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

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Fourier transform of a compactly supported distribution is a smooth polynomially bounded multiplier

Statement

Assume Countable Choice. Let $v \in D^{'} (R^{n})$ have compact support, let $\tilde{v} \in S^{'}$ be its canonical extension, and let $χ \in D$ equal one on a neighborhood of $supp v$ . Then $F \tilde{v}$ is the regular tempered distribution represented by

$V (ξ) = ⟨ v_{x}, χ (x) e^{- 2 π i x \cdot ξ} ⟩ .$

The function $V$ is independent of $χ$ , is smooth, and for every multi-index $α$ there are $C_{α}, m_{α}$ with $∣ \partial^{α} V (ξ) ∣ \leq C_{α} (1 + ∣ ξ ∣)^{m_{α}}$ . Consequently multiplication by $V$ is continuous on $S$ and, by transpose, on $S^{'}$ in both dual topologies.

Facts & Assumptions

Given: Countable Choice, a compactly supported distribution $v$ , and a cutoff $χ$ as in the statement.

[F1]

The extension $\tilde{v}$ is tempered and its pairing with a smooth function is computed using any cutoff equal to one near the support (Compactly supported distributions are tempered, Compactly supported distributions extend to smooth functions).

[F2]

Compactly supported distribution pairings with smooth parameter families differentiate in the parameter (Distribution pairing with smooth parameter families).

[F3]

The Fourier transform is defined by bilinear transposition (Fourier transform of a tempered distribution), and the local Schwartz integral lemma permits a seminorm-dominated integral to cross a tempered pairing (Schwartz parameter pairing and integral interchange).

[F4]

A smooth function whose derivatives grow polynomially is a continuous Schwartz multiplier, as is its transpose (Smooth polynomially bounded multipliers on schwartz space).

Proof

technique · compact finite-order estimate and pairing interchange

1.1

If $χ_{1}$ and $χ_{2}$ are both one near $supp v$ , then $(χ_{1} - χ_{2}) e^{- 2 π i x \cdot ξ}$ vanishes near that support, so [F1] makes its pairing with $v$ zero. Thus $V$ is cutoff-independent.

1.2

Apply smooth parameter differentiation from [F2].

$\partial_{ξ}^{α} V (ξ) = ⟨ v_{x}, χ (x) (- 2 π i x)^{α} e^{- 2 π i x \cdot ξ} ⟩ .$

On one fixed compact containing $supp χ$ , the finite-order estimate for $v$ differentiates the displayed test in $x$ only finitely many times. Each resulting term is bounded by a constant times $(1 + ∣ ξ ∣)^{m}$ . Hence $V$ is smooth and every derivative has the claimed polynomial bound. [F1, F2, algebra]

1.3

Let $φ \in S$ and set $H (ξ, x) = χ (x) e^{- 2 π i x \cdot ξ} φ (ξ)$ . As an $x$ -Schwartz family, $H (ξ, \cdot)$ is continuous in $ξ$ , and every $x$ -Schwartz seminorm has an integrable majorant $C (1 + ∣ ξ ∣)^{q} ∣ φ (ξ) ∣$ . Thus [F3] applies.

$\begin{aligned} ⟨ F \tilde{v}, φ ⟩ & = ⟨ \tilde{v}, F φ ⟩ \\ = \int ⟨ v_{x}, χ (x) e^{- 2 π i x \cdot ξ} ⟩ φ (ξ) d ξ = \int V (ξ) φ (ξ) d ξ . \end{aligned}$

[F1, F3]

2.1

Equality in step 1.3 identifies $F \tilde{v}$ with the regular distribution $u_{V}$ . The derivative bounds from step 1.2 satisfy [F4], which proves both multiplier assertions. For $v = 0$ , $V = 0$ ; empty support causes no exception. Countable Choice enters only through the published Fourier and Lebesgue-interchange clauses.

F4step 1.2step 1.3∎

Depends on

Used by

Fourier transform converts allowed tempered convolutions to products Theorem

Dependency tree · two levels

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Sources

Semyon Dyatlov, Lecture notes for 18.155 (2022) (standard reference, not scraped)