1 Introduction
Throughout this paper, we use following notation: \(\mathbb{N}\) is the set of all natural numbers, \(\mathbb{R}\) is the set of all real numbers, and \(\mathbb{R}^{+}\) is the set of all nonnegative real numbers.
Definition 1.1
([2])
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A partial metric on a nonempty set X is a function \(p:X\times X\to \mathbb{R}^{+}\) such that, for all \(x,y,z\in X\):
(P1)
\(x=y\) if and only if \(p(x,x)=p(x,y)=p(y,y)\);
(P2)
\(p(x,x)\leq p(x,y)\);
(P3)
\(p(x,y)=p(y,x)\);
(P4)
\(p(x,y)\leq p(x,z)+p(z,y)-p(z,z)\).
The pair \((X,p)\) is called a partial metric space. Many fixed point results in partial metric spaces have been proved; see [3‐17]. Recently, Beg and Pathak [1] introduced a weaker form of partial metrics called a weak partial metric.
Definition 1.2
([1])
Let X be a nonempty set. A function \(q:X\times X\rightarrow \mathbb{R}^{+}\) is called a weak partial metric on X if for all \(x,y,z\in X\), the following conditions hold:
The pair \((X, q)\) is called a weak partial metric space.
\((WP1)\)
\(q(x,x)=q(x,y)\) if and only if \(x=y\);
\((WP2)\)
\(q(x, x) \le q(x, y)\);
\((WP3)\)
\(q(x, y) = q(y, x)\);
\((WP4)\)
\(q(x,y)\leq q(x,z)+q(z,y)\).
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Examples of weak partial metric spaces [1] are: Notice that
(1)
\((\mathbb{R}^{+},q)\), where \(q:\mathbb{R}^{+}\times \mathbb{R} ^{+}\rightarrow \mathbb{R}^{+}\) is defined as \(q(x,y)=\vert x-y \vert +1\) for \(x,y\in \mathbb{R}^{+}\).
(2)
\((\mathbb{R}^{+},q)\), where \(q:\mathbb{R}^{+}\times \mathbb{R}^{+}\rightarrow \mathbb{R}^{+}\) is defined as \(q(x,y)=\frac{1}{4}\vert x-y \vert +\max \{x,y\}\) for \(x,y\in \mathbb{R}^{+}\).
(3)
\((\mathbb{R}^{+},q)\), where \(q:\mathbb{R}^{+}\times \mathbb{R}^{+}\rightarrow \mathbb{R}^{+}\) is defined as \(q(x,y)=\max \{x,y\}+e^{\vert x-y \vert }+1\) for \(x,y\in \mathbb{R} ^{+}\).
-
If \(q(x,y)=0\), then \((WP1)\) and \((WP2)\) imply that \(x=y\), but the converse need not be true.
-
\((P1)\) implies \((WP1)\), but the converse need not be true.
-
\((P4)\) implies \((WP4)\), but the converse need not be true.
Example 1.1
([1])
If \(X=\{[a,b]:a,b\in \mathbb{R},a\leq b\}\), then \(q([a,b],[c,d])=\max \{b,d\}-\min \{a,c\}\) is a weak partial metric.
Each weak partial metric q on X generates a \(T_{0}\) topology \(\tau _{q}\) on X. Topology \(\tau _{q}\) has as a base the family of open q-balls \(\{B_{q}(x,\epsilon ):x\in X,\epsilon >0\}\), where \(B_{q}(x,\epsilon )=\{y\in X:q(x,y)< q(x,x)+\epsilon \}\) for all \(x\in X\) and \(\epsilon >0\).
If q is a weak partial metric on X, then the function \(q^{s}:X\times X\to [0,\infty )\) given by \(q^{s}(x,y)=q(x,y)-\frac{1}{2}[q(x,x)+q(y,y)]\) defines a metric on X.
Definition 1.3
Let \((X,q)\) be a weak partial metric space.
(i)
A sequence \(\{x_{n}\}\) in \((X, q)\) converges to a point \(x\in X\), with respect to \(\tau _{q}\) if \(q(x,x)=\lim_{n\rightarrow \infty }q(x,x_{n})\);
(ii)
A sequence \(\{x_{n}\}\) in X is said to be a Cauchy sequence if \(\lim_{n, m\rightarrow \infty } q(x_{n}, x_{m})\) exists and is finite;
(iii)
\((X, q)\) is called complete if every Cauchy sequence \(\{x_{n}\}\) in X converges to \(x\in X\) with respect to topology \(\tau _{q}\).
Clearly, we also have the following:
Lemma 1.1
Let
\((X,q)\)
be a weak partial metric space. Then
(a)
A sequence
\(\{x_{n}\}\)
in
X
is Cauchy sequence in
\((X,q)\)
if and only if it is a Cauchy sequence in the metric space
\((X, q^{s})\);
(b)
\((X,q)\)
is complete if and only if the metric space
\((X,q^{s})\)
is complete. Furthermore, a sequence
\(\{x_{n}\} \)
converges in
\((X,q^{s})\)
to a point
\(x\in X\)
if and only if
$$ \lim_{ n, m \to \infty } q(x_{n}, x_{m}) = \lim_{n \to \infty } q(x_{n}, x) = q(x, x). $$
(1.1)
Let \((X,q)\) be a weak partial metric space. Let \(CB^{q}(X)\) be the family of all nonempty closed bounded subsets of \((X,q)\). Here, the boundedness is given as follows: E is a bounded subset in \((X,q)\) if there exist \(x_{0}\in X\) and \(M\geq 0\) such that, for all \(a\in E\), we have \(a\in B_{q}(x_{0},M)\), that is, \(q(x_{0},a)< q(a,a)+M\).
