Quasi-partial b-metric spaces and some related fixed point theorems

Abstract

In this paper, the quasi-partial b-metric space is defined and general fixed point theorems on this space are discussed with examples.

1 Introduction

A generalization of the metric space can be obtained as a partial-metric space by replacing the condition $d(x,x)=0$ with the condition $d(x,x)\le d(x,y)$ for all x, y in the definition of the metric. In the year 1993, Czerwik [1] introduced the concept of a b-metric space as another generalization of the concept of metric space. Several authors have focused on fixed point theorems for a metric space, a partial-metric space, quasi-partial metric space and a partial b-metric space. For further information on the subject see [2–16].

The concept of a quasi-partial-metric space was introduced by Karapınar et al. [17]. He studied some fixed point theorems on these spaces whereas Shatanawi and Pitea [18] studied some coupled fixed point theorems on quasi-partial-metric spaces.

The aim of this paper is to introduce the concept of quasi-partial b-metric spaces which is a generalization of the concept of quasi-partial-metric spaces. The fixed point results are proved in setting of such spaces and some examples are given to verify the effectiveness of the main results.

2 Preliminaries

We begin the section with some basic definitions and concepts.

Definition 2.1

([17])

A quasi-partial metric on a non-empty set X is a function $q:X\times X\rightarrow \mathbb{R}^{+}$, satisfying

(QPM₁):

If $q(x,x)=q(x,y)=q(y,y)$, then $x=y$.

(QPM₂):

$q(x,x)\le q(x,y)$.

(QPM₃):

$q(x,x)\le q(y,x)$.

(QPM₄):

$q(x,y)+q(z,z)\le q(x,z)+q(z,y)$ for all $x,y,z\in X$.

A quasi-partial-metric space is a pair $(X, q)$ such that X is a non-empty set and q is a quasi-partial metric on X.

Let q be a quasi-partial metric on the set X. Then

$$d_{q} (x,y)=q(x,y)+q(y,x)-q(x,x)-q(y,y) \mbox{ is a metric on }X. $$

Lemma 2.1

([17])

For a quasi-partial metric q on X,

$$p_{q} (x,y)=\frac{1}{2} \bigl[q(x,y)+q(y,x)\bigr] \quad\textit{for all } x,y\in X \textit{ is a partial metric on }X. $$

Lemmas 2.2

([19–21])

• (A) A sequence $\{x_{n} \}$ is Cauchy in a partial-metric space $(X,p)$ if and only if $\{x_{n} \}$ is Cauchy in the (corresponding) metric space $(X,d_{p})$.

• (B) A partial-metric space $(X,p)$ is complete if and only if the (corresponding) metric space $(X,d_{p})$ is complete. Moreover,

  $$\lim_{n\to\infty} d_{p} (x,x_{n})=0\quad \Leftrightarrow\quad p(x,x)=\lim_{n\to\infty} p(x,x_{n})=\lim _{n,m\to\infty} p(x_{n} ,x_{m}). $$

Lemma 2.3

([17])

Let $(X,q)$ be a quasi-partial metric space, let $(X,p_{q})$ be the corresponding partial-metric space, and let $(X,d_{p_{q} })$ be the corresponding metric space. Then the following statements are equivalent:

• (A) The sequence $\{x_{n} \}$ is Cauchy in $(X,q)$ and $(X,q)$ is complete.

• (B) The sequence $\{x_{n} \}$ is Cauchy in $(X,p_{q})$ and $(X,p_{q})$ is complete.

• (C) The sequence $\{x_{n} \}$ is Cauchy in $(X,d_{p_{q} })$ and $(X,d_{p_{q} })$ is complete.

Also,

$$\begin{aligned} &\lim_{n\to\infty} d_{q} (x,x_{n})=0 \quad\Leftrightarrow\quad p_{q} (x,x)=\lim_{n\to\infty} p_{q} (x,x_{n})=\lim_{n,m\to\infty}p_{q} (x_{n} ,x_{m})\\ &\hphantom{\lim_{n\to\infty} d_{q} (x,x_{n})=0}\quad\Leftrightarrow\quad q(x,x)=\lim_{n\to\infty}q(x,x_{n})= \lim_{n,m\to\infty}q(x_{n} ,x_{m})\\ &\hphantom{\lim_{n\to\infty} d_{q} (x,x_{n})=0\quad\Leftrightarrow\quad q(x,x)} =\lim_{n\to\infty} q(x_{n},x)=\lim _{n,m\to\infty} q(x_{m},x_{n}). \end{aligned}$$

Definition 2.2

([17])

If $T:X\to X$ is any map on X, $O(x)=\{x,Tx,T^{2} x,\ldots\}$ is called the orbit of x. A mapping $G:X\to{\mathbb{R}}^{+}$ is T-orbitally lower semi-continuous at x if $\{x_{n} \}$ is a sequence in $O(x)$ and $\lim x_{n} =z$ implies $G(z)\le\liminf G(x_{n})$.

3 Quasi-partial b-metric space

We introduce the concept of quasi-partial b-metric space here.

Definition 3.1

A quasi-partial b-metric on a non-empty set X is a mapping $qp_{b} :X\times X\to{\mathbb{R}}^{+}$ such that for some real number $s \geq1$ and all $x, y, z\in X$:

(QPb₁):

$qp_{b} (x,x)=qp_{b} (x,y)=qp_{b} (y,y)\Rightarrow x=y$,

(QPb₂):

$qp_{b} (x,x)\le qp_{b} (x,y)$,

(QPb₃):

$qp_{b} (x,x)\le qp_{b} (y,x)$,

(QPb₄):

$qp_{b} (x,y)\le s[qp_{b} (x,z)+qp_{b} (y,z)]-qp_{b} (z,z)$.

