Applied Mathematics E-Notes, 15(2015), 147-161 c
Available free at mirror sites of http://www.math.nthu.edu.tw/ amen/
ISSN 1607-2510
Some Fixed Point Theorems For Mappings Involving
Rational Type Expressions In Partial Metric Spaces
Amarjeet Singh Salujay, Mohammad Saeed Khanz,
Pankaj Kumar Jhadex, Brian Fisher{
Received 13 October 2014
Abstract
The aim of this paper is to establish some …xed point theorems for mappings
involving rational expressions in a complete partial metric space using a class of
pairs of functions satisfying certain assumptions. Our result extends and generalizes some well known results of [6, 7, 9] in (usual) metric spaces.
1
Introduction
It is well known that, there are lots of literature dealing with …xed points and common
…xed points for Banach contraction principle in di¤erent spaces. One of the most
interesting space is Partial metric space introduced by Matthews ([16, 17]) in 1994.
In fact, a partial metric space is a generalization of usual metric spaces in which the
self distance for any point need not be equal to zero. The partial metric space has
wide applications in many branches of mathematics as well as in the …eld of computer
domain and semantics. After this remarkable contribution, many authors focused on
partial metric spaces and its topological properties (see, e.g. [1-3, 11-17, 22, 24]).
On the other hand, Banach contraction principle has been generalized in various
ways either by using contractive conditions or by imposing some additional conditions
on the ambient spaces. Das and Gupta [4] were the pioneers in proving …xed point
theorems using contractive conditions involving rational expressions. They prove the
following …xed point theorem in the setting of complete metric space.
THEOREM 1 [6]. Let (X; d) be a complete metric space and T : X ! X a mapping
such that there exists ; > 0 with + < 1 satisfying
d(T x; T y)
d(x; y) +
d(y; T y)[1 + d(x; T x)]
;
1 + d(x; y)
(1)
Mathematics Subject Classi…cations: 47H10, 54H25, 46J10, 46J15.
of Mathematics, J H Government PG College, Betul, India-460001
z Corresponding author, Department of Mathematics & Statistics, Sultan Qaboos University, P. O.
Box 36, Al-Khoud 123, Muscat, Sultanate of Oman
x Department of Mathematics, NRI Institute of Information Science & Technology, Bhopal, MP,
India-462021
{ Department of Mathematics, The University of Leicester, LEI 7RH, England
y Department
147
148
Some Fixed Point Theorems
for all x; y 2 X. Then T has a unique …xed point.
Recently, one of the most important ingredients of a contractivity condition is to
study the kind of involved functions, like altering distance functions introduced by
Khan et al. [17] as follows.
DEFINITION 1 [17]. An altering distance functions is a continuous, nondecreasing
function : [0; 1) ! [0; 1) such that (t) = 0 if and only if t = 0.
In [5], Berzig et al. introduced the notion of pair of generalized altering distance
functions as follows.
DEFINITION 2 [5]. The pair ('; ), where '; : [0; 1) ! [0; 1), is called a pair
of generalized altering distance functions if the following conditions are satis…ed:
(a1) ' is continuous;
(a2) ' is nondecreasing;
(a3) limn!1 (tn ) = 0 ) limn!1 tn = 0.
The condition (a3) was introduced by Popescu in [20] and Moradi and Farajzadeh
in [18]. Notice that the above conditions do not determine the values of '(0) and (0).
In the recent work, Agarwal et al. [4] introduced the following family of function.
';
DEFINITION 3 [4]. We will denote by F the family of all pairs ('; ), where
: [0; 1) ! [0; 1) are functions satisfying the following three conditions.
(F1) ' is nondecreasing;
(F2) if there exists t0 2 [0; 1) such that (t0 ) = 0, then t0 = 0 and '
1
(0) = f0g;
(F3) if fak g; fbk g [0; 1) are sequences such that fak g ! L; fbk g ! L and verifying
L < bk and '(bk ) ('
)(ak ) for all k, then L = 0.
Recently, Karapinar et al. [12] studied the existence and uniqueness of a …xed point
for mappings satisfying rational type contractive condition using auxiliary functions.
The main purpose of this paper is to present some …xed point theorems for contractions
of rational type by using a class of pairs of functions satisfying certain assumptions in
the setting of partial metric spaces.
Saluja et al.