For \(E,F\in \mathit{CB}^{q}(X)\) and \(x\in X\), define and Now, \(q(x,E)=0\) implies \(q^{s}(x,E)=0\), where \(q^{s}(x,E)=\inf \{q^{s}(x,a),a\in E\}\).
$$ q(x,E)=\inf \bigl\{ q(x,a),a\in E \bigr\} ,\qquad \delta _{q}(E,F)=\sup \bigl\{ q(a,F):a\in E \bigr\} $$
$$ \delta _{q}(F,E)=\sup \bigl\{ q(b,E):b\in F \bigr\} . $$
Remark 1.1
([1])
Let \((X,q)\) be a weak partial metric space, and let E be a nonempty set in (\(X,q\)). Then where E̅ denotes the closure of E with respect to the weak partial metric q.
$$ a\in \overline{E} \quad \mbox{if and only if}\quad q(a,E)=q(a,a), $$
(1.2)
Note that E is closed in \((X,q)\) if and only if \(E=\overline{E}\).
First, we study properties of the mapping \(\delta _{q}:\mathit{CB}^{q}(X)\times \mathit{CB}^{q}(X)\rightarrow[ 0,\infty )\).
Proposition 1.1
([1])
Let (\(X, q\)) be a weak partial metric space,We have the following:
(i)
\(\delta _{q}(E,E)=\sup \{q(a,a):a\in E\}\);
(ii)
\(\delta _{q}(E,E)\leq \delta _{q}(E,F)\);
(iii)
\(\delta _{q}(E,F)=0\)
implies
\(E\subseteq F\);
(iv)
\(\delta _{q}(E,F)\leq \delta _{q}(E,H)+\delta _{q}(H,F)\)
for all
\(E ,F, H\in \mathit{CB}^{q}(X)\).
Definition 1.4
([1])
Let \((X,q)\) be a weak partial metric space. For \(E,F\in \mathit{CB}^{q}(X)\), define
$$ \mathcal{H}_{q}^{+}(E,F)=\frac{1}{2} \bigl\{ \delta _{q}(E,F)+\delta _{q}(F,E) \bigr\} . $$
(1.3)
The following proposition is a consequence of Proposition 1.1.
Proposition 1.2
([1])
Let
\((X, q)\)
be a weak partial metric space. Then, for all
\(E, F, H\in \mathit{CB}^{q}(X)\), we have
(wh1)
\(\mathcal{H}_{q}^{+}(E,E)\leq \mathcal{H}_{q}^{+}(E,F)\);
(wh2)
\(\mathcal{H}_{q}^{+}(E,F)=\mathcal{H}_{q}^{+}(F,E)\);
(wh3)
\(\mathcal{H}_{q}^{+}(E,F)\leq \mathcal{H}_{q}^{+}(E,H)+\mathcal{H}_{q}^{+}(H,F)\).
The mapping \(\mathcal{H}_{q}^{+}:\mathit{CB}^{q}(X)\times \mathit{CB}^{q}(X)\rightarrow[ 0,+\infty )\), is called the \(\mathcal{H}^{+}\)-type Pompeiu–Hausdorff metric induced by q.
Definition 1.5
([1])
Let \((X,q)\) be a complete weak partial metric space. A multivalued map \(T:X\rightarrow \mathit{CB}^{q}(X)\) is called an \(\mathcal{H}_{q}^{+}\)-contraction if
\((1^{\circ })\)
there exists k in \((0, 1)\) such that
$$ \mathcal{H}_{q}^{+} \bigl(Tx\setminus \{x\}, Ty\setminus \{y \} \bigr) \le k q(x, y) \quad \mbox{for every } x, y \in X, $$
(1.4)
\((2^{\circ })\)
for all x in \(X, y\) in Tx, and \(\epsilon > 0\), there exists z in Ty such that
$$ q(y, z) \leq \mathcal{H}_{q}^{+}(Ty, Tx) + \epsilon . $$
(1.5)
Beg and Pathak [1] proved the following fixed point theorem.
Theorem 1.1
([1])
Let
\((X,q)\)
be a complete weak partial metric space. Every
\(\mathcal{H}_{q}^{+}\)-type multivalued contraction mapping
\(T:X\rightarrow \mathit{CB}^{q}(X)\)
with Lipschitz constant
\(k<1\)
has a fixed point.
In this paper, we generalize the concept of \(\mathcal{H}_{q}^{+}\)-type multivalued contractions by introducing \(\mathcal{H}_{q}^{+}\)-type Suzuki mult-valued contraction mappings.
2 Fixed point results
First, let \(\psi : [0, 1)\rightarrow (0, 1]\) be the nonincreasing function Now, we state a fixed point result for \(\mathcal{H}_{q}^{+}\)-type Suzuki multivalued contraction mappings.
$$ \psi (r)= \textstyle\begin{cases} 1& \mbox{if } 0\leq r < \frac{1}{2}, \\ 1-r& \mbox{if } \frac{1}{2}\leq r < 1. \end{cases} $$
(2.1)
Theorem 2.1
Let
\((X,q)\)
be a complete weak partial metric space, and let
\(F : X \to \mathcal{CB}^{q}(X)\)
be a multivalued mapping. Let
\(\psi : [0, 1)\rightarrow (0, 1]\)
be the nonincreasing function defined by (2.1). Suppose that there exists
\(0 \leq s < 1\)
such that
T
satisfies the condition
for all
\(x,y \in X\). Suppose also that, for all
x
in
\(X, y\)
in
Fx, and
\(t > 1\), there exists
z
in
Fy
such that
Then
F
has a fixed point.
