==== TABLE OF CONTENTS ====
The-mean-analytic-rank-of-quadratic-twists-of-elliptic-curves//main.tex
The-mean-analytic-rank-of-quadratic-twists-of-elliptic-curves//sections/algebraic.tex
The-mean-analytic-rank-of-quadratic-twists-of-elliptic-curves//sections/analytic.tex
The-mean-analytic-rank-of-quadratic-twists-of-elliptic-curves//sections/introduction.tex
The-mean-analytic-rank-of-quadratic-twists-of-elliptic-curves//sections/model.tex
The-mean-analytic-rank-of-quadratic-twists-of-elliptic-curves//sections/setup.tex
The-mean-analytic-rank-of-quadratic-twists-of-elliptic-curves//sections/tail.tex
The-mean-analytic-rank-of-quadratic-twists-of-elliptic-curves//sections/weights.tex
The-mean-analytic-rank-of-quadratic-twists-of-elliptic-curves//references.bib
===========================

==== BEGIN The-mean-analytic-rank-of-quadratic-twists-of-elliptic-curves//main.tex ====
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\hypersetup{colorlinks=true,allcolors=linkblue,
  pdftitle={The mean analytic rank of quadratic twists of elliptic curves},
  pdfauthor={OpenAI}}
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\title{The mean analytic rank of quadratic twists of elliptic curves}
\author{OpenAI}
\date{September 23, 2026}

\begin{document}
\maketitle
\begin{abstract}
For every elliptic curve over \(\Q\), we prove that the average analytic
rank of its quadratic twists tends to \(1/2\) when signed squarefree twist
parameters are ordered by absolute value. This resolves Goldfeld's mean
analytic-rank conjecture in this counting convention.
\end{abstract}

\input{sections/introduction}
\input{sections/setup}
\input{sections/analytic}
\input{sections/weights}
\input{sections/model}
\input{sections/tail}
\input{sections/algebraic}

\bibliographystyle{plainnat}
\bibliography{references}
\end{document}
==== END The-mean-analytic-rank-of-quadratic-twists-of-elliptic-curves//main.tex ====

==== BEGIN The-mean-analytic-rank-of-quadratic-twists-of-elliptic-curves//sections/algebraic.tex ====
\section{Algebraic-rank moments}\label{sec:algebraic}

We now prove Corollary~\ref{cor:algebraic-moments}.
The inputs are the companion density and rank-equality theorem and
the exponential-moment estimate of Koymans and Smith.
The latter bounds the contribution of the density-zero exceptional set
for every fixed exponential parameter or moment order; the analytic
tail theorem is not needed.

\begin{proof}[Proof of Corollary~\ref{cor:algebraic-moments}]
Put \(N(Y)=\#\mathcal D(Y)\), \(r_d=r(E^{(d)})\), and let
\(N_j^{\mathrm{alg}}(Y)\) count the parameters with \(r_d=j\).
The rank-equality conclusion of the companion density theorem
\cite[Theorem~1.2]{OpenAIAnalyticDensityV2}, together with its analytic
rank-zero and rank-one densities, gives
\[
 \frac{N_j^{\mathrm{alg}}(Y)}{N(Y)}\longrightarrow\frac12
 \qquad(j=0,1).
\]
Indeed, analytic and algebraic ranks agree outside a density-zero set,
so their rank-\(j\) counts differ by \(o(N(Y))\). Consequently
\[
 \mathcal B(Y)=\{d\in\mathcal D(Y):r_d\ge2\}
 \quad\text{satisfies}\quad
 \#\mathcal B(Y)=o(N(Y)).
\]

For the next bound, \(E^{(v)}\) for any nonzero integer \(v\) denotes
the twist determined by its squareclass, so the integer sum may repeat
squareclasses. Theorem~1.4 of Koymans and Smith
\cite{KoymansSmith2026}, specialized to \(F=\Q\), \(A=E\),
one variable, and \(P(u)=u\), gives a constant \(C_E>0\) such that
for every fixed \(\kappa>0\),
\[
 \frac1{2Y}\sum_{\substack{v\in\Z\\0<|v|\le Y}}
 \exp\!\bigl(\kappa\,\operatorname{rank}_{\Z}E^{(v)}(\Q)\bigr)
 \le \exp\!\bigl(\exp(C_E\kappa)\bigr)
\]
for all sufficiently large \(Y\), with the threshold allowed to depend
on \(\kappa\). Their integer box already includes both signs, and
\(P(v)\ne0\) excludes exactly \(v=0\). Restricting this nonnegative
sum to \(\mathcal D(Y)\), and using
\(N(Y)\sim2Y/\zeta(2)\), gives for each fixed \(\kappa>0\)
\begin{equation}\label{eq:algebraic-exp-bound}
 \limsup_{Y\to\infty}\frac1{N(Y)}
 \sum_{d\in\mathcal D(Y)}e^{\kappa r_d}
 \le \zeta(2)\exp\!\bigl(\exp(C_E\kappa)\bigr)<\infty.
\end{equation}

For fixed \(t>0\), Cauchy--Schwarz and
\eqref{eq:algebraic-exp-bound} at \(\kappa=2t\) give
\[
 \frac1{N(Y)}\sum_{d\in\mathcal B(Y)}e^{t r_d}
 \le
 \left(\frac{\#\mathcal B(Y)}{N(Y)}\right)^{1/2}
 \left(\frac1{N(Y)}
       \sum_{d\in\mathcal D(Y)}e^{2t r_d}\right)^{1/2}
 \longrightarrow0.
\]
For fixed \(t<0\), the same exceptional average is at most
\(\#\mathcal B(Y)/N(Y)\), since \(r_d\ge0\); for \(t=0\) the full
average is identically one. The contributions from ranks zero and one
are \(N_0^{\mathrm{alg}}(Y)/N(Y)\) and
\(e^tN_1^{\mathrm{alg}}(Y)/N(Y)\), proving the exponential limit.

Finally, for each fixed positive integer \(m\), one has
\(x^{2m}\ll_m e^x\) for \(x\ge0\). A second Cauchy--Schwarz application
and \eqref{eq:algebraic-exp-bound} at \(\kappa=1\) therefore give
\[
 \frac1{N(Y)}\sum_{d\in\mathcal B(Y)}r_d^m
 \le
 \left(\frac{\#\mathcal B(Y)}{N(Y)}\right)^{1/2}
 \left(\frac1{N(Y)}\sum_{d\in\mathcal D(Y)}r_d^{2m}\right)^{1/2}
 \longrightarrow0.
\]
The rank-zero contribution is zero and the rank-one contribution is
\(N_1^{\mathrm{alg}}(Y)/N(Y)\), proving the power-moment limit without
differentiating a limiting exponential-moment formula.
\end{proof}
==== END The-mean-analytic-rank-of-quadratic-twists-of-elliptic-curves//sections/algebraic.tex ====

==== BEGIN The-mean-analytic-rank-of-quadratic-twists-of-elliptic-curves//sections/analytic.tex ====
\section{Poisson summation and logarithmic second moments}
\label{analytic:section}
\label{sec:analytic}

We retain the finite family $\mathcal F$, the fixed set $S$, and the
unitary normalization from Lemmas~\ref{setup:reduction} and
\ref{setup:local-factors}. In particular, $|\lambda_F(n)|\leq\tau(n)$,
all local parameters have modulus at most one, and reduction of a
squarefree twisting parameter to an admissible one does not increase its
absolute value. All constants in this section are uniform in
$F\in\mathcal F$.

Our first aim is to bound the second moments of the twist functions and
their localized Dirichlet polynomials with only logarithmic losses.
For a compactly supported smooth function $G$ on $(0,\infty)$, write
\[
 P_d(N,t;G)=\sum_{n\geq1}
 \lambda_F(n)\chi_d(n)n^{-1/2-it}G(n/N).
\]
All dyadic parameters below belong to $\{1,2,4,\ldots\}$.

\begin{proposition}[Logarithmic second moments]\label{analytic:moment}
There are fixed positive integers $A,B$ such that, for every dyadic
$M\geq1$, every $F\in\mathcal F$, every $t\in\mathbb R$, and
$1/2\leq\sigma\leq2$,
\begin{equation}\label{analytic:L-moment}
 \frac1M\sum_{\substack{d\ \mathrm{admissible}\\M\leq|d|<2M}}
 |L_{F^{(d)}}(\sigma+it)|^2
 \ll_{\mathcal F}\log^B(3M)(1+|t|)^A.
\end{equation}
For arbitrary signed squarefree twisting parameters and $H\geq1$,
\begin{equation}\label{analytic:prefix-moment}
 \sum_{\substack{0<|h|\leq H\\h\ \mathrm{squarefree}}}
 |L_{F^{(h)}}(\sigma+it)|^2
 \ll_{\mathcal F}H\log^B(3H)(1+|t|)^A.
\end{equation}
\end{proposition}

\begin{proposition}[Dirichlet-polynomial second moments]
\label{analytic:polynomial-moment}
With the same fixed integers $A,B$, every fixed smooth $G$ of compact
support in $(0,\infty)$ satisfies
\begin{equation}\label{analytic:polynomial-bound}
 \frac1M\sum_{\substack{d\ \mathrm{admissible}\\M\leq|d|<2M}}
 |P_d(N,t;G)|^2
 \ll_{\mathcal F,G}
 \log^B(3M)(1+|t|)^A(1+N/M)^A
 \qquad(N>0).
\end{equation}
The bound is uniform for fixed uniformly smooth families of weights
with common compact support.
\end{proposition}

We prove these estimates by induction on the twisting height $M$.
Poisson summation turns a polynomial mean square into a sum of products
of twist $L$-functions at dual squarefree parameters. To put those
parameters below the current height, we first enlarge the family by
replacing $d$ with $dp^2$, where $p\asymp P$. The new averaging height is
$U=MP^2$. A weight with compact Fourier support then restricts the dual
parameters to size $O(N^2/(MP^2))$, which is at most $M/2$ for a
sufficiently large $P$. Recovering the original squarefree average costs
$O(P\log(2P))$. The nonzero-frequency estimate gains $P^{-2}$, so after
recovery its recursive coefficient contains $N\log(2P)/(MP)$.
We choose $P$ as a sufficiently large fixed power of
$\log(3M)(1+N/M)(1+|t|)$, with a large fixed leading constant, to make
this coefficient small. The approximate functional equation and a strip
estimate then close the induction for the $L$-functions themselves.
The compact Fourier support construction and prime-square inflation
are related to Li's treatment of quadratic twists \cite{Li2022}.

The arithmetic preparation below keeps its arithmetic exponents independent of the
number of weight derivatives. Once the induction is closed, we sum all
dual frequencies to compare short polynomial averages with the integer
model. The polynomial moment bound will also control the unmollified
terminal series in Section~\ref{sec:weights}.

\subsection{The normalized Gauss sums}

Put $e(x)=\exp(2\pi i x)$. For an odd positive integer $l$, set
\[
 \tau_h(l)=\sum_{a\bmod l}\left(\frac a l\right)e(ah/l),\qquad
 \epsilon_l=\begin{cases}1&l\equiv1\pmod4,\\i&l\equiv3\pmod4,\end{cases}
 \qquad B_h(l)=\frac{\tau_h(l)}{\epsilon_l\sqrt l}.
\]
We use $B_h(1)=1$. The square root is positive.
These are the normalized quadratic Gauss sums of
\cite[Section~2.2, Lemma~2.3]{Soundararajan2000}: in that notation,
$B_h(l)=G_h(l)/\sqrt l$. We give the formulas and their proof in the
present normalization.

\begin{lemma}[Exact Gauss-sum formulas]\label{analytic:gauss}
For fixed $h$, the function $l\mapsto B_h(l)$ is multiplicative on odd
positive integers, and
\begin{equation}\label{analytic:phase}
 B_{-h}(l)=\left(\frac{-1}{l}\right)B_h(l),\qquad
 \epsilon_l B_h(l)=\frac{1+i}{2}B_h(l)+\frac{1-i}{2}B_{-h}(l).
\end{equation}
If $h\ne0$, $p$ is odd, $t=v_p(h)$, and $v\geq1$, then
\begin{equation}\label{analytic:prime-powers}
 B_h(p^v)=
 \begin{cases}
 p^{v/2}(1-p^{-1}),&v\text{ even},\ v\leq t,\\
 -p^{v/2-1},&v\text{ even},\ v=t+1,\\
 p^{(v-1)/2}\left(\dfrac{h/p^{v-1}}p\right),
       &v\text{ odd},\ v=t+1,\\
 0,&\text{otherwise}.
 \end{cases}
\end{equation}
Consequently $B_h(p^v)=0$ for $v>t+1$ and
$|B_h(p^v)|\leq p^{v/2}$. If $t=0$, only $v=1$ can survive and
$B_h(p)=(h/p)$. If $t=1$, only $v=2$ can survive and
$B_h(p^2)=-1$.
\end{lemma}

\begin{proof}
For coprime odd $l_1,l_2$, the Chinese remainder theorem, followed by a
change of variable in each Gauss sum, gives
\[
 \tau_h(l_1l_2)=
 \left(\frac{l_1}{l_2}\right)\left(\frac{l_2}{l_1}\right)
 \tau_h(l_1)\tau_h(l_2).
\]
Quadratic reciprocity says that the displayed product of symbols is
$\epsilon_{l_1l_2}/(\epsilon_{l_1}\epsilon_{l_2})$. This proves
multiplicativity. Substituting $a\mapsto-a$ proves the first identity
in \eqref{analytic:phase}; the second follows by considering the two
residue classes of $l$ modulo four.

For even $v$, the Jacobi symbol modulo $p^v$ is the indicator of the
units. Thus
\[
 \tau_h(p^v)=\sum_{a\bmod p^v}e(ah/p^v)
             -\sum_{b\bmod p^{v-1}}e(bh/p^{v-1}),
\]
which equals $p^v-p^{v-1}$ when $p^v\mid h$, equals $-p^{v-1}$ when
$p^{v-1}\Vert h$, and otherwise vanishes. Here $\epsilon_{p^v}=1$.
For odd $v$, write $a=y+pz$ with $y$ modulo $p$ and $z$ modulo
$p^{v-1}$. The $z$-sum vanishes unless $p^{v-1}\mid h$; in that case
\[
 \tau_h(p^v)=p^{v-1}\sum_{y\bmod p}
       \left(\frac yp\right)e\big(y(h/p^{v-1})/p\big).
\]
This is zero if $p^v\mid h$. Otherwise the quadratic Gauss-sum formula
makes it
$p^{v-1}\epsilon_p\sqrt p\,\bigl(\frac{h/p^{v-1}}p\bigr)$.
Since $\epsilon_{p^v}=\epsilon_p$ for odd $v$, division by
$\epsilon_{p^v}p^{v/2}$ gives \eqref{analytic:prime-powers}.
\end{proof}

For later use, fix once and for all the arithmetic exponent
\begin{equation}\label{analytic:arithmetic-exponent}
 C_{\rm ar}=256.
\end{equation}
This exponent will not change when we require more derivatives of a
weight or more decay in an imaginary direction.

\begin{lemma}[Two-variable Euler factorization]\label{analytic:euler}
Let $r$ be positive and odd, and write $h=h_1h_2^2\ne0$, where $h_1$ is
signed squarefree and $h_2\geq1$. For $\nu\in\{1,-1\}$, the series
\[
 \mathcal G_{h,r}^{\nu}(s_1,s_2)=
 \sum_{\substack{n_1,n_2\geq1\\n_1,n_2\ \mathrm{odd}}}
 \frac{\lambda_F(n_1)\lambda_F(n_2)B_{\nu h}(rn_1n_2)}
      {n_1^{s_1}n_2^{s_2}}
\]
initially converges in a right half-plane and continues to
$\Re s_1,\Re s_2>1/2$ as
\begin{equation}\label{analytic:euler-factorization}
 \mathcal G_{h,r}^{\nu}(s_1,s_2)=
 L_{F^{(\nu h_1)}}(s_1)L_{F^{(\nu h_1)}}(s_2)
 \mathcal Q_{h,r}^{\nu}(s_1,s_2).
\end{equation}
For $0<b\leq1/4$ and $\Re s_i\geq1/2+b$,
\begin{equation}\label{analytic:euler-bound}
 |\mathcal Q_{h,r}^{\nu}(s_1,s_2)|
 \ll_{\mathcal F} b^{-C_{\rm ar}}r^{C_{\rm ar}}
                       \tau(h_2)^{C_{\rm ar}}.
\end{equation}
The function $\mathcal Q$ is holomorphic in these open half-planes.
\end{lemma}

\begin{proof}
For an odd prime put $e_p=v_p(r)$ and $t_p=v_p(h)$. Multiplicativity
expresses the series as the product of the local sums
\[
 T_p(s_1,s_2)=\sum_{a,b'\geq0}
 \lambda_F(p^a)\lambda_F(p^{b'})
 B_{\nu h}(p^{e_p+a+b'})p^{-as_1-b's_2}.
\]
The symbol $b'$ here is an integer exponent, distinct from the real
shift $b$. No division by $B_{\nu h}(p^{e_p})$ is made; that quantity
may vanish. Write $D_p^*(s)$ for the inverse local factor of the
elliptic curve $F^{(\nu h_1)}$. The correction factor at $p$ is
$T_p(s_1,s_2)D_p^*(s_1)D_p^*(s_2)$. At two it is just
$D_2^*(s_1)D_2^*(s_2)$, because the summation indices are odd.

There are three types of odd prime. If $p\notin S$ and $p\nmid rh$,
put $\chi=(\nu h/p)=(\nu h_1/p)$ and $x_i=p^{-s_i}$. The local
formulas give exactly
\[
 T_p=1+\lambda_F(p)\chi(x_1+x_2),\qquad
 D_p^*(s_i)=1-\lambda_F(p)\chi x_i+x_i^2.
\]
Multiplication cancels both terms of total degree one. The sum of the
absolute values of all polynomial coefficients before cancellation is
at most $5\cdot4\cdot4=80$. Therefore the correction is
$1+O(80p^{-1-2b})$, uniformly in the imaginary parts.

If $p\notin S$, $p\nmid r$, and $t_p=1$, the twisting curve has a
trivial local factor at $p$. By \eqref{analytic:prime-powers},
\[
 T_p=1-\lambda_F(p^2)(x_1^2+x_2^2)-\lambda_F(p)^2x_1x_2
     =1+O(10p^{-1-2b}).
\]
In particular these primes do not produce a cost for every prime
factor of the squarefree part $h_1$.

The remaining odd primes lie in $S$ or divide $rh_2$. The support in
\eqref{analytic:prime-powers} gives $e_p+a+b'\leq t_p+1$ for every
nonzero summand, except that $B_{\nu h}(1)=1$ is already covered when
$e_p=a=b'=0$. Hence
\[
 |T_p|\leq p^{e_p/2}
       \sum_{a+b'\leq t_p+1}(a+1)(b'+1)
 \leq p^{e_p/2}(t_p+3)^4.
\]
Each inverse local factor has modulus at most
$(1+p^{-1/2})^2$, so the product of the two costs at most $9$.
Writing $u_p=v_p(h_2)$, we have $t_p\leq2u_p+1$.
If $u_p\geq1$, the bound
$9(2u_p+4)^4\leq(u_p+1)^{17}$ absorbs this cost into a fixed divisor
power, with $p^{e_p/2}$ absorbed by $r$.
If $u_p=0$ and $e_p\geq1$, the cost is at most
$9\cdot4^4p^{e_p/2}\leq p^{8e_p}$ for $p\geq3$.
If $u_p=e_p=0$, this is one of the fixed primes of $S$, whose cost
is a fixed constant. The factor at two is bounded by
$(1+2^{-1/2})^4$.

