Fixed point theorems of contractive mappings of integral type

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.

1 Introduction

Branciari [1] 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 $(X,d)$ into itself satisfying

$${\int }_0^{d(fx,fy)}\phi (t)dt\le c{\int }_0^{d(x,y)}\phi (t)dt,\mathrm{\forall }x,y\in X,$$

where $c\in (0,1)$ 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., [1–8] and the references therein. In particular, in 2011, Liu et al. [5] extended the result of Branciari [1] and deduced the following fixed point theorems.

Theorem 1.2 Let f be a mapping from a complete metric space $(X,d)$ into itself satisfying

$${\int }_0^{d(fx,fy)}\phi (t)dt\le \alpha (d(x,y)){\int }_0^{d(x,y)}\phi (t)dt,\mathrm{\forall }x,y\in X,$$

where $\phi \in {\mathrm{\Phi }}_1$ and $\alpha :{\mathbb{R}}^{+}\to [0,1)$ is a function with

$$\underset{s\to t}{lim\hspace{0.17em}sup}\alpha (s)<1,\mathrm{\forall }t>0.$$

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 $(X,d)$ into itself satisfying

$${\int }_0^{d(fx,fy)}\phi (t)dt\le \alpha (d(x,y)){\int }_0^{d(x,fx)}\phi (t)dt+\beta (d(x,y)){\int }_0^{d(y,fy)}\phi (t)dt,\mathrm{\forall }x,y\in X,$$

where $\phi \in {\mathrm{\Phi }}_1$ and $\alpha ,\beta :{\mathbb{R}}^{+}\to [0,1)$ are two functions with

$$\alpha (t)+\beta (t)<1,\mathrm{\forall }t\in {\mathbb{R}}^{+},\underset{s\to 0^{+}}{lim\hspace{0.17em}sup}\beta (s)<1,\underset{s\to t^{+}}{lim\hspace{0.17em}sup}\frac{\alpha (s)}{1-\beta (s)}<1,\mathrm{\forall }t>0.$$

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 [9] proved the following result.

Theorem 1.4 Let f be a mapping from a complete metric space $(X,d)$ into itself satisfying

$$\psi (d(fx,fy))\le \psi (d(x,y))-\phi (d(x,y)),\mathrm{\forall }x,y\in X,$$

where $\psi ,\phi :{\mathbb{R}}^{+}\to {\mathbb{R}}^{+}$ are both continuous and monotone nondecreasing functions with $\psi (t)=\phi (t)=0$ 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:

$$\psi ({\int }_0^{d(fx,fy)}\phi (t)dt)\le \psi ({\int }_0^{d(x,y)}\phi (t)dt)-\varphi ({\int }_0^{d(x,y)}\phi (t)dt),\mathrm{\forall }x,y\in X,$$

(1.1)

where $(\phi ,\varphi ,\psi )\in {\mathrm{\Phi }}_1\times {\mathrm{\Phi }}_2\times {\mathrm{\Phi }}_3$;

$$\psi ({\int }_0^{d(fx,fy)}\phi (t)dt)\le \alpha (d(x,y))\psi ({\int }_0^{d(x,y)}\phi (t)dt),\mathrm{\forall }x,y\in X,$$

(1.2)

where $(\phi ,\psi ,\alpha )\in {\mathrm{\Phi }}_1\times {\mathrm{\Phi }}_3\times {\mathrm{\Phi }}_5$;

$$\begin{array}{rl}\psi ({\int }_0^{d(fx,fy)}\phi (t)dt)\le & \alpha (d(x,y))\varphi ({\int }_0^{d(x,fx)}\phi (t)dt)\\ +\beta (d(x,y))\psi ({\int }_0^{d(y,fy)}\phi (t)dt),\mathrm{\forall }x,y\in X,\end{array}$$

(1.3)

where $(\phi ,\psi ,\varphi )\in {\mathrm{\Phi }}_1\times {\mathrm{\Phi }}_3\times {\mathrm{\Phi }}_4$ and $(\alpha ,\beta )\in {\mathrm{\Phi }}_6$.

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 ${\mathbb{R}}^{+}=[0,+\mathrm{\infty })$, ${\mathbb{N}}_0=\mathbb{N}\cup \{0\}$, ℕ denotes the set of all positive integers, $(X,d)$ is a metric space, $f:X\to X$ is a self-mapping and

$$d_n=d(f^{n}x,f^{n+1}x),\mathrm{\forall }(n,x)\in {\mathbb{N}}_0\times X,$$

${\mathrm{\Phi }}_1$ = {$\phi :\phi :{\mathbb{R}}^{+}\to {\mathbb{R}}^{+}$ is Lebesgue integrable, summable on each compact subset of ${\mathbb{R}}^{+}$ and ${\int }_0^{\epsilon }\phi (t)dt>0$ for each $\epsilon >0$};

${\mathrm{\Phi }}_2$ = {$\phi :\phi :{\mathbb{R}}^{+}\to {\mathbb{R}}^{+}$ satisfies that ${lim\hspace{0.17em}inf}_{n\to \mathrm{\infty }}\phi (a_n)>0\iff {lim\hspace{0.17em}inf}_{n\to \mathrm{\infty }}{a}_n>0$ for each ${\{a_n\}}_{n\in \mathbb{N}}\subset {\mathbb{R}}^{+}$};

${\mathrm{\Phi }}_3$ = {$\phi :\phi :{\mathbb{R}}^{+}\to {\mathbb{R}}^{+}$ is nondecreasing continuous and $\phi (t)=0\iff t=0$};

${\mathrm{\Phi }}_4$ = {$\phi :\phi :{\mathbb{R}}^{+}\to {\mathbb{R}}^{+}$ satisfies that $\phi (0)=0$};

${\mathrm{\Phi }}_5$ = {$\phi :\phi :{\mathbb{R}}^{+}\to [0,1)$ satisfies that ${lim\hspace{0.17em}sup}_{s\to t}\phi (s)<1$ for each $t>0$};

${\mathrm{\Phi }}_6$ = {$(\alpha ,\beta ):\alpha ,\beta :{\mathbb{R}}^{+}\to [0,1)$ satisfy that ${lim\hspace{0.17em}sup}_{s\to 0^{+}}\beta (s)<1$, ${lim\hspace{0.17em}sup}_{s\to t^{+}}\frac{\alpha (s)}{1-\beta (s)}<1$ and $\alpha (t)+\beta (t)<1$ for each $t>0$}.

The following lemmas play important roles in this paper.

