CHAPTER-4

Some Common fixed point theorems in L-Spaces

CHAPTER-4

Some Common fixed point theorems in L-Spaces

4.1 In tro d u c t io n and main resu its :

In this chapter we develop the concept of common fixed

points of commuting .compatible mapping of type(A),iteration,semi

continuity provide the framework for the main results.We further

illustrate the use of a new contractive type condition in L-space for

fixed point.

Kasahara [80 ] established the several known generalizations

of Banach contraction principle derived easily without using the

notion of metric space,in particular the axiom of triangular

inequality is not required necessarily in their proffs.He introduced

the general idea of L-space in the fixed point theory and then

Yeh[155], gives few fixed point theorems in L-space.After that

lseki[67],used the fundamental idea of Kasahara to investigate the

generalization of some known theorems in L-space.

During past few years many great mathematicians worked on

L-space.The names them are S ingh[136], Pachpatte[103],

Park[107] , Som and Mukharjee [145], Jungck [71-72], Pathak and

Dubey [104-105], Sharma and Agrawal [134] are worth

mentioning.Recall the definitions of L-space as below :

Let N be a set of all natural numbers and X be a non empty

set.A pair (X,-» )of a set of X and a subset -> of the set XN x X is

called an L-Space if

(i) If x n = x , x e x for all neN then ({xn}, neN ,X )e-).

(ii) If ( { x n},neN ,X)e_> then({xni} i^N .X )e

*for every subsequence {x ni} i e N of { x n},neN

D e f i n i t i o n a l .2 ):

L-space (X,-»)is said to be separated if each sequence in X

converges to at most one point of X .

D e f i n i t i o n a l .3 ):

A mapping T of an L-space (X,->) into an L-space (X,->) is

said to be c o n t in u o u s if xn-> x implies Txn->Tx for some

subsequence {x nj } , i e N of {x„},neN

When d be non negative extended real valued function on

XxX ,

0 < d(x,y) < oo, V x .y ^X .

D e f in i t io n s . 1.4):

An L-space (X, —>) is said to be d-com p le te if each sequence

{x n}, neN in X with J d (x m,x m+i) < °° converges to at most one0

point of X. When d be non negative extended real valued function

on XxX,

0<d(x,y)< oo V x,y eX .

D e f in i t io n ^ . 1.5):

Let A and B be a mappings from L-space (X,-> ) into itself ,

then A and B are said to be com patib le m appings if

limd(AB x n , BA x n )=0,n —►<»

by d-completeness {x n},neN is a sequence in X , such that :

limA x n = lim B x n = t , for some t in X .w — ► co n —><»

Example(4.1.1):

Let X = [0,1] A ,B :X -^X be Euclidean metric

Ax= (Vi) x 1/2 Bx= x1/2

by d-completeness {x„ },n eN is a sequence in X , such that :

lim A x n = lim B x n = t , for some t in X .rt— ► » W-feo

Then A and B are compatible mappings.

Definit»on(4.1.6):

Let A and B be a mappings from L-space (X, —> ) into itself ,

then A and B are said to be com patib le m a p p in g so f type(S ) if

limd(BAB x n , BBA x n ) = limd(ABA x n , AAB x n ) = 0,n-»cc n~*co

by d-completeness {x „ },n eN is a sequence in X , such that :

lim A x n = lim B x n = t , for some t in X .o n—

Exam ple(4.1.2):

Let X = [0,1] A . B ^ - ^ X be Euclidean metric

Ax= 1/8 x1/2 Bx= 1/2 x1/2

by d-completeness {x„},neN is a sequence in X , such that :

limA x n = lim B x n = t , for some t in X .n —>00 n-*oo

Then A and B are compatible mappings of type(S).

D e fin it io n (4 .1 .7 ) :

Let A and B be a mappings from L-space (X, —> ) into itself ,

then A and B are said to be com patib le m a p p in g so f type(A ) iff

limd(AB x n , BB x n ) = limd(BA x n , AA x n ) = 0,n - > 00 a —>ao

by d-completeness {x n},neN is a sequence in X , such that :

lim A x n = lim B x n = t , for some t in X .n—>oo oo

E xam p le (4 .1 .3):

Let X = [0,1] A .B iX -^X be Euclidean metric

Ax= (1/4) x1/2 Bx= x 1/2

by d-completeness {x n},neN is a sequence in X , such that :

lim A x n = lim B x n = t , for some t in X .n—>00 a —>oo

Then A and B are compatible mappings of type(A).

