---

Statement

Setting (p. 29). For positive integers $a_1\le a_2\le\cdots\le a_n$ put

$$M(a_1,\ldots,a_n)=\max_{\lvert z\rvert=1} \Bigl\lvert\prod_{i=1}^n(1-z^{a_i})\Bigr\rvert, \qquad f(n)=\min_{a_1,\ldots,a_n}M(a_1,\ldots,a_n).$$

The exponents need not be distinct.

Theorem 2 (p. 33).

$$\lim_{n\to\infty}f(n)^{1/n}=1.$$

Since $f(n)\ge1$ for $n\ge1$ (by Theorem 3), the theorem says that $\log f(n)=o(n)$. The paper remarks (p. 29), without proof, that a refinement of its method might give $f(n)<\exp(n^{1-c})$ for some $c<1$, and it says the determination of $f(n)$ seems to be a very difficult question.

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 2 on p. 33, its proof on pp. 33--34. The edition read is identified on the source card.

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

Proof pointer

Pages 33--34. For $m^2\le n<(m+1)^2$ the proof takes $n-m^2$ exponents equal to $1$ and the $m^2$ exponents $2^kl$, $1\le k\le m$, $1\le l\le m$. The factor $(1-z)^{n-m^2}$ is at most $2^{2\sqrt n}$ in modulus, so it suffices that the remaining product is at most $(1+2\varepsilon)^{m^2}$ on the circle for $m$ large, display (17). Writing $z=e^{2\pi i\alpha}$, for a fixed $q\le A$ the points $2^k\alpha$, $1\le k\le m$, can satisfy the exceptional inequalities (10) of Theorem 1 (with $m$ in place of $n$) only for $o(m)$ values of $k$, because doubling moves the error out of the window after boundedly many steps. Those $k$ contribute at most $2^{o(m^2)}$, display (18), and Theorem 1 bounds each of the other inner products over $l$ by $(1+\varepsilon)^m$, display (19).

Dependencies

Theorem 1 of the same paper.

Bears on

• Problem 256: the problem asks to estimate $f(n)$, defined there as here, and whether $\log f(n)\gg n^c$ for some $c>0$. Theorem 2 gives the upper estimate $\log f(n)=o(n)$. It does not determine the order of $f(n)$, and it does not settle whether $\log f(n)\gg n^c$; the sharper bound $f(n)<\exp(n^{1-c})$ is only suggested in the paper, not proved.