Uniform Beta Smoothing ofWidder Sums for the Riemann ξ-Function and a Schur–Triple Application to Simple Critical Zeros

1. Introduction

Let G be the entire function determined by $G\left(y^2\right)=\xi (1/2+y)$, and put $F_{\xi }=2G^{\prime }/G$. For $m\ge 1$ define

$${\mathcal{W}}_m\left[F\right]\left(x\right)={(-1)}^{m-1}{\left(x^{m}F\left(x\right)\right)}^{(2m-1)},A_m\left(x\right)={\displaystyle \frac{{\int }_0^x{u}^{m-1}{\mathcal{W}}_m\left[F_{\xi }\right]\left(u\right)du}{2(2m-1)!B(m,m)}}.$$

The pointwise Widder expressions have zero weights concentrated around ordinate $\sqrt{x}$. Their normalized integrals instead have decreasing beta weights and can therefore be compared with cumulative zero counts.

Our main result, Theorem 1, is the uniform formula

$$A_m\left(x\right)={\displaystyle \frac{c_m\sqrt{x}}{2\pi }}\left(log{\displaystyle \frac{\sqrt{x}}{2\pi }}-1+{\displaystyle \frac{1}{2m-1}}\right)+O_L(logx),1\le m\le Lx,$$

where $c_m=\Gamma (m-1/2)\Gamma (m+1/2)/\Gamma {\left(m\right)}^2$. The main term follows by integrating the Riemann–von Mangoldt formula against the real beta kernel. The genuinely uniform step is the replacement of each complex reflected zero by its ordinate: the vertical displacement is only $O\left(x^{-1/2}\right)$, but the derivative of the beta kernel may have size $O\left(\sqrt{m}\right)$ when m grows with x. Lemma 3 proves that the total replacement error is nevertheless only $O_L(logx)$. Corollary A5 then gives a direct approximation to the ordinary zero count. Proposition 1 isolates the elementary positive-kernel argument used to transfer cumulative density bounds.

1.0.0.1. Contribution and scope.

The core contribution is the uniform beta-smoothed zero-count asymptotic in the full range $m\le Lx$, its explicit-formula derivative refinement, and the resulting count-recovery and density-transfer statements. The derivative theorem is proved from a fully stated Guinand–Weil formula with the Fourier normalization used in this paper. A sharp shifted-kernel saddle estimate gives pointwise Widder positivity, for each fixed $\epsilon >0$, uniformly through $m\le (4-\epsilon )xloglogx$ for all sufficiently large x; in particular it covers every fixed linear range $m\le Lx$. These results are contained in Section 3 and do not depend on the later RH-equivalent reformulations.

Section 4 gives a separate quantitative application. The logarithmic beta densities compactify to polynomial probability windows on $[-1/2,1/2]$. An explicit three-beta mixture is inserted into the general window framework of Alpöge–Furman [11]. Motivated by the Schur–Jensen observation of Shi [12], we reprove the needed spectral rank–trace inequality and strengthened zero-side seam in our own notation, track all fixed-bandwidth mixed, tail, trace, and boundary errors, and then apply the resulting Schur remainder on disjoint triples. Together with two independent interval-arithmetic certificates, this gives

$$\underset{T\to \infty }{lim\; inf}{\displaystyle \frac{N_0^s(T,2T)}{N(T,2T)}}\ge 0.672625773769468\dots .$$

This improves the Montgomery–Taylor/Lamzouri baseline inside the present three-point beta-window argument. It is a method-specific statement, not a numerical-record claim: recent non-peer-reviewed multi-point preprints report larger constants, and none of those claims is used in our proof.

Appendix A records complementary fixed-height consequences. There a nonreal reflected zero is shown to force infinitely many negative high orders at one fixed point. The same mechanism yields a prime-side root-growth criterion, two order-generating functions, and a semicircular logarithmic-derivative formula. These are downstream deductions from the beta–Widder hierarchy and are separated from the main smoothing theorem so that their logical status is explicit. In particular, the continuation formulas are stated only on domains on which the square-root branches are fixed and the compositions with $-{\zeta }^{\prime }/\zeta$ are meromorphic. The verified RH height then gives an unconditional fixed-height transfer for a large range of T.

The finite-height Stieltjes calculation in Appendix B is a separate application of known finite Stieltjes theory, a published rigorous zero verification, and an explicit zero-free region. It is included because it gives a concrete global finite-order consequence, but it is not presented as the source of the paper’s main novelty.

The integrated smoothing theorem uses the critical strip, the genus-zero product, and the classical zero-count remainder. The stronger pointwise derivative estimate additionally uses the Guinand–Weil explicit formula. It does not require a lower bound on the number of critical-line zeros. Such a bound enters only the density applications. Lamzouri [13] gives the Montgomery–Taylor value $3/2-2^{-1/2}cot\left(2^{-1/2}\right)$, while Section 4 derives a slightly larger value from the Alpöge–Furman finite compression combined with a beta-window triple Schur remainder. The literature comparison in this manuscript is frozen at 15 September 2026. In particular, Wang [14] has since developed a short-interval extension of the Lamzouri approach; that work does not enter any proof here and addresses a different interval-length question.

For context, Bondesson–Simon ([1], Definition 1, Theorem 1 and Section 4.1) establish the finite Stieltjes classes, their integral representations, and the quadratic angular threshold used below. Our finite-height application combines these ingredients with the zero verification of Platt–Trudgian [5] and, for the fully published numerical theorem, the zero-free region of Mossinghoff–Trudgian–Yang [3]. The stronger Bellotti–Trudgian–Yang preprint region [2] is used only for the separately labelled enhancement. The contribution here is the height-to-order conversion. The quadratic examples and finite Stieltjes representation are recalled consequences of the existing theory, not separate originality claims.

Section 2 establishes the zero-sum identities. Section 3 contains the uniform smoothing theorem, the exact explicit-formula normalization, and its density consequences. Section 4 develops the compactified beta window and the Schur–triple simple-zero application. Appendix A collects the fixed-height and high-order generating-function consequences. Appendix B treats finite-height positivity. Appendix C explains the relation with existing criteria. The appendices are intentionally separated from the four-section core so that the principal analytic theorem and the simple-zero application can be read and refereed without relying on any of the later RH-equivalent reformulations or finite-height applications. A reader concerned only with the two main theorem chains may therefore stop after Section 4; no result in the appendices is used retroactively in the core. The precise analytic contribution investigated here is the uniform beta-smoothed zero-count statement and its explicit-formula derivative refinement. The sparse RH criterion recorded in Appendix C is an elementary consequence of Widder’s classical criterion and a downward recurrence for the normalized sums. More generally, analytic-continuation, logarithmic-derivative, and prime-side reformulations of RH are classical in broad form; the later sections claim only the explicit formulas and deductions arising from the particular beta–Widder hierarchy developed here. Recent work of Polson [7,8] develops related Thorin/GGC, Stieltjes, zero-measure, reciprocal-$\xi$, and explicit-formula viewpoints for the Riemann $\xi$-function. In particular, the public abstract of [8] formulates RH through positivity of a Thorin measure attached to a reciprocal-$\xi$ transform and also discusses finite truncations of derivative-type positivity conditions. We do not use those results, and we do not identify those derivative conditions with the Widder orders studied here without a separate term-by-term comparison. The novelty claims of the present paper are therefore deliberately restricted to the specific beta-regularized sums $A_m$, their joint uniform asymptotics in $(m,x)$, the sharp shifted-kernel factor $M_{\sigma }$, and the resulting $(4-\epsilon )xloglogx$ positivity wedge. The separate Section 4 additionally claims the explicit compactified beta-window construction and its three-point Schur bootstrap, with its computer-assisted finite inequality stated transparently. No broad priority claim is made for Stieltjes/GGC formulations, reciprocal-$\xi$ positivity, prime–zero explicit-formula dictionaries, or RH-equivalent reformulations.

1.0.0.2. Logical dependency map.

To reduce the verification burden, the proof architecture is summarized here. The first three rows form the four-section core together with the Introduction; the last row is logically downstream.

Part	Principal output	External input actually used
Section 2	Exact zero-sum/Widder identities	Classical properties and genus-zero product of $\xi$; no finite-compression input.
Section 3	Uniform beta smoothing, count recovery, density transfer, and the pointwise positivity wedge	Riemann–von Mangoldt and, only for the derivative refinement, the explicitly normalized Guinand–Weil formula.
Section 4	The beta-window Schur–triple simple-zero bound	Version-locked analytic estimates from Alpöge–Furman v2. The strengthened Schur seam and triple inequality are proved here; the latter has two independent interval certificates.
Appendices Appendix A–Appendix C	Fixed-height, high-order, generating-function, and comparison consequences	Not used to prove either the smoothing theorem or the beta–Schur theorem.

2. Zero Sums and Widder Expressions

We use the standard completed zeta function

$$\xi \left(s\right)={\displaystyle \frac{1}{2}}s(s-1){\pi }^{-s/2}\Gamma \left({\displaystyle \frac{s}{2}}\right)\zeta \left(s\right),$$

which is entire and satisfies

$$\xi \left(s\right)=\xi (1-s),\overline{\xi \left(s\right)}=\xi \left(\overline{s}\right).$$

Put

$$H\left(y\right)=\xi \left({\displaystyle \frac{1}{2}}+y\right).$$

Then H is even. Consequently there is an entire function G such that

$$H\left(y\right)=G\left(y^2\right),G\left(x\right)=\xi \left({\displaystyle \frac{1}{2}}+\sqrt{x}\right).$$

(1)

Since H has order one, G has order $1/2$. In particular G has a genus-zero canonical product.

Define, for $x>0$,

$$F_{\xi }\left(x\right):=2{\displaystyle \frac{G^{\prime }\left(x\right)}{G\left(x\right)}}={\displaystyle \frac{1}{\sqrt{x}}}{\displaystyle \frac{{\xi }^{\prime }}{\xi }}\left({\displaystyle \frac{1}{2}}+\sqrt{x}\right).$$

(2)

The function $F_{\xi }$ is positive on $(0,\infty )$. For completeness, we record an explicit form of the classical Riemann kernel. Put

$$\Phi \left(u\right)={\displaystyle \sum _{n=1}^{\infty }}\left(4{\pi }^2{n}^4{e}^{9u/2}-6\pi n^2{e}^{5u/2}\right)e^{-\pi n^2{e}^{2u}},u\ge 0.$$

(3)

Then the standard Fourier representation of the Riemann $\Xi$-function gives, after the substitution $t=-iy$,

$$H\left(y\right)=\xi \left({\displaystyle \frac{1}{2}}+y\right)=2{\int }_0^{\infty }\Phi \left(u\right)cosh\left(yu\right)du,y\in \mathbb{R},$$

(4)

see Titchmarsh ([17], Ch. II). Every summand in (3) is strictly positive for $u\ge 0$, since

$$4{\pi }^2{n}^4{e}^{9u/2}-6\pi n^2{e}^{5u/2}=2\pi n^2{e}^{5u/2}\left(2\pi n^2{e}^{2u}-3\right)>0.$$

The series and all differentiated integrands have the standard rapid decay needed for local uniform convergence. Hence differentiation under the integral gives

$$H^{\prime }\left(y\right)=2{\int }_0^{\infty }u\Phi \left(u\right)sinh\left(yu\right)du>0(y>0),$$

while $H\left(y\right)>0$ for $y\ge 0$. Since

$$F_{\xi }\left(x\right)={\displaystyle \frac{H^{\prime }\left(\sqrt{x}\right)}{\sqrt{x}H\left(\sqrt{x}\right)}},$$

we obtain $F_{\xi }\left(x\right)>0$ for every $x>0$.

Let ${\mathcal{Z}}_{+}$ denote the multiset of nontrivial zeros $\rho =\beta +i\gamma$ with $\gamma >0$, counted with multiplicity. The functional equation pairs the zero $\rho$ with $1-\rho$ in the lower half-plane, so the zeros of the genus-zero function G in (1) are precisely

$$z_{\rho }={(\rho -1/2)}^2,\rho \in {\mathcal{Z}}_{+}.$$

It is convenient to reflect them:

$$w_{\rho }:=-z_{\rho }=-{(\rho -1/2)}^2.$$

(5)

Writing

$$\delta =\beta -{\displaystyle \frac{1}{2}},$$

we obtain the exact identity

$$w_{\rho }=-{(\delta +i\gamma )}^2={(\gamma -i\delta )}^2.$$

(6)

Thus

$$|arg{w}_{\rho }|=2arctan{\displaystyle \frac{\left|\delta \right|}{\gamma }}.$$

(7)

For a critical-line zero, $\delta =0$ and $w_{\rho }={\gamma }^2>0$.

Since the number of zeta zeros up to height T is $O(TlogT)$,

$$\sum _{\rho \in {\mathcal{Z}}_{+}}{\displaystyle \frac{1}{|w_{\rho }|}}<\infty .$$

Therefore

$$G\left(x\right)=G\left(0\right)\prod _{\rho \in {\mathcal{Z}}_{+}}\left(1+{\displaystyle \frac{x}{w_{\rho }}}\right)$$

(8)

and

$$F_{\xi }\left(x\right)=2\sum _{\rho \in {\mathcal{Z}}_{+}}{\displaystyle \frac{1}{x+w_{\rho }}}.$$

(9)

For an off-line zero $\rho =1/2+\delta +i\gamma$, the symmetry $1-\overline{\rho }$ is also an upper-half-plane zero and gives $w_{1-\overline{\rho }}=\overline{w_{\rho }}$. Hence the nonreal terms in (9) occur in conjugate pairs, as used in the sector argument below.

For $m\ge 1$, define

$${\mathcal{W}}_m\left[F\right]\left(x\right):={(-1)}^{m-1}{\displaystyle \frac{d^{2m-1}}{d{x}^{2m-1}}}\left(x^{m}F\left(x\right)\right).$$

(10)

Lemma 1 

(Elementary derivative identity). For $m\ge 1$, $w\ne 0$, and $x\ne -w$,

$${(-1)}^{m-1}{\displaystyle \frac{d^{2m-1}}{d{x}^{2m-1}}}\left({\displaystyle \frac{x^m}{x+w}}\right)=(2m-1)!{\displaystyle \frac{w^m}{{(x+w)}^{2m}}}.$$

(11)

Proof. 

Divide $x^m$ by $x+w$. The quotient has degree $m-1$, hence its $(2m-1)$st derivative vanishes. The remainder is ${(-w)}^m/(x+w)$. Differentiating the latter $2m-1$ times gives (11).    □

For a genus-zero product $G\left(x\right)=G\left(0\right){\prod }_j(1+x/w_j)$ with ${\sum }_j{\left|w_j\right|}^{-1}<\infty$, put $F=2G^{\prime }/G$. It follows that

$${\mathcal{W}}_m\left[F\right]\left(x\right)=2(2m-1)!\sum _j{\displaystyle \frac{w_j^m}{{(x+w_j)}^{2m}}}.$$

(12)

For later use we record the convergence explicitly. Fix a compact interval $I⋐(0,\infty )$. Under the sector bound $|arg{w}_j|\le \vartheta <\pi /2$ there is a constant $c_I>0$ such that $|x+w_j|\ge c_I(1+|w_j\left|\right)$ for $x\in I$. Hence

$$\left|{\displaystyle \frac{w_j^m}{{(x+w_j)}^{2m}}}\right|\le C_{I,m}min\{1,|w_j{|}^{-m}\}.$$

Only finitely many $w_j$ lie in a bounded set, while ${\sum }_j{\left|w_j\right|}^{-1}<\infty$; consequently the series in (12) converges absolutely and locally uniformly, for every $m\ge 1$. The same estimates, applied before differentiation to finitely many derivative orders, justify term-by-term differentiation of the logarithmic-derivative series.

3. Uniform Beta Smoothing and Density Transfer

A sector argument for pointwise Widder positivity uses a bound on every reflected zero. A cumulative lower density of critical-line zeros is a different input: it does not directly control the weight

$${\displaystyle \frac{t^{2m}}{{(x+t^2)}^{2m}}},$$

which increases up to $t=\sqrt{x}$ and then decreases. We introduce an integrated Widder expression whose zero weight is decreasing. This permits a rigorous transfer of cumulative density information. The uniformity in the order is recorded explicitly.

For $m\ge 1$ put $B_m=B(m,m)=\Gamma {\left(m\right)}^2/\Gamma \left(2m\right)$ and define

$$A_m\left(x\right)={\displaystyle \frac{1}{2(2m-1)!B_m}}{\int }_0^x{u}^{m-1}{\mathcal{W}}_m\left[F_{\xi }\right]\left(u\right)du.$$

(13)

The expression at zero is understood by analytic continuation of $F_{\xi }$ to a neighborhood of zero. Introduce the rational function

$$P_m(x,w)={\displaystyle \sum _{j=m}^{2m-1}}\left({\displaystyle \genfrac{}{}{0pt}{}{2m-1}{j}}\right){\displaystyle \frac{x^j{w}^{2m-1-j}}{{(x+w)}^{2m-1}}}.$$

(14)

For positive real w, this is the regularized incomplete beta function $I_{x/(x+w)}(m,m)$. Formula (14), rather than a choice of branch for the incomplete beta function, defines its complex extension here. Differentiating gives

$${\partial }_x{P}_m(x,w)={\displaystyle \frac{x^{m-1}{w}^m}{B_m{(x+w)}^{2m}}},P_m(0,w)=0.$$

Consequently, by (12) and local uniform convergence,

$$A_m\left(x\right)=\sum _{\rho \in {\mathcal{Z}}_{+}}{P}_m(x,w_{\rho }).$$

(15)

Indeed, for fixed m and bounded x, the summands are $O_m\left(\right|w_{\rho }{|}^{-m})$. The sum is real by conjugation symmetry. In particular,

$$A_1\left(x\right)={\displaystyle \frac{x{F}_{\xi }\left(x\right)}{2}},P_2(x,w)={\displaystyle \frac{x^2(x+3w)}{{(x+w)}^3}}.$$

Let $N_0^{\mathrm{s}}\left(t\right)$ count simple critical-line zeros with positive ordinate at most t. Define the positive simple-zero contribution

$$S_m\left(x\right)=\sum _{\begin{array}{c}\rho \in {\mathcal{Z}}_{+}\\ \beta =1/2\\ \rho \mathrm{simple}\end{array}}{P}_m(x,{\gamma }^2).$$

(16)

Here and below a simple zero is counted once, whereas $N\left(t\right)$ and unrestricted zero sums count multiplicities.

Theorem 1 

(Uniform smoothed asymptotic and density transfer). Set

$$c_m={\displaystyle \frac{\Gamma (m-1/2)\Gamma (m+1/2)}{\Gamma {\left(m\right)}^2}}.$$

For every fixed $L>0$, as $x\to \infty$, uniformly over integers $1\le m\le Lx$,

$$A_m\left(x\right)={\displaystyle \frac{c_m\sqrt{x}}{2\pi }}\left(log{\displaystyle \frac{\sqrt{x}}{2\pi }}-1+{\displaystyle \frac{1}{2m-1}}\right)+O_L(logx).$$

(17)

In particular $A_m\left(x\right)>0$ throughout this range once x is sufficiently large (depending on L).

Suppose, for some $p\in [0,1]$, that

$$\underset{t\to \infty }{lim\; inf}{\displaystyle \frac{N_0^{\mathrm{s}}\left(t\right)}{N\left(t\right)}}\ge p.$$

(18)

Then

$$\underset{x\to \infty }{lim\; inf}\underset{\begin{array}{c}m\in \mathbb{N}\\ 1\le m\le Lx\end{array}}{inf}{\displaystyle \frac{S_m\left(x\right)}{A_m\left(x\right)}}\ge p\mathrm{for}\mathrm{every}\mathrm{fixed}L>0.$$

(19)

We give details of the uniform estimates; a fixed-order asymptotic by itself would not establish this theorem. The homogeneity of (14) will be used repeatedly: for $T>0$,

$$P_m(T^2,{(\gamma -i\delta )}^2)=P_m\left(1,{\left({\displaystyle \frac{\gamma -i\delta }{T}}\right)}^2\right)=k_m\left({\displaystyle \frac{\gamma -i\delta }{T}}\right),$$

(20)

where on the right $k_m$ denotes the rational continuation defined by $P_m(1,z^2)$.

Lemma 2 

(Uniform real-kernel estimates). Put

$$k_m\left(y\right)=P_m(1,y^2)=I_{1/(1+y^2)}(m,m),y\ge 0.$$

Then $k_m\left(0\right)=1$, $k_m(\infty )=0$, and

$$h_m\left(y\right):=-k_m^{\prime }\left(y\right)={\displaystyle \frac{2y^{2m-1}}{B_m{(1+y^2)}^{2m}}}>0(y>0).$$

(21)

Uniformly in $m\ge 1$,

$${\int }_0^{\infty }{h}_m\left(y\right)dy=1,\underset{y\ge 0}{sup}{h}_m\left(y\right)\ll \sqrt{m},{\int }_0^{\infty }{h}_m\left(y\right)log(2+y)dy\ll 1,$$

and $h_m\left(y\right)\ll y^{-3}$ for $y\ge 2$. Moreover,

$$\begin{array}{cc}\hfill {\int }_0^{\infty }{k}_m\left(y\right)dy& =c_m,\hfill \end{array}$$

(22)

$$\begin{array}{cc}\hfill {\int }_0^{\infty }{k}_m\left(y\right)logydy& =c_m\left(-1+{\displaystyle \frac{1}{2m-1}}\right).\hfill \end{array}$$

(23)

One has $1\le c_m\le \pi /2$ and $c_m=1+O\left(m^{-1}\right)$.

Proof. 

The derivative follows either from the incomplete beta integral or directly from (14). The beta integral and Stirling’s formula give $B_m^{-1}\ll 4^m\sqrt{m}$. The function $h_m$ increases to its unique maximum at $y^2=(2m-1)/(2m+1)$ and decreases thereafter. At that maximum $y\ge 1/\sqrt{3}$ and $4y^2/{(1+y^2)}^2\le 1$, which proves the supremum estimate. For $y\ge 2$, set $r\left(y\right)=4y^2/{(1+y^2)}^2\le 16/25$. Then

$$h_m\left(y\right)\ll {\displaystyle \frac{\sqrt{m}r{\left(y\right)}^m}{y}}\ll {\displaystyle \frac{r\left(y\right)}{y}}\ll y^{-3}.$$

The asserted logarithmic moment of $h_m$ follows from this tail bound and its total mass one.

For a in a neighborhood of zero, Tonelli’s theorem applied to the beta integral gives

$${\int }_0^{\infty }{y}^a{k}_m\left(y\right)dy={\displaystyle \frac{B(m-(a+1)/2,m+(a+1)/2)}{(a+1)B_m}}.$$

Evaluation and differentiation at $a=0$ prove (22), using $\psi (m+1/2)-\psi (m-1/2)=1/(m-1/2)$. Finally $c_1=\pi /2$, $c_{m+1}/c_m=1-1/\left(4m^2\right)$, and $c_m\to 1$ by Stirling’s formula. The same recurrence gives

$$0\le log{c}_m=-{\displaystyle \sum _{j=m}^{\infty }}log\left(1-{\displaystyle \frac{1}{4j^2}}\right)\ll {\displaystyle \sum _{j=m}^{\infty }}{j}^{-2}\ll m^{-1},$$

which also proves $c_m=1+O\left(m^{-1}\right)$.    □

Lemma 3 

(Replacing reflected zeros by their ordinates). For $T\to \infty$ and each fixed $L>0$,

$$\sum _{\rho \in {\mathcal{Z}}_{+}}\left|P_m(T^2,{(\gamma -i\delta )}^2)-k_m(\gamma /T)\right|{\ll }_{L}logT(1\le m\le L{T}^2).$$

(24)

Proof. 

