Given linearly independent vectors $w_1,w_2\in\mathbb{R}^2$, $\exists\, k > 0$ such that $|m w_1 + n w_2| \geq k(|m|+|n|)\, \forall$ integers $m,n$ I am unable to see the correctness of this statement. It seems the author has considered this statement trivial and hence has not given any proof of this statement. But I am unable to prove it.
 A: The mapping $Ax = x_1w_1+x_2 w_2$ is linear and injective. Hence
$k=\min_{\|x||=1} \|Ax\| >0$ (for any norm $\|\cdot\|$), hence using the $\|\cdot\|_1$ norm we have
$\|Ax \|_1 \ge k \|x\|_1$, for all $x$.
In particular, we have
$\|m w_1 + n w_2\|_1 \ge k (|m|+|n|)$ (for all $m,n \in \mathbb{R}$, not just integers).
Since all norms on $\mathbb{R}^2$ are equivalent, you can replace the norm on
the left by whatever norm you want.
A: By changing $\omega_1$ to $-\omega_1$ we can reduce this to the case when $m$ and $n$ have the same sign. We may as well assume that $m$ and $n$ are non-negative integers.
Let $f(t)=\frac {|t+1|} {|t\omega_1+\omega_2|}$ defined on $\mathbb R$. This is a ratio of two continuous functions and the denominator never vanishes. Hence, $f$ is continuous. I will let you check that $|f(t)| \to \frac 1 {|\omega_1|}$ as $|t| \to \infty$. [$\omega_ \neq 0$ by hypthesis. It follows that $f$ is a  bounded function. If $f(t) \leq M$ for all $t$ we get $|t+1|\leq |t\omega_1+\omega_2|$ and putting $t=m/n$ and multiplying by $n$ finishes the proof.
