= Solution
If unit vectors $f_n$ satisfy $(L-\lambda I)f_n\to0$, a bounded inverse would give $1\le\|(L-\lambda I)^{-1}\|\|(L-\lambda I)f_n\|\to0$. Hence $\lambda$ belongs to the <spectrum of a bounded operator>.
Conversely a spectral point is real by the preceding argument. Put $A=L-\lambda I$, again a <self-adjoint operator>. If $\inf_{\|f\|=1}\|Af\|>0$, then $A$ is injective with closed range. Its range is dense by <image-kernel orthogonality for an adjoint>, so it is bijective with a bounded inverse, a contradiction. Thus this infimum is zero; choose unit $f_n$ with $\|Af_n\|<1/n$. We have proved
$$
\boxed{\lambda\in\Sigma(L)\iff\exists(f_n):\ \|f_n\|=1,\quad(L-\lambda I)f_n\to0.}
$$
Such a sequence is a <spectral Weyl sequence>. No weak convergence is required here; nonreal $\lambda$ are ruled out by the lower bound in the preceding solution.
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