A COMMON FIXED POINT THEOREM UNDER AN AUXILIARY FUNCTION

A generalization of a result of Badshah and Singh [1] was proved in [5] for a pair of compatible maps and dropping the continuity of one of the self-maps. A generalization of the result of [5] is obtained in this paper, by employing an auxiliary function. AMS Subject Classification: 54H25


Introduction
Badshah and Singh [1] proved the following result for commuting self-maps: Theorem 1.1.Let f and g be self-maps on a complete metric space X satisfying the inclusion f (X) ⊂ g(X) and the inequality f or all x, y ∈ X, where (a) α and β are nonnegative constants with α + 2β < 1, (b) (f, g) is a commuting pair, (c) f and g are continuous.
Then f and g have a unique common fixed point.
A generalization of Theorem 1.1 was obtained in [5], by dropping the continuity of f and using a compatible pair1 (f, g) in (b) with the choice: for some t ∈ X.
It is easy to observe that every commuting pair of self-maps is necessarily compatible.Converse is not true.For instance, see [2], [3] and [4].
The generalization proved in [5] is the following: Theorem 1.2.Let f and g be self-maps on a complete metric space X satisfying the inclusion (1) and the inequality (2), where α and β are nonnegative constants with α + 2β < 1.If g is continuous, and (f, g) is a compatible pair, then f and g have a unique common fixed point.
We prove a generalization of Theorem 1.2 by replacing (2) with a general inequality involving an auxiliary function.

Preliminary Notations
Several fixed point theorems in metric space setting have been proved through contraction conditions involving different types of auxiliary functions.Given a positive integer α, a generalized class Φ α of auxiliary functions was introduced in [6] as follows: ( It is obvious that, for α = 1, Φ α reduces to the class Ψ of all contractive moduli ψ [7] such that ψ(0) = 0 and psi(t) < t for t > 0.
Definition 2.1.A mapping φ ∈ Φ α is said to be upper semicontinuous at Our main result is Theorem 2.1.Let f and g be self-maps on a complete metric space X satisfying the inclusion (1), and the inequality where φ ∈ Φ 2 is nondecreasing and upper semicontinuous.If g is continuous, and (f, g) is a compatible pair, then f and g have a unique common fixed point.
Proof.Let x 0 ∈ X be arbitrary.
In view of (1), we can choose points x 1 , x 2 , . . ., x n , . . . in X inductively such that Writing x = x n−1 and y = x n in (6) and using (7), we get We now prove that If possible, suppose that d(y m , y m−1 ) < d(y m+1 , y m ) for some m ≥ 2. Then d(y m+1 , y m ) > 0. Since φ is nondecreasing, from (8) it follows that which is a contradiction.This proves (9).In other words, d(y n+1 , y n ) ∞ n=1 is a decreasing sequence of nonnegative real numbers and hence converges to some t ≥ 0. Now using (9) in (8), we get Taking the limit superior as n → ∞ in this and then using the upper semicontinuity of φ, we obtain that t ≤ φ(2t). ( If t > 0 in (10), then the choice of φ implies that t ≤ φ(2t) < t, which is a contradiction.Thus We now prove that y n ∞ n=1 is a Cauchy sequence in X.
If possible we suppose that y n ∞ n=1 is not Cauchy.Then for some ǫ > 0, we choose sequences Suppose that m k is the smallest integer exceeding n k which satisfies (12).That is Now by triangle inequality of d, we see that and from (11), we see that Using ( 15) in ( 14), we get Again, by the triaangle inequality of d, we get As k → ∞ this in view of ( 16) and ( 17), gives On the other hand, writing x = x m k −1 , y = x n k −1 in (6), we have Since φ is nondecreasing, proceeding the limit as n → ∞ in this, and then using upper semicontinuity of φ, (13), (15), ( 16),( 17) and (18) we get which is a contradiction.Hence y n ∞ n=1 must be a G-Cauchy sequence in X.
But the use of ( 6) yields Applying the limit as n → ∞ in this, and using ( 22) and (23), we obtain that [d(f x, f y)] 2 ≤ α [d(f x, gx)d(f y, gy) + d(f y, gx)d(f x, gy)] +β [d(f x, gx)d(f x, gy) + d(f y, gx)d(f y, gy)] Since (X, G) is G-Complete, there exists a point p ∈ X such that y n∞ n=1 is G-convergent to p.That is lim n→∞ y n−1 = lim n→∞ y n = p.(20)Now the compatibility of f and g, and (20) imply thatlim n→∞ d(f gx n , gf x n ) = 0,(21)while the sequential property of the continuiy of g and (20) give lim n→∞ gf x n = lim n→∞ g 2 x n = gz.(22) Hence it follows from (21) and (22), that lim n→∞ d(f gx n , gz) = 0 or lim n→∞ f gx n = gz.