15-ijaest-volume-no-2-issue-no-1-related-fixed-points-theorems-on-three-metric-spaces-(fpttms)-104-1

Related Fixed Points Theorems on Three Metric Spaces ( FPTTMS)

Manish Kumar Mishra and Deo Brat Ojha

[email protected], [email protected]

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.

Mathematics Subject Classification: 54H25

Keywords- Three metric space, fixed point, integral type inequality.

.

I. INTRODUCTION 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:

( , '), ( , ),( , ') max

( ' '), ( , ')

( , '), ( , ),( , ') max

( ' '), ( , ')

d x x d x RSxd RSx RSx c

d x RSx d Sx Sx

z z z SRzSRz SRz c

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

Z into X satisfying the inequalities:

( , ) max{ ( , ), ( , ), ( , ), ( , )}( , ) max{ ( , ), ( , ), ( , ), ( , )}( , ) max{ ( , ), ( , ), ( , ), ( , )}

d RSTx RSy c d x RSy d x RSTx y Tx Sy STx

TRSy STz c y TRz x TRSy z Sy d Rz RSy

STRz STx c z STx z STRz d x Rz Tx TRz

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

STR has a unique fixed point w in Z. Further Tu = v, Sv = w and

Rw = u. The next theorem was proved in [3].

Theorem 1.3 : Let ( , )X d , ( , )Y and ( , )Z be complete metric


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

Z into X satisfying the inequalities:

( , '), ( , ),( , ') max ( ', '), ( , '),

, '

( , '), ( , ),( , ') max ( ', '), ( , '),

, '

( , '), ( , ),( , ') max ( ', '

d x x d x RSTx

d RSTx RSTx c d x RSTx Tx Tx

STx STx

y y y TRSy

TRSy TRSy c y TRSy Sy Sy

d RSy RSy

z z z STRz

STRz STRz c 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 a

mapping such that for each x , y X ,

,)()(),(

0

),(

0

dttdtt

fyfxd yxd

Where RR ; is a

lebesgue integrable mapping which is summable, nonnegative and

such that, for each 0

0, ( ) 0t dt

. Then f has a unique

Manish Kumar Mishra et al. / (IJAEST) INTERNATIONAL JOURNAL OF ADVANCED ENGINEERING SCIENCES AND TECHNOLOGIES

Vol No. 2, Issue No. 1, 104 - 107

ISSN: 2230-7818 @ 2011 http://www.ijaest.iserp.org. All rights Reserved. 04

IJAEST

common fixed z X such that for each x X , .lim zxf n

n

Rhoades(2003)[4], extended this result by replacing the above

condition by the following

( , ) 1max{ ( , ), ( , ), ( , ), [ ( , ) ( , )]

20

0

( ) ( )d fx fy

d x y d x fx d y fy d x fy d y fx

t dt t dt

Ojha (2010) [5] 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

1

1

1

1

max

0

( , ) ( , ),( , ) ( , ),( , ) ( , ),

( , ) ( , )( , )

0( ) ( )

P

P

P

PP

ad fx fy d fx Fx

ad fx fy d fy Fy

ad fx Fx d fy Fy

cd fx Fy d fy FxJ Fx Fy

t dt t dt

for all x , y in X ,where 2p is an integer 0a and

0 1c then f and F have unique common fixed point in

X .

for all x , y in X ,where 2p is an integer 0a and 0 1c

then f and F have unique common fixed point in X .

II. MAIN RESULTS

We now prove the following related fixed point theorem.

Theorem 1.4 : Let ( , )X d , ( , )Y and ( , )Z be complete metric


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

into X satisfying the inequalities:

2

2

max{ ( , ') ( , '),( , ) ( , '),

', ' , ' ,( , ') , ' ', }

0 0max{ ( , ') ( , '),( , ) ( , '),

, ' ( ', '( , ')

0 0

.......... 1

c d x RSTx y yd x RSTx d RSTx RSTxd x RSTx z z

d RSTx RSTx d x x d x RSTx

c y TRSy z zy TRSy TRSy TRSy

d x x y TRSyTRSy TRSy

t dt t dt

t dt t dt

2

),( ', ) ( , ')}

max{ ( ', ') , ' ,( ', ') ( , ),( ', ') , ' ,

( , ') ', ( , ')}

0 0

......... 2

.......... 3

y TRSy y y

c z STRz d x xz STRz z STRzz STRz y y

STRz STRz z STRz z zt dt t dt

for all , 'x x in X, , 'y y in Y and , 'z z in Z where 0 1c . 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.

Proof : Let 0x be an arbitrary point in X. Define the sequence

,n nx y nz,

0 1, ,n

n n n n nx RST x y Tx z Sy for 1,2...........n .

