TheoremStatement: Literature-sourcedProof: AI-adaptedSession-authored (Fable 5 assisted)precheck passaudited 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.

Differentiable convex functions are characterized by the gradient inequality

Statement

Let $U \subseteq R^{n}$ be open and convex, and let $f : U \to R$ be differentiable. Then $f$ is convex if and only if

$f (y) \geq f (x) + ⟨ \nabla f (x), y - x ⟩ (x, y \in U) .$

Equivalently, $\nabla f (x)$ is a subgradient at every $x$ (Subgradients and the subdifferential of a convex function, The Jacobian matrix of partial derivatives and the gradient in the scalar-valued case).

Facts & Assumptions

Given: The domain and differentiable function in the Statement, with convexity from Convex and strictly convex functions on Euclidean convex sets.

[L1]

The total derivative of a composite is the composite of the total derivatives (The chain rule for total derivatives: $D (g \circ f) (a) = D g (f (a)) \circ D f (a)$ ).

Proof

technique · direct

1.1

For the forward implication, fix $x, y \in U$ and put $d = y - x$ . For $0 < t \leq 1$ , convexity gives $f (x + t d) \leq (1 - t) f (x) + t f (y)$ , hence $\frac{f (x + t d) - f (x)}{t} \leq f (y) - f (x) .$ By [L1] the left side tends to $⟨ \nabla f (x), d ⟩$ as $t ↓ 0$ , giving the displayed gradient inequality.

L1givenalgebra

2.1

For the reverse implication, assume the gradient inequality and take $z = (1 - t) x + t y$ . Apply it at $z$ toward $x$ and toward $y$ , multiply the results by $1 - t$ and $t$ , and add. The gradient terms cancel because $(1 - t) (x - z) + t (y - z) = 0$ , leaving $f (z) \leq (1 - t) f (x) + t f (y)$ . Thus $f$ is convex.

assume-hypalgebra∎

Depends on

Used by

Nothing in the library uses this result yet.

Dependency tree · two levels

13 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, §3.1.3 (standard reference, not scraped)
D. Bertsekas, MIT 6.253 Convex Analysis and Optimization, Lecture 3 (standard reference, not scraped)