∎

11institutetext: Talat Nazir, 22institutetext: Department of Mathematical
Sciences, University of South Africa, Florida 0003, South Africa.
22email<EMAIL_ADDRESS>33institutetext: Sergei Silvestrov
44institutetext: Division of Mathematics and Physics, School of Education,
Culture and Communication, Mälardalen University, Box 883, 72123 Västerås,
Sweden.
44email<EMAIL_ADDRESS>
# Unique common fixed points of four generalized contractive mappings in
ordered partial metric spaces

Talat Nazir  Sergei Silvestrov

###### Abstract

The existence and uniqueness of the common fixed point for generalized
contractive mappings in order partial metric spaces is investigated. The
existence of nonnegative solution of implicit nonlinear integral equations is
also studied. Some examples demonstrating the validity of our main results are
constructed. The presented results extend and unify various comparable results
in the existing literature.
MSC 2020: 47H09, 47H10, 54C60, 54H25

###### Keywords:

Common fixed point, generalized contractive mapping, partially ordered set,
partial metric space

##  1 Introduction

Fixed point theory is a one of the powerful tool, noteworthy and stimulating
themes of nonlinear functional analysis that blends topology, analysis and
applied mathematics. In the fixed-point technique, the controllability problem
is converted to a fixed-point problem for an applicable nonlinear operator in
a function space. An essential part of this approach is to guarantee the
solvability of the equations for an invariant subset for this operator.

Alber and Guerre-Delabrere Alber  introduced the concept of weakly contractive
mappings and proved that weakly contractive mapping defined on a Hilbert space
is a Picard operator. Later, Rhoades Rhoades  proved that the corresponding
result is also valid when Hilbert space is replaced by a complete metric
space. Dutta et al. Dutta  generalized the weak contractive condition and
proved a fixed point theorem for a self map, which in turn generalizes
(Rhoades, , Theorem 1) and the corresponding result in Alber . The study of
common fixed points of mappings satisfying certain contractive conditions has
been at the center of rigorous research activity. The study of common fixed
point theory for sets of several single valued maps, started with the
assumption that all of the maps commuted. Sessa Sessa  generalized the concept
of commuting maps and introducing the weakly commuting maps. Then, Jungck
generalized this idea, first to compatible mappings Jungck1  and then to
weakly compatible mappings Jungck2 . There are examples that show that each of
these generalizations of commutativity is a proper extension of the previous
definition. On the other hand, Beg and Abbas Beg  obtained a common fixed
point theorem extending weak contractive condition for two maps. In this
direction, Zhang and Song Zhang  introduced the concept of a generalized
$\varphi$-weak contraction condition and obtained a common fixed point for two
maps. Doric Doric  proved a common fixed point theorem for generalized
$(\psi,\varphi)$-weak contractions. Abbas and Doric Abbas  obtained a common
fixed point theorem for four maps that satisfy contractive condition which is
more general than that given in Zhang .

In 2004, Ran and Reurings Ran  investigated the existence of fixed points in
partially ordered metric spaces, and then by Nieto and Lopez Neito . Further
results in this direction under weak contractive condition were proved (see
for example ATS ; Amini ; Ciric ; Harjani ; Hussain ; Nashine ; Petrusel ;
Radenovic ; Regan ).

In 2011, Abbas et al. ATS  presented some common fixed point theorems for
generalized $\left(\psi,\varphi\right)$-weakly contractive mappings in
partially ordered metric spaces. Further, Radenović and Kadelburg Radenovic
proved a result for generalized weak contractive mappings in partially ordered
metric spaces.

Partial metric space is a generalized metric space in which each object does
not necessarily have to have a zero distance from itself Matthews . A
motivation behind introducing the concept of a partial metric was to obtain
appropriate mathematical models in the theory of computation Aydi14 ; Heckmann
; Matthews2 ; Schellekens . Altun and Erduran Altun , Oltra and Valero Oltra
and Valero Valero  established some further generalizations of the results in
Matthews , and Romaguera Romaguera  proved a Caristi type fixed point theorem
on partial metric spaces. Karapinar Kara  proved some fixed point theorems for
weak $\varphi$-contraction on partial metric spaces in partially ordered sets.
Further results in the direction of partial metric space were proved in AT ;
ATR ; Aydi ; Bukatin ; Bukatin2 ; Romaguera1 ; WasfiSametAbbas .

It was shown that, in some cases, the results of fixed point in partial metric
spaces can be obtained directly from their induced metric counterparts HRS13 ;
JKS13 ; SVV13 . However, some conclusions important for the application of
partial metrics in information sciences cannot be obtained in this way. For
example, if $u$ is a fixed point of map $f$, then, by using the method from
HRS13 , we cannot conclude that $p(fu,fu)=0=p(u,u)$. For further details, we
refer the reader to NK13 ; NKR13 .

Our aim is to study the unique common fixed point results for four mappings
satisfying generalized contractive conditions in the setup of ordered partial
metric spaces.

In the sequel, $\mathbb{R},$ $\mathbb{R}_{\geq 0}$ and $\mathbb{Z}_{\geq 0}$
will denote the set of all real numbers, the set of all nonnegative real
numbers and the set of all non-negative integers, respectively. The usual
order on $\mathbb{R}$ (respectively, on $\mathbb{R}_{\geq 0})$ will be
indistinctly denoted by $\leq$ or $\geq$.

Consistent with Altun  and Matthews , the following definitions and results
will be needed in the sequel.

###### Definition 1

Let $X$ be a nonempty set. A function $p:X\times X\rightarrow\mathbb{R}_{\geq
0}$ is said to be a partial metric on $X$ if for any $x,y,z\in X,$ the
following conditions hold true:

  1. 1)

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

  2. 2)

$p(x,x)\leq p(x,y);$

  3. 3)

$p(x,y)=p(y,x);$

  4. 4)

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

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

If $p(x,y)=0$, then (${1}$) and (${2}$) imply that $x=y$. But the converse
does not always hold.

A trivial example of a partial metric space is the pair $(\mathbb{R}_{\geq
0},p)$, where the partial metric $p:\mathbb{R}_{\geq 0}\times\mathbb{R}_{\geq
0}\rightarrow\mathbb{R}_{\geq 0}$ is defined as $p(x,y)=\max\\{x,y\\}$.

###### Example 1

(Matthews ) If $X=\\{[a,b]:a,b\in\mathbb{R},a\leq b\\},$ then

$p([a,b],[c,d])=\max\\{b,d\\}-\min\\{a,c\\}$

defines a partial metric $p$ on $X$.

For some more examples of partial metric spaces, we refer to AT ; ATR ; Altun
; Bukatin2 ; Romaguera ; Schellekens .

Each partial metric $p$ on $X$ generates a $T_{0}$ topology $\tau_{p}$ on $X$
which has as a base the family open $p$-balls $\\{B_{p}(x,\varepsilon):x\in
X,\varepsilon>0\\},$ where $B_{p}(x,\varepsilon)=\\{y\in
X:p(x,y)<p(x,x)+\varepsilon\\},$ for all $x\in X$ and $\varepsilon>0$.

Observe (see (Matthews, , p. 187)) that a sequence $\\{x_{n}\\}$ in a partial
metric space $X$ converges to a point $x\in X$, with respect to $\tau_{p},$ if
and only if $p(x,x)=\lim\limits_{n\rightarrow\infty}p(x,x_{n})$.

If $p$ is a partial metric on $X$, then the function $p^{S}:X\times
X\rightarrow\mathbb{R}_{\geq 0}$ given by $p^{S}(x,y)=2p(x,y)-p(x,x)-p(y,y)$
defines a metric on $X$.

Furthermore, a sequence $\\{x_{n}\\}$ converges in $(X,p^{S})$ to a point
$x\in X$ if and only if

$\lim_{n,m\rightarrow\infty}p(x_{n},x_{m})=\lim_{n\rightarrow\infty}p(x_{n},x)=p(x,x).$

######  Definition 2 (Matthews )

A sequence $\\{x_{n}\\}$ in a partial metric space $X$ is said to be a Cauchy
sequence if $\lim\limits_{n,m\rightarrow\infty}p(x_{n},x_{m})$ exists and is
finite.

A partial metric space $X$ is said to be complete if every Cauchy sequence
$\\{x_{n}\\}$ in $X$ converges with respect to $\tau_{p}$ to a point $x\in X$
such that $\lim\limits_{n\rightarrow\infty}p(x,x_{n})=p(x,x)$. In this case,
we say that the partial metric $p$ is complete.

######  Lemma 1 (Altun ; Matthews )

Let $X$ be a partial metric space.

  1. (i)

A sequence $\\{x_{n}\\}$ in $X$ is a Cauchy sequence in $X$ if and only if it
is a Cauchy sequence in metric space $(X,p^{S})$.

  2. (ii)

A partial metric space $(X,p)$ is complete if and only if the metric space
$(X,p^{S})$ is complete.

Two self maps $f$ and $g$ on $X$ are said to be compatible if, whenever
$\\{x_{n}\\}$ in $X$ such that
$\lim\limits_{n\rightarrow\infty}p^{S}(fx_{n},x)=0$ and$\
\lim\limits_{n\rightarrow\infty}p^{S}(gx_{n},x)=0$ for some $x\in X,$ then
$\lim\limits_{n\rightarrow\infty}p^{S}(fgx_{n},gfx_{n})=0.$

If $fx=gx$ for some $x$ in $X$, then $x$ is called a coincidence point of $f$
and $g$. Furthermore, if the mappings are commuting on their coincidence
point, then such mappings are called weakly compatible, Jungck2 .

###### Definition 3

Let $X$ be a nonempty set. Then $(X,\preceq,p)$ is called an ordered partial
metric space if

  1. (i)

$p$ is a partial metric on $X,$

  2. (ii)

$\preceq$ is a partial order on $X$.

We say that the elements $x,y\in X$ are called comparable if either $x\preceq
y$ or $y\preceq x$ holds.

######  Definition 4 (ATS )

Let $(X,\preceq)$ be a partially ordered set and $f\ $and $g$ be two self-maps
of $X.$ Mapping $f$ is said to be dominated if $fx\preceq x$ for each $x$ in
$X$. A mapping $g$ is said to be dominating if $x\preceq gx$ for each $x$ in
$X.$

###### Example 2

Let $X=[0,1]$ be endowed with usual ordering. Let $f,g:X\rightarrow X$ defined
by $fx=\dfrac{x}{k}$ and $gx=kx$ for any positive real number $k\geq 1.$ It is
easy to see that $f$ is dominated and $g$ is a dominating map.

Zhang and Song Zhang  obtained the following common fixed point result in
metric spaces for a generalized $\varphi$-weak contraction.

######  Theorem 1.1 (Zhang )

Let $(X,d)$ be a complete metric space, and let $f,g:X\rightarrow X$ be two
self-mappings such that for all $x,y\in X,$ $d(fx,gy)\leq
M(x,y)-\varphi(M(x,y))$ holds, where $\varphi:[0,\infty)\rightarrow[0,\infty)$
is a lower semi-continuous function with $\varphi\left(t\right)>0$ for
$t\in\left(0,\infty\right)$, $\varphi(0)=0,$ and

$M(x,y)=\max\\{d(x,y),d(fx,x),d(gy,y),\dfrac{d(x,gy)+d(fx,y)}{2}\\}.$

Then there exists a unique point $u\in X$ such that $u=fu=gu$.

Aydi in Aydi  obtained the following result in partial metric spaces endowed
with a partial order.