2
149
Preliminaries
The de…nition of partial metric space is given by Matthews [19, 20] as follows:
DEFINITION 4. Let X be a nonempty set and let p : X
X ! [0; 1) satis…es
(1) x = y , p(x; x) = p(y; y) = p (x; y) ;
(2) p (x; x)
p (x; y) ;
(3) p (x; y) = p (y; x) ;
(4) p (x; y)
p (x; z) + p (z; y)
p (z; z)
for all x; y; z 2 X. Then the pair (X; p) is called a partial metric space and p is called
a partial metric on X.
It is clear that if p (x; y) = 0, then x = y. But if x = y then p (x; y) may need not
be zero. Each partial metric p on X generates a T0 topology p on X which has as a
base the family of open p-balls fBp (x; ") : x 2 X; " > 0g where
Bp (x; ") = fy 2 X : p (x; y) < p (x; x) + "g
for all x 2 X and " > 0. Similarly, closed p-balls is de…ned as
Bp [x; "] = fy 2 X : p (x; y)
p (x; x) + "g :
If p is a partial metric on X, then the function dp : X
dp (x; y) = 2p(x; y)
p(x; x)
X ! R+ is given by
p(y; y);
(2)
is a (usual) metric on X.
EXAMPLE 1. Let X = R and p(x; y) = emaxfx;yg for all x; y 2 X. Then (X; p) is
a partial metric space.
DEFINITION 5 [18, 19].
(1) A sequence fxn g in a partial metric space (X; p) converges to x 2 X, if and only
if p (x; x) = limn!1 p (x; xn ).
(2) A sequence fxn g in a partial metric space (X; p) is called a Cauchy sequence if
and only if limn;m!1 p (xn ; xm ) exist and …nite.
(3) A partial metric space (X; p) is said to be complete if every Cauchy sequence
fxn g 2 X converges, with respect to p , to a point x 2 X such that p (x; x) =
limn;m!1 p (xn ; xm ).
(4) A mapping f : X ! X is said to be continuous at x0 2 X, if for each " > 0 ,
there exists > 0 such that f (B (x0 ; )) B (f (x0 ) ; ").
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Some Fixed Point Theorems
The following lemmas of [12] and [17] will be used in the sequel.
LEMMA 1. The following statements hold.
(1) A sequence fxn g is Cauchy in a partial metric space (X; p) if and only if fxn g is
Cauchy in a metric space (X; dp ).
(2) A partial metric space (X; p) is said to be complete if a metric space (X; dp ) is
complete, i.e.
lim dp (x; xn ) = 0 , p (x; x) = lim p (x; xn ) =
n!1
n!1
lim p (xn ; xm ) :
n;m!1
LEMMA 2. Let (X; p) be a partial metric space.
(1) If p (x; y) = 0 then x = y.
(2) If x 6= y then p (x; y) > 0.
LEMMA 3. Let xn ! z as n ! 1 in a partial metric space (X; p), where p (z; z) =
0. Then limn!1 p (xn ; y) = p (z; y) for every y 2 X.
3
Main Results
We start this section presenting the following class of pairs of functions F . A pair of
functions ('; ) is said to belong to the class F , if they satisfy the following conditions:
(i) ';
: [0; 1) ! [0; 1);
(ii) for t; s 2 [0; 1), if ' (t)
(s), then t
s;
(iii) for ftn g and fsn g sequences in [0; 1) such that limn!1 tn = limn!1 sn = a , if
' (tn )
(sn ) for any n 2 N, then a = 0.
Notice that, if a pair '; veri…es (F1) and (F2), then the pair ('; = '
)
satis…es (i) and (ii). Furthermore, if ('; = '
) satis…es (iii), then ('; ) satis…es
(F3).
EXAMPLE 2. The conditions (i)–(iii) of the above de…nition are ful…lled for the
functions '; : [0; 1) ! [0; 1) de…ned by
'(t) = ln
1 + 2t
1+t
and (t) = ln
for all t 2 [0; 1):
2
2
REMARK 1. Note that, if ('; ) 2F and ' (t)
(t), then t = 0. Since we can
take tn = sn = t for any n 2 N and by condition (iii), we deduce that t = 0.
Saluja et al.
151
Now we present some interesting examples of pairs of functions belonging to the
class F .