The nonexceptional product is bounded by
\[
 \exp\left(80\sum_p p^{-1-2b}\right)
 \leq\zeta(1+2b)^{80}\ll b^{-80}.
\]
Combining the estimates proves \eqref{analytic:euler-bound} with the
stated $C_{\rm ar}$. The nonexceptional correction product converges
normally on compact subsets of $\Re s_i>1/2$; all exceptional factors
are polynomials. Multiplication by the entire twist $L$-functions
therefore gives the claimed continuation. This argument never divides
by a global $L$-value.
\end{proof}

\subsection{A weighted Poisson estimate}

The Euler factorization identifies the twist functions that occur at
nonzero frequencies. We now turn it into a weighted comparison formula:
the zero frequency gives the integer residue average, and the remaining
frequencies are bounded by those twist functions. This is a quadratic
Poisson method used in \cite[Section~2.4, Lemma~2.6]{Soundararajan2000},
with the weights and normalization specified below.

Fix a compact interval $[a_0,a_1]\subset(0,\infty)$ for the first two
scaled variables. The third variable ranges over $\mathbb R$. For an
integer $D\geq0$, a convenient finite seminorm is
\[
 \|W\|_{D,*}=\max_{\alpha+\beta+\gamma\leq D}
 \sup_{y_1,y_2,x}(1+|x|)^D
 \left|\partial_{y_1}^{\alpha}\partial_{y_2}^{\beta}
                    \partial_x^\gamma W(y_1,y_2,x)\right|.
\]
The functions under consideration vanish outside $[a_0,a_1]^2$ in the
first two variables and have the indicated finite decay and smoothness
in the third. Requiring a larger fixed $D$ below causes no change to
$C_{\rm ar}$.

\begin{lemma}[Poisson estimate]\label{analytic:poisson}
Let $U>0$, let $r$ be odd and positive, and let $N_i$ be bounded below
by a fixed positive constant. Fix $0<b\leq1/4$. In the expression
\begin{equation}\label{analytic:poisson-average}
 \frac1U\sum_{m\in\mathbb Z}
 \sum_{\substack{n_1,n_2\geq1\\n_1,n_2\ \mathrm{odd}}}
 \frac{\lambda_F(n_1)\lambda_F(n_2)}
 {(n_1n_2)^{1/2}n_1^{it_1}n_2^{it_2}}
 \left(\frac m{rn_1n_2}\right)
 W(n_1/N_1,n_2/N_2,m/U),
\end{equation}
Poisson summation gives the zero frequency, namely the complete-residue
average integrated in $x$, together with the nonzero frequencies.
For every fixed $J_1>0$, a sufficiently large fixed seminorm order $D$
bounds the absolute contribution of frequency $h=h_1h_2^2\ne0$ by
\begin{align}
 &C_{J_1,\mathcal F}\|W\|_{D,*}
 (N_1N_2)^{-1/2+b}b^{-C_{\rm ar}}r^{C_{\rm ar}}
 \tau(h_2)^{C_{\rm ar}}
 \left(1+\frac{|h|U}{rN_1N_2}\right)^{-J_1}
 \nonumber\\[-2pt]
 &\hspace{8mm}\cdot
 \sum_{\nu=\pm1}\int_{\mathbb R^2}
 \frac{\prod_{i=1}^2
 |L_{F^{(\nu h_1)}}(1/2+b+i(t_i+v_i))|}
 {(1+|v_1|+|v_2|)^{J_1}}\,dv_1\,dv_2.
 \label{analytic:nonzero-mode}
\end{align}
If the Fourier transform of $W$ in $x$ is supported in a fixed interval
$[-C_W,C_W]$, all modes with
$|h|>C_Wa_1^2rN_1N_2/U$ vanish exactly.
\end{lemma}

\begin{proof}
Our Fourier convention is
$\widehat W(y_1,y_2,\xi)=\int_{\mathbb R}W(y_1,y_2,x)e(-x\xi)\,dx$.
Applying Poisson summation in each residue class modulo $l=rn_1n_2$
gives exactly
\[
 \frac1U\sum_m\left(\frac ml\right)W(y_1,y_2,m/U)
 =\sum_{h\in\mathbb Z}\frac{\tau_h(l)}l
                         \widehat W(y_1,y_2,hU/l).
\]
Thus the normalized frequency factor is
\[
 \frac{1}{\sqrt{n_1n_2}}\frac{\tau_h(rn_1n_2)}{rn_1n_2}
 =\frac{\epsilon_{rn_1n_2}B_h(rn_1n_2)}
       {\sqrt r\,n_1n_2}.
\]
Use \eqref{analytic:phase} to replace the numerator by a fixed linear
combination of $B_h$ and $B_{-h}$.

Put $\rho=hU/(rN_1N_2)$ and
\[
 V_h(y_1,y_2)=\widehat W(y_1,y_2,\rho/(y_1y_2)),\qquad
 \widetilde V_h(z_1,z_2)=
 \int_0^\infty\!\int_0^\infty
 V_h(y_1,y_2)y_1^{z_1}y_2^{z_2}
                    \frac{dy_1dy_2}{y_1y_2}.
\]
On any fixed bounded range of real parts, Fourier integration by parts
in $x$, followed by Mellin integration by parts in $y_1,y_2$, yields
\begin{equation}\label{analytic:mellin-kernel}
 |\widetilde V_h(z_1,z_2)|
 \ll_{J_1}\|W\|_{D,*}(1+|\rho|)^{-J_1}
                 (1+|\Im z_1|+|\Im z_2|)^{-J_1}.
\end{equation}
To see why differentiation causes no arithmetic loss, each derivative
of $\rho/(y_1y_2)$ contributes a bounded multiple of $\rho$ on the
fixed compact support. For any fixed number of such derivatives, take
that many additional integrations by parts in $x$. The powers of
$\rho$ are then absorbed by its Fourier decay. All other factors are
bounded functions of $y_1,y_2$.

Mellin inversion first on right lines expresses the $\nu$ part of the
mode as
\[
 \frac1{\sqrt r(2\pi i)^2}
 \int\!\int \widetilde V_h(z_1,z_2)N_1^{z_1}N_2^{z_2}
 \mathcal G_{h,r}^{\nu}(1+it_1+z_1,1+it_2+z_2)
 \,dz_1\,dz_2.
\]
Move both lines to $\Re z_i=-1/2+b$. Lemma~\ref{analytic:euler}
and the entire continuation of the twist functions show that no pole
is crossed. Polynomial growth in fixed strips, together with a larger
fixed order in \eqref{analytic:mellin-kernel}, makes the horizontal
segments tend to zero. The scale factor is
$(N_1N_2)^{-1/2+b}$. Applying
\eqref{analytic:euler-bound} and \eqref{analytic:mellin-kernel}, and
discarding $r^{-1/2}\leq1$, proves
\eqref{analytic:nonzero-mode}. The final assertion follows directly
from the support of $\widehat W$ before Mellin inversion.
\end{proof}

Whenever unrestricted positive indices occur, write $n=2^a n'$ with
$n'$ odd. For $z_m(2)=c$ the factor taken outside an odd-index sum is
$\lambda_F(2^a)c^a2^{-a(1/2+it)}$. Its absolute values have the
uniformly convergent majorant
\begin{equation}\label{analytic:two-sum}
 \sum_{a\geq0}(a+1)2^{-a/2}<\infty.
\end{equation}
The new length is $N/2^a$; a nonempty localized sum has this length
bounded below by a constant depending only on $[a_0,a_1]$.

\subsection{Closing a logarithmic second-moment induction}

\begin{proof}[Proof of Propositions~\ref{analytic:moment} and~\ref{analytic:polynomial-moment}]
We give a simultaneous induction for all $F\in\mathcal F$.
The induction hypothesis at smaller dyadic heights is
\begin{equation}\label{analytic:induction-hypothesis}
 \frac1{M'}\sum_{\substack{d\ \mathrm{admissible}\\M'\leq|d|<2M'}}
 |L_{F^{(d)}}(\sigma+it)|^2
 \leq K\log^B(3M')(1+|t|)^A \qquad(M'<M).
\end{equation}
The constant $K$, common to the finite family, will be selected last.
Since the reduction in Lemma~\ref{setup:reduction} does not increase
height and has bounded multiplicity, this hypothesis implies
\begin{equation}\label{analytic:inductive-prefix}
 \sum_{\substack{0<|h_1|\leq H\\h_1\ \mathrm{squarefree}}}
 |L_{F^{(\nu h_1)}}(1/2+b+iv)|^2
 \leq C_{\mathcal F}KH\log^B(3M)(1+|v|)^A
 \quad(1\leq H\leq M/2).
\end{equation}
Indeed all admissible images lie in earlier dyadic blocks, whose
lengths have sum at most $2H$. If $H<1$, the sum is empty. In
particular the arguments below at $M=1$ require no induction input.

\smallskip
\noindent\emph{The polynomial average at inflated height.}
First we prove a polynomial estimate with a small coefficient in
front of $K$. Put $L=\log(3M)$, $T=1+|t|$, and let $N>0$.
Choose $P$ sufficiently large, with
\begin{equation}\label{analytic:inflation-condition}
 P\geq C_{\mathcal F,G}(1+N/M).
\end{equation}
Let $\mathcal P$ consist of the primes $p\in[P,2P]$ outside $S$.
For a sufficiently large fixed lower threshold on $P$, the elementary
dyadic prime estimate of Erd\H{o}s \cite[\S6, equation~(10)]{Erdos1932}
gives $|\mathcal P|\gg P/\log(2P)$.

Choose a fixed nonnegative Schwartz function $\Phi$ that is at least
one on $[-8,8]$ and has compactly supported Fourier transform. For
completeness, take a nonzero even real smooth bump $g$ in frequency
space with $\int g>0$. Its inverse Fourier transform $f$ is real,
Schwartz, and nonzero near zero. A sufficiently small dilation makes
$f(\alpha x)$ nonzero throughout $[-8,8]$; a fixed multiple of
$f(\alpha x)^2$ then has the required properties. Its Fourier
transform is a compactly supported convolution.

Put $U=MP^2$. For $c=\pm1$ use the formal character $z_m$ with
$z_m(2)=c$, and let $P_m^c(N',t;G)$ be the corresponding polynomial.
We claim, uniformly for $0<N'\leq N$,
\begin{align}
 \frac1U\sum_{m\in\mathbb Z}\Phi(m/U)|P_m^c(N',t;G)|^2
 &\leq C_G\log^{C_0}(3M+3N)
 \left(1+C_{A,B,\mathcal F,G}K\frac{N}{MP^2}L^BT^A\right),
 \label{analytic:inflated-moment}
\end{align}
where we may fix $C_0=512$, independently of $A,B$ and of the required
smoothness orders.

To prove the claim, apply Lemma~\ref{analytic:poisson} with $r=1$ to
the expansion of the square, first separating powers of two by
\eqref{analytic:two-sum}. Take
$b=1/(10\log(3M+3N))$. For the zero mode the complete-residue mean
vanishes unless the odd part of $n_1n_2$ is a square, and its absolute
value is at most one. Inserting $(n_1n_2)^{-b}$ at the bounded cost
$(C_GN)^{2b}\ll_G1$ majorizes the resulting sum by an Euler product.
At an odd prime this product has factor
\[
 \sum_{\substack{a,b'\geq0\\a+b'\ \mathrm{even}}}
       (a+1)(b'+1)p^{-(a+b')(1/2+b)}
 =1+O(100p^{-1-2b}).
\]
For example this follows by putting $z=p^{-1/2-b}\leq3^{-1/2}$ in
$\tfrac12((1-z)^{-4}+(1+z)^{-4})$ and bounding its terms of degree
at least two. The prime two costs a bounded factor by
\eqref{analytic:two-sum}. The zero mode is therefore $O_G(b^{-100})$.

For the nonzero modes, denote the two odd lengths by $N_1,N_2$.
The Fourier support restricts the frequencies to
\[
 0<|h|\leq H_0=C_\Phi N_1N_2/(MP^2).
\]
Increasing the constant in \eqref{analytic:inflation-condition}
ensures $H_0\leq M/2$ for every such pair of lengths. For fixed
$h_2$, Cauchy--Schwarz and \eqref{analytic:inductive-prefix} bound
the sum over $h_1$ occurring in \eqref{analytic:nonzero-mode} by
\[
 \ll_{\mathcal F}K\frac{H_0}{h_2^2}L^B
 (1+|t+v_1|)^{A/2}(1+|-t+v_2|)^{A/2}.
\]
This formula is only used when $H_0/h_2^2\geq1$; otherwise that sum
is empty. Choose $J_1>A+4$. Since
$1+|\pm t+v|\leq T(1+|v|)$, the integral in $v_1,v_2$ is
$O_A(T^A)$. Also
\begin{equation}\label{analytic:square-factor-sum}
 \sum_{h_2\geq1}\frac{\tau(h_2)^{C_{\rm ar}}}{h_2^2}<\infty.
\end{equation}
One proof of this convergence is its Euler product: its local factor
is $1+O_{C_{\rm ar}}(p^{-2})$, since
$\sum_{a\geq1}(a+1)^{C_{\rm ar}}p^{-2a}
\ll_{C_{\rm ar}}p^{-2}$ uniformly for $p\geq2$.
Consequently the nonzero contribution is bounded by
\[
 \ll_{A,B,\mathcal F,G}b^{-C_{\rm ar}}KL^BT^A
 (N_1N_2)^{-1/2+b}H_0
 \ll_{A,B,\mathcal F,G}
 \log^{C_0}(3M+3N)K\frac{N}{MP^2}L^BT^A.
\]
Here $(N_1N_2)^b\ll_G1$ and
$\sqrt{N_1N_2}\ll_GN$. The sums of the factors from two converge by
\eqref{analytic:two-sum}. This proves
\eqref{analytic:inflated-moment}. If the Fourier support contains no
nonzero frequency, the same proof uses only its zero-mode part.

\smallskip
\noindent\emph{Recovering the squarefree average.}
Now inflate an admissible parameter by setting $m=dp^2$. If
$c=\chi_d(2)$, multiplicativity gives the exact identity
\begin{equation}\label{analytic:inflation-identity}
 P_d(N,t;G)=\sum_{a\geq0}
 \lambda_F(p^a)\chi_d(p)^a p^{-a(1/2+it)}
 P_{dp^2}^c(N/p^a,t;G).
\end{equation}
Indeed $z_{dp^2}(n)$ deletes all positive powers of $p$ and agrees
with $\chi_d(n)$ on integers prime to $p$; reinstating their Euler
coefficients gives the original sum. If $p\mid d$, the terms with
$a\geq1$ vanish, so the identity also covers that case.

Equip pairs $(d,p)$, with $M\leq|d|<2M$, $\chi_d(2)=c$, and
$p\in\mathcal P$, with squared norm
$(M|\mathcal P|)^{-1}\sum_{d,p}|\cdot|^2$.
The map $(d,p)\mapsto dp^2$ is injective: in the prime factorization
of its absolute value the only exponent exceeding one occurs at
$p$, and is either two or three. Moreover $|dp^2|<8U$.
Thus the $a=0$ term has norm at most
\[
 \left(\frac{U}{M|\mathcal P|}\right)^{1/2}
 \sup_{0<N'\leq N}
 \left(\frac1U\sum_m\Phi(m/U)|P_m^c(N',t;G)|^2\right)^{1/2}
 \ll(P\log(2P))^{1/2}\mathcal B^{1/2},
\]
where $\mathcal B$ denotes the right side of
\eqref{analytic:inflated-moment}.
For $a\geq1$ fix $p$ before summing in $d$. The weaker bound
$\sum_d|P_{dp^2}^c(N/p^a,t;G)|^2\leq U\mathcal B$ gives pair norm
at most
$(a+1)P^{1-a/2}\mathcal B^{1/2}$. Minkowski's inequality and
\[
 \sum_{a\geq1}(a+1)P^{-a/2}\ll P^{-1/2}
\]
show that these terms together have norm
$O(P^{1/2}\mathcal B^{1/2})$. Summing the two choices of $c$ proves
\begin{align}
 \frac1M\sum_d|P_d(N,t;G)|^2
 &\ll_G\log^{C_0}(3M+3N)P\log(2P)
 \left(1+C_{A,B,\mathcal F,G}K\frac{N}{MP^2}L^BT^A\right).
 \label{analytic:deinflated-moment}
\end{align}

Here is an explicit noncircular choice of exponents. Set
\begin{equation}\label{analytic:exponent-choices}
 \begin{split}
 C_0&=512,\qquad C_1=C_0+32,\qquad C_2=C_0+C_1+4,\\
 R&=C_2+4,\qquad C_3=3C_2+10,\qquad
 A=B=2(C_3+20).
 \end{split}
\end{equation}
The letter $R$ in this proof denotes a fixed contour abscissa; it is
unrelated to a rank truncation threshold used later in the paper.
For any prescribed $\varepsilon>0$, choose
\begin{equation}\label{analytic:choice-P}
 P=P_0[L(1+N/M)T]^{C_1}.
\end{equation}
Once $A,B$ and the fixed weight family have been selected, a sufficiently
large $P_0$ ensures \eqref{analytic:inflation-condition} and gives
\begin{align}
 \frac1M\sum_d|P_d(N,t;G)|^2
 &\leq C_*[L(1+N/M)T]^{C_2}
 \nonumber\\
 &\quad+\varepsilon K L^{B-10}T^{A-4}(1+N/M)^{-4}.
 \label{analytic:small-recursion}
\end{align}
The constant $C_*$ can depend on $P_0$ and the chosen fixed data, but
not on $K,M,N,t$.
For explicit verification, put $q=1+N/M$ and $D_0=LqT$.
Then $\log(3M+3N)\ll Lq$ and
\[
 \frac{\log(2P)}P
 \leq\frac{\log(2P_0)+C_1}{P_0}D_0^{-C_1+1}.
\]
The recursive coefficient in \eqref{analytic:deinflated-moment} is
therefore at most a fixed multiple of
\[
 \frac{\log(2P_0)+C_1}{P_0}
 K L^{B+C_0-C_1+1}T^{A-C_1+1}q^{C_0-C_1+2}.
\]
Our value of $C_1$ leaves at least the losses $10,4,4$ displayed in
\eqref{analytic:small-recursion}. Its prefactor can be made smaller
than $\varepsilon$. The nonrecursive term is bounded by
$C_*D_0^{C_0+C_1+1}$ and hence by the stated $C_2$ power. In
particular $C_2$ was fixed before $A,B,\varepsilon,P_0,K$.

\smallskip
\noindent\emph{Reconstructing the central-line values.}
We next reconstruct $L$ on the central line. Choose a fixed \mbox{smooth}
dyadic partition $G_0$ with common compact support in $(0,\infty)$,
so that
$\sum_{N=1,2,4,\ldots}G_0(n/N)=1$ for every integer $n\geq1$.
Let $C_d=\sqrt{q_F}|d|/(2\pi)$ and define
\[
 I_d(t)=\frac1{2\pi i}\int_{(R)}
 L_{F^{(d)}}(1/2+it+w)C_d^w
 \frac{\Gamma(1+it+w)}{\Gamma(1+it)}\frac{e^{w^2}}w\,dw.
\]
Moving the contour to $\Re w=-R$ crosses only the residue
$L_{F^{(d)}}(1/2+it)$, because the completed function is entire.
Apply its functional equation on the new line and replace $w$ by
$-w$. One obtains the exact approximate functional equation
\begin{equation}\label{analytic:afe-identity}
 L_{F^{(d)}}(1/2+it)=I_d(t)+\omega_d(t)I_d(-t),\qquad
 \omega_d(t)=\epsilon_d C_d^{-2it}
                       \frac{\Gamma(1-it)}{\Gamma(1+it)},
 \quad |\omega_d(t)|=1.
\end{equation}
The Gaussian makes the contour displacement legitimate. Initially,
on $\Re w=R$, the Dirichlet series is absolutely convergent, so we
may insert the dyadic partition.

