Multivalued fixed point theorems in cone b-metric spaces

Abstract

In this paper we extend the Banach contraction for multivalued mappings in a cone b-metric space without the assumption of normality on cones and generalize some attractive results in literature.

MSC:47H10, 54H25.

1 Introduction

The analysis on existence of linear and nonlinear operators was developed after the Banach contraction theorem [1] presented in 1922. Many generalizations are available with applications in the literature [2–13]. Nadler [14] gave its set-valued form in his classical paper in 1969 on multivalued contractions. A real generalization of Nadler’s theorem was presented by Mizoguchi and Takahashi [15] as follows.

Theorem 1.1 [15]

Let $(X,d)$ be a complete metric space and let $T:X\to 2^X$ be a multivalued map such that Tx is a closed bounded subset of X for all $x\in X$. If there exists a function $\phi :(0,\mathrm{\infty })\to [0,1)$ such that ${lim}_{r\to t^{+}}sup\phi (r)<1$ for all $t\in [0,\mathrm{\infty })$ and if

$$H(Tx,Ty)\le \phi (d(x,y))(d(x,y))\mathit{\text{for all }}x,y\in X,$$

then T has a fixed point in X.

Huang and Zhang [10] introduced a cone metric space with normal cone as a generalization of a metric space. Rezapour and Hamlbarani [16] presented the results of [10] for the case of a cone metric space without normality in cone. Many authors worked on it (see [17]). Cho and Bae [18] presented the result of [15] for multivalued mappings in cone metric spaces with normal cone.

Recently Hussain and Shah [19] introduced the notion of cone b-metric spaces as a generalization of b-metric and cone metric spaces. In [20] the authors presented some fixed point results in cone b-metric spaces without assumption of normality on cone.

In this article we present the generalized form of Cho and Bae [18] for the case of cone b-metric spaces without normality on cone. We also give an example to support our main theorem.

2 Preliminaries

Let $\mathbb{E}$ be a real Banach space and P be a subset of $\mathbb{E}$. By θ we denote the zero element of $\mathbb{E}$ and by intP the interior of P. The subset P is called a cone if and only if:

• (i) P is closed, nonempty, and $P=\{\theta \}$;

• (ii) $a,b\in \mathbb{R}$, $a,b\ge 0$, $x,y\in P\Rightarrow ax+by\in P$;

• (iii) $P\cap (-P)=\{\theta \}$.

For a given cone $P\subseteq \mathbb{E}$, we define a partial ordering ⪯ with respect to P by $x⪯y$ if and only if $y-x\in P$; $x\prec y$ will stand for $x⪯y$ and $x\ne y$, while $x\ll y$ will stand for $y-x\in intP$, where intP denotes the interior of P. The cone P is said to be solid if it has a nonempty interior.

Definition 2.1 [19]

Let X be a nonempty set and $r\ge 1$ be a given real number. A function $d:X\times X\to \mathbb{E}$ is said to be a cone b-metric if the following conditions hold:

(C1) $\theta ⪯d(x,y)$ for all $x,y\in X$ and $d(x,y)=\theta$ if and only if $x=y$;

(C2) $d(x,y)=d(y,x)$ for all $x,y\in X$;

(C3) $d(x,z)⪯r[d(x,y)+d(y,z)]$ for all $x,y,z\in X$.

The pair $(X,d)$ is then called a cone b-metric space.

Example 2.1 [20]

Let $X=l^p$ with $0<p<1$, where $l^p=\{\{x_n\}\subset \mathbb{R}:{\sum }_{n=1}^{+\mathrm{\infty }}|x_n{|}^p<\mathrm{\infty }\}$. Let $d:X\times X\to \mathbb{R}$ be defined as

$$d(x,y)={(\sum _{n=1}^{+\mathrm{\infty }}|x_n-y_n{|}^p)}^{\frac{1}{p}},$$

where $x=\{x_n\},y=\{y_n\}\in l^p$. Then $(X,d)$ is a b-metric space. Put $E=l^1$, $P=\{\{x_n\}\in E:x_n\ge 0\text{ for all }n\ge 1\}$. Letting the map $d^{\prime }:X\times X\to E$ be defined by $d^{\prime }(x,y)={\{\frac{d(x,y)}{2^n}\}}_{n\ge 1}$, we conclude that $(X,d^{\prime })$ is a cone b-metric space with the coefficient $r=2^{\frac{1}{p}}>1$, but is not a cone metric space.

