Manish Kumar Mishra et al. / (IJAEST) INTERNATIONAL JOURNAL OF ADVANCED ENGINEERING SCIENCES AND TECHNOLOGIES Vol No. 2, Issue No. 1, 104 - 107
Related Fixed Points Theorems on Three Metric Spaces ( FPTTMS) Manish Kumar Mishra and Deo Brat Ojha
mkm2781@rediffmail.com, deobratojha@rediffmail.com
Department of Mathematics R.K.G.Institute of Technology Ghaziabad,U.P.,INDIA
Abstract— We obtained related fixed point theorem on three metric spaces satisfying integral type inequality.
Theorem 1.3 : Let ( X , d ) , (Y , ) and ( Z , ) be complete metric spaces and suppose T is a continuous mapping of X into Y, S is a continuous mapping of Y into Z and R is a continuous mapping of
Mathematics Subject Classification: 54H25
.
INTRODUCTION
Z into X satisfying the inequalities:
ES
Keywords- Three metric space, fixed point, integral type inequality.
I.
T
Rw = u. The next theorem was proved in [3].
The following fixed point theorem was proved by Fisher [1].
Theorem 1.1: Let ( X , d ) and ( Z , ) be complete metric spaces. If S is a continuous mapping of X into Z, and R is a continuous mapping of Z into X satisfying the inequalities:
IJ A
d ( x, x '), d ( x, RSx), d ( RSx, RSx ') c max d ( x ' RSx '), d ( Sx, Sx ')
( z, z '), ( z, SRz ), ( SRz , SRz ') c max ( z ' SRz '), ( Rz, Rz ')
for all x, x' in X, and z, z' in Z, where 0≤ c < 1, then RS has a unique fixed point u in X and RS has a unique fixed point w in Z. Further Su = w and Rw = u. The next theorem was proved in [2]. Theorem 1.2: Let ( X , d ) , (Y , ) and ( Z , ) be complete metric spaces and suppose T is a continuous mapping of X into Y, S is a continuous mapping of Y into Z and R is a continuous mapping of
d ( x, x '), d ( x, RSTx), d ( RSTx, RSTx ') c max d ( x ', RSTx '), (Tx, Tx '), STx, STx ' ( y, y '), ( y, TRSy ), (TRSy, TRSy ') c max ( y ', TRSy '), ( Sy, Sy '), d RSy, RSy '
( z , z '), ( z, STRz ), ( STRz , STRz ') c max ( z ', STRz '), d Rz , Rz ' , (TRz , TRz ') for all x, x
Then RST has a unique fixed point u in X, TRS has a unique fixed point v in Y and STR has a unique fixed point w in Z. Further, Tu = v, Sv = w and Rw = u. In recently Ansari , Sharma[6] generate Related Fixed Points Theorems on Three Metric Spaces. Now We obtained related fixed point theorem on three metric spaces satisfying integral type inequality. Let
(X,d ) be a complete metric space, [0,1], f : X X
Z into X satisfying the inequalities:
mapping
d ( RSTx, RSy ) c max{d ( x, RSy ), d ( x, RSTx), ( y, Tx), ( Sy, STx)} (TRSy, STz ) c max{ ( y, TRz ), ( x, TRSy), ( z, Sy), d ( Rz, RSy)} ( STRz, STx) c max{ ( z, STx), ( z, STRz ), d ( x, Rz ), (Tx, TRz )}
d ( fx , fy )
such
that
(t )dt
for
each
x
,
y X
a ,
d ( x, y )
0
(t )dt , 0
Where
; R R is a
lebesgue integrable mapping which is summable, nonnegative and for all x in X, y in Y and z in Z, where 0≤ c < 1. Then RST has a unique fixed point u in X, TRS has a unique fixed point v in Y and
such that, for each
0, (t )dt 0 .
