Bad approximation, alternative definiton

Question

I read Complex dynamics of Carleson and Gamelin. They state without a proof the following

$\theta$ is called bad approximate if there exists $c>0$ and $\mu<\infty$ such that $$ \Big\vert \theta - \frac{p}{q}\Big \vert > \frac{c}{q^{\mu}}$$ for all integer $p$ and $q$ ( $ q \neq 0$). This is supposedly only occures if and only if:

$\lambda = e^{2\pi i \theta}$ satisfying $$ \Big \vert \lambda^n -1 \Big \vert > c n^{1-\mu} , \quad \forall n\geq 1$$ For some $c>0$ and $\mu<\infty$

I can not se any way to prove this statement, grateful for help.

Answer 1

Not quite sure if this is along the lines of what you're looking for, but I thought I'd give it a shot anyway. We have the exponential map \begin{align*} \exp : \mathbb{R} / \mathbb{Z} &\longrightarrow \mathbb{T},\\ \left[x \right] &\longmapsto e^{2 \pi i x} \end{align*} from $\mathbb{R}/\mathbb{Z}$ to the (complex) unit circle. The topology on $\mathbb{R}/\mathbb{Z}$ is induced by the following metric: For $\left[ x \right], \left[ y \right] \in \mathbb{R}/\mathbb{Z}$, \begin{align*} d_1 ( \left[ x \right], \left[ y \right] ) = \min_{k \in \mathbb{Z}} |x-k\cdot y|. \end{align*} The topology on $\mathbb{T}$ is induced by the metric \begin{align*} d_2 ( e^{2 \pi i x}, e^{2 \pi i y} ) = \text{length of the shortest path on $\mathbb{T}$ from $e^{2 \pi i x}$ to $e^{2 \pi i y}$}. \end{align*} These metrics are related in the following way: $2 \pi \cdot d_1 (\left[x \right], \left[ y \right]) = d_2 (\exp (x), \exp (y))$.

Now, if there is $c > 0$ and $\mu \geqslant 1$ such that $| \theta - p/q | > c/q^{\mu}$ for all $p/q \in \mathbb{Q}$, then for any fixed integer $p$, $|n \theta - p | > c/n^{\mu - 1}$ for all $n \geqslant 1$. This implies that \begin{align*} d_1 (\left[ p \right], \left[n \theta \right]) > \frac{c}{n^{\mu - 1}}, \end{align*} which implies that \begin{align*} d_2 ( \exp (p), \exp(n \theta)) = d_2 (1, e^{2 \pi i n \theta}) > \frac{2 \pi c}{n^{\mu - 1}}. \end{align*} So the shortest arc from $1$ to $e^{2 \pi i n \theta}$ has length at least $2 \pi c / n^{\mu - 1}$. But since \begin{align*} \text{chordal length} \geqslant \frac{2}{\pi} \cdot \text{arc length}, \end{align*} we have \begin{align*} |e^{2 \pi i n \theta} - 1| > \frac{2}{\pi} \cdot \frac{2 \pi c}{n^{\mu - 1}} = \frac{4c}{n^{\mu - 1}}. \end{align*} It is also possible to prove the other direction this way, although possibly with a smaller $c$. But that shouldn't matter since being badly approximable just means that there is some $c > 0$ such that the relevant inequality is satisfied.

Answer 2

I have thought a bit and think i have solved one direction myself. The proof is based on proving following inequality.

$$ \vert \lambda^n -1 \vert=\vert e^{2\pi i n \theta}-1\vert < 2 \pi \vert \theta n -p\vert,\quad \forall p,n \in \mathbb{Z}$$.

We have

$$ \begin{align} \vert e^{2 \pi i n \theta}-1 \vert& = \vert e^{\pi i n\theta}\vert \vert e^{2 \pi i n \theta}-1 \vert\\ & = \vert e^{ \pi i n \theta}- e^{-\pi i n \theta} \vert \\ & =2 \vert \frac{e^{ \pi i n \theta}- e^{-\pi i n \theta}}{2i} \vert \\ & = 2 \vert\sin{\pi n\theta}\vert \\ & = \{\vert \sin{x}\vert \text{ is } \pi\text{-periodic}\}\\ & = 2 \vert \sin{\pi n \theta - \pi p}\vert\\ & = \{ \text{ Use that } \vert \sin{x}\vert \leq \vert x \vert \quad \forall x \in \mathbb{R} \}\\ & = 2\vert \pi (n\theta -p)\vert\\ & = 2\pi\vert n \theta -p\vert \end{align} $$

So given this inequality we can first prove the first direction. Assume $\theta$ is not bad approximal. Then for all $c>0$ and $ \mu < \infty$ we have some $p\in \mathbb{Z},q\in \mathbb{Z}^*$ such that

$$\vert \theta - \frac{p}{q}\vert < \frac{c}{q^{\mu}}$$

But then we have that

$$ \begin{align} \vert \lambda^n-1 \vert & < 2 \pi \vert n \theta -p\vert\\ & = 2\pi n\vert \theta -\frac{p}{n}\vert\\ & < 2 \pi n \frac{c}{n^{\mu }}\\ & = 2 \pi cn^{1-\mu} \end{align} $$

For the converse i can not see if one can apply the same metod.