###### Theorem 1.2

Let $(X,\leq_{X})$ be a partially ordered set and let $p$ be a partial metric
on $X$ such that $(X,p)$ is complete. Let $f:X\rightarrow X$ be a
nondecreasing map with respect to $\leq_{X}$. Suppose that the following
conditions hold for $y\leq_{X}x$:

  1. (i)

the inequality holds

$p(fx,fy)\leq p(x,y)-\varphi(p(x,y)),$

where $\varphi:[0,\infty)\rightarrow[0,\infty)$ is a continuous and non-
decreasing function such that it is positive in $(0,\infty)$, $\varphi(0)=0$
and $\underset{t\rightarrow\infty}{\lim}\varphi(t)=\infty;$

  2. (ii)

there exist $x_{0}\in X$ such that $x_{0}\leq_{X}fx_{0}$;

  3. (iii)

$f$ is continuous in $(X,p)$, or;
if a non-decreasing sequence $\\{x_{n}\\}$ converges to $x\in X$ , then
$x_{n}\leq_{X}x$ for all $n$.

Then $f$ has a fixed point $u\in X$. Moreover, $p(u,u)=0$.

######  Definition 5 (Doric )

The control functions $\psi$ and $\varphi$ are defined as

  1. 1)

$\psi:[0,\infty)\rightarrow[0,\infty)$ is a continuous nondecreasing function
with $\psi(t)=0$ if and only if $t=0$,

  2. 2)

$\varphi:[0,\infty)\rightarrow[0,\infty)$ is a lower semi-continuous function
with $\varphi(t)=0$ if and only if $t=0.$

A subset $W$ of a partially ordered set $X$ is said to be well ordered if
every two elements of $W$ are comparable.

Recently, Abbas et al. AT  obtained the following result in partial metric
spaces.

###### Theorem 1.3

Let $(X,\preceq)$ be a partially ordered set such that there exist a complete
partial metric $p$ on $X$ and $f$ a nondecreasing self map on $X$. Suppose
that for every two elements $x,y\in X$ with $y\preceq x,$ we have

$\displaystyle\psi(p(fx,fy))\leq\psi(M(x,y))-\phi(M(x,y)),$ $\displaystyle
M(x,y)=\max\\{p(x,y),p(fx,x),p(fy,y),\dfrac{p(x,fy)+p(y,fx)}{2}\\},$

where $\psi$ and $\phi$ are control functions. If there exists $x_{0}\in X$
with $x_{0}\preceq fx_{0}$ and one of the following two conditions is
satisfied:

  1. (i)

$f$ is continuous self map on $(X,p^{S})$,

  2. (ii)

for any nondecreasing sequence $\\{x_{n}\\}$ in $(X,\preceq)$ with
$\lim\limits_{n\rightarrow\infty}p^{S}(z,x_{n})=0$ it follows $x_{n}\preceq z$
for all $n\in\mathbb{Z}_{\geq 0},$

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.

##  2 Common Fixed Point Results

In this section, we obtain common fixed point theorems for four mappings
defined on an ordered partial metric space. We start with the following
result.

###### Theorem 2.1

Let $(X,\preceq,p)$ be an ordered complete partial metric space. Let $f,g,S$
and $T$ be self maps on $X$, $(f,g)$ be the pair of dominated and $(S,T)$ be
the pair of dominating maps with $f(X)\subseteq T(X)$ and $g(X)\subseteq
S(X)$. Suppose that, there exists control functions $\psi$ and $\varphi$ such
that, for every two comparable elements $x,y\in X$,

$\psi(p(fx,gy))\leq\psi(M_{p}(x,y))-\varphi(M_{p}(x,y)),$ (1)

is satisfied where

$M_{p}(x,y)=\max\\{p(Sx,Ty),p(fx,Sx),p(gy,Ty),\dfrac{p(Sx,gy)+p(fx,Ty)}{2}\\}.$

If for any nonincreasing sequence $\\{x_{n}\\}$ in $(X,\preceq)$ with
$x_{n}\preceq y_{n}$ for all $n$ and
$\lim\limits_{n\rightarrow\infty}p^{S}(x_{n},u)=0$ it holds that $u\preceq
y_{n}$ for all $n\in\mathbb{Z}_{\geq 0}$ and either of the following
conditions hold:

  1. (i)

$\\{f,S\\}$ are compatible, $f$ or $S$ is continuous on $(X,p^{S})$ and
$\\{g,T\\}$ are weakly compatible;

  2. (ii)

$\\{g,T\\}$ are compatible, $g$ or $T$ is continuous on $(X,p^{S})$ and
$\\{f,S\\}$ are weakly compatible,

then $f$,$g$,$S$ and $T$ have a common fixed point. Moreover, the set of
common fixed points of $f$, $g$, $S$ and $T$ is well ordered if and only if
$f$, $g$, $S$ and $T$ have one and only one common fixed point.

###### Proof

∎Let $x_{0}$ be an arbitrary point in $X.$ We construct sequences
$\\{x_{n}\\}$ and $\\{y_{n}\\}$ in $X$ such that $y_{2n+1}=fx_{2n}=Tx_{2n+1},$
and $y_{2n+2}=gx_{2n+1}=Sx_{2n+2}.$ By given assumptions, $x_{2n+2}\preceq
Sx_{2n+2}=gx_{2n+1}\preceq x_{2n+1},$ and $x_{2n+1}\preceq
Tx_{2n+1}=fx_{2n}\preceq x_{2n}.$ Thus, for all $n\in\mathbb{Z}_{\geq 0}$ we
have $x_{n+1}\preceq x_{n}.$ We suppose that $p(y_{2n},y_{2n+1})>0,$ for every
$n.$ If not, then $y_{2n}=y_{2n+1},$ for some $n$. Further, since $x_{2n}$ and
$x_{2n+1}$ are comparable, so from (1), we have

$\psi(p(y_{2n+1},y_{2n+2}))=\psi(p(fx_{2n},gx_{2n+1}))\\\
\leq\psi(M_{p}(x_{2n},x_{2n+1}))-\varphi(M_{p}(x_{2n},x_{2n+1})),$

where

$\displaystyle
M_{p}(x_{2n},x_{2n+1})=\max\\{p(Sx_{2n},Tx_{2n+1}),p(fx_{2n},Sx_{2n}),p(gx_{2n+1},Tx_{2n+1}),$
$\displaystyle\hskip
113.81102pt\dfrac{p(Sx_{2n},gx_{2n+1})+p(fx_{2n},Tx_{2n+1})}{2}\\}$
$\displaystyle=$
$\displaystyle\max\\{p(y_{2n},y_{2n+1}),p(y_{2n+1},y_{2n}),p(y_{2n+2},y_{2n+1}),$
$\displaystyle\hskip
85.35826pt\dfrac{p(y_{2n},y_{2n+2})+p(y_{2n+1},y_{2n+1})}{2}\\}$
$\displaystyle=$ $\displaystyle p(y_{2n+1},y_{2n+2}).$

Hence,
$\psi(p(y_{2n+1},y_{2n+2}))\leq\psi(p(y_{2n+1},y_{2n+2}))-\varphi(p(y_{2n+1},y_{2n+2}))$
implies that $\varphi(p(y_{2n+1},y_{2n+2}))=0.$ As, $\varphi(t)=0$ if and only
if $t=0$, it follows that $y_{2n+1}=y_{2n+2}.$ Following the similar
arguments, we get $y_{2n+2}=y_{2n+3}$ and so on. Thus $y_{2n}$ is the common
fixed point of $f,$ $g,$ $S$ and $T$ as $\\{y_{n}\\}$ became a constant
sequence in $X$.

Taking $p(y_{2n},y_{2n+1})>0$ for each $n.$ Since $x_{2n}$ and $x_{2n+1}$ are
comparable, from (1), we obtain that

$\psi(p(y_{2n+2},y_{2n+1}))=\psi(p(y_{2n+1},y_{2n+2}))=\psi(p(fx_{2n},gx_{2n+1}))$
$\leq\psi(M_{p}(x_{2n},x_{2n+1}))-\varphi(M_{p}(x_{2n},x_{2n+1})),$ (2)

where

$\displaystyle
M_{p}(x_{2n},x_{2n+1})=\max\\{p(Sx_{2n},Tx_{2n+1}),p(fx_{2n},Sx_{2n}),$
$\displaystyle\hskip
56.9055ptp(gx_{2n+1},Tx_{2n+1}),\dfrac{p(Sx_{2n},gx_{2n+1})+p(fx_{2n},Tx_{2n+1})}{2}\\}$
$\displaystyle=$ $\displaystyle\max\\{p(y_{2n},y_{2n+1}),p(y_{2n+1},y_{2n}),$
$\displaystyle\hskip
56.9055ptp(y_{2n+2},y_{2n+1}),\dfrac{p(y_{2n},y_{2n+2})+p(y_{2n+1},y_{2n+1})}{2}\\}$
$\displaystyle\leq$
$\displaystyle\max\\{p(y_{2n+1},y_{2n}),p(y_{2n+2},y_{2n+1}),\dfrac{p(y_{2n},y_{2n+1})+p(y_{2n+1},y_{2n+2})}{2}\\}$
$\displaystyle=$
$\displaystyle\max\\{p(y_{2n+1},y_{2n}),p(y_{2n+2},y_{2n+1})\\}.$

If $\max\\{p(y_{2n+1},y_{2n}),p(y_{2n+2},y_{2n+1})\\}=p(y_{2n+2},y_{2n+1}),$
then

$M_{p}(x_{2n},x_{2n+1})\leq p(y_{2n+2},y_{2n+1}).$

But $M_{p}(x_{2n},x_{2n+1})\geq p(y_{2n+2},y_{2n+1}),$ and so

$M_{p}(x_{2n},x_{2n+1})=p(y_{2n+2},y_{2n+1}),$

and (2) give

$\displaystyle\psi(p(y_{2n+2},y_{2n+1}))$ $\displaystyle\leq$
$\displaystyle\psi(M_{p}(x_{2n},x_{2n+1}))-\varphi(M_{p}(x_{2n},x_{2n+1}))$
$\displaystyle=$
$\displaystyle\psi(p(y_{2n+2},y_{2n+1}))-\varphi(p(y_{2n+2},y_{2n+1})),$

a contradiction. Hence $p(y_{2n+2},y_{2n+1})\leq p(y_{2n+1},y_{2n})$. Moreover
$M_{p}(x_{2n},x_{2n+1})\leq p(y_{2n},y_{2n+1}).$ But, since
$M_{p}(x_{2n},x_{2n+1})\geq p(y_{2n},y_{2n+1}),$ it follows that

$M_{p}(x_{2n},x_{2n+1})=p(y_{2n},y_{2n+1}).$

Similarly, $p(y_{2n+3},y_{2n+2})\leq p(y_{2n+2},y_{2n+1}).$ Thus, the sequence
$\\{p(y_{2n+1},y_{2n})\\}$ is nonincreasing. Hence, there exists $c\geq 0$
such that $\lim\limits_{n\rightarrow\infty}p(y_{2n+1},y_{2n})=c.$ Suppose that
$c>0.$ Then,
$\psi(p(y_{2n+2},y_{2n+1}))\leq\psi(M_{p}(x_{2n+1},x_{2n}))-\varphi(M_{p}(x_{2n+1},y_{2n})),$
and by lower semicontinuity of $\varphi,$ we have

