Showing that (on $\mathbb{R}^n$) the $\|\cdot\|_\infty$ norm is weaker than any other norm.

I am doing past papers and the question is this: "Prove that any norm on $\mathbb{R}^n$ is weaker than the $\|\cdot\|_\infty$ norm. Use this fact to show that on $\mathbb{R}^n$ any two norms are equivalent"

(You can find all the needed definitions here, save me typing them out)

I have shown that $\exists C>0\forall x\in\mathbb{R}^n$ such that $\|x\|\le C\|x\|_\infty$ is true. (using the triangle inequality property of $\|\cdot\|$)

I must now show that $\exists C>0\forall x\in\mathbb{R}^n$ such that $\|x\|_\infty\le C\|x\|$ - I sense I'll need the reverse triangle inequality (simply because that's all I've got) you can find the proof that given the triangle inequality we have the reverse here.

Not sure how though.

Forward case

Simply $\|x\|\le\sum^n_{i=1}|a_i|\|e_i\|$
Let $M_e=\max_{1\le i\le n}(\{\|e_i\|\})$ then
$\sum^n_{i=1}|a_i|\|e_i\|\le\sum^n_{i=1}|a_i|M_e$ so
$\|x\|\le M_e\sum^n_{i=1}|a_i|$
Let $M_a=\max_{1\le i\le n}(\{|a_i|\})$ then
$M_e\sum^n_{i=1}|a_i|\le M_e\sum^n_{i=1}M_a=nM_eM_a$

But wait! $M_a=\|x\|_\infty$ so we conclude:

$\|x\|\le nM_e\|x\|_\infty$ the proof follows by choosing $C=nM_e$

  • $\begingroup$ Please don't use the definition of equivalence of norms given in your link. It is lacking teh condition that $c>0$. $\endgroup$ – Hagen von Eitzen Apr 21 '15 at 18:40
  • $\begingroup$ @HagenvonEitzen edited but even without that the "Note also that if ∥⋅∥1 is both weaker and stronger than ∥⋅∥2 they are equivalent " should make it clear $\endgroup$ – Alec Teal Apr 21 '15 at 18:43
  • $\begingroup$ @HagenvonEitzen how would you write it, $\exists c,C\in\mathbb{R}\forall x\in V[c,C>0\wedge c\|x\|_1\le \|x\|_2\le C\|x\|_1]$ or how it is written, they both look pretty bad. $\endgroup$ – Alec Teal Apr 21 '15 at 18:55

Here is a proof by contradiction:

You have shown that there is some $C>0$ such that $\|x\| \le C \|x\|_\infty$.

Suppose for all $n$ there is some $x_n$ such that $\|x_n\|_\infty > n \|x_n\|$. Without loss of generality we may assume $\|x_n\|_\infty = 1$. Since the unit ball (with $\|\cdot\|_\infty$) is compact, we can find a subsequence and $x$ such that $\|x_{n_k} - x\|_\infty \to 0$. We have $\|x\|_\infty = 1$ and so $x \neq 0$. Also, we have $\|x_{n_k}\| < {1 \over n_k}$.

We have $\|x\| \le \|x_{n_k}\| + \|x_{n_k} -x \| \le {1 \over n_k} + C \|x_{n_k} -x \|_\infty$ which implies $x = 0$, a contradiction.

  • $\begingroup$ While this is good - PLEASE do an edit where you answer the question, although sequential compactness is good. I was really hoping for a ... nicer proof, by that I mean using norms and finite sums - because as you know any two norms on ANY finite vec space are equiv. $\endgroup$ – Alec Teal Apr 21 '15 at 19:08
  • $\begingroup$ I don't understand your comment. The above does answer the question. It shows that there exists some $n$ such that $\|x\|_\infty \le n\|x\|$ for all $x$. Can you elaborate what it is that you wanted? $\endgroup$ – copper.hat Apr 21 '15 at 19:31
  • $\begingroup$ Good point, sorry for that I came across not how I meant and rather rudely infact. I was hoping for a proof that could easily be applied to any finite vector space, that'd probably end up using the reverse triangle inequality and without mentioning compactness. $\endgroup$ – Alec Teal Apr 21 '15 at 19:32
  • $\begingroup$ In a normed space, the unit ball is compact iff the space is finite dimensional, so I suspect that compactness is inextricably involved in an equivalence proof. Since we are dealing with the $\sup$ norm (for one of the norms), we really only need to consider sequential compactness of $[-1,1]$. $\endgroup$ – copper.hat Apr 21 '15 at 19:37
  • $\begingroup$ I didn't know that (that explains my difficulty) - please do give this some thought though, - I've added my +1 $\endgroup$ – Alec Teal Apr 21 '15 at 19:38

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