# 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).

## 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

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

2. (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 $$x, y \in X$$.

3. (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  introduced a condition as follows:

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

2. (v)

Ciric  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 ).

In 2008, Suzuki  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  generalized the Kannan mapping resulting in the following theorem.

### Theorem 1.1

(Kikkawa and Suzuki )

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 )

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  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  proved a multi-valued extension of the Banach contraction theorem as follows.

### Theorem 1.3

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  is a generalization of Nadler.

### Theorem 1.4

(Kikkawa and Suzuki )

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  generalized the result of Kannan (iii) and Nadler.

### Theorem 1.5

(Damjanović and Dorić )

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 ) 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 .

## 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  and therefore the Kannan fixed point theorem  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. □

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## Acknowledgements

The authors would like to thank the Thailand Research Fund under the project RTA5780007 and Science Achievement Scholarship of Thailand, which provides funding for our research.

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Correspondence to Chakkrid Klin-eam.

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All authors contributed equally to the writing of this paper. All authors read and approved the final manuscript.

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