1 Introduction

The notion of a partial metric space (PMS) was introduced by Matthews [1] in 1992 (see also [2]). The PMS is a generalization of the usual metric space in which \(d(x,x)\) is no longer necessarily zero. Recently, many authors have focused on the PMS and its topological properties(see for example [3–15]). Partial metric spaces have extensive application potential in the research area of computer domains and semantics (see [15–18]).

A partial metric is a function \(p:X\times X\rightarrow[0,\infty)\) satisfying the following conditions:

  1. (a)

    \(p(x,y)=p(y,x)\) (symmetry),

  2. (b)

    \(x=y\Longleftrightarrow p(x,x)=p(x,y)=p(y,y)\) (equality),

  3. (c)

    \(p(x,x)\leq p(x,y)\) (small self-distances),

  4. (d)

    \(p(x,y)\leq p(x,z)+p(y,z)-p(z,z)\) (triangularity),

for all \(x,y,z\in X\). Then \((X,p)\) is called a partial metric space.

Each partial metric p on X generates a \(T_{0}\) topology \(\tau_{p}\) on X with a base of the family of open p-balls \(\lbrace B_{p}(x,\varepsilon):x\in X,\varepsilon>0\rbrace\), where

$$B_{p}(x,\varepsilon)=\bigl\lbrace y\in X:p(x,y)< p(x,x)+\varepsilon \bigr\rbrace , $$

for all \(x\in X\) and \(\varepsilon>0\). For a partial metric p on X, the function \(d_{p}:X\times X\rightarrow[0,\infty)\) given by

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

is a (usual) metric on X. Another metric on X induced by p is defined in [19] as \(d(x,y)=p(x,y)\) whenever \(x\neq y\) and \(d(x,y)=0\) whenever \(x=y\).

Some topological concepts and basic results on a PMS are defined as follows.

A sequence \(\{ x_{n} \}_{n\geq1}\) in a PMS \((X,p)\) converges to \(x\in X\) if and only if \(p(x,x)=\lim_{n\rightarrow\infty} p(x, x_{n})\).

A sequence \(\{ x_{n} \}_{n\geq1}\) is called a Cauchy sequence if and only if \(\lim_{n,m\rightarrow\infty}p(x_{n},x_{m})\) exists and is finite.

A PMS \((X,p)\) is said to be complete whenever every Cauchy sequence \(\{ x_{n} \}_{n\geq1}\) in X converges to a point x with respect to \(\tau_{p}\), that is, \(p(x,x)=\lim _{n,m\rightarrow\infty}p(x_{n},x_{m})\).

Lemma 1.1

([3])

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

  1. (i)

    A sequence \(\{ x_{n} \}_{n\geq1}\) is Cauchy in a PMS \((X,p)\) if and only if \(\{ x_{n} \}_{n\geq1}\) is Cauchy in the metric space \((X,d_{p})\).

  2. (ii)

    A PMS \((X,p)\) is complete if and only if the metric space \((X,d_{p})\) is complete. Moreover,

    $$\lim_{n\rightarrow\infty} d_{p}(x,x_{n})=0\quad \Longleftrightarrow\quad p(x,x)=\lim_{n\rightarrow\infty} p(x,x_{n})= \lim_{n,m\rightarrow\infty} p(x_{n},x_{m}). $$

An interesting property of partial metric spaces is the nonuniqueness of limits of sequences. To emphasize this property we consider the following example.

Example 1.1

Let \(X=[0,\infty)\) and define a partial metric p on X as

$$p(x,y)=\max\{x,y\}. $$

Consider the sequence \(\{x_{n}\}=\{1+\frac{1}{n}\}\). Notice that

$$\lim_{n\to\infty}p(x_{n},1)=\lim_{n\to\infty} \max\biggl\{ 1+\frac{1}{n},1\biggr\} =\lim_{n\to\infty} 1+ \frac{1}{n}=1. $$

Also

$$\lim_{n\to\infty}p(x_{n},2)=\lim_{n\to\infty} \max\biggl\{ 1+\frac{1}{n},2\biggr\} =\lim_{n\to\infty} 2=2. $$

