# A new approach on generalized quasimetric spaces induced by partial metric spaces

## Abstract

In this paper, we introduce the concept of generalized quasimetric spaces by a new approach, and present some examples in the partial metric spaces. Furthermore, we obtain some results on (strong) complete partial metric spaces.

## 1 Introduction

In 1931, Wilson  initiated the notion of quasimetric spaces, which was defined without the symmetric condition comparing to the axioms of the standard metric. Later, Matthews  defined the concept of partial metric space in 1994, in which the distance of each object to itself is not necessarily zero. Additionally, he constructed quasimetric q and weighted metric $$p^{m}$$ by partial metric p, where $$q(x,y)=p(x,y)-p(x,x)$$ and $$p^{m}(x,y)= 2p(x,y)-p(x,x)-p(y,y)$$, respectively. Over the past few decades, these methods of construction have appeared in many papers on partial metric spaces, and the fixed-pointed theory has been one of the most important topics in topology ().

The object of this paper tries to give a generalized quasimetric , i.e., $$d_{p}$$  is its special case,  and $$p^{s}$$  are equivalent. Furthermore, we obtain some results on (strong) complete partial metric spaces.

## 2 Preliminaries

Throughout this paper, X is always a nonempty set, the letters $$\mathbb{R}$$, $$\mathbb{R}\mathbbm{^{+}}$$, $$\mathbb{N}\mathbbm{^{+}}$$ always denote the set of real numbers, of all positive real numbers and of all positive integers, respectively.

### Definition 2.1

()

A quasimetric is a function $$d: X \times X \rightarrow [0,+\infty )$$ satisfying the following conditions: $$\forall x,y,z\in X$$,

1. (M1)

$$x=y \Leftrightarrow d(x,y)=d(y,x)=0$$;

2. (M2)

$$d(x,z)\le d(x,y)+d(y,z)$$.

A quasimetric d is called a metric if it also satisfies

1. (M3)

$$d(x,y)=d(y,x)$$.

A (quasi)metric space is a pair $$(X,d)$$ such that d is a (quasi)metric on X.

### Definition 2.2

()

A partial metric is a function $$p: X \times X \rightarrow [0,+\infty )$$ satisfying the following conditions: $$\forall x,y,z\in X$$,

1. (P1)

$$x=y\Leftrightarrow p(x,x)=p(x,y)=p(y,y)$$;

2. (P2)

$$p(x,x)\le p(x,y)$$;

3. (P3)

$$p(x,y)=p(y,x)$$;

4. (P4)

$$p(x,z)\le p(x,y)+p(y,z)-p(y,y)$$.

A partial metric space is a pair $$(X,p)$$ such that p is a partial metric on X.

Apparently, each metric is precisely a partial metric on X, and a partial metric p is a metric if and only if $$p(x,x)=0$$ for all $$x\in X$$. Similar to the definition of open balls in metric spaces, that is $$B^{d}_{\varepsilon}(x)= \{{y\in X:d(x,y)<{\varepsilon}}\}$$, Matthews used $$B^{p}_{\varepsilon}(x)= \{{y\in X:p(x,y)<{\varepsilon}}\}$$ to denote open p-balls for all $$x\in X$$ and $$\varepsilon >0$$, we can see that some open p-balls may be empty (see more details in ).

### Lemma 2.3

For each partial metric p: $$X \times X \rightarrow [0,+\infty )$$, set $$\hat {p}(x,y)= p(x,y)-[\alpha p(x,x)+\beta p(y,y)]$$, where $$0\le \alpha ,\beta \le 1$$, $$\alpha +\beta =1$$. Then, the following statements hold:

1. (1)

is a quasimetric.

2. (2)

is a metric if and only if $$\alpha =\beta =\frac{1}{2}$$, where we denote

$$\tilde {p}(x,y)=p(x,y)-\frac{p(x,x)+ p(y,y)}{2}.$$
3. (3)

is a metric, where $$\hat {q}(x,y)=\max \{{\hat{p}(x,y),\hat{p}(y,x)}\}$$.

### Proof

(1) We verify the conditions (M1) and (M2) one by one.

1. (M1):

() Suppose that $$x=y$$. It is clear that $$\hat {p}(x,y)=\hat {p}(y,x)=0$$.

() Suppose that $$\hat {p}(x,y)=\hat {p}(y,x)=0$$. Then, $$p(x,y)=\alpha p(x,x)+\beta p(y,y)$$, and $$p(y,x)=\alpha p(y,y)+\beta p(x,x)$$, which implies $$p(x,y)+p(y,x)= p(x,x)+p(y,y)$$. Since $$p(x,x)\le p(x,y)$$ by (P2), we have $$p(x,x)+p(y,x)\le p(x,x)+p(y,y)$$, namely $$p(y,x)\le p(y,y)$$. By (P2) and (P3), we have $$p(y,x)= p(y,y)$$. Analogously, we can deduce $$p(x,y)= p(x,x)$$. Hence, $$p(x,y)=p(x,x)=p(y,y)$$, which implies $$x=y$$ by (P1).

1. (M2):

By (P4), we have

\begin{aligned} \hat {p}(x,y)+\hat {p}(y,z)&={p(x,y)- \bigl[\alpha p(x,x)+\beta p(y,y) \bigr]}+{p(y,z)- \bigl[ \alpha p(y,y)+\beta p(z,z) \bigr]} \\ &=p(x,y)+{p(y,z)-p(y,y)- \bigl[\alpha p(x,x)+\beta p(z,z) \bigr]} \\ &\ge p(x,z)- \bigl[\alpha p(x,x)+\beta p(z,z) \bigr]=\hat {p}(x,z), \end{aligned}

for all $$x,y,z\in X$$. Therefore, is a quasimetric.

(2) and (3) are trivial in that and satisfy (M1)–(M3). □

### Remark 2.4

1. (1)

If $$\alpha =1$$ and $$\beta =0$$, then is q (see ).

