Suzuki type fixed point theorems for generalized multi-valued mappings in b-metric spaces

Abstract

In this paper, we obtained a new condition for a multi-valued mapping in a b-metric space, which guarantees the existence of its fixed point.

MSC:47H10, 54H25, 54E50.

1 Introduction

Let $(X,d)$ be a metric space, and let $CB(X)$ be a collection of all non-empty closed and bounded subsets of X. For every $A,B\in CB(X)$, a Hausdorff metric H induced by the metric d of X is given by

$$H(A,B)=max\{\underset{a\in A}{sup}d(a,B),\underset{b\in B}{sup}d(b,A)\},$$

where $d(a,B)=inf\{d(a,x),x\in B\}$.

For a multi-valued mapping $T:X\to 2^X$, a point $x\in X$ is called a fixed point of T if $x\in Tx$. We denote the set of fixed points of T by $Fix(T)$.

Banach’s fixed point theorem is extended to the following result of Nadler [1] from the single-valued mappings to the multi-valued contractive mappings.

Theorem 1.1 [1]

Let $(X,d)$ be a complete metric space, and let $T:X\to CB(X)$ be a set-valued α-contraction, that is, a mapping, for which there exists a constant $\alpha \in (0,1)$ such that $H(Tx,Ty)\le \alpha d(x,y)$, $\mathrm{\forall }x,y\in X$. Then T has at least one fixed point.

The following remarkable generalization of the classical Banach contraction theorem due to Suzuki [2], states the following.

Theorem 1.2 [2]

For a metric space $(X,d)$, define a nonincreasing function θ from $[0,1)$ onto $(1/2,1]$ by

$$\theta (r)=\{\begin{array}{cc}1\hfill & \mathit{\text{if }}0\le r\le (\sqrt{5}-1)/2,\hfill \\ (1-r)r^{-2}\hfill & \mathit{\text{if }}(\sqrt{5}-1)/2\le r\le 2^{-1/2},\hfill \\ {(1+r)}^{-1}\hfill & \mathit{\text{if }}{2}^{-1/2}\le r<1.\hfill \end{array}$$

The following are equivalent:

• (i) X is complete.

• (ii) Every mapping T on X such that there exists $r\in [0,1)$, $\theta (r)d(x,Tx)\le d(x,y)$ implies that $d(Tx,Ty)\le rd(x,y)$ for all $x,y\in X$ has a fixed point.

Theorem 1.2 has been generalized to multi-valued mappings by Kikkawa and Suzuki [3], Mot and Petrusel [4], Dhompongsa and Yingtaweesittikul [5], Singh and Mishra [6], Shahzad and Bassindowa [7], and Aleomraninejad et al. [8].

The concept of a b-metric space was introduced by Czerwik (see [9] and [10]). We recall from [9] the following definition.

Definition 1.3 [9]

Let X be a set, and let $s\ge 1$ be a given real number. A function $d:X\times X\to R^{+}$ is said to be a b-metric if and only if for all $x,y,z\in X$, the following conditions are satisfied:

• 1. $d(x,y)=0$ if and only if $x=y$.

• 2. $d(x,y)=d(y,x)$.

• 3. $d(x,z)\le s[d(x,y)+d(y,z)]$.

A pair $(X,d)$ is called a b-metric space.

We remark that a metric space is evidently a b-metric space. However, Czerwik (see [9, 10]) has shown that a b-metric on X need not be a metric on X.

We cite the following lemmas from Czerwik [9–11] and Singh et al. [6]

Lemma 1.4 Let $(X,d)$ be a b-metric space. For any $A,B,C\in CB(X)$ and any $x,y\in X$,

• 1. $d(x,B)\le d(x,b)$ for any $b\in B$,

• 2. $d(x,B)\le H(A,B)$ for any $x\in A$,

• 3. $d(x,A)\le s[d(x,y)+d(y,A)]$.

Lemma 1.5 Let $(X,d)$ be a b-metric space, and let $A,B\in C(X)$. Then for each $\alpha >0$ and for all $b\in B$, there exists $a\in A$ such that

$$d(a,b)\le H(A,B)+\alpha .$$

Some examples of b-metric spaces and some fixed point theorems for single-valued and multi-valued mappings in b-metric spaces can also be found in Czerwik [9], Boriceanu et al. [12], Boriceanu et al. [13], Aydi and Bota [14], Bota et al. [15], and Bota [16].

