$b_2$-Metric spaces and some fixed point theorems

Abstract

The aim of this paper is to establish the structure of $b_2$-metric spaces, as a generalization of 2-metric spaces. Some fixed point results for various contractive-type mappings in the context of ordered $b_2$-metric spaces are presented. We also provide examples to illustrate the results presented herein, as well as an application to integral equations.

MSC:47H10, 54H25.

1 Introduction

The concept of metric spaces has been generalized in many directions.

The notion of a b-metric space was studied by Czerwik in [1, 2] and many fixed point results were obtained for single and multivalued mappings by Czerwik and many other authors.

On the other hand, the notion of a 2-metric was introduced by Gähler in [3], having the area of a triangle in ${\mathbb{R}}^2$ as the inspirative example. Similarly, several fixed point results were obtained for mappings in such spaces. Note that, unlike many other generalizations of metric spaces introduced recently, 2-metric spaces are not topologically equivalent to metric spaces and there is no easy relationship between the results obtained in 2-metric and in metric spaces.

In this paper, we introduce a new type of generalized metric spaces, which we call $b_2$-metric spaces, as a generalization of both 2-metric and b-metric spaces. Then we prove some fixed point theorems under various contractive conditions in partially ordered $b_2$-metric spaces. These include Geraghty-type conditions, conditions using comparison functions and almost generalized weakly contractive conditions. We illustrate these results by appropriate examples, as well as an application to integral equations.

2 Mathematical preliminaries

The notion of a b-metric space was studied by Czerwik in [1, 2].

Definition 1 [1]

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.

Note that a b-metric is not always a continuous function of its variables (see, e.g., [[4], Example 2]), whereas an ordinary metric is.

On the other hand, the notion of a 2-metric was introduced by Gähler in [3].

Definition 2 [3]

Let X be a nonempty set and let $d:X^3\to \mathbb{R}$ be a map satisfying the following conditions:

• 1. For every pair of distinct points $x,y\in X$, there exists a point $z\in X$ such that $d(x,y,z)\ne 0$.

• 2. If at least two of three points x, y, z are the same, then $d(x,y,z)=0$.

• 3. The symmetry: $d(x,y,z)=d(x,z,y)=d(y,x,z)=d(y,z,x)=d(z,x,y)=d(z,y,x)$ for all $x,y,z\in X$.

• 4. The rectangle inequality: $d(x,y,z)\le d(x,y,t)+d(y,z,t)+d(z,x,t)$ for all $x,y,z,t\in X$.

Then d is called a 2-metric on X and $(X,d)$ is called a 2-metric space.

Definition 3 [3]

Let $(X,d)$ be a 2-metric space, $a,b\in X$ and $r\ge 0$. The set $B(a,b,r)=\{x\in X:d(a,b,x)<r\}$ is called a 2-ball centered at a and b with radius r.

The topology generated by the collection of all 2-balls as a subbasis is called a 2-metric topology on X.

Note that a 2-metric is not always a continuous function of its variables, whereas an ordinary metric is.

Remark 1

• 1. [5] It is straightforward from Definition 2 that every 2-metric is non-negative and every 2-metric space contains at least three distinct points.

• 2. A 2-metric $d(x,y,z)$ is sequentially continuous in each argument. Moreover, if a 2-metric $d(x,y,z)$ is sequentially continuous in two arguments, then it is sequentially continuous in all three arguments; see [6].

• 3. A convergent sequence in a 2-metric space need not be a Cauchy sequence; see [6].

• 4. In a 2-metric space $(X,d)$, every convergent sequence is a Cauchy sequence if d is continuous; see [6].

• 5. There exists a 2-metric space $(X,d)$ such that every convergent sequence in it is a Cauchy sequence but d is not continuous; see [6].

For some fixed point results on 2-metric spaces, the readers may refer to [5–15].

Now, we introduce new generalized metric spaces, called $b_2$-metric spaces, as a generalization of both 2-metric and b-metric spaces.

Definition 4 Let X be a nonempty set, $s\ge 1$ be a real number and let $d:X^3\to \mathbb{R}$ be a map satisfying the following conditions:

• 1. For every pair of distinct points $x,y\in X$, there exists a point $z\in X$ such that $d(x,y,z)\ne 0$.

• 2. If at least two of three points x, y, z are the same, then $d(x,y,z)=0$.

• 3. The symmetry: $d(x,y,z)=d(x,z,y)=d(y,x,z)=d(y,z,x)=d(z,x,y)=d(z,y,x)$ for all $x,y,z\in X$.

• 4. The rectangle inequality: $d(x,y,z)\le s[d(x,y,t)+d(y,z,t)+d(z,x,t)]$ for all $x,y,z,t\in X$.

Then d is called a $b_2$-metric on X and $(X,d)$ is called a $b_2$-metric space with parameter s.

Obviously, for $s=1$, $b_2$-metric reduces to 2-metric.

Definition 5 Let $\{x_n\}$ be a sequence in a $b_2$-metric space $(X,d)$.

• 1. $\{x_n\}$ is said to be $b_2$-convergent to $x\in X$, written as ${lim}_n{x}_n=x$, if for all $a\in X$, ${lim}_{n}d(x_n,x,a)=0$.

• 2. $\{x_n\}$ is said to be a $b_2$-Cauchy sequence in X if for all $a\in X$, ${lim}_{n}d(x_n,x_m,a)=0$.

• 3. $(X,d)$ is said to be $b_2$-complete if every $b_2$-Cauchy sequence is a $b_2$-convergent sequence.

The following are some easy examples of $b_2$-metric spaces.

Example 1 Let $X=[0,+\mathrm{\infty })$ and $d(x,y,z)={[xy+yz+zx]}^p$ if $x\ne y\ne z\ne x$, and otherwise $d(x,y,z)=0$, where $p\ge 1$ is a real number. Evidently, from convexity of function $f(x)=x^p$ for $x\ge 0$, then by Jensen inequality we have

$${(a+b+c)}^p\le 3^{p-1}(a^p+b^p+c^p).$$

So, one can obtain the result that $(X,d)$ is a $b_2$-metric space with $s\le 3^{p-1}$.

Example 2 Let a mapping $d:{\mathbb{R}}^3\to [0,+\mathrm{\infty })$ be defined by

$$d(x,y,z)=min\{|x-y|,|y-z|,|z-x|\}.$$

Then d is a 2-metric on ℝ, i.e., the following inequality holds:

$$d(x,y,z)\le d(x,y,t)+d(y,z,t)+d(z,x,t),$$

for arbitrary real numbers x, y, z, t. Using convexity of the function $f(x)=x^p$ on $[0,+\mathrm{\infty })$ for $p\ge 1$, we obtain that

$$d_p(x,y,z)={[min\{|x-y|,|y-z|,|z-x|\}]}^p$$

is a $b_2$-metric on ℝ with $s<3^{p-1}$.

Definition 6 Let $(X,d)$ and $(X^{\prime },d^{\prime })$ be two $b_2$-metric spaces and let $f:X\to X^{\prime }$ be a mapping. Then f is said to be $b_2$-continuous at a point $z\in X$ if for a given $\epsilon >0$, there exists $\delta >0$ such that $x\in X$ and $d(z,x,a)<\delta$ for all $a\in X$ imply that $d^{\prime }(fz,fx,a)<\epsilon$. The mapping f is $b_2$-continuous on X if it is $b_2$-continuous at all $z\in X$.

Proposition 1 Let $(X,d)$ and $(X^{\prime },d^{\prime })$ be two $b_2$-metric spaces. Then a mapping $f:X\to X^{\prime }$ is $b_2$-continuous at a point $x\in X$ if and only if it is $b_2$-sequentially continuous at x; that is, whenever $\{x_n\}$ is $b_2$-convergent to x, $\{f{x}_n\}$ is $b_2$-convergent to $f(x)$.