2. (2)

If $$\alpha =\beta =\frac{1}{2}$$, then and $$p^{s}$$ are equivalent (see ).

### Proposition 2.5

Let X be a nonempty set, p be a partial metric, and $$(X,\hat{p})$$ be the corresponding quasimetric space defined in Lemma 2.3, i.e., $$\hat{p}(x,y)=p(x,y)-[\alpha p(x,x)+\beta p(y,y)]$$ for any $$x,y\in X$$. Then, the following statements hold.

1. (1)

The set of all open p-balls $$B^{p}_{\varepsilon}(x)$$ is the basis of a topology $$\mathcal{T}(p)$$ on X, where $$B^{p}_{\varepsilon}(x)=\{{y\in X:p(x,y)<\varepsilon }\}$$ for any $$\varepsilon >0$$. We call $$\mathcal{T}(p)$$ the topology generated by the partial metric p on X.

2. (2)

The set of all open -balls $$B^{\hat {p}}_{\varepsilon}(x)$$ is the basis of a topology $$\mathcal{T}(\hat {p})$$ on X, where $$B^{\hat {p}}_{\varepsilon}(x)=\{{y\in X: \hat {p}(x,y)<\varepsilon}\}$$ for any $$\varepsilon >0$$. We call $$\mathcal{T}(\hat {p})$$ the topology generated by the quasimetric on X.

### Proof

(1) It is trivial by Theorem 3.1 in .

(2) It is not difficult to prove that $$X = \bigcup_{x\in X}B^{\hat {p}}_{\varepsilon}(x)$$, where $$\varepsilon >0$$.

Moreover, we have $$B^{\hat {p}}_{\varepsilon}(x)\cap B^{\hat {p}}_{\delta}(y) = \bigcup\{{B^{\hat {p}}_{\eta}(z):z\in B^{\hat {p}}_{\varepsilon}(x)\cap B^{ \hat {p}}_{\delta}(y)}\}$$, where $$\eta =\beta p(z,z)+\min \{{\varepsilon -p(x,z)+\alpha p(x,x),\delta -p(y,z)+ \alpha p(y,y)}\}$$. □

### Theorem 2.6

Let X be a nonempty set, p be a partial metric and $$\hat{p}(x,y)=p(x,y)-[\alpha p(x,x)+\beta p(y,y)]$$, where $$0\le \alpha ,\beta \le 1$$, $$\alpha +\beta =1$$ and $$\alpha \ne 1/2$$, $$\beta \ne 1/2$$, for any $$x,y\in X$$. The following statements hold:

1. (1)

Each partial metric p on X generates a $$T_{0}$$ topology $$\mathcal{T}(p)$$ on X.

2. (2)

Each quasimetric on X generates a $$T_{0}$$ topology $$\mathcal{T}(\hat {p})$$ on X.

3. (3)

$$\mathcal{T}(p)=\mathcal{T}(\hat {p})$$.

4. (4)

$$(X,\mathcal{T}(p))$$ and $$(X,\mathcal{T}(\hat {p}))$$ are first countable.

### Proof

(1) It is trivial by Theorem 3.3 in .

(2) By Lemma 2.3(1), we know that is a quasimetric. Suppose that $$x\ne y$$. By (P2) and (P3), we have $$\alpha p(x,x)+\beta p(y,y)\le \alpha p(x,y)+\beta p(x,y)=p(x,y)$$. Set $$\varepsilon =\frac{p(x,y)-[\alpha p(x,x)+\beta p(y,y)]}{2}$$. Then, $$x\in B^{\hat {p}}_{\varepsilon}(x)$$ and $$y\notin B^{\hat {p}}_{\varepsilon}(x)$$. Therefore, $$(X,\mathcal{T}(\hat {p}))$$ is a $$T_{0}$$ topology space.

(3) For any $$x\in X$$ and $$\varepsilon >0$$, suppose $$y\in B^{p}_{\varepsilon}(x)$$, namely, $$p(x,y)<\varepsilon$$. Since $$\alpha p(x,x)+\beta p(y,y)\le p(x,y)$$, we have $$\alpha p(x,x)+\beta p(y,y)<\varepsilon$$. Set $$\delta =\varepsilon -[\alpha p(x,x)+\beta p(y,y)]$$. We can deduce $$p(x,y)<\delta +[\alpha p(x,x)+\beta p(y,y)]$$, which implies $$y\in B^{\hat {p}}_{\delta}(x)$$. Therefore, $$B^{p}_{\varepsilon}(x)\subseteq B^{\hat {p}}_{\delta}(x)$$.

On the other hand, for any $$x\in X$$ and $$\varepsilon >0$$, suppose $$y\in B^{\hat {p}}_{\varepsilon}(x)$$. We have $$p(x,y)-[\alpha p(x,x)+\beta p(y,y)]<\varepsilon$$. Set $$\eta =\varepsilon +[\alpha p(x,x)+\beta p(y,y)]$$. Then, we can deduce $$p(x,y)<\eta$$, which implies $$y\in B^{p}_{\eta}(x)$$, thus $$B^{\hat {p}}_{\varepsilon}(x)\subseteq B^{ p}_{\eta}(x)$$. Hence, $$\mathcal{T}(p)=\mathcal{T}(\hat {p})$$.

(4) Set $$\varepsilon \in \mathbb{Q}\mathbbm{^{+}}$$, where $$\mathbb{Q}\mathbbm{^{+}}$$ denotes the set of all positive rational numbers. For any $$x\in X$$, $$B^{ p}_{\varepsilon}(x)$$ and $$B^{\hat {p}}_{\varepsilon}(x)$$ are countable neighborhoods at x in $$(X,\mathcal{T}(p))$$ and $$(X,\mathcal{T}(\hat {p}))$$, respectively. □

## 3 Some results on (strong) complete partial metric spaces

### Definition 3.1

Let $$(X,p)$$ be a partial metric space and $$\{{x_{n}}\}$$ be a sequence in X.

