TheoremStatement: AI-adaptedProof: AI-adaptedPipeline-generatedjudge pass (gpt-5.6-terra)audited 2026-09-14

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.

Tempered distributions embed continuously in distributions

Statement

Restriction to compactly supported tests defines an injective complex-linear map

$ρ : S^{'} (R^{n}) ⟶ D^{'} (R^{n}) .$

It is continuous from weak to weak topology and from strong to strong topology. No assertion is made that the strong topology on $S^{'}$ is the subspace topology inherited from $D^{'}$ .

Facts & Assumptions

Given: The continuous dense inclusion $j : D ↪ S$ (Test function inclusion in schwartz space is continuous).

[F1]

Weak and strong topologies on $S^{'}$ are generated by point tests and bounded Schwartz-test sets (Weak and strong topologies on tempered distributions).

[F2]

The corresponding topologies on $D^{'}$ use point tests and bounded LF-test sets (Weak and strong topologies on distributions).

Proof

technique · transpose the dense continuous inclusion

1.1

Define $ρ (u) (ψ) = u (j ψ)$ . Since $j$ and $u$ are continuous complex-linear maps, $ρ (u)$ is a distribution, and $ρ$ is complex-linear. If $ρ (u) = 0$ , then $u$ vanishes on the dense image of $j$ ; continuity makes $u = 0$ on $S$ . Thus $ρ$ is injective.

given

1.2

For each $ψ \in D$ , $p_{ψ} (ρ u) = p_{j ψ} (u)$ . Hence every weak seminorm on the target pulls back to a weak seminorm on the source, proving weak continuity.

F1F2

2.1

If $B \subseteq D$ is bounded, continuity and linearity of $j$ imply that $j (B)$ is bounded in $S$ .

F1F2step 1.1

$p_{B} (ρ u) = \sup_{ψ \in B} ∣ u (j ψ) ∣ = p_{j (B)} (u),$

which proves strong continuity. Density was essential only for injectivity; continuity alone would not prove that conclusion. [F1, F2, given] ∎

Depends on

Used by

Dependency tree · two levels

10 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 (2022) (standard reference, not scraped)
Radu Gelca, Functional Analysis (standard reference, not scraped)