On multivalued weakly Picard operators in partial Hausdorff metric spaces

Abstract

We discuss multivalued weakly Picard operators on partial Hausdorff metric spaces. First, we obtain Kikkawa-Suzuki type fixed point theorems for a new type of generalized contractive conditions. Then, we prove data dependence of a fixed points set theorem. Finally, we present sufficient conditions for well-posedness of a fixed point problem. Our results generalize, complement and extend classical theorems in metric and partial metric spaces.

1 Introduction and preliminaries

In 1937, von Neumann [1] initiated the fixed point theory for multivalued mappings in the study of game theory. Indeed, the fixed point theorems for multivalued mappings are quite useful in control theory and have been frequently used in solving many problems of economics. In 1969, Nadler [2] initiated the development of the metric fixed point theory for multivalued mappings. Nadler used the concept of Hausdorff metric to establish the multivalued contraction principle containing the Banach contraction principle as a special case. Also, for the basic problems of fixed point theory for multivalued mappings, we refer to [3].

Let $(X,d)$ be a metric space and let $\operatorname{CB}(X)$ be the family of all nonempty, closed and bounded subsets of X. For $A,B\in \operatorname{CB}(X)$, $x \in X$, let

$$ H(A,B)=\max\Bigl\{ \sup_{x\in A}d(x,B),\sup_{y\in B}d(y,A) \Bigr\} , $$

where $d(x,B)=\inf\{d(x,y):y\in B\}$ and $H:\operatorname{CB}(X) \times \operatorname{CB}(X) \to \mathbb{R}^{+}$ is the Hausdorff metric induced by d.

Now on, the letters ℝ, $\mathbb{R}^{+}$ and ℕ will denote the set of all real numbers, the set of all non-negative real numbers and the set of all positive integers, respectively. Also, $\operatorname{CL}(X)$ is the collection of nonempty closed subsets of X.

Definition 1.1

Let X be a nonempty set. If $T : X \to \operatorname{CB}(X)$ is a multivalued operator, then an element $x \in X$ is called

• (i) fixed point of T if $x \in Tx$;

• (ii) strict fixed point of T if $\{x\} = T x$.

In the sequel, we denote by $\operatorname{Fix}(T):= \{x \in X: x \in T x\}$ the set of all fixed points of T and by $S\operatorname{Fix}(T):= \{x \in X: \{x\} = T x\}$ the set of all strict fixed points of T.

Definition 1.2

([4])

Let $(X, d)$ be a metric space and $T: X \to \operatorname{CL}(X)$ be a multivalued operator. T is called a multivalued weakly Picard operator (briefly MWP operator) if for all $x \in X$ and all $y \in Tx$, there exists a sequence $\{x_{n}\}$ such that:

• (i) $x_{0} = x$, $x_{1} = y$;

• (ii) $x_{n+1} \in Tx_{n}$ for all $n \in\mathbb{N}\cup\{0\}$;

• (iii) the sequence $\{x_{n}\}$ is convergent and its limit is a fixed point of T.

A sequence $\{x_{n}\}$ satisfying (i) and (ii) is also called a sequence of successive approximations (briefly s.s.a.) of T starting from $x_{0}$.

For interested readers, the theory of MWP operators was presented in [4–7].

In 2008 Suzuki [8] introduced a new type of mappings in order to generalize the well-known Banach contraction principle. This result has led to some important contributions in metric fixed point theory (see, for instance, [9] and the references therein).

As we mentioned above, Nadler proved the following multivalued version of the Banach contraction principle.

Theorem 1.3

([2])

Let $(X,d)$ be a complete metric space and $T:X\rightarrow \operatorname{CB}(X)$ be a multivalued mapping satisfying $H(Tx,Ty)\leq kd(x,y)$ for all $x,y\in X$ and $k\in(0,1)$. Then T has a fixed point.

In the last decades, a number of fixed point results (see [10–17]) have been obtained in attempts to generalize Theorem 1.3.

One of the most significant fixed point theorems for multivalued mappings appeared in [9], Theorem 2.1. This theorem merges the ideas of Suzuki [8] and Nadler [2] into a consistent framework.

Theorem 1.4

([9])

Let $(X,d)$ be a complete metric space and $T:X\rightarrow \operatorname{CB}(X)$. Consider the non-increasing function $\psi:[0,1)\rightarrow(0,1]$ defined by

$$ \psi(r)= \begin{cases} 1 & \textit{if }0\leq r< \frac{1}{2}, \\ 1-r & \textit{if }\frac{1}{2}\leq r<1. \end{cases} $$

Assume that there exists $r \in[0,1)$ such that

$$ \psi(r)d(x,Tx)\leq d(x,y)\quad \Longrightarrow\quad H(Tx,Ty)\leq rM_{d}(x,y) $$

for all $x,y\in X$, where

$$ M_{d}(x,y)=\max \biggl\{ d(x,y),d(x,Tx),d(y,Ty), \frac {d(x,Ty)+d(y,Tx)}{2} \biggr\} . $$

Then T has a fixed point.

We remark that the right-hand side of the above implication is known as Ćirić type contractive condition, see [10, 11, 18]. Also, for our further use, we recall the following refinement of Nadler’s theorem, see [14].

Theorem 1.5

Let $\eta: [0, 1) \to(\frac{1}{2}, 1]$ be a function defined by $\eta(r) = \frac{1}{1+r}$. Let $(X, d)$ be a complete metric space and $T: X \to \operatorname{CB}(X)$ be an r-KS multivalued operator, that is, there exists $r \in[0, 1)$ such that

$$\eta(r)d(x,T x) \leq d(x, y) \quad \Longrightarrow\quad H(Tx,Ty) \leq rd(x, y) $$

for all $x,y \in X$. Then T is an MWP operator.

The other basic notion for the development of our work is the concept of partial metric space, which was introduced by Matthews [19] as a part of the study of denotational semantics of dataflow networks. Matthews presented a modified version of the Banach contraction principle, more suitable in this context, see also [20, 21]. For more reading on interesting approaches to partial metric spaces and related contexts, we refer to [22–24]. Now, the (complete) partial metric spaces constitute a suitable framework to model several distinguished examples of the theory of computation and also to model metric spaces via domain theory, see [18, 19, 25–37]. In this direction, Aydi et al. [38] introduced the concept of partial Hausdorff metric and extended Nadler’s fixed point theorem in the setting of partial metric spaces.

Consistent with [19, 38–40], the following definitions and results will be needed in the sequel.

Definition 1.6

([19])

Let X be any nonempty set. A function $p:X\times X\rightarrow \mathbb{R}^{+}$ is said to be a partial metric if and only if for all $x,y,z\in X$ the following conditions are satisfied:

• (P1) $p(x,x)=p(y,y)=p(x,y)$ if and only if $x=y$;

• (P2) $p(x,x)\leq p(x,y)$;

• (P3) $p(x,y)=p(y,x)$;

• (P4) $p(x,z)\leq p(x,y)+p(y,z)-p(y,y)$.

The pair $(X,p)$ is called a partial metric space. If $p(x,y)=0$, then (P1) and (P2) imply that $x=y$, but the converse does not hold in general. A trivial example of a partial metric space is the pair $(\mathbb{R}^{+},p)$, where $p:\mathbb{R}^{+} \times\mathbb{R}^{+} \rightarrow \mathbb{R}^{+}$ is given by $p(x,y)=\max\{x,y\}$, see also [41].

Example 1.7

([19])

Let $X=\{[a,b]:a,b\in \mathbb{R},a\leq b\}$. It is easy to show that the function $p:X\times X\rightarrow\mathbb{R}^{+}$ given by $p([a,b],[c,d])=\max\{b,d\}-\min\{a,c\}$ defines a partial metric on X.

For further examples, we refer to [34, 35, 39, 42, 43]. Note that each partial metric p on X generates a $T_{0}$ topology $\tau_{p}$ on X which has as a base the family of the open balls (p-balls) $\{B_{p}(x,\epsilon): x \in X, \epsilon>0\}$, where for all $x\in X$ and $\epsilon>0$,

$$ B_{p}(x,\epsilon)=\bigl\{ y\in X: p(x,y)< p(x,x)+\epsilon\bigr\} . $$

A sequence $\{x_{n}\}$ in a partial metric space $(X,p)$ is called convergent to a point $x\in X$, with respect to $\tau_{p}$, if and only if $p(x,x)=\lim_{n\rightarrow +\infty}p(x,x_{n})$, see [19] for details. If p is a partial metric on X, then the function

$$p^{S}(x,y)=2p(x,y)-p(x,x)-p(y,y) $$

defines a metric on X. Further a sequence $\{x_{n}\}$ converges in the metric space $(X,p^{S})$ to a point $x\in X$ if and only if

$$ p(x,x)=\lim_{n\rightarrow+\infty }p(x,x_{n})=\lim _{n,m\rightarrow+\infty}p(x_{n},x_{m}). $$

Definition 1.8

([19])

Let $(X,p)$ be a partial metric space. Then:

• (i) A sequence $\{x_{n}\}$ in X is called Cauchy if and only if $\lim_{n,m\rightarrow+\infty}p(x_{n},x_{m})$ exists and is finite.