D e f i n i t i o n a l .8):

Let A and B be a mappings from L-space (X,-> ) into itself ,

then A and B are said to be com patib le m app ings o f type(P ) if

limd(BBx n , AAx n ) =0.n —► oo

and

limd(AAx n , BB x n )=0n —>oo

by d-completeness {x n} , n GN is a sequence in X , such that :

limA x n - lim B x n = t , for some t in Xn—>00 n-* oo

E xam p le (4 .1.4):

Let X = [0,1] A .B iX ^ X be Euclidean metric

Ax= % x 1/2 Bx= x1/2

by d-completeness {x n},nGN is a sequence in X , such that :

lim A x n = lim B x n = t , for some t in X .n—><» w—>oo*

Then A and B are compatible mappings of type(P)

D efin it ion (4 .1 .9 ):

Let A and B be a mappings from L-space (X,-> ) into itself ,

then A and B are said to be weakly com m utings if

d(ABx , BAx) £ d(Ax , Bx ) for all x in X .

by d-completeness {x „ },n eN is a sequence in X , such that :

lim A x n = lim B x n = t , for some t in X .rt—>oo n—>oo

Exam ple(4.1.5):

Let X=[0,1] A ,B :X->X be Euclidean metric space .such that

Ax=x/10 Bx=[0,x/x+10]

by d-completeness { x n},neN is a sequence in X , such that :

lim A x n = lim B x n = t , for some t in X .>oo n -x o

Then A and B are weakly commutings.

Let A and B be a mappings from L-space (X, ->) into itself.

The mappings A and B are said to be weakly com m utings pa ir o f

m app ings w ith respec t to T if

(i) d(TA x,B x )£ d(AT x,B x)

(ii) d(A x.TB x)s d(A x,BT x) for all x in X.♦

by d-completeness {x n},neN is a sequence in X , such that :

lim A x n = lim B x n = lim T x n =t , for some t in X .n—>°o n-¥ oo n~>»

Exam ple(4 .1 .6):

Let Ax = 0 if x£ 0 Bx =0 if x£ 0 Tx =0 if xs 0

= x/(1+2x) if 0<xs 1 =x if 0 < x s 1 =x if x>0

= 1/3 if x>1 =1 if x>1

by d-completeness { x n},neN is a sequence in X , such that :

lim A x n = lim B x n = limTxn = t , for some t in X .w -> oo >co n-*<o

Then A and B are weakly commutings pair of mappings with

respect to T

P ro p o s it io n in L -sp a ce :

The following propositions show that definition of compatible

mappings and compatible mappings of type(A) are equivalent

under some conditions.

P ro p o s it io n (4 .1 .1 ) :

Let A and B be a sequentially continous mappings of L-space

(X,-> ) on X into i t s e l f ,if A and B are compatible mappings .then

they are compatible mappings of type(A)

Proof: Suppose that A and B are compatible mappings by d-

completeness {x n},neN be a sequence in X.such that :♦

A xn ,B x „ -» t for some t in X.

By compability of A and B we have

limd(ABx n , BAx n ) =0n-> oo

then we have

ABx n = BAx n

=> At = Bt as « - > G0

=> d(At , Bt ) = 0

^ limd(ABx n , BBx n ) =0n-*oo

Therefore A and B are compatible mappings of type(A).

E xam p le (4 .1 .7):

Let X = [0,1] A ,B :X->X be Euclidean metric

Ax= (%) x 1/2 Bx= x1'2

by d-completeness {x n} ,n GN is a sequence in X , such that :

lim A x n = lim B x n =t , for some t in X .n - t oo n~¥ oo

Then A,B are compatible mappings of type(A).

Let A and B be a compatible mappings of type(A) from

L-space (X ,—» ) into itself . If one of A or B is sequentially

continous , then A and B are compatible mappings .

Proof: Let A is continuous .To show that A and B are

compatible mappings , by d-completeness {x n},neN is a sequence9

in X.such that :

A xn ,B xn -» t for some t in X.

Then BA xn -> Bt since B is continuous ,by compatible

mappings of type(A),we have

limd(ABx n , BBx n ) =0n-> co

by d-completeness we have

ABx n = BBx n

=> At = Bt

=> d(At.Bt) =0

^ lim d(ABx n , BAx n ) =0

so A and B are compatible mappings.

