# Existence of a bounded operator which satisfies the discrete product rule

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

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.

• Thank you for your help. I was not aware of the general theorem, but this simple proof is quite elegant.
– Chr
Mar 12, 2019 at 10:53