Partial quasi-metric completeness via Kannan-type fixed points

TL;DR

This paper extends Kannan's fixed point theorem to partial quasi-metric spaces and uses these results to characterize a specific form of completeness, supported by an illustrative example.

Contribution

It introduces partial quasi-metric versions of Kannan's fixed point theorem and applies them to characterize a type of completeness in these spaces.

Findings

Established fixed point theorems for self-mappings in partial quasi-metric spaces.

Characterized a specific type of completeness using fixed point results.

Provided an example supporting the theoretical findings.

Abstract

In this short note, we obtain partial quasi-metric versions of Kannan's fixed point theorem for self-mappings. Moreover, we use these fixed points results to characterize a certain type of completeness in partial quasi-metric spaces. We have reported an example to support our result.

Equations

$$d(Tx,Ty)\le K\{d(x,Tx)+d(y,Ty)\}$$

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

$$p(x,ϵ)=\{y\in X:p(x,y)<ϵ+p(x,x)\}$$

$$\underset{n\le m,n,m\to \infty }{\lim }p(x_n,x_m)$$

$$\underset{n,m\to \infty }{\lim }{p}^{+}(x_n,x_m)$$

$$\underset{n<m,n,m\to \infty }{\lim }p(x_n,x_m)=\underset{n\to \infty }{\lim }p(x,x_n)=p(x,x).$$

$$\underset{n,m\to \infty }{\lim }{p}^{+}(x_n,x_m)=\underset{n\to \infty }{\lim }p(x,x_n)=p(x,x).$$

$$\underset{n,m\to \infty }{\lim }\text{ means “simultaneously", i.e. }\underset{n\to \infty ,m\to \infty }{\lim }$$

$$\tau (p)\text{-Smyth complete}⟹\text{left }p\text{-sequentially complete}⟹p\text{-sequentially complete}.$$

$$p(Tx,Ty)\le \lambda [p(x,Tx)+p(y,Ty)]$$

$$p^{+}(Tx,Ty)\le \gamma [p^{+}(x,Tx)+p^{+}(y,Ty)],\text{ for all }x,y\in X.$$

$$\underset{n,m\to \infty }{\lim }{p}^{+}(x_n,x_m)=0.$$

$$p(Tx,Ty)\le \lambda [p(x,Tx)+p(y,Ty)]\text{and}p(Ty,Tx)\le \lambda [p(y,Ty)+p(x,Tx)],$$

$$p^{+}(Tx,Ty)\le 2\lambda [p(x,Tx)+p(y,Ty)],$$

$$p^{+}(Tx,Ty)\le 2\lambda [p^{+}(x,Tx)+p^{+}(y,Ty)]=\gamma [p^{+}(x,Tx)+p^{+}(y,Ty)],$$

$$\underset{n,m\to \infty }{\lim }{p}^{+}(x_n,x_m)=0.$$

$$0=\underset{n,m\to \infty }{\lim }{p}^{+}(x_n,x_m)=\underset{n\to \infty }{\lim }p(z,x_n)=p(z,z).$$

$$p(z,Tz)$$

$$\le p(z,T^n{x}_0)+p(T^n{x}_0,Tz)$$

$$\le p(z,T^n{x}_0)+\lambda [d(T^{n-1}{x}_0,T^n{x}_0)+p(z,Tz)],$$

$$p^{+}(Tz,T^{2}z)\le 2\lambda (p(z,Tz)+p(Tz,T^{2}z)),$$

$$p^{+}(u,Tz)=p^{+}(Tu,T^{2}z)\le \lambda [p(u,Tu)+p(Tz,T^{2}z)].$$

$$p^{+}(Tx,Ty)=p^{+}\left(0,\frac{y}{8}\right)=\frac{y}{4}\le \frac{2}{15}\left(2x+\frac{15}{8}y\right)=\frac{2}{15}(p(x,Tx)+p(y,Ty).$$

$$p^{+}(Tx,Ty)=p^{+}\left(\frac{x}{8},\frac{y}{8}\right)=\frac{y}{4}\le \frac{2}{15}\left(\frac{15}{8}x+\frac{15}{8}y\right)=\frac{2}{15}(p(x,Tx)+p(y,Ty).$$

$$p^{+}(x_n,x_m)<\frac{1}{4}p(x,C_x),\text{ for all }m,n\ge n(x).$$

$$p^{+}(Tx,Ty)$$

$$\le \frac{1}{8}p(x,x_{n(x)})=\frac{1}{8}p(x,Tx).$$

Full text

1. Introduction and preliminaries

In 1968, Kannan[3] proved the following fixed point theorem:

Theorem 1.1**.**

Let $(X,d)$ be a complete metric space and $T$ be a self-mapping on $X$ satisfying

[TABLE]

for all $x,y\in X$ and $K\in[0,\frac{1}{2})$. Then $T$ has a unique fixed point $z\in X$, and for any $x\in X$ the sequence of iterates $(T^{n}x)$ converges to $z$.

