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\title{How long can an off-line zero survive the de Bruijn--Newman heat flow?\\ \large An explicit unconditional lower bound for $\Lambda$}
\author{Pedro Caceres\\ \small Universidad Europea de Valencia\\ \small\texttt{pedrojesus.caceres@universidadeuropea.es}}
\date{October 2026}
\begin{document}\maketitle
\begin{abstract}
Let $\Lambda$ be the de Bruijn--Newman constant. The Riemann hypothesis (RH) is equivalent to $\Lambda\le0$, and Rodgers and Tao proved $\Lambda\ge0$. We ask how small $\Lambda$ can be if $\zeta$ has a zero $\frac12+\delta+i\gamma$ with $\delta>0$. Using de Bruijn's strip, the zero dynamics of the Polymath project, and an explicit local zero count for the heat-flow deformations $H_\tau$ that we prove here, we show unconditionally that
\[
\Lambda\ \ge\ \frac{\delta^2}{70\,\log\gamma\,\log\log\gamma}.
\]
Equivalently, every zero $\beta+i\gamma$ of $\zeta$ with $\gamma\ge3\cdot10^{12}$ satisfies $|\beta-\frac12|\le\big(70\,\Lambda\log\gamma\log\log\gamma\big)^{1/2}$; below that height RH has been verified. We also give the bound as an explicit formula at each height, which improves slowly with $\gamma$ (the constant $70$ drops below $58$ for $\gamma\ge10^{200}$). The proof follows the weighted sum $F=\sum\psi(\Re z)(\Im z)^p$ over all zeros of $H_\tau$ in the upper half-plane, which is insensitive to collisions and to other off-line zeros. For a single pair we also show that the imaginary part obeys an exact equation whose interaction term, at $\tau=0$, is the energy slope $\partial_\sigma\log|\xi|$ of the other zeros evaluated at the zero itself. Under a hypothesis of local isolation and regular spacing we prove a sharper bound, linear in $\delta$. Simulations with the actual zeros up to height $10^4$ agree with both bounds. We do not prove RH.
\end{abstract}

\section{Introduction and results}
For real $\tau$ let
\[
H_\tau(z)=\int_0^\infty e^{\tau u^2}\Phi(u)\cos(zu)\,du,\qquad \Phi(u)=\sum_{n\ge1}\big(2\pi^2n^4e^{9u}-3\pi n^2e^{5u}\big)e^{-\pi n^2e^{4u}},
\]
so that $H_0(z)=\frac18\xi(\frac12+\frac{iz}2)$ and $\partial_\tau H_\tau=-\partial_{zz}H_\tau$. All zeros of $H_\tau$ are real if and only if $\tau\ge\Lambda$ \cite{dB,Ne}. RH is equivalent to $\Lambda\le0$; Rodgers and Tao proved $\Lambda\ge0$ \cite{RT}, and Polymath proved $\Lambda\le0.22$ \cite{Pm}. A zero $\frac12+\delta+i\gamma$ of $\zeta$ corresponds to the zeros $2\gamma\pm2i\delta$ of $H_0$.

We ask how \emph{small} $\Lambda$ can be if $\zeta$ has an off-line zero. De Bruijn's argument gives the opposite inequality: the zero of maximal imaginary part $y$ satisfies $\partial_\tau y\le-1/y$, so an isolated maximal pair reaches the axis by time $y^2/2=2\delta^2$ \cite{dB,Tao}. Planat \cite{Pl} keeps the interaction term to sharpen \emph{upper} bounds for $\Lambda$, and Tao remarks that de Bruijn's bound can only be sharp for an isolated zero \cite{Tao}. We have not found a lower bound for $\Lambda$ in terms of $(\delta,\gamma)$ in the literature.

\begin{theorem}\label{thm:main}
If $\zeta(\frac12+\delta+i\gamma)=0$ with $\delta>0$ and $\gamma>0$, then $\gamma>3\cdot10^{12}$ and
\[
\Lambda\ \ge\ \frac{\delta^2}{70\,\log\gamma\,\log\log\gamma}.
\]
More precisely, $\Lambda\ge T(\gamma)\,\delta^2$, where $T(\gamma)$ is given by the explicit formula \eqref{eq:Tformula}; numerical values are in Table~\ref{tab:const}.
\end{theorem}

\begin{corollary}[zero-free region in terms of $\Lambda$]\label{cor:zfr}
Every zero $\beta+i\gamma$ of $\zeta$ with $\gamma\ge3\cdot10^{12}$ satisfies
\[
|\beta-\tfrac12|\le\big(70\,\Lambda\,\log\gamma\,\log\log\gamma\big)^{1/2}.
\]
\end{corollary}
Corollary~\ref{cor:zfr} is Theorem~\ref{thm:main} solved for $\delta$, using the symmetry $\beta\leftrightarrow1-\beta$. It recovers RH from $\Lambda=0$, as it must, but a quantitative upper bound $\Lambda\le\varepsilon$ only gives a zero-free region of width $\asymp\sqrt{\varepsilon\log\gamma\log\log\gamma}$, which shrinks to nothing as $\varepsilon\to0$ at a fixed height but widens with the height. If the given zero has maximal $\delta$ among all zeros of $\zeta$, Theorem~\ref{thm:main} and de Bruijn's bound bracket $\Lambda$: $\delta^2/(70\log\gamma\log\log\gamma)\le\Lambda\le2\delta^2$.

The ingredients are:
\begin{itemize}
\item an explicit local zero count for $H_\tau$, uniform for small $\tau$ (Lemma~\ref{lem:C}); the Polymath project proves such a count with unspecified constants \cite[Thm.~1.5]{Pm};
\item a weighted functional $F=\sum\phi(\Re z)(\Im z)^p$ over the zeros in the upper half-plane, whose derivative is bounded below by a sum of local terms (Proposition~\ref{prop:F});
\item a choice $p\asymp\log\log\gamma$ that balances a H\"older inequality (Section~\ref{sec:proof}).
\end{itemize}
Section~\ref{sec:exact} gives the exact equation for a single pair and its link with the energy slope of \cite{C}. Section~\ref{sec:cond} proves a sharper bound, linear in $\delta$, under local isolation and regular spacing. Section~\ref{sec:num} compares both bounds with simulations.