EXAMPLE 3. Let ' : [0; 1) ! [0; 1) be a continuous and increasing function such
that ' (t) = 0 if and only if t = 0 (these functions are known as the altering distance
function in the literature).
Let : [0; 1) ! [0; 1) be a non-decreasing function such that (t) = 0 if and only
if t = 0 and suppose that
'. Then the pair ('; '
) 2F . In fact, it is clear that
('; '
) satis…es condition (i).
Now we will prove condition (ii). For this, let us suppose that t; s 2 [0; 1) and
' (t) ('
) (s). Then from ' (t) ' (s)
(s) ' (s) and since ' is increasing,
we can deduce that t s. Finally, to prove (iii), we assume that
' (tn )
' (sn )
(sn ) ;
(3)
for any n 2 N, where tn ; sn 2 [0; 1) and
lim tn = lim sn = a:
n!1
n!1
Taking limit as n ! 1 in (3), we infer that limn!1 (sn ) = 0.
Now, let us suppose that a > 0. Since limn!1 sn = a > 0, we can …nd " > 0 and
a sequence fsnk g of fsn g such that snk > " for any k 2 N. As is non-decreasing,
we have (snk ) > (") for any k 2 N and consequently, limk!1 (snk )
("). This
contradicts the fact that limk!1 (snk ) = 0. Therefore, a = 0. This proves that
('; '
) 2F .
An interesting particular case is, when ' is the identity mapping, ' = 1[0;1) and
: [0; 1) ! [0; 1) is a non-decreasing function such that (t) = 0 if and only if t = 0
and (t) t for any t 2 [0; 1).
EXAMPLE 4. Let S be the class of functions de…ned by
n
o
S=
: [0; 1) ! [0; 1) : If lim (tn ) = 1 ) lim tn = 0 :
n!1
n!1
2 S and 1[0;1)
Let us consider the pair of functions 1[0;1) ; 1[0;1) , where
is de…ned by
1[0;1) (t) = (t) t for t 2 [0; 1). Then it is easy to prove that
1[0;1) ; 1[0;1) 2 F for 2 S.
REMARK 2. Suppose that g : [0; 1) ! [0; 1) is an increasing function and
('; ) 2F . Then it is easily seen that the pair (g '; g
) 2F .
Now we are in a position to prove our main results.
THEOREM 2. Let (X; p) be a complete partial metric space and T : X ! X a self
map such that there exists a pair of functions ('; ) 2 F such that
' (p (T x; T y))
max
(p (x; y)) ;
p (y; T y)
1 + p (x; T x)
1 + p (x; y)
152
Some Fixed Point Theorems
for all x; y 2 X . Then T has a unique …xed point.
PROOF. Let x0 2 X be arbitrary. We construct a sequence fxn g in X as follows:
xn+1 = T xn for n
0:
(4)
Now applying contractive condition, we have
'(p(xn+1 ; xn )) = '(p(T xn ; T xn
1 ))
max
(p(xn ; xn
1 ));
p(xn
max
(p(xn ; xn
1 ));
p(xn
1 + p(xn ; T xn )
1 + p(xn ; xn 1 )
1 + p(xn ; xn+1 )
:
1 ; xn )
1 + p(xn ; xn 1 )
1 ; T xn 1 )
(5)
Now, let us assume that there exists n0 2 N such that p(xn0 +1 ; xn0 ) = 0. In this case,
xn0 +1 = xn0 and consequently, by (4)
T xn0 = xn0 +1 = xn0 ;
i.e. xn0 would be the …xed point of T . So we may assume that p(xn+1 ; xn ) 6= 0 for any
n 2 N. Now we consider two cases.
Case 1. Consider
max
(p(xn ; xn
1 ));
p(xn
1 ; xn )
1 + p(xn ; xn+1 )
1 + p(xn ; xn 1 )
= (p(xn ; xn
1 )):
(6)
Then from (5), we have
'(p(xn+1 ; xn ))
(p(xn ; xn
and since ('; ) 2 F , we deduce that p(xn+1 ; xn )
Case 2. If
max
=
(p(xn ; xn
p(xn
1 ; xn )
1 ));
p(xn
1 ));
p(xn ; xn
1 ; xn )
1 + p(xn ; xn+1 )
1 + p(xn ; xn 1 )
(7)
1 ).
