Some common fixed point results in ordered partial b-metric spaces

Abstract

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 $(\psi ,\phi )$-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.

MSC: 47H10, 54H25.

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., [3–8]).

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., [12–26]).

Definition 1 [10]

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

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

• (b₂) $d(x,y)=d(y,x)$,

• (b₃) $d(x,z)\le s[d(x,y)+d(y,z)]$.

In this case, the pair $(X,d)$ 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 [28–33].

Definition 2 [27]

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

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

• (p₂) $p(x,x)\le p(x,y)$,

• (p₃) $p(x,y)=p(y,x)$,

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

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

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

Definition 3 [27]

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

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

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

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

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

Example 1 [32]

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

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

Then, clearly, $\{x_n\}$ is a convergent sequence and for every $x\ge 1$, we have ${lim}_{n\to \mathrm{\infty }}p(x_n,x)=p(x,x)$. But ${lim}_{n,m\to \mathrm{\infty }}p(x_n,x_m)$ does not exist, that is, $\{x_n\}$ 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 $p_b:X\times X\to {\mathbb{R}}^{+}$ such that for all $x,y,z\in X$:

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

• (${\text{p}}_{b2}$) $p_b(x,x)\le p_b(x,y)$,

• (${\text{p}}_{b3}$) $p_b(x,y)=p_b(y,x)$,

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

A partial b-metric space is a pair $(X,p_b)$ 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 $(X,p_b)$.

In a partial b-metric space $(X,p_b)$, if $x,y\in X$ and $p_b(x,y)=0$, 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 $p_b:X\times X\to {\mathbb{R}}^{+}$ be defined by

$$p_b(x,y)={[max\{x,y\}]}^q+|x-y{|}^q\text{for all }x,y\in X.$$

Then $(X,p_b)$ 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 $p_b:X\times X\to {\mathbb{R}}^{+}$, defined by $p_b(x,y)=p(x,y)+d(x,y)$ for all $x,y\in X$, is a partial b-metric on X with the coefficient s.

Proposition 2 [34]

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

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

Definition 5 [35]

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

• 1. ψ is continuous and nondecreasing;

• 2. $\psi (t)=0$ 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, 36–41]).

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 $(\psi ,\phi )$-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 $p_b:X\times X\to {\mathbb{R}}^{+}$ is a partial b-metric if, for all $x,y,z\in X$, the following conditions are satisfied:

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

• (${\text{p}}_{b2}$) $p_b(x,x)\le p_b(x,y)$,

• (${\text{p}}_{b3}$) $p_b(x,y)=p_b(y,x)$,

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

The pair $(X,p_b)$ is called a partial b-metric space.

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

$$p_b(x,y)\le s(p_b(x,z)+p_b(z,y)-p_b(z,z))\le s(p_b(x,z)+p_b(z,y))-p_b(z,z).$$

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 $(X,d)$ be a metric space and $p_b(x,y)=d{(x,y)}^q+a$, 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 $f(x)=x^q$ ($x>0$) implies that ${(a+b)}^q\le 2^{q-1}(a^q+b^q)$ holds for $a,b\ge 0$. Thus, for each $x,y,z\in X$, we obtain

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

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

Note that $(X,p_b)$ is not necessarily a partial metric space. For example, if $X=\mathbb{R}$ is the set of real numbers, $d(x,y)=|x-y|$, $q=2$ and $a=3$, then $p_b(x,y)={(x-y)}^2+3$ 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 $p_b(2,5)=12$, $p_b(2,\frac{5}{2})=\frac{13}{4}$ and $p_b(\frac{5}{2},5)=\frac{37}{4}$, hence $p_b(2,5)=12\nleqq \frac{38}{4}=p_b(2,\frac{5}{2})+p_b(\frac{5}{2},5)-p_b(\frac{5}{2},\frac{5}{2})$.

Also, $p_b$ is not a b-metric since $p_b(x,x)\ne 0$ for $x\in X$.

Proposition 3 Every partial b-metric $p_b$ defines a b-metric $d_{p_b}$, where

$$d_{p_b}(x,y)=2p_b(x,y)-p_b(x,x)-p_b(y,y)$$

for all $x,y\in X$.

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

$$\begin{array}{r}{d}_{p_b}(x,y)\\ =2p_b(x,y)-p_b(x,x)-p_b(y,y)\\ \le 2[s(p_b(x,z)+p_b(z,y)-p_b(z,z))+\left(\frac{1-s}{2}\right)(p_b(x,x)+p_b(y,y))]\\ -p_b(x,x)-p_b(y,y)\\ =2s{p}_b(x,z)+2s{p}_b(z,y)-2s{p}_b(z,z)+(1-s)(p_b(x,x)+p_b(y,y))\\ -p_b(x,x)-p_b(y,y)\\ =2s{p}_b(x,z)+2s{p}_b(z,y)-2s{p}_b(z,z)-s{p}_b(x,x)-s{p}_b(y,y)\\ =s[2p_b(x,z)-p_b(x,x)-p_b(z,z)+2p_b(z,y)-p_b(z,z)-p_b(y,y)]\\ =s[d_{p_b}(x,z)+d_{p_b}(z,y)].\end{array}$$

 □

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 $(X,p_b)$ 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

$$B_{p_b}(x,ϵ)=\{y\in X\mid p_b(x,y)<p_b(x,x)+ϵ\}.$$

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

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

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

Proof Let $y\in B_{p_b}(x,r)$. If $y=x$, then we choose $\delta =r$. Suppose that $y\ne x$. Then we have $p_b(x,y)\ne 0$. Now, we consider two cases.

Case 1. If $p_b(x,y)=p_b(x,x)$, then for $s=1$ we choose $\delta =r$. If $s>1$, then we consider the set

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

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 $p_b(x,x)\le r/(2s^m(s-1))$ and we choose $\delta =r/(2s^{m+1})$. Let $z\in B_{p_b}(y,\delta )$; by the property (${\text{p}}_{b4}$), we have

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

Hence, $B_{p_b}(y,\delta )\subseteq B_{p_b}(x,r)$.

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

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

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

For $s>1$, we consider the set

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

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

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

Hence, $B_{p_b}(y,\delta )\subseteq B_{p_b}(x,r)$. □

Thus, from the above proposition the family of all $p_b$-balls

$$\mathrm{\Delta }=\{B_{p_b}(x,r)\mid x\in X,r>0\}$$

is a base of a $T_0$ topology ${\tau }_{p_b}$ on X which we call the $p_b$-metric topology.