We will need the following simple lemma about the $b_2$-convergent sequences in the proof of our main results.

Lemma 1 Let $(X,d)$ be a $b_2$-metric space and suppose that $\{x_n\}$ and $\{y_n\}$ are $b_2$-convergent to x and y, respectively. Then we have

$$\frac{1}{s^2}d(x,y,a)\le \underset{n\to \mathrm{\infty }}{lim\hspace{0.17em}inf}d(x_n,y_n,a)\le \underset{n\to \mathrm{\infty }}{lim\hspace{0.17em}sup}d(x_n,y_n,a)\le s^{2}d(x,y,a),$$

for all a in X. In particular, if $y_n=y$ is constant, then

$$\frac{1}{s}d(x,y,a)\le \underset{n\to \mathrm{\infty }}{lim\hspace{0.17em}inf}d(x_n,y,a)\le \underset{n\to \mathrm{\infty }}{lim\hspace{0.17em}sup}d(x_n,y,a)\le sd(x,y,a),$$

for all a in X.

Proof Using the rectangle inequality in the given $b_2$-metric space, it is easy to see that

$$\begin{array}{rl}d(x,y,a)& =d(x,a,y)\le sd(x,a,x_n)+sd(a,y,x_n)+sd(y,x,x_n)\\ \le sd(x,a,x_n)+s^2[d(a,y,y_n)+d(y,x_n,y_n)+d(x_n,a,y_n)]+sd(y,x,x_n)\end{array}$$

and

$$\begin{array}{rl}d(x_n,y_n,a)& =d(x_n,a,y_n)\le sd(x_n,a,x)+sd(a,y_n,x)+sd(y_n,x,x_n)\\ \le sd(x_n,a,x)+s^2[d(a,y_n,y)+d(y_n,x,y)+d(x,a,y)]+sd(y_n,x,x_n).\end{array}$$

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 desired result.

If $y_n=y$, then

$$d(x,y,a)\le sd(x,y,x_n)+sd(y,a,x_n)+sd(a,x,x_n)$$

and

$$d(x_n,y,a)\le sd(x_n,y,x)+sd(y,a,x)+sd(a,x_n,x).$$

 □

3 Main results

3.1 Results under Geraghty-type conditions

In 1973, Geraghty [16] proved a fixed point result, generalizing the Banach contraction principle. Several authors proved later various results using Geraghty-type conditions. Fixed point results of this kind in b-metric spaces were obtained by Ðukić et al. in [17].

Following [17], for a real number $s\ge 1$, let ${\mathcal{F}}_s$ denote the class of all functions $\beta :[0,\mathrm{\infty })\to [0,\frac{1}{s})$ satisfying the following condition:

$$\beta (t_n)\to \frac{1}{s}\text{as }n\to \mathrm{\infty }\text{implies}{t}_n\to 0\text{as }n\to \mathrm{\infty }.$$

Theorem 1 Let $(X,⪯)$ be a partially ordered set and suppose that there exists a $b_2$-metric d on X such that $(X,d)$ is a $b_2$-complete $b_2$-metric space. Let $f:X\to X$ be an increasing mapping with respect to ⪯ such that there exists an element $x_0\in X$ with $x_0⪯f{x}_0$. Suppose that

$$sd(fx,fy,a)\le \beta (d(x,y,a))M(x,y,a)$$

(3.1)

for all $a\in X$ and for all comparable elements $x,y\in X$, where

$$M(x,y,a)=max\{d(x,y,a),\frac{d(x,fx,a)d(y,fy,a)}{1+d(fx,fy,a)}\}.$$

If f is $b_2$-continuous, then f has a fixed point. Moreover, the set of fixed points of f is well ordered if and only if f has one and only one fixed point.

Proof Starting with the given $x_0$, put $x_n=f^n{x}_0$. Since $x_0⪯f{x}_0$ and f is an increasing function we obtain by induction that

$$x_0⪯f{x}_0⪯f^2{x}_0⪯\dots ⪯f^n{x}_0⪯f^{n+1}{x}_0⪯\cdots .$$

Step I: We will show that ${lim}_{n}d(x_n,x_{n+1},a)=0$. Since $x_n⪯x_{n+1}$ for each $n\in \mathbb{N}$, then by (3.1) we have

$$\begin{array}{rl}sd(x_n,x_{n+1},a)& =sd(f{x}_{n-1},f{x}_n,a)\le \beta (d(x_{n-1},x_n,a))M(x_{n-1},x_n,a)\\ \le \frac{1}{s}d(x_{n-1},x_n,a)\le d(x_{n-1},x_n,a),\end{array}$$

(3.2)

because

$$\begin{array}{rl}M(x_{n-1},x_n,a)& =max\{d(x_{n-1},x_n,a),\frac{d(x_{n-1},f{x}_{n-1},a)d(x_n,f{x}_n,a)}{1+d(f{x}_{n-1},f{x}_n,a)}\}\\ =max\{d(x_{n-1},x_n,a),\frac{d(x_{n-1},x_n,a)d(x_n,x_{n+1},a)}{1+d(x_n,x_{n+1},a)}\}\\ =d(x_{n-1},x_n,a).\end{array}$$

Therefore, the sequence $\{d(x_n,x_{n+1},a)\}$ is decreasing. Then there exists $r\ge 0$ such that ${lim}_{n}d(x_n,x_{n+1},a)=r$. Suppose that $r>0$. Then, letting $n\to \mathrm{\infty }$, from (3.2) we have

$$\frac{1}{s}r\le sr\le \underset{n}{lim}\beta (d(x_{n-1},x_n,a))r\le r.$$

So, we have ${lim}_n\beta (d(x_{n-1},x_n,a))\ge \frac{1}{s}$ and since $\beta \in {\mathcal{F}}_s$ we deduce that ${lim}_{n}d(x_{n-1},x_n,a)=0$ which is a contradiction. Hence, $r=0$, that is,

$$\underset{n}{lim}d(x_n,x_{n+1},a)=0.$$

(3.3)

Step II: As $\{d(x_n,x_{n+1},a)\}$ is decreasing, if $d(x_{n-1},x_n,a)=0$, then $d(x_n,x_{n+1},a)=0$. Since from part 2 of Definition 4, $d(x_0,x_1,x_0)=0$, we have $d(x_n,x_{n+1},x_0)=0$ for all $n\in \mathbb{N}$. Since $d(x_{m-1},x_m,x_m)=0$, we have

$$d(x_n,x_{n+1},x_m)=0$$

(3.4)

for all $n\ge m-1$. For $0\le n<m-1$, we have $m-1\ge n+1$, and from (3.4) we have

$$d(x_{m-1},x_m,x_{n+1})=d(x_{m-1},x_m,x_n)=0.$$

(3.5)

It implies that

$$\begin{array}{rl}d(x_n,x_{n+1},x_m)& \le sd(x_n,x_{n+1},x_{m-1})+sd(x_{n+1},x_m,x_{m-1})+sd(x_m,x_n,x_{m-1})\\ =sd(x_n,x_{n+1},x_{m-1}).\end{array}$$

Since $d(x_n,x_{n+1},x_{n+1})=0$, from the above inequality, we have

$$d(x_n,x_{n+1},x_m)\le s^{m-n-1}d(x_n,x_{n+1},x_{n+1})=0$$

(3.6)

for all $0\le n<m-1$. From (3.4) and (3.6), we have

$$d(x_n,x_{n+1},x_m)=0$$

(3.7)

for all $n,m\in \mathbb{N}$.