STR has a unique fixed point w in Z. Further Tu = v, Sv = w and
ISSN: 2230-7818
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Then f has a unique
0
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Manish Kumar Mishra et al. / (IJAEST) INTERNATIONAL JOURNAL OF ADVANCED ENGINEERING SCIENCES AND TECHNOLOGIES Vol No. 2, Issue No. 1, 104 - 107
common fixed z X such that for each x X ,
lim f n x z. n
Rhoades(2003)[4], extended this result by replacing the above condition by the following
0
(t )dt
(t )dt
Let ( X , d ) be a metric space and let
f : X X , F : X CB( X ) be a single and a
multi-
valued map respectively, suppose that f and F are occasionally weakly commutative (OWC) and satisfy the inequality
J P ( Fx , Fy )
0
(t ) dt
ad ( fx , fy ) d P 1 ( fx , Fx ), ad ( fx , fy ) d P 1 ( fy , Fy ), max P 1 , Fx ) d ( fy , Fy ), ad P(fx 1 cd ( fx , Fy ) d ( fy , Fx )
n xn RST x0 , yn Txn 1 , zn Syn for n 1, 2........... .
0
d 2 ( RSTxn1 , RSTxn )
0
0
d 2 ( xn , xn1 )
0
c max{d ( xn , xn1 ) ( yn1 , yn ), d ( xn1 , xn ) d ( xn , xn1 ), d xn , xn1 zn1 , zn , d xn1 , xn d xn , xn }
t dt
0
d ( xn , xn1 )
0
t dt
c max{ ( yn1 , yn ), zn1 , zn , d xn1 , xn }
0
2 (TRSyn1 ,TRSyn )
0
2 ( yn , yn1 )
t dt
0
(t ) dt
c max{ ( yn ,TRSyn ) ( zn1 , zn ), ( yn1 ,TRSyn1 ) (TRSyn1 ,TRSyn ), d xn1 , xn ( yn ,TRSyn ), ( yn ,TRSyn1 ) ( yn1 , yn )}
0
t dt.
t dt.
t dt........ 4
2 ( yn , yn1 )
( yn , yn1 )
f and F have unique common fixed point in X .
IJ A
We now prove the following related fixed point theorem.
( X , d ) , (Y , ) and ( Z , ) be complete metric
t dt
t dt
t dt
c max{ ( yn , yn1 ) ( zn1 , zn ), ( yn1 , yn ) ( yn , yn1 ), d xn1 , xn ( yn , yn1 ), ( yn , yn ) ( yn1 , yn )}
ES
p 2 is an integer a 0 and 0 c 1 y for all x , in X ,where
MAIN RESULTS
t dt
0
0
II.
d 2 ( xn , xn1 )
c max{d ( xn , RSTxn ) ( yn1 , yn ), d ( xn1 , RSTxn1 ) d ( xn , RSTxn ), d xn , RSTxn zn1 , zn , d xn1 , xn d xn , RSTxn1 }
0 c 1 then f and F have unique common fixed point in X.
t dt
A Applying inequality (2) we have
for all x , y in X ,where p 2 is an integer a 0 and
then
,
Applying inequality (1) we have
1 max{d ( x , y ), d ( x , fx ), d ( y , fy ), [ d ( x , fy ) d ( y , fx )] 2 0
Ojha (2010) [5]
zn
T
d ( fx , fy )
xn , yn
0
0
t dt
t dt
c max ( zn1 , zn ), d xn1 , xn , ( yn1 , yn )
0
t dt...... 5
Now, applying inequality (3) we have
2 ( STRzn1 , STRzn )
0
t dt
2 ( zn , zn1 )
0
t dt
c max{ ( zn , STRzn ) d xn1 , xn , ( zn , STRzn ) ( zn1 , STRzn1 ), ( zn , STRzn ) yn1 , yn zn , STRzn1 ( zn1 , zn )}
0
t dt
c max{ ( zn , zn1 ) d xn1 , xn , ( zn , zn1 ) ( zn1 , zn ), n , zn1 ) yn1 , yn zn , zn ( zn1 , zn )}
2 ( zn , zn1 )
continuous mapping of Y into Z and R is a continuous mapping of Z
( zn , zn1 )
into X satisfying the inequalities:
follows easily by induction on using inequalities (4), (5) and (6) that
Theorem 1.4 : Let
spaces and suppose T is a continuous mapping of X into Y, S is a
d 2 ( RSTx , RSTx ')
0
0
2
(TRSy ,TRSy ')
0
c max{ ( y ,TRSy ') ( z , z '), ( y ,TRSy ) (TRSy ,TRSy '), d x , x ' ( y ',TRSy '), ( y ',TRSy ) ( y , y ')}
t dt
0
2 ( STRz , STRz ')
0
t dt
c max{d ( x , RSTx ') ( y , y '), d ( x , RSTx ) d ( RSTx , RSTx '), d x ', RSTx ' z , z ' , d x , x ' d x ', RSTx }
t dt
t dt.......... 3
RST has a unique fixed point u in X, TRS has a unique fixed point v in Y and STR has a unique fixed point w in Z. Further, Tu = v, Sv = w and Rw = u.