The topological space $(X,p_b)$ is $T_0$, but need not be $T_1$.

Definition 8 A sequence $\{x_n\}$ in a partial b-metric space $(X,p_b)$ is said to be:

• (i) $p_b$-convergent to a point $x\in X$ if ${lim}_{n\to \mathrm{\infty }}{p}_b(x,x_n)=p_b(x,x)$;

• (ii) a $p_b$-Cauchy sequence if ${lim}_{n,m\to \mathrm{\infty }}{p}_b(x_n,x_m)$ exists (and is finite).

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

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

Lemma 1

• (1) A sequence $\{x_n\}$ is a $p_b$-Cauchy sequence in a partial b-metric space $(X,p_b)$ if and only if it is a b-Cauchy sequence in the b-metric space $(X,d_{p_b})$.

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

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

Proof First, we show that every $p_b$-Cauchy sequence in $(X,p_b)$ is a b-Cauchy sequence in $(X,d_{p_b})$. Let $\{x_n\}$ be a $p_b$-Cauchy sequence in $(X,p_b)$. Then, there exists $\alpha \in \mathbb{R}$ such that, for arbitrary $\epsilon >0$, there is $n_{\epsilon }\in \mathbb{N}$ with

$$|p_b(x_n,x_m)-\alpha |<\frac{\epsilon }{4}$$

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

$$\begin{array}{r}|d_{p_b}(x_n,x_m)|\\ =2p_b(x_n,x_m)-p_b(x_n,x_n)-p_b(x_m,x_m)\\ =|p_b(x_n,x_m)-\alpha +\alpha -p_b(x_n,x_n)+p_b(x_m,x_n)-\alpha +\alpha -p_b(x_m,x_m)|\\ \le |p_b(x_n,x_m)-\alpha |+|\alpha -p_b(x_n,x_n)|+|p_b(x_m,x_n)-\alpha |+|\alpha -p_b(x_m,x_m)|\\ <\epsilon \end{array}$$

for all $n,m\ge n_{\epsilon }$. Hence, we conclude that $\{x_n\}$ is a b-Cauchy sequence in $(X,d_{p_b})$.

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

$$\underset{n\to \mathrm{\infty }}{lim}[p_b(x_n,y)-p_b(y,y)+p_b(y,x_n)-p_b(x_n,x_n)]=0,$$

therefore, ${lim}_{n\to \mathrm{\infty }}[p_b(x_n,y)-p_b(y,y)]=0$. Further, we have

$$\underset{n\to \mathrm{\infty }}{lim}[p_b(y,x_n)-p_b(x_n,x_n)]=0.$$

Consequently,

$$\underset{n\to \mathrm{\infty }}{lim}{p}_b(x_n,y)=p_b(y,y)=\underset{n\to \mathrm{\infty }}{lim}{p}_b(x_n,x_n).$$

On the other hand,

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

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

$$p_b(y,y)\le \underset{n,m\to \mathrm{\infty }}{lim}{p}_b(x_n,y)=\underset{n,m\to \mathrm{\infty }}{lim}{p}_b(x_n,x_n)\le \underset{n,m\to \mathrm{\infty }}{lim}{p}_b(x_n,x_m).$$

Hence, we obtain that $\{x_n\}$ is a $p_b$-convergent sequence in $(X,p_b)$.

Now, we prove that every b-Cauchy sequence $\{x_n\}$ in $(X,d_{p_b})$ is a $p_b$-Cauchy sequence in $(X,p_b)$. Let $\epsilon =\frac{1}{2}$. Then there exists $n_0\in \mathbb{N}$ such that $d_{p_b}(x_n,x_m)<\frac{1}{2}$ for all $n,m\ge n_0$. Since

$$p_b(x_n,x_{n_0})-p_b(x_{n_0},x_{n_0})\le d_{p_b}(x_n,x_{n_0})<\frac{1}{2},$$

hence

$$p_b(x_n,x_n)\le p_b(x_n,x_{n_0})\le d_{p_b}(x_n,x_{n_0})+p_b(x_{n_0},x_{n_0})<\frac{1}{2}+p_b(x_{n_0},x_{n_0}).$$

Consequently, the sequence $\{p_b(x_n,x_n)\}$ is bounded in ℝ, and so there exists $a\in \mathbb{R}$ such that a subsequence $\{p_b(x_{n_k},x_{n_k})\}$ of $\{p_b(x_n,x_n)\}$ is convergent to a, i.e.,

$$\underset{k\to \mathrm{\infty }}{lim}{p}_b(x_{n_k},x_{n_k})=a.$$

Now, we prove that $\{p_b(x_n,x_n)\}$ is a Cauchy sequence in ℝ. Since $\{x_n\}$ is a b-Cauchy sequence in $(X,d_{p_b})$ for given $\epsilon >0$, there exists $n_{\epsilon }\in \mathbb{N}$ such that $d_{p_b}(x_n,x_m)<\epsilon$ for all $n,m\ge n_{\epsilon }$. Thus, for all $n,m\ge n_{\epsilon }$,

$$\begin{array}{rl}{p}_b(x_n,x_n)-p_b(x_m,x_m)& \le p_b(x_n,x_m)-p_b(x_m,x_m)\\ \le d_{p_b}(x_m,x_n)<\epsilon .\end{array}$$

Therefore, ${lim}_{n\to \mathrm{\infty }}{p}_b(x_n,x_n)=a$.

On the other hand,

$$\begin{array}{rl}|p_b(x_n,x_m)-a|& =|p_b(x_n,x_m)-p_b(x_n,x_n)+p_b(x_n,x_n)-a|\\ \le d_{p_b}(x_m,x_n)+|p_b(x_n,x_n)-a|\end{array}$$

for all $n,m\ge n_{\epsilon }$. Hence, ${lim}_{n,m\to \mathrm{\infty }}{p}_b(x_n,x_m)=a$, and consequently, $\{x_n\}$ is a $p_b$-Cauchy sequence in $(X,p_b)$.

