Alphabeta Math
LemmaStatement: AI-adaptedProof: AI-adaptedPipeline-generatedjudge pass (gpt-5.6-terra)audited 2026-09-13
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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.

Compact support continuous primitive representation

Statement

Assume AC. Let u∈D′(Ω) have compact support K and global order at most m. For every open V with K⊆V⊆Ω, there are finitely many continuous functions gβ on Ω, each compactly supported in V, such that u=∑0≤βi≤m+2∂βugβ. Zero functions may fill unused indices. Only the AC use in the local representation theorem is needed beyond the stated elementary operations.

Facts & Assumptions

[F1]

Under AC, local representation by a continuous function can use the multi-index α=(m+2,…,m+2) when the localized distribution has order bound m (Local structure of distributions as derivatives of continuous functions).

[F2]

Distributional Leibniz's rule holds for smooth multipliers (Leibniz rule for distributions).

[F3]

A compact set inside an open set admits a smooth compact cutoff equal to one near it (Test function cutoffs and euclidean localization).

[F4]

A compactly supported distribution pairs with smooth functions by cutoff, and vanishes on smooth functions zero near its support (Compactly supported distributions extend to smooth functions).

[F5]

The assumed axiom is The Axiom of Choice.

Proof

Given: u,K,m,V and AC.

1.1givenF1F3F4F5

Use F3 in V to choose θ∈D(V) equal to one near K, and put L=supp⁡θ. By F1 and F5 there is continuous f on Ω with u=∂αuf on DL, where each αi=m+2. For every test φ, F4 gives u(φ)=u(θφ), and θφ∈DL, hence u=θ∂αuf globally on Ω.

2.1step 1.1F2algebra

For any distribution w, F2 implies the inverse product identity [step 1.1, F2, algebra] θ∂αw=∑β≤α(−1)∣α−β∣(αβ)∂β((∂α−βθ)w). Indeed expand each derivative on the right by F2. The coefficient of (∂α−γθ)∂γw is (αγ)∑δ≤α−γ(−1)∣α−γ−δ∣(α−γδ). The sum is the product of the binomial expansions of (1−1)αi−γi, so it is zero unless γ=α, when it is one. Thus precisely the left side remains.

3.1step 2.1step 1.1F4

Apply step 2.1 to w=uf and define gβ=(−1)∣α−β∣(αβ)(∂α−βθ)f. These functions are continuous and supported in L⊆V. Multiplying a regular functional by the smooth factor yields its pointwise product by the defining integral, so step 1.1 and the identity give the required sum. There are only ∏i(m+3) indices. If K is empty, u=0 by support locality and all functions may be zero. For m=0 the bound is still βi≤2. No further infinite selection occurs. □

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Sources