## 1 Introduction

Fixed points theorems in partially ordered metric spaces were firstly obtained in 2004 by Ran and Reurings [1], and then by Nieto and Lopez [2]. In this direction several authors obtained further results under weak contractive conditions (see, e.g., [38]).

The concept of b-metric space was introduced by Bakhtin [9] and extensively used by Czerwik in [10, 11]. After that, several interesting results about the existence of a fixed point for single-valued and multi-valued operators in (ordered) b-metric spaces have been obtained (see, e.g., [1226]).

Definition 1 [10]

Let X be a (nonempty) set and $s\ge 1$ be a given real number. A function $d:X×X\to {\mathbb{R}}^{+}$ is a b-metric on X if, for all $x,y,z\in X$, the following conditions hold:

• (b1) $d\left(x,y\right)=0$ if and only if $x=y$,

• (b2) $d\left(x,y\right)=d\left(y,x\right)$,

• (b3) $d\left(x,z\right)\le s\left[d\left(x,y\right)+d\left(y,z\right)\right]$.

In this case, the pair $\left(X,d\right)$ is called a b-metric space.

On the other hand, Matthews [27] introduced the notion of a partial metric space as a part of the study of denotational semantics of dataflow networks. In partial metric spaces, self-distance of an arbitrary point need not be equal to zero. Several authors obtained many useful fixed point results in these spaces - we mention just [2833].

Definition 2 [27]

A partial metric on a nonempty set X is a mapping $p:X×X\to {\mathbb{R}}^{+}$ such that for all $x,y,z\in X$:

• (p1) $x=y$ if and only if $p\left(x,x\right)=p\left(x,y\right)=p\left(y,y\right)$,

• (p2) $p\left(x,x\right)\le p\left(x,y\right)$,

• (p3) $p\left(x,y\right)=p\left(y,x\right)$,

• (p4) $p\left(x,y\right)\le p\left(x,z\right)+p\left(z,y\right)-p\left(z,z\right)$.

In this case, $\left(X,p\right)$ is called a partial metric space.

It is clear that if $p\left(x,y\right)=0$, then from (p1) and (p2), $x=y$. But if $x=y$, $p\left(x,y\right)$ may not be 0. A basic example of a partial metric space is the pair $\left({\mathbb{R}}^{+},p\right)$, where $p\left(x,y\right)=max\left\{x,y\right\}$ for all $x,y\in {\mathbb{R}}^{+}$.

Each partial metric p on a set X generates a ${T}_{0}$ topology ${\tau }_{p}$ on X which has as a base the family of open p-balls $\left\{{B}_{p}\left(x,\epsilon \right):x\in X,\epsilon >0\right\}$, where ${B}_{p}\left(x,\epsilon \right)=\left\{y\in X:p\left(x,y\right) for all $x\in X$ and $\epsilon >0$.

Definition 3 [27]

Let $\left(X,p\right)$ be a partial metric space, and let $\left\{{x}_{n}\right\}$ be a sequence in X and $x\in X$. Then:

1. (i)

The sequence $\left\{{x}_{n}\right\}$ is said to converge to x with respect to ${\tau }_{p}$ if ${lim}_{n\to \mathrm{\infty }}p\left({x}_{n},x\right)=p\left(x,x\right)$.

2. (ii)

The sequence $\left\{{x}_{n}\right\}$ is said to be Cauchy in $\left(X,p\right)$ if ${lim}_{n,m\to \mathrm{\infty }}p\left({x}_{n},{x}_{m}\right)$ exists and is finite.

3. (iii)

$\left(X,p\right)$ is said to be complete if every Cauchy sequence $\left\{{x}_{n}\right\}$ in X converges, with respect to ${\tau }_{p}$, to a point $x\in X$ such that ${lim}_{n,m\to \mathrm{\infty }}p\left({x}_{n},{x}_{m}\right)={lim}_{n\to \mathrm{\infty }}p\left({x}_{n},x\right)=p\left(x,x\right)$.

The following example shows that a convergent sequence $\left\{{x}_{n}\right\}$ in a partial metric space $\left(X,p\right)$ may not be Cauchy. In particular, it shows that the limit may not be unique.

Example 1 [32]

Let $X=\left[0,\mathrm{\infty }\right)$ and $p\left(x,y\right)=max\left\{x,y\right\}$. Let

${x}_{n}=\left\{\begin{array}{cc}0,\hfill & n=2k,\hfill \\ 1,\hfill & n=2k+1.\hfill \end{array}$

Then, clearly, $\left\{{x}_{n}\right\}$ is a convergent sequence and for every $x\ge 1$, we have ${lim}_{n\to \mathrm{\infty }}p\left({x}_{n},x\right)=p\left(x,x\right)$. But ${lim}_{n,m\to \mathrm{\infty }}p\left({x}_{n},{x}_{m}\right)$ does not exist, that is, $\left\{{x}_{n}\right\}$ is not a Cauchy sequence.

As a generalization and unification of partial metric and b-metric spaces, Shukla [34] introduced the concept of partial b-metric space as follows.

Definition 4 [34]

A partial b-metric on a nonempty set X is a mapping ${p}_{b}:X×X\to {\mathbb{R}}^{+}$ such that for all $x,y,z\in X$:

• (${\text{p}}_{b1}$) $x=y$ if and only if ${p}_{b}\left(x,x\right)={p}_{b}\left(x,y\right)={p}_{b}\left(y,y\right)$,

• (${\text{p}}_{b2}$) ${p}_{b}\left(x,x\right)\le {p}_{b}\left(x,y\right)$,

• (${\text{p}}_{b3}$) ${p}_{b}\left(x,y\right)={p}_{b}\left(y,x\right)$,

• (${\text{p}}_{b4}$) ${p}_{b}\left(x,y\right)\le s\left[{p}_{b}\left(x,z\right)+{p}_{b}\left(z,y\right)\right]-{p}_{b}\left(z,z\right)$.

A partial b-metric space is a pair $\left(X,{p}_{b}\right)$ such that X is a nonempty set and ${p}_{b}$ is a partial b-metric on X. The number $s\ge 1$ is called the coefficient of $\left(X,{p}_{b}\right)$.

In a partial b-metric space $\left(X,{p}_{b}\right)$, if $x,y\in X$ and ${p}_{b}\left(x,y\right)=0$, then $x=y$, but the converse may not be true. It is clear that every partial metric space is a partial b-metric space with the coefficient $s=1$ and every b-metric space is a partial b-metric space with the same coefficient and zero self-distance. However, the converse of these facts need not hold.

Example 2 [34]

Let $X={\mathbb{R}}^{+}$, $q>1$ be a constant and ${p}_{b}:X×X\to {\mathbb{R}}^{+}$ be defined by

Then $\left(X,{p}_{b}\right)$ is a partial b-metric space with the coefficient $s={2}^{q-1}>1$, but it is neither a b-metric nor a partial metric space.

Note that in a partial b-metric space the limit of a convergent sequence may not be unique (see [[34], Example 2]).

Some more examples of partial b-metrics can be constructed with the help of the following propositions.

Proposition 1 [34]

Let X be a nonempty set, and let p be a partial metric and d be a b-metric with the coefficient $s\ge 1$ on X. Then the function ${p}_{b}:X×X\to {\mathbb{R}}^{+}$, defined by ${p}_{b}\left(x,y\right)=p\left(x,y\right)+d\left(x,y\right)$ for all $x,y\in X$, is a partial b-metric on X with the coefficient s.

Proposition 2 [34]

Let $\left(X,p\right)$ be a partial metric space and $q\ge 1$. Then $\left(X,{p}_{b}\right)$ is a partial b-metric space with the coefficient $s={2}^{q-1}$, where ${p}_{b}$ is defined by ${p}_{b}\left(x,y\right)={\left[p\left(x,y\right)\right]}^{q}$.

Altering distance functions were introduced by Khan et al. in [35].

Definition 5 [35]

A function $\psi :\left[0,\mathrm{\infty }\right)\to \left[0,\mathrm{\infty }\right)$ is called an altering distance function if the following properties are satisfied:

1. 1.

ψ is continuous and nondecreasing;

2. 2.

$\psi \left(t\right)=0$ if and only if $t=0$.

So far, many authors have studied fixed point theorems which are based on altering distance functions (see, e.g., [12, 28, 3641]).

In this paper, we introduce a modified version of ordered partial b-metric spaces. We demonstrate a fundamental lemma for the convergence of sequences in such spaces. Using this lemma, we prove some fixed point and common fixed point results for $\left(\psi ,\phi \right)$-weakly contractive mappings in the setup of ordered partial b-metric spaces. Finally, examples are presented to verify the effectiveness and applicability of our main results.

## 2 Definition and basic properties of partial b-metric spaces

In the following definition, we modify Definition 4 in order to obtain that each partial b-metric ${p}_{b}$ generates a b-metric ${d}_{{p}_{b}}$.

Definition 6 Let X be a (nonempty) set and $s\ge 1$ be a given real number. A function ${p}_{b}:X×X\to {\mathbb{R}}^{+}$ is a partial b-metric if, for all $x,y,z\in X$, the following conditions are satisfied:

• (${\text{p}}_{b1}$) $x=y⟺{p}_{b}\left(x,x\right)={p}_{b}\left(x,y\right)={p}_{b}\left(y,y\right)$,

• (${\text{p}}_{b2}$) ${p}_{b}\left(x,x\right)\le {p}_{b}\left(x,y\right)$,

• (${\text{p}}_{b3}$) ${p}_{b}\left(x,y\right)={p}_{b}\left(y,x\right)$,

• (${\text{p}}_{b{4}^{\prime }}$) ${p}_{b}\left(x,y\right)\le s\left({p}_{b}\left(x,z\right)+{p}_{b}\left(z,y\right)-{p}_{b}\left(z,z\right)\right)+\left(\frac{1-s}{2}\right)\left({p}_{b}\left(x,x\right)+{p}_{b}\left(y,y\right)\right)$.

The pair $\left(X,{p}_{b}\right)$ is called a partial b-metric space.

Since $s\ge 1$, from (${\text{p}}_{b{4}^{\prime }}$) we have

${p}_{b}\left(x,y\right)\le s\left({p}_{b}\left(x,z\right)+{p}_{b}\left(z,y\right)-{p}_{b}\left(z,z\right)\right)\le s\left({p}_{b}\left(x,z\right)+{p}_{b}\left(z,y\right)\right)-{p}_{b}\left(z,z\right).$

Hence, a partial b-metric in the sense of Definition 6 is also a partial b-metric in the sense of Definition 4.

It should be noted that the class of partial b-metric spaces is larger than the class of partial metric spaces, since a partial b-metric is a partial metric when $s=1$. We present an example which shows that a partial b-metric on X (in the sense of Definition 6) might be neither a partial metric, nor a b-metric on X.

Example 3 Let $\left(X,d\right)$ be a metric space and ${p}_{b}\left(x,y\right)=d{\left(x,y\right)}^{q}+a$, where $q>1$ and $a\ge 0$ are real numbers. We will show that ${p}_{b}$ is a partial b-metric with $s={2}^{q-1}$.

Obviously, conditions (${\text{p}}_{b1}$)-(${\text{p}}_{b3}$) of Definition 6 are satisfied.

Since $q>1$, the convexity of the function $f\left(x\right)={x}^{q}$ ($x>0$) implies that ${\left(a+b\right)}^{q}\le {2}^{q-1}\left({a}^{q}+{b}^{q}\right)$ holds for $a,b\ge 0$. Thus, for each $x,y,z\in X$, we obtain

$\begin{array}{rl}{p}_{b}\left(x,y\right)& =d{\left(x,y\right)}^{q}+a\le {\left(d\left(x,z\right)+d\left(z,y\right)\right)}^{q}+a\\ \le {2}^{q-1}\left(d{\left(x,z\right)}^{q}+d{\left(z,y\right)}^{q}\right)+a\\ ={2}^{q-1}\left(d{\left(x,z\right)}^{q}+a+d{\left(z,y\right)}^{q}+a-a\right)+a-{2}^{q-1}a\\ ={2}^{q-1}\left({p}_{b}\left(x,z\right)+{p}_{b}\left(z,y\right)-{p}_{b}\left(z,z\right)\right)+\left(\frac{1-{2}^{q-1}}{2}\right)\left({p}_{b}\left(x,x\right)+{p}_{b}\left(y,y\right)\right).\end{array}$

Hence, condition (${\text{p}}_{b{4}^{\prime }}$) of Definition 6 is fulfilled and ${p}_{b}$ is a partial b-metric on X.

Note that $\left(X,{p}_{b}\right)$ is not necessarily a partial metric space. For example, if $X=\mathbb{R}$ is the set of real numbers, $d\left(x,y\right)=|x-y|$, $q=2$ and $a=3$, then ${p}_{b}\left(x,y\right)={\left(x-y\right)}^{2}+3$ is a partial b-metric on X with $s={2}^{2-1}=2$, but it is not a partial metric on X. Indeed, the ordinary (partial) triangle inequality does not hold. To see this, let $x=2$, $y=5$ and $z=\frac{5}{2}$. Then ${p}_{b}\left(2,5\right)=12$, ${p}_{b}\left(2,\frac{5}{2}\right)=\frac{13}{4}$ and ${p}_{b}\left(\frac{5}{2},5\right)=\frac{37}{4}$, hence ${p}_{b}\left(2,5\right)=12\nleqq \frac{38}{4}={p}_{b}\left(2,\frac{5}{2}\right)+{p}_{b}\left(\frac{5}{2},5\right)-{p}_{b}\left(\frac{5}{2},\frac{5}{2}\right)$.

