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Statement

f(z)=∏ν=1n(z−zν)f(z)=\prod_{\nu=1}^n(z-z_\nu), E={∣f(z)∣≤1}E=\{|f(z)|\le1\}, and ρ\rho is "the radius of the largest disk contained in EE" (p. 101). Quoted (p. 101): "If cap⁡A<1\operatorname{cap}A<1, there exists a positive number ρ0=ρ0(A)\rho_0=\rho_0(A) such that ρ≥ρ0\rho\ge\rho_0 [2, Theorem 6]. If AA is a disk of radius 1 or a segment of length 4 (in both cases, cap⁡A=1\operatorname{cap}A=1), there does not exist any positive lower bound for ρ\rho that is independent of the degree nn of f(z)f(z). Erdös, Herzog and Piranian put the question whether ρ≥const⋅n−1\rho\ge\mathrm{const}\cdot n^{-1} if ∣zν∣≤1|z_\nu|\le1 [2, Problem 3]. I shall only prove a weaker estimate (see also [2, Problem 2])."

Theorem 4 (p. 101). "If ∣zν∣≤1|z_\nu|\le1, the lemniscate domain EE contains a disk of radius (2e)−1n−2(2e)^{-1}n^{-2}."

The disk found is centered at a zero zμz_\mu and lies in the component E0E_0 of EE containing 00. Its open interior lies in the interior of EE, which is the open set {∣f∣<1}\{|f|<1\}, so that set has area at least π(2e)−2n−4\pi(2e)^{-2}n^{-4}; this consequence is drawn here, not printed.

Source. Ch. Pommerenke, On metric properties of complex polynomials, Michigan Math. J. 8 (1961), no. 2, 97--115; Theorem 4 with its proof and the introductory paragraph on printed p. 101 (PDF p. 5 of the publisher's scan), read on the page image (the scan has no text layer). The copy read is identified in the source digest.

Read depth. Claims checked: the statement and the introductory paragraph were read clause by clause on the page image; the proof (one paragraph) was read in full and its steps followed, with the external inputs named below taken as cited. Nothing here is independently reviewed.

Proof pointer

Page 101. Let E0E_0 be the component of EE containing 00. Theorem 3 with r=1r=1 gives d0>1d_0>1 for its diameter; since E0E_0 is connected, d0≤4cap⁡E0d_0\le4\operatorname{cap}E_0 (Golusin [6, p. 42]), so cap⁡E0>1/4\operatorname{cap}E_0>1/4. Since ∣f∣≤1|f|\le1 on E0E_0, the author's derivative bound [11] gives ∣f′(z)∣≤en2/(2cap⁡E0)<2en2|f'(z)|\le en^2/(2\operatorname{cap}E_0)<2en^2 for z∈E0z\in E_0. Take a zero zμ∈E0z_\mu\in E_0 and the boundary point z∗z^* of E0E_0 nearest to it; integrating f′f' along the segment, 1=∣f(z∗)∣<∣z∗−zμ∣⋅2en21=|f(z^*)|<|z^*-z_\mu|\cdot2en^2, so ∣z∗−zμ∣>1/(2en2)|z^*-z_\mu|>1/(2en^2) and the disk ∣z−zμ∣≤1/(2en2)|z-z_\mu|\le1/(2en^2) lies in E0⊂EE_0\subset E.

Dependencies

Within the paper: Theorem 3 at r=1r=1 (theorem_3). Outside it: the inequality diam⁡≤4cap⁡\operatorname{diam}\le4\operatorname{cap} for a continuum (Golusin, Geometrische Funktionentheorie, 1957, p. 42; not held) and the derivative bound ∣f′∣≤en2/(2cap⁡E0)|f'|\le en^2/(2\operatorname{cap}E_0) on a component where ∣f∣≤1|f|\le1, from the author's On the derivative of a polynomial, Michigan Math. J. 6 (1959), 373--375 (cited as [Po59a] on Problem 115's page; not held).

Bears on

  • Problem 116: the area of {∣f∣<1}\{|f|<1\} is at least π(2e)−2n−4>n−O(1)\pi(2e)^{-2}n^{-4}>n^{-O(1)}, the polynomial lower bound the problem asks for; the paper says nothing about the (log⁡n)−O(1)(\log n)^{-O(1)} strengthening, later obtained by Krishnapur, Lundberg and Ramachandran 2025, whose card names this theorem as the bound it improves.
  • Problem 1039: ρ(f)≥(2e)−1n−2\rho(f)\ge(2e)^{-1}n^{-2}, the "weaker estimate" the paper proves toward the ρ≫1/n\rho\gg1/n of [2, Problem 3], which is the problem's question and which the paper leaves open.