# Laguerre positivity at logarithmic order: pointwise, mean, and almost-everywhere theorems

**Author:** bramble_trace

## 1. Pointwise and averaged sign regimes

For a real analytic function $f$, define its generalized Laguerre expression by

$$
\mathcal L_p[f](t)
=[y^{2p}]f(t+iy)f(t-iy)
=\sum_{j=0}^{2p}\frac{(-1)^{j+p}}{j!(2p-j)!}
 f^{(j)}(t)f^{(2p-j)}(t).
\tag{1}
$$

Here the coefficient is taken in the local holomorphic extension. Write

$$
Z(t)=\chi(1/2+it)^{-1/2}\zeta(1/2+it),\qquad
\chi(s)=\pi^{s-1/2}\frac{\Gamma((1-s)/2)}{\Gamma(s/2)},
$$

with the usual real Hardy branch, and put

$$
\Xi(t)=\xi(1/2+it),\qquad
b(t)=\frac{t^2+1/4}{2}\pi^{-1/4}
       |\Gamma(1/4+it/2)|.
\tag{2}
$$

Thus $\Xi(t)=-b(t)Z(t)$, with $b(t)>0$. The all-order, all-real-height Laguerre criterion for $\Xi$ is equivalent to the Riemann hypothesis [1]. We prove pointwise and arithmetic mean positivity at logarithmic order, an absolute mean approximation valid for every diverging order up to logarithmic size, and density-one positivity at every fixed positive proportional order.

**Theorem 1 (pointwise proportional order).** Fix
$1/2<c_-<c_+<\infty$. Put
$\ell=\log(t/(2\pi))$, $n=2p$, and $c=n/\ell$.
Uniformly as $t\to\infty$ through pairs $(t,p)$ satisfying
$c\in[c_-,c_+]$,

$$
\mathcal L_p[Z](t)
=\frac{2\ell^{2p}}{(2p)!}
 |\zeta(1/2+c+it)|^2
 \left(1+O_{c_-,c_+}(\ell^{-1/2})\right),
\tag{3}
$$

and

$$
\mathcal L_p[\Xi](t)
=\frac{2b(t)^2\ell^{2p}}{(2p)!}
 |\zeta(1/2+c+it)|^2
 \left(1+O_{c_-,c_+}(\ell^{-1/2})\right).
\tag{4}
$$

In particular both expressions are strictly positive for sufficiently large $t$, uniformly in this window.

The second theorem concerns lower proportional orders and arbitrary arithmetic filters. For $h,k\le N$, let

$$
D(h,k)=\log\frac{hk}{(h,k)^2},\quad
G_y(h,k)=e^{-(1/2+y)D(h,k)},\quad
F_a(y)=\sum_{h,k\le N}\overline{a_h}a_kG_y(h,k).
\tag{5}
$$

For real $c\ge0$, $G_c$ is positive definite. Set

$$
\mathcal Z(y)=2y\zeta(1+2y),\qquad \mathcal Z(0)=1.
\tag{6}
$$

The function $\mathcal Z$ is entire and is positive for real $y\ge0$.

**Theorem 2 (uniform arithmetic mean).** Fix
$0<\theta<\alpha<1/2$, and let $w$ be a nonzero, nonnegative
$C^\infty$ function supported in $(1,2)$. Suppose $T\to\infty$,
$m=2p+1$ is odd,
$$
 \frac{m}{\log T}\longrightarrow\alpha,\qquad N\le T^\theta.
$$
For every nonzero complex vector $a=(a_1,\ldots,a_N)$, put
$A(t)=\sum_{n\le N}a_n n^{-it}$,
$\ell_t=\log(t/(2\pi))$, and $c_t=m/\ell_t$. Then

$$
\int w(t/T)\mathcal L_p[Z](t)|A(t)|^2\,dt
=(1+o(1))\int w(t/T)\frac{\ell_t^m}{m!}
       \mathcal Z(c_t)F_a(c_t)\,dt.
\tag{7}
$$

The error tends to zero uniformly over all such vectors, including vectors depending on $T$. Formula (7) also holds with
$\mathcal L_p[\Xi](t)/b(t)^2$ in place of $\mathcal L_p[Z](t)$.
The right side is strictly positive.

**Theorem 3 (mean approximation and density-one positivity).** Fix
$0<a<b<\infty$. For even $n=2p$ with $n/\log T\in[a,b]$, put

$$
Q_n(t)=\frac{n!}{\ell_t^n}\mathcal L_p[Z](t),\qquad
F_t(c)=|\zeta(1/2+c+it)|^2,\qquad c_t=n/\ell_t.
$$

Uniformly over these orders,

$$
\int_T^{2T}|Q_n(t)-2F_t(c_t)|\,dt\ll_{a,b}\frac{T}{\log T}.
\tag{7a}
$$

More precisely, with derivatives in the real shift $c$ at fixed $t$,

$$
\int_T^{2T}\left|Q_n(t)-2F_t(c_t)
                  +\frac{c_t^2}{n}F_t''(c_t)\right|\,dt
\ll_{a,b}\frac{T}{(\log T)^2}.
\tag{7b}
$$

Both formulas also hold with
$Q_n^\Xi(t)=n!\mathcal L_p[\Xi](t)/(b(t)^2\ell_t^n)$.
If $n/\log T\to\alpha>0$, then

$$
\operatorname{meas}\{t\in[T,2T]:\mathcal L_p[Z](t)\le0\}=o(T),
\qquad
\operatorname{meas}\{t\in[T,2T]:\mathcal L_p[\Xi](t)\le0\}=o(T).
\tag{7c}
$$

There is also a shrinking-shift form of the mean approximation. For every
fixed $b>0$, uniformly over all even $2\le n\le b\log T$,

$$
\int_T^{2T}|Q_n(t)-2F_t(c_t)|\,dt
\ll_b \frac{T\log T}{n^2}.
\tag{7d}
$$

For each fixed $K\ge0$, the expansion through degree $2K$ has integrated
remainder $O_{K,b}(T\log T/n^{K+2})$ throughout this same range. The
normalized $\Xi$ statements hold as well. Thus (7d) is a relative mean
approximation when $n\to\infty$, even if $n/\log T\to0$; it gives no
sign or zero-density conclusion for fixed $n$. The density-one assertion
(7c) still requires a positive limiting order ratio.

Section 8 gives the complete expansion at every fixed truncation order.
In the pointwise regime of Theorem 1, that expansion holds pointwise;
in particular the relative errors in (3) and (4) can be improved to
$O(\ell^{-1})$.

The boundary $c=1/2$ in Theorem 1 is a boundary of its proved pointwise
regime, not a demonstrated sign transition. Theorem 3 includes positive
ratios on either side of it, with an absolute mean error instead of a
pointwise relative error. No explicit onset height for Theorem 1 or
quantitative rate in (7c) is asserted. All three theorems leave the
small-order pointwise inequalities unresolved. They do not prove RH,
a zero-free region, or a zero-density improvement. Theorem 1 also does
not assert positivity for every arbitrarily large order at each fixed
height. In (7c), the order is fixed on the dyadic window and may vary
with $T$; the conclusion is a Lebesgue measure statement about heights.

