# Mathematics Prelims

## July 16, 2008

### Equivalence of Norms in n dimensions

Filed under: Analysis,Functional Analysis — cjohnson @ 4:08 pm

We say that two norms, $\| \cdot \|_1$ and $\| \cdot \|_2$, on the same vector space $X$ are equivalent if there exist $\alpha, \beta > 0$ such that for every $x \in X$,

$\displaystyle \alpha \| x \|_1 \leq \| x \|_2 \leq \beta \| x \|_1$.

In a finite dimensional normed space, all norms are equivalent.

Proof: Suppose $X$ is an n-dimensional space with basis elements $\{ e_1, ..., e_n \}$ and that $\| \cdot \|_1$ and $\| \cdot \|_2$ are norms on $X$.  We know (from an earlier lemma) there exists a $c > 0$ such that for any $x \in X$ we choose

$\displaystyle \| x \|_1 \geq c (| \alpha_1 | + ... + | \alpha_n |)$

(Where $x = \alpha_1 e_1 + ... + \alpha_n e_n$.)

Now consider $\| x \|_2$ and apply the triangle inequality.

$\displaystyle \| x \|_2 \leq k \sum_{i=1}^n | \alpha_i |$

Where $k = \max \{ \| e_1 \|_2, ..., \| e_n \|_2 \}$.  If we then apply our earlier inequality we’re left with

$\displaystyle \| x \|_2 \leq \frac{k}{c} \| x \|_1$.

If we repeat this process but with $\| \cdot \|_1$ and $\| \cdot \|_2$ reversed, we achieve the other inequality.  Since our choice of $x \in X$ was arbitrary, and since our $k$ and $c$ don’t depend on our choice of $x$, we see that $\| \cdot \|_1$ and $\| \cdot \|_2$ are equivalent.  Since these were arbitrary norms, all norms on $X$ are equivalent.