= Solution
The intended <self-adjoint operator> identity is $\langle Lf,g\rangle=\langle f,Lg\rangle$, with the <inner product> linear in its first argument. The printed second $Lf$ must be $Lg$. Taken literally, the printed identity forces $L=0$: put $g=0$ and then vary $g$. Its spectral conclusion is then trivial, but it does not define self-adjointness.
For the corrected identity, $\langle Lv,v\rangle$ is real. If $\operatorname{Im}\lambda\ne0$, the <Cauchy-Schwarz inequality> gives
$$
\|(L-\lambda I)v\|\|v\|\ge|\operatorname{Im}\langle(L-\lambda I)v,v\rangle|=|\operatorname{Im}\lambda|\|v\|^2.
$$
Thus $L-\lambda I$ is injective and bounded below, so its range is closed: a convergent image sequence has a <Cauchy sequence> of preimages. Its range is dense because its <orthogonal complement> is $\ker(L-\bar\lambda I)$, also zero by the same lower bound. Hence it is surjective with inverse norm at most $1/|\operatorname{Im}\lambda|$. Therefore
$$
\boxed{\Sigma(L)\subset\mathbb R.}
$$
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