Contour perturbation projection

[The supplied spectral support gives the gap estimate] Apply [F1] to the arbitrary Hilbert space and self-adjoint $H_0$. Its PVM is concentrated on $\sigma (H_0)$, its resolvent multiplier has the exact domain, and its nonzero-neighborhood characterization identifies the spectral support with that spectrum. Its bound applies equally after changing from $(H_0-z{)}^{-1}$ to $R_0(z)=(z-H_0{)}^{-1}$, which changes only the sign. For $z\in \mathrm{\Gamma }$, the distinguished spectral point $E$ has distance $g\mathrm{/}2$ from $z$, while every other $t\in \sigma (H_0)$ satisfies $\mathrm{\mid }t-z\mathrm{\mid }\ge \mathrm{\mid }t-E\mathrm{\mid }-\mathrm{\mid }z-E\mathrm{\mid }\ge g\mathrm{/}2$. Therefore $\mathrm{dist}(z,\sigma (H_0))\ge g\mathrm{/}2$ and $\mathrm{\parallel }{R}_0(z)\mathrm{\parallel }\le 2\mathrm{/}g$ on $\mathrm{\Gamma }$. The only spectral point in the open interior is $E$. No separability or arbitrary global spectral decomposition is used.

[Correct Neumann factors and exact domains] The bounded perturbation is closed on $D(H_0)$: if $x_n\to x$ and $H_{\lambda }{x}_n\to y$, then $H_0{x}_n=H_{\lambda }{x}_n-\lambda V{x}_n\to y-\lambda Vx$ and closedness of $H_0$ gives the claim. This domain is dense. On $D(H_0)$ factor $z-H_{\lambda }=(I-\lambda V{R}_0(z))(z-H_0)$. Step 1.1 gives $\mathrm{\parallel }\lambda V{R}_0(z)\mathrm{\parallel }\le q<1$, so the displayed inverse series converges uniformly and maps $H$ into $D(H_0)$. Multiplying in the stated order gives both $(z-H_{\lambda })R_{\lambda }=I_H$ and $R_{\lambda }(z-H_{\lambda })=I_{D(H_0)}$. The geometric-series bounds give (a), with resolvent difference bound $(2\mathrm{/}g)q\mathrm{/}(1-q)$. Local inverse Neumann series also make $z\mapsto R_{\lambda }(z)$ norm holomorphic near the circle, and the uniform coupling power series makes $\lambda \mapsto R_{\lambda }(z)$ norm holomorphic throughout the full disk. A bounded perturbation adjoint test gives $D(H_{\lambda }^{\ast })=D(H_0)$ and $H_{\lambda }^{\ast }=H_0+\bar{\lambda }V$: subtract the bounded functional arising from $\lambda V$ in the defining adjoint identity to identify the domain with $D(H_0^{\ast })$. In particular $H_{\lambda }$ is self-adjoint for real $\lambda$.

[Idempotence without a complex PVM] For each fixed $\lambda$, the inverse norm bound in step 2.1 and another Neumann series give an open annulus around $\mathrm{\Gamma }$ consisting of resolvent points. Within this annulus, circle integrals of a holomorphic operator function are independent of radius: differentiating ${\int }_0^{2\pi }f(E+r{e}^{it})ir{e}^{it}dt$ in $r$ gives the integral of the $t$ derivative of $f(E+r{e}^{it})e^{it}$, which is zero by periodicity. Choose concentric outer and inner circles in the annulus. The inverse equations give $R_{\lambda }(z)R_{\lambda }(w)=[R_{\lambda }(w)-R_{\lambda }(z)]\mathrm{/}(z-w)$ on the two circles. In the double integral of this identity, integrating the first term in the outer variable gives its inner projection integral because $\oint dz\mathrm{/}(z-w)=2\pi i$; integrating the second term in the inner variable gives zero because $\oint dw\mathrm{/}(z-w)=0$. Interchanging continuous compact integrals is justified by norm Riemann sums. Both circle projections equal $P_{\lambda }$ by annular independence, so $P_{\lambda }^2=P_{\lambda }$. Resolvents commute by the same identity, and thus commute with $P_{\lambda }$. Also $H_{\lambda }{R}_{\lambda }(z)=z{R}_{\lambda }(z)-I$. Integrating in the graph norm of the closed $H_{\lambda }$ shows $P_{\lambda }H\subseteq D(H_0)$ and $H_{\lambda }{P}_{\lambda }=(2\pi i{)}^{-1}\oint z{R}_{\lambda }(z)dz$, a bounded operator; the inverse equations on $D(H_0)$ give $P_{\lambda }{H}_{\lambda }=H_{\lambda }{P}_{\lambda }$ there. The uniform series makes $P_{\lambda }$ and this bounded $H_{\lambda }{P}_{\lambda }$ holomorphic in $\lambda$.

