# If a product of polynomials converges, does some product of their zeros also converge?

Suppose $$\{p_k\}$$ is a sequence of polynomials with $$p_k(0)=1$$. Let $$a_1,a_2,\ldots$$ be an enumeration of all of the zeros of the $$p_k$$. Suppose that $$\prod_{k=1}^\infty p_k(z)$$ converges uniformly on compact subsets of $$\mathbb{C}$$. Is there some permutation $$\sigma:\mathbb{N}\to\mathbb{N}$$ such that $$f_\sigma (z)=\prod_{k=1}^\infty \left(1-\frac{z}{a_{\sigma(k)}}\right)$$ converges uniformly on compact subsets of $$\mathbb{C}$$?

This is a follow-up to Factoring a convergent infinite product of polynomials., in which an example of such $$\{p_k\}$$ is given along with a permutation $$\sigma$$ for which the product $$f_\sigma (z)$$ does $$\underline{\text{not converge}}$$.

• Since you have $a_{\sigma(k)}$ in a denominator, you probably don't want to include $0$ in your enumeration of the zeros, even though it is a zero of all the $p_k$. Dec 6, 2018 at 15:56
• Thank you. I meant $p_k(0)=1$. Dec 6, 2018 at 15:57
• My intuition is telling me that the answer is definitely "no". But I need to think more. Dec 9, 2018 at 3:00
• I think this is true, but possibly not that easy to prove. It might be provable with a product version of the "polygonal confinement theorem" by Steinitz: sites.math.washington.edu/~morrow/335_17/levy.pdf Dec 11, 2018 at 23:45
• @LukasGeyer Do you know if there is an analog of something like Lemma 3.1 in the paper you linked for holomorphic functions? Something like: if $f_1,...,f_m$ are analytic on $|z|\leq R$, $\left\| \sum_{n=1}^m f_n \right\|_{|z|\leq R}< \epsilon$ and $\|f_n\|_{|z|\leq R} < \epsilon$, then there exists a permutation $\sigma$ of $1,2,\ldots,m$ such that whenever $1\leq j\leq m$, $$\left\| \sum_{n=1}^j f_{\sigma(n)} \right\|_R< 4\epsilon.$$ Jan 2, 2019 at 22:16