Lemma 2.1 ([5])

Let $\phi \in {\mathrm{\Phi }}_1$ and ${\{r_n\}}_{n\in \mathbb{N}}$ be a nonnegative sequence with ${lim}_{n\to \mathrm{\infty }}{r}_n=a$. Then

$$\underset{n\to \mathrm{\infty }}{lim}{\int }_0^{r_n}\phi (t)dt={\int }_0^a\phi (t)dt.$$

Lemma 2.2 ([5])

Let $\phi \in {\mathrm{\Phi }}_1$ and ${\{r_n\}}_{n\in \mathbb{N}}$ be a nonnegative sequence. Then

$$\underset{n\to \mathrm{\infty }}{lim}{\int }_0^{r_n}\phi (t)dt=0$$

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

Lemma 2.3 Let $\phi \in {\mathrm{\Phi }}_2$. Then $\phi (t)>0$ 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 $t={lim\hspace{0.17em}inf}_{n\to \mathrm{\infty }}{a}_n>0$, which together with $\phi \in {\mathrm{\Phi }}_2$ ensures that

$$\phi (t)=\underset{n\to \mathrm{\infty }}{lim\hspace{0.17em}inf}\phi (a_n)>0.$$

Conversely, suppose that $\phi (t)>0$ for some $t\in {\mathbb{R}}^{+}$. Set $a_n=t$ for each $n\in \mathbb{N}$. It is clear that $\phi (t)={lim\hspace{0.17em}inf}_{n\to \mathrm{\infty }}\phi (a_n)>0$, which together with $\phi \in {\mathrm{\Phi }}_2$ guarantees that

$$t=\underset{n\to \mathrm{\infty }}{lim\hspace{0.17em}inf}{a}_n>0.$$

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 $(X,d)$ 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

$$d_n\le d_{n-1},\mathrm{\forall }n\in \mathbb{N}.$$

(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

$${\int }_0^{d_{n_0}}\phi (t)dt>0.$$

(3.3)

Using (1.1), (3.2) and $(\phi ,\varphi ,\psi )\in {\mathrm{\Phi }}_1\times {\mathrm{\Phi }}_2\times {\mathrm{\Phi }}_3$, we conclude immediately that

$$\begin{array}{rl}\psi ({\int }_0^{d_{n_0-1}}\phi (t)dt)& \le \psi ({\int }_0^{d_{n_0}}\phi (t)dt)\\ =\psi ({\int }_0^{d(f^{n_0}x,f^{n_0+1}x)}\phi (t)dt)\\ \le \psi ({\int }_0^{d(f^{n_0-1}x,f^{n_0}x)}\phi (t)dt)-\varphi ({\int }_0^{d(f^{n_0-1}x,f^{n_0}x)}\phi (t)dt)\\ =\psi ({\int }_0^{d_{n_0-1}}\phi (t)dt)-\varphi ({\int }_0^{d_{n_0-1}}\phi (t)dt)\\ \le \psi ({\int }_0^{d_{n_0-1}}\phi (t)dt),\end{array}$$

which yields that

$$\psi ({\int }_0^{d_{n_0}}\phi (t)dt)=\psi ({\int }_0^{d_{n_0-1}}\phi (t)dt)$$

(3.4)

and

$$\varphi ({\int }_0^{d_{n_0-1}}\phi (t)dt)=0.$$

(3.5)

Combining (3.5) and Lemma 2.3, we get that

$${\int }_0^{d_{n_0-1}}\phi (t)dt=0,$$

which together with $\psi \in {\mathrm{\Phi }}_3$ and (3.4) means that

$$\psi ({\int }_0^{d_{n_0}}\phi (t)dt)=\psi ({\int }_0^{d_{n_0-1}}\phi (t)dt)=\psi (0)=0,$$

that is,

$${\int }_0^{d_{n_0}}\phi (t)dt=0,$$

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 ${\{d_n\}}_{n\in {\mathbb{N}}_0}$ 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

$$\begin{array}{rl}\psi ({\int }_0^{d_n}\phi (t)dt)& =\psi ({\int }_0^{d(f^{n}x,f^{n+1}x)}\phi (t)dt)\\ \le \psi ({\int }_0^{d(f^{n}x,f^{n-1}x)}\phi (t)dt)-\varphi ({\int }_0^{d(f^{n}x,f^{n-1}x)}\phi (t)dt)\\ =\psi ({\int }_0^{d_{n-1}}\phi (t)dt)-\varphi ({\int }_0^{d_{n-1}}\phi (t)dt),\mathrm{\forall }n\in \mathbb{N}.\end{array}$$

(3.7)

Taking upper limit in (3.7) and using Lemma 2.1 and $(\phi ,\varphi ,\psi )\in {\mathrm{\Phi }}_1\times {\mathrm{\Phi }}_2\times {\mathrm{\Phi }}_3$, we conclude that

$$\begin{array}{rl}\psi ({\int }_0^c\phi (t)dt)& =\underset{n\to \mathrm{\infty }}{lim\hspace{0.17em}sup}\psi ({\int }_0^{d_n}\phi (t)dt)\\ \le \underset{n\to \mathrm{\infty }}{lim\hspace{0.17em}sup}[\psi ({\int }_0^{d_{n-1}}\phi (t)dt)-\varphi ({\int }_0^{d_{n-1}}\phi (t)dt)]\\ \le \underset{n\to \mathrm{\infty }}{lim\hspace{0.17em}sup}\psi ({\int }_0^{d_{n-1}}\phi (t)dt)-\underset{n\to \mathrm{\infty }}{lim\hspace{0.17em}inf}\varphi ({\int }_0^{d_{n-1}}\phi (t)dt)\\ =\psi ({\int }_0^c\phi (t)dt)-\underset{n\to \mathrm{\infty }}{lim\hspace{0.17em}inf}\varphi ({\int }_0^{d_{n-1}}\phi (t)dt)\\ <\psi ({\int }_0^c\phi (t)dt),\end{array}$$

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

Thirdly, we show that ${\{f^{n}x\}}_{n\in \mathbb{N}}$ is a Cauchy sequence. Suppose that ${\{f^{n}x\}}_{n\in \mathbb{N}}$ 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 $m(k)$ and $n(k)$ with $m(k)>n(k)>k$ satisfying

$$d(f^{m(k)}x,f^{n(k)}x)>\epsilon .$$

(3.8)

For each positive integer k, let $m(k)$ denote the least integer exceeding $n(k)$ and satisfying (3.8). It follows that

$$d(f^{m(k)}x,f^{n(k)}x)>\epsilon \text{and}d(f^{m(k)-1}x,f^{n(k)}x)\le \epsilon ,\mathrm{\forall }k\in \mathbb{N}.$$

(3.9)

