Extending $T$ by $T^{**}$

Question

Let $T:D → H$ be an unbounded operator, $T^*:D(T^*) → H$ is closed and $T^{**}:D(T^{**}) → H$ is also closed

I have proved that $T^{**}$ is unbounded operator ie. $D(T^{**})$ is dense in $H$ , how can I prove that $T^{**}$ is an extension of $T$?

Answer 1

I'll assume $T$ is densely-defined and closable so that the adjoints make sense. The graph $\mathcal{G}(T)$ of $T$ is a subspace of $H\times H$, and $\mathcal{G}(T^*)$ is $J(\mathcal{G}(T))^{\perp}$, where $J(x,y)=(-y,x)$ acts on $H\times H$. Likewise $$ \mathcal{G}(T^{**})=J(\mathcal{G}(T^*))^{\perp} = J(J(\mathcal{G}(T)^{\perp})^{\perp}=\mathcal{G}(T)^{\perp\perp}. $$ And $\mathcal{G}(T)^{\perp\perp}$ is the closed subspace generated by $\mathcal{G}(T)$, which means $T^{**}$ is the closure of $T$.