Alphabeta Math
LemmaStatement: 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.

Compact support continuous primitive representation

Statement

Assume AC. Let uD(Ω) have compact support K and global order at most m. For every open V with KVΩ, there are finitely many continuous functions gβ on Ω, each compactly supported in V, such that u=0βim+2βugβ. 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.

1.1

Use F3 in V to choose θD(V) equal to one near K, and put L=suppθ. 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 θφDL, hence u=θαuf globally on Ω.

givenF1F3F4F5
2.1

For any distribution w, F2 implies the inverse product identity [step 1.1, F2, algebra] θαw=βα(1)αβ(αβ)β((αβθ)w). Indeed expand each derivative on the right by F2. The coefficient of (αγθ)γw is (αγ)δαγ(1)αγδ(αγδ). The sum is the product of the binomial expansions of (11)αiγi, so it is zero unless γ=α, when it is one. Thus precisely the left side remains.

step 1.1F2algebra
3.1

Apply step 2.1 to w=uf and define gβ=(1)αβ(αβ)(αβθ)f. These functions are continuous and supported in LV. 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 βi2. No further infinite selection occurs.

step 2.1step 1.1F4

Depends on

Used by

Dependency tree · two levels

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Sources