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CorollaryStatement: Literature-sourcedProof: AI-adaptedSession-authored (Fable 5 assisted)precheck passaudited 2026-08-24
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For one regular constraint, the objective gradient is a scalar multiple of the constraint gradient

Statement

Let f,G:URmR be C1, and suppose a is a local maximum or minimum of f subject to G(x)=c. If G(a)0, then there is a unique scalar λ such that f(a)=λG(a).

Facts & Assumptions

Given: The functions, constrained local extremum, and nonzero constraint gradient.

[L1]

For a scalar function, the Jacobian is the row G(a)T (The Jacobian matrix of partial derivatives and the gradient in the scalar-valued case), so DG(a):RmR is surjective exactly when G(a)0.

[L2]

At a constrained local extremum with DG(a) surjective, there is a unique λR such that f(a)=DG(a)Tλ (Lagrange multipliers for a regular vector-valued level-set constraint).

Proof

technique · direct
1.1

By [L1], the nonzero-gradient hypothesis makes DG(a) surjective.

givenL1
2.1

Apply [L2]. Since the transpose of the one-row matrix DG(a) sends λ to λG(a), its conclusion is the displayed equation.

step 1.1L2algebra
3.1

Uniqueness is part of [L2] and also follows directly from G(a)0.

step 2.1

Depends on

Used by

Dependency tree · two levels

12 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