Let ${T: D(T) \subset H \to H}$ be a linear operator. $T$ is symmetric if:
- $(\phi, T\psi)= (T\phi, \psi)$ holds $\forall \phi, \psi \in D(T) $ and
- $D(T)$ is dense in $H$.
The point 2. is necessary to guarantee the uniqueness of the adjoint $T^*$, defined as the operator for which $(\phi, T\psi)= (T^*\phi, \psi)$ holds $\forall \phi \in D(T^*), \forall \psi \in D(T)$.
We can hence conclude that $T$ symmetric $\implies T^* \supset T$, which is an abbreviation for "$T$ is extended by its adjoint $T^*$".
Guess. Can we say that $D(T^*) \supset D(T)$ is dense? Or, more generally, that any extension of a dense set is dense?
If this is true, it follows that $T \subset T^{**} = {\overline T }\subset T^{*}$. Since $\overline T$ is the smallest closed extension for $T$, and $T^*$ is always closed
The problem is that the adjoint of a (densely defined) symmetric operator is not symmetric (unless the given operator was already self-adjoint). Sure, the domain of the adjoint contains the domain of the original, so is still dense, and/but that's not the issue.