Generalized multi-valued mappings satisfy some inequalities conditions on metric spaces

Abstract

In this paper, we prove a condition of the existence for generalized multi-valued mappings satisfying some inequalities in metric spaces. These results are improved versions of results of Boško Damjanović and Dragan Dorić (Filomat 25:125-131, 2011).

1 Introduction and preliminaries

Let $(X,d)$ be a metric space. We denote by $\operatorname {CB}(X)$ the family of all non-empty closed bounded subsets of X. Let $H(\cdot,\cdot)$ be the Hausdorff metric, i.e.,

$$H(A,B) = \max\Bigl\{ \sup_{a\in A}d(a,B),\sup_{b\in B}d(A,b) \Bigr\} , $$

for $A, B\in \operatorname {CB}(X)$, where

$$d(x,B)=\inf_{y\in B} d(x,y). $$

• (i) Let T be a self-mapping on X. Then T is called a Banach contraction mapping if there exists $r\in[0, 1)$ such that

  $$d(Tx,Ty)\leq r d(x,y) $$

  for all $x, y \in X$.

• (ii) T is called a Kannan mapping if there exists $a\in[0, \frac{1}{2})$ such that

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

  for all $x, y \in X$.

• (iii) If T is a mapping such that

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

  such that $r\in[0, 1)$ and all $x, y \in X$, then T is called a generalized Kannan mapping.

In 1973, Hardy and Rogers [1] introduced a condition as follows:

• (iv) Let $x, y \in X$. Then there exists $a_{i}\geq 0$ such that

  $$d(Tx, Ty) \leq a_{1}d(x,y)+a_{2}d(x,Tx)+a_{3}d(y, Ty)+a_{4}d(x,Ty)+a_{5}d(y, Tx), $$

  where $\sum_{i=1}^{5}a_{i}<1$.

• (v) Ciric [2] defined the following condition which generalizes the Banach contraction and Kannan mapping, that is,

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

  such that $r\in[0, 1)$ and all $x, y \in X$.

If X is complete and at least one of (i), (ii), (iii), (iv), and (v) holds, then T has a unique fixed point (see [1–5]).

In 2008, Suzuki [6] introduced the condition C as follows. T is said to satisfy condition C if

$$\frac{1}{2}d(x, Tx) \leq d(x, y) \quad \mbox{implies}\quad d(Tx, Ty) \leq d(x, y), $$

for all $x, y \in C$.

In the same year, Kikkawa and Suzuki [7] generalized the Kannan mapping resulting in the following theorem.

Theorem 1.1

(Kikkawa and Suzuki [7])

Let T be a mapping on complete metric space $(X, d)$ and let φ be a non-increasing function from $[0,1)$ into $(\frac{1}{2},1]$ defined by

$$\varphi(r)= \textstyle\begin{cases} 1, &\textit{if }0\leq r< \frac{1}{\sqrt{2}},\\ \frac{1}{1+r}, &\textit{if }\frac{1}{\sqrt{2}}\leq r < \frac {1}{2}. \end{cases} $$

Let $\alpha\in[0,\frac{1}{2})$ and put $r = \frac{\alpha}{1-\alpha} \in [0,1)$. Suppose that

$$\begin{aligned} \varphi(r)d(x, T x) \leq d(x,y) \quad \textit{implies} \quad d(T x, T y) \leq\alpha d(x,Tx)+\alpha d(y,Ty) \end{aligned}$$

(1.1)

for all $x, y\in X$. Then T has a unique fixed point z and $\lim_{n\to\infty} T^{n}x = z$ holds for every $x \in X$.

Theorem 1.2

(Kikkawa and Suzuki [7])

Let T be a mapping on a complete metric space $(X, d)$ and let θ be a non-increasing function from $[0,1)$ into $(\frac{1}{2},1]$ defined by

$$\theta(r)= \textstyle\begin{cases} 1, &\textit{if }0\leq r< \frac{1}{2}(\sqrt{5}-1),\\ \frac{1-r}{r^{2}}, &\textit{if }\frac{1}{2}(\sqrt{5}-1)\leq r < \frac {1}{\sqrt{2}},\\ \frac{1}{r+1}, &\textit{if }\frac{1}{\sqrt{2}}\leq r < 1. \end{cases} $$

Suppose that $r\in[0,1)$ such that

$$\begin{aligned} \theta(r)d(x, T x) \leq d(x,y)\quad \textit{implies}\quad d(T x, T y) \leq r\max\bigl\{ d(x,Tx),d(y,Ty)\bigr\} \end{aligned}$$

(1.2)

for all $x, y\in X$. Then T has a unique fixed point z and $\lim_{n\to\infty} T^{n}x = z$ holds for every $x \in X$.

In 2011, Karapinar and Tas [8] stated some new conditions which are modifications of Suzuki’s condition C, as follows. T is said to satisfy condition $SCC$ if

$$\frac{1}{2}d(x,Tx) \leq d(x,y)\quad \mbox{implies}\quad d(Tx,Ty) \leq M(x,y) $$

for all $x, y \in K$, where

$$M(x, y) = \max \bigl\{ d(x, y), d(x, Tx), d(y,Ty ), d(y,Tx ), d(x, Ty) \bigr\} . $$

In 1969, Nadler [9] proved a multi-valued extension of the Banach contraction theorem as follows.

Theorem 1.3

(Nadler [10])

Let $(X, d)$ be a complete metric space and let T be a mapping from X into $\operatorname {CB}(X)$. Assume that there exists $r \in[0, 1)$ such that

$$H(Tx, Ty) \leq rd(x, y) $$

for all $x, y \in X$. Then there exists $z \in X$ such that $z \in Tz$.

Next, the result of Kikkawa and Suzuki [9] is a generalization of Nadler.

Theorem 1.4

(Kikkawa and Suzuki [9])

Let $(X, d)$ be a complete metric space and let T be a mapping from X into $\operatorname {CB}(X)$. Define a strictly decreasing function η from $[0, 1)$ onto $(\frac{1}{2},1]$ by

$$\eta(r)=\frac{r}{1+r} $$

and assume that there exists $r \in[0, 1)$ such that

$$\eta(r)d(x,Tx) \leq d(x, y) \quad \textit{implies}\quad H(Tx, Ty) \leq rd(x, y) $$

for all $x, y \in X$. Then there exists $z \in X$ such that $z \in Tz$.

In 2011, Damjanović and Dorić [11] generalized the result of Kannan (iii) and Nadler.

Theorem 1.5

(Damjanović and Dorić [11])

Define a non-increasing function φ from $[0,1)$ into $(0,1]$ by

$$\varphi(r)= \textstyle\begin{cases} 1, &\textit{if }0\leq r< \frac{\sqrt{5}-1}{2},\\ 1-r, &\textit{if }\frac{\sqrt{5}-1}{2}\leq r < 1. \end{cases} $$

Let $(X, d)$ be a complete metric space and let T be a mapping from X into $\operatorname {CB}(X)$. Assume that

$$\begin{aligned} \varphi(r)d(x, T x) \leq d(x,y) \quad \textit{implies} \quad H(T x, T y) \leq r\max\bigl\{ d(x,Tx),d(y,Ty)\bigr\} \end{aligned}$$

(1.3)

for all $x, y\in X$. Then there exists $z\in X$ such that $z \in T z$.

Corollary 1.6

(Damjanović and Dorić [11]) Let $(X, d)$ be a complete metric space and let T be a mapping from X into $\operatorname {CB}(X)$. Let $\alpha\in[0, \frac{1}{2})$ and put $r = 2\alpha$. Suppose that

$$\begin{aligned} \varphi(r)d(x, T x) \leq d(x,y) \quad\textit{implies}\quad H(T x, T y) \leq\alpha d(x,Tx)+\alpha d(y,Ty) \end{aligned}$$

(1.4)

for all $x, y\in X$, where the function φ is defined as in Theorem  1.5. Then there exists $z\in X$ such that $z \in T z$.

In this paper, we prove a condition of the existence for generalized multi-valued mappings under SCC conditions in metric spaces. These results are improved versions of results of Boško Damjanović and Dragan Dorić [11].

