Fixed point theorems for Kannan-type maps

Abstract

We introduce the new classes of Kannan-type maps with respect to u-distance and prove some fixed point theorems for these mappings. Then we present several examples to illustrate the main theorems.

1 Introduction

A mapping T on a metric space $(X, d)$ is called Kannan if there exists $\alpha\in[0, \frac{1}{2})$ such that

$$ d(Tx, Ty) \leq\alpha d(x,Tx) + \alpha d(y, Ty) $$

(1.1)

for all $x, y \in X$. Kannan [1] proved that if X is complete, then a Kannan mapping has a fixed point. It is interesting that Kannan’s theorem is independent of the Banach contraction principle [2]. Also, Kannan’s fixed point theorem is very important because Subrahmanyam [3] proved that Kannan’s theorem characterizes the metric completeness. That is, a metric space X is complete if and only if every Kannan mapping on X has a fixed point.

Using the concept of Hausdorff metric, Nadler [4] proved the fixed point theorem for multi-valued contraction maps, which is a generalization of the Banach contraction principle [2]. Since then various fixed point results concerning multi-valued contractions have appeared; for example, see [5–7] and the references cited there.

Without using the concept of Hausdorff metric, most recently Dehaish and Latif [8] generalized fixed point theorems of Latif and Abdou [9], Suzuki [10], Suzuki and Takahashi [11].

In 1996, Kada et al. [12] introduced the notion of w-distance and improved several classical results including Caristi’s fixed point theorem. Suzuki and Takahashi [11] introduced single-valued and multi-valued weakly contractive maps with respect to w-distance and proved fixed point results for such maps. Generalizing the concept of w-distance, in 2001, Suzuki [10] introduced the notion of τ-distance on a metric space and improved several classical results including the corresponding results of Suzuki and Takahashi [11]. In 2010, Ume [13] introduced the new concept of a distance called u-distance, which generalizes w-distance, Tataru’s distance and τ-distance. Then he proved a new minimization theorem and a new fixed point theorem by using u-distance on a complete metric space.

Distances in uniform spaces were given by Vályi [14]. More general concepts of distances were given by Wlodarczyk and Plebaniak [15–18] and Wlodarczyk [19].

In this paper, we introduce the new classes of Kannan-type multi-valued maps without using the concept of Hausdorff metric and Kannan-type single-valued maps with respect to u-distance and prove some fixed point theorems for these mappings. Then we present several examples to illustrate the main theorems.

2 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.

Ume [13] introduced u-distance as follows: Let X be metric space with metric d. Then a function $p:X \times X \rightarrow R_{+}$ is called 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 $x, y,z \in X$:

($u_{1}$):

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

($u_{2}$):

$\theta(x, y, 0, 0) = 0$ and $\theta(x, y, s, t) \geq\min\{s,t\}$ for all $x,y\in X$ and $s, t \in R_{+}$, 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 $|\theta(x,y,s,t) -\theta(x,y,s_{0} , t_{0} ) |< \varepsilon$.

($u_{3}$):

$\lim_{n \rightarrow\infty} x_{n} =x$ 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 $y\in X$.

($u_{4}$):

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

($u_{5}$):

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

We recall remark, examples, definition and lemmas which will be useful in what follows.

Remark 2.1

([13])

(a) Suppose that θ from $X \times X \times R_{+} \times R_{+}$ into $R_{+}$ is a mapping satisfying $(u_{2}) {\sim}(u_{5})$. 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 $(u_{2})_{\eta}{\sim}(u_{5})_{\eta}$, where $(u_{2})_{\eta}{\sim}(u_{5})_{\eta}$ stand for substituting η for θ in $(u_{2}) {\sim}(u_{5})$, respectively.

(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 $(u_{2}) {\sim}(u_{5})$.

(c) Each τ-distance p on a metric space $(X,d)$ is also a u-distance on X.

We present some examples of u-distance which are not τ-distance (for details, see [13]).