Moreover, for any \(a\geq1\) we have

$$\lim_{n\to\infty}p(x_{n},a)=a. $$

In what follows, we introduce the notions, notations, and assumptions used in the discussion. Throughout this paper, we suppose that \((X,p)\) is a partial metric space. We denote the family of all nonempty subsets of X by \(2^{X}\), the family of all closed subsets of X by \(C(X)\) and the family of all closed and bounded subsets of X by \(\mathit{CB}(X)\). The partial Hausdorff distance \(H_{p}\) on \(\mathit{CB}(X)\) was introduced by Aydi et al. [20] as follows:

$$ H_{p}(A,B)=\max\Bigl\{ \sup_{a\in A} p(a,B), \sup _{b\in B} p(b, A)\Bigr\} , $$
(1)

for all \(A,B\in\mathit{CB}(X)\), where

$$ p(x,A)=\inf_{a\in A} p(x,a). $$
(2)

Let \(T:X\rightarrow2^{X}\) be a multi-valued function (multifunction). We denote the set of fixed points of T by \(F(T)\), i.e.,

$$ F(T)=\{x\in X: x\in Tx\}. $$
(3)

Lemma 1.2

([4])

Let \((X, p)\) be a partial metric space, \(A \subseteq X\), and \(x \in X\). Then \(x \in\overline{A}\) if and only if \(p(x,A) = p(x,x)\).

In 2012, Aydi et al. [20] proved the following fixed point theorem on partial metric space.

Theorem 1.3

Let \((X,p)\) be a complete partial metric space and \(T:X\rightarrow \mathit{CB}(X)\) a multifunction. Suppose that there exist \(k\in(0,1)\) such that

$$ H_{p}(Tx,Ty)\leq kp(x,y), $$
(4)

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

Some fixed point theorems for multifunctions on metric space are given next (see [6, 21, 22]).

Theorem 1.4

Let \((X, d)\) be a complete metric space and \(T : X \rightarrow \mathit{CB}(X)\) a multifunction. Assume that there exists \(r \in[0, 1)\) such that

$$ \frac{1}{1+r} d(x, Tx) \leq d(x, y)\quad\textit{implies} \quad H(Tx, Ty) \leq rd(x, y), $$
(5)

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

Theorem 1.5

Let \((X,d)\) be a complete metric space and \(T : X \rightarrow C(X)\) a multifunction. Assume that there exist \(a, b, c\in[0,1)\) such that \(a+b+c<1 \) and

$$ \frac{(1-b-c)}{1+a}d(x, Tx)\leq d(x,y) $$
(6)

implies

$$ H(Tx, Ty) \leq ad(x,y)+bd(x,Tx)+cd(y, Ty), $$
(7)

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

The aim of this paper is to provide a new, more general condition for the multifunction T which guarantees the existence of its fixed point. Our results generalize some of the existing ones.

In what follows, we consider two classes of functions, namely, \(R_{1}\) and \(R_{2}\) as defined below.

Definition 1.2

Let \(R_{1}\) be the set of all continuous functions \(g:\left.[0,\infty )\right.^{5}\rightarrow[0,\infty)\), satisfying the conditions:

  1. (i)

    \(g(t,t,t,2t,t)< t\), for all \(t\in[0,\infty)\),

  2. (ii)

    g is subhomogeneous, i.e., \(g(\alpha x_{1},\alpha x_{2},\alpha x_{3},\alpha x_{4},\alpha x_{5})\leq\alpha g(x_{1},x_{2},x_{3},x_{4},x_{5})\), for all \(\alpha\geq0\),

  3. (iii)

    if \(x_{i},y_{i}\in[0,\infty)\), \(x_{i}\leq y_{i} \) for \(i=1,\ldots ,5 \) we have \(g(x_{1},x_{2},x_{3},x_{4},x_{5})\leq g(y_{1},y_{2},y_{3},y_{4},y_{5})\).