Also, ${p}_{b}$ is not a b-metric since ${p}_{b}\left(x,x\right)\ne 0$ for $x\in X$.

Proposition 3 Every partial b-metric ${p}_{b}$ defines a b-metric ${d}_{{p}_{b}}$, where

${d}_{{p}_{b}}\left(x,y\right)=2{p}_{b}\left(x,y\right)-{p}_{b}\left(x,x\right)-{p}_{b}\left(y,y\right)$

for all $x,y\in X$.

Proof Let $x,y,z\in X$. Then we have

$\begin{array}{r}{d}_{{p}_{b}}\left(x,y\right)\\ \phantom{\rule{1em}{0ex}}=2{p}_{b}\left(x,y\right)-{p}_{b}\left(x,x\right)-{p}_{b}\left(y,y\right)\\ \phantom{\rule{1em}{0ex}}\le 2\left[s\left({p}_{b}\left(x,z\right)+{p}_{b}\left(z,y\right)-{p}_{b}\left(z,z\right)\right)+\left(\frac{1-s}{2}\right)\left({p}_{b}\left(x,x\right)+{p}_{b}\left(y,y\right)\right)\right]\\ \phantom{\rule{2em}{0ex}}-{p}_{b}\left(x,x\right)-{p}_{b}\left(y,y\right)\\ \phantom{\rule{1em}{0ex}}=2s{p}_{b}\left(x,z\right)+2s{p}_{b}\left(z,y\right)-2s{p}_{b}\left(z,z\right)+\left(1-s\right)\left({p}_{b}\left(x,x\right)+{p}_{b}\left(y,y\right)\right)\\ \phantom{\rule{2em}{0ex}}-{p}_{b}\left(x,x\right)-{p}_{b}\left(y,y\right)\\ \phantom{\rule{1em}{0ex}}=2s{p}_{b}\left(x,z\right)+2s{p}_{b}\left(z,y\right)-2s{p}_{b}\left(z,z\right)-s{p}_{b}\left(x,x\right)-s{p}_{b}\left(y,y\right)\\ \phantom{\rule{1em}{0ex}}=s\left[2{p}_{b}\left(x,z\right)-{p}_{b}\left(x,x\right)-{p}_{b}\left(z,z\right)+2{p}_{b}\left(z,y\right)-{p}_{b}\left(z,z\right)-{p}_{b}\left(y,y\right)\right]\\ \phantom{\rule{1em}{0ex}}=s\left[{d}_{{p}_{b}}\left(x,z\right)+{d}_{{p}_{b}}\left(z,y\right)\right].\end{array}$

□

Hence, the advantage of our definition of partial b-metric is that by using it we can define a dependent b-metric which we call the b-metric associated with ${p}_{b}$. This allows us to readily transport many concepts and results from b-metric spaces into a partial b-metric space.

Now, we present some definitions and propositions in a partial b-metric space.

Definition 7 Let $\left(X,{p}_{b}\right)$ be a partial b-metric space. Then, for $x\in X$ and $ϵ>0$, the ${p}_{b}$-ball with center x and radius ϵ is

${B}_{{p}_{b}}\left(x,ϵ\right)=\left\{y\in X\mid {p}_{b}\left(x,y\right)<{p}_{b}\left(x,x\right)+ϵ\right\}.$

For example, let $\left(X,{p}_{b}\right)$ be the partial b-metric space from Example 3 (with $X=\mathbb{R}$, $q=2$ and $a=3$). Then

$\begin{array}{rl}{B}_{{p}_{b}}\left(1,4\right)& =\left\{y\in X\mid {p}_{b}\left(1,y\right)<{p}_{b}\left(1,1\right)+4\right\}=\left\{y\in X\mid {\left(y-1\right)}^{2}+3<3+4\right\}\\ =\left\{y\in X\mid {\left(y-1\right)}^{2}<4\right\}=\left(-1,3\right).\end{array}$

Proposition 4 Let $\left(X,{p}_{b}\right)$ be a partial b-metric space, $x\in X$ and $r>0$. If $y\in {B}_{{p}_{b}}\left(x,r\right)$, then there exists $\delta >0$ such that ${B}_{{p}_{b}}\left(y,\delta \right)\subseteq {B}_{{p}_{b}}\left(x,r\right)$.

Proof Let $y\in {B}_{{p}_{b}}\left(x,r\right)$. If $y=x$, then we choose $\delta =r$. Suppose that $y\ne x$. Then we have ${p}_{b}\left(x,y\right)\ne 0$. Now, we consider two cases.

Case 1. If ${p}_{b}\left(x,y\right)={p}_{b}\left(x,x\right)$, then for $s=1$ we choose $\delta =r$. If $s>1$, then we consider the set

$A=\left\{n\in \mathbb{N}|\frac{r}{2{s}^{n+1}\left(s-1\right)}<{p}_{b}\left(x,x\right)\right\}.$

By the Archimedean property, A is a nonempty set; then by the well ordering principle, A has the least element m. Since $m-1\notin A$, we have ${p}_{b}\left(x,x\right)\le r/\left(2{s}^{m}\left(s-1\right)\right)$ and we choose $\delta =r/\left(2{s}^{m+1}\right)$. Let $z\in {B}_{{p}_{b}}\left(y,\delta \right)$; by the property (${\text{p}}_{b4}$), we have

$\begin{array}{rl}{p}_{b}\left(x,z\right)& \le s\left({p}_{b}\left(x,y\right)+{p}_{b}\left(y,z\right)-{p}_{b}\left(y,y\right)\right)\\ \le s\left({p}_{b}\left(x,x\right)+\delta \right)\\ \le {p}_{b}\left(x,x\right)+\frac{r}{2{s}^{m}}+\frac{r}{2{s}^{m}}\\ ={p}_{b}\left(x,x\right)+\frac{r}{{s}^{m}}\\ <{p}_{b}\left(x,x\right)+r.\end{array}$

Hence, ${B}_{{p}_{b}}\left(y,\delta \right)\subseteq {B}_{{p}_{b}}\left(x,r\right)$.

Case 2. If ${p}_{b}\left(x,y\right)\ne {p}_{b}\left(x,x\right)$, then from the property (${\text{p}}_{b2}$) we have ${p}_{b}\left(x,x\right)<{p}_{b}\left(x,y\right)$ and for $s=1$ we consider the set

$B=\left\{n\in \mathbb{N}|\frac{r}{{2}^{n+3}}<{p}_{b}\left(x,y\right)-{p}_{b}\left(x,x\right)\right\}.$

Similarly, by the well ordering principle, there exists an element m such that ${p}_{b}\left(x,y\right)-{p}_{b}\left(x,x\right)\le r/\left({2}^{m+2}\right)$, and we choose $\delta =r/\left({2}^{m+2}\right)$. One can easily obtain that ${B}_{{p}_{b}}\left(y,\delta \right)\subseteq {B}_{{p}_{b}}\left(x,r\right)$.

For $s>1$, we consider the set

$C=\left\{n\in \mathbb{N}|\frac{r}{2{s}^{n+2}}<{p}_{b}\left(x,y\right)-\frac{1}{s}{p}_{b}\left(x,x\right)\right\}$

and by the well ordering principle, there exists an element m such that ${p}_{b}\left(x,y\right)-\frac{1}{s}{p}_{b}\left(x,x\right)\le \frac{r}{2{s}^{m+1}}$ and we choose $\delta =\frac{r}{2{s}^{m+1}}$. Let $z\in {B}_{{p}_{b}}\left(y,\delta \right)$. By the property (${\text{p}}_{b4}$), we have

$\begin{array}{rl}{p}_{b}\left(x,z\right)& \le s\left({p}_{b}\left(x,y\right)+{p}_{b}\left(y,z\right)-{p}_{b}\left(y,y\right)\right)\\ \le s\left({p}_{b}\left(x,y\right)+\delta \right)\\ \le {p}_{b}\left(x,x\right)+\frac{r}{2{s}^{m}}+\frac{r}{2{s}^{m}}\\ ={p}_{b}\left(x,x\right)+\frac{r}{{s}^{m}}\\ <{p}_{b}\left(x,x\right)+r.\end{array}$

Hence, ${B}_{{p}_{b}}\left(y,\delta \right)\subseteq {B}_{{p}_{b}}\left(x,r\right)$. □

Thus, from the above proposition the family of all ${p}_{b}$-balls

$\mathrm{\Delta }=\left\{{B}_{{p}_{b}}\left(x,r\right)\mid x\in X,r>0\right\}$

is a base of a ${T}_{0}$ topology ${\tau }_{{p}_{b}}$ on X which we call the ${p}_{b}$-metric topology.

The topological space $\left(X,{p}_{b}\right)$ is ${T}_{0}$, but need not be ${T}_{1}$.

Definition 8 A sequence $\left\{{x}_{n}\right\}$ in a partial b-metric space $\left(X,{p}_{b}\right)$ is said to be:

1. (i)

${p}_{b}$-convergent to a point $x\in X$ if ${lim}_{n\to \mathrm{\infty }}{p}_{b}\left(x,{x}_{n}\right)={p}_{b}\left(x,x\right)$;

2. (ii)

a ${p}_{b}$-Cauchy sequence if ${lim}_{n,m\to \mathrm{\infty }}{p}_{b}\left({x}_{n},{x}_{m}\right)$ exists (and is finite).

3. (iii)

A partial b-metric space $\left(X,{p}_{b}\right)$ is said to be ${p}_{b}$-complete if every ${p}_{b}$-Cauchy sequence $\left\{{x}_{n}\right\}$ in X ${p}_{b}$-converges to a point $x\in X$ such that ${lim}_{n,m\to \mathrm{\infty }}{p}_{b}\left({x}_{n},{x}_{m}\right)={lim}_{n,m\to \mathrm{\infty }}{p}_{b}\left({x}_{n},x\right)={p}_{b}\left(x,x\right)$.

The following lemma shows the relationship between the concepts of ${p}_{b}$-convergence, ${p}_{b}$-Cauchyness and ${p}_{b}$-completeness in two spaces $\left(X,{p}_{b}\right)$ and $\left(X,{d}_{{p}_{b}}\right)$ which we state and prove according to Lemma 2.2 of [31].

Lemma 1

1. (1)

A sequence $\left\{{x}_{n}\right\}$ is a ${p}_{b}$-Cauchy sequence in a partial b-metric space $\left(X,{p}_{b}\right)$ if and only if it is a b-Cauchy sequence in the b-metric space $\left(X,{d}_{{p}_{b}}\right)$.

2. (2)

A partial b-metric space $\left(X,{p}_{b}\right)$ is ${p}_{b}$-complete if and only if the b-metric space $\left(X,{d}_{{p}_{b}}\right)$ is b-complete. Moreover, ${lim}_{n\to \mathrm{\infty }}{d}_{{p}_{b}}\left(x,{x}_{n}\right)=0$ if and only if

$\underset{n\to \mathrm{\infty }}{lim}{p}_{b}\left(x,{x}_{n}\right)=\underset{n,m\to \mathrm{\infty }}{lim}{p}_{b}\left({x}_{n},{x}_{m}\right)={p}_{b}\left(x,x\right).$

Proof First, we show that every ${p}_{b}$-Cauchy sequence in $\left(X,{p}_{b}\right)$ is a b-Cauchy sequence in $\left(X,{d}_{{p}_{b}}\right)$. Let $\left\{{x}_{n}\right\}$ be a ${p}_{b}$-Cauchy sequence in $\left(X,{p}_{b}\right)$. Then, there exists $\alpha \in \mathbb{R}$ such that, for arbitrary $\epsilon >0$, there is ${n}_{\epsilon }\in \mathbb{N}$ with

$|{p}_{b}\left({x}_{n},{x}_{m}\right)-\alpha |<\frac{\epsilon }{4}$

for all $n,m\ge {n}_{\epsilon }$. Hence,

$\begin{array}{r}|{d}_{{p}_{b}}\left({x}_{n},{x}_{m}\right)|\\ \phantom{\rule{1em}{0ex}}=2{p}_{b}\left({x}_{n},{x}_{m}\right)-{p}_{b}\left({x}_{n},{x}_{n}\right)-{p}_{b}\left({x}_{m},{x}_{m}\right)\\ \phantom{\rule{1em}{0ex}}=|{p}_{b}\left({x}_{n},{x}_{m}\right)-\alpha +\alpha -{p}_{b}\left({x}_{n},{x}_{n}\right)+{p}_{b}\left({x}_{m},{x}_{n}\right)-\alpha +\alpha -{p}_{b}\left({x}_{m},{x}_{m}\right)|\\ \phantom{\rule{1em}{0ex}}\le |{p}_{b}\left({x}_{n},{x}_{m}\right)-\alpha |+|\alpha -{p}_{b}\left({x}_{n},{x}_{n}\right)|+|{p}_{b}\left({x}_{m},{x}_{n}\right)-\alpha |+|\alpha -{p}_{b}\left({x}_{m},{x}_{m}\right)|\\ \phantom{\rule{1em}{0ex}}<\epsilon \end{array}$

for all $n,m\ge {n}_{\epsilon }$. Hence, we conclude that $\left\{{x}_{n}\right\}$ is a b-Cauchy sequence in $\left(X,{d}_{{p}_{b}}\right)$.

