---

Statement

Printed p. 137: "Suppose $x_0=0,x_1,x_2,\ldots$ is a strictly increasing sequence of reals with $x_n\to\infty$ as $n\to\infty$ and with (1) $\lim_{n\to\infty}x_{n+1}/x_n=1$. For any $\lambda$ in $0<\lambda\le1$ let $M(\lambda)$ be the set of numbers $y$ which lie in one of the intervals (2) $x_n\le y<x_n+\lambda(x_{n+1}-x_n)$ ($n=0,1,2,\ldots$). Let $\alpha$ be a positive number, and $F(N,\lambda)$ the number of positive integers $k\le N$ for which $k\alpha$ lies in $M(\lambda)$. If (3) $\lim_{N\to\infty}F(N,\lambda)/N=\lambda$ for every $\lambda$, we say that the sequence (4) $\alpha,2\alpha,3\alpha,\ldots$ is uniformly distributed relative to the sequence $x_n$. This concept is due to LeVeque [3]. Davenport and Erdős [1] conjectured that the sequence (4) is uniformly distributed relative to the sequence $x_n$ for almost all $\alpha>0$." The test function (5) is $f(x)=1$ if $x\in M(1/2)$ and $f(x)=-1$ otherwise: "If the conjecture were true, we would have that $\sum_{n=1}^Nf(\alpha n)=o(N)$ for almost every $\alpha>0$." As printed on p. 137:

Theorem 1. "There is a function $f(x)$ of the type considered above such that

$$\limsup_{N\to\infty}\Bigl|N^{-1}\sum_{n=1}^Nf(\alpha n)\Bigr|=1 \tag{6}$$

for almost every $\alpha>0$."

That is, there is a sequence $x_0=0<x_1<\cdots$ with $x_n\to\infty$ and $x_{n+1}/x_n\to1$ relative to which (4) is not uniformly distributed for almost every $\alpha>0$. The sequence the proof constructs (pp. 140--141) is real, not integral: its gaps satisfy $x_{n+1}-x_n\le1/k$ in the $k$-th block (p. 141), so they tend to zero.

Source. W. M. Schmidt, Disproof of some conjectures on Diophantine approximations, Studia Sci. Math. Hungar. 4 (1969), 137--144 (received April 2, 1968; the paper's own running header misprints "3 (1968)"); Theorem 1 on printed p. 137, which is physical p. 139 of the repository's 488-page scan of the whole volume (printed p. $n$ is physical p. $n+2$ for this paper), the proof on printed pp. 138--141 (physical pp. 140--143), read on the rendered page images (the scan's OCR text layer garbles the formulas). The edition read is identified in the source digest.

Read depth. Claims checked: the definitions (1)--(5), the account of the earlier positive results and Theorem 1 were read clause by clause on the page image of p. 137; Lemma 1 (pp. 138--139) was read as a statement; the proof of Theorem 1 (pp. 139--141) was read for its structure and not checked.

Proof pointer

Lemma 1 (pp. 138--140): for $N>1$ and $\varepsilon>0$ there is a subdivision $0=x_0<x_1<\cdots<x_h=1$ of the unit interval with $x_{i+1}-x_i<\varepsilon$ and a set $\sigma_\varepsilon$ of measure less than $\varepsilon$ such that the test function (5) of this subdivision satisfies (11) $f(x)=f(mx)$ whenever $x,mx$ lie in the unit interval, $x\notin\sigma_\varepsilon$ and $m$ is an integer with $1\le m\le N$; proved with Dirichlet's theorem on simultaneous approximation of $\log m$, $1\le m\le N$, by $p_m/q$ (12), the auxiliary function $f^*$ with breakpoints $e^{t/q}$ and the subdivision (14) $x_1=e^{-q}$, $x_2=e^{-q+1/q}$, ..., $x_{q^2+1}=1$. Section 3 (pp. 140--141): integers $N_1<N_2<\cdots$ with (15) $N_k\ge2k^2N_{k-1}$ and $\varepsilon_k<\min(1/(2k^2),1/(kN_k))$ (16); $f$ is built block by block, $f(x)=f_k(x/N_k)$ on $M_k\le x<N_k$ (18), with the single subdivision interval $[N_{k-1},M_k)$ as a bridge (17), $f=1$ on its lower half and $-1$ on its upper half, where $f_k$ is the function of Lemma 1 with $N=N_k$, $\varepsilon=\varepsilon_k$; the resulting subdivision has $x_{n+1}-x_n\le\varepsilon_kN_k\le1/k$, so (1) holds; for $\alpha\in[b^{-1},b]$ outside a set of measure $\le\varepsilon_kN_k$ the values $f(m\alpha)$ for $N_k/k\le m<N_k/b$ (19) are all equal, whence $\bigl|\sum_{m\le N_k/b}f(m\alpha)\bigr|\ge(N_k/b)(1+O(1/k))$, and (6) follows for almost all $\alpha$ in $[b^{-1},b]$. Not reconstructed here.

Dependencies

Lemma 1 (pp. 138--139), which rests on Dirichlet's theorem on simultaneous approximation; otherwise self-contained.

Bears on

• Problem 492: the disproof the site cites. With $a_i=x_i$ for $i\ge1$ (so that $[a_i,a_{i+1})=[x_i,x_{i+1})$), the problem's $f(x)=(x-a_i)/(a_{i+1}-a_i)$ on $[a_i,a_{i+1})$ is the fractional position within the interval, so on $x\ge a_1$ the set $M(1/2)$ is where the problem's $f<1/2$ and Schmidt's test function is $f_{\mathrm{Sch}}=2\cdot\mathbf 1[f<1/2]-1$; if $(f(\alpha n))_{n\ge1}$ were uniformly distributed in $[0,1)$ the proportion of $n\le N$ with $f(\alpha n)<1/2$ would tend to $1/2$ and $N^{-1}\sum_{n\le N}f_{\mathrm{Sch}}(\alpha n)\to0$, against (6) (an authored translation; Schmidt's subdivision also has the interval $[x_0,x_1)=[0,a_1)$, which holds only the finitely many $\alpha n<a_1$ for fixed $\alpha>0$). The theorem answers no for real sequences, the form in which LeVeque, Davenport and Erdős posed the question and the problem page's corrected Statement; it says nothing about the site's restriction to $A\subseteq\mathbb N$, since the constructed gaps tend to zero.