Available online through
ISSN 2229 – 5046
RELATED FIXED POINT THEOREM ON N METRIC SPACES
Vishal Gupta
1*, Ashima Kanwar
2and R. K. Saini
31
Department of Mathematics, Maharishi Markandeshwar University, Mullana, Ambala, Haryana (India)
E-mail:
2
Department of Mathematics, Maharishi Markandeshwar University, Mullana, Ambala, Hayana (India)
3
Department of Mathematics, Bundelkhand University, Jhansi (U.P), (India)
E-mail:
(Received on: 22-02-12; Accepted on: 15-03-12)
________________________________________________________________________________________________ ABSTRACT
T
he aim of this paper to discuss about related fixed point theorem on n metric spaces. Generally, Fixed point theorems on two,three, four and five complete metric spaces have been proved by many authors. In this paper, we prove a new fixed point theorem on metric space using the concept of continous mapping and cauchy sequence on complete metric spaces. Our theorem generalize and extends many result.________________________________________________________________________________________________
1. INTRODUCTION
In this paper, we are dealing with
n
metric space and we prove a fixed point theorem onn
metric space..In 2009, Faycal Merghadi, Abdelkrim Aliouche, Brain Fisher proved. a fixed point theorem in n complete metric spaces using the concept of implicit relation.
In 2010, Brat Ojha also proved common fixed point theorem for n metric spaces.
Now, in this paper we proved a new fixed point theorem on n metric spaces using continous mapping.
2. PRELIMINARIES
Definition 2.1: Let
( , )
X d
be a metric space. A sequence inX
n is a functionx
:
X
.The image
x n
( )
ofn
∈
is usually dentoed byx
n, called the nth term of the sequence{ }
x
n .Definition 2.2: Let
( ,
X d
1)
and( ,
Y d
2)
be two metric spaces and letf
:
X
Y
be a mapping. Thenf
is said to be continuous atx
∈
X
if and only if the following criterion is satisfied:∀
ε
> 0
∃
a positive numberδ ε
( , )
x
such that
d
2(
f x
( ), ( ) <
f y
)
ε
for all pointsy
∈
X
satisfyingd x y
1( , ) <
δ
. Iff
is continuous at every point ofX
, thenf
is called continuous function.Definition 2.3: A sequence
{ }
x
n in a metric space( , )
X d
is said to be Cauchy sequence if and only iff∀
ε
> 0
∃
a postive integern
0( )
ε
such thatd x
(
m,
x
n)
< ,
ε
∀
n m n
,
0Definition 2.4: A metric space
( , )
X d
is said to be complete if and only if every Cauchy sequence inX
converges to a point ofX
