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.

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Compactly supported distributions are tempered

Statement

Every compactly supported distribution $v \in D^{'} (R^{n})$ has a unique extension $\tilde{v} \in S^{'} (R^{n})$ . Restricting $\tilde{v}$ to $D$ recovers $v$ . No choice axiom is required.

Facts & Assumptions

Given: A distribution $v$ with compact support in $R^{n}$ .

[F1]

Such a distribution extends uniquely to a continuous linear functional on $C^{\infty}$ , with $\tilde{v} (f) = v (χ f)$ for any fixed cutoff equal to one near the support (Compactly supported distributions extend to smooth functions).

[F2]

A finite Schwartz-seminorm estimate proves temperateness (Finite seminorm bound characterizes tempered distributions).

[F3]

The inclusion $D \to S$ is continuous with dense image (Test function inclusion in schwartz space is continuous).

Proof

technique · cutoff extension and density

1.1F1

Restrict the $C^{\infty}$ extension from [F1] to $S$ . Its continuity gives a compact $K$ , an integer $m$ , and $C \geq 0$ satisfying the following estimate.

$∣ \tilde{v} (φ) ∣ \leq C \max_{∣ β ∣ \leq m} \sup_{x \in K} ∣ \partial^{β} φ (x) ∣ \leq C \max_{∣ β ∣ \leq m} p_{0, β} (φ) .$

Thus the restriction is tempered by [F2]. [F1, F2]

1.2F1

For $ψ \in D$ , the extension property in [F1] gives $\tilde{v} (ψ) = v (ψ)$ ; this includes the zero distribution and empty support. Hence $\tilde{v}$ really extends $v$ .

2.1F3step 1.2∎

If $w \in S^{'}$ is another extension, then $w - \tilde{v}$ vanishes on $D$ . This difference is continuous on $S$ , and $D$ is dense there, so it vanishes on all of $S$ . Therefore the extension is unique.

Compactly supported distributions are tempered

Statement

Facts & Assumptions

Proof

Depends on

Dependency tree · 2 levels

Sources