Let \(R_{2}\) be the set of all continuous function \(g:\left.[0,\infty )\right.^{5}\rightarrow[0,\infty)\) satisfying the following conditions:

  1. (a)

    \(g(t,t,t,t,t)< t\), for all \(t\in[0,\infty)\),

  2. (b)

    g is subhomogeneous,

  3. (c)

    if \(x_{i},y_{i}\in[0,\infty)\), \(x_{i}\leq y_{i} \) for \(i=1,\ldots ,5 \) we have \(g(x_{1},x_{2},x_{3},x_{4},x_{5})\leq g(y_{1},y_{2},y_{3},y_{4},x_{5})\),

  4. (d)

    for all \(0\leq a\leq x_{4}\), \(g(x_{1},x_{2},x_{3},x_{4}-a,a)= g(x_{1},x_{2},x_{3},x_{4}, 0)\).

Remark 1.1

It is easy to see that if \(g\in R_{1}\), then \(g(1,1,1,2,1)=h\in(0,1)\). Indeed, if \(g\in R_{1}\), the conditions (i) and (ii) give

$$ g(t,t,t,2t,t)\leq tg(1,1,1,2,1)< t, $$
(8)

which implies \(g(1,1,1,2,1)=h\in(0,1)\). In addition, if \(g\in R_{2}\), then by a similar argument we observe that \(g(1,1,1,1,1)=h\in(0,1)\)

Examples of functions from both classes are given below.

Example 1.3

The function \(g(x_{1},x_{2},x_{3},x_{4},x_{5})=k \max\{x_{i}\} _{i=1}^{5}\) for \(k\in(0,\frac{1}{2})\) is in class \(R_{1}\).

Example 1.4

The function \(g(x_{1},x_{2},x_{3},x_{4},x_{5})=k \max\{x_{1},x_{2},x_{3},\frac {x_{4}+x_{5}}{2}\}\) for \(k\in(0,1)\) belongs to \(R_{2}\).

The following results are quite trivial.

Proposition 1.6

If \(g \in{R_{1}} \) and \(u,v \in[0,\infty)\) are such that

$$ u\leq\max\bigl\{ g(v,u,v,u,v),g(v,u,v,v+u,v)\bigr\} , $$
(9)

then \(u\leq hv\), where \(h=g(1,1,1,2,1)\).

Proof

From (iii) it is clear that \(g(v,u,v,u,v)\leq g(v,u,v,v+u,v)\) and hence \(u\leq\max\{g(v,u,v, u,v),g(v,u,v,v+u,v)\}=g(v,u,v,v+u,v)\). If \(v < u\), then

$$u \leq g(v, u, v, v + u, v) \leq g(u, u, u, 2u, u) \leq u g(1, 1, 1, 2, 1) = hu < u, $$

which is a contradiction. Thus \(u \leq v\), which implies

$$u \leq g(v,u,v,v+u, v) \leq g(v, v, v, 2v, v) \leq v g(1, 1, 1, 2, 1) = hv. $$

 □

Proposition 1.7

If \(g \in{R_{2}} \) and \(u,v \in[0,\infty)\) are such that

$$ u\leq\max\bigl\{ g(v, u, v, u+v, 0),g(v, u, v, u, v)\bigr\} , $$
(10)

then \(u\leq hv\) where \(h=g(1,1,1,1,1)\).

Proof

Let \(\max\{g(v, u, v, u+v, 0),g(v, u, v, u, v)\}=g(v, u, v, u+v, 0)\). If \(v < u\), then (d) implies

$$u \leq g(v, u, v, v + u, 0) \leq g(u, u, u, 2u, 0)\leq ug(1, 1, 1, 2, 0) = u g(1,1,1,1,1)=hu < u, $$

which is a contradiction. Thus \(u \leq v\), and hence,

$$u \leq g(v, u, v, v+u, 0) \leq g(v, v, v, 2v, 0) \leq vg(1, 1, 1, 2, 0) = v g(1,1,1,1,1)=hv. $$

Let \(\max\{g(v, u, v, u+v, 0),g(v, u, v, u, v)\}=g(v, u, v, u, v)\). If \(v < u\), then

$$u \leq g(v, u, v, u, v) \leq g(u, u, u, u, u)\leq u g(1,1,1,1,1)=hu < u. $$

This contradicts our assumption, that is, we should have \(u \leq v\). Then

$$u \leq g(v, u, v, u, v) \leq g(v,v,v,v,v) \leq v g(1,1,1,1,1)=hv, $$

which completes the proof. □

2 Main results

We state and proof our main results in this section.

