For bounded linear operators $S$ and $T$ such that $ST-TS=I$, show that $\|T^n\|=0$ for some $n \in \mathbb{N}$

Question

I found in the first part of the question that $ST^{n+1}-T^{n+1}S=(n+1)T^n$ for any non negative integer $n$, any clues on how to proceed?

Thanks.

Answer 1

From $ST^{n+1}-T^{n+1}S=(n+1)T^n$ we get

$(n+1)||T^n|| \le 2||S||\cdot||T^{n+1}||\le 2||S||\cdot ||T|| \cdot||T^n||.$

Now suppose that $||T^n|| \ne 0$ for all $n$. Then we get

$n+1 \le 2||S||\cdot ||T||$ for all $n$, which is absurd.