TITLE: What is the meaning of the last part of the proof of this theorem?
QUESTION [5 upvotes]: I am trying to understand the last part of the proof of this theorem but I am really confused as to what the author has done. I have an exam in a couple of days and I don't like memorising things without understanding them but just got lost between the meanings of $y_m, y_{1,m}, y_{2,m}, \beta^{(m)}_{1}$ and $\beta^{(m)}_{2}$.
Thanks.
Any help is appreciated.

REPLY [2 votes]: Another way to look at this: The function
$$
        F(\alpha_1,\alpha_2,\cdots,\alpha_n)=\|\alpha_1 x_1+\alpha_2 x_2+\cdots+\alpha_n x_n\|
$$
is a jointly continuous function of the scalars $\alpha_j$. The subset $\mathscr{C}$ defined by
$$
      |\alpha_1|+|\alpha_2|+\cdots+|\alpha_n|=1
$$
is a compact subset of $\mathbb{C}^n$ (or $\mathbb{R}^n$ if you are working over the reals.) And $F$ is never $0$ on this set. Therefore, $F$ achieves a minimum value on $\mathscr{C}$ that is non-zero. Let $c$ be that minimum non-zero value. Then, for any $\alpha_j$ for which $|\alpha_1|+\cdots+|\alpha_n| > 0$,
$$
   F(\frac{\alpha_1}{|\alpha_1|+\cdots+|\alpha_n|},\cdots,\frac{\alpha_n}{|\alpha_1|+\cdots+|\alpha_n|}) \ge c > 0\\
   \implies F(\alpha_1,\cdots,\alpha_n) \ge c(|\alpha_1|+\cdots+|\alpha_n|),
$$
which proves the inequality.