Available online through
ISSN 2229 – 5046Theorem on Fixed Points in n- Metric Spaces
Vishal Gupta*, Ashima Kanwar and Richa Sharma
Department of mathematics, Maharishi Markandeshwar University, Mullana, Ambala, Haryana, India
(Received on: 05-06-12; Accepted on: 30-06-12)
ABSTRACT
I
n this paper , we prove a fixed point theorem on n- metric spaces. Here we are dealing with n-metric spaces and n mapping. This paper result extends the result of L.Kikina [1] from three metric spaces to n-metric spaces and also it generalizes the results of S.Č.Nešić [2] from two metric spaces to n-metric spaces.Keywords: Complete metric space, fixed point.
2010 AMS Mathematics Subject Classification: 47H10, 54H25.
1. INTRODUCTION
The following Fixed point theorem on two complete metric spaces proved by B. Fisher [3].
Theorem1.1: Let (X, d) and (Y, e) be complete metric spaces. If T is a mapping of X into Y and S is mapping of Y into X satisfying the Inequality
e Tx TSy
(
,
)
≤
c
max{ ( ,
d x Sy e y Tx e y TSy
), ( ,
), ( ,
)}
d Sy STx
(
,
)
≤
c
max{ ( ,
e y Tx d x Sy d x STx
), ( ,
), ( ,
)}
For all x in X and y in Y. where 0 ≤ c <1, then ST has a unique fixed point z in X and TS has unique fixed point w in y. Further, Tz = w and Sw = z.
In 1991, V. Popa [5] gave a result on two complete metric spaces.
Theorem1.2: Let (X, d) and (Y, ρ) be complete metric spaces, if T is a mapping of X into Y and S is a mapping of Y into X satisfying the inequalities
(
)
(
) (
) (
) (
) (
) (
)
2
1
d
Sy,STx
≤
c max{
ρ
y, Tx d x,Sy ,
ρ
y, Tx d x,STx , d x,Sy d x,STx }
(
)
(
) (
) (
) (
) (
) (
)
2
2
Tx, TSy
c max{d x,Sy
y, Tx , d x,Sy ,
y, TSy , y, Tx
y, TSy }
ρ
≤
ρ
ρ
ρ
ρ
For all x in X and y in Y where 0 ≤ c1.c2<1, then ST has a unique fixed point z in X and TS has a unique fixed point w in Y. Further Tz = w and Sw = z.
The following Fixed point theorem on three complete metric spaces proved by R.K .Jain [4].
Theorem1.3: Let (X, d1), (Y, d2) and (Z, d3) be complete metric spaces. If T is continuous mapping X into Y, S is continuous mapping Y into Z and R is mapping of Z into X satisfying the following inequalities
(
)
(
) (
) (
) (
) (
)
1 1 1 1 2 3
d RSTx, RST x '
≤
cmax{d x, x ' , d x, RSTx , d x ', RSTx ' , d
Tx, Tx ' , d STx,STx ' }
(
)
(
) (
) (
) (
) (
)
2 2 2 2 3 1
d
TRSy ,TRSy '
≤
c max{d
y ,y ' , d
y ,TRSy , d
y ', TRS y ' , d Sy,Sy ' , d RSy, RSy ' }
(
)
(
) (
) (
) (
) (
)
3 3 3 3 1 2
d STRz,STRz '
≤
c max{ d
z, z ' , d
z,STRz , d
z ',STR z ' , d Rz, Rz ' , d
TRz, TRz ' }
for all x,x' in X , y,y' in Y and z,z'in Z, where 0 ≤ c <1, then RST has unique fixed point u in X , TRS has unique fixed point v in Y and STR has fixed point w in Z . Further, Tu = v, Sv = w and Rw = u.
*Corresponding author:Vishal Gupta*
Page: 2393-2399
2. PRELIMINARIES
Definition 2.1 A Metric space is an ordered pair (X, d) where X is a non empty Set and d is a metric on X or also called distance function or simple distance, i.e a function d:XxX→R such that
i. d(x,y) ≥ 0 (non -negative) ii. d(x,y)= 0 iff x=y
iii. d(x,y)= d(y, x) (symmetric)
iv. d(x,y)≤ d(x, z)+ d(z, y) (triangle inequality)
Definition2.2: Let (X, d) be a Metric Space. A Sequence {xn} in X is said to converge to point x in X iff the following criterion is satisfied.