1. (1)

A sequence $$\{{x_{n}}\}$$ converges to a point $$x\in X$$ if $$p(x,x)= \lim_{n\to +\infty}p(x,x_{n})$$;

2. (2)

A sequence $$\{{x_{n}}\}$$ is called a Cauchy sequence if $$\lim_{n,m\to +\infty}p(x_{n},x_{m})$$ exists and is finite;

3. (3)

$$(X,p)$$ is said to be complete if every Cauchy sequence $$\{{x_{n}}\}$$ in X converges, with respect to $$\mathcal{T}(p)$$, to a point $$x\in X$$ such that $$p(x,x)=\lim_{n,m\to +\infty}p(x_{n},x_{m})=\lim_{n\to +\infty}p(x_{n},x)$$.

### Lemma 3.2

Let $$(X,p)$$ be a partial metric space and $$(X,\tilde {p})$$ be the corresponding metric space defined in Lemma 2.3(2), i.e., $$\tilde {p}(x,y)=p(x,y)-\frac{p(x,x)+ p(y,y)}{2}$$, for all $$x,y\in X$$. Let $$(X,\hat{q})$$ be the corresponding metric space, where $$\hat {q}(x,y) = \max \{{\hat{p}(x,y), \hat{p}(y,x)}\}$$, and $$\hat{p}(x,y) = p(x,y)-[\alpha p(x,x)+\beta p(y,y)]$$, $$0\le \alpha ,\beta \le 1$$, $$\alpha +\beta =1$$ and $$\alpha \ne \frac{1}{2}$$, $$\beta \ne \frac{1}{2}$$, for all $$x,y\in X$$. The following statements hold:

1. (1)

A sequence is a Cauchy sequence in $$(X,p)$$ if and only if it is a Cauchy sequence in $$(X,\hat {q})$$.

2. (2)

$$(X,p)$$ is complete if and only if $$(X,\hat {q})$$ is complete.

3. (3)

$$(X,p)$$ is complete if and only if $$(X,\tilde {p})$$ is complete.

4. (4)

$$\lim_{n\to +\infty}\tilde{p}(x_{n},x)=0$$ if and only if

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

### Proof

(1) () Let $$\{{x_{n}}\}$$ be a Cauchy sequence in $$(X,p)$$. There exists $$\eta \in [0,+\infty )$$ such that $$\lim_{n,m\to +\infty}p(x_{n},x_{m})=\eta$$. Then, for any $$\varepsilon >0$$, there exists $$n_{\varepsilon}\in \mathbb{N}\mathbbm{^{+}}$$ such that

$$\bigl\vert p(x_{n},x_{m})-\eta \bigr\vert < \frac{\varepsilon}{2},\quad \forall n,m>n_{ \varepsilon}.$$

Then, we have that

\begin{aligned} \bigl\vert \hat {p}(x_{n},x_{m}) \bigr\vert &= \bigl\vert p(x_{n},x_{m})- \bigl[\alpha p(x_{n},x_{n})+ \beta p(x_{m},x_{m}) \bigr] \bigr\vert \\ &\leqslant \bigl\vert p(x_{n},x_{m})-\eta \bigr\vert + \alpha \bigl\vert p(x_{n},x_{n})-\eta \bigr\vert + \beta \bigl\vert p(x_{m},x_{m})-\eta \bigr\vert \\ &< \frac{\varepsilon}{2} +\alpha \cdot \frac{\varepsilon}{2}+\beta \cdot \frac{\varepsilon}{2}=\varepsilon . \end{aligned}

This implies that $$\{{x_{n}}\}$$ is a Cauchy sequence in $$(X,\hat{q})$$.

() Suppose $$\{{x_{n}}\}$$ is a Cauchy sequence in $$(X,\hat{q})$$ and let $$\varepsilon >0$$. Then, there exists $$n_{\varepsilon}\in \mathbb{N}\mathbbm{^{+}}$$, such that

$$\hat{q}(x_{n},x_{m})< \frac{ \vert \alpha -\beta \vert \varepsilon}{2},\quad \forall n,m>n_{ \varepsilon}.$$

Set $$\varepsilon =1$$. Then, there exists $$n_{0}\in \mathbb{N}\mathbbm{^{+}}$$ such that

$$\hat{q}(x_{n},x_{m})< \frac{ \vert \alpha -\beta \vert }{2},\quad \forall n,m>n_{0}.$$

We prove that $$\{{x_{n}}\}$$ is a Cauchy sequence in $$(X,{p})$$ in the following steps.

Step 1: Since $$p(x_{n},x_{n_{0}})=p(x_{n_{0}},x_{n})$$ by (P3) for all $$n\geq n_{0}$$, we have

\begin{aligned} \hat{p}(x_{n},x_{n_{0}})+ \bigl[ \alpha p(x_{n},x_{n})+\beta p(x_{n_{0}},x_{n_{0}}) \bigr]= \hat{p}(x_{n_{0}},x_{n})+ \bigl[\alpha p(x_{n_{0}},x_{n_{0}})+\beta p(x_{n},x_{n}) \bigr].\end{aligned}

Thus, we have $$(\alpha -\beta ) p(x_{n},x_{n})=\hat{p}(x_{n_{0}},x_{n})+(\alpha - \beta ) p(x_{n_{0}},x_{n_{0}})-\hat {p}(x_{n},x_{n_{0}})$$, which implies that

$$p(x_{n},x_{n})=\frac{1}{\alpha -\beta} \bigl[ \hat{p}(x_{n_{0}},x_{n})- \hat {p}(x_{n},x_{n_{0}}) \bigr]+ p(x_{n_{0}},x_{n_{0}}).$$

Then, we have

\begin{aligned} \bigl\vert p(x_{n},x_{n}) \bigr\vert &\le \frac{1}{ \vert \alpha -\beta \vert } \bigl[ \bigl\vert \hat{p}(x_{n_{0}},x_{n}) \bigr\vert + \bigl\vert \hat {p}(x_{n},x_{n_{0}}) \bigr\vert \bigr]+ p(x_{n_{0}},x_{n_{0}}) \\ &\le \frac{2}{ \vert \alpha -\beta \vert } \bigl\vert \hat{q}(x_{n},x_{n_{0}}) \bigr\vert +p(x_{n_{0}},x_{n_{0}}) \\ &< 1+p(x_{n_{0}},x_{n_{0}}), \end{aligned}

for all $$n\ge n_{0}$$, which implies that the sequence $$\{{{p}(x_{n},x_{n})}\}$$ is bounded in $$\mathbb{R}$$. Hence, the sequence $$\{{p(x_{n},x_{n})}\}$$ exists with a subsequence $$\{{p(x_{n_{k}},x_{n_{k}})}\}$$ that is convergent and we denote $$\lim_{n_{k}\to +\infty}p(x_{n_{k}},x_{n_{k}})=a$$.

