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

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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.
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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

1.1given

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.

1.2F1F2

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.

2.1F1F2step 1.1

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

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

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

Tempered distributions embed continuously in distributions

Statement

Facts & Assumptions

Proof

Depends on

Dependency tree · 2 levels

Sources