Next, we prove that b-completeness of $\left(X,{d}_{{p}_{b}}\right)$ implies ${p}_{b}$-completeness of $\left(X,{p}_{b}\right)$. Indeed, if $\left\{{x}_{n}\right\}$ is a ${p}_{b}$-Cauchy sequence in $\left(X,{p}_{b}\right)$, then according to the above discussion, it is also a b-Cauchy sequence in $\left(X,{d}_{{p}_{b}}\right)$. Since the b-metric space $\left(X,{d}_{{p}_{b}}\right)$ is b-complete, we deduce that there exists $y\in X$ such that ${lim}_{n\to \mathrm{\infty }}{d}_{{p}_{b}}\left(y,{x}_{n}\right)=0$. Hence,

$\underset{n\to \mathrm{\infty }}{lim}\left[{p}_{b}\left({x}_{n},y\right)-{p}_{b}\left(y,y\right)+{p}_{b}\left(y,{x}_{n}\right)-{p}_{b}\left({x}_{n},{x}_{n}\right)\right]=0,$

therefore, ${lim}_{n\to \mathrm{\infty }}\left[{p}_{b}\left({x}_{n},y\right)-{p}_{b}\left(y,y\right)\right]=0$. Further, we have

$\underset{n\to \mathrm{\infty }}{lim}\left[{p}_{b}\left(y,{x}_{n}\right)-{p}_{b}\left({x}_{n},{x}_{n}\right)\right]=0.$

Consequently,

$\underset{n\to \mathrm{\infty }}{lim}{p}_{b}\left({x}_{n},y\right)={p}_{b}\left(y,y\right)=\underset{n\to \mathrm{\infty }}{lim}{p}_{b}\left({x}_{n},{x}_{n}\right).$

On the other hand,

$\begin{array}{rl}\underset{n,m\to \mathrm{\infty }}{lim}{p}_{b}\left({x}_{n},{x}_{m}\right)\le & \underset{n,m\to \mathrm{\infty }}{lim}s{p}_{b}\left({x}_{n},y\right)+\underset{n,m\to \mathrm{\infty }}{lim}s{p}_{b}\left({x}_{m},y\right)-s{p}_{b}\left(y,y\right)\\ +\left(\frac{1-s}{2}\right)\left({p}_{b}\left({x}_{n},{x}_{n}\right)+{p}_{b}\left({x}_{m},{x}_{m}\right)\right)\\ =& {p}_{b}\left(y,y\right).\end{array}$

Also, from (${\text{p}}_{b2}$),

${p}_{b}\left(y,y\right)\le \underset{n,m\to \mathrm{\infty }}{lim}{p}_{b}\left({x}_{n},y\right)=\underset{n,m\to \mathrm{\infty }}{lim}{p}_{b}\left({x}_{n},{x}_{n}\right)\le \underset{n,m\to \mathrm{\infty }}{lim}{p}_{b}\left({x}_{n},{x}_{m}\right).$

Hence, we obtain that $\left\{{x}_{n}\right\}$ is a ${p}_{b}$-convergent sequence in $\left(X,{p}_{b}\right)$.

Now, we prove that every b-Cauchy sequence $\left\{{x}_{n}\right\}$ in $\left(X,{d}_{{p}_{b}}\right)$ is a ${p}_{b}$-Cauchy sequence in $\left(X,{p}_{b}\right)$. Let $\epsilon =\frac{1}{2}$. Then there exists ${n}_{0}\in \mathbb{N}$ such that ${d}_{{p}_{b}}\left({x}_{n},{x}_{m}\right)<\frac{1}{2}$ for all $n,m\ge {n}_{0}$. Since

${p}_{b}\left({x}_{n},{x}_{{n}_{0}}\right)-{p}_{b}\left({x}_{{n}_{0}},{x}_{{n}_{0}}\right)\le {d}_{{p}_{b}}\left({x}_{n},{x}_{{n}_{0}}\right)<\frac{1}{2},$

hence

${p}_{b}\left({x}_{n},{x}_{n}\right)\le {p}_{b}\left({x}_{n},{x}_{{n}_{0}}\right)\le {d}_{{p}_{b}}\left({x}_{n},{x}_{{n}_{0}}\right)+{p}_{b}\left({x}_{{n}_{0}},{x}_{{n}_{0}}\right)<\frac{1}{2}+{p}_{b}\left({x}_{{n}_{0}},{x}_{{n}_{0}}\right).$

Consequently, the sequence $\left\{{p}_{b}\left({x}_{n},{x}_{n}\right)\right\}$ is bounded in ℝ, and so there exists $a\in \mathbb{R}$ such that a subsequence $\left\{{p}_{b}\left({x}_{{n}_{k}},{x}_{{n}_{k}}\right)\right\}$ of $\left\{{p}_{b}\left({x}_{n},{x}_{n}\right)\right\}$ is convergent to a, i.e.,

$\underset{k\to \mathrm{\infty }}{lim}{p}_{b}\left({x}_{{n}_{k}},{x}_{{n}_{k}}\right)=a.$

Now, we prove that $\left\{{p}_{b}\left({x}_{n},{x}_{n}\right)\right\}$ is a Cauchy sequence in ℝ. Since $\left\{{x}_{n}\right\}$ is a b-Cauchy sequence in $\left(X,{d}_{{p}_{b}}\right)$ for given $\epsilon >0$, there exists ${n}_{\epsilon }\in \mathbb{N}$ such that ${d}_{{p}_{b}}\left({x}_{n},{x}_{m}\right)<\epsilon$ for all $n,m\ge {n}_{\epsilon }$. Thus, for all $n,m\ge {n}_{\epsilon }$,

$\begin{array}{rl}{p}_{b}\left({x}_{n},{x}_{n}\right)-{p}_{b}\left({x}_{m},{x}_{m}\right)& \le {p}_{b}\left({x}_{n},{x}_{m}\right)-{p}_{b}\left({x}_{m},{x}_{m}\right)\\ \le {d}_{{p}_{b}}\left({x}_{m},{x}_{n}\right)<\epsilon .\end{array}$

Therefore, ${lim}_{n\to \mathrm{\infty }}{p}_{b}\left({x}_{n},{x}_{n}\right)=a$.

On the other hand,

$\begin{array}{rl}|{p}_{b}\left({x}_{n},{x}_{m}\right)-a|& =|{p}_{b}\left({x}_{n},{x}_{m}\right)-{p}_{b}\left({x}_{n},{x}_{n}\right)+{p}_{b}\left({x}_{n},{x}_{n}\right)-a|\\ \le {d}_{{p}_{b}}\left({x}_{m},{x}_{n}\right)+|{p}_{b}\left({x}_{n},{x}_{n}\right)-a|\end{array}$

for all $n,m\ge {n}_{\epsilon }$. Hence, ${lim}_{n,m\to \mathrm{\infty }}{p}_{b}\left({x}_{n},{x}_{m}\right)=a$, and consequently, $\left\{{x}_{n}\right\}$ is a ${p}_{b}$-Cauchy sequence in $\left(X,{p}_{b}\right)$.

Conversely, let $\left\{{x}_{n}\right\}$ be a b-Cauchy sequence in $\left(X,{d}_{{p}_{b}}\right)$. Then $\left\{{x}_{n}\right\}$ is a ${p}_{b}$-Cauchy sequence in $\left(X,{p}_{b}\right)$, and so it is convergent to a point $x\in X$ with

$\underset{n\to \mathrm{\infty }}{lim}{p}_{b}\left(x,{x}_{n}\right)=\underset{n,m\to \mathrm{\infty }}{lim}{p}_{b}\left({x}_{m},{x}_{n}\right)={p}_{b}\left(x,x\right).$

Then, for given $\epsilon >0$, there exists ${n}_{\epsilon }\in \mathbb{N}$ such that

${p}_{b}\left(x,{x}_{n}\right)-{p}_{b}\left(x,x\right)<\frac{\epsilon }{4}$

and

${p}_{b}\left({x}_{n},{x}_{n}\right)-{p}_{b}\left(x,x\right)\le {p}_{b}\left({x}_{m},{x}_{n}\right)-{p}_{b}\left(x,x\right)<\frac{\epsilon }{4}.$

Therefore,

$\begin{array}{rl}|{d}_{{p}_{b}}\left({x}_{n},x\right)|& =|{p}_{b}\left({x}_{n},x\right)-{p}_{b}\left({x}_{n},{x}_{n}\right)+{p}_{b}\left({x}_{n},x\right)-{p}_{b}\left(x,x\right)|\\ \le |{p}_{b}\left({x}_{n},x\right)-{p}_{b}\left(x,x\right)|+|{p}_{b}\left(x,x\right)-{p}_{b}\left({x}_{n},{x}_{n}\right)|+|{p}_{b}\left({x}_{n},x\right)-{p}_{b}\left(x,x\right)|\\ <\epsilon ,\end{array}$

whenever $n\ge {n}_{\epsilon }$. Therefore, $\left(X,{d}_{{p}_{b}}\right)$ is complete.

Finally, let ${lim}_{n\to \mathrm{\infty }}{d}_{{p}_{b}}\left({x}_{n},x\right)=0$. So,

$\underset{n\to \mathrm{\infty }}{lim}\left[{p}_{b}\left({x}_{n},x\right)-{p}_{b}\left({x}_{n},{x}_{n}\right)\right]+\underset{n\to \mathrm{\infty }}{lim}\left[{p}_{b}\left({x}_{n},x\right)-{p}_{b}\left(x,x\right)\right]=0.$

On the other hand,

$\begin{array}{r}\underset{n,m\to \mathrm{\infty }}{lim}\left[{p}_{b}\left({x}_{n},{x}_{m}\right)-{p}_{b}\left(x,x\right)\right]\\ \phantom{\rule{1em}{0ex}}\le \underset{n\to \mathrm{\infty }}{lim}\left[s{p}_{b}\left({x}_{n},x\right)+s{p}_{b}\left(x,{x}_{m}\right)-s{p}_{b}\left(x,x\right)\\ \phantom{\rule{2em}{0ex}}+\left(\frac{1-s}{2}\right)\left({p}_{b}\left({x}_{n},{x}_{n}\right)+{p}_{b}\left({x}_{m},{x}_{m}\right)\right)-{p}_{b}\left(x,x\right)\right]\\ \phantom{\rule{1em}{0ex}}=0.\end{array}$

□

Definition 9 Let $\left(X,{p}_{b}\right)$ and $\left({X}^{\prime },{p}_{b}^{\prime }\right)$ be two partial b-metric spaces, and let $f:\left(X,{p}_{b}\right)\to \left({X}^{\prime },{p}_{b}^{\prime }\right)$ be a mapping. Then f is said to be ${p}_{b}$-continuous at a point $a\in X$ if for a given $\epsilon >0$, there exists $\delta >0$ such that $x\in X$ and ${p}_{b}\left(a,x\right)<\delta +{p}_{b}\left(a,a\right)$ imply that ${p}_{b}^{\prime }\left(f\left(a\right),f\left(x\right)\right)<\epsilon +{p}_{b}^{\prime }\left(f\left(a\right),f\left(a\right)\right)$. The mapping f is ${p}_{b}$-continuous on X if it is ${p}_{b}$-continuous at all $a\in X$.

Proposition 5 Let $\left(X,{p}_{b}\right)$ and $\left({X}^{\prime },{p}_{b}^{\prime }\right)$ be two partial b-metric spaces. Then a mapping $f:X\to {X}^{\prime }$ is ${p}_{b}$-continuous at a point $x\in X$ if and only if it is ${p}_{b}$-sequentially continuous at x; that is, whenever $\left\{{x}_{n}\right\}$ is ${p}_{b}$-convergent to x, $\left\{f\left({x}_{n}\right)\right\}$ is ${p}_{b}^{\prime }$-convergent to $f\left(x\right)$.

Definition 10 A triple $\left(X,⪯,{p}_{b}\right)$ is called an ordered partial b-metric space if $\left(X,⪯\right)$ is a partially ordered set and ${p}_{b}$ is a partial b-metric on X.

## 3 Fixed point results in partial b-metric spaces

The following crucial lemma is useful in proving our main results.