Step 2: By Step 1, we have

\begin{aligned} \bigl\vert p(x_{n},x_{n})-p(x_{m},x_{m}) \bigr\vert &=\frac{1}{ \vert \alpha -\beta \vert } \bigl\vert \hat{p}(x_{m},x_{n})- \hat{p}(x_{n},x_{m}) \bigr\vert \\ &\le \frac{1}{ \vert \alpha -\beta \vert } \bigl[ \bigl\vert \hat{p}(x_{m},x_{n}) \bigr\vert + \bigl\vert \hat{p}(x_{n},x_{m}) \bigr\vert \bigr] \\ &< \frac{2}{ \vert \alpha -\beta \vert } \bigl\vert \hat{q}(x_{m},x_{n}) \bigr\vert < \varepsilon , \end{aligned}

for all $$n,m>n_{\varepsilon}$$. In addition, since

$$p(x_{n},x_{n})=\frac{1}{\alpha -\beta} \bigl[ \hat{p}(x_{m},x_{n})-\hat {p}(x_{n},x_{m}) \bigr]+ p(x_{m},x_{m}),$$

we have

$$\lim_{n\to +\infty}p(x_{n},x_{n})=\lim _{m\to +\infty}p(x_{m},x_{m})=a,$$

for all $$n,m>n_{1}$$, where $$n_{1}=\max \{{n_{\varepsilon},n_{0}}\}$$.

Furthermore,

\begin{aligned} & \bigl\vert p(x_{n},x_{m})-a \bigr\vert \\ &\quad = \bigl\vert p(x_{n},x_{m})- \bigl[\alpha p(x_{n},x_{n})+\beta p(x_{m},x_{m}) \bigr]+ \bigl[ \alpha p(x_{n},x_{n})+\beta p(x_{m},x_{m}) \bigr]-a \bigr\vert \\ &\quad \le \bigl\vert p(x_{n},x_{m})- \bigl[\alpha p(x_{n},x_{n})+\beta p(x_{m},x_{m}) \bigr] \bigr\vert + \bigl\vert \alpha p(x_{n},x_{n})+ \beta p(x_{m},x_{m})-a \bigr\vert \\ &\quad =\hat{p}(x_{n},x_{m})+\alpha \bigl\vert p(x_{n},x_{n})-a \bigr\vert +\beta \bigl\vert p(x_{m},x_{m})-a \bigr\vert \\ &\quad < \frac{ \vert \alpha -\beta \vert }{2}\cdot \varepsilon +\alpha \cdot \varepsilon +\beta \cdot \varepsilon =\frac{ \vert \alpha -\beta \vert +2}{2} \cdot \varepsilon , \end{aligned}
(1)

for all $$n,m>n_{1}$$. This implies that $$\{{x_{n}}\}$$ is a Cauchy sequence in $$(X,p)$$.

(2) () First, without loss of generality, we claim that $$0\le \beta <\frac{1}{2}$$ (in fact, by $$0\le \alpha ,\beta \le 1$$, $$\alpha +\beta =1$$ and $$\alpha \ne \frac{1}{2}$$, $$\beta \ne \frac{1}{2}$$, then, we have $$\alpha <\frac{1}{2}$$ or $$\beta <\frac{1}{2}$$).

Step 1: Let $$\{{x_{n}}\}$$ be a Cauchy sequence in $$(X,p)$$. It is clear that $$\{{x_{n}}\}$$ is a Cauchy sequence in $$(X,\hat{q})$$ by Lemma 3.2(1). Since $$(X,\hat{q})$$ is complete, there exists $$x\in X$$ such that $$\lim_{n\to +\infty}\hat{q}(x,x_{n})=0$$, i.e., for any $$\varepsilon >0$$, there exists $$n_{0}\in \mathbb{N}\mathbbm{^{+}}$$ such that

$$\bigl\vert \hat{q}(x,x_{n}) \bigr\vert < \frac{1-2\beta}{2}\cdot \varepsilon$$

for all $$n>n_{0}$$. Since $$\hat {q}(x,y) = \max \{{\hat{p}(x,y), \hat{p}(y,x)}\}$$, we have $$\lim_{n\to +\infty}\hat{p}(x,x_{n})=0$$. This shows that $$\{{x_{n}}\}$$ is a convergent sequence in $$(X,\hat{p})$$.