For each ε >0, there exit a positive integer n0(ε), Such that
d ( xn, x) < ε, for all n ≥ n0
Definition2.3: A sequence {xn} in a metric space (X, d) is said to be a Cauchy Sequence iff for each ε>0, there exist a positive integer number no(ε) such that
d ( xm , xn ) < ε, for all n, m ≥ n0
Definition2.4: A metric space (X, d) is said to be complete iff every Cauchy sequence in X converges to point of X.
Let
(
X d
i,
i)
be complete metric spaces wherei
= 1, 2, 3,
,
n
. IfA
i is mapping ofX
i toX
i+1 where= 1, 2, 3,
,
1
i
n
−
andA
n is mapping ofx
n tox
1.we denote
1 2 1 2 1 1 2 1
1
( ,
)
{
1( ,
1 2),
1( ,
1 2 1),
2(
,
1)}
p p p
n n n n
M x x
=
d
x A A
−
A x
d
x A A
−
A A x
d
x A x
(2.1)
2 3 2 3 2 2 3 2
2
(
,
)
{
2(
,
1 3),
2(
,
1 2),
3(
,
2)}
p p p
n n
M
x x
=
d
x
A A
A x
d
x
A A
A x
d
x A x
(2.2)
(2.3)
And so on,
1 1 1
1 2 1 1 1 1
(
n,
)
{
p(
n,
),
p(
n,
n),
p( ,
n)}
n n n n n n n
M
x x
=
d
x
A
−
A A x
d
x
A
−
A A x
d
x A x
(2.4)
Let
F
: 0,
[ ]
∞ →
R
+be continuous mapping in 0 withF
(0)
=
0
In this paper we prove a fixed point theorem on n metric spaces and some corollaries which extend the result of L.Kikina [1] from three metric spaces to n-metric spaces and a version of Vishal Gupta et al [6-7].
3. MAIN RESULT
Theorem 3.1: Let
(
X d
i,
i)
be complete metric spaces wherei
= 1, 2, 3,
,
n
. IfA
i is mapping ofX
i toX
i+1 wherei
= 1, 2, 3,
,
n
−
1
andA
n is mapping ofx
n tox
1 satisfying the following inequalities.2 1
1
(
1 2,
1 2 1)
p
n n n n
d
A A
−
A x
A A
−
A A x
1 21 1
max
( ,
)
(min
)
c
M x x
F
M
≤
+
(3.1)
3 2
2
(
1 4 3,
1 3 2)
p
n n
d
A A
A A x A A
A A x
≤
c
max
M
2(
x x
2,
3)
+
F
(min
M
2)
(3.2)4 3
3
(
2 1 5 4,
2 1 4 3)
p
n n
d
A A A
A A x A A A
A A x
≤
c
max
M x x
3(
3,
4)
+
F
(min
M
3)
(3.3)So continuously like above.
1
1 2 2 1 1 2 2 1
(
,
)
p n
n n n n n n
d
A
−A
−
A A x A
−A
−
A A A x
≤
c
max
M
n(
x x
n,
1)
+
F
(min
M
n)
(3.4)
3 4 3 4 3 3 4 3
3
(
,
)
{
3(
,
2 1 4),
3(
,
2 1 3),
4(
,
3)}
p p p
n n
1 2
1
,
2,
,
n n
x
X x
X
x
X
∀ ∈
∈
∈
, where 0≤c<1. ThenA A
n n−1
A A
2 1 has a unique fixed pointβ
1∈
X
1 ,1 n 3 2
A A
A A
has a unique fixed pointβ
2∈
X
2,A A
2 1
A A
4 3 has a unique fixed pointβ
3∈
X
3 and so on1 2 1
n n n
A A
− −
A A
has a unique fixed pointβ
n∈
X
n . Further,A
1( )
β
1=
β
2,
A
2( )
β
2=
β
3,
A
3( )
β
3=
β
4,
(
)
( )
1 1 1
,
A
n−β
n−=
β
n,
A
nβ
n=
β
Proof: Let
