Erdős Problem 998: bounded remainder intervals
The displayed statement needs an endpoint correction
The current page asks whether bounded discrepancy for $[u,v)$ under an irrational rotation forces
both $u$ and $v$ to belong to the orbit of zero. That literal condition is false. The correct Hecke--Ostrowski--Kesten theorem is \[
\sup_{N\ge1}\left|\sum_{m=1}^{N}1_{[u,v)}(\{m\alpha\})-N(v-u)\right|<\infty
\quad\Longleftrightarrow\quad
v-u\in\mathbb Z\alpha+\mathbb Z.
\tag{1}
\] We prove $1$ and then give an explicit counterexample to the extra endpoint restriction. The necessity is the classical bounded-coboundary/eigenvalue argument; no novelty is claimed. Standard Fourier completeness on the circle is the only analytic background used.
Sufficiency by a telescoping identity
Let $T(x)=x+\alpha$ on $\mathbb R/\mathbb Z$, and first suppose $0<\ell=v-u<1$ and $\ell\equiv k\alpha\pmod1$, $k\in\mathbb Z\setminus\{0\}$. With $g(x)=\{x-u\}$ we have the exact half-open interval identity \[
1_{[u,v)}(x)-\ell=g(T^{-k}x)-g(x).
\] Indeed $\{y-\ell\}-\{y\}$ equals $1-\ell$ for $0\le\{y\}<\ell$ and $-\ell$ otherwise. Summing along $T^m x$ cancels all but two blocks of at most $|k|$ terms, each in $[0,1)$, giving discrepancy at most $|k|$ for every starting point and every $N$. The empty and full intervals have identically zero discrepancy and also satisfy the right side of $1$.
Bounded sums along one orbit give a bounded $L^2$ family
Conversely let $f=1_{[u,v)}-\ell$ and suppose its sums along the orbit of zero are bounded by $C$ for every $N$; an eventual bound can be enlarged to cover the finite initial segment. Every shifted orbit block then has modulus at most $2C$. Fix $N$. The function $S_Nf(x)=\sum_{j=0}^{N-1}f(T^jx)$ is constant on each interval between its finitely many discontinuities. The orbit of zero is dense, so its block bound implies $|S_Nf(x)|\le2C$ away from those finitely many points. In particular \[
\|S_Nf\|_2\le2C\qquad(N\ge1).
\tag{2}
\] This argument is what justifies passing from the one specified starting point to an almost-everywhere statement.
Constructing a transfer function
For $h\ne0$, the $h$-th Fourier coefficient of $S_Nf$ is \[
\widehat f(h)\frac{1-e(Nh\alpha)}{1-e(h\alpha)}.
\] Average Parseval's identity for $1\le N\le M$ and let $M\to\infty$. Since $\alpha$ is irrational, the average of $|1-e(Nh\alpha)|^2$ tends to $2$ for every nonzero $h$. Fatou's lemma for the nonnegative Fourier sum, together with $2$, gives \[
\sum_{h\ne0}\frac{|\widehat f(h)|^2}{|1-e(h\alpha)|^2}\le2C^2.
\] There is therefore a real $L^2$ function $g$ whose nonzero Fourier coefficients are $\widehat f(h)/(1-e(h\alpha))$, and whose mean is zero. The reality follows from conjugate symmetry. Fourier completeness yields \[
f(x)=g(x)-g(Tx)\quad\hbox{for almost every }x.
\tag{3}
\]
Exponentiation forces the interval length
Put $H(x)=e(g(x))$, so $|H|=1$ and $H$ is a nonzero $L^2$ function. Because $f(x)=1_{[u,v)}(x)-\ell$, equation $3$ implies $H(Tx)=e(\ell)H(x)$ almost everywhere. Some Fourier coefficient $\widehat H(h)$ is nonzero. Comparing that coefficient on the two sides gives \[
e(h\alpha)\widehat H(h)=e(\ell)\widehat H(h),
\] so $\ell\equiv h\alpha\pmod1$. This proves necessity and completes $1$.
A counterexample to the literal endpoint condition
Take $\alpha=\sqrt2-1$, \[
u=(1-\alpha)/2=1-\sqrt2/2,\qquad v=(1+\alpha)/2=\sqrt2/2.
\] Then $v-u=\alpha$, and the telescoping proof gives discrepancy at most $1$ for every $N$. Neither endpoint equals $\{k\alpha\}$ for any integer $k$: comparison of rational and $\sqrt2$ coefficients would require respectively $k=-1/2$ or $k=1/2$. Thus the page's endpoint assertion is disproved, while its intended interval-length theorem is proved.
Sources and scope
Sources:
https://www.erdosproblems.com/998; H. Kesten,
On a conjecture of Erdős and Szüsz related to uniform distribution mod 1, Acta Arith. 12 $1966$, 193--212,
https://doi.org/10.4064/aa-12-2-193-212; the bounded-remainder/eigenvalue formulation recorded as Theorems 5--6 in
https://www.numdam.org/item/10.5802/aif.2079.pdf. The full proof above addresses both formulations explicitly, without inheriting the erroneous endpoint condition.
COMPLETION ESTIMATE
COMPLETION: 100%