For $0\leq c\leq R$, Stirling's formula uniformly gives
\begin{equation}\label{analytic:gamma-quotient}
 \left|\frac{\Gamma(1+c+i(t+v))}{\Gamma(1+it)}\right|
 \ll_R T^c(1+|v|)^{R+1/2}e^{\pi|v|/2}.
\end{equation}
Indeed the quotient of the exponential factors is at most
$e^{\pi|v|/2}$, and
\[
 \frac{(1+|t+v|)^{c+1/2}}{(1+|t|)^{1/2}}
 \leq T^c(1+|v|)^{c+1/2}.
\]
The same estimates with $1+|t+v|$ also cover bounded ordinates, including
$v$ near $-t$. Thus the bound is uniform as $c$ tends to zero.
For a fixed dyadic piece the polynomial is finite, so its contour can
be moved to any $0<c\leq R$. Multiplying
\eqref{analytic:gamma-quotient} by $|e^{(c+iv)^2}|$ absorbs its
$v$-factors into $O_R(e^{-v^2/2})$ and yields
\begin{equation}\label{analytic:afe-pieces}
 |L_{F^{(d)}}(1/2+it)|
 \ll_{R,\mathcal F}\sum_{\nu=\pm1}\sum_N
 \int_{\mathbb R}\frac{e^{-v^2/2}}{|c_N+iv|}
      \left(\frac{MT}N\right)^{c_N}
 |P_d(N,\nu t+v;G_0(y)y^{-c_N})|\,dv.
\end{equation}
There is no use of the desired moment to justify this representation:
on the initial fixed line $R$, the elementary bound
$|P_d(N,t;G)|\ll_G\sqrt N\log(2+N)$ makes the sum of the long
pieces converge, since $R>1/2$. Finitely many shorter pieces can
then be moved separately. The moment estimates below also show
convergence after taking the family norm.

Use
\[
 c_N=\frac1{\log(3MT)}\quad(N\leq MT^2),\qquad
 c_N=R\quad(N>MT^2).
\]
Apply \eqref{analytic:small-recursion} uniformly to the fixed smooth
family $G_0(y)y^{-c}$, $0\leq c\leq R$, and use Minkowski in
\eqref{analytic:afe-pieces}. We spell out the two contributions.
For the nonrecursive square-root term and $N\leq MT^2$,
\[
 [L(1+N/M)(1+|\nu t+v|)]^{C_2/2}
 \ll L^{C_2/2}T^{3C_2/2}(1+|v|)^{C_2/2}.
\]
There are $O(\log(3MT^2))$ such dyadic pieces,
$(MT/N)^{c_N}\leq e$ for $N\geq1$, and
$1/|c_N+iv|\leq\log(3MT)$.
Since both logarithms are $O(LT)$, their squared total cost gives
at most $C_{**}(LT)^{3C_2+4}$ in the mean square.
For the long pieces put $q'=N/M>T^2$. Then
\[
 \sum_{q'>T^2}\left(\frac T{q'}\right)^R(q')^{C_2/2}
 \ll_R T^{-R+C_2},
\]
where $q'$ runs over a dyadic progression. Including the outside
factor $L^{C_2/2}T^{C_2/2}$ again fits within
$C_{**}(LT)^{C_3}$ by \eqref{analytic:exponent-choices}.

For the recursive square-root term, the short dyadic sum satisfies
\[
 \sum_{N\leq MT^2}(1+N/M)^{-2}\ll\log(3M)=L.
\]
The Gaussian absorbs $(1+|v|)^{(A-4)/2}$, so the resulting norm is
at most
\[
 C_{A,R,\mathcal F}\sqrt{\varepsilon K}\,
 L^{B/2-5}T^{A/2-2}\,L\log(3MT)
 \ll C_{A,R,\mathcal F}\sqrt{\varepsilon K}\,
 L^{B/2}T^{A/2}.
\]
For the long pieces we have the stronger estimate
\begin{equation}\label{analytic:long-recursive-sum}
 \sum_{q'>T^2}\left(\frac T{q'}\right)^R(1+q')^{-2}
 \ll_R T^{-R-4},
\end{equation}
which gives the same bound. Squaring the total norm now proves
\begin{equation}\label{analytic:central-recursion}
 \frac1M\sum_d|L_{F^{(d)}}(1/2+it)|^2
 \leq C_{**}(LT)^{C_3}
       +C_{\dagger}\varepsilon K L^BT^A.
\end{equation}
Here $C_{\dagger}$ depends on $A,R,\mathcal F$ but not on $P_0,K,M,t$;
$C_{**}$ may depend on $P_0$. This independence is essential when
choosing the constants.

\smallskip
\noindent\emph{Closing the induction in the strip.}
It remains to extend the estimate to the strip without a factor for
the number of twists. Form the vector
\[
 \mathbf V_F(s)=\big(M^{-1/2}L_{F^{(d)}}(s)\big)_d
\]
in the finite-dimensional Euclidean space indexed by the current
admissible block. On $\Re s=2$ its squared norm is bounded by an
absolute constant: each $L$-value has modulus at most
$\sum_n\tau(n)n^{-2}=\zeta(2)^2$, and the number of parameters is
$O(M)$. On $\Re s=1/2$, \eqref{analytic:central-recursion} gives the
stated vector norm bound.

For each constant unit vector $u$ apply the scalar strip principle to
\[
 f_u(s)=\frac{\langle\mathbf V_F(s),u\rangle}{(1+s)^{A/2}}.
\]
Here $A$ is even by \eqref{analytic:exponent-choices}. On both
boundaries $|1+s|\asymp1+|\Im s|$, with constants independent of the
number of twists. The scalar boundary bound is therefore at most
\[
 C_A\big(C_{**}L^{C_3}+C_{\dagger}\varepsilon K L^B+1\big)^{1/2}.
\]
For clarity, finite-order growth suffices for the strip principle:
multiply $f_u(s)$ by
$\exp\{-\rho\cos(s-5/4)\}$, with $\rho>0$. Its modulus is at most
one on the strip boundaries and it decays faster than every
finite-order growth bound on horizontal sides, since
$\cos(\Re s-5/4)>0$ for $1/2\leq\Re s\leq2$. Apply the maximum
principle on rectangles and then let their heights tend to infinity
and $\rho$ decrease to zero. Taking the supremum over unit vectors
at the desired point gives
\begin{equation}\label{analytic:strip-recursion}
 \|\mathbf V_F(\sigma+it)\|_2^2
 \leq C_{\rm st}
       \big(C_{**}L^{C_3}+C_{\dagger}\varepsilon K L^B+1\big)T^A,
 \qquad 1/2\leq\sigma\leq2.
\end{equation}
This argument has no dimension-dependent loss.

The choice order is now explicit. Fix the family, the elementary
arithmetic exponent, $C_0,C_1,C_2,R,C_3,A,B$ as above, and then all
needed fixed kernel orders. Next choose
$0<\varepsilon<(4C_{\rm st}C_{\dagger})^{-1}$, choose $P_0$ to obtain
\eqref{analytic:small-recursion} for the entire weight family, and
finally choose $K$ so large that
$C_{\rm st}(C_{**}L^{C_3}+1)\leq K L^B/2$ for every $L\geq\log3$.
This is possible because $B>C_3$ and $C_{**}$ is independent of $K$.
Equation~\eqref{analytic:strip-recursion} proves
\eqref{analytic:induction-hypothesis} at height $M$. At $M=1$, all
dual sums used above are empty, so this also establishes the initial
step. The simultaneous induction is complete.

The induction just completed uses the fixed weight family
$G_0(y)y^{-c}$, $0\leq c\leq R$. For any other fixed smooth compactly
supported $G$, repeat the polynomial argument through
\eqref{analytic:small-recursion}, using the established $L$-function
bound at smaller heights. The constants $P_0$ and $C_*$ may now depend
on $G$, while $K,A,B$ remain fixed. Since $A,B>C_2$, this proves
\eqref{analytic:polynomial-bound}, with the asserted uniformity for
fixed uniformly smooth weight families.
Finally use Lemma~\ref{setup:reduction}, bounded multiplicity, and a
sum over admissible dyadic blocks of total length $O(H)$ to obtain
\eqref{analytic:prefix-moment}. No analytic-density input has entered
the argument.
\end{proof}

\subsection{Short polynomials and auxiliary multipliers}

The second moments are now available at every height. We use them to
bound the complete sum of nonzero Poisson frequencies, retaining the
zero frequency as an exact independent-model average. The resulting
comparison permits a short polynomial multiplier, as required for the
mollifiers below.

Fix a nonnegative smooth function $\Psi$ compactly supported in
$\{x:0<|x|<\infty\}$, with $\Psi(x)\geq1$ when $1\leq|x|\leq2$.
Fix $c\in\{1,-1\}$. The formal character $z_m$ and the independent
variables $Z$ always have value $c$ at two. At odd primes their model
law is
\[
 \mathbb P(Z(p)=0)=\frac1p,\qquad
 \mathbb P(Z(p)=1)=\mathbb P(Z(p)=-1)=\frac{1-1/p}{2}.
\]
They are extended completely multiplicatively. This is a model for
complete integer residue averages.

\begin{lemma}[Summation of all dual frequencies]\label{analytic:frequency-sum}
For fixed $B\geq0$, $J_1>1$, and $Q>0$,
\begin{equation}\label{analytic:dyadic-frequency-bound}
 \sum_{T=1,2,4,\ldots}T\log^B(3T)(1+T/Q)^{-J_1}
 \ll_{B,J_1}
 \begin{cases}
 Q^{J_1},&0<Q<1,\\
 Q\log^B(3Q),&Q\geq1.
 \end{cases}
\end{equation}
In particular the first bound is at most a constant times $Q$.
\end{lemma}

\begin{proof}
If $Q<1$, bound $(1+T/Q)^{-J_1}$ by $(Q/T)^{J_1}$.
The remaining series is
$\sum_{j\geq0}2^{-(J_1-1)j}\log^B(3\cdot2^j)<\infty$.
For $Q\geq1$, the part $T\leq Q$ is bounded by
$\log^B(3Q)\sum_{T\leq Q}T\ll Q\log^B(3Q)$.
For the remaining part choose the first dyadic $T_0>Q$, so
$Q<T_0\leq2Q$, and write $T=2^jT_0$. The summand is at most
\[
 C_B Q\,2^{-(J_1-1)j}
       \big(\log^B(3Q)+(j+1)^B\big).
\]
Both resulting geometric series converge. This estimates the infinite
frequency tail before any logarithm is replaced by $\log X$.
\end{proof}

\begin{proposition}[Comparison with the independent model]
\label{analytic:comparison}
Let $\ell=\log X$, where $X$ is sufficiently large, and let
\[
 c_1\ell^{-1/10}\leq\Delta\leq\frac12
\]
for a fixed $c_1>0$. There are fixed constants $\eta,c_2>0$ for which
the following holds. Let
\[
 S_V(x)=\sum_{n\geq1}\frac{\lambda_F(n)V(n)}{\sqrt n}\,
                         w_x(\ell^{-1}\log n),\qquad
 H_V=\sum_{r'\leq R_0}h(r')V(r'),
\]
where $V=z_m$ or $Z$. Suppose that $S_V$ has length at most
$X^{1-\Delta}$, that
\begin{equation}\label{analytic:multiplier-conditions}
 1\leq R_0\leq X^{\eta\Delta},\qquad
 \sum_{r'\leq R_0}|h(r')|\leq C_H R_0^2,
\end{equation}
and that $w_x$ and a function $A(x)$ have bounds by a fixed power of
$\ell$ for all derivatives up to a sufficiently large fixed order.
For $w_x$ these bounds are required after smooth dyadic localization
in $n$, on the scaled variables $n/N$ and $x$ in a fixed neighborhood
of $\operatorname{supp}\Psi$. They may in particular follow from
bounds in $u=\ell^{-1}\log n\geq0$ and $x$, with a smooth extension
at $n=1$. Then
\begin{align}
 &\frac1X\sum_{m\in\mathbb Z}\Psi(m/X)
       |H_{z_m}S_{z_m}(m/X)-A(m/X)|^2
 \nonumber\\
 &\hspace{8mm}=
 \int_{\mathbb R}\Psi(x)\,
       \mathbb E|H_ZS_Z(x)-A(x)|^2\,dx
       +O\big(\exp(-c_2\Delta\ell)\big).
 \label{analytic:model-comparison}
\end{align}
The constants and the sufficiently large $X$ threshold are uniform
when the stated support, coefficient, and derivative bounds are
uniform. They do not depend on $\Delta$ or on an index parametrizing
such weights. One may take
\begin{equation}\label{analytic:comparison-constants}
 C'=C_{\rm ar}+1=257,\qquad
 \eta=\frac1{4(4+2C')},\qquad c_2=\frac12.
\end{equation}
\end{proposition}

\begin{proof}
Put $L=X^{1-\Delta}$. If the length hypothesis specifies only that
the integer coefficients vanish for $n>L$, multiply the interpolating
weight by $\rho_+(n/L)$, where $\rho_+$ is a fixed smooth function equal
to one on $(-\infty,1]$ and zero on $[2,\infty)$. This preserves every
integer coefficient and gives real support in $n\leq2L$. On a dyadic
scale meeting the transition, $N/L$ is bounded, so derivatives of
$\rho_+(Ny/L)$ add only fixed constants to the assumed scaled seminorm
bounds. Thus all nonempty localized lengths satisfy $N_i\ll L$.

Expand the square on the left of \eqref{analytic:model-comparison}.
A term involving two multiplier indices $r_1',r_2'$ has character
factor $z_m(r_1'r_2'n_1n_2)$. Its odd multiplier part is an integer
$r\leq R_0^2$. Powers of two contribute only fixed signs. Split
powers of two out of $n_1,n_2$, and insert smooth dyadic partitions
in their odd parts. Each resulting term has exactly the form
\eqref{analytic:poisson-average} with $U=X$, $t_1=t_2=0$, and a
weight $W$ whose required seminorm is $O(\ell^{D_1})$ for a fixed
$D_1$. The independent factors coming from the powers of two satisfy
\eqref{analytic:two-sum}. Cross terms with $A(x)$ are covered by
using an index supported just at one: choose a fixed smooth function
supported in $(1/2,3/2)$ and equal to one at one. The identity
$\lambda_F(1)=1$ then puts them in the same form. The term $|A|^2$
is covered by two such indices. Its derivatives obey the same
fixed polynomial bounds.

The zero frequency is exactly the model expression. In fact, for
an odd prime and $v\geq1$,
\[
 \frac1{p^v}\sum_{a\bmod p^v}\left(\frac a{p^v}\right)
 =\begin{cases}0&v\text{ odd},\\1-p^{-1}&v\text{ even},\end{cases}
 =\mathbb E Z(p)^v.
\]
For $v=0$ both sides are one. Chinese remaindering gives independence
at distinct odd primes, and the value at two is already fixed.
The integral of the zero-frequency smooth amplitude is thus the
corresponding term on the right of
\eqref{analytic:model-comparison}.

It remains to estimate the nonzero frequencies. Fix an odd multiplier
$r$ and localized odd lengths $N_1,N_2$, and set
\[
 b=\frac1{10\ell},\qquad Q=\frac{rN_1N_2}{X}.
\]
Only nonempty lengths need be considered, so $N_i$ is bounded below
by a fixed positive constant. Also $N_i\ll X^{1-\Delta}$, whence
$(N_1N_2)^b\ll1$. Proposition~\ref{analytic:moment} is available
at every height. For a dyadic frequency range $T\leq|h|<2T$,
write $h=h_1h_2^2$. For fixed $h_2$, the parameter $h_1$ lies in
$0<|h_1|<2T/h_2^2$. Cauchy--Schwarz and
\eqref{analytic:prefix-moment} imply
\begin{align*}
 &\sum_{\substack{T\leq|h|<2T\\h=h_1h_2^2}}
 \tau(h_2)^{C_{\rm ar}}
 \prod_{i=1}^2
 |L_{F^{(\nu h_1)}}(1/2+b+iv_i)|\\
 &\quad\ll
 T\log^B(3T)(1+|v_1|)^{A/2}(1+|v_2|)^{A/2}
 \sum_{h_2\leq\sqrt{2T}}\frac{\tau(h_2)^{C_{\rm ar}}}{h_2^2}\\
 &\quad\ll T\log^B(3T)
               (1+|v_1|)^{A/2}(1+|v_2|)^{A/2}.
\end{align*}
This inequality applies separately to each $\nu$; it includes both
signs of $h$. Choose $J_1>A+4$ in
\eqref{analytic:nonzero-mode}, and then fix the corresponding
smoothness order. Its vertical integral is convergent. The remaining
sum is bounded, by Lemma~\ref{analytic:frequency-sum}, by
\[
 \ll \ell^{D_2}r^{C_{\rm ar}}(N_1N_2)^{-1/2}
 \begin{cases}Q^{J_1}&Q<1,\\Q\log^B(3Q)&Q\geq1.
 \end{cases}
\]
When $Q\geq1$ we have $Q\ll X$ using $r\leq R_0^2$,
\eqref{analytic:multiplier-conditions}, and $\eta<1$; hence
$\log(3Q)\ll\ell$. When $Q<1$ use $Q^{J_1}\leq Q$.
In both regimes the error for this localized pair is therefore
\begin{equation}\label{analytic:localized-comparison-error}
 \ll\ell^{D_3}r^{C'}\frac{\sqrt{N_1N_2}}X,
 \qquad C'=C_{\rm ar}+1.
\end{equation}
The increase from $C_{\rm ar}$ to $C'$ is only the factor $r$
in $Q$, not a smoothness loss. Notice that this argument controls
arbitrarily large dual frequencies as well as the case $Q<1$.

There are $O(\ell)$ possible nonempty dyadic pieces for each main
index. Their cost is a fixed logarithmic power. Summing the
coefficients from powers of two by \eqref{analytic:two-sum} costs
only a constant. The two multiplier coefficient sums and the
bound $r\leq R_0^2$ cost at most
\[
 (1+\textstyle\sum|h(r')|)^2R_0^{2C'}\ll_{C_H}R_0^{4+2C'}.
\]
Since $\sqrt{N_1N_2}/X\ll X^{-\Delta}$, including the terms with an
index at one, the full error is
\begin{equation}\label{analytic:total-comparison-error}
 \ll\ell^{D_4}X^{-\Delta}R_0^{4+2C'}
 \leq\ell^{D_4}\exp(-3\Delta\ell/4).
\end{equation}
Our choices in \eqref{analytic:comparison-constants} give the last
inequality. The power $D_4$ is fixed before choosing $X$.
Because $\Delta\ell\geq c_1\ell^{9/10}$, for one sufficiently large
$X$ the factor $\ell^{D_4}$ is at most
$\exp(\Delta\ell/4)$. This proves
\eqref{analytic:model-comparison} with $c_2=1/2$.
\end{proof}

The endpoint at $n=1$ in this proposition needs no extension to a fixed
negative interval in $u$.
Extend a dyadically localized weight only to, say, $n\geq1/2$.
In the coordinate $\log n=\ell u$ this is a fixed interval, and
smooth cutoffs there introduce at most fixed powers of $\ell$ in
$u$ derivatives. On a dyadic scale, repeated differentiation of
$w_x(\log(Ny)/\ell)$ instead introduces powers of $1/\ell$ and
bounded powers of $1/y$. Thus the required finite seminorm bounds
are precisely those in the proposition, even at the first piece.
==== END The-mean-analytic-rank-of-quadratic-twists-of-elliptic-curves//sections/analytic.tex ====

==== BEGIN The-mean-analytic-rank-of-quadratic-twists-of-elliptic-curves//sections/introduction.tex ====
\section{Introduction}\label{sec:introduction}

Goldfeld's mean-rank conjecture predicts that the average central order
of vanishing in a quadratic-twist family is \(1/2\)
\cite[p.~113, Conjecture~(B)]{Goldfeld1979}.
The issue addressed here is the contribution of rare twists of large
analytic rank: knowing the densities of ranks zero and one does not
determine the mean without control of this contribution.

