\phi :\phi :{\mathbb{R}}^{+}\to {\mathbb{R}}^{+}$ is Lebesgue integrable, summable on each compact subset of ${\mathbb{R}}^{+}$ and ${\int }_{0}^{\epsilon }\phi \left(t\right)\phantom{\rule{0.2em}{0ex}}dt>0$ for each $\epsilon >0$};

${\mathrm{\Phi }}_{2}$ = {$\phi :\phi :{\mathbb{R}}^{+}\to {\mathbb{R}}^{+}$ satisfies that ${lim inf}_{n\to \mathrm{\infty }}\phi \left({a}_{n}\right)>0⇔{lim inf}_{n\to \mathrm{\infty }}{a}_{n}>0$ for each ${\left\{{a}_{n}\right\}}_{n\in \mathbb{N}}\subset {\mathbb{R}}^{+}$};

${\mathrm{\Phi }}_{3}$ = {$\phi :\phi :{\mathbb{R}}^{+}\to {\mathbb{R}}^{+}$ is nondecreasing continuous and $\phi \left(t\right)=0⇔t=0$};

${\mathrm{\Phi }}_{4}$ = {$\phi :\phi :{\mathbb{R}}^{+}\to {\mathbb{R}}^{+}$ satisfies that $\phi \left(0\right)=0$};

${\mathrm{\Phi }}_{5}$ = {$\phi :\phi :{\mathbb{R}}^{+}\to \left[0,1\right)$ satisfies that ${lim sup}_{s\to t}\phi \left(s\right)<1$ for each $t>0$};

${\mathrm{\Phi }}_{6}$ = {$\left(\alpha ,\beta \right):\alpha ,\beta :{\mathbb{R}}^{+}\to \left[0,1\right)$ satisfy that ${lim sup}_{s\to {0}^{+}}\beta \left(s\right)<1$, ${lim sup}_{s\to {t}^{+}}\frac{\alpha \left(s\right)}{1-\beta \left(s\right)}<1$ and $\alpha \left(t\right)+\beta \left(t\right)<1$ for each $t>0$}.

The following lemmas play important roles in this paper.

Lemma 2.1 ()

Let $\phi \in {\mathrm{\Phi }}_{1}$ and ${\left\{{r}_{n}\right\}}_{n\in \mathbb{N}}$ be a nonnegative sequence with ${lim}_{n\to \mathrm{\infty }}{r}_{n}=a$. Then
$\underset{n\to \mathrm{\infty }}{lim}{\int }_{0}^{{r}_{n}}\phi \left(t\right)\phantom{\rule{0.2em}{0ex}}dt={\int }_{0}^{a}\phi \left(t\right)\phantom{\rule{0.2em}{0ex}}dt.$

Lemma 2.2 ()

Let $\phi \in {\mathrm{\Phi }}_{1}$ and ${\left\{{r}_{n}\right\}}_{n\in \mathbb{N}}$ be a nonnegative sequence. Then
$\underset{n\to \mathrm{\infty }}{lim}{\int }_{0}^{{r}_{n}}\phi \left(t\right)\phantom{\rule{0.2em}{0ex}}dt=0$

if and only if ${lim}_{n\to \mathrm{\infty }}{r}_{n}=0$.

Lemma 2.3 Let $\phi \in {\mathrm{\Phi }}_{2}$. Then $\phi \left(t\right)>0$ if and only if $t>0$.

Proof Let $t>0$. Put ${a}_{n}=t$ for each $n\in \mathbb{N}$. It is easy to see that $t={lim inf}_{n\to \mathrm{\infty }}{a}_{n}>0$, which together with $\phi \in {\mathrm{\Phi }}_{2}$ ensures that
$\phi \left(t\right)=\underset{n\to \mathrm{\infty }}{lim inf}\phi \left({a}_{n}\right)>0.$
Conversely, suppose that $\phi \left(t\right)>0$ for some $t\in {\mathbb{R}}^{+}$. Set ${a}_{n}=t$ for each $n\in \mathbb{N}$. It is clear that $\phi \left(t\right)={lim inf}_{n\to \mathrm{\infty }}\phi \left({a}_{n}\right)>0$, which together with $\phi \in {\mathrm{\Phi }}_{2}$ guarantees that
$t=\underset{n\to \mathrm{\infty }}{lim inf}{a}_{n}>0.$

This completes the proof. □

## 3 Main results

In this section we show the existence, uniqueness and iterative approximations of fixed points for contractive mappings (1.1)-(1.3) of integral type, respectively.

Theorem 3.1 Let f be a mapping from a complete metric space $\left(X,d\right)$ into itself satisfying (1.1). Then f has a unique fixed point $a\in X$ such that ${lim}_{n\to \mathrm{\infty }}{f}^{n}x=a$ for each $x\in X$.

Proof Let x be an arbitrary point in X. Firstly, we show that
${d}_{n}\le {d}_{n-1},\phantom{\rule{1em}{0ex}}\mathrm{\forall }n\in \mathbb{N}.$
(3.1)
Suppose that (3.1) does not hold. It follows that there exists some ${n}_{0}\in \mathbb{N}$ satisfying
${d}_{{n}_{0}}>{d}_{{n}_{0}-1}.$
(3.2)
Note that (3.2) and $\phi \in {\mathrm{\Phi }}_{1}$ imply that
${\int }_{0}^{{d}_{{n}_{0}}}\phi \left(t\right)\phantom{\rule{0.2em}{0ex}}dt>0.$
(3.3)
Using (1.1), (3.2) and $\left(\phi ,\varphi ,\psi \right)\in {\mathrm{\Phi }}_{1}×{\mathrm{\Phi }}_{2}×{\mathrm{\Phi }}_{3}$, we conclude immediately that
$\begin{array}{rl}\psi \left({\int }_{0}^{{d}_{{n}_{0}-1}}\phi \left(t\right)\phantom{\rule{0.2em}{0ex}}dt\right)& \le \psi \left({\int }_{0}^{{d}_{{n}_{0}}}\phi \left(t\right)\phantom{\rule{0.2em}{0ex}}dt\right)\\ =\psi \left({\int }_{0}^{d\left({f}^{{n}_{0}}x,{f}^{{n}_{0}+1}x\right)}\phi \left(t\right)\phantom{\rule{0.2em}{0ex}}dt\right)\\ \le \psi \left({\int }_{0}^{d\left({f}^{{n}_{0}-1}x,{f}^{{n}_{0}}x\right)}\phi \left(t\right)\phantom{\rule{0.2em}{0ex}}dt\right)-\varphi \left({\int }_{0}^{d\left({f}^{{n}_{0}-1}x,{f}^{{n}_{0}}x\right)}\phi \left(t\right)\phantom{\rule{0.2em}{0ex}}dt\right)\\ =\psi \left({\int }_{0}^{{d}_{{n}_{0}-1}}\phi \left(t\right)\phantom{\rule{0.2em}{0ex}}dt\right)-\varphi \left({\int }_{0}^{{d}_{{n}_{0}-1}}\phi \left(t\right)\phantom{\rule{0.2em}{0ex}}dt\right)\\ \le \psi \left({\int }_{0}^{{d}_{{n}_{0}-1}}\phi \left(t\right)\phantom{\rule{0.2em}{0ex}}dt\right),\end{array}$
which yields that
$\psi \left({\int }_{0}^{{d}_{{n}_{0}}}\phi \left(t\right)\phantom{\rule{0.2em}{0ex}}dt\right)=\psi \left({\int }_{0}^{{d}_{{n}_{0}-1}}\phi \left(t\right)\phantom{\rule{0.2em}{0ex}}dt\right)$
(3.4)
and
$\varphi \left({\int }_{0}^{{d}_{{n}_{0}-1}}\phi \left(t\right)\phantom{\rule{0.2em}{0ex}}dt\right)=0.$
(3.5)
Combining (3.5) and Lemma 2.3, we get that
${\int }_{0}^{{d}_{{n}_{0}-1}}\phi \left(t\right)\phantom{\rule{0.2em}{0ex}}dt=0,$
which together with $\psi \in {\mathrm{\Phi }}_{3}$ and (3.4) means that
$\psi \left({\int }_{0}^{{d}_{{n}_{0}}}\phi \left(t\right)\phantom{\rule{0.2em}{0ex}}dt\right)=\psi \left({\int }_{0}^{{d}_{{n}_{0}-1}}\phi \left(t\right)\phantom{\rule{0.2em}{0ex}}dt\right)=\psi \left(0\right)=0,$
that is,
${\int }_{0}^{{d}_{{n}_{0}}}\phi \left(t\right)\phantom{\rule{0.2em}{0ex}}dt=0,$

which contradicts (3.3). Hence (3.1) holds.