$\underset{n\rightarrow\infty}{\lim\sup}\
\psi(p(y_{2n+2},y_{2n+1}))\leq\underset{n\rightarrow\infty}{\lim\sup}\
\psi(p(y_{2n},y_{2n+1}))-\underset{n\rightarrow\infty}{\lim\inf}\
\varphi(p(y_{2n},y_{2n+1})),$

which implies that $\psi(c)\leq\psi(c)-\varphi(c),$ a contradiction. Therefore
$c=0.$ So we conclude that

$\lim_{n\rightarrow\infty}p(y_{2n+1},y_{2n})=0.$ (3)

Now, we show that $\lim\limits_{n,m\rightarrow\infty}p(y_{2n},y_{2m})=0$. If
not, there is $\varepsilon>0,$ and there exist even integers $2n_{k}$ and
$2m_{k}$ with $2m_{k}>2n_{k}>k$ such that

$p(y_{2m_{k}},y_{2n_{k}})\geq\varepsilon,$ (4)

and $p(y_{2m_{k}-2},y_{2n_{k}})<\varepsilon.$ Since

$\displaystyle\varepsilon$ $\displaystyle\leq$ $\displaystyle
p(y_{2m_{k}},y_{2n_{k}})$ $\displaystyle\leq$ $\displaystyle
p(y_{2n_{k}},y_{2m_{k}-2})+p(y_{2m_{k}-2},y_{2m_{k}})-p(y_{2m_{k}-2},y_{2m_{k}-2})$
$\displaystyle\leq$ $\displaystyle
p(y_{2n_{k}},y_{2m_{k}-2})+p(y_{2m_{k}-2},y_{2m_{k}-1})+p(y_{2m_{k}-1},y_{2m_{k}})$
$\displaystyle-p(y_{2m_{k}-1},y_{2m_{k}-1})-p(y_{2m_{k}-2},y_{2m_{k}-2}),$

from (3) and (4), we have

$\lim_{k\rightarrow\infty}p(y_{2m_{k}},y_{2n_{k}})=\varepsilon.$ (5)

Also (5) and inequality

$p(y_{2m_{k}},y_{2n_{k}})\leq
p(y_{2m_{k}},y_{2m_{k}-1})+p(y_{2m_{k}-1},y_{2n_{k}})-p(y_{2m_{k}-1},y_{2m_{k}-1})$

give that
$\varepsilon\leq\lim\limits_{k\rightarrow\infty}p(y_{2m_{k}-1},y_{2n_{k}}),$
while inequality

$p(y_{2m_{k}-1},y_{2n_{k}})\leq
p(y_{2m_{k}-1},y_{2m_{k}})+p(y_{2m_{k}},y_{2n_{k}})-p(y_{2m_{k}},y_{2m_{k}})$

yields
$\lim\limits_{k\rightarrow\infty}p(y_{2m_{k}-1},y_{2n_{k}})\leq\varepsilon,$
and hence

$\lim_{k\rightarrow\infty}p(y_{2m_{k}-1},y_{2n_{k}})=\varepsilon.$ (6)

Now (6) and inequality

$p(y_{2m_{k}-1},y_{2n_{k}})\leq
p(y_{2m_{k}-1},y_{2n_{k}+1})+p(y_{2n_{k}+1},y_{2n_{k}})-p(y_{2n_{k}+1},y_{2n_{k}+1})$

give
$\varepsilon\leq\lim\limits_{k\rightarrow\infty}p(y_{2m_{k}-1},y_{2n_{k}+1}),$
while inequality

$p(y_{2m_{k}-1},y_{2n_{k}+1})\leq
p(y_{2m_{k}-1},y_{2n_{k}})+p(y_{2n_{k}},y_{2n_{k}+1})-p(y_{2n_{k}},y_{2n_{k}})$

yields
$\lim\limits_{k\rightarrow\infty}p(y_{2m_{k}-1},y_{2n_{k}+1})\leq\varepsilon,$
and so

$\lim_{k\rightarrow\infty}p(y_{2m_{k}-1},y_{2n_{k}+1})=\varepsilon.$

As

$\displaystyle
M_{p}(x_{2n_{k}},x_{2m_{k}-1})=\max\\{p(Sx_{2n_{k}},Tx_{2m_{k}-1}),p(fx_{2n_{k}},Sx_{2n_{k}}),$
$\displaystyle\hskip
28.45274ptp(gx_{2m_{k}-1},Tx_{2m_{k}-1}),\frac{p(Sx_{2n_{k}},gx_{2m_{k}-1})+p(fx_{2n_{k}},Tx_{2m_{k}-1})}{2}\\}$
$\displaystyle=\max\\{p(y_{2n_{k}},y_{2m_{k}-1}),p(y_{2n_{k}+1},y_{2n_{k}}),$
$\displaystyle\hskip
28.45274ptp(y_{2m_{k}},y_{2m_{k}-1}),\frac{p(y_{2n_{k}},y_{2m_{k}})+p(y_{2n_{k}+1},y_{2m_{k}-1})}{2}\\}.$

So,
$\lim\limits_{k\rightarrow\infty}M_{p}(x_{2n_{k}},x_{2m_{k}-1})=\max\\{\varepsilon,0,0,\varepsilon\\}=\varepsilon.$
From (1), we obtain

$\psi(p(y_{2n_{k}+1},y_{2m_{k}}))=\psi(p(fx_{2n_{k}},gx_{2m_{k}-1}))\\\
\leq\psi(M_{p}(x_{2n_{k}},x_{2m_{k}-1}))-\varphi(M_{p}(x_{2n_{k}},x_{2m_{k}-1})).$

Taking upper limit as $k\rightarrow\infty$ implies that
$\psi(\varepsilon)\leq\psi(\varepsilon)-\varphi(\varepsilon),$ a contradiction
as $\varepsilon>0$. Thus, we obtain
$\lim\limits_{n,m\rightarrow\infty}p(y_{2n},y_{2m})=0,$ and it follows that
$\\{y_{2n}\\}$ is a Cauchy sequence in $(X,p),$ and hence Cauchy in
$(X,p^{S})$ by Lemma 1. Since $(X,p)$ is complete, it follows from Lemma 1,
$(X,p^{S})$ is also complete, so the sequence $\\{y_{2n}\\}$ is convergent in
the metric space $(X,p^{S}).$ Therefore, there exists a point $z$ in $X$ such
that $\lim\limits_{n\rightarrow\infty}p^{S}(y_{2n},z)=0.$ Hence,

$\displaystyle\lim_{n\rightarrow\infty}y_{2n+1}=\lim_{n\rightarrow\infty}Tx_{2n+1}=\lim_{n\rightarrow\infty}fx_{2n}=z,$
$\displaystyle\lim_{n\rightarrow\infty}y_{2n+2}=\lim_{n\rightarrow\infty}Sx_{2n+2}=\lim_{n\rightarrow\infty}gx_{2n+1}=z.$

Equivalently, we have
$\lim\limits_{n,m\rightarrow\infty}p(y_{2n},y_{2m})=\lim\limits_{n\rightarrow\infty}p(y_{2n},z)=p(z,z).$

Assume that $S$ is continuous on $(X,p^{S})$. Then
$\lim\limits_{n\rightarrow\infty}SSx_{2n+2}=\lim\limits_{n\rightarrow\infty}Sfx_{2n+2}=Sz.$
Also, since $\\{f,S\\}$ are compatible, we have

$\lim_{n\rightarrow\infty}fSx_{2n+2}=\lim_{n\rightarrow\infty}Sfx_{2n+2}=Sz.$

As, $Sx_{2n+2}=gx_{2n+1}\preceq x_{2n+1},$ so from (1), we have

$\psi(p(fSx_{2n+2},gx_{2n+1}))\leq\psi\left(M_{p}(Sx_{2n+2},x_{2n+1})\right)-\varphi(M_{p}(Sx_{2n+2},x_{2n+1})),$
(7)

where

$M_{p}(Sx_{2n+2},x_{2n+1})=\max\\{p(SSx_{2n+2},Tx_{2n+1}),p(fSx_{2n+2},SSx_{2n+2}),\\\
p(gx_{2n+1},Tx_{2n+1}),\frac{p(SSx_{2n+2},gx_{2n+1})+p(fSx_{2n+2},Tx_{2n+1})}{2}\\}.$

Now we show that
$\lim\limits_{n\rightarrow\infty}p(fSx_{2n+2},gx_{2n+1})=p(Sz,z).$ Indeed,

$p^{S}(fSx_{2n+2},gx_{2n+1})=\\\
2p(fSx_{2n+2},gx_{2n+1})-p(fSx_{2n+2},fSx_{2n+2})-p(gx_{2n+1},gx_{2n+1}),$

implies

$p(fSx_{2n+2},fSx_{2n+2})+p(gx_{2n+1},gx_{2n+1})+p^{S}(fSx_{2n+2},gx_{2n+1})\\\
=2p(fSx_{2n+2},gx_{2n+1}),$

which on taking limit as $n\rightarrow\infty,$ implies that

$p(Sz,Sz)+p(z,z)+p^{S}(Sz,z)=2\lim_{n\rightarrow\infty}p(fSx_{2n+2},gx_{2n+1}).$

This further implies that

$p(Sz,Sz)+p(z,z)+[2p(Sz,z)-p(Sz,Sz)-p(z,z)]=2\lim_{n\rightarrow\infty}p(fSx_{2n+2},gx_{2n+1}),$

that is,

$p(Sz,z)=\lim_{n\rightarrow\infty}p(fSx_{2n+2},gx_{2n+1}).$

From (7), on taking upper limit as $n\rightarrow\infty,$ we obtain

$\psi(p(Sz,z))\leq\psi\left(p(Sz,z)\right)-\varphi(p(Sz,z)),$

and $Sz=z.$

Now, as $gx_{2n+1}\preceq x_{2n+1}$ and $gx_{2n+1}\rightarrow z$ as
$n\rightarrow\infty,$ it follows that $z\preceq x_{2n+1}$. Hence from (1), we
have

$\psi(p(fz,gx_{2n+1}))\leq\psi\left(M_{p}(z,x_{2n+1})\right)-\varphi(M_{p}(z,x_{2n+1})),$

where

$\displaystyle M_{p}(z,x_{2n+1})$
$\displaystyle=\max\\{p(Sz,Tx_{2n+1}),p(fz,Sz),p(gx_{2n+1},Tx_{2n+1}),$
$\displaystyle\hskip 56.9055pt\frac{p(Sz,gx_{2n+1})+p(fz,Tx_{2n+1})}{2}\\}$
$\displaystyle=\max\\{p(z,Tx_{2n+1}),p(fz,z),p(gx_{2n+1},Tx_{2n+1}),$
$\displaystyle\hskip 56.9055pt\frac{p(z,gx_{2n+1})+p(fz,Tx_{2n+1})}{2}\\}.$

On taking upper limit as $n\rightarrow\infty,$ we have
$\psi(p(fz,z))\leq\psi\left(p(fz,z)\right)-\varphi(p(fz,z)),$ and $fz=z.$

Since $f(X)\subseteq T(X),$ there exists a point $w\in X$ such that $fz=Tw.$
Suppose that $gw\neq Tw.$ Since $w\preceq Tw=fz\preceq z\ $implies $w\preceq
z.$ From (1), we obtain

