Does for an operator on an Hilbert space symmetric, densley defined and bijective imply self adjoint?

Question

I have an operator $L \colon D(L) \subset H \to H$, where $H$ is an Hilbert space and $D(L)$ is a dense subset of $H$. If $L$ is now symmetric and bijective, is it then automatically self-adjoint?

Answer 1

Yes, it is.

Symmetry implies $L \preceq L^*$ (graph inclusion.) To show that $L^*\preceq L$, suppose $x \in \mathcal{D}(L^*)$. Then there exists $y\in \mathcal{D}(L)$ such that $Ly=L^*x$ because $L$ is surjective. Then, for all $z\in\mathcal{D}(L)$, $$ \langle x,Lz\rangle= \langle L^*x, z \rangle = \langle Ly,z\rangle=\langle y,Lz\rangle. $$ Because $L$ is surjective, then $x=y$, which means an arbitrary $x\in\mathcal{D}(L^*)$ is in $\mathcal{D}(L)$.