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Is there a bounded self-adjoint operator $H$ acting on $\ell^2(\mathbb{Z})$ such that for all sequences $u,v\in \ell^2(\mathbb{Z})$ $$ H(uv)=(Hu)v+u(Hv)$$ where uv is the pointwise product. This is for instance not the case of $i\partial$ where $[\partial u](n)=u(n+1)-u(n-1)$

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Assuming that $uv$ means the pointwise product, the only such $H$ is the zero operator. (Self-adjointness is unnecessary.)

Proof. Let $n\in {\bf Z}$. Taking $u=v=e_n$ we get $H(e_n) = H(e_ne_n)= 2e_n\cdot (He_n)$.

Multiply both sides by $e_n$ to get $e_n\cdot H(e_n) = 2e_n\cdot H(e_n)$ and deduce that $e_n\cdot H(e_n)=0$.

Hence $H(e_n)=2e_n\cdot H(e_n)=0$.

So $H$ annihilates each standard basis vector $e_n$ and by continuity it vanishes on all of ${\ell^2}({\bf Z})$. QED

(The motivation for this trick is that we have $(2e_n-{\bf 1})\cdot (He_n)=0$ where ${\bf 1}$ is the vector of all 1s, i.e. the constant function ${\bf Z}\to \{1\}$, and since $e_n$ is an idempotent $2e_n-1$ is an involution. One can make this rigorous by working in the multiplier algebra of $\ell^2({\bf Z})$ but it is quicker to use the argument above, which is also standard folklore for those studying derivations on commutative Banach algebras.)


Edit: of course, what I was really doing above was digging out of my memory a proof of the following general result, which as I said is folklore. (The particular trick of avoiding the unitization was pointed out to me by Herb Kamowitz after a conference talk.)

Theorem. Let $A$ be an algebra (not necessarily with identity), let $X$ be an $A$-bimodule, and let $D:A\to X$ be a derivation. Then $D(e)=0$ for every $e\in Z(A)$ which satisfies $e^2=e$.

The proof is essentially just the same as the argument above.

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  • $\begingroup$ Thank you for your help. I was not aware of the general theorem, but this simple proof is quite elegant. $\endgroup$ – Chr Mar 12 at 10:53

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