On the other hand, we have

\begin{aligned} & \bigl\vert p(x_{n},x_{n})-p(x,x) \bigr\vert \\ &\quad = \bigl\vert (\alpha +\beta ) p(x,x)-(\alpha +\beta )p(x_{n},x_{n}) \bigr\vert \\ &\quad = \bigl\vert \bigl[\alpha p(x,x)+\beta p(x_{n},x_{n})-p(x,x_{n}) \bigr]+ \bigl[p(x,x_{n})- \alpha p(x_{n},x_{n})- \beta p(x,x) \bigr] \\ &\qquad {} -2\beta \bigl[p(x_{n},x_{n})-p(x,x) \bigr] \bigr\vert \\ &\quad \le \bigl\vert p(x,x_{n})-\alpha p(x,x)-\beta p(x_{n},x_{n}) \bigr\vert + \bigl\vert p(x,x_{n})- \alpha p(x_{n},x_{n})- \beta p(x,x) \bigr\vert \\ &\qquad {} +2\beta \bigl\vert p(x_{n},x_{n})-p(x,x) \bigr\vert . \end{aligned}
(1)

Then,

\begin{aligned} & (1-2\beta ) \bigl\vert p(x_{n},x_{n})-p(x,x) \bigr\vert \\ &\quad \le \bigl\vert p(x,x_{n})- \bigl[\alpha p(x,x)+\beta p(x_{n},x_{n}) \bigr] \bigr\vert + \bigl\vert p(x_{n},x)- \bigl[ \alpha p(x_{n},x_{n})+\beta p(x,x) \bigr] \bigr\vert \\ &\quad =\hat{p}(x,x_{n})+\hat{p}(x_{n},x) \\ &\quad < 2\hat{q}(x,x_{n}) < (1-2\beta )\cdot \varepsilon , \end{aligned}

for all $$n>n_{0}$$. Therefore, we can deduce $$|p(x_{n},x_{n})-p(x,x)|<\varepsilon$$, which implies $$\lim_{n\to +\infty}p(x_{n}, x_{n})= p(x,x)$$.

Step 2: Since $$\hat{p}(x,y)=p(x,y)-[\alpha p(x,x)+\beta p(y,y)]$$, by Step 1, we have $$\lim_{n\to +\infty}\hat{p}(x,x_{n})=\lim_{n\to +\infty}{p}(x,x_{n})- \alpha{p}(x,x) -\beta \lim_{n\to +\infty}p(x_{n},x_{n})$$. Then, we can deduce $$\lim_{n\to +\infty}p(x,x_{n})=\lim_{n\to +\infty}p(x_{n},x)= p(x,x)$$.

In addition, by (P4) we have $$p(x_{n},x_{m})\le p(x_{n},x)+p(x,x_{m})-p(x,x)$$. Hence, $$\lim_{n,m\to +\infty}p(x_{n},x_{m})\le p(x,x)$$. Moreover, by (P2), we have $$p(x_{n},x_{m})\ge p(x_{n},x_{n})$$, which implies $$\lim_{n,m\to +\infty}p(x_{n},x_{m})\ge p(x,x)$$. Then, we have

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

Therefore, $$(X,p)$$ is complete.

() Let $$\{{x_{n}}\}$$ be a Cauchy sequence in $$(X,\hat{q})$$. Then, $$\{{x_{n}}\}$$ is a Cauchy sequence in $$(X,p)$$ by Lemma 3.2(2). There exists a point $$x\in X$$, such that $$\lim_{n,m\to +\infty}p(x_{n},x_{m})=\lim_{n\to +\infty}p(x,x_{n})=p(x,x)$$. Therefore, for any $$\varepsilon >0$$, there exists $$n_{0}\in \mathbb{N}\mathbbm{^{+}}$$ such that

$$\bigl\vert p(x,x_{n})-p(x,x) \bigr\vert < \varepsilon$$

and

$$\bigl\vert p(x_{n},x_{n})-p(x,x) \bigr\vert < \varepsilon ,$$

for all $$n\ge n_{0}$$. Then, we have

\begin{aligned} \bigl\vert \hat{p}(x,x_{n}) \bigr\vert =& \bigl\vert p(x,x_{n}) - \bigl[\alpha p(x,x)+\beta p(x_{n},x_{n}) \bigr] \bigr\vert \\ =&\alpha \bigl\vert \bigl[p(x,x_{n})- p(x,x) \bigr] \bigr\vert + \beta \bigl\vert \bigl[p(x,x_{n})-p(x_{n},x_{n}) \bigr] \bigr\vert \\ < &\alpha \cdot \varepsilon + \beta \bigl\vert p(x,x_{n}) -p(x,x) \bigr\vert +\beta \bigl\vert p(x_{n},x_{n})-p(x,x) \bigr\vert \\ < &\alpha \cdot \varepsilon +2\beta \cdot \varepsilon =\beta \cdot \varepsilon . \end{aligned}

Therefore, we have $$\lim_{n\to +\infty}\hat{p}(x,x_{n})=0$$.

Analogously, we have

\begin{aligned} \bigl\vert \hat{p}(x_{n},x) \bigr\vert &= \bigl\vert p(x_{n},x)- \bigl[\alpha p(x_{n},x_{n})+ \beta p(x,x) \bigr] \bigr\vert \\ &< \beta \cdot \varepsilon + \alpha \bigl\vert p(x,x_{n}) -p(x,x) \bigr\vert +\alpha \bigl\vert p(x_{n},x_{n})-p(x,x) \bigr\vert \\ &< \beta \cdot \varepsilon +2\alpha \cdot \varepsilon =\alpha \cdot \varepsilon . \end{aligned}

This implies $$\lim_{n\to +\infty}\hat{p}(x_{n},x)=0$$.

Furthermore, by (M2), we have $$\hat {p}(x_{n},x_{m})\le \hat {p}(x_{n},x)+\hat {p}(x,x_{m})$$. Therefore, $$\lim_{n,m\to +\infty} \hat {p}(x_{n}, x_{m})=0$$, which implies $$(X,\hat {p})$$ is complete. It is not difficult to show $$(X,\hat {q})$$ is complete.

(3) It is trivial by Lemma 3.2 in .

(4) It is trivial by Lemma 2.1 in . □

### Corollary 3.3

Let $$(X,p)$$ be a partial metric space. Then, $$\lim_{n\to +\infty}\hat {q}(x,x_{n})=0$$ if and only if $$\lim_{n,m\to +\infty}p(x_{n},x_{m})=\lim_{n\to +\infty}p(x,x_{n})=p(x,x)$$.

