Let $H$ be a Hilbert space. Let $A$ be a closed unbounded operator, and let $B\in B(H)$ be a bounded operator.

Definition:$A$ and $B$strong-commuteif the partial isometry in the polar decomposition of $A$ commutes with $B$, and all the spectral projections of $|A|$ also commute with $B$.

In this question, I learned that, when $A$ is self-adjoint, we have $$ BA \subset AB \quad\Longrightarrow\quad \text{$A$ and $B$ strong-commute}. $$

When $A$ is not self-adjoint, then $BA \subset AB$ is not enough to ensure the strong-commuting of $A$ and $B$ (see below for a counterexample). One should think of strong-commuting as saying, roughly, that $B$ commutes with both $A$ and $A^*$.

Question:Is it true that $$ (\,\,\,BA \subset AB \quad\text{and}\quad BA^* \subset A^*B\,\,\,) \quad\Longrightarrow\quad \text{$A$ and $B$ strong-commute}. $$

Ok, now the counterexample:

Let $D \subset E$ be two closed operators (both densely defined) on some Hilbert space $K$.
Then I take $A := ...\oplus D\oplus D\oplus D\oplus E\oplus E\oplus E...$ on the Hilbert space $H:=\ell^2(\mathbb Z)\otimes K$, and I take $B$ to be the shift operator.

verystrong condition --- for "typical" non-normal $A$ it would imply that $B$ is a scalar multiple of the identity. Is that really what you want? $\endgroup$ – Nik Weaver Apr 17 '18 at 22:46