Existence and data dependence of the fixed points of generalized contraction mappings with applications

Abstract

The aim of this paper is to introduce a new type of generalized multivalued contraction mappings and to present some results regarding fixed points of new class of multivalued contractions. As applications we obtain some basic results in fixed point theory like characterization of metric completeness, data dependence of fixed points and homotopy result. We prove the existence and uniqueness of bounded solution of functional equation arising in dynamic programming. Our results generalize, extend and unify various comparable results in the existing literature.

1 Introduction and preliminaries

The Hausdorff metric $H$ induced by the metric $d$ of $X$ is given by

$$\begin{aligned} H(A,B)=\max \left\{ \sup _{x\in A}d(x,B),\sup _{y\in B}d(y,A)\right\} \end{aligned}$$

for every $A,B\in CB(X),$ where $CB(X)$ denotes the collection of closed and bounded subsets of $X.$ It is well known that if $(X,d)$ is a complete metric space, then the pair $(CB(X),H)$ is a complete metric space. In 1969, Nadler [16] obtained the following multivalued version of Banach contraction principle.

Theorem 1.1

Let $(X,d)$ be a complete metric space and $T:X\longrightarrow CB(X)$ a multivalued mapping such that

$$\begin{aligned} H(Tx,Ty)\le kd(x,y) \end{aligned}$$

for all $x,y\in X$ and for some $k\in (0,1).$ Then there exists a fixed point $x\in X$ of $T,$ i.e., $x\in Tx$.

A number of fixed point theorems (see [5, 6, 8, 9, 12, 14, 19, 21]) have been proved in the context of generalization of Theorem 1.1. Kikkawa and Suzuki [13] refined Nadler’s result by proving the following result.

Theorem 1.2

Let $(X,d)$ be a complete metric space and $T:X\rightarrow CB(X)$ a multivalued mapping. Define the mapping $\beta :[0,1)\rightarrow ( \frac{1}{2},1]$ by $\beta (b)=\dfrac{1}{1+b}.$ If there exists a $b\in [0,1)$ such that

$$\begin{aligned} \beta (b)d(x,Tx)\le d(x,y)\text { implies }H(Tx,Ty)\le bd(x,y) \end{aligned}$$

for all $x,y\in X.$ Then $T$ has a fixed point. In this case, we call $T$ as $b$-KS multivalued operator.

Theorem 1.2 has further been generalized in [7, 10, 11, 15, 23].

Definition 1.3

[20] Let $(X,d)$ be a metric space. A mapping $T:X\rightarrow CB(X)$ is called a multivalued weakly Picard operator (MWP operator), if for all $x\in X$ and $y\in Tx$, there exists a sequence $\{x_{n}\}_{n\ge 0}$ satisfying (a) $x_{0}=x,$ $x_{1}=y$ (b) $x_{n+1}\in Tx_{n}$ for all $n\ge 0$ (c) the sequence $\{x_{n}\}_{n\ge 0}$ converges to a fixed point of $T.$

The sequence $\{x_{n}\}$ satisfying (a) and (b) is called a sequence of successive approximations (briefly s.s.a.) of $T$ starting from $x_{0}$.

Let $(X,d)$ be a metric space and $T:X\longrightarrow CB(X)$ a multivalued mapping. We define

$$\begin{aligned} M_{T}(x,y)=\max \left\{ d(x,y),d(x,Tx),d(y,Ty),\frac{d(x,Ty)+d(y,Tx)}{2}\right\} \end{aligned}$$

(1)

for all $x,y\in X$.

Recently Popescu [18] introduced the following class of multivalued operators.

Definition 1.4

[18] Let $(X,d)$ be a complete metric space. A mapping $T:X\longrightarrow CB(X)$ is called an $(s,r)$-contractive multivalued operator if $r\in [0,1)$, $s\ge r$ and $x,y\in X$ with $d(y,Tx)\le sd(y,x)$ implies $H(Tx,Ty)\le rM_{T}(x,y)$.

Theorem 1.5

[18] Let $(X,d)$ be a complete metric space and $T:X\longrightarrow CB(X)$ an $(s,r)$-contractive multivalued operator with $s>r.$ Then $T\,$is a MWP operator.

In this paper, we introduce a new type of generalized multivalued contraction in metric spaces. As a result we generalize results given in [5, 13, 15, 16, 18].

2 Main results

Let $\psi :[0,1)\rightarrow (0,\dfrac{1}{2}]$ be a strictly decreasing mapping defined by

$$\begin{aligned} \psi (s)=\left\{ \begin{array}{ll} \frac{1}{2(1+s)}&{}\quad \text {if }0\le s<\frac{1}{2} \\ \frac{1-s}{2}&{}\quad \text {if }\frac{1}{2}\le s<1 \end{array} \right. . \end{aligned}$$

(2)

We define $(\psi ,r)$-contractive multivalued operators as follows:

Definition 2.1

Let $(X,d)$ be a metric space. A mapping $T:X\rightarrow CB(X)$ is said to be a $(\psi ,r)$-contractive multivalued operator if $r\in [0,1)$, $s\ge r$ and $x,y\in X$ with

$$\begin{aligned} \psi (s)(d(x,Tx)+d(y,Tx))\le d(x,y) \end{aligned}$$

(3)

implies

$$\begin{aligned} H(Tx,Ty)\le rM_{T}(x,y). \end{aligned}$$

(4)

Theorem 2.2

Let $(X,d)$ be a complete metric space and $T:X\longrightarrow CB(X)$ a $(\psi ,r)$-contractive multivalued operator. Then $T$ is a MWP operator and has a fixed point.

Proof

Let $r_{1}$ be a real number such that $0\le r<r_{1}<1$ and $r_{1}\le s.$ Let $u_{1}$ be a given point in $X$. We can arbitrary choose $u_{2}\in Tu_{1}.$ If $h=\frac{1}{\sqrt{r}},$ then there exists $u_{3}\in Tu_{2}$ such that $d(u_{2},u_{3})\le \frac{1}{\sqrt{r}}H(Tu_{1},Tu_{2}).$ As $\psi (s)\le 1,$ so we have

$$\begin{aligned} \psi (s)(d(u_{1},Tu_{1})+d(u_{2},Tu_{1})) \le d(u_{1},Tu_{1})+d(u_{2},Tu_{1}) \\ \quad \le d(u_{1},Tu_{1})\le d(u_{1},u_{2}), \end{aligned}$$

which implies that $\psi (s)(d(u_{1},Tu_{1})+d(u_{2},Tu_{1}))\le d(u_{1},u_{2}).$ Now by (4), we have

$$\begin{aligned} d(u_{2},u_{3})&\le \frac{1}{\sqrt{r}}H(Tu_{1},Tu_{2})\le r\frac{1}{\sqrt{ r}}M_{T}(u_{1},u_{2}) \\&= \sqrt{r}\max \left\{ d(u_{1},u_{2}),d(u_{1},Tu_{1}),d(u_{2},Tu_{2}),\dfrac{ d(u_{1},Tu_{2})+d(u_{2},Tu_{1})}{2}\right\} \\&\le \sqrt{r}\max \left\{ d(u_{1},u_{2}),d(u_{2},u_{3}),\dfrac{ d(u_{1},u_{2})+d(u_{2},u_{3})}{2}\right\} . \end{aligned}$$

Thus

$$\begin{aligned} d(u_{2},u_{3})\le \sqrt{r}\max \{d(u_{1},u_{2}),d(u_{2},u_{3})\}. \end{aligned}$$

