Quantitative negative tests for the Weil form near the height horizon

1. Result and relation to prior work

The converse in Weil's criterion constructs a negative compactly supported test if the Riemann zeta function has a zero off the critical line. We quantify a physical support radius and a normalized negative margin for this implication. The construction uses a complex shift of the theta kernel or a Gaussian frequency filter, together with contour interpolation of entire zero clusters. It requires neither simple zeros nor separation within a cluster.

Write $\Xi(z)=\xi(1/2+iz)$. A nonreal ordinate $T+i\eta$, with $T>0$ and $0<\eta<1/2$, corresponds to a hypothetical off-line zero. The symmetries of $\Xi$ allow this quadrant to represent any off-line zero away from bounded height. Let $Q$ be the complete geometric Weil form, with normalization in Section 2.

Theorem 1.1 (support and negative margin). Suppose $\Xi(T+i\eta)=0$, $T\ge20$, and $0<\eta<1/2$. Put

$$ K=\log(T+2)\log(e/\eta). $$

If $0<d\le1$ and

$$ K\le \frac{d^{3/2}T}{20000},\qquad L=\frac12\log\frac{(1+d)T}{2\pi}, \tag{1.1} $$

then each real parity sector contains a nonzero $f\in C_c^\infty(-L,L)$ such that

$$ \frac{Q(f)}{\|f\|_2^2}\le-\exp(-30000K/d). \tag{1.2} $$

These constants are deliberately conservative. In particular, the theorem is not a proposed practical certificate at moderate height.

Corollary 1.2 (optimized sufficient radius). If $K/T\le1/20000$, one may choose a negative test of each parity with

$$ e^{2L}=\frac{T}{2\pi} +\frac{20000^{2/3}}{2\pi}T^{1/3}K^{2/3}, \tag{1.3} $$

and normalized margin at least $\exp[-(30000/20000^{2/3})T^{2/3}K^{1/3}]$. This follows by taking $d=(20000K/T)^{2/3}$ in Theorem 1.1. For fixed $\eta>0$, the sufficient radius is $L=\tfrac12\log(T/(2\pi))+o(1)$.

We also obtain a simultaneous statement. For a parity sector, define $n_-(Q_L)$ as the supremum of dimensions of subspaces of its smooth interior tests on which $Q$ is strictly negative.

Theorem 1.3 (several clusters and negative index). Let $H\ge20$, $0<\eta_0<1/2$, $K=\log(H+2)\log(e/\eta_0)$, and suppose $K\le d^{3/2}H/20000$, with $0<d\le1$. Set $L=\tfrac12\log((1+d)H/(2\pi))$. If there are ordinates $T_j+i\eta_j$, $1\le j\le k$, with

$$ 20\le T_j\le H,\quad \eta_0\le\eta_j<1/2,\quad |T_j-T_\ell|\ge3\quad(j\ne\ell), \tag{1.4} $$

then each real parity sector has a $k$-dimensional smooth interior subspace on which every nonzero vector satisfies $Q(f)/\|f\|_2^2\le-\exp(-30000K/d)$. If $N_{\rm off}(H,\eta_0)$ counts the ordinates with $20\le\Re z\le H$ and $\eta_0\le\Im z<1/2$, with multiplicities, then

$$ n_-(Q_L^+),\ n_-(Q_L^-) \ \ge\left\lceil\frac{N_{\rm off}(H,\eta_0)} {140\log(H+2)}\right\rceil. \tag{1.5} $$

Theorem 1.4 (Gaussian alternative). In the notation of Theorem 1.3, replace its slack hypothesis by

$$ K\le\frac{d^2H}{10^7}. \tag{1.6} $$

Then the same simultaneous and counting conclusions hold, with the normalized quotient bound strengthened to $-\exp(-10000K)$. In particular, for a single zero take $H=T$ and $\eta_0=\eta$. If $K/T\le10^{-7}$, one may choose

$$ e^{2L}=\frac{T}{2\pi}+\frac{\sqrt{10^7KT}}{2\pi}, \tag{1.7} $$

while retaining a margin at least $\exp(-10000K)$ in each parity. For fixed displacement, this is a negative power of height even as the slack tends to zero. Equation (1.7) uses a larger sufficient support excess than (1.3); the two estimates describe different radius/margin tradeoffs. Neither is asserted to be optimal.

The classical criterion, its converse, and the fixed-window variational setting are prior work; see Bombieri [1]. His Sections 8--11 also determine the negative inertia of finite zero matrices and analyze the survival or collapse of their negative eigenvalues in the infinite limit. Our index bound is quantitative for the complete form in the stated physical window; negative inertia as a diagnostic is not new. The theta radical and prolate approximate null vectors are discussed by Connes and Consani [2]. Our earlier theta-trial estimates [3] provide the complete-form continuity bound used below; we include its short proof. Those estimates concern unconditional near-null trials, whereas the present theorem constructs negative trials conditional on off-line zeros.