Theorem 1.6 [9]

Let $(X,d)$ be a b-complete metric space, and let $T:X\to CB(X)$ be a multi-valued mapping such that T satisfies the inequality

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

where $0<r<\frac{1}{s}$. Then T has a fixed point.

Theorem 1.7 [16]

Let $(X,d)$ be a b-complete metric space, and let $T:X\to CB(X)$ be a multi-valued mapping. Suppose that there exist $a,b,c>0$ with $c<1$ and $\frac{a+b}{1-c}<\frac{1}{s}$ such that T satisfies the inequality

$$H(Tx,Ty)\le ad(x,y)+bd(x,Tx)+cd(y,Ty)$$

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

Theorem 1.8 [14]

Let $(X,d)$ be a b-complete metric space, and let $T:X\to CB(X)$ be a multi-valued mapping such that for all $x,y\in X$,

$$H(Tx,Ty)\le rmax\{d(x,y),d(x,Tx),d(y,Ty),d(x,Ty),d(y,Tx)\},$$

where $r<\frac{1}{s^2+s}$. Then T has a fixed point.

In 2011, Aleomraninejad et al. [17] gave a new condition for multi-valued mappings in a metric space, which guarantees the existence of its fixed point.

Consider a continuous function $g:{[0,\mathrm{\infty })}^5\to [0,\mathrm{\infty })$ satisfying the following conditions:

• (i) $g(1,1,1,2,0)=g(1,1,1,0,2)=h\in (0,1)$.

• (ii) g is subhomogeneous, that is, for all $(x_1,x_2,x_3,x_4,x_5)\in {[0,\mathrm{\infty })}^5$, $\alpha >0$

  $$g(\alpha x_1,\alpha x_2,\alpha x_3,\alpha x_4,\alpha x_5)\le \alpha g(x_1,x_2,x_3,x_4,x_5).$$

• (iii) If $x_i,y_i\in [0,\mathrm{\infty })$, $x_i<y_i$ for $i=1,\dots ,4$, then

  $$g(x_1,x_2,x_3,x_4,0)<g(y_1,y_2,y_3,y_4,0)\text{and}g(x_1,x_2,x_3,0,x_4)<g(y_1,y_2,y_3,0,y_4).$$

Theorem 1.9 [17]

Let $(X,d)$ be a complete metric space, and let $F,G:X\to CB(X)$ be two multi-valued mappings. Suppose that there exist $\alpha \in (0,1)$ and $g\in R$ such that $\alpha (h+1)\le 1$ and $\alpha d(x,Fx)\le d(x,y)$ or $\alpha d(y,Gy)\le d(x,y)$ implies that

$$H(Fx,Gy)\le g(d(x,y),d(x,Fx),d(y,Gy),d(x,Gy),d(y,Fx))$$

for all $x,y\in X$. Then $Fix(F)=Fix(G)$ and $Fix(F)$ is non-empty.

The aim of this paper is to apply the concept of this function g to b-metric spaces.

Let $s\ge 1$ be fixed, and let $R_s$ be the set of all continuous functions $g:{[0,\mathrm{\infty })}^5\to [0,\mathrm{\infty })$ satisfying the conditions (ii), (iii) and

• (iv) $g(1,1,1,2s,0)=g(1,1,1,0,2s)=h_s\in (0,1/s)$.

Following the proofs in [18] and [17] with minor modification, we get the following results, respectively.

Lemma 1.10 If $g\in R_s$ and $u,v\in [0,\mathrm{\infty })$ are such that

$$\begin{array}{rl}u\le & max\{g(v,v,u,s(v+u),0),g(v,v,u,0,s(v+u)),\\ g(v,u,v,s(v+u),0),g(v,u,v,0,s(v+u))\},\end{array}$$

then $u\le h_{s}v$.

Proof Without loss of generality, we can suppose that $u\le g(v,v,u,s(v+u),0)$.

