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

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Fourier differentiation and multiplication identities on tempered distributions

Statement

Assume Countable Choice. For $u \in S^{'} (R^{n})$ and every multi-index $α$ ,

$F (\partial^{α} u) = (2 π i ξ)^{α} F u, F (x^{α} u) = {(- \frac{1}{2 π i})}^{∣ α ∣} \partial_{ξ}^{α} F u .$

Every operation and equality is in $S^{'} (R^{n})$ .

Facts & Assumptions

Given: Countable Choice, $u \in S^{'} (R^{n})$ , and a multi-index $α$ .

[F1]

Derivatives and polynomial products on $S^{'}$ use the bilinear transpose conventions (Differentiation and polynomial multiplication preserve tempered distributions).

[F2]

The distributional Fourier transform is the bilinear transpose of the Schwartz transform (Fourier transform of a tempered distribution).

[F3]

On Schwartz tests, $F (\partial_{j} φ) = 2 π i ξ_{j} F φ$ and $\partial_{j} F φ = F (- 2 π i x_{j} φ)$ (Fourier transform acts continuously on Schwartz space), and all test operations involved are continuous (Basic operations are continuous on Schwartz space).

Proof

technique · first-order test calculation and iteration

1.1F1F2F3

Fix a coordinate $j$ and a Schwartz test $φ$ ; transposition gives the following calculation.

$\begin{aligned} ⟨ F (\partial_{j} u), φ ⟩ & = - ⟨ u, \partial_{j} (F φ) ⟩ \\ = 2 π i ⟨ u, F (ξ_{j} φ) ⟩ = ⟨ 2 π i ξ_{j} F u, φ ⟩ . \end{aligned}$

The first minus sign is the distributional derivative sign; the second identity is the second formula in [F3]. [F1, F2, F3]

1.2F1F2F3

The first formula in [F3] gives $x_{j} F φ = (2 π i)^{- 1} F (\partial_{j} φ)$ , so transposition yields the second calculation.

$⟨ F (x_{j} u), φ ⟩ = \frac{1}{2 π i} ⟨ F u, \partial_{j} φ ⟩ = - \frac{1}{2 π i} ⟨ \partial_{j} F u, φ ⟩ .$

[F1, F2, F3]

2.1step 1.1step 1.2∎

Coordinate derivatives and coordinate multipliers commute among themselves in the relevant families. Iterating steps 1.1 and 1.2 $∣ α ∣$ times therefore gives the two multi-index identities. For $α = 0$ both reduce to $F u = F u$ . Countable Choice is used only through the published Schwartz Fourier identity.

Fourier differentiation and multiplication identities on tempered distributions

Statement

Facts & Assumptions

Proof

Depends on

Dependency tree · 2 levels

Sources