Secondly, we show that
$\underset{n\to \mathrm{\infty }}{lim}{d}_{n}=0.$
(3.6)
In view of (3.1), we deduce that the nonnegative sequence ${\left\{{d}_{n}\right\}}_{n\in {\mathbb{N}}_{0}}$ is nonincreasing, which means that there exists a constant c with ${lim}_{n\to \mathrm{\infty }}{d}_{n}=c\ge 0$. Suppose that $c>0$. It follows from (1.1) that
$\begin{array}{rl}\psi \left({\int }_{0}^{{d}_{n}}\phi \left(t\right)\phantom{\rule{0.2em}{0ex}}dt\right)& =\psi \left({\int }_{0}^{d\left({f}^{n}x,{f}^{n+1}x\right)}\phi \left(t\right)\phantom{\rule{0.2em}{0ex}}dt\right)\\ \le \psi \left({\int }_{0}^{d\left({f}^{n}x,{f}^{n-1}x\right)}\phi \left(t\right)\phantom{\rule{0.2em}{0ex}}dt\right)-\varphi \left({\int }_{0}^{d\left({f}^{n}x,{f}^{n-1}x\right)}\phi \left(t\right)\phantom{\rule{0.2em}{0ex}}dt\right)\\ =\psi \left({\int }_{0}^{{d}_{n-1}}\phi \left(t\right)\phantom{\rule{0.2em}{0ex}}dt\right)-\varphi \left({\int }_{0}^{{d}_{n-1}}\phi \left(t\right)\phantom{\rule{0.2em}{0ex}}dt\right),\phantom{\rule{1em}{0ex}}\mathrm{\forall }n\in \mathbb{N}.\end{array}$
(3.7)
Taking upper limit in (3.7) and using Lemma 2.1 and $\left(\phi ,\varphi ,\psi \right)\in {\mathrm{\Phi }}_{1}×{\mathrm{\Phi }}_{2}×{\mathrm{\Phi }}_{3}$, we conclude that
$\begin{array}{rl}\psi \left({\int }_{0}^{c}\phi \left(t\right)\phantom{\rule{0.2em}{0ex}}dt\right)& =\underset{n\to \mathrm{\infty }}{lim sup}\psi \left({\int }_{0}^{{d}_{n}}\phi \left(t\right)\phantom{\rule{0.2em}{0ex}}dt\right)\\ \le \underset{n\to \mathrm{\infty }}{lim sup}\left[\psi \left({\int }_{0}^{{d}_{n-1}}\phi \left(t\right)\phantom{\rule{0.2em}{0ex}}dt\right)-\varphi \left({\int }_{0}^{{d}_{n-1}}\phi \left(t\right)\phantom{\rule{0.2em}{0ex}}dt\right)\right]\\ \le \underset{n\to \mathrm{\infty }}{lim sup}\psi \left({\int }_{0}^{{d}_{n-1}}\phi \left(t\right)\phantom{\rule{0.2em}{0ex}}dt\right)-\underset{n\to \mathrm{\infty }}{lim inf}\varphi \left({\int }_{0}^{{d}_{n-1}}\phi \left(t\right)\phantom{\rule{0.2em}{0ex}}dt\right)\\ =\psi \left({\int }_{0}^{c}\phi \left(t\right)\phantom{\rule{0.2em}{0ex}}dt\right)-\underset{n\to \mathrm{\infty }}{lim inf}\varphi \left({\int }_{0}^{{d}_{n-1}}\phi \left(t\right)\phantom{\rule{0.2em}{0ex}}dt\right)\\ <\psi \left({\int }_{0}^{c}\phi \left(t\right)\phantom{\rule{0.2em}{0ex}}dt\right),\end{array}$

which is a contradiction. Hence $c=0$.

Thirdly, we show that ${\left\{{f}^{n}x\right\}}_{n\in \mathbb{N}}$ is a Cauchy sequence. Suppose that ${\left\{{f}^{n}x\right\}}_{n\in \mathbb{N}}$ is not a Cauchy sequence, which means that there is a constant $\epsilon >0$ such that for each positive integer k, there are positive integers $m\left(k\right)$ and $n\left(k\right)$ with $m\left(k\right)>n\left(k\right)>k$ satisfying
$d\left({f}^{m\left(k\right)}x,{f}^{n\left(k\right)}x\right)>\epsilon .$
(3.8)
For each positive integer k, let $m\left(k\right)$ denote the least integer exceeding $n\left(k\right)$ and satisfying (3.8). It follows that
$d\left({f}^{m\left(k\right)}x,{f}^{n\left(k\right)}x\right)>\epsilon \phantom{\rule{1em}{0ex}}\text{and}\phantom{\rule{1em}{0ex}}d\left({f}^{m\left(k\right)-1}x,{f}^{n\left(k\right)}x\right)\le \epsilon ,\phantom{\rule{1em}{0ex}}\mathrm{\forall }k\in \mathbb{N}.$
(3.9)
Note that
$\begin{array}{r}d\left({f}^{m\left(k\right)}x,{f}^{n\left(k\right)}x\right)\le d\left({f}^{n\left(k\right)}x,{f}^{m\left(k\right)-1}x\right)+{d}_{m\left(k\right)-1},\phantom{\rule{1em}{0ex}}\mathrm{\forall }k\in \mathbb{N};\\ |d\left({f}^{m\left(k\right)}x,{f}^{n\left(k\right)+1}x\right)-d\left({f}^{m\left(k\right)}x,{f}^{n\left(k\right)}x\right)|\le {d}_{n\left(k\right)},\phantom{\rule{1em}{0ex}}\mathrm{\forall }k\in \mathbb{N};\\ |d\left({f}^{m\left(k\right)+1}x,{f}^{n\left(k\right)+1}x\right)-d\left({f}^{m\left(k\right)}x,{f}^{n\left(k\right)+1}x\right)|\le {d}_{m\left(k\right)},\phantom{\rule{1em}{0ex}}\mathrm{\forall }k\in \mathbb{N};\\ |d\left({f}^{m\left(k\right)+1}x,{f}^{n\left(k\right)+1}x\right)-d\left({f}^{m\left(k\right)+1}x,{f}^{n\left(k\right)+2}x\right)|\le {d}_{n\left(k\right)+1},\phantom{\rule{1em}{0ex}}\mathrm{\forall }k\in \mathbb{N}.\end{array}$
(3.10)
In light of (3.9) and (3.10), we get that
$\begin{array}{rl}\epsilon & =\underset{k\to \mathrm{\infty }}{lim}d\left({f}^{n\left(k\right)}x,{f}^{m\left(k\right)}x\right)=\underset{k\to \mathrm{\infty }}{lim}d\left({f}^{m\left(k\right)}x,{f}^{n\left(k\right)+1}x\right)\\ =\underset{k\to \mathrm{\infty }}{lim}d\left({f}^{m\left(k\right)+1}x,{f}^{n\left(k\right)+1}x\right)=\underset{k\to \mathrm{\infty }}{lim}d\left({f}^{m\left(k\right)+1}x,{f}^{n\left(k\right)+2}x\right).\end{array}$
(3.11)
In view of (1.1), we deduce that
$\begin{array}{r}\psi \left({\int }_{0}^{d\left({f}^{m\left(k\right)+1}x,{f}^{n\left(k\right)+2}x\right)}\phi \left(t\right)\phantom{\rule{0.2em}{0ex}}dt\right)\\ \phantom{\rule{1em}{0ex}}\le \psi \left({\int }_{0}^{d\left({f}^{m\left(k\right)}x,{f}^{n\left(k\right)+1}x\right)}\phi \left(t\right)\phantom{\rule{0.2em}{0ex}}dt\right)-\varphi \left({\int }_{0}^{d\left({f}^{m\left(k\right)}x,{f}^{n\left(k\right)+1}x\right)}\phi \left(t\right)\phantom{\rule{0.2em}{0ex}}dt\right),\phantom{\rule{1em}{0ex}}\mathrm{\forall }k\in \mathbb{N}.\end{array}$
(3.12)
Taking upper limit in (3.12) and using (3.11), $\left(\phi ,\varphi ,\psi \right)\in {\mathrm{\Phi }}_{1}×{\mathrm{\Phi }}_{2}×{\mathrm{\Phi }}_{3}$ and Lemma 2.1, we deduce that
$\begin{array}{r}\psi \left({\int }_{0}^{\epsilon }\phi \left(t\right)\phantom{\rule{0.2em}{0ex}}dt\right)\\ \phantom{\rule{1em}{0ex}}=\underset{k\to \mathrm{\infty }}{lim sup}\psi \left({\int }_{0}^{d\left({f}^{m\left(k\right)+1}x,{f}^{n\left(k\right)+2}x\right)}\phi \left(t\right)\phantom{\rule{0.2em}{0ex}}dt\right)\\ \phantom{\rule{1em}{0ex}}\le \underset{k\to \mathrm{\infty }}{lim sup}\left[\psi \left({\int }_{0}^{d\left({f}^{m\left(k\right)}x,{f}^{n\left(k\right)+1}x\right)}\phi \left(t\right)\phantom{\rule{0.2em}{0ex}}dt\right)-\varphi \left({\int }_{0}^{d\left({f}^{m\left(k\right)}x,{f}^{n\left(k\right)+1}x\right)}\phi \left(t\right)\phantom{\rule{0.2em}{0ex}}dt\right)\right]\\ \phantom{\rule{1em}{0ex}}\le \underset{k\to \mathrm{\infty }}{lim sup}\psi \left({\int }_{0}^{d\left({f}^{m\left(k\right)}x,{f}^{n\left(k\right)+1}x\right)}\phi \left(t\right)\phantom{\rule{0.2em}{0ex}}dt\right)-\underset{k\to \mathrm{\infty }}{lim inf}\varphi \left({\int }_{0}^{d\left({f}^{m\left(k\right)}x,{f}^{n\left(k\right)+1}x\right)}\phi \left(t\right)\phantom{\rule{0.2em}{0ex}}dt\right)\\ \phantom{\rule{1em}{0ex}}=\psi \left({\int }_{0}^{\epsilon }\phi \left(t\right)\phantom{\rule{0.2em}{0ex}}dt\right)-\underset{k\to \mathrm{\infty }}{lim inf}\varphi \left({\int }_{0}^{d\left({f}^{m\left(k\right)}x,{f}^{n\left(k\right)+1}x\right)}\phi \left(t\right)\phantom{\rule{0.2em}{0ex}}dt\right)\\ \phantom{\rule{1em}{0ex}}<\psi \left({\int }_{0}^{\epsilon }\phi \left(t\right)\phantom{\rule{0.2em}{0ex}}dt\right),\end{array}$

which is impossible. Thus ${\left\{{f}^{n}x\right\}}_{n\in \mathbb{N}}$ is a Cauchy sequence.