Let \(E/\Q\) be an elliptic curve. For a nonzero squarefree integer \(d\),
write \(E^{(d)}\) for its quadratic twist, and put
\[
 \mathcal D(Y)=\{d\in\Z:0<|d|\le Y,\ d\text{ squarefree}\},
 \qquad
 a(E^{(d)})=\ord_{s=1}L(E^{(d)},s).
\]
Both signs of \(d\) are included. We call \(a\) the analytic rank and
write \(r(E^{(d)})=\operatorname{rank}_{\Z}E^{(d)}(\Q)\) for the
algebraic or Mordell--Weil rank. Outside the explicitly algebraic
statements below, rank means analytic order of vanishing.

\begin{theorem}\label{thm:mean}
For every elliptic curve \(E/\Q\),
\[
 \lim_{Y\to\infty}\frac{1}{\#\mathcal D(Y)}
     \sum_{d\in\mathcal D(Y)}a(E^{(d)})=\frac12.
\]
\end{theorem}

Theorem~\ref{thm:mean} resolves positively Goldfeld's mean analytic-rank
conjecture in this signed squarefree counting convention. It applies
without restrictions on complex multiplication, rational torsion,
rational isogenies, or reduction type.

The analytic density theorem
\cite[Theorem~1.2]{OpenAIAnalyticDensityV2}
states that, for this same family and every \(E/\Q\),
\begin{equation}\label{eq:density-input}
 \frac{\#\{d\in\mathcal D(Y):a(E^{(d)})=j\}}{\#\mathcal D(Y)}
       \longrightarrow\frac12
       \qquad(j=0,1).
\end{equation}
For the analytic mean theorem, \eqref{eq:density-input} is the only input
from the companion manuscript identified in the bibliography.
A zero-density exceptional family can still contribute
to a rank-weighted average. The new assertion needed for the mean is the
following tail estimate, whose proof is independent of
\eqref{eq:density-input}.

\begin{theorem}\label{thm:tail}
For every elliptic curve \(E/\Q\), there are constants \(C_E>0\) and
an integer \(R_E\ge1\) such that, for every integer \(R\ge R_E\),
\[
 \limsup_{Y\to\infty}\frac1Y
 \sum_{\substack{d\in\mathcal D(Y)\\a(E^{(d)})>R}}a(E^{(d)})
       \le \frac{C_E}{R}.
\]
\end{theorem}

Together, \eqref{eq:density-input} and Theorem~\ref{thm:tail} show that
\[
 \sum_{\substack{d\in\mathcal D(Y)\\a(E^{(d)})\ge2}}a(E^{(d)})=o_E(Y).
\]
For each fixed sufficiently large integer \(R\), \eqref{eq:density-input}
makes the complement of ranks zero and one have density zero, so ranks
\(2\) through \(R\) contribute \(o_{E,R}(Y)\).
For the remaining ranks, use the height limsup in Theorem~\ref{thm:tail}
and then let \(R\) tend to infinity. This order of limits will be
made explicit in Section~\ref{sec:tail}.

\subsection{Algebraic-rank moments}

The same signed squarefree family has the following algebraic-rank
moment limits. This consequence uses the companion's density conclusions
and its density-one equality of analytic and algebraic ranks, together
with the exponential-moment bound of Koymans and Smith. It does not use
the analytic tail estimate of Theorem~\ref{thm:tail}.

\begin{corollary}[Fixed algebraic-rank moments]\label{cor:algebraic-moments}
Fix an elliptic curve \(E/\Q\). For every fixed real \(t\),
\[
 \lim_{Y\to\infty}\frac1{\#\mathcal D(Y)}
 \sum_{d\in\mathcal D(Y)}\exp\!\bigl(t\,r(E^{(d)})\bigr)
 =\frac{1+e^t}{2}.
\]
For every fixed positive integer \(m\),
\[
 \lim_{Y\to\infty}\frac1{\#\mathcal D(Y)}
 \sum_{d\in\mathcal D(Y)}r(E^{(d)})^m=\frac12.
\]
\end{corollary}

The proof appears in Section~\ref{sec:algebraic}.

All parameters \(t\) and moment orders \(m\) in
Corollary~\ref{cor:algebraic-moments} are fixed before \(Y\to\infty\).
The corollary concerns algebraic ranks only: it gives no higher
analytic-rank moments, no uniformity for parameters varying with \(Y\),
and no moment limit for thin polynomial subfamilies.

\subsection{History and significance}

Goldfeld's original question concerns the change of rank when a fixed
elliptic curve is viewed over quadratic fields. Under
Birch--Swinnerton-Dyer, that change is the central order of the associated
quadratic twist. His Conjecture~(B) formulates the average directly in
terms of this analytic order, using quadratic-field discriminants
ordered by absolute value \cite[pp.~113--114]{Goldfeld1979}.
Here the counting parameter is instead the signed squarefree integer
\(d\) in \(\mathcal D(Y)\).
The rank-zero/rank-one density prediction describes the smallest orders
allowed by the two functional-equation signs. The mean prediction also
requires the larger orders, however sparse, to have negligible total
contribution.

Conditional analytic estimates illustrate both aspects of the problem.
Heath-Brown proved an upper bound \(3/2+o(1)\) for smooth averages over
fundamental discriminants coprime to the conductor, separately for each
functional-equation sign \cite[Theorem~3]{HeathBrown2003}.
His nonnegative compactly supported weight selects one sign of the
discriminant, and his Riemann-hypothesis assumption concerns every
quadratic twist, including those not coprime to the conductor.
Under the same twistwise Riemann-hypothesis assumption, Miller and Wong
bounded higher analytic-rank moments and obtained exponentially
decreasing large-rank counting tails
\cite[Theorem~1.1 and Corollary~1.2]{MillerWong2012}.
Their weighted sums run over integer parameters, with possible repeated
squareclasses; the weight selects one parameter sign without fixing the
functional-equation sign. Their tail estimate fixes the rank threshold
before taking the height limit. Fiorilli obtained the exact mean \(1/2\) over
signed squarefree parameters coprime to the conductor, assuming the
Riemann hypothesis for elliptic-curve \(L\)-functions and an additional
averaged cancellation hypothesis for nonreal zeros
\cite[Theorem~1.1 and Hypothesis~M]{Fiorilli2016}.
Thus this exact conditional mean uses more than the Riemann hypothesis.

A different route to densities comes from Selmer groups.
Smith proved that, for every rational elliptic curve, the full
\(2\)-power Selmer corank is zero or one with density \(1/2\) each,
using signed integer twist parameters
\cite[Theorem~1.1]{Smith2025}. His analytic density corollary assumes
Birch--Swinnerton-Dyer \cite[Corollary~1.2]{Smith2025}.
The companion's \(2\)-converse supplies the analytic implication needed
for \eqref{eq:density-input} in the signed squarefree convention.
Theorem~\ref{thm:tail} addresses the additional rank-mass question
independently of that density argument.

Koymans and Smith prove exponential moment bounds for Mordell--Weil rank
in polynomial quadratic-twist families
\cite[Theorem~1.4]{KoymansSmith2026}. The specialization in the proof of
Corollary~\ref{cor:algebraic-moments} supplies the needed algebraic
tail control. Such bounds do not supply an analytic-rank identity or
higher analytic-rank moments.

Hanners claims the full Birch--Swinnerton-Dyer conjecture for every
rational elliptic curve \cite[Theorems~28.1, 39.4 and~39.5]{Hanners2026}. We do not use
this claim. The unresolved point is how the bridge conditions supported
by tests on eighteen curves in Section~39.1 are established for every
rational elliptic curve, as required by the transfer in Section~39.2.
This is not a refutation of the claimed theorem.

Modularity provides the analytic continuation and functional equation
for every curve in the argument
\cite[Theorem~A]{BreuilConradDiamondTaylor2001}. Our estimates use these analytic properties
and the local coefficient bounds. The proof invokes neither the
Birch--Swinnerton-Dyer conjecture nor the generalized Riemann hypothesis.

Earlier work on a fixed elliptic curve already connected derivative
moments to nonvanishing and analytic-rank averages. Perelli and
Pomyka\l a proved first-derivative nonvanishing results and bounds for
the sum of analytic ranks \cite{PerelliPomykala1997};
Pomyka\l a extended this nonvanishing approach to higher derivatives of
fixed order in a congruence-restricted twisting family \cite{Pomykala1997}.
In the family of weight-two newforms of varying prime level,
Kowalski, Michel, and VanderKam combined high completed derivatives,
mollification, and consecutive derivative orders to control both
functional-equation signs. Their passage from these nonvanishing counts
to rank-weighted tails also uses a separate bound for the second moment
of analytic rank
\cite[Sections~2, 4, and~8]{KowalskiMichelVanderKam2000}.
In the present quadratic-twist argument the derivative order grows with
height. This uniformity, together with a pointwise rank bound, controls
the rank mass beyond the available counting range without requiring a
bounded second analytic-rank moment.

Second moments of the twist \(L\)-functions themselves supply the
analytic comparison used below. For full-level holomorphic Hecke forms
of weight divisible by four, Soundararajan and Young obtained the
second-moment asymptotic over positive discriminants \(8d\), with \(d\)
odd and squarefree, under GRH for the twists, the Riemann zeta function,
and the symmetric-square \(L\)-function
\cite[Theorem~1.2]{SoundararajanYoung2010}.
Li proved the asymptotic unconditionally in this full-level setting
\cite[Theorem~1.1]{Li2022}. We adapt the Fourier-cutoff and prime-square
inflation mechanism in Li's proof, establishing the needed logarithmic
bounds for our finite family of arbitrary elliptic curves with its
conductor and bad-prime factors retained. Neither asymptotic is an input
to the argument.

\subsection{The analytic argument}

The starting point is the exact identity of Lemma~\ref{weights:vanishing},
forced by the vanishing of a derivative of order \(k\) of the completed \(L\)-function.
Its Dirichlet-series weight is obtained by exponentially smoothing
a \(k\)-th power of a truncated logarithm.
Near the square root of the conductor, at lengths of order \(X\) for
twists of height \(X\), this weight suppresses the late terms.
The approximation \((1-u)^k\approx e^{-ku}\) for \(u=O(1/k)\)
motivates mollification: at \(u=\log n/\log X\), the exponential
corresponds to a small positive shift from the center of the
\(L\)-function. A short approximation to its inverse Euler product
then makes the first weighted piece close to a positive main term.
We split the weighted series using smooth cutoffs into finite pieces
and a terminal series, use progressively shorter mollifiers as the
lengths of the finite pieces increase, and estimate the terminal series
without mollification.

Two features make the resulting estimates uniform for
\(k\) as large as a fixed power of \(\log X\).
First, a Poisson argument and inflation give unmollified second moments
with only powers of logarithms
(Propositions~\ref{analytic:moment} and~\ref{analytic:polynomial-moment}).
The inflation mechanism is related
to Li's work on quadratic twists
\cite[Lemmas~2.7 and~3.1, Proposition~3.2]{Li2022}; we prove the estimate
needed here for the entire fixed finite family of curves, including its
local factors at bad primes.
Second, Proposition~\ref{model:main-model-estimates} gives mean-square
bounds for the mollified finite pieces in a model that replaces
odd-prime character values by independent variables with their
complete-residue distributions. The mean-square estimates use the whole
probability space, without discarding exceptional outcomes.
Proposition~\ref{analytic:comparison} transfers these bounds to integer averages.
Separately, Proposition~\ref{model:good-event} compares the mollifiers
with positive reciprocal Euler products on most integer parameters
and relates these products to one another. We can then remove the
mollifiers and compare every piece with the same positive product:
the first piece dominates the later pieces and
the terminal series. The resulting nonzero sum contradicts the
completed-derivative identity whenever the rank exceeds the derivative
order of matching parity. Consecutive orders cover both signs.

For admissible twists of height \(X\), Proposition~\ref{tail:count} proves a bound
\(O_E(X/k^2)\) for the number of ranks exceeding \(k\), uniformly up to
\(k=(\log X)^{3/5}\).
A conductor-uniform Jensen bound (Lemma~\ref{tail:jensen}) gives the pointwise estimate
\(a(E^{(d)})=O_E(\log X)\).
The count at the largest permissible \(k\) then controls the remaining
rank mass, and summation of the integer tails proves
Theorem~\ref{thm:tail}.
The numerical exponents used in the length partition leave ample
room between the comparison error and the later thresholds; their
optimization is not needed.

The sections follow the inputs required by this argument.
Section~\ref{sec:setup} fixes the finite auxiliary family and the integer
residue model. Section~\ref{sec:analytic} proves the moments and the
comparison with that model. Section~\ref{sec:weights} constructs the
completed-derivative weights and bounds their terminal series.
Section~\ref{sec:model} constructs the short mollifiers and proves the
model estimates. Section~\ref{sec:tail} combines them into the counting
and rank-weighted tail bounds, and uses the companion density theorem
only in the final deduction of the mean. Section~\ref{sec:algebraic}
proves the separate algebraic-rank moment corollary.

\subsection{Conventions}

Implied constants may depend on the original curve \(E\), on fixed
smooth functions, and on a specified fixed order of differentiation.
They do not depend on the varying height, derivative order, or scale
index unless this is explicitly stated.
We write \(\tau(n)\) for the divisor function and use the Fourier kernel
\(e^{-2\pi ixy}\) when applying Poisson summation.
All sufficiently large lower thresholds for the integer \(k\) are
chosen after the fixed analytic exponents and smoothness orders.
==== END The-mean-analytic-rank-of-quadratic-twists-of-elliptic-curves//sections/introduction.tex ====

==== BEGIN The-mean-analytic-rank-of-quadratic-twists-of-elliptic-curves//sections/model.tex ====
\section{Block mollifiers and independent-model mean squares}
\label{sec:model}
\label{model:section}

We seek polynomial multipliers that make the first short piece close
to the positive quantity $A_x$ from \eqref{weights:Ax}, while keeping
the later pieces small. The construction has two separate tasks:
control the multipliers on most integers, and prove mean-square bounds
on the entire independent model. Their combination will allow the
first piece to dominate the weighted-series identity.

Fix a member $F$ of the finite family $\mathcal F$ and a value
$c\in\{-1,1\}$ at the prime $2$, and write $\lambda=\lambda_F$ and $\xi_p=\xi_{F,p}$.
The constants in this section are uniform in these two choices.  We use
the integer characters $z_m$ and the independent variables $Z$ defined
above; in particular, at an odd prime
\[
 \mathbb P(Z(p)=0)=p^{-1},\qquad
 \mathbb P(Z(p)=1)=\mathbb P(Z(p)=-1)=(1-p^{-1})/2,
 \qquad Z(2)=c.
\]
All model norms below are norms on this entire probability space.
We use the parameters and weights of Section~\ref{sec:weights}:
\[
 \ell=\log X,\qquad k_0\le k\le\ell^{3/5},\qquad
 J=\left\lfloor\frac{\log\ell}{10\log2}\right\rfloor,
 \qquad \delta_j=2^{-j},\qquad s=k/\ell.
\]
Here $k$ is an integer.  We always take $X$ sufficiently large that
$J\ge1$ and $0<s\le1$.  The scaled variable $x$ ranges over a fixed
compact subset $\mathcal X$ of $\mathbb R\setminus\{0\}$ containing the
support of the averaging weight.  Constants may depend on this compact
set, the fixed cutoff $f_*$, and $\mathcal F$.

The shift $s$ comes from the weight of the first piece. Replacing the
truncated power in \eqref{weights:W} by its exponential reference gives
the exact integral
\[
 \int_0^\infty e^{-v}
       \exp\bigl(-ku+s\log(C_F|x|v)\bigr)\,dv
       =A_xe^{-ku}.
\]
At $u=\log n/\ell$, the reference weight is therefore $A_xn^{-s}$.
This suggests a multiplier approximating the inverse Euler product at
the real shift $s$. Lemma~\ref{model:first-Fourier-error} and the final
Fourier argument will quantify the replacement in the required norm;
the reference calculation alone does not replace the weighted series.

\subsection{Elementary prime estimates and local factors}

We shall repeatedly use the elementary estimate
$\pi(y)\ll y/\log(2y)$.  One proof starts with
\[
 \sum_{n<p\le2n}\log p\le\log\binom{2n}{n}\le2n\log2.
\]
Summing on dyadic intervals gives $\sum_{p\le y}\log p\ll y$;
separating $p\le\sqrt y$ then proves the asserted estimate for $\pi(y)$.
Partial summation gives the following consequences, with absolute
constants:
\begin{align}
 \sum_{p\le y}\frac1p&\ll1+\log\log(3y),\nonumber\\
 \sum_{a<p\le b}\frac1p&\ll
 1+\log\frac{\log b}{\log a}\quad(2\le a\le b),
 \label{model:prime-mass}\\
 \sum_{p>Y}p^{-1-2a}
 &\ll\int_{2a\log Y}^{\infty}\frac{e^{-v}}v\,dv
 \ll 1+\log^+\frac1{a\log Y}\qquad(0<a\le1,\ Y\ge2).
 \label{model:prime-tail}
\end{align}
Here $\log^+t=\max(0,\log t)$.  If $a\log Y\ge1/2$, we also have
\[
 \int_{2a\log Y}^{\infty}\frac{e^{-v}}v\,dv
 \ll\frac{e^{-2a\log Y}}{a\log Y}.
\]
For clarity, the first bound in \eqref{model:prime-tail} follows by
integrating $t^{-1-2a}$ against $d\pi(t)$, discarding the nonpositive
boundary term at $Y$, and using $1+2a\le3$.
We also need, for $0<a\le1$ and $h\ge a$,
\begin{equation}
 \sum_p\frac{\min(1,h\log p)}p p^{-2a}
 \ll 1+\log(1+h/a).
 \label{model:prime-transition}
\end{equation}
Here is a direct verification.  If $h\le1$, primes with
$\log p\le1/h$ contribute at most
\[
 h\sum_{p\le e^{1/h}}\frac{\log p}{p}\ll1.
\]
Between $1/h$ and $1/a$ the reciprocal-prime mass is
$O(1+\log(h/a))$ by \eqref{model:prime-mass}; primes beyond $e^{1/a}$
contribute $O(1)$ by \eqref{model:prime-tail}.  Empty ranges are omitted.
If $h>1$, discard the minimum and use the same estimates to obtain
$O(1+\log(1/a))$, which is bounded by the right side of
\eqref{model:prime-transition}.

For any real character value $V(p)\in\{-1,0,1\}$ define
\begin{equation}
 D_p(b,V)=1-\lambda(p)V(p)p^{-1/2-b}
                +\xi_p V(p)^2p^{-1-2b},\qquad \Re b\ge0.
 \label{model:inverse-local}
\end{equation}
The local parameter bound gives a factorization
$D_p(b,V)=\prod_{\nu=1}^2(1-\alpha_{p,\nu}V(p)p^{-1/2-b})$,
where $|\alpha_{p,\nu}|\le1$ and zero parameters are permitted.
Consequently
\begin{equation}
 (1-p^{-1/2})^2\le |D_p(b,V)|\le(1+p^{-1/2})^2.
 \label{model:local-bounds}
\end{equation}
For real $b\ge0$, the value of $D_p(b,V)$ is positive: at a good
prime its two factors are complex conjugates (or both positive real),
and at a bad prime its nontrivial factor is positive.  These facts apply
equally to $V=Z$ and $V=z_m$.

\subsection{Construction, length, and the simultaneous product event}

Radziwi\l\l{} and Soundararajan use disjoint prime blocks, truncated
exponential polynomials, and relative truncation estimates in their
study of central values of quadratic twists
\cite[Sections~3, 8, and~9]{RadziwillSoundararajan2015}.
Here we truncate products of inverse local Euler factors by total
degree and estimate shifted norms over the full integer model defined
in Section~\ref{sec:setup}.