Conversely, let $\{x_n\}$ be a b-Cauchy sequence in $(X,d_{p_b})$. Then $\{x_n\}$ is a $p_b$-Cauchy sequence in $(X,p_b)$, and so it is convergent to a point $x\in X$ with

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

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

$$p_b(x,x_n)-p_b(x,x)<\frac{\epsilon }{4}$$

and

$$p_b(x_n,x_n)-p_b(x,x)\le p_b(x_m,x_n)-p_b(x,x)<\frac{\epsilon }{4}.$$

Therefore,

$$\begin{array}{rl}|d_{p_b}(x_n,x)|& =|p_b(x_n,x)-p_b(x_n,x_n)+p_b(x_n,x)-p_b(x,x)|\\ \le |p_b(x_n,x)-p_b(x,x)|+|p_b(x,x)-p_b(x_n,x_n)|+|p_b(x_n,x)-p_b(x,x)|\\ <\epsilon ,\end{array}$$

whenever $n\ge n_{\epsilon }$. Therefore, $(X,d_{p_b})$ is complete.

Finally, let ${lim}_{n\to \mathrm{\infty }}{d}_{p_b}(x_n,x)=0$. So,

$$\underset{n\to \mathrm{\infty }}{lim}[p_b(x_n,x)-p_b(x_n,x_n)]+\underset{n\to \mathrm{\infty }}{lim}[p_b(x_n,x)-p_b(x,x)]=0.$$

On the other hand,

$$\begin{array}{r}\underset{n,m\to \mathrm{\infty }}{lim}[p_b(x_n,x_m)-p_b(x,x)]\\ \le \underset{n\to \mathrm{\infty }}{lim}[s{p}_b(x_n,x)+s{p}_b(x,x_m)-s{p}_b(x,x)\\ +\left(\frac{1-s}{2}\right)(p_b(x_n,x_n)+p_b(x_m,x_m))-p_b(x,x)]\\ =0.\end{array}$$

 □

Definition 9 Let $(X,p_b)$ and $(X^{\prime },p_b^{\prime })$ be two partial b-metric spaces, and let $f:(X,p_b)\to (X^{\prime },p_b^{\prime })$ 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 $p_b(a,x)<\delta +p_b(a,a)$ imply that $p_b^{\prime }(f(a),f(x))<\epsilon +p_b^{\prime }(f(a),f(a))$. The mapping f is $p_b$-continuous on X if it is $p_b$-continuous at all $a\in X$.

Proposition 5 Let $(X,p_b)$ and $(X^{\prime },p_b^{\prime })$ 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 $\{x_n\}$ is $p_b$-convergent to x, $\{f(x_n)\}$ is $p_b^{\prime }$-convergent to $f(x)$.

Definition 10 A triple $(X,⪯,p_b)$ is called an ordered partial b-metric space if $(X,⪯)$ 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 $(X,p_b)$ be a partial b-metric space with the coefficient $s>1$ and suppose that $\{x_n\}$ and $\{y_n\}$ are convergent to x and y, respectively. Then we have

$$\begin{array}{rl}\frac{1}{s^2}{p}_b(x,y)-\frac{1}{s}{p}_b(x,x)-p_b(y,y)& \le \underset{n\to \mathrm{\infty }}{lim\hspace{0.17em}inf}{p}_b(x_n,y_n)\le \underset{n\to \mathrm{\infty }}{lim\hspace{0.17em}sup}{p}_b(x_n,y_n)\\ \le s{p}_b(x,x)+s^2{p}_b(y,y)+s^2{p}_b(x,y).\end{array}$$

In particular, if $p_b(x,y)=0$, then we have ${lim}_{n\to \mathrm{\infty }}{p}_b(x_n,y_n)=0$.

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

$$\begin{array}{rl}\frac{1}{s}{p}_b(x,z)-p_b(x,x)& \le \underset{n\to \mathrm{\infty }}{lim\hspace{0.17em}inf}{p}_b(x_n,z)\le \underset{n\to \mathrm{\infty }}{lim\hspace{0.17em}sup}{p}_b(x_n,z)\\ \le s{p}_b(x,z)+s{p}_b(x,x).\end{array}$$

In particular, if $p_b(x,x)=0$, then we have

$$\frac{1}{s}{p}_b(x,z)\le \underset{n\to \mathrm{\infty }}{lim\hspace{0.17em}inf}{p}_b(x_n,z)\le \underset{n\to \mathrm{\infty }}{lim\hspace{0.17em}sup}{p}_b(x_n,z)\le s{p}_b(x,z).$$

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

$$p_b(x,y)\le s{p}_b(x,x_n)+s^2{p}_b(x_n,y_n)+s^2{p}_b(y_n,y)$$

and

$$p_b(x_n,y_n)\le s{p}_b(x_n,x)+s^2{p}_b(x,y)+s^2{p}_b(y,y_n).$$

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 $p_b(x,y)=0$, then by the triangle inequality we get $p_b(x,x)=0$ and $p_b(y,y)=0$. Therefore, we have ${lim}_{n\to \mathrm{\infty }}{p}_b(x_n,y_n)=0$. Similarly, using again the triangle inequality, the other assertions follow. □

Let $(X,⪯,p_b)$ be an ordered partial b-metric space, and let $f:X\to X$ be a mapping. Set

$$M_s^f(x,y)=max\{p_b(x,y),p_b(x,fx),p_b(y,fy),\frac{p_b(x,fy)+p_b(y,fx)}{2s}\}.$$

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

$$\psi (s{p}_b(fx,fy))\le \psi (M_s^f(x,y))-\phi (M_s^f(x,y))$$

(3.1)

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

First, we prove the following result.