If $v<u$, then

$$u\le g(v,v,u,s(v+u),0)\le g(u,u,u,2us,0)\le ug(1,1,1,2s,0)=h_{s}u<u$$

which is a contradiction. Thus $u\le v$. So,

$$u\le g(v,v,u,s(u+v),0)\le g(v,v,v,2vs,0)\le vg(1,1,1,2s,0)=h_{s}v.$$

 □

Lemma 1.11 Let $(X,d)$ be a b-complete metric space, and let $F,G:X\to CB(X)$ be two multi-valued mappings. Suppose that there exist $\alpha \in (0,\mathrm{\infty })$ and $g\in R_s$ such that $\alpha d(x,Fx)\le d(x,y)$ or $\alpha d(y,Gy)\le d(x,y)$ implies that

$$H(Fx,Gy)\le g(d(x,y),d(x,Fx),d(y,Gy),d(x,Gy),d(y,Fx))$$

for all $x,y\in X$. Then $Fix(F)=Fix(G)$.

Proof Let $x\in Fix(F)$, then $\alpha d(x,Fx)=0=d(x,x)$. Thus,

$$\begin{array}{rl}d(x,Gx)& \le H(Fx,Gx)\\ \le g(d(x,x),d(x,Fx),d(x,Gx),d(x,Gx),d(x,Fx))\\ \le g(0,0,d(x,Gx),d(x,Gx),0)\\ \le g(0,0,d(x,Gx),sd(x,Gx),0).\end{array}$$

Using Lemma 1.10, we have $d(x,Gx)\le h_{s}0=0$. So, $x\in Fix(G)$.

Hence $Fix(F)\subseteq Fix(G)$. Similarly, we can obtain $Fix(G)\subseteq Fix(F)$. □

2 Main results

Theorem 2.1 Let $(X,d)$ be a b-complete metric space, and let $F,G:X\to CB(X)$ be two multi-valued mappings. Suppose that there exist $\alpha \in (0,1)$ and $g\in R_s$ such that $s\alpha (h_s+1)\le 1$ and $\alpha d(x,Fx)\le d(x,y)$ or $\alpha d(y,Gy)\le d(x,y)$ implies that

$$H(Fx,Gy)\le g(d(x,y),d(x,Fx),d(y,Gy),d(x,Gy),d(y,Fx))$$

for all $x,y\in X$. Then $Fix(F)=Fix(G)$ and $Fix(F)$ is non-empty.

Proof The main idea of the proof follows from Theorem 1.9.

By Lemma 1.11, $Fix(F)=Fix(G)$. Let $r\in (h_s,\frac{1}{s})$ and $x_0\in X$. If $x_0$ is not a fixed point, choose $x_1\in F{x}_0$ such that $\alpha d(x_0,F{x}_0)<d(x_0,x_1)$. Thus,

$$\begin{array}{rl}d(x_1,G{x}_1)& \le H(F{x}_0,G{x}_1)\\ \le g(d(x_0,x_1),d(x_0,F{x}_0),d(x_1,G{x}_1),d(x_0,G{x}_1),d(x_1,F{x}_0))\\ \le g(d(x_0,x_1),d(x_0,x_1),d(x_1,G{x}_1),s[d(x_0,x_1)+d(x_1,G{x}_1)],0).\end{array}$$

By Lemma 1.10, we have $d(x_1,G{x}_1)\le h_{s}d(x_0,x_1)<rd(x_0,x_1)$. If $x_1$ is not a fixed point, there exists $x_2\in G{x}_1$ such that $d(x_1,x_2)<rd(x_0,x_1)$. Since $\alpha d(x_1,G{x}_1)<d(x_1,x_2)$,

$$\begin{array}{rl}d(x_2,F{x}_2)& \le H(F{x}_2,G{x}_1)\\ \le g(d(x_1,x_2),d(x_2,F{x}_2),d(x_1,G{x}_1),d(x_2,G{x}_1),d(x_1,F{x}_2))\\ \le g(d(x_1,x_2),d(x_2,F{x}_2),d(x_1,x_2),0,s[d(x_1,x_2)+d(x_2,F{x}_2)]).\end{array}$$

By Lemma 1.10, we have $d(x_2,F{x}_2)\le h_{s}d(x_1,x_2)<rd(x_1,x_2)$.

Similarly, there exists $x_3\in F{x}_2$ such that $d(x_2,x_3)<rd(x_1,x_2)<r^{2}d(x_0,x_1)$.

