Multivalued Contraction Fixed-Point Theorem in b-Metric Spaces

Abstract

The authors explore fixed-point theory in b-metric spaces and strong b-metric spaces. They wish to prove some new extensions of the Covitz and Nadler fixed-point theorem in b-metric spaces. In so doing, they wish to answer a question proposed by Kirk and Shahzad about Nadler’s theorem holding in strong b-metric spaces. In addition, they offer an improvement to the fixed-point theorem proven by Dontchev and Hager.

1. Introduction

Fixed-point theory is a major and important tool in the study of nonlinear phenomena. This theory has been applied in such diverse fields as topology, differential equations and inclusions, economics, game theory, engineering, physics, optimal control, and nonlinear functional analyses. Many authors are interested in fixed-point theorems in metric spaces. The concept of a b-metric space is an old notion that is used in many areas of mathematics. In 1970, Coifman and Guzffian [1] introduced a weaker notion of a metric space called a quasi-metric space; some researchers have used the notion of the b-distance in an attempt to include b-distance functions such as

$$d(x,y)={|x-y|}^n,x,y\in {\mathbb{R}}^n,$$

to resolve some central questions in harmonic analyses (see also [2,3,4]). The actual definition of a b-metric was introduced in 1979 by Madas and Segovia [5]. The notion of a b-metric was first used in fixed-point theory by Bakhtin [6] and extended by Czerwik [7]. Chapter 12, and in particular Section 12.1, of the monograph by Kirk and Shahzad [8] presents a nice introduction to the origin and history of this type of metric space as well as some elementary examples of such spaces.

Our aim in this work is to prove some new versions of the Covitz and Nadler fixed-point theorem [9,10] and to answer a question proposed by Kirk and Shahzad [8], namely, does Nadler’s theorem hold in strong b-metric spaces [8] (page 128) (see Theorem 3 below)?

2. Preliminaries

We begin with some essential concepts and results. In what follows, $\mathcal{P}\left(X\right)$ denotes the set of all nonempty subsets of X so that ${\mathcal{P}}_{cl,b}\left(X\right)$ is the set of all nonempty closed and bounded subsets of X, and ${\mathcal{P}}_{cp}\left(X\right)$ is the set of all nonempty compact subsets of X.

Definition 1.

Let A, $B\in \mathcal{P}\left(X\right)$ and define:

• ${\displaystyle H_d^{*}(A,B)=sup\{d(a,B):a\in A\}}$;
• ${\displaystyle H_d^{*}(B,A)=sup\{d(A,b):b\in B\}}$;
• ${\displaystyle H_d(A,B)=max(H_d^{*}(A,B),H_d^{*}(B,A))}$ (the b-Hausdorff distance between A and B).

Remark 1.

For $ϵ>0$, let

$$A_{ϵ}=\{x\in X:d(x,A)<ϵ\}.$$

Then,

$${\displaystyle H_d^{*}(A,B)=inf\{ϵ>0:A\subset B_{ϵ}\},H_d^{*}(B,A)=inf\{ϵ>0:B\subset A_{ϵ}\}.}$$

Next, we define what is meant by a b-metric space and a strong b-metric space.

Definition 2.

Let X be a nonempty set and $s\ge 1$. By a b-metric on X, we mean a map $d:X\times X\to [0,\infty )$ with the following properties for all x, y, $z\in X$:

(i)

$d(x,y)=0$ if and only if $x=y$;

(ii)

(Symmetry) $d(x,y)=d(y,x)$;

(iii)

(s-relaxed triangle inequality) $d(x,y)\le s\left[d\right(x,z)+d(z,y\left)\right]$.

The triple $(X,d,s)$ is called a b-metric space.

Definition 3.

Let X be a nonempty set and $s\ge 1$. By a strong b-metric on X, we mean a map $d:X\times X\to [0,\infty )$ with the following properties for all x, y, $z\in X$:

(i) 

$d(x,y)=0$ if and only if $x=y$;

(ii) 

(Symmetry) $d(x,y)=d(y,x)$;

(iii) 

(s-relaxed triangle inequality) $d(x,y)\le d(x,z)+sd(z,y)$.

The triple $(X,d,s)$ is called a strong b-metric space.

A useful generalization of the s-relaxed triangle inequality is given in the following lemma.

Lemma 1.

