Journal of mathematics and computer science
12 (2014), 85-98
Fixed Point Results in Partial Metric Spaces Using Generalized Weak
Contractive Conditions
M. Akram*, W. Shamaila †
*
†
Department of Mathematics GC University, Lahore, Pakistan.
[email protected]
Department of Mathematics, Kinnaird College for Women, Lahore, Pakistan.
[email protected]
Article history:
Received July 2014
Accepted September 2014
Available online October 2014
Abstract
In this work, fixed point results using generalized weakly contractive conditions on partial metric
spaces are presented. These results generalize many previously obtained results. Some examples are also
given to show the usability of these results.
Keywords: Partial Metric, Generalized Weak contractive condition, Fixed point.
1. Introduction
The concept of partial metric spaces was given by Steve Matthews [1, 2] in 1992 to study the
denotational semantics of data flow network. He presented partial metric space as a generalization of
metric space in the sense that the self distance of any point need not be zero. Recently many authors have
focused on the fixed point results in partial metric spaces (see e.g.[3-9]).
The notion of  -contraction was introduced by Boyd and Wong [10] and the weak  -contraction was
introduced by Alber and Guerre-Delabriere [11] as a generalization of  -contraction. Later on  contractions and weak  -contractions have been studied by many authors (see e.g. [5-9, 12, 14]) in
metric spaces as well as in partial metric spaces.
Consistent with Matthews [1, 2], Karapinar [3] and Altun and Erduron [4] some important definitions
and results which are used in this paper are given in the following.
M. Akram, W. Shamaila / J. Math. Computer Sci. 12 (2014), 85-98
Definition 1.1 [1, 2] A partial metric " p" on X is a function from X  X to R  such that for every
element x, y and z of X it satisfies following axioms.
p1 : 0  px, x   px, y .
p2 : px, x  = px, y  = p( y, y) if and only if x = y.
p3 : px, y  = p y, x . (symmetry)
p4 : p( x, z)  px, y   p y, z   p y, y . (triangular inequality)
If " p" is a partial metric on X then  X , p  is called a partial metric space (PMS).
For
a
partial
metric
p
X,
on
the
function
d p : X  X  R
defined
by
d p x, y  = 2 px, y   px, x   p y, y  for all x, y, z  X is a metric on X . Each partial metric " p" on
X generates a T0 topology  p on X for which the collection Bp x,   : x  X ,  > 0 of all open balls
forms a base. Where Bp x,   = y  X : px, y  < px, x     for each  > 0 and x  X .
Definition 1.2 [1, 2, 4]
1. A sequence yn  in a partial metric space  X , p  converges to the limit y  X if and
only if lim p y, yn  = p y, y  .
n
2. A sequence yn  in a partial metric space  X , p  is called Cauchy if and only if
lim p ym , yn  exists and is finite.
m , n 
3.
A partial metric space  X , p  is said to be complete if every Cauchy sequence yn  in
X converges, with respect to  p , to a point y  X such that lim p ym , yn  = p y, y .
m , n
4. The mapping f : X  X is said to be continuous at y0  X , if for every  > 0, there


exists  > 0 such that f Bp  y0 ,    Bp  f  y0 ,  .
The following lemma will be frequently used in the proofs of the main results.
Lemma 1.3 [1, 4] A sequence yn  is a Cauchy sequence in a partial metric space  X , p  if and only if


it is a Cauchy sequence in the metric space X , d p .
I.


A partial metric space  X , p  is complete if and only if the metric space X , d p is
complete. Moreover, lim d p  y, yn  = 0, if and only if
n 
px, x  = lim p y, yn  = lim p yn , ym . Where y is the limit of yn  in X , d p .
n
n , m
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M. Akram, W. Shamaila / J. Math. Computer Sci. 12 (2014), 85-98
II.
Let X be a complete partial metric space. Then
(a) If px, y  = 0, then x = y.
(b) If x  y, then px, y  > 0.
Let X be a partial metric space. Assume that the sequence yn  is converging to z as
III.
n  . such that pz, z  = 0. Then lim p yn , y  = pz, y  for all elements y of X .
n
2. Main results
In the following theorem a generalized form of weak  -contraction is used.
Theorem 2.1 Let  X , p  be a complete partial metric space and T : X  X be a self map such that for
all x, y  X
p(Tx, Ty)  M ( x, y)   (M ( x, y)) (1)
where
1
1


