A Fixed Point Theorem on Semi-Metric Space using Occasionally Weakly Compatible Mappings

A Fixed Point Theorem on Semi-Metric Space using Occasionally Weakly Compatible Mappings

B. Vijayabasker Reddy

Department of Mathematics, Sreenidhi Institute of Science and Technology,Ghatkesar, Hyderabad, India-501 301

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 semi-metric space with X=[1/2,5] and d(x ,y)=(x-y)2. Define two self mappings A and B as fololws

1. INTRODUCTION

The concept of semi-metric 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 semi-metric 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 semi-metric 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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