Let $(X,d,s)$ be a strong b-metric space. Then, for $x_0,x_1,\dots ,x_n\in X$, we have

$$d(x_0,x_n)\le \sum _{i=0}^{n-2}{s}^{i+1}d(x_i,x_{i+1})+s^{n-1}d(x_{n-1},x_n).$$

The next two lemmas will be used in our proofs.

Lemma 2.

Let $(X,d,s)$ be a strong b-metric space. Then, d is a continuous mapping.

Proof. 

For any x, y, $x_0$, $y_0\in X$,

$$\begin{array}{cc}\hfill d(x,y)& \le sd(x,x_0)+d(x_0,y)\hfill \\ \hfill & \le sd(x,x_0)+d(x_0,y_0)+sd(y_0,y).\hfill \end{array}$$

Hence,

$$d(x,y)-d(x_0,y_0)\le sd(x,x_0)+sd(y_0,y).$$

Similarly,

$$d(x_0,y_0)-d(x,y)\le sd(x,x_0)+sd(y_0,y).$$

(1)

This implies that

$$|d(x,y)-d(x_0,y_0)|\le s[d(x,x_0)+d(y_0,y)],$$

and therefore d is continuous. □

Lemma 3

([11]). Let $(X,d,s)$ be a b-metric space. Then, every sequence ${\left(x_n\right)}_{n\in \mathbb{N}}\subset X$ for which there exists $\gamma \in (0,1)$ such that

$$d(x_n,x_{n+1})\le \gamma d(x_n,x_{n-1}),n\in \mathbb{N},$$

is a Cauchy sequence.

Lemma 4.

Let $(X,d,s)$ be a b-metric space and A, $B\in {\mathcal{P}}_{cp}\left(X\right)$, which is the set of all nonempty compact subsets of X. If d is a continuous b-metric, then for any $x\in A$, there exists $y\in B$ such that

$$d(x,y)\le d(x,B).$$

Proof. 

Let $x\in A$; then, for every $n\in \mathbb{N}$, there exists $y_n\in B$ with

$$d(x,y_n)\le d(x,B)+\frac{1}{n}.$$

(2)

Since B is compact, there exists a subsequence ${\left(y_{n_k}\right)}_{k\in \mathbb{N}}$ of ${\left(y_n\right)}_{n\in \mathbb{N}}$ converging to $y\in B$. Since d is continuous, letting $n\to \infty$ in $\left(2\right)$, we obtain

$$d(x,y)\le d(x,B)\le H_d(A,B),$$

which proves the lemma. □

3. Covitz–Nadler-Type Fixed-Point Theorems

In this section, we give versions of the Covitz and Nadler fixed-point theorem in b-metric spaces. They proved their classical fixed-point theorem in metric spaces for contraction multi-valued operators in 1970 (see [9,10]) (also see Deimling [12] (Theorem 11.1)).

Definition 4.

A mapping $F:X\to \mathcal{P}\left(X\right)$ is a multivalued map if for each $x\in X$, $F\left(x\right)\in \mathcal{P}\left(X\right)$. The point p is a fixed point of a multivalued map F if $p\in F\left(p\right)$. We will denote the set of fixed points of the mapping F by $FixF$.

We also have the notion of a contraction for multivalued maps.

Definition 5.

If the mapping F has a Lipschitz constant $c<1$, then f is called a multivalued contraction mapping.

The following lemma is referred to as the Covitz and Nadler fixed-point theorem [9].

Lemma 5.

Let $(X,d)$ be a complete metric space. If $F:X\to P_{cl}\left(X\right)$ is a contraction, then $FixX\ne \varnothing$.

Our first result is contained in the following theorem.

Theorem 1.

Let $(X,d,s)$ be a complete b-metric space and d be continuous. If $F:X\to {\mathcal{P}}_{cp}\left(X\right)$ is a contraction, then $FixF\ne \varnothing$.

Proof. 

