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Statement

Setting (p. 29). For positive integers $a_1\le\cdots\le a_n$, $M(a_1,\ldots,a_n)$ is the maximum of $\lvert\prod_{i=1}^n(1-z^{a_i})\rvert$ over $\lvert z\rvert=1$, and $f(n)$ is the minimum of $M(a_1,\ldots,a_n)$ over all such exponents, as on the page of Theorem 2.

Theorem 3 (p. 34).

$$f(n)\ge\sqrt{2n}.$$

The paper calls this lower bound nearly trivial and says it is unable at present to improve it (p. 29).

Source. P. Erdős and G. Szekeres, On the product $\prod_{k=1}^n(1-z^{a_k})$, Acad. Serbe Sci. Publ. Inst. Math. 13 (1959), 29--34: the setting and the remark on p. 29, Theorem 3 and its proof on p. 34. The edition read is identified on the source card.

Read depth. Claims checked: the statement was read on the printed page. The proof was read for its structure only; no step was checked, and nothing here is independently reviewed.

Proof pointer

Page 34. Expand the product as $\sum_i x^{b_i}-\sum_i x^{c_i}$ with $b_1<b_2<\cdots$ and $c_1<c_2<\cdots$. Since $x=1$ is a root of order $n$, the derivatives of order $p<n$ vanish at $1$, which gives the power-sum identities $\sum_ib_i^p=\sum_ic_i^p$ for $p=0,1,\ldots,n-1$, display (20). The paper concludes from (20) that at least $n$ of the $b$'s and $n$ of the $c$'s are present, and Parseval's identity on the unit circle then bounds the square of the maximum modulus below by the number of nonzero coefficients, at least $2n$.

Dependencies

None beyond Parseval's identity.

Bears on

• Problem 256: the problem asks to estimate $f(n)$, defined there as here. Theorem 3 is a lower bound for $f(n)$; it does not determine the order of $f(n)$ and does not bear on whether $\log f(n)\gg n^c$.