# Well distributed sequence uniformly over small intervals

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

• I would also be happy if it holds only for some class of irrational numbers Oct 20 '19 at 9:57
• 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 $\frac1n \sum\ldots > cu$ ? Oct 21 '19 at 2:33
• 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. Oct 21 '19 at 6:34

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