## Archive for June, 2011

### Equivalence of norms

June 17, 2011

It is well-known that all norms on a finite dimensional ${{\mathbb R}}$-vector space are equivalent: the proof is immediate, because the unit ball is compact. A similar argument works with ${{\mathbb R}}$ replaced by any locally compact field. It is less known that the same is true over any complete field, without assuming local compactness. The compactness argument no longer works, but the proof is still not very difficult. However, the arguments I saw in the literature so far are somewhat bulky (see, for instance, Section 144 of van der Waerden’s “Algebra”).

Recently our master student Martin Djukanovic showed me a wonderful short proof of this statement. We argue by induction in the dimension. There is nothing to prove in dimension ${1}$. Now let ${V}$ be an ${m}$-dimensional vector space over some complete field, ${m>1}$. We fix a basis ${e_1,\ldots, e_m}$ of ${V}$ and show that any norm ${\|\cdot\|}$ on ${V}$ is equivalent to the sup-norm ${\|\cdot\|_\infty}$ with respect to this basis:

$\displaystyle \|x_1e_1+\cdots+x_me_m\|_\infty = \max\{|x_1|, \ldots, |x_m|\}.$

By induction, this holds in dimension ${m-1}$. In particular, in dimension ${m-1}$ any normed space over a complete field is itself complete, because this is obviously true for the sup-norm.

For any ${v\in V}$ we have ${\|v\|\le C\|v\|_\infty}$ with ${C=\|e_1\|+\cdots+\|e_m\|}$. It remains to show that ${\|v\|_\infty\le C\|v\|}$ with some other ${C>0}$. Assuming the contrary, we find an infinite sequence of non-zero vectors ${v_n\in V}$ such that ${\|v_n\|/\|v_n\|_\infty\rightarrow 0}$ as ${n\rightarrow \infty}$. Taking a subsequence, we may assume that the sup-norm of each ${v_n}$ is attained at its ${m}$-th coordinate, and after re-scaling we may assume that the ${m}$-th coordinate (and thereby the sup-norm) of each ${v_n}$ is ${1}$. Thus, each ${v_n}$ can be written as ${u_n+e}$, where ${u_n\in U=\langle e_1, \ldots, e_{m-1}\rangle}$ and ${e=e_m}$.

Since ${\|v_n\|_\infty=1}$, we have ${\|v_n\|=\|u_n+e\|\rightarrow 0}$ as ${n\rightarrow \infty}$. Thus, the sequence ${(u_n)}$ converges to ${-e}$. Hence it is a Cauchy sequence. But ${U}$ is complete by the induction, which implies that ${(u_n)}$ converges in ${U}$. We obtain ${-e\in U}$, a contradiction.