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TheoremStatement: Literature-sourcedProof: AI-adaptedSession-authored (Fable 5 assisted)precheck passaudited 2026-08-02
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The multivariable second-derivative test by definiteness of the Hessian

Statement

Let aa be a critical point of a C2C^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 C2C^2 scalar field and a critical point aa.

[L1]

The second-order Taylor expansion has quadratic term 12Hf(a)h,h\tfrac12\langle H_f(a)h,h\rangle and remainder o(h2)o(\|h\|^2) (Second-order Taylor expansion f(a+h)=f(a)+f(a)h+12hTHf(a)h+o(h2)f(a+h)=f(a)+\nabla f(a)\cdot h+\tfrac12h^TH_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=ruh=ru with r=h2r=\|h\|_2 and u2=1\|u\|_2=1 when h0h\ne0. The sign bound in [L2] dominates the o(r2)o(r^2) remainder for sufficiently small rr.

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 00 the one-variable functions x4x^4, x4-x^4, and x3x^3 all have zero Hessian, but respectively have a strict local minimum, a strict local maximum, and neither.

step 2.1algebra

Depends on

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Dependency tree · next 3 levels

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