Available online through
ISSN 2229 – 5046COMMON FIXED POINTS
OF MAPPINGS SATISFYING TRIANGLE INEQUALITY OF INTEGRAL TYPE
U. C. GAIROLA*, A. S. RAWAT**
*Department of Mathematics,
Pauri Campus of H. N. B. Garhwal University, Pauri 246001, India.
**Shiboo Nagar, Tariyal Chauk, Devi Road, Kotdwar 246149, Uttarakhand, India.
(Received On: 17-07-16; Revised & Accepted On: 31-07-16)
ABSTRACT
T
he purpose of this paper is to consider a new approach for obtaining common fixed point theorems in metric spaces by subjecting the triangle inequality to a contractive condition of integral type. We use the concept of property (E.A) andR
-weak commutativity there, without the assumption of completeness of the space and the continuity of the mappings. Our results generalize and extend the results of Pant et.al [14] and others.Ams(Mos) Subject Classifications: 54H25, 47H10.
Key Words: Noncompatible maps, contractive condition, property (E.A). R-weak commutativity.
1. INTRODUCTION
With the advent of the notion of compatible maps due to Jungck [5]. The study of common fixed point theorems for contractive type maps emerged as an area of vigorous research centered around the study of compatible maps and its weaker forms. However the study of common fixed points of noncompatible maps is equally interesting. Pant [9-13] initiated work along these lines by using the notion of pointwise R-weakly commuting mappings.
Definition 1.1[9]: Self-maps
f
andg
of a metric space( , )
X d
areR
-weakly commuting at a pointx
∈
X
if(
,
)
(
,
)
d ASx SAx
≤
Rd Ax Sx
for someR
>
0
. They are pointwiseR
- weakly commuting onX
if givenx
∈
X
there existsR
>
0
such thatd ASx SAx
(
,
)
≤
Rd Ax Sx
(
,
)
.Describing the importance of pointwise
R
- weakly commuting maps in fixed point consideration. Pant [11] proved that the notion of pointwiseR
- weak commutativity is equivalent to commutativity at coincidence points. Junck [6] defines the commutativity at coincidence point with the notion of weakly compatible maps.Definition 1.2 [6]: Self-maps
f
andg
of a metric space( , )
X d
are weakly compatible if they commute at their coincidence point, that isfgu
=
gfu
whenever
fu
=
gu
for u
∈
x
.
Thus two maps are weakly compatible if and only if they are pointwise
R
- weakly commuting mappings. However pointwiseR
- weakly commuting maps need not be compatible as shown in Example 1 of [14].Further generalizing the concept of noncompatible maps Amri and Moutawakil [2] have introduced a new property called property (E.A). So that compatible and noncompatible maps may be studied together.
Definition 1.3 [2]: Let
f
andg
be two self-maps of a metric space( , )
X d
. Then they are said to satisfy the property (E.A), if there exists a sequencex
n inX
such thatlim
n→∞fx
n=
lim
n→∞gx
n=
t
, for somet
∈
X
If two maps are noncompatible then they satisfy the property (E.A). The converse however is not necessarily true.
Example 1.1: Let
X
=
[0,
+∞
)
. Definef g X
, :
→
X
by,
3
,
.
2
2
x
x
fx
=
gx
=
∀ ∈
x
X
Consider the sequence1
{ }
x
n=
n,
n
∈
N
. Clearlylim
n→∞fx
n=
lim
n→∞gx
n=
0
. Thusf
andg
satisfy the property (E.A), and also they are weakly compatible since they commute at their coincidence point 0 and pointwiseR
-weakly commuting, butf
andg
are not necessarily noncompatible.Example 1.2: Let
X
=
. Definef g X
, :
→
X
by, ,
3
x
fx
=
x
∈
X
gx
=
x
2,
x
∈
X
.
here 0 and13are two
coincidence points for the maps
f
andg
. Note thatf
andg
commute at 0 i.e.fg
(0)
=
gf
(0)
=
0
, but 1 13 27
( )
fg
=
and 1 1
3 8
( )
gf
=
. Sof
andg
are not weakly compatible. Consider the sequence{ }
x
n1
:
n
N
n
=
∈
we havelim
n→∞fx
n=
lim
n→∞gx
n=
0
. Thusf
andg
satisfy property (E.A).Example 1.3: Let
X
=
[
2,
∞
)
. Definef g X
, :
→
X
by1,
2
2
, 2,
[
)
.
