Suppose the radius of convergence for $\sum a_nz^n$ and $\sum b_n z^n$ have radii $R_1$ and $R_2$. Prove that $\sum a_nb_n z^n$ is at least $R_1 R_2$.

My work so far:

I am thinking of using the property that $sup(AB) \geq supAsupB $ for this proof. (1)

We define $R_1$ as $\frac{1}{\alpha_1}$, where $\alpha_1 = lim sup |a_n|^{-n}$.

And $R_2$ as $\frac{1}{\alpha_1}$, where $\alpha_2 = lim sup |b_n|^{-n}$

Then the radius of convergence for $\sum a_nb_n z^n$ is $R_3 = \frac{1}{\alpha_3}$

$$ \alpha_3 = limsup \sqrt[n]{|a_nb_n|} $$ $$ = limsup \sqrt[n]{|a_n|}\sqrt[n]{|b_n|} $$ $$ = sup (A^*B^*) $$

Where $A^*$ and $B^*$ are the set of subsequential limits of $\{\sqrt[n]{a_n}\}$ & $\{\sqrt[n]{b_n}\}$ respectively.

Is it possible to now impose the property of I mentioned above now?


Given two analytic functions in a neighbourhood of the origin, let us introduce the notation $f\ll g$ with the following meaning: given $f(x)=\sum_{n\geq 0} f_n x^n$ and $g(x)=\sum_{n\geq 0}g_n x^n$, we say that $f\ll g$ ("$f$ is dominated by $g$", or "$g$ dominates $f$") if for some positive constant $C$ the inequality $|f_n|\leq C|g_n|$ holds for every large enough $n$. If the radius of convergence of $f(x)$ at the origin equals $\rho>0$ then $f$ is dominated by $\frac{1}{1-(\rho+\varepsilon)x}$ for any $\varepsilon >0$. It is pretty trivial that if $f$ is dominated by $\frac{1}{1-(r+\varepsilon)x}$ and $g$ is dominated by $\frac{1}{1-(R+\varepsilon)x}$ then $h(x)=\sum_{n\geq 0}f_n g_n x^n$ is dominated by $\frac{1}{1-(r+\varepsilon)(R+\varepsilon)x}$, hence the radius of convergence of $h(x)$ at the origin is at least $rR$.

This notation is pretty useful for showing, in a similar fashion, that the radius of convergence of $f(x)\cdot g(x)$ equals $\min(r,R)$.


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