I started to read the following article but I stuck at the beginning:
The authur say that "clearly" Eq. (1.1) holds when ... and when ...
Question: Why Eq. (1.1) holds for each of the two cases mentioned (underlined red and green)?
Simple clear explanation would be much appreciated. My knowledge is Rudin's Principles of Real Analysis snd Hardy's Number Theory book.
For the first "when":
If $1,\alpha$ and $\beta$ are linearly dependent over the rationals then this means that we can write: $$ r=s\alpha+t\beta,$$ for some $r,s,t\in\mathbb{Q}$, where $r,s$ and $t$ are not all $0$. We will assume that neither $\alpha$ nor $\beta$ are rational numbers, since for any $\frac{u}{v}\in\mathbb{Q}$, we have: $$\inf\limits_{q\geq{1}}\left\|q\frac{u}{v}\right\|=\left\|v\frac{u}{v}\right\|=\|u\|=0.$$
In this case $s\neq{}0$, since otherwise this would imply that:
$$r=t\beta,$$
and so, either $\beta\in\mathbb{Q}$ or $t=0$. Similarly, we can conclude that $t\neq{0}$. By rearranging the top equation, dividing through by $t$ and relabelling, we have:
$$\beta={s}'\alpha+r',$$
with $s',r'\in\mathbb{Q}$ and $r'\neq{0}$. To simplify future steps, we will write $r'=\frac{u}{v}$ and $s'=\frac{w}{v}$, where $u,w\in\mathbb{Z}$ and $v\in\mathbb{Z}_{\geq{1}}$.
The first thing to note is that if $y=x+m$ for some $m\in\mathbb{Z}$, then: $$\|y\|=\|x+m\|=\|x\|,$$ by definition of the nearest integer function.
Secondly, if for some $x,y\in\mathbb{R}$ there is an $m\in\mathbb{Z}$ such that $y=mx$, then the distance to the nearest integer function satisfies the following property:
\begin{align} \|y\|=\|mx\| &= \min\left\{|mx-r|: n\in\mathbb{Z}\right\}\\ &= \min\left\{|m|\cdot\left|x-\frac{r}{m}\right|: r\in\mathbb{Z}\right\}\\ &= |m|\cdot\min\left\{\left|x-\frac{r}{m}\right|: r\in\mathbb{Z}\right\}\\ &\leq|m|\cdot\min\left\{\left|x-r\right|:r\in\mathbb{Z}\right\}\\ &=|m|\cdot\|x\| \end{align}
Piecing this together, if there is some $\frac{u}{v},\frac{w}{v}\in\mathbb{Q}$ with $\beta=\frac{u}{v}\alpha+\frac{w}{v}$, we have the following:
\begin{align} vq\cdot\|vq\alpha\|\cdot\|vq\beta\| &=vq\cdot\|vq\alpha\|\cdot\left\|v\left(q\frac{u}{v}\alpha+\frac{w}{v}\right)\right\|\\ &= vq\cdot\|vq\alpha\|\cdot\left\|uq\alpha+w\right\|\\ &= v^2\cdot{q}\cdot\|q\alpha\|\cdot\|uq\alpha\|\\ &\leq{} v^2\cdot|u|\cdot{q}\cdot\|q\alpha\|^2\end{align}
where $C\in\mathbb{N}$ is some constant (and assuming $v\geq{1}$).
Here, we can use Theorem 163 and 171. of Hardy and Wright to see that if $\alpha$ has continued fraction expansion $\overline{\alpha}:=[a_0;a_1,a_2,\ldots]$ and $\frac{p_k}{q_k}$ is the $k^{th}$ convergent of $\alpha$ , then we have:
\begin{align} \|q_k\alpha\|=|q_k\alpha-p_k| &=q_k\cdot\left|\alpha-\frac{p_k}{q_k}\right|\\ &\leq{} q_k\cdot\frac{1}{a_{k+1}q_k^2}\\ &\leq{} \frac{1}{a_{k+1}q_k} \end{align}
Plugging this back into the above equation we get:
\begin{align} C\cdot{q_k}\cdot\|q_k\alpha\|^2 &\leq{} C\cdot{q_k}\cdot\left(\frac{1}{a_{k+1}q_k}\right)^2\\ &\leq\frac{C}{a_{k+1}^2q_k} \end{align}
However, since $\alpha$ is not rational (by assumption), the convergent denominators are unbounded, therefore taking the subsequence $q=nq_k$, we find that:
$$\inf_{k\in\mathbb{N}} nq_k\cdot\|nq_k\alpha\|\cdot|nq_k\beta\| \leq\inf_{k\in\mathbb{N}} \frac{C}{a^2_{k+1}q_k}=0$$ As required.
The second "when" comes from the fact that if $\alpha$ has bounded partial quotients if and only if $\alpha\in\textbf{Bad}$. This result can be found in most introductory textbooks. Since $\textbf{Bad}$ can also be defined as:
$$\textbf{Bad}:=\{\alpha\in\mathbb{R}:\inf_{q\geq{1}} q\cdot\|q\alpha\|>0\},$$
we see that $\alpha\not\in\textbf{Bad}$ implies that:
$$\inf_{q\geq1} q\cdot\|q\alpha\|=0.$$
Since the distance to the nearest integer function is bounded (i.e. $\|x\|\in[0,\frac{1}{2})$), it is easy to see that for any $\beta\in\mathbb{R}$, we have:
$$\inf_{q\geq1} q\cdot\|q\alpha\|\cdot\|q\beta\|=0.$$