A FOCUS ON FIXED POINT THEOREM IN BANACH SPACE

IJESM Volume 2, Issue 2 ISSN: 2320-0294
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2013
A FOCUS ON FIXED POINT THEOREM IN BANACH
SPACE
Neena B. Gupta
Abstract
In this paper we present a fixed point theorem with the help of self mapping which satisfy the
contractive type of condition in Banach Space. Its purpose is to change the contractive condition
by D.P. Shukla & Shivkant Tiwari [4]
Keywords :
Fixed point, self maps, contraction mapping, Banach Space, Cauchy Sequence, Convex Set.
Introduction :
In (1979) Fisher gave the contractive condition for a mapping :S X X
2
[ ( , )] ( , ) ( , ) ( , ) ( , )d Sx Ty d x Sx d y Sy d x Sy d y Sx  
For all x, y X and 0 1  and 0 
In 2012 D.P. Shukla [4] established a fixed point theorem satisfying the following condition
1
{ ( , ) ( , ) ( , ) ( , )},
5
1
{ ( , ) ( , ) ( , ) ( , )},
2 5
1
{ ( , ) ( , ) ( , ) ( , )}
5
[ ( , ] .min
d x Sx d x Sy d x Sy d y Sx
d x Sx d x Sy d x Sx d y Sx
d x Sy d y Sx d x Sx d y Sx
d Sx Sy 



 
 
 
 
 
 
 
 

Asst. Prof. of Mathematics, Career College, Bhopal, M.P., India

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Definition and Preliminaries :
 Banach Space : A normed linear space which is complete as a
metric space is called a Banach Space.
 Contacting Mapping : If (X, d) be a complete metric space. A
mapping S: X X is
called a contacting mapping if there exists a real number  with
0 a 1 such that d(Sx,Sy) d(x,y) ( , ), for every x,y Xd x y    
Thus in a contracting mapping the distance between the images of any two pints is less then the
distance between the points.
 Fixed point : Let X be a non empty set and Let S : X X, for all
x  X
Such that Sx = x, for all x  X
That is S maps X to itself.
Then x is called fixed point of the mapping S
 Convex Set : A non empty subset X of a Bonach Space is said to be
convex
if (1-) x+y  X, for all x,y  X
where  is any real such that 0 ≤ <1
Theorem :
Let X be a closed and convex subset of a Banach Space and let S be a self mapping of X into
itself which satisfies the following condition :
1
( , ) ( , ),
4
1
( , ) ( , ),
2 4
1
( , ) ( , ),
4
[ ( , )] max , For all x X
d x Sx d x STx
d x Sx d Tx Sx
d x STx d Tx Sx
d Sx Sy 
 
 
 
  
 
 
 
 
And { , , }y Sx Tx STx & 0 1 
Where T is self mapping in X such that

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2013
........................................(2)
2
x Sx
Tx


Then S has a fixed point
Proof- By the definition of metric space
   
( , )
2 ( )
2
2
2
2
2
2 , (2)
2d x,Tx 3
d x Sx x Sx
x x Sx x
x Sx x
Sx x
x
x Tx
x Tx by
 
   
 


 
 
 
 
 Now d Sx, Tx
, (2)
2
2
1
2
1
( , )...................(4)
2
Sx Tx
x Sx
Sx by
Sx x
x Sx
d x Sx
 

 


 

 
1
Again Sx,Tx ( , )
2
1
[2 ( , )], (3)
2
d d x Sx
d x Tx by


 Sx,Tx ( , )............................................(5)d d x Tx
Now taking A =2[(Tx-STx)+STx]
x+Sx
=2 -STx
2
= x+Sx-STx ............................................(6)
STx
 
  

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2
Now d (A, ) =
= , by(6)
2 2 , 2
= , by(6)
2 2 , 2
2
2 ( , )
2 .2 ( , ), by(3)
4 ( , )............................................(7)
STx A STx
x Sx STx STx
Tx STx by
x Sx STx STx
Tx STx by
Tx STx
d Tx STx
d Tx TTx
d Tx T x

  
 
  
 
 



( , ) ( , ) ( , ),
( , )
( , ), (6)
( ) ( ) ( , )
( , )
( , ) ( , ) ( , )
( , ) 2 ( , )
Now d A STx d A Sx d Sx STx by triangular Inequality
A Sx d Sx STx
x Sx STx Sx d Sx STx by
x Sx Sx ST d Sx STx
x Sx Sx STx d Sx STx
d x Sx d Sx STx d Sx STx
d x Sx d Sx STx
 
  
    
    
    
  
 
( , ) ( , ) 2 ( , )............................................(8)d A STx d x Sx d Sx STx  
By (7) & (8)
2
4 ( , ) ( , ) 2 ( , )d Tx T x d x Sx d Sx STx 
2 ( , ) 2 ( , ), (3)d x Tx d Sx STx By 
2
2
4 ( , ) 2 ( , ) 2 ( , )
2 ( , ) ( , ) ( , )............................................(9)
d Tx T x d x Tx d Sx STx
d Tx T x d x Tx d Sx STx
  
 
Now from (1)
2
1
( , ). ( , ),
4
1
[ ( , )] .max ( , ). ( , ),
4
1
( , ). ( , )
4
d x Sx d x STx
d Sx Sy d x Sx d Tx Sx
d x STx d Tx Sx

 
 
 
 
 
 
 
  

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 
1
( , )[ ( , ) ( , ),
4
1 1
( , ) . max ( , ) ( , ), ByTriangular Inequality & By 4 &
4 2
1
[ ( , )( ( , ) ( , )
4
d x Sx d x Sx d Sx STx
d Sx STx d x Sx x Sx y Tx
d Sx Tx d x Sx d Sx STx

