Every half-turn sends the endpoint bill up by a factorial

Let a finite complex measure live in the shallow endpoint window \([-r,0]\), and write

$$ B(z)=\int_{[-r,0]}e^{i\tau z}\,d\mu(\tau). $$

Suppose \(B\) never vanishes on a horizontal interval of length \(S\), and its phase advances by \(V\). Put

$$ n=\left\lfloor\frac V\pi\right\rfloor, \qquad a_*=\min |B|, \qquad M_y=\int_{[-r,0]}e^{-y\tau}\,d|\mu|(\tau). $$

Then

$$ \boxed{ \frac{M_y}{a_*} \ge \frac{(n+1)n!}{(rS)^n}. } $$

Every half-turn sends the endpoint bill up by another order of divided difference. The nodes do not need to be equally spaced. The phase does not need to move monotonically. They may bunch up horribly. Bunching only makes the barycentric denominators smaller, which makes the lower bound stronger.

I keep staring at the \(n!\). It is exactly the right kind of rude.

The half-turn nodes align themselves#

Choose a continuous phase \(\theta\) along the interval. Every time it advances by \(\pi\), continuity supplies another ordered point

$$ x_0<x_1<\cdots<x_n $$

with

$$ B(x_j+iy) = (-1)^je^{i\theta(x_0)}|B(x_j+iy)|. $$

For increasing real nodes,

$$ \operatorname{sgn} \prod_{k\ne j}(x_j-x_k) = (-1)^{n-j}. $$

That sign cancels the alternating phase. Every term in the divided difference

$$ [B(\cdot+iy);x_0,\ldots,x_n] = \sum_{j=0}^n \frac{B(x_j+iy)} {\prod_{k\ne j}(x_j-x_k)} $$

lies on one complex ray. There is no cancellation left:

$$ \left|[B;x_0,\ldots,x_n]\right| \ge \frac{(n+1)a_*}{S^n}. $$

On the other hand, Hermite--Genocchi turns the same divided difference into an average of the \(n\)-th derivative. Endpoint support gives

$$ |B^{(n)}(x+iy)|\le r^nM_y, $$

and therefore

$$ \left|[B;x_0,\ldots,x_n]\right| \le \frac{r^n}{n!}M_y. $$

The two estimates collide and produce the factorial bill.

If the front carries \(n\sim cLS\) half-turns while \(r=o(L)\), Stirling gives

$$ \boxed{ \log\frac{M_y}{a_*} \ge (c-o(1))LS\log\frac Lr. } $$

A shallow spectrum may imitate depth-\(L\) phase rotation, but the signed variation needed to do it grows superoscillatorily in the number of phase cells.

Curvature has two ways to pay#

A Stokes front is usually not horizontal. There are now two complementary ways to transfer its winding into the same endpoint cost.

The robust version uses the argument principle. Put the curved graph above a straight comparison boundary. If the projective identity forces phase advance \(V\) along the graph, then the combined winding on the lower and vertical sides differs from it only by \(2\pi\) times the number of interior zeros. Zeros do not erase the debt. They increase it.

So one straight side carries at least a third of the phase advance. Rolle's theorem gives the factorial derivative cost there, while harmonic measure transfers the amplitude comparison back to the curved front:

$$ \frac{M_+}{A_\gamma} \ge \left(\frac{n}{erT_*}\right)^{\omega n}, \qquad n=\left\lfloor\frac{V}{6\pi}\right\rfloor-1. $$

This version is fussy but forgiving. It allows an arbitrary graph arc and does not need slow curvature. The price is a factor six in phase density and the harmonic-measure exponent \(\omega\).

The sharp version assumes a long scaled-\(C^2\) component

$$ \gamma_L(t)=t+iy_L(t), $$

with

$$ \|y_L'\|_\infty=O(S_L^{-1}), \qquad \|y_L''\|_\infty=O(S_L^{-2}), $$

and uniform projective phase locking. The curve slightly rotates every Vandermonde denominator. Instead of pretending those rotations are zero, one solves for nodes whose boundary phases absorb them exactly.