Assume that $H_d(F\left(x\right),F\left(y\right))\le Ld(x,y)$ for every x, $y\in X$, where $L\in [0,1)$, and let $x\in X$. Since $F\left(x\right)$ is compact, by Lemma 4, we can choose $x_1\in F\left(x\right)$ such that

$$d(x,x_1)\le d(x,F\left(x\right)).$$

Then, we may choose $x_2\in F\left(x_1\right)$ such that

$$d(x_1,x_2)\le d(x_1,F\left(x_1\right))\mathrm{implies}d(x_1,x_2)\le H_d(F\left(x\right),F\left(x_1\right)).$$

This means that

$$d(x_1,x_2)\le Ld(x,F\left(x\right)).$$

Continuing this way, we can find a sequence $\{x_n:n\in \mathbb{N}\}\subset X$ with

$$d(x_n,x_{n+1})\le d(x_n,F\left(x_n\right)).$$

Hence,

$$\begin{array}{cc}\hfill d(x_n,x_{n+1})& \le d(x_n,F\left(x_n\right))\le H_d(F\left(x_{n-1}\right),x_n)\hfill \\ \hfill & \le Ld(x_{n-1},x_n)\le L^{n}d(x,F\left(x\right)).\hfill \end{array}$$

By Lemma 3, ${\left\{x_n\right\}}_{n\in \mathbb{N}}$ is a Cauchy sequence. Since X is complete, we let ${\displaystyle \tilde{x}=\underset{n\to \infty }{lim}{x}_n}$. Then, $x_{n+1}\in F\left(x_n\right)$ for every $n\in \mathbb{N}$, and

$$0\le d(\tilde{x},F\left(\tilde{x}\right))\le s[d(x_{n+1},\tilde{x})+d(x_{n+1},F\left(\tilde{x}\right))]\le s[d(x_{n+1},\tilde{x})+Ld(x_n,\tilde{x})].$$

Letting $n\to +\infty$ gives $\tilde{x}\in F\left(\tilde{x}\right)$ as claimed, and this proves the theorem. □

As a direct consequence of Theorem 1, we are able to obtain the following generalization of Nadler’s fixed-point theorem to strong b-metric spaces.

Corollary 1.

Let $(X,d,s)$ be a complete strong b-metric space. If $F:X\to {\mathcal{P}}_{cp}\left(X\right)$ is an L-contraction, then $FixF\ne \varnothing$.

Proof. 

Since $(X,d,s)$ is a complete strong b-metric space, it is complete. By Lemma 2, d is continuous. By Theorem 1, F has at least one fixed point, and this completes the proof. □

Our next result on the existence of a fixed point is contained in the following theorem.

Theorem 2.

Let $(X,d,s)$ be a complete b-metric space and $F:X\to {\mathcal{P}}_{cl,b}\left(X\right)$ be an L-contraction multi-valued mapping. Then, F has a fixed point in X.

Proof. 

We will employ a standard iterative procedure for contracting mappings. Let $L\in (0,1)$ be such that

$$H_d(F\left(x\right),F\left(y\right))\le Ld(x,y)\mathrm{for}\mathrm{all}x,y\in X.$$

Let $x_0\in X$ be fixed and choose $x_1\in F\left(x_0\right)$ such that

$$d(x_1,x_0)\le d(x_0,F\left(x_0\right))+L.$$

From the definition of the Hausdorff distance, we can find $x_2\in F\left(x_1\right)$ with

$$d(x_1,x_2)\le d(x_1,F\left(x_1\right))+L,\mathrm{which} \mathrm{implies}d(x_1,x_2)\le H_d(F\left(x_0\right),F\left(x_1\right))+L.$$

Similarly, we can find $x_3\in F\left(x_2\right)$, with

$$d(x_3,x_2)\le H_d(F\left(x_2\right),F\left(x_1\right))+L^2.$$

Continuing this process, we obtain a sequence ${\left(x_n\right)}_{n\in \mathbb{N}}$ in X such that $x_{i+1}\in \left(x_n\right)$ and

$$d(x_{i+1},x_i)\le H_d(F\left(x_i\right),F\left(x_{i-1}\right))+L^i.$$

For fixed $m\in \mathbb{N}$,

$$\begin{array}{cc}\hfill d(x_m,x_{m+1})& \le H_d(F\left(x_m\right),F\left(x_{m-1}\right))+L^m\hfill \\ \hfill & \le Ld(x_m,x_{m-1})+L^m\hfill \\ \hfill & \le L{H}_d(F\left(x_{m-1}\right),F\left(x_{m-2}\right))+2L^m\hfill \\ \hfill & \le L^{2}d(x_{m-1},x_{m-2})+2L^m\hfill \\ \hfill & \le L^2(H_d(F\left(x_{m-2}\right),F\left(x_{m-3}\right))+L^{m-2})+2L^m\hfill \\ \hfill & \le L^{3}d(x_{m-2},x_{m-3})+3L^m\hfill \\ \hfill & ⋮\hfill \\ \hfill & \le L^{m}d(x_1,x_0)+m{L}^m.\hfill \end{array}$$

