CorollaryStatement: Literature-sourcedProof: AI-adaptedPipeline-generatedjudge pass (gpt-5.6-terra)audited 2026-09-12

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Fourier transform is a topological automorphism of Schwartz space

Statement

Assume countable choice. The Fourier transform is a topological automorphism of $S (R^{n})$ , with $F^{2} = R$ and $F^{- 1} = R F$ , where $R f (x) = f (- x)$ .

Facts & Assumptions

Given: The Axiom of Countable Choice ( ${A C}_{ω}$ ).

[F1]

Inversion holds everywhere on Schwartz space (Fourier inversion on Schwartz space).

[F2]

Fourier transformation is continuous on Schwartz space (Fourier transform acts continuously on Schwartz space).

[F3]

Reflection is continuous on Schwartz space (Basic operations are continuous on Schwartz space).

Proof

technique · direct

1.1F1F2F3algebra

Evaluate [F1] at $- x$ . Its right-hand side is $F (\hat{f}) (x)$ , so $F^{2} f = R f$ . Also $R^{2} = I$ directly. By associativity, $F R = F F^{2} = F^{2} F = R F$ . All compositions are defined by [F2] and [F3].

2.1step 1.1F2F3∎

Consequently $(R F) F = R^{2} = I$ and $F (R F) = R F^{2} = I$ . These two identities prove both injectivity and surjectivity and the asserted inverse. Both the map and its inverse are continuous by [F2], [F3] and composition, establishing the topological automorphism.

Depends on

Dependency tree · 2 levels

Sources

Semyon Dyatlov, MIT 18.155 (2022)