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Statement

Setting (p. 3). The minimal partition $p_*$ of the setup recorded on Lemma 4.1, with $K_S$, $n_S$, $c_S$ and $r(S)$ as in Claim 3.1. For a block $S$ the note picks $z_S\in K_S$ with $\lvert z_S-c_S\rvert=r(S)$, which exists since $K_S$ is compact, and argues that $\lvert f(z_S)\rvert=1$ when $r(S)>0$.

Lemma 5.1 (Internal product upper bound, p. 3). Let $S\in p_*$ and let $\alpha_1,\ldots,\alpha_{n_S}$ be the roots belonging to $S$, counted with multiplicity. If $z_S\in K_S$ satisfies $\lvert z_S-c_S\rvert=r(S)$, then

$$\prod_{j=1}^{n_S}\lvert z_S-\alpha_j\rvert\le\bigl(\sqrt2\,r(S)\bigr)^{n_S}.$$

Inequality (1) (p. 4). Write $P_S(z)=\prod_{\alpha\in S}(z-\alpha)^{m(\alpha)}$ and $Q_S=f/P_S$, so $f=P_SQ_S$. When $\lvert f(z_S)\rvert=1$, Lemma 5.1 gives

$$r(S)\ge\frac{1}{\sqrt2\,\lvert Q_S(z_S)\rvert^{1/n_S}},$$

and summing, $\sum_{S\in p_*}\lvert Q_S(z_S)\rvert^{-1/n_S}\le\sqrt2\sum_{S\in p_*}r(S)$.

Proof pointer

P. 3. Normalize to $c_S=0$ and $z_S=r(S)=r$ and set $\nu_j=\alpha_j/r$. Then $\lvert\nu_j\rvert\le1$ and the $\nu_j$ have mean $0$, so the mean of $\lvert1-\nu_j\rvert^2$ is at most $2$, and the arithmetic-geometric mean inequality bounds $\prod_j\lvert1-\nu_j\rvert^2$ by $2^{n_S}$.

Read depth

Claims checked: Lemma 5.1 and inequality (1) were read clause by clause on the page images of the print, and the proofs on pp. 3--4 were followed. Nothing here is independently reviewed.

Source. Boon Qing Hong, Strategy Proposal on Covering Lemniscates for Erdős Problem #509, unpublished note (2026), 11 pp.; the edition read is named on the source card.

Bears on

• Problem 509: a lower bound on the radius of each block of a hypothetical minimal counterexample to Claim 3.1 in terms of the roots outside the block; the note does not turn it into a proof of any case of the problem.