\documentclass[11pt]{amsart} \usepackage[utf8]{inputenc} \usepackage{amsmath,amssymb,amsthm,mathtools} \usepackage{booktabs} \usepackage{graphicx} \usepackage[margin=1in]{geometry} \usepackage[colorlinks=true,linkcolor=blue!50!black,citecolor=blue!50!black,urlcolor=blue!50!black]{hyperref} \usepackage{xcolor} \newtheorem{theorem}{Theorem} \newtheorem{proposition}[theorem]{Proposition} \newtheorem{lemma}[theorem]{Lemma} \newtheorem{corollary}[theorem]{Corollary} \theoremstyle{remark} \newtheorem{remark}[theorem]{Remark} \newtheorem{heuristic}[theorem]{Heuristic} \newcommand{\re}{\operatorname{Re}} \newcommand{\im}{\operatorname{Im}} \newcommand{\Ei}{\operatorname{Ei}} \newcommand{\Ci}{\operatorname{Ci}} \newcommand{\Ein}{\operatorname{Ein}} \newcommand{\doi}[1]{\href{https://doi.org/#1}{doi:#1}} \title[An Euler-product mollifier and an energy criterion for RH]{An Euler-product mollifier for the Jensen moment of $\zeta$\\ and an energy criterion for the Riemann hypothesis} \author{Pedro Caceres} \address{Universidad Europea de Valencia, Valencia, Spain} \email{pedrojesus.caceres@universidadeuropea.es} \subjclass[2020]{Primary 11M26; Secondary 11M06, 30H15} \keywords{Riemann hypothesis, Jensen formula, Poisson--Jensen formula, mollifier, explicit formula, Euler product, Davenport--Heilbronn function, Beurling primes} \date{October 2026} \begin{document} \begin{abstract} The Jensen moment of Balazard, Saias and Yor expresses the off-line zeros of $\zeta$ through the mean of $\log|\zeta|$ on the critical line, and Jensen's inequality turns any mollifier into an upper bound for them. With Dirichlet polynomials as mollifiers this bound cannot fall below $\frac12\log(1+c_0)=0.0386$, a barrier caused by the pole at $s=1$. We use instead the reciprocal of a smoothed Euler factor $E^*_X$, built from the prime powers up to a constant multiple of $X$ and an explicit pole term, for which $\zeta/E^*_X$ is a product over the nontrivial zeros alone. We prove that the resulting bound $B_X(s_0)$, at any centre $s_0$ of the half-plane $\re s>\frac12$, tends to $0$ under RH, so that RH is equivalent to $B_X(s_0)\to0$; and that, if the supremum $\Theta$ of the real parts of the zeros is attained and exceeds $\frac12$, it grows at least like $X^r/\log X$ for some $r>0$. Hence, when $\Theta$ is attained, RH holds if and only if the single mean square $\mathcal E(X)=\int|\zeta/E^*_X(\frac12+it)|^2\,dt/(1+t^2)$ stays bounded as $X\to\infty$. Numerically the bound falls to $0.0015$ at $X=10^4$ for the centre $s_0=1$ of Balazard, Saias and Yor, and at $s_0=2$ it decays like $0.037/\log X$ up to $X=10^6$, close to the value predicted by a heuristic with the constant $\kappa=\int_0^\infty(e^{2\Ci(v)}-1)\,dv=1.828\ldots$; for the Davenport--Heilbronn function it grows. We describe the geometry of the factorization and discuss, with two control functions, what a proof of the boundedness of $\mathcal E(X)$ would have to use. \end{abstract} \maketitle \section{Introduction} Let $F(s)=(s-1)\zeta(s)/s$ and let $\mu$ be the probability measure $dt/(2\pi(\frac14+t^2))$ on the critical line. Balazard, Saias and Yor [BSY] proved \begin{equation}\label{eq:bsy} \int_{-\infty}^{\infty}\log|\zeta(\tfrac12+it)|\,d\mu(t)=\sum_{\re\rho>\frac12}\log\Big|\frac{\rho}{1-\rho}\Big|, \end{equation} so that RH holds if and only if the left side vanishes. The identity is the Poisson--Jensen formula for $F$ in the half-plane $\re s>\frac12$ at the point $s=1$; $\mu$ is the Poisson measure of that point. For any function $M$ analytic in the half-plane, Jensen's inequality $\int\log|FM|\,d\mu\le\frac12\log\int|FM|^2d\mu$ then bounds the right side of \eqref{eq:bsy} by \[ \tfrac12\log\int|\zeta M(\tfrac12+it)|^2\,d\mu(t)-\log|M(1)|. \] For Dirichlet polynomials $M(s)=\sum_{n\le N}a_nn^{-s}$ this quantity is at least $\frac12\log(1+c_0)=0.0386$ with $c_0=\sum_{k\ge1}(1-k(k+1)\log^2(1+\frac1k))=0.0803\ldots$ [C, Prop.~3.5$''$]: the normalization at $s=1$, where $F$ carries the pole of $\zeta$, forces $M$ to approximate a step function, and the sawtooth costs $c_0$ whatever the length of $M$. The barrier is a property of Dirichlet polynomials, not of $\zeta$. In this paper the mollifier is $1/P$, where $P(s)=(s-1)E^*_g(s)/s$ and $E^*_g$ is the smoothed Euler factor of the explicit formula in [C, Prop.