By the s-relaxed triangle inequality in b-metric spaces, for every $p\in \mathbb{N}$ and $q=\left[{log}_{2}p\right]$,

$$\begin{array}{cc}\hfill d(x_{m+1},x_{m+p})& \le sd(x_{m+1},x_{m+2})+sd(x_{m+2},x_{m+p})\hfill \\ \hfill & \le sd(x_{m+1},x_{m+2})+s^{2}d(x_{m+2},x_{m+2^2})+s^{2}d(x_{m+2^2},x_{m+p})\hfill \\ \hfill & ⋮\hfill \\ \hfill & \le \sum _{n=1}^q{s}^{n}d(x_{m+2^{n-1}},x_{m+2^n})+s^{q+1}d(x_{m+2^q},x_{m+p}).\hfill \end{array}$$

By Lemma 1, we obtain

$$\begin{array}{cc}\hfill d(x_{m+1},x_{m+p})& \le \sum _{n=1}^q{s}^{2n}\sum _{i=m}^{m+2^{n-1}-1}d(x_{2^{n-1}+i},x_{m+2^{n-1}+i+1})\hfill \\ \hfill & +s^{2(q+1)}\sum _{i=m}^{m+p-2^q-1}d(x_{2^q+i},x_{2^q+i+1}).\hfill \end{array}$$

Consequently,

$$\begin{array}{cc}\hfill d(x_{m+1},x_{m+p})& \le \sum _{n=1}^q{s}^{2n}\sum _{i=m}^{m+2^{n-1}-1}(L^{2^{n-1}+i}d(x_0,x_1)+(2^{n-1}+i)L^{2^{n-1}+i})\hfill \\ \hfill & +s^{2(q+1)}\sum _{i=m}^{m+p-2^q-1}(L^{2^q+i}d(x_0,x_1)+(2^q+i)L^{2^q+i})\hfill \\ \hfill & \le \sum _{n=1}^{q+1}{s}^{2n}\sum _{i=0}^{2^{n-1}-1}[L^{2^{n-1}+i+m}d(x_0,x_1)\hfill \\ \hfill & +(2^{n-1}+i+m)L^{2^{n-1}+i+m}]\hfill \\ \hfill & \le L^m\sum _{n=1}^{q+1}{s}^{2n}\sum _{i=0}^{2^{n-1}-1}{L}^{2^{n-1}+i}d(x_0,x_1)\hfill \\ \hfill & +L^m\sum _{n=1}^{q+1}{s}^{2n}\sum _{i=0}^{2^{n-1}-1}(2^{n-1}+i)L^{2^{n-1}+i}\hfill \\ \hfill & +m{L}^m\sum _{n=1}^{q+1}{s}^{2n}\sum _{i=0}^{2^{n-1}-1}(2^{n-1}+i)L^{2^{n-1}+i}.\hfill \end{array}$$

Using simple calculations, we can see that

$$L^m\sum _{n=1}^{q+1}{s}^{2n}\sum _{i=0}^{2^{n-1}-1}{L}^{2^{n-1}+i}d(x_0,x_1)\le \frac{L^{m}d(x_0,x_1)}{1-L}\sum _{n=1}^{q+1}{L}^{2n{log}_{L}s+2^{n-1}},$$

(3)

and

$$\begin{array}{cc}\hfill L^m\sum _{n=1}^{q+1}{s}^{2n}& \sum _{i=0}^{2^{n-1}-1}(2^{n-1}+i)L^{2^{n-1}+i}\hfill \\ \hfill & \le L^m\sum _{n=1}^{q+1}{s}^{2n}{2}^{n-1}{L}^{2^{n-1}}\sum _{i=0}^{2^{n-1}-1}{L}^i+L^m\sum _{n=1}^{q+1}{s}^{2n}{L}^{2^{n-1}}\sum _{i=0}^{2^{n-1}-1}i{L}^i\hfill \\ \hfill & \le \frac{2L^m}{1-L}\sum _{n=1}^{q+1}{\left(2s\right)}^{2n}{L}^{2^{n-1}}+L^m\sum _{i=0}^{\infty }i{L}^i\sum _{n=1}^{q+1}{s}^{2n}{L}^{2^{n-1}}.\hfill \end{array}$$

