On $({\alpha }^{\ast },\psi )$-contractive multi-valued mappings

Abstract

In this paper, we generalize the contractive condition for multi-valued mappings given by Asl, Rezapour and Shahzad in 2012. We establish some fixed point theorems for multi-valued mappings from a complete metric space to the space of closed or bounded subsets of the metric space satisfying generalized $({\alpha }^{\ast },\psi )$-contractive condition.

MSC:47H10, 54H25.

1 Introduction

Samet et al. [1] introduced the notion of α-ψ-contractive self-mappings of a metric space. Recently, Asl et al. [2] introduced the notion of ${\alpha }^{\ast }$-ψ-contractive mappings to extend the notion α-ψ-contractive mappings. In this paper, we generalize the notion of ${\alpha }^{\ast }$-ψ-contractive mappings and prove some fixed point theorems for such mappings.

Let Ψ be a family of nondecreasing functions, $\psi :[0,\mathrm{\infty })\to [0,\mathrm{\infty })$ such that ${\sum }_{n=1}^{\mathrm{\infty }}{\psi }^n(t)<\mathrm{\infty }$ for each $t>0$, where ${\psi }^n$ is the n th iterate of ψ. It is known that for each $\psi \in \mathrm{\Psi }$, we have $\psi (t)<t$ for all $t>0$ and $\psi (0)=0$ for $t=0$ [1]. Let $(X,d)$ be a metric space. A mapping $G:X\to X$ is called α-ψ-contractive if there exist two functions $\alpha :X\times X\to [0,\mathrm{\infty })$ and $\psi \in \mathrm{\Psi }$ such that $\alpha (x,y)d(Gx,Gy)\le \psi (d(x,y))$ for each $x,y\in X$. A mapping $G:X\to X$ is called α-admissible [1] if $\alpha (x,y)\ge 1\Rightarrow \alpha (Gx,Gy)\ge 1$. We denote by $N(X)$ the space of all nonempty subsets of X, by $B(X)$ the space of all nonempty bounded subsets of X and by $\mathit{CL}(X)$ the space of all nonempty closed subsets of X. For $A\in N(X)$ and $x\in X$, $d(x,A)=inf\{d(x,a):a\in A\}$. For every $A,B\in B(X)$, $\delta (A,B)=sup\{d(a,b):a\in A,b\in B\}$. When $A=\{x\}$, we denote $\delta (A,B)$ by $\delta (x,B)$. For every $A,B\in \mathit{CL}(X)$, let

$$H(A,B)=\{\begin{array}{cc}max\{{sup}_{x\in A}d(x,B),{sup}_{y\in B}d(y,A)\}\hfill & \text{if the maximum exists};\hfill \\ \mathrm{\infty }\hfill & \text{otherwise}.\hfill \end{array}$$

Such a map H is called generalized Hausdorff metric induced by d. Let $(X,⪯,d)$ be an ordered metric space and $A,B\subseteq X$. We say that $A{\prec }_{r}B$ if for each $a\in A$ and $b\in B$, we have $a⪯b$. We give a few definitions and the result due to Asl et al. [2] for convenience.

Definition 1.1 [2]

Let $(X,d)$ be a metric space and let $\alpha :X\times X\to [0,\mathrm{\infty })$ be a mapping. A mapping $G:X\to \mathit{CL}(X)$ is ${\alpha }^{\ast }$-admissible if $\alpha (x,y)\ge 1\Rightarrow {\alpha }^{\ast }(Gx,Gy)\ge 1$, where ${\alpha }^{\ast }(Gx,Gy)=inf\{\alpha (a,b):a\in Gx,b\in Gy\}$.

Definition 1.2 [2]

Let $(X,d)$ be a metric space. A mapping $G:X\to \mathit{CL}(X)$ is called ${\alpha }^{\ast }$-ψ-contractive if there exist two functions $\alpha :X\times X\to [0,\mathrm{\infty })$ and $\psi \in \mathrm{\Psi }$ such that

$${\alpha }^{\ast }(Gx,Gy)H(Gx,Gy)\le \psi (d(x,y))$$

(1.1)

for all $x,y\in X$.