• (ii) A partial metric space $(X,p)$ is said to be complete if every Cauchy sequence $\{x_{n}\}$ in X converges, with respect to $\tau _{p}$, to a point $x\in X$ such that $p(x,x)=\lim_{n,m \rightarrow + \infty}p(x_{n},x_{m})$.

Lemma 1.9

([19, 39])

Let $(X,p)$ be a partial metric space. Then:

• (i) A sequence $\{x_{n}\}$ in X is Cauchy in $(X,p)$ if and only if it is Cauchy in $(X,p^{S})$.

• (ii) A partial metric space $(X,p)$ is complete if and only if the metric space $(X,p^{S})$ is complete.

Consistent with [38], let $(X,p)$ be a partial metric space and let $\operatorname{CB}^{p}(X)$ be the family of all nonempty, closed and bounded subsets of the partial metric space $(X,p)$, induced by the partial metric p. Note that the closedness is taken from $(X,\tau_{p})$ ($\tau_{p}$ is the topology induced by p) and the boundedness is given as follows: A is a bounded subset in $(X,p)$ if there exist $x_{0}\in X$ and $M\geq0$ such that for all $a\in A$, we have $a\in B_{p}(x_{0},M)$, that is, $p(x_{0},a)< p(x_{0},x_{0})+M$. For $A,B\in \operatorname{CB}^{p}(X)$, $x\in X$, $\delta _{p}:\operatorname{CB}^{p}(X)\times \operatorname{CB}^{p}(X)\rightarrow\mathbb{R}^{+}$ define

$$\begin{aligned}& p(x,A) = \inf\bigl\{ p(x,a):a\in A\bigr\} , \\& p(A,B) = \inf\bigl\{ p(x,y):x\in A, y \in B\bigr\} , \\& \delta_{p}(A,B) = \sup\bigl\{ p(a,B):a\in A\bigr\} , \\& H_{p}(A,B) = \max\bigl\{ \delta_{p}(A,B), \delta_{p}(B,A)\bigr\} , \end{aligned}$$

where $H_{p}:\operatorname{CB}^{p}(X)\times \operatorname{CB}^{p}(X)\rightarrow\mathbb{R}^{+}$ is called the partial Hausdorff metric induced by p. Also, it is easy to show that $p(x,A)=0$ implies that $p^{S}(x,A)=0$, where

$$ p^{S}(x,A)=\inf\bigl\{ p^{S}(x,a):a\in A\bigr\} . $$

Lemma 1.10

([39])

Let $(X,p)$ be a partial metric space and A be any nonempty subset of X, then $a\in\overline{A}$ if and only if $p(a,A)=p(a,a)$.

Lemma 1.11

Let $(X,p)$ be a partial metric space and A be any nonempty subset of X. If A is closed in $(X,p)$, then A is closed in $(X,p^{S})$.

Proof

Let $\{x_{n}\}$ be a sequence converging to some $x \in X$ in $(X,p^{S})$. Then we have

$$p(x,x)=\lim_{n\rightarrow+\infty }p(x,x_{n}), $$

which implies, by definition, that $\{x_{n}\}$ converges to $x \in X$ also in $(X,p)$. Now, since A is closed in $(X,p)$, then $x \in A$ and so we deduce that A is closed also in $(X,p^{S})$. □

Proposition 1.12

([38])

Let $(X,p)$ be a partial metric space. For any $A,B,C\in \operatorname{CB}^{p}(X)$, we have:

• (i) $\delta_{p}(A,A)=\sup\{p(a,a):a\in A\}$;

• (ii) $\delta_{p}(A,A)\leq\delta_{p}(A,B)$;

• (iii) $\delta_{p}(A,B)=0\Longrightarrow A\subseteq B$;

• (iv) $\delta_{p}(A,B)\leq\delta_{p}(A,C)+\delta _{p}(C,B)-\inf_{c\in C}p(c,c)$.

Proposition 1.13

([38])

Let $(X,p)$ be a partial metric space. For any $A,B,C\in \operatorname{CB}^{p}(X)$, we have:

• (h1) $H_{p}(A,A)\leq H_{p}(A,B)$;

• (h2) $H_{p}(A,B)=H_{p}(B,A)$;

• (h3) $H_{p}(A,B)\leq H_{p}(A,C)+H_{p}(C,B)-\inf_{c\in C}p(c,c)$;

• (h4) $H_{p}(A,B)=0\Longrightarrow A=B$.

Notice that each Hausdorff metric is a partial Hausdorff metric but the converse is not true, see Example 2.6 in [38].

Lemma 1.14

([38])

Let $(X,p)$ be a partial metric space, $A,B\in \operatorname{CB}^{p}(X)$ and $q>1$, then for any $a\in A$, there exists $b(a)\in B$ such that $p(a,b(a))\leq qH_{p}(A,B)$.

Theorem 1.15

([38])

Let $(X,p)$ be a partial metric space. If $T:X\rightarrow \operatorname{CB}^{p}(X)$ is a multivalued mapping such that for all $x,y\in X$, we have $H_{p}(Tx,Ty)\leq kp(x,y)$, where $k\in(0,1)$, then T has a fixed point.

In view of the above considerations and following the ideas in [44], the aim of this paper is to discuss multivalued weakly Picard operators on partial Hausdorff metric spaces, see also [45] for other interesting results. First, we obtain Kikkawa-Suzuki type fixed point theorems for a new type of generalized contractive conditions. Then, we prove data dependence of a fixed points set theorem. Finally, we present sufficient conditions for well-posedness of a fixed point problem. The presented results extend and unify some recently obtained comparable results for multivalued mappings (see [9] and the references therein).

2 Fixed point theorems in partial Hausdorff metric spaces

In this section we present several theorems which characterize MWP operators, defined in the previous section, in terms of different contractive conditions. Results of this section are generalizations of Theorem 1.4, Theorem 1.15 (and so Nadler’s Theorem 1.3), Ćirić’s theorem in [10] and others.

2.1 Result - I

To provide the first theorem we introduce the notion of $(s,r)$-contractive multivalued operator in partial Hausdorff metric spaces as follows.

Definition 2.1

Let $p: X \times X \to\mathbb{R}^{+}$ be a partial metric and $T:X\rightarrow \operatorname{CB}^{p}(X)$ be a multivalued mapping. T is called an $(s,r)$-contractive multivalued operator if there exist $r \in[0,1)$ and $s \geq r$ such that

$$ p(y,Tx) \leq s p(y,x)\quad \Longrightarrow \quad H_{p}(Tx,Ty) \leq r M_{p}(x,y) $$

(1)

for all $x,y \in X$, where

$$M_{p}(x,y)=\max \biggl\{ p(x,y),p(x,Tx),p(y,Ty),\frac {p(x,Ty)+p(y,Tx)}{2} \biggr\} . $$

Now we state and prove our theorem.

Theorem 2.2

Let $(X,p)$ be a complete partial metric space and $T:X\rightarrow \operatorname{CB}^{p}(X)$ be an $(s,r)$-contractive multivalued operator with $s \geq1$. Then T is an MWP operator.

Proof

Let $r_{1}$ be a real number such that $0\leq r< r_{1}<1$. Let $u_{1}\in X$. As $Tu_{1}$ is nonempty, we can choose $u_{2}\in Tu_{1}$. Clearly, if $u_{2}=u_{1}$ the proof is finished and so we assume $u_{2} \neq u_{1}$. Then we get

$$ p(u_{2},Tu_{1})=p(u_{2},u_{2}) \leq p(u_{2},u_{1}) \leq s p(u_{2},u_{1}). $$

(2)

Next, we choose $u_{3}\in Tu_{2}$ such that

$$p(u_{3},u_{2}) \leq\frac{r_{1}}{r}H_{p}(Tu_{2},Tu_{1}). $$

Now, from (2) and by using condition (1) we write

$$\begin{aligned} p(u_{3},u_{2}) \leq& \frac{r_{1}}{r}H_{p}(Tu_{2},Tu_{1}) \leq r_{1}M_{p}(u_{1},u_{2}) \\ = & r_{1} \max \biggl\{ p(u_{1},u_{2}),p(u_{1},Tu_{1}),p(u_{2},Tu_{2}), \frac{p(u_{1},Tu_{2})+p(u_{2},Tu_{1})}{2} \biggr\} \\ \leq& r_{1} \max \biggl\{ p(u_{1},u_{2}),p(u_{1},u_{2}),p(u_{2},u_{3}), \frac {p(u_{1},u_{3})+p(u_{2},u_{2})}{2} \biggr\} \\ \leq& r_{1} \max \biggl\{ p(u_{1},u_{2}),p(u_{2},u_{3}), \frac {p(u_{1},u_{2})+p(u_{2},u_{3})}{2} \biggr\} \\ \leq& r_{1} \max \bigl\{ p(u_{1},u_{2}),p(u_{2},u_{3}) \bigr\} . \end{aligned}$$

Thus, we obtain

$$ p(u_{2},u_{3})\leq r_{1} \max\bigl\{ p(u_{1},u_{2}),p(u_{2},u_{3})\bigr\} . $$

If $\max\{p(u_{1},u_{2}),p(u_{2},u_{3})\}=p(u_{2},u_{3})$, then $p(u_{2},u_{3})\leq r_{1} p(u_{2},u_{3})$ implies that $p(u_{2},u_{3})=0$, and we obtain $p^{S}(u_{2},u_{3})\leq2p(u_{2},u_{3})=0$, which further implies that $p^{S}(u_{2},u_{3})=0$. Hence $u_{2}=u_{3}\in Tu_{2}$ and the proof is finished. On the contrary, if $\max \{p(u_{1},u_{2}),p(u_{2},u_{3})\}=p(u_{1},u_{2})$, then we have

$$ p(u_{2},u_{3})\leq r_{1} p(u_{1},u_{2}). $$

Continuing this process, we can construct a sequence $\{u_{n}\}$ in X such that $u_{n+1}\in Tu_{n}$, $u_{n+1} \neq u_{n}$ and

$$ p(u_{n},u_{n+1})\leq r_{1}^{n-1}p(u_{1},u_{2}) $$

for every $n>1$. This shows that $\lim_{n\rightarrow+\infty }p(u_{n},u_{n+1})=0$.

