# $AB$ is compact iff $BA$ is

Is it in general true or false that for $A, B \in \mathcal L (\mathcal H)$, $A B \in \mathcal S^\infty (\mathcal H) \Longleftrightarrow B A \in \mathcal S^\infty (\mathcal H )$?

I think it holds only for $A, B$ normal. Could someone tell me if I'm right?

Proof (sketch) for normal operators: $A B \in \mathcal S^\infty (\mathcal H)$, then $\psi_n \rightharpoonup 0$ implies $A B \psi_n \to 0$, because $A, B$ normal, this implies $A^* B^* \psi_n = (BA)^* \psi_n \to 0$, i.e. $(BA)^* \in \mathcal S^\infty(\mathcal H)$ and this is equivalent to $BA \in \mathcal S^\infty (\mathcal H)$.

Counterexample for the general case (I think this is one): $\mathcal H = \ell^2(\mathbb N)$ and denote projections in Bra-Ket notation like $\vert n \rangle \langle n \vert$. $$A := \operatorname{s-}\lim_{N\to\infty} \sum_{n=1}^N \frac{1}{2n+1} \vert 2n+1\rangle \langle 2n+1\vert + \operatorname{s-}\lim_{N\to\infty} \sum_{n=1}^N \vert 2n \rangle \langle 2n \vert$$ $$B := \operatorname{s-}\lim_{N\to\infty} \sum_{n=1}^N \vert 2n+1\rangle \langle n \vert$$ Then $AB$ is compact but $BA$ is not (since it maps norm-preserving on the infinite-dimensional subspace of even $n$s)...

Yes, you are right. A slightly simplified version of your example is, with $\{E_{kj}\}$ denoting the matrix units corresponding to an orthonormal basis $\{e_n\}$, $$A=\sum_n E_{2n+1,n},\ \ \ B=\sum_n E_{n,2n}.$$ Then $$AB=\sum_{n} E_{2n+1,2n},$$ with range containing the closed span of $\{e_{2n+1}\}$, while $BA=0$.