reviewed_theorem_4.md.txt

---
name: library/discrepancy/kesten_1966_bounded_remainder/theorem_4
desc: "Kesten Theorem 4: bounded discrepancy depends on interval length; exact domains and transfers."
review_status: unreviewed
created: 2026-09-06T06:28:01Z
updated: 2026-09-06T06:44:03Z
---

***

The new local reconstruction on this page is unreviewed and covers only
irrational anchored necessity: bounded discrepancy for $[0,b)$, $0<b<1$,
forces $b=\{j\xi\}$. It does not cover the arbitrary-translate/Bohl
reduction, rational rotations, or the already accepted Ostrowski sufficiency
proof. The full published statement is retained separately below.

## Statement and source

Harry Kesten, *On a conjecture of Erdős and Szüsz related to uniform distribution mod 1*, Acta Arithmetica **12** (1966), 193–212, Theorem 4 on printed p. 193 (the right-hand leaf of PDF sheet 1). The selected journal header gives 1966; the inherited “1966/67” citation is retained as a bibliographic variant.

For $\xi\in[0,1]$, $0\le a<b\le1$, and $b-a<1$, put

$$
N(M,\xi,a,b)=\#\{m:1\le m\le M,\ a\le\{m\xi\}<b\},
\qquad
R(M,\xi,a,b)=N(M,\xi,a,b)-M(b-a).
$$

For fixed $\xi,a,b$, Theorem 4 states that $R(M,\xi,a,b)$ is bounded as $M$ ranges over the positive integers if and only if

$$
b-a=\{j\xi\}\qquad\text{for some }j\in\mathbb Z.
$$

The source includes rational $\xi$ (its footnote calls that case trivial). The application to [[problems/irrationality/E0998]] concerns irrational $\alpha$, with $\xi=\{\alpha\}$. This theorem constrains the **length**, not the separate endpoints, and permits every non-wrapping translate $[a,b)\subseteq[0,1]$ of the given proper length.

## Exact elementary transfers

For $0<\ell<1$, the condition $\ell=\{j\alpha\}$ is equivalent to $\ell\in\mathbb Z\alpha+\mathbb Z$. One implication follows from $\{j\alpha\}=j\alpha-\lfloor j\alpha\rfloor$. Conversely, if $\ell=j\alpha+k$ with integers $j,k$ and $0<\ell<1$, then the unique representative of $j\alpha$ modulo $1$ in $[0,1)$ is $\ell$. For irrational $\alpha$, the integer $j$ here is nonzero.

The full interval $[0,1)$, omitted by the theorem's strict length restriction, has $N(M)=M$ and $R(M)=0$. Its length $1$ belongs to $\mathbb Z\alpha+\mathbb Z$, but is never a fractional part. It must be treated separately when the theorem is restated using group membership. An empty interval similarly has identically zero discrepancy and lies outside the displayed hypothesis $a<b$.

The imported problem asks for a uniform bound for all sufficiently large $n$. Such a bound is equivalent to boundedness for all positive $n$: if $|R(n)|\le C$ for $n\ge n_0$, replace $C$ by the maximum of $C$ and the finitely many values $|R(1)|,\ldots,|R(n_0-1)|$. The reverse implication is immediate. Thus there is no eventual-versus-all-indices gap in applying the theorem.

For irrational $\alpha$, if both endpoints are orbit points and $0<b-a<1$, subtracting their two representations shows $b-a\in\mathbb Z\alpha+\mathbb Z$. This proves that the endpoint condition is sufficient. Necessity for the endpoints does not follow. A complete counterexample for every irrational $\alpha$ is given in a companion page not part of this release, using the complete short source proof in [[library/irrationality/ostrowski_1927_mathematische_miszellen/equation_3]].

## Anchored necessity: selected subdirection

The proof below reconstructs only the following proper subdirection of
Theorem 4. For irrational $0<\xi<1$ and fixed $0<b<1$, if

$$
R(M)=\#\{1\le k\le M:\{k\xi\}<b\}-Mb
$$

is bounded for all positive integers $M$, then $b=\{j\xi\}$ for some
$j\in\mathbb Z$. Braces denote fractional parts in $[0,1)$.

This is an author-recorded reconstruction, not independently accepted proof
coverage. It excludes arbitrary starting endpoints, the Bohl reduction cited
on p. 205, rational rotations, and the already reconstructed Ostrowski
sufficiency direction. The full translated Theorem 4 remains a named
published interface outside this selected proof.