Now, since

$$ p(u_{n},u_{n})+p(u_{n+1},u_{n+1}) \leq2p(u_{n},u_{n+1}), $$

then we get

$$ \lim_{n\rightarrow+\infty}p(u_{n},u_{n})=0\quad \text{and}\quad \lim_{n\rightarrow+\infty}p(u_{n+1},u_{n+1})=0. $$

(3)

Let $\epsilon>0$ and pick $N \in\mathbb{N}$ large enough so that for $n \geq N$ we have

$$ 2r_{1}^{n-1}\frac{1}{1- r_{1}}p(u_{1},u_{2})< \epsilon. $$

Then, for every positive integer $k > n \geq N$, there is some $m \in \mathbb{N}$ such that $k=n+m$, and we have

$$\begin{aligned} p^{S}(u_{n},u_{n+m}) \leq&2p(u_{n},u_{n+m}) \\ \leq&2\bigl[p(u_{n},u_{n+1})+\cdots+p(u_{n+m-1},u_{n+m}) \bigr] \\ \leq&2\bigl[r_{1}^{n-1}p(u_{1},u_{2})+ \cdots+ r_{1}^{n+m-2}p(u_{1},u_{2})\bigr] \\ \leq&2 r_{1}^{n-1}\frac{1}{1-r_{1}}p(u_{1},u_{2})< \epsilon. \end{aligned}$$

It is immediate to deduce that $\{u_{n}\}$ is a Cauchy sequence in $(X,p^{S})$, but by Lemma 1.9 $\{u_{n}\}$ is Cauchy also in $(X,p)$. Moreover, since $(X,p)$ is complete, again by Lemma 1.9 we deduce the completeness of $(X,p^{S})$. It follows that there exists $z\in X$ such that $\lim_{n\rightarrow+\infty}u_{n}=z$ in $(X,p^{S})$. Therefore $\lim_{n\rightarrow+\infty}p^{S}(u_{n},z)=0$ implies

$$ p(z,z)=\lim_{n\rightarrow+\infty }p(u_{n},z)=\lim _{m,n\rightarrow+\infty}p(u_{n},u_{m}). $$

Now, since $\{u_{n}\}$ is a Cauchy sequence in $(X,p^{S})$, then we have

$$ \lim_{m,n\rightarrow+\infty}p^{S}(u_{n},u_{m})=0 $$

and so

$$ \lim_{m,n\rightarrow+\infty }2p(u_{n},u_{m})-\lim _{m\rightarrow+\infty }p(u_{m},u_{m})-\lim _{n\rightarrow+\infty}p(u_{n},u_{n})=0. $$

It follows from (3) that

$$ \lim_{m\rightarrow+\infty}p(u_{m},u_{m})=\lim _{n\rightarrow +\infty}p(u_{n},u_{n})=0, $$

which further implies that

$$ \lim_{m,n\rightarrow+\infty}p(u_{n},u_{m})=0 $$

and hence

$$ p(z,z)=\lim_{n\rightarrow+\infty }p(u_{n},z)=\lim _{n\rightarrow+\infty}p(u_{n},u_{n})=0. $$

Next, we will show that there exists a subsequence $\{u_{n(k)}\}$ of $\{ u_{n}\}$ such that

$$p(z,Tu_{n(k)}) \leq sp(z,u_{n(k)}) $$

for all $k \in\mathbb{N}$. Reasoning by contradiction, we assume that there exists a positive integer N such that $p(z,Tu_{n}) > sp(z,u_{n})$ for all $n \geq N$. This implies $p(z,u_{n+1}) > s p(z,u_{n})$ for all $n \geq N$. By induction, for all $n \geq N$ and $m' \geq1$, we get that

$$ p(z,u_{n+m'}) > s^{m'} p(z,u_{n}). $$

(4)

Since

$$\begin{aligned} p(u_{n+m'},u_{n}) \leq& p(u_{n},u_{n+1}) + p(u_{n+1},u_{n+2})+ \cdots+ p(u_{n+m'-1},u_{n+m'}) - \sum_{i=n+1}^{m'-1} p(u_{i},u_{i}) \\ \leq& p(u_{n},u_{n+1}) + p(u_{n+1},u_{n+2})+ \cdots+ p(u_{n+m'-1},u_{n+m'}) \\ \leq& p(u_{n},u_{n+1}) \bigl(1+r_{1} + r_{1}^{2} + r_{1}^{3}+ \cdots+ r_{1}^{m'-1}\bigr) \\ \leq& p(u_{n},u_{n+1}) \biggl[\frac{1-r_{1}^{m'}}{1-r_{1}} \biggr] \end{aligned}$$

for all $n \geq N$ and $m' \geq1$, then we get

$$\begin{aligned} p(z,u_{n}) \leq& p(z, u_{n+m'}) + p(u_{n+m'},u_{n}) - p(z,z) \\ \leq& p(z, u_{n+m'}) + p(u_{n},u_{n+1}) \biggl[ \frac {1-r_{1}^{m'}}{1-r_{1}} \biggr]. \end{aligned}$$

Passing to the limit as $m' \to+\infty$, we have

$$p(z,u_{n}) \leq\frac{1}{1-r_{1}} p(u_{n},u_{n+1}) \quad \text{for all } n \geq N. $$

Then we obtain

$$ p(z,u_{n+m'}) \leq\frac{1}{1-r_{1}} p(u_{n+m'},u_{n+m'+1}) \leq\frac {r_{1}^{m'}}{1-r_{1}} p(u_{n},u_{n+1}) $$

(5)

for all $n \geq N$ and $m' \geq1$. By (4) and (5) we get

$$ p(z,u_{n}) < \frac{(\frac{r_{1}}{s})^{m'}}{1-r_{1}} p(u_{n},u_{n+1}) $$

for all $n \geq N$ and $m' \geq1$. Next, passing to the limit as $m' \to+\infty$, we obtain that $p(z,u_{n}) = 0$ for all $n \geq N$. This contradicts (4) and therefore there exists a subsequence $\{ u_{n(k)}\}$ of $\{u_{n}\}$ such that $p(z,Tu_{n(k)}) \leq s p(z,u_{n(k)})$ for all $k \in\mathbb{N}$.

Also, by hypothesis we have

$$\begin{aligned}& H_{p}(Tu_{n(k)},Tz) \\& \quad \leq r\max \biggl\{ p(u_{n(k)},z),p(u_{n(k)},Tu_{n(k)}),p(z,Tz), \frac{ p(u_{n(k)},Tz)+p(z,Tu_{n(k)})}{2} \biggr\} \\& \quad \leq r\max \biggl\{ p(u_{n(k)},z),p(u_{n(k)},u_{n(k)+1}),p(z,Tz), \frac {p(u_{n(k)},z)+ p(z,Tz)+p(z,u_{n(k)+1})}{2} \biggr\} . \end{aligned}$$

Therefore, we write

$$\begin{aligned}& p(z,Tz) \\& \quad \leq p(z,u_{n(k)+1}) + H_{p}(Tu_{n(k)},Tz) - p(u_{n(k)+1},u_{n(k)+1}) \\& \quad \leq p(z,u_{n(k)+1}) \\& \qquad {} + r\max \biggl\{ p(u_{n(k)},z),p(u_{n(k)},u_{n(k)+1}),p(z,Tz), \frac{p(u_{n(k)},z)+ p(z,Tz)+p(z,u_{n(k)+1})}{2} \biggr\} . \end{aligned}$$

On passing to the limit as $k\rightarrow+\infty$, we get

$$ p(z,Tz)\leq r\max \biggl\{ p(z,Tz),\frac{p(z,Tz)}{2} \biggr\} =r p(z,Tz). $$

Since $r<1$, it follows that

$$ p(z,Tz)=0. $$

Therefore $p(z,Tz)=0=p(z,z)$ and hence by Lemma 1.10 we deduce that $z\in Tz$, that is, z is a fixed point of T. □

The following example illustrates the use of Theorem 2.2.

