Alphabeta Math
LemmaStatement: Literature-sourcedProof: AI-adaptedPipeline-generatedverified 2026-10-15 (gpt-6-luna, gpt-6.1-sol, and delegated Codex review/repair agents; see completed run evidence)
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.

Fenchel inequality and its equality condition

Statement

Let f:Rd→R‾ be proper, let x∈dom⁡f, and let p∈Rd. Then f∗(p)≥⟨p,x⟩−f(x). Equality holds exactly when p∈∂f(x), equivalently when x attains the supremum defining f∗(p); in that case the supremum is finite.

Facts & Assumptions

Given: The Fenchel conjugate and subdifferential from Fenchel conjugate, biconjugate, and subdifferential.

[F1]

Fenchel conjugate, biconjugate, and subdifferential: f∗(p) is the supremum of ⟨p,y⟩−f(y) over y∈dom⁡f, and p∈∂f(x) means f(y)≥f(x)+⟨p,y−x⟩ for every y.

Proof

technique · direct
1.1F1given

Since x∈dom⁡f, the value ⟨p,x⟩−f(x) is one of the values over which f∗(p) is the supremum; the stated inequality follows.

2.1F1algebra∎

Equality means every y∈dom⁡f satisfies ⟨p,y⟩−f(y)≤⟨p,x⟩−f(x), which rearranges to the subgradient inequality; for y∉dom⁡f that inequality is automatic because f(y)=+∞. Conversely the subgradient inequalities make the value at x a finite upper bound attained there, so f∗(p)=⟨p,x⟩−f(x).

Depends on

Dependency tree · 2 levels

Sources