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

1.1

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

F1F2F3

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

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

1.2

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

F1F2F3

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

[F1, F2, F3]

2.1

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.

step 1.1step 1.2∎

Depends on

Used by

Dependency tree · two levels

29 results within two dependency steps of this one, each drawn at its shortest distance from it. An arrow runs from a result to what uses it, so the chart reads left to right and ends at this result, which carries a heavier outline. Every node is a link to that result. Click elsewhere on the chart to enlarge it.

Sources

Semyon Dyatlov, Lecture notes for 18.155 (2022) (standard reference, not scraped)
Radu Gelca, Functional Analysis (standard reference, not scraped)