For an irrational number $\alpha \in \Bbb{R} \setminus \Bbb{Q}$, the Dirichlet approximation theorem states that there are infinitely many irreducible fractions $\frac{p}{q}$ such that $|\alpha – \frac{p}{q}| < \frac{1}{q^2}$.
I would like to know if anything is known about how "sporadic" these "good denominators" $q$ are. Specifically, I would like to know whether the following holds:
Are there $B,C > 1$ such that for every $n \in \Bbb{N}$, there is $q \in \Bbb{N} \cap [B^n, B^{n+1}]$ and $p \in \Bbb{Z}$ such that $\frac{p}{q}$ is irreducible and $|\alpha – \frac{p}{q}| \leq \frac{C}{q^2}$?
In the above question, $B,C$ are allowed to depend on $\alpha$ if need be. I would also be willing to relax the desired estimate to $|\alpha – \frac{p}{q}| \leq \frac{C}{q^{2-\epsilon}}$ for arbitrary $\epsilon \in (0,2)$ and $C,B$ depending on $\alpha$ and $\epsilon$.
Motivation:
The idea here is that for any given $K \geq 1$, I want to be able to choose $q \sim K$ satisfying the estimate $|\alpha – \frac{p}{q}| < \frac{C}{q^2} \sim K^{-2}$. My ultimate goal is to prove a kind of "quantitative equidistribution theorem", i.e., I want to estimate $\sup_{f \in \mathcal{F}} \big|\frac{1}{N} \sum_{\ell=1}^N f(\alpha \ell) – \int_0^1 f(x) \, dx \big| \lesssim N^{-\beta}$ for a certain class of periodic functions $\mathcal{F}$, and the above estimate would be a big help for doing so.
Disclaimer: I have next to no knowledge about number theory, so it might be that my question is somewhat stupid. In any case I am looking forward to learning something.
Best Answer
Technical remark to start: the interval $[X,B^2 X]$ contains two powers of $B$, so one can replace the assumption that the integer $q \in [B^m,B^{m+1}]$ by the weaker condition that for any $X$ there is a $q \in [X,B^2 X]$ satisfying the conditions.
Let's start with the bad news; certainly some $\alpha$ do not have any such approximation, note even the weaker bound:
$$\left| \alpha - \frac{p}{q} \right| < \frac{1}{q^{1 + \epsilon}}$$
for any fixed $\epsilon > 0$. To see this, let
$$\alpha = \sum_{m=1}^{\infty} \frac{1}{10^{m!}}$$
This is Liouville number (transcendental) which has extremely good rational approximations. For example, we can let
$$\frac{p_n}{q_n} = \sum_{m=1}^{n} \frac{1}{10^{m!}}$$
and then
$$\left| \alpha - \frac{p_n}{q_n} \right| \le \frac{2}{q^{n+1}_n} = \frac{2}{q_{n+1}}.$$
The key point, however, is that very good approximations tend to repell other good approximations. The triangle inequality says that
$$\left| \frac{p}{q} - \frac{p_n}{q_n} \right| \le \left| \alpha - \frac{p}{q} \right| + \left| \alpha - \frac{p_n}{q_n} \right| \le \frac{1}{q^{1 + \epsilon}} + \frac{2}{q_{n+1}}.$$
Now choose $X$ such that:
$$3 \cdot 10^{n!/\epsilon} < X < B^2 X < 10^{(n+1)!/(1 + \epsilon)},$$
As long as $\epsilon > 0$, this is possible to do for $n$ large enough. Now if $q \in [X,B^2 X]$ then $q^{1 + \epsilon} < q_{n+1}$, so the error term above is at must $3/q^{1 + \epsilon}$. But then we have:
$$\frac{3}{q^{1 + \epsilon}} \ge \left| \frac{p}{q} - \frac{p_n}{q_n} \right| \ge \frac{1}{q_n q},$$
the last inequality following from using a common denominator and assuming $p/q \ne p_n/q_n$. But this gives:
$$q^{\epsilon} < 3q_n,$$
which is a contradition. The same argument more or less should even work with inequalities of the form
$$\left| \alpha - \frac{p}{q} \right| \le \frac{1}{q f(q)}$$
for any monotonically increasing function $f(q) \rightarrow \infty$, after replacing $\alpha$ by another Liouville number with even more drastic convergents.
Now the second bad news. Suppose instead of $p/q$ such that:
$$\left| \alpha - \frac{p}{q} \right| < \frac{1}{q^2},$$
for all $\alpha$, you only ask about it for almost all $\alpha$. Here I am particular in making the RHS of the form $1/q^2$. Even this is too much to ask. It turns out that the inequality above implies that $p/q$ is a convergent to $\alpha$. But the denominators $q_n$ of the convergent turn out to grow exponentially, and given the continued fraction $\alpha = [a_0,a_1,a_2,a_3, \ldots]$ there's a formula: $$q_n = a_{n} q_{n-1} + q_{n-2} \in [a_n q_{n-1},(a_n + 1) q_{n-1}].$$ Hence requiring that there is a $q \in [B^m,B^{m+1}]$ is would imply that the integers $a_i$ bounded. But it turns out that almost all real numbers have unbounded $a_i$ (by the Gauss-Kuzmin Theorem).
The continued fraction at least tells you roughly what to expect -- by Khinchin's theorem the geometric mean of the $a_n$ converge (almost always) to a fixed constant $K$, which implies that (a.a.) the $q_n$ also grow at most exponentially, which implies that you can find a $B$ such that the desired approximation holds for $q \in [B^n,B^{n+1}]$ for "many" $n$. You could also consider the relaxed inequality with $1/q^{2 - \epsilon}$ and then think about what happens for almost all $\alpha$, although I haven't thought much about that.