$$ \psi (s)q(x,Fx)\leq q(x,y) \quad \textit{implies } \mathcal{H}_{q}^{+} \bigl(Fx\setminus \{x\},Fy\setminus \{y\} \bigr)\leq s q(x,y) $$
(2.2)
$$ q(y, z) \le t \mathcal{H}_{q}^{+}(Fy, Fx). $$
(2.3)
Proof
Let \(s_{1}\in (0, 1)\) be such that \(0\leq s \leq s_{1} < 1\) and \(w_{0}\in X\). Since \(Fw_{0}\) is nonempty, it follows that if \(w_{0} \in Fw_{0}\), then the proof is completed. Let \(w_{0}\notin Fw_{0}\). Then there exists \(w_{1}\in Fw_{0}\) such that \(w_{1}\neq w_{0}\).
Similarly, there exists \(w_{2}\in Fw_{1}\) such that \(w_{1}\neq w_{2}\), and from (2.3) we have Since from (2.2) and (2.4) we get By repeating this process n times we obtain Hence Now we prove that \(\{w_{n}\}\) is a Cauchy sequence in \((X,q^{s})\). For all \(m\in N\), we have Hence This implies that \(\{w_{n}\}\) is a Cauchy sequence in the complete metric space \((X,q^{s})\). It follows that there exists \(u\in X \) such that From \((WP2)\) we obtain By taking the limit as \(n\to \infty \) from (2.6) we get Also, from (2.7) and (2.10) we find Therefore Now, we prove that Since \(\lim_{n\to \infty } q(w_{n},u)=0\), there exists \(n_{0}\in \mathbb{N}\) such that Then This implies that Since \(w_{n+1}\in Fw_{n}\), we have By taking the limit as \(n\to \infty \) we get Also, since and we have From (2.14) and (2.15) we find that We claim that If \(x=u\), then at that point, this clearly holds. So, let \(x\neq u\). Then for every positive integer \(n\in \mathbb{N}\), there exists \(y_{n} \in Fx \) such that Therefore From (2.16) and (2.17) we get Hence This implies that Finally, we show that \(u\in Fu\). For this, We deduce that \(q(u,u)=q(u,Fu)=0\). Since Fu is closed, \(u\in \overline{Fu}=Fu\). □
$$ q(w_{1},w_{2})\leq \frac{1}{\sqrt{s_{1}}}H^{+}_{q}(Fw_{0},Fw_{1}). $$
(2.4)
$$ \psi (s)q(w_{1},Fw_{1})\leq q(w_{1},Fw_{1}) \leq q(w_{1},w_{2}), $$
$$\begin{aligned} q(w_{1},w_{2})&\leq \frac{1}{\sqrt{s_{1}}}H^{+}_{q}(Fw_{0},Fw_{1})\leq \frac{1}{\sqrt{s_{1}}}H^{+}_{q} \bigl(Fw_{0} \setminus \{w_{0}\},Fw_{1}\setminus \{w_{1}\} \bigr) \\ & \leq \frac{1}{\sqrt{s_{1}}}. s .q(w_{0},w_{1})< \sqrt{s_{1}}. q(w_{0},w_{1}). \end{aligned}$$
$$ q(w_{n},w_{n+1})\leq (\sqrt{s_{1}})^{n} \cdot q(w_{0},w_{1}). $$
(2.5)
$$ \lim_{n\to \infty }q(w_{n},w_{n+1})=0. $$
(2.6)
$$\begin{aligned} q^{s}(w_{n},w_{n+m})&= q(w_{n},w_{n+m})- \frac{1}{2} \bigl[q(w_{n},w_{n})+q(w_{n+m},w_{n+m}) \bigr] \\ &\leq q(w_{n},w_{n+m}) \\ &\leq q(w_{n},w_{n+1})+q(w_{n+1},w_{n+2})+ \cdots+q(w_{n+m-1},w_{n+m}) \\ & \leq \bigl[(\sqrt{s_{1}})^{n}+(\sqrt{s_{1}})^{n+1}+ \cdots+(\sqrt{s_{1}})^{n+m-1} \bigr]q(w_{0},w_{1}) \\ &\leq (\sqrt{s_{1}})^{n}\frac{1}{1-\sqrt{s_{1}}}q(w_{0},w_{1}). \end{aligned}$$
$$ \lim_{n\to \infty }q^{s}(w_{n},w_{n+m})=0. $$
(2.7)
$$ \lim_{n\to \infty }q(w_{n},u)=\lim _{n,m\to \infty }q(w_{n},w_{m})=q(u,u). $$
(2.8)