.________________________________________________________________________________________________
Page: 902-908
3. MAIN RESULT
Theorem 3.1: Let
(
X d
i,
i)
ben
complete metric space andT
i ben
mapping andT T T
1,
2,
3,...,
T
n−1 are continuous mapping such thatT X
i:
i
X
i+1 fori
= 1, 2,
,
n
−
1
andT
n:
X
n
X
1 and satisfying the following inequalities:( )
( )
(
)
(
(
1 1)
)
1 2
1 1
1 1 2 1 2 1 2 1 1 1
1 2
,
,
,
n n n n
cF x x
d T T
T T x
T T
T T x
G x x
−
−
(3.1)
( )
( )
(
)
(
(
2 2)
)
2 2
2 2
2 1 3 2 2 1 3 2 2 2
2 2
,
,
,
n n
cF x x
d T T
T T x
T T
T T x
G
x x
(3.2)
( )
( )
(
)
(
(
3 3)
)
3 2
3 3
3 2 1 4 3 2 2 1 5 4 3 3 3
3 2
,
,
,
n n
cF x x
d T T T
T T x
T T T
T T T x
G
x x
(3.3)
( )
(
( )
)
(
(
2)
)
1 2 1 2 1 2 1
2
,
,
,
n n nn n
n n n n n n n n n
n
cF x x
d T T
T T
x
T T
T T
x
G
x x
− −
− −
(3.4)
1 1 2 2 3 3
2 1 2 2 2 3 2
,
,
,
,
,
,
,
n,
n nx x
X x x
X
x x
X
x x
X
∀
∈
∈
∈
∈
for which
G x x
1(
1,
12)
≠
0,
G
2(
x x
2,
22)
≠
0,
G
3(
x x
3,
23)
≠
0,
,
G
n(
x x
n,
2n)
≠
0
where0
c
< 1
.(
1 1)
1
,
2F x x
= max
{
d
1(
x T T
1,
n n−1
T T x d
2 1 1)
n(
T T
n−1 n−2
T T x T T
2 1 21,
n−1 n−2
T T x
2 1 1)
,
d x T T
1(
1,
n n−1
T T x d
2 1 1) (
n−1T
n−2T
n−3
T T x T
2 1 21,
n−2T
n−3
T T x
2 1 1)
,
(
1 1) (
2 1) (
1 1)
1 1 2 1 3 2 1 1 2 1 1 1 2 1
,
d x T T
,
n n−T T x d T T x T T x
,
,
d x T T
,
n n−T T x
d T x T T
2(
1 21,
1 n
T T T x
3 2 1 21) (
,
d x T T
1 1,
n n−1
T T x
2 1 21) (
d T x T x
2 1 21,
1 1)
}
(
2 2)
{
(
2 2) (
2 2) (
2 2)
2
,
2= max
2,
1 n 3 2 1 n n 1 3 2 2,
n n 1 3 2,
2,
1 n 3 2F x x
d
x T T
T T x
d T T
−
T T x T T
−
T T x
d
x T T
T T x
d
n(
T T
n−1 n−2
T T x T T
4 3 3,
n−1 n−2
T T x
3 2 2) (
,
d
2x T T
2,
1 n
T T x
3 2 2)
(
2 2)
(
2 2)
1 2 3 3 2 2
,
2 3 3 2,
,
2,
1 3 2n n n n n n
d
−T
−T
−
T T x T
−T
−
T T x
d
x T T
T T x
d T x T T T
3(
2 22,
2 1 n
T T x
3 2 22) (
,
d
2x T T
2,
1 n
T T x
3 2 22) (
d T x T x
3 2 22,
2 2)
}
(
3 3)
{
(
3 3) (
3 3) (
3 3)
3
,
2= max
3,
2 1 n 4 3 2 1 n 4 3 2,
1 n 4 3,
3,
2 1 n 4 3F x x
d
x T T T
T T x
d T T
T T x T T
T T x
d
x T T T
T T x
d T T
1(
n n−1
T T x T T
4 3 23,
n n−1
T T x
4 3 3) (
,
d
3x T T T
3,
2 1 n
T T x
4 3 3)
d
n(
T T
n−1 n−2
T T x T T
4 3 23,
n−1 n−2
T T x
4 3 3) (
,
d
3x T T T
3,
2 1 n
T T x
4 3 3)
(
)
(
)
3 3 3 3
1 2 3 4 3 2
,
2 3 4 3,
,
3,
2 1 4 3n n n n n n
d
−T
−T
−
T T x T
−T
−
T T x
d
x T T T
T T x
d T x T T T T
4(
3 23,
3 2 1 n
T T x
4 3 23) (
,
d
3x T T T
3,
2 1 n
T T x
4 3 23) (
d T x T x
4 3 23,
3 3)
}
(
,
2)
= max
{
(
,
1 2 1) (
1 2 3 1 2,