Put
\[
 Y_j=X^{\delta_j^2/\sqrt k},\qquad
 P_j^{\mathrm{eu}}(V)=\prod_{p\le Y_j}D_p(s,V)^{-1}
 \quad(0\le j\le J).
\]
Since $s\log Y_0=\sqrt k$, the prime-tail estimate
\eqref{model:prime-tail} makes $\sum_{p>Y_0}p^{-1-2s}$ exponentially
small. This sum will control the residual Euler product in mean square.
For the later pieces, the available
multiplier length decreases as their lengths approach $X$.
The smaller cutoffs $Y_j$ will keep the truncated products within
the comparison allowance $X^{\eta\delta_j/4}$.
In the estimates for these later pieces, the weight factor $\delta_j^k$
will absorb fixed polynomial losses in $k$ and $2^j$.
In the stated range $\log Y_J\ge\ell^{1/2}$, so every $Y_j$ exceeds
$2$ for a single sufficiently large $X$ threshold.  For $i\ge0$ let
\[
 I_{ji}=\{p:Y_j^{2^{-i-1}}<p\le Y_j^{2^{-i}}\},\qquad
 m_{ji}=\lceil k^{1/4}+j+i\rceil.
\]
We omit empty blocks.  They form a disjoint partition of the primes at
most $Y_j$, and \eqref{model:prime-mass} gives
\begin{equation}
 \sum_{p\in I_{ji}}p^{-1}\le C
 \label{model:block-mass}
\end{equation}
uniformly in $j,i,k,X$.  Indeed, when the lower endpoint is at least
$2$, the ratio of the logarithms of the endpoints is $2$; otherwise a
nonempty block has upper endpoint less than $4$.
Set
\begin{align}
 K_{ji}(V,z)&=\prod_{p\in I_{ji}}
  \left(1-\lambda(p)V(p)p^{-1/2-s}z
          +\xi_pV(p)^2p^{-1-2s}z^2\right),
 \nonumber\\
 T_{ji}(V)&=\sum_{b=0}^{m_{ji}}[z^b]K_{ji}(V,z),
 &H_j(V)&=\prod_i T_{ji}(V).
 \label{model:mollifier-definition}
\end{align}
The coefficient extraction in this definition is followed by evaluation
at $z=1$.

\begin{lemma}[Length and coefficients]\label{model:mollifier-length}
The polynomial $H_j$ has an expansion
\[
 H_j(V)=\sum_{r\le R_j}h_j(r)V(r),\qquad
 |h_j(r)|\le\tau(r)r^{-1/2},\qquad
 \sum_r|h_j(r)|\ll R_j^2,
\]
where, for a fixed constant $C_{\rm len}>0$, one may take
\begin{equation}
 R_j=\exp\left(C_{\rm len}(k^{1/4}+j+1)
                     \ell\delta_j^2/\sqrt k\right).
 \label{model:length-bound}
\end{equation}
For every fixed $\eta>0$, a sufficiently large fixed $k_0$ ensures
$R_j\le X^{\eta\delta_j/4}$ for all $0\le j\le J$ and all $k$ in
the stated range.  Thus $H_j$ satisfies the multiplier hypotheses of
Proposition~\ref{analytic:comparison}, with $\Delta=\delta_j/4$.
\end{lemma}

\begin{proof}
In a block, a chosen monomial has prime exponents $e_p\in\{0,1,2\}$
whose sum is at most $m_{ji}$.  Its integer index is therefore at most
$\exp(m_{ji}2^{-i}\log Y_j)$.  Since
\[
 \sum_{i\ge0}m_{ji}2^{-i}
 \le\sum_{i\ge0}(k^{1/4}+j+i+1)2^{-i}
 =2k^{1/4}+2j+4,
\]
we obtain \eqref{model:length-bound}.
Unique factorization shows that each index occurs in exactly one
prime-exponent pattern.  The truncations only delete patterns.  A
surviving coefficient is a product of local coefficients $1$,
$-\lambda(p)p^{-1/2-s}$, or $\xi_pp^{-1-2s}$, and hence is bounded by
$\tau(r)r^{-1/2-s}\le\tau(r)r^{-1/2}$.  For example,
\[
 \sum_{r\le R}\tau(r)r^{-1/2}
 =\sum_{ab\le R}(ab)^{-1/2}
 \le2\sqrt R\sum_{a\le R}\frac1a
 \ll\sqrt R\log(2R)\ll R^2\qquad(R\ge1).
\]
Finally,
\[
 \frac{\log R_j}{\ell\delta_j/4}
 \le4C_{\rm len}\bigl(k^{-1/4}+(j+1)2^{-j}k^{-1/2}\bigr)
 \le8C_{\rm len}k^{-1/4}.
\]
Here $(j+1)2^{-j}\le1$. Thus the fixed choice
$k_0\ge(8C_{\rm len}/\eta)^4$ suffices simultaneously for all $j$.
The coefficient estimate and this length bound verify both multiplier
conditions of Proposition~\ref{analytic:comparison}.
\end{proof}

\begin{lemma}[Prime-sum moments for the integer model]
\label{model:integer-moments}
Let $\mathcal I$ be any set of primes at most $Y_0$, and put
\[
 B_{\mathcal I}(V)=\sum_{p\in\mathcal I}
                 \lambda(p)V(p)p^{-1/2-s}.
\]
If $k_0$ is a sufficiently large absolute constant, then
\begin{equation}
 \frac1X\sum_{X\le|m|<2X}|B_{\mathcal I}(z_m)|^{64}
 \ll\left(1+\sum_{p\in\mathcal I}\frac1p\right)^{32}
 \label{model:64th-moment}
\end{equation}
uniformly in $\mathcal I,k,X$.
\end{lemma}

\begin{proof}
First omit the deterministic prime $2$ and write
$a_p=\lambda(p)p^{-1/2-s}$.  On expanding the model moment, any odd
prime that occurs just once has zero expectation.  Group the remaining
terms by their partition of the $64$ positions into $q\le32$ parts of
sizes at least two.  Since $|a_p|\le2$, for every part of size $r\ge2$
we have $|a_p|^r\le2^{r-2}|a_p|^2$.  Summing the distinct-prime choices
and then discarding distinctness bounds the moment by
\[
 C\sum_{q=1}^{32}\left(\sum_{p\in\mathcal I}|a_p|^2\right)^q
 \ll\left(1+\sum_{p\in\mathcal I}\frac1p\right)^{32}.
\]
Restoring the prime $2$ changes only the constant, by
$|u+v|^{64}\le2^{63}(|u|^{64}+|v|^{64})$ and $|a_2c|\le\sqrt2$.

For completeness the passage to integers needs no estimate for primes
in progressions.  For each ordered $64$-tuple of primes, the product
of its character values is periodic modulo the product of its odd
primes, which is at most $Y_0^{64}$.  Its absolute value is at most one.
The arithmetic mean over the two integer intervals $X\le|m|<2X$ differs from its
complete-residue mean by $O(Y_0^{64}/X)$.  The complete-residue mean is
exactly the independent-model mean, by the Chinese remainder theorem
and the counts of $0,1,-1$ for the Legendre symbol.  There are at most
$Y_0^{64}$ tuples and each coefficient has absolute value at most
$2^{64}$.  Thus the discrepancy between that arithmetic mean and the
model moment is
\[
 O(Y_0^{128}/X)=O(X^{-1+128/\sqrt k})=O(X^{-1/2})
\]
after requiring $\sqrt{k_0}\ge256$.  Since the number of integers in
these intervals is $O(X)$, this proves \eqref{model:64th-moment} with
its normalization by $X$.  If endpoints are not integers, the same
period-counting argument changes only its absolute constant.
\end{proof}

\begin{proposition}[Simultaneous product event]\label{model:good-event}
Except for $O(Xk^{-2})$ integers in $X\le|m|<2X$, the following
inequalities hold simultaneously:
\begin{align}
 \frac12\le H_j(z_m)P_j^{\mathrm{eu}}(z_m)&\le2
                         &&(0\le j\le J),
 \label{model:good-mollifier}\\
 \frac{P_0^{\mathrm{eu}}(z_m)}{P_j^{\mathrm{eu}}(z_m)}
   &\ge2^{-jk/8} &&(1\le j\le J),
 \label{model:good-ratio}\\
 P_0^{\mathrm{eu}}(z_m)&\ge2^{-Jk/8}.
 \label{model:good-full-product}
\end{align}
The implied constant and the lower threshold $k_0$ do not depend on
$k,j,X$.
\end{proposition}

\begin{proof}
For a block $I=I_{ji}$ write $B=B_I(z_m)$.  The elementary expansion
$\log(1+w)=w+O(|w|^2)$ at all sufficiently large primes, together with
\eqref{model:block-mass}, gives
\begin{equation}
 \sup_{|z|=2}|K_{ji}(z_m,z)|\le e^{2|B|+C},
 \qquad K_{ji}(z_m,1)\ge e^{-|B|-C}.
 \label{model:pointwise-block}
\end{equation}
The finitely many smaller primes are included using
\eqref{model:local-bounds} at $z=1$ and the direct upper bound
$\prod_{\nu}(1+2p^{-1/2})$ on $|z|=2$; they change $C$ by an absolute
amount.  Thus no nonvanishing assertion on the radius-two circle is
required.  Cauchy's coefficient estimate now gives, with $m=m_{ji}$,
\[
 \frac{|T_{ji}(z_m)-K_{ji}(z_m,1)|}{K_{ji}(z_m,1)}
 \le C2^{-m}e^{3|B|}.
\]
If every block satisfies $|B_{I_{ji}}(z_m)|\le m_{ji}/100$, the last
quantity is at most $Ce^{-c m_{ji}}$ for a fixed $c>0$.  The sum over
$i$ is $O(e^{-c(k^{1/4}+j)})$.  By enlarging $k_0$, the product over
blocks of the ratios $T_{ji}(z_m)/K_{ji}(z_m,1)$ lies between $1/2$
and $2$.
Since $\prod_iK_{ji}(z_m,1)=(P_j^{\mathrm{eu}}(z_m))^{-1}$,
this proves \eqref{model:good-mollifier} on the specified event.
The number excluded, divided by $X$, is at most
\[
 C\sum_{j=0}^J\sum_{i\ge0}(k^{1/4}+j+i)^{-64}
 \ll (k^{1/4})^{-62}=k^{-31/2}\ll k^{-2},
\]
by Lemma~\ref{model:integer-moments} and
\eqref{model:block-mass}.  The sum may be extended to all $i$ because
only nonempty blocks impose an event.

The real logarithm is available by positivity, and the same local
expansion, including the finitely many small primes separately, gives
for every prime set $\mathcal I$
\begin{equation}
 \log\prod_{p\in\mathcal I}D_p(s,z_m)^{-1}
 =B_{\mathcal I}(z_m)+O\left(\sum_{p\in\mathcal I}p^{-1}\right).
 \label{model:log-product}
\end{equation}
For $Y_j<p\le Y_0$, \eqref{model:prime-mass} bounds this reciprocal
mass by $C(1+j)$, since $\log Y_0/\log Y_j=4^j$.
For $p\le Y_0$ it is at most $C(1+J)$: indeed
$\log\log Y_0\le\log\ell$, and
$\log\ell<10(J+1)\log2$.
Require that the corresponding prime sums have absolute values at
most $(\log2)jk/16$ and $(\log2)Jk/16$, respectively.
Once $k_0$ dominates the fixed drift constants in
\eqref{model:log-product}, these requirements imply
\eqref{model:good-ratio} and \eqref{model:good-full-product}.
Their total exceptional proportion is at most
\[
 C\sum_{j=1}^J\frac{(1+j)^{32}}{(jk)^{64}}
   +C\frac{(1+J)^{32}}{(Jk)^{64}}
 \ll k^{-64}.
\]
Combining the exceptional sets proves the proposition.
\end{proof}

\subsection{Euler-product norms on the full model}

The preceding proposition controls the multipliers on most integers.
We now prove the model norms needed for arithmetic comparison, including
the contribution of every event. We first estimate exact Euler products,
and then replace each complete block by its degree truncation with a
summable relative error.

To avoid any limiting interchange, fix for now a finite real cutoff
$Q\ge Y_0$ and put
\begin{equation}
 \mathcal L_Q(b)=\prod_{p\le Q}D_p(b,Z)^{-1},\qquad \Re b>0.
 \label{model:finite-Euler}
\end{equation}
All bounds in this subsection are independent of $Q$.  At fixed $Q$
the product has the absolutely convergent expansion
\begin{equation}
 \mathcal L_Q(b)=
 \sum_{\substack{n\ge1\\p\mid n\Rightarrow p\le Q}}
       \lambda(n)Z(n)n^{-1/2-b}.
 \label{model:finite-expansion}
\end{equation}
Absolute convergence follows by multiplying the finitely many local
geometric series; their absolute sums are bounded by
$\prod_{p\le Q}(1-p^{-1/2-\Re b})^{-2}$.

\begin{lemma}[Exact shifted products]\label{model:exact-shifts}
Let $0<s,\alpha\le1$, $b=\alpha+it$, and $Y_j\ge2$.  If
$h=\alpha+s+|t|$ and $a=\min(\alpha,s)$, then
\begin{equation}
 \|(P_j^{\mathrm{eu}})^{-1}\mathcal L_Q(b)\|_2
 \ll (1+h/a)^C
       \left(1+\frac1{\alpha\log Y_j}\right)^C.
 \label{model:general-exact-norm}
\end{equation}
At the real point $b=s$, with $j=0$ and the parameters of this section,
\begin{equation}
 \|(P_0^{\mathrm{eu}})^{-1}\mathcal L_Q(s)-1\|_2^2
 \ll \frac{e^{-2\sqrt k}}{\sqrt k}.
 \label{model:exact-distance}
\end{equation}
\end{lemma}

\begin{proof}
We first give the local cancellation with its dependence on $b$.
Write $r=p^{-1/2}$, $u=p^{-s}$, $v=p^{-b}$, and
$M=\min(1,h\log p)$.  The mean-value formula for the exponential and
the trivial bound imply
\begin{equation}
 |u-v|\ll M(u+|v|).
 \label{model:local-difference}
\end{equation}
By \eqref{model:local-bounds}, uniformly for every prime,
\[
 D_p(b,Z)^{-1}-1=O(r|v|).
\]
Subtract the two inverse polynomials and use this last estimate to
obtain
\begin{equation}
 \frac{D_p(s,Z)}{D_p(b,Z)}
 =1+\lambda(p)Z(p)r(v-u)
      +O\big(r^2|v-u|(u+|v|)\big).
 \label{model:local-ratio-expansion}
\end{equation}
For odd $p$, the displayed linear term has expectation zero.  Expanding
the squared absolute value, and using $M^2\le M$ and
$(u+|v|)^2\le2(u^2+|v|^2)$, therefore gives
\begin{equation}
 \mathbb E\left|\frac{D_p(s,Z)}{D_p(b,Z)}\right|^2
 \le \exp\left\{\frac Cp\min(1,h\log p)
                    (p^{-2s}+p^{-2\alpha})\right\}.
 \label{model:local-shift-cost}
\end{equation}
The constants in the error in \eqref{model:local-ratio-expansion} are
uniform even at the small primes, because
$(1-2^{-1/2})^2$ is a common positive lower bound for the denominator.
At the deterministic prime $2$ the local ratio has a fixed upper bound
by \eqref{model:local-bounds}.

For odd primes outside the mollifier, the expansion
\[
 D_p(b,Z)^{-1}=1+\lambda(p)Z(p)p^{-1/2-b}
                       +O(p^{-1-2\alpha})
\]
and centering give
\begin{equation}
 \mathbb E|D_p(b,Z)^{-1}|^2
       \le \exp(Cp^{-1-2\alpha}).
 \label{model:unmatched-local-cost}
\end{equation}
Independence multiplies the local second moments exactly.  Sum
\eqref{model:local-shift-cost} by
\eqref{model:prime-transition}, using both $a=s$ and $a=\alpha$
there, and sum \eqref{model:unmatched-local-cost} by
\eqref{model:prime-tail}.  Taking a square root proves
\eqref{model:general-exact-norm}.

For the distance estimate there is exact cancellation below $Y_0$.
Set
\[
 U=(P_0^{\mathrm{eu}})^{-1}\mathcal L_Q(s)
       =\prod_{Y_0<p\le Q}D_p(s,Z)^{-1},\qquad
 \Xi=\sum_{p>Y_0}p^{-1-2s}.
\]
This is a positive real random variable.  The same expansion gives
\[
 \mathbb E D_p(s,Z)^{-1}=1+O(p^{-1-2s}),\qquad
 \mathbb E D_p(s,Z)^{-2}=1+O(p^{-1-2s}).
\]
The first error may have either sign.  Nevertheless, the elementary
product bound $|\prod(1+e_p)-1|\le\exp(\sum|e_p|)-1$ and
independence show that
\begin{equation}
 \mathbb EU=1+O(\Xi),\qquad \mathbb EU^2=1+O(\Xi).
 \label{model:residual-moments}
\end{equation}
Indeed \eqref{model:prime-tail} and $s\log Y_0=\sqrt k$ give
\[
 \Xi\ll\int_{2\sqrt k}^{\infty}e^{-v}\frac{dv}v
       \ll e^{-2\sqrt k}/\sqrt k,
\]
so $\Xi$ is uniformly small after choosing $k_0$.  Finally,
\[
 \|U-1\|_2^2=\mathbb EU^2-2\mathbb EU+1\ll\Xi.
\]
This uses both moment estimates in \eqref{model:residual-moments} and
proves \eqref{model:exact-distance}.
\end{proof}

\begin{lemma}[Relative truncation in mean square]
\label{model:relative-truncation}
For every $b$ with $\Re b>0$, uniformly in $Q\ge Y_0$,
\begin{equation}
 \|(H_j-(P_j^{\mathrm{eu}})^{-1})\mathcal L_Q(b)\|_2
 \le \left(\exp\left(C\sum_i2^{-m_{ji}}\right)-1\right)
       \|(P_j^{\mathrm{eu}})^{-1}\mathcal L_Q(b)\|_2.
 \label{model:relative-error}
\end{equation}
In particular the relative factor is
$O(2^{-k^{1/4}-j})$.
\end{lemma}

\begin{proof}
For a block $I=I_{ji}$ let
$\mathcal L_I(b)=\prod_{p\in I}D_p(b,Z)^{-1}$.  We claim there are
fixed positive constants $c_*,C_*$ such that
\begin{equation}
 \|K_{ji}(Z,1)\mathcal L_I(b)\|_2\ge c_*,\qquad
 \sup_{|z|=2}\|K_{ji}(Z,z)\mathcal L_I(b)\|_2\le C_*.
 \label{model:block-two-norms}
\end{equation}
For the first assertion, at a large odd prime the exact local ratio is
\[
 \frac{D_p(s,Z)}{D_p(b,Z)}
 =1+\lambda(p)Z(p)p^{-1/2}(p^{-b}-p^{-s})+O(p^{-1}).
\]
The centered linear term shows that its local second moment is
$1+O(p^{-1})$; for all sufficiently large primes it is at least
$1-C/p\ge1/2$.  Their product is bounded below by
$\exp(-2C\sum_{p\in I}1/p)$, which is a fixed positive constant by
\eqref{model:block-mass}.  At each of the finitely many smaller primes,
including $2$, \eqref{model:local-bounds} gives the pointwise lower bound
\[
 \left|\frac{D_p(s,Z)}{D_p(b,Z)}\right|
 \ge\frac{(1-p^{-1/2})^2}{(1+p^{-1/2})^2}>0.
\]
Their finite product is also bounded below.  This proves the first
assertion of \eqref{model:block-two-norms}.

For the second, on $|z|=2$ the local quotient is
\[
 \frac{1-\lambda(p)Z(p)p^{-1/2-s}z
                +\xi_p Z(p)^2p^{-1-2s}z^2}{D_p(b,Z)}
 =1+\lambda(p)Z(p)p^{-1/2}(p^{-b}-zp^{-s})+O(p^{-1})
\]
at large odd primes, uniformly in $z$ and $b$.  Its second moment is
$1+O(p^{-1})$, and therefore its product is bounded above, again by
\eqref{model:block-mass}.  At small primes use the direct numerator
upper bound and \eqref{model:local-bounds}.  The numerator is permitted
to vanish; only an upper bound has been used on this circle.