2 Main results

Theorem 2.1

Define a non-increasing function φ from $[0,\frac{1}{2})$ into $(0,1]$ by

$$\varphi(r)= \textstyle\begin{cases} 1, &\textit{if }0\leq r< \frac{\sqrt{5}-1}{\sqrt{5}+1},\\ \frac{1-2r}{1-r}, &\textit{if }\frac{\sqrt{5}-1}{\sqrt{5}+1}\leq r < \frac {1}{2}. \end{cases} $$

Let $(X, d)$ be a complete metric space and let T be a mapping from X into $\operatorname {CB}(X)$. Assume that

$$\begin{aligned} \varphi(r)d(x, T x) \leq d(x,y) \quad \textit{implies} \quad H(T x, T y) \leq r M(x,y) \end{aligned}$$

(2.1)

where $M(x,y)=\max\{d(x,y),d(x,Tx),d(y,Ty),d(x,Ty),d(y,Tx)\}$, for all $x, y\in X$. Then there exists $z\in X$ such that $z \in T z$.

Proof

Let $r_{1}$ be a real number such that $0 \leq r < r_{1} < \frac {1}{2}$. Let $u_{1} \in X$ and $u_{2} \in T u_{1}$ be arbitrary. Since $u_{2} \in T u_{1}$, we have $d(u_{2},T u_{2}) \leq H(T u_{1},T u_{2})$ and

$$\varphi(r)d(u_{1},T u_{1}) \leq d(u_{1}, T u_{1})\leq d(u_{1}, u_{2}). $$

Thus from the assumption (2.1),

$$d(u_{2},T u_{2}) \leq H(T u_{1}, T u_{2}) \leq r M(u_{1},u_{2}) $$

where $M(u_{1},u_{2})=\max\{ d(u_{1},u_{2}),d(u_{1},Tu_{1}),d(u_{2},Tu_{2}),d(u_{1},Tu_{2}),d(u_{2},Tu_{1})\}$. Consider

$$\begin{aligned} d(u_{2},T u_{2}) &\leq r \max\bigl\{ d(u_{1},u_{2}),d(u_{1},Tu_{1}),d(u_{2},Tu_{2}),d(u_{1},Tu_{2}),d(u_{2},Tu_{1}) \bigr\} \\ &= r \max\bigl\{ d(u_{1},u_{2}),d(u_{1},Tu_{2}) \bigr\} . \end{aligned}$$

If $\max\{d(u_{1},u_{2}),d(u_{1},Tu_{2})\}=d(u_{1},Tu_{2})$, then

$$\begin{aligned} d(u_{2},T u_{2}) &\leq r d(u_{1},Tu_{2}) \\ &\leq r d(u_{1},u_{2})+rd(u_{2},Tu_{2}) \end{aligned}$$

and then

$$d(u_{2},T u_{2})\leq\biggl(\frac{r}{1-r} \biggr)d(u_{1},u_{2}). $$

If $\max\{d(u_{1},u_{2}),d(u_{1},Tu_{2})\}=d(u_{1},u_{2})$, then

$$d(u_{2},T u_{2})\leq r d(u_{1},u_{2}) \leq \biggl(\frac{r}{1-r}\biggr)d(u_{1},u_{2}). $$

So

$$d(u_{2},T u_{2})\leq \biggl(\frac{r}{1-r} \biggr)d(u_{1},u_{2}). $$

So there exists $u_{3}\in Tu_{2}$ such that $d(u_{2},u_{3})\leq (\frac {r_{1}}{1-r_{1}})d(u_{1},u_{2})$. Thus, we can construct a sequence $\{x_{n}\}$ in X such that $u_{n+1} \in Tu_{n}$ and

$$d(u_{n+1}, u_{n+2}) \leq\biggl(\frac{r_{1}}{1-r_{1}} \biggr)d(u_{n}, u_{n+1}). $$

Hence, by induction,

$$d(u_{n}, u_{n+1}) \leq\biggl(\frac{r_{1}}{1-r_{1}} \biggr)^{n-1} d(u_{1}, u_{2}). $$

Then by the triangle inequality, we have

$$\sum^{\infty}_{n=1}d(u_{n}, u_{n+1}) \leq\sum^{\infty}_{n=1}\biggl( \frac{r_{1}}{1-r_{1}}\biggr)^{n-1} d(u_{1}, u_{2})< \infty. $$

Hence we conclude that $\{u_{n}\}$ is a Cauchy sequence. Since X is complete, there is some point $z \in X$ such that

$$\lim_{n\to\infty}u_{n} = z. $$

Now, we will show that $d(z,Tx)\leq rd(x,Tx)$ for all $x\in X\setminus \{z\}$.

Let $x\in X\setminus\{z\}$. Since $u_{n}\to z$, there exists $n_{0} \in N$ such that $d(z, u_{n})\leq (\frac{1}{3})d(z, x)$ for all $n\geq n_{0}$. Then we have

$$\begin{aligned} \varphi(r) d(u_{n}, Tu_{n}) &\leq d(u_{n},Tu_{n}) \\ &\leq d(u_{n},u_{n+1}) \\ &\leq d(u_{n}, z) + d(z, u_{n+1}) \\ &\leq\biggl(\frac{2}{3}\biggr)d(z, x) \\ &=d(z, x)-\frac{1}{3}d(z, x) \\ &\leq d(z, x)-d(z, u_{n}) \\ &\leq d(x, u_{n}). \end{aligned}$$

(2.2)

Then from (2.1) we have

$$H(Tu_{n}, Tx) \leq r \max\bigl\{ d(u_{n}, x),d(u_{n}, Tu_{n}), d(x, Tx), d(u_{n}, Tx),d(x, Tu_{n})\bigr\} . $$

Since $u_{n+1} \in Tu_{n}$, $d(u_{n+1}, Tx) \leq H(Tu_{n}, Tx)$, so that

$$d(u_{n+1}, Tx) \leq r \max\bigl\{ d(u_{n}, x),d(u_{n}, u_{n+1}), d(x, Tx), d(u_{n}, Tx),d(x, u_{n+1})\bigr\} $$

for all $n\geq n_{0}$. Letting $n \to\infty$, we obtain

$$d(z, Tx) \leq r \max\bigl\{ d(z, x), d(x, Tx), d(z, Tx)\bigr\} . $$

It follows that

$$\begin{aligned} d(z, Tx) \leq\biggl( \frac{r}{1-r}\biggr)d(x, Tx) \quad \mbox{for all } x\in X\setminus\{z\}. \end{aligned}$$

(2.3)

Next, we show that $z \in Tz$. Suppose that z is not an element in Tz.

Case (i): $0\leq r<\frac{\sqrt{5}-1}{\sqrt{5}+1}$. Let $a \in Tz$. Then $a\neq z$ and so by (2.3), we have

$$d(z, Ta) \leq\biggl( \frac{r}{1-r}\biggr)d(a, Ta). $$

On the other hand, since $\varphi(r) d(z, Tz) = d(z, Tz)\leq d(z, a)$, from (2.1) we have

$$\begin{aligned} H(Tz, Ta) \leq r \max\bigl\{ d(z,a),d(z, Tz), d(a, Ta),d(z, Ta), d(a, Tz)\bigr\} . \end{aligned}$$

So

$$\begin{aligned} d(a, Ta)\leq H(Tz, Ta) \leq r \max\bigl\{ d(z,a),d(z, Tz),d(z, Ta) \bigr\} . \end{aligned}$$

(2.4)

It implies that

$$d(a, Ta)\leq r \max\bigl\{ d(z,a),d(z, Tz),d(z, Ta)\bigr\} . $$

Since $d(z,a)\leq d(z, Tz)+d( Tz,a)=d(z, Tz)$, we have

$$\begin{aligned} d(a, Ta)\leq\biggl(\frac{r}{1-r}\biggr) d(z, Tz). \end{aligned}$$

(2.5)

Using (2.3), (2.4), and (2.5), we have

$$\begin{aligned} d(z, Tz)&\leq d(z, Ta)+H( Ta,Tz) \\ &\leq\biggl(\frac{r}{1-r}\biggr)d(a, Ta)+r \max\bigl\{ d(z,a), d(z, Tz),d(z, Ta)\bigr\} \\ &\leq\biggl(\frac{r}{1-r}\biggr)d(a, Ta)+r \max\biggl\{ d(z,a), d(z, Tz), \biggl(\frac {r}{1-r}\biggr)d(a, Ta)\biggr\} \\ &\leq\biggl(\frac{r}{1-r}\biggr)d(a, Ta)+r \max\bigl\{ d(z,a), d(z, Tz)\bigr\} \\ &\leq\biggl(\frac{r}{1-r}\biggr)d(a, Ta)+rd(z, Tz) \\ &\leq\biggl(\frac{r}{1-r}\biggr)^{2}d(z, Tz)+rd(z, Tz) \\ &\leq\biggl(\frac{r}{1-r}\biggr)^{2}d(z, Tz)+\biggl( \frac{r}{1-r}\biggr)d(z, Tz) \\ &\leq\biggl[\biggl(\frac{r}{1-r}\biggr)^{2}+\biggl( \frac{r}{1-r}\biggr)\biggr]d(z, Tz) \\ &\leq\bigl[k^{2}+k\bigr]d(z, Tz), \end{aligned}$$

where $k=\frac{r}{1-r}$. Since $r<\frac{\sqrt{5}-1}{\sqrt{5}+1}$, we have $k^{2}+k<1$ and so $d(z, Tz)< d(z, Tz)$, a contradiction. Thus $z\in Tz$.

