Checking of continuity of the operators

Question

Let $X$ be a normed vector space and $S,T : X → X$ be linear and such that $ST −TS = Id$. Show that either $S$ or $T$ is not continuous.

Answer 1

Let me give an answer that also works if $\mathcal L(X)$ is not a Banach algebra, i.e. if $X$ is real or noncomplete.

We show by induction that $ST^{n+1}-T^{n+1}S = (n+1)T^n$:

The base case is clear. Now if this is true for some $n\in \mathbb N$, then

\begin{align} &ST^{n+1}-T^{n+1}S = (n+1)T^n \\\ \Rightarrow\quad &ST^{n+2}-T^{n+1}ST = (n+1)T^{n+1} \\\ \Rightarrow\quad &ST^{n+2}-T^{n+1}(ST-Id)-T^{n+1} = (n+1)T^{n+1}\\\ \Rightarrow\quad &ST^{n+2}-T^{n+1}(TS) = (n+2)T^{n+1} \\\ \Rightarrow\quad &ST^{n+2}-T^{n+2}S = (n+2)T^{n+1}. \end{align}

Inductively it follows aswell that $T^n\neq 0$ for each $n\in\mathbb N$.

But now $(n+1)\|T^n\| \leq 2\|S\|\|T^{n+1}\|\leq 2\|S\|\|T\|\|T^n\|$. Dividing by $\|T^n\|$ (assuming it is finite) yields $$(n+1)\leq 2\|S\|\|T\|$$ for each $n\in\mathbb N$, hence one of $S,T$ must be unbounded.