# Proof of the lemma used in proving that a finite-dimensional normed space is complete

I'm trying to understand the proof for the lemma:

$$\|\alpha _1 e_1 + \alpha _2 e_2 + \cdots + \alpha_n e_n\| \geq c (|\alpha_1|+|\alpha_2|+\cdots+|\alpha_n|)$$ where $c>0$ and the $e_i$s are finite and linearly independent.

The proof I'm referring to is found in many places, like Kreyszig’s book and on page 16 of this PDF file; and in other locations.

The proof is based on proving that no sequence of the form $(\alpha_1 e_1 + \alpha _2 e_2 + \cdots + \alpha _n e_n)$ could converge to a sequence with zero norm.

What I don't quite get is the following:

1. Why the proof is not simply stated as: If the lemma is not true, then $\|\alpha _1 e_1 + \cdots + \alpha_n e_n\| = 0$ which is not possible since the $e_i$s are independent. I mean, why do they involve the sequence convergence business here?

2. Even when they assume that a sequence has to converge to a zero normed sequence, why does it have to be with the constraint that $|\alpha_1^{(m)}|+|\alpha_2^{(m)}|+\cdots+|\alpha_n^{(m)}| = 1$? Shouldn’t we consider that any possible sequence converges to zero normed sequence without the restriction that the sum should always be one?

Thanks a lot!

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I tried to clean this up slightly. Let me know if I goofed on anything. – Dylan Moreland Jul 12 '12 at 19:54
Thanks Dylan, looks better now. – Khalid Salman Jul 13 '12 at 18:21

1- The lemma they are trying to prove is that there exists some $c>0$ such that, for any $\alpha_1,\ldots,\alpha_n$, we have $\|\alpha_1e_1+\cdots+\alpha_ne_n\|\geq c(|\alpha_1|+\cdots+|\alpha_n|)$. The key point here is that $c$ does not depend on $\alpha_1,\ldots,\alpha_n$. What you propose (showing that the norm is not $0$) only allows us to say we can pick some $c>0$ dependent on $\alpha_1,\ldots,\alpha_n$, which is a weaker statement.

2- It suffices to consider the case $|\alpha_1|+\cdots+|\alpha_n|=1$ to prove that if $|\alpha_1|+\cdots+|\alpha_n|=1$, then $\|\alpha_1e_1+\cdots+\alpha_ne_n\|\geq c$. The full lemma can then be proved by dividing both sides by $|\alpha_1|+\cdots+|\alpha_n|$.

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Thanks Alex. I should have mentioned that I'm an engineer, and therefore can't grasp such mathematical subtleties easily :). Based on your comment, how does the proof show the existence of c based on the basis but not based on the chosen alphas? I mean the proof only shows that convergence to zero vector is not possible and does not mention c itself anywhere. Can you please help me understand? – Khalid Salman Jul 13 '12 at 18:33
@KhalidSalman Suppose no such $c$ existed. Then for any $c>0$, we have some term (which I'll call $x_n$) such that $\|x_n\|<c$. Thus we have some $x_{n_1}$ such that $\|x_{n_1}\|<1/1$, some $x_{n_2}$ such that $\|x_{n_2}\|<1/2$, etc. The sequence $x_{n_k}$ then converges to $0$ as $k\to \infty$. – Alex Becker Jul 13 '12 at 20:55
Thanks Alex, all is clear now. – Khalid Salman Jul 16 '12 at 17:43

The lemma states that there exists $c\in\mathbb{R},c>0$ such that $\|\alpha_1e_1+\cdots+\alpha_ne_n\|\ge c|\alpha_1+\cdots+\alpha_n|$. The negation is: "for each $c\in\mathbb{R},c>0$, there exists $\alpha_1,\cdots,\alpha_n$ such that $\|\alpha_1e_1+\cdots+\alpha_n\|<c|\alpha_1+\cdots+\alpha_n|$". Maybe this example clarify a little: consider $l_2(\mathbb{N})$. Then $M=\left\{\frac{1}{n}e_n:n\in\mathbb{N}\right\}$ form a linearly independent set, but for every $c>0$, you can find $m\in\mathbb{N}$ such that $1/m<c$, so, $\left\|1\cdot \frac{1}{m}e_m\right\|_2=\frac{1}{m}<c|1|$. The problem here is that $M$ is infinite.

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Thanks Yuki, I believe you and Alex are in agreement that this proof is 'all about the c value' which I still don't quite get as in my comment to Alex. Your example is very useful BTW. – Khalid Salman Jul 13 '12 at 18:36