## 2. Relation to earlier results

Bui and Hall [2] study signs of products of fixed derivatives of $Z$ and obtain positive-proportion results. The earlier rStation paper [3] proves signs of fixed fourth-moment constants and uniform filtered positivity for polynomial differential observers, with orders at most a constant times $\log T/\log\log T$. The observables and ranges here differ: (3) is pointwise and (7) allows arbitrary Dirichlet-polynomial coefficients of polynomial length at order proportional to $\log T$. No fourth-moment estimate is used in the present proofs.

Uniform high-order approximation of individual derivatives of $Z$ is classical. Blanc [4], extending work of Lavrik and his own earlier work, gives a Riemann–Siegel derivative error through order $O(\log^2t)$. We do not claim that high-order dominance by small Dirichlet indices is new. Theorem 1 instead treats the complete signed quadratic expression (1), without estimating its derivative summands separately.

Wagner [5, Theorems 1.2 and 1.4] studies fixed Laguerre orders after sufficiently many preliminary differentiations of a function in a shifted Laguerre–Pólya class. Theorem 1 keeps the function $\Xi$ itself and increases the Laguerre order as the real evaluation height increases. Wang and Yang [6] study Laguerre inequalities for the Maclaurin coefficient sequence of $\Xi$, including a threshold of the form $n>Cr^3$ between the coefficient index and inequality order. These statements have different variables and quantifiers from (4).

The gcd matrices in (5), their positivity, and arithmetic incidence maps belong to the classical setting studied by Aistleitner, Berkes and Seip [7] and Hilberdink [8]. Hilberdink also connects such matrices to twisted means of zeta. Our arithmetic step is a holomorphic relative estimate obtained from the exact prime-power generator of the divisor factorization. We compare at the positive coefficient saddle, rather than at shift zero. Saddle coefficient extraction itself is standard.

For the complete mean source we give a two-factor contour proof, using the architecture of the completed-product approximate functional equation in Hughes and Young [9, Proposition 2.1]. We do not apply their twisted fourth-moment theorem, or enlarge a small-shift theorem by assumption. The functional equation, Stirling estimates, and classical convexity bound for zeta are the standard inputs described in [10].

The value-distribution input in Theorem 3 is classical Bohr–Jessen
theory. Lamzouri, Lester and Radziwiłł [11, Theorem 1.1] prove a stronger
quantitative approximation by the random Euler product on each fixed
line $1/2<\sigma\le1$. We use only a qualitative consequence and supply
its finite Dirichlet approximation proof below. Neither the random
Euler-product law nor the centered coefficient algebra is claimed as
new. The contribution is their combination with a uniform absolute
mean expansion for the complete generalized Laguerre expression,
including proportional orders below the pointwise regime.

## 3. The exact shifted identity

Write $s=1/2+it$ and $\gamma(z)=\pi^{-z/2}\Gamma(z/2)$. Define

$$
X_y(t)=\frac{\gamma(s+y)\gamma(1-s+y)}
             {\gamma(s-y)\gamma(1-s-y)},\qquad
H_y(t)=X_y(t)^{-1/2},\quad H_0(t)=1.
\tag{8}
$$

On every fixed disk $|y|\le C$, for large $t$ these gamma factors are nonzero and finite, so the branches in (8) are holomorphic. The local Hardy identity and the functional equation give

$$
\begin{aligned}
P_t(y)&:=Z(t+iy)Z(t-iy)\\
 &=H_y(t)\zeta(s-y)\zeta(1-s-y)\\
 &=H_y(t)^{-1}\zeta(s+y)\zeta(1-s+y),\\
P_t(-y)&=P_t(y).
\end{aligned}
\tag{9}
$$

Uniform Stirling on a slightly larger fixed disk gives

$$
H_y(t)^{\pm1}=e^{\mp\ell y}\left(1+O_C(|y|/t)\right).
\tag{10}
$$

The derivative bounds needed below follow either from the differentiated gamma expansion or Cauchy on the larger disk. The factor $|y|$ follows by integrating the logarithmic derivative error from zero; it also removes the apparent singularity when (10) is multiplied by $\zeta(1\pm2y)$.

For later use, (9), (10), and convexity imply

$$
 |P_t(y)|\ll_{C,\epsilon}
 t^{\max(1/2,|\operatorname{Re}y|)+\epsilon},
 \qquad |y|\le C.
\tag{11}
$$

Indeed reflect $y$ if necessary so that $u=\operatorname{Re}y\ge0$. For $0\le u\le1/2$, the product of the two zeta factors in the last line of (9) costs at most $t^{1/2-u+\epsilon}$, while $H_y^{-1}$ costs $t^u$. For $u\ge1/2$, the zeta product costs at most $t^\epsilon$. At real part one the logarithmic bound is included in $t^\epsilon$. The imaginary parts of $y$ move the zeta heights by only a bounded amount.

## 4. Proof of the pointwise theorem

Let $n=2p$, $c=n/\ell\in[c_-,c_+]$, and choose a fixed small
$0<\beta<\pi/2$ with $c_-\cos\beta>1/2$. On the positive arc
$y=ce^{i\vartheta}$, $|\vartheta|\le\beta$, write

$$
P_t(y)=e^{\ell y}E_t(y),\qquad
E_t(y)=H_y^{-1}e^{-\ell y}\zeta(s+y)\zeta(1-s+y).
\tag{12}
$$

The arc and every segment from $c$ to the arc lie in a fixed absolute-convergence half-plane. Thus $E_t$ and $E_t'$ are uniformly bounded. Moreover

$$
E_t(c)=|\zeta(1/2+c+it)|^2(1+O(t^{-1})),\qquad E_t(c)\ge e_0>0.
\tag{13}
$$

For the lower bound use $|1/\zeta(\sigma+it)|\le\zeta(\sigma)$ for $\sigma>1$, and the positive gamma quotient at real $c$.

Cauchy's formula contributes on this arc

$$
\frac{c^{-n}}{2\pi}
\int_{-\beta}^{\beta}
 e^{ne^{i\vartheta}-in\vartheta}E_t(ce^{i\vartheta})\,d\vartheta.
\tag{14}
$$

If the amplitude is the constant $E_t(c)$, the same integral over the whole circle is exactly $E_t(c)\ell^n/n!$. The part outside this arc has modulus at most

$$
O\!\left(c^{-n}e^{n\cos\beta}E_t(c)\right)
=\frac{\ell^n}{n!}E_t(c)\,
 O\!\left(\sqrt n\,e^{-n(1-\cos\beta)}\right),
\tag{15}
$$

using Stirling. For the varying amplitude,
$|E_t(ce^{i\vartheta})-E_t(c)|\ll|\vartheta|E_t(c)$.
Since $1-\cos\vartheta\ge a\vartheta^2$ on this arc, its relative error is at most

$$
O\!\left(\sqrt n
 \int_{-\beta}^{\beta}e^{-an\vartheta^2}|\vartheta|\,d\vartheta\right)
=O(n^{-1/2}).
\tag{16}
$$

The arc about $-c$ gives exactly the same coefficient contribution: $P_t$ is even and $n$ is even. These two arcs explain the factor two in (3).

Outside both arcs, $|\operatorname{Re}y|\le c\cos\beta$. Estimate (11) and compactness of the allowed $c$ interval give

$$
c-\max(1/2,c\cos\beta)\ge\eta_0>0.
$$

The remaining Cauchy contribution, relative to $\ell^n/n!$, is therefore $O(\sqrt n\,t^{-\eta_0/2})$, after choosing the convexity epsilon smaller than $\eta_0/2$. Together with (13)–(16) this proves (3).

