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CounterexampleConstruction: Literature-sourcedVerification: AI-adaptedSession-authored (Fable 5 assisted)precheck passaudited 2026-08-02
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The degenerate constraint x2+y2=0x^2+y^2=0 defeats the multiplier conclusion

Statement refuted

Refuted: every constrained local extremum satisfies ∇f=λ∇g\nabla f=\lambda\nabla g, even when the constraint gradient vanishes.

Facts & Assumptions

Given: g(x,y)=x2+y2g(x,y)=x^2+y^2 and f(x,y)=xf(x,y)=x.

[L1]

The Jacobian and gradient use the convention of The Jacobian matrix of partial derivatives and the gradient in the scalar-valued case.

Counterexample

Proof

technique · direct
1.1

The constraint set {g=0}\{g=0\} is the singleton (0,0)(0,0), so ff has both a constrained local maximum and a constrained local minimum there.

givenalgebra
1.2

At the origin, ∇g=(0,0)\nabla g=(0,0) while ∇f=(1,0)\nabla f=(1,0).

givenL1algebra
2.1

No scalar λ\lambda can satisfy ∇f(0,0)=λ∇g(0,0)\nabla f(0,0)=\lambda\nabla g(0,0). Thus a regularity hypothesis is necessary for the usual multiplier conclusion.

step 1.1step 1.2algebra∎

Depends on

Used by

Nothing in the library uses this result yet.

Dependency tree · next 3 levels

Direct dependencies and their dependencies through the next three levels: 51 results over 17 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