= Solution
If $|z|>\|L\|$, the <Neumann series> gives
$$
(L-zI)^{-1}
=-\frac1z\sum_{n=0}^\infty\left(\frac Lz\right)^n.
$$
The series converges in <operator norm> because $\|L/z\|<1$. Hence
$$
\boxed{\Sigma(L)\subset\{z:|z|\leq\|L\|\}.}
$$
The same argument applied to small perturbations of an invertible operator shows that the <resolvent set> is open and the <spectrum> is closed.
Suppose $L=L^*$ and $\operatorname{Im}z\ne0$. The quadratic form $\langle Lh,h\rangle$ is real, so the <Cauchy-Schwarz inequality> implies
$$
|\operatorname{Im}z|\,\|h\|^2
=|\operatorname{Im}\langle(L-zI)h,h\rangle|
\leq\|(L-zI)h\|\,\|h\|.
$$
Thus $L-zI$ is bounded below by $|\operatorname{Im}z|$: it is injective and has closed range. Its <adjoint operator> $L-\bar zI$ has the same lower bound and trivial kernel. Since
$$
\operatorname{ran}(L-zI)^\perp=\ker(L-\bar zI)=\{0\},
$$
the range is dense as well as closed, and therefore is all of $H$. The inverse has norm at most $|\operatorname{Im}z|^{-1}$. \b[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.
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