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Suppose $\{\alpha n\}$ is the fractional part of $\alpha n$. Put $$A_{\alpha}(n) = \#\{\{\alpha k\} < 1/\sqrt{k} : k \leq n\}.$$ If $\alpha$ is irrational, can I find some constant $K$ such that $A_{\alpha}(n) < K \sqrt{n}$ for all $n$? Does the order of $A_{\alpha}(n)$ depend on $\alpha$? Suppose $1/\sqrt{k}$ is replaced by some function $f(k)$. What can I say about the number of $\{\alpha n\}$ less than $f(n)$ as $n$ tends to infinity?

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up vote 5 down vote accepted

The Weyl equidistribution theorem says that for irrational $\alpha$ and sufficiently many $k$, the fractional parts $\{\alpha k\}$ will be equidistributed in the interval $[0,1]$.

To apply this to your particular problem, consider a large interval $k\in[N,2N]$. Because of equidistribution, the numbers $k$ will "hit" the condition $\{\alpha k\} < 1/\sqrt{N}$ roughly $1/\sqrt{N}$ of the time, i.e. there will be roughly $N·1/\sqrt{N} = \sqrt{N}$ numbers $k$ from the interval that fulfill the condition.

Of course, we are actually interested in the condition $\{\alpha k\} < 1/\sqrt{k}$, so we have overestimated things a bit, but it will still work out.

Now, piecing intervals together by choosing $N=2^M$ as a sequence of powers of two, we obtain an estimate along the lines of

$$A_\alpha(n=2^M) \lesssim \sqrt{1} + \sqrt{2^1} + \sqrt{2^2} + .. + \sqrt{2^{M-1}} \leq K \sqrt{2}^M = K\sqrt{n} $$

as desired. You might have to fill in some epsilons and stuff to make the proof precise, but this is the core argument.

In the general case, a similar argument will yield a good bound as long as the function $f(k)$ doesn't vanish too fast. If it does vanish very fast, then it can only get better, though a better bound might be harder to prove.

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Thanks for the answer. What could I read to learn more? – yjj Feb 13 '11 at 23:04
Unfortunately, I don't know a good reference myself. Results of this kind are usually attributed to the topic of diophantine approximation and are often proven with methods from ergodic theory, but the latter topic is so vast that there may well be books on ergodic theory that don't mention Weyl's theorem at all. Maybe the recent Ergodic Theory with a view towards Number Theory is suitable, though. – Greg Graviton Feb 14 '11 at 13:20
Thanks for the references. – yjj Feb 15 '11 at 3:35

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