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.

Vanishing data errors with persistent small-singular-value reconstruction error

Example

For every integer $n \geq 1$ , let $X_{n} = Y_{n} = R^{n}$ with the Euclidean inner product and let $A_{n} = A_{h, n} = diag (1, 1 / 2, \dots, 1 / n) .$ Set $x_{n}^{†} = 0$ , exact data $y_{n} = A_{n} x_{n}^{†} = 0$ , noise $e_{n} = (0, \dots, 0, 1 / n)$ , noise bound $δ_{n} = 1 / n$ , and regularization coefficient $λ_{n} = n^{- 4}$ . The observed data are $y_{n, δ} = y_{n} + e_{n}$ , and the returned output is the exact Tikhonov output ${\tilde{x}}_{n} = (A_{n}^{*} A_{n} + λ_{n} I)^{- 1} A_{n}^{*} y_{n, δ} .$ The smallest singular value of $A_{n}$ is $1 / n$ , while $∥ e_{n} ∥ = δ_{n} \to 0$ . Nevertheless $∥ {\tilde{x}}_{n} - x_{n}^{†} ∥ = \frac{n^{2}}{n^{2} + 1} ⟶ 1.$ At each resolution the discrete operator equals the forward operator and the discretized datum is formed exactly before adding noise, so the operator, pre-inversion, and algebraic-solve channels vanish; only data perturbation remains. Thus vanishing noise alone does not force vanishing reconstruction error when the retained singular values shrink and $λ_{n}$ is chosen as above.

Example references

Five-channel error-budget convention

Clason, Regularization of Inverse Problems, Theorem 8.8 and its proof

Hansen, Introduction to Inverse Problems, Chapter 5, slides 4–6

Van Leeuwen and Brune, 10 Lectures on Inverse Problems and Imaging, §3.4.1

Facts & Assumptions

Given: For each $n \geq 1$ , the real finite-dimensional spaces $X_{n} = Y_{n} = R^{n}$ , the diagonal map $A_{n} = A_{h, n}$ , $x_{n}^{†} = 0$ , $y_{n} = 0$ , $e_{n} = (0, \dots, 0, 1 / n)$ , $δ_{n} = 1 / n$ , and $λ_{n} = n^{- 4}$ .

[F1]

Error budget for computed Tikhonov inversion: The budget uses $y = A x^{†}$ , $y_{δ} = y + e$ , $y_{h, δ} = A_{h} x^{†} + e$ , $d_{h} = (A - A_{h}) x^{†}$ , and the five bias, data, operator, pre-inversion, and solve vectors; its shifted normal operators are assumed bijective.

[F2]

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

Verification

Given: The finite matrices and data in the Example, with the standard coordinate vectors $q_{1}, \dots, q_{n}$ in $R^{n}$ .

1.1givenF1F2constructalgebra

The map $A_{n}$ is bounded and has range $R^{n}$ , which has the finite ordered basis $q_{1}, \dots, q_{n}$ . Hence $A_{n}$ is compact by [F2]. Its adjoint is $A_{n}^{*} = A_{n}$ , and $A_{n}^{*} A_{n} + λ_{n} I$ is diagonal with positive entries $j^{- 2} + n^{- 4}$ for $1 \leq j \leq n$ ; therefore the adjoints and bijective shifted normal operator required by [F1] are supplied in this finite model.

2.1givenF1step 1.1algebra

Let $q_{n}$ be the last coordinate vector. Then $e_{n} = q_{n} / n$ , $A_{n}^{*} y_{n, δ} = q_{n} / n^{2}$ , and $(A_{n}^{*} A_{n} + λ_{n} I) q_{n} = (n^{- 2} + n^{- 4}) q_{n}$ . Solving this one diagonal coordinate gives ${\tilde{x}}_{n} = \frac{n^{2}}{n^{2} + 1} q_{n}$ .

3.1givenF1step 2.1algebra

Since $x_{n}^{†} = y_{n} = 0$ and $A_{h, n} = A_{n}$ , the bias, pre-inversion, and operator channels are zero. Since ${\tilde{x}}_{n}$ is the exact discrete Tikhonov output, the solve channel is zero. The only nonzero channel is the data vector $R_{λ_{n}} y_{n} - R_{λ_{n}} y_{n, δ} = - {\tilde{x}}_{n}$ . Therefore $∥ {\tilde{x}}_{n} - x_{n}^{†} ∥ = n^{2} / (n^{2} + 1)$ , while $δ_{n} = 1 / n \to 0$ and the smallest singular value is $1 / n \to 0$ .

4.1givenstep 3.1algebra∎

The scalar data-noise allowance in the general budget is $δ_{n} / (2 \sqrt{λ_{n}}) = n / 2$ , so it does not force the exact errors in step 3.1 to tend to zero. The sequence has resolved forward data at every level and asserts no convergence theorem for a fixed infinite-dimensional discretization.

Depends on

Used by

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Dependency tree · two levels

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Sources

Christian Clason, Regularization of Inverse Problems: Lecture Notes, arXiv:2001.00617v2, Chapter 8 (standard reference, not scraped)
Per Christian Hansen, Introduction to Inverse Problems, Chapter 5: Regularization Parameter Choice (standard reference, not scraped)
Tristan van Leeuwen and Christoph Brune, 10 Lectures on Inverse Problems and Imaging, Chapter 3 (standard reference, not scraped)