Any two norms on finite dimensional space are equivalent

Question

Any two norms on a finite dimensional linear space are equivalent.

Suppose not, and that $||\cdot||$ is a norm such that for any other norm $||\cdot||'$ and any constant $C$, $C||x||'<||x||$ for all $x$. Define $||\cdot||''=\sum |x_i|\cdot||e_i||$ (*). This is a norm and attains at least as large values as $||\cdot||$ for all $x$.

Could this be used as part of a proof? That two norms $||\cdot||_1,||\cdot||_2$ are equivalent means that there are $m,M$ such that $m||x||_1 \leq ||x||_2 \leq M||x||_1$ for all $x$. In the above I only say that there cannot be a norm such that there does NOT exist an $M$ such that $||x||\leq M||x||'$ for any other norm $||\cdot||'$, but I'm really not sure that proves the entire assertion.

If this is utter nonsense, some hints would be appreciated, thank you.

(*The $e_i$ form a base and sloppily assumed the space to be real.)

Normally you prove this directly by proving everything is equivalent to the Euclidean norm. — Tim Seguine, Dec 09 '13 at 14:09
The negation of two norms being equivalent in the above is wrong. You can not claim to have a constant C... — Sergio Parreiras, Dec 09 '13 at 14:20

score 12 · Answer 1 · edited Feb 26 '18 at 02:03

12

Note that it is sufficient to prove that any norm $\|\cdot\|$ is equivalent to the sup norm $\|\cdot\|_\infty$.

First write $a=\sum_{i=1}^n a_ie_i$ as a linear combination of basis elements of the $n$-dimensional vector space, say $K$. Then $\| a\|=\left\|\sum_{i=1}^n a_i e_i\right\|$ and use triangle inequality to get $c$ such that $\| a\|\leq c\|a\|_\infty.$

For the other direction consider the identity map $$id:(K,\|\cdot\|_\infty)\longrightarrow(K,\|\cdot\|)$$

Let $S$ be the unit sphere in $(K,\|\cdot\|_\infty )$. Then

Show that $S$ is compact.
Show that $id$ is continuous.

edited Feb 26 '18 at 02:03

answered Dec 09 '13 at 14:34

pritam

10,157

same directions? – Lookout Apr 07 '16 at 01:19
1

@user166445: ??? Could you be more explicit? – Alex M. Sep 06 '16 at 19:41
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@pritam, couldn't we use the compacity of $S$ and continuity of $id$ to prove also that $||b||\leq c$? – rmdmc89 Jun 09 '18 at 21:34
At the end, you really want to use that ${|x| : x \in S}$ is compact, hence closed, so there exists $c>0$ such that $|x| > c$ for $x \in S$. – Bart Michels May 15 '19 at 18:04

score 5 · Answer 2 · answered Apr 01 '21 at 04:55

5

This lecture note answers this question quite well. https://math.mit.edu/~stevenj/18.335/norm-equivalence.pdf

answered Apr 01 '21 at 04:55

Ariel So

359

Tim Seguine · Answer 3 · 2013-12-09T14:26:52.467

1

The negation of the statement that all norms on a finite dimensional space are equivalent is different than what you claim. All you can say is that at least one norm exists for which the inequality doesn't hold, not that it doesn't hold for all norms.

So no, your attempt doesn't work.

edited Dec 09 '13 at 14:26

answered Dec 09 '13 at 14:20

Tim Seguine

1,270

Any two norms on finite dimensional space are equivalent

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