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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 $E$.

Marc van Leeuwen
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anonymous67
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  • related: http://math.stackexchange.com/questions/99175/solutions-to-the-matrix-equation-mathbfab-ba-i-over-general-fields – Ehsan M. Kermani Aug 31 '14 at 04:32
  • I'm not sure if this is a valid example. If $E = C^{\infty}[0,1]$ (set of infinitely differentiable functions on $[0,1]$), and we defined $u(x(t)) = x'(t)$ and $v(x(t)) = tx(t)$, then $u(v(x(t)))-v(u(x(t))) = x(t)$ for any function $x(t) \in C^{\infty}[0,1]$. – JimmyK4542 Aug 31 '14 at 04:38
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    @EhsanM.Kermani: Well only sort-of related, as the question there asks for (1) finite dimension (2) arbitrary field, (3) no norm condition. Some answers do give examples that ignore (1) (so infinite dimensional), however none address a norm. My guess is that functional analysis treats this question. – Marc van Leeuwen Aug 31 '14 at 04:39
  • Jimmy's example occurred to me, too (it's standard). Can we make them both continuous w.r.t. some norm? – Jyrki Lahtonen Aug 31 '14 at 04:39
  • @JyrkiLahtonen No. Let's restrict $u$ and $v$ to $\mathbb{R}[x]$. Then for every $n=1,2,\ldots$, $n\Vert x^{n-1}\Vert=\Vert u(x^n)\Vert\leq\Vert u\Vert\Vert x^n\Vert=\Vert u\Vert\Vert v(x^{n-1})\Vert\leq\Vert u\Vert \Vert v\Vert \Vert x^{n-1}\Vert$, hence $n\leq \Vert u\Vert \Vert v\Vert$ for every $n$, a contradiction. – Luiz Cordeiro Aug 31 '14 at 05:37
  • @LuizCordeiro: I knew that. Restricting the domain to $[0,1]$ allows us to bound the norm of $v$, but can we use a different norm (or use a different space) that also allows us to bound the norm of $u$? – Jyrki Lahtonen Aug 31 '14 at 05:45
  • Hmm. On second thought your argument does show that if the space includes all the monomials, $u$ cannot be bounded. Sorry :-) If both $u$ and $v$ are continuous, so are $uv$ and $vu$, and ... – Jyrki Lahtonen Aug 31 '14 at 05:47

1 Answers1

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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?

anonymous67
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