A quasi-partial b-metric space is a pair $(X,qp_{b})$ such that X is a non-empty set and $(X,qp_{b})$ is a quasi partial b-metric on X. The number s is called the coefficient of $(X,qp_{b})$.

For a quasi-partial b-metric space $(X,qp_{b})$, the function $d_{qp_{b} } :X\times X\to{\mathbb{R}}^{+}$ defined by

$$d_{qp_{b} } (x,y)=qp_{b} (x,y)+qp_{b} (y,x)-qp_{b} (x,x)-qp_{b} (y,y) \mbox{ is a }b \mbox{-metric on }X. $$

Example 3.1

Let $X=[0,1]$.

Define $qp_{b}(x,y)=|x-y|+x$. Here

$$\begin{aligned} qp_{b}(x,x)=qp_{b}(x,y)=qp_{b}(y,y)\quad \Rightarrow\quad x=y \quad\mbox{as } x=|x-y|+x=y \mbox{ gives } x=y . \end{aligned}$$

Again, $qp_{b}(x,x)\le qp_{b}(x,y)$ as $x\le|x-y|+x$ and similarly, $qp_{b}(x,x)\le qp_{b}(y,x)$ as $x\le|y-x|+y$ for $0< x<y$.

Also $qp_{b}(x,y)+qp_{b}(z,z)\le s[qp_{b}(x,z)+qp_{b}(z,y)]$ as

$$|x-y|+x+z\le s\bigl[|x-z|+x+|z-y|+z\bigr] \quad\mbox{for all } s\ge1. $$

It can be observed that

$$\begin{aligned} |x-y|+x+z=|x-z+z-y|+x+z\le|x-z|+|z-y|+x+z . \end{aligned}$$

So $(X,qp_{b})$ is a quasi-partial b-metric space with $s\ge1$.

Example 3.2

Let $X=[1, \infty)$.

Define $qp_{b} :X\times X\to{\mathbb{R}}^{+}$ as $qp_{b} (x,y)=\ln (xy)$. Then $(X,qp_{b})$ is a quasi-partial b-metric space.

Let $qp_{b} (x,x)=qp_{b} (x,y)=qp_{b} (y,y) \Rightarrow \ln(x^{2})=\ln(xy)=\ln(y^{2}) \Rightarrow x=y$.

Let $x,y\in X$. Without loss of generality $x \leq y \Rightarrow\ln x\le\ln y \Rightarrow2\ln x\le\ln x+\ln y \Rightarrow\ln(x^{2})\le\ln x+\ln y$.

Thus, $qp_{b} (x,x)\le qp_{b} (x,y)$.

Similarly $qp_{b} (x,x)\le qp_{b} (y,x)$.

For (QPb₄) we have

$$\begin{aligned} qp_{b} (x,y) &=\ln x+\ln y \\ &\le s\ln x+s\ln y\quad\mbox{since }s\ge1\mbox{ and also }\ln x\ge0\mbox{ and }\ln y \ge0 \\ &\le s\ln x+s\ln y+2\ln z(s-1)\quad\mbox{since }\ln z\ge0\mbox{ and }s-1\ge 0 \\ &=s\bigl\{ qp_{b} (x,z)+qp_{b} (z,y)\bigr\} -qp_{b} (z,z). \end{aligned}$$

Example 3.3

Let $X= [0, \frac{\pi}{4} ]$ and define $qp_{b}:X\times X\to \mathbb{R}^{+}$ as

$$qp_{b} (x,y)=\sin x+\sin y. $$

Then $(X,qp_{b})$ is a quasi-partial b-metric space.

Lemma 3.4

Let $(X,qp_{b})$ be a quasi-partial b-metric space. Then the following hold:

• (A) If $qp_{b} (x,y)=0$ then $x = y$.

• (B) If $x\ne y$, then $qp_{b} (x,y)>0$ and $qp_{b} (y,x)>0$.

The proof is similar to the case of quasi-partial-metric space [17].

Lemma 3.5

Every quasi-partial space is a quasi-partial b-metric space. But the converse does not need to be true.

Definition 3.2

Let $(X,qp_{b})$ be a quasi-partial b-metric. Then:

• (i) A sequence $\{x_{n} \} \subset X$ converges to $x\in X$ if and only if

  $$qp_{b} (x,x)=\lim_{n\to\infty} qp_{b} (x,x_{n})=\lim_{n\to\infty} qp_{b} (x_{n} ,x). $$

• (ii) A sequence $\{x_{n} \} \subset X$ is called a Cauchy sequence if and only if

  $$\lim_{n,m\to\infty} qp_{b} (x_{n} ,x_{m}) \quad\mbox{and}\quad \lim_{n,m\to\infty}qp_{b} (x_{m} ,x_{n}) \mbox{ exist (and are finite)}. $$

• (iii) The quasi-partial b-metric space $(X,qp_{b})$ is said to be complete if every Cauchy sequence $\{x_{n} \}\subset X$ converges with respect to $\tau_{qp_{b} }$ to a point $x\in X$ such that

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

• (iv) A mapping $f:X\to X$ is said to be continuous at $x_{0} \in X$ if, for every $\varepsilon>0$, there exists $\delta>0$ such that $f(B(x_{0} ,\delta))\subset B(f(x_{0}),\varepsilon)$.