Since $\left(X,d\right)$ is complete, it follows that there exists a point $a\in X$ satisfying ${lim}_{n\to \mathrm{\infty }}{f}^{n}x=a$. By virtue of (1.1), we infer that
$\psi \left({\int }_{0}^{d\left({f}^{n+1}x,fa\right)}\phi \left(t\right)\phantom{\rule{0.2em}{0ex}}dt\right)\le \psi \left({\int }_{0}^{d\left({f}^{n}x,a\right)}\phi \left(t\right)\phantom{\rule{0.2em}{0ex}}dt\right)-\varphi \left({\int }_{0}^{d\left({f}^{n}x,a\right)}\phi \left(t\right)\phantom{\rule{0.2em}{0ex}}dt\right),\phantom{\rule{1em}{0ex}}\mathrm{\forall }n\in \mathbb{N},$
which together with $\left(\phi ,\varphi ,\psi \right)\in {\mathrm{\Phi }}_{1}×{\mathrm{\Phi }}_{2}×{\mathrm{\Phi }}_{3}$ and Lemmas 2.1 and 2.2 gives that
$\begin{array}{rl}\psi \left({\int }_{0}^{d\left(a,fa\right)}\phi \left(t\right)\phantom{\rule{0.2em}{0ex}}dt\right)& =\underset{n\to \mathrm{\infty }}{lim sup}\psi \left({\int }_{0}^{d\left({f}^{n+1}x,fa\right)}\phi \left(t\right)\phantom{\rule{0.2em}{0ex}}dt\right)\\ \le \underset{n\to \mathrm{\infty }}{lim sup}\left[\psi \left({\int }_{0}^{d\left({f}^{n}x,a\right)}\phi \left(t\right)\phantom{\rule{0.2em}{0ex}}dt\right)-\varphi \left({\int }_{0}^{d\left({f}^{n}x,a\right)}\phi \left(t\right)\phantom{\rule{0.2em}{0ex}}dt\right)\right]\\ \le \underset{n\to \mathrm{\infty }}{lim sup}\psi \left({\int }_{0}^{d\left({f}^{n}x,a\right)}\phi \left(t\right)\phantom{\rule{0.2em}{0ex}}dt\right)-\underset{n\to \mathrm{\infty }}{lim inf}\varphi \left({\int }_{0}^{d\left({f}^{n}x,a\right)}\phi \left(t\right)\phantom{\rule{0.2em}{0ex}}dt\right)\\ =\psi \left(0\right)-0\\ =0,\end{array}$
which together with $\psi \in {\mathrm{\Phi }}_{3}$ yields that
${\int }_{0}^{d\left(a,fa\right)}\phi \left(t\right)\phantom{\rule{0.2em}{0ex}}dt=0,$

that is, $a=fa$.

Finally, we show that a is a unique fixed point of f in X. Suppose that f has another fixed point $b\in X\setminus \left\{a\right\}$. It follows from (1.1) and $\left(\phi ,\varphi ,\psi \right)\in {\mathrm{\Phi }}_{1}×{\mathrm{\Phi }}_{2}×{\mathrm{\Phi }}_{3}$ that
$\begin{array}{rl}\psi \left({\int }_{0}^{d\left(a,b\right)}\phi \left(t\right)\phantom{\rule{0.2em}{0ex}}dt\right)& =\psi \left({\int }_{0}^{d\left(fa,fb\right)}\phi \left(t\right)\phantom{\rule{0.2em}{0ex}}dt\right)\\ \le \psi \left({\int }_{0}^{d\left(a,b\right)}\phi \left(t\right)\phantom{\rule{0.2em}{0ex}}dt\right)-\varphi \left({\int }_{0}^{d\left(a,b\right)}\phi \left(t\right)\phantom{\rule{0.2em}{0ex}}dt\right)\\ <\psi \left({\int }_{0}^{d\left(a,b\right)}\phi \left(t\right)\phantom{\rule{0.2em}{0ex}}dt\right),\end{array}$

which is a contradiction. This completes the proof. □

Theorem 3.2 Let f be a mapping from a complete metric space $\left(X,d\right)$ into itself satisfying (1.2). Then f has a unique fixed point $a\in X$ such that ${lim}_{n\to \mathrm{\infty }}{f}^{n}x=a$ for each $x\in X$.

Proof Let x be an arbitrary point in X. Suppose that (3.2) holds for some ${n}_{0}\in \mathbb{N}$. Using (1.2), (3.2) and $\left(\phi ,\psi ,\alpha \right)\in {\mathrm{\Phi }}_{1}×{\mathrm{\Phi }}_{3}×{\mathrm{\Phi }}_{5}$, we get that
$\psi \left({\int }_{0}^{{d}_{{n}_{0}}}\phi \left(t\right)\phantom{\rule{0.2em}{0ex}}dt\right)>0$
and
$\begin{array}{rl}\psi \left({\int }_{0}^{{d}_{{n}_{0}-1}}\phi \left(t\right)\phantom{\rule{0.2em}{0ex}}dt\right)& \le \psi \left({\int }_{0}^{{d}_{{n}_{0}}}\phi \left(t\right)\phantom{\rule{0.2em}{0ex}}dt\right)=\psi \left({\int }_{0}^{d\left({f}^{{n}_{0}}x,{f}^{{n}_{0}+1}x\right)}\phi \left(t\right)\phantom{\rule{0.2em}{0ex}}dt\right)\\ \le \alpha \left(d\left({f}^{{n}_{0}-1}x,{f}^{{n}_{0}}x\right)\right)\psi \left({\int }_{0}^{d\left({f}^{{n}_{0}-1}x,{f}^{{n}_{0}}x\right)}\phi \left(t\right)\phantom{\rule{0.2em}{0ex}}dt\right)\\ =\alpha \left({d}_{{n}_{0}-1}\right)\psi \left({\int }_{0}^{{d}_{{n}_{0}-1}}\phi \left(t\right)\phantom{\rule{0.2em}{0ex}}dt\right)<\psi \left({\int }_{0}^{{d}_{{n}_{0}-1}}\phi \left(t\right)\phantom{\rule{0.2em}{0ex}}dt\right),\end{array}$
which is a contradiction, and hence (3.2) does not hold. Consequently, (3.1) is true. Notice that the nonnegative sequence ${\left\{{d}_{n}\right\}}_{n\in {\mathbb{N}}_{0}}$ is nonincreasing, which implies that there exists a constant $c\ge 0$ with ${lim}_{n\to \mathrm{\infty }}{d}_{n}=c$. Suppose that $c>0$. In light of (1.2), we infer that
$\begin{array}{rl}\psi \left({\int }_{0}^{{d}_{n}}\phi \left(t\right)\phantom{\rule{0.2em}{0ex}}dt\right)& =\psi \left({\int }_{0}^{d\left({f}^{n}x,{f}^{n+1}x\right)}\phi \left(t\right)\phantom{\rule{0.2em}{0ex}}dt\right)\\ \le \alpha \left(d\left({f}^{n-1}x,{f}^{n}x\right)\right)\psi \left({\int }_{0}^{d\left({f}^{n-1}x,{f}^{n}x\right)}\phi \left(t\right)\phantom{\rule{0.2em}{0ex}}dt\right)\\ =\alpha \left({d}_{n-1}\right)\psi \left({\int }_{0}^{{d}_{n-1}}\phi \left(t\right)\phantom{\rule{0.2em}{0ex}}dt\right),\phantom{\rule{1em}{0ex}}\mathrm{\forall }n\in \mathbb{N}.\end{array}$
(3.13)
Taking upper limit in (3.13) and using Lemma 2.1 and $\left(\phi ,\psi ,\alpha \right)\in {\mathrm{\Phi }}_{1}×{\mathrm{\Phi }}_{3}×{\mathrm{\Phi }}_{5}$, we know that
$\begin{array}{rl}\psi \left({\int }_{0}^{c}\phi \left(t\right)\phantom{\rule{0.2em}{0ex}}dt\right)& =\underset{n\to \mathrm{\infty }}{lim sup}\psi \left({\int }_{0}^{{d}_{n}}\phi \left(t\right)\phantom{\rule{0.2em}{0ex}}dt\right)\\ \le \underset{n\to \mathrm{\infty }}{lim sup}\left[\alpha \left({d}_{n-1}\right)\psi \left({\int }_{0}^{{d}_{n-1}}\phi \left(t\right)\phantom{\rule{0.2em}{0ex}}dt\right)\right]\\ \le \underset{n\to \mathrm{\infty }}{lim sup}\alpha \left({d}_{n-1}\right)\cdot \underset{n\to \mathrm{\infty }}{lim sup}\psi \left({\int }_{0}^{{d}_{n-1}}\phi \left(t\right)\phantom{\rule{0.2em}{0ex}}dt\right)\\ <\psi \left({\int }_{0}^{c}\phi \left(t\right)\phantom{\rule{0.2em}{0ex}}dt\right),\end{array}$

which is a contradiction, and hence $c=0$, that is, (3.6) holds.