Example 2.2

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 2.3

Let X be a normed space with $\|\cdot \|$, 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 examples and (c) of Remark 2.1 that u-distance is a proper extension of τ-distance.

Definition 2.4

([13])

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 $(u_{2}) {\sim}(u_{5})$ 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 2.5

([13])

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

Lemma 2.6

([13])

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

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

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

Lemma 2.7

([13])

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

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

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

3 Main result

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,

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

(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 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 ($u_{2}$), for this $a_{k_{1}} \in X$ and $\varepsilon=1$, there exists $\delta_{1} >0$ such that

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

(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 ($u_{2}$) 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 ($u_{4}$), 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 ($u_{5}$), 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, $2^{X}$ be a set of all nonempty subsets of X and $\operatorname{Cl}(X)$ be a set of all nonempty closed subsets of X. Let $T: X\rightarrow2^{X}$. Then an element $z \in X$ is a fixed point of T if $z \in Tz$.

A mapping $T: X\rightarrow2^{X}$ is called Kannan-type multi-valued p-contractive mapping if there exist a u-distance p on X and $r\in[0, \frac{1}{2})$ such that

• (i) $p(y_{1}, y_{2})\leq r[p(x_{1}, y_{1})+p(x_{2}, y_{2})]$ for any $x_{1}, x_{2} \in X$, $y_{1} \in Tx_{1}$ and $y_{2} \in Tx_{2}$,

• (ii) $Ty \subseteq Tx$ for all $x,y \in X$ with $y \in Tx$.

In the next example we shall show that if $(X,d)$ is a complete metric space with a u-distance p and a mapping $T:X\rightarrow \operatorname{Cl}(X)$ is not Kannan-type multi-valued p-contractive, in general, T may have no fixed point in X.

Example 3.3

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

$$\begin{aligned}& p(x,y) =\left \{ \begin{array}{l@{\quad}l} 2, & x=0, \\ x, & x\neq0, \end{array} \right . \end{aligned}$$

(3.18)

$$\begin{aligned}& Tx= \left \{ \begin{array}{l@{\quad}l} \{\frac{1}{4}\}, & x=0, \\ {[ \frac{x}{8(1+x)}, \frac{x}{4(1+x)} ]} , & x \neq0. \end{array} \right . \end{aligned}$$

(3.19)

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

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

(3.20)

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

From (3.18) and (3.20) easily we can obtain that p is a u-distance on X.

In terms of (3.18) and (3.19), we have

$$ p(y_{1},y_{2})\leq\frac{1}{4} \bigl[p(x_{1}, y_{1}) + p(x_{2}, y_{2}) \bigr] $$

(3.21)

for all $x_{1},x_{2} \in X$, $y_{1} \in Tx_{1}$ and $y_{2} \in Tx_{2}$.

To show that (3.21) is satisfied, we need to consider several possible cases.

$$\begin{aligned}& \begin{aligned} &\mbox{Case 1. Let } x_{1}=x_{2}=0. \mbox{ Then }y_{1} \in Tx_{1}=\biggl\{ \frac{1}{4} \biggr\} , y_{2} \in Tx_{2}=\biggl\{ \frac{1}{4} \biggr\} , \\ &p(y_{1},y_{2})=y_{1}=\frac{1}{4}, p(x_{1},y_{1})=2 \mbox{ and }p(x_{2}, y_{2})=2 \mbox{ and } \\ &\frac{1}{4} \bigl[p(x_{1},y_{1})+p(x_{2},y_{2}) \bigr]=\frac{1}{4}[2+2]=1 \geq \frac{1}{4} = p(y_{1},y_{2}). \mbox{ Thus } \\ &p(y_{1},y_{2}) \leq\frac{1}{4}\bigl[p(x_{1},y_{1})+p(x_{2},y_{2}) \bigr]. \end{aligned} \end{aligned}$$

