Page 1

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

@ 2011 http://www.ijaest.iserp.org. All rights Reserved.

Then f has a unique

0

Page 104


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 ( RSTxn1 , RSTxn )

0

0



d 2 ( xn , xn1 )

0

c max{d ( xn , xn1 )  ( yn1 , yn ), d ( xn1 , xn ) d ( xn , xn1 ), d  xn , xn1   zn1 , zn  , d  xn1 , xn  d  xn , xn }

  t  dt  

0

d ( xn , xn1 )

0

  t  dt  

c max{  ( yn1 , yn ),   zn1 , zn , d  xn1 , xn }

0

 2 (TRSyn1 ,TRSyn )

0

 2 ( yn , yn1 )

  t  dt  

0

 (t ) dt 

c max{  ( yn ,TRSyn ) ( zn1 , zn ),  ( yn1 ,TRSyn1 )  (TRSyn1 ,TRSyn ), d  xn1 , xn   ( yn ,TRSyn ),  ( yn ,TRSyn1 )  ( yn1 , yn )}

0

  t  dt.

  t  dt.

  t  dt........  4 

 2 ( yn , yn1 )

 ( yn , yn1 )

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 , yn1 ) ( zn1 , zn ),  ( yn1 , yn )  ( yn , yn1 ), d  xn1 , xn   ( yn , yn1 ),  ( yn , yn )  ( yn1 , 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 , xn1 )

c max{d ( xn , RSTxn )  ( yn1 , yn ), d ( xn1 , RSTxn1 ) d ( xn , RSTxn ), d  xn , RSTxn   zn1 , zn , d  xn1 , xn  d  xn , RSTxn1 }

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 ( zn1 , zn ), d  xn1 , xn ,  ( yn1 , yn )

0

  t  dt......  5

Now, applying inequality (3) we have

 2 ( STRzn1 , STRzn )

0

  t  dt  

 2 ( zn , zn1 )

0



  t  dt

c max{ ( zn , STRzn ) d  xn1 , xn  , ( zn , STRzn )  ( zn1 , STRzn1 ),  ( zn , STRzn )   yn1 , yn   zn , STRzn1  ( zn1 , zn )}

0

  t  dt

c max{ ( zn , zn1 ) d  xn1 , xn , ( zn , zn1 )  ( zn1 , zn ), n , zn1 )   yn1 , yn   zn , zn  ( zn1 , zn )}

 2 ( zn , zn1 )

continuous mapping of Y into Z and R is a continuous mapping of Z

 ( zn , zn1 )

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 , xn1 )

0

 ( yn , yn1 )

0

 ( zn , zn1 )

0

c max{d  xn1 , xn ,   yn1 , yn  ( zn1 , zn )}

  t  dt  

  t  dt

  t  dt.........  6 

cn1 max{  ( yn1 , yn ),  zn1 , zn , d  xn1 , xn }

0

  t  dt  

cn1 max{ ( zn1 , zn ), d  xn1 , xn ,  ( yn1 , yn )}

0

  t  dt  

c n1 max{d  xn1 , xn ,   yn1 , yn  ( zn1 , 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

@ 2011 http://www.ijaest.iserp.org. All rights Reserved.

Page 105


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 ( xn1 , RSTxn1 )  ( yn1 , v ), d ( u , RSTu ) d ( RSTu , RSTxn1 ),

  d  xn1 , RSTxn1   w, zn1 , d u , xn1  d  xn1 , 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 ( xn1 , xn )  ( yn1 , v ), d ( u , RSTu ) d ( RSTu , xn ), d  xn1 , xn   w , zn1  , d  u , xn1  d  xn1 , RSTu }

by

  t  dt.

that 0

  t  dt  

cd 2 ( RSTu , u )

0



d ( RSTu , u )

0

d ( RSTxn1 , RSTxn ) max{ d ( xn , xn1 ), d ( xn , RSTxn1 )}

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 , xn1 )

  t  dt  

d ( xn , xn1 )

0

  t  dt  

  t  dt

or

d 2 ( xn , xn1 )

d ( xn , xn1 )

0

  t  dt

IJ A

point of TRS

that

0

 ( zn , zn1 ) max{ ( zn , zn1 ), ( zn , zn )}

 2 ( zn , zn1 )

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 =

ISSN: 2230-7818

0

  t  dt  

c  ( yn , yn1 ) max{ ( zn , zn1 ),  ( yn , yn1 )}

0

  t  dt  

c  ( yn , yn1 ) ( zn , zn1 )

  t  dt  

c  ( yn , yn1 )

0

 2 ( zn , zn1 )

0

  t  dt  

  t  dt

  t  dt

it

  t  dt

c  2 ( yn , yn1 )

0

  t  dt

follows

 ( zn , zn1 )

0

  t  dt.....  9 

implies

  t  dt..........................(10)

0

 ( zn , zn1 )

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 , zn1 )

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 , zn1 )

  t  dt

either

d ( xn , xn1 )

Rw  RSTRw  RST Rwand so Rw is a fixed point of RST. Since

0

c 2 ( zn , zn1 )

0

  t  dt

  t  dt

0

  t  dt  

and w is a unique fixed point of STR.

  t  dt  

c ( zn , zn1 )

  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 , zn1 ) d ( xn , xn1 )

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 , zn1 ) max{ ( zn , zn1 ), d ( xn , xn1 )}

0

We now prove the uniqueness of the fixed point u. Suppose

d 2 ( RSTu , RSTu ')

d ( xn , xn1 ) max{ d ( xn , xn1 ), d ( xn , xn )}



Hence v and w are fixed points of TRS and STR respectively.

inequality (1), we have

c ( STxn1 , 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 , RSTxn1 )

  t  dt  

as b ( yn , yn1 )

0

above

  t  dt.........................(11)

applying inequality (8) we have

 ( yn , yn1 ) max{  ( yn , yn1 ),  ( yn , yn )}

0



  t  dt

cd ( xn1 , xn ) max{d ( xn1 , xn ),  ( yn , yn1 )}

0

@ 2011 http://www.ijaest.iserp.org. All rights Reserved.

  t  dt

Page 106

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 , yn1 )

0

follows

  t  dt  

bd ( xn1 , xn )

0

as

above

that

  t  dt.........(12)

RSTRw  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 , xn1 )

0

  t  dt  

b ( zn , zn1 )

0



b2  ( yn , yn1 )

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 ( xn1 , xn ), d ( xn1 , RSTu )}

0



c ( STu , xn ) max{ ( zn1 , 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.

ISSN: 2230-7818

@ 2011 http://www.ijaest.iserp.org. All rights Reserved.

Page 107

15-IJAEST-Volume-No-2-Issue-No-1-Related-Fixed-Points-Theorems-on-Three-Metric-Spaces-(FPTTMS)-104-1  

)(X,d be a complete metric space,   RR  0, () 0 tdt for all x, x Then RST has a unique fixed point u in X, TRS has a unique fixed point v...

Advertisement