Gen. Math. Notes, Vol. 7, No. 2, December 2011, pp. 24-37
ISSN 2219-7184; Copyright © ICSRS Publication, 2011
www.i-csrs.org
Available free online at http://www.geman.in
A Fixed Point Theorem for Six Mappings
in Metric Space
Rajesh Shrivastava1, Sarvesh Agrawal2, Ramakant Bhardwaj3
and R.N. Yadava4
1
Prof. and Head (Mathematics), Govt. Science & Commerce College,
Benezeer, Bhopal, India
E-mail: rajeshraju0101@rediffmail.com
2
Department of Mathematics, Truba Group of Institutions,
Bhopal, India
E-mail: agrawalsarvesh1962@gmail.com
3
Department of Mathematics, Truba Group of Institutions,
Bhopal, India
E-mail: rkbhardwaj100@gmail.com
4
Ex-Director & Head, Resource Development Centre (AMPRI),
Bhopal, India
E-mail: dryadava@gmail.com
(Received: 1-6-11/ Accepted: 10-8-11)
Abstract
In present paper we prove a common fixed point theorem for six mappings
using compatibility, weak compatibility and commutativity. Our results improve
one of the result of Imdad and Khan [3], fisher [2].
Keywords: Fixed Points, Metric
Commutativity, Coincidence Point.
Space,
Weak
Compatibility
and
A Fixed Point Theorem for Six Mappings…
2
25
Preliminaries
Before starting the result first we discuss some definitions.
Definition 2.1 [15]. A pair of self-mapping ( A, B ) on a metric space
said to be weakly commuting if d ( ABx, BAx ) ≤ d ( Bx, Ax ) .For all x ∈ X .
( X , d ) is
Definition 2.2. A pair of self mapping ( A, B ) of a metric space ( X , d ) is said to be
compatible if lim d ( ABxn , BAxn ) = 0
n →∞
Whenever lim Axn = lim Bxn = t , for all t ∈ X .
n →∞
n →∞
Theorem 2.1 [2]. Let S and T be two self-mapping of a complete metric space
( X , d ) such that for all x, y in X either
b  d ( x, Ty )  + c  d ( y, Sx ) 
d ( Sx, Ty ) ≤ 
d ( x, Ty ) + d ( y, Sx )
2
(i)
2
if d ( x, Ty ) + d ( y, Sx ) ≠ 0, 0 ≤ b, c, b + c < 1
(ii)
d ( Sx, Ty ) = 0 if d ( x, Ty ) + d ( y, Sx ) = 0
if one of S or T is continuous then S and T have a unique common fixed point.
3
Main Result
Theorem 3.1. Let F , G , H , R, S and T be self mappings of a complete metric
space ( X , d ) satisfying the conditions:
FG ( X ) ⊂ T ( X ) , HR ( X ) ⊂ S ( X ) and for each x, y ∈ X either
 d ( FGx, Ty )2 + d ( HRy, Sx )2 

( 3.1.1) d ( FGx, HRy ) ≤ α 
 d ( FGx, Ty ) + d ( HRy, Sx ) 
 d ( FGx, Ty ) + d ( HRy, Sx ) + d ( FGx, Sx ) + d ( HRy, Ty ) 
+β 
1 +  d ( FGx, Ty ) .d ( HRy, Sx ) .d ( FGx, Sx ) .d ( HRy, Ty ) 
+ γ  d ( FGx, Sx ) + d ( HRy, Ty )  + δ d ( Sx, Ty )
if d ( FGx, Ty ) + d ( HRy, Sx ) ≠ 0, α , β , γ , δ ≥ 0 and 2α + 4β + 2γ + δ < 1
26
Rajesh Shrivastava et al.
or
( 3.1.2 ) d ( FGx, HRy ) = 0 if d ( FGx, Ty ) + d ( HRy, Sx ) = 0 if either
(i) ( FG, S ) are compatible, S or FG is continuous and ( HR, T ) are
weakly
compatible or
(ii) ( HR, T ) are compatible, T or HR is continuous and
weakly
( FG, S ) are
compatible, then FG , HR, S and T have a unique common fixed point.
Further more if the pairs ( F , G ) , ( F , S ) , ( G, S ) , ( H , R ) , ( H , T ) and ( R, T ) are
commuting mappings then F , G , H , R, S and T have a unique common fixed
point.
Proof. Let x0 be an arbitrary point in X. Since GF ( X ) ⊂ T ( X ) , we can find a
point x1 in X such that FGx0 = Tx1. Also since HR ( X ) ⊂ S ( X ) we can choose a
point x2 such that HRx1 = Sx2 . Using this argument repeatedly one can constract a
sequence { pn } such that
p2 n = FGx2 n = Tx2 n +1 , p2 n +1 = HRx2 n +1 = Sx2 n + 2 , for n = 0,1, 2, 3,......
d( p2n+1, p2n+2 ) =d( HRx2n+1,GFx2n+2 )
d( FGx2n+2,Tx2n+1)2 +d( HRx2n+1, Sx2n+2 )2 

