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 [1] initiated the notion of quasimetric spaces, which was defined without the symmetric condition comparing to the axioms of the standard metric. Later, Matthews [2] 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 ([3–12]).

The object of this paper tries to give a generalized quasimetric p̂, i.e., $d_{p}$ [6] is its special case, p̃ [13] and $p^{s}$ [9] 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

([1])

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

• (M1) $x=y \Leftrightarrow d(x,y)=d(y,x)=0$;

• (M2) $d(x,z)\le d(x,y)+d(y,z)$.

A quasimetric d is called a metric if it also satisfies

• (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

([2])

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

• (P1) $x=y\Leftrightarrow p(x,x)=p(x,y)=p(y,y)$;

• (P2) $p(x,x)\le p(x,y)$;

• (P3) $p(x,y)=p(y,x)$;

• (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 [2]).

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) p̂ is a quasimetric.

• (2) p̃ 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) q̂ 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.

• (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).

• (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, p̂ is a quasimetric.

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

Remark 2.4

• (1) If $\alpha =1$ and $\beta =0$, then p̂ is q (see [2]).

• (2) If $\alpha =\beta =\frac{1}{2}$, then p̃ and $p^{s}$ are equivalent (see [4]).

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) 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) The set of all open p̂-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 p̂ on X.

Proof

(1) It is trivial by Theorem 3.1 in [2].

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

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

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

• (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].

(2) By Lemma 2.3(1), we know that p̂ 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) A sequence $\{{x_{n}}\}$ converges to a point $x\in X$ if $p(x,x)= \lim_{n\to +\infty}p(x,x_{n})$;

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

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

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

• (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 [14].

(4) It is trivial by Lemma 2.1 in [4]. □

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

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

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

• (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, $\hat{p}(x_{0},x^{*})<1$. □