Lemma 2.1

Let \((X,p)\) be a partial metric space and \(T,S:X\rightarrow C(X)\) be two multifunctions. Suppose that there exist \(\alpha\in(0,\infty)\) and \(g\in{R_{1}\cup R_{2}}\) such that \(\alpha p(x,Tx)\leq p(x,y)\) or \(\alpha p(y,Sy)\leq p(x,y)\) implies

$$ H_{p}(Tx,Sy)\leq g\bigl(p(x,y),p(y,Sy),p(x,Tx),p(x,Sy),p(y,Tx) \bigr), $$
(11)

for all \(x,y\in X\). Then for every \(x\in F(T)\cup F(S)\) we have \(p(x,x)=0\).

Proof

Without loss of generality, we can suppose that \(x\in Tx\). Then \(p(x,Tx)=p(x,x)\) and hence

$$\begin{aligned} p(x,Sx) \leq&H_{p}(Tx,Sx)\leq g\bigl(p(x,x), p(x,Sx), p(x,Tx), p(x,Sx), p(x,Tx)\bigr) \\ \leq& g\bigl(p(x,x), p(x,Sx), p(x,x), p(x,Sx), p(x,x)\bigr). \end{aligned}$$
(12)

By using Proposition 1.6 if \(g\in R_{1}\) or Proposition 1.7 if \(g\in R_{2}\), we have

$$p(x, x)\leq p(x, Sx) \leq h p(x, x). $$

However, since \(h<1\) we have \(p(x, x)=0\). □

Lemma 2.2

Let \((X, p)\) be a partial metric space and \(T,S:X\rightarrow C(X)\) be two multifunctions. Suppose that there exist \(\alpha\in(0,\infty)\) and \(g\in{R_{1} \cup R_{2}}\) such that \(\alpha p(x,Tx)\leq p(x,y)\) or \(\alpha p(y,Sy)\leq p(x,y)\) implies

$$H_{p}(Tx,Sy)\leq g\bigl(p(x,y),p(y,Sy),p(x,Tx),p(x,Sy),p(y,Tx)\bigr), $$

for all \(x,y\in X\). Then \({F}(T)={F}(S)\).

Proof

If \(x\in Tx\), then \(p(x,Tx)=p(x,x)=0\) by Lemma 2.1. Hence,

$$\begin{aligned} p(x,Sx) \leq& H_{p}(Tx,Sx)\leq g\bigl(p(x,x),p(x,Sx),p(x,Tx),p(x,Sx),p(x,Tx) \bigr) \\ \leq&g\bigl(p(x,x),p(x,Sx),p(x,x),p(x,Sx),p(x,x)\bigr) \\ \leq&g\bigl(0,p(x,Sx),0,p(x,Sx),0\bigr). \end{aligned}$$
(13)

By using Proposition 1.6 whenever \(g\in R_{1}\) or Proposition 1.7 in case \(g\in R_{2}\), we have \(p(x,Sx)\leq h0=0\), and thus \(x\in F(S)\). Thus, \(F(T)\subseteq F(S)\). Similarly, we can show that \(F(S)\subseteq F(T)\), which completes the proof. □

In what follows, we state our main existence result.

Theorem 2.3

Let \((X,p)\) be a complete partial metric space and \(T,S:X\rightarrow C(X)\) be two multifunctions. Suppose that there exist \(g\in{R_{1}\cup R_{2}} \) and \(\alpha\in(0,1)\), such that \(\alpha (h+1)\leq1\) where \(h=g(1,1,1,2,1)\) if \(g\in R_{1}\) and \(h=g(1,1,1,1,1)\) if \(g\in R_{2}\). Suppose also that \(\alpha p(x,Tx)\leq p(x,y)\) or \(\alpha p(y,Sy)\leq p(x,y)\) implies

$$ H_{p}(Tx,Sy)\leq g\bigl(p(x,y),p(y,Sy),p(x,Tx),p(x,Sy),p(y,Tx) \bigr), $$
(14)

for all \(x,y\in X\). Then \(F(T)=F(S)\) and \(F(T)\) is nonempty.