x
10 be an arbitrary point inX
1 , let define sequence{
x
m1}
,{
x
m2},
,{
x
mn}
inX X
1,
2,
,
X
n respectively by(
)
1 11 3 2 1 0
=
,
m
n n m
A A
−
A A A
x
x
x
m2=
A x
1( )
1m−1,
x
m3=
A x
2( )
m2
x
mn=
A
n−1( )
x
mn−1
x
1m=
A x
n( )
mnfor
m
= 1, 2, 3,
We will assume that
x
m1≠
x
1m+1,
x
m2≠
x
m2+1and so onx
mn≠
x
mn+1for all m. Otherwise, ifx
m1=
x
m1+1for some m, then2 2 1
m m
x
=
x
+ ,x
m3=
x
m3+1and so onx
mn=
x
mn+1, we could putx
1m=
β
1,x
m2+1=
β
2and so onx
mn+1=
β
n.First, we prove the sequences
{
x
1m}
,{
x
m2},
,{
x
mn}
are cauchy sequences.Taking
x
1=
x
1m,x
2=
x
m2 in (2.1) and (3.1) , we obtain1 2 1 2 1 1 2 1
1
(
,
)
{
1(
,
1 2),
1(
,
1 2 1),
2(
,
1)}
p p p
m m m n n m m n n m m m
M x
x
=
d
x
A A
−
A x
d
x
A A
−
A A x
d
x
A x
=
{
d
1p(
x
m1,
x
1m),
d
1p(
x
m1,
x
1m+1),
d
2p(
x
m2,
x
m2+1)}
{0,
1p(
1,
1 1),
2p(
2,
2 1)}
m m m m
d
x
x
+d
x
x
+=
1 1 1
(
,
1)
p
m m
d
x
x
+=
1(
1 2 2,
1 2 1 1)
p
n n m n n m
d
A A
−
A x
A A
−
A A x
≤
c
max
M x
1(
1m,
x
m2)
+
F
(min
M x
1(
1m,
x
m2))
=
c
max{0,
d
1p(
x
1m,
x
1m+1),
d
2p(
x
m2,
x
m2+1)}
+
F
(0)
=
cd
2p(
x
m2,
x
m2+1)
Since, if
max
M x
1(
m1,
x
m2)
=
d
1p(
x
1m,
x
m1+1)
, then1 1 1 1
1
(
,
1)
1(
,
1)
p p
m m m m
d
x
x
+≤
cd
x
x
+It follows
x
1m=
x
1m+1since0
≤ <
c
1
, so1 1 1
(
,
1)
p m m
d
x
x
+≤
cd
2p(
x
m2,
x
m2+1)
(3.5)Taking
x
2=
x
m2,
x
3=
x
m3−1 in (2.2) and (3.2), we get2 3 2 3 2 2 3 2
2
(
,
1)
{
2(
,
1 3 1),
2(
,
1 2),
3(
1,
2)}
p p p
m m m n m m n m m m
M
x
x
−=
d
x
A A
A x
−d
x
A A
A x
d
x
−A x
=
{
d
2p(
x
m2,
x
m2),
d
2p(
x
m2,
x
m2+1),
d
3p(
x
m3−1,
x
m3)}
{0,
2p(
2,
2 1),
3p(
3 1,
3)}
m m m m
Page: 2393-2399
2 2 2
(
,
1)
p m m
d
x
x
+ =d
2p(
A A
1 n
A A x
4 3 3m−1,
A A
1 n
A A x
3 2 m2)
≤
c
max
M
2(
x
m2,
x
m3−1)
+
F
(min
M
2(
x
m2,
x
m3−1))
=
c
max{0,
d
2p(
x
m2,
x
m2+1),
d
3p(
x
m3−1,
x
m3)}
+
F
(0)
Since
0
≤ <
c
1
, we get2 2 3 3
2
(
,
1)
3(
1,
)
p p
m m m m
d
x
x
+≤
cd
x
−x
(3.6)Taking
x
3=
x
m3,
x
4=
x
m4−1 in (2.3) and (3.3), we get3 4 3 4 3 3 4 3
3
(
,
1)
{
3(
,
2 1 4 1),
3(
,
2 1 3),
4(
1,
3)}
p p p
m m m n m m n m m m
M x
x
−=
d
x
A A A
A x
−d
x
A A A
A x
d
x
−A x
=
{
d
3p(
x
m3,
x
m3),
d
3p(
x
m3,
x
m3+1),
d
4p(
x
m4−1,
x
m4)}
=
{0,
d
3p(
x
m3,
x
m3+1),
d
4p(
x
m4−1,
x
m4)}
3 3 4 3
3
(
,
1)
3(
2 1 5 4 1,
2 1 4 3)
p p
m m n m n m
d
x
x
+=
d
A A A
A A x
−A A A
A A x
≤
c
max
M x
3(
m3,
x
m4−1)
+
F
(min