$\psi(p(Tw,gw))=\psi(p(fz,gw))\leq\psi\left(M_{p}(z,w)\right)-\varphi(M_{p}(z,w)),$
(8) $\displaystyle\text{where }M_{p}(z,w)$ $\displaystyle=$
$\displaystyle\max\\{p(Sz,Tw),p(fz,Sz),p(gw,Tw),\frac{p(Sz,gw)+p(fz,Tw)}{2}\\}$
$\displaystyle=$
$\displaystyle\max\\{p(z,z),p(z,z),p(gw,Tw),\frac{p(Tw,gw)+p(Tw,Tw)}{2}\\}$
$\displaystyle=$ $\displaystyle p(Tw,gw).$

Now (8) becomes
$\psi(p(Tw,gw))\leq\psi\left(p(Tw,gw)\right)-\varphi(p(Tw,gw)),$ a
contradiction. Hence, $Tw=gw.$ Since $g$ and $T$ are weakly compatible,
$gz=gfz=gTw=Tgw=Tfz=Tz.$ Thus $z$ is a coincidence point of $g$ and $T.$

Now, $fx_{2n}\preceq x_{2n}$ and $x_{2n}\rightarrow z$ as
$n\rightarrow\infty,$ imply that $z\preceq fx_{2n}.$ Hence from (1), we get
$\psi(p(fx_{2n},gz))\leq\psi\left(M_{p}(x_{2n},z)\right)-\varphi(M_{p}(x_{2n},z)),$
where

$\displaystyle M_{p}(x_{2n},z)$ $\displaystyle=$
$\displaystyle\max\\{p(Sx_{2n},Tz),p(fx_{2n},Sx_{2n}),p(gz,Tz),\frac{p(Sx_{2n},gz)+p(fx_{2n},Tz)}{2}\\}$
$\displaystyle=$
$\displaystyle\max\\{p(Sx_{2n},gz),p(fx_{2n},Sx_{2n}),p(gz,gz),\frac{p(Sx_{2n},gz)+p(fx_{2n},gz)}{2}\\}$
$\displaystyle=$ $\displaystyle p(z,gz)\text{ as }n\rightarrow\infty.$

On taking upper limit as $n\rightarrow\infty,$ we have
$\psi(p(z,gz))\leq\psi\left(p(z,gz)\right)-\varphi(p(z,gz)),$ and $z=gz.$
Therefore $fz=gz=Sz=Tz=z.$ The proof is similar when $f$ is continuous.

Similarly, the result follows when (ii) holds.

Now, suppose that the set of common fixed points of $f$, $g$, $S$ and $T$ is
well ordered. We are to show that the common fixed point of $f$, $g$, $S$ and
$T$ is unique. Suppose that $u$ and $v$ be two fixed points of $f,$ $g,$ $S$
and $T$ i.e., $fu=gu=Su=Tu=u$ and $fv=gv=Sv=Tv=v$ with $u\neq v.$ Then from
(1), we have

$\psi(p(u,v))=\psi(p(fu,gv))\leq\psi(M_{p}(u,v))-\varphi(M_{p}(u,v)),$

where

$\displaystyle\text{ }M_{p}(u,v)$ $\displaystyle=$
$\displaystyle\max\\{p(Su,Tv),p(fu,Su),p(gv,Tv),\dfrac{p(Su,gv)+p(fu,Tv)}{2}\\}$
$\displaystyle=$
$\displaystyle\max\\{p(u,v),p(u,u),p(v,v),\dfrac{p(u,v)+p(u,v)}{2}\\}$
$\displaystyle=$ $\displaystyle p(u,v).$

Thus $\psi(p(u,v))\leq\psi(p(u,v))-\varphi(p(u,v)),$ a contradiction. Hence
$u=v$. Conversely, if $f,$ $g,$ $S$ and $T$ have only one common fixed point
then the set of common fixed point of $f,$ $g,$ $S\ $and $T$ is well ordered
being singleton. ∎

###### Example 3

Let $X=[0,k]$ for a real number $k\geq 9/10$ endowed with usual order $\leq.$
Let $p:X\times X\rightarrow\mathbb{R}_{\geq 0}$ be defined by
$p(x,y)=\left|x-y\right|$ if $x,y\in[0,1),\,$and $p(x,y)=\max\\{x,y\\}$
otherwise. It is easily seen that $(X,p)$ is a complete partial metric space
AT . Consider $\psi(t)=\left\\{\begin{array}[]{l}3t,\text{ if }0\leq
t\leq\frac{1}{3}\\\ 1,\text{ \ if }x\in(\frac{1}{3},1]\\\ t,\text{ \
otherwise}\end{array}\right.$ and
$\varphi(t)=\left\\{\begin{array}[]{l}0,\text{ \ if }t=0\\\ \frac{t}{3},\text{
\ if }0<t\leq\frac{1}{3}\\\ \frac{1}{9},\text{ \
otherwise}\end{array}\right..$ Define the self mappings $f$, $g$, $S$ and $T$
on $X$ by

$\begin{array}[]{ll}f(x)=\left\\{\begin{array}[]{l}\frac{1}{6}x,\text{ if
}x\leq\frac{1}{3}\\\ \frac{1}{18},\text{ if
}x\in(\frac{1}{3},k]\end{array}\right.,&\text{
}gx=\left\\{\begin{array}[]{l}0,\text{ if }x\leq\frac{1}{3}\\\
\frac{1}{3},\text{ if }x\in(\frac{1}{3},k]\end{array}\right.,\\\
T(x)=\left\\{\begin{array}[]{l}0,\text{ \ if }x=0\\\ x,\text{ \ if
}x\in(0,\frac{1}{3}]\\\ k,\text{ \ if
}x\in(\frac{1}{3},k]\end{array}\right.,&\text{
}Sx=\left\\{\begin{array}[]{l}0,\text{ if }x=0\\\ \frac{1}{3},\text{ if
}x\in(0,\frac{1}{3}]\\\ k,\text{ if
}x\in(\frac{1}{3},k]\end{array}\right..\end{array}$

Then $f(X)\subseteq T(X)$ and $g(X)\subseteq S(X)$ with $f$ and $g$ are
dominated and $S$ and $T$ are dominating mappings as

$\text{for each }x\text{ in }X$ | $fx\leq x$ | $gx\leq x$ | $x\leq Sx$ | $x\leq Tx$
---|---|---|---|---
$x=0$ | $f\left(0\right)=0$ | $g\left(0\right)=0$ | $0=S(0)$ | $0=T(0)$
$x\in(0,\frac{1}{3}]$ | $fx=\frac{1}{6}x\leq x$ | $gx=0<x$ | $x\leq\frac{1}{3}=S(x)$ | $x=T(x)$
$x\in(\frac{1}{3},k]$ | $fx=\frac{1}{18}<x$ | $gx=\frac{1}{3}<x$ | $x\leq k=S(x)$ | $x\leq k=T(x)$

Also note that $\\{f,S\\}$ are compatible, $\\{g,T\\}$ are weakly compatible
with $f$ is a continuous map.

To show that $f,$ $g,$ $S$ and $T$ satisfy (1) for all $x,y\in X,$ we consider
the following cases:

  1. (i)

If $x=0\ $and $y\in[0,\frac{1}{3}],$ then $p(fx,gy)=0$ and (1) is satisfied.

  2. (ii)

For $x=0\ $and $y\in(\frac{1}{3},k],$ we have

$\displaystyle\psi(p(fx,gy))$ $\displaystyle=$
$\displaystyle\psi(p(0,\frac{1}{3}))=\psi(\frac{1}{3})$ $\displaystyle=$
$\displaystyle 1<k-\frac{1}{9}=\psi(k)-\varphi(k)$ $\displaystyle=$
$\displaystyle\psi(p(0,k))-\varphi(p(0,k))$ $\displaystyle=$
$\displaystyle\psi(p(Sx,Ty))-\varphi(p(Sx,Ty))$ $\displaystyle=$
$\displaystyle\psi(M_{p}(x,y))-\varphi(M_{p}(x,y)).$

  3. (iii)

When $x=(0,\frac{1}{3}]\ $and $y\in[0,\frac{1}{3}],$ then

$\displaystyle\psi(p(fx,gy))$ $\displaystyle=$
$\displaystyle\psi(p(\frac{1}{6}x,0))=\psi(\frac{1}{6}x)=\frac{1}{2}x$
$\displaystyle\leq$ $\displaystyle
3\max\\{(\frac{1}{3}-\frac{1}{6}x),y\\}-\frac{1}{3}\max\\{(\frac{1}{3}-\frac{1}{6}x),y\\}$
$\displaystyle=$
$\displaystyle\psi(\max\\{(\frac{1}{3}-\frac{1}{6}x),y\\})-\varphi(\max\\{(\frac{1}{3}-\frac{1}{6}x),y\\})$
$\displaystyle=$
$\displaystyle\psi(\max\\{p(fx,Sx),p(gy,Ty)\\})-\varphi(\max\\{p(fx,Sx),p(gy,Ty)\\})$
$\displaystyle=$ $\displaystyle\psi(M_{p}(x,y))-\varphi(M_{p}(x,y)).$

  4. (iv)

If $x=(0,\frac{1}{3}]\ $and $y\in(\frac{1}{3},k],$ then

$\displaystyle\psi(p(fx,gy))$ $\displaystyle=$
$\displaystyle\psi(p(\frac{1}{6}x,\frac{1}{3}))=\psi(\frac{1}{3}(1-\frac{x}{2}))=1-\frac{1}{2}x$
$\displaystyle<$ $\displaystyle
k-\frac{1}{9}=\psi(\max\\{\frac{1}{3},k\\})-\varphi(\max\\{\frac{1}{3},k\\})$
$\displaystyle=$ $\displaystyle\psi(p(gy,Ty))-\varphi(p(gy,Ty))$
$\displaystyle=$ $\displaystyle\psi(M_{p}(x,y))-\varphi(M_{p}(x,y)).$

  5. (v)

For $x\in(\frac{1}{3},k]\ $and $y\in[0,\frac{1}{3}],$ we obtain

$\displaystyle\psi(p(fx,gy))$ $\displaystyle=$
$\displaystyle\psi(p(\frac{1}{18},0))=\psi(\frac{1}{18})=\frac{1}{6}<k-\frac{1}{9}$
$\displaystyle=$
$\displaystyle\psi(\max\\{\frac{1}{18},k\\})-\varphi(\max\\{\frac{1}{18},k\\})$
$\displaystyle=$ $\displaystyle\psi(p(fx,Sx))-\varphi(p(fx,Sx))$
$\displaystyle=$ $\displaystyle\psi(M_{p}(x,y))-\varphi(M_{p}(x,y)).$

  6. (vi)

Finally, when $x,y\in(\frac{1}{3},k],$ then we have

$\displaystyle\psi(p(fx,gy))$ $\displaystyle=$
$\displaystyle\psi(p(\frac{1}{18},\frac{1}{3}))=\psi(\frac{5}{18})=\frac{5}{6}<k-\frac{1}{9}$
$\displaystyle=$
$\displaystyle\psi(\max\\{\frac{1}{3},k\\})-\varphi(\max\\{\frac{1}{3},k\\})$
$\displaystyle=$ $\displaystyle\psi(p(gy,Ty))-\varphi(p(gy,Ty))$
$\displaystyle=$ $\displaystyle\psi(M_{p}(x,y))-\varphi(M_{p}(x,y)).$

The mappings $f,g,S$ and $T$ satisfy (1). Thus all the conditions given in
Theorem 2.1 are satisfied. Moreover, $0$ is the unique common fixed point of
$f,$ $g,$ $S$ and $T$.

