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# Fixed point theorems for Ćirić type mapping and application to integral equation

## Abstract

In this paper, we introduce a new class of Ćirić type single-valued mapping with respect to u-distance and prove some fixed point theorems for this mapping. An example is given to show that our results are a proper extension of many well-known results. As an application, we establish the existence of a solution for an integral equation.

## Introduction

The Banach contraction principle is a remarkable result in metric fixed point theory. Over the years, it has been generalized in different directions and spaces by several mathematicians, see [117] and the references therein. In 1974, Ćirić [5] proved the following fixed point theorem on a complete metric space, which generalizes the Banach contraction principle: Let X be a complete metric space and let $T:X\rightarrow X$ be a quasi-contractive mapping; i.e., there exists a constant $q \in[0,1)$ such that, for all $x, y \in X$,

$$d(Tx, Ty) \leq q \cdot\max\bigl\{ d(x,y), d(x,Tx), d(y, Ty), d(x,Ty), d(y, Tx) \bigr\} .$$

Then:

1. (1)

T has a unique fixed point w in X.

2. (2)

$\lim_{n \rightarrow\infty} T^{n} x= w$ for every $x \in X$.

3. (3)

$d(T^{n} x, w) \leq[ \frac{q^{n}}{1-q} ] d(x, Tx)$ for every x in X.

Recently, Ume [15] generalized the notion of τ-distance [18] by introducing u-distance as follows.

Let X be metric space with metric d. Then a function $p:X \times X \rightarrow R_{+}$ is called a u-distance on X if there exists a function $\theta: X \times X \times R_{+} \times R_{+} \rightarrow R_{+}$ such that the following hold for $x, y,z \in X$:

(u1):

$p(x, z) \leq p(x, y) + p(y, z)$;

(u2):

$\theta(x, y, 0, 0) = 0$ and $\theta(x, y, s, t) \geq\min\{s,t\}$ for each $s, t \in R_{+}$, and for any $x \in X$ and for every $\varepsilon> 0$, there exists $\delta > 0$ such that $| s- s_{0} |< \delta$, $|t - t_{0} |< \delta$, $s, s_{0}, t, t_{0} \in R_{+}$ and $y \in X$ imply $|\theta(x,y,s,t) -\theta(x,y,s_{0} , t_{0} ) |< \varepsilon$;

(u3):

$\lim_{n \rightarrow\infty} x_{n} =x$ and $\lim_{n \rightarrow \infty} \sup\{ \theta(w_{n} , z_{n} , p(w_{n} , x_{m}), p(z_{n} ,x_{m} )) : m\geq n \} = 0$ imply $p(y,x) \leq\lim_{n \rightarrow\infty} \inf p(y, x_{n} )$ for all $y\in X$;

(u4):

$\lim_{n \rightarrow\infty} \sup\{p(x_{n} , w_{m}):m \geq n \} =0$, $\lim_{n \rightarrow\infty} \sup\{p(y_{n} , z_{m}):m \geq n \} =0$, $\lim_{n \rightarrow\infty} \theta(x_{n} , w_{n} , s_{n} ,t_{n} )=0$, $\lim_{n \rightarrow\infty} \theta(y_{n} , z_{n} , s_{n} ,t_{n} )=0$ imply $\lim_{n \rightarrow\infty} \theta(w_{n} , z_{n} , s_{n} ,t_{n} )=0$ or $\lim_{n \rightarrow\infty} \sup\{p(w_{m} , x_{n} ): m \geq n \} =0$, $\lim_{n \rightarrow\infty} \sup\{p(z_{m} , y_{n}): m \geq n \} =0$, $\lim_{n \rightarrow\infty} \theta(x_{n} , w_{n} , s_{n} ,t_{n} )=0$, $\lim_{n \rightarrow\infty} \theta(y_{n} , z_{n} , s_{n} ,t_{n} )=0$ imply $\lim_{n \rightarrow\infty} \theta(w_{n} , z_{n} , s_{n} ,t_{n} )=0$;

(u5):

$\lim_{n \rightarrow\infty} \theta(w_{n} , z_{n} , p(w_{n} , x_{n}), p(z_{n} ,x_{n} ))=0$, $\lim_{n \rightarrow\infty} \theta(w_{n} , z_{n} , p(w_{n} , y_{n}), p(z_{n} ,y_{n} ))=0$ imply $\lim_{n \rightarrow\infty} d(x_{n}, y_{n})=0$ or $\lim_{n \rightarrow\infty} \theta(a_{n} , b_{n} , p(x_{n} , a_{n}), p(x_{n} ,b_{n} ))=0$, $\lim_{n \rightarrow\infty} \theta(a_{n} , b_{n} , p(y_{n} , a_{n}), p(y_{n} ,b_{n} ))=0$ imply $\lim_{n \rightarrow\infty} d(x_{n}, y_{n})=0$.

### Remark 1.1

([15])

1. (a)

Suppose that θ from $X \times X \times R_{+} \times R_{+}$ into $R_{+}$ is a mapping satisfying $(\mathrm{u}_{2}){\sim}(\mathrm{u}_{5})$. Then there exists a mapping η from $X \times X \times R_{+} \times R_{+}$ into $R_{+}$ such that η is nondecreasing in its third and fourth variable, satisfying $(\mathrm{u}_{2})_{\eta}{\sim}(\mathrm{u}_{5})_{\eta}$, where $(\mathrm{u}_{2})_{\eta}{\sim}(\mathrm{u}_{5})_{\eta}$ stand for substituting η for θ in $(\mathrm{u}_{2}){\sim}(\mathrm{u}_{5})$, respectively.

2. (b)

On account of (a), we may assume that θ is nondecreasing in its third and fourth variables, respectively, for a function θ from $X \times X \times R_{+} \times R_{+}$ into $R_{+}$ satisfying $(\mathrm{u}_{2}){\sim}(\mathrm{u}_{5})$.

3. (c)

Each τ-distance p on a metric space $(X,d)$ is also a u-distance on X. We present some example of u-distance which are not τ-distance. (For details, see [15].)

### Example 1.2

Let $X=R_{+}$ with the usual metric. Define $p: X \times X \rightarrow R_{+}$ by $p(x,y) = (\frac{1}{4})x^{2}$. Then p is a u-distance on X but not a τ-distance on X.

### Example 1.3

Let X be a normed space with norm $\| \cdot\|$. Then a function $p:X \times X \rightarrow R_{+}$ defined by $p(x,y)=\| x \|$ for every $x, y \in X$ is a u-distance on X but not a τ-distance.

It follows from the above example and Remark 1.1(c) that u-distance is a proper extension of τ-distance. Other useful examples on u-distance are given in [15].

## Preliminaries

Throughout this paper we denote by N the set of all positive integers, by R the set of all real numbers and by $R_{+}$ the set of all nonnegative real numbers.

### Definition 2.1

([15])

Let X be a metric space with a metric d and let p be a u-distance on X. Then a sequence $\{x_{n}\}$ in X is called p-Cauchy if there exists a function θ from $X \times X \times R_{+} \times R_{+}$ into $R_{+}$ satisfying $(\mathrm{u}_{2}){\sim}(\mathrm{u}_{5})$ and a sequence $\{z_{n}\}$ of X such that

\begin{aligned}& \lim_{n \rightarrow\infty} \sup\bigl\{ \theta\bigl(z_{n} , z_{n} , p(z_{n} , x_{m}), p(z_{n} ,x_{m} )\bigr): m \geq n \bigr\} =0\quad \mbox{or} \\& \lim_{n \rightarrow\infty} \sup\bigl\{ \theta\bigl(z_{n} , z_{n} , p(x_{m} , z_{n}), p(x_{m} ,z_{n} )\bigr): m \geq n \bigr\} =0. \end{aligned}

### Lemma 2.2

([15])

Let X be a metric space with a metric d and let p a u-distance on X. If $\{x_{n}\}$ is a p-Cauchy sequence, then $\{x_{n} \}$ is a Cauchy sequence.

### Lemma 2.3

([15])

Let X be a metric space with a metric d and let p be a u-distance on X.

