Alphabeta Math
TheoremStatement: Literature-sourcedProof: AI-adaptedSession-authored (Fable 5 assisted)precheck passaudited 2026-08-02
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A constrained local extremum annihilates every velocity of a differentiable parametrization

Statement

Let f:U→Rf:U\to\mathbb R be differentiable at a∈U⊆RNa\in U\subseteq\mathbb R^N, and let γ:(−η,η)→U\gamma:(-\eta,\eta)\to U be differentiable at 00 with γ(0)=a\gamma(0)=a. If f∘γf\circ\gamma has a local maximum or minimum at 00, then Df(a)γ′(0)=0Df(a)\gamma'(0)=0.

Facts & Assumptions

Given: The hypotheses of the statement.

[L1]

The total-derivative chain rule is D(f∘γ)(0)=Df(γ(0))Dγ(0)D(f\circ\gamma)(0)=Df(\gamma(0))D\gamma(0) (The chain rule for total derivatives: D(g∘f)(a)=Dg(f(a))∘Df(a)D(g\circ f)(a)=Dg(f(a))\circ Df(a)).

[L2]

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

Proof

technique · direct
1.1

The composite g=f∘γg=f\circ\gamma is differentiable at 00 by [L1], and 00 is an interior local extremum of gg by the hypothesis.

givenL1algebra
2.1

Fermat's theorem gives g′(0)=0g'(0)=0.

step 1.1L2
3.1

The chain-rule identity in [L1] and γ(0)=a\gamma(0)=a give g′(0)=Df(a)γ′(0)g'(0)=Df(a)\gamma'(0). Combining with step 2.1 proves the conclusion.

L1step 2.1algebra∎

Depends on

Used by

Dependency tree · next 3 levels

Direct dependencies and their dependencies through the next three levels: 51 results over 17 levels. An arrow runs from a result to what uses it, and this result sits at the bottom with a heavier outline. Click the chart to enlarge it.

Sources