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Statement

Setting (§1, p. 7). For positive integers $a_1\le a_2\le\cdots\le a_n$ put

$$M(a_1,\ldots,a_n)=\max_{\theta}\prod_{k=1}^{n}\bigl|1-\exp(a_ki\theta)\bigr|,$$

the maximum over all real $\theta$, and let $f(n)$ be the greatest lower bound of $M(a_1,\ldots,a_n)$ over all such sets of positive integers. Write $g(n)=\log f(n)$. The paper records that $g$ is subadditive, $g(m+n)\le g(m)+g(n)$, with $g(1)=\log2$, and quotes from Erdős and Szekeres the bounds $g(n)=o(n)$ as $n\to\infty$ (its (3)) and $g(n)\ge\frac12\log(2n)$ (its (4)).

Inequality (5) (p. 7). The aim of the note is to improve (3) to

$$g(n)\le n^{1/2}\Bigl(\tfrac12\log n+4\log2\Bigr).$$

The print states (5) with no restriction on $n$; the deduction on p. 11 applies to every positive integer $n$. Equivalently, $f(n)\le\exp\bigl(n^{1/2}(\frac12\log n+4\log2)\bigr)$.

Triangular case (§5, p. 11). On the way the paper proves, for every positive integer $p$,

$$g\bigl(\tfrac12p(p+1)\bigr)\le\tfrac12(p+1)(\log p+2\log2),$$

obtained from the exponents in which each $k=1,\ldots,p$ occurs $p+1-k$ times.

Source. Inequality (5), stated on p. 7 and proved on p. 11, of F. V. Atkinson, On a problem of Erdős and Szekeres, Canad. Math. Bull. 4 (1961), 7–12, DOI 10.4153/CMB-1961-002-5, as identified on the source card.

Read depth. Claims checked: the setting, (5), the triangular case and the deduction of §5 were read clause by clause on pp. 7–11. Nothing here is independently reviewed.

Proof pointer

§§2 and 5, pp. 8 and 11. Taking logarithms and grouping equal exponents, $g(n)$ is the minimum over $1\le p\le n$ of the greatest lower bound of $N(c_1,\ldots,c_p)$, the maximum over $\theta$ of $\sum_{k=1}^pc_k\log|1-e^{ki\theta}|$, over non-negative integers $c_1,\ldots,c_p$ with sum $n$ (p. 8, (6)). Lemma 2 is applied with $c_0=\frac12(p+1)$ and $c_k=p+1-k$, whose cosine polynomial is a non-negative Fejér-kernel expression, and with $M=p$; this gives the triangular case. For general $n$, write $n=\frac12p(p+1)+q$ with $p$ largest and $0\le q\le p$, use subadditivity and $g(q)\le q\log2$, and bound $\frac12(p+1)\le\sqrt n$ and $q\le p<\sqrt{2n}$.

Dependencies

Lemma 2, which rests on Lemma 1.

Bears on

• Problem 256: the problem asks to estimate $f(n)$ and whether $\log f(n)\gg n^c$ for some constant $c>0$. Inequality (5) gives $\log f(n)\ll n^{1/2}\log n$, so no constant $c>\frac12$ works; it does not decide the question for $0<c\le\frac12$.