Theorem 1 Let $(X,⪯,p_b)$ 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 ${(\psi ,\phi )}_s$-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 $(x_n)$ 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

$$\begin{array}{rl}\psi (p_b(x_n,x_{n+1}))& \le \psi (s{p}_b(x_n,x_{n+1}))\\ =\psi (s{p}_b(f{x}_{n-1},f{x}_n))\\ \le \psi (M_s^f(x_{n-1},x_n))-\phi (M_s^f(x_{n-1},x_n)),\end{array}$$

(3.2)

where

$$\begin{array}{rl}{M}_s^f(x_{n-1},x_n)=& max\{p_b(x_{n-1},x_n),p_b(x_{n-1},f{x}_{n-1}),p_b(x_n,f{x}_n),\\ \frac{p_b(x_{n-1},f{x}_n)+p_b(x_n,f{x}_{n-1})}{2s}\}\\ =& max\{p_b(x_{n-1},x_n),p_b(x_n,x_{n+1}),\frac{p_b(x_{n-1},x_{n+1})+p_b(x_n,x_n)}{2s}\}\\ \le & max\{p_b(x_{n-1},x_n),p_b(x_n,x_{n+1}),\\ \frac{s{p}_b(x_{n-1},x_n)+s{p}_b(x_n,x_{n+1})+(1-s)p_b(x_n,x_n)}{2s}\}\\ =& max\{p_b(x_{n-1},x_n),p_b(x_n,x_{n+1})\}.\end{array}$$

So, we have

$$M_s^f(x_{n-1},x_n)=max\{p_b(x_{n-1},x_n),p_b(x_n,x_{n+1})\}.$$

(3.3)

From (3.2), (3.3) we get

$$\begin{array}{rl}\psi (p_b(x_n,x_{n+1}))\le & \psi (max\{p_b(x_{n-1},x_n),p_b(x_n,x_{n+1})\})\\ -\phi (max\{p_b(x_{n-1},x_n),p_b(x_n,x_{n+1})\}).\end{array}$$

(3.4)

If

$$max\{p_b(x_{n-1},x_n),p_b(x_n,x_{n+1})\}=p_b(x_n,x_{n+1}),$$

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

$$\begin{array}{rl}\psi (p_b(x_n,x_{n+1}))& \le \psi (p_b(x_n,x_{n+1}))-\phi (p_b(x_n,x_{n+1}))\\ <\psi (p_b(x_n,x_{n+1})),\end{array}$$

which gives a contradiction. Thus,

$$\psi (p_b(x_n,x_{n+1}))\le \psi (p_b(x_{n-1},x_n))-\phi (p_b(x_{n-1},x_n)).$$

(3.5)

Therefore, $\{p_b(x_n,x_{n+1}):n\in \mathbb{N}\cup \{0\}\}$ is a nonincreasing sequence of positive numbers. So, there exists $r\ge 0$ such that

$$\underset{n\to \mathrm{\infty }}{lim}{p}_b(x_n,x_{n+1})=r.$$

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

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

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

$$\underset{n\to \mathrm{\infty }}{lim}{p}_b(x_n,x_{n+1})=0.$$

(3.6)

Next, we show that $\{x_n\}$ is a $p_b$-Cauchy sequence in X. For this, we have to show that $\{x_n\}$ is a b-Cauchy sequence in $(X,d_{p_b})$ (see Lemma 1). Suppose the contrary; that is, $\{x_n\}$ is not a b-Cauchy sequence. Then there exists $\epsilon >0$ for which we can find two subsequences $\{x_{m_i}\}$ and $\{x_{n_i}\}$ of $\{x_n\}$ such that $n_i$ is the smallest index for which

$$n_i>m_i>i,d_{p_b}(x_{m_i},x_{n_i})\ge \epsilon .$$

(3.7)

This means that

$$d_{p_b}(x_{m_i},x_{n_i-1})<\epsilon .$$

(3.8)

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

$$\epsilon \le d_{p_b}(x_{m_i},x_{n_i})\le s{d}_{p_b}(x_{m_i},x_{n_i-1})+s{d}_{p_b}(x_{n_i-1},x_{n_i}).$$

(3.9)

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

$$\frac{\epsilon }{s}\le \underset{i\to \mathrm{\infty }}{lim\hspace{0.17em}inf}{d}_{p_b}(x_{m_i},x_{n_i-1})\le \underset{i\to \mathrm{\infty }}{lim\hspace{0.17em}sup}{d}_{p_b}(x_{m_i},x_{n_i-1})\le \epsilon .$$

(3.10)

Also, from (3.9) and (3.10),

$$\epsilon \le \underset{i\to \mathrm{\infty }}{lim\hspace{0.17em}sup}{d}_{p_b}(x_{m_i},x_{n_i})\le s\epsilon .$$

Further,

$$d_{p_b}(x_{m_i+1},x_{n_i})\le s{d}_{p_b}(x_{m_i+1},x_{m_i})+s{d}_{p_b}(x_{m_i},x_{n_i}),$$

and hence

$$\underset{i\to \mathrm{\infty }}{lim\hspace{0.17em}sup}{d}_{p_b}(x_{m_i+1},x_{n_i})\le s^2\epsilon .$$

Finally,

$$d_{p_b}(x_{m_i+1},x_{n_i-1})\le s{d}_{p_b}(x_{m_i+1},x_{m_i})+s{d}_{p_b}(x_{m_i},x_{n_i-1}),$$

and hence

$$\underset{i\to \mathrm{\infty }}{lim\hspace{0.17em}sup}{d}_{p_b}(x_{m_i+1},x_{n_i-1})\le s\epsilon .$$

On the other hand, by the definition of $d_{p_b}$ and (3.6),

$$\underset{i\to \mathrm{\infty }}{lim\hspace{0.17em}sup}{d}_{p_b}(x_{m_i},x_{n_i-1})=2\underset{i\to \mathrm{\infty }}{lim\hspace{0.17em}sup}{p}_b(x_{m_i},x_{n_i-1}).$$

Hence, by (3.10),

$$\frac{\epsilon }{2s}\le \underset{i\to \mathrm{\infty }}{lim\hspace{0.17em}inf}{p}_b(x_{m_i},x_{n_i-1})\le \underset{i\to \mathrm{\infty }}{lim\hspace{0.17em}sup}{p}_b(x_{m_i},x_{n_i-1})\le \frac{\epsilon }{2}.$$

(3.11)

Similarly,

$$\underset{i\to \mathrm{\infty }}{lim\hspace{0.17em}sup}{p}_b(x_{m_i},x_{n_i})\le \frac{s\epsilon }{2},$$

(3.12)

$$\frac{\epsilon }{2s}\le \underset{i\to \mathrm{\infty }}{lim\hspace{0.17em}sup}{p}_b(x_{m_i+1},x_{n_i}),$$

(3.13)

$$\underset{i\to \mathrm{\infty }}{lim\hspace{0.17em}sup}{p}_b(x_{m_i+1},x_{n_i-1})\le \frac{s\epsilon }{2}.$$

(3.14)

From (3.1), we have

$$\begin{array}{rl}\psi (s{p}_b(x_{m_i+1},x_{n_i}))& =\psi (s{p}_b(f{x}_{m_i},f{x}_{n_i-1}))\\ \le \psi (M_s^f(x_{m_i},x_{n_i-1}))-\phi (M_s^f(x_{m_i},x_{n_i-1})),\end{array}$$