Similar arguments for compatible mappings of type(P).

As a direct consequence of proposition of (4.1.1) and (4.1.2),we

have

Let A and B be a sequentially continous mappings of L-space

(X, —> ) into itself ,if A and B are compatible if and only if they are

compatible of type(A).

Similar arguments for compatible mappings of type(P).

The following examples show that proposition(4.1.3) is not true if A

and B are not sequentially continuous

E xam p le (4 .1 .8):

Let X=R , the set of real numbers with the usual metric

d(x,y)= \x -y \

Define A,B : (X,->) -> (X,->) as follows :x2

Ax = - if x / 0 ; Ax = if x * 0 ;

Then A and T are not continuous at x=0 .

by d-completeness {x n} ,n GN is a sequence in X .Consider a

X x2

= 1 if x = 0 . = 2 if x = 0 .

sequence {xn } in X defined by x„ = n2

for n = 1,2,3 ..........

Then we have as « -* ° °

A x n = 0 B X n = 0n2 n4

by d-completeness we have

limd(ABx n , BAx n ) = 0

limd(ABx n , BBx n ) = 00

limd(BAx n , AAx n ) = °°

limd(BBx n , AAx n ) =°°n->oo

Therefore A and B are compatible mappings but not compatible

mappings of type(A) and not compatible mappings of type(P).

E xam ple(4 .1 .9):

Let X =[0,1] with the usual metric

d(x,y)= \x -y \ .Define A,B:[0,1] ->[0,1] by

Ax = x if x e [o I ) Bx =1-x if x e [0, - )

= 1 if X € (1 ,1 ] =1 if X 6 ( 1 , 1 ]

by d-completeness {x n},neN is a sequence in X.

Then A and B are not continuous a t x = - .2

Now , we assert that A and B are not compatible mappings

but are compatible mappings of type(A) as well as not compatible

mappings of type(P).To see this , suppose that {x n } — [0,1] and

that A xn ,B xn -> t .

By definition of A and B , t e {1 ,1 } since A and B agree on

[^ ,1 ] we need only consider t ,so we can suppose that

x n -> 1/ 2 and that x n < 1A for all n .Then by d-completeness

we have

limd(ABx n , BAx n ) * 0n-*°o

limd(ABx n , BBx n ) = 0

lim d( BAx n , AAx n ) = 0o

limd(BBx n , AAx n ) = 0n->oo

Therefore A and B are compatible mappings of type(A) as well as

compatible mappings of type(P) ,but not compatible mappings.

P ro p o s it io n (4 .1 .4):

Let A and B be compatible mappings of type(A) from

L-space (X,-» ) into itself and A(t)=B(t) for some t in X then

AB(t) = BB(t) = BA(t)=AA(t).

Proof: Suppose that {x n} ,n GN be a sequence in X defined by

xn = t , n = 1 ,2 ,.......... and At=Bt .

Then we have A xn ,B xn ->At as «->°°

Since A and B are compatible mappings of type(A),we have

d(ABt , BBt) = limd(ABx n , BBx n ) =0n-¥ eo

by d-completeness ,we have

ABt = BBt .

Similarly ,we have

BAt = AAt ;

But Bt = At

=> BBt = BAt .

Therefore ABt =BBt = BAt =AAt .

Let A and B be a compatible mappings of type(A) from

L-space (X,-> ) into itself and by d-completeness {xn},neN is a

sequence in X .

A(xn),B (xn) -> t for some t in X .

Then we have f

(i) limBA(Xn) = A(t), if A is continuous./j—>00

(ii) AB(t) =BA(t)

and

A(t)=B(t) , if A and B are continuous at t.

Proof: Using the argunts of proposition^. 1.4).

P ro p o s it io n (4 .1 .6 ):

Let A and B be a sequentially continous mappings L-space

(X,-> ) into itse lf .if A and B are compatible mappings if and only if

they are compatible mappings of type(P).

