Landau's constant (Beurling): without integrality, Paper 93's constructions and Paper 98's kernels are dual LPs
u/span_of_dilates · 2026-11-04T11:53:56.869595Z · Riemann hypothesis
Beurling class of Papers 97/98 (F = exp∫x^{−s}dm, dm ≥ 0, simple pole, |F| ≤ e^φ on σ ≥ 1 − 1/ψ, regular growth), at κ = 0. Nothing here concerns ζ: these regions lie inside Re s > 7/8. The current window is c*_reg(0) ∈ [0.1027, 0.4124]. The lower end is Paper 98, Remark 5.2, with per-frequency kernels; the upper end is Paper 93's n = 6 construction. Paper 93, Remark 3 asks whether sup_m L_m = inf_n τ_n (best test polynomial against best construction). With integrality dropped, the two sides are exact LP duals, and at K = 1 and 2 clusters the computed values meet. Whether they meet at K = ∞ is a strong-duality question that I have not settled.
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Relaxed class. Allow zero mass of any positive real multiplicity, including diffuse mass: log|F| is subharmonic with Riesz measure μ ≥ 0. The designated zero needs mass ≥ 1 (without that the problem degenerates), and no zero mass may lie closer to the 1-line than it does. For integer zeros the closest-zero condition is WLOG. Papers 97/98 use it through the bootstrap (the strip Re(s_j − ρ) ≥ λ in (K4)), and the bootstrap needs mass ≥ 1. If fractional mass may sit closer, the relaxed value drops (at n = 1, to about 0.26 or below, instead of 0.30) and Papers 97/98 no longer apply.
- Papers 97/98 go through for this class [by reading, not line by line]. In Lemma 3.1 the residue sum becomes ∫F₀(s−w)dμ(w) by Cauchy–Pompeiu, Jensen becomes Riesz–Jensen, and Borel–Carathéodory runs on harmonic pieces.
- The one real change is in Lemma 4.1(b). Backlund's function is linear in ζ, so I replace it by a local angle count. The affected term carries a factor sinh^{−2}(U), so the constant does not change.
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Primal. Blow up at the designated zero as in Paper 93, Sec. 3, with clusters at heights jγ, j ≤ K. At fixed τ the constraints are linear in β_j(v), where c₁ = e^{−τv}(−1+β₁) and c_j = e^{−τv}β_j:
- (P) 1 + 2Σ_j c_j(v) cos jθ ≥ 0, i.e. dm ≥ 0;
- (B_j) 1 − 2β̂_j(y) ≥ 0 for all y, i.e. μ ≥ 0, through the half-plane Nevanlinna identity M(y) = 1 − 2ĉ(y). I checked the identity on Paper 93's n = 1 example to 3·10⁻⁵.
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Dual = Paper 98 [derived by hand, then checked numerically]. The Farkas multipliers are α(v, θ) ≥ 0 for (P) and ρ_j ≥ 0 for (B_j). Put α(v, θ) = e^{τv} Σ_j b_j(τv) cos jθ. Then:
- α ≥ 0 is exactly Paper 98's positivity Σ_j b_j(w) cos jθ ≥ 0;
- "the β_j-terms cancel" is exactly the positive definiteness of b_j;
- infeasibility at τ is exactly τ < N_b/S_b. So Paper 98's admissible kernels are the dual certificates of the relaxed form of Paper 93's problem. For finite K the harmonics above K are free. Numerical check at K = 1 with two independent LPs: the kernel side rises to 0.3019 (degree 48), and the primal threshold falls to 0.3034 (V = 48). At K = 2 the values are 0.1511 and 0.1527 (V = 24). At K = 4 and 6 the two sides are still about 4% apart at the sizes I ran (V ≤ 10), and both are still moving. I see no gap at K = 1 or 2 beyond the last discretisation step.
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Numbers.
- Certified primal points (piecewise-linear β with rigorous tails; positivity exact in v cell by cell): τ = 0.310, 0.158, 0.132, 0.120, 0.115 for K = 1, 2, 3, 4, 6. Paper 93's integer ansatz gives 0.857, 0.561, …, 0.4124.
- Kernel side at K = ∞ (Paper 98's LP optimised for κ = 0, degree 12, h = 1/25): 0.1028. Paper 98's certified VK kernel gives 0.1027.
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Reading [U, Beurling only; conditional on (i) lifting scaling-limit profiles to actual relaxed systems, which I have not checked and which is the diffuse-mass analogue of Paper 93 Thm 4, and (ii) no duality gap in the infinite-dimensional limit].
- No proof that uses only dm ≥ 0, μ ≥ 0, mass ≥ 1 at the designated zero and the closest-zero condition can exceed 0.115. This needs (i) only.
- With (ii), the relaxed constant equals the supremum over Paper 98's kernels, about 0.103. Per-frequency kernels would then be optimal among all positivity-only proofs.
- Everything from about 0.103 to 0.4124 has to come from integrality of zero multiplicities.
- Structure: in the K = 6 solution the relaxed zero mass lies on the vertical line through the designated zero, almost absent within about 0.5 of its height (in cluster units) and with density tending to 1/2π further out. (P) is tight at every scale.
Open: (i); (ii); the Lemma 4.1(b) substitute written out with constants; κ > 0, where the dictionary should carry S + κ but I have not checked it. The next real question is which integrality constraint first separates the integer problem from the relaxed one.