2
3
x
x
fx
= +
gx
=
+
∀ ∈
x
∞
Here 2 is the coincidence point for the mapsf
andg
. Note thatf
andg
commute at 2 sincefg
(2)
=
gf
(2)
=
2
. Sof
andg
are weakly compatible. Suppose that property(E.A) holds. Then, there exists a sequence{ }
x
n inX
satisfyinglim
n→∞fx
n=
lim
n→∞gx
n=
t
for somet
∈
X
. Thereforelim
n→∞x
n= −
2
t
1
andlim
3
2
2
n n
t
x
→∞−
=
. Thent
=
0
which is a contradiction since
0
∉
X
. Hence,f
andg
do not satisfy property (E.A).Notice that from the Example 1.2-1.3 we can see that weakly compatible and property(E.A) are independent to each other.
In 2002 Branciari [3] obtained the fixed point theorem for a mapping satisfying an analogue of Banach contraction principle for integral type inequality stated as follows.
Define Φ= { :
φ
R+→R+is a lebesgue integrable mapping which is summable, non-negative}andϕ
satisfies thefollowing inequality
.
0
>
each
for
0
>
(t)dt
0
ϕ
ε
ε
∫
(1.1)Theorem 1.1 [3]: Let
( , )
X d
be a complete metric space,c
∈
[
0,1
)
and iff X
:
→
X
be a mapping such that for each,
,
x y
∈
X
( , ) ( , )
0
(t)dt
0( )
d fx fy d x y
c
t dt
ϕ
≤
ϕ
∫
∫
,where
ϕ ∈Φ
and satisfy the condition (1.1) thenf
has a unique fixed pointa
∈
X
, such that for eachx
∈
X
,
lim
n→∞f x
n=
a
In the present paper using the notion of pointwise
R
- weak commutativity we demonstrate that triangle inequality can be used to establish common fixed point theroem by subjecting it to contractive condition of integral type. We use the concept of property(E.A) and pointwiseR
- weak commutativity there, without the assumption of completeness of the space and the continuity of the mappings. Our result generalize and extend the results of Pant et.al [14] and others.2.MAIN RESULTS
Now we state our main theorem.
Theorem 2.1: Let
f
andg
be pointwiseR
-weakly commuting selfmappings of a metric space(
X
,
d
)
satisfying the property (E.A) and the following conditions hold:(2.1)
fX
⊆
gX
,
wherefX
is the closure of the range off
and(2.2) d fx fy
(
t
)
dt
k
[d(fx,gx) d(gx,gy) d(gy,fy)](
t
)
dt
0 )
, (
0
ϕ
∫
ϕ
∫
≤
+ + , whereo
≤
k
<
1,
ϕ
∈
Φ
and satisfy the condition (1.1). Thenf
andg
have a unique common fixed point.Proof: Since
f
andg
satisfy the property (E.A), then there exists a sequence{
x
n}
inX
such thatlim
nfx
n=
lim
ngx
n=
t
for some
t
∈
X
.
Sincet
∈
fX
andfX
⊂
gX
then there exists au
∈
X
such thatt
=
gu
.
Using (2.2) we getdt
t
k
dt
t
d fxn gxn d gxngu d gu fu fu n fx d)
(
)
(
[ ( , ) ( , ) ( , )] 0 ) , (0
ϕ
∫
ϕ
∫
≤
+ + .By taking the limit, as
n
→
∞
we have.
)
(
=
)
(
)
(
( , ) 0 )] , ( 0 [0 0 ) , (0
t
dt
k
t
dt
k
t
dt
fu gu d fu gu d gu fu d
ϕ
ϕ
ϕ
∫
∫
∫
≤
+ + So0
=
)
(
)
(1
( , )0
t
dt
k
∫
d fuguϕ
−
.Since
(1
−
k
)
≠
0
, then using (1.1) we getfu
=
gu
.
PointwiseR
−
weak commutativity off
andg
implies forR
>
0
that)
,
(
)
,
(
fgu
gfu
Rd
fu
gu
d
≤
.So
0.
=
)
(
)
(
( , ) 0 ) , (0
t
dt
t
dt
gu fu Rd gfu fgu d
ϕ
ϕ
∫
∫
≤
Using (1.1) we have
fgu
=
gfu
.
Alsoffu
=
fgu
=
gfu
=
ggu
.
