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J. Comp. & Math. Sci. Vol.4 (3), 161-166 (2013)

A Common Fixed Point Theorem Using Weaker Conditions V. SRINIVAS, B.V.B. REDDY and R. UMAMAHESHWAR RAO Department of Mathematics, Sreenidhi Institue of Science and Technology, Ghatkesar, R.R. District, Andhra Pradesh, INDIA. (Received on: May 14, 2013) ABSTRACT The purpose of this paper is to present a common fixed point theorem in a metric space which generalizes the result of Brian Fisher using the weaker conditions such as Weakly compatible mappings and Associated sequence in place of compatibility and completeness of the metric space. More over the condition of continuity of any one of the mapping is being dropped. Keywords: Fixed point, self maps, compatible mappings, weakly compatible mappings, associated sequence. AMS (2000) Mathematics Classification: 54H25, 47H10

1. INTRODUCTION

1.1. 2 Weakly compatible

G. Jungck1 introduced the concept of compatible maps which is weaker than weakly commuting maps. After wards Jungck and Rhoades4 defined weaker class of maps known as weakly compatible maps.

Two self maps S and T of a metric space (X,d) are said to be weakly compatible if they commute at their coincidence point. i.e if Su=Tu for some u∈X then STu=TSu. It is clear that every compatible pair is weakly compatible but its converse need not be true. Brian Fisher proved the following theorem.

1.1 Definitions and Preliminaries 1.1.1 Compatible mappings Two self maps S and T of a metric space (X,d) are said to be compatible mappings if lim d(STxn,TSxn)=0, whenever n →∞

<xn> is a sequence in X such that

lim Txn= t n →∞

for some t∈X.

lim Sxn= n →∞

1.1.3 Theorem (A): Suppose S,I,T and J be self mappings from a complete metric space (X,d) into itself satisfying the following conditions S(X) ⊆ J(X) and T(X) ⊆ I(X)

Journal of Computer and Mathematical Sciences Vol. 4, Issue 3, 30 June, 2013 Pages (135-201)

(1.1.4)

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V. Srinivas, et al., J. Comp. & Math. Sci. Vol.4 (3), 161-166 (2013)

One of A,B,S or T is continuous

(1.1.5)

[ d (Sx,Ty)] ≤ c Max {d ( Ix, Jy), d (Ix, Sx), d ( Jy, Ty)} (1.1.6) The pairs (A,S) and (B,T) are commuting on X (1.1.7) Further if X is a complete metric space (1.1.8) then A,B,S and T have a unique common fixed point in X. Now we generalize the theorem for extending the result to six self maps using weaker conditions like is weakly compatible mappings and Associated Sequence. 1.1.9 Associated Sequence: Let A,B,P,Q,S and T are self maps of a metric space (X,d). Then for an arbitrary x0∈X, P(X) ⊆ Ab(X) gives there exists x0∈X such that Px0 = ABx1 and for this point x1, there exist a point x2 in X.

Since Q(X) ⊆ St(X) gives there exists x1∈X such that Qx1= STx2 and so on. Proceeding in the similar manner, we can define a sequence <yn> in X such that y2n=Px2n= ABx2n+1 and y2n+1=QX2n+1 = STx2n+2 for n ≥ 0. We shall call this sequence as an “associated sequence of x0 “relative to the four self maps A,B,P,Q,S and T. Now we prove a lemma which plays an important role in our main Theorem. 1.1.10 Lemma: Let A,B,P,Q,S and T are self maps of a metric space (X,d) into itself satisfying the conditions P(X) ⊆ AB(X) and P(X) ⊆ AB(X) Then the associated sequence {yn} relative to four self maps is a Cauchy sequence in X. Proof: From the conditions P(X) ⊆ AB(X) and P(X) ⊆ AB(X) and from the definition of associated sequence we have

[ d ( Px, Qy )] ≤ c.Max d ( Px, ABy) , d ( STx, ABy), [ d ( STx, Px) + d ( Px, Qy)] Forall x, y ∈ X

 where 0 ≤ c < 1

1 2

[ d ( Px2n , Qx2n+1 )] ≤ c.Max d ( Px2n , ABx2n+1 ) , d ( STx2n , ABx2n+1 ), [ d ( STx2n , Px2n ) + d ( Px2n , Qx2n+1 )] 1 2

[ d ( Px2n , Qx2n+1 )] ≤ c.Max d ( Px2n , Px2 n ) , d (Qx2n−1 , Px2n ), [ d (Qx2n−1 , Px2n ) + d ( Px2n , Qx2n+1 )] 