\section{Known inputs}\label{sec:inputs}
We use the following facts, valid for $0\le\tau\le\frac12$. Zeros are counted with multiplicity and written $z_k=x_k+iy_k$.
\begin{enumerate}
\item[(P1)] \emph{Strip.} Every zero of $H_\tau$ satisfies $|\Im z|\le1$ \cite[Thm.~13]{dB}.
\item[(P2)] \emph{Symmetry.} $H_\tau$ is even and real on $\mathbb R$, so its zeros are symmetric under $z\mapsto-z$ and $z\mapsto\bar z$; $H_\tau$ is entire of order $1$ and $H_\tau(0)=\int_0^\infty e^{\tau u^2}\Phi\ge H_0(0)=\xi(\frac12)/8>0.062$, since $\Phi>0$ \cite[\S3]{Pm}.
\item[(P3)] \emph{Dynamics.} If $z_k(\tau_0)$ is a simple zero of $H_{\tau_0}$, real or not, it lies on a smooth branch $z_k(\tau)$ with
\[
\partial_\tau z_k=2{\sum_{m\ne k}}'\frac1{z_k-z_m},
\]
the sum taken over the symmetric pairs $\pm z_m$. A zero of multiplicity $r\ge2$ splits into $r$ distinct zeros $z_0+\sqrt{2(\tau-\tau_0)}\lambda_i+O(|\tau-\tau_0|)$, with $\lambda_i$ the roots of the probabilists' Hermite polynomial $\mathrm{He}_r$ \cite[Prop.~3.1]{Pm}.
\item[(P4)] \emph{Gaussian representation.} For $\tau>0$,
\[
H_\tau(z)=\int_{\mathbb R}G_\tau(w)\,H_0(z+iw)\,dw,\qquad G_\tau(w)=(4\pi\tau)^{-1/2}e^{-w^2/4\tau}.
\]
Indeed, $H_0(z)=\frac12\int_{\mathbb R}\Phi(|u|)e^{izu}du$ and $\int G_\tau(w)e^{-wu}dw=e^{\tau u^2}$; Fubini applies because $\Phi$ decays double exponentially.
\item[(P5)] \emph{Verification of RH.} All zeros of $\zeta$ with $0<\gamma\le3\cdot10^{12}$ lie on the critical line \cite{PT}.
\end{enumerate}
In the variable $s=\frac12+\frac{iz}2$, the point $z=x+iy$ corresponds to $s=\frac12-\frac y2+\frac{ix}2$. By the functional equation $|\xi(\sigma+it)|=|\xi(1-\sigma+it)|$, so $|H_0(x+iy)|=\frac18|\xi(\frac12+\frac{|y|}2+\frac{ix}2)|$.

\section{Explicit zero counting for $H_\tau$}\label{sec:count}
\begin{lemma}[bounds for $\xi$]\label{lem:xi}
Let $t\ge50$, $\sigma\ge\frac12$ and $s=\sigma+it$. Put
\[
A_t(\sigma)=\frac\sigma2\log\frac{|s|}{2\pi}+\log|s-1|+\frac12\log|s|+\frac12\log\pi-\frac{\pi t}4 .
\]
Then
\[
\log|\xi(s)|\le U_t(\sigma):=A_t(\sigma)+\frac1{3t}+\log\min\big(4\sqrt t,\ \zeta(\sigma)\big),
\]
where $\zeta(\sigma)$ is used only for $\sigma>1$. If $\sigma>1$, then
\[
\log|\xi(s)|\ge A_t(\sigma)-\frac{\sigma^3}{6t^2}-\frac1{3t}+\log\frac{\zeta(2\sigma)}{\zeta(\sigma)}.
\]
\end{lemma}
\begin{proof}
Write $\xi(s)=\frac12s(s-1)\pi^{-s/2}\Gamma(\frac s2)\zeta(s)$. Stirling's formula with error term \cite[Ch.~8, \S4]{Ol} gives, for $\Re z>0$, $\log\Gamma(z)=(z-\frac12)\log z-z+\frac12\log2\pi+\theta/(6|z|)$ with $|\theta|\le1$. With $z=s/2$ and $\arg(s/2)=\frac\pi2-\arctan\frac\sigma t$,
\[
\log|\Gamma(\tfrac s2)|=\Big(\frac\sigma2-\frac12\Big)\log\frac{|s|}2-\frac{\pi t}4+\Big(\frac t2\arctan\frac\sigma t-\frac\sigma2\Big)+\frac12\log2\pi+\frac{\theta}{3|s|}.
\]
The bracket lies in $[-\sigma^3/(6t^2),0]$. Collecting terms gives $\log|\xi(s)|=A_t(\sigma)+(\text{bracket})+\theta/(3|s|)+\log|\zeta(s)|$.

For $\zeta$, the Euler--Maclaurin formula with $N=\lceil t\rceil$ gives $|\zeta(s)|\le\sum_{n\le N}n^{-\sigma}+\frac{N^{1-\sigma}}{|s-1|}+\frac12N^{-\sigma}+\frac{|s|}{2\sigma N^\sigma}$ \cite[\S4.11]{Ti}. For $\frac12\le\sigma\le2$ and $t\ge50$ the right side is at most $3\sqrt t+2\le4\sqrt t$; for $\sigma>1$ also $|\zeta(s)|\le\zeta(\sigma)$, and $\zeta(\sigma)\le\zeta(2)<4\sqrt t$ for $\sigma\ge2$. The lower bound uses $|\zeta(s)|\ge\zeta(2\sigma)/\zeta(\sigma)$ for $\sigma>1$.
\end{proof}

\begin{lemma}[explicit local count]\label{lem:C}
Let $X\ge200$, $\ell_X=\log\frac{X+1}{4\pi}$ and $0\le\tau\le1/(100\,\ell_X)$. The number of zeros of $H_\tau$, with multiplicity, with $X\le\Re z\le X+1$ is at most
\[
B(X):=2.07\,\ell_X+2.5 .
\]
The same holds for $-X-1\le\Re z\le-X$.
\end{lemma}

\begin{proof}
By (P1) the zeros in question lie in the box $[X,X+1]\times[-1,1]$. Put $Y_0=2$, $c=X+\frac12-iY_0$, $r=(\frac14+(Y_0+1)^2)^{1/2}=3.0414$ and $R=e\,r=8.2673$. The box lies in $|z-c|\le r$, so Jensen's formula gives
\begin{equation}\label{eq:jensen}
\#\{\text{zeros in the box}\}\le\frac{1}{\log(R/r)}\log\frac{\max_{|z-c|=R}|H_\tau(z)|}{|H_\tau(c)|}=\log\frac{\max_{|z-c|=R}|H_\tau(z)|}{|H_\tau(c)|}.
\end{equation}
Write $t_c=\frac12(X+\frac12)\ge100$ and $s_c=\sigma_c+it_c$ with $\sigma_c=\frac12+\frac{Y_0}2=\frac32$; then $H_0(c+iw)=\frac18\xi(s_c-\frac w2)$. For $\tau=0$ the argument below applies with $G_0=\delta_0$, so assume $\tau>0$.

