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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.

Variational approximation, residual certificates and propagation errors — Examples

1 · Prerequisites

2 · Summary

3 · Logical flowchart

4 · Definitions, postulates, theorems and proofs

None yet.

5 · Examples, experiments and counterexamples

ExampleConstruction: AI-generatedVerification: AI-generatedprecheck passverified 2026-10-22 (gpt-6.1-sol)Open item page →

Hermite Ritz exactness

Example

Assume Choice and Countable Choice. Fix m,ω,ℏ>0 and let Hosc be the self-adjoint coefficient-domain realization on L2(R,dx) of −ℏ22md2dx2+mω22x2 with normalized Hermite eigenfunctions (ψj)j≥0, where Hoscψj=Ejψj and Ej=ℏω(j+12). For n≥1, let Vn=span⁡{ψ0,…,ψn−1}⊆Q(Hosc). The n Ritz values of the associated form on Vn are exactly E0,…,En−1 in increasing order, and each continuum residual (Hosc−EjI)ψj is zero.

Facts & Assumptions

Given: Choice and Countable Choice; m,ω,ℏ>0; the oscillator Hosc; its normalized Hermite eigenfunctions ψj; and Vn as in the Example.

[A1]

Choice and Countable Choice are the assumptions of the form min–max result and the exact oscillator realization used here (The Axiom of Choice, The Axiom of Countable Choice (ACω)).

[F1]

The oscillator realization is self-adjoint, has the stated coefficient domain, satisfies Hosc≥ℏω2I>0, its Hermite functions form a complete orthonormal family, and Hoscψj=ℏω(j+12)ψj (Oscillator scaling and exact realization).

[F2]

In the spectral-form definition of Conforming Ritz eigenvalue upper bounds, take A=Hosc and c=0. Since Hosc≥0, this gives Q(Hosc)=D(S) and q(f,g)=⟨Sf,Sg⟩ for S=Hosc1/2 defined by the Borel functional calculus.

[F3]

The spectral/Borel calculus gives D(Hosc)={f:∫x2 dμf(x)<∞} and Q(Hosc)=D(Hosc1/2)={f:∫x dμf(x)<∞} because Hosc≥0; x≤1+x2 on [0,∞) therefore gives D(Hosc)⊆Q(Hosc) (Spectral theorem for unbounded self-adjoint operators (PVM form), Unbounded Borel functional calculus: domains, products, spectral mapping).

[F4]

The Ritz values on a finite-dimensional conforming form subspace are the ordered eigenvalues of its exact Gram/form generalized eigenproblem (Conforming Ritz eigenvalue upper bounds).

Proof

1.1A1F1F2F3algebra

By [F1], each ψj∈D(Hosc), the family is orthonormal, and Hoscψj=Ejψj. By [F3], each ψj lies in Q(Hosc)=D(S). The product rule gives S2=Hosc with D(S2)=D(Hosc); since S is self-adjoint, for i,j<n we have q(ψj,ψi)=⟨Sψj,Sψi⟩=⟨S2ψj,ψi⟩=⟨Hoscψj,ψi⟩=Ejδij. Thus the Gram matrix on Vn is the identity and its form matrix is diagonal with entries E0,…,En−1.

2.1F1F4step 1.1algebra

Since ℏω>0, E0<⋯<En−1. The generalized matrix eigenproblem therefore returns exactly these n ordered Ritz values by [F4], and the oscillator eigenvalue equations in [F1] give (Hosc−EjI)ψj=0 for each j<n.

3.1step 2.1∎

Hence this conforming Hermite trial space attains the first n oscillator eigenvalues exactly and every corresponding continuum residual vanishes.

CounterexampleConstruction: AI-generatedVerification: AI-generatedprecheck passverified 2026-10-22 (gpt-6.1-sol)Open item page →

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 E0, every normalized u∈D(H), and every real λ, if (H−λI)u=0, then λ=E0 and u belongs to the ground eigenspace ker⁡(H−E0I).

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 (ACω)).

[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=C2 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 E0=0, and (T−100I)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≠E0. Thus the zero residual verifies a spectral point, as in [F1], but refutes the asserted ground-state label.

Sources