1 + p(xn ; xn+1 )
1 + p(xn ; xn 1 )
;
(8)
1 + p(xn ; xn+1 )
1 + p(xn ; xn 1 )
(9)
then from (5), we obtain
'(p(xn+1 ; xn ))
p(xn
1 ; xn )
and since ('; ) 2 F , we deduce that
p(xn+1 ; xn )
which implies that p(xn+1 ; xn )
p(xn
p(xn ; xn
1 ; xn )
1 ).
1 + p(xn ; xn+1 )
;
1 + p(xn ; xn 1
Saluja et al.
153
Therefore, from both the cases we conclude that fp(xn+1 ; xn )g is a decreasing sequence of non-negative real numbers.
Put r = limn!1 p(xn+1 ; xn ), where r 0, and denote
A = fn 2 N : n satis…es (6)g and B = fn 2 N : n satis…es (8)g:
Now, we make the following remark.
(1). If Card A = 1, then from (5), we can …nd in…nitely many natural numbers
satisfying (7) and since limn!1 p(xn+1 ; xn ) = limn!1 p(xn ; xn 1 ) = r and ('; ) 2 F ,
we deduce that r = 0.
(2). If Card B = 1, then from (5), we can …nd in…nitely many natural numbers
satisfying (9) and since ('; ) 2 F , and using the same argument one used in Case 2,
we obtain
1 + p(xn ; xn+1 )
:
p(xn+1 ; xn ) p(xn 1 ; xn )
1 + p(xn ; xn 1
Taking limit as n ! 1 in the last inequality and taking into account that
lim p(xn+1 ; xn ) = r;
n!1
one can deduce that r = 0.
Therefore,
lim p(xn+1 ; xn ) = 0:
(10)
n!1
Due to inequality (2), we have
dp (xn+1 ; xn )
2p(xn+1 ; xn );
therefore,
lim dp (xn+1 ; xn ) = 0:
(11)
n!1
Now, we prove that fxn g is a Cauchy sequence in X, i.e. we prove that
lim p(xn ; xm ) = 0:
n;m!1
In the contrary case, since limn!1 p(xn+1 ; xn ) = 0, we can …nd " > 0 and subsequences
fxm(k) g and fxn(k) g of fxn g such that n(k) is the smallest index for which
n(k) > m(k) > k;
p(xn(k) ; xm(k) )
":
(12)
This means that
p(xn(k)
1 ; xm(k) )
":
(13)
From (12) and (13), we have
"
p(xn(k) ; xm(k) )
p(xn(k) ; xn(k)
1)
p(xn(k) ; xn(k)
+ p(xn(k)
1)
+ p(xn(k)
1 ; xm(k) )
1 ; xm(k) )
< " + p(xn(k) ; xn(k)
p(xn(k)
1 ):
1 ; xn(k) 1 )
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Some Fixed Point Theorems
Taking the limit as k ! 1 and using (10), we get
lim p(xn(k) ; xm(k) ) = ":
(14)
k!1
By the triangular inequality, we have
p(xn(k) ; xm(k) )
p(xn(k) ; xn(k)
1)
+ p(xn(k)
1 ; xm(k) )
p(xn(k) ; xn(k)
1 ) + p(xn(k)
1 ; xm(k) )
p(xn(k) ; xn(k)
1)
1 ; xm(k) 1 )
+ p(xm(k)
1 ; xm(k) )
1 ; xm(k) 1 )
+ p(xm(k)
1 ; xm(k) )
p(xm(k)
+ p(xn(k)
p(xn(k)
1 ; xn(k) 1 )
1 ; xm(k) 1 )
p(xn(k) ; xn(k)
1)
+ p(xn(k)
and
P (xn(k)
1 ; xm(k) 1 )
p(xn(k)
1 ; xn(k) )
+ p(xn(k) ; xm(k)
1)
p(xn(k)
1 ; xn(k) ) + p(xn(k) ; xm(k)
1)
p(xn(k)
1 ; xn(k) )
p(xn(k) ; xn(k) )
+ p(xn(k) ; xm(k) )
p(xm(k) ; xm(k)
1)
+ p(xn(k) ; xm(k) )
p(xm(k) ; xm(k)
1 ):
p(xm(k) ; xm(k) )
p(xn(k)
1 ; xn(k) )
Taking the limit as k ! 1 in the above two inequalities and using (10) and (14), we
get
lim p(xn(k) 1 ; xm(k) 1 ) = ":
(15)
n!1
Now applying the contractive condition, we have
'(p(xm(k) ; xn(k) )) = '(p(T xm(k) 1 ; T xn(k) 1 ))
n
max (p(xm(k) 1 ; xn(k) 1 ));
1 + p(xm(k) 1 ; T xm(k) 1 ) o
p(xn(k) 1 ; T xn(k) 1 )
1 + p(xm(k) 1 ; xn(k) 1 )
n
= max (p(xm(k) 1 ; xn(k) 1 ));
1 + p(xm(k) 1 ; xm(k) ) o
p(xn(k) 1 ; xn(k) )
for k 2 N:
1 + p(xm(k) 1 ; xn(k) 1 )
(16)
Let us put
C = fk 2 N : '(p(xm(k) ; xn(k) ))
D=
k 2 N : '(p(xm(k) ; xn(k) ))
p(xn(k)
(p(xm(k)
1 ; xn(k) )
1 ; xn(k) 1 ))g;
1 + p(xm(k) 1 ; xm(k) )
1 + p(xm(k) 1 ; xn(k) 1 )
:
By (16), we have Card C = 1 or Card D = 1.
