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Let $a$ an irrational number. Can we say that there is $c>0$ such that for all integer $k,n$ and $1>u>0$, $$ \frac {1}{n}\sum_{i=k}^{k+n} 1_{( (ia) \in [0,u])}< cu ? $$ where (x) is the fractional part of a real number $x$.

It might be related to the fact that the sequence $\{an\}$ is "well-distributed": for all interval $I$ in $[0,1]$, $$

\frac 1n \sum_{i=k}^{k+n} 1_{(ia)\in I}\to |I| $$ "uniformly over k", as $n\to \infty$. I think the "uniformly over k" means that for $I$ fixed, the distance to the limit can be bounded only by means of n. The problem for me is that it is not "uniformly over $ I$" (or at least over $I=[0,u]$).

EDIT: I found a partial answer (with a power of u larger than 1 on the RHS) but I'm still open to a solution! I also posted the following related question: Very badly approximable numbers

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  • $\begingroup$ I would also be happy if it holds only for some class of irrational numbers $\endgroup$
    – kaleidoscop
    Oct 20, 2019 at 9:57
  • $\begingroup$ As the question is written, I think $c=1$ works trivially. Do you want to require $c<1$? Maybe you meant the inequality $\frac1n \sum\ldots < cu$ should be reversed to $\frac1n \sum\ldots > cu$ ? $\endgroup$
    – Noam D. Elkies
    Oct 21, 2019 at 2:33
  • $\begingroup$ For a trivial bound with c=1 i would imagine to bound the indicator by 1, in which case 1<u has to be true, and it is obviously false for u<1. I want to prove that the portion of fractional parts of ia smaller than u for i in a set of integers of length n is bounded by u, up to a uniforml constant. $\endgroup$
    – kaleidoscop
    Oct 21, 2019 at 6:34

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This is certainly false for Liouville numbers, and more generally for any $a$ with unbounded digits in the continued fraction expansion. See Theorem 5 in https://en.wikipedia.org/wiki/Continued_fraction

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