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

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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.

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The subdifferential of a differentiable convex function is its gradient singleton

Statement

Let $f : U \to R$ be convex on an open convex set and differentiable at $a \in U$ (The total (Fréchet) derivative $D f (a)$ as the linear first-order approximation with $o (∥ h ∥_{2})$ remainder, The Jacobian matrix of partial derivatives and the gradient in the scalar-valued case). Then $\partial f (a) = {\nabla f (a)}$ .

Facts & Assumptions

Given: The function and point in the Statement and the subdifferential convention Subgradients and the subdifferential of a convex function.

[F1]

If $f$ is convex, then $f ((1 - t) x + t y) \leq (1 - t) f (x) + t f (y)$ for $x, y$ in its convex domain and $t \in [0, 1]$ (Convex and strictly convex functions on Euclidean convex sets).

Proof

technique · direct

1.1

Fix $y \in U$ and put $d = y - a$ . For $0 < t \leq 1$ , [F1] gives $f (a + t d) \leq (1 - t) f (a) + t f (y)$ , hence $\frac{f (a + t d) - f (a)}{t} \leq f (y) - f (a) .$ Differentiability at $a$ makes the left side tend to $⟨ \nabla f (a), d ⟩$ as $t ↓ 0$ . Thus $f (y) \geq f (a) + ⟨ \nabla f (a), y - a ⟩$ for every $y \in U$ , so $\nabla f (a) \in \partial f (a)$ .

F1givenalgebra

2.1

Let $v \in \partial f (a)$ . For each coordinate vector $e_{i}$ and sufficiently small positive and negative $t$ , apply the subgradient inequality at $a + t e_{i}$ . Dividing by $t$ with the appropriate reversal and taking the two one-sided limits gives $v_{i} \leq \partial_{i} f (a)$ and $v_{i} \geq \partial_{i} f (a)$ . Thus $v = \nabla f (a)$ , proving the singleton claim.

step 1.1givenalgebra∎

Depends on

Used by

The Euclidean norm and its square are convex, with a ball of subgradients at zero for the norm Example
FALSE: every convex function is differentiable False statement

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

D. Bertsekas, MIT 6.253 Convex Analysis and Optimization, Lecture 12 (standard reference, not scraped)
D. Drusvyatskiy, Convex Analysis and Nonsmooth Optimization, §3.5 (standard reference, not scraped)