Fixed point of a pair of contractive dominated mappings on a closed ball in an ordered dislocated metric space

Abstract

Common fixed point results for mappings satisfying locally contractive conditions on a closed ball in an ordered complete dislocated metric space have been established. The notion of dominated mappings is applied to approximate the unique solution of nonlinear functional equations. Our results improve several well-known conventional results.

MSC:46S40, 47H10, 54H25.

1 Introduction and preliminaries

Let $T:X⟶X$ be a mapping. A point $x\in X$ is called a fixed point of T if $x=Tx$. Let $x_0$ be an arbitrarily chosen point in X. Define a sequence $\{x_n\}$ in X by a simple iterative method given by

$$x_{n+1}=T{x}_n,\text{where }n\in \{0,1,2,\dots \}.$$

Such a sequence is called a Picard iterative sequence, and its convergence plays a very important role in proving the existence of a fixed point of a mapping T. A self-mapping T on a metric space X is said to be a Banach contraction mapping if

$$d(Tx,Ty)\le kd(x,y)$$

holds for all $x,y\in X$, where $0\le k<1$.

Fixed points results of mappings satisfying a certain contractive condition on the entire domain have been at the center of rigorous research activity (for example, see [1–12]) and they have a wide range of applications in different areas such as nonlinear and adaptive control systems, parameter estimation problems, computing magnetostatic fields in a nonlinear medium and convergence of recurrent networks (see [13–15]).

From the application point of view, the situation is not yet completely satisfactory because it frequently happens that a mapping T is a contraction not on the entire space X but merely on a subset Y of X. However, if Y is closed and a Picard iterative sequence $\{x_n\}$ in X converges to some x in X, then by imposing a subtle restriction on the choice of $x_0$, one may force the Picard iterative sequence to stay eventually in Y. In this case, the closedness of Y coupled with some suitable contractive condition establishes the existence of a fixed point of T. Azam et al. [16] proved a significant result concerning the existence of fixed points of a mapping satisfying contractive conditions on a closed ball of a complete metric space. Recently, many results related to the fixed point theorem in complete metric spaces endowed with a partial ordering ⪯ appeared in literature. Ran and Reurings [17] proved an analogue of Banach’s fixed point theorem in a metric space endowed with a partial order and gave applications to matrix equations. In this way, they weakened the usual contractive condition. Subsequently, Nieto et al. [18] extended this result in [17] for non-decreasing mappings and applied it to obtain a unique solution for a first-order ordinary differential equation with periodic boundary conditions. Thereafter, many works related to fixed point problems have also been considered in partially ordered metric spaces (see [17–23]). Indeed, they all deal with monotone mappings (either order-preserving or order-reversing) such that for some $x_0\in X$, either $x_0⪯f{x}_0$ or $f{x}_0⪯x_0$, where f is a self-map on a metric space. To obtain a unique solution, they used an additional restriction that each pair of elements has a lower bound and an upper bound. We have not used these conditions in our results. In this paper we introduce a new condition of partial order.

On the other hand, the notion of a partial metric space was introduced by Matthews in [24]. In partial metric spaces, the distance of a point from itself may not be zero. He also proved a partial metric version of the Banach fixed point theorem. Karapınar et al. [25] have proved a common fixed point in partial metric spaces. Partial metric spaces have applications in theoretical computer science (see [26]). Altun et al. [20], Aydi [27], Samet et al. [28] and Paesano et al. [29] used the idea of a partial metric space and partial order and gave some fixed point theorems for the contractive condition on ordered partial metric spaces. Further useful results can be seen in [28]. To generalize a partial metric, Hitzler and Seda [30] introduced the concept of dislocated topologies and its corresponding generalized metric, named a dislocated metric, and established a fixed point theorem in complete dislocated metric spaces to generalize the celebrated Banach contraction principle. The notion of dislocated topologies has useful applications in the context of logic programming semantics (see [31]). Further useful results can be seen in [32–35]. The dominated mapping, which satisfies the condition $fx⪯x$, occurs very naturally in several practical problems. For example, if x denotes the total quantity of food produced over a certain period of time and $f(x)$ gives the quantity of food consumed over the same period in a certain town, then we must have $fx⪯x$. In this paper, we exploit this concept for contractive mappings [36] to generalize, extend and improve some classical fixed point results for two, three and four mappings in the framework of an ordered complete dislocated metric space X. Our results not only extend some primary theorems to ordered dislocated metric spaces, but also restrict the contractive conditions on a closed ball only. The concept of a dominated mapping has been applied to approximate the unique solution of nonlinear functional equations.

Consistent with [30, 32, 34] and [35], the following definitions and results will be needed in the sequel.

Definition 1.1 Let X be a nonempty set and let $d_l:X\times X\to [0,\mathrm{\infty })$ be a function, called a dislocated metric (or simply $d_l$-metric), if the following conditions hold for any $x,y,z\in X$:

• (i) if $d_l(x,y)=0$, then $x=y$,

• (ii) $d_l(x,y)=d_l(y,x)$,

• (iii) $d_l(x,y)\le d_l(x,z)+d_l(z,y)$.

The pair $(X,d_l)$ is then called a dislocated metric space. It is clear that if $d_l(x,y)=0$, then from (i), $x=y$. But if $x=y$, $d_l(x,y)$ may not be 0.

Recently Sarma and Kumari [34] proved the results that establish the existence of a topology induced by a dislocated metric and the fact that this topology is metrizable. This topology has as a base the family of sets $\{B(x,\epsilon )\cup \{x\}:x\in X,\epsilon >0\}$, where $B(x,\epsilon )$ is an open ball and $B(x,\epsilon )=\{y\in X:d_l(x,y)<\epsilon \}$ for some $x\in X$ and $\epsilon >0$. Also, $\overline{B(x,\epsilon )}=\{y\in X:d_l(x,y)\le \epsilon \}$ is a closed ball.