Then,

$$\begin{array}{cc}\hfill L^m\sum _{n=1}^{q+1}{s}^{2n}& \sum _{i=0}^{2^{n-1}-1}(2^{n-1}+i)L^{2^{n-1}+i}\hfill \\ \hfill & \le \frac{2L^m}{1-L}\sum _{n=1}^{q+1}{L}^{2n{log}_{L}2s+2^{n-1}}+L^m\sum _{i=0}^{\infty }i{L}^i\sum _{n=1}^{q+1}{L}^{2n{log}_{L}s+2^{n-1}}.\hfill \end{array}$$

Hence,

$$L^m\sum _{n=1}^{q+1}{s}^{2n}\sum _{i=0}^{2^{n-1}-1}{2}^{n-1}{L}^{2^{n-1}+i}\le \frac{2L^m}{1-L}\sum _{n=1}^{q+1}{L}^{2n{log}_{L}2s+2^{n-1}},$$

(4)

and

$$L^m\sum _{n=1}^{q+1}{s}^{2n}\sum _{i=0}^{2^{n-1}-1}i{L}^{2^{n-1}+i}\le L^m\sum _{i=0}^{\infty }i{L}^i\sum _{n=1}^{q+1}{L}^{2n{log}_{L}s+2^{n-1}}.$$

(5)

We observe that

$$\underset{n\to \infty }{lim}(2n{log}_{L}s+2^{n-1}-n)=\underset{n\to \infty }{lim}(2n{log}_{L}2s+2^{n-1}-n)=\infty .$$

For a fixed $M>0$, there exist $n_0\in \mathbb{N}$ such that

$$2n{log}_{L}s+2^{n-1}-n\ge M,\mathrm{and}2n{log}_{L}2s+2^{n-1}-n\ge M,\mathrm{for}\mathrm{all}n\ge n_0.$$

Then,

$$L^{2n{log}_{L}s+2^{n-1}}\le L^M{L}^n\mathrm{and}{L}^{2n{log}_{L}2s+2^{n-1}}\le L^n{L}^M,$$

and since ${lim}_{n\to \infty }\frac{(n+1)L^{n+1}}{n{L}^n}=L\in (0,1)$, we conclude that

$$L_1:=\sum _{n=1}^{\infty }{L}^{2n{log}_{L}s+2^{n-1}},L_2:=\sum _{n=1}^{\infty }{L}^{2n{log}_{L}2s+2^{n-1}},L_3:=\sum _{n=1}^{\infty }n{L}^n$$

(6)

are convergent series. Using (3)–(6), we obtain

$$d(x_{m+1},x_{m+p})\le \frac{L^m{L}_{1}d(x_1,x_0)}{1-L}+\frac{(2+(1-L)L_1)L_2(1+m)L^m}{1-L}.$$

Thus, ${\left(x_n\right)}_{n\in \mathbb{N}}$ is a Cauchy sequence, and so $x_n\to x$ for some $x\in X$.

Next, we prove that $x\in F\left(x\right)$. For all $n\in \mathbb{N}$,

$$\begin{array}{cc}\hfill 0\le d(x,F(x\left)\right)& \le s[d(x,x_n)+d(x_n,F\left(x\right))]\hfill \\ \hfill & \le s[d(x,x_n)+H_d(F\left(x_{n-1}\right),F\left(x\right))]\hfill \\ \hfill & \le s[d(x,x_n)+Ld(x_{n-1},x)].\hfill \end{array}$$

Letting $n\to \infty$, we see that

$$d(x,F(x\left)\right)=0,$$

which implies $x\in F\left(x\right)$, and so x is a fixed point of F. This proves the theorem. □

Remark 2.

In [13], Czerwik obtained the result in Theorem 2 for b-metric spaces, but with the more restrictive condition that $sL\in (0,1)$. Kirk and Shahzad [8] (Theorem 12.5) relaxed the result for strong b-metric spaces with $L\in (0,1)$. Theorem 2 is an extension of the results of Czerwik and Kirk and Shahzad.

4. Local Version of the Covitz–Nadler Theorem

For the next result, we give a version of the fixed-point theorems proved by Beer and Dontchev [14] (see Theorem 4) and Dontchev and Hager [15] in a strong b-metric space. Hence, we obtain a partial answer to the question raised by Kirk and Shahzad [8] (p. 128).