≤α 
d( FGx2n+2,Tx2n+1) +d( HRx2n+1, Sx2n+2 )
d( FGx2n+2,Tx2n+1) +d( HRx2n+1, Sx2n+2 ) +d( FGx2n+2, Sx2n+2 ) +d( HRx2n+1,Tx2n+1)
+β 
1+d( FGx2n+2,Tx2n+1) .d( HRx2n+1, Sx2n+2 ) .d( FGx2n+2, Sx2n+2 ) .d( HRx2n+1,Tx2n+1)
+γ d( FGx2n+2, Sx2n+2 ) +d( HRx2n+1,Tx2n+1) +δd( Sx2n+2,Tx2n+1)
≤ α d ( p2n+2 , p2n+1 ) + d ( p2n+1 , p2n ) 
+2β d ( p2n+2 , p2n+1 ) + d ( p2n+1 , p2n ) 
+γ d ( p2n+2 , p2n+1 ) + d ( p2n+1 , p2n )  + δ d ( p2n+1 , p2n )
≤ (α + 2β + γ ) d ( p2n+2 , p2n+1 ) + (α + 2β + γ + δ ) d ( p2n+1 , p2 n )
A Fixed Point Theorem for Six Mappings…
27
 α + 2 β + γ + δ 
d ( p 2 n + 2 , p 2 n +1 ) ≤ 
 d ( p 2 n +1 , p 2 n )
 1 − (α + 2 β + γ ) 
d ( p 2 n + 2 , p 2 n +1 ) ≤ k d ( p 2 n +1 , p 2 n ) ,
where k =
α + 2β + γ + δ
<1
1 − (α + 2 β + γ )
Similarly, d ( p2 n , p2 n +1 ) ≤ k d ( p2 n −1 , p2 n )
Thus for every n, we have
(3.1.3) d ( pn , pn +1 ) ≤ k d ( pn −1 , pn )
Which shows that
( X ,d ) .
{ pn } is
a Cauchy Sequence in the complete metric space
Hence the sequence FGxn = Tx2 n +1 and HRx2 n +1 = Sx2 n + 2 which are
subsequences also converges to the point p.
Let us now assume that S is continuous so that the sequences {S 2 x2n } and
{SFGx2n } converge to Sp. Also the compatibility of {S , FG} , {FGSx2 n } converges
to Sp.
Now
2
2
2


 d ( F G S x 2 n , T x 2 n + 1 ) + d ( H R x 2 n + 1 , S x 2 n ) 

d ( F G S x 2 n , H R x 2 n +1 ) ≤ α
 d ( F G S x 2 n , T x 2 n +1 ) + d ( H R x 2 n +1 , S 2 x 2 n )


 d ( F G S x 2 n , T x 2 n +1 ) + d ( H R x 2 n +1 , S 2 x 2 n ) + d ( F G S x 2 n , S 2 x 2 n )


 + d ( H R x 2 n + 1 , T x 2 n + 1 )

+β
 d ( F G S x 2 n , T x 2 n + 1 ) .d ( H R x 2 n + 1 , S 2 x 2 n ) .d ( F G S x 2 n , S 2 x 2 n ) 

1+ 
 .d ( H R x 2 n + 1 , T x 2 n + 1 )

