Symmetric operator

Remarks

Example: on $H = L^2(0, 1)$ let $D(T)$ be the smooth functions with compact support in $(0, 1)$ and $Tf = -i f'$. For $f, g \in D(T)$, integration by parts with vanishing boundary terms gives $\int_0^1 -i f' \bar g = \int_0^1 f \, \overline{-i g'}$, so $T$ is symmetric.

Non-example: on the same domain, $Sf = f'$ satisfies $\ip{Sf}{g} = -\ip{f}{Sg}$, so $\ip{Sf}{g} \ne \ip{f}{Sg}$ whenever $\ip{f}{Sg} \ne 0$; it is skew-symmetric, not symmetric.

An equivalent form: $T$ is symmetric if and only if its adjoint $T^*$ extends $T$, written $T \subseteq T^*$. $T$ is self-adjoint when moreover $T^* = T$, including equality of domains. By the Hellinger–Toeplitz theorem, a symmetric operator with $D(T) = H$ is bounded, and then it is self-adjoint. Some authors, and Mathlib, call a linear map $T \colon E \to E$ on an inner product space symmetric when the identity holds on all of $E$, without a domain.