Now we show that ${\left\{{f}^{n}x\right\}}_{n\in \mathbb{N}}$ is a Cauchy sequence. Suppose that ${\left\{{f}^{n}x\right\}}_{n\in \mathbb{N}}$ is not a Cauchy sequence. As in the proof of Theorem 3.1, we conclude that there exist $\epsilon >0$ and $\left\{m\left(k\right),n\left(k\right):k\in \mathbb{N}\right\}\subseteq \mathbb{N}$ with $m\left(k\right)>n\left(k\right)>k$ for each $k\in \mathbb{N}$ satisfying (3.8)-(3.11). By means of (1.2), (3.11), Lemma 2.1 and $\left(\phi ,\psi ,\alpha \right)\in {\mathrm{\Phi }}_{1}×{\mathrm{\Phi }}_{3}×{\mathrm{\Phi }}_{5}$, we get that
$\begin{array}{rl}\psi \left({\int }_{0}^{\epsilon }\phi \left(t\right)\phantom{\rule{0.2em}{0ex}}dt\right)& =\underset{k\to \mathrm{\infty }}{lim sup}\psi \left({\int }_{0}^{d\left({f}^{m\left(k\right)+1}x,{f}^{n\left(k\right)+2}x\right)}\phi \left(t\right)\phantom{\rule{0.2em}{0ex}}dt\right)\\ =\underset{k\to \mathrm{\infty }}{lim sup}\left[\alpha \left(d\left({f}^{m\left(k\right)}x,{f}^{n\left(k\right)+1}x\right)\right)\psi \left({\int }_{0}^{d\left({f}^{m\left(k\right)}x,{f}^{n\left(k\right)+1}x\right)}\phi \left(t\right)\phantom{\rule{0.2em}{0ex}}dt\right)\right]\\ \le \underset{k\to \mathrm{\infty }}{lim sup}\alpha \left(d\left({f}^{m\left(k\right)}x,{f}^{n\left(k\right)+1}x\right)\right)\cdot \underset{k\to \mathrm{\infty }}{lim sup}\psi \left({\int }_{0}^{d\left({f}^{m\left(k\right)}x,{f}^{n\left(k\right)+1}x\right)}\phi \left(t\right)\phantom{\rule{0.2em}{0ex}}dt\right)\\ <\psi \left({\int }_{0}^{\epsilon }\phi \left(t\right)\phantom{\rule{0.2em}{0ex}}dt\right),\end{array}$

which is a contradiction. Hence ${\left\{{f}^{n}x\right\}}_{n\in \mathbb{N}}$ is a Cauchy sequence.

It follows from completeness of $\left(X,d\right)$ that there exists $a\in X$ with ${lim}_{n\to \mathrm{\infty }}{f}^{n}x=a$. In view of (1.2), we have
$\psi \left({\int }_{0}^{d\left({f}^{n+1}x,fa\right)}\phi \left(t\right)\phantom{\rule{0.2em}{0ex}}dt\right)\le \alpha \left(d\left({f}^{n}x,a\right)\right)\psi \left({\int }_{0}^{d\left({f}^{n}x,a\right)}\phi \left(t\right)\phantom{\rule{0.2em}{0ex}}dt\right),\phantom{\rule{1em}{0ex}}\mathrm{\forall }n\in {\mathbb{N}}_{0}.$
(3.14)
Taking upper limit in (3.14) and making use of $\left(\phi ,\psi ,\alpha \right)\in {\mathrm{\Phi }}_{1}×{\mathrm{\Phi }}_{3}×{\mathrm{\Phi }}_{5}$ and Lemmas 2.1 and 2.2, we get that
$\begin{array}{rl}\psi \left({\int }_{0}^{d\left(a,fa\right)}\phi \left(t\right)\phantom{\rule{0.2em}{0ex}}dt\right)& =\underset{n\to \mathrm{\infty }}{lim sup}\psi \left({\int }_{0}^{d\left({f}^{n+1}x,fa\right)}\phi \left(t\right)\phantom{\rule{0.2em}{0ex}}dt\right)\\ \le \underset{n\to \mathrm{\infty }}{lim sup}\left[\alpha \left(d\left({f}^{n}x,a\right)\right)\psi \left({\int }_{0}^{d\left({f}^{n}x,a\right)}\phi \left(t\right)\phantom{\rule{0.2em}{0ex}}dt\right)\right]\\ \le \underset{n\to \mathrm{\infty }}{lim sup}\alpha \left(d\left({f}^{n}x,a\right)\right)\cdot \underset{n\to \mathrm{\infty }}{lim sup}\psi \left({\int }_{0}^{d\left({f}^{n}x,a\right)}\phi \left(t\right)\phantom{\rule{0.2em}{0ex}}dt\right)\\ =0,\end{array}$
which means that
$\psi \left({\int }_{0}^{d\left(a,fa\right)}\phi \left(t\right)\phantom{\rule{0.2em}{0ex}}dt\right)=0,$

that is, $fa=a$.

Next we prove that a is a unique fixed point of f in X. Suppose that f has another fixed point $b\in X\setminus \left\{a\right\}$. It follows from (1.2) and $\left(\phi ,\psi ,\alpha \right)\in {\mathrm{\Phi }}_{1}×{\mathrm{\Phi }}_{3}×{\mathrm{\Phi }}_{5}$ that
$\begin{array}{rl}\psi \left({\int }_{0}^{d\left(a,b\right)}\phi \left(t\right)\phantom{\rule{0.2em}{0ex}}dt\right)& =\psi \left({\int }_{0}^{d\left(fa,fb\right)}\phi \left(t\right)\phantom{\rule{0.2em}{0ex}}dt\right)\le \alpha \left(d\left(a,b\right)\right)\psi \left({\int }_{0}^{d\left(a,b\right)}\phi \left(t\right)\phantom{\rule{0.2em}{0ex}}dt\right)\\ <\psi \left({\int }_{0}^{d\left(a,b\right)}\phi \left(t\right)\phantom{\rule{0.2em}{0ex}}dt\right),\end{array}$

which is a contradiction. This completes the proof. □

Theorem 3.3 Let f be a mapping from a complete metric space $\left(X,d\right)$ into itself satisfying (1.3) and
$\varphi \left(t\right)\le \psi \left(t\right),\phantom{\rule{1em}{0ex}}\mathrm{\forall }t\in {\mathbb{R}}^{+}.$
(3.15)

Then f has a unique fixed point $a\in X$ such that ${lim}_{n\to \mathrm{\infty }}{f}^{n}x=a$ for each $x\in X$.

Proof Let x be an arbitrary point in X. If there exists ${n}_{0}\in {\mathbb{N}}_{0}$ satisfying ${d}_{{n}_{0}}=0$, it is clear that ${f}^{{n}_{0}}x$ is a fixed point of f and ${lim}_{n\to \mathrm{\infty }}{f}^{n}x={f}^{{n}_{0}}x$. Now we assume that ${d}_{n}\ne 0$ for all $n\in {\mathbb{N}}_{0}$. Suppose that (3.2) holds for some ${n}_{0}\in \mathbb{N}$. It follows from (1.3) that
$\begin{array}{rl}\psi \left({\int }_{0}^{{d}_{{n}_{0}}}\phi \left(t\right)\phantom{\rule{0.2em}{0ex}}dt\right)=& \psi \left({\int }_{0}^{d\left({f}^{{n}_{0}}x,{f}^{{n}_{0}+1}x\right)}\phi \left(t\right)\phantom{\rule{0.2em}{0ex}}dt\right)\\ \le & \alpha \left(d\left({f}^{{n}_{0}-1}x,{f}^{{n}_{0}}x\right)\right)\varphi \left({\int }_{0}^{d\left({f}^{{n}_{0}-1}x,{f}^{{n}_{0}}x\right)}\phi \left(t\right)\phantom{\rule{0.2em}{0ex}}dt\right)\\ +\beta \left(d\left({f}^{{n}_{0}-1}x,{f}^{{n}_{0}}x\right)\right)\psi \left({\int }_{0}^{d\left({f}^{{n}_{0}}x,{f}^{{n}_{0}+1}x\right)}\phi \left(t\right)\phantom{\rule{0.2em}{0ex}}dt\right)\\ =& \alpha \left({d}_{{n}_{0}-1}\right)\varphi \left({\int }_{0}^{{d}_{{n}_{0}-1}}\phi \left(t\right)\phantom{\rule{0.2em}{0ex}}dt\right)+\beta \left({d}_{{n}_{0}-1}\right)\psi \left({\int }_{0}^{{d}_{{n}_{0}}}\phi \left(t\right)\phantom{\rule{0.2em}{0ex}}dt\right),\end{array}$
which together with (3.2), (3.15), $\left(\phi ,\psi ,\varphi \right)\in {\mathrm{\Phi }}_{1}×{\mathrm{\Phi }}_{3}×{\mathrm{\Phi }}_{4}$ and $\left(\alpha ,\beta \right)\in {\mathrm{\Phi }}_{6}$ implies that
$\begin{array}{rl}0& <\psi \left({\int }_{0}^{{d}_{{n}_{0}-1}}\phi \left(t\right)\phantom{\rule{0.2em}{0ex}}dt\right)\le \psi \left({\int }_{0}^{{d}_{{n}_{0}}}\phi \left(t\right)\phantom{\rule{0.2em}{0ex}}dt\right)\\ \le \frac{\alpha \left({d}_{{n}_{0}-1}\right)}{1-\beta \left({d}_{{n}_{0}-1}\right)}\varphi \left({\int }_{0}^{{d}_{{n}_{0}-1}}\phi \left(t\right)\phantom{\rule{0.2em}{0ex}}dt\right)\\ \le \frac{\alpha \left({d}_{{n}_{0}-1}\right)}{1-\beta \left({d}_{{n}_{0}-1}\right)}\psi \left({\int }_{0}^{{d}_{{n}_{0}-1}}\phi \left(t\right)\phantom{\rule{0.2em}{0ex}}dt\right)\\ <\psi \left({\int }_{0}^{{d}_{{n}_{0}-1}}\phi \left(t\right)\phantom{\rule{0.2em}{0ex}}dt\right),\end{array}$

which is a contradiction, and hence (3.2) does not hold. Consequently, (3.1) holds.

