Alphabeta Math
CounterexampleConstruction: Literature-sourcedVerification: AI-adaptedSession-authored (Fable 5 assisted)precheck passaudited 2026-08-02
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A zero Hessian occurs at a strict minimum, a strict maximum, and a saddle

Statement refuted

Refuted: a zero Hessian determines the local type of a critical point.

Facts & Assumptions

Given: f+(x,y)=x4+y4f_+(x,y)=x^4+y^4, f−(x,y)=−(x4+y4)f_-(x,y)=-(x^4+y^4), and fs(x,y)=x4−y4f_s(x,y)=x^4-y^4.

[L1]

The second-derivative test gives no conclusion for a semidefinite but not definite Hessian (The multivariable second-derivative test by definiteness of the Hessian).

Counterexample

Proof

technique · calculation
1.1

Each displayed function has zero gradient and zero Hessian at (0,0)(0,0).

givenalgebra
1.2

f+f_+ is positive off the origin, so the origin is a strict local minimum; f−f_- is negative off the origin, so it is a strict local maximum.

givenalgebra
1.3

The values fs(t,0)=t4f_s(t,0)=t^4 and fs(0,t)=−t4f_s(0,t)=-t^4 have opposite signs for t≠0t\ne0, so the origin is a saddle.

givenalgebra
2.1

These three different local types share the same zero Hessian, exactly as the inconclusive clause [L1] permits.

step 1.1step 1.2step 1.3L1algebra∎

Depends on

Used by

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Dependency tree · next 3 levels

Direct dependencies and their dependencies through the next three levels: 30 results over 10 levels. An arrow runs from a result to what uses it, and this result sits at the bottom with a heavier outline. Click the chart to enlarge it.

Sources