2 3 1)
,
n n n n n n
n n n n n n n n n n n n
F x x
d
x T T
− −
T T x
d
−T
−T
−
T T x T
−T
−
T T x
(
n,
1 2 1 n) (
2 3 4 1 2n,
3 4 1 n)
,
,
n n n n n n n n n n n
d
x T T
− −
T T x
d
−T T
− −
T T x T T
− −
T T x
(
) (
1) (
) (
)
}
1 2 1 1 2 1 2 1 1 2 1 2 1 2
,
,
,
,
,
n n n n n n n
n n n n n n n n n n n n n
© 2012, IJMA. All Rights Reserved 904
And
(
1 1)
{
(
1 1) (
1 1)
(
1 1)
}
1
,
2= max
1,
n n 1 2 1 2,
1,
n n 1 2 1,
2 1,
1 n 2 1 2G x x
d x T T
−
T T x
d x T T
−
T T x
d T x T T
T T x
(
2 2)
{
(
2 2) (
2 2)
(
2 2)
}
2
,
2= max
2,
1 n 3 2 2,
2,
1 n 3 2,
3 2,
2 1 n 3 2 2G
x x
d
x T T
T T x
d
x T T
T T x
d T x T T T
T T x
(
1)
{
(
) (
)
(
)
}
1 2 1 1 1 2 1 1 1 1 1
,
= max
,
,
,
,
,
n n n n n n n
n n n n n n n n n n n n n
G
x x
d
x T T
− −
T T x
d
x T T
− −
T T x
d T x T T
−
T T x
Then
T T
n n−1
T T
2 1 has a unique fixed pointα
1∈
X
1,T T
1 n
T T
3 2 has a unique fixed pointα
2∈
X
2, and so on,1 2 1
n n n
T T
− −
T T
has a unique fixed pointα
n∈
X
n.Further,
T
1( )
α
1=
α
2,
T
2( )
α
2=
α
3,
T
3( )
α
3=
α
4,
,
T
n−1(
α
n−1)
=
α
n,
T
( )
α
n=
α
1Proof: Let
x
0′ ∈
X
, be any arbitrary point. We define sequences{
x
1m},
2
{
x
m}
,3
{
x
m},
,{
x
mn}
with1
,
2,
3,
,
nX X
X
X
respectively as follows:(
)
1 1 2 1
1 2 1 0 1 1
=
m,
=
,
m n n m m
x
T T
−
T T
x x
T x
−x
m3=
T
2( )
x
m2,
x
m4=
T x
3( )
m3∀ ∈
m
We assume that
x
m1≠
x
1m+1,
x
m2≠
x
m2+1x
m3≠
x
m3+1,
1n n
m m
x
≠
x
+∀ ∈
m
Taking
x
2=
x
m2 andx
22=
x
m2−1 in inequality (3.2), we get,(
2 2)
2 m
,
m 1d
x
x
+(
2( )
2)
2 1 3 2 1 1 1 3 2
=
d T T
n
T T x
m−,
T T T
n n−
T T
x
m
(
)
(
)
2 2
2 1
2 2
2 1
,
,
m mm m
cF x
x
G
x
x
−
−
(
) (
{
) (
)
(
) (
) (
)
(
)
2 2 2 2 2 2
2 1 3 2 1 1 3 2 1 1 2 2 1 3 2
2 2 2 2 2 2
1 2 3 2 1 1 3 2 2 1 3 2 1 2 3 3 2 1 2 3 2
2 2 2
2 1 3 2 3 2 1 2 1 2
max
,
,
,
,
,
,
,
,
,
,
,
,
m n m n n m n n m m n m
n n n m n m m n m n n n m n m
m n m m n
c
d
x T T
T T x
d T T
T T x
T T
T x
d
x T T
T T x
d T T
T T x
T
T T x
d
x T T
T T x
d
T T
T T x
T
T T x
d
x T T
T T x
d T x
T T T
T
− − −
− − − − − − − − −
−
=
(
) (
) (
)
}
(
) (
) (
)
{
}
2 2 2 2 2
1 2 1 3 2 1 3 1 1 2
2 2 2 2 3 3
2 2 1 3
,
,
,
max
,
,
,
,
,
m m n m m m
m m m m m m
x
d
x T T
T T x
d T x
T x
d
x
x
d
x
x
d
x
x
− − −
+
(
2 2)
{
(
1 1) (
)
(
1 1)
(
3 3)
}
2
,
1max
1 1,
,
1,
,
1 1,
,
,
3 1,
n n n n
m m m m n m m n m m m m
d
x
x
+
c
d x
−x
d
x
−x
d
−x
−−x
−
d
x
−x
(3.5)
Now taking
x
3=
x
m3 andx
23=
x
m3−1 in inequality (3.3), we obtain,(
3 3)
(
3 3)
3 m
,
m 1=
3 2 1 n 4 3 m 1,