Example 2.2 [20]

Let $X=\{1,2,3,4\}$, $E={\mathbb{R}}^2$, $P=\{(x,y)\in E:x\ge 0,y\ge 0\}$. Define $d:X\times X\to E$ by

$$d(x,y)=\{\begin{array}{ll}(|x-y{|}^{-1},|x-y{|}^{-1})& \text{if }x\ne y,\\ \theta & \text{if }x=y.\end{array}$$

Then $(X,d)$ is a cone b-metric space with coefficient $r=\frac{6}{5}$. But it is not a cone metric space, because

$$\begin{array}{c}d(1,2)>d(1,4)+d(4,2),\hfill \\ d(3,4)>d(3,1)+d(1,4).\hfill \end{array}$$

Remark 2.1 [19]

The class of cone b-metric spaces is larger than the class of cone metric spaces since any cone metric space must be a cone metric b-metric space. Therefore, it is obvious that cone b-metric spaces generalize b-metric spaces and cone metric spaces.

Definition 2.2 [19]

Let $(X,d)$ be a cone b-metric space, $x\in X$, let $\{x_n\}$ be a sequence in X. Then

• (i) $\{x_n\}$ converges to x whenever for every $c\in \mathbb{E}$ with $\theta \ll c$ there is a natural number $n_0$ such that $d(x_n,x)\ll c$ for all $n\ge n_0$. We denote this by ${lim}_{n\to \mathrm{\infty }}{x}_n=x$;

• (ii) $\{x_n\}$ is a Cauchy sequence whenever for every $c\in \mathbb{E}$ with $\theta \ll c$ there is a natural number $n_0$ such that $d(x_n,x_m)\ll c$ for all $n,m\ge n_0$;

• (iii) $(X,d)$ is complete cone b-metric if every Cauchy sequence in X is convergent.

Remark 2.2 [17]

The results concerning fixed points and other results, in case of cone spaces with non-normal solid cones, cannot be provided by reducing to metric spaces, because in this case neither of the conditions of Lemmas 1-4 in [10] hold. Further, the vector cone metric is not continuous in the general case, i.e., from $x_n\to x$, $y_n\to y$ it need not follow that $d(x_n,y_n)\to d(x,y)$.

Let $\mathbb{E}$ be an ordered Banach space with a positive cone P. The following properties hold [17, 19]:

(PT1) If $u⪯v$ and $v\ll w$, then $u\ll w$.

(PT2) If $u\ll v$ and $v⪯w$, then $u\ll w$.

(PT3) If $u\ll v$ and $v\ll w$, then $u\ll w$.

(PT4) If $\theta ⪯u\ll c$ for each $c\in intP$, then $u=\theta$.

(PT5) If $a⪯b+c$ for each $c\in intP$, then $a⪯b$.

(PT6) Let $\{a_n\}$ be a sequence in $\mathbb{E}$. If $c\in intP$ and $a_n\to \theta$ (as $n\to \mathrm{\infty }$), then there exists $n_0\in \mathbb{N}$ such that for all $n\ge n_0$, we have $a_n\ll c$.

3 Main result

According to [18], we denote by Λ a family of nonempty closed and bounded subsets of X, and

$$\begin{array}{c}s(p)=\{q\in \mathbb{E}:p⪯q\}\text{for }q\in \mathbb{E},\hfill \\ s(a,B)=\bigcup _{b\in B}s(d(a,b))=\bigcup _{b\in B}\{x\in \mathbb{E}:d(a,b)⪯x\}\text{for }a\in X\text{ and }B\in \mathrm{\Lambda }.\hfill \end{array}$$

For $A,B\in \mathrm{\Lambda }$, we define

$$s(A,B)=(\bigcap _{a\in A}s(a,B))\cap (\bigcap _{b\in B}s(b,A)).$$

Remark 3.1 Let $(X,d)$ be a cone b-metric space. If $\mathbb{E}=\mathbb{R}$ and $P=[0,+\mathrm{\infty })$, then $(X,d)$ is a b-metric space. Moreover, for $A,B\in CB(X)$, $H(A,B)=infs(A,B)$ is the Hausdorff distance induced by d.

