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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 $A{v}_n\to 0$, 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. ThereforeFor nonreal $\lambda$, the resolvent lower bound excludes such a sequence, so the equivalence covers all $\lambda \in \mathbb{C}$.

For a bounded self-adjoint operator $L$, every bounded sequence whose images converge has a norm-convergent subsequence exactly when its kernel is finite-dimensional and its range is closed, or equivalently $0\notin {\Sigma }_{\mathrm{e}}(L)$ for the essential spectrum of a bounded self-adjoint operator. Split the sequence into kernel and kernel complement: finite dimensionality gives a subsequence on the first part and the closed-range bound on the kernel complement makes the second part a Cauchy sequence. Infinite kernel or approximate null unit vectors in its complement obstruct the property.

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.

If $L$ is bounded and self-adjoint on a complex Hilbert space and $K$ is compact and self-adjoint, thenA singular Weyl sequence for $L$ stays singular for $L+K$, because compact operators send weak convergence to norm convergence and therefore $K{f}_n\to 0$. Applying the same argument with $-K$ proves the reverse inclusion. Finite-multiplicity isolated eigenvalues can move under such perturbations; the essential spectrum of a bounded self-adjoint operator is unchanged.