Now, for all $i,j,k\in \mathbb{N}$ with $i<j$, we have

$$d(x_{j-1},x_j,x_i)=d(x_{j-1},x_j,x_k)=0.$$

(3.8)

Therefore, from (3.8) and triangular inequality

$$\begin{array}{rl}d(x_i,x_j,x_k)& \le s[d(x_i,x_j,x_{j-1})+d(x_j,x_k,x_{j-1})+d(x_k,x_i,x_{j-1})]\\ =sd(x_i,x_{j-1},x_k)\le \cdots \le s^{j-i}d(x_i,x_i,x_k)=0.\end{array}$$

This proves that for all $i,j,k\in \mathbb{N}$

$$d(x_i,x_j,x_k)=0.$$

(3.9)

Step III: Now, we prove that the sequence $\{x_n\}$ is a $b_2$-Cauchy sequence. Using the rectangle inequality and by (3.1) we have

$$\begin{array}{rl}d(x_n,x_m,a)\le & sd(x_n,x_m,x_{n+1})+sd(x_m,a,x_{n+1})+sd(a,x_n,x_{n+1})\\ \le & sd(x_n,x_{n+1},x_m)+s^2[d(x_m,x_{m+1},a)+d(x_{n+1},x_{m+1},a)\\ +d(x_m,x_{m+1},x_{n+1})]+sd(x_n,x_{n+1},a)\\ \le & sd(x_n,x_{n+1},x_m)+s^{2}d(x_m,x_{m+1},a)+s\beta (d(x_n,x_m,a))M(x_n,x_m,a)\\ +s^{2}d(x_m,x_{m+1},x_{n+1})+sd(x_n,x_{n+1},a).\end{array}$$

Letting $m,n\to \mathrm{\infty }$ in the above inequality and applying (3.3) and (3.7) we have

$$\underset{m,n\to \mathrm{\infty }}{lim}d(x_n,x_m,a)\le s\underset{m,n\to \mathrm{\infty }}{lim}\beta (d(x_n,x_m,a))\underset{m,n\to \mathrm{\infty }}{lim}M(x_n,x_m,a).$$

(3.10)

Here

$$\begin{array}{rl}d(x_n,x_m,a)& \le M(x_n,x_m,a)\\ =max\{d(x_n,x_m,a),\frac{d(x_n,f{x}_n,a)d(x_m,f{x}_m,a)}{1+d(f{x}_n,f{x}_m,a)}\}\\ =max\{d(x_n,x_m,a),\frac{d(x_n,x_{n+1},a)d(x_m,x_{m+1},a)}{1+d(x_{n+1},x_{m+1},a)}\}.\end{array}$$

Letting $m,n\to \mathrm{\infty }$ in the above inequality we get

$$\underset{m,n\to \mathrm{\infty }}{lim}M(x_n,x_m,a)=\underset{m,n\to \mathrm{\infty }}{lim}d(x_n,x_m,a).$$

(3.11)

Hence, from (3.10) and (3.11), we obtain

$$\underset{m,n\to \mathrm{\infty }}{lim}d(x_n,x_m,a)\le s\underset{m,n\to \mathrm{\infty }}{lim}\beta (d(x_n,x_m,a))\underset{m,n\to \mathrm{\infty }}{lim}d(x_n,x_m,a).$$

(3.12)

Now we claim that ${lim}_{m,n\to \mathrm{\infty }}d(x_n,x_m,a)=0$. If, to the contrary, ${lim}_{m,n\to \mathrm{\infty }}d(x_n,x_m,a)\ne 0$, then we get

$$\frac{1}{s}\le \underset{m,n\to \mathrm{\infty }}{lim}\beta (d(x_n,x_m,a)).$$

Since $\beta \in {\mathcal{F}}_s$ we deduce that

$$\underset{m,n\to \mathrm{\infty }}{lim}d(x_n,x_m,a)=0,$$

(3.13)

which is a contradiction. Consequently, $\{x_n\}$ is a $b_2$-Cauchy sequence in X. Since $(X,d)$ is $b_2$-complete, the sequence $\{x_n\}$ $b_2$-converges to some $z\in X$, that is, ${lim}_{n}d(x_n,z,a)=0$.

Step IV: Now, we show that z is a fixed point of f.

Using the rectangle inequality, we get

$$d(fz,z,a)\le sd(fz,f{x}_n,z)+sd(z,a,f{x}_n)+sd(a,fz,f{x}_n).$$

Letting $n\to \mathrm{\infty }$ and using the continuity of f, we have $fz=z$. Thus, z is a fixed point of f.

Step V: Finally, suppose that the set of fixed point of f is well ordered. Assume, to the contrary, that u and v are two distinct fixed points of f. Then by (3.1), we have

$$\begin{array}{rl}sd(u,v,a)& =sd(fu,fv,a)\le \beta (d(u,v,a))M(u,v,a)\\ =\beta (d(u,v,a))d(u,v,a)<\frac{1}{s}d(u,v,a),\end{array}$$

(3.14)

because

$$\begin{array}{rl}M(u,v,a)& =max\{d(u,v,a),\frac{d(u,fu,a)d(v,fv,a)}{1+d(fu,fv,a)}\}\\ =max\{d(u,v,a),0\}=d(u,v,a).\end{array}$$

Thus, we get $sd(u,v,a)<\frac{1}{s}d(u,v,a)$, a contradiction. Hence, f has a unique fixed point. The converse is trivial. □

Note that the continuity of f in Theorem 1 can be replaced by certain property of the space itself.

Theorem 2 Under the hypotheses of Theorem  1, without the $b_2$-continuity assumption on f, assume that whenever $\{x_n\}$ is a nondecreasing sequence in X such that $x_n\to u$, one has $x_n⪯u$ for all $n\in \mathbb{N}$. Then f has a fixed point. Moreover, the set of fixed points of f is well ordered if and only if f has one and only one fixed point.

Proof Repeating the proof of Theorem 1, we construct an increasing sequence $\{x_n\}$ in X such that $x_n\to z\in X$. Using the assumption on X we have $x_n⪯z$. Now, we show that $z=fz$. By (3.1) and Lemma 1,

$$\begin{array}{rl}s[\frac{1}{s}d(z,fz,a)]& \le s\underset{n\to \mathrm{\infty }}{lim\hspace{0.17em}sup}d(x_{n+1},fz,a)\\ \le \underset{n\to \mathrm{\infty }}{lim\hspace{0.17em}sup}\beta (d(x_n,z,a))\underset{n\to \mathrm{\infty }}{lim\hspace{0.17em}sup}M(x_n,z,a),\end{array}$$

where

$$\begin{array}{rl}\underset{n\to \mathrm{\infty }}{lim}M(x_n,z,a)& =\underset{n}{lim}max\{d(x_n,z,a),\frac{d(x_n,f{x}_n,a)d(z,fz,a)}{1+d(f{x}_n,fz,a)}\}\\ =\underset{n}{lim}max\{d(x_n,z,a),\frac{d(x_n,x_{n+1},a)d(z,fz,a)}{1+d(x_{n+1},fz,a)}\}=0\left(\text{see }\text{(3.3)}\right).\end{array}$$

Therefore, we deduce that $d(z,fz,a)\le 0$. As a is arbitrary, hence, we have $z=fz$.

The proof of uniqueness is the same as in Theorem 1. □

If in the above theorems we take $\beta (t)=r$, where $0\le r<\frac{1}{s}$, then we have the following corollary.

Corollary 1 Let $(X,⪯)$ be a partially ordered set and suppose that there exists a $b_2$-metric d on X such that $(X,d)$ is a $b_2$-complete $b_2$-metric space. Let $f:X\to X$ be an increasing mapping with respect to ⪯ such that there exists an element $x_0\in X$ with $x_0⪯f{x}_0$. Suppose that for some r, with $0\le r<\frac{1}{s}$,

$$sd(fx,fy,a)\le rM(x,y,a)$$

holds for each $a\in X$ and all comparable elements $x,y\in X$, where

$$M(x,y,a)=max\{d(x,y,a),\frac{d(x,fx,a)d(y,fy,a)}{1+d(fx,fy,a)}\}.$$

If f is continuous, or, for any nondecreasing sequence $\{x_n\}$ in X such that $x_n\to u\in X$ one has $x_n⪯u$ for all $n\in \mathbb{N}$, then f has a fixed point. Additionally, the set of fixed points of f is well ordered if and only if f has one and only one fixed point.

