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


1 Answer 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.

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

Your Answer

By clicking “Post Your Answer”, you agree to our terms of service, privacy policy and cookie policy

Not the answer you're looking for? Browse other questions tagged or ask your own question.