1. (1)

If sequences $\{x_{n}\}$ and $\{y_{n}\}$ of X satisfy $\lim_{n \rightarrow\infty} p(z,x_{n}) =0$ and $\lim_{n \rightarrow\infty} p(z,y_{n}) =0$ for some $z \in X$, then $\lim_{n\rightarrow\infty} d(x_{n}, y_{n}) =0$.

2. (2)

If $p(z,x)=0$ and $p(z,y)=0$, then $x=y$.

3. (3)

Suppose that sequences $\{x_{n}\}$ and $\{y_{n}\}$ of X satisfy $\lim_{n \rightarrow\infty} p(x_{n},z) =0$ and $\lim_{n \rightarrow\infty} p(y_{n} ,z) =0$ for some $z \in X$, then $\lim_{n\rightarrow\infty} d(x_{n}, y_{n}) =0$.

4. (4)

If $p(x,z)=0$ and $p(y,z)=0$, then $x=y$.

### Lemma 2.4

([15])

Let X be a metric space with a metric d and let p be a u-distance on X. Suppose that a sequence $\{x_{n}\}$ of X satisfies

\begin{aligned}& \lim_{n \rightarrow\infty} \sup\bigl\{ p(x_{n} ,x_{m} ): m \geq n \bigr\} =0 \quad \textit{or} \\& \lim_{n \rightarrow\infty} \sup\bigl\{ p(x_{m} ,x_{n} ): m \geq n \bigr\} =0. \end{aligned}

Then:

1. (i)

$\{ x_{n} \}$ is a p-Cauchy sequence.

2. (ii)

If $\{x_{n}\}$ is a p-Cauchy sequence, then $\{x_{n}\}$ is a Cauchy sequence.

## Fixed point theorems

The following lemma plays an important role in proving our theorems.

### Lemma 3.1

Let $(X,d)$ be a metric space with a u-distance p on X and $\{a_{n}\}$ and $\{b_{n}\}$ be sequences of X such that

\begin{aligned}& \lim_{n \rightarrow\infty} \sup\bigl\{ p(a_{n} ,a_{m} ): m > n \bigr\} =0 \quad \textit{and} \\& \lim_{n \rightarrow\infty} \sup\bigl\{ p(a_{n} ,b_{m} ): m > n \bigr\} =0. \end{aligned}

Then there exist a subsequence $\{a_{k_{n}}\}$ of $\{a_{n}\}$ and a subsequence $\{b_{k_{n}}\}$ of $\{b_{n}\}$ such that $\lim_{n \rightarrow\infty} d(a_{k_{n}}, b_{k_{n}})=0$.

### Proof

Since p is a u-distance on X,

\begin{aligned}& \mbox{there exists a mapping } \theta: X \times X \times R_{+} \times R_{+} \rightarrow R_{+} \\& \mbox{such that } \theta \mbox{ is nondecreasing in its third and} \\& \mbox{fourth variable respectively, satisfying } (\mathrm{u}_{2}) { \sim}(\mathrm{u}_{5}). \end{aligned}
(3.1)

For each $n\in N$, let

$$\alpha_{n}=\sup\bigl\{ p(a_{n}, a_{m}):m>n\bigr\} \quad \mbox{and}\quad \beta_{n}=\sup\bigl\{ p(a_{n}, b_{m}) : m>n\bigr\} .$$
(3.2)

By the hypotheses and (3.2), we have

$$\lim_{n \rightarrow\infty}(\alpha_{n} + \beta_{n} )=0.$$
(3.3)

Let $k_{1} \in N$ be an arbitrary and fixed element. Then, by (u2), for this $a_{k_{1}} \in X$ and $\varepsilon=1$, there exists $\delta_{1} >0$ such that

$$| s |=s < \delta_{1}, \qquad | t |=t < \delta_{1},\qquad y\in X \quad \mbox{imply} \quad \theta(a_{k_{1}}, y, s, t)< 1.$$
(3.4)

By virtue of (3.3) and (3.4), for this $\delta_{1}>0$, there exists $M_{1} \in N$ such that

$$n\geq M_{1} \quad \mbox{implies}\quad \alpha_{n} + \beta_{n}< \delta_{1}.$$
(3.5)

Let $k_{2} \in N$ be such that

$$k_{2}\geq\max\{1+k_{1}, M_{1}\}.$$
(3.6)

Due to (3.6), we have

$$k_{1}< k_{2} \quad \mbox{and} \quad k_{2}\geq M_{1} .$$
(3.7)

From (3.4), (3.5), (3.6), and (3.7) we get

$$\theta(a_{k_{1}}, a_{k_{2}}, \alpha_{k_{2}}+ \beta_{k_{2}}, \alpha_{k_{2}}+\beta_{k_{2}}) < 1.$$
(3.8)

In terms of (u2) and (3.6), for this $a_{k_{2}} \in X$ and $\varepsilon=\frac{1}{2}$, there exists $\delta_{2} >0$ such that $| s |=s < \delta_{2}$, $| t |=t < \delta_{2}$, $y \in X$ imply

$$\theta(a_{k_{2}}, y, s, t) < \frac{1}{2}.$$
(3.9)

In view of (3.3) and (3.9), for this $\delta_{2}>0$, there exists $M_{2}\in N$ such that

$$n\geq M_{2} \quad \mbox{implies}\quad \alpha_{n} +\beta_{n} < \delta_{2} .$$
(3.10)

Let $k_{3} \in N$ be such that

$$k_{3} \geq\max\{1+k_{2}, M_{2}\}.$$
(3.11)

On account of (3.9), (3.10), (3.11), we obtain

$$k_{2} < k_{3} \quad \mbox{and}\quad \theta(a_{k_{2}}, a_{k_{3}}, \alpha_{k_{3}} + \beta_{k_{3}}, \alpha_{k_{3}} + \beta_{k_{3}}) < \frac{1}{2}.$$
(3.12)

Continuing this process, there exist a subsequence $\{a_{k_{n}}\}$ of $\{a_{n}\}$, and a subsequence $\{b_{k_{n}}\}$ of $\{b_{n}\}$ such that, for all $n \in N$,

$$\theta(a_{k_{n}}, a_{k_{n+1}}, \alpha_{k_{n+1}} + \beta_{k_{n+1}}, \alpha_{k_{n+1}} + \beta_{k_{n+1}}) < \frac{1}{n}.$$
(3.13)

Using (3.2), (3.3), and (3.13), we know that

\begin{aligned} &\lim_{n\rightarrow\infty} \bigl\{ \sup \bigl[p(a_{k_{n}},a_{k_{m+1}}): m\geq n\bigr]\bigr\} \\ &\quad \leq \lim_{n\rightarrow\infty} \bigl\{ \sup\bigl[p(a_{k_{n}},a_{l}): l>k_{n}\bigr]\bigr\} \\ &\quad = \lim_{n\rightarrow\infty} \alpha_{k_{n}} = 0\quad \mbox{and} \\ &\lim_{n\rightarrow\infty} \theta(a_{k_{n}}, a_{k_{n+1}}, \alpha _{k_{n+1}} + \beta_{k_{n+1}}, \alpha_{k_{n+1}} + \beta_{k_{n+1}}) =0. \end{aligned}
(3.14)

Using (3.1), (3.2), (3.14) and putting $x_{n} =y_{n} = a_{k_{n}}$, $w_{m} =z_{m} = a_{k_{m+1}}$ and $s_{n} =t_{n} = \alpha_{k_{n+1}} + \beta_{k_{n+1}}$ in (u4) we deduce

\begin{aligned} &\lim_{n\rightarrow\infty} \theta \bigl(a_{k_{n+1}},a_{k_{n+1}}, p(a_{k_{n+1}}, a_{k_{n+2}}), p(a_{k_{n+1}}, a_{k_{n+2}})\bigr)=0\quad \mbox{and} \\ &\lim_{n\rightarrow\infty} \theta\bigl(a_{k_{n+1}}, a_{k_{n+1}}, p(a_{k_{n+1}}, b_{k_{n+2}}), p(a_{k_{n+1}}, b_{k_{n+2}}) \bigr)=0. \end{aligned}
(3.15)

Using (3.15) and putting $w_{n} =z_{n} = a_{k_{n+1}}$, $x_{n}= a_{k_{n+2}}$, and $y_{n}=b_{k_{n+2}}$ in (u5), we have

$$\lim_{n\rightarrow\infty} d(a_{k_{n+2}},b_{k_{n+2}})=0.$$
(3.16)

Due to (3.13) and (3.16), there exist a subsequence $\{a_{k_{n}}\}$ of $\{a_{n}\}$ and a subsequence $\{b_{k_{n}}\}$ of $\{b_{n}\}$ such that

$$\lim_{n\rightarrow\infty} d(a_{k_{n}},b_{k_{n}})=0.$$
(3.17)

□

### Definition 3.2

Let $(X,d)$ be a metric space with a u-distance p on X and let T be a selfmapping on X. For $A\subseteq X$, let $\delta(A)= \sup\{ p(x,y) : x,y \in A \}$ and for each $x, y \in X$, $n \in N$, let

$$O(x,y,n) = \bigl\{ T^{i} x, T^{j} y :0\leq i,j \leq n, i,j \in N \cup\{0\} \bigr\} ,$$

where $T^{0} x =x$ and $T^{i}$ is the i times repeated composition of T with itself. Let

$$O(x,y, \infty) = \bigl\{ T^{i} x, T^{j} y : i,j \in N \cup \{0\} \bigr\}$$

for each $x, y \in X$.