By (20), the summand is the change of the rational function $k_m$ along the vertical segment joining $y=\gamma /T$ to $y-i\delta /T$. Since $\left|\delta \right|\le 1/2$, every such segment has length at most $1/\left(2T\right)$. The only poles of the rational continuation of $k_m$ are at $\pm i$, and differentiation of (14) gives

$$k_m^{\prime }\left(z\right)=-{\displaystyle \frac{2z^{2m-1}}{B_m{(1+z^2)}^{2m}}}.$$

(25)

We first treat $y\ge 1/2$. Put $z=y-iϵ$, with $\left|ϵ\right|\le 1/\left(2T\right)$. From (25) and (21),

$${\displaystyle \frac{|k_m^{\prime }\left(z\right)|}{h_m\left(y\right)}}={\left({\displaystyle \frac{\left|z\right|}{y}}\right)}^{2m-1}{\left({\displaystyle \frac{1+y^2}{|1+z^2|}}\right)}^{2m}.$$

(26)

Now

$${\displaystyle \frac{\left|z\right|}{y}}={\left(1+{\displaystyle \frac{{ϵ}^2}{y^2}}\right)}^{1/2}\le exp\left({\displaystyle \frac{{ϵ}^2}{2y^2}}\right),$$

while, for $T\ge 4$,

$$|1+z^2|\ge \Re (1+z^2)=1+y^2-{ϵ}^2>0.$$

Hence

$${\displaystyle \frac{1+y^2}{|1+z^2|}}\le {\left(1-{\displaystyle \frac{{ϵ}^2}{1+y^2}}\right)}^{-1}.$$

Since $y\ge 1/2$ and $\left|ϵ\right|\le 1/8$, the quantity $u={ϵ}^2/(1+y^2)$ satisfies $0\le u\le 1/80$; thus $-log(1-u)\le (80/79)u$. Substituting these two estimates into (26) yields, with room to spare,

$$|k_m^{\prime }(y-iϵ)|\le e^{8m{ϵ}^2}{h}_m\left(y\right)\le e^{2L}{h}_m\left(y\right),y\ge {\displaystyle \frac{1}{2}},$$

(27)

for $1\le m\le L{T}^2$.

It remains to control the low-y rectangle. If $0\le y<1/2$ and $\left|ϵ\right|\le 1/8$, a direct calculation gives

$$r\left(z\right):={\displaystyle \frac{{4\left|z\right|}^2}{|1+z^2{|}^2}}\le {\displaystyle \frac{4352}{6241}}<{\displaystyle \frac{3}{4}}.$$

The point that is useful near $z=0$ is to factor one power of z before using the beta normalization. From (25),

$$\begin{array}{cc}\hfill |k_m^{\prime }\left(z\right)|& ={\displaystyle \frac{2\left|z\right|}{B_m{|1+z^2|}^2}}{\left({\displaystyle \frac{{\left|z\right|}^2}{|1+z^2{|}^2}}\right)}^{m-1}\hfill \\ \hfill & \ll \sqrt{m}{\displaystyle \frac{\left|z\right|}{|1+z^2{|}^2}}r{\left(z\right)}^{m-1}.\hfill \end{array}$$

Here we used $B_m^{-1}\ll 4^m\sqrt{m}$, absorbing the remaining fixed factor 4. On the closed rectangle $0\le y\le 1/2$, $\left|ϵ\right|\le 1/8$, the factor $\left|z\right|/|1+z^2{|}^2$ is uniformly bounded. Therefore

$$|k_m^{\prime }(y-iϵ)|\ll \sqrt{m}{(3/4)}^{m-1}\ll 1,0\le y<{\displaystyle \frac{1}{2}}.$$

(28)

This explicit factorization avoids any spurious singular factor $1/\left|z\right|$ at $z=0$.

Integrating along the vertical segments and splitting at $\gamma =T/2$ now bounds the left side of (24) by

$${\displaystyle \frac{C}{T}}N(T/2)+{\displaystyle \frac{C_L}{T}}\sum _{\gamma \ge T/2}{h}_m(\gamma /T).$$

(29)

Write the classical Riemann–von Mangoldt formula as

$$N\left(t\right)=M\left(t\right)+E\left(t\right),M\left(t\right)={\displaystyle \frac{t}{2\pi }}log{\displaystyle \frac{t}{2\pi }}-{\displaystyle \frac{t}{2\pi }},E\left(t\right)=O(log(t+2)).$$

(30)

Here $M\left(0\right)=0$ by continuity; the bound for E extends to all $t\ge 0$ by adjusting the constant (see [17], Ch. IX). Let $a=T/2$. For the sum over ordinates $\gamma >a$, Stieltjes integration gives

$$\sum _{\gamma >a}{h}_m(\gamma /T)={\int }_{(a,\infty )}{h}_m(t/T)dM\left(t\right)+{\int }_{(a,\infty )}{h}_m(t/T)dE\left(t\right).$$

Since $M^{\prime }\left(t\right)={\left(2\pi \right)}^{-1}log(t/\left(2\pi \right))$ for $t>0$,

$$\left|{\int }_a^{\infty }{h}_m(t/T)dM\left(t\right)\right|\ll T{\int }_{1/2}^{\infty }{h}_m\left(y\right)\left(logT+log(2+y)\right)dy\ll TlogT.$$

Integration by parts against $dE$ bounds the second contribution by

$$C{h}_m(1/2)log(T+2)+C{\int }_{1/2}^{\infty }\left|h_m^{\prime }\left(y\right)\right|log(Ty+2)dy.$$

(31)

The boundary term at infinity vanishes because $h_m\left(y\right)\ll y^{-3}$ there. On $[1/2,2]$, unimodality gives

$${\int }_{1/2}^2\left|h_m^{\prime }\left(y\right)\right|dy\le 2sup{h}_m\ll \sqrt{m}.$$

On $[2,\infty )$, $h_m$ is decreasing, and another integration by parts gives

$$\begin{array}{cc}\hfill {\int }_2^{\infty }\left|h_m^{\prime }\left(y\right)\right|log(Ty+2)dy& =h_m\left(2\right)log(2T+2)+{\int }_2^{\infty }{\displaystyle \frac{T{h}_m\left(y\right)}{Ty+2}}dy\hfill \\ \hfill & \ll logT+{\int }_2^{\infty }{y}^{-4}dy\ll logT.\hfill \end{array}$$

Thus (31) is $O(\sqrt{m}logT)$. If a itself is a zero ordinate, its multiplicity is $O(log(T+2\left)\right)$, obtained by comparing (30) at $a-1$ and $a+1$. Restoring that possible boundary atom therefore costs at most $O(\sqrt{m}logT)$. We have proved

$$\sum _{\gamma \ge T/2}{h}_m(\gamma /T)\ll (T+\sqrt{m})logT.$$

(32)

Substituting this into (29), using $N(T/2)\ll TlogT$ and $\sqrt{m}/T\le \sqrt{L}$, proves (24).    □

Proposition 1 

(Transfer by a decreasing kernel). Let N and Q be nonnegative, nondecreasing counting functions on $[0,\infty )$ with $N\left(0\right)=Q\left(0\right)=0$. Suppose

$$Q\left(t\right)\ge qN\left(t\right)-C(t\ge 0)$$

for constants $q,C\ge 0$. Let k be an absolutely continuous decreasing function with $k\left(0\right)=1$, $k(\infty )=0$, and $h=-k^{\prime }\ge 0$. For any $T>0$ for which the integrals below are finite and the boundary products vanish,

$${\int }_0^{\infty }k(t/T)dQ\left(t\right)\ge q{\int }_0^{\infty }k(t/T)dN\left(t\right)-C.$$

Proof. 

Integration by parts writes the two integrals as ${\int }_0^{\infty }Q\left(Ty\right)h\left(y\right)dy$ and ${\int }_0^{\infty }N\left(Ty\right)h\left(y\right)dy$. Integrate the assumed inequality and use ${\int }_0^{\infty }h\left(y\right)dy=1$.    □

Proof 

(Proof of Theorem 1). Write

$$R_m\left(T\right)=\sum _{\rho \in {\mathcal{Z}}_{+}}{k}_m(\gamma /T).$$

Partial summation in (30) gives

$$R_m\left(T\right)={\displaystyle \frac{T}{2\pi }}{\int }_0^{\infty }{k}_m\left(y\right)log{\displaystyle \frac{Ty}{2\pi }}dy+O(logT),$$

uniformly in all $m\ge 1$. More explicitly, $R_m\left(T\right)={\int }_0^{\infty }{k}_m(t/T)dN\left(t\right)$; the $dM$ part becomes the displayed integral after $t=Ty$, while integration by parts in the $dE$ part gives an error bounded by a constant multiple of

$${\int }_0^{\infty }{h}_m\left(y\right)log(Ty+2)dy=O(logT)$$

by Lemma 2. The boundary terms vanish: at zero one uses $M\left(0\right)=E\left(0\right)=N\left(0\right)=0$, and at infinity $k_m(t/T)=O_m\left(t^{-2m}\right)$ for each fixed m occurring in the Stieltjes integral. The resulting error bound is uniform because only the normalized mass and logarithmic moment of $h_m$ enter the estimate. Lemma 2 evaluates the integral, while Lemma 3 gives

$$A_m\left(T^2\right)=R_m\left(T\right)+O_L(logT)(1\le m\le L{T}^2).$$

(33)

This proves (17). Since $1\le c_m\le \pi /2$, the main term is bounded below by a positive constant times $TlogT$ for sufficiently large T, uniformly in this range.

For the density assertion, first take $p>0$ and fix $0<\epsilon <p$. Assumption (18) implies

$$N_0^{\mathrm{s}}\left(t\right)\ge (p-\epsilon )N\left(t\right)-C_{\epsilon }(t\ge 0)$$

for some finite $C_{\epsilon }$. Proposition 1, with $q=p-\epsilon$, gives

$$\begin{array}{cc}\hfill S_m\left(T^2\right)& ={\int }_0^{\infty }{N}_0^{\mathrm{s}}\left(Ty\right)h_m\left(y\right)dy\hfill \\ \hfill & \ge (p-\epsilon )R_m\left(T\right)-C_{\epsilon }.\hfill \end{array}$$

Combine this with (33) and the uniform lower bound $A_m\left(T^2\right)\gg TlogT$. Let $T\to \infty$ and then $\epsilon \downarrow 0$. If $p=0$, the conclusion follows directly from $S_m\ge 0$ and eventual positivity of $A_m$.    □

Proposition 2 

(Guinand–Weil formula in our Fourier convention). Let h be even and holomorphic in a strip $|\Im z|\le 1/2+\epsilon$ for some $\epsilon >0$, and suppose that, uniformly in that strip,

$$\left|h\left(z\right)\right|\ll {(1+|z\left|\right)}^{-1-\eta }$$

for some $\eta >0$. Define

$$g\left(u\right):={\displaystyle \frac{1}{2\pi }}{\int }_{-\infty }^{\infty }h\left(r\right)e^{-iur}dr.$$

(34)

Then, with the nontrivial zeros of ζ counted with multiplicity and written as $\rho =1/2+i{\Gamma }_{\rho }$ (where ${\Gamma }_{\rho }$ need not be real),

$$\begin{array}{cc}\hfill \sum _{\rho }h\left({\Gamma }_{\rho }\right)& =2h(i/2)+{\displaystyle \frac{1}{2\pi }}{\int }_{-\infty }^{\infty }h\left(r\right)\left\{\Re {\displaystyle \frac{{\Gamma }^{\prime }}{\Gamma }}\left({\displaystyle \frac{1}{4}}+{\displaystyle \frac{ir}{2}}\right)-log\pi \right\}dr\hfill \\ \hfill & -2{\displaystyle \sum _{n=2}^{\infty }}{\displaystyle \frac{\Lambda \left(n\right)}{\sqrt{n}}}g(logn).\hfill \end{array}$$

(35)

This is the Guinand–Weil explicit formula with the $e^{-iur}/\left(2\pi \right)$ Fourier normalization. In the more common convention $\widehat{h}\left(\xi \right)=\int h\left(r\right)e^{-2\pi ir\xi }dr$, one has $\widehat{h}\left(\xi \right)=2\pi g\left(2\pi \xi \right)$. Formula (35) is therefore exactly the standard formula; see Conrey ([16], pp. 345–346) and, with the hypotheses stated in this form, Chirre–Gonçalves ([9],Proposition 5). No assumption of RH is involved.

Lemma 4 

(Admissibility of the beta test functions). Fix $T\ge 2$ and $m\ge 1$, and set

$$h_{m,T}\left(r\right)=p_m(r/T),p_m\left(z\right)={\displaystyle \frac{z^{2m}}{B_m{(1+z^2)}^{2m}}}.$$

Then $h_{m,T}$ is even, real-valued on $\mathbb{R}$, holomorphic in the closed strip $|\Im r|\le 3/4$, and satisfies the strip-uniform bound

$$\underset{\left|v\right|\le 3/4}{sup}|h_{m,T}(u+iv)|\le C_{m,T}{(1+|u\left|\right)}^{-2m}(u\in \mathbb{R}).$$

(36)

In particular $h_{m,T}$ satisfies the hypotheses of Proposition 2 (take $\epsilon =1/4$), its Fourier transform is absolutely convergent, and the vertical sides of the rectangular contours used to shift a Fourier line inside this strip tend to zero. For these test functions the zero sum in (35) is absolutely convergent as well.

Proof. 

The only poles of $h_{m,T}$ are at $r=\pm iT$, each of order $2m$, and hence lie outside $|\Im r|\le 3/4$. Evenness and reality on the real axis are immediate. Write $r=u+iv$. If $\left|u\right|\ge 2T$, then

$${|1+(r/T)}^2|={\displaystyle \frac{|r-iT||r+iT|}{T^2}}\ge {\displaystyle \frac{u^2}{T^2}},$$

whereas $\left|r\right|\le {(1+9/\left(64T^2\right))}^{1/2}\left|u\right|\ll \left|u\right|$ uniformly for $\left|v\right|\le 3/4$. Consequently

$$|h_{m,T}(u+iv)|\le {\displaystyle \frac{1}{B_m}}{\left({\displaystyle \frac{CT}{\left|u\right|}}\right)}^{2m}\left(\right|u|\ge 2T,|v|\le 3/4).$$

On the compact rectangle $\left|u\right|\le 2T$, $\left|v\right|\le 3/4$, the denominator is bounded away from zero, so enlarging the constant proves (36). This is stronger than the strip decay required in Proposition 2.

Finally, a nontrivial zero has $|\Im {\Gamma }_{\rho }|=|\beta -1/2|<1/2$ and the classical zero count is $N\left(R\right)=O(RlogR)$. The bound (36) therefore gives absolute convergence of the zero sum already for $m=1$.    □

Lemma 5 

(Sharp shifted $L^1$ bound for the beta kernel). Let

$$p_m\left(z\right)={\displaystyle \frac{z^{2m}}{B_m{(1+z^2)}^{2m}}},m\ge 2.$$

There is an absolute constant $C>0$ such that, uniformly for $0\le \sigma \le 1/8$ and $m\ge 2$,

$${\int }_{-\infty }^{\infty }\left|p_m(y-i\sigma )\right|dy\le C{M}_{\sigma }^m,M_{\sigma }:={\displaystyle \frac{2}{1+\sqrt{1-4{\sigma }^2}}}.$$

(37)

Moreover,

$$log{M}_{\sigma }={\sigma }^2+{\displaystyle \frac{3}{2}}{\sigma }^4+O\left({\sigma }^6\right)(\sigma \to 0),$$

(38)

and in particular

$$log{M}_{\sigma }\le {\sigma }^2+C{\sigma }^4(0\le \sigma \le 1/8).$$

(39)

Proof. 

Put $q={\sigma }^2$ and

$${\mathcal{R}}_{\sigma }\left(y\right):={\displaystyle \frac{4(y^2+q)}{{(1+y^2-q)}^2+4q{y}^2}}.$$

Then

$$|p_m(y-i\sigma )|={\displaystyle \frac{{\mathcal{R}}_{\sigma }{\left(y\right)}^m}{4^m{B}_m}}.$$

(40)

Set

$$r_{\sigma }:=\sqrt{1-4q},u:=y^2+q,y_{\sigma }^2:=r_{\sigma }-q.$$

Since

$${(1+y^2-q)}^2+4q{y}^2={(1+u)}^2-4q=u^2+2u+r_{\sigma }^2,$$

differentiation in u shows that the maximum occurs at $u=r_{\sigma }$, i.e. at $y=\pm y_{\sigma }$, and

$$\underset{y\in \mathbb{R}}{max}{\mathcal{R}}_{\sigma }\left(y\right)=M_{\sigma }={\displaystyle \frac{2}{1+r_{\sigma }}}.$$

More importantly, the same algebra gives the exact identity

$${\displaystyle \frac{{\mathcal{R}}_{\sigma }\left(y\right)}{M_{\sigma }}}={\left(1+{\displaystyle \frac{{(u-r_{\sigma })}^2}{2u(1+r_{\sigma })}}\right)}^{-1}.$$

(41)

This turns the uniform Laplace estimate into a quantitative calculation.

We give the details. Because $0\le q\le 1/64$, $r_{\sigma }\ge \sqrt{15}/4$ and $y_{\sigma }^2\ge \sqrt{15}/4-1/64>{(15/16)}^2$. For $1/2\le y\le 3/2$, let

$$X_{\sigma }\left(y\right):={\displaystyle \frac{{(u-r_{\sigma })}^2}{2u(1+r_{\sigma })}}.$$

Using $u-r_{\sigma }=y^2-y_{\sigma }^2$, $1/4\le u\le 145/64$, and $y_{\sigma }>15/16$, we obtain

$$X_{\sigma }\left(y\right)\ge {\displaystyle \frac{{(y-y_{\sigma })}^2{(y+y_{\sigma })}^2}{145/16}}\ge {\displaystyle \frac{1}{5}}{(y-y_{\sigma })}^2.$$

On the same interval, $|u-r_{\sigma }|<4/3$ and $2u(1+r_{\sigma })>31/32$, so $X_{\sigma }\left(y\right)<2$. Hence $log(1+X)\ge X/(1+X)\ge X/3$ gives

$${\left({\displaystyle \frac{{\mathcal{R}}_{\sigma }\left(y\right)}{M_{\sigma }}}\right)}^m\le exp\left(-{\displaystyle \frac{m}{15}}{(y-y_{\sigma })}^2\right),{\displaystyle \frac{1}{2}}\le y\le {\displaystyle \frac{3}{2}}.$$

(42)

Therefore the contribution of this interval is at most $\sqrt{15\pi /m}$.

The complementary regions have a uniform geometric gap. If $0\le y\le 1/2$, then $u\le 17/64$ while $r_{\sigma }\ge 15/16$, and (41) gives

$${\displaystyle \frac{{\mathcal{R}}_{\sigma }\left(y\right)}{M_{\sigma }}}\le {\displaystyle \frac{5}{7}}.$$

If $3/2\le y\le 3$, then $u\ge 9/4$ and

$$X_{\sigma }\left(y\right)\ge {\displaystyle \frac{{(u-1)}^2}{4u}}\ge {\displaystyle \frac{25}{144}},{\displaystyle \frac{{\mathcal{R}}_{\sigma }\left(y\right)}{M_{\sigma }}}\le {\displaystyle \frac{144}{169}}<{\displaystyle \frac{6}{7}}.$$

Finally, for $y\ge 3$ we have $M_{\sigma }\ge 1$ and

$${\displaystyle \frac{{\mathcal{R}}_{\sigma }\left(y\right)}{M_{\sigma }}}\le {\mathcal{R}}_{\sigma }\left(y\right)={\displaystyle \frac{4u}{u^2+2u+r_{\sigma }^2}}\le {\displaystyle \frac{4}{u}}\le {\displaystyle \frac{4}{y^2}}.$$

Using evenness, these estimates imply

$$\begin{array}{cc}\hfill {\int }_{-\infty }^{\infty }{\left({\displaystyle \frac{{\mathcal{R}}_{\sigma }\left(y\right)}{M_{\sigma }}}\right)}^{m}dy& \le 2\sqrt{{\displaystyle \frac{15\pi }{m}}}+{\left({\displaystyle \frac{5}{7}}\right)}^m+3{\left({\displaystyle \frac{6}{7}}\right)}^m+{\displaystyle \frac{6}{2m-1}}{\left({\displaystyle \frac{4}{9}}\right)}^m\hfill \\ \hfill & \ll m^{-1/2},\hfill \end{array}$$

(43)

uniformly for $0\le \sigma \le 1/8$ and $m\ge 2$. Equivalently,

$${\int }_{-\infty }^{\infty }{\mathcal{R}}_{\sigma }{\left(y\right)}^{m}dy\ll {\displaystyle \frac{M_{\sigma }^m}{\sqrt{m}}}.$$

(44)

The beta normalization is equally explicit:

$${\displaystyle \frac{1}{4^m{B}_m}}={\displaystyle \frac{m}{2}}{\displaystyle \frac{\left(\genfrac{}{}{0pt}{}{2m}{m}\right)}{4^m}}\ll \sqrt{m}$$

by the Wallis bound for the central binomial coefficient. Combining this with (40) and (44) proves (37). Thus the $m^{-1/2}$ saddle width exactly cancels the $\sqrt{m}$ beta normalization; no hidden power of m remains.

The expansion (38) follows by Taylor expansion of $2/(1+\sqrt{1-4{\sigma }^2})$. Since $[log{M}_{\sigma }-{\sigma }^2]/{\sigma }^4$ extends continuously to $3/2$ at $\sigma =0$, it is bounded on $[0,1/8]$, proving (39) with an absolute constant.    □

Theorem 2 

(Explicit-formula derivative asymptotic with refined order dependence). There is an absolute constant $C>0$ such that, as $T\to \infty$, uniformly for every integer $m\ge 2$, with

$$\lambda :={\displaystyle \frac{m}{T^2}},$$

one has

$$A_m^{\prime }\left(T^2\right)={\displaystyle \frac{c_m}{4\pi T}}\left(log{\displaystyle \frac{T}{2\pi }}+{\displaystyle \frac{1}{2m-1}}\right)+O\left({\displaystyle \frac{(1+\lambda )exp\left(\lambda /4+C\lambda /T^2\right)}{T}}\right).$$

(45)

Consequently, for every fixed $\epsilon \in (0,4)$,

$$A_m^{\prime }\left(T^2\right)>0\left(2\le m\le (4-\epsilon )T^{2}loglogT\right)$$

(46)

for all sufficiently large T, uniformly throughout the displayed range. Equivalently, for every fixed $\epsilon \in (0,4)$,

$${\mathcal{W}}_m\left[F_{\xi }\right]\left(x\right)>0\left(2\le m\le (4-\epsilon )xloglogx\right)$$

(47)

for all sufficiently large x.

Proof. 

