ExampleConstruction: AI-adaptedVerification: AI-adaptedPipeline-generatedjudge pass (gpt-5.6-terra)audited 2026-09-14

How statement and proof provenance work

The first chip identifies the source of the statement or construction; the second identifies the source of its local proof or verification.

Literature-sourced: the exact statement appears in a cited source; only wording and notation differ.
AI-adapted: a semantically identical restatement of literature-sourced material, modulo indexing, notation, and boundary cases adopted by the library.
AI-generated: a genuinely novel statement formulated by AI, with no source for the claim itself.

These labels describe origin, not correctness: citations and verification chips remain separate evidence.

Dirac comb and poisson summation

Example

Assume Countable Choice. Fourier invariance of the unit-lattice comb is equivalent, on Schwartz tests, to

$\sum_{k \in Z^{n}} φ (k) = \sum_{k \in Z^{n}} \hat{φ} (k) .$

For $g_{t} (x) = e^{- π t ∣ x ∣^{2}}$ , $t > 0$ , this gives the theta transformation

$\sum_{k \in Z^{n}} e^{- π t ∣ k ∣^{2}} = t^{- n / 2} \sum_{k \in Z^{n}} e^{- π ∣ k ∣^{2} / t} .$

Facts & Assumptions

Given: Countable Choice, $n \geq 1$ , and $t > 0$ .

[F1]

The unit-lattice comb is Fourier invariant (Dirac comb is fourier invariant).

[F2]

The $2 π$ -normalized Gaussian formula is ${\hat{g}}_{t} (ξ) = t^{- n / 2} e^{- π ∣ ξ ∣^{2} / t}$ (Euclidean Gaussian transform with the 2π normalization).

[F3]

The defining lattice sum for the comb converges absolutely on every Schwartz test (Dirac comb), and the Fourier transform sends Schwartz tests to Schwartz tests (Fourier transform is a topological automorphism of Schwartz space).

Verification

technique · evaluate one distributional identity on two classes of tests

1.1

Evaluate comb invariance on an arbitrary $φ \in S$ .

F1F3

$\sum_{k} \hat{φ} (k) = ⟨ III, \hat{φ} ⟩ = ⟨ F III, φ ⟩ = ⟨ III, φ ⟩ = \sum_{k} φ (k) .$

This is Poisson summation at the origin, with no rearrangement of a conditionally convergent series. [F1, F3]

1.2

Conversely, suppose the displayed lattice-sum identity holds for every $φ \in S$ .

F3def. equality in tempered distributions

Then [F3] and the definition of the distributional Fourier transform give

$⟨ F III, φ ⟩ = ⟨ III, \hat{φ} ⟩ = ⟨ III, φ ⟩ .$

Thus $F III = III$ in $S^{'}$ , which proves the asserted equivalence. [F3, def. equality in tempered distributions]

2.1

Apply step 1.1 to $g_{t}$ and substitute [F2]. The left and right lattice sums become exactly the two sides of the theta transformation. At $t = 1$ the Gaussian is itself Fourier invariant; as $t$ varies, the formula exchanges $t$ and $1 / t$ with the dimension factor $t^{- n / 2}$ . Countable Choice is used only through [F1]–[F3].

F2step 1.1∎

Depends on

Used by

Nothing in the library uses this result yet.

Dependency tree · two levels

30 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)