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CounterexampleConstruction: Literature-sourcedVerification: AI-adaptedSession-authored (Fable 5 assisted)precheck passaudited 2026-08-21
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A strictly convex function can have a singular Hessian

Statement refuted

Every twice differentiable strictly convex function has a positive-definite Hessian at every point.

The counterexample below establishes: For every n1, the function f:RnR given by f(x)=j<nxj4 is strictly convex, but its Hessian at the origin is the zero matrix.

Facts & Assumptions

Given: Fix n1. Strict convexity is as in Convex and strictly convex functions on Euclidean convex sets, and the Hessian convention is The Hessian matrix and critical points of a scalar field.

Counterexample

technique · direct
1.1

Let r<s and 0<t<1, and put c=(1t)r+ts. Apply [L2] on [r,c] and [c,s]. The two secant slopes equal 4u3 and 4v3 for some r<u<c<v<s by [L1], so the first is strictly smaller; rearranging gives c4<(1t)r4+ts4. Thus x4 is strictly convex. For distinct vectors, at least one coordinate gives this strict inequality and every other coordinate gives the corresponding weak one; adding shows that f(x)=j<nxj4 is strictly convex for n1.

L1L2givenalgebra
2.1

Differentiating again with [L1] and [L3], the Hessian is diagonal with entries 12xj2 and zero off-diagonal entries. At the origin it is the zero matrix, hence singular and not positive definite because n1. Thus f(x)=j<nxj4 is strictly convex, but its Hessian at the origin is the zero matrix.

L1L3step 1.1givenalgebra

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