Alphabeta Math
CounterexampleConstruction: Literature-sourcedVerification: AI-adaptedprecheck passaudited 2026-08-02
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The degenerate constraint x2+y2=0 defeats the multiplier conclusion

Statement refuted

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

Facts & Assumptions

Given: g(x,y)=x2+y2 and f(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} is the singleton (0,0), so f has both a constrained local maximum and a constrained local minimum there.

givenalgebra
1.2

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

givenL1algebra
2.1

No scalar λ can satisfy ∇f(0,0)=λ∇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 · two levels

8 results within two dependency steps of this one, each drawn at its shortest distance from it. An arrow runs from a result to what uses it, so the chart reads left to right and ends at this result, which carries a heavier outline. Every node is a link to that result. Click elsewhere on the chart to enlarge it.

Sources