 
 
 
   
 
 
 
  
2
2
2
2
2
1
[( ( , ) ( , ) ( , )],
4
1
.max [ ( , )]
8
1 1
[ ( , )[ ( , ) ( , )]
4 2
1
[( ( , )] ( , ) ( , ),
4
1 1
.max [ ( ( , )) ],
2 4
1
{[ ( , )] ( , ) ( , )}
8
d x Sx
d x Sx


 
 
 
 
 
 
 
  
 
 
 
 
 
 
 
 
2
2
2
1 1
2 ( ( , )] ( , ) ( , )] ,
8 8
1
max ( ( , )] ,
8
1 1
( ( , ) ( , ) ( , )
8 8
d x Sx

  
    
 
  
 
 
 
 
21 1
.2 [ ( , )] ( , ) ( , )
8 8
d x Sx d x Sx d Sx STx
 
   
2
[ ( , )] ( , ) ( , )
4

   
2 2
4[ ( , )] [ ( , )] ( , ) ( , )d Sx STx d x Sx d x Sx d Sx STx     
2 2
4[ ( , )] ( , ) ( , ) [ ( , )] 0d Sx STx d x Sx d x STx d x Sx    
Which is quadratic equation
Then by the solution for equation ax2
+bx+c=0 is given by
2
4
2
b b ac
x
a
  


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 
2 2 2
( , ) [ ( , )] 16 ( [( , )]
Hence d S,STx - 0
2 4
x Sx d x Sx d x Sx   


 
2
( , ) ( , ) 16
d S,STx - 0
8
x Sx d x Sx   

 
2
2 ( , )[ 16
d S,STx - 0
8
d x Tx    

 
2
( , )[ 16
d S,STx - 0, (3) , ( )
4
d x Tx
by on taking ve sign
   
 
 d Sx,STx - d(x,Tx) 0 
( , ) ( , )............................................(10)d Sx STx d x Tx
2
16
Where =
4
  

 
Where 0 1 
2
+ 16
Because if 1 then 1
4
  


 
2
+ 16 4    
2
16 4     
2 2
16 (4 )     
2 2
16 16 8       
16
24 16 0.66 <1 => 1
24
         
Hence 1 
0 0 Hence 0 1And also       
Now by (9) & (10)
2
2 ( , ) ( , ) ( , ) (1 ) ( , )d Tx T x d x Tx d x Tx d x Tx    
2 (1 )
( , ) ( , )
2
d Tx T x d x Tx


Similarly

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2 3 2
2
(1 )
( , ) ( , )
2
(1 ) (1 ) 1
( , ) ( , )
2 2 2
d T x T x d Tx T x
d x Tx d x Tx

  


   
   
 
Similarly we can find
3
3 4 1
( , ) ( , )
2
d T x T x d x Tx
 
  
 
4
4 5 1
( , ) ( , )
2
and d T x T x d x Tx
 
  
 
- - - - - - - - - - - - - - - - - - - - - - -
- - - - - - - - - - - - - - - - - - - - - - -
1 1
( , ) ( , )............................................(11)
2
n
n n
d T x T x d x Tx
  
  
 
(1 )
1 1 2 1
2

 

     
1
lim 0
2
n
n
Then


 
 
 
1
(11) ( , ) 0n n
from euqtion d T x T x as n
   
1
( , ) for 0n n
d T x T x as n 
   
1{ } a Cauchy Sequence in X. (By the defination of Cauchy Sequence)n
nT x is

But X is a Banach Space. Then by the property of completeness
1{ } , converges to a fixed point.n
nT x is a convergent sequence in X which

Let there exists a point in X such that lim ............................................(12)n
n
T x 


1 1
Now consider d( , ) ( , ) ( , ) triangular,n n
S d T d T S by Inequality      
 
1
( , ) ( , )n n
d T d TT S   
 
1
( , )n n
d T TT S   
  

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1 1
( , ) ( ) , by(2)
2
n n n
d T T ST S    
   
1 1 1 1 1
( , )
2 2 2 2
n n n
d T T S ST S     
    
1 1 1
( , )
2 2
n n n
d T ST S T S     
    
1 1 1
( , ) ( , ) ( , )
2 2
n n n
d T d ST S d T S     
  
 1 1
( , ) ( , ) ( , )
2
n n n
d T d T S d S ST     
  
 1 1
( , ) ( , )
2
n n n
d T d T ST   
 
1
( , ) ( , ), (3)n n n
d T d T TT by   
 
1 1
( , ) ( , )n n n
d T d T T    
 
1 1
( , ) ( , )n n n
d T d T T    
 
( , )n
d T 
( , ) .n
d T X isconvex  
( , ) ( , )n
d S d T    
( , ) 0, (12)n
As n then d T by   
( , ) 0d S S      
point in XS has a fixed 

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References :
[1] Bajaj N., 2001. A fixed point theorem in Banach space:
Vikram Mathematical Journal, 21:7-10.
[2] Chaubey A.K. and Sahu D.P., 2011. Fixed point of contraction type mapping in Banach space.
Napier Indian Advanced Research Journal of Science, 6:43-46.
[3] Ciric Lj., 1977. Quasi-Contractions in Banach space: Publ. L’Instt. Math. (N.S.), 21(35):41-48.
[4] D.P., Shukla, 2012, Fixed point theorem in Banach Space, Indian J.Sci. Res. 3(1):177-178
[5] Fisher B., 1979. Mappings with a common fixed point: Math Sem. Notes, Kobe Unive., 7:81-84.

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