If $\max \{d(u_{1},u_{2}),d(u_{2},u_{3})\}=d(u_{1},u_{2}),$ then we have $d(u_{2},u_{3})\le \sqrt{r}d(u_{1},u_{2}).$ If $\max \{d(u_{1},u_{2}),d(u_{2},u_{3})\}=d(u_{2},u_{3}),$ then we get $d(u_{2},u_{3})\le \sqrt{r}d(u_{2},u_{3})$ which implies that $d(u_{2},u_{3})=0$, that is, $u_{2}=u_{3}\in Tu_{2}.$ Hence the result follows. So we assume that $\max \{d(u_{1},u_{2}),d(u_{2},u_{3})\}=d(u_{1},u_{2}).$ Thus

$$\begin{aligned} d(u_{2},u_{3})\le \sqrt{r}d(u_{1},u_{2})\le \sqrt{r_{1}}d(u_{1},u_{2}). \end{aligned}$$

By continuing this way, we can obtain a sequence $\{u_{n}\}$ in $X$ such that $u_{n+1}\in Tu_{n}$, we have

$$\begin{aligned} d(u_{n},u_{n+1})\le (\sqrt{r_{1}})^{n-1}d(u_{1},u_{2}), \end{aligned}$$

(5)

which implies that $\lim _{n\rightarrow \infty }d(u_{n},u_{n+1})=0.$ Now we show that $\{u_{n}\}$ is a Cauchy sequence. For a positive integer $p,$ we have

$$\begin{aligned} d(u_{n},u_{n+p})&\le (d(u_{n},u_{n+1})+\ldots +d(u_{n+p-1},u_{n+p})) \\&\le ((\sqrt{r_{1}})^{n-1}d(u_{1},u_{2})+\ldots +(\sqrt{r_{1}} )^{n+p-2}d(u_{1},u_{2}))\\&\le (\sqrt{r_{1}})^{n-1}\frac{1}{1-\sqrt{r_{1}}} d(u_{1},u_{2}), \end{aligned}$$

which on taking limit as $n$ tends to infinity implies that

$$\begin{aligned} \underset{n\rightarrow \infty }{\lim }d(u_{n},u_{n+p})=0. \end{aligned}$$

(6)

Therefore $\{u_{n}\}$ is a Cauchy sequence in $(X,d).$ Since $(X,d)$ is complete, there exists an element $z\in X$ such that ${\lim }_{n\rightarrow \infty }u_{n}=z$, that is, ${\lim }_{n\rightarrow \infty } d(u_{n},z)=0.$ Next we show that

$$\begin{aligned} d(z,Tx)\le r\max \{d(z,x),d(x,Tx)\} \end{aligned}$$

(7)

for all $x\ne z.$ As $\lim _{n\rightarrow \infty }d(u_{n},z)=0,$ so there exists a positive integer $n_{0}$ such that $d(z,u_{n})<\frac{1}{9} d(z,x)$ for all $n\ge n_{0}.$ Using $u_{n+1}\in Tu_{n},$ we obtain

$$\begin{aligned} 2\psi (s)(d(u_{n},Tu_{n})+d(x,Tu_{n}))&\le d(u_{n},Tu_{n})+d(x,Tu_{n})\le d(u_{n},u_{n+1})+d(x,u_{n+1}) \\&\le d(u_{n},z)+d(z,u_{n+1})+d(x,z) +d(z,u_{n+1})\le \dfrac{4}{3}d(z,x) \\&= 2[d(z,x)-\dfrac{1}{3}d(z,x)]\le 2[d(z,x)-\dfrac{1}{9}d(z,x)] \\&\le 2[d(z,x)-d(u_{n},z)]\le 2d(u_{n},x). \end{aligned}$$

So for any $n\ge n_{0}$,

$$\begin{aligned} \psi (s)(d(u_{n},Tu_{n})+d(x,Tu_{n}))\le d(u_{n},x). \end{aligned}$$

Also from (4), we have

$$\begin{aligned} d(u_{n+1},Tx)&\le H(Tu_{n},Tx) \\&\le r\max \left\{ d(u_{n},x),d(u_{n},Tu_{n}),d(x,Tx),\frac{ d(u_{n},Tx)+d(x,Tu_{n})}{2}\right\} \\&\le r\max \left\{ d(u_{n},x),d(u_{n},u_{n+1}),d(x,Tx),\frac{ d(u_{n},Tx)+d(x,u_{n+1})}{2}\right\} . \end{aligned}$$

On taking limit as $n\rightarrow \infty$ on both sides of above inequality, we have

$$\begin{aligned} d(z,Tx)\le r\max \left\{ d(z,x),d(x,Tx),\frac{d(z,Tx)+d(x,z)}{2}\right\} . \end{aligned}$$

Now we claim that

$$\begin{aligned} d(z,Tx)\le r\max \{d(z,x),d(x,Tx)\} \end{aligned}$$

holds for all $x\ne z.$ Indeed, if we suppose that

$$\begin{aligned} \max \left\{ d(z,x),d(x,Tx),\frac{d(z,Tx)+d(x,z)}{2}\right\} =\frac{d(z,Tx)+d(x,z)}{2}, \end{aligned}$$

then we have $d(z,Tx)\le r\frac{d(z,Tx)+d(x,z)}{2}.$ As $r<1,$ so we have $d(z,Tx)\le \frac{2r}{2-r}d(x,z)<rd(x,z)\le r\max \{d(z,x),d(x,Tx)\}.$ Thus

$$\begin{aligned} d(z,Tx)\le r\max \{d(z,x),d(x,Tx)\} \end{aligned}$$

holds for all $x\ne z.$ If $x=z$ then $d(z,Tz)\le r\max \{d(z,z),d(z,Tz)\}$ implies that $d(z,Tz)=0,$ that is, $z\in Tz$ . Now we prove that $z\in Tz,$ given that

$$\begin{aligned} d(z,Tx)\le r\max \{d(z,x),d(x,Tx)\} \end{aligned}$$

holds for all $x\ne z.$ For this we consider the case for $0\le r\le s<1/2$. Assume on contrary that $z\notin Tz,$ we can choose $a\in Tz$ such that

$$\begin{aligned} d(a,z)<d(z,Tz)+\left( \frac{1}{2r}-1\right) d(z,Tz) \end{aligned}$$

that is

$$\begin{aligned} 2rd(a,z)<d(z,Tz). \end{aligned}$$

(8)

As $a\in Tz$ and $z\notin Tz$, so $a\ne z$, and hence we have

$$\begin{aligned} \psi (s)(d(z,Tz)+d(a,Tz))\le d(z,Tz)\le d(z,a). \end{aligned}$$

Thus

$$\begin{aligned} \psi (s)(d(z,Tz)+d(a,Tz))\le d(z,a). \end{aligned}$$

By (4), we have

$$\begin{aligned} H(Tz,Ta)&\le r\max \left\{ d(z,a),d(z,Tz),d(a,Ta),\dfrac{d(z,Ta)+d(a,Tz)}{2}\right\} \nonumber \\&\le r\max \left\{ d(z,a),d(a,Ta),\dfrac{d(z,a)+d(a,Ta)}{2}\right\} \nonumber \\&= r\max \left\{ d(z,a),d(a,Ta)\right\} . \end{aligned}$$

(9)

Clearly, $d(a,Ta)\le H(Tz,Ta)$. By (9), we obtain $H(Tz,Ta)\le r\max \left\{ d(z,a),H(Tz,Ta)\right\}$. Now $r<1$ implies that

$$\begin{aligned} H(Tz,Ta)\le rd(z,a). \end{aligned}$$

(10)

Hence $d(a,Ta)\le d(z,a).$ Now by (7), (9) and (10), we have

$$\begin{aligned} d(z,Tz)&\le d(z,Ta)+H(Tz,Ta)\le r\max \{d(z,a),d(a,Ta)\}+rd(z,a) \\&= rd(z,a)+rd(z,a)=2rd(z,a)<d(z,Tz), \end{aligned}$$

a contradiction. Hence $z\in Tz.$ If $\dfrac{1}{2}\le r\le s<1$ and $r\le s$, then first we show that

$$\begin{aligned} H(Tx,Tz)\le r\max \left\{ d(x,z),d(x,Tx),d(z,Tz),\dfrac{d(x,Tz)+d(z,Tx)}{2}\right\} \end{aligned}$$