### Lemma 3.3

Let $(X,d)$ be a metric space with a u-distance p on X. Let $T : X \rightarrow X$ and $\varphi: R_{+} \rightarrow R_{+}$ be mappings that satisfy the following conditions:

\begin{aligned} \begin{aligned}[b] (\mathrm{i}) &\quad p(Tx,Ty) \leq\varphi\bigl(\max \bigl\{ p(x,y),p(x,Tx),p(y,Ty),p(x,Ty),p(y,Tx), \\ &\hphantom{\quad p(Tx,Ty) \leq{}}p(y,x),p(Tx,x),p(Ty,y),p(Ty,x),p(Tx,y)\bigr\} \bigr) \end{aligned} \end{aligned}
(3.18)

for all $x,y \in X$;

\begin{aligned} (\mathrm{ii})&\quad \varphi\textit{ is nondecreasing and }\varphi(t)< t\textit{ for all }t>0; \\ (\mathrm{iii})&\quad I-\varphi\textit{ is nondecreasing and bijective}, \textit{where }I\textit{ is identity mapping on }R_{+}; \\ (\mathrm{iv})&\quad \sum_{n=1}^{\infty} \varphi^{n} (t)< \infty\quad \textit{for each } t \in (0, \infty), \end{aligned}
(3.19)

where $\varphi^{n}$ is n-times repeated composition of φ with itself.

Then:

1. (1)

For each $x,y \in X$ and $n\in N$,

\begin{aligned}& \max\bigl\{ p\bigl(T^{i} x, T^{j} x\bigr), p \bigl(T^{i} x, T^{j} y\bigr), p\bigl(T^{i} y, T^{j} x\bigr), p\bigl(T^{i} y, T^{j} y\bigr)\mid i,j \in N, i,j \leq n \bigr\} \\& \quad \leq\varphi\bigl(\delta\bigl(O(x,y,n)\bigr)\bigr). \end{aligned}
2. (2)

For each $x,y \in X$, $n \in N$ and for each $i\in N$ with $i \leq8$, there exists $l_{i} \in N$ with $l_{i} \leq n$ such that

\begin{aligned} \begin{aligned} \delta\bigl(O(x,y,n)\bigr) ={}& \max\bigl\{ p(x,x),p(x,y),p(y,x),p(y,y),p \bigl(x,T^{l_{1}} x\bigr), p\bigl(x,T^{l_{2}} y\bigr), \\ &p\bigl(y, T^{l_{3}} x\bigr),p\bigl(y, T^{l_{4}} y\bigr), p\bigl(T^{l_{5}} x, x\bigr), p \bigl(T^{l_{6}} x ,y \bigr),p\bigl(T^{l_{7}} y , x\bigr), p \bigl(T^{l_{8}} y , y\bigr) \bigr\} . \end{aligned} \end{aligned}
3. (3)

For each $x,y \in X$,

$$\delta\bigl(O(x,y,\infty)\bigr) \leq(I-\varphi)^{-1} \bigl(b(x,y) \bigr),$$

where $b(x,y)=p(x,x)+p(y,y)+p(x,y)+p(y,x)+p(x,Tx)+p(Tx,x) +p(y,Ty)+ p(Ty,y)$.

4. (4)

For each $x \in X$, $\{T^{n} x \}$ is a Cauchy sequence.

5. (5)

For each $x,y \in X$ and $n \in N$,

$$p\bigl(T^{n} x, T^{n} y\bigr) \leq\varphi^{n-1} \bigl((I-\varphi)^{-1} \bigl(b(x,y)\bigr)\bigr).$$
6. (6)

For each $x,y \in X$, $\lim_{n\rightarrow\infty} p(T^{n} x, T^{n} y)=0$.

### Proof

Let $x,y \in X$ and $n \in N$, and let i and j be natural numbers with $i,j \leq n$. Then $T^{i-1} x, T^{i} x, T^{j-1} x, T^{j} x, T^{i-1} y, T^{i} y, T^{j-1} y, T^{j} y \in O(x,y,n)$.

From (3.18) and hypothesis (ii), we have

\begin{aligned}& p\bigl(T^{i} x, T^{j} x\bigr) = p\bigl(TT^{i-1} x, TT^{j-1} x\bigr) \\& \hphantom{p\bigl(T^{i} x, T^{j} x\bigr)}\leq\varphi\bigl(\max\bigl\{ p\bigl(T^{i-1} x, T^{j-1} x\bigr), p\bigl(T^{i-1} x, T^{i} x\bigr), p \bigl(T^{j-1} x, T^{j} x\bigr), p\bigl(T^{i-1} x, T^{j} x\bigr), \\& \hphantom{p\bigl(T^{i} x, T^{j} x\bigr)={}}p\bigl(T^{j-1} x, T^{i} x\bigr), p \bigl(T^{j-1} x, T^{i-1} x\bigr), p\bigl(T^{i} x, T^{i-1}x\bigr), \\& \hphantom{p\bigl(T^{i} x, T^{j} x\bigr)={}}p\bigl(T^{j} x, T^{j-1} x\bigr), p \bigl(T^{j} x, T^{i-1} x\bigr),p\bigl(T^{i} x, T^{j-1} x\bigr) \bigr\} \bigr) \\& \hphantom{p\bigl(T^{i} x, T^{j} x\bigr)}\leq \varphi\bigl(\delta\bigl(O(x,y,n)\bigr)\bigr), \\& p\bigl(T^{i} x, T^{j} y\bigr) = p\bigl(TT^{i-1} x, TT^{j-1} y\bigr) \\& \hphantom{p\bigl(T^{i} x, T^{j} y\bigr)}\leq\varphi\bigl(\max\bigl\{ p\bigl(T^{i-1} x, T^{j-1} y\bigr), p\bigl(T^{i-1} x, T^{i} x\bigr), p \bigl(T^{j-1} y, T^{j} y\bigr), p\bigl(T^{i-1} x, T^{j} y\bigr), \\& \hphantom{p\bigl(T^{i} x, T^{j} y\bigr) ={}}p\bigl(T^{j-1} y, T^{i} x\bigr), p \bigl(T^{j-1} y, T^{i-1} x\bigr), p\bigl(T^{i} x, T^{i-1} x\bigr), \\& \hphantom{p\bigl(T^{i} x, T^{j} y\bigr) ={}}p\bigl(T^{j} y, T^{j-1} y\bigr), p \bigl(T^{j} y, T^{i-1} x\bigr),p\bigl(T^{i} x, T^{j-1} y\bigr) \bigr\} \bigr) \\& \hphantom{p\bigl(T^{i} x, T^{j} y\bigr)}\leq \varphi\bigl(\delta\bigl(O(x,y,n)\bigr)\bigr), \\& p\bigl(T^{i} y, T^{j} x\bigr) = p\bigl(TT^{i-1} y, TT^{j-1} x\bigr) \\& \hphantom{p\bigl(T^{i} y, T^{j} x\bigr)}\leq\varphi\bigl(\max\bigl\{ p\bigl(T^{i-1} y, T^{j-1} x\bigr), p\bigl(T^{i-1} y, T^{i} y\bigr), p \bigl(T^{j-1} x, T^{j} x\bigr), p\bigl(T^{i-1} y, T^{j} x\bigr), \\& \hphantom{p\bigl(T^{i} y, T^{j} x\bigr) ={}}p\bigl(T^{j-1} x, T^{i} y\bigr), p \bigl(T^{j-1} x, T^{i-1} y\bigr), p\bigl(T^{i} y, T^{i-1} y\bigr), \\& \hphantom{p\bigl(T^{i} y, T^{j} x\bigr) ={}}p\bigl(T^{j} x, T^{j-1} x\bigr), p \bigl(T^{j} x, T^{i-1} y\bigr),p\bigl(T^{i} y, T^{j-1} x\bigr) \bigr\} \bigr) \\& \hphantom{p\bigl(T^{i} y, T^{j} x\bigr)}\leq \varphi\bigl(\delta\bigl(O(x,y,n)\bigr)\bigr), \\& p\bigl(T^{i} y, T^{j} y\bigr) = p\bigl(TT^{i-1} y, TT^{j-1} y\bigr) \\& \hphantom{p\bigl(T^{i} y, T^{j} y\bigr)}\leq\varphi\bigl(\max\bigl\{ p\bigl(T^{i-1} y, T^{j-1} y\bigr), p\bigl(T^{i-1} y, T^{i} y\bigr), p \bigl(T^{j-1} y, T^{j} y\bigr), p\bigl(T^{i-1} y, T^{j} y\bigr), \\& \hphantom{p\bigl(T^{i} y, T^{j} y\bigr) ={}}p\bigl(T^{j-1} y, T^{i} y\bigr), p \bigl(T^{j-1} y, T^{i-1} y\bigr), p\bigl(T^{i} y, T^{i-1} y\bigr), \\& \hphantom{p\bigl(T^{i} y, T^{j} y\bigr) ={}}p\bigl(T^{j} y, T^{j-1} y\bigr), p \bigl(T^{j} y, T^{i-1} y\bigr),p\bigl(T^{i} y, T^{j-1} y\bigr) \bigr\} \bigr) \\& \hphantom{p\bigl(T^{i} y, T^{j} y\bigr)}\leq \varphi\bigl(\delta\bigl(O(x,y,n)\bigr)\bigr), \end{aligned}