The selected
[[library/discrepancy/kesten_1966_bounded_remainder/kesten_1966_bounded_remainder.pdf|1966 journal PDF]]
has SHA-256
<removed: the digest of the held PDF, which Git LFS records>.
The proof consumes the definitions on printed pp. 193–194, the needed parts
of Theorem 1 on pp. 196–199, and Section 4 on pp. 204–212. The elementary
inputs and the consumed partition geometry are proved below; no external
continued-fraction theorem is left as an unproved premise.
Labels (A)–(Q) below belong to this reconstruction, not to the source.

### Continued-fraction identities

Apply the continued-fraction algorithm to $\xi$: put
$T_1=1/\xi$, $a_i=\lfloor T_i\rfloor$, and
$T_{i+1}=1/(T_i-a_i)$. Irrationality makes every step defined, with
$a_i\ge1$ and $T_i=a_i+1/T_{i+1}>1$. Set

$$
p_{-1}=1,\quad p_0=0,\quad q_{-1}=0,\quad q_0=1,
\qquad
p_n=a_np_{n-1}+p_{n-2},\quad
q_n=a_nq_{n-1}+q_{n-2}\quad(n\ge1).
$$

Induction in these recurrences gives

$$
p_nq_{n-1}-p_{n-1}q_n=(-1)^{n-1},\qquad
\xi=\frac{p_nT_{n+1}+p_{n-1}}{q_nT_{n+1}+q_{n-1}}.
$$

For the first identity the determinant changes sign at each step. For the
second, the case $n=0$ is $\xi=1/T_1$; substituting
$T_{n+1}=a_{n+1}+1/T_{n+2}$ gives the next case. In particular
$\gcd(p_n,q_n)=1$.

Write

$$
\sigma_n=(-1)^n,\quad A_n=T_{n+1},\quad
Q_n=A_nq_n+q_{n-1},\quad
\delta_n=Q_n^{-1},\quad
\varepsilon_n=q_n\xi-p_n=\sigma_n\delta_n.
$$

The error formula follows by subtracting $p_n/q_n$ in the preceding
fraction. Here $A_n$ and $Q_n$ are Kesten's $a'_{n+1}$ and $q'_{n+1}$,
respectively. Since $A_n=a_{n+1}+1/A_{n+1}$, direct substitution gives

$$
Q_{n+1}=A_{n+1}Q_n,\qquad
\delta_{n+1}=\frac{\delta_n}{A_{n+1}},\qquad
\delta_{n-1}=a_{n+1}\delta_n+\delta_{n+1}.
\tag{A}
$$

For the last identity one can also use the recurrence for
$\varepsilon_n$ and their alternating signs. For $n\ge2$ we have
$0<q_{n-1}<q_n$, $q_{n+2}\ge2q_n$, and, with $a=a_{n+1}$,

$$
q_{n+1}<Q_n<(a+2)q_n.
\tag{B}
$$

Thus $q_n\to\infty$ and $\delta_n\to0$. Telescoping (A) gives

$$
\sum_{j=s}^{\infty}a_{j+1}\delta_j
=\delta_{s-1}+\delta_s\longrightarrow0.
\tag{C}
$$

Only these identities are needed, not existence or uniqueness of a general
Ostrowski expansion of an arbitrary integer.

### The consumed partition geometry, with both parities

Fix $n\ge2$ and abbreviate $q=q_n$, $v=q_{n-1}$, $Q=Q_n$,
$A=A_n$, $\delta=\delta_n$, $\sigma=\sigma_n$. Use the oriented circle
coordinate

$$
y_n(k)=\{\sigma k\xi\}.
$$

For $r=0,\ldots,q-1$, let $\lambda_r$ be the unique integer in $[1,q]$
such that $\sigma p_n\lambda_r\equiv r\pmod q$. The error formula gives

$$
Y_r:=y_n(\lambda_r)=\frac rq+\frac{\lambda_r}{qQ}
\in\left(\frac rq,\frac{r+1}{q}\right).
\tag{D}
$$

Indeed $0<\lambda_r/(qQ)<1/q$ by (B). These are exactly the first $q$
orbit points, in increasing oriented order. Put $Y_q=Y_0+1$ when measuring
the interval that crosses zero.