$$ \frac{1}{2} \bigl[q(w_{n},w_{n})+q(w_{n+1},w_{n+1}) \bigr]\leq q(w_{n},w_{n+1}). $$
(2.9)
$$ \lim_{n\to \infty }q(w_{n},w_{n})= \lim_{n\to \infty }q(w_{n+1},w_{n+1})=\lim _{n\to \infty }q(w_{n},w_{n+1})=0. $$
(2.10)
$$ \lim_{n\to \infty }q^{s}(w_{n},w_{n+m})=0= \lim_{n\to \infty }q(w_{n},w_{n+m})-\frac{1}{2} \lim_{n\to \infty } \bigl[q(w_{n},w_{n})+q(w_{n+m},w_{n+m}) \bigr]. $$
(2.11)
$$ \lim_{n\to \infty }q(w_{n},w_{n+m})=0= \lim_{n\to\infty }q(w_{n},u)=q(u,u). $$
(2.12)
$$ q(u,Fx)\leq 2s q(u,x)\quad \mbox{for all } x\in X\setminus \{u\}. $$
(2.13)
$$ q(w_{n},u)\leq \frac{1}{3}q(x,u)\quad \mbox{for all } n\geq n_{0}. $$
$$\begin{aligned} \psi (s)q(w_{n},Fw_{n}) &\leq q(w_{n},Fw_{n}) \\ &\leq q(w_{n},w_{n+1}) \\ &\leq q(w_{n},u)+q(u,w_{n+1}) \\ &\leq \frac{1}{3}q(u,x)+ \frac{1}{3}q(u,x) \\ &\leq q(u,x)-\frac{1}{3}q(u,x) \\ &\leq q(u,x)-q(u,w_{n})\leq q(x,w_{n}). \end{aligned}$$
$$ H^{+}_{q}(Fw_{n},Fx)\leq s q(w_{n},x). $$
$$\begin{aligned} q(w_{n+1},Fx)&\leq \delta _{q}(Fw_{n},Fx) \\ &\leq 2H^{+}_{q}(Fw_{n},Fx) \\ &\leq 2 s q(w_{n},x) \\ &\leq 2s \bigl[q(w_{n},u)+q(u,x) \bigr]. \end{aligned}$$
$$ \lim_{n\to \infty }q(w_{n+1},Fx)\leq 2s q(u,x). $$
(2.14)
$$ q(u,Fx)\leq q(u,w_{n+1})+q(w_{n+1},Fx) $$
$$ q(w_{n+1},Fx)\leq q(w_{n+1},w_{n})+q(w_{n},u)+q(u,Fx), $$
$$ \lim_{n\to \infty }q(w_{n+1},Fx)= q(u,Fx). $$
(2.15)
$$ q(u,Fx)\leq 2s q(u,x)\quad \mbox{for all } x\in X\setminus \{u \}. $$
(2.16)
$$ H^{+}_{q}(Fx,Fu)\leq s q(u,u) \quad \mbox{for all } x\in X. $$
$$ q(u,y_{n})\leq q(u,Fx)+\frac{1}{n} q(u,x). $$
$$\begin{aligned} q(x,Fx)&\leq q(x,y_{n}) \\ &\leq q(x,u)+q(u,y_{n}) \\ &\leq q(x,u)+q(u,Fx)+\frac{1}{n} q(x,u). \end{aligned}$$
(2.17)
$$\begin{aligned} q(x,Fx)&\leq q(u,x)+2s q(u,x)+\frac{1}{n} q(x,u) \end{aligned}$$
(2.18)
$$\begin{aligned} &= \biggl[1+2s+\frac{1}{n} \biggr]q(x,u). \end{aligned}$$
(2.19)
$$\begin{aligned} \frac{1}{1+2s+\frac{1}{n}} q(x,Fx)\leq q(u,x). \end{aligned}$$
$$\begin{aligned} H^{+}_{q}(Fu,Fx)\leq s q(u,x). \end{aligned}$$
$$\begin{aligned} q(u,Fu)&=\lim_{n\to \infty }q(w_{n+1},Fu) \\ &\leq \lim _{n\to \infty }\delta _{q}(Fw_{n},Fu) \\ &\leq 2\lim_{n\to \infty }H^{+}_{q}(Fw_{n},Fu) \\ &\leq 2 s\lim_{n\to \infty } q(w_{n},u)=0. \end{aligned}$$
We provide the following example.
Example 2.1
Let \(X=\{0,\frac{1}{2},1\}\) and define a weak partial metric \(q:X\times X \to [0,\infty ) \) as follows: \(q(0,0)=0\), \(q(\frac{1}{2}, \frac{1}{2})=\frac{1}{3}\), \(q(1,1)=\frac{1}{4}\), \(q(0,\frac{1}{2})=q(\frac{1}{2},0)=\frac{1}{2}\), \(q(\frac{1}{2},1)=q(1,\frac{1}{2})=\frac{3}{4}\), and \(q(1,0)=q(0,1)=1\). It is clear that \((X,q)\) is a weak partial metric space. Note that Then \((X,q)\) is not a partial metric space. Define the mapping \(F:X\to \mathit{CB}^{q}(X)\) by \(F(0)=F(\frac{1}{2})=\{0\}\) and \(F(1)=\{0,\frac{1}{3}\}\). Choose \(s=0.5\). From the definition of ψ we have \(\psi (s)=1\).