Let us suppose that Card C = 1. Then there exists in…nitely many k 2 N such
that
'(p(xm(k) ; xn(k) ))
(p(xm(k) 1 ; xn(k) 1 ));
Saluja et al.
155
and since ('; ) 2 F , we get
p(xm(k) ; xn(k) )
p(xm(k)
1 ; xn(k) 1 ):
Taking the limit as k ! 1 in the above inequality, we get " = 0 a contradiction.
Let us suppose that Card D = 1. In this case, we can …nd in…nitely many k 2 N
such that
'(p(xm(k) ; xn(k) ))
p(xn(k)
1 ; xn(k) )
1 + p(xm(k) 1 ; xm(k) )
1 + p(xm(k) 1 ; xn(k) 1 )
;
and since ('; ) 2 F , we infer
p(xm(k) ; xn(k) )
p(xn(k)
1 ; xn(k) )
1 + p(xm(k) 1 ; xm(k) )
:
1 + p(xm(k) 1 ; xn(k) 1 )
Taking limit as k ! 1 and in view of (10) and (15), it follows that " 0 and we get
a contradiction.
Therefore, in both possibilities, we obtain a contradiction and so we have
lim p(xn ; xm ) = 0:
n;m!1
Since limn;m!1 p(xn ; xm ) exists and …nite, we conclude that fxn g is a Cauchy sequence
in (X; p). Due to inequality (2), we have dp (xn ; xm ) 2p(xn ; xm ). Therefore
lim dp (xn ; xm ) = 0:
n;m!1
Thus, by Lemma 1, fxn g is a Cauchy sequence in (X; dp ) and (X; p). Since (X; p) is a
complete partial metric space, there exists x 2 X such that limn!1 p(xn ; x) = p(x; x).
Since limn;m!1 p(xn ; xm ) = 0, then again by Lemma 1, we have p(x; x) = 0.
Next, we will prove that x is a …xed point of T . Suppose that T x 6= x. Now
applying contractive condition and Lemma 3, we have
'(p(T x; T xn ))
max
(p(x; xn ));
p(xn ; T xn )
1 + p(x; T x)
1 + p(x; xn )
:
We can distinguish two cases again.
Case 1. There exist in…nitely many n 2 N such that
'(p(T x; T xn ))
(p(x; xn )):
Since ('; ) 2 F , we obtain
p(T x; T xn )
Letting n ! 1, we get
p(x; xn ):
lim T xn = T x;
n!1
(17)
156
Some Fixed Point Theorems
where to simplify our consideration, we will denote the subsequence by the same symbol
fT xn g. By (4)
T x = lim T xn = lim xn+1 :
(18)
n!1
n!1
Since xn ! x in X, this means that lim sup p(xn ; x) ! 0 and consequently,
lim xn+1 = x:
n!1
From this last result and (18), we deduce that T x = x, which means that x is a …xed
point of T .