Note that

$$\begin{array}{r}d(f^{m(k)}x,f^{n(k)}x)\le d(f^{n(k)}x,f^{m(k)-1}x)+d_{m(k)-1},\mathrm{\forall }k\in \mathbb{N};\\ |d(f^{m(k)}x,f^{n(k)+1}x)-d(f^{m(k)}x,f^{n(k)}x)|\le d_{n(k)},\mathrm{\forall }k\in \mathbb{N};\\ |d(f^{m(k)+1}x,f^{n(k)+1}x)-d(f^{m(k)}x,f^{n(k)+1}x)|\le d_{m(k)},\mathrm{\forall }k\in \mathbb{N};\\ |d(f^{m(k)+1}x,f^{n(k)+1}x)-d(f^{m(k)+1}x,f^{n(k)+2}x)|\le d_{n(k)+1},\mathrm{\forall }k\in \mathbb{N}.\end{array}$$

(3.10)

In light of (3.9) and (3.10), we get that

$$\begin{array}{rl}\epsilon & =\underset{k\to \mathrm{\infty }}{lim}d(f^{n(k)}x,f^{m(k)}x)=\underset{k\to \mathrm{\infty }}{lim}d(f^{m(k)}x,f^{n(k)+1}x)\\ =\underset{k\to \mathrm{\infty }}{lim}d(f^{m(k)+1}x,f^{n(k)+1}x)=\underset{k\to \mathrm{\infty }}{lim}d(f^{m(k)+1}x,f^{n(k)+2}x).\end{array}$$

(3.11)

In view of (1.1), we deduce that

$$\begin{array}{r}\psi ({\int }_0^{d(f^{m(k)+1}x,f^{n(k)+2}x)}\phi (t)dt)\\ \le \psi ({\int }_0^{d(f^{m(k)}x,f^{n(k)+1}x)}\phi (t)dt)-\varphi ({\int }_0^{d(f^{m(k)}x,f^{n(k)+1}x)}\phi (t)dt),\mathrm{\forall }k\in \mathbb{N}.\end{array}$$

(3.12)

Taking upper limit in (3.12) and using (3.11), $(\phi ,\varphi ,\psi )\in {\mathrm{\Phi }}_1\times {\mathrm{\Phi }}_2\times {\mathrm{\Phi }}_3$ and Lemma 2.1, we deduce that

$$\begin{array}{r}\psi ({\int }_0^{\epsilon }\phi (t)dt)\\ =\underset{k\to \mathrm{\infty }}{lim\hspace{0.17em}sup}\psi ({\int }_0^{d(f^{m(k)+1}x,f^{n(k)+2}x)}\phi (t)dt)\\ \le \underset{k\to \mathrm{\infty }}{lim\hspace{0.17em}sup}[\psi ({\int }_0^{d(f^{m(k)}x,f^{n(k)+1}x)}\phi (t)dt)-\varphi ({\int }_0^{d(f^{m(k)}x,f^{n(k)+1}x)}\phi (t)dt)]\\ \le \underset{k\to \mathrm{\infty }}{lim\hspace{0.17em}sup}\psi ({\int }_0^{d(f^{m(k)}x,f^{n(k)+1}x)}\phi (t)dt)-\underset{k\to \mathrm{\infty }}{lim\hspace{0.17em}inf}\varphi ({\int }_0^{d(f^{m(k)}x,f^{n(k)+1}x)}\phi (t)dt)\\ =\psi ({\int }_0^{\epsilon }\phi (t)dt)-\underset{k\to \mathrm{\infty }}{lim\hspace{0.17em}inf}\varphi ({\int }_0^{d(f^{m(k)}x,f^{n(k)+1}x)}\phi (t)dt)\\ <\psi ({\int }_0^{\epsilon }\phi (t)dt),\end{array}$$

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

Since $(X,d)$ 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

$$\psi ({\int }_0^{d(f^{n+1}x,fa)}\phi (t)dt)\le \psi ({\int }_0^{d(f^{n}x,a)}\phi (t)dt)-\varphi ({\int }_0^{d(f^{n}x,a)}\phi (t)dt),\mathrm{\forall }n\in \mathbb{N},$$

which together with $(\phi ,\varphi ,\psi )\in {\mathrm{\Phi }}_1\times {\mathrm{\Phi }}_2\times {\mathrm{\Phi }}_3$ and Lemmas 2.1 and 2.2 gives that

$$\begin{array}{rl}\psi ({\int }_0^{d(a,fa)}\phi (t)dt)& =\underset{n\to \mathrm{\infty }}{lim\hspace{0.17em}sup}\psi ({\int }_0^{d(f^{n+1}x,fa)}\phi (t)dt)\\ \le \underset{n\to \mathrm{\infty }}{lim\hspace{0.17em}sup}[\psi ({\int }_0^{d(f^{n}x,a)}\phi (t)dt)-\varphi ({\int }_0^{d(f^{n}x,a)}\phi (t)dt)]\\ \le \underset{n\to \mathrm{\infty }}{lim\hspace{0.17em}sup}\psi ({\int }_0^{d(f^{n}x,a)}\phi (t)dt)-\underset{n\to \mathrm{\infty }}{lim\hspace{0.17em}inf}\varphi ({\int }_0^{d(f^{n}x,a)}\phi (t)dt)\\ =\psi (0)-0\\ =0,\end{array}$$

which together with $\psi \in {\mathrm{\Phi }}_3$ yields that

$${\int }_0^{d(a,fa)}\phi (t)dt=0,$$

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 $b\in X\setminus \{a\}$. It follows from (1.1) and $(\phi ,\varphi ,\psi )\in {\mathrm{\Phi }}_1\times {\mathrm{\Phi }}_2\times {\mathrm{\Phi }}_3$ that

$$\begin{array}{rl}\psi ({\int }_0^{d(a,b)}\phi (t)dt)& =\psi ({\int }_0^{d(fa,fb)}\phi (t)dt)\\ \le \psi ({\int }_0^{d(a,b)}\phi (t)dt)-\varphi ({\int }_0^{d(a,b)}\phi (t)dt)\\ <\psi ({\int }_0^{d(a,b)}\phi (t)dt),\end{array}$$

which is a contradiction. This completes the proof. □

Theorem 3.2 Let f be a mapping from a complete metric space $(X,d)$ 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 $(\phi ,\psi ,\alpha )\in {\mathrm{\Phi }}_1\times {\mathrm{\Phi }}_3\times {\mathrm{\Phi }}_5$, we get that

$$\psi ({\int }_0^{d_{n_0}}\phi (t)dt)>0$$

and

$$\begin{array}{rl}\psi ({\int }_0^{d_{n_0-1}}\phi (t)dt)& \le \psi ({\int }_0^{d_{n_0}}\phi (t)dt)=\psi ({\int }_0^{d(f^{n_0}x,f^{n_0+1}x)}\phi (t)dt)\\ \le \alpha \left(d(f^{n_0-1}x,f^{n_0}x)\right)\psi ({\int }_0^{d(f^{n_0-1}x,f^{n_0}x)}\phi (t)dt)\\ =\alpha (d_{n_0-1})\psi ({\int }_0^{d_{n_0-1}}\phi (t)dt)<\psi ({\int }_0^{d_{n_0-1}}\phi (t)dt),\end{array}$$