Case (ii): $\frac{\sqrt{5}-1}{\sqrt{5}+1}\leq r <\frac{1}{2}$. Let $x\in X$. If $x = z$, then $H(T x, T z) \leq r \max\{ d(x,z),d(x,Tx),d(z,Tz), d(x,Tz),d(z,Tx)\}$ holds. If $x \neq z$, then for all $n\in\mathbb{N}$, there exists $y_{n} \in Tx$ such that

$$d(z, y_{n}) \leq d(z,Tx) + \biggl(\frac{1}{n}\biggr)d(x, z). $$

We consider

$$\begin{aligned} d(x, Tx)&\leq d(x, y_{n}) \\ &\leq d(x,z)+d(z, y_{n}) \\ &\leq d(x,z)+d(z,Tx) + \biggl(\frac{1}{n}\biggr)d(x, z) \\ &\leq d(x,z)+\biggl( \frac{r}{1-r}\biggr)d(x, Tx)+ \biggl( \frac{1}{n}\biggr)d(x, z). \end{aligned}$$

Thus, $( \frac{1-2r}{1-r})d(x, Tx)\leq(1+\frac{1}{n})d(x,z)$. Take $n\to\infty$, we obtain

$$\biggl( \frac{1-2r}{1-r}\biggr)d(x, Tx)\leq d(x,z), $$

by using (2.1), implies $H(T x, T z) \leq r \max\{ d(x,z),d(x,Tx),d(z,Tz),d(x,Tz),d(z,Tx)\}$. Hence, as $u_{n+1}\in Tu_{n}$, it follows that with $x=u_{n}$

$$\begin{aligned} d(z, Tz)&=\lim_{n\to\infty}d(u_{n+1}, Tz) \\ &\leq H(T u_{n}, T z) \\ &\leq\lim_{n\to\infty}r \max\bigl\{ d(u_{n},z),d(u_{n},Tu_{n}),d(z,Tz),d(u_{n},Tz),d(z,Tu_{n}) \bigr\} \\ &\leq\lim_{n\to\infty}r \max\bigl\{ d(u_{n},z),d(u_{n},u_{n+1}),d(z,Tz),d(u_{n},Tz),d(z,u_{n+1}) \bigr\} \\ &\leq rd(z,Tz). \end{aligned}$$

Therefore, $(1-r)d(z, Tz)\leq0$, which implies $d(z, Tz)=0$. Since Tz is closed, we have $z\in Tz$. This completes the proof. □

Example 2.2

Let $X=[0,\infty)$ be endowed with the usual metric d. Define $T: X\to \operatorname {CB}(X)$ by

$$\begin{aligned} T(x)= \textstyle\begin{cases} [0,x^{2}], & 0\leq x\leq\frac{1}{2},\\ {[0,\frac{x}{3}]}, &\frac{1}{2}< x < 1,\\ {[0,\log(x)]}, & 1\leq x. \end{cases}\displaystyle \end{aligned}$$

(2.6)

Proof

We show that T satisfies (2.1). Let $x,y\in X$. We prove by cases.

Case (i): Suppose that $x,y \in[0,\frac{1}{2}]$. Thus, if $x^{2}\leq y$, then

$$\varphi\biggl(\frac{1}{4}\biggr)d(x, T x)=\bigl\vert x-x^{2} \bigr\vert \geq \vert x-y\vert =d(x,y). $$

But if $x^{2}> y$, then

$$\varphi\biggl(\frac{1}{4}\biggr)d(x, T x)=\bigl\vert x-x^{2} \bigr\vert \leq \vert x-y\vert =d(x,y) $$

and

$$\begin{aligned} H(T x, T y)&=\bigl\vert x^{2}-y^{2}\bigr\vert \\ &\leq\frac{1}{4}\bigl\vert (2x)^{2}-(2y)^{2}\bigr\vert \\ &\leq\frac{1}{4}\bigl\vert x-2y^{2}\bigr\vert \\ &\leq\frac{1}{4}\bigl\vert x-y^{2}\bigr\vert \\ &=\frac{1}{4}\max\bigl\{ \vert x-y\vert ,\bigl\vert x-x^{2}\bigr\vert ,\bigl\vert y-y^{2}\bigr\vert ,\bigl\vert x-y^{2}\bigr\vert ,\bigl\vert y-x^{2}\bigr\vert \bigr\} \\ &= \frac{1}{4}\max\bigl\{ d(x,y),d(x,Tx),d(y,Ty),d(x,Ty),d(y,Tx)\bigr\} \\ &= r M(x,y), \end{aligned}$$

(2.7)

where $r=\frac{1}{4}$. Hence T satisfies (2.1).

Case (ii): Suppose that $x,y \in(\frac{1}{2},1)$. Thus, if $\frac {x}{3}\leq y$, then

$$\varphi\biggl(\frac{1}{3}\biggr)d(x, T x)=\biggl\vert x- \frac{x}{3}\biggr\vert \geq \vert x-y\vert =d(x,y). $$

But if $\frac{x}{3}> y$, then

$$\varphi\biggl(\frac{1}{3}\biggr)d(x, T x)=\biggl\vert x- \frac{x}{3}\biggr\vert \leq \vert x-y\vert =d(x,y) $$

and

$$\begin{aligned} H(T x, T y)&=\frac{1}{3}\vert x-y\vert \\ &\leq\frac{1}{3}\biggl\vert x-\frac{y}{3}\biggr\vert \\ &=\frac{1}{3}\max\biggl\{ \vert x-y\vert ,\biggl\vert x- \frac{x}{3}\biggr\vert ,\biggl\vert y-\frac{y}{3}\biggr\vert , \biggl\vert x-\frac {y}{3}\biggr\vert ,\biggl\vert y- \frac{x}{3}\biggr\vert \biggr\} \\ &= \frac{1}{3}\max\bigl\{ d(x,y),d(x,Tx),d(y,Ty),d(x,Ty),d(y,Tx)\bigr\} \\ &= r M(x,y), \end{aligned}$$

(2.8)

where $r=\frac{1}{3}$. Hence T satisfies (2.1).

Case (iii): Suppose that $x,y \in[1,\infty]$. Thus, if $\log(x) \leq y$, then

$$\varphi\biggl(\frac{1}{3}\biggr)d(x, T x)=\bigl\vert x-\log(x)\bigr\vert \geq \vert x-y\vert =d(x,y). $$

But if $\log(x)> y$, then

$$\varphi\biggl(\frac{1}{3}\biggr)d(x, T x)=\bigl\vert x-\log(x)\bigr\vert \leq \vert x-y\vert =d(x,y) $$

and

$$\begin{aligned} H(T x, T y)&=\bigl\vert \log(x)-\log(y)\bigr\vert \\ &=\frac{1}{3}\bigl(3\log(x)-3\log(y)\bigr) \\ &\leq\frac{1}{3}\bigl\vert x-\log(y)\bigr\vert \\ &=\frac{1}{3}\max\bigl\{ \vert x-y\vert ,\bigl\vert x-\log(x)\bigr\vert ,\bigl\vert y-\log(y)\bigr\vert ,\bigl\vert x-\log(y)\bigr\vert , \bigl\vert y-\log (x)\bigr\vert \bigr\} \\ &= \frac{1}{3}\max\bigl\{ d(x,y),d(x,Tx),d(y,Ty),d(x,Ty),d(y,Tx)\bigr\} \\ &= r M(x,y), \end{aligned}$$

(2.9)

where $r=\frac{1}{3}$. Hence T satisfies (2.1).

Case (iv): Suppose that $x \in[0,\frac{1}{2}]$ and $y \in(\frac {1}{2},1)$. Then $x^{2}< x< y$. Thus, $\varphi(\frac{1}{3})d(x, T x)=\vert x-x^{2}\vert \geq \vert x-y\vert =d(x,y)$. Hence T satisfies (2.1).