Lemma 2 Let $\left(X,{p}_{b}\right)$ be a partial b-metric space with the coefficient $s>1$ and suppose that $\left\{{x}_{n}\right\}$ and $\left\{{y}_{n}\right\}$ are convergent to x and y, respectively. Then we have

$\begin{array}{rl}\frac{1}{{s}^{2}}{p}_{b}\left(x,y\right)-\frac{1}{s}{p}_{b}\left(x,x\right)-{p}_{b}\left(y,y\right)& \le \underset{n\to \mathrm{\infty }}{lim inf}{p}_{b}\left({x}_{n},{y}_{n}\right)\le \underset{n\to \mathrm{\infty }}{lim sup}{p}_{b}\left({x}_{n},{y}_{n}\right)\\ \le s{p}_{b}\left(x,x\right)+{s}^{2}{p}_{b}\left(y,y\right)+{s}^{2}{p}_{b}\left(x,y\right).\end{array}$

In particular, if ${p}_{b}\left(x,y\right)=0$, then we have ${lim}_{n\to \mathrm{\infty }}{p}_{b}\left({x}_{n},{y}_{n}\right)=0$.

Moreover, for each $z\in X$, we have

$\begin{array}{rl}\frac{1}{s}{p}_{b}\left(x,z\right)-{p}_{b}\left(x,x\right)& \le \underset{n\to \mathrm{\infty }}{lim inf}{p}_{b}\left({x}_{n},z\right)\le \underset{n\to \mathrm{\infty }}{lim sup}{p}_{b}\left({x}_{n},z\right)\\ \le s{p}_{b}\left(x,z\right)+s{p}_{b}\left(x,x\right).\end{array}$

In particular, if ${p}_{b}\left(x,x\right)=0$, then we have

$\frac{1}{s}{p}_{b}\left(x,z\right)\le \underset{n\to \mathrm{\infty }}{lim inf}{p}_{b}\left({x}_{n},z\right)\le \underset{n\to \mathrm{\infty }}{lim sup}{p}_{b}\left({x}_{n},z\right)\le s{p}_{b}\left(x,z\right).$

Proof Using the triangle inequality in a partial b-metric space, it is easy to see that

${p}_{b}\left(x,y\right)\le s{p}_{b}\left(x,{x}_{n}\right)+{s}^{2}{p}_{b}\left({x}_{n},{y}_{n}\right)+{s}^{2}{p}_{b}\left({y}_{n},y\right)$

and

${p}_{b}\left({x}_{n},{y}_{n}\right)\le s{p}_{b}\left({x}_{n},x\right)+{s}^{2}{p}_{b}\left(x,y\right)+{s}^{2}{p}_{b}\left(y,{y}_{n}\right).$

Taking the lower limit as $n\to \mathrm{\infty }$ in the first inequality and the upper limit as $n\to \mathrm{\infty }$ in the second inequality, we obtain the first desired result. If ${p}_{b}\left(x,y\right)=0$, then by the triangle inequality we get ${p}_{b}\left(x,x\right)=0$ and ${p}_{b}\left(y,y\right)=0$. Therefore, we have ${lim}_{n\to \mathrm{\infty }}{p}_{b}\left({x}_{n},{y}_{n}\right)=0$. Similarly, using again the triangle inequality, the other assertions follow. □

Let $\left(X,⪯,{p}_{b}\right)$ be an ordered partial b-metric space, and let $f:X\to X$ be a mapping. Set

${M}_{s}^{f}\left(x,y\right)=max\left\{{p}_{b}\left(x,y\right),{p}_{b}\left(x,fx\right),{p}_{b}\left(y,fy\right),\frac{{p}_{b}\left(x,fy\right)+{p}_{b}\left(y,fx\right)}{2s}\right\}.$

Definition 11 Let $\left(X,{p}_{b}\right)$ be an ordered partial b-metric space. We say that a mapping $f:X\to X$ is a generalized ${\left(\psi ,\phi \right)}_{s}$-weakly contractive mapping if there exist two altering distance functions ψ and φ such that

$\psi \left(s{p}_{b}\left(fx,fy\right)\right)\le \psi \left({M}_{s}^{f}\left(x,y\right)\right)-\phi \left({M}_{s}^{f}\left(x,y\right)\right)$
(3.1)

for all comparable $x,y\in X$.

First, we prove the following result.

Theorem 1 Let $\left(X,⪯,{p}_{b}\right)$ be a ${p}_{b}$-complete ordered partial b-metric space. Let $f:X\to X$ be a nondecreasing, with respect to ⪯, continuous mapping. Suppose that f is a generalized ${\left(\psi ,\phi \right)}_{s}$-weakly contractive mapping. If there exists ${x}_{0}\in X$ such that ${x}_{0}⪯f{x}_{0}$, then f has a fixed point.

Proof Let ${x}_{0}\in X$ be such that ${x}_{0}⪯f{x}_{0}$. Then we define a sequence $\left({x}_{n}\right)$ in X such that ${x}_{n+1}=f{x}_{n}$ for all $n\ge 0$. Since ${x}_{0}⪯f{x}_{0}={x}_{1}$ and f is nondecreasing, we have ${x}_{1}=f{x}_{0}⪯{x}_{2}=f{x}_{1}$. Again, as ${x}_{1}⪯{x}_{2}$ and f is nondecreasing, we have ${x}_{2}=f{x}_{1}⪯{x}_{3}=f{x}_{2}$. By induction, we have

${x}_{0}⪯{x}_{1}⪯\cdots ⪯{x}_{n}⪯{x}_{n+1}⪯\cdots .$

If ${x}_{n}={x}_{n+1}$ for some $n\in \mathbb{N}$, then ${x}_{n}=f{x}_{n}$ and hence ${x}_{n}$ is a fixed point of f. So, we may assume that ${x}_{n}\ne {x}_{n+1}$ for all $n\in \mathbb{N}$. By (3.1), we have

$\begin{array}{rl}\psi \left({p}_{b}\left({x}_{n},{x}_{n+1}\right)\right)& \le \psi \left(s{p}_{b}\left({x}_{n},{x}_{n+1}\right)\right)\\ =\psi \left(s{p}_{b}\left(f{x}_{n-1},f{x}_{n}\right)\right)\\ \le \psi \left({M}_{s}^{f}\left({x}_{n-1},{x}_{n}\right)\right)-\phi \left({M}_{s}^{f}\left({x}_{n-1},{x}_{n}\right)\right),\end{array}$
(3.2)

where

$\begin{array}{rl}{M}_{s}^{f}\left({x}_{n-1},{x}_{n}\right)=& max\left\{{p}_{b}\left({x}_{n-1},{x}_{n}\right),{p}_{b}\left({x}_{n-1},f{x}_{n-1}\right),{p}_{b}\left({x}_{n},f{x}_{n}\right),\\ \frac{{p}_{b}\left({x}_{n-1},f{x}_{n}\right)+{p}_{b}\left({x}_{n},f{x}_{n-1}\right)}{2s}\right\}\\ =& max\left\{{p}_{b}\left({x}_{n-1},{x}_{n}\right),{p}_{b}\left({x}_{n},{x}_{n+1}\right),\frac{{p}_{b}\left({x}_{n-1},{x}_{n+1}\right)+{p}_{b}\left({x}_{n},{x}_{n}\right)}{2s}\right\}\\ \le & max\left\{{p}_{b}\left({x}_{n-1},{x}_{n}\right),{p}_{b}\left({x}_{n},{x}_{n+1}\right),\\ \frac{s{p}_{b}\left({x}_{n-1},{x}_{n}\right)+s{p}_{b}\left({x}_{n},{x}_{n+1}\right)+\left(1-s\right){p}_{b}\left({x}_{n},{x}_{n}\right)}{2s}\right\}\\ =& max\left\{{p}_{b}\left({x}_{n-1},{x}_{n}\right),{p}_{b}\left({x}_{n},{x}_{n+1}\right)\right\}.\end{array}$

So, we have

${M}_{s}^{f}\left({x}_{n-1},{x}_{n}\right)=max\left\{{p}_{b}\left({x}_{n-1},{x}_{n}\right),{p}_{b}\left({x}_{n},{x}_{n+1}\right)\right\}.$
(3.3)

From (3.2), (3.3) we get

$\begin{array}{rl}\psi \left({p}_{b}\left({x}_{n},{x}_{n+1}\right)\right)\le & \psi \left(max\left\{{p}_{b}\left({x}_{n-1},{x}_{n}\right),{p}_{b}\left({x}_{n},{x}_{n+1}\right)\right\}\right)\\ -\phi \left(max\left\{{p}_{b}\left({x}_{n-1},{x}_{n}\right),{p}_{b}\left({x}_{n},{x}_{n+1}\right)\right\}\right).\end{array}$
(3.4)

If

$max\left\{{p}_{b}\left({x}_{n-1},{x}_{n}\right),{p}_{b}\left({x}_{n},{x}_{n+1}\right)\right\}={p}_{b}\left({x}_{n},{x}_{n+1}\right),$

then by (3.4) and properties of φ, we have

$\begin{array}{rl}\psi \left({p}_{b}\left({x}_{n},{x}_{n+1}\right)\right)& \le \psi \left({p}_{b}\left({x}_{n},{x}_{n+1}\right)\right)-\phi \left({p}_{b}\left({x}_{n},{x}_{n+1}\right)\right)\\ <\psi \left({p}_{b}\left({x}_{n},{x}_{n+1}\right)\right),\end{array}$

$\psi \left({p}_{b}\left({x}_{n},{x}_{n+1}\right)\right)\le \psi \left({p}_{b}\left({x}_{n-1},{x}_{n}\right)\right)-\phi \left({p}_{b}\left({x}_{n-1},{x}_{n}\right)\right).$
(3.5)

Therefore, $\left\{{p}_{b}\left({x}_{n},{x}_{n+1}\right):n\in \mathbb{N}\cup \left\{0\right\}\right\}$ is a nonincreasing sequence of positive numbers. So, there exists $r\ge 0$ such that

$\underset{n\to \mathrm{\infty }}{lim}{p}_{b}\left({x}_{n},{x}_{n+1}\right)=r.$

Letting $n\to \mathrm{\infty }$ in (3.5), we get

$\psi \left(r\right)\le \psi \left(r\right)-\phi \left(r\right)\le \psi \left(r\right).$

Therefore, $\phi \left(r\right)=0$, and hence $r=0$. Thus, we have

$\underset{n\to \mathrm{\infty }}{lim}{p}_{b}\left({x}_{n},{x}_{n+1}\right)=0.$
(3.6)

Next, we show that $\left\{{x}_{n}\right\}$ is a ${p}_{b}$-Cauchy sequence in X. For this, we have to show that $\left\{{x}_{n}\right\}$ is a b-Cauchy sequence in $\left(X,{d}_{{p}_{b}}\right)$ (see Lemma 1). Suppose the contrary; that is, $\left\{{x}_{n}\right\}$ is not a b-Cauchy sequence. Then there exists $\epsilon >0$ for which we can find two subsequences $\left\{{x}_{{m}_{i}}\right\}$ and $\left\{{x}_{{n}_{i}}\right\}$ of $\left\{{x}_{n}\right\}$ such that ${n}_{i}$ is the smallest index for which

${n}_{i}>{m}_{i}>i,\phantom{\rule{1em}{0ex}}{d}_{{p}_{b}}\left({x}_{{m}_{i}},{x}_{{n}_{i}}\right)\ge \epsilon .$
(3.7)

This means that

${d}_{{p}_{b}}\left({x}_{{m}_{i}},{x}_{{n}_{i}-1}\right)<\epsilon .$
(3.8)

From (3.7) and using the triangular inequality, we get

$\epsilon \le {d}_{{p}_{b}}\left({x}_{{m}_{i}},{x}_{{n}_{i}}\right)\le s{d}_{{p}_{b}}\left({x}_{{m}_{i}},{x}_{{n}_{i}-1}\right)+s{d}_{{p}_{b}}\left({x}_{{n}_{i}-1},{x}_{{n}_{i}}\right).$
(3.9)

Taking the upper limit as $i\to \mathrm{\infty }$ and using (3.8), we get

$\frac{\epsilon }{s}\le \underset{i\to \mathrm{\infty }}{lim inf}{d}_{{p}_{b}}\left({x}_{{m}_{i}},{x}_{{n}_{i}-1}\right)\le \underset{i\to \mathrm{\infty }}{lim sup}{d}_{{p}_{b}}\left({x}_{{m}_{i}},{x}_{{n}_{i}-1}\right)\le \epsilon .$
(3.10)

Also, from (3.9) and (3.10),

$\epsilon \le \underset{i\to \mathrm{\infty }}{lim sup}{d}_{{p}_{b}}\left({x}_{{m}_{i}},{x}_{{n}_{i}}\right)\le s\epsilon .$

Further,

${d}_{{p}_{b}}\left({x}_{{m}_{i}+1},{x}_{{n}_{i}}\right)\le s{d}_{{p}_{b}}\left({x}_{{m}_{i}+1},{x}_{{m}_{i}}\right)+s{d}_{{p}_{b}}\left({x}_{{m}_{i}},{x}_{{n}_{i}}\right),$

and hence

$\underset{i\to \mathrm{\infty }}{lim sup}{d}_{{p}_{b}}\left({x}_{{m}_{i}+1},{x}_{{n}_{i}}\right)\le {s}^{2}\epsilon .$

