$\nexists \ (u, v)$ 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.
 A: The finite dimensional case is easy, since if $E$ is over the real numbers, say, then linear functions $u,v : E \to E$ (with choice of basis) are just square matrices with dimension equal to dimension of $E$ and real entries. But the tracial property asserts that $tr(uv-vu) = 0$, so if $(uv-vu) = I$  then the traces of both sides have to match, and this does not happen.
However, this does not happen in infinite dimensions : for example, one can see that if we take the vector space of infinitely differentiable functions on $\mathbb R$ with compact support (as an  exercise, show that this is not finite dimensional, by finding explicitly, infinitely many linearly independent functions in this vector space), then the operators given by $(\mathbf x(\psi))(y) = y \psi(y)$ and $(\mathbf p(\psi))(y) = -\frac{\partial \psi}{\partial x}(y)$ actually satisfy the relation $\mathbf{xp} -\mathbf{px} = I$. This is seen by taking a function $\psi$, and noting : 
$$
(\mathbf{px})(\psi)= \mathbf{p}(y \psi) = -\psi - y(-\mathbf p \psi) = -\psi + (\mathbf{xp})(\psi) \implies I = \mathbf{xp} - \mathbf{px}
$$
That is, the tracial property fails to hold here.
However, there is still a resolution of the problem for sufficiently regular vector spaces in infinite dimensions : it's called the Stone Von-Neumann theorem, and it states that roughly, the above, is upto some conditions of equivalence, the only such example on these kind of vector space.
I cannot go on to explain the content of the result, unfortunately, since it involves some fairly difficult terminology to be known a priori.
EDIT : This answer by Mariano Suarez-Alvarez will be helpful to understand the content of the theorem : Solutions to the matrix equation $\mathbf{AB-BA=I}$ over general fields
A: 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 $\forall \ n \in \mathbb N, \ u^{n+1} \neq 0$, there is a contradiction with the last inequality.
So $\exists  \ n \in \mathbb N, \ u^{n} \neq 0$ and $u^{n+1} = 0 \ $  contradiction with the first equality.