Now, we start with the main result of this paper.

Theorem 3.1 Let $(X,d)$ be a complete cone b-metric space with the coefficient $r\ge 1$ and cone P, and let $T:X\to \mathrm{\Lambda }$ be a multivalued mapping. If there exists a function $\phi :P\to [0,\frac{1}{r})$ such that

$$\underset{n\to \mathrm{\infty }}{lim}sup\phi (a_n)<\frac{1}{r}$$

(a)

for any decreasing sequence $\{a_n\}$ in P. If for all $x,y\in X$,

$$\phi (d(x,y))d(x,y)\in s(Tx,Ty),$$

(b)

then T has a fixed point in X.

Proof Let $x_0$ be an arbitrary point in X, then $T{x}_0\in \mathrm{\Lambda }$, so $T{x}_0\ne \varphi$. Let $x_1\in T{x}_0$ and consider

$$\phi (d(x_0,x_1))d(x_0,x_1)\in s(T{x}_0,T{x}_1).$$

By definition we have

$$\phi (d(x_0,x_1))d(x_0,x_1)\in (\bigcap _{x\in T{x}_0}s(x,T{x}_1))\cap (\bigcap _{y\in T{x}_1}s(y,T{x}_0)),$$

which implies

$$\phi (d(x_0,x_1))d(x_0,x_1)\in s(x,T{x}_1)\text{for all }x\in T{x}_0.$$

Since $x_1\in T{x}_0$, so we have

$$\phi (d(x_0,x_1))d(x_0,x_1)\in s(x_1,T{x}_1).$$

We have

$$\phi (d(x_0,x_1))d(x_0,x_1)\in \bigcup _{x\in T{x}_1}s(d(x_1,x)).$$

So there exists some $x_2\in T{x}_1$ such that

$$\phi (d(x_0,x_1))d(x_0,x_1)\in s(d(x_1,x_2)).$$

It gives

$$d(x_1,x_2)⪯\phi (d(x_0,x_1))d(x_0,x_1).$$

By induction we can construct a sequence $\{x_n\}$ in X such that

$$d(x_n,x_{n+1})⪯\phi (d(x_{n-1},x_n))d(x_{n-1},x_n),x_{n+1}\in T{x}_n\text{ for }n\in \mathbb{N}.$$

(c)

If $x_n=x_{n+1}$ for some $n\in \mathbb{N}$, then T has a fixed point. Assume that $x_n\ne x_{n+1}$, then from (c) the sequence $\{d(x_n,x_{n+1})\}$ is decreasing in P. Hence from (a) there exists $a\in (0,\frac{1}{r})$ such that

$$\underset{n\to \mathrm{\infty }}{lim}sup\phi (d(x_n,x_{n+1}))<a.$$

Thus, for any $k\in (a,\frac{1}{r})$, there exists some $n_0\in \mathbb{N}$ such that for all $n\ge n_0$, implies $\phi (d(x_n,x_{n+1}))<k$. Now consider, for all $n\ge n_0$,

$$\begin{array}{rcl}d(x_n,x_{n+1})& ⪯& \phi (d(x_{n-1},x_n))d(x_{n-1},x_n)\prec kd(x_{n-1},x_n)\prec k^{n-n_0}d(x_{n_0},x_{n_0+1})\\ =& k^n{v}_0,\end{array}$$

where $v_0=k^{-n_0}d(x_{n_0},x_{n_0+1})$.

