CounterexampleConstruction: AI-generatedVerification: AI-generatedprecheck passaudited 2026-08-02

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Literature-sourced: the exact statement appears in a cited source; only wording and notation differ.
AI-adapted: a semantically identical restatement of literature-sourced material, modulo indexing, notation, and boundary cases adopted by the library.
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$x^{2} + y^{2} (1 - x)^{3}$ 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: $f (x, y) = x^{2} + y^{2} (1 - x)^{3}$ on $R^{2}$ .

[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 $f_{x} = 2 x - 3 y^{2} (1 - x)^{2}$ and $f_{y} = 2 y (1 - x)^{3}$ . Their simultaneous vanishing forces $(x, y) = (0, 0)$ .

givenalgebra

1.2

At $x = 2$ , $f (2, y) = 4 - y^{2}$ , which is negative for $∣ y ∣ > 2$ , whereas $f (0, 0) = 0$ .

givenalgebra

2.1

At the origin the Hessian is $diag (2, 2)$ , 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

Used by

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x2+y2(1−x)3x2+y2(1−x)3 has a unique critical point, a strict local but nonglobal minimum