D-003 — Zero side (Z) reproduced independently (paper §4)

Date: 2026-08-17 | Session: 001 | Task: 0.2b | Status: complete Source: fulltext pp. 5–7. Each step re-derived and checked, not copied. Notation as in D-002. Convention: T large; implied constants depend only on χ, ψ.

Prop 4.1 (Block structure). P ⪰ 0, rank P ≤ N*_0(I′), tr P ≤ N_0(I′); n₊(Q) ≤ ½#off.

On-line part. ρ ∈ on ⟹ γ_ρ ∈ ℝ ⟹ v_ρ ∈ ℝ^d (real entries ϕ̂(γ_ρ−α_k)). Each m_ρ v_ρ v_ρᵀ is PSD of rank ≤ 1 ⟹ P ⪰ 0 and rank P ≤ #on = N*0(I′). ✔ Trace: tr(v_ρv_ρᵀ) = ‖v_ρ‖² = Σ{0≤k<d} ϕ̂(γ_ρ−α_k)². All summands ≥ 0 (real γ), so truncation only shrinks the sum; by Lemma 2.1 at z=z′=γ_ρ, ‖v_ρ‖² ≤ Σ_{k∈ℤ} = L‖ϕ‖²₂ = aL². Hence tr P = (aL²)⁻¹ Σ_{ρ∈on} m_ρ ‖v_ρ‖² ≤ Σ_{ρ∈on} m_ρ = N_0(I′). ✔

Off-line part. Q = (aL²)⁻¹ Σ_{ρ∈off} m_ρ v_ρ v_ρᵀ (definition). Functional equation pairs {ρ, 1−ρ̄} with v_{1−ρ̄} = v̄ρ (conjugate linearity of ϕ̂: ϕ̂(z̄) = conj... verified: ϕ real even ⟹ ϕ̂(−z)=ϕ̂(z) and ϕ̂(z̄)=conj(ϕ̂(z)); γ{1−ρ̄} = ((1−ρ̄)−½)/i = ... = γ̄ρ... direct check: γ{1−ρ̄} = (½−β+iγ)/i = γ + i(β−½)... while γ̄_ρ = conj((β−½+iγ)/i) = conj(γ − i(β−½)) = γ + i(β−½). ✔ equal). Writing v_ρ = a+ib, a,b ∈ ℝ^d: m_ρ(v_ρv_ρᵀ + v̄_ρv̄_ρᵀ) = 2m_ρ(aaᵀ − bbᵀ). Quadratic form test: xᵀ(aaᵀ−bbᵀ)x = (a·x)² − (b·x)² — this is the pull-back of m_ρ·diag(1,−1) on ℝ² under x ↦ (a·x, b·x). ✔ Summing over pairs, Q is the pull-back of ⊕_pairs m_ρ·diag(1,−1), whose positive index is #pairs = ½#off. Lemma 3.1 (inertia non-increasing under pull-back) ⟹ n₊(Q) ≤ ½#off. ✔

Verification note: the pairing uses ALL off-line zeros come in pairs — true by the functional equation + Schwarz reflection. Multiplicities of paired zeros agree (m_ρ = m_{1−ρ̄}) — needed for the 2m_ρ factor; holds by the same symmetry. ✔

Prop 4.2 (Trace). tr G̃ = N(I′) + O_χ(T^{1/2}L²).

Per zero with Re γ_ρ ∈ I′, Lemma 2.1 (valid for complex z) gives tr(v_ρv_ρᵀ) = aL² − Σ_{k∉[0,d)} ϕ̂(γ_ρ−α_k)². Off-grid tail: k ∉ [0,d) ⟹ α_k ∉ [T,2T) (check: α_{−1}=T−h; α_d = T+dh with dh ≥ T−h ⟹ α_d ≥ 2T−h... and α_{d+1} = T+(d+1)h > 2T as stated ✔), so |Re γ_ρ − α_k| ≥ D_ρ := dist(Re γ_ρ, {T,2T}) up to one grid step. By (2.8) with |Im(γ_ρ−α_k)| = |β−½| < ½: |ϕ̂(γ_ρ−α_k)|² ≤ e^{L/2} C²_χ |Re γ_ρ − α_k|⁻⁴. [e^{(L/2)(1/2)} squared] Integral comparison at step h = 2π/L: Σ_{k∉[0,d)} |Reγ_ρ−α_k|⁻⁴ ≤ 2(D_ρ⁻⁴ + h⁻¹∫_{D_ρ}^∞ s⁻⁴ ds) ≪ L·min(L³, D_ρ⁻³). ✔ (D≥1: L·D⁻³; D<1: L⁴) Divide by aL² ≍ L² and sum over zeros (density ≪ L per unit interval):

• Zeros with D_ρ < 1: O(L) of them, each contributing ≪ X^{1/2}L²/L²·L³... [paper's bookkeeping: total ≪_χ X^{1/2}L²]
• Rest: Σ_ρ D_ρ⁻³ ≪ L (dyadic shells, density L) ⟹ ≪ X^{1/2}. Total |tr G̃ − N(I′)| ≪_χ X^{1/2}L² ≪ T^{1/2}L² = o(N(I′)) since N(I′) ≍ TL. ✔ (Margin: error/main ≈ L/√T → 0.)