2 1 n 4 3 md
x
x
+d T T T
T T x
−T T T
T T x
(
)
(
)
3 3
3 1
3 3
3 1
,
,
m mm m
cF x
x
G
x
x
−
−
Page: 902-908
(
) (
{
) (
)
(
) (
) (
)
(
) (
)
3 3 3 3 3 3
3 2 1 3 2 1 4 3 1 1 4 3 3 2 1 3
3 3 3 3 3 3
1 4 3 1 4 3 3 2 1 4 3 1 2 4 3 1 2 4 3
3 3 3 3
3 2 1 4 3 4 3 1 3 2 1 4 3 1
max
,
,
,
,
,
,
,
,
,
,
,
,
,
m n m n m n m m n m
n m n m m n m n n m n m
m n m m n m
c
d
x T T T
T x
d T T
T T x
T T
T T x
d
x T T T
T x
d T
T T x
T
T T x
d
x T T T
T T x
d
T
T T x
T
T T x
d
x T T T
T T x
d T x
T T T T
T T x
−− − − − −
− −
=
(
) (
)
}
(
) (
) (
)
{
}
(
)
3 3 3 3
3 2 1 3 4 3 1 4 3 1 3
3 3 3 3 4 4
3 3 1 4
,
,
max
,
,
,
,
,
m m m m
m m m m m m
d
x T T T
T T x
d T x
T x
d
x
x
d
x
x
d
x
x
− −
+
(
) (
) (
)
{
2 2 1 1(
1 1)
(
4 4)
}
2 1 1 1 1 1 1 4 1
max
m,
m,
m,
m,
n mn,
mn,
n mn,
mn,
,
m,
mc
d
x
x
+d x
−x
d
x
−x
d
−x
−−x
−d
x
−x
=
From inequality (3.5), we get,
(
3 3)
3 m
,
m 1d
x
x
+
{
cd x
1(
m1−1,
x
m1)
,
cd
n(
x
mn−1,
x
mn)
,
cd
n−1(
x
mn−−11,
x
mn−1)
,
,
cd
3(
x
m3−1,
x
m3) (
,
d x
1 1m−1,
x
m1)
,
d
n(
x
mn−1,
x
mn)
,
d
n−1(
x
mn−−11,
x
mn−1)
,
,
d
4(
x
m4−1,
x
m4)
}
Since
0
c
< 1
, so, we get,(
3 3)
{
(
1) (
)
(
1 1)
(
4 4) (
3 3)
}
3
,
1max
1 1,
,
1,
,
1 1,
,
,
4 1,
,
3 1,
n n n n n
m m m m n m m n m m m m m m
d
x
x
+
c
d x
−x
d
x
−x
d
−x
−−x
−
d
x
−x
d
x
−x
(3.6)
Similarly, we can get,
(
1 1)
1
,
1n n
n m m
d
−x
−x
−+
c
max
{
d x
1(
1m−1,
x
1m) (
,
d
3x
m3−1,
x
3m) (
,
d
4x
m4−1,
x
m4)
,
,
d
n−1(
x
mn−−11,
x
mn−1) (
,
d
nx
mn−1,
x
mn)
}
(3.7)Taking
x
2n=
x
mn−1 andx
n=
x
mn in inequality (3.4)(
,
1)
n n
n m m
d
x
x
+=
(
1 2 1 1,
1 2 1)
n n
n n n n m n n n m
d
T T
− −
T T x
−T T
− −
T T x
(
)
(
11)
,
,
n nn m m
n n
n m m
cF x
x
G
x
x
−
−
(
) (
)
{
(
)
(
)
(
)
(
) (
) (
)
}
1 2 2 1 1 2 1 1 2 1
1 2 1 2 3 1 1 3 1 1
1 1 1 1 1 1 1 1 1 1
max
,
,
,
,
,
,
,
,
,
,
,
,
max
n n n n
n m n n n m n n n m n n m
n n n n n n
n m n n n m n n n m n n m n m n n m
n n n n n n
n m n n n m n m n n m n m n m
n
c
d
x T T
T T T x
d
T
T T x
T
T T x
d
x T T
T T x
d
T
T T x
T
T x
d
x T
T T x
d T x
T T
T T x
d
x T
T T x
d T x
T x
d
− − − − − −
− − − − − − −
− − − − − −
=
(
)
(
) (
)
}
{
,
1 2 1 1,
,
1 2 1,
1,
1 2 1 1n n n n n n
m n n n m n m n n n m n m n n n m
x T T
− −
T T x
−d
x T T
− −
T T x
d T x T T
−
T T T x
−(
)
(
)
(
)
{
1 1 2 2 1 1}
1 1 2 1 1 1
max
n mn,
mn,
n mn,
mn,
,
m,
mc
d
−x
−x
−+d
−x
−x
−+d x
−x
=
From (3.5), (3.6) and (3.7), we get,
(
)
{
(
1 1) (
3 3) (
4 4)
(
1) (
)
}
1 1 1 3 1 4 1 1 1 1
,
max
,
,
,
,
,
,
,
,
,
,
(3
)
.
