Some Common Fixed Point Theorem for Cone Metric Space

Shailesh T.Patel^{1}, Ramakant Bhardwaj^{2} ^{1}Research Scholar of Singhania University, Pacheri Bari, Jhunjhunu (RJ) INDIA ^{2}Truba Institutions of Engineering & I.T. Bhopal, (MP) INDIA ^{1}Corresponding Address: [email protected]

Research Article

Abstract: In this paper, we proof some fixed point and common fixed point theorem for Cone metric space.
Let X be a Real banach space and P a subset of X.P is called a cone if P satisfy followings conditions:

P is closed, non empty and P≠0

Ax + By Є P for all x,y Є P and non negative real numbers a,b

P ∩ (-P) = { 0 }

Given a cone P X, we define a partial ordering ≤ on X with respect to P by y-x Є P.
We shall write x << y if (y-x) Є int P, denoted by the norm on X. the cone P is called normal if there is a number k > 0 such that for all x,y Є X
0 ≤ x ≤ y ………………..(A)
The least positive number k satisfying the above condition (A) is called the normal constant of P.
The authors showed that there is no normal cones with normal constant M < 1 and for each k > 1
There are cone with normal constant M > k.
The cone P is called regular if every increasing sequence which is bounded from the above is convergent, that is if {xn}n≥1 is a sequence such that x1 ≤ x2 ≤ …..≤ y for some y Є X,
Then there is x Є X .
The cone P is regular iff every decreasing sequence which is bounded from below is convergent.

Definition:1 Let X be a nonempty set and X is a real Banach Space, d is a mapping from X into itself such that, d satisfying following conditions,
.d1 : d(x,y) ≥ 0 x,y Є X
.d2 : d(x,y) = 0 x=y
.d3 : d(x,y) = d(y,x)
.d4 : d(x,y) ≤ d(x,z) + d(z,y)
Then d is called a cone metric on X and (X,d) is called cone metric space.
Definition:2 Let A and S be two mapping of a cone metric space (X,d) then it is said to be compatible if
, whenever {xn} is a sequence in X such that and
For some t Є X.
Let A and S be two self mapping of a cone metric space (X,d) then it is said to be weakly compatible, if they commute at coincidence point,that is Ax = Sx implies that, ASx = SAx for x Є X.
It is easy to see that compatible mapping commute at there coincidence points.It is note that a compatible maps are weakly compatible but converges need not be true.

Theorem:1.1Let (X,d) be a complete cone metric space and P a normal cone with normal constant k. Suppose that the mapping T, from X into itself satisfy the condition,

…………………………………(1)
For all x,yЄX and such that < 1.Then T has unique fixed point in X. Proof: For any arbitrary X0, in X, we choose X1,X2Є X such that,
Tx0=x1 and Tx1=x2
In general we can define a sequence of elements of X such that,
X2n+1=TX2n and X2n+2=TX2n+1
Now , from (1)
d( X2n+1,X2n+2)=d(TX2n,TX2n+1)
+

+
By using triangle inequality,we get
d( X2n+1,X2n+2)d( X2n,,X2n+1)
Similarly we can show that,
d( X2n,,X2n+1)d( X2n-1,,X2n)
In general we can write,
d( X2n+1,X2n+2)d( X0,,,X1)
On taking
d( X2n+1,X2n+2)d( X0,,,X1)
For n≤m, we have
d( X2n,X2m)
d( X2n,X2m)
d( X2n,X2m)
as n→∞

In this way
as n→∞
Hence {xn} is a Cauchy sequence which converges to uЄX
Hence (X,d) is complete cone metric space.
Thus xn→u as n→∞, TX2n→u and T2n+1→u as n→∞,
u is fixed point of T in X. Uniqueness: Let us assume that , v is another fixed point of T in X different from v.then,
Tu=u andTv=v
From (1)
d(u,v)=d(Tu,Tv)

d(Tu,Tv)≤()d(u,v)
Which contradiction,
u is unique fixed point of T in X. Theorem:2 Let (X,d) be a complete cone metric space and P a normal cone with normal constant k. Suppose that S and T, the mapping from X into itself satisfies the condition,

……………………………………………………………(2)
For all x,yЄX and such that < 1.Then S and T has unique fixed point in X.Further more if, ST=TS then it have unique common fixed point in X. Proof: For any arbitrary X0, in X, we choose X1,X2Є X such that,
Sx0=x1 and Tx1=x2
In general we can define a sequence of elements of X such that,
X2n+1=SX2n and X2n+2=TX2n+1
Now , from (1)
d( X2n+1,X2n+2)=d(SX2n,TX2n+1)