$$ q(1,0)=1\nleq q \biggl(1,\frac{1}{2} \biggr)+q \biggl(\frac{1}{2},0 \biggr)-q \biggl(\frac{1}{2},\frac{1}{2} \biggr)=\frac{3}{4}+ \frac{1}{2}-\frac{1}{3}. $$
To prove the contraction condition (2.2), we need the following cases:
Case 1. At \(x=0\), we have For \(y=0\), we have For \(y=\frac{1}{2}\), we get If f \(y=1\), then
$$ \psi (s)q \bigl(0,F(0) \bigr)=q(0,0)=0\leq q(0,y) \quad \mbox{for all } x\in X. $$
$$ H^{+}_{q} \bigl(F(0)\setminus \{0\},F(0)\setminus \{0\}\bigr)=H^{+}_{q}(\phi ,\phi )=0\leq s q(0,0). $$
$$ H^{+}_{q} \biggl(F(0)\setminus \{0\},F \biggl(\frac{1}{2} \Bigm\backslash \biggl\{ \frac{1}{2} \biggr\} \biggr)=H^{+}_{q} \bigl(\phi ,\{0\} \bigr)=0\leq s q \biggl(0, \frac{1}{2} \biggr). $$
$$ H^{+}_{q} \bigl(F(0)\setminus \{0\},F(1)\setminus \{1\} \bigr)=H^{+}_{q} \biggl(\phi , \biggl\{ 0,\frac{1}{2} \biggr\} \biggr)=0\leq s q(0,1). $$
Case 2. At \(x=\frac{1}{2}\), we have Similarly, if \(y=0, then\)
If \(y=1\), then
$$ \psi (s)q \biggl(\frac{1}{2},F \biggl(\frac{1}{2} \biggr) \biggr)=q \biggl(\frac{1}{2},0 \biggr)=\frac{1}{2}\leq q \biggl( \frac{1}{2},y \biggr)\quad\mbox{for all } y\in X\Bigm\backslash \biggl\{ \frac{1}{2} \biggr\} . $$
$$ H^{+}_{q} \biggl(F \biggl(\frac{1}{2} \Bigm\backslash \biggl\{ \frac{1}{2} \biggr\} ,F(0)\setminus \{0\} \biggr)=H^{+}_{q} \bigl(\{0\},\phi \bigr)=0\leq s q \biggl(\frac{1}{2},0 \biggr), $$
$$ H^{+}_{q} \biggl(F \biggl(\frac{1}{2} \biggr)\Bigm\backslash \biggl\{ \frac{1}{2} \biggr\} ,F(1)\setminus \{1\} \biggr)=H^{+}_{q}\biggl(\{0\}, \biggl\{ 0,\frac{1}{2} \biggr\} \biggr)=\frac{1}{4}< s q \biggl(\frac{1}{2},1 \biggr)=\frac{3}{8}. $$
Case 3. At \(x=1\), we have Again, if \(y=0\), then If \(y=\frac{1}{2}\), then Finally, we will enquire the condition (2.3) with \(t=2\). For this, we discuss the following situations:
$$ \psi (s)q \bigl(1,F(1) \bigr)=q \biggl(1,\frac{1}{2} \biggr)= \frac{3}{4} \leq q(1,y)\quad\mbox{for all } y\in X\setminus \{1\}. $$
$$ H^{+}_{q} \bigl(F(1)\setminus \{1\},F(0)\setminus \{0\} \bigr)=H^{+}_{q} \biggl( \biggl\{ 0,\frac{1}{2} \biggr\} , \phi \biggr)=0\leq s q(1,0). $$
$$ H^{+}_{q} \biggl(F(1)\setminus \{1\},F \biggl( \frac{1}{2} \Bigm\backslash \biggl\{ \frac{1}{2} \biggr\} \biggr)=H^{+}_{q} \biggl( \biggl\{ 0,\frac{1}{2} \biggr\} ,\{0\} \biggr)=\frac{1}{4}< s q \biggl(1,\frac{1}{2} \biggr)= \frac{3}{8}. $$
(i)
If \(x=0\) or \(x=\frac{1}{2}\), then \(y\in F(0)=F(\frac{1}{2})=\{0\}\). This yields that \(y=0\), so there exists \(z\in F(y)\) such that
$$ 0=q(y,z)\leq 2 H^{+}_{q} \bigl(F(x),F(y) \bigr). $$
(ii)
If \(x=1\), then \(y\in F(1)=\{0,\frac{1}{2}\}\). If \(y=0\), then \(z=0\), and condition (2.3) is satisfied.
Also, If \(y=\frac{1}{2}\), then \(z=0\), so that Therefore all conditions of Theorem 2.1 are satisfied, and the function F has a fixed point \(u=0\).
$$ \frac{1}{2}=q(y,z)= 2 H^{+}_{q} \biggl(F(1),F \biggl( \frac{1}{2} \biggr) \biggr)=\frac{1}{2}. $$
On the other hand, the result of Beg and Pathak [1] is not applicable. Indeed,
$$ H^{+}_{q} \bigl(F(1)\setminus \{1\},F(1)\setminus \{1\} \bigr)= \frac{1}{3}>\frac{1}{2} q(1,1)=\frac{1}{8}. $$
3 Applications
First, we present an application concerning a homotopy result for complete weak partial metric spaces.
Theorem 3.1
Let
\((X, q)\)
be a complete weak partial metric space, let
D
be an open subset of
X, and let
W
be a closed subset of
X
with
\(D\subset W\). Let
\(F : W \times [0, 1] \to \mathit{CB}^{q} (X)\)
be an operator satisfying:
(i)
\(x\notin F (x, t)\)
for each
\(x \in W \setminus D\)
and each
\(t\in [0, 1]\);
(ii)
there exists
\(s \in (0, \frac{1}{2})\)
such that, for each
\(t \in [0, 1]\)
and each
\(x, y \in W\), we have
$$ \psi (s) q \bigl(x,F(x,t) \bigr)\leq q(x,y)\Rightarrow H^{+}_{q} \bigl( F (x, t) \setminus \{x\}, F (y, t) \setminus \{y\} \bigr) \leq s q(x, y); $$
(iii)
for all
\(x \in W\), \(y \in F (x, t)\), and
\(h > 1\), there exists
\(z \in F (y, t)\)
such that
$$ q(y, z) \leq h H^{+}_{q} \bigl(F (y, t), F (x, t) \bigr); $$
(iv)
there exists a continuous function
\(\eta : [0, 1] \to \mathbb{R}\)
such that
for all
\(t_{1}, t_{2} \in [0, 1]\)
and
\(x \in W\);
$$ H^{+}_{q} \bigl( F (x, t_{1}) \setminus \{x\}, F (x, t_{2}) \setminus \{x\} \bigr) \leq s\bigl\vert \eta (t_{1}) -\eta (t_{2}) \bigr\vert $$
(v)
if
\(x \in F (x, t)\), then
\(F (x, t) = \{x\}\). Then
\(F (\cdot , 0)\)
has a fixed point if and only if
\(F (\cdot , 1)\)
has a fixed point.
