\section{The margin: overview}\label{ce:sec:intro}

For $N>1$ put
\[
C(N)=2\sum_{n<N}\frac{\Lambda(n)}{\sqrt n}\log\frac Nn ,
\]
the Ces\`aro (Riesz) mean of order one of $\sum\Lambda(n)n^{-1/2}$, and
\begin{equation}\label{ce:eq:closed}
R(N)=8\sqrt N-\Big(\log\pi+\gamma_E+\frac\pi2+3\log2\Big)\log N-16+\frac{\pi^2}2+4G-2\sum_{k\ge1}\frac{N^{-(2k+1/2)}}{(2k+\frac12)^2},
\end{equation}
with $\gamma_E$ Euler's constant and $G$ Catalan's constant. The three pieces of $R$ are the pole of $\zeta$, the conductor with the $\Gamma$ factor at $s=\frac12$, and the trivial zeros. The object of this paper is the \emph{margin} $M(N)=R(N)-C(N)$. Both $R$ and $C$ are of size $8\sqrt N$. By Suzuki's theorem \cite{Su1,Su2} (Theorem~\ref{ce:thm:landau}), RH holds if and only if $M(N)\ge0$ for all large $N$, and then $0\le M(N)\le0.18484$ for every $N>1$. So RH is equivalent to confining a sum of size $8\sqrt N$ over the prime powers below $N$ to a window of fixed width, from both sides. This is as hard as RH, and we do not prove it.

\emph{What the paper does.} It studies this one-variable inequality from three sides and records exactly where each side stops.
\begin{enumerate}
\item \emph{The margin} (Section~\ref{ce:sec:margin}). The margin is a Fej\'er sum over the zeros (Proposition~\ref{ce:prop:fejer}). The window width is $4(\log\xi)''(\frac12)$, computable without the zeros (Corollary~\ref{ce:cor:sandwich}). The two-sided inequality holds for all real $1<N\le10^{12}$, by computations from the primes with rigorous error bounds, and, from the certified first $4520$ zeros, up to $2.15\cdot10^{29}$; without any hypothesis it holds up to an explicit error $E(N)$ (Proposition~\ref{ce:prop:verified}, Theorem~\ref{ce:thm:explicit}, Corollary~\ref{ce:cor:twosided}).
\item \emph{The minima and the slope} (Section~\ref{ce:sec:minima}). $M$ is convex between prime powers, so its local minima form an explicit countable set, and RH is equivalent to their nonnegativity (Corollary~\ref{ce:cor:discrete}). Under RH they accumulate at $0$ (Proposition~\ref{ce:prop:nomargin}). The slope of the margin is the fluctuation $\alpha$ of $\sum\Lambda(n)n^{-1/2}$ (Proposition~\ref{ce:prop:slope}). The minima are the downward zeros of $\alpha$, and they follow from one another by a recurrence (Proposition~\ref{ce:prop:betarec}).
\item \emph{Energy} (Section~\ref{ce:sec:energy}). Under RH, $M$ is the variogram of $\alpha$, and the window is the inequality $0\le\langle|\Delta f|^2\rangle\le4\langle|f|^2\rangle$ (Proposition~\ref{ce:prop:variogram}). RH is equivalent to finite energy of $\alpha$ (Proposition~\ref{ce:prop:energy}), and to a one-sided bound for the $L$-average of $M$ (Proposition~\ref{ce:prop:average}). Without hypothesis we prove an energy identity (Theorem~\ref{ce:thm:energyidentity}) and a lower bound for the energy (Theorem~\ref{ce:thm:energylower}). Block energies to $10^{10}$ agree with the prediction.
\item \emph{Limits} (Section~\ref{ce:sec:limits}). Theorem~\ref{ce:thm:nogo} plants an off-line zero in positive weights that satisfy every structural input of the energy arguments, so these arguments cannot prove RH. Proposition~\ref{ce:prop:mobiusrep} carries the identity $\Lambda=\mu*\log$, which excludes the planted weights, through the slope exactly. The result is a type~I sum of $\mu$ against the sawtooth (Remark~\ref{ce:rem:mobius}). Section~\ref{ce:sec:windows} summarizes the companion results on Weil's form in windows \cite{P20,P21,P22}.
\end{enumerate}

\emph{What is due to others.} Weil's explicit formula and criterion \cite{We52,Bo,IK}; the explicit formula for Riesz means \cite{In}; the one-variable criterion, its bounds and its Laplace transform are Suzuki's \cite{Su1,Su2} (we found his papers after our proofs were written); Landau's theorem on Laplace transforms \cite{Wi,In}; the evaluation of $\sum_\rho(\rho-\frac12)^{-2}$ \cite{Vo}; the limiting distributions of prime error terms \cite{RuSa,ANS}; Beurling systems with large oscillation \cite{DMV}; the distribution of the Mertens function \cite{Ng}. What we believe is new: the verification to $10^{12}$, the exact description of the minima and their recurrence, the variogram reading of the margin, the energy identity and the energy lower bound, the planted-zero theorem, and the M\"obius representation of the slope.

\emph{Controls.} A statement that holds for the Davenport--Heilbronn function \cite{DH,T}, which has the functional equation of $\zeta$ but no Euler product and zeros off the line, cannot be the reason for RH. We use it, a Beurling system \cite{DMV} and planted weights (Section~\ref{ce:subsec:nogo}) as controls throughout.

\emph{Notation.} The nontrivial zeros are $\rho=\frac12+i\gamma_\rho$, with $\gamma_\rho$ real exactly when $\rho$ is on the line; $m_\gamma$ is the multiplicity. We write $T_0=3\cdot10^{12}$ for the height to which RH is verified \cite{PT}. For a test function $g$, $\widehat g(z)=\int g(x)e^{izx}dx$. The explicit formula, for $G$ even, continuous, of bounded variation and compact support \cite{Bo,IK}, reads
\begin{multline}\label{ce:eq:weilzeta}
\sum_\rho\widehat G(\gamma_\rho)=\widehat G(\tfrac i2)+\widehat G(-\tfrac i2)-G(0)\log\pi+\frac1{2\pi}\int_{-\infty}^\infty\widehat G(r)\,\re\psi\big(\tfrac14+\tfrac{ir}2\big)dr\\-\sum_{n\ge2}\frac{\Lambda(n)}{\sqrt n}\big(G(\log n)+G(-\log n)\big),
\end{multline}

the sum over the zeros taken symmetrically. Weil's criterion says that RH holds if and only if the right side is nonnegative for every $G=g*\tilde g$, where $\tilde g(x)=\overline{g(-x)}$. The right side is \emph{Weil's quadratic form} $Q(g)$.

\section{The margin}\label{ce:sec:margin}
The margin is the explicit formula for one test function, the autocorrelation of the indicator of $[0,\log N]$.

\begin{proposition}[the margin over the zeros]\label{ce:prop:fejer}
Let $N>1$ and $L=\log N$. Without any hypothesis,
\[
M(N)=\sum_\rho F_L(\gamma_\rho),\qquad F_L(z)=\frac{4\sin^2(Lz/2)}{z^2}=\Big|\int_0^Le^{izu}du\Big|^2\ (z\in\mathbb R),
\]
the sum over all nontrivial zeros with multiplicity, absolutely convergent. More generally, for a window $\varphi\ge0$ on $[0,L]$ with autocorrelation $A(u)=\int_0^{L-u}\varphi(v)\varphi(v+u)dv$ and $t\in\mathbb R$, let $C^\varphi_N(t)=2\sum_{n<N}\Lambda(n)n^{-1/2}A(\log n)\cos(t\log n)$ and let $R^\varphi_N(t)$ be the right side of \eqref{ce:eq:weilzeta} without the prime term for $G(u)=A(|u|)\cos(tu)$. Then $R^\varphi_N(t)-C^\varphi_N(t)=\sum_\rho\frac12\big(\widehat\varphi(\gamma_\rho-t)\overline{\widehat\varphi(\bar\gamma_\rho-t)}+\widehat\varphi(\gamma_\rho+t)\overline{\widehat\varphi(\bar\gamma_\rho+t)}\big)$, a two-parameter family \cite[Proposition~10]{P20}; the case $\varphi=1$, $t=0$ is the first statement.
\end{proposition}

\begin{proof}
Take $G(u)=(L-|u|)_+$, which is $g*\tilde g$ for $g=1_{[0,L]}$, so $\widehat G=|\widehat g|^2=F_L$ on $\mathbb R$; $G$ is even, continuous, of bounded variation, supported in $[-L,L]$, and $\widehat G(z)=O(|z|^{-2})$ on horizontal lines, so \eqref{ce:eq:weilzeta} applies. Its prime side is $\sum_n\Lambda(n)n^{-1/2}\cdot2(L-\log n)_+=C(N)$. The pole terms are $\int G(u)(e^{u/2}+e^{-u/2})du=8(\sqrt N+N^{-1/2}-2)$. For the archimedean term, Gauss's integral $\psi(z)=\int_0^\infty(\frac{e^{-x}}x-\frac{e^{-zx}}{1-e^{-x}})dx$ and $\frac1{2\pi}\int\widehat G(r)e^{-irx/2}dr=G(\frac x2)$ give, after $x=2u$,
\[
-L\log\pi+\int_0^\infty\Big[\frac{Le^{-2u}}u-\frac{2e^{-u/2}G(u)}{1-e^{-2u}}\Big]du ,
\]
and Proposition~\ref{ce:prop:closed} evaluates the total as $R(N)$. The general window is the same computation with $G(u)=A(|u|)\cos(tu)=\frac12\sum_\pm(k_\pm*\tilde k_\pm)(u)$, $k_\pm(u)=\varphi(u)e^{\pm itu}$.
\end{proof}

\begin{proposition}[the remainder in closed form]\label{ce:prop:closed}
The right side of \eqref{ce:eq:weilzeta} for $G(u)=(L-|u|)_+$, without the prime term, is $R(N)$ given by \eqref{ce:eq:closed}.
\end{proposition}

\begin{proof}
The pole term is $8(\sqrt N+N^{-1/2}-2)$. Extending the archimedean integral to $(0,\infty)$,
\[
L\int_0^\infty\Big[\frac{e^{-2u}}u-\frac{2e^{-u/2}}{1-e^{-2u}}\Big]du+\int_0^\infty\frac{2u\,e^{-u/2}}{1-e^{-2u}}du-\int_L^\infty\frac{2(u-L)e^{-u/2}}{1-e^{-2u}}du .
\]
The first integral is $\psi(\frac14)=-\gamma_E-\frac\pi2-3\log2$. Expanding $(1-e^{-2u})^{-1}=\sum_ke^{-2ku}$, the second is $2\sum_{k\ge0}(2k+\frac12)^{-2}=\frac12\psi'(\frac14)=\frac{\pi^2}2+4G$, and the third is $2\sum_{k\ge0}N^{-(2k+1/2)}(2k+\frac12)^{-2}$, whose term $k=0$ cancels the $8N^{-1/2}$ of the pole.
\end{proof}

A second derivation, independent of Proposition~\ref{ce:prop:fejer}, integrates von Mangoldt's explicit formula for $\sum_{n<N}\Lambda(n)n^{-1/2}$ twice in $\log N$ \cite{In,MV}: it gives $C(N)=R(N)-M(N)$ with
\[
R(N)=8(\sqrt N-1)-2\frac{\zeta'}\zeta(\tfrac12)\log N+2\sum_{m\ge1}\frac{1-N^{-(2m+1/2)}}{(2m+\frac12)^2},\qquad M(N)=2\sum_\rho\frac{1-N^{i\gamma_\rho}}{\gamma_\rho^2},
\]
and comparing with \eqref{ce:eq:closed} gives $\log\pi+\gamma_E+\frac\pi2+3\log2=2\frac{\zeta'}\zeta(\frac12)$, which is $\xi'(\frac12)=0$. So the remainder is the smoothed sum of the smooth part of the partial sum (the pole, the constant $\frac{\zeta'}\zeta(\frac12)$ and the trivial zeros), and the margin is the smoothed sum of its oscillating part.

