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ISSN 2229 – 5046

Theorem 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*

(2)

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 where

i

= 1, 2, 3,

,

n

. If

A

i is mapping of

X

i to

X

i+1 where

= 1, 2, 3,

,

1

i

n

and

A

n is mapping of

x

n to

x

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 with

F

(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 where

i

= 1, 2, 3,

,

n

. If

A

i is mapping of

X

i to

X

i+1 where

i

= 1, 2, 3,

,

n

1

and

A

n is mapping of

x

n to

x

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 2

1 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

(3)

1 2

1

,

2

,

,

n n

x

X x

X

x

X

∀ ∈

, where 0≤c<1. Then

A A

n n1

A A

2 1 has a unique fixed point

β

1

X

1 ,

1 n 3 2

A A

A A

has a unique fixed point

β

2

X

2,

A A

2 1

A A

4 3 has a unique fixed point

β

3

X

3 and so on

1 2 1

n n n

A A

A A

has a unique fixed point

β

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

n

β

n

=

β

Proof: Let

x

10 be an arbitrary point in

X

1 , let define sequence

{

x

m1

}

,

{

x

m2

},

,{

x

mn

}

in

X X

1

,

2

,

,

X

n respectively by

(

)

1 1

1 3 2 1 0

=

,

m

n n m

A A

A A A

x

x

x

m2

=

A x

1

( )

1m1

,

x

m3

=

A x

2

( )

m2

x

mn

=

A

n1

( )

x

mn−1

x

1m

=

A x

n

( )

mn

for

m

= 1, 2, 3,

We will assume that

x

m1

x

1m+1

,

x

m2

x

m2+1and so on

x

mn

x

mn+1for all m. Otherwise, if

x

m1

=

x

m1+1for some m, then

2 2 1

m m

x

=

x

+ ,

x

m3

=

x

m3+1and so on

x

mn

=

x

mn+1, we could put

x

1m

=

β

1,

x

m2+1

=

β

2and so on

x

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 obtain

1 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

)

, then

1 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+1since

0

≤ <

c

1

, so

1 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

m31 in (2.2) and (3.2), we get

2 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

m31

,

x

m3

)}

{0,

2p

(

2

,

2 1

),

3p

(

3 1

,

3

)}

m m m m

(4)

Page: 2393-2399

2 2 2

(

,

1

)

p m m

d

x

x

+ =

d

2p

(

A A

1 n

A A x

4 3 3m1

,

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

m31

,

x

m3

)}

+

F

(0)

Since

0

≤ <

c

1

, we get

2 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

m41 in (2.3) and (3.3), we get

3 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

m41

,

x

m4

)}

=

{0,

d

3p

(

x

m3

,

x

m3+1

),

d

4p

(

x

m41

,

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

m41

)

+

F

(min

M x

3

(

m3

,

x

m41

))

=

c

max{0,

d

3p

(

x

m3

,

x

3m+1

),

d

4p

(

x

m41

,

x

m4

)}

+

F

(0)

=

cd

4p

(

x

m41

,

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

m11in (2.4) and (3.4) we obtain

1 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

m11

,

x

m1

)}

=

{0,

d

np

(

x

mn

,

x

mn+1

),

d

1p

(

x

m11

,

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

1m1

)

+

F

(min

M

n

(

x

mn

,

x

m11

))

=

c

max{0,

d

np

(

x

mn

,

x

mn+1

),

d

1p

(

x

m11

,

x

m1

)}

+

F

(0)

=

cd

1p

(

x

1m1

,

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 3

3

(

1

,

)

p

m m

c d

x

x

. . .

c

n 1

d

np

(

x

m nn 2

,

x

m nn 3

)

c d

n 1p

(

x

m nn 1

,

x

m nn 2

)

− + − + − + − +

c

2n

d

1p

(

x

mn2n+2

,

x

mn2n+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

=

+

≤ ≤ 

=

(5)

Since

0

≤ <

c

1

, the sequences

{

x

m1

}

,

{

x

m2

},

,{

x

mn

}

are cauchy sequences. Since

(

X d

i

,

i

)

be complete metric spaces where

i

= 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

1mand

x

2

=

β

2in the inequality (3.1) we obtain

1 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 n1

A

2

β

2

),

d

1p

(

x

1m

,

A A

n n1

A A x

2 1 1m

),

d

2p

(

β

2

,

A x

1 m1

)}

=

{

d

1p

(

x

m1

,

A A

n n1

A

2

β

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 n1

A

2

β β

2

,

1

)

cd

1p

(

β

1

,

A A

n n1

A

2

β

2

)

So we get,

A A

n n1

A

2

β

2

=

β

1

In same way, we obtain

A A

1 n

A A

4 3

β

3

=

β

2 ,

A A A

2 1 n

A A

5 4

β

4

=

β

3, . . . ,

A

n1

A

n2

A A

2 1

β

1

=

β

n

Using (3.1), taking

x

1

=

β

1

,

x

2

=

x

m2 we get

d

1p

(

x

1m

,

A A

n n1

A A

2 1

β

1

)

