\section{Three unconditional families}\label{uncond:section} We count zeros of the derivative of the uncompleted $L$-function. In particular, differentiating a gamma factor would give a different problem. For a real analytic function $F$, let $R^{\rm dist}(F)$ and $R^{\rm mult}(F)$ denote its numbers of distinct zeros and zeros counted with multiplicity in $[1/2,1]$, and let $S^-(F)$ denote its strict changes of sign in $(1/2,1)$. Put \[ c_*={e^{-1/2}\over \pi\sqrt2}. \] All the probability measures in this section are the original uniform counting measures. Restricting an expectation to a good event always means inserting its indicator, without renormalizing the measure. \begin{theorem}\label{uncond:quadratic} Let $\mathcal F(X)$ be the set of fundamental discriminants $D$ with $1<|D|\le X$. For each fixed $\eta>0$, all but $o_\eta(X)$ members satisfy, simultaneously for $C=R^{\rm dist},R^{\rm mult},S^-$, \[ \big|C(L'(\cdot,\chi_D))-c_*\log\log|D|\big| \le\eta\log\log|D|. \] Both signs of the discriminant and every possible conductor at $2$ are included. \end{theorem} \begin{theorem}\label{uncond:gaussian} Let $K=\mathbb Q(i)$ and \[ \mathcal D_X=\{d\in\mathbb Z[i]:(d,2)=1,\ (d)\text{ square-free}, \ X0$, all but $o_\eta(X)$ labels satisfy \[ \max_{C=R^{\rm dist},R^{\rm mult},S^-} \big|C(L'_d)-c_*\log\log Q_d\big|\le\eta\log\log Q_d. \] The same assertion holds with uniform measure on the distinct primitive characters occurring in this family. \end{theorem} \begin{theorem}\label{uncond:weight} Let $k$ tend to infinity through all even integers and let $H_k$ be the normalized level-one Hecke eigenbasis of weight $k$. Use the unitary normalization \[ L(s,f)=\prod_p(1-\lambda_f(p)p^{-s}+p^{-2s})^{-1}. \] For each fixed $\eta>0$, all but $o_\eta(|H_k|)$ forms satisfy \[ \max_{C=R^{\rm dist},R^{\rm mult},S^-} \big|C(L'(\cdot,f))-c_*\log\log(k^2)\big| \le\eta\log\log(k^2). \] The measure is unweighted, and the assertion includes both root numbers $i^k=1$ and $i^k=-1$. \end{theorem} No theorem above assumes the Riemann hypothesis, nonvanishing at the central point, or simplicity of a zero. We give the arithmetic and analytic inputs in some detail, since the passage from a mollified moment to the very short interval next to the centre is sensitive to the family and to the range of the shift. \subsection{The arithmetic probability spaces}\label{uncond:arithmetic} For the rational quadratic family the independent model has variables $X_p\in\{0,1,-1\}$ with \[ \mathbb P(X_p=1)=\mathbb P(X_p=-1)={p\over2(p+1)},\qquad \mathbb P(X_p=0)={1\over p+1}. \] This includes $p=2$. For every positive integer $n$, \begin{equation}\label{uncond:qorth} \E_{\mathcal F(X)}\chi_D(n) =\mathbf1_{n=\square}\prod_{p\mid n}{p\over p+1} +O(X^{-1/2}n). \end{equation} Here and below sufficiently large implied constants also cover bounded parameters. One elementary derivation, which will also fix the normalization needed for the mollifier, is as follows. Write $D=\epsilon cd$, where $d>0$ is odd and square-free, $\epsilon=\pm1$, $c\in\{1,4,8\}$, and $d\equiv r\pmod8$. The allowed classes are \[ c=1:\ \epsilon r\equiv1\pmod4;\qquad c=4:\ \epsilon r\equiv3\pmod4;\qquad c=8:\ r\text{ arbitrary odd}. \] Expanding $\mu^2(d)=\sum_{\alpha^2\mid d}\mu(\alpha)$ shows that each odd class contains $Y/\pi^2+O(\sqrt Y)$ elements up to $Y$. For odd nonsquare $n$ the complete character sum in the progression modulo $8$ vanishes; summing the incomplete-period error over $\alpha\le\sqrt Y$ costs $O(\sqrt Y n)$. For square $n$, inclusion-exclusion for $(d,n)=1$ gives the local factors $p/(p+1)$ and the same error. The classes with $c=1,4,8$ contribute respectively $4X/\pi^2,X/\pi^2,X/\pi^2$. In the first group $\chi_D(2)=(-1)^{(r^2-1)/8}$; in the other two it is zero. Summing the classes proves \eqref{uncond:qorth}, including the factor $2/3$ at $2$, and \begin{equation}\label{uncond:qsize} |\mathcal F(X)|={6X\over\pi^2}+O(\sqrt X). \end{equation} For $K=\mathbb Q(i)$, use independent variables $X_{\mathfrak p}$ with the same ternary law with $p$ replaced by $N\mathfrak p$, except that $X_{(1+i)}=0$. Distinct prime ideals of the same norm are distinct independent coordinates. We have \begin{equation}\label{uncond:korth} |\mathcal D_X|={2\pi X\over3\zeta_K(2)}+O(\sqrt X\log(2X)),\qquad \left|\E_{\mathcal D_X}\chi_d(\mathfrak n)-\E X_{\mathfrak n}\right| \ll X^{-1/3}N\mathfrak n. \end{equation} For completeness, the number of points of one residue class modulo an ideal $\mathfrak q$ in a norm annulus $(Y,2Y]$ is $\pi Y/N\mathfrak q+O(\sqrt{Y/N\mathfrak q}+1)$. This follows by covering the boundary circle with fundamental squares of the rotated lattice $\mathfrak q$. A nonprincipal character modulo $\mathfrak n$ therefore has annular sum $O(\sqrt{YN\mathfrak n}+N\mathfrak n)$, also with the oddness condition. Apply this after the ideal-M\"obius expansion and sum $N\mathfrak a\le\sqrt{2X}$; the error is $O(\sqrt X N\mathfrak n\log(2X))$. In the square-ideal case the main Euler product is \[ {\pi X\over2}\prod_{\mathfrak p\mid\mathfrak n}(1-(N\mathfrak p)^{-1}) \prod_{\mathfrak p\nmid2\mathfrak n}(1-(N\mathfrak p)^{-2}) ={2\pi X\over3\zeta_K(2)} \prod_{\mathfrak p\mid\mathfrak n}{N\mathfrak p\over N\mathfrak p+1}. \] The elementary bound $2^{\omega(\mathfrak n)}\le N\mathfrak n$ suffices for inclusion-exclusion. Ideals divisible by $(1+i)$ have character value zero. Division by the first formula in \eqref{uncond:korth} proves the second. Gao's family definition and functional equation give \begin{equation}\label{uncond:kcompletion} \Lambda_d(s)=(128N(d))^{s/2}(2\pi)^{-s}\Gamma(s)L_d(s) =\Lambda_d(1-s) \end{equation} and primitivity of conductor $((1+i)^5d)$ \cite[\S2.3, (2.1)--(2.4)]{GaoReal2020}. Equal primitive characters must have equal conductor ideals, so their labels differ by a unit. The supplementary quadratic law gives $(-1/\mathfrak n)=1$ for odd ideals, whereas $(i/n)=-1$ for a primary $n\equiv-1+2i\pmod4$ \cite[(2.2)]{GaoFourth2020}. The Chinese remainder theorem permits the additional condition $n\equiv1\pmod{\operatorname{rad}(d)}$. Consequently the fibers of $d\mapsto\chi_d$ are exactly $\{d,-d\}$. Thus passing to distinct characters divides both the numerator and the denominator of every character-dependent event by two. The weight family has a different local law. If $U_j=U_j(x/2)$ are the Chebyshev polynomials, set \begin{equation}\label{uncond:plancherel} d\mu_p(x)={1+p^{-1}\over(1+p^{-1})^2-p^{-1}x^2} {\sqrt{4-x^2}\over2\pi}\,dx,\quad -2\le x\le2. \end{equation} The identity $\sum_{j\ge0}p^{-j}U_{2j}(x/2) =(1+p^{-1})/((1+p^{-1})^2-p^{-1}x^2)$ and Sato--Tate orthogonality show that $\E U_j(X_p/2)=p^{-j/2}$ for even $j$ and zero otherwise. The density relative to Sato--Tate lies between $2/3$ and $6$. Put $A_1(x)=x$ and $A_j(x)=U_j(x/2)-U_{j-2}(x/2)$ for $j\ge2$. Then $|A_j|\le2$, $\E X_p^2=1+p^{-1}$, and \begin{equation}\label{uncond:weightdrift} \E A_{2j}(X_p)=-(1-p^{-1})p^{-j+1},\qquad \E A_{2j-1}(X_p)=0. \end{equation} Hough's unweighted coefficient estimate says that, for fixed $0<\epsilon<2$ and $n0$, \begin{equation}\label{uncond:hecketransfer} |\E_{H_k}P-\E P(X)|\ll k^{-\kappa}\|P\|_{\rm H} \end{equation} for every finite Hecke polynomial, with no support restriction. In the two quadratic families the analogous submultiplicative norm is $\sum|a_n|n$ or $\sum|a_{\mathfrak n}|N\mathfrak n$, and the exponents are $1/2$ and $1/3$ by \eqref{uncond:qorth} and \eqref{uncond:korth}. \subsection{A central mollified bound for all signed discriminants} \label{uncond:cbcsub} The following extension of the quadratic mollifier calculation is needed on the complete family, rather than only on $\chi_{-8d}$. \begin{proposition}\label{uncond:cbc} Write $L=\log X$, $H=\log L$, $M=X^{1/100}$, and let $Q(v)=0$ for $v\le0$, $Q(v)=3(2v)^2-2(2v)^3$ for $0V$. With $Y_l=Y/l^2$, the large sieve \cite[Lemma 3.4]{CS2002} bounds the fourth mollifier sum by $Y^\epsilon(Y_l+M^2)$, and the product approximate functional equation and \cite[Lemma 3.5]{CS2002} bound the second $L$-sum by $Y^\epsilon Y_l$. The principal case $m=1$ is bounded separately and is included in $M^2$. Two applications of Cauchy--Schwarz and summation over $l$ give $Y^\epsilon(V^{-1}+MY^{-1/2}\log Y)$ after normalization. Summing \eqref{uncond:twistederror} against the two mollifier coefficients therefore leaves at most \begin{equation}\label{uncond:cbcerror} Y^\epsilon\{MY^{-1/4}+MV^{-1}+M^2VY^{-1/2} +V^{-1}+MY^{-1/2}\}, \end{equation} a fixed negative power when $.009\le\log M/\log Y\le.02$. It remains to bound the main terms uniformly through colliding shifts. We give the cancellation estimate used for this purpose. For a generic term $(w,v)$ in \eqref{uncond:fourmain}, its mollifier sum is \[ \mathcal A(w,v)=\sum_{r,s\ \mathrm{odd}}{\lambda(r)\lambda(s)\over r^{1/2+\delta_1}s^{1/2+\delta_2}} {r_v(l_1)\over l_1^{1/2+w}}\eta_v(1+2w;rs),\quad \lambda(n)=\mu(n)Q(\log(M/n)/\log M). \] The square-free rearrangement of the Euler factors gives \begin{equation}\label{uncond:flatfactor} \mathcal A(w,v)=\eta_v(1+2w;1) \sum_{\gamma\ \mathrm{odd}}{H_{w,v}(\gamma)\over \gamma^{1+2\tau}h_{w,v}(\gamma)} S_\gamma(\delta_1+w,v)S_\gamma(\delta_2+w,v), \end{equation} where \begin{align*} h_{w,v}(p)&=1+p^{-1}+p^{-1-2w} -(p^{2v}+p^{-2v})p^{-2-2w}+p^{-3-4w},\\ H_{w,v}(p)&=1+p^{-1-2w}-r_v(p)^2p^{-1-2w}/h_{w,v}(p),\\ S_\gamma(u,v)&=\sum_{a\le M/\gamma\atop a\ \mathrm{odd}} {r_v(a)\mu(a\gamma)\over a^{1+u}h_{w,v}(a)} Q(\log(M/(a\gamma))/\log M). \end{align*} Let $E_\gamma=\prod_{p\mid\gamma}(1+C/\sqrt p)$. The residue formula of \cite[Lemma 6.1]{CS2002}, applied to these explicit Euler factors, gives \begin{equation}\label{uncond:flatbound} |S_\gamma(u,v)|\ll E_\gamma H^6 (|u+v|+L^{-1})(|u-v|+L^{-1}). \end{equation} Here are the estimates within that application. Its generating series is $\mu(\gamma)G(1+s,\gamma;u,v)J(s+u+v)J(s+u-v)$, with $J(z)=1/\zeta(1+z)=z+O(z^2)$. The derivatives of $G$ through order three are $O(E_\gamma H^{2j})$: local ratios are $1+O(p^{-1})$ and their logarithmic derivatives cost $\sum_{p\mid\gamma}(\log p)^j/p\ll1+H^{j+1}$. The cubic residues are divided by $(\log M)^j$, and the error is $O(E_\gamma/L^2)$. For $\gamma>M/2$ use $Q(v)=O(v^2)$ directly. For $\gamma\le M^{1/2}$ apply the residue formula at $M$ and at $M^{1/2}$ to the two pieces of the flat cutoff; their derivative residues cancel. These observations prove \eqref{uncond:flatbound} also across the join of $Q$. Finally \[ \sum_{\gamma\le M}{|H_{w,v}(\gamma)/h_{w,v}(\gamma)|E_\gamma^2\over\gamma} \ll\prod_{p\le M}(1+p^{-1}+Cp^{-3/2})\ll L. \] For $\delta_1=it$, $\delta_2=-it$, $|t|\ge4/L$, the two primary residues must be taken together on $|w|=1/L$. There $\mathcal A\ll H^{12}(|t|+L^{-1})^2/L$ and $Z(1+2w;it)\ll L/(|t|+L^{-1})^2$; the kernel $2/w$ and the contour length cancel. Each secondary term at $w=\pm it$, $v=0$ is $O(H^{12}/(|t|L))$. Thus the main contribution is $O(H^{12})$. For $|t|\le4/L$, apply two-variable Cauchy to the entire mollified moment on $|\delta_1|=10/L$, $|\delta_2|=20/L$. The boundary shifts are separated, $Z=O(L^3)$, and $\mathcal A=O(H^{12}L^{-3})$. The Cauchy kernels have bounded total norm. This proves the same bound inside. Bounding individual primary residues at $\tau=0$ would not be legitimate. Divide by $(cY/\pi)^\tau\Gamma_{a,\delta}(\tau)$, take $\Psi(x)=x^{-\tau}\Phi(x)$, and normalize by the actual class mass $\check\Phi(0)/\pi^2+O(Y^{-1/2})$. Equations \eqref{uncond:cbcerror}--\eqref{uncond:flatbound} prove the weighted bound, weakened to $H^{20}$. Fixed plateaus which are at least one on $[6/5,9/5]$, at scales $5Y/6$ and $10Y/9$, remove the weight. For $|D|\ge X^{.9}$ the single global choice $M=X^{.01}$ remains in the compact interval $.009\le\log M/\log Y\le.02$. The sum of the dyadic conductor masses is $O(X)$, so no logarithm is lost in combining the blocks. For $|D|\sigma$ and $|\gamma|\le T$. We first record the logarithmic fourth-moment bound needed to turn Proposition~\ref{uncond:cbc} into a zero detector: \begin{equation}\label{uncond:ss4} \E_{\mathcal F(X)}|L(1/2+it,\chi_D)|^4 \ll (\log X)^{10}(1+|t|)^8. \end{equation} Here is a derivation with a harmless, but explicit, height exponent. For the fixed smooth partition function $G$ in \cite[Lemma 2.9(ii)]{ShenStucky2024}, Proposition 1.3 of that paper bounds the dyadic conductor sum of $|\sum\chi_D(n)n^{-1/2-it}G(n/N)|^4$ by \[ C(1+|t|)^2\{M+N^2\log(2+N^2/M)\}\log^6(2+MN). \] Put $T=1+|t|$. If $N\le\sqrt{MT}$ this is $O(MT^4\log^6(2+M))$, since $\log^7(2+T)\ll T$. For $N>\sqrt{MT}$ set $N_1=MT/N$. The functional-equation Mellin transform has the gamma quotient \[ {\Gamma(1/4+(s-it+a)/2)\over \Gamma(1/4-(s-it-a)/2)},\qquad a=0,1. \] Decompose the transformed series into dyadic blocks $N_0$ and insert $V(x)=G(2x)+G(x)+G(x/2)$, which equals one on the support of $G$. For $N_0\le N_1$ move the Mellin line only to $\Re s=-1/8$; the first possible numerator pole has real part $-1/2$ and is not crossed. For $N_0>N_1$ move it to $\Re s=4$. Stirling's formula and rapid Mellin decay give the pointwise majorant \begin{align*} C\sum_{N_0\le N_1}^{\rm dyad}(N_0/N_1)^{1/8} \int_{\mathbb R}{|A_D(N_0,v-t)|\over(1+|v|)^9}\,dv +C\sum_{N_0>N_1}^{\rm dyad}(N_1/N_0)^4 \int_{\mathbb R}{|A_D(N_0,v-t)|\over(1+|v|)^9}\,dv, \end{align*} where $A_D(N_0,w)=\sum\chi_D(n)n^{-1/2-iw}V(n/N_0)$. The lower-line bound uses $T\le(1+|v|)(1+|v-t|)$; two Mellin factors with decay of order $24$ leave the displayed integrable weight. All contour shifts can first be made on finite blocks. The high-block tail is absolutely summable using $|A_D|\ll\sqrt{N_0}$, so passage to the full series is justified. Weighted H\"older and Proposition 1.3 now give \[ CT^2\left\{\sum_{N_0\le N_1}(N_0/N_1)^{1/8}F_M(N_0) +\sum_{N_0>N_1}(N_1/N_0)^4F_M(N_0)\right\}, \] where $F_M(v)=\{M+v^2\log(2+v^2/M)\}\log^6(2+Mv)$. For $N_1\ge1$ this is $O(T^2F_M(N_1))$ by geometric summation; for $N_1<1$ it is $O(MT^2\log^6(2+M))$. Thus the same $MT^4\log^6(2+M)$ bound holds for every $N$. Summing conductor blocks and dividing by \eqref{uncond:qsize} bounds each smooth block in $L^4$ by $CL^{3/2}T$. Take a partition cutoff $Y\asymp X^4T^4$. Periodicity and the vanishing complete sum of a nonprincipal primitive character show by partial summation that the remaining tail is $O(XTY^{-1/2})=O(1)$. There are $O(L+\log(2+T))$ blocks. Minkowski proves \eqref{uncond:ss4}. This argument avoids moving a localized even-character gamma quotient past its pole at $-1/2+it$. \begin{lemma}\label{uncond:qdetector} For the rational family let $r=L/\ell\to\infty$, $\ell\to\infty$, and $0\le T_*\le L^{-5/6}/32$. With probability at least \begin{equation}\label{uncond:qdetectorerror} 1-CH^{20}(1+T_*\ell)e^{-cr}-Ce^{-KH}, \end{equation} there is no zero with $\beta>2/\ell$ and $|\gamma|\le T_*+\ell^{-1}e^{3\ell\beta}$. Here $K$ is any fixed constant, and the constants in the estimate may depend on $K$. For $\nu=A\sqrt H$, the probability of a zero with $\beta>\nu/(2L)$ and $|\gamma|\le2\beta$ is \begin{equation}\label{uncond:qwedgebound} \ll H^{20}e^{-c\nu}+e^{-KH}. \end{equation} There is also a coarse detector, valid for $8\le\ell\le c_0L$, $\alpha\ge1/L$, and $T_*\le1$, with error $CL^{C_G}e^{-c_G\alpha L}$ for the same region with $\beta>\alpha$. \end{lemma} \begin{proof} Put $B_D(s)=L(s,\chi_D)M_D(s)-1$. The mollifier fourth moment is $O(L^6)$: the square diagonal is bounded by $\prod_{p\le M}(1+6/p+1/p^2)\ll L^6$, and the error from \eqref{uncond:qorth} is $O(X^{-1/2}M^6)$. Cauchy--Schwarz and \eqref{uncond:ss4} give $\E|B_D(1/2+it)|^2\ll L^8(1+|t|)^C$. The flat part of the mollifier annihilates coefficients $1a_c$ cost $O(e^{-KH})$ by the coarse detector. Below $a_c$ use bins $j/\ell<\beta\le(j+1)/\ell$, $j\ge2$, and height $2(T_*+\ell^{-1}e^{3(j+1)})$ in \eqref{uncond:qlocalband}. If $2/\ell\ge a_c$ there are no such bins. Otherwise their nonconstant heights have logarithm $-H+\log r+3A_cH/r+O(1)=-H+o(H)$ and hence are $o(h)$. The sum is bounded by $CH^{20}(1+T_*\ell)\sum_{j\ge2}(1+e^{3(j+1)})e^{-crj}$, which proves \eqref{uncond:qdetectorerror}. For the wedge use the same high-zero cutoff and, below it, the bands $\sigma=2^{j-1}\nu/L$, $T=4\sigma$. Their geometric sum proves \eqref{uncond:qwedgebound}. \end{proof} We next derive the Hecke-family detector from the precise range of Gao's mollified moment. Fix $\kappa=10^{-3}$, $m=1/2-5\kappa$, $b=1/2$, $M=X^m$, and the cubic flat cutoff $Q$ used above, with join at $b$. The mollifier is the sum over primary Gaussian integers $n\equiv1\pmod{(1+i)^3}$, with coefficient $\mu_K(n)Q(\log(M/Nn)/\log M)$ and denominator $(Nn)^s$. For a fixed nonnegative weight $\Phi$ on $[1,2]$, let $W(z)=\E_\Phi|L_d(1/2+z)M_d(1/2+z)|^2$. Gao's Proposition 3.5 \cite{GaoReal2020} states, with $z=\tau+it$, $L_0=\log(32X/\pi^2)$, \begin{align} W(z)=1+&\left\{{1-e^{-2\tau L_0}\over2\tau mL} -e^{-\tau L_0}{\sin(tL_0)\over tmL}\right\} \int_0^b e^{-2\tau mL(1-v)} \left|Q'(v)+{Q''(v)\over2zmL}\right|^2dv\notag\\ &+O_\Phi(X^{-\kappa-\epsilon} +e^{-2m(1-b)\tau L}|z|^6L^5), \label{uncond:gaomoment} \end{align} for fixed $\epsilon>0$, $\tau\ge-1/(\epsilon L)$ and $\epsilon/L\le|z|\le\kappa$. Quotients at zero are interpreted by continuity. This is the second moment of $LM$, not of $LM-1$. Put $\eta_0=m(1-b)$. For $\tau\ge1/L$, the prefactor is nonnegative and bounded: use $\sinh y/y\ge1$ and $\sin y/y\le1$. Since $|z|L\ge1$, the integral is $O(e^{-2\eta_0\tau L})$, and the $\tau^6L^5$ part of the error is absorbed into that exponential. Consequently \begin{equation}\label{uncond:gaodecay} |W(\tau+it)-1|\ll(1+|t|^6L^5)e^{-\eta_0\tau L}+X^{-\kappa}. \end{equation} In particular the loss is logarithm-free only at height $|t|\le L^{-5/6}$. \begin{lemma}\label{uncond:kdensity} There are fixed positive $\sigma_0,\tau_0,c$ such that \begin{equation}\label{uncond:kdensitybound} \E_{\mathcal D_X}N(\sigma,T) \ll(1+TL)(1+T^6L^5)e^{-c\sigma L} \quad(4/L\le\sigma\le\sigma_0, 0\le T\le\tau_0). \end{equation} For every fixed $\alpha>0$ there is $\theta_\alpha>0$ such that \begin{equation}\label{uncond:kfar} \mathbb P_{\mathcal D_X}\{N(\alpha,X^{\theta_\alpha})>0\} \ll X^{-c_\alpha}. \end{equation} Also, for $8\le\nu=o(L)$, the probability of a zero with $\beta>\nu/(2L)$, $|\gamma|\le2\beta$ is \begin{equation}\label{uncond:kwedge} \ll \nu e^{-c\nu}+X^{-c_0}. \end{equation} \end{lemma} \begin{proof} We spell out the application of Selberg's rectangle lemma, because discarding its signed logarithm would lose \eqref{uncond:kdensitybound}. Take $\lambda=\eta_0/100$, $S=\pi/(2\lambda)$, and rectangle half-height $h=S/L$, with left side $1/2+\sigma/2$ and right side $2$. A zero with $\beta>\sigma$ and $|\gamma-t_0|\le h/2$ contributes at least $4h\cos(\pi/4)\sinh(\lambda\sigma L/2)$. On each boundary average the signed $\log|LM|$ and use $\E_\Phi\log|LM|\le\frac12\log W\le\frac12|W-1|$. The left side is bounded by $Ch\{(1+(T+2h)^6L^5)e^{-\eta_0\sigma L/2}+X^{-\kappa}\}$. On the horizontal sides split the offset $u$ into $[\sigma/2,\kappa/2]$, $[\kappa/2,.6]$, and $[.6,1.5]$. The first uses \eqref{uncond:gaodecay} and integrates to $C(1+(T+2h)^6L^5)L^{-1}e^{-\eta_0\sigma L/2}$ plus a fixed power error. On the second use Gao's Proposition 3.2 with a fixed $\epsilon_0<\eta_0\kappa/100$: \[ |W(u+it)-1|\ll X^{\epsilon_0} \{e^{-2\eta_0uL}+e^{(\eta_0(1/2-u)-1/2)L}\}. \] After multiplication by the boundary weight $e^{\lambda uL}$ both exponents are negative fixed multiples of $L$. For the last interval the flat mollifier gives directly $|LM(1/2+u+it)-1|\ll Le^{-\eta_0(u-1/2)L}$. It also makes $LM$ zero-free near the right edge and gives $|\log LM(2+it)|\ll Le^{-\eta_0L}$. Thus the last interval and the right edge contribute fixed negative powers of $X$. Divide by the zero weight and cover $[-T,T]$ with $O(1+TL)$ rectangles. Shrinking fixed $\sigma_0,\tau_0$ absorbs all these fixed-power errors and proves the weighted form of \eqref{uncond:kdensitybound}. Two fixed plateaus at scales $5X/6$ and $10X/9$ remove the weight; using $4/L$ in the stated lower endpoint allows for the bounded change of logarithmic scale. The denominator is still $|\mathcal D_X|$. For \eqref{uncond:kfar} a polynomial height bound is needed, rather than the compact-height statement of Proposition 3.2 alone. Its proof, using Gao's Lemmas 2.7--2.8, decomposes $B=LM-1$ as $T_1-T_2$. The large sieve gives $\E|T_1|^2\ll X^\epsilon e^{-2\eta_0uL}$ uniformly in the phases $n^{-it}$. The second term is a Mellin integral on $\Re w=-u$; the inverse Mellin weight is $e^{-Nn/X}$. For $\alpha/2\le u\le.7$, the fourth-moment bounds \[ \E|L(1/2+iv)|^4\ll X^\epsilon(1+v^2)^{1+\epsilon},\qquad \E|M(1/2+iv)|^4\ll X^\epsilon(2+M^2/X) \] and gamma decay imply $\E|T_2|^2\ll X^{-2u+\epsilon}(1+|t|)^4$. Hence $\E|B(1/2+u+it)|^2\ll X^\epsilon(1+|t|)^4e^{-2\eta_0uL}$. Cover $\beta\ge\alpha$, $|\gamma|\le T$ by $O_\alpha(T+1)$ submean disks of fixed radius $\alpha/8$. Since $B=-1$ at an $L$-zero, the probability is at most $C_\alpha X^{\epsilon-\eta_0\alpha}(1+T)^5$. Take $\epsilon=\theta_\alpha=\eta_0\alpha/100$. For larger fixed $u$ the deterministic right-hand estimate applies. This proves \eqref{uncond:kfar}. Finally take $\beta_0<\min(\sigma_0/2,\tau_0/16)$. The part $\beta>\beta_0$ of the wedge is controlled by \eqref{uncond:kfar}. Below it use the bands $b_j=2^{j-1}\nu/L$, with height $8b_j$, in \eqref{uncond:kdensitybound}. Writing $v_j=b_jL$, their bound is $Cv_j(1+v_j^6/L)e^{-cv_j}\ll Cv_je^{-cv_j/2}$. Its geometric sum proves \eqref{uncond:kwedge}. \end{proof} For the weight family the needed zero density is a direct literature input. Hough's Main Theorem 1.1 \cite{HoughDensity2012} states, for some sufficiently small fixed positive $\theta,\delta$, \begin{equation}\label{uncond:hdensity} \E_{H_k}N(\sigma,T)\ll T\log k\,e^{-\theta\sigma\log k}, \quad {2\over\log k}<\sigma<\tfrac12,\quad {10\over\log k}2/\ell, |\gamma|\le T_*+\ell^{-1}e^{3\ell\beta}\} \ll(T_*L+r)e^{-cr}. \end{equation} Indeed the $j$th bin