(3.22)

$$\begin{aligned}& \begin{aligned} &\mbox{Case 2. Let } x_{1}=0 \mbox{ and } x_{2} \neq0. \mbox{ Then } y_{1} \in Tx_{1}= \biggl\{ \frac{1}{4}\biggr\} , \\ &y_{2} \in Tx_{2}=\biggl[\frac{x_{2}}{8(1+x_{2})}, \frac{x_{2}}{4(1+x_{2})}\biggr], p(y_{1},y_{2})=y_{1}= \frac{1}{4}, \\ &p(x_{1},y_{1})=2\mbox{ and }p(x_{2},y_{2})=x_{2}. \mbox{ Thus } \\ &\frac{1}{4}\bigl[p(x_{1},y_{1})+p(x_{2},y_{2}) \bigr]=\frac{1}{4}[2+x_{2}] \geq \frac{1}{4} = p(y_{1},y_{2}). \end{aligned} \end{aligned}$$

(3.23)

$$\begin{aligned}& \begin{aligned} &\mbox{Case 3. Let } x_{1} \neq0\mbox{ and }x_{2}=0. \mbox{ Then } y_{1} \in Tx_{1}= \biggl[\frac{x_{1}}{8(1+x_{1})},\frac{x_{1}}{4(1+x_{1})} \biggr], \\ &y_{2} \in Tx_{2}=\biggl\{ \frac{1}{4}\biggr\} , p(y_{1},y_{2})=y_{1} \leq \frac{x_{1}}{4(1+x_{1})}, p(x_{1},y_{1})=x_{1} \\ &\mbox{and } p(x_{2},y_{2})=2.\mbox{ Thus } \\ &\frac{1}{4}\bigl[p(x_{1},y_{1})+p(x_{2},y_{2}) \bigr]=\frac{1}{4}[x_{1}+2] \geq \frac{x_{1}}{4(1+x_{1})} \geq p(y_{1},y_{2}). \end{aligned} \end{aligned}$$

(3.24)

$$\begin{aligned}& \begin{aligned} &\mbox{Case 4. Let } x_{1} \neq0\mbox{ and } x_{2} \neq0. \mbox{ Then } y_{1} \in Tx_{1}=\biggl[\frac{x_{1}}{8(1+x_{1})},\frac{x_{1}}{4(1+x_{1})}\biggr], \\ &y_{2} \in Tx_{2}=\biggl[\frac{x_{2}}{8(1+x_{2})}\frac{x_{2}}{4(1+x_{2})} \biggr], p(y_{1},y_{2})=y_{1} \leq \frac{x_{1}}{4(1+x_{1})}, \\ &p(x_{1},y_{1})=x_{1} \mbox{ and } p(x_{2},y_{2})=x_{2}.\mbox{ Thus } \\ &\frac{1}{4}\bigl[p(x_{1},y_{1})+p(x_{2},y_{2}) \bigr]=\frac{1}{4}[x_{1}+x_{2}] \geq\max\biggl\{ \frac{x_{1}}{4(1+x_{1})},\frac{x_{2}}{4(1+x_{2})}\biggr\} \geq p(y_{1},y_{2}). \end{aligned} \end{aligned}$$

(3.25)

From (3.18)∼(3.25), we have

$$ p(y_{1},y_{2}) \leq \frac{1}{4} \bigl[p(x_{1},y_{1})+p(x_{2},y_{2}) \bigr] $$

(3.26)

for all $x_{1}, x_{2} \in X$, $y_{1} \in Tx_{1}$ and $y_{2} \in Tx_{2}$.

But there exist $x=0 \in X$ and $y=\frac{1}{4} \in X$ with $y \in Tx$ such that $Ty=T \frac{1}{4}= [ \frac{1}{40}, \frac{1}{20} ] \nsubseteq\{ \frac{1}{4} \} =T0$. Therefore T is not Kannan-type multi-valued p-contractive and T does not have a fixed point.

Using Lemma 3.1, we have the following main theorem.

Theorem 3.4

Let $(X,d)$ be a complete metric space and let $T: X \rightarrow \operatorname{Cl}(X)$ be a Kannan-type multi-valued p-contractive mapping. Then T has a unique fixed point in X.