which proves (1).

From (1), it follows that for each $x,y \in X$, $n\in N$ and for each $i \in N$ with $i \leq8$, there exists $l_{i} \in N$ with $l_{i} \le n$ such that

\begin{aligned} \delta\bigl(O(x,y,n)\bigr) =& \max\bigl\{ p(x,x),p(x,y),p(y,x),p(y,y),p \bigl(x,T^{l_{1}} x\bigr), p\bigl(x,T^{l_{2}} y\bigr), p\bigl(y, T^{l_{3}} x\bigr), \\ &p\bigl(y, T^{l_{4}} y\bigr), p\bigl(T^{l_{5}} x, x\bigr), p \bigl(T^{l_{6}} x ,y \bigr),p\bigl(T^{l_{7}} y , x\bigr), p \bigl(T^{l_{8}} y , y\bigr) \bigr\} , \end{aligned}

which proves (2).

Applying the triangle inequality, hypothesis (iii), (1), and (2), we have

\begin{aligned}& p\bigl(x, T^{l_{1}} x\bigr) \leq p(x, Tx)+p\bigl(Tx, T^{l_{1}} x \bigr) \leq p(x, Tx) + \varphi\bigl(\delta\bigl(O(x,y,n)\bigr)\bigr), \\& p\bigl(x, T^{l_{2}} y\bigr) \leq p(x, Tx)+p\bigl(Tx, T^{l_{2}} y \bigr) \leq p(x, Tx) + \varphi\bigl(\delta\bigl(O(x,y,n)\bigr)\bigr), \\& p\bigl(y, T^{l_{3}} x\bigr) \leq p(y, Ty)+p\bigl(Ty, T^{l_{3}} x \bigr) \leq p(y, Ty) + \varphi\bigl(\delta\bigl(O(x,y,n)\bigr)\bigr), \\& p\bigl(y, T^{l_{4}} x\bigr) \leq p(y, Ty)+p\bigl(Ty, T^{l_{4}} x \bigr) \leq p(y, Ty) + \varphi\bigl(\delta\bigl(O(x,y,n)\bigr)\bigr), \\& p\bigl(T^{l_{5}} x, x\bigr) \leq p\bigl(T^{l_{5}} x, Tx \bigr)+p(Tx, x) \leq p(Tx, x) + \varphi\bigl(\delta\bigl(O(x,y,n)\bigr)\bigr), \\& p\bigl(T^{l_{6}} x, y\bigr) \leq p\bigl(T^{l_{6}} x, Ty \bigr)+p(Ty, y) \leq p(Ty, y) + \varphi\bigl(\delta\bigl(O(x,y,n)\bigr)\bigr), \\& p\bigl(T^{l_{7}} y, x\bigr) \leq p\bigl(T^{l_{7}} y, Tx \bigr)+p(Tx, x) \leq p(Tx, x) + \varphi\bigl(\delta\bigl(O(x,y,n)\bigr)\bigr), \\& p\bigl(T^{l_{8}} y, y\bigr) \leq p\bigl(T^{l_{8}} y, Ty \bigr)+p(Ty, y) \leq p(Ty, y) + \varphi\bigl(\delta\bigl(O(x,y,n)\bigr)\bigr). \end{aligned}

Therefore $\delta(O(x,y,n)) \leq(I- \varphi)^{-1} (b(x,y))$.

Since n is arbitrary, the proof of (3) is complete.

To prove (4), let x be an arbitrary point of X and define $x_{n} = T^{n} x$ for every $n \in N$. On account of (3.18) and hypothesis (ii), we have