The determinant identity implies
$\sigma p_nv\equiv-1\pmod q$. Consequently the label of the next point,
including the cyclic last-to-first pair, satisfies

$$
\lambda_{r+1}=
\begin{cases}
\lambda_r-v,&\lambda_r>v,\\
\lambda_r+q-v,&\lambda_r\le v.
\end{cases}
\tag{E}
$$

Here $\lambda_q=\lambda_0$; representatives are always in $[1,q]$.
Subtracting the two instances of (D), using the lift at the cyclic pair,
shows that the interval from $Y_r$ to $Y_{r+1}$ has length

$$
L_r=
\begin{cases}
A\delta,&\lambda_r>v\quad\text{(short)},\\
(A+1)\delta,&\lambda_r\le v\quad\text{(long)}.
\end{cases}
\tag{F}
$$

This proves the needed long/short classification for odd as well as even
$n$: reflection is built into $y_n$, rather than left as an omitted case.
All these lengths are less than $2/q$, and hence tend uniformly to zero.

Now include all indices through $q_{n+1}=aq+v$, where $a=a_{n+1}$.
Every such index has a unique form $\lambda_r+sq$, with

$$
0\le s\le m_r,\qquad
m_r=
\begin{cases}
a-1,&\lambda_r>v,\\
a,&\lambda_r\le v.
\end{cases}
$$

Since $\lambda_r+sq\le q_{n+1}<Q$, the same calculation as (D) yields

$$
y_n(\lambda_r+sq)=Y_r+s\delta
<\frac{r+1}{q}.
\tag{G}
$$

Thus there are no additional points between consecutive displayed points
in a column, or between its last displayed point and $Y_{r+1}$. The column
interval is split into $m_r$ pieces of length $\delta_n$ and a last piece
of length

$$
L_r-m_r\delta_n
=(A_n-a_{n+1}+1)\delta_n
=\delta_n+\delta_{n+1}.
\tag{H}
$$

At level $n+1$ the orientation reverses. Its short length is
$A_{n+1}\delta_{n+1}=\delta_n$, and its long length is
$\delta_n+\delta_{n+1}$. Hence the regular pieces in (G) become short
intervals and the last piece becomes a long interval. This proves all the
refinement information used from Theorem 1; its other general-$N$
statements and corollaries are not required.

### Locating the endpoint and deriving the counting identity

Assume for contradiction that $b$ is not $\{j\xi\}$ for any integer $j$.
In particular no positive orbit point is $b$ or zero. Put

$$
z_n=
\begin{cases}
b,&n\ \text{even},\\
1-b,&n\ \text{odd}.
\end{cases}
$$

Choose $n$ sufficiently large that the partition interval containing $z_n$
does not cross zero; this is possible by (F) and
$\min(b,1-b)>0$. Write its index as $r_n$, its initial label as
$\lambda_n=\lambda_{r_n}$, its length as $L_n$, and

$$
h_n=z_n-Y_{r_n},\qquad 0<h_n<L_n.
$$

The inequalities are strict because its endpoints are orbit points. Define
$d_n$ as the largest integer $d$ with $0\le d\le m_{r_n}$ and
$d\delta_n<h_n$. Call the case $d_n=m_{r_n}$ terminal. If it is not
terminal, then

$$
d_n\delta_n<h_n<(d_n+1)\delta_n.
\tag{I}
$$

The partition refinement gives the exact transition rules. In the
nonterminal case the next interval is short and

$$
\lambda_{n+1}=\lambda_n+(d_n+1)q_n,\qquad
h_{n+1}=(d_n+1)\delta_n-h_n.
\tag{J}
$$

In the terminal case it is long; its initial point in the reversed
orientation is the far endpoint of the old interval. Thus

$$
\lambda_{n+1}=
\begin{cases}
\lambda_n-q_{n-1},&\text{old interval short},\\
\lambda_n+q_n-q_{n-1},&\text{old interval long},
\end{cases}
\qquad h_{n+1}=L_n-h_n.
\tag{K}
$$

These are statements about positive integer labels, not a choice of
fractional-part representatives.