Let $m>n\ge n_0$. Applying (C3) to triples $\{x_n,x_{n+1},x_m\},\{x_{n+1},x_{n+2},x_m\},\dots ,\{x_{m-2},x_{m-1},x_m\}$, we obtain

$$\begin{array}{rcl}d(x_n,x_m)& ⪯& r[d(x_n,x_{n+1})+d(x_{n+1},x_m)]\\ ⪯& rd(x_n,x_{n+1})+r^2[d(x_{n+1},x_{n+2})+d(x_{n+2},x_m)]\\ ⪯& \cdots \\ ⪯& rd(x_n,x_{n+1})+r^{2}d(x_{n+1},x_{n+2})+\cdots +r^{m-n-1}[d(x_{m-2},x_{m-1})+d(x_{m-1},x_m)]\\ ⪯& rd(x_n,x_{n+1})+r^{2}d(x_{n+1},x_{n+2})+\cdots +r^{m-n-1}d(x_{m-2},x_{m-1})+r^{m-n}d(x_{m-1},x_m).\end{array}$$

Now $d(x_n,x_{n+1})⪯k^n{v}_0$ and $kr<1$ imply that

$$\begin{array}{rcl}d(x_n,x_m)& ⪯& (r{k}^n+r^2{k}^{n+1}+\cdots +r^{m-n}{k}^{m-1})v_0\\ =& r{k}^n(1+(rk)+\cdots +{(rk)}^{m-n-1})v_0\\ ⪯& \frac{r{k}^n}{1-rk}{v}_0\to \theta \text{when }n\to \mathrm{\infty }.\end{array}$$

Now, according to (PT6) and (PT1), we obtain that for a given $\theta \ll c$ there exists $m_0\in \mathbb{N}$ such that

$$d(x_n,x_m)\ll c\text{for all }m,n>m_0,$$

that is, $\{x_n\}$ is Cauchy sequence in $(X,d)$. Since $(X,d)$ is a complete cone b-metric space, so there exists some $u\in X$ such that $x_n\to u$. Take $k_0\in \mathbb{N}$ such that $d(x_n,u)\ll \frac{c}{2r}$ for all $n\ge k_0$. Now we will prove $u\in Tu$. For this let us consider

$$\phi (d(x_n,u))d(x_n,u)\in s(T{x}_n,Tu).$$

By definition we have

$$\phi (d(x_n,u))d(x_n,u)\in (\bigcap _{x\in T{x}_n}s(x,Tu))\cap (\bigcap _{v\in Tu}s(y,T{x}_n)),$$

which implies

$$\begin{array}{c}\phi (d(x_n,u))d(x_n,u)\in (\bigcap _{x\in T{x}_n}s(x,Tu)),\hfill \\ \phi (d(x_n,u))d(x_n,u)\in s(x,Tu)\text{for all }x\in T{x}_n.\hfill \end{array}$$

Since $x_{n+1}\in T{x}_n$, so we have

$$\phi (d(x_n,u))d(x_n,u)\in s(x_{n+1},Tu).$$

So there exists some $v_n\in Tu$ such that

$$\phi (d(x_n,u))d(x_n,u)\in s(d(x_{n+1},v_n)).$$

It gives

$$d(x_{n+1},v_n)⪯\phi (d(x_n,u))d(x_n,u)⪯d(x_n,u).$$

(d)

Now consider

$$\begin{array}{rcl}d(u,v_n)& ⪯& r[d(u,x_{n+1})+d(x_{n+1},v_n)]\\ ⪯& rd(u,x_{n+1})+rd(x_n,u)\\ \ll & \frac{c}{2}+\frac{c}{2}=c\text{for all }n\ge k_0,\end{array}$$

which means $v_n\to u$, since Tu is closed so $u\in Tu$. □

Corollary 3.1 [18]

Let $(X,d)$ be a complete cone metric space with a normal cone P, and let $T:X\to CB(X)$ be a multivalued mapping. If there exists a function $\phi :P\to [0,1)$ such that

$$\underset{n\to \mathrm{\infty }}{lim}sup\phi (a_n)<1$$

for any decreasing sequence $\{a_n\}$ in P. If for all $x,y\in X$,

$$\phi (d(x,y))d(x,y)\in s(Tx,Ty),$$

then T has a fixed point in X.

Corollary 3.2 [15]

Let $(X,d)$ be a complete metric space and let $T:X\to 2^X$ be a multivalued map such that Tx is a closed bounded subset of X for all $x\in X$. If there exists a function $\phi :(0,\mathrm{\infty })\to [0,1)$ such that ${lim\hspace{0.17em}sup}_{r\to t^{+}}\phi (r)<1$ for all $t\in [0,\mathrm{\infty })$ and if

$$H(Tx,Ty)\le \phi (d(x,y))(d(x,y))\mathit{\text{for all }}x,y\in X,$$

then T has a fixed point in X.