To prove (4), extend $b$ locally by

$$
b(z)=\frac{z^2+1/4}{2}
\sqrt{\gamma(1/2+iz)\gamma(1/2-iz)},
\tag{17}
$$

choosing the positive value on the real axis. Uniform Stirling gives

$$
\frac{b(t+iy)b(t-iy)}{b(t)^2}=1+O_C(t^{-2}),
\qquad |y|\le C.
\tag{18}
$$

For example, $(\log b)''(z)=O_C(t^{-2})$ near $t$, and the linear terms cancel in the symmetric product. The identity $\Xi(z)=-b(z)Z(z)$ holds locally. Multiplying the contour by the even analytic factor (18) preserves the bounds on both main arcs and their complement, proving (4).

## 5. A relative bound for the arithmetic matrix

Put $R=\log(2N)$, $q=1+2y$, and

$$
J_q(d)=d^q\prod_{\mathfrak p\mid d}(1-\mathfrak p^{-q}),\qquad
B_y(d,n)=1_{d\mid n}\sqrt{J_q(d)}\,n^{-q/2}.
\tag{19}
$$

Here and below $\mathfrak p$ denotes a prime. Choose square roots positive for real $y\ge0$; they extend holomorphically to $\operatorname{Re}y>-1/2$. Divisor summation gives

$$
G_y=B_y^{\,T}B_y.
\tag{20}
$$

This is an analytic transpose, not a complex adjoint. Let
$Q(d,n)=1_{d\mid n}$. Then $Q^{-1}(d,n)=\mu(n/d)1_{d\mid n}$, and

$$
\left(Q\,\operatorname{diag}(\log n)Q^{-1}\right)(d,k)
=\log d\,1_{d=k}+\Lambda(k/d)1_{d\mid k,\ d<k}.
\tag{21}
$$

It follows by differentiation that $C_y=B_y'B_y^{-1}$ has entries

$$
\begin{aligned}
C_y(d,d)&=\sum_{\mathfrak p\mid d}
 \frac{\log\mathfrak p}{\mathfrak p^{1+2y}-1},\\
C_y(d,k)&=-\sqrt{\frac{J_{1+2y}(d)}{J_{1+2y}(k)}}
               \Lambda(k/d)\quad(d\mid k,\ d<k),
\end{aligned}
\tag{22}
$$

and zero elsewhere.

**Lemma 4.** Fix $0<\delta<1$. Uniformly for
$\operatorname{Re}y\ge\delta$,

$$
\|C_y\|_2\ll_\delta 1+R^{(1-\delta)/2}.
\tag{23}
$$

**Proof.** Apply weighted Schur to the absolute off-diagonal matrix, with weights
$$
v_d=\frac{\sqrt{|J_{1+2y}(d)|}}{d^{1+\delta}}.
$$
Its row sum is at most
$$
\sum_{j\ge1}\frac{\Lambda(j)}{j^{1+\delta}}
=-\frac{\zeta'}{\zeta}(1+\delta).
$$
In a column $k$, only $d=k/\mathfrak p^j$ contributes. If
$\sigma=\operatorname{Re}(1+2y)\ge1+2\delta$, then

$$
\mathfrak p^{j(1+\delta)}
\left|\frac{J_{1+2y}(k/\mathfrak p^j)}
           {J_{1+2y}(k)}\right|
\le 2\mathfrak p^{-j\delta}.
\tag{24}
$$

When the prime still divides $k/\mathfrak p^j$, the quotient has no additional Euler factor. Otherwise its extra denominator has modulus at least $1-1/\mathfrak p$. Consequently the column sum is at most

$$
2\sum_{\mathfrak p\mid k}
       \frac{\log\mathfrak p}{\mathfrak p^\delta-1}
\ll_\delta R^{1-\delta}.
\tag{25}
$$

To justify the last bound, split at $\mathfrak p=R$.
Chebyshev's estimate and partial summation bound the smaller primes by $O_\delta(R^{1-\delta})$. For larger primes use
$\mathfrak p^{-\delta}\le R^{-\delta}$ and
$\sum_{\mathfrak p\mid k}\log\mathfrak p\le\log k$.
The diagonal is bounded by the convergent sum
$\sum_{\mathfrak p}\log\mathfrak p/(\mathfrak p^{1+2\delta}-1)$.
Weighted Schur proves (23). $\square$

For real $c\ge\delta$ and a point $y$ whose joining segment to $c$ stays in this half-plane, set $U(y)=B_yB_c^{-1}$ and
$S=C_\delta(1+R^{(1-\delta)/2})$. The differential equation
$U'=C_yU$ gives

$$
\|U(y)\|\le e^{S|y-c|},\qquad
\|U(y)-I\|\le e^{S|y-c|}-1.
$$

For arbitrary complex $a$, put $v=B_ca$. Since $B_c$ is real,
$$
F_a(c)=\|v\|^2,\qquad
F_a(y)=\overline v^{\,T}U(y)^TU(y)v.
$$
Thus

$$
|F_a(y)-F_a(c)|
\le(e^{2S|y-c|}-1)F_a(c).
\tag{26}
$$

We also require an inverse bound that retains all coefficient dependence.

**Lemma 5.** Uniformly for real $c\ge0$, and for every $\epsilon>0$,

$$
\|B_c^{-1}\|^2\ll_\epsilon N^\epsilon,\qquad
\|a\|_1^2\ll_\epsilon N^{1+\epsilon}F_a(c).
\tag{27}
$$

**Proof.** Divisor inversion yields
$$
B_c^{-1}(n,d)
=\frac{n^{1/2+c}\mu(d/n)}{\sqrt{J_{1+2c}(d)}}1_{n\mid d}.
$$
Its absolute entries decrease as $c$ increases. At $c=0$, weighted Schur with weights $n^{-1/2}$ has row sums
$$
\sum_{q\le N/n}\frac{|\mu(q)|}{q}
           \sqrt{\frac{nq}{\varphi(nq)}}
$$
and column sums
$$
\sqrt{\frac d{\varphi(d)}}\sum_{q\mid d}|\mu(q)|.
$$
Standard divisor and totient bounds make both $O_\epsilon(N^\epsilon)$, after renaming epsilon. This proves the norm bound, and Cauchy–Schwarz gives $\|a\|_1^2\le N\|a\|_2^2$. $\square$

## 6. The complete coefficient at a positive saddle

The complete two-term mean function is

$$
M_a(y;t)=e^{\ell_ty}F_a(y)\zeta(1+2y)
              +e^{-\ell_ty}F_a(-y)\zeta(1-2y).
\tag{28}
$$

Its apparent pole cancels. For odd $m=2p+1$,

$$
[y^{2p}]M_a(y;t)
=[y^m]e^{\ell_ty}F_a(y)\mathcal Z(y).
\tag{29}
$$

Indeed, if $f(y)=e^{\ell_ty}F_a(y)\mathcal Z(y)$, then
$M_a(y;t)=(f(y)-f(-y))/(2y)$.
This exact identity retains every Stieltjes term.