Theorem 1.3 [2]

Let $(X,d)$ be a complete metric space, let $\alpha :X\times X\to [0,\mathrm{\infty })$ be a function, let $\psi \in \mathrm{\Psi }$ be a strictly increasing map and T be a closed-valued, ${\alpha }^{\ast }$-admissible and ${\alpha }^{\ast }$-ψ-contractive multi-function on X. Suppose that there exist $x_0\in X$ and $x_1\in G{x}_0$ such that $\alpha (x_0,x_1)\ge 1$. Assume that if $\{x_n\}$ is a sequence in X such that $\alpha (x_n,x_{n+1})\ge 1$ for all n and $x_n\to x$, then $\alpha (x_n,x)\ge 1$ for all n. Then G has a fixed point.

2 Main results

We begin this section by introducing the following definition.

Definition 2.1 Let $(X,d)$ be a metric space and let $G:X\to \mathit{CL}(X)$ be a mapping. We say that G is generalized $({\alpha }^{\ast },\psi )$-contractive if there exists $\psi \in \mathrm{\Psi }$ such that

$${\alpha }^{\ast }(Gx,Gy)d(y,Gy)\le \psi (d(x,y))$$

(2.1)

for each $x\in X$ and $y\in Gx$, where ${\alpha }^{\ast }(Gx,Gy)=inf\{\alpha (a,b):a\in Gx,b\in Gy\}$.

Note that an ${\alpha }^{\ast }$-ψ-contractive mapping is generalized $({\alpha }^{\ast },\psi )$-contractive. In case when $\psi \in \mathrm{\Psi }$ is strictly increasing, generalized $({\alpha }^{\ast },\psi )$-contractive is called strictly generalized $({\alpha }^{\ast },\psi )$-contractive. The following lemma is inspired by [[3], Lemma 2.2].

Lemma 2.2 Let $(X,d)$ be a metric space and $B\in \mathit{CL}(X)$. Then, for each $x\in X$ with $d(x,B)>0$ and $q>1$, there exists an element $b\in B$ such that

$$d(x,b)<qd(x,B).$$

(2.2)

Proof It is given that $d(x,B)>0$. Choose

$$ϵ=(q-1)d(x,B).$$

Then, by using the definition of $d(x,B)$, it follows that there exists $b\in B$ such that

$$d(x,b)<d(x,B)+ϵ=qd(x,B).$$

 □

Lemma 2.3 Let $(X,d)$ be a metric space and $G:X\to \mathit{CL}(X)$. Assume that there exists a sequence $\{x_n\}$ in X such that ${lim}_{n\to \mathrm{\infty }}d(x_n,G{x}_n)=0$ and $x_n\to x\in X$. Then x is a fixed point of G if and only if the function $f(\xi )=d(\xi ,G\xi )$ is lower semi-continuous at x.

Proof Suppose $f(\xi )=d(\xi ,G\xi )$ is lower semi-continuous at x, then

$$d(x,Gx)\le \underset{n}{lim\hspace{0.17em}inf}f(x_n)=\underset{n}{lim\hspace{0.17em}inf}d(x_n,G{x}_n)=0.$$

By the closedness of G it follows that $x\in Gx$. Conversely, suppose that x is a fixed point of G, then $f(x)=0\le {lim\hspace{0.17em}inf}_{n}f(x_n)$. □

Theorem 2.4 Let $(X,d)$ be a complete metric space and let $G:X\to \mathit{CL}(X)$ be an ${\alpha }^{\ast }$-admissible strictly generalized $({\alpha }^{\ast },\psi )$-contractive mapping. Assume that there exist $x_0\in X$ and $x_1\in G{x}_0$ such that $\alpha (x_0,x_1)\ge 1$. Then x is a fixed point of G if and only if $f(\xi )=d(\xi ,G\xi )$ is lower semi-continuous at x.

