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Fourier transform on tempered distributions is well defined and continuous
Statement
Assume Countable Choice. The Fourier transform is well-defined and complex-linear. It is continuous for both the weak topology and the strong topology .
Facts & Assumptions
Given: Countable Choice and a tempered distribution .
The distributional transform is the bilinear transpose of the Schwartz transform (Fourier transform of a tempered distribution).
The Schwartz transform is a continuous linear automorphism (Fourier transform is a topological automorphism of Schwartz space).
Weak dual seminorms use single tests and strong dual seminorms use bounded test sets (Weak and strong topologies on tempered distributions).
Proof
By [F2], is a Schwartz test and depends continuously and linearly on . Thus is a continuous complex-linear functional. This proves well-definedness, and linearity in follows directly from the pairing.
For a single test , . Every target weak seminorm therefore pulls back to a source weak seminorm, so is weakly continuous.
If is bounded, continuity and linearity of the Schwartz transform imply that is bounded: each output seminorm is bounded by finitely many input seminorms.
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
Dependency tree · two levels
13 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)