=

d

1p

(

A A

n n1

A x

2 m2

,

A A

n n1

A A

2 1

β

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 n1

A x

2 m2

),

d

1p

(

β

1

,

A A

n n1

A A

2 1

β

1

),

d

2p

(

x

m2

,

A

1

β

1

)}

=

{

d

1p

(

β

1

,

A A

n n1

A x

2 1m

),

d

1p

(

β

1

,

A A

n n1

A A

2 1

β

1

),

d

2p

(

x

m2

,

A

1

β

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 n1

A A

2 1

β

1

)

1

(

1

,

1 2 1 1

)

1 2 1 1 1

p

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 n1

A A

2 1

β

1

)

cd

2p

(

β

2

,

A

1

β

1

)

1

(

1

,

1 2 1 1

)

p

n n

(6)

Page: 2393-2399

This can be also written in following form

d

1p

(

β

1

,

A

n

β

n

)

cd

2p

(

β

2

,

A

1

β

1

)

(3.10)

Since,

A A

n1 n2

A A

2 1

β

1

=

β

n

Taking

x

3

=

x

m3

,

x

2

=

β

2in inequality (3.2) in the same way as above, we obtain

d

2p

(

β

2

,

A

1

β

1

)

cd

3p

(

β

3

,

A

2

β

2

)

(3.11)

Continuously like above, we get

d

np

(

β

n

,

A

n1

β

n1

)

cd

1p

(

β

1

,

A

n

β

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

≤ ≤

1

1 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

n

β

n

)

c d

n 1p

(

β

1

,

A

n

β

n

)

A

n

β

n

=

β

1 , since

0

≤ <

c

1

Thus again

d

1p

(

β

1

,

A A

n n1

A A

2 1

β

1

)

=

d

1p

(

β

1

,

A

n

β

n

)

=

0

A A

n n1

A A

2 1

β

1

=

β

1

So, we proved that

β

1 is a fixed point of

A A

n n1

A A

2 1

β

1

In same way it can be shown that

A A

1 n

A A

3 2 has a unique fixed point

β

2

X

2,

A A

2 1

A A

4 3 has a unique fixed point

β

3

X

3 and so on

A A

n1 n2

A A

1 n has a unique fixed point

β

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

n1

(

β

n1

)

=

β

n

,

A

n

( )

β

n

=

β

1

Now let assume now that

β

11

X

1is another fixed point of

A A

n n−1

A A

2 1 , different from

β

1.

Using (3.1), if we take

x

1

=

β

1

,

x

2

=

A

1

β

11, we get

d

1p

(

β β

11

,

1

)

=

d

1p

(

A A

n n1

A A

2 1

β

11

,

A A

n n1

A A

2 1

β

1

)

c

max

M

1

(

β

1

,

A

1

β

11

)

+

F

(min

M

1

(

β

1

,

A

1

β

11

))

Where,

M

1

(

β

1

,

A

1

β

11

)

=

{

d

1p

(

β

1

,

A A

n n1

A A

2 1

β

11

),

d

1p

(

β

1

,

A A

n n1

A A

2 1

β

1

),

d

2p

(

A

1

β

11

,

A

1

β

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

1

β β

11

,

2

)

(3.13)

Taking

x

3

=

β

3

,

x

2

=

A

1

β

11 in inequality (3.2) we obtain

2p

(

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

(7)

Where,

M

2

(

A

1

β β

11

,

3

)

=

{

d

2p

(

A

1

β

11

,

A A

1 n

A

3

β

3

),

d

2p

(

A

1

β

11

,

A A

1 n

A A

2 1

β

11

),

d

3p

(

β

3

,

A A

2 1

β

11

)}

=

d

2p

(

A

1

β β

11

,

2

),

d

2p

(

A

1

β

11

,

A

1

β

11

),

d

3p

(

β

3

,

A A

2 1

β

11

)}

Then,

d

2p

(

β

2

,

A

1

β

11

)

cd

3p

(

β

3

,

A A

2 1

β

11

)

(3.14)

Continuously like above, using (3.4), we get

d

np

(

A

n1

A

n2

A A

2 1

β 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

1

β β

11

,

2

)

c d

2 3p

(

β

3

,

A A

2 1

β

11

)

≤ ≤

..

c

n−1

d

np

(

A

n1

A

n2

A A

2 1

β 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

)

where

0

≤ <

c

1

.

It follows that

β

11

=

β

1.

Thus we proved

β

1is the fixed point of

A 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 point

β

2

X

2,

A A

2 1

A A

4 3 has a unique fixed point

β

3

X

3 and so on

A A

n1 n2

A A

1 n has a unique fixed point

β

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.

References

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