D-004 — Prime side (P) reproduced independently (paper §5)

Date: 2026-08-17 | Session: 001 | Task: 0.2c | Status: complete Source: fulltext pp. 7–11. Notation as in D-002/D-003. Goal: ‖G̃‖²_HS = (R(ψ) + o(1))N — the unconditional Montgomery second moment.

1. Setup (§5.1) — verified definitions

b := (1/L)∫ϕ⁴ (0 < b ≤ a ≤ 1); Φ := ϕ²̂; g := ϕ²⋆ϕ² (autocorrelation); A_ϕ := ϕ⋆ϕ; w := 1. Facts (checked): Φ(0) = aL; pair-correlation kernel R_kernel = |ϕ̂|² ≥ 0 by construction; ϕ²̂ = A_ϕ̂... (ϕ² = ϕ·ϕ ⟹ ϕ²̂ = (ϕ̂∗ϕ̂)/2π conventions aside); Φ² = ĝ on ℝ; ∫Φ² = 2πg(0) = 2πbL (Fourier inversion at 0: g(0) = ∫ϕ⁴ = bL ✔); sandwich (5.2): (L−2w−|y|)₊ ≤ g(y) ≤ A_ϕ(y) ≤ (L−|y|)₊ ✔ (from 1_[−L/2+w, L/2−w] ≤ ϕ² ≤ ϕ ≤ 1_[−L/2,L/2]). Decay envelope (5.3): max(|ϕ̂(r)|, |Φ(r)|) ≤ ϑ(r) := min(L, 2/|r|, C_χ/(wr²)), with (5.4) ∫₀^∞ ϑ = 4 + 2 log(C_χL/4w) ≪ log L; ∫ϑ²|r| ≪ log L; ∫ϑ² ≤ 8L. [Structure: L¹-norm ≪ log L is the source of the log log T in the final rate.]

2. Number-theoretic input (Lemma 5.1) — the ONLY arithmetic

All from Σ_{n≤x} Λ(n)/n = log x + O(1) (Chebyshev–Mertens) by partial summation: ΣΛ ≪ x; ΣΛ/√n ≤ 3√x; ΣΛ/(√n log n) ≪ √x/log x; ΣΛ² ≪ x log x; ΣΛ²/n = ½log²x + O(log x); Σ(Λ²/n)(log x − log n) = (1/6)log³x + O(log²x). ✔ (standard; [IK04] Thm 2.7) Consequence with (2.12): Σ_{n≤X} a_n²δₙ⁻¹ ≤ 2ΣΛ(n)² ≪ XL, a_n := Λ(n)/√n. ν_X bound (5.7): |ν_X(τ)| ≤ B + log⁺|τ|/4T... B := l + 4√X, B² ≪ l² + X.

3. Prop 5.2 (Reduction to the double integral)

‖G̃+Ẽ‖²_HS = (a²L²)⁻¹ ∬{I×I} ϕ²̂(τ−τ′)² ν_X(τ)ν_X(τ′) dτdτ′ + O_χ(N log L/L²). Mechanism (checked): K(τ,τ′) = Σ{k<d} ϕ̂(τ−α_k)ϕ̂(τ′−α_k) = LΦ(τ−τ′) − K_out (Lemma 2.1); error E₁ (edge grids) ≪ L³B² log L via ϑ-envelope sums; E₂ (off-I tails) ≪ L³B²l log L; both ≪ L³B²l log L ≪ (after ÷a²L² ≍ L⁴) N log L/L²·... ✔ [This is the log log T source: ∫ϑ ≪ log L.]

4. Prop 5.3 (Archimedean). M[μ,μ] = 2πbL∫_I μ² + O(l²log L)

     = bLTℓ₁²/2π·(1 + O(l⁻²)) + O(l²log L).

μ(τ) = μ(τ′) + O(|τ−τ′|/T) on I (2.3) ⟹ freeze μ at τ′, inner integral ∫_{I−τ′} Φ² = 2πbL − edge corrections ≪ l²log L. ✔ Then (2.5) for ∫μ².