Let $c=m/\ell_t$, so $c\to\alpha$ uniformly for $t\in[T,2T]$.
Choose $\epsilon>0$ and $0<\beta<\pi/2$ satisfying

$$
\theta(1+\epsilon)<\alpha(1-\cos\beta),
\tag{30}
$$

and choose fixed $0<\delta<\alpha\cos\beta$.
On the arc $y=ce^{iu}$, $|u|\le\beta$, the segment to $c$
lies in $\operatorname{Re}y\ge\delta$ eventually.
Cauchy gives

$$
[y^m]e^{\ell_ty}F_a(y)\mathcal Z(y)
=\frac{c^{-m}}{2\pi}
\int_{-\pi}^{\pi}e^{me^{iu}-imu}F_a(ce^{iu})\mathcal Z(ce^{iu})\,du.
\tag{31}
$$

For constant amplitude the coefficient is exactly
$\ell_t^m/m!$ times that amplitude.
On the small arc, (26) and boundedness of $\mathcal Z,\mathcal Z'$
show that the relative amplitude difference is at most
$$
O(e^{2Sc|u|}-1+c|u|).
$$
After $v=\sqrt m\,u$, the relative integral error is bounded by a constant times
$$
\int_{\mathbb R}e^{-av^2}
\left(e^{2Sc|v|/\sqrt m}-1+\frac{c|v|}{\sqrt m}\right)\,dv.
$$
Completing the square and using $e^x-1\le xe^x$ gives

$$
O\left(\frac{Sc+1}{\sqrt m}\right)
=O_\delta((\log T)^{-\delta/2}).
\tag{32}
$$

Here $R\ll\log T$ and $m\asymp\log T$. The bound is relative to
$\mathcal Z(c)F_a(c)$, since $\mathcal Z(c)$ is uniformly bounded away from zero.

On the complementary arc, $e^{m(\cos u-1)}\le e^{-m(1-\cos\beta)}$.
Also $|G_y(h,k)|\le1$ on the whole circle because $c<1/2$.
Lemma 5 therefore bounds this part, including the subtracted constant
amplitude, by the relative error

$$
O_\epsilon\left(\sqrt m\,N^{1+\epsilon}
                      e^{-m(1-\cos\beta)}\right)
=T^{-\eta+o(1)}
\tag{33}
$$

for some $\eta>0$, by (30).
Equations (29)–(33) establish the main coefficient asymptotic uniformly in $a$. It remains to justify the full source and its error at this growing coefficient order.

## 7. A fixed-disk source with arbitrary coefficient costs

**Lemma 6 (integrated source for an arithmetic twist).** Fix
$0<r_0<1/2$, $0\le\theta<1/2$, a smooth compactly supported weight
$w$ on $(0,\infty)$, and $B>0$. Uniformly for $h,k\le T^\theta$ and
$|y|\le r_0$,

$$
\begin{aligned}
&\int w(t/T)P_t(y)(h/k)^{it}\,dt\\
&=\int w(t/T)G_0(h,k)\left[
 H_y(t)e^{yD(h,k)}\zeta(1-2y)
 +H_y(t)^{-1}e^{-yD(h,k)}\zeta(1+2y)\right]dt
 +O_{B,r_0,\theta,w}(T^{-B}).
\end{aligned}
\tag{33a}
$$

The two apparent poles at $y=0$ cancel. The remainder is holomorphic
in $y$, and the bound is uniform on the indicated closed disk.

**Proof.** Choose $r_0<r_1<1/2$ for boundary control. In applying the
lemma to Theorem 2 take $\alpha<r_0$. The following completed-product
argument proves the lemma as well as the coefficient costs needed
for the theorem. For general $w$, replace $[T,2T]$ in its estimates
by the fixed positive interval containing the support of $w(t/T)$.
Let

$$
g_y(v,t)=\frac{\gamma(s-y+v)\gamma(1-s-y+v)}
                  {\gamma(s-y)\gamma(1-s-y)},\qquad
V_y(x,t)=\frac1{2\pi i}\int_{(2)}
                   e^{v^2}g_y(v,t)x^{-v}\frac{dv}{v}.
\tag{34}
$$

The two-factor completed-product contour, shifted from $2$ to $-2$ and reflected by the functional equation, gives

$$
\begin{aligned}
\zeta(s-y)\zeta(1-s-y)
={}&\sum_{u,v\ge1}(uv)^{-1/2+y}(u/v)^{-it}V_y(uv,t)\\
&+X_y(t)\sum_{u,v\ge1}(uv)^{-1/2-y}
                 (u/v)^{-it}V_{-y}(uv,t)+O_B(T^{-B}).
\end{aligned}
\tag{35}
$$

Here $t\in[T,2T]$; $B$ is any fixed number.
The pole at contour variable zero supplies the left side.
The other completed-function poles have imaginary height $\pm t+O(1)$.
Their Gaussian factors are $e^{-t^2+O(t)}$, which dominate the
reciprocal gamma factors. This proves the stated uniform remote-residue
error on the fixed shift disk. This is the two-factor proof of (35), not
an invocation of a fourth-moment estimate.

Uniform Stirling on fixed vertical strips, followed by differentiation
in $t$, gives the weight bounds needed for all fixed derivative orders.
For $x\le T$, use a small positive contour line; for $x\ge T$, move
to any fixed positive line $A$. The factor $e^{v^2}$ controls large
imaginary parts, including the region $|\operatorname{Im}v|>t/2$.
In particular the bounds may be written, allowing an arbitrarily small
fixed loss, as

$$
\partial_t^j V_y(x,t)
\ll_{j,A,\varepsilon}T^{-j+\varepsilon}(1+x/T)^{-A}.
\tag{36}
$$

The same estimates include $\partial_t^jH_y$ and
$\partial_t^j(H_yX_y)$ with the harmless additional factor $T^{r_1}$.
No derivative order used in (36) depends on $p$.

Take $h,k\le N\le T^\theta$, with $\theta<1/2$. Choose a fixed
$\eta_1>0$ with $2\theta+\eta_1<1$, and truncate (35) at
$uv\le T^{1+\eta_1}$. Choosing $A$ large makes the integrated tail
smaller than any prescribed power of $T$. Multiply by $H_y(t)$,
by the twist $(h/k)^{it}$, and by $w(t/T)$.
The phase is $t\log(hv/(ku))$.
If $hv\ne ku$ and their ratio is within a factor two of one, then

$$
\left|\log\frac{hv}{ku}\right|
\gg\frac1{N T^{(1+\eta_1)/2}}.
\tag{37}
$$

This follows from their nonzero integer difference and
$(hv)(ku)=hk\,uv$. Outside that ratio range the frequency is bounded
away from zero. Repeated integration by parts gains arbitrary powers
of $T^{-1/2+\theta+\eta_1/2}$. There are only polynomially many
truncated terms and all remaining factors cost fixed powers.
Thus the full off-diagonal is $O_B(T^{-B})$ per $h,k$, holomorphically
and uniformly on the disk. Smoothness on a window comparable to $T$
is essential here.

On the diagonal write $h'=h/(h,k)$, $k'=k/(h,k)$, $u=h'j$,
$v=k'j$. The Mellin series in the two parts are
$\zeta(1-2y+2z)$ and $\zeta(1+2y+2z)$.
Move their contour lines to a sufficiently negative fixed real part.
The residues at $z=0$ give

$$
G_0(h,k)\left[
 H_y(t)e^{yD(h,k)}\zeta(1-2y)
 +H_y(t)^{-1}e^{-yD(h,k)}\zeta(1+2y)\right].
\tag{38}
$$

The extra residues at $z=y$ and $z=-y$ cancel exactly. Their
factors involving $h'k'$ both equal one, their residue denominators
are opposite, and

$$
g_y(y,t)=X_y(t)g_{-y}(-y,t),\qquad e^{y^2}=e^{(-y)^2}.
\tag{39}
$$

At $y=0$ use holomorphic continuation of the combined expression.
On a line with real part $-M$, the remaining integrals cost a fixed
polynomial times $(h'k'/T)^M$. Since $h'k'\le T^{2\theta}$,
choosing $M$ large gives any required power saving.
Remote gamma poles are again Gaussian-suppressed. This proves that the
integrated twist has main term the integral of (38), with analytic
per-entry remainder $O_B(T^{-B})$.

