COMMON FIXED POINT THEOREM OF THREE MAPPINGS IN COMPLETE METRIC SPACE Latpate V.V. and Dolhare U.P. ACS College Gangakhed DSM College Jintur Abstract:-In this paper we prove common fixed point theorem of weakly compatible mappings in complete Metric space. AMS Subject classification:-47h0,47h09 Keywords:-Fixed point, Common fixed point, contraction mapping, weakly compatible mapping. I. INTRODUCTION The study of common fixed point of mappings satisfying contractive type conditions has been a very active field of research activity during the last some decades. The fixed point theory has several applications in many fields of science and engineering field.s. Banach [ ] derived a well known theorem for a contraction mapping in a complete Metric space, which states that, A contraction has a unique fixed point theorem in a complete Metric space. After that many authors proved fixed point theorems for mappings satisfying certain contraction conditions. In 968 R. Kanan [ 5 ] introduced another type of map called as Kanan map and investigated unique fixed point theorem in complete Metric space. In 973 B.K. Das and Sattya Gupta [ ] Generalized Banach Contraction Principle in terms of rational expression.various fixed point thereoms proved by several authors. Recently V.V. Latpate and Dolhare U.P.[ 6 ] proved fixed point theorems for uniformly locally contractive mappings. Sesa [ 8 ] Introduced a concept of weakly commuting mappings and obtained some common fixed point theorems in complete Metric space. S.T. Patil [ 7 ] Proved some common fixed point theorems for weakly commuting mappings satisfying a contractive conditions in complete Metric space. In 986 G. Jungck [ 4 ] defined compatible mappings and proved some common fixed point theorems in complete Metric space. Also he proved weak commuting mappings are compatible. Also we prove the common fixed point theorem of weakly compatible mappings satisfying the inequality similar to C- Contraction. II. PRELIMINARIES Definition.:- Let X be a non-empty set. A mapping d : X X R is said to be a Metric or a distance function if it satisfies following conditions.. d( x, y is non-negative.. d( x, y =0 if and only if x and y coincides i.e. x = y. 3. d( x, y d( y, x (Symmetry 4. d( x, y d( x, z d( z, y (Triangle inequality Then the function d is referred to as metric on X. And (X,d or simply X is said to as Metric space. Definition.:- A Metric space (X,d is said to be a complete Metric space if every Cauchy sequence in X converges to a point of X. Definition.3:- If (X,d be a complete Metric space and a function F : X X is said to be a contraction map if d( F( x, F( y d( x, y DOI:0.63/IJAPSA.07.308.QZLGT Page
Volume 03, Issue 8, [August- 07] e-issn: 394-553, p-issn: 394-83X For all x, yòx and for 0 Definition.4:- Let F : X X, then x ò X is said to be a fixed point of F if F( x x Definition.5:-Let X be a Metric space and if F and F be any two maps. An element a ò X is said to be a common fixed point of F and F if F( a F( a For ex:- If F( x sin( x and F( x tan( x Then 0 is the common fixed point F and F Since F (0 sin(0 and F (0 tan(0 0 Definition.6:-( S.K. Chatterjea [ 3 ] A mapping F : X X where (X,d is a Metric space is said to be C-Contraction if there is a some s.t. 0 s.t. the following inequality holds d F, F ( d( x, F d( y, F ( x y y x If ( Xd, be a complete Metric space, then any C-contraction on X has a unique fixed point. Definition.7:- Let F and G be two self mappings of a Metric space ( Xd,. F and G are said to be weakly compatible if for all xx ò Fx Gx FGx GFx Theorem.:-Suppose P,Q,R,S be four self maps of a Metric space (X,d which satisfies the conditions given below.. P( X S( X and R( X Q( X.. Pair of mappings (P,Q and (R,S are Commuting. 