Theorem 3.

Let $(X,d,s)$ be a complete strong b-metric space and $F:X\to {\mathcal{P}}_{cp}\left(X\right)$. Assume there exist $x_0\in X$, $r>0$, and $sL\in (0,1)$ such that

$\left(i\right)$

$d(x_0,F\left(x_0\right))<r(1-sL)$;

$\left(ii\right)$

$H_d^{*}(F\left(x\right)\cap \overline{B}(x_0,r),F\left(y\right))\le Ld(x,y)$ for all x, $y\in \overline{B}(x_0,r)$.

Then, F has a fixed point in $\overline{B}(x_0,r).$

Proof. 

Since $F\left(x_0\right)\in {\mathcal{P}}_{cp}\left(X\right)$, there exists $x_1\in F\left(x_0\right)$ with $x_1\in B(x_0,r)$ such that

$$d(x_1,x_0)<r(1-sL)$$

(7)

and

$$H_d^{*}(F\left(x_0\right)\cap \overline{B}(x_0,r),F\left(x_1\right))\le Ld(x_1,x_0).$$

Since $x_1\in F\left(x_0\right)\cap \overline{B}(x_0,r)$,

$$d(x_1,F\left(x_1\right))\le H_d^{*}(F\left(x_0\right)\cap \overline{B}(x_0,r),F\left(x_1\right))\le Ld(x_1,x_0)<r(1-sL)L.$$

Then, there exists $x_2\in F\left(x_1\right)$ with

$$d(x_1,x_2)<r(1-sL)L,$$

so we have

$$d(x_0,x_2)\le d(x_0,x_1)+sd(x_1,x_2)<r(1-sL)+sr(1-sL)L,$$

that is,

$$d(x_1,x_2)<r(1-sL)L,d(x_0,x_2)<r(1-{\left(sL\right)}^2),\mathrm{and}{x}_2\in \overline{B}(x_0,r).$$

(8)

Hence,

$$d(x_2,F\left(x_2\right))\le H_d^{*}(F\left(x_1\right)\cap \overline{B}(x_0,r),F\left(x_2\right))\le Ld(x_1,x_2)<r{L}^2(1-sL).$$

Then, there exists $x_3\in F\left(x_2\right)$ such that

$$d(x_2,x_3)<r{L}^2(1-sL),$$

and so

$$\begin{array}{cc}\hfill d(x_0,x_3)\le d(x_0,x_2)+sd(x_2,x_3)& \le r(1-{\left(sL\right)}^2)+sr{L}^2(1-sL)\hfill \\ \hfill & \le r(1-{\left(sL\right)}^2)+sr{L}^2(1-{\left(sL\right)}^2)\hfill \end{array}$$

since $sL<1$. We then have

$$d(x_2,x_3)<r{L}^2(1-sL),d(x_0,x_3)<r(1-{\left(sL\right)}^4),\mathrm{and}{x}_3\in \overline{B}(x_0,r).$$

(9)

From (7)–(9), we can proceed by induction, so that there exist ${\left(x_n\right)}_{n\in \mathbb{N}}\subset \overline{B}(x_0,r)$ with $x_n\in F\left(x_{n-1}\right)$, $n\in \mathbb{N}$, such that

$$d(x_n,x_{n+1})<r{L}^n(1-sL),n\in {\mathbb{N}}_0.$$

By the s-relaxed triangular inequality, for $n\ge m$, we have

$$d(x_m,x_n)\le s\sum _{i=m}^{n-1}d(x_i,x_{i+1})\le rs(1-sL)\sum _{i=m}^{n-1}{L}^i\le rs(1-sL)L^m\sum _{i=0}^{\infty }{L}^i.$$

Therefore,

$$d(x_m,x_n)\le \frac{rs(1-sL)L^m}{1-L}\to 0\mathrm{as}m\to \infty ,$$

which implies that ${\left(x_n\right)}_{n\in \mathbb{N}}$ is a Cauchy sequence in X. Since X is complete, there exists $x\in X$ such that ${\displaystyle \underset{n\to \infty }{lim}{x}_n=x\in \overline{B}(x_0,r)}$. By condition $\left(ii\right)$,

$$d(x_n,F\left(x\right))\le H_d^{*}(F\left(x_{n-1}\right)\cap \overline{B}(x_0,r),F\left(x\right))\le Ld(x_{n-1},x).$$

The s-relaxed triangle inequality implies that

$$d(x,F\left(x\right))\le sd(x,x_n)+d(x_n,F\left(x\right))\le sd(x,x_n)+Ld(x_{n-1},x)\to 0$$

as $n\to \infty$. Therefore, $d(x,F(x\left)\right)=0$, and hence, x is a fixed point of F. This proves the theorem. □

A second result in the same direction is contained in the following theorem.

