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Lemma: Let $\{x_1, \ldots, x_n \}$ be a linearly independent set of vectors in a normed space $X$ (of any dimension). Then there is a number $c > 0$ such that for every choice of scalars $\alpha_1, \ldots, \alpha_n$, we have $$\Vert \alpha_1 x_1 + \ldots + \alpha_n x_n \Vert \geq c (\lvert\alpha_1\rvert + \ldots + \lvert\alpha_n\rvert).$$

I have understood the proof until the second page where the author writes each sequence $(\beta_j^{(m)})$ is bounded. I don't understand next exactly how $(y_{n, m})$ is a subsequence and how we obtain the subsequence $(y_{n,m})=(y_{n, 1}, y_{n, 2}, \cdots)$ of $(y_m)$

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banana.space
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  • Do you know about singular value decomposition? If you know about SVD and want to see why this is true in a linear-algebra-centric way, it might be helpful to think in terms of the SVD of X. Note that I'm not sure about the connection because on the right hand side, you don't have the norm of $\begin{bmatrix} \alpha_1 \ \alpha_2 \. \. \. \ \alpha_n \end{bmatrix} $ – AnlamK Apr 17 '15 at 12:52
  • Wait I figured something out. For instance, you can easily show why $(2)$ (above) is true by considering the singular values of $X$, where $X$ is the matrix whose columns are $x_k$. $c$ will just be the minimum singular value of $X$. – AnlamK Apr 17 '15 at 13:03
  • Also, this question is related – AnlamK Apr 17 '15 at 13:05
  • I've added my alternative proof to the linked question. – AnlamK Apr 17 '15 at 14:19
  • I posted a different proof here: http://math.stackexchange.com/questions/1043581/lemma-2-4-1-in-erwin-kreyszigs-introductory-functional-analysis-with-applicati/1043748#1043748 – Disintegrating By Parts Apr 18 '15 at 02:05

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Suppose you have got a convergent subsequence of $(\beta_1^{(m)})$, say, $(\beta_1^{(m_i)})$. Now consider the subsequence of $(\beta_2^{(m_i)})$ of $(\beta_2^{(m)})$. Now apply the same procedure to get a convergent subsequence of $(\beta_2^{(m_i)})$, say, $(\beta_2^{(m_{i_j})})$. Now go back to $(\beta_1^{(m)})$ and consider the subsequence $(\beta_1^{(m_{i_j})})$ of $(\beta_1^{(m)})$ . So, now you have subsequences $(\beta_1^{(m_{i_j})})$ and $(\beta_2^{(m_{i_j})})$ (with same indices) of $(\beta_1^{(m)})$ and $(\beta_2^{(m)})$ converging to $\beta_1$ and $\beta_2$ respectively. Repeat this process and you are done.

Urban PENDU
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