Is there a bounded selfadjoint 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 $[\partial u](n)=u(n+1)u(n1)$
1 Answer
Assuming that $uv$ means the pointwise product, the only such $H$ is the zero operator. (Selfadjointness 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 $2e_n1$ 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.

$\begingroup$ Thank you for your help. I was not aware of the general theorem, but this simple proof is quite elegant. $\endgroup$– ChrMar 12, 2019 at 10:53