This result by Kannan is an extension of Banach contraction principle in the sense that it proves that there exists a contractive map accompanied with fixed point which is not necessarily continuous. Kannan’s theorem is important because Subrahmanyam [7] proved that Kannan’s theorem characterizes the metric completeness. That is, a metric space $X$ is complete if and only if every mapping satisfying (1.1) on $X$ with constant $K<\frac{1}{2}$ has a fixed point. However contractions (in the sense of Banach) do not have this property.

On the other hand, and motivated in part by the fact that partial quasi-metric spaces provide suitable frameworks in several areas of asymmetric functional analysis, domain theory, complexity analysis and in modelling partially defined information, which often appears in computer science, the development of the fixed point theory for theses spaces appears to be an interesting focus for current research. In this setting, the problem of characterizing partial quasi-metric completeness via fixed point theorems arises in a natural way. This problem is indeed more interesting due to the fact that asymmetric structures present a natural inclination for different type of completeness.

Partial metrics were introduced by Matthews[5] in 1992. They generalize the concept of a metric space in the sense that the self-distance from a point to itself need not be equal to zero. In [4], by dropping the symmetry condition in the definition of a partial metric, Künzi et al. studied another variant of partial metrics, namely partial quasi-metrics.

In concluding this introductory part, we recall some pertinent notions and properties on partial quasi-metric spaces which will be useful later on.

Definition 1.2**.**

(Compare [5]) A partial metric type on a set $X$ is a function $p:X\times X\to[0,\infty)$ such that:

1. (pm1)

$x=y$ iff $(p(x,x)=p(x,y)=p(y,y)$ whenever $x,y\in X$, 2. (pm2)

$p(x,x)\leq p(x,y)$ whenever $x,y\in X$, 3. (pm3)

$p(x,y)=p(y,x);$ whenever $x,y\in X$, 4. (pm4)

[TABLE]

for any points $x,y,z\in X$.

The pair $(X,p)$ is called a partial metric space.

It is clear that, if $p(x,y)=0$ , then, from (pm1) and (pm2), $x=y$.

Definition 1.3**.**

([4, Definition 1.]) A partial quasi-metric on a set $X$ is a function $p:X\times X\to[0,\infty)$ such that:

1. (1a)

$p(x,x)\leq p(x,y)$ whenever $x,y\in X$, 2. (1b)

$p(x,x)\leq p(y,x)$ whenever $x,y\in X$, 3. (2)

$p(x,z)+p(y,y)\leq(p(x,y)+p(y,z))$ whenever $x,y,z\in X$, for some $K\geq 1,$ 4. (3)

$x=y$ iff $(p(x,x)=p(x,y)$ and $p(y,y)=p(y,x))$ whenever $x,y\in X$.

The pair $(X,p)$ will be called partial quasi-metric space.

If $p$ satisfies all these conditions except possibly (1b), we shall speak of a lopsided partial quasi-metric type or a lopsided partial quasi-metric.

Remark 1.4**.**

If $p$ is a partial quasi-metric on $X$ satisfying $(4)$ $p(x,y)=p(y,x)$ whenever $x,y\in X$, then $p$ is called a partial metric on $X$ in the sense of [5].

Lemma 1.5**.**

([4, Lemma 2.])

1. (a)

Each quasi-metric $p$ on X is a partial quasi-metric on $X$ with $p(x,x)=0$ whenever $x\in X$. 2. (b)

If $p$ is a partial quasi-metric on $X$, then so is its conjugate $p^{-1}(x,y)=p(y,x)$ whenever$x,y\in X$. 3. (c)

If $p$ is a partial quasi-metric on $X$, then $p^{+}$ defined by $p^{+}(x,y)=p(x,y)+p^{-1}(x,y)$ is a partial $K$-metric on $X$.

The notions such as convergence, completeness, Cauchy sequence in the setting of partial metric spaces, can be found in [2, 5] and references therein.

For every $K$-partial quasi-metric space $(X,p,K)$, the collection of balls

[TABLE]

yields a base for a $T_{0}$-Topology $\tau(p)$ on $X$.

Now, we define Cauchy sequence and convergent sequence in $K$-partial quasi-pseudometric spaces.