which is a contradiction, and hence (3.2) does not hold. Consequently, (3.1) is true. Notice that the nonnegative sequence ${\{d_n\}}_{n\in {\mathbb{N}}_0}$ 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

$$\begin{array}{rl}\psi ({\int }_0^{d_n}\phi (t)dt)& =\psi ({\int }_0^{d(f^{n}x,f^{n+1}x)}\phi (t)dt)\\ \le \alpha \left(d(f^{n-1}x,f^{n}x)\right)\psi ({\int }_0^{d(f^{n-1}x,f^{n}x)}\phi (t)dt)\\ =\alpha (d_{n-1})\psi ({\int }_0^{d_{n-1}}\phi (t)dt),\mathrm{\forall }n\in \mathbb{N}.\end{array}$$

(3.13)

Taking upper limit in (3.13) and using Lemma 2.1 and $(\phi ,\psi ,\alpha )\in {\mathrm{\Phi }}_1\times {\mathrm{\Phi }}_3\times {\mathrm{\Phi }}_5$, we know that

$$\begin{array}{rl}\psi ({\int }_0^c\phi (t)dt)& =\underset{n\to \mathrm{\infty }}{lim\hspace{0.17em}sup}\psi ({\int }_0^{d_n}\phi (t)dt)\\ \le \underset{n\to \mathrm{\infty }}{lim\hspace{0.17em}sup}[\alpha (d_{n-1})\psi ({\int }_0^{d_{n-1}}\phi (t)dt)]\\ \le \underset{n\to \mathrm{\infty }}{lim\hspace{0.17em}sup}\alpha (d_{n-1})\cdot \underset{n\to \mathrm{\infty }}{lim\hspace{0.17em}sup}\psi ({\int }_0^{d_{n-1}}\phi (t)dt)\\ <\psi ({\int }_0^c\phi (t)dt),\end{array}$$

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

Now we show that ${\{f^{n}x\}}_{n\in \mathbb{N}}$ is a Cauchy sequence. Suppose that ${\{f^{n}x\}}_{n\in \mathbb{N}}$ is not a Cauchy sequence. As in the proof of Theorem 3.1, we conclude that there exist $\epsilon >0$ and $\{m(k),n(k):k\in \mathbb{N}\}\subseteq \mathbb{N}$ with $m(k)>n(k)>k$ for each $k\in \mathbb{N}$ satisfying (3.8)-(3.11). By means of (1.2), (3.11), Lemma 2.1 and $(\phi ,\psi ,\alpha )\in {\mathrm{\Phi }}_1\times {\mathrm{\Phi }}_3\times {\mathrm{\Phi }}_5$, we get that

$$\begin{array}{rl}\psi ({\int }_0^{\epsilon }\phi (t)dt)& =\underset{k\to \mathrm{\infty }}{lim\hspace{0.17em}sup}\psi ({\int }_0^{d(f^{m(k)+1}x,f^{n(k)+2}x)}\phi (t)dt)\\ =\underset{k\to \mathrm{\infty }}{lim\hspace{0.17em}sup}\left[\alpha \left(d(f^{m(k)}x,f^{n(k)+1}x)\right)\psi ({\int }_0^{d(f^{m(k)}x,f^{n(k)+1}x)}\phi (t)dt)\right]\\ \le \underset{k\to \mathrm{\infty }}{lim\hspace{0.17em}sup}\alpha \left(d(f^{m(k)}x,f^{n(k)+1}x)\right)\cdot \underset{k\to \mathrm{\infty }}{lim\hspace{0.17em}sup}\psi ({\int }_0^{d(f^{m(k)}x,f^{n(k)+1}x)}\phi (t)dt)\\ <\psi ({\int }_0^{\epsilon }\phi (t)dt),\end{array}$$

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

It follows from completeness of $(X,d)$ that there exists $a\in X$ with ${lim}_{n\to \mathrm{\infty }}{f}^{n}x=a$. In view of (1.2), we have

$$\psi ({\int }_0^{d(f^{n+1}x,fa)}\phi (t)dt)\le \alpha \left(d(f^{n}x,a)\right)\psi ({\int }_0^{d(f^{n}x,a)}\phi (t)dt),\mathrm{\forall }n\in {\mathbb{N}}_0.$$

(3.14)

Taking upper limit in (3.14) and making use of $(\phi ,\psi ,\alpha )\in {\mathrm{\Phi }}_1\times {\mathrm{\Phi }}_3\times {\mathrm{\Phi }}_5$ and Lemmas 2.1 and 2.2, we get that

$$\begin{array}{rl}\psi ({\int }_0^{d(a,fa)}\phi (t)dt)& =\underset{n\to \mathrm{\infty }}{lim\hspace{0.17em}sup}\psi ({\int }_0^{d(f^{n+1}x,fa)}\phi (t)dt)\\ \le \underset{n\to \mathrm{\infty }}{lim\hspace{0.17em}sup}\left[\alpha \left(d(f^{n}x,a)\right)\psi ({\int }_0^{d(f^{n}x,a)}\phi (t)dt)\right]\\ \le \underset{n\to \mathrm{\infty }}{lim\hspace{0.17em}sup}\alpha \left(d(f^{n}x,a)\right)\cdot \underset{n\to \mathrm{\infty }}{lim\hspace{0.17em}sup}\psi ({\int }_0^{d(f^{n}x,a)}\phi (t)dt)\\ =0,\end{array}$$

which means that

$$\psi ({\int }_0^{d(a,fa)}\phi (t)dt)=0,$$

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 $b\in X\setminus \{a\}$. It follows from (1.2) and $(\phi ,\psi ,\alpha )\in {\mathrm{\Phi }}_1\times {\mathrm{\Phi }}_3\times {\mathrm{\Phi }}_5$ that

$$\begin{array}{rl}\psi ({\int }_0^{d(a,b)}\phi (t)dt)& =\psi ({\int }_0^{d(fa,fb)}\phi (t)dt)\le \alpha (d(a,b))\psi ({\int }_0^{d(a,b)}\phi (t)dt)\\ <\psi ({\int }_0^{d(a,b)}\phi (t)dt),\end{array}$$

which is a contradiction. This completes the proof. □

Theorem 3.3 Let f be a mapping from a complete metric space $(X,d)$ into itself satisfying (1.3) and