Finally,

${d}_{{p}_{b}}\left({x}_{{m}_{i}+1},{x}_{{n}_{i}-1}\right)\le s{d}_{{p}_{b}}\left({x}_{{m}_{i}+1},{x}_{{m}_{i}}\right)+s{d}_{{p}_{b}}\left({x}_{{m}_{i}},{x}_{{n}_{i}-1}\right),$

and hence

$\underset{i\to \mathrm{\infty }}{lim sup}{d}_{{p}_{b}}\left({x}_{{m}_{i}+1},{x}_{{n}_{i}-1}\right)\le s\epsilon .$

On the other hand, by the definition of ${d}_{{p}_{b}}$ and (3.6),

$\underset{i\to \mathrm{\infty }}{lim sup}{d}_{{p}_{b}}\left({x}_{{m}_{i}},{x}_{{n}_{i}-1}\right)=2\underset{i\to \mathrm{\infty }}{lim sup}{p}_{b}\left({x}_{{m}_{i}},{x}_{{n}_{i}-1}\right).$

Hence, by (3.10),

$\frac{\epsilon }{2s}\le \underset{i\to \mathrm{\infty }}{lim inf}{p}_{b}\left({x}_{{m}_{i}},{x}_{{n}_{i}-1}\right)\le \underset{i\to \mathrm{\infty }}{lim sup}{p}_{b}\left({x}_{{m}_{i}},{x}_{{n}_{i}-1}\right)\le \frac{\epsilon }{2}.$
(3.11)

Similarly,

$\underset{i\to \mathrm{\infty }}{lim sup}{p}_{b}\left({x}_{{m}_{i}},{x}_{{n}_{i}}\right)\le \frac{s\epsilon }{2},$
(3.12)
$\frac{\epsilon }{2s}\le \underset{i\to \mathrm{\infty }}{lim sup}{p}_{b}\left({x}_{{m}_{i}+1},{x}_{{n}_{i}}\right),$
(3.13)
$\underset{i\to \mathrm{\infty }}{lim sup}{p}_{b}\left({x}_{{m}_{i}+1},{x}_{{n}_{i}-1}\right)\le \frac{s\epsilon }{2}.$
(3.14)

From (3.1), we have

$\begin{array}{rl}\psi \left(s{p}_{b}\left({x}_{{m}_{i}+1},{x}_{{n}_{i}}\right)\right)& =\psi \left(s{p}_{b}\left(f{x}_{{m}_{i}},f{x}_{{n}_{i}-1}\right)\right)\\ \le \psi \left({M}_{s}^{f}\left({x}_{{m}_{i}},{x}_{{n}_{i}-1}\right)\right)-\phi \left({M}_{s}^{f}\left({x}_{{m}_{i}},{x}_{{n}_{i}-1}\right)\right),\end{array}$
(3.15)

where

$\begin{array}{r}{M}_{s}^{f}\left({x}_{{m}_{i}},{x}_{{n}_{i}-1}\right)\\ \phantom{\rule{1em}{0ex}}=max\left\{{p}_{b}\left({x}_{{m}_{i}},{x}_{{n}_{i}-1}\right),{p}_{b}\left({x}_{{m}_{i}},f{x}_{{m}_{i}}\right),{p}_{b}\left({x}_{{n}_{i}-1},f{x}_{{n}_{i}-1}\right),\\ \phantom{\rule{2em}{0ex}}\frac{{p}_{b}\left({x}_{{m}_{i}},f{x}_{{n}_{i}-1}\right)+{p}_{b}\left(f{x}_{{m}_{i}},{x}_{{n}_{i}-1}\right)}{2s}\right\}\\ \phantom{\rule{1em}{0ex}}=max\left\{{p}_{b}\left({x}_{{m}_{i}},{x}_{{n}_{i}-1}\right),{p}_{b}\left({x}_{{m}_{i}},{x}_{{m}_{i}+1}\right),{p}_{b}\left({x}_{{n}_{i}-1},{x}_{{n}_{i}}\right),\\ \phantom{\rule{2em}{0ex}}\frac{{p}_{b}\left({x}_{{m}_{i}},{x}_{{n}_{i}}\right)+{p}_{b}\left({x}_{{m}_{i}+1},{x}_{{n}_{i}-1}\right)}{2s}\right\}.\end{array}$
(3.16)

Taking the upper limit as $i\to \mathrm{\infty }$ in (3.16) and using (3.6), (3.11), (3.12) and (3.14), we get

$\begin{array}{rl}\underset{i\to \mathrm{\infty }}{lim sup}{M}_{s}^{f}\left({x}_{{m}_{i}},{x}_{{n}_{i}-1}\right)=& max\left\{\underset{i\to \mathrm{\infty }}{lim sup}{p}_{b}\left({x}_{{m}_{i}},{x}_{{n}_{i}-1}\right),0,0,\\ \frac{{lim sup}_{i\to \mathrm{\infty }}{p}_{b}\left({x}_{{m}_{i}},{x}_{{n}_{i}}\right)+{lim sup}_{i\to \mathrm{\infty }}{p}_{b}\left({x}_{{m}_{i}+1},{x}_{{n}_{i}-1}\right)}{2s}\right\}\\ \le & max\left\{\frac{\epsilon }{2},\frac{\frac{\epsilon s+\epsilon s}{2}}{2s}\right\}=\frac{\epsilon }{2}.\end{array}$
(3.17)

Now, taking the upper limit as $i\to \mathrm{\infty }$ in (3.15) and using (3.13) and (3.17), we have

$\begin{array}{rl}\psi \left(s\frac{\epsilon }{2s}\right)& \le \psi \left(s\underset{i\to \mathrm{\infty }}{lim sup}{p}_{b}\left({x}_{{m}_{i}+1},{x}_{{n}_{i}}\right)\right)\\ \le \psi \left(\underset{i\to \mathrm{\infty }}{lim sup}{M}_{s}^{f}\left({x}_{{m}_{i}},{x}_{{n}_{i}-1}\right)\right)-\underset{i\to \mathrm{\infty }}{lim inf}\phi \left({M}_{s}^{f}\left({x}_{{m}_{i}},{x}_{{n}_{i}-1}\right)\right)\\ \le \psi \left(\frac{\epsilon }{2}\right)-\phi \left(\underset{i\to \mathrm{\infty }}{lim inf}{M}_{s}^{f}\left({x}_{{m}_{i}},{x}_{{n}_{i}-1}\right)\right),\end{array}$

which further implies that

$\phi \left(\underset{i\to \mathrm{\infty }}{lim inf}{M}_{s}^{f}\left({x}_{{m}_{i}},{x}_{{n}_{i}-1}\right)\right)=0,$

so ${lim inf}_{i\to \mathrm{\infty }}{M}_{s}^{f}\left({x}_{{m}_{i}},{x}_{{n}_{i}-1}\right)=0$, and by (3.16) we get ${lim inf}_{i\to \mathrm{\infty }}{d}_{{p}_{b}}\left({x}_{{m}_{i}},{x}_{{n}_{i}-1}\right)=0$, a contradiction with (3.11).

Thus, we have proved that $\left\{{x}_{n}\right\}$ is a b-Cauchy sequence in the b-metric space $\left(X,{d}_{{p}_{b}}\right)$. Since $\left(X,{p}_{b}\right)$ is ${p}_{b}$-complete, then from Lemma 1, $\left(X,{d}_{{p}_{b}}\right)$ is a b-complete b-metric space. Therefore, the sequence $\left\{{x}_{n}\right\}$ converges to some $z\in X$, that is, ${lim}_{n\to \mathrm{\infty }}{d}_{{p}_{b}}\left({x}_{n},z\right)=0$. Again, from Lemma 1,

$\underset{n\to \mathrm{\infty }}{lim}{p}_{b}\left(z,{x}_{n}\right)=\underset{n\to \mathrm{\infty }}{lim}{p}_{b}\left({x}_{n},{x}_{n}\right)={p}_{b}\left(z,z\right).$

On the other hand, thanks to (3.6) and condition (${\text{p}}_{b2}$), ${lim}_{n\to \mathrm{\infty }}{p}_{b}\left({x}_{n},{x}_{n}\right)=0$, which yields that

$\underset{n\to \mathrm{\infty }}{lim}{p}_{b}\left(z,{x}_{n}\right)=\underset{n\to \mathrm{\infty }}{lim}{p}_{b}\left({x}_{n},{x}_{n}\right)={p}_{b}\left(z,z\right)=0.$

Using the triangular inequality, we get

${p}_{b}\left(z,fz\right)\le s{p}_{b}\left(z,f{x}_{n}\right)+s{p}_{b}\left(f{x}_{n},fz\right).$

Letting $n\to \mathrm{\infty }$ and using the continuity of f, we get

${p}_{b}\left(z,fz\right)\le s\underset{n\to \mathrm{\infty }}{lim}{p}_{b}\left(z,f{x}_{n}\right)+s\underset{n\to \mathrm{\infty }}{lim}{p}_{b}\left(f{x}_{n},fz\right)=s{p}_{b}\left(fz,fz\right).$
(3.18)

Note that from (3.1), we have

$\psi \left(s{p}_{b}\left(fz,fz\right)\right)\le \psi \left({M}_{s}^{f}\left(z,z\right)\right)-\phi \left({M}_{s}^{f}\left(z,z\right)\right),$
(3.19)

where

${M}_{s}^{f}\left(z,z\right)=max\left\{{p}_{b}\left(z,z\right),{p}_{b}\left(z,fz\right),{p}_{b}\left(z,fz\right),\frac{{p}_{b}\left(z,fz\right)+{p}_{b}\left(z,fz\right)}{2s}\right\}={p}_{b}\left(fz,z\right).$

Hence, as ψ is nondecreasing, we have $s{p}_{b}\left(fz,fz\right)\le {p}_{b}\left(fz,z\right)$. Thus, by (3.18) we obtain that $s{p}_{b}\left(fz,fz\right)={p}_{b}\left(fz,z\right)$. But then, using (3.19), we get that $\phi \left({M}_{s}^{f}\left(z,z\right)\right)=0$.

Hence, we have ${p}_{b}\left(fz,z\right)=0$ and $fz=z$. Thus, z is a fixed point of f. □

We will show now that the continuity of f in Theorem 1 is not necessary and can be replaced by another assumption.

Theorem 2 Under the hypotheses of Theorem  1, without the continuity assumption on f, assume that whenever $\left\{{x}_{n}\right\}$ is a nondecreasing sequence in X such that ${x}_{n}\to x\in X$, one has ${x}_{n}⪯x$ for all $n\in \mathbb{N}$. Then f has a fixed point in X.

Proof Following similar arguments as those given in Theorem 1, we construct an increasing sequence $\left\{{x}_{n}\right\}$ in X such that ${x}_{n}\to z$ for some $z\in X$. Using the assumption on X, we have ${x}_{n}⪯z$ for all $n\in \mathbb{N}$. Now, we show that $fz=z$. By (3.1), we have

$\begin{array}{rl}\psi \left(s{p}_{b}\left({x}_{n+1},fz\right)\right)& =\psi \left(s{p}_{b}\left(f{x}_{n},fz\right)\right)\\ \le \psi \left({M}_{s}^{f}\left({x}_{n},z\right)\right)-\phi \left({M}_{s}^{f}\left({x}_{n},z\right)\right),\end{array}$
(3.20)

where

$\begin{array}{rl}{M}_{s}^{f}\left({x}_{n},z\right)& =max\left\{{p}_{b}\left({x}_{n},z\right),{p}_{b}\left({x}_{n},f{x}_{n}\right),{p}_{b}\left(z,fz\right),\frac{{p}_{b}\left({x}_{n},fz\right)+{p}_{b}\left(f{x}_{n},z\right)}{2s}\right\}\\ =max\left\{{p}_{b}\left({x}_{n},z\right),{p}_{b}\left({x}_{n},{x}_{n+1}\right),{p}_{b}\left(z,fz\right),\frac{{p}_{b}\left({x}_{n},fz\right)+{p}_{b}\left({x}_{n+1},z\right)}{2s}\right\}.\end{array}$
(3.21)

Letting $n\to \mathrm{\infty }$ in (3.21) and using Lemma 2, we get

$\begin{array}{rl}\frac{{p}_{b}\left(z,fz\right)}{2{s}^{2}}& =min\left\{{p}_{b}\left(z,fz\right),\frac{\frac{{p}_{b}\left(z,fz\right)}{s}}{2s}\right\}\le \underset{i\to \mathrm{\infty }}{lim inf}{M}_{s}^{f}\left({x}_{n},z\right)\le \underset{i\to \mathrm{\infty }}{lim sup}{M}_{s}^{f}\left({x}_{n},z\right)\\ \le max\left\{{p}_{b}\left(z,fz\right),\frac{s{p}_{b}\left(z,fz\right)}{2s}\right\}={p}_{b}\left(z,fz\right).\end{array}$
(3.22)

Again, taking the upper limit as $n\to \mathrm{\infty }$ in (3.20) and using Lemma 2 and (3.22), we get

$\begin{array}{rl}\psi \left({p}_{b}\left(z,fz\right)\right)& =\psi \left(s\frac{1}{s}{p}_{b}\left(z,fz\right)\right)\le \psi \left(s\underset{n\to \mathrm{\infty }}{lim sup}{p}_{b}\left({x}_{n+1},fz\right)\right)\\ \le \psi \left(\underset{n\to \mathrm{\infty }}{lim sup}{M}_{s}^{f}\left({x}_{n},z\right)\right)-\underset{n\to \mathrm{\infty }}{lim inf}\phi \left({M}_{s}^{f}\left({x}_{n},z\right)\right)\\ \le \psi \left({p}_{b}\left(z,fz\right)\right)-\phi \left(\underset{n\to \mathrm{\infty }}{lim inf}{M}_{s}^{f}\left({x}_{n},z\right)\right).\end{array}$

Therefore, $\phi \left({lim inf}_{n\to \mathrm{\infty }}{M}_{s}^{f}\left({x}_{n},z\right)\right)\le 0$, equivalently, ${lim inf}_{n\to \mathrm{\infty }}{M}_{s}^{f}\left({x}_{n},z\right)=0$. Thus, from (3.22) we get $z=fz$, and hence z is a fixed point of f. □

Corollary 1 Let $\left(X,⪯,{p}_{b}\right)$ be a ${p}_{b}$-complete ordered partial b-metric space. Let $f:X\to X$ be a continuous mapping, nondecreasing with respect to ⪯. Suppose that there exists $k\in \left[0,1\right)$ such that

${p}_{b}\left(fx,fy\right)\le \frac{k}{s}max\left\{{p}_{b}\left(x,y\right),{p}_{b}\left(x,fx\right),{p}_{b}\left(y,fy\right),\frac{{p}_{b}\left(x,fy\right)+{p}_{b}\left(y,fx\right)}{2s}\right\}$

for all comparable elements $x,y\in X$. If there exists ${x}_{0}\in X$ such that ${x}_{0}⪯f{x}_{0}$, then f has a fixed point.