Corollary 2 Let $(X,⪯)$ be a partially ordered set and suppose that there exists a $b_2$-metric d on X such that $(X,d)$ is a $b_2$-complete $b_2$-metric space. Let $f:X\to X$ be an increasing mapping with respect to ⪯ such that there exists an element $x_0\in X$ with $x_0⪯f{x}_0$. Suppose that

$$d(fx,fy,a)\le \alpha d(x,y,a)+\beta \frac{d(x,fx,a)d(y,fy,a)}{1+d(fx,fy,a)}$$

for each $a\in X$ and all comparable elements $x,y\in X$, where $\alpha ,\beta \ge 0$ and $\alpha +\beta \le \frac{1}{s}$.

If f is continuous, or, for any nondecreasing sequence $\{x_n\}$ in X such that $x_n\to u\in X$ one has $x_n⪯u$ for all $n\in \mathbb{N}$, then f has a fixed point. Moreover, the set of fixed points of f is well ordered if and only if f has one and only one fixed point.

Proof Since

$$\begin{array}{r}\alpha d(x,y,a)+\beta \frac{d(x,fx,a)d(y,fy,a)}{1+d(fx,fy,a)}\\ \le (\alpha +\beta )max\{d(x,y,a),\frac{d(x,fx,a)d(y,fy,a)}{1+d(fx,fy,a)}\}.\end{array}$$

Putting $r=\alpha +\beta$, the conditions of Corollary 1 are satisfied and f has a fixed point. □

Example 3 Let $X=\{(\alpha ,0):\alpha \in [0,+\mathrm{\infty })\}\cup \{(0,2)\}\subset {\mathbb{R}}^2$ and let $d(x,y,z)$ denote the square of the area of triangle with vertices $x,y,z\in X$, e.g.,

$$d((\alpha ,0),(\beta ,0),(0,2))={(\alpha -\beta )}^2.$$

It is easy to check that d is a $b_2$-metric with parameter $s=2$. Introduce an order ⪯ in X by

$$(\alpha ,0)⪯(\beta ,0)\iff \alpha \ge \beta ,$$

with all other pairs of distinct points in X incomparable.

Consider the mapping $f:X\to X$ given by

$$f(\alpha ,0)=(\frac{\alpha }{3},0)\text{for }\alpha \in [0,+\mathrm{\infty })\text{ and }f(0,2)=(0,2),$$

and the function $\beta \in {\mathcal{F}}_2$ given as

$$\beta (t)=\frac{1+t}{2+4t}\text{for }t\in [0,+\mathrm{\infty }).$$

Then f is an increasing mapping with $(\alpha ,0)⪯f(\alpha ,0)$ for each $\alpha \ge 0$. If $\{x_n\}=\{({\alpha }_n,0)\}$ is a nondecreasing sequence in X, converging to some $z=(\gamma ,0)$, then $({\alpha }_n,0)⪯(\gamma ,0)$ for all $n\in \mathbb{N}$. Finally, in order to check the contractive condition (3.1), only the case when $x=(\alpha ,0)$, $y=(\beta ,0)$, $a=(0,2)$ is nontrivial. But then $d(x,y,a)={(\alpha -\beta )}^2$ and

$$\begin{array}{rl}sd(fx,fy,a)& =2d((\frac{1}{3}\alpha ,0),(\frac{1}{3}\beta ,0),(0,2))=2\cdot \frac{1}{9}{(\alpha -\beta )}^2\le \frac{1}{4}{(\alpha -\beta )}^2\\ \le \beta (d(x,y,a))d(x,y,a)\le \beta (d(x,y,a))M(x,y,a).\end{array}$$

All the conditions of Theorem 2 are satisfied and f has two fixed points, $(0,0)$ and $(0,2)$. Note that the condition (stated in Theorem 1 and Theorem 2) for the uniqueness of a fixed point is here not satisfied.

3.2 Results using comparison functions

Let Ψ denote the family of all nondecreasing and continuous functions $\psi :[0,\mathrm{\infty })\to [0,\mathrm{\infty })$ such that ${lim}_n{\psi }^n(t)=0$ for all $t>0$, where ${\psi }^n$ denotes the n th iterate of ψ. It is easy to show that, for each $\psi \in \Psi$, the following are satisfied:

• (a) $\psi (t)<t$ for all $t>0$;

• (b) $\psi (0)=0$.

Theorem 3 Let $(X,⪯)$ be a partially ordered set and suppose that there exists a $b_2$-metric d on X such that $(X,d)$ is a $b_2$-complete $b_2$-metric space. Let $f:X\to X$ be an increasing mapping with respect to ⪯ such that there exists an element $x_0\in X$ with $x_0⪯f{x}_0$. Suppose that

$$sd(fx,fy,a)\le \psi (M(x,y,a)),$$

(3.15)

where

$$M(x,y,a)=max\{d(x,y,a),\frac{d(x,fx,a)d(y,fy,a)}{1+d(fx,fy,a)}\},$$

for some $\psi \in \Psi$ and for all elements $x,y,a\in X$, with x, y comparable. If f is $b_2$-continuous, then f has a fixed point. In addition, the set of fixed points of f is well ordered if and only if f has one and only one fixed point.

Proof Since $x_0⪯f{x}_0$ and f is an increasing function, we obtain by induction that

$$x_0⪯f{x}_0⪯f^2{x}_0⪯\cdots ⪯f^n{x}_0⪯f^{n+1}{x}_0⪯\cdots .$$

By letting $x_n=f^n{x}_0$, we have

$$x_0⪯x_1⪯x_2⪯\cdots ⪯x_n⪯x_{n+1}⪯\cdots .$$

If there exists $n_0\in \mathbb{N}$ such that $x_{n_0}=x_{n_0+1}$, then $x_{n_0}=f{x}_{n_0}$ and so we have nothing to prove. Hence, we assume that $x_n\ne x_{n+1}$ for all $n\in \mathbb{N}$.

Step I. We will prove that ${lim}_{n}d(x_n,x_{n+1},a)=0$. Using condition (3.15), we obtain

$$d(x_{n+1},x_n,a)\le sd(x_{n+1},x_n,a)=sd(f{x}_n,f{x}_{n-1},a)\le \psi (M(x_n,x_{n-1},a)).$$

Here

$$\begin{array}{rl}M(x_{n-1},x_n,a)& =max\{d(x_{n-1},x_n,a),\frac{d(x_{n-1},f{x}_{n-1},a)d(x_n,f{x}_n,a)}{1+d(f{x}_{n-1},f{x}_n,a)}\}\\ =d(x_{n-1},x_n,a).\end{array}$$

Hence,

$$d(x_n,x_{n+1},a)\le sd(x_n,x_{n+1},a)\le \psi (d(x_{n-1},x_n,a))<d(x_{n-1},x_n,a).$$

(3.16)

By induction, we get

$$d(a,x_{n+1},x_n)\le \psi (d(a,x_n,x_{n-1}))\le {\psi }^2(d(a,x_{n-1},x_{n-2}))\le \cdots \le {\psi }^n(d(a,x_1,x_0)).$$

As $\psi \in \Psi$, we conclude that

$$\underset{n}{lim}d(x_n,x_{n+1},a)=0.$$

(3.17)

From similar arguments as in Theorem 1, since $\{d(x_n,x_{n+1},a)\}$ is decreasing, we can conclude that

$$d(x_i,x_j,x_k)=0$$

(3.18)

for all $i,j,k\in \mathbb{N}$.

