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Statement

Notation (p. 34). For real $y$, $R(y)$ is the reduced value of $y$ modulo $1$: $0\le R(y)<1$ and $y-R(y)$ is an integer. For an interval $J$ and an integer $x>0$, $N(J,x)=N(J,x,\alpha)$ counts the points $R(\alpha),R(2\alpha),\ldots,R(x\alpha)$ that lie in $J$, and $(J)$ is the length of $J$. Intervals are taken modulo $1$, so an interval may contain the point $0$ in its interior (p. 35), and an interval always contains its initial point and never its endpoint (footnote 3, p. 35).

Satz I (p. 35, quoted; the footnote mark after Teilintervall is omitted).

I. Ist $\alpha$ eine beliebige reelle Zahl, $J$ ein Teilintervall des Intervalles $0\ldots1$ von der Länge $R(\nu\alpha)$, wo $\nu\gtrless0$ ganz ist, so ist für alle ganzen $x>0$

$$|N(J,x)-(J)x|<|\nu|. \tag{2}$$

In words: for every real $\alpha$, every nonzero integer $\nu$ and every interval $J$ of length $R(\nu\alpha)$ modulo $1$, in any position, the count of $R(m\alpha)$, $1\le m\le x$, in $J$ differs from $x\,R(\nu\alpha)$ by less than $|\nu|$, for every integer $x>0$. The paper stresses that $\alpha$ need not be irrational, so the theorem also says something about the residues of $pa$ modulo $q$ for integers $a,p,q$ (p. 35). It credits the theorem to the author's 1927 note (Jber. Deutsch. Math.-Verein. 36, p. 179) and calls it a generalization of Hecke's special case, intervals starting at $0$ and irrational $\alpha$ (Abh. Math. Sem. Hamburg 1 (1922), 73--74).

Further statements of Section III (pp. 39--41). The proof yields more than (2).

• The left side of (2) is $0$ for every position of $J$ exactly when $x\alpha$ is an integer (p. 39).
• Statement 1 (p. 40): under arbitrary translations of $J$ the difference $N(J,x)-(J)x$ ranges over an interval of length at most $|\nu|$, whereas (2) alone gives only $2|\nu|$.
• Statement 2 (p. 40): if the difference does not vanish for every position of $J$, it takes both positive and negative values.
• Formulas (6), (6') (p. 40). For $\nu>0$, with $\xi$ the distance of the endpoint of $J$ from $1$,

$$(J)x-N(J,x)=\sum_{n=1}^{\nu}\bigl[R(n\alpha+\xi+x\alpha)-R(n\alpha+\xi)\bigr];$$

for $\nu<0$, with $\xi'$ the distance of the initial point of $J$ from $1$,

$$(J)x-N(J,x)=\sum_{n=1}^{|\nu|}\bigl[R(n\alpha+\xi')-R(n\alpha+\xi'+x\alpha)\bigr].$$

A footnote says the 1927 note proved only (6), and (6') follows by applying (6) to the complementary interval.

• Bounds (7) (p. 40). Each summand of (6) equals $R(x\alpha)$ or $R(x\alpha)-1$, so

$$\nu R(x\alpha)\ge(J)x-N(J,x)\ge\nu\bigl(R(x\alpha)-1\bigr)\quad(\nu>0),$$ $$\nu\bigl(R(x\alpha)-1\bigr)\ge(J)x-N(J,x)\ge\nu R(x\alpha)\quad(\nu<0).$$

The paper notes that (7) gives Statement 1 at once and is an essential sharpening of it, and that for irrational $\alpha$ and fixed $\nu$ there are arbitrarily large $x$ with $N(J,x)=[(J)x]+1$, and others with $N(J,x)=[(J)x]$ (pp. 40--41).

• Reciprocity (6*), (6*') (p. 41). With $S(x)=\sum_{n=1}^{x}R(n\alpha+\xi)$, (6) reads $(J)x-N(J,x)=S(x+\nu)-S(x)-S(\nu)$ for $\nu>0$, which is symmetric in $x$ and $\nu$. Hence $(J)x-N(J,x)=(J^*)\nu-N(J^*,\nu)$, where $J^*$ has length $R(x\alpha)$ and its endpoint at distance $\xi$ from $1$; for $\nu<0$ the same holds with $|\nu|$ in place of $\nu$.

Source. Alexander Ostrowski, Mathematische Miszellen. XVI. Zur Theorie der linearen Diophantischen Approximationen, Jber. Deutsch. Math.-Verein. 39 (1930), 34--46; notation on p. 34, Satz I and equation (2) on p. 35, its proof and Statements 1 and 2 on pp. 39--40, (6), (6') and (7) on p. 40, (6*) on p. 41. The edition read is identified on the source card.

Read depth. Claims checked: the statement, the endpoint convention and the Section III statements 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. 39--40. If the first case of Satz II holds for the length $\zeta=R(\nu\alpha)$, the difference in (2) is $0$. Otherwise Satz II gives intervals $J_+$, $J_-$ of that length with positive and negative difference; choose them with the largest count $N_+$ and the smallest count $N_-$, so that it suffices to show $N_+-N_-\le|\nu|$ (the paper's (5), printed as $N_+-N_-\le\nu$). Slide $J_+$ forward until it reaches $J_-$. When its initial point passes an orbit point $R(k\alpha)$, its endpoint passes $R((k+\nu)\alpha)$ at the same moment, so the count drops only for the $k$ with $\nu+k$ outside $(0,x]$, and there are exactly $|\nu|$ such $k$. A second route to Statement 1 is formula (6), carried over from the 1927 proof.

Dependencies

Satz II of the same paper (p. 36), for the first proof; the 1927 formula (6), reconstructed in the corpus as the 1927 note's equation (3) and its proof, for the second.

Bears on

• Problem 998: for irrational $\alpha$ and $0\le u<v\le1$ with $v-u=R(j\alpha)$ for a nonzero integer $j$, Satz I bounds the discrepancy of $[u,v)$ by $|j|$ at every $n$; this is the sufficiency direction of the problem's corrected Statement, the same bound as the 1927 note's equation (3). It says nothing about the converse, which the problem asks.