\subsection{RH in one variable}
\begin{theorem}[Suzuki \cite{Su1,Su2}]\label{ce:thm:landau}
RH holds if and only if $M(N)\ge0$ for all sufficiently large $N$. Under RH,
\begin{equation}\label{ce:eq:wave}
M(N)=4\sum_{\gamma>0}\frac{1-\cos(\gamma\log N)}{\gamma^2}\in\Big[0,\ 8\sum_{\gamma>0}\gamma^{-2}\Big],
\end{equation}
so then $M(N)\ge0$ for every $N>1$.
\end{theorem}

\begin{proof}
By Proposition~\ref{ce:prop:fejer}, $M(e^L)=\sum_\rho F_L(\gamma_\rho)$. Under RH every term is $\ge0$, and pairing $\pm\gamma$ gives \eqref{ce:eq:wave}. Conversely, suppose $M(e^L)\ge0$ for $L\ge L_0$. For $\re s>\frac12$, since $|F_L(\gamma)|\le(2+2e^{L/2})|\gamma|^{-2}$ and $\sum|\gamma_\rho|^{-2}<\infty$, Fubini gives
\[
\int_0^\infty M(e^L)e^{-sL}dL=\sum_\rho\frac2{s(s^2+\gamma_\rho^2)}=:\mathcal G(s),
\]
meromorphic on $\mathbb C$ with poles at $0$ and at $s=\pm i\gamma_\rho$. A pole $s$ fixes $\gamma^2=-s^2$, so only $\pm\gamma$ meet there, with equal nonzero residues, and no cancellation occurs. An off-line zero $\frac12+\delta+i\gamma_0$, $\delta>0$, has $\gamma_0\ne0$ because $\zeta$ has no real zeros in $(0,1)$, and gives a pole at $\delta+i\gamma_0$ off the real axis; the zeros on the line give poles on the imaginary axis. So $\mathcal G$ is analytic on $(0,\infty)$. By Landau's theorem \cite{In,MV}, the Laplace integral of a function that is $\ge0$ on $[L_0,\infty)$ has a singularity at the real point of its abscissa of convergence; it differs from $\mathcal G$ by an entire function, so the abscissa is $\le0$, and $\mathcal G$ has no pole with $\re s>0$.
\end{proof}

Status: proved. It is Landau's oscillation method for a Riesz mean of $\psi(x)-x$ weighted by $n^{-1/2}$, and the theorem is Suzuki's: with $M=2\Psi(\log N)$, RH holds if and only if $\Psi\ge0$, and if and only if $\Psi=O(1)$ \cite[Theorems 1.6, 1.7]{Su1}; the criterion with the main term alone is \cite[Theorem 1]{Su2}. Our proof is independent. The constants in \eqref{ce:eq:wave} are values of $\xi$ at the centre: Hadamard's product gives $(\log\xi)''(s)=-\sum_\rho(s-\rho)^{-2}$, and at $s=\frac12$
\begin{equation}\label{ce:eq:xi2}
(\log\xi)''(\tfrac12)=\sum_\rho\frac1{\gamma_\rho^2}=-8+\frac{\pi^2}4+2G+\Big(\frac{\zeta''}\zeta-\Big(\frac{\zeta'}\zeta\Big)^2\Big)(\tfrac12)=0.0462100,
\end{equation}
unconditionally; under RH it is $2\sum_{\gamma>0}\gamma^{-2}$, and with the $4520$ zeros below $5000$ and the tail the two agree to $6\cdot10^{-10}$.

\begin{corollary}[the sandwich]\label{ce:cor:sandwich}
The following are equivalent: (1) RH; (2) $M(N)\ge0$ for all large $N$; (3) $M(N)\le B$ for some $B$ and all large $N$; (4) for every $N>1$,
\begin{equation}\label{ce:eq:sandwich}
R(N)-4(\log\xi)''(\tfrac12)\;\le\;C(N)\;\le\;R(N),\qquad4(\log\xi)''(\tfrac12)=0.18484 .
\end{equation}
\end{corollary}

\begin{proof}
(1)$\Rightarrow$(4) is \eqref{ce:eq:wave} with \eqref{ce:eq:xi2}; (4)$\Rightarrow$(2),(3) are immediate; (2)$\Rightarrow$(1) is Theorem~\ref{ce:thm:landau}; for (3)$\Rightarrow$(1) apply the same argument to $B-M(e^L)\ge0$, whose Laplace integral has the same poles off the real axis.
\end{proof}

Status: proved; the equivalences (1)--(3) and the bounds are in \cite{Su1}, and the evaluation \eqref{ce:eq:xi2} of $\sum_\rho\gamma_\rho^{-2}$ is classical \cite{Vo}. Writing the window this way makes \eqref{ce:eq:sandwich} checkable without the zeros, which is what the verification below uses. The two sides are not redundant in general: a real zero $\beta\in(\frac12,1)$ would push $M$ upwards only, and only the upper side sees it. For $\zeta$, which has no such zeros, either side alone is equivalent to RH; for the Beurling system and the fake curve of \cite{P20} the lower side never fails and the upper side fails for all large $N$.

\subsection{What is known unconditionally}
\begin{proposition}[unconditional bound, crude form]\label{ce:prop:uncond}
For every $N>1$, $M(N)\ge-8(3+\sqrt N)S\ge-2.29\cdot10^{-11}(3+\sqrt N)$, with $S=\sum_{\im\rho>T_0}(\im\rho)^{-2}\le2.854\cdot10^{-12}$.
\end{proposition}

\begin{proof}
In Proposition~\ref{ce:prop:fejer} the zeros on the line contribute $\ge0$. By Platt and Trudgian \cite{PT} every zero with $0<\im\rho\le T_0$ is on the line, so an off-line zero has $\gamma_\rho=\gamma_0+id$ with $|\gamma_0|>T_0$, $|d|<\frac12$. With $F_L(z)=(2-e^{iLz}-e^{-iLz})/z^2$, $|F_L(z)|\le(2+N^{|d|}+N^{-|d|})/|z|^2\le(3+\sqrt N)/\gamma_0^2$, and summing over the off-line ordinates, which come in quadruples $\pm\gamma_0\pm id$, gives at most $8(3+\sqrt N)S$ (a conservative constant). Stieltjes integration and Trudgian's bound for $N(u)$ \cite{Tr} give $S\le\int_{T_0}^\infty2N(u)u^{-3}du\le2.854\cdot10^{-12}$.
\end{proof}

The factor $\sqrt N$ is the price of a hypothetical zero near $\re s=1$. A zero-density estimate says that such zeros are rare, and it improves the bound by many orders of magnitude. Let $N(\sigma,T)$ be the number of zeros $\beta+i\gamma$ with $\beta>\sigma$, $0<\gamma<T$. Kadiri, Lumley and Ng \cite{KLN} prove, for $T\ge H_0=3.0610046\cdot10^{10}$ and each $\sigma$ in their Table~1 ($\sigma=0.60,0.65,\dots,0.85,0.86,\dots,0.99$), with constants $A_\sigma,B_\sigma$ and $k_\sigma\le1$ listed there,
\begin{equation}\label{ce:eq:kln}
N(\sigma,T)\le A_\sigma(\log k_\sigma T)^{2\sigma}(\log T)^{5-4\sigma}T^{\frac83(1-\sigma)}+B_\sigma(\log T)^2\le A_\sigma(\log T)^{5-2\sigma}T^{\frac83(1-\sigma)}+B_\sigma(\log T)^2 .
\end{equation}

\begin{theorem}[explicit unconditional lower bound]\label{ce:thm:explicit}
Let $T_0=3\cdot10^{12}$, $\widetilde N(t)=\frac t{2\pi}\log\frac t{2\pi}$, and let $\frac12=\sigma_0<\sigma_1<\dots<\sigma_J<\sigma_{J+1}=1$ be $\frac12$, the twenty values of $\sigma$ in \cite[Table 1]{KLN}, and $1$. Put
\[
S_{1/2}=\int_{T_0}^\infty\widetilde N(t)\,t^{-3}\,dt,\qquad S_j=2\int_{T_0}^\infty\min\Bigl(N_{\rm KLN}(\sigma_j,t),\tfrac12\widetilde N(t)\Bigr)t^{-3}\,dt\quad(1\le j\le J),
\]
with $N_{\rm KLN}$ the right side of \eqref{ce:eq:kln}. Then for every $N>1$
\begin{equation}\label{ce:eq:EN}
M(N)\ge-E(N),\qquad E(N)=4\Bigl[(3+N^{\sigma_1-\frac12})S_{1/2}+\sum_{j=1}^{J}\bigl(N^{\sigma_{j+1}-\frac12}-N^{\sigma_j-\frac12}\bigr)S_j\Bigr].
\end{equation}
One may take $S_{1/2}=1.4822\cdot10^{-12}$, $S_j=S_{1/2}$ to four digits for $\sigma_j\le0.75$, $S(0.80)=4.71\cdot10^{-13}$, $S(0.85)=8.63\cdot10^{-15}$, $S(0.90)=1.57\cdot10^{-16}$, $S(0.95)=2.85\cdot10^{-18}$, $S(0.99)=1.14\cdot10^{-19}$, and $E(N)$ takes the values of Table~\ref{ce:tab:EN}. In particular $|\min(M(N),0)|<10^{-3}$ for $N<5\cdot10^{24}$ and $E(N)<0.043$ for $N<2.3\cdot10^{29}$.
\end{theorem}

\begin{proof}
As in Proposition~\ref{ce:prop:uncond}, $M(N)\ge-\sum'|F_L(\gamma_\rho)|$, the sum over the off-line zeros, all of which have $|\im\rho|>T_0$, and $|F_L(\gamma_0+id)|\le(3+N^{|d|})/\gamma_0^2$. A zero $\rho=\beta+i\gamma$ with $\beta>\frac12$, $\gamma>T_0$, brings the quadruple $\rho,\bar\rho,1-\rho,1-\bar\rho$, so
\[
\sum{}'|F_L|\le4\sum_{\beta>\frac12,\ \gamma>T_0}\frac{3+N^{\beta-1/2}}{\gamma^2}.
\]
Write $\Sigma(\sigma)=\sum_{\beta>\sigma,\gamma>T_0}\gamma^{-2}$. Since $N^{\beta-1/2}\le N^{\sigma_{j+1}-1/2}$ when $\sigma_j<\beta\le\sigma_{j+1}$, Abel summation over the bins gives
\[
\sum_{\beta>\frac12,\gamma>T_0}\frac{N^{\beta-1/2}}{\gamma^2}\le N^{\sigma_1-\frac12}\Sigma(\tfrac12)+\sum_{j=1}^J\bigl(N^{\sigma_{j+1}-\frac12}-N^{\sigma_j-\frac12}\bigr)\Sigma(\sigma_j).
\]
It remains to show $\Sigma(\frac12)\le S_{1/2}$ and $\Sigma(\sigma_j)\le S_j$. Since $N(\sigma,T_0)=0$, Stieltjes integration gives $\Sigma(\sigma)=2\int_{T_0}^\infty N(\sigma,t)t^{-3}dt$. By the symmetry $\beta\leftrightarrow1-\beta$, at most half of the zeros up to height $t$ lie to the right of the line, so $N(\sigma,t)\le\frac12N(t)$; and $N(t)\le\widetilde N(t)$ for $t\ge T_0$, since $N(t)=\frac t{2\pi}\log\frac t{2\pi e}+\frac78+\theta(0.112\log t+0.278\log\log t+2.51+0.2/t)$ with $|\theta|\le1$ \cite{Tr}, and the term $-\frac t{2\pi}$ dominates. This gives $\Sigma(\frac12)\le S_{1/2}$. For $\sigma_j\ge0.6$, \eqref{ce:eq:kln} applies since $T_0>H_0$, and $\Sigma(\sigma_j)\le S_j$. The values used come from adaptive quadrature in $u=\log t$ (\texttt{ebound.py}); they are rigorous upper bounds, because each integrand $g(u)$ is positive and decreasing for $t\ge T_0$ (both $N_{\rm KLN}(\sigma,t)t^{-2}$ and $\widetilde N(t)t^{-2}$ decrease), so a left Riemann sum of step $10^{-4}$ on $[\log T_0,\log T_0+120]$ plus an explicit tail bounds the integral from above, and every quadrature value exceeds that bound (\texttt{ebound\_rig.py}; the ratio is at most $0.9984$). For instance $S_{1/2}=(\log\frac{T_0}{2\pi}+1)/(2\pi T_0)=1.47970\cdot10^{-12}$ exactly, the Riemann sum gives $1.47977\cdot10^{-12}$ and the value used is $1.48222\cdot10^{-12}$.
\end{proof}

\begin{corollary}[unconditional two-sided bound]\label{ce:cor:twosided}
For every $N>1$,
\[
-E(N)\le M(N)\le4(\log\xi)''(\tfrac12)+E(N)=0.18484+E(N).
\]
\end{corollary}

\begin{proof}
The lower side is Theorem~\ref{ce:thm:explicit}. For the upper side split $M=\sum_\rho F_L(\gamma_\rho)$ into zeros on and off the line. A pair $\pm\gamma$ on the line contributes $4(1-\cos\gamma L)/\gamma^2\le8/\gamma^2$, and the off-line zeros contribute at most $E(N)$ in absolute value, as in the proof of Theorem~\ref{ce:thm:explicit}. Identity \eqref{ce:eq:xi2} holds without RH, with $\gamma_\rho=(\rho-\frac12)/i$ complex for an off-line zero. A quadruple $\pm\gamma_0\pm id$ contributes to it $4(\gamma_0^2-d^2)/(\gamma_0^2+d^2)^2>0$, since $|d|<\frac12<\gamma_0$. Hence $\sum_{\text{on the line}}\gamma^{-2}\le(\log\xi)''(\frac12)$, and the zeros on the line contribute at most $4(\log\xi)''(\frac12)$.
\end{proof}

So without any hypothesis \eqref{ce:eq:sandwich} holds up to $E(N)$ on both sides; RH removes $E(N)$.

\begin{table}[ht]
\centering
\caption{The off-line contribution: crude bound of Proposition~\ref{ce:prop:uncond} against $E(N)$ of Theorem~\ref{ce:thm:explicit}.}
\label{ce:tab:EN}
\begin{tabular}{lcccccc}
\toprule
$N$ & $10^{10}$ & $10^{15}$ & $10^{20}$ & $10^{25}$ & $10^{27}$ & $10^{30}$\\
\midrule
$8(3+\sqrt N)S$ & $2.3\cdot10^{-6}$ & $7.2\cdot10^{-4}$ & $0.23$ & $72$ & $720$ & $2.3\cdot10^4$\\
$E(N)$ & $1.0\cdot10^{-8}$ & $4.7\cdot10^{-7}$ & $2.4\cdot10^{-5}$ & $1.3\cdot10^{-3}$ & $6.3\cdot10^{-3}$ & $0.072$\\
\bottomrule
\end{tabular}
\end{table}

The crude bound is below the least margin observed ($0.0427$, Proposition~\ref{ce:prop:verified}) as long as $N<3.5\cdot10^{18}$; the explicit bound \eqref{ce:eq:EN} stays below it up to $2.3\cdot10^{29}$. Beyond that it grows like $N^{0.35}$ (the bin $0.80<\beta\le0.85$ dominates) and says nothing. This is the familiar fact that verifying RH to a height $H$, together with zero-density estimates, gives bounds of RH quality only in a range polynomial in $H$ \cite{Jn}; it comes from the zeros, not from the primes. Theorem~\ref{ce:thm:explicit} is not a step towards RH: it holds whether or not RH is true, and it is silent for $N>10^{30}$.

