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Dependent equality constraints have nonunique multiplier vectors

Statement refuted

Refuted: that the multiplier vector produced by the Lagrange multiplier rule is unique without any independence hypothesis on the constraint derivatives — equivalently, that the conclusion of The multiplier vector is unique when the constraint gradients are independent remains true when surjectivity of $D G (u)$ is dropped.

The witness is the finite-dimensional problem $X = R^{2}$ , $G (x, y) = (x, 2 x)$ and $I (x, y) = x^{2} + y^{2}$ (The Lagrange multiplier rule for finitely many regular constraints, Fréchet derivative between Banach spaces). On the level set $G = 0$ , which is the $y$ -axis, the point $u = (0, 0)$ is the strict global minimiser of $I$ , and $D I (u) = 0$ , while $D G_{1} (u) = (1, 0)$ and $D G_{2} (u) = (2, 0)$ are linearly dependent, so $D G (u)$ is not surjective. The stationarity equation $D I (u) = λ_{1} D G_{1} (u) + λ_{2} D G_{2} (u)$ holds exactly for the pairs with $λ_{1} + 2 λ_{2} = 0$ , a one-parameter family of multiplier vectors. Hence surjectivity of $D G (u)$ , equivalently independence of the constraint gradients, is what makes the multiplier unique.

Facts & Assumptions

Given: The maps $G : R^{2} \to R^{2}$ , $G (x, y) = (x, 2 x)$ , and $I : R^{2} \to R$ , $I (x, y) = x^{2} + y^{2}$ , with components $G_{1} (x, y) = x$ , $G_{2} (x, y) = 2 x$ , and the point $u = (0, 0)$ .

[F1]

Fréchet derivative between Banach spaces: $I$ and the components $G_{1}, G_{2}$ are differentiable everywhere with $D I (x, y) = (2 x, 2 y)$ as a linear functional, $D G_{1} \equiv (1, 0)$ and $D G_{2} \equiv (2, 0)$ , and $D G (x, y) : R^{2} \to R^{2}$ is the linear map $(ξ, η) \mapsto (ξ, 2 ξ)$ .

[F2]

The Lagrange multiplier rule for finitely many regular constraints, The multiplier vector is unique when the constraint gradients are independent: the multiplier rule asserts the existence of multipliers when $D G (u)$ is surjective, and the uniqueness lemma shows that surjectivity is precisely the hypothesis that rules out the degeneracy exhibited here.

Counterexample

technique · direct

Given: The maps and point above.

1.1givenF1

The level set ${G = G (u)} = {(x, y) : x = 0}$ is the $y$ -axis, and $I (0, y) = y^{2} \geq 0$ with equality only for $y = 0$ ; hence $u = (0, 0)$ is the strict global minimiser of $I$ on the level set.

1.2givenF1

The derivatives at $u$ are $D I (u) = (0, 0) = 0$ , $D G_{1} (u) = (1, 0)$ and $D G_{2} (u) = (2, 0)$ by [F1]; the map $D G (u) : (ξ, η) \mapsto (ξ, 2 ξ)$ has image ${(a, 2 a) : a \in R} \neq R^{2}$ , so $D G (u)$ is not surjective, and $D G_{2} (u) = 2 D G_{1} (u)$ shows that the two constraint gradients are linearly dependent.

2.1step 1.2F1algebra

A pair $(λ_{1}, λ_{2}) \in R^{2}$ satisfies $λ_{1} D G_{1} (u) + λ_{2} D G_{2} (u) = D I (u)$ exactly when $(λ_{1} + 2 λ_{2}, 0) = (0, 0)$ , that is, exactly when $λ_{1} + 2 λ_{2} = 0$ ; the solution set is the line of all pairs $(- 2 t, t)$ , $t \in R$ , a one-parameter family.

3.1step 1.1step 1.2step 2.1F2∎

The stationarity equation therefore holds for infinitely many multiplier vectors although the constrained minimiser is unique, so the claim that uniqueness of the multiplier follows from stationarity alone is false; the uniqueness statement of [F2] genuinely requires the surjectivity, equivalently the independence, hypothesis.

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Sources

Riccardo Cristoferi, Calculus of Variations: Lecture Notes, Carnegie Mellon University 2016 (complete 133-page notes) (standard reference, not scraped)