M x
3(
m3,
x
m4−1))
=
c
max{0,
d
3p(
x
m3,
x
3m+1),
d
4p(
x
m4−1,
x
m4)}
+
F
(0)
=
cd
4p(
x
m4−1,
x
m4)
Replacing m with m-1 we obtain
3 3 4 4
3
(
1,
)
4(
2,
1)
p p
m m m m
d
x
−x
≤
cd
x
−x
− (3.7)Continuously like above, taking
x
n=
x
mn,
x
1=
x
m1−1in (2.4) and (3.4) we obtain1 1 1 1
1 1 2 1 1 1 1 1 1 1
(
n,
)
{
p(
n,
),
p(
n,
n),
p(
,
)}
n m m n m n m n m n n m m n m
M
x
x
−=
d
x
A
−
A A x
−d
x
A
−
A A x
d
x
−A x
−
=
{
d
np(
x
mn,
x
mn),
d
np(
x
mn,
x
mn+1),
d
1p(
x
m1−1,
x
m1)}
=
{0,
d
np(
x
mn,
x
mn+1),
d
1p(
x
m1−1,
x
m1)}
1
1 1 2 2 1 1 1 2 2 1
(
,
)
(
,
)
p n n p n
n m m n n n m n n n m
d
x
x
+=
d
A
−A
−
A A x
−A
−A
−
A A A x
≤
c
max
M
n(
x
mn,
x
1m−1)
+
F
(min
M
n(
x
mn,
x
m1−1))
=
c
max{0,
d
np(
x
mn,
x
mn+1),
d
1p(
x
m1−1,
x
m1)}
+
F
(0)
=
cd
1p(
x
1m−1,
x
1m)
Replacing m with m-n+2, we get
1 1
2 3 1 1 2
(
,
)
(
,
)
p n n p
n m n m n m n m n
d
x
− +x
− +≤
cd
x
− +x
− + (3.8)Using (3.5), (3.6), (3.7) and (3.8), we get
1 1 1
(
,
1)
p m m
d
x
x
+ 2(
2,
2 1)
p
m m
cd
x
x
+≤
2 3 33
(
1,
)
p
m m
c d
x
−x
≤
≤
. . .c
n 1d
np(
x
m nn 2,
x
m nn 3)
c d
n 1p(
x
m nn 1,
x
m nn 2)
−
− + − + − + − +
≤
≤
≤
c
2nd
1p(
x
mn−2n+2,
x
mn−2n+3)
1 1 1 1 2 1 1 1 0 1
( ,
), ,
(
1)
1
...
( ,
), ,
(
1)
nk p
nk p
c d
x x
m
n
k
c d
x x
m
n
k
=
−
+
≤ ≤
=
−
Since
0
≤ <
c
1
, the sequences{
x
m1}
,{
x
m2},
,{
x
mn}
are cauchy sequences. Since(
X d
i,
i)
be complete metric spaces wherei
= 1, 2, 3,
,
n
.
1
1 1 2
2 2 3
3 3
lim
lim
lim
.
.
.
lim
m m
m m
m m
n
m m n n
x
X
x
X
x
X
x
X
β
β
β
β
→∞
→∞
→∞
→∞
=
∈
=
∈
=
∈
=
∈
Now taking
x
1=
x
1mandx
2=
β
2in the inequality (3.1) we obtain1 1
(
1 2 2,
1)
p
n n m
d
A A
−
A
β
x
+ =1 1
(
1 2 2,
1 2 1)
p
n n n n m
d
A A
−
A
β
A A
−
A A x
≤
c
max
M x
1(
m1,
β
2)
+
F
(min
M x
1(
1m,
β
2))
(3.9)
Where
M x
1(
m1,
β
2)
=
{
d
1p(
x
1m,
A A
n n−1
A
2β
2),
d
1p(
x
1m,
A A
n n−1
A A x
2 1 1m),
d
2p(
β
2,
A x
1 m1)}
=
{
d
1p(
x
m1,
A A
n n−1
A
2β
2),
d
1p(
x
m1,
x
m1+1),
d
2p(
β
2,
x
m2+1)}
As m tend to infinity in (3.9) and F is continuous in 0 we get
d
1p(
A A
n n−1
A
2β β
2,
1)
≤
cd
1p(
β
1,
A A
n n−1
A
2β
2)
So we get,
A A
n n−1
A
2β
2=
β
1In same way, we obtain
A A
1 n
A A
4 3β
3=
β
2 ,A A A
2 1 n
A A
5 4β
4=
β
3, . . . ,A
n−1A
n−2
A A
2 1β
1=
β
nUsing (3.1), taking
x
1=
β
1,
x
2=
x
m2 we get
d
1p(
x
1m,
A A
n n−1
A A
2 1β
1)
=
d
1p(
A A
n n−1
A x
2 m2,
A A
n n−1
A A
2 1β