+ γ  d ( F G S x 2 n , S 2 x 2 n ) + d ( H R x 2 n + 1 , T x 2 n + 1 )  + δ d ( S 2 x 2 n , T x 2 n + 1 )
which on letting n → ∞ , we have
 d ( Sp , p )2 + d ( p , Sp )2 
 + β  d ( SP , P ) + d ( P , SP ) + d ( SP , SP ) + d ( P , p ) 
d ( SP , P ) ≤ α 
1 +  d ( SP , P ) .d ( P , SP ) .d ( SP , SP ) .d ( P , p ) 
 d ( SP , P ) + d ( p , SP ) 
+ γ  d ( SP , SP ) + d ( P , P )  + δ d ( SP , P )
≤ α d ( SP , P ) + 2 β d ( SP , P ) + δ d ( SP , P )
≤ (α + 2 β + δ ) d ( SP , P )
Which is possible only when SP = P , since α + 2 β + δ < 1
Now
28
Rajesh Shrivastava et al.
d ( FGP,Tx2n+1 )2 + d ( HRx2n+1, SP)2 

d ( FGP, HRx2n+1 ) ≤α 
d ( FGP,Tx2n+1 ) + d ( HRx2n+1, SP)
d ( FGP,Tx2n+1 ) + d ( HRx2n+1, SP) + d ( FGP, SP) + d ( HRx2n+1,Tx2n+1 )
+β 
1+d ( FGP,Tx2n+1 ) .d ( HRx2n+1, SP) .d ( FGP, SP) .d ( HRx2n+1,Tx2n+1 ) 
+γ d ( FGP, SP) + d ( HRx2n+1,Tx2n+1 ) +δ d ( SP,Tx2n+1 )
On letting n → ∞ and using SP = P , we obtain
 d ( FGP, P )2 + d ( P, P ) 2 

d ( FGP, P ) ≤ α 
 d ( FGP, P ) + d ( P, P ) 
 d ( FGP, P ) + d ( P, P ) + d ( FGP, P ) + d ( P, P ) 
+β 
1 +  d ( FGP, P ) .d ( P, P ) .d ( FGP, P ) .d ( P, P ) 
+ γ  d ( FGP, P ) + d ( P, P )  + δ d ( P, P )
≤ α d ( FGP, P ) + 2 β d ( FGP, P ) + γ d ( FGP, P )
≤ (α + 2 β + γ ) d ( FGP, P )
which is possible only when FGP = P , since α + 2 β + γ < 1
Since FG ( X ) ⊂ T ( X ) , a point q Then Tq = P so HRP = HR ( Tq ) Now
d ( P , HRq ) = d ( FGP , HRq )
 d ( FGP , Tq ) 2 + d ( HRq , SP )2 

≤α 
 d ( FGP , Tq ) + d ( HRq , SP ) 
 d ( FGP , Tq ) + d ( HRq , SP ) + d ( FGP , SP ) + d ( HRq , Tq ) 
+β 
1 +  d ( FGP , Tq ) .d ( HRq , SP ) .d ( FGP , SP ) .d ( HRq , Tq ) 
+ γ  d ( FGP , SP ) + d ( HRq , Tq )  + δ d ( SP , Tq )
A Fixed Point Theorem for Six Mappings…
 d ( P , P )2 + d
≤ α 
 d ( P , P ) + d
29
( H R q , P ) 
( H R q , P ) 
2
 d ( P , P ) + d ( H R q , P ) + d ( P , P ) + d ( H R q , P ) 
+ β 
1 +  d ( P , P ) . d ( H R q , P ) . d ( P , P ) . d ( H R q , P ) 
+ γ  d
(P , P ) +
d
( H R q , P ) 
+δ d
(P , P )
≤ α d (P, H Rq ) + 2β d (P, H Rq ) + γ d (P, H Rq )
≤ (α + 2 β + γ
)
d (P , H Rq )
Which is possible only when HRq = P = Tq , since α + 2 β + γ < 1 .
Which shows that q is a point of HR and T .
Now by weak compatibility of ( HR, T ) , we have
HRP = HR ( Tq ) = T ( HRq ) = TP
Which shows that P is also a point of ( HR, T ) .
Now
d (P , H R P ) = d (F G P , H R P )
 d ( F G P , T P )2 + d
≤ α 
 d ( F G P , T P ) + d
( H R P , S P ) 
( H R P , S P ) 
2
 d ( F G P , T P ) + d ( H R P , S P ) + d ( F G P , S P ) + d ( H R P , T P ) 
+ β 
1 +  d ( F G P , T P ) . d ( H R P , S P ) . d ( F G P , S P ) . d ( H R P , T P ) 
+ γ  d
(F G P , SP ) +
d
( H R P , T P ) 
+δ d
(S P ,T P )
≤ α d (P , H Rq ) + 2β d (P , H Rq ) + γ d (P , H Rq )
≤ (α + 2 β + γ
)
d (P , H Rq )
Which is possible only when HRq = P = Tq, since α + 2 β + γ < 1
Which shows that q is a point of HR and T .
Now by weak compatibility of ( HR, T ) , we have
HRP = HR ( Tq ) = T ( HRq ) = TP
Which shows that P is also a point of ( HR, T ) .
30
Rajesh Shrivastava et al.
Now
d
(P , H RP ) =
d
(F G P , H R P )
 d ( F G P , T P )2 + d
≤α 
 d ( F G P , T P ) + d
( H R P , S P ) 
( H R P , S P ) 
2
 d ( F G P , T P ) + d ( H R P , S P ) + d ( F G P , S P ) + d ( H R P , T P ) 
+ β 
1 +  d ( F G P , T P ) .d ( H R P , S P ) .d ( F G P , S P ) . d ( H R P , T P ) 
+ γ  d
(FG P , SP ) +
d
( H R P , T P )  + δ
d (SP ,T P
)
 d ( P, P ) 2 + d ( HRP, P )2 

