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Statement

Notation (p. 125): $f(z)=\prod_{\nu=1}^n(z-z_\nu)$ is the paper's polynomial (1), written $\prod(x-x_\nu)$ when only real variables occur; $E=E(f)$ is the set where $|f(z)|<1$; $L$ is the real axis and $I=[-1,1]$.

Theorem 1 (p. 126). "Let the zeros $x_\nu$ of the polynomial (1) lie in $I$, and let their centroid $\bar x$ lie in $[0,1]$. Then the set $E\cap L$ contains an interval $J$ which contains the open interval $(0,1)$; moreover, the interval $J$ contains at least $n/2$ of the $x_\nu$, and $|J|\geq\sqrt2$. On the other hand, the set $E$ does not meet the interval $(-\infty,-\sqrt2]$."

The paper places the theorem after two earlier facts it cites (p. 126): $|E\cap I|\ge1$ for zeros on $I$, with equality only for $(x\pm1)^n$ (its reference [2], p. 957), and the result of Steinberg and others (its [7]) that $E$ contains one of the open halves $(-1,0)$, $(0,1)$ of $I$. Theorem 1 names the half: the one on the side of the centroid. The paper remarks (p. 126) that the half of $I$ holding at least half of the zeros need not lie in $E$, by the example $(x-1)(x+1/4)^2$.

Source. P. Erdős, F. Herzog, G. Piranian, Metric properties of polynomials, J. Analyse Math. 6 (1958), 125--148, doi:10.1007/BF02790232; Theorem 1 on p. 126, its proof on pp. 126--128. The copy read is identified on the source card.

Read depth. Claims checked: the statement was read clause by clause on the page image of p. 126 on 2026-10-08; the proof was read for structure, not checked. Nothing here is independently reviewed.

Proof pointer

Pages 126--128. The function $F(x)=\frac1n\sum|x-x_\nu|$ is convex with $F(0)\le1$ and $F(1)=1-\bar x\le1$; outside the trivial case $f=(x^2-1)^p$ (where $E\cap L$ is two open intervals of length $\sqrt2$ each), $F<1$ on $(0,1)$, and the inequality of the arithmetic and geometric means gives $|f|<1$ there. With the zeros ordered decreasingly, $F$ is least at some $x_h$ with $h>n/2$, so $[x_h,1)\subset E$ too, and $J$ is the component of $E\cap L$ holding both. The bound on $|J|$ comes from two comparisons that can only shrink $J$: the zeros inside $J$ are first merged at their centroid and then moved to $1$, and the resulting polynomial is below $1$ at $\sqrt2$ because more than half of its zeros sit at $1$. The last clause uses the paper's inequality (2), $|f(x)|\ge|x-1|^\lambda|x+1|^{n-\lambda}$ for $|x|>1$ with $\lambda/n=(1+\bar x)/2$, a consequence of the concavity of $\log$, evaluated at $x=-\sqrt2$ with $\lambda\ge n/2$.

Dependencies

None within the paper. Inequality (2) of this proof is reused for Theorem 2 (p. 129), and Theorem 1 itself in the proof of Theorem 3 (p. 132).

Bears on

• #1038: for zeros in $[-1,1]$, $J\subset E\cap L$ gives $|E\cap L|\ge\sqrt2$ when the centroid is in $[0,1]$, and the substitution $x\mapsto-x$ (with the sign $(-1)^n$ that keeps the polynomial monic) covers the other case, so the infimum the problem asks for is at least $\sqrt2$. This consequence is drawn here; the paper does not state it. The paper's own remarks on that infimum are on Problem 1.