ExampleConstruction: AI-generatedVerification: AI-generatedPipeline-generatedjudge 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.

An added variable is not identifiable without a penalty

Example

The original scalar model is $F (c) = c$ with exact datum $y \in R$ . Relax it by adding an unconstrained variable $z \in R$ and using $\tilde{F} (c, z) = c + z$ , with penalty coordinate $P (c, z) = z$ . The relaxed map satisfies the range-invariance factorization at $(0, 0)$ with $K (a, b) = a + b$ and $r (c, z) = (c, z)$ . For the exact datum, the unpenalized least-squares objective $J_{0} (c, z) = (c + z - y)^{2}$ has every pair with $c + z = y$ as a minimizer. For any fixed $λ > 0$ , adding $λ ∣ P (c, z) ∣^{2}$ gives $J_{λ} (c, z) = (c + z - y)^{2} + λ z^{2}$ , whose unique minimizer is $(y, 0)$ . This finite-dimensional example shows why an enlarged variable can require an explicit penalty; it asserts no general uniqueness result for other models.

Example references

The range-invariance factorization

Facts & Assumptions

Given: The original map $F (c) = c$ , the relaxed variable $(c, z) \in R^{2}$ , the exact datum $y \in R$ , the relaxed map $\tilde{F} (c, z) = c + z$ , and $P (c, z) = z$ .

[F1]

Differential range invariance for Banach-space maps: range invariance at $(0, 0)$ means $\tilde{F} (c, z) - \tilde{F} (0, 0) = K r (c, z)$ for fixed bounded linear $K$ and a map $r$ .

Verification

1.1F1givenconstructalgebra

With $Z = R^{2}$ , $K (a, b) = a + b$ is bounded because $∣ a + b ∣ \leq \sqrt{2} \sqrt{a^{2} + b^{2}}$ . Taking $r (c, z) = (c, z)$ gives $K r (c, z) = c + z = \tilde{F} (c, z) - \tilde{F} (0, 0)$ , so the relaxed map satisfies [F1].

1.2givenalgebra

For every $c \in R$ , the pair $(c, y - c)$ makes $J_{0} (c, y - c) = 0$ . Since $J_{0} \geq 0$ , all these infinitely many pairs are global minimizers, so the relaxed variables are not identifiable from the exact datum alone.

2.1givenalgebra∎

The pair $(y, 0)$ makes $J_{λ} (y, 0) = 0$ and $J_{λ} \geq 0$ . If another pair minimizes, both nonnegative terms must vanish: $z = 0$ because $λ > 0$ , and $c + z = y$ from the data term. Hence $c = y$ , so $(y, 0)$ is the unique minimizer.

Depends on

Differential range invariance for Banach-space maps

Used by

Nothing in the library uses this result yet.

Dependency tree · two levels

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