Also, Harandi [37] defined the concept of a metric-like space which is similar to a dislocated metric space. Each metric-like σ on X generates a topology ${\tau }_{\sigma }$ on X whose base is the family of open σ-balls

$$B_{\sigma }(x,\epsilon )=\{y\in X:|\sigma (x,y)-\sigma (x,x)|<\epsilon \}.$$

Definition 1.2 Let $p:X\times X\to R^{+}$, where X is a nonempty set. p is said to be a partial metric on X if for any $x,y,z\in X$:

(P₁) $p(x,x)=p(y,y)=p(x,y)$ if and only if $x=y$,

(P₂) $p(x,x)\le p(x,y)$,

(P₃) $p(x,y)=p(y,x)$,

(P₄) $p(x,z)\le p(x,y)+p(y,z)-p(y,y)$.

The pair $(X,p)$ is then called a partial metric space.

Each partial metric p on X induces a $T_0$ topology p on X which has as a base the family of open 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 $\epsilon >0$.

It is clear that any partial metric is a $d_l$-metric. A basic example of a partial metric space is the pair $(R^{+},p)$, where $p(x,y)=max\{x,y\}$ for all $x,y\in R^{+}$. It is also a $d_l$-metric. An example of a $d_l$-metric space which is not a partial metric is given below.

Example 1.3 If $X=R^{+}\cup \{0\}$, then $d_l(x,y)=x+y$ defines a dislocated metric $d_l$ on X. Note that this metric is not a partial metric as (P₂) is not satisfied.

From the examples and definitions, it is clear that any partial metric is a $d_l$-metric, whereas a $d_l$-metric may not be a partial metric. We also remark that for those $d_l$-metrics which are also partial metrics, we have $B_{d_l}(x,\epsilon )\subseteq B_p(x,\epsilon )$. Also, for any $d_l$-metric, $B_{d_l}(x,\epsilon )\subseteq B_{\sigma }(x,\epsilon )$. Thus it is better to find a fixed point on a closed ball defined by Hitzler in a $d_l$-metric because we restrict ourselves to applying the contractive condition on the smallest closed ball. In this way, we also weaken the contractive condition.

Definition 1.4 [30]

A sequence $\{x_n\}$ in a $d_l$-metric space $(X,d_l)$ is called a Cauchy sequence if given $\epsilon >0$, there corresponds $n_0\in N$ such that for all $n,m\ge n_0$, we have $d_l(x_m,x_n)<\epsilon$ or ${lim}_{n,m\to \mathrm{\infty }}{d}_l(x_n,x_m)=0$.

Definition 1.5 [30]

A sequence $\{x_n\}$ in a $d_l$-metric space converges with respect to $d_l$ if there exists $x\in X$ such that $d_l(x_n,x)\to 0$ as $n\to \mathrm{\infty }$. In this case, x is called the limit of $\{x_n\}$, and we write $x_n\to x$.

Definition 1.6 [30]

A $d_l$-metric space $(X,d_l)$ is called complete if every Cauchy sequence in X converges to a point in X.

In Harandi’s sense, a sequence $\{x_n\}$ in the metric-like space $(X,\sigma )$ converges to a point $x\in X$ if and only if ${lim}_{n\to \mathrm{\infty }}\sigma (x_n,x)=\sigma (x,x)$. The sequence ${\{x_n\}}_{n=0}^{\mathrm{\infty }}$ of elements of X is called σ-Cauchy if the limit ${lim}_{n,m\to \mathrm{\infty }}\sigma (x_n,x_m)$ exists and is finite. The metric-like space $(X,\sigma )$ is called complete if for each σ-Cauchy sequence ${\{x_n\}}_{n=0}^{\mathrm{\infty }}$, there is some $x\in X$ such that

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

Romaguera [38] has given the idea of a 0-Cauchy sequence and a 0-complete partial metric space. Using his idea, we can observe the following:

• (a) Every Cauchy sequence with respect to Hitzler is a Cauchy sequence with respect to Harandi.

• (b) Every complete metric space with respect to Harandi is complete with respect to Hitzler. The following example shows that the converse assertions of (a) and (b) do not hold.

Example 1.7 Let $X=Q^{+}\cup \{0\}$ and let $d_l:X\times X\to X$ be defined by $d_l(x,y)=x+y$. Note that $\{x_n\}={(1+\frac{1}{n})}^n$ is a Cauchy sequence with respect to Harandi, but it is not a Cauchy sequence with respect to Hitzler. Also, every Cauchy sequence (with respect to Hitzler) in X converges to a point ‘0’ in X. Hence X is complete with respect to Hitzler, but X is not complete with respect to Harandi as ${lim}_{n\to \mathrm{\infty }}{(1+\frac{1}{n})}^n=e\notin X$.

Definition 1.8 Let X be a nonempty set. Then $(X,⪯,d_l)$ is called an ordered dislocated metric space if (i) $d_l$ is a dislocated metric on X and (ii) ⪯ is a partial order on X.

Definition 1.9 Let $(X,⪯)$ be a partial ordered set. Then $x,y\in X$ are called comparable if $x⪯y$ or $y⪯x$ holds.

Definition 1.10 [39]

Let $(X,⪯)$ be a partially ordered set. A self-mapping f on X is called dominated if $fx⪯x$ for each x in X.

