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CounterexampleConstruction: AI-generatedVerification: AI-generatedSession-authored (Fable 5 assisted)precheck passaudited 2026-08-02
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x2+y2(1−x)3x^2+y^2(1-x)^3 has a unique critical point, a strict local but nonglobal minimum

Statement refuted

Refuted: a unique critical point which is a strict local minimum must be a global minimum.

Facts & Assumptions

Given: f(x,y)=x2+y2(1−x)3f(x,y)=x^2+y^2(1-x)^3 on R2\mathbb R^2.

[L1]

A positive definite Hessian at a critical point gives a strict local minimum (The multivariable second-derivative test by definiteness of the Hessian).

Proof

technique · calculation
1.1

The partial derivatives are fx=2x−3y2(1−x)2f_x=2x-3y^2(1-x)^2 and fy=2y(1−x)3f_y=2y(1-x)^3. Their simultaneous vanishing forces (x,y)=(0,0)(x,y)=(0,0).

givenalgebra
1.2

At x=2x=2, f(2,y)=4−y2f(2,y)=4-y^2, which is negative for ∣y∣>2|y|>2, whereas f(0,0)=0f(0,0)=0.

givenalgebra
2.1

At the origin the Hessian is diag⁡(2,2)\operatorname{diag}(2,2), so [L1] makes it a strict local minimum.

step 1.1L1algebra
3.1

Thus the unique critical point is a strict local minimum but not a global one.

step 1.1step 2.1step 1.2algebra∎

Depends on

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