Proof By the hypothesis, there exist $x_0\in X$ and $x_1\in G{x}_0$ such that $\alpha (x_0,x_1)\ge 1$. If $x_0=x_1$, then we have nothing to prove. Let $x_0\ne x_1$. If $x_1\in G{x}_1$, then $x_1$ is a fixed point. Let $x_1\notin G{x}_1$. Since G is ${\alpha }^{\ast }$-admissible, so ${\alpha }^{\ast }(G{x}_0,G{x}_1)\ge 1$, we have

$$0<d(x_1,G{x}_1)\le {\alpha }^{\ast }(G{x}_0,G{x}_1)d(x_1,G{x}_1).$$

(2.3)

For given $q>1$ by Lemma 2.2, there exists $x_2\in G{x}_1$ such that

$$0<d(x_1,x_2)<qd(x_1,G{x}_1).$$

(2.4)

It follows from (2.3), (2.4) and (2.1) that

$$0<d(x_1,x_2)<q\psi (d(x_0,x_1)).$$

(2.5)

It is clear that $x_1\ne x_2$ and $\alpha (x_1,x_2)\ge 1$. Thus ${\alpha }^{\ast }(G{x}_1,G{x}_2)\ge 1$. Since ψ is strictly increasing, by (2.5), we have

$$\psi (d(x_1,x_2))<\psi \left(q\psi (d(x_0,x_1))\right).$$

Put $q_1=\frac{\psi (q\psi (d(x_0,x_1)))}{\psi (d(x_1,x_2))}$, then $q_1>1$. If $x_2\in G{x}_2$, then $x_2$ is a fixed point. Let $x_2\notin G{x}_2$, then by Lemma 2.2, there exists $x_3\in G{x}_2$ such that

$$\begin{array}{rcl}0& <& d(x_2,x_3)<q_{1}d(x_2,G{x}_2)\le q_1{\alpha }^{\ast }(G{x}_1,G{x}_2)d(x_2,G{x}_2)\\ \le & q_1\psi (d(x_1,x_2))=\psi \left(q\psi (d(x_0,x_1))\right).\end{array}$$

It is clear that $x_2\ne x_3$, $\alpha (x_2,x_3)\ge 1$ and $\psi (d(x_2,x_3))<{\psi }^2(q\psi (d(x_0,x_1)))$. Now put $q_2=\frac{{\psi }^2(q\psi (d(x_0,x_1)))}{\psi (d(x_2,x_3))}$. Then $q_2>1$. If $x_3\in G{x}_3$, then $x_3$ is a fixed point. Let $x_3\notin G{x}_3$. Then by Lemma 2.2 there exists $x_4\in G{x}_3$ such that

$$\begin{array}{rcl}0& <& d(x_3,x_4)<q_{2}d(x_3,G{x}_3)\le q_2{\alpha }^{\ast }(G{x}_2,G{x}_3)d(x_3,G{x}_3)\\ \le & q_2\psi (d(x_2,x_3))={\psi }^2\left(q\psi (d(x_0,x_1))\right).\end{array}$$

By continuing the same process, we get a sequence $\{x_n\}$ in X such that $x_{n+1}\in G{x}_n$. Also, $x_n\ne x_{n+1}$, $\alpha (x_n,x_{n+1})\ge 1$ and $0<d(x_n,x_{n+1})<{\psi }^{n-1}(q\psi (d(x_0,x_1)))$ or

$$0<d(x_n,G{x}_n)<{\psi }^{n-1}\left(q\psi (d(x_0,x_1))\right).$$

(2.6)

For each $m>n$, we have

$$d(x_n,x_m)\le \sum _{i=n}^{m-1}d(x_i,x_{i+1})<\sum _{i=n}^{m-1}{\psi }^{i-1}\left(q\psi (d(x_0,x_1))\right).$$