By continuing this process, we obtain a sequence $\{x_n\}$ in X such that

$$\begin{array}{c}{x}_{2n-1}\in F{x}_{2n-2},x_{2n}\in G{x}_{2n-1},d(x_n,x_{n+1})\le r^{n}d(x_0,x_1)\hfill \\ d(x_{2n},F{x}_{2n})\le hd(x_{2n-1},x_{2n})\text{and}d(x_{2n-1},G{x}_{2n-1})\le hd(x_{2n-2},x_{2n-1}).\hfill \end{array}$$

We prove next that the sequence $\{x_n\}$ is Cauchy,

$$\begin{array}{rl}d(x_n,x_{n+p})\le & s[d(x_n,x_{n+1})+d(x_{n+1},x_{n+p})]=sd(x_n,x_{n+1})+sd(x_{n+1},x_{n+p})\\ \le & sd(x_n,x_{n+1})+s^2[d(x_{n+1},x_{n+2})+d(x_{n+2},x_{n+p})]\\ =& sd(x_n,x_{n+1})+s^{2}d(x_{n+1},x_{n+2})+s^{2}d(x_{n+2},x_{n+p})\\ \le & sd(x_n,x_{n+1})+s^{2}d(x_{n+1},x_{n+2})+s^{3}d(x_{n+2},x_{n+3})+\cdots \\ +s^{p-1}d(x_{n+p-2},x_{n+p-1})+s^{p-1}d(x_{n+p-1},x_{n+p})\\ \le & s{r}^{n}d(x_0,x_1)+s^2{r}^{n+1}d(x_0,x_1)+s^3{r}^{n+2}d(x_0,x_1)+\cdots \\ +s^{p-1}{r}^{n+p-2}d(x_0,x_1)+s^{p-1}{r}^{n+p-1}d(x_0,x_1),\\ d(x_n,x_{n+p})\le & s{r}^n(1+sr+s^2{r}^2+\cdots +s^{p-2}{r}^{p-2}+s^{p-2}{r}^{p-1})d(x_0,x_1)\\ \le & s{r}^n(1+sr+s^2{r}^2+\cdots +s^{p-2}{r}^{p-2}+s^{p-1}{r}^{p-1})d(x_0,x_1)\\ =& s{r}^n\left[\frac{1-{(sr)}^p}{1-sr}\right]d(x_0,x_1).\end{array}$$

Notice that

$$s{r}^n\left[\frac{1-{(sr)}^p}{1-sr}\right]d(x_0,x_1)\to 0\text{as }n\to \mathrm{\infty }.$$

So $\{x_n\}$ is Cauchy, and $x_n\to x$ for some $x\in X$.

Now, we claim that for each $n\ge 1$,

$$\alpha d(x_{2n},F{x}_{2n})\le d(x_{2n},x)\text{or}\alpha d(x_{2n+1},G{x}_{2n+1})\le d(x_{2n+1},x).$$

If $\alpha d(x_{2n},F{x}_{2n})>d(x_{2n},x)$ and $\alpha d(x_{2n+1},G{x}_{2n+1})>d(x_{2n+1},x)$ for some $n\ge 1$, then

$$\begin{array}{rl}d(x_{2n},x_{2n+1})& \le s[d(x_{2n},x)+d(x_{2n+1},x)]\\ <s[\alpha d(x_{2n},F{x}_{2n})+\alpha d(x_{2n+1},G{x}_{2n+1})]\\ \le s[\alpha d(x_{2n},x_{2n+1})+\alpha hd(x_{2n},x_{2n+1})]\\ =s(\alpha +\alpha h)d(x_{2n},x_{2n+1})=s\alpha (h_s+1)d(x_{2n},x_{2n+1}).\end{array}$$

Thus, we get $s\alpha (h_s+1)>1$, which is a contradiction. By using the assumption, for each $n\ge 1$, either

$$H(F{x}_{2n},Gx)\le g(d(x_{2n},x),d(x_{2n},F{x}_{2n}),d(x,Gx),d(x_{2n},Gx),d(x,F{x}_{2n}))$$

or

$$H(Fx,G{x}_{2n+1})\le g(d(x,x_{2n+1}),d(x,Fx),d(x_{2n+1},G{x}_{2n+1}),d(x,G{x}_{2n+1}),d(x_{2n+1},Fx)).$$

Therefore, one of the following cases holds.