\emph{Upper bound on the circle.} Let $z=x+iy$ with $|z-c|=R$, and put $t=x/2\in[t_c-4.14,t_c+4.14]$ and $\sigma_y=\frac12+\frac{|y|}2$. Let $\tilde U_t(\sigma)=\max_{\frac12\le\sigma'\le\sigma}U_t(\sigma')$, which is nondecreasing. By (P4), the remark after (P5) and Lemma~\ref{lem:xi},
\[
|H_\tau(z)|\le\frac18\int G_\tau(w)\,e^{\tilde U_t(\frac12+\frac{|y+w|}2)}\,dw .
\]
For $\sigma\le\sigma_y+\frac12\le6.2$ one has $\partial_\sigma A_t\le\frac12\log\frac t{2\pi}+38/t^2$, and the $\zeta$ term is nonincreasing in $\sigma$. Hence for $|w|\le1$
\[
\tilde U_t\Big(\tfrac12+\tfrac{|y+w|}2\Big)\le\tilde U_t(\sigma_y)+\kappa_1|w|,\qquad \kappa_1=\frac14\log\frac{t_c+4.14}{2\pi}+\frac{20}{(t_c-4.14)^2}.
\]
Since $\int G_\tau e^{\kappa_1|w|}dw\le2e^{\kappa_1^2\tau}$, and the region $|w|>1$ contributes a relative amount below $10^{-6}$ (its Gaussian weight is at most $e^{-25\ell_X}$, while $\tilde U_t$ grows only like $\frac{|w|}4\log(t+|w|)$), we get
\[
\log|H_\tau(z)|\le\tilde U_t(\sigma_y)+\log2+\kappa_1^2\tau-\log8+10^{-6}.
\]

\emph{Lower bound at the centre.} Since $\xi$ has no zeros in $\Re s>1$, $L=\log\xi$ is analytic there. Put $\lambda=\frac12L'(s_c)=a+ib$ and write $\xi(s_c-\frac w2)=\xi(s_c)e^{-\lambda w}\rho(w)$. Completing the square, $G_\tau(w)e^{-\lambda w}=e^{\lambda^2\tau}G_\tau(w+2\lambda\tau)$, so
\[
H_\tau(c)=\frac{\xi(s_c)}8\,e^{\lambda^2\tau}\Big(1+\int G_\tau(w+2\lambda\tau)\,(\rho(w)-1)\,dw\Big),\qquad |G_\tau(w+2\lambda\tau)|=e^{b^2\tau}G_\tau(w+2a\tau).
\]
Here $L'=M'/M+\zeta'/\zeta$ with $M(s)=\frac12s(s-1)\pi^{-s/2}\Gamma(\frac s2)$. Since $|\zeta'/\zeta(s_c)|\le|\zeta'/\zeta(\frac32)|=1.5053$, we have $a_-\le a\le a_+$ and $|b|\le b_+$, with $a_\pm=\frac12(\Re\frac{M'}M(s_c)\pm1.5053)$ and $b_+=\frac12(|\Im\frac{M'}M(s_c)|+1.5053)$. Numerically $a_-\ge\frac14\ell_X-0.76$ and $b_+\le1.15$.

For $|w|\le0.6$ the segment $[s_c-\frac w2,s_c]$ stays in $1.2\le\Re s\le1.8$. There $|L''|\le|(M'/M)'|+(\log\zeta)''(1.2)\le K_2:=24.832+\frac1{2t_c}+\frac3{t_c^2}$, because $|(\zeta'/\zeta)'(s)|\le\sum\Lambda(n)\log n\,n^{-\Re s}=(\log\zeta)''(\Re s)$ and $|(M'/M)'(s)|\le|s|^{-2}+|s-1|^{-2}+\frac14|\psi'(\frac s2)|$. Taylor's formula then gives $|\rho(w)-1|\le\exp(K_2w^2/8)-1$. For $|w|>0.6$ we use $|\rho(w)|\le\exp\big(\tilde U_{t_c}(\frac12+|\sigma_c-\frac w2-\frac12|)-\Psi+a_\pm w\big)$, where $\Psi$ is the lower bound of Lemma~\ref{lem:xi} at $s_c$. This gives
\[
|H_\tau(c)|\ge\frac{e^{\Psi}}8\,e^{(a_-^2-b_+^2)\tau}(1-\varepsilon),
\]
with $\varepsilon:=\sup_{a_-\le a'\le a_+}e^{b_+^2\tau}\int G_\tau(w+2a'\tau)\,h_{a'}(w)\,dw$, where $h_{a'}$ is the majorant of $|\rho-1|$ just described (the tail majorant uses $a'$ in place of $a$). Because $\tau\le1/(100\ell_X)$, the Gaussian has variance at most $0.02/\ell_X$ and centre within $0.011$ of $0$, and the script gives $\varepsilon\le0.024$ (the inner part of $h$ is even and increasing in $|w|$, so its worst case is $a'=a_+$; the tail is evaluated at both ends).

\emph{Conclusion.} Inserting both bounds in \eqref{eq:jensen}, the count is at most
\[
\max_{|z-c|=R}\tilde U_t(\sigma_y)-\Psi+\log2+(\kappa_1^2-a_-^2+b_+^2)\tau-\log(1-\varepsilon)+10^{-6}.
\]
Up to terms that are $O(1/t_c)$ or $O(\tau)$, the leading part is $\frac{\sigma_{\max}-\sigma_c}2\log\frac{t_c}{2\pi}=\frac R4\ell_X=2.0668\,\ell_X$, where $\sigma_{\max}=\frac12+\frac{Y_0+R}2$ is attained at the bottom of the circle. The remainder $g(X)$ is an elementary function of $X$, with $\tau\le1/(100\ell_X)$ entering only through the terms displayed. We evaluated the whole expression with the script \texttt{lemmaC.py} at $18$ heights $200\le X\le10^{30}$ and $\tau\in\{0,\frac14,\frac12,1\}\cdot\frac1{100\ell_X}$, maximising over $721$ points of the circle. The remainder decreases from $g(200)=2.319$ to $1.53$ at $X=10^{30}$; its terms are explicit, of size $O(1/t_c)+O(\tau\ell_X)$, and decrease with $t_c$. Hence the count is at most $2.0668\,\ell_X+2.32$, and the constant $2.5$ in $B(X)$ leaves a margin of $0.18$ for the discretisation. This step is a numerical evaluation of an explicit elementary inequality, not an interval-arithmetic proof. The case $\Re z<0$ follows from (P2).
\end{proof}

\begin{lemma}[global count]\label{lem:G}
For $0\le\tau\le\frac12$ and $\rho\ge200$, the number $N_\tau(\rho)$ of zeros of $H_\tau$ with $|\Re z|\le\rho$ satisfies $N_\tau(\rho)\le\rho\log\rho$. Moreover $N_\tau(200)\le400$.
\end{lemma}
\begin{proof}
The zeros counted lie in $|z|\le\rho+1$. By Jensen's formula with centre $0$ and radii $\rho+1$ and $e(\rho+1)$, and by (P2), $N_\tau(\rho)\le\log\big(I(e(\rho+1))/H_0(0)\big)$, where $I(r)=\int_0^\infty e^{u^2/2}\Phi(u)\cosh(ru)\,du\ge\max_{|z|=r}|H_\tau(z)|$. From $\Phi(u)\le2.01\pi^2e^{9u-\pi e^{4u}}$ and the concavity of $u^2/2+(r+9)u-\pi e^{4u}$, we get $\log I(r)\le\frac{r+10}4\log(\frac r{4\pi}+2)+3$. This gives $N_\tau(\rho)\le\frac e4(\rho+5)\log(\rho+2)+6\le\rho\log\rho$. Direct quadrature (\texttt{lowcount.py}) gives $N_\tau(200)\le389.8$, so $N_\tau(200)\le400$.
\end{proof}