Proof

By Lemma 2.2 we already have \(F(T)=F(S)\). Fix arbitrary \(1>r>h\) and \(x_{0}\in X\) and choose \(x_{1}\in Tx_{0}\) such that \(\alpha p(x_{0},Tx_{0})< p(x_{0},x_{1})\). Then by the hypothesis of the theorem and condition (iii) or (c) in Definition 1.2, respectively, we have

$$\begin{aligned} p(x_{1},Sx_{1}) \leq& H_{p}(Tx_{0},Sx_{1}) \leq g\bigl(p(x_{0},x_{1}),p(x_{1},Sx_{1}),p(x_{0},Tx_{0}),p(x_{0},Sx_{1}),p(x_{1},Tx_{0}) \bigr) \\ \leq& g\bigl(p(x_{0},x_{1}),p(x_{1},Sx_{1}),p(x_{0},x_{1}),p(x_{0},Sx_{1}),p(x_{1},x_{1}) \bigr) \\ \leq& g\bigl(p(x_{0},x_{1}),p(x_{1},Sx_{1}),p(x_{0},x_{1}),p(x_{0},x_{1})+p(x_{1},Sx_{1})-p(x_{1},x_{1}),p(x_{1},x_{1}) \bigr) \\ \leq& g\bigl(p(x_{0},x_{1}),p(x_{1},Sx_{1}),p(x_{0},x_{1}),p(x_{0},x_{1})+p(x_{1},Sx_{1}),p(x_{1},x_{1}) \bigr), \end{aligned}$$

where obviously \(p(x_{0},Sx_{1})\leq p(x_{0},x_{1})+p(x_{1},Sx_{1})-p(x_{1},x_{1})\) due to triangle inequality in PMS. Suppose that \(g\in{R_{1}}\). Since

$$p(x_{1},Sx_{1})\leq g\bigl(p(x_{0},x_{1}),p(x_{1},Sx_{1}),p(x_{0},x_{1}),p(x_{0},x_{1})+p(x_{1},Sx_{1}),p(x_{0},x_{1}) \bigr), $$

then, by Proposition 1.6, we have

$$ p(x_{1},Sx_{1})\leq hp(x_{0},x_{1})< rp(x_{0},x_{1}). $$
(15)

Now let \(g\in{R_{2}}\). Since

$$p(x_{1},Sx_{1})\leq g\bigl(p(x_{0},x_{1}),p(x_{1},Sx_{1}),p(x_{0},x_{1}),p(x_{0},x_{1})+p(x_{1},Sx_{1})-p(x_{1},x_{1}),p(x_{1},x_{1}) \bigr), $$

and obviously \(0\leq p(x_{1},x_{1})\leq p(x_{0},x_{1})+p(x_{1},Sx_{1})\), we let \(a=p(x_{1},x_{1})\) and employ condition (d) in Definition 1.2 to get

$$p(x_{1},Sx_{1})\leq g\bigl(p(x_{0},x_{1}),p(x_{1},Sx_{1}),p(x_{0},x_{1}),p(x_{0},x_{1})+p(x_{1},Sx_{1}),0 \bigr). $$

Now by Proposition 1.7, we have

$$ p(x_{1},Sx_{1})\leq hp(x_{0},x_{1})< rp(x_{0},x_{1}). $$
(16)