$$\varphi (t)\le \psi (t),\mathrm{\forall }t\in {\mathbb{R}}^{+}.$$

(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

$$\begin{array}{rl}\psi ({\int }_0^{d_{n_0}}\phi (t)dt)=& \psi ({\int }_0^{d(f^{n_0}x,f^{n_0+1}x)}\phi (t)dt)\\ \le & \alpha \left(d(f^{n_0-1}x,f^{n_0}x)\right)\varphi ({\int }_0^{d(f^{n_0-1}x,f^{n_0}x)}\phi (t)dt)\\ +\beta \left(d(f^{n_0-1}x,f^{n_0}x)\right)\psi ({\int }_0^{d(f^{n_0}x,f^{n_0+1}x)}\phi (t)dt)\\ =& \alpha (d_{n_0-1})\varphi ({\int }_0^{d_{n_0-1}}\phi (t)dt)+\beta (d_{n_0-1})\psi ({\int }_0^{d_{n_0}}\phi (t)dt),\end{array}$$

which together with (3.2), (3.15), $(\phi ,\psi ,\varphi )\in {\mathrm{\Phi }}_1\times {\mathrm{\Phi }}_3\times {\mathrm{\Phi }}_4$ and $(\alpha ,\beta )\in {\mathrm{\Phi }}_6$ implies that

$$\begin{array}{rl}0& <\psi ({\int }_0^{d_{n_0-1}}\phi (t)dt)\le \psi ({\int }_0^{d_{n_0}}\phi (t)dt)\\ \le \frac{\alpha (d_{n_0-1})}{1-\beta (d_{n_0-1})}\varphi ({\int }_0^{d_{n_0-1}}\phi (t)dt)\\ \le \frac{\alpha (d_{n_0-1})}{1-\beta (d_{n_0-1})}\psi ({\int }_0^{d_{n_0-1}}\phi (t)dt)\\ <\psi ({\int }_0^{d_{n_0-1}}\phi (t)dt),\end{array}$$

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 ${\{d_n\}}_{n\in \mathbb{N}}$ 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

$$\begin{array}{rl}\psi ({\int }_0^{d_n}\phi (t)dt)=& \psi ({\int }_0^{d(f^{n}x,f^{n+1}x)}\phi (t)dt)\\ \le & \alpha \left(d(f^{n-1}x,f^{n}x)\right)\varphi ({\int }_0^{d(f^{n-1}x,f^{n}x)}\phi (t)dt)\\ +\beta \left(d(f^{n-1}x,f^{n}x)\right)\psi ({\int }_0^{d(f^{n}x,f^{n+1}x)}\phi (t)dt)\\ =& \alpha (d_{n-1})\varphi ({\int }_0^{d_{n-1}}\phi (t)dt)+\beta (d_{n-1})\psi ({\int }_0^{d_n}\phi (t)dt),\mathrm{\forall }n\in \mathbb{N},\end{array}$$

which means that

$$\psi ({\int }_0^{d_n}\phi (t)dt)\le \frac{\alpha (d_{n-1})}{1-\beta (d_{n-1})}\varphi ({\int }_0^{d_{n-1}}\phi (t)dt),\mathrm{\forall }n\in \mathbb{N}.$$

(3.16)

Taking upper limit in (3.16) and using (3.15), $(\phi ,\psi ,\varphi )\in {\mathrm{\Phi }}_1\times {\mathrm{\Phi }}_3\times {\mathrm{\Phi }}_4$, $(\alpha ,\beta )\in {\mathrm{\Phi }}_6$ and Lemma 2.1, we arrive at

$$\begin{array}{rl}\psi ({\int }_0^c\phi (t)dt)& =\underset{n\to \mathrm{\infty }}{lim\hspace{0.17em}sup}\psi ({\int }_0^{d_n}\phi (t)dt)\\ \le \underset{n\to \mathrm{\infty }}{lim\hspace{0.17em}sup}\left[\frac{\alpha (d_{n-1})}{1-\beta (d_{n-1})}\varphi ({\int }_0^{d_{n-1}}\phi (t)dt)\right]\\ \le \underset{n\to \mathrm{\infty }}{lim\hspace{0.17em}sup}\frac{\alpha (d_{n-1})}{1-\beta (d_{n-1})}\cdot \underset{n\to \mathrm{\infty }}{lim\hspace{0.17em}sup}\psi ({\int }_0^{d_{n-1}}\phi (t)dt)\\ \le \underset{s\to c^{+}}{lim\hspace{0.17em}sup}\frac{\alpha (s)}{1-\beta (s)}\cdot \psi ({\int }_0^c\phi (t)dt)\\ <\psi ({\int }_0^c\phi (t)dt),\end{array}$$

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

Next we show that ${\{f^{n}x\}}_{n\in \mathbb{N}}$ is a Cauchy sequence. Suppose that ${\{f^{n}x\}}_{n\in \mathbb{N}}$ is not a Cauchy sequence. As in the proof of Theorem 3.1, we conclude that there exist $\epsilon >0$ and $\{m(k),n(k):k\in \mathbb{N}\}\subseteq \mathbb{N}$ with $m(k)>n(k)>k$ for each $k\in \mathbb{N}$ satisfying (3.8)-(3.11). By means of (3.12), we deduce that

$$\begin{array}{r}\psi ({\int }_0^{d(f^{m(k)+1}x,f^{n(k)+2}x)}\phi (t)dt)\\ \le \alpha \left(d(f^{m(k)}x,f^{n(k)+1}x)\right)\varphi ({\int }_0^{d(f^{m(k)}x,f^{m(k)+1}x)}\phi (t)dt)\\ +\beta \left(d(f^{m(k)}x,f^{n(k)+1}x)\right)\psi ({\int }_0^{d(f^{n(k)+1}x,f^{n(k)+2}x)}\phi (t)dt)\\ =\alpha \left(d(f^{m(k)}x,f^{n(k)+1}x)\right)\varphi ({\int }_0^{d_{m(k)}}\phi (t)dt)\\ +\beta \left(d(f^{m(k)}x,f^{n(k)+1}x)\right)\psi ({\int }_0^{d_{n(k)}}\phi (t)dt),\mathrm{\forall }k\in \mathbb{N}.\end{array}$$

(3.17)