1)
≤
c
max
M
1(
β
1,
x
m2)
+
F
(min
M
1(
β
1,
x
m2))
Where,
M
1(
β
1,
x
m2)
=
{
d
1p(
β
1,
A A
n n−1
A x
2 m2),
d
1p(
β
1,
A A
n n−1
A A
2 1β
1),
d
2p(
x
m2,
A
1β
1)}
=
{
d
1p(
β
1,
A A
n n−1
A x
2 1m),
d
1p(
β
1,
A A
n n−1
A A
2 1β
1),
d
2p(
x
m2,
A
1β
1)}
Now letting m tend to infinity we get
1 1 1 1 1 1 2 1 1 2 2 1 1
max{
p(
,
),
p(
,
n n),
p(
,
)}
c
d
β β
d
β
A A
−A A
β
d
β
A
β
≤
1 1 1 2 1 1 2 2 1 1
max{
p(
,
n n),
p(
,
)}
c
d
β
A A
−A A
β
d
β
A
β
=
From which it follows, or
d
1p(
β
1,
A A
n n−1
A A
2 1β
1)
1(
1,
1 2 1 1)
1 2 1 1 1p
n n n n
cd
β
A A
−A A
β
A A
−A A
β
β
≤
⇔
=
Or
d
1p(
β
1,
A A
n n−1
A A
2 1β
1)
≤
cd
2p(
β
2,
A
1β
1)
1
(
1,
1 2 1 1)
p
n n
Page: 2393-2399
This can be also written in following form
d
1p(
β
1,
A
nβ
n)
≤
cd
2p(
β
2,
A
1β
1)
(3.10)
Since,
A A
n−1 n−2
A A
2 1β
1=
β
nTaking
x
3=
x
m3,
x
2=
β
2in inequality (3.2) in the same way as above, we obtain
d
2p(
β
2,
A
1β
1)
≤
cd
3p(
β
3,
A
2β
2)
(3.11)
Continuously like above, we get
d
np(
β
n,
A
n−1β
n−1)
≤
cd
1p(
β
1,
A
nβ
n)
(3.12)
By (3.10), (3.11) and (3.12), we obtain
1
(
1,
)
2(
2,
1 1)
p p
n n
d
β
A
β
≤
cd
β
A
β
2 3(
3,
2 2)
...
p
c d
β
A
β
≤
≤ ≤
11 1
(
,
)
n p
n n n n
c
−d
β
A
−β
− 1(
1,
)
n p
n n
c d
β
A
β
≤
⇒
d
1p(
β
1,
A
nβ
n)
≤
c d
n 1p(
β
1,
A
nβ
n)
⇔
A
nβ
n=
β
1 , since0
≤ <
c
1
Thus again
d
1p(
β
1,
A A
n n−1
A A
2 1β
1)
=
d
1p(
β
1,
A
nβ
n)
=
0
⇔
A A
n n−1
A A
2 1β
1=
β
1So, we proved that
β
1 is a fixed point ofA A
n n−1
A A
2 1β
1In same way it can be shown that
A A
1 n
A A
3 2 has a unique fixed pointβ
2∈
X
2,A A
2 1
A A
4 3 has a unique fixed pointβ
3∈
X
3 and so onA A
n−1 n−2
A A
1 n has a unique fixed pointβ
n∈
X
n.Further, we also showed that
A
1( )
β
1=
β
2,
A
2( )
β
2=
β
3,
A
3( )
β
3=
β
4,
,
A
n−1(
β
n−1)
=
β
n,
A
n( )
β
n=
β
1Now let assume now that
β
11∈
X
1is another fixed point ofA A
n n−1
A A
2 1 , different fromβ
1.Using (3.1), if we take
x
1=
β
1,
x
2=
A
1β
11, we get
d
1p(
β β
11,
1)
=
d
1p(
A A
n n−1
A A
2 1β
11,
A A
n n−1
A A
2 1β
1)
≤
c
max
M
1(
β
1,
A
1β
11)
+
F
(min
M
1(
β
1,
A
1β
11))
Where,
M
1(
β
1,
A
1β
11)
=
{
d
1p(
β
1,
A A
n n−1
A A
2 1β
11),
d
1p(
β
1,
A A
n n−1
A A
2 1β
1),
d
2p(
A
1β
11,
A
1β
1)}
1 1
1 1 1 1 1 1 2 1 1 2
{
d
p(
β β
,
),
d
p(
β β
,
),
d
p(
A
β β
,
)}
=
From which it follows
d
1p(
β β
11,
1)
≤
cd
2p(
A
1β β
11,
2)
(3.13)
Takingx
3=
β
3,
x
2=
A
1β
