The first assertion is obvious because, for $a\in k^{\times}$, $av\in\mathscr{V}$ if and only if $\|v\|\leq|a|^{-1}$.

For $v\in V$, let $a\in k^{\times}$ with $av\in\mathscr{V}$. Then $\|av\|\leq 1$, that is, $\|v\|\leq|a|^{-1}$, and hence $\|v\|\leq\|v\|_{\mathscr{V}}$.

Finally we consider the second inequality, that is, $\|v\|_{\mathscr{V}}\leq|\alpha|\|v\|$ for $v\in V$. Clearly we may assume that $v\not=0$. As $|\alpha|^{-1}<1$, there is $\epsilon>0$ with $|\alpha|^{-1}e^{\epsilon}<1$. By the first assertion, we can choose $b\in k^{\times}$ such that $\|v\|\leq|b|\leq e^{\epsilon}\|v\|_{\mathscr{V}}$. If $\|v\|<|b\alpha^{-1}|$, then

$$\|v\|_{\mathscr{V}}\leq|b||\alpha|^{-1}\leq e^{\epsilon}\|v\|_{\mathscr{V}}|\alpha|^{-1}.$$

Thus $1\leq e^{\epsilon}|\alpha|^{-1}$. This is a contradiction, so that $\|v\|\geq|b\alpha^{-1}|$. Therefore,

$$\|v\|_{\mathscr{V}}\leq|b|\leq|\alpha|\|v\|,$$

as required. ∎