# ${b}_{2}$-Metric spaces and some fixed point theorems

- Zead Mustafa
^{1}Email author, - Vahid Parvaneh
^{2}, - Jamal Rezaei Roshan
^{3}and - Zoran Kadelburg
^{4}

**2014**:144

https://doi.org/10.1186/1687-1812-2014-144

© Mustafa et al.; licensee Springer. 2014

**Received: **27 January 2014

**Accepted: **30 June 2014

**Published: **22 July 2014

## 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.

## Keywords

*b*-metric space2-metric spacepartially ordered setfixed pointgeneralized contractive map

## 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:

(b_{1}) $d(x,y)=0$ if and only if $x=y$,

(b_{2}) $d(x,y)=d(y,x)$,

(b_{3}) $d(x,z)\le s[d(x,y)+d(y,z)]$.

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]

*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

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*is a 2-metric on ℝ,

*i.e.*, the following inequality holds:

*x*,

*y*,

*z*,

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

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*

*for all*

*a*

*in*

*X*.

*In particular*,

*if*${y}_{n}=y$

*is constant*,

*then*

*for all* *a* *in* *X*.

*Proof*Using the rectangle inequality in the given ${b}_{2}$-metric space, it is easy to see that

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.

□

## 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].

**Theorem 1**

*Let*$(X,\u2aaf)$

*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*${x}_{0}\u2aaff{x}_{0}$.

*Suppose that*

*for all*$a\in X$

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

*where*

*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 ${x}_{0}\u2aaff{x}_{0}$ and

*f*is an increasing function we obtain by induction that

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

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*.

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

*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

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* ${x}_{n}\u2aafu$ *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 ${x}_{n}\u2aafz$. Now, we show that $z=fz$. By (3.1) and Lemma 1,

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*$(X,\u2aaf)$

*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*${x}_{0}\u2aaff{x}_{0}$.

*Suppose that for some*

*r*,

*with*$0\le r<\frac{1}{s}$,

*holds for each*$a\in X$

*and all comparable elements*$x,y\in X$,

*where*

*If* *f* *is continuous*, *or*, *for any nondecreasing sequence* $\{{x}_{n}\}$ *in* *X* *such that* ${x}_{n}\to u\in X$ *one has* ${x}_{n}\u2aafu$ *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*$(X,\u2aaf)$

*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*${x}_{0}\u2aaff{x}_{0}$.

*Suppose that*

*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* ${x}_{n}\u2aafu$ *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

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*is a ${b}_{2}$-metric with parameter $s=2$. Introduce an order ⪯ in

*X*by

with all other pairs of distinct points in *X* incomparable.

*f*is an increasing mapping with $(\alpha ,0)\u2aaff(\alpha ,0)$ for each $\alpha \ge 0$. If $\{{x}_{n}\}=\{({\alpha}_{n},0)\}$ is a nondecreasing sequence in

*X*, converging to some $z=(\gamma ,0)$, then $({\alpha}_{n},0)\u2aaf(\gamma ,0)$ 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

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

*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*$(X,\u2aaf)$

*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*${x}_{0}\u2aaff{x}_{0}$.

*Suppose that*

*where*

*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 ${x}_{0}\u2aaff{x}_{0}$ and

*f*is an increasing function, we obtain by induction that

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 $n\in \mathbb{N}$.

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

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*.

*f*, we get

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* ${x}_{n}\u2aafu$ *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 ${x}_{n}\u2aafz$. Now, we show that $z=fz$. By (3.15) we have

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

**Corollary 3**

*Let*$(X,\u2aaf)$

*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*${x}_{0}\u2aaff{x}_{0}$.

*Suppose that*

*where*$0\le r<1$

*and*

*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* ${x}_{n}\u2aafu$ *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 $A\u2ab0B\u2ab0C$, with all other pairs of distinct points incomparable. Define $d:{X}^{3}\to \mathbb{R}$ by

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 $s=\frac{4}{3}$.