### Lemma 3.4

()

Let $$(X,d)$$ be a complete metric space, $$\varphi :X\rightarrow [0,+\infty )$$ be a lower semicontinuous function, and $$T:X\rightarrow X$$ be a given mapping. The following statements hold:

1. (1)

Suppose that for any $$0< a< b<+\infty$$, there exists $$0<\gamma (a,b)<1$$ such that for all $$x,y\in X$$, $$a\le d(x,y)+\frac{\varphi (x)+\varphi (y)}{2}\le b$$ implies $$d(Tx,Ty)+\frac{\varphi (Tx)+\varphi (Ty)}{2}\le \gamma (a,b)[d(x,y)+ \frac{\varphi (x)+\varphi (y)}{2}]$$. Then, T has a unique fixed point $$x^{*}\in X$$. Moreover, we have $$\varphi (x^{*})=0$$.

2. (2)

Suppose that for all $$x,y\in X$$, there exist $$a,b,c\in [0,+\infty )$$ with $$a+b+c<1$$ such that $$d(Tx,Ty)+\varphi{Tx}+\varphi{Ty)}\le a[d(x,y)+\varphi (x)+\varphi (y)]+b[d(x,Tx)+ \varphi (x)+\varphi (Tx)]+c[d(y,Ty)+\varphi (y)+\varphi (Ty)]$$. Then, T has a unique fixed point $$x^{*}\in X$$. Moreover, we have $$\varphi (x^{*})=0$$.

### Theorem 3.5

Let $$(X,p)$$ be a complete partial metric space and $$T:X\rightarrow X$$ be a given mapping. The following statements hold:

1. (1)

Suppose for any $$a,b\in (0,+\infty )$$, there exists $$0<\gamma (a,b)<1$$ such that for all $$x,y\in X$$, $$a\le p(x,y)\le b$$ implies $$p(Tx,Ty)\le \gamma (a,b)p(x,y)$$. Then, T has a unique fixed point $$x^{*}\in X$$. Moreover, we have $$p(x^{*},x^{*})=0$$.

2. (2)

Suppose for all $$x,y\in X$$, there exist $$a,b,c\in (0,+\infty )$$ and $$a+b+c<1$$ such that $$p(Tx,Ty)\le a p(x,y)+b p(x,Tx)+c p(y,Ty)$$. Then, T has a unique fixed point $$x^{*}\in X$$. Moreover, we have $$p(x^{*},x^{*})=0$$.

### Proof

(1) We have $$\tilde {p}(x,y)=p(x,y)-\frac{p(x,x)+ p(y,y)}{2}$$ by Lemma 2.3(2). Then,

$$p(x,y)=\tilde {p}(x,y)+\frac{p(x,x)+ p(y,y)}{2} ,$$

for all $$x,y\in X$$. Since $$(X,p)$$ is complete, we have that $$(X,\tilde{p})$$ is complete by Lemma 3.2(3). Define a function $$\varphi :X\rightarrow [0,+\infty )$$. Set $$\varphi (x)=p(x,x)$$ for all $$x\in X$$. Since $$p(x,y)=\tilde {p}(x,y)+\frac{p(x,x)+ p(y,y)}{2}$$, there exists $$0<\gamma (a,b)<1$$ for any $$a,b,c\in (0,+\infty )$$. From Lemma 3.4, we can deduce that $$a\le \tilde {p}(x,y)+\frac{p(x,x)+ p(y,y)}{2}\le b$$ implies $$\tilde {p}(Tx,Ty)+\frac{\varphi (Tx)+\varphi (Ty)}{2}\le \gamma (a,b)[ \tilde{p}(x,y)+\frac{\varphi (x)+\varphi (y)}{2}]$$.

On the other hand, let $$\{{x_{n}}\}$$ be a sequence in X such that $$\lim_{n\to +\infty}\tilde {p}(x_{n},x)=0$$, where $$x\in X$$. Then, we have $$\lim_{n\to +\infty}{p}(x_{n},x)= p(x,x)$$ by Lemma 3.2(4), i.e., $$\lim_{n\to +\infty}\varphi (x_{n})=\varphi (x)$$, so φ is continuous. By Lemma 3.4(1), the result follows.

(2) It is not difficult to show that

\begin{aligned} & 2\tilde {p}(Tx,Ty)+p(Tx,Tx)+p(Ty,Ty) \\ &\quad \le a \bigl[2\tilde {p}(x,y)+p(x,x)+ p(y,y) \bigr]+ b \bigl[2\tilde {p}(x,Tx)+p(x,x)+p(Tx,Tx) \bigr] \\ & \qquad {}+c \bigl[2\tilde{p}(y,Ty)+p(y,y)+p(Ty,Ty) \bigr]. \end{aligned}
(1)

Set $$d=2\tilde {p}$$ and $$\varphi (x)=p(x,x)$$. By Lemma 3.4(2), then this statement holds. □

### Example 3.6

Let $$X=[0,+\infty )$$. Define $$p: X \times X\times \rightarrow [0,+\infty )$$ as follows: $$p(x, y)=\max {\{x,y\}}$$ for all $$x,y\in X$$. It is clear that $$(X,p)$$ is a partial metric space. Define a mapping T: $$X\rightarrow X$$ by $$Tx=\frac{x}{1+x}$$ for all $$x\in X$$, and taking $$\gamma (a,b)=\frac{a+b}{1+a+b}$$ for all $$a,b\in (0,+\infty )$$. Thus, all the conditions of Theorem 3.5(1) are satisfied and obviously $$x=0$$ is a fixed point of T.

### Definition 3.7

Let p be a partial metric and $$(X,\hat{p})$$ be the corresponding quasimetric space defined in Theorem 2.6, i.e., $$\hat{p}(x,y) = p(x,y)-[\alpha p(x,x)+\beta p(y,y)]$$, $$0\le \alpha ,\beta \le 1,\alpha +\beta =1$$ and $$\alpha \ne \frac{1}{2}$$, $$\beta \ne \frac{1}{2}$$, for all $$x,y\in X$$. $$(X,p)$$ is said to be a strong complete partial metric space if $$\lim_{m>n\to +\infty}\hat{p}(x_{n},x_{m})=0$$ can imply $$\lim_{n\to +\infty}x_{n}=x$$ for some $$x\in X$$.

