**Theorem 2.1** *Let* (X,\u2aaf,{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)\phantom{\rule{1em}{0ex}}\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\u2aaf{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\u2aafx *and* z\u2aafy, *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}\u2aaf{x}_{0}, so {x}_{1}\u2aaf{x}_{0} and let {x}_{2}=T{x}_{1}. Now T{x}_{1}\u2aaf{x}_{1} gives {x}_{2}\u2aaf{x}_{1}. Continuing this process, we construct a sequence {x}_{n} of points in *X* such that

{x}_{2i+1}=S{x}_{2i},\phantom{\rule{2em}{0ex}}{x}_{2i+2}=T{x}_{2i+1}\phantom{\rule{1em}{0ex}}\text{and}\phantom{\rule{1em}{0ex}}{x}_{2i+1}=S{x}_{2i}\u2aaf{x}_{2i},\phantom{\rule{1em}{0ex}}\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}\u2aaf{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}\u2aaf{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})\phantom{\rule{1em}{0ex}}\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\phantom{\rule{1em}{0ex}}\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})\phantom{\rule{1em}{0ex}}\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})\phantom{\rule{1em}{0ex}}\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})\u27f60\phantom{\rule{1em}{0ex}}\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}\u2aaf{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}\u2aaf{x}^{\ast} and {z}_{0}\u2aafy. Choose a point {z}_{1} in *X* such that {z}_{1}=T{z}_{0}. As T{z}_{0}\u2aaf{z}_{0}, so {z}_{1}\u2aaf{z}_{0}and let {z}_{2}=S{z}_{1}. Now S{z}_{1}\u2aaf{z}_{1} gives {z}_{2}\u2aaf{z}_{1}. Continuing this process and having chosen {z}_{n} in *X* such that

{z}_{2i+1}=T{z}_{2i},\phantom{\rule{2em}{0ex}}{z}_{2i+2}=S{z}_{2i+1}\phantom{\rule{1em}{0ex}}\text{and}\phantom{\rule{1em}{0ex}}{z}_{2i+1}=T{z}_{2i}\u2aaf{z}_{2i},\phantom{\rule{1em}{0ex}}\text{where}i=0,1,2,\dots ,

we obtain that {z}_{n+1}\u2aaf{z}_{n}\u2aaf\cdots \u2aaf{z}_{0}\u2aaf{x}^{\ast}. As {z}_{0}\u2aaf{x}^{\ast} and {z}_{0}\u2aafy, it follows that {z}_{n}\u2aafT{x}^{\ast} and {z}_{n}\u2aafTy 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}\u2aaf{x}^{\ast} and {z}_{0}\u2aafy, it follows that {z}_{n}\u2aaf{T}^{n}{x}^{\ast}, {z}_{n}\u2aaf{S}^{n}{x}^{\ast}, {z}_{n}\u2aaf{S}^{n}y and {z}_{n}\u2aaf{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)\\ \vdots \\ \le & {k}^{n+1}{d}_{l}({x}^{\ast},{z}_{0})+{k}^{n+1}{d}_{l}({z}_{0},y)\u27f60\phantom{\rule{1em}{0ex}}\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})\u2aaf({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+y1\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+y1.\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{3{y}_{1}}{11}+\frac{4{x}_{2}}{15}+\frac{2{y}_{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\phantom{\rule{1em}{0ex}}\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)\phantom{\rule{1em}{0ex}}\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,\u2aaf,{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)\phantom{\rule{1em}{0ex}}\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\u2aaf{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,\u2aaf,{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)\phantom{\rule{1em}{0ex}}\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\u2aaf{x}_{n}, *and for any two points* *x*, *y* *in* *X*, *there exists a point* z\in X *such that* z\u2aafx *and* z\u2aafy, *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\u2aaf{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)\phantom{\rule{1em}{0ex}}\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,\u2aaf,{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\u2aaffx, Sx\u2aaffx, *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\u2aaf{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\u2aafz *and* y\u2aafx, *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\u2aaffx implies that g(fx)\u2aaffx and Tx\u2aaffx implies that h(fx)\u2aaffx, 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})\u2aaf{y}_{0}, so {y}_{1}\u2aaf{y}_{0} and let {y}_{2}=g({y}_{1}). Now g({y}_{1})\u2aaf{y}_{1} gives {y}_{2}\u2aaf{y}_{1}. Continuing this process and having chosen {y}_{n} in *fX* such that

{y}_{2i+1}=h({y}_{2i})\phantom{\rule{1em}{0ex}}\text{and}\phantom{\rule{1em}{0ex}}{y}_{2i+2}=g({y}_{2i+1}),\phantom{\rule{1em}{0ex}}\text{where}i=0,1,2,\dots ,

then {y}_{n+1}\u2aaf{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,\u2aaf,{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\u2aaffx, Sx\u2aaffx, *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\u2aaf{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\u2aafz *and* y\u2aafx; *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,\u2aaf,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)\phantom{\rule{1em}{0ex}}\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\u2aaf{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\u2aafx *and* z\u2aafy, *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,\u2aaf,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\u2aaffx. *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*:

*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\u2aaf{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\u2aafz *and* y\u2aafx; *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.