Proof Follows from Theorem 1 by taking $\psi \left(t\right)=t$ and $\phi \left(t\right)=\left(1-k\right)t$, for all $t\in \left[0,+\mathrm{\infty }\right)$. □

Corollary 2 Under the hypotheses of Corollary  1, without the continuity assumption on f, for any nondecreasing sequence $\left\{{x}_{n}\right\}$ in X such that ${x}_{n}\to x\in X$, let us have ${x}_{n}⪯x$ for all $n\in \mathbb{N}$. Then f has a fixed point in X.

Now, in order to support the usability of our results, we present the following example.

Example 4 Let $X=\left[0,+\mathrm{\infty }\right)$ be equipped with the partial order ⪯ defined by

$x⪯y\phantom{\rule{1em}{0ex}}⇔\phantom{\rule{1em}{0ex}}x=y\vee \left(x,y\in \left[0,1\right]\wedge x\le y\right),$

and with the partial b-metric ${p}_{b}$ given by ${p}_{b}\left(x,y\right)={\left[max\left\{x,y\right\}\right]}^{2}$ (with $s=2$). Consider the mapping $f:X\to X$ given by

$fx=\left\{\begin{array}{cc}\frac{x}{\sqrt{2}\sqrt{1+x}},\hfill & x\in \left[0,1\right],\hfill \\ \frac{x}{2},\hfill & x>1.\hfill \end{array}$

Then f is continuous and increasing, and $0⪯f0$. Take altering distance functions

$\psi \left(t\right)=t,\phantom{\rule{2em}{0ex}}\phi \left(t\right)=\left\{\begin{array}{cc}\frac{t\sqrt{t}}{1+\sqrt{t}},\hfill & 0\le t\le 1,\hfill \\ \frac{t}{2},\hfill & t>1.\hfill \end{array}$

In order to check the contractive condition (3.1) of Theorem 1, without loss of generality, we may take $x,y\in X$ such that $y⪯x$. Consider the following two possible cases.

Case 1. $0\le y\le x\le 1$. Then

${p}_{b}\left(fx,fy\right)={\left[max\left\{\frac{x}{\sqrt{2}\sqrt{1+x}},\frac{y}{\sqrt{2}\sqrt{1+y}}\right\}\right]}^{2}=\frac{{x}^{2}}{2\left(1+x\right)}$

and

${M}_{s}^{f}\left(x,y\right)=max\left\{{x}^{2},{x}^{2},{y}^{2},\frac{{x}^{2}+{max}^{2}\left\{y,\frac{x}{\sqrt{2}\sqrt{1+x}}\right\}}{2s}\right\}={x}^{2}.$

Thus, (3.1) reduces to

$\psi \left(2\cdot \frac{{x}^{2}}{2\left(1+x\right)}\right)=\frac{{x}^{2}}{1+x}\le {x}^{2}-\frac{{x}^{3}}{1+x}=\psi \left({x}^{2}\right)-\phi \left({x}^{2}\right).$

Case 2. $x=y\ge 1$. Then ${p}_{b}\left(fx,fy\right)=\frac{{x}^{2}}{4}$ and ${M}_{s}^{f}\left(x,y\right)={x}^{2}$, so (3.1) reduces to

$\psi \left(2\cdot \frac{{x}^{2}}{4}\right)=\frac{{x}^{2}}{2}\le {x}^{2}-\frac{{x}^{2}}{2}=\psi \left({x}^{2}\right)-\phi \left({x}^{2}\right).$

Hence, all the conditions of Theorem 1 are satisfied and f has a fixed point (which is $z=0$).

## 4 Common fixed point results in partial b-metric spaces

Let $\left(X,⪯,{p}_{b}\right)$ be an ordered partial b-metric space with the coefficient $s\ge 1$, and let $f,g:X\to X$ be two mappings. Set

${M}_{s}^{f,g}\left(x,y\right)=max\left\{{p}_{b}\left(x,y\right),{p}_{b}\left(x,fx\right),{p}_{b}\left(y,gy\right),\frac{{p}_{b}\left(x,gy\right)+{p}_{b}\left(y,fx\right)}{2s}\right\}.$

Now, we present the following definition.

Definition 12 Let $\left(X,⪯,{p}_{b}\right)$ be an ordered partial b-metric space, and let ψ and φ be altering distance functions. We say that a pair $\left(f,g\right)$ of self-mappings $f,g:X\to X$ is a generalized ${\left(\psi ,\phi \right)}_{s}$-contraction pair if

$\psi \left({s}^{2}{p}_{b}\left(fx,gy\right)\right)\le \psi \left({M}_{s}^{f,g}\left(x,y\right)\right)-\phi \left({M}_{s}^{f,g}\left(x,y\right)\right)$
(4.1)

for all comparable $x,y\in X$.

Definition 13 [42]

Let $\left(X,⪯\right)$ be a partially ordered set. Then two mappings $f,g:X\to X$ are said to be weakly increasing if $fx⪯gfx$ and $gx⪯fgx$ for all $x\in X$.

Theorem 3 Let $\left(X,⪯,{p}_{b}\right)$ be a ${p}_{b}$-complete ordered partial b-metric space with the coefficient $s\ge 1$, and let $f,g:X\to X$ be two weakly increasing mappings with respect to ⪯. Suppose that $\left(f,g\right)$ is a generalized ${\left(\psi ,\phi \right)}_{s}$-contraction pair for some altering distance functions ψ and φ. If f and g are continuous, then f and g have a common fixed point.

Proof Let us divide the proof into two parts as follows.

First part. We prove that $u\in X$ is a fixed point of f if and only if it is a fixed point of g. Suppose that u is a fixed point of f, that is, $fu=u$. As $u⪯u$, by (4.1), we have

$\begin{array}{r}\psi \left({s}^{2}{p}_{b}\left(u,gu\right)\right)=\psi \left({s}^{2}{p}_{b}\left(fu,gu\right)\right)\\ \phantom{\rule{1em}{0ex}}\le \psi \left(max\left\{{p}_{b}\left(u,u\right),{p}_{b}\left(u,fu\right),{p}_{b}\left(u,gu\right),\frac{1}{2s}\left({p}_{b}\left(u,gu\right)+{p}_{b}\left(u,fu\right)\right)\right\}\right)\\ \phantom{\rule{2em}{0ex}}-\phi \left(max\left\{{p}_{b}\left(u,u\right),{p}_{b}\left(u,fu\right),{p}_{b}\left(u,gu\right),\frac{1}{2s}\left({p}_{b}\left(u,gu\right)+{p}_{b}\left(u,fu\right)\right)\right\}\right)\\ \phantom{\rule{1em}{0ex}}\le \psi \left({p}_{b}\left(u,gu\right)\right)-\phi \left(max\left\{{p}_{b}\left(u,u\right),{p}_{b}\left(u,fu\right),{p}_{b}\left(u,gu\right),\frac{1}{2s}\left({p}_{b}\left(u,gu\right)+{p}_{b}\left(u,fu\right)\right)\right\}\right)\\ \phantom{\rule{1em}{0ex}}\le \psi \left({s}^{2}{p}_{b}\left(u,gu\right)\right)\\ \phantom{\rule{2em}{0ex}}-\phi \left(max\left\{{p}_{b}\left(u,u\right),{p}_{b}\left(u,fu\right),{p}_{b}\left(u,gu\right),\frac{1}{2s}\left({p}_{b}\left(u,gu\right)+{p}_{b}\left(u,fu\right)\right)\right\}\right).\end{array}$

Therefore, ${p}_{b}\left(u,gu\right)=0$ and hence $gu=u$. Similarly, we can show that if u is a fixed point of g, then u is a fixed point of f.

Second part (construction of a sequence by iterative technique).

Let ${x}_{0}\in X$. We construct a sequence $\left\{{x}_{n}\right\}$ in X such that ${x}_{2n+1}=f{x}_{2n}$ and ${x}_{2n+2}=g{x}_{2n+1}$ for all nonnegative integers n. As f and g are weakly increasing with respect to ⪯, we have

$\begin{array}{rl}{x}_{1}& =f{x}_{0}⪯gf{x}_{0}={x}_{2}=g{x}_{1}⪯fg{x}_{1}={x}_{3}⪯\cdots \\ ⪯{x}_{2n+1}=f{x}_{2n}⪯gf{x}_{2n}={x}_{2n+2}⪯\cdots .\end{array}$

If ${x}_{2n}={x}_{2n+1}$ for some $n\in \mathbb{N}$, then ${x}_{2n}=f{x}_{2n}$. Thus ${x}_{2n}$ is a fixed point of f. By the first part, we conclude that ${x}_{2n}$ is also a fixed point of g.

If ${x}_{2n+1}={x}_{2n+2}$ for some $n\in \mathbb{N}$, then ${x}_{2n+1}=g{x}_{2n+1}$. Thus, ${x}_{2n+1}$ is a fixed point of g. By the first part, we conclude that ${x}_{2n+1}$ is also a fixed point of f. Therefore, we assume that ${x}_{n}\ne {x}_{n+1}$ for all $n\in \mathbb{N}$. Now, we complete the proof in the following steps.

Step 1: We will prove that

$\underset{n\to \mathrm{\infty }}{lim}{p}_{b}\left({x}_{n},{x}_{n+1}\right)=0.$

As ${x}_{2n+1}$ and ${x}_{2n+2}$ are comparable, by (4.1), we have

$\begin{array}{rl}\psi \left({p}_{b}\left({x}_{2n+1},{x}_{2n+2}\right)\right)& \le \psi \left({s}^{2}{p}_{b}\left({x}_{2n+1},{x}_{2n+2}\right)\right)\\ =\psi \left({s}^{2}{p}_{b}\left(f{x}_{2n},g{x}_{2n+1}\right)\right)\\ \le \psi \left({M}_{s}^{f,g}\left({x}_{2n},{x}_{2n+1}\right)\right)-\phi \left({M}_{s}^{f,g}\left({x}_{2n},{x}_{2n+1}\right)\right),\end{array}$

where

$\begin{array}{rl}{M}_{s}^{f,g}\left({x}_{2n},{x}_{2n+1}\right)=& max\left\{{p}_{b}\left({x}_{2n},{x}_{2n+1}\right),{p}_{b}\left({x}_{2n},f{x}_{2n}\right),{p}_{b}\left({x}_{2n+1},g{x}_{2n+1}\right),\\ \frac{{p}_{b}\left(f{x}_{2n},{x}_{2n+1}\right)+{p}_{b}\left({x}_{2n},g{x}_{2n+1}\right)}{2s}\right\}\\ =& max\left\{{p}_{b}\left({x}_{2n},{x}_{2n+1}\right),{p}_{b}\left({x}_{2n+1},{x}_{2n+2}\right),\\ \frac{{p}_{b}\left({x}_{2n+1},{x}_{2n+1}\right)+{p}_{b}\left({x}_{2n},{x}_{2n+2}\right)}{2s}\right\}\\ \le & max\left\{{p}_{b}\left({x}_{2n},{x}_{2n+1}\right),{p}_{b}\left({x}_{2n+1},{x}_{2n+2}\right),\\ \frac{s{p}_{b}\left({x}_{2n},{x}_{2n+1}\right)+s{p}_{b}\left({x}_{2n+1},{x}_{2n+2}\right)}{2s}\right\}\\ =& max\left\{{p}_{b}\left({x}_{2n},{x}_{2n+1}\right),{p}_{b}\left({x}_{2n+1},{x}_{2n+2}\right)\right\}.\end{array}$