Apply Cauchy's coefficient formula to the polynomial
$z\mapsto K_{ji}(Z,z)\mathcal L_I(b)$ with values in $L^2$.
For $m=m_{ji}$, Minkowski's inequality and
\eqref{model:block-two-norms} give
\[
 \|(T_{ji}(Z)-K_{ji}(Z,1))\mathcal L_I(b)\|_2
 \le C_*\sum_{h>m}2^{-h}\le C_*2^{-m}.
\]
Dividing by the positive lower bound in
\eqref{model:block-two-norms} makes this a relative error
$\varepsilon_i=C2^{-m_{ji}}$.

To see explicitly how relative errors combine, put
$A_i=K_{ji}(Z,1)\mathcal L_I(b)$ and
$B_i=T_{ji}(Z)\mathcal L_I(b)$.  Then
$\|B_i-A_i\|_2\le\varepsilon_i\|A_i\|_2$ and
$\|B_i\|_2\le(1+\varepsilon_i)\|A_i\|_2$.
Variables belonging to distinct blocks and to the residual primes
$Y_j<p\le Q$ are independent.  In the telescoping identity for
$\prod B_i-\prod A_i$, the norm of each product of factors on distinct
blocks is the product of their norms.  Thus its relative norm, also
including the independent residual product, is at most
\[
 \sum_i\varepsilon_i\prod_{h<i}(1+\varepsilon_h)
   =\prod_i(1+\varepsilon_i)-1
   \le\exp\left(\sum_i\varepsilon_i\right)-1.
\]
This proves \eqref{model:relative-error}, without a factor equal to
the number of blocks.  Finally
$\sum_i2^{-m_{ji}}\le2^{1-k^{1/4}-j}$.
\end{proof}

\begin{proposition}[The shifted norms]\label{model:shifted-norms}
There are fixed $C,c_0>0$ such that, uniformly for every real $\theta$
and every cutoff $Q\ge Y_0$,
\begin{align}
 \|H_0(Z)\mathcal L_Q(s/2+i\theta s)\|_2
       &\ll(1+|\theta|)^C,
 \label{model:half-shift}\\
 \|H_0(Z)\mathcal L_Q(s)-1\|_2
       &\ll e^{-c_0 k^{1/4}},
 \label{model:mollified-distance}\\
 \|H_j(Z)\mathcal L_Q((1+i\theta)/\ell)\|_2
       &\ll (1+|\theta|)^C(k2^j)^C
          \qquad(0\le j\le J).
 \label{model:unit-shift}
\end{align}
\end{proposition}

\begin{proof}
In \eqref{model:general-exact-norm}, first take
$\alpha=s/2$, $t=\theta s$, and $j=0$.
Then $h/\min(\alpha,s)=3+2|\theta|$ and
$\alpha\log Y_0=\sqrt k/2$.  This proves
\eqref{model:half-shift} with the exact multiplier; use
Lemma~\ref{model:relative-truncation} to replace it by $H_0$.
For \eqref{model:unit-shift}, take $\alpha=1/\ell$ and
$t=\theta/\ell$.  Since $k\ge1$,
\[
 h/\min(\alpha,s)=1+k+|\theta|,
 \qquad \alpha\log Y_j=\delta_j^2/\sqrt k.
\]
The bound follows from \eqref{model:general-exact-norm} after increasing
the fixed exponent $C$, and then from relative truncation.
Finally, \eqref{model:exact-distance} bounds the distance from one for
the exact multiplier and also bounds its norm.  The triangle inequality
and \eqref{model:relative-error} give
\[
 \|H_0\mathcal L_Q(s)-1\|_2
 \ll 2^{-k^{1/4}}+e^{-\sqrt k}k^{-1/4}
 \ll e^{-c_0k^{1/4}}.
\]
\end{proof}

\subsection{Fourier representations and the first weight}

We have obtained bounds for the mollified finite Euler products at the
three shifts needed below. Fourier inversion now converts these bounds
into estimates for the actual weights. For the first piece, the main
term is $A_x$. Its estimate combines the power-to-exponential comparison
with the cutoff tail and the exponentially small mollifier error.

We use the Fourier convention
\[
 \widehat g(\theta)=\int_{\mathbb R}g(y)e^{i\theta y}\,dy,
 \qquad
 g(y)=\frac1{2\pi}\int_{\mathbb R}\widehat g(\theta)e^{-i\theta y}
 \,d\theta.
\]
The following elementary bound includes negative real arguments, which
occur inside the gamma integral.

\begin{lemma}[Global approximation of the truncated power]
\label{model:global-power-approximation}
For each fixed nonnegative integer $r$, if $k\ge2r+4$, then for every
$z\in\mathbb R$,
\begin{equation}
 \left|\frac{d^r}{dz^r}
           \left((1-z/k)_+^k-e^{-z}\right)\right|
 \le\frac{C_r}{k}(1+|z|)^2e^{-z}.
 \label{model:global-power-bound}
\end{equation}
\end{lemma}

\begin{proof}
For $z<k$ the derivative of the power is
\[
 (-1)^r\frac{(k)_r}{k^r}(1-z/k)^{k-r},\qquad
 (k)_r=k(k-1)\cdots(k-r+1),
\]
and it is zero for $z\ge k$.  The formula is continuous through $z=k$
in the derivative orders under consideration.  For $z<k$, the logarithm
of its ratio in absolute value to $e^{-z}$ is
\[
 \log((k)_r/k^r)+z+(k-r)\log(1-z/k).
\]
The last two terms have derivative $(r-z)/(k-z)$, so their maximum
over $z<k$ is at $z=r$.  It follows at once that the absolute ratio is
at most $e^r$.  For $|z|\ge\sqrt k$ this bound and
\mbox{$k^{-1}(1+|z|)^2\ge1$} prove \eqref{model:global-power-bound}.
For $|z|<\sqrt k$ we have $|z/k|\le1/2$.  Taylor's formula for
$\log(1-z/k)$ and
$\log((k)_r/k^r)=O_r(k^{-1})$ bound the displayed logarithm by
\[
 O_r\big(k^{-1}(1+|z|)^2\big).
\]
This quantity is bounded in this range; applying
$|e^w-1|\le |w|e^{|w|}$ proves the result there too.  All estimates
hold for negative $z$ as well as positive $z$.
\end{proof}

\begin{lemma}[Fourier error for the first piece]
\label{model:first-Fourier-error}
Define
\[
 A_x=(C_F|x|)^s\Gamma(1+s),\qquad
 E_x(y)=w_{0,x}(y/k)-A_xe^{-y}\quad(y\ge0).
\]
For every required fixed integer $r$, there is an extension $g_x$ of
$e^{y/2}E_x(y)$ from $y\ge0$ to $\mathbb R$ such that
\begin{equation}
 |\widehat g_x(\theta)|\ll_r k^{-1}(1+|\theta|)^{-r}.
 \label{model:first-Fourier-decay}
\end{equation}
The constants are uniform in $x,k,X$ after choosing $k_0$ depending on
$r$.  The quantities $A_x$ are bounded above and below by positive
fixed constants.
\end{lemma}

\begin{proof}
Write $q=C_F|x|$, which lies in a fixed compact subinterval of
$(0,\infty)$.  The exact weight definition gives
\[
 W_{k,x}(y/k)=\int_0^\infty e^{-v}
 \left(1-\frac{y-s\log(qv)}k\right)_+^k\,dv,
\]
whereas
\[
 \int_0^\infty e^{-v}e^{-y+s\log(qv)}\,dv=A_xe^{-y}.
\]
For every fixed $M$, the logarithmic moments
\begin{equation}
 \int_0^\infty e^{-v}(qv)^s
           (1+|s\log(qv)|)^M\,dv\le C_M
 \label{model:gamma-log-moments}
\end{equation}
are uniform for $0<s\le1$: on $0<v\le1$, bound $v^s\le1$ and
$s|\log v|\le|\log v|$; on $v\ge1$, use $v^s\le v$ and the
exponential decay.  Lemma~\ref{model:global-power-approximation} and
differentiation under this dominated integral now imply, for
$0\le h\le r$ and $y\ge-2$,
\begin{equation}
 \left|\frac{d^h}{dy^h}
       \left(W_{k,x}(y/k)-A_xe^{-y}\right)\right|
 \ll_r k^{-1}(1+y^2)e^{-y}.
 \label{model:gamma-power-error}
\end{equation}
In particular the proof uses \eqref{model:global-power-bound} when
$y-s\log(qv)$ is negative and arbitrarily large in absolute value.

The cutoff $F_0(y/k)$ is one for $y\le k/2$, zero for
$y\ge3k/4$, and its $h$th derivative in $y$ is $O_h(k^{-h})$.
Write
\[
 E_x(y)=F_0(y/k)\big(W_{k,x}(y/k)-A_xe^{-y}\big)
                    +(F_0(y/k)-1)A_xe^{-y}
\]
also for $y\ge-2$.  By \eqref{model:gamma-power-error}, after
multiplication by $e^{y/2}$ the first term and its first $r$ derivatives
have $L^1([-2,\infty))$ norm $O_r(k^{-1})$.  The second term is
supported in $y\ge k/2$ and has these norms $O_r(e^{-k/4})$.
Choose a fixed smooth function $\rho$ equal to zero for $y\le-2$ and
one for $y\ge-1$, and extend by
\[
 g_x(y)=\rho(y)e^{y/2}E_x(y),
\]
with zero value to the left of $-2$.  All derivatives through order
$r$ have $L^1(\mathbb R)$ norm $O_r(k^{-1})$.  Integration by parts
$r$ times and the bound without integration by parts give
\eqref{model:first-Fourier-decay}.  Lastly, continuity and positivity
of $q^s\Gamma(1+s)$ on the compact parameter set $q\in[q_-,q_+]$,
$0\le s\le1$, give the two bounds for $A_x$.
\end{proof}

\begin{proposition}[Independent-model estimates for the weight pieces]
\label{model:main-model-estimates}
With the weights and sums $S_{j,Z}(x)$ defined in
Section~\ref{sec:weights}, there is a fixed exponent $C_5$ such
that
\begin{align}
 \|H_0(Z)S_{0,Z}(x)-A_x\|_2&\ll k^{-1},
 \label{model:first-piece-norm}\\
 \|H_j(Z)S_{j,Z}(x)\|_2&\ll(k2^j)^{C_5}\delta_j^k
                    \qquad(1\le j\le J).
 \label{model:later-piece-norm}
\end{align}
These bounds are uniform in $x\in\mathcal X$ and over the entire
range $k_0\le k\le\ell^{3/5}$, with one fixed $k_0$ and one
sufficiently large $X$ threshold.
\end{proposition}

\begin{proof}
Choose a finite cutoff $Q$ exceeding $Y_0$ and the lengths of all the
short polynomials $S_{j,Z}$, $0\le j\le J$.  This is the only
requirement on $Q$; Proposition~\ref{model:shifted-norms} is uniform
in its choice.
For $j\ge1$, set $g_{j,x}(u)=e^u w_{j,x}(u)$.  The support is
contained in
$[1-\delta_j,1-\delta_j/4]\subset[1/2,1]$.
The fixed-order derivative bounds in Lemma~\ref{weights:derivatives}
and integration by parts give, for each required fixed $r$,
\begin{equation}
 |\widehat g_{j,x}(\theta)|
 \ll_r (k/\delta_j)^{C_r}\delta_j^k(1+|\theta|)^{-r}.
 \label{model:later-Fourier-decay}
\end{equation}
Fourier inversion and \eqref{model:finite-expansion} give the exact
identity
\begin{equation}
 S_{j,Z}(x)=\frac1{2\pi}\int_{\mathbb R}
      \widehat g_{j,x}(\theta)
      \mathcal L_Q((1+i\theta)/\ell)\,d\theta.
 \label{model:later-Fourier-representation}
\end{equation}
Indeed every nonzero weight coefficient has index below the chosen
cutoff and hence occurs in \eqref{model:finite-expansion}; all other
indices contribute zero after inversion.  At this fixed finite $Q$,
absolute convergence of the Dirichlet expansion and the integrable
kernel justify interchanging the sum and integral.  Multiply by $H_j$
and use Minkowski's inequality, \eqref{model:unit-shift}, and
\eqref{model:later-Fourier-decay}, taking $r>C+2$.  Their integral
is $O((k/\delta_j)^{C_r}(k2^j)^C\delta_j^k)$, which proves
\eqref{model:later-piece-norm} after fixing $C_5$.

For the first piece, Lemma~\ref{model:first-Fourier-error} and the
same finite-product expansion give
\begin{equation}
 S_{0,Z}(x)-A_x\mathcal L_Q(s)
 =\frac1{2\pi}\int_{\mathbb R}\widehat g_x(\theta)
                    \mathcal L_Q(s/2+i\theta s)\,d\theta.
 \label{model:first-Fourier-representation}
\end{equation}
To check the scaling, at an integer index $n$ put
$y=k\log n/\ell=s\log n\ge0$.  Fourier inversion for $g_x$ then
reads
\[
 w_{0,x}(\log n/\ell)-A_x n^{-s}
 =\frac1{2\pi}\int\widehat g_x(\theta)
                          n^{-s/2-i\theta s}\,d\theta,
\]
which is exactly the coefficient identity needed in
\eqref{model:first-Fourier-representation}.
Use \eqref{model:half-shift} and
\eqref{model:first-Fourier-decay} with $r>C+2$ to bound the norm of
\eqref{model:first-Fourier-representation} after multiplication by
$H_0$ by $O(k^{-1})$.  By \eqref{model:mollified-distance},
\[
 \|H_0S_{0,Z}-A_x\|_2
 \le \|H_0(S_{0,Z}-A_x\mathcal L_Q(s))\|_2
        +A_x\|H_0\mathcal L_Q(s)-1\|_2
 \ll k^{-1}+e^{-c_0k^{1/4}}\ll k^{-1}.
\]
This proves \eqref{model:first-piece-norm}.

All smoothness orders used here are fixed after the absolute exponent
in Proposition~\ref{model:shifted-norms}, and before choosing $k_0$.
The derivative estimates for the weights, the inequality
$s\le\ell^{-2/5}$, and the lower bound
$\delta_J\ge\ell^{-1/10}$ then give a single $X$ threshold for the
whole parameter range.  We have not used the event of
Proposition~\ref{model:good-event} anywhere in these model norm
arguments.  Nor is a passage to an infinite Euler cutoff required:
the coefficient identities hold for every sufficiently large finite
$Q$, and all norm estimates are uniform in that cutoff.
\end{proof}
==== END The-mean-analytic-rank-of-quadratic-twists-of-elliptic-curves//sections/model.tex ====

==== BEGIN The-mean-analytic-rank-of-quadratic-twists-of-elliptic-curves//sections/setup.tex ====
\section{Twist families and the integer model}\label{sec:setup}

We first reduce upper bounds for arbitrary twists to a fixed finite
family of curves with coprime twisting parameters. This gives exact
conductor formulas while preserving the bounds for nonnegative sums
needed later. We then define the probability model for complete integer
residue averages.

We use unitary normalization:
\[
 L_F(s)=L(F,s+1/2)=\sum_{n\ge1}\lambda_F(n)n^{-s},
 \qquad q_F=\operatorname{cond}(F),
 \qquad C_F=\frac{\sqrt{q_F}}{2\pi}.
\]
The series converges absolutely in \(\Re s>1\).
Modularity and the local rules for the Hasse--Weil \(L\)-function give
real multiplicative coefficients and inverse Euler factors
\begin{equation}\label{eq:setup-euler}
 1-\lambda_F(p)p^{-s}+\xi_{F,p}p^{-2s},
 \qquad \xi_{F,p}=\mathbf 1_{p\nmid q_F}.
\end{equation}
At a good prime the two factor parameters have modulus one by Hasse's
bound \cite[Chapter~V, Theorem~2.3.1(a)]{Silverman2009}.
At a multiplicative prime there is one parameter, of modulus
\(p^{-1/2}\), and at an additive prime the factor is one
\cite[Definitions~17--18 and Theorem~19]{DokchitserAnni2014}.
In particular,
\begin{equation}\label{eq:setup-coefficients}
 |\lambda_F(n)|\le\tau(n).
\end{equation}
The completed function
\[
 \Lambda_F(s)=C_F^s\Gamma(s+1/2)L_F(s)
\]
is entire of finite order and satisfies
\begin{equation}\label{eq:setup-fe}
 \Lambda_F(s)=\epsilon_F\Lambda_F(1-s),\qquad \epsilon_F\in\{1,-1\}.
\end{equation}
Analytic continuation and the functional equation follow from
modularity \cite[Theorem~A]{BreuilConradDiamondTaylor2001}
and the standard completion
\cite[Appendix~C, Theorems~16.3--16.4]{Silverman2009}.
The standard finite-order bounds follow, for example, from the split
Mellin integral of the modular cusp form and its Fricke transform.
They also give polynomial growth in every fixed vertical strip for each
fixed curve \(F\), independently of the averaged estimates proved below.
Indeed, enclose the strip in \(a\le\Re s\le b\), with \(a\le-1\) and
\(b\ge2\). Absolute convergence bounds \(L_F\) on the right boundary;
\eqref{eq:setup-fe} and Stirling's formula give
\(L_F(a+it)\ll_{F,a}(1+|t|)^{1-2a}\) on the left boundary.
Take \(A=1-a\) and an integer \(N>1-2a\), so that
\(L_F(s)/(s+A)^N\) is bounded on both boundaries. For \(\varepsilon>0\) and \(0<\kappa<\pi/(b-a)\), multiply
this quotient by
\[
 \exp\left\{-\varepsilon\cos\left(\kappa
              \left(s-\frac{a+b}{2}\right)\right)\right\}.
\]
The multiplier has modulus at most one on the vertical boundaries and
at most \(\exp(-c\varepsilon\cosh(\kappa\Im s))\) on the horizontal
sides, for a fixed \(c>0\).
This dominates the finite-order growth. The maximum principle on
expanding rectangles, followed by \(\varepsilon\downarrow0\), therefore
gives \(L_F(\sigma+it)\ll_{F,a,b}(1+|t|)^N\) on the strip.
These individual bounds justify contour displacements. Their constants
may depend on the fixed curve; estimates uniform in the varying twist
parameters will be proved separately.

\subsection{A finite family closed under reduction}

Fix the primes
\[
 S=\{p:p\mid 2q_E\},\qquad Q_S=\prod_{p\in S}p.
\]
Let \(\mathcal F\) be the finite set, up to \(\Q\)-isomorphism, of curves
obtained from \(E\) by twisting by signed squarefree products of primes
in \(S\). The curve \(E\) is included. Every \(F\in\mathcal F\) has good
reduction outside \(S\).