\begin{aligned}& p(x_{n}, x_{n+1})= p(Tx_{n-1},Tx_{n}) \\& \hphantom{p(x_{n}, x_{n+1})}\leq\varphi\bigl(\max\bigl\{ p(x_{n-1},x_{n}), p(x_{n-1},x_{n}), p(x_{n}, x_{n+1}),p(x_{n-1}, x_{n+1}),p(x_{n}, x_{n}), \\& \hphantom{p(x_{n}, x_{n+1})={}}p(x_{n},x_{n-1}),p(x_{n},x_{n-1}), p(x_{n+1},x_{n}),p(x_{n+1},x_{n-1}),p(x_{n},x_{n}) \bigr\} \bigr), \end{aligned}
(3.20)
\begin{aligned}& p(x_{n+1},x_{n})= p(Tx_{n},Tx_{n-1}) \\& \hphantom{p(x_{n+1},x_{n})}\leq\varphi\bigl(\max\bigl\{ p(x_{n},x_{n-1}), p(x_{n},x_{n+1}), p(x_{n-1},x_{n}),p(x_{n}, x_{n}),p(x_{n-1}, x_{n+1}), \\& \hphantom{p(x_{n+1},x_{n})={}}p(x_{n-1},x_{n}),p(x_{n+1},x_{n}), p(x_{n},x_{n-1}),p(x_{n},x_{n}),p(x_{n+1},x_{n-1}) \bigr\} \bigr), \end{aligned}
(3.21)
\begin{aligned}& p(x_{n-1},x_{n})= p(Tx_{n-2},Tx_{n-1}) \\& \hphantom{p(x_{n-1},x_{n})}\leq\varphi\bigl(\max\bigl\{ p(x_{n-2},x_{n-1}), p(x_{n-2},x_{n-1}),p(x_{n-1}, x_{n}),p(x_{n-2}, x_{n}),p(x_{n-1}, x_{n-1}), \\& \hphantom{p(x_{n-1},x_{n})={}}p(x_{n-1},x_{n-2}),p(x_{n-1},x_{n-2}), p(x_{n},x_{n-1}),p(x_{n}, x_{n-2}),p(x_{n-1},x_{n-1}) \bigr\} \bigr), \end{aligned}
(3.22)
\begin{aligned}& p(x_{n}, x_{n-1})= p(Tx_{n-1},Tx_{n-2}) \\& \hphantom{p(x_{n}, x_{n-1})}\leq\varphi\bigl(\max\bigl\{ p(x_{n-1},x_{n-2}), p(x_{n-1},x_{n}),p(x_{n-2},x_{n-1}),p(x_{n-1}, x_{n-1}),p(x_{n-2}, x_{n}), \\& \hphantom{p(x_{n}, x_{n-1})={}}p(x_{n-2},x_{n-1}),p(x_{n}, x_{n-1}), p(x_{n-1},x_{n-2}),p(x_{n-1},x_{n-1}),p(x_{n}, x_{n-2}) \bigr\} \bigr), \end{aligned}
(3.23)
\begin{aligned}& p(x_{n-1},x_{n+1})= p(Tx_{n-2},Tx_{n}) \\& \hphantom{p(x_{n-1},x_{n+1})}\leq\varphi\bigl(\max\bigl\{ p(x_{n-2},x_{n}), p(x_{n-2},x_{n-1}),p(x_{n}, x_{n+1}),p(x_{n-2},x_{n+1}), \\& \hphantom{p(x_{n-1},x_{n+1})={}}p(x_{n},x_{n-1}),p(x_{n},x_{n-2}),p(x_{n-1},x_{n-2}), \\& \hphantom{p(x_{n-1},x_{n+1})={}}p(x_{n+1},x_{n}),p(x_{n+1}, x_{n-2}),p(x_{n-1},x_{n}) \bigr\} \bigr), \end{aligned}
(3.24)
\begin{aligned}& p(x_{n+1}, x_{n-1})= p(Tx_{n},Tx_{n-2}) \\& \hphantom{p(x_{n+1}, x_{n-1})}\leq\varphi\bigl(\max\bigl\{ p(x_{n},x_{n-2}), p(x_{n},x_{n+1}),p(x_{n-2},x_{n-1}),p(x_{n}, x_{n-1}), \\& \hphantom{p(x_{n+1}, x_{n-1})={}}p(x_{n-2},x_{n+1}),p(x_{n-2},x_{n}),p(x_{n+1},x_{n}), \\& \hphantom{p(x_{n+1}, x_{n-1})={}}p(x_{n-1},x_{n-2}),p(x_{n-1},x_{n}),p(x_{n+1}, x_{n-2}) \bigr\} \bigr), \end{aligned}
(3.25)
\begin{aligned}& p(x_{n}, x_{n})= p(Tx_{n-1},Tx_{n-1}) \\& \hphantom{p(x_{n}, x_{n})}\leq\varphi\bigl(\max\bigl\{ p(x_{n-1},x_{n-1}), p(x_{n-1},x_{n}),p(x_{n},x_{n-1}) \bigr\} \bigr). \end{aligned}
(3.26)

Substituting (3.21)(3.26) into (3.20), proceeding in this manner and by hypotheses, (1), (2), and (3) of Lemma 3.3, we have

\begin{aligned} p(x_{n}, x_{n+1}) \leq&\varphi\bigl(\max\bigl\{ p(x_{i},x_{j}): n-1 \leq i,j \leq n+1\bigr\} \bigr) \\ \leq&\varphi^{2} \bigl(\max\bigl\{ p(x_{i},x_{j}): n-2 \leq i,j \leq n+1\bigr\} \bigr) \\ &\vdots \\ \leq&\varphi^{n-1} \bigl(\max\bigl\{ p(x_{i},x_{j}): 1 \leq i,j \leq n+1\bigr\} \bigr) \\ \leq&\varphi^{n-1}\bigl(\delta\bigl(O(x,x, \infty)\bigr)\bigr) \\ \leq&\varphi^{n-1}\bigl((I-\varphi)^{-1}\bigl(a(x)\bigr) \bigr), \end{aligned}
(3.27)

where $a(x)=4[p(x,x)+p(x, Tx) +p(Tx,x)]$.

If $n< m$, then, by (3.27),

\begin{aligned} p(x_{n}, x_{m}) \leq& p(x_{n},x_{n+1})+p(x_{n+1},x_{n+2})+ \cdots+p(x_{m-1},x_{m}) \\ =& \sum_{k=n}^{m-1} p(x_{k},x_{k+1}) \\ \leq&\sum_{k=n}^{m-1} \varphi^{k-1} \bigl((I-\varphi)^{-1}\bigl(a(x)\bigr)\bigr) \\ \leq& \sum_{k=n-1}^{m} \varphi^{k} \bigl((I-\varphi)^{-1}\bigl(a(x)\bigr)\bigr). \end{aligned}
(3.28)

Combining (3.19) and (3.28), we get

$$\lim_{n\rightarrow\infty} \sup\bigl\{ p(x_{n}, x_{m}): m>n\bigr\} =0.$$
(3.29)

By means of Lemma 2.4 and (3.29),

$$\{x_{n}\} \mbox{ is a Cauchy sequence}, \textit{i.e.}, \bigl\{ T^{n} x \bigr\} \mbox{ is a Cauchy sequence}$$

for each $x \in X$. This is the proof of (4).

To prove (5), let $x,y \in X$ and define $x_{n} = T^{n} x$ and $y_{n}=T^{n} y$ for every $n \in N$. By the same method as in (3.18)(3.27), we get

\begin{aligned} p(x_{n}, y_{n}) \leq&\varphi\bigl(\max\bigl\{ p(x_{i},x_{j}),p(x_{i},y_{j}),p(y_{i},x_{j}),p(y_{i},y_{j}) \mid n-1 \leq i,j \leq n\bigr\} \bigr) \\ \leq&\varphi^{2} \bigl(\max\bigl\{ p(x_{i},x_{j}),p(x_{i},y_{j}),p(y_{i},x_{j}),p(y_{i},y_{j}) \mid n-2 \leq i,j \leq n\bigr\} \bigr) \\ &\vdots \\ \leq&\varphi^{n-1} \bigl(\max\bigl\{ p(x_{i},x_{j}),p(x_{i},y_{j}),p(y_{i},x_{j}),p(y_{i},y_{j}) \mid1 \leq i,j \leq n\bigr\} \bigr) \\ \leq&\varphi^{n-1}\bigl(\delta\bigl(O(x,y,n)\bigr)\bigr) \\ \leq&\varphi^{n-1}\bigl((I-\varphi)^{-1}\bigl(b(x,y)\bigr) \bigr), \end{aligned}
(3.30)

which proves (5).

By virtue of (3.19) and (3.30), we deduce that

$$\lim_{n \rightarrow\infty} p\bigl(T^{n} x, T^{n}y\bigr)=0$$
(3.31)

for each $x,y\in X$. This is the proof of (6). □

### Definition 3.4

Let $(X,d)$ be a metric space, a mapping $T:X\rightarrow X$ is called Ćirić type φ-generalized single-valued p-contractive if it satisfies the following:

1. (c1)

There exist a u-distance p on X and $\varphi: [0, \infty) \rightarrow[0, \infty)$ such that

\begin{aligned} p(Tx,Ty) \leq&\varphi\bigl(\max\bigl[p(x,y),p(x,Tx),p(y,Ty),p(x,Ty),p(y,Tx), \\ &p(y,x),p(Tx,x),p(Ty,y),p(Ty,x),p(Tx,y)\bigr]\bigr) \end{aligned}

for all $x,y \in X$.

2. (c2)

For each $x \in X$ with $\lim_{n \rightarrow\infty} T^{n}x=c_{x} \in X$, there exists $y \in X$ such that $\lim_{n \rightarrow \infty} T^{n}y=Tc_{x}$.

### Theorem 3.5

Let $(X,d)$ be a complete metric space with a u-distance p. Let $T:X\rightarrow X$ be Ćirić type φ-generalized single-valued p-contractive satisfying (ii)(iv) of Lemma  3.3. Then:

1. (1)

$\lim_{n \rightarrow\infty} T^{n} x=z$ for each $x \in X$.

2. (2)

$p(T^{n} x, z) \leq\sum_{k=n-1}^{\infty} \varphi^{k} ((I-\varphi)^{-1} (a(x)))$ for each $x \in X$, where $a(x) = [p(x,x)+p(x,Tx)+p(Tx,x)] \times4$.