###### Corollary 1

Let $(X,$$\preceq,p)$ be an ordered complete partial metric space. Let $f,g,S$
and $T$ be self maps on $X$, $(f,g)$ be the pair of dominated and $(S,T)$ be
the pair of dominating maps with $f\left(X\right)\subseteq T\left(X\right)$
and $g\left(X\right)\subseteq S\left(X\right)$. Suppose that, there exists
control functions $\psi$ and $\varphi$ such that for every two comparable
elements $x,y\in X,$

$\psi(p(fx,gy))\leq\psi(p(Sx,Ty))-\varphi(p(Sx,Ty))$

is satisfied.

If for any nonincreasing sequence $\\{x_{n}\\}$ in $(X,\preceq)$ with
$x_{n}\preceq y_{n}$ for all $n$ and
$\lim\limits_{n\rightarrow\infty}p^{S}(x_{n},u)=0$ it holds that $u\preceq
y_{n}$ for all $n\in\mathbb{Z}_{\geq 0}$, and either of the following
conditions hold:

  1. 1)

$\\{f,S\\}$ are compatible, $f$ or $S$ is continuous on $(X,p^{S})$ and
$\\{g,T\\}$ are weakly compatible,

  2. 2)

$\\{g,T\\}$ are compatible, $g$ or $T$ is continuous on $(X,p^{S})$ and
$\\{f,S\\}$ are weakly compatible,

then $f,g,S$ and $T$ have a common fixed point. Moreover, the set of common
fixed points of $f$, $g$, $S$ and $T$ is well ordered if and only if $f$, $g$,
$S$ and $T$ have one and only one common fixed point.

Consistent with the terminology in Radenovic , we denote $\Upsilon$ the set of
all functions $\phi:\mathbb{R}_{\geq 0}\rightarrow\mathbb{R}_{\geq 0},$ where
$\phi$ is a Lebesgue integrable mapping with finite integral on each compact
subset of $\mathbb{R}_{\geq 0},$ nonnegative, and for each $\varepsilon>0,$
$\int_{0}^{\varepsilon}\phi(t)dt>0$ (see also, Branciari ). As a consequence
of Theorem 2.1, we obtain following fixed point result for a mapping
satisfying contractive conditions of integral type in a complete partial
metric space $X.$

###### Corollary 2

Let $(X,\preceq,p)$ be an ordered complete partial metric space. Let $f,g,S$
and $T$ be self maps on $X$, $(f,g)$ be the pair of dominated and $(S,T)$ be
the pair of dominating maps with $f\left(X\right)\subseteq T\left(X\right)$
and $g\left(X\right)\subseteq S\left(X\right)$. Suppose that, there exists
control functions $\psi$ and $\varphi$ such that for every two comparable
elements $x,y\in X,$

$\int_{0}^{\psi(p(fx,gy))}\phi(t)dt\leq\int\limits_{0}^{\psi(M_{p}(x,y))}\phi(t)dt-\int\limits_{0}^{\varphi(M_{p}(x,y))}\phi(t)dt,$
(9)

is satisfied, where $\phi\in\Upsilon$ and

$M_{p}(x,y)=\max\\{p(Sx,Ty),p(fx,Sx),p(gy,Ty),\dfrac{p(Sx,gy)+p(fx,Ty)}{2}\\}.$

If for any nonincreasing sequence $\\{x_{n}\\}$ in $(X,\preceq)$ with
$x_{n}\preceq y_{n}$ for all $n$ and
$\lim\limits_{n\rightarrow\infty}p^{S}(x_{n},u)=0$, it holds that $u\preceq
y_{n}$ for all $n\in\mathbb{Z}_{\geq 0}$, and either of the following
conditions hold:

  1. 1)

$\\{f,S\\}$ are compatible, $f$ or $S$ is continuous on $(X,p^{S})$ and
$\\{g,T\\}$ are weakly compatible

  2. 2)

$\\{g,T\\}$ are compatible, $g$ or $T$ is continuous on $(X,p^{S})$ and
$\\{f,S\\}$ are weakly compatible,

then $f,g,S$ and $T$ have a common fixed point. Moreover, the set of common
fixed points of $f,$ $g$, $S$ and $T$ is well ordered if and only if $f$, $g$,
$S$ and $T$ have one and only one common fixed point.

###### Proof

∎Define $\Psi:\mathbb{R}_{\geq 0}\rightarrow\mathbb{R}_{\geq 0}$ by
$\Psi(x)=\int\limits_{0}^{x}\phi(t)dt,$ then from (9), we have

$\Psi\left(\psi(p(fx,gy))\right)\leq\Psi\left(\psi(M_{p}(x,y))\right)-\Psi\left(\varphi(M_{p}(x,y))\right),$

which can be written as

$\psi_{1}(p(fx,gy))\leq\psi_{1}(M_{p}(x,y))-\varphi_{1}(M_{p}(x,y)),$

where $\psi_{1}=\Psi\circ\psi$ and $\varphi_{1}=\Psi\circ\varphi$. Clearly,
$\psi_{1},\varphi_{1}:\mathbb{R}_{\geq 0}\rightarrow\mathbb{R}_{\geq 0},$
$\psi_{1}$ is continuous and nondecreasing, $\varphi_{1}$ is a lower
semicontinuous, and $\psi_{1}(t)=\varphi_{1}(t)=0$ if and only if $t=0.$ Hence
by Theorem 2.1, $f,g,S$ and $T$ have a unique common fixed point. ∎

###### Remark 1

We have the following remarks.

  1. 1)

If we take $f=g$ and $S=T=I$ (an identity map) in Corollary 1, then it extends
(Dutta, , Theorem 2.1) to ordered partial metric spaces.

  2. 2)

We can not apply Corollary 1 in the setup of ordered metric space to the
mappings given in Example 3. Indeed, if we take $x,y\in(\frac{1}{3},2]$
contractive condition in the Corollary 1 in the setup of ordered metric space
is not satisfied.

  3. 3)

Theorem 2.1 generalizes (AT, , Theorem 2.1), (ATR, , Theorem 2.1) and (Doric,
, Theorem 2.1) for four maps in the setup of ordered partial metric spaces.

##  3 Application for Solutions of Implicit Integral Equations

Let $\Omega=[0,1]$ be bounded open set in $\mathbb{R}$, and $L^{2}(\Omega)$ be
the set of comparable functions on $\Omega$ whose square is integrable on
$\Omega.$ Consider an integral equation

$F(t,x(t))=\int\limits_{\Omega}\kappa(t,s,x(s))ds$ (10)

where $F:\Omega\times\mathbb{R}_{\geq 0}\rightarrow\mathbb{R}_{\geq 0}$ and
$\kappa:\Omega\times\Omega\times\mathbb{R}_{\geq 0}\rightarrow\mathbb{R}_{\geq
0}$ be two mappings. Feckan Fec  obtained the nonnegative solutions of
implicit integral equation (10) as an application of fixed point theorem. We
shall study the sufficient condition for existence of solution of integral
equation in framework of ordered complete partial metric space. Define
$p:X\times X\rightarrow\mathbb{R}_{\geq 0}$ by

$p(x,y)=\max\left(\underset{t\in\Omega}{\sup}\
x(t),\underset{t\in\Omega}{\sup}\ y(t)\right).$

Then $\left(X,p\right)$ is a complete partial metric space. We assume the
following that there exists a positive number $h\in[0,\frac{1}{4})$:

  1. (i)

$F(s,u(t))\leq hu(t)$ for each $s,t\in\Omega.$

  2. (ii)

$\int\limits_{\Omega}\kappa(t,s,v(s))ds\leq 2hv(t)\ $for each $s,t\in\Omega$.

  3. (iii)

The control functions $\psi$ and $\varphi$ are connected with relation that

$\psi(a)+\phi(2a)\leq\psi(2a),$

for every $a\in\mathbb{R}_{\geq 0}$.

Then integral equation (10) has a solution in $L^{2}(\Omega)$.

###### Proof

∎Define $(fx)(t)=F(t,x(t))$ and
$(gx)(t)=\int\limits_{\Omega}\kappa(t,s,x(s))ds.$ Now

$\displaystyle\psi(p(fx,gy))$ $\displaystyle=$
$\displaystyle\psi\left(\max\left(\underset{t\in\Omega}{\sup}\
\left(fx\right)(t),\underset{t\in\Omega}{\sup}\
\left(fy\right)(t)\right)\right)$ $\displaystyle=$
$\displaystyle\psi\left(\max\left(\underset{t\in\Omega}{\sup}\
F(t,x(t)),\underset{t\in\Omega}{\sup}\
\int\limits_{\Omega}\kappa(t,s,y(t))dt\right)\right)$ $\displaystyle\leq$
$\displaystyle\psi\left(\max\left(\underset{t\in\Omega}{\sup}\
hx(t),\underset{t\in\Omega}{\sup}\ 2hy(t)\right)\right)$ $\displaystyle\leq$
$\displaystyle\psi\left(2h\max\left(\underset{t\in\Omega}{\sup}\
x(t),\underset{t\in\Omega}{\sup}\ y(t)\right)\right)$ $\displaystyle\leq$
$\displaystyle\psi\left(\frac{1}{2}\max\left(\underset{t\in\Omega}{\sup}\
x(t),\underset{t\in\Omega}{\sup}\ y(t)\right)\right)$ $\displaystyle=$
$\displaystyle\psi\left(\max\left(\underset{t\in\Omega}{\sup}\
x(t),\underset{t\in\Omega}{\sup}\
y(t)\right)\right)-\phi\left(\max\left(\underset{t\in\Omega}{\sup}\
x(t),\underset{t\in\Omega}{\sup}\ y(t)\right)\right)$ $\displaystyle=$
$\displaystyle\psi\left(p(x,y)\right)-\phi\left(p(x,y)\right)$
$\displaystyle=$ $\displaystyle\psi(M_{p}(x,y))-\varphi(M_{p}(x,y)).$

Thus for every comparable elements $x,y\in X,$

$\psi(p(fx,gy))\leq\psi(M_{p}(x,y))-\varphi(M_{p}(x,y)),$

is satisfies where

$M_{p}(x,y)=\max\\{p(x,y),p(fx,x),p(gy,y),\dfrac{p(fx,y)+p(gy,x)}{2}\\}.$

Now we can apply Theorem 2.1 by taking $S$ and $T$ as identity maps to obtain
the solution of integral equation (10) in $L^{2}(\Omega).$ ∎

##  4 Fractals in Partial Metric Spaces

Consistent with AY , let $CB^{p}(X)$ be the family of all non-empty, closed
and bounded subsets of the partial metric space $(X,p)$, induced by the
partial metric $p$. Note that closedness is taken from $(X,\tau_{p})$
($\tau_{p}$ is the topology induced by $p$) and boundedness is given as
follows: $A$ is a bounded subset in $(X,p)$ if there exists an $x_{0}\in X$
and $M\geq 0$ such that for all $a\in A,$ we have $a\in B_{p}(x_{0},M)$, that
is, $p(x_{0},a)<p(a,a)+M$. For $A,B\in CB^{p}(X)$ and $x\in X$, define
$\delta_{p}:CB^{p}(X)\times CB^{p}(X)\rightarrow[0,\infty)$ and

