TheoremStatement: Literature-sourcedProof: AI-adaptedprecheck passaudited 2026-08-02

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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.

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A constrained local extremum annihilates every velocity of a differentiable parametrization

Statement

Let $f : U \to R$ be differentiable at $a \in U \subseteq R^{N}$ , and let $γ : (- η, η) \to U$ be differentiable at $0$ with $γ (0) = a$ . If $f \circ γ$ has a local maximum or minimum at $0$ , then $D f (a) γ^{'} (0) = 0$ .

Facts & Assumptions

Given: The hypotheses of the statement.

[L1]

The total-derivative chain rule is $D (f \circ γ) (0) = D f (γ (0)) D γ (0)$ (The chain rule for total derivatives: $D (g \circ f) (a) = D g (f (a)) \circ D f (a)$ ).

[L2]

A differentiable one-variable function with an interior local extremum has derivative zero (Fermat's interior extremum theorem: if $f$ has a local extremum at a point $c$ interior to its domain and is differentiable at $c$ , then $f^{'} (c) = 0$ ).

Proof

technique · direct

1.1

The composite $g = f \circ γ$ is differentiable at $0$ by [L1], and $0$ is an interior local extremum of $g$ by the hypothesis.

givenL1algebra

2.1

Fermat's theorem gives $g^{'} (0) = 0$ .

step 1.1L2

3.1

The chain-rule identity in [L1] and $γ (0) = a$ give $g^{'} (0) = D f (a) γ^{'} (0)$ . Combining with step 2.1 proves the conclusion.

L1step 2.1algebra∎

Depends on

Used by

Lagrange multipliers for a regular graph constraint y=ψ(x) Corollary
The degenerate constraint x²+y²=0 defeats the multiplier conclusion Counterexample
Lagrange multipliers for a regular vector-valued level-set constraint Theorem

Dependency tree · two levels

18 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

Lagrange Multipliers (standard reference, not scraped)