3. (3)

T has a unique fixed point z in X and $p(z,z)=0$.

### Proof

Let $x,y \in X$ and let $x_{n} = T^{n} x$ and $y_{n}=T^{n} y$ for every $n \in N$. Then, by (4) of Lemma 3.3, $\{ x_{n} \}$ is a Cauchy sequence.

Since X is complete, $\{ x_{n} \}$ converges to some $z \in X$. This is the proof of (1). Due to (3.28), (iv) of Lemma 3.3, Lemma 2.4, Definition 2.1, and (u3), we have

$$p(x_{n},z) \leq\lim_{m \rightarrow\infty}\inf p(x_{n},x_{m}) \leq\sum_{k=n-1}^{\infty} \varphi^{k} \bigl((I-\varphi)^{-1} \bigl(a(x)\bigr)\bigr),$$

which proves (2).

By (1) and (c2) of Definition 3.4, there exists $y \in X$ such that

$$\lim_{n \rightarrow\infty} T^{n} y= Tz.$$
(3.32)

In view of (3.29) and (3.31), we get

\begin{aligned}& \lim_{n \rightarrow\infty} \sup\bigl\{ \sup\bigl[p \bigl(T^{n} x, T^{m} y\bigr):m>n\bigr] \bigr\} \\& \quad \leq\lim_{n \rightarrow\infty} \sup\bigl\{ \sup\bigl[p \bigl(T^{n} x, T^{m} x\bigr)+ p\bigl(T^{m} x, T^{m} y\bigr):m>n\bigr] \bigr\} \\& \quad \leq\lim_{n \rightarrow\infty} \sup\bigl\{ \sup\bigl[p \bigl(T^{n} x, T^{m} x\bigr):m>n\bigr] \bigr\} + \lim _{n \rightarrow\infty} \sup\bigl\{ \sup\bigl[p\bigl(T^{m} x, T^{m} y\bigr):m>n\bigr] \bigr\} \\& \quad =0. \end{aligned}
(3.33)

Due to (3.33), we obtain

$$\lim_{n \rightarrow\infty} \sup\bigl\{ p\bigl(T^{n} x, T^{m} y\bigr):m>n \bigr\} =0.$$
(3.34)

In terms of (3.29), (3.34), and Lemma 3.1, there exist a subsequence $\{x_{k_{n}}\}$ of $\{x_{n}\}$ and a subsequence $\{y_{k_{n}}\}$ of $\{y_{n}\}$ such that

$$\lim_{n \rightarrow\infty}d(x_{k_{n}} , y_{k_{n}})=0.$$
(3.35)

From (1), (3.32), and (3.35), we have

$$d(z,Tz)=0.$$

Thus z is a fixed point of T.

To prove the unique fixed point of T, let $z=Tz$ and $w=Tw$.

Then, by hypothesis, we obtain

\begin{aligned} &p(w,z)= p(Tw,Tz)\leq\varphi\bigl(\max \bigl\{ p(w,z), p(w,w), p(z,z), p(z,w)\bigr\} \bigr), \\ &p(z,w)= p(Tz,Tw)\leq\varphi\bigl(\max\bigl\{ p(w,z), p(w,w), p(z,z), p(z,w) \bigr\} \bigr), \\ &p(z,z) = p(Tz,Tz)\leq\varphi\bigl(\max\bigl\{ p(w,z), p(w,w), p(z,z), p(z,w) \bigr\} \bigr), \\ &p(w,w)= p(Tw,Tw)\leq\varphi\bigl(\max\bigl\{ p(w,z), p(w,w), p(z,z), p(z,w) \bigr\} \bigr). \end{aligned}
(3.36)

By (3.36) and the hypothesis

$$\max\bigl\{ p(w,z), p(w,w), p(z,z), p(z,w)\bigr\} =0 .$$
(3.37)

From Lemma 2.3 and (3.37), we have

$$w=z.$$

□

From Theorem 3.5, we have the following corollary.

### Corollary 3.6

Let $(X,d)$ be a complete metric space with a u-distance p on X. Let $T:X \rightarrow X$ be a mapping that satisfies the following conditions:

\begin{aligned} \begin{aligned}[b] (1)&\quad p(Tx,Ty) \leq k \bigl(\max \bigl[p(x,y),p(x,Tx),p(y,Ty),p(x,Ty),p(y,Tx), \\ &\hphantom{\quad p(Tx,Ty) \leq{}}p(y,x),p(Tx,x),p(Ty,y),p(Ty,x),p(Tx,y)\bigr]\bigr) \end{aligned} \end{aligned}
(3.38)

for all $x, y \in X$ and for some $k \in(0,1)$;

\begin{aligned} (2)&\quad \textit{for each }x \in X\textit{ with }\lim_{n \rightarrow\infty} T^{n}x=c_{x} \in X, \textit{there exists }y \in X, \\ \hphantom{(2)}&\quad \textit{such that }\lim_{n \rightarrow \infty} T^{n}y=Tc_{x}. \end{aligned}

Then T has a unique fixed point z in X and $p(z,z)=0$.

### Proof

Let $\varphi: R_{+} \rightarrow R_{+}$ be defined by

$$\varphi{(t)} = kt, \quad 0< k< 1.$$
(3.39)

Then, by (3.39), all the conditions of Theorem 3.5 are satisfied.

Thus T has a unique fixed point z in X and $p(z,z)=0$. □

### Lemma 3.7

Let $(X,d)$ be a complete metric space with a u-distance p on X and let $T:X \rightarrow X$ be a mapping satisfying (3.38) and

$$\inf\bigl\{ p(x,y) + p(x,Tx) : x \in X \bigr\} >0$$
(3.40)

for every $y \in X$ with $y \neq Ty$.

Then, for each $x \in X$ with $\lim_{n \rightarrow\infty} T^{n}x=c_{x} \in X$, there exists $y \in X$ such that $\lim_{n \rightarrow\infty} T^{n}y=Tc_{x}$.

### Proof

Suppose that there exists some $x \in X$ with $\lim_{n \rightarrow\infty} T^{n}x=c_{x} \in X$ such that

$$\lim_{n \rightarrow\infty} T^{n}y\neq Tc_{x}\quad \mbox{for all } y \in X .$$
(3.41)

From (3.41) we get

$$\lim_{n \rightarrow\infty} T^{n} x=c_{x} \in X \quad \mbox{and}\quad \lim_{n \rightarrow\infty} T^{n} (Tx)= \lim_{n \rightarrow\infty} T^{n+1} x \neq Tc_{x} .$$
(3.42)

Then, by (3.42), the same method as in Theorem 3.5 and simple calculations, we have

$$c_{x} \neq Tc_{x},\quad \lim_{n \rightarrow\infty} p\bigl(T^{n} x, c_{x}\bigr) =0 \quad \mbox{and} \quad \lim _{n \rightarrow\infty} p\bigl(T^{n} x,T^{n+1} x\bigr)=0.$$
(3.43)

On account of (3.43) and the hypotheses of Lemma 3.7, we obtain

\begin{aligned} 0 < & \inf\bigl\{ p(x,c_{x}) + p(x,Tx) : x \in X \bigr\} \\ \leq&\inf\bigl\{ p\bigl(T^{n} x,c_{x}\bigr) + p \bigl(T^{n}x,T^{n+1}x\bigr) : n \in N \bigr\} =0 . \end{aligned}

From Corollary 3.6 and Lemma 3.7 we have the following corollary.

### Corollary 3.8

([15])

Let $(X,d)$ be a complete metric space with a u-distance p on X. Let $T:X\rightarrow X$ be a mapping satisfying (3.38) and (3.40). Then T has a unique fixed point z in X and $p(z,z)=0$.

### Proof

Since all the conditions of Corollary 3.8 satisfy all the conditions of Corollary 3.6, we obtain result of Corollary 3.8. □

In the next example we shall show that all the conditions of Theorem 3.5 are satisfied, but condition (3.38) in Corollary 3.6 and condition (3.40) in Lemma 3.7 are not satisfied.

