 Open Access
 Total Downloads : 290
 Authors : B. Vijayabasker Reddy, V. Srinivas
 Paper ID : IJERTV3IS090372
 Volume & Issue : Volume 03, Issue 09 (September 2014)
 Published (First Online): 17092014
 ISSN (Online) : 22780181
 Publisher Name : IJERT
 License: This work is licensed under a Creative Commons Attribution 4.0 International License
A Fixed Point Theorem on SemiMetric Space using Occasionally Weakly Compatible Mappings
B. Vijayabasker Reddy
Department of Mathematics, Sreenidhi Institute of Science and Technology,Ghatkesar, Hyderabad, India501 301
V. Srinivas
Department of Mathematics,University college of Science,Saifabad,Osmania UniversityHyderabad, India.
Abstract The aim of this paper is to prove common fixed point theorem for four self mappings in semi metric space using the concept of occasionally weakly compatible. This theorem generalizes the result of Bijendra Singh and M.S Chauhan[1].
Keywords Semi metric space ,coincidence point, weakly compatible, occasionally weakly compatible, Fixedpoint.
Remark 1: Weakly compatible mappings are occasionally compatible mappings but converse is not true.
Example 1: Let (X, d) be semimetric space with X=[1/2,5] and d(x ,y)=(xy)2. Define two self mappings A and B as fololws

INTRODUCTION
The concept of semimetric space is introduced by
x2
A x
if 1 x 1
2
and
Menger, which is a generalization of metric space.Cicchese introduced the notion of a contractive mappings in semi
2x 1 if x 1
metric space and proved the fixed point theorem.In 2006 Jungck and Rhoades introduced the concept of
2x if
B x
1 x 1
2
Occasionally weakly compatible mappings which generalizes weakly compatible mappings.
x2
if x 1
II PRELIMINARIES
Definition 1 : (X ,d) is said to be Semi metric space if and only if it satisfies the following conditions:
M1: d(x ,y)=0 if and only of x=y.
M2:d(x ,y)=d(y ,x) if and only if x=y for any x,yX.
Definition 2 : Let A and B be two self mappings of a semi metric space (X, d) then A and B are said to be weakly compatible mappings if they commute at their coincidence points.
Definition 3 : Let A and B be two self maps of a semi metric space (X ,d) then A and B are said to be occasionally weakly compatible mappings if there is a coincidence point xX of A and B at which A and B are commute.
Clearly, X=1/2 and x=1 are two coincidence points. If x=1 then A(1)=1=B(1) which gives AB(1)=1=BA(1).If x=1/2 then A(1/2)=B(1/2)=1/4 but AB(1/2)BA(1/2).Therefore A and B are occasionally weakly compatible but not weakly compatible
Lemma 1: Let (X ,d) be a semimetric space, A,B are occasionally weakly compatible mappings of X. If the self mappings A and B on X have a unique point of coincidence w=Ax=Bx. Then w is unique common fixed point of A and B.
Proof: Since A and B are occasionally weakly compatible mappings, there exists a point xX such that Ax=Bx=w and ABx=BAx.Thus AAx=ABx=BAx Which gives Ax is also point of coincidence of A and B. since the point of coincidence w=Ax is unique then, BAx=AAx=Ax ,and w=Ax is a common fixed point of A and B.If z is any common fixed point of and A and B then z=Az=Bz=w by the uniqueness of the point of coincidence.
III MAIN RESULT
Theorem 1: Let A,B,S,T,P and Q be self maps on a semi metric space (X ,d) If
(i) (AP,S) and (BQ,T) are occasionally weakly compatible mappings.
d (w, u)2 d ( APw, BQu)2
1 d (BQu, Sw)d ( APw,Tu)
k d ( APw, Sw) d (BQu,Tu)
k
d ( APw, Sw)d ( APw,Tu)
2 d (BQu,Tu)d (BQu, Sw)
(ii)
2 d ( APx, Sx) d (BQy,Ty)
1
d (w, u)2 k d (u, w)d (w, u)
d ( APx, BQy)
k1 d (BQy, Sx)d ( APx,Ty)
d (w, u)2 (1 k ) 0
2 d (BQy,Ty)d (BQy, Sx)
k d ( APx, Sx)d ( APx,Ty)
Where x,yX and k1+2k21,k1,k20
then AP,BQ,S and T have a Common fixed point. Further if AP=PA,BQ=QB Then A,B,P,Q,S and T have a common fixed point ,
Proof: (AP,S) and (BQ,T) are occasionally weakly compatible, then there exists some x,yX such that
1
This is contradiction. There fore u=w.Hence w is unique common fixed point of AP,BQ,S and T.
If AP=PA and BQ=QB then Aw=A(APw)=A(Paw)=AP(Aw).
Put x=w and y=Aw in (ii)
d ( APw, BQ( Aw))2
k
d ( APw, Sw) d (BQ( Aw),T ( Aw))
1 d (BQ( Aw), Sw)d ( APw,T ( Aw))
APx=Sx and BQy=Ty. Using (ii) we claim APx=BQy.
k d ( APw, Sw)d ( APw,T ( Aw))
2 d ( APx, APx) d (BQy, BQy)
2 d (BQ( Aw),T ( Aw))d (BQ( Aw), Sw)
d ( APx, BQy)
k1 d (BQy, APx)d ( APx, BQy)

k d ( APx, APx)d ( APx, BQy)
2 d (BQy, BQy)d (BQy, APx) 2
d (w, Aw k1 d ( Aw, w)d (w, Aw)
d ( APx, BQy)2 k d (BQy, APx)d (APx, BQy) 2 2
1
1
d ( APx, BQy)2 (1 k ) 0
d (w, Aw
d (w, Aw2
k1 d (w, Aw
(1 k1 ) 0
This is contradiction. So APx=BQy. Therefore APx=BQy=Sx=Ty.
if there is another point of coincident say,w such that APz=Sz=w then APz=Sz=BQy=Ty. Which gives APz=APx implies z=x.
Hence w=APx=Sx for wX is the unique point of coincidence of AP and S.By lemma (1.1) w is a fixed point of AP and S Hence APw=Sw=w.Similarly there
exists a common fixed point uX such that u=BQu=Tu.
Which gives w Aw
Pw= A(Pw)=P(Aw)=w.
Similarly we have Bw=Qw=w.
Hence A,B,S,T,P and Q have unique fixed point.
Example 2 :
Let X, d be the semimetric Space with X 0,1/ 2 and d x y2 .
Define Self mappings A, B,T , S, P and Q as
A(x) 2x 1, B(x) 4x 1,T (x) 4x 3 ,
4 6 10
Suppose uw
Put x=w and y=u in (ii)
S(x) 6x 1, P(x) 2 6x and Q(x) 2x 3 .
8 10 8
.
Also the mappings satisfy all the conditions of theorem 1. Here the common fixed point is 1/2
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