Step II. We will prove that $\{x_n\}$ is a $b_2$-Cauchy sequence. Suppose the contrary. Then there exist $a\in X$ and $\epsilon >0$ 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

$$n_i>m_i>i\text{and}d(x_{m_i},x_{n_i},a)\ge \epsilon .$$

(3.19)

This means that

$$d(x_{m_i},x_{n_i-1},a)<\epsilon .$$

(3.20)

From (3.19) and using the rectangle inequality, we get

$$\epsilon \le d(x_{m_i},x_{n_i},a)\le sd(x_{m_i},x_{n_i},x_{m_i+1})+sd(x_{m_i+1},x_{n_i},a)+sd(x_{m_i+1},x_{m_i},a).$$

Taking the upper limit as $i\to \mathrm{\infty }$, from (3.17) and (3.18) we get

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

(3.21)

From the definition of $M(x,y,a)$ we have

$$\begin{array}{rl}M(x_{m_i},x_{n_i-1},a)& =max\{d(x_{m_i},x_{n_i-1},a),\frac{d(x_{m_i},f{x}_{m_i},a)d(x_{n_i-1},f{x}_{n_i-1},a)}{1+d(f{x}_{m_i},f{x}_{n_i-1},a)}\}\\ =max\{d(x_{m_i},x_{n_i-1},a),\frac{d(x_{m_i},a,x_{m_i+1})d(x_{n_i-1},a,x_{n_i})}{1+d(x_{m_i+1},x_{n_i},a)}\}\end{array}$$

and if $i\to \mathrm{\infty }$, by (3.17) and (3.20) we have

$$\underset{i\to \mathrm{\infty }}{lim\hspace{0.17em}sup}M(x_{m_i},x_{n_i-1},a)\le \epsilon .$$

Now, from (3.15) we have

$$sd(x_{m_i+1},x_{n_i},a)=sd(f{x}_{m_i},f{x}_{n_i-1},a)\le \psi (M(x_{m_i},x_{n_i-1},a)).$$

Again, if $i\to \mathrm{\infty }$ by (3.21) we obtain

$$\epsilon =s\cdot \frac{\epsilon }{s}\le s\underset{i\to \mathrm{\infty }}{lim\hspace{0.17em}sup}d(x_{m_i+1},x_{n_i},a)\le \psi (\epsilon )<\epsilon ,$$

which is a contradiction. Consequently, $\{x_n\}$ is a $b_2$-Cauchy sequence in X. Therefore, the sequence $\{x_n\}$ $b_2$-converges to some $z\in X$, that is, ${lim}_{n}d(x_n,z,a)=0$ for all $a\in X$.

Step III. Now we show that z is a fixed point of f.

Using the rectangle inequality, we get

$$d(z,fz,a)\le sd(z,fz,f{x}_n)+sd(f{x}_n,fz,a)+sd(f{x}_n,z,a).$$

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

$$d(z,fz,a)\le 0.$$

Hence, we have $fz=z$. Thus, z is a fixed point of f.

The uniqueness of the fixed point can be proved in the same manner as in Theorem 1. □

Theorem 4 Under the hypotheses of Theorem  3, without the $b_2$-continuity assumption on f, assume that whenever $\{x_n\}$ is a nondecreasing sequence in X such that $x_n\to u\in X$, one has $x_n⪯u$ for all $n\in \mathbb{N}$. Then f has a fixed point. In addition, the set of fixed points of f is well ordered if and only if f has one and only one fixed point.

Proof Following the proof of Theorem 3, we construct an increasing sequence $\{x_n\}$ in X such that $x_n\to z\in X$. Using the given assumption on X we have $x_n⪯z$. Now, we show that $z=fz$. By (3.15) we have

$$sd(fz,x_n,a)=sd(fz,f{x}_{n-1},a)\le \psi (M(z,x_{n-1},a)),$$

(3.22)

where

$$M(z,x_{n-1},a)=max\{d(z,x_{n-1},a),\frac{d(z,fz,a)d(x_{n-1},f{x}_{n-1},a)}{1+d(fz,f{x}_{n-1},a)}\}.$$

Letting $n\to \mathrm{\infty }$ in the above relation, we get

$$\underset{n\to \mathrm{\infty }}{lim\hspace{0.17em}sup}M(z,x_{n-1},a)=0.$$

(3.23)

Again, taking the upper limit as $n\to \mathrm{\infty }$ in (3.22) and using Lemma 1 and (3.23) we get

$$\begin{array}{rl}s[\frac{1}{s}d(z,fz,a)]& \le s\underset{n\to \mathrm{\infty }}{lim\hspace{0.17em}sup}d(x_n,fz,a)\\ \le \underset{n\to \mathrm{\infty }}{lim\hspace{0.17em}sup}\psi (M(z,x_{n-1},a))=0.\end{array}$$

So we get $d(z,fz,a)=0$, i.e., $fz=z$. □

Corollary 3 Let $(X,⪯)$ be a partially ordered set and suppose that there exists a $b_2$-metric d on X such that $(X,d)$ is a $b_2$-complete $b_2$-metric space. Let $f:X\to X$ be an increasing mapping with respect to ⪯ such that there exists an element $x_0\in X$ with $x_0⪯f{x}_0$. Suppose that

$$sd(fx,fy,a)\le rM(x,y,a),$$

where $0\le r<1$ and

$$M(x,y,a)=max\{d(x,y,a),\frac{d(x,fx,a)d(y,fy,a)}{1+d(fx,fy,a)}\},$$

for all elements $x,y,a\in X$ with x, y comparable. If f is continuous, or, whenever $\{x_n\}$ is a nondecreasing sequence in X such that $x_n\to u\in X$, one has $x_n⪯u$ for all $n\in \mathbb{N}$, then f has a fixed point. Moreover, the set of fixed points of f is well ordered if and only if f has one and only one fixed point.

Example 4 Let $X=\{A,B,C,D\}$ be ordered by $A⪰B⪰C$, with all other pairs of distinct points incomparable. Define $d:X^3\to \mathbb{R}$ by

$$d(A,B,C)=0,d(A,B,D)=1,d(A,C,D)=4,d(B,C,D)=2,$$

with symmetry in all variables and with $d(x,y,z)=0$ when at least two of the arguments are equal. Then it is easy to check that $(X,d)$ is a complete $b_2$-metric space with $s=\frac{4}{3}$.

Consider the mapping $f:X\to X$ given as

$$f=\left(\begin{array}{cccc}A& B& C& D\\ A& A& B& D\end{array}\right)$$

and a comparison function $\psi (t)=\frac{2}{3}t$. Then f is a nondecreasing mapping w.r.t. ⪯ and there exists $x_0\in X$ such that $x_0⪯f{x}_0$. The only nontrivial cases for checking the contractive condition (3.15) are when $a=D$ and $x=A$, $y=C$ or $x=B$, $y=C$ (or vice versa). Then we have

$$sd(fA,fC,D)=\frac{4}{3}d(A,B,D)=\frac{4}{3}<\frac{2}{3}\cdot 4=\psi (4)=\psi (d(A,C,D))\le \psi (M(A,C,D)),$$

resp.

$$sd(fB,fC,D)=\frac{4}{3}d(A,B,D)=\frac{4}{3}=\frac{2}{3}\cdot 2=\psi (2)=\psi (d(B,C,D))\le \psi (M(B,C,D)).$$

Hence, all the conditions of Theorem 3 are fulfilled. The mapping f has two fixed points (A and D).

3.3 Results for almost generalized weakly contractive mappings

Berinde in [18–21] initiated the concept of almost contractions and obtained many interesting fixed point theorems. Results with similar conditions were obtained, e.g., in [22] and [23]. In this section, we define the notion of almost generalized ${(\psi ,\phi )}_{s,a}$-contractive mapping and we prove some new results. In particular, we extend Theorems 2.1, 2.2 and 2.3 of Ćirić et al. in [24] to the setting of $b_2$-metric spaces.

Recall that Khan et al. introduced in [25] the concept of an altering distance function as follows.