Case (v): Suppose that $x \in(\frac{1}{2},1)$ and $y \in[0,\frac {1}{2}]$. So $x>y$. Thus, if $\frac{x}{3} \leq y$, then

$$\varphi\biggl(\frac{1}{3}\biggr)d(x, T x)=\biggl\vert x- \frac{x}{3}\biggr\vert \geq \vert x-y\vert =d(x,y). $$

But if $\frac{x}{3}> y$, then

$$\varphi\biggl(\frac{1}{3}\biggr)d(x, T x)=\biggl\vert x- \frac{x}{3}\biggr\vert \leq \vert x-y\vert =d(x,y) $$

and

$$\begin{aligned} H(T x, T y)&=\biggl\vert \frac{x}{3}-y^{2}\biggr\vert \\ &\leq\frac{1}{3}\bigl\vert x-3y^{2}\bigr\vert \\ &\leq\frac{1}{3}\bigl\vert x-y^{2}\bigr\vert \\ &=\frac{1}{3}\max\biggl\{ \vert x-y\vert ,\biggl\vert x- \frac{x}{3}\biggr\vert ,\bigl\vert y-y^{2}\bigr\vert ,\bigl\vert x-y^{2}\bigr\vert ,\biggl\vert y-\frac {x}{3}\biggr\vert \biggr\} \\ &= \frac{1}{3}\max\bigl\{ d(x,y),d(x,Tx),d(y,Ty),d(x,Ty),d(y,Tx)\bigr\} \\ &= r M(x,y), \end{aligned}$$

(2.10)

where $r=\frac{1}{3}$. Hence T satisfies (2.1).

Case (vi): Suppose that $x \in[0,\frac{1}{2}]$ and $y \in[1,\infty]$.

$$\varphi\biggl(\frac{1}{3}\biggr)d(x, T x)=\bigl\vert x-x^{2} \bigr\vert \leq \vert x-y\vert =d(x,y) $$

and

$$\begin{aligned} H(T x, T y)&=\bigl\vert x^{2}-\log(y)\bigr\vert \\ &=\frac{1}{3}\bigl\vert 3x^{2}-3\log(y)\bigr\vert = \frac{1}{3}\bigl\vert 3\log(y)-3x^{2}\bigr\vert \\ &\leq\frac{1}{3}\max\bigl\{ \bigl\vert y-\log(y)\bigr\vert ,\bigl\vert y-x^{2}\bigr\vert \bigr\} \\ &=\frac{1}{3}\max\bigl\{ \vert x-y\vert ,\bigl\vert x-x^{2}\bigr\vert ,\bigl\vert y-\log(y)\bigr\vert ,\bigl\vert x-\log(y)\bigr\vert ,\bigl\vert y-x^{2}\bigr\vert \bigr\} \\ &= \frac{1}{3}\max\bigl\{ d(x,y),d(x,Tx),d(y,Ty),d(x,Ty),d(y,Tx)\bigr\} \\ &= r M(x,y), \end{aligned}$$

(2.11)

where $r=\frac{1}{3}$. Hence T satisfies (2.1).

Case (vii): Suppose that $x \in[1,\infty]$ and $y \in[0,\frac {1}{2}]$. Thus, if $\log(x) \leq y$, then

$$\varphi\biggl(\frac{1}{4}\biggr)d(x, T x)=\bigl\vert x-\log(x)\bigr\vert \geq \vert x-y\vert =d(x,y). $$

But if $\log(x) > y$, then

$$\varphi\biggl(\frac{1}{4}\biggr)d(x, T x)=\bigl\vert x-\log(x)\bigr\vert \leq \vert x-y \vert =d(x,y) $$

and

$$\begin{aligned} H(T x, T y)&=\bigl\vert \log(x) -y^{2}\bigr\vert \\ &=\frac{1}{4}\bigl\vert 4\log(x) -4y^{2}\bigr\vert \\ &\leq\frac{1}{4}\bigl\vert x-y^{2}\bigr\vert \\ &=\frac{1}{4}\max\bigl\{ \vert x-y\vert ,\bigl\vert x-\log(x) \bigr\vert ,\bigl\vert y-y^{2}\bigr\vert ,\bigl\vert x-y^{2}\bigr\vert ,\bigl\vert y-\log(x) \bigr\vert \bigr\} \\ &= \frac{1}{4}\max\bigl\{ d(x,y),d(x,Tx),d(y,Ty),d(x,Ty),d(y,Tx)\bigr\} \\ &= r M(x,y), \end{aligned}$$

(2.12)

where $r=\frac{1}{4}$. Hence T satisfies (2.1).

Case (viii): Suppose that $x \in(\frac{1}{2},1)$ and $y \in[1,\infty]$.

$$\varphi\biggl(\frac{1}{3}\biggr)d(x, T x)=\biggl\vert x- \frac{x}{3}\biggr\vert \leq \vert x-y\vert =d(x,y) $$

and

$$\begin{aligned} H(T x, T y)&=\biggl\vert \frac{x}{3}-\log(y)\biggr\vert \\ &=\frac{1}{3}\bigl\vert x-3\log(y)\bigr\vert =\frac{1}{3}\bigl\vert 3\log(y)-x\bigr\vert \\ &\leq\frac{1}{3}\max\biggl\{ \bigl\vert y-\log(y)\bigr\vert ,\biggl\vert y-\frac{x}{3}\biggr\vert \biggr\} \\ &=\frac{1}{3}\max\biggl\{ \vert x-y\vert ,\biggl\vert x- \frac{x}{3}\biggr\vert ,\bigl\vert y-\log(y)\bigr\vert ,\bigl\vert x-\log (y)\bigr\vert ,\biggl\vert y-\frac{x}{3}\biggr\vert \biggr\} \\ &= \frac{1}{3}\max\bigl\{ d(x,y),d(x,Tx),d(y,Ty),d(x,Ty),d(y,Tx)\bigr\} \\ &= r M(x,y), \end{aligned}$$

(2.13)

where $r=\frac{1}{3}$. Hence T satisfies (2.1).

Case (ix): Suppose that $x \in[1,\infty]$ and $y \in(\frac{1}{2},1)$. Thus, if $\log(x) \leq y$, then

$$\varphi\biggl(\frac{1}{3}\biggr)d(x, T x)=\bigl\vert x-\log(x)\bigr\vert \geq \vert x-y\vert =d(x,y). $$

But if $\log(x) > y$, then

$$\varphi\biggl(\frac{1}{3}\biggr)d(x, T x)=\bigl\vert x-\log(x)\bigr\vert \leq \vert x-y \vert =d(x,y) $$

and

$$\begin{aligned} H(T x, T y)&=\biggl\vert \log(x) -\frac{y}{3}\biggr\vert \\ &=\frac{1}{3}\bigl\vert 3\log(x) -y\bigr\vert \\ &\leq\frac{1}{3}\vert x-y\vert \\ &=\frac{1}{3}\max\biggl\{ \vert x-y\vert ,\bigl\vert x-\log(x) \bigr\vert ,\biggl\vert y-\frac{y}{3}\biggr\vert ,\biggl\vert x- \frac {y}{3}\biggr\vert ,\bigl\vert y-\log(x) \bigr\vert \biggr\} \\ &= \frac{1}{3}\max\bigl\{ d(x,y),d(x,Tx),d(y,Ty),d(x,Ty),d(y,Tx)\bigr\} \\ &= r M(x,y), \end{aligned}$$

(2.14)

where $r=\frac{1}{3}$. Hence T satisfies (2.1).

Thus we see that T satisfies condition (2.1) and satisfies Theorem 2.1. So there exists $z\in X$ such that $z \in T z$. Moreover, $0\in T(0)$. □

Theorem 2.3

Define a non-increasing function φ from $[0,\frac{1}{5})$ into $(0,1]$ by

$$\varphi(r)= \textstyle\begin{cases} 1, &\textit{if }0\leq r< \frac{\sqrt{5}-1}{4+2\sqrt{5}},\\ \frac{1-5r}{1-2r}, &\textit{if }\frac{\sqrt{5}-1}{4+2\sqrt{5}}\leq r < \frac{1}{5}. \end{cases} $$

Let $(X, d)$ be a complete metric space and let T be a mapping from X into $\operatorname {CB}(X)$. Assume that

$$\begin{aligned} \varphi(r)d(x, T x) \leq d(x,y) \quad \textit{implies} \quad H(T x, T y) \leq S(x,y) \end{aligned}$$

(2.15)

where $S(x,y)=r d(x,y)+r d(x,Tx)+r d(y,Ty)+r d(x,Ty)+r d(y,Tx)$ for all $x, y\in X$. Then there exists $z\in X$ such that $z \in T z$.