Since the statement is asymptotic as $T\to \infty$, we may assume $T\ge 6$ throughout the proof. For $m\ge 2$ put

$$h_{m,T}\left(r\right)=p_m(r/T),p_m\left(z\right)={\displaystyle \frac{z^{2m}}{B_m{(1+z^2)}^{2m}}}.$$

Differentiating (15) term by term gives, for every fixed m,

$$T^2{A}_m^{\prime }\left(T^2\right)=\sum _{\rho \in {\mathcal{Z}}_{+}}{p}_m\left({\displaystyle \frac{\gamma -i(\beta -1/2)}{T}}\right).$$

(48)

For later use, the beta integral gives

$$\begin{array}{cc}\hfill {\int }_0^{\infty }{p}_m\left(y\right)dy& ={\displaystyle \frac{c_m}{2}},\hfill \end{array}$$

(49)

$$\begin{array}{cc}\hfill {\int }_0^{\infty }{p}_m\left(y\right)logydy& ={\displaystyle \frac{c_m}{2(2m-1)}}.\hfill \end{array}$$

(50)

Indeed, for a near zero,

$${\int }_0^{\infty }{y}^a{p}_m\left(y\right)dy={\displaystyle \frac{B(m+(a+1)/2,m-(a+1)/2)}{2B_m}}.$$

By Lemma 4, the test function $h_{m,T}$ is admissible for the Guinand–Weil explicit formula. Write a nontrivial zero as $\rho =1/2+i{\Gamma }_{\rho }$, where

$${\Gamma }_{\rho }=\gamma -i(\beta -1/2).$$

If

$$g_{m,T}\left(u\right)={\displaystyle \frac{1}{2\pi }}{\int }_{-\infty }^{\infty }{h}_{m,T}\left(r\right)e^{-iur}dr,$$

then Proposition 2 gives

$$\begin{array}{cc}\hfill \sum _{\rho }{h}_{m,T}\left({\Gamma }_{\rho }\right)& =2h_{m,T}(i/2)\hfill \\ \hfill & +{\displaystyle \frac{1}{2\pi }}{\int }_{-\infty }^{\infty }{h}_{m,T}\left(r\right)\left\{\Re {\displaystyle \frac{{\Gamma }^{\prime }}{\Gamma }}\left({\displaystyle \frac{1}{4}}+{\displaystyle \frac{ir}{2}}\right)-log\pi \right\}dr\hfill \\ \hfill & -2{\displaystyle \sum _{n=2}^{\infty }}{\displaystyle \frac{\Lambda \left(n\right)}{\sqrt{n}}}{g}_{m,T}(logn).\hfill \end{array}$$

(51)

The reflected zero set is symmetric under $\Gamma \mapsto -\Gamma$, while $h_{m,T}$ is even. Therefore (48) gives

$$\sum _{\rho }{h}_{m,T}\left({\Gamma }_{\rho }\right)=2T^2{A}_m^{\prime }\left(T^2\right).$$

(52)

We first evaluate the archimedean term. The standard uniform digamma estimate gives

$$\Re {\displaystyle \frac{{\Gamma }^{\prime }}{\Gamma }}\left({\displaystyle \frac{1}{4}}+{\displaystyle \frac{ir}{2}}\right)-log\pi =log{\displaystyle \frac{\left|r\right|}{2\pi }}+O\left({\displaystyle \frac{1}{\left|r\right|}}\right)\left(\right|r|\ge 1).$$

On $\left|r\right|\le 1$ the difference between the left side and $log\left(\right|r|/(2\pi \left)\right)$ is $O(1+|log\left|r\right|\left|\right)$. This contribution is uniformly $O\left(1\right)$. Indeed, by the Wallis bound used in Lemma 5,

$$p_m\left(y\right)\ll \sqrt{m}{\left({\displaystyle \frac{4y^2}{{(1+y^2)}^2}}\right)}^m,$$

and for $T\ge 6$ and $\left|r\right|\le 1$ the base is at most $1/9$. Hence the product with $1+|log|r\left|\right|$ is dominated by an integrable function with an absolute bound because ${sup}_{m\ge 2}\sqrt{m}{9}^{-m}<\infty$. Moreover

$${\int }_{-\infty }^{\infty }{\displaystyle \frac{p_m\left(y\right)}{\left|y\right|}}dy=1,$$

(53)

which follows from the beta integral ${\int }_0^{\infty }{p}_m\left(y\right)y^{-1}dy=1/2$. Thus the $O(1/|r\left|\right)$ error contributes $O\left(1\right)$. Using (49)–(), now on both halves of the real line, gives

$$\begin{array}{cc}\hfill & {\displaystyle \frac{1}{2\pi }}{\int }_{-\infty }^{\infty }{h}_{m,T}\left(r\right)\left\{\Re {\displaystyle \frac{{\Gamma }^{\prime }}{\Gamma }}\left({\displaystyle \frac{1}{4}}+{\displaystyle \frac{ir}{2}}\right)-log\pi \right\}dr\hfill \\ \hfill & ={\displaystyle \frac{c_{m}T}{2\pi }}\left(log{\displaystyle \frac{T}{2\pi }}+{\displaystyle \frac{1}{2m-1}}\right)+O\left(1\right).\hfill \end{array}$$

(54)

The endpoint term is also uniformly bounded, with no hidden dependence on m. Indeed, putting $y=1/\left(2T\right)\le 1/4$ and again using the Wallis bound,

$$|h_{m,T}(i/2)|={\displaystyle \frac{y^{2m}}{B_m{(1-y^2)}^{2m}}}\ll \sqrt{m}{\left({\displaystyle \frac{4y^2}{{(1-y^2)}^2}}\right)}^m\le C\sqrt{m}{\left({\displaystyle \frac{64}{225}}\right)}^m=O\left(1\right).$$

It remains to control the prime sum. Every contour displacement used below has $\sigma =a/T\le 1/8$. For $u>0$ shift the Fourier contour from $\Im r=0$ to $\Im r=-a$, where $1/2<a\le 3/4$. No pole is crossed, and the strip-uniform decay from Lemma 4 gives

$$|g_{m,T}\left(u\right)|\le {\displaystyle \frac{e^{-au}}{2\pi }}{\int }_{-\infty }^{\infty }|h_{m,T}(r-ia)|dr={\displaystyle \frac{T{e}^{-au}}{2\pi }}{\int }_{-\infty }^{\infty }\left|p_m(y-ia/T)\right|dy.$$

(55)

Lemma 5 therefore gives

$$|g_{m,T}\left(u\right)|\ll T{e}^{-au}{M}_{a/T}^m.$$

(56)

Let $\lambda =m/T^2$. If $0<\lambda \le 4$, take $a=3/4$. Since $log{M}_{a/T}\ll a^2/T^2$, we have $M_{a/T}^m\ll 1$, and therefore

$${\displaystyle \sum _{n=2}^{\infty }}{\displaystyle \frac{\Lambda \left(n\right)}{\sqrt{n}}}\left|g_{m,T}(logn)\right|\ll T.$$

(57)

If $\lambda >4$, choose instead

$$a={\displaystyle \frac{1}{2}}+{\displaystyle \frac{1}{\lambda }},$$

so that $1/2<a<3/4$. The absolutely convergent Dirichlet series gives

$${\displaystyle \sum _{n=2}^{\infty }}{\displaystyle \frac{\Lambda \left(n\right)}{n^{1/2+a}}}=-{\displaystyle \frac{{\zeta }^{\prime }}{\zeta }}\left(1+{\displaystyle \frac{1}{\lambda }}\right)\ll 1+\lambda ,$$

where the last bound follows from the Laurent expansion at $s=1$ (and is uniform for $\lambda \ge 4$). By (39),

$$\begin{array}{cc}\hfill mlog{M}_{a/T}& \le \lambda a^2+C{\displaystyle \frac{\lambda a^4}{T^2}}\hfill \\ \hfill & ={\displaystyle \frac{\lambda }{4}}+1+{\displaystyle \frac{1}{\lambda }}+O\left({\displaystyle \frac{\lambda }{T^2}}\right).\hfill \end{array}$$

Together with (56), this yields

$${\displaystyle \sum _{n=2}^{\infty }}{\displaystyle \frac{\Lambda \left(n\right)}{\sqrt{n}}}\left|g_{m,T}(logn)\right|\ll T(1+\lambda )exp\left({\displaystyle \frac{\lambda }{4}}+C{\displaystyle \frac{\lambda }{T^2}}\right).$$

(58)

After enlarging the implied constant, (58) also contains the case $0<\lambda \le 4$ by (57).

Substitution of (54) and (58) into (51)–(52), followed by division by $2T^2$, proves (45).

Now fix $\epsilon \in (0,4)$ and suppose $\lambda \le (4-\epsilon )loglogT$. Since then $\lambda /T^2=o\left(1\right)$,

$$(1+\lambda )exp\left({\displaystyle \frac{\lambda }{4}}+C{\displaystyle \frac{\lambda }{T^2}}\right)\ll (loglogT){(logT)}^{1-\epsilon /4+o\left(1\right)}=o(logT).$$

The main term in (45) is $\asymp (logT)/T$, uniformly in $m\ge 2$ because $1\le c_m\le \pi /2$. This proves (46). Finally,

$$A_m^{\prime }\left(x\right)={\displaystyle \frac{x^{m-1}}{2(2m-1)!B_m}}{\mathcal{W}}_m\left[F_{\xi }\right]\left(x\right)$$

has positive prefactor. Since $loglogx=loglogT+log2$ for $x=T^2$, write $L=loglogT$. For every fixed $\epsilon \in (0,4)$ and all sufficiently large T,

$$(4-\epsilon )(L+log2)\le (4-\epsilon /2)L.$$

Applying the T-statement with $\epsilon /2$ therefore proves (47).    □

Remark 1 

(Where a hypothetical obstruction can remain). Theorem 2 does not prove RH: it controls a large asymptotic region of the two-parameter Widder family, not every pair $(m,x)$. It shows more precisely that, for every fixed $\epsilon >0$, a sequence of negative Widder values with $x\to \infty$ must eventually escape the region

$$m\le (4-\epsilon )xloglogx.$$

The constant 4 has a natural origin in the absolute-convergence contour method. If $\lambda =m/T^2$ is large, the prime Dirichlet series forces the shift parameter toward $a=1/2$, while $log{M}_{a/T}\sim a^2/T^2$. Thus the least absolute-value amplification available from this method is asymptotically $exp(\lambda /4)$, and balancing it against the archimedean factor $logT$ gives the transition $\lambda \sim 4loglogT$. Going materially beyond that boundary would therefore require genuine cancellation on the prime side rather than a sharper absolute-value estimate alone.

The quantitative bound (45) also shows that positivity holds whenever

$${\displaystyle \frac{\lambda }{4}}+log(1+\lambda )+C{\displaystyle \frac{\lambda }{T^2}}\le loglogT-\omega \left(1\right).$$

For example, it is enough that

$$\lambda \le 4loglogT-4logloglogT-\omega \left(1\right).$$

4. A beta–Widder Schur–triple Application to Simple Critical Zeros

This section is logically separate from the uniform smoothing theorem of Section 3. Its purpose is to record a second use of the same beta family. After a compactifying change of variables, the beta densities become positive polynomial windows on $[-1/2,1/2]$. We combine one such window with the analytic finite-compression framework of Alpöge–Furman [11]. The positive Schur–Jensen remainder is motivated by Shi [12], but the rank–trace inequality, the zero-side seam, and its identification with the atoms used below are all reproved here in our own notation. Thus Shi is no longer used as a black-box input. Throughout this section, Alpöge–Furman means arXiv:2608.13637v2, and Shi means the preprint archived at the DOI in the bibliography. The external interface is version-locked: the only analytic inputs in the transfer step are the explicitly cited Poisson–Gabor, trace, tail, reduction, Archimedean, prime-square, and mixed-term estimates of Alpöge–Furman. Their fixed-$\lambda$ normalization errors are tracked below, and the limit $\lambda \uparrow 1$ is taken only after $T\to \infty$.

We make no claim that the numerical constant below is the strongest value obtainable from recent multi-point computations. The point here is that the window is an explicit beta–Widder mixture, the local Schur estimate uses only three consecutive simple critical zeros, and the finite verification is small and reproducible.

4.1. Compactifying the Beta Densities

Recall the beta density

$$p_m\left(y\right)={\displaystyle \frac{y^{2m}}{B(m,m){(1+y^2)}^{2m}}},y>0.$$

The change of variables

$$s={\displaystyle \frac{y^2-1}{2(y^2+1)}},-{\displaystyle \frac{1}{2}}<s<{\displaystyle \frac{1}{2}},$$

(59)

converts the logarithmic beta mass into a compactly supported polynomial probability density.

Lemma 6 

(Beta compactification). For $m\ge 1$ define

$$v_m\left(s\right):={\displaystyle \frac{4^{1-m}}{B(m,m)}}{(1-4s^2)}^{m-1},\left|s\right|\le {\displaystyle \frac{1}{2}}.$$

(60)

Then

$${\displaystyle \frac{2p_m\left(y\right)}{y}}dy=v_m\left(s\right)ds$$

(61)

under (59), and ${\int }_{-1/2}^{1/2}{v}_m\left(s\right)ds=1$.

Proof. 

Since

$${\displaystyle \frac{ds}{dy}}={\displaystyle \frac{2y}{{(1+y^2)}^2}}$$

and

$$1-4s^2={\displaystyle \frac{4y^2}{{(1+y^2)}^2}},$$

we obtain

$${\displaystyle \frac{2p_m\left(y\right)}{y}}{\displaystyle \frac{dy}{ds}}={\displaystyle \frac{1}{B(m,m)}}{\left({\displaystyle \frac{y^2}{{(1+y^2)}^2}}\right)}^{m-1}={\displaystyle \frac{4^{1-m}}{B(m,m)}}{(1-4s^2)}^{m-1}.$$

Integrating (61) over $y>0$ gives the normalization; equivalently it follows directly from the beta integral.    □

We use the following explicit three-beta mixture:

$${\psi }_{†}\left(s\right):={\displaystyle \frac{49621}{60000}}{v}_1\left(s\right)+{\displaystyle \frac{1}{6}}{v}_2\left(s\right)+{\displaystyle \frac{379}{60000}}{v}_3\left(s\right).$$

(62)

The coefficients are positive and sum to 1. Since

$$v_1\left(s\right)=1,v_2\left(s\right)={\displaystyle \frac{3}{2}}(1-4s^2),v_3\left(s\right)={\displaystyle \frac{15}{8}}{(1-4s^2)}^2,$$

we have the exact polynomial form

$${\psi }_{†}\left(s\right)={\displaystyle \frac{379}{2000}}{s}^4-{\displaystyle \frac{4379}{4000}}{s}^2+{\displaystyle \frac{522653}{480000}},\left|s\right|\le {\displaystyle \frac{1}{2}}.$$

(63)

In particular, ${\psi }_{†}$ is even, $C^{\infty }$ and strictly positive on $[-1/2,1/2]$, with

$${\int }_{-1/2}^{1/2}{\psi }_{†}\left(s\right)ds=1,{\psi }_{†}(\pm 1/2)={\displaystyle \frac{49621}{60000}},\underset{\left|s\right|\le 1/2}{max}{\psi }_{†}\left(s\right)={\psi }_{†}\left(0\right)={\displaystyle \frac{522653}{480000}}.$$

To remain literally inside the bounded-window normalization used in ([11], Section 2.2), we feed the compression the scaled window

$${\psi }_{†}^{\mathrm{AF}}\left(s\right):=c_{†}{\psi }_{†}\left(s\right),c_{†}:={\displaystyle \frac{480000}{522653}},$$

(64)

so that $0<{\psi }_{†}^{\mathrm{AF}}\le 1$. This scaling changes neither $R_{\lambda }$, which is homogeneous of degree zero in the window, nor the normalized correlation kernel: dividing the Fourier transform of ${\psi }_{†}^{\mathrm{AF}}$ by its mass $c_{†}$ gives exactly the $K_{†}$ below. We therefore keep the probability-normalized ${\psi }_{†}$ in all exact rational calculations.

Let

$$K_{†}\left(u\right):={\int }_{-1/2}^{1/2}{\psi }_{†}\left(s\right)e^{2\pi ius}ds.$$

(65)

Since the window is even and normalized, $K_{†}$ is real and even on $\mathbb{R}$ and $K_{†}\left(0\right)=1$. Writing $x=\pi u$, elementary integration gives

$$K_{†}\left(u\right)={\displaystyle \frac{(49621x^4+24315x^2+17055)sinx-(30000x^3+17055x)cosx}{60000x^5}},x=\pi u,$$

(66)

with the removable value $K_{†}\left(0\right)=1$.

4.2. The Window Constant

For an even positive window $\psi$ on $[-1/2,1/2]$, Alpöge–Furman ([11], Lemma 5.6) define

$$R\left(\psi \right):={\displaystyle \frac{{\displaystyle {\int }_{-1/2}^{1/2}\psi {\left(s\right)}^{2}ds+{\int }_{-1/2}^{1/2}{\int }_{-1/2}^{1/2}|s-t|\psi \left(s\right)\psi \left(t\right)dsdt}}{{\left({\displaystyle {\int }_{-1/2}^{1/2}\psi \left(s\right)ds}\right)}^2}}.$$

(67)

Their finite-compression argument gives the baseline proportion $2-R\left(\psi \right)$ for simple zeros on the critical line. Direct rational integration of (63) gives

$$\begin{array}{cc}\hfill {\int }_{-1/2}^{1/2}{\psi }_{†}{\left(s\right)}^{2}ds& ={\displaystyle \frac{8451863641}{8400000000}},\hfill \end{array}$$

(68)

$$\begin{array}{cc}\hfill {\int }_{-1/2}^{1/2}{\int }_{-1/2}^{1/2}|s-t|{\psi }_{†}\left(s\right){\psi }_{†}\left(t\right)dsdt& ={\displaystyle \frac{17814265853}{55440000000}}.\hfill \end{array}$$

(69)

Hence

$$R\left({\psi }_{†}\right)={\displaystyle \frac{183991414709}{138600000000}}=1.327499384624819\dots ,$$

(70)

and

$$p_{†}:=2-R\left({\psi }_{†}\right)={\displaystyle \frac{93208585291}{138600000000}}=0.672500615375180\dots .$$

(71)

This is slightly below the Montgomery–Taylor optimum among certificates of the form $2-R\left(\psi \right)$, as it must be; see ([11], Theorem A and Section 7.2). The gain below comes instead from retaining a positive Schur remainder.

4.3. A Three-Point Schur Remainder

Following Shi [12], put

$$g\left(t\right):=t^2-2t-{(t-2)}_{+}^2=\left\{\begin{array}{cc}{t}^2-2t,\hfill & 0\le t\le 2,\hfill \\ 2t-4,\hfill & t\ge 2.\hfill \end{array}\right.$$

(72)

Let M be a $3\times 3$ correlation Gram matrix, so $M⪰0$ and $M_{jj}=1$. Write

$$E\left(M\right):=|M_{12}{|}^2+|M_{13}{|}^2+{\left|M_{23}\right|}^2$$

and

$${\mathcal{S}}_3\left(M\right):=trg\left(M\right)-3g\left(1\right).$$

(73)

Lemma 7 

(Universal three-point Schur bound). Every $3\times 3$ correlation Gram matrix satisfies

$${\mathcal{S}}_3\left(M\right)\ge {\displaystyle \frac{5}{3}}E\left(M\right).$$

(74)

Moreover, if ${\lambda }_{max}\left(M\right)<2$, then

$${\mathcal{S}}_3\left(M\right)=2E\left(M\right).$$

(75)

Proof. 

Let the eigenvalues be ${\lambda }_1,{\lambda }_2,{\lambda }_3\ge 0$. Since $\sum {\lambda }_j=3$ and $tr{M}^2=3+2E\left(M\right)$, formula (72) gives

$${\mathcal{S}}_3\left(M\right)=2E\left(M\right)-{\displaystyle \sum _{j=1}^3}{({\lambda }_j-2)}_{+}^2.$$

(76)

At most one eigenvalue can exceed 2. If none does, this proves (75). Otherwise write $L={\lambda }_{max}\in (2,3]$. The other two eigenvalues have sum $3-L$, so

$$E\left(M\right)={\displaystyle \frac{1}{2}}\left(\sum _j{\lambda }_j^2-3\right)\ge {\displaystyle \frac{3}{4}}{(L-1)}^2.$$

For $2\le L\le 3$,

$${\displaystyle \frac{3}{4}}{(L-1)}^2\ge 3{(L-2)}^2,$$

whence ${(L-2)}^2\le E\left(M\right)/3$. Substitution into (76) proves (74).    □

The block form needed below is a direct extension of the matching argument in ([12], Lemma 2.2).

Lemma 8 

(Disjoint-block Schur–Horn bound). Let $v_1,\dots ,v_s$ be vectors with $\parallel v_j\parallel \le 1$, let $P={\sum }_j{v}_j{v}_j^{*}$, and let $M=\left(\langle v_i,v_j\rangle \right)$. Define

$$\mathcal{S}\left(P\right):=trg\left(P\right)-{\displaystyle \sum _{j=1}^s}g(\parallel v_j{\parallel }^2).$$

For any partition of a subset of $\{1,\dots ,s\}$ into disjoint blocks B, one has

$$\mathcal{S}\left(P\right)\ge \sum _B\left(trg\left(M_B\right)-\sum _{j\in B}g\left(M_{jj}\right)\right),$$

(77)

where $M_B$ is the corresponding principal Gram block.

Proof. 

For each selected block choose a unitary matrix diagonalizing $M_B$, and let U be the direct sum of these block unitaries and the identity on the unselected coordinates. The nonzero eigenvalues of P and M agree, so $trg\left(P\right)=trg\left(M\right)$. By Schur–Horn and the convexity of g,

$$trg\left(M\right)\ge \sum _{j}g\left({\left(U^{*}MU\right)}_{jj}\right).$$

On each selected block the diagonal entries of $U^{*}MU$ are the eigenvalues of $M_B$, while on all other coordinates the diagonal remains $M_{jj}$. Subtracting ${\sum }_{j}g\left(M_{jj}\right)$ gives (77).    □

4.4. An Affine Triple-Energy Inequality

For $x,y\ge 0$ let

$$M_{†}(x,y):=\left(\begin{array}{ccc}1& K_{†}\left(x\right)& K_{†}(x+y)\\ K_{†}\left(x\right)& 1& K_{†}\left(y\right)\\ K_{†}(x+y)& K_{†}\left(y\right)& 1\end{array}\right)$$

(78)

and write ${\mathcal{S}}_{†}(x,y):={\mathcal{S}}_3\left(M_{†}(x,y)\right)$. Because (78) is the Gram matrix of three translates of $\sqrt{{\psi }_{†}}$, it is positive semidefinite.

Lemma 9 

(Affine triple-energy certificate). For every $x,y\ge 0$,

$${\mathcal{S}}_{†}(x,y)+{\displaystyle \frac{x+y}{800}}\ge {\displaystyle \frac{171}{40000}}.$$

(79)

Proof. 

Put

$$E_{†}(x,y):=K_{†}{\left(x\right)}^2+K_{†}{\left(y\right)}^2+K_{†}{(x+y)}^2.$$

If (79) failed, then by Lemma 7,

$$E_{†}(x,y)<{\displaystyle \frac{3}{5}}{\displaystyle \frac{171}{40000}}=0.002565.$$

Hence

$$\parallel M_{†}{(x,y)-I\parallel }_{\mathrm{op}}\le {\parallel M_{†}(x,y)-I\parallel }_{\mathrm{F}}=\sqrt{2E_{†}(x,y)}<0.072,$$

so all eigenvalues of $M_{†}(x,y)$ are $<2$. Therefore (75) applies, and it suffices to prove

$$2E_{†}(x,y)+{\displaystyle \frac{x+y}{800}}\ge {\displaystyle \frac{171}{40000}}.$$

(80)

The remaining statement is a finite interval-arithmetic verification of the explicit function (66). We record the reduction so that the trust base is transparent. Define

$$q\left(u\right):=2K_{†}{\left(u\right)}^2+{\displaystyle \frac{u}{1600}}.$$

Then the left side of (80) is $q\left(x\right)+q\left(y\right)+q(x+y)$. If $x+y\ge 3.42$, the linear term alone proves (80). For $0\le u\le 0.48$, positivity of ${\psi }_{†}$ and its normalization give

$$K_{†}\left(u\right)={\int }_{-1/2}^{1/2}{\psi }_{†}\left(s\right)cos\left(2\pi us\right)ds\ge cos\left(\pi u\right),$$

so again $q\left(u\right)>171/40000$.

