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Let $\alpha \in \mathbb{R}$ be a fixed (positive) number. For each $k \in \mathbb{N}$ we choose $\varepsilon_k >0$ with the property that $\lim_k \varepsilon_k =0$.

If $\alpha \in \mathbb{R} \setminus \mathbb{Q}$, we know that there exist infinitely many $q_k \in \mathbb{Z}$ such that $$ \frac{1}{2} - \varepsilon_k \leq \{q_k \alpha \} \leq \frac{1}{2}, $$ since $\{q \alpha\}$ is dense in $[0,1)$. Here $\{ z \}$ denotes the fractional part of a number $z$, i.e. $\{z\} = z -[z]$. Can we also choose $q_k$ such that $q_k = O(\varepsilon_k^{-1})$? In other words, can we select $q_k$ so that $\limsup_k \varepsilon_k q_k < \infty$?

It seems to me that the answer is not trivial, since we are, roughly speaking, trying to single out a slow sequence of approximation. Maybe this holds only for some particular (kind of) irrational number $\alpha$.

Edit. It seems that there is a positive result if one requires the weaker condition $k/\varepsilon_k \geq q_k$, see here. Of course $q_k$ can be much larger than we need.

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  • $\begingroup$ What does $\lim_k$ mean? $\endgroup$
    – joro
    Commented May 8, 2016 at 11:10
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    $\begingroup$ @joro $\lim_k = \lim_{k \to +\infty}$, the only possible limit as $k \in \mathbb{N}$. $\endgroup$
    – Siminore
    Commented May 8, 2016 at 11:18
  • $\begingroup$ I think this should be true using some form of the pigeonhole principle, like Dirichlet's approximation theorem. $\endgroup$
    – benblumsmith
    Commented May 8, 2016 at 11:29
  • $\begingroup$ @benblumsmith Do you think it is so easy? Take for instance $\varepsilon_k = 2^{-k}$. Then we want $q_k \lesssim 2^k$. I don't see how a pigeonhole principle can be invoked. How can we bound, a priori, the largeness of the integer $q_k$? In some sense there are too few integers below $2^k$... $\endgroup$
    – Siminore
    Commented May 8, 2016 at 11:32

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For each positive integer $N$, let $A_N$ be the set of $\alpha$ such that for every positive integer $k$ there is a positive integer $q_k$ with $q < N \epsilon_k^{-1}$ and $1/2 - \epsilon_k \le \{q_k \alpha\} \le 1/2$.

Note that if $\alpha = b/c$ is rational with $c$ odd, $1/2 - \epsilon_k \le \{q_k \alpha\} \le 1/2$ is impossible unless $2 c \epsilon_k \ge 1$. So $A_N$ is a closed set that contains no rationals with odd denominator, and thus is nowhere dense. By the Baire category theorem, the union of these is a meagre set. In particular, there exist irrationals not in that union. For such $\alpha$, you can't have $q_k = O(\epsilon_k^{-1})$.

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  • $\begingroup$ It is very interesting. I suspected that this could be true only for most irrationals. However I am unable to make a reasonable guess. $\endgroup$
    – Siminore
    Commented May 9, 2016 at 19:15
  • $\begingroup$ Another thing I suspect is that the constant $C$ in the upper bound $q_k \leq C \varepsilon_k^{-1}$ has a role. There seem to be (partial) results with a small $C$. $\endgroup$
    – Siminore
    Commented May 10, 2016 at 8:52

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