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We have just come across the lemma that every norm on $\mathbb{R}^n$ is equivalent to $\|\cdot \|_\infty$. My question is that do all norms on $\mathbb{R}^n$ take the form $\|\cdot \|_1, \|\cdot \|_2, \dots, \|\cdot \|_\infty$ etc. ? And if not how can you possibly prove that any norm on $\mathbb{R}^n$ is equivalent to $\|\cdot \|_\infty$ if you don't know what 'form' the norm takes?

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A proof of the lemma can be found here: with further links. Of course, not every norm is a $p$-norm. Every symmetric convex set $C$ with non-empty interior gives a norm (by declaring that $C$ is the unit ball of the norm); symmetry means $x \in C$ if and only if $-x \in C$. For example, a regular hexagon in the plane gives a norm that certainly isn't a $p$-norm. – t.b. Apr 6 '12 at 11:01
For that particular proof why can you write $x = \sum_m x_m e_m$ with $e_m$ a basis? – user26069 Apr 6 '12 at 11:09
My lecture notes say that $e_m$ is the standard basis, i.e. $e_1 = (1,0,\dots,0), e_2 = (0,1,\dots,0), \dots$ etc. I assume this is not the case? – user26069 Apr 6 '12 at 11:32
The proof works with any basis, so in particular for the standard basis. But no, it isn't assumed that the basis in Fabian's proof is the standard basis. – t.b. Apr 6 '12 at 11:36
up vote 1 down vote accepted

For the sake of completeness, here's an answer to your questions:

1) Consider, on $\mathbb{R}^2$, the norm $$ \|(x,y)\|=2|x|+3|y| $$ (of course, any other choice of positive coefficients will to, too). You can also do the same kind of variation with the $p$-norms. Notice that if you take any subadditive positive function on $\mathbb{R}$ ($f(t)>0$ for all $t\ne0$, $f(0)=0$, $f(s+t)\leq f(s)+f(t)$), then $$ \|(x,y)\|_f=f(x)+f(y) $$ defines a norm, and even (for $p\geq1$) $$ \|(x,y)\|_{f,p}=\left(f(x)^p+f(y)^p\right)^{1/p}. $$

2) The answer to your second question can be seen from the proof of Fabian quoted above: in a finite-dimensional space any norm is bounded above by a multiple of the infinity norm; this together with the compactness of any closed and bounded set (again a consequence of the finite-dimensionality) guarantees the equivalence of your norm with the infinity norm.

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