Proof

Let $a_{1} \in X$ be arbitrary, $a_{2} \in Ta_{1}$ and $a_{3} \in Ta_{2}$ be chosen. Since T is Kannan-type p-contractive,

$$ p(a_{2},a_{3}) \leq r\bigl[p(a_{1},a_{2})+ p(a_{2},a_{3})\bigr], $$

(3.27)

where $r \in[0,\frac{1}{2})$.

From (3.27), we get

$$ p(a_{2},a_{3}) \leq k p(a_{1},a_{2}), $$

(3.28)

where $k=\frac{r}{1-r} \in[0,1)$.

By (3.27) and (3.28), we obtain a sequence $\{a_{n}\}$ in X such that

$$ a_{n+1}\in Ta_{n} \quad \mbox{and} \quad p(a_{n+1}, a_{n+2}) \leq k p(a_{n}, a_{n+1}) $$

(3.29)

for all $n \in N$.

By repeated application of (3.29), we have

$$ p(a_{n}, a_{n+1}) \leq k^{n-1}p(a_{1}, a_{2}) $$

(3.30)

for all $n \in N$.

Now we shall know that $\{a_{n}\}$ is a Cauchy sequence.

Let $n,m \in N$ be such that $n< m$. Then, by virtue of (3.30), we deduce

$$\begin{aligned} p(a_{n},a_{m}) \leq& p(a_{n},a_{n+1})+p(a_{n+1}, a_{n+2})+\cdots+p(a_{m-1},a_{m}) \\ =&\sum_{i=n}^{m-1} p(a_{i}, a_{i+1}) \leq\sum_{i=n}^{m-1} k^{i-1} p(a_{1},a_{2}) \\ \leq&\biggl(\frac{k^{n-1}}{1-k}\biggr)p(a_{1}, a_{2}). \end{aligned}$$

(3.31)

In view of (3.31), we get

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

(3.32)

On account of Lemma 2.5, Lemma 2.7 and (3.32), $\{a_{n}\}$ is a Cauchy sequence in X. Since X is complete, there exists $b_{1} \in X$ such that

$$ \lim_{n \rightarrow\infty} a_{n} =b_{1}. $$

(3.33)

By the same method as that in (3.27)∼(3.33), there exists a sequence $\{ b_{n} \}$ in X such that

$$ b_{n+1} \in Tb_{n}\quad \mbox{and}\quad p(b_{n}, b_{n+1}) \leq k^{n-1} p(b_{1}, b_{2}) $$

(3.34)

for all $n \in N$.

Combining the hypothesis, (3.27), (3.28), (3.30), (3.31) and (3.34), we have

$$\begin{aligned} p(a_{n}, b_{m}) \leq& p(a_{n}, a_{m}) + p(a_{m}, b_{m}) \\ \leq& p(a_{n}, a_{m})+r\bigl[p(a_{m-1}, a_{m}) + p(b_{m-1}, b_{m})\bigr] \\ \leq&\biggl(\frac{k^{n-1}}{1-k}\biggr)p(a_{1}, a_{2})+ r \bigl[k^{m-2}p(a_{1},a_{2})+k^{m-2}p(b_{1},b_{2}) \bigr] \\ \leq&\biggl(\frac{k^{n-1}}{1-k}\biggr)p(a_{1}, a_{2})+ k^{n-1}p(a_{1},a_{2})+k^{n-1}p(b_{1},b_{2}) \\ =&k^{n-1} \biggl\{ \biggl(\frac{1}{1-k}\biggr) p(a_{1},a_{2}) +p(a_{1},a_{2})+p(b_{1},b_{2}) \biggr\} \end{aligned}$$

(3.35)

for all $n,m \in N$ with $m>n$.

By (3.35), we have

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

(3.36)

Due to Lemma 3.1, (3.32) and (3.36), there exist a subsequence $\{a_{k_{n}}\}$ of $\{a_{n}\}$ and a subsequence $\{b_{k_{n}}\}$ of $\{b_{n}\}$ such that

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

(3.37)

On account of the hypothesis and (3.34), we obtain

$$ b_{n+1} \in Tb_{1} $$

(3.38)

for all $n \in N$.

