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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 $\lambda \in \mathbb{C}$.

Let $(f_n)$ be a singular Weyl sequence for $L$ at $\lambda \in {\Sigma }_{\mathrm{e}}(L)$. Since $K$ is compact and $f_n\rightharpoonup 0$, the preceding result gives $K{f}_n\to 0$. ThereforeThe norms remain one and the weak limit remains zero. The singular Weyl sequence criterion yields $\lambda \in {\Sigma }_{\mathrm{e}}(L+K)$. Apply the same argument to $L+K$ and the compact self-adjoint operator $-K$ for the reverse inclusion. Both operators are bounded and self-adjoint. Thus the Weyl theorem for compact self-adjoint perturbations isThe corrected shifted-range definition is necessary. For the diagonal example in the preceding solutions, a rank-one perturbation changing the entry $2$ to $0$ removes $2$ from the spectrum. The unshifted printed definition had classified $2$ as essential merely because the original range was not closed, so it would make this invariance false.

For the subsequent essential spectrum arguments, the discrete-spectrum definition must use $\mathrm{ran}(L-\lambda I)$ closed, rather than the printed unshifted range. With that correction, $0\notin {\Sigma }_{\mathrm{e}}(L)$ means exactly that $N=\ker L$ is finite-dimensional and $\mathrm{ran}L$ is closed; this includes the case $0$ is in the resolvent set.

A closed-range bound on the kernel complement supplies the useful equivalenceFor the forward implication, $L:N^{\perp }\to \mathrm{ran}L$ is a bounded bijection between Banach spaces, so the bounded inverse theorem applies. For the reverse implication, any Cauchy sequence of image points has a Cauchy sequence of preimages in $N^{\perp }$, and completeness gives a preimage of its limit.

Now decompose a bounded sequence as $f_n=p_n+v_n$ with $p_n\in N$, $v_n\in N^{\perp }$. If $L{f}_n$ converges, the lower bound makes $(v_n)$ a Cauchy sequence. The finite-dimensional vector space $N$ makes the bounded $p_n$ have a convergent subsequence. Their sum has a norm-convergent subsequence.

Conversely, if every bounded sequence with convergent images has a norm-convergent subsequence, the kernel cannot be infinite-dimensional: an orthonormal sequence in it would have zero images and no convergent subsequence. If the range were not closed, the lower-bound equivalence would provide unit vectors $v_n\in N^{\perp }$ with $L{v}_n\to 0$. Any norm limit would lie in both $N$ and $N^{\perp }$, hence be zero, contradicting its unit norm. This proves the required sequential properness for a self-adjoint operator equivalence.

The spectral-shift repair is essential for later parts. On ${\ell }^2$, take $L=\mathrm{diag}(2,1,1/2,1/3,\dots )$. Its range is not closed, but $2$ is an isolated eigenvalue with one-dimensional eigenspace and closed shifted range. The printed definition would incorrectly place $2$ in the essential spectrum, although no singular Weyl sequence exists there: on the complement of that eigenspace, $\Vert (L-2I)v\Vert \ge \Vert v\Vert$.

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.