Applying inequality (1) we have

2 21 1

1 1 11 1 1

( , ) ( , )

0 0max{ ( , ) ( , ), ( , ) ( , ),

, , , , , }

0.

n n n n

n n n n n n n n

n n n n n n n n

d RSTx RSTx d x x

c d x RSTx y y d x RSTx d x RSTxd x RSTx z z d x x d x RSTx

t dt t dt

t dt

1 11 1

1 121 1

max{ ( , ) ( , ),( , ) ( , ),

, , ,( , ) , , }

0 0.

n n n n

n n n n

n n n nn n

n n n n

c d x x y yd x x d x xd x x z z

d x x d x x d x xt dt t dt

1

11 1

max{ ( , ),( , ) , , , }

0 0........ 4

n nn n

n n n n

c y yd x x

z z d x xt dt t dt

A Applying inequality (2) we have

2 21 1

11 1 11

1 1

( , ) ( , )

0 0max{ ( , ) ( , ),( , ) ( , ),

, ( , ),( , ) ( , )}

0

n n n n

n n n n

n n n n

n n n n

n n n n

TRSy TRSy y y

c y TRSy z zy TRSy TRSy TRSy

d x x y TRSyy TRSy y y

t dt t dt

t dt

1 11 1

2 1 11

1

max{ ( , ) ( , ),( , ) ( , ),

, ( , ),( , ) ( , ) ( , )}

0 0

n n n n

n n n n

n n n nn n

n n n n

c y y z zy y y y

d x x y yy y

y y y yt dt t dt

1 1 1 1( , ) max ( , ), , , ( , )

0 0...... 5n n n n n n n ny y c z z d x x y y

t dt t dt

Now, applying inequality (3) we have

2 21 1

1 1 1

1 1 1

( , ) ( , )

0 0

max{ ( , ) , , ( , ) ( , ),( , ) , , ( , )}

0

n n n n

n n n n n n n n

n n n n n n n n

STRz STRz z z

c z STRz d x x z STRz z STRz

z STRz y y z STRz z z

t dt t dt

t dt

2 1 1 1 1

11 1 1

max{ ( , ) , , ( , ) ( , ),( , ) ( , ) , , ( , )}0 0

n n n n n n n nn n

n n n n n n n n

c z z d x x z z z zz z

z z y y z z z zt dt t dt

1 1 1 1( , ) max{ , , , ( , )}

0 0......... 6n n n n n n n nz z c d x x y y z z

t dt t dt

It

follows easily by induction on using inequalities (4), (5) and (6) that

1

1 1 1 1( , ) max{ ( , ), , , , }

0 0

nn n n n n n n nd x x c y y z z d x x

t dt t dt

1

1 1 1 1( , ) max{ ( , ), , , ( , )}

0 0

nn n n n n n n ny y c z z d x x y y

t dt t dt

1

1 1 1 1( , ) max{ , , , ( , )}

0 0

nn n n n n n n nz z c d x x y y z z

t dt t dt

Since c < 1, it follows that ,n nx y nz are Cauchy

sequences with limits u, v and w in X, Y and Z respectively. Since T

and S are continuous, we have

Using inequality (1) again we have

lim lim , lim lim .n n n nn n n n

y Tx Tu v z Sy Sv w


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IJAEST

2

2 2

2 21

1 1 1 11 1 1 1 1

2

1 1

( , ) ( , )

0 0max{ ( , ) ( , ), ( , ) ( , ),

, , , , , }

0

( , )

0max{ ( , ) ( , ), ( , )

0

.

n n

n n n n

n n n n n

n

n n n

d RSTu RSTx d RSTu x

c d x RSTx y v d u RSTu d RSTu RSTxd x RSTx w z d u x d x RSTu

d RSTu x

c d x x y v d u RSTu d

t dt t dt

t dt

t dt

t dt

1 1 1 1

( , ),, , , , , } .

n

n n n n n

RSTu xd x x w z d u x d x RSTu

Since T and S are continuous, it follows on letting n tend to infinity

that

2 2( , ) ( , )

0 0

( , ) ( , )

0 0

d RSTu u cd RSTu u

d RSTu u cd RSTu u

t dt t dt

t dt t dt

Thus RSTu = u, as 1, 1c c and so u is a fixed point of RST.

Now, we have TRSv TRSTu Tu v

And so STRw STRSv 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 (1), we have

2 2

2 2

( , ') ( , ')

0 0max{ ( ', ') ( , '), ( , ) ( , '),

', ' , ' , , ' ', }

0

( , ') ( , ')

0 0

( , ') ( , ')

0 0

. .

1, 1,

d u u d RSTu RSTu

c d u RSTu Tu Tu d u RSTu d RSTu RSTud u RSTu STu STu d u u d u RSTu

d u u c d u u

d u u c d u u

t dt t dt

t dt

i e t dt t dt

t dt t dt

as c c henc

'.e 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.

Rw RSTRw RST Rwand so Rw is a fixed point of RST. Since

u is a unique fixed point of RST, it follows that Rw = u. We now

prove the following another theorem.