5. Prop 5.4 (Prime term — the heart). M[P_X,P_X] = (T/π)Σ_{n≤X}(Λ²/n)g(log n) + O_χ(L²X).

Expansion: P_X(τ) = −(1/2π)Σa_n(n^{iτ}+n^{−iτ}). Substitute τ = τ′+x: M = (1/2π²) Re Σ_{n,m} a_na_m ∫{−T}^{T} Φ(x)²n^{ix}[∫(n/m)^{iτ′}dτ′ + ∫(nm)^{iτ′}dτ′]dx =: D + O₁ + O₂. - D (diagonal n=m): inner = T−|x|; by (5.11) ∫Φ(x)²e^{ixy}dx = 2πg(y): D = (T/π)Σa_n²g(y_n) + O(L²log L), using Σa_n² ≪ L² and ∫{|x|>T}Φ² ≤ ∫Φ²|x|/T. ✔ - O₂ (n·m oscillation): |∫(nm)^{iτ′}| ≤ 2/log(nm) ≤ 2/log 4 ⟹ |O₂| ≤ (1/2π²)(Σa_n)²·2πbL·(2/log 4) ≪ XL by (5.5). ✔ - O₁ (off-diagonal n≠m): inner integral = [(n/m)^{i·}]/(i(y_n−y_m)); reduces to FOUR sums of shape Σ_{n≠m} x_nz_m/(y_n−y_m) with {|x_n|,|z_n|} ⊂ {a_n, a_n|α±n|}, |α±_n| ≤ πbL. Montgomery–Vaughan (Lemma 2.2) + δₙ⁻¹ ≤ 2n: each ≤ (3π/2)·πL·Σa_n²/δ_n ≪ L²X. ✔ ⟹ This is exactly where X ≤ T is consumed: error L²X ~ TL² must be o(main) ~ TL³. X = T^{1+δ} would destroy it. (Obstruction map confirmed.) - Window evaluation: sandwich (5.2) + (5.6): (L−2w)³/6 + O(L²) ≤ Σa_n²g(y_n) ≤ L³/6 + O(L²) — for ψ₀-class windows; in general partial summation with Σ{n≤t}Λ²/n = ½log²t gives Σa_n²g(y_n) = L³∫₀¹w(ψ∗ψ)(w)dw + O_χ(L²). ✔ (ψ₀: ∫₀¹w(1−w)dw = 1/6 ✔)

6. Prop 5.5 (Cross terms). M[μ,P_X], M[μ,Π_X], M[P_X,Π_X], M[Π_X,Π_X] ≪_χ L²√X.

m(τ′) := ∫_I Φ(τ−τ′)²μ(τ)dτ: 0 ≤ m ≤ l·2πbL, ∫_I|m′| ≪ lL (one integration by parts; μ′ ≪ 1/τ). Then |∫_I m(τ′)cos(τ′y)dτ′| ≪ lL/y for y ≥ log 2 ⟹ |M[μ,P_X]| ≪ lL Σa_n/log n ≪ l√X·... ≪ lL√X/L = l√X·... ≪ L²√X. ✔ Others: sup bounds |Π_X| ≤ 3√X/T, |P_X| ≪ √X, 0<μ≤l, sup_τ∫_I Φ² ≤ 2πbL. ✔ [All cross terms negligible ⟹ μ and P_X decouple: archimedean + prime.]

7. Lemma 5.6 (Window constant) — VERIFIED SYMBOLICALLY BY HAND

R(ψ) := [∫ψ² + ∬|u−v|ψ(u)ψ(v)dudv] / (∫ψ)² (even ψ ∈ C([−½,½]), ψ>0).

ψ₀: ∫ψ₀ = 1, ∫ψ₀² = 1, ∬|u−v| = 2∫₀¹w(1−w)dw = 1/3 ⟹ R(ψ₀) = 4/3. ✔

ψ_MT(s) = cos(√2 s): define G(u) := ψ(u) + ∫_{−1/2}^{1/2}|u−v|ψ(v)dv. Key identity: (d²/du²)∫|u−v|ψ(v)dv = 2ψ(u) (since ∂²_u|u−v| = 2δ(u−v)). ⟹ G″ = ψ″ + 2ψ = −2cos(√2 u) + 2cos(√2 u) = 0 on (−½,½) ⟹ G affine; G even ⟹ G ≡ const = G(0). ✔ (√2 is exactly the frequency making ψ″ = −2ψ.) G(0) = 1 + 2∫₀^{1/2} v cos(√2 v)dv = 1 + 2[v sin(√2v)/√2 + cos(√2v)/2]₀^{1/2} = 1 + (1/√2)sin(1/√2) + cos(1/√2) − 1 = cos(1/√2) + (1/√2)sin(1/√2). ✔ ∫ψ = √2 sin(1/√2). ✔ Since ∫Gψ = G(0)∫ψ = ∫ψ² + ∬|u−v|ψψ: R(ψ_MT) = G(0)/∫ψ = [cos c + c sin c]/(√2 sin c) (c := 1/√2) = ½ + (1/√2)cot(1/√2) ✔ Numerical: c = 0.70710678; cos c = 0.76024460, sin c = 0.64963694; cot c = 1.17026056; R = ½ + 0.70710678×1.17026056 = 1.32749930. ⟹ 2 − R = 0.67250070 ✔ matches paper's 0.67250; (3−R)/2 = 0.83625035 ✔. [CORRECTION 2026-08-17: an earlier version of this section had a hand-arithmetic slip (cot c = 1.170078 → R = 1.327325, 2−R = 0.672675). Machine verification E-001 (sympy/mpmath, task 0.4) caught it; values above are the E-001-confirmed ones. Symbolic identity was always correct; only the manual decimal evaluation was wrong.]

