TheoremStatement: Literature-sourcedProof: AI-adaptedSession-authored (Fable 5 assisted)precheck passjudge pass (deepseek-v4-pro + gpt-5.6-terra)audited 2026-08-21

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.

A function is convex exactly when its epigraph is convex

Statement

Let $C \subseteq R^{n}$ be convex and let $f : C \to R$ . The function $f : C \to R$ is convex if and only if its epigraph is a convex subset of $R^{n + 1}$ . This includes the empty-domain convention.

Facts & Assumptions

Given: The function and convex domain in the Statement, with convex subsets interpreted by A convex subset of $R^{m}$ contains every line segment between two of its points.

[F1]

The function $f : C \to R$ is convex when $f ((1 - t) x + t y) \leq (1 - t) f (x) + t f (y)$ for all $x, y \in C$ and $t \in [0, 1]$ (Convex and strictly convex functions on Euclidean convex sets).

[F2]

The epigraph of $f : A \to R$ is $epi f = {(x, s) \in A \times R : f (x) \leq s}$ (The epigraph and hypograph of a real-valued function).

Proof

technique · direct

1.1

For the forward implication, take $(x, r), (y, s) \in epi f$ and $t \in [0, 1]$ . By [F1], $f ((1 - t) x + t y) \leq (1 - t) f (x) + t f (y) \leq (1 - t) r + t s,$ so [F2] puts the convex combination in the epigraph.

F1F2

2.1

For the reverse implication, assume the epigraph convex and apply its convexity to $(x, f (x))$ and $(y, f (y))$ . By [F2], membership of their convex combination is exactly the inequality in [F1]. Thus $f$ is convex; if $C$ is empty, both conditions are vacuous.

step 1.1F1F2∎

Depends on

Used by

A convex function has a subgradient at every interior point of its domain Theorem

Dependency tree · two levels

6 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

S. Boyd and L. Vandenberghe, Convex Optimization, §3.1.7 (standard reference, not scraped)