Proof

Let $r_{1}$ be a real number such that $0 \leq r < r_{1} < 1$. Let $u_{1} \in X$ and $u_{2} \in T u_{1}$ be arbitrary. Since $u_{2} \in T u_{1}$, we have $d(u_{2},T u_{2}) \leq H(T u_{1},T u_{2})$ and

$$\varphi(r)d(u_{1},T u_{1}) \leq d(u_{1}, T u_{1})\leq d(u_{1}, u_{2}). $$

Thus, from the assumption (2.15),

$$d(u_{2},T u_{2}) \leq H(T u_{1}, T u_{2}) \leq S(u_{1},u_{2}) $$

where $S(u_{1},u_{2})=r d(u_{1},u_{2})+r d(u_{1},Tu_{1})+r d(u_{2},Tu_{2})+r d(u_{1},Tu_{2})+r d(u_{2},Tu_{1})$. Consider

$$\begin{aligned} d(u_{2},T u_{2}) &\leq r d(u_{1},u_{2})+r d(u_{1},Tu_{1})+r d(u_{2},Tu_{2})+rd(u_{1},Tu_{2})+r d(u_{2},Tu_{1}) \\ &\leq3r d(u_{1},u_{2})+2r d(u_{2},Tu_{2}). \end{aligned}$$

So

$$\begin{aligned} d(u_{2},T u_{2}) &\leq\biggl(\frac{3r}{1-2r} \biggr)d(u_{1},u_{2}). \end{aligned}$$

So there exists $u_{3}\in Tu_{2}$ such that $d(u_{2},u_{3})\leq( \frac {3r_{1}}{1-2r_{1}})d(u_{1},u_{2})$. Thus, we can construct a sequence $\{x_{n}\}$ in X such that $u_{n+1} \in Tu_{n}$ and

$$d(u_{n+1}, u_{n+2}) \leq\biggl(\frac{3r_{1}}{1-2r_{1}} \biggr)d(u_{n}, u_{n+1}). $$

Hence, by induction,

$$d(u_{n}, u_{n+1}) \leq\biggl(\frac{3r_{1}}{1-2r_{1}} \biggr)^{n-1} d(u_{1}, u_{2}). $$

Then by the triangle inequality, we have

$$\sum^{\infty}_{n=1}d(u_{n}, u_{n+1}) \leq\sum^{\infty}_{n=1}\biggl( \frac {3r_{1}}{1-2r_{1}}\biggr)^{n-1} d(u_{1}, u_{2})< \infty. $$

Hence we conclude that $\{u_{n}\}$ is a Cauchy sequence. Since X is complete, there is some point $z \in X$ such that

$$\lim_{n\to\infty}u_{n} = z. $$

Now, we will show that $d(z,Tx)\leq( \frac{3r}{1-2r})d(x,Tx)$ for all $x\in X\setminus\{z\}$.

Let $x\in X\setminus\{z\}$. Since $u_{n}\to z$, there exists $n_{0} \in N$ such that $d(z, u_{n})\leq (\frac{1}{3})d(z, x)$ for all $n\geq n_{0}$. By using (2.2), we get

$$\varphi(r) d(u_{n}, Tu_{n}) \leq d(x, u_{n}). $$

Then from (2.15) we have

$$H(Tu_{n}, Tx) \leq r\bigl[d(u_{n}, x)+d(u_{n}, Tu_{n})+ d(x, Tx)+d(u_{n}, Tx)+d(x, Tu_{n})\bigr]. $$

Since $u_{n+1} \in Tu_{n}$, $d(u_{n+1}, Tx) \leq H(Tu_{n}, Tx)$, so that

$$d(u_{n+1}, Tx) \leq r \bigl[d(u_{n}, x)+d(u_{n}, u_{n+1})+ d(x, Tx)+d(u_{n}, Tx)+d(x, u_{n+1})\bigr] $$

for all $n\geq n_{0}$. Letting $n \to\infty$, we obtain

$$\begin{aligned} d(z, Tx) &\leq r\bigl[2d(z, x)+d(x, Tx)+d(z, Tx)\bigr] \\ & \leq r 3d(z, x)+r2d(z, Tx). \end{aligned}$$

It follows that

$$\begin{aligned} d(z, Tx) \leq\biggl( \frac{3r}{1-2r}\biggr)d(x, Tx) \quad \mbox{for all } x\in X\setminus\{z\}. \end{aligned}$$

(2.16)

Next, we show that $z \in Tz$. Suppose that z is not an element in Tz.

Case (i): $0\leq r<\frac{\sqrt{5}-1}{4+2\sqrt{5}}$. Let $a \in Tz$. Then $a\neq z$ and so by (2.16), we have

$$d(z, Ta) \leq \biggl(\frac{3r}{1-2r}\biggr)d(a, Ta). $$

On the other hand, since $\varphi(r) d(z, Tz) = d(z, Tz)\leq d(z, a)$, from (2.15) we have

$$\begin{aligned} H(Tz, Ta) \leq r \bigl[d(z,a)+d(z, Tz)+d(a, Ta)+d(z, Ta)+d(a, Tz)\bigr]. \end{aligned}$$

So

$$\begin{aligned} d(a, Ta)&\leq H(Tz, Ta) \leq r \bigl[2d(z,a)+d(a, Ta)+d(z, Ta) \bigr] \\ &\leq r \bigl[3d(z,a)+2d(a, Ta)\bigr]. \end{aligned}$$

(2.17)

Since $d(z,a)\leq d(z, Tz)+d( Tz,a)=d(z, Tz)$, we have

$$\begin{aligned} d(a, Ta)\leq\biggl(\frac{3r}{1-2r}\biggr) d(z, Tz). \end{aligned}$$

Using (2.15), (2.16), and (2.17), we have

$$\begin{aligned} d(z, Tz)&\leq d(z, Ta)+H( Ta,Tz) \\ &\leq\biggl(\frac{3r}{1-2r}\biggr)d(a, Ta)+S(a,z) \\ &\leq\biggl(\frac{3r}{1-2r}\biggr)d(a, Ta)+ r \bigl[d(z,a)+d(z, Tz)+d(a, Ta)+d(z, Ta)+d(a, Tz)\bigr] \\ &\leq\biggl(\frac{3r}{1-2r}\biggr)d(a, Ta)+3r d(z,a) \\ &\leq\biggl(\frac{3r}{1-2r}\biggr)^{2}d(z, Tz)+\biggl( \frac{3r}{1-2r}\biggr) d(z, Tz) \\ &\leq\bigl(k^{2}+k\bigr)d(z, Tz), \end{aligned}$$

where $k=\frac{3r}{1-2r}$.

Since $0\leq r<\frac{\sqrt{5}-1}{4+2\sqrt{5}}$, we have $0\leq k^{2}+k<1$ and so, $d(z, Tz)< d(z, Tz)$, a contradiction. Thus $z\in Tz$.

Case (ii): $\frac{\sqrt{5}-1}{4+2\sqrt{5}}\leq r <\frac{1}{5}$. Let $x\in X$.

If $x = z$, then $H(T x, T z) \leq r [d(x,z)+d(x,Tx)+d(z,Tz)+d(x,Tz)+d(z,Tx)]$ holds. If $x \neq z$, then for all $n\in\mathbb{N}$, there exists $y_{n} \in Tx$ such that

$$d(z, y_{n}) \leq d(z,Tx) + \biggl(\frac{1}{n}\biggr)d(x, z). $$

We consider

$$\begin{aligned} d(x, Tx)&\leq d(x, y_{n}) \\ &\leq d(x,z)+d(z, y_{n}) \\ &\leq d(x,z)+d(z,Tx) + \biggl(\frac{1}{n}\biggr)d(x, z) \\ &\leq d(x,z)+\biggl(\frac{3r}{1-2r}\biggr)d(x, Tx)+ \biggl(\frac{1}{n} \biggr)d(x, z) . \end{aligned}$$

Thus, $(\frac{1-5r}{1-2r})d(x, Tx)\leq(1+\frac{1}{n})d(x,z)$. Take $n\to\infty$, we obtain

$$\biggl( \frac{1-5r}{1-2r}\biggr)d(x, Tx)\leq d(x,z), $$

by using (2.15), implies $H(T x, T z) \leq S(x,z)$, where $S(x,z)=r [d(x,z)+d(x,Tx)+d(z,Tz)+d(x,Tz)+d(z,Tx)]$.