Definition 7 [25]

A function $\phi :[0,+\mathrm{\infty })\to [0,+\mathrm{\infty })$ is called an altering distance function, if the following properties hold:

• 1. φ is continuous and nondecreasing.

• 2. $\phi (t)=0$ if and only if $t=0$.

Let $(X,d)$ be a $b_2$-metric space and let $f:X\to X$ be a mapping. For $x,y,a\in X$, set

$$M_a(x,y)=max\{d(x,y,a),d(x,fx,a),d(y,fy,a),\frac{d(x,fy,a)+d(y,fx,a)}{2s}\}$$

and

$$N_a(x,y)=min\{d(x,fx,a),d(x,fy,a),d(y,fx,a),d(y,fy,a)\}.$$

Definition 8 Let $(X,d)$ be a $b_2$-metric space. We say that a mapping $f:X\to X$ is an almost generalized ${(\psi ,\phi )}_{s,a}$-contractive mapping if there exist $L\ge 0$ and two altering distance functions ψ and φ such that

$$\psi (sd(fx,fy,a))\le \psi (M_a(x,y))-\phi (M_a(x,y))+L\psi (N_a(x,y))$$

(3.24)

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

Now, let us prove our new result.

Theorem 5 Let $(X,⪯)$ be a partially ordered set and suppose that there exists a $b_2$-metric d on X such that $(X,d)$ is a $b_2$-complete $b_2$-metric space. Let $f:X\to X$ be a continuous mapping, nondecreasing with respect to ⪯. Suppose that f satisfies condition (3.24), for all elements $x,y,a\in X$, with x, y comparable. If there exists $x_0\in X$ such that $x_0⪯f{x}_0$, then f has a fixed point. Moreover, the set of fixed points of f is well ordered if and only if f has one and only one fixed point.

Proof Starting with the given $x_0$, 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 $x_1=f{x}_0⪯x_2=f{x}_1$, and by induction

$$x_0⪯x_1⪯\cdots ⪯x_n⪯x_{n+1}⪯\cdots .$$

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 (3.24), we have

$$\begin{array}{rl}\psi (d(x_n,x_{n+1},a))& \le \psi (sd(x_n,x_{n+1},a))\\ =\psi (sd(f{x}_{n-1},f{x}_n,a))\\ \le \psi (M_a(x_{n-1},x_n))-\phi (M_a(x_{n-1},x_n))+L\psi (N_a(x_{n-1},x_n)),\end{array}$$

(3.25)

where

$$\begin{array}{r}{M}_a(x_{n-1},x_n)\\ =max\{d(x_{n-1},x_n,a),d(x_{n-1},f{x}_{n-1},a),d(x_n,f{x}_n,a),\\ \frac{d(x_{n-1},f{x}_n,a)+d(x_n,f{x}_{n-1},a)}{2s}\}\\ =max\{d(x_{n-1},x_n,a),d(x_n,x_{n+1},a),\frac{d(x_{n-1},x_{n+1},a)}{2s}\}\\ \le max\{d(x_{n-1},x_n,a),d(x_n,x_{n+1},a),\\ \frac{d(x_{n-1},x_{n+1},x_n)+d(x_{n+1},a,x_n)+d(a,x_{n-1},x_n)}{2}\}\end{array}$$

(3.26)

and

$$\begin{array}{r}{N}_a(x_{n-1},x_n)\\ =min\{d(x_{n-1},f{x}_{n-1},a),d(x_{n-1},f{x}_n,a),d(x_n,f{x}_{n-1},a),d(x_n,f{x}_n,a)\}\\ =min\{d(x_{n-1},x_n,a),d(x_{n-1},x_{n+1},a),0,d(x_n,x_{n+1},a)\}=0.\end{array}$$

(3.27)

From (3.25)–(3.27) and the properties of ψ and φ, we get

$$\begin{array}{r}\psi (d(x_n,x_{n+1},a))\\ \le \psi (max\{d(x_{n-1},x_n,a),d(x_n,x_{n+1},a),\\ \frac{d(x_{n-1},x_{n+1},x_n)+d(x_{n+1},a,x_n)+d(a,x_{n-1},x_n)}{2}\left\}\right)\\ -\phi (max\{d(x_{n-1},x_n,a),d(x_n,x_{n+1},a),\frac{d(x_{n-1},x_{n+1},a)}{2s}\}).\end{array}$$

(3.28)

If

$$\begin{array}{r}max\{d(x_{n-1},x_n,a),d(x_n,x_{n+1},a),\frac{d(x_{n-1},x_{n+1},x_n)+d(x_{n+1},a,x_n)+d(a,x_{n-1},x_n)}{2}\}\\ =d(x_n,x_{n+1},a),\end{array}$$

then by (3.28) we have

$$\begin{array}{rl}\psi (d(x_n,x_{n+1},a))\le & \psi (d(x_n,x_{n+1},a))\\ -\phi (max\{d(x_{n-1},x_n,a),d(x_n,x_{n+1},a),\frac{d(x_{n-1},x_{n+1},a)}{2s}\}),\end{array}$$

which gives a contradiction.

If $d(x_{n-1},x_{n+1},x_n)=0$, then

$$\begin{array}{r}max\{d(x_{n-1},x_n,a),d(x_n,x_{n+1},a),\frac{d(x_{n-1},x_{n+1},x_n)+d(x_{n+1},a,x_n)+d(a,x_{n-1},x_n)}{2}\}\\ =d(x_{n-1},x_n,a),\end{array}$$

therefore (3.28) becomes

$$\begin{array}{rl}\psi (d(x_n,x_{n+1},a))\le & \psi (d(x_n,x_{n-1},a))\\ -\phi (max\{d(x_{n-1},x_n,a),d(x_n,x_{n+1},a),\frac{d(x_{n-1},x_{n+1},a)}{2s}\})\\ \le & \psi (d(x_n,x_{n-1},a)).\end{array}$$

(3.29)

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

$$\underset{n}{lim}d(x_n,x_{n+1},a)=r.$$

Letting $n\to \mathrm{\infty }$ in (3.29), we get

$$\psi (r)\le \psi (r)-\phi (max\{r,r,\underset{n}{lim}\frac{d(x_{n-1},x_{n+1},a)}{2s}\})\le \psi (r).$$

Therefore,

$$\phi (max\{r,r,\underset{n}{lim}\frac{d(x_{n-1},x_{n+1},a)}{2s}\})=0,$$

and hence $r=0$. Thus, we have

$$\underset{n}{lim}d(x_n,x_{n+1},a)=0,$$

(3.30)

for each $a\in X$.

Note that if $d(x_{n-1},x_{n+1},x_n)\ne 0$ and

$$\begin{array}{r}max\{d(x_{n-1},x_n,a),d(x_n,x_{n+1},a),\frac{d(x_{n-1},x_{n+1},x_n)+d(x_{n+1},a,x_n)+d(a,x_{n-1},x_n)}{2}\}\\ =\frac{d(x_{n-1},x_{n+1},x_n)+d(x_{n+1},a,x_n)+d(a,x_{n-1},x_n)}{2}.\end{array}$$

Then, by (3.28) and taking $a=x_{n-1}$, we have

$$\begin{array}{r}\psi (d(x_n,x_{n+1},x_{n-1}))\\ \le \psi \left(\frac{d(x_{n-1},x_{n+1},x_n)+d(x_{n+1},x_{n-1},x_n)+d(x_{n-1},x_{n-1},x_n)}{2}\right)\\ -\phi (max\{d(x_{n-1},x_n,x_{n-1}),d(x_n,x_{n+1},x_{n-1}),\frac{d(x_{n-1},x_{n+1},x_{n-1})}{2s}\}),\end{array}$$

which gives $d(x_{n-1},x_{n+1},x_n)=0$, a contradiction.

