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Definition

If $V$ is a finite dimensional vector space then we say that the norm $||\cdot||_1$ and $||\cdot||_2$ are equivalent if and only if there exist two positive constant $m$ and $M$ such that $$ m||v||_1\le||v||_2\le M||v||_2 $$ for any $v\in V$.

Lemma

If $V$ is a finite dimensional vector space then the function $||\cdot||_\infty:V\rightarrow\Bbb R^+_0$ defined through the condition $$ ||v||_\infty:=\max_{i=1,...,n}|v_i| $$ is a norm in $V$.

Theorem

If $V$ is a finite dimensional vector space then all norm in $V$ are equivalent.

Clearly if $||\cdot||$ is a norm in $V$ then by triangle inequality $$ ||v||\le M\cdot||v||_\infty $$ where $M:=\max_{i=1,...,n}||e_i||$ for any basis $\mathcal{B}:=\{e_1,...,e_n\}$.

Unfortunately I can't prove the other inequality. So could someone help me, please?

Antonio Maria Di Mauro
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1 Answers1

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Method 1

Suppose There are no $D>0$ s.t. for all $x\in V$, $\|x\|\geq D\|x\|_\infty $. This mean that $$\forall n\in\mathbb N^*, \exists x_n\in V: \|x_n\|\leq \frac{\|x_n\|_\infty }{n}.$$

Instead of considering $y_n=\frac{x_n}{\|x_n\|_\infty }$, we can suppose WLOG that $\|x_n\|_\infty =1$ for all $n$. Therefore, by Bolzano-Weierstrass, there is a subsequence (still denoted $(x_n)$) that converges to $x$. Since a norm is continuous, you have $$\|x\|=\lim_{n\to \infty }\|x_n\|=0,$$ and thus $x=0$, but on the other hand, $$\|x\|_\infty =\lim_{n\to \infty }\|x_n\|_\infty =1,$$ which is a contradiction.

Method 2

You have that $\|\cdot \|$ is continuous on $$S=\{x\in V\mid \|x\|_\infty =1\},$$ which is compact. Therefore, there are $x_0,x_1\in S$, s.t. for all $x\in S$, $$0<\|x_0\|\leq \|x\|\leq \|x_1\|.$$ Since $$S=\left\{\frac{y}{\|y\|_\infty }\mid y\in V-\{0\}\right\},$$ you get the wished result.

Surb
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  • Sorry, I don't understend why $|x|\infty =\lim{n\to \infty }|x_n|_\infty =1$; then whe without loos of generality we can suppose that $||x_n||=1$ for all $n\in\Bbb N$? Could you explain this to me, please? – Antonio Maria Di Mauro Jul 22 '20 at 10:21
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    because $|x_n|\infty =1$ for all $n$. Therefore (and by continuity of $|\cdot |\infty $) you get $|x|=\lim_{n\to \infty }|x_n|_\infty =1$. – Surb Jul 22 '20 at 10:37
  • Okay, and then why we can suppose without loss of generality that $||x_n||_\infty=1$ for all $n\in\Bbb N$? – Antonio Maria Di Mauro Jul 22 '20 at 10:38
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    Because we have $$|x_n|\leq \frac{1}{n}|x_n|\infty \iff \left|\frac{x_n}{|x_n|\infty }\right|\leq \frac{1}{n}.$$ Thus, either you set $y_n=\frac{x_n}{|x_n|\infty }$ which is s.t. $|y_n|\infty =1$, or you just suppose that $|x_n|_\infty =1$, and do the whole prove with $(x_n)$ instead of $y_n$ (what I did). – Surb Jul 22 '20 at 10:41
  • Okay, finally if we use Bolzano-Weierstrass theorem then the the subsequence $x_n$ is a subsequence of $y_n$, right? Forgive my confusion: unfortunately I'm confused by notation. – Antonio Maria Di Mauro Jul 22 '20 at 10:48
  • not at all. $(x_n)$ is a bounded sequence for the supremum norm (because $|x_n|\infty =1$). Therefore there is a subsequence $(x{n_k})$ that converges for $|\cdot |\infty $. Instead of working with $(x{n_k})$ I just supposed directly that $(x_n)$ converges (but you can rather work with $(x_{n_k})$ if this make thing more clear for you). – Surb Jul 22 '20 at 10:54
  • Okay, I understand: anyway I know that Bolzano-Weierstrass theorem is valid for the space $\Bbb R^n$ but I suppose that it can be generalise to metric spaces, right? – Antonio Maria Di Mauro Jul 22 '20 at 11:03
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    No, only normed vector space of finite dimension (because all normed space $(V,|\cdot |)$ of dimension $n$ is homeomorphic to $(\mathbb R^n,|\cdot |)$ – Surb Jul 22 '20 at 11:07
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    Okay, thansk too much for your assistance! – Antonio Maria Di Mauro Jul 22 '20 at 11:08
  • Perhaps could you help me here? It seems you know the proof of the general Bolzano-Weierstarss theorem. – Antonio Maria Di Mauro Jul 22 '20 at 11:48
  • However I'd like your second proof but I don't understand why the set $S$ is compact: could you explain this to me? – Antonio Maria Di Mauro Jul 22 '20 at 13:37
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    $S$ is compact because it's closed, ($S=|\cdot |\infty ^{-1}({1})$ i.e. the converse image of a closed set by a continuous function) and obviously bounded (since $x\in S\implies |x|\infty \leq1$). And in normed vector space of finite dimension, closed and bounded sets are compact (Heine-Borel-Theorem). – Surb Jul 22 '20 at 14:34
  • Okay. Clearly if $X$ is a normed space then any compact set $S$ is closed and bounded since $X$ is a metric space too. However if $S$ is closed and bounded how prove that it is compact? Is possible to do this without use the homeomorphism between $X$ and $\Bbb R^n$? – Antonio Maria Di Mauro Jul 23 '20 at 09:45