Call a signed squarefree integer \(d\) \emph{admissible} if
\[
 d\equiv1\pmod4,\qquad (d,Q_S)=1.
\]
For a dyadic \(X\), put
\[
 \mathcal A(X)=\{d:d\text{ admissible},\ X\le|d|<2X\}.
\]

\begin{lemma}\label{setup:reduction}
For every \(F\in\mathcal F\) and signed squarefree \(b\), there are
\(F'\in\mathcal F\) and an admissible \(d\) such that
\[
 F^{(b)}\simeq (F')^{(d)},\qquad |d|\le |b|.
\]
The correspondence can be chosen to have bounded multiplicity, uniformly
in \(b\). Every sum of nonnegative functions of these twists can
therefore be bounded by a fixed multiple of the corresponding sums over
\(\mathcal F\) and admissible parameters of no greater height.
\end{lemma}

\begin{proof}
Write \(b=b_S b_0\), where \(b_S>0\) is the product of the primes of \(S\)
dividing \(b\). Then \(b_0\) is odd and prime to \(Q_S\).
Choose \(\varepsilon\in\{1,-1\}\) so that
\(d=\varepsilon b_0\equiv1\pmod4\), and put
\(F'=F^{(\varepsilon b_S)}\).
Quadratic twisting depends only on the square class and composes by
multiplication of square classes. Thus \(F'\) is in \(\mathcal F\) and
\((F')^{(d)}\simeq F^{(b)}\).
Moreover \(|d|=|b|/b_S\le |b|\).
There are at most \(2^{|S|+1}\) choices of the signed \(S\)-part.
Possible coincidences among the finitely many curves only change this
fixed multiplicity.
\end{proof}

\begin{lemma}\label{setup:local-factors}
For \(F\in\mathcal F\) and admissible \(d\),
\begin{equation}\label{eq:setup-twist}
 L_{F^{(d)}}(s)=\sum_{n\ge1}\lambda_F(n)\chi_d(n)n^{-s},
 \qquad q_{F^{(d)}}=q_F|d|^2,
 \qquad \chi_d(n)=\left(\frac dn\right).
\end{equation}
Here the symbol is the Kronecker symbol and \(d=1\) has the untwisted
convention. For an arbitrary signed squarefree \(b\), the same good-prime
twisting rule holds at primes outside \(S\): the local trace is multiplied
by \((b/p)\) when \(p\nmid b\), and the local factor is one when \(p\mid b\).
\end{lemma}

\begin{proof}
We compute the inverse local Euler factors using geometric Frobenius
on the inertia invariants of the dual of the rational Tate module,
and the conductor as the Artin conductor
\cite[Definitions~17--18 and Theorem~19]{DokchitserAnni2014}.
The rational Tate-module representation of a quadratic twist is the
original representation tensored with the corresponding quadratic
character. This follows from the isomorphism over the twisting extension,
whose conjugate differs by the scalar automorphism \([-1]\).

An unramified scalar twist preserves the conductor and multiplies
Frobenius eigenvalues on inertia invariants by the character value.
Since an admissible \(d\) is an odd fundamental discriminant, its character
is unramified at every prime of \(S\).
At a prime \(p\mid d\), the representation of \(F\) is unramified and the
quadratic character is tamely ramified. The tensor product has no inertia
invariants, Swan conductor zero, and Artin conductor exponent two.
At all other primes the scalar twist is unramified.
Computing the inertia-invariant Euler factors and the conductor gives
\eqref{eq:setup-twist}.
For an arbitrary squarefree \(b\), these same computations apply at odd
primes outside \(S\), regardless of the character's behavior at two.
\end{proof}

The constants in what follows are chosen uniformly over the finite set
\(\mathcal F\).
The nonincreasing height in Lemma~\ref{setup:reduction} is useful in the
second-moment induction: a dual squarefree parameter of absolute value
at most \(M/2\) remains in a smaller dyadic block after reduction.

\subsection{Complete integer averages}

Fix \(c\in\{1,-1\}\). For \(m\in\Z\), define a completely multiplicative
formal character on positive integers by
\[
 z_m(2)=c,\qquad z_m(p)=\left(\frac mp\right)
 \quad(p\text{ odd}).
\]
Let \(Z\) be the completely multiplicative random model with \(Z(2)=c\)
and independent odd-prime values distributed by
\begin{equation}\label{eq:setup-model}
 \Prob(Z(p)=0)=\frac1p,\qquad
 \Prob(Z(p)=1)=\Prob(Z(p)=-1)=\frac{1-1/p}{2}.
\end{equation}
For every polynomial involving only finitely many prime values,
its expectation is the average over a complete residue system for the
corresponding odd-prime periods. This follows from the Chinese remainder
theorem and the equal numbers of nonzero quadratic residues and
nonresidues modulo each odd prime.

We use this model for averages over all integers \(m\).
Nonnegative estimates can subsequently be restricted to admissible
squarefree \(d\), separated into the two classes \(\chi_d(2)=c\);
on either class, \(z_d(n)=\chi_d(n)\) for every \(n\).
==== END The-mean-analytic-rank-of-quadratic-twists-of-elliptic-curves//sections/setup.tex ====

==== BEGIN The-mean-analytic-rank-of-quadratic-twists-of-elliptic-curves//sections/tail.tex ====
\section{Nonvanishing and the rank-weighted tail}\label{sec:tail}

We now combine the transferred model estimates, the integer product
event, and the terminal mean square to count large orders of vanishing.
A separate bound for each individual rank then converts this count into
the rank-weighted tail. The companion density theorem enters only in
the final subsection, after the tail theorem is proved.
The constants and the fixed
lower threshold \(k_0\) are increased finitely many times in this section.
All choices depend only on the fixed family and the already fixed
smoothness orders.

\Needspace{11\baselineskip}
\subsection{Counting large orders of vanishing}

\begin{proposition}\label{tail:count}
There are \(k_0\), \(X_0\), and \(C\), depending on \(E\), such that for
every \(F\in\mathcal F\), every dyadic \(X\ge X_0\), and every integer
\[
 k_0+1\le k\le(\log X)^{3/5},
\]
one has
\[
 \#\{d\in\mathcal A(X):a(F^{(d)})>k\}\le C Xk^{-2}.
\]
\end{proposition}

\begin{proof}
Write \(\ell=\log X\), and first consider the functional-equation sign
\((-1)^k\), with \(k_0\le k\le\ell^{3/5}\).
Fix \(c=\pm1\), and use the weights and partition of
Section~\ref{sec:weights}. We write \(P_j^{\rm eu}\) for the positive
Euler products of Section~\ref{sec:model}.

Corollary~\ref{weights:comparison-interface} and
Lemma~\ref{model:mollifier-length} permit
Proposition~\ref{analytic:comparison} to be applied to \(H_jS_j\),
for \(0\le j\le J\), with \(\Delta=\delta_j/4\).
By the first assertion of Proposition~\ref{model:main-model-estimates},
outside \(O(X/k^2)\) integers in \(X\le|m|<2X\),
\begin{equation}\label{eq:tail-main-good}
 |H_0(z_m)S_{0,z_m}(m/X)-A_{m/X}|<\frac12 A_{m/X}.
\end{equation}
Indeed \(A_x=(C_F|x|)^s\Gamma(1+s)\) is bounded above and below by positive
constants on the fixed \(x\)-range, uniformly for \(s=k/\ell\le\ell^{-2/5}\).
Chebyshev's inequality applies to the transferred \(O(k^{-2})\)
mean-square error. The exponentially small comparison error is absorbed
uniformly.

For \(1\le j\le J\), the second model estimate gives
\[
 \frac1X\sum_m\Psi(m/X)|H_j(z_m)S_{j,z_m}(m/X)|^2
 \ll (k2^j)^{2C_5}2^{-2jk}+e^{-c\delta_j\ell}
 \ll 2^{-3jk/2}.
\]
The last inequality follows first by increasing fixed \(k_0\) to absorb
the polynomial factor. For the comparison error, use
\[
 \delta_j\ell\ge\ell^{9/10},\qquad
 jk\ll\ell^{3/5}\log\ell,
\]
and then increase one \(X_0\) for all the permitted \(k,j\).
The terminal estimate of Lemma~\ref{weights:terminal-small} gives
\[
 \frac1X\sum_{d\in\mathcal A(X)}
       |S_{*,\chi_d}(d/X)|^2\ll 2^{-3Jk/2}.
\]
Chebyshev and summation of the resulting geometric probabilities imply,
outside \(O(X/k^2)\) further admissible parameters,
\begin{equation}\label{eq:tail-other-good}
 |H_j(\chi_d)S_{j,\chi_d}(d/X)|\le2^{-jk/2}\quad(1\le j\le J),
 \qquad
 |S_{*,\chi_d}(d/X)|\le2^{-Jk/2}.
\end{equation}
Here \(\sum_{j\ge1}2^{-jk/2}\ll k^{-2}\) after fixing \(k_0\), and \(J\ge1\)
for sufficiently large \(X\).

Intersect these events with the event in
Proposition~\ref{model:good-event}. Its additional exceptional set is
also \(O(X/k^2)\). On the intersection,
\[
 \frac12\le H_jP_j^{\rm eu}\le2,\qquad
 P_j^{\rm eu}\le 2^{jk/8}P_0^{\rm eu},\qquad
 1\le 2^{Jk/8}P_0^{\rm eu}.
\]
Choose a fixed \(a_*>0\) with \(A_x\ge a_*\) for all the parameters.
Equation~\eqref{eq:tail-main-good} yields
\[
 |S_{0,\chi_d}(d/X)|\ge\frac{a_*}{4}P_0^{\rm eu}.
\]
Equation~\eqref{eq:tail-other-good} yields
\[
 \sum_{j=1}^J|S_{j,\chi_d}(d/X)|+|S_{*,\chi_d}(d/X)|
 \le P_0^{\rm eu}
     \left(2\sum_{j=1}^J2^{-3jk/8}+2^{-3Jk/8}\right).
\]
For a sufficiently large fixed \(k_0\), the last parenthesis is less
than \(a_*/4\). The entire weighted sum is therefore nonzero.
Lemma~\ref{weights:vanishing} excludes the simultaneous conditions
\(a(F^{(d)})>k\) and \(\epsilon_{F^{(d)}}=(-1)^k\).
Including both values of \(c\), we have proved
\begin{equation}\label{eq:tail-parity-count}
 \#\{d\in\mathcal A(X):
       a(F^{(d)})>k,\ \epsilon_{F^{(d)}}=(-1)^k\}
       \ll Xk^{-2}.
\end{equation}
For the other functional-equation sign, apply
\eqref{eq:tail-parity-count} at \(k-1\). A rank exceeding \(k\) also exceeds
\(k-1\), and \((k-1)^{-2}\ll k^{-2}\).
This proves the proposition.
\end{proof}

\subsection{A uniform bound for an individual rank}

\begin{lemma}\label{tail:jensen}
Uniformly for \(F\in\mathcal F\) and \(d\in\mathcal A(X)\),
\[
 a(F^{(d)})\ll_E\log(3X).
\]
\end{lemma}

\begin{proof}
Put \(L=L_{F^{(d)}}\) and \(C_d=C_F|d|\).
On \(\Re s=5\), absolute convergence and
\eqref{eq:setup-coefficients} give a uniform bound.
On \(\Re s=-1\), the functional equation is
\[
 L(s)=\epsilon_{F^{(d)}} C_d^{\,1-2s}
          \frac{\Gamma(3/2-s)}{\Gamma(1/2+s)}L(1-s).
\]
At \(s=-1+it\), the modulus of the gamma quotient equals
\[
 |(-1/2+it)(1/2+it)(3/2+it)|.
\]
Since \(L(2-it)\) is uniformly bounded,
\[
 |L(-1+it)|\ll_E X^3(1+|t|)^3.
\]
Apply the strip principle to \(L(s)/(s+2)^4\) on
\(-1\le\Re s\le5\). To see that the possibly conductor-dependent
finite-order growth in the strip introduces no new constant, first
multiply by
\[
 \exp\!\left(-\varepsilon
       \cos\frac{\pi(s-2)}{12}\right),\qquad \varepsilon>0.
\]
Its modulus is at most one on this strip and decays faster than any
finite-order growth as \(|\Im s|\to\infty\).
The maximum principle on expanding rectangles gives the bound from
the two vertical sides. Letting \(\varepsilon\) decrease to zero shows
that \(L(s)/(s+2)^4\ll_E X^3\) throughout the strip.
In particular, the maximum of \(|L|\) on the closed disk
\(|s-2|\le5/2\) is \(O_E(X^3)\).

At the center, the Euler factors and their parameter bounds give
\[
 |L(2)|\ge \prod_p(1+p^{-2})^{-2}
          =\left(\frac{\zeta(4)}{\zeta(2)}\right)^2>0.
\]
Jensen's formula, with inner radius \(2\) and outer radius \(5/2\),
therefore bounds the number of zeros in the inner disk, counted with
multiplicity, by \(O_E(\log(3X))\).
The point \(1/2\) lies in that inner disk. This proves the result.
\end{proof}

\subsection{Summing the integer tails}

\begin{proof}[Proof of Theorem~\ref{thm:tail}]
We first work in \(\mathcal A(X)\) for a fixed \(F\in\mathcal F\).
Let
\[
 T_{F,X}(k)=\#\{d\in\mathcal A(X):a(F^{(d)})>k\},
 \qquad K=\lfloor(\log X)^{3/5}\rfloor.
\]
For an integer \(R\ge k_0+1\) and \(X\) large enough that \(K>R\),
the elementary integer-tail bound gives
\begin{align}
 \sum_{\substack{d\in\mathcal A(X)\\a(F^{(d)})>R}}a(F^{(d)})
 &\le R T_{F,X}(R)
       +\sum_{k=R}^{K-1}T_{F,X}(k)
       +C_E\log X\,T_{F,X}(K) \notag\\
 &\ll_E \frac X R+X(\log X)^{-1/5}.
 \label{eq:tail-dyadic-mass}
\end{align}
For ranks between \(R+1\) and \(K\), the first two terms count the rank
exactly. For ranks exceeding \(K\), the last term bounds the remaining
mass by Lemma~\ref{tail:jensen}.
Proposition~\ref{tail:count} supplies every counting estimate used here.
In particular the last term is
\(O_E(X\log X/K^2)=O_E(X(\log X)^{-1/5})\).
The constant in \eqref{eq:tail-dyadic-mass} is independent of \(R\).

For completeness, fix \(R\) and cover \(0<|d|\le Y\) by dyadic blocks.
Enlarge the last partial block to a full block, using nonnegativity.
The sum of their left endpoints is less than \(2Y\).
The finitely many blocks below the threshold for
\eqref{eq:tail-dyadic-mass} contribute a finite number \(B_{E,R}\).
For the error term in the remaining blocks, split at \(X=\sqrt Y\).
The smaller blocks contribute \(O_E(\sqrt Y)\), and in the larger ones
\(\log X\ge\frac12\log Y\).
Consequently
\[
 \sum_{\substack{d\ {\rm admissible}\\0<|d|\le Y\\a(F^{(d)})>R}}
     a(F^{(d)})
 \le B_{E,R}+\frac{C_EY}{R}
       +O_E\bigl(\sqrt Y+Y(\log Y)^{-1/5}\bigr).
\]
After division by \(Y\) and passage to the limsup, the desired
\(C_E/R\) bound holds for each \(F\in\mathcal F\).
Lemma~\ref{setup:reduction}, its bounded multiplicity, and the finite
maximum of these constants prove the same bound for every original
signed squarefree parameter of \(E\).
\end{proof}

\subsection{The mean}

\begin{proof}[Proof of Theorem~\ref{thm:mean}]
Write \(N(Y)=\#\mathcal D(Y)\) and let \(N_j(Y)\) count the parameters
of analytic rank \(j\).
The elementary squarefree sieve gives
\[
 N(Y)=2\sum_{n\le Y}\mu(n)^2
     =\frac{2Y}{\zeta(2)}+O(\sqrt Y).
\]
Indeed, expand \(\mu(n)^2=\sum_{r^2\mid n}\mu(r)\), interchange the
finite sums, and bound the omitted tail of
\(\sum_r\mu(r)/r^2=1/\zeta(2)\).

By the analytic density theorem \eqref{eq:density-input},
\(N_0(Y)/N(Y)\to1/2\) and \(N_1(Y)/N(Y)\to1/2\).
Thus \(N(Y)-N_0(Y)-N_1(Y)=o(Y)\).
For every fixed integer \(R\ge R_E\),
\[
 \sum_{\substack{d\in\mathcal D(Y)\\2\le a(E^{(d)})\le R}}
        a(E^{(d)})
 \le R\bigl(N(Y)-N_0(Y)-N_1(Y)\bigr)=o_{E,R}(Y).
\]
Theorem~\ref{thm:tail} now implies
\[
 \limsup_{Y\to\infty}\frac1Y
 \sum_{\substack{d\in\mathcal D(Y)\\a(E^{(d)})\ge2}}a(E^{(d)})
 \le\frac{C_E}{R}.
\]
Let \(R\to\infty\). The left side is zero by nonnegativity.
The total rank sum is consequently \(N_1(Y)+o_E(Y)\).
Dividing by \(N(Y)\) proves the limit \(1/2\).
\end{proof}
==== END The-mean-analytic-rank-of-quadratic-twists-of-elliptic-curves//sections/tail.tex ====

==== BEGIN The-mean-analytic-rank-of-quadratic-twists-of-elliptic-curves//sections/weights.tex ====
\section{Completed derivatives and their weights}
\label{sec:weights}

A large analytic rank forces a completed derivative to vanish. For the
matching functional-equation sign, we express that vanishing as an exact
weighted-series identity, partition the series into short pieces, and
bound the remaining infinite piece using the second moments of
Section~\ref{sec:analytic}.
The short pieces will be treated by mollification in
Section~\ref{sec:model}.

Fix a curve \(F\) in the finite family \(\mathcal F\), write
\(\lambda=\lambda_F\), and put \(C_F=\sqrt{q_F}/(2\pi)\).
The variable \(x\) will range over a fixed compact subset
\(\mathcal K\) of \(\mathbb R\setminus\{0\}\), chosen large enough to
contain the support of the averaging weight in
Proposition~\ref{analytic:comparison}. Derivative bounds in \(x\) are
understood on a fixed compact neighborhood of \(\mathcal K\) disjoint
from zero. All constants below may depend on this compact set and on
the fixed finite family.

The order of choices is as follows. First fix the exponents in
Proposition~\ref{analytic:polynomial-moment}, and all the finite orders of
differentiation needed for Proposition~\ref{analytic:comparison}, the
Mellin inversion below, and the subsequent Fourier estimates. Next
choose a sufficiently large fixed integer \(k_0\). Finally choose one
lower bound for \(X\), valid simultaneously for all the integers
\begin{equation}
 \ell=\log X,\qquad
 k_0\le k\le \lfloor\ell^{3/5}\rfloor .
 \label{weights:range}
\end{equation}
The estimates in this section respect this order. In particular, the
number of derivatives is never allowed to increase with \(k\) or \(X\).

\subsection{The exact completed derivative identity}

For \(k\ge1\), \(x\ne0\), and real \(u\), define
\begin{equation}
 W_{k,x}(u)=
 \int_0^\infty e^{-v}
 \left(1-u+\frac{\log(C_F|x|v)}{\ell}\right)_+^k\,dv,
 \qquad y_+=\max(y,0).
 \label{weights:W}
\end{equation}
For every fixed choice of the parameters the integral is finite.
The weighted sum below is an infinite, absolutely convergent smoothed
Dirichlet series; its individual short pieces will be finite sums.

\begin{lemma}[Completed derivative identity]
\label{weights:vanishing}
Let \(d\) be admissible, \(X\le |d|<2X\), and \(x=d/X\). Set
\[
 g_d(z)=L_{F^{(d)}}(1/2+z)(C_F|d|)^z\Gamma(1+z),
 \qquad \epsilon_d=\operatorname{sign}(F^{(d)}).
\]
Then \(g_d\) is entire, \(g_d(-z)=\epsilon_d g_d(z)\), and
\begin{equation}
 \begin{split}
 \mathcal S_d
 &:=
 \sum_{n\ge1}\frac{\lambda(n)\chi_d(n)}{\sqrt n}
                 W_{k,x}(\ell^{-1}\log n)\\
 &=\frac{k!}{\ell^k}\frac{1}{2\pi i}
       \int_{(2)}g_d(z)\frac{dz}{z^{k+1}}.
 \end{split}
 \label{weights:completed-integral}
\end{equation}
More precisely,
\begin{equation}
 \bigl(1+\epsilon_d(-1)^k\bigr)\mathcal S_d
       =\ell^{-k}g_d^{(k)}(0).
 \label{weights:exact-derivative}
\end{equation}
Consequently, if \(a(F^{(d)})>k\) and
\(\epsilon_d=(-1)^k\), then
\begin{equation}
 \sum_{n\ge1}\frac{\lambda(n)\chi_d(n)}{\sqrt n}
                  W_{k,d/X}(\ell^{-1}\log n)=0.
 \label{weights:zero}
\end{equation}
\end{lemma}

\begin{proof}
The completed function is
\[
 \Lambda_d(s)=(C_F|d|)^s\Gamma(s+1/2)L_{F^{(d)}}(s).
\]
Thus \(g_d(z)=(C_F|d|)^{-1/2}\Lambda_d(1/2+z)\).
Its entireness and reflection law follow from those of \(\Lambda_d\).

For \(T>0\), Mellin inversion gives
\[
 \frac{1}{2\pi i}\int_{(2)}\frac{T^z}{z^{k+1}}\,dz
       =\frac{(\log T)_+^k}{k!}.
\]
Insert the absolutely convergent Dirichlet series for
\(L_{F^{(d)}}(1/2+z)\), and use
\(\Gamma(1+z)=\int_0^\infty e^{-v}v^z\,dv\).
On this line all exchanges are absolutely convergent: the series is
dominated by \(\sum_n\tau(n)n^{-5/2}\), and the other two factors to be
integrated are dominated by
\(e^{-v}v^2\) and \(|2+it|^{-k-1}\).
The result is \eqref{weights:completed-integral}, because
\[
 \ell^{-1}\log(C_F|d|v/n)
       =1-\ell^{-1}\log n+\ell^{-1}\log(C_F|x|v).
\]
This argument, or the same computation with absolute coefficient
values, also proves absolute convergence of the series.

Let \(I_+\) and \(I_-\) denote the integrals of
\(g_d(z)z^{-k-1}/(2\pi i)\) on the upward oriented lines
\(\operatorname{Re}z=2\) and \(\operatorname{Re}z=-2\).
Shifting between these lines crosses only the pole at zero, and gives
\[
 I_+-I_-=\frac{g_d^{(k)}(0)}{k!}.
\]
The horizontal integrals tend to zero: in this fixed strip the
uncompleted \(L\)-function has polynomial growth, while the gamma
factor has exponential decay on horizontal segments tending to
infinity. Apparent gamma poles on the real axis are removable in
\(g_d\). Changing \(z\) to \(-z\), with orientations included, gives
\[
 I_-=(-1)^{k+1}\epsilon_d I_+.
\]
Multiplication by \(k!/\ell^k\) proves
\eqref{weights:exact-derivative}. Finally, the completing factor is
holomorphic and nonzero at zero. Hence \(a(F^{(d)})>k\) implies
\(g_d^{(k)}(0)=0\), and the stipulated sign makes the coefficient on
the left of \eqref{weights:exact-derivative} equal to two.
\end{proof}

\subsection{Partition and derivative estimates}

Fix a \(C^\infty\), nondecreasing function \(f_*\colon\mathbb R\to[0,1]\)
which is zero on \((-\infty,1/4]\) and one on \([1/2,\infty)\).
Define
\begin{equation}
 J=\left\lfloor\frac{\log\ell}{10\log2}\right\rfloor,\qquad
 \delta_j=2^{-j},\qquad
 F_j(u)=f_*((1-u)/\delta_j)\quad(0\le j\le J).
 \label{weights:scales}
\end{equation}
In particular, for sufficiently large \(X\),
\begin{equation}
 J\ge1,\qquad
 \ell^{-1/10}\le\delta_J<2\ell^{-1/10},\qquad
 \frac{k}{\ell\delta_j}\le\ell^{-3/10}
       \quad(0\le j\le J).
 \label{weights:scale-bounds}
\end{equation}
Set
\begin{equation}
 \begin{aligned}
 w_{0,x}(u)&=F_0(u)W_{k,x}(u),\\
 w_{j,x}(u)&=(F_j(u)-F_{j-1}(u))W_{k,x}(u)
                    &&(1\le j\le J),\\
 w_{*,x}(u)&=(1-F_J(u))W_{k,x}(u).
 \end{aligned}
 \label{weights:partition}
\end{equation}
These weights sum exactly to \(W_{k,x}\). For \(V=\chi_d,z_m\), or \(Z\),
write
\begin{equation}
 \begin{aligned}
 S_{j,V}(x)&=\sum_{n\ge1}
       \frac{\lambda(n)V(n)}{\sqrt n}w_{j,x}(\ell^{-1}\log n),\\
 S_{*,V}(x)&=\sum_{n\ge1}
       \frac{\lambda(n)V(n)}{\sqrt n}w_{*,x}(\ell^{-1}\log n).
 \end{aligned}
 \label{weights:S}
\end{equation}
The first definition includes \(j=0\). All the formal character
values in these definitions have modulus at most one. The sums
through \(j=J\) have length at most \(X^{1-\delta_j/4}\), and
\begin{equation}
 \operatorname{supp} w_{j,x}
       \subset[1-\delta_j,1-\delta_j/4]\quad(1\le j\le J).
 \label{weights:short-support}
\end{equation}
Indeed \(F_j-F_{j-1}\) vanishes when \(1-u\le\delta_j/4\) or
\(1-u\ge\delta_j\). The terminal weight is supported in
\(u\ge1-\delta_J/2\).

\begin{lemma}[Fixed-order weight bounds]
\label{weights:derivatives}
Fix an integer \(r\ge0\), and then take \(k_0\ge r+2\).
For \(a,b\ge0\), \(a+b\le r\), uniformly in \eqref{weights:range},
\(x\in\mathcal K\), and the indicated arguments, one has
\begin{align}
 |\partial_u^a\partial_x^b w_{j,x}(u)|
   &\le C_r(k/\delta_j)^r\delta_j^k
                  &&(1\le j\le J,\ u\in\mathbb R), \label{weights:short-derivatives}\\
 |\partial_u^a\partial_x^b w_{0,x}(u)|
   &\le C_r k^r
                  &&(u\ge-\log2/\ell). \label{weights:main-derivatives}
\end{align}
There is also \(c>0\), independent of \(r,k,X\), such that, with
\(\delta=\delta_J\) and \(n=e^{\ell u}>0\),
\begin{equation}
 |\partial_u^a\partial_x^b w_{*,x}(u)|
       \le C_r(k/\delta)^r\delta^k e^{-cn/X}.
 \label{weights:terminal-derivatives}
\end{equation}
Changing the fixed compact neighborhood of \(\mathcal K\) changes
only the constants.
\end{lemma}

\begin{proof}
It is convenient first to differentiate using
\(D_x=x\partial_x\). For
\[
 h=1-u+\ell^{-1}\log(C_F|x|v)
\]
and \(a+b\le r\), differentiation under the integral gives
\begin{equation}
 \partial_u^aD_x^b h_+^k
    =(-1)^a\ell^{-b}(k)_{a+b}h_+^{k-a-b},
 \label{weights:power-derivative}
\end{equation}
where \((k)_t=k(k-1)\cdots(k-t+1)\).
The choice \(k\ge r+2\) ensures enough continuous derivatives at
\(h=0\); domination for differentiating follows from the estimates
below. Ordinary \(x\)-derivatives are fixed linear combinations of
\(x^{-b}D_x^i\), \(i\le b\), and therefore satisfy the same bounds on
our compact set.

Suppose \(1-u\le\delta'\), where \(\delta'>0\). For \(0\le t\le r\),
write \(m=k-t>0\) and \(a_t=m/(\ell\delta')\). The elementary
inequality \((1+q)_+^m\le e^{mq}\), valid for every real \(q\), gives
\begin{equation}
 h_+^m
    \le(\delta')^m(C_F|x|v)^{a_t}.
 \label{weights:gamma-majorant}
\end{equation}
If \(\delta'=\delta_j\), then \(0<a_t\le\ell^{-3/10}\) by
\eqref{weights:scale-bounds}, so
\[
 \int_0^\infty e^{-v}(C_F|x|v)^{a_t}\,dv
       =(C_F|x|)^{a_t}\Gamma(1+a_t)\le C.
\]
The bound is uniform once \(X\) is large enough that \(a_t\le1\).
Thus, on \(1-u\le\delta_j\), a derivative of \(W_{k,x}\) involving at
most \(r\) differentiations is bounded by a constant times
\(k^r\delta_j^{k-r}\).
The cutoff derivatives of order \(q\) cost \(O_r(\delta_j^{-q})\).
More precisely, each Leibniz term with \(t\) derivatives hitting
\(W\) and \(q\) hitting a cutoff has \(t+q\le r\), and is at most
\[
 C_r k^t\delta_j^{k-t-q}
       \le C_r(k/\delta_j)^r\delta_j^k.
\]
This proves \eqref{weights:short-derivatives}.

For \(u\ge-\log2/\ell\), use instead
\(\delta'=1+\log2/\ell\). Now
\((\delta')^k\le\exp(k\log2/\ell)\le2\) for large \(X\), and
\(a_t\le k/\ell\le\ell^{-2/5}\).
The same argument and the fixed derivative bounds for \(F_0\)
prove \eqref{weights:main-derivatives}.

On the support of the terminal cutoff, or any nonzero derivative
of it, \(1-u\le\delta_J/2\le\delta_J\).
Moreover \(h>0\) implies
\[
 v>\frac{n}{C_F|x|X}.
\]
Retain half the exponential in the integral. Uniformly on our
compact set,
\begin{equation}
 \begin{split}
 \int_{\{h>0\}}e^{-v}(C_F|x|v)^{a_t}\,dv
 &\le e^{-n/(2C_F|x|X)}
       (C_F|x|)^{a_t}
       \int_0^\infty e^{-v/2}v^{a_t}\,dv\\
 &=e^{-n/(2C_F|x|X)}
       (C_F|x|)^{a_t}2^{1+a_t}\Gamma(1+a_t)
 \le C e^{-cn/X}.
 \end{split}
 \label{weights:retained-exponential}
\end{equation}
Here \(a_t=(k-t)/(\ell\delta_J)\le\ell^{-3/10}\), so
\(2^{1+a_t}\Gamma(1+a_t)\) is bounded independently of \(k\).
The same Leibniz calculation now proves
\eqref{weights:terminal-derivatives}. All integrands used for
differentiation are dominated on compact sets by the displayed
integrable majorants, which also justifies
\eqref{weights:power-derivative} under the integral.
\end{proof}

\begin{corollary}[Comparison interface]
\label{weights:comparison-interface}
For every fixed derivative order \(r\), the short weights satisfy
the smoothness assumptions of Proposition~\ref{analytic:comparison}
with derivative bounds \(\ell^{C_r}\), uniformly in \(j,k,X\).
They have length \(X^{1-\Delta_j}\), with
\[
 \Delta_j=\delta_j/4,\qquad
 \tfrac14\ell^{-1/10}\le\Delta_j\le\tfrac14 .
\]
The deterministic term
\begin{equation}
 s=k/\ell,\qquad A_x=(C_F|x|)^s\Gamma(1+s)
 \label{weights:Ax}
\end{equation}
has uniformly bounded derivatives of each fixed order on
\(\mathcal K\), and is bounded above and below by positive constants.
The endpoint enlargement at \(n=1\) can be made inside the fixed
range \(n\ge1/2\).
\end{corollary}

\begin{proof}
The bounds \(k\le\ell^{3/5}\) and
\(\delta_j^{-1}\le\ell^{1/10}\) make
\eqref{weights:short-derivatives} polynomial in \(\ell\).
Choose a fixed smooth function \(\chi\) equal to zero on
\((-\infty,-\log2]\) and equal to one on \([0,\infty)\).
For the comparison alone, replace \(w_{0,x}(u)\) on the real line
by
\[
 \widetilde w_{0,x}(u)=\chi(\ell u)w_{0,x}(u).
\]
It agrees with the original weight at every positive integer
\(n=e^{\ell u}\), and its extra derivatives cost only fixed powers
of \(\ell\), by \eqref{weights:main-derivatives}. Its left support
is \(u\ge-\log2/\ell\), exactly the enlargement \(n\ge1/2\).
For \(j\ge1\), the weights already vanish on a neighborhood of
\(u\le0\). After a fixed smooth dyadic localization \(n=Ny\),
the operator \(y\partial_y\) acts as \(\ell^{-1}\partial_u\), so
the same bounds give all the required seminorms in the scaled
variables. The endpoint extension has therefore taken place in
\(\log n\); no extension to a fixed negative value of \(u\) is used.

Finally \(0<s\le\ell^{-2/5}\), while \(C_F|x|\) lies in a fixed
positive compact interval. The formula for \(A_x\), its derivatives
in \(x\), and the continuity and positivity of \(\Gamma(1+s)\)
give the assertions concerning \(A_x\).
\end{proof}

\subsection{The terminal mean square}

We use the following consequence of Proposition~\ref{analytic:polynomial-moment}.
For general $X$, cover $[X,2X)$ by at most two dyadic blocks whose
heights are comparable to $X$; the normalization and polynomial factors
in that proposition then change by fixed constants only.
For any fixed smooth compactly supported function \(G\) on
\((0,\infty)\), there are fixed
exponents \(A,B\) such that
\begin{equation}
 \frac1X\sum_{\substack{d\ {\rm admissible}\\X\le|d|<2X}}
 \left|\sum_{n\ge1}\frac{\lambda(n)\chi_d(n)}{\sqrt n}\,
                         n^{-it}G(n/N)\right|^2
 \ll_G \ell^B(1+|t|)^A(1+N/X)^A.
 \label{weights:moment-interface}
\end{equation}
Its applicability includes the fixed enlarging bump used below.

\begin{proposition}[Terminal mean square]
\label{weights:terminal}
There is a fixed exponent \(C_T\), chosen before \(k_0\), such that
uniformly in \eqref{weights:range},
\begin{equation}
 \frac1X\sum_{\substack{d\ {\rm admissible}\\X\le|d|<2X}}
       |S_{*,\chi_d}(d/X)|^2
       \ll \ell^{C_T}\delta_J^{2k}.
 \label{weights:terminal-square}
\end{equation}
The same estimate holds after restricting the summation to either
value of \(\chi_d(2)\).
\end{proposition}

\begin{proof}
Choose a smooth dyadic partition with a fixed
\(\rho\in C_c^\infty((1/2,2))\) satisfying
\[
 \sum_{N\in\{1,2,4,\ldots\}}\rho(n/N)=1\qquad(n\ge1).
\]
For example one may start with a smooth nonincreasing function
equal to one on \((0,1]\) and zero on \([2,\infty)\), and take
its difference at arguments \(y\) and \(2y\); choosing its
transition strictly inside \((1,2)\) gives the asserted compact
support. Fix also \(G\in C_c^\infty((0,\infty))\) equal to one
on \(\operatorname{supp}\rho\). Put
\[
 T_N(y,x)=\rho(y)w_{*,x}(\ell^{-1}\log(Ny)),\qquad
 \widehat T_N(t,x)=\int_0^\infty T_N(y,x)y^{it}\frac{dy}{y}.
\]
Select once and for all integers \(r>A/2+2\) and \(D>A/2+2\),
and then choose \(k_0\ge r+2\). If derivatives required elsewhere
are of higher order, increase this fixed \(r\) first.
By Lemma~\ref{weights:derivatives}, differentiation on the
\(y\)-scale through order \(r\) gives
\[
 \left|(y\partial_y)^a T_N(y,x)\right|
       \ll_r \ell^{C_r}\delta_J^k e^{-c'N/X}
       \qquad(0\le a\le r),
\]
uniformly in \(x\in\mathcal K\).
There is no boundary term when integrating by parts in \(\log y\).
Consequently,
\begin{equation}
 \sup_{x\in\mathcal K}|\widehat T_N(t,x)|
   \ll_{r,D}\ell^{C_r}\delta_J^k
       (1+N/X)^{-D}(1+|t|)^{-r}.
 \label{weights:terminal-kernel}
\end{equation}
In passing from the exponential to \((1+N/X)^{-D}\), the constant
depends on the fixed \(D\), not on \(k\).

Mellin inversion now expresses the \(N\)-piece of the terminal sum as
\[
 \frac1{2\pi}\int_{\mathbb R}
   \widehat T_N(t,d/X)N^{it}
   P_d(N,t)\,dt,\qquad
 P_d(N,t)=\sum_{n\ge1}\frac{\lambda(n)\chi_d(n)}{\sqrt n}\,
                        n^{-it}G(n/N).
\]
Indeed \(G(n/N)T_N(n/N,x)=T_N(n/N,x)\).
Define the normalized family norm by
\[
 \|B_d\|_{2,X}
   =\left(\frac1X\sum_{\substack{d\ {\rm admissible}\\X\le|d|<2X}}
                      |B_d|^2\right)^{1/2}.
\]
We bound the kernel uniformly before applying the family moment.
For each fixed \(t\),
\[
 \|\widehat T_N(t,d/X)P_d(N,t)\|_{2,X}
   \le \sup_{x\in\mathcal K}|\widehat T_N(t,x)|
                           \|P_d(N,t)\|_{2,X}.
\]
Minkowski's inequality, followed by
\eqref{weights:moment-interface} and
\eqref{weights:terminal-kernel}, therefore yields
\[
 \begin{split}
 \|S_{*,\chi_d}(d/X)\|_{2,X}
 &\ll \ell^{C_r+B/2}\delta_J^k
   \sum_{N=1,2,4,\ldots}(1+N/X)^{-D+A/2}
   \int_{\mathbb R}(1+|t|)^{-r+A/2}\,dt\\
 &\ll \ell^{C_r+B/2+1}\delta_J^k.
 \end{split}
\]
The integral converges by the fixed choice of \(r\).
There are \(O(\ell)\) dyadic terms with \(N\le X\), and those
with \(N>X\) form a convergent geometric tail by the choice of
\(D\). All sums and integrals can first be truncated; the displayed
integrable majorants justify passage to the limit. Squaring proves
\eqref{weights:terminal-square}, for example with
\(C_T=2C_r+B+2\). This exponent depends only on choices made before
\(k_0\). Restricting the family can only decrease the nonnegative
sum.
\end{proof}

\begin{lemma}[Uniform absorption of the terminal loss]
\label{weights:terminal-small}
After increasing the fixed \(k_0\), one sufficiently large lower
bound on \(X\) makes
\begin{equation}
 \frac1X\sum_{\substack{d\ {\rm admissible}\\X\le|d|<2X}}
       |S_{*,\chi_d}(d/X)|^2
       \ll 2^{-3Jk/2}
 \label{weights:terminal-small-square}
\end{equation}
simultaneously for all \(k\) in \eqref{weights:range}.
Moreover, for any fixed \(c,L>0\), the comparison errors obey
\begin{equation}
 e^{-c\Delta_j\ell}
       \le k^{-L}2^{-Ljk}\qquad(0\le j\le J)
 \label{weights:comparison-absorption}
\end{equation}
for one sufficiently large lower bound on \(X\).
\end{lemma}

\begin{proof}
The ratio of the right side of
\eqref{weights:terminal-square} to \(2^{-3Jk/2}\), apart from its
fixed implied constant, is
\[
 \ell^{C_T}2^{-Jk/2}.
\]
For \(J\ge1\) this decreases with \(k\), so throughout the permitted
range it is at most its value at the fixed integer \(k_0\).
The floor in the definition of \(J\) gives
\[
 \ell^{C_T}2^{-Jk/2}
       \le \ell^{C_T}2^{-Jk_0/2}
       \le 2^{k_0/2}\ell^{C_T-k_0/20}.
\]
Choose \(k_0>20C_T\) after \(C_T\) is fixed. The last expression
then tends to zero with \(X\). In particular the choice of a
single \(X\)-threshold works for the entire \(k\)-range.

For the second assertion,
\(\Delta_j\ell\ge\ell^{9/10}/4\), whereas uniformly in the same
range
\[
 L\log k+Ljk\log2
       =O_L(\ell^{3/5}\log\ell)=o(\ell^{9/10}).
\]
Taking logarithms proves \eqref{weights:comparison-absorption}.
\end{proof}
==== END The-mean-analytic-rank-of-quadratic-twists-of-elliptic-curves//sections/weights.tex ====

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@misc{WeiZhai2026,
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}
==== END The-mean-analytic-rank-of-quadratic-twists-of-elliptic-curves//references.bib ====