Extreme case bounds on Diophantine approximation

Question

I am wondering about the possible best case approximation and worst case approximation of irrational numbers. I think the appropriate formulation is whether there are functions $\hat{b}(q)$ and $\check{b}(q)$ such that for any $\alpha\in [0,1)\setminus \mathbb{Q}$ we have for only finitely many $q's$

$$ \check{b}(q) \leq \Big\vert \alpha-\frac{p}{q} \Big\vert \quad \text{and} \quad \Big\vert \alpha-\frac{p}{q} \Big\vert \leq \hat{b}(q) \quad \text{for some}\;\; p\in \mathbb{Z}. $$

Later edit: If I try to phrase it alternatively, for all irrational $\alpha$ we have functions $\check{b}(\alpha,q),\hat{b}(\alpha,q)$ satisfying for infinitely many $q$'s that $$ \check{b}(\alpha,q) \leq \Big\vert \alpha-\frac{p}{q} \Big\vert \leq \hat{b}(\alpha,q). $$

Given $f:\mathbb{N}\to [1,\infty)$ satisfying that $f(n)\to \infty$, can we find $\alpha$ such that $\hat{b}(\alpha,q)=o\big(\frac{1}{f(q)})$? Can we find $\alpha$ irrational such that $\frac{1}{qf(q)}=O\big(\check{b}(\alpha,q)\big)$?

End of later edit.

By the Piegonhole principle, we know that $\check{b}(q)\geq \frac{1}{2q}$. Also, by Liouville's theorem we can see that $\hat{b}(q)=o(q^{-n})$ for all $n$. Are there more explicit asymptotic bounds ? Can we say something like $\hat{b}(q)=o(e^{-q})$ ? Can we say that $\check{b}(q)=\Theta(\frac{1}{q})$?

I tried to ask this questions on Math Stackexchange, but I had no response. I assume something like this is known and I have not found this in a short search online. I am unused to these notions so I apologize if this is a silly question.

I would appreciate any insights on this.

Later edit: By Oleg Eroshkin's comments, it seems like there is no effective bound on how small $\hat{b}(\alpha,q)$ can be.

Answer 1

If $\alpha$ is a real irrational number, then there are infinitely many coprime integers $p,q$ with $q > 0$ such that

$$\displaystyle \left \lvert \alpha - \frac{p}{q} \right \rvert < \frac{1}{q^2}$$

by Dirichlet's theorem.

One can easily construct a real number such that one can replace the right hand side with a function of $q$ that goes to zero faster than $q^{-2}$ by an arbitrary amount. That is, for any function $f$ with $\lim_{x \rightarrow \infty} f(x) = 0$ we can find a real number $\alpha_f$ such that there exist infinitely many integers $p,q$ with $\gcd(p,q) = 1, q > 0$ satisfying

$$\displaystyle \left \lvert \alpha - \frac{p}{q} \right \rvert < \frac{f(q)}{q^2}.$$

One can do this in at least two ways. First, one can simply choose the sequence of partial quotients in the continued fraction expansion of $\alpha$ to tend to infinity arbitrarily fast. Essentially equivalently, one can choose

$$\displaystyle \alpha = \sum_{n=1}^\infty a_n$$

with the sequence $\{a_n\}_{n \geq 1} \subset \mathbb{R}_{>0}$ tending to zero arbitrarily fast. For example, Liouville's original construction of a transcendental number had the choice $a_n = 10^{-n!}$.

Conversely, the badly approximable numbers are those whose sequence of partial quotients is bounded. The absolute worst approximable number is therefore the unique real number whose sequence of partial quotients consist of only 1's. This is the Golden ratio, and the statement that this is the worst approximable irrational number is due to Hurwitz. In particular, Hurwitz proved that one can improve the constant 1 in the numerator of Dirichlet's theorem by any number greater than $1/\sqrt{5}$, but no smaller.