We choose a number μ such that \(\inf_{y\in Sx_{1}} p(x_{1},y)=p(x_{1},Sx_{1})<\mu<r p(x_{0},x_{1})\). Thus there exists \(x_{2}\in Sx_{1}\) such that \(p(x_{1},x_{2})<\mu<rp(x_{0},x_{1})\). Since \(\alpha p(x_{1},Sx_{1})< p(x_{1},x_{2})\), by using (14) and the properties of the function g we have

$$\begin{aligned} p(x_{2},Tx_{2}) \leq&H_{p}(Sx_{1},Tx_{2}) \leq g\bigl(p(x_{1},x_{2}),p(x_{2},Tx_{2}),p(x_{1},Sx_{1}),p(x_{1},Tx_{2}),p(x_{2},Sx_{1}) \bigr) \\ \leq& g\bigl(p(x_{1},x_{2}),p(x_{2},Tx_{2}),p(x_{1},x_{2}),p(x_{1},x_{2})+p(x_{2},Tx_{2})-p(x_{2},x_{2}),p(x_{2},x_{2}) \bigr) \\ \leq&g\bigl(p(x_{1},x_{2}),p(x_{2},Tx_{2}),p(x_{1},x_{2}),p(x_{1},x_{2})+p(x_{2},Tx_{2}),p(x_{1},x_{2}) \bigr). \end{aligned}$$

Now, if \(g\in R_{1}\), using Proposition 1.6 and mimicking the proof of (15) we obtain

$$ p(x_{2},Tx_{2})\leq hp(x_{1},x_{2})< rp(x_{1},x_{2}). $$
(17)

If \(g\in R_{2}\), letting \(a=p(x_{2},x_{2})\) we get

$$\begin{aligned} p(x_{2},Tx_{2}) \leq& g\bigl(p(x_{1},x_{2}),p(x_{2},Tx_{2}),p(x_{1},x_{2}),p(x_{1},x_{2})+p(x_{2},Tx_{2})- p(x_{2},x_{2}),p(x_{2},x_{2})\bigr) \\ \leq&g\bigl(p(x_{1},x_{2}),p(x_{2},Tx_{2}),p(x_{1},x_{2}),p(x_{1},x_{2})+p(x_{2},Tx_{2}),0 \bigr), \end{aligned}$$

and hence Proposition 1.7 yields

$$ p(x_{2},Tx_{2})\leq hp(x_{1},x_{2})< rp(x_{1},x_{2}). $$
(18)

In a similar way, we can choose \(x_{3}\in Tx_{2}\) such that

$$ p(x_{2},x_{3})< rp(x_{1},x_{2})< r^{2}p(x_{0},x_{1}). $$
(19)

By continuing this process, we obtain a sequence \(\{x_{n}\}_{n\geq1}\) in X such that

$$ x_{2n-1}\in Tx_{2n-2}, \qquad x_{2n} \in Sx_{2n-1}, $$
(20)

which satisfies

$$ p(x_{n},x_{n+1})\leq r^{n} p(x_{0},x_{1}). $$
(21)

Then \(p(x_{2n},Tx_{2n})\leq h p(x_{2n-1},x_{2n})\) and \(p(x_{2n-1},Sx_{2n-1})\leq hp(x_{2n-2},x_{2n-1})\).

If \(x_{m}=x_{m+1}\) for some \(m\geq1\) where \(m=2k\), then

$$p(x_{2k},x_{2k})\leq p(x_{2k},Tx_{2k}) \leq p(x_{2k},x_{2k+1})=p(x_{2k},x_{2k}), $$

so \(p(x_{2k},Tx_{2k})=p(x_{2k},x_{2k})\), and hence \(x_{2k}\in Tx_{2k}\). Thus T and S have a fixed point. If \(m=2k+1\) in a similar way we find that T and S have a fixed point.

Suppose that \(x_{n}\neq x_{n+1}\), for all \(n\geq1\). Repeated application of the triangle inequality implies

$$\begin{aligned} p(x_{n},x_{n+m}) \leq&p(x_{n},x_{n+1})+p(x_{n+1},x_{n+2})+ \cdots+p(x_{n+m-1},x_{n+m}) \\ \leq&r^{n} p(x_{0},x_{1})+r^{n+1} p(x_{0},x_{1})+\cdots+r^{n+m-1} p(x_{0},x_{1}) \\ \leq&r^{n} p(x_{0},x_{1}) \bigl(1+r+r^{2}+ \cdots+r^{m-1}\bigr) \\ \leq& \frac{ r^{n}}{1-r} p(x_{0},x_{1}). \end{aligned}$$