(11)

for all $x\in X$ with $x\ne z$. Now for each $n\in N$, there exists $y_{n}\in Tx$ such that

$$\begin{aligned} d(z,y_{n})<d(z,Tx)+\dfrac{1}{n}d(x,z). \end{aligned}$$

So we have

$$\begin{aligned} d(x,Tx)+d(z,Tx)&\le d(x,y_{n})+d(z,Tx)\le d(x,z)+d(z,y_{n})+d(z,Tx) \\&< d(x,z)+2d(z,Tx)+\frac{1}{n}d(x,z). \end{aligned}$$

Hence by (7), we have

$$\begin{aligned} d(x,Tx)+d(z,Tx)<d(x,z)+2r\max \{d(z,x),d(x,Tx)\}+\dfrac{1}{n}d(x,z). \end{aligned}$$

(12)

If $\max \{d(z,x),d(x,Tx)\}=d(x,z),$ then by (12), we have

$$\begin{aligned} d(x,Tx)+d(z,Tx)&< d(x,z)+2rd(z,x)+\dfrac{1}{n}d(x,z) \\&< \left( (1+2r)+\dfrac{1}{n}\right) d(x,z)\le \left( (1+2s)+\dfrac{1}{n}\right) d(x,z), \end{aligned}$$

which implies that

$$\begin{aligned} \psi (s)(d(x,Tx)+d(z,Tx))&= \frac{1-s}{2}(d(x,Tx)+d(z,Tx)) \\&\le \frac{1}{1+2s}(d(x,Tx)+d(z,Tx))<\left( 1+\frac{1}{(1+2s)n}\right) d(x,z). \end{aligned}$$

On taking limit as $n$ tends to $\infty ,$ we obtain that

$$\begin{aligned} \psi (s)[d(x,Tx)+d(z,Tx)]\le d(x,z). \end{aligned}$$

Now by (4) with $y=z,$ we get (11). If $\max \{d(z,x),d(x,Tx)\}=d(x,Tx),$ then by (8), we have

$$\begin{aligned} d(x,Tx)\le d(x,z)+d(z,Tx)\le d(x,z)+rd(x,Tx). \end{aligned}$$

Hence

$$\begin{aligned} d(x,Tx)\le \frac{1}{1-r}d(x,z). \end{aligned}$$

Now by (12), we have

$$\begin{aligned} d(x,Tx)+d(z,Tx)&\le d(x,z)+2rd(x,Tx)+\dfrac{1}{n}d(x,z) \\&\le d(x,z)+\frac{2r}{1-r}d(x,z)+\dfrac{1}{n}d(x,z)\le \frac{2}{1-r} d(x,z)+\dfrac{1}{n}d(x,z). \end{aligned}$$

As $\dfrac{1}{2}\le r<1$ and $r\le s,$ so we have

$$\begin{aligned} \psi (s)(d(x,Tx)+d(z,Tx))&= \frac{1-s}{2}(d(x,Tx)+d(z,Tx)) \\&\le \frac{1-r}{2}(d(x,Tx)+d(z,Tx))\le d(z,x)+\frac{1-r}{2n}d(z,x), \end{aligned}$$

which on taking limit as $n$ tends to $\infty$ gives that

$$\begin{aligned} \psi (s)(d(x,Tx)+d(z,Tx))\le d(x,z). \end{aligned}$$

We get (11). Now by (11) with $x=u_{n}$ and $y=z,$ we have

$$\begin{aligned} d(u_{n+1},Tz)&\le H(Tu_{n},Tz) \\&\le r\max \left\{ d(u_{n},z),d(u_{n},Tu_{n}),d(z,Tz),\dfrac{ d(u_{n},Tz)+d(z,Tu_{n})}{2}\right\} \\&\le r\max \left\{ d(u_{n},z),\ d(u_{n},u_{n+1}),d(z,Tz),\dfrac{ d(u_{n},Tz)+d(z,u_{n+1})}{2}\right\} , \end{aligned}$$

which on taking limit as $n$ tends to $\infty$ implies that

$$\begin{aligned} d(z,Tz)\le rd(z,Tz). \end{aligned}$$

As $r<1,$ so we have $d(z,Tz)=0,$ that is, $z\in Tz.$ $\square$

Remark 2.3

Let $(X,d)$ be a complete metric space and $T:X\longrightarrow CB(X)$. We show that every $(s,r)$-contractive multivalued operator is $(\psi ,r)$-contractive multivalued operators. We consider the case when $0\le r\le s<\frac{1}{2}.$ If $d(y,Tx)\le sd(y,x)$ then we have

$$\begin{aligned} d(x,Tx)-d(y,x)\le d(y,Tx)\le sd(y,x), \end{aligned}$$

which implies that

$$\begin{aligned} d(x,Tx)\le (1+s)d(y,x), \end{aligned}$$

(13)

that is $\frac{1}{1+s}d(x,Tx)\le d(y,x).$ As $\frac{1}{1+s}\le 1$ and $\psi (s)\le \frac{1}{2},$ so we have

$$\begin{aligned} 2\psi (s)(d(x,Tx)+d(y,Tx))\le d(x,Tx)+d(y,Tx)\le (1+2s)d(y,x)\le 2d(x,y). \end{aligned}$$

Hence

$$\begin{aligned} \psi (s)(d(x,Tx)+d(y,Tx))\le d(x,y). \end{aligned}$$

If $\frac{1}{2}\le r\le s<1,$ then $1-s\le \frac{1}{2}$ and $\frac{1}{1+s} <1.$ Then we have

$$\begin{aligned} 2\psi (s)(d(x,Tx)+d(y,Tx))&= (1-s)(d(x,Tx)+d(y,Tx)) \\&\le \frac{1}{2}d(x,Tx)+\frac{1}{2}d(y,Tx)\le \frac{1+s}{2}d(x,y)+\frac{s }{2}d(y,x) \\&\le \frac{1+2s}{2}d(x,y)\le \frac{3}{2}d(x,y)\le 2d(x,y). \end{aligned}$$

Thus

$$\begin{aligned} \psi (s)(d(x,Tx)+d(y,Tx))\le d(x,y). \end{aligned}$$

Remark 2.4

Theorem 2.2 extends and generalizes results in [5, 13, 15, 16, 18].

Example 2.5

Let $X=\{0,1,2\}$ and $d$ be the metric on $X\$defined by:

$$\begin{aligned} d(0,0)&= d(1,1)=d(2,2)=0,\,d(0,1)=d(1,0)=\frac{1}{4}, \\ d(0,2)&= d(2,0)=\frac{1}{3},\,d(2,1)=d(1,2)=\frac{1}{2}. \end{aligned}$$

Define the mapping $T:X\longrightarrow CB(X)$ by

$$\begin{aligned} Tx=\left\{ \begin{array}{ll} \{0\},&{}\quad \text {when }x\ne 2 \\ \{0,1\},&{}\quad \text {when }x=2 \end{array} \right. . \end{aligned}$$

Note that, for all $x,y\in X,$ and any $s\in [0,1),$ we have

$$\begin{aligned} \psi (s)(d(x,Tx)+d(y,Tx))\le d(x,y). \end{aligned}$$

If $s=\dfrac{4}{5}>\dfrac{3}{4}=r,$ then $\psi (s)=\dfrac{1}{10}.$ Note that

$$\begin{aligned} H(Tx,Ty)\le rM_{T}(x,y) \end{aligned}$$

is satisfied for all $x,y\in X.$ Thus, all the conditions of Theorem 2.24 are satisfied.