Theorem 4.

Let $(X,d,s)$ be a complete strong b-metric space and $F:X\to {\mathcal{P}}_{cl}\left(X\right)$. Assume there exist $x_0\in X$, $r>0$, and $L\in (0,1)$ such that

$\left(i\right)$

$d(x_0,F\left(x_0\right))<\frac{r}{s}(1-L)$;

$\left(ii\right)$

$H_d^{*}(F\left(x\right)\cap \overline{B}(x_0,r),F\left(y\right))\le Ld(x,y)$ for all $x,y\in \overline{B}(x_0,r)$.

Then, F has a fixed point in $\overline{B}(x_0,r).$

Proof. 

Since $F\left(x_0\right)\in {\mathcal{P}}_{cp}\left(X\right)$, there exists $x_1\in F\left(x_0\right)$ with $x_1\in \overline{B}(x_0,r)$ such that

$$d(x_1,x_0)<\frac{r}{s}(1-L)$$

and

$$H_d^{*}(F\left(x_0\right)\cap \overline{B}(x_0,r),F\left(x_1\right))\le Ld(x_1,x_0).$$

Since $x_1\in F\left(x_0\right)\cap \overline{B}(x_0,r)$,

$$d(x_1,F\left(x_1\right))\le H_d^{*}(F\left(x_0\right)\cap \overline{B}(x_0,r),F\left(x_1\right))\le Ld(x_1,x_0)<\frac{r}{s}(1-L)L,$$

and so there exists $x_2\in F\left(x_1\right)$ such that

$$d(x_1,x_2)<\frac{r}{s}(1-L)L.$$

Hence, we have

$$d(x_0,x_2)\le s[d(x_0,x_1)+d(x_1,x_2)]<r(1-L)+r(1-L)L=r(1-L^2),$$

which means

$$d(x_1,x_2)<\frac{r}{s}(1-L)L,d(x_0,x_2)<r(1-L^2),\mathrm{and}{x}_2\in \overline{B}(x_0,r).$$

Thus,

$$d(x_2,F\left(x_2\right))\le H_d^{*}(F\left(x_1\right)\cap \overline{B}(x_0,r),F\left(x_2\right))\le Ld(x_1,x_2)<\frac{r}{s}{L}^2(1-L),$$

and so then there exists $x_3\in F\left(x_2\right)$ such that

$$d(x_2,x_3)<\frac{r}{s}{L}^2(1-L).$$

This implies

$$\begin{array}{cc}\hfill d(x_0,x_3)\le d(x_0,x_2)+sd(x_2,x_3)& \le r(1-L^2)+r{L}^2(1-L)\hfill \\ \hfill & \le r(1-L^2)+r{L}^2(1-L^2)\hfill \end{array}$$

since $L<1$. Thus, we have

$$d(x_2,x_3)<\frac{r}{s}{L}^2(1-L),d(x_0,x_3)<r(1-L^4),\mathrm{and}{x}_3\in \overline{B}(x_0,r).$$

Proceeding by induction, there exists ${\left(x_n\right)}_{n\in \mathbb{N}}\subset \overline{B}(x_0,r)$ with $x_n\in F\left(x_{n-1}\right)$, $n\in \mathbb{N}$, such that

$$d(x_n,x_{n+1})<\frac{r}{s}{L}^n(1-L),n\in {\mathbb{N}}_0.$$

As in the proof of Theorem 3, we again see that ${\left(x_n\right)}_{n\in {\mathbb{N}}_0}$ is a Cauchy sequence. Since X is complete, there exists $x\in \overline{B}(x_0,r)$ such that ${\displaystyle \underset{n\to \infty }{lim}{x}_n=x}$ and $x\in F\left(x\right)$, which proves the theorem. □

The next result is our improvement of Dontchev and Hager’s [15] (Lemma) fixed-point theorem.

Theorem 5.

