Weak Partial b-Metric Spaces and Nadler’s Theorem

Abstract

The purpose of this paper is to define the notions of weak partial b-metric spaces and weak partial Hausdorff b-metric spaces along with the topology of weak partial b-metric space. Moreover, we present a generalization of Nadler’s theorem by using weak partial Hausdorff b-metric spaces in the context of a weak partial b-metric space. We present a non-trivial example which show the validity of our result and an application to nonlinear Volterra integral inclusion for the applicability purpose.

1. Introduction

The famous Banach contraction principle has been generalized in many directions, whether by generalizing the contractive condition or by extending the domain of the function. Bakhtin [1] and Czerwik [2] introduced b-metric spaces generalizing the ordinary metric space and considering the problem of convergence of measurable functions with respect to measure; Czerwik [2] proved the variant of Banach contraction in b-metric spaces. Later on, many authors proved fixed point results for both single and multivalued mapping in the context of b-metric spaces (see also [2,3,4,5,6,7,8,9,10,11,12,13]).

Matthews [14] established the notion of a partial metric space and proved an analogue of Banach’s principle in such spaces. The concept of partial Hausdorff metric was given by Aydi et al. [6] and they established a fixed point theorem for multivalued mappings in partial metric spaces. Excluding the idea of small self-distance, Heckmann [15] generalized the partial metric space to weak partial metric spaces (see more [16,17,18,19,20,21,22]).

Shukla [23] introduced the concept of the partial b-metric and proved some fixed point results. Beg [7] presented the idea of the almost partial Hausdorff metric and extended Nadler’s theorem (Nadler [19]) to weak partial metric spaces.

The aim of this paper is to introduce the notion of the weak partial b-metric space, the ${\mathcal{H}}^{+}$-type partial Hausdorff b-metric and prove Nadler’s theorem to weak partial b-metric spaces. An example and application to Volterra type integral inclusion to support our result will be given.

2. Preliminaries

Consistent with Beg [7], notion of weak partial metric and related concepts are as follows:

Definition 1.

[7] Let M be a nonempty set. A function $\varrho :M\times M\to {\mathbb{R}}^{+}$ is called weak partial metric if for all $s,t,z\in M$, following assertions hold:

(WP1) 

$\varrho (s,s)=\varrho (s,t)iffs=t;$

(WP2) 

$\varrho (s,s)\le \varrho (s,t);$

(WP3) 

$\varrho (s,t)=\varrho (t,s);$

(WP4) 

$\varrho (s,t)\le \varrho (s,z)+\varrho (z,t).$

The pair $(M,\varrho )$ is called weak partial metric space.

We refer [7] to readers for detail work in weak partial metric space.

Let $C{B}^{\varrho }\left(M\right)$ be the family of nonempty, closed and bounded subsets of a weak partial metric space $(M,\varrho )$. Define

$$\varrho (x,U)=inf\{\varrho (x,u),u\in U\},{\delta }_{\varrho }(U,V)=sup\{\varrho (u,V):u\in U\}$$

and

$${\delta }_{\varrho }(V,U)=sup\{\varrho (v,U):v\in V\},$$

where $U,V\in C{B}^{\varrho }\left(M\right)$ and $s\in M$. Also

$$\varrho (x,U)=0\Rightarrow {\varrho }^s(x,U)=0,$$

where ${\varrho }^s(x,U)=inf\{{\varrho }^s(x,u),u\in U\}$.

Remark 1.

[7] If $\varphi \ne U\subseteq M$, then

$$u\in \overline{U}\mathit{if}\mathit{and}\mathit{only}\mathit{if}\varrho (u,U)=\varrho (u,u).$$

Definition 2.

[7] Let $(M,\varrho )$ be a weak partial metric space. For $U,V\in C{B}^{\varrho }\left(M\right)$, define

$${\mathcal{H}}_{\varrho }^{+}(U,V)=\frac{1}{2}\{{\delta }_{\varrho }(U,V)+{\delta }_{\varrho }(V,U)\}.$$

The mapping ${\mathcal{H}}_{\varrho }^{+}:C{B}^{\varrho }\left(M\right)\times C{B}^{\varrho }\left(M\right)\to [0,\infty )$, is called ${\mathcal{H}}_{\varrho }^{+}$-type Hausdorff metric induced by ϱ.

Proposition 1.

[7] Let $(M,\varrho )$ be a weak partial metric space. For any $U,V,Y\in C{B}^{\varrho }\left(M\right)$, we have:

(wh1) 

${\mathcal{H}}_{\varrho }^{+}(U,U)\le {\mathcal{H}}_{\varrho }^{+}(U,V)$;

(wh2) 

${\mathcal{H}}_{\varrho }^{+}(U,V)={\mathcal{H}}_{\varrho }^{+}(V,U)$;

(wh3) 

${\mathcal{H}}_{\varrho }^{+}(U,V)\le {\mathcal{H}}_{\varrho }^{+}(U,Y)+{\mathcal{H}}_{\varrho }^{+}(Y,V)$.

Definition 3.

[7] Let $(M,\varrho )$ be a weak partial metric space. A multivalued mapping $\mathcal{T}:M\to C{B}^{\varrho }\left(M\right)$ is called ${\mathcal{H}}_{\varrho }^{+}$-contraction if

$\left(1^o\right)$

∃ $k\in (0,1)$ such that

$${\mathcal{H}}_{\varrho }^{+}(\mathcal{T}s\backslash \left\{s\right\},\mathcal{T}t\backslash \left\{t\right\})\le k\varrho (s,t)\mathit{for}\mathit{every}s,t\in M,$$

$\left(2^o\right)$

for every $s\in M$, t in $\mathcal{T}s$ and $ϵ>0$, there exists z in $\mathcal{T}t$ such that

$$\varrho (t,z)\le {\mathcal{H}}_{\varrho }^{+}(\mathcal{T}s,\mathcal{T}t)+ϵ.$$

Beg [7] gave the following variant of Nadler’s fixed point theorem.

Theorem 1.

[7] Every ${\mathcal{H}}_{\varrho }^{+}$-type multivalued contraction on a complete weak partial metric space $(M,\varrho )$ has a fixed point.

3. Weak Partial $\mathit{b}$-Metric Space

We now define weak partial b-metric space and related concepts:

Definition 4.