Since $\psi \in \mathrm{\Psi }$, it follows that $\{x_n\}$ is a Cauchy sequence in X. Thus there is $x\in X$ such that $x_n\to x$. Letting $n\to \mathrm{\infty }$ in (2.6), we have

$$\underset{n\to \mathrm{\infty }}{lim}d(x_n,G{x}_n)=0.$$

(2.7)

The rest of the proof follows from Lemma 2.3. □

Example 2.5 Let $X=\mathbb{R}$ be endowed with the usual metric d. Define $G:X\to \mathit{CL}(X)$ and $\alpha :X\times X\to [0,\mathrm{\infty })$ by

$$Gx=\{\begin{array}{cc}[x,\mathrm{\infty })\hfill & \text{if }x\ge 0,\hfill \\ (-\mathrm{\infty },-x^2]\hfill & \text{if }x<0\hfill \end{array}$$

(2.8)

and

$$\alpha (x,y)=\{\begin{array}{cc}1\hfill & \text{if }x,y\ge 0,\hfill \\ 0\hfill & \text{otherwise}.\hfill \end{array}$$

(2.9)

Let $\psi (t)=\frac{t}{2}$ for all $t\ge 0$. For each $x\in X$ and $y\in Gx$, we have

$${\alpha }^{\ast }(Gx,Gy)d(y,Gy)=0\le \frac{1}{2}d(x,y).$$

Hence G is a strictly generalized $({\alpha }^{\ast },\psi )$-contractive mapping. Clearly, G is ${\alpha }^{\ast }$-admissible. Also, we have $x_0=1$ and $x_1=1\in G{x}_0$ such that $\alpha (x_0,x_1)=1$. Therefore, all conditions of Theorem 2.4 are satisfied and G has infinitely many fixed points. Note that Theorem 1.3 in Section 1 is not applicable here. For example, take $x=1$ and $y=-1$.

Corollary 2.6 Let $(X,⪯,d)$ be a complete ordered metric space, $\psi \in \mathrm{\Psi }$ be a strictly increasing map and $G:X\to \mathit{CL}(X)$ be a mapping such that for each $x\in X$ and $y\in Gx$ with $x⪯y$, we have

$$d(y,Gy)\le \psi (d(x,y)).$$

(2.10)

Also, assume that

• (i) there exist $x_0\in X$ and $x_1\in G{x}_0$ such that $x_0⪯x_1$,

• (ii) if $x⪯y$, then $Gx{\prec }_{r}Gy$.

Then x is a fixed point of G if and only if $f(\xi )=d(\xi ,G\xi )$ is lower semi-continuous at x.

Proof Define $\alpha :X\times X\to [0,\mathrm{\infty })$ by

$$\alpha (x,y)=\{\begin{array}{cc}1\hfill & \text{if }x⪯y,\hfill \\ 0\hfill & \text{otherwise}.\hfill \end{array}$$

By using condition (i) and the definition of α, we have $\alpha (x_0,x_1)=1$. Also, from condition (ii), we have $x⪯y$ implies $Gx{\prec }_{r}Gy$; by using the definitions of α and ${\prec }_r$, we have $\alpha (x,y)=1$ implies ${\alpha }^{\ast }(Gx,Gy)=1$. Moreover, it is easy to check that G is a strictly generalized $({\alpha }^{\ast },\psi )$-contractive mapping. Therefore, by Theorem 2.4, x is a fixed point of G if and only if $f(\xi )=d(\xi ,G\xi )$ is lower semi-continuous at x. □

Definition 2.7 Let $(X,d)$ be a metric space and $G:X\to B(X)$ be a mapping. We say that G is a generalized $({\alpha }^{\ast },\psi ,\delta )$-contractive mapping if there exists $\psi \in \mathrm{\Psi }$ such that

$${\alpha }^{\ast }(Gx,Gy)\delta (y,Gy)\le \psi (d(x,y))$$

(2.11)

for each $x\in X$ and $y\in Gx$, where ${\alpha }^{\ast }(Gx,Gy)=inf\{\alpha (a,b):a\in Gx,b\in Gy\}$.