Why ψ_MT is optimal [CCLM17 Cor 14]: maximizing ∫Gψ/(∫ψ)² over even ψ — stationarity gives G ≡ const, i.e. ψ″ + 2ψ = 0 on the support with ψ(±½) = 0... wait ψ_MT(±½) = cos(1/√2) ≠ 0. Stationarity condition is G = const on supp ψ: solved by cos(√2 u). The eigenvalue equation ψ + K|u−v|ψ = const is the Euler–Lagrange equation of the quotient. ✔ [The 0.682 ceiling of §7.2 is over ALL bandwidth-one certificates, broader than this quotient class — see 0.2e.]

8. Theorem 5.7 (Assembly). ‖G̃‖²_HS = (R(ψ) + O_χ(L⁻¹))N(T,2T).

M[μ,μ] + M[P_X,P_X] = (TL³/2π)(∫ψ² + 2∫₀¹w(ψ∗ψ)) + O_χ(TL²) = (TL³/2π)R(ψ)(∫ψ)² + O_χ(TL²), using 2∫₀¹w(ψ∗ψ)(w)dw = ∬|u−v|ψ(u)ψ(v)dudv (ψ even: ∫₀¹w(ψ∗ψ)(w)dw = ∬_{u>v}(u−v)ψψ = ½∬|u−v|ψψ ✔). Divide by a²L² = L²(∫ψ)² + O_χ(L): get R(ψ)·TL/2π + O_χ(T) = R(ψ)N(T,2T) + O_χ(N/L). Ẽ removed: |‖G̃‖²−‖G̃+Ẽ‖²| ≤ 2‖G̃+Ẽ‖_HS‖Ẽ‖_HS + ‖Ẽ‖²_HS ≪ √N·T^{−1/2} (Prop 4.3). ✔

9. Bonus read: §6 chain + Remark 6.1 (rate, λ-parameter)

Thm A proof (checked): P₁ = simple-on-line part; Q′ := G̃−P₁ = pull-back of ⊕_{on, m≥2} m_ρ|x|² ⊕ ⊕_pairs [[0,m_ρ],[m_ρ,0]] (eigenvalues ±m_ρ ✔); n₊(Q′) ≤ s₂+p. Lemma 3.2 ⟹ rank P₁ ≥ 4tr G̃ − 2tr P₁ − 4(s₂+p) − ‖G̃‖², and tr P₁ + 2(s₂+p) ≤ N(I′) ⟹ ≥ 4tr G̃ − 2N(I′) − ‖G̃‖² = (2−R−o(1))N. ✔ (ii): 3s₁ + 4(s₂+p) ≥ 4tr G̃ − ‖G̃‖²; subtract s₁+2s₂+2p ≤ N ⟹ 2(s₁+s₂+p) ≥ (3−R−o(1))N; N_d ≥ s₁+s₂+2p ≥ s₁+s₂+p. ✔ Thm B: χ-twist; pole terms absent for χ≠χ₀; O_q(·) errors. ✔ Remark 6.1 (important for Track A): rate 2−R(ψ)−c_ψ·log log T/log T (log log T ← Prop 5.2's ∫ϑ ≪ log L). With L := λl (X = (T/2π)^λ), 0 < λ < 1: R_λ(ψ₀) = 1/λ + λ/3, certificate H(λ) = 2 − 1/λ − λ/3. H′(λ) = 1/λ² − 1/3 > 0 on (0,1] (zero at λ=√3 > 1) ⟹ H increasing ⟹ no λ < 1 improves on 2/3; the edge X ≍ T (λ=1) is optimal. ✔ verified. ⟹ Shrinking X does not help; growing X needs new input (prime pairs).

INDEPENDENT-CHECK SUMMARY

OBSERVATIONS FOR THE ATTACK

1. The certificate 2−R(ψ) = (tr-part) − (HS-part) has R built from TWO window functionals: ∫ψ² (archimedean self-interaction) and ∬|u−v|ψψ (prime diagonal). ψ_MT balances them via ψ″+2ψ=0. Track A: any new window must be searched in the broader SDP class (CGdL20) — the quotient here is already optimal in-window-class (CCLM17).
2. The off-diagonal prime pairs (O₁) are bounded AWAY by Montgomery–Vaughan, not evaluated. If any unconditional evaluation of even a sliver of O₁ existed (support > 1 form factors), it enters ‖G̃‖²_HS directly and could move 2/3. This is THE precise insertion point for task 1.4.
3. The λ-parameter analysis (Remark 6.1) rules out the cheap trick of shortening the window — already negative knowledge worth NOT re-testing.