≤α 
 d ( P, P ) + d ( HRP, P ) 
 d ( P, P ) + d ( HRP, P ) + d ( P, P ) + d ( HRP, P ) 
+β 
1 +  d ( P, P ) .d ( HRP, P ) .d ( P, P ) .d ( HRP, P ) 
+ γ  d ( P, P ) + d ( HRP, P )  + δ d ( P, P )
≤ (α + 2 β + γ ) d ( HRP, P )
Which is possible only when HRP = P = TP , since α + 2 β + γ < 1 .
Hence P is a common fixed point of FG , S , HR and T .
Now suppose that FG is continuous so that the sequence
{FGSx2n } converges to FGP . Since ( FG, S ) are
{SFGx2n } also converges to FGP . Thus
{FG x } and
2
2n
compatible it follows that
2
2
 d ( F G 2 x , Tx
2n
2 n + 1 ) + d ( H Rx 2 n + 1 , SFG x 2 n ) 



d ( F G x 2 n , H R x 2 n +1 ) ≤ α
 d ( F G 2 x 2 n , Tx 2 n + 1 ) + d ( H Rx 2 n + 1 , SFG x 2 n ) 


2
 d ( F G 2 x 2 n , Tx 2 n + 1 ) + d ( H Rx 2 n + 1 , SFG x 2 n ) 


 + d ( FG 2 x 2 n , SF G x 2 n ) + d ( H Rx 2 n + 1 , Tx 2 n + 1 ) 

+β 
 d ( F G 2 x 2 n , Tx 2 n + 1 ) .d ( H R x 2 n +1 , SF G x 2 n ) 

1+ 
 .d ( F G 2 x 2 n , SF G x 2 n ) .d ( H R x 2 n + 1 , Tx 2 n + 1 ) 


+ γ  d ( F G 2 x 2 n , SFG x 2 n ) + d ( H R x 2 n + 1 , T x 2 n + 1 )  + δ d ( SF G x 2 n , Tx 2 n + 1 )
A Fixed Point Theorem for Six Mappings…
31
On making n → ∞ , we obtain
 d ( F G P , P )2 + d
d (F G P , P ) ≤ α 
 d ( F G P , P ) + d
( P , F G P ) 
( P , F G P ) 
2
 d ( F G P , P ) + d ( P , F G P ) + d ( F G P , F G P ) + d ( P , P ) 
+ β 
1 +  d ( F G P , P ) . d ( P , F G P ) . d ( F G P , F G P ) . d ( P , P ) 
+ γ  d
(F G P , F G P ) +
≤ (α + 2 β + δ
)
d
d
( P , P ) 
+ δ d
(F G P , P )
(F G P , P )
Which is possible only when FGP = P , since α + 2 β + δ < 1.
As earlier there exists q in X such that FGP = P = Tq. Then
d ( FG2 x ,Tq)2 + d ( HRq, SFGx )2 
2n
2n 


d ( FG2 x2n , HRq) ≤ α 
2
d ( FG x2n ,Tq) + d ( HRq, SFGx2n ) 


d ( FG2 x2n ,Tq) + d ( HRq, SFGx2n ) + d ( FG2 x2n , SFGx2n ) + d ( HRq,Tq) 