Example 1.11 [39]

Let $X=[0,1]$ be endowed with the usual ordering and $f:X\to X$ be defined by $fx=x^n$ for some $n\in \mathbb{N}$. Since $fx=x^n\le x$ for all $x\in X$, therefore f is a dominated map.

Definition 1.12 Let X be a nonempty set and $T,f:X\to X$. A point $y\in X$ is called a point of coincidence of T and f if there exists a point $x\in X$ such that $y=Tx=fx$. The mappings T, f are said to be weakly compatible if they commute at their coincidence point (i.e., $Tfx=fTx$ whenever $Tx=fx$).

For $A\subset X$, we denote by $D(A)$ the set of all limit points of A and $\overline{A}$ closure of A in X. We state without proof the following simple facts due to [34].

Lemma 1.13 A subset of A of a dislocated metric space is closed if and only if $\overline{A}=A$.

Lemma 1.14 The topology induced by a dislocated metric is a Hausdorff topology.

Lemma 1.15 Every closed ball in a complete dislocated metric space is complete.

We also need the following results for subsequent use.

Lemma 1.16 [40]

Let X be a nonempty set and let $f:X\to X$ be a function. Then there exists a subset $E\subset X$ such that $fE=fX$ and $f:E\to X$ is one-to-one.

Lemma 1.17 [1]

Let X be a nonempty set and let the mappings $S,T,f:X\to X$ have a unique point of coincidence v in X. If $(S,f)$ and $(T,f)$ are weakly compatible, then S, T, f have a unique common fixed point.

Theorem 1.18 [[36], p.303]

Let $(X,d)$ be a complete metric space, let $S:X\to X$ be a mapping, let $r>0$ and $x_0$ be an arbitrary point in X. Suppose that there exists $k\in [0,1)$ with

$$d(Sx,Sy)\le kd(x,y)\mathit{\text{for all}}x,y\in Y=\overline{B(x_0,r)}$$

and $d(x_0,S{x}_0)<(1-k)r$. Then there exists a unique point $x^{\ast }$ in $\overline{B(x_0,r)}$ such that $x^{\ast }=S{x}^{\ast }$.

2 Fixed points of contractive mappings

Theorem 2.1 Let $(X,⪯,d_l)$ be an ordered complete dislocated metric space, let $S,T:X\to X$ be dominated maps and let $x_0$ be an arbitrary point in X. Suppose that for $k\in [0,1)$ and for $S\ne T$, we have

$$d_l(Sx,Ty)\le k{d}_l(x,y)\mathit{\text{for all comparable elements}}x,y\mathit{\text{in}}\overline{B(x_0,r)}$$

(2.1)

and

$$d_l(x_0,S{x}_0)\le (1-k)r.$$

(2.2)

If for a non-increasing sequence $\{x_n\}$ in $\overline{B(x_0,r)}$, $\{x_n\}\to u$ implies that $u⪯x_n$, then there exists $x^{\ast }\in \overline{B(x_0,r)}$ such that $d_l(x^{\ast },x^{\ast })=0$ and $x^{\ast }=S{x}^{\ast }=T{x}^{\ast }$. Also if, for any two points x, y in $\overline{B(x_0,r)}$, there exists a point $z\in \overline{B(x_0,r)}$ such that $z⪯x$ and $z⪯y$, that is, every pair of elements has a lower bound, then $x^{\ast }$ is a unique common fixed point in $\overline{B(x_0,r)}$.

Proof Choose a point $x_1$ in X such that $x_1=S{x}_0$. As $S{x}_0⪯x_0$, so $x_1⪯x_0$ and let $x_2=T{x}_1$. Now $T{x}_1⪯x_1$ gives $x_2⪯x_1$. Continuing this process, we construct a sequence $x_n$ of points in X such that

$$x_{2i+1}=S{x}_{2i},x_{2i+2}=T{x}_{2i+1}\text{and}{x}_{2i+1}=S{x}_{2i}⪯x_{2i},\text{where }i=0,1,2,\dots .$$

First we show that $x_n\in \overline{B(x_0,r)}$ for all $n\in N$. Using inequality (2.2), we have

$$d_l(x_0,x_1)\le (1-k)r\le r.$$

It follows that

$$x_1\in \overline{B(x_0,r)}.$$

Let $x_2,\dots ,x_j\in \overline{B(x_0,r)}$ for some $j\in N$. If $j=2i+1$, then $x_{2i+1}⪯x_{2i}$, where $i=0,1,2,\dots ,\frac{j-1}{2}$. So, using inequality (2.1), we obtain

$$\begin{array}{rcl}{d}_l(x_{2i+1},x_{2i+2})& =& d_l(S{x}_{2i},T{x}_{2i+1})\le k[d_l(x_{2i},x_{2i+1})]\\ \le & k^2[d_l(x_{2i-1},x_{2i})]\le \cdots \le k^{2i+1}{d}_l(x_0,x_1).\end{array}$$

(2.3)

If $j=2i+2$, then as $x_1,x_2,\dots ,x_j\in \overline{B(x_0,r)}$ and $x_{2i+2}⪯x_{2i+1}$ ($i=0,1,2,\dots ,\frac{j-2}{2}$). We obtain

$$d_l(x_{2i+2},x_{2i+3})\le k^{2(i+1)}{d}_l(x_0,x_1).$$

(2.4)

Thus from inequalities (2.3) and (2.4), we have

$$d_l(x_j,x_{j+1})\le k^j{d}_l(x_0,x_1).$$

(2.5)