Proof: Let {x n } is a sequence in X . Suppose that the

mappings A and B are compatible mappings then by the triangle

inequality.we have

d(AAx n ,BBx n ) * d(AAx „ .AB x n )+d(ABx „ ,BB x n )

5 d(AAx n ,ABx n )+d(ABx n , B A x „ )

+d(BAx n .BBx n )

by d-completeness {x n} ,n GN is a sequence in X , such that :

lim A x n - lim B x n = t , for some t in X .Since A and B aren—>oo n—>co

compatible mappings and sequentially continuous.then we have

Hence limd(AA x n ,BB x „ )=0 .

Conversely suppose that A and B are compatible mappings of

type(P),then by triangle inequality,we have

d(ABx n'.BAx n ) £ d(ABx n ,AA x n )+d(AA x n ,BA x „ )

£ d(ABx n ,AA x n )+d(AAx „ ,BB x n )

+d(BBx n ,BA x n )

by d-completeness {xn},neN is a sequence in X , such that :

lim A x n = lim B x n = t , for some t in X. Since A and B are compatiblen-> oo n-+ oo

mappings of type(P) and sequentially continuous.then we have

limd(AB x n ,BA x n )=0 .n—>00

P ro p o s it io n M .1 .7 ) :

Let A and B be a compatible mappings of type(A) from

L-space (X,-> ) into itself.If one of A or B is sequentially

continuous then A and B are compatible mappings of type(P).

Proof: If A is sequentially continuous then by triangle inequality,

we have

d(AAx n ,BBx n ) ^ d(AAx n ,AB x „ )+d(ABx n ,BB x n )

£ d(AAx n ,AB x n )+d(ABx n ,BA x n )

+d(BA x n ,BB x n )

by d-completeness {x n},neN is a sequence in X , such that

lim A x n = lim B x n = t , for some t in X .Then we haven—* co n—>co

limd(AA x n ,BB x n )=0 .rt-+CO

Similarly limd(BB x n ,AA x n )=0 ,by considering B is sequentially

continuous instead of A . Therefore A and B are compatible*

mappings of type(P).

E xam p le (4 .1 .10):

Let X = [0,1] A ,B :X-»X be Euclidean metric

Ax= (V*) x1/2 Bx= x1/2

by d-completeness {x n},neN is a sequence in X , such that :

lim A x n = lim B x n = t , for some t in X.Take a sequence { x n } in X.n -* co « —>oo

Then A and B are compatible mappings of type(A) and type(P).

Using the propositions(4.1.3),(4.1.6) and (4.1.7),we have

P ro p o s i t io n s . 1.8)

Let A and B are sequentially continous mappings from

L-space (X,-» ) into itself,then

(i) A and B are compatible mappings of type(A) if and only if A

and B are compatible mappings of type(P).

(ii) A and B are compatible mappings if and only if A and B are

compatible mappings of type(P)

Let A and B be a compatible mappings of type(P) from

L-space (X,-> ) into itself .If At = Bt for some t in X .

Then ABt = AAt = BBt = BAt .

Proof: Let {x n } is a sequence in X , defined by

x n = t for n=1,2...............

and At=Bt .

by d-completeness (x n},neN is a sequence in X , such that :

lim A x n = lim B x n = At .>00 n-> oo

Since A and B are compatible mappings of type(P),we have

d(AAt , BBt ) = lim d(AA x „ , BB x „ ) =0n -¥ oo

by d-completeness , we have

AAt = BBt ;

Therefore ABt = AAt = BBt = BAt .

P r o p o s i t io n ^ . 1.10):

Let A and B be a compatible mappings of type(P) from

L-space (X,-> ) into itself. Suppose {x n } is a sequence in X , such

that :

lim Axn = lim Bxn = t for some t in Xn—>oo n -* oo

Then we have the following:

(i) lim BBxn = At if A is sequentially continuous at t ;>00

(ii) lim A Axn = Bt if B is sequentially continuous at t ;n —>oo

(iii) ABt BAt and At = Bt if A and B are sequentially continuous

at t ;

Proof, (i) Let limAxn = lim Bxn = t for some t in X .Since A isn-*ao

sequentially continuous we have

limABXn = At ;17—»«>

Also byf triangle inequality

d(BBx n ,At ) £ d(BBx n ,AA x n )+d(AAx n ,At)

by d-completeness {x n},neN is a sequence in X .Since A and B

are compatible mappings of type(P) then

lim BBxn = Atn—>oo

(ii) Using the similar argument(i) we have

limAAXn = Btn—► oo

(iii) As B is sequentially continuous at t ,we have

BBxn = Bt by (i)

If A is sequentially continuous at t ,we have

BBxn = At .