$\displaystyle p(x,A)$ $\displaystyle=$ $\displaystyle\inf\\{p(x,a):a\in
A\\},$ $\displaystyle\delta_{p}(A,B)$ $\displaystyle=$
$\displaystyle\sup\\{p(a,B):a\in A\\},$ $\displaystyle H_{p}(A,B)$
$\displaystyle=$ $\displaystyle\max\\{\delta_{p}(A,B),\delta_{p}(B,A)\\}.$

It can be verified that $p(x,A)=0$ implies $p^{S}(x,A)=0,$ where

$p^{S}(x,A)=\inf\\{p^{S}(x,a):a\in A\\}.$

######  Lemma 2 (AL1 )

Let $(X,p)$ be a partial metric space and $A$ be a non-empty subset of $X,$
then $a\in\overline{A}$ if and only if $p(a,A)=p(a,a).$

######  Proposition 1 (AY )

Let $(X,p)$ be a partial metric space. For any $A,B,C\in CB^{p}(X)$,

  1. (i)

$\delta_{p}(A,A)=\sup\\{p(a,a):a\in A\\};$

  2. (ii)

$\delta_{p}(A,A)\leq\delta_{p}(A,B);$

  3. (iii)

$\delta_{p}(A,B)=0$ implies $A\subseteq B;$

  4. (iv)

$\delta_{p}(A,B)\leq\delta_{p}(A,C)+\delta_{p}(C,B)-\inf\limits_{c\in
C}p(c,c).$

######  Proposition 2 (AY )

Let $(X,p)$ be a partial metric space. For any $A,B,C\in CB^{p}(X),$

  1. 1)

$H_{p}(A,A)\leq H_{p}(A,B);$

  2. 2)

$H_{p}(A,B)=H_{p}(B,A);$

  3. 3)

$H_{p}(A,B)\leq H_{p}(A,C)+H_{p}(C,B)-\inf\limits_{c\in C}p(c,c);$

  4. 4)

$H_{p}(A,B)=0$ implies that $A=B.$

The mapping $H_{p}:CB^{p}(X)\times CB^{p}(X)\rightarrow[0,\infty)$ is called
partial Hausdorff metric induced by partial metric $p.$ Every Hausdorff metric
is partial Hausdorff metric but converse is not true (see (AY, , Example
2.6)).

######  Theorem 4.1 (AY )

Let $(X,p)$ be a partial metric space. If $T:X\rightarrow CB^{p}(X)$ be a
multi-valued mapping such that for all $x,y\in X,$ we have $H_{p}(Tx,Ty)\leq
kp(x,y),$ where $k\in(0,1).$ Then $T$ has a fixed point.

###### Definition 6

Let $(X,p)$ be a partial metric space and and $\mathcal{H}_{p}(X)$ denotes the
set of all non-empty compact subsets of $X.$ Let $\\{f_{n}:n=1,\dots,N\\}$ be
a finite family of self-mappings on $X$ that satisfy

$\psi(p(f_{i}x,f_{i}y))\leq\psi(M_{p}(x,y))-\varphi(M_{p}(x,y)),$

where

$\displaystyle M_{p}(x,y)$ $\displaystyle=$
$\displaystyle\max\\{p(x,y),p(f_{i}x,x),p(f_{i}y,y),p(f_{i}^{2}x,f_{i}x),$
$\displaystyle
p(f_{i}^{2}y,y),p(f_{i}^{2}y,f_{i}y),\dfrac{p(f_{i}x,y)+p(f_{i}y,x)}{2}\\}$

for every $x,y\in X.$ We call these maps as a family of generalized
$\left(\psi,\phi\right)$-contraction mappings. Define
$T:\mathcal{H}_{p}(X)\rightarrow\mathcal{H}_{p}(X)$ by

$\displaystyle T(A)$ $\displaystyle=$ $\displaystyle f_{1}(A)\bigcup
f_{2}(A)\bigcup\cdot\cdot\cdot\bigcup f_{N}(A)$ $\displaystyle=$
$\displaystyle\bigcup_{n=1}^{N}f_{n}(A),\text{ for each
}A\in\mathcal{H}_{p}(X).$

If $f_{n}:X\rightarrow X$, $n=1,\dots,N$ are generalized
$\left(\psi,\phi\right)$-contraction mappings, then
$(X;f_{1},f_{2},\dots,f_{N})$ is called generalized
$\left(\psi,\phi\right)$-iterated function system
($\left(\psi,\phi\right)$-IFS).

###### Definition 7

A nonempty compact set $A\subseteq X$ is said to be an attractor of the
generalized $\left(\psi,\phi\right)$-IFS if

  1. 1)

$T(A)=A$ and

  2. 2)

there is an open set $U\subseteq X$ such that $A\subseteq U$ and
$\lim\limits_{k\rightarrow\infty}T^{k}(B)=A$ for any compact set $B\subseteq
U$, where the limit is taken with respect to the partial Hausdorff metric.

The largest open set $U$ satisfying 2 is called a basin of attraction.

###### Theorem 4.2

Let $(X,p)$ be a complete partial metric space and $(X;f_{n},n=1,\dots,k)$ a
generalized $\left(\psi,\phi\right)$-iterated function system. Let
$T:\mathcal{H}_{p}(X)\rightarrow\mathcal{H}_{p}(X)$ be a mapping defined by

$T(A)=\bigcup_{n=1}^{k}f_{n}(A),\text{ for all }A\in\mathcal{H}_{p}(X).$

Suppose that, there exists control functions $\psi$ and $\varphi$ such that
for every $A$, $B\in\mathcal{H}_{p}\left(X\right),$

$\psi(H_{p}(T\left(A\right),T\left(B\right)))\leq\psi(M_{T}(A,B))-\phi(M_{T}(A,B))$
(11)

is satisfied, where

$\displaystyle M_{T}(A,B)$ $\displaystyle=$
$\displaystyle\max\\{H_{p}(A,B),H_{p}(A,T\left(A\right)),H_{p}(B,T\left(B\right)),H_{p}(T^{2}\left(A\right),T\left(A\right)),$
$\displaystyle
H_{p}(T^{2}\left(A\right),B),H_{p}(T^{2}\left(A\right),T\left(B\right)),\dfrac{H_{p}(A,T\left(B\right))+H_{p}(B,T\left(A\right))}{2}\\}.$

Then $T$ has a unique fixed point $U\in\mathcal{H}_{p}\left(X\right),$ that is

$U=T\left(U\right)=\bigcup_{n=1}^{k}f_{n}(U).$

Moreover, for any initial set $A_{0}\in\mathcal{H}_{p}\left(X\right)$, the
sequence $\\{A_{0},T\left(A_{0}\right),T^{2}\left(A_{0}\right),\dots\\}$ of
compact sets converges to a fixed point of $T$.

###### Proof

∎Let $A_{0}\ $be an arbitrary element in $\mathcal{H}_{p}\left(X\right).$ If
$A_{0}=T\left(A_{0}\right),$ then the proof is finished. So we assume that
$A_{0}\neq T\left(A_{0}\right).$ Define, for $m\in\mathbb{Z}_{\geq 0}$,

$A_{1}=T(A_{0}),\text{
}A_{2}=T\left(A_{1}\right),\dots,A_{m+1}=T\left(A_{m}\right).$

We may assume that $A_{m}\neq A_{m+1}$ for all $m\in\mathbb{Z}_{\geq 0}.$ If
not, then $A_{k}=A_{k+1}$ for some $k$ implies $A_{k}=T(A_{k})\ $and this
completes the proof. Take $A_{m}\neq A_{m+1}$ for all $m\in\mathbb{Z}_{\geq
0}$. From (11), we have

$\displaystyle\psi\left(H_{p}(A_{m+1},A_{m+2})\right)$ $\displaystyle=$
$\displaystyle\psi\left(H_{p}(T\left(A_{m}\right),T\left(A_{m+1}\right))\right)$
$\displaystyle\leq$
$\displaystyle\psi\left(M_{T}\left(A_{m},A_{m+1}\right)\right)-\phi\left(M_{T}\left(A_{m},A_{m+1}\right)\right),$

where

$\displaystyle
M_{T}\left(A_{m},A_{m+1}\right)=\max\\{H_{p}(A_{m},A_{m+1}),H_{p}\left(A_{m},T\left(A_{m}\right)\right),H_{p}\left(A_{m+1},T\left(A_{m+1}\right)\right),$
$\displaystyle
H_{p}(T^{2}\left(A_{m}\right),T\left(A_{m}\right)),H_{p}\left(T^{2}\left(A_{m}\right),A_{m+1}\right),H_{p}\left(T^{2}\left(A_{m}\right),T\left(A_{m+1}\right)\right),$
$\displaystyle\displaystyle{\frac{H_{p}\left(A_{m},T\left(A_{m+1}\right)\right)+H_{p}\left(A_{m+1},T\left(A_{m}\right)\right)}{2}}\\}$
$\displaystyle=\max\\{H_{p}(A_{m},A_{m+1}),H_{p}\left(A_{m},A_{m+1}\right),H_{p}\left(A_{m+1},A_{m+2}\right),$
$\displaystyle
H_{p}(A_{m+2},A_{m+1}),H_{p}\left(A_{m+2},A_{m+1}\right),H_{p}\left(A_{m+2},A_{m+2}\right),$
$\displaystyle\displaystyle{\frac{H_{p}\left(A_{m},A_{m+2}\right)+H_{p}\left(A_{m+1},A_{m+1}\right)}{2}}\\}$
$\displaystyle\leq\max\\{H_{p}(A_{m},A_{m+1}),H_{p}\left(A_{m+1},A_{m+2}\right),\displaystyle{\frac{H_{p}\left(A_{m},A_{m+1}\right)+H_{p}\left(A_{m+1},A_{m+2}\right)}{2}}\\}$
$\displaystyle=\max\\{H_{p}\left(A_{m},A_{m+1}\right),H_{p}\left(A_{m+1},A_{m+2}\right)\\}.$

As
$\max\\{H_{p}\left(A_{m},A_{m+1}\right),H_{p}\left(A_{m+1},A_{m+2}\right)\\}\leq
M_{T}\left(A_{m},A_{m+1}\right).$ Therefore,

$M_{T}\left(A_{m},A_{m+1}\right)=\max\\{H_{p}\left(A_{m},A_{m+1}\right),H_{p}\left(A_{m+1},A_{m+2}\right)\\}.$

Now if $M_{T}\left(A_{m},A_{m+1}\right)=H_{p}\left(A_{m+1},A_{m+2}\right),$
then (11) gives that

$\psi\left(H_{p}(A_{m+1},A_{m+2})\right)\leq\psi(H_{p}\left(A_{m+1},A_{m+2}\right))-\phi\left(H_{p}\left(A_{m+1},A_{m+2}\right)\right),$

a contradiction. Hence
$M_{T}\left(A_{m},A_{m+1}\right)=H_{p}\left(A_{m+1},A_{m+2}\right)$ and

$\displaystyle\psi\left(H_{p}(A_{m+1},A_{m+2})\right)$ $\displaystyle\leq$
$\displaystyle\psi(H_{p}\left(A_{m},A_{m+1}\right))-\phi\left(H_{p}\left(A_{m},A_{m+1}\right)\right)$
$\displaystyle\leq$ $\displaystyle\psi(H_{p}\left(A_{m},A_{m+1}\right)),$

that is, $H_{p}\left(A_{m+1},A_{m+2}\right)\leq
H_{p}\left(A_{m},A_{m+1}\right).$ Thus the sequence
$\\{H_{p}\left(A_{m},A_{m+1}\right)\\}$is nonincreasing. Hence there exists
$c\geq 0$ such that $\lim\limits_{n\rightarrow\infty}H_{p}(A_{n},A_{n+1})=c.$
Suppose that $c>0.$ Then,
$\psi(H_{p}(A_{n+2},A_{n+1}))\leq\psi(H_{p}(A_{n+1},A_{n}))-\varphi(H_{p}(A_{n+1},A_{n})),$
and by lower semicontinuity of $\varphi,$ we have