(3.15)

where

$$\begin{array}{r}{M}_s^f(x_{m_i},x_{n_i-1})\\ =max\{p_b(x_{m_i},x_{n_i-1}),p_b(x_{m_i},f{x}_{m_i}),p_b(x_{n_i-1},f{x}_{n_i-1}),\\ \frac{p_b(x_{m_i},f{x}_{n_i-1})+p_b(f{x}_{m_i},x_{n_i-1})}{2s}\}\\ =max\{p_b(x_{m_i},x_{n_i-1}),p_b(x_{m_i},x_{m_i+1}),p_b(x_{n_i-1},x_{n_i}),\\ \frac{p_b(x_{m_i},x_{n_i})+p_b(x_{m_i+1},x_{n_i-1})}{2s}\}.\end{array}$$

(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

$$\begin{array}{rl}\underset{i\to \mathrm{\infty }}{lim\hspace{0.17em}sup}{M}_s^f(x_{m_i},x_{n_i-1})=& max\{\underset{i\to \mathrm{\infty }}{lim\hspace{0.17em}sup}{p}_b(x_{m_i},x_{n_i-1}),0,0,\\ \frac{{lim\hspace{0.17em}sup}_{i\to \mathrm{\infty }}{p}_b(x_{m_i},x_{n_i})+{lim\hspace{0.17em}sup}_{i\to \mathrm{\infty }}{p}_b(x_{m_i+1},x_{n_i-1})}{2s}\}\\ \le & max\{\frac{\epsilon }{2},\frac{\frac{\epsilon s+\epsilon s}{2}}{2s}\}=\frac{\epsilon }{2}.\end{array}$$

(3.17)

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

$$\begin{array}{rl}\psi \left(s\frac{\epsilon }{2s}\right)& \le \psi (s\underset{i\to \mathrm{\infty }}{lim\hspace{0.17em}sup}{p}_b(x_{m_i+1},x_{n_i}))\\ \le \psi (\underset{i\to \mathrm{\infty }}{lim\hspace{0.17em}sup}{M}_s^f(x_{m_i},x_{n_i-1}))-\underset{i\to \mathrm{\infty }}{lim\hspace{0.17em}inf}\phi (M_s^f(x_{m_i},x_{n_i-1}))\\ \le \psi \left(\frac{\epsilon }{2}\right)-\phi (\underset{i\to \mathrm{\infty }}{lim\hspace{0.17em}inf}{M}_s^f(x_{m_i},x_{n_i-1})),\end{array}$$

which further implies that

$$\phi (\underset{i\to \mathrm{\infty }}{lim\hspace{0.17em}inf}{M}_s^f(x_{m_i},x_{n_i-1}))=0,$$

so ${lim\hspace{0.17em}inf}_{i\to \mathrm{\infty }}{M}_s^f(x_{m_i},x_{n_i-1})=0$, and by (3.16) we get ${lim\hspace{0.17em}inf}_{i\to \mathrm{\infty }}{d}_{p_b}(x_{m_i},x_{n_i-1})=0$, a contradiction with (3.11).

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

$$\underset{n\to \mathrm{\infty }}{lim}{p}_b(z,x_n)=\underset{n\to \mathrm{\infty }}{lim}{p}_b(x_n,x_n)=p_b(z,z).$$

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

$$\underset{n\to \mathrm{\infty }}{lim}{p}_b(z,x_n)=\underset{n\to \mathrm{\infty }}{lim}{p}_b(x_n,x_n)=p_b(z,z)=0.$$

Using the triangular inequality, we get

$$p_b(z,fz)\le s{p}_b(z,f{x}_n)+s{p}_b(f{x}_n,fz).$$

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

$$p_b(z,fz)\le s\underset{n\to \mathrm{\infty }}{lim}{p}_b(z,f{x}_n)+s\underset{n\to \mathrm{\infty }}{lim}{p}_b(f{x}_n,fz)=s{p}_b(fz,fz).$$

(3.18)

Note that from (3.1), we have

$$\psi (s{p}_b(fz,fz))\le \psi (M_s^f(z,z))-\phi (M_s^f(z,z)),$$

(3.19)

where

$$M_s^f(z,z)=max\{p_b(z,z),p_b(z,fz),p_b(z,fz),\frac{p_b(z,fz)+p_b(z,fz)}{2s}\}=p_b(fz,z).$$

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

Hence, we have $p_b(fz,z)=0$ 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 $\{x_n\}$ 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 $\{x_n\}$ 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

$$\begin{array}{rl}\psi (s{p}_b(x_{n+1},fz))& =\psi (s{p}_b(f{x}_n,fz))\\ \le \psi (M_s^f(x_n,z))-\phi (M_s^f(x_n,z)),\end{array}$$

(3.20)

where

$$\begin{array}{rl}{M}_s^f(x_n,z)& =max\{p_b(x_n,z),p_b(x_n,f{x}_n),p_b(z,fz),\frac{p_b(x_n,fz)+p_b(f{x}_n,z)}{2s}\}\\ =max\{p_b(x_n,z),p_b(x_n,x_{n+1}),p_b(z,fz),\frac{p_b(x_n,fz)+p_b(x_{n+1},z)}{2s}\}.\end{array}$$

(3.21)

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

$$\begin{array}{rl}\frac{p_b(z,fz)}{2s^2}& =min\{p_b(z,fz),\frac{\frac{p_b(z,fz)}{s}}{2s}\}\le \underset{i\to \mathrm{\infty }}{lim\hspace{0.17em}inf}{M}_s^f(x_n,z)\le \underset{i\to \mathrm{\infty }}{lim\hspace{0.17em}sup}{M}_s^f(x_n,z)\\ \le max\{p_b(z,fz),\frac{s{p}_b(z,fz)}{2s}\}=p_b(z,fz).\end{array}$$

(3.22)

