How can we show that:
$||x||_\infty \leq ||x||_2 \leq \sqrt{n} ||x||_\infty$
$||A||_\infty \leq \sqrt{n}||A||_2 \leq n||A||_\infty$
Where A is $n\times n$ and $x$ is $n$ vector.
I think this implies the equivalence of these norms for finite $n$, that is boundedness in one norm also means boundedness in the other.