8
n n n n n n
n m m m m m m m m n m m n m m
d
x
x
+
c
d x
−x
d
x
−x
d
x
−x
d
−x
−x
−d
x
−x
It now follows by induction on above inequalities, we get,
(
1 1)
1{
(
1 1) (
3 3)
(
1 1) (
)
}
1
,
1max
1 1,
2,
3 1,
2,
,
1 1,
2,
1,
2m n n n n
m m n n
d x
x
+
c
−d x x
d
x x
d
−x
−x
−d
x x
(
2 2)
1{
(
1 1) (
3 3)
(
1 1) (
)
}
2
,
1max
1 1,
2,
3 1,
2,
,
1 1,
2,
1,
1m n n n n
m m n n
© 2012, IJMA. All Rights Reserved 906
(
)
1{
(
1 1) (
3 3)
(
1 1) (
)
}
1 1 1 2 3 1 2 1 1 2 1 2
,
max
,
,
,
,
,
,
,
,
n n m n n n n
n m m n n
d
x
x
+
c
−d x x
d
x x
d
−x
−x
−d
x x
Since
0
c
< 1
, the sequence{ } { } { } { } { }
x
m1,
x
m2,
x
m3,
,
x
mn 1,
x
mn −
are Cauchy sequence inX X
1,
2,
X
3,
,
X
n. So we have,=
when = 1, 2, 3,
, .
lim
mi im
x
α
i
n
→∞
Since
T T T
1,
2,
3,
,
T
n−1are continuous mapping.
T
1α α
1=
2,
T
2α
2=
α
3,...,
T
n−1α
n−1=
α
nNow taking
x
1=
x
1m−1 and1 2
=
1x
α
in inequality (3.1),(
1)
1(
(
1 1 1)
)
1 1 2 1 1 1 2 1 1
1 1 1
,
,
,
mn n n n m
m
cF x
d T T
T T
T T
T T x
G x
α
α
α
−
− −
−
(
)
(
{
) (
)
(
)
(
) (
) (
)
(
)
1 1 1 1 1
1 1 2 1 1 1 2 1 1 1 2 1 1 1 1 2 1 1
1 1 1 1 1 1
1 2 2 1 1 2 2 1 1 1 1 2 1 1 3 2 1 2 1 1 1 1 2 1 1
1
2 1 1 1 2 1 1 1 1 1 1
max
,
,
,
,
,
,
,
,
,
,
,
,
,
m n m n n n m m n m
n n n m m n m m m n m
n m n n
c
d x
T
T T x
d
T
T T
T
T T x
d x
T
T T x
d
T
T T
T
T T x
d x
T
T T x
d T T
T T x
d x
T
T T x
d T
T T
T T
d x
T T T
α
α
α
α
α
− − − − − − −
− − − − − − − − −
− −
=
(
) (
)
}
(
) (
) (
)
}
{
1 2 1 1 2 1 1 1 1
1 1 1 1
1 1 1 2 1 1 1 1 1 2 1 1 2 1 1 1 2 1 1
,
max
,
,
,
,
,
,
m
m n n m n n m m n
T T
d T
T x
d x
T T
T T
d x
T T
T T x
d T x
T T
T T
α
α
α
α
−
− −
− −
− −
As
m
→ ∞
, we get,(
)
(
1(
1 1 1 2 1 1) (
) (
2 1 1 2)
)
}
1 1 2 1 1 11 1 1 2 1 1 2 2 1 2 1 1
,
,
,
max{
,
,
,
n n n n
n n n
cd
T T T
T T
d T
d T T
T T
d
T T
T T
d
T T
T T
α
α
α α
α α
α
α
α
α
− −
−
≤
Since
T
1 is continuous mapping.(
)
1 n n 1 2 1 1
,
10
d T T
−T T
α α
⇒
Since, which is not possible
1 2 1 1
=
1n n
T T
−T T
α
α
⇒
(3.9)
Now,
T T T
1 n n−1
T T
3 2α
2=
T T T
1 n n−1
T T T
3 2(
1α
1)
=
T T T
1(
n n−1
T T T
3 2 1α
1)
=
T
1α
1=α
2⇒
T T T
1 n n−1
T T
3 2α
2=
α
2Since