d ( xn , xn1 )
0
( yn , yn1 )
0
( zn , zn1 )
0
c max{d xn1 , xn , yn1 , yn ( zn1 , zn )}
t dt
t dt
t dt......... 6
cn1 max{ ( yn1 , yn ), zn1 , zn , d xn1 , xn }
0
t dt
cn1 max{ ( zn1 , zn ), d xn1 , xn , ( yn1 , yn )}
0
t dt
c n1 max{d xn1 , xn , yn1 , yn ( zn1 , zn )}
0
Since c < 1, it follows that
xn , yn
It
t dt
t dt
t dt
zn
are Cauchy
sequences with limits u, v and w in X, Y and Z respectively. Since T and S are continuous, we have
lim yn lim Txn Tu v, lim zn lim Syn Sv w. n
Proof : Let x0 be an arbitrary point in X. Define the sequence
t dt
0
t dt.......... 1
for all x, x ' in X, y, y ' in Y and z, z ' in Z where 0 c 1 . Then
ISSN: 2230-7818
0
0
t dt......... 2
c max{ ( z ', STRz ') d x , x ' , ( z ', STRz ') ( z , STRz ), ( z ', STRz ') y , y ' , z ', STRz ( z , z ')}
0
0
t dt ( z
n
n
n
Using inequality (1) again we have
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Manish Kumar Mishra et al. / (IJAEST) INTERNATIONAL JOURNAL OF ADVANCED ENGINEERING SCIENCES AND TECHNOLOGIES Vol No. 2, Issue No. 1, 104 - 107
0
t dt
d 2 ( RSTu , xn )
0
w and Rw = u.
t dt
c max{d ( xn1 , RSTxn1 ) ( yn1 , v ), d ( u , RSTu ) d ( RSTu , RSTxn1 ),
d xn1 , RSTxn1 w, zn1 , d u , xn1 d xn1 , RSTu } 0
d 2 ( RSTu , xn )
0
Proof : Let x0 be an arbitrary point in X. Define the sequence
t dt.
xn , yn
t dt
c max{d ( xn1 , xn ) ( yn1 , v ), d ( u , RSTu ) d ( RSTu , xn ), d xn1 , xn w , zn1 , d u , xn1 d xn1 , RSTu }
by
t dt.
that 0
t dt
cd 2 ( RSTu , u )
0
d ( RSTu , u )
0
d ( RSTxn1 , RSTxn ) max{ d ( xn , xn1 ), d ( xn , RSTxn1 )}
c 1, c 1 and so u is a fixed point of RST.
TRSv TRSTu Tu v
0
t dt
0
0
i.e.
d 2 ( u , u ')
0
d ( u , u ')
0
t dt
c d 2 ( u ,u ')
0
t dt
c d ( u ,u ')
0
d 2 ( xn , xn1 )
t dt
d ( xn , xn1 )
0
t dt
t dt
or
d 2 ( xn , xn1 )
d ( xn , xn1 )
0
t dt
IJ A
point of TRS
that
0
( zn , zn1 ) max{ ( zn , zn1 ), ( zn , zn )}
2 ( zn , zn1 )
prove the following another theorem.
