LemmaStatement: Literature-sourcedProof: AI-generatedPipeline-generatedprecheck passjudge pass (gpt-6-luna)verified 2026-11-04 (gpt-6.1-sol)

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.

Compactly Supported Tests Determine Continuous Local Coefficients

Statement

Let $n, m \geq 1$ , let $U \subseteq R^{n}$ be open, let $w : U \to (0, \infty)$ be continuous, and let $A_{i} : U \to R$ , $i = 1, \dots, m$ , be continuous. For a smooth compactly supported test $ψ = (ψ^{1}, \dots, ψ^{m}) : U \to R^{m}$ , define the chart integral of $\sum_{i} A_{i} ψ^{i} w$ by extending its compactly supported continuous integrand by zero to a containing rectangle. If

\int_{U} \sum_{i = 1}^{m} A_{i} (x) ψ^{i} (x) w (x) d^{n} x = 0

for every such $ψ$ , then $A_{i} (x) = 0$ for every $i$ and $x$ . The lemma assumes no metric, field equation, or choice principle.

Facts & Assumptions

Given: The integers $n, m \geq 1$ , open $U \subseteq R^{n}$ , positive continuous $w$ , continuous $A_{i}$ , and the stated vanishing pairing for all smooth compactly supported tests.

[F1]

For $0 < r < R$ , there is a smooth $ρ : R^{n} \to [0, 1]$ equal to $1$ on ${\overline{B}}_{r} (0)$ and supported in $B_{R} (0)$ (A smooth bump between concentric Euclidean balls).

[F2]

Every continuous real function on a closed nondegenerate rectangle in $R^{n}$ is Riemann integrable (Every continuous function on a closed nondegenerate rectangle in $R^{m}$ is Riemann integrable).

[F3]

The lower Darboux integral is the supremum of the lower sums, and each lower sum is $\sum_{j} (\inf_{Q_{j}} f) vol (Q_{j})$ (The lower and upper Darboux integrals over a nondegenerate rectangle in $R^{m}$ , Lower and upper Darboux sums over a grid partition in $R^{m}$ ).

[F4]

For a nondegenerate rectangle its volume is the product of its positive side lengths; in particular it is positive (Axis-parallel rectangles in $R^{m}$ and their volume).

[F5]

A finite grid is obtained by partitioning each coordinate interval; inserting the endpoints of a closed subrectangle makes it a cell of a grid (Grid partitions of a rectangle in $R^{m}$ , their cells, refinements and mesh).

Proof

technique · contradiction; isolate one nonzero component with a nonnegative smooth bump and obtain a strictly positive lower Darboux sum

1.1givenassume-contra

Suppose $A_{j} (p) \neq 0$ for some $j$ and $p \in U$ . Choose $ε \in {- 1, 1}$ so that $ε A_{j} (p) w (p) > 0$ . By continuity of $x \mapsto ε A_{j} (x) w (x)$ and openness of $U$ , there are $R > 0$ and $c > 0$ such that ${\overline{B}}_{R} (p) \subset U$ and $ε A_{j} (x) w (x) > c$ throughout ${\overline{B}}_{R} (p)$ .

2.1step 1.1F1construct

Choose $0 < r < R$ with ${\overline{B}}_{r} (p)$ containing a nondegenerate closed coordinate cube $Q$ centered at $p$ . By [F1], after translation there is $0 \leq ρ \leq 1$ with $ρ = 1$ on ${\overline{B}}_{r} (p)$ and support in $B_{R} (p)$ . Set $ψ^{j} = ε ρ$ and $ψ^{i} = 0$ for $i \neq j$ ; this is a smooth compactly supported test in $U$ . Its pairing integrand $F = ε ρ A_{j} w$ is nonnegative everywhere, is continuous and compactly supported, and satisfies $F > c$ on $Q$ .

3.1step 2.1F2F5

Extend $F$ by zero to a closed rectangle containing its support. It is continuous there and Riemann integrable by [F2]. By [F5], choose a grid that includes the faces of $Q$ , so $Q$ is one cell.

4.1step 1.1step 2.1step 3.1F3F4

On every other grid cell $F \geq 0$ , while on $Q$ , $\inf_{Q} F \geq c$ ; [F3] and [F4] therefore give a lower sum at least $c vol (Q) > 0$ . The integral, being the lower Darboux integral, is positive.

5.1step 4.1givendischarge-contradiction∎

Step 4.1 makes the pairing for the admissible $ψ$ strictly positive, contrary to its assumed value zero. Thus no component $A_{j}$ is nonzero at any point, as required.

Compactly Supported Tests Determine Continuous Local Coefficients

Statement

Facts & Assumptions

Proof

Depends on

Dependency tree · 2 levels

Sources