If

$$ \chi_j = \sum_{k\ne j} \arg\frac{\gamma_L(t_j)-\gamma_L(t_k)}{t_j-t_k}, $$

the node equation is

$$ \varphi_L(t_j)-\chi_j=\alpha+\pi j. $$

The scaled curvature bounds make this a contraction on the ordered gap box. At its fixed point, all complex barycentric terms again lie on one ray. Chebyshev's extremal bound then recovers the sharp density

$$ n_L=(1+o(1))\frac{L\ell_L}{\pi}, $$

and hence

$$ \boxed{ \log\frac{M^-_{L,r}}{A_{*,L}} \ge (1-o(1)) \frac{L\ell_L}{\pi} \log\frac L{r_L}. } $$

The curvature correction is not thrown away as an error. It is built into the phase quantization. That is the part I find cutest.

Fragmentation does not make the phases private#

The sharp curved theorem originally wanted one long component. A front can try to evade it by breaking into many short intervals, each carrying only a piece of the total winding.

That works only if the fragmentation itself reaches the phase scale.

Let

$$ G=\bigcup_{\nu=1}^J I_\nu $$

have total length \(\ell\) inside a span \(S\), with every interval lying on one shared scaled-\(C^2\) graph. Suppose the sum \(V\) of the positive componentwise phase advances obeys

$$ V\ge \rho L\ell, \qquad J\le \frac{\rho L\ell}{32\pi}. $$

First-passage boxes of phase width \(4\pi\) are cut out separately on every component. Discard the long boxes, then keep alternating short boxes so that their endpoints cannot collide. This still leaves

$$ N=\left\lfloor\frac{\rho L\ell}{64\pi}\right\rfloor $$

ordered boxes spread across the fragments.

The cute point is that the node equations are global. Each node sees the Vandermonde phase contributed by every other node, including nodes on other components. Small scaled curvature keeps each of these global secant-phase corrections inside a half-plane, and Poincare--Miranda chooses one node in every box simultaneously.

All barycentric terms again land on one ray. With \(n=N-1\),

$$ \boxed{ \frac{M_Y}{A_*} \ge \frac{(n+1)n!} {\left(rS\sqrt{1+K^2/S^2}\right)^n}. } $$

So a front may be fragmented, but its phases are not allowed to pretend they belong to separate universes. If the fragments live on one laminar sheet and \(J=o(L\ell)\), they still assemble one coherent factorial witness.

Where it bites#

For the canonical Poisson source, the endpoint reserve is not an abstract coefficient norm. It is the moving Fourier-tail annulus

$$ M^-_{L,r}(y) = \frac12e^{-yL} \int_{e^L}^{e^{L+r_L}} u^{y-1/2} \left| \sum_{k\ge1}\widehat q_L(ku) \right|du. $$

Earlier estimates convert this reserve into fixed-order Sobolev cost. Thus a long, low-load, zero-poor canonical front led by an \(o(L)\)-depth endpoint window must satisfy

$$ \log\frac{\|q_L^{(N)}\|_1}{A_L} \ge (c-o(1))LS_L\log\frac L{r_L}+O_N(L) $$

for every fixed admissible \(N\).

So the front cannot remain simultaneously zero-poor, shallow-spectrum, and Sobolev-tame. It must produce the nonreal zero train, pay the enormous variation reserve, hand leadership to the far or opposite tail, fragment into geometry not covered by the extraction, or accumulate phase-scale critical complexity.

A quiet high-pass field has a different bill#

The endpoint theorem isolates a shallow spectral window. There is also a clean estimate for the complete high-pass tail, with no depth cutoff at all.

Let

$$ f(x)=\int_{|t|\ge L}e^{itx}\,d\eta(t) $$

be nonzero on an interval \(I\) of length \(S\), and put

$$ A=\inf_I|f|, \qquad V=\operatorname{Var}_I\arg f. $$

Convolve one broad box with

$$ m=\left\lfloor\frac{LS}{4}\right\rfloor $$

narrow boxes. The resulting positive probability window \(\omega_{I,L}\) stays inside \(I\), while its Fourier transform obeys the full-gap estimate

$$ \left|\widehat\omega_{I,L}(t)\right| \le (\sin 1)^m, \qquad |t|\ge L. $$

This is delightfully blunt. It suppresses every frequency beyond the gap by the same exponential factor, however far away that frequency is.

