Alphabeta Math
LemmaStatement: AI-adaptedProof: AI-adaptedPipeline-generatedjudge pass (gpt-5.6-terra)audited 2026-09-14
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Fourier transform on tempered distributions is well defined and continuous

Statement

Assume Countable Choice. The Fourier transform F:S(Rn)S(Rn) is well-defined and complex-linear. It is continuous for both the weak topology σ(S,S) and the strong topology β(S,S).

Facts & Assumptions

Given: Countable Choice and a tempered distribution u.

[F1]

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

[F2]

The Schwartz transform is a continuous linear automorphism (Fourier transform is a topological automorphism of Schwartz space).

[F3]

Weak dual seminorms use single tests and strong dual seminorms use bounded test sets (Weak and strong topologies on tempered distributions).

Proof

technique · transpose seminorm calculation
1.1

By [F2], Fφ is a Schwartz test and depends continuously and linearly on φ. Thus φu(Fφ) is a continuous complex-linear functional. This proves well-definedness, and linearity in u follows directly from the pairing.

F1F2
1.2

For a single test φ, pφ(Fu)=u(Fφ)=pFφ(u). Every target weak seminorm therefore pulls back to a source weak seminorm, so F is weakly continuous.

F1F3
2.1

If BS is bounded, continuity and linearity of the Schwartz transform imply that F(B) is bounded: each output seminorm is bounded by finitely many input seminorms.

F2step 1.1

pB(Fu)=supφBu(Fφ)=pF(B)(u).

Thus every target strong seminorm pulls back to a strong seminorm and the map is strongly continuous. Countable Choice was used only in [F2], not in these transpose calculations. [F2, F3] ∎

Depends on

Used by

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Sources