If $X$ is a nonzero normed space. $S,T \in Lin(X)$ satisfies $ST-TS=I$, where $I$ is the identity function.
Prove the two linear operators cannot be both bounded.
My attempt:
Suppose $S,T$ are bounded, then
- $\Vert Tx\Vert \leq a_1\Vert x\Vert,\Vert Sx\Vert \leq a_2\Vert x\Vert$
- $\Vert x\Vert =\Vert (ST-TS)x\Vert$
I am not sure how to combine the two conditions to reach a contradiction.
Any help would be appreciated.
Best Answer
Suppose $[S,T]=ST-TS=I$ and $S$ and $T$ are both bounded operators. I write $\Vert S\Vert$ for the operator norm of $S$, that is, $\Vert S \Vert:= \inf \left\{M\mid \forall v\in X:\Vert S(v)\Vert\leq M\Vert v\Vert\right\}$.
There is a theorem by Wielandt that shows that in any normed algebra $A$, the identity cannot be expressed by a commutator. Clearly this theorem shows that $S$ and $T$ are not both bounded as they do belong to the normed algebra of bounded operators on $X$.
Here's the proof: By induction you can easily show that $[S,T^n]=nT^{n-1}$.
Note that $n \Vert T^{n-1}\Vert=\Vert n T^{n-1}\Vert=\Vert ST^n-T^nS\Vert\leq 2\Vert S\Vert \Vert T\Vert \Vert T^{n-1}\Vert$. Thus $n\leq \Vert S\Vert \Vert T\Vert$. As this identity holds for any $n\geq 1$, we arrive at a contradiction.
That's how easy the proof of the theorem is. You can wonder how one comes up with this proof. Note that you probably wrote down the following: $\Vert v\Vert= \Vert (ST-TS)v\Vert\leq 2 \Vert S\Vert \Vert T\Vert \Vert v\Vert$. From this you already find that the product of the operator norms of $S$ and $T$ needs to be at least $\frac{1}{2}$. It's strange that you get such a specific restriction on an otherwise fairly general setting. The induction above allows to find even more restrictions to the point it becomes rediculous!