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Source. Alexander Ostrowski, Mathematische Miszellen. IX. Notiz zur Theorie der Diophantischen Approximationen, Jahresbericht der Deutschen Mathematiker-Vereinigung 36 (1927), 178--180. The selected theorem is equation (3), printed p. 179; its complete short proof runs through equation (4) on printed p. 180.

Selected translated-interval theorem

Write $\{t\}=t-\lfloor t\rfloor$. Let $\alpha\in\mathbb R$ and $j\in\mathbb Z\setminus\{0\}$, and suppose $\beta=\{j\alpha\}\in(0,1)$. For any $u\in\mathbb R$, let $J$ be the circle interval represented by $[u,u+\beta)$ modulo $1$, including its initial endpoint and excluding its final endpoint. Set

$$N_J(n)=\#\{1\le m\le n:\{m\alpha\}\in J\}.$$

Then, for every positive integer $n$,

$$|N_J(n)-n\beta|<|j|. \tag{O3}$$

This is the nonempty proper-interval case of the source's (3), retaining arbitrary translation and the source's integer sampling parameter. The source allows real $\alpha$; irrationality is not needed for this bound. A wrapped circle interval may appear as two intervals in $[0,1)$.

Complete source proof in modern notation

Suppose first that $j>0$, and put $\xi=1-u-\beta$. For any real $t$, membership of $\{t\}$ in $J$ is equivalent to

$$\{t+\xi\}\in[1-\beta,1).$$

Indeed, adding $1-u-\beta$ sends the half-open arc from $u$ to $u+\beta$ to the half-open arc from $1-\beta$ to $1$ modulo $1$. The chosen endpoint convention is preserved, including points exactly at an endpoint.

Since $j\alpha$ differs from $\beta$ by an integer, if $r=\{t+\xi\}$ then

$$\{t+j\alpha+\xi\}-\{t+\xi\} =\{r+\beta\}-r =\begin{cases} \beta,&0\le r<1-\beta,\\ \beta-1,&1-\beta\le r<1. \end{cases}$$

Summing this identity for $t=m\alpha$, $1\le m\le n$, gives

$$\begin{aligned} S_\xi(n) &=\sum_{m=1}^{n} \bigl(\{(m+j)\alpha+\xi\}-\{m\alpha+\xi\}\bigr)\\ &=n\beta-N_J(n). \tag{O4} \end{aligned}$$

Reindexing finite sums also gives

$$S_\xi(n)=\sum_{h=1}^{j} \bigl(\{(n+h)\alpha+\xi\}-\{h\alpha+\xi\}\bigr).$$

For completeness, if $A(k)=\sum_{m=1}^{k}\{m\alpha+\xi\}$, each side equals $A(n+j)-A(n)-A(j)$. Thus this equality also holds when $n<j$. Each of its $j$ summands has absolute value strictly below $1$, because both fractional parts belong to $[0,1)$. The triangle inequality gives $|S_\xi(n)|<j$, proving (O3) for $j>0$.

If $j<0$, the complementary circle interval $J^c$ has length $1-\beta=\{-j\alpha\}$ and positive index $-j$. Its half-open convention partitions the circle with $J$, so $N_{J^c}(n)=n-N_J(n)$. Hence

$$N_{J^c}(n)-n(1-\beta)=-(N_J(n)-n\beta).$$

Applying the positive-index case to $J^c$ proves the same strict bound $|j|$. This is the source's complement reduction for the negative index. Every finite-sum and endpoint step needed for the selected result is now included. The proof uses no external equidistribution or Kesten theorem.

The zero-length case is outside the displayed selected statement; if $\{j\alpha\}=0$ with $j\ne0$ and $J$ is interpreted as the empty arc, both discrepancy terms vanish and the same bound is immediate. No $j=0$ strict bound is asserted.

Relation to the endpoint question

The source first discusses intervals whose endpoints are rotation-orbit points and then explicitly extends the bound to arbitrary translations. Kesten's Theorem 4 supplies the separate necessity of the length condition; its proof is not included here.

The complete selected source proof survived whole-proof review by the independent source reviewer. A distinct grader passed the report contract and independence; the [[irrationality/ostrowski_1927_mathematische_miszellen/evidence/verify/translated_interval_review|retained review and grade]] identify the exact frozen subject and its unchanged current mathematics. This coverage grants no full Kesten proof, formal verification or community-acceptance finding.

Bears on. Problem 998: (O3) is the direction of the length criterion opposite to the one the corrected statement asks for, giving bounded discrepancy to every translate of an interval of length $\{j\alpha\}$, $j\ne0$. It does not decide the corrected statement, whose direction is Kesten's necessity.