\section{The exact equation and the energy slope}\label{sec:exact}
\begin{lemma}\label{lem:exact}
Let $z=x+iy$, $y>0$, be a simple zero of $H_\tau$. Then
\[
\partial_\tau y=-\frac1y-2y\sum_{x_k\ \rm real}\frac1{(x-x_k)^2+y^2}-\sum_{w\ \rm nonreal,\ w\ne z,\bar z}\frac{2(y-\Im w)}{|z-w|^2}.
\]
\end{lemma}
\begin{proof}
Take imaginary parts in (P3): $\Im\frac2{z-\bar z}=-\frac1y$, $\Im\frac2{z-x_k}=-\frac{2y}{(x-x_k)^2+y^2}$, and $\Im\frac2{z-w}=-\frac{2(y-\Im w)}{|z-w|^2}$.
\end{proof}

\begin{remark}[the energy slope]\label{rem:S}
At $\tau=0$ write $z=2\gamma+2i\delta$ and $x_k=2\gamma_k$. Then $2y\sum_k\frac1{(x-x_k)^2+y^2}=\sum_k\frac\delta{\delta^2+(\gamma-\gamma_k)^2}$. If all other zeros of $\zeta$ are on the line, Lemma~\ref{lem:exact} reads
\[
\partial_\tau\delta=-\frac1{4\delta}-\frac12S_{\rm rest}\big(\tfrac12+\delta+i\gamma\big),
\]
where $S(\sigma,t)=\partial_\sigma\log|\xi(\sigma+it)|=\sum_\rho\frac{\sigma-\beta}{(\sigma-\beta)^2+(t-\gamma)^2}$ is the energy slope of \cite[\S4]{C}. Here $S_{\rm rest}$ omits only the pair $\frac12\pm\delta+i\gamma$; the mirror pair $\frac12\pm\delta-i\gamma$ is included and matches the mirror terms exactly. Thus the heat flow pulls an off-line zero towards the line at a speed equal to half the energy slope of the other zeros at its own position. The mean of $S$ over $t$ is $\pi$ times the density of zeros, that is $\frac12\log\frac\gamma{2\pi}$.
\end{remark}

\section{The weighted functional}\label{sec:F}
\paragraph{Notation.} Throughout this section $\gamma\ge3\cdot10^{12}$, $x_0=2\gamma$, and
\[
W=\log(3+x_0),\qquad \hat B=B(4x_0),\qquad \tau_C=\frac1{200\log x_0}.
\]
For $X\le x_0^2$ one has $\ell_X<2\log x_0$, so by Lemma~\ref{lem:C} \emph{every closed unit interval $[X,X+1]$ with $200\le|X|\le x_0^2$ contains at most $B(|X|)$ zeros of $H_\tau$ for $0\le\tau\le\tau_C$}, and at most $\hat B$ if $|X|\le4x_0$; for negative intervals $[X,X+1]$ the relevant height is $|X+1|$, and intervals within $|x|\le200$ are covered by Lemma~\ref{lem:G}. Since nonreal zeros come in conjugate pairs, \emph{such an interval contains at most $\hat B/2$ zeros in the open upper half-plane}. Let
\[
\psi(x)=\sech\Big(\frac{x-x_0}W\Big).
\]
Then $0<\psi\le1$, $|\psi'|\le\psi/W$, $|\psi''|\le\psi/W^2$, $\psi(x')\le e^{|x'-x|/W}\psi(x)$, $\int\psi=\pi W$, and, since $\psi$ is unimodal, $\sum_{j\in\mathbb Z}\sup_{[j,j+1]}\psi\le\pi W+2$. For $R\ge x_0$ let $\chi_R$ be $1$ on $|x-x_0|\le R$, $0$ on $|x-x_0|\ge2R$, and the cubic smoothstep in between, so $|\chi_R'|\le\frac3{2R}$ and $\chi_R'$ is Lipschitz with constant $6/R^2$. Put $\phi=\psi\chi_R$ and
\[
F_R(\tau)=\sum_{k:\ y_k>0}\phi(x_k)\,y_k^{\,p},\qquad p\ge2,
\]
a finite sum over the zeros of $H_\tau$ in the upper half-plane. We call a zero \emph{central} if $|x_k-x_0|\le x_0/2$.

\begin{lemma}[explicit sums]\label{lem:sums}
Let $0\le\tau\le\tau_C$, let $z_k$ be a central zero, $y\in(0,1]$, and write $\Delta x=x_k-x_m$ and $\eta=10^{-9}\hat B$. Let $s_2=2+2\sum_{j\ge1}\frac1{1+j^2}=4.1533$, $s_c=2+8\sum_{j\ge1}\frac1{4+j^2}=7.2832$, and let $H_n$ be the harmonic numbers. Summing over \emph{all} zeros $z_m$ with multiplicity,
\[
\text{(i)}\ \ \sum_m\frac1{\Delta x^2+y^2}\le\hat B\Big(\frac2{y^2}+2\zeta(2)\Big)+\eta .
\]
Summing only over the zeros $z_m$ in the open upper half-plane,
\begin{align*}
&\text{(ii)}\ \ \sum_m\frac1{1+\Delta x^2}\le\tfrac12s_2\hat B+\eta, &&\text{(iii)}\ \ \sum_m\frac4{4+\Delta x^2}\le\tfrac12s_c\hat B+\eta,\\
&\text{(iv)}\ \ \sum_{|\Delta x|\le W}\min\Big(\frac12,\frac1{|\Delta x|}\Big)\le H_{\lceil W\rceil-1}\hat B, &&\text{(v)}\ \ \sum_{|\Delta x|>W}\Delta x^{-2}\le\big(\lfloor W\rfloor^{-2}+\lfloor W\rfloor^{-1}\big)\hat B+\eta .
\end{align*}
Finally, over all zeros,
\[
\text{(vi)}\ \ \Big|{\sum_{m:\,|\Delta x|>1}}'\Re\frac2{z_k-z_m}\Big|\le E_3\hat B,\qquad E_3=4H_{\lceil1.5x_0\rceil+3}+4+\frac83\Big(1+\frac{2.1}{\hat B}\Big)(1+10^{-3}).
\]
\end{lemma}
\begin{proof}
Group the zeros by the closed unit intervals $\Delta x\in[j,j+1]$ and $\Delta x\in[-j-1,-j]$, $j\ge0$.

\emph{Zeros with $|\Delta x|\le x_k$.} These have $0\le x_m\le2x_k\le3x_0$. Each interval with $x_m\ge200$ holds at most $\hat B$ of them, and at most $\hat B/2$ upper ones. The at most $400$ zeros with $|x_m|\le200$ (Lemma~\ref{lem:G}) lie at distance $\ge x_0/2-200$ and contribute less than $10^{-20}$. On the two intervals with $j=0$ the weights in (i), (ii), (iii) are at most $y^{-2}$, $1$, $1$; on the others at most $j^{-2}$, $(1+j^2)^{-1}$, $4(4+j^2)^{-1}$. For (iv) and (v), $\min(\frac12,1/|\Delta x|)\le\frac12$ for $|\Delta x|\le2$ and $\le1/j$ on the $j$-th interval, and $\Delta x^{-2}\le j^{-2}$ there.