Case 2 There exist in…nitely many n 2 N such that
'(p(T x; T xn ))
p(xn ; T xn )
1 + p(x; T x)
1 + p(x; xn )
:
To simplify our consideration, we will denote the subsequence by the same symbol
fT xn g. Since ('; ) 2 F and T xn = xn+1 , we have
p(T x; T xn )
p(xn ; T xn )
1 + p(x; T x)
for any n 2 N:
1 + p(x; xn )
Taking the limit as n ! 1 and by (11), limn!1 p(xn ; xn+1 ) = 0, we infer (17). From
the above Case 1, we deduce that x is a …xed point of T .
Therefore, in both the cases we have shown that x is a …xed point of T .
Finally, we will prove the uniqueness of the …xed point. Suppose that y is another
…xed point of T such that x 6= y. Now using contractive condition, we get
'(p(x; y)) = '(p(T x; T y))
max
(p(x; y));
p(y; T y)
1 + p(x; T x)
1 + p(x; y)
maxf (p(x; y)); (0)g:
(19)
Now there are two cases.
Case 1. Consider maxf (p(x; y)); (0)g = (p(x; y)). In this case, from (19)
'(p(x; y))
(p(x; y)):
Since ('; ) 2 F and by Remark 1, we deduce that p(x; y) = 0, that is, x = y.
Case 2. Consider maxf (p(x; y)); (0)g = (0). Then from (19), we obtain
'(p(x; y))
(0):
Since ('; ) 2 F , we infer that p(x; y) 0. Therefore, p(x; y) = 0, that is, x = y.
Hence in both the cases x = y, that is, the …xed point is unique. This completes
the proof of the Theorem 2.
From Theorem 2, we obtain the following corollaries.
Saluja et al.
157
COROLLARY 1. Let (X; p) be a complete partial metric space and T : X ! X be
a self map such that there exists a pair of functions ('; ) 2 F satisfying
'(p(T x; T y))
(p(x; y));
for all x; y 2 X. Then T has a unique …xed point in X.
COROLLARY 2. Let (X; p) be a complete partial metric space and T : X ! X a
self map such that there exists a pair of functions ('; ) 2 F satisfying
'(p(T x; T y))
p(y; T y)
1 + p(x; T x)
1 + p(x; y)
;
for all x; y 2 X. Then T has a unique …xed point in X.
REMARK 3. Notice that the contractive condition appearing in Theorem 1,
d(T x; T y)
for all x; y 2 X, where ;
d(T x; T y)
d(x; y) +
> 0 with
+
d(y; T y)[1 + d(x; T x)]
;
1 + d(x; y)
< 1, implies that
d(y; T y)[1 + d(x; T x)]
1 + d(x; y)
d(y; T y)[1 + d(x; T x)]
max ( + )d(x; y); ( + )
1 + d(x; y)
( + ) max d(x; y);
for all x; y 2 X.
This condition is a particular case of the contractive condition appearing in Theorem
2 with the pair of functions ('; ) 2 F given by ' = 1[0;1) and = ( + )1[0;1) .
Therefore, Theorem 1 is a particular case of the following corollary and considered as
an extension and generalizations of Theorem 1 in the setting of complete partial metric
spaces.
COROLLARY 3. Let (X; p) be a complete partial metric space and T : X ! X be
a self map such that there exists a pair of functions ('; ) 2 F satisfying
d(T x; T y)
max ( + )d(x; y); ( + )
d(y; T y)[1 + d(x; T x)]
1 + d(x; y)
for all x; y 2 X. Then T has a unique …xed point in X.
Taking into account Example 3, we have the following corollary.
COROLLARY 4. Let (X; p) be a complete partial metric space and T : X ! X be
a self map such that there exists two functions '; : [0; 1) ! [0; 1) such that
n
'(p(T x; T y)) max '(p(x; y))
(p(x; y));
1 + p(x; T x)
1 + p(x; T x) o
' p(y; T y)
p(y; T y)
1 + p(x; y)
1 + p(x; y)
158
Some Fixed Point Theorems
for any x; y 2 X, where ' is a continuous and increasing function satisfying '(t) = 0
if and only if t = 0 and is a nondecreasing function such that (t) = 0 if and only if
t = 0 and
'. Then T has a unique …xed point in X.
Corollary 4 has the following consequences.