\subsection{Verification}
\begin{proposition}[the sandwich verified]\label{ce:prop:verified}
\begin{enumerate}
\item For every real $N$ with $1<N\le10^{10}$, $0<M(N)<0.14$; more precisely $M(N)\ge0.0148\log N$ for $1<N\le e^{1/15}$, $M(N)\ge0.0276$ for $e^{1/30}\le N\le10^{10}$, $M(N)\ge0.0429$ for $10^3\le N\le10^{10}$, and $M(N)\le0.1379$ throughout.
\item For every real $N$ with $10^{10}\le N\le10^{12}$, $0.04268\le M(N)\le0.13122$. The least value of $M$ on $(10^3,10^{12}]$ is attained near $N=1.056\cdot10^{11}$.
\item $M(N)\ge0.0415$ for every real $N$ with $10^8\le N\le10^{15}$, $M(N)\ge0.0481$ for $10^{15}\le N\le10^{27}$, and $M(N)>0$ for $10^{15}\le N\le2.15\cdot10^{29}$; $M(N)\le0.1377$ for $10^8\le N\le10^{27}$ and $M(N)<0.18484$ for $10^8\le N\le4.28\cdot10^{29}$; so \eqref{ce:eq:sandwich} holds on $[10^8,2.15\cdot10^{29}]$. Here the $4520$ zeros below $5000$ are certified: with ball arithmetic, $N(5000)=4520$ and each ordinate used lies within $9.1\cdot10^{-12}$ of the true one.
\end{enumerate}
Since $0.14<0.18484$, parts (1) and (3) give the two-sided inequality \eqref{ce:eq:sandwich} for every real $N$ with $1<N\le2.15\cdot10^{29}$; part (2) checks the range $[10^{10},10^{12}]$ independently of the zeros.
\end{proposition}

\begin{proof}
(1) Write $m(L)=M(e^L)$ and $r(L)=R(e^L)$. \emph{Near $N=1$}, no prime power lies in $(1,2)$ and $M=R$; elementary bounds on the series in \eqref{ce:eq:closed} give $r(L)\ge0.01488L$ for $L\le\frac1{15}$ and $r(L)\le0.1016$ for $L\le\frac1{30}$, and on $[\frac1{30},\log2]$ monotonicity of each piece on $3300$ intervals gives $0.0670\le M\le0.1282$. \emph{From $2$ to $10^{10}$}: between consecutive prime powers $q<q'$, $m(L)=r(L)-2AL+2B$ with $A=\sum_{n\le q}\Lambda(n)n^{-1/2}$ and $B=\sum_{n\le q}\Lambda(n)n^{-1/2}\log n$ constant, and $m''=r''=2e^{L/2}-2e^{-L/2}/(e^{2L}-1)>0$ for $N\ge2$, so $m$ is convex there and its minimum on $[\log q,\log q']$ is bounded below by the tangent lines at the two ends; its maximum is at an end. \emph{Arithmetic}: Python's \texttt{decimal} at $34$ digits, whose $+,-,\times,/,\sqrt{\ },\log,\exp$ are correctly rounded; $\pi,\gamma_E,G$ entered to $48$ digits; a margin of $10^{-15}$ is subtracted from every lower bound and added to every upper bound. The computation ran over all $455\,062\,595$ prime powers below $10^{10}$, from a segmented sieve; $8$ intervals below $N=100$ needed subdivision. The least lower bound for $N\ge10^3$ is $0.042997$, on $[34\,186\,367,34\,186\,499]$, where the minimum of $M$ below $10^{10}$ is attained; the largest upper bound is $0.137816$ at $N=319\,439$.

(2) Same convexity argument, over all $37\,607\,992\,088$ prime powers below $10^{12}$ (the count equals $\pi(10^{12})=37\,607\,912\,018$ plus the $80\,070$ powers $p^k$, $k\ge2$, which checks the sieve). On $[\log q,\log q']$, $\min m\ge\max\bigl(m(a)-(b-a)\max(0,-m'(a)),\,m(b)-(b-a)\max(0,m'(b))\bigr)$. The arithmetic is IEEE double precision. The operations $+,-,\times,/,\sqrt{\ }$ are correctly rounded; the library logarithm was checked at run time to be within $0.5$ ulp of the correctly rounded value on $2\cdot10^4$ test points, and an error of $2$ ulp is assumed in the bounds. The sums $A$ and $B$ are accumulated in segments of length $5\cdot10^7$, each total exactly rounded (\texttt{math.fsum}) and the totals added exactly in rational arithmetic. Every lower bound is lowered, and every upper bound raised, by an explicit bound for all rounding errors (the error in $A(10^{12})=1\,999\,997.3576058$ is at most $2.1\cdot10^{-9}$, and the error of one evaluation of $m$ is of order $10^{-7}$). The least lower bound is $0.042686$, near $N=105\,606\,345\,001$; the largest upper bound on $[10^{10},10^{12}]$ is $0.131211$, near $N=742\,864\,907\,663$. The same program reproduces part (1) on $[10^3,10^{10}]$ ($0.042997$, $0.137816$, $455\,062\,595$ prime powers). The minimum near $1.056\cdot10^{11}$ is also visible from the zeros: there the sum over the $4520$ zeros below $5000$ in part (3) is $0.041711$, and the difference $0.00098$ is the mean contribution $4\sum_{\gamma>5000}\gamma^{-2}=0.00098$ of the omitted zeros.

(3) Unconditionally $M=\sum_\rho F_L(\gamma_\rho)$; the zeros up to $3\cdot10^{12}$ are on the line and contribute $\ge0$, so dropping those not in the list and bounding the rest as in Proposition~\ref{ce:prop:uncond},
$M(N)\ge4\sum_{\gamma\in\text{list}}(1-\cos\gamma L)/\gamma^2-8(3+\sqrt N)S$. The sum is Lipschitz with constant $4\sum1/\gamma=14.132$; on a grid of $805\,906$ points covering $[\log10^8,\log10^{15}]$, the least value of the sum on a cell, minus the Lipschitz loss and minus $8(3+\sqrt N)S$ at the right end of the cell (at most $7.2\cdot10^{-4}$), is $0.04156$; it is attained near $N=1.06\cdot10^{11}$, where the sum is $0.041711$, the Lipschitz loss $1.41\cdot10^{-4}$ and the tail term $7\cdot10^{-6}$. For $10^{15}\le N\le10^{30}$ the same computation with $E(N)$ of Theorem~\ref{ce:thm:explicit} in place of $8(3+\sqrt N)S$, on a grid of step $2\cdot10^{-5}$ in $L$ ($1\,726\,941$ points), gives the least lower bound $0.048104$ up to $10^{27}$, $0.047549$ up to $10^{28}$ and $0.034188$ up to $10^{29}$; the first cell where $W-E\le0$ is at $N=2.159\cdot10^{29}$, where $E=0.0418$ and the sum is near its least value on the range, $0.038908$ at $N=2.18\cdot10^{29}$ (Figure~\ref{ce:fig:explicit}). For the upper side, the zeros on the line above $5000$ contribute at most $8\sum_{\gamma>5000}\gamma^{-2}$, and by Stieltjes integration with $N(5000)=4520$ and Trudgian's bound for $N(t)$ this is at most $8\cdot2.446\cdot10^{-4}=1.957\cdot10^{-3}$. (As a check, $2(\sum_{\rm list}4\gamma^{-2}+4\cdot2.446\cdot10^{-4})=0.184841$, against $4(\log\xi)''(\frac12)=0.184840$.) So $M\le W+1.957\cdot10^{-3}+E(N)$. The largest upper bound on the cells, the maximum of $W$ plus the Lipschitz gain, is $0.137298$ up to $10^{20}$, $0.137626$ up to $10^{27}$ and $0.161553$ up to $10^{29}$, and it first reaches $0.18484$ at $N=4.29\cdot10^{29}$; the maximum of $W$ on $[10^8,10^{30}]$ is $0.135632$. An error $\delta$ in each ordinate changes the sum by at most $4\delta L\sum\gamma^{-2}\le3.2\,\delta$ for $N\le10^{30}$.
\end{proof}

Status: (1) is computer-assisted, with correctly rounded arithmetic, and was checked in three independent ways (an extended-precision scan of $A$ and $B$ to $10^{10}$, $45$-digit values below $2\cdot10^6$, and an independent sieve on three windows). (2) is computer-assisted in double precision with rigorous error bounds; it relies on the stated accuracy of the library logarithm, which was tested but not proved. (3) uses the zeros below $5000$, certified with Arb \cite{Ar} through python-flint: \texttt{acb.zeta\_zeros} returns balls guaranteed to contain the $n$-th zeros, isolated by sign changes of Hardy's $Z$ with the count checked by Turing's method. The $4520$-th zero lies in $[4999.32968137297,4999.32968137298]$ and the $4521$-st above $5000.2343169313$, so the list is complete, and it is accurate to $9.1\cdot10^{-12}$, which changes $W$ by less than $3\cdot10^{-11}$. The grid evaluation of $W$ in double precision has error below $10^{-10}$ (each term $4(1-\cos\gamma L)/\gamma^2$ to a few units in the last place of $\gamma L\le3.5\cdot10^5$). Neither reaches the statement for all large $N$: the primes cost one evaluation per prime power, and the zeros stop near $2\cdot10^{29}$ by Theorem~\ref{ce:thm:explicit}.

\begin{figure}[htbp]\centering\includegraphics[width=\textwidth]{fig/CE_explicit.png}
\caption{(a) The bound on how far off-line zeros could move $M(N)$: the crude bound of Proposition~\ref{ce:prop:uncond} and $E(N)$ of Theorem~\ref{ce:thm:explicit}, which uses the zero density of \cite{KLN}. (b) For $10^{15}\le N\le10^{30}$: the contribution $W$ of the $4520$ zeros below $5000$ (grey), the bound $E(N)$ (red) the certified lower bound $W-E$ for $M$ (blue), positive up to $2.16\cdot10^{29}$ (dashed), and the upper bound $W+1.96\cdot10^{-3}+E$ (green), below $0.18484$ up to $4.29\cdot10^{29}$.}
\label{ce:fig:explicit}
\end{figure}

\emph{The wave.} Formula \eqref{ce:eq:wave} says what the primes must produce: an almost periodic function of $\log N$ with mean $4\sum_{\gamma>0}\gamma^{-2}=0.0924$, frequencies at the ordinates and amplitudes $4/\gamma^2$. From the $5\,762\,859$ prime powers below $10^8$, at $6000$ values of $\log N$, $M$ agrees with the right side of \eqref{ce:eq:wave} truncated at the zeros below $5000$ to $5.4\cdot10^{-5}$; its spectrum in $\log N$, from the primes alone, has peaks at $14.132$, $21.023$, $25.014$, $30.425,\dots$ with heights agreeing with $4/\gamma^2$ to $10^{-5}$ (Figure~\ref{ce:fig:wave}). Over $4\cdot10^5$ values of $\log N$ in $[\log2,\log10^8]$ the mean is $0.092367$ (predicted $2(\log\xi)''(\frac12)=0.092420$) and the standard deviation $0.017219$ (predicted $\sqrt{8\sum\gamma^{-4}}=0.017245$ for independent phases).

\begin{figure}[htbp]\centering\includegraphics[width=\textwidth]{fig/CE20_wave.png}
\caption{Top: the margin $M(N)$ computed from the primes below $10^8$ (black) and the right side of \eqref{ce:eq:wave} from the $4520$ zeros below $5000$ (dashed); mean $0.0924$, least value $0.043$. Bottom: its spectrum in the variable $\log N$, from the primes, against $4/\gamma^2$ at the zeros.}\label{ce:fig:wave}
\end{figure}

A least-squares fit of $m(L)=M(e^L)$ on the first $40$ ordinates, from the prime powers below $10^7$, gives cosine amplitudes equal to $-4/\gamma_k^2$ to within $0.4\%$ and sine amplitudes below $10^{-5}$, as \eqref{ce:eq:wave} predicts under RH. The decay $4/\gamma^2$ is the regularity of $m$: it is continuous, and its slope jumps by $-2\Lambda(q)/\sqrt q$ at $L=\log q$ (Proposition~\ref{ce:prop:minima}).

\emph{Where the difficulty sits.} By blocks at $N=0.999\cdot10^8$: $R(N)=79853.63$, the primes give $79665.63$, the prime squares $130.68$, the higher powers $57.21$, and $M=0.114$. The squares are controlled without RH; the whole difficulty is in the sum over primes, which has to be known to $O(1)$ out of $8\sqrt N$. Without RH the best bound for that sum is of size $\sqrt N\exp(-c(\log N)^{3/5}(\log\log N)^{-1/5})$, and closing the gap to $O(1)$ is RH. For the Davenport--Heilbronn function the same construction gives its own window $w_f=0.78456$; its margin computed from its coefficients up to $2\cdot10^7$ stays inside, and the model with its zeros predicts the first exit near $N\approx2.3\cdot10^8$ \cite{P20}: in one variable the criterion is the simplest, but the slowest to detect an off-line zero.