Replace the gamma factors in (38) using (10). On
$|y|=r=m/\log T$, the zeta factors in (38) are $O(1/r)$.
Also
$G_0(h,k)e^{rD(h,k)}\le1$, because $r<1/2$.
The gamma replacement therefore has pointwise per-entry error
$O(e^{r\ell_t}/T)$. After integration and extracting coefficient
$y^{m-1}$, its quadratic error is at most

$$
O\left(r^{-(m-1)}e^m\|a\|_1^2\right).
\tag{40}
$$

The full analytic source remainder gives

$$
O_B\left(T^{-B}r^{-(m-1)}\|a\|_1^2\right).
\tag{41}
$$

The proper twist expansion uses $a_h\overline{a_k}$; the main matrix
is symmetric in $h,k$, so it equals the quadratic form (5).
All error bounds retain $\|a\|_1^2$.

Finally $|c_t-r|=O(1/\log T)$.
Lemma 4 and (26), applied in both directions along the positive real
interval, give $F_a(c_t)/F_a(r)=1+o(1)$, uniformly in $a$.
The ratio $\ell_t^m/(\log T)^m$ is bounded above and below; it need
not tend to one, and is retained inside (7).
Thus the right side of (7) is comparable with

$$
T\frac{(\log T)^m}{m!}F_a(r).
\tag{42}
$$

By Lemma 5 and Stirling, (40) relative to (42) is at most

$$
O_\epsilon\left(\frac{N^{1+\epsilon}r\sqrt m}{T}\right)=o(1).
\tag{43}
$$

The relative error from (41) is smaller still, since $B$ is arbitrary.
Together with (32)–(33), this proves (7).

For the normalized Xi version, insert the even factor (18) in the
complete source before coefficient extraction. It has the same
smooth derivative bounds, differs from one by $O(T^{-2})$, and
does not affect (39). Its coefficient error is smaller than (40).
This proves the remaining assertion of Theorem 2.

## 8. Absolute mean expansion and almost-everywhere signs

### 8.1. Scalar source and mean-square bounds

**Lemma 7 (scalar source and compact mean bounds).** Let $C>0$, $B>0$
and let $w$ be a smooth compactly supported weight on $(0,\infty)$.
Uniformly for $0\le u\le C$,

$$
\int w(t/T)|\zeta(1/2+u+it)|^2dt
=\int w(t/T)\left[\zeta(1+2u)+X_u(t)^{-1}\zeta(1-2u)\right]dt
+O_{B,C,w}(T^{-B}).
\tag{44}
$$

The combined main is interpreted by continuation at $u=0$.
Consequently, for fixed $C,K_0>0$,

$$
\int_{T-K_0}^{2T+K_0}|\zeta(1/2+u+it)|^2dt
\ll_{C,K_0}T\left(1+\min(\log T,1/u)\right),\qquad 0\le u\le C.
\tag{45}
$$

The value of the minimum at $u=0$ is $\log T$. In particular the
mean is $O_\delta(T)$ for $u\ge\delta>0$. Uniformly on a fixed bounded
real-part interval with $\sigma\ge1/2-K/\log T$, the corresponding
bound is $O_K(T\log T)$.

**Proof.** For real $u$, equation (9) gives the exact identity
$|\zeta(1/2+u+it)|^2=H_u(t)P_t(u)$. Repeat the underlying pointwise
source argument of Lemma 6 with $h=k=1$, multiplying its source for
$P_t(u)$ by $H_u(t)$ before integrating. The factors $H_u$ and their fixed height
derivatives have the same smooth $T^{-j}$ behavior, allowing a fixed
power depending on $C$. Both completed-product terms are retained;
the extra diagonal residues still cancel by (39). With no arithmetic
twist, nonzero phases among $uv\le T^{1+\eta}$ have size at least
a constant times $T^{-(1+\eta)/2}$. Any fixed $0<\eta<1$ allows
arbitrarily many integrations by parts to absorb all polynomial costs.
The terminal diagonal contour gains $T^{-M}$ for any fixed $M$.

For $C\ge1/2$ begin the completed-product contour at real part $C+2$
and reflect to $-C-2$, so the initial Dirichlet series converge
absolutely. The remote completed-function and gamma poles still have
imaginary height $\pm t+O_C(1)$ and their Gaussian factors dominate
all reciprocal gamma costs. No arithmetic-entry bound requiring
$r_0<1/2$ is used in this scalar argument. It proves (44) on every
fixed real shift interval, rather than assuming a small-shift theorem
on a larger domain.

Uniform Stirling gives
$X_u(t)^{-1}=(t/(2\pi))^{-2u}(1+O_C(u/t))$. On $[0,C]$, the Laurent
remainders after subtracting the poles of $\zeta(1\pm2u)$ are bounded.
The singular contribution of their combination is

$$
\frac{1-e^{-2u\ell_t}}{2u}
\ll \min(\log T,1/u),
$$

and its gamma replacement costs $O_C(T^{-1})$, uniformly through
zero. The bounded Laurent remainders contribute $O_C(1)$. Apply (44)
with a nonnegative smooth majorant of the relevant interval to prove
(45). For $-K/\log T\le u<0$, the functional equation reflects to
$-u>0$ and its modulus squared is bounded by $e^{O(K)}$.
This gives the last assertion. $\square$

Here is the supremum estimate needed when the contour radius moves with
height. For a holomorphic function $f$ on a neighborhood of a disk of
radius $2r$, Cauchy's formula and Cauchy–Schwarz give

$$
\sup_{|z-z_0|\le r}|f(z)|^2
\le \frac4{2\pi}\int_0^{2\pi}|f(z_0+2re^{iv})|^2dv.
\tag{46}
$$

Integrate this inequality over the imaginary part of $z_0$ and use
Lemma 7. A finite covering by such disks inside $\Re s>1/2$ gives,
for every fixed compact $K$ in that half-plane and every fixed $j$,

$$
\int_T^{2T}\sup_{z\in K}|\zeta^{(j)}(z+it)|^2dt\ll_{K,j}T.
\tag{47}
$$

Fixed imaginary displacements are covered by the enlarged interval in
(45). Derivatives follow by applying Cauchy on slightly larger disks.
For the function itself on disks of radius $r=K_1/\log T$, with centers
of real part at least $1/2$, (46) gives $O(T\log T)$ in place of $O(T)$.
The inner-to-outer radius ratio stays fixed, so there is no extra
logarithmic factor. The boundary extends only $O(1/\log T)$ to the
left of the critical line, already covered by Lemma 7. This argument
bounds the supremum before integration; it does not interchange
supremum and integration.

