Simple-level reduced-resolvent coefficients

[The actual complementary inverse] The restricted PVM of $H_Q$ has zero projection at $E$, and its spectral support is contained in $\sigma (H_0)\setminus \{E\}$. Indeed its representing measures are restrictions to $QH$, and the neighborhood characterization of [F1] shows that no spectral point can occur in the gap $\mathrm{\mid }t-E\mathrm{\mid }<g$. The multiplier $b(t)=1\mathrm{/}(t-E)$ on this restricted support has $\mathrm{\mid }b\mathrm{\mid }\le 1\mathrm{/}g$ and bounded $tb(t)$, so its calculus maps $QH$ into exactly $D(H_Q)$ and gives the two-sided inverse $S_Q=(H_Q-E{)}^{-1}$ with norm at most $1\mathrm{/}g$. Extend it by zero on $P_{0}H$ to get the stated $S$. It is self-adjoint because its multiplier is real, and $S{P}_0=P_{0}S=0$, $(H_0-E)S=Q$ on $H$ and $S(H_0-E)=Q$ on $D(H_0)$. This uses the whole complementary spectral support, including any continuous part.

[Spectral inclusion and an exact Q-block inverse] Put $\delta =\mathrm{\mid }\lambda \mathrm{\mid }v$ and $e_{\lambda }=E_{\lambda }-E$. Whenever $\mathrm{dist}(z,\sigma (H_0))>\delta$, [F1] makes $\mathrm{\parallel }\lambda V(z-H_0{)}^{-1}\mathrm{\parallel }<1$ and the correct Neumann factor $(z-H_{\lambda })=[I-\lambda V(z-H_0{)}^{-1}](z-H_0)$ gives a two-sided inverse on the actual domain. Hence every spectral point of $H_{\lambda }$ has distance at most $\delta$ from $\sigma (H_0)$. Since $E_{\lambda }$ lies inside the $g\mathrm{/}2$ circle and $\delta <g\mathrm{/}4$, all spectral points of $H_0$ other than $E$ have distance greater than $g\mathrm{/}2$ from it; therefore $\mathrm{\mid }{e}_{\lambda }\mathrm{\mid }\le \delta$. On $QH$ consider $T_{\lambda }=H_Q-E-e_{\lambda }+\lambda QVQ$, on $D(H_Q)$. The bounded perturbation $D_{\lambda }=\lambda QVQ-e_{\lambda }I$ has norm at most $2\delta =qg<g$. The factorization $T_{\lambda }=(I+D_{\lambda }{S}_Q)(H_Q-E)$ yields the two-sided inverse $B_{\lambda }=S_Q(I+D_{\lambda }{S}_Q{)}^{-1}$, mapping $QH$ into the exact $D(H_Q)$, with $\mathrm{\parallel }{B}_{\lambda }\mathrm{\parallel }\le 1\mathrm{/}[g(1-q)]$. These inverses are holomorphic in $\lambda$ because $e_{\lambda }$ is and the geometric series converges locally uniformly.

[The gauge denominator and full-radius norm bound] If $u\mathrm{\ne }0$ is an eigenvector in $P_{\lambda }H$ and $P_{0}u=0$, the projected eigenvalue equation is $T_{\lambda }u=0$, contradicting step 2.1; thus every nonzero eigenvector has a nonzero ${\psi }_0$ coefficient. Also $P_{\lambda }{\psi }_0\mathrm{\ne }0$ because $\mathrm{\parallel }{P}_{\lambda }-P_0\mathrm{\parallel }<1$ by [F2]. It is an eigenvector, so its gauge denominator $⟨P_{\lambda }{\psi }_0,{\psi }_0⟩$ is nonzero throughout the full $q<1\mathrm{/}2$ disk. Division gives the holomorphic eigenvector of the Statement, with $⟨\psi (\lambda ),{\psi }_0⟩=1$. Write $\psi ={\psi }_0+\chi$, $\chi \in QH$. The Q-block equation is $T_{\lambda }\chi =-\lambda QV{\psi }_0$, so $\chi =-\lambda B_{\lambda }QV{\psi }_0$ and $\mathrm{\parallel }\chi \mathrm{\parallel }\le q\mathrm{/}[2(1-q)]<1\mathrm{/}2$, in particular $\mathrm{\parallel }Q\psi \mathrm{\parallel }\le q$. Orthogonality gives $\mathrm{\parallel }\psi {\mathrm{\parallel }}^2=1+\mathrm{\parallel }\chi {\mathrm{\parallel }}^2<5\mathrm{/}4$, so the original public bound $\mathrm{\parallel }\psi \mathrm{\parallel }\le 3$ holds on the entire stated complex disk. Graph analyticity also holds: $H_0{P}_{\lambda }=H_{\lambda }{P}_{\lambda }-\lambda V{P}_{\lambda }$ is bounded holomorphic by [F2], and division by the same nonzero scalar gives holomorphy of both $\psi$ and $H_0\psi$. Thus differentiating the unbounded eigenvalue equation is justified in the exact graph domain.

[The holomorphic gauge energy identity] Pair the eigenvalue equation with the fixed ${\psi }_0$ in the second argument. Self-adjointness of $H_0$ and the gauge give $⟨H_0\psi ,{\psi }_0⟩=⟨\psi ,H_0{\psi }_0⟩=E$ and $⟨E_{\lambda }\psi ,{\psi }_0⟩=E_{\lambda }$. Consequently $E_{\lambda }=E+\lambda ⟨V\psi (\lambda ),{\psi }_0⟩$, a holomorphic scalar identity with a fixed second vector. Since $\psi (0)={\psi }_0$, its derivative is $E^{\mathrm{\prime }}(0)=⟨V{\psi }_0,{\psi }_0⟩$. The gauge gives $⟨{\psi }^{\mathrm{\prime }}(0),{\psi }_0⟩=0$. No unit-norm claim for the gauge vector, or holomorphic pairing with a moving second vector, is used.

[Vector derivative and second energy coefficient] Differentiate the eigenvalue equation at zero using graph analyticity in step 3.1: $(H_0-E){\psi }^{\mathrm{\prime }}(0)=-V{\psi }_0+E^{\mathrm{\prime }}(0){\psi }_0$. Apply the actual $S$ and use the gauge, so $Q{\psi }^{\mathrm{\prime }}(0)={\psi }^{\mathrm{\prime }}(0)$; step 1.1 gives ${\psi }^{\mathrm{\prime }}(0)=-SV{\psi }_0$. Differentiating the holomorphic scalar identity of step 4.1 twice gives $E^{\mathrm{\prime }\mathrm{\prime }}(0)=2⟨V{\psi }^{\mathrm{\prime }}(0),{\psi }_0⟩=-2⟨VSV{\psi }_0,{\psi }_0⟩$. All quantities are finite because $V,S$ are bounded. The second coefficient is real because $VSV$ is self-adjoint, but no real-part operation is inserted into a complex derivative. Taylor expansion gives exactly the first two terms and $O({\lambda }^3)$ of the Statement; the companion remainder item supplies its explicit bound. The public analytic radius and all coefficient claims are retained under CC on arbitrary $H$.