Theorem 1.5 : 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 Z into X satisfying the inequalities:

( , ') ( , ')max{ ( , '), ( ', )} max{ ( ', '), ( ', ')

0 0( , ') ( , ')

max{ ( ', ), ( ', ')} max{ ( , ), ( ', ')}

0 0

0

...... 7

..... 8

d RSTx RSTx c STx STxd x RSTx d x RSTx z STRz d x RSTx

TRSy TRSy cd RSy RSyy TRSy y TRSy d x RSTx y TRSy

t dt t dt

t dt t dt

t dt

( , ') ( , ')

max{ ( ', '), ( ', ')} max{ ( ', '), ( ', ')}

0..... 9

STRz STRz c TRz TRzz STRz z STRz z STRz y TRSy

t dt

for all , 'x x in X, , 'y y in Y and , 'z z in Z where 0 1c . 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.

Proof : Let 0x be an arbitrary point in X. Define the sequence

,n nx y nzX, Y and Z respectively

by 0 1, ,n

n n n n nx RST x y Tx z Sy for 1,2...........n .


1 1 1

1

( , ) max{ ( , ), ( , )}

0( , ) max{ ( , ), ( , )

0

n n n n n n

n n n n n n

d RSTx RSTx d x x d x RSTx

c STx STx z STRz d x RSTx

t dt

t dt

1 1

1 1 1

( , ) max{ ( , ), ( , )}

0( , ) max{ ( , ), ( , )}

0

n n n n n n

n n n n n n

d x x d x x d x x

c z z z z d x x

t dt

t dt

and so either

2

1 1 1( , ) ( , ) ( , )

0 0

n n n n n nd x x c z z d x x

t dt t dt

which implies that

1 1( , ) ( , )

0 0

n n n nd x x c z z

t dt t dt

or

which implies that

1 1( , ) ( , )

0 0

n n n nd x x b z z

t dt t dt

where 1b c c .

Thus either case implies

that

1 1( , ) ( , )

0 0..........................(10)n n n nd x x b z z

t dt t dt


1 1 1 1 1( , )max{ ( , ), ( , )} ( , )max{ ( , ), ( , )}

0 0

n n n n n n n n n n n nz z z z z z c y y z z y y

t dt t dt

and so either

2

1 1 1( , ) ( , ) ( , )

0 0

n n n n n nz z c y y z z

t dt t dt

which implies that

1 1( , ) ( , )

0 0

n n n nz z c y y

t dt t dt

or

2 2

1 1( , ) ( , )

0 0

n n n nz z c y y

t dt t dt

it follows as above that

1 1( , ) ( , )

0 0.........................(11)n n n nz z b y y

t dt t dt

applying inequality (8) we have

1 1

1 1 1

( , ) max{ ( , ), ( , )}

0( , ) max{ ( , ), ( , )}

0

n n n n n n

n n n n n n

y y y y y y

cd x x d x x y y

t dt

t dt

2 2

1 1( , ) ( , )

0 0

n n n nd x x c z z

t dt t dt


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IJAEST

and it follows as above that

1 1( , ) ( , )

0 0.........(12)n n n ny y bd x x

t dt t dt

It now follows from inequalities (10), (11) and (12) that

1 1

21

30 1

( , ) ( , )

0 0

( , )

0

( , )

0...............

n n n n

n n

n

x x b z z

b y y

b d x x

t dt t dt

t dt

t dt

0 1,b it follows that ,n nx y nz

sequences with limits u, v and w in X, Y and Z respectively. Since T

and S are continuous, we have

lim lim ,..................................... 13

lim lim ...................................... 14

n nn n

n nn n

y Tx Tu v

z Sy Sw w

inequality (7) again we have

1 1

1 1

( , ) max{ ( , ), ( , )}

0( , ) max{ ( , ), ( , )

0

n n n n

n n n n n

d RSTu x d x x d x RSTu

c STu x z z d x x

t dt

t dt

Since T and S are continuous, it follows on letting n tend to infinity

that

Thus RSTu = u, and so u is a fixed point of RST.

It now follows from equalities (13), (14)

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

Hence v and w are .

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

( , ') max{ ( ', '), ( ', )}

0( , ') max{ ( ', '), ( ', ')}

0

d RSTu RSTu d u RSTu d u RSTu

c STu STu STu STRu d u RSTu

t dt

t dt

which implies that

2 ( , ')

00

d u u

t dt

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.

We finally prove that we also have Rw = u. To do this note that Rw

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.

This completes the proof.

REFERENCES

[1] B. Fisher: Related fixed points on two metric spaces, Math.Sem.Notes,Kobe Univ., 10(1982), 17-26.

[2] N.P. Nung: A fixed point theorem in three metric spaces, Math.Sem.Notes,Kobe Univ., 11(1983), 77-79.

[3] .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.

[4] 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.

[5] 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.

[6] 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.

2 ( , )

00

d RSTu u

t dt

TRSv TRSTu T(RSTu) Tu vAnd so

( )STRw STRSv S TRSv Sv w


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