Proof
Define the set Since \(F (\cdot , 0)\) has a fixed point, from condition (i), we get \(0\in \Delta \), so \(\Delta \neq\phi \). First, we want to show that Δ is an open set. Let \(t_{1}\in \Delta \) and \(x_{1}\in D\) be such that \(x_{1}\in F(x_{1},t_{1} )\). Since D is open in \((X,q)\), there exists \(r>0\) such that \(B(x_{1},r)\subset D\). Consider \(\epsilon =(\frac{1-2s}{2})(q(x_{1},x_{1})+r)>0\). Since η is continuous at \(t_{1}\), there exists \(\delta (\epsilon )>0\) such that \(\vert \eta (t)-\eta (t_{1}) \vert < \epsilon \) for all \(t\in (t_{1}-\delta (\epsilon ),t_{1}+\delta (\epsilon ))\).
$$ \Delta := \bigl\{ t\in [0,1] ;x\in F(x, t) \mbox{ for some } x\in D \bigr\} . $$
Let \(t\in (t_{1}-\delta (\epsilon ),t_{1}+\delta (\epsilon ))\) and \(x\in B(x_{1},r)=\{x\in X ;q(x_{1},x)\leq q(x_{1},x_{1})+r\}\). Since \(x_{1}\in F(x_{1},t_{1} )\), from \((WP2)\) we have Thus Therefore \(F(x,t)\subset B(x_{1},r)\). Since \(F(\cdot ,t):B(x_{1},r)\to \mathit{CB}^{q}(X)\) for each fixed \(t\in (t_{1}-\delta (\epsilon ),t-1+\delta (\epsilon ))\) and (ii) holds, all the hypotheses of Theorem 2.1 are satisfied. We conclude that \(F(\cdot ,t)\) has a fixed point in \(B(x_{1},r)\subset W\). This fixed point must be in D due to (i). Hence \((t_{1}-\delta (\epsilon ),t-1+\delta (\epsilon ))\subset \Delta \), and therefore Δ is open in \([0,1]\).
$$ \psi (s)q \bigl(x_{1},F(x_{1},t_{1} ) \bigr) \leq q(x_{1},x_{1})\leq q(x_{1},x)\quad\mbox{for all } x\in X. $$
$$\begin{aligned} q \bigl(x_{1},F(x,t) \bigr)&\leq 2H^{+}_{q} \bigl(F(x,t),F(x_{1},t_{1}) \bigr) \\ &\leq 2 \bigl[H^{+}_{q} \bigl(F(x,t),F(x,t_{1}) \bigr)+H^{+}_{q} \bigl(F(x,t_{1}),F(x_{1},t_{1}) \bigr) \bigr] \\ &=2 \bigl[H^{+}_{q} \bigl(F(x,t)\setminus \{x \},F(x,t_{1})\setminus \{x\} \bigr) +H^{+}_{q} \bigl(F(x,t_{1})\setminus \{x \},F(x_{1},t_{1})\setminus \{x_{1}\} \bigr) \bigr] \\ &\leq 2 \bigl[\bigl\vert \eta (t) - \eta (t_{1}) \bigr\vert +s q(x,x_{1}) \bigr] \\ &\leq 2 \bigl[\epsilon +s \bigl(q(x_{1},x_{1})+r \bigr) \bigr] \\ &\leq 2 \biggl[ \biggl(\frac{1-2s}{2} \biggr) \bigl(q(x_{1},x_{1})+r \bigr)+s \bigl(q(x_{1},x_{1})+r \bigr) \biggr] \\ &\leq q(x_{1},x_{1})+r. \end{aligned}$$
Second, we prove that Δ is closed in \([0,1]\). To show this, choose a sequence \(\{t_{n}\}\) in Δ such that \(t_{n}\to t^{*}\in [0,1]\) as \(n\to \infty \). We must show that \(t^{*} \in \Delta \). By the definition of Δ there exists \(x_{n} \in D\) with \(x_{n} \in F(x_{n},t_{n})\). Then This implies that, for all positive integers \(m, n\in \mathbb{N}\), using (v) and \((Wh3)\), we have This implies that Hence \(\lim_{n,m\to \infty }q(x_{n},x_{m})=0\). Therefore \(\{x_{n}\}\) is a Cauchy sequence in \((X, q)\). Since \((X, q)\) is complete, there exists \(x^{*}\in W\) such that On the other hand, we have Taking the limit as \(n\to \infty \) in the above inequality, we get It follows that \(x^{*}\in F(x^{*} ,t^{*} )\). Thus \(t^{*}\in \Delta \), and hence Δ is closed in \([0,1]\). By the connectedness of \([0,1]\) we have \(\Delta =[0,1]\).
