= Solution
Use the corrected <essential spectrum of a bounded self-adjoint operator> and put $A=L-\lambda I$. Essential spectral points are real. If $\ker A$ is infinite-dimensional, choose an <orthonormal sequence> in the kernel. It converges weakly to zero by the <Bessel inequality>, and its residuals vanish.
If $\ker A$ is finite-dimensional, membership in the <essential spectrum> means the range is not closed. Choose unit $v_n\in(\ker A)^\perp$ with $Av_n\to0$, using the <closed-range bound on the kernel complement>. A bounded <Hilbert space> sequence has a weakly convergent subsequence. Its weak limit $v$ satisfies $Av=0$ because bounded operators preserve <weak convergence>, and $v\in(\ker A)^\perp$; hence $v=0$. This subsequence is a <singular Weyl sequence>.
Conversely a <singular Weyl sequence> first places $\lambda$ in the <spectrum of a bounded operator>. If it were not essential, the <sequential properness for a self-adjoint operator> equivalence for $A$ would yield a norm-convergent subsequence. Its weak limit is zero, whereas norm convergence of unit vectors gives a unit norm limit, a contradiction. Therefore
$$
\boxed{\lambda\in\Sigma_{\mathrm e}(L)\iff\exists(f_n):\ \|f_n\|=1,\ f_n\rightharpoonup0,\ (L-\lambda I)f_n\to0.}
$$
For nonreal $\lambda$, the resolvent lower bound excludes such a sequence, so the equivalence covers all $\lambda\in\mathbb C$.
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