Lemma 3.6

Let $(X,qp_{b})$ be a quasi-partial b-metric space and $(X,d_{qp_{b} })$ be the corresponding b-metric space. Then $(X,d_{qp_{b} })$ is complete if $(X,qp_{b})$ is complete.

Proof

Since $(X,qp_{b})$ is complete, every Cauchy sequence $\{x_{n} \}$ in X converges with respect to $\tau_{qp_{b} }$ to a point $x\in X$ such that

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

(1)

Consider a Cauchy sequence $\{x_{n} \}$ in $(X,d_{qp_{b} })$. We will show that $\{x_{n} \}$ is Cauchy in $(X,qp_{b})$. Since $\{x_{n} \}$ is Cauchy in $(X,d_{qp_{b} })$, $\lim_{n,m\to\infty} d_{qp_{b} } (x_{n} ,x_{m})$ exists and is finite.

Also, $d_{qp_{b} } (x_{n} ,x_{m})=qp_{b} (x_{n} ,x_{m})+qp_{b} (x_{m} ,x_{n})-qp_{b} (x_{n} ,x_{n})-qp_{b} (x_{m} ,x_{m})$.

Clearly, $\lim_{n,m\to\infty} qp_{b} (x_{n} ,x_{m})$ and $\lim_{n,m\to\infty} qp_{b} (x_{m} ,x_{n})$ exist and are finite.

Therefore, $\{x_{n} \}$ is a Cauchy sequence in $(X,qp_{b})$. Now, since $(X,qp_{b})$ is complete, the sequence $\{x_{n} \}$ converges with respect to $\tau_{qp_{b} }$to a point $x\in X$ such that (1) holds.

For $\{x_{n} \}$ to be convergent in $(X,d_{qp_{b} })$ we will show that $d_{qp_{b} } (x,x)=\lim_{n\to\infty} d_{qp_{b} } (x,x_{n})$.

If follows from the definition of $d_{qp_{b} }$ that $d_{qp_{b} } (x,x)=0$. Also,

$$\begin{aligned} \lim_{n\to\infty} d_{qp_{b} } (x,x_{n}) &=\lim _{n\to\infty} qp_{b} (x,x_{n})+\lim _{n\to\infty } qp_{b} (x_{n} ,x) -\lim _{n\to\infty} qp_{b} (x_{n},x_{n})-\lim _{n\to\infty}qp_{b} (x,x) \\ &= 0 \quad\mbox{by (1) and definition of convergence in }(X,qp_{b}). \end{aligned}$$

Hence, $d_{qp_{b} } (x,x)=\lim_{n\to\infty} d_{qp_{b} } (x,x_{n})$. □

In [17] Karapınar et al. proved a fixed point theorem on quasi-partial-metric space. Motivated by this, we have generalized the results on a quasi-partial b-metric space.

4 The main results

Theorem 4.1

Let $(X,qp_{b})$ be a quasi-partial b-metric space, and let $T:X\to X$. Then the following hold:

• (A) There exists $\phi:X\to{\mathbb{R}}^{+}$ such that

  $$\begin{aligned} &qp_{b} (x,Tx)\le\phi(x)-\phi(Tx) \quad\textit{for all } x \in X \quad\textit{if and only if} \\ &\quad\sum_{n=0}^{\infty} qp_{b} \bigl(T^{n} x,T^{n+1} x\bigr) \textit{ converges for all } x\in X. \end{aligned}$$

• (B) There exists $\phi:X\to{\mathbb{R}}^{+}$ such that

  $$\begin{aligned} &qp_{b} (x,Tx)\le\phi(x)-\phi(Tx) \quad\textit{for all }x\in O(x)\quad\textit{if and only if} \\ &\quad\sum_{n=0}^{\infty} qp_{b} \bigl(T^{n} x,T^{n+1} x\bigr) \textit{ converges for all }x\in O(x). \end{aligned}$$

Proof

(A) Let $x\in X$, and let

$$qp_{b} (x,Tx)\le\phi(x)-\phi(Tx). $$

Define the sequence $\{x_{n} \} _{n=1}^{\infty}$ in the following way:

$$x_{0} =x \quad\mbox{and} \quad x_{n+1} =Tx_{n} =T^{n+1} x_{0}, \quad\mbox{for all } n = 0, 1, 2, \ldots. $$

Set $z_{n} (x)=\sum_{k=0}^{n} qp_{b} (x_{k} ,x_{k+1})=\sum_{k=0}^{n} qp_{b} (T^{k} x_{0} ,T^{k+1} x_{0})$. Then

$$\begin{aligned} z_{n} (x) &\le\sum_{k=0}^{n} \bigl[\phi\bigl(T^{k} x_{0}\bigr)-\phi\bigl(T^{k+1} x_{0}\bigr)\bigr] \\ &=\bigl[\phi(x_{0})-\phi(Tx_{0})\bigr]+\cdots+\bigl[\phi \bigl(T^{n} x_{0}\bigr)-\phi\bigl(T^{n+1} x_{0}\bigr)\bigr] \\ &=\bigl[\phi(x_{0})-\phi\bigl(T^{n+1} x_{0}\bigr) \bigr]\le\phi(x_{0})=\phi(x). \end{aligned}$$

(2)

Thus, (2) implies that $\{z_{n} (x)\}$ is bounded. Also $\{ z_{n} (x)\}$ is non-decreasing and hence convergent. Therefore, $\sum_{n=0}^{\infty} qp_{b} (T^{n} x,T^{n+1} x)$ converges.