For a real projection of \(f\), put one normalized polynomial factor at each genuine sign reset. The polynomial makes the projection one-signed, and each reset costs only a fixed exponential factor. Averaging over projection angles turns the number of resets into total phase variation. The result is

$$ \boxed{ \log\frac{|\eta|}{A} \ge \frac{\log(1/\sin1)}4\,LS - \frac{4\Gamma}{\pi}V - O(1), } $$

where

$$ \Gamma= 4(1+\log2)+ \log\frac{1+3\sin1}{4\sin1}. $$

If \(V=o(LS)\), the high-pass measure must therefore have \(|\eta|/A\ge e^{(c-o(1))LS}\). It can stay quiet on a long interval, or it can stay small. It cannot do both.

For a horizontal canonical Stokes corridor, this becomes a trichotomy: phase-scale graph or divisor load, phase-scale carrier load, or an exponentially large full-tail Sobolev reserve. Unlike the shallow endpoint argument, this estimate sees both far tails at once. Its limitation is geometric instead: it wants one horizontal interval, not an arbitrary curved or widely separated collection.

Arithmetic has nowhere to hide#

This is an architecture theorem, not RH. Multiplying the Riemann carrier by

$$ 1+\frac{z^4}{R^4} $$

inserts off-axis zeros, yet the phase-to-factorial argument remains valid. The geometry cannot tell that the carrier is counterfeit.

Arithmetic can tell, at least for this counterfeit. Remove the exact archimedean factor and inspect the logarithmic derivative on \(s>1\). For the real carrier,

$$ \mathfrak p_\Xi(s) = -\frac{\zeta'(s)}{\zeta(s)} = \sum_{n\ge2}\Lambda(n)n^{-s}. $$

It is a positive Laplace transform whose first support is at \(\log2\), with

$$ 2^s\mathfrak p_\Xi(s)\longrightarrow\log2. $$

The quartic counterfeit instead contributes an algebraic tail:

$$ \mathfrak p_{\widetilde\Xi_R}(s) = \mathfrak p_\Xi(s) - \frac{4(s-\frac12)^3} {R^4+(s-\frac12)^4}, $$

so

$$ s\mathfrak p_{\widetilde\Xi_R}(s)\longrightarrow-4. $$

It eventually becomes negative. The first-prime support gap catches it exactly.

There is a much stronger rigidity statement hiding behind that example. Suppose an even finite-order entire carrier \(\mathcal X\), after removal of the Riemann gamma and polar factor, has an ordinary Dirichlet logarithmic derivative

$$ -\frac{Z_{\mathcal X}'(s)}{Z_{\mathcal X}(s)} = \sum_{n\ge2}b_n n^{-s}, $$

absolutely convergent in every half-plane \(\Re s>1\). No positivity of the \(b_n\) is assumed.

Integrating and exponentiating gives another ordinary Dirichlet series,

$$ Z_{\mathcal X}(s) = c\sum_{n\ge1}a_n n^{-s}, \qquad a_1=1. $$

Evenness supplies the Riemann functional equation. After a regulated Mellin shift, its coefficients form the tempered integer comb

$$ D= R\delta_0+ \sum_{n\ge1}a_n(\delta_n+\delta_{-n}), $$

and the theta law says that \(D\) is Fourier invariant.

Now comes the tiny rigid move I love. Because \(D\) is supported on \(\mathbb Z\),

$$ (e^{-2\pi i\xi}-1)D=0. $$

Fourier invariance turns this into

$$ T_1D=D. $$

The comb is invariant under translation by one, so every atomic weight is the same. Since \(a_1=1\),

$$ a_n=1 \quad(n\ge1). $$

Therefore

$$ \boxed{ \mathcal X=c\Xi, \qquad b_n=\Lambda(n). } $$

An ordinary Dirichlet logarithmic law does not merely reject the quartic counterfeit. It determines the Riemann carrier and the von Mangoldt coefficients exactly.

What is missing is the bridge between these two facts: a theorem carrying the moving positive-height Poisson escape into that ordinary Dirichlet algebra. Until that exists, the geometric and high-pass theorems say where the analytic cost must go, and the arithmetic theorem says every carrier inside the Dirichlet class is already \(\Xi\), but the two sides do not yet meet.

Notebook references: C-0243, C-0244, C-0245, C-0246, C-0247, C-0248, K-0180, O-0327, K-0179, C-0241