Next, we show that $\{x_n\}$ is a $b_2$-Cauchy sequence in X. For this purpose, we use the following relation (see (3.9) and (3.18)):

$$d(x_i,x_j,x_k)=0,$$

(3.31)

for all $i,j,k\in N$ (note that this can obtained as $\{d(x_n,x_{n+1},a):n\in \mathbb{N}\cup \{0\}\}$ is a nonincreasing sequence of positive numbers).

Suppose the contrary, that is, $\{x_n\}$ is not a $b_2$-Cauchy sequence. Then there exist $a\in X$ and $\epsilon >0$ 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

$$n_i>m_i>i,d(x_{m_i},x_{n_i},a)\ge \epsilon .$$

(3.32)

This means that

$$d(x_{m_i},x_{n_i-1},a)<\epsilon .$$

(3.33)

Using (3.33) and taking the upper limit as $i\to \mathrm{\infty }$, we get

$$\underset{n\to \mathrm{\infty }}{lim\hspace{0.17em}sup}d(x_{m_i},x_{n_i-1},a)\le \epsilon .$$

(3.34)

On the other hand, we have

$$d(x_{m_i},x_{n_i},a)\le sd(x_{m_i},x_{n_i},x_{m_i+1})+sd(x_{n_i},a,x_{m_i+1})+sd(a,x_{m_i},x_{m_i+1}).$$

Using (3.30), (3.31), (3.32), and taking the upper limit as $i\to \mathrm{\infty }$, we get

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

(3.35)

Again, using the rectangular inequality, we have

$$d(x_{m_i+1},x_{n_i-1},a)\le sd(x_{m_i+1},x_{n_i-1},x_{m_i})+sd(x_{n_i-1},a,x_{m_i})+sd(a,x_{m_i+1},x_{m_i}),$$

and

$$d(x_{m_i},x_{n_i},a)\le sd(x_{m_i},x_{n_i},x_{n_i-1})+sd(x_{n_i},a,x_{n_i-1})+sd(a,x_{m_i},x_{n_i-1}).$$

Taking the upper limit as $i\to \mathrm{\infty }$ in the first inequality above, and using (3.30), (3.31), and (3.34) we get

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

(3.36)

Similarly, taking the upper limit as $i\to \mathrm{\infty }$ in the second inequality above, and using (3.30), (3.31), and (3.33), we get

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

(3.37)

From (3.24), we have

$$\begin{array}{r}\psi (sd(x_{m_i+1},x_{n_i},a))\\ =\psi (sd(f{x}_{m_i},f{x}_{n_i-1},a))\\ \le \psi (M_a(x_{m_i},x_{n_i-1}))-\phi (M_a(x_{m_i},x_{n_i-1}))+L\psi (N_a(x_{m_i},x_{n_i-1})),\end{array}$$

(3.38)

where

$$\begin{array}{r}{M}_a(x_{m_i},x_{n_i-1})\\ =max\{d(x_{m_i},x_{n_i-1},a),d(x_{m_i},f{x}_{m_i},a),d(x_{n_i-1},f{x}_{n_i-1},a),\\ \frac{d(x_{m_i},f{x}_{n_i-1},a)+d(f{x}_{m_i},x_{n_i-1},a)}{2s}\}\\ =max\{d(x_{m_i},x_{n_i-1},a),d(x_{m_i},x_{m_i+1},a),d(x_{n_i-1},x_{n_i},a),\\ \frac{d(x_{m_i},x_{n_i},a)+d(x_{m_i+1},x_{n_i-1},a)}{2s}\},\end{array}$$

(3.39)

and

$$\begin{array}{r}{N}_a(x_{m_i},x_{n_i-1})\\ =min\{d(x_{m_i},f{x}_{m_i},a),d(x_{m_i},f{x}_{n_i-1},a),d(x_{n_i-1},f{x}_{m_i},a),d(x_{n_i-1},f{x}_{n_i-1},a)\}\\ =min\{d(x_{m_i},x_{m_i+1},a),d(x_{m_i},x_{n_i},a),d(x_{n_i-1},x_{m_i+1},a),d(x_{n_i-1},x_{n_i},a)\}.\end{array}$$

(3.40)

Taking the upper limit as $i\to \mathrm{\infty }$ in (3.39) and (3.40) and using (3.30), (3.34), (3.36), and (3.37), we get

$$\begin{array}{r}\underset{n\to \mathrm{\infty }}{lim\hspace{0.17em}sup}{M}_a(x_{m_i-1},x_{n_i-1})\\ =max\{\underset{n\to \mathrm{\infty }}{lim\hspace{0.17em}sup}d(x_{m_i},x_{n_i-1},a),0,0,\\ \frac{{lim\hspace{0.17em}sup}_{n\to \mathrm{\infty }}d(x_{m_i},x_{n_i},a)+{lim\hspace{0.17em}sup}_{n\to \mathrm{\infty }}d(x_{m_i+1},x_{n_i-1},a)}{2s}\}\\ \le max\{\epsilon ,\frac{\epsilon s+\epsilon s}{2s}\}=\epsilon .\end{array}$$

(3.41)

So, we have

$$\underset{n\to \mathrm{\infty }}{lim\hspace{0.17em}sup}{M}_a(x_{m_i-1},x_{n_i-1})\le \epsilon ,$$

(3.42)

and

$$\underset{n\to \mathrm{\infty }}{lim\hspace{0.17em}sup}{N}_a(x_{m_i},x_{n_i-1})=0.$$

(3.43)

Now, taking the upper limit as $i\to \mathrm{\infty }$ in (3.38) and using (3.35), (3.42), and (3.43) we have

$$\begin{array}{rl}\psi (s\cdot \frac{\epsilon }{s})& \le \psi (s\underset{n\to \mathrm{\infty }}{lim\hspace{0.17em}sup}d(x_{m_i+1},x_{n_i},a))\\ \le \psi (\underset{n\to \mathrm{\infty }}{lim\hspace{0.17em}sup}{M}_a(x_{m_i},x_{n_i-1}))-\underset{n\to \mathrm{\infty }}{lim\hspace{0.17em}inf}\phi (M_a(x_{m_i},x_{n_i-1}))\\ \le \psi (\epsilon )-\phi (\underset{n\to \mathrm{\infty }}{lim\hspace{0.17em}inf}{M}_a(x_{m_i},x_{n_i-1})),\end{array}$$

which further implies that

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

so ${lim\hspace{0.17em}inf}_{n\to \mathrm{\infty }}{M}_a(x_{m_i},x_{n_i-1})=0$, a contradiction to (3.32). Thus, $\{x_{n+1}=f{x}_n\}$ is a $b_2$-Cauchy sequence in X.

As X is a $b_2$-complete space, there exists $u\in X$ such that $x_n\to u$ as $n\to \mathrm{\infty }$, that is,

$$\underset{n}{lim}{x}_{n+1}=\underset{n}{lim}f{x}_n=u.$$

Now, using continuity of f and the rectangle inequality, we get

$$d(u,fu,a)\le sd(u,fu,f{x}_n)+sd(fu,a,f{x}_n)+sd(a,u,f{x}_n).$$

Letting $n\to \mathrm{\infty }$, we get

$$d(u,fu,a)\le s\underset{n}{lim}d(u,fu,f{x}_n)+s\underset{n}{lim}d(fu,a,f{x}_n)+s\underset{n\to \mathrm{\infty }}{lim}d(a,u,f{x}_n)=0.$$

Therefore, we have $fu=u$. Thus, u is a fixed point of f.

The uniqueness of fixed point can be proved as in Theorem 1. □

Note that the continuity of f in Theorem 5 can be replaced by a property of the space.

Theorem 6 Under the hypotheses of Theorem  5, 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. Moreover, the set of fixed points of f is well ordered if and only if f has one and only one fixed point.

