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

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Fourier transform is a topological automorphism of tempered distributions

Statement

Assume Countable Choice. Fourier transformation is a topological automorphism of $S^{'} (R^{n})$ for both the weak and strong dual topologies. If

$⟨ R u, φ ⟩ = ⟨ u, φ (- \cdot) ⟩,$

then $F^{2} u = R u$ and $F^{- 1} = R F = F R$ .

Facts & Assumptions

Given: Countable Choice and $u \in S^{'} (R^{n})$ .

[F1]

Fourier transformation on $S^{'}$ is weakly and strongly continuous (Fourier transform on tempered distributions is well defined and continuous).

[F2]

On Schwartz space, $F^{2} = R$ , $R^{2} = I$ , and $F^{- 1} = R F$ (Fourier transform is a topological automorphism of Schwartz space).

Proof

technique · transpose the Schwartz identities

1.1F2

Evaluate the second transform on an arbitrary Schwartz test $φ$ .

$⟨ F^{2} u, φ ⟩ = ⟨ u, F^{2} φ ⟩ = ⟨ u, R φ ⟩ = ⟨ R u, φ ⟩ .$

Thus $F^{2} u = R u$ . This is a direct test calculation and uses no density assertion about $S$ inside its dual. [F2]

2.1step 1.1algebra

Reflection on the dual satisfies $R^{2} = I$ . Since step 1.1 gives $R = F^{2}$ as operators on $S^{'}$ , associativity gives the following two-sided inverse calculation.

$(R F) F = R^{2} = I, F (R F) = F^{4} = R^{2} = I .$

Hence $F^{- 1} = R F$ ; also $R F = F^{3} = F R$ . [step 1.1, algebra]

3.1F1step 2.1∎

The inverse $R F = F^{3}$ is a composition of weakly continuous maps and also of strongly continuous maps. Therefore $F$ is a topological automorphism for both topologies. Countable Choice is used only through [F1]–[F2].

Fourier transform is a topological automorphism of tempered distributions

Statement

Facts & Assumptions

Proof

Depends on

Dependency tree · 2 levels

Sources