Conversely, define

$$\phi(x)=\sum_{n=0}^{\infty} qp_{b} \bigl(T^{n} x,T^{n+1} x\bigr) \quad\mbox{and}\quad z_{n} (x)=\sum_{k=0}^{n} qp_{b} \bigl(T^{k} x, T^{k+1} x\bigr). $$

Then

$$\phi(Tx)=\sum_{n=0}^{\infty} qp_{b} \bigl(T^{n+1} x,T^{n+2} x\bigr) \quad\mbox{and}\quad z_{n} (Tx)=\sum_{k=0}^{n} qp_{b} \bigl(T^{k+1} x,T^{k+2} x\bigr). $$

Using these definitions, we get

$$ \begin{aligned}[b] z_{n} (x)-z_{n} (Tx) &=\sum _{k=0}^{n} qp_{b} \bigl(T^{k} x,T^{k+1} x\bigr)-\sum_{k=0}^{n} qp_{b} \bigl(T^{k+1} x,T^{k+2} x\bigr)\\ &=qp_{b} (x,Tx)-qp_{b} \bigl(T^{n+1} x,T^{n+2} x\bigr) . \end{aligned} $$

(3)

Since $\sum_{n=0}^{\infty} qp_{b} (T^{n} x, T^{n+1} x)$ converges for all $x \in X$,

$$\lim_{n\to\infty}z_{n} (x)=\phi(x) \quad\mbox{and}\quad \lim_{n\to\infty}qp_{b} \bigl(T^{n} x,T^{n+1} x\bigr)=0. $$

Letting $n\to\infty$ in (3) gives $qp_{b} (x,Tx)=\phi (x)-\phi(Tx)$.

(B) It can easily be proved using part (A). □

Example 4.1

Let $X=[0, 1]$. Define $qp_{b} (x,y)=|x-y| + |x|$.

Then $qp_{b} (x,y)$ satisfies all conditions of quasi-partial b-metric space. It is also quasi-partial metric. But for $x\ne y$, $qp_{b} (x,y)\ne qp_{b} (y,x)$ and $qp_{b} (x,x)\ne0$ for $x\ne0$. So $qp_{b}$ is not a partial metric or a quasi-metric. Define $T:X\to X$ as $Tx=\frac{x}{3}$ for all $x\in X$. Then the series $\sum_{n=0}^{\infty} qp_{b} (T^{n} x,T^{n+1} x)$ is convergent. Indeed,

$$\begin{aligned} \sum_{n=0}^{\infty} qp_{b} \bigl(T^{n} x,T^{n+1} x\bigr) &=\sum _{n=0}^{\infty} qp_{b} \biggl(\frac{x}{3^{n} }, \frac {x}{3^{n+1} } \biggr) =\sum_{n=0}^{\infty} \biggl\vert \frac {x}{3^{n} } -\frac{x}{3^{n+1} } \biggr\vert +\biggl\vert \frac{x}{3^{n} } \biggr\vert \\ &=\sum_{n=0}^{\infty} \biggl\vert \frac{2x}{3^{n+1} } \biggr\vert +\biggl\vert \frac{x}{3^{n} } \biggr\vert = \sum_{n=0}^{\infty} \frac {5x}{3^{n+1} } = \frac{5x}{3}\cdot\frac{1}{1-\frac{1}{3} } =\frac {5x}{2} . \end{aligned}$$

Then the conditions of Theorem 4.1 are satisfied for $\phi(x)=\frac {5x}{2}$. Indeed

$$\begin{aligned} qp_{b} (x,Tx) &=qp_{b} \biggl(x,\frac{x}{3} \biggr)= \biggl\vert x-\frac{x}{3} \biggr\vert +|x|=\biggl\vert \frac{2x}{3} \biggr\vert +|x|=\frac{5x}{3}=\phi(x)-\phi(Tx). \end{aligned}$$

The next result gives conditions for the existence of fixed points of operators on quasi-partial b-metric space.

Theorem 4.2

Let $(X, qp_{b})$ and $(Y, qp_{b})$ be complete quasi-partial b-metric spaces. Let also $T:X\to X$, $R: X\to Y$, and $\phi:R(X)\to{\mathbb{R}}^{+}$. If there exist $x\in X$ and $c>0$ such that

$$ \max\bigl\{ qp_{b} (y, Ty), cqp_{b} (Ry, RTy)\bigr\} \le\phi(Ry)-\phi(RTy) $$

(4)

for all $y\in O(x)$, then the following hold:

• (A) $\lim_{n\to\infty} T^{n} x=z$ exists.

• (B) $Tz=z$ if and only if $G(x)=qp_{b} (x, Tx)$ is T-orbitally lower semi-continuous at x.

• (C) $qp_{b} (x, T^{n} x)\le s^{n-1} \phi(Rx)$.

• (D) For $m > n$, $qp_{b} (T^{n} x, T^{m} x)\le s^{m-n} [\phi(RT^{n} x)]$.

Proof

(A) Let $x\in X$. Define the sequence $\{x_{n} \} _{n=1}^{\infty}$ as follows:

$$x_{0} =x \quad\mbox{and}\quad x_{n+1} =Tx_{n} =T^{n+1} x_{0}, \quad\mbox{for all } n = 0, 1, 2,\ldots . $$

We will show that $\{x_{n} \} _{n=1}^{\infty}$ is Cauchy.