Proof Following similar arguments to those given in the proof of Theorem 5, we construct an increasing sequence $\{x_n\}$ in X such that $x_n\to u$, for some $u\in X$. Using the assumption on X, we have $x_n⪯u$, for all $n\in \mathbb{N}$. Now, we show that $fu=u$. By (3.24), we have

$$\begin{array}{rl}\psi (sd(x_{n+1},fu,a))& =\psi (sd(f{x}_n,fu,a))\\ \le \psi (M_a(x_n,u))-\phi (M_a(x_n,u))+L\psi (N_a(x_n,u)),\end{array}$$

(3.44)

where

$$\begin{array}{r}{M}_a(x_n,u)\\ =max\{d(x_n,u,a),d(x_n,f{x}_n,a),d(u,fu,a),\frac{d(x_n,fu,a)+d(f{x}_n,u,a)}{2s}\}\\ =max\{d(x_n,u,a),d(x_n,x_{n+1},a),d(u,fu,a),\frac{d(x_n,fu,a)+d(x_{n+1},u,a)}{2s}\}\end{array}$$

(3.45)

and

$$\begin{array}{rl}{N}_a(x_n,u)& =min\{d(x_n,f{x}_n,a),d(x_n,fu,a),d(u,f{x}_n,a),d(u,fu,a)\}\\ =min\{d(x_n,x_{n+1},a),d(x_n,fu,a),d(u,x_{n+1},a),d(u,fu,a)\}.\end{array}$$

(3.46)

Letting $n\to \mathrm{\infty }$ in (3.45) and (3.46) and using Lemma 1, we get

$$\begin{array}{rl}\frac{\frac{1}{s}d(u,fu,a)}{2s}& \le \underset{n\to \mathrm{\infty }}{lim\hspace{0.17em}inf}{M}_a(x_n,u)\le \underset{n\to \mathrm{\infty }}{lim\hspace{0.17em}sup}{M}_a(x_n,u)\\ \le max\{d(u,fu,a),\frac{sd(u,fu,a)}{2s}\}=d(u,fu,a),\end{array}$$

(3.47)

and

$$N_a(x_n,u)\to 0.$$

Again, taking the upper limit as $i\to \mathrm{\infty }$ in (3.44) and using Lemma 1 and (3.47) we get

$$\begin{array}{rl}\psi (d(u,fu,a))& =\psi (s\cdot \frac{1}{s}d(u,fu,a))\le \psi (s\underset{n\to \mathrm{\infty }}{lim\hspace{0.17em}sup}d(x_{n+1},fu,a))\\ \le \psi (\underset{n\to \mathrm{\infty }}{lim\hspace{0.17em}sup}{M}_a(x_n,u))-\underset{n\to \mathrm{\infty }}{lim\hspace{0.17em}inf}\phi (M_a(x_n,u))\\ \le \psi (d(u,fu,a))-\phi (\underset{n\to \mathrm{\infty }}{lim\hspace{0.17em}inf}{M}_a(x_n,u)).\end{array}$$

Therefore, $\phi ({lim\hspace{0.17em}inf}_{n\to \mathrm{\infty }}{M}_a(x_n,u))\le 0$, equivalently, ${lim\hspace{0.17em}inf}_{n\to \mathrm{\infty }}{M}_a(x_n,u)=0$. Thus, from (3.47) we get $u=fu$ and hence u is a fixed point of f. □

Corollary 4 Let $(X,⪯)$ be a partially ordered set and suppose that there exists a $b_2$-metric d on X such that $(X,d)$ is a $b_2$-complete $b_2$-metric space. Let $f:X\to X$ be a nondecreasing continuous mapping with respect to ⪯. Suppose that there exist $k\in [0,1)$ and $L\ge 0$ such that

$$\begin{array}{rl}d(fx,fy,a)\le & \frac{k}{s}max\{d(x,y,a),d(x,fx,a),d(y,fy,a),\frac{d(x,fy,a)+d(y,fx,a)}{2s}\}\\ +\frac{L}{s}min\{d(x,fx,a),d(y,fx,a)\},\end{array}$$

for all elements $x,y,a\in X$ with x, y comparable. If there exists $x_0\in X$ such that $x_0⪯f{x}_0$, then f has a fixed point. Moreover, the set of fixed points of f is well ordered if and only if f has one and only one fixed point.

Proof Follows from Theorem 5 by taking $\psi (t)=t$ and $\phi (t)=(1-k)t$, for all $t\in [0,+\mathrm{\infty })$. □

Corollary 5 Under the hypotheses of Corollary  4, without the continuity assumption of f, let for any nondecreasing sequence $\{x_n\}$ in X such that $x_n\to x\in X$ we have $x_n⪯x$, for all $n\in \mathbb{N}$. Then f has a fixed point in X.

4 An application to integral equations

As an application of our results, inspired by [26], we will consider the following integral equation:

$$x(t)=h(t)+{\int }_0^{T}g(t,s)F(s,x(s))ds,t\in I=[0,T].$$

(4.1)

Consider the set $X=C_{\mathbb{R}}(I)$ of all real continuous functions on I, ordered by the natural relation

$$x⪯y\iff x(t)\le y(t)\text{for all }t\in I,$$

and take arbitrary real $p>1$. We will use the following assumptions.

• (I) $h:I\to \mathbb{R}$, $g:I\times \mathbb{R}\to [0,+\mathrm{\infty })$ and $F:I\times \mathbb{R}\to \mathbb{R}$ are continuous functions;

• (II) for $x,y\in X$,

  $$x⪯y\Rightarrow {\int }_0^{T}g(\cdot ,s)F(s,x(s))ds\le {\int }_0^{T}g(\cdot ,s)F(s,y(s))ds;$$

• (III) for some $0\le r<1$ and all $x,y,a\in X$, with x and y comparable (w.r.t. ⪯),

  $$\begin{array}{r}{3}^{p-1}[\underset{0\le t\le T}{max}min\{|{\int }_0^{T}g(t,s)[F(s,x(s))-F(s,y(s))]ds|,\\ |h(t)+{\int }_0^{T}g(t,s)F(s,y(s))ds-a(t)|,\\ {|h(t)+{\int }_0^{T}g(t,s)F(s,x(s))ds-a(t)|\}]}^p\\ \le r{[\underset{0\le t\le T}{max}min\{|x(t)-y(t)|,|y(t)-a(t)|,|x(t)-a(t)|\}]}^p;\end{array}$$

• (IV) there exists $x_0\in X$ such that $x_0(t)\le h(t)+{\int }_0^{T}g(t,s)F(s,x_0(s))ds$ for all $t\in I$.

Let $d:X\times X\times X\to [0,\mathrm{\infty })$ be defined by

$$d(x,y,z)={[\underset{0\le t\le T}{max}min\{|x(t)-y(t)|,|y(t)-z(t)|,|x(t)-z(t)|\}]}^p.$$

Then $(X,d)$ is a $b_2$-complete $b_2$-metric space, with $s\le 3^{p-1}$ (similarly as in Example 2). We have the following result.

Theorem 7 Let the functions h, g, F satisfy conditions (I)-(IV) and let the space $(X,⪯,d)$ satisfy the requirement that if $\{x_n\}$ is a sequence in X, nondecreasing w.r.t. ⪯, and converging (in d) to some $u\in X$, then $x_n⪯u$ for all $n\in \mathbb{N}$. Then the integral equation (4.1) has a solution in X.

Proof Define the mapping $f:X\to X$ by

$$fx(t)=h(t)+{\int }_0^{T}g(t,s)F(s,x(s))ds,t\in I.$$

Then all the conditions of Corollary 3 are fulfilled. In particular, condition (III) implies that, for all $x,y,a\in X$, with x, y comparable, we have

$$sd(fx,fy,a)\le 3^{p-1}d(fx,fy,a)\le rd(x,y,a)\le rM(x,y,a).$$

Hence, using Corollary 3, we conclude that there exists a fixed point $x\in X$ of f, which is obviously a solution of (4.1). □