M ( x, y ) = max  px, y ,  px, Tx  p y, Ty ,  px, Ty   p y, Tx
2
2


and  : 0,   0,  is continuous non-decreasing function with  t  = 0 if and only if t = 0. Then
T has a unique fixed point.
Proof: Let y0  X be fixed. Define a sequence yn  in X by yn1 = Tyn , for all n  0. If there exist a

positive integer n0 such that p yn
0 1



, yn = 0 or p Tyn , yn = 0, then Tyn = yn , this shows that yn
0
0
0
0
0
0
is the fixed point of T . Hence we assume that pTyn , yn  = p yn1 , yn   0, for all n  0 . By
substituting x = yn and y = yn 1 in (1) , we have
pTyn , Tyn1  = p yn1 , yn2   M  yn , yn1    M  yn , yn1 
where
1


 p yn , yn 1 , 2  p yn , yn 1   p yn 1 , yn  2 ,
M ( yn , yn 1 ) = max 
.
1
  p yn , yn  2   p yn 1 , yn 1 

2

By p4 , p yn , yn2   p yn1 , yn1   p yn , yn1   p yn1 , yn2 
If p yn , yn 1  <
1
 p yn , yn1   p yn1 , yn2 , then
2
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M. Akram, W. Shamaila / J. Math. Computer Sci. 12 (2014), 85-98
M  yn , yn1  =
1
p yn , yn1   p yn1 , yn2 
2
From (2) we have
p yn1 , yn  2  
1
 p yn , yn1   p yn1 , yn2     1  p y n , y n1   p y n1 , y n2 
2
2

<
(3)
1
 p yn , yn1   p yn1 , yn2 
2
Which implies
p yn1 , yn2   p yn , yn1 
(4)
If
1
 p yn , yn1   p yn1 , yn2  < p yn , yn1 ,
2
then M  yn , yn1  = p yn , yn1  and again from (2) , we have
p yn1 , yn2   p yn , yn1     p yn , yn1 
(5)
< p yn , yn1 
Hence
p yn1 , yn2   p yn , yn1 
(6)
Thus in both cases we have p yn1 , yn 2   p yn , yn1  for all n. Hence p yn , yn1  is monotone
decreasing sequence of non-negative real numbers so there exists a real number r  0 , such that
lim p yn , yn1  = r.
n
(7)
Letting n  , in (3) or in (5) , using (7) and regarding the continuity of  we have r  r   r ,
which forces r = 0. Hence, in both cases
lim p yn , yn1  = 0.
n
Now consider,
88
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M. Akram, W. Shamaila / J. Math. Computer Sci. 12 (2014), 85-98
pTym1 , Tym1  = p ym , ym   M ( ym1 , ym1 )   M ( ym1 , ym1 )
1


= max  p ym1 , ym1 ,  p ym1 , ym   p ym1 , ym 
2



1


   max  p y m1 , y m1 ,  p y m1 , y m   p y m1 , y m  
2



= maxp y m1 , y m1 , p y m1 , y m    maxp y m1 , y m1 , p y m1 , y m 
= p ym1 , ym     p ym1 , ym 
by p1 
Hence, p y m , y m   p y m1 , y m     p y m1 , y m  .
Also by P1 0  p y m , y m   p y m1 , y m     p y m1 , y m . Let m   , using (8) and continuity of
 we have
lim p ym , ym  = 0.
(9)
m


Now, in order to show that yn  is a Cauchy sequence in the complete metric space, X , d p . Assume

that yn  is not Cauchy. Then there exists some  > 0 for which we can find the subsequences ymk 
 

and yn k  of yn  with n(k ) > m(k ) > k such that
d p  ymk  , xnk     .
(10)
Further, we can choose nk  corresponding to mk , in such a way that it is the smallest integer
satisfying (10) hence
d p ymk  , ynk 1  <  .
(11)
From (10)
  d p ymk  , ynk   d p ymk  , ynk 1   d p ynk 1 , ynk  
<   d p ynk 1 , ynk  .
Hence,
  d p ymk  , ynk   <   d p ynk 1 , ynk  .
89
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M. Akram, W. Shamaila / J. Math. Computer Sci. 12 (2014), 85-98
We know that,
d p ynk 1 , ynk   = 2 pynk 1 , ynk    pynk  , ynk    py nk 1 , y nk 1 .
Let k  , using (8) and (9), we get
lim d p  yn k 1 , ynk   = 0.
(13)
lim d p ymk  , ynk   =  .
(14)
k 
Using (13) in (12), we have
k 
We, also know that
d p ymk  , ynk   = 2 pymk  , ynk    pymk  , ymk    pynk  , ynk  
Let k   , using (9) and (14) we get
lim d p ymk  , ynk   = 2 lim pymk  , ynk  .
k 
k 
Therefore, we get
lim p ymk  , yn k  =
k 