Hence by the uniqueness of the limit.we have

At=Bt

And so by proposition (4.1.9) we have

ABt = BAt .

We showing that a similar investigation and extension of

Naimpally and Singh[100], and others on a common fixed point

theorem for three and four mappings in L-spaces gives

considerable information about the special space may have.

Theorem[128]Sharma and Agrawal proved the following

theorem:

Let (X, -»)be a separated L-space which is d-complete for a

non negative real valued function d on XxX with d(x,x)= 0 for each

x in X Let E,F and T be threecontinuous self mappings of X

satisfying the following conditions:

(i)ET=TE, FT=TF,

(ii)E(X) c T(X), F (X )c T (X ) ,

(iii)d(Ex.Fy)) Sa + p [d(Tx,Ty)l + d(Tx,Ty)

For all x .yeX with T x * T y , and a, ft ^0, a + /? <1.

Then E,F and T has a unique common fixed point.

We want to prove the following:

Theorem 4.2:Let (X, —>) be a separated L-space which is d-

complete for non negative real valued function d on XxX with

d(x,x)= 0 for each x in X .

Suppose that A,B, and T are self mapping on X and

continuous.Also (T,A) , (T,B) are commuting pair of mapping for

a i+2a2+2a3<1 ai , a2, a3>0and satisfying the conditions.

(i) A a(X) <= T(X) , Bb(X) c T(X) (4.2.1)

\ + d(TxJy)

+ a3[d(Tx,Ty) + d(Aay,Bbx)]

+a2[d(Tx,Aax)+d(Ty,Bby)]

(4.2.2)

Then A,B and T have a unique common fixed point.

Proof : Let x0 be an arbitrary point in X. We construct a

sequence {xn} of points X such that

Consider Qm = d(Txm , Txm+i ) for m=0,1,2, ................

Now using this fact, we see that

d(Txn+i,Txn+2) [1+ d(Txn ,Txn+i)]d(Tx n+1 >Tx n + 2 )) — a i-------------------------------------------------

1+ d(Tx n ,T X n+1 )

+ a2 [d(Tx n ,Tx n +1 )+ d(Tx n +i ,TX n + 2 ) ]

+ a3 [d(Tx n ,Txn+i ) + d (T x n+2 ,Txn+i )

On simplification and we write

Q n + 1 < , ..-~2+^ ----- Q n (4-2.4)1 — ax - a2 - a3

< k Qn

Continuing this process we get

Q n+1 < k n Q 0 as n —>00

If Q n+1 = 0 then by d-compleeness of X, the sequence {Tnx0},

ue N converges to u, so {Anx0} ,ne N, {Bnx0},ne N .Also converges

to the point u.

A x n - Tx n+1 1 B x n+1 = T x n+2 for n — 0,1,2,3.......(4.2.3)

=> Q n + 1 = o (4.2.5)

Since A, B and T are continuous so Aa,Bb and T are also

continous , therefore the subsequences of {Tnx0}, n^N is

converges to the same point u.Now we write :

A axn = TAaxn = Tu , since A commutes with T

Bbxn+i = TBbxn+i = Tu , since B commutes with T

TTxn = Tu , since T is continuous

Suppose Tu= Bbu , and if Tu * Bbu ,

Then we observe that

d(AaTx n , Bbu )< [a i[d(Tu ,Bbu)+a2 [d(TTx n . AaTx n)+d(Tu,Bbu)]

+ a3[d(TTx n , Tu)+d(Aau,BbTx n )]

=> d(Tu,Bbu)<a1d(TuIBbu)+a2[d(Tu,Tu)+d(TuITu)]

+ a3 [d(Tu,Tu)+d(Tu,Bbu)]

=> d(Tu,Bbu) < ad(Tu,Bbu) wchich contradict our hypothesis

Thus Tu= Bbu

Similarly Tu = Au

Therefore Tu = Au = Bbu

So that T nx0 = u

Hence T( limT nXo ) = Tun-*co

This shows that by d- completeness

Tu=u so u is also common fixed point of A8,Bb and T .hence for

also A,B and T.