Next we show that ${lim}_{n\to \mathrm{\infty }}{d}_{n}=0$. Note that the nonnegative sequence ${\left\{{d}_{n}\right\}}_{n\in \mathbb{N}}$ is nonincreasing, which implies that there exists a constant $c\ge 0$ with ${lim}_{n\to \mathrm{\infty }}{d}_{n}=c$. Suppose that $c>0$. It follows from (1.3) that
$\begin{array}{rl}\psi \left({\int }_{0}^{{d}_{n}}\phi \left(t\right)\phantom{\rule{0.2em}{0ex}}dt\right)=& \psi \left({\int }_{0}^{d\left({f}^{n}x,{f}^{n+1}x\right)}\phi \left(t\right)\phantom{\rule{0.2em}{0ex}}dt\right)\\ \le & \alpha \left(d\left({f}^{n-1}x,{f}^{n}x\right)\right)\varphi \left({\int }_{0}^{d\left({f}^{n-1}x,{f}^{n}x\right)}\phi \left(t\right)\phantom{\rule{0.2em}{0ex}}dt\right)\\ +\beta \left(d\left({f}^{n-1}x,{f}^{n}x\right)\right)\psi \left({\int }_{0}^{d\left({f}^{n}x,{f}^{n+1}x\right)}\phi \left(t\right)\phantom{\rule{0.2em}{0ex}}dt\right)\\ =& \alpha \left({d}_{n-1}\right)\varphi \left({\int }_{0}^{{d}_{n-1}}\phi \left(t\right)\phantom{\rule{0.2em}{0ex}}dt\right)+\beta \left({d}_{n-1}\right)\psi \left({\int }_{0}^{{d}_{n}}\phi \left(t\right)\phantom{\rule{0.2em}{0ex}}dt\right),\phantom{\rule{1em}{0ex}}\mathrm{\forall }n\in \mathbb{N},\end{array}$
which means that
$\psi \left({\int }_{0}^{{d}_{n}}\phi \left(t\right)\phantom{\rule{0.2em}{0ex}}dt\right)\le \frac{\alpha \left({d}_{n-1}\right)}{1-\beta \left({d}_{n-1}\right)}\varphi \left({\int }_{0}^{{d}_{n-1}}\phi \left(t\right)\phantom{\rule{0.2em}{0ex}}dt\right),\phantom{\rule{1em}{0ex}}\mathrm{\forall }n\in \mathbb{N}.$
(3.16)
Taking upper limit in (3.16) and using (3.15), $\left(\phi ,\psi ,\varphi \right)\in {\mathrm{\Phi }}_{1}×{\mathrm{\Phi }}_{3}×{\mathrm{\Phi }}_{4}$, $\left(\alpha ,\beta \right)\in {\mathrm{\Phi }}_{6}$ and Lemma 2.1, we arrive at
$\begin{array}{rl}\psi \left({\int }_{0}^{c}\phi \left(t\right)\phantom{\rule{0.2em}{0ex}}dt\right)& =\underset{n\to \mathrm{\infty }}{lim sup}\psi \left({\int }_{0}^{{d}_{n}}\phi \left(t\right)\phantom{\rule{0.2em}{0ex}}dt\right)\\ \le \underset{n\to \mathrm{\infty }}{lim sup}\left[\frac{\alpha \left({d}_{n-1}\right)}{1-\beta \left({d}_{n-1}\right)}\varphi \left({\int }_{0}^{{d}_{n-1}}\phi \left(t\right)\phantom{\rule{0.2em}{0ex}}dt\right)\right]\\ \le \underset{n\to \mathrm{\infty }}{lim sup}\frac{\alpha \left({d}_{n-1}\right)}{1-\beta \left({d}_{n-1}\right)}\cdot \underset{n\to \mathrm{\infty }}{lim sup}\psi \left({\int }_{0}^{{d}_{n-1}}\phi \left(t\right)\phantom{\rule{0.2em}{0ex}}dt\right)\\ \le \underset{s\to {c}^{+}}{lim sup}\frac{\alpha \left(s\right)}{1-\beta \left(s\right)}\cdot \psi \left({\int }_{0}^{c}\phi \left(t\right)\phantom{\rule{0.2em}{0ex}}dt\right)\\ <\psi \left({\int }_{0}^{c}\phi \left(t\right)\phantom{\rule{0.2em}{0ex}}dt\right),\end{array}$

which is impossible. Therefore $c=0$, that is, ${lim}_{n\to \mathrm{\infty }}{d}_{n}=0$.

Next we show that ${\left\{{f}^{n}x\right\}}_{n\in \mathbb{N}}$ is a Cauchy sequence. Suppose that ${\left\{{f}^{n}x\right\}}_{n\in \mathbb{N}}$ is not a Cauchy sequence. As in the proof of Theorem 3.1, we conclude that there exist $\epsilon >0$ and $\left\{m\left(k\right),n\left(k\right):k\in \mathbb{N}\right\}\subseteq \mathbb{N}$ with $m\left(k\right)>n\left(k\right)>k$ for each $k\in \mathbb{N}$ satisfying (3.8)-(3.11). By means of (3.12), we deduce that
$\begin{array}{r}\psi \left({\int }_{0}^{d\left({f}^{m\left(k\right)+1}x,{f}^{n\left(k\right)+2}x\right)}\phi \left(t\right)\phantom{\rule{0.2em}{0ex}}dt\right)\\ \phantom{\rule{1em}{0ex}}\le \alpha \left(d\left({f}^{m\left(k\right)}x,{f}^{n\left(k\right)+1}x\right)\right)\varphi \left({\int }_{0}^{d\left({f}^{m\left(k\right)}x,{f}^{m\left(k\right)+1}x\right)}\phi \left(t\right)\phantom{\rule{0.2em}{0ex}}dt\right)\\ \phantom{\rule{2em}{0ex}}+\beta \left(d\left({f}^{m\left(k\right)}x,{f}^{n\left(k\right)+1}x\right)\right)\psi \left({\int }_{0}^{d\left({f}^{n\left(k\right)+1}x,{f}^{n\left(k\right)+2}x\right)}\phi \left(t\right)\phantom{\rule{0.2em}{0ex}}dt\right)\\ \phantom{\rule{1em}{0ex}}=\alpha \left(d\left({f}^{m\left(k\right)}x,{f}^{n\left(k\right)+1}x\right)\right)\varphi \left({\int }_{0}^{{d}_{m\left(k\right)}}\phi \left(t\right)\phantom{\rule{0.2em}{0ex}}dt\right)\\ \phantom{\rule{2em}{0ex}}+\beta \left(d\left({f}^{m\left(k\right)}x,{f}^{n\left(k\right)+1}x\right)\right)\psi \left({\int }_{0}^{{d}_{n\left(k\right)}}\phi \left(t\right)\phantom{\rule{0.2em}{0ex}}dt\right),\phantom{\rule{1em}{0ex}}\mathrm{\forall }k\in \mathbb{N}.\end{array}$
(3.17)
Taking upper limit in (3.17) and making use of (1.3), (3.11), Lemma 2.1, $\left(\phi ,\psi ,\varphi \right)\in {\mathrm{\Phi }}_{1}×{\mathrm{\Phi }}_{3}×{\mathrm{\Phi }}_{4}$ and $\left(\alpha ,\beta \right)\in {\mathrm{\Phi }}_{6}$, we deduce that
$\begin{array}{rl}0<& \psi \left({\int }_{0}^{\epsilon }\phi \left(t\right)\phantom{\rule{0.2em}{0ex}}dt\right)=\underset{k\to \mathrm{\infty }}{lim sup}\psi \left({\int }_{0}^{d\left({f}^{m\left(k\right)+1}x,{f}^{n\left(k\right)+2}x\right)}\phi \left(t\right)\phantom{\rule{0.2em}{0ex}}dt\right)\\ \le & \underset{k\to \mathrm{\infty }}{lim sup}\left[\alpha \left(d\left({f}^{m\left(k\right)}x,{f}^{n\left(k\right)+1}x\right)\right)\varphi \left({\int }_{0}^{{d}_{m\left(k\right)}}\phi \left(t\right)\phantom{\rule{0.2em}{0ex}}dt\right)\\ +\beta \left(d\left({f}^{m\left(k\right)}x,{f}^{n\left(k\right)+1}x\right)\right)\psi \left({\int }_{0}^{{d}_{n\left(k\right)}}\phi \left(t\right)\phantom{\rule{0.2em}{0ex}}dt\right)\right]\\ \le & \underset{k\to \mathrm{\infty }}{lim sup}\alpha \left(d\left({f}^{m\left(k\right)}x,{f}^{n\left(k\right)+1}x\right)\right)\cdot \underset{k\to \mathrm{\infty }}{lim sup}\psi \left({\int }_{0}^{{d}_{m\left(k\right)}}\phi \left(t\right)\phantom{\rule{0.2em}{0ex}}dt\right)\\ +\underset{k\to \mathrm{\infty }}{lim sup}\beta \left(d\left({f}^{m\left(k\right)}x,{f}^{n\left(k\right)+1}x\right)\right)\cdot \underset{k\to \mathrm{\infty }}{lim sup}\psi \left({\int }_{0}^{{d}_{n\left(k\right)}}\phi \left(t\right)\phantom{\rule{0.2em}{0ex}}dt\right)\\ \le & \underset{s\to \epsilon }{lim sup}\alpha \left(s\right)\cdot \psi \left({\int }_{0}^{0}\phi \left(t\right)\phantom{\rule{0.2em}{0ex}}dt\right)+\underset{s\to \epsilon }{lim sup}\beta \left(s\right)\cdot \psi \left({\int }_{0}^{0}\phi \left(t\right)\phantom{\rule{0.2em}{0ex}}dt\right)\\ =& 0,\end{array}$

which is a contradiction. Hence ${\left\{{f}^{n}x\right\}}_{n\in \mathbb{N}}$ is a Cauchy sequence.