Example 2.3

Let $X=\{0,1,2\}$ and $p:X\times X\rightarrow \mathbb{R}^{+}$ be defined by $p(0,0) =0$, $p(1,1)=p(2,2)=\frac{1}{4}$, $p(1,0)=\frac{1}{3}$, $p(2,0)=\frac{3}{5}$, $p(2,1)=\frac{2}{5}$ and $p(y,x)=p(x,y)$ for all $x,y\in X$. Then $(X,p)$ is a complete partial metric space. Also define $T:X\rightarrow \operatorname{CB}^{p}(X)$ by

$$ Tx= \begin{cases} \{0\} & \text{if }x \neq2, \\ \{0,1\} & \text{if }x = 2. \end{cases} $$

Therefore, we get

$$\begin{aligned}& H_{p}(T0,T0) = H_{p}(T0,T1) =H_{p}(T1,T1) =p(0,T0)=p(0,T1)=p(0,0), \\& H_{p}(T0,T2) = H_{p}(T1,T2)=p(1,T0)=p(1,T1)=p(1,0), \\& p(2,T2) = \min\bigl\{ p(2,0),p(2,1)\bigr\} =p(2,1), \\& H_{p}(T2,T2) = p(2,2). \end{aligned}$$

It follows easily that the inequality

$$ p(y,Tx)\leq s p(x,y) $$

holds for all $x,y\in X$ with $s \geq\frac{8}{5}$. Also, for all $x,y\in X$ with $r \in[\frac{5}{6},1)$, we get

$$ H_{p}(Tx,Ty)\leq r M_{p}(x,y). $$

Thus all the conditions of Theorem 2.2 are satisfied and 0 is a fixed point of T.

From Theorem 2.2 we deduce some corollaries.

Corollary 2.4

Let $(X,p)$ be a complete partial metric space and $T:X\rightarrow \operatorname{CB}^{p}(X)$ be a multivalued mapping. Assume that there exist $r \in[0,1)$ and $s \geq1$ such that

$$ p(y,Tx)\leq s p(x,y) \quad \Longrightarrow \quad H_{p}(Tx,Ty)\leq r \max \bigl\{ p(x,y),p(x,Tx),p(y,Ty)\bigr\} $$

for all $x,y \in X$. Then T is an MWP operator.

Corollary 2.5

Let $(X,p)$ be a complete partial metric space and $T:X\rightarrow \operatorname{CB}^{p}(X)$ be a multivalued mapping. Assume that there exist $r \in[0,1)$ and $s \geq1$ such that

$$ p(y,Tx)\leq s p(x,y)\quad \Longrightarrow\quad H_{p}(Tx,Ty)\leq \frac{r}{3} \bigl[ p(x,y)+p(x,Tx)+p(y,Ty)\bigr] $$

for all $x,y \in X$. Then T is an MWP operator.

In the case of single-valued mappings, Theorem 2.2 reduces to the following significant corollary.

Corollary 2.6

Let $(X,p)$ be a complete partial metric space and $T:X\rightarrow X$ be a single-valued mapping. Assume that there exist $r \in[0,1)$ and $s \geq1$ such that

$$ p(y,Tx) \leq s p(x,y) \quad \Longrightarrow \quad p(Tx,Ty) \leq r M_{p}(x,y), $$

(6)

where

$$M_{p}(x,y)=\max \biggl\{ p(x,y),p(x,Tx),p(y,Ty),\frac {p(x,Ty)+p(y,Tx)}{2} \biggr\} . $$

Then T has a unique fixed point.

Proof

The existence part of the proof follows easily by Theorem 2.2. Thus, we need to prove uniqueness of the fixed point. Suppose to the contrary that, for $s \geq1$, there exist $x,y \in \operatorname{Fix}(T)$ with $x \neq y$. It follows immediately that

$$p(y,T x) = p(y, x) \leq s p(y,x). $$

Thus, by hypothesis on T, we would have

$$\begin{aligned} p(Tx,Ty) \leq& r \max \biggl\{ p(x,y),p(x,Tx),p(y,Ty),\frac {p(x,Ty)+p(y,Tx)}{2} \biggr\} \\ =& r \max \biggl\{ p(x,y),p(x,x),p(y,y),\frac{p(x,y)+p(y,x)}{2} \biggr\} \\ =& r p(x,y). \end{aligned}$$

We deduce that $p(x,y) = 0$, which further implies that $p^{S} (x, y) \leq2 p(x, y) = 0$ and hence $x = y$, a contradiction. This completes the proof. □

The following two examples, adapted from [46], show the validity of Corollary 2.6.

Example 2.7

Let $X=[0,1]$ be endowed with the partial metric $p(x,y)=\max\{x,y\}$ for all $x,y \in X$. Then $(X,p)$ is a complete partial metric space. Also define $T:X\to X$ by

$$Tx=\frac{x^{2}}{1+x}. $$

Take arbitrary elements $x,y \in X$ with $y\leq x$. Then we have

$$p(y,Tx) = \max \biggl\{ y, \frac{x^{2}}{1+x} \biggr\} \leq sx = s p(x,y) $$

for all $x, y \in X$ and $s \geq1$. On the other hand, we get

$$p(Tx,Ty)=\max \biggl\{ \frac{x^{2}}{1+x},\frac{y^{2}}{1+y} \biggr\} = \frac {x^{2}}{1+x} $$

and

$$\begin{aligned} M_{p}(x,y)&=\max \biggl\{ p(x,y),p(x,Tx),p(y,Ty),\frac {p(x,Ty)+p(y,Tx)}{2} \biggr\} \\ &=\max \biggl\{ p(x,y),p\biggl(x,\frac{x^{2}}{1+x}\biggr),p\biggl(y, \frac {y^{2}}{1+y}\biggr),\frac{1}{2}\biggl(p\biggl(x,\frac{y^{2}}{1+y} \biggr)+p\biggl(y,\frac {x^{2}}{1+x}\biggr)\biggr) \biggr\} \\ &=\max\biggl\{ x,x,y,\frac{1}{2}\biggl(x+\max\biggl\{ y,\frac{x^{2}}{1+x} \biggr\} \biggr)\biggr\} =x. \end{aligned}$$

Thus the inequality $p(Tx,Ty)\leq r M_{p}(x,y)$ holds for all $x, y \in X$ with $y \leq x$ and for any $r \in[\frac{1}{2},1)$. Note that we obtain the same conclusion if we assume that $x \leq y$. Hence, all the conditions of Corollary 2.6 are satisfied and 0 is a fixed point of T.

The following example underlines the crucial role of the right-hand side of (6) in establishing existence of the fixed point.

Example 2.8

Let $X=\{1,2,3,4\}$ and $p:X\times X\rightarrow \mathbb{R}^{+}$ be defined by $p(x,x)=\frac{1}{2}$ for each $x\in X$, $p(1,2)=p(3,4)=2$, $p(1,3)=p(2,4)=1$, $p(1,4)=p(2,3)=\frac{3}{2}$ and $p(y,x)=p(x,y)$ for all $x,y\in X$. Then $(X,p)$ is a complete partial metric space. Also define $T:X\to X$ by

$$T=\left(\begin{array}{cccc}1& 2& 3& 4\\ 2& 1& 1& 2\end{array}\right).$$

Trivially T has no fixed points, but we try to apply Corollary 2.6. It is easy to check that the inequality $p(y,Tx) \leq s p(y,x)$ certainly holds for all $x,y\in X$ with $s \geq 2$. Now we note that, for $x=3$ and $y=1$, we get

$$p(T3,T1)=p(1,2)=2 $$

and

$$\begin{aligned} M_{p}(3,1)&=\max \biggl\{ p(3,1),p(3,T3),p(1,T1),\frac {p(3,T1)+p(1,T3)}{2} \biggr\} \\ &=\max \biggl\{ 1,1,2,\frac{1}{2} \biggl(\frac{3}{2}+ \frac{1}{2} \biggr) \biggr\} =2. \end{aligned}$$

It follows that

$$p(T3,T1)=2 \nleq2r=r M_{p}(3,1), $$

whatever $r \in[0,1)$ is chosen. We conclude that Corollary 2.6 is not applicable in this case.

2.2 Result - II

Another interesting characterization of MWP operators is provided by the following theorem.

Theorem 2.9

Let $(X,p)$ be a complete partial metric space and $T:X\rightarrow \operatorname{CB}^{p}(X)$ be a multivalued operator. Assume that there exist $r,s \in[0,1)$, with $r < s$, such that

$$ \frac{1}{1+r} p(x,Tx) \leq p(x,y) \leq\frac{1}{1-s} p(x,Tx)\quad \Longrightarrow\quad H_{p}(Tx,Ty) \leq r M_{p}(x,y) $$

(7)

for all $x,y \in X$, where

$$M_{p}(x,y)=\max \biggl\{ p(x,y),p(x,Tx),p(y,Ty),\frac {p(x,Ty)+p(y,Tx)}{2} \biggr\} . $$

Then T is an MWP operator.