~6.1$'$] (Lemma~\ref{lem:factor} below): $\log E^*_g$ is a finite sum over prime powers up to $e^aX$, plus a term that carries the pole and terms for the trivial zeros. It is not a Dirichlet polynomial, $P$ is outer (Lemma~\ref{lem:outer}), and $F/P=Z_g$ is a product over the nontrivial zeros alone. Throughout, $g_0\in C_c^\infty(\mathbb R)$ is fixed with $g_0\ge0$, $\int g_0=1$, $\operatorname{supp}g_0\subset[-a,a]$, and for $X>e^{a}$ we put $g_X(u)=g_0(u-\log X)$, $L=\log X$, and write $E^*_X$, $Z_X$ for $E^*_{g_X}$, $Z_{g_X}$. Let $\Theta=\sup\re\rho$. \begin{theorem}[Jensen moment at a general centre]\label{thm:A} Let $s_0=\sigma_0+it_0$ with $\sigma_0>\frac12$, let $P_{s_0}(t)=(\sigma_0-\frac12)/\big(\pi((\sigma_0-\frac12)^2+(t-t_0)^2)\big)$ be the Poisson kernel of $\re s>\frac12$ at $s_0$, and $b_\rho(s)=(s-\rho)/(s-1+\bar\rho)$. Then \[ \int\log|\zeta(\tfrac12+it)|\,P_{s_0}(t)\,dt=\log\Big|\frac{(s_0-1)\zeta(s_0)}{s_0}\Big|+\sum_{\re\rho>\frac12}\log\frac1{|b_\rho(s_0)|}, \] and, if $\zeta(s_0)\neq0$, for every $X>e^a$ \[ \sum_{\re\rho>\frac12}\log\frac1{|b_\rho(s_0)|}\ \le\ B_X(s_0):=\frac12\log\int\Big|\frac{\zeta}{E^*_X}\big(\tfrac12+it\big)\Big|^2P_{s_0}(t)\,dt-\log\Big|\frac{\zeta}{E^*_X}(s_0)\Big|. \] \end{theorem} \begin{theorem}[an off-line zero makes the bound explode]\label{thm:B} Suppose that $\Theta>\frac12$ and that some zero has real part $\Theta$. Then for every $s_0$ with $\re s_0>\Theta$ (in particular every $s_0$ with $\re s_0>1$) there are $r\in(0,\Theta-\frac12)$, $C$ and $X_0$ with \[ B_X(s_0)\ \ge\ \tfrac12\Ei\big(r(\log X-a)\big)-C\log\log X\ \ge\ \frac{X^r}{3re^{ra}\log X}\qquad(X\ge X_0). \] \end{theorem} \begin{theorem}[the converse under RH]\label{thm:C} Let \[ \mathcal E(X)=\int_{-\infty}^{\infty}\Big|\frac{\zeta}{E^*_X}\big(\tfrac12+it\big)\Big|^2\frac{dt}{1+t^2}, \] which is finite for every $X>e^a$. Assume RH. Then $\mathcal E(X)\to\pi$ as $X\to\infty$, and $B_X(s_0)\to0$ for every $s_0$ with $\re s_0>\frac12$. \end{theorem} \begin{corollary}[criteria]\label{cor:D} \textup{(a)} For every $s_0$ with $\re s_0>\frac12$ and $\zeta(s_0)\neq0$, \[\textup{RH}\iff\lim_{X\to\infty}B_X(s_0)=0 .\] \textup{(b)} If $\Theta$ is attained, \[\textup{RH}\iff\limsup_{X\to\infty}\mathcal E(X)<\infty\iff\liminf_{X\to\infty}\mathcal E(X)<\infty .\] \end{corollary} Part (a) is unconditional, but it is a Jensen-type reformulation: by Theorem~\ref{thm:A}, $B_X(s_0)$ is bounded below by the off-line contribution, which does not depend on $X$. The new content is part (b), which rests on Theorem~\ref{thm:B}: a single weighted mean square, with no normalization at a point, detects an off-line zero by blowing up. Numerically the Dirichlet barrier disappears: at the centre $s_0=1$ of [BSY] the bound $B_X(1)$ is $0.0049$, $0.0026$, $0.0018$, $0.0015$ for $X=10,\dots,10^4$, against $0.0384$ for the optimal Dirichlet polynomial of length $150$; at $s_0=2$ it decays to $0.0026$ at $X=10^6$ (Section~\ref{sec:numerics}). The decay is close to $\kappa W(s_0)/\log X$ with $W(s_0)=\sum_\rho P_{s_0}(\gamma)$ and \[ \kappa=\int_0^\infty\big(e^{2\Ci(v)}-1\big)\,dv=1.828\ldots, \] a constant that comes from the profile $e^{\Ci(|v|)}$, $v=(t-\gamma)\log X$, of $|\zeta/E^*_X|$ at a zero on the line. We derive this law only heuristically (Heuristic~\ref{heu:kappa}); Theorem~\ref{thm:C} proves $B_X\to0$ without a rate. \textbf{Scope.} We do not prove that $\mathcal E(X)$ is bounded. $\mathcal E(X)$ compares $|\zeta|$ on the critical line with a finite, zero-free factor built from the primes, and Corollary~\ref{cor:D}(b) says that, when $\Theta$ is attained, its boundedness is equivalent to RH; without that hypothesis, boundedness gives only ``RH or $\Theta$ not attained''. Numerically $\mathcal E(X)$ grows for the Davenport--Heilbronn function, which has a functional equation and no Euler product; we expect it to grow, by the mechanism of Theorem~\ref{thm:B}, for a Beurling system with a positive prime measure and an off-line zero, which has an Euler product and no functional equation. This suggests that a proof must use the positivity of $\Lambda(n)$ and the exact spacing of the integers $1,2,3,\dots$ at the same