On the remaining interval, outward interval arithmetic at 50 decimal digits certifies

$$q\left(u\right)\ge {\displaystyle \frac{171}{40000}}$$

outside

$$J_1=[1.01,1.11],J_2=[1.94,2.13],J_3=[2.895,3.153].$$

(81)

Thus a counterexample to (80) would have $x,y,x+y\in J_1\cup J_2\cup J_3$. Since $x+y<3.42$, the only additive possibilities are, up to exchanging x and y,

$$(x,y)\in J_1\times J_1\mathrm{or}(x,y)\in J_1\times J_2.$$

The same outward interval calculation certifies (80) on both full rectangles. The supplied script beta_widder_affine_certificate.py performs these checks using mpmath.iv; it closes the four one-dimensional complement intervals with $15,39,29,13$ boxes, respectively, and the two rectangles with 5053 and 3333 boxes, with no unresolved boxes. The subdivision endpoints are exact rationals, and the script compares outward interval endpoints directly. As an independent implementation, the supplied standard-library script beta_widder_affine_certificate_decimal.py uses directed decimal rounding, exact rational subdivision, Taylor’s theorem with explicit Lagrange remainders for sine and cosine, and the Fourier-moment bounds $|K_{†}|\le 1$, $|K_{†}^{\prime }|\le \pi$, $|K_{†}^{\prime \prime }|\le {\pi }^2$. In separate fresh-process runs it closes the same four one-dimensional regions with $23,69,99,51$ boxes and the two rectangles with 2309 and 5351 boxes, respectively; the corresponding output log is included in the source package.    □

Remark 2 

(Computational status). The interval verification in Lemma 9 is small and fully reproducible from the source package. Two algorithmically independent interval implementations are supplied. The first usesmpmath.ivwithmpmath==1.3.0at 50 decimal digits and outward intervals. The second uses only Python’s standard-librarydecimalmodule with directed rounding and rigorous Taylor remainders; it does not importmpmathor any external interval package. Both use exact rational subdivision coordinates and certify the same fixed rational target $171/40000$ on the same analytic reduction. The exact rational window integrals and bootstrap fixed point are separately checked, using onlyfractions.Fraction, by verifybetaschurrationals.py. The exact commands, expected box counts, software assumptions, and the SHA-256 hashes of the scripts and archived logs are recorded in REPRODUCIBILITY.md and SHA256SUMS.txt. The recorded outputs are betaschurcertificateoutput.txt and betaschurcertificatedecimaloutput.txt. Timing is not part of the certificate. An Arb/FLINT replay would be a useful third archival implementation, but the proof package does not rely on such a replay because the directed-Decimal/Taylor implementation already supplies an independent transcendental enclosure path. These certificates verify only the finite inequality in Lemma 9; they do not certify the external analytic estimates imported from Alpöge–Furman.

4.5. The Fixed-Bandwidth Finite-Compression Interface

Remark 3 

(Version-locked Alpöge–Furman interface). For auditability, the external dependency is frozen at arXiv:2608.13637v2. We use exactly the Fourier convention of that paper, $\widehat{f}\left(\xi \right)={\int }_{\mathbb{R}}f\left(u\right)e^{-iu\xi }du$. The only change of scale is

$$\ell :=log(T/2\pi ),L:=\lambda \ell ,X:=e^L={(T/2\pi )}^{\lambda },0<\lambda <1,$$

with λ fixed while $T\to \infty$. No assertion at the endpoint $\lambda =1$ is imported.

The following ledger records the exact external formulas used. Equation and proposition numbers in the first column refer to Alpöge–Furman v2.

No rank–trace inequality, Schur remainder, triple packing lemma, or three-point certificate is imported from this list. More importantly, the general-window formula for $R_{\lambda }\left(\psi \right)$ isnotquoted from Remark 6.1: Proposition 3 re-derives it from the raw terms in the ledger. Remark 6.1 is used only as a check on the special case $\psi ={\mathbf{1}}_{[-1/2,1/2]}$.

There is also no hidden use of $L=\ell$ in this re-scaling. In the cited proofs, the Archimedean density is naturally expressed in ℓ, whereas window support and the prime cutoff are expressed in $L=logX$; Proposition 5.4 only requires the same diagonal regime $X\le T$, which is automatic for fixed $\lambda <1$, and the mixed-term proof changes only by the fixed ratio $\ell /L=1/\lambda$. Proposition 5.2 already keeps ℓ and X separate through its quantity B. This is exactly the deformation summarized for the indicator window in their Remark 6.1.   

External item	Quantity retained before the specialization $L=\ell$
(2.7)–(2.11), Lemma 2.1	The tapered test family, grid spacing $2\pi /L$, normalized matrix, and the exact Poisson–Gabor identity. These give the atom norm bound and the local Gram limit used below.
Proposition 4.2, proof	$tr\widehat{A}=N\left(I^{\prime }\right)+O_{\chi }\left(X^{1/2}{L}^2\right)$ for the finite zero-side compression.
Proposition 4.3, proof	$\parallel \widehat{E}{\parallel }_1{\ll }_{\chi }{X}^{1/2}/T$ for the omitted zero-side tail.
(5.7), Proposition 5.2, proof	With $B^2\ll {\ell }^2+X$, the two raw reduction errors satisfy $|{\mathcal{E}}_1|\ll L^3{B}^{2}logL$ and $|{\mathcal{E}}_2|\ll L^3{B}^2\ell logL$.
Proposition 5.3	$\mathcal{M}[\mu ,\mu ]=bLT{\ell }_1^2/\left(2\pi \right)(1+O\left({\ell }^{-2}\right))+O({\ell }^{2}logL)$, where $b=L^{-1}\int {\varphi }^4$ and ${\ell }_1=\ell +O\left(1\right)$.
Proposition 5.4	$\mathcal{M}[P_X,P_X]=(T/\pi ){\sum }_{n\le X}\Lambda {\left(n\right)}^2{n}^{-1}g(logn)+O_{\chi }\left(L^{2}X\right)$; the same partial summation as in Theorem 5.7 gives the $T{L}^{3}D/\left(2\pi \right)$ main term below.
Proposition 5.5	The four mixed terms are $O_{\lambda ,\psi ,\chi }\left(L^2{X}^{1/2}\right)$ after fixing $\lambda$ (the displayed $\mu$–prime term is smaller).
Remark 6.1	Independent consistency check: for the indicator window the fixed-$\lambda$ constant is $R_{\lambda }={\lambda }^{-1}+\lambda /3$.

For clarity we now state exactly the part of the recent finite-compression machinery that is used below and record the uniform normalization estimate needed for triples. Put

$$\ell :=log(T/2\pi ),L:=\lambda \ell ,0<\lambda <1$$

with $\lambda$ fixed while $T\to \infty$. Let $\chi$ be the fixed endpoint taper used in ([11], Section 2.2), and set

$${\varphi }_T\left(u\right):=\chi \left({\displaystyle \frac{L}{2}}+u\right)\chi \left({\displaystyle \frac{L}{2}}-u\right){\left({\psi }_{†}^{\mathrm{AF}}(u/L)\right)}^{1/2},a_T:={\displaystyle \frac{1}{L}}{\int }_{\mathbb{R}}{\varphi }_T{\left(u\right)}^{2}du.$$

(82)

Since $\int {\psi }_{†}^{\mathrm{AF}}=c_{†}$, one has $a_T=c_{†}+O\left(L^{-1}\right)$. Put

$${\tau }_k:=T+{\displaystyle \frac{2\pi k}{L}},d:=⌊{\displaystyle \frac{LT}{2\pi }}⌋.$$

Put

$$I:=(T,2T],I^{\prime }:=(T-T^{1/2},2T+T^{1/2}],I^{\circ }:=[T+T^{1/2},2T-T^{1/2}].$$

For every simple critical-line zero $1/2+i\gamma$ with $\gamma \in I^{\prime }$, define its normalized finite atom by

$$v_{\gamma }:={\displaystyle \frac{1}{\sqrt{a_T{L}^2}}}{\left(\widehat{{\varphi }_T}(\gamma -{\tau }_k)\right)}_{0\le k<d}.$$

(83)

These are exactly the simple-line vectors entering the positive block of the normalized finite zero-side matrix. The full-lattice Poisson–Gabor identity and truncation immediately give $\parallel v_{\gamma }{\parallel }^2\le 1$ for every $\gamma \in I^{\prime }$. The stronger normalization $\parallel v_{\gamma }\parallel =1+o\left(1\right)$ is needed only for the selected triples, which will lie in the deep core $I^{\circ }$.

Lemma 10 

(Uniform atom normalization and local Gram limit). For every fixed $U>0$ there is a quantity ${\eta }_T\left(U\right)=o\left(1\right)$, depending on $\lambda ,U$ and the fixed taper but not on the chosen core zeros, such that

$$\begin{array}{cc}\hfill 0\le 1-\parallel v_{\gamma }{\parallel }^2& \le {\eta }_T\left(U\right),\hfill \end{array}$$

(84)

$$\begin{array}{cc}\hfill \left|{\displaystyle \frac{\langle v_{\gamma },v_{{\gamma }^{\prime }}\rangle }{\parallel v_{\gamma }\parallel \parallel v_{{\gamma }^{\prime }}\parallel }}-K_{†}\left({\displaystyle \frac{L(\gamma -{\gamma }^{\prime })}{2\pi }}\right)\right|& \le {\eta }_T\left(U\right)\hfill \end{array}$$

(85)

whenever $|L(\gamma -{\gamma }^{\prime })/\left(2\pi \right)|\le U$. More precisely one may take

$${\eta }_T\left(U\right){\ll }_U{L}^{-1}+{\left(L{T}^{3/2}\right)}^{-1}.$$

Proof. 

The Poisson–Gabor identity ([11], Lemma 2.1) gives, on the full sampling lattice,

$$\sum _{k\in \mathbb{Z}}\widehat{{\varphi }_T}(\gamma -{\tau }_k)\overline{\widehat{{\varphi }_T}({\gamma }^{\prime }-{\tau }_k)}=L\widehat{{\varphi }_T^2}(\gamma -{\gamma }^{\prime }).$$

At ${\gamma }^{\prime }=\gamma$ the right side is $a_T{L}^2$. Two integrations by parts give $|\widehat{{\varphi }_T}{\left(r\right)|\ll |r|}^{-2}$ away from $r=0$, uniformly for fixed $\lambda$. A core ordinate is at distance $T^{1/2}-O\left(L^{-1}\right)$ from the omitted part of the lattice, hence comparison with ${\int }_{T^{1/2}}^{\infty }{r}^{-4}dr$ gives

$$\sum _{k\notin [0,d)}{\left|\widehat{{\varphi }_T}(\gamma -{\tau }_k)\right|}^2\ll L{T}^{-3/2}.$$

After division by $a_T{L}^2\asymp L^2$ this proves (84), with error $O\left({\left(L{T}^{3/2}\right)}^{-1}\right)$.

Now put $u=L(\gamma -{\gamma }^{\prime })/\left(2\pi \right)$. Uniformly for $\left|u\right|\le U$, rescaling (82) gives

$${\displaystyle \frac{\widehat{{\varphi }_T^2}(2\pi u/L)}{a_{T}L}}={\displaystyle \frac{1}{c_{†}}}{\int }_{-1/2}^{1/2}{\psi }_{†}^{\mathrm{AF}}\left(s\right)e^{2\pi ius}ds+O_U\left(L^{-1}\right)=K_{†}\left(u\right)+O_U\left(L^{-1}\right).$$

The omitted cross terms satisfy the same $O\left(L{T}^{-3/2}\right)$ estimate by Cauchy–Schwarz. Dividing by $a_T{L}^2$ and then by the two norms, which are $1+o\left(1\right)$ uniformly by the first part, proves ().    □

The preceding lemma lets us keep the original finite atoms in the global Schur remainder; no renormalization of the finite-compression seam is needed. The next elementary stability statement makes this explicit.

Lemma 11 

(Stability of a three-point Schur block). For a positive semidefinite $3\times 3$ Gram matrix M with diagonal $r_1,r_2,r_3\in (0,1]$, let

$$C=D^{-1}M{D}^{-1},D=diag(\sqrt{r_1},\sqrt{r_2},\sqrt{r_3}).$$

Then C is a correlation matrix and, as ${max}_j|r_j-1|\to 0$,

$$trg\left(M\right)-{\displaystyle \sum _{j=1}^3}g\left(r_j\right)={\mathcal{S}}_3\left(C\right)+O\left(\underset{j}{max}|r_j-1|\right),$$

(86)

uniformly over all such Gram matrices. Moreover the left side of (86) is always nonnegative.

Proof. 

All eigenvalues of M and C lie in $[0,3]$. On this interval the convex function g is 2-Lipschitz. Since $M=DCD$ and $D=I+O\left(\eta \right)$ when $\eta ={max}_j|r_j-1|$, one has ${\parallel M-C\parallel }_{\mathrm{F}}=O\left(\eta \right)$. The Hoffman–Wielandt inequality therefore gives

$$|trg\left(M\right)-trg\left(C\right)|=O\left(\eta \right),\left|\sum _{j}g\left(r_j\right)-3g\left(1\right)\right|=O\left(\eta \right),$$

which proves (86). Finally, the vector of diagonal entries of a Hermitian matrix is majorized by its eigenvalue vector. Convexity of g and Schur–Horn therefore give $trg\left(M\right)\ge {\sum }_{j}g\left(M_{jj}\right)$.    □

4.6. General-window Scaling of the Prime-Side Second Moment

We next make explicit the fixed-$\lambda$ scaling and the error bookkeeping needed in the bootstrap. This removes the phrase “by the same estimates” from the transfer argument and makes clear why the ordered limit $T\to \infty$ first, $\lambda \uparrow 1$ second, is harmless.

Proposition 3 

(Fixed-bandwidth window constant with an explicit error). Let ψ be an even positive $C^2$ window on $[-1/2,1/2]$, and put

$$J:={\int }_{-1/2}^{1/2}\psi \left(s\right)ds,I_2:={\int }_{-1/2}^{1/2}\psi {\left(s\right)}^{2}ds,D:=\underset{{[-1/2,1/2]}^2}{\int \int }|s-t|\psi \left(s\right)\psi \left(t\right)dsdt.$$

Use the Alpöge–Furman compression with $L=\lambda \ell$, $X=e^L={(T/2\pi )}^{\lambda }$, where $0<\lambda <1$ is fixed. Let $\widehat{A}$ denote the normalized finite zero-side block, $\widehat{E}$ the omitted zero-side tail, and put $\widehat{G}:=\widehat{A}+\widehat{E}$ for the full compression evaluated by the explicit formula. Then

$$\parallel \widehat{G}{\parallel }_{\mathrm{HS}}^2=\left(R_{\lambda }\left(\psi \right)+O_{\lambda ,\psi ,\chi }\left({\epsilon }_{\lambda }\left(T\right)\right)\right)N(T,2T),$$

(87)

where

$$R_{\lambda }\left(\psi \right)={\displaystyle \frac{{\lambda }^{-1}{I}_2+\lambda D}{J^2}},{\epsilon }_{\lambda }\left(T\right):={\displaystyle \frac{1}{\ell }}+{\displaystyle \frac{\ell log\ell }{T}}+{\displaystyle \frac{T^{\lambda -1}log\ell }{\ell }}+{\displaystyle \frac{T^{\lambda /2-1}}{\ell }}.$$

(88)

In particular ${\epsilon }_{\lambda }\left(T\right)\to 0$ for every fixed $0<\lambda <1$.

Proof. 

We spell out the normalization of every term used from ([11], Sections 4–5). For the tapered window built from $\psi$ one has

$$a_T=J+O_{\psi ,\chi }\left(L^{-1}\right),b_T:=L^{-1}{\parallel {\varphi }_T\parallel }_4^4=I_2+O_{\psi ,\chi }\left(L^{-1}\right).$$

The normalization is worth writing before estimating anything. In the notation of the proof of ([11], Proposition 5.2), the exact reduction has the form

$${\left(a_T{L}^2\right)}^2{\parallel \widehat{G}\parallel }_{\mathrm{HS}}^2=L^2\mathcal{M}+{\mathcal{E}}_1+{\mathcal{E}}_2,$$

(89)

up to the harmless notation change from their full compression to $\widehat{G}$. Thus the main double integral is divided by $a_T^2{L}^2$, while the raw errors are divided by $a_T^2{L}^4$; these two powers of L should not be conflated. Their proof gives

$$|{\mathcal{E}}_1|{\ll }_{\psi ,\chi }{L}^3{B}^{2}logL,\left|{\mathcal{E}}_2\right|{\ll }_{\psi ,\chi }{L}^3{B}^2\ell logL,B^2\ll {\ell }^2+X.$$

Using $L=\lambda \ell$, $a_T{\asymp }_{\psi }1$, and $N(T,2T)\asymp T\ell$, division of the two errors in (89) by $a_T^2{L}^{4}N(T,2T)$ gives

$$O_{\lambda ,\psi ,\chi }\left({\displaystyle \frac{\ell log\ell }{T}}+{\displaystyle \frac{T^{\lambda -1}log\ell }{\ell }}\right).$$

(90)

The first raw error alone is smaller; the displayed envelope is chosen to cover both and to keep the later bookkeeping one-line. Their Proposition 5.3 gives

$$\mathcal{M}[\mu ,\mu ]={\displaystyle \frac{b_{T}LT{\ell }_1^2}{2\pi }}\left(1+O\left({\ell }^{-2}\right)\right)+O({\ell }^{2}logL),{\ell }_1=\ell +O\left(1\right).$$

Since $N(T,2T)\sim T\ell /\left(2\pi \right)$ and $L=\lambda \ell$, its normalized contribution is

$${\displaystyle \frac{\mathcal{M}[\mu ,\mu ]}{a_T^2{L}^{2}N(T,2T)}}={\displaystyle \frac{1}{\lambda }}{\displaystyle \frac{I_2}{J^2}}+O_{\lambda ,\psi ,\chi }\left({\ell }^{-1}\right).$$

(91)

For the prime square, Proposition 5.4 and the partial-summation calculation used in Theorem 5.7 give

$$\mathcal{M}[P_X,P_X]={\displaystyle \frac{T{L}^3}{2\pi }}D+O_{\psi ,\chi }(T{L}^2+L^{2}X).$$

Consequently

$${\displaystyle \frac{\mathcal{M}[P_X,P_X]}{a_T^2{L}^{2}N(T,2T)}}=\lambda {\displaystyle \frac{D}{J^2}}+O_{\lambda ,\psi ,\chi }\left({\displaystyle \frac{1}{\ell }}+{\displaystyle \frac{T^{\lambda -1}}{\ell }}\right).$$

(92)

The four mixed terms are bounded in Proposition 5.5 by $O_{\chi }\left(L^2\sqrt{X}\right)$ (with the first one in fact smaller). After the same normalization their total contribution is

$$O_{\lambda ,\psi ,\chi }\left({\displaystyle \frac{\sqrt{X}}{N(T,2T)}}\right)=O_{\lambda ,\psi ,\chi }\left({\displaystyle \frac{T^{\lambda /2-1}}{\ell }}\right).$$

(93)

For reference, the normalized contributions per zero are therefore

$$\begin{array}{cc}\mathrm{term}& \mathrm{contribution}\mathrm{after}\mathrm{division}\mathrm{by}N(T,2T)\\ \mathrm{Archimedean}& {\lambda }^{-1}{I}_2/J^2+O\left({\ell }^{-1}\right)\\ \mathrm{prime}\mathrm{square}& \lambda D/J^2+O({\ell }^{-1}+T^{\lambda -1}/\ell )\\ \mathrm{mixed}\mathrm{terms}& O(T^{\lambda /2-1}/\ell )\\ \mathrm{finite}/\mathrm{full}\mathrm{reduction}& O(\ell log\ell /T+T^{\lambda -1}log\ell /\ell ).\end{array}$$

All implied constants may depend on the fixed $\lambda$, window, and taper. Combining (90), (91), (92), and (93) proves (87). For later use in the zero-side seam, we also record that the proof of ([11], Proposition 4.3), with $X=e^L$, gives

$$\parallel \widehat{E}{\parallel }_1{\ll }_{\psi ,\chi }{\displaystyle \frac{X^{1/2}}{T}}{\ll }_{\lambda ,\psi ,\chi }{T}^{\lambda /2-1}.$$

(94)

This estimate is not absorbed into (87); it is kept separately because it is precisely what compares the finite zero-side matrix $\widehat{A}$ with the full explicit-formula matrix $\widehat{G}$. For $\psi ={\mathbf{1}}_{[-1/2,1/2]}$ the formula reads $R_{\lambda }={\lambda }^{-1}+\lambda /3$, in agreement with ([11], Remark 6.1). At the level of the main term, $R_{\lambda }\left(\psi \right)\to R\left(\psi \right)$ as $\lambda \uparrow 1$; no uniformity at the endpoint is asserted or needed.    □

For the normalized window ${\psi }_{†}$ we shall therefore use

$$R_{\lambda }\left({\psi }_{†}\right)={\displaystyle \frac{1}{\lambda }}{\int }_{-1/2}^{1/2}{\psi }_{†}{\left(s\right)}^{2}ds+\lambda \int \int |s-t|{\psi }_{†}\left(s\right){\psi }_{†}\left(t\right)dsdt,$$

(95)

and in particular $R_{\lambda }\left({\psi }_{†}\right)\to R\left({\psi }_{†}\right)$ as $\lambda \uparrow 1$.

4.7. A Self-Contained Strengthened Finite-Compression Seam

We now reproduce the finite-dimensional refinement needed for the Schur remainder. It is the mechanism isolated by Shi [12], but the proof below is included so that the transfer theorem does not require transporting notation or an unstated hypothesis from that preprint.

Lemma 12 

(Spectral rank–trace inequality). Let $P,Q$ be Hermitian $d\times d$ matrices, with $P⪰0$ and $n_{+}\left(Q\right)\le b$. Then

$${\parallel P+Q\parallel }_{\mathrm{HS}}^2\ge 2trP+trg\left(P\right)+4trQ-4b.$$

(96)

Proof. 

Write $Q=Q_{+}-Q_{-}$ for the positive and negative parts. Since $tr\left(P{Q}_{+}\right)\ge 0$ and $Q_{+}{Q}_{-}=0$,

$${\parallel P+Q\parallel }_{\mathrm{HS}}^2\ge \parallel P-Q_{-}{\parallel }_{\mathrm{HS}}^2+{\parallel Q_{+}\parallel }_{\mathrm{HS}}^2.$$

Let $p_j,n_j\ge 0$ be the eigenvalues of $P,Q_{-}$ in decreasing order. Von Neumann’s trace inequality gives

$$\parallel P-Q_{-}{\parallel }_{\mathrm{HS}}^2\ge \sum _j{(p_j-n_j)}^2.$$

For $p,n\ge 0$ one has

$${(p-n)}^2\ge 2p+g\left(p\right)-4n.$$

(97)

Indeed the difference is $n(n+4-2p)\ge 0$ when $p\le 2$ and ${(p-n-2)}^2\ge 0$ when $p\ge 2$. Thus

$$\parallel P-Q_{-}{\parallel }_{\mathrm{HS}}^2\ge 2trP+trg\left(P\right)-4tr{Q}_{-}.$$

If $q_1,\dots ,q_r$ are the positive eigenvalues of Q, then $r\le b$ and $q_j^2\ge 4q_j-4$, whence $\parallel Q_{+}{\parallel }_{\mathrm{HS}}^2\ge 4tr{Q}_{+}-4b$. Adding the last two estimates proves (96).    □

Let $\widehat{A}$ be the normalized finite zero-side matrix associated with $I^{\prime }$ and the same window and sampling grid as above. The simple critical-line zeros in $I^{\prime }$ contribute exactly the atoms (83). Write

$$\widehat{A}=P_1+Q^{\prime },P_1:={\displaystyle \sum _{j=1}^{s_1}}{v}_j{v}_j^{*},$$

(98)

where $s_1$ is the number of simple critical-line zeros in $I^{\prime }$. Let $s_2$ be the number of distinct multiple critical-line zeros in $I^{\prime }$ and let p be the number of off-line functional-equation pairs. The block-structure argument of ([11], Proposition 4.1), or equivalently Sylvester inertia applied pair by pair, gives

$$n_{+}\left(Q^{\prime }\right)\le s_2+p,N\left(I^{\prime }\right)\ge s_1+2s_2+2p.$$

(99)

Define the global simple-atom Schur remainder

$$\mathcal{S}:=trg\left(P_1\right)-{\displaystyle \sum _{j=1}^{s_1}}g(\parallel v_j{\parallel }^2).$$

(100)

By Schur–Horn and convexity of g, $\mathcal{S}\ge 0$.