Let $M\ne \varphi$ and $s\ge 1$, a function ${\varrho }_b:M\times M\to {\mathbb{R}}^{+}$ is called weak partial b-metric on M if for all $s,t,z\in M$, following conditions are satisfied:

(WPB1) 

${\varrho }_b(s,s)={\varrho }_b(s,t)\iff s=t$;

(WPB2) 

${\varrho }_b(s,s)\le {\varrho }_b(s,t)$;

(WPB3) 

${\varrho }_b(s,t)={\varrho }_b(t,s)$;

(WPB4) 

${\varrho }_b(s,t)\le s[{\varrho }_b(s,z)+{\varrho }_b(z,t)].$

The pair $(M,{\varrho }_b)$ is a weak partial b-metric space.

Example 1.

(i) 

$({\mathbb{R}}^{+},{\varrho }_b)$, where ${\varrho }_b:{\mathbb{R}}^{+}\times {\mathbb{R}}^{+}\to {\mathbb{R}}^{+}$ is defined as

$${\varrho }_b(s,t)={|s-t|}^2+1\mathit{for}\mathit{all}s,t\in {\mathbb{R}}^{+}.$$

(ii) 

$({\mathbb{R}}^{+},{\varrho }_b)$, where ${\varrho }_b:{\mathbb{R}}^{+}\times {\mathbb{R}}^{+}\to {\mathbb{R}}^{+}$ is defined as

$${\varrho }_b(s,t)=\frac{1}{2}{|s-t|}^2+max\{s,t\}\mathit{for}\mathit{all}s,t\in {\mathbb{R}}^{+}.$$

Definition 5.

A sequence $\left\{s_n\right\}$ in $(M,{\varrho }_b)$ is said to converges a point $s\in X$, if and only if

$${\varrho }_b(s,s)=\underset{n\to \infty }{lim}{\varrho }_b(s,s_n).$$

Remark 2.

If ${\varrho }_b$ is a weak partial b-metric on M, the function ${{\varrho }_b}^s:M\times M\to {\mathbb{R}}^{+}$ given by ${{\varrho }_b}^s(s,t)={\varrho }_b(s,t)-\frac{1}{2}[{\varrho }_b(s,s)+{\varrho }_b(t,t)]$, defines a b-metric on M. Further, a sequence $\left\{s_n\right\}$ in $(M,{{\varrho }_b}^s)$ converges to a point $s\in M$, iff

$$\underset{n,m\to \infty }{lim}{\varrho }_b(s_n,s_m)=\underset{n\to \infty }{lim}{\varrho }_b(s_n,s)={\varrho }_b(s,s).$$

(1)

Definition 6.

Let $(M,{\varrho }_b)$ be a weak partial b-metric space. Then

(1) 

A Cauchy sequence in metric space $(M,{\varrho }_b^s)$ is Cauchy in M.

(2) 

If the metric space $(M,{\varrho }_b^s)$ is complete, so is weak partial b-metric space $(M,{\varrho }_b)$.

Let $(M,{\varrho }_b)$ be a weak partial b-metric space and $C{B}^{{\varrho }_b}\left(M\right)$ be class of all nonempty, closed and bounded subsets of $(M,{\varrho }_b)$. For $U,V\in C{B}^{{\varrho }_b}\left(M\right)$ and $s\in M$, define

$${\varrho }_b(s,U)=inf\{{\varrho }_b(s,u),u\in U\},{\delta }_{{\varrho }_b}(U,V)=sup\{{\varrho }_b(u,V):u\in U\}$$

and

$${\delta }_{{\varrho }_b}(V,U)=sup\{{\varrho }_b(v,U):v\in V\}.$$

Now ${\varrho }_b(s,U)=0\Rightarrow {{\varrho }_b}^s(s,U)=0$, where ${{\varrho }_b}^s(s,U)=inf\{{{\varrho }_b}^s(s,u),u\in U\}$.

Remark 3.

Let $(M,{\varrho }_b)$ be a weak partial b-metric space and U a nonempty subset of M, then

$$u\in \overline{U}\iff {\varrho }_b(u,U)={\varrho }_b(u,u).$$

Proposition 2.

Let $(M,{\varrho }_b)$ be a weak partial b-metric space. For any $U,V,Y\in C{B}^{{\varrho }_b}\left(M\right)$, we have the following:

(i) 

${\delta }_{{\varrho }_b}(U,U)=sup\{{\varrho }_b(u,u):u\in U\}$;

(ii) 

${\delta }_{{\varrho }_b}(U,U)\le {\delta }_{{\varrho }_b}(U,V)$;

(iii) 

${\delta }_{{\varrho }_b}(U,V)=0\Rightarrow U\subseteq V$;

(iv) 

${\delta }_{{\varrho }_b}(U,V)\le s[{\delta }_{{\varrho }_b}(U,Y)+{\delta }_{{\varrho }_b}(Y,V)]$.

Proof. 

(i) 

If $U\in C{B}^{{\varrho }_b}\left(M\right)$, then for all $u\in U$, we have ${\varrho }_b(u,U)={\varrho }_b(u,u)$ as $\overline{U}=U$. This implies that ${\delta }_{{\varrho }_b}(U,U)=sup\{{\varrho }_b(u,U):u\in U\}=sup\{{\varrho }_b(u,u):u\in U\}$.

(ii) 

Let $u\in U$. Since ${\varrho }_b(u,u)\le {\varrho }_b(u,w)$ for all $w\in U$, therefore we have $\varrho {}_b(u,u)\le inf\{\varrho {}_b(u,v)$ $:v\in V\}={\varrho }_b(u,V)\le sup\{{\varrho }_b(u,V):u\in U\}={\delta }_{{\varrho }_b}(U,V)$.

(iii) 

If ${\delta }_{{\varrho }_b}(U,V)=0$, then ${\varrho }_b(u,V)=0$ for all $u\in U$. From (i) and (ii), it follows that ${\varrho }_b(u,u)\le$ ${\delta }_{{\varrho }_b}(U,V)=0$ for all $u\in U$. Hence ${\varrho }_b(u,V)={\varrho }_b(u,u)$ for all $u\in U$. By Remark 3, we have $u\in$ $\overline{V}=V$, so $U\subseteq V$.

