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Fourier coefficients of (1 − β|z|/m)^T are nonnegative and small

Proved
QAlgorithms.Arunachalam.fourier_weightPow_bound

by Goku · Oct 13, 2026 · Mathlib 0df444a (Lean v4.33.1)

arunachalamlecture-notesquantum-information

Let m≥10m\ge10m≥10, β∈(0,1]\beta\in(0,1]β∈(0,1] and let TTT be an integer with 1≤T≤m/(e3β)1\le T\le m/(e^3\beta)1≤T≤m/(e3β) (the setting of Theorem 7.5.1). Let f:{0,1}m→Rf:\{0,1\}^m\to\mathbb Rf:{0,1}m→R, f(z)=(1−β∣z∣/m)Tf(z)=(1-\beta|z|/m)^Tf(z)=(1−β∣z∣/m)T, with Fourier coefficients f^(S)=Ez[f(z)(−1)S⋅z]\hat f(S)=\mathbb E_z[f(z)(-1)^{S\cdot z}]f^​(S)=Ez​[f(z)(−1)S⋅z]. Then for every S∈{0,1}mS\in\{0,1\}^mS∈{0,1}m of Hamming weight ∣S∣=q|S|=q∣S∣=q,

0 ≤ f^(S) ≤ 4e(1−β2)T(Tβm)qe22T2β2/m.0\ \le\ \hat f(S)\ \le\ 4e\left(1-\frac\beta2\right)^T\left(\frac{T\beta}{m}\right)^q e^{22T^2\beta^2/m}.0 ≤ f^​(S) ≤ 4e(1−2β​)T(mTβ​)qe22T2β2/m.

Combined with Claim 7.5.3 this bounds the diagonal of A\sqrt AA​ in Theorem 7.5.1.

Formalization Note The conditions m≥10m\ge10m≥10 and 1≤T≤m/(e3β)1\le T\le m/(e^3\beta)1≤T≤m/(e3β) are those of Theorem 7.5.1, which the proof of the lemma uses.

Preamble
import Definitions.Def_QAlgorithms_Defs_LearningHardInstances
set_option autoImplicit false
Formal statement
namespace QAlgorithms.Arunachalam

open QAlgorithms.Learning





/-- Arunachalam p.172–175, Lemma 7.5.4, in the setting of Theorem 7.5.1 (`m ≥ 10`,
`T ∈ [1, m/(e³β)]` a natural number, which the proof uses): for `β ∈ (0, 1]`, the Fourier
coefficients of `f(z) = (1 − β|z|/m)^T` on `{0,1}^m` satisfy
`0 ≤ f̂(S) ≤ 4e (1 − β/2)^T (Tβ/m)^q e^{22T²β²/m}` for all `S` with `|S| = q`. -/
theorem fourier_weightPow_bound {m T : ℕ} {β : ℝ} (hm : 10 ≤ m) (hβ : 0 < β) (hβ1 : β ≤ 1)
    (hT : 1 ≤ T) (hTm : (T : ℝ) ≤ (m : ℝ) / (Real.exp 3 * β)) (S : Fin m → ZMod 2) :
    0 ≤ boolFourier (weightPowFun m β T) S ∧
      boolFourier (weightPowFun m β T) S ≤
        4 * Real.exp 1 * (1 - β / 2) ^ T * ((T : ℝ) * β / (m : ℝ)) ^ hammingNorm S *
          Real.exp (22 * (T : ℝ) ^ 2 * β ^ 2 / (m : ℝ)) := by
  sorry


end QAlgorithms.Arunachalam
Source
Srinivasan Arunachalam, Quantum Algorithms and Learning Theory — Part II, 2018 thesis, ILLC DS-2018-08, PDF p. 172, Lemma 7.5.4. https://ir.cwi.nl/pub/27536/Thesis_S_Arunachalam
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What the Lean code literally says, in plain math · claude-opus-5-5

Let m,T∈Nm, T \in \mathbb{N}m,T∈N and β∈R\beta \in \mathbb{R}β∈R. Assume:

  • m≥10m \ge 10m≥10;
  • 0<β≤10 < \beta \le 10<β≤1;
  • T≥1T \ge 1T≥1;
  • T≤m/(e3β)T \le m / (e^3\beta)T≤m/(e3β).

Let S∈F2mS \in \mathbb{F}_2^mS∈F2m​ be arbitrary. Define w(u)=(1−β∣u∣/m)Tw(u) = (1 - \beta|u|/m)^Tw(u)=(1−β∣u∣/m)T on F2m\mathbb{F}_2^mF2m​, where ∣u∣|u|∣u∣ is the Hamming weight. Let w^(S)=2−m∑u∈F2mw(u)(−1)S⋅u\hat w(S) = 2^{-m}\sum_{u \in \mathbb{F}_2^m} w(u)(-1)^{S\cdot u}w^(S)=2−m∑u∈F2m​​w(u)(−1)S⋅u be its Boolean Fourier transform.

The theorem asserts

0≤w^(S)≤4e (1−β2)T(Tβm)∣S∣exp⁡ ⁣(22 T2β2m),0 \le \hat w(S) \le 4e\,\Big(1 - \frac{\beta}{2}\Big)^T \Big(\frac{T\beta}{m}\Big)^{|S|} \exp\!\Big(\frac{22\,T^2\beta^2}{m}\Big),0≤w^(S)≤4e(1−2β​)T(mTβ​)∣S∣exp(m22T2β2​),

where ∣S∣|S|∣S∣ is the Hamming weight of SSS.

Human review
  • Endorsed by Shuze Chen · Oct 13, 2026

    Confirmed by the moderator at approval.

  • Endorsed by Goku · Oct 13, 2026

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

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