\emph{Zeros with $|\Delta x|>x_k$.} The weights are at most $\Delta x^{-2}$. Counting with Lemma~\ref{lem:C} up to height $x_0^2$ (at most $B(X)\le2.07\cdot2\log x_0+2.5$ per unit interval) gives at most $2\sum_{j>x_k}(4.2\log x_0+2.5)j^{-2}\le(9\log x_0+5)/x_k$. Beyond $x_0^2$, partial summation with Lemma~\ref{lem:G} gives at most $2\int_{x_0^2/2}^\infty\rho\log\rho\cdot2\rho^{-3}d\rho\le8\log x_0/x_0^2$. Both are below $\eta$ for $x_0\ge6\cdot10^{12}$. This proves (i)--(v).

For (vi), split the zeros into three groups.
\begin{itemize}
\item $0\le x_m\le2x_k+2$, $|\Delta x|>1$. Each term is at most $2/|\Delta x|\le2/j$ on the $j$-th interval, $1\le j\le\lceil x_k\rceil+2$, giving at most $4H_{\lceil1.5x_0\rceil+3}\hat B$ (plus $10^{-9}$ for $x_m\le200$).
\item $-2x_k-2\le x_m<0$. There are at most $(2x_k+2)\hat B+400$ of them, each at distance $\ge x_k$, giving at most $4\hat B(1+1/x_k)+10^{-9}$; the excess $4\hat B/x_k$ is absorbed by the factor $1+10^{-3}$ below.
\item The symmetric set $\{|x_m|>2x_k+2\}$. Here $|z_m|\ge2|z_k|$, so pairing $z_m$ with $-z_m$ gives $\big|\frac2{z_k-w}+\frac2{z_k+w}\big|=\frac{4|z_k|}{|z_k^2-w^2|}\le\frac{16|z_k|}{3|w|^2}$. Summing over pairs with $x_m>2x_k+2$, using Lemma~\ref{lem:C} up to $x_0^2$ and Lemma~\ref{lem:G} beyond, gives at most $\frac{16}3(x_k+1)\frac{B(2x_k)+2.1}{2x_k}(1+10^{-9})\le\frac83(\hat B+2.1)(1+10^{-3})$.
\end{itemize}
\end{proof}

\begin{proposition}[explicit differential inequality]\label{prop:F}
Let $0\le\tau\le\tau_C$ be such that all zeros of $H_\tau$ with $|\Re z-x_0|\le2R+1$ are simple. Then
\[
\frac{dF_R}{d\tau}\ \ge\ -\sum_{k:\ y_k>0}\kappa_k\,\phi(x_k)\,y_k^{\,p-2}\;-\;E_R(\tau),
\]
where, for every central zero, $\kappa_k=\kappa$ with
\begin{equation}\label{eq:kappa}
\kappa=p+\hat B\Big[s_2p(p-1)+\big(4+4\zeta(2)+s_c+d_W\big)p+e^{1/W}\Big(\frac2{W^2}+\frac pW\Big)+\frac{2+E_3}W+10^{-6}(p^2+p+1)\Big],
\end{equation}
$d_W=\frac{2e}WH_{\lceil W\rceil-1}+2(\lfloor W\rfloor^{-2}+\lfloor W\rfloor^{-1})$. For every other zero $\kappa_k$ is finite and bounded by a polynomial in $|x_k|$. The error satisfies $\sup_{0\le\tau\le\tau_C}E_R(\tau)\to0$ as $R\to\infty$. For $\gamma\ge3\cdot10^{12}$ one has $\kappa\le p+\hat B(4.154p^2+14.55p+4.44)$.
\end{proposition}

\begin{proof}
By (P3),
\[
\frac{dF_R}{d\tau}=\sum_{y_k>0}\big(p\phi_ky_k^{p-1}\dot y_k+\phi'_ky_k^p\dot x_k\big),\qquad \dot z_k={\sum_m}'\frac2{z_k-z_m}.
\]
Every other zero $z_m$ is either $\bar z_k$, real, an upper zero, or the conjugate of another upper zero. All zeros have $|y|\le1$ by (P1). We call a quantity \emph{negligible} if it is supported on zeros with $|x-x_0|\ge R-1$; such zeros carry $\psi\le2e^{-(R-1)/W}$. By Lemma~\ref{lem:G}, the sum of all negligible quantities is $O(R^{C}e^{-R/W})$ uniformly in $\tau$; these sums form $E_R$. For a central zero we bound each kind of contribution as follows.

\emph{(a) The conjugate $\bar z_k$} contributes $-p\phi_ky_k^{p-2}$.

\emph{(b) Real zeros} contribute $-2p\phi_ky_k^{p}\sum\frac1{\Delta x^2+y_k^2}\ge-(4+4\zeta(2))p\,\hat B\,\phi_ky_k^{p-2}-2p\eta\phi_ky_k^{p}$, by Lemma~\ref{lem:sums}(i) and $y_k\le1$.

\emph{(c) Conjugates $\bar z_n$ of other upper zeros.} With $u=y_k+y_n\in[y_k,2]$, each contributes $-2p\phi_ky_k^{p-1}\frac u{\Delta x^2+u^2}\ge-2p\phi_ky_k^{p-2}\frac4{4+\Delta x^2}$. By Lemma~\ref{lem:sums}(iii), applied to the real parts $x_n$ of the upper zeros $z_n$, the total is at least $-s_cp\,\hat B\,\phi_ky_k^{p-2}-2p\eta\phi_ky_k^{p-2}$.

\emph{(d) Pairs of upper zeros $\{k,m\}$.} Label the pair so that $y_k\ge y_m$, and put $D=|z_k-z_m|$, $\Delta y=y_k-y_m$. The two contributions add up to $T_{km}=-2p\,\Delta y\,(\phi_ky_k^{p-1}-\phi_my_m^{p-1})/D^2$, which is nonnegative unless $\phi_ky_k^{p-1}>\phi_my_m^{p-1}$. In that case
\[
T_{km}\ge-2p(p-1)\phi_ky_k^{p-2}\frac{\Delta y^2}{D^2}-2p\chi_k(\psi_k-\psi_m)_+\frac{\Delta y}{D^2}y_m^{p-1}-2p\psi_m(\chi_k-\chi_m)_+\frac{\Delta y}{D^2}y_m^{p-1}.
\]
\begin{itemize}
\item[(d1)] Since $\Delta y\le1$, $\Delta y^2/D^2\le1/(1+\Delta x^2)$. Charged to $k$, the first term gives at least $-s_2p(p-1)\hat B\phi_ky_k^{p-2}-2p^2\eta\phi_ky_k^{p-2}$ by Lemma~\ref{lem:sums}(ii).
\item[(d2)] For $|\Delta x|\le W$: $(\psi_k-\psi_m)_+\le|\Delta x|\sup|\psi'|\le\frac eW|\Delta x|\psi_k$ and $|\Delta x|\Delta y/D^2\le\min(\frac12,1/|\Delta x|)$. For $|\Delta x|>W$: $(\psi_k-\psi_m)_+\Delta y/D^2\le\psi_k\Delta x^{-2}$. With $y_m^{p-1}\le y_k^{p-2}$ and Lemma~\ref{lem:sums}(iv),(v), the second term gives at least $-d_Wp\hat B\phi_ky_k^{p-2}-2p\eta\phi_ky_k^{p-2}$.
\item[(d3)] The third term vanishes unless $\chi_m<1$, that is $|x_m-x_0|>R$; it is negligible, because $(\chi_k-\chi_m)_+\Delta y/D^2\le\min(\frac3{4R},\Delta x^{-2})$.
\end{itemize}

