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.
This is definitely a useful trick to understand.
We start by looking at the sequence $(\beta_1^{(m)})_{m \in \Bbb N}$. It has a convergent subsequence, $(\beta_1^{k_m})_{m \in \Bbb N}$. We will equivalently label this sequence as $(\beta_1^{k_{1,m}})_{m \in \Bbb N}$; more on that in a bit. We can define a corresponding subsequence $(y_{1,m})_{m \in \Bbb N}$ by $$ y_{1,m} = \beta_1^{(k_{1,m})}x_1 + \beta_2^{(k_{1,m})}x_2 + \cdots + \beta_n^{(k_{1,m})}x_n $$ From there, we look at the sequence $(\beta^{(k_{1,m})}_2)_{m \in \Bbb N}$ (which doesn't necessarily converge). We can select a subsequence (of this subsequence!) that converges. That is, we look at $(\beta^{(k_{2,m})}_2)_{m \in \Bbb N}$, which satisfies $\lim_{m \to \infty} \beta^{(k_{2,m})}_2 \to \beta_2$. Notice that because we've taken a subsequence, we also have $\beta^{(k_{2,m})}_1 \to \beta_1$. The corresponding subsequence of $y_n$ is $$ y_{2,m} = \beta_1^{(k_{2,m})}x_1 + \beta_2^{(k_{2,m})}x_2 + \cdots + \beta_n^{(k_{2,m})}x_n $$ the process continues until we reach $y_{n,m}$, for which each sequence of coefficients converges.
Note: If we have the Heine-Borel theorem, then we can reach the convergent subsequence $(y_{n,m})_{m \in \Bbb N}$ in one step. In fact, this trick is how one would prove Heine-Borel using Bolzano-Weierstrass.
The first point is that if $(b_i^{(m)})_{m\in \mathbb N}$ is a bounded sequence for each $i=1,...,n$ then there is a strictly increasing $f:\mathbb N\to \mathbb N$ such that $(b_i^{(f(m))})_m$ converges to a limit $b_i$ for each $i=1,...,n.$ ...(In the Lemma, every $|b_i^{(m)}|\leq 1.$ )
The second point is that, in the Lemma, not all $b1,...,b_n$ can be $0$ because $\sum_{i=1}^n|b_i|=\lim_{m\to \infty}\sum_{i=1}^n|b_i^{(f(m))}|=\lim_{m\to \infty} 1=1.$
The third point is that $y_{f(m)}=\sum_{i=1}^nb_i^{(f(m))}x_i$ converges to $y=\sum_{i=1}^nb_ix_i$ as $m\to \infty$ because $$ \|y_{f(m)}-y\|\leq \sum_{i=1}^n|b_i^{(f(m))}-b_i|\cdot \|x_i\|$$ and each $b_i^{(f(m)}\to b_i$ as $m\to \infty.$
The final point is that is that the third point implies $\|y\|=\lim_{m\to \infty}\|y_{f(m)}\|=0,$ so $0=y=\sum_{i=1}^nb_ix_i,$ with $b_1,...,b_n$ not all $0$ (second point), contradicting the linear independence of $x_1,...,x_n.$
NOTES. (1). The first point is a useful general tool. (2). The fact that, in a normed space, $\lim_{n\to \infty}\|v_n-v\|=0$ implies $\lim_{n\to \infty}\|v_n\|=\|v\|,$ while easily proven, is a good little time-saver.
Two consequences of this lemma are: (i). A finite-dimensional vector-subspace of a normed linear space is closed. (ii). Any two norms on $\mathbb R^n$ (or on $\mathbb C^n$), for finite $n$, that make $\mathbb R^n$ (or $\mathbb C^n$) a normed linear space, will generate the same topology, and, regarding the metrics defined by the norms, the norms' metrics are uniformly equivalent.
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.