TheoremStatement: AI-adaptedProof: AI-generatedPipeline-generatedprecheck passverified 2026-10-22 (gpt-6.1-sol)

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Rank-one resolvent and bound-state equation

Statement

Assume the Axiom of Choice. Let $H_{0}$ be a self-adjoint operator on a complex Hilbert space $H$ , with inner product linear in its first variable and $\inf σ (H_{0}) > - \infty$ . Let $g \in H$ , $λ \in R$ , and define $P_{g} : H \to H$ by $P_{g} f = ⟨ f, g ⟩ g$ . Put $H_{λ} = H_{0} + λ P_{g}$ on $D (H_{λ}) = D (H_{0})$ and $R_{0} (z) = (H_{0} - z I)^{- 1}$ . Then $H_{λ}$ is self-adjoint. For every $z \in ρ (H_{0})$ such that $d_{λ} (z) : = 1 + λ ⟨ R_{0} (z) g, g ⟩ \neq 0$ , one has $z \in ρ (H_{λ})$ and $R_{λ} (z) : = (H_{λ} - z I)^{- 1} = R_{0} (z) - \frac{λ R_{0} (z) P_{g} R_{0} (z)}{1 + λ ⟨ R_{0} (z) g, g ⟩} .$ For real $E < \inf σ (H_{0})$ and $g \neq 0$ , $E$ is an eigenvalue of $H_{λ}$ if and only if $d_{λ} (E) = 0$ ; in that case the eigenspace is one-dimensional.

Statement references

Kato-Rellich theorem

Resolvent and spectrum of an unbounded operator

Facts & Assumptions

Given: The self-adjoint lower-semibounded operator $H_{0}$ , its domain, $g \in H$ , real $λ$ , and the rank-one map $P_{g}$ defined in the Statement.

[A1]

The resolvent set consists of those $z$ for which $H_{0} - z I : D (H_{0}) \to H$ is bijective with bounded inverse; for self-adjoint $H_{0}$ , each real $E < \inf σ (H_{0})$ belongs to the resolvent set (Resolvent and spectrum of an unbounded operator, Resolvent of a self-adjoint operator: nonreal resolvents and the estimate).

[F1]

Kato–Rellich: a symmetric perturbation with relative operator bound less than one preserves self-adjointness on the original operator domain (Kato-Rellich theorem).

[F2]

Cauchy–Schwarz gives $∣ ⟨ f, g ⟩ ∣ \leq ∥ f ∥ ∥ g ∥$ , and the inner product is linear in its first argument (Cauchy–Schwarz: $∣ ⟨ x, y ⟩ ∣ \leq ∥ x ∥ ∥ y ∥$ , with equality exactly for dependent pairs, Real and complex inner-product spaces and their induced length).

[A2]

Kato–Rellich assumes the Axiom of Choice; the self-adjoint resolvent input assumes Countable Choice. The stated AC hypothesis supplies the latter: for any sequence $(X_{n})_{n \in N}$ of nonempty sets, apply AC to $F = {X_{n} : n \in N}$ to obtain a choice function $c$ ; then $f (n) = c (X_{n}) \in X_{n}$ for every $n$ , which is Countable Choice (The Axiom of Choice).

Proof

Given: [A1], [A2], $P_{g} f = ⟨ f, g ⟩ g$ , and the operators in the Statement.

1.1A2F1F2given

For every $f \in H$ , [F2] gives $∥ P_{g} f ∥ = ∣ ⟨ f, g ⟩ ∣ ∥ g ∥ \leq ∥ g ∥^{2} ∥ f ∥$ . Also $⟨ P_{g} f, h ⟩ = ⟨ f, g ⟩ ⟨ g, h ⟩ = ⟨ f, P_{g} h ⟩$ , so $P_{g}$ is bounded and self-adjoint, including $g = 0$ . Thus $λ P_{g}$ is symmetric on $D (H_{0})$ with relative bound zero; [F1] implies that $H_{λ}$ is self-adjoint on $D (H_{0})$ .

1.2A1givenalgebra

Fix $z \in ρ (H_{0})$ and write $R_{0} = R_{0} (z)$ , $m = ⟨ R_{0} g, g ⟩$ , and $X = R_{0} P_{g}$ . Then $X f = ⟨ f, g ⟩ R_{0} g$ and $X^{2} f = ⟨ f, g ⟩ ⟨ R_{0} g, g ⟩ R_{0} g = m X f$ ; hence $X^{2} = m X$ . If $d = 1 + λ m \neq 0$ , direct multiplication gives $(I + λ X)^{- 1} = I - λ X / d$ .

2.1step 1.2A1algebra

Define $S = (I - λ X / d) R_{0}$ . Its range is contained in $D (H_{0})$ . Since $P_{g} R_{0} P_{g} = m P_{g}$ , on $D (H_{0})$ one has $S (H_{λ} - z I) = (I - λ X / d) (I + λ X) = I$ . On $H$ one has $(H_{λ} - z I) S = I - λ P_{g} R_{0} / d + λ P_{g} R_{0} - λ^{2} m P_{g} R_{0} / d = I$ , using $d = 1 + λ m$ . Thus $S$ is a bounded two-sided inverse from $H$ onto $D (H_{λ})$ , and its expansion is the stated resolvent formula.

3.1A1step 1.1step 1.2algebra∎

Let $E < \inf σ (H_{0})$ be real. By [A1], $R_{0} (E)$ exists. If $H_{λ} u = E u$ with $u \neq 0$ , then $u = - λ ⟨ u, g ⟩ R_{0} (E) g$ . The coefficient $⟨ u, g ⟩$ cannot be zero, because that would give $u = 0$ ; applying $⟨ \cdot, g ⟩$ therefore yields $1 + λ ⟨ R_{0} (E) g, g ⟩ = 0$ . Conversely, if this scalar is zero, then $u = R_{0} (E) g \neq 0$ and $(H_{λ} - E I) u = g + λ ⟨ R_{0} (E) g, g ⟩ g = 0$ . Every eigenvector satisfies the same scalar-multiple formula for $R_{0} (E) g$ , so the eigenspace is one-dimensional.

Rank-one resolvent and bound-state equation

Statement

Statement references

Facts & Assumptions

Proof

Depends on

Dependency tree · 2 levels

Sources