3. One of the function P, Q, R, S is continuous. 4. d( Px, Rx ( x, y where ( x, y max d( Qx, Sy, d( Qx, Px, d( Sy, Ry For all x, y ò X and 0 and 5. X is complete. Then P,Q,R and S have Unique Common Fixed point zx ò.further more z is the unique common fixed point of (P,Q and (R,S. Theorem.:- Let (X,d be a Complete Metric space. Suppose that the mappings P,Q,R and S are four self maps of X which satisfies the following, S(X P( x and R( X Q( X ; d(rx,sx ( ( xy, Where is an upper semi continuous, contractive modulus and ( x, y max { d( Px, Qy, d( Px, Rx, d( Qy, Sy, ( d( Px, Sy d( Qy, Rx} 3. The pairs (R,P and (S,Q are weakly compatible. Then P,Q,R and S have a unique common fixed point. We obtain Common Fixed point theorems for three maps which satisfies contraction condition. III. MAIN RESULT Theorem 3.:- Let ( Xd, be a complete Metric space and Let A be a non empty closed subset of X. Let P, Q : A A be s.t. d( P, Q ( d( R, Q d( R, P d( S, R ( d( R, Q d( R, P (. x y x y y x x y x y y x @IJAPSA-07, All rights Reserved Page
Volume 03, Issue 8, [August- 07] e-issn: 394-553, p-issn: 394-83X For any ( x, y ò X X, where a function :[0, [0, is a continuous and ( xy, 0 iff x y 0 and R : A X which satisfies the following condition. (i PA RA and QA RA (ii The pair of mappings (P,R and (Q,R are weakly compatible. (iii R(A is closed subset of X. Then P,R and Q have unique common fixed point. Proof:- Let x 0 be any arbitrary element of A as PA RA and QA RA. y0 Px0 Rx, y Qx Rx, y Px Rx3,... Let { x n } and{ y n } be two sequences s.t.... y Px Rx, y Qx Rx,... First we shall prove that d( yn, yn 0 as n Let n=k by inequality (, we have d( y, y d( Px, Qx n k k k n n n n n n ( d ( Rx, Qx d( Rx, Px d ( Sx (,,,, k k k k k Rx k d ( Rx Qx d k k ( Rx Px k k ( d( y, y d( y, (, ( y d y y ( d y, y, d( y, y k k k k k k k k k k ( y y d ( k, k (. ( d( y k, y k d ( y k,y k This gives d( y, y d( y, y k k k k For n=k+, similarly, we can show that d( yk, yk d( y k, yk (.3 d( yn, yn is a non-increasing sequence of non-negative real numbers and hence it is convergent. Let l lim d( yn, yn From (. we have n d( yn, yn d( yn, yn and by triangle inequality d( yn, yn ( d( yn, yn d( yn, yn (.4 Letting n, we have lim d( yn, yn lim d( yn, yn lim d( yn, yn n n n l lim d ( yn, yn l n lim d( y, y =l n n n @IJAPSA-07, All rights Reserved Page 3
Consider d( y, y d( Px, Qx k k k k Volume 03, Issue 8, [August- 07] e-issn: 394-553, p-issn: 394-83X ( d( y k, y k d( y k, y k d( y k, y k ( d( y k, y k, d( yk, y k (.5 Letting k and Since is given to be continuous we obtain l l ( l,0 This gives ( l,0 0 By definition of, ( x, y 0 if x y 0 l 0, l 0 l lim d( yn, yn 0 (.5 n Now our claim is that { y n } is a Cauchy sequence From (.3 we have d( y, y d( y, y, n n n n To prove { y n } is a Cauchy sequence we only prove that the subsequence { y n } is a Cauchy. If possible suppose that { y n } is not a Cauchy sequence. There exists 0 for which we can find two subsequence s { y nk ( } and { y mk ( } of { y n } Such that n k is the