Is it true that $U^{-1}\overline{A}U=\overline{U^{-1}AU}?$

Question

Let $X_0$ and $X_1$ be Banach spaces and suppose that $A:D(A)\subseteq X_0\times X_1$ is a closable linear operator. If $U: X_0\to X_1$ is a unitary operator, is it always true that $$U^{-1}\overline{A}U=\overline{U^{-1}AU}?$$ That is to ask, is it always true that a unitary transformation of the closure of a linear operator coincides with the closure of the unitary transformation? If so, how would one prove this, or is there a standard reference with a proof?

Answer 1

Yes, it's straightforward. First, $U^{-1}AU\subset U^{-1}\overline{A}U$, so $\overline{U^{-1}AU}\subset U^{-1}\overline{A}U$. Now, let $x\in\operatorname{dom}U^{-1}\overline{A}U$. Then $Ux\in\operatorname{dom}\overline{A}$, so there exists $(y_n)\subset\operatorname{dom}A$ such that $y_n\to Ux$ and $Ay_n\to\overline{A}Ux$. Set $x_n := U^{-1}y_n$. Then $x_n\to x$ and $U^{-1}AUx_n\to U^{-1}\overline{A}Ux$. Hence, $x\in\operatorname{dom}\overline{U^{-1}AU}$ and $\overline{U^{-1}AU}\,x = U^{-1}\overline{A}Ux$.