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 $[\partial u](n)=u(n+1)-u(n-1)$ 



Existence of a bounded operator which satisfies the discrete product rule
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.)




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**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. 