Taking upper limit in (3.17) and making use of (1.3), (3.11), Lemma 2.1, $(\phi ,\psi ,\varphi )\in {\mathrm{\Phi }}_1\times {\mathrm{\Phi }}_3\times {\mathrm{\Phi }}_4$ and $(\alpha ,\beta )\in {\mathrm{\Phi }}_6$, we deduce that

$$\begin{array}{rl}0<& \psi ({\int }_0^{\epsilon }\phi (t)dt)=\underset{k\to \mathrm{\infty }}{lim\hspace{0.17em}sup}\psi ({\int }_0^{d(f^{m(k)+1}x,f^{n(k)+2}x)}\phi (t)dt)\\ \le & \underset{k\to \mathrm{\infty }}{lim\hspace{0.17em}sup}[\alpha \left(d(f^{m(k)}x,f^{n(k)+1}x)\right)\varphi ({\int }_0^{d_{m(k)}}\phi (t)dt)\\ +\beta \left(d(f^{m(k)}x,f^{n(k)+1}x)\right)\psi ({\int }_0^{d_{n(k)}}\phi (t)dt)]\\ \le & \underset{k\to \mathrm{\infty }}{lim\hspace{0.17em}sup}\alpha \left(d(f^{m(k)}x,f^{n(k)+1}x)\right)\cdot \underset{k\to \mathrm{\infty }}{lim\hspace{0.17em}sup}\psi ({\int }_0^{d_{m(k)}}\phi (t)dt)\\ +\underset{k\to \mathrm{\infty }}{lim\hspace{0.17em}sup}\beta \left(d(f^{m(k)}x,f^{n(k)+1}x)\right)\cdot \underset{k\to \mathrm{\infty }}{lim\hspace{0.17em}sup}\psi ({\int }_0^{d_{n(k)}}\phi (t)dt)\\ \le & \underset{s\to \epsilon }{lim\hspace{0.17em}sup}\alpha (s)\cdot \psi ({\int }_0^0\phi (t)dt)+\underset{s\to \epsilon }{lim\hspace{0.17em}sup}\beta (s)\cdot \psi ({\int }_0^0\phi (t)dt)\\ =& 0,\end{array}$$

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

Completeness of $(X,d)$ 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), $(\phi ,\psi ,\varphi )\in {\mathrm{\Phi }}_1\times {\mathrm{\Phi }}_3\times {\mathrm{\Phi }}_4$, $(\alpha ,\beta )\in {\mathrm{\Phi }}_6$ and Lemma 2.1, we infer that

$$\begin{array}{rl}\psi ({\int }_0^{d(a,fa)}\phi (t)dt)=& \underset{n\to \mathrm{\infty }}{lim\hspace{0.17em}sup}\psi ({\int }_0^{d(f^{n+1}x,fa)}\phi (t)dt)\\ \le & \underset{n\to \mathrm{\infty }}{lim\hspace{0.17em}sup}[\alpha \left(d(f^{n}x,a)\right)\varphi ({\int }_0^{d(f^{n}x,f^{n+1}x)}\phi (t)dt)\\ +\beta \left(d(f^{n}x,a)\right)\psi ({\int }_0^{d(a,fa)}\phi (t)dt)]\\ \le & \underset{n\to \mathrm{\infty }}{lim\hspace{0.17em}sup}\alpha \left(d(f^{n}x,a)\right)\cdot \underset{n\to \mathrm{\infty }}{lim\hspace{0.17em}sup}\psi ({\int }_0^{d_n}\phi (t)dt)\\ +\underset{n\to \mathrm{\infty }}{lim\hspace{0.17em}sup}\beta \left(d(f^{n}x,a)\right)\cdot \psi ({\int }_0^{d(a,fa)}\phi (t)dt)\\ \le & \underset{s\to 0^{+}}{lim\hspace{0.17em}sup}\beta (s)\cdot \psi ({\int }_0^{d(a,fa)}\phi (t)dt),\end{array}$$

which together with $(\alpha ,\beta )\in {\mathrm{\Phi }}_6$ yields that

$$\psi ({\int }_0^{d(a,fa)}\phi (t)dt)=0,$$

which gives that $d(fa,a)=0$, 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 $b\in X\setminus \{a\}$. It follows from (1.3) and $(\phi ,\psi ,\varphi )\in {\mathrm{\Phi }}_1\times {\mathrm{\Phi }}_3\times {\mathrm{\Phi }}_4$ and $(\alpha ,\beta )\in {\mathrm{\Phi }}_6$ that

$$\begin{array}{rl}0& \le \psi ({\int }_0^{d(fa,fb)}\phi (t)dt)\\ \le \alpha (d(a,b))\varphi ({\int }_0^{d(a,fa)}\phi (t)dt)+\beta (d(a,b))\psi ({\int }_0^{d(b,fb)}\phi (t)dt)\\ =0,\end{array}$$

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 $X=[0,\frac{1}{2}]\cup \{1\}\cup \{3\}$ 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

$$\begin{array}{c}f(x)=\{\begin{array}{cc}\frac{x}{2},\hfill & \mathrm{\forall }x\in [0,\frac{1}{2}],\hfill \\ 0,\hfill & x=1,\hfill \\ 1,\hfill & x=3,\hfill \end{array}\phi (t)=\{\begin{array}{cc}\frac{t}{2},\hfill & \mathrm{\forall }t\in [0,1],\hfill \\ 1,\hfill & \mathrm{\forall }t\in (1,+\mathrm{\infty }),\hfill \end{array}\hfill \\ \varphi (t)=\{\begin{array}{cc}\frac{t^2}{4},\hfill & \mathrm{\forall }t\in [0,1],\hfill \\ \frac{t^2}{8},\hfill & \mathrm{\forall }t\in (1,+\mathrm{\infty }),\hfill \end{array}\psi (t)=\{\begin{array}{cc}t,\hfill & \mathrm{\forall }t\in [0,1],\hfill \\ \frac{t^2+1}{2},\hfill & \mathrm{\forall }t\in (1,+\mathrm{\infty }).\hfill \end{array}\hfill \end{array}$$

Clearly, $(X,d)$ is a complete metric and $(\phi ,\varphi ,\psi )\in {\mathrm{\Phi }}_1\times {\mathrm{\Phi }}_2\times {\mathrm{\Phi }}_3$. Let $x,y\in X$ with $x<y$. In order to verify (1.1), we have to consider the following four cases.

Case 1. Let $x,y\in [0,\frac{1}{2}]$. Note that

$$\begin{array}{rl}\psi ({\int }_0^{d(fx,fy)}\phi (t)dt)& =\psi ({\int }_0^{\frac{1}{2}|x-y|}\phi (t)dt)=\psi \left(\frac{{|x-y|}^2}{16}\right)=\frac{{|x-y|}^2}{16}\\ \le \frac{{|x-y|}^2}{4}-\frac{{|x-y|}^4}{16}\\ =\psi \left(\frac{{|x-y|}^2}{4}\right)-\varphi \left(\frac{{|x-y|}^2}{4}\right)\\ =\psi ({\int }_0^{|x-y|}\phi (t)dt)-\varphi ({\int }_0^{|x-y|}\phi (t)dt)\\ =\psi ({\int }_0^{d(x,y)}\phi (t)dt)-\varphi ({\int }_0^{d(x,y)}\phi (t)dt).\end{array}$$

