Show $\liminf_n |\sum^n a_kb_{n-k}|^{-\frac{1}{n}} = \liminf_n |a_n|^{-\frac{1}{n}}$?

Question

Suppose for some complex $a_n$ and $b_n$ such that $\liminf_n |a_n|^{-\frac{1}{n}} = \liminf_n |b_n|^{-\frac{1}{n}}= R$. Show that $\liminf_n |\sum^n a_kb_{n-k}|^{-\frac{1}{n}} = R$?

I'm somehow trying to make a bound like $$\liminf_n |a_n|^{-\frac{1}{n}} \geq \liminf_n |\sum^n a_kb_{n-k}|^{-\frac{1}{n}}\geq \liminf_n |b_n|^{-\frac{1}{n}}$$

for the left inequality, I think intuitively as long as $\limsup_n|\sum^n a_kb_{n-k}| -|a_n| \geq 0$ it should hold. But then I'm not pretty sure how to proceed. The inequality would not hold if $b_{n-k}$ is small, but then that would violate the assumption that $\liminf_n |b_n|^{-\frac{1}{n}}= R$. Any hints to formulate this or have some other better ways?

Answer 1

Let $f(z)=\sum_na_nz^n$ and $g(z)=\sum_nb_nz^n$. The assumption means that $f$ and $g$ have the same radius of convergence, say $\rho$. Then $$\rho=\frac{1}{\limsup_n\sqrt[n]{|ca_n|}}=\frac{1}{\limsup_n\sqrt[n]{|b_n|}}$$ and both $f$ and $g$ converge absolutely and uniformly in compact subsets of $\{z:|z|<\rho\}$.

Let $h=f\cdot g$ Then $$ h(z)=\sum_nc_nz^n,\qquad c_n=\sum^n_{k=0}a_kb_{n-k}$$ The radius of convergence of $h$ is $\rho_c:=\frac{1}{\limsup_n\sqrt[n]{|c_n|}}$. By definition of radius of convergence, $h$ converges in $\{z:|z|<\rho_c\}$ and diverges in $\{z:|z|>\rho_c\}$.

Since both $f(z)$ and $g(z)$ converge in $\{z:|z|<\rho\}$, so does $h(z)$. Hence $\rho\leq \rho_c$.

Edit: As @MartinR observed the inequality in the opposite direction may not hold, his example $f(z)=\frac{z-1}{z+1}$ and $g(z)=\frac{z+1}{z-1}$ satisfies the assumptions of the OP (their powers series around $0$ have both radius of convergence equal to $1$); however, $h(z)=f(z)g(z)=1$ has radius $\infty$