Now

$$\begin{array}{rcl}{d}_l(x_0,x_{j+1})& \le & d_l(x_0,x_1)+\cdots +d_l(x_j,x_{j+1})\\ \le & d_l(x_0,x_1)+\cdots +k^j{d}_l(x_0,x_1)\text{(by (2.5))}\\ \le & d_l(x_0,x_1)[1+\cdots +k^{j-1}+k^j]\\ \le & \frac{(1-k^{j+1})}{1-k}{d}_l(x_0,x_1)\\ \le & \frac{(1-k^{j+1})}{1-k}(1-k)r\text{(by (2.2))}\\ \le & (1-k^{j+1})r\le r.\end{array}$$

Thus $x_{j+1}\in \overline{B(x_0,r)}$. Hence $x_n\in \overline{B(x_0,r)}$ for all $n\in N$. It implies that

$$d_l(x_n,x_{n+1})\le k^n{d}_l(x_0,x_1)\text{for all }n\in N.$$

(2.6)

It implies that

$$\begin{array}{rcl}{d}_l(x_n,x_{n+i})& \le & d_l(x_n,x_{n+1})+\cdots +d_l(x_{n+i-1},x_{n+i})\\ \le & k^n{d}_l(x_0,x_1)+\cdots +k^{n+i-1}{d}_l(x_0,x_1)\text{(by (2.6))}\\ \le & k^n{d}_l(x_0,x_1)[1+\cdots +k^{i-2}+k^{i-1}]\\ \le & \frac{k^n(1-k^i)}{1-k}{d}_l(x_0,x_1)⟶0\text{as }n\to \mathrm{\infty }.\end{array}$$

Notice that the sequence $\{x_n\}$ is a Cauchy sequence in $(\overline{B(x_0,r)},d_l)$. Therefore there exists a point $x^{\ast }\in \overline{B(x_0,r)}$ with ${lim}_{n\to \mathrm{\infty }}{x}_n=x^{\ast }$. Also,

$$\underset{n\to \mathrm{\infty }}{lim}{d}_l(x_n,x^{\ast })=0.$$

(2.7)

Now,

$$d_l(x^{\ast },S{x}^{\ast })\le d_l(x^{\ast },x_{2n+2})+d_l(x_{2n+2},S{x}^{\ast }).$$

On taking limit as $n\to \mathrm{\infty }$ and using the fact that $x^{\ast }⪯x_n$ when $x_n\to x^{\ast }$, we have

$$d_l(x^{\ast },S{x}^{\ast })\le \underset{n\to \mathrm{\infty }}{lim}[d_l(x^{\ast },x_{2n+2})+k{d}_l(x_{2n+1},x^{\ast })].$$

By equation (2.7), we obtain

$$d_l(x^{\ast },S{x}^{\ast })\le 0,$$

and hence $x^{\ast }=S{x}^{\ast }$. Similarly, by using

$$d_l(x^{\ast },T{x}^{\ast })\le d_l(x^{\ast },x_{2n+1})+d_l(x_{2n+1},T{x}^{\ast }),$$

we can show that $x^{\ast }=T{x}^{\ast }$. Hence S and T have a common fixed point in $\overline{B(x_0,r)}$. Now,

$$d_l(x^{\ast },x^{\ast })=d_l(S{x}^{\ast },T{x}^{\ast })\le k{d}_l(x^{\ast },x^{\ast }).$$

This implies that

$$d_l(x^{\ast },x^{\ast })=0.$$

For uniqueness, assume that y is another fixed point of T and S in $\overline{B(x_0,r)}$. If $x^{\ast }$ and y are comparable, then

$$\begin{array}{rcl}{d}_l(x^{\ast },y)& =& d_l(S{x}^{\ast },Ty)\\ \le & k{d}_l(x^{\ast },y).\end{array}$$

This shows that $x^{\ast }=y$. Now if $x^{\ast }$ and y are not comparable, then there exists a point $z_0\in \overline{B(x_0,r)}$ such that $z_0⪯x^{\ast }$ and $z_0⪯y$. Choose a point $z_1$ in X such that $z_1=T{z}_0$. As $T{z}_0⪯z_0$, so $z_1⪯z_0$and let $z_2=S{z}_1$. Now $S{z}_1⪯z_1$ gives $z_2⪯z_1$. Continuing this process and having chosen $z_n$ in X such that

$$z_{2i+1}=T{z}_{2i},z_{2i+2}=S{z}_{2i+1}\text{and}{z}_{2i+1}=T{z}_{2i}⪯z_{2i},\text{where }i=0,1,2,\dots ,$$

we obtain that $z_{n+1}⪯z_n⪯\cdots ⪯z_0⪯x^{\ast }$. As $z_0⪯x^{\ast }$ and $z_0⪯y$, it follows that $z_n⪯T{x}^{\ast }$ and $z_n⪯Ty$ for all $n\in N$. We will prove that $z_n\in \overline{B(x_0,r)}$ for all $n\in N$ by using mathematical induction. For $n=1$,

$$\begin{array}{rcl}{d}_l(x_0,z_1)& \le & d_l(x_0,x_1)+d_l(x_1,z_1)\\ \le & (1-k)r+k{d}_l(x_0,z_0)\\ \le & (1-k)r+kr=r.\end{array}$$

It follows that $z_1\in \overline{B(x_0,r)}$. Let $z_2,z_3,\dots ,z_j\in \overline{B(x_0,r)}$ for some $j\in N$. Note that if j is odd, then

$$d_l(x_{j+1},z_{j+1})=d_l(T{x}_j,S{z}_j)\le k{d}_l(x_j,z_j)\le \cdots \le k^{j+1}{d}_l(x_0,z_0),$$

and if j is even, then

$$d_l(x_{j+1},z_{j+1})=d_l(S{x}_j,T{z}_j)\le k{d}_l(x_j,z_j)\le \cdots \le k^{j+1}{d}_l(x_0,z_0).$$