$$ \psi (s)q \bigl(x_{n},F(x_{n},t_{n} ) \bigr)\leq q(x_{n},x_{n})\leq q(x_{n},x)\quad\mbox{for all } x \in X. $$
$$\begin{aligned} q(x_{n},x_{m}) &\leq 2H^{+}_{q} \bigl(F(x_{n},t_{n}), F (x_{m},t_{m}) \bigr) \\ & \leq 2H^{+}_{q} \bigl(F(x_{n},t_{n}), F (x_{n},t_{m}) \bigr)+2H^{+}_{q} \bigl(F (x_{n},t_{m}), F (x_{m},t_{m}) \bigr) \\ & =2H^{+}_{q} \bigl(F(x_{n},t_{n}) \setminus \{x_{n}\},F(x_{n},t_{m})\setminus \{x_{n}\} \bigr) \\ &\quad {} +2H^{+}_{q} \bigl(F(x_{n},t_{m}) \setminus \{x_{n}\},F(x_{m},t_{m})\setminus \{x_{m}\} \bigr) \\ & \leq 2s\bigl\vert \eta (t_{n})-\eta (t_{m}) \bigr\vert +2s q(x_{n},x_{m}). \end{aligned}$$
$$ q(x_{n},x_{m})\leq \frac{2s}{1-2s} \bigl(\bigl\vert \eta(t_{n})-\eta (t_{m}) \bigr\vert \bigr). $$
$$ q \bigl(x^{*} ,x^{*} \bigr) = \lim_{n\to \infty }q \bigl(x^{*},x_{n} \bigr)=\lim_{n,m\to \infty } q(x_{n},x_{m})=0. $$
$$\begin{aligned} q \bigl(x_{n},F \bigl(x^{*} ,t^{*} \bigr) \bigr)& \leq 2H^{+}_{q} \bigl(F (x_{n},t_{n}), F \bigl(x^{*} ,t^{*} \bigr) \bigr) \\ & \leq 2H^{+}_{q} \bigl(F (x_{n},t_{n}), F \bigl(x_{n},t^{*} \bigr)+2H^{+}_{q} \bigl(F \bigl(x_{n},t^{*} \bigr) \bigr), F \bigl(x^{*} ,t^{*} \bigr) \bigr) \\ &=2 H^{+}_{q} \bigl( F(x_{n},t_{n}) \setminus \{x_{n}\},F \bigl(x_{n},t^{*} \bigr) \setminus \{x_{n}\} \bigr) \\ &\quad {} +2H^{+}_{q} \bigl( F \bigl(x_{n},t^{*} \bigr)\setminus \{x_{n}\},F \bigl(x^{*} ,t^{*} \bigr)\setminus \bigl\{ x^{*} \bigr\} \bigr) \\ & \leq 2s\bigl\vert \eta (t_{n})-\eta \bigl(t^{*} \bigr) \bigr\vert +2sq \bigl(x_{n},x^{*} \bigr). \end{aligned}$$
$$ q \bigl(x^{*} ,F \bigl(x^{*} ,t^{*} \bigr) \bigr)= \lim_{n\to \infty } q \bigl(x_{n},F \bigl(x^{*} ,t^{*} \bigr) \bigr)=0. $$
The reverse implication easily follows by applying the same strategy. This completes the proof. □
Now, we give another application to the solvability of integral inclusions of Fredholm type. Let \(I=[0,1]\), and let \(C(I,\mathbb{R})\) be the space of all continuous functions \(f :I\to R\). Consider the weak partial metric on X given by for all \(x,y\in C(I,R)\) and \(\alpha >0\). We have \(q^{s}(x,y)=\sup_{t\in I} \vert x(t)-y(t) \vert \), so by Lemma 1.1
\((C(I,\mathbb{R}),q)\) is a complete weak partial metric space. Denote by \(P_{cv}(\mathbb{R})\) the family of all nonempty compact and convex subsets of \(\mathbb{R}\) and by \(P_{cl}(\mathbb{R})\) the family of all nonempty closed subsets of \(\mathbb{R}\).
$$ q(x,y)=\sup_{t\in I} \bigl\vert x(t)-y(t) \bigr\vert +\alpha $$
Theorem 3.2
Consider the integral inclusion of Fredholm type
Suppose that:
$$ h(t)\in f(t)+ \int ^{1}_{0} K \bigl(t,u,h(u) \bigr)\,du, \quad t\in [0,1]. $$
(3.1)
(i)
\(K:I\times I\times R\to P_{cv}(\mathbb{R})\)
is such that
\(K_{h}(t,u):=K(t,u,h(u))\)
is a lower semicontinuous for all
\((t,u)\in I\times I\)
and
\(h\in C(I,\mathbb{R})\),
(ii)
\(f \in C(I,R)\);
(iii)
for each
\(t\in I\), there exists
\(l(t,\cdot )\in L^{1}(I)\)
such that
\(\sup_{ t\in I} \int ^{1}_{0} l(t,u)\,du = \frac{s}{2}\)
with
\(s\in [0,1)\)
and
for all
\(t,u\in I\)
and all
\(h,r\in C(I,\mathbb{R})\).
$$ H^{+}_{q} \bigl( K \bigl(t,u,h(u) \bigr),K \bigl(t,u,r(u) \bigr) \bigr) \leq l(t,u) \Bigl(\sup_{u\in I}\bigl\vert h(u)-r(u) \bigr\vert +\alpha \Bigr) $$
Proof
Consider the multivalued operator \(T:C(I,R)\to P_{CL}(C(I,R))\) defined by for \(x\in C(I,\mathbb{R})\). For each \(K_{x}(t,u):I\times I \to P_{cv}(\mathbb{R})\), by the Michael selection theorem there exists a continuous operator \(k_{x}: I\times I\to \mathbb{R}\) such that \(k_{x}(t,u)\in K_{x}(t,u)\) for all \(t,u\in I\). This implies that \(f(t)+ \int ^{1}_{0} k_{x}(t,u)\,du\in Tx \), and so \(Tx\neq \emptyset \). It is easy to prove that Tx is closed, and so we omit the details (see also [18]). This implies that Tx is closed in \((C(I,\mathbb{R}),q)\).