The following is Nadler’s theorem for multivalued mappings in a complete metric space.

Corollary 3.3 [14]

Let $(X,d)$ be a complete metric space and let $T:X\to 2^X$ be a multivalued map such that Tx is a closed bounded subset of X for all $x\in X$. If there exists $k\in [0,1)$ such that

$$H(Tx,Ty)\le kd(x,y)\mathit{\text{for all }}x,y\in X,$$

then T has a fixed point in X.

Example 3.1 Let $X=[0,1]$ and $\mathbb{E}$ be the set of all real-valued functions on X which also have continuous derivatives on X. Then $\mathbb{E}$ is a vector space over ℝ under the following operations:

$$(x+y)(t)=x(t)+y(t),(\alpha x)(t)=\alpha x(t)$$

for all $x,y\in \mathbb{E}$, $\alpha \in \mathbb{R}$. That is, $E=C_R^1[0,1]$ with the norm $\parallel f\parallel ={\parallel f\parallel }_{\mathrm{\infty }}+{\parallel f^{\prime }\parallel }_{\mathrm{\infty }}$ and

$$P=\{x\in \mathbb{E}:\theta ⪯x\},\text{where }\theta (t)=0\text{ for all }t\in X,$$

then P is a non-normal cone. Define $d:X\times X\to \mathbb{E}$ as follows:

$$(d(x,y))(t)=|x-y{|}^p{e}^t\text{for }p>1.$$

Then $(X,d)$ is a cone b-metric space but not a cone metric space. For $x,y,z\in X$, set $u=x-z$, $v=z-y$, so $x-y=u+v$. From the inequality

$${(a+b)}^p\le {(2max\{a,b\})}^p\le 2^p(a^p+b^p)\text{for all }a,b\ge 0,$$

we have

$$\begin{array}{c}|x-y{|}^p=|u+v{|}^p\le {(|u|+|v|)}^p\le 2^p(|u{|}^p+|v{|}^p)=2^p(|x-z{|}^p+|z-y{|}^p),\hfill \\ |x-y{|}^p{e}^t\le 2^p(|x-z{|}^p{e}^t+|z-y{|}^p{e}^t),\hfill \end{array}$$

which implies that

$$d(x,y)\preccurlyeq r[d(x,z)+d(y,z)]\text{with }r=2^p>1.$$

But

$$|x-y{|}^p{e}^t\le |x-z{|}^p{e}^t+|z-y{|}^p{e}^t$$

is impossible for all $x>z>y$. Indeed, taking advantage of the inequality

$${(a+b)}^p>a^p+b^p,$$

we have

$$\begin{array}{c}|x-y{|}^p>|x-z{|}^p+|z-y{|}^p,\hfill \\ |x-y{|}^p{e}^t>|x-z{|}^p{e}^t+|z-y{|}^p{e}^t\hfill \end{array}$$

for all $x>z>y$. Thus the triangular inequality in a cone metric space is not satisfied, so $(X,d)$ is not a cone metric space but is a cone b-metric space.

Let $T:X\to \mathrm{\Lambda }$ be such that

$$Tx=[0,\frac{x}{30}],$$

then we have, for $x<y$,

$$s(Tx,Ty)=s\left(\right|\frac{x}{30}-\frac{y}{30}{|}^p{e}^t).$$

Since

$$|\frac{x}{30}-\frac{y}{30}{|}^p{e}^t\le \frac{1}{3^p}|x-y{|}^p{e}^t,$$

so

$$\frac{1}{3^p}(|x-y{|}^p{e}^t)\in s\left(\right|\frac{x}{30}-\frac{y}{30}{|}^p{e}^t).$$

Hence, for $\phi (d(x,y))=\frac{1}{3^p}$, we have

$$\phi (d(x,y))d(x,y)\in s(Tx,Ty).$$

All conditions of our main theorems are satisfied, so T has a fixed point.