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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 $\underline{\text{not converge}}$.

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  • $\begingroup$ 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 $p_k$. $\endgroup$ – Robert Israel Dec 6 '18 at 15:56
  • $\begingroup$ Thank you. I meant $p_k(0)=1$. $\endgroup$ – user122916 Dec 6 '18 at 15:57
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    $\begingroup$ My intuition is telling me that the answer is definitely "no". But I need to think more. $\endgroup$ – mathworker21 Dec 9 '18 at 3:00
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    $\begingroup$ 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 $\endgroup$ – Lukas Geyer Dec 11 '18 at 23:45
  • $\begingroup$ @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 $1\leq j\leq m$, $$\left\| \sum_{n=1}^j f_{\sigma(n)} \right\|_R< 4\epsilon.$$ $\endgroup$ – user122916 Jan 2 at 22:16

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