Proposition 4 

(Self-contained Schur seam). With the notation above,

$$s_1\ge 4tr\widehat{A}-{\parallel \widehat{A}\parallel }_{\mathrm{HS}}^2-2N\left(I^{\prime }\right)+\mathcal{S}.$$

(101)

Consequently,

$$N_0^s\left(I\right)\ge 4tr\widehat{A}-{\parallel \widehat{A}\parallel }_{\mathrm{HS}}^2-2N\left(I\right)-3N(I^{\prime }\setminus I)+\mathcal{S}.$$

(102)

Proof. 

Apply Lemma 12 to $P_1+Q^{\prime }$. If $x_j={\parallel v_j\parallel }^2$, then by (100),

$$\parallel \widehat{A}{\parallel }_{\mathrm{HS}}^2\ge 2tr{P}_1+\sum _{j}g\left(x_j\right)+\mathcal{S}+4tr{Q}^{\prime }-4(s_2+p).$$

Using $\widehat{A}=P_1+Q^{\prime }$ this becomes

$$\parallel \widehat{A}{\parallel }_{\mathrm{HS}}^2\ge 4tr\widehat{A}-2tr{P}_1+\sum _{j}g\left(x_j\right)+\mathcal{S}-4(s_2+p).$$

Here $0\le x_j\le 1$, so $g\left(x_j\right)=x_j^2-2x_j\ge -1$, while $tr{P}_1={\sum }_j{x}_j\le s_1$. Therefore

$$\parallel \widehat{A}{\parallel }_{\mathrm{HS}}^2\ge 4tr\widehat{A}-3s_1+\mathcal{S}-4(s_2+p).$$

From (99), $4(s_2+p)\le 2N\left(I^{\prime }\right)-2s_1$, and rearrangement proves (101). Finally $N_0^s\left(I\right)\ge s_1-N(I^{\prime }\setminus I)$ and $N\left(I^{\prime }\right)\le N\left(I\right)+N(I^{\prime }\setminus I)$, which gives (102).    □

Corollary 1 

(Analytic insertion into the seam). Fix $0<\lambda <1$. For the beta–Widder shape ${\psi }_{†}$, implemented in the compression by the harmless scaling (64),

$${\displaystyle \frac{N_0^s(T,2T)}{N(T,2T)}}\ge 2-R_{\lambda }\left({\psi }_{†}\right)+{\displaystyle \frac{\mathcal{S}}{N(T,2T)}}-O_{\lambda ,\chi }\left({\delta }_{\lambda }\left(T\right)\right),$$

(103)

where, for example, one may take

$${\delta }_{\lambda }\left(T\right):={\displaystyle \frac{1}{\ell }}+{\displaystyle \frac{\ell log\ell }{T}}+{\displaystyle \frac{T^{\lambda -1}log\ell }{\ell }}+T^{\lambda /2-1}\ell +T^{-1/2},{\delta }_{\lambda }\left(T\right)⟶0.$$

(104)

Proof. 

Let $\widehat{G}$ be the normalized full finite-compression matrix and $\widehat{E}$ the omitted zero tail, so that $\widehat{A}=\widehat{G}-\widehat{E}$. The proof of ([11], Proposition 4.2) gives, before the harmless specialization $X\le T$,

$$tr\widehat{A}=N\left(I^{\prime }\right)+O_{\chi }\left(X^{1/2}{L}^2\right).$$

(105)

Since $\widehat{G}=\widehat{A}+\widehat{E}$ and (94) gives $B_0:={\parallel \widehat{E}\parallel }_1\ll X^{1/2}/T$, we also have

$$tr\widehat{G}=N\left(I^{\prime }\right)+O_{\chi }\left(X^{1/2}{L}^2\right),$$

(106)

with the tail contribution absorbed in the displayed error. Hence

$$\begin{array}{cc}\hfill 4tr\widehat{A}-{\parallel \widehat{A}\parallel }_{\mathrm{HS}}^2& \ge 4tr\widehat{G}-{\parallel \widehat{G}\parallel }_{\mathrm{HS}}^2\hfill \end{array}$$

(107)

$$\begin{array}{c}\hfill -B_0\left(4+2\parallel \widehat{G}{\parallel }_{\mathrm{HS}}+B_0\right).\end{array}$$

(108)

Indeed this follows from $|tr\widehat{E}|\le \parallel \widehat{E}{\parallel }_1$ and Frobenius Cauchy–Schwarz. Proposition 3 implies $\parallel \widehat{G}{\parallel }_{\mathrm{HS}}=O\left(\sqrt{N(T,2T)}\right)$, so the last line of (108) is $o\left(N\right(T,2T\left)\right)$ for every fixed $\lambda <1$.

Now insert (106), (87), and $N(I^{\prime }\setminus I)=O(T^{1/2}\ell )$ into (102). Dividing by $N(T,2T)\asymp T\ell$ gives (103). The normalized trace error is $O(T^{\lambda /2-1}\ell )$, the boundary error is $O\left(T^{-1/2}\right)$, and the Hilbert–Schmidt errors are bounded by (88); these are all dominated by (104). This proves the claimed fixed-$\lambda$ error with no appeal to an external Schur seam.    □

4.8. Triple Packing and the Transfer Theorem

We now combine Lemma 9 with the self-contained seam of Proposition 4 and the analytic insertion in Corollary 1. The elementary spacing step is stated separately.

Lemma 13 

(Three-offset packing). Let $0<y_1<\dots <y_m$ lie in an interval of length X. Among the three partitions into consecutive triples obtained by the three possible starting offsets, one has

$$\sum _{B}span\left(B\right)\le {\displaystyle \frac{2X}{3}},$$

(109)

where the sum runs over its complete triple blocks. The number of complete blocks is $m/3+O\left(1\right)$.

Proof. 

Write $d_j=y_{j+1}-y_j$. In a fixed offset partition a gap contributes to a block span exactly when it lies inside a triple rather than between two triples. Across the three offsets, every gap away from incomplete end blocks is counted exactly twice, while an end gap is counted at most twice. Hence the sum of the three total block-spans is at most $2{\sum }_j{d}_j=2(y_m-y_1)\le 2X$. One offset therefore satisfies (109).    □

Let

$$A_{*}:={\displaystyle \frac{171}{40000}},B_{*}:={\displaystyle \frac{1}{800}},{\displaystyle \frac{A_{*}}{B_{*}}}=3.42.$$

Proposition 5 

(Rigorous three-point transfer). Fix $0<\lambda <1$. Suppose

$$\underset{T\to \infty }{lim\; inf}{\displaystyle \frac{N_0^s(T,2T)}{N(T,2T)}}\ge a.$$

(110)

Then the Schur remainder $\mathcal{S}$ in the strengthened finite-compression seam satisfies

$${\displaystyle \frac{\mathcal{S}}{N(T,2T)}}\ge {\displaystyle \frac{A_{*}}{3}}a-{\displaystyle \frac{2B_{*}}{3}}\lambda -o\left(1\right).$$

(111)

Consequently,

$$\underset{T\to \infty }{lim\; inf}{\displaystyle \frac{N_0^s(T,2T)}{N(T,2T)}}\ge 2-R_{\lambda }\left({\psi }_{†}\right)+{\displaystyle \frac{A_{*}}{3}}a-{\displaystyle \frac{2B_{*}}{3}}\lambda .$$

(112)

Proof. 

Fix $\epsilon >0$. For all sufficiently large T, after removing the two boundary strips of width $T^{1/2}$, the core contains

$$m\ge (a-\epsilon -o(1\left)\right)N(T,2T)$$

simple critical-line zeros. Order their ordinates and put $y_j=L{\gamma }_j/\left(2\pi \right)$. By Riemann–von Mangoldt,

$$y_m-y_1\le (\lambda +o\left(1\right))N(T,2T).$$

(113)

Choose the offset in Lemma 13. Thus the number of complete blocks is ${\displaystyle \frac{1}{3}}(a-\epsilon -o\left(1\right))N(T,2T)$ and

$$\sum _{B}span\left(B\right)\le \left({\displaystyle \frac{2\lambda }{3}}+o\left(1\right)\right)N(T,2T).$$

(114)

For a selected block $B=({\gamma }_1,{\gamma }_2,{\gamma }_3)$ let $M_B$ be the Gram block of the original normalized finite atoms (83), and write

$$x={\displaystyle \frac{L({\gamma }_2-{\gamma }_1)}{2\pi }},y={\displaystyle \frac{L({\gamma }_3-{\gamma }_2)}{2\pi }},s_B=x+y=span\left(B\right).$$

If $s_B\ge A_{*}/B_{*}=3.42$, Lemma 11 gives

$$trg\left(M_B\right)-\sum _{j\in B}g\left({\left(M_B\right)}_{jj}\right)\ge 0\ge A_{*}-B_{*}{s}_B.$$

Thus no asymptotic kernel approximation is needed for long blocks.

If $s_B<3.42$, then $x,y,x+y$ all belong to the fixed compact interval $[0,3.42]$. Lemma 10, followed by Lemma 11, gives uniformly over all such blocks

$$trg\left(M_B\right)-\sum _{j\in B}g\left({\left(M_B\right)}_{jj}\right)={\mathcal{S}}_{†}(x,y)+o\left(1\right)\ge A_{*}-B_{*}{s}_B-o\left(1\right),$$

where the last step is Lemma 9. Since there are $O\left(N\right(T,2T\left)\right)$ blocks and the $o\left(1\right)$ is uniform on the compact span range, the accumulated error is $o\left(N\right(T,2T\left)\right)$. Applying Lemma 8 with $P=P_1$ from (98) and summing over the disjoint blocks therefore yields

$$\mathcal{S}\ge A_{*}\#\left\{B\right\}-B_{*}\sum _B{s}_B-o\left(N\right)\ge \left({\displaystyle \frac{A_{*}}{3}}(a-\epsilon )-{\displaystyle \frac{2B_{*}}{3}}\lambda -o\left(1\right)\right)N.$$

Letting $\epsilon \downarrow 0$ proves (111).

Corollary 1 applies to this very same $\mathcal{S}$, because the vectors in the selected triple blocks are principal subfamilies of the simple atoms in $P_1$ from (98). Hence

$${\displaystyle \frac{N_0^s(T,2T)}{N(T,2T)}}\ge 2-R_{\lambda }\left({\psi }_{†}\right)+{\displaystyle \frac{\mathcal{S}}{N(T,2T)}}-o\left(1\right).$$

Insert (111) to obtain (112). This closes the transfer using only the explicitly stated Alpöge–Furman analytic estimates and the finite-dimensional argument proved above.    □

Letting $\lambda \uparrow 1$ only after $T\to \infty$ in (112) gives the bootstrap map

$$\mathcal{F}\left(a\right)=p_{†}+{\displaystyle \frac{171}{120000}}a-{\displaystyle \frac{1}{1200}}.$$

(115)

Its slope is $171/120000<1$. Notice that this order of limits requires no endpoint $\lambda =1$ tail estimate.

Theorem 3 

(Computer-assisted beta–Widder Schur–triple lower bound). Let $N_0^s(T,2T)$ denote the number of simple zeros of $\zeta \left(s\right)$ on the critical line with ordinates in $(T,2T]$, and let $N(T,2T)$ count all nontrivial zeros there with multiplicity. Then

$$\underset{T\to \infty }{lim\; inf}{\displaystyle \frac{N_0^s(T,2T)}{N(T,2T)}}\ge {\displaystyle \frac{93093085291}{138402495000}}=0.672625773769468\dots .$$

(116)

The same lower bound holds for the cumulative ratio $N_0^s(0,T)/N(0,T)$.

Proof. 

Alpöge–Furman ([11], Theorem A) supply an unconditional starting value exceeding $2/3$, so the bootstrap is applicable. If a lower bound a is known, Proposition 5, followed by $\lambda \uparrow 1$, gives the improved lower bound $\mathcal{F}\left(a\right)$ in (115). Iterating the contraction yields the unique fixed point

$$a_{*}={\displaystyle \frac{p_{†}-1/1200}{1-171/120000}}={\displaystyle \frac{93093085291}{138402495000}}.$$

The cumulative form follows by the standard dyadic summation used in ([11], Theorem A and Remark 6.1); no additional finite-dimensional input is needed.    □

Corollary 2 

(Improved smoothed simple-zero density). In the notation of the smoothing section, for every fixed $L>0$,

$$\underset{x\to \infty }{lim\; inf}\underset{\begin{array}{c}m\in \mathbb{N}\\ 1\le m\le Lx\end{array}}{inf}{\displaystyle \frac{S_m\left(x\right)}{A_m\left(x\right)}}\ge {\displaystyle \frac{93093085291}{138402495000}}=0.672625773769468\dots .$$

(117)

Proof. 

Apply Proposition 1 with the cumulative density bound furnished by Theorem 3.    □

Remark 4 

(Relation to the recent literature). The literature status in this remark was checked on 15 September 2026. The Montgomery–Taylor value $3/2-2^{-1/2}cot\left(2^{-1/2}\right)=0.672500703679\dots$ appears in Alpöge–Furman [11] and in Lamzouri’s shorter proof [13]. Shi [12] retains a positive Schur remainder and selects a matching of pairs. Wang [14] develops a short-interval extension of the Lamzouri method. The present argument instead uses the Alpöge–Furman finite-compression analytic input, an affine estimate on disjoint triples, and a window chosen from the compactified beta–Widder family. The value in (116) is therefore presented only as the output of this beta–Widder three-point mechanism, not as a record claim. A non-peer-reviewed Zenodo preprint of Devine [15], for example, reports the larger computer-assisted value $0.673399$. We neither use nor validate that separate claim here; it is cited only to delimit the numerical-priority statement.

Appendix A.1. Superdiagonal Localization and High-Order Witnesses

The preceding estimate leaves open the regime in which the beta kernel becomes microscopic. In that regime one should not expect a further unconditional positivity theorem: nonreal reflected zeros are exponentially amplified by high Widder orders. The following proposition makes this precise for the genus-zero framework used above.

For a reflected parameter w and $x>0$ put

$$Q_w\left(x\right):=\left|{\displaystyle \frac{w}{{(x+w)}^2}}\right|.$$

If $w=r{e}^{i\varphi }$ with $\left|\varphi \right|<\pi /2$, then

$$Q_w\left(x\right)={\displaystyle \frac{r}{x^2+2xrcos\varphi +r^2}}.$$

(A1)

For a positive real parameter one always has $Q_w\left(x\right)\le {\left(4x\right)}^{-1}$, whereas for a nonreal parameter

$$Q_w\left(r\right)={\displaystyle \frac{1}{2r(1+cos\varphi )}}={\displaystyle \frac{1}{4r{cos}^2(\varphi /2)}}>{\displaystyle \frac{1}{4r}}.$$

(A2)

Theorem A1 

(A nonreal zero forces arbitrarily high negative orders). Let

$$G\left(x\right)=G\left(0\right)\prod _j(1+x/w_j),\sum _j{\left|w_j\right|}^{-1}<\infty ,$$

be a real genus-zero product whose reflected parameters satisfy $\Re w_j>0$ and occur in conjugate pairs. Put $F=2G^{\prime }/G$. If at least one $w_j$ is nonreal, then there exists $x_{*}>0$ and infinitely many integers $m\to \infty$ for which

$${\mathcal{W}}_m\left[F\right]\left(x_{*}\right)<0.$$

(A3)

More precisely, one can choose $x_{*}$, a nonreal conjugate pair $w_{*},\overline{w_{*}}$ of multiplicity $\nu \ge 1$, a number $Q_{*}>0$, an angle ${\theta }_{*}\neg \equiv 0(mod2\pi )$, and $0<\rho <1$ such that

$${\displaystyle \frac{{\mathcal{W}}_m\left[F\right]\left(x_{*}\right)}{2(2m-1)!}}=2\nu Q_{*}^{m}cos\left(m{\theta }_{*}\right)+O\left({\left(\rho Q_{*}\right)}^m\right).$$

(A4)

Proof. 

Choose one nonreal parameter $w_0=r_0{e}^{i{\varphi }_0}$. At $x_0=r_0$, (A2) shows that $Q_{w_0}\left(x_0\right)>{\left(4x_0\right)}^{-1}$, while every positive real parameter has $Q_w\left(x_0\right)\le {\left(4x_0\right)}^{-1}$. Moreover $Q_{w_j}\left(x_0\right){\ll }_{x_0}{\left|w_j\right|}^{-1}$ as $|w_j|\to \infty$; hence the maximum of $Q_{w_j}\left(x_0\right)$ is attained, and every maximizing class is nonreal.

On a sufficiently small compact interval about $x_0$, only finitely many conjugate-pair classes can compete with this maximum. Indeed the tail is uniformly small there, by ${\sum }_j{\left|w_j\right|}^{-1}<\infty$. For two distinct classes the functions $Q_w\left(x\right)$ in (A1) are distinct real-analytic functions: equality identically would force the same r and the same $\Re w$, hence the same conjugate pair. Their tie points are therefore discrete. We may consequently choose $x_{*}$ arbitrarily near $x_0$ so that one nonreal conjugate pair $w_{*},\overline{w_{*}}$ is the unique maximizing class. We also choose $x_{*}\ne \left|w_{*}\right|$.

Write

$${\displaystyle \frac{w_{*}}{{(x_{*}+w_{*})}^2}}=Q_{*}{e}^{i{\theta }_{*}}.$$

For $w_{*}=r{e}^{i\varphi }$ with $0<\left|\varphi \right|<\pi /2$, the argument is zero exactly when $x_{*}=r$: indeed $arg(x_{*}+w_{*})=\varphi /2$ is equivalent to $x_{*}=r$. Thus our choice gives ${\theta }_{*}\ne 0$. Uniqueness of the maximizing class and uniform control of the tail give a number $\rho <1$ such that the sum of all remaining terms in (12) is $O\left({\left(\rho Q_{*}\right)}^m\right)$. The dominant conjugate pair contributes $2\nu Q_{*}^{m}cos\left(m{\theta }_{*}\right)$, proving (A4).

Finally, if ${\theta }_{*}/\left(2\pi \right)$ is irrational, the sequence $m{\theta }_{*}(mod2\pi )$ is dense; if it is rational and nonzero, it is periodic and contains an angle with negative cosine. In either case there are infinitely many m for which $cos\left(m{\theta }_{*}\right)\le -c$ for some $c>0$. For all sufficiently large such m, the exponentially smaller remainder in (A4) cannot change the sign. This proves (A3).    □

For the Riemann xi-function all reflected parameters satisfy $\Re w_{\rho }>0$, since $\gamma >14$ and $|\beta -1/2|<1/2$. Hence Theorem A1 applies directly.

Corollary A1 

(Eventual-in-order criterion for RH). The following are equivalent:

1.

the Riemann hypothesis holds;

2.

for every $x>0$ there exists an integer $M_x$ such that

$${\mathcal{W}}_m\left[F_{\xi }\right]\left(x\right)\ge 0(m\ge M_x);$$

3.

for every $x>0$ there exists an integer $M_x$ such that

$$A_m^{\prime }\left(x\right)\ge 0(m\ge M_x).$$

If RH is false, there is in fact a single $x_{*}>0$ at which both inequalities fail for infinitely many arbitrarily large orders.

Proof. 

Under RH, $F_{\xi }$ is Stieltjes and every Widder inequality is nonnegative. Conversely, if RH is false then some reflected parameter $w_{\rho }$ is nonreal, so Theorem A1 gives a fixed $x_{*}$ and infinitely many negative Widder orders. The equivalence with the derivative statement uses

$$A_m^{\prime }\left(x\right)={\displaystyle \frac{x^{m-1}}{2(2m-1)!B_m}}{\mathcal{W}}_m\left[F_{\xi }\right]\left(x\right),$$

whose prefactor is positive.    □

Remark A1 

(Interpretation of the superdiagonal barrier). The beta density $p_m\left(y\right)$ concentrates at $y=1$ on the scale $m^{-1/2}$. Indeed, uniformly for z in compact subsets of $\mathbb{C}$,

$${\displaystyle \frac{1}{\sqrt{m}}}{p}_m\left(1+{\displaystyle \frac{z}{\sqrt{m}}}\right)⟶{\displaystyle \frac{1}{2\sqrt{\pi }}}{e}^{-z^2}.$$

(A5)

Thus at $x=T^2$ the ordinate window has width $T/\sqrt{m}$, while a horizontal displacement $\delta =\beta -1/2$ enters on the amplified scale $\sqrt{m}\delta /T$. Theorem 2 controls the broad kernel regime. Theorem A1 shows why the opposite, microscopic regime cannot be made unconditionally positive without resolving RH itself.

Appendix A.2. An Exact Prime-Side High-Order Criterion

The preceding high-order witness theorem can be transported through the Guinand–Weil formula to a criterion involving only prime powers and values of $-{\zeta }^{\prime }/\zeta$ in its absolutely convergent half-plane. This also identifies quantitatively the cancellation that is missing from a purely absolute-value estimate on the prime side.

For $m\ge 1$ put

$${\Phi }_m\left(v\right):={\displaystyle \frac{1}{2\pi }}{\int }_{-\infty }^{\infty }{p}_m\left(y\right)e^{-ivy}dy(v\in \mathbb{R}).$$

Since $p_m$ is even, ${\Phi }_m$ is real and even. For $v\ge 0$, closing the contour in the lower half-plane shows that

$${\Phi }_m\left(v\right)=e^{-v}{R}_m\left(v\right),$$

(A6)

where $R_m$ is the real polynomial of degree $2m-1$ defined by

$$R_m\left(v\right):=-i{e}^v\underset{z=-i}{Res}\left({\displaystyle \frac{z^{2m}{e}^{-ivz}}{B_m{(1+z^2)}^{2m}}}\right).$$

(A7)

For example,

$$R_1\left(v\right)={\displaystyle \frac{1-v}{4}},R_2\left(v\right)={\displaystyle \frac{v^3-6v^2+3v+3}{16}}.$$

By scaling,

$$g_{m,T}\left(u\right)=T{\Phi }_m\left(T\right|u\left|\right)=T{e}^{-T\left|u\right|}{R}_m\left(T\right|u\left|\right).$$

(A8)

For fixed $T\ge 2$ define the prime functional

$${\mathcal{P}}_m\left(T\right):=2T{\displaystyle \sum _{n=2}^{\infty }}{\displaystyle \frac{\Lambda \left(n\right)}{n^{T+1/2}}}{R}_m(Tlogn).$$

(A9)

The sum converges absolutely for each fixed m. If

$$Q\left(s\right):=-{\displaystyle \frac{{\zeta }^{\prime }}{\zeta }}\left(s\right)={\displaystyle \sum _{n=2}^{\infty }}{\displaystyle \frac{\Lambda \left(n\right)}{n^s}},\Re s>1,$$

and $D=-d/ds$, then equivalently

$${\mathcal{P}}_m\left(T\right)=2T{\left[R_m\left(TD\right)Q\left(s\right)\right]}_{s=T+1/2}.$$

(A10)

Thus every quantity in ${\mathcal{P}}_m\left(T\right)$ is evaluated strictly inside the ordinary Dirichlet-series half-plane.

Proposition A1 

(Spectral radius of the fixed-T Widder sequence). For fixed $T>0$ put

$$\mathcal{R}\left(T\right):=4T^2\underset{\rho \in {\mathcal{Z}}_{+}}{max}\left|{\displaystyle \frac{w_{\rho }}{{(T^2+w_{\rho })}^2}}\right|.$$

(A11)

Then the maximum is attained and

$$\underset{m\to \infty }{lim\; sup}{\left|T^2{A}_m^{\prime }\left(T^2\right)\right|}^{1/m}=\mathcal{R}\left(T\right).$$

(A12)

Under RH, $\mathcal{R}\left(T\right)\le 1$ for every $T>0$. If RH is false, then $\mathcal{R}\left(T\right)>1$ on a nonempty open interval of T-values.

Proof. 