time (Section~\ref{sec:controls}). Section~\ref{sec:open} lists the open points. \section{The factorization}\label{sec:factor} Let $g\ge0$ be piecewise continuous with compact support in $(0,\infty)$ and $\int g=1$; put \[ \psi_g(v)=\int_v^\infty g(u)\,du,\qquad K_g(w)=\int_0^\infty g(u)\,E_1(wu)\,du, \] with $E_1(z)=\int_z^\infty e^{-\tau}\tau^{-1}d\tau$ on the principal branch, $|\arg z|<\pi$. We use $E_1(z)=-\gamma_0-\log z+\Ein(z)$ with $\Ein$ entire, and $E_1(z)=e^{-z}\varphi(z)$ with $\varphi(z)=\int_0^\infty e^{-\tau}(z+\tau)^{-1}d\tau$. For $\tau\ge0$ the distance from $z+\tau$ to $0$ is at least $d(z):=\operatorname{dist}(z,(-\infty,0])$, so \begin{equation}\label{eq:phi} |\varphi^{(k)}(z)|\le k!\,d(z)^{-k-1}\qquad(k\ge0,\ z\notin(-\infty,0]). \end{equation} Across the cut $E_1$ jumps by $2\pi i$, so $K_g$ jumps by $2\pi i\int g=2\pi i$ and $e^{\pm K_g}$ is single-valued; $\re K_g$ is continuous across the cut. \begin{lemma}[{[C, Prop.~6.1$'$]}]\label{lem:factor} For $t\ne0$ and $s$ neither a zero of $\zeta$ nor of the form $\rho-r$ with $r>0$, \[ \log\zeta(s)=\sum_{n\ge2}\frac{\Lambda(n)\,\psi_g(\log n)}{n^s\log n}+K_g(s-1)-\sum_\rho K_g(s-\rho)-\sum_{k\ge1}K_g(s+2k), \] the zeros counted with multiplicity and in order of increasing $|\im\rho|$. Consequently $\zeta=E^*_gZ_g$ as meromorphic functions on $\mathbb C$, with $Z_g(s)=\prod_\rho e^{-K_g(s-\rho)}$ and \[ \log E^*_g(s)=\sum_{n\ge2}\frac{\Lambda(n)\,\psi_g(\log n)}{n^s\log n}+K_g(s-1)-\sum_{k\ge1}K_g(s+2k), \] and $\log|Z_g(s)|=-\sum_\rho\re K_g(s-\rho)$ for every $s$ that is not a zero. \end{lemma} \begin{proof}[Sketch] For $x>1$ not a prime power the truncated explicit formula with the term $\frac12\log(1-x^{-2})$ of $\psi(x)$ kept exactly gives $\sum_{n\le x}\Lambda(n)n^{-s}/\log n+E_1((s-1)\log x)=\log\zeta(s)+\sum_\rho E_1((s-\rho)\log x)+\sum_kE_1((s+2k)\log x)$; multiply by $g(\log x)$ and integrate [C]. Each factor $e^{-K_g(s-\rho)}$ is entire with a single simple zero at $\rho$, since $e^{-E_1(z)}=z\,e^{\gamma_0}e^{-\Ein(z)}$. For $g=g_X$ the product converges absolutely and locally uniformly by Lemma~\ref{lem:kernel} below (for general $g$, in the stated order), so both sides of $\zeta=E^*_gZ_g$ are meromorphic and the identity extends from the open set of the statement to $\mathbb C$. The formula for $\log|Z_g|$ follows, $\re K_g$ being continuous across the cut. \end{proof} The prime sum is finite when $g$ has compact support, and $e^{K_g(s-1)}$ has a simple pole at $s=1$. Hence $P(s)=(s-1)E^*_g(s)/s$ has neither zeros nor poles in $\re s\ge\frac12$, and $E^*_g$ itself has no zeros: the zeros of $\zeta$ enter only through $Z_g$. \begin{lemma}[$P$ is outer]\label{lem:outer} $\log|P|$ is continuous and bounded on $\re s\ge\frac12$. Hence $P$ is an outer function in $\re s>\frac12$, $Z_g=F/P$ lies in the Smirnov class $N^+$ of the half-plane, and $\log|Z_g(s)|\le\int\log|Z_g(\frac12+it)|P_s(t)\,dt$ for $\re s>\frac12$. \end{lemma} \begin{proof} Write $\log|P(s)|=-\log|s|+\big(\log|s-1|+\re K_g(s-1)\big)+\re D(s)-\sum_k\re K_g(s+2k)$, $D$ the finite prime sum. $|\re D|\le\sum\Lambda(n)n^{-1/2}/\log n$ on $\re s\ge\frac12$. Since $\log|z|+\re E_1(z)=-\gamma_0+\re\Ein(z)$, the bracket equals $-\gamma_0-\int g\log u+\re\int g(u)\Ein((s-1)u)\,du$ and is continuous; it is bounded on $|s-1|\le R$, and for $|s-1|\ge R$ with $R$ large, $z=(s-1)u$ satisfies $\re z\ge-u/2$ and either $\re z\ge0$, $d(z)=|z|\ge Ru$, or $|\im z|\ge\sqrt{R^2-1}\,u$, so $|E_1(z)|\le e^{u/2}/d(z)$ by \eqref{eq:phi} and $K_g(s-1)$ is bounded, as is $\log|(s-1)/s|$. Finally $|E_1(z)|\le e^{-\re z}/|z|$ for $\re z>0$ makes $\sum_k K_g(s+2k)$ bounded. A zero-free analytic function with $\log|P|$ bounded, harmonic in the half-plane and continuous up to the line, is the Poisson integral of its boundary values, i.e.