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Lemma:

Let $\left \{ v_{1},\cdot \cdot \cdot ,v_{n} \right \}$ be a linearly independent set of vectors in a normed space $V,\left \| \cdot \right \|$. Then, there exists a constant $c>0$ such that for any scalars $\alpha_{1},\cdot \cdot \cdot ,\alpha_{n}$:

$\left \| \alpha^{1}v_{1}+\cdot \cdot \cdot +\alpha^{n}v_{n} \right \|\leq c\left ( \left | \alpha^{1} \right |+\cdot \cdot \cdot + \left |\alpha^{n} \right | \right )$.

If $s=0$, the proof is trivial.

So suppose $s\neq 0$ and define $\beta^{i}:= \frac{\alpha^{i}}{s}$.

With some manipulation we get $\left \| \sum_{i}^{n} \beta^{i}v_{i}\right \| \leq c$

Suppose this statement is false. Then there exists a sequence $\left ( \beta_{m} \right )=\left ( b_{m}^{1}v_{1}+\cdot \cdot \cdot +\beta_{m}^{n}v_{n} \right )$ in $V $ with $\sum_{i=1}^{n}\left | \beta_{m}^{i} \right |=1$ such that $\lim_{m\rightarrow \infty}\left \| \beta_{m} \right \| = 0$

I'll like to know why the limit of $\beta_{m}$ tends to $0$.

Any help is appreciated.

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  • $\begingroup$ We speak of "a proof" and "to prove", not "a prove" or "to proof". $\endgroup$
    – Hans Hüttel
    Nov 2, 2016 at 6:15
  • $\begingroup$ Yes that is correct @Masacroso $\endgroup$
    – Mathematicing
    Nov 2, 2016 at 6:24
  • $\begingroup$ @EwanDelanoy that was what I meant. Note that the superscripts are not exponents but rather are indices. $\endgroup$
    – Mathematicing
    Nov 2, 2016 at 6:25
  • $\begingroup$ The index notation is standard in this context but it can cause some confusion. To solve this the common solution is to write $x^{(n)}$ instead of $x^n$. $\endgroup$
    – Masacroso
    Nov 2, 2016 at 6:27
  • $\begingroup$ I agree with you. Unfortunately, my notes are rather inconsistent in the use of notation. I'd much prefer it if $\left ( x_{n} \right )$ were used as a notation for sequences. In fact, the limit notation in the OP was properly notated by me. The note uses $\left \| \beta_{m} \right \|\rightarrow 0$ $\endgroup$
    – Mathematicing
    Nov 2, 2016 at 6:29

1 Answer 1

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$$||\alpha^1v_1+...+\alpha^nv_n||\le ||\alpha^1v_1||+...+||\alpha^nv_n||=|\alpha^1| ||v_1||+...+|\alpha^n| ||v_n||\le max(||v_i||)(|\alpha^1|+...|\alpha^n|)$$

Any $c\ge max(||v_i||)$ is ok

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