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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.

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 ([312]).

The object of this paper tries to give a generalized quasimetric , i.e., \(d_{p}\) [6] is its special case, [13] and \(p^{s}\) [9] are equivalent. Furthermore, we obtain some results on (strong) complete partial metric spaces.

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: \(\forall x,y,z\in X\),

  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

([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 [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. (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. (2)

    If \(\alpha =\beta =\frac{1}{2}\), then 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. (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].

(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].

(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. □

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 [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. (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, \(\hat{p}(x_{0},x^{*})<1\). □

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Acknowledgements

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

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

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Wu, Y. A new approach on generalized quasimetric spaces induced by partial metric spaces. J Inequal Appl 2022, 61 (2022). https://doi.org/10.1186/s13660-022-02800-5

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MSC

  • 47H10
  • 54A05
  • 54H25

Keywords

  • Partial metric
  • Quasimetric
  • Fixed-point theorem
  • Strong complete partial metric spaces