\ outer. By [BSY], $F$ is a Blaschke product times an outer function (no singular inner factor, including at $\infty$, since $F(\sigma)\to1$ as $\sigma\to+\infty$); a quotient of outer functions is outer, so $Z_g\in N^+$, and the last inequality is the subharmonic inequality for functions in $N^+$. \end{proof} \section{The Jensen moment at a general centre}\label{sec:general} \begin{proof}[Proof of Theorem~\ref{thm:A}] The Möbius map $s\mapsto(s-s_0)/(s-1+\bar s_0)$, which sends $\re s>\frac12$ to the unit disc and $s_0$ to $0$, carries the Poisson measure of $s_0$ to normalized arc length. By [BSY] the Poisson--Jensen formula for $F$ holds without singular part: $\int\log|F(\frac12+it)|P_{s_0}(t)\,dt=\log|F(s_0)|+\sum_{\re\rho>\frac12}\log(1/|b_\rho(s_0)|)$, and $|F|=|\zeta|$ on the line. Since $P$ is outer (Lemma~\ref{lem:outer}), $\log|P(s_0)|=\int\log|P|\,P_{s_0}$; subtracting, $Z_X=F/P$ satisfies the same formula with $\log|Z_X(s_0)|$ on the right. Jensen's inequality $\int\log|Z_X|\,P_{s_0}\le\frac12\log\int|Z_X|^2P_{s_0}$ and $|Z_X|=|\zeta/E^*_X|$ on the line give the bound. \end{proof} For $s_0=1$ the identity is \eqref{eq:bsy}. Moving the centre to $s_0=\sigma_0+it_0$ concentrates the sensitivity near height $t_0$: an off-line zero at height close to $t_0$ contributes of order $(\beta-\frac12)(\sigma_0-\frac12)/|s_0-\rho|^2$ instead of $(\beta-\frac12)/|\rho|^2$. Table~\ref{tab:identity} checks the identity numerically, together with its analogue for the Davenport--Heilbronn function $f$, whose eight off-line zeros below height $400$ are listed in [C]. \begin{table}[h]\centering\small \begin{tabular}{llccc} \toprule function & $s_0$ & line mean & interior (and pole) & excess\\ \midrule $\zeta$ & $2+20i$ & $-0.03463$ & $-0.03408$ & $-0.0006$\\ $\zeta$ & $1.5+100i$ & $0.27150$ & $0.27144$ & $+0.0001$\\ $\zeta$ & $1+85.7i$ & $0.13177$ & $0.13158$ & $+0.0002$\\ $f$ & $1+85.7i$ & $-0.05385$ & $-1.49465$ & $+1.4408$ (off-line zeros: $+1.4407$)\\ \bottomrule \end{tabular} \caption{Theorem~\ref{thm:A}, first identity, with the window $|t-t_0|<4000$; the remaining kernel mass is below $3\cdot10^{-4}$.}\label{tab:identity} \end{table} \section{Craters: the constant $\kappa$}\label{sec:kappa} \begin{heuristic}\label{heu:kappa} Assume RH and simple zeros separated at scale $1/\log X$. Then for fixed $s_0$ \[ B_X(s_0)\approx\frac{\kappa\,W(s_0)}{\log X},\qquad \kappa=\int_0^\infty\big(e^{2\Ci(v)}-1\big)\,dv=1.8280\ldots,\qquad W(s_0)=\sum_\rho P_{s_0}(\gamma). \] \end{heuristic} \emph{Derivation.} On the line, under RH, $\re(s-\rho)=0$ and $\re E_1(ix)=-\Ci(|x|)$, so $\log|Z_X(\frac12+it)|=\sum_\gamma\int g_X(u)\Ci(|t-\gamma|u)\,du$. As $X\to\infty$ the weight concentrates at $u=\log X$ in relative terms, and near a zero $|Z_X|$ takes the profile $e^{\Ci(|v|)}$ with $v=(t-\gamma)\log X$ (Figure~\ref{fig:geom}b): a well of width of order $1/\log X$, with an overshoot of height $e^{\max\Ci}=e^{\Ci(\pi/2)}=e^{0.4720}=1.60$. Since $\int_0^\infty\Ci(v)\,dv=0$, the mean of $\log|Z_X|$ is unchanged to first order, while each zero adds $\frac2{\log X}\int_0^\infty(e^{2\Ci(v)}-1)\,dv$ to the energy $\int|Z_X|^2P_{s_0}$; and $\log|Z_X(s_0)|\to0$. Interactions between neighbouring wells enter through the tail $\Ci(v)\sim\sin v/v$; they are neglected, which is reasonable because $P_{s_0}$ weights low heights, where the zeros are sparse. We do not make this rigorous; the numbers below support it. For $s_0=1$, $W=\frac1{2\pi}\sum_\rho|\rho|^{-2}=0.00735$ and $\kappa W/\log10^4=0.00146$, against the measured $0.0015$; for $s_0=1+100i$, $W=0.2305$ and the ratio of the measured value to $\kappa W/\log X$ is $1.07$ at $X=10^3$ and $1.05$ at $X=10^4$; for $s_0=2$, $W=0.0220$ and $B\log X$ tends to $0.037$ against $\kappa W=0.040$ (Table~\ref{tab:largeX}). \section{Volcanoes: proof of Theorem~\ref{thm:B}}\label{sec:explosion} \begin{lemma}[the kernel]\label{lem:kernel} Let $w=-x+iy$, $|x|\le x_{\max}$, $X\ge e^{a+1}$, $L=\log X$. For every integer $A\ge0$ there is $C$ depending on $A$, $x_{\max}$ and $g_0$ only, such that: \begin{enumerate} \item[(L1)] if $\re w\ge\eta>0$, then $|K_X(w)|\le C_\eta\,X^{-\re w}(1+|y|)^{-A-1}/L$, with $C_\eta$ depending also on $\eta$; \item[(L2)] if $\re w\ge0$, $w\ne0$, then $|K_X(w)|\le C(1+|\log(|w|L)|)$, and if moreover $|y|\ge1$, $|K_X(w)|\le C|y|^{-A-1}/L$; \item[(L3)] if $x>0$ and $y\ne0$, then $|K_X(w)|\le e^{x(L+a)}/(|y|(L-a))$, and if moreover $|y|\ge1$, $|K_X(w)|\le C\,e^{xa}X^{x}|y|^{-A-1}/L$; \item[(L4)] for $x>0$, $-\re K_X(-x)=\int g_X(u)\Ei(xu)\,du\ge\Ei\big(x(L-a)\big)$. \end{enumerate} \end{lemma} \begin{proof} Write $K_X(w)=e^{-wL}\int g_0(v)\,e^{-wv}\,\varphi(w(v+L))\,dv$ and $u=v+L\in[L-a,L+a]$; note $d(wu)=u\,d(w)$, with $d(w)=|w|$ if $x\le0$ and $d(w)=|y|$ if $x>0$. With $A=0$ and \eqref{eq:phi}, $|K_X(w)|\le e^{x(L+a)}/(d(w)(L-a))$; for $x>0$ this is the first bound in (L3). Integrating by parts $A$ times against $e^{-wv}$ (no boundary terms, $g_0\in C_c^\infty$), \[ K_X(w)=\frac{e^{-wL}}{w^{A}}\int e^{-wv}\,\frac{d^{A}}{dv^{A}}\Big[g_0(v)\varphi(w(v+L))\Big]dv, \] and by Leibniz and \eqref{eq:phi} the $v$-derivative of order $m$ of $\varphi(w(v+L))$ is at most $m!