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A C² local minimum has vanishing gradient and positive semidefinite Hessian

Statement

Let $n \geq 1$ , let $Ω \subseteq R^{n}$ be open, let $U : Ω \to R$ be of class $C^{2}$ ( $C^{k}$ Euclidean maps and diffeomorphisms), and let $q_{0} \in Ω$ be a local minimum of $U$ : there is $ρ > 0$ with $\overline{B} (q_{0}, ρ) \subseteq Ω$ and $U (q) \geq U (q_{0})$ for every $q \in \overline{B} (q_{0}, ρ)$ . Then

$\nabla U (q_{0}) = 0, h^{T} D^{2} U (q_{0}) h \geq 0 for every h \in R^{n},$

where $D^{2} U (q_{0})$ is the Hessian matrix, which is symmetric (The Hessian of a $C^{2}$ scalar field is symmetric, The Hessian matrix and critical points of a scalar field).

Facts & Assumptions

Given: The $C^{2}$ function $U$ and the local minimum $q_{0}$ .

[F1]

Fermat's theorem in Euclidean space: if a differentiable function has a local extremum at an interior point, its gradient vanishes there (Fermat's theorem: an interior differentiable local extremum has zero gradient, The Jacobian matrix of partial derivatives and the gradient in the scalar-valued case, The total (Fréchet) derivative $D f (a)$ as the linear first-order approximation with $o (∥ h ∥_{2})$ remainder).

[F2]

Second-order Taylor expansion: for $C^{2}$ $f$ near $a$ , $f (a + h) = f (a) + \nabla f (a) \cdot h + \frac{1}{2} ⟨ H_{f} (a) h, h ⟩ + o (∥ h ∥^{2})$ , and the Hessian is symmetric (Second-order Taylor expansion $f (a + h) = f (a) + \nabla f (a) \cdot h + \frac{1}{2} h^{T} H_{f} (a) h + o (∥ h ∥^{2})$ , The Hessian of a $C^{2}$ scalar field is symmetric, The Euclidean inner product $⟨ x, y ⟩ = \sum_{k < n} x_{k} y_{k}$ on $R^{n}$ ).

[F3]

The Peano remainder is a limit statement: writing $f (q_{0} + t h) = f (q_{0}) + t \nabla f (q_{0}) \cdot h + \frac{1}{2} t^{2} h^{T} H_{f} (q_{0}) h + r (t h)$ for small $t$ , one has $r (t h) / ∥ t h ∥_{2}^{2} \to 0$ as $t ↓ 0$ ; by the definition of the function limit (The $ε$ - $δ$ limit $\lim_{x \to c} f (x) = L$ of $f : A \to R$ at a limit point $c$ of $A$ ), for every $ε > 0$ there is $δ > 0$ with $∣ r (t h) ∣ \leq ε t^{2} ∥ h ∥_{2}^{2}$ for $0 < t < δ$ (Second-order Taylor expansion $f (a + h) = f (a) + \nabla f (a) \cdot h + \frac{1}{2} h^{T} H_{f} (a) h + o (∥ h ∥^{2})$ , The total (Fréchet) derivative $D f (a)$ as the linear first-order approximation with $o (∥ h ∥_{2})$ remainder).

Proof

technique · direct; apply Fermat's theorem and expand along a line

1.1givenF1F2algebra

The restriction $φ_{h} (t) : = U (q_{0} + t h)$ is defined and differentiable for $∣ t ∣$ small, with $φ_{h}^{'} (0) = \nabla U (q_{0}) \cdot h$ by the chain rule; it has a local minimum at $t = 0$ because $U$ does. By the one-variable Fermat theorem (or [F1] applied in one dimension) $φ_{h}^{'} (0) \geq 0$ and $φ_{- h}^{'} (0) \geq 0$ ; since $φ_{- h}^{'} (0) = - \nabla U (q_{0}) \cdot h$ , both inequalities give $\nabla U (q_{0}) \cdot h = 0$ . As this holds for every $h$ , $\nabla U (q_{0}) = 0$ .

2.1step 1.1F2F3algebra∎

Fix $h \in R^{n}$ , assume $h \neq 0$ (the case $h = 0$ is trivial), and apply [F2] at $a = q_{0}$ to the vector $t h$ with $t > 0$ small: $U (q_{0} + t h) = U (q_{0}) + t \nabla U (q_{0}) \cdot h + \frac{1}{2} t^{2} h^{T} D^{2} U (q_{0}) h + o (t^{2}) .$ By step 1.1 the linear term vanishes, and the left side is $\geq U (q_{0})$ ; hence, using the notation of [F3], $\frac{1}{2} t^{2} h^{T} D^{2} U (q_{0}) h + r (t h) \geq 0$ for all sufficiently small $t > 0$ . Suppose for contradiction that $h^{T} D^{2} U (q_{0}) h = - c < 0$ . Choose $ε : = c / (4 ∥ h ∥_{2}^{2})$ in [F3]; then for $0 < t < δ$ , $r (t h) \leq ε t^{2} ∥ h ∥_{2}^{2} = c t^{2} / 4$ , so the left side is at most $- \frac{1}{2} c t^{2} + \frac{1}{4} c t^{2} = - \frac{1}{4} c t^{2} < 0$ , a contradiction. Hence $h^{T} D^{2} U (q_{0}) h \geq 0$ .

Remarks

The converse is false without a strictness hypothesis: a $C^{2}$ function with $\nabla U (q_{0}) = 0$ and positive semidefinite Hessian need not have a local minimum at $q_{0}$ . The lemma records only the necessary conditions, which is what the mechanical energy-stability theorem and the linearisation proposition use.
Nothing is asserted about strict positivity of the Hessian: a strict minimum may have a degenerate Hessian (for example $U (q) = ∣ q ∣^{4}$ at $q_{0} = 0$ ), and the energy-stability theorem is stated for strict minima without any Hessian hypothesis.

A C² local minimum has vanishing gradient and positive semidefinite Hessian

Statement

Facts & Assumptions

Proof

Remarks

Depends on

Dependency tree · 2 levels

Sources