( X , d ) , (Y , ) and ( Z , ) be
complete
0
0
is a continuous mapping of Y into Z and R is a continuous mapping
or
0
0
cd ( RSy , RSy ') max{ d ( x , RSTx ), ( y ',TRSy ')}
t dt
0
( STRz , STRz ') max{ ( z ', STRz '), ( z ', STRz ')}
t dt
t dt..... 8
c (TRz ,TRz ') max{ ( z ', STRz '), ( y ',TRSy ')}
0
for all x, x ' in X, y, y '
t dt...... 7
in Y and z, z ' in Z where 0 c 1 . Then
RST has a unique fixed point u in X, TRS has a unique fixed point v in Y and STR has a unique fixed point w in Z. Further, Tu = v, Sv =
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0
t dt
c ( yn , yn1 ) max{ ( zn , zn1 ), ( yn , yn1 )}
0
t dt
c ( yn , yn1 ) ( zn , zn1 )
t dt
c ( yn , yn1 )
0
2 ( zn , zn1 )
0
t dt
t dt
t dt
it
t dt
c 2 ( yn , yn1 )
0
t dt
follows
( zn , zn1 )
0
t dt..... 9
implies
t dt..........................(10)
0
( zn , zn1 )
c ( STx , STx ') max{ ( z ', STRz '), d ( x ', RSTx ')
case
which implies that
metric spaces and suppose T is a continuous mapping of X into Y, S of Z into X satisfying the 2inequalities: 2
t dt
b ( zn , zn1 )
Applying inequality (9) we have
u is a unique fixed point of RST, it follows that Rw = u. We now
0
t dt
0
and so either
(TRSy ,TRSy ') max{ ( y ',TRSy ), ( y ',TRSy ')}
b ( zn , zn1 )
t dt
either
d ( xn , xn1 )
Rw RSTRw RST Rwand so Rw is a fixed point of RST. Since
0
c 2 ( zn , zn1 )
0
t dt
t dt
0
t dt
and w is a unique fixed point of STR.
t dt
c ( zn , zn1 )
t dt
Thus
Similarly, it can be proved that v is a unique fixed
d ( RSTx , RSTx ') max{ d ( x , RSTx '), d ( x ', RSTx )}
and so either
where b c c 1 .
This shows that u is a unique fixed point of RST.
1.5 : Let
c ( zn , zn1 ) d ( xn , xn1 )
0
which implies that
as c 1, c 1, hence u u '.
Theorem
n 1, 2........... .
t dt
t dt
0
0
t dt
respectively
which implies that
t dt
c max{d ( u ', RSTu ') (Tu ,Tu '), d ( u , RSTu ) d ( RSTu , RSTu '), d u ', RSTu ' STu , STu ' , d u ,u ' d u ', RSTu }
0
ES
that RST has a second fixed point u', and then using 2
for
Z
t dt
c ( zn , zn1 ) max{ ( zn , zn1 ), d ( xn , xn1 )}
0
We now prove the uniqueness of the fixed point u. Suppose
d 2 ( RSTu , RSTu ')
d ( xn , xn1 ) max{ d ( xn , xn1 ), d ( xn , xn )}
Hence v and w are fixed points of TRS and STR respectively.