Example 2.6

Let $X=[0,10]$ be a usual metric space. Define $T:X\rightarrow CB(X)$, where $Tx=[0,ke^{-\frac{1}{2}}x^{2}+1)],$ where $k\in (0,\frac{1}{20}).$ Fix $x,y\in X$ such that $\psi (s)(d(x,Tx)+d(y,Tx))\le d(x,y).$ Note that

$$\begin{aligned} H(Tx,Ty)&= ke^{-\frac{1}{2}}\left| x^{2}-y^{2}\right| =ke^{-\frac{1 }{2}}\left| x-y\right| \left| x+y\right| \le 20ke^{-\frac{1 }{2}}\left| x^{2}-y^{2}\right| \\&\le e^{-\frac{1}{2}}\left| x-y\right| =e^{-\frac{1}{2} }d(fx,fy)\le rM_{T}(x,y) \end{aligned}$$

for all $x,y\in X,$ where $M_{T}(x,y)$ is defined in (1) and $r=e^{- \frac{1}{2}}$. Then for any $0<r<s<1$ $T$ is $(\psi ,r)$-contractive multivalued mapping. Note that every $x\le 10\sqrt{e}-2e\sqrt{5(5e-\sqrt{e}) }$ is such that $x\in Tx.$

Corollary 2.7

Let $(X,d)$ be a complete metric space and $T:X\longrightarrow CB(X)$ a multivalued mapping. Let $\psi$ be the same as defined in Theorem 2.2 and $\psi _{1}(s)=\frac{\psi (s)}{2}$. If there exist $0\le r\le s<1$ such that

$$\begin{aligned} \psi _{1}(s)(d(x,Tx)+d(y,Tx))&\le d(x,y)\text { implies that} \nonumber \\ H(Tx,Ty)&\le rM_{T}(x,y) \end{aligned}$$

(14)

for all $x,y\in X$ whenever $x\ne y.$ Then $T$ has a fixed point.

Corollary 2.8

Let $(X,d)$ be a complete metric space and $T:X\longrightarrow CB(X)$ a multivalued mapping. Let $\psi$ be the same as defined in Theorem 2.2 and $\psi _{1}(s)=\frac{\psi (s)}{2}$. If there exist $0\le r\le s<1$ such that

$$\begin{aligned} \psi _{1}(s)(d(x,Tx)+d(y,Tx))&\le d(x,y)\text { implies } \\ H(Tx,Ty)&\le r\max \{d(x,y),d(x,Tx),d(y,Ty)\} \end{aligned}$$

for all $x,y\in X$ whenever $x\ne y.$ Then $T$ has a fixed point.

Remark 2.9

Let $(X,d)$ be a complete metric space and $T:X\longrightarrow CB(X)$ a multivalued mapping. Let $\psi$ be the same as defined in Theorem 2.2 and $\psi _{1}(s)=\frac{\psi (s)}{2}$. Suppose that there exists $0\le r\le s<1$ satisfying

$$\begin{aligned} \frac{1}{1+r}d(x,Tx)&\le d(x,y)\le \frac{1}{1-s}d(x,Tx)\text { implies } \end{aligned}$$

(15)

$$\begin{aligned} H(Tx,Ty)&\le r\max \{d(x,y),d(x,Tx),d(y,Ty)\}. \end{aligned}$$

(16)

Above contraction condition [18, Theorem 2.7] was employed to prove the existence of fixed points of $T.$ Now if $0\le r\le s<\frac{1}{2},$ then $4\psi _{1}(s)<1$ and we have

$$\begin{aligned} 4\psi _{1}(s)(d(x,Tx)+d(y,Tx))&\le \frac{1}{1+s}d(x,Tx)+\frac{1}{1+s} d(y,Tx) \\&\le \frac{2}{1+s}d(x,Tx)+\frac{1}{1+s}d(y,x) \\&\le \frac{2(1+r)}{1+s}d(x,y)+\frac{1}{1+s}d(y,x) \\&\le 4d(x,y). \end{aligned}$$

Thus

$$\begin{aligned} \psi _{1}(s)(d(x,Tx)+d(y,Tx))\le d(x,y). \end{aligned}$$

When $\frac{1}{2}\le r\le s<1.$ Then

$$\begin{aligned} 4\psi _{1}(s)(d(x,Tx)+d(y,Tx))&\le (1-s)d(x,Tx)+(1-s)d(y,Tx) \\&\le 2(1-s)d(x,Tx)+(1-s)d(y,x) \\&\le 2(1-s)d(x,Tx)+d(x,Tx) \\&\le (3-2s)d(x,Tx)\le 2(1+r)d(x,y)\le 4d(x,y). \end{aligned}$$

Hence we obtain

$$\begin{aligned} \psi _{1}(s)(d(x,Tx)+d(y,Tx))\le d(x,y). \end{aligned}$$

Corollary 2.8 can be viewed as a generalization of results in [18, Theorem 2.7] which in turn generalize the results in [13, Theorem 1.6].

Corollary 2.10

Let $(X,d)$ be a complete metric space and $T:X\longrightarrow CB(X)$ a multivalued mapping. Let $\psi$ be the same as defined in Theorem 2.2 and $\psi _{1}(s)=\frac{\psi (s)}{2}$. If there exist $0\le r\le s<1$ and $\alpha \in [0,\frac{1}{3})$ such that

$$\begin{aligned} \psi _{1}(s)(d(x,Tx)+d(y,Tx))&\le d(x,y)\text { implies } \\ H(Tx,Ty)&\le \alpha [d(x,y)+d(x,Tx)+d(y,Ty)] \end{aligned}$$

for all $x,y\in X$ whenever $x\ne y$ and $r=3\alpha .$ Then $T$ has a fixed point.

For single valued mappings, Theorem 2.2 reduces to the following corollary:

Corollary 2.11

Let $(X,d)$ be a complete metric space and $T:X\longrightarrow X$ a single valued mapping. Let $\psi (s)$ be given as in Theorem 2.2. If there exist $0\le r\le s<1$ such that

$$\begin{aligned}&\psi _{1}(s)(d(x,Tx)+d(y,Tx)) \le d(x,y)\text { implies } \\&\quad d(Tx,Ty) \le r\max \left\{ d(x,y),d(x,Tx),d(y,Ty),\dfrac{d(x,Ty)+d(y,Tx)}{2}\right\} \end{aligned}$$

for all $x,y\in X$ whenever $x\ne y.$ Then $T$ has a unique fixed point.

Proof

Existence of fixed point follows from Theorem 2.2. We prove the uniqueness. If there exist $z_{1}\ne z_{2}$ such that $z_{1}=Tz_{1}$ and $z_{2}=Tz_{2}.$ Then

$$\begin{aligned} \psi (s)(d(z_{1},Tz_{1})+d(z_{2},Tz_{1}))&\le d(z_{1},Tz_{1})+d(z_{2},Tz_{1}) \\&= d(z_{1},z_{1})+d(z_{2},z_{1})\le d(z_{1},z_{2}), \end{aligned}$$

which implies that

$$\begin{aligned} \psi (s)(d(z_{1},Tz_{1})+d(z_{2},Tz_{1}))\le d(z_{1},z_{2}). \end{aligned}$$

It follows that

$$\begin{aligned} d(z_{1},z_{2})&= d(Tz_{1},Tz_{2}) \\&\le r\max \{d(z_{1},z_{2}),d(z_{1},Tz_{1}),d(z_{2},Tz_{2}),\frac{ d(z_{2},Tz_{1})+d(z_{1},Tz_{2})}{2}\} \\&\le r\max \{d(z_{1},z_{2}),d(z_{1},z_{1}),d(z_{2},z_{2})\} \le rd(z_{1},z_{2}). \end{aligned}$$

Hence $d(z_{1},z_{2})=0,$ that is, $z_{1}=z_{2}.$ $\square$

3 Characterization of metric completeness for multivalued mappings

Motivated by the work of Suzuki [24] we prove the characterization of metric space completeness for the class of $(\psi ,r)$-contractive multivalued mappings.