Theorem 4.3 : Let (X -> ) be a separated L-Space which is

d-complete for a non negative real valued fuction d on XxX with

d(x,x) =0 for each x in X . A,B,S and T be four self mapping of X

and continuous for a+2b+2c<1 satisfying the following conditions:

(i) A(X) c T(X) , B(X) c S(X) (4.3.1)

(ii) d(Ax,By)) < a d(Sx,Ty)+ b[d(Sx,Ax)+d(Ty,By) ]

+c [d(Sx,By)+ d(Ty.Ax) ]

where (T,A),(S,B) are commuting pairs of mapping. (4.3.2)

Then A,B,S and T have a unique common fixed point.

Proof: Let x0 be an arbitrary point in X . We construct a

sequence

{x n} of points X such that

Ax n =TX n+1 i

Bx n+i = S x n+2 for n = 0 ,1 ,2 ,3 ..........................

Consider Qm = d(T x m , Sx m+i ) fo rm = 0 ,1 ,2 ..................

and using this fact of theorem (4.3) we see that :

d(Ax n+2 iBx n + i))<[{ad(Sxn+2,Txn+i)+b[d(Sx n+2 ,Ax n+2)

+ d(TX n+i, BX n+l)]+C [d(SX n+2 ,Bxn+l)

+ d(TX n+1 iAXn+2 )]

On simplification,we write

Q n+1 < k Q n , Where a+2b+2c < 1

Continuing this process ,we get

Q n+i < k n Q o , as n

Q n+1 = 0 ,

by d-completeness of X ,it implies that the sequence {Tnxo},u^N

converges to some uex.so {Anx0} ,n^N ,{Bnx0} ,n e N also

converges to the same point u. Since A,B,S and T are

continuous,therefore the subsequences of {Tnx0}{Snx0 } ne N are

also converges to the point u e x .Also we have

Axn = TAxn = Tu , since A commutes with T

Bkn+i = SBXn+i = Su , since B commutes with S

and TTxn = Tu , since T is continuous

Suppose Tu= Bu ,and if Tu * Bu then we observe that

d(ATx n , Bu )< [a [d(Sxn , Tu) +b [d(S x n , AT x n)

+ d(Tu,Bu)]+c[d(Sx „ , Bu)+d( Tu.ATx n )]

=> d(Tu,Bu) < ad(Tu,Tu)+b[d(Tu,Tu)+d(Tu,Tu)]

+ c[d(Tu,Bu)+d(Tu,Tu)]

==> d(Tu,Bu) < a.O+b.O+c.O

which contradicts our hypothesis . Therefore

Tu= Bu

Similarly Tu = Au

And Tu = Au = Bu = Su

Hence Tnx0 = u

=> T(Iim TnXo ) = Tu

=> S (lim S n x0 ) = Sun -* oo

This implies by d- completeness Tu=Su=Au=Bu=u

i.e u is common fixed point of A,B ,S and T .

The unicity of common fixed point can be easily seen by

using the fact of the equation (4.3.2 ) .

Thus the proof of the theorem is complete .

Theorem 4.4: Let (X -> ) be a separated L-Space which is

d-complete for a non negative real valued fuction d on XxX with

d(x,x) =0 for each x in X . A,B,S and T be four self mappings of X

satisfying the following conditions :

(i) Aa(X) c T(X) ,Bb(X) c S(X) (4.4.1)

where A,B,S and T are continuous and (A,S) and (B,T) are two

commuting mapping and a,b>0 such that :

(ii) d(Aax,Bby) < k max [c d(Sx,Ty),d(Sx,Aax)+d(Ty,Bby),

d(Sx,Bby)+d(Ty,Aax)] ..(4.4.2)

for all x,y in X , 0<k <1 ,c > 0.

And

S(X)3 (1 -a i)S (X )+a1Aa(X) ,T(X) 3 (1-a2)T(X)+a2 Bb(X) ..... (4.4.3)

where 0<ai , a2 < 1 . If Xo in X the sequence {xn} defined in

accordance with Mann iteration associated as given below

Sx2n+i= (1 -a i)S x 2n+aiAax2n

Tx2n+2= (1 -a 2)Tx2n+a2Bbx2n+i ..........(4.4 .4)

converge to a point u in X .If S,T are continuous at u and

max {kc,2k} < 1. Then A.B.S and T have a unique common fixed

point

Proof : We want to prove that

A au = Bbu=Su=Tu .