Now

$$\begin{array}{rcl}{d}_l(x_0,z_{j+1})& \le & d_l(x_0,x_1)+d_l(x_1,x_2)+\cdots +d_l(x_{j+1},z_{j+1})\\ \le & d_l(x_0,x_1)+k{d}_l(x_0,x_1)+\cdots +k^{j+1}{d}_l(x_0,z_0)\\ \le & d_l(x_0,x_1)[1+k+\cdots +k^j]+k^{j+1}r\\ \le & (1-k)r\frac{(1-k^{j+1})}{1-k}+k^{j+1}r,\\ d_l(x_0,z_{j+1})& \le & r,\end{array}$$

which implies that

$$d_l(x_{j+1},z_{j+1})=d_l(S{x}_j,T{z}_j)\le k{d}_l(x_j,z_j)\le \cdots \le k^{j+1}{d}_l(x_0,z_0).$$

Thus $z_{j+1}\in \overline{B(x_0,r)}$. Hence $z_n\in \overline{B(x_0,r)}$ for all $n\in N$. As $z_0⪯x^{\ast }$ and $z_0⪯y$, it follows that $z_n⪯T^n{x}^{\ast }$, $z_n⪯S^n{x}^{\ast }$, $z_n⪯S^{n}y$ and $z_n⪯T^{n}y$ for all $n\in N$ as $S^n{x}^{\ast }=T^n{x}^{\ast }=x^{\ast }$ and $S^{n}y=T^{n}y=y$ for all $n\in N$. If n is odd, then

$$\begin{array}{rcl}{d}_l(x^{\ast },y)& =& d_l(T^n{x}^{\ast },T^{n}y)\\ \le & d_l(T^n{x}^{\ast },S{z}_n)+d_l(S{z}_n,T^{n}y)\\ \le & k{d}_l(T^{n-1}{x}^{\ast },z_n)+k{d}_l(z_n,T^{n-1}y)\\ =& k{d}_l(S^{n-1}{x}^{\ast },T{z}_{n-1})+k{d}_l(T{z}_{n-1},S^{n-1}y)\\ \le & k^2{d}_l(S^{n-2}{x}^{\ast },z_{n-1})+k^2{d}_l(z_{n-1},S^{n-2}y)\\ ⋮\\ \le & k^{n+1}{d}_l(x^{\ast },z_0)+k^{n+1}{d}_l(z_0,y)⟶0\text{as }n\to \mathrm{\infty }.\end{array}$$

So, $x^{\ast }=y$. Similarly, we can show that $x^{\ast }=y$ if n is even. Hence $x^{\ast }$ is a unique common fixed point of T and S in $\overline{B(x_0,r)}$. □

Theorem 2.1 extends Theorem 1.18 to ordered complete dislocated metric spaces.

Example 2.2 Let $X=Q^{+}\cup \{0\}$ be endowed with the order $(x_1,y_1)⪯(x_2,y_2)$ if $x_1\le x_2$, $y_1\le y_2$. Let $S,T:X^2\to X^2$ be defined by

$$S(x,y)=\{\begin{array}{ll}(\frac{x}{7},\frac{3y}{11})& \text{if }x+y\le 1,\\ (x-\frac{1}{3},y-\frac{3}{8})& \text{if }x+y>1\end{array}$$

and

$$T(x,y)=\{\begin{array}{ll}(\frac{4x}{15},\frac{2y}{7})& \text{if }x+y\le 1,\\ (x-\frac{1}{4},y-\frac{1}{5})& \text{if }x+y>1.\end{array}$$

Clearly, S and T are dominated mappings. Let $d_l:X^2\times X^2\to X$ be defined by $d_l((x_1,y_1),(x_2,y_2))=x_1+y_1+x_2+y_2$. Then it is easy to prove that $(X^2,d_l)$ is a complete dislocated metric space. Let $(x_0,y_0)=(\frac{3}{7},\frac{4}{7})$, $r=2$, then

$$\overline{B((x_0,y_0),r)}=\{(x,y)\in X:x+y\le 1\}$$

with $k=\frac{3}{10}\in [0,1)$,

$$\begin{array}{c}(1-k)r=(1-\frac{3}{10})2=\frac{7}{5},\hfill \\ d_l((x_0,y_0),S(x_0,y_0))=\frac{656}{539}<\frac{7}{5}.\hfill \end{array}$$

Also, for all comparable elements $(x_1,y_1),(x_2,y_2)\in X^2$ such that $x_1+y_1>1$ and $x_2+y_2>1$, we have

$$\begin{array}{c}{d}_l(S(x_1,y_1),T(x_2,y_2))=x_1-\frac{1}{3}+y_1-\frac{3}{8}+x_2-\frac{1}{4}+y_2-\frac{1}{5}\hfill \\ \phantom{d_l(S(x_1,y_1),T(x_2,y_2))}\ge \frac{3}{10}\{x_1+y_1+x_2+y_2\},\hfill \\ d_l(Sx,Ty)\ge k{d}_l[(x_1,y_1),(x_2,y_2)].\hfill \end{array}$$

So, the contractive condition does not hold on $X^2$. Now if $(x_1,y_1),(x_2,y_2)\in \overline{B((x_0,y_0),r)}$, then

$$\begin{array}{rcl}{d}_l(S(x_1,y_1),T(x_2,y_2))& =& \frac{x_1}{7}+\frac{3y_1}{11}+\frac{4x_2}{15}+\frac{2y_2}{7}\\ \le & \frac{3}{10}\{x_1+y_1+x_2+y_2\}=k{d}_l[(x_1,y_1),(x_2,y_2)].\end{array}$$

Therefore, all the conditions of Theorem 2.1 are satisfied. Moreover, $(0,0)$ is the common fixed point of S and T. Also, note that for any metric d on $X^2$, the respective condition does not hold on $\overline{B((x_0,y_0),r)}$ since

$$\begin{array}{rcl}d(S(\frac{2}{5},\frac{3}{5}),T(\frac{2}{5},\frac{3}{5}))& =& d((\frac{2}{35},\frac{9}{55}),(\frac{8}{75},\frac{6}{35}))\\ >& kd((\frac{2}{5},\frac{3}{5}),(\frac{2}{5},\frac{3}{5}))=0\text{for any }k\in [0,1).\end{array}$$

Moreover, $X^2$ is not complete for any metric d on $X^2$.