Using (QPb₄), we get

$$\begin{aligned} qp_{b} (x_{n} ,x_{n+2}) &\le s\bigl\{ qp_{b} (x_{n},x_{n+1})+qp_{b} (x_{n+1},x_{n+2})\bigr\} -qp_{b} (x_{n+1},x_{n+1}) \\ &\le s\bigl\{ qp_{b} (x_{n},x_{n+1})+qp_{b} (x_{n+1},x_{n+2})\bigr\} \end{aligned}$$

(5)

and, similarly,

$$\begin{aligned} qp_{b} (x_{n},x_{n+3}) &\le s\bigl\{ qp_{b} (x_{n},x_{n+2})+qp_{b} (x_{n+2},x_{n+3})\bigr\} -qp_{b} (x_{n+2},x_{n+2}) \\ &\le s^{2} \bigl\{ qp_{b} (x_{n},x_{n+1})+qp_{b} (x_{n+1},x_{n+2})\bigr\} + s\bigl\{ qp_{b} (x_{n+2},x_{n+3})\bigr\} . \end{aligned}$$

(6)

Now,

$$\begin{aligned} qp_{b} (x_{n},x_{n+4}) \le{}& s\bigl\{ qp_{b} (x_{n},x_{n+3})+qp_{b} (x_{n+3},x_{n+4})\bigr\} -qp_{b} (x_{n+3},x_{n+3}) \\ \ \le{}& s^{3} \bigl\{ qp_{b} (x_{n},x_{n+1})+qp_{b} (x_{n+1},x_{n+2})\bigr\} +s^{2} \bigl\{ qp_{b} (x_{n+2},x_{n+3})\bigr\} \\ &{} +s\bigl\{ qp_{b} (x_{n+3},x_{n+4})\bigr\} . \end{aligned}$$

On generalization, we get

$$\begin{aligned} qp_{b} (x_{n},x_{m}) \le{}& s^{m-n-1} \bigl\{ qp_{b} (x_{n},x_{n+1})+qp_{b} (x_{n+1},x_{n+2})\bigr\} \\ &{} +s^{m-n-2} \bigl\{ qp_{b} (x_{n+2},x_{n+3}) \bigr\} +\cdots+s\bigl\{ qp_{b} (x_{m-1},x_{m})\bigr\} \\ \le{}& s^{m-n-1} \bigl\{ qp_{b} \bigl(T^{n} x, T^{n+1} x\bigr)+qp_{b} \bigl(T^{n+1} x, T^{n+2} x\bigr)\bigr\} \\ &{} +s^{m-n-2} \bigl\{ qp_{b} \bigl(T^{n+2} x, T^{n+3} x\bigr)\bigr\} +\cdots+s\bigl\{ qp_{b} \bigl(T^{m-1} x, T^{m} x\bigr)\bigr\} \\ ={}&\sum_{k=n+1}^{m-1} s^{m-k} \bigl\{ qp_{b} \bigl(T^{k} x, T^{k+1} x\bigr)\bigr\} +s^{m-n-1} qp_{b} (x_{n},x_{n+1}) \\ ={}&\sum_{k=n}^{m-1} s^{m-k} \bigl\{ qp_{b} \bigl(T^{k} x, T^{k+1} x\bigr)\bigr\} +s^{m-n-1} qp_{b} (x_{n},x_{n+1}) -s^{m-n} qp_{b} (x_{n},x_{n+1}) \\ ={}&\sum_{k=n}^{m-1} s^{m-k} \bigl\{ qp_{b} \bigl(T^{k} x, T^{k+1} x\bigr)\bigr\} -s^{m-n} qp_{b} (x_{n},x_{n+1}) \biggl[1- \frac{1}{s} \biggr] \\ \le{}&\sum_{k=n}^{m-1} s^{m-k} \bigl\{ qp_{b} \bigl(T^{k} x, T^{k+1} x\bigr)\bigr\} \quad \mbox{for } m > n. \end{aligned}$$

(7)

Set $z_{n} (x)=\sum_{k=0}^{n} s^{m-k} \{qp_{b} (T^{k} x, T^{k+1} x)\}$.

From (4) we have

$$\begin{aligned}& \begin{aligned}[b] s^{m-k} \bigl\{ qp_{b} \bigl(T^{k} x, T^{k+1} x\bigr)\bigr\} &\le s^{m-k} \max\bigl\{ qp_{b} \bigl(T^{k} x, T^{k+1} x\bigr), cqp_{b} \bigl(RT^{k} x, RT^{k+1} x\bigr)\bigr\} \\ & \le s^{m-k} \bigl\{ \phi\bigl(RT^{k} x\bigr)-\phi \bigl(RT^{k+1} x\bigr)\bigr\} \quad\mbox{for all } k = 0, 1, \ldots \end{aligned} \end{aligned}$$

(8)

$$\begin{aligned}& \begin{aligned}[b] \quad\Rightarrow\quad z_{n} (x)\le{}&\sum _{k=0}^{n} s^{m-k} \bigl\{ \phi \bigl(RT^{k} x\bigr)-\phi\bigl(RT^{k+1} x\bigr)\bigr\} \\ \le{}& s^{m} \phi(Rx)-s^{m} \phi(RTx)+s^{m} \phi(RTx)-s^{m-1} \phi \bigl(RT^{2} x\bigr)+\cdots \\ &{} +s^{m-n+1} \phi\bigl(RT^{n} x\bigr)-s^{m-n} \phi \bigl(RT^{n+1} x\bigr) \\ ={}&s^{m} \phi(Rx)-s^{m-n} \phi\bigl(RT^{n+1} x \bigr) \\ \le{}& s^{m} \phi(Rx). \end{aligned} \end{aligned}$$

(9)

Thus, $\sum_{k=0}^{\infty} s^{m-k} \{ qp_{b} (T^{k} x, T^{k+1} x)\}$ is convergent.