Theorem 3.1

Let $(X,d)$ be a metric space then the following statements are equivalent:

• (a) $(X,d)$ is complete;

• (b) For each $r\in [0,1)$ and $s\ge r,$ every mapping $T:X\longrightarrow CB(X)$ such that $\psi (s)((d(x,Tx)+d(y,Tx))\le d(x,y)$ implies

  $$\begin{aligned} H(Tx,Ty)\le rM_{T}(x,y) \end{aligned}$$

  (17)

  for all $x,y\in X$ has a fixed point.

Proof

By Theorem (2.1) $(a)\Rightarrow (b)$. Now we prove that $(b)\Rightarrow (a).$ Suppose on contrary that $(X,d)$ is not complete. That is there exists a Cauchy sequence $\{u_{n}\}$ which does not converge. Define a function $f:X\rightarrow [0,\infty )$ by $f(x)=\lim _{n\rightarrow \infty }d(x,u_{n})$ for $x\in X.$ Since $f(x)>0$ and $\lim _{n\rightarrow \infty }f(u_{n})=0$ therefore for every $x\in X$ there exists $\upsilon \in \mathbb {N}$ such that $f(u_{\upsilon })\le \frac{\psi (s)r}{4+r+\psi (s)r}f(x).$ We put $T(x)=\{u_{n}:f(u_{n})\le \frac{\psi (s)r}{4+r+\psi (s)r}f(x)\}.$ Define $g(x)=\sup _{y\in Tx}f(y),$ then $g(x)\le \frac{\psi (s)r}{ 4+r+\psi (s)r}f(x)$ for all $x\in X.$ Since $f(y)<f(x)$ for all $y\in Tx,$ therefore $T$ has no fixed point. By the definition of mapping $f$ we have

$$\begin{aligned} f(x)-f(y)&\le d(x,y)\le f(x)+f(y)\text { for all }y\in Tx, \end{aligned}$$

(18)

$$\begin{aligned} f(y)-f(x)&\le d(x,y)\le f(x)+f(y)\text { for all }y\in Tx. \end{aligned}$$

(19)

This implies

$$\begin{aligned} f(x)-g(x)&\le d(x,Tx)\le f(x)+g(x)\text {,} \end{aligned}$$

(20)

$$\begin{aligned} H(Tx,Ty)&\le g(x)+g(y). \end{aligned}$$

(21)

Now fix $x,y\in X$ such that $\psi (s)((d(x,Tx)+d(y,Tx))\le d(x,y),$ we need to show that 17 holds. Observe that

$$\begin{aligned} \left\{ \begin{array}{l} d(x,y)\ge \psi (s)((d(x,Tx)+d(y,Tx))\ge \psi (s)(d(x,Tx) \\ \ge \psi (s)f(x)-g(x)\ge \psi (s)\left( 1-\frac{\psi (s)r}{4+r+\psi (s)r} \right) f(x)=\left( \frac{(4+r)\psi (s)}{4+r+\psi (s)r}\right) f(x). \end{array} \right. \nonumber \\ \end{aligned}$$

(22)

Case (1) when $f(y)\ge f(x),$ then by

$$\begin{aligned} H(Tx,Ty)&\le g(x)+g(y)=\frac{4+r}{4}(g(x)+g(y))-\frac{r}{4}(g(x)+g(y)) \\&\le \frac{4+r}{4}\dfrac{\psi (s)r}{4+r+\psi (s)r}(f(x)+f(y))-\frac{r}{4} (g(x)+g(y))+\frac{r}{4}(f(y)-f(x)) \\&\le \frac{r}{4}\dfrac{(4+r)\psi (s)}{4+r+\psi (s)r}(f(x)+f(y))-\frac{r}{4} (g(x)+g(y)) \\&+\frac{r}{4}(f(y)-f(x))+\frac{r}{4}(f(x)-f(y)) \\&\le \frac{r}{4}(f(x)+f(y))-\frac{r}{4}(g(x)+g(y))+\frac{r}{4}(f(y)-f(x))+ \frac{r}{4}(f(x)-f(y)) \\&\le \frac{r}{4}d(x,Tx)+\frac{r}{4}d(y,Ty)+\frac{r}{4}d(x,y)+\frac{r}{4} d(x,y) \\&\le \frac{r}{4}(4M_{T}(x,y))=rM_{T}(x,y). \end{aligned}$$

Case (2) when $f(y)<f(x),$ then

$$\begin{aligned} H(Tx,Ty)&\le g(x)+g(y)\le \dfrac{\psi (s)r}{4+r+\psi (s)r}(f(x)+f(y)) \\&= \dfrac{\psi (s)r}{4+r+\psi (s)r}f(x)+\dfrac{\psi (s)r}{4+r+\psi (s)r} f(y)\\&\le \dfrac{\psi (s)r}{4+r+\psi (s)r}f(x)+\dfrac{\psi (s)r}{4+r+\psi (s)r }f(x) \\&\le \dfrac{\psi (s)r}{4+r}d(x,Tx)+\dfrac{r}{4+r}d(x,y)\le \dfrac{r}{4} d(x,Tx)+\dfrac{r}{4}d(x,y)\\&\le \frac{r}{4}(2M_{T}(x,y))\le rM_{T}(x,y). \end{aligned}$$

Hence $\psi (s)((d(x,Tx)+d(y,Tx))\le d(x,y)$ implies

$$\begin{aligned} H(Tx,Ty)\le rM_{T}(x,y) \end{aligned}$$

for all $x,y\in X.$ this implies that $T$ has a fixed point, a contradiction. Hence $X$ is complete and consequently $(b)\Rightarrow (a).$ $\square$

4 Data dependence of the fixed point set

Let $(X,d)$ be a metric space and and $T:X\longrightarrow P(X)$ (the collection of all the subsets of $X$) be a MWP operator. Define a multivalued operator $T^{\infty }:G(T)\rightarrow P(Fix(T))$ by

$$\begin{aligned} T^{\infty }(x,y)&= \{z\in Fix(T):\text { there exists a sequence of successive approximations } \\&\text { of }T\text { starting from }(x,y)\text { that converges to }z\}. \end{aligned}$$

Further

$$\begin{aligned} G(T)=\{(x,y):x\in X,y\in Tx\} \end{aligned}$$

is called graph of multivalued mapping $T.$ A selection for $T$ is a single valued mapping $t:X\rightarrow X$ such that $tx\in Tx$ for all $x\in X.$

Definition 4.1

[20] Let $(X,d)$ be a metric space and $T:X\longrightarrow P(X)$ a MWP operator. Then $T$ is called $c$-multivalued weakly Picard (briefly $c$-MWP) operator if $c>0$ and there exists a selection $t^{\infty }$ of $T^{\infty }$ such that

$$\begin{aligned} d(x,t^{\infty }(x,y))\le cd(x,y) \end{aligned}$$

(23)

for all $(x,y)\in G(T).$

One of the main result concerning $c$-MWP operators is the following:

Theorem 4.2

[20] Let $(X,d)$ be a metric space and $T_{1},T_{2}:X\rightarrow P(X)$ two multivalued operators. Suppose that:

• (i) $T_{i}$ is a $c_{i}$-MWP operator for each $i\in \{1,2\};$

• (ii) There exists $\lambda >0$ such that $H(T_{1}x,T_{2}x)\le \lambda ,$ for all $x\in X.$

Then

$$\begin{aligned} H(Fix(T_{1}),Fix(T_{2}))\le \lambda \max \{c_{1},c_{2}\}. \end{aligned}$$

Moţ and Petruşel [15] proved the following result.

Theorem 4.3

[15] Let $(X,d)$ be a metric space and $T_{1},T_{2}:X\rightarrow P(X)$ two multivalued operators. If

• (i) $T_{i}$ is a $b_{i}$-KS multivalued operator for each $i\in \{1,2\};$

• (ii) There exists $\lambda >0$ such that $H(T_{1}x,T_{2}x)\le \lambda ,$ for all $x\in X.$

Then:

• (a) $Fix(T_{i})\in CB(X),$ $i\in \{1,2\};$

• (b) Each $T_{i}$ is a MWP operator and

  $$\begin{aligned} H(Fix(T_{1}),Fix(T_{2}))\le \frac{\lambda }{1-\max \{b_{1},b_{2}\}}. \end{aligned}$$

  (24)

Recently Popescu [18] proved the following theorem.