1 2

[ d ( Px2 n , Qx2n +1 )] ≤ c.Max {d (Qx2 n−1 , Px2 n ), d ( Px2 n , Qx2 n+1 )} [ d ( y2n , y2 n+1 )] ≤ c.Max {d ( y2 n−1 , y2n ), d ( y2 n , y2 n+1 )} This implies For every int eger p > 0, we get d ( yn , yn + p ) ≤ d ( yn , yn +1 ) + d ( yn +1 , yn + 2 ) + ............ + d ( yn + p −1 , yn + p ) ≤ h n d ( y0 , y1 ) + h n +1d ( y0 , y1 ) + ............. + h n + p −1d ( y0 , y1 )

≤ ( h n + h n +1 + ............. + h n + p −1 ) d ( y0 , y1 )

≤ h n (1 + h + h 2 + ............. + h p −1 ) d ( y0 , y1 ) Journal of Computer and Mathematical Sciences Vol. 4, Issue 3, 30 June, 2013 Pages (135-201)

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V. Srinivas, et al., J. Comp. & Math. Sci. Vol.4 (3), 161-166 (2013)

Since h<1, , hn → 0 as n→ ∞, so that d(yn,yn+1)→ 0. This shows that the sequence {yn} is a Cauchy sequence in X and since X is a complete metric space, it converges to a limit, say z∈X. The converse of the Lemma is not true, that is A,B,S and T are self maps of a metric space (X,d) satisfying (1.1.4) and (1.1.6), even if for x0∈X and for associated sequence of x0 converges, the metric space (X,d) need not

be complete. The establishes this.

following

example

Now we generalize the above Theorem (A) in the following form. 1.1.12. Theorem (B): Let A,B,P,Q,S and T are self maps of a metric space (X,d) satisfying the conditions P(X) ⊆ AB(X) and Q(X) ⊆ ST(X)

[ d ( Px, Qy)] ≤ c.Max d ( Px, ABy) , d (STx, ABy ), [ d (STx, Px) + d ( Px, Qy)] 1 2

 where 0 ≤ c < 1

Forall x, y ∈ X



P(ST)x2n→Pz and (ST)Px2n→STz as n→∞ From the compatibility of the pair (P,ST) gives, ŕ˘”â†’âˆž ŕŤ›â€Ť ܖ‏, ŕŤ›â€Ť ܖ‏ D (Pz,STz)=0 Pz=STz Since P(X) ⊆ Ab(X) implies there exists u ∈ X such that ABu= z. And Q(X) ⊆ ST(X) implies there exists v ∈ X such that z=STv. To prove Pz=z Put x=z, y=x2n+1

the pairs (P,ST) is Reciprocally continuous and compatible and (Q,AB) are weakly compatible AB=BA,ST=TS,TP=PT,QA=AQ Then A,B,P,Q,S and T have a unique common fixed point z in X. Proof: Px2n→z, ABx2n+1 →z, Qx2n+1 →z STx2n+2→z Also since (P,ST) is reciprocally continuous

[ d ( Px2 n , Qu )] ≤ c.Max d ( Px2 n , ABu ) , d ( STx2 n , ABu ), [ d ( STx2 n , Px2 n ) + d ( Px2 n , Qu )] 1 2





Letting n → ∞

[ d ( z , Qu )] ≤ c.Max d ( z, z ), d ( z , z ), [ d ( z , z ) + d ( z , Qu )] 

1 2



1 2

[ d ( z , Qu )] ≤ c. [ d ( z , Qu )] [ d ( z , Qu )] 1 − c

1 ≤0 2   This gives d ( Qu , z ) = 0 or Qu = z Hence Qu = ABu = z Journal of Computer and Mathematical Sciences Vol. 4, Issue 3, 30 June, 2013 Pages (135-201)

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V. Srinivas, et al., J. Comp. & Math. Sci. Vol.4 (3), 161-166 (2013)

Since (Q,AB) is weakly compatible implies Q[(AB)]u=[AB(Q)]u⇒Qz=ABz To prove z=Qz put x=x2n ,y=z

[ d ( Px2 n , Qz )] ≤ c.Max d ( Px2 n , ABz ), d ( STx2 n , ABz ), [ d ( STx2 n , Px2 n ) + d ( Px2 n , Qz ) ] 1 2

Letting n → ∞

[ d ( z, Qz )] ≤ c.Max d ( z , Qz ) , d ( z , Qz ), [ d ( z , z ) + d ( z, Qz )] 1 2

 [ d ( z, Qz )] ≤ c.d ( z, Qz )

[ d ( z, Qz )][1 − c ] ≤ c

This gives d ( Qz , z ) = 0 or Qz = z Hence Qz = ABz = z

To prove Tz=Z, put x=Tz ,y= x2n+1

[ d ( PTz, Qx2 n +1 )] ≤ c.Max d ( PTz, ABx2 n +1 ) , d ( STTz, ABx2 n+1 ), [ d ( STTz, PTz ) + d ( PTz, Qx2n +1 )] 1 2