(iv) 

Let $u\in U,v\in V$ and $t\in Y$. By (WPB4), we have ${\varrho }_b(u,v)\le s[{\varrho }_b(u,t)+{\varrho }_b(t,v)].$ Since $v\in V$ is arbitrary, therefore ${\varrho }_b(u,V)\le s[{\varrho }_b(u,t)+{\varrho }_b(t,V)]$ and ${\varrho }_b(u,V)\le s[{\varrho }_b(u,t)+{sup}_{t\in Y}{\varrho }_b(t,V)],$ so that ${\varrho }_b(u,V)\le s[{\varrho }_b(u,t)+{\delta }_{{\varrho }_b}(Y,V)].$ Since $t\in Y$ is arbitrary, therefore ${\varrho }_b(u,V)\le s$ $[{\varrho }_b(u,Y)+{\delta }_{{\varrho }_b}(Y,V)].$ Since $u\in U$ is arbitrary, we have ${\delta }_{{\varrho }_b}(U,V)\le s[{\delta }_{{\varrho }_b}(U,Y)+{\delta }_{{\varrho }_b}(Y,V)].$

 ☐

Definition 7.

Let $(M,{\varrho }_b)$ be a weak partial b-metric space. For $U,V\in C{B}^{{\varrho }_b}\left(M\right)$, the mapping ${\mathcal{H}}_{{\varrho }_b}^{+}:C{B}^{{\varrho }_b}\left(M\right)\times C{B}^{{\varrho }_b}\left(M\right)\to [0,\infty )$ define by

$${\mathcal{H}}_{{\varrho }_b}^{+}(U,V)=\frac{1}{2}\{{\delta }_{{\varrho }_b}(U,V)+{\delta }_{{\varrho }_b}(V,U)\}$$

is called ${\mathcal{H}}_{{\varrho }_b}^{+}$-type Hausdorff metric induced by ${\varrho }_b$.

Proposition 3.

Let $(M,{\varrho }_b)$ be a weak partial b-metric space. For any $U,V,Y\in C{B}^{{\varrho }_b}\left(M\right)$, we have:

(whb1) 

${\mathcal{H}}_{{\varrho }_b}^{+}(U,U)\le {\mathcal{H}}_{{\varrho }_b}^{+}(U,V)$;

(whb2) 

${\mathcal{H}}_{{\varrho }_b}^{+}(U,V)={\mathcal{H}}_{{\varrho }_b}^{+}(V,U)$;

(whb3) 

${\mathcal{H}}_{{\varrho }_b}^{+}(U,V)\le s[{\mathcal{H}}_{{\varrho }_b}^{+}(U,Y)+{\mathcal{H}}_{{\varrho }_b}^{+}(Y,V)]$.

Proof. 

From (ii) of Proposition 2, we have

$${\mathcal{H}}_{{\varrho }_b}^{+}(U,U)={\delta }_{{\varrho }_b}(U,U)\le {\delta }_{{\varrho }_b}(U,V)\le {\mathcal{H}}_{{\varrho }_b}^{+}(U,V).$$

Also (whb2) obviously holds by definition. Now for (whb3), from (iv) of Proposition 2, we have

$$\begin{array}{ccc}\hfill {\mathcal{H}}_{{\varrho }_b}^{+}(U,V)& =& \frac{1}{2}\{{\delta }_{{\varrho }_b}(U,V)+{\delta }_{{\varrho }_b}(V,U)\}\hfill \\ & \le & \frac{1}{2}\{s[{\delta }_{{\varrho }_b}(U,Y)+{\delta }_{{\varrho }_b}(Y,V)]+s[{\delta }_{{\varrho }_b}(V,Y)+{\delta }_{{\varrho }_b}(Y,U)]\}\hfill \\ & =& s[\frac{1}{2}\{{\delta }_{{\varrho }_b}(U,Y)+{\delta }_{{\varrho }_b}(Y,U)\}+\frac{1}{2}\{{\delta }_{{\varrho }_b}(Y,V)+{\delta }_{{\varrho }_b}(V,Y)\}]\hfill \\ & =& s[{\mathcal{H}}_{{\varrho }_b}^{+}(U,Y)+{\mathcal{H}}_{{\varrho }_b}^{+}(Y,V)].\hfill \end{array}$$

 ☐

Following lemma is essential:

Lemma 1.

Let $\left(M,{\varrho }_b\right)$ be weak partial b-metric space with $s\ge 1$ and $\mathcal{T}:M\to C{B}^{{\varrho }_b}\left(M\right)$ be a multivalued mapping. If $\left\{u_n\right\}$ is a sequence in M such that $u_n\in \mathcal{T}{u}_{n-1}$ and

$${\varrho }_b\left(u_n,u_{n+1}\right)\le \lambda {\varrho }_b\left(u_{n-1},u_n\right)$$

for each where $\lambda \in (0,1)$, then $\left\{u_n\right\}$ is Cauchy.

Proof. 

Let $u_0\in M$ and $u_n\in \mathcal{T}{u}_{n-1}$ for all $n\in \mathbb{N}.$ We divide the proof into two cases:

Case I.

Let $\lambda \in [0,\frac{1}{s})$ $\left(s>1\right).$ By the hypotheses, we have

$${\varrho }_b\left(u_n,u_{n+1}\right)\le \lambda {\varrho }_b\left(u_{n-1},u_n\right)\le {\lambda }^2{\varrho }_b\left(u_{n-2},u_{n-1}\right)\le ···\le {\lambda }^n{\varrho }_b\left(u_0,u_1\right).$$