$$\begin{aligned} \Rightarrow \quad\sum_{n=0}^{\infty} s^{m-n} \bigl\{ qp_{b} \bigl(T^{n} x, T^{n+1} x\bigr)\bigr\} \mbox{ is convergent}. \end{aligned}$$

Taking the limit as $n,m\to\infty$ in (7), we get

$$ \lim_{m,n\to\infty} qp_{b} (x_{n},x_{m})=\lim_{m,n\to \infty} \bigl(z_{m-1} (x)-z_{n-1} (x)\bigr)=0. $$

(10)

Using similar arguments,

$$\begin{aligned} \lim_{m,n\to\infty} qp_{b} (x_{m} ,x_{n})=0. \end{aligned}$$

(11)

Thus the sequence $\{x_{n} \}$ is Cauchy in $(X,qp_{b})$. Since $(X,qp_{b})$is complete, $(X,d_{qp_{b} })$ is also complete by Lemma 2.3, and hence $\lim_{n\to\infty} d_{qp_{b} } (T^{n} x,z)=0$, $\lim_{n\to\infty} T^{n} x=z$.

Further, $\lim_{n\to\infty} qp_{b} (T^{n} x, T^{n+1} x)=0$ and hence $\lim_{n\to\infty} qp_{b} (T^{n} x, T^{n+1} x)=qp_{b} (z,z)=0$.

(B) Assume that $Tz=z$ and that $x_{n}$ is a sequence in $O(x)$ with $x_{n} \to z$.

By Lemma 3.6,

$$\begin{aligned} \lim_{n\to\infty} d_{qp_{b} } (z,x_{n})=0 \quad \Leftrightarrow\quad qp_{b} (z,z)=\lim_{n\to\infty} qp_{b} (z,x_{n}) =\lim_{n,m\to\infty} qp_{b} (x_{n} ,x_{m}). \end{aligned}$$

(12)

Then $G(z)=qp_{b} (z,Tz)=qp_{b} (z,z)\le\lim_{n\to\infty} \inf qp_{b} (x_{n} ,Tx_{n})=\lim_{n\to\infty} \inf G(x_{n})$.

Thus G is T-orbitally lower semi-continuous at x.

Conversely, suppose that $x_{n} =T^{n} x\to z$ and that G is T-orbitally lower semi-continuous at x. Then

$$\begin{aligned} 0&\le qp_{b} (z, Tz)=G(z) \le\lim_{n\to\infty} \inf G(x_{n})=\lim_{n\to\infty } \inf qp_{b} \bigl(T^{n} x, T^{n+1} x\bigr) \\ &=\lim_{n\to\infty} \inf qp_{b} (x_{n},x_{n+1})= qp_{b} (z,z)=0. \end{aligned}$$

(13)

By Lemma 3.4, we have $Tz=z$.

(C) We have, from (QPb₄) and (4),

$$\begin{aligned}& \begin{aligned}[b] qp_{b} \bigl(x, T^{2} x\bigr) &\le s \bigl\{ qp_{b} (x, Tx)+qp_{b} \bigl(Tx, T^{2} x \bigr)\bigr\} -qp_{b} (Tx, Tx) \\ &\le s\bigl\{ qp_{b} (x, Tx)+qp_{b} \bigl(Tx, T^{2} x\bigr)\bigr\} , \end{aligned} \\& \begin{aligned}[b] qp_{b} \bigl(x, T^{3} x\bigr)&\le s \bigl\{ qp_{b} \bigl(x, T^{2} x\bigr)+qp_{b} \bigl(T^{2} x, T^{3} x\bigr)\bigr\} -qp_{b} \bigl(T^{2} x, T^{2} x\bigr) \\ &\le s\bigl[s\bigl\{ qp_{b} (x, Tx)+qp_{b} \bigl(Tx, T^{2} x\bigr)\bigr\} +qp_{b} \bigl(T^{2} x, T^{3} x\bigr)\bigr] \\ &\le s^{2} \bigl\{ qp_{b} (x, Tx)+qp_{b} \bigl(Tx, T^{2} x\bigr)\bigr\} +s\bigl\{ qp_{b} \bigl(T^{2} x, T^{3} x\bigr)\bigr\} . \end{aligned} \end{aligned}$$

On generalization, we get

$$\begin{aligned} &qp_{b} \bigl(x, T^{n} x\bigr) \\ &\quad\le s^{n-1} \bigl\{ qp_{b} (x, Tx)+qp_{b} \bigl(Tx, T^{2} x\bigr)\bigr\} +s^{n-2} \bigl\{ qp_{b} \bigl(T^{2} x, T^{3} x\bigr)\bigr\} + \cdots \\ &\qquad{} +s\bigl\{ qp_{b} \bigl(T^{n-1} x, T^{n} x \bigr)\bigr\} \\ &\quad\le s^{n-1} \bigl\{ qp_{b} (x, Tx)\bigr\} +s^{n-1} \bigl\{ qp_{b} \bigl(Tx, T^{2} x\bigr)\bigr\} +s^{n-2} \bigl\{ qp_{b} \bigl(T^{2} x, T^{3} x\bigr)\bigr\} +\cdots \\ &\qquad{} +s\bigl\{ qp_{b} \bigl(T^{n-1} x, T^{n} x \bigr)\bigr\} \\ &\quad\le s^{n-1} \bigl\{ \phi(Rx)-\phi(RTx)\bigr\} +s^{n-1} \bigl\{ \phi(RTx)-\phi \bigl(RT^{2} x\bigr)\bigr\} \\ &\qquad{} +s^{n-2} \bigl\{ \phi\bigl(RT^{2} x\bigr)-\phi \bigl(RT^{3} x\bigr)\bigr\} +\cdots+s\bigl\{ \phi \bigl(RT^{n-1} x\bigr)-\phi\bigl(RT^{n} x\bigr)\bigr\} \\ &\quad\le s^{n-1} \phi(Rx)-s^{n-1} \phi\bigl(RT^{2} x\bigr)+s^{n-2} \phi\bigl(RT^{2} x\bigr) -s^{n-2} \phi\bigl(RT^{3} x\bigr)+\cdots \\ &\qquad{} +s\phi\bigl(RT^{n-1} x\bigr)-s\phi\bigl(RT^{n} x \bigr) \\ &\quad\le s^{n-1} \phi(Rx)-s\phi\bigl(RT^{2} x\bigr)-s\phi \bigl(RT^{n-1} x\bigr)-s\phi \bigl(RT^{n} x\bigr) \\ &\quad\le s^{n-1} \phi(Rx). \end{aligned}$$

