Proving Analyticity of Product of Analytic Functions

[MathematicalPhysicist]

i need to prove that if f and and g are analytic functions in (-a,a) then so is fg.

well basically i need to find the radius of convergence of fg, which its coefficient is: $$c_n=\sum_{i=0}^{n} b_i*a_{n-i}$$, by using cauchy hadamard theorom for finding the radius of convergence, and to show that it's not greater than a.
well limsup |c_n|^1/n, then $$c_n=a_0b_n+...+a_nb_0$$
now i think that $$(c_n)^{\frac{1}{n}}<= ((n+1)(\max_{n \in N}(|a_n|,||b_n|))^2)^{1/n}$$
i think this inequality also applies to alternating sequences.
anyway i don't know how bound it below.

 

[MathematicalPhysicist]

wait a minute i think that's enough iv'e shown that the radius cannot be greater than a.
stupid me. (-;

 

[matt grime]

But this isn't true. If f is analytic in some region, then gf is analytic in that region if f maps into some region in which g is analytic. And the converse is certainly false.

 

[AKG]

He's talking about the product of f and g, not the composition.

 

[matt grime]

Ah! Now why didn't I think of that? (Suggestions not necessary.)

 

[MathematicalPhysicist]

yes I am talking of the prodcut.
anyway, i think this appraoch of mine isn't correct.
i think that bassically in order to prove it you need to use here cauchy's product.
i.e if two partial sums f_n and g_n converges absolutely then also their cauchy's product converges absolutely.
bassically here we assume that both g and f converges in 0<r<a, in [-r,r] they converge absolutely so also their cauchy product converges absolutely at [-r,r].
is this better than my other obvoiusly faulty answer?