\section{The local minima and the slope}\label{ce:sec:minima}
The proof of Proposition~\ref{ce:prop:verified} uses that the margin is convex between consecutive prime powers. This has a sharper consequence: the minima of $M$ are a discrete set, one at most between two prime powers, with an explicit location and value. The criterion of Theorem~\ref{ce:thm:landau}, a statement about every real $N$, becomes a statement about countably many points given by sums over primes. We keep the notation of the proof of Proposition~\ref{ce:prop:verified}: $m(L)=M(e^L)$, $r(L)=R(e^L)$, and for consecutive prime powers $q<q'$
\[
A_q=\sum_{n\le q}\frac{\Lambda(n)}{\sqrt n},\qquad B_q=\sum_{n\le q}\frac{\Lambda(n)\log n}{\sqrt n},\qquad m(L)=r(L)-2A_qL+2B_q\quad(\log q\le L\le\log q').
\]
Write $c_0=\log\pi+\gamma_E+\frac\pi2+3\log2=5.3721834$ and $K_0=-16+\frac{\pi^2}2+4G=-7.4013354$, so that by \eqref{ce:eq:closed}
\begin{equation}\label{ce:eq:rL}
r(L)=8e^{L/2}-c_0L+K_0-\tau(L),\qquad \tau(L)=2\sum_{k\ge1}\frac{e^{-(2k+\frac12)L}}{(2k+\frac12)^2}.
\end{equation}

\begin{proposition}[the local minima of the margin]\label{ce:prop:minima}
\begin{enumerate}
\item On $L>L_c=0.2824$ ($N>1.326$), $r''(L)=2e^{L/2}-2e^{-L/2}/(e^{2L}-1)>0$; so $m$ is strictly convex on every interval between consecutive prime powers.
\item At a prime power $q=p^k$ the derivative of $m$ jumps down: $m'(\log q^+)-m'(\log q^-)=-2\Lambda(q)/\sqrt q<0$. Hence no prime power is a local minimum of $m$.
\item For $N\ge2$, the local minima of $M$ are exactly the points $N^*_q=e^{L^*_q}$ with $r'(L^*_q)=2A_q$ and $q<N^*_q<q'$, one at most between two consecutive prime powers; at such a point
\begin{equation}\label{ce:eq:minval}
M(N^*_q)=2B_q+K_0-8u_q\big(\log u_q-1\big)+\epsilon_q,\qquad u_q=\frac{2A_q+c_0}4,
\end{equation}
with $N^*_q=u_q^2(1+O(u_q^{-5}))$ and $\epsilon_q$ coming from $\tau$: $|\epsilon_q|\le0.035$, and $|\epsilon_q|\le3\cdot10^{-6}$ for $N^*_q>100$.
\item On $1<N<2$, where $M=R$, the only local minimum is at $N=1.5904$, with $M=0.0793$, and $M(N)\to0$ as $N\to1^+$.
\end{enumerate}
\end{proposition}

\begin{proof}
(1) Differentiating \eqref{ce:eq:rL}, $r''(L)=2e^{L/2}-2\sum_{k\ge1}e^{-(2k+\frac12)L}=2e^{L/2}-2e^{-L/2}/(e^{2L}-1)$, which is positive exactly when $e^L(e^{2L}-1)>1$, that is for $L>0.2824$. Between prime powers $m''=r''$. (2) $C(e^L)=2\sum_n\Lambda(n)n^{-1/2}(L-\log n)_+$ gains the slope $2\Lambda(q)/\sqrt q$ at $L=\log q$, and $r$ is smooth. A local minimum at $\log q$ would need $m'(\log q^-)\le0\le m'(\log q^+)$, which the jump forbids. (3) By (1) and (2), for $L\ge\log2$ every local minimum is an interior critical point of a strictly convex piece, so it is the unique solution of $m'=r'-2A_q=0$ in that piece, if it lies there. With $r'(L)=4e^{L/2}-c_0+\tau_1(L)$, $\tau_1=-\tau'$, the equation is $4e^{L/2}=2A_q+c_0-\tau_1(L)$, so $e^{L^*/2}=u_q-\tau_1(L^*)/4$. Inserting $e^{L^*/2}=u_q$ and $L^*=2\log u_q$ in $m=r-2A_qL+2B_q$ gives $8u_q-2\log u_q\,(c_0+2A_q)+K_0+2B_q=2B_q+K_0-8u_q(\log u_q-1)$; the difference with the exact value comes from $\tau$ and $\tau_1$, of order $N^{-5/2}$, and its size is the computed one. (4) On $(1,2)$ the sum over primes is empty, $r''<0$ for $N<1.326$ and $r''>0$ after, $r'$ vanishes once on $(1.326,2)$, at $N=1.5904$, and $r(0)=0$ with $r(L)\ge0.0148L$ near $0$ (Proposition~\ref{ce:prop:verified}).
\end{proof}

Status: proved; the bounds for $\epsilon_q$ and the values in (4) are numerical, from the exact solution of $r'(L)=2A_q$ by Newton's method. Part (1) is the convexity already used in Proposition~\ref{ce:prop:verified}; parts (2) and (3) are new to this paper as far as we know.

\begin{corollary}[a discrete criterion]\label{ce:cor:discrete}
RH holds if and only if $M(N^*_q)\ge0$ for every prime power $q$ with $q<N^*_q<q'$, that is, if and only if
\begin{equation}\label{ce:eq:discrete}
2B_q\;\ge\;8u_q\big(\log u_q-1\big)-K_0-\epsilon_q\qquad\text{whenever }q<N^*_q<q'.
\end{equation}
\end{corollary}

\begin{proof}
If RH holds, $M\ge0$ everywhere by Theorem~\ref{ce:thm:landau}. Conversely, assume \eqref{ce:eq:discrete} and suppose $M(N_0)<0$ for some $N_0\ge2$ ($M>0$ on $(1,2)$ by Proposition~\ref{ce:prop:verified}). Let $L_0=\log N_0$. If $m'(L_0^-)>0$, follow $m$ to the left: on the convex piece containing $L_0$ it decreases towards the left end unless it meets a critical point, which would be a local minimum with value $<m(L_0)<0$; at a prime power the left derivative is larger than the right one by (2), so $m$ keeps decreasing to the left. Either we meet a local minimum with negative value, contradicting \eqref{ce:eq:discrete}, or we reach $L=\log2$ with $m(\log2)<0$, contradicting Proposition~\ref{ce:prop:verified}. If $m'(L_0^+)<0$, follow $m$ to the right in the same way: either a local minimum with negative value appears, or $m$ decreases for all $L>L_0$; in the second case $M\le M(N_0)$ for all $N\ge N_0$, so $M$ is bounded above, and Corollary~\ref{ce:cor:sandwich}(3) gives RH, hence $M\ge0$, a contradiction. One of the two cases occurs, since $m'(L_0^-)\ge m'(L_0^+)$ and they cannot be both $\le0$ and $\ge0$ unless $L_0$ is itself a critical point, which is then a local minimum with negative value.
\end{proof}

So RH is equivalent to countably many inequalities between two explicit sums over the prime powers below $q$, at the prime powers $q$ for which the stationary point $u_q^2$ falls before the next prime power. The reduction removes the need to look at every real $N$; it does not make the inequality easier. Both sides of \eqref{ce:eq:discrete} are of size $4\sqrt q\log q$, and their difference is $M(N^*_q)\in[0,0.185]$ under RH.

\begin{proposition}[no margin]\label{ce:prop:nomargin}
If RH holds, then $\liminf_{N\to\infty}M(N)=0$, and the local minima of Proposition~\ref{ce:prop:minima} accumulate at $0$: $\liminf M(N^*_q)=0$ over the prime powers $q$ with $q<N^*_q<q'$.
\end{proposition}

\begin{proof}
Under RH, $m(L)=4\sum_{\gamma>0}(1-\cos\gamma L)/\gamma^2$ by \eqref{ce:eq:wave}, an absolutely convergent trigonometric series, so $m$ is uniformly almost periodic with mean $2(\log\xi)''(\frac12)=0.0924$. Given $\varepsilon\in(0,0.09)$, choose $T$ with $8\sum_{\gamma>T}\gamma^{-2}<\varepsilon/2$. By Dirichlet's simultaneous approximation theorem there are arbitrarily large $L_0$ with $\|\gamma L_0/2\pi\|$ so small for every $0<\gamma\le T$ that $4\sum_{\gamma\le T}(1-\cos\gamma L_0)/\gamma^2<\varepsilon/2$; then $m(L_0)<\varepsilon$. Since $m$ is almost periodic with mean $0.0924>\varepsilon$, the set $\{m>\varepsilon\}$ is relatively dense, so there are $a<L_0<b$ with $m(a),m(b)>\varepsilon$. The minimum of $m$ on $[a,b]$ is $<\varepsilon$, is not at $a$ or $b$, and is not at a prime power by Proposition~\ref{ce:prop:minima}(2); so it is a local minimum $N^*_q$ with $M(N^*_q)<\varepsilon$, and $N^*_q\ge e^a$ is arbitrarily large.
\end{proof}

Status: proved, conditional on RH, and needs no linear independence of the ordinates. So no proof of $M\ge0$ can rest on a uniform positive lower bound: under RH the margin is positive without margin, as the least eigenvalues of the windows are in Section~\ref{ce:sec:windows}.

\emph{The minima from the primes.} Below $10^7$ there are $665\,134$ prime powers and $10\,341$ of the intervals between them, $1.6\%$, contain a local minimum. The least is $M(N^*_q)=0.055041$ at $N^*_q=3099.71$, between $q=3089$ and $q'=3109$; the largest is $0.137816$ at $N^*_q=319\,433.67$, next to the maximum of $M$ found in Proposition~\ref{ce:prop:verified}; the mean is $0.091059$, against the mean $0.0924$ of $M$. The first ones are $M=0.0718$, $0.0754$, $0.0862$, $0.0652$ at $N^*_q=2.448$, $3.598$, $4.298$, $5.934$. The minimum $0.042997$ of $M$ below $10^{10}$, near $N=3.4\cdot10^7$, is necessarily one of these points.

Both sides of \eqref{ce:eq:discrete} are of size $4\sqrt q\log q$, and their difference is at most $0.185$. In this subsection we remove the common smooth part of the two sums $A$ and $B$ and look at what remains. The criterion becomes a condition on the fluctuation of one weighted Chebyshev sum, $B$, at the points where the fluctuation of the other, $A$, vanishes. The identities below are exact and elementary, and they contain no information beyond Corollary~\ref{ce:cor:discrete}. What they do is separate the two roles that the primes play at a minimum, where it lies and how deep it is, and show the size of the problem.

For real $x\ge1$ let $A(x)=\sum_{n\le x}\Lambda(n)n^{-1/2}$ and $B(x)=\sum_{n\le x}\Lambda(n)n^{-1/2}\log n$, so that $A_q=A(q)$ and $B_q=B(q)$, and define their fluctuations
\begin{equation}\label{ce:eq:alphabeta}
\alpha(x)=A(x)-\Big(2\sqrt x-\frac{c_0}2\Big),\qquad \beta(x)=B(x)-\big(2\sqrt x\log x-4\sqrt x\big).
\end{equation}
The smooth parts are the integrals $\int_0^x t^{-1/2}\,dt=2\sqrt x$ and $\int_0^x t^{-1/2}\log t\,dt=2\sqrt x\log x-4\sqrt x$, that is, the two sums with $d\psi(t)$ replaced by $dt$; the constant $-c_0/2$ in $\alpha$ is the one that the closed form \eqref{ce:eq:rL} imposes. By the prime number theorem and partial summation, $\alpha(x)=o(\sqrt x)$ and $\beta(x)=o(\sqrt x\log x)$; under RH, $\psi(x)-x=O(\sqrt x\log^2x)$ gives $\alpha(x)=O(\log^2x)$ and $\beta(x)=O(\log^3x)$. Between two consecutive prime powers $A$ and $B$ are constant, so $\alpha$ decreases, with $d\alpha/d\log x=-\sqrt x$, and at a prime power $q$ it jumps up by $\Lambda(q)/\sqrt q$. It is a sawtooth (Figure~\ref{ce:fig:floors}(a)).

\begin{proposition}[the two sums give the slope and the value of the margin]\label{ce:prop:slope}
Let $N\ge2$ not be a prime power, $L=\log N$, and $\tau_1=-\tau'$, that is, $\tau_1(L)=2\sum_{k\ge1}e^{-(2k+\frac12)L}/(2k+\frac12)$. Then
\begin{align}
\alpha(N)&=\frac{\tau_1(L)-m'(L)}2,\label{ce:eq:slope}\\
\beta(N)-L\,\alpha(N)&=\frac{m(L)-K_0+\tau(L)}2.\label{ce:eq:value}
\end{align}
Equivalently, $M(N)=K_0+2\big(\beta(N)-\alpha(N)\log N\big)-\tau(\log N)$ and $\beta(N)=\frac12\big(m-Lm'-K_0+\tau+L\tau_1\big)$.
\end{proposition}

\begin{proof}
Let $q<N<q'$ be the consecutive prime powers around $N$. On $(\log q,\log q')$ we have $A(e^L)=A_q$, $B(e^L)=B_q$ and $m(L)=r(L)-2A_qL+2B_q$. Differentiating \eqref{ce:eq:rL}, $r'(L)=4e^{L/2}-c_0+\tau_1(L)$, so
\[
m'(L)=4e^{L/2}-c_0+\tau_1(L)-2A_q=\tau_1(L)-2\alpha(N),
\]
which is \eqref{ce:eq:slope}. Write $A_q=2e^{L/2}-\frac{c_0}2+\alpha$ and $B_q=2Le^{L/2}-4e^{L/2}+\beta$, with $\alpha=\alpha(N)$, $\beta=\beta(N)$. Inserting these and \eqref{ce:eq:rL} in $m=r-2A_qL+2B_q$ gives
\[
m=8e^{L/2}-c_0L+K_0-\tau-2L\Big(2e^{L/2}-\frac{c_0}2+\alpha\Big)+2\big(2Le^{L/2}-4e^{L/2}+\beta\big)=K_0-\tau+2(\beta-L\alpha).
\]
The terms in $e^{L/2}$, $Le^{L/2}$ and $L$ cancel, and what remains is \eqref{ce:eq:value}.
\end{proof}

Status: proved, elementary. So $\alpha$ is, up to the factor $-\frac12$ and the trivial-zero term $\tau_1$, the slope of the margin in the variable $\log N$. The combination $\beta-\alpha\log N$ is its value, up to the constant $K_0$ and the factor $2$. The trivial-zero terms are small: $\tau(\log2)=0.0615$ and $\tau_1(\log2)=0.165$, and both are $O(N^{-5/2})$.

\begin{corollary}[the minima as zeros of $\alpha$]\label{ce:cor:beta}
For $N\ge2$, the local minima of $M$ are the points $N^*_q$ of an interval $(q,q')$ between consecutive prime powers at which the sawtooth $\alpha$ crosses the level $\frac12\tau_1(\log N)$, a quantity below $0.083$ that tends to $0$. At such a point
\begin{equation}\label{ce:eq:floorbeta}
M(N^*_q)=K_0+2\,\beta(N^*_q)-\tau(L^*_q)-L^*_q\,\tau_1(L^*_q),\qquad L^*_q=\log N^*_q.
\end{equation}
Consequently RH holds if and only if
\begin{equation}\label{ce:eq:betacrit}
\beta(N^*_q)\;\ge\;-\frac{K_0}2+\frac{\tau(L^*_q)+L^*_q\tau_1(L^*_q)}2\qquad\text{at every local minimum }N^*_q,
\end{equation}
where $-K_0/2=8-\pi^2/4-2G=3.7006677$ and the correction is below $1.8\cdot10^{-5}$ for $N^*_q>100$.
\end{corollary}

\begin{proof}
By Proposition~\ref{ce:prop:minima}(3) the local minima are the solutions of $m'(L)=0$ inside the intervals $(\log q,\log q')$, and by \eqref{ce:eq:slope} these are the points where $\alpha=\frac12\tau_1$. On $L\ge\log2$ the function $\alpha-\frac12\tau_1$ is strictly decreasing between prime powers, since its derivative in $L$ is $-e^{L/2}+\sum_{k\ge1}e^{-(2k+\frac12)L}<0$; so there is at most one such point in each interval, and $\alpha$ crosses the level from above. At that point \eqref{ce:eq:value} gives $m=K_0-\tau+2\beta-2L\alpha=K_0-\tau+2\beta-L\tau_1$, which is \eqref{ce:eq:floorbeta}. The equivalence with RH is Corollary~\ref{ce:cor:discrete}. The bound for the correction is the computed maximum of $\frac12(\tau+L\tau_1)$ over the minima with $N^*_q>100$.
\end{proof}

In words: a minimum of the margin lies where the staircase $A$ comes down to its smooth curve $2\sqrt x-\frac{c_0}2$, inside an interval free of prime powers. Its depth is twice the excess of the staircase $B$ over its smooth curve at that point, minus $7.40$. The large term $\alpha\log N$ in \eqref{ce:eq:value} is switched off exactly at the minima, which is why the value of a minimum depends on $\beta$ alone. One function of the primes locates the minima and the other measures them.