Let $(X,d,s)$ be a complete strong b-metric space and $F:X\to \mathcal{P}\left(X\right)$. Assume there exist $x_0\in X$, $r>0$, and $L\in (0,1)$ such that

$\left(i\right)$

The set $\mathcal{G}r\left(F\right)\cap \overline{B}(x_0,r)\times \overline{B}(x_0,r)$ is a closed set;

$\left(ii\right)$

$d(x_0,F\left(x_0\right))<\frac{r}{s}(1-L)$;

$\left(iii\right)$

$H_d^{*}(F\left(x\right)\cap \overline{B}(x_0,r),F\left(y\right))\le Ld(x,y)$ for all x, $y\in \overline{B}(x_0,r)$.

Then, F has a fixed point in $\overline{B}(x_0,r).$

Proof. 

Since $d(x_0,F\left(x_0\right))<\frac{r}{s}(1-L),$ there exists $x_1\in F\left(x_0\right)$ with $x_1\in \overline{B}(x_0,r)$ such that

$$d(x_1,x_0)<\frac{r}{s}(1-L)$$

(10)

and

$$H_d^{*}(F\left(x_0\right)\cap \overline{B}(x_0,r),F\left(x_1\right))\le Ld(x_1,x_0).$$

Since $x_1\in F\left(x_0\right)\cap \overline{B}(x_0,r)$,

$$d(x_1,F\left(x_1\right))\le H_d^{*}(F\left(x_0\right)\cap \overline{B}(x_0,r),F\left(x_1\right))\le Ld(x_1,x_0)<\frac{r}{s}(1-L)L,$$

and so there exists $x_2\in F\left(x_1\right)$ such that

$$d(x_1,x_2)<\frac{r}{s}(1-L)L$$

and

$$d(x_0,x_2)\le s[d(x_0,x_1)+d(x_1,x_2)]<r(1-L)+r(1-L)L=r(1-L^2).$$

That is,

$$d(x_1,x_2)<\frac{r}{s}(1-L)L,d(x_0,x_2)<r(1-L^2),\mathrm{and}{x}_2\in \overline{B}(x_0,r).$$

(11)

Hence,

$$d(x_2,F\left(x_2\right))\le H_d^{*}(F\left(x_1\right)\cap \overline{B}(x_0,r),F\left(x_2\right))\le Ld(x_1,x_2)<\frac{r}{s}{L}^2(1-L),$$

so there exists $x_3\in F\left(x_2\right)$ such that

$$d(x_2,x_3)<\frac{r}{s}{L}^2(1-L).$$

It follows that

$$\begin{array}{cc}\hfill d(x_0,x_3)\le d(x_0,x_2)+sd(x_2,x_3)& \le r(1-L^2)+r{L}^2(1-L)\hfill \\ \hfill & \le r(1-L^2)+r{L}^2(1-L^2),\hfill \end{array}$$

that is,

$$d(x_2,x_3)<\frac{r}{s}{L}^2(1-L),d(x_0,x_3)<r(1-L^4),\mathrm{and}{x}_3\in \overline{B}(x_0,r).$$

(12)

By induction, there exists

$${\left(x_n\right)}_{n\in \mathbb{N}}\subset \overline{B}(x_0,r),x_n\in F\left(x_{n-1}\right),n\in \mathbb{N},$$

(13)

with

$$d(x_n,x_{n+1})<\frac{r}{s}{L}^n(1-L),n\in {\mathbb{N}}_0.$$

As in the proof of Theorem 3, ${\left(x_n\right)}_{n\in {\mathbb{N}}_0}$ is a Cauchy sequence, and since X is complete, there exists $x\in \overline{B}(x_0,r)$ such that ${\displaystyle \underset{n\to \infty }{lim}{x}_n=x}$. Hence, $(x_{n-1},x_n)\to (x,x)$ as $n\to \infty$. From (13) and condition $\left(i\right)$, we have

$${\left\{(x_{n-1},x_n)\right\}}_{n\in \mathbb{N}}\subset \mathcal{G}r\left(F\right)\cap \overline{B}(x_0,r)\times \overline{B}(x_0,r),$$

and so

$$(x,x)\in \mathcal{G}r\left(F\right)\cap \overline{B}(x_0,r)\times \overline{B}(x_0,r).$$

Therefore, $x\in F\left(x\right)$ and this completes the proof of the theorem. □