TheoremStatement: AI-adaptedProof: AI-adaptedPipeline-generatedjudge pass (gpt-5.6-terra)audited 2026-09-14

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Dirac comb is fourier invariant

Statement

Assume Countable Choice. Under $\hat{f} (ξ) = \int f (x) e^{- 2 π i x \cdot ξ} d x$ , the unit-lattice Dirac comb satisfies

$F {III}_{Z^{n}} = {III}_{Z^{n}}$

in $S^{'} (R^{n})$ .

Facts & Assumptions

Given: Countable Choice and $n \geq 1$ .

[F1]

The comb pairs with a Schwartz test by its absolutely convergent lattice sum (Dirac comb).

[F2]

Fourier transformation on $S^{'}$ is defined by transposition (Fourier transform of a tempered distribution).

[F3]

Poisson summation at $x = 0$ says $\sum_{k \in Z^{n}} \hat{φ} (k) = \sum_{k \in Z^{n}} φ (k)$ , with both sums absolutely convergent (Poisson summation for Schwartz functions).

Proof

technique · test against Poisson summation

1.1F1F2

Let $φ \in S (R^{n})$ . Apply the defining transpose and the comb pairing.

$⟨ F {III}_{Z^{n}}, φ ⟩ = ⟨ {III}_{Z^{n}}, \hat{φ} ⟩ = \sum_{k \in Z^{n}} \hat{φ} (k) .$

All terms and sums are defined by [F1]–[F2]. [F1, F2]

2.1F1F3step 1.1∎

Poisson summation changes the last sum to $\sum_{k} φ (k) = ⟨ {III}_{Z^{n}}, φ ⟩$ . Equality on every Schwartz test proves the stated identity. Countable Choice is used only through the published Fourier and Poisson suppliers.

Depends on

Used by

Dependency tree · two levels

23 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)