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The convolution module property

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QAlgorithms.Papers.Independent.Intakethird100.P38.convolution_module

by Gohan · Nov 1, 2026 · Mathlib 0df444a (Lean v4.33.1)

prove2quantum

Let n≥1n\ge1n≥1, let S=S(Rn;C)\mathcal S=\mathcal S(\mathbb R^n;\mathbb C)S=S(Rn;C) have its full Schwartz Fréchet topology, and write ψ^(u)=∫Rne−iu⋅xψ(x) dx\widehat\psi(u)=\int_{\mathbb R^n}e^{-iu\cdot x}\psi(x)\,dxψ​(u)=∫Rn​e−iu⋅xψ(x)dx. Let C\mathcal CC consist of the regular positive Borel measures σ\sigmaσ for which ∫(1+∣u∣2)−p dσ(u)<∞\int(1+|u|^2)^{-p}\,d\sigma(u)<\infty∫(1+∣u∣2)−pdσ(u)<∞ for some nonnegative integer ppp.

Let σ∈C\sigma\in\mathcal Cσ∈C. Suppose a regular positive Borel measure τ\tauτ belongs to Cb\mathcal C_bCb​: for each integer p≥1p\ge1p≥1 there are an integer q≥1q\ge1q≥1 and C>0C>0C>0, both independent of v∈Rnv\in\mathbb R^nv∈Rn, with

∫(1+∣u+v∣2)−q dτ(u)≤C(1+∣v∣2)−p(∀v∈Rn).\int(1+|u+v|^2)^{-q}\,d\tau(u)\le C(1+|v|^2)^{-p}\qquad(\forall v\in\mathbb R^n).∫(1+∣u+v∣2)−qdτ(u)≤C(1+∣v∣2)−p(∀v∈Rn).

Then σ⋆τ∈C\sigma\star\tau\in\mathcal Cσ⋆τ∈C, where convolution is the pushforward of the product measure under addition. This is the module inclusion C⋆Cb⊂C\mathcal C\star\mathcal C_b\subset\mathcal CC⋆Cb​⊂C in every positive dimension.

Preamble
import Definitions.Def_QAlgorithms_Papers_Independent_Intakethird100_P38
Formal statement
noncomputable section

namespace QAlgorithms.Papers.Independent.Intakethird100.P38

open MeasureTheory Filter

open scoped Topology ComplexConjugate ENNReal

/-- Theorem 11.1, PDF pp.25--26: C is a convolution module over C_b.
In (11.1) both q and C are uniform in v; the proof's missing q is restored. -/
theorem convolution_module (n : ℕ) (hn : 0 < n)
    (σ τ : Measure (SpectralDomain n)) (hσ : InC σ) (hτ : InCb τ) :
    InC (measureConvolution σ τ) := by sorry

end QAlgorithms.Papers.Independent.Intakethird100.P38

end
Source
Stochastic processes induced by singular operators, arXiv:1109.5273v1, PDF p. 25, Theorem 11.1: the convolution module property
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What the Lean code literally says, in plain math · gpt-6.1-sol

For every positive integer nnn, put X=RFin⁡nX=\mathbb R^{\operatorname{Fin}n}X=RFinn, and let σ,τ\sigma,\tauσ,τ be measures on XXX. Assume that σ\sigmaσ is regular and that ∫X(1+∥u∥2)−r dσ(u)<∞\int_X(1+\|u\|^2)^{-r}\,d\sigma(u)<\infty∫X​(1+∥u∥2)−rdσ(u)<∞ for some natural number rrr, with r=0r=0r=0 allowed. Assume also that τ\tauτ is regular and that, for every natural number p≥1p\ge1p≥1, there exist a natural number q≥1q\ge1q≥1 and a real number C>0C>0C>0 such that, for every v∈Xv\in Xv∈X,

∫X ⁣ ⁣−(1+∥u+v∥2)−q dτ(u)≤C(1+∥v∥2)−p,\int_X^{\!\!-}(1+\|u+v\|^2)^{-q}\,d\tau(u) \le C(1+\|v\|^2)^{-p},∫X−​(1+∥u+v∥2)−qdτ(u)≤C(1+∥v∥2)−p,

where the nonnegative expression on the right is viewed as an extended nonnegative real number. Define σ∗τ\sigma*\tauσ∗τ as the pushforward of σ⊗τ\sigma\otimes\tauσ⊗τ under (u,v)↦u+v(u,v)\mapsto u+v(u,v)↦u+v. Then σ∗τ\sigma*\tauσ∗τ is regular, and there exists a natural number sss, possibly zero, such that

∫X ⁣ ⁣−(1+∥u∥2)−s d(σ∗τ)(u)<∞.\int_X^{\!\!-}(1+\|u\|^2)^{-s}\,d(\sigma*\tau)(u)<\infty.∫X−​(1+∥u∥2)−sd(σ∗τ)(u)<∞.
Human review
  • Endorsed by Shuze Chen · Nov 1, 2026

    Confirmed by the moderator at approval.

  • Endorsed by Gohan · Nov 1, 2026

    Confirmed by the mission captain (proposal self-audit).

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