(14)

(D) From (7) we get

$$qp_{b} (x_{n},x_{m})\le\sum _{k=n}^{m-1} s^{m-k} \bigl\{ qp_{b} \bigl(T^{k} x, T^{k+1} x\bigr)\bigr\} \quad\mbox{for } m >n. $$

Note that

$$\begin{aligned} &\sum_{k=n}^{m-1} s^{m-k} qp_{b} \bigl(T^{k} x, T^{k+1} x \bigr) \\ &\quad\le\sum_{k=n}^{m-1} s^{m-k} \bigl[\phi\bigl(RT^{k} x\bigr)-\phi\bigl(RT^{k+1} x\bigr)\bigr] \\ &\quad =s^{m-n} \phi\bigl(RT^{n} x\bigr)-s^{m-n} \phi\bigl(RT^{n+1} x\bigr)+s^{m-n-1} \phi\bigl(RT^{n+1} x \bigr) \\ &\qquad{} -s^{m-n-1} \phi\bigl(RT^{n+2} x\bigr)+\cdots+s\phi \bigl(RT^{m-1} x\bigr)-s\phi\bigl(RT^{m} x\bigr) \\ &\quad =s^{m-n} \phi\bigl(RT^{n} x\bigr)-s\phi \bigl(RT^{n+1} x\bigr)-s\phi\bigl(RT^{m-1} x\bigr)-s\phi \bigl(RT^{m} x\bigr) \\ & \quad\le s^{m-n} \phi\bigl(RT^{n} x\bigr). \end{aligned}$$

(15)

Here, $0\le qp_{b} (x_{n},x_{m}) =qp_{b} (T^{n} x, T^{m} x)\le s^{m-n} \phi(RT^{n} x)$ for $m>n$. □

Example 4.2

Let $X=Y=[0,1]$. Define $qp_{b}(x,y)=|x-y|+x$. Then $qp_{b}$ is a quasi-partial b-metric with $s=1$. Also define $T:X\to X$ as $T(x)=\frac{x}{3}$; $R:X\to Y$ as $R(x)=3x$, and $\phi:R(X)\to \mathbb{R}^{+}$ as $\phi(x)=3x$. Then for $c=1$ and $x\in [0,1]$ we have

$$\begin{aligned} \max\bigl\{ qp_{b}(y,Ty),cqp_{b}(Ry,RTy)\bigr\} &=\max \biggl\{ qp_{b} \biggl(y,\frac{y}{3} \biggr),qp_{b}(3y,y) \biggr\} \\ &=\max \biggl\{ \biggl\vert y-\frac{y}{3}\biggr\vert +y,|3y-y|+3y \biggr\} \\ &=\max \biggl\{ \frac{5y}{3},5y \biggr\} =5y< 6y=\phi(3y)-\phi(y) \\ &=\phi(Ry)-\phi(RTy). \end{aligned}$$

We now prove that (A), (B), (C), and (D) of the above theorem hold:

(A) $\lim_{n\to\infty}T^{n}x=\lim_{n\to\infty}\frac {x}{3^{n}}=0=z$ (say).

So $\lim_{n\to\infty}T^{n}x=z$ exists.

(B) By (A) part above, $z=0$.

Therefore $T(z)=T(0)=0=z$ holds trivially.

Hence whenever $G(x)=qp_{b}(x,Tx)$ is T-orbitally lower semi-continuous at x then $Tz=z$.

Conversely, let $Tz=z$ and we show that G is T-orbitally lower semi-continuous at x, i.e.,

$$G(z)\le\lim\inf G(x_{n}) \quad\forall \{x_{n}\}\subseteq O(x), x_{n}\to z. $$

Let $\{x_{n}\}\subseteq O(x)$ be a sequence converging to z. Then

$$\begin{aligned} G(z) &=qp_{b}(z,Tz)=qp_{b}(z,z)=z \\ &=\frac{5z}{3} \ (\mbox{as }z=0) =\lim\inf\frac{5x_{n}}{3} \\ &=\lim\inf\frac{2x_{n}}{3}+x_{n}=\lim\inf\biggl\vert x_{n}-\frac{x_{n}}{3}\biggr\vert +x_{n} \\ &=\lim\inf qp_{b} \biggl(x_{n},\frac{x_{n}}{3} \biggr)= \lim\inf qp_{b} (x_{n},Tx_{n})=\lim\inf G(x_{n}) . \end{aligned}$$

Hence $G(z)=\lim\inf G(x_{n})$.