COROLLARY 5. Let (X; p) be a complete partial metric space and T : X ! X be
a self map such that there exists two functions '; : [0; 1) ! [0; 1) such that
'(p(T x; T y))
'(p(x; y))
(p(x; y))
for any x; y 2 X, where ' is an increasing function and is a nondecreasing function
and they satisfy '(t) = (t) = 0 if and only if t = 0 and ' is continuous with
'.
Then T has a unique …xed point in X.
Corollary 5 can be considered as an extension of the following result about …xed
point theorems which appears in [7] in the setting of complete partial metric space.
THEOREM 3. Let (X; d) be a complete metric space and T : X ! X a mapping
satisfying
'(d(T x; T y)) '(d(x; y))
(d(x; y))
for any x; y 2 X, where ' and
has a unique …xed point.
satisfy the same conditions as in Corollary 5. Then T
COROLLARY 6. Let (X; p) be a complete partial metric space and T : X ! X
be a self map such that there exists two functions '; : [0; 1) ! [0; 1) satisfying the
same conditions as in Corollary 5 such that
'(p(T x; T y))
' p(y; T y)
1 + p(x; T x)
1 + p(x; y)
p(y; T y)
1 + p(x; T x)
1 + p(x; y)
for any x; y 2 X. Then T has a unique …xed point.
Taking into account Example 4, we have the following corollary.
COROLLARY 7. Let (X; p) be a complete partial metric space and T : X ! X be
a self map such that there exist 2 S (see Example 4) satisfying
n
p(T x; T y) max (p(x; y))p(x; y);
1 + p(x; T x) o
1 + p(x; T x)
p(y; T y)
p(y; T y)
1 + p(x; y)
1 + p(x; y)
for any x; y 2 X. Then T has a unique …xed point.
Corollary 7 has the following consequence.
Saluja et al.
159
COROLLARY 8. Let (X; p) be a complete partial metric space and T : X ! X be
a self map such that there exist 2 S (see Example 4) satisfying
p(T x; T y)
(p(x; y))p(x; y)
for any x; y 2 X. Then T has a unique …xed point.
Corollary 8 can be considered as an extension of the following result about …xed
point theorems which appears in [9] in the setting of complete partial metric spaces.
THEOREM 4. Let (X; d) be a complete metric space and T : X ! X a mapping
satisfying
d(T x; T y)
(d(x; y))d(x; y)
for any x; y 2 X, where
4
2 S (see Example 4). Then T has a unique …xed point.
Example
In this section, we give an example in support of our main result.
EXAMPLE 5. Let X = [0; 1] and p(x; y) = maxfx; yg, then (X; p) is a partial
metric space. Suppose T : X ! X such that T x = x2 for all x 2 X. De…ne the function
'; : [0; 1) ! [0; 1) as follows:
'(x) = ln
5x
1
+
12 12
Without loss of generality, assume that x
'(p(T x; T y)) = ln
= ln
1
3x
+
12 12
and (x) = ln
for all x 2 [0; 1):
y. Then we have
1
5
+ p(T x; T y)
12 12
1
1
+ x :
12 12
= ln
1
5
+
maxfx; yg
12 12
On the other hand,
3
1
+ p(x; y)
12 12
1
3
= ln
+ x
12 12
(p(x; y)) = ln
= ln
1
3
+
maxfx; yg
12 12
and
p(y; T y)
1 + p(x; T x)
1
3
1 + p(x; T x)
= ln
+
p(y; T y)
1 + p(x; y)
12 12
1 + p(x; y)
1
3
1+x
= ln
+
y
12 12 1 + x
1
3
= ln
+ y :
12 12
160
Some Fixed Point Theorems
Therefore,
max
n
(p(x; y));
p(y; T y)
n
1 + p(x; T x) o
= max ln
1 + p(x; y)
1
= ln
+
12
1
3
1
3 o
+ x ; ln
+ y
12 12
12 12
3
x :
12
Combining the observations above, we get
1
1
1
3
+ x
ln
+ x
12 12
12 12
n
1 + p(x; T x) o
:
= max (p(x; y)); p(y; T y)
1 + p(x; y)
'(p(T x; T y)) = ln
Thus all the conditions of Theorem 2 are satis…ed. Hence T has a unique …xed point,
indeed x = 0 is the required …xed point.
Acknowledgment. The authors are thankful to the anonymous referees for their
valuable comments and suggestions which improves the presentation of the paper.
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