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The weighted Fourier seminorm separates Schwartz functions

Statement

Assume Countable Choice. For every $n \geq 1$ , real $s$ , and $u \in S (R^{n})$ , $q_{s} (u) = 0$ implies that $u = 0$ as an actual smooth function. Consequently $Q_{s}$ from Weighted Fourier candidate norm on Schwartz space is a positive-definite inner product, and its induced norm is $q_{s} (u) = ∥ ⟨ ξ ⟩^{s} \hat{u} ∥_{L^{2}}$ .

Facts & Assumptions

Given: Countable Choice, $n \geq 1$ , $s \in R$ , and $u \in S (R^{n})$ .

[F1]

The form and candidate seminorm satisfy $Q_{s} (u, v) = (⟨ ξ ⟩^{s} \hat{u}, ⟨ ξ ⟩^{s} \hat{v})_{L^{2}}$ and $q_{s} (u) = ∥ ⟨ ξ ⟩^{s} \hat{u} ∥_{2}$ (Weighted Fourier candidate norm on Schwartz space).

[F2]

The repository Fourier transform extends to a unitary map on complex $L^{2}$ and preserves the $L^{2}$ norm of Schwartz functions (Plancherel theorem).

[F3]

A Schwartz function is an actual continuous smooth function, not only an almost-everywhere class (Schwartz space and its seminorms).

Proof

technique · Weighted $L^2$ separation and continuity

1.1F1given

Suppose $q_{s} (u) = 0$ . By [F1], $∥ w_{s} \hat{u} ∥_{2} = 0$ , so $w_{s} \hat{u} = 0$ almost everywhere.

2.1F2step 1.1algebra

Since $w_{s} (ξ) = ⟨ ξ ⟩^{s} > 0$ at every $ξ$ , step 1.1 implies $\hat{u} = 0$ almost everywhere; Plancherel [F2] then gives $∥ u ∥_{2} = ∥ \hat{u} ∥_{2} = 0$ .

3.1F3step 2.1

If $u (x_{0}) \neq 0$ , continuity from [F3] gives a ball on which $∣ u ∣ > ∣ u (x_{0}) ∣ / 2$ ; its positive Lebesgue measure contradicts $∥ u ∥_{2} = 0$ . Therefore the actual Schwartz function vanishes everywhere.

4.1F1step 3.1

By [F1], $Q_{s}$ is the complex $L^{2}$ inner product of the weighted Fourier images, hence is linear in the first variable, conjugate symmetric, and nonnegative on the diagonal; step 3.1 makes it positive definite, and [F1] gives $Q_{s} (u, u)^{1 / 2} = q_{s} (u)$ .

5.1F1∎

Conversely, if $u = 0$ , its Fourier transform vanishes and the defining formula [F1] gives $q_{s} (u) = 0$ ; thus the kernel is exactly ${0}$ .

Depends on

Used by

Real-order Bessel-potential completion Hˢ Definition

Dependency tree · two levels

19 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, current revision (standard reference, not scraped)
Richard B. Melrose, Differential Analysis, Chapter 3 (standard reference, not scraped)