$$ Tx(t)= \biggl\{ h\in C(I,\mathbb{R}) \mbox{ such that } h(t)\in f(t)+ \int ^{1}_{0} K \bigl(t,u,x(u) \bigr)\,du,t \in I \biggr\} $$
Now, we will show that T is \(H^{+}_{q}\)-type Suzuki multivalued contraction mapping. Let \(x_{1}, x_{2}\in C(I,\mathbb{R})\) and \(h\in Tx\). Then there exists \(k_{x_{1}}(t,u)\in K_{x_{1}}(t,u)\) with \(t,u\in I\) such that \(h(t)=f(t)+ \int ^{1}_{0} k_{x}(t,u)\,du,t\in I\). Also, by hypothesis (iii), Then there exists \(z(t,u)\in K_{x_{2}}(t,u)\) such that for all \(t,u\in I\). Now, we define the multivalued operator \(M(t,u)\) by for \(t,u\in I\). Since M is a lower semicontinuous operator, there exists a continuous operator \(k_{x_{2}} : I\times I \to \mathbb{R}\) such that \(k_{x_{2}}(t,u)\in M(t,u)\) for all \(t,u\in I\) and Therefore Since \(h(t)\in Tx_{1}\) is arbitrary, we have Similarly, we can get From (3.2) and (3.3) we have In particular, the previous inequality holds for any \(t\in I\), so that Thus all conditions of Theorem 2.1 are satisfied, and hence a solution of (3.1) exists. □
$$ H^{+}_{q} \bigl( K \bigl(t,u,x_{1}(u) \bigr),K \bigl(t,u,x_{2}(u) \bigr) \bigr) \leq l(t,u) \Bigl(\sup _{u\in I}\bigl\vert x_{1}(u)-x_{2}(u) \bigr\vert +\alpha \Bigr)\quad \forall t,u\in I. $$
$$ \bigl\vert k_{x_{1}}(t,u)-z(t,u) \bigr\vert +n \leq l(t,u) \bigl[ \bigl\vert x_{1}(u)-x_{2}(u) \bigr\vert + \alpha \bigr] $$
$$ M(t,u)=K_{x_{2}}(t,u)\cap \bigl\{ m\in \mathbb{R}, \bigl\vert k_{x_{1}}(t,u)-m \bigr\vert + \alpha \leq l(t,u) \bigl(\bigl\vert x_{1}(u)-x_{2}(u) \bigr\vert +\alpha \bigr) \bigr\} $$
$$ w(t)=f(t)+ \int ^{1}_{0} k_{x_{2}}(t,u)\,du\in f(t)+ \int ^{1}_{0} K \bigl(t,u,x_{2}(u) \bigr)\,du. $$
$$\begin{aligned} q \bigl(h(t),Tx_{2}(t) \bigr)&\leq q \bigl(h(t),w(t) \bigr) \\ &=\sup _{t\in I}\bigl\vert h(t)-w(t) \bigr\vert +\alpha \\ & =\sup_{t\in I}\biggl\vert \int ^{1}_{0} \bigl[k_{x_{1}}(t,u)-k_{x_{2}}(t,u) \bigr]\,du \biggr\vert +\alpha \\ & \leq \sup_{t\in I}\int ^{1}_{0} \bigl(\bigl\vert k_{x_{1}}(t,u)-k_{x_{2}}(t,u) \bigr\vert +\alpha -\alpha \bigr)\,du+\alpha \\ & \leq \sup_{t\in I}\int ^{1}_{0}l(t,u) \bigl[\bigl\vert x_{1}(u)-x_{2}(u) \bigr\vert + \alpha \bigr]\,du- \int ^{1}_{0}\alpha \,du+\alpha \\ & = \Bigl(\sup_{t\in I}\bigl\vert x_{1}(u)-x_{2}(u)\bigr\vert + \alpha \Bigr)\int ^{1}_{0}l(t,u)\,du \\ & \leq s q \bigl(x_{1}(t),x_{2}(t) \bigr). \end{aligned}$$
$$\begin{aligned} \delta _{q}(Tx_{1},Tx_{2})\leq s q(x_{1},x_{2}). \end{aligned}$$
(3.2)
$$\begin{aligned} \delta _{q}(Tx_{2},Tx_{1})\leq s q(x_{1},x_{2}). \end{aligned}$$
(3.3)
$$ H^{+}_{q}(Tx_{1},Tx_{2})= \frac{\delta _{q}(Tx_{1},Tx_{2})+\delta _{q}(Tx_{2},Tx_{1})}{2}\leq s q(x_{1},x_{2}). $$
$$ \psi (s)q(x_{1},Tx_{1})\leq q(x_{1},x_{2}). $$
4 Perspectives
In 2010, Romaguera [19] introduced the notions of 0-Cauchy sequences and 0-complete partial metric spaces and proved some characterizations of partial metric spaces in terms of completeness and 0-completeness. Adapting the same concepts, we introduce the concepts of 0-Cauchy sequences and 0-complete weak partial metric spaces.
Definition 4.1
Let \((X,q)\) be a weak partial metric space.
(i)
A sequence \(\{x_{n}\}\)in X is said to be 0-Cauchy if \(\displaystyle \lim_{n,m\rightarrow \infty } q(x_{n}, x_{m})=0\);
(iii)
\((X,q)\) is called 0-complete if every 0-Cauchy sequence \(\{x_{n}\}\) in X converges to \(x\in X\) such that \(q(x,x)=0\).
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