Completeness of $\left(X,d\right)$ implies that there exists a point $a\in X$ such that ${lim}_{n\to \mathrm{\infty }}{f}^{n}x=a$. In view of (1.3), $\left(\phi ,\psi ,\varphi \right)\in {\mathrm{\Phi }}_{1}×{\mathrm{\Phi }}_{3}×{\mathrm{\Phi }}_{4}$, $\left(\alpha ,\beta \right)\in {\mathrm{\Phi }}_{6}$ and Lemma 2.1, we infer that
$\begin{array}{rl}\psi \left({\int }_{0}^{d\left(a,fa\right)}\phi \left(t\right)\phantom{\rule{0.2em}{0ex}}dt\right)=& \underset{n\to \mathrm{\infty }}{lim sup}\psi \left({\int }_{0}^{d\left({f}^{n+1}x,fa\right)}\phi \left(t\right)\phantom{\rule{0.2em}{0ex}}dt\right)\\ \le & \underset{n\to \mathrm{\infty }}{lim sup}\left[\alpha \left(d\left({f}^{n}x,a\right)\right)\varphi \left({\int }_{0}^{d\left({f}^{n}x,{f}^{n+1}x\right)}\phi \left(t\right)\phantom{\rule{0.2em}{0ex}}dt\right)\\ +\beta \left(d\left({f}^{n}x,a\right)\right)\psi \left({\int }_{0}^{d\left(a,fa\right)}\phi \left(t\right)\phantom{\rule{0.2em}{0ex}}dt\right)\right]\\ \le & \underset{n\to \mathrm{\infty }}{lim sup}\alpha \left(d\left({f}^{n}x,a\right)\right)\cdot \underset{n\to \mathrm{\infty }}{lim sup}\psi \left({\int }_{0}^{{d}_{n}}\phi \left(t\right)\phantom{\rule{0.2em}{0ex}}dt\right)\\ +\underset{n\to \mathrm{\infty }}{lim sup}\beta \left(d\left({f}^{n}x,a\right)\right)\cdot \psi \left({\int }_{0}^{d\left(a,fa\right)}\phi \left(t\right)\phantom{\rule{0.2em}{0ex}}dt\right)\\ \le & \underset{s\to {0}^{+}}{lim sup}\beta \left(s\right)\cdot \psi \left({\int }_{0}^{d\left(a,fa\right)}\phi \left(t\right)\phantom{\rule{0.2em}{0ex}}dt\right),\end{array}$
which together with $\left(\alpha ,\beta \right)\in {\mathrm{\Phi }}_{6}$ yields that
$\psi \left({\int }_{0}^{d\left(a,fa\right)}\phi \left(t\right)\phantom{\rule{0.2em}{0ex}}dt\right)=0,$

which gives that $d\left(fa,a\right)=0$, that is, $fa=a$.

Finally, we prove that a is a unique fixed point of f in X. Suppose that f has another fixed point $b\in X\setminus \left\{a\right\}$. It follows from (1.3) and $\left(\phi ,\psi ,\varphi \right)\in {\mathrm{\Phi }}_{1}×{\mathrm{\Phi }}_{3}×{\mathrm{\Phi }}_{4}$ and $\left(\alpha ,\beta \right)\in {\mathrm{\Phi }}_{6}$ that
$\begin{array}{rl}0& \le \psi \left({\int }_{0}^{d\left(fa,fb\right)}\phi \left(t\right)\phantom{\rule{0.2em}{0ex}}dt\right)\\ \le \alpha \left(d\left(a,b\right)\right)\varphi \left({\int }_{0}^{d\left(a,fa\right)}\phi \left(t\right)\phantom{\rule{0.2em}{0ex}}dt\right)+\beta \left(d\left(a,b\right)\right)\psi \left({\int }_{0}^{d\left(b,fb\right)}\phi \left(t\right)\phantom{\rule{0.2em}{0ex}}dt\right)\\ =0,\end{array}$

which is a contradiction. This completes the proof. □

## 4 Three examples

Now we construct three examples to explain Theorems 3.1-3.3.

Example 4.1 Let $X=\left[0,\frac{1}{2}\right]\cup \left\{1\right\}\cup \left\{3\right\}$ be endowed with the Euclidean metric $d=|\cdot |$. Assume that $f:X\to X$ and $\phi ,\varphi ,\psi :{\mathbb{R}}^{+}\to {\mathbb{R}}^{+}$ are defined by
$\begin{array}{c}f\left(x\right)=\left\{\begin{array}{cc}\frac{x}{2},\hfill & \mathrm{\forall }x\in \left[0,\frac{1}{2}\right],\hfill \\ 0,\hfill & x=1,\hfill \\ 1,\hfill & x=3,\hfill \end{array}\phantom{\rule{2em}{0ex}}\phi \left(t\right)=\left\{\begin{array}{cc}\frac{t}{2},\hfill & \mathrm{\forall }t\in \left[0,1\right],\hfill \\ 1,\hfill & \mathrm{\forall }t\in \left(1,+\mathrm{\infty }\right),\hfill \end{array}\hfill \\ \varphi \left(t\right)=\left\{\begin{array}{cc}\frac{{t}^{2}}{4},\hfill & \mathrm{\forall }t\in \left[0,1\right],\hfill \\ \frac{{t}^{2}}{8},\hfill & \mathrm{\forall }t\in \left(1,+\mathrm{\infty }\right),\hfill \end{array}\phantom{\rule{2em}{0ex}}\psi \left(t\right)=\left\{\begin{array}{cc}t,\hfill & \mathrm{\forall }t\in \left[0,1\right],\hfill \\ \frac{{t}^{2}+1}{2},\hfill & \mathrm{\forall }t\in \left(1,+\mathrm{\infty }\right).\hfill \end{array}\hfill \end{array}$

Clearly, $\left(X,d\right)$ is a complete metric and $\left(\phi ,\varphi ,\psi \right)\in {\mathrm{\Phi }}_{1}×{\mathrm{\Phi }}_{2}×{\mathrm{\Phi }}_{3}$. Let $x,y\in X$ with $x. In order to verify (1.1), we have to consider the following four cases.

Case 1. Let $x,y\in \left[0,\frac{1}{2}\right]$. Note that
$\begin{array}{rl}\psi \left({\int }_{0}^{d\left(fx,fy\right)}\phi \left(t\right)\phantom{\rule{0.2em}{0ex}}dt\right)& =\psi \left({\int }_{0}^{\frac{1}{2}|x-y|}\phi \left(t\right)\phantom{\rule{0.2em}{0ex}}dt\right)=\psi \left(\frac{{|x-y|}^{2}}{16}\right)=\frac{{|x-y|}^{2}}{16}\\ \le \frac{{|x-y|}^{2}}{4}-\frac{{|x-y|}^{4}}{16}\\ =\psi \left(\frac{{|x-y|}^{2}}{4}\right)-\varphi \left(\frac{{|x-y|}^{2}}{4}\right)\\ =\psi \left({\int }_{0}^{|x-y|}\phi \left(t\right)\phantom{\rule{0.2em}{0ex}}dt\right)-\varphi \left({\int }_{0}^{|x-y|}\phi \left(t\right)\phantom{\rule{0.2em}{0ex}}dt\right)\\ =\psi \left({\int }_{0}^{d\left(x,y\right)}\phi \left(t\right)\phantom{\rule{0.2em}{0ex}}dt\right)-\varphi \left({\int }_{0}^{d\left(x,y\right)}\phi \left(t\right)\phantom{\rule{0.2em}{0ex}}dt\right).\end{array}$
Case 2. Let $x\in \left[0,\frac{1}{2}\right]$ and $y=1$. It follows that
$\begin{array}{rl}\psi \left({\int }_{0}^{d\left(fx,fy\right)}\phi \left(t\right)\phantom{\rule{0.2em}{0ex}}dt\right)& =\psi \left({\int }_{0}^{\frac{x}{2}}\phi \left(t\right)\phantom{\rule{0.2em}{0ex}}dt\right)=\psi \left(\frac{{x}^{2}}{16}\right)=\frac{{x}^{2}}{16}\le \frac{{\left(1-x\right)}^{2}}{4}-\frac{{\left(1-x\right)}^{4}}{16}\\ =\psi \left(\frac{{\left(1-x\right)}^{2}}{4}\right)-\varphi \left(\frac{{\left(1-x\right)}^{2}}{4}\right)\\ =\psi \left({\int }_{0}^{|x-1|}\phi \left(t\right)\phantom{\rule{0.2em}{0ex}}dt\right)-\varphi \left({\int }_{0}^{|x-1|}\phi \left(t\right)\phantom{\rule{0.2em}{0ex}}dt\right)\\ =\psi \left({\int }_{0}^{d\left(x,y\right)}\phi \left(t\right)\phantom{\rule{0.2em}{0ex}}dt\right)-\varphi \left({\int }_{0}^{d\left(x,y\right)}\phi \left(t\right)\phantom{\rule{0.2em}{0ex}}dt\right).\end{array}$
Case 3. Let $x\in \left[0,\frac{1}{2}\right]$ and $y=3$. It follows that
$\begin{array}{rl}\psi \left({\int }_{0}^{d\left(fx,fy\right)}\phi \left(t\right)\phantom{\rule{0.2em}{0ex}}dt\right)& =\psi \left({\int }_{0}^{\frac{2-x}{2}}\phi \left(t\right)\phantom{\rule{0.2em}{0ex}}dt\right)=\psi \left(\frac{{\left(2-x\right)}^{2}}{16}\right)=\frac{{\left(2-x\right)}^{2}}{16}<\frac{1}{2}\\ \le \frac{1}{2}\left[{\left(\frac{9}{4}-x\right)}^{2}+1\right]-\frac{1}{8}{\left(\frac{9}{4}-x\right)}^{2}=\psi \left(\frac{9}{4}-x\right)-\varphi \left(\frac{9}{4}-x\right)\\ =\psi \left({\int }_{0}^{1}\phi \left(t\right)\phantom{\rule{0.2em}{0ex}}dt+{\int }_{1}^{3-x}\phi \left(t\right)\phantom{\rule{0.2em}{0ex}}dt\right)-\varphi \left({\int }_{0}^{1}\phi \left(t\right)\phantom{\rule{0.2em}{0ex}}dt+{\int }_{1}^{3-x}\phi \left(t\right)\phantom{\rule{0.2em}{0ex}}dt\right)\\ =\psi \left({\int }_{0}^{3-x}\phi \left(t\right)\phantom{\rule{0.2em}{0ex}}dt\right)-\varphi \left({\int }_{0}^{3-x}\phi \left(t\right)\phantom{\rule{0.2em}{0ex}}dt\right)\\ =\psi \left({\int }_{0}^{d\left(x,y\right)}\phi \left(t\right)\phantom{\rule{0.2em}{0ex}}dt\right)-\varphi \left({\int }_{0}^{d\left(x,y\right)}\phi \left(t\right)\phantom{\rule{0.2em}{0ex}}dt\right).\end{array}$
Case 4. Let $x=1$ and $y=3$. Note that
$\begin{array}{rl}\psi \left({\int }_{0}^{d\left(fx,fy\right)}\phi \left(t\right)\phantom{\rule{0.2em}{0ex}}dt\right)& =\psi \left({\int }_{0}^{1}\phi \left(t\right)\phantom{\rule{0.2em}{0ex}}dt\right)=\psi \left(\frac{1}{4}\right)=\frac{1}{4}<\frac{139}{128}\\ =\frac{1}{2}\left(\frac{25}{16}+1\right)-\frac{1}{8}\cdot \frac{25}{16}=\psi \left(\frac{5}{4}\right)-\varphi \left(\frac{5}{4}\right)\\ =\psi \left({\int }_{0}^{1}\phi \left(t\right)\phantom{\rule{0.2em}{0ex}}dt+{\int }_{1}^{2}\phi \left(t\right)\phantom{\rule{0.2em}{0ex}}dt\right)-\varphi \left({\int }_{0}^{1}\phi \left(t\right)\phantom{\rule{0.2em}{0ex}}dt+{\int }_{1}^{2}\phi \left(t\right)\phantom{\rule{0.2em}{0ex}}dt\right)\\ =\psi \left({\int }_{0}^{2}\phi \left(t\right)\phantom{\rule{0.2em}{0ex}}dt\right)-\varphi \left({\int }_{0}^{2}\phi \left(t\right)\phantom{\rule{0.2em}{0ex}}dt\right)\\ =\psi \left({\int }_{0}^{d\left(x,y\right)}\phi \left(t\right)\phantom{\rule{0.2em}{0ex}}dt\right)-\varphi \left({\int }_{0}^{d\left(x,y\right)}\phi \left(t\right)\phantom{\rule{0.2em}{0ex}}dt\right).\end{array}$