Thus, for $n>m,$ we have

$$\begin{array}{ccc}\hfill {\varrho }_b\left(u_m,u_n\right)& \le & s\left[{\varrho }_b\left(u_m,u_{m+1}\right)+{\varrho }_b\left(u_{m+1},u_n\right)\right]\hfill \\ & \le & s{\varrho }_b\left(u_m,u_{m+1}\right)+s^2\left[{\varrho }_b\left(u_{m+1},u_{m+2}\right)+{\varrho }_b\left(u_{m+2},u_n\right)\right]\hfill \\ & \le & s{\varrho }_b\left(u_m,u_{m+1}\right)+s^2{\varrho }_b\left(u_{m+1},{\varrho }_{um+2}\right)+s^3\left[{\varrho }_b\left(u_{m+2},u_{m+3}\right)+{\varrho }_b\left(u_{m+3},u_n\right)\right]\hfill \\ & \le & s{\varrho }_b\left(u_m,u_{m+1}\right)+s^2{\varrho }_b\left(u_{m+1},u_{m+2}\right)+s^3{\varrho }_b\left(u_{m+2},u_{m+3}\right)\hfill \\ & & +···+s^{n-m-1}{\varrho }_b\left(u_{n-2},u_{n-1}\right)+s^{n-m-1}{\varrho }_b\left(u_{n-1},u_n\right)\hfill \\ & \le & s{\lambda }^m{\varrho }_b\left(u_0,u_1\right)+s^2{\lambda }^{m+1}{\varrho }_b\left(u_0,u_1\right)+s^3{\lambda }^{m+2}{\varrho }_b\left(u_0,u_1\right)\hfill \\ & & +···+s^{n-m-1}{\lambda }^{n-2}{\varrho }_b\left(u_0,u_1\right)+s^{n-m-1}{\lambda }^{n-1}{\varrho }_b\left(u_0,u_1\right)\hfill \\ & \le & s{\lambda }^m\left(1+\left(s\lambda \right)+{\left(s\lambda \right)}^2+···+{\left(s\lambda \right)}^{n-m-2}+\frac{{\left(s\lambda \right)}^{n-m-1}}{s}\right){\varrho }_b\left(u_0,u_1\right)\hfill \\ & \le & s{\lambda }^m\left(\frac{1}{1-s\lambda }+\frac{{\left(s\lambda \right)}^{n-m-1}}{s}\right){\varrho }_b\left(u_0,u_1\right)\hfill \\ & =& \left(\frac{s{\lambda }^m}{1-s\lambda }+{\left(s\lambda \right)}^{n-1}\right){\varrho }_b\left(u_0,u_1\right)\to 0\left(n,m\to \infty \right).\hfill \end{array}$$

Using (1) and the definition of ${\varrho }_b^s$, we get that ${\varrho }_b^s(u_m,u_n)\le {\varrho }_b(u_m,u_n)$ tends to 0 as $m,n$ tends $to+\infty$ which implies that $\left\{u_n\right\}$ is Cauchy in b-metric space $(M,{\varrho }_b^s)$. Since $(M,{\varrho }_b)$ is complete, therefore $(M,{\varrho }_b^s)$ is a complete b-metric space. Consequently, the sequence $\left\{u_n\right\}$ converges to a point (say) $u^{*}\in M$ w.r.t b-metric ${\varrho }_b^s$, that is, $\underset{n\to +\infty }{lim}{\varrho }_b^s(u_n,u^{*})=0$. Again, from (1) we get

$${\varrho }_b(u^{*},u^{*})=\underset{n\to +\infty }{lim}{\varrho }_b(u_n,u^{*})=\underset{n\to +\infty }{lim}{\varrho }_b(u_n,u_n)=0.$$

Thus $\left\{u_n\right\}$ is a Cauchy sequence in $(M,{\varrho }_b)$.

Case II.

Let $\lambda \in [\frac{1}{s},1)$$\left(s>1\right).$ In this case, we have ${\lambda }^n\to 0$ as $n\to \infty ,$ then there is $k\in \mathbb{N}$ such that ${\lambda }^k<\frac{1}{s}.$ Thus, by Case-I, we have that

$$\left\{u_k,u_{k+1},u_{k+2},...,u_{k+n},...\right\},$$

is a Cauchy sequence. Since

$${\left\{u_n\right\}}_{n=0}^{\infty }=\left\{u_0,u_{1,}...,u_{k-1}\right\}\cup \left\{u_k,u_{k+1},u_{k+2},...,u_{k+n},...\right\},$$

we obtain that $u_n\in {\mathcal{T}}^n{u}_0$, $n=1,2,...$ is a Cauchy sequence in $M.$  ☐

Definition 8.

Let $(M,{\varrho }_b)$ be a complete weak partial b-metric space. A multivalued mapping $\mathcal{T}:M\to C{B}^{{\varrho }_b}\left(M\right)$ is called ${\mathcal{H}}_{{\varrho }_b}^{+}$-contraction if

$\left(1^{\prime }\right)$

for every $s,t\in M,$ ∃ $k\in (0,1)$ such that

$${\mathcal{H}}_{{\varrho }_b}^{+}(Ts\backslash \left\{s\right\},Tt\backslash \left\{t\right\})\le k{\varrho }_b(s,t);$$

$\left(2^{\prime }\right)$

for every $s\in X$, t in $Ts$ and $ϵ>0$, ∃ z in $\mathcal{T}t$ such that

$${\varrho }_b(t,z)\le {\mathcal{H}}_{{\varrho }_b}^{+}(\mathcal{T}s,\mathcal{T}t)+ϵ.$$

4. Fixed Point Result

Our main result is the following:

Theorem 2.

Every ${\mathcal{H}}_{{\varrho }_b}^{+}$-type multivalued contraction on a complete weak partial b-metric space $(M,{\varrho }_b)$ has a fixed point.

Proof. 

Let $u_0\in M$ be arbitrary. If $u_0\in \mathcal{T}{u}_0$ then $u_0$ is the fixed point. Therefore, we assume that $u_0\notin \mathcal{T}{u}_0$. Let $u_1\in \mathcal{T}{u}_0$ and $u_0\ne u_1$ such that $u_1\notin \mathcal{T}{u}_1$. From $\left(2^{\prime }\right)$, we have $u_2\in \mathcal{T}{u}_1$ such that $u_2\ne u_1$ and

$${\varrho }_b(u_1,u_2)\le {\mathcal{H}}_{{{\varrho }_b}_b}^{+}(\mathcal{T}{u}_0,\mathcal{T}{u}_1)+ϵ.$$

Continuing this process we get $u_{n+1}\in \mathcal{T}{u}_n$ such that $u_{n+1}\ne u_n$ and

$${\varrho }_b(u_n,u_{n+1})\le {\mathcal{H}}_{{\varrho }_b}^{+}(\mathcal{T}{u}_{n-1},\mathcal{T}{u}_n)+ϵ.$$

(2)

Choosing $ϵ=\left(\frac{1}{\sqrt{k}}-1\right){\mathcal{H}}_{{\varrho }_b}^{+}(\mathcal{T}{u}_{n-1},\mathcal{T}{u}_n)$ in (2), we have

$${\varrho }_b(u_n,u_{n+1})\le {\mathcal{H}}_{{\varrho }_b}^{+}(\mathcal{T}{u}_{n-1},\mathcal{T}{u}_n)+\left(\frac{1}{\sqrt{k}}-1\right){\mathcal{H}}_{{\varrho }_b}^{+}(\mathcal{T}{u}_{n-1},\mathcal{T}{u}_n)=\frac{1}{\sqrt{k}}{\mathcal{H}}_{{\varrho }_b}^{+}(\mathcal{T}{u}_{n-1},\mathcal{T}{u}_n).$$