For this let u ^ x and lim x „ = un~+<x>

Consider d(Sx,Bby)<d(Su,Sx 2n +1 )+d (Sx 2n+i ,Bbu ) ..(4.4.5)

Now using the equation (4.4.4) and (4.4.2) respectively.we see

that :♦

d(Sx2n+i ,Bbu ) =d((1-ai) Sx2n + Aa x 2n ,Bbu)

<(1-a!)d( Sx2n ,Bbu )+ai d( A x 2n ,Bbu ) ....(4.4.6)

d(Ax2n ,Bbu ) < k max{ c d( Sx2n ,Tu) +d(S x 2n, Aa x 2n )

+d(Tu,Bbu) ,d(S x 2n ,Bbu ) + d(Tu, Aa x 2n)

Since S is continuous at u then we have

Sx 2n -» Su as n -> oo .

And applying equation (4.4.3),we observe that

d(Aa x 2n ,S x n ) -> 0 as n -> oo .

Therefore

d(Aax2n,Bbu)<kmax{cd(Su,Tu),d(Tu1Bbu)1d(Su,Bbu)+d(Tu,Su)},

Also with the help of (4.4.5) and (4.4.6), we have

d( A ax 2n,Sxn )< (1-ai) d(Su,Tu)+a1k {max {c d(Su,Tu) ,

d(Tu,Bbu),d(Su,Bbu)+d(Tu,Su)} ..... (4.4.7)

Similarly

d(Su,Bbu)<k{max{cd(Su,Tu),d(Tu,Bbu),d(Su, Bbu)+d(Tu,Su) (4.4.8)

And

d(Aau,Tu)<kmax{cd(Su,Tu),d(Tu,Aau),d(Su,Aau)+d(Tu,Aau) (4.4.9)

Since S and T are continuous so by virtue of the equations (4 4 2)

and (4 .4.4),we observe that

d(Aax2n ,Bbx2n+i )< k max{c d(Sx2n ,Tx2n+i) ,d(Sx 2n ,Aax 2n)

+d(Tx2n+1,Bbx2n+i),d(Sx 2n ,Bbx2n+i)

+ d(Tx2n+1, Aa x 2n)

And

d(Su,Tu) < k max {c d(Su,Tu) ,0 , d(Su,Tu) +d(Tu,Su) }

< k max {c d(Su,Tu) ,0 , 2d(Su,Tu) }

< d(Su,Tu)

contradicts the hypothesis as 0<k < 1, max{kc,2k} < 1

Thus Su = Tu .

Hence from (4.4.8) and (4.4.9) we conclude that

Aau=Su=Tu=Bbu. ..... (4.4.10)

Again Suppose the pair (Aa,S) and (Bb,T) are commuting , so using

the equations (4.4.2) and (4.4.10) we get

d(AaSu,Su) = d(AaSu ,Bbu)

<kmax{cd(SSu,Tu),d(SSu,AaSu)+d(Tu,Bbu)1

d(SSu,Bbu)+d(Tu,AaSu) }

<kmax{cd(SSu,AaSu)+d(AaSu1Tu),d(SSu,AaSu)

+d(Tu,Bbu),d(SSu,AaSu)+d(AaSu,Bbu)

+d(Tu,AaSu)}

< k max {c d(AaSu,Su) ,0 ,2d(AaSu,Su)}

=> d(AaSu,Su) < d(AaSu,Su) is a contradiction

Therefore A aSu = Su

Similarly AaSu = SSu = Su

And BbSu = Su = TSu

Now it is obvious to show that :

d(Su,u) < d(Su,u), which is again contradiction.and

therefore

Su = u .

This implies by d-completeness that

Tu=u=Su=Aau=Bbu

So . u is also a common fixed point of Aa,Bb,S and T,so also of

A,B,S and T..

The unicity of common fixed point can be easily shown will the

help of the equation (4.4.2) as

d(Su ! ,Su) = d(AaSu,BbSu)

On simplification we have

d(Su i.Su) < d(Su i,Su) which is again contradiction .

Hence Sui = Su .

Thus the proof of the theorem is complete

Theorem 4.5[*]:Let(X,-»)be a separated L-space which is d-

complete for a non negative real valued function d on XxX with

d(x,x)= 0 for each x in X.