Hence, we have

$\begin{array}{rl}\psi \left({p}_{b}\left({x}_{2n+1},{x}_{2n+2}\right)\right)\le & \psi \left(max\left\{{p}_{b}\left({x}_{2n},{x}_{2n+1}\right),{p}_{b}\left({x}_{2n+1},{x}_{2n+2}\right)\right\}\right)\\ -\phi \left(max\left\{{p}_{b}\left({x}_{2n},{x}_{2n+1}\right),{p}_{b}\left({x}_{2n+1},{x}_{2n+2}\right)\right\}\right).\end{array}$
(4.2)

If

$max\left\{{p}_{b}\left({x}_{2n},{x}_{2n+1}\right),{p}_{b}\left({x}_{2n+1},{x}_{2n+2}\right)\right\}={p}_{b}\left({x}_{2n+1},{x}_{2n+2}\right),$

then (4.2) becomes

$\begin{array}{rl}\psi \left({p}_{b}\left({x}_{2n+1},{x}_{2n+2}\right)\right)& \le \psi \left({p}_{b}\left({x}_{2n+1},{x}_{2n+2}\right)\right)-\phi \left({p}_{b}\left({x}_{2n+1},{x}_{2n+2}\right)\right)\\ <\psi \left({p}_{b}\left({x}_{2n+1},{x}_{2n+2}\right)\right),\end{array}$

$max\left\{{p}_{b}\left({x}_{2n},{x}_{2n+1}\right),{p}_{b}\left({x}_{2n+1},{x}_{2n+2}\right)\right\}={p}_{b}\left({x}_{2n},{x}_{2n+1}\right),$

and (4.2) becomes

$\begin{array}{rl}\psi \left({p}_{b}\left({x}_{2n+1},{x}_{2n+2}\right)\right)& \le \psi \left({p}_{b}\left({x}_{2n},{x}_{2n+1}\right)\right)-\phi \left({p}_{b}\left({x}_{2n},{x}_{2n+1}\right)\right)\\ \le \psi \left({p}_{b}\left({x}_{2n},{x}_{2n+1}\right)\right).\end{array}$
(4.3)

Similarly, we can show that

$\psi \left({p}_{b}\left({x}_{2n+1},{x}_{2n}\right)\right)\le \psi \left({p}_{b}\left({x}_{2n-1},{x}_{2n}\right)\right)-\phi \left({p}_{b}\left({x}_{2n-1},{x}_{2n}\right)\right)\le \psi \left({p}_{b}\left({x}_{2n-1},{x}_{2n}\right)\right).$
(4.4)

By (4.3) and (4.4), we get that $\left\{{p}_{b}\left({x}_{n},{x}_{n+1}\right):n\in \mathbb{N}\right\}$ is a nonincreasing sequence of positive numbers. Hence, there is $r\ge 0$ such that

$\underset{n\to \mathrm{\infty }}{lim}{p}_{b}\left({x}_{n},{x}_{n+1}\right)=r.$

Letting $n\to \mathrm{\infty }$ in (4.3), we get

$\psi \left(r\right)\le \psi \left(r\right)-\phi \left(r\right)\le \psi \left(r\right),$

which implies that $\phi \left(r\right)=0$ and hence $r=0$. So, we have

$\underset{n\to \mathrm{\infty }}{lim}{p}_{b}\left({x}_{n},{x}_{n}\right)\le \underset{n\to \mathrm{\infty }}{lim}{p}_{b}\left({x}_{n},{x}_{n+1}\right)=0.$
(4.5)

Step 2. We will prove that $\left\{{x}_{n}\right\}$ is a ${p}_{b}$-Cauchy sequence. Because of (4.5), it is sufficient to show that $\left\{{x}_{2n}\right\}$ is a ${p}_{b}$-Cauchy sequence. By Lemma 1, we should show that $\left\{{x}_{2n}\right\}$ is b-Cauchy in $\left(X,{d}_{{p}_{b}}\right)$. Suppose the contrary, i.e., that $\left\{{x}_{2n}\right\}$ is not a b-Cauchy sequence in $\left(X,{d}_{{p}_{b}}\right)$. Then there exists $\epsilon >0$ for which we can find two subsequences $\left\{{x}_{2{m}_{i}}\right\}$ and $\left\{{x}_{2{n}_{i}}\right\}$ of $\left\{{x}_{2n}\right\}$ such that ${n}_{i}$ is the smallest index for which

${n}_{i}>{m}_{i}>i,\phantom{\rule{1em}{0ex}}{d}_{{p}_{b}}\left({x}_{2{m}_{i}},{x}_{2{n}_{i}}\right)\ge \epsilon .$
(4.6)

This means that

${d}_{{p}_{b}}\left({x}_{2{m}_{i}},{x}_{2{n}_{i}-2}\right)<\epsilon .$
(4.7)

From (4.6) and using the triangular inequality, we get

$\epsilon \le {d}_{{p}_{b}}\left({x}_{2{m}_{i}},{x}_{2{n}_{i}}\right)\le s{d}_{{p}_{b}}\left({x}_{2{m}_{i}},{x}_{2{m}_{i}+1}\right)+s{d}_{{p}_{b}}\left({x}_{2{m}_{i}+1},{x}_{2{n}_{i}}\right).$

Using (4.5) and taking the upper limit as $i\to \mathrm{\infty }$, we get

$\frac{\epsilon }{s}\le \underset{i\to \mathrm{\infty }}{lim sup}{d}_{{p}_{b}}\left({x}_{2{m}_{i}+1},{x}_{2{n}_{i}}\right).$

On the other hand, we have

${d}_{{p}_{b}}\left({x}_{2{m}_{i}},{x}_{2{n}_{i}-1}\right)\le s{d}_{{p}_{b}}\left({x}_{2{m}_{i}},{x}_{2{n}_{i}-2}\right)+s{d}_{{p}_{b}}\left({x}_{2{n}_{i}-2},{x}_{2{n}_{i}-1}\right).$

Using (4.5), (4.7) and taking the upper limit as $i\to \mathrm{\infty }$, we get

$\underset{i\to \mathrm{\infty }}{lim sup}{d}_{{p}_{b}}\left({x}_{2{m}_{i}},{x}_{2{n}_{i}-1}\right)\le \epsilon s.$
(4.8)

Again, using the triangular inequality, we have

$\begin{array}{rl}{d}_{{p}_{b}}\left({x}_{2{m}_{i}},{x}_{2{n}_{i}}\right)& \le s{d}_{{p}_{b}}\left({x}_{2{m}_{i}},{x}_{2{n}_{i}-2}\right)+s{d}_{{p}_{b}}\left({x}_{2{n}_{i}-2},{x}_{2{n}_{i}}\right)\\ \le s{d}_{{p}_{b}}\left({x}_{2{m}_{i}},{x}_{2{n}_{i}-2}\right)+{s}^{2}{d}_{{p}_{b}}\left({x}_{2{n}_{i}-2},{x}_{2{n}_{i}-1}\right)+{s}^{2}{d}_{{p}_{b}}\left({x}_{2{n}_{i}-1},{x}_{2{n}_{i}}\right)\end{array}$

and

${d}_{{p}_{b}}\left({x}_{2{m}_{i}+1},{x}_{2{n}_{i}-1}\right)\le s{d}_{{p}_{b}}\left({x}_{2{m}_{i}+1},{x}_{2{m}_{i}}\right)+s{d}_{{p}_{b}}\left({x}_{2{m}_{i}},{x}_{2{n}_{i}-1}\right).$

Taking the upper limit as $i\to \mathrm{\infty }$ in the above inequalities and using (4.5), (4.7) and (4.8), we get

$\underset{i\to \mathrm{\infty }}{lim sup}{d}_{{p}_{b}}\left({x}_{2{m}_{i}},{x}_{2{n}_{i}}\right)\le \epsilon s$

and

$\underset{i\to \mathrm{\infty }}{lim sup}{d}_{{p}_{b}}\left({x}_{2{m}_{i}+1},{x}_{2{n}_{i}-1}\right)\le \epsilon {s}^{2}.$

From the definition of ${d}_{{p}_{b}}$ and (4.5), we have the following relations:

$\frac{\epsilon }{2s}\le \underset{i\to \mathrm{\infty }}{lim sup}{p}_{b}\left({x}_{2{m}_{i}+1},{x}_{2{n}_{i}}\right),$
(4.9)
$\frac{\epsilon }{2s}\le \underset{i\to \mathrm{\infty }}{lim inf}{p}_{b}\left({x}_{2{m}_{i}},{x}_{2{n}_{i}-1}\right)\le \underset{i\to \mathrm{\infty }}{lim sup}{p}_{b}\left({x}_{2{m}_{i}},{x}_{2{n}_{i}-1}\right)\le \frac{s\epsilon }{2},$
(4.10)
$\underset{i\to \mathrm{\infty }}{lim sup}{p}_{b}\left({x}_{2{m}_{i}},{x}_{2{n}_{i}}\right)\le \frac{s\epsilon }{2},$
(4.11)
$\underset{i\to \mathrm{\infty }}{lim sup}{p}_{b}\left({x}_{2{m}_{i}+1},{x}_{2{n}_{i}-1}\right)\le \frac{{s}^{2}\epsilon }{2}.$
(4.12)

Since ${x}_{2{m}_{i}}$ and ${x}_{2{n}_{i}-1}$ are comparable, using (4.1) we have

$\begin{array}{rl}\psi \left({s}^{2}{p}_{b}\left({x}_{2{m}_{i}+1},{x}_{2{n}_{i}}\right)\right)& =\psi \left({s}^{2}{p}_{b}\left(f{x}_{2{m}_{i}},g{x}_{2{n}_{i}-1}\right)\right)\\ \le \psi \left({M}_{s}^{f,g}\left({x}_{2{m}_{i}},{x}_{2{n}_{i}-1}\right)\right)-\phi \left({M}_{s}^{f,g}\left({x}_{2{m}_{i}},{x}_{2{n}_{i}-1}\right)\right),\end{array}$
(4.13)

where

$\begin{array}{rl}{M}_{s}^{f,g}\left({x}_{2{m}_{i}},{x}_{2{n}_{i}-1}\right)=& max\left\{{p}_{b}\left({x}_{2{m}_{i}},{x}_{2{n}_{i}-1}\right),{p}_{b}\left({x}_{2{m}_{i}},{x}_{2{m}_{i}+1}\right),{p}_{b}\left({x}_{2{n}_{i}-1},{x}_{2{n}_{i}}\right),\\ \frac{{p}_{b}\left({x}_{2{m}_{i}},{x}_{2{n}_{i}}\right)+{p}_{b}\left({x}_{2{m}_{i}+1},{x}_{2{n}_{i}-1}\right)}{2s}\right\}.\end{array}$
(4.14)

Taking the upper limit in (4.14) and using (4.5) and (4.10)-(4.12), we get

$\begin{array}{rl}\underset{i\to \mathrm{\infty }}{lim sup}{M}_{s}^{f,g}\left({x}_{2{m}_{i}},{x}_{2{n}_{i}-1}\right)=& max\left\{\underset{i\to \mathrm{\infty }}{lim sup}{p}_{b}\left({x}_{2{m}_{i}},{x}_{2{n}_{i}-1}\right),0,0,\\ \frac{{lim sup}_{i\to \mathrm{\infty }}{p}_{b}\left({x}_{2{m}_{i}},{x}_{2{n}_{i}}\right)+{lim sup}_{i\to \mathrm{\infty }}{p}_{b}\left({x}_{2{m}_{i}+1},{x}_{2{n}_{i}-1}\right)}{2s}\right\}\\ \le & max\left\{\frac{s\epsilon }{2},\frac{\frac{\epsilon s+\epsilon {s}^{2}}{2}}{2s}\right\}=\frac{s\epsilon }{2}.\end{array}$
(4.15)

Now, taking the upper limit as $i\to \mathrm{\infty }$ in (4.13) and using (4.9) and (4.15), we have

$\begin{array}{rl}\psi \left(\frac{s\epsilon }{2}\right)& =\psi \left({s}^{2}\frac{\epsilon }{2s}\right)\le \psi \left({s}^{2}\underset{i\to \mathrm{\infty }}{lim sup}{p}_{b}\left({x}_{2{m}_{i}+1},{x}_{2{n}_{i}}\right)\right)\\ \le \psi \left(\underset{i\to \mathrm{\infty }}{lim sup}{M}_{s}^{f,g}\left({x}_{2{m}_{i}},{x}_{2{n}_{i}-1}\right)\right)-\phi \left(\underset{i\to \mathrm{\infty }}{lim inf}{M}_{s}^{f,g}\left({x}_{2{m}_{i}},{x}_{2{n}_{i}-1}\right)\right)\\ \le \psi \left(\frac{s\epsilon }{2}\right)-\phi \left(\underset{i\to \mathrm{\infty }}{lim inf}{M}_{s}^{f,g}\left({x}_{2{m}_{i}},{x}_{2{n}_{i}-1}\right)\right),\end{array}$

which implies that $\phi \left({lim inf}_{i\to \mathrm{\infty }}{M}_{s}^{f,g}\left({x}_{2{m}_{i}},{x}_{2{n}_{i}-1}\right)\right)=0$. By (4.14), it follows that