\emph{The numbers.} On the $10\,341$ local minima below $10^7$ the identities hold to the precision of the computation: $|\alpha(N^*_q)-\frac12\tau_1|\le5.5\cdot10^{-12}$, and \eqref{ce:eq:floorbeta} holds to $2.2\cdot10^{-10}$. The values of $\beta$ at the minima lie in $[3.72819,3.79680]$, with mean $3.74621$, against the threshold $3.70067$ of \eqref{ce:eq:betacrit} (Figure~\ref{ce:fig:floors}(c)). The least value is at $N^*_q=3099.71$, the least minimum of $M$ below $10^7$, and its excess over the threshold, $0.02752$, is half of $M(N^*_q)=0.05504$. The least value of $M$ below $10^{10}$, $0.042997$, corresponds to $\beta=3.72217$.

\emph{The minima one after another.} Let $N^*_1<N^*_2<\cdots$ be the local minima of $M$ on $[2,\infty)$, and write $L_j=\log N^*_j$, $u_j=\sqrt{N^*_j}$ and $\beta_j=\beta(N^*_j)$.

\begin{proposition}[the recurrence of the minima]\label{ce:prop:betarec}
The first minimum is $N^*_1=2.44777$, with $\beta_1=(\log2)^2/\sqrt2-4u_1(\log u_1-1)=3.796801$ and $M(N^*_1)=0.071795$. For $j\ge1$, $N^*_{j+1}$ is the stationary point $u^2$, $4u=2A_q+c_0-\tau_1(2\log u)$, of the first interval $(q,q')$ after $N^*_j$ that contains it. With $dA$ the measure of mass $\Lambda(n)/\sqrt n$ at each prime power $n$,
\begin{equation}\label{ce:eq:betamoment}
\beta_{j+1}-\beta_j=\int_{N^*_j}^{N^*_{j+1}}\log\frac x{N^*_j}\,\Big(dA(x)-\frac{dx}{\sqrt x}\Big)+\frac{L_j}2\big(\tau_1(L_{j+1})-\tau_1(L_j)\big),
\end{equation}
and the two measures have equal mass on $(N^*_j,N^*_{j+1})$ up to $\frac12(\tau_1(L_{j+1})-\tau_1(L_j))$. RH holds if and only if the running sum $\beta_J$ never falls below the threshold of \eqref{ce:eq:betacrit}.
\end{proposition}

\begin{proof}
The position follows from Proposition~\ref{ce:prop:minima}(3) and Corollary~\ref{ce:cor:beta}. Subtracting $\alpha(N^*_j)=\frac12\tau_1(L_j)$ from $\alpha(N^*_{j+1})=\frac12\tau_1(L_{j+1})$ gives $A(N^*_{j+1})-A(N^*_j)-2(u_{j+1}-u_j)=\frac12(\tau_1(L_{j+1})-\tau_1(L_j))$, which is the statement about the masses. With $S(u)=4u(\log u-1)$ we have $\beta_{j+1}-\beta_j=(B(N^*_{j+1})-B(N^*_j))-(S(u_{j+1})-S(u_j))$. Write $\log n=\log(n/N^*_j)+L_j$ in the first difference and $\int x^{-1/2}\log x\,dx=\int x^{-1/2}\log(x/N^*_j)\,dx+L_j\int x^{-1/2}dx$ in the second. The terms in $L_j$ combine to $L_j$ times the mass difference, which gives \eqref{ce:eq:betamoment}. The last statement is Corollary~\ref{ce:cor:beta}.
\end{proof}

Status: proved, elementary. Between two minima, the primes and the smooth density $x^{-1/2}dx$ contribute the same mass, and $\beta$ changes by the difference of their first moments in $\log(x/N^*_j)$. RH says that, starting from $\beta_1=3.796801$, these mass-balanced moments never lose more than $\beta_1+K_0/2=0.0961$. Run forward with the prime powers alone, the recurrence finds the same $10\,341$ minima below $10^7$, with $\beta$ correct to $1.7\cdot10^{-11}$; $5\,258$ steps go up and $5\,082$ go down. Like Corollary~\ref{ce:cor:discrete}, it is exact but contains no information beyond Landau's form of the criterion \cite{In}.

\begin{figure}[htbp]\centering\includegraphics[width=\textwidth]{fig/CE_fondos.png}
\caption{The minima read from the slope. (a) The sawtooth $\alpha(x)$ on $[2950,3250]$; its downward zeros (red) are the local minima $N^*_q$, among them the least minimum below $10^7$. (b) $\alpha(e^L)$ from the primes and $-2\sum_{\gamma\le\gamma_{1200}}\sin(\gamma L)/\gamma$ from the zeros. (c) $\beta(N^*_q)$ at the $10\,341$ minima below $10^7$; the red line is $M=0$ by \eqref{ce:eq:floorbeta}. (d) Share of the variance of the minima reproduced by the first $K$ zeros.}\label{ce:fig:floors}
\end{figure}

\section{The margin as an energy}\label{ce:sec:energy}
By Proposition~\ref{ce:prop:slope}, $\alpha$ is, up to the factor $-\frac12$ and the term $\tau_1$, the slope of the margin. Under RH, $\alpha$ is almost periodic in $\log N$, and the margin turns out to be the mean square of its increments. This makes both sides of the sandwich of Corollary~\ref{ce:cor:sandwich} visible: $M\ge0$ because it is a mean of squares, and $M\le4(\log\xi)''(\frac12)$ because the variance of $\alpha$ is $(\log\xi)''(\frac12)$. The same identity turns RH into a statement about energy: $\alpha$ has finite mean square in $\log x$.

Write $\langle f\rangle=\lim_{U\to\infty}\frac1U\int_0^Uf(u)\,du$ when the limit exists. $B^2$ is the Besicovitch space of almost periodic functions, the closure of trigonometric polynomials in the seminorm $\langle|f|^2\rangle^{1/2}$. For $f\in B^2$ with Fourier series $\sum_\omega a_\omega e^{i\omega u}$, Parseval's identity $\langle|f|^2\rangle=\sum|a_\omega|^2$ holds \cite{Bes}.

\begin{proposition}[the margin is the variogram of $\alpha$]\label{ce:prop:variogram}
Assume RH. Then $u\mapsto\alpha(e^u)$ belongs to $B^2$, has mean $0$, and has the Fourier series
\[
\alpha(e^u)\sim\sum_{\gamma}\frac{i\,m_\gamma}{\gamma}\,e^{i\gamma u}=-2\sum_{\gamma>0}\frac{m_\gamma}{\gamma}\sin(\gamma u),
\]
where $\gamma$ runs over the distinct ordinates of the zeros (both signs in the first sum) and $m_\gamma$ is the multiplicity. Consequently, for every $L>0$,
\begin{align}
V(L)&:=\bigl\langle|\alpha(e^{u+L})-\alpha(e^u)|^2\bigr\rangle=4\sum_{\gamma>0}\frac{m_\gamma^2\,(1-\cos\gamma L)}{\gamma^2}\ \ge\ M(e^L),\label{ce:eq:variogram}\\
\langle\alpha^2\rangle&=2\sum_{\gamma>0}\frac{m_\gamma^2}{\gamma^2}\ \ge\ (\log\xi)''(\tfrac12)=0.046210\ldots,\label{ce:eq:energy}
\end{align}
with equality in both, for all $L$, if and only if all zeros are simple. In that case $M(e^L)=V(L)$ is a mean of squares, and the sandwich $0\le M\le4(\log\xi)''(\frac12)$ is the elementary inequality
\[
0\le\langle|f(u+L)-f(u)|^2\rangle\le2\langle|f(u+L)|^2\rangle+2\langle|f(u)|^2\rangle=4\langle|f|^2\rangle .
\]
\end{proposition}

\begin{proof}
Let $\phi(u)=e^{-u/2}(\psi(e^u)-e^u)$. Under RH, $\phi\in B^2$ with Fourier series $-\sum_\rho e^{i\gamma_\rho u}/\rho$ \cite{RuSa,ANS}. This is the explicit formula $\psi(x)=x-\sum_\rho x^\rho/\rho-\log2\pi-\frac12\log(1-x^{-2})$, with the series converging in $B^2$ because $\sum_\rho|\rho|^{-2}<\infty$.

Partial summation with $A(x)=\sum_{n<x}\Lambda(n)n^{-1/2}$ gives, for $x=e^u$ not a prime power,
\[
A(x)=\frac{\psi(x)}{\sqrt x}+\frac12\int_1^x\frac{\psi(t)}{t^{3/2}}dt,\qquad \alpha(e^u)=\phi(u)+\frac12\int_0^u\phi(v)\,dv+\frac{c_0}2-1 .
\]
Integrating the explicit formula termwise, which is justified as in \cite[Ch.~IV]{In}, gives
\[
\frac12\int_0^u\phi=-\frac12\sum_\rho\frac{e^{i\gamma_\rho u}-1}{i\gamma_\rho\,\rho}+\kappa+o(1)
\]
for a constant $\kappa$. Here the series converges absolutely and uniformly, since $\sum|\gamma\rho|^{-1}<\infty$. Hence $\alpha(e^u)\in B^2$, and its coefficient at $e^{i\gamma u}$ is
\[
-m_\gamma\Big(\frac1\rho+\frac1{2i\gamma\rho}\Big)=-m_\gamma\,\frac{2i\gamma+1}{2i\gamma\rho}=-\frac{m_\gamma}{i\gamma}=\frac{i\,m_\gamma}{\gamma}.
\]

The mean is $0$. Indeed, for $\Re s>\frac12$, the Laplace transform of $\alpha(e^u)$ is
\[
\mathcal L[\alpha](s)=-\frac{\zeta'}{\zeta}\bigl(s+\tfrac12\bigr)\frac1s-\frac2{s-\frac12}+\frac{c_0}{2s}.
\]
By the Abelian theorem \cite{Wi}, the mean equals $\lim_{\sigma\to0^+}\sigma\mathcal L[\alpha](\sigma)=-\frac{\zeta'}{\zeta}(\frac12)+\frac{c_0}2$, which is $0$ by the identity $c_0=2\frac{\zeta'}{\zeta}(\frac12)$ of Section~\ref{ce:sec:margin}.

Translation and difference preserve $B^2$. So $\alpha(e^{u+L})-\alpha(e^u)$ has coefficients $\frac{im_\gamma}{\gamma}(e^{i\gamma L}-1)$, and Parseval gives \eqref{ce:eq:variogram}, because $|e^{i\gamma L}-1|^2=2(1-\cos\gamma L)$ and each $\gamma>0$ appears with its negative. Parseval for $\alpha$ itself gives \eqref{ce:eq:energy}.

For the comparisons, \eqref{ce:eq:wave} counts zeros with multiplicity, so $M(e^L)=4\sum_{\gamma>0}m_\gamma(1-\cos\gamma L)/\gamma^2$. Similarly $(\log\xi)''(\frac12)=-\sum_\rho(\frac12-\rho)^{-2}=2\sum_{\gamma>0}m_\gamma/\gamma^2$. Since $m^2\ge m$, with equality iff $m=1$, both inequalities and the equality cases follow, because $1-\cos\gamma L$ does not vanish identically in $L$.
\end{proof}

\begin{proposition}[RH as finite energy]\label{ce:prop:energy}
The following are equivalent:
\begin{enumerate}
\item[(i)] the Riemann hypothesis;
\item[(ii)] $\alpha$ has bounded mean energy in $\log x$: $\frac1U\int_0^U\alpha(e^u)^2\,du=O(1)$;
\item[(iii)] $\int_0^U\alpha(e^u)^2\,du=e^{o(U)}$, that is, $\int_1^X\alpha(x)^2\frac{dx}x=X^{o(1)}$.
\end{enumerate}
\end{proposition}

\begin{proof}
(i)$\Rightarrow$(ii) is Proposition~\ref{ce:prop:variogram}. In substance it is Cram\'er's theorem that, under RH, $\int_2^X(\psi(x)-x)^2x^{-2}dx\sim c\log X$ \cite{In}.