### Remark 3.8

A strong complete partial metric space is a complete partial metric space, but the converse may not be true.

In fact, by (P4), we have

\begin{aligned} & p(x_{n},x_{m})-p(x,x) \\ &\quad \le p(x_{n},x)+p(x,x_{m})-2p(x,x) \\ &\quad =\hat{p}(x_{n},x)+\hat{p}(x,x_{m})+\alpha p(x_{n},x_{n})+\beta p(x_{m},x_{m})-p(x,x), \end{aligned}

for all $$n,m\in \mathbb{N}\mathbbm{^{+}}$$. Since $$(X,p)$$ is a strong complete partial metric space, we have $$\lim_{m>n\to +\infty}\hat{p}(x_{n},x_{m})=0$$ and $$\lim_{n\to +\infty}x_{n}=x$$, which implies that $$\lim_{m,n\to +\infty}[p(x_{n},x_{m})-p(x,x)]=0$$, namely, $$(X,p)$$ is complete.

The following example shows that a complete partial metric space may not be a strong complete partial metric space.

### Example 3.9

Let $$A=\{{a_{i}:a_{i}=2i, i\in \mathbb{N}\mathbbm{^{+}} }\}$$ and $$B =\{{b_{i}:b_{i}=2i+1, i\in \mathbb{N}\mathbbm{^{+}}}\}$$ be two disjoint infinitely countable sets, and $$X=A\cup B$$. Define a function $$p: X \times X \rightarrow [0,+\infty )$$ by

$$P{(x,y)}= \textstyle\begin{cases} 1,&x=y\in A\text{ or }x=y\in B; \\ 1+\frac{1}{i}+\frac{1}{j},& x\ne y\text{ and } \{{x,y}\}\in \{{\{{a_{i},a_{j}}\},\{{a_{i},b_{j}}\},\{{b_{i},b_{j}} \}}\}. \end{cases}$$

It is not difficult to prove that $$(X,p)$$ is a complete partial metric space. Set $$x_{n}=2n$$, $${x_{m}=2m+1}$$ for all $$m>n$$, where $$n,m\in \mathbb{N}\mathbbm{^{+}}$$. Then, we have $$\hat{p}(x_{n},x_{m})= p(x_{n},x_{m})-[\alpha p(x_{n},x_{n})+\beta p(x_{m},x_{m})]= \frac{1}{n}+\frac{1}{m}$$, and we can deduce $$\lim_{m>n\to +\infty}\hat {p}(x_{n},x_{m})=0$$. However, $$\lim_{n\to +\infty}x_{n}$$ does not exist.

### Theorem 3.10

Let p be a partial metric and $$(X,\hat{p})$$ be the corresponding metric space defined in Theorem 2.6, i.e., $$\hat{p}(x,y) = p(x,y)-[\alpha p(x,x)+\beta p(y,y)]$$, $$0\le \alpha ,\beta \le 1,\alpha +\beta =1$$ and $$\alpha \ne \frac{1}{2}$$, $$\beta \ne \frac{1}{2}$$, for all $$x,y\in X$$, and satisfies the following conditions:

1. (1)

$$(X,\hat{p})$$ is a strong complete partial metric space.

2. (2)

$$f:X\rightarrow \mathbb{R}$$ is a lower semicontinuous function bounded from below.

3. (3)

Let $$\varepsilon >0$$, there exists $$x_{0}\in X$$ such that $$f(x_{0})\le \inf_{x\in X}f(x)+\varepsilon$$ for all $$x\in X$$.

Then, there exists a point $$x^{*}$$ such that $$\hat{p}(x_{0},x^{*})<1$$.

### Proof

For any $$\varepsilon >0$$, we denote $$T(x,y)=\varepsilon p(x,y)$$. Suppose

$$S_{n}=\bigl\{ {t:f(t)+T(t,x_{n})\le f(x_{n})+ \alpha T(t,t)+\beta T(x_{n},x_{n})} \bigr\} .$$

It is easy to see that $$x_{n}\in S_{n}$$, namely, $$S_{n}\ne \varnothing$$. Take $$x_{n+1}\in S_{n}$$, such that

$$f(x_{n+1})-\inf_{S_{n}}f\le \frac {f(x_{n})-\inf_{S_{n}}f}{2}.$$

We have that the sequence $$\{{f(x_{n})}\}$$ is nonincreasing and bounded from below. Hence, $$\{{f(x_{n})}\}$$ is a Cauchy sequence. We prove in the following steps:

Step 1: By (P4), we can deduce that

\begin{aligned} f(t)+T(t,x_{n}) &= f(t)+\varepsilon p(t,x_{n}) \\ &\le f(t)+\varepsilon \bigl[p(t,x_{n+1})+p(x_{n+1},x_{n})-p(x_{n+1},x_{n+1}) \bigr] \\ &= f(t)+T(t,x_{n+1})+T(x_{n+1},x_{n})-T(x_{n+1},x_{n+1}). \end{aligned}

For any $$t\in S_{n+1}$$, we have

$$f(t)+T(t,x_{n+1})\le f(x_{n+1})+\alpha T(t,t)+\beta T(x_{n+1},x_{n+1}) .$$

Then,

\begin{aligned} & f(t)+T(t,x_{n}) \\ &\quad \le f(x_{n+1})+\alpha T(t,t)+\beta T(x_{n+1},x_{n+1})+T(x_{n+1},x_{n})-T(x_{n+1},x_{n+1}). \end{aligned}
(1)

Since $$x_{n+1}\in S_{n}$$, we have that

$$f(x_{n+1})+T(x_{n+1},x_{n})\le f(x_{n})+ \alpha T(x_{n+1},x_{n+1})+ \beta T(x_{n},x_{n}).$$

Then,

\begin{aligned} & f(t)+T(t,x_{n}) \\ &\quad \le f(x_{n})+\alpha T(x_{n+1},x_{n+1})+\beta T(x_{n},x_{n})+\alpha T(t,t)+ \beta T(x_{n+1},x_{n+1}) \\ &\qquad {} -T(x_{n+1},x_{n+1}) \\ &\quad =f(x_{n})+\alpha T(t,t)+\beta T(x_{n},x_{n}), \end{aligned}
(1)

which implies that $$t\in S_{n}$$, then $$S_{n+1}\subset S_{n}$$.