Remark 2.3 If we impose a Banach-type contractive condition for a pair of mappings $S,T:X\to X$ on a metric space $(X,d)$, that is,

$$d(Sx,Ty)\le kd(x,y)\text{for all }x,y\in X,$$

then it follows that $Sx=Tx$ for all $x\in X$ (that is, S and T are equal). Therefore the above condition fails to find common fixed points of S and T. However, the same condition in a dislocated metric space does not assert that $S=T$, which is seen in Example 2.2. Hence Theorem 2.1 cannot be obtained from a metric fixed point theorem.

Theorem 2.4 Let $(X,⪯,d_l)$ be an ordered complete dislocated metric space, let $S:X\to X$ be a dominated map and let $x_0$ be an arbitrary point in X. Suppose that there exists $k\in [0,1)$ with

$$d_l(Sx,Sy)\le k{d}_l(x,y)\mathit{\text{for all comparable elements}}x,y\mathit{\text{in}}\overline{B(x_0,r)}$$

and

$$d_l(x_0,S{x}_0)\le (1-k)r.$$

If, for a non-increasing sequence $\{x_n\}$ in $\overline{B(x_0,r)}$, $\{x_n\}\to u$ implies that $u⪯x_n$, and also, for any two points x, y in $\overline{B(x_0,r)}$, there exists a point $z\in \overline{B(x_0,r)}$ such that every pair of elements has a lower bound, then there exists a unique fixed point $x^{\ast }$ of S in $\overline{B(x_0,r)}$. Further, $d_l(x^{\ast },x^{\ast })=0$.

Proof By following similar arguments to those we have used to prove Theorem 2.1, one can easily prove the existence of a unique fixed point $x^{\ast }$ of S in $\overline{B(x_0,r)}$. □

In Theorem 2.1, condition (2.2) is imposed to restrict condition (2.1) only for x, y in $\overline{B(x_0,r)}$ and Example 2.2 explains the utility of this restriction. However, the following result relaxes condition (2.2) but imposes condition (2.1) for all comparable elements in the whole space X.

Theorem 2.5 Let $(X,⪯,d_l)$ be an ordered complete dislocated metric space, let $S,T:X\to X$ be the dominated map and let $x_0$ be an arbitrary point in X. Suppose that for $k\in [0,1)$ and for $S\ne T$, we have

$$d_l(Sx,Ty)\le k{d}_l(x,y)\mathit{\text{for all comparable elements}}x,y\mathit{\text{in}}X.$$

Also, if for a non-increasing sequence $\{x_n\}$ in X, $\{x_n\}\to u$ implies that $u⪯x_n$, and for any two points x, y in X, there exists a point $z\in X$ such that $z⪯x$ and $z⪯y$, then there exists a unique point $x^{\ast }$ in X such that $x^{\ast }=S{x}^{\ast }=T{x}^{\ast }$. Further, $d_l(x^{\ast },x^{\ast })=0$.

In Theorem 2.1, the condition ‘for a non-increasing sequence, $\{x_n\}\to u$ implies that $u⪯x_n$’ and the existence of z or a lower bound is imposed to restrict condition (2.1) only for comparable elements. However, the following result relaxes these restrictions but imposes condition (2.1) for all elements in $\overline{B(x_0,r)}$. In Theorem 2.1, it may happen that S has more fixed points, but these fixed points of S are not the fixed points of T, because a common fixed point of S and T is unique, whereas without order we can obtain a unique fixed point of S and T separately, which is proved in the following theorem.

Theorem 2.6 Let $(X,d_l)$ be a complete dislocated metric space, let $S,T:X\to X$ be self-maps and let $x_0$ be an arbitrary point in X. Suppose that for $k\in [0,1)$ and for $S\ne T$, we have

$$d_l(Sx,Ty)\le k{d}_l(x,y)\mathit{\text{for all elements}}x,y\mathit{\text{in}}\overline{B(x_0,r)}$$

and

$$d_l(x_0,S{x}_0)\le (1-k)r.$$

Then there exists a unique $x^{\ast }\in \overline{B(x_0,r)}$ such that $d_l(x^{\ast },x^{\ast })=0$ and $x^{\ast }=S{x}^{\ast }=T{x}^{\ast }$. Further, S and T have no fixed point other than $x^{\ast }$.

Proof By Theorem 2.1, $x^{\ast }=S{x}^{\ast }=T{x}^{\ast }$. Let y be another point such that $y=Ty$. Then

$$d_l(x^{\ast },y)=d_l(S{x}^{\ast },Ty)\le k{d}_l(x^{\ast },y).$$

This shows that $x^{\ast }=y$. Thus T has no fixed point other than $x^{\ast }$. Similarly, S has no fixed point other than $x^{\ast }$. □

Now we apply our Theorem 2.1 to obtain a unique common fixed point of three mappings on a closed ball in an ordered complete dislocated metric space.