uses $T_j=2(T_*+\ell^{-1}e^{3(j+1)})$, $j\ge2$. Its smallest height is greater than $10/L$, its smallest offset is greater than $2/L$, and its largest height is $e^{3\ell/2+O(1)}=k^{o(1)}$. Thus every bin is within \eqref{uncond:hdensity}, and summing $C(T_*L+re^{3(j+1)})e^{-crj}$ proves the assertion. The same estimate on the bands $2^{j-1}\nu/L$, with height eight times the offset, proves the wedge bound \begin{equation}\label{uncond:hwedge} \mathbb P\{\exists\rho:\beta>\nu/(2L), |\gamma|\le2\beta\} \ll\nu e^{-c\nu},\qquad \nu\to\infty,\quad\nu=o(L). \end{equation} \subsection{A zero-free complex approximation}\label{uncond:appsub} Write $\mathcal L=-L'/L$. For the rational and weight families use \[ K_\ell(z)={e^{\ell z}(e^{\ell z}-1)^2\over\ell^2z^3},\qquad P_\ell(s)=\sum_{n\le e^{3\ell}}\Lambda_L(n)w_\ell(n)n^{-s}, \] where $\Lambda_L(p^j)=\chi_D(p)^j\log p$ in the rational family and $\Lambda_L(p^j)=A_j(\lambda_f(p))\log p$ in the weight family, and \[ w_\ell(n)={ (3\ell-\log n)_+^2-2(2\ell-\log n)_+^2 +(\ell-\log n)_+^2\over2\ell^2}. \] The weight lies in $[0,1]$, is one for $n\le e^\ell$, and is zero for $n>e^{3\ell}$. The arithmetic norms of $P_\ell$ are at most $Ce^{9\ell/2}$ and $Ce^{7\ell}$, respectively. In the second case expand $A_j$ as $e_{p^j}-e_{p^{j-2}}$ and use $\tau(n)\le2\sqrt n$ to bound the norm by $C\sum_{n\le e^{3\ell}}n\log n$. The Gaussian Hecke family needs a kernel with stronger vertical decay. Choose a fixed $\tau>0$ smaller than $\tau_0/10$, put $a=4/\tau$, $k_\ell=\lfloor\ell/a\rfloor$, and define \begin{equation}\label{uncond:dampedkernel} K_\ell^*(z)=K_\ell(z) \left({\sinh(az/2)\over az/2}\right)^{k_\ell}. \end{equation} If $U_1,U_2$ are uniform on $[0,\ell]$ and the $V_j$ are uniform on $[-a/2,a/2]$, then $K_\ell^*(z)=\E e^{Vz}/z$ for $V=\ell+U_1+U_2+\sum V_j\in[\ell/2,7\ell/2]$. Thus its Dirichlet weight $w_\ell^*(n)=\mathbb P(V\ge\log n)$ is nonnegative, at most one, equals one up to $e^{\ell/2}$, and vanishes above $e^{7\ell/2}$. The corresponding ideal polynomial $P_\ell^*$ has arithmetic norm at most $Ce^{8\ell}$. The kernel satisfies \begin{equation}\label{uncond:kernelbounds} |K_\ell^*(u+iv)|\le{4\over\ell^2|u+iv|^3} \begin{cases}e^{7\ell u/2},&u\ge0,\\e^{\ell u/2},&u\le0. \end{cases} \end{equation} Moreover, for $\beta\le\beta_0=\tau/1000$, $\delta\ge0$, $|v|\ge\tau$, \begin{equation}\label{uncond:verticaldamping} |K_\ell^*(\beta-\delta+iv)| \le {4e^{-\ell\delta/2-\tau\ell/20}\over\ell^2|v|^3} (\tau/|v|)^{k_\ell}. \end{equation} Indeed the extra factor is at most $\{\tfrac12 e^{a|u|/2}\tau/|v|\}^{k_\ell}$ and $k_\ell\ge\tau\ell/8$ for large $\ell$. The loss $7\beta_0\ell/2$ is smaller than the gain from $2^{-k_\ell}$. We use the following basic analytic facts in their unconditional form. The primitive completed functions are entire of order one, their zeros are in $0\le\Re s\le1$, and their real functional equations pair $\rho$ with $1-\rho$ and $\bar\rho$. These facts follow from the usual theta-integral continuation for the two degree-one families, and from the Mellin integral of the cusp form in the weight family. Euler coefficients bound the function on a right-hand line; the functional equation, Stirling's formula and the maximum principle give polynomial bounds on every fixed vertical strip. Jensen at $2+iT$, with inner radius $3$ and outer radius $4$, then gives \begin{equation}\label{uncond:unitzeros} \#\{\rho:|\Im\rho-T|\le1\}\ll\log(Q(|T|+2)) \end{equation} in the degree-one families. The lower bound at the centre of the Jensen disk is supplied by the absolutely convergent Euler product. For weight $k$, the same bound is $O(\log(k+|T|+2))$; alternatively take the real part of the completed logarithmic derivative at $2+iT$, where each zero in the indicated band has weight at least $1/5$ and the Euler logarithmic derivative is bounded. Writing $z=s-1/2$, the order-one product and functional equation give \begin{equation}\label{uncond:pairedhadamard} \mathcal L(1/2+z)=B(z)-{m_0\over z} -\sum_{\alpha\in\mathcal P}{2z\over z^2-\alpha^2}. \end{equation} Here $m_0$ is the central multiplicity, $\mathcal P$ contains one representative of each pair of noncentral completed zeros, and \[ B(z)=\begin{cases} \tfrac12\log(|D|/\pi)+\tfrac12\psi((1/2+a_D+z)/2),&\chi_D,\\ \tfrac12\log Q_d+\psi(1/2+z)-\log(2\pi),&K,\\ \psi(k/2+z)-\log(2\pi),&H_k. \end{cases} \] In particular $B=O(L)$ near $z=0$. The paired series and its inverse-square tail converge by \eqref{uncond:unitzeros}. For root number $-1$, the completed function is $z^{m_0}$ times an even nonzero function at zero, with $m_0$ odd; its logarithmic derivative is still odd after the $m_0/z$ term is retained. No assumption $m_0=1$ has been made. The contour formula is \begin{equation}\label{uncond:selbergidentity} \mathcal L(s)=P_\ell(s)+\sum_{\rho}K_\ell(\rho-s)+E_{\rm tr}(s), \end{equation} with starred versions for $K$. Start the convolution integral on $\Re z=2$, so that its Dirichlet series is absolutely convergent. Move left across the completed zeros and the gamma poles, using horizontal edges a distance $\gg1/\log(Q(2+T))$ from zero ordinates. The local count \eqref{uncond:unitzeros} and the order-one product bound the logarithmic derivative there by a fixed power of the logarithm. The kernels have vertical decay $|z|^{-3}$, or $|z|^{-3-k_\ell}$. The functional equation on left vertical edges, chosen a fixed distance from gamma poles, and exponential kernel decay let the edges tend to infinity. The residue of $-L'/L$ at a zero is its negative multiplicity; consequently the sign in \eqref{uncond:selbergidentity} is positive when the argument is $\rho-s$. This also fixes the sign if the denominator is instead written $(s-\rho)^3$. The trivial-zero contribution is $O(\ell^{-2}e^{-\ell\Re s})$ in the unstarred cases, and $O(\ell^{-2}e^{-\ell\Re s/2})$ in the starred case. For the latter the gamma poles are $0,-1,-2,\ldots$; a completed zero at a boundary point, if present, is counted separately with its multiplicity. \begin{lemma}\label{uncond:approximation} For the rational and weight families, outside the detector in Lemma~\ref{uncond:qdetector} or \eqref{uncond:hdetector}, the function $L$ is zero-free at $s=1/2+\delta+it$, simultaneously for $|t|\le T_*$, $4/\ell\le\delta\le3/2$, and \begin{equation}\label{uncond:app} \mathcal L(s)=P_\ell(s)+ O\left(e^{-\delta\ell} \{L+|P_\ell(1/2+4/\ell+it)|\}\right). \end{equation} For $K$, outside a set of probability \begin{equation}\label{uncond:kappbad} C(T_*L+r)(1+T_*^6L^5)e^{-cr}+CX^{-c_0}, \qquad r=L/\ell\to\infty,\quad T_*\le\tau/4, \end{equation} the analogous assertion holds for $8/\ell\le\delta\le3/2$, with $P_\ell^*$ and $e^{-\delta\ell/2}$ in place of $P_\ell$ and $e^{-\delta\ell}$. If $T_*\le L^{-5/6}$ the factor $1+T_*^6L^5$ in \eqref{uncond:kappbad} is bounded. \end{lemma} \begin{proof} At the anchor $s_0=1/2+a+it$, set $S_0=\sum_\rho a/|s_0-\rho|^2$ and $T_0=\sum_\rho(a-\beta)/|s_0-\rho|^2$. The ratio of the contribution of a horizontal pair to $T_0$ and to $S_0$ is \[ {a^2+(\gamma-t)^2-\beta^2\over a^2+(\gamma-t)^2+\beta^2}\ge3/5 \] whenever $|\beta|\le a/2$ or $|\gamma-t|>2|\beta|$. For $a=4/\ell$, the unstarred detector gives exactly these alternatives. In the variables $A=\ell a=4$, $B=\ell\beta$, $C=\ell(\gamma-t)$, the sum of the three exponential terms