For a nonterminal $d=d_n$, set $t=d+1$ and $M_n=tq_n$. Then
$1\le t\le a_{n+1}$, so $M_n<q_{n+1}$. The first $M_n$ points have exactly
$t$ representatives in each grid cell $(r/q_n,(r+1)/q_n)$, namely those
in (G) with $0\le s<t$. By (I), all $t$ points in cell $r_n$ are below
$z_n$. Moreover

$$
z_n<Y_{r_n}+t\delta_n
=\frac{r_n}{q_n}
 +\frac{\lambda_n+tq_n}{q_nQ_n}
<\frac{r_n+1}{q_n},
$$

where nonterminality ensures $\lambda_n+tq_n\le q_{n+1}<Q_n$.
All earlier cells contribute $t$ points and all later cells contribute
none. Consequently

$$
D_n(M_n):=\sigma_nR(M_n)
=t(r_n+1)-tq_nz_n
=t\left(1-\frac{\lambda_n}{Q_n}-q_nh_n\right).
\tag{L}
$$

For odd $n$ this uses the exact identity

$$
\#\{1\le k\le M:\{-k\xi\}<1-b\}
=M-\#\{1\le k\le M:\{k\xi\}<b\}.
$$

It holds because neither equality $k\xi\in\mathbb Z$ nor
$\{k\xi\}=b$ occurs. Thus (L) includes the sign and endpoint conventions
in the source's even and odd counting formulas.

### Why separated discrepancy blocks add

For every fixed positive integer $M$ there is $\eta(M)>0$ such that moving
each of the first $M$ orbit points by any circle distance less than
$\eta(M)$ leaves its membership in $[0,b)$ unchanged. Take less than the
minimum of their positive circle distances to the two boundary points
$0,b$. This is a finite positive minimum.

Suppose infinitely many indices $n$ have block lengths
$M_n=c_nq_n$, where $c_n$ is an integer with $1\le c_n\le a_{n+1}$, and
$\sigma_nR(M_n)\ge c$ for one fixed $c>0$. One parity contains infinitely
many of them. Choose increasing indices $n_1,n_2,\ldots$ of that parity,
so separated that

$$
\delta_{n_{i+1}-1}+\delta_{n_{i+1}}<\eta(M_{n_i}).
$$

This is possible by (C). For any finite terminal index $u>i$, put
$S_i=\sum_{j=i+1}^uM_{n_j}$. Its rotation differs from an integer by the
signed sum $\sum_{j=i+1}^u c_{n_j}\varepsilon_{n_j}$, whose absolute
value is at most the tail in (C). Hence

$$
N(S_i+M_{n_i},\xi,0,b)-N(S_i,\xi,0,b)
=N(M_{n_i},\xi,0,b).
$$

Use the empty-count convention $N(0,\xi,0,b)=R(0)=0$.
The same equality is trivial for $i=u$, with $S_u=0$. Subtracting
$M_{n_i}b$ and summing in reverse block order gives

$$
R\left(\sum_{i=1}^uM_{n_i}\right)
=\sum_{i=1}^uR(M_{n_i}).
\tag{M}
$$

The common parity makes all summands have the same sign and magnitude at
least $c$. Thus $|R|$ is unbounded. Under our boundedness hypothesis, every
class of blocks with such a uniform positive signed discrepancy is finite.
This proves the accumulation step (4.19)–(4.20) without assuming arbitrary
blocks add, or confusing a fractional part with a small signed error.

### Excluding all nonterminal digits

In this paragraph abbreviate $a=a_{n+1}$, $B=a_{n+2}$, $q=q_n$,
$v=q_{n-1}$, $Q=Q_n$, $A=A_n$, and $t=d_n+1$. Whenever
$d_n\le a-2$ it is nonterminal. From (I), (L) and $\lambda_n\le q$,

$$
D_n(M_n)>
\frac{t}{Q}\bigl((A-d_n-2)q+v\bigr).
\tag{N}
$$

If $d_n\le a-3$, then $a\ge3$ and $1\le t\le a-2$. Using (B) and
$A>a$ gives

$$
D_n(M_n)>
\frac{t(a-1-t)}{a+2}
\ge\frac{a-2}{a+2}\ge\frac15.
$$

The middle inequality follows by minimizing the concave quadratic on the
integer interval $1\le t\le a-2$, whose two endpoint values are $a-2$.
If $d_n=a-2\ge0$ and $B\le6$, then $A-a=1/A_{n+1}>1/7$, so

$$
D_n(M_n)>
\frac{a-1}{7(a+2)}\ge\frac1{28}.
$$

The accumulation argument therefore excludes both cases for all sufficiently
large $n$. After enlarging the starting index, every $n$ satisfies

$$
d_n\ge a_{n+1}-1
\quad\text{or}\quad
d_n=a_{n+1}-2\ge0,\quad a_{n+2}\ge7.
\tag{O}
$$

