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-commute** if 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](https://mathoverflow.net/questions/278409/commuting-with-self-adjoint-operator/278439), 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.



Commuting with an unbounded operator
Yes, your statement is true.


First observe that $$M := \{B \in \mathcal{B}(\mathcal{H}) : BA \subset AB \,\,\mbox{ and }\,\, BA^\* \subset A^\*B\}$$ is a von Neumann algebra (on the nose, without needing to take a strong closure). To see that M is closed under adjoints, it may be helpful to note that $BA^\* \subset A^\*B$ is equivalent to $B^\*A \subset AB^\*$ when $B$ is bounded. You wish to show that $M \subset N:=\{E\_S , V\}'$ where $E\_S$ and $V$ are the spectral projections and polar part of A, respectively. It suffices to show that $U \in N$ for all unitary $U \in M$.


If $U \in M$, then $U$ maps $D(A)$ onto itself, and similarly $D(A^\*)$. One can check that $$A = UAU^\* = (UVU^\*)(U|A|U^\*),$$ that $U|A|U^\*$ is positive self-adjoint, and that $UVU^\*$ is a partial isometry with the correct initial and final spaces, so by the uniqueness of the polar decomposition (see Reed and Simon, Functional Analysis, Thm. VIII.32) $U$ commutes with $V$ and $U$ commutes (strongly, by the spectral theorem/your earlier question) with $|A|$.