in $K_\ell$ divided by the anchor summand is, for $B\le2$, at most \[ {2\over A^2}(e^{-A/2}+2e^{-A}+e^{-3A/2})<1/32. \] For $B>2$, $|C|>e^{3B}$ and it is at most $(e^{-2A}+2e^{-5A/2}+e^{-3A})/A<1/32$. The real part of \eqref{uncond:pairedhadamard} and \eqref{uncond:selbergidentity} therefore absorb the zero sum with coefficient $3/5-1/32=91/160$, giving $S_0\ll L+|P_\ell(s_0)|$. At a rightward point with $A_1=\ell\delta\ge4$, the same comparison gives $Ce^{-A_1}$ times the anchor sum: if $B\le2$ use $|s-\rho|\ge|s_0-\rho|$ and $\ell|s-\rho|\ge A_1/2$; if $B>2$ use $|C|>e^{3B}$ to remove the positive exponential. This proves \eqref{uncond:app}. The inequalities hold up to $\delta=3/2$; this endpoint is useful for the first complex disks. For the starred kernel exclude the following two sets of zeros: \[ \begin{array}{ll} 4/\ell<\beta\le\beta_0,& |\gamma|\le T_*+\min(\tau,\ell^{-1}e^{4\ell\beta}),\\ \beta>\beta_0,&|\gamma|\le Y=X^{\theta_{\beta_0}}. \end{array} \] The second has probability $O(X^{-c_0})$ by \eqref{uncond:kfar}. Apply \eqref{uncond:kdensitybound} to the first in bins $j/\ell<\beta\le(j+1)/\ell$, $j\ge4$. Their heights are at most $2(T_*+\min(\tau,\ell^{-1}e^{4(j+1)}))$. The two prefactors are bounded by $C(T_*L+r)e^{4(j+1)}$ and $C(1+T_*^6L^5)(1+r^6/L)e^{24(j+1)}$. The geometric sum and $r\to\infty$ absorb $1+r^6/L$, proving \eqref{uncond:kappbad}. Now take $a=8/\ell$. The pair ratio is again at least $3/5$. Zeros with $\beta\le\beta_0$, $|\gamma-t|>\tau$ contribute, by \eqref{uncond:verticaldamping} and \eqref{uncond:unitzeros}, $O(Le^{-\ell\delta/2-\tau\ell/20})$. Those with $\beta>\beta_0$ have $|\gamma-t|>Y/2$ and contribute \[ O\left(\ell^{-2}e^{7\ell/4-\ell\delta/2} Y^{-2}\log(QY)\right) \ll Le^{-\ell\delta/2}X^{-c}, \] because $\ell=o(L)$. For the remaining zeros put $A=8$, and use \eqref{uncond:kernelbounds}. If $B\le4$ the kernel-to-anchor ratio is at most $8A^{-2}e^{-A/4}=e^{-2}/8<1/32$. If $B>4$, then $|C|>e^{4B}$, and the ratio is at most $4A^{-1}e^{-9A/4}<1/32$. The same absorption works. At $A_1=\ell\delta\ge8$ the ratio is $Ce^{-A_1/2}$, using $|s-\rho|\ge|s_0-\rho|$ in the first case and $|C|>e^{4B}$ in the second. This proves the starred assertion, including zero-freeness throughout the stated complex region. \end{proof} \subsection{Model bounds and the central interval}\label{uncond:central} In all three families set \[ G_D(w)=e^{-w}\mathcal L_D(1/2+e^{-w}),\qquad G_E(w)=e^{-w}\sum_{p,j\ge1}a_{p,j}(\log p)p^{-j(1/2+e^{-w})}, \] with prime ideals and their norms in the Hecke case. In the quadratic cases $a_{p,j}=X_p^j$; in the weight case $a_{p,j}=A_j(X_p)$. The analytic construction, moment bounds and Gaussian comparison of Lemmas~\ref{random:analytic-construction}, \ref{random:general-model} and \ref{random:gaussian-comparison} apply. Here their parameters are \begin{equation}\label{uncond:modelparameters} \begin{aligned} \E G_E(t)&=\mu+O(e^{-t}),\\ \operatorname{Cov}(G_E(t),G_E(s)) &={1\over4}\operatorname{sech}^2((t-s)/2)+O(e^{-t-s}),\\ \mu&=\begin{cases}1/2,&\text{quadratic cases},\\-1/2,&H_k.\end{cases} \end{aligned} \end{equation} We indicate the checks that distinguish these applications. The prime sum is an analytic $L^2$ martingale; Cauchy's formula on a slightly larger fixed sector gives local uniform convergence and derivative moments. Higher prime powers converge absolutely there. The prime number theorem, or the prime-ideal theorem for the fixed field $K$, gives $\sum(\log p)p^{-1-z}=z^{-1}+O(1)$ and $\sum(\log p)^2p^{-1-z}=z^{-2}+O(1)$ in the required small sectors. The second prime powers have mean $1+O(p^{-1})$ in the quadratic cases and $-1+O(p^{-1})$ by \eqref{uncond:weightdrift}; these give the two means in \eqref{uncond:modelparameters}. The first powers give the covariance. All remaining covariance terms are absolutely summable. Whole-prime blocks have range at most $C|a|(\log p)p^{-1/2-\Re a}/(1-p^{-1/2-\Re a})$. The squared ranges sum to $O(1)$ in each fixed sector, and their cubes sum to $O(|a|^3)$. Thus the moment and Lindeberg estimates also hold for the continuous local laws \eqref{uncond:plancherel}. In particular, for a fixed finite number of points and every $q\ge2$, the normalized model polynomials have $L^q$ norm $O(\sqrt q)$. The same holds for the actual polynomials for fixed $q$, by the arithmetic norm estimates above. At an anchor $a_0=4/\ell$ or $8/\ell$, the precise bound is \begin{equation}\label{uncond:anchormoment} \|\ell^{-1}P_\ell(1/2+a_0+it)\|_q\ll_q1, \end{equation} uniformly in the height $t$; the unnormalized norm is $O_q(\ell)$. Phases do not affect the range estimates. Finite-word transfer bounds the error for a product of degree $2m$ by $X^{-\kappa}\exp(Cm\ell+O(m\log m))$, with $X$ replaced by $k$ in the weight family. Taylor expansion of $e^{ivY}$ to order $2m-1$ then gives characteristic-function error \begin{equation}\label{uncond:cfbudget} 2(C|v|/\sqrt m)^{2m} +X^{-\kappa}\exp(Cm(\ell+\log(2+|v|)+\log m)). \end{equation} The actual even remainder is transferred before it is bounded; this is why the estimate is valid on the original family. This is the argument of Proposition~\ref{transfer:characteristic-functions}, with the actual arithmetic norms verified here. The ternary models have a uniform negative one-point moment, by Lemma~\ref{random:joint-small} and Corollary~\ref{random:ideal-onepoint-negative}: \begin{equation}\label{uncond:ternarynegative} \E|G_E(t)|^{-a_0}\le C\qquad(t\ge\log2) \end{equation} for some $a_0>0$. The prime-ideal version is obtained by precisely the same prime-window argument: the fixed-field prime-ideal theorem counts every logarithmic window, the active local variances are at least $3/4$, and prime ideals of equal norm remain independent. The single inactive ideal $(1+i)$ is omitted. No assertion about independence of equal norms is used. For the weight model the following substitutes suffice: \begin{equation}\label{uncond:weightsmall} \mathbb P(|G_E(t)-v|\le h)\ll e^th,\qquad \mathbb P(|G_E(t)-v|\le h)\ll h+e^{-t/4}. \end{equation} For the first inequality condition on all primes except $2$. If $a=e^{-t}$ and $r=2^{-1/2-a}$, its exact block is $a\log2\,(rx-2r^2)/(1-rx+r^2)$, whose derivative divided by $a$ is $\log2\,r(1-r^2)/(1-rx+r^2)^2\gg1$ on $[-2,2]$. The density in \eqref{uncond:plancherel} is bounded, giving density $O(a^{-1})$. For the second inequality smooth an interval at width $e^{-t/4}$ and use the Gaussian comparison; its first three derivative costs are absorbed by $e^{-t}$, $e^{-2t}$ and $e^{-3t}$, respectively. The Gaussian marginal density is bounded. Take now $\nu=A\sqrt H$, $\Delta=\nu/L$, with fixed $A>0$, and $\ell=L/\sqrt\nu$. On an interval $c\Delta\le\delta\le C\Delta$ the detectors and Lemma~\ref{uncond:approximation} imply \[ \|\mathbf1_{\rm good}\,\delta(\mathcal L-P_\ell) (1/2+\delta)\|_1 \ll\nu e^{-c\sqrt\nu}=o(1). \] Here \eqref{uncond:anchormoment} was used with its normalization. The detector probabilities tend to zero, including the factor $H^{20}$ in the rational case. In \eqref{uncond:cfbudget} choose $m=\lfloor\nu^{1/8}\rfloor$ and $|v|\le\nu^{1/64}$. Then $|v|/\sqrt m\ll\nu^{-3/64}$ and $m\ell/L\ll\nu^{-3/8}$. The model tail tends to zero, so uniformly in these real shifts \begin{equation}\label{uncond:anchorlaw} \delta\mathcal L_D(1/2+\delta) \Longrightarrow\ N(\mu,1/4). \end{equation} For $U$ uniform on $[1,2]$, the same statement implies \begin{equation}\label{uncond:anchoravoid} \mathbb P\bigl(|\Delta\mathcal L_D(1/2+U\Delta)|2|\beta|$. For a horizontal pair its real resolvent at $\Delta$ is at least \[ {6\over5}{\Delta\over\Delta^2+\beta^2+\gamma^2}. \] This follows by writing $D_0=\Delta^2+\beta^2+\gamma^2$: the numerator is at least $3D_0/5$ and the denominator at most $D_0^2$. Zeros on the symmetry line give nonnegative terms directly. Thus the positive statistic \[ K_D=\sum_\rho{\Delta^2\over\Delta^2+|\rho-1/2|^2} \le{5\over3}\Delta\Re\{B(\Delta)-\mathcal L_D(1/2+\Delta)\} \] has restricted expectation $O(\nu+1)$ by the anchor estimate and $B=O(L)$. If $N_D$ counts the near zeros $|\rho-1/2|\le12\Delta$ and $W_D=\Delta^2\sum_{|\alpha|>12\Delta}|\alpha|^{-2}$ is over the paired representatives in \eqref{uncond:pairedhadamard}, then $N_D\le145K_D$ and $W_D\le145K_D/288$. All hypotheses of Lemma~\ref{analytic:local-jensen} now hold with the original denominator and its good-set indicator. Using \eqref{uncond:anchoravoid}, that lemma bounds the probability of a count exceeding $\eta H$ by \[ o(1)+{C\{\nu+1+\log(\nu+2)+\sqrt H\}\over\eta H}=o(1). \] The central pole $m_0/z$ is included in $K_D$ and in $N_D$. At a central zero of order $m_0$, the derivative has order $m_0-1$; the lemma counts this contribution. The argument consequently applies unchanged to every root-number $-1$ form, without a rank-one assumption. Finally $L'\not\equiv0$: a quadratic character has an active prime, and for a level-one form $\lambda_f(2)=0$ implies $\lambda_f(4)=-1$. \end{proof} \subsection{The short bridge}\label{uncond:bridgesub} We will need a growing-dimensional small-ball test for the weight family as well as for the ternary models. The following variant makes the necessary hypothesis on continuous blocks explicit. \begin{lemma}\label{uncond:rangeblock} Let $Y=\mu+\sum\xi_p\in\mathbb R^r$ be a finite sum of independent centred blocks, with covariance $cI\le\Sigma\le CI$ and $\|\mu\|^2\le Cr$. Suppose $\|\xi_p\|\le\beta_p\le1$ and, for every $h\in\mathbb R^r$, the directional ranges satisfy \[ \sum_p\bigl(\mathop{\rm ess\,sup}h\cdot\xi_p -\mathop{\rm ess\,inf}h\cdot\xi_p\bigr)^2 \le C\|h\|^2. \] For all sufficiently large $r$, and a sufficiently small fixed $\epsilon_0>0$, put $A_0=\operatorname{tr}\Sigma+\|\mu\|^2$ and $m=\lfloor\epsilon_0r\rfloor$ and $\mathcal Q(y)=((A_0-\|y\|^2)/(A_0-1))^{2m}$. Then \[ \mathbf1_{\{\|y\|\le1\}}\le\mathcal Q(y),\qquad \E\mathcal Q(Y)\le e^{-cr}+e^{Cr}\sum_p\beta_p^3. \] \end{lemma} \begin{proof} For sufficiently large $r$, $A_0-1\ge cr/2$ and $m\ge1$. The Gaussian calculation is that in Lemma~\ref{random:finite-block-test}: its centred quadratic norm has $L^q$ norm $O(\sqrt{rq}+q)$, so the expected polynomial is at most $2^{-2m}$ when $\epsilon_0$ is small. For completeness, the replacement argument uses only the displayed directional-range condition. Hoeffding bounds every mixed partial sum, including Gaussian replacement blocks, by $\E e^{h\cdot W}\le e^{C\|h\|^2}$. A $1/2$-net of the sphere with at most $5^r$ points gives $\|W\|_{L^q}\ll\sqrt r+\sqrt q$. Write $p=2m$ and $q(y)=A_0-\|y\|^2$. The third directional derivative is \[ (A_0-1)^{-p}\{p(p-1)(p-2)q(y)^{p-3}(-2y\cdot h)^3 +3p(p-1)q(y)^{p-2}(-2y\cdot h)(-2\|h\|^2)\}. \] Its absolute value is bounded by $e^{Cr}\|h\|^3(1+\|y\|/\sqrt r)^{2p+3}$. Taylor's formula to order two cancels the matching means and covariances. Independence of the omitted block and the partial sum, the preceding norm bounds, and $\|g_p\|_{L^q}\ll\sqrt q\,\beta_p$ for its Gaussian replacement bound each remainder by $e^{Cr}\beta_p^3$. For example the expansion of the power in the derivative reduces to $r^{-j/2}(j+3)^{(j+3)/2}\le C^j(Cr)^{3/2}$ for $j=O(r)$; the polynomial factor is absorbed in $e^{Cr}$. Sum over the finite blocks. The even power makes $\mathcal Q$ nonnegative, and $A_0-\|y\|^2\ge A_0-1>0$ on the unit ball. \end{proof} Put $U=H^{1/4}$ and choose \begin{equation}\label{uncond:cutoffs} u=\begin{cases}\lfloor\sqrt H\rfloor,&\mathcal F(X),\\ \lfloor H^{1/8}\rfloor,&\mathcal D_X,H_k, \end{cases} \qquad d={H^8e^{2u}\over L},\qquad \Delta={AU^2\over L}. \end{equation} Here $A$ will be a sufficiently large fixed constant. The bridge is $[1/2+\Delta,1/2+e^5d]$. \begin{proposition}\label{uncond:bridge} For this choice of $A$ the multiplicity count on the bridge is $o_{\mathbb P}(H)$ in all three families. \end{proposition} \begin{proof} Cover its logarithmic image by half-step cells, with centres $t_j$ and $\delta_j=e^{-t_j}$. Extend the endpoint centres by at most $1/2$. The number of cells is $n=O(u+\log H)$ and, writing $S_j=\delta_jL$, \begin{equation}\label{uncond:bridgerange} e^{-1/2}AU^2\le S_j\le e^{11/2}H^8e^{2u}. \end{equation} The required heights are $T_*=3\delta_j$. In the rational and Hecke families these are $o(L^{-5/6})$, because $\delta_jL^{5/6}\le\exp(-H/6+O(u+\log H))$. Thus every detector is used within its actual height range. First use $\ell_j=L/\sqrt{S_j}$ to control large maxima. The detector errors are bounded respectively by \[ CH^{20}(1+3\sqrt{S_j})e^{-c\sqrt{S_j}}+Ce^{-4H},\quad C(S_j+\sqrt{S_j})e^{-c\sqrt{S_j}}+CX^{-c_0},\quad C(S_j+\sqrt{S_j})e^{-c\sqrt{S_j}}. \] Their sum over the cells is $o(1)$. The normalized approximation error has every fixed restricted $L^q$ norm at most $C_qS_je^{-c\sqrt{S_j}}$, by \eqref{uncond:anchormoment}. Cauchy's formula on fixed disks and the polynomial moment bounds therefore show that the sum of the positive logarithms of the large disk maxima has expectation $O(n)=o(H)$. The small disk maximum requires a longer polynomial. Choose $r=\lfloor\epsilon U\rfloor$, with fixed $\epsilon\log64<1/4$, put $v_j=r+\log(2+S_j)$, and take \begin{equation}\label{uncond:bridgecoefficientcutoff} \ell'_j=C_yv_j/\delta_j. \end{equation} The ratio $L/\ell'_j=S_j/(C_yv_j)$ tends to infinity uniformly and is $\gg U$ after $A$ is chosen. More precisely, \[ \sup_j{\log H+\log S_j\over S_j/(C_yv_j)}\longrightarrow0. \] This follows from the monotonicity of $S/(r+\log(2+S))$ for large $S$ and \eqref{uncond:bridgerange}. It is needed in the rational case, where $\log S_j$ is not $o(r)$. The detector errors for \eqref{uncond:bridgecoefficientcutoff}, including the $H^{20}$ factor, again sum to $o(1)$. Use the Cayley map $a(z)=\delta_j(1+z)/(1-z)$ and define \[ \mathcal H_D(z)={\delta_j\over(1-z)^2} \mathcal L_D(1/2+a(z)),\qquad Z_k={2\over\sqrt{k+1}}[z^k]\mathcal H_D(z),\quad0\le ke^{b_0v_j}$ has probability at most $Cv_j^4e^{-2b_0v_j}\le e^{-b_0v_j}$. Choose $C_y$ after $b_0$. The normalized analytic error is then at most $Ce^{-C_yv_j/6}(e^{b_0v_j}+S_j)\le e^{-30v_j}$ (the denominator $6$ allows the starred kernel). Another Cauchy estimate, costing $C\sqrt r\,2^r$, gives $\|Z-Y_D\|<1/4$. Thus a small actual disk maximum implies $\|Y_D\|\le1$ outside the stated exceptional set. We verify the model-vector hypotheses instead of assuming independent Taylor coefficients. Expansion of \eqref{uncond:modelparameters} under the Cayley map gives \begin{equation}\label{uncond:cayleycov} \E Y_k={2\mu\,\mathbf1_{2\mid k}\over\sqrt{k+1}}+O(\delta_j2^k), \qquad \operatorname{Cov}(Y_k,Y_l) =\mathbf1_{k=l}+O(\delta_j^2 2^{k+l}). \end{equation} The squared polynomial-tail contribution is $O(r4^re^{-\delta_j\ell'_j/6})$. Also $\delta_j^2r4^r\to0$ uniformly. Hence $\frac12I\le\Sigma\le2I$ and $\|\mu_Y\|^2\le Cr$. For a ternary prime block write $\xi_p=a_pX_p+b_p(X_p^2-\theta_p)$. Odd and even parts are uncorrelated, $\theta_p\ge2/3$ in the rational case and $\theta_p\ge3/4$ for active ideals. Thus $\sum a_pa_p^{\mathsf T}\ll I$, while the higher-power estimate gives $\sum\|b_p\|^2=o(1)$. Moreover $\max\|\xi_p\|\le\delta_je^{Cr}=o(1)$ and $\sum\|\xi_p\|^3\le\delta_j^3e^{Cr}$. Lemma~\ref{random:finite-block-test} applies. For the weight model write instead $\xi_p=a_pX_p+\eta_p(X_p)$, with $\eta_p$ the centred higher powers. Cauchy's coefficient estimate on $|z|=1/2$ gives \[ \|\eta_p\|_\infty\le b_p=C\delta_j\sqrt r\,2^r (\log p)p^{-1-2\delta_j/3},\qquad \|a_p\|\le C\delta_j\sqrt r\,2^r (\log p)p^{-1/2-\delta_j/3}. \] Here $\sum b_p^2=o(1)$. The $L^2$ triangle inequality, used separately at each prime, implies \[ \sum_p\E X_p^2(h\cdot a_p)^2 \le2\operatorname{Var}(h\cdot Y)+2\sum_p\E|h\cdot\eta_p|^2 \ll\|h\|^2. \] Since $\E X_p^2\ge1$, the linear Gram matrix is bounded. The directional range is at most $4|h\cdot a_p|+2\|h\|b_p$, proving the hypothesis of Lemma~\ref{uncond:rangeblock}. The block envelopes have $\sum\beta_p^3\le\delta_j^3e^{Cr}$ and maximum $o(1)$. Correlation between $X_p$ and its higher powers has not been discarded. It follows in every family that $\E\mathcal Q(Y_E)\le e^{-cr}+\delta_j^3e^{Cr}$. The polynomial $\mathcal Q$ has degree $4m$, not $2m$. The norm of each coefficient coordinate is bounded by $C2^r\exp(C_N\ell'_j)$ with $C_N=9/2,8,7$ respectively. Consequently its arithmetic norm has logarithm at most \begin{equation}\label{uncond:bridgebudget} 4C_Nm\ell'_j+O(mr+m\log r+m). \end{equation} Choose constants in this order: the arithmetic exponent $\kappa$, then $\epsilon$, then the small $\epsilon_0$ in the polynomial test, then $b_0$, then $C_y$, and finally $A$. Indeed, \[ {m\ell'_j\over L}\ll {r(r+\log(2+S_j))\over S_j} \ll {\epsilon^2\over A}+o(1). \] Increasing $A$ makes the leading term in \eqref{uncond:bridgebudget} smaller than $\kappa L/4$, and its remaining terms are $o(L)$. The arithmetic error is therefore at most $X^{-\kappa/2}$ (or $k^{-\kappa/2}$). This proves, for every cell, the small-maximum probability bound $Ce^{-cU}+e^{-H}+X^{-\kappa/2}$, including the analytic exceptions already summed. Its sum over $n$ tends to zero. On the retained set all depths $\log^+(1/m_j)$ are at most $U$. Their sum is at most $nU=o(H)$: in the rational case this is $O(H^{3/4}+H^{1/4}\log H)$, and in the other cases it is smaller. The large-log sum has expectation $O(n)$, as proved at the start. The shifted Jensen inequality \eqref{random:shifted-jensen} on each cell now proves the assertion by Markov. The approximation supplies zero-freeness of $L$ on the surrounding complex disks, so the zeros and multiplicities of $G_D$ there are exactly those of $L'_D$. \end{proof} \subsection{The bulk count with retained local moments}\label{uncond:bulksub} We provide the transfer and counting argument on the unconditional good sets just constructed. In particular an approximation under GRH is not being used for this step. Let $d,u$ be as in \eqref{uncond:cutoffs}, and set \begin{equation}\label{uncond:bulkparameters} b_H=-\log d=H-8\log H-2u,\qquad \ell_B={C_yH\over d}={C_yL\over H^7e^{2u}},\qquad a_H=H^{1/4}. \end{equation} Consider the disks $|w-t|\le.9$ about half-step centres. For $a=e^{-w}$, $\Re a\ge c_+e^{-t}$ with $c_+=e^{-.9}\cos(.9)>0$, and $|a|<3/2$ for $t\ge\log2$. The rational coarse detector, with $T_*=1$, and the weight detector therefore give one common zero-free event on all disks with centres $\log2\le t\le b_H$. For the Hecke family take first $t\ge t_0=\log(16e^{.9}/\tau)$ and use the starred detector with $T_*=\tau/4$. In all cases \[ {L\over\ell_B}={H^7e^{2u}\over C_y},\qquad {L/\ell_B\over H}\longrightarrow\infty. \] This pays any fixed power of $L$ in the detector, including the $L^5$ factor in \eqref{uncond:kappbad} at fixed height. The anchor line segment is covered by $O(\ell_B+1)$ disks of radius $1/(2\ell_B)$ and slightly larger disks of radius $1/\ell_B$. Equation \eqref{uncond:anchormoment}, finite-moment transfer, and Cauchy's formula show that \[ \mathbb P\left(\sup_{|t|\le T_*} |P(1/2+a_0+it)|>Le^{4H}\right)\ll e^{-7H}. \] Indeed the second moment costs $O((\ell_B+1)\ell_B^2)$ and division by $L^2e^{8H}$ gives the assertion since $\ell_B\le L$. For each prescribed fixed $E>0$, choose $C_y$ sufficiently large. Lemma~\ref{uncond:approximation} then gives, on this common event, \begin{equation}\label{uncond:commonapp} \sup_{t,w}|G_D(w)-e^{-w}P(1/2+e^{-w})|\le e^{-EH}. \end{equation} In the starred case the exponent in the kernel is halved, which is allowed in the choice of $C_y$. The squared random tail is $O(e^{-c\Re(a)\ell_B})$; its $L^2$ norm has half this exponent, again paid by the same choice. For every fixed $q$ the normalized actual polynomial moments are bounded, since $q\ell_B=o(L)$. Applying Cauchy's formula to each good path and then taking expectations bounds all fixed moments of maxima and fixed derivatives on the smaller disks. These are restricted moments with the original denominator. In the Hecke family the fixed interval $[\log2,t_0+2]$ is treated without using a shrinking-height detector. Let $a_*>0$ be a lower bound for $\Re(e^{-w})$ on its disks, and take $\alpha=a_*/2$ in \eqref{uncond:kfar}. Outside probability $O(X^{-c})$, zeros with $\beta\ge\alpha$ have ordinate larger than $Y=X^{\theta_\alpha}$. Use the exact starred identity with $\ell=C_yH$. Zeros with $\beta<\alpha$ contribute at most $C\ell^{-2}Le^{-a_*\ell/4}$, by \eqref{uncond:unitzeros} and the negative-real kernel bound. The others contribute at most $C\ell^{-2}e^{7\ell/4}Y^{-2}\log(QY)=O(X^{-c})$. Thus \eqref{uncond:commonapp} and its moment consequences hold on these finitely many disks as well, with constants depending on $t_0$. We next control the lower tails of their small-disk maxima. Write $m_j=\max_{|w-t_j|\le1/8}|G_D(w)|$ and $v_j=\log^+(1/m_j)$. A convenient nonnegative majorant is \[ g_n(x)=\exp(1-e^{2n}x^2/4), \qquad \mathbf1_{\{|x|\le e^{-n}\}}\le g_n(x). \] Its Fourier transform has bounded normalized $L^1$ norm, and its tail outside $|v|\le C_E\sqrt H e^n$ is $O(e^{-EH})$. For the two quadratic models \eqref{uncond:ternarynegative} gives $\E g_n(G_E(t))\ll e^{-a_0n}$. Use $m=\lceil H^4e^{2u}\rceil$ in \eqref{uncond:cfbudget}, for every $n\le u$. Uniformly over the disks, \begin{equation}\label{uncond:depthbudget} {|v|\over\sqrt m}=O(H^{-3/2}),\qquad {m\ell_B\over L}=O(H^{-3}),\qquad m(u+\log H)=o(L). \end{equation} Thus the finite Taylor error and the arithmetic norm error are smaller than every fixed negative power of $e^H$. The Lipschitz cost $O(e^n)$ in \eqref{uncond:commonapp} and in the random tail is also absorbed. Positivity permits us to drop the good-set indicator when bounding the polynomial majorant, giving \begin{equation}\label{uncond:depthtail} \mathbb P({\rm good},v_j\ge n) \ll e^{-a_0n}+O(e^{-EH}),\qquad 1\le n\le u. \end{equation} For the finitely many Hecke front disks, the same calculation uses $m\ell/L=O(H^5e^{2u-H})\to0$. Retain only samples with all these depths below $u$. The discarded probability is $O(He^{-a_0u})+o(1)=o(1)$. Summing the