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Lagrange multipliers for a regular vector-valued level-set constraint

Statement

Let $U \subseteq R^{m}$ be open, let $f : U \to R$ and $G : U \to R^{q}$ be $C^{1}$ , and suppose $a$ is a local maximum or minimum of $f$ subject to $G (x) = c$ . If $D G (a)$ is surjective, then there is a unique $λ \in R^{q}$ such that $D f (a) v = ⟨ λ, D G (a) v ⟩ (v \in R^{m}),$ or equivalently $\nabla f (a) = D G (a)^{T} λ .$ This is a necessary condition, not a sufficient condition for a constrained extremum.

Facts & Assumptions

Given: The maps $f, G$ , the regular constrained point $a$ , and $c = G (a)$ .

[L1]

If $c$ is a regular value of a $C^{1}$ map on an open set, then at every point $a$ of its fibre a vector lies in the tangent space exactly when it is the velocity at zero of a $C^{1}$ curve through $a$ inside that fibre (Tangent vectors to a regular level set are exactly its curve velocities).

[L4]

For a $C^{1}$ map the locus where the derivative has rank at least $r$ is open, so the submersion locus is open (Differential rank is lower semicontinuous); a value is regular when every point of its fibre is a submersion point (Regular and critical points, regular and critical values, and level sets).

[L2]

A local extremum is defined by the objective inequality on a neighbourhood, and if a differentiable function restricted to a differentiable curve has a local extremum, then its derivative along the curve is zero (Local and strict local extrema for scalar fields on Euclidean open sets, A constrained local extremum annihilates every velocity of a differentiable parametrization).

[L3]

A linear functional vanishing on the kernel of a surjection is a unique transpose multiple; $D f (a) v = ⟨ \nabla f (a), v ⟩$ (A linear functional annihilating the kernel of a surjection is a unique transpose multiple, For a differentiable scalar field, $D_{v} f (a) = ⟨ \nabla f (a), v ⟩$ and the unit direction of steepest ascent is the normalized gradient, The Jacobian matrix of partial derivatives and the gradient in the scalar-valued case).

Proof

technique · direct

1.1givenL1L2L4choose

By [L4] the set $W$ of points of $U$ at which $D G$ is surjective is open, and $a \in W$ ; on $W$ every point of every fibre is a submersion point, so $c$ is a regular value of $G ∣_{W}$ and $a$ is a local extremum of $f$ subject to $G ∣_{W} = c$ . Fix $v \in \ker D G (a)$ . By [L1] applied to $G ∣_{W}$ , choose a level-set curve $γ$ with $γ (0) = a$ and $γ^{'} (0) = v$ inside $G ∣_{W}^{- 1} (c) \subseteq G^{- 1} (c)$ . The constrained local extremum condition in [L2] makes $f \circ γ$ locally extremal at $0$ .

2.1step 1.1L2

By [L2], $D f (a) v = 0$ . Since $v$ was arbitrary, the functional $D f (a)$ vanishes on $\ker D G (a)$ .

3.1step 2.1L3

Apply [L3] to the surjection $D G (a)$ . It gives a unique $λ$ with $D f (a) v = ⟨ λ, D G (a) v ⟩$ for all $v$ , and the gradient representation turns this equality of functionals into $\nabla f (a) = D G (a)^{T} λ$ .

4.1step 3.1∎

The argument derives the multiplier equation from a constrained extremum and makes no converse assertion, as claimed.

Depends on

Used by

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Sources

University of Toronto MAT237 notes, Section 2.8 (standard reference, not scraped)
J. M. Lee, Introduction to Smooth Manifolds, Lagrange multipliers discussion (standard reference, not scraped)