That is, (1.1) holds. Thus Theorem 3.1 guarantees that f has a unique fixed point $0\in X$ such that ${lim}_{n\to \mathrm{\infty }}{f}^{n}x=0$ for each $x\in X$.

Example 4.2 Let $X=\left[0,1\right]\cup \left[4,5\right]$ be endowed with the Euclidean metric $d=|\cdot |$. Assume that $f:X\to X$ and $\phi ,\psi :{\mathbb{R}}^{+}\to {\mathbb{R}}^{+}$ and $\alpha :{\mathbb{R}}^{+}\to \left[0,1\right)$ are defined by
$\begin{array}{c}f\left(x\right)=\left\{\begin{array}{cc}\frac{{x}^{2}}{4},\hfill & \mathrm{\forall }x\in \left[0,1\right],\hfill \\ \frac{{x}^{2}}{26},\hfill & \mathrm{\forall }x\in \left[4,5\right],\hfill \end{array}\phantom{\rule{2em}{0ex}}\phi \left(t\right)=\left\{\begin{array}{cc}4{t}^{3},\hfill & \mathrm{\forall }t\in \left[0,1\right],\hfill \\ 2t,\hfill & \mathrm{\forall }t\in \left[4,5\right],\hfill \end{array}\hfill \\ \psi \left(t\right)={t}^{\frac{1}{2}},\phantom{\rule{1em}{0ex}}\mathrm{\forall }t\in {\mathbb{R}}^{+},\phantom{\rule{2em}{0ex}}\alpha \left(t\right)=\left\{\begin{array}{cc}\frac{1}{3}+\frac{{t}^{2}}{2},\hfill & \mathrm{\forall }t\in \left[0,1\right],\hfill \\ \frac{1}{2t},\hfill & \mathrm{\forall }t\in \left(1,3\right),\hfill \\ \frac{1}{\sqrt{t}},\hfill & \mathrm{\forall }t\in \left(3,+\mathrm{\infty }\right).\hfill \end{array}\hfill \end{array}$

Obviously, $\left(\phi ,\psi ,\alpha \right)\in {\mathrm{\Phi }}_{1}×{\mathrm{\Phi }}_{3}×{\mathrm{\Phi }}_{5}$. Put $x,y\in X$ with $x. In order to verify (1.2), we have to consider three possible cases as follows.

Case 1. Let $x,y\in \left[0,1\right]$. It is clear that
$\begin{array}{rl}\psi \left({\int }_{0}^{d\left(fx,fy\right)}\phi \left(t\right)\phantom{\rule{0.2em}{0ex}}dt\right)& ={\left({\int }_{0}^{\frac{{y}^{2}-{x}^{2}}{4}}4{t}^{3}\phantom{\rule{0.2em}{0ex}}dt\right)}^{\frac{1}{2}}=\frac{{\left(x+y\right)}^{2}}{16}{|x-y|}^{2}\le \frac{1}{4}{|x-y|}^{2}\\ \le \left(\frac{1}{3}+\frac{1}{2}{|x-y|}^{2}\right){|x-y|}^{2}\\ =\alpha \left(d\left(x,y\right)\right)\psi \left({\int }_{0}^{d\left(x,y\right)}\phi \left(t\right)\phantom{\rule{0.2em}{0ex}}dt\right).\end{array}$
Case 2. Let $x,y\in \left[4,5\right]$. It follows that
$\begin{array}{rl}\psi \left({\int }_{0}^{d\left(fx,fy\right)}\phi \left(t\right)\phantom{\rule{0.2em}{0ex}}dt\right)& ={\left({\int }_{0}^{\frac{{y}^{2}-{x}^{2}}{26}}4{t}^{3}\phantom{\rule{0.2em}{0ex}}dt\right)}^{\frac{1}{2}}={\left(\frac{x+y}{26}\right)}^{2}{|x-y|}^{2}\le \frac{25}{169}{|x-y|}^{2}\\ \le \left(\frac{1}{3}+\frac{1}{2}{|x-y|}^{2}\right){|x-y|}^{2}\\ =\alpha \left(d\left(x,y\right)\right)\psi \left({\int }_{0}^{d\left(x,y\right)}\phi \left(t\right)\phantom{\rule{0.2em}{0ex}}dt\right).\end{array}$
Case 3. Let $x\in \left[0,1\right]$ and $y\in \left[4,5\right]$. It follows that
$\begin{array}{rl}\psi \left({\int }_{0}^{d\left(fx,fy\right)}\phi \left(t\right)\phantom{\rule{0.2em}{0ex}}dt\right)& ={\left({\int }_{0}^{\frac{{y}^{2}}{26}-\frac{{x}^{2}}{4}}4{t}^{3}\phantom{\rule{0.2em}{0ex}}dt\right)}^{\frac{1}{2}}={\left(\frac{{y}^{2}}{26}-\frac{{x}^{2}}{4}\right)}^{2}\le {\left(\frac{25}{26}\right)}^{2}<1<\sqrt{|x-y|}\\ =\alpha \left(|x-y|\right)|x-y|=\alpha \left(|x-y|\right){\left({\int }_{0}^{1}4{t}^{3}\phantom{\rule{0.2em}{0ex}}dt+{\int }_{1}^{|x-y|}2t\phantom{\rule{0.2em}{0ex}}dt\right)}^{\frac{1}{2}}\\ =\alpha \left(d\left(x,y\right)\right){\left({\int }_{0}^{|x-y|}\phi \left(t\right)\phantom{\rule{0.2em}{0ex}}dt\right)}^{\frac{1}{2}}\\ =\alpha \left(d\left(x,y\right)\right)\psi \left({\int }_{0}^{d\left(x,y\right)}\phi \left(t\right)\phantom{\rule{0.2em}{0ex}}dt\right).\end{array}$

That is, (1.2) holds. Consequently, the conditions of Theorem 3.2 are satisfied. It follows from Theorem 3.2 that f has a unique fixed point $0\in X$ such that ${lim}_{n\to \mathrm{\infty }}{f}^{n}x=0$ for each $x\in X$.