+β 
2
2


1+ d ( FG x2n ,Tq) .d ( HRq, SFGx2n ) .d ( FG x2n , SFGx2n ) .d ( HRq,Tq)
+ γ d ( FG2 x2n , SFGx2n ) + d ( HRq,Tq)  +δ d ( SFGx2n ,Tq)
which on letting n → ∞ , we obtain
d ( P, P)2 + d ( HRq, P)2 
 + β d ( P, P) + d ( HRq, P) + d ( P, P) + d ( HRq, P) 
d ( P, HRq) ≤ α 
d ( P, P) + d ( HRq, P)
1+ d ( P, P) .d ( HRq, P) .d ( P, P) .d ( HRq, P) 
+γ d ( P, P) + d ( HRq, P) +δ d ( P, P)
≤ (α + 2β + γ ) d ( HRq, P)
which given HRq = P = Tq , since α + 2 β + γ < 1.
( HR, T ) . Since the pair ( HR, T ) is weakly
HRP = HR ( Tq ) = T ( HRq ) = TP which shows that HRP = TP.
Thus q is a coincidence point of
compatible one has
Further
32
Rajesh Shrivastava et al.
d ( FGx2n , TP)2 + d ( HRP, Sx2n )2 

d ( FGx2n , HRP) ≤ α 
d ( FGx2n , TP ) + d ( HRP, Sx2n ) 
d ( FGx2n , TP) + d ( HRP, Sx2n ) + d ( FGx2n , Sx2n ) + d ( HRP, TP) 
+β 
1 + d ( FGx2n , TP) .d ( HRP, Sx2n ) .d ( FGx2n , Sx2n ) .d ( HRP, TP) 
+ γ d ( FGx2n , Sx2n ) + d ( HRP, TP)  + δ d ( Sx2n , TP)
making n → ∞ . We obtain
 d ( P, P )2 + d ( HRP, P )2 

d ( P, HRP ) ≤ α 
 d ( P, P ) + d ( HRP, P ) 
 d ( P, P ) + d ( HRP, P ) + d ( P, P ) + d ( HRP, P ) 
+β 
1 +  d ( P, P ) .d ( HRP, P ) .d ( P, P ) .d ( HRP, P ) 
+ γ  d ( P, P ) + d ( HRP, P )  + δ d ( P, P )
≤ (α + 2β + γ ) d ( HRP, P )
It follows that HRP = P = TP , since α + 2 β + γ < 1 .
The point P therefore is in the range of HR and since HR ( X ) ⊂ S ( X ) . There
exists a point w in X such that Sw = P. Thus
d ( FGw, P ) = d ( FGw, HRP )
 d ( FGw, TP )2 + d ( HRP, Sw ) 2 

≤α 
 d ( FGw, TP ) + d ( HRP, Sw ) 
 d ( FGw, TP ) + d ( HRP, Sw ) + d ( FGw, Sw ) + d ( HRP, TP ) 
+β 
1 +  d ( FGw, TP ) .d ( HRP, Sw ) .d ( FGw, Sw ) .d ( HRP, TP ) 
+ γ  d ( FGw, Sw ) + d ( HRP, TP )  + δ d ( Sw, TP )
A Fixed Point Theorem for Six Mappings…
33
d ( FGw, P)2 + d ( P, P)2 
 + β d ( FGw, P) + d ( P, P) + d ( FGw, P) + d ( P, P) 
≤α 
d ( FGw, P) + d ( P, P)
1+ d ( FGw, P) .d ( P, P) .d ( FGw, P) .d ( P, P) 
+ γ d ( FGw, P) + d ( P, P) + δ d ( P, P)
≤ (α + 2β + γ ) d ( FGw, P)
Hence FGw = P , since α + 2 β + γ < 1.
Also FGP = SP = P.
Thus we have proved that P is common fixed point of FG , HR, S and T .
Let v be another fixed point of S , T , FG and HR , then
d ( P, v) = d ( FGP, HRv)
d ( FGP,Tv)2 + d ( HRv, SP)2 