(ii)$\Rightarrow$(iii) is trivial.

(iii)$\Rightarrow$(i): let $I(U)=\int_0^U\alpha(e^u)^2du=e^{o(U)}$ and $\sigma>0$. Integration by parts against $dI$ shows that $\int_0^\infty\alpha(e^u)^2e^{-2\sigma'u}du<\infty$ for every $\sigma'>0$. By Cauchy--Schwarz with $0<\sigma'<\sigma$,
\[
\int_0^\infty|\alpha(e^u)|e^{-\sigma u}du\le\Big(\int_0^\infty\alpha^2e^{-2\sigma'u}du\Big)^{1/2}\Big(\int_0^\infty e^{-2(\sigma-\sigma')u}du\Big)^{1/2}<\infty .
\]
So $\mathcal L[\alpha](s)$ converges absolutely in $\Re s>0$ and is holomorphic there. For $\Re s>\frac12$ it equals $-\frac{\zeta'}{\zeta}(s+\frac12)\frac1s-\frac2{s-\frac12}+\frac{c_0}{2s}$, by partial summation from $\sum_n\Lambda(n)n^{-1/2-s}=-\frac{\zeta'}\zeta(s+\frac12)$.

By analytic continuation, $-\frac{\zeta'}\zeta(w)/(w-\frac12)-2/(w-1)$ is holomorphic in $\Re w>\frac12$. The pole of $-\frac{\zeta'}\zeta$ at $w=1$ is cancelled, since its residue $1$ divided by $w-\frac12=\frac12$ gives $\frac2{w-1}$. So $\zeta$ has no zeros in $\Re w>\frac12$.
\end{proof}

\begin{remark}[what Proposition~\ref{ce:prop:energy} changes, and what it does not]\label{ce:rem:energy}
Proposition~\ref{ce:prop:energy} replaces the positivity $M\ge0$ by an energy bound for the slope, $\int_1^X\alpha(x)^2\frac{dx}x=X^{o(1)}$. Positivity is then automatic: under (ii), Proposition~\ref{ce:prop:variogram} applies and $M$ is a mean of squares. The bound in (iii) is weak: it allows sub-exponential growth in $\log X$. Yet it is equivalent to RH.

Unconditionally, $\alpha(x)\ll x^{\Theta-1/2+\varepsilon}$, where $\Theta$ is the supremum of the real parts of the zeros. This gives only $\int_1^X\alpha^2\frac{dx}x\ll X^{2\Theta-1+\varepsilon}$, and the gap between $X^{2\Theta-1}$ and $X^{o(1)}$ is exactly the gap between the known zero-free regions and RH.

By the Paley--Wiener theorem, (ii) says that $\mathcal L[\alpha]$ belongs to the Hardy space $H^2$ of every half-plane $\Re s>\sigma>0$. Its holomorphy there is the content of RH. Its $L^2$ size on vertical lines is a mean-square bound for $\frac{\zeta'}{\zeta}$ to the right of the critical line, which is known under RH \cite{MV}. So an unconditional proof of (iii) would have to control $\frac{\zeta'}\zeta$ in mean square on every line $\Re w=\frac12+\sigma$ and show that it extends holomorphically. No method we know does the first without the second. We do not claim progress on it.
\end{remark}

Numerically, with the prime powers below $10^8$, $M(e^L)$ and the variogram of $\alpha$ agree to three digits at $L=0.2,0.6,\dots,3$, and the empirical variance of $\alpha(e^u)$ is $0.04625$, against $(\log\xi)''(\frac12)=0.04621$.

\subsection{The energy programme: dyadic blocks and averages in \texorpdfstring{$L$}{L}}\label{ce:subsec:programme}
Proposition~\ref{ce:prop:energy} reduces RH to an energy bound for $\alpha$. This subsection follows the two natural ways of attacking that bound: in dyadic blocks, using the recurrence of the minima, and by averaging the margin in $L$. Each route gives exact statements and a clear account of where it stops. Neither route proves RH.

\emph{An exact identity: the energy of the slope is a C-transform of the margin.} Write $m(L)=M(e^L)$, so that $m'=\tau_1-2\alpha$ by Proposition~\ref{ce:prop:slope}.

\begin{theorem}[energy identity]\label{ce:thm:energyidentity}
For $1<X<Y$, neither of them a prime power, and without any hypothesis,
\[
\int_{\log X}^{\log Y}m'(L)^2\,dL=2\sum_{X<q<Y}\frac{\Lambda(q)}{\sqrt q}\,M(q)\;-\;\int_{\log X}^{\log Y}m(L)\,r''(L)\,dL\;+\;\Big[m\,m'\Big]_{\log X}^{\log Y},
\]
where $q$ runs over prime powers and $r''(L)=2e^{L/2}-2\sum_{k\ge1}e^{-(2k+\frac12)L}$.
\end{theorem}

\begin{proof}
Between consecutive prime powers, $m=r-2A_qL+2B_q$ is smooth with $m''=r''$. At a prime power $q$, $m$ is continuous, and $m'=r'-2A$ jumps by $-2\Lambda(q)/\sqrt q$. On each gap, $\int m'^2=[mm']-\int m\,r''$. Summing over the gaps, the interior boundary terms combine into $\sum_q m(\log q)\bigl(m'(\log q^-)-m'(\log q^+)\bigr)=2\sum_q\frac{\Lambda(q)}{\sqrt q}M(q)$.
\end{proof}

The prime measure $\frac{\Lambda(q)}{\sqrt q}\delta_{\log q}$ and its smooth density $e^{L/2}dL=\frac12r''dL+O(e^{-5L/2})$ both look at the margin. Theorem~\ref{ce:thm:energyidentity} says that the energy of the slope is twice the difference between what the primes see and what the continuum sees, which is the C-transform of $M$ with the weight $\Lambda(n)/\sqrt n$.

\begin{corollary}[the rate is the window width]\label{ce:cor:rate}
Assume RH and that the zeros are simple. Then
\[
2\sum_{q<Y}\frac{\Lambda(q)}{\sqrt q}M(q)-\int_0^{\log Y}m\,r''\,dL=4\langle\alpha^2\rangle\log Y+o(\log Y)+O\big(1+|\alpha(Y)|\big),
\]
with $4\langle\alpha^2\rangle=4(\log\xi)''(\tfrac12)=0.18484\ldots$
In particular, the left side divided by $\log Y$ tends to $0.18484$ along every sequence with $\alpha(Y)=o(\log Y)$, and so on a set of $Y$ of logarithmic density $1$. In words: per unit of $\log N$, the primes see the margin more than the continuum does by exactly the width of the window of Corollary~\ref{ce:cor:sandwich}.
\end{corollary}

\begin{proof}
Under RH, $0\le m\le0.18484$ (Corollary~\ref{ce:cor:sandwich}), and $m'=\tau_1-2\alpha$ with $\tau_1(L)=O(e^{-5L/2})$. So the boundary term of Theorem~\ref{ce:thm:energyidentity} is $O(1+|\alpha(Y)|)$, and $\int_0^{\log Y}m'^2=4\int_0^{\log Y}\alpha^2+O(1)=4\langle\alpha^2\rangle\log Y+o(\log Y)$ by Proposition~\ref{ce:prop:variogram}. For simple zeros, \eqref{ce:eq:energy} gives the value. Finally, $\alpha(e^u)$ has a limiting distribution under RH \cite{ANS}, so $|\alpha(Y)|\le K$ on a set of logarithmic density tending to $1$ as $K\to\infty$. (The boundary term cannot be dropped altogether: $\alpha$ is unbounded, by Littlewood's theorem.)
\end{proof}

\begin{theorem}[a lower bound for the energy, without hypothesis]\label{ce:thm:energylower}
Unconditionally,
\[
\limsup_{X\to\infty}\frac1{\log X}\int_1^X\alpha(x)^2\frac{dx}x\ \ge\ (\log\xi)''(\tfrac12)=0.046210\ldots,
\]
and the limit exists and equals $(\log\xi)''(\frac12)$ if and only if RH holds and all zeros are simple.
\end{theorem}

\begin{proof}
If RH fails, Proposition~\ref{ce:prop:energy} shows that $\int_1^X\alpha^2\frac{dx}x$ is not $X^{o(1)}$, so the $\limsup$ is infinite. If RH holds, Proposition~\ref{ce:prop:variogram} gives the limit $\langle\alpha^2\rangle=2\sum_{\gamma>0}m_\gamma^2/\gamma^2$. This is $\ge2\sum_{\gamma>0}m_\gamma/\gamma^2=(\log\xi)''(\frac12)$, with equality iff every $m_\gamma=1$.
\end{proof}

So the energy of the slope is a computable test of ``RH and simple zeros'' together, with an excess $2\sum(m_\gamma^2-m_\gamma)/\gamma^2$ for multiple zeros. The empirical value is $0.04625$ up to $10^8$, and $0.0465$ from the block energies to $10^{10}$ below.

\emph{Dyadic blocks.} Let $E_k=\int_{k\log2}^{(k+1)\log2}\alpha(e^u)^2du$ be the energy in the block $[2^k,2^{k+1})$; by Proposition~\ref{ce:prop:energy}, RH is equivalent to $E_k=2^{o(k)}$. We computed $E_k$ and $W_k=\int m'^2$ for every block below $10^{10}$, exactly between consecutive prime powers, from all $455\,062\,595$ prime powers (Figure~\ref{ce:fig:energy}(a)). The block energies do not grow: the mean of $E_k/\log2$ over $6\le k\le32$ is $0.0465$, against $0.0462$, and $W_k/\log2$ has mean $0.1858$, against the rate $0.1848$ of Corollary~\ref{ce:cor:rate}. Theorem~\ref{ce:thm:energyidentity} holds block by block to $4\cdot10^{-11}$ for $k\ge15$.

Why this does not give a proof: on a block, $\alpha(x)=\alpha(2^k)+D_k(x)$, where $D_k$ depends only on the prime powers of the block. Its energy is the energy of the normalized error $\bigl(\psi(x)-\psi(2^k)-(x-2^k)\bigr)/2^{k/2}$ of the prime number theorem on an interval as long as its position. A bound $2^{o(k)}$ for it is already RH. The recurrence of Proposition~\ref{ce:prop:betarec} moves $\alpha$ exactly from one prime power to the next and reproduces $D_k$, so it gives nothing more. Finally, no computation can see an off-line zero. Such a zero has $\gamma>T_0$, so its term in $\alpha$ is at most $10^{-12}x^{\beta-1/2}$, which reaches the size $0.2$ of $\alpha$ only beyond $x=e^{54}$ even for $\beta=1$.

\begin{figure}[htbp]\centering\includegraphics[width=\textwidth]{fig/CE_energy.png}
\caption{(a) Energy per dyadic block of the slope, $E_k/\log2$, and of the margin's derivative, $W_k/(4\log2)$, from all prime powers below $10^{10}$, against $(\log\xi)''(\frac12)$. (b) The same block energy, below $2^{26}$, for the true weights and for the weights \eqref{ce:eq:planted} with a planted zero: (A) $\delta=0.15$, $\gamma_0=10$; (B) $\delta=0.25$, $\gamma_0=1000$ (Theorem~\ref{ce:thm:nogo}).}\label{ce:fig:energy}
\end{figure}

\emph{Averages in $L$.} Averaging the margin in $L$ is a second-order Riesz mean:
\[
M_2(N):=\int_0^{\log N}M(e^L)\,dL=R_2(N)-\sum_{n<N}\frac{\Lambda(n)}{\sqrt n}\log^2\frac Nn,\qquad R_2(N)=\int_0^{\log N}R(e^L)\,dL .
\]
Under RH and simple zeros, $M_2(N)=4\sum_{\gamma>0}\gamma^{-2}\log N-4\sum_{\gamma>0}\sin(\gamma\log N)/\gamma^3=2\langle\alpha^2\rangle\log N+O(1)$, which is the identity $\langle M\rangle=2\langle\alpha^2\rangle$.

\begin{proposition}[averaging in $L$]\label{ce:prop:average}
The following are equivalent:
\begin{enumerate}
\item[(i)] RH;
\item[(ii)] for every $\varepsilon>0$, $M_2(N)\ge-C_\varepsilon N^\varepsilon$;
\item[(iii)] $|M_2(N)|=N^{o(1)}$.
\end{enumerate}
\end{proposition}

\begin{proof}
(i)$\Rightarrow$(ii),(iii): under RH, $0\le M\le0.18484$ by Corollary~\ref{ce:cor:sandwich}, so $0\le M_2(N)\le0.18484\log N$.

(iii)$\Rightarrow$(i): integrating by parts, $\int_0^\infty M(e^L)e^{-sL}dL=s\int_0^\infty M_2(e^L)e^{-sL}dL$ converges for $\Re s>0$, so the function $\mathcal G$ of the proof of Theorem~\ref{ce:thm:landau} is holomorphic in $\Re s>0$. But each zero $\rho=\frac12+\delta+it$ with $\delta>0$ gives $\mathcal G$ a pole at $s=\delta+it$.

(ii)$\Rightarrow$(i): apply Landau's theorem \cite{Wi} to $M_2(e^L)+C_\varepsilon e^{\varepsilon L}\ge0$. Its Laplace transform is $\mathcal G(s)/s+C_\varepsilon/(s-\varepsilon)$, which is holomorphic at every real point $s>\varepsilon$, because zeros of $\zeta$ are not real and $\zeta(\sigma)\ne0$ for $\frac12<\sigma<1$. So the integral converges in $\Re s>\varepsilon$, and $\mathcal G$ has no pole there.
\end{proof}

Proposition~\ref{ce:prop:average} is weaker than Theorem~\ref{ce:thm:landau} in its hypothesis, since it needs only a one-sided bound on an average and allows $N^\varepsilon$. It is equivalent all the same, and it is the same statement as Proposition~\ref{ce:prop:energy}, because $M_2=2\langle\alpha^2\rangle\log N+O(1)$ under RH. The least eigenvalue of a window (Section~\ref{ce:sec:windows}) is an infimum of such energies over test functions, so it enters through the same door.

\begin{remark}[where both routes stop]\label{ce:rem:programmestop}
Every statement in this subsection that is equivalent to RH is equivalent through a Landau-type argument. The analytic content is always the same: the Laplace transform of $m$, of $\alpha$, or of the kernel's remainder is $-\frac{\zeta'}\zeta$ shifted by $\frac12$, divided by a polynomial in $s$, plus explicit terms. Each bound in the programme (energy, average, one-sided bound) is equivalent to the holomorphy of $\frac{\zeta'}\zeta$ in $\Re w>\frac12$.