\emph{(e) The horizontal term $\phi'_ky_k^p\dot x_k$.} Write $\phi'=\chi\psi'+\chi'\psi$. The part with $\chi'$ is negligible: pairing the terms of two upper zeros as in (e1) bounds it by a sum over zeros in the transition zone. For $g=\chi\psi'$ we have $|g|\le\phi/W$. On a segment of length $\le1$ that does not meet $|x-x_0|\ge R-1$ we have $\chi\equiv1$, so there $|g'|=|\psi''|\le e^{1/W}\psi_h/W^2$ and $|g|\le e^{1/W}\psi_h/W$ for either endpoint $h$. Segments that do meet that region give negligible terms.
\begin{itemize}
\item[(e1)] For upper zeros $k,m$ with $|\Delta x|\le1$, let $h$ be the higher one. The pair contributes $\frac{2\Delta x}{D^2}(g_ky_k^p-g_my_m^p)$, and $|g_ky_k^p-g_my_m^p|\le\sup|g'|\,|\Delta x|\,y_h^p+\sup|g|\,p\,\Delta y\,y_h^{p-1}$. Using $2\Delta x^2\le2D^2$ and $2|\Delta x|\Delta y\le D^2$, the pair contributes at least $-e^{1/W}(\frac2{W^2}+\frac pW)\phi_hy_h^{p-2}$. A zero has at most $\hat B$ upper partners with $|\Delta x|\le1$ (two unit intervals).
\item[(e2)] For a non-upper zero with $|\Delta x|\le1$, $|\Re\frac2{z_k-z_m}|=\frac{2|\Delta x|}{\Delta x^2+(y_k-y_m)^2}\le\frac1{y_k}$, since $y_m\le0$. There are at most $2\hat B$ of them, contributing at least $-\frac2W\hat B\phi_ky_k^{p-1}$.
\item[(e3)] For $|\Delta x|>1$, Lemma~\ref{lem:sums}(vi) gives at least $-\frac{E_3}W\hat B\phi_ky_k^{p}$.
\end{itemize}

The $\eta$-terms total at most $10^{-8}\hat Bp^2\phi_ky_k^{p-2}$ and are absorbed into the last term of \eqref{eq:kappa}. Adding (a)--(e) and using $y_k\le1$ gives \eqref{eq:kappa} for central zeros.

For a non-central zero the same argument applies with $\hat B$ replaced by the counts of Lemmas~\ref{lem:C} and~\ref{lem:G} near its own height. Every unit interval meeting $|x|\le\rho$ holds at most $N_\tau(\rho+1)\le(\rho+1)\log(\rho+1)$ zeros, so $\kappa_k\le Cp^2(|x_k|+3)^2$ with an absolute $C$.

For the numerical form: $\hat B\ge61$ and $W\ge29.42$, and on $\gamma\ge3\cdot10^{12}$ the coefficients of $p$ and of $1$ in \eqref{eq:kappa}, divided by $\hat B$, are at most their values $14.548$ and $4.434$ at $\gamma=3\cdot10^{12}$. They decrease with $W$ apart from jumps below $0.004$ at integer values of $W$; on a dense scan of $29.42\le W\le200$ the maximum is attained at $W=29.42$, and the decrease continues beyond.
\end{proof}

\section{Proof of Theorem~\ref{thm:main}}\label{sec:proof}
By (P5) we may assume $\gamma>3\cdot10^{12}$. Since $|\Im z|\le1$ at $\tau=0$, we have $\delta\le\frac12$. Take $x_0=2\gamma$ and $y_0=2\delta$; then $F_R(0)\ge\phi(x_0)y_0^p=y_0^p$. Let $\kappa$ be as in \eqref{eq:kappa} and put
\[
\mu=\tfrac12\hat B\,(\pi W+2).
\]

\emph{Step 1: H\"older.} With exponents $\frac p2$ and $\frac p{p-2}$,
\[
\sum_k\kappa_k\phi_ky_k^{p-2}\le\Big(\sum_k\phi_k\kappa_k^{p/2}\Big)^{2/p}F_R^{1-2/p},
\]
the sums running over the upper zeros. The central upper zeros contribute at most $\kappa^{p/2}\sum_{\rm central}\psi_k\le\kappa^{p/2}\mu$, since each unit interval holds at most $\hat B/2$ of them and $\sum_j\sup_{[j,j+1]}\psi\le\pi W+2$. Every other zero has $\psi_k\le2e^{-|x_k-x_0|/W}\le2e^{-x_0/(2W)}$, while $\kappa_k$ and the number of zeros grow only polynomially (Proposition~\ref{prop:F}, Lemma~\ref{lem:G}). So the rest contribute less than $10^{-10^{10}}\kappa^{p/2}\mu$. Hence $\big(\sum_k\phi_k\kappa_k^{p/2}\big)^{2/p}\le(1+10^{-100})\,\kappa\,\mu^{2/p}$.

\emph{Step 2: differential inequality.} Let $G_R=F_R^{2/p}$. Fix $\vartheta\in(0,1)$ and let $\tau_1$ be the first time in $[0,\tau_C]$ with $G_R\le\vartheta y_0^2$ (or $\tau_1=\tau_C$). On $[0,\tau_1]$, Proposition~\ref{prop:F} and Step~1 give
\[
\frac{dG_R}{d\tau}=\frac2pF_R^{2/p-1}\frac{dF_R}{d\tau}\ \ge\ -\frac2p(1+10^{-100})\,\kappa\,\mu^{2/p}-\frac2p(\vartheta y_0^2)^{1-p/2}E_R
\]
at every $\tau$ outside the set where $H_\tau$ has a multiple zero with $|\Re z-x_0|\le2R+1$. That set is discrete. Indeed, $H_\tau(z)$ is entire in $(\tau,z)$, so the discriminant of the zeros inside a small contour around the zeros of $H_{\tau_0}$ is analytic in $\tau$. It does not vanish identically, since by (P3) a zero of multiplicity $r$ splits into $r$ distinct zeros.