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

$$\begin{array}{rl}\psi (p_b(z,fz))& =\psi (s\frac{1}{s}{p}_b(z,fz))\le \psi (s\underset{n\to \mathrm{\infty }}{lim\hspace{0.17em}sup}{p}_b(x_{n+1},fz))\\ \le \psi (\underset{n\to \mathrm{\infty }}{lim\hspace{0.17em}sup}{M}_s^f(x_n,z))-\underset{n\to \mathrm{\infty }}{lim\hspace{0.17em}inf}\phi (M_s^f(x_n,z))\\ \le \psi (p_b(z,fz))-\phi (\underset{n\to \mathrm{\infty }}{lim\hspace{0.17em}inf}{M}_s^f(x_n,z)).\end{array}$$

Therefore, $\phi ({lim\hspace{0.17em}inf}_{n\to \mathrm{\infty }}{M}_s^f(x_n,z))\le 0$, equivalently, ${lim\hspace{0.17em}inf}_{n\to \mathrm{\infty }}{M}_s^f(x_n,z)=0$. Thus, from (3.22) we get $z=fz$, and hence z is a fixed point of f. □

Corollary 1 Let $(X,⪯,p_b)$ 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 $k\in [0,1)$ such that

$$p_b(fx,fy)\le \frac{k}{s}max\{p_b(x,y),p_b(x,fx),p_b(y,fy),\frac{p_b(x,fy)+p_b(y,fx)}{2s}\}$$

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 $\psi (t)=t$ and $\phi (t)=(1-k)t$, for all $t\in [0,+\mathrm{\infty })$. □

Corollary 2 Under the hypotheses of Corollary  1, without the continuity assumption on f, for any nondecreasing sequence $\{x_n\}$ 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 $X=[0,+\mathrm{\infty })$ be equipped with the partial order ⪯ defined by

$$x⪯y\iff x=y\vee (x,y\in [0,1]\wedge x\le y),$$

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

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

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

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

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

$$p_b(fx,fy)={[max\{\frac{x}{\sqrt{2}\sqrt{1+x}},\frac{y}{\sqrt{2}\sqrt{1+y}}\}]}^2=\frac{x^2}{2(1+x)}$$

and

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

Thus, (3.1) reduces to

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

Case 2. $x=y\ge 1$. Then $p_b(fx,fy)=\frac{x^2}{4}$ and $M_s^f(x,y)=x^2$, so (3.1) reduces to

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

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 $(X,⪯,p_b)$ be an ordered partial b-metric space with the coefficient $s\ge 1$, and let $f,g:X\to X$ be two mappings. Set

$$M_s^{f,g}(x,y)=max\{p_b(x,y),p_b(x,fx),p_b(y,gy),\frac{p_b(x,gy)+p_b(y,fx)}{2s}\}.$$

Now, we present the following definition.

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

$$\psi (s^2{p}_b(fx,gy))\le \psi (M_s^{f,g}(x,y))-\phi (M_s^{f,g}(x,y))$$

(4.1)

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

Definition 13 [42]

Let $(X,⪯)$ 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 $(X,⪯,p_b)$ 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 $(f,g)$ is a generalized ${(\psi ,\phi )}_s$-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

$$\begin{array}{r}\psi (s^2{p}_b(u,gu))=\psi (s^2{p}_b(fu,gu))\\ \le \psi (max\{p_b(u,u),p_b(u,fu),p_b(u,gu),\frac{1}{2s}(p_b(u,gu)+p_b(u,fu))\})\\ -\phi (max\{p_b(u,u),p_b(u,fu),p_b(u,gu),\frac{1}{2s}(p_b(u,gu)+p_b(u,fu))\})\\ \le \psi (p_b(u,gu))-\phi (max\{p_b(u,u),p_b(u,fu),p_b(u,gu),\frac{1}{2s}(p_b(u,gu)+p_b(u,fu))\})\\ \le \psi (s^2{p}_b(u,gu))\\ -\phi (max\{p_b(u,u),p_b(u,fu),p_b(u,gu),\frac{1}{2s}(p_b(u,gu)+p_b(u,fu))\}).\end{array}$$

Therefore, $p_b(u,gu)=0$ 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 $\{x_n\}$ 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

$$\underset{n\to \mathrm{\infty }}{lim}{p}_b(x_n,x_{n+1})=0.$$

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

$$\begin{array}{rl}\psi (p_b(x_{2n+1},x_{2n+2}))& \le \psi (s^2{p}_b(x_{2n+1},x_{2n+2}))\\ =\psi (s^2{p}_b(f{x}_{2n},g{x}_{2n+1}))\\ \le \psi (M_s^{f,g}(x_{2n},x_{2n+1}))-\phi (M_s^{f,g}(x_{2n},x_{2n+1})),\end{array}$$

where

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

Hence, we have

$$\begin{array}{rl}\psi (p_b(x_{2n+1},x_{2n+2}))\le & \psi (max\{p_b(x_{2n},x_{2n+1}),p_b(x_{2n+1},x_{2n+2})\})\\ -\phi (max\{p_b(x_{2n},x_{2n+1}),p_b(x_{2n+1},x_{2n+2})\}).\end{array}$$

(4.2)

If

$$max\{p_b(x_{2n},x_{2n+1}),p_b(x_{2n+1},x_{2n+2})\}=p_b(x_{2n+1},x_{2n+2}),$$

then (4.2) becomes

$$\begin{array}{rl}\psi (p_b(x_{2n+1},x_{2n+2}))& \le \psi (p_b(x_{2n+1},x_{2n+2}))-\phi (p_b(x_{2n+1},x_{2n+2}))\\ <\psi (p_b(x_{2n+1},x_{2n+2})),\end{array}$$

which gives a contradiction. Hence,

$$max\{p_b(x_{2n},x_{2n+1}),p_b(x_{2n+1},x_{2n+2})\}=p_b(x_{2n},x_{2n+1}),$$

and (4.2) becomes

$$\begin{array}{rl}\psi (p_b(x_{2n+1},x_{2n+2}))& \le \psi (p_b(x_{2n},x_{2n+1}))-\phi (p_b(x_{2n},x_{2n+1}))\\ \le \psi (p_b(x_{2n},x_{2n+1})).\end{array}$$

(4.3)

Similarly, we can show that

$$\psi (p_b(x_{2n+1},x_{2n}))\le \psi (p_b(x_{2n-1},x_{2n}))-\phi (p_b(x_{2n-1},x_{2n}))\le \psi (p_b(x_{2n-1},x_{2n})).$$

(4.4)

