Definition 2.1.2 ([X.2.2, Con85]).label Let $H$ be a Hilbert space and $T: D(T) \to H$ be densely defined linear map, then the following are equivalent:
- (1)
$T^{*}$ is an extension of $T$.
- (2)
$\angles{Tx, y}= \angles{x, Ty}$ for all $x, y \in D(T)$.
- (3)
$\angles{Tx, x}\in \real$ for all $x \in D(T)$.
If the above holds, then $T$ is a symmetric operator.
If $T = T^{*}$, then $T$ is self-adjoint.
Proof. $(1) \Rightarrow (2)$: Let $x, y \in D(T)$, then
\[\angles{Tx, y}_{H} = \angles{x, T^*y}_{H} = \angles{x, Ty}_{H}\]
$(2) \Rightarrow (3)$: For any $x, y \in D(T)$,
\[\angles{Tx, x}_{H} = \overline{\angles{Tx, x}_H}\quad \angles{Tx, x}_{H} \in \real\]
$(3) \Rightarrow (1)$: By the polarisation identity, for any $x, y \in D(T)$,
\begin{align*}4\angles{Tx, y}_{H}&= \sum_{k = 0}^{3}i^{k}\angles{T(x + i^ky), x + i^ky}_{H} \\&= \sum_{k = 0}^{3}i^{k}\angles{x + i^ky, T(x + i^ky)}_{H} \\&= 4\angles{x, Ty}_{H}\end{align*}
Hence $T^{*}$ is an extension of $T$.$\square$