Definition 1.6**.**

Let $(X,p)$ be a partial quasi-metric space. Let $(x_{n})_{n\geq 1}$ be any sequence in $X$ and $x\in X$. Then:

1. (1)

The sequence $(x_{n})_{n\geq 1}$ is said to be convergent with respect to $\tau(p)$ (or $\tau(p)$-convergent) and converges to $x$, if $\lim\limits_{n\to\infty}p(x,x_{n})=p(x,x)$. 2. (2)

The sequence $(x_{n})_{n\geq 1}$ is said to be convergent with respect to $\tau(p^{+})$ (or $\tau(p^{+})$-convergent) and converges to $x$, if $\lim\limits_{n\to\infty}p^{+}(x_{n},x)=p^{+}(x,x)$. 3. (3)

The sequence $(x_{n})_{n\geq 1}$ is said to be a left $p$-Cauchy sequence if

[TABLE]

exists and is finite. 4. (4)

The sequence $(x_{n})_{n\geq 1}$ is said to be $\tau(p^{+})$-Cauchy sequence if

[TABLE]

exists and is finite. 5. (5)

$(X,p)$ is said to be $\tau(p)$-complete if for every $\tau(p)$-Cauchy sequence $(x_{n})_{n\geq 1}\subseteq X$, there exists $x\in X$ such that:

[TABLE] 6. (6)

$(X,p)$ is said to be left $p$-sequentially complete if every left $p$-Cauchy sequence converges for the topology $\tau(p)$. 7. (7)

$(X,p)$ is said to be $p$-sequentially complete if every $\tau(p^{+})$-Cauchy sequence is $\tau(p)$-convergent, i.e. there exists $x\in X$ such that:

[TABLE] 8. (8)

$(X,p)$ is said to be $\tau(p)$-Smyth complete if every left $p$-Cauchy sequence is $\tau(p^{+})$-convergent.

Remark 1.7**.**

It is worthwhile here to point out the fact that, in this manuscript, the notation

[TABLE]

Indeed, in other contexts it means $\lim\limits_{n\to\infty}\lim\limits_{m\to\infty}$, for instance, which is clearly a different type of convergence in general .

Also, the following implications are easy to check:

[TABLE]

2. Main results

Our proofs are inspired by the recent work of Romaguera et al. [1].

Definition 2.1**.**

Let $(X,p)$ be a partial quasi-metric space. By a $p$-Kannan mapping on $X$, we mean a self-mapping $T$ on $X$ such that there exists a constant $0\leq\lambda<1/4$ satisfying

[TABLE]

for all $x,y\in X.$

Lemma 2.2**.**

Let $T$ be a $p$-Kannan mapping on the partial quasi-metric space $(X,p,K)$ with $0\leq\lambda<1/4$. Then:

1. (a)

$p^{+}(Tx,Ty)\leq 2\lambda(p(x,Tx)+p(y,Ty))$, for all $x,y\in X.$ 2. (b)

$T$* is a Kannan mapping on the partial metric space $(X,p^{+})$ with $0\leq\gamma<1/2$, i.e.*

[TABLE] 3. (c)

For any $x_{0}\in X$, the sequence $(T^{n}x_{0})_{n\geq 1}$, is $\tau(p^{+})$-Cauchy sequence and

[TABLE]

Proof.

(a) Given $x,y\in X$, we have

[TABLE]

so

[TABLE]

(b) Since $p(x,Tx)\leq p^{+}(x,Tx)$ and $p(y,Ty)\leq p^{+}(y,Ty)$ for any $x,y\in X$,

[TABLE]

where $\gamma=2\lambda.$ It follows that $T$ is a Kannan mapping on the partial metric space $(X,p^{+})$ with $0\leq\gamma<1/2$.

(c) Since $T$ is a Kannan mapping on the partial $K$-metric space $(X,p^{+})$ (here $K=1$), the classical proof of Shukla’s fixed point theorem [6] shows that for any $x_{0}\in X$, the sequence $(T^{n}x_{0})_{n\geq 1}$, is $\tau(p^{+})$-Cauchy sequence and

[TABLE]

∎

Theorem 2.3**.**

Let $(X,p)$ be a $p$-sequentially complete partial quasi-metric space. Then every $p$-Kannan mapping on $(X,p,K)$ with constant $0\leq\lambda<1/4$ has a unique fixed point $x^{*}$ and $p(x^{*},x^{*})=0.$

Proof.

Let $T$ be a $p$-Kannan mapping on $(X,p)$. Then, there exists $c\in[0,1/4)$ such that the contraction condition (2.1) follows for all $x,y\in X.$ For any $x_{0}\in X$ fixed, Lemma 2.2 guarantees that the sequence $(T^{n}x_{0})_{n\geq 1}$ is a $\tau(p^{+})$-Cauchy sequence in the partial metric space $(X,p^{+})$. Since $(X,p)$ is $p$-sequentially complete, there exists $z\in X$ such that

[TABLE]

Next we show that $Tz$ is the unique fixed point of $T$. To this end, we first show that $p(z,Tz)=0$. Indeed, we have

[TABLE]

for all $n\in\mathbb{N}$. Using (2.2), we deduce that $p(z,Tz)\leq\lambda p(z,Tz)$. Consequently, $p(z,Tz)=0$.

