LemmaStatement: Literature-sourcedProof: AI-generatedPipeline-generatedprecheck passjudge pass (gpt-6-luna)

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 H1 functions lie in H1_0

Statement

Assume Countable Choice. Let $U \subset R^{d}$ be open and let $z \in H^{1} (U)$ vanish almost everywhere outside a compact set $K \subset U$ . Then $z \in H_{0}^{1} (U)$ , where $H_{0}^{1} (U)$ is the $H^{1}$ closure of $C_{c}^{\infty} (U)$ .

Facts & Assumptions

Given: Countable Choice; open $U \subset R^{d}$ ; compact $K \subset U$ ; and $z \in H^{1} (U)$ with $z = 0$ almost everywhere on $U ∖ K$ .

[A1]

The Axiom of Countable Choice ( ${A C}_{ω}$ ) is used through the zero-extension, Meyers--Serrin density and Sobolev closure interfaces [F1], [F2], and [F4]; the cutoff existence [F3] itself is choice-free.

[F1]

Compactly supported Sobolev functions extend by zero in every integer order: the zero extension $E_{0} z$ belongs to $H^{1} (R^{d})$ and its Sobolev norm equals that of $z$ .

[F2]

Meyers–Serrin density on an arbitrary open set: under Countable Choice, smooth functions in $C^{\infty} (R^{d}) \cap H^{1} (R^{d})$ are dense in $H^{1} (R^{d})$ .

[F3]

Test function cutoffs and euclidean localization: for compact $K \subset U$ there is $χ \in C_{c}^{\infty} (U)$ equal to one on a neighborhood of $K$ .

[F4]

Zero-boundary Sobolev space as a norm closure: $H_{0}^{1} (U)$ is the closure of $C_{c}^{\infty} (U)$ in the $H^{1}$ norm.

Proof

1.1A1F1F2F3given

By [F1], $E_{0} z \in H^{1} (R^{d})$ and $E_{0} z = z$ on $U$ , $E_{0} z = 0$ off $U$ . By [F2], choose $q_{n} \in C^{\infty} (R^{d}) \cap H^{1} (R^{d})$ with $q_{n} \to E_{0} z$ in $H^{1} (R^{d})$ . Choose $χ$ from [F3].

2.1F1F3F4step 1.1algebra∎

Multiplication by $χ$ is bounded on $H^{1} (R^{d})$ . Indeed, for a weak derivative and a test function $ψ$ , the weak identity gives $\int χ q \partial_{j} ψ = \int q \partial_{j} (χ ψ) - \int q (\partial_{j} χ) ψ = - \int (χ \partial_{j} q + q \partial_{j} χ) ψ,$ so $D_{j} (χ q) = χ D_{j} q + q \partial_{j} χ$ ; the boundedness of $χ$ and its derivatives gives the $H^{1}$ norm bound. Since $χ E_{0} z = z$ almost everywhere, $χ q_{n} \to z$ in $H^{1} (U)$ . Every $χ q_{n}$ is smooth and compactly supported in $U$ , so [F4] proves $z \in H_{0}^{1} (U)$ .

Depends on

Used by

Dependency tree · two levels

27 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

Juha Kinnunen, Sobolev Spaces, Chapter 1 §1.3 and Chapter 2, complete lecture notes (standard reference, not scraped)
John K. Hunter, Notes on Partial Differential Equations, §3.4 (standard reference, not scraped)
Stefan Takacs, Numerical Analysis of Isogeometric Methods, complete JKU lecture notes, 2018/19 (standard reference, not scraped)