\,|w|^m/(u\,d(w))^{m+1}\le m!\,(|w|/d(w))^m/(u\,d(w))$. If $x\le0$, $|w|/d(w)=1$; if $x>0$ and $|y|\ge1$, $|w|/d(w)\le1+x_{\max}$. Hence $|K_X(w)|\le C\,e^{x(L+a)}|w|^{-A}/(d(w)(L-a))$, with $e^{x(L+a)}=X^{x}e^{xa}$, which gives the second bounds in (L2), (L3), and (L1) (there $|w|\ge\max(\eta,|y|)$). For the first bound in (L2): if $|wu|\ge1$, $|E_1(wu)|\le|\varphi(wu)|\le1$ since $\re(wu)\ge0$; if $|wu|<1$, $|E_1(wu)|\le\gamma_0+|\log|wu||+\pi+e$; and $\log|wu|=\log(|w|L)+O(1)$. For (L4), $E_1(-\xi\pm i0)=-\Ei(\xi)\mp i\pi$ for $\xi>0$, and $\Ei$ is increasing. \end{proof} \begin{proof}[Proof of Theorem~\ref{thm:B}] Write $Z=Z_X$ and fix $s_0$ with $\re s_0>\Theta$. (a) \emph{Reduction to a point.} By Lemma~\ref{lem:outer} and Jensen's inequality, for every $s_1$ with $\re s_1>\frac12$, $\log|Z(s_1)|\le\int\log|Z|\,P_{s_1}\le\frac12\log\int|Z|^2P_{s_1}$ (also when the last integral is infinite). Since $P_{s_1}\le H\,P_{s_0}$ with $H=\sup_tP_{s_1}(t)/P_{s_0}(t)<\infty$, independent of $X$, \[ B_X(s_0)\ \ge\ \log|Z(s_1)|-\tfrac12\log H-\log|Z(s_0)|. \] (b) \emph{Choice of $s_1$.} Take a zero $\rho_0=\Theta+i\gamma_0$, and let $d\in(0,1]$ be at most the distance from $\gamma_0$ to the ordinates of the other zeros of real part $\Theta$ (finitely many lie in any bounded window); $d$ depends only on $\rho_0$. Let $S=\sum'\min\big(1/|y|,\ C|y|^{-3}\big)$ over the zeros $\rho\ne\rho_0$ with $\re\rho=\Theta$, $y=\gamma_0-\gamma$, with $C$ the constant of (L3) for $A=2$, $x_{\max}=\frac12$; $S<\infty$ and $S$ does not depend on $r$. Choose $r\in(0,\Theta-\frac12)$ with $2re^{2ra}S\le\frac14$ and such that $s_1=\rho_0-r$ is not a zero (this excludes a countable set of $r$). Let $\delta>0$ be the distance from $s_1$ to the zeros. (c) \emph{Lower bound at $s_1$.} By Lemma~\ref{lem:factor}, $\log|Z(s_1)|=-\sum_\rho\re K_X(s_1-\rho)$, and $w=s_1-\rho$ has $x=r-(\Theta-\beta)\le r$. \begin{itemize} \item $\rho=\rho_0$: $w=-r$, and by (L4) the term is at least $\Ei(r(L-a))$. \item $\re\rho=\Theta$, $\rho\ne\rho_0$: $x=r$, $|y|\ge d$. By (L3) the sum of $|K_X|$ is at most $X^re^{ra}\frac{L}{L-a}S/L\le\frac12\cdot\frac{X^re^{-ra}}{r(L-a)}$ for $X\ge X_0$, by the choice of $r$. \item $\Theta-r<\beta<\Theta$: $00$ for every zero, so by (L1) with $A=2$, $|\log|Z(s_0)||\le C_\eta X^{-\eta}L^{-1}\sum_\rho(1+|t_0-\gamma|)^{-3}\to0$. Combining (a), (c) and (d) gives the theorem. \end{proof} \begin{remark}[Davenport--Heilbronn] The Davenport--Heilbronn function $f$ has an analogue $E^*_f$, built from the coefficients $b(n)$ of $-f'/f$ and its trivial zeros at $-1,-3,\dots$. Theorem~\ref{thm:B} does \emph{not} apply to it as stated: $f$ has infinitely many zeros in $\re s>1$ [DH], so $\Theta_f>1$, its attainment is unknown, and the zero $\rho_f=0.808517+85.699348i$ studied below is not of maximal real part; we also do not prove the factorization and the Smirnov property for $f$. Numerically, at $s_1=\rho_f-r$ the dominant term (L4) is nevertheless visible for moderate $X$: $\log|Z_f(s_1)|$ is $1.0332$ and $1.6702$ for $X=10^3,10^4$ with $r=0.1$, against $\int g\Ei(ru)\,du=1.0373$ and $1.6800$, and $2.9489$, $4.3721$ with $r=0.2$, against $2.9564$, $4.3976$. \end{remark} \section{The converse under RH}\label{sec:converse} \begin{proof}[Proof of Theorem~\ref{thm:C}] \emph{Finiteness.} On the line $s-1=-\frac12+it$, so by the proof of Lemma~\ref{lem:outer} $\log|E^*_X|$ is bounded below there; with $|\zeta(\frac12+it)|\ll(|t|+2)^{1/4}$ this makes $\mathcal E(X)$ finite. \emph{Convergence.