2
.
(15)
From the triangular inequality
d p ynk  , ymk    d p ynk  , ynk 1  d p ynk 1 , ymk 1  d p ymk 1 , ymk  
and
d p ynk 1 , ymk 1   d p ynk 1 , ynk   d p ynk  , ymk   d p ymk  , ymk 1 .
Let k  , and using (13) and (14) we get
lim d p ynk  , ymk    lim d p ynk 1 , ymk 1 
k 
k 
and
lim d p ynk 1 , ymk 1  lim d p ynk  , ymk  
k 
k 
Hence,
lim d p ynk 1 , ymk 1  = lim d p ynk  , ymk  =  .
k 
k 
90
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M. Akram, W. Shamaila / J. Math. Computer Sci. 12 (2014), 85-98
By definition of d p ,
d p ymk 1 , ynk 1  = 2 pymk 1 , ynk 1   pymk 1 , ymk 1   py nk 1 , y nk 1 .
Let k  , and using (9) we get
lim d p ymk 1 , ynk 1  = 2lim pymk 1 , ynk 1  =  .
k 
k 
Which gives
lim p ymk 1 , yn k 1  =
k 

2
.
(17)
Now, consider
d p ymk  , ynk    d p ymk  , ynk 1  d p ynk 1 , ynk  
and
d p ymk  , ynk 1   d p ymk  , ynk   d p ynk  , ynk 1 .
Let k  , in the above inequalities and using (13) and (14) we get
  lim d p y mk  , y nk 1 
k 
lim d p y mk  , y nk 1    .
and
k 
Therefore,
lim d p  ymk  , ynk 1  =  .
(18)
lim d p ynk  , ymk 1  =  .
(19)
k 
Similarly, we can show that
k 
Again by using the definition of d p , we have
d p ymk  , ynk 1 = 2 pymk  , ynk 1   pymk  , ymk   pynk 1 , ynk 1 
Letting k  , and using (9) and (18) we get
lim d p  ymk  , ynk 1 = 2 lim p ymk  , yn k 1 ,
k 
k 
which gives
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M. Akram, W. Shamaila / J. Math. Computer Sci. 12 (2014), 85-98
lim p ymk  , yn k 1  =

k 
2
lim p yn k  , ymk 1  =

.
(20)
.
(21)
Similarly, we can show that
k 
2
Now by substituting x = ymk  and y = yn k  in (1) we have
1


 p ym k  , yn k  , 2  p ym k  , ym k 1   p yn k  , yn k 1 ,
p(Tymk  , Tynk  ) = p( ymk 1 , ynk 1 )  max 

1
  p ym k  , yn k 1   p yn k  , ym k 1 

2


1











p
y
,
y
,
p
y
,
y

p
y
,
y
,
m

k

n

k

m

k

m

k


1
n

k

n

k


1


2
 max 

 
1
  p  y m k  , y n k 1   p  y n k  , y m k 1 


2





Letting k  , and using (8) , (15) , (17) , (20) , (21) and using the continuity of  we get

   
   
 max  ,0,     max  ,0,  ,
2
2 2 
 2 2 

hence

2


 
   .
2
2

A contradiction. Thus yn  is a Cauchy sequence in X , d p . Which gives
lim d p  yn , ym  = 0.
m , n


(22)
Since X , d p is complete so there exists z  X such that lim d p  yn , z  = 0, if and only if
n
pz, z  = lim p yn , z  = lim p yn , ym  =
n 
m , n 
1
lim d p  yn , ym  = 0.
2 m,n
(by Lemma 1.3(II) and (22)). This gives,
pz, z  = lim p yn , z  = lim p yn , ym  = 0.
n
m , n
Now, applying (1) with x = yn and y = z , we have
p(Tyn , Tz) = p( yn1 , Tz)
92
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M. Akram, W. Shamaila / J. Math. Computer Sci. 12 (2014), 85-98
1
1


 max  p yn , z ,  p yn , yn 1   pz, Tz ,  p yn , Tz   pz, yn 1 
2
2



1
1


  max  p yn , z ,  p yn , yn 1   pz , Tz ,  p yn , Tz   pz, yn 1  
2
2



Letting k  , using (8) , (23) , Lemma 1.3 (IV) and the continuity of  we get
pz, Tz  
1
1
 1
pz, Tz     pz, Tz  < pz, Tz .
2
2
 2
Which is possible only if pz,Tz  = 0 and hence Tz = z.
To show the uniqueness of z consider z * as another fixed point of T then by (1),
 

 

1
1



 
p( z, z * )  max  p z, z * , pz, z   p z * , z *     max  p z, z * , pz, z   p z * , z *  .
2
2



 






 

(24)

By using P1 we have pz, z   p z, z * and p z * , z *  p z, z * .
Adding above two inequalities, we get



 


p z , z   p z * , z *  2 p z , z * . .