Example 4.3 Let $X=\left[\frac{1}{2},1\right]\cup \left[\frac{3}{2},2\right]$ be endowed with the Euclidean metric $d=|\cdot |$. Assume that $f:X\to X$, $\phi ,\varphi ,\psi :{\mathbb{R}}^{+}\to {\mathbb{R}}^{+}$ and $\alpha ,\beta :{\mathbb{R}}^{+}\to \left[0,1\right)$ are defined by
$\begin{array}{c}f\left(x\right)=\left\{\begin{array}{cc}1,\hfill & \mathrm{\forall }x\in \left[\frac{1}{2},1\right],\hfill \\ \frac{x}{2},\hfill & \mathrm{\forall }x\in \left[\frac{3}{2},2\right],\hfill \end{array}\phantom{\rule{2em}{0ex}}\varphi \left(t\right)=\left\{\begin{array}{cc}0,\hfill & t\in \left[0,\frac{9}{16}\right)\hfill \\ \frac{32{t}^{2}}{9},\hfill & t\in \left[\frac{9}{16},+\mathrm{\infty }\right),\hfill \end{array}\hfill \\ \phi \left(t\right)=2t,\phantom{\rule{2em}{0ex}}\psi \left(t\right)=4{t}^{2},\phantom{\rule{2em}{0ex}}\alpha \left(t\right)=\frac{t}{{\left(\frac{1}{2}+t\right)}^{2}},\phantom{\rule{2em}{0ex}}\beta \left(t\right)=\frac{{t}^{2}}{{\left(\frac{1}{2}+t\right)}^{2}},\phantom{\rule{1em}{0ex}}\mathrm{\forall }t\in {\mathbb{R}}^{+}.\hfill \end{array}$

It is easy to see that $\left(\phi ,\psi ,\varphi \right)\in {\mathrm{\Phi }}_{1}×{\mathrm{\Phi }}_{3}×{\mathrm{\Phi }}_{4}$, $\left(\alpha ,\beta \right)\in {\mathrm{\Phi }}_{6}$ and (3.15) holds. In order to verify (1.3), we have to consider the five possible cases below.

Case 1. Let $x,y\in \left[\frac{3}{2},2\right]$ with $x\ge y$. Note that
$\begin{array}{rl}\psi \left({\int }_{0}^{d\left(fx,fy\right)}\phi \left(t\right)\phantom{\rule{0.2em}{0ex}}dt\right)& =\psi \left({\int }_{0}^{\frac{|x-y|}{2}}\phi \left(t\right)\phantom{\rule{0.2em}{0ex}}dt\right)=\psi \left(\frac{{\left(x-y\right)}^{2}}{4}\right)=\frac{{\left(x-y\right)}^{4}}{4}\le \frac{x-y}{2}\\ \le \frac{x-y}{2}\cdot \frac{{x}^{4}}{{\left(\frac{1}{2}+x-y\right)}^{2}}\le \frac{x-y}{{\left(\frac{1}{2}+x-y\right)}^{2}}\cdot \frac{32}{9}{\left(\frac{{x}^{2}}{4}\right)}^{2}\\ =\alpha \left(d\left(x,y\right)\right)\varphi \left({\int }_{0}^{d\left(x,fx\right)}\phi \left(t\right)\phantom{\rule{0.2em}{0ex}}dt\right)\\ \le \alpha \left(d\left(x,y\right)\right)\varphi \left({\int }_{0}^{d\left(x,fx\right)}\phi \left(t\right)\phantom{\rule{0.2em}{0ex}}dt\right)d+\beta \left(d\left(x,y\right)\right)\psi \left({\int }_{0}^{d\left(y,fy\right)}\phi \left(t\right)\phantom{\rule{0.2em}{0ex}}dt\right).\end{array}$
Case 2. Let $x,y\in \left[\frac{3}{2},2\right]$ with $y>x$. Note that
$\begin{array}{rl}\psi \left({\int }_{0}^{d\left(fx,fy\right)}\phi \left(t\right)\phantom{\rule{0.2em}{0ex}}dt\right)& =\psi \left({\int }_{0}^{\frac{|y-x|}{2}}\phi \left(t\right)\phantom{\rule{0.2em}{0ex}}dt\right)=\frac{{\left(y-x\right)}^{4}}{4}\le \frac{{\left(y-x\right)}^{2}}{4}\cdot \frac{{y}^{4}}{{\left(\frac{1}{2}+y-x\right)}^{2}}\\ =\frac{{\left(y-x\right)}^{2}}{{\left(\frac{1}{2}+y-x\right)}^{2}}\cdot \frac{{y}^{4}}{4}=\beta \left(d\left(x,y\right)\right)\psi \left(\frac{{y}^{2}}{4}\right)\\ =\beta \left(d\left(x,y\right)\right)\psi \left({\int }_{0}^{\frac{y}{2}}\phi \left(t\right)\phantom{\rule{0.2em}{0ex}}dt\right)=\beta \left(d\left(x,y\right)\right)\psi \left({\int }_{0}^{d\left(y,fy\right)}\phi \left(t\right)\phantom{\rule{0.2em}{0ex}}dt\right)\\ \le \alpha \left(d\left(x,y\right)\right)\varphi \left({\int }_{0}^{d\left(x,fx\right)}\phi \left(t\right)\phantom{\rule{0.2em}{0ex}}dt\right)+\beta \left(d\left(x,y\right)\right)\psi \left({\int }_{0}^{d\left(y,fy\right)}\phi \left(t\right)\phantom{\rule{0.2em}{0ex}}dt\right).\end{array}$
Case 3. Let $x\in \left[\frac{3}{2},2\right]$ and $y\in \left[\frac{1}{2},1\right]$. It follows that
$\begin{array}{rl}\psi \left({\int }_{0}^{d\left(fx,fy\right)}\phi \left(t\right)\phantom{\rule{0.2em}{0ex}}dt\right)& =\psi \left({\int }_{0}^{\frac{x-2}{2}}\phi \left(t\right)\phantom{\rule{0.2em}{0ex}}dt\right)=\psi \left(\frac{{\left(x-2\right)}^{2}}{4}\right)=\frac{{\left(x-2\right)}^{4}}{4}\le \frac{1}{64}<\frac{27}{64}\\ =\frac{3}{8}\cdot \frac{2}{9}\cdot \frac{81}{16}\le \frac{x-y}{{\left(\frac{1}{2}+x-y\right)}^{2}}\cdot \frac{2}{9}\cdot {x}^{4}=\alpha \left(d\left(x,y\right)\right)\varphi \left(\frac{{x}^{2}}{4}\right)\\ =\alpha \left(d\left(x,y\right)\right)\varphi \left({\int }_{0}^{d\left(x,fx\right)}\phi \left(t\right)\phantom{\rule{0.2em}{0ex}}dt\right)\\ \le \alpha \left(d\left(x,y\right)\right)\varphi \left({\int }_{0}^{d\left(x,fx\right)}\phi \left(t\right)\phantom{\rule{0.2em}{0ex}}dt\right)+\beta \left(d\left(x,y\right)\right)\psi \left({\int }_{0}^{d\left(y,fy\right)}\phi \left(t\right)\phantom{\rule{0.2em}{0ex}}dt\right).\end{array}$
Case 4. Let $x\in \left[\frac{1}{2},1\right]$ and $y\in \left[\frac{3}{2},2\right]$. Note that
$\begin{array}{rl}\psi \left({\int }_{0}^{d\left(fx,fy\right)}\phi \left(t\right)\phantom{\rule{0.2em}{0ex}}dt\right)& =\psi \left({\int }_{0}^{\frac{|y-2|}{2}}2t\phantom{\rule{0.2em}{0ex}}dt\right)=\frac{{\left(y-2\right)}^{4}}{4}\le \frac{1}{64}<\frac{1}{4}\cdot \frac{81}{64}\\ \le \frac{{\left(y-x\right)}^{2}}{{\left(\frac{1}{2}+y-x\right)}^{2}}\cdot \frac{{y}^{4}}{4}=\beta \left(d\left(x,y\right)\right)\psi \left(\frac{{y}^{2}}{4}\right)\\ =\beta \left(d\left(x,y\right)\right)\psi \left({\int }_{0}^{d\left(y,fy\right)}\phi \left(t\right)\phantom{\rule{0.2em}{0ex}}dt\right)\\ \le \alpha \left(d\left(x,y\right)\right)\varphi \left({\int }_{0}^{d\left(x,fx\right)}\phi \left(t\right)\phantom{\rule{0.2em}{0ex}}dt\right)+\beta \left(d\left(x,y\right)\right)\psi \left({\int }_{0}^{d\left(y,fy\right)}\phi \left(t\right)\phantom{\rule{0.2em}{0ex}}dt\right).\end{array}$
Case 5. Let $x,y\in \left[\frac{1}{2},1\right]$. Notice that $fx=fy=1$. It follows that
$\begin{array}{r}\psi \left({\int }_{0}^{d\left(fx,fy\right)}\phi \left(t\right)\phantom{\rule{0.2em}{0ex}}dt\right)\\ \phantom{\rule{1em}{0ex}}=0\le \alpha \left(d\left(x,y\right)\right)\varphi \left({\int }_{0}^{d\left(x,fx\right)}\phi \left(t\right)\phantom{\rule{0.2em}{0ex}}dt\right)+\beta \left(d\left(x,y\right)\right)\psi \left({\int }_{0}^{d\left(y,fy\right)}\phi \left(t\right)\phantom{\rule{0.2em}{0ex}}dt\right).\end{array}$

That is, (1.3) holds. Thus all the conditions of Theorem 3.3 are satisfied. It follows from Theorem 3.3 that f has a unique fixed point $1\in X$ such that ${lim}_{n\to \mathrm{\infty }}{f}^{n}x=1$ for each $x\in X$.

## Declarations

### Acknowledgements

This research was supported by the Science Research Foundation of Educational Department of Liaoning Province (L2012380).

## Authors’ Affiliations

(1)
Department of Mathematics, Liaoning Normal University, Dalian, Liaoning, 116029, People’s Republic of China
(2)
Department of Mathematics and RINS, Gyeongsang National University, Jinju, 660-701, Korea

## References

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