ExampleConstruction: AI-generatedVerification: AI-generatedPipeline-generatedprecheck passjudge pass (gpt-6-luna)verified 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.

Early stopping leaves a nonzero Tikhonov normal residual

Example

Let $X = Y = R$ with the usual inner product, let $A = A_{h} = I$ , let $x^{†} = 0$ , $y = y_{h, δ} = 0$ , $e = 0$ , and set $λ = 1$ . The exact Tikhonov output is $R_{λ, h} y_{h, δ} = 0$ , and its normal equation is $2 x = 0$ . Apply one Richardson step with step size $1 / 4$ from $x_{0} = 1 / 2$ : ${\tilde{x}}_{h} = x_{1} = x_{0} - \frac{1}{4} (2 x_{0}) = \frac{1}{4} .$ The returned output has exact normal-equation residual $r_{h, h} = 2 {\tilde{x}}_{h} = \frac{1}{2} \neq 0,$ and reconstruction error $∣ x^{†} - {\tilde{x}}_{h} ∣ = 1 / 4$ . The bias, data, operator, and pre-inversion channels are zero; only the algebraic solve channel is nonzero.

Example references

Five-channel Tikhonov error budget

Facts & Assumptions

Given: The one-dimensional real Hilbert spaces $X = Y = R$ , the identity maps $A = A_{h} = I$ , $x^{†} = 0$ , $y = y_{h, δ} = 0$ , $e = 0$ , $δ = 0$ , $λ = 1$ , initial iterate $x_{0} = 1 / 2$ , and Richardson step size $1 / 4$ .

[F1]

Error budget for computed Tikhonov inversion: The solve vector is $b_{s o l v e} = R_{λ, h} y_{h, δ} - {\tilde{x}}_{h}$ , and the exact normal residual is $r_{h, h} = (A_{h}^{*} A_{h} + λ I_{X}) {\tilde{x}}_{h} - A_{h}^{*} y_{h, δ}$ .

[F2]

Bounded finite rank operators are compact: Every bounded linear operator whose range has a finite ordered basis is compact.

Verification

Given: The scalar data and iteration in the Example.

1.1givenF1F2constructalgebra

The identity is bounded and has the finite range basis ${1}$ , so it is compact by [F2]. The shifted normal operator is $A_{h}^{*} A_{h} + λ I = 2 I$ , which is bijective with inverse $(1 / 2) I$ ; hence the finite scalar instance satisfies [F1].

1.2givenF1algebra

One Richardson step for $2 x = 0$ from $x_{0} = 1 / 2$ gives ${\tilde{x}}_{h} = 1 / 2 - (1 / 4) (2 \cdot 1 / 2) = 1 / 4$ . Substitution into the residual definition gives $r_{h, h} = 2 (1 / 4) - 0 = 1 / 2$ , while the exact regularized output is $R_{λ, h} 0 = 0$ .

2.1givenF1step 1.2algebra∎

The truth bias and data perturbation vanish because $x^{†} = y = e = 0$ . Since $A_{h} = A$ , $d_{h} = 0$ and the regularized-operator difference vanishes; the pre-inversion channel is therefore also zero. The solve channel is $0 - {\tilde{x}}_{h} = - 1 / 4$ , so its norm equals the reconstruction error $1 / 4$ and is at most the residual allowance $∣ r_{h, h} ∣ / λ = 1 / 2$ .

Depends on

Used by

Nothing in the library uses this result yet.

Dependency tree · two levels

13 results within two dependency steps of this one, each drawn at its shortest distance from it. An arrow runs from a result to what uses it, so the chart reads left to right and ends at this result, which carries a heavier outline. Every node is a link to that result. Click elsewhere on the chart to enlarge it.