*f*is a nondecreasing mapping w.r.t. ⪯ and there exists ${x}_{0}\in X$ such that ${x}_{0}\u2aaff{x}_{0}$. 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

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]

- 1.
*φ*is continuous and nondecreasing. - 2.
$\phi (t)=0$ if and only if $t=0$.

**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

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

Now, let us prove our new result.

**Theorem 5** *Let* $(X,\u2aaf)$ *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* ${x}_{0}\u2aaff{x}_{0}$, *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 ${x}_{0}\u2aaff{x}_{0}={x}_{1}$ and

*f*is nondecreasing, we have ${x}_{1}=f{x}_{0}\u2aaf{x}_{2}=f{x}_{1}$, and by induction

*f*. So, we may assume that ${x}_{n}\ne {x}_{n+1}$, for all $n\in \mathbb{N}$. By (3.24), we have

*ψ*and

*φ*, we get

which gives a contradiction.

for each $a\in X$.

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

*X*. For this purpose, we use the following relation (see (3.9) and (3.18)):

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).

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*.

*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,

*f*and the rectangle inequality, we get

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* ${x}_{n}\u2aafx$, *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 ${x}_{n}\u2aafu$, for all $n\in \mathbb{N}$. Now, we show that $fu=u$. By (3.24), we have

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*$(X,\u2aaf)$

*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*

*for all elements* $x,y,a\in X$ *with* *x*, *y* *comparable*. *If there exists* ${x}_{0}\in X$ *such that* ${x}_{0}\u2aaff{x}_{0}$, *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* ${x}_{n}\u2aafx$, *for all* $n\in \mathbb{N}$. *Then* *f* *has a fixed point in* *X*.

## 4 An application to integral equations

*I*, ordered by the natural relation

- (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 $x,y\in X$,$x\u2aafy\phantom{\rule{1em}{0ex}}\Rightarrow \phantom{\rule{1em}{0ex}}{\int}_{0}^{T}g(\cdot ,s)F(s,x(s))\phantom{\rule{0.2em}{0ex}}ds\le {\int}_{0}^{T}g(\cdot ,s)F(s,y(s))\phantom{\rule{0.2em}{0ex}}ds;$
- (III)for some $0\le r<1$ and all $x,y,a\in X$, with
*x*and*y*comparable (w.r.t. ⪯),$\begin{array}{r}{3}^{p-1}[\underset{0\le t\le T}{max}min\{|{\int}_{0}^{T}g(t,s)[F(s,x(s))-F(s,y(s))]\phantom{\rule{0.2em}{0ex}}ds|,\\ \phantom{\rule{2em}{0ex}}|h(t)+{\int}_{0}^{T}g(t,s)F(s,y(s))\phantom{\rule{0.2em}{0ex}}ds-a(t)|,\\ {\phantom{\rule{2em}{0ex}}|h(t)+{\int}_{0}^{T}g(t,s)F(s,x(s))\phantom{\rule{0.2em}{0ex}}ds-a(t)|\}]}^{p}\\ \phantom{\rule{1em}{0ex}}\le r{[\underset{0\le t\le T}{max}min\{|x(t)-y(t)|,|y(t)-a(t)|,|x(t)-a(t)|\}]}^{p};\end{array}$ - (IV)
there exists ${x}_{0}\in X$ such that ${x}_{0}(t)\le h(t)+{\int}_{0}^{T}g(t,s)F(s,{x}_{0}(s))\phantom{\rule{0.2em}{0ex}}ds$ for all $t\in I$.

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* $(X,\u2aaf,d)$ *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* ${x}_{n}\u2aafu$ *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

*x*,

*y*comparable, we have

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). □

## Declarations

### Acknowledgements

The authors are highly indebted to the referees of this paper who helped us to improve it in several places. The fourth author is thankful to the Ministry of Education, Science and Technological Development of Serbia.

## Authors’ Affiliations

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