TheoremStatement: Literature-sourcedProof: AI-adaptedprecheck passaudited 2026-08-02

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The multivariable second-derivative test by definiteness of the Hessian

Statement

Let $a$ be a critical point of a $C^{2}$ scalar field. A positive definite Hessian gives a strict local minimum, a negative definite Hessian gives a strict local maximum, and an indefinite Hessian gives neither. If the Hessian is semidefinite but not definite, the Hessian test gives no conclusion in general.

Facts & Assumptions

Given: A $C^{2}$ scalar field and a critical point $a$ .

[L1]

The second-order Taylor expansion has quadratic term $\frac{1}{2} ⟨ H_{f} (a) h, h ⟩$ and remainder $o (∥ h ∥^{2})$ (Second-order Taylor expansion $f (a + h) = f (a) + \nabla f (a) \cdot h + \frac{1}{2} h^{T} H_{f} (a) h + o (∥ h ∥^{2})$ ).

[L2]

A definite quadratic form has a uniform signed bound on the unit sphere (A definite quadratic form has a uniform signed bound on the Euclidean unit sphere).

Proof

technique · direct

1.1

At a critical point the linear term in [L1] vanishes.

L1given

2.1

Write $h = r u$ with $r = ∥ h ∥_{2}$ and $∥ u ∥_{2} = 1$ when $h \neq 0$ . The sign bound in [L2] dominates the $o (r^{2})$ remainder for sufficiently small $r$ .

step 1.1L2algebra

3.1

This gives the strict minimum and maximum conclusions in the definite cases; two unit directions of opposite quadratic sign give neither extremum in the indefinite case.

step 2.1algebra

4.1

For a semidefinite Hessian which is not definite, its quadratic form has a nonzero null direction, so step 2.1 supplies no signed quadratic bound. No universal conclusion is possible: at $0$ the one-variable functions $x^{4}$ , $- x^{4}$ , and $x^{3}$ all have zero Hessian, but respectively have a strict local minimum, a strict local maximum, and neither.

step 2.1algebra∎

Depends on

Used by

The two-variable Hessian determinant test Corollary
A zero Hessian occurs at a strict minimum, a strict maximum, and a saddle Counterexample
x²+y²(1-x)³ has a unique critical point, a strict local but nonglobal minimum Counterexample
The monkey saddle x³-3xy² has an indefinite higher-order critical point Example

Dependency tree · two levels

16 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

Analysis, Convexity, and Optimization (standard reference, not scraped)