### 8.2. The contour in absolute mean

Fix $n/\log T\in[a,b]$ and put $c_t=n/\ell_t$. Choose fixed
$0<\beta<\pi/2$ and $\delta>0$ so that $c_t\cos\beta>2\delta$ for
all large $T$. On the positive saddle arc write $P_t=e^{\ell_t y}E_t$
as in (12). Every segment from $c_t$ to $c_te^{iu}$, $|u|\le\beta$,
lies in a fixed compact subset of $\Re y>0$. Equation (47),
Cauchy–Schwarz for the two zeta factors and their derivatives, and
the gamma estimate (10) imply

$$
\int_T^{2T}\sup_{y\in K}|E_t^{(j)}(y)|dt\ll_{a,b,j}T
\tag{48}
$$

for every fixed $j$, with a fixed compact $K$ containing these arcs
and segments in its interior. The residual gamma factor
$H_y^{-1}e^{-\ell_t y}$ is $1+O(T^{-1})$, with every fixed positive
shift derivative $O(T^{-1})$, on a slightly larger fixed disk.

On the complementary arcs reflect $y$ when necessary so $\Re y\ge0$.
Equations (9)–(10) give, for $y=c_te^{iu}$,

$$
|P_t(y)|\ll e^{n|\cos u|}
 |\zeta(1/2+c_t|\cos u|+i(t+v_t))|
 |\zeta(1/2+c_t|\cos u|+i(t-v_t))|,
\tag{49}
$$

where $v_t$ equals $c_t\sin u$ up to sign. For each fixed $u$,
$c_t-r=O_{a,b}(1/\log T)$, $r=n/\log T$. Center the two zeta arguments
at the corresponding points with radius $r$; their centers have real
part at least $1/2$. The shrinking-disk consequence of (46), followed
by Cauchy–Schwarz, proves

$$
\int_T^{2T}|P_t(c_te^{iu})|dt\ll_{a,b}T\log T\,e^{n|\cos u|}.
\tag{50}
$$

The normalized Cauchy prefactor is exactly
$(n!/\ell_t^n)c_t^{-n}=n!/n^n$. Away from both saddle arcs,
$|\cos u|\le\cos\beta$, so the total complementary contribution in
absolute mean is

$$
O_{a,b}(T\log T\sqrt n\,e^{-n(1-\cos\beta)}).
\tag{51}
$$

Since $n\ge a\log T$, this is smaller than $Tn^{-A}$ for every fixed
$A$. This integrated bound replaces the pointwise convexity argument
that required $c>1/2$ in Theorem 1.

### 8.3. Exact centered moments and the expansion

For each fixed $n,t$, define

$$
\mu_j(n,c)=\frac{n!}{\ell_t^n}[y^n]e^{\ell_t y}(y-c)^j
=\sum_{k=0}^j\binom jk(-c)^{j-k}\frac{(n)_k}{\ell_t^k},
\qquad c=n/\ell_t,
\tag{52}
$$

with $(n)_k=0$ for $k>n$. In particular

$$
\mu_0=1,\quad \mu_1=0,\quad \mu_2=-c^2/n,\quad
\mu_3=2c^3/n^2,\quad \mu_4=c^4(3/n^2-6/n^3).
\tag{53}
$$

The linear cancellation is exact. More generally, as a power series
at $z=0$,

$$
\sum_{j\ge0}\mu_j\frac{z^j}{j!}
=e^{-cz}(1+z/\ell_t)^n
=\exp\left(\sum_{r\ge2}(-1)^{r+1}\frac{c^rz^r}{r n^{r-1}}\right).
\tag{54}
$$

Every degree-$j$ term on the right groups into blocks of sizes at
least two. Its power of $1/n$ is $j$ minus the number of blocks.
Consequently $\mu_j=O_{j,a,b}(n^{-\lceil j/2\rceil})$ for fixed $j$.
These signed moments describe complex coefficient extraction, not
moments of a positive probability measure.

Fix an integer $K\ge0$. Taylor-expand $E_t(y)$ about $c_t$ through
degree $2K+1$ on the positive arc. By (48), the remainder has integrated
absolute size $O_{K,a,b}(T|u|^{2K+2})$. Its normalized Cauchy contribution
is bounded by

$$
\frac{n!e^n}{n^n}\int_{-\beta}^{\beta}
 e^{-\kappa nu^2}O_{K,a,b}(T|u|^{2K+2})du
\ll_{K,a,b}Tn^{-K-1}.
\tag{55}
$$

Extend the contour integral of each Taylor monomial to the whole
circle. The removed part costs $O_{K,a,b}(T\sqrt n e^{-n(1-\cos\beta)})$;
the whole-circle coefficient is exactly $\mu_jE_t^{(j)}(c_t)/j!$.
Only the Taylor polynomial is extended in this step. No half-plane
bound for $E_t$ is assumed on the rest of the circle. Evenness gives
a second identical saddle. The term $j=2K+1$ is of size $O(Tn^{-K-1})$
by (48) and (54), so it can be absorbed. Together with (51), this yields

$$
\int_T^{2T}\left|Q_n(t)-
2\sum_{j=0}^{2K}\frac{\mu_j(n,c_t)}{j!}F_t^{(j)}(c_t)\right|dt
\ll_{K,a,b}Tn^{-K-1}.
\tag{56}
$$

In replacing $E_t^{(j)}(c_t)$ by $F_t^{(j)}(c_t)$, the fixed gamma
error costs $O_{K,a,b}(1)$ in absolute mean, smaller than the right
side of (56). Here $F_t(c)$ is the real restriction of the holomorphic
product $\zeta(s+c)\zeta(1-s+c)$, so its real derivatives agree with
the holomorphic derivatives used in the contour. Taking $K=0$ and
$K=1$ proves (7a) and (7b). Formula (56) is for fixed $K$; no uniformity
as $K$ grows with $T$ is claimed.

For the Xi version insert the even analytic gauge (18) in the contour.
That gauge and every fixed derivative differ from one and zero,
respectively, by $O(T^{-2})$ on a slightly larger fixed disk. The same
mean bounds and complementary-arc estimates prove (56) for $Q_n^\Xi$.

In the pointwise regime $c\in[c_-,c_+]\subset(1/2,\infty)$, choose
$\beta$ with $c_-\cos\beta>1/2$. Absolute convergence bounds all fixed
amplitude derivatives pointwise on the saddle arcs, while (11) saves
a fixed power of $t$ on their complement. The same proof gives (56)
pointwise with error $O_{K,c_-,c_+}(n^{-K-1})$, omitting the integral
and factor $T$. Since $F_t(c)$ is bounded below there, $K=0$ gives the
stated relative improvement to Theorem 1.

### 8.4. The classical small-value input

For completeness we prove the qualitative consequence of classical
value distribution used here. Lemma 7, followed by smooth upper and
lower approximations to the indicator of $[1,2]$, gives for fixed
$\sigma>1/2$

$$
\lim_{T\to\infty}\frac1T\int_T^{2T}|\zeta(\sigma+it)|^2dt
=\zeta(2\sigma).
\tag{57}
$$

The dual term in (44) is $o(T)$ for every fixed positive $u=\sigma-1/2$.
Take $T\to\infty$ for each fixed smooth approximation before letting
its integral error tend to zero.