• (a) There exists an infinite subset $I\subseteq N$ such that

  $$\begin{array}{rl}d(x_{2n+1},Gx)& \le H(F{x}_{2n},Gx)\\ \le g(d(x_{2n},x),d(x_{2n},F{x}_{2n}),d(x,Gx),d(x_{2n},Gx),d(x,F{x}_{2n}))\end{array}$$

  for all $n\in I$.

• (b) There exists an infinite subset $J\subseteq N$ such that

  $$\begin{array}{rl}d(Fx,x_{2n+2})& \le H(Fx,G{x}_{2n+1})\\ \le g(d(x,x_{2n+1}),d(x,Fx),d(x_{2n+1},G{x}_{2n+1}),d(x,G{x}_{2n+1}),d(x_{2n+1},Fx))\end{array}$$

  for all $n\in J$.

In case (a), we obtain

$$\begin{array}{rl}d(x,Gx)\le & s[d(x,x_{2n+1})+d(x_{2n+1},Gx)]\\ \le & s[d(x,x_{2n+1})+g(d(x_{2n},x),d(x_{2n},F{x}_{2n}),d(x,Gx),d(x_{2n},Gx),d(x,F{x}_{2n}))]\\ \le & s[d(x,x_{2n+1})\\ +g(d(x_{2n},x),d(x_{2n},x_{2n+1}),d(x,Gx),s[d(x_{2n},x)+d(x,Gx)],d(x,x_{2n+1}))]\end{array}$$

for all $n\in I$. Since g is continuous, $d(x,Gx)\le s(g(0,0,d(x,Gx),sd(x,Gx),0))$. Using Lemma 1.10, $d(x,Gx)=0$. We have $x\in Gx$.

In case (b), we obtain

$$\begin{array}{rl}d(x,Fx)\le & s[d(x,x_{2n+2})+d(x_{2n+2},Fx)]\\ \le & s[d(x,x_{2n+2})\\ +g(d(x,x_{2n+1}),d(x,Fx),d(x_{2n+1},G{x}_{2n+1}),d(x,G{x}_{2n+1}),d(x_{2n+1},Fx))]\\ \le & s[d(x,x_{2n+2})\\ +g(d(x,x_{2n+1}),d(x,Fx),d(x_{2n+1},x_{2n+2}),d(x,x_{2n+2}),s[d(x_{2n+1},x)+d(x,Fx)])]\end{array}$$

for all $n\in J$. Since g is continuous, $d(x,Fx)\le s(g(0,d(x,Fx),0,0,sd(x,Fx)))$. Using Lemma 1.10, $d(x,Fx)=0$. We have $x\in Fx$. This completes the proof. □

Remark 2.2 Taking $s=1$ in Theorem 2.1 (case of metric spaces), we recover Theorem 1.9.

The following result is a consequence of Theorem 2.1.

Corollary 2.3 Let $(X,d)$ be a b-complete metric space, and let $T:X\to CB(X)$ be a multi-valued mapping. Suppose that there exist $\alpha \in (0,1)$ and $g\in R_s$ such that $s\alpha (h_s+1)\le 1$ and $\alpha d(x,Tx)\le d(x,y)$ implies that

$$H(Tx,Ty)\le g(d(x,y),d(x,Tx),d(y,Ty),d(x,Ty),d(y,Tx))$$

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

Corollary 2.4 Let $(X,d)$ be a b-complete metric space, and let $T:X\to CB(X)$ be a multi-valued mapping. Suppose that there exists $r\in (0,\frac{1}{s})$ such that $\frac{1}{s(r+1)}d(x,Tx)\le d(x,y)$ implies

$$H(Tx,Ty)\le rmax\{d(x,y),d(x,Tx),d(y,Ty)\}$$

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

Proof Let $g\in R_s$ by $g(x_1,x_2,x_3,x_4,x_5)=rmax\{x_1,x_2,x_3\}$, where $r<\frac{1}{s}$. Put $\alpha =\frac{1}{s(r+1)}$. Since $h_s=r<\frac{1}{s}$ and $s\alpha (h_s+1)\le 1$, by using Corollary 2.3, T has a fixed point. □

Remark 2.5 Corollary 2.4 is an extension of Theorem 1.6.