truncated tails in \eqref{uncond:depthtail} gives uniformly bounded restricted fourth moments of $v_j$. The shifted Jensen inequality, the positive maximum moments, and Cauchy's formula therefore give \begin{equation}\label{uncond:retainedmoments} \E[\mathbf1_{\rm retained}N_j^4]\ll1,\qquad \E[\mathbf1_{\rm retained}\sup_{[t_j-1/4,t_j+1/4]}|G_D''|^8]\ll1, \end{equation} where $N_j$ is the multiplicity count on the half-cell. In particular the fixed Hecke front interval has count $O_{\mathbb P}(1)$. The weight model requires a separate early interval because the first bound in \eqref{uncond:weightsmall} grows with $t$. On $[\log2,a_H+2]$ use $d_F=e^{-a_H-3}$, $\ell_F=C_yH/d_F$, and cap $v_F=4a_H$. Use $m=\lceil H^4e^{2v_F}\rceil$ for its depth tests. Then $m\ell_F/L=O(H^5e^{9a_H+3-H})\to0$, and the other two Taylor budgets are as in \eqref{uncond:depthbudget}. The first estimate in \eqref{uncond:weightsmall} gives \[ \mathbb P({\rm good},v_j\ge n)\ll\min(1,e^{t_j-n})+e^{-EH}. \] The union of the early cap failures is $O(a_He^{-3a_H+3})+o(1)$, and the restricted expected depth is $O(1+t_j)$. Jensen summed over the early cells gives expected count $O(a_H^2)=O(H^{1/2})=o(H)$. On the late interval $[a_H-2,b_H-2]$ use $\ell_B$ and cap $u$. The second estimate in \eqref{uncond:weightsmall} gives instead $\mathbb P({\rm good},v_j\ge n) \ll e^{-n}+e^{-t_j/4}+e^{-EH}$. Its union of cap failures is $O(H(e^{-u}+e^{-a_H/4}+e^{-EH}))=o(1)$, and its restricted fourth depth moment is bounded by $C+Cu^4(e^{-a_H/4}+e^{-EH})=O(1)$. Thus \eqref{uncond:retainedmoments} holds on the late interval also in the weight family. No uniform negative moment for the weight model has been assumed. \begin{proposition}\label{uncond:bulk} For each family and each fixed $\eta>0$, outside a proportion $o_\eta(1)$, both the distinct and the multiplicity counts on $[1/2+e^2d,1]$ are between $(c_*-\eta)H$ and $(c_*+\eta)H$. \end{proposition} \begin{proof} Work on the retained set just constructed, putting the real process equal to $1$ on its complement whenever a bounded sign test is used. Its exceptional probability tends to zero. The counts on the time interval before $a_H$ are $o_{\mathbb P}(H)$: for the quadratic cases this follows from \eqref{uncond:retainedmoments} and $O(a_H)$ cells; for weight it was proved separately above. For at most four time points in $[a_H,b_H-2]$, Taylor transfer with $m=\lceil H^4\rceil$ and Fourier cutoff $H$, followed by fixed smooth Fourier approximation, gives uniform convergence of bounded smooth tests to the Gaussian process \[ \mathcal G(t)=\mu+\tfrac12 g(t),\qquad \E g(t)g(s)=\operatorname{sech}^2((t-s)/2). \] The budgets are $H/\sqrt m=H^{-1}$ and $m\ell_B/L=O(H^{-3}e^{-2u})$. The model Gaussian comparison is uniform since the smallest time tends to infinity. Repeated and arbitrarily close points are allowed: no inverse covariance matrix enters the smooth comparison. To pass to signs, truncate values at height $K$, costing $O(K^{-2})$, and smooth the sign at width $\varepsilon$, costing $O(\varepsilon)+o(1)$ by the Gaussian marginal density. Let $X$ or $k$ tend to infinity first, then $K\to\infty$ and $\varepsilon\to0$. Thus two-point and four-point strict-sign tests converge uniformly. Fix a small mesh $h>0$. Choose a deterministic offset, depending on the finite arithmetic family if necessary, so that no retained function vanishes at an internal mesh point. This is possible because there are finitely many functions and each has only finitely many zeros on the compact interval. Let $S_h$ count crossing cells. The two-point convergence and stationarity give \begin{equation}\label{uncond:signmean} {\E S_h\over H}\longrightarrow {p(h)\over h},\qquad p(h)=\mathbb P(\mathcal G(0)\mathcal G(h)<0). \end{equation} For pairs of cells separated by at most $R$ there are only $O_h(H(R+1))$ pairs. For the others the off-diagonal Gaussian covariance block is $O(e^{-R})$. Interpolate this block for smoothed sign tests. Gaussian integration by parts bounds the difference from the product of expectations by $O(\varepsilon^{-2}e^{-R})$; the four marginal smoothing strips cost $O(\varepsilon)$. With $\varepsilon=e^{-R/3}$ the total is $O(e^{-R/3})$. The uniform four-point arithmetic convergence now implies \[ \limsup_{X\to\infty}\operatorname{Var}(S_h/H)\ll_h e^{-R/3}. \] Let $R\to\infty$. The same statement holds with $k$ as the family parameter. Therefore $S_h/H-p(h)/h\to0$ in probability. It remains to recover all multiplicities from mesh signs. If a cell of length $h$ contains at least two zeros counted with multiplicity, Rolle's theorem, including a repeated zero, gives \[ |G_D(t_{\mathrm{left}})|\le h^2\sup_I|G_D''|. \] Set $K=h^{-1/4}$. One-point transfer and the second bound in \eqref{uncond:retainedmoments} imply \[ \mathbb P({\rm retained},N_I\ge2) \ll h^{7/4}+h^2+o(1). \] H\"older and the first bound in \eqref{uncond:retainedmoments} give $\E[\mathbf1_{\rm retained}N_I\mathbf1_{N_I\ge2}] \ll h^{21/16}+o(1)$, uniformly over cells. Summing $O(H/h)$ cells bounds the expected multiplicity excess by $CHh^{5/16}+o(H)$. Cells meeting the two ends cost $O_{\mathbb P}(1)$ by the same local moments. A cell with one simple zero has a strict sign change, while all failures of this correspondence are bounded by the preceding excess. Hence the distinct and multiplicity counts, divided by $H$, differ from $S_h/H$ by a quantity with expectation at most $Ch^{5/16}+o(1)$ on the retained set. Finally the Gaussian calculation \eqref{random:crossing-constant}, with the parameters of \eqref{uncond:modelparameters}, gives $p(h)=c_*h+O(h^3)$. The sign change $\mu=1/2$ to $-1/2$ does not alter this probability. Choose $h$ sufficiently small for the prescribed $\eta$, then let the arithmetic parameter tend to infinity. This proves the proposition. The interval ends at $b_H-2$ in time, which is exactly the physical lower offset $e^2d$. \end{proof} \subsection{Assembly and normalization}\label{uncond:assembly} The three physical intervals \[ [1/2,1/2+\Delta],\qquad [1/2+\Delta,1/2+e^5d],\qquad [1/2+e^2d,1] \] cover $[1/2,1]$, with overlap between the bridge and bulk. Propositions~\ref{uncond:centralnegligible} and \ref{uncond:bridge} make the first two multiplicity counts $o_{\mathbb P}(H)$, while Proposition~\ref{uncond:bulk} gives both bulk counts $c_*H+o_{\mathbb P}(H)$. An upper bound uses the sum of these three counts; a lower bound uses just the bulk. Thus on the full closed interval \[ R^{\rm dist}(L')=c_*H+o_{\mathbb P}(H),\qquad R^{\rm mult}(L')=c_*H+o_{\mathbb P}(H). \] The deterministic multiplicity/sign inequality of Lemma~\ref{analytic:sign-sandwich} is \[ 2R^{\rm dist}(L')-R^{\rm mult}(L')-2 \le S^-(L')\le R^{\rm dist}(L'). \] The subtraction of two allows for the closed endpoints, and gives the same asymptotic for strict changes of sign on the open interval. For the rational family discard $|D|\le\sqrt X$, which is a proportion $O(X^{-1/2})$ by \eqref{uncond:qsize}. On the remaining family, $H-\log2\le\log\log|D|\le H$. Given the requested tolerance, use a smaller fixed tolerance in the $H$-normalized statement; the bounded difference is then absorbed for all sufficiently large $X$. This proves Theorem~\ref{uncond:quadratic} with the full cumulative denominator, not a conditional dyadic denominator. For the Gaussian family $128X