inequality (1), we have
c ( STxn1 , STxn ) max{ ( zn , STRzn ), d ( xn , RSTxn )
0
And so STRw STRSv Sv w
d 2 ( u , u ')
and
t dt
0
t dt
0
Now, we have
t dt
cd ( RSTu , u )
t dt
Thus RSTu = u, as
xn RST x0 , yn Txn 1 , zn Syn
0
d 2 ( RSTu , u )
Y
Applying inequality (7) we have
Since T and S are continuous, it follows on letting n tend to infinity
X,
n
0
zn
T
d 2 ( RSTu , RSTxn1 )
t dt
as b ( yn , yn1 )
0
above
t dt.........................(11)
applying inequality (8) we have
( yn , yn1 ) max{ ( yn , yn1 ), ( yn , yn )}
0
t dt
cd ( xn1 , xn ) max{d ( xn1 , xn ), ( yn , yn1 )}
0
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t dt
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that
Manish Kumar Mishra et al. / (IJAEST) INTERNATIONAL JOURNAL OF ADVANCED ENGINEERING SCIENCES AND TECHNOLOGIES Vol No. 2, Issue No. 1, 104 - 107
it
( yn , yn1 )
0
follows
t dt
bd ( xn1 , xn )
0
as
above
that
t dt.........(12)
RSTRw RST Rw and so Rw is a fixed point of RST. Since u is a uniquefixed point of RST, it follows that Rw = u.
It now follows from inequalities (10), (11) and (12) that
( xn , xn1 )
0
t dt
b ( zn , zn1 )
0
b2 ( yn , yn1 )
0
b3 n d ( x0 , x1 )
0
t dt REFERENCES
follows that
[1]
t dt
xn , yn
zn
sequences with limits u, v and w in X, Y and Z respectively. Since T and S are continuous, we have
lim yn lim Txn Tu v,..................................... 13
n
n
n
n
This completes the proof.
t dt
...............
0 b 1, it
We finally prove that we also have Rw = u. To do this note that Rw
lim zn lim Syn Sw w...................................... 14
[2]
[3]
[4]
inequality (7) again we have d ( RSTu , xn ) max{ d ( xn1 , xn ), d ( xn1 , RSTu )}
0
c ( STu , xn ) max{ ( zn1 , zn ), d ( xn 1 , xn )
0
[5]
t dt
t dt
ES
[6]
Since T and S are continuous, it follows on letting n tend to infinity that
d 2 ( RSTu ,u )
0
t dt 0
B. Fisher: Related fixed points on two metric spaces, Math.Sem.Notes,Kobe Univ., 10(1982), 17-26. N.P. Nung: A fixed point theorem in three metric spaces, Math.Sem.Notes,Kobe Univ., 11(1983), 77-79. .K. Jain, H.K. Sahu and B. Fisher: Related fixed point theorem for three metric spaces, NOVI SAD J.Math.VOL.26, No.1, (1996), 11-17. Deo Brat Ojha, Manish Kumar Mishra and Udayana Katoch,A Common Fixed Point Theorem Satisfying Integral Type for Occasionally Weakly Compatible Maps, Research Journal of Applied Sciences, Engineering and Technology 2(3): 239-244, 2010. Rhoades, B.E.,Two fixed point theorem for mapping satisfying a general contractiv condition of integral type. Int. J. Math. Sci., 3: 2003 4007-4013. K. Ansari , Manish Sharma and Arun Garg, “ Related Fixed Points Theorems on Three Metric Spaces” , Int. J. Contemp. Math. Sciences, Vol. 5, 2010, no. 42, 2059 – 2064.
T
and
Thus RSTu = u, and so u is a fixed point of RST. It now follows from equalities (13), (14)
IJ A
TRSv TRSTu T(RSTu) Tu v And so STRw STRSv S (TRSv) Sv w
Hence v and w are fixed points of TRS and STR respectively
We now prove the uniqueness of the fixed point u. Suppose that RST has a second fixed point u', and then using inequality (7), we have
d ( RSTu , RSTu ') max{d ( u ', RSTu '), d ( u ', RSTu )}
0
t dt
c ( STu , STu ') max{ ( STu ', STRu '), d ( u ', RSTu ')}
0
t dt
which implies that
d 2 ( u ,u ')
0
t dt 0
hence u u ' ' This shows that u is a unique fixed point of RST. Similarly, it can be proved that v is a unique fixed point of TRS and w is a unique fixed point of STR.it can be proved that v is a unique fixed point of TRS and w is a unique fixed point of STR.
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