$\underset{n\rightarrow\infty}{\lim\sup}\
\psi(H_{p}(A_{n+2},A_{n+1}))\leq\underset{n\rightarrow\infty}{\lim\sup}\
\psi(H_{p}(A_{n+1},A_{n}))-\underset{n\rightarrow\infty}{\lim\inf}\
\varphi(H_{p}(A_{n+1},A_{n})),$

which implies that $\psi(c)\leq\psi(c)-\varphi(c),$ a contradiction. Therefore
$c=0.$ So we conclude that

$\lim_{n\rightarrow\infty}H_{p}(A_{n+1},A_{n})=0.$ (12)

Now, we show that $\lim\limits_{n,m\rightarrow\infty}H_{p}(A_{n},A_{m})=0$ .
If not, there is $\varepsilon>0,$ and there exist even integers $n_{k}$ and
$m_{k}$ with $m_{k}>n_{k}>k$ such that

$H_{p}(A_{m_{k}},A_{n_{k}})\geq\varepsilon,$ (13)

and $H_{p}(A_{m_{k}-2},A_{n_{k}})<\varepsilon.$ Since

$\displaystyle\varepsilon$ $\displaystyle\leq$ $\displaystyle
H_{p}(A_{m_{k}},A_{n_{k}})$ $\displaystyle\leq$ $\displaystyle
H_{p}(A_{n_{k}},A_{m_{k}-2})+H_{p}(A_{m_{k}-2},A_{m_{k}})-\inf_{a_{1}\in
A_{m_{k}-2}}p(a_{1},a_{1})$ $\displaystyle\leq$ $\displaystyle
H_{p}(A_{n_{k}},A_{m_{k}-2})+H_{p}(A_{m_{k}-2},A_{m_{k}-1})+H_{p}(A_{m_{k}-1},A_{m_{k}})$
$\displaystyle-\inf_{a_{2}\in A_{m_{k}-1}}p(a_{2},a_{2})-\inf_{a_{1}\in
A_{m_{k}-2}}p(a_{1},a_{1}).$

From (12) and (13), we have

$\lim_{k\rightarrow\infty}H_{p}(A_{m_{k}},A_{n_{k}})=\varepsilon.$ (14)

Also (13) and inequality

$H_{p}(A_{m_{k}},A_{n_{k}})\leq
H_{p}(A_{m_{k}},A_{m_{k}-1})+H_{p}(A_{m_{k}-1},A_{n_{k}})-\inf\limits_{a_{2}\in
A_{m_{k}-1}}p(a_{2},a_{2})$

give that
$\varepsilon\leq\lim\limits_{k\rightarrow\infty}H_{p}(A_{m_{k}-1},A_{n_{k}}),$
while inequality

$H_{p}(A_{m_{k}-1},A_{n_{k}})\leq
H_{p}(A_{m_{k}-1},A_{m_{k}})+H_{p}(A_{m_{k}},A_{n_{k}})-\inf\limits_{a_{3}\in
A_{m_{k}}}p(a_{3},a_{3})$

yields
$\lim\limits_{k\rightarrow\infty}H_{p}(A_{m_{k}-1},A_{n_{k}})\leq\varepsilon,$
and hence

$\lim_{k\rightarrow\infty}H_{p}(A_{m_{k}-1},A_{n_{k}})=\varepsilon.$ (15)

Now (15) and inequality

$H_{p}(A_{m_{k}-1},A_{n_{k}})\leq
H_{p}(A_{m_{k}-1},A_{n_{k}+1})+H_{p}(A_{n_{k}+1},A_{n_{k}})-\inf_{a_{4}\in
A_{n_{k}+1}}p(a_{4},a_{4})$

give

$\varepsilon\leq\lim\limits_{k\rightarrow\infty}H_{p}(A_{m_{k}-1},A_{n_{k}+1}),$

while inequality

$H_{p}(A_{m_{k}-1},A_{n_{k}+1})\leq
H_{p}(A_{m_{k}-1},A_{n_{k}})+H_{p}(A_{n_{k}},A_{n_{k}+1})-\inf\limits_{a_{5}\in
A_{n_{k}}}p(a_{5},a_{5})$

yields
$\lim\limits_{k\rightarrow\infty}H_{p}(A_{m_{k}-1},A_{n_{k}+1})\leq\varepsilon,$
and so

$\lim_{k\rightarrow\infty}H_{p}(A_{m_{k}-1},A_{n_{k}+1})=\varepsilon.$ (16)

As

$\displaystyle M_{T}(A_{n_{k}},A_{m_{k}-1})$ $\displaystyle=$
$\displaystyle\max\\{H_{p}(A_{n_{k}},A_{m_{k}-1}),H_{p}(A_{n_{k}},A_{n_{k}}),$
$\displaystyle
H_{p}(A_{m_{k}-1},A_{m_{k}-1}),H_{p}(A_{n_{k}+2},A_{n_{k}}),H_{p}(A_{n_{k}+2},A_{m_{k}-1}),$
$\displaystyle
H_{p}(A_{n_{k}+2},A_{m_{k}}),\frac{H_{p}(A_{n_{k}},A_{m_{k}-1})+H_{p}(A_{n_{k}},A_{m_{k}-1})}{2}\\}.$

So,
$\lim\limits_{k\rightarrow\infty}M_{T}(x_{n_{k}},x_{m_{k}-1})=\max\\{\varepsilon,0,0,0,\varepsilon,\varepsilon,\varepsilon\\}=\varepsilon.$
From (16), we obtain

$\displaystyle\psi(H_{p}(A_{n_{k}+1},A_{m_{k}}))$ $\displaystyle=$
$\displaystyle\psi(H_{p}(A_{n_{k}},A_{m_{k}-1}))$ $\displaystyle\leq$
$\displaystyle\psi(M_{T}(A_{n_{k}},A_{m_{k}-1}))-\varphi(M_{T}(A_{n_{k}},A_{m_{k}-1})).$

Taking upper limit as $k\rightarrow\infty$ implies that
$\psi(\varepsilon)\leq\psi(\varepsilon)-\varphi(\varepsilon),$ a contradiction
as $\varepsilon>0$. Therefore $\\{A_{n}\\}$ is a Cauchy sequence in $X.$ Since
$(\mathcal{H}_{p}(X),p)$ is complete as $(X,p)$ is complete, so
$\lim\limits_{n\rightarrow\infty}H_{p}(A_{n},U)=H_{p}\left(U,U\right)$ for
some $U\in\mathcal{H}_{p}(X),$ that is, we have $A_{n}\rightarrow U$ as
$n\rightarrow\infty.$

In order to show that $U$ is the fixed point of $T,$ we contrary assume that
$H_{p}\left(U,T\left(U\right)\right)\neq 0$. Now

$\begin{array}[]{ll}\psi\left(H_{p}(A_{n+1},T\left(U\right))\right)&=\psi(H_{p}(T\left(A_{n}\right),T\left(U\right)))\\\
&\leq\psi\left(M_{T}\left(A_{n},U\right)\right)-\phi\left(M_{T}\left(A_{n},U\right)\right),\end{array}$
(17)

where

$\displaystyle M_{T}\left(A_{n},U\right)$ $\displaystyle=$
$\displaystyle\max\\{H_{p}(A_{n},U),H_{p}(A_{n},T\left(A_{n}\right)),H_{p}(U,T\left(U\right)),H_{p}(T^{2}\left(A_{n}\right),T\left(A_{n}\right)),$
$\displaystyle
H_{p}(T^{2}\left(A_{n}\right),U),H_{p}(T^{2}\left(A_{n}\right),T\left(U\right)),\frac{H_{p}(A_{n},T\left(U\right))+H_{p}(U,T\left(A_{n}\right))}{2}\\}$
$\displaystyle=$
$\displaystyle\max\\{H_{p}(A_{n},U),H_{p}(A_{n},A_{n+1}),H_{p}(U,T\left(U\right)),H_{p}(A_{n+2},A_{n+1}),$
$\displaystyle
H_{p}(A_{n+2},U),H_{p}(A_{n+2},T\left(U\right)),\frac{H_{p}(A_{n},T\left(U\right))+H_{p}(U,A_{n+1})}{2}\\}.$

Now we consider the following cases:

  1. 1)

If $M_{T}\left(A_{n},U\right)=H_{p}(A_{n},U,),$ then on taking upper limit as
$n\rightarrow\infty$ in (17), we have

$\psi\left(H_{p}(T\left(U\right),U)\right)\leq\psi\left(H_{p}\left(U,U\right)\right)-\psi\left(H_{p}\left(U,U\right)\right),$

a contradiction.

  2. 2)

When $M_{T}\left(A_{n},U\right)=H_{p}(A_{n},A_{n+1}),$ then on taking upper
limit as $n\rightarrow\infty$ in (17), implies

$\psi\left(H_{p}(T\left(U\right),U)\right)\leq\psi\left(H_{p}\left(U,U\right)\right)-\phi\left(H_{p}\left(U,U\right)\right),$

gives a contradiction.

  3. 3)

In case $M_{T}\left(A_{n},U\right)=H_{p}(U,T\left(U\right)),$ then on taking
upper limit as $n\rightarrow\infty$ in (17), we get

$\psi\left(H_{p}(T\left(U\right),U)\right)\leq\psi\left(H_{p}\left(U,T\left(U\right)\right)\right)-\phi\left(H_{p}\left(U,T\left(U\right)\right)\right),$

a contradiction.

  4. 4)

If
$M_{T}\left(A_{n},U\right)=\dfrac{H_{p}(A_{n},T\left(U\right))+H_{p}(U,A_{n+1})}{2},$
then on upper taking limit as $n\rightarrow\infty,$ we have

$\psi\left(H_{p}(T\left(U\right),U)\right)\leq\psi(\frac{H_{p}\left(U,T\left(U\right)\right)+H_{p}\left(U,U\right)}{2})\\\
-\phi(\frac{H_{p}\left(U,T\left(U\right)\right)+H_{p}\left(U,U\right)}{2})\\\
=\psi(\frac{H_{p}\left(U,T\left(U\right)\right)}{2})-\phi(\frac{H_{p}\left(U,T\left(U\right)\right)}{2}),$

a contradiction.

  5. 5)

When $M_{T}\left(A_{n},U\right)=H_{p}(A_{n+2},A_{n+1}),$ then on taking upper
limit as $n\rightarrow\infty$ in (17), we get

$\psi\left(H_{p}(T\left(U\right),U)\right)\leq\psi\left(H_{p}\left(U,U\right)\right)-\phi\left(H_{p}\left(U,U\right)\right),$

gives a contradiction.

  6. 6)

In case $M_{T}\left(A_{n},U\right)=H_{p}(A_{n+2},U),$ then on taking upper
limit as $n\rightarrow\infty$ in (17), we get

$\psi\left(H_{p}(T\left(U\right),U)\right)\leq\psi\left(H_{p}\left(U,U\right)\right)-\phi\left(H_{p}\left(U,U\right)\right),$

a contradiction.