Proof

Let $r_{1}$ be a real number such that $0\leq r< r_{1}<s$. Also, let $u_{1}\in X$ and $u_{2}\in Tu_{1}$ be such that

$$p(u_{1},u_{2}) \leq\frac{1-r_{1}}{1-s}p(u_{1},Tu_{1}). $$

Then we have

$$\frac{1}{1+r} p(u_{1},Tu_{1}) \leq p(u_{1},Tu_{1}) \leq p(u_{1},u_{2}) \leq \frac{1}{1-s} p(u_{1},Tu_{1}) $$

and so, by using condition (7), we obtain

$$\begin{aligned} p(u_{2},Tu_{2}) \leq& H_{p}(Tu_{1},Tu_{2}) \\ \leq& r \max \biggl\{ p(u_{1},u_{2}),p(u_{1},Tu_{1}),p(u_{2},Tu_{2}), \frac {p(u_{1},Tu_{2})+p(u_{2},Tu_{1})}{2} \biggr\} \\ \leq& r \max \biggl\{ p(u_{1},u_{2}),p(u_{2},Tu_{2}), \frac {p(u_{1},Tu_{2})+p(u_{2},u_{2})}{2} \biggr\} \\ \leq& r \max \biggl\{ p(u_{1},u_{2}),p(u_{2},Tu_{2}), \frac {p(u_{1},u_{2})+p(u_{2},Tu_{2})}{2} \biggr\} . \end{aligned}$$

Thus

$$ p(u_{2},Tu_{2})\leq r \max\bigl\{ p(u_{1},u_{2}),p(u_{2},Tu_{2}) \bigr\} . $$

Now, if $\max\{p(u_{1},u_{2}),p(u_{2},Tu_{2})\}=p(u_{2},Tu_{2})$, then $p(u_{2},Tu_{2})\leq r p(u_{2},Tu_{2})$ implies that $p(u_{2},Tu_{2})=0$ and so we obtain

$$p^{S}(u_{2},Tu_{2})\leq2p(u_{2},Tu_{2})=0, $$

which further implies that $p^{S}(u_{2},Tu_{2})=0$. In view of Lemma 1.11, $u_{2}\in Tu_{2}$, and the proof is finished. On the contrary, if $\max \{p(u_{1},u_{2}),p(u_{2},u_{3})\}=p(u_{1},u_{2})$, then we have

$$ p(u_{2},Tu_{2})\leq r p(u_{1},u_{2}). $$

It follows that there exists $u_{3} \in Tu_{2}$ such that

$$p(u_{2},u_{3})\leq r_{1} p(u_{1},u_{2}) \quad \mbox{and}\quad p(u_{2},u_{3}) \leq \frac{1-r_{1}}{1-s}p(u_{2},Tu_{2}). $$

This implies

$$\frac{1}{1+r} p(u_{2},Tu_{2}) \leq p(u_{2},u_{3}) \leq\frac{1}{1-s} p(u_{2},Tu_{2}). $$

Now, by using condition (7), we get $p(u_{3},Tu_{3})\leq r p(u_{2},u_{3})$. Continuing this process, we can construct a sequence $\{u_{n}\}$ in X with the following properties:

$$\begin{cases} u_{n+1}\in Tu_{n}, \\ p(u_{n+1},Tu_{n+1})\leq r p(u_{n},u_{n+1}), \\ p(u_{n+1},u_{n+2})\leq r_{1} p(u_{n},u_{n+1}), \\ p(u_{n+1},u_{n+2}) \leq\frac{1-r_{1}}{1-s}p(u_{n+1},Tu_{n+1}) \end{cases} $$

for every $n \in\mathbb{N}$. Next, from $p(u_{n+1},u_{n+2})\leq r_{1} p(u_{n},u_{n+1})$ we deduce that

$$\lim_{n\rightarrow+\infty }p(u_{n},u_{n+1})=0. $$

Since

$$ p(u_{n},u_{n})+p(u_{n+1},u_{n+1}) \leq2p(u_{n},u_{n+1}), $$

then we get

$$ \lim_{n\rightarrow+\infty}p(u_{n},u_{n})=0\quad \text{and}\quad \lim_{n\rightarrow+\infty}p(u_{n+1},u_{n+1})=0. $$

Proceeding as in the proof of Theorem 2.2, one can show that the sequence $\{u_{n}\}$ is a Cauchy sequence in $(X,p)$ converging to some $z\in X$ with $p(z,z)=0$.

Now, since

$$p(u_{n},u_{n+1}) \leq\frac{1-r_{1}}{1-s} p(u_{n},Tu_{n}), $$

then we have

$$p(z,u_{n}) \leq\frac{1}{1-s} p(u_{n},Tu_{n}) $$

for all $n \geq1$. Then we assume that there exists a positive integer N such that

$$p(z, u_{n}) < \frac{1}{1+r} p(u_{n}, Tu_{n}) $$

holds for every $n \geq N$. Consequently, we have

$$\begin{aligned} p(u_{n},u_{n+1}) \leq& p(z,u_{n})+ p(z,u_{n+1}) - p(z,z) \\ < & \frac{1}{1+r} \bigl[p(u_{n},Tu_{n}) + p(u_{n+1},Tu_{n+1})\bigr] \\ <& \frac{1}{1+r} \bigl[p(u_{n},Tu_{n}) + r p(u_{n},u_{n+1})\bigr] \\ \leq& \frac{1}{1+r} \bigl[p(u_{n},u_{n+1}) + r p(u_{n},u_{n+1})\bigr]=p(u_{n},u_{n+1}), \end{aligned}$$

which is a contradiction. Hence, there exists a subsequence $\{u_{n(k)}\}$ of $\{u_{n}\}$ such that

$$p(z,u_{n(k)}) \geq\frac{1}{1+r} p(u_{n(k)},Tu_{n(k)}) $$

holds for every $k \geq N$. Since $p(z,u_{n}) \leq\frac{1}{1-s} p(u_{n},Tu_{n})$ for all $n \geq1$, by condition (7), we have $H_{p}(Tz, Tu_{n(k)}) \leq r M_{p}(z, u_{n(k)})$. This implies

$$\begin{aligned}& p(z,Tz) \\& \quad \leq p(u_{n(k)+1},z) + H_{p}(Tu_{n(k)},Tz) \\& \quad \leq p(u_{n(k)+1},z) \\& \qquad{} + r\max \biggl\{ p(u_{n(k)},z),p(u_{n(k)},Tu_{n(k)}),p(z,Tz), \frac{p(u_{n(k)},Tz)+p(z,Tu_{n(k)})}{2} \biggr\} \\& \quad \leq r\max \biggl\{ p(u_{n(k)},z),p(u_{n(k)},u_{n(k)+1}),p(z,Tz), \frac {p(u_{n(k)},Tz)+p(z,u_{n(k)+1})}{2} \biggr\} . \end{aligned}$$

On passing to the limit as $k \rightarrow+\infty$, we get

$$ p(z,Tz)\leq r\max \biggl\{ p(z,Tz),\frac{p(z,Tz)}{2} \biggr\} . $$

Since $r<1$, it follows that

$$ p(z,Tz)=0. $$

Therefore $p(z,Tz)=0=p(z,z)$ and hence by Lemma 1.10 we have $z\in Tz$, that is, z is a fixed point of T. □

In the case of single-valued mappings, Theorem 2.9 reduces to the following corollary.

Corollary 2.10

Let $(X,p)$ be a complete partial metric space and $T:X\rightarrow X$ be a single-valued mapping. Assume that there exists $r \in[0,1)$ such that

$$ \frac{1}{1+r} p(x,Tx) \leq p(x,y) \leq\frac{1}{1-r} p(x,Tx) \quad \Longrightarrow\quad p(Tx,Ty) \leq r M_{p}(x,y) $$

for all $x,y \in X$, where

$$M_{p}(x,y)=\max\biggl\{ p(x,y),p(x,Tx),p(y,Ty),\frac{p(x,Ty)+p(y,Tx)}{2} \biggr\} . $$

Then T has a fixed point.