+

+
By using triangle inequality,we get
d( X2n+1,X2n+2)d( X2n,,X2n+1)
Similarly we can show that,
d( X2n,,X2n+1)d( X2n-1,,X2n)
In general we can write,
d( X2n+1,X2n+2)d( X0,,,X1)
On taking
d( X2n+1,X2n+2)d( X0,,,X1)
For n≤m, we have
d( X2n,X2m)
d( X2n,X2m)
d( X2n,X2m)
as n→∞

In this way
as n→∞
Hence {xn} is a Cauchy sequence which converges to uЄX
Hence (X,d) is complete cone metric space.
Thus xn→u as n→∞, SX2n→u and T2n+1→u as n→∞,
u is fixed point of S and T in X.
Since ST=TS this give,
u=Tu=TSu=STu=Su=u
u is common fixed point of S and T. Uniqueness: Let us assume that , v is another fixed point of T in X different from v.then,
Tu=u andTv=v also Su=u and Sv=v
From (2)
d(u,v)=d(Su,Tv)

d(Tu,Tv)≤()d(u,v)
Which contradiction,
u is unique fixed point of S and T in X. Theorem:3 Let (X,d) be a complete cone metric space and P a normal cone with normal constant k. Suppose that S,R and T,be the mapping from X into itself satisfies the condition,

……………………………………………………………(3)
For all x,yЄX and such that < 1.Then S,R and T has unique fixed point in X.Further more either SR=RS or TR=RT then it have unique common fixed point in X. Proof: For any arbitrary X0, in X, we choose X1,X2Є X such that,
SRx0=x1 and TRx1=x2
In general we can define a sequence of elements of X such that,
X2n+1=SRX2n and X2n+2=TRX2n+1
Now ,
d( X2n+1,X2n+2)=d(SRX2n,TRX2n+1)

+

+
By using triangle inequality,we get
d( X2n+1,X2n+2)d( X2n,,X2n+1)
Similarly we can show that,
d( X2n,,X2n+1)d( X2n-1,,X2n)
In general we can write,
d( X2n+1,X2n+2)d( X0,,,X1)
On taking
d( X2n+1,X2n+2)d( X0,,,X1)
For n≤m, we have
d( X2n,X2m)
d( X2n,X2m)
d( X2n,X2m)
as n→∞

In this way
as n→∞
Hence {xn} is a Cauchy sequence which converges to uЄX
Hence (X,d) is complete cone metric space.
Thus xn→u as n→∞, SRX2n→u and TR2n+1→u as n→∞,
u is fixed point of S and T in X.
Since ST=TS this give,
u=Tu=TSu=STu=Su=u
u is common fixed point of S and T. Uniqueness: Let us assume that , v is another fixed point of T in X different from v.then,
Tu=u andTv=v also Su=u and Sv=v
From (3)
d(u,v)=d(Su,Tv)

d(Tu,Tv)≤()d(u,v)
Which contradiction,
u is unique fixed point of S and T in X. Theorem:4 Let (X,d) be a complete cone metric space and P a normal cone with normal constant k. Suppose that A,B S and T,be the mapping from X into itself satisfies the condition,
(1)A(X)T(X), B(X)S(X).
(2){A,S} and {B,T} are weakly compatible.
(3)S or T is continuous.
(4)
…………………………..(4)
For all x,yЄX and such that < 1.Then A,B,S,R and T have unique fixed point in X. Proof: For any arbitrary X0, in X, we define the sequence {xn} and {yn} in X, such that,
AX 2n=TX2n+1=y2nand BX2n+1=SX2n+2=y2n+1for all n=0,1,2,…….
Now ,
d( y2n,y2n+1)=d(AX2n,BX2n+1)

+

+
By using triangle inequality,we get
d( y2n,y2n+1)d( y2n-1,,y2n)
In general we can write,
d( y2n,y2n+1)d( y0,,,y1)
On taking
d( y2n,y2n+1)d( y0,,,y1)
For n≤m, we have
d( y2n,y2m)
d( y2n,y2m)
as n→∞

Hence {yn} is a Cauchy sequence which converges to uЄX,By the continuity of S and T,{xn} is also convergent sequence which converges to uЄX,
Hence (X,d) is complete cone metric space.
u is fixed point of A,B,S and T.
Since {A,S} and {B,T} are weakly compatible, implies that u is common fixed point of A,B,S and T Uniqueness: Let us assume that , v is another fixed point of A,B,S and T in X different from v.then,
Au=u and Av=v also Bu=u and Bv=v
From (4)
d(u,v)=d(Au,Bv)

d(Au,Bv)≤()d(u,v)
Which contradiction,
u is unique fixed point of A,B, S and T in X.

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