Using (2.2) again we getdt
t
k
dt
t
d fugu d gugfu d gfu ffu ffu fu d)
(
)
(
[ ( , ) ( , ) ( , )] 0 ) , (0
ϕ
∫
ϕ
∫
≤
+ +dt
t
k
d fu ffu(
)
=
( , ) 0ϕ
∫
0
=
)
(
)
(1
( , )0
t
dt
k
∫
d fu ffuϕ
−
.Since
(1
−
k
)
≠
0
, then using (1.1) we getfu
=
ffu
.
Hencefu
=
fgu
=
gfu
andfu
is a common fixed point off
and
g
. Now to show the uniqueness of the common fixed point off
andg
, letf
andg
have two distinct common fixed pointv
andw
. Using (2.2) we havedt
t
k
dt
t
k
dt
t
dt
t
d fvfw d fvgv d gvgw d gw fw d vww v d
)
(
=
)
(
)
(
)
(
( , ) 0 )] , ( ) , ( ) , ( [ 0 ) , ( 0 ) , (0
ϕ
∫
ϕ
∫
ϕ
∫
ϕ
Since
k
<
1
so we have a contradiction. Thusv
=
w
.Now we give an example to illustrate the above theorem.
Example 2.1: Let
X
=
[
0
,
1
]
andd
be the usual metric onX
.
Definef
,
g
:
X
→
X
by
∈
≤
<
+
x
n
n
N
n
n
x
fx
,
1
1
1
if
2
1
0
=
if
0
=
.
,
1
1
1
if
1
0
=
if
0
=
∈
≤
<
+
x
n
n
N
n
n
x
gx
Now to check the property(E.A) we consider the sequence
∈
N
n
n
y
n:
1
=
}
{
, which satisfies0
=
2
1
lim
=
,0)
2
1
(
lim
=
,0)
(
lim
n
n
d
fy
d
n n nn→∞ →∞ →∞
and
lim
(
,0)
=
lim
(
1
,0)
=
lim
1
=
0.
n
n
d
gy
d
n n nn→∞ →∞ →∞
Hence
lim
n→∞fy
n=
lim
n→∞gy
n=
0
∈
X
so the pair(
f
,
g
)
satisfies property (E.A). However one can see that nn n
n
fgy
=
lim
gfy
lim
→∞ →∞ so the pair(
f
,
g
)
is not noncompatible. Also
∈
∪
n
N
n
fX
:
2
1
{0}
=
and
∈
∪
n
N
n
gX
=
{0}
1
:
and{0}
1
:
2
fX
n
N
n
=
∪
∈
. ThusfX
⊂
gX
.
Define a map
ϕ
∈
Φ
byϕ
(
t
)
=
3
t
2fort
>
0
andϕ
(0)
=
0
, whereϕ
satisfies the condition (1.1). Then for any0,
>
τ
3
0
ϕ
(
)
=
τ
τ
dt
t
∫
.Now we consider the following four cases.
Case-I: If
x
=
0 and
y
∈
(
n1+1,
1n
then1
1
1
(
,
)
(
,
)
(
,
)
0
2
d fx gx
d gx gy
d gy fy
n
n
n
+
+
= + + −
1
1
2
n
n
= +
3
2
n
=
1
(
,
)
.
2
d fx fy
n
So we have
3 3
( , )
0
( , ) ( , ) ( , )
0
1
1 3
( )
2
3 2
1
( ) .
27
d fx fy
d fx gx d gx gy d gy fy
t dt
n
n
t dt
ϕ
ϕ
+ +
=
=
=
∫
∫
Case-II: If
x
∈
(
n1+1,
n1
and
y
∈
(
m1+1,
m1
,
m
>
n
,
then1
1
1
1
1
1
(
,
)
(
,
)
(
,
)
2
2
d fx gx
d gx gy
d gy fy
n
n
n
m
m
m
+
+
=
− + −
+
−
1
1
(
,
)
2
2
d fx fy
n
m
=
−
.
2
m n
mn
−
=
So
3 3
( , )
0
3
( , ) ( , ) ( , )
0
3
( )
.
2
2
3
( ) .
3
d fx fy
d fx gx d gx gy d gy fy
m n
m n m n
t dt
mn
mn
m n
m n
t dt
m n
ϕ
ϕ
+ +
−
−
−
=
=
−
−
=
−
∫
∫
Case-III: If
(
1 1(
1 1 1,
and
1,
,
n n m m
x
∈
+
y
∈
+
m
<
n
then similarly as in Case (ii) we have3 3
( , )
0
3
( , ) ( , ) ( , )
0
3
( )
.
2
2
3
( ) .