By (4.3) and (4.4), we get that $\{p_b(x_n,x_{n+1}):n\in \mathbb{N}\}$ is a nonincreasing sequence of positive numbers. Hence, there is $r\ge 0$ such that

$$\underset{n\to \mathrm{\infty }}{lim}{p}_b(x_n,x_{n+1})=r.$$

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

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

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

$$\underset{n\to \mathrm{\infty }}{lim}{p}_b(x_n,x_n)\le \underset{n\to \mathrm{\infty }}{lim}{p}_b(x_n,x_{n+1})=0.$$

(4.5)

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

$$n_i>m_i>i,d_{p_b}(x_{2m_i},x_{2n_i})\ge \epsilon .$$

(4.6)

This means that

$$d_{p_b}(x_{2m_i},x_{2n_i-2})<\epsilon .$$

(4.7)

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

$$\epsilon \le d_{p_b}(x_{2m_i},x_{2n_i})\le s{d}_{p_b}(x_{2m_i},x_{2m_i+1})+s{d}_{p_b}(x_{2m_i+1},x_{2n_i}).$$

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

$$\frac{\epsilon }{s}\le \underset{i\to \mathrm{\infty }}{lim\hspace{0.17em}sup}{d}_{p_b}(x_{2m_i+1},x_{2n_i}).$$

On the other hand, we have

$$d_{p_b}(x_{2m_i},x_{2n_i-1})\le s{d}_{p_b}(x_{2m_i},x_{2n_i-2})+s{d}_{p_b}(x_{2n_i-2},x_{2n_i-1}).$$

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

$$\underset{i\to \mathrm{\infty }}{lim\hspace{0.17em}sup}{d}_{p_b}(x_{2m_i},x_{2n_i-1})\le \epsilon s.$$

(4.8)

Again, using the triangular inequality, we have

$$\begin{array}{rl}{d}_{p_b}(x_{2m_i},x_{2n_i})& \le s{d}_{p_b}(x_{2m_i},x_{2n_i-2})+s{d}_{p_b}(x_{2n_i-2},x_{2n_i})\\ \le s{d}_{p_b}(x_{2m_i},x_{2n_i-2})+s^2{d}_{p_b}(x_{2n_i-2},x_{2n_i-1})+s^2{d}_{p_b}(x_{2n_i-1},x_{2n_i})\end{array}$$

and

$$d_{p_b}(x_{2m_i+1},x_{2n_i-1})\le s{d}_{p_b}(x_{2m_i+1},x_{2m_i})+s{d}_{p_b}(x_{2m_i},x_{2n_i-1}).$$

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

$$\underset{i\to \mathrm{\infty }}{lim\hspace{0.17em}sup}{d}_{p_b}(x_{2m_i},x_{2n_i})\le \epsilon s$$

and

$$\underset{i\to \mathrm{\infty }}{lim\hspace{0.17em}sup}{d}_{p_b}(x_{2m_i+1},x_{2n_i-1})\le \epsilon s^2.$$

From the definition of $d_{p_b}$ and (4.5), we have the following relations:

$$\frac{\epsilon }{2s}\le \underset{i\to \mathrm{\infty }}{lim\hspace{0.17em}sup}{p}_b(x_{2m_i+1},x_{2n_i}),$$

(4.9)

$$\frac{\epsilon }{2s}\le \underset{i\to \mathrm{\infty }}{lim\hspace{0.17em}inf}{p}_b(x_{2m_i},x_{2n_i-1})\le \underset{i\to \mathrm{\infty }}{lim\hspace{0.17em}sup}{p}_b(x_{2m_i},x_{2n_i-1})\le \frac{s\epsilon }{2},$$

(4.10)

$$\underset{i\to \mathrm{\infty }}{lim\hspace{0.17em}sup}{p}_b(x_{2m_i},x_{2n_i})\le \frac{s\epsilon }{2},$$

(4.11)

$$\underset{i\to \mathrm{\infty }}{lim\hspace{0.17em}sup}{p}_b(x_{2m_i+1},x_{2n_i-1})\le \frac{s^2\epsilon }{2}.$$

(4.12)

Since $x_{2m_i}$ and $x_{2n_i-1}$ are comparable, using (4.1) we have

$$\begin{array}{rl}\psi (s^2{p}_b(x_{2m_i+1},x_{2n_i}))& =\psi (s^2{p}_b(f{x}_{2m_i},g{x}_{2n_i-1}))\\ \le \psi (M_s^{f,g}(x_{2m_i},x_{2n_i-1}))-\phi (M_s^{f,g}(x_{2m_i},x_{2n_i-1})),\end{array}$$

(4.13)

where

$$\begin{array}{rl}{M}_s^{f,g}(x_{2m_i},x_{2n_i-1})=& max\{p_b(x_{2m_i},x_{2n_i-1}),p_b(x_{2m_i},x_{2m_i+1}),p_b(x_{2n_i-1},x_{2n_i}),\\ \frac{p_b(x_{2m_i},x_{2n_i})+p_b(x_{2m_i+1},x_{2n_i-1})}{2s}\}.\end{array}$$

(4.14)

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

$$\begin{array}{rl}\underset{i\to \mathrm{\infty }}{lim\hspace{0.17em}sup}{M}_s^{f,g}(x_{2m_i},x_{2n_i-1})=& max\{\underset{i\to \mathrm{\infty }}{lim\hspace{0.17em}sup}{p}_b(x_{2m_i},x_{2n_i-1}),0,0,\\ \frac{{lim\hspace{0.17em}sup}_{i\to \mathrm{\infty }}{p}_b(x_{2m_i},x_{2n_i})+{lim\hspace{0.17em}sup}_{i\to \mathrm{\infty }}{p}_b(x_{2m_i+1},x_{2n_i-1})}{2s}\}\\ \le & max\{\frac{s\epsilon }{2},\frac{\frac{\epsilon s+\epsilon s^2}{2}}{2s}\}=\frac{s\epsilon }{2}.\end{array}$$

(4.15)