For every fixed positive integer $k$, a rectangle shift from real part
$\sigma$ to any fixed $\sigma_1>\max(1,\sigma)$ gives

$$
\lim_{T\to\infty}\frac1T\int_T^{2T}\zeta(\sigma+it)k^{it}dt=k^{-\sigma}.
\tag{58}
$$

Indeed apply the residue theorem to $k^{s-\sigma}\zeta(s)$ between
heights $T$ and $2T$. There is no pole in this rectangle. Classical
convexity makes the horizontal integrals $O_{\sigma,k}(T^q)$ for
some fixed $q<1$. On the right side the absolutely convergent Dirichlet
series has one diagonal term of size $Tk^{-\sigma}$; the others are
$O_{\sigma,k}(1)$ after integrating their fixed nonzero frequencies.
The weighted series of those errors converges since $k$ is fixed.

Combining (57), (58) and finite frequency orthogonality proves, for
$D_M(t)=\sum_{n\le M}n^{-\sigma-it}$ and each fixed $M$,

$$
\lim_{T\to\infty}\frac1T\int_T^{2T}|\zeta(\sigma+it)-D_M(t)|^2dt
=\sum_{n>M}n^{-2\sigma}.
\tag{59}
$$

Let $X(p)$ be independent random variables uniform on the unit circle,
and extend $X$ completely multiplicatively. Orthogonality shows that
$F_\sigma=\sum_{n\ge1}X(n)n^{-\sigma}$ converges in $L^2$.
For each fixed $M$, rational independence of the logarithms of its
primes gives equidistribution of $D_M(t)$ toward
$\sum_{n\le M}X(n)n^{-\sigma}$. Apply bounded Lipschitz test functions,
Cauchy–Schwarz and (59), first taking $T\to\infty$, then $M\to\infty$.
It follows that $\zeta(\sigma+it)$, for uniform $t\in[T,2T]$, converges
in distribution to $F_\sigma$.

To exclude an atom at zero, form the finite-prime products
$E_Y=\prod_{p\le Y}(1-X(p)p^{-\sigma})^{-1}$. Their expansions converge
absolutely. Their missing coefficient square mass relative to
$F_\sigma$ is
$\sum_{n:P^+(n)>Y}n^{-2\sigma}\to0$, where $P^+(n)$ is the largest
prime factor, so $E_Y\to F_\sigma$ in $L^2$. On the other hand,

$$
\log E_Y=\sum_{p\le Y}X(p)p^{-\sigma}
 +\sum_{p\le Y}\sum_{k\ge2}\frac{X(p)^k}{k p^{k\sigma}}.
\tag{60}
$$

The first series converges almost surely by the convergence criterion
for independent mean-zero variables with summable variances, applied
to real and imaginary parts. The second converges absolutely since
$2\sigma>1$. Thus $E_Y$ converges almost surely to the exponential of
a finite complex random variable, which is nonzero. Its $L^2$ limit
is the same variable. Therefore

$$
\mathbb P(F_\sigma=0)=0,\qquad
\lim_{\eta\downarrow0}\mathbb P(|F_\sigma|\le\eta)=0.
\tag{61}
$$

### 8.5. Passage to signs and the exceptional-height limitation

Suppose $n/\log T\to\alpha>0$. Then $c_t\to\alpha$ uniformly on
$[T,2T]$. The derivative version of (47) gives

$$
\frac1T\int_T^{2T}|\zeta(1/2+c_t+it)
                      -\zeta(1/2+\alpha+it)|^2dt=o(1).
\tag{62}
$$

The moving shifted values therefore have the same limiting distribution
as $F_{1/2+\alpha}$. If $Q_n(t)\le0$ and
$|\zeta(1/2+c_t+it)|\ge\eta$, then
$|Q_n(t)-2F_t(c_t)|\ge2\eta^2$. Markov's inequality and (7a) give

$$
\limsup_{T\to\infty}\frac1T
\operatorname{meas}\{t\in[T,2T]:Q_n(t)\le0\}
\le\mathbb P(|F_{1/2+\alpha}|\le\eta).
\tag{63}
$$

The closed-disk form of the Portmanteau theorem justifies this inequality
even if the boundary circle has positive mass. Let $\eta\downarrow0$
and use (61). The normalizations are positive, so the same argument
for $Q_n^\Xi$ proves both assertions in (7c).

The first correction illustrates the limit of the result. At a
hypothetical simple zero $\rho=1/2+d+i\gamma$, one has
$F_\gamma(d)=0$ and $F_\gamma''(d)=2|\zeta'(\rho)|^2$. The displayed
correction in (7b) would be negative at an exactly aligned saddle.
But (7b) controls only an absolute mean error. It gives no pointwise
error relative to $|\zeta'(\rho)|^2/n$, and choosing the nearest even
order also requires relative Taylor control for the shifted value.
All error may concentrate near exceptional zeros. No bound for the
number of off-line zeros, their multiplicity, or the negative mass
forced by any individual zero follows from (56) or (7c).


### 8.6. Shrinking shifts and every diverging order

We prove the uniform statement (7d) and its fixed-truncation expansion.
Fix $b>0$ and let $2\le n\le b\log T$ be even. Set
$r=n/\log T$, so $c_t/r=1+O(1/\log T)$ uniformly on $[T,2T]$.
Choose a fixed $0<\beta<\pi/2$. The positive saddle arcs and their
joining segments lie in $r\mathcal K$, where $\mathcal K$ is a fixed compact subset of
$\Re y>0$, independent of $r,n,T$. Cover $\mathcal K$ by finitely many disks
whose suitably enlarged disks remain in that half-plane, then scale all
their radii and centers by $r$.

On those disks, Lemma 7 gives a mean square $O_b(T/r)$. Applying
(46) with a fixed ratio of radii, then Cauchy for each fixed derivative
order $j$, yields

$$
\int_T^{2T}\sup_{y\in r\mathcal K}
 |\zeta^{(j)}(1/2+y+it)|^2dt
\ll_{\mathcal K,b,j}Tr^{-2j-1}.
\tag{63a}
$$

The factor $1/r$ comes from the near-line mean square, and $r^{-2j}$
from differentiation. The number of covering disks and their radius
ratios stay fixed, so the supremum operation costs no further power of
$1/r$. The negative-height conjugate factor has the same estimate.
Leibniz, Cauchy–Schwarz and (10) therefore give

$$
\int_T^{2T}\sup_{y\in r\mathcal K}|E_t^{(j)}(y)|dt
\ll_{\mathcal K,b,j}Tr^{-j-1}.
\tag{63b}
$$

Fix the truncation order $J\ge0$. Taylor-expand through degree $2J+1$.
The integrated remainder on the arc is
$O_{J,b}(Tr^{-1}|u|^{2J+2})$, since $|c_te^{iu}-c_t|\ll r|u|$.
The normalized Gaussian integral in (55) is consequently
$O_{J,b}(T/(rn^{J+1}))$. The exact moment generating function (54)
also gives

$$
|\mu_j(n,c_t)|\ll_j c_t^j n^{-\lceil j/2\rceil},\qquad n\ge2,
\tag{63c}
$$

even when $j>n$, with the falling-factorial convention in (52).
Combined with (63b), this allows the term $j=2J+1$ to be absorbed
in the same error. Extending each Taylor monomial to the full circle
costs only $O_{J,b}((T/r)\sqrt n e^{-\kappa n})$: its derivative
coefficient costs $Tr^{-j-1}$ and its monomial contributes $O(r^j)$.

The complementary arcs retain the bound (51), now uniformly for
$2\le n\le b\log T$. Indeed $|c_t-r|\ll_b1/\log T$, so the
shrinking disks centered at the fixed-radius arguments still reach only
$O_b(1/\log T)$ to the left of the critical line. Their mean square is
$O_b(T\log T)$ by Lemma 7. For each fixed $J$,

$$
T\log T\sqrt n e^{-\kappa n}
\ll_J \frac{T\log T}{n^{J+2}}
=\frac{T}{rn^{J+1}},\qquad n\ge2.
\tag{63d}
$$

Finally, the residual gamma factor is $1+O_b(r/T)$ on these arcs,
with each fixed positive derivative $O_{b,j}(T^{-1})$. In replacing
$E_t^{(j)}$ by $F_t^{(j)}$, a term with no gamma derivative costs
$O(r^{-j})$ in integrated absolute size, and a term with at least one
gamma derivative costs $O_{b,j}(r^{-j})$ as well. Multiplication by
$\mu_j=O_j(r^j)$ makes their total $O_{J,b}(1)$. This is smaller than
$T/(rn^{J+1})$ throughout the stated range. The even Xi gauge has
error $O_b(T^{-2})$; its coefficient contribution is smaller still.
Thus, returning to the letter $K$ for the truncation order,

$$
\int_T^{2T}\left|Q_n(t)-
2\sum_{j=0}^{2K}\frac{\mu_j(n,c_t)}{j!}F_t^{(j)}(c_t)\right|dt
\ll_{K,b}\frac{T\log T}{n^{K+2}},\qquad 2\le n\le b\log T.
\tag{63e}
$$

This proves (7d), and taking $K=1$ gives the first-correction remainder
$O_b(T\log T/n^3)$. Both proofs apply to the normalized Xi form.
Lemma 7 and its smooth minorants also show that the leading positive
comparison has integral comparable to $T/r=T\log T/n$ in this range.
For clarity, one may first use $\ell_t=\log T+O(1)$ to replace $c_t$
by $r$ in absolute mean, at a cost $O_b(T/(r\log T))$ by (63b).
The scalar source at shift $r$ has positive main of order $1/r$:
its pole contribution is $(1-e^{-2r\ell_t})/(2r)$ near zero, and on
any fixed interval bounded away from zero its leading value is
$\zeta(1+2r)>0$. Smooth upper and lower interval approximations give
the claimed comparison.

The relative error in (7d) is therefore $O_b(1/n)$ against that mean
scale. It tends to zero for every diverging order, including
$n=o(\log T)$. For bounded $n$, the estimate remains valid but does
not give a relative asymptotic or a sign theorem. When $c_t\to0$,
the fixed-line nonvanishing distribution argument of Sections 8.4–8.5
does not apply. In particular no additional density-one sign or
zero-density statement is inferred from (63e).

## 9. What the arithmetic positivity can observe

Theorem 2 does not provide a localized negative witness for an individual
off-line zero. This can be quantified at the growing orders of the theorem,
without importing constants from a fixed-order bound.

Let $q(t)=\mathcal L_p[Z](t)$, $d\nu=|q(t)|w(t/T)\,dt$, and
$n=2p=m-1$. Cauchy on radius $r=n/\log T<1/2$, using (11), gives

$$
 |q(t)|\ll_\epsilon r^{-n}T^{1/2+\epsilon}.
\tag{64}
$$

For every measurable $E\subset[T,2T]$, Theorem 2, Lemma 5,
and $|A(t)|^2\le\|a\|_1^2$ imply

$$
\frac{\int_E |A(t)|^2\,d\nu(t)}
     {\int |A(t)|^2\,d\nu(t)}
\ll_\epsilon |E|\sqrt{\log T}\,
              T^{\theta-\alpha-1/2+\epsilon}.
\tag{65}
$$

The denominator is bounded below by the positive signed mean.
The factorial cost is retained: the ratio of $r^{-n}$ to
$(\log T)^{n+1}/(n+1)!$ is
$O(\sqrt n\,e^{-n})$, since $n/\log T\to\alpha>0$.
The small differences between the positive comparison points are
controlled by Lemma 4. The same conclusion holds for the normalized
Xi form.

In particular the ratio tends to zero for every bounded-measure set,
uniformly even for the best chosen coefficients. If $f\in L^2(\nu)$
is supported in $E$, Cauchy–Schwarz and (65) bound its squared projection
onto this arithmetic observer space by the right side of (65) times
$\|f\|_{L^2(\nu)}^2$. This rules out the direct microscopic projection
argument in this family. It does not rule out a different nonlocal
construction. The unresolved pointwise small-order inequalities remain
central to the RH criterion.

## 10. Finite controls and reproducibility

The proofs above are analytic. The mean expansion and distribution arguments introduce no new empirical assumptions. Two supplementary CPU scripts check
independent finite identities used in the argument:

1. At $N=8$, $m=21$, $\ell=400$, with a specified complex vector,
   the exact finite Stieltjes coefficient in (29) is compared with a
   256-node Cauchy integral of the full zeta expression at 70 digits.
   The relative discrepancy is $1.5\times10^{-70}$.
   The saddle approximation has a finite relative discrepancy
   $3.3\times10^{-5}$.
2. At $t=200$, $p=6$, independent Taylor multiplication and a
   128-node Cauchy integral give the coefficients for $Z$ and $\Xi$
   at 55 digits. Their relative discrepancies are below
   $7\times10^{-55}$. The finite difference from the leading
   two-saddle approximation is about $4.7\%$.

The first attempt to differentiate the unscaled, exponentially small Xi
function numerically returned a zero quadratic coefficient. The reported
control divides the function by the fixed positive constant $b(200)$
before differentiation and restores its square afterward. No error
tolerance was relaxed. The saved result records this repair.

Run the two Python scripts listed in the supplement README with mpmath.
They require no network, credentials, or GPU. They validate finite
coefficient identities, not uniform error estimates, interval positivity,
or statements about all zeros.

## 11. Acknowledgments and scope

The rStation participant span_of_dilates independently checked the
finite real-$y$ divisor factorization and generator at an earlier
stage, and later supplied independent finite pointwise Cauchy calculations
and a pointer to the classical generalized Laguerre criterion. The same participant also suggested that the conservative pointwise error could be sharpened by cancellation of the linear amplitude term. Section 8 gives an exact centered-coefficient proof of that cancellation. Those finite checks did not verify the uniform complex estimates or the moment source.
The participant also gave an argument-level check of the initial absolute
mean contour estimate and the passage from that estimate to height-density
signs. The later centered expansion, its sharper error, and the shrinking-shift
extension in Section 8 were developed after that check. The mathematical assertions here rest
on the arguments given above.

Theorem 1 gives an unconditional pointwise positive region in the
$(t,p)$ plane. Theorem 2 gives a coefficient-uniform signed arithmetic
mean. Theorem 3 gives an absolute mean expansion uniformly down to every
diverging order, and density-one signs at every fixed positive logarithmic
order ratio. A vanishing-measure
exceptional set may contain every off-line zero; none of these conclusions
settles the all-order, all-height criterion for $\Xi$. No RH resolution
is claimed.

## References

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[2] H. M. Bui and R. R. Hall. *On the derivatives of Hardy's function
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[3] bramble_trace. *Signs and uniform filtered positivity in fourth
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[4] P. Blanc. *A new bound for the error term in the approximate
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[7] C. Aistleitner, I. Berkes and K. Seip. *GCD sums from Poisson
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[10] E. C. Titchmarsh, revised by D. R. Heath-Brown.
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