  7. 7)

Finally if $M_{T}\left(A_{n},U\right)=H_{p}(A_{n+2},T\left(U\right)),$ then on
taking upper limit as $n\rightarrow\infty,$ we have

$\psi\left(H_{p}(T\left(U\right),U)\right)\leq\psi(H_{p}(U,T\left(U\right)))-\phi\left(H_{p}\left(U,U\right)\right),$

a contradiction.

Thus, $U$ is the fixed point of $T$.

To show the uniqueness of fixed point of $T$, assume that $U$ and $V$ are two
fixed points of $T$ with $H_{p}\left(U,V\right)$ is not zero. From (11), we
obtain that

$\displaystyle\psi(H_{p}(U,V))$ $\displaystyle=$
$\displaystyle\psi(H_{p}(T\left(U\right),T\left(V\right)))$
$\displaystyle\leq$
$\displaystyle\psi\left(M_{T}\left(U,V\right)\right)-\phi\left(M_{T}\left(U,V\right)\right),$

where

$\displaystyle
M_{T}\left(U,V\right)=\max\\{H_{p}(U,V),H_{p}(U,T\left(U\right)),H_{p}(V,T\left(V\right)),$
$\displaystyle\frac{H_{p}(U,T\left(V\right))+H_{p}(V,T\left(U\right))}{2},H_{p}(T^{2}\left(U\right),U),H_{p}(T^{2}\left(U\right),V),H_{p}(T^{2}\left(U\right),T\left(V\right))\\}$
$\displaystyle=\max\\{H_{p}\left(U,V\right),H_{p}(U,U),H_{p}(V,V),$
$\displaystyle\hskip
56.9055pt\frac{H_{p}(U,V)+H_{p}(V,U)}{2},H_{p}\left(U,U\right),H_{p}(U,V),H_{p}(U,V)\\}$
$\displaystyle=H_{p}(U,V),$

that is,

$\psi(H_{p}(U,V))\leq\psi\left(H_{p}\left(U,V\right)\right)-\phi\left(H_{p}\left(U,V\right)\right),$

a contradiction. Thus $T$ has a unique fixed point $U\in\mathcal{H}_{p}(X)$. ∎

###### Remark 2

In Theorem 4.2, if we take $\mathcal{S}(X)$ the collection of all singleton
subsets of $X,$ then clearly $\mathcal{S}(X)\subseteq\mathcal{H}_{p}(X).$
Moreover, consider $f_{n}=f$ for each $n,$ where $f=f_{1}$ then the mapping
$T$ becomes

$T(x)=f(x).$

With this setting, we obtain the following fixed point result.

###### Corollary 3

Let $(X,p)$ be a complete partial metric space and $\\{X:f_{n},n=1,\dots,k\\}$
a generalized iterated function system. Let $f:X\rightarrow X$ be a mapping
defined as in Remark 2. Suppose that, there exists control functions $\psi$
and $\varphi$ such that for any $x,y\in\mathcal{H}_{p}\left(X\right)$, the
following holds:

$\psi\left(p\left(fx,fy\right)\right)\leq\psi(M_{p}(x,y))-\phi(M_{p}(x,y)),$

where

$\displaystyle M_{p}(x,y)=$
$\displaystyle\max\\{p(x,y),p(x,fx),p(y,fy),p(f^{2}x,y),$ $\displaystyle
p(f^{2}x,fx),p(f^{2}x,fy),\dfrac{p(x,fy)+p(y,fx)}{2}\\}.$

Then $f$ has a unique fixed point $x\in X,$ Moreover, for any initial set
$x_{0}\in X$, the sequence of compact sets
$\\{x_{0},fx_{0},f^{2}x_{0},\dots\\}$ converges to a fixed point of $f$.

###### Corollary 4

Let $(X,p)$ be a complete partial metric space and $(X;f_{n},n=1,\dots,k)$ be
iterated function system where each $f_{i}$ for $i=1,\dots,k$ is a contraction
self-mapping on $X$. Then $T:\mathcal{H}(X)\rightarrow\mathcal{H}(X)$ defined
in Theorem 4.2 has a unique fixed point in $\mathcal{H}\left(X\right).$
Furthermore, for any set $A_{0}\in\mathcal{H}\left(X\right)$, the sequence of
compact sets $\\{A_{0},T\left(A_{0}\right),T^{2}\left(A_{0}\right),\dots\\}$
converges to a fixed point of $T$.

###### Proof

∎It follows from Theorem 1.2 that if each $f_{i}$ for $i=1,\dots,k$ is a
contraction mapping on $X,$ then the mapping
$T:\mathcal{H}(X)\rightarrow\mathcal{H}(X)$ defined by

$T(A)=\bigcup_{n=1}^{k}f_{n}(A),\text{ for all }A\in\mathcal{H}(X)$

is contraction on $\mathcal{H}\left(X\right)$. Using Theorem 2.1, the result
follows. ∎

###### Corollary 5

Let $(X,p)$ be a complete partial metric space and $(X;f_{n},n=1,\dots,k)$ an
iterated function system where each $f_{i}$ for $i=1,\dots,k$ is a mapping on
$X$ satisfying

$d\left(f_{i}x,f_{i}y\right)e^{d\left(f_{i}x,f_{i}y\right)-d\left(x,y\right)}\leq
e^{-\tau}d\left(x,y\right),\text{ for all }x,y\in X,\text{ }f_{i}x\neq
f_{i}y,$

where $\tau>0.$ Then the mapping $T:\mathcal{H}(X)\rightarrow\mathcal{H}(X)$
defined in Theorem 4.2 has a unique fixed point in
$\mathcal{H}\left(X\right).\ $Furthermore, for any set
$A_{0}\in\mathcal{H}\left(X\right)$ , the sequence of compact sets
$\\{A_{0},T\left(A_{0}\right),T^{2}\left(A_{0}\right),\dots\\}$ converges to a
fixed point of $T$.

###### Proof

∎Take $F\left(\lambda\right)=\ln\left(\lambda\right)+\lambda,$ $\lambda>0$ in
Theorem 1.3, then each mapping $f_{i}$ for $i=1,\dots,k$ on $X$ satisfies

$d\left(f_{i}x,f_{i}y\right)e^{d\left(f_{i}x,f_{i}y\right)-d\left(x,y\right)}\leq
e^{-\tau}d\left(x,y\right),\text{ for all }x,y\in X,\text{ }f_{i}x\neq
f_{i}y,$

where $\tau>0.$ Again from Theorem 1.3, the mapping
$T:\mathcal{H}(X)\rightarrow\mathcal{H}(X)$ defined by

$T(A)=\bigcup_{n=1}^{k}f_{n}(A),\text{ for all }A\in\mathcal{H}(X)$

satisfies

$H\left(T\left(A\right),T\left(B\right)\right)e^{H\left(T\left(A\right),T\left(B\right)\right)-H\left(A,B\right)}\leq
e^{-\tau}H\left(A,B\right),$

for all $A,B\in\mathcal{H}(X),$
$H\left(T\left(A\right),T\left(B\right)\right)\neq 0$. Using Theorem 2.1, the
result follows. ∎

###### Corollary 6

Let $(X,p)$ be a complete partial metric space and $(X;f_{n},n=1,\dots,k)$ be
iterated function system such that each $f_{i}$ for $i=1,\dots,k$ is a mapping
on $X$ satisfying

$d\left(f_{i}x,f_{i}y\right)(d\left(f_{i}x,f_{i}y\right)+1)\leq
e^{-\tau}d\left(x,y\right)(d\left(x,y\right)+1),\text{ for all }x,y\in X,\
f_{i}x\neq f_{i}y,$

where $\tau>0.$ Then the mapping $T:\mathcal{H}(X)\rightarrow\mathcal{H}(X)$
defined in Theorem 2.1 has a unique fixed point in
$\mathcal{H}\left(X\right).$ Furthermore, for any set
$A_{0}\in\mathcal{H}\left(X\right)$, the sequence of compact sets
$\\{A_{0},T\left(A_{0}\right),T^{2}\left(A_{0}\right),\dots\\}$ converges to a
fixed point of $T$.

###### Proof

∎By taking
$F\left(\lambda\right)=\ln\left(\lambda^{2}+\lambda\right)+\lambda,$
$\lambda>0$ in Theorem 1.3, we obtain that each mapping $f_{i}$ for
$i=1,\dots,k$ on $X$ satisfies

$d\left(f_{i}x,f_{i}y\right)(d\left(f_{i}x,f_{i}y\right)+1)\leq
e^{-\tau}d\left(x,y\right)(d\left(x,y\right)+1),\text{ for all }x,y\in X,\
f_{i}x\neq f_{i}y,$

where $\tau>0.$ Again it follows from Theorem 1.3 that the mapping
$T:\mathcal{H}(X)\rightarrow\mathcal{H}(X)$ defined by

$T(A)=\bigcup_{n=1}^{k}f_{n}(A),\text{ for all }A\in\mathcal{H}(X)$

satisfies

$H\left(T\left(A\right),T\left(B\right)\right)(H\left(T\left(A\right),T\left(B\right)\right)+1)\leq
e^{-\tau}H\left(A,B\right)(H\left(A,B\right)+1),$

for all $A,B\in\mathcal{H}(X),$
$H\left(T\left(A\right),T\left(B\right)\right)\neq 0$. Using Theorem 2.1, the
result follows. ∎

###### Corollary 7

Let $(X,p)$ be a complete partial metric space and $(X;f_{n},n=1,\dots,k)$ be
iterated function system such that each $f_{i}$ for $i=1,\dots,k$ is a mapping
on $X$ satisfying

$d\left(f_{i}x,f_{i}y\right)\leq\frac{1}{(1+\tau\sqrt{d\left(x,y\right)})}d\left(x,y\right),\text{
for all }x,y\in X,\text{ }f_{i}x\neq f_{i}y,$

where $\tau>0.$ Then the mapping $T:\mathcal{H}(X)\rightarrow\mathcal{H}(X)$
defined in Theorem 4.2 has a unique fixed point $\mathcal{H}\left(X\right).$
Furthermore, for any set $A_{0}\in\mathcal{H}\left(X\right)$, the sequence of
compact sets $\\{A_{0},T\left(A_{0}\right),T^{2}\left(A_{0}\right),\dots\\}$
converges to a fixed point of $T$.

###### Proof

∎Take $F\left(\lambda\right)=-1/\sqrt{\lambda},$ $\lambda>0$ in Theorem 1.3,
then each mapping $f_{i}$ for $i=1,\dots,k$ on $X$ satisfies

$d\left(f_{i}x,f_{i}y\right)\leq\frac{1}{(1+\tau\sqrt{d\left(x,y\right)})^{2}}d\left(x,y\right),\text{
for all }x,y\in X,\ f_{i}x\neq f_{i}y,$

where $\tau>0.$ Again it follows from Theorem 1.3 that the mapping
$T:\mathcal{H}(X)\rightarrow\mathcal{H}(X)$ defined by

$T(A)=\bigcup\limits_{n=1}^{k}f_{n}(A),\text{ for all }A\in\mathcal{H}(X)$

satisfies

$H\left(T\left(A\right),T\left(B\right)\right)\leq\frac{1}{(1+\tau\sqrt{H\left(A,B\right)})^{2}}H\left(A,B\right),$

for all $A,B\in\mathcal{H}(X),$
$H\left(T\left(A\right),T\left(B\right)\right)\neq 0$. Using Theorem 2.1, the
result follows. ∎

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