Theorem 2.7 Let $(X,⪯,d_l)$ be an ordered dislocated metric space, let S, T be self-mappings and let f be a dominated mapping on X such that $SX\cup TX\subset fX$, $Tx⪯fx$, $Sx⪯fx$, and let $x_0$ be an arbitrary point in X. Suppose that for $k\in [0,1)$ and for $S\ne T$, we have

$$d_l(Sx,Ty)\le k{d}_l(fx,fy)$$

(2.8)

for all comparable elements $fx,fy\in \overline{B(f{x}_0,r)}\subseteq fX$; and

$$d_l(f{x}_0,T{x}_0)\le (1-k)r.$$

(2.9)

If for a non-increasing sequence, $\{x_n\}\to u$ implies that $u⪯x_n$, and for any two points z and x in $\overline{B(f{x}_0,r)}$, there exists a point $y\in \overline{B(f{x}_0,r)}$ such that $y⪯z$ and $y⪯x$, that is, every pair of elements in $\overline{B(f{x}_0,r)}$ has a lower bound in $\overline{B(f{x}_0,r)}$; if fX is a complete subspace of X and $(S,f)$ and $(T,f)$ are weakly compatible, then S, T and f have a unique common fixed point fz in $\overline{B(f{x}_0,r)}$. Also, $d_l(fz,fz)=0$.

Proof By Lemma 1.16, there exists $E\subset X$ such that $fE=fX$ and $f:E\to X$ is one-to-one. Now, since $SX\cup TX\subset fX$, we define two mappings $g,h:fE\to fE$ by $g(fx)=Sx$ and $h(fx)=Tx$, respectively. Since f is one-to-one on E, then g, h are well defined. As $Sx⪯fx$ implies that $g(fx)⪯fx$ and $Tx⪯fx$ implies that $h(fx)⪯fx$, therefore g and h are dominated maps. Now $f{x}_0\in \overline{B(f{x}_0,r)}\subseteq fX$. Then $f{x}_0\in fX$. Let $y_0=f{x}_0$, choose a point $y_1$ in fX such that $y_1=h(y_0)$. As $h(y_0)⪯y_0$, so $y_1⪯y_0$ and let $y_2=g(y_1)$. Now $g(y_1)⪯y_1$ gives $y_2⪯y_1$. Continuing this process and having chosen $y_n$ in fX such that

$$y_{2i+1}=h(y_{2i})\text{and}{y}_{2i+2}=g(y_{2i+1}),\text{where }i=0,1,2,\dots ,$$

then $y_{n+1}⪯y_n$ for all $n\in N$. Following similar arguments of Theorem 2.1, $y_n\in \overline{B(f{x}_0,r)}$. Also, by inequality (2.9),

$$d_l(f{x}_0,h(f{x}_0))\le (1-k)r.$$

Note that for $fx,fy\in \overline{B(f{x}_0,r)}$, where fx, fy are comparable. Then by using inequality (2.8), we have

$$d_l(g(fx),h(fy))\le k{d}_l(fx,fy).$$

As fX is a complete space, all the conditions of Theorem 2.1 are satisfied, we deduce that there exists a unique common fixed point $fz\in \overline{B(f{x}_0,r)}$ of g and h. Also, $d_l(fz,fz)=0$. Now $fz=g(fz)=h(fz)$ or $fz=Sz=Tz=fz$. Thus fz is the point of coincidence of S, T and f. Let $v\in \overline{B(f{x}_0,r)}$ be another point of coincidence of f, S and T, then there exists $u\in \overline{B(f{x}_0,r)}$ such that $v=fu=Su=Tu$, which implies that $fu=g(fu)=h(fu)$, a contradiction as $fz\in \overline{B(f{x}_0,r)}$ is a unique common fixed point of g and h. Hence $v=fz$. Thus S, T and f have a unique point of coincidence $fz\in \overline{B(f{x}_0,r)}$. Now, since $(S,f)$ and $(T,f)$ are weakly compatible, by Lemma 1.17 fz is a unique common fixed point of S, T and f. □

In a similar way, we can apply our Theorems 2.5 and 2.6 to obtain a unique common fixed point of three mappings in an ordered complete dislocated metric space and a unique common fixed point of three mappings on a closed ball in a complete dislocated metric space, respectively.

In the following theorem, we use Theorem 2.6 to establish the existence of a unique common fixed point of four mappings on a closed ball in a complete dislocated metric space. One cannot prove the following theorem for an ordered dislocated metric space in a way similar to that of Theorem 2.7. In order to prove the unique common fixed point of four mappings on a closed ball in an ordered dislocated metric space, we should prove that S and T have no fixed point other than $x^{\ast }$ in Theorem 2.1.

Theorem 2.8 Let $(X,d_l)$ be a dislocated metric space and let S, T, g and f be self-mappings on X such that $SX,TX\subset fX=gX$. Assume that for $x_0$, an arbitrary point in X, and for $k\in [0,1)$ and for $S\ne T$, the following conditions hold:

$$d_l(Sx,Ty)\le k{d}_l(fx,gy)$$

(2.10)

for all elements $fx,gy\in \overline{B(f{x}_0,r)}\subseteq fX$; and

$$d_l(f{x}_0,S{x}_0)\le (1-k)r.$$

(2.11)

If fX is a complete subspace of X, then there exists $fz\in X$ such that $d_l(fz,fz)=0$. Also, if $(S,f)$ and $(T,g)$ are weakly compatible, then S, T, f and g have a unique common fixed point fz in $\overline{B(f{x}_0,r)}$.

