Some Common fixed point theorems in...
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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.
*