By virtue of the hypothesis, (3.33), (3.37) and (3.38), we have

$$ b_{1} \in Tb_{1}. $$

(3.39)

Due to (3.39), $b_{1}$ is a fixed point of T. To prove the unique fixed point of T, let $c_{1}$ be another fixed point of T. Then

$$ c_{1} \in Tc_{1}. $$

(3.40)

Since T is Kannan-type multi-valued p-contractive, by (3.39) and (3.40), we have

$$\begin{aligned}& p(b_{1},b_{1}) \leq r\bigl[p(b_{1},b_{1})+p(b_{1},b_{1}) \bigr], \end{aligned}$$

(3.41)

$$\begin{aligned}& p(c_{1},c_{1}) \leq r\bigl[p(c_{1},c_{1})+p(c_{1},c_{1}) \bigr], \end{aligned}$$

(3.42)

$$\begin{aligned}& p(b_{1},c_{1}) \leq r\bigl[p(b_{1},b_{1})+p \bigl(c_{1},c_{1}' \bigr)\bigr]. \end{aligned}$$

(3.43)

Since $r \in[0, \frac{1}{2} )$, from (3.41), (3.42) and (3.43), we get

$$ p(b_{1},b_{1})=p(c_{1},c_{1})=p(b_{1},c_{1})=0. $$

(3.44)

By virtue of Lemma 2.6 and (3.44), we have

$$ b_{1}=c_{1}. $$

(3.45)

On account of (3.39), (3.40) and (3.45), T has a unique fixed point. □

Now we give an example to support Theorem 3.4.

Example 3.5

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

$$ \begin{aligned} &p(x,y)=x, \\ &Tx=\biggl[0,\frac{1}{4}x\biggr]. \end{aligned} $$

(3.46)

Let $\theta: X \times X \times R_{+} \times R_{+} \rightarrow R_{+}$ be as in (3.20). Then, due to (3.46), we easily can obtain that p is a u-distance on X.

From (3.46), we have

$$ p(y_{1},y_{2}) \leq\frac{1}{4} \bigl[p(x_{1},y_{1})+p(x_{2},y_{2}) \bigr] $$

(3.47)

for all $x_{1}, x_{2} \in X$, $y_{1} \in Tx_{1}$ and $y_{2} \in Tx_{2}$. To show that (3.47) is satisfied, let $x_{1},x_{2} \in X$, $y_{1} \in Tx_{1}$ and $y_{2} \in Tx_{2}$. Then $p(y_{1},y_{2})=y_{1} \leq\frac{1}{4}x_{1}$ and $\frac{1}{4}[p(x_{1},y_{1})+p(x_{2},y_{2})]=\frac{1}{4}(x_{1}+x_{2}) \geq \frac{1}{4}x_{1}$. Thus (3.47) is satisfied. Let $x,y \in X$ be such that $y \in Tx$. Then $0 \leq y \leq\frac{1}{4}x$ and $Ty \subseteq[0,\frac{1}{16}x] \subseteq[0, \frac{1}{4}x] =Tx$. Thus $Ty \subseteq Tx$ for all $x,y \in X$ with $y\in Tx$. Therefore all the conditions of Theorem 3.4 are satisfied and T has a unique fixed point 0 in X.

Definition 3.6

Let $(X,d)$ be a metric space. A mapping $T:X \rightarrow X$ is called Kannan-type single-valued p-contractive mapping if there exist a u-distance p on X and $r \in[0, \frac{1}{2})$ such that

• (iii) $p(Tx_{1},Tx_{2}) \leq r[p(x_{1},Tx_{1})+p(x_{2},Tx_{2})]$ for any $x_{1}, x_{2} \in X$,

• (iv) if $\{x_{n}\}$ is a sequence in X such that $x_{n+1}=Tx_{n}$ for each $n \in N$ and $\lim_{n \to\infty} x_{n} =c \in X$, then $p(Tc,c) \leq r [p(Tc, Tc)+p(c, Tc) ]$ and $p(c,Tc) \leq r [p(Tc, Tc)+p(Tc, c) ]$.