Proof By Lemma 1.16, there exist $E_1,E_2\subset X$ such that $f{E}_1=fX=gX=g{E}_2$, $f:E_1\to X$, $g:E_2\to X$ are one-to-one. Now define the mappings $A,B:f{E}_1\to f{E}_1$ by $A(fx)=Sx$ and $B(gx)=Tx$, respectively. Since f, g are one-to-one on $E_1$ and $E_2$, respectively, then the mappings A, B are well defined. As fX is a complete space, all the conditions of Theorem 2.6 are satisfied, we deduce that there exists a unique common fixed point $fz\in \overline{B(f{x}_0,r)}$ of A and B. Further, A and B have no fixed point other than fz. Also, $d_l(fz,fz)=0$. Now $fz=A(fz)=B(fz)$ or $fz=Sz=fz$. Thus fz is a point of coincidence of f and S. Let $w\in \overline{B(f{x}_0,r)}$ be another point of coincidence of S and f, then there exists $u\in \overline{B(f{x}_0,r)}$ such that $w=fu=Su$, which implies that $fu=A(fu)$, a contradiction as $fz\in \overline{B(f{x}_0,r)}$ is a unique fixed point of A. Hence $w=fz$. Thus S and f have a unique point of coincidence $fz\in \overline{B(f{x}_0,r)}$. Since $(S,f)$ are weakly compatible, by Lemma 1.17 fz is a unique common fixed point of S and f. As $fX=gX$, then there exists $v\in X$ such that $fz=gv$. Now, as $A(fz)=B(fz)=fz\Rightarrow A(gv)=B(gv)=gv\Rightarrow Tv=gv$, thus gv is the point of coincidence of T and g. Now, if $Tx=gx\Rightarrow B(gx)=gx$, a contradiction. This implies that $gv=gx$. As $(T,g)$ are weakly compatible, we obtain gv, a unique common fixed point for T and g. But $gv=fz$. Thus S, T, g and f have a unique common fixed point $fz\in \overline{B(f{x}_0,r)}$. □

Corollary 2.9 Let $(X,⪯,d_l)$ be an ordered dislocated metric space, let S, T be self-mappings and let f be a dominated mapping on X such that $SX\cup TX\subset fX$, $Tx⪯fx$, $Sx⪯fx$, and let $x_0$ be an arbitrary point in X. Suppose that for $k\in [0,1)$ and for $S\ne T$, we have

$$d_l(Sx,Ty)\le k{d}_l(fx,fy)$$

for all comparable elements $fx,fy\in \overline{B(f{x}_0,r)}\subseteq fX$; and

$$d_l(f{x}_0,T{x}_0)\le (1-k)r.$$

If for a non-increasing sequence, $\{x_n\}\to u$ implies that $u⪯x_n$, and for any two points z and x in $\overline{B(f{x}_0,r)}$, there exists a point $y\in \overline{B(f{x}_0,r)}$ such that $y⪯z$ and $y⪯x$; if fX is a complete subspace of X, then S, T and f have a unique point of coincidence $fz\in \overline{B(f{x}_0,r)}$. Also, $d_l(fz,fz)=0$.

In a similar way, we can obtain a coincidence point result of four mappings as a corollary of Theorem 2.8.

A partial metric version of Theorem 2.1 is given below.

Theorem 2.10 Let $(X,⪯,p)$ be an ordered complete partial metric space, let $S,T:X\to X$ be dominated maps and let $x_0$ be an arbitrary point in X. Suppose that for $k\in [0,1)$ and for $S\ne T$,

$$p(Sx,Ty)\le kp(x,y)\mathit{\text{for all comparable elements}}x,y\mathit{\text{in}}\overline{B(x_0,r)}$$

and

$$p(x_0,S{x}_0)\le (1-k)[r+p(x_0,x_0)].$$

Then there exists $x^{\ast }\in \overline{B(x_0,r)}$ such that $p(x^{\ast },x^{\ast })=0$. Also, if for a non-increasing sequence $\{x_n\}$ in $\overline{B(x_0,r)}$, $\{x_n\}\to u$ implies that $u⪯x_n$, and for any two points x, y in $\overline{B(x_0,r)}$, there exists a point $z\in \overline{B(x_0,r)}$ such that $z⪯x$ and $z⪯y$, then there exists a unique point $x^{\ast }$ in $\overline{B(x_0,r)}$ such that $x^{\ast }=S{x}^{\ast }=T{x}^{\ast }$.

A partial metric version of Theorem 2.7 is given below.

Theorem 2.11 Let $(X,⪯,p)$ be an ordered partial metric space, let S, T be self-mappings and let f be a dominated mapping on X such that $SX\cup TX\subset fX$ and $Tx,Sx⪯fx$. Assume that for $x_0$, an arbitrary point in X, and for $k\in [0,1)$ and for $S\ne T$, the following conditions hold:

$$p(Sx,Ty)\le kp(fx,fy)$$

for all comparable elements $fx,fy\in \overline{B(f{x}_0,r)}\subseteq fX$; and

$$p(f{x}_0,T{x}_0)\le (1-k)[r+p(f{x}_0,f{x}_0)].$$

If for a non-increasing sequence, $\{x_n\}\to u$ implies that $u⪯x_n$, also for any two points z and x in $\overline{B(f{x}_0,r)}$, there exists a point $y\in \overline{B(f{x}_0,r)}$ such that $y⪯z$ and $y⪯x$; if fX is complete subspace of X and $(S,f)$ and $(T,f)$ are weakly compatible, then S, T and f have a unique common fixed point fz in $\overline{B(f{x}_0,r)}$. Also, $p(fz,fz)=0$.

Remark 2.12 We can obtain a partial metric version as well as a metric version of other theorems in a similar way.