The two routes make this concrete. The block computation shows that ``local'' in $\log x$ means a prime number theorem on intervals as long as their position. Averaging shows that averaging changes the order of the Riesz mean but not the abscissa. Theorem~\ref{ce:thm:energyidentity} is the one exact link between the energy and the primes. To be useful it would need an independent lower bound for the continuum term or an upper bound for the prime term, and each of these is again a statement about $\frac{\zeta'}\zeta$ to the right of the line.

We record the programme as a reformulation with exact identities and verified numerics. It does not prove RH.
\end{remark}

\section{The margin at large $N$: a probabilistic projection}\label{ce:sec:projection}
Proposition~\ref{ce:prop:nomargin} says that under RH the margin comes arbitrarily close to $0$. This section asks how fast. The answer is a projection, not a theorem: it assumes RH and the linear independence of the ordinates (LI), as in the work of Rubinstein and Sarnak \cite{RuSa} and of Akbary, Ng and Shahabi \cite{ANS} on limiting distributions.

\emph{The limiting distribution.} Under RH, \eqref{ce:eq:wave} writes $M(e^L)=\sum_{\gamma>0}a_\gamma(1-\cos\gamma L)$ with $a_\gamma=4/\gamma^2$. Under LI, by the Kronecker--Weyl theorem the phases $\gamma L$ modulo $2\pi$ become equidistributed on the infinite torus as $L\to\infty$, so the proportion of $L\in[0,T]$ with $M(e^L)<x$ tends to $\mathbb P(X<x)$, where
\begin{equation}\label{ce:eq:Xlimit}
X=\sum_{\gamma>0}a_\gamma(1-\cos\theta_\gamma),\qquad \mathbb E\,e^{itX}=\prod_{\gamma>0}e^{ita_\gamma}J_0(ta_\gamma),
\end{equation}
with independent $\theta_\gamma$ uniform on $[0,2\pi)$. Its mean is $2(\log\xi)''(\frac12)=0.0924$ and its standard deviation $(8\sum\gamma^{-4})^{1/2}=0.01724$. Fourier inversion of \eqref{ce:eq:Xlimit} with the $4520$ certified zeros (the rest contribute their mean $0.00098$ and a variance below $3\cdot10^{-11}$) gives its density, a smooth bump on $[0.04,0.145]$ with two shallow maxima. For the lower tail we use the saddle point: with $K(s)=\log\mathbb E e^{sX}=\sum_\gamma\bigl(sa_\gamma+\log I_0(sa_\gamma)\bigr)$ and $K'(\hat s)=x$, the Lugannani--Rice formula gives
\[
\begin{gathered}\mathbb P(X<0.04)=3.7\cdot10^{-6},\quad \mathbb P(X<0.03)=1.0\cdot10^{-11},\\ \mathbb P(X<0.02)=4.2\cdot10^{-27},\quad \mathbb P(X<0.01)=1.7\cdot10^{-97},\end{gathered}
\]
in agreement with Fourier inversion to $0.6\%$ where both apply ($x\ge0.03$). Numerically $\log\mathbb P(X<x)\approx-C/x^2$: halving $x$ multiplies the exponent by about $3.7$.

\emph{How often the margin dips.} The rate of downcrossings of a level $x$, per unit of $L=\log N$, is given by Rice's formula $\nu(x)=f_X(x)\,\mathbb E\bigl[|X'|\,\big|\,X=x\bigr]/2$, with $X'=\sum_\gamma\gamma a_\gamma\sin\theta_\gamma$. Conditioning on $X=x$ by exponential tilting turns each $\theta_\gamma$ into a von Mises variable of concentration $\kappa_\gamma=|\hat s|a_\gamma$, for which $\mathbb E\sin^2\theta=I_1(\kappa)/(\kappa I_0(\kappa))$; taking $X'$ conditionally Gaussian,
\begin{equation}\label{ce:eq:rice}
\nu(x)=\frac{f_X(x)}{\sqrt{2\pi}}\Bigl(\sum_{\gamma>0}\frac{16}{\gamma^2}\frac{I_1(\kappa_\gamma)}{\kappa_\gamma I_0(\kappa_\gamma)}\Bigr)^{1/2},\qquad f_X(x)\approx\frac{e^{K(\hat s)-\hat sx}}{\sqrt{2\pi K''(\hat s)}} .
\end{equation}
We tested the model against the function itself, computed from the zeros on $7\le L\le3000$: for $x=0.060,0.055,0.050,0.045$ the observed proportion of time below $x$ is $2.12\cdot10^{-2}$, $6.1\cdot10^{-3}$, $1.17\cdot10^{-3}$, $1.2\cdot10^{-4}$ against $2.14\cdot10^{-2}$, $6.3\cdot10^{-3}$, $1.16\cdot10^{-3}$, $1.1\cdot10^{-4}$, and the observed number of excursions per unit of $L$ on $[1000,3000]$ is $0.686$, $0.259$, $0.068$, $0.0085$ against $0.683$, $0.262$, $0.065$, $0.0083$ from \eqref{ce:eq:rice}.

\emph{The record low.} If dips below $x$ arrive as a Poisson process of rate $\nu(x)$, then $\mathbb P\bigl(\min_{N'\le N}M(N')\ge x\bigr)\approx e^{-\nu(x)\log N}$. The median value of $\log N$ at which $M$ first drops below $x$ is $\log2/\nu(x)$:

\begin{center}\footnotesize\setlength{\tabcolsep}{4pt}
\begin{tabular}{lccccccc}
\toprule
$x$ & $0.040$ & $0.035$ & $0.030$ & $0.025$ & $0.020$ & $0.015$ & $0.010$\\
\midrule
$1/x$ & $25$ & $28.6$ & $33.3$ & $40$ & $50$ & $66.7$ & $100$\\
median $\log_{10}N$ & $7.5\cdot10^2$ & $7.9\cdot10^4$ & $1.3\cdot10^8$ & $2.7\cdot10^{13}$ & $1.2\cdot10^{23}$ & $5.8\cdot10^{42}$ & $6.4\cdot10^{92}$\\
\bottomrule
\end{tabular}
\end{center}
Since $\log\mathbb P(X<x)\approx-C/x^2$, the record low decreases like $(\log\log N)^{-1/2}$. Fitting the median gives
\begin{equation}\label{ce:eq:recordlaw}
\max_{10^3\le N'\le N}\frac1{M(N')}\approx10.4\sqrt{\log_{10}\log N}\qquad(N\to\infty),
\end{equation}
which reproduces the median to within $3\%$ for $\log_{10}\log N\ge13$; for smaller $N$ the median lies above the law (Figure~\ref{ce:fig:inverse}(b)).

\emph{The actual record, under RH only.} Under RH alone, $M(e^L)=W(L)+T(L)$, where $W$ is the sum \eqref{ce:eq:wave} over the $4520$ certified zeros and $0\le T\le2\cdot0.00098$, $T$ having mean $0.00098$. We followed the record low of $W+0.00098$ for $7\le L\le10^5$: a coarse pass with the first $100$ zeros, whose partial sum is a pointwise lower bound for $W$, selects the candidates, which are then refined with all $4520$ zeros. The records are
\[
\begin{gathered}0.0430\ (N=3.4\cdot10^7),\quad 0.04269\ (1.06\cdot10^{11}),\quad 0.0399\ (2.2\cdot10^{29}),\\ 0.0385\ (10^{1521}),\quad 0.0369\ (10^{8274}),\quad 0.0362\ (10^{24461}),\end{gathered}
\]
the first two agreeing with the primes (Proposition~\ref{ce:prop:verified}) to $10^{-5}$. The last three fall inside the $10$--$90\%$ band of the model; the first three come early (each had probability $2$--$6\%$ of arriving so soon), so the region $N<10^{30}$ is somewhat atypically low.

\begin{figure}[htbp]\centering\includegraphics[width=\textwidth]{fig/CE_inverse.png}
\caption{The reciprocal $1/M$. (a) $1/M(N)$ for $10^3<N<10^{130}$, from the certified zeros under RH: it stays between $7$ and $15$ most of the time and spikes when many waves are in phase. (b) The record of $1/M$ against $\log_{10}\log N$: actual records for $\log N\le10^5$ (red), the median record and $10$--$90\%$ band of the model \eqref{ce:eq:rice} under RH and LI (blue), and the law \eqref{ce:eq:recordlaw} (dashed).}
\label{ce:fig:inverse}
\end{figure}

\begin{remark}[status]\label{ce:rem:projection}
Formula \eqref{ce:eq:Xlimit} for the limiting distribution is a theorem under RH and LI; the saddle-point values are numerical approximations checked against Fourier inversion; the rate \eqref{ce:eq:rice}, the Poisson approximation and the law \eqref{ce:eq:recordlaw} are heuristics, checked against the function itself where it can be computed. Nothing here bears on the truth of RH: the projection describes how the margin behaves if RH holds. Read through the reciprocal, it says that $1/M$ is positive and unbounded, and that its spikes grow like $\sqrt{\log\log N}$, so that $1/M$ reaches $100$ only for $N$ with about $10^{93}$ digits. If RH failed, $1/M$ would eventually change sign.
\end{remark}

\section{The walk of the primes against the curve}\label{ce:sec:walk}
Proposition~\ref{ce:prop:minima} can be read as a geometric reformulation of the whole problem. Write $P_q=(A_q,B_q)$ for the pair of prime sums at the prime power $q$, and
\begin{equation}\label{ce:eq:Dq}
D_q=2B_q+K_0-\Phi(u_q),\qquad \Phi(u)=8u(\log u-1),\qquad u_q=\frac{2A_q+c_0}4=\sqrt q+\frac{\alpha(q)}2,
\end{equation}
the last equality because $A_q=2\sqrt q+\alpha(q)-\frac{c_0}2$. By \eqref{ce:eq:minval}, $D_q$ is, up to $|\epsilon_q|\le3\cdot10^{-6}$, the value of the local minimum of $M$ attached to the block $[q,q')$ when that minimum is interior. Every local minimum of $M$ on $N\ge2$ is of this form (Proposition~\ref{ce:prop:minima}(3)), and a sign change of the continuous function $M$, positive near $N=1$, would happen at a negative local minimum. So RH is the statement that \emph{the walk $q\mapsto P_q$, pushed forward one prime power at a time, never crosses the fixed curve} $2B+K_0=\Phi\bigl(\frac{2A+c_0}4\bigr)$ beyond some point; $D_q$ is the height of the walk above the curve (Figure~\ref{ce:fig:walk}).

\subsection{The exact step}
\begin{proposition}[the step of the walk]\label{ce:prop:step}
Let $q<q'$ be consecutive prime powers, $\lambda=\Lambda(q')q'^{-1/2}$. Then, for some $\xi\in(u_q,u_{q'})$,
\begin{equation}\label{ce:eq:step}
D_{q'}-D_q=2\lambda\Bigl(\log\frac{q'}q-2\log\Bigl(1+\frac{\alpha(q)}{2\sqrt q}\Bigr)\Bigr)-\frac{\lambda^2}\xi .
\end{equation}
In particular the step is $2\lambda\,\log(q'/N^*_q)+O(\lambda^2/\sqrt q)$: the prime power $q'$ \emph{lowers} the floor when it arrives before the pending minimum $N^*_q$, and \emph{raises} it when it arrives after; and no single prime power moves the floor appreciably: for $q\ge10^6$,
\[
|D_{q'}-D_q|\le\frac{2\log q'}{\sqrt{q'}}\Bigl(\log\frac{q'}q+\frac{|\alpha(q)|+1}{\sqrt q}\Bigr)\le\frac{2\log q\,(|\alpha(q)|+2)}{q}.
\]
\end{proposition}

\begin{proof}
$A_{q'}=A_q+\lambda$, $B_{q'}=B_q+\lambda\log q'$, $u_{q'}=u_q+\frac\lambda2$. So $D_{q'}-D_q=2\lambda\log q'-\bigl(\Phi(u_q+\frac\lambda2)-\Phi(u_q)\bigr)$, and by Taylor with $\Phi'(u)=8\log u$, $\Phi''(u)=8/u$, the bracket is $4\lambda\log u_q+\lambda^2/\xi$. Now $2\log u_q=\log q+2\log(1+\frac{\alpha(q)}{2\sqrt q})$, which gives \eqref{ce:eq:step}; and $2\log u_q=\log N^*_q+O(u_q^{-5})$ by Proposition~\ref{ce:prop:minima}(3). The bound follows from $\Lambda(q')\le\log q'$ and the Bertrand bound $q'<2q$, and for $q\ge10^6$ from the known prime gaps ($q'-q<1500$ below $4\cdot10^{18}$).
\end{proof}

Identity \eqref{ce:eq:step} was checked against the exact $D_q$ over $10^5$-long stretches at $q\sim10^6$, agreeing to $1.9\cdot10^{-11}$; the largest single step there is $1.6\cdot10^{-5}$, falling to $9\cdot10^{-8}$ at $q\sim10^8$ and $3.3\cdot10^{-9}$ at $q\sim10^9$. The two terms of the step are the two clocks of the problem: the gap $\log(q'/q)$, what the primes do, against $\alpha(q)/\sqrt q$, where the running account stands. Their accumulated difference is $-2\int\alpha\,d\log x$ (Section~\ref{ce:sec:minima}), so no individual prime matters and only the \emph{area} under the slope does.