$\underset{i\to \mathrm{\infty }}{lim inf}{p}_{b}\left({x}_{2{m}_{i}},{x}_{2{n}_{i}-1}\right)=0,$

which is in contradiction with (4.10). Thus, we have proved that $\left\{{x}_{n}\right\}$ is a b-Cauchy sequence in the metric space $\left(X,{d}_{{p}_{b}}\right)$. Since $\left(X,{p}_{b}\right)$ is ${p}_{b}$-complete, then from Lemma 1, $\left(X,{d}_{{p}_{b}}\right)$ is a b-complete b-metric space. Therefore, the sequence $\left\{{x}_{n}\right\}$ converges to some $z\in X$, that is, ${lim}_{n\to \mathrm{\infty }}{d}_{{p}_{b}}\left({x}_{n},z\right)=0$. Again, from Lemma 1,

$\underset{n\to \mathrm{\infty }}{lim}{p}_{b}\left(z,{x}_{n}\right)=\underset{n\to \mathrm{\infty }}{lim}{p}_{b}\left({x}_{n},{x}_{n}\right)={p}_{b}\left(z,z\right).$

On the other hand, from (4.5) we get that

$\underset{n\to \mathrm{\infty }}{lim}{p}_{b}\left(z,{x}_{n}\right)=\underset{n\to \mathrm{\infty }}{lim}{p}_{b}\left({x}_{n},{x}_{n}\right)={p}_{b}\left(z,z\right)=0.$

Step 3 (Existence of a common fixed point). Using the triangular inequality, we get

$\begin{array}{r}{p}_{b}\left(z,fz\right)\le s{p}_{b}\left(z,f{x}_{2n}\right)+s{p}_{b}\left(f{x}_{2n},fz\right),\\ {p}_{b}\left(z,gz\right)\le s{p}_{b}\left(z,g{x}_{2n+1}\right)+s{p}_{b}\left(g{x}_{2n+1},gz\right).\end{array}$

Letting $n\to \mathrm{\infty }$ and using the continuity of f and g, we get

$\begin{array}{rl}{p}_{b}\left(z,fz\right)& \le s\underset{n\to \mathrm{\infty }}{lim}{p}_{b}\left(z,f{x}_{2n}\right)+s\underset{n\to \mathrm{\infty }}{lim}{p}_{b}\left(f{x}_{2n},fz\right)=s{p}_{b}\left(fz,fz\right),\\ {p}_{b}\left(z,gz\right)& \le s\underset{n\to \mathrm{\infty }}{lim}{p}_{b}\left(z,g{x}_{2n+1}\right)+s\underset{n\to \mathrm{\infty }}{lim}{p}_{b}\left(g{x}_{2n+1},gz\right)=s{p}_{b}\left(gz,gz\right).\end{array}$

Therefore,

$max\left\{{p}_{b}\left(z,fz\right),{p}_{b}\left(z,gz\right)\right\}\le max\left\{s{p}_{b}\left(fz,fz\right),s{p}_{b}\left(gz,gz\right)\right\}\le {s}^{2}{p}_{b}\left(gz,fz\right).$
(4.16)

From (4.1), we have

$\psi \left({s}^{2}{p}_{b}\left(fz,gz\right)\right)\le \psi \left({M}_{s}^{f,g}\left(z,z\right)\right)-\phi \left({M}_{s}^{f,g}\left(z,z\right)\right),$
(4.17)

where

$\begin{array}{rl}{M}_{s}^{f,g}\left(z,z\right)& =max\left\{{p}_{b}\left(z,z\right),{p}_{b}\left(z,fz\right),{p}_{b}\left(z,gz\right),\frac{{p}_{b}\left(z,gz\right)+{p}_{b}\left(z,fz\right)}{2s}\right\}\\ =max\left\{{p}_{b}\left(z,fz\right),{p}_{b}\left(z,gz\right)\right\}.\end{array}$

As ψ is nondecreasing, we have ${s}^{2}{p}_{b}\left(fz,gz\right)\le max\left\{{p}_{b}\left(z,fz\right),{p}_{b}\left(z,gz\right)\right\}$. Hence, by (4.16) we obtain that ${s}^{2}{p}_{b}\left(fz,gz\right)=max\left\{{p}_{b}\left(z,fz\right),{p}_{b}\left(z,gz\right)\right\}$. But then, using (4.17), we get that $\phi \left({M}_{s}^{f,g}\left(z,z\right)\right)=0$. Thus, we have $fz=gz=z$ and z is a common fixed point of f and g. □

The continuity of functions f and g in Theorem 3 can be replaced by another condition.

Theorem 4 Under the hypotheses of Theorem  3, without the continuity assumption on the functions f and g, for any nondecreasing sequence $\left\{{x}_{n}\right\}$ in X such that ${x}_{n}\to x\in X$, let us have ${x}_{n}⪯x$ for all $n\in \mathbb{N}$. Then f and g have a common fixed point in X.

Proof Reviewing the proof of Theorem 3, we construct an increasing sequence $\left\{{x}_{n}\right\}$ in X such that ${x}_{n}\to z$ for some $z\in X$. Using the given assumption on X, we have ${x}_{n}⪯z$ for all $n\in \mathbb{N}$. Now, we show that $fz=gz=z$. By (4.1), we have

$\begin{array}{rl}\psi \left({s}^{2}{p}_{b}\left({x}_{2n+1},gz\right)\right)& =\psi \left({s}^{2}{p}_{b}\left(f{x}_{2n},gz\right)\right)\\ \le \psi \left({M}_{s}^{f,g}\left({x}_{2n},z\right)\right)-\phi \left({M}_{s}^{f,g}\left({x}_{2n},z\right)\right),\end{array}$
(4.18)

where

$\begin{array}{rl}{M}_{s}^{f,g}\left({x}_{2n},z\right)& =max\left\{{p}_{b}\left({x}_{2n},z\right),{p}_{b}\left({x}_{2n},f{x}_{2n}\right),{p}_{b}\left(z,gz\right),\frac{{p}_{b}\left({x}_{2n},gz\right)+{p}_{b}\left(f{x}_{2n},z\right)}{2s}\right\}\\ =max\left\{{p}_{b}\left({x}_{2n},z\right),{p}_{b}\left({x}_{2n},{x}_{2n+1}\right),{p}_{b}\left(z,gz\right),\frac{{p}_{b}\left({x}_{2n},gz\right)+{p}_{b}\left({x}_{2n+1},z\right)}{2s}\right\}.\end{array}$
(4.19)

Letting $n\to \mathrm{\infty }$ in (4.19) and using Lemma 2, we get

$\begin{array}{rl}\frac{{p}_{b}\left(z,gz\right)}{{s}^{2}}& \le max\left\{{p}_{b}\left(z,gz\right),\frac{\frac{{p}_{b}\left(z,gz\right)}{s}}{2s}\right\}\le \underset{n\to \mathrm{\infty }}{lim inf}{M}_{s}^{f,g}\left({x}_{2n},z\right)\\ \le \underset{n\to \mathrm{\infty }}{lim sup}{M}_{s}^{f,g}\left({x}_{2n},z\right)\le max\left\{{p}_{b}\left(z,gz\right),\frac{s{p}_{b}\left(z,gz\right)}{2s}\right\}={p}_{b}\left(z,gz\right).\end{array}$
(4.20)

Again, taking the upper limit as $n\to \mathrm{\infty }$ in (4.18) and using Lemma 2 and (4.20), we get

$\begin{array}{rl}\psi \left({p}_{b}\left(z,gz\right)\right)& =\psi \left({s}^{2}\frac{1}{{s}^{2}}{p}_{b}\left(z,gz\right)\right)\le \psi \left({s}^{2}\underset{n\to \mathrm{\infty }}{lim sup}d\left({x}_{2n+1},gz\right)\right)\\ \le \psi \left(\underset{n\to \mathrm{\infty }}{lim sup}{M}_{s}^{f,g}\left({x}_{2n},z\right)\right)-\phi \left(\underset{n\to \mathrm{\infty }}{lim inf}{M}_{s}^{f,g}\left({x}_{2n},z\right)\right)\\ \le \psi \left({p}_{b}\left(z,gz\right)\right)-\phi \left(\underset{n\to \mathrm{\infty }}{lim inf}{M}_{s}^{f,g}\left({x}_{2n},z\right)\right).\end{array}$

Therefore, $\phi \left({lim inf}_{n\to \mathrm{\infty }}{M}_{s}^{f,g}\left({x}_{2n},z\right)\right)\le 0$, equivalently, ${lim inf}_{n\to \mathrm{\infty }}{M}_{s}^{f,g}\left({x}_{2n},z\right)=0$. Thus, from (4.20) we get $z=gz$ and hence z is a fixed point of g. On the other hand, similar to the first part of the proof of Theorem 3, we can show that $fz=z$. Hence, z is a common fixed point of f and g. □

Also, we have the following results.

Corollary 3 Let $\left(X,⪯,{p}_{b}\right)$ be a ${p}_{b}$-complete ordered partial b-metric space with the coefficient $s\ge 1$, and let $f,g:X\to X$ be two weakly increasing mappings with respect to ⪯. Suppose that there exists $k\in \left[0,1\right)$ such that

${p}_{b}\left(fx,gy\right)\le \frac{k}{{s}^{2}}max\left\{{p}_{b}\left(x,y\right),{p}_{b}\left(x,fx\right),{p}_{b}\left(y,gy\right),\frac{{p}_{b}\left(x,gy\right)+{p}_{b}\left(fx,y\right)}{2s}\right\}$

for all comparable elements $x,y\in X$. If f and g are continuous, then f and g have a common fixed point.

Corollary 4 Under the hypotheses of Corollary  3, without the continuity assumption on the functions f and g, assume that whenever $\left\{{x}_{n}\right\}$ is a nondecreasing sequence in X such that ${x}_{n}\to x\in X$, then ${x}_{n}⪯x$ for all $n\in \mathbb{N}$. Then f and g have a common fixed point in X.

Remark 1 Recall that a subset W of a partially ordered set X is said to be well ordered if every two elements of W are comparable. Note that in Theorems 1 and 2, it can be proved in a standard way that f has a unique fixed point provided that the fixed points of f are comparable. Similarly, in Theorems 3 and 4, the set of common fixed points of f and g is well ordered if and only if f and g have one and only one common fixed point.

The usability of these results is demonstrated by the following example.

Example 5 Let $X=\left\{0,1,2,3,4\right\}$ be equipped with the following partial order ⪯:

$⪯:=\left\{\left(0,0\right),\left(1,1\right),\left(1,2\right),\left(2,2\right),\left(3,3\right),\left(4,2\right),\left(4,4\right)\right\}.$

Define a partial b-metric ${p}_{b}:X×X\to {\mathbb{R}}^{+}$ by

It is easy to see that $\left(X,{p}_{b}\right)$ is a ${p}_{b}$-complete partial b-metric space, with $s=49/25$.

Define self-maps f and g by

$f=\left(\begin{array}{ccccc}0& 1& 2& 3& 4\\ 0& 2& 2& 1& 2\end{array}\right),\phantom{\rule{2em}{0ex}}g=\left(\begin{array}{ccccc}0& 1& 2& 3& 4\\ 0& 2& 2& 1& 1\end{array}\right).$

We see that f and g are weakly increasing mappings with respect to ⪯ and that f and g are continuous.

Define $\psi ,\phi :\left[0,\mathrm{\infty }\right)\to \left[0,\mathrm{\infty }\right)$ by $\psi \left(t\right)=\sqrt{t}$ and $\phi \left(t\right)=\frac{t}{300}$. In order to check that $\left(f,g\right)$ is a generalized ${\left(\psi ,\phi \right)}_{s}$-contractive pair, only the case $x=2$, $y=4$ is nontrivial (when x and y are comparable and the left-hand side of condition (4.1) is positive). Then

$\psi \left({s}^{2}{p}_{b}\left(f2,g4\right)\right)=\sqrt{{s}^{2}\cdot {3}^{2}}=\frac{147}{25}=\sqrt{36}-\frac{36}{300}=\psi \left({M}_{s}^{f,g}\left(2,4\right)\right)-\phi \left({M}_{s}^{f,g}\left(2,4\right)\right).$

Thus, all the conditions of Theorem 3 are satisfied and hence f and g have common fixed points. Indeed, 0 and 2 are two common fixed points of f and g. Note that the ordered set $\left(\left\{0,2\right\},⪯\right)$ is not well ordered.

Note that if the same example is considered in the space without order, then the contractive condition is not satisfied. For example,

$\begin{array}{rl}\psi \left({s}^{2}{p}_{b}\left(f1,g4\right)\right)& =\sqrt{{s}^{2}\cdot {3}^{2}}=\frac{147}{25}\\ >\frac{59}{12}=\sqrt{25}-\frac{25}{300}=\psi \left({M}_{s}^{f,g}\left(1,4\right)\right)-\phi \left({M}_{s}^{f,g}\left(1,4\right)\right).\end{array}$