Corollary 2.6 Let $(X,d)$ be a b-complete metric space, and let $T:X\to CB(X)$ be a multi-valued mapping. Suppose that there exists $a,b\in [0,1)$ and $a+2b<\frac{1}{s}$ such that $\frac{1}{s(1+a+2b)}d(x,Tx)\le d(x,y)$ implies that

$$H(Tx,Ty)\le ad(x,y)+bd(x,Tx)+bd(y,Ty)$$

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

Proof Let $g\in R_s$ be $g(x_1,x_2,x_3,x_4,x_5)=a{x}_1+b(x_2+x_3)$, where $a+2b<\frac{1}{s}$. Put $\alpha =\frac{1}{s(1+a+2b)}$. Since $h_s=a+2b<\frac{1}{s}$ and $s\alpha (h_s+1)\le 1$, by using Corollary 2.3, T has a fixed point. □

The following examples show that we can apply Corollary 2.3 but cannot apply Theorem 1.8.

Example 2.7 Let $X=[0,1]$ and $d(x,y)={|x-y|}^2$ for all $x,y\in X$. It is obvious that d is a b-metric on X with $s=2$ and $(X,d)$ is complete. Also, d is not a metric on X. Define $T:X\to CB(X)$ by

$$Tx=\{\begin{array}{cc}\{\frac{1}{3},\frac{2}{3}\}\hfill & \text{if }0\le x<1,\hfill \\ \{\frac{1}{3}\}\hfill & \text{if }x=1.\hfill \end{array}$$

Let $x,y\in X$. Without loss of generality, take $x\le y$.

If $x=y$ or $x,y<1$, then $Tx=Ty$. Hence $H(Tx,Ty)=0$.

If $x<1$ and $y=1$, then

$$H(Tx,Ty)=\frac{1}{9}\le \frac{4}{27}=\frac{1}{3}\cdot \frac{4}{9}=\frac{1}{3}d(y,Ty)\le rmax\{d(x,y),d(x,Tx),d(y,Ty)\},$$

where $r=\frac{1}{3}<\frac{1}{2}=\frac{1}{s}$. So all the conditions of Corollary 2.4 are satisfied. Moreover, $\frac{1}{3}$ and $\frac{2}{3}$ are the two fixed points of T.

On the other hand, if we choose $x=\frac{1}{3}$ and $y=1$, then

$$\begin{array}{rl}H(Tx,Ty)& =\frac{1}{9}>\frac{1}{6}\cdot \frac{4}{9}=\frac{1}{s^2+s}max\{\frac{4}{9},0,\frac{4}{9},0,\frac{1}{9}\}\\ =\frac{1}{s^2+s}max\{d(x,y),d(x,Tx),d(y,Ty),d(x,Ty),d(y,Tx)\}.\end{array}$$

So we could not apply Theorem 1.8.

Example 2.8 Let $X=[1,\mathrm{\infty })$ and $d(x,y)={|x-y|}^2$ for all $x,y\in X$. Then $(X,d)$ is a complete b-metric space with $s=2$. Define $T:X\to CB(X)$ by

$$Tx=[1,1+\frac{x}{2}]\text{for all }x\in X.$$

Consider $H(Tx,Ty)=\frac{1}{4}{(x-y)}^2=\frac{1}{4}d(x,y)\le rmax\{d(x,y),d(x,Tx),d(y,Ty)\}$, where $r=\frac{1}{4}<\frac{1}{2}=\frac{1}{s}$ for all $x,y\in X$. So all the conditions of Corollary 2.4 are satisfied. Moreover, 1 and 2 are the two fixed points of T.

On the other hand, if we choose $x=1$ and $y=2$, then

$$\begin{array}{rl}H(Tx,Ty)& =\frac{1}{4}>\frac{1}{6}=\frac{1}{s^2+s}max\{1,0,0,0,\frac{1}{4}\}\\ =\frac{1}{s^2+s}max\{d(x,y),d(x,Tx),d(y,Ty),d(x,Ty),d(y,Tx)\}.\end{array}$$

So we could not apply Theorem 1.8.