\subsection{What it takes to cross}
\begin{proposition}[local crossing bound]\label{ce:prop:block}
Let $x<y$ and suppose $M$ changes sign in $[x,y]$, with $M(x)>0$. Then
\[
\max_{t\in[x,y]}\alpha(t)\ \ge\ \frac{M(x)}{2\log(y/x)},
\]
and the maximum is attained at a prime power of $[x,y]$. Moreover $\alpha(t)-\alpha(x)$ depends only on the prime powers in $(x,t]$: $\alpha(t)-\alpha(x)=\sum_{x<n\le t}\Lambda(n)n^{-1/2}-2(\sqrt t-\sqrt x)$.
\end{proposition}

\begin{proof}
With $m(L)=M(e^L)$, $m'=-2\alpha(e^L)+\tau_1(L)$ and $\tau_1>0$ (Section~\ref{ce:sec:minima}), so for $L\in[\log x,\log y]$, $m(L)\ge m(\log x)-2(L-\log x)\max_{[x,y]}\alpha\ge M(x)-2\log(y/x)\max\alpha$. If $M$ changes sign the left side is $<0$ somewhere. Between prime powers $\alpha(t)=A-2\sqrt t+\frac{c_0}2$ decreases, and it jumps up at each prime power, so its maximum on a closed block sits at a prime power.
\end{proof}

\begin{table}[ht]
\centering
\caption{What a first sign change in a dyadic block $[x,2x]$ would require, from Proposition~\ref{ce:prop:block} with the verified lower bounds for $M(x)$: a surplus $\alpha\ge M(x)/(2\log2)$, i.e.\ about $\alpha\sqrt x/\log x$ prime powers above the smooth count, present at some prime power of the block.}
\label{ce:tab:crossing}
\begin{tabular}{lcccc}
\toprule
$x$ & $10^{12}$ & $10^{20}$ & $10^{27}$ & $10^{29}$\\
\midrule
$M(x)\ge$ & $0.0427$ & $0.0481$ & $0.0481$ & $0.0342$\\
needs $\alpha\ge$ & $0.031$ & $0.035$ & $0.035$ & $0.025$\\
extra prime powers & $1.1\cdot10^{3}$ & $7.5\cdot10^{6}$ & $1.8\cdot10^{10}$ & $1.2\cdot10^{11}$\\
\bottomrule
\end{tabular}
\end{table}

Status: both propositions are proved, and they answer a natural question -- can one added prime, or one added zero, flip the sign of $M$? -- precisely. A zero on the line only adds the nonnegative window $4(1-\cos\gamma L)/\gamma^2$; an off-line zero flips the sign for large $N$, but that is RH itself; a single prime power beyond $10^6$ moves the floor by less than $3\cdot10^{-5}$. What Proposition~\ref{ce:prop:block} does \emph{not} give is RH: the required surplus, about $0.15$ standard deviations of $\alpha$ (whose mean square is $(\log\xi)''(\frac12)$ by Theorem~\ref{ce:thm:energylower}), occurs in a positive proportion of blocks. The content is the exact bookkeeping: a sign change cannot start in a block where the primes run behind their smooth count, and the surplus must be present at a prime power, locally auditable through the second identity of Proposition~\ref{ce:prop:block}. The obstruction to converting this into a proof is the one of Theorem~\ref{ce:thm:nogo}: the bound uses $|\alpha|$, not the identity $\Lambda=\mu*\log$.

\subsection{Anatomy of the close approaches}
The walk has come closest to the curve at the record minima of Section~\ref{ce:sec:projection}. Two complementary mechanisms show up, one in the primes and one in the zeros.

\emph{In the primes.} At the record $N=34\,186\,433$ ($M=0.042997$), the surplus was assembled one octave earlier: on $(8.5\cdot10^6,\,1.7\cdot10^7]$ the weighted count of prime powers exceeds $2\sqrt x$ by $+0.223$ in the slope -- about $49$ extra prime powers -- while the neighbouring octaves run near balance ($+0.017$) or behind ($-0.193$, $-0.203$). The final descent is short: $M$ crosses its mean $0.0924$ for the last time at $N=3.09\cdot10^7$ and reaches the record a factor $1.106$ later, paying $2\int\alpha\,d\log x=0.0504$ with $\alpha$ averaging $+0.25$ on the stretch; the bottom is reached exactly when the surplus is spent, $\alpha(N_1)=0.0149$ (the minimum is the zero of $\alpha$, Proposition~\ref{ce:prop:minima}). Figure~\ref{ce:fig:anatomy}(a) shows $\alpha$ from the primes against $\alpha=-2\sum_\gamma\sin(\gamma\log x)/\gamma$ from the $4520$ certified zeros: the two are indistinguishable at this scale. At the next record, $N=1.056\cdot10^{11}$, the primes (Proposition~\ref{ce:prop:verified}(2)) give $0.042686$ and the zeros give $0.042689$: the same event seen from both sides, to $3\cdot10^{-6}$.

\emph{In the zeros.} At every record the low-lying zeros conspire: of the first $20$ phases $\cos(\gamma_k\log N)$, between $13$ and $16$ exceed $\frac12$ simultaneously (Figure~\ref{ce:fig:anatomy}(b)), the first $10$ zeros carry $74$--$92\%$ of the depth $0.0924-M$, and the first $100$ carry $99\%$. Independent phases would align like this with probability of order $3^{-14}\approx10^{-7}$ per sample, which is the rarity that the rates of Section~\ref{ce:sec:projection} quantify; the next deeper conspiracies are predicted near $N\sim10^{10^5}$.

\begin{figure}[htbp]\centering\includegraphics[width=\textwidth]{fig/CE_walk.png}
\caption{The walk against the curve. (a) The height $D_q$ of \eqref{ce:eq:Dq} over the first prime powers: each prime power moves the point, down when it arrives before the pending minimum $N^*_q$, up when after \eqref{ce:eq:step}. (b) $D_q$ for all prime powers below $10^{10}$; the closest approach $0.042997$ at $q=34\,186\,367$ is the global minimum of $M$ beyond $N=10^3$.}
\label{ce:fig:walk}
\end{figure}

\begin{figure}[htbp]\centering\includegraphics[width=\textwidth]{fig/CE_anatomy.png}
\caption{Anatomy of the record minima. (a) The slope $\alpha$ around the record at $N=3.42\cdot10^7$, from the primes and from the zeros: a surplus of about $49$ prime powers accumulates on $(8.5\cdot10^6,1.7\cdot10^7]$, the margin pays for it on the shaded stretch, and the record sits where $\alpha$ returns to $0$. (b) The phases $\cos(\gamma_k\log N)$ of the first $20$ zeros at three records: the dips are conspiracies of the low zeros.}
\label{ce:fig:anatomy}
\end{figure}

\section{The margin as a one-sided mean of the prime-counting error}\label{ce:sec:mean}
The attempt to push the verification beyond the computed range with explicit prime-counting bounds fails, and it fails for a reason worth recording exactly. Let
\[
K_N(t)=t^{-3/2}\Bigl(1+\tfrac12\log\tfrac Nt\Bigr),\qquad
E^*(t)=\psi(t)-t+\log2\pi+\tfrac12\log(1-t^{-2}),\qquad
\widehat B=1+\frac\gamma2-\frac12\log4\pi ,
\]
so that $E^*$ is the remainder of $\psi$ against the full smooth model of the explicit formula, $\widehat B=\sum_\rho\rho^{-1}=0.0230957\ldots$ is the Hadamard constant of $\xi$, and $\int_1^NK_N(t)\,dt=\log N$ exactly, since $-t^{-1/2}\log(N/t)$ is an antiderivative of $K_N$.

\begin{proposition}[the margin as a weighted mean]\label{ce:prop:mean}
For every $N>1$, exactly and unconditionally,
\begin{equation}\label{ce:eq:mean}
M(N)=-2\int_1^NE^*(t)\,K_N(t)\,dt-2\widehat B\log N .
\end{equation}
\end{proposition}

\begin{proof}
All three steps are elementary. \emph{First}, $C(N)=2\int_1^N\psi(t)K_N(t)\,dt$: by the antiderivative above, $2\int_q^NK_N=2q^{-1/2}\log(N/q)$, so the right side is $\sum_{q<N}2\Lambda(q)q^{-1/2}\log(N/q)=C(N)$, with no boundary terms. \emph{Second}, with $S(t)=t-\log2\pi-\frac12\log(1-t^{-2})$, termwise integration with $\int_1^Nt^{a-1}(1+\frac12\log(N/t))\,dt=\frac{N^a-1}a+\frac{N^a-1-a\log N}{2a^2}$, $a=-(2k+\frac12)$ for the $k$-th term of $-\frac12\log(1-t^{-2})=\frac12\sum_kt^{-2k}/k$, gives
\[
2\int_1^NS(t)K_N(t)\,dt=8\sqrt N+\Bigl(2-5\log2-2\log\pi-\frac\pi2\Bigr)\log N+K_0-\tau(\log N):
\]
the $N^{-2k-\frac12}$ coefficients are $\frac1k\bigl(\frac1a+\frac1{2a^2}\bigr)=-\frac2{(2k+\frac12)^2}$, which is $-\tau$ term by term; the $\log N$ coefficient uses $\sum_{k\ge1}\frac1{k(4k+1)}=\psi(\tfrac54)+\gamma=4-3\log2-\frac\pi2$; and the constant uses $\sum_{k\ge1}\frac1{k(4k+1)^2}=8-3\log2-\frac\pi2-\frac{\pi^2}4-2G$, which follows from the trigamma value $\psi_1(\frac14)=\pi^2+8G$, and comes out as exactly $K_0=-16+\frac{\pi^2}2+4G$. \emph{Third}, comparing with \eqref{ce:eq:closed},
\[
R(N)-2\int_1^NS(t)K_N(t)\,dt=\bigl(-c_0-2+5\log2+2\log\pi+\tfrac\pi2\bigr)\log N=(\log4\pi-\gamma-2)\log N=-2\widehat B\log N .
\]
Since $\psi=S+E^*$, $M=R-C=-2\int E^*K_N-2\widehat B\log N$.
\end{proof}

Identity \eqref{ce:eq:mean} was checked to machine precision against \eqref{ce:eq:closed} for $1.5\le N\le10^{15}$ (\texttt{ident.py}). The proof never mentions the zeros; the Hadamard constant $\widehat B=\sum_\rho\rho^{-1}$ materializes on the prime side out of the digamma values at $\frac14$.

\begin{corollary}[RH as one inequality for one mean]\label{ce:cor:onesided}
Let $\mu_K(N)=\frac1{\log N}\int_1^NE^*(t)K_N(t)\,dt$, the $K_N$-weighted mean of $E^*$, so that $M(N)=-2\log N\,(\mu_K(N)+\widehat B)$. Then RH holds if and only if
\[
\mu_K(N)\le-\widehat B\qquad\text{for all sufficiently large }N,
\]
and in that case $-\widehat B-\frac{0.18484}{2\log N}\le\mu_K(N)\le-\widehat B$ for $N>1$, so $\mu_K(N)\to-\widehat B$ \emph{one-sidedly}. Unconditionally, $\mu_K(N)\le-\widehat B+E(N)/(2\log N)$ by Theorem~\ref{ce:thm:explicit}, and the verification gives $\mu_K(N)\le-\widehat B-0.0171/\log N$ for $10^3\le N\le2.15\cdot10^{29}$ (Figure~\ref{ce:fig:mean}).
\end{corollary}

Three remarks on what \eqref{ce:eq:mean} says, and what it cannot do.

\emph{Where the weight sits.} The measure $K_N(t)\,dt$ has total mass $\log N$, of which the interval $[1,2]$ -- before the first prime -- carries the share $1-2^{-1/2}+O(1/\log N)$, about $30\%$, and $[1,4]$ about $51\%$; the oscillation at height $t$ enters with density $\sim t^{-1/2}$ per unit. The mean is dominated by the first integers.

\emph{The head start and the payback.} On $[1,2]$, $\psi=0$ and the integrand is forced: that interval alone contributes $d\log N+0.115$ to $M(N)$, with $d=\int_1^2S(t)t^{-3/2}dt=0.06344$. Since the total drift in \eqref{ce:eq:mean} is $-2\widehat B\log N=-0.04619\log N$, the primes from $2$ on must pay back at the exact asymptotic rate $d-2\widehat B=0.01725$ per unit of $\log N$ -- they must run, in this weighted sense, slightly \emph{ahead} of the smooth model forever -- and RH is the statement that they never overpay: the accumulated account $\mu_K(N)+\widehat B$ reaches $0$ from below at most.

\emph{Why explicit prime bounds are powerless here.} The sharpest explicit estimates of the $\psi$-remainder, such as $|\psi(t)-t|\le0.94\sqrt t$ for $11<t\le10^{19}$ \cite{Bu}, give only $|\mu_K(N)|\le0.94\,(1+\frac14\log N)+O(1)$, which does not even keep $\mu_K$ bounded; the criterion needs the \emph{signed} mean accurate to the absolute constant $\widehat B=0.0231$. At $N=10^{12}$ the gap is a factor of about $300$. This is the quantitative form of the obstruction of Section~\ref{ce:sec:limits}: positivity of $M$ is a statement about cancellation in $E^*$ at the scale $\sqrt t$, which no absolute-value estimate can see, and Theorem~\ref{ce:thm:nogo} forbids any route that uses only $|E^*|$-information. We record \eqref{ce:eq:mean} because it compresses RH into a single one-sided inequality for a single weighted mean, with an explicit classical constant on the other side.

\begin{figure}[htbp]\centering\includegraphics[width=\textwidth]{fig/CE_mean.png}
\caption{The one-sided convergence that is RH: $\mu_K(N)+\widehat B=-M(N)/(2\log N)$, computed from the primes up to $10^{10}$, against the mean $-0.0924/(2\log N)$ and the extreme $-0.1848/(2\log N)$ allowed under RH. RH says the blue curve approaches the axis from below and never touches it.}
\label{ce:fig:mean}
\end{figure}