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Statement

Notation (pp. 125, 128): $f$ is a monic polynomial (1), $E$ the set where $|f|<1$, $L$ the real axis and $I_r=[-r,r]$; $\delta(r)$ is the supremum of $\operatorname{diam}(E\cap L)$ over the polynomials whose zeros lie on $I_r$.

Theorem 2 (p. 128).

$$\delta(r)=2\sqrt{1+r^2}\quad(0\leq r\leq3/4),\qquad \delta(r)=1+2r\quad(3/4\leq r<\infty).$$

The paper notes (p. 128) that $2\sqrt{1+r^2}$ is attained by $|E\cap L|$ for $f=x^2-r^2$, that $1+2r$ is approached by $f=(x-r)^m(x+r)$ with $m$ large, and that the theorem stays valid when the zeros are only required to lie in the closed disk $\bar D_r$ (referring to the proof of Theorem 9).

Source. P. Erdős, F. Herzog, G. Piranian, Metric properties of polynomials, J. Analyse Math. 6 (1958), 125--148, doi:10.1007/BF02790232; Theorem 2 on p. 128, its proof on pp. 129--131. The copy read is identified on the source card.

Read depth. Claims checked: the statement and the remarks after it were read on the page image of p. 128 on 2026-10-08; the proof was read for structure, not checked. Nothing here is independently reviewed.

Proof pointer

Pages 129--131. An inequality like (2) of Theorem 1 reduces the question to the functions $g(x)=|x-r|^m|x+r|$ with real $m>0$. For $r\le3/4$ the diameter of $E(g)\cap L$ is shown largest at $m=1$: writing the endpoints as $s-\sqrt{1+r^2}$ and $t+\sqrt{1+r^2}$, the claim $s>t$ for $m>1$ becomes the positivity of a power series in $s$ whose coefficients are checked to be positive through the inequality (3) on p. 130. For $r\ge3/4$ the paper shows $g(r+s-(1+2r))\ge1$, using that the relevant expression increases with $r$ and the case $r=3/4$ from the first part (p. 131).

Dependencies

The comparison inequality (2) from the proof of Theorem 1. The first part of Theorem 9 rests on this theorem.

Bears on

• #1038: a set of reals has measure at most its diameter, so for zeros in $[-1,1]$ the case $r=1$ gives $|E\cap L|\le\delta(1)=3$, an upper bound for the supremum the problem asks for. This consequence is drawn here; the paper does not state it, and its own conjecture for that supremum is $2\sqrt2$ (see Problem 1).