 Solution

If $|z|>\Vert L\Vert$, the Neumann series givesThe series converges in operator norm because $\Vert L/z\Vert <1$. HenceThe same argument applied to small perturbations of an invertible operator shows that the resolvent set is open and the spectrum is closed.

Suppose $L=L^{\ast }$ and $\mathrm{Im}z\ne 0$. The quadratic form $⟨Lh,h⟩$ is real, so the Cauchy-Schwarz inequality impliesThus $L-zI$ is bounded below by $|\mathrm{Im}z|$: it is injective and has closed range. Its adjoint operator $L-\bar{z}I$ has the same lower bound and trivial kernel. Sincethe range is dense as well as closed, and therefore is all of $H$. The inverse has norm at most $|\mathrm{Im}z{|}^{-1}$. For a bounded self-adjoint operator, $\Sigma (L)\subset \mathbb{R}$. The closed-range and adjoint steps are essential: merely excluding nonreal eigenvalues would not exclude all nonreal spectral points.