$\nexists \ (u, v)$ such that $v \circ u - u \circ v = Id_E $ Let $E$ be a normed non-zero vector-space. _Show there exist no continuous linear functions $u$ and $v$ such that $v \circ u - u \circ v = Id_E $.

$\nexists \ (u, v)$ such that $v \circ u - u \circ v = Id_E $ Let $E$ be a normed non-zero vector-space. _Show there exist no continuous linear functions $u$ and $v$ such that $v \circ u - u \circ v = Id_E $._ See the answer below.

Answer to the question:

By induction $\forall \ n \in \mathbb N\ \ \ v \circ u^{n+1} - u^{n+1} \circ v = (n+1)u^n $.

Define $\| u\|$, algebra norm.

$\forall \ n \in \mathbb N, \ (n+1)\| u^n\|≤2\| v\| \| u^{n+1}\|$

Hence $\forall \ n \in \mathbb N, \ (n+1)\| u^{n+1}\|≤2\| v\|\| u\| \| u^{n+1}\| $

If $\forall \ n \in \mathbb N, \ u^{n+1} \
eq 0$, there is a contradiction with the last inequality.

So $\exists \ n \in \mathbb N, \ u^{n} \
eq 0$ and $u^{n+1} = 0 \ $ contradiction with the first equality.