Continuum contribution in perturbation

[Explicit operator and spectral measure] The $L^2((1,2))$ completeness and pairing in [F1] make it a complex Hilbert space. If $c_n\oplus f_n$ is Cauchy in the stated direct-sum norm, each component is Cauchy; completeness of $\mathbb{C}$ and this $L^2$ space gives limits $c,f$, and the norm identity gives convergence to $c\oplus f$. Thus $H$ is complete before any Hilbert-space theorem or PVM definition is used. Since $1<x<2$, $M_x$ is bounded on all $L^2(1,2)$ with norm $2$, and real multiplication gives $⟨M_{x}f,h⟩=⟨f,M_{x}h⟩$. A bounded symmetric operator on the whole Hilbert space equals its adjoint directly by the defining pairing, so $H_0$ is self-adjoint with domain $H$. For every real Borel $B$ define $\mathcal{E}(B)(c\oplus f)={\mathbf{1}}_B(0)c\oplus {\mathbf{1}}_B(x)f(x)$. These are orthogonal projections, $\mathcal{E}(\mathbb{R})=I$, and their products are $\mathcal{E}(B\cap C)$. On a fixed vector, disjoint countable additivity follows from pointwise indicator sums and the integrable squared-norm majorant $\mathrm{\mid }f{\mathrm{\mid }}^2$ (with the scalar component evaluated at $0$), so this is a PVM. Its scalar measures are $\mathrm{\mid }c{\mathrm{\mid }}^2{\delta }_0(B)+{\int }_{B\cap (1,2)}\mathrm{\mid }f(x){\mathrm{\mid }}^{2}dx$. Integrating bounded simple approximations to the coordinate function on $\{0\}\cup [1,2]$ yields $H_0=\int td\mathcal{E}(t)$ on all $H$. Thus the representing PVM and its bounded multiplier realization are explicit.

[Spectrum, gap and absence of complementary eigenvectors] For $z\mathrm{\notin }\{0\}\cup [1,2]$, $(z-H_0{)}^{-1}(c\oplus f)=c\mathrm{/}z\oplus f\mathrm{/}(z-x)$ is a bounded two-sided inverse on $H$. For any $t\in [1,2]$ and $\epsilon >0$, the interval $J=(1,2)\cap (t-\epsilon ,t+\epsilon )$ has positive length. The vector $0\oplus {\mathbf{1}}_J\mathrm{/}\mathrm{\mid }J{\mathrm{\mid }}^{1\mathrm{/}2}$ is unit and has $\mathrm{\parallel }(H_0-t)u\mathrm{\parallel }\le \epsilon$, excluding a bounded inverse at $t$. Finally $\ker H_0=\mathbb{C}\oplus \{0\}$, so $0$ is an isolated simple eigenvalue and its distance to the remaining spectrum is $g=1$. The equation $(x-t)f=0$ forces $f=0$ in $L^2(1,2)$ because a singleton has Lebesgue measure zero. Consequently the complementary block has no eigenvectors and all its spectral measures are absolutely continuous with densities $\mathrm{\mid }f{\mathrm{\mid }}^2$ on $(1,2)$, including zero atom at the endpoints. In particular $P_0=\mathcal{E}(\{0\})$ and ${\psi }_0=1\oplus 0$ has norm one.

[Perturbation hypotheses and the restricted inverse] Let $I(f)={\int }_1^{2}f(x)dx=⟨f,\mathbf{1}⟩$. Cauchy-Schwarz and the interval's length one give $\mathrm{\mid }I(f)\mathrm{\mid }\le \mathrm{\parallel }f{\mathrm{\parallel }}_2$. Therefore $\mathrm{\parallel }V(c\oplus f){\mathrm{\parallel }}^2=\mathrm{\mid }I(f){\mathrm{\mid }}^2+\mathrm{\mid }c{\mathrm{\mid }}^2\le \mathrm{\parallel }c\oplus f{\mathrm{\parallel }}^2$; equality at ${\psi }_0$ gives $v=\mathrm{\parallel }V\mathrm{\parallel }=1>0$. For $c\oplus f,d\oplus h$, the first-variable-linear pairing gives $⟨V(c\oplus f),d\oplus h⟩=I(f)\bar{d}+c\overline{I(h)}=⟨c\oplus f,V(d\oplus h)⟩$, so $V=V^{\ast }$. Moreover $V{\psi }_0=0\oplus \mathbf{1}$ and the first-order pairing is zero. On $QH=0\oplus L^2(1,2)$, $H_Q-E=M_x$ has inverse $M_{1\mathrm{/}x}$, bounded on that exact whole complementary domain with norm one. Hence $S(c\oplus f)=0\oplus (f\mathrm{/}x)$, $S{P}_0=P_{0}S=0$ and $(H_0-E)S=S(H_0-E)=Q$ on $H$. No global inverse of $H_0-E$ is used. The preceding operator, spectral and perturbation computations verify every concrete hypothesis of [F2], with inherited CC and $q=2\mathrm{\mid }\lambda \mathrm{\mid }$.

[Exact coefficient and its continuous origin] The explicit reduced inverse gives $SV{\psi }_0=0\oplus (1\mathrm{/}x)$ and $VSV{\psi }_0=({\int }_1^{2}dx\mathrm{/}x)\oplus 0$. Pairing with the unit ${\psi }_0$ and applying [F2], [F3] yields $E^{\mathrm{\prime }\mathrm{\prime }}(0)\mathrm{/}2=-{\int }_1^{2}dx\mathrm{/}x=-\log 2<0$. The scalar spectral measure of $V{\psi }_0=0\oplus \mathbf{1}$ is exactly ${\mathbf{1}}_{(1,2)}(x)dx$ by step 1.1, so the entire coefficient integrates $1\mathrm{/}x$ over the continuous complementary spectrum. Step 2.1 shows that the only bound eigenvector belongs to $P_{0}H$, with zero coupling matrix element; there are no complementary bound eigenvectors. A sum restricted to bound states therefore gives zero and omits the exact negative coefficient. This proves all original example claims, its normalization and class, with the correct restricted inverse and explicit inherited CC.