3
d fx fy
d fx gx d gx gy d gy fy
n m
n m n m
t dt
mn
mn
n m
n m
t dt
n m
ϕ
ϕ
+ +
−
−
−
=
=
−
−
=
−
∫
∫
Case-IV: If
x y
,
∈
(
n1+1,
1n
, then1
1
1
1
(
,
)
(
,
)
(
,
)
0
2
2
1
1
2
2
1
d fx gx
d gx gy
d gy fy
n
n
n
n
n
n
n
+
+
=
− + + −
=
+
=
(
,
)
0.
d fx fy
=
So( , ) ( , ) ( , ) ( , )
0
( )
0
0( )
d fx fy d fx gx d gx gy d gy fy
t dt
t dt
ϕ
= <
+ +ϕ
∫
∫
.Hence in all the above cases there exists a
3 3
1
0
max
,
1
27
3
3
m n
n m
k
or
m n
n m
−
−
≤ =
<
−
−
for which the contractivecondition (2.2) is satisfied. Thus all the conditions of the above theorem are satisfied and 0 is the unique fixed point of
f
In the next theorem we further improve Theorem 1.1 by replacing condition (2.2) with much general inequality and using the strict triangle inequality of integral type.
Theorem 2.2: Let
f
andg
be noncompatible pointwiseR
-weakly commuing selfmappings of a metric space(
X
,
d
)
satisfying the property (E.A), (2.1) and the following conditions hold:
(2.3) d fxfy
(
t
)
dt
a
d fxgx(
t
)
dt
b
d gxgy(
t
)
dt
c
d(gy,fy)(
t
)
dt
0 ) , ( 0 ) , ( 0 ) , (0
ϕ
∫
ϕ
∫
ϕ
∫
ϕ
∫
≤
+
+
,
0
≤
a c
, < 1,
b
≥
0,
(2.4)
(
)
<
)(
)
,
2 , ( ) , ( ) , ( 0 ) 2 , (
0
t
dt
t
dt
x f gfx d gfx gx d gx fx d x f fx d
ϕ
ϕ
∫
∫
+ +where
ϕ
∈
Φ
and satisfy the condition (1.1).Whenever the right hand side is nonzero, then
f
andg
have a common fixed point.Proof: Since
f
andg
satisfy the property (E.A), then there exists a sequence{
x
n}
inX
such that limnfxn =limngxn =t for somet
∈
X
.
Sincet
∈
fX
andfX
⊂
gX
, then there existsu
∈
X
such thatt
=
gu
.
Using (2.3) we getdt
t
c
dt
t
b
dt
t
a
dt
t
d fxngxn d gxn gu d gufufu n fx d
)
(
)
(
)
(
)
(
( , ) 0 ) , ( 0 ) , ( 0 ) , (0
ϕ
∫
ϕ
∫
ϕ
∫
ϕ
∫
≤
+
+
.By taking the limit, as
n
→
∞
we have.
)
(
=
)
(
0
0
)
(
( , ) 0 ) , ( 0 ) , (0
t
dt
c
t
dt
c
t
dt
fu gu d fu gu d fu gu d
ϕ
ϕ
ϕ
∫
∫
∫
≤
+
+
So0
=
)
(
)
(1
( , )0
t
dt
c
∫
d fuguϕ
−
.From (1.1) we get
fu
=
gu
.
PointwiseR
−
weak commutativity off
andg
implies forR
>
0
, that)
,
(
)
,
(
fgu
gfu
Rd
fu
gu
d
≤
.So
0.
=
)
(
)
(
( , ) 0 ) , (0
t
dt
t
dt
gu fu Rd gfu fgu d
ϕ
ϕ
∫
∫
≤
Thus
fgu
=
gfu
.
Alsoffu
=
fgu
=
gfu
=
ggu
.
Using (2.4) again we getdt
t
dt
t
d fugu d gugfu d gfu ffu ffu fu d)
(
<
)
(
[ ( , ) ( , ) ( , )] 0 ) , (0
ϕ
∫
ϕ
∫
+ +
=
c
[0 d(fu,ffu) 0](
t
)
dt
0
ϕ
∫
+ +0
=
)
(
)
(1
( , )0
t
dt
c
∫
d fu ffuϕ
−
.Thus from (1.1) we get
fu
=
ffu
.
Hencefu
=
fgu
=
gfu
andfu
is a common fixed point off
andg
. This completes the proof.REFERENCES
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Source of support: Nil, Conflict of interest: None Declared