Lemma 2.8 Let $(X,d)$ be a metric space and $G:X\to B(X)$. Assume that there exists a sequence $\{x_n\}$ in X such that ${lim}_{n\to \mathrm{\infty }}\delta (x_n,G{x}_n)=0$ and $x_n\to x\in X$. Then $\{x\}=Gx$ if and only if the function $f(\xi )=\delta (\xi ,G\xi )$ is lower semi-continuous at x.

Proof Suppose that $f(\xi )=\delta (\xi ,G\xi )$ is lower semi-continuous at x, then

$$\delta (x,Gx)\le \underset{n}{lim\hspace{0.17em}inf}f(x_n)=\underset{n}{lim\hspace{0.17em}inf}\delta (x_n,G{x}_n)=0.$$

Hence, $\{x\}=Gx$ because $\delta (A,B)=0$ implies $A=B=\{a\}$. Conversely, suppose that $\{x\}=Gx$. Then $f(x)=0\le {lim\hspace{0.17em}inf}_{n}f(x_n)$. □

Theorem 2.9 Let $(X,d)$ be a complete metric space and let $G:X\to B(X)$ be an ${\alpha }^{\ast }$-admissible generalized $({\alpha }^{\ast },\psi ,\delta )$-contractive mapping. Assume that there exist $x_0\in X$ and $x_1\in G{x}_0$ such that $\alpha (x_0,x_1)\ge 1$. Then there exists $x\in X$ such that $\{x\}=Gx$ if and only if $f(\xi )=\delta (\xi ,G\xi )$ is lower semi-continuous at x.

Proof By the hypothesis of the theorem, there exist $x_0\in X$ and $x_1\in G{x}_0$ such that $\alpha (x_0,x_1)\ge 1$. Assume that $x_0\ne x_1$, for otherwise, $x_0$ is a fixed point. Let $x_1\notin G{x}_1$. As G is ${\alpha }^{\ast }$-admissible, we have ${\alpha }^{\ast }(G{x}_0,G{x}_1)\ge 1$. Then

$$\delta (x_1,G{x}_1)\le {\alpha }^{\ast }(G{x}_0,G{x}_1)\delta (x_1,G{x}_1)\le \psi (d(x_0,x_1)).$$

(2.12)

Since $G{x}_1\ne \mathrm{\varnothing }$, there is $x_2\in G{x}_1$. Then

$$0<d(x_1,x_2)\le \delta (x_1,G{x}_1).$$

(2.13)

From (2.12) and (2.13), we have

$$0<d(x_1,x_2)\le \psi (d(x_0,x_1)).$$

(2.14)

Since ψ is nondecreasing, we have

$$\psi (d(x_1,x_2))\le {\psi }^2(d(x_0,x_1)).$$

(2.15)

As $x_2\in G{x}_1$, we have $\alpha (x_1,x_2)\ge 1$. Since $G{x}_2\ne \mathrm{\varnothing }$, there is $x_3\in G{x}_2$. Assume that $x_2\ne x_3$, for otherwise, $x_2$ is a fixed point of G. Then

$$\begin{array}{rcl}0<d(x_2,x_3)& \le & \delta (x_2,G{x}_2)\le {\alpha }^{\ast }(G{x}_1,G{x}_2)\delta (x_2,G{x}_2)\\ \le & \psi (d(x_1,x_2))\le {\psi }^2(d(x_0,x_1)).\end{array}$$

(2.16)

Since ψ is nondecreasing, we have

$$\psi (d(x_2,x_3))\le {\psi }^3(d(x_0,x_1)).$$

(2.17)

By continuing in this way, we get a sequence $\{x_n\}$ in X such that $x_{n+1}\in G{x}_n$ and $x_n\ne x_{n+1}$ for $n=0,1,2,3,\dots$ . Further we have

$$0<d(x_n,x_{n+1})\le \delta (x_n,G{x}_n)\le {\psi }^n(d(x_0,x_1)).$$

(2.18)