Then we get

$$\lim_{n\rightarrow\infty} p(x_{n},x_{n+m})\rightarrow0, $$

and hence \(\{x_{n}\}_{n\geq1}\) is a Cauchy sequence in \((X,p)\). Regarding Lemma 1.1, \(\{x_{n}\}_{n\geq1}\) is also a Cauchy sequence in \((X,d_{p})\). Since \((X,p)\) is a complete partial metric space, by Lemma 1.1, \((X,d_{p})\) is also complete. Thus \(\{x_{n}\} _{n\geq1}\) converges to a limit, say, \(x\in X\), that is,

$$ \lim_{n\rightarrow\infty} d_{p}(x_{n},x)=0. $$
(22)

Notice that Lemma 1.1 yields

$$ p(x,x)=\lim_{n\rightarrow\infty} p(x_{n},x)=\lim _{n,m\rightarrow\infty} p(x_{n},x_{m})=0. $$
(23)

Now, we claim that for each \(n\geq1\) one of the relations

$$ \alpha p(x_{2n}, Tx_{2n})\leq p(x_{2n},x)\quad\text{or} \quad\alpha p(x_{2n+1},Sx_{2n+1}) \leq p(x_{2n+1},x) $$
(24)

holds. If for some \(n\geq1\) we have \(\alpha p(x_{2n}, Tx_{2n})> p(x_{2n},x)\) and \(\alpha p(x_{2n+1},Sx_{2n+1})> p(x_{2n+1},x)\) then

$$\begin{aligned} p(x_{2n},x_{2n+1}) \leq&p(x_{2n},x)+p(x,x_{2n+1}) \\ < & \alpha p(x_{2n},Tx_{2n})+\alpha p(x_{2n+1}, Sx_{2n+1}) \\ \leq& \alpha p(x_{2n}, x_{2n+1})+\alpha h p(x_{2n},x_{2n+1}). \end{aligned}$$

This results in \(\alpha(h+1)> 1\), which contradicts the initial assumption. Hence, our claim is proved. Observe that by the assumption of the theorem, for each \(n\geq1\) we have either

$$H_{p}(Tx_{2n},Sx)\leq g\bigl(p(x_{2n},x),p(x,Sx),p(x_{2n},Tx_{2n}),p(x_{2n},Sx),p(x,Tx_{2n}) \bigr) $$

or

$$H_{p}(Sx_{2n+1},Tx)\leq g\bigl(p(x_{2n+1},x),p(x,Tx),p(x_{2n+1},Sx_{2n+1}),p(x_{2n+1},Tx),p(x,Sx_{2n+1}) \bigr). $$

Therefore, one of the following cases holds.

Case (i). There exists an infinite subset \(I\subseteq\mathbb{N}\) such that

$$\begin{aligned} p(x_{2n+1},Sx) \leq&H_{p}(Tx_{2n},Sx) \\ \leq& g\bigl(p(x_{2n},x),p(x,Sx),p(x_{2n},Tx_{2n}),p(x_{2n},Sx),p(x,Tx_{2n}) \bigr), \end{aligned}$$

for all \(n\in I\).

Case (ii). There exists an infinite subset \(J\subseteq\mathbb{N}\) such that

$$\begin{aligned} p(x_{2n+2},Tx) \leq&H_{p}(Sx_{2n+1},Tx) \\ \leq&g\bigl(p(x_{2n+1},x),p(x,Tx),p(x_{2n+1},Sx_{2n+1}),p(x_{2n+1},Tx),p(x,Sx_{2n+1}) \bigr), \end{aligned}$$

for all \(n\in J\).

In Case (i), we get

$$\begin{aligned} p(x,Sx) \leq& p(x,x_{2n+1})+p(x_{2n+1},Sx) \\ \leq& p(x, x_{2n+1})+ g\bigl(p(x_{2n},x),p(x,Sx),p(x_{2n},Tx_{2n}),p(x_{2n},Sx),p(x,Tx_{2n}) \bigr) \\ \leq& p(x, x_{2n+1}) \\ &{}+g\bigl(p(x_{2n},x),p(x,Sx),p(x_{2n},x_{2n+1}),p(x_{2n},x)+p(x,Sx)-p(x,x),p(x,x_{2n+1}) \bigr), \end{aligned}$$

for all \(n\in I\). Continuity of g implies

$$ p(x,Sx)\leq g\bigl(0,p(x,Sx),0,0+p(x,Sx)-0,0\bigr). $$
(25)

Now by using Propositions 1.6 and 1.7, we have \(p(x,Sx)=0\), and thus \(x\in Sx\).