Letting n → ∞

[ d (Tz, z )] ≤ c.Max d ( Tz, z ) , d (Tz, z ), [ d (Tz, Tz ) + d (Tz, z )] 1 2

 ( , ) . ( d Tz z ≤ c d Tz [ ] { , z) }

This gives d ( Tz , z ) = 0 or Tz = z STz = z Im plies Sz = z

To prove Az=z, put x=z,y=Az

[ d ( Pz , QAz )] ≤ c.Max d ( Pz, ABAz ) , d (STz, ABAz ), [ d (STz, Pz ) + d ( Pz, QAz )] 1 2

Letting n → ∞

[ d ( z, Az )] ≤ c.Max d ( z, Az ) , d ( z , Az ), [ d ( z, z ) + d ( z, Az )]  This gives d ( Az , z ) = 0 or Az = z Now ABz = z ⇒ BAz = z ⇒ Bz = z

1 2

We get z in a common fixed point of A,B,P,Q,S and T. The uniqueness of the fixed point can be easily proved. Journal of Computer and Mathematical Sciences Vol. 4, Issue 3, 30 June, 2013 Pages (135-201)

V. Srinivas, et al., J. Comp. & Math. Sci. Vol.4 (3), 161-166 (2013)

165

1.1.11. Example: Let X=[-1,1] with d(x,y)= x − y

1  5 Px = Qx =  1  6

1 6

if − 1 < x < if

  x if Ax = Sx =   x if  1 if  5 ABx =  1 − x  3

1 ≤ x <1 6 −1 ≤ x <

1 6

1 ≤ x <1 6 −1 < x < if

1 6

1 ≤ x <1 6

1 1  5 if − 1 < x < 6 Bx =   1 − x if 1 ≤ x < 1  3 6 1 1  5 if − 1 < x < 6 Tx =   6 x + 5 if 1 ≤ x < 1  36 6 1 1 if − 1 < x <  5 6 STx =  + 6 x 5 1  if ≤ x < 1  36 6

 1  1 −2   1  1 11  ∪  ,  ST(X) =  ∪  ,   5  6 3   5  6 36 

1 1  5 6 

Then P(X) =Q(X)=  ,  while AB(X) = 

so that the conditions P(X) ⊆ AB(X and Q(X) ⊆ ST(X) are satisfied. Clearly (X,d) is a complete metric space. 1.1.15. Remark From the example given above, clearly the pairs (P,ST) is reciprocally continuous and compatible and (Q,AB) are weakly compatible as they commute at coincident points

1 . But the 6

pair (Q,AB) are not compatible. For this, take a sequence xn=  1 + 1  for n≥1, then 6 n 

also

lim n →∞

Clearly

1 is the unique common fixed 6

point of A,B,P,Q,S and T.

1 1 and lim P(ST)xn= n → ∞ 6 6 1 (ST)Pxn= . So that lim n →∞ 6

lim Pxn= lim STxn= n →∞ n →∞

d(P(ST)xn,(ST)Pxn)=0. Also note that none of the mappings are continuous and the rational inequality holds for the values of 0 ≤ k1 + 2k2 < 1, where k1 , k2 ≥ 0 .

REFERENCES 1. Jungck. G, Compatible mappings and common fixed points, Internat. J. Math. & Math.Sci.9, 771- 778 (1986).

Journal of Computer and Mathematical Sciences Vol. 4, Issue 3, 30 June, 2013 Pages (135-201)

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V. Srinivas, et al., J. Comp. & Math. Sci. Vol.4 (3), 161-166 (2013)

2. R.P. Pant, A Common fixed point theorem under a new condition, Indian J. of Pure and App. Math., 30(2), 147152 (1999). 3. Jungck,G , Compatible mappings and common fixed points(2), Internat. J. Math. & Math. Sci.11, 285-288 (1988). 4. Jungck.G. and Rhoades.B.E., Fixed point for set valued functions without continuity, Indian J. Pure. Appl. Math., 29 (3), 227-238 (1998). 5. Bijendra Singh and S.Chauhan On common fixed points of four mappings,

Bull. Cal. Math. Soc., 88,301-308, (1998). 6. B.Fisher, Common fixed points of four mappings, Bull. Inst. Math. Acad. Sinica, 11,103(1983). 7. Srinivas.V and Umamaheshwar Rao.R, A fixed point theorem for four self maps under weakly compatible, Proceeding of world congress on Engineering, Vol. II,WCE 2008, London,U.K. (2008). 8. Srinivas.V &Umamaheshwar Rao.R, Common Fixed Point of Four Self Maps� Vol.3,No.2, 113-118 (2011).

Journal of Computer and Mathematical Sciences Vol. 4, Issue 3, 30 June, 2013 Pages (135-201)

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