### Example 3.9

Let $k \in(0,1)$ and let $X=[0,1]$ be closed interval with the usual metric, and $p:X \times X \rightarrow R_{+}$, $T:X \rightarrow X$ and $\varphi: R_{+} \rightarrow R_{+}$ be mappings defined as follows:

\begin{aligned}& p(x,y)=\biggl(\frac{1-k}{1+k}\biggr)x, \end{aligned}
(3.44)
\begin{aligned}& Tx=\biggl(\frac{1+k}{2}\biggr)x, \end{aligned}
(3.45)
\begin{aligned}& \varphi(t)= \left \{ \textstyle\begin{array}{l@{\quad}l} ( \frac{1+k}{2})t, & 0 \leq t \leq\frac{1-k}{1+k} , \\ \frac{t}{1+t}, & \frac{1-k}{1+k}< t. \end{array}\displaystyle \displaystyle \right . \end{aligned}
(3.46)

Define $\theta: X \times X \times R_{+} \times R_{+} \rightarrow R_{+}$ by

$$\theta(x,y,s,t)=s$$
(3.47)

for all $x,y \in X$ and $s,t \in R_{+}$.

Then, by (3.44)(3.47) and simple calculations, we know that p is a u-distance on X and φ satisfies (ii) and (iii) in Lemma 3.3. We now show that φ satisfies (iv) in Lemma 3.3.

On account of (3.46), if $0 \leq t \leq\frac{1-k}{1+k}$, then $\varphi^{n} (t) = (\frac{1+k}{2})^{n} t$ for all $n \in N$ and so (iv) holds for all $t \in [0,\frac{1-k}{1+k}]$.

If $t > \frac{1-k}{1+k}$, then there exists $M \in N$ such that

$$\varphi^{M} (t) \leq\frac{1-k}{1+k} .$$
(3.48)

Suppose that $\varphi^{n} (t)> \frac{1-k}{1+k}$ for all $n \in N$ and $t \in(\frac{1-k}{1+k}, \infty)$.

Then, by (3.46), $\frac{1-k}{1+k} < \varphi^{n}(t) = \frac{t}{1+nt}$ for all $n \in N$ and $t \in(\frac{1-k}{1+k}, \infty)$. Thus $0< \frac{1-k}{1+k} \leq \lim_{n \rightarrow \infty} \varphi^{n}(t) = \lim_{n \rightarrow\infty} \frac{t}{1+nt} =0$, a contradiction.

Hence (3.48) holds.

By virtue of (3.46) and (3.48), we get

$$\varphi^{n} (t)= \varphi^{n-M} \bigl(\varphi^{M}(t) \bigr)= \biggl(\frac{1+k}{2}\biggr)^{n-M}\cdot\varphi^{M} (t)$$

for all $n \in N$ with $n >M$. Thus φ satisfies (iv) in Lemma 3.3 for $t \in(\frac{1-k}{1+k}, \infty)$. Therefore (iv) in Lemma 3.3 holds. Using (3.44)(3.46), we have

\begin{aligned}& \varphi\bigl(\max\bigl[p(x,y),p(x,Tx),p(y,Ty),p(x,Ty),p(y,Tx), \\& \qquad p(y,x),p(Tx,x),p(Ty,y),p(Ty,x),p(Tx,y)\bigr]\bigr) \\& \quad = \varphi\biggl(\max\biggl[\biggl(\frac{1-k}{1+k}\biggr)x, \biggl( \frac{1-k}{1+k}\biggr)y\biggr]\biggr), \\& p(Tx,Ty)=\biggl(\frac{1-k}{1+k}\biggr)Tx=\biggl(\frac{1-k}{1+k}\biggr) \biggl(\frac{1+k}{2}\biggr)x=\biggl(\frac{1-k}{2}\biggr)x \quad \mbox{and} \\& \varphi \biggl(\biggl(\frac{1-k}{1+k}\biggr)x\biggr)=\biggl(\frac{1+k}{2} \biggr) \biggl(\frac{1-k}{1+k}\biggr)x=\biggl(\frac{1-k}{2}\biggr)x \end{aligned}
(3.49)

for all $x, y \in X$.

By (3.49), (c1) of Definition 3.4 is satisfied.

Due to (3.45), since $\lim_{n \rightarrow\infty} T^{n} x = \lim_{n \rightarrow\infty} (\frac{1+k}{2})^{n} x=0$ for each $x \in X$, there exists $y = \frac{x}{2} \in X$ such that $\lim_{n \rightarrow\infty} T^{n} y =\lim_{n \rightarrow\infty} (\frac{1+k}{2})^{n} \cdot\frac{x}{2}=0=T0$.

This implies (c2) of Definition 3.4.

Therefore all the conditions of Theorem 3.5 are satisfied.

By means of (3.44) and (3.45), there exist $a=1 \in X$ and $b= 0 \in X$ such that

\begin{aligned}& k \cdot\max\bigl\{ p(a,b),p(a,Ta),p(b,Tb),p(a,Tb),p(b,Ta), \\& \qquad p(b,a),p(Ta,a),p(Tb,b),p(Tb,a),p(Ta,b)\bigr\} \\& \quad = k \cdot\biggl(\frac{1-k}{1+k}\biggr) \quad \mbox{and} \\& p(Ta,Tb)= \biggl(\frac{1-k}{1+k}\biggr)Ta=\biggl(\frac{1-k}{1+k}\biggr) \biggl( \frac{1+k}{2}\biggr)=\biggl(\frac {1-k}{2}\biggr)\quad \mbox{and} \\& \frac{1-k}{2}> k\biggl(\frac{1-k}{1+k}\biggr) \quad \mbox{for } k \in (0,1). \end{aligned}
(3.50)

On account of (3.50), (3.38) in Corollary 3.6 is not satisfied.

In terms of (3.44) and (3.45) we obtain

\begin{aligned} 0 \leq&\inf\bigl\{ p(x,y) + p(x,Tx) : x \in X \bigr\} \\ \leq&\inf\bigl\{ p\bigl(T^{n} x, y\bigr) + p\bigl(T^{n} x, T^{n+1} x\bigr): n \in N \bigr\} \\ =&\inf\biggl\{ \biggl(\frac{1-k}{1+k}\biggr)T^{n} x + \biggl( \frac{1-k}{1+k}\biggr)T^{n} x: n \in N \biggr\} \\ =&\inf\biggl\{ 2 \biggl(\frac{1-k}{1+k}\biggr)\cdot\biggl(\frac{1+k}{2} \biggr)^{n} x : n \in N \biggr\} =0 \end{aligned}

for all $y \in X$ with $y \neq Ty$. This means that (3.40) in Lemma 3.7 is not satisfied.

### Remark 3.10

It follows from Lemma 3.7 and Example 3.9 that Theorem 3.5 is a proper extension of Corollary 3.6 and Corollary 3.8, the results of Ćirić [5], Kannan [12] and Ume [15].

The following theorem is a generalization of Suzuki’s fixed point theorem [18].

### Theorem 3.11

Let $(X,d)$ be a complete metric space with a u-distance p on X. Let $\varphi: R_{+}\rightarrow R_{+}$ be a mapping satisfying conditions (ii)(iv) of Lemma  3.3.

Let $T: X \rightarrow X$ be a mapping that satisfies the following conditions:

\begin{aligned} \begin{aligned} (\mathrm{i})&\quad p\bigl(Tx, T^{2} x\bigr) \leq\varphi \bigl(p(x,Tx)\bigr)\quad \textit{for all } x \in X; \\ (\mathrm{ii})&\quad \textit{If } \lim_{n \rightarrow\infty} \sup\bigl\{ p(x_{n}, x_{m}):m>n \bigr\} =0, \lim_{n \rightarrow\infty}p(x_{n}, Tx_{n})=0 \textit{ and} \\ \hphantom{(\mathrm{ii})}&\quad\lim_{n \rightarrow\infty}p(x_{n}, y) =0, \textit{then } Ty=y. \end{aligned} \end{aligned}
(3.51)

Then there exists $z \in X$ such that $Tz=z$ and $p(z,z)=0$.