Hence, for any $$x_{m}\in S_{m}$$, we have

$$f(x_{m})+T(x_{m},x_{n})\le f(x_{n})+ \alpha T(x_{m},x_{m})+\beta T(x_{n},x_{n}),$$

for all $$m\ge n$$.

Step 2: By step 1, we have

$$T(x_{m},x_{n})-\alpha T(x_{m},x_{m})- \beta T(x_{n},x_{n})\le f(x_{n})-f(x_{m}),$$

for all $$m\ge n$$. Since $$\{{f(x_{n})}\}$$ is a Cauchy sequence, we have

$$\lim_{m>n\to +\infty }\bigl[T(x_{m},x_{n})-\alpha T(x_{m},x_{m})-\beta T(x_{n},x_{n}) \bigr]=0,$$

which implies that $$\lim_{m>n\to +\infty }[p(x_{m},x_{n})-\alpha p(x_{m},x_{m})-\beta p(x_{n},x_{n})]=0$$, namely, $$\lim_{m>n\to +\infty }\hat {p}(x_{m},x_{n})=0$$. Since $$(X,\hat{p})$$ is strong complete, there exists some point $$x^{*}$$, such that $$\lim_{m\to +\infty }x_{m}=x^{*}$$, and $$\lim_{m\to +\infty }f(x_{m})=f(x^{*})$$.

Furthermore, for any $$m\ge n$$, we have

$$f(x_{m})+T(x_{m},x_{n})\le f(x_{n})+ \alpha T(x_{m},x_{m})+\beta T(x_{n},x_{n}),$$

which implies that

$$f\bigl(x^{*}\bigr)+T\bigl(x^{*},x_{n}\bigr)\le f(x_{n})+ \alpha T\bigl(x^{*},x^{*}\bigr)+\beta T(x_{n},x_{n}),$$

so $$x^{*}\in S_{n}$$.

Step 3: From step 1 and step 2, we have

\begin{aligned} & T \bigl(x^{*},x_{0} \bigr)- \alpha T \bigl(x^{*},x^{*} \bigr)-\beta T(x_{0},x_{0}) \\ &\quad =\lim_{n\to +\infty } \bigl[T(x_{n},x_{0})-\alpha T(x_{n},x_{n})-\beta T(x_{0},x_{0}) \bigr] \\ &\quad = f(x_{0})-\alpha f \bigl(x^{*} \bigr)-\beta f \bigl(x^{*} \bigr) \\ &\quad = f(x_{0})-f \bigl(x^{*} \bigr) \\ &\quad \le f(x_{0})-\inf_{x\in X}f \bigl(x^{*} \bigr)< \varepsilon , \end{aligned}
(1)

which implies $$p(x^{*},x_{0})-\alpha p(x^{*},x^{*})-\beta p(x_{0},x_{0})<1$$, namely, $$\hat{p}(x_{0},x^{*})<1$$. □

## Availability of data and materials

The data used to support the findings of this study are included within the article.

## References

1. Wilson, W.A.: On quasi-metric spaces. Am. J. Math. 53, 675–684 (1931)

2. Matthews, S.G.: Partial metric topology. Ann. N.Y. Acad. Sci. 728, 183–197 (1994)

3. Bakhtin, I.A.: The contraction mapping principle in almost metric spaces. Funct. Anal. 30, 26–37 (1989)

4. Samet, B., Verto, C., Vetro, F.: From metric spaces to partial metric spaces. Fixed Point Theory Appl. 2013, 5 (2013)

5. Romaguera, S., Schellekens, M.: Quasi-metric properties of complexity spaces. Topol. Appl. 98, 311–322 (1999)

6. Valero, O.: On Banach fixed point theorems for partial metric spaces. Appl. Gen. Topol. 6, 229–240 (2005)

7. Paluszyński, M., Stempak, K.: On quasi-metric and metric spaces. Proc. Am. Math. Soc. 137, 4307–4312 (2009)

8. Bukatin, M., Kopperman, R., Matthews, S., Pajoohesh, H.: Partial metric spaces. Am. Math. Mon. 116, 708–718 (2009)

9. Altun, I., Sola, F., Simsek, H.: Generalized contractions on partial metric spaces. Topol. Appl. 157, 2778–2785 (2010)

10. Abdeljawad, T., Karapinar, E., Tas, K.: Existence and uniqueness of a common fixed point on partial metric spaces. Appl. Math. Lett. 24, 1900–1904 (2011)

11. Karapinar, E., Erhan, I.M.: Fixed point theorems for operators on partial metric spaces. Appl. Math. Lett. 24, 1894–1899 (2011)

12. Parvaneh, V., Kadelburg, Z.: Extended partial b-metric spaces and some fixed point results. Filomat 32, 2837–2850 (2018)

13. Han, S.Z., Wu, J.F., Zhang, D.: Properties and principles on partail metric spaces. Topol. Appl. 230, 77–98 (2017)

14. Reilly, I.L., Subrahmanyam, P.V., Vamanamurthy, M.K.: Cauchy sequences in quasi-pseudo-metric spaces. Monatshefte Math. 93, 127–140 (1982)

## Acknowledgements

The author thanks the editor and the referees for constructive and pertinent suggestions, which have improved the quality of the manuscript greatly.

## Funding

This work was supported by Program through the guiding science and technology plan project of Suqian (No. Z2021131).

## Author information

Authors

### Corresponding author

Correspondence to Yaoqiang Wu.

## Ethics declarations

### Competing interests

The author declares that they have no competing interests.

## Rights and permissions 