Hence, as $u_{n+1}\in Tu_{n}$, it follows that with $x=u_{n}$

$$\begin{aligned} d(z, Tz)&=\lim_{n\to\infty}d(u_{n+1}, Tz) \\ &\leq H(T u_{n}, T z) \\ &\leq\lim_{n\to\infty}r\bigl[d(u_{n},z)+d(u_{n},Tu_{n})+d(z,Tz)+d(u_{n},Tz)+d(z,Tu_{n}) \bigr] \\ &\leq\lim_{n\to\infty} \bigl[r d(u_{n},z)+r d(u_{n},u_{n+1})+r d(z,Tz)+r d(u_{n},Tz)+r d(z,u_{n+1})\bigr] \\ &\leq(2r)d(z,Tz). \end{aligned}$$

(2.18)

Using (2.18), we have $(1-2r)d(z, Tz)\leq0$, which implies $d(z, Tz)=0$. Since Tz is closed, we have $z\in Tz$. This completes the proof. □

Example 2.4

Let $X=[0,\frac{1}{2}]$ with the metric $d(x,y)=\frac {\vert x-y\vert }{\vert x-y\vert +1}$ for all $x,y \in X$. Define $T: X\to \operatorname {CB}(X)$ by

$$T(x)= \bigl[0,x^{2}\bigr]. $$

Proof

We show that T satisfies (2.15). Let $x,y\in X$. Thus, if $x^{2}\leq y$, then

$$\varphi\biggl(\frac{1}{6}\biggr)d(x, T x)=\frac{\vert x-x^{2}\vert }{\vert x-x^{2}\vert +1}\geq \frac {\vert x-y\vert }{\vert x-y\vert +1}=d(x,y). $$

But if $x^{2}> y$, then

$$\varphi\biggl(\frac{1}{6}\biggr)d(x, T x)=\frac{\vert x-x^{2}\vert }{\vert x-x^{2}\vert +1}\leq \frac {\vert x-y\vert }{\vert x-y\vert +1}=d(x,y) $$

and

$$\begin{aligned} &H(T x, T y) \\ &\quad =\frac{\vert x^{2}-y^{2}\vert }{\vert x^{2}-y^{2}\vert +1} \\ &\quad =\frac{1}{6}\frac{6\vert x^{2}-y^{2}\vert }{\vert x^{2}-y^{2}\vert +1} \\ &\quad =\frac{1}{6}\biggl\{ \frac{\vert x^{2}-y^{2}\vert }{\vert x^{2}-y^{2}\vert +1}+\frac {\vert x^{2}-y^{2}\vert }{\vert x^{2}-y^{2}\vert +1}+ \frac{\vert x^{2}-y^{2}\vert }{\vert x^{2}-y^{2}\vert +1}+\frac {2\vert x^{2}-y^{2}\vert }{\vert x^{2}-y^{2}\vert +1}+\frac{\vert x^{2}-y^{2}\vert }{\vert x^{2}-y^{2}\vert +1}\biggr\} \\ &\quad < \frac{1}{6}\biggl\{ \frac{\vert x-y\vert }{\vert x-y\vert +1}+\frac{\vert x-x^{2}\vert }{\vert x-x^{2}\vert +1}+ \frac {\vert y-y^{2}\vert }{\vert y-y^{2}\vert +1}+\frac{\vert x-y^{2}\vert }{\vert x-y^{2}\vert +1}+\frac {\vert y-x^{2}\vert }{\vert y-x^{2}\vert +1}\biggr\} \\ &\quad = \frac{1}{6}\bigl\{ d(x,y)+d(x,Tx)+d(y,Ty)+d(x,Ty)+d(y,Tx)\bigr\} \\ &\quad = \frac{1}{6} S(x,y), \end{aligned}$$

(2.19)

where $r= \frac{1}{6}$.

Thus we see that T satisfies condition (2.15) and satisfies Theorem 2.3. So there exists $z\in X$ such that $z \in T z$. Moreover, $0\in T(0)$. □

Theorem 2.5

Define a non-increasing function φ from $[0,1)$ into $(0,1]$ by

$$\varphi(r)= \textstyle\begin{cases} 1, &if 0\leq r< \frac{\sqrt{5}-1}{2},\\ 1-r, &if \frac{\sqrt{5}-1}{2}\leq r < 1. \end{cases} $$

Let $\alpha\in[0,\frac{1}{2})$ and $r=\frac{\alpha}{1-\alpha}$, and let $(X, d)$ be a complete metric space and let T be a mapping from X into $\operatorname {CB}(X)$.

Assume that

$$\begin{aligned} \varphi(r)d(x, T x) \leq d(x,y)\quad \textit{implies} \quad H(T x, T y) \leq\alpha M(x,y) \end{aligned}$$

(2.20)

where $M(x,y)=\max\{d(x,y),d(x,Tx),d(y,Ty),d(x,Ty),d(y,Tx)\}$, for all $x, y\in X$. Then there exists $z\in X$ such that $z \in T z$.

Proof

Let $r_{1}$ be a real number such that $0 \leq r < r_{1} < \frac {1}{2}$. Let $u_{1} \in X$ and $u_{2} \in T u_{1}$ be arbitrary. Since $u_{2} \in T u_{1}$, we have $d(u_{2},T u_{2}) \leq H(T u_{1},T u_{2})$ and

$$\varphi(r)d(u_{1},T u_{1}) \leq d(u_{1}, T u_{1})\leq d(u_{1}, u_{2}). $$

Thus, from the assumption (2.20),

$$d(u_{2},T u_{2}) \leq H(T u_{1}, T u_{2}) \leq\alpha M(u_{1},u_{2}) $$

where $M(u_{1},u_{2})=\max\{ d(u_{1},u_{2}),d(u_{1},Tu_{1}),d(u_{2},Tu_{2}),d(u_{1},Tu_{2}),d(u_{2},Tu_{1})\}$.

Consider

$$\begin{aligned} d(u_{2},T u_{2}) &\leq\alpha\max\bigl\{ d(u_{1},u_{2}),d(u_{1},Tu_{1}),d(u_{2},Tu_{2}),d(u_{1},Tu_{2}),d(u_{2},Tu_{1}) \bigr\} \\ &= \alpha\max\bigl\{ d(u_{1},u_{2}),d(u_{1},Tu_{2}) \bigr\} . \end{aligned}$$

If $\max\{d(u_{1},u_{2}),d(u_{1},Tu_{2})\}=d(u_{1},Tu_{2})$, then

$$\begin{aligned} d(u_{2},T u_{2}) &\leq\alpha d(u_{1},Tu_{2}) \\ &\leq\alpha d(u_{1},u_{2})+\alpha d(u_{2},Tu_{2}) \end{aligned}$$

and then

$$d(u_{2},T u_{2})\leq\biggl(\frac{\alpha}{1-\alpha} \biggr)d(u_{1},u_{2})= rd(u_{1},u_{2}), $$

where $r=\frac{\alpha}{1-\alpha}$.

So there exists $u_{3}\in Tu_{2}$ such that $d(u_{2},u_{3})\leq r_{1}d(u_{1},u_{2})$. Thus, we can construct a sequence $\{x_{n}\}$ in X such that $u_{n+1} \in Tu_{n}$ and

$$d(u_{n+1}, u_{n+2}) \leq r_{1}d(u_{n}, u_{n+1}). $$

Hence, by induction

$$d(u_{n}, u_{n+1}) \leq(r_{1})^{n-1} d(u_{1}, u_{2}). $$

Then by the triangle inequality, we have

$$\sum^{\infty}_{n=1}d(u_{n}, u_{n+1}) \leq\sum^{\infty}_{n=1}(r_{1})^{n-1} d(u_{1}, u_{2})< \infty. $$

Hence we conclude that $\{u_{n}\}$ is a Cauchy sequence. Since X is complete, there is some point $z \in X$ such that

$$\lim_{n\to\infty}u_{n} = z. $$

Now, we will show that $d(z,Tx)\leq rd(x,Tx)$ for all $x\in X\setminus \{z\}$.