} Under RH, $\re(s-\rho)=0$ for $s=\frac12+it$, so by Lemma~\ref{lem:factor}, for $t$ not an ordinate, \[ \log|Z_X(\tfrac12+it)|=\sum_\gamma c_X(t-\gamma),\qquad c_X(y)=\int g_X(u)\Ci(|y|u)\,du, \] with multiplicity. Three bounds: (c1) $c_X(y)\le\max\Ci=\Ci(\frac\pi2)=0.4720$, since $g_X\ge0$ and $\int g_X=1$; (c2) $|c_X(y)|\le2/(|y|(L-a))$, since $|\Ci(x)|\le2/x$ (indeed $\le1/x$) from $\Ci(x)=\frac{\sin x}x-\int_x^\infty\frac{\sin\tau}{\tau^2}d\tau$; (c3) $|c_X(y)|\le C|y|^{-3}/L$ for $|y|\ge1$, by (L2). Under RH, $S(T)\ll\log T/\log\log T$ [T, Thm.~14.13], so an interval of length $h\le1$ centred at $t$ contains at most $\frac h{2\pi}\log(|t|+3)+C\log(|t|+3)/\log\log(|t|+3)$ ordinates. By (c1) the ordinates with $|t-\gamma|<1/L$ contribute at most $0.472\big(\frac1{\pi L}+\frac C{\log\log(|t|+3)}\big)\log(|t|+3)$; by (c2) on dyadic shells $2^k/L\le|t-\gamma|<2^{k+1}/L$ the ordinates with $|t-\gamma|<1$ contribute $\ll\big(\frac{\log L}L+\frac{1}{\log\log(|t|+3)}\big)\log(|t|+3)$; by (c3) the others contribute $\ll\log(|t|+3)/L$. Treating $|t|\le T_\varepsilon$ separately, for every $\varepsilon>0$ there are $X_\varepsilon$, $C_\varepsilon$ with \begin{equation}\label{eq:upper} \log|Z_X(\tfrac12+it)|\le\varepsilon\log(|t|+3)+C_\varepsilon\qquad(t\in\mathbb R,\ X\ge X_\varepsilon). \end{equation} For each $t$ that is not an ordinate, (c2) and (c3) give $\sum_\gamma|c_X(t-\gamma)|\le(L-a)^{-1}\big(\sum_{|t-\gamma|<1}2/|t-\gamma|+C\sum_{|t-\gamma|\ge1}|t-\gamma|^{-3}\big)\to0$, so $\log|Z_X(\frac12+it)|\to0$ almost everywhere. By \eqref{eq:upper} with $\varepsilon=\frac14$, $|Z_X|^2\le C(|t|+3)^{1/2}$, integrable against $dt/(1+t^2)$; dominated convergence gives $\mathcal E(X)\to\pi$, and in the same way $\int|Z_X|^2P_{s_0}\to1$. Finally $\re(s_0-\rho)=\re s_0-\frac12>0$ for every zero, so $\log|Z_X(s_0)|\to0$ by (L1), and $B_X(s_0)\to0$. \end{proof} \begin{proof}[Proof of Corollary~\ref{cor:D}] (a) If RH fails, the right side of Theorem~\ref{thm:A} is a positive number independent of $X$, so $B_X(s_0)$ does not tend to $0$; if RH holds, $B_X(s_0)\to0$ by Theorem~\ref{thm:C}. (b) RH implies $\mathcal E(X)\to\pi$ by Theorem~\ref{thm:C}. Conversely, suppose $\Theta>\frac12$ is attained, and fix $s_0$ with $\re s_0>\max(\Theta,1)$. Since $P_{s_0}(t)\le C/(1+t^2)$, $B_X(s_0)\le\frac12\log(C\,\mathcal E(X))-\log|Z_X(s_0)|$, and $\log|Z_X(s_0)|\to0$ by part (d) of the proof of Theorem~\ref{thm:B}. As $B_X(s_0)\to\infty$ by Theorem~\ref{thm:B}, $\mathcal E(X)\to\infty$; so $\liminf\mathcal E(X)<\infty$ already excludes $\Theta>\frac12$. \end{proof} Numerically, with the weight of Section~\ref{sec:numerics}, the ratio $R_X=\max_{t<600}\log|Z_X(\frac12+it)|/\log(t+2)$ is $0.185$, $0.169$, $0.170$ for $X=10^2,10^3,10^4$, and on $200\le t<600$ it decreases, $0.179$, $0.161$, $0.146$, consistent with \eqref{eq:upper}. \section{The geometry of $Z_X$}\label{sec:geometry} Figure~\ref{fig:geom} shows the factorization $\zeta=E^*_XZ_X$. On the critical line the truncated Euler product $\prod_{p\le X}(1-p^{-s})^{-1}$ does not converge and oscillates around $|\zeta|$, while $E^*_X$ follows $|\zeta(\frac12+it)|$ closely and never vanishes: at $\gamma_1$ it equals $0.097$ for $X=10^2$ and $0.048$ for $X=10^4$. $E^*_X$ is built from the prime powers up to $e^aX$ and the pole alone; it is a smooth envelope that passes over each zero without touching it, and the zero is carried entirely by $Z_X$ (panels a, b). In the strip (panel c) $\log|Z_X|$ is flat to the right of the line. The factor $e^{-K_X(s-\rho)}$ grows like $\exp\big(X^{\beta-\sigma}/((\beta-\sigma)\log X)\big)$ where $\re(s-\rho)<0$, so every zero has a \emph{volcano} to its left and is seen from the right only as a \emph{crater}; for a zero on the line the volcano lies in $\sigma<\frac12$, outside the half-plane of Theorem~\ref{thm:A}. For the Davenport--Heilbronn function (panel d) the volcano of the off-line zero starts at $\sigma=0.81$ and crosses the critical line. In these terms, Theorems~\ref{thm:B} and \ref{thm:C} say that, when $\Theta$ is attained, RH holds if and only if only craters reach the critical line. \begin{figure}[htbp]\centering\includegraphics[width=\textwidth]{Geometry_ZX.png} \caption{The geometry of $Z_X$. (a) $|\zeta(\frac12+it)|$, the truncated Euler product $|E_X|$ with $X=100$, and $|E^*_X|$ for $X=10^2,10^4$; dots: zeros. (b) $|Z_X|=|\zeta/E^*_X|$ on the line: wells at the zeros with the overshoot $e^{0.472}$. (c) $\log|Z_X(\sigma+it)|$ for $\zeta$, $X=10^3$: craters on the line, volcanoes to the left. (d) The same for the Davenport--Heilbronn function near its off-line zero (stars: the zero and its reflection); its volcano crosses the line.