Alphabeta Math
LemmaStatement: Literature-sourcedProof: AI-adaptedPipeline-generatedjudge pass (gpt-6.1-sol)
How statement and proof provenance work

The first chip identifies the source of the statement or construction; the second identifies the source of its local proof or verification.

  • 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.
  • AI-generated: a genuinely novel statement formulated by AI, with no source for the claim itself.

These labels describe origin, not correctness: citations and verification chips remain separate evidence.

The Hessian is negative semidefinite at an interior local maximum

Statement

Let n≥1, let U⊆Rn be open, let f∈C2(U), and let a∈U be a local maximum of f. Then hTHf(a)h≤0for every h∈Rn; in particular ∇f(a)=0 and Δf(a)=tr⁡Hf(a)≤0.

Facts & Assumptions

Given: An open U⊆Rn, f∈C2(U), an interior local maximum a∈U, and an arbitrary h∈Rn.

[F1]

At an interior local extremum of a differentiable scalar field the gradient vanishes: ∇f(a)=0 (Fermat's theorem: an interior differentiable local extremum has zero gradient).

[F2]

For a C2 scalar field, f(a+h)=f(a)+∇f(a)⋅h+12⟨Hf(a)h,h⟩+o(∥h∥2) (Second-order Taylor expansion f(a+h)=f(a)+∇f(a)⋅h+12hTHf(a)h+o(∥h∥2)).

[F3]

The Hessian Hf(a) is the matrix of second partial derivatives, and Δf=∑i∂i∂if is the trace of the Hessian (The Hessian matrix and critical points of a scalar field, The Laplacian of a C2 function and of a C2 vector field).

[F4]

Local maximality means f(a)≥f(x) for all x in some Euclidean neighbourhood of a (Local and strict local extrema for scalar fields on Euclidean open sets).

Proof

Given: An open U⊆Rn, f∈C2(U), an interior local maximum a∈U, and h∈Rn.

1.1F1F4given

Since a is an interior point of U at which f has a local maximum, [F1] applies and gives ∇f(a)=0.

2.1step 1.1F2F4given

Suppose hTHf(a)h=:c>0; then h≠0 and [F2] and step 1.1 give f(a+th)=f(a)+t22hTHf(a)h+o(t2)=f(a)+t2(c2+o(1))>f(a) for every sufficiently small t≠0, and a+th lies in the neighbourhood of a on which the local maximum is attained for small t; this contradicts [F4]. Hence hTHf(a)h≤0, and since h was arbitrary the quadratic form of the Hessian is negative semidefinite.

3.1step 2.1F3given∎

Taking h=ei in step 2.1 gives Hii(a)≤0 for every coordinate index i, and by the trace formula [F3] the Laplacian is Δf(a)=∑iHii(a)≤0; together with step 1.1 this is the stated conclusion.

Depends on

Used by

Dependency tree · two levels

14 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