A FIXED POINT THEOREM USING E.A PROPERTY ON MULTIPLICATIVE METRIC SPACE

The emphasis of this paper is to establish a common fixed point theorem on a multiplicative metric space using the conditions weakly compatible mappings and EA-property. Further some examples are discussed to substantiate our result.


INTRODUCTION
In the recent past, the notion of multiplicative metric space (MMS) was introduced by Bashirove et.al. [1]. Many authors [3], [4], [5], [7], [8] and [9] proved fixed point theorems on multiplicative metric space. Jungck and Rhoades [10] defined the weaker class of mappings as weakly compatible mappings. Aamri and Moutawakil [2] developed the notion of E.A 1789 E.A PROPERTY ON MULTIPLICATIVE METRIC SPACE property .Further Ozavsar et.al. [7] designed the notion of convergence and proved unique common fixed point results in multiplicative metric space. In this paper we generate a common fixed point theorem using the concept of weakly compatible mappings with EA property. Our presentation is also supported by the provision of a suitable example.
iii. An MMS is complete if every multiplicative Cauchy sequence is convergent in it.

Definition:
We define mappings G and I of a MMS as compatible if ( , ) = 1 as k → ∞, when ever {α k } is a sequence in X such that = = as k → ∞ for some ∈ X .

Definition:
1790 V. SRINIVAS, T. THIRUPATHI, K. MALLIAH The mappings G and Hof a MMS in which if µ = µ for some µ such that GIµ = µ holds then we say that and I are weakly compatible mappings.

Definition:
Mappings G and I of a MMS (X,d) are said to hold EA property if Now we discuss an example for E.A property.

MAIN RESULTS
Now we prove our main theorem on MMS.

Theorem
Suppose in a complete MMS (X, δ), there are four mappings G, H, I and J holding the conditions , ( Then the above mappings will be having a common fixed point.
Continuing this process, it is possible to construct a Sequence   k  in X Such that . 0 for 2 2 1 2 1 2 and We now prove {βk} is a Cauchy sequence in MMS.
Hence for k<l, on using the multiplicative triangle inequality we get (2) and (3) it gives In the inequality (C2), by putting Nowthe pair (G,I) is weakly compatible with Gαk=Iαk gives GIαk=IGαk and this inturn implies Gµ=Iµ. Now we show that Gµ = µ.

Example:
Showing that the compatibility condition is not fulfilled.
We now establish that the mappings G,H,I and J satisfy the Condition(C2) .   Hence the condition (C2) is satisfied.
Similarly we can prove other cases.