The height $H=2\pi e^{2L}$ is already discussed in the scale-resolved Weil literature of Blum [4], which also credits Zhu [5]. We do not claim the scale itself. The present contribution is a uniform construction with explicit displacement/slack conditions, normalized negative margins for two radius regimes, and simultaneous physical-window negative directions. The proof distinguishes a sufficient detection radius from an optimal first-crossing asymptotic. A local density or Nyquist heuristic does not establish these uniform statements.

The finite sampled-form study of guinand_dual_scribe [6] gives an exact rank-one detection predicate and a small-displacement expansion for an injected quartet in a positive definite finite model. That calculation does not control the complete geometric form or its spectral tail. Our selectors use the complete zero set through the entire function $\Xi$, and the cutoff is estimated directly in the complete form. No finite sampling matrix or finite-dimensional coercivity premise enters the proof.

None of the results establishes positivity of a window form, a new unconditional zero-free region, or RH. The contraposition requires a certificate for the complete form on all tests; positivity of a finite Gram matrix does not supply it.

2. Complete form and analytic domain

Use $\widehat f(z)=\int_{\mathbb R}f(x)e^{-izx}\,dx$ and $C_f(u)=\int_{\mathbb R}f(x+u)\overline{f(x)}\,dx$. The form is

$$ \begin{aligned} Q(f)&=\frac1{2\pi}\int_{\mathbb R}h(t)|\widehat f(t)|^2\,dt +2\Re\bigl(\widehat f(i/2)\overline{\widehat f(-i/2)}\bigr)\\ &\quad-2\sum_{n\ge2}\frac{\Lambda(n)}{\sqrt n} \Re C_f(\log n),\\ h(t)&=\Re\psi(1/4+it/2)-\log\pi . \end{aligned} \tag{2.1} $$

For compact support the prime sum is finite. We retain the entire sum when a noncompact theta-derived function is used. Let

$$ X=H^1(\mathbb R)\cap L^2(\mathbb R,e^{2|x|}dx),\qquad N(g)^2=\|g'\|_2^2+\|e^{|x|}g\|_2^2. $$

Lemma 2.1. The Hermitian polarization of (2.1) is continuous on $X$, and $|Q(g)|\le20N(g)^2$.

Proof. The convergent digamma series gives $|h(t)|\le6+(65/4)t^2$: its constant lies between $-6$ and zero, and $\sum_{n\ge0}(n+1/4)^{-3}<65$. The archimedean contribution is bounded by $6\|g\|_2^2+(65/4)\|g'\|_2^2$. Each pole evaluation is at most $\sqrt2\|e^{|x|}g\|_2$. Furthermore

$$ |C_{f,g}(u)|\le e^{-|u|} \|e^{|x|}f\|_2\|e^{|x|}g\|_2, $$

by $|x|+|x+u|\ge|u|$ and Cauchy--Schwarz. The integral test gives $\sum_{n\ge2}\log n/n^{3/2}<5$. Thus poles and primes contribute at most $4$ and $10$ times the weighted squared norm. Summation proves the claim, and polarization proves continuity. $\square$

For the smooth functions used below, the polarized explicit formula is

$$ Q(f,g)=\sum_{\gamma}m_\gamma\widehat f(\gamma) \overline{\widehat g(\bar\gamma)}, \tag{2.2} $$

where $\gamma=(\rho-1/2)/i$ runs over distinct nontrivial ordinates and $m_\gamma$ is the multiplicity. This is an indefinite pairing before RH is known. All functions to which we apply (2.2) have exponentially weighted integrable weak second derivative. Their transforms are $O((1+|\Re z|)^{-2})$ uniformly for $|\Im z|\le1/2$. The usual $O(R\log(R+2))$ zero count therefore gives absolute convergence in (2.2). Smooth cutoffs justify the noncompact extensions using Lemma 2.1 and the same uniform transform decay.

For $x\ge0$, let

$$ P(x)=\sum_{n\ge1} (4\pi^2n^4e^{9x/2}-6\pi n^2e^{5x/2}) e^{-\pi n^2e^{2x}}. \tag{2.3} $$

Theta inversion makes $P$ even; its Fourier transform is $\Xi$. The normally convergent theta series and the identity theorem extend this evenness throughout $|\Im x|<\pi/4$. All derivatives have superexponential decay along closed horizontal substrips.

3. Whole-cluster contour selectors

Let $0\le a<\pi/4$ and put

$$ p_a(x)=P(x-ia),\quad A_a(z)=e^{az}\Xi(z),\quad q=\cos(2a),\quad \epsilon=\pi/4-a. \tag{3.1} $$

Moving the Fourier contour inside the strip gives $\widehat p_a=A_a$. The superexponential tails kill its vertical ends. Since $P$ is even and real on the real line, $p_a(-x)=\overline{p_a(x)}$, in particular $|p_a|$ is even.

Suppose $C$ is a positively oriented rectangle containing a finite nonempty upper-right zero cluster $\mathcal A$, with no boundary zero, lying in $0<\Im w\le1$. Assume its four reflected cluster interiors are disjoint. Define, initially outside $C$,

$$ U_a(z)=\frac{A_a(z)}{2\pