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Updated 08-18-2026

D-026 — Q-S3a: exact factorization of W(a,θ;z); Cor 1.6 reduced to a Wiener–Hopf remainder estimate

Date: 2026-08-18 | Session: 006 cont. | Task: 2.7 (Q-S3a) Status: COMPLETE (structural answer). The vague question "does v±(a,·) → A_∞^{−1}e^{±·}?" is replaced by an EXACT identity with an explicit leading term and a precisely-posed remainder problem.

1. The identity [DERIVED — unconditional]

Setup (Suzuki 2606.09096, §6.2, §8.3): with T_a = A_a − λI (λ = 0 when A_a > 0, e.g. under RH; our numerics use λ = −1), the deficiency vectors v± = v±(a,·) ∈ L²(−a,a) satisfy the first-kind integral equation with CONTINUOUS kernel (g∗v±)(x) = h±(x), x ∈ [−a,a], h±(x) := e^{±x} ± i (equation after (8.2)ff. of the paper; kernel g = screw function (1.3), the same g our E-009–E-016 Galerkin machinery uses). Extend v± by zero outside [−a,a] and define the OUT-OF-INTERVAL TAIL τ±(x) := (g∗v±)(x), |x| > a. Then g∗v± = h±·1_{[−a,a]} + τ± on ℝ. The screw function's Fourier transform (convention v̂(z) = ∫v(x)e^{izx}dx) is ĝ(z) = z^{−2} ξ′(½−iz)/ξ(½−iz) [Suzuki 2023, (1.2), (1.8)] so, as tempered distributions (g grows ~ t log t; ĝ is meromorphic; v̂±, ĥ± entire by Paley–Wiener; τ̂± an analytic functional carried by |x| ≥ a), v̂±(z) = (ĥ±(z) + τ̂±(z)) · z²ξ(½−iz)/ξ′(½−iz). Substituting into W(a,θ;z) = (z−i)v̂+(z) + e^{iθ}(z+i)v̂−(z) (Thm 1.5): W(a,θ;z) = F(z) · M(a,θ;z), F(z) := z²ξ(½−iz)/ξ′(½−iz), M(a,θ;z) := R₀(a,θ;z) + R_τ(a,θ;z), R₀ := (z−i)ĥ+(z) + e^{iθ}(z+i)ĥ−(z) — EXPLICIT (below), R_τ := (z−i)τ̂+(z) + e^{iθ}(z+i)τ̂−(z) — the tail transform. F is EXACTLY the Cor 1.6 target. Hence Cor 1.6 ⟺ e^{ϕ(a,z)} M(a,θ;z) → 1 uniformly on compacts. [Honesty: the FT step is at the tempered-distribution level; the resulting identity of meromorphic functions is rigorous once ĝv̂± = ĥ± + τ̂± is established — standard Wiener–Hopf/Paley–Wiener–Schwartz bookkeeping. Flagged for the Lean/Phase-5 pass, not a gap in substance.]

2. R₀ in closed form — the lattice skeleton [DERIVED]

ĥ±(z) = ∫_{−a}^{a}(e^{±x} ± i)e^{izx}dx: ĥ+ = 2 sinh(a(1+iz))/(1+iz) + 2i sin(az)/z, ĥ− = 2 sinh(a(iz−1))/(iz−1) − 2i sin(az)/z. Using (1±iz) = i(z∓i) the (z∓i) factors cancel EXACTLY: R₀(a,θ;z) = −2i[sinh(a(1+iz)) + e^{iθ} sinh(a(iz−1))] + (2i sin(az)/z)[(z−i) − e^{iθ}(z+i)]. For z in a compact K ⊂ {|Im z| < 1} and a → ∞ the sinh terms are one-sided, giving the clean asymptotic skeleton R₀(a,θ;z) = 2 e^{a+iθ/2} · sin(az − θ/2) + o(e^a). Special cases: θ = 0: R₀ ≈ 2e^a sin(az); θ = π: R₀ ≈ −2i e^a cos(az).

3. Instant reconciliation with ALL measured phenomenology

4. The remainder problem — precise form of the "thinning"

Cor 1.6 ⟺ e^{ϕ}M → 1. With R₀ = 2e^{a+iθ/2} sin(az − θ/2): M = 2e^{a+iθ/2} sin(az − θ/2) · (1 + R_τ/R₀) + o(e^a). The sine's zeros get DENSER (spacing π/a → 0) — the OPPOSITE of the required density collapse. Therefore the thinning is equivalent to: (★) R_τ(a,θ;z) = −2e^{a+iθ/2} sin(az − θ/2) + e^{−ϕ(a,z)} + o(1) i.e. near-total cancellation between the tail transform and the sine skeleton, leaving a ZERO-FREE remainder. M's zeros (= W's zeros, all real by Thm 1.5) must flow out to ±∞ along ℝ — consistent with D-023's confinement argument.

5. Structural interpretation (the RH gate) [ANALYSIS]

The equation for v± is a FINITE-INTERVAL Wiener–Hopf equation: symbol ĝ(z) = z^{−2}ξ′(½−iz)/ξ(½−iz), data h± on the interval, complementary unknown τ± outside. As a → ∞, finite-interval WH solutions are governed by the half-line problem (Krein, Gohberg–Feldman theory), whose solution is built from the FACTORIZATION of the symbol. The symbol's poles/zeros are at the ξ-zeros (poles of ξ′/ξ): the factorization problem for ĝ "knows" the zero locations — this is exactly where RH must enter (and must fail for Davenport–Heilbronn/Epstein: off-line zeros destroy the factorization ⇒ (★) impossible ⇒ consistent with the sanity gate). This also explains Suzuki's remark that A_∞ "is only a formal symbol": the whole-line equation g∗v = e^{±·} has no L²/FT solution; the rigorous limit object is the WH factorization, not an operator inverse. The original Q-S3a question (v± → A_∞^{−1}e^{±·} in some norm) is thereby REPLACED by the correct one: the (★) remainder estimate.

6. What this buys us (status vs tasks)

7. Honesty notes