### Proof

By (3.51), the same methods in Theorem 3.5 and simple calculations, we deduce that

\begin{aligned} &\lim_{n \rightarrow\infty} \sup\bigl\{ p\bigl(T^{n} x, T^{m} x\bigr):m>n \bigr\} =0,\qquad \lim _{n \rightarrow\infty} T^{n} x=z, \\ &\lim_{n \rightarrow\infty}p\bigl(T^{n} x, z\bigr)=0\quad \mbox{and} \quad \lim_{n \rightarrow\infty} p\bigl(T^{n} x, T^{n+1} x \bigr) =0 . \end{aligned}
(3.52)

By means of (3.52) and hypotheses (i), (ii), we obtain

$$Tz=z \quad \mbox{and}\quad p(z,z)=0 \quad \mbox{and} \quad z \mbox{ is a unique fixed point of }T.$$

□

### Corollary 3.12

([18])

Let $(X,d)$ be a complete metric space with a τ-distance p on X. Let $T: X \rightarrow X$ be a mapping satisfying (ii) of Theorem  3.11 and

$$p\bigl(Tx, T^{2} x\bigr) \leq k p(x,Tx)$$

for all $x \in X$ and some $k \in(0,1)$.

Then T has a unique fixed point z in X and $p(z,z)=0$.

### Proof

Let $\varphi: R_{+}\rightarrow R_{+}$ be defined by

$$\varphi(t) =kt,\quad 0< k< 1.$$

Since p is a τ-distance, p is a u-distance. Thus all the conditions of Theorem 3.11 are satisfied.

Therefore T has a unique fixed point z in X and $p(z,z)=0$. □

## Existence of a solution for an integral equation

In what follows, we assume that $X=C([0,1])$ is the set of all continuous functions defined on $[0,1]$ and $\varphi: R_{+} \rightarrow R_{+}$ satisfy conditions (ii), (iii), and (iv) of Lemma 3.3. Let $d,p:X \times X \rightarrow R_{+}$ and $\theta:X \times X \times R_{+} \times R_{+} \rightarrow R_{+}$ be mappings defined as follows:

$$d(x,y)=\sup_{t \in[0,1]} \bigl\vert x(t)-y(t)\bigr\vert , \qquad p(x,y)=\sup_{t \in [0,1]} \bigl\vert x(t)\bigr\vert$$

and

$$\theta(x,y,s,t)=s$$

for all $x,y \in X$ and $s,t \in R_{+}$. Then clearly $(X,d)$ is a complete metric space and p is a u-distance on X. Now we prove the existence theorem for a solution of the following integral equation by using Theorem 3.5:

$$x(t)=r(x,t)+ \int_{0}^{1} G(t,s)f \bigl(s,x (s)\bigr)\, ds,$$
(4.1)

where $x \in X$, $r: X \times R \rightarrow R$, $G: [0,1] \times[0,1] \rightarrow R$ and $f : [0,1] \times R \rightarrow R$ are given mappings.

### Theorem 4.1

Suppose that the following hypotheses hold:

(I1):

$r: X \times R \rightarrow R$ is a continuous mapping such that

$$\bigl\vert r(x,t)\bigr\vert \leq{{1}\over {2}} \varphi\bigl(\bigl\vert x(t)\bigr\vert \bigr) \quad \textit{for all } x \in X \textit{ and } t \in R .$$
(I2):

$G: [0,1] \times[0,1] \rightarrow R$ is a continuous mapping such that

$$\bigl\vert G(t,s)\bigr\vert \leq{{1}\over {2}} \quad \textit{for all } t,s \in [0,1].$$
(I3):

$f: [0,1] \times R \rightarrow R$ is a continuous mapping such that

$$\bigl\vert f\bigl(s,x(s)\bigr)\bigr\vert \leq \varphi\bigl(\bigl\vert x(s) \bigr\vert \bigr) \quad \textit{for all } x \in X \textit{ and } s \in [0,1] .$$
(I4):

For each $x \in X$ with $\lim_{n \rightarrow\infty} {T^{n} x=c_{x} \in X}$, there exists $y \in X$ such that $\lim_{n \rightarrow\infty} T^{n} y=Tc_{x}$.

Then the integral equation (4.1) has a solution $x \in X$.

### Proof

Let $T: X\rightarrow X$ be a mapping defined by

$$(Tx ) (t)=r(x,t)+ \int_{0}^{1} {G(t,s)f \bigl(s,x(s)\bigr)\, ds}$$

for all $x \in X$ and $t \in[0,1]$. By conditions (I1), (I2), and (I3), we have

\begin{aligned} \bigl\vert (Tx ) (t)\bigr\vert =& \biggl\vert r(x,t)+ \int _{0}^{1} G(t,s)f\bigl(s,x(s)\bigr)\, ds \biggr\vert \\ \leq&\bigl\vert r(x,t)\bigr\vert +\biggl\vert \int_{0}^{1} {G(t,s)f\bigl(s,x(s)\bigr)\, ds} \biggr\vert \\ \leq&\bigl\vert r(x,t)\bigr\vert + \int_{0}^{1} \bigl\vert G(t,s)\bigr\vert \bigl\vert f\bigl(s,x(s)\bigr)\bigr\vert \, ds \\ \leq&\bigl\vert r(x,t)\bigr\vert + \frac{1}{2} \int _{0}^{1} {\bigl\vert f\bigl(s,x(s)\bigr)\bigr\vert \, ds} \\ \leq&\bigl\vert r(x,t)\bigr\vert + \frac{1}{2} \int _{0}^{1} {\varphi\bigl(\bigl\vert x(s)\bigr\vert \bigr)\, ds} \\ \leq&\frac{1}{2} \varphi\bigl(\bigl\vert x(t)\bigr\vert \bigr)+ \frac{1}{2}\int_{0}^{1} {\varphi\Bigl(\sup _{t \in[0,1]} \bigl\vert x(t)\bigr\vert \Bigr)\, ds} \\ \leq&\frac{1}{2} \varphi\bigl(\bigl\vert x(t)\bigr\vert \bigr)+ \frac{1}{2} {\varphi\Bigl(\sup_{t \in[0,1]} \bigl\vert x(t) \bigr\vert \Bigr)} \end{aligned}

for all $x \in X$ and $t \in[0,1]$. Then

\begin{aligned} p(Tx, Ty) =& \sup_{t \in[0,1]}\bigl\vert (Tx) (t)\bigr\vert \leq \sup_{t \in[0,1]}\biggl\{ \frac {1}{2} \varphi\bigl(\bigl\vert x(t)\bigr\vert \bigr)+ \frac{1}{2} {\varphi\Bigl(\sup _{t \in [0,1]}\bigl\vert x(t)\bigr\vert \Bigr)} \biggr\} \\ \leq& \frac{1}{2} \varphi\Bigl(\sup_{t \in[0,1]}\bigl\vert x(t)\bigr\vert \Bigr)+ \frac{1}{2} {\varphi\Bigl(\sup _{t \in [0,1]}\bigl\vert x(t)\bigr\vert \Bigr)} \\ =& \varphi\Bigl(\sup_{t \in[0,1]} \bigl\vert x(t)\bigr\vert \Bigr) \\ \leq& \varphi \Bigl(\max\Bigl\{ \sup_{t \in[0,1]}\bigl(\bigl\vert x(t)\bigr\vert \bigr), \sup_{t \in[0,1]}\bigl\vert y(t)\bigr\vert , \sup_{t \in[0,1]}\bigl\vert (Tx) (t)\bigr\vert , \sup_{t \in[0,1]}\bigl\vert (Ty) (t)\bigr\vert \Bigr\} \Bigr) \\ =& \varphi\bigl( \max\bigl\{ p(x, y), p(x, Tx), p(y, Ty), p(x, Ty), p(y, Tx), \\ &p( y,x), p(Tx, x), p(Ty, y), p( Ty,x), p( Tx,y) \bigr\} \bigr) \end{aligned}

for all $x,y \in X$.

Thus all of the hypotheses of Theorem 3.5 are satisfied. Hence the mapping T has a fixed point that is a solution in $X=C([0,1])$ of the integral equation (4.1). □

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## Acknowledgement

This research was supported by Basic Science Research Program through the National Research Foundation of Korea (NRF) funded by the Ministry of Science, ICT & Future Planning (2013R1A1A2057665).

## Author information

Correspondence to Jeong Sheok Ume.

### Competing interests

The author declares that he has no competing interests.

### Author’s contributions

The author completed the paper himself. The author read and approved the final manuscript.