Unbounded self-adjoint operator ($A=A^{\ast },D(A)=D(A^{\ast })$)

A densely defined linear operator on a complex Hilbert space is self-adjoint when its adjoint operator has exactly the same domain and agrees there. Symmetry only requires $A\subseteq A^{\ast }$ and does not guarantee self-adjointness. A self-adjoint operator is a closed operator with real spectrum. Domains matter for differential operators and unbounded multiplication operators.

Positivity makes $A-z$ bounded below with dense closed range for every $z<0$, excluding negative spectral values. Conversely if the spectrum is nonnegative, $R=(I+A{)}^{-1}$ has spectrum in $[0,1]$ by the resolvent spectral mapping identity. The numerical-range endpoint approximate-eigenvalue theorem gives $0\le R\le I$, and positive-form Cauchy-Schwarz inequality gives $\Vert Rf{\Vert }^2\le ⟨Rf,f⟩$. For $u=Rf$, $⟨Au,u⟩=⟨Rf,f⟩-\Vert Rf{\Vert }^2\ge 0$.

Symmetry gives $\Vert (A-z)u\Vert \ge |\mathrm{Im}z|\Vert u\Vert$ when $z$ is nonreal. Closedness makes the range closed; its orthogonal complement is $\ker (A-\overline{z})=0$, so it is also dense and hence surjective. This proves the resolvent estimate and reality of the spectrum. Equality of the adjoint domains is needed in the range argument.