A DISCUSSION ON A COMMON FIXED POINT THEOREM ON SEMICOMPATIBLE MAPPINGS

A DISCUSSION ON A COMMON FIXED POINT THEOREM ON SEMICOMPATIBLE MAPPINGS RAVI S. AND V. SRINIVAS Department of Mathematics, University Post Graduate College, Secunderabad Osmania University, Hyderabad-500003(Telangana), India Department of Mathematics, University College of Science, Saifabad Osmania University, Hyderabad-500004(Telangana), India Copyright © 2018 Ravi and Srinivas. This is an open access article distributed under the Creative Commons Attribution License, which permits unrestricted use, distribution, and reproduction in any medium, provided the original work is properly cited. Abstract: In this paper, we prove a common fixed point theorem which is a generalization of Bijendra Singh and M.S. Chauhan using some weaker conditions namely semi compatible and associated sequence instead of compatibility and completeness of the metric space. Also, we give suitable example to validate our theorem.


Introduction
A contraction mapping defined on complete metric space is having unique fixed point, this is known as Banach contraction principle and which is first ever result in fixed point theory.This result was further generalized and extended in various ways by many authors.S.Sessa [8] defined weak commutativity and proved common fixed point theorems for weakly commuting maps.Afterwards G.Jungck [1] introduced the concept of compatible mappings which is weaker than weakly commuting mappings.Thereafter Jungck and Rhoades [4] defined weaker class of maps known as weakly compatible maps.
The concept of semi compatible mappings was introduced by Y.J.Cho, B.K. Sharma and D.R.Sharma [6].In this paper we prove a common fixed point theorem for four self maps using semicompatible mappings.

Definitions and Preliminaries
2.1 Definition [3].A and S are two self maps of a metric space (X,d)   for some tX  .
2.2 Example.Let (X,d) be a metric space where X [0, 2]  and d(x, y) | x y |  .We define self maps A and S as  for some tX  .

Example. Let X [0, 2]
 with the usual metric.Define A and S by .Hence (S,A) is not semi compatible.
Further the mappings A and S are not compatible.
2.5 Remark.Semi compatibility of the pair (A,S) does not imply the semi compatibility of the pair (S,A) and semi compatible mappings need not be compatible.
(2.6.4) the pairs (A,S) and (B,T) are compatible on X .
Further if X is a complete metric space then A,B,S and T have a unique common fixed point in X.Now we generalize the above theorem using semi compatible mappings and an associated sequence.

Associated sequence [7]
: Suppose A, B, S and T are four self maps of a metric space (X,d) satisfying the condition(2.6.1),then for an arbitrary 0 xX  such that 0 ( ) Bx Sx  and so on.Proceeding in this similar manner, we can define  for 0 n  .We shall call this sequence as an "Associated sequence" connected to the four self maps A, B, S and T. Now we prove a lemma with an example which plays an important role in our main theorem.

Lemma:
Suppose A, B, S and T are self maps of a complete metric space (X, d) into itself satisfying the conditions (2.6.1) and (2.6.3),then the associated sequence {} n y relative to four self maps given in (2.7) is a Cauchy sequence in X.
Proof: Using the conditions (2.6.1),(2.6.3) and from the definition of associated sequence, we have Now for every integer 0 m  , we get n n m d y y   .This shows that the sequence {} n y is a Cauchy sequence in X and since X is a complete metric space, it converges to some point, 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 the (2.6.1) and (2.6.3),though if for any x0X and for the associated sequence , , , ,...., , ,... nn Ax Bx Ax Bx Ax Bx  converges, the metric space (X,d) need not be complete.
For this we give an example.

Example:
x y  .Define self maps of A, B, S and T of X by showing the conditions ( ) ( ) . Also it is easy to show that the associated sequence , but X is not a complete metric space.2), we have

Theorem
] letting n and using the conditions (3.1.6)and Sz z  ,we get   Hence Bz Tz z  … (3.1.12),showing that z is a common fixed point of B andT .
From the conditions (3.1.9)and (3.1.12),we have Az Sz Bz Tz z     .
Since Bz Tz Az Sz z     , we get z is a common fixed point of A, B, S and T.

Conclusion:
From the example (2.6), we prove that the pairs (A, S) and (B, T) are semicompatible and S and T are continuous.
For this, take a sequence which is a common fixed point of A, B, S and T. We observe that 1 2 is the unique common fixed point of A, B, S and T.

2 [ 2 (
 and using the conditions (3.1.6),(3.1.7)and (3.1.8),we get 1 Sz d z z d z Sz d Sz z k d Sz Sz d Sz z d z Sz d z z z  .Since the distance function can never be negative, we get ] d Az z k d Az z d z z d z z d Az z k d Az z d Az z d z z d z z d Az z  , this gives ]