Case 2. Let $x\in [0,\frac{1}{2}]$ and $y=1$. It follows that

$$\begin{array}{rl}\psi ({\int }_0^{d(fx,fy)}\phi (t)dt)& =\psi ({\int }_0^{\frac{x}{2}}\phi (t)dt)=\psi \left(\frac{x^2}{16}\right)=\frac{x^2}{16}\le \frac{{(1-x)}^2}{4}-\frac{{(1-x)}^4}{16}\\ =\psi \left(\frac{{(1-x)}^2}{4}\right)-\varphi \left(\frac{{(1-x)}^2}{4}\right)\\ =\psi ({\int }_0^{|x-1|}\phi (t)dt)-\varphi ({\int }_0^{|x-1|}\phi (t)dt)\\ =\psi ({\int }_0^{d(x,y)}\phi (t)dt)-\varphi ({\int }_0^{d(x,y)}\phi (t)dt).\end{array}$$

Case 3. Let $x\in [0,\frac{1}{2}]$ and $y=3$. It follows that

$$\begin{array}{rl}\psi ({\int }_0^{d(fx,fy)}\phi (t)dt)& =\psi ({\int }_0^{\frac{2-x}{2}}\phi (t)dt)=\psi \left(\frac{{(2-x)}^2}{16}\right)=\frac{{(2-x)}^2}{16}<\frac{1}{2}\\ \le \frac{1}{2}[{(\frac{9}{4}-x)}^2+1]-\frac{1}{8}{(\frac{9}{4}-x)}^2=\psi (\frac{9}{4}-x)-\varphi (\frac{9}{4}-x)\\ =\psi ({\int }_0^1\phi (t)dt+{\int }_1^{3-x}\phi (t)dt)-\varphi ({\int }_0^1\phi (t)dt+{\int }_1^{3-x}\phi (t)dt)\\ =\psi ({\int }_0^{3-x}\phi (t)dt)-\varphi ({\int }_0^{3-x}\phi (t)dt)\\ =\psi ({\int }_0^{d(x,y)}\phi (t)dt)-\varphi ({\int }_0^{d(x,y)}\phi (t)dt).\end{array}$$

Case 4. Let $x=1$ and $y=3$. Note that

$$\begin{array}{rl}\psi ({\int }_0^{d(fx,fy)}\phi (t)dt)& =\psi ({\int }_0^1\phi (t)dt)=\psi \left(\frac{1}{4}\right)=\frac{1}{4}<\frac{139}{128}\\ =\frac{1}{2}(\frac{25}{16}+1)-\frac{1}{8}\cdot \frac{25}{16}=\psi \left(\frac{5}{4}\right)-\varphi \left(\frac{5}{4}\right)\\ =\psi ({\int }_0^1\phi (t)dt+{\int }_1^2\phi (t)dt)-\varphi ({\int }_0^1\phi (t)dt+{\int }_1^2\phi (t)dt)\\ =\psi ({\int }_0^2\phi (t)dt)-\varphi ({\int }_0^2\phi (t)dt)\\ =\psi ({\int }_0^{d(x,y)}\phi (t)dt)-\varphi ({\int }_0^{d(x,y)}\phi (t)dt).\end{array}$$

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 $X=[0,1]\cup [4,5]$ be endowed with the Euclidean metric $d=|\cdot |$. Assume that $f:X\to X$ and $\phi ,\psi :{\mathbb{R}}^{+}\to {\mathbb{R}}^{+}$ and $\alpha :{\mathbb{R}}^{+}\to [0,1)$ are defined by

$$\begin{array}{c}f(x)=\{\begin{array}{cc}\frac{x^2}{4},\hfill & \mathrm{\forall }x\in [0,1],\hfill \\ \frac{x^2}{26},\hfill & \mathrm{\forall }x\in [4,5],\hfill \end{array}\phi (t)=\{\begin{array}{cc}4t^3,\hfill & \mathrm{\forall }t\in [0,1],\hfill \\ 2t,\hfill & \mathrm{\forall }t\in [4,5],\hfill \end{array}\hfill \\ \psi (t)=t^{\frac{1}{2}},\mathrm{\forall }t\in {\mathbb{R}}^{+},\alpha (t)=\{\begin{array}{cc}\frac{1}{3}+\frac{t^2}{2},\hfill & \mathrm{\forall }t\in [0,1],\hfill \\ \frac{1}{2t},\hfill & \mathrm{\forall }t\in (1,3),\hfill \\ \frac{1}{\sqrt{t}},\hfill & \mathrm{\forall }t\in (3,+\mathrm{\infty }).\hfill \end{array}\hfill \end{array}$$

Obviously, $(\phi ,\psi ,\alpha )\in {\mathrm{\Phi }}_1\times {\mathrm{\Phi }}_3\times {\mathrm{\Phi }}_5$. Put $x,y\in X$ with $x<y$. In order to verify (1.2), we have to consider three possible cases as follows.

Case 1. Let $x,y\in [0,1]$. It is clear that

$$\begin{array}{rl}\psi ({\int }_0^{d(fx,fy)}\phi (t)dt)& ={\left({\int }_0^{\frac{y^2-x^2}{4}}4t^{3}dt\right)}^{\frac{1}{2}}=\frac{{(x+y)}^2}{16}{|x-y|}^2\le \frac{1}{4}{|x-y|}^2\\ \le (\frac{1}{3}+\frac{1}{2}{|x-y|}^2){|x-y|}^2\\ =\alpha (d(x,y))\psi ({\int }_0^{d(x,y)}\phi (t)dt).\end{array}$$

Case 2. Let $x,y\in [4,5]$. It follows that

$$\begin{array}{rl}\psi ({\int }_0^{d(fx,fy)}\phi (t)dt)& ={\left({\int }_0^{\frac{y^2-x^2}{26}}4t^{3}dt\right)}^{\frac{1}{2}}={\left(\frac{x+y}{26}\right)}^2{|x-y|}^2\le \frac{25}{169}{|x-y|}^2\\ \le (\frac{1}{3}+\frac{1}{2}{|x-y|}^2){|x-y|}^2\\ =\alpha (d(x,y))\psi ({\int }_0^{d(x,y)}\phi (t)dt).\end{array}$$

Case 3. Let $x\in [0,1]$ and $y\in [4,5]$. It follows that

$$\begin{array}{rl}\psi ({\int }_0^{d(fx,fy)}\phi (t)dt)& ={\left({\int }_0^{\frac{y^2}{26}-\frac{x^2}{4}}4t^{3}dt\right)}^{\frac{1}{2}}={(\frac{y^2}{26}-\frac{x^2}{4})}^2\le {\left(\frac{25}{26}\right)}^2<1<\sqrt{|x-y|}\\ =\alpha (|x-y|)|x-y|=\alpha (|x-y|){({\int }_0^{1}4t^{3}dt+{\int }_1^{|x-y|}2tdt)}^{\frac{1}{2}}\\ =\alpha (d(x,y)){({\int }_0^{|x-y|}\phi (t)dt)}^{\frac{1}{2}}\\ =\alpha (d(x,y))\psi ({\int }_0^{d(x,y)}\phi (t)dt).\end{array}$$