For any nonterminal digit, (J) and
$h_{n+1}>d_{n+1}\delta_{n+1}$ give a stronger bound than (N):

$$
D_n(M_n)>
\frac{t}{Q}
\left(Q-\lambda_n-tq+
      \frac{d_{n+1}q}{A_{n+1}}\right).
\tag{P}
$$

In the second case of (O), $t=a-1$, $a\ge2$, $B\ge7$, and (O) at
the next index gives $d_{n+1}\ge B-2$. Since
$Q=(a+1/A_{n+1})q+v$ and $\lambda_n\le q$, the parenthesis in (P)
is at least $(B-2)q/A_{n+1}$. Hence

$$
D_n(M_n)>
\frac{a-1}{a+2}\frac{B-2}{B+1}
\ge\frac14\frac58=\frac5{32}.
$$

This class is also finite by (M). We have therefore proved that eventually
$d_n\ge a_{n+1}-1$ at every index, not merely at one parity.

The only remaining nonterminal possibility is $d_n=a-1$ in a long
interval: a short interval has maximal permitted digit $a-1$.
For a long interval $\lambda_n\le v$. At sufficiently large indices,
$d_{n+1}\ge B-1$. Now (P), with $t=a$, gives

$$
D_n(aq)>
\frac{a}{Q}
\left(v-\lambda_n+\frac{(1+d_{n+1})q}{A_{n+1}}\right)
\ge\frac{aqB}{QA_{n+1}}
>\frac{a}{a+2}\frac{B}{B+1}\ge\frac16.
$$

This last class is finite too. This combines the transition and counting
steps of source (4.24)–(4.31) in the common oriented coordinates; it does not
leave the reversed-parity calculation implicit.

Thus eventually every digit is terminal. By (K), a terminal step is
followed by a long interval. At all sufficiently large indices the interval
is consequently long and its terminal digit is

$$
d_n=a_{n+1}.
\tag{Q}
$$

### The terminal label forces an orbit endpoint

In (Q) the long-interval case of (K) applies at every step:

$$
\lambda_{n+1}=\lambda_n+q_n-q_{n-1}.
$$

Therefore $\lambda_n-q_{n-1}=j$ is one fixed integer once $n$ is
sufficiently large. The initial point of the interval containing $b$ in
the original circle is $P_n=\{\lambda_n\xi\}$. Its distance from $b$
tends to zero by (F). Since
$\lambda_n\xi=j\xi+p_{n-1}+\varepsilon_{n-1}$ and
$\varepsilon_{n-1}\to0$, the circle distance from $b-j\xi$ to an integer
is zero. Thus $b-j\xi\in\mathbb Z$, and $0<b<1$ implies
$b=\{j\xi\}$. This contradicts the assumed absence of such an orbit point
and proves the selected anchored necessity direction.

The integer-label invariant is an equivalent presentation of the final
telescoping argument on pp. 211–212. It avoids taking an unjustified
ordinary real limit through a fractional-part discontinuity.

## Current proof coverage and remaining obligations

The anchored necessity argument above is complete as an author-recorded
reconstruction and has review status unreviewed. Its continued-fraction
identities, consumed Theorem 1 geometry, parity transfer, finite-prefix
stability, block accumulation, digit exclusions and final consequence are
written out. The source was visually inspected at the listed pages.
The signed-coordinate presentation and integer-label invariant are local
reorganizations of Kesten's argument, not an author-issued correction or a
claim of a new theorem.

Independent whole-proof review of this exact reconstruction remains due.
No mathematical code, Lean proof, numerical tier, or new resolution of the
endpoint problem is claimed. Theorem 1 outside the consumed geometry, the
paper's Farey and metric results, rational rotations, and the Bohl
arbitrary-translate reduction are not reconstructed here. In particular,
this partial direction must not be described as full local proof coverage
of Theorem 4.

The
[[library/irrationality/ostrowski_1927_mathematische_miszellen/evidence/verify/translated_interval_review|retained independent review and distinct grade]]
cover the previously recorded Theorem 4 statement, exact domains and
elementary transfers, not the new anchored proof. The endpoint disproof
does not depend on the necessity direction. Those completed earlier scopes
and their acceptance are unchanged.