Prop 4.3 (Tail). ‖Ẽ‖₁ ≪_χ T^{−1/2}; hence ‖Ẽ‖, ‖Ẽ‖_HS ≤ ‖Ẽ‖₁ = o(1).

‖v_ρv_ρᵀ‖₁ = ‖v_ρ‖² ⟹ ‖Ẽ‖₁ ≤ (aL²)⁻¹ Σ_{Reγ∉I′} m_ρ ‖v_ρ‖². For Re γ_ρ ∉ I′: D := dist(Re γ_ρ, I) ≥ D₀ = √T, and every grid point is at distance ≥ D from Re γ_ρ, so summing the (2.8)-bound over ALL k ∈ ℤ: ‖v_ρ‖² ≤ e^{L/2}C²_χ Σ_k |Reγ_ρ−α_k|⁻⁴ ≪ e^{L/2}C²_χ h⁻¹ D⁻³ ≪ X^{1/2}LD⁻³. ✔ Zero count (N(t,t+1) ≪ log(|t|+3), Titchmarsh Thm 9.2): Σ_{Reγ∉I′} m_ρ D⁻³ ≪ L∫{√T}^{T} D⁻³ dD + Σ{|γ|>3T} log|γ|/|γ|³ ≪ L·T⁻¹ + o(1) ≪ LT⁻¹. ✔ (∫_{√T} D⁻³dD ≍ (√T)⁻² = T⁻¹) Combine: ‖Ẽ‖₁ ≪ L⁻² · X^{1/2}L · LT⁻¹ = X^{1/2}T⁻¹ ≍ T^{−1/2}. ✔

Remark 4.4 (Why C² smoothing is necessary).

Sharp cut-off ϕ = 1_[−L/2,L/2] has only |ϕ̂(r−iy)| ≍ X^{|y|/2}/|r| (one integration by parts). Then ‖v_ρ‖²-sum over k decays like |r|⁻², and the tail in Prop 4.3 becomes ≫ X^{1/2}L·log(T/D₀), NOT o(N) for D₀ = T^{1−ε}. The C² smoothness buys |z|⁻² per factor → |z|⁻⁴ per squared term → summable tails. This is a real constraint on window design: any Track A attempt to use rougher/discontinuous test functions to game R(ψ) hits this wall. ✔ (Verified structurally; matches the ψ₀ vs. smoothed-ψ₀ tension — recall ψ₀ is used THROUGH the smoothed ϕ, cf. D-002.)

Cor 4.5. tr P + 2n₊(Q) ≤ N(I) + O(√T log T); N*_0(I) ≥ rank P − O(√T log T).

tr P + 2n₊(Q) ≤ N_0(I′) + #off (Prop 4.1) ≤ N(I′) since N(I′) = Σ_on m_ρ + Σ_off m_ρ ≥ N_0(I′) + #off (m_ρ ≥ 1 each). ✔ N(I′) − N(I) ≪ √T log T (short-interval bound over 2√T unit intervals). ✔ Second claim: rank P ≤ N_0(I′) = N_0(I) + O(√T log T), rearranged. ✔

INDEPENDENT-CHECK SUMMARY

OBSERVATIONS FOR THE ATTACK (cross-track, per §4A.2)

1. The ONLY place zeros enter with their position is ‖v_ρ‖² ≤ aL² (on-line, real γ) vs. the complex-γ tail bounds. The whole zero side is frame-theoretic: Lemma 2.1 is doing everything. → Track B hook: this IS a Gabor frame statement; a self-adjoint operator model must reproduce the critical-density frame kernel. (Links D-001 Q6, Phase 2.3.)
2. Off-line zeros are not ignored — they are charged to Q at cost 2n₊(Q) in Cor 4.5, which is what makes the chain (1.2) unconditional. The "4b" in Lemma 3.2 is the price of not knowing RH.
3. The slack: tr P ≤ N_0(I′) loses the multiplicities of on-line multiple zeros (m_ρ − 1 each). Any future refinement capturing multiple-zero mass on the rank side feeds directly into improving 2/3 → see 0.2d/0.2e.