Thus

$$\sqrt{k}{\varrho }_b(u_n,u_{n+1})\le {\mathcal{H}}_{{\varrho }_b}^{+}(\mathcal{T}{u}_{n-1},\mathcal{T}{u}_n)={\mathcal{H}}_{{\varrho }_b}^{+}\left(\mathcal{T}{u}_{n-1}\backslash \left\{u_{n-1}\right\},\mathcal{T}{u}_n\backslash \left\{u_n\right\}\right).$$

From $\left(1^{\prime }\right)$, we get

$$\sqrt{k}{\varrho }_b(u_n,u_{n+1})\le k{\varrho }_b(u_{n-1},u_n)={\left(\sqrt{k}\right)}^2{\varrho }_b(u_{n-1},u_n).$$

Thus for all $n\in \mathbb{N}$,

$$\begin{array}{c}\hfill {\varrho }_b(u_n,u_{n+1})\le \sqrt{k}{\varrho }_b(u_{n-1},u_n).\end{array}$$

(3)

Taking $\sqrt{k}=\lambda$, we obtained by Lemma 1 that $\left\{u_n\right\}$ is a Cauchy sequence. Since $(M,{\varrho }_b)$ is complete. Therefore, there exists $u^{*}\in M$ such that $\underset{n\to +\infty }{lim}{u}_n=u^{*}$. To show that $u^{*}\in \mathcal{T}$. On contrary suppose that $u^{*}\notin \mathcal{T}{u}^{*}$. Since

$$\begin{array}{ccc}\hfill \frac{1}{2}[{\delta }_{{\varrho }_b}(\mathcal{T}{u}_n,\mathcal{T}{u}^{*})+{\delta }_{{\varrho }_b}(\mathcal{T}{u}^{*},\mathcal{T}{u}_n)]& =& {\mathcal{H}}_{{\varrho }_b}^{+}(\mathcal{T}{u}_n,\mathcal{T}{u}^{*})\hfill \\ & =& {\mathcal{H}}_{{\varrho }_b}^{+}(\mathcal{T}{u}_n\backslash \left\{u_n\right\},\mathcal{T}{u}^{*}\backslash \left\{u^{*}\right\})\hfill \\ & \le & k{\varrho }_b(u_n,u^{*}),\hfill \end{array}$$

hence

$$\underset{n\to +\infty }{lim}inf[{\delta }_{{\varrho }_b}(\mathcal{T}{u}_n,\mathcal{T}{u}^{*})+{\delta }_{{\varrho }_b}(\mathcal{T}{u}^{*},\mathcal{T}{u}_n)]=0.$$

Since

$$\underset{n\to +\infty }{lim}inf{\delta }_{{\varrho }_b}(\mathcal{T}{u}_n,\mathcal{T}{u}^{*})+\underset{n\to +\infty }{lim}inf{\delta }_{{\varrho }_b}(\mathcal{T}{u}^{*},\mathcal{T}{u}_n)\le \underset{n\to +\infty }{lim}inf[{\delta }_{{\varrho }_b}(\mathcal{T}{u}_n,\mathcal{T}{u}^{*})+{\delta }_{{\varrho }_b}(\mathcal{T}{u}^{*},\mathcal{T}{u}_n)],$$

we have

$$\underset{n\to +\infty }{lim}inf{\delta }_{{\varrho }_b}(\mathcal{T}{u}_n,\mathcal{T}{u}^{*})+\underset{n\to +\infty }{lim}inf{\delta }_{{\varrho }_b}(\mathcal{T}{u}^{*},\mathcal{T}{u}_n)=0.$$

This implies that

$$\underset{n\to +\infty }{lim}inf{\delta }_{{\varrho }_b}(\mathcal{T}{u}_n,\mathcal{T}{u}^{*})=0.$$

Since

$${\varrho }_b(u^{*},\mathcal{T}{u}^{*})\le {\delta }_{{\varrho }_b}(\mathcal{T}{u}_n,\mathcal{T}{u}^{*})+{\varrho }_b(u_{n+1},u^{*}),$$

therefore

$$\begin{array}{ccc}\hfill {\varrho }_b(u^{*},\mathcal{T}{u}^{*})& \le & \underset{n\to +\infty }{lim}inf[{\delta }_{{\varrho }_b}(\mathcal{T}{u}_n,\mathcal{T}{u}^{*})+{\varrho }_b(u_{n+1},u^{*})]\hfill \\ & =& \underset{n\to +\infty }{lim}inf{\delta }_{{\varrho }_b}(\mathcal{T}{u}_n,\mathcal{T}{u}^{*})+\underset{n\to +\infty }{lim}{\varrho }_b(u_{n+1},u^{*}).\hfill \end{array}$$

This implies ${\varrho }_b(u^{*},\mathcal{T}{u}^{*})=0$, therefore from (1), we obtain

$${\varrho }_b(u^{*},u^{*})={\varrho }_b(u^{*},\mathcal{T}{u}^{*}),$$

which implies $u^{*}\in \overline{\mathcal{T}{u}^{*}}=\mathcal{T}{u}^{*}$, as $\mathcal{T}{u}^{*}$ is closed.  ☐

Example 2.

