Symmetric, self-adjoint and essentially self-adjoint operators

Assume Countable Choice. Let $T$ be a densely defined linear operator on $H$.

• $T$ is symmetric when $T\subseteq T^{\ast }$, that is, $D(T)\subseteq D(T^{\ast })$ and $Tx=T^{\ast }x$ for all $x\in D(T)$; equivalently $⟨Tx,y⟩=⟨x,Ty⟩$ for all $x,y\in D(T)$.
• $T$ is self-adjoint when $T=T^{\ast }$, that is, the domains and the values agree.
• $T$ is essentially self-adjoint when its closure $\bar{T}$ is self-adjoint; by Closability is equivalent to density of the adjoint domain this presupposes that $T$ is closable.

Consequences, with proofs. These are part of the content of the definition.

1. A symmetric $T$ is closable. $T\subseteq T^{\ast }$ and $T^{\ast }$ is closed (The adjoint is well defined, closed, and reverses inclusions), so $T^{\ast }$ is a closed extension of $T$. In particular the closure $\bar{T}$ exists and $\bar{T}\subseteq T^{\ast }$.
2. The closure of a symmetric operator is symmetric. If $T\subseteq T^{\ast }$, the inclusion reversal of The adjoint is well defined, closed, and reverses inclusions applied to $T\subseteq T^{\ast }$ gives $(T^{\ast }{)}^{\ast }\subseteq T^{\ast }$, that is $T^{\ast \ast }\subseteq T^{\ast }$. Applying it once more to the inclusion $T^{\ast \ast }\subseteq T^{\ast }$ gives $(T^{\ast }{)}^{\ast }\subseteq (T^{\ast \ast }{)}^{\ast }$, that is $T^{\ast \ast }\subseteq T^{\ast \ast \ast }$. Since $\bar{T}=T^{\ast \ast }$ and $(\bar{T}{)}^{\ast }=T^{\ast \ast \ast }$ (Closability is equivalent to density of the adjoint domain), this reads $\bar{T}\subseteq (\bar{T}{)}^{\ast }$: the closure is symmetric. (Each application is legitimate because the adjoint is defined once its operator is densely defined, and $D(T^{\ast \ast })$ is dense because $T\subseteq T^{\ast \ast }=\bar{T}$, so that $D(T^{\ast \ast })$ contains the dense domain $D(T)$.)
3. A self-adjoint operator has no proper symmetric extension. If $T=T^{\ast }$ and $T\subseteq S\subseteq S^{\ast }$, then $S^{\ast }\subseteq T^{\ast }$ by inclusion reversal, so $T\subseteq S\subseteq S^{\ast }\subseteq T^{\ast }=T$ and all inclusions are equalities.
4. A self-adjoint operator is closed, being equal to the adjoint $T^{\ast }$ of the densely defined $T$, and an essentially self-adjoint operator has exactly one self-adjoint extension, namely $\bar{T}$: a self-adjoint extension $S$ of $T$ is closed, hence contains the least closed extension $\bar{T}$, and then $S\subseteq S^{\ast }\subseteq (\bar{T}{)}^{\ast }=\bar{T}$ by item 2 and symmetry, so $S=\bar{T}$.