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I think some version of the following definition given in "Diophantine Approximations" by W.M. Schmidt:

Let $\sigma \in (0,1)$ and $x \in \mathbb{R}$. Then $x$ is $\sigma$-improvement for Dirichlet theorem if there is $Q_0$ so that for all $Q > Q_0$ there are $p,q \in \mathbb{Z}$, so that $ 0 < q < \sigma Q$ and $\| x - \frac{p}{q} \| \leq \frac{\sigma}{qQ}$.

I'm having a bit of trouble understanding the following question, making use of the above mentioned definition:

If $x$ is an irrational $\sigma$-improvement for Dirichlet theorem for some $\sigma \in (0,1)$, then $x \in BA$ (there is $c > 0$ so that for all $p,q \in \mathbb{Z}$, $\|x - \frac{p}{q}\| \geq \frac{c}{q^2}$).

I think it's something that has to do with continued fractions, but I'm not sure how to use it here. And even before that, I'm not sure where is the contradiction if I assume the opposite.

And the other direction (kind of):

If $x$ is $\sigma$-improvement for Dirichlet theorem for all $\sigma \in (0,1)$, then $x$ is rational.

Intuitively, this one seems true, as we approximate $x$ as tightly as we want. I'm not sure what theorem makes it true, though.

I'll appreciate any clue.

Pavel
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1 Answers1

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You are saying that this is homework, so I won't develop a complete solution yet, but I think I can set you in the right direction.

First of all, note that the second question is, as you almost wrote, really easy given the first, as being a $\sigma$-improvement of Dirichlet's theorem implies the existence of infinitely many high quality approximations: $$\left|x-\frac{p}{q}\right| \le \frac{\sigma}{q^2}$$ If those exist for all $\sigma\in(0,1)$, then $x$ certainly can't be badly-approximable, but the first part proved that if it's irrational it must be BA, so $x$ can't be irrational altogether.

As for the first part, I think your continued fraction intuition is correct. Recall that a number $x$ is BA if and only if its continued fraction approximation coefficients are bounded. Thus, if you assume $x\notin BA$, there must be infinitely many arbitrarily large coefficients.

If you observe that $q_n=a_nq_{n-1} + q_{n-2}$, where $q_k$ is the denominator of the $k$th continued fraction approximation of $x$ and $a_n$ is the $n$th coefficient, you can see that if $a_n > \frac{1}{\sigma}$, then $q_{n-1}<\sigma q_n$, and then the $\sigma$-improvement property will apply for the previous continued fraction approximation, as it is the best approximation with denominator lower than $q_n$.

I believe that you can then a lower bound on $\left|x-\frac{p_n}{q_n}\right|$ using the condition imposed by the $\sigma$-improvement and the equality $$\left|x-\frac{p_k}{q_k}\right|=\frac{1}{x_{k+1}q_k^2+q_kq_{k-1}}$$ where $x_{n+1}$ denotes the tail $[a_{n+1},a_{n+2},\ldots]$ of the approximation of $x$, for a few values of $k$ around $n$.

Alfonso Fernandez
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  • Thanks, it clears things a bit. Though I still have a hard time understanding a couple of things - $x_{n+1}$ can be as large as we want? It seems like I can't find a constant large enough to bound $\left|x-\frac{p_n}{q_n}\right|$ from below. And one more thing I don't understand - we did get an improvement for some $n$ and the ones below it, but will it hold for every $n$? Even for those where $a_n \leq \frac{1}{\sigma}$? – Pavel Jul 10 '13 at 08:22
  • $x_{n+1}$ will have an upper bound depending on $\sigma$ that is based on the lower bound of $x_n$ with the relation $x_n=\frac{1}{a_{n+1}+x_{n+1}}$. Note that because $x_{n+1}$ is in the denominator, it needs to be small for $\left|x-\frac{p_n}{q_n}\right|$ to be large. – Alfonso Fernandez Jul 10 '13 at 20:31
  • If we indeed manage to prove that CF approximations where $a_n>\frac{1}{\sigma}$ are worse than some $\frac{c}{q^2}$, we can use the following reasoning: Either there are only finitely many $a_n$ larger than $\frac{1}{\sigma}$, in which case the CF representation is bounded and $x$ is BA, or there are infinitely many "bad" CF approximations - which, because CF approximations are the best, means that $x$ is BA. – Alfonso Fernandez Jul 10 '13 at 20:36
  • The last part is what I missed. Thought that I need to find a constant to bound $x_{n+1}$ for all $n$. Thank you very much. – Pavel Jul 11 '13 at 07:13