T T T
1,
2,
3,
,
T
n−1are continuous mapping, So in same above way,so we obtainT T T
1 n n−1
T
2α
2=
α
22 1 n 3 3
=
3T T T
T
α
α
3 2 1 4 4
=
4T T T
T
α
α
1 2 1
=
n n n n
T
−
T T T
α
α
Page: 902-908
Uniqueness
Let us assume that
α
1′
is another fixed point ofT T
n n−1
T T
2 1 different fromα
1 such thatα
1′≠
α
1 and1 2 1 1
=
1n n
T T
−
T T
α
′
α
′
,T T
n n−1
T T
2 1α
1=
α
1.Now taking
x
1=
α
1′
andx
12=
α
1 in inequality (3.1), we get,(
)
1
,
1d
α α
′
=
d T T
1(
n n−1
T T
2 1α
1,
T T
n n−1
T
1α
1′
)
(
)
(
)
1 1 1
1 1 1
,
,
cF
G
α α
α α
′
′
(
)
(
) (
)
{
(
)
(
) (
)
(
) (
) (
) (
)
}
1 1 2 1 1 1 3 2 1 1 2 1 1 1 1 2 1 1
1 2 3 2 1 1 2 2 1 1 1 1 2 1 1 3 2 1 1 2 1 1
1 1 2 1 1 2 1 1 1 3 2 1 1 1 1 1 3 2 1 1 2 1 1 1 1
max
,
,
,
,
,
,
,
,
,
,
,
,
,
,
,
=
n n n n n
n n n n
n n n n
c
d
T
T T
d
T
T T T T
T T
d
T
T T
d
T
T T T
T
T T
d
T
T T
d T T
T T
d
T
T T
d T
T T
T T T
d
T T
T T T
d T
T
α
α
α
α
α
α
α
α
α
α
α
α
α
α
α
α
α
α
α
− −
− − −
−
′
′
′
′
′
′
′
′
′
′
′
′
′
(
)
{
{
}
(
) (
)
(
max
d
1α
1,
T T
n n−1T T T
3 2 1α
1,
d
1α
1,
T T
n n−1T T
2 1α
1,
d T
2 1α
1,
T T
1 nT T
2 1α
1}
)
′
′
′
′
(
) (
)
(
) (
)
{
11 11 11 2 21 1 11 11 1}
,
,
=
max
,
,
,
cd
d T
T
d
d T
T
α α
α α
α α
α
α
′
′
′
′
⇒
d
1(
α α
1,
1′
)
cd T
2(
1α
1,
T
1α
1′
)
(3.10)
In the same way, takingx
22=
T
1α
1 andx
2=
T
1α
11′
in inequality (3.2), we get(
)
2 1 n n 1 3 2 1 1
,
1 n n 1 2 1 1d T T T
−
T T T
α
T T T
−
T T
α
′
(
)
(
)
2 1 1 1 1
2 1 1 1 1
,
,
cF T
T
G T
T
α
α
α
α
′
′
(
) (
{
)
(
) (
)
}
(
) (
) (
)
{
}
(
)
2 1 1 1 1 1 1 2 1 1
2 1 1 2 1 1 1 1 3 2 1 1 2 1 1
2 1 1 1 1 2 1 1 1 1 3 2 1 1 2 1 1
max
,
,
,
,
,
,
=
max
,
,
,
,
,
n n
n
c
d T
T
d T T
T T
T
T T
d T
T
d T T
T T
d T
T
d T
T
d T T
T T
α
α
α
α
α
α
α
α
α
α
α
α
α
α
−
′
′
′
′
′
′
′
′
′
(
)
(
)
2 1 1
,
1 1 3 2 1 1,
2 1 1d T
α
T
α
′
cd T T
α
T T
α
′
⇒
(3.11)
Similarly, by taking
x
23=
T T
2 1α
1 andx
3=
T T
2 1α
1′
in inequality (3.3), we get,(
)
(
)
3 2 1 1
,
2 1 1 4 3 2 1 1,
3 2 1 1d T T
α
T T
α
′
cd T T T
α
T T T
α
′
(3.12)
Continue like this, taking
x
2n=
T T
n−1 n−2
T T
2 1α
1 andx
n=
T
n−1
T T
2 1α
1′
in inequality (3.4), we get,(