Theorem 4.4

Let $(X,d)$ be a metric space and $T_{1},T_{2}:X\rightarrow P(X)$ two multivalued operators. If

• (i) $T_{i}$ is an $(1,r_{i})$-contractive multivalued operator for each $i\in \{1,2\};$

• (ii) There exists $\lambda >0$ such that $H(T_{1}x,T_{2}x)\le \lambda ,$ for all $x\in X.$

Then:

• (a) $Fix(T_{i})\in CB(X),$ $i\in \{1,2\};$

• (b) Each $T_{i}$ is a MWP operator and

  $$\begin{aligned} H(Fix(T_{1}),Fix(T_{2}))\le \frac{\lambda }{1-\max \{r_{1},r_{2}\}}. \end{aligned}$$

  (25)

  Now we prove the following result for $(\psi ,r)$-contractive multivalued operators.

Theorem 4.5

Let $(X,d)$ be a complete metric space and $T_{1},T_{2}:X\rightarrow P(X)$ two multivalued operators. If

• (i) $T_{i}$ is $(\psi ,r_{i})$-contractive multivalued operators for each $i\in \{1,2\};$

• (ii) There exists $\lambda >0$ such that $H(T_{1}x,T_{2}x)\le \lambda ,$ for all $x\in X.$

Then:

• (a) $Fix(T_{i})\in CB(X),$ $i\in \{1,2\};$

• (b) Each $T_{i}$ is a MWP operator and

  $$\begin{aligned} H(Fix(T_{1}),Fix(T_{2}))\le \frac{\lambda }{1-\max \{r_{1},r_{2}\}}. \end{aligned}$$

  (26)

Proof

From Theorem 2.2, $Fix(T_{i})$ is nonempty for each $i\in \{1,2\}.$ Let $x_{n}\in Fix(T_{1})$ be such that $x_{n}\rightarrow z$ as $n\rightarrow \infty ,$ that is,

$$\begin{aligned} \lim _{n\rightarrow \infty }d(x_{n},z)=0. \end{aligned}$$

(27)

Note that

$$\begin{aligned} \psi (s)(d(x_{n},T_{1}x_{n})+d(z,T_{1}x_{n}))=\psi (s)d(z,x_{n})\le d(z,x_{n}). \end{aligned}$$

Thus

$$\begin{aligned} d(z,T_{1}z)&\le d(z,x_{n})+d(x_{n},T_{1}z)\le d(z,x_{n})+H(T_{1}x_{n},T_{1}z) \\&\le d(z,x_{n})\!+\!r_{1}\max \left\{ p(z,x_{n}),p(z,T_{1}z),p(T_{1}x_{n},x_{n}),\frac{p(z,T_{1}x_{n})\!+\!p(x_{n},T_{1}z)}{2}\right\} \\&\le d(z,x_{n})+r_{1}d(z,x_{n}). \end{aligned}$$

Taking limit as $n\rightarrow \infty$, we obtain that $d(z,T_{1}z)=0,$ that is, $z\in T_{1}z.$ Hence $Fix(T_{1})$ is closed. In the same way, we can prove that $Fix(T_{2})$ is closed. Using arguments as in proof of the Theorem 2.2, each $T_{i}$ is a MWP operator. To prove

$$\begin{aligned} H(Fix(T_{1}),Fix(T_{2}))\le \dfrac{\lambda }{1-\max \{r_{1},r_{2}\}}. \end{aligned}$$

$(C_{1})$ A “Classical” proof: Let $a>1.$ Then for an arbitrary $x_{0}\in Fix(T_{1}),$ there exists $x_{1}\in T_{2}x_{0}$ such that

$$\begin{aligned} d(x_{0},x_{1})\le aH(T_{1}x_{0},T_{2}x_{0}). \end{aligned}$$

As $x_{1}\in T_{2}x_{0},$ so there exists $x_{2}\in T_{2}x_{1}$ such that

$$\begin{aligned} \psi (s)(d(x_{0},T_{2}x_{0})+d(x_{1},T_{2}x_{0}))\le \psi (s)d(x_{0},x_{1})\le d(x_{0},x_{1}), \end{aligned}$$

which implies that

$$\begin{aligned} d(x_{1},x_{2})&\le aH(T_{2}x_{0},T_{2}x_{1}) \\&\le ar_{2}\max \left\{ d(x_{0},x_{1}),d(x_{0},Tx_{0}),d(x_{1},Tx_{1}),\frac{d(x_{0},Tx_{1})+d(x_{1},Tx_{0})}{2}\right\} \\&\le ar_{2}d(x_{0},x_{1}). \end{aligned}$$

Continuing this way, we can obtain a sequence $\{x_{n}\}$ in $X$ such that $x_{n+1}\in T_{2}x_{n}$ and

$$\begin{aligned} d(x_{n},x_{n+1})\le ar_{2}d(x_{n},x_{n+1})\le \ldots \le (ar_{2})^{n}d(x_{0},x_{1}). \end{aligned}$$

Thus

$$\begin{aligned} d(x_{n},x_{n+p})&\le d(x_{n},x_{n+1})+d(x_{n+1},x_{n+2})+\ldots +d(x_{n+p-1},x_{n+p}) \nonumber \\&\le (ar_{2})^{n}d(x_{0},x_{1})+\ldots +(ar_{2})^{n+p-1}d(x_{0},x_{1})\le \frac{(ar_{2})^{n}}{1-ar_{2}}d(x_{0},x_{1}). \end{aligned}$$

(28)

Chose $1<a<\min \left\{ \frac{1}{r_{1}},\frac{1}{r_{2}}\right\} .$ This implies that $\{x_{n}\}$ is Cauchy sequence in $X.$ Then there exists $u$ in $X$ such that $x_{n}\rightarrow u$ as $n\rightarrow \infty .$ Following arguments similar to those given in Theorem 2.2, it follows that $u\in T_{2}u.$ By (28), we obtain that

$$\begin{aligned} d(x_{n},u)\le \frac{(ar_{2})^{n}}{1-ar_{2}}d(x_{0},x_{1}) \end{aligned}$$

(29)

Thus, in particular

$$\begin{aligned} d(x_{0},u)\le \frac{1}{1-ar_{2}}d(x_{0},x_{1})\le \frac{\lambda }{1-ar_{2}}. \end{aligned}$$

(30)

In a similar way, we conclude that for each $z_{0}\in Fix(T_{2}),$ there is an $x\in Fix(T_{1})$ such that

$$\begin{aligned} d(z_{0},x)\le \frac{1}{1-ar_{1}}d(z_{0},z_{1})\le \frac{\lambda }{1-ar_{1}}. \end{aligned}$$

(31)

By (30) and (31), we obtain that

$$\begin{aligned} H(Fix(T_{1}),Fix(T_{2}))\le \frac{\lambda }{1-\max \{ar_{1},ar_{2}\}}. \end{aligned}$$

Letting $a\searrow 1$ we get the conclusion.