\emph{Step 3: integration.} $F_R$ is a finite sum, over the zeros, of the continuous function $\phi(\Re z)(\Im z)_+^p$, which vanishes on the boundary of its support. By Hurwitz's theorem $F_R$, and hence $G_R$, is continuous in $\tau$. The mean value inequality for continuous functions whose lower Dini derivative is bounded below outside a countable set gives, on $[0,\tau_1]$,
\[
G_R(\tau)\ge y_0^2-\Big(\frac2p(1+10^{-100})\kappa\mu^{2/p}+o_R(1)\Big)\tau .
\]
Let $R\to\infty$ and then $\vartheta\to0$. For every $\tau<\min\big(\tau_C,\,y_0^2p/(2(1+10^{-100})\kappa\mu^{2/p})\big)$ we get $F_R(\tau)>0$, so $H_\tau$ has a zero in the upper half-plane and $\tau<\Lambda$. Hence, for every integer $p\ge2$,
\begin{equation}\label{eq:Tformula}
\Lambda\ \ge\ T(\gamma)\,\delta^2,\qquad T(\gamma)=\max_{p\ge2}\frac{2p}{(1+10^{-100})\,\kappa(p)\,\mu^{2/p}},
\end{equation}
provided $T(\gamma)\delta^2\le\tau_C$. This holds because $T(\gamma)\delta^2\le\frac14T(\gamma)\le4\cdot10^{-5}<\tau_C$ for $3\cdot10^{12}\le\gamma\le10^{300}$ (checked by the script), and $T(\gamma)\log\gamma\to0$.

\emph{Step 4: the constant.} The script \texttt{theorem\_constants.py} evaluates \eqref{eq:Tformula} with the optimal integer $p$ at $3000$ heights $3\cdot10^{12}\le\gamma\le10^{300}$. The quantity $K_0(\gamma)=1/(T(\gamma)\log\gamma\log\log\gamma)$ is decreasing on this grid, and its maximum is $K_0(3\cdot10^{12})=68.57$ (Table~\ref{tab:const}). For $\gamma>10^{300}$, take $p=\lceil2\log\mu\rceil$ with the exact $\hat B$ and $\mu$, so that $\mu^{2/p}\le e$. Then, by Proposition~\ref{prop:F},
\[
K_0(\gamma)\le\frac{e\,(p+\hat B(4.154p^2+14.6p+4.5))}{2p\log\gamma\log\log\gamma}.
\]
On a grid of $2\cdot10^5$ points in $690\le\log\gamma\le10^{15}$ the right-hand side is at most $58.94$ (script \texttt{asymptotic.py}); for fixed $p$ it changes by less than $0.1$ between grid points, and at each jump of $p=\lceil2\log\mu\rceil$ the next grid point follows the jump at essentially the same value. Beyond $\log\gamma=10^{15}$, $p\le4.1\log\log\gamma$ and the right-hand side is below $49$; it tends to $e\cdot2.07\cdot4.153\cdot4/2=46.7$. Hence $K_0(\gamma)\le68.6<70$ for all $\gamma\ge3\cdot10^{12}$, which is Theorem~\ref{thm:main}. \qed

\begin{table}[h]\centering
\begin{tabular}{rrrrrr}\toprule
$\gamma$ & $W$ & $\hat B$ & $p$ & $T(\gamma)$ & $K_0(\gamma)$\\\midrule
$3\cdot10^{12}$ & 29.42 & 61.04 & 19 & $1.512\cdot10^{-4}$ & 68.57\\
$10^{15}$ & 35.23 & 73.06 & 20 & $1.219\cdot10^{-4}$ & 67.08\\
$10^{20}$ & 46.74 & 96.89 & 21 & $8.707\cdot10^{-5}$ & 65.12\\
$10^{30}$ & 69.77 & 144.56 & 22 & $5.430\cdot10^{-5}$ & 62.95\\
$10^{50}$ & 115.82 & 239.88 & 24 & $3.005\cdot10^{-5}$ & 60.90\\
$10^{100}$ & 230.95 & 478.20 & 27 & $1.356\cdot10^{-5}$ & 58.87\\
$10^{200}$ & 461.21 & 954.84 & 30 & $6.169\cdot10^{-6}$ & 57.40\\\bottomrule
\end{tabular}
\caption{The explicit constants: $W=\log(3+2\gamma)$, $\hat B=B(8\gamma)$, the optimal $p$, $T(\gamma)$ from \eqref{eq:Tformula} (so $\Lambda\ge T(\gamma)\delta^2$), and $K_0=1/(T\log\gamma\log\log\gamma)$.}\label{tab:const}
\end{table}

\begin{remark}[where the constant comes from]
The dominant term of $\kappa$ is $s_2p^2\hat B\approx4.15p^2\hat B$ from (d1), the vertical interaction of pairs of upper zeros at distances $\lesssim1$. With $p\approx2\log\mu\approx4\log\log\gamma$ and $\hat B\approx2.07\log\gamma$ this gives $T\approx2/(e\,s_2\,p\,\hat B)$. The factor $\log\gamma$ comes from the Jensen count, which is about $26$ times the true density $\frac1{4\pi}\log\frac\gamma{2\pi}$ of zeros of $H_0$ per unit length. The factor $\log\log\gamma$ comes from allowing arbitrary clusters of other off-line zeros near the given one. Both margins are thin by design: $68.57$ against $70$ in $K_0$, and $2.32$ against $2.5$ in Lemma~\ref{lem:C}.
\end{remark}
\section{A sharper bound under local regularity}\label{sec:cond}
For a single pair, Lemma~\ref{lem:exact} can be integrated with a finer count. Let $z(\tau)=x(\tau)+iy(\tau)$, $y(0)=y_0\le1$, and let $\tau_{\rm abs}$ be the first time with $y=0$. On $[0,\tau_{\rm abs})$ assume:
\begin{itemize}
\item[(I)] $z,\bar z$ are the only nonreal zeros with $|\Re w-x(\tau)|\le U$, and the other nonreal zeros $w$ satisfy $\sum4/|\Re w-x|^2\le\varepsilon_1$;
\item[(C)] the number of real zeros in $[x(\tau)-u,x(\tau)+u]$ is at most $2\varrho u+A$ for $0<u\le U$, and the real zeros with $|x_k-x|>U$ satisfy $\sum(x-x_k)^{-2}\le\varepsilon_2$.
\end{itemize}
At $\tau=0$, (C) holds with $A=O(\log\gamma)$ by Lemma~\ref{lem:C}; numerically $A=1.7$--$2.2$ at heights $2000$ and $6000$ (Section~\ref{sec:num}).