In the following example we show that if $(X,d)$ is a complete metric space and a mapping $T:X \rightarrow X$ is not Kannan-type single-valued p-contractive, in general, T may have no fixed point in X.

Example 3.7

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

$$\begin{aligned}& p(x,y) =\left \{ \begin{array}{l@{\quad}l} 2, & x=0, \\ x, & x \neq0, \end{array} \right . \end{aligned}$$

(3.48)

$$\begin{aligned}& Tx = \left \{ \begin{array}{l@{\quad}l} \frac{1}{4},& x=0, \\ \frac{x}{4(1+x)}, & x \neq0. \end{array} \right . \end{aligned}$$

(3.49)

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

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

(3.50)

By the same methods as in Example 3.3, we know that p is a u-distance on X and T is not Kannan-type single-valued p-contractive and T has no fixed point in X.

Theorem 3.8

Let $(X,d)$ be a metric space with a u-distance p on X.

Let $T: X \rightarrow X$ be a Kannan-type single-valued p-contractive mapping such that there exist a sequence $\{x_{n}\}$ of X and $c \in X$ satisfying $x_{n+1}=Tx_{n}$ for each $n \in N$ and $\lim_{n \to\infty} x_{n} =c \in X$. Then c is a fixed point of T, i.e., $Tc=c$.

Proof

By hypotheses, we obtain

$$\begin{aligned}& p(Tc,Tc) \leq r\bigl[p(c,Tc)+p(c,Tc)\bigr] = 2rp(c,Tc), \end{aligned}$$

(3.51)

$$\begin{aligned}& p(Tc, c) \leq r\bigl[p(Tc,Tc)+p(c, Tc)\bigr] \\& \hphantom{p(Tc, c)} \leq r\bigl[2rp(c,Tc)+p(c,Tc)\bigr] \\& \hphantom{p(Tc, c)} = r(2r+1)p(c,Tc), \end{aligned}$$

(3.52)

$$\begin{aligned}& p(c, Tc) \leq r\bigl[p(Tc,Tc)+p(Tc,c)\bigr] \\& \hphantom{p(c, Tc)} \leq r \bigl\{ 2rp(c,Tc)+r(2r+1)p(c,Tc)\bigr\} \\& \hphantom{p(c, Tc)} = \bigl(2r^{3} + 3r^{2}\bigr)p(c,Tc). \end{aligned}$$

(3.53)

Since $r \in[0, \frac{1}{2})$, $2r, r(2r+1), (2r^{3}+3r^{2}) \in[0,1)$ and thus, by (3.51), (3.52) and (3.53), we have

$$ p(Tc,Tc)= p(Tc,c) = p(c, Tc). $$

(3.54)

In view of Lemma 2.6 and (3.54),

$$ Tc=c. $$

(3.55)

This means that c is a fixed point of T. □

Theorem 3.9

Let $(X,d)$ be a complete metric space and let $T: X \to X$ be a Kannan-type single-valued p-contractive mapping.

Then T has a unique fixed point in X.

Proof

Since T is Kannan type single-valued p-contractive, there exists a sequence $\{ x_{n} \}$ of X such that

$$ x_{n+1} = Tx_{n} \quad \mbox{and}\quad p(x_{n+1},x_{n+2}) \leq kp(x_{n},x_{n+1}) $$

(3.56)

for all $n\in N$, where $k \in[0,1)$.

By repeated application of (3.56), we have

$$ p(x_{n},x_{n+1}) \leq k^{n-1}p(x_{1},x_{2}) $$

(3.57)

for all $n \in N$.