Set

$$q_{\rho }\left(T\right):={\displaystyle \frac{w_{\rho }}{{(T^2+w_{\rho })}^2}}.$$

For fixed T, one has ${\sum }_{\rho }\left|q_{\rho }\left(T\right)\right|<\infty$ and $q_{\rho }\left(T\right)\to 0$, so $Q\left(T\right):={max}_{\rho }\left|q_{\rho }\left(T\right)\right|$ is attained. Put $S_m\left(T\right)={\sum }_{\rho }{q}_{\rho }{\left(T\right)}^m$. The generating function

$$\sum _{m\ge 1}{S}_m\left(T\right)z^m=\sum _{\rho }{\displaystyle \frac{q_{\rho }\left(T\right)z}{1-q_{\rho }\left(T\right)z}}$$

converges normally for ${\left|z\right|<Q\left(T\right)}^{-1}$. At a point $z=q_{*}^{-1}$ corresponding to a maximizing value $q_{*}$, the finitely many identical maximizing terms give a genuine pole; all other terms are analytic there after separating the uniformly small tail. Hence the radius of convergence is exactly $Q{\left(T\right)}^{-1}$. Cauchy–Hadamard therefore gives

$$\underset{m\to \infty }{lim\; sup}{\left|S_m\left(T\right)\right|}^{1/m}=Q\left(T\right).$$

Since

$$T^2{A}_m^{\prime }\left(T^2\right)={\displaystyle \frac{T^{2m}}{B_m}}{S}_m\left(T\right),B_m^{-1/m}⟶4,$$

we obtain (A12).

Under RH every $w_{\rho }>0$, and $4T^2{w}_{\rho }/{(T^2+w_{\rho })}^2\le 1$ by the arithmetic–geometric mean inequality. Conversely, if $w=r{e}^{i\varphi }$ is nonreal, then at $T^2=r$, (A2) gives $4T^2{Q}_w\left(T^2\right)>1$. Continuity preserves the strict inequality on a neighborhood, proving the last assertion.    □

Lemma A1 

(Fixed-T control of the nonprime explicit-formula terms). Fix $T\ge 2$. Define

$$\begin{array}{cc}\hfill {\mathcal{H}}_m\left(T\right)& :=2h_{m,T}(i/2)-g_{m,T}\left(0\right)log\pi \hfill \\ \hfill & +{\displaystyle \frac{1}{2\pi }}{\int }_{-\infty }^{\infty }{h}_{m,T}\left(r\right){\displaystyle \frac{{\Gamma }^{\prime }}{\Gamma }}\left({\displaystyle \frac{1}{4}}+{\displaystyle \frac{ir}{2}}\right)dr.\hfill \end{array}$$

(A13)

Then, for every $m\ge 1$,

$$2T^2{A}_m^{\prime }\left(T^2\right)={\mathcal{H}}_m\left(T\right)-{\mathcal{P}}_m\left(T\right),$$

(A14)

and

$$\underset{m\ge 1}{sup}\left|{\mathcal{H}}_m\left(T\right)\right|<\infty .$$

(A15)

Consequently, as $m\to \infty$,

$$4^m{B}_m{\mathcal{H}}_m\left(T\right)=O_T\left(m^{-1/2}\right).$$

(A16)

Proof. 

Because $h_{m,T}$ is even and real on $\mathbb{R}$, while $\psi \left(\overline{z}\right)=\overline{\psi \left(z\right)}$ for $\psi ={\Gamma }^{\prime }/\Gamma$, pairing r with $-r$ shows that

$${\int }_{-\infty }^{\infty }{h}_{m,T}\left(r\right){\displaystyle \frac{{\Gamma }^{\prime }}{\Gamma }}\left({\displaystyle \frac{1}{4}}+{\displaystyle \frac{ir}{2}}\right)dr={\int }_{-\infty }^{\infty }{h}_{m,T}\left(r\right)\Re {\displaystyle \frac{{\Gamma }^{\prime }}{\Gamma }}\left({\displaystyle \frac{1}{4}}+{\displaystyle \frac{ir}{2}}\right)dr.$$

Thus the identity (A14) is exactly (51)–(52), because (A8) identifies the prime term with ${\mathcal{P}}_m\left(T\right)$.

It remains to prove a bound uniform in the order. First,

$$g_{m,T}\left(0\right)={\displaystyle \frac{T}{2\pi }}{\int }_{-\infty }^{\infty }{p}_m\left(y\right)dy={\displaystyle \frac{T{c}_m}{2\pi }}=O_T\left(1\right),$$

since $1\le c_m\le \pi /2$. Next put $u={\left(2T\right)}^{-1}\le 1/4$. Then

$$|h_{m,T}(i/2)|={\displaystyle \frac{1}{4^m{B}_m}}{\left({\displaystyle \frac{4u^2}{{(1-u^2)}^2}}\right)}^m.$$

The quantity in parentheses is strictly less than 1, while ${\left(4^m{B}_m\right)}^{-1}=O\left(\sqrt{m}\right)$ by Stirling’s formula. Thus $h_{m,T}(i/2)=O_T\left(1\right)$ (in fact it decays exponentially with m).

For the gamma-factor term we use the standard bound

$$\left|{\displaystyle \frac{{\Gamma }^{\prime }}{\Gamma }}\left({\displaystyle \frac{1}{4}}+{\displaystyle \frac{ir}{2}}\right)\right|\ll 1+log(2+|r\left|\right),r\in \mathbb{R}.$$

After $r=Ty$ it is therefore enough to bound

$$T{\int }_{-\infty }^{\infty }{p}_m\left(y\right)\left(1+log(2+T|y\left|\right)\right)dy$$

uniformly in m. The part $\left|y\right|\le 1$ is $O_T\left(1\right)$ because ${\int }_{\mathbb{R}}{p}_m\left(y\right)dy=c_m=O\left(1\right)$. For $m\ge 2$ and $y\ge 1$, $log(2+Ty){\ll }_{T}y$, and the beta integral gives

$${\int }_0^{\infty }y{p}_m\left(y\right)dy={\displaystyle \frac{B(m+1,m-1)}{2B(m,m)}}={\displaystyle \frac{m}{2(m-1)}}\le 1.$$

The case $m=1$ is a single fixed integrable function and is absorbed into the constant. This proves (A15). Finally $4^m{B}_m=2\sqrt{\pi }{m}^{-1/2}(1+O\left(m^{-1}\right))$, so (A16) follows.    □

Theorem A2 

(Prime-side root-growth criterion for RH). The Riemann hypothesis is equivalent to

$$\underset{m\to \infty }{lim\; sup}{\left|{\mathcal{P}}_m\left(T\right)\right|}^{1/m}\le 1\mathrm{for}\mathrm{every}\mathrm{fixed}T\ge 2.$$

(A17)

If RH is false, one may choose a fixed $T>14$ and constants $R_{*}>1$, ${\theta }_{*}\neg \equiv 0(mod2\pi )$, $\nu \ge 1$, and $0<\rho <1$ such that, along all integers $m\ge 1$,

$${\mathcal{P}}_m\left(T\right)=-{\displaystyle \frac{4\nu }{B_m}}{\left({\displaystyle \frac{R_{*}}{4}}\right)}^{m}cos\left(m{\theta }_{*}\right)+O\left({\displaystyle \frac{1}{B_m}}{\left({\displaystyle \frac{\rho R_{*}}{4}}\right)}^m\right)+O_T\left(1\right).$$

(A18)

In particular

$$\underset{m\to \infty }{lim\; sup}{\left|{\mathcal{P}}_m\left(T\right)\right|}^{1/m}=R_{*}>1.$$

(A19)

Proof. 

By Lemma A1, the Guinand–Weil formula gives the exact identity

$$2T^2{A}_m^{\prime }\left(T^2\right)={\mathcal{H}}_m\left(T\right)-{\mathcal{P}}_m\left(T\right),{\mathcal{H}}_m\left(T\right)=O_T\left(1\right),$$

with a constant independent of m.

Assume RH. Then every ${\Gamma }_{\rho }=\gamma$ is real and

$$p_m(\gamma /T)={\displaystyle \frac{1}{B_m{4}^m}}{\left({\displaystyle \frac{4{(\gamma /T)}^2}{{(1+{(\gamma /T)}^2)}^2}}\right)}^m.$$

The quantity in parentheses lies in $(0,1]$, and its sum over positive zero ordinates is finite, since it is $O_T\left({\gamma }^{-2}\right)$. Moreover ${\left(B_m{4}^m\right)}^{-1}=O\left(\sqrt{m}\right)$. Hence

$$T^2{A}_m^{\prime }\left(T^2\right)=O_T\left(\sqrt{m}\right),$$

and (A14) implies ${\mathcal{P}}_m\left(T\right)=O_T\left(\sqrt{m}\right)$. This proves (A17).

Conversely suppose RH is false. In the proof of Theorem A1 choose $x_{*}=T^2$ sufficiently close to the modulus of the selected nonreal reflected parameter so that, in addition to unique dominance,

$$R_{*}:=4T^2{Q}_{*}>1.$$

For the xi-function T may be taken larger than 14. Since

$$p_m(\Gamma /T)={\displaystyle \frac{1}{B_m}}{\left(T^2{\displaystyle \frac{w}{{(T^2+w)}^2}}\right)}^m,$$

the dominant-pair argument gives

$$T^2{A}_m^{\prime }\left(T^2\right)={\displaystyle \frac{2\nu }{B_m}}{\left({\displaystyle \frac{R_{*}}{4}}\right)}^{m}cos\left(m{\theta }_{*}\right)+O\left({\displaystyle \frac{1}{B_m}}{\left({\displaystyle \frac{\rho R_{*}}{4}}\right)}^m\right).$$

Substitution into (A14) proves (A18). Finally $B_m^{-1}=4^m\sqrt{m}{\left(2\sqrt{\pi }\right)}^{-1}(1+O\left(m^{-1}\right))$, while $|cos(m{\theta }_{*}\left)\right|$ is bounded below by a positive constant for infinitely many m. This gives (A19) and completes the converse.    □

The next proposition quantifies the horizontal-contour bound obtained without using cancellation between prime powers.

Proposition A2 

(Optimized absolute-convergence bound). For every fixed $T\ge 2$,

$$\underset{m\to \infty }{lim\; sup}{\left|{\mathcal{P}}_m\left(T\right)\right|}^{1/m}\le \mathcal{M}\left(T\right):={\displaystyle \frac{2}{1+\sqrt{1-T^{-2}}}}.$$

(A20)

In particular

$$\mathcal{M}\left(T\right)=1+{\displaystyle \frac{1}{4T^2}}+O\left(T^{-4}\right).$$

(A21)

Thus the absolute-value method leaves a strictly positive exponential gap between the unconditional bound and the RH threshold 1.

Proof. 

Fix $a>1/2$ sufficiently close to $1/2$ that $a/T\le 1/3$, and put $\sigma =a/T<1/2$. Shifting the Fourier contour to $\Im r=-a$ as in (55) gives

$$|g_{m,T}\left(u\right)|\le {\displaystyle \frac{T{e}^{-au}}{2\pi }}{\int }_{-\infty }^{\infty }\left|p_m(y-i\sigma )\right|dy.$$

Set

$${\mathcal{R}}_{\sigma }\left(y\right):={\displaystyle \frac{4(y^2+{\sigma }^2)}{{(1+y^2-{\sigma }^2)}^2+4y^2{\sigma }^2}}.$$

Then

$$|p_m(y-i\sigma )|={\displaystyle \frac{{\mathcal{R}}_{\sigma }{\left(y\right)}^m}{B_m{4}^m}}.$$

For $0\le \sigma \le 1/3$ a one-variable maximization, with $t=y^2$, gives

$$\underset{y\in \mathbb{R}}{sup}{\mathcal{R}}_{\sigma }\left(y\right)={\displaystyle \frac{2}{1+\sqrt{1-4{\sigma }^2}}}.$$

(A22)

Indeed the interior critical point satisfies $t+{\sigma }^2=\sqrt{1-4{\sigma }^2}$, and substitution yields (A22). Since ${\mathcal{R}}_{\sigma }\left(y\right)=O_{\sigma }\left(y^{-2}\right)$ at infinity, choose a fixed Y so large that the tail is bounded by $C_{\sigma }{y}^{-2}$ with $C_{\sigma }/Y^2$ strictly smaller than the maximum in (A22). Splitting at $\left|y\right|=Y$ then shows

$${\int }_{-\infty }^{\infty }{\mathcal{R}}_{\sigma }{\left(y\right)}^{m}dy{\ll }_{\sigma }{\left({\displaystyle \frac{2}{1+\sqrt{1-4{\sigma }^2}}}\right)}^m.$$

Together with ${\left(B_m{4}^m\right)}^{-1}=O\left(\sqrt{m}\right)$ this yields

$$|g_{m,T}\left(u\right)|{\ll }_{T,a}T\sqrt{m}{e}^{-au}{\left({\displaystyle \frac{2}{1+\sqrt{1-4a^2/T^2}}}\right)}^m.$$

Since $a>1/2$,

$${\displaystyle \sum _{n=2}^{\infty }}{\displaystyle \frac{\Lambda \left(n\right)}{n^{1/2+a}}}<\infty .$$

Applying this to (A9) and taking mth roots gives

$$\underset{m\to \infty }{lim\; sup}{\left|{\mathcal{P}}_m\left(T\right)\right|}^{1/m}\le {\displaystyle \frac{2}{1+\sqrt{1-4a^2/T^2}}}.$$

Letting $a\downarrow 1/2$ proves (A20); the expansion (A21) is immediate.    □

Remark A2 

(What prime cancellation would have to prove). Theorem A2 and Proposition A2 isolate the remaining arithmetic problem sharply. Absolute convergence reaches the boundary $\mathcal{M}\left(T\right)>1$, corresponding to shifting just beyond the line required to sum $\Lambda \left(n\right)n^{-1/2-a}$ absolutely. Proving RH through this route would require cancellation in the oscillatory prime functional strong enough to replace $\mathcal{M}\left(T\right)$ by 1 for every fixed $T\ge 2$. This is not supplied by the present argument and should not be interpreted as a proof of RH.

Appendix A.3. Summing the Widder Order: a Prime Generating-Function Criterion

The preceding root-growth criterion can be packaged into a single generating function. Besides being more economical, this formulation identifies the absolute-convergence barrier geometrically.

Put

$$q\left(y\right):={\displaystyle \frac{y^2}{{(1+y^2)}^2}}.$$

Since

$${\displaystyle \frac{1}{B(m,m)}}={\displaystyle \frac{m}{2}}\left({\displaystyle \genfrac{}{}{0pt}{}{2m}{m}}\right),$$

the central-binomial generating function gives, for $\left|4zq\right(y\left)\right|<1$,

$$\sum _{m\ge 1}{p}_m\left(y\right)z^m={\displaystyle \frac{zq\left(y\right)}{{\left(1-4zq\left(y\right)\right)}^{3/2}}}.$$

(A23)

We denote the right-hand side by

$${\mathcal{K}}_z\left(y\right):={\displaystyle \frac{zq\left(y\right)}{{\left(1-4zq\left(y\right)\right)}^{3/2}}},$$

where the branch is chosen to equal 1 at $z=0$.

For fixed $T\ge 2$ define the two Taylor germs

$${\mathfrak{A}}_T\left(z\right):=\sum _{m\ge 1}{T}^2{A}_m^{\prime }\left(T^2\right)z^m,{\mathfrak{P}}_T\left(z\right):=\sum _{m\ge 1}{\mathcal{P}}_m\left(T\right)z^m.$$

(A24)

If

$$a_{\rho }\left(T\right):={\displaystyle \frac{T^2{w}_{\rho }}{{(T^2+w_{\rho })}^2}},$$

then, in the disk of absolute convergence,

$${\mathfrak{A}}_T\left(z\right)=\sum _{\rho \in {\mathcal{Z}}_{+}}{\displaystyle \frac{z{a}_{\rho }\left(T\right)}{{\left(1-4z{a}_{\rho }\left(T\right)\right)}^{3/2}}}.$$

(A25)

In particular Proposition A1 may be restated as

$$rad\left({\mathfrak{A}}_T\right)=\mathcal{R}{\left(T\right)}^{-1}.$$

(A26)

The same summation can be performed directly on the prime side. For $0<a\le 1/3$ set

$$r\left(a\right):={\displaystyle \frac{1+\sqrt{1-4a^2}}{2}}.$$

(A27)

The maximization in (A22) says exactly that, in this range,

$$\underset{y\in \mathbb{R}}{sup}4\left|q(y-ia)\right|=r{\left(a\right)}^{-1}.$$

(A28)

The restriction $a\le 1/3$ is more than sufficient below and keeps us inside the range in which the interior maximizer used in (A22) is valid.

Theorem A3 

(Prime generating germ and the unit-disk RH criterion). Fix $T\ge 2$ and put

$$r_{\mathrm{abs}}\left(T\right):={\displaystyle \frac{1+\sqrt{1-T^{-2}}}{2}}=\mathcal{M}{\left(T\right)}^{-1}.$$

(A29)

Then the following hold.

(i)The prime Taylor germ ${\mathfrak{P}}_T$ is holomorphic for $\left|z\right|<r_{\mathrm{abs}}\left(T\right)$. More precisely, if

$${\displaystyle \frac{1}{2T}}<a\le {\displaystyle \frac{1}{3}},\left|z\right|<r\left(a\right),$$

then

$${\mathfrak{P}}_T\left(z\right)={\displaystyle \frac{T}{\pi }}{\int }_{-\infty }^{\infty }{\mathcal{K}}_z(y-ia)Q\left({\displaystyle \frac{1}{2}}+aT+iTy\right)dy,$$

(A30)

where $Q\left(s\right)=-{\zeta }^{\prime }\left(s\right)/\zeta \left(s\right)$ and the integral is absolutely convergent. Thus the integral representation in (A30) uses Q only in the half-plane $\Re s>1$.

(ii)RH is equivalent to the assertion that, for every fixed $T\ge 2$, the germ ${\mathfrak{P}}_T$ extends holomorphically to the full unit disk $\left|z\right|<1$. It is already enough to require this for every rational $T>14$.

(iii)If RH is false, one can choose a rational $T>14$ for which ${\mathfrak{P}}_T$ has a pair of conjugate algebraic singularities inside the unit disk. After choosing T so that one nonreal reflected pair is uniquely dominant, these singularities occur at

$$z_{*}={\displaystyle \frac{1}{4a_{*}\left(T\right)}},\overline{z_{*}},\left|z_{*}\right|<1,$$

(A31)

and are of local type ${(1-z/z_{*})}^{-3/2}$ and its conjugate.

Proof. 

The identity (A23) follows from

$$\sum _{m\ge 0}\left({\displaystyle \genfrac{}{}{0pt}{}{2m}{m}}\right)t^m={(1-4t)}^{-1/2}$$

by applying ${\displaystyle \frac{1}{2}}td/dt$ and then taking $t=zq\left(y\right)$. Summing (48) therefore gives (A25) wherever the interchange is absolutely convergent. Equation (A26) also follows directly from Proposition A1 and Cauchy–Hadamard.

For the prime side, let

$${\widehat{\mathcal{K}}}_z\left(v\right):={\displaystyle \frac{1}{2\pi }}{\int }_{-\infty }^{\infty }{\mathcal{K}}_z\left(y\right)e^{-ivy}dy.$$

We justify the contour deformation in the full strip, not merely on its terminal line. Let

$${\mathcal{S}}_a:=\{y\in \mathbb{C}:-a\le \Im y\le 0\}.$$

For $0\le \sigma \le a$, formula (A28) gives

$$\underset{t\in \mathbb{R}}{sup}4\left|q(t-i\sigma )\right|=r{\left(\sigma \right)}^{-1}\le r{\left(a\right)}^{-1},$$

because $r\left(\sigma \right)$ is decreasing on $[0,1/3]$. Hence, if $\left|z\right|<r\left(a\right)$,

$$\underset{y\in {\mathcal{S}}_a}{sup}\left|4zq\left(y\right)\right|\le {\displaystyle \frac{\left|z\right|}{r\left(a\right)}}<1.$$

(A32)

The poles of q at $\pm i$ lie outside ${\mathcal{S}}_a$ since $a<1/2$. Moreover (A32) places $1-4zq\left(y\right)$ in the open disk centered at 1 of radius $\left|z\right|/r\left(a\right)<1$; thus it never meets the nonpositive real axis or 0. The principal branch of ${(1-4zq\left(y\right))}^{-3/2}$ is therefore holomorphic throughout the strip and agrees with the branch normalized to 1 at $z=0$. Uniformly for $-a\le \Im y\le 0$ one also has ${\mathcal{K}}_z\left(y\right)=O_{z,a}{\left(\right|\Re y|}^{-2})$ as $|\Re y|\to \infty$. The vertical sides of the usual rectangular contour consequently vanish, and for $v>0$ Cauchy’s theorem gives the rigorous shift

$${\widehat{\mathcal{K}}}_z\left(v\right)={\displaystyle \frac{e^{-av}}{2\pi }}{\int }_{-\infty }^{\infty }{\mathcal{K}}_z(y-ia)e^{-ivy}dy.$$

Using

$${\mathcal{P}}_m\left(T\right)=2T\sum _{n\ge 2}{\displaystyle \frac{\Lambda \left(n\right)}{\sqrt{n}}}{\Phi }_m(Tlogn)$$

and summing first over m now yields

$${\mathfrak{P}}_T\left(z\right)=2T\sum _{n\ge 2}{\displaystyle \frac{\Lambda \left(n\right)}{\sqrt{n}}}{\widehat{\mathcal{K}}}_z(Tlogn).$$

Because $aT>1/2$, the Dirichlet series

$$\sum _{n\ge 2}\Lambda \left(n\right)n^{-1/2-aT-iTy}=Q\left({\displaystyle \frac{1}{2}}+aT+iTy\right)$$

is absolutely convergent. Fubini’s theorem therefore gives (A30). Given any $\left|z\right|<r_{\mathrm{abs}}\left(T\right)$, one may choose $a\in (1/(2T),1/3]$ sufficiently close to $1/\left(2T\right)$ so that $\left|z\right|<r\left(a\right)$. (The interval is nonempty because $T\ge 2$.) This proves (i).

Assume RH. Theorem A2 gives

$$\underset{m\to \infty }{lim\; sup}{\left|{\mathcal{P}}_m\left(T\right)\right|}^{1/m}\le 1,$$

so the Taylor series in (A24) is holomorphic throughout $\left|z\right|<1$ for every $T\ge 2$.

Conversely, suppose that all the stated prime germs are holomorphic in the unit disk. Their Taylor series at the origin then have radius at least one, so Cauchy–Hadamard gives

$$\underset{m\to \infty }{lim\; sup}{\left|{\mathcal{P}}_m\left(T\right)\right|}^{1/m}\le 1.$$

Theorem A2 yields RH if this holds for every $T\ge 2$. If RH were false, Proposition A1 and the proof of Theorem A2 give a nonempty open interval of $T>14$ on which the prime root growth is larger than one. That interval contains a rational number, proving the countable version of (ii).

Finally suppose RH is false and choose, as in the high-order witness proof, a rational $T>14$ at which one nonreal conjugate reflected pair is uniquely dominant. Write $a_{*}=a_{{\rho }_{*}}\left(T\right)$ for one member of that pair and let $\nu \ge 1$ be its multiplicity. Then

$$z_{*}={\left(4a_{*}\right)}^{-1},\left|z_{*}\right|=\mathcal{R}{\left(T\right)}^{-1}<1.$$

Because $a_{\rho }\left(T\right)\to 0$ along the zero set, unique dominance implies that there is $\eta >0$ such that every nonidentical reflected parameter $a_{\rho }\left(T\right)$ satisfies $|4a_{\rho }\left(T\right)z_{*}|\le 1-\eta$. After removing the finitely many terms corresponding to the selected multiplicity, the remaining series in (A25) is normally convergent in a sufficiently small closed disk about $z_{*}$. Indeed ${\sum }_{\rho }\left|a_{\rho }\left(T\right)\right|<\infty$, and for the tail the factors ${(1-4z{a}_{\rho }\left(T\right))}^{-3/2}$ are uniformly bounded there. Hence the remainder is holomorphic near $z_{*}$. The singular part is exactly

$$\nu {\displaystyle \frac{z{a}_{*}}{{(1-4z{a}_{*})}^{3/2}}}=\nu z{a}_{*}{(1-z/z_{*})}^{-3/2}.$$

Since $z_{*}{a}_{*}=1/4$, its analytic numerator is nonzero at $z_{*}$, so this is a genuine algebraic branch singularity of local type ${(1-z/z_{*})}^{-3/2}$. The conjugate argument applies at $\overline{z_{*}}$.