For each $m>n$, we have

$$d(x_n,x_m)\le \sum _{i=n}^{m-1}d(x_i,x_{i+1})\le \sum _{i=n}^{m-1}{\psi }^i(d(x_0,x_1)).$$

Since $\psi \in \mathrm{\Psi }$, it follows that $\{x_n\}$ is a Cauchy sequence in X. As X is complete, there exists $x\in X$ such that $x_n\to x$. Letting $n\to \mathrm{\infty }$ in (2.18), we have

$$\underset{n\to \mathrm{\infty }}{lim}\delta (x_n,G{x}_n)=0.$$

(2.19)

The rest of the proof follows from Lemma 2.8. □

Example 2.10 Let $X=\{0,2,4,6,8,10,\dots \}$ be endowed with the usual metric d. Define $G:X\to B(X)$ and $\alpha :X\times X\to [0,\mathrm{\infty })$ by

$$Gx=\{\begin{array}{cc}\{(x-2),x\}\hfill & \text{if }x\ne 0,\hfill \\ \{0\}\hfill & \text{if }x=0\hfill \end{array}$$

and

$$\alpha (x,y)=\{\begin{array}{cc}0\hfill & \text{if }x=y\ne 0,\hfill \\ 1\hfill & \text{if }x=y=0,\hfill \\ \frac{1}{4}\hfill & \text{otherwise}.\hfill \end{array}$$

Let $\psi (t)=\frac{t}{2}$ for all $t\ge 0$. For each $x\in X$ and $y\in Gx$, we have

$${\alpha }^{\ast }(Gx,Gy)\delta (y,Gy)\le \frac{1}{2}(d(x,y)).$$

Hence G is a generalized $({\alpha }^{\ast },\psi ,\delta )$-contractive mapping. Clearly, G is ${\alpha }^{\ast }$-admissible. Also, we have $x_0=0\in X$ and $x_1=0\in G0$ such that $\alpha (x_0,x_1)=1$. Therefore, all conditions of Theorem 2.9 are satisfied and G has infinitely many fixed points.

Corollary 2.11 Let $(X,⪯,d)$ be a complete ordered metric space, $\psi \in \mathrm{\Psi }$ and $G:X\to B(X)$ be a mapping such that for each $x\in X$ and $y\in Gx$ with $x⪯y$, we have

$$\delta (y,Gy)\le \psi (d(x,y)).$$

(2.20)

Also, assume that

• (i) there exists $x_0\in X$ such that $\{x_0\}{\prec }_{1}G{x}_0$, i.e., there exists $x_1\in G{x}_0$ such that $x_0⪯x_1$,

• (ii) if $x⪯y$, then $Gx{\prec }_{r}Gy$.

Then there exists $x\in X$ such that $\{x\}=Gx$ if and only if $f(\xi )=\delta (\xi ,G\xi )$ is lower semi-continuous at x.

Proof Define $\alpha :X\times X\to [0,\mathrm{\infty })$ by

$$\alpha (x,y)=\{\begin{array}{cc}1\hfill & \text{if }x⪯y,\hfill \\ 0\hfill & \text{otherwise}.\hfill \end{array}$$

By using condition (i) and the definition of α, we have $\alpha (x_0,x_1)=1$. Also, from condition (ii), we have $x⪯y$ implies $Gx{\prec }_{r}Gy$, by using the definitions of α and ${\prec }_r$, we have $\alpha (x,y)=1$ implies ${\alpha }^{\ast }(Gx,Gy)=1$. Moreover, it is easy to check that G is a generalized $({\alpha }^{\ast },\psi ,\delta )$-contractive mapping. Therefore, by Theorem 2.9, there exists $x\in X$ such that $\{x\}=Gx$ if and only if $f(\xi )=\delta (\xi ,G\xi )$ is lower semi-continuous at x. □