Show that if $ A B - B A = \Id $, $ A , B : X \to X $ ($ X $ is a normed space) linear operators, then either $ A $ or $ B $ is unbounded.
I am not sure how to approach this, because I've never worked with commutators. I have to show that $ \frac { \| B x \| _ X } { \| x \| _ X } $ is unbounded. My Idea: $ 1 = \frac { \| \Id ( x ) \| _ X } { \| x \| _ X } = \frac { \| ( A B - B A ) x \| _ X } { \| x \| _ X } $ But this leads to nothing. I am very thankful for hints.
Greetings.
Since, there is a bit of interest. I decided to post my solution.
Given $AB-BA=\mathbb{1}_X$, assume $\|A\|,\|B\| \le C$ , one can show that $AB^n-B^nA=nB^{n-1}$ per induction.
Taking norm on the equation
$$|n|\|B^{n-1}\|=\|AB^n-B^nA\| \le \|A\|\|B^n\|+\|B^n\|\|A\|= 2 \|A\| \|B^n\|$$ since (showing via induction) $B^{n-1} \ne 0 $ $$\implies n \le 2 \|A\|\|B\|$$ for arbitary $n$ this is impossible due to Archimedes' principle. Therefore one Operator is unbounded.