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.)
Best Answer
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
Then $S$ is compact and hence bounded in $\|\cdot\|$. Now take any $a\in K$ and let $b=a/\|a\|_\infty$. Then $\| b\|\geq c^\prime$ for some positive constant $c^\prime$ and you are done.