TheoremStatement: 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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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.
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A convex function has a subgradient at every interior point of its domain

Statement

Assume the Axiom of Choice (The Axiom of Choice) and the Axiom of Countable Choice (The Axiom of Countable Choice ( ${A C}_{ω}$ )). Let $C \subseteq R^{n}$ be convex and let $f : C \to R$ be convex. Then $\partial f (a)$ is nonempty for every $a \in int C$ .

Facts & Assumptions

Given: Fix $a \in int C$ and assume the choice principles in the Statement (Interior, closure, boundary, limit point, isolated point and dense subset of a metric space). The restriction of $f$ to the open convex set $int C$ is continuous by A convex function on an open convex set is continuous, and subgradients have the convention of Subgradients and the subdifferential of a convex function, The epigraph and hypograph of a real-valued function.

[A1]

AC and ${A C}_{ω}$ supply the choice functions asserted in The Axiom of Choice and The Axiom of Countable Choice ( ${A C}_{ω}$ ).

[L1]

The function $f : C \to R$ is convex if and only if its epigraph is a convex subset of $R^{n + 1}$ (A function is convex exactly when its epigraph is convex).

[L2]

At every boundary point of a nonempty convex set there is a nonzero supporting normal $u$ whose inner product with every displacement into the set is nonpositive (Every boundary point belonging to a nonempty Euclidean convex set has a supporting hyperplane).

Proof

technique · direct

1.1

Choose a closed ball $B$ centred at $a$ and contained in $int C$ . The restricted epigraph $E = {(x, s) : x \in B, f (x) \leq s}$ is closed by continuity and convex by [L1]. The point $(a, f (a))$ is on its boundary, so [A1] licenses the hypotheses of [L2], which gives a supporting normal $(u, μ) \neq 0$ . Since the epigraph contains every upward vertical ray, $μ \leq 0$ ; if $μ = 0$ , the ball contains small displacements from $a$ in both directions and forces $u = 0$ , impossible. Thus $μ < 0$ , and rescaling to $μ = - 1$ gives $f (x) \geq f (a) + ⟨ u, x - a ⟩$ on $B$ .

A1L1L2givenalgebra

2.1

Let $y \in C$ . Choose $0 < t \leq 1$ so that $z = a + t (y - a) \in B$ . The local inequality from step 1.1 gives $f (z) \geq f (a) + t ⟨ u, y - a ⟩$ , while convexity gives $f (z) \leq (1 - t) f (a) + t f (y)$ . Combining and dividing by $t > 0$ yields $f (y) \geq f (a) + ⟨ u, y - a ⟩$ . Thus $u \in \partial f (a)$ .

step 1.1givenalgebra∎

Depends on

Used by

x↦-√1-‖x‖₂² has empty subdifferential on the unit-sphere boundary Counterexample

Dependency tree · two levels

28 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, Lectures 7 and 12 (standard reference, not scraped)
D. Drusvyatskiy, Convex Analysis and Nonsmooth Optimization, Theorem 3.36 (standard reference, not scraped)