[The initial projection and full-disk rank argument] At $\lambda =0$, [F1] evaluates the scalar contour integrals against the spectral measures of $H_0$: their support misses the circle, and the circle integral of $(z-t{)}^{-1}$ is one for $\mathrm{\mid }t-E\mathrm{\mid }<g\mathrm{/}2$ and zero outside. Hence $P_{\lambda =0}=E_{H_0}(\{E\})=P_0$. The circle length is $\pi g$, so the difference estimate in step 2.1 gives $\mathrm{\parallel }{P}_{\lambda }-P_0\mathrm{\parallel }\le q\mathrm{/}(1-q)$. For bounded idempotents $P,Q$ with $\mathrm{\parallel }P-Q\mathrm{\parallel }<1$, $Qx=0$, $x\in \mathrm{ran}P$, implies $\mathrm{\parallel }x\mathrm{\parallel }=\mathrm{\parallel }(P-Q)x\mathrm{\parallel }<\mathrm{\parallel }x\mathrm{\parallel }$ unless $x=0$. Thus $Q{\mathrm{\mid }}_{\mathrm{ran}P}$ is injective, and the reverse restriction is injective by the same argument. For fixed $\lambda$ and $s,t\in [0,1]$, the inverse equations give $R_{s\lambda }(z)-R_{t\lambda }(z)=(s-t)\lambda R_{s\lambda }(z)V{R}_{t\lambda }(z)$, so $\mathrm{\parallel }{P}_{s\lambda }-P_{t\lambda }\mathrm{\parallel }\le q\mathrm{\mid }s-t\mathrm{\mid }\mathrm{/}(1-q{)}^2$. Choose an integer $N>q\mathrm{/}(1-q{)}^2$ and compare the finite sequence $P_{j\lambda \mathrm{/}N}$, $0\le j\le N$. Consecutive distances are less than one, so the two injections preserve the one-dimensional rank at every step. Therefore $P_{\lambda }$ has rank one for every $q<1$, even when its distance from $P_0$ is not less than one.

[The entire interior spectrum is one simple eigenvalue] The invariant one-dimensional range of $P_{\lambda }$ supports a scalar $E_{\lambda }$ with $H_{\lambda }{P}_{\lambda }=E_{\lambda }{P}_{\lambda }$. Since $R_{\lambda }(z)P_{\lambda }=(z-E_{\lambda }{)}^{-1}{P}_{\lambda }$, the projection integral and $P_{\lambda }\mathrm{\ne }0$ imply that $E_{\lambda }$ lies strictly inside the circle. For any $w$ inside it define the bounded operator $B(w)=(2\pi i{)}^{-1}\oint R_{\lambda }(z)\mathrm{/}(z-w)dz$. Its range lies in $D(H_0)$ by graph-norm integration, and the inverse equations give $(w-H_{\lambda })B(w)=I-P_{\lambda }$ on $H$ and $B(w)(w-H_{\lambda })=I-P_{\lambda }$ on $D(H_0)$. It commutes with $P_{\lambda }$. Its product with $P_{\lambda }$ is zero: the scalar integral of $[(z-E_{\lambda })(z-w){]}^{-1}$ is zero, both when the two interior poles are distinct by partial fractions and when equal by the double-pole monomial integral. Thus $B(w)$ is the two-sided inverse of $w-H_{\lambda }$ on $\ker P_{\lambda }$, including the exact restricted domain. For $w\mathrm{\ne }{E}_{\lambda }$, $B(w)+(w-E_{\lambda }{)}^{-1}{P}_{\lambda }$ is the two-sided inverse on all $H$, proving absence of any other interior spectrum. At $w=E_{\lambda }$, the complementary inverse excludes complementary eigenvectors and generalized eigenvectors, while the rank-one range has no nontrivial Jordan chain. The eigenspace and algebraic multiplicity are therefore precisely those asserted in (b).

[Holomorphic eigenvalue and the real-coupling interpretation] Around any coupling ${\lambda }_0$ choose a nonzero $x\in \mathrm{ran}{P}_{{\lambda }_0}$ and the continuous complex-linear functional $\mathrm{\ell }(y)=⟨y,x⟩$. Continuity makes $\mathrm{\ell }(P_{\lambda }x)\mathrm{\ne }0$ nearby. There $E_{\lambda }=\mathrm{\ell }(H_{\lambda }{P}_{\lambda }x)\mathrm{/}\mathrm{\ell }(P_{\lambda }x)$ is holomorphic by step 3.1. The uniqueness of step 5.1 makes these locally defined functions agree, so $E_{\lambda }$ is holomorphic throughout the disk. For real $\lambda$, self-adjointness from step 2.1 and [F1] imply $E_{\lambda }\in \mathbb{R}$; the same scalar contour evaluation as in step 4.1, now applied to $H_{\lambda }$, identifies $P_{\lambda }$ with its orthogonal PVM projection on the interior, hence on $\{E_{\lambda }\}$. For complex coupling the valid construction is the Riesz projection of steps 3.1–5.1. All claims follow on arbitrary $H$ under CC without adding separability or AC.