Proof

It is easy to prove that for every $u_{1} \in X$ the sequence $\{u_{n}\}$ defined by $u_{n+1}=Tu_{n}$ satisfies the relationship $p(u_{n+1},u_{n+2}) \leq r p(u_{n},u_{n+1})$. Consequently, the sequence $\{u_{n}\}$ is Cauchy, and there is some point $z \in X$ such that $\lim_{n \to+\infty} u_{n} = z$. Proceeding as in the proof of Theorem 2.9, we can show that

$$ p(z,z)=\lim_{n\rightarrow+\infty}p(u_{n},z)=\lim _{n\rightarrow+\infty}p(u_{n},u_{n})=0. $$

Also in view of Theorem 2.9, for all $n \geq N$, we can assume $p(z,u_{n}) \leq\frac{1}{1+r} p(u_{n},u_{n+1})$ for leading to contradiction. Consequently, there exists a subsequence $\{u_{n(k)}\}$ of $\{u_{n}\}$ such that

$$p(z,u_{n(k)}) \geq\frac{1}{1+r} p(u_{n(k)},u_{n(k)+1}) $$

holds for every $k \geq N$. Therefore, we obtain that

$$\begin{aligned}& p(u_{n(k)+1},Tz) \\& \quad \leq r\max \biggl\{ p(u_{n(k)},z),p(u_{n(k)},Tu_{n(k)}),p(z,Tz), \frac{ p(u_{n(k)},Tz)+p(z,Tu_{n(k)})}{2} \biggr\} \\& \quad \leq r\max \biggl\{ p(u_{n(k)},z),p(u_{n(k)},u_{n(k)+1}),p(z,Tz), \frac {p(u_{n(k)},z)+ p(z,Tz)+p(z,u_{n(k)+1})}{2} \biggr\} . \end{aligned}$$

On passing to the limit as $k\rightarrow+\infty$, we get

$$ p(z,Tz)\leq r\max \biggl\{ p(z,Tz),\frac{p(z,Tz)}{2} \biggr\} . $$

Since $r<1$, it follows that

$$ p(z,Tz)=0 $$

and so $z= Tz$, that is, z is a fixed point of T. □

The following example shows that Theorem 2.9 is proper extension of the respective result in standard metric spaces.

Example 2.11

Let $X=\{0,1,2\}$ and $p:X\times X\rightarrow \mathbb{R}^{+}$ be defined by $p(0,0) =p(1,1)=0$, $p(2,2)=\frac{1}{4}$, $p(1,0)=\frac{1}{3}$, $p(2,0)=\frac{2}{5}$, $p(2,1)=\frac{11}{15}$ and $p(y,x)=p(x,y)$ for all $x,y\in X$. Then $(X,p)$ is a complete partial metric space. Also define $T:X\rightarrow \operatorname{CB}^{p}(X)$ by

$$ Tx= \begin{cases} \{0\} & \text{if }x\neq2, \\ \{0,1\} & \text{if }x=2. \end{cases} $$

Therefore, we get

$$\min\bigl\{ p(x,Tx):x \in X \setminus\{0\}\bigr\} =\min\bigl\{ p(x,y) :x,y \in X\text{ and }x\neq y\bigr\} =\frac{1}{3} $$

and

$$\max\bigl\{ p(x,Tx):x \in X\bigr\} =\frac{2}{5}. $$

It follows easily that the inequalities

$$ \frac{1}{1+r} p(x,Tx) \leq p(x,y) \leq\frac{1}{1-s} p(x,Tx) $$

hold for all $x,y \in X$ with $x \neq y$ and for some $1>s>r \geq\frac {1}{5}$. Also the above inequalities hold for $x=y=2$ with $1>s>r \geq \frac{3}{5}$. On the other hand, the above inequalities are not applicable for $x=y \in\{0,1\}$. Clearly we have

$$\begin{aligned}& H_{p}(T0,T0) = H_{p}(T0,T1) =H_{p}(T1,T1) =p(0,T0)=p(0,T1)=p(0,0), \\& H_{p}(T0,T2) = H_{p}(T1,T2)=p(1,T0)=p(1,T1)=p(1,0), \\& p(2,T2) = \min\bigl\{ p(2,0),p(2,1)\bigr\} =p(2,0), \\& H_{p}(T2,T2) = p(2,2). \end{aligned}$$

Finally, by routine calculations and taking $1>s>r\geq\frac{5}{6}$, one can show that the inequality

$$ H_{p}(Tx,Ty)\leq r M_{p}(x,y) $$

holds true as for all $x,y \in X$ with $x \neq y$, as for $x=y=2$. Thus all the conditions of Theorem 2.9 are satisfied and 0 is a fixed point of T.

Next, we consider the metric $p^{S}$ induced by the partial metric p. Indeed, we have $p^{S}(x,x)=0$ for all $x \in X$, $p^{S}(1,0)=\frac{2}{3}$, $p^{S}(2,1)=\frac{73}{60}$, $p^{S}(2,0)=\frac{11}{20}$ and $p^{S}(x,y)=p^{S}(y,x)$ for all $x,y \in X$.

We show that Theorem 2.9 is not applicable in this case. Indeed, for $x=2$ and $y=0$, the inequalities

$$ \frac{1}{1+r} p^{S}(2,T2) = \frac{1}{1+r} p^{S} \bigl(2,\{0,1\}\bigr) = \frac {1}{1+r}p^{S}(2,0) \leq p^{S}(2,0) \leq\frac{1}{1-s} p^{S}(2,T2) $$

hold true for all $r \in[0,1)$ with $r< s$. Therefore, we need to have

$$ H_{p^{S}}(T0,T2)\leq rM_{p^{S}}(0,2). $$

Unfortunately, this is not the case because

$$ H_{p^{S}}(T0,T2)=H_{p^{S}}\bigl(\{0\},\{0,1\}\bigr)= \frac{2}{3} $$

and

$$\begin{aligned} M_{p^{S}}(0,2) =&\max \biggl\{ p^{S}(0,2),p^{S}(0,T0),p^{S}(2,T2), \frac{p^{S}(0,T2)+p^{S}(2,T0)}{2} \biggr\} \\ =&\max \biggl\{ \frac{11}{20},0,\frac{11}{20},\frac{0+\frac{11}{20 }}{2} \biggr\} =\frac{11}{20}< \frac{2}{3}. \end{aligned}$$

Consequently, for any $r \in[0,1)$ we have

$$ H_{p^{S}}(T0,T2)\nleqslant rM_{p^{S}}(0,2). $$

3 Data dependence theorem in partial Hausdorff metric spaces

The aim of this section is to discuss data dependence of a fixed points set for MWP operators on partial metric spaces. Also, this section is motivated by Popescu [44], see also [4]. Precisely, we will prove a result for $(1, r)$-contractive multivalued operators in partial Hausdorff metric spaces.

First, we need the following auxiliary lemma.

Lemma 3.1

Let $(X, p)$ be a partial metric space and $T : X \to \operatorname{CB}^{p}(X)$ be a $(1,r)$-contractive multivalued operator. If $z \in Tz$, then $p(z,z)=0$.

Proof

Since $p(z,Tz)=p(z,z)$, then we have

$$H_{p}(Tz,Tz) \leq r \max \biggl\{ p(z,z),p(z,Tz),p(z,Tz), \frac {p(z,Tz)+p(z,Tz)}{2} \biggr\} =rp(z,z). $$

Thus, by definition, we write

$$p(z,z) \leq p(z,Tz) \leq H_{p}(Tz,Tz)\leq rp(z,z)< p(z,z) $$

and hence we deduce that $p(z,z)=0$. □

We recall the following concept.

Definition 3.2

([47])

Let $(X,p)$ be a partial metric space and let $\phi:X \to \mathbb{R}^{+}$ be a function on X. Then the function ϕ is called p-lower semi-continuous on X whenever

$$\lim_{n \to+\infty}p(x_{n}, x)= p(x,x)\quad \Longrightarrow \quad \phi(x) \leq \lim_{n \to+\infty}\inf\phi(x_{n}) = \sup _{n \geq1}\inf_{m \geq n}\phi(x_{m}). $$

Now we state and prove our theorem.

Theorem 3.3

Let $(X, p)$ be a partial metric space and $T_{1},T_{2} : X \to \operatorname{CB}^{p}(X)$ be two multivalued operators. We suppose that:

• (i) $T_{i}$ is a $(1,r_{i})$-contractive multivalued operator for $i \in{1, 2}$;

• (ii) there exists $\lambda> 0$ such that $H_{p}(T_{1}x,T_{2}x) \leq \lambda$ for all $x \in X$;

• (iii) the function $\phi:X \to \mathbb{R}^{+}$ defined by $\phi(x)=p(x,x)$ is p-lower semi-continuous.

