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Fourier transform on tempered distributions is well defined and continuous

Statement

Assume Countable Choice. The Fourier transform $F : S^{'} (R^{n}) \to S^{'} (R^{n})$ 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

1.1F1F2

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

1.2F1F3

For a single test $φ$ , $p_{φ} (F u) = ∣ u (F φ) ∣ = p_{F φ} (u)$ . Every target weak seminorm therefore pulls back to a source weak seminorm, so $F$ is weakly continuous.

2.1F2step 1.1

If $B \subseteq S$ 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.

$p_{B} (F u) = \sup_{φ \in B} ∣ u (F φ) ∣ = p_{F (B)} (u) .$

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

Dependency tree · 2 levels

Sources

Semyon Dyatlov, Lecture notes for 18.155 (2022)