≤α 
d ( FGP,Tv) + d ( HRv, SP) 
d ( FGP,Tv) + d ( HRv, SP) + d ( FGP, SP) + d ( HRv, Tv) 
+β 
1+ d ( FGP,Tv) .d ( HRv, SP) .d ( FGP, SP) .d ( HRv,Tv)
+ γ d ( FGP, SP) + d ( HRv,Tv)  + δ d ( SP,Tv)
d ( P, v)2 + d ( v, P)2  d ( P, v) + d ( v, P) + d ( P, P) + d ( v, v) 
 +β 

≤α 
d ( P, v) + d ( v, P) 
1+ d ( P, v) .d ( v, P) .d ( P, P) .d ( v,v)
+γ d ( P, P) + d ( v, v)  +δ d ( P, v)
≤ (α + 2β +δ ) d ( P, v)
Yielding thereby P = v , since α + 2 β + δ < 1 .
Finally, we prove that P is also a common fixed point of F , G , H , R, S and T .
34
Rajesh Shrivastava et al.
For this P be the unique common fixed point of both the pairs
( HR, T ) . Then
( FG, S ) and
FP = F ( FGP ) = F ( GFP ) = FG ( FP ) , FP = F ( SP ) = S ( FP ) .
GP = G ( FGP ) = G ( FGP ) = GF ( GP ) = FG ( GP ) , GP = G ( SP ) = S ( GP ) .
Which shows that FP and GP is a common fixed point of
( FG, S ) .
Yielding
thereby FP = P = GP = SP = FGP in the view of uniqueness of the common fixed
point of the pair ( FG, S ) .
Similarly using the commutativity of ( H , R ) , ( H , T ) and ( R, T ) it can be shown
that HP = P = RP = TP = HRP .
Now we show that FP = HP, GP = RP . Also remains a common fixed point of
both the pairs ( FG, S ) and ( HR, T ) . For this
d ( FP , H P ) = d ( F ( G FP ) , H ( RH P ) )
= d ( FG ( FP ) , H R ( H P ) )
 d ( FG ( F P ) , T ( H P ) )2 + d ( H R ( H P ) , S ( FP ) )2 

≤α 
 d ( FG ( FP ) , T ( H P ) ) + d ( H R ( H P ) , S ( FP ) ) 
 d ( FG ( F P ) , T ( H P ) ) + d ( H R ( H P ) , S ( FP ) ) + d ( FG ( F P ) , S ( FP ) ) 


 + d ( H R ( H P ) , T ( H P ) )

+β
 d ( F G ( FP ) , T ( H P ) ) .d ( H R ( H P ) , S ( F P ) ) .d ( FG ( FP ) , S ( F P ) ) 

1+ 
 .d ( H R ( H P ) , T ( H P ) )

