TheoremStatement: AI-adaptedProof: AI-adaptedPipeline-generatedjudge pass (gpt-5.6-terra)audited 2026-09-13

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.

Sequential convergence of smooth multipliers and distributions

Statement

Assume Dependent Choice. If $u_{j} \to u$ weakly in $D^{'} (Ω)$ and $a_{j}, a \in C^{\infty} (Ω)$ satisfy uniform convergence of every derivative on every compact set, then $a_{j} u_{j} \to a u$ weakly. Dependent Choice is used only through the uniform finite-order bound for the sequence of distributions.

Facts & Assumptions

[F1]

Under Dependent Choice, a pointwise bounded family of distributions has a common finite-order estimate on each $D_{K}$ ; a pointwise convergent sequence has a pointwise bounded range (Uniform finite order bounds for pointwise bounded distributions).

[F2]

Smooth multiplication acts by $(a v) (φ) = v (a φ)$ (Multiplication of a distribution by a smooth function).

[F3]

The assumed axiom is The axiom of dependent choice: a relation in which every element is related to something admits an $N$ -indexed chain.

Proof

Given: the sequences and limits in the statement, and Dependent Choice.

1.1

Fix a test $φ$ with compact support $K$ . By F1 and F3 there are $C, m$ such that $∣ u_{j} (ψ) ∣ \leq C p_{m} (ψ)$ for every $j$ and $ψ \in D_{K}$ . Every $(a_{j} - a) φ$ is supported in $K$ , and the finite product rule gives $p_{m} ((a_{j} - a) φ) \leq 2^{m} q_{K, m} (a_{j} - a) p_{m} (φ) \to 0$ , where $q_{K, m}$ is the maximum of derivatives through degree $m$ on $K$ .

givenF1F2F3algebra

2.1

By F2 the difference of pairings is $u_{j} ((a_{j} - a) φ) + (u_{j} - u) (a φ)$ . The first term tends to zero by step 1.1 and the second by weak convergence applied to the fixed test $a φ$ . Therefore $(a_{j} u_{j}) (φ) \to (a u) (φ)$ . Since the test was arbitrary this is weak convergence. Zero multipliers and the zero test satisfy the same bounds, and on the empty domain all distributions are zero. The finite initial part of the sequence is covered by F1, with no unsupported arbitrary-net uniform bound. $□$

step 1.1givenF1F2

Depends on

Used by

Nothing in the library uses this result yet.

Dependency tree · two levels

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Sources

Razvan Gelca, Functional Analysis (standard reference, not scraped)