Symmetric operator : Is it $(\overline{T})^{*}= \overline{T^{*}}$ true?

Question

I have known that if the operator $T$ is closable, then $$ T^{*}=(\overline{T})^{*}.$$

If the closure and dual can commutative? In other words, if $T$ is symmetric and unbounded, then $$(\overline{T})^{*}=\overline{T^{*}}.$$

Is it true? Obviously, $T \subset T^{*}$, and $\overline{T} \subset \overline{T^{*}}$.

Moreover, is it true if T is essentially self-adjoint? ( If $T$ is self-adjoint, the proposition is true.)

Actually, I want to ask why it is true if $T$ is symmetric and essential self-adjoint $$ker(T^{*}+iI)=ker((\overline{T})^{*}+iI)?$$

Answer 1

If $J$ is the isometric transpose map $J(x,y)=(y,-x)$ from $H\times H$ to itself, then $$ \mathcal{G}(T^*)= J\mathcal{G}(T)^{\perp}. $$ This is well-defined if $T$ is densely-defined. The graph of the closure of $T$ is $\overline{\mathcal{G}(T)}$. Basic properties of orthogonal complement give $$ \mathcal{G}(T^*) = J\overline{\mathcal{G}(T)}^{\perp} = J\mathcal{G}(\overline{T})^{\perp} = \mathcal{G}(\overline{T}^*)$$

Answer 2

$T^{\ast}$ is a closed operator, so $\overline{T^{\ast}}=T^{\ast}$. In general, we have $(\overline{T})^{\ast}\subseteq T^{\ast}$.