$(C_{2})$ Proof based on MWP operator technique: Suppose that $T$ is a $(\psi ,r)$-contractive multivalued operators. Now we show that $T$ is $c$-MWP operator with $c=\frac{1}{1-r}.$ Then the conclusion will follow from Theorem 4.2. Let $a>1,$ $x\in X$ and $y\in Tx$ be arbitrary chosen. By a similar approach to $(C_{1}),$ we obtain a sequence of successive approximations $\{x_{n}\}$ starting from $(x=x_{0},y=x_{1})\in G(T)$ such that

$$\begin{aligned} d(x_{n},x_{n+p})\le \frac{(ar_{2})^{n}}{1-ar_{2}}d(x_{0},x_{1}), \end{aligned}$$

for each $n\in \mathbb {N}$ and $p\rightarrow +\infty$ in the above estimation we get that $d(x_{n},u)\le \frac{(ar_{2})^{n}}{1-ar_{2}}d(x_{0},x_{1}),$ for each $n\in \mathbb {N} .$ For $n=0$ we obtain that $d(x,u)\le \frac{1}{1-ar_{2}}d(x,y).$ Letting $a\searrow 1$ we obtain $d(x,u)\le \frac{1}{1-r}d(x,y).$ Thus $T$ is a $\frac{1}{1-r}$-MWP operator. $\square$

5 Application in dynamic programming

A dynamic process consists of a state space (a set of initial states, actions and transitions) and a decision space (set of possible input and output actions). We assume $U$ and $V$ are Banach spaces where $W\subseteq U$ is state space and $D\subseteq V$ is decision space$.$ Now define the mappings as

$$\begin{aligned} \tau :W\times D\longrightarrow W,\,g:W\times D\longrightarrow \mathbb {R} ,G:W\times D\times \mathbb {R} \longrightarrow \mathbb {R}, \end{aligned}$$

where $\mathbb {R}$ is the field of real numbers. Dynamic programming provides tools for mathematical optimization and computer programing as well. It is well known that the problem of dynamic programming related to multistage process reduces to the problem of solving the functional equation:

$$\begin{aligned} q(x):=\underset{y\in D}{\sup }\{g(x,y)+G(x,y,q(\tau (x,y)))\},\,\,x\in W. \end{aligned}$$

(32)

For the detailed background of the problem (see [1–4, 17, 22]). Let $B(W)$ be the set of all bounded real-valued functions on $W$. For an arbitrary $h\in B(W)$, define $\left\| h\right\| =\sup \nolimits _{x\in W}\left| hx\right|$. Then $(B(W),\left\| \cdot \right\| )$ is a Banach space endowed with the metric $d$ defined by

$$\begin{aligned} d(h,k)=\sup \limits _{x\in W}\left| hx-kx\right| \end{aligned}$$

(33)

where $h,k\in B(W)$. Suppose that the following conditions hold:

$(DT-1)$ functions $G$ and $g$ are bounded.

$(DT-2)$ For $h,k\in B(W)$ and $x,z\in W,$ define $T$ by

$$\begin{aligned} T(hx):=\underset{y\in D}{\sup }\{g(x,y)+G(x,y,h(\tau (x,y)))\}. \end{aligned}$$

(34)

Moreover, there exist $0\le r\le s<1$ such that

$$\begin{aligned} \left| G(x,y,hz)-G(x,y,kz)\right| \le rM_{T}(hz,kz) \end{aligned}$$

for all $h,k\in B(W)$ and $x,z\in W,$ where

$$\begin{aligned} M_{T}(hz,kz)=\max \left\{ d(hz,kz),d(hz,Thz),d(kz,Tkz),\dfrac{d(hz,Tkz)+d(kz,Thz)}{2}\right\} . \end{aligned}$$

Theorem 5.1

If conditions $(DT-1)$ and $(DT-2)$ are satisfied, then the functional Eq. (32) has a unique bounded solution.

Proof

Note that $(B(W),d)$ is a complete metric space and $T$ is a self map of $B(W).$ Let $\lambda$ be an arbitrary positive number and $h_{1},h_{2}\in B(W)$. Choose $x\in W$ and $y_{1},y_{2}\in D$ such that

$$\begin{aligned} Th_{1}x&< g(x,y_{1})+G(x,y_{1},h_{1}(\tau (x,y_{1})))+\lambda , \end{aligned}$$

(35)

$$\begin{aligned} Th_{2}x&< g(x,y_{2})+G(x,y_{2},h_{2}(\tau (x,y_{2})))+\lambda , \end{aligned}$$

(36)

$$\begin{aligned} Th_{1}x&\ge g(x,y_{2})+G(x,y_{2},h_{1}(\tau (x,y_{2}))), \end{aligned}$$

(37)

$$\begin{aligned} Th_{2}x&\ge g(x,y_{1})+G(x,y_{1},h_{2}(\tau (x,y_{1}))). \end{aligned}$$

(38)

By (35) and (38), we have

$$\begin{aligned} Th_{1}x-Th_{2}x&< G(x,y_{1},h_{1}(\tau (x,y_{1})))-G(x,y_{1},h_{2}(\tau (x,y_{2})))+\lambda \\&\le \left| G(x,y_{1},h_{1}(\tau (x,y_{1})))-G(x,y_{1},h_{2}(\tau (x,y_{2})))\right| +\lambda \\&\le rM_{T}(h_{1}x,h_{2}x)+\lambda . \end{aligned}$$

That is

$$\begin{aligned} Th_{1}x-Th_{2}x\le rM_{T}(h_{1}x,h_{2}x)+\lambda . \end{aligned}$$

(39)

By (36) and (37), we obtain

$$\begin{aligned} Th_{2}x-Th_{1}x\le rM_{T}(h_{1}x,h_{2}x)+\lambda . \end{aligned}$$

(40)

Finally, by (39) and (40), we have

$$\begin{aligned} \left| Th_{1}x-Th_{2}x\right| \le rM_{T}(h_{1}x,h_{2}x)+\lambda , \end{aligned}$$

(41)

that is

$$\begin{aligned} d(Th_{1},Th_{2})\le rM_{T}(h_{1}x,h_{2}x). \end{aligned}$$

As above inequality is true for any $x\in W$, $\psi$ and $\lambda >0$, so

$$\begin{aligned} \psi (s)(d(h,Th)+d(k,Th))\le d(h,k)\text { implies }d(Th_{1},Th_{2})\le rM_{T}(h_{1}x,h_{2}x). \end{aligned}$$

Thus all the conditions of Corollary 2.11 are satisfied for the mapping $T.$ So functional Eq. 32 has a unique bounded solution. $\square$

Example 5.2

Let $U=V= \mathbb {R} ,$ $W=[0,20]$ and $D=[0,10].$ Consider the functional equation

$$\begin{aligned} q(x)=\max _{y\in D}\left\{ 2x^{2}y+\frac{x}{30+30x^{2}y}\cos (q(\frac{x+y}{2}))\right\} ,\,x\in W. \end{aligned}$$

(42)

For each $h,k\in B(W),$ define the functional

$$\begin{aligned} Th(x)=\max _{y\in D}\left\{ 2x^{2}y+\frac{x}{30+30x^{2}y}\cos (h(\frac{x+y}{2}))\right\} , \,x\in W. \end{aligned}$$

Suppose that $g(x,y)=2x^{2}y,$ $G(x,y,z)=\frac{x}{2+2x^{2}y}\cos (hz).$ Clearly $g,G$ are bounded and

$$\begin{aligned} \left| G(x,y,hz)-G(x,y,kz)\right|&= \left| \frac{x}{30+30x^{2}y} \cos (hz)-\frac{x}{30+30x^{2}y}\cos (kz)\right| \\&\le \frac{x}{30+30x^{2}y}\left| \cos (hz)-\cos (kz)\right| \le \frac{x}{30+30x^{2}y}\left| hz-kz\right| \\&\le \frac{2}{3}\sup _{z\in W}\left| hz-kz\right| =\frac{2}{3} d(h,k)\le rM_{T}(h,y) \end{aligned}$$

for all $\in B(W),$ where $r=\frac{2}{3}.$ Hence all the conditions of Theorem 5.1 are satisfiend and consequently functional Eq. (42) has a unique and bounded solution.

6 Homotopy result

Following is the local fixed point result for $(\psi ,r)$-contractive multivalued mappings.