On the other hand Lemma A1 gives

$${\mathfrak{P}}_T\left(z\right)={\mathfrak{H}}_T\left(z\right)-2{\mathfrak{A}}_T\left(z\right),{\mathfrak{H}}_T\left(z\right):=\sum _{m\ge 1}{\mathcal{H}}_m\left(T\right)z^m,$$

and (A15) makes ${\mathfrak{H}}_T$ holomorphic in $\left|z\right|<1$. A holomorphic function cannot cancel the nonintegral local branch term just displayed. Thus ${\mathfrak{P}}_T$ has the asserted conjugate branch points. This proves (iii).    □

Remark A3 

(The cancellation annulus). The gap left by absolute convergence is now visible directly in the generating variable. The half-plane representation (A30) reaches only

$$\left|z\right|<r_{\mathrm{abs}}\left(T\right)=1-{\displaystyle \frac{1}{4T^2}}+O\left(T^{-4}\right),$$

whereas RH is equivalent to holomorphy throughout $\left|z\right|<1$. Thus this route must continue the prime germ across the thin annulus

$$r_{\mathrm{abs}}\left(T\right)\le \left|z\right|<1.$$

The thickness is $1/\left(4T^2\right)+O\left(T^{-4}\right)$, exactly dual to the exponential gap $\mathcal{M}\left(T\right)-1$ in Proposition A2. Crossing that annulus cannot be justified by absolute convergence of the prime Dirichlet series alone; it requires additional cancellation or equivalent information about zeros.

Appendix A.4. Rational Normalization and a Semicircular Formula

There is an even simpler order-generating kernel if the beta normalization is removed coefficient by coefficient. Define

$${\tilde{\mathfrak{A}}}_T\left(z\right):=\sum _{m\ge 1}{4}^m{B}_m{T}^2{A}_m^{\prime }\left(T^2\right)z^m,{\tilde{\mathfrak{P}}}_T\left(z\right):=\sum _{m\ge 1}{4}^m{B}_m{\mathcal{P}}_m\left(T\right)z^m.$$

(A33)

Then

$${\mathcal{L}}_z\left(y\right):=\sum _{m\ge 1}{4}^m{B}_m{p}_m\left(y\right)z^m={\displaystyle \frac{4zq\left(y\right)}{1-4zq\left(y\right)}}.$$

(A34)

Thus the algebraic $3/2$-singularities in (A23) become simple poles after normalization. On the zero side, put

$$r_{\rho }\left(T\right):={\displaystyle \frac{4T^2{w}_{\rho }}{{(T^2+w_{\rho })}^2}}.$$

(A35)

Then, in the disk of convergence,

$${\tilde{\mathfrak{A}}}_T\left(z\right)=\sum _{\rho \in {\mathcal{Z}}_{+}}{\displaystyle \frac{z{r}_{\rho }\left(T\right)}{1-z{r}_{\rho }\left(T\right)}}.$$

(A36)

The Fourier transform of (A34) is elementary. For real $0<z<1$, write

$$a=\sqrt{z},b=\sqrt{1-z}.$$

Then

$${\mathcal{L}}_z\left(y\right)={\displaystyle \frac{4a^2{y}^2}{\left({(y-a)}^2+b^2\right)\left({(y+a)}^2+b^2\right)}}.$$

Closing in the lower half-plane for $v>0$ and taking the residues at $a-ib$ and $-a-ib$ gives

$${\widehat{\mathcal{L}}}_z\left(v\right):={\displaystyle \frac{1}{2\pi }}{\int }_{-\infty }^{\infty }{\mathcal{L}}_z\left(y\right)e^{-ivy}dy={\displaystyle \frac{a}{b}}{e}^{-bv}\left(acos\left(av\right)-bsin\left(av\right)\right).$$

(A37)

Definition A1 

(Branch and continuation convention). Let

$$\mathcal{D}:=\mathbb{C}\setminus \left((-\infty ,0]\cup [1,\infty )\right).$$

On $\mathcal{D}$ let

$$a\left(z\right)=\sqrt{z},b\left(z\right)=\sqrt{1-z}$$

be the branches that are positive for $0<z<1$, and put

$$s_{\pm }\left(z\right)={\displaystyle \frac{1}{2}}+Tb\left(z\right)\pm iTa\left(z\right).$$

Since $Q\left(s\right)=-{\zeta }^{\prime }\left(s\right)/\zeta \left(s\right)$ is meromorphic on $\mathbb{C}$, the expression

$${\mathcal{Q}}_T\left(z\right):=T{\displaystyle \frac{a\left(z\right)}{b\left(z\right)}}\left((a\left(z\right)-ib\left(z\right))Q\left(s_{+}\left(z\right)\right)+(a\left(z\right)+ib\left(z\right))Q\left(s_{-}\left(z\right)\right)\right)$$

(A38)

is meromorphic on $\mathcal{D}$, with poles only where one of the composed arguments meets a pole of Q. Whenever we refer below to continuation of the prime generating germ by the semicircular formula, we mean the unique meromorphic continuation obtained from (A38) along a path in $\mathcal{D}$ starting in the absolute-convergence region and avoiding its isolated poles. This convention makes no assertion that a pole can be crossed holomorphically.

Theorem A4 

(Semicircular logarithmic-derivative formula). Fix $T\ge 2$ and let $r_{\mathrm{abs}}\left(T\right)$ be as in (A29). For real

$$0<z<r_{\mathrm{abs}}\left(T\right),$$

(A39)

the Taylor series defining ${\tilde{\mathfrak{P}}}_T\left(z\right)$ converges absolutely. Let $a=a\left(z\right)=\sqrt{z}$, $b=b\left(z\right)=\sqrt{1-z}$ as in Definition A1, and write

$$s_{\pm }\left(z\right):={\displaystyle \frac{1}{2}}+Tb\pm iTa.$$

(A40)

Then $\Re s_{\pm }>1$ and

$$\begin{array}{|c|}\hline {\displaystyle {\tilde{\mathfrak{P}}}_T\left(z\right)=T\frac{a}{b}\left((a-ib)Q\left(s_{+}\left(z\right)\right)+(a+ib)Q\left(s_{-}\left(z\right)\right)\right).}\\ \hline\end{array}$$

(A41)

For real z this is equivalently

$${\tilde{\mathfrak{P}}}_T\left(z\right)=2T{\displaystyle \frac{a}{b}}\Re \left((a-ib)Q\left({\displaystyle \frac{1}{2}}+Tb+iTa\right)\right).$$

(A42)

The meromorphic function ${\mathcal{Q}}_T$ in (A38) is the continuation of this Taylor germ. On the larger real interval

$$0<z<1-{\displaystyle \frac{1}{4T^2}},$$

(A43)

one still has $\Re s_{\pm }>1$, so ${\mathcal{Q}}_T\left(z\right)$ is represented by the absolutely convergent prime Dirichlet series for Q. Beyond that endpoint the same formula is only a meromorphic continuation and no absolute prime-series claim is made.

As real z increases from 0 to 1, the points $s_{\pm }\left(z\right)$ trace the two halves of the right semicircle

$$|s-1/2|=T,\Re s>1/2.$$

(A44)

Proof. 

Equation (A34) is the geometric series applied to $4zq\left(y\right)$. Formula (A36) follows by summing (48) after multiplication by $4^m{B}_m{z}^m$.

For (A37), the lower-half-plane poles are

$$y_1=a-ib,y_2=-a-ib.$$

A direct residue calculation gives

$${\widehat{\mathcal{L}}}_z\left(v\right)={\displaystyle \frac{a}{2b}}\left((a-ib)e^{-iv(a-ib)}-(-a-ib)e^{-iv(-a-ib)}\right),$$

which simplifies to (A37).

For $0<z<r_{\mathrm{abs}}\left(T\right)$, the root-growth bound (A20) and ${\left(4^m{B}_m\right)}^{1/m}\to 1$ show that the Taylor series defining ${\tilde{\mathfrak{P}}}_T\left(z\right)$ converges absolutely. The same shifted-contour majorant used in Proposition A2, now summed in m through the geometric identity (A34), justifies summing the prime-side formula in this range. Thus

$$\begin{array}{cc}\hfill {\tilde{\mathfrak{P}}}_T\left(z\right)& =2T{\displaystyle \frac{a}{b}}\sum _{n\ge 2}{\displaystyle \frac{\Lambda \left(n\right)}{n^{1/2+Tb}}}\left(acos(Talogn)-bsin(Talogn)\right)\hfill \\ \hfill & =T{\displaystyle \frac{a}{b}}\left((a-ib)Q\left(s_{+}\left(z\right)\right)+(a+ib)Q\left(s_{-}\left(z\right)\right)\right),\hfill \end{array}$$

which proves (A41) on the Taylor-germ interval. Since $r_{\mathrm{abs}}\left(T\right)<1-1/\left(4T^2\right)$, the right-hand side is holomorphic in a complex neighborhood of every real point up to the latter endpoint and is represented there by the absolutely convergent Dirichlet series for Q. The identity theorem therefore identifies (A38) with the meromorphic continuation of the Taylor germ along every admissible path described in Definition A1. Finally $a^2+b^2=1$ gives the semicircle identity.    □

Corollary A2 

(Real pole detector for an off-critical zero). Suppose RH is false and let

$$\rho ={\displaystyle \frac{1}{2}}+\delta +i\gamma ,\delta >0,$$

be a nontrivial zero. Put

$$T_{\rho }:=\sqrt{{\gamma }^2+{\delta }^2},z_{\rho }:={\displaystyle \frac{{\gamma }^2}{{\gamma }^2+{\delta }^2}}.$$

(A45)

Then $0<z_{\rho }<1$,

$$s_{+}\left(z_{\rho }\right)=\rho ,r_{\rho }\left(T_{\rho }\right)=z_{\rho }^{-1}>1,$$

(A46)

and ${\tilde{\mathfrak{A}}}_{T_{\rho }}$ has a simple pole at the real point $z=z_{\rho }$. The meromorphic continuation of ${\tilde{\mathfrak{P}}}_{T_{\rho }}$ specified in Definition A1 has a nonremovable simple pole there as well. Equivalently, $s_{+}\left(z_{\rho }\right)=\rho$ makes the $Q\left(s_{+}\left(z\right)\right)$ term in (A38) singular with a nonzero prefactor.

Consequently RH is equivalent to the absence of such interior real poles for every $T>14$. This is only a reformulation of the zero-free condition, not an independent proof of it.

Proof. 

With $a=\gamma /T_{\rho }$ and $b=\delta /T_{\rho }$, one has $a^2+b^2=1$, $z_{\rho }=a^2$, and (A40) gives $s_{+}\left(z_{\rho }\right)=\rho$. Moreover

$$w_{\rho }={(\gamma -i\delta )}^2=T_{\rho }^2{(a-ib)}^2$$

and

$$T_{\rho }^2+w_{\rho }=2T_{\rho }^{2}a(a-ib).$$

Substitution into (A35) yields

$$r_{\rho }\left(T_{\rho }\right)={\displaystyle \frac{1}{a^2}}={\displaystyle \frac{1}{z_{\rho }}}>1.$$

The corresponding term in (A36) is therefore $z/z_{\rho }{(1-z/z_{\rho })}^{-1}$, giving a simple pole whose residue at $z=z_{\rho }$ is $-z_{\rho }$. The upper-half-plane partner $1-\overline{\rho }$ has the conjugate reflected parameter and, at the same $(T_{\rho },z_{\rho })$, produces the identical real denominator and the same residue $-z_{\rho }$ (multiplied by the zero multiplicity, if necessary). Hence the paired principal parts add rather than cancel. All other zero summands whose reflected parameters differ from this value are holomorphic at $z_{\rho }$; summands with the same parameter merely add another copy of the same-sign principal part. Thus cancellation at the real pole is impossible.

There are two equivalent ways to see the same pole on the prime side. First, multiplying (A14) by $4^m{B}_m{z}^m$ and summing shows that the difference between ${\tilde{\mathfrak{P}}}_T$ and $-2{\tilde{\mathfrak{A}}}_T$ is holomorphic for $\left|z\right|<1$, because $4^m{B}_m{\mathcal{H}}_m\left(T\right)=O_T\left(m^{-1/2}\right)$. Second, in the continuation formula (A38), the map $z\mapsto s_{+}\left(z\right)$ is holomorphic near $z_{\rho }$ with nonzero derivative, while Q has a pole at $s=\rho$ and the coefficient multiplying $Q\left(s_{+}\left(z\right)\right)$ is nonzero. Thus the pole is nonremovable. The last assertion follows from the functional equation, which pairs every zero with $\Re \rho <1/2$ with one having $\Re \rho >1/2$.    □

Remark A4 

(Where the arithmetic barrier sits on the semicircle). The absolutely convergent prime formula (A41) reaches exactly the portion of the semicircle with $\Re s>1$. Its real endpoint is

$$z=1-{\displaystyle \frac{1}{4T^2}}.$$

On the other hand an off-critical zero in the critical strip has $0<\delta <1/2$ and therefore

$$z_{\rho }=1-{\displaystyle \frac{{\delta }^2}{T_{\rho }^2}}>1-{\displaystyle \frac{1}{4T_{\rho }^2}}.$$

Thus every possible off-critical pole occurs precisely in the narrow terminal part of the semicircle that is inaccessible to absolute convergence of the prime Dirichlet series. Reaching that terminal arc is equivalent to obtaining new information about $Q\left(s\right)$ in $1/2<\Re s\le 1$; algebraic resummation alone does not cross it.

Appendix A.5. Verified-height Transfer and the Terminal Zero-Free Collar

The semicircular formulation also converts existing zero information into a sharp statement about the order-generating variable. The first observation uses only the rigorous verification of RH through a finite height.

Theorem A5 

(Verified-height transfer to the exact root threshold). Let

$$H_{*}:=3·{10}^{12},$$

so that every nontrivial zero with $0<\gamma \le H_{*}$ is known to lie on the critical line by Platt–Trudgian [5]. Then, for every fixed T with

$$0<T\le H_{*}-{\displaystyle \frac{1}{2}},$$

(A47)

one has

$$\mathcal{R}\left(T\right)\le 1.$$

(A48)

Consequently

$$\underset{m\to \infty }{lim\; sup}{\left|T^2{A}_m^{\prime }\left(T^2\right)\right|}^{1/m}\le 1,$$

(A49)

and, for $2\le T\le H_{*}-1/2$,

$$\underset{m\to \infty }{lim\; sup}{\left|{\mathcal{P}}_m\left(T\right)\right|}^{1/m}\le 1.$$

(A50)

Equivalently, the normalized zero generating function ${\tilde{\mathfrak{A}}}_T\left(z\right)$ is holomorphic for $\left|z\right|<1$ throughout the range (A47).

Proof. 

Critical-line zeros have $\delta =0$, and hence $|r_{\rho }\left(T\right)|\le 1$ by the arithmetic–geometric mean inequality. Consider a hypothetical off-critical zero

$$\rho ={\displaystyle \frac{1}{2}}+\delta +i\gamma ,0<\delta <{\displaystyle \frac{1}{2}}.$$

By the verified-height input, necessarily $\gamma >H_{*}$. A direct calculation from (A35) gives

$$\begin{array}{cc}\hfill & |T^2+w_{\rho }{|}^2-4T^2\left|w_{\rho }\right|\hfill \end{array}$$

(A51)

$$\begin{array}{cc}\hfill & =\left(-T^2-2T\delta +{\delta }^2+{\gamma }^2\right)\left(-T^2+2T\delta +{\delta }^2+{\gamma }^2\right).\hfill \end{array}$$

(A52)

Since $\gamma >T$, the inequality $|r_{\rho }\left(T\right)|>1$ would force the first factor in (A51) to be negative, and therefore

$${\gamma }^2<T^2+2T\delta -{\delta }^2<{(T+\delta )}^2.$$

Thus $\gamma <T+\delta <T+1/2\le H_{*}$, a contradiction. Hence every zero satisfies $|r_{\rho }\left(T\right)|\le 1$, proving (A48). Equation (A49) follows from Proposition A1. The exact relation (A14), together with ${\mathcal{H}}_m\left(T\right)=O_T\left(1\right)$, then gives (A50). Finally, ${\sum }_{\rho }\left|r_{\rho }\left(T\right)\right|<\infty$ for fixed T, so (A36) converges normally on every compact subset of $\left|z\right|<1$.    □

The next statement quantifies the remaining real-axis collar using any classical zero-free region.

Proposition A3 

(Zero-free penetration of the terminal semicircle). Assume that, for some $R>0$ and $t_0\ge 2$,

$$\zeta (\sigma +it)\ne 0\left(\left|t\right|\ge t_0,\sigma \ge 1-{\displaystyle \frac{1}{Rlog\left|t\right|}}\right).$$

(A53)

For $T>0$ put

$$z_R\left(T\right):=1-{\displaystyle \frac{1}{T^2}}{\left({\displaystyle \frac{1}{2}}-{\displaystyle \frac{1}{RlogT}}\right)}^2.$$

(A54)

If $T\ge \sqrt{t_0^2+1/4}$ and $1/(RlogT)<1/2$, then the right-hand side of (A41) analytically continues the real generating germ throughout

$$0<z<z_R\left(T\right).$$

(A55)

Moreover, if an off-critical zero $\rho =1/2+\delta +i\gamma$ gives the real pole $(T_{\rho },z_{\rho })$ of Corollary A2 and $\gamma \ge t_0$, then

$$z_{\rho }>z_R\left(T_{\rho }\right).$$

(A56)

Proof. 

Write $a=\sqrt{z}$, $b=\sqrt{1-z}$. For $z\le 1-1/\left(4T^2\right)$ one has $\Re s_{\pm }\ge 1$, so continuation from the original Dirichlet-series range is standard. On the remaining interval up to $z_R\left(T\right)$,

$$Tb\ge {\displaystyle \frac{1}{2}}-{\displaystyle \frac{1}{RlogT}},\Re s_{\pm }\ge 1-{\displaystyle \frac{1}{RlogT}}.$$

Also $|\Im s_{\pm }|=Ta\le T$. Since the terminal interval begins at $z=1-1/\left(4T^2\right)$, its imaginary ordinate is at least $\sqrt{T^2-1/4}\ge t_0$. Because $t\mapsto 1/(Rlogt)$ is decreasing,

$$1-{\displaystyle \frac{1}{RlogT}}\ge 1-{\displaystyle \frac{1}{Rlog|\Im s_{\pm }|}},$$

so (A53) excludes zeros throughout this part of the arc. Hence $Q\left(s_{\pm }\left(z\right)\right)$ is analytic there and (A41) provides the claimed continuation.

For an off-critical zero, (A53) gives

$$\delta <{\displaystyle \frac{1}{2}}-{\displaystyle \frac{1}{Rlog\gamma }}.$$

Since $T_{\rho }=\sqrt{{\gamma }^2+{\delta }^2}>\gamma$, one has $1/(Rlog\gamma )>1/(Rlog{T}_{\rho })$. Therefore

$$\delta <{\displaystyle \frac{1}{2}}-{\displaystyle \frac{1}{Rlog{T}_{\rho }}},$$

and substitution into $z_{\rho }=1-{\delta }^2/T_{\rho }^2$ proves (A56).    □

Corollary A3 

(Current explicit collar). Using the Bellotti–Trudgian–Yang zero-free region

$$\sigma \ge 1-{\displaystyle \frac{1}{4.896log\left|t\right|}},\left|t\right|\ge 3,$$

Proposition A3 applies for $T\ge 4$ with

$$z_{\mathrm{BTY}}\left(T\right)=1-{\displaystyle \frac{1}{T^2}}{\left({\displaystyle \frac{1}{2}}-{\displaystyle \frac{1}{4.896logT}}\right)}^2.$$

(A57)

Relative to the absolute-convergence endpoint $z_{\mathrm{abs}}\left(T\right)=1-1/\left(4T^2\right)$, the unconditional gain is

$$z_{\mathrm{BTY}}\left(T\right)-z_{\mathrm{abs}}\left(T\right)={\displaystyle \frac{1}{T^2}}\left({\displaystyle \frac{1}{4.896logT}}-{\displaystyle \frac{1}{4.{896}^2{(logT)}^2}}\right).$$

(A58)

The published Mossinghoff–Trudgian–Yang region gives the same statements with $4.896$ replaced by $5.558691$.

Corollary A4 

(Reduction to the unverified terminal collar). With $H_{*}=3·{10}^{12}$, the prime-side criterion of Theorem A2 may be reduced to

$$\mathrm{RH}⟺\underset{m\to \infty }{lim\; sup}{\left|{\mathcal{P}}_m\left(T\right)\right|}^{1/m}\le 1\mathrm{for}\mathrm{every}\mathrm{fixed}T>H_{*}-{\displaystyle \frac{1}{2}}.$$

(A59)

Likewise, using the Bellotti–Trudgian–Yang region, RH is equivalent to the absence of real poles of ${\tilde{\mathfrak{A}}}_T$ in the restricted set

$$T>H_{*},z_{\mathrm{BTY}}\left(T\right)<z<1.$$

(A60)

The same statement with the published constant uses the corresponding $z_R\left(T\right)$ with $R=5.558691$.

Proof. 

Theorem A5 supplies the root-growth bound for every $2\le T\le H_{*}-1/2$, so Theorem A2 leaves only the stated range. If RH is false, choose an off-critical zero with $\delta >0$. The verified-height result forces $\gamma >H_{*}$ and hence $T_{\rho }=\sqrt{{\gamma }^2+{\delta }^2}>H_{*}$. Corollary A3 and Proposition A3 force its real pole to satisfy $z_{\rho }>z_{\mathrm{BTY}}\left(T_{\rho }\right)$. Conversely every such real pole comes from an off-critical zero by Corollary A2, so its absence is equivalent to RH.    □

Remark A5 

(Quantitative control below the verified height). A recent explicit estimate of Leong ([4], Corollary 3 and Table 2) uses the verified RH height to show, unconditionally, that

$$\left|{\displaystyle \frac{{\zeta }^{\prime }}{\zeta }}(\sigma +it)\right|\le 31447logt(\sigma \ge 0.51,15\le t\le H_{*}).$$

Consequently, on the real semicircle and in the same ordinate range, (A42) gives

$$|{\tilde{\mathfrak{P}}}_T\left(z\right)|\le 62894T{\displaystyle \frac{\sqrt{z}}{\sqrt{1-z}}}logT$$

(A61)

whenever $z\le 1-{10}^{-4}/T^2$ and $T\sqrt{z}\ge 15$. Thus, for $16\le T\le H_{*}$, an explicit logarithmic-derivative estimate penetrates $99.96\%$ of the real z-gap between $1-1/\left(4T^2\right)$ and 1; the remaining issue is not absolute convergence but control arbitrarily close to the critical line. This is a quantitative finite-height statement and does not extend RH beyond the verified height.

Corollary A5 

(Recovery of the ordinary zero count). For fixed $a,L>0$, uniformly for integers $a\sqrt{x}\le m\le Lx$ as $x\to \infty$,

$$A_m\left(x\right)=N\left(\sqrt{x}\right)+O_{a,L}(logx).$$

Proof. 

Put $T=\sqrt{x}$. Theorem 1 and $c_m=1+O\left(m^{-1}\right)$ give

$$A_m\left(T^2\right)=M\left(T\right)+O_L\left(logT+{\displaystyle \frac{TlogT}{m}}\right).$$

If $m\ge aT$, the second error is $O_a(logT)$. Apply (30) to replace $M\left(T\right)$ by $N\left(T\right)$.    □

Remark A6 

(Meaning of the density ratio). Although $S_m$ is a positive sum, individual off-line blocks in $A_m$ need not be nonnegative in the range of Theorem 1. The ratio $S_m/A_m$ is therefore not a probability or a literal partition of positive mass. Nevertheless, $0\le S_m\left(T^2\right)\le R_m\left(T\right)$ and (33) imply

$$0\le {\displaystyle \frac{S_m\left(T^2\right)}{A_m\left(T^2\right)}}\le 1+O_L\left(T^{-1}\right)$$

uniformly for $1\le m\le L{T}^2$, once T is sufficiently large. The density theorem is an asymptotic comparison with an eventually positive normalizing sum.