Since by Lemma 2.2(a),

[TABLE]

we deduce that $d^{+}(Tz,T2z)\leq cp(Tz,T^{2}z),$ so $p^{+}(Tz,T^{2}z)=0$, i.e., $Tz$ is a fixed point of $T$.

Now, let us show that if $u\in X$ is a fixed point of $T$, that is, $Tu=u$, then $p(u,u)=0.$

Indeed, from (2.1), $p^{+}(u,u)=p^{+}(Tu,Tu)\leq\lambda[p(u,Tu)+p(u,Tu)]=2\lambda p(u,u)<p(u,u),$ a contradiction. Therefore, we must have $p(u,u)=0.$

Finally, if $Tu=u$ , it follows from Lemma 2.2(a) that

[TABLE]

Since $p(u,Tu)=p(Tz,T^{2}z)=0$, we deduce that $p^{+}(u,Tz)=0$, i.e., $u=Tz$ and the fixed point is unique.

This concludes the proof.

∎

The following examples illustrate Theorem 2.3.

Example 2.4**.**

Let $X=[0,\infty)$ and let $p$ be the partial quasi-metric on X given by $p(x,y)=\max\{x-y,0\}+x$ for all $x,y\in X$. $(X,p)$ is $p$-sequentially complete (in fact, it is left $p$-sequentially complete because every left $p$-Cauchy sequence in $X$ converges to 0 for $\tau(p)$).

Now define $T:X\to X$ as $Tx=0$ if $x\in[0,1]$ and $Tx=x/8$ if $x\in(1,\infty)$.

Let $x,y\in X$, and assume, without loss of generality, that $x\leq y.$ We distinguish the three following cases:

• •

Case 1: If $x,y\in[0,1],$ $p^{+}(Tx,Ty)=p^{+}(0,0)=0.$

• •

Case 2: If $x\in[0,1]$ and $y\in(1,\infty)$, we have

[TABLE]

• •

Case 3: If $x,y\in(1,\infty)$, we have

[TABLE]

Therefore $T$ is a $p$-Kannan mapping on $(X,p)$ for $c=2/15$. Thus, all conditions of Theorem 2.3 are satisfied. In fact $z=0$ is the unique fixed point of $T$.

Theorem 2.5**.**

A partial quasi-metric space (X, d) is $p$-sequentially complete if and only if every $p$-Kannan mapping on $(X,p)$ has a fixed point.

Proof.

Suppose that $(X,p)$ is $p$-sequentially complete. Then, every $p$-Kannan mapping on $(X,p)$ has a (unique) fixed point by Theorem 2.3.

For the converse suppose that $(X,p)$ is not $p$-sequentially complete. Then there exists a $\tau(p^{+})$-Cauchy sequence $(x_{n})_{n\geq 1}$ such that $\lim\limits_{n,m\to\infty}p^{+}(x_{n},x_{m})=0$ that does not converge for $\tau(p)$. Then, for each $x\in X$ there exists $n_{x}\in\mathbb{N}$ such that $p(x,x_{n})>0,$ for all $n\geq n_{x}$ (indeed, otherwise there is $x\in X$ such that for each $n\in\mathbb{N}$ we can find $m_{n}\geq n$ for which $p(x,x_{m_{n}})=0;$ since $(x_{n}))_{n\geq 1}$ is a $\tau(p^{+})$-Cauchy sequence it follows that $(x_{n})_{n\geq 1}$ converges to $x$ for $\tau(p)$, a contradiction).

Now, for each $x\in X$ put $C_{x}=\{xn:n\geq n_{x}\}$ and observe that $p(x,C_{x}):=\inf\{p(x,y),y\in C_{x}\}>0.$ Since $(x_{n})_{n\geq 1}$ is a $\tau(p^{+})$-Cauchy sequence such that $\lim\limits_{n,m\to\infty}p^{+}(x_{n},x_{m})=0$, for each $x\in X$ there exists $n(x)\geq n_{x}$ such that

[TABLE]

Define $T:X\to X$ as $Tx=x_{n(x)}$ for all $x\in X$. Since $n(x)\geq n_{x}$, we have that $p(x,x_{n(x)})>0$, and hence $T$ has no fixed point.

To complete the proof, we shall show that, $T$ is a $p$-Kannan mapping on $(X,p)$ for $c=1/8$. Indeed, let $x,y\in X$ and suppose, without loss of generality, that $n(x)\leq n(y)$. Then

[TABLE]

Since $p(Tx,Ty)\leq p^{+}(Tx,Ty)$ and $p(Ty,Tx)\leq p^{+}(Tx,Ty)$, we conclude that $T$ is a $p$-Kannan mapping on $(X,p)$ for $c=1/8$. This contradiction finishes the proof.

∎