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If unit vectors $f_n$ satisfy $(L-\lambda I)f_n\to 0$, a bounded inverse would give $1\le \Vert (L-\lambda I{)}^{-1}\Vert \Vert (L-\lambda I)f_n\Vert \to 0$. 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 ${\inf }_{\Vert f\Vert =1}\Vert Af\Vert >0$, 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 $\Vert A{f}_n\Vert <1/n$. We have provedSuch 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.

A singular Weyl sequence for a bounded self-adjoint operator $L$ at $\lambda$ is a spectral Weyl sequence that also converges weakly to zero. Such a sequence exists exactly when $\lambda$ is in the essential spectrum of a bounded self-adjoint operator. An infinite-dimensional shifted kernel gives a weakly null orthonormal sequence. If that kernel is finite-dimensional but the shifted range is not closed, choose approximate null vectors in the kernel complement and extract a weakly convergent subsequence; its limit belongs to both kernel and complement, so is zero.