Theorem 6.1

Let $(X,d)$ be a complete metric space, $x_{0}\in X$ and $a>0.$ Suppose that $T:B(x_{0},a)\rightarrow CB(X)$ be $(\psi ,r)$-contractive multivalued mappings and $d(x_{0},Tx_{0})<(1-s)a.$ Then $T$ has a fixed point in $B(x_{0},a).$

Proof

Let $0<a_{1}<a$ be such that $\widetilde{B}(x_{0},a_{1})\subset B(x_{0},a)$ and $d(x_{0},Tx_{0})<(1-s)a_{1}<(1-s)a.$ Let $x_{1}\in Tx_{0}$ be such that $d(x_{0},x_{1})<(1-s)a_{1}.$ Then for $h=\frac{1}{\sqrt{r}}>1$ and $x_{1}\in Tx_{0}$ there exists $x_{2}\in Tx_{1}$ such that

$$\begin{aligned} d(x_{1},x_{2})\le hH(Tx_{0},Tx_{1}) \end{aligned}$$

Since $\psi (s)(d(x_{0},Tx_{0})+d(x_{1},Tx_{0}))=\psi (s)d(x_{0},Tx_{0})\le \psi (s)d(x_{0},x_{1})\le d(x_{0},x_{1}),$ therefore we obtain

$$\begin{aligned} d(x_{1},x_{2})&\le hH(Tx_{0},Tx_{1})=\frac{1}{\sqrt{r}} H(Tx_{0},Tx_{1})\le \sqrt{r}M_{T}(x_{0},x_{1}) \\&\le \sqrt{r}\max \left\{ d(x_{0},x_{1}),d(x_{0},Tx_{0}),d(x_{1},Tx_{1}),\frac{ d(x_{0},Tx_{1})+d(Tx_{0},x_{1})}{2}\right\} \\&\le \sqrt{r}\max \left\{ d(x_{0},x_{1}),d(x_{0},x_{1}),d(x_{1},x_{2}),\frac{ d(x_{0},x_{1})+d(x_{1},x_{2})}{2}\right\} \\&\le \sqrt{r}\max \left\{ d(x_{0},x_{1}),d(x_{0},x_{1}),d(x_{1},x_{2}),\frac{ d(x_{0},x_{1})+d(x_{1},x_{2})}{2}\right\} \\&\le \sqrt{r}d(x_{0},x_{1})<\sqrt{r}(1-s)a_{1}\le \sqrt{r}(1-r)a_{1}. \end{aligned}$$

Also, we have $x_{2}\in B(x_{0},a)$ because

$$\begin{aligned} d(x_{0},x_{2})&\le d(x_{0},x_{1})+d(x_{1},x_{2})<(1-s)a_{1}+\sqrt{r} (1-r)a_{1} \\&\le (1-r)a_{1}+\sqrt{r}(1-r)a_{1}=(1-r)(1+\sqrt{r})a_{1}. \end{aligned}$$

In this way, we obtain inductively a sequence $(x_{n})_{n\in \mathbb {N} }$ satisfying (i) $x_{n}\in B(x_{0},a)$; for each $n\in \mathbb {N} ,$ (ii) $x_{n+1}\in Tx_{n},$ for all $n\in \mathbb {N} ,$ (iii) $d(x_{n},x_{n+1})\le (\sqrt{r})^{n}(1-r)s$ for each $n\in \mathbb {N}.$ From (iii) the sequence $(x_{n})_{n\in \mathbb {N} }$ is Cauchy and hence, it converges to a certain $u\in B(x_{0},a)$. Following similar arguments to those given in Theorem 2.2, we obtain $u\in Tu$. $\square$

Now we present a homotopy result for $(\psi ,r)$-contractive multivalued mappings.

Theorem 6.2

Let $(X,d)$ be a complete metric space and $U$ an open subset of $X.$ Let $G:\overline{U}\times [0,1]\rightarrow P(X)$ be a multivalued operator such that the following conditions are satisfied:

h-1:

$x\notin G(x,t)$, for each $x\in \partial U$ (boundary of $U$) and each $t\in [0,1];$

h-2:

$G(.,t):\overline{U}\rightarrow P(X)$ is a $(\psi ,r)$-contractive multivalued mappings for each $t\in [0,1];$

h-3:

there exists a continuous increasing function $\rho :[0,1]\rightarrow \mathbb {R}$ such that

$$\begin{aligned} H(G(x,t),G(x,s))\le \left| \rho (t)-\rho (s)\right| \text { for all } t,s\in [0,1]\text { and each }x\in \overline{U;} \end{aligned}$$

h-4:

$G:\overline{U}\times [0,1]\rightarrow P(X)$ is closed.

Then $G(.,0)$ has a fixed point if and only if $G(.,1)$ has a fixed point.

Proof

Let $G(.,0)$ has a fixed point $z,$ then (h-1) implies that $z\in U$. Define

$$\begin{aligned} \Delta =\{(t,x)\in [0,1]\times U\mid x\in G(x,t)\}. \end{aligned}$$

Since $(0,z)\in \Delta$ therefore $\Delta \ne \emptyset ,$ as. Now we define a partial order on $\Delta$, that is

$$\begin{aligned} (t,x)\le (s,y)\text { if and only if }t\le s\text { and }d(x,y)\le \frac{2}{ 1-r}[\rho (s)-\rho (t)] \end{aligned}$$

where $0\le r<1.$ Let $M$ be a totally ordered subset of $\Delta$ and $t^{*}:=\sup \{t\mid (t,x)\in M\}.$ Consider a sequence $(t_{n},x_{n})_{n\in \mathbb {N} }\subset M$ such that $(t_{n},x_{n})\le (t_{n+1},x_{n+1})$ and $t_{n}\rightarrow t^{*}$ as $n\rightarrow \infty .$ Then

$$\begin{aligned} d(x_{m},x_{n})\le \frac{2}{1-r}[\rho (t_{m})-\rho (t_{n})],\text { for each } m,n\in \mathbb {N} ,m>n. \end{aligned}$$

Taking limit as $m,n\rightarrow \infty ,$ we obtain $d(x_{m},x_{n}) \rightarrow 0$. Thus $(x_{n})_{n\in \mathbb {N} }$ is Cauchy sequence which converges to (say) $x^{*}$ in $X$. As $x_{n}\in G(x_{n},t_{n}),$ $n\in \mathbb {N}$ and $G$ is closed, so $x^{*}\in G(x^{*},t^{*})$. Also, from (h-1) we have $x^{*}\in U$. Hence $(t^{*},x^{*})\in \Delta .$ Since $M$ is totally ordered, therefore $(t,x)\le (t^{*},x^{*}),$ for each $(t,x)\in M.$ That is, $(t^{*},x^{*})$ is an upper bound of $M$. By Zorn’s Lemma $\Delta$ have a maximal element $(t_{0},x_{0})\in \Delta .$ We claim that $t_{0}=1.$ Suppose that $t_{0}<1.$ Choose $a>0$ and $t\in (t_{0},1]$ such that $B(x_{0},a)\subset U$ and $a=\frac{2}{1-r}[\rho (t)-\rho (t_{0})].$ Note that

$$\begin{aligned} d(x_{0},G(x_{0},t))&\le d(x_{0},G(x_{0},t_{0}))+H(G(x_{0},t_{0}),G(x_{0},t)) \\&\le [\rho (t)-\rho (t_{0})]=\frac{(1-r)a}{2}<(1-r)a. \end{aligned}$$

Thus $G(.,t):B(x_{0},a)\rightarrow CL(X)$ satisfies, for all $t\in [0,1]$, the assumptions of Theorem 6.1. Hence, for all $t\in [0,1]$, there exists $x\in B(x_{0},a)$ such that $x\in G(x,t)$ which implies that $(t,x)\in \Delta .$ Now

$$\begin{aligned} d(x_{0},x)\le a=\frac{2}{1-r}[\rho (t)-\rho (t_{0})], \end{aligned}$$

gives $(t_{0},x_{0})<(t,x),$ a contradiction to the maximality of $(t_{0},x_{0}).$ Conversely if $G(.,1)$ has a fixed point, then by a similar approach we obtain that $G(.,0)$ has a fixed point.