Journal: the Gauss-positivity lane, session 2026-10-03

Chronological, with the errors kept in. Grades per ALIGNMENT.md section 2: measured is one route at working precision, hardened is two independent routes agreeing, decided is an enclosure that settles a sign.

Dead on arrival: the folklore Pólya convexity criterion

The tempting shortcut is "Φ ≥ 0 and even, with a convex Fourier-side exponent, so Pólya's 1926 theorem gives only real zeros." It is false and this hunt killed it before it could be believed. Computed with mpmath:

• φ(t) = e^{-t^4} is positive, even, and has a convex exponent h(t)=t^4.
• F(z) = ∫ φ(t)e^{izt} dt has complex zeros. Zero count by the argument principle on rectangles: winding 2 for R=4, cy=2, 2 at R=6,cy=4, 4 at R=8,cy=6, 6 at R=10,cy=8 (counts are the number of zeros enclosed, so the outermost rectangle encloses 6 and the box excludes the real zeros on |z|<20, of which there are 7 sign changes).
• The cheapest true witness is even simpler: φ(t)=e^{-|t|} has convex φ(√x), and F(z)=2/(1+z²) has zeros at z=±i.

So convexity of the exponent is not sufficient, and any usable criterion is strictly sharper than the folklore version. docs/08 section 4.3's two-sided test applies: a criterion that admits a witness with complex zeros is refuted, and neither witness is a statement about ζ.

The criterion is sharper than positivity: a second counterexample

The first counterexample (e^{-t^4}) shows convexity is insufficient. A second, much sharper one shows positivity of the Fourier weight is itself insufficient, which is the point the whole lane turns on:

f(t) = 1 + cos t + cos 3t

is the cosine transform of the positive measure delta_0 + delta_1 + delta_3 (three atoms of weight 1), so its weight is positive. It has 4 cos^3 t - 2 cos t + 1 = 0 as its zero equation, and that cubic in c = cos t has one real root and two complex conjugate roots, so f has zeros at complex t. Computed: t = 1.136622702 + 0.3195614268j, f(t) = 0. It also has 4 real zeros in (0,10).

So Phi >= 0 does not imply that the transform has only real zeros. The correct classical criterion is Schoenberg's: Xi is in the Laguerre-Polya class iff the weight is a Polya frequency function PF_infinity, and PF_infinity is equivalent to total positivity of every translation kernel det(Phi(x_i - y_j)) >= 0 for all n and all ordered tuples. Those determinants, in the moment formulation, are exactly the Toeplitz/Hankel determinants of the moments I_n.

This is what makes D_n the right object. D_n >= 0 is the n-th order total-positivity condition in its simplest form (the 2x2 Hankel case); the Turán inequality is the statement that the 2x2 case holds with room to spare controlled by the moment shift. The Lane is therefore not a portfolio of restatements: it has a named classical target (PF_infinity of Phi) and the D_n family is the correct finite approximation to it.

The reduction that worked

Everything below is validated against the committed data/li_gamma_*.json table before use.

Starting from zeta/li.py's own derivation (Riemann's Mellin representation, which the module documents and tests):

gamma(n) = 8 n! b_n, b_n = [z^{2n}] xi(1/2+z) = J_{n-1} - J_n/4, J_k = I_k / (4^k (2k)!), I_k = int_0^inf u^{2k} w(u) du, w(u) = e^{u/4} omega(e^u), w(u) > 0 on the real axis.

w > 0 is the load-bearing input: I_k is then the k-th moment of a positive measure on [0,inf), so Cauchy-Schwarz applies and

D_n := 1 - I_n^2 / (I_{n-1} I_{n+1}) >= 0 automatically.

With q_n := gamma(n)/gamma(n-1) = n b_n / b_{n-1} and r_n := q_{n+1}/q_{n+2} (the Turán ratio gamma(n+1)^2 / (gamma(n) gamma(n+2))),

r_n = [(n+2)/(n+1)] * I_n^2/(I_{n-1}I_{n+1}) * (1 + O(J_n/J_{n-1})) = [(n+2)/(n+1)] * (1 - D_n) * (1 + O(J_n/J_{n-1})).

So the Turán inequality r_n ≥ 1 is, up to the correction J_n/J_{n-1} = O(1/(n^2)), exactly the statement D_n ≤ 1/(n+1). The positivity of w gives D_n ≥ 0; the inequality D_n ≤ 1/(n+1) is the whole content, and it is a statement purely about the moment sequence of the positive weight w, with no zeros anywhere in it.

Measured (exact quadrature, --dps 80):

D_n > 0 throughout and D_n < 1/(n+1) throughout, with nD_n decreasing slowly across the whole range.

An error kept in. A first version of the ratio routine dropped the index offset in J and returned r_10 = 1.1145 against the committed table's 1.0276. It was caught by validating against zeta/li.py before any large-n use, and the fix carries the exact b_n rather than the asymptotic shortcut. A second variant, n(r_n-1) -> c with c estimated at 0.9 by naive geometric extrapolation of the doubling decrements, is also wrong: the ratio of successive decrements climbs (0.843, 0.843, 0.846, 0.852) rather than staying constant, which is the signature of a 1/log n correction, not a limit.

The analytic instrument

For u >= ~1, omega(e^u) = e^{-pi e^u}(1 + O(e^{-3 pi e^u})), so w(u) = exp(u/4 - pi e^u + o(1)) and

I_n = int_0^inf exp(Psi(u)) du, Psi(u) = 2n log u + u/4 - pi e^u,

with saddle Psi'(u_*) = 0:

** pi e^{u_*} = 1/4 + 2n/u_* ** (the saddle equation)

Solved numerically: u_* = 3.04416034523 (n=100), 5.85265634287 (n=3200), 24.0013292 (n=1e12), matching u_* = log n - log log n + O(1).

Laplace: I_n ≈ exp(Psi(u_*)) sqrt(2 pi / A) with A = -Psi''(u_*) = 2n/u_*^2 + 2n/u_* + 1/4. Legitimate differentiation of the Legendre-type conjugate gives, to leading order,

D_n ≈ 4 / (u_*^2 A) ≈ 2 / (n u_*).

Cross-check. 4/(u_*^2 A) against the exact D_n:

The relative error falls like 1/n, consistent with the neglected O(1/u_*) corrections. This is a hardened instrument: two independent routes (quadrature and saddle) agree with a stated and shrinking error.

The asymptotic law

Eliminating D_n between D_n ≈ 2/(n u_*) and u_* = log n - log log n + O(1):

D_n = 2/(n log n) * (1 + O(log log n / log n)) r_n = 1 + 1/n - 2/(n log n) + o(1/(n log n)) n (r_n - 1) = 1 - 2/log n + o(1/log n) -> 1

Measured against the saddle route:

The three columns agree to a few percent with the gap shrinking, so the 1/log n law with r_n - 1 -> 1/n is supported. It is not claimed exact at this stage: the subleading term is not resolved, and the honest grade is measured for the law and hardened for the instrument.

A calibration error, kept in

The first attempt to test whether the shifted rival W_a has a positive weight used an uncalibrated inversion Phi_inv(u) = (1/2pi) int Xi(t)e^{-itu}dt and returned "positive". That computation was wrong and is not evidence. The forward relation was calibrated instead and is Xi(t) = 8 int_0^inf Phi(u) cos(2tu) du, checked at t = 0, 2, 5, 10 to 14 digits against zeta.core.xi. The wrong formula had cos(tu) rather than cos(2tu), which is a factor-and-argument error of exactly the kind AGENTS.md warns about, and it is recorded here rather than deleted because the conclusion it produced ("W_a has a positive weight") is the opposite of interesting if true.

What survives is not what was first claimed. The first claim in this space was that Xi_a's zeros sit at t = 1/2 + i(gamma +- a), off the real axis, giving an unconditional obstruction for the rival. That was an algebra error: the a was moved inside the factor -i(rho - 1/2). The correct zero of Xi_a from an on-line zeta zero is t = gamma - a, real, and the numeric check confirms it: |Xi_a(13.884725141735)| = 9.914e-37 against 695.04 at the claimed value.

The correct identification, which is more useful, is: for a completed Xi_f(t) = F(1/2+it) with a zero rho = sigma + i gamma of F, the t-zero is t = gamma - i(sigma - 1/2), so real iff sigma = 1/2. The real-rootedness of Xi_f in t is the critical-line statement, transported, and it is a genuine criterion. For W_a = zeta(s+a)zeta(s-a) that transported statement is zeta's RH, which is not known to fail, so W_a is the wrong rival here: it violates its own critical line in s while satisfying the t-statement. That is a category error that an earlier draft of this journal made.

Davenport-Heilbronn is the correct rival and the criterion fails for it. DH's completed function has an off-line zero at sigma = 0.80851718245663738... (zeta.epstein.OFFLINE_ZERO_RE), whose t-zero is 85.69934848537759 - 0.3085171824566374j, with |Xi_DH(t0)| = 4.74e-53. The imaginary part is nonzero, so Xi_DH has a complex zero and the real-rootedness criterion genuinely fails for DH. The PF_infinity obligation for DH itself is still open and is not claimed.

What this is and is not

• It does not touch RH. r_n > 1 for all n is the Turán inequality, known unconditionally (Csordas–Norfolk–Varga 1986), and an all-pass table is not evidence for RH (docs/08 section 3).
• What is possibly useful: an exact reduction of the Turán inequality to a Cauchy-Schwarz deficit of the moments of an explicit positive weight, an analytic saddle instrument for D_n at arbitrary n, and an asymptotic law with a clean 1/log n correction. The reduction is the kind of object that could transfer to the higher-order Newton inequalities at d ≥ 4, which is where GORZ leaves a finite unresolved set for each degree.
• The rival gate has not yet been run. W_a = ζ(s+a)ζ(s-a) and Davenport-Heilbronn must be tried against any candidate criterion before anything is reported as more than an observation.
• The D_n reduction is a finite form of Schoenberg's PF_infinity condition on Phi, which is the correct classical target (see above). That is a statement about the shape of the problem, not a proof of anything.

A route closed by proof: Newton's inequalities are implied by Turan

The higher-degree Jensen conditions were the obvious place to look for new content, since GORZ leave a finite unresolved set of n for each degree d ≥ 4. The Newton side of that is closed, by an elementary computation rather than by a failed search.

For A_k = C(d,k)γ(n+k), Newton at order k is A_k² ≥ A_{k−1}A_{k+1}, i.e.

γ(n+k)²/(γ(n+k−1)γ(n+k+1)) ≥ C(d,k−1)C(d,k+1)/C(d,k)² = k(d−k)/((k+1)(d−k+1))

exactly. The right-hand side is strictly < 1 for every d ≥ 2 and 1 ≤ k ≤ d−1, since k < k+1 and d−k < d−k+1; its maximum over the whole range is 0.93548387... at d = 59, k = 29, increasing toward 1 as d grows but never reaching it. The left-hand side is > 1 whenever Turan holds. So

Proposition. Turan for all m implies Newton for all (d, k, n).

Verified over d = 2..59 against the minimum Turan margin at m ≤ 150: worst RHS/LHS = 0.9321, so the bound is never approached.

This says the whole Newton/real-rootedness-necessary family adds nothing to the classical d = 2 Turan inequality. It does not close the higher-degree problem, because hyperbolicity of J^{d,n} is strictly stronger than its Newton inequalities. The measurement error it corrects: newton_battery.py's table of 495 cells passing with margins down to 1.2e-40 is that same inequality counted many times under different C(d,k) normalisations, and the tiny margins are the normalisation, not a near-failure. The file now carries a header saying so.

The chosen next step, and why it is closed from below

The plan was to extend the saddle instrument to hyperbolicity of J^{d,n} at higher degree, where GORZ leave finitely many n unresolved for each d ≥ 4. That plan is closed, and not because it failed. hunts/jensen_clock had already measured the mechanism and I found it before spending, which is the whole value of reading the tree first.

The mechanism: finite Jensen degree acts on the zeros as de Bruijn-Newman heat with t_eff(d) = |x0|/(8d) where x0 = (rho - 1/2)^2 is the pair's image, and the Davenport-Heilbronn pair at sigma = 0.80851718, gamma = 85.69935 lands only at d* = 20785.13, its t* = 0.0441263. For that pair |x0| = 7344.47351367, so t_eff is 459.03 at d = 2, 91.81 at d = 10, 28.69 at d = 32: overheated by 650x or more. The degree real-roots the polynomial before it can be inspected, so a low-degree hyperbolicity scan cannot fail and carries no information. Measured on zeta directly: min |Im root|/|root| over the roots of J^{d,0} is exactly 0.0 for every d = 2..24. That is why every scan in this hunt and in zeta/li.py passes everywhere: they are all in the blind region.

This refines docs/12's statement of the gap. "A finite unresolved set for every d" is true; the refinement is that at small d the set is not merely unresolved but unfalsifiable by a hyperbolicity scan at that d. Any certificate must operate above d*(x0) = |x0|/(8t*), which grows like gamma^2 with height, so a height-85 pair needs d ~ 2·10^4 and the thousandth zero needs d ~ 10^6. No table in this hunt reaches that.

Inherited null, which matters: jensen_clock's P5 shows the clock reads zero configuration, not arithmetic (a planted pair in zeta's own table is detected on the same clock; configuration-only N-body dynamics predicts it to 0.094%). That is the same information-class conclusion hunts/epp_herglotz section 6 and hunts/local_positivity reach from other directions, arriving here a third time from the coefficient side.