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

$$\begin{array}{rl}\psi \left(\frac{s\epsilon }{2}\right)& =\psi \left(s^2\frac{\epsilon }{2s}\right)\le \psi (s^2\underset{i\to \mathrm{\infty }}{lim\hspace{0.17em}sup}{p}_b(x_{2m_i+1},x_{2n_i}))\\ \le \psi (\underset{i\to \mathrm{\infty }}{lim\hspace{0.17em}sup}{M}_s^{f,g}(x_{2m_i},x_{2n_i-1}))-\phi (\underset{i\to \mathrm{\infty }}{lim\hspace{0.17em}inf}{M}_s^{f,g}(x_{2m_i},x_{2n_i-1}))\\ \le \psi \left(\frac{s\epsilon }{2}\right)-\phi (\underset{i\to \mathrm{\infty }}{lim\hspace{0.17em}inf}{M}_s^{f,g}(x_{2m_i},x_{2n_i-1})),\end{array}$$

which implies that $\phi ({lim\hspace{0.17em}inf}_{i\to \mathrm{\infty }}{M}_s^{f,g}(x_{2m_i},x_{2n_i-1}))=0$. By (4.14), it follows that

$$\underset{i\to \mathrm{\infty }}{lim\hspace{0.17em}inf}{p}_b(x_{2m_i},x_{2n_i-1})=0,$$

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

$$\underset{n\to \mathrm{\infty }}{lim}{p}_b(z,x_n)=\underset{n\to \mathrm{\infty }}{lim}{p}_b(x_n,x_n)=p_b(z,z).$$

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

$$\underset{n\to \mathrm{\infty }}{lim}{p}_b(z,x_n)=\underset{n\to \mathrm{\infty }}{lim}{p}_b(x_n,x_n)=p_b(z,z)=0.$$

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

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

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

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

Therefore,

$$max\{p_b(z,fz),p_b(z,gz)\}\le max\{s{p}_b(fz,fz),s{p}_b(gz,gz)\}\le s^2{p}_b(gz,fz).$$

(4.16)

From (4.1), we have

$$\psi (s^2{p}_b(fz,gz))\le \psi (M_s^{f,g}(z,z))-\phi (M_s^{f,g}(z,z)),$$

(4.17)

where

$$\begin{array}{rl}{M}_s^{f,g}(z,z)& =max\{p_b(z,z),p_b(z,fz),p_b(z,gz),\frac{p_b(z,gz)+p_b(z,fz)}{2s}\}\\ =max\{p_b(z,fz),p_b(z,gz)\}.\end{array}$$

As ψ is nondecreasing, we have $s^2{p}_b(fz,gz)\le max\{p_b(z,fz),p_b(z,gz)\}$. Hence, by (4.16) we obtain that $s^2{p}_b(fz,gz)=max\{p_b(z,fz),p_b(z,gz)\}$. But then, using (4.17), we get that $\phi (M_s^{f,g}(z,z))=0$. 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 $\{x_n\}$ 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 $\{x_n\}$ 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

$$\begin{array}{rl}\psi (s^2{p}_b(x_{2n+1},gz))& =\psi (s^2{p}_b(f{x}_{2n},gz))\\ \le \psi (M_s^{f,g}(x_{2n},z))-\phi (M_s^{f,g}(x_{2n},z)),\end{array}$$

(4.18)

where

$$\begin{array}{rl}{M}_s^{f,g}(x_{2n},z)& =max\{p_b(x_{2n},z),p_b(x_{2n},f{x}_{2n}),p_b(z,gz),\frac{p_b(x_{2n},gz)+p_b(f{x}_{2n},z)}{2s}\}\\ =max\{p_b(x_{2n},z),p_b(x_{2n},x_{2n+1}),p_b(z,gz),\frac{p_b(x_{2n},gz)+p_b(x_{2n+1},z)}{2s}\}.\end{array}$$

(4.19)

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

$$\begin{array}{rl}\frac{p_b(z,gz)}{s^2}& \le max\{p_b(z,gz),\frac{\frac{p_b(z,gz)}{s}}{2s}\}\le \underset{n\to \mathrm{\infty }}{lim\hspace{0.17em}inf}{M}_s^{f,g}(x_{2n},z)\\ \le \underset{n\to \mathrm{\infty }}{lim\hspace{0.17em}sup}{M}_s^{f,g}(x_{2n},z)\le max\{p_b(z,gz),\frac{s{p}_b(z,gz)}{2s}\}=p_b(z,gz).\end{array}$$

(4.20)

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

$$\begin{array}{rl}\psi (p_b(z,gz))& =\psi (s^2\frac{1}{s^2}{p}_b(z,gz))\le \psi (s^2\underset{n\to \mathrm{\infty }}{lim\hspace{0.17em}sup}d(x_{2n+1},gz))\\ \le \psi (\underset{n\to \mathrm{\infty }}{lim\hspace{0.17em}sup}{M}_s^{f,g}(x_{2n},z))-\phi (\underset{n\to \mathrm{\infty }}{lim\hspace{0.17em}inf}{M}_s^{f,g}(x_{2n},z))\\ \le \psi (p_b(z,gz))-\phi (\underset{n\to \mathrm{\infty }}{lim\hspace{0.17em}inf}{M}_s^{f,g}(x_{2n},z)).\end{array}$$

Therefore, $\phi ({lim\hspace{0.17em}inf}_{n\to \mathrm{\infty }}{M}_s^{f,g}(x_{2n},z))\le 0$, equivalently, ${lim\hspace{0.17em}inf}_{n\to \mathrm{\infty }}{M}_s^{f,g}(x_{2n},z)=0$. 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 $(X,⪯,p_b)$ 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 $k\in [0,1)$ such that

$$p_b(fx,gy)\le \frac{k}{s^2}max\{p_b(x,y),p_b(x,fx),p_b(y,gy),\frac{p_b(x,gy)+p_b(fx,y)}{2s}\}$$

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 $\{x_n\}$ 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 $X=\{0,1,2,3,4\}$ be equipped with the following partial order ⪯:

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

Define a partial b-metric $p_b:X\times X\to {\mathbb{R}}^{+}$ by

$$p_b(x,y)=\{\begin{array}{cc}0\hfill & \text{if }x=y,\hfill \\ {(x+y)}^2\hfill & \text{if }x\ne y.\hfill \end{array}$$

It is easy to see that $(X,p_b)$ is a $p_b$-complete partial b-metric space, with $s=49/25$.

Define self-maps f and g by

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

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

Define $\psi ,\phi :[0,\mathrm{\infty })\to [0,\mathrm{\infty })$ by $\psi (t)=\sqrt{t}$ and $\phi (t)=\frac{t}{300}$. In order to check that $(f,g)$ is a generalized ${(\psi ,\phi )}_s$-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

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

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 $(\{0,2\},⪯)$ 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,

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