LemmaStatement: AI-adaptedProof: AI-adaptedPipeline-generatedprecheck passjudge pass (gpt-6-luna)

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 regular level set realizes every compatible second-order jet

Statement

Let $H : U \subseteq R^{n} \to R^{q}$ be $C^{2}$ , let $x \in U$ , and suppose $D H (x)$ is surjective. If $d \in \ker D H (x)$ and $D H (x) z + D^{2} H (x) [d, d] = 0,$ then there is a $C^{2}$ curve $γ$ on the level $H = H (x)$ near $0$ with $γ (0) = x$ , $γ^{'} (0) = d$ , and $γ^{''} (0) = z$ . For $q = 0$ the level is all of $U$ and the statement is immediate.

Facts & Assumptions

Given: The $C^{2}$ map, regular point, vectors $d, z$ , and compatibility equation in the Statement.

[F1]

Full row rank of $D H (x)$ persists on a neighborhood, where the level is a $C^{2}$ graph $x + u + ψ (u)$ over $\ker D H (x)$ with $ψ (0) = 0$ and $D ψ (0) = 0$ (Differential rank is lower semicontinuous, A regular level set is locally a $C^{k}$ graph of dimension $m - n$ ).

[F2]

A $C^{2}$ Euclidean map has its second-order Taylor expansion with a Peano remainder (Second-order Taylor expansion $f (a + h) = f (a) + \nabla f (a) \cdot h + \frac{1}{2} h^{T} H_{f} (a) h + o (∥ h ∥^{2})$ ).

Proof

1.1givenconstruct

If $q = 0$ , then $H$ is the unique map into $R^{0}$ and the level is all of $U$ . The curve $γ (t) = x + t d + \frac{1}{2} t^{2} z$ has the required value and first two derivatives.

1.2F1F2givenalgebra

Now suppose $q > 0$ and take the graph in [F1]. Put $v = D^{2} ψ (0) [d, d]$ . Since $H (x + t d + ψ (t d)) = H (x)$ for small $t$ , [F2] and $ψ (t d) = \frac{1}{2} t^{2} v + o (t^{2})$ give $D H (x) d = 0$ and $D H (x) v + D^{2} H (x) [d, d] = 0$ .

2.1step 1.2F1F2construct∎

The compatibility equation for $z$ and step 1.2 imply $e = z - v \in \ker D H (x)$ . Define $u (t) = t d + \frac{1}{2} t^{2} e$ and $γ (t) = x + u (t) + ψ (u (t))$ . This curve stays on $H = H (x)$ by [F1]. Using [F2] and $D ψ (0) = 0$ gives $γ (t) = x + t d + \frac{1}{2} t^{2} (e + v) + o (t^{2}) = x + t d + \frac{1}{2} t^{2} z + o (t^{2})$ , so its first two derivatives at $0$ are $d$ and $z$ .

Depends on

Used by

Second-order necessity on the critical cone under LICQ Theorem

Dependency tree · two levels

20 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.

Sources

Boyd and Vandenberghe, Convex Optimization (standard reference, not scraped)
Nocedal and Wright, Numerical Optimization, second edition (standard reference, not scraped)