Let A,B,S and T are self mapping X,t>0,b s i when as2 satisfying

the following conditions:

(i) A(X) c T(X), B(X) c S(X) (4.5.1)

(ii) d2(ATx,BSy)<<|>{d2(Sx,Ty),d(Sx,ATx).d (Ty.BSy),

d(Sx,Ty).dSx,ATx),(Sx1Ty).d(Ty,BSy),

d(Sx,Ty).d(Sx,BSy),d(Sx,Ty).d(Ty,ATx),

d(Sx,BSy).d(Ty,ATx),d(Sx,ATx).dTy,ATx),

d(Sx,BSy).d(Ty,BSy)} (4.5.2)

(iii) <t>(t,t,t,t,at,0 ,0 ,0,at) < bt ,

and <|>(t,t,t,t,0,at,0,at,0)< bt , (4.5.3)

(iv) <(> ( t,0, 0,0, t , t , t , 0 , 0 ) < t for some t (4.5.4)

where <|> s F (The family of all functions from : R+9 -> R+ which

are upper semi continuous and non decresing in each variable

coordinate and R+ be the set of non negative real numbers )

Then A,B,S and T have a unique common fixed point

[*][Sao,G.S.:Com m on fixed point theorem for nonlinear contraction

mapping on L-space,Acta Ciencia Indica.vol.XXXI, M(2005) No.4,

1165-1167]

Proof. Suppose a point Xo in X ,we construct a sequence

{xn} of points of X such that :

ATX n =SX n+1 ,

BSx n+i = T x n+2 , n = 0,1,2,3, .......................

Consider Qm= d(Sxm , Txm+1 a ) form=0,1,2, .. (4.5.5)

Then we have

Qn ^ <j) {Q n -1 > Qn-1. Qn ,Q n-1-Q n-1. Qn-1-Qn .

Q n - 1 - ( Q n - 1 + Q n ) , 0 , 0 , 0 , ( Q n - i + Q n ) - Q n }

Suppose Q n > Q n -1

then

Q n -i + Q n = a Q n for a < 2 , Since <|> is non decreasing in each

variable and b< 1 for a < 2 ,we have

Qn2 < (t> { Qn2 , Qn2 , Qn" , Qn" ,3 Qn2 ,0,0,0,3 Qn2 }

=> Qn2 < b Qn2 for b < 1 which is a contradiction

Therefore Q n ^ Q n-i for n = 1,2 .............

Continuing this process ,we have

d( Sx n+i , Tx n+2 ) = 0 as n ->«>

by d-completeness of X,This shows that {Txn} is Cauchy

sequence with respect to x^X, that the sequence {Txn} converges

to z in T(x).S ince {ATxn} and { BSxn+i} are sub- sequences of

{ Sxn}, {Txn+1} so these sub - sequences are also converges to a

point z .So that we write :

Tz = z .

Now putting x = x n and y = z in equation (4.5.2) when n -x»,

d2(ATxn ,BSz) < <j) {d2(Sxn Tz),d(Sx n ,ATx n ) d (Tz, BSz),

d(Sxn ,Tz).d(Sxn . ATXn), d(Sxn ,Tz).d(Tz ,BSz) ,

d(Sxn ,Tz).d(SXn,BSz),d(Sxn,Tz).d(Tz ,ATxn).

d(Sxn ,BSz).d(Tz,ATXn),d(Sxn,ATXn).d(Tz,ATXn),

d(Sxn ,BSz).d(Tz,BSz)}

Taking n -»a>, then by d-completeness we have

d2(Sz ,BSz )<<t>{d2(Sz,BSz),0,0,0,d2(Sz,BSz),d2(Sz,BSz),

d2(Sz ,Bz),0,0}

On simplification , we observe that

d (Sz ,BSz) < <(> {d (Sz ,BSz )

Therefore Sz = BSz

Similarly Tz = Az

And Tz = Az = Bz

Thus by d-completeness

d (ASz - SSz ) = 0

=> ASz - SSz

And d (SAz - AAz ) = 0

=> SAz = AAz

Therefore SSZ= ASz = AAz = SAz

Similarly TTZ = BTz = BBz = TBz by d-completeness

Thus BBz = Bz and SSz = Sz .,

This implies by d- completeness we write

Sz = Az = Bz as a common fixed point of A.B.S and T.

Hence z is a unique common fixed point of A,B , S and T This

c o m p l e t e s the proof of the theorem.

*