Then:

• (a) $\operatorname{Fix}(T_{i} ) \in \operatorname{CL}^{p}(X)$, $i \in\{1, 2\}$;

• (b) $T_{1}$ and $T_{2}$ are MWP operators and

  $$H_{p}\bigl(\operatorname{Fix}(T_{1}), \operatorname{Fix}(T_{2}) \bigr) \leq\frac{\lambda}{1- \max\{r_{1}, r_{2}\}}. $$

Proof

(a) From Theorem 2.2 we have that $\operatorname{Fix}(T_{i})$ is a nonempty set, $i \in\{1, 2\}$. Let us prove that the fixed point set of a $(1,r_{i})$-contractive multivalued operator $T_{i}$ is closed. Let $x_{n} \in \operatorname{Fix}(T_{i})$, with $n \geq1$, be such that $\lim_{n\rightarrow+\infty}x_{n}=z$ in $(X,p)$. In view of (iii) and Lemma 3.1, we have

$$\lim_{n \to+\infty}p(x_{n}, z)= p(z,z) \quad \Longrightarrow \quad p(z,z) \leq \lim_{n \to+\infty}\inf p(x_{n},x_{n}) = 0. $$

It follows that

$$ p(z,z)=\lim_{n\rightarrow+\infty}p(x_{n},z)=\lim _{n\rightarrow+\infty}p(x_{n},x_{n})=0. $$

Also, since $x_{n} \in T_{i}x_{n}$, we have $p(z,T_{i}x_{n}) \leq p(z,x_{n})$ and then

$$\begin{aligned} p(z,T_{i}z) \leq& p(z,x_{n})+ p(x_{n},T_{i}z) - p(x_{n},x_{n}) \\ \leq& p(z,x_{n})+H_{p}(T_{i}x_{n},T_{i} z) \\ \leq& p(z,x_{n})+r_{i} p(x_{n},z). \end{aligned}$$

Passing to the limit as $n\to+\infty$, we obtain that $p(z,T_{i} z) = 0$. Therefore

$$p(z,T_{i}z)=0=p(z,z) $$

and hence by Lemma 1.10 and $T_{i}z \in \operatorname{CL}^{p}(X)$ we get $z \in T_{i}z$, that is, $z \in \operatorname{Fix}(T_{i})$.

(b) From the proof of Theorem 2.2 we immediately get that a $(1,r_{i})$-contractive multivalued operator is an MWP operator. For the second conclusion, let q be a real number such that $q>1$, and $x_{0} \in \operatorname{Fix}(T_{1})$ be arbitrary. Then, by Lemma 1.14, there exists $x_{1} \in T_{2}x_{0}$ such that

$$p(x_{0},x_{1}) \leq q H_{p}(T_{1}x_{0},T_{2}x_{0}). $$

Next, for $x_{1} \in T_{2}x_{0}$, there exists $x_{2} \in T_{2}x_{1}$ such that

$$p(x_{1},x_{2}) \leq q H_{p}(T_{2}x_{0},T_{2}x_{1}). $$

Since $x_{1} \in T_{2}x_{0}$, $p(x_{1},T_{2}x_{0}) = p(x_{1},x_{1}) \leq p(x_{0}, x_{1})$, then we have

$$p(x_{1},x_{2}) \leq q H_{p}(T_{2}x_{0},T_{2}x_{1}) \leq qr_{2} p(x_{0}, x_{1}). $$

Iterating this process allows us to construct a sequence of successive approximations for $T_{2}$ starting from $x_{0}$, satisfying the following assertions:

$$x_{n+1} \in T_{2}x_{n} \quad \text{and}\quad p(x_{n}, x_{n+1}) \leq(qr_{2})^{n} p(x_{0}, x_{1})\quad \text{for all } n \in\mathbb{N}. $$

Hence, for all $n \geq N$ and $m \geq1$, we write

$$\begin{aligned} p(x_{n+m},x_{n}) \leq& p(x_{n},x_{n+1}) + p(x_{n+1},x_{n+2})+ \cdots+ p(x_{n+m-1},x_{n+m})- \sum_{i=n+1}^{m-1} p(x_{i},x_{i}) \\ \leq& p(x_{n},x_{n+1}) + p(x_{n+1},x_{n+2})+ \cdots+ p(x_{n+m-1},x_{n+m}) \\ \leq& \frac{(qr_{2})^{n}}{1 - qr_{2}} p(x_{0}, x_{1}). \end{aligned}$$

Consequently, we get

$$ p^{S}(x_{n},x_{n+m}) \leq 2p(x_{n},x_{n+m}) \leq\frac{2(qr_{2})^{n}}{1 - qr_{2}} p(x_{0}, x_{1}). $$

(8)

Now, choosing $1 < q < \min\{\frac{1}{r_{1}}, \frac{1}{r_{2}}\}$ and passing to the limit as $n \to+\infty$, we deduce easily that the sequence $\{x_{n}\}$ is Cauchy in $(X,p^{S})$ and, by Lemma 1.9, $\{x_{n}\}$ is Cauchy in $(X,p)$. Then there exists $u \in X$ such that

$$\lim_{n\rightarrow+\infty}x_{n}=u \quad \mbox{in } \bigl(X,p^{S}\bigr). $$

Therefore, (8) and $\lim_{n\rightarrow+\infty}p^{S}(x_{n},u)=0$ imply

$$ p(u,u)=\lim_{n\rightarrow+\infty}p(x_{n},u)=\lim _{n,m\rightarrow+\infty}p(x_{n},x_{m})=0. $$

We will prove that u is a fixed point for $T_{2}$. To this aim, suppose that there exists a positive integer N such that

$$p(u,T_{2}x_{n}) > p(u,x_{n}) \quad \mbox{for all }n \geq N. $$

This implies that $p(u,x_{n+1}) > p(u,x_{n})$ for all $n \geq N$, which leads to contradiction since $x_{n} \to u$ as $n \to+\infty$. Hence, there exists a subsequence $\{x_{n(k)}\}$ such that

$$p(u,T_{2}x_{n(k)}) \leq p(u,x_{n(k)})\quad \mbox{for all } k \in\mathbb{N}. $$

Next, from

$$ p(u,T_{2}u) \leq p(x_{n(k)+1},u) + H_{p}(T_{2}x_{n(k)},T_{2}u), $$

on passing to the limit as $k \to+\infty$, we get

$$\begin{aligned} p(u,T_{2}u) \leq& \lim_{k \to+\infty} p(x_{n(k)+1},u) + \lim_{k \to +\infty} H_{p}(T_{2}x_{n(k)},T_{2}u) \\ \leq& \lim_{k \to+\infty} r_{2} p(x_{n(k)},u) = 0 \end{aligned}$$

and so $u \in \operatorname{Fix}(T_{2})$.

By (8), passing to the limit as $m \to+\infty$, we get

$$\begin{aligned} p(x_{n},u) \leq& \lim_{m \to+\infty} \bigl[ p(x_{n},x_{n+m}) + p(x_{n+m},u) -p(u,u)\bigr] \\ \leq& \lim_{m \to+\infty} \biggl[\frac{2(qr_{2})^{n}}{1 - qr_{2}} p(x_{0}, x_{1}) + p(x_{n+m},u) \biggr] \\ \leq& \frac{(qr_{2})^{n}}{1-qr_{2}} p(x_{0},x_{1})\quad \text{for each } n \in \mathbb{N}. \end{aligned}$$

Then

$$p(x_{0},u) \leq\frac{1}{1-qr_{2}} p(x_{0},x_{1}) \leq\frac{q\lambda}{1-qr_{2}}. $$

Analogously, one can show that, for each $u_{0} \in \operatorname{Fix}(T_{2})$, there exists $x \in \operatorname{Fix}(T_{1})$ such that

$$p(u_{0}, x) \leq\frac{1}{1-qr_{1}} p(u_{0},u_{1}) \leq\frac{q\lambda}{1-qr_{1}} $$

and hence

$$H_{p}\bigl(\operatorname{Fix}(T_{1}),\operatorname{Fix}(T_{2}) \bigr) \leq\frac{q \lambda}{1- \max\{qr_{1}, qr_{2}\}}. $$

Finally, passing to the limit as $q \to1^{+}$, we obtain the assertion. This concludes the proof. □

4 Well-posedness of fixed point problems in partial Hausdorff metric spaces

According to [48, 49], we get the notions of well-posedness of a fixed point problem in the setting of partial metric spaces.

Definition 4.1

(see [48, 49])

Let $(X, p)$ be a partial metric space and $T : X \to \operatorname{CB}^{p}(X)$ be a multivalued operator. Then the fixed point problem is well posed for T with respect to p if:

(a₁):

$\operatorname{Fix}(T ) = \{z\}$;

(b₁):

if $x_{n} \in X$, $n \in\mathbb{N}$ and $\lim_{n\rightarrow+\infty}p(x_{n},Tx_{n}) = 0$, then $\lim_{n\rightarrow+\infty}p(x_{n}, z) = 0$.

Definition 4.2

(see [48, 49])

Let $(X, p)$ be a partial metric space and $T : X \to \operatorname{CB}^{p}(X)$ be a multivalued operator. Then the fixed point problem is well posed for T with respect to $H_{p}$ if:

(a₂):

$S\operatorname{Fix}(T) = \{z\}$;

(b₂):

if $x_{n} \in X$, $n \in\mathbb{N}$ and $\lim_{n\rightarrow+\infty}H_{p}(x_{n},Tx_{n}) = 0$, then $\lim_{n\rightarrow+\infty}p(x_{n},z) = 0$.

Clearly, (b₂) of Definition 4.2 implies (b₁) of Definition 4.1. Moreover, from (a₁) and (a₂), that is, $\operatorname{Fix}(T) = S\operatorname{Fix}(T) = \{z\}$, we deduce that if the fixed point problem is well posed for T with respect to p, then it is well posed for T with respect to $H_{p}$.

Motivated by the above facts, we will prove the following theorem for $(s, r)$-contractive multivalued operators, with $s >1$, in partial metric spaces.

