# About the commutatvity of a bounded self-adjoint operator with an unbounded symmetric one?

Let $$B\in B(H)$$ be self-adjoint and let $$A$$ be a densely defined symmetric (and closed if needed) operator such that $$A^2$$ is densely defined. If $$BA^2\subset A^2B$$ say, is there a result which gives $$BA\subset AB$$?

Notice that I already have a counterexample when $$A^2$$ is not densely defined.

Cheers,

Hichem

• I don't know what kind of result you are looking for, but without further conditions, this can already fail for bounded operators ($2\times 2$ matrices even). Jan 29 at 18:28
• you may add a positive $A$ to avoid trivialities. Jan 29 at 19:10