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.



  • $\begingroup$ 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). $\endgroup$
    – MaoWao
    Jan 29 at 18:28
  • $\begingroup$ you may add a positive $A$ to avoid trivialities. $\endgroup$ Jan 29 at 19:10


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