ExampleConstruction: Literature-sourcedVerification: AI-adaptedPipeline-generatedprecheck passjudge pass (gpt-6-sol)audited 2026-09-30

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.

Every Schwartz function belongs to every real-order H^s

Statement

Assume Countable Choice. For every $n \geq 1$ , real $s$ , and actual Schwartz function $u \in S (R^{n})$ , its canonical completion class is sent by $E_{s}$ to the usual regular distribution of $u$ . In particular $S (R^{n}) \subset H^{s} (R^{n})$ for every real $s$ , with $∥ u ∥_{H^{s}} = ∥ ⟨ ξ ⟩^{s} \hat{u} ∥_{2} < \infty .$ This inclusion does not assert that the intersection of all real-order $H^{s}$ spaces is exactly Schwartz space.

Facts & Assumptions

Given: Countable Choice, $n \geq 1$ , $s \in R$ , and $u \in S (R^{n})$ .

[A1]

Countable Choice is used by the metric completion construction defining $H^{s}$ (The Axiom of Countable Choice ( ${A C}_{ω}$ )).

[F1]

Multiplication by $⟨ ξ ⟩^{s}$ maps Schwartz space continuously to itself (Real powers of the Japanese bracket act on Schwartz space).

[F2]

Fourier transformation is an automorphism of Schwartz space (Fourier transform is a topological automorphism of Schwartz space).

[F3]

Every Schwartz function determines a complex $L^{2}$ class (Schwartz space is dense in L2).

[F4]

The candidate norm is $q_{s} (u) = ∥ ⟨ ξ ⟩^{s} \hat{u} ∥_{2}$ (Weighted Fourier candidate norm on Schwartz space).

[F5]

The canonical constant-sequence map embeds Schwartz space linearly and isometrically into its completion $H^{s}$ (Real-order Bessel-potential completion H^s).

[F6]

The canonical embedding sends a Schwartz class to its usual regular distribution (The Bessel completion embeds canonically in tempered distributions).

Proof

technique · Apply the weighted Fourier norm directly to a Schwartz function

1.1F1F2F3F4given

By [F2], $\hat{u} \in S$ ; then [F1] gives $f (ξ) = ⟨ ξ ⟩^{s} \hat{u} (ξ) \in S$ . Thus [F3] gives $f \in L^{2}$ and the defining integral calculation is $q_{s} (u)^{2} = \int_{R^{n}} ∣ ⟨ ξ ⟩^{s} \hat{u} (ξ) ∣^{2} d ξ = ∥ f ∥_{2}^{2} < \infty$ by [F4].

2.1A1F5F6step 1.1∎

Under the stated Countable Choice assumption [A1], [F5] places the constant sequence $i (u) = [(u, u, \dots)]$ in $H^{s}$ with $∥ i (u) ∥_{H^{s}} = q_{s} (u)$ ; [F6] gives $E_{s} (i (u)) = u_{u}$ , the regular distribution of $u$ . Step 1.1 supplies the finite norm, proving the asserted inclusion and formula for every real $s$ .

Depends on

Used by

Nothing in the library uses this result yet.

Dependency tree · two levels

28 results within two dependency steps of this one, each drawn at its shortest distance from it. An arrow runs from a result to what uses it, so the chart reads left to right and ends at this result, which carries a heavier outline. Every node is a link to that result. Click elsewhere on the chart to enlarge it.

Sources

Semyon Dyatlov, Lecture Notes for 18.155, current revision (standard reference, not scraped)
Richard B. Melrose, Differential Analysis, Chapter 3 (standard reference, not scraped)