+ γ  d ( FG ( F P ) , S ( FP ) ) + d ( H R ( H P ) , T ( H P ) )  + δ d ( S ( FP ) , T ( H P ) )
d ( FP, HP ) ≤ 0
⇒ d ( FP, HP ) = 0
⇒ FP = HP
Similarly, it can be shown that GP = RP .
Thus P is the unique common fixed point of F , G , H , R, S and T .
Now, if d ( FGx, Ty ) + d ( HRy, Sx ) = 0 ⇒ d ( FGx, HRy ) = 0 . Then suppose that
there exists n such that Pn = Pn +1 . Then, also Pn +1 = Pn + 2 , suppose not. Then from
(3.1.3) we have 0 < d ( Pn +1 , Pn + 2 ) < k d ( Pn +1 , Pn ) yielding thereby Pn +1 = Pn + 2 . Thus
A Fixed Point Theorem for Six Mappings…
35
Pn = Pn + k for k=1,2,3,… . It then follows that there exists two points z1 and z2 such
that v1 = FGz1 = Tz2 and v2 = HRz2 = Sz1 . Since d ( FGz1 , Tz2 ) + d ( HRz2 , Sz1 ) = 0 ,
from
(3.1.2)
d ( FGz1 , HRz2 ) = 0 i.e.
Sv1 = I ( FGz1 ) = FG ( Iz1 ) = FGv1 .
v1 = FGz1 = HRz2 = v2 .
Similarly
HRv2 = Tv2 .
Also
Define
y1 = FGv1 , y2 = HRv2 . Since d ( FGv1 , Tv2 ) + d ( HRv2 , Sv1 ) = 0 , it follows from
(3.1.2) that d ( FGv1 , HRv2 ) = 0 i.e. y1 = y2 . Thus FGv1 = Sv1 = HRv2 = Tv2 .
But v1 = v2 .
Therefore FG , S , HR and T have a common coincidence point. Define w = FGv1 ,
it then follows that w is also a common coincidence point of FG , S , HR and T . If
FGw ≠ FGv1 = HRv1 ,
d ( FGw, HRv1 ) > 0 .
then
d ( FGw, Tv1 ) + d ( HRv1 , Sw ) = 0 ,
it
follows
from
But,
(3.1.2)
Since
that
d ( FGw, HRv1 ) = 0 i.e. FGw = HRv1 , which is a contradiction.
Hence FGw = HGv1 = w and w is a common fixed point of FG , HR, S and T . The
rest of the proof is identical to the previous calculation, hence it is omitted.
This completes the proof.
Corollary 3.2. Theorem 3.1 remains true if contraction condition (3.1.1) and
(3.1.2) are replaced by any one of the following condition:
(a) By changing β = 0, δ = 0. Then
 d ( FGx, Ty ) 2 + d ( HRy, Sx )2 

d ( FGx, HRy ) ≤ α 
 d ( FGx, Ty ) + d ( HRy, Sx ) 
+ γ  d ( FGx, Sx ) + d ( HRy, Ty ) 
if d ( FGx, Ty ) + d ( HRy, Sx ) ≠ 0, α , γ > 0, 2α + 2γ < 1
or d ( FGx, HRy ) = 0 if d ( FGx, Ty ) + d ( HRy, Sx ) = 0.
Which is a results of Fisher [2] and Kannan [9].
(b) By choosing β = 0, γ = 0. Then
 d ( FGx, Ty ) 2 + d ( HRy, Sx )2 
 + δ d Sx, Ty
d ( FGx, HRy ) ≤ α 
(
)
 d ( FGx, Ty ) + d ( HRy, Sx ) 
36
Rajesh Shrivastava et al.
if d ( FGx, Ty ) + d ( HRy, Sx ) ≠ 0, α , δ > 0, 2α + δ < 1
or d ( FGx, HRy ) = 0 if d ( FGx, Ty ) + d ( HRy, Sx ) = 0.
Which is a results of Fisher [2].
(c) By choosing β = 0, γ = 0, δ = 0. Then
 d ( F G x , T y )2 + d ( H R y , S x )2 

d ( FG x, H Ry ) ≤ α 
 d ( F G x , T y ) + d ( H R y , S x ) 
if d
(F G x,T y )+
or d
(F G x, H R y ) =
d
(H R y , S x ) ≠
0 if d
0,α > 0,α < 1 / 2
(F G x,T y ) +
d
(H R y , Sx ) =
0.
Which extends a theorem of Fisher [2].
(d)
d ( FGx,Ty)2 + d ( HRy, Sx)2  d ( FGx,Ty) + d ( HRy, Sx) 2
≤
 ≤ d FGx,Ty + d HRy, Sx
If 
(
) (
)
d ( FGx,Ty) + d ( HRy, Sx)  d ( FGx,Ty) + d ( HRy, Sx) 
and β = 0, then condition ( 3.1.1) be
d ( FGx, HRy) ≤α d ( FGx,Ty) + d ( HRy, Sx)  +γ d ( FGx, Sx) + d ( HRy,Ty)  +δ d ( Sx,Ty)
If 2α + 2γ + δ < 1 .
Which extends a theorem of Hardy-Rogers. [14].
4
Remarks
(i)
If we put F = A, G = B, H = S , R = T , S = I and T = J and
α = α1 , β = 0, γ = α 2 , δ = α 3 , then we get the result of Imdad and Khan.
Our result is motivated by Fisher and J. Rhoades, Kannan, Hardy-Roger.
(ii)
Reference
[1]
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A Fixed Point Theorem for Six Mappings…
[3]
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[8]
[9]
[10]
[11]
[12]
[13]
[14]
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