}\label{fig:geom}\end{figure} \section{Numerical results}\label{sec:numerics} All computations use $\zeta$ and Hurwitz zeta functions by Euler--Maclaurin summation in double precision. For $\zeta$ the weight is either the Gaussian of width $0.4$ around $\log X$ (Tables~\ref{tab:barrier}, \ref{tab:dh}) or, for large $X$, the $C_c^\infty$ bump $g_0(v)\propto\exp(-1/(1-(v/1.2)^2))$ on $[-1.2,1.2]$, for which the prime sum is exactly finite (Table~\ref{tab:largeX}). With the Gaussian truncated at $4.5$ standard deviations the neglected tail of the prime sum grows like $X^{1/2}$ and inflates $B$ from $X=10^5$ on; this is why the compactly supported weight is used there. The theorems are stated for compactly supported weights. \begin{table}[h]\centering\small \begin{tabular}{lcccc} \toprule $s_0$ & $X=10$ & $10^2$ & $10^3$ & $10^4$\\ \midrule $1$ (Euler--Cáceres) & $0.0049$ & $0.0026$ & $0.0018$ & $0.0015$\\ $1$ (best Dirichlet polynomial, $N\le150$) & \multicolumn{4}{c}{$0.0384$ (floor $0.0386$)}\\ $1+100i$ (Euler--Cáceres) & $0.131$ & $0.101$ & $0.066$ & $0.048$\\ $1+100i$ (best Dirichlet polynomial, $N=500$) & \multicolumn{4}{c}{$0.080$}\\ \bottomrule \end{tabular} \caption{The bound of Theorem~\ref{thm:A} with the mollifier $1/P$ against the optimal Dirichlet polynomial (computed exactly as a quadratic form).}\label{tab:barrier} \end{table} \begin{table}[h]\centering\small \begin{tabular}{lcccccc} \toprule $X$ & $10$ & $10^2$ & $10^3$ & $10^4$ & $10^5$ & $10^6$\\ \midrule $B_X(2)$ & $0.0138$ & $0.0074$ & $0.0052$ & $0.0039$ & $0.0032$ & $0.0026$\\ $B_X(2)\log X$ & $0.032$ & $0.034$ & $0.036$ & $0.036$ & $0.036$ & $0.037$\\ \bottomrule \end{tabular} \caption{The limit $X\to\infty$ at the single centre $s_0=2$ with the compactly supported weight; $\kappa W(2)=0.040$. The Jensen identity holds to $1.3\cdot10^{-4}$.}\label{tab:largeX} \end{table} \begin{table}[h]\centering\small \begin{tabular}{lcccc} \toprule & $X=10$ & $10^2$ & $10^3$ & $10^4$\\ \midrule $\zeta$, $s_0=1+100i$ & $0.130$ & $0.101$ & $0.065$ & $0.048$\\ $\zeta$, $s_0=2+85.7i$ & $0.241$ & $0.121$ & $0.090$ & $0.071$\\ $f$, $s_0=1+85.7i$ & $1.65$ & $2.55$ & $4.49$ & $7.58$\\ $f$, $s_0=1+100i$ & $0.34$ & $0.31$ & $1.26$ & $5.91$\\ $f$, $s_0=2+85.7i$ & $0.775$ & $1.779$ & $3.822$ & $7.081$\\ \bottomrule \end{tabular} \caption{$B_X(s_0)$ for $\zeta$ and for the Davenport--Heilbronn function $f$ (Gaussian weight).}\label{tab:dh} \end{table} \begin{figure}[htbp]\centering\includegraphics[width=\textwidth]{Euler_Caceres_barrier.png} \caption{(a) Profile $e^{\Ci(|v|)}$ of $|Z_X|$ at a zero on the line, $v=(t-\gamma)\log X$. (b) $B(s_0)$ for $\zeta$ at $s_0=1$ and $1+100i$, the heuristic law $\kappa W(s_0)/\log X$ (dashed), and the floor $0.0386$ of Dirichlet mollifiers (dotted). (c) $B(s_0)$ for the Davenport--Heilbronn function, which grows with $X$; dashed: its off-line excess.}\label{fig:barrier}\end{figure} \section{What a proof of the bound would need}\label{sec:controls} \textbf{Positivity alone.} For each $n$, $3+4\cos\theta+\cos2\theta\ge0$, so $\Lambda(n)\ge0$ gives $3\re\Sigma(\frac12)+4\re\Sigma(\frac12+it)+\re\Sigma(\frac12+2it)\ge0$ for the prime sum $\Sigma$ of $\log E^*_X$, and hence an upper bound for $\log|Z_X(\frac12+it)|$. The bound is useless: the $a_0$ term sits at $t=0$, on the pole, where the prime sum is of size $2\sqrt X/\log X$ ($6.6$, $14.1$, $31.0$ for $X=10^2,10^3,10^4$). At $t=20,30,40$ the resulting bound for $\log|Z_X|$ is about $5$, $10$ and $23$ for these $X$, while the true values lie between $-0.2$ and $0.5$. Every nonnegative trigonometric polynomial has $a_0>0$, so every argument of this type pays a correction of size $X^{1/2}$ on the critical line [C, \S6, \emph{Positivity and the pole}]. \textbf{The two controls.} Two functions isolate the two halves of the structure of $\zeta$ (Table~\ref{tab:controls}). \begin{table}[h]\centering\small \begin{tabular}{lccc} \toprule & $\Lambda\ge0$, one Euler product & integers exactly $1,2,3,\dots$ & off-line zeros\\ \midrule $\zeta$ & yes & yes (functional equation) & none known\\ Beurling system [DMV], [C] & yes & no & yes\\ Davenport--Heilbronn $f$ [DH] & no & yes (functional equation) & yes\\ \bottomrule \end{tabular} \caption{The two controls.