Is there $u,v\in L(E): uv-vu=id_E$ Let $E$ be a normed vector space over $\mathbb{R}$. Is there continuous linear transformations $u$ and $v$ such that: $$uv-vu=id_E$$ (.ie $\forall x\in E:u(v(x))-v(u(x))=x$) I suspect that the answer is no. When $E$ is finite dimensional we can use Trace Operator to prove that there is indeed no satisfied transformations. I don't know how to process in the case of infinite dimensional $E$.

I have a following reasoning. Is it acceptable?

Let's assume that there exist $u$ and $v$ with that property. Define norm of linear transformation $\|u\|=\sup_{\|x\|=1}\|u(x)\|$

We can prove that $uv^n-v^nu=nv^{n-1}$ by induction. Take $n>(\|uv\|+\|vu\|)$. Then we have:

$$n\|v^{n-1}\|=\|uv^n-v^nu\|\leq\|v^{n-1}\|(\|uv\|+\|vu\|)$$

Or $(\|uv\|+\|vu\|)\ge n$, contradiction.

The only thing I'm not sure is the norm of linear operator. Does it always exist in a normed vector space?