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 $X=[\frac{1}{2},1]\cup [\frac{3}{2},2]$ be endowed with the Euclidean metric $d=|\cdot |$. Assume that $f:X\to X$, $\phi ,\varphi ,\psi :{\mathbb{R}}^{+}\to {\mathbb{R}}^{+}$ and $\alpha ,\beta :{\mathbb{R}}^{+}\to [0,1)$ are defined by

$$\begin{array}{c}f(x)=\{\begin{array}{cc}1,\hfill & \mathrm{\forall }x\in [\frac{1}{2},1],\hfill \\ \frac{x}{2},\hfill & \mathrm{\forall }x\in [\frac{3}{2},2],\hfill \end{array}\varphi (t)=\{\begin{array}{cc}0,\hfill & t\in [0,\frac{9}{16})\hfill \\ \frac{32t^2}{9},\hfill & t\in [\frac{9}{16},+\mathrm{\infty }),\hfill \end{array}\hfill \\ \phi (t)=2t,\psi (t)=4t^2,\alpha (t)=\frac{t}{{(\frac{1}{2}+t)}^2},\beta (t)=\frac{t^2}{{(\frac{1}{2}+t)}^2},\mathrm{\forall }t\in {\mathbb{R}}^{+}.\hfill \end{array}$$

It is easy to see that $(\phi ,\psi ,\varphi )\in {\mathrm{\Phi }}_1\times {\mathrm{\Phi }}_3\times {\mathrm{\Phi }}_4$, $(\alpha ,\beta )\in {\mathrm{\Phi }}_6$ and (3.15) holds. In order to verify (1.3), we have to consider the five possible cases below.

Case 1. Let $x,y\in [\frac{3}{2},2]$ with $x\ge y$. Note that

$$\begin{array}{rl}\psi ({\int }_0^{d(fx,fy)}\phi (t)dt)& =\psi ({\int }_0^{\frac{|x-y|}{2}}\phi (t)dt)=\psi \left(\frac{{(x-y)}^2}{4}\right)=\frac{{(x-y)}^4}{4}\le \frac{x-y}{2}\\ \le \frac{x-y}{2}\cdot \frac{x^4}{{(\frac{1}{2}+x-y)}^2}\le \frac{x-y}{{(\frac{1}{2}+x-y)}^2}\cdot \frac{32}{9}{\left(\frac{x^2}{4}\right)}^2\\ =\alpha (d(x,y))\varphi ({\int }_0^{d(x,fx)}\phi (t)dt)\\ \le \alpha (d(x,y))\varphi ({\int }_0^{d(x,fx)}\phi (t)dt)d+\beta (d(x,y))\psi ({\int }_0^{d(y,fy)}\phi (t)dt).\end{array}$$

Case 2. Let $x,y\in [\frac{3}{2},2]$ with $y>x$. Note that

$$\begin{array}{rl}\psi ({\int }_0^{d(fx,fy)}\phi (t)dt)& =\psi ({\int }_0^{\frac{|y-x|}{2}}\phi (t)dt)=\frac{{(y-x)}^4}{4}\le \frac{{(y-x)}^2}{4}\cdot \frac{y^4}{{(\frac{1}{2}+y-x)}^2}\\ =\frac{{(y-x)}^2}{{(\frac{1}{2}+y-x)}^2}\cdot \frac{y^4}{4}=\beta (d(x,y))\psi \left(\frac{y^2}{4}\right)\\ =\beta (d(x,y))\psi ({\int }_0^{\frac{y}{2}}\phi (t)dt)=\beta (d(x,y))\psi ({\int }_0^{d(y,fy)}\phi (t)dt)\\ \le \alpha (d(x,y))\varphi ({\int }_0^{d(x,fx)}\phi (t)dt)+\beta (d(x,y))\psi ({\int }_0^{d(y,fy)}\phi (t)dt).\end{array}$$

Case 3. Let $x\in [\frac{3}{2},2]$ and $y\in [\frac{1}{2},1]$. It follows that

$$\begin{array}{rl}\psi ({\int }_0^{d(fx,fy)}\phi (t)dt)& =\psi ({\int }_0^{\frac{x-2}{2}}\phi (t)dt)=\psi \left(\frac{{(x-2)}^2}{4}\right)=\frac{{(x-2)}^4}{4}\le \frac{1}{64}<\frac{27}{64}\\ =\frac{3}{8}\cdot \frac{2}{9}\cdot \frac{81}{16}\le \frac{x-y}{{(\frac{1}{2}+x-y)}^2}\cdot \frac{2}{9}\cdot x^4=\alpha (d(x,y))\varphi \left(\frac{x^2}{4}\right)\\ =\alpha (d(x,y))\varphi ({\int }_0^{d(x,fx)}\phi (t)dt)\\ \le \alpha (d(x,y))\varphi ({\int }_0^{d(x,fx)}\phi (t)dt)+\beta (d(x,y))\psi ({\int }_0^{d(y,fy)}\phi (t)dt).\end{array}$$

Case 4. Let $x\in [\frac{1}{2},1]$ and $y\in [\frac{3}{2},2]$. Note that

$$\begin{array}{rl}\psi ({\int }_0^{d(fx,fy)}\phi (t)dt)& =\psi \left({\int }_0^{\frac{|y-2|}{2}}2tdt\right)=\frac{{(y-2)}^4}{4}\le \frac{1}{64}<\frac{1}{4}\cdot \frac{81}{64}\\ \le \frac{{(y-x)}^2}{{(\frac{1}{2}+y-x)}^2}\cdot \frac{y^4}{4}=\beta (d(x,y))\psi \left(\frac{y^2}{4}\right)\\ =\beta (d(x,y))\psi ({\int }_0^{d(y,fy)}\phi (t)dt)\\ \le \alpha (d(x,y))\varphi ({\int }_0^{d(x,fx)}\phi (t)dt)+\beta (d(x,y))\psi ({\int }_0^{d(y,fy)}\phi (t)dt).\end{array}$$

Case 5. Let $x,y\in [\frac{1}{2},1]$. Notice that $fx=fy=1$. It follows that

$$\begin{array}{r}\psi ({\int }_0^{d(fx,fy)}\phi (t)dt)\\ =0\le \alpha (d(x,y))\varphi ({\int }_0^{d(x,fx)}\phi (t)dt)+\beta (d(x,y))\psi ({\int }_0^{d(y,fy)}\phi (t)dt).\end{array}$$

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 $x\in X$.