Let $H$ be a Hilbert space. Let $S∈B(H)$ and let $T$ be a densely defined closed operator such that $TS\subset ST$. Assume further that $T$ is boundedly invertible.

Is it true that $ST$ is closed?

In my problem, $S$ and $T$ are both self adjoint and positive.

  • $\begingroup$ What does boundedly invertible mean? $T: D(T) \to H$ is injective and $T^{-1}: T(D(T)) \to H$ is bounded? $\endgroup$ Jan 15 at 20:22
  • $\begingroup$ Are you also sure that you don't mean $ST \subseteq TS$? $\endgroup$ Jan 15 at 20:30
  • $\begingroup$ Yes. It is not the usual commutativity... $\endgroup$ Jan 15 at 20:34
  • $\begingroup$ Any ideas or suggestions? $\endgroup$ Jan 16 at 12:22


Your Answer

By clicking “Post Your Answer”, you agree to our terms of service, privacy policy and cookie policy