11 in inequality (3.2) we obtain2p
(
2,
1 11)
2p(
1 4 3 3,
1 3 2 1 11)
n n
d
β
A
β
=
d
A A
A A
β
A A
A A A
β
Where,
M
2(
A
1β β
11,
3)
=
{
d
2p(
A
1β
11,
A A
1 n
A
3β
3),
d
2p(
A
1β
11,
A A
1 n
A A
2 1β
11),
d
3p(
β
3,
A A
2 1β
11)}
=
d
2p(
A
1β β
11,
2),
d
2p(
A
1β
11,
A
1β
11),
d
3p(
β
3,
A A
2 1β
11)}
Then,
d
2p(
β
2,
A
1β
11)
≤
cd
3p(
β
3,
A A
2 1β
11)
(3.14)
Continuously like above, using (3.4), we get
d
np(
A
n−1A
n−2
A A
2 1β β
11,
n)
≤
cd
1p(
β β
11,
1)
(3.15)
By using (3.13), (3.14) and (3.15)
1 1
(
1,
1)
p
d
β β
≤
cd
2p(
A
1β β
11,
2)
≤
c d
2 3p(
β
3,
A A
2 1β
11)
≤ ≤
..
c
n−1d
np(
A
n−1A
n−2
A A
2 1β β
11,
n)
≤
c d
n 1p(
β β
11,
1)
⇔
d
1p(
β β
11,
1)
≤
c d
n 1p(
β β
11,
1)
where0
≤ <
c
1
.It follows that
β
11=
β
1.Thus we proved
β
1is the fixed point ofA A
n n−1
A A
2 1.In the same way, it can be shown that
A A
1 n
A A
3 2 has a unique fixed pointβ
2∈
X
2,A A
2 1
A A
4 3 has a unique fixed pointβ
3∈
X
3 and so onA A
n−1 n−2
A A
1 n has a unique fixed pointβ
n∈
X
n.Corollary3.2: if we take n=3 ,then we obtain [1] proved by Luljeta kikina.
Corollary3.3: if we take n=2 ,then we obtain [2] proved by S.Č.Nešić.
Corollary3.4: if we take n=3, p=2 and F(t)=0 for all t belongs to R+, then we obtain a gerernalization of theorem 2[5], extended to three metric spaces.
REFERENCES
[1] L.Kikina, Fixed point theorem in three metric spaces, Int.Journal of math.analysis. ,Vol.3 (2009), no.13, 619-626.
[2] S.Č.Nešić, A fixed point theorem in two metric spaces, Bull. Math. Soc. Sci.Math. Roumanie (N.S.).,Vol. 44(2001), no.92, 253-257.
[3] B. Fisher, Fixed points theorem on two metric spaces, Glasnik Mat., Vol. 16 (1981) no.36, 333-337.
[4] R. K. Jain, H. K. Sahu, and B. Fisher, Related fixed point theorems for three metric spaces, Novi Sad J. Math., Vol. 26 (1996), no.1 ,11–17.
[5] V. Popa, Fixed points on two metric spaces, Zb. Rad, Prirod, Mat. Fak (N.S.)Ser. Mat.,Vol. 21 (1991), 83-93.
[6] Vishal Gupta, Ashima Kanwar and S.B. Singh, “A Fixed Point Theorem For n-Mapping on n-Metric Spaces” International journal of Emerging Trends in Engineering and Development, Vol.1, Issue 2, p.p.257-268, 2012.
[7] Vishal Gupta, Ashima Kanwar and R.K. Saini, “Related Fixed Point Theorem on n Metric Spaces”, International Journal of Mathematical Archive Vol.3 , No. 3, pp. 902-908, 2012, ISSN: 2229 – 5046.