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Statement

Conventions (p. 29 and p. 31). For real $\alpha$ write $\langle\alpha\rangle=\lvert1-e^{2\pi i\alpha}\rvert$, so that $\langle t\alpha\rangle=\lvert1-z^t\rvert$ for $z=e^{2\pi i\alpha}$. The paper takes $0\le\alpha<1$ throughout and writes $c_1,c_2,\ldots$ for positive absolute constants.

Theorem 1 (pp. 31--32). For every $\varepsilon$ there are $n_0(\varepsilon)$, $A=A(\varepsilon)$ and $B=B(\varepsilon)$ with the following property. Let $n>n_0(\varepsilon)$, and let $\alpha$ be such that there are no integers $p,q$ with $0\le p<q\le A$ and

$$\frac1{Bn}<\Bigl\lvert\alpha-\frac pq\Bigr\rvert\le\frac1{\varepsilon n}. \qquad(10)$$

Then

$$\prod_{t=1}^n\langle t\alpha\rangle<(1+\varepsilon)^n.\qquad(11)$$

The print phrases the exception as every $\alpha$ "which does not satisfy one of the inequalities" (10); the reading above is the one the paper itself gives on p. 32, that (11) holds unless $\alpha$ can be approximated well but not too well by rationals with small denominators. The statement leaves the range of $\varepsilon$ implicit; the proof takes $\varepsilon$ small.

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 conventions on pp. 29 and 31, Lemma 1 on p. 31, Theorem 1 and its proof on pp. 31--32. The edition read is identified on the source card.

Read depth. Claims checked: the statement, its quantifiers and the inequalities (10) and (11) 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 31--32. Lemma 1 (p. 31): if $\alpha=p/q+\theta/q^2$ with $(p,q)=1$ and $\lvert\theta\rvert<1$, then every block of $q$ consecutive factors $\langle t\alpha\rangle$, $l+1\le t\le l+q$, has product less than $q^{c_1}$; the points $e^{2\pi it\alpha}$ in the block sit close to the $q$-th roots of unity shifted by half a step, whose distances from $1$ multiply to $2$. In the proof of the theorem, if $\lvert\alpha-p/q\rvert\ge1/(\varepsilon n)$ for every $q\le A$, Dirichlet's theorem gives a $q\le\varepsilon n$ with $\lvert\alpha-p/q\rvert<1/(q\varepsilon n)$, necessarily $q>A$; splitting $1,\ldots,n$ into blocks of length $q$ and applying Lemma 1 to each gives (11) once $A$ is large. If instead $\alpha$ lies within $1/(Bn)$ of some $p/q$ with $q\le A$, each full block of $q$ consecutive factors has product below $1/2$ for $B$ large, and the whole product is below $1$.

Dependencies

Lemma 1 of the same paper (p. 31), summarized above, and Dirichlet's approximation theorem.

Bears on

• Problem 256: only as the tool for Theorem 2, whose page states the relation. Theorem 1 itself bounds a product with exponents $1,\ldots,n$ at every point of the circle outside the exceptional set, and gives no bound for $f(n)$.