## 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 [216].

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

([1921])

1. (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 $$(X,d_{p})$$.

2. (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:

1. (A)

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

2. (B)

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

3. (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:

1. (A)

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

2. (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:

1. (i)

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

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

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

$$\lim_{n,m\to\infty} qp_{b} (x_{n} ,x_{m}) \quad\mbox{and}\quad \lim_{n,m\to\infty}qp_{b} (x_{m} ,x_{n}) \mbox{ exist (and are finite)}.$$
3. (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

$$qp_{b} (x,x)=\lim_{n,m\to\infty}qp_{b} (x_{m} ,x_{n})=\lim_{n,m\to\infty} qp_{b} (x_{n} ,x_{m}).$$
4. (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:

1. (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}
2. (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:

1. (A)

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

2. (B)

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

3. (C)

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

4. (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:

1. (A)

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

2. (B)

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

3. (C)

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

4. (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:

1. (A)

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

2. (B)

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

3. (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:

1. (A)

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

2. (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}

□