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Statement

Notation as on the Satz I page: $R(y)$ is $y$ reduced modulo $1$ into $[0,1)$, intervals are taken modulo $1$, and each contains its initial point and not its endpoint. Fix a real $\alpha$ and consider the points

$$R(\alpha),\ R(2\alpha),\ \ldots,\ R(x\alpha). \tag{3}$$

Satz II (p. 36). Let $\zeta$ be any positive real number and $x$ any positive integer. Then some interval $J_+$ of length $\zeta$ contains, after reduction modulo $1$, at least $\zeta x$ of the numbers (3), and some interval $J_-$ of length $\zeta$ contains at most $\zeta x$ of them.

More precisely (p. 36), for fixed $x$ and $\alpha$ exactly one of the following holds:

1. every interval of length $\zeta$ contains the same number of the numbers (3), namely exactly $\zeta x$;
2. some interval of length $\zeta$ contains more than $\zeta x$ of them, and some contains fewer than $\zeta x$.

The paper adds that the first case can occur only for rational $\zeta$ (p. 36).

When the first case occurs (pp. 38--39). The proof reduces to $0<\zeta<1$ (p. 37). The paper then states and proves that the count is the same for every position of an interval of length $\zeta$ if and only if $\zeta=p/q$ with $(p,q)=1$, $\alpha=r/(qq')$ with $(r,qq')=1$, so that $\alpha$ is rational with denominator divisible by $q$, and $x\alpha$ is an integer; the proof shows that then $x$ is a multiple $sqq'$ of $qq'$, and each interval holds $x\zeta=spq'$ of the points. The print writes this count as $sr=x\zeta$; since $x\zeta=spq'$, the $sr$ appears to be a misprint (an observation of this page).

Context (pp. 36--37). The paper says Satz II, unlike Satz I, carries over to several dimensions, while Satz I, even in Hecke's narrower form, fails there; a footnote on p. 37 gives a two-dimensional example with $\xi=\sqrt2$, $\eta=\sqrt2+\varepsilon/100$ and a rectangle holding more than $42$ of the first $100$ points while its area is below $1/4$.

Source. Alexander Ostrowski, Mathematische Miszellen. XVI. Zur Theorie der linearen Diophantischen Approximationen, Jber. Deutsch. Math.-Verein. 39 (1930), 34--46; Satz II and display (3) on p. 36, its proof in Section II on pp. 37--39. The edition read is identified on the source card.

Read depth. Claims checked: the statement, the dichotomy and the characterization of the first case were read clause by clause on the page images; the proof was followed but not checked step by step. Nothing here is independently reviewed.

Proof pointer

Pp. 37--39. For irrational $\zeta$, choose $\varepsilon>0$ with no point of (3) in $[1-\varepsilon,1)$, and $q$ with $R(q\zeta)>1-\varepsilon$ and $[x\zeta]<\frac xq[q\zeta+1]<[x\zeta+1]$. Laying $q$ intervals of length $\zeta$ end to end from $0$ covers $[0,1)$ exactly $[q\zeta+1]$ times, except on a piece of $[1-\varepsilon,1)$ free of the points, so the average count lies strictly between the consecutive integers $[x\zeta]$ and $[x\zeta+1]$, forcing counts above and below $x\zeta$. For $\zeta=p/q$, if one interval has a count other than $x\zeta$, the $q$ translates from its initial point cover $[0,1)$ exactly $p$ times, so the counts average $x\zeta$ and some count lies on each side. The characterization follows by sliding an interval of length $\zeta$ past an orbit point and comparing multiplicities.

Dependencies

None beyond the definitions.

Bears on

None recorded. Satz II is the step from which Section III of the paper proves Satz I, which bears on Problem 998; on its own it bounds no discrepancy and bears on neither direction of that problem.