In Case (ii), we have

$$\begin{aligned} p(x,Tx) \leq&p(x,x_{2n+2})+p(x_{2n+2},Tx) \\ \leq& p(x,x_{2n+2})+g\bigl(p(x_{2n+1},x),p(x,Tx),p(x_{2n+1},Sx_{2n+1}),p(x_{2n+1},Tx),p(x,Sx_{2n+1}) \bigr) \\ \leq& p(x,x_{2n+2}) \\ &{}+ g\bigl(p(x_{2n+1},x),p(x,Tx),p(x_{2n+1},x_{2n+2}), \\ &p(x_{2n+1},x)+p(x,Tx)-p(x,x),p(x,x_{2n+2}) \bigr), \end{aligned}$$

for all \(n\in J\). Since g is continuous, we obtain

$$ p(x,Tx)\leq g\bigl(0,p(x,Tx),0,0+p(x,Tx)-0,0\bigr). $$
(26)

Again, by using Propositions 1.6 and 1.7, we have \(p(x,Tx)=0\), which gives \(x\in Tx\). This completes the proof. □

The following results are consequences of Theorem 2.3.

Theorem 2.4

Let \((X,p)\) be a complete partial metric space and \(T:X\rightarrow C(X)\) be a multifunction. Suppose that there exist \(\alpha\in(0,1)\) and \(g\in{R}\) with \(h=g(1,1,1,2,0)\) such that \(\alpha (h+1)\leq1\) and \(\alpha p(x,Tx)\leq p(x,y)\) implies

$$ H_{p} (Tx,Ty)\leq g\bigl(p(x,y),p(y,Ty),p(x,Tx),p(x,Ty),p(y,Tx) \bigr), $$
(27)

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

Corollary 2.5

Theorem  1.3 introduced in [20] is a special case of Theorem  2.4.

Proof

Define \(g\in{R_{1}}\) by \(g(x_{1},x_{2},x_{3},x_{4},x_{5})=kx_{1}\). □

Now we provide the partial metric versions of Theorems 1.4 and 1.5.

Theorem 2.6

Let \((X, p)\) be a complete partial metric space and \(T : X \rightarrow\mathit{CB}(X)\) be a multifunction. Assume that there exists \(r \in[0, 1) \) such that

$$ \frac{1}{1+r} p(x, Tx) \leq p(x, y)\quad\textit{implies} \quad H_{p}(Tx, Ty) \leq rp(x, y), $$
(28)

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

Proof

Define \(g\in{R_{1}}\) by \(g(x_{1},x_{2},x_{3},x_{4},x_{5})=rx_{1}\). Let \(\alpha =\frac{1}{1+r}\). Since \(h=r\) and \(\alpha(1+h)\leq1\), by using Theorem 2.3, T has a fixed point. □

Theorem 2.7

Let \((X,p)\) be a complete partial metric space and \(T : X \rightarrow C(X)\) be a multifunction. Assume that there exist \(a, b, c\in[0,1)\) such that \(a+ b+c<1 \) and

$$ \begin{aligned} &\frac{(1 - b - c)}{1+a}p(x, Tx)\leq p(x,y)\quad\textit {implies} \\ &H_{p}(Tx, Ty) \leq ap(x,y)+bp(x,Tx)+cp(y, Ty). \end{aligned} $$
(29)

Then T has a fixed point.

Proof

Define \(g \in{ R_{1}}\) by \(g(x_{1}, x_{2}, x_{3}, x_{4}, x_{5}) = ax_{1} + cx_{2} + bx_{3}\). Let \(\alpha=\frac{1-b-c}{1+a}\). Since \(h = a + b + c\) and \(\alpha(1 + h) \leq1\), by Theorem 2.3, T has a fixed point. □