$\begin{array}{ll} (\mathrm{C})\ \displaystyle qp_{b}\bigl(x,T^{n}x\bigr) &\displaystyle=qp_{b} \biggl(x, \frac{x}{3^{n}} \biggr)=\biggl \vert x-\frac{x}{3^{n}}\biggr \vert +x=x \biggl(2-\frac{1}{3^{n}} \biggr)< x(9) \quad \forall n\in N\\ \displaystyle&\displaystyle=\phi(3x) =s^{n-1}\phi(Rx) \quad\mbox{where } s=1. \end{array}$

(D) Let $m>n$ then

$$\begin{aligned} qp_{b}\bigl(T^{n}x,T^{m}x\bigr) &=qp_{b} \biggl(\frac{x}{3^{n}},\frac{x}{3^{m}} \biggr)=\biggl\vert \frac {x}{3^{n}}-\frac{x}{3^{m}}\biggr\vert +\frac{x}{3^{n}} \\ &=\frac{x}{3^{n}} \biggl[2-\frac{1}{3^{m-n}} \biggr]< \frac{x}{3^{n}}(9) \quad \forall n\in N \\ &=\phi \biggl(\frac{x}{3^{n-1}} \biggr)=\phi\bigl(3T^{n} x \bigr)=s^{m-n}\bigl[\phi\bigl(RT^{n} x\bigr)\bigr] \quad \mbox{where }s=1. \end{aligned}$$

Corollary 4.3

Let $(X, qp_{b})$ be a complete quasi-partial b-metric space. Let $T:X\to X$ and $\phi:X\to{\mathbb{R}}^{+}$. Suppose that there exists $x\in X$ such that

$$qp_{b} (y, Ty)\le\phi(y)-\phi(Ty) \quad\textit{for all } y\in O(x). $$

Then the following hold:

• (A) $\lim_{n\to\infty} T^{n} x=z$ exists.

• (B) $Tz=z$ if and only if $G(x)=qp_{b} (x, Tx)$ is T-orbitally lower semi-continuous at x.

• (C) $qp_{b} (x, T^{n} x)\le s^{n-1} \phi(x)$.

• (D) For $m > n$, $qp_{b} (T^{n} x, T^{m} x)\le s^{m-n} \phi(T^{n} x)$.

Proof

Take $Y = X$, $R = I$, and $c = 1$ in Theorem 4.2. □

Corollary 4.4

Let $(X, qp_{b})$ be a complete quasi-partial b-metric space, and let $0< k<1$. Suppose that $T:X\to X$ and that there exists $x\in X$ such that

$$\begin{aligned} qp_{b} \bigl(Ty, T^{2} y\bigr)\le kqp_{b} (y, Ty) \quad\textit{for all } y\in O(x). \end{aligned}$$

(16)

Then the following hold:

• (A) $\lim_{n\to\infty} T^{n} x=z$ exists.

• (B) $Tz=z$ if and only if $G(x)=qp_{b} (x, Tx)$ is T-orbitally lower semi-continuous at x.

• (C) $qp_{b} (x, T^{n} x)\le\frac{s^{n-1} }{1-k} qp_{b} (x, Tx)$.

Proof

Set $\phi(y)=\frac{1}{1-k} qp_{b} (y, Ty)$ for $y\in O(x)$.

Let $y=T^{n} x$ in (16). Then

$$qp_{b} \bigl(T^{n+1} x, T^{n+2} x\bigr)\le kqp_{b} \bigl(T^{n} x, T^{n+1} x\bigr) $$

and

$$qp_{b} \bigl(T^{n} x, T^{n+1} x \bigr)-kqp_{b} \bigl(T^{n} x, T^{n+1} x\bigr)\le qp_{b} \bigl(T^{n} x, T^{n+1} x\bigr)-qp_{b} \bigl(T^{n+1} x, T^{n+2} x\bigr). $$

Thus, $qp_{b} (T^{n} x, T^{n+1} x)\le\frac{1}{1-k} [qp_{b} (T^{n} x, T^{n+1} x)-qp_{b} (T^{n+1} x, T^{n+2} x)]$ or $qp_{b} (y,Ty)\le[\phi(y)-\phi(Ty)]$.

(A)-(C) follow immediately from Corollary 4.3. □

Corollary 4.5

Let $(X,qp_{b})$ be a complete quasi-partial b-metric space where $qp_{b}$ is continuous. Let $T:X\to X$ and $\phi:X\to{\mathbb{R}}^{+}$ is continuous. Suppose that there exists $x\in X$ such that

$$qp_{b} (y, Ty)\le\phi(y)-\phi(Ty) \quad\textit{for all } y\in O(x). $$

Then the following hold:

• (A) $\lim_{n\to\infty} T^{n} x=z$ exists.

• (B) $q_{p} (z,z)\le s\phi(z)$.

Proof

In Theorem 4.2(D) taking $m = n + 1$, $R = I$, $c = 1$, and $Y = X$,

$$qp_{b} \bigl(T^{n} x, T^{n+1} x\bigr)\le s \bigl[ \phi\bigl(RT^{n} x\bigr)\bigr]. $$

Now taking $\lim n\to\infty$

$$\begin{aligned} &\lim_{n\to\infty} qp_{b} \bigl(T^{n} x, T^{n+1} x\bigr)\le\lim_{n\to\infty} s \bigl[\phi \bigl(T^{n} x\bigr)\bigr], \\ &qp_{b} (z,z)\le s\phi(z). \end{aligned}$$

 □