Common Fixed Points of Compatible Mappings

In this paper we prove two common fixed point theorems by considering four mappings in complete metric space. In the first theorem we consider two pairs of compatible mappings of type (A) and in the second theorem we consider two pairs of compatible mappings of type (B). Our results modify and extend some earlier results in the literature.


Introduction
The first important result in the theory of fixed point of compatible mappings was obtained by Jungck [1] as a generalization of commuting mappings.In 1993 Jungck et al. [2] introduced the concept of compatible mappings of type (A) by generalizing the definition of weakly uniformly contraction maps.The concept of compatible mappings of type (B) was introduced by Pathak and Khan [3] in the year 1995.Recently, Nema and Qureshi [4] proved two common fixed point theorems of compatible mappings of type (P).Koireng et.al. [5] also proved another theorem of compatible mappings of type (R).
The aim of this paper is to prove some common fixed point theorems of compatible mappings of type (A) and type (B) in metric spaces by considering four self mappings.Our results extend and modify the results in [4][5][6].

Preliminaries
We recall definitions of various types of compatible mappings and other results which will be needed in the sequel.Definition 1.1 [1]: Let S and T be mappings from a complete metric space X into itself.The mappings S and T are said to be compatible if ) , ( lim Definition 1.2 [2]: Let S and T be mappings from a complete metric space X into itself.
The mappings S and T are said to be compatible of type (A) if ) , ( lim for some t ∈ X.
Definition 1.3 [5]: Let S and T be mappings from a complete metric space X into itself.
The mappings S and T are said to be compatible of type (B) if (iii) STz=TSz and Sz=Tz if S and T are continuous at z. Proposition 2.8 [3]: If S and T be continuous from a metric space X into itself then (i) S and T are compatible if and only if they are compatible of type (B).
(ii) S and T are compatible of type (A) if and only if they are compatible of type (B).
Remark [3]: Proposition 2.8 is not true if S and T are not continuous.Following example will illustrate the remark.
Example: Let X = R with the metric d and define two mappings S and T: X→X as follows: Both S and T are discontinuous at z = 0. Consider a sequence {x n } in X defined by Lemma 2.9 [6] Let A, B, S and T be mapping from a metric space (X, d) into itself satisfying the following conditions: (1) A (X) ⊆ T(X) and B(X) ⊆ S(X) (3) Let x 0 ∈ X then by (1) there exists x 1 ∈ X such that Tx 1 = Ax 0 and for x 1 there exists x 2 ∈ X such that Sx 2 = Bx 1 and so on.Continuing this process we can define a sequence {y n } in X such that y 2n+1 =Tx 2n+1 =Ax 2n and y 2n = Sx 2n = Bx 2n-1 then the sequence {y n } is Cauchy sequence in X.

Proposition 2 . 4 [ 2 ]
: Let S and T be mappings from a complete metric space (X, d) into itself.If a pair {S, T} is compatible of type (A) on X and Sz = Tz for z ∈ X, then STz = TSz = SSz = TTz.Proposition 2.5[2]: Let S and T be mappings from a complete metric space (X, d) into itself.If a pair {S, T} is compatible of type (A) on X and z ∈ X, then we have (i) d(TSx n , Sz) →0 as n → ∞ if S is continuous, (ii) STz=TSz and Sz=Tz if S and T are continuous at z. Proposition 2.6[5]: Let S and T be mappings from a complete metric space (X, d) into itself.If a pair {S, T} is compatible of type (B) on X and Sz = Tz for z ∈ X, then STz = TSz = SSz = TTz.Proposition 2.7[5]: Let S and T be mappings from a complete metric space (X, d) into itself.If a pair {S, T} is compatible of type (B) on X and z ∈ X, then we have