Let $x\in X\setminus\{z\}$. Since $u_{n}\to z$, there exists $n_{0} \in N$ such that $d(z, u_{n})\leq (\frac{1}{3})d(z, x)$ for all $n\geq n_{0}$. By using (2.2), we get

$$\varphi(r) d(u_{n}, Tu_{n}) \leq d(x, u_{n}). $$

Then from (2.20), we have

$$H(Tu_{n}, T_{x}) \leq\alpha\max\bigl\{ d(u_{n}, x),d(u_{n}, Tu_{n}), d(x, Tx), d(u_{n}, Tx),d(x, Tu_{n})\bigr\} . $$

Since $u_{n+1} \in Tu_{n}$, we have $d(u_{n+1}, T_{x}) \leq H(Tu_{n}, T_{x})$, so that

$$d(u_{n+1}, Tx) \leq\alpha\max\bigl\{ d(u_{n}, x),d(u_{n}, u_{n+1}), d(x, Tx), d(u_{n}, Tx),d(x, u_{n+1})\bigr\} $$

for all $n\geq n_{0}$. Letting $n \to\infty$, we obtain

$$\begin{aligned}& d(z, Tx) \leq\alpha\max\bigl\{ d(z, x), d(x, Tx), d(z, Tx)\bigr\} . \end{aligned}$$

We obtain

$$\begin{aligned} d(z, Tx) \leq\biggl( \frac{\alpha}{1-\alpha}\biggr)d(x, Tx)=rd(x, Tx) \quad \mbox{for all } x\in X\setminus\{z\}. \end{aligned}$$

(2.21)

Next, we show that $z \in Tz$. Suppose that z is not an element in Tz.

Case (i): $0\leq r<\frac{\sqrt{5}-1}{2}$. Let $a \in Tz$. Then $a\neq z$ and so by (2.21), we have

$$d(z, Ta) \leq rd(a, Ta). $$

On the other hand, since $\varphi(r) d(z, Tz) = d(z, Tz)\leq d(z, a)$, from (2.20) we have

$$H(Tz, Ta) \leq\alpha\max\bigl\{ d(z,a),d(z, Tz), d(a, Ta),d(z, Ta), d(a, Tz)\bigr\} . $$

So

$$\begin{aligned} d(a, Ta)\leq H(Tz, Ta) \leq\alpha\max\bigl\{ d(z,a),d(z, Tz),d(z, Ta)\bigr\} . \end{aligned}$$

(2.22)

It implies that

$$d(a, Ta)\leq\alpha\max\bigl\{ d(z,a),d(z, Tz),d(z, Ta)\bigr\} . $$

Since $d(z,a)\leq d(z, Tz)+d( Tz,a)=d(z, Tz)$, we have

$$\begin{aligned} d(a, Ta)\leq r d(z, Tz). \end{aligned}$$

(2.23)

Using (2.20), (2.21), (2.22), and (2.23), we have

$$\begin{aligned} d(z, Tz)&\leq d(z, Ta)+H( Ta,Tz) \\ &\leq rd(a, Ta)+\alpha\max\bigl\{ d(z,a), d(z, Tz),d(z, Ta)\bigr\} \\ &\leq rd(a, Ta)+\alpha\max\bigl\{ d(z,a), d(z, Tz),rd(a, Ta)\bigr\} \\ &\leq rd(a, Ta)+\alpha\max\bigl\{ d(z,a), d(z, Tz)\bigr\} \\ &\leq rd(a, Ta)+\alpha d(z, Tz) \\ &\leq(r)^{2}d(z, Tz)+rd(z, Tz) \\ &\leq\bigl(r^{2}+r\bigr)d(z, Tz), \end{aligned}$$

where $r=\frac{\alpha}{1-\alpha}$.

Since $r<\frac{\sqrt{5}-1}{2}$, we have $r^{2}+r<1$ and so $d(z, Tz)< d(z, Tz)$, a contradiction. Thus $z\in Tz$.

Case (ii) $\frac{\sqrt{5}-1}{2}\leq r <1$. Let $x\in X$. If $x = z$, then $H(T x, T z) \leq\alpha\max\{ d(x,z),d(x,Tx),d(z,Tz), d(x,Tz),d(z,Tx)\}$ holds. If $x \neq z$, then for all $n\in\mathbb{N}$, there exists $y_{n} \in Tx$ such that

$$d(z, y_{n}) \leq d(z,Tx) + \biggl(\frac{1}{n}\biggr)d(x, z). $$

We consider

$$\begin{aligned} d(x, Tx)&\leq d(x, y_{n}) \\ &\leq d(x,z)+d(z, y_{n}) \\ &\leq d(x,z)+d(z,Tx) + \biggl(\frac{1}{n}\biggr)d(x, z) \\ &\leq d(x,z)+ rd(x, Tx)+ \biggl(\frac{1}{n}\biggr)d(x, z) . \end{aligned}$$

Thus, $( 1-r)d(x, Tx)\leq(1+\frac{1}{n})d(x,z)$. Take $n\to\infty$, we obtain

$$( 1-r)d(x, Tx)\leq d(x,z), $$

by using (2.20), this implies $H(T x, T z) \leq\alpha\max\{ d(x,z),d(x,Tx),d(z,Tz),d(x,Tz),d(z,Tx)\}$. Hence, as $u_{n+1}\in Tu_{n}$, it follows that with $x=u_{n}$

$$\begin{aligned} d(z, Tz)&=\lim_{n\to\infty}d(u_{n+1}, Tz) \\ &\leq H(T u_{n}, T z) \\ &\leq\lim_{n\to\infty}\alpha\max\bigl\{ d(u_{n},z),d(u_{n},Tu_{n}),d(z,Tz),d(u_{n},Tz),d(z,Tu_{n}) \bigr\} \\ &\leq\lim_{n\to\infty}\alpha\max\bigl\{ d(u_{n},z),d(u_{n},u_{n+1}),d(z,Tz),d(u_{n},Tz),d(z,u_{n+1}) \bigr\} \\ &\leq\alpha d(z,Tz). \end{aligned}$$

Therefore, $(1-\alpha)d(z, Tz)\leq0$, which implies $d(z, Tz)=0$. Since Tz is closed, we have $z\in Tz$. This completes the proof. □

Corollary 2.6

Let be $(X, d)$ a complete metric space and let T be a mapping from X into $\operatorname {CB}(X)$. Let $\alpha\in[0,\frac{1}{5})$ and $r=5\alpha$. Assume that

$$\varphi(r)d(x, T x) \leq d(x,y) \quad \textit{implies} \quad H(T x, T y) \leq S(x,y) $$

where $S(x,y)=\alpha d(x,y)+\alpha d(x,Tx)+\alpha d(y,Ty)+\alpha d(x,Ty)+\alpha d(y,Tx)$ for all $x, y\in X$, where the function φ is defined as Theorem  2.5. Then there exists $z\in X$ such that $z \in T z$.

Remark 2.7

We see that Theorem 2.5 is a multi-valued mapping generalization of Theorem 2.3 of Kikkawa and Suzuki [7] and therefore the Kannan fixed point theorem [6] for generalized Kannan mappings.

Theorem 2.8

Define a non-increasing function φ from $[0,1)$ into $(0,1]$ by

$$\varphi(r)= \textstyle\begin{cases} 1, &\textit{if }0\leq r< \frac{1}{2},\\ 1-r, &\textit{if }\frac{1}{2}\leq r < 1. \end{cases} $$

Let $\alpha\in[0,\frac{1}{5})$ and $r=\frac{3\alpha}{1-2\alpha}$, and let be $(X, d)$ a complete metric space and let T be a mapping from X into $\operatorname {CB}(X)$.

Assume that

$$\begin{aligned} \varphi(r)d(x, T x) \leq d(x,y) \quad \textit{implies} \quad H(T x, T y) \leq S(x,y) \end{aligned}$$

(2.24)

where $S(x,y)=\alpha d(x,y)+\alpha d(x,Tx)+\alpha d(y,Ty)+\alpha d(x,Ty)+\alpha d(y,Tx)$ for all $x, y\in X$. Then there exists $z\in X$ such that $z \in T z$.

