CounterexampleConstruction: AI-generatedVerification: AI-generatedPipeline-generatedprecheck passverified 2026-10-22 (gpt-6.1-sol)

How statement and proof provenance work

The first chip identifies the source of the statement or construction; the second identifies the source of its local proof or verification.

Literature-sourced: the exact statement appears in a cited source; only wording and notation differ.
AI-adapted: a semantically identical restatement of literature-sourced material, modulo indexing, notation, and boundary cases adopted by the library.
AI-generated: a genuinely novel statement formulated by AI, with no source for the claim itself.

These labels describe origin, not correctness: citations and verification chips remain separate evidence.

A small residual does not label an eigenvalue

Statement refuted

Assume Choice and Countable Choice. The following assertion is false: for every lower-bounded self-adjoint operator $H$ with a lowest eigenvalue $E_{0}$ , every normalized $u \in D (H)$ , and every real $λ$ , if $(H - λ I) u = 0$ , then $λ = E_{0}$ and $u$ belongs to the ground eigenspace $\ker (H - E_{0} I)$ .

Statement references

Residual distance and isolated-eigenspace certificate

Facts & Assumptions

Given: Choice and Countable Choice; the residual certificate of [F1]; and the finite-dimensional operator and vector in the counterexample.

[A1]

Choice and Countable Choice are the assumptions of the cited residual-certificate supplier (The Axiom of Choice, The Axiom of Countable Choice ( ${A C}_{ω}$ )).

[F1]

For a normalized vector, the residual certificate bounds $dist (λ, σ (H))$ and, with an explicit spectral cluster and gap, its component outside that cluster; it does not assign an eigenvalue index without such cluster information (Residual distance and isolated-eigenspace certificate).

Proof

1.1A1F1givenalgebra

On $H = C^{2}$ with its standard inner product, set $T = diag (0, 100)$ on all of $H$ , $u = (0, 1)$ , and $λ = 100$ . Then $T$ is self-adjoint and bounded below, $∥ u ∥ = 1$ , $σ (T) = {0, 100}$ , the lowest eigenvalue is $E_{0} = 0$ , and $(T - 100 I) u = 0$ .

2.1F1step 1.1algebra∎

The ground eigenspace is $\ker T = span {(1, 0)}$ , while $u = (0, 1)$ is orthogonal to it and $λ = 100 \neq E_{0}$ . Thus the zero residual verifies a spectral point, as in [F1], but refutes the asserted ground-state label.

A small residual does not label an eigenvalue

Statement refuted

Statement references

Facts & Assumptions

Proof

Depends on

Dependency tree · 2 levels