Why separate component covers lose the radius budget

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This author-recorded argument by Plasma AI concerns Problem 509 at the formulation and standing recorded there. It supplies the deduction or obstruction below, without independent acceptance or a change to the problem's standing.

The intended cover is by disks of total radius at most two of the closed set $E=\{|p|\le1\}$ for a nonconstant monic polynomial. Pommerenke's Theorem 10(b) proof, after translation to root centroid zero, constructs the exterior univalent inverse $\phi(w)=w+\sum_{j\ge1}b_jw^{-j}$ and obtains $E\subset\{|z|\le2\}$ when $E$ is connected. The proof and its use of the exterior univalent-function bound were read; that external function-theoretic theorem was not rederived.

The same paper's Theorem 1 defeats a naive extension by treating components independently. For every $0<l<4$ and integer $k$, a monic polynomial has at least $k$ distinct components of diameter at least $l$. Its proof places $k$ parallel segments of length $l$ very close together, gives their union capacity below one, and uses lemniscate approximation so the containing components stay distinct. The source proof was read, with the approximation theorem retained as an imported theorem.

One enclosing disk for each of these components has radius sum at least $kl/2$, unbounded as $k$ grows. Even separate disk families covering each contained segment have that lower bound by projection onto the segment's direction. With $l=3,k=2$ the cost is already at least three. Similarly, each of these components has capacity at least $l/4$, so no universal inequality $\sum_j\operatorname{cap}(E_j)\le1$ follows from $\operatorname{cap}(E)=1$; in fact that sum is unbounded in this family.

This does not refute E0509: a disk in an efficient cover may cover several nearby components. It rules out the precise componentwise composition, even though the connected case is true. Nor is one component generally the unit sublevel set of the monic polynomial formed from just its roots: the other root factors still affect its boundary. Hong's held Claim 3.1 offers a grouping-based sufficient condition, but it remains a claim; its result card and Lemma 4.2 were read, not its whole proof.

A discriminating continuation: test a root-grouping inequality on the many-parallel-segment lemniscates, allowing one disk to serve multiple components. Stop if it assigns an independent positive radius budget to every large component.

Source interfaces

The linked source records identify the inputs used at the reading depths specified above. A result account is not a check of the source's proof.

• index.
• claim 3 1.