Consider a set $M=\{0,\frac{1}{2},1\}$ and ${\varrho }_b:M\times M\to {\mathbb{R}}^{+}$ a weak partial b-metric given by

$${\varrho }_b(u,v)=\frac{1}{2}{|u-v|}^2+\frac{1}{2}max\{u,v\}\mathrm{for}\mathrm{all}u,v\in M.$$

Since ${\varrho }_b\left(\frac{1}{2},\frac{1}{2}\right)=\frac{1}{4}\ne 0$ and ${\varrho }_b(1,1)=\frac{1}{2}\ne 0$. Also

$$\begin{array}{ccc}\hfill u\in \overline{\left\{0\right\}}& \iff & {\varrho }_b(u,\left\{0\right\})={\varrho }_b(u,u)\hfill \\ & \iff & \frac{1}{2}{u}^2+\frac{1}{2}u=\frac{1}{2}u\iff u=0\hfill \\ & \iff & u\in \left\{0\right\}.\hfill \end{array}$$

Also

$$\begin{array}{ccc}\hfill u\in \overline{\{0,1\}}& \iff & {\varrho }_b(u,\{0,1\})={\varrho }_b(u,u)\hfill \\ & \iff & min\left\{\frac{1}{2}{u}^2+\frac{1}{2}u,\frac{1}{2}{|u-1|}^2+\frac{1}{2}max\{u,1\}\right\}=\frac{1}{2}u\hfill \\ & \iff & u\in \{0,1\}\hfill \end{array}$$

and

$$\begin{array}{ccc}\hfill u\in \overline{\left\{0,\frac{1}{2}\right\}}& \iff & {\varrho }_b\left(u,\left\{0,\frac{1}{2}\right\}\right)={\varrho }_b(u,u)\hfill \\ & \iff & min\left\{\frac{1}{2}{u}^2+\frac{1}{2}u,\frac{1}{2}{\left|u-\frac{1}{2}\right|}^2+\frac{1}{2}max\left\{u,\frac{1}{2}\right\}\right\}=\frac{1}{2}u\hfill \\ & \iff & u\in \left\{0,\frac{1}{2}\right\}.\hfill \end{array}$$

Hence, $\left\{0\right\}$, $\{0,1\}$ and $\left\{0,\frac{1}{2}\right\}$ are closed w.r.t weak partial b-metric ${\varrho }_b$.

Define $\mathcal{T}:X\to C{B}^{{\varrho }_b}\left(M\right)$ by

$$\mathcal{T}\left(0\right)=\left\{0\right\},\mathcal{T}\left(\frac{1}{2}\right)=\left\{0,\frac{1}{2}\right\}\mathrm{and}\mathcal{T}\left(1\right)=\{0,1\}.$$

To show that for all $u,v\in M$, the contractive condition $\left(1^{\prime }\right)$ holds for all $k\in (0,1)$, we consider the following cases:

• For $u=v=0$, we have

  $${\mathcal{H}}_{{\varrho }_b}^{+}\left(\mathcal{T}\left(0\right)\backslash \left\{0\right\},\mathcal{T}\left(0\right)\backslash \left\{0\right\}\right)={\mathcal{H}}_{{\varrho }_b}^{+}\left(\left\{0\right\}\backslash \left\{0\right\},\left\{0\right\}\backslash \left\{0\right\}\right)={\mathcal{H}}_{{\varrho }_b}^{+}\left(\varnothing ,\varnothing \right)=0,$$

  so $\left(1^{\prime }\right)$ satisfied.
• For $u=0,v=\frac{1}{2}$, we have

  $${\mathcal{H}}_{{\varrho }_b}^{+}\left(\mathcal{T}\left(0\right)\backslash \left\{0\right\},\mathcal{T}\left(\frac{1}{2}\right)\backslash \left\{\frac{1}{2}\right\}\right)={\mathcal{H}}_{{\varrho }_b}^{+}\left(\left\{0\right\}\backslash \left\{0\right\},\left\{0,\frac{1}{2}\right\}\backslash \left\{\frac{1}{2}\right\}\right)={\mathcal{H}}_{{\varrho }_b}^{+}\left(\varnothing ,\left\{0\right\}\right)=0,$$

  so $\left(1^{\prime }\right)$ satisfied.
• For $u=v=\frac{1}{2}$, we have

  $$\begin{array}{c}{\mathcal{H}}_{{\varrho }_b}^{+}\left(\mathcal{T}\left(\frac{1}{2}\right)\backslash \left\{\frac{1}{2}\right\},\mathcal{T}\left(\frac{1}{2}\right)\backslash \left\{\frac{1}{2}\right\}\right)={\mathcal{H}}_{{\varrho }_b}^{+}\left(\left\{0,\frac{1}{2}\right\}\backslash \left\{\frac{1}{2}\right\},\left\{0,\frac{1}{2}\right\}\backslash \left\{\frac{1}{2}\right\}\right)\hfill \\ ={\mathcal{H}}_{{\varrho }_b}^{+}\left(\left\{0\right\},\left\{0\right\}\right)={\varrho }_b(0,0)=0,\hfill \end{array}$$

  so $\left(1^{\prime }\right)$ satisfied.
• For $u=0,v=1$, we have

  $${\mathcal{H}}_{{\varrho }_b}^{+}\left(\mathcal{T}\left(0\right)\backslash \left\{1\right\},\mathcal{T}\left(1\right)\backslash \left\{1\right\}\right)={\mathcal{H}}_{{\varrho }_b}^{+}\left(\left\{0\right\}\backslash \left\{0\right\},\{0,1\}\backslash \left\{0\right\}\right)={\mathcal{H}}_{{\varrho }_b}^{+}\left(\varnothing ,\left\{0\right\}\right)=0,$$

  so $\left(1^{\prime }\right)$ satisfied.
• For $u=\frac{1}{2},v=1$, we have

  $$\begin{array}{c}{\mathcal{H}}_{{\varrho }_b}^{+}\left(\mathcal{T}\left(\frac{1}{2}\right)\backslash \left\{\frac{1}{2}\right\},\mathcal{T}\left(1\right)\backslash \left\{1\right\}\right)={\mathcal{H}}_{{\varrho }_b}^{+}\left(\left\{0,\frac{1}{2}\right\}\backslash \left\{0\right\},\{0,1\}\backslash \left\{1\right\}\right)={\mathcal{H}}_{{\varrho }_b}^{+}\left(\left\{0\right\},\left\{0\right\}\right)\hfill \\ ={\varrho }_b(0,0)=0,\hfill \end{array}$$

  so $\left(1^{\prime }\right)$ satisfied.
• For $u=v=1$, we have

  $${\mathcal{H}}_{{\varrho }_b}^{+}\left(\mathcal{T}\left(1\right)\backslash \left\{1\right\},\mathcal{T}\left(1\right)\backslash \left\{1\right\}\right)={\mathcal{H}}_{{\varrho }_b}^{+}\left(\{0,1\}\backslash \left\{1\right\},\{0,1\}\backslash \left\{1\right\}\right)={\mathcal{H}}_{{\varrho }_b}^{+}\left(\left\{0\right\},\left\{0\right\}\right)={\varrho }_b(0,0)=0,$$

  so $\left(1^{\prime }\right)$ satisfied.

