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

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Fourier transform converts allowed tempered convolutions to products

Statement

Assume Countable Choice. Let $u \in S^{'} (R^{n})$ and $φ \in S (R^{n})$ . Then

$F (u * φ) = (F u) (F φ), F (φ u) = (F φ) * (F u) .$

In the second formula the convolution means $(F u) * (F φ)$ under the distribution-first convention. If $v \in D^{'} (R^{n})$ has compact support, then

$F (u * v) = (F u) (F v) .$

Here $F v$ is the smooth polynomially bounded function representing the transform of the canonical tempered extension of $v$ . No product of two arbitrary distributions and no convolution of two arbitrary tempered distributions occurs.

Facts & Assumptions

Given: Countable Choice, $u \in S^{'}$ , $φ \in S$ , and, for the last formula, compactly supported $v$ .

[F1]

The convolution $u * φ$ is a regular tempered distribution (Tempered convolution is smooth with polynomial growth).

[F2]

Schwartz multipliers act on $S^{'}$ , and $F v$ is a smooth polynomially bounded multiplier (Smooth polynomially bounded multipliers on schwartz space, Fourier transform of a compactly supported distribution is a smooth polynomially bounded multiplier).

[F3]

Seminorm-dominated Schwartz integrals commute with tempered pairings (Schwartz parameter pairing and integral interchange).

[F4]

Compact-distribution convolution preserves $S^{'}$ and agrees with the support-conditioned distribution convolution (Compact distribution convolution preserves schwartz and tempered spaces).

[F5]

Fourier transformation or inversion is available on $S^{'}$ , with $F^{2} = R$ (Fourier transform is a topological automorphism of tempered distributions).

[F6]

Products and convolutions of two Schwartz functions satisfy the same $2 π$ -normalized transform laws (Schwartz convolution and product laws).

Proof

technique · test-pairing interchange

1.1F3

Let $ψ \in S$ . The family $x \mapsto [y \mapsto φ (x - y) \hat{ψ} (x)]$ is dominated in every $y$ -Schwartz seminorm by an integrable polynomial weight times $\hat{ψ} (x)$ . Therefore [F3] applies.

$⟨ F (u * φ), ψ ⟩ = ⟨ u_{y}, \int φ (x - y) \hat{ψ} (x) d x ⟩ .$

[F1, F3]

1.2F6

Absolute scalar interchange, or equivalently [F6] on Schwartz functions, identifies the inner integral.

$F (\hat{φ} ψ) (y) .$

Indeed, inserting $\hat{ψ} (x) = \int ψ (ξ) e^{- 2 π i x \cdot ξ} d ξ$ and translating $x - y$ produces $\hat{φ} (ξ) e^{- 2 π i y \cdot ξ}$ . Hence step 1.1 equals $⟨ F u, \hat{φ} ψ ⟩$ , which is $⟨ (F u) (F φ), ψ ⟩$ . [F2, F3, F6, step 1.1]

1.3F4

Put $V = F v$ . Evaluate the compact-factor convolution on an arbitrary $ψ \in S$ .

$⟨ F (u * v), ψ ⟩ = ⟨ u_{x}, ⟨ v_{y}, \hat{ψ} (x + y) ⟩ ⟩ .$

The compact support of $v$ and [F3] permit its pairing to cross the rapidly convergent Fourier integral, giving

$⟨ v_{y}, \hat{ψ} (x + y) ⟩ = \int V (ξ) ψ (ξ) e^{- 2 π i x \cdot ξ} d ξ = F (V ψ) (x) .$

Thus the outer pairing is $⟨ F u, V ψ ⟩ = ⟨ (F u) (F v), ψ ⟩$ . [F2, F3, F4]

2.1F5step 1.2

Apply the first identity to $F u$ and the Schwartz function $F φ$ , then use Fourier squaring.

$F ((F u) * (F φ)) = (F^{2} u) (F^{2} φ) = (R u) (R φ) = R (φ u) .$

Since $F^{- 1} R = F$ , inversion gives $(F u) * (F φ) = F (φ u)$ . Zero factors are included. Countable Choice is used only through the published Fourier/Lebesgue suppliers. [F5, step 1.2] ∎

Fourier transform converts allowed tempered convolutions to products

Statement

Facts & Assumptions

Proof

Depends on

Dependency tree · 2 levels

Sources