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Local minima and nonnegative second variation

Statement

Let $U$ be a real-valued $C^{2}$ function on an open set $O \subseteq R^{m}$ , and let $q \in O$ be a local minimum. Thus $U$ has continuous partial derivatives through order two, and $U (q) \leq U (x)$ for all $x$ in some neighbourhood of $q$ . Write $H = \nabla^{2} U (q)$ for the matrix of second partial derivatives. Then $\nabla U (q) = 0$ and $v^{T} H v \geq 0$ for every $v \in R^{m}$ . The matrix $H$ is real symmetric, and $z^{*} H z \geq 0$ for every $z \in C^{m}$ , where $z^{*}$ is the conjugate transpose. Consequently, if $H$ has a strictly negative eigenvalue, $q$ cannot be a local minimum. An eigenvalue here is a scalar $λ$ for which $H w = λ w$ for some nonzero vector $w$ . This is a necessary condition; no converse is asserted.

Facts & Assumptions

Given: The function $U$ , open set $O$ , and local minimum $q$ in the statement.

Goal: Prove the vanishing gradient, the real and complex quadratic-form inequalities, and the negative-eigenvalue exclusion.

[F1]

The local-minimum and Hessian conventions are those of Local and strict local extrema for scalar fields on Euclidean open sets and The Hessian matrix and critical points of a scalar field.

[F2]

At an interior differentiable local minimum the gradient vanishes, by Fermat's theorem: an interior differentiable local extremum has zero gradient.

[F3]

For a $C^{2}$ scalar function, the second-order Taylor expansion at $q$ has remainder $o (∥ h ∥^{2})$ , by Second-order Taylor expansion f(a+h)=f(a)+∇ f(a)· h+1/2h^TH_f(a)h+o(‖h‖²).

[F4]

Continuous second partial derivatives commute, so $H$ is symmetric, by Clairaut--Schwarz theorem for continuous second partial derivatives.

Proof

technique · direct

1.1givenF2

Since $O$ is open, $q$ is an interior point; $U$ is differentiable there. Hence $\nabla U (q) = 0$ by [F2].

2.1givenF1F3step 1.1

Fix $v \in R^{m}$ . For sufficiently small real $t$ , the local-minimum property gives $U (q + t v) - U (q) \geq 0$ . Taylor expansion and step 1.1 give $U (q + t v) - U (q) = \frac{1}{2} t^{2} v^{T} H v + o (t^{2})$ as $t \to 0$ when $v \neq 0$ ; the case $v = 0$ is immediate. Dividing by $t^{2} > 0$ and taking the limit yields $v^{T} H v \geq 0$ .

3.1F4step 2.1algebra

By [F4], $H = H^{T}$ . For $z = x + i y$ with $x, y \in R^{m}$ , direct expansion gives $z^{*} H z = x^{T} H x + y^{T} H y + i (x^{T} H y - y^{T} H x) = x^{T} H x + y^{T} H y \geq 0$ .

4.1step 3.1algebra∎

If $H w = λ w$ for a nonzero complex vector $w$ and a real $λ < 0$ , then $w^{*} H w = λ w^{*} w < 0$ , contradicting step 3.1. Thus a strictly negative eigenvalue excludes a local minimum.

Depends on

Used by · 4

Dependency tree · 2 levels

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