On account of (3.57), we get

$$\begin{aligned} p(x_{n},x_{m}) \leq& p(x_{n},x_{n+1})+ p(x_{n+1},x_{n+2})+ \cdots+ p(x_{m-1},x_{m}) \\ \leq&\sum_{i=n}^{m-1} p(x_{i},x_{i+1}) \leq\sum_{i=n}^{m-1} k^{i-1} p(x_{1},x_{2}) \\ \leq&\biggl(\frac{k^{n-1}}{1-k}\biggr)p(x_{1}, x_{2}) \end{aligned}$$

(3.58)

for all $n, m \in N$ with $n< m$.

In view of (3.58), we deduce that

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

(3.59)

By virtue of Lemma 2.5, Lemma 2.7 and (3.59), we know that $\{ x_{n} \}$ is a Cauchy sequence.

Since X is complete, there exists $c \in X$ such that

$$ \lim_{n \rightarrow\infty} x_{n}=c. $$

(3.60)

On account of the hypothesis, Theorem 3.8, (3.56) and (3.60), we know that c is a fixed point, i.e.,

$$ Tc=c. $$

(3.61)

By the same method as that in (3.40)∼(3.45), we can prove that T has a unique fixed point X. □

From Theorem 3.9, we have the following corollary.

Corollary 3.10

([1])

Let $(X,d)$ be a complete metric space and let $T: X \rightarrow X$ be a mapping such that

$$d(Tx,Ty) \leq r\bigl[d(x,Tx)+d(y,Ty)\bigr] $$

for all $x,y \in X$ and some $r \in[0, \frac{1}{2})$.

Then T has a unique fixed point in X.

Proof

By the same methods as in (3.56)∼(3.59), we deduce that

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

(3.62)

where $x_{n+1} =Tx_{n}$ for all $n \in N$.

Since X is complete, there exists $c \in X$ such that

$$ \lim_{n \rightarrow\infty} x_{n}=c. $$

(3.63)

Due to the hypothesis and (3.62), we get

$$\begin{aligned} d(x_{n+1}, Tc) =& d(Tx_{n}, Tc) \\ \leq& r\bigl[d(x_{n}, Tx_{n})+d(c, Tc)\bigr] \\ =&r\bigl[d(x_{n},x_{n+1})+d(c,Tc)\bigr] \end{aligned}$$

(3.64)

for all $n \in N$ and some $r \in[0, \frac{1}{2})$.

Taking the limit as $n \rightarrow\infty$ in (3.64), we obtain

$$ d(c,Tc) \leq r d(c,Tc) $$

(3.65)

for some $r \in[0, \frac{1}{2})$.

From (3.65), we have

$$ d(c,Tc)=0. $$

(3.66)

Since metric d is a u-distance, by view of (3.62), (3.63), (3.66) and the hypothesis, conditions of Corollary 3.10 satisfy all conditions of Theorem 3.9.

Therefore T has a unique fixed point. □

Finally we shall present an example to show that all conditions of Theorem 3.9 are satisfied, but all conditions of Corollary 3.10 are not satisfied.

Example 3.11

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

$$ \begin{aligned} &p(x,y)=x, \\ &Tx= \frac{1}{4} x. \end{aligned} $$

(3.67)

Then, due to (3.67), we easily can obtain that p is a u-distance on X, but not metric and

$$ p(Tx,Ty) \leq\frac{1}{4}\bigl[p(x,Tx)+p(y,Ty)\bigr] $$

(3.68)

for all $x,y \in X$.

Suppose that $\{x_{n} \}$ is a sequence of X such that $x_{n+1}=Tx_{n}$ for all $x \in N$.

Then, by (3.67), we have

$$ x_{n+1} = Tx_{n} = \frac{1}{4} x_{n} $$

(3.69)

for all $n \in N$.

By virtue of (3.69),

$$ \lim_{n \rightarrow\infty} x_{n}= \lim _{n \rightarrow \infty}\biggl({\frac{1}{4}}\biggr)^{n-1} x_{1} =0. $$

(3.70)

On account of (3.67)∼(3.70), all conditions of Theorem 3.9 are satisfied, but all conditions of Corollary 3.10 are not satisfied since p is not metric.