Further, we show that for every $u\in M,v\in \mathcal{T}u$ and $ϵ>0$, ∃ $w\in \mathcal{T}v$ such that

$${\varrho }_b(v,w)\le {\mathcal{H}}_{{\varrho }_b}^{+}\left(\mathcal{T}u,\mathcal{T}v\right)+ϵ.$$

So,

(a) 

If $u=0,v\in \mathcal{T}\left(0\right)=\left\{0\right\},ϵ>0$, ∃ $w\in \mathcal{T}v=\left\{0\right\}$

$$0={\varrho }_b(v,w)\le {\mathcal{H}}_{{\varrho }_b}^{+}(\mathcal{T}v,\mathcal{T}u)+ϵ.$$

(b) 

If $u=\frac{1}{2},v\in \mathcal{T}u=\mathcal{T}\left(\frac{1}{2}\right)=\{0,\frac{1}{2}\}$, for $v=0,ϵ>0$, ∃ $w\in \mathcal{T}v=\left\{0\right\}$ such that

$$0={\varrho }_b(v,w)<\frac{3}{16}+ϵ\le {\mathcal{H}}_{{\varrho }_b}^{+}(\mathcal{T}v,\mathcal{T}u)+ϵ$$

and for $v=\frac{1}{2},ϵ>0$, ∃ $w\in Tv=\{0,\frac{1}{2}\}$ such that

$$\frac{1}{4}={\varrho }_b(v,w)<\frac{1}{4}+ϵ\le {\mathcal{H}}_{{\varrho }_b}^{+}(\mathcal{T}v,\mathcal{T}u)+ϵ.$$

(c) 

If $u=1,v\in \mathcal{T}u=\mathcal{T}\left(1\right)=\{0,1\}$, for $v=0,ϵ>0$, ∃ $w\in Tv=\left\{0\right\}$ such that

$$0={\varrho }_b(v,w)<\frac{3}{4}+ϵ\le {\mathcal{H}}_{{\varrho }_b}^{+}(\mathcal{T}v,\mathcal{T}u)+ϵ$$

and for $v=1,ϵ>0$, ∃ $w\in \mathcal{T}v=\{0,1\}$ such that

$$\frac{1}{2}={\varrho }_b(v,w)<\frac{1}{2}+ϵ\le {\mathcal{H}}_{{\varrho }_b}^{+}(\mathcal{T}v,\mathcal{T}u)+ϵ.$$

Thus condition $\left(2^{\prime }\right)$ is satisfied.

Hence Theorem 2 can be applied and we conclude that $u\in \{0,\frac{1}{2},1\}$ is fixed points of $\mathcal{T}$.

5. Application

We now apply our main result to show the existence of solution of nonlinear integral inclusion of Volterra type. Suppose $l=(0,1)$, and $M=C[l,\mathbb{R})$, the space of all continuous functions $f:l\to \mathbb{R}$. Consider weak partial b-metric on M by

$${\varrho }_b(x,y)=\underset{t\in l}{sup}{e}^{-\beta t}{|x\left(t\right)-y\left(t\right)|}^p+\alpha ,$$

∀$x,y\in C(l,\mathbb{R})$, $p>1$ and $\alpha >0$. We have ${\varrho }_b^s(x,y)={sup}_{t\in l}{e}^{-\beta t}{|x\left(t\right)-y\left(t\right)|}^p$, so by Definition 6, $(C(l,\mathbb{R}),{\varrho }_b)$ is complete partial b-metric space. Denote by $P_{cl}\left(\mathbb{R}\right)$ the class of all nonempty closed subsets of $\mathbb{R}$.

Theorem 3.

Assume the integral equation inclusion of Volterra type

$$y\left(t\right)\in f\left(t\right)+{\int }_0^{t}K(t,s,y\left(s\right))ds,t\in l.$$

(4)

Suppose

(a) 

$K:l\times l\times \mathbb{R}\to P_{cl}\left(\mathbb{R}\right)$ is such that $K_y(t,s):=K(t,s,y\left(s\right))$ is continuous for all $(t,s)\in l\times l$ and $y\in C(l,\mathbb{R})$;

(b) 

$f\in C(l,\mathbb{R})$;

(c) 

for each $t\in l$, there exist $y\in C(l,\mathbb{R})$, such that

$${\mathcal{H}}_{{\varrho }_b}^{+}(K(t,x,y\left(x\right)),K(t,x,h\left(x\right)))\le \frac{1}{t^{p-1}}\left(\underset{x\in l}{sup}{|y\left(x\right)-h\left(x\right)|}^p+\alpha \right),$$

for all $t,x\in l$ and all $y,h\in C(l,\mathbb{R})$.

Then there is at least one solution of (4) in $C(l,\mathbb{R})$.

Proof. 

Define $\mathcal{T}:C(l,\mathbb{R})\to P_{cl}\left(C(l,\mathbb{R})\right)$ by

$$\mathcal{T}x\left(t\right)=\left\{y\in C(l,\mathbb{R})\mathrm{such}\mathrm{that}y\left(t\right)\in f\left(t\right)+{\int }_0^{t}K(t,s,x\left(s\right))ds,t\in l\right\}$$

for each $x\in C(l,\mathbb{R})$. For each $K_x:l\times l\to P_{cl}\left(\mathbb{R}\right)$ there exists $k_x:l\times l\to \mathbb{R}$ such that $k_x(t,s)\in K_x(t,s)$ for all $t,s\in l$. This implies that $f\left(t\right)+{\int }_0^t{k}_x(t,s)ds\in \mathcal{T}x$, and so $\mathcal{T}x\ne \varnothing$. It is easy to prove that $\mathcal{T}x$ is closed.