Corollary A6 

(Application of Lamzouri’s stated proportion theorem). Using Theorem 1.1 of Lamzouri [13], Theorem 1 applies with the Montgomery–Taylor constant

$$C_0={\displaystyle \frac{3}{2}}-{\displaystyle \frac{1}{\sqrt{2}}}cot{\displaystyle \frac{1}{\sqrt{2}}}=0.6725007036794116457\dots .$$

Thus for every fixed $L>0$,

$$\underset{x\to \infty }{lim\; inf}\underset{\begin{array}{c}m\in \mathbb{N}\\ 1\le m\le Lx\end{array}}{inf}{\displaystyle \frac{S_m\left(x\right)}{A_m\left(x\right)}}\ge C_0.$$

The dependence on this external theorem is confined to the numerical value of p; the smoothed asymptotic and the general transfer theorem do not use it.

Remark A7 

(What this extension establishes). The construction links the Widder expressions to cumulative zero counts through an exact beta-kernel identity. The estimates are uniform for $m\le Lx$, rather than only for fixed m. The density transfer itself is an Abel-summation consequence of (18); it does not improve $C_0$. Positivity of $A_m\left(x\right)$ alone is positivity of an integral of a Widder expression and does not imply pointwise positivity. The derivative refinement in Theorem 2 is what yields pointwise positivity, for every fixed $\epsilon >0$, throughout the asymptotic region $m\le (4-\epsilon )xloglogx$ for sufficiently large x. Neither result supplies the full infinite hierarchy required for RH, and no explicit onset $x_0\left(\epsilon \right)$ is asserted.

Appendix B.5.5.3. Attribution of the quadratic examples.

Bondesson–Simon ([1], Section 4.1, Proposition 3 and its proof) use $f_{\alpha }\left(x\right)={(1+2cos\left(\pi \alpha \right)x+x^2)}^{-1}$ and $g_{\alpha }=-{(log{f}_{\alpha })}^{\prime }$. Their calculation gives the finite threshold $2\alpha k\le 1$. Below $F_{\varphi }=2g_{\alpha }$ with $\varphi =\pi \alpha$. The examples and threshold are therefore recorded as known building blocks; the extension to products uses dilation, summation and convergence.

Proposition A4 

(Exact universal sector threshold). Let $0\le \vartheta <\pi /2$ and $m\ge 1$. The inequality ${\mathcal{W}}_m[2G^{\prime }/G]\left(x\right)>0$ for all $x>0$ holds for every nonconstant real genus-zero product (A62) with $|arg{w}_j|\le \vartheta$ if and only if $m\vartheta \le \pi /2$.

Proof. 

Sufficiency is Theorem A6. If $m\vartheta >\pi /2$, choose

$${\displaystyle \frac{\pi }{2m}}<\varphi <min\{\vartheta ,\pi /m\}.$$

The interval is nonempty. The quadratic

$$G_{\varphi }\left(x\right)=(1+x{e}^{i\varphi })(1+x{e}^{-i\varphi })=1+2xcos\varphi +x^2$$

has reflected zeros $e^{i\varphi },e^{-i\varphi }$ in the sector. Its logarithmic derivative $F_{\varphi }=2G_{\varphi }^{\prime }/G_{\varphi }$ is positive on $(0,\infty )$. Formula (12), which here extends continuously to $x=0$, gives

$${\mathcal{W}}_m\left[F_{\varphi }\right]\left(0\right)=4(2m-1)!cos\left(m\varphi \right)<0.$$

Continuity gives a negative value for all sufficiently small positive x.    □

Proposition A5 

(Finite positivity with nonreal poles). For every integer $N\ge 1$, there is a positive rational function $F_{\varphi }=2G_{\varphi }^{\prime }/G_{\varphi }$ in ${\mathcal{S}}_N\setminus {\mathcal{S}}_{N+1}$ whose poles are nonreal.

Proof. 

Choose

$${\displaystyle \frac{\pi }{2(N+1)}}<\varphi <{\displaystyle \frac{\pi }{2N}}$$

and use the preceding quadratic $G_{\varphi }$. It has nonreal zeros, and

$$F_{\varphi }\left(x\right)={\displaystyle \frac{4(x+cos\varphi )}{x^2+2xcos\varphi +1}}>0.$$

Since $N\varphi <\pi /2$, Theorem A6 proves all the first N Widder inequalities. On the other hand,

$${\displaystyle \frac{\pi }{2}}<(N+1)\varphi <{\displaystyle \frac{(N+1)\pi }{2N}}\le \pi ,$$

so ${\mathcal{W}}_{N+1}\left[F_{\varphi }\right]\left(0\right)<0$ and the same holds for small positive x. Thus $F_{\varphi }\notin {\mathcal{S}}_{N+1}$. Its nonreal poles also exclude an ordinary Stieltjes representation.    □

These examples establish a limitation of the general finite hierarchy. They do not assert the existence of an off-line zero of $\xi$, or optimality of the numerical cutoff for $F_{\xi }$ itself.

Appendix B.1. Transfer from a Verified Height

We now combine a finite verification of RH with any explicit two-sided zero-free strip above the verified height.

Theorem A7 

(Finite-height transfer principle). Let $H>1$ and let $d:[H,\infty )\to [0,\infty )$ be finite-valued with $d\left(t\right)/t$ nonincreasing. Assume that every nontrivial zero with $0<\gamma \le H$ lies on the critical line, and that every zero with $\gamma >H$ satisfies

$$|\beta -1/2|\le d\left(\gamma \right).$$

(A68)

Set $b\left(t\right)=min\left\{d\right(t),1/2\}$ and

$${\vartheta }_H=2arctan{\displaystyle \frac{b\left(H\right)}{H}},N_H=\left\{\begin{array}{cc}⌊{\displaystyle \frac{\pi }{2{\vartheta }_H}}⌋,\hfill & b\left(H\right)>0,\hfill \\ +\infty ,\hfill & b\left(H\right)=0.\hfill \end{array}\right.$$

(A69)

Then ${\mathcal{W}}_m\left[F_{\xi }\right]\left(x\right)>0$ for every $x>0$ and every integer $1\le m\le N_H$. When $N_H=+\infty$, this means all positive integer orders; in that case the assumptions themselves imply RH.

Proof. 

The critical strip and (A68) give $|\beta -1/2|\le b\left(\gamma \right)$ above H. Moreover,

$${\displaystyle \frac{b\left(t\right)}{t}}=min\left\{{\displaystyle \frac{d\left(t\right)}{t}},{\displaystyle \frac{1}{2t}}\right\}$$

is nonincreasing, since both entries are nonincreasing. For zeros above H,

$$|arg{w}_{\rho }|\le 2arctan{\displaystyle \frac{b\left(\gamma \right)}{\gamma }}\le 2arctan{\displaystyle \frac{b\left(H\right)}{H}}={\vartheta }_H.$$

Zeros below H give positive real reflected parameters. Since $H>1$ and $b\left(H\right)\le 1/2$, one has $0\le {\vartheta }_H<\pi /2$. If $b\left(H\right)>0$, Theorem A6 applies for $m\le \lfloor \pi /\left(2{\vartheta }_H\right)\rfloor$, including a possible integer endpoint. If $b\left(H\right)=0$, monotonicity and nonnegativity force $b\left(t\right)=0$ for all $t\ge H$. All remaining zeros then lie on the critical line, and every Widder order is strictly positive.    □

Even without a zero-free region, the critical strip gives $|\beta -1/2|<1/2$, so finite verification to height H alone yields the simpler sector

$${\vartheta }_H^{\left(0\right)}=2arctan{\displaystyle \frac{1}{2H}}.$$

The explicit zero-free region sharpens this slightly but significantly at the trillion-scale order considered here.

Appendix B.2. An Explicit Numerical Application

Platt and Trudgian [5] rigorously verified, using interval arithmetic and Turing’s method, that every nontrivial zero with

$$0<\gamma \le H,H=3·{10}^{12},$$

(A70)

lies on the critical line, and their published result also states that all such zeros are simple. Only the critical-line location is needed for the finite-order positivity theorem below; simplicity is not used in its proof.

Bellotti, Trudgian and Yang ([2], Theorem 1) establish the explicit classical zero-free region

$$\zeta (\sigma +it)\ne 0\mathrm{if}t\ge 3,\sigma >1-{\displaystyle \frac{1}{Rlogt}},R=4.896.$$

(A71)

By conjugation and the functional equation, a nontrivial zero at height $t>H$ obeys

$${\displaystyle \frac{1}{Rlogt}}\le \beta \le 1-{\displaystyle \frac{1}{Rlogt}}.$$

(A72)

Thus

$$\left|\beta -{\displaystyle \frac{1}{2}}\right|\le d\left(t\right):={\displaystyle \frac{1}{2}}-{\displaystyle \frac{1}{Rlogt}}.$$

(A73)

Moreover,

$${\displaystyle \frac{d}{dt}}\left({\displaystyle \frac{d\left(t\right)}{t}}\right)={\displaystyle \frac{1}{t^2}}\left[{\displaystyle \frac{1}{R{(logt)}^2}}-{\displaystyle \frac{1}{2}}+{\displaystyle \frac{1}{Rlogt}}\right]<0$$

for $t\ge H$, since $logt>28$ and $R>4$. Here $0<d\left(t\right)<1/2$, so $b=d$ in Theorem A7. Although (A73) is weak, $\gamma >H$ and strict decrease of $d\left(t\right)/t$ give a strict angular bound above H. Consequently all reflected zero parameters lie in the sector

$$|arg{w}_{\rho }|<{\vartheta }_H,{\vartheta }_H=2arctan\left[{\displaystyle \frac{1}{H}}\left({\displaystyle \frac{1}{2}}-{\displaystyle \frac{1}{RlogH}}\right)\right].$$

(A74)

Appendix B.3. Arithmetic Verification

The source package contains the script verify_cutoff.py, which evaluates (A74) using the interval context of mpmath at 80 decimal digits. The endpoints are outward rounded. It gives the enclosures

$$\begin{array}{cc}\hfill 0.4928906727372165928820573& <d\left(H\right)<0.4928906727372165928820574,\hfill \end{array}$$

(A75)

$$\begin{array}{cc}\hfill 3.2859378182481106192137157\times {10}^{-13}& <{\vartheta }_H<3.2859378182481106192137158\times {10}^{-13},\hfill \end{array}$$

(A76)

and

$$\begin{array}{cc}\hfill 4,780,359,257,170.4922923019926& <{\displaystyle \frac{\pi }{2{\vartheta }_H}}<4,780,359,257,170.4922923019927.\hfill \end{array}$$

(A77)

The script also checks these displayed enclosures directly, starting from the exact decimal inputs H and R. Therefore

$$N_H=4,780,359,257,170,N_H{\vartheta }_H<{\displaystyle \frac{\pi }{2}}<(N_H+1){\vartheta }_H.$$

(A78)

For comparison with a fully published zero-free input, let $R_0=5.558691$, the constant in Mossinghoff–Trudgian–Yang [3], and define

$$d_0\left(H\right)={\displaystyle \frac{1}{2}}-{\displaystyle \frac{1}{R_{0}logH}},{\vartheta }_{H,0}=2arctan{\displaystyle \frac{d_0\left(H\right)}{H}}.$$

The same interval-arithmetic script gives

$$4,772,153,270,068<{\displaystyle \frac{\pi }{2{\vartheta }_{H,0}}}<4,772,153,270,069.$$

(A79)

Theorem A8 

(Published-input finite-order consequence). Let $F_{\xi }$ be defined by (2). Using only the published Platt–Trudgian zero verification and the published Mossinghoff–Trudgian–Yang zero-free region, one has

$${(-1)}^{m-1}{\displaystyle \frac{d^{2m-1}}{d{x}^{2m-1}}}\left[x^m{\displaystyle \frac{1}{\sqrt{x}}}{\displaystyle \frac{{\xi }^{\prime }}{\xi }}\left({\displaystyle \frac{1}{2}}+\sqrt{x}\right)\right]>0$$

(A80)

for every $x>0$ and every integer

$$1\le m\le \begin{array}{|c|}\hline 4,772,153,270,068\\ \hline\end{array}.$$

(A81)

In particular, $F_{\xi }\in {\mathcal{S}}_{4,772,153,270,068}$ in the sense of Bondesson–Simon. Membership requires nonnegative inequalities; the displayed strict inequalities give a stronger conclusion.

Proof. 

Apply Theorem A7 with the published constant $R_0=5.558691$ and the verified height (A70). The enclosure (A79) and Theorem A6 give the stated range. □

Corollary A7 

(Preprint-enhanced finite-order consequence). If one additionally invokes Theorem 1 of the 2026 Bellotti–Trudgian–Yang preprint [2], with $R=4.896$, then the range in Theorem A8 improves to

$$1\le m\le \begin{array}{|c|}\hline 4,780,359,257,170\\ \hline\end{array}.$$

(A82)

Proof. 

For the preprint constant, (A77) gives (A78). Apply Theorem A7 with (A70), (A73), and (A74), followed by Theorem A6. □

Remark A8 

(Status of the two numerical cutoffs). The cutoff in Theorem A8 depends only on peer-reviewed published inputs. Corollary A7 records the slightly larger value obtained from the stronger 2026 Bellotti–Trudgian–Yang preprint theorem, which is still a preprint as of 12 September 2026. Neither integer is claimed to be optimal for $F_{\xi }$.

Appendix B.4. A Prescribed Order and a Positive Margin

The transfer principle can be inverted. If one uses only the critical-strip bound $|\beta -1/2|<1/2$, a sufficient condition for the mth Widder inequality is

$$2marctan{\displaystyle \frac{1}{2H}}\le {\displaystyle \frac{\pi }{2}}.$$

Equivalently,

$$H\ge {\displaystyle \frac{1}{2}}cot{\displaystyle \frac{\pi }{4m}}.$$

(A83)

Since

$${\displaystyle \frac{1}{2}}cot{\displaystyle \frac{\pi }{4m}}={\displaystyle \frac{2m}{\pi }}+O\left(m^{-1}\right),$$

we obtain the following simple rule.

Corollary A8 

(Linear height-order law). For $H>1$, a rigorous verification of RH through height

$$H\ge {\displaystyle \frac{1}{2}}cot{\displaystyle \frac{\pi }{4m}}$$

is sufficient, even without an additional zero-free region, to prove the mth Widder inequality globally on $(0,\infty )$. Asymptotically, height H certifies all orders

$$m\le {\displaystyle \frac{\pi }{4arctan(1/(2H\left)\right)}}={\displaystyle \frac{\pi }{2}}H+{\displaystyle \frac{\pi }{24H}}+O\left(H^{-3}\right).$$

Thus a successful verification of RH through the indicated finite height suffices for each prescribed order. This is a sufficient implication, not a guarantee that arbitrarily high successful verifications can be obtained. The zero-free region improves the constant at a fixed verified height by reducing the maximum possible horizontal displacement of all as-yet-unverified zeros.

For comparison, if one ignored the explicit zero-free region and used only $|\beta -1/2|<1/2$ above $H=3·{10}^{12}$, one would still obtain

$$1\le m\le 4,712,388,980,384.$$

The 2026 zero-free region increases the certified order by exactly $67,970,276,786$ relative to the critical-strip-only cutoff above.

In the range of Theorem A8, retaining the first critical-line zero ${\gamma }_1=14.1347251417\dots$ in (12) gives

$${\mathcal{W}}_m\left[F_{\xi }\right]\left(x\right)\ge 2(2m-1)!{\displaystyle \frac{{\gamma }_1^{2m}}{{(x+{\gamma }_1^2)}^{2m}}}>0.$$

This is a pointwise margin, not a uniform lower bound on the entire half-line. Only the previously published zero verification is computationally substantial; the accompanying script checks the elementary cutoff arithmetic.

Appendix C.1. The Infinite Hierarchy and Finite Representations

Recall that a Stieltjes function has the form

$$f\left(x\right)=a+{\int }_{[0,\infty )}{\displaystyle \frac{d\mu \left(t\right)}{x+t}},a\ge 0,\mu \ge 0,$$

(A86)

with ${\int }_{[0,\infty )}{(1+t)}^{-1}d\mu \left(t\right)<\infty$. Widder’s real-variable criterion, in the form recalled in ([1], (4)), states that a nonnegative smooth function is Stieltjes if and only if

$${(-1)}^{m-1}{\displaystyle \frac{d^{2m-1}}{d{x}^{2m-1}}}\left(x^{m}f\left(x\right)\right)\ge 0(m\ge 1,x>0).$$

(A87)

Bondesson and Simon call a nonnegative function satisfying (A87) for $m=1,\dots ,k$ a k-Stieltjes function.

Proposition A6 

(Stieltjes formulation). The Riemann hypothesis is equivalent to $F_{\xi }$ being a Stieltjes function on $(0,\infty )$.

Proof. 

Assume RH. Write the upper-half-plane nontrivial zeros as $\rho =1/2+i\gamma$, $\gamma >0$, with multiplicity. Since G has order $1/2$ and genus zero,

$$G\left(x\right)=G\left(0\right)\prod _{\gamma >0}\left(1+{\displaystyle \frac{x}{{\gamma }^2}}\right),$$

with locally uniform convergence. Hence

$$F_{\xi }\left(x\right)=2\sum _{\gamma >0}{\displaystyle \frac{1}{x+{\gamma }^2}}.$$

Because ${\sum }_{\gamma >0}{\gamma }^{-2}<\infty$, this is a Stieltjes representation with positive discrete measure $2{\sum }_{\gamma >0}{\delta }_{{\gamma }^2}$.

Conversely, suppose $F_{\xi }$ is Stieltjes. Its representation extends to a holomorphic function $S\left(z\right)$ on the slit plane

$$\Omega =\mathbb{C}\setminus (-\infty ,0].$$

On $(0,\infty )$ it agrees with the meromorphic function $M\left(z\right)=2G^{\prime }\left(z\right)/G\left(z\right)$. Suppose, toward a contradiction, that G has a zero $z_0\in \Omega$. Choose a small disk D centered at $z_0$ containing no other zero of G, and join a point of $(0,\infty )$ to a point of the punctured disk $D\setminus \left\{z_0\right\}$ by a path in $\Omega$ that avoids the discrete zero set of G. Analytic continuation of the identity $S=M$ along this path shows that $S=M$ on $D\setminus \left\{z_0\right\}$. But if $z_0$ has multiplicity $r\ge 1$, then

$$M\left(z\right)={\displaystyle \frac{2r}{z-z_0}}+O\left(1\right)(z\to z_0),$$

whereas S is holomorphic at $z_0$. This contradiction shows that every zero of G lies on $(-\infty ,0)$.

If $\rho$ is a nontrivial zero of $\zeta$, then ${(\rho -1/2)}^2$ is a zero of G, so

$${(\rho -1/2)}^2<0.$$

Therefore $\rho -1/2$ is purely imaginary, and RH follows. □

Thus RH asks for the entire infinite Widder hierarchy (A87). A single finite level does not suffice, as Proposition A5 shows.

Proposition A7 

(Downward propagation of Widder positivity). Put

$$U_m\left(x\right)={\displaystyle \frac{{\mathcal{W}}_m\left[F_{\xi }\right]\left(x\right)}{(2m-1)!}}.$$

Then for every $m\ge 1$ and $x>0$,

$$U_{m+1}\left(x\right)=-{\displaystyle \frac{U_m^{\prime }\left(x\right)}{2m}}-{\displaystyle \frac{x{U}_m^{\prime \prime }\left(x\right)}{2m(2m+1)}},$$

(A88)

and hence

$${\left(x^{2m+1}{U}_m^{\prime }\left(x\right)\right)}^{\prime }=-2m(2m+1)x^{2m}{U}_{m+1}\left(x\right).$$

(A89)

If $U_{m+1}\left(x\right)\ge 0$ for all $x>0$, then $U_m\left(x\right)\ge 0$ for all $x>0$. Consequently positivity at one order M implies all lower Widder inequalities $1\le m\le M$.

Proof. 

From the zero-sum identity one has

$$U_m\left(x\right)=2\sum _{\rho \in {\mathcal{Z}}_{+}}{\displaystyle \frac{w_{\rho }^m}{{(x+w_{\rho })}^{2m}}}.$$

Termwise differentiation on compact subsets of $(0,\infty )$ gives (A88); multiplying by $2m(2m+1)x^{2m}$ yields (A89).

Assume $U_{m+1}\ge 0$. Since $F_{\xi }$ is analytic at $x=0$, so is $U_m$, and therefore

$$\underset{x\downarrow 0}{lim}{x}^{2m+1}{U}_m^{\prime }\left(x\right)=0.$$

Equation (A89) shows that $x^{2m+1}{U}_m^{\prime }\left(x\right)$ is nonincreasing, hence $U_m^{\prime }\left(x\right)\le 0$ on $(0,\infty )$. For fixed m, the zero sum gives $U_m\left(x\right)\to 0$ as $x\to \infty$. Indeed, $\Re w_{\rho }>0$ implies $|x+w_{\rho }|\ge |w_{\rho }|$, and hence

$$\left|{\displaystyle \frac{w_{\rho }^m}{{(x+w_{\rho })}^{2m}}}\right|\le {\left|w_{\rho }\right|}^{-m}.$$

Since ${\sum }_{\rho }{\left|w_{\rho }\right|}^{-1}<\infty$, dominated convergence applies. Thus $U_m$ decreases to zero and must be nonnegative everywhere. Iteration proves the final assertion. □

Corollary A9 

(Sparse Widder criteria for RH). Let $1\le n_1<n_2<\dots$ be any unbounded sequence of integers. Then

$$\mathrm{RH}⟺{\mathcal{W}}_{n_j}\left[F_{\xi }\right]\left(x\right)\ge 0\mathrm{for}\mathrm{every}x>0\mathrm{and}\mathrm{every}j.$$

(A90)

Equivalently, with $A_m$ defined by (13),

$$\mathrm{RH}⟺A_{n_j}\mathrm{is}\mathrm{nondecreasing}\mathrm{on}(0,\infty )\mathrm{for}\mathrm{every}j.$$

(A91)

In particular one may take $n_j=2^j$.

Proof. 

Under RH, Proposition A6 and Widder’s criterion give all the inequalities. Conversely, fix $m\ge 1$ and choose j with $n_j\ge m$. Proposition A7 propagates the $n_j$th inequality down to the mth one. Hence the full Widder hierarchy holds, so $F_{\xi }$ is Stieltjes and Proposition A6 gives RH. Finally,

$$A_m^{\prime }\left(x\right)={\displaystyle \frac{x^{m-1}}{2(2m-1)!B_m}}{\mathcal{W}}_m\left[F_{\xi }\right]\left(x\right),$$

whose prefactor is positive, proving (A91). □

Bondesson and Simon ([1], Theorem 1) define ${\mathcal{S}}_k$ by nonnegativity together with the first k Widder conditions. Their Theorem 1 states that, for every $k\ge 2$, $f\in {\mathcal{S}}_k$ if and only if there exist $a_k\ge 0$ and a nonnegative measure ${\mu }_k$ on $[0,\infty )$, integrating ${(1+t)}^{-1}$, such that

$$f\left(x\right)=a_k+{\int }_0^{\infty }{\Phi }_{k-1}(x,t)d{\mu }_k\left(t\right).$$

The kernels ${\Phi }_{k-1}$ are their finite-type approximants to the ordinary Stieltjes kernel.

In particular, Theorem A8 yields this representation for $F_{\xi }$ at $k=4,772,153,270,068$ using published inputs; under the additional preprint input of Corollary A7, the same representation holds at $k=4,780,359,257,170$. This is an immediate application of their representation theorem. Its kernel is not the ordinary Stieltjes kernel, so this conclusion does not imply RH.