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

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.

Gaussian summability kernels

Statement

Assume countable choice. For $t > 0$ put $k_{t} (x) = t^{- n / 2} e^{- π ∣ x ∣^{2} / t}$ on $R^{n}$ , $n \geq 1$ . Then $k_{t} \geq 0$ , $\int k_{t} = ∥ k_{t} ∥_{1} = 1$ , $\hat{k_{t}} (ξ) = e^{- π t ∣ ξ ∣^{2}}$ , and $(k_{t})_{t > 0}$ is an $L^{1}$ approximate identity as $t ↓ 0$ . Directly, $k_{t} (z) = \int e^{- π t ∣ ξ ∣^{2}} e^{2 π i z \cdot ξ} d ξ .$

Facts & Assumptions

Given: $n \geq 1$ , $t > 0$ and The Axiom of Countable Choice ( ${A C}_{ω}$ ).

[F1]

The Gaussian transform is $F (e^{- π s ∣ x ∣^{2}}) (ξ) = s^{- n / 2} e^{- π ∣ ξ ∣^{2} / s}$ for $s > 0$ (Euclidean Gaussian transform with the 2π normalization).

[F2]

The complex Lebesgue substitution formula uses the absolute Jacobian (A C^1 diffeomorphism satisfies the change-of-variables formula for L^1 functions).

[F3]

Dominated convergence passes integrable tails to zero (Dominated convergence).

[F4]

An approximate identity has unit mass, uniformly bounded $L^{1}$ norm and vanishing absolute tails outside every fixed radius (An $L^{1}$ approximate identity on $R^{n}$ ).

Proof

1.1F1given

Apply F1 with $s = 1 / t$ and multiply by $t^{- n / 2}$ to obtain $\hat{k_{t}} (ξ) = e^{- π t ∣ ξ ∣^{2}}$ . At $ξ = 0$ this gives mass one; positivity gives the same $L^{1}$ norm. Applying F1 with $s = t$ at frequency $- z$ proves the displayed inverse integral without an inversion theorem.

2.1F1F2F3F4step 1.1∎

Write $y = x / \sqrt{t}$ . F2 gives $\int_{∣ x ∣ > δ} k_{t} (x) d x = \int_{∣ y ∣ > δ / \sqrt{t}} e^{- π ∣ y ∣^{2}} d y$ . This tends to zero: F3 applied to the explicit integer-radius tails with majorant the integrable Gaussian gives their convergence to zero, and monotonicity bounds every sufficiently small t-tail by any fixed integer tail. All three conditions in F4 now hold with norm bound one. Countable choice is inherited from the Gaussian and substitution results.

Depends on

Dependency tree · 2 levels

Sources

Gerald Teschl, Topics in Real and Functional Analysis (2017)