least index for which n k > m k >k and d( y mk (, y nk ( This gives Using triangle inequality d ( y, y m( k n( k d( y, y d( y, y m( k n( k m( k n( k n( k n( k n( k n( k (.6 d( y, y d( y, y (.7 Now as k and from (.6, we have lim k d( y m( k, y n( k (.8 d( y, y d( y, y d( y, y (.9 Also And m( k n( k m( k n( k n( k n( k d( y, y d( y, y d( y, y (.0 n( k m( k n( k m( k m( k m( k d( y, y d( y, y d( y, y (. n( k m( k n( k m( k m( k m( k From (.6,(.9,(.0 and (., We have lim d( y, y lim d( y, y k m( k n( k k m( k n( k Inequality (. gives d(, y d( Px, Qx y m( k n( k n( k m( k = lim d( y, y ò k m( k n( k (. @IJAPSA-07, All rights Reserved Page 4
Volume 03, Issue 8, [August- 07] e-issn: 394-553, p-issn: 394-83X ( d ( Rx (, Qx ( d ( Rx (, Px ( d ( Sx (, Rx n k m k m k n k n k m( k ( d( Rx, Qx, d( Rx, Px n( k m( k m( k n( k d( y n( k, y m( k d( y m( k, y n( k d( y n( k, y m( k ( d( y, y, d( y, y n( k m( k m( k n( k ( d( y, ( m( k y m( k d y m( k, y m( k (.3 Letting k in above inequality and from (.3 and is continuous, We have ( ( this gives (, 0. By assumption of ( x, y 0 if x y 0 0. But 0 therefore which is contradiction. { y n } is a Cauchy Sequence. To prove P,Q,R have a Common fixed point. Given (X,d be complete and { Sequence, there is px ò y p y } s.t. lim n as A is closed and { n n closed.so there is ua ò s.t. p= Ru for every nn ò d( Pu, y d( Pu, Qx n n n n y n } be Cauchy A, pa ò, By hypothesis R(A is ( d ( Ru, Qx d ( Rx, Pu d ( Su, Rx n n n (.4 - ( d( Ru, Qx, d( Rx, Pu When n ( d ( p, y d ( y, Pu d ( Su n n, y n ( d( Ru, Qx, d( Rx, Pu n n d( Pu, p ( d ( p, p d ( p, Pu d ( Su, p ( d ( Ru, p, d ( p, Pu And hence (0, d( p, Pu ( d( Pu, p d( Su, p 0, d( p, pu 0Pu p Simillarly we can show that Su p. Pu Qu Ru p. Given pairs (R,P and (R,Q are weakly compatible, Pp Qp Rp Now consider d( Pp, y d( Pp, Qx n n ( d ( Rp, Qx d ( Rx, Pp d ( Sp, Rx n n n (.5 - ( d( Rp, Qx, d( Rx, Pp n n @IJAPSA-07, All rights Reserved Page 5
Volume 03, Issue 8, [August- 07] e-issn: 394-553, p-issn: 394-83X ( d ( Rp, y d ( y, Pp d ( Sp, y n n n ( d( Rp, y, d( y, Pp n n As n, Since Pp Qp Rp, We have d ( Pp, p ( d ( Pp, p d ( p, Pp d ( Sp, p ( d ( Pp, p, d (, P p p (.6 Hence ( d( Pp, p, d( p, Pp 0 and so d( Pp, p 0 Pp p and from Pp Qp Rp We have Pp Qp Rp p. Thus Uniqueness of common fixed point is easily obtained from inequality (. IV. CONCLUSION Thus we proved common fixed point theorem for weakly compatible mappings. We are thankful to all authors. V. ACKNOWLEDGEMENT BIBLIOGRAPHY [] S.Banach,Sur les operations dans les ensembles abstraits et leur application aux equations integrals, Fund. Math. 3,(9 338 (French [] B.K. Das and S. Gupta, An extension of Banach Contraction principle through Rational expression. [3] S.K.Chattterjea, Fixed point theorems, C.R. Acad.Buglare Sci.,5 (97 [4] G. Jungck, Compatible mappings and common fixed points, Internat. J. Math. Sci. 9, [5] (986, 77-779. [6] Kanan R. (968 Some results on fixed points Bull. Calcutta Math. Soc., 60,7-76. [7] Latpate V.V. and Dolhare U.P., Uniformly Locally contractive mapping and fixed point theorems in generalized Metric space.ijapsa,06,73-78. [8] S.T. Patil, Results on fixed points and its applications,ph.d. thesis, Marathwada University, Aurangabad, India 990. [9] Sesa,S., on weakly commutativity condition in a fixed point consideration, Publ. Int. Math. 3(46 (986 @IJAPSA-07, All rights Reserved Page 6