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

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Fourier inversion on Schwartz space

Statement

Assume countable choice. For every $f \in S (R^{n})$ and every $x \in R^{n}$ , $f (x) = \int_{R^{n}} \hat{f} (ξ) e^{2 π i x \cdot ξ} d ξ .$ The integral is absolutely convergent.

Facts & Assumptions

Given: $f \in S$ and The Axiom of Countable Choice ( ${A C}_{ω}$ ).

[F1]

The Fourier transform preserves Schwartz space (Fourier transform acts continuously on Schwartz space).

[F2]

Schwartz functions are integrable (Schwartz derivatives are integrable).

[F3]

If $f, \hat{f} \in L^{1}$ , inversion gives its value at every Lebesgue point (L1 Fourier inversion with an integrable transform).

Proof

technique · direct

1.1F1F2given

By [F1] and [F2], both $f$ and $\hat{f}$ are integrable, and the displayed integral is absolutely convergent since the exponential has modulus one. Fix $x$ . Smoothness implies continuity, so for every $ε > 0$ some $δ > 0$ gives $∣ f (x - y) - f (x) ∣ < ε$ for $∣ y ∣ < δ$ . Averaging over any ball of radius $0 < r < δ$ bounds its mean oscillation by $ε$ . Thus $x$ is a Lebesgue point with specified value $f (x)$ .

2.1step 1.1F3∎

Apply [F3] at this arbitrary point. This proves the formula everywhere, inheriting exactly the countable-choice assumption of these three suppliers. The proof never exchanges an undamped double Fourier integral.

Depends on

Dependency tree · 2 levels

Sources

Semyon Dyatlov, MIT 18.155 (2022)