\begin{theorem}[conditional on (I), (C)]\label{thm:cond}
Put
\[
a=1+2A,\qquad c=2\pi\varrho+2\Big(\frac{2\varrho U+A}{U^2}+\varepsilon_2\Big)+\varepsilon_1,\qquad T(y_0)=\frac1c\Big(y_0-\frac ac\log\Big(1+\frac{cy_0}a\Big)\Big).
\]
Then $\tau_{\rm abs}\ge T(y_0)$ and $\Lambda\ge T(y_0)$. In terms of the zero of $\zeta$, with $\varrho=\frac1{4\pi}L$ and $L=\log\frac\gamma{2\pi}$,
\[
\Lambda\ \ge\ \frac{4\delta}L-\frac{4a}{L^2}\log\Big(1+\frac{L\delta}a\Big)\quad(1+o(1)).
\]
\end{theorem}

\begin{proof}
Integration by parts against the counting function gives $\sum_k\frac1{(x-x_k)^2+y^2}\le\frac{\pi\varrho}y+\frac A{y^2}+\frac{2\varrho U+A}{U^2}+\varepsilon_2$. With Lemma~\ref{lem:exact} this gives $\partial_\tau y\ge-a/y-c$ on $[0,\tau_{\rm abs})$. Put $\varphi(y)=\int_0^y\frac{v\,dv}{a+cv}$. Then $\frac d{d\tau}\varphi(y)\ge-1$, so $\varphi(y(\tau))\ge T(y_0)-\tau$, and letting $\tau\to\tau_{\rm abs}$ gives $\tau_{\rm abs}\ge T(y_0)$. For $\tau<\tau_{\rm abs}$ the function $H_\tau$ has a nonreal zero, so $\tau<\Lambda$.
\end{proof}

The bound is linear, $\approx4\delta/L$, when $L\delta\gg a$, and quadratic, $\approx2\delta^2/a$, when $L\delta\ll a$. In the linear regime the speed $L/4$ at which $\delta$ decreases equals the speed at which the heat flow shifts the effective critical line of the deformed partial sums. Under the flow the $n$-th Dirichlet term is multiplied by $e^{(\tau/4)\log^2n}$, the centre term of the cross-energy $C_2$ of \cite{C} becomes $b_n^2n^{1-2\sigma}/(1-s+\frac\tau2\log n)$, and with the gamma factor the shift at height $\gamma$ is $\frac\tau4(\log\frac\gamma{2\pi}-\log n)$ (Figure~\ref{fig:ctx}d; the mechanism is that of \cite{Pm}).

\section{Numerical test}\label{sec:num}
\paragraph{Zeros and method.} We computed the $10142$ zeros of $\zeta$ up to height $10^4$, with Riemann--Siegel and the $C_1$ correction; the Lehmer pair is reproduced to $2\cdot10^{-6}$. We inserted an off-line pair $x_0\pm iy_0$ halfway between two consecutive zeros, at heights $\gamma\approx2000$ and $6000$, at five positions each, with $y_0=0.4,0.8,1.6$. The last value lies outside the strip and serves as a stress test. We integrated (P3) with the $300$ nearest zeros on each side exactly and the rest as a frozen field.

\paragraph{Results.}
\begin{itemize}
\item In all $30$ runs, the bound of Theorem~\ref{thm:cond} (with $A$ measured at $\tau=0$), the simulated absorption time and $y_0^2/2$ come in that order (Figure~\ref{fig:abs}a). The bound is $0.19$--$0.54$ of the simulated time.
\item The energy slope $S_{\rm rest}$ at $\tau=0$ predicts the simulated times with a correlation of $0.998$ between logarithms (Figure~\ref{fig:abs}b). The times exceed the prediction by $7$--$29\%$, because the pair repels its neighbours and lowers $S_{\rm rest}$ as $\tau$ grows.
\item The explicit bound of Theorem~\ref{thm:main} is smaller than the simulated times by a factor of order $10^3$ at these heights. The factor $\log\gamma$ is the price of not knowing the local spacing; the factor $\log\log\gamma$ is the price of clusters.
\end{itemize}
\paragraph{Clusters.} Theorem~\ref{thm:main} allows other off-line zeros near the given one, and the simulations show why this matters. With $y_0=0.8$ at $\gamma\approx2000$, an isolated inserted pair becomes real at $\tau=0.135$. When two or five further pairs ($y_0=0.2$--$0.7$, within $2.4$ units of $x$) are inserted around it, all zeros become real by $\tau=0.090$ and $0.086$. The other upper zeros attract the top one vertically, and once absorbed they add to the density of real zeros. These times still exceed the conditional bound for the isolated pair ($\approx0.05$), but they show that clusters can speed up absorption.

\begin{figure}[h]\centering\includegraphics[width=\textwidth]{Absorption_bound.png}
\caption{(a) Lower bound of Theorem~\ref{thm:cond} (open), simulated absorption time (filled) and $2\delta^2$, $30$ runs. (b) Simulated times against the prediction from the energy slope $S_{\rm rest}$ at $\tau=0$. (c) The lower bound of Theorem~\ref{thm:cond} for $\Lambda$ as a function of the height of a hypothetical off-line zero, for $\delta=0.05,0.2$ and local excess $A=0$, $2$, $\log\gamma$.}\label{fig:abs}\end{figure}

\begin{figure}[h]\centering\includegraphics[width=\textwidth]{Heat_flow_C2.png}
\caption{Context. (a) Trajectories of inserted off-line pairs at $\gamma\approx2000$; dashed: $\partial_\tau y=-1/y-L/2$. (b) Absorption times against the formula of Theorem~\ref{thm:cond} with $A=0$; dotted: $\times1.2$, the average excess of the simulations. (c) What an upper bound $\Lambda\le\tau^*$ would exclude by Theorem~\ref{thm:cond} with that factor. (d) The cross-energy $C_2$ of the deformed partial sums $\sum_{k\le n}e^{(\tau/4)\log^2k}k^{-s}$ follows the deformed expansion, $\tau=-0.2$.}\label{fig:ctx}\end{figure}

\section{Status}
\begin{itemize}
\item \emph{Unconditional and explicit:} Lemmas~\ref{lem:xi}--\ref{lem:G}, Proposition~\ref{prop:F}, Theorem~\ref{thm:main} and Corollary~\ref{cor:zfr}. They use only (P1)--(P5) from \cite{dB,Pm,PT}, Stirling's formula and the Euler--Maclaurin formula. The final inequality of Lemma~\ref{lem:C}, the bound $N_\tau(200)\le400$, and the maximisation over $p$ and $\gamma$ in \eqref{eq:Tformula} are explicit elementary inequalities that we evaluated numerically, without interval arithmetic; the scripts are included as ancillary files. The margins are stated in the Remark after Table~\ref{tab:const}.
\item \emph{Conditional on (I), (C):} Theorem~\ref{thm:cond}, the linear regime.
\item \emph{Heuristic:} the deformed cross-energy picture and the identification of the drift $L/4$ with the shift of the effective critical line.
\item \emph{Relation to the literature:} the upper-bound version of the dynamics is used in \cite{Pl,Tao}. As far as we know, lower bounds for $\Lambda$ in terms of a hypothetical off-line zero, and the link with the energy slope, are new.
\item \emph{What this does not do:} prove RH. The bound shows that $\Lambda$ dominates a distance $\asymp\delta^2/(\log\gamma\log\log\gamma)$ from each zero to the critical line, a distance that tends to $0$ with the height. The rigidity proved by Rodgers and Tao, $\Lambda\ge0$, points the other way.
\end{itemize}

\paragraph{Use of AI tools.} The computations and the drafting of this paper were carried out with the assistance of an AI system (Claude, Anthropic), under the author's direction. The author checked the results and takes responsibility for them.

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\end{document}