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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.

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    The finite-dimensional case is easy: https://math.stackexchange.com/questions/284901/ab-ba-i-having-no-solutions – lhf Mar 04 '18 at 10:58
  • Yes, but take $\mathbb F = \mathbb Z / p \mathbb Z$. –  Mar 04 '18 at 11:06
  • Then it might be false as $I$ can be traceless ; as stated in the link provided by @lhf. – krirkrirk Mar 04 '18 at 11:09

2 Answers2

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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

Sarvesh Ravichandran Iyer
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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.