Theorem 4.3

Let $(X, p)$ be a partial metric space and $T : X \to \operatorname{CB}^{p}(X)$ be a multivalued operator. We suppose that:

• (1) T is an $(s,r)$-contractive multivalued operator with $s \geq1$;

• (2) $S\operatorname{Fix}(T) \neq\emptyset$.

Then:

• (a) $\operatorname{Fix}(T) = S\operatorname{Fix}(T) = \{z \}$;

• (b) the fixed point problem is well posed for T with respect to $H_{p}$ if $s >1$.

Proof

(a) Suppose $z \in S\operatorname{Fix}(T)$. Clearly, $z \in \operatorname{Fix}(T)$. We show that $\operatorname{Fix}(T) = \{z\}$. To this aim, let $u \in \operatorname{Fix}(T)$ with $u \neq z$. Since $p(u,Tz) = p(u, z) \leq s p(u, z)$, we get $H_{p}(Tz,Tu) \leq r p(z,u)$ and therefore

$$p(z,u) = p(Tz,u) \leq H_{p}(Tz,Tu) \leq r p(z,u), $$

which leads to contradiction. Thus, $\operatorname{Fix}(T ) = \{z\}$ and so the assertion (a) holds true.

(b) Now, let $x_{n} \in X$, with $n \in\mathbb{N}$, be such that $\lim_{n\rightarrow+\infty}p(x_{n},Tx_{n}) = 0$. We have to show that $\lim_{n\rightarrow+\infty}p(x_{n},z) = 0$. Suppose this is not the case; suppose that $p(x_{n},z)$ does not converge to zero. Consequently, there exist $\epsilon>0$ and a subsequence $\{ x_{n(k)}\}$ such that

$$p(x_{n(k)},z) \geq\epsilon\quad \mbox{for all }k \in\mathbb{N}. $$

Now, assume that there exists a subsequence $\{x_{n(k(j))}\}$ of $\{x_{n(k)}\}$ with

$$p(z,Tx_{n(k(j))}) \leq s p(z,x_{n(k(j))}). $$

Then we get

$$H_{p}(Tz,Tx_{n(k(j))}) \leq r p(z,x_{n(k(j))}), $$

and so we write

$$\begin{aligned} p(z,x_{n(k(j))}) =& p(x_{n(k(j))},Tz) \\ \leq& p(x_{n(k(j))},Tx_{n(k(j))})+H_{p}(Tz,Tx_{n(k(j))}) \\ \leq& p(x_{n(k(j))},Tx_{n(k(j))})+ r p(z,x_{n(k(j))}). \end{aligned}$$

The above inequality leads to the following:

$$\epsilon\leq p(z,x_{n(k(j))}) \leq\frac{1}{1- r}p(x_{n(k(j))},Tx_{n(k(j))}). $$

Passing to the limit as $j \to+\infty$, since $\lim_{n\rightarrow+\infty}p(x_{n},Tx_{n}) = 0$, we get the contradiction $\epsilon=0$. Consequently, we deduce that there exists $k_{1} \in \mathbb{N}$ such that

$$p(z,Tx_{n(k)}) > sp(z,x_{n(k)})\quad \mbox{for all }k \geq k_{1}. $$

Again, since $\lim_{n\rightarrow+\infty}p(x_{n},Tx_{n}) = 0$, there exists $k_{2} \geq k_{1}$ such that

$$p(x_{n(k)},Tx_{n(k)}) < (s-1) \epsilon\quad \mbox{for all }k \geq k_{2}. $$

Finally, for all $k \geq k_{2}$, we write

$$\begin{aligned} (s - 1)\epsilon \leq& (s - 1)p(z, x_{n(k)}) \\ =&sp(z, x_{n(k)})-p(z, x_{n(k)}) \\ < & p(z,T x_{n(k)}) - p(z,x_{n(k)}) \\ \leq& p(x_{n(k)},Tx_{n(k)}) \quad \mbox{(by (P4) of Definition 1.6)} \\ <& (s-1)\epsilon, \end{aligned}$$

which leads to contradiction. Consequently, we conclude that $\lim_{n\rightarrow+\infty}p(x_{n},z) = 0$. □

5 Application to integral equations

The literature is rich with papers focusing on the study of integral operators of various types: Fredholm, Urysohn, Volterra and others. It is well known that integral operators provide an important subject of numerous mathematical investigations and are often applicable in many scientific disciplines as physics, biology and economics. The papers we refer to essentially present a fixed point approach based on the Banach contraction principle and its constructive proof. These results give sufficient conditions for establishing the existence (and uniqueness) of solution of certain integral operators, see [50–52].

Here, following this line of research, we prove an existence theorem for the solution of integral equations by using Corollary 2.6. Precisely, we consider the following integral equation:

$$ u(t)= \int_{0}^{\mathcal{T}} K\bigl(t,s,u(s) \bigr)\, ds +g(t), $$

(9)

$t\in I=[0,\mathcal{T}]$, where $\mathcal{T}>0$. Also we denote

$$C(I):=\{u:I\rightarrow\mathbb{R}\mid u \mbox{ is continuous on } I\}, $$

and define $d:C(I)\times C(I)\to\mathbb{R}$ by

$$d(u,v)=\max_{t\in I}\bigl\vert u(t)-v(t)\bigr\vert $$

for all $u,v\in C(I)$, so that $(C(I),d)$ is a complete metric space.

Finally, we define the operator $T: C(I)\to C(I)$ by

$$T x(t)=\int_{0}^{\mathcal{T}} K\bigl(t,s,x(s)\bigr)\, ds+g(t) $$

for all $x \in C(I)$ and $t\in I$.

Then we consider the following hypotheses:

(i) $K: I\times I\times\mathbb{R}\rightarrow \mathbb{R}$ and $g: I\rightarrow\mathbb{R}$ are continuous;

(ii) there exist $\alpha\in[0,1)$, $\beta\geq1$ and a continuous function $G:I\times I\to \mathbb{R}^{+}$ such that

$$\max_{t\in I}\int_{0}^{\mathcal{T}} G(t,s) \, ds = \alpha, $$

and the inequality

$$ \max_{t\in I}\biggl\vert \int_{0}^{\mathcal{T}} K\bigl(t,s,u(s)\bigr) \, ds+g(t) - v(t)\biggr\vert \leq\beta\max _{t\in I}\bigl\vert u(t) - v(t)\bigr\vert $$

implies

$$\begin{aligned}& \bigl\vert K\bigl(t,s,u(t)\bigr)-K\bigl(t,s,v(t)\bigr)\bigr\vert \\& \quad \leq G(t,s) \max \biggl\{ \bigl\vert u(t) - v(t)\bigr\vert , \\& \qquad \biggl\vert \int_{0}^{\mathcal{T}} K \bigl(t,s,u(s)\bigr) \, ds+g(t) - u(t)\biggr\vert , \biggl\vert \int _{0}^{\mathcal{T}} K\bigl(t,s,v(s)\bigr) \, ds+g(t) - v(t) \biggr\vert , \\& \qquad \frac{1}{2} \biggl(\biggl\vert \int_{0}^{\mathcal{T}} K\bigl(t,s,v(s)\bigr) \, ds+g(t) - u(t)\biggr\vert + \biggl\vert \int _{0}^{\mathcal{T}} K\bigl(t,s,u(s)\bigr) \, ds+g(t) - v(t) \biggr\vert \biggr) \biggr\} \end{aligned}$$

for all $u,v\in C(I)$ and $t,s\in I$.

We will prove the following result.

Theorem 5.1

Suppose that hypotheses (i) and (ii) hold. Then the integral equation (9) has a unique solution $x^{*}\in C(I)$.

Proof

First we note that the space $(C(I),d)$ is trivially a complete partial metric space with zero self-distance.

Next, for all $u,v\in C(I)$, by (ii), we deduce that

$$\begin{aligned}& \bigl\vert {T}u(t)- {T} v(t)\bigr\vert \\& \quad \leq \int_{0}^{T} \bigl\vert K \bigl(t,s,u(s)\bigr)-K\bigl(t,s,v(s)\bigr)\bigr\vert \, ds \\& \quad \leq\int_{0}^{T} G(t,s)\, ds\max \biggl\{ \bigl\vert u(t) - v(t)\bigr\vert , \bigl\vert u(t) - T\bigl(u(t)\bigr)\bigr\vert , \\& \qquad \bigl\vert v(t) - T\bigl(v(t) \bigr)\bigr\vert , \frac{|u(t)-T(v(t))|+ |v(t)- T(u(t))|}{2} \biggr\} , \end{aligned}$$

which, on routine calculations, leads to

$$ d(Tu, {T}v)\leq \alpha M_{d}(u,v). $$

Therefore, without loss of generality, we can write that

$$d(v,Tu) \leq\beta d(u,v) \quad \Longrightarrow\quad d(Tu, Tv) \leq \alpha M_{d}(u,v) $$

for all $u,v\in C(I)$. Thus Corollary 2.6 is applicable in this case, and hence the operator T has a unique fixed point $x^{*}\in C(I)$. Clearly, $x^{*}\in C(I)$ is the unique solution of (9). □