LemmaStatement: AI-adaptedProof: AI-adaptedPipeline-generatedprecheck passverified 2026-10-15 (gpt-6-luna, gpt-6.1-sol, and delegated Codex review/repair agents; see completed run evidence)

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.

Interior rays from a unique supporting plane

Statement

Use the Euclidean distance $∥ q - z ∥$ ; $int K$ consists of points containing a ball in $K$ , closure consists of points whose every ball meets $K$ , and $\partial K = \overline{K} ∖ int K$ . Closed means equal to closure. A supporting hyperplane at $x \in K$ is ${q : n \cdot (q - x) = 0}$ with $n \neq 0$ and $K$ contained in one closed half-space. Let $K \subset R^{d}$ be closed and convex with nonempty interior, $d \geq 1$ , and let $x \in \partial K$ . Suppose $K$ has exactly one supporting hyperplane at $x$ . Choose its inward unit normal $n$ , so $n \cdot (q - x) \geq 0$ for every $q \in K$ . Then for every $v$ with $n \cdot v > 0$ , there is $δ > 0$ such that $x + t v \in int K$ for every $0 < t < δ$ .

Facts & Assumptions

Given: The closed convex set, boundary point, unique supporting hyperplane, and Euclidean inner product.

[F1]

Strictly separate a point from a nonempty closed convex Euclidean set: a point outside a nonempty closed convex Euclidean set admits a strict separating linear functional.

[F2]

The Euclidean inner product $⟨ x, y ⟩ = \sum_{k < n} x_{k} y_{k}$ on $R^{n}$ : Euclidean linear functionals are represented by vectors.

[F4]

A convex subset of $R^{m}$ contains every line segment between two of its points: finite convex combinations of points in a convex set remain in it.

Proof

1.1givenF2

Let $T = \overline{{a (q - x) : a \geq 0, q \in K}}$ . This is a closed convex cone. If a nonzero functional $ℓ$ is nonnegative on $T$ , then it is nonnegative on $K - x$ , so its zero plane through $x$ supports $K$ . Uniqueness in the hypothesis makes that plane the plane with normal $n$ ; the inward sign forces $ℓ = c n \cdot (\cdot)$ for some $c > 0$ .

2.1F1step 1.1algebra

The support at $x$ gives $T \subseteq {w : n \cdot w \geq 0}$ . If $v$ with $n \cdot v > 0$ were outside $T$ , strict separation [F1] of $v$ from the cone would give a nonzero functional $ℓ$ nonnegative on $T$ and negative at $v$ : the separating bound is zero because $0 \in T$ and $T$ is a cone. Step 1.1 then makes $ℓ$ a positive multiple of $n$ , contradicting $n \cdot v > 0$ . Thus the open half-space $n \cdot w > 0$ lies in $T$ , and each such $v$ is an interior point of $T$ .

3.1step 2.1constructF2algebra

Fix such $v$ . Since $v \in int T$ , choose $d + 1$ affinely independent vectors in a small ball about $v$ whose simplex contains a neighborhood of $v$ . Approximate each vector by a nonzero cone generator $g_{j} = a_{j} (q_{j} - x)$ with $a_{j} > 0$ , $q_{j} \in K$ . Sufficiently close approximations preserve affine independence and keep $v$ in the simplex interior, so a neighborhood $N$ of $v$ lies in the simplex of the $g_{j}$ . For each $w \in N$ , write $w = \sum_{j} θ_{j} (w) a_{j} (q_{j} - x)$ with $θ_{j} (w) \geq 0$ and $\sum_{j} θ_{j} (w) = 1$ . Put $A = \max_{j} a_{j} > 0$ . The nonnegative barycentric weights sum to one, so $\sum_{j} θ_{j} (w) a_{j} \leq A$ for every $w \in N$ .

4.1step 3.1givenF4algebra∎

Choose $δ > 0$ with $δ A < 1$ . For $0 < t < δ$ and each $w \in N$ , $x + t w = (1 - t \sum_{j} θ_{j} a_{j}) x + \sum_{j} t θ_{j} a_{j} q_{j}$ is a convex combination of points of $K$ , hence lies in $K$ by [F4]. Therefore $x + t N \subset K$ , so a neighborhood of $x + t v$ lies in $K$ and $x + t v \in int K$ .

Interior rays from a unique supporting plane

Statement

Facts & Assumptions

Proof

Depends on

Dependency tree · 2 levels

Sources