Proof

Let $r_{1}$ be a real number such that $0 \leq r < r_{1} < 1$. Let $u_{1} \in X$ and $u_{2} \in T u_{1}$ be arbitrary. Since $u_{2} \in T u_{1}$, we have $d(u_{2},T u_{2}) \leq H(T u_{1},T u_{2})$ and

$$\varphi(r)d(u_{1},T u_{1}) \leq d(u_{1}, T u_{1})\leq d(u_{1}, u_{2}). $$

Thus, from the assumption (2.24),

$$d(u_{2},T u_{2}) \leq H(T u_{1}, T u_{2}) \leq S(u_{1},u_{2}) $$

where $S(u_{1},u_{2})=\alpha d(u_{1},u_{2})+\alpha d(u_{1},Tu_{1})+\alpha d(u_{2},Tu_{2})+\alpha d(u_{1},Tu_{2})+\alpha d(u_{2},Tu_{1})$. Consider

$$\begin{aligned} d(u_{2},T u_{2}) &\leq\alpha d(u_{1},u_{2})+ \alpha d(u_{1},Tu_{1})+\alpha d(u_{2},Tu_{2})+ \alpha d(u_{1},Tu_{2})+\alpha d(u_{2},Tu_{1}) \\ &\leq3\alpha d(u_{1},u_{2})+2\alpha d(u_{2},Tu_{2}). \end{aligned}$$

Then

$$\begin{aligned} d(u_{2},T u_{2}) &\leq\biggl(\frac{3\alpha}{1-2\alpha } \biggr)d(u_{1},u_{2})=rd(u_{1},u_{2}), \end{aligned}$$

where $r=\frac{3\alpha}{1-2\alpha}$.

So there exists $u_{3}\in Tu_{2}$ such that $d(u_{2},u_{3})\leq r_{1}d(u_{1},u_{2})$. Thus, we can construct a sequence $\{x_{n}\}$ in X such that $u_{n+1} \in Tu_{n}$ and

$$d(u_{n+1}, u_{n+2}) \leq r_{1}d(u_{n}, u_{n+1}). $$

Hence, by induction

$$d(u_{n}, u_{n+1}) \leq(r_{1})^{n-1} d(u_{1}, u_{2}). $$

Then by the triangle inequality, we have

$$\sum^{\infty}_{n=1}d(u_{n}, u_{n+1}) \leq\sum^{\infty}_{n=1}(r_{1})^{n-1} d(u_{1}, u_{2})< \infty. $$

Hence we conclude that $\{u_{n}\}$ is a Cauchy sequence. Since X is complete, there is some point $z \in X$ such that

$$\lim_{n\to\infty}u_{n} = z. $$

Now, we will show that $d(z,Tx)\leq rd(x,Tx)$ for all $x\in X\setminus \{z\}$.

Let $x\in X\setminus\{z\}$. Since $u_{n}\to z$, there exists $n_{0} \in N$ such that $d(z, u_{n})\leq (\frac{1}{3})d(z, x)$ for all $n\geq n_{0}$. By using (2.2), we get

$$\begin{aligned} \varphi(r) d(u_{n}, Tu_{n}) &\leq d(x, u_{n}). \end{aligned}$$

Then from (2.1), we have

$$H(Tu_{n}, T_{x}) \leq\alpha\bigl[d(u_{n}, x)+d(u_{n}, Tu_{n})+ d(x, Tx)+d(u_{n}, Tx)+d(x, Tu_{n})\bigr] . $$

Since $u_{n+1} \in Tu_{n}$, $d(u_{n+1}, T_{x}) \leq H(Tu_{n}, T_{x})$, so that

$$d(u_{n+1}, Tx) \leq\alpha\bigl[d(u_{n}, x)+d(u_{n}, u_{n+1})+ d(x, Tx)+d(u_{n}, Tx)+d(x, u_{n+1})\bigr] $$

for all $n\geq n_{0}$. Letting $n \to\infty$, we obtain

$$\begin{aligned} d(z, Tx) &\leq\alpha\bigl[2d(z, x)+d(x, Tx)+d(z, Tx)\bigr] \\ & \leq\alpha3d(z, x)+\alpha2d(z, Tx). \end{aligned}$$

It follows that

$$\begin{aligned} d(z, Tx) \leq\biggl( \frac{3\alpha}{1-2\alpha }\biggr)d(x, Tx)=rd(x, Tx) \quad \mbox{for all } x\in X\setminus\{z\}. \end{aligned}$$

(2.25)

Next, we show that $z \in Tz$. Suppose that z is not element in Tz.

Case (i): $0\leq r<\frac{1}{2}$. Let $a \in Tz$. Then $a\neq z$ and so by (2.25), we have

$$d(z, Ta) \leq rd(a, Ta). $$

On the other hand, since $\varphi(r) d(z, Tz) = d(z, Tz)\leq d(z, a)$, from (2.24) we have

$$\begin{aligned} H(Tz, Ta) \leq\alpha\bigl[d(z,a)+d(z, Tz)+d(a, Ta)+d(z, Ta)+d(a, Tz)\bigr]. \end{aligned}$$

So

$$\begin{aligned} d(a, Ta)&\leq H(Tz, Ta) \leq\alpha\bigl[2d(z,a)+d(a, Ta)+d(z, Ta) \bigr] \\ &\leq\alpha\bigl[3d(z,a)+2d(a, Ta)\bigr]. \end{aligned}$$

(2.26)

Since $d(z,a)\leq d(z, Tz)+d( Tz,a)=d(z, Tz)$, we have

$$\begin{aligned} d(a, Ta)\leq\biggl(\frac{3\alpha}{1-2\alpha}\biggr) d(z, Tz)=rd(z, Tz). \end{aligned}$$

(2.27)

Using (2.24), (2.26), and (2.27), we have

$$\begin{aligned} d(z, Tz)&\leq d(z, Ta)+H( Ta,Tz) \\ &\leq rd(a, Ta)+S(a,z) \\ &\leq rd(a, Ta)+ \alpha\bigl[d(z,a)+d(z, Tz)+d(a, Ta)+d(z, Ta)+d(a, Tz)\bigr] \\ &\leq(r+2\alpha)d(a, Ta)+3\alpha d(z,a) \\ &\leq(r+2\alpha)rd(z, Tz)+3\alpha d(z, Tz) \\ &\leq(r+r)rd(z, Tz)+r d(z, Tz) \\ &\leq\bigl(2r^{2}+r\bigr)d(z, Tz). \end{aligned}$$

Since $0\leq r<\frac{1}{2}$, we have $0\leq2r^{2}+r<1$ and so, $d(z, Tz)< d(z, Tz)$, a contradiction. Thus $z\in Tz$.

Case (ii): $\frac{1}{2}\leq r <1$. Let $x\in X$. If $x = z$, then $H(T x, T z) \leq\alpha [d(x,z)+d(x,Tx)+d(z,Tz)+d(x,Tz)+d(z,Tx)]$ holds. If $x \neq z$, then for all $n\in\mathbb{N}$, there exists $y_{n} \in Tx$ such that

$$d(z, y_{n}) \leq d(z,Tx) + \biggl(\frac{1}{n}\biggr)d(x, z). $$

We consider

$$\begin{aligned} d(x, Tx)&\leq d(x, y_{n}) \\ &\leq d(x,z)+d(z, y_{n}) \\ &\leq d(x,z)+d(z,Tx) + \biggl(\frac{1}{n}\biggr)d(x, z) \\ &\leq d(x,z)+rd(x, Tx)+ \biggl(\frac{1}{n}\biggr)d(x, z) . \end{aligned}$$

Thus, $(1-r)d(x, Tx)\leq(1+\frac{1}{n})d(x,z)$. Take $n\to\infty$, we obtain

$$( 1-r)d(x, Tx)\leq d(x,z), $$

by using (2.24), this implies $H(T x, T z) \leq S(x,z)$, where $S(x,z)=\alpha[d(x,z)+d(x,Tx)+d(z,Tz)+d(x,Tz)+d(z,Tx)]$. Hence, as $u_{n+1}\in Tu_{n}$, it follows that with $x=u_{n}$

$$\begin{aligned} d(z, Tz)&=\lim_{n\to\infty}d(u_{n+1}, Tz) \\ &\leq H(T u_{n}, T z) \\ &\leq\lim_{n\to\infty}\alpha\bigl[d(u_{n},z)+d(u_{n},Tu_{n})+d(z,Tz)+d(u_{n},Tz)+d(z,Tu_{n}) \bigr] \\ &\leq\lim_{n\to\infty} \bigl[\alpha d(u_{n},z)+\alpha d(u_{n},u_{n+1})+\alpha d(z,Tz)+\alpha d(u_{n},Tz)+ \alpha d(z,u_{n+1})\bigr] \\ &\leq(2\alpha) d(z,Tz). \end{aligned}$$

(2.28)

Therefore, $(1-2\alpha)d(z, Tz)\leq0$, which implies $d(z, Tz)=0$. Since Tz is closed, we have $z\in Tz$. This completes the proof. □