CounterexampleConstruction: Literature-sourcedVerification: 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.

$x \mapsto - \sqrt{1 - ∥ x ∥_{2}^{2}}$ has empty subdifferential on the unit-sphere boundary

Statement refuted

Every convex real-valued function on a closed convex domain has a subgradient at every domain point.

Facts & Assumptions

Given: Let $B = {x \in R^{n} : ∥ x ∥_{2} \leq 1}$ for $n \geq 1$ (The $p$ -norms $∥ x ∥_{p}$ for rational $p \geq 1$ , and $∥ x ∥_{\infty}$ ) and define $f (x) = - \sqrt{1 - ∥ x ∥_{2}^{2}}$ on $B$ , with convexity as in Convex and strictly convex functions on Euclidean convex sets. For the comparison with the interior-point theorem, assume the Axiom of Choice and the Axiom of Countable Choice (The Axiom of Choice, The Axiom of Countable Choice ( ${A C}_{ω}$ )).

[F1]

A vector $v$ is a subgradient of $f$ at $a$ when $f (y) \geq f (a) + ⟨ v, y - a ⟩$ for every $y$ in the domain (Subgradients and the subdifferential of a convex function).

[L1]

Assuming the Axiom of Choice and the Axiom of Countable Choice, a convex function has nonempty subdifferential at every interior point of its domain (A convex function has a subgradient at every interior point of its domain).

[L2]

The Euclidean norm is a norm on $R^{n}$ and satisfies the triangle inequality (Cauchy-Schwarz $∣ ⟨ x, y ⟩ ∣ \leq ∥ x ∥_{2} ∥ y ∥_{2}$ with its equality case, the triangle inequality for $∥ \cdot ∥_{2}$ , the parallelogram law and polarisation).

Counterexample

technique · direct

1.1

Put $s_{x} = \sqrt{1 - ∥ x ∥_{2}^{2}}$ . The lifted vectors $(x, s_{x})$ have norm one in $R^{n + 1}$ . By the triangle inequality in [L2], the convex combination $((1 - t) x + t y, (1 - t) s_{x} + t s_{y})$ has norm at most one, so its nonnegative last coordinate is at most $s_{(1 - t) x + t y}$ . Negating gives the convexity inequality for $f$ on the whole closed ball.

L2givenalgebra

2.1

Fix a unit vector $u$ and suppose $v$ satisfied [F1] at $u$ . Testing $y = α u$ for $0 \leq α < 1$ gives $⟨ v, u ⟩ \geq \frac{\sqrt{1 - α^{2}}}{1 - α} = \sqrt{\frac{1 + α}{1 - α}} .$ The right side is unbounded as $α$ approaches $1$ , impossible for the fixed vector $v$ . Hence $\partial f (u) = \emptyset$ .

step 1.1F1algebra

3.1

Step 2.1 applies to every sphere point, while [L1] gives nonempty subdifferentials at every interior point. Thus the interior hypothesis in the existence theorem is sharp.

step 2.1L1∎

Depends on

Used by

Nothing in the library uses this result yet.

Dependency tree · two levels

37 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

CUHK ENGG 5501 Convex Analysis notes (standard reference, not scraped)

x↦−1−∥x∥22x↦−1−∥x∥22​​ has empty subdifferential on the unit-sphere boundary