We show that $\mathcal{T}$ is ${\mathcal{H}}_{{\varrho }_b}^{+}$-type multivalued contraction. Let $u_1,u_2\in C(l,\mathbb{R})$ and $y\in \mathcal{T}x$. Then ∃ $k_{u_1}(t,s)\in K_{u_1}(t,s)$, $t,s\in l$ such that $y\left(t\right)=f\left(t\right)+{\int }_0^t{k}_x(t,s)ds,t\in l$. Also by hypothesis (iii),

$${\mathcal{H}}_{{\varrho }_b}^{+}(K(t,s,u_1\left(s\right)),K(t,s,u_2\left(s\right)))\le \frac{1}{t^{p-1}}\left(\underset{s\in l}{sup}{|u_1\left(s\right)-u_2\left(s\right)|}^p+\alpha \right)\forall t,s\in l.$$

Then there exist $g(t,s)\in K_{u_1}(t,s)$ such that

$$|k_{u_1}{(t,s)-g(t,s)|}^p+\xi \le \frac{1}{t^{p-1}}\left[|u_1\left(s\right)-u_2{\left(s\right)|}^p+\alpha \right]$$

for all $t,s\in l$. Define a multivalued operator $Q(t,s)$ by

$$Q(t,s)=K_{u_2}(t,s)\cap \{\eta \in \mathbb{R},|k_{u_1}{-\eta |}^p+\alpha \le \frac{1}{t^{p-1}}|u_1\left(s\right)-u_2\left(s\right){|}^p+\alpha \}$$

for all $t,s\in l$. Since Q is continuous operator, there exists a continuous operator $k_{u_2}:l\times l\to \mathbb{R}$ such that $k_{u_2}(t,s)\in Q(t,s)$ for all $t,s\in l$ and

$$h\left(t\right)=f\left(t\right)+{\int }_0^t{k}_{u_2}(t,s)ds\in f\left(t\right)+{\int }_0^{t}K(t,s,u_2\left(s\right))ds.$$

Therefore, let $q>1$ such that $\frac{1}{p}+\frac{1}{q}=1$.

$$\begin{array}{ccc}\hfill {\varrho }_b(y\left(t\right),\mathcal{T}{u}_2\left(t\right))& \le & {\varrho }_b(y\left(t\right),h\left(t\right))\hfill \\ & =& \underset{t\in l}{sup}{e}^{-\beta t}{|y\left(t\right)-h\left(t\right)|}^p+\alpha \hfill \\ & =& \underset{t\in l}{sup}{e}^{-\beta t}{\left|{\int }_0^t[k_{u_1}(t,s)-k_{u_2}(t,s)]ds\right|}^p+\alpha \hfill \\ & \le & \underset{t\in l}{sup}{e}^{-\beta t}{\left[{\left({\int }_0^{t}ds\right)}^{\frac{1}{q}}{\left({\int }_0^t{|k_{u_1}(t,s)-k_{u_2}(t,s)|}^{p}ds\right)}^{\frac{1}{p}}\right]}^p+\alpha \hfill \\ & \le & \underset{t\in l}{sup}{e}^{-\beta t}{\left(t\right)}^{\frac{p}{q}}\left({\int }_0^t{|k_{u_1}(t,s)-k_{u_2}(t,s)|}^{p}ds\right)+\alpha \hfill \\ & =& \underset{t\in l}{sup}{e}^{-\beta t}{\left(t\right)}^{p-1}\left({\int }_0^t{e}^{\beta s}{e}^{-\beta s}{|k_{u_1}(t,s)-k_{u_2}(t,s)|}^{p}ds\right)+\alpha \hfill \\ & =& \underset{t\in l}{sup}{e}^{-\beta t}{\left(t\right)}^{p-1}\left({\int }_0^t{e}^{\beta s}{e}^{-\beta s}{|k_{u_1}(t,s)-k_{u_2}(t,s)|}^{p}ds\right)+\alpha \hfill \\ & =& e^{-\beta t}{\left(t\right)}^{p-1}\left({\int }_0^t\left(e^{\beta s}\underset{t\in l}{sup}\{e^{-\beta s}|k_{u_1}(t,s)-k_{u_2}(t,s){|}^p+\alpha \}-\alpha \right)ds\right)+\alpha \hfill \\ & \le & e^{-\beta t}{\left(t\right)}^{p-1}\left({\int }_0^t\left(e^{\beta s}\underset{t\in l}{sup}\{\frac{1}{t^{p-1}}|u_1\left(t\right)-u_2\left(t\right){|}^p+\alpha \}-\alpha \right)ds\right)+\alpha \hfill \\ & =& e^{-\beta t}{\left(t\right)}^{p-1}\frac{1}{t^{p-1}}{\varrho }_b(u_1\left(t\right),u_2\left(t\right)){\int }_0^t{e}^{\beta s}ds-e^{-\beta t}{\left(t\right)}^{p-1}{\int }_0^t\alpha ds+\alpha \hfill \\ & =& e^{-\beta t}{\varrho }_b(u_1\left(t\right),u_2\left(t\right))(e^{\beta t}-1)-e^{-\beta t}{\left(t\right)}^{p-1}\alpha t+\alpha \hfill \\ & =& (1-e^{-\beta t}){\varrho }_b(u_1\left(t\right),u_2\left(t\right))+(1-e^{-\beta t}{t}^p)\alpha \hfill \\ & \le & (1-e^{-\beta t}){\varrho }_b(u_1\left(t\right),u_2\left(t\right))\hfill \\ & =& k.{\varrho }_b(u_1\left(t\right),u_2\left(t\right)),\hfill \end{array}$$

where $k=(1-e^{-\beta t})<1$. Since $y\left(t\right)$ is arbitrary, we have

$${\delta }_{{\varrho }_b}(\mathcal{T}{u}_1,\mathcal{T}{u}_2)\le k.{\varrho }_b(u_1,u_2).$$

(5)

Similarly, we get

$${\delta }_{{\varrho }_b}(\mathcal{T}{u}_2,\mathcal{T}{u}_1)\le k.{\varrho }_b(u_2,u_1).$$

(6)

From (5) and (6), we get

$${\mathcal{H}}_{{\varrho }_b}^{+}(\mathcal{T}{u}_1,\mathcal{T}{u}_2)=k.\frac{{\delta }_{{\varrho }_b}(\mathcal{T}{u}_1,\mathcal{T}{u}_2)+{\delta }_{{\varrho }_b}(\mathcal{T}{u}_2,\mathcal{T}{u}_1))}{2}\le k.{\varrho }_b(u_2,u_1).$$

Hence, $\mathcal{T}$ is ${\mathcal{H}}_{{\varrho }_b}^{+}$-type multivalued contraction. Thus all the assertions of Theorem 2 are satisfied and hence (4) has a solution.  ☐

6. Conclusions

In this paper, we present the concept of weak partial b-metric spaces with their topology and weak partial Hausdorff b-metric spaces and generalized the famous Nadler’s theorem in weak partial b-metric space by using weak partial Hausdorff b-metric spaces. We give an example to show the validity and an application to nonlinear Volterra integral inclusion for the usability of our result.