}\label{tab:controls} \end{table} For $f$ the bound fails numerically (Table~\ref{tab:dh}), so an argument that used only the functional equation and mean-value estimates valid for every Dirichlet series with a functional equation would, if it applied to $f$, prove something false; this is a heuristic constraint, since we have not proved that $\mathcal E_f(X)$ is unbounded. For a Beurling system with an off-line zero of maximal real part, the mechanism of Theorem~\ref{thm:B} (one dominant term (L4) at $s_1=\rho_0-r$) should again force growth; Diamond, Montgomery and Vorhauer [DMV] constructed such systems with zeros as close to $\sigma=1$ as the region of de la Vallée Poussin allows, and [C] gives a simplified example with a zero at $0.75+20i$. We do not prove the analogue of Lemmas~\ref{lem:factor}--\ref{lem:outer} for these functions. Taken together, the two controls suggest that a proof of the bound must use the positivity of $\Lambda(n)$ and the exact spacing of the integers at the same time. \textbf{A quantity in which $\zeta$ cancels.} For $s^*=1-\bar s$ the functional equation $|\zeta(s^*)|=|\zeta(s)|/|\chi(s)|$ gives \[ D_X(s):=\log|Z_X(s)|-\log|Z_X(s^*)|=\log|\chi(s)|-\log|E^*_X(s)|+\log|E^*_X(s^*)|, \] a quantity built from the primes and the gamma factor alone (checked to $2\cdot10^{-13}$, also for $f$). It might seem to combine the two ingredients, but reindexing the zeros by $\rho\mapsto1-\bar\rho$ (legitimate by absolute convergence) gives $D_X(s)=\sum_\rho h(s-\rho)$ with $h(w)=-\re[K_X(w)-K_X(-\bar w)]$, the same kernel for every zero; $D_X$ looks alike for $\zeta$, $f$ and a Beurling-type toy (Figure~\ref{fig:reflect}), and its prime side contains $E^*_X$ at $\sigma<\frac12$, where the pole barrier reappears. \begin{figure}[htbp]\centering\includegraphics[width=\textwidth]{Reflection_asymmetry.png} \caption{The reflection asymmetry $D_X(s)$, $X=10^3$, computed from the primes and $\chi$ only. (a) $\zeta$. (b) The Davenport--Heilbronn function. (c) A Beurling-type toy with the nonnegative prime measure $(1-2\re x^{\rho_B-1})\,dx/\log x$ on $[16,\infty)$, $\rho_B=0.75+20i$, and no functional equation. (d) Difference between the primes-only formula and the direct one on $\sigma=0.7$.}\label{fig:reflect}\end{figure} \section{Open points}\label{sec:open} \textbf{The bound.} Prove $\limsup_X\mathcal E(X)<\infty$. By Corollary~\ref{cor:D}(b) this is equivalent to RH when $\Theta$ is attained, so it is at least as hard as RH in that case. It is a mean-value statement related to the hybrid Euler--Hadamard products of Gonek, Hughes and Keating [GHK, BGM], uniform in the length of the Euler factor at fixed weight. Unconditional mean-value methods with Dirichlet mollifiers are limited by the length of the mollifier relative to the height (the exponent $\frac47$ of Conrey [Co]); here the weight $1/(1+t^2)$ is fixed and the length tends to infinity, the opposite regime. \textbf{The case $\Theta$ not attained.} Here is a heuristic outline only. For each $X$ one can choose $\rho^*$ maximizing $\beta\log X-A\log(2+|\gamma|)$; the zeros at distance at least $1$ from $\rho^*$ then contribute at most a small multiple of the main term, and a vertical shift of $s_1$ aligns the common phase $e^{-i\tau\log X}$. What remains is the cluster amplitude $\sum_{|\gamma-\gamma^*|<1}X^{\beta-\beta^*}e^{i(\gamma-\gamma^*)\log X}\hat g_0(w)/w$, a sum of Cauchy kernels, which can be small only if off-line zeros of nearly maximal real part close to $\rho^*$ cancel one another. For two such zeros at distance $\Delta$ we expect a remainder of order $X^r\Delta/(r^2\log X)$, so a lower bound $\Delta\ge X^{-o(1)}$ for their separation would plausibly close this case; no such bound is known. \section*{Data and code} All tables and figures are reproduced by the ancillary Python scripts distributed with this paper (NumPy, SciPy, Matplotlib), which are also in the folder \texttt{anc/euler\_mollifier/} of the source of [C]. 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