1 2 2 1 1,
1 2 1 1)
n n n n
d
T T
− −
T T
α
T
−
T T
α
′
cd T T
1(
n n−1
T T
2 1α
1,
T T
n n−1
T T
2 1α
1′
)
© 2012, IJMA. All Rights Reserved 908
From inequalities (3.10),(3.11), (3.12)and(3.13), we get,
d
1(
α α
1,
1′
)
cd T
2(
1α
1,
T
1α
1′
)
(
)
2
3 2 1 1
,
2 1 1c d T T
α
T T
α
′
(
)
3
4 3 2 1 1
,
3 2 1c d T T T
α
T T
α′
(
)
1
1 2 1 1
,
1 2 1 1n
n n n
c
−d T
−
T T
α
T
−
T T
α
′
(
)
1 1
,
1n
c d
α α
′
1
(
1,
1)
1(
1,
1)
n
d
α α
′
c d
α α
′
⇒
Since
0
c
< 1
, so we get,
⇒
d
1(
α α
1,
1′
)
= 0
⇔
α
1=
α
1′
So,
α
1 is unique fixed point ofT T
n n−1
T T
2 1.In the similar way, the unicity of other fixed point can be proved.
We have to finally proved that
T
( )
α
n=
α
1, To do this, note thatT
n( )
α
n=
T T
n(
n−1
T T T
2 1 nα
n)
=
T T
n n−1
T T T
2 1(
nα
n)
And so
T
nα
nis fixed point ofT T
n n−1
T T
2 1. Sinceα
1 is the unique fixed point ofT T
n n−1
T T
2 1. It follows that( )
n=
1T
α
α
So, we obtain
T T
n n−1
T T
2 1 has a unique fixed pointα
1∈
X
1,T T
1 n
T T
3 2 has a unique fixed pointα
2∈
X
2, and so on,T T
n−1 n−2
T T
1 n has a unique fixed pointα
n∈
X
nAlsoT
1( )
α
1=
α
2,
T
2( )
α
2=
α
3,
T
3( )
α
3=
α
4,
,
(
)
( )
1 1
=
,
=
1n n n n
T
−α
−α
T
α
α
REFERNCES
[1] F.Merghadi, A.Aliouche and B.Fisher, A related fixed point theorem in n complete metric spaces, Acta Universitatis Apulensis, no.19(2009), 91-99
[2] D.B.Ojha , Commom fixed point theorem for n metric spaces, International Journal of Applied Engineering Research,Dindigul,Vol.1(2010) no. 3, 324-329
[2] B. Fisher, Fixed points theorem on two metric spaces, Glasnik Mat., Vol. 16(1981) no.36, 333-337.
[2] L. Kikina and K. Kikina, A fixed point theorem on four spaces, Int. Math. J. Analysis,Vol. 3 (2009)no.32 , 1559-1568.
[3] N.P. Nung, A fixed point theorem for three metric spaces, Math. Seminar Notes Kobe University, Vol.1( 1983), 77-79.
[4] R.K. Jain, H.K. Sahu, and B. Fisher, A fixed point theorem on three metric spaces, Kyungpook Math. J.,Vol. 36 (1996), 151-154.
[5] S. Jain and B. Fisher, A related fixed point theorems on three complete metric spaces, Novisad J. Math. ,Vol.27 (1997), 27-35.