 
(25)
Using (25) in (24), we have p z, z *  p z, z *   p z, z * . Further by using the property of  we


deduce that p z, z * = 0 and hence, z = z * . Thus z is the unique fixed point of T .
Example 2.2 Suppose X = R  and px, y  = maxx, y; Then  X , p  is a complete partial metric
x2
space. Let T : X  X be defined by Tx =
for all x  X and  : 0,   0,  is such that
1 x
t
 t  =
. Assume that x  y. Then from the contractive condition of Theorem 3.1, we have
1 t


1   x2 
y 2  1  
y2 

  p y,
,  p x,
 
pTx, Ty   max  px, y ,  p x,
2   1 x 

 1  y  2   1  y 


x 2  
 
p y,
 1  x  




1   x2 
y 2  1  
y2 
x 2   

  p y,
,  p x,
  p y,
 
  max  px, y ,  p x,

2
1

x
1

y
2
1

y
1

x








  





93
M. Akram, W. Shamaila / J. Math. Computer Sci. 12 (2014), 85-98
 x2
x2
y2 
x
x2
  x   x  = x 
= p
,
=
. All the conditions of Theorem 3.1 are true,
1 x
1 x 1 x
1 x 1 y 
thus T has a unique fixed point namely, 0.
As
In the following theorem we shall obtain a fixed point theorem on partial metric space for a
generalized weak contractive type mapping.
Theorem 2.3 Let  X , p  be a complete partial metric space and T : X  X be a self-mapping such
that for all x, y  X ,



  pTx, Ty     max  px, y ,

1
 px, Tx  p y, Ty , 1  px, Ty   p y, Tx 
2
2

  maxpx, y , px, Tx ,
(26)
where  : 0,   0,  is a continuous function with  t  = 0 if and only if t = 0,  : 0,   0, 
is monotone non-decreasing and continuous function with  t  = 0 if and only if t = 0 . Then T has a
unique fixed point.
Proof: Let y0  X be fixed. Define a sequence of iterates yn  in X , by yn1 = Tyn for all n  0. If
for some positive integer m, p ym1 , ym  = 0, then by p1 and p2 , ym is the fixed point of T . Hence,
assume that p yn1 , yn   0 for all n  0. Put x = yn , y = yn 1 in (26) we have

1



 p yn , yn 1 , 2  p yn , yn 1   p yn 1 , yn  2 , 
  p yn 1 , yn  2     max 

1

  p yn , yn  2   p yn 1 , yn 1 
 

2


  maxp y n , y n1 , p y n , y n1 .
(27)
From p4 , we have p yn , yn 2   p yn1 , yn1   p yn , yn1   p yn1 , yn2 . ,
therefore (27) becomes



  p yn1 , yn 2     max  p yn , yn1 ,

1
 p yn , yn1   p yn1 , yn2      p yn , yn1 
2

94
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M. Akram, W. Shamaila / J. Math. Computer Sci. 12 (2014), 85-98
If p yn , yn 1  
1
 p yn , yn1   p yn1 , yn2  . Then (28) becomes
2
1
 p yn , yn1   p yn1 , yn2     p yn , yn1 
2

  p yn1 , yn 2    
1

<    p y n , y n1   p y n1 , y n 2 
2

Since  is monotone increasing so p yn 1 , yn  2 
because
  p y n , y n1  > 0
1
 p yn , yn1   p yn1 , yn2 , which gives
2
p yn1 , yn2   p yn , yn1 
If
(29)
1
 p yn , yn1   p yn1 , yn2   p yn , yn1 
2
then (28) becomes   p yn1 , yn 2     p yn , yn1     p yn , yn1 
<   p yn , yn1 
  p yn , yn1  > 0
Hence, we get
p yn1 , yn2   p yn , yn1 
(30)
Since  is monotone non-decreasing, so in both cases p yn , yn1  is monotone decreasing sequence
of non-negative real numbers. Hence, there exists a real number r  0 such that
lim p yn , yn1  = r.
n


If max  p yn , yn 1 ,
(31)
1
 p yn , yn1   p yn1 , yn2  = p yn , yn1 .
2

Then from (28) we have
  p yn1 , yn2     p yn , yn1     p yn , yn1 .
Taking limit as n  , using (31) and the continuity of  and  we get,
 r    r    r , which is a contradiction unless r = 0.
If
95
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M. Akram, W. Shamaila / J. Math. Computer Sci. 12 (2014), 85-98
1

 1
max  p yn , yn1 ,  p yn , yn1   p yn 1 , yn 2  =  p yn , yn 1   p yn 1 , yn  2 ,
2

 2
then from (28) we have
1
 p yn , yn1   p yn1 , yn2     p yn , yn1 .
2

  p yn1 , yn 2    
(33)
Taking the limit as n  , using (31) and the continuity of  and  we get
 r    r    r , which is a contradiction unless r = 0. Thus lim p yn , yn1  = 0.
(34)
n
Also from p1 , we get
lim p yn , yn  = 0.
(35)
n 


Next, we prove that yn  is a Cauchy sequence in the complete metric space X , d p .
Following the steps (10  21) in the proof of Theorem 3.1 and applying (26) with x = yn k  and
y = ymk  we obtain
  pynk 1 , ymk 1 

1



 p yn k  , ym k  , 2 p yn k  , yn k 1   p ym k  , ym k 1  , 
   max 

1

 p yn k  , ym k 1   p ym k  , yn k 1 
 

2







  max pynk  , ymk  , pynk  , ynk 1 .
 
2
 
2
(36)
 
2
Taking limit as k  , in (36) we have           , which is impossible since
 
   > 0. Hence, yn  is a Cauchy sequence in the complete metric space X , d p . So there exists
2
some z  X such that lim d p  yn , z  = 0 if and only if
n 
pz, z  = lim p yn , z  = lim p yn , ym  =
n 
m , n 
Hence
pz, z  = lim p yn , z  = lim p yn , ym  = 0.
n
m, n
(37)
96
1
lim d p  yn , ym  = 0.
2 m,n
M. Akram, W. Shamaila / J. Math. Computer Sci. 12 (2014), 85-98
To show that z is the fixed point of T put x = yn and y = z in (26).



  p yn1 , Tz     max  p yn , z ,

1
 p yn , yn1   pz, Tz , 1  p yn , Tz   pz, yn1  
2
2

  maxp yn , z , p yn , yn1 .
(38)
1

pz, Tz . Which forces pz,Tz  = 0 and hence
2

Letting n  , in (38) we have   pz, Tz    
Tz = z. Thus z is the fixed point of T . In order to prove the uniqueness of z consider z * as another
fixed point of T then for taking x = z and y = z * in (26) we have

  pTz, Tz*  =   pz, z *     max  pz, z * ,







  

1
1

pz, z   p z * , z * , p z, z *  p z * , z  
2
2


 

  max p z, z * , pz, z  .

  
  



Thus  p z, z *   p z, z *   p z, z * . Which is possible only if p z, z * = 0. By using p1 and
p2 we get z = z *. Thus T has a unique fixed point.
Now, we consider an example to support the usability of Theorem 2.3.
Example 2.4 In Example 2.2 if we define  : 0,   0,  by  t  = t , for all t  0, . Then the
contractive condition of Theorem 2.3 is satisfied and we have 0 as the unique fixed point of T .
Corollary 2.5 Let X be a complete partial metric space. Let T : X  X be a self-mapping such that all
elements x, y of X , satisfy

 
 


1




p
x
,
y
,
p x, T m x  p y , T m y

m
m
2
 p T x, T y    max 
1

 p x, T m y  p y , T m x

2








  max px, y , p x, T m x ,
, 

 

(39)
where m is a positive integer and  ,  are as defined in Theorem 2.3. Then T has a unique fixed point
in X .
Proof: Put S = T m in (39) we have,
97
M. Akram, W. Shamaila / J. Math. Computer Sci. 12 (2014), 85-98

1

 px, Sx   p y, Sy , 1  px, Sy   p y, Sx  
2
2



  max px, y , px, Sx 
  pSx, Sy     max  px, y ,
Hence by Theorem 2.3, S has a unique fixed point z that is, Sz = z since S = T m so T m z = z which
 
gives T T m z = T m1 z = T m Tz  = S Tz  = Tz, which shows that Tz is also a fixed point of S . Since S
has a unique fixed point so Tz = z. Hence, z is the fixed point of T . Condition (39) implies the
uniqueness of z.
Conclusion: We have generalized the theorem proved by Rhoades[13] for a self map on a complete
partial metric space and we obtain Matthews’ generalization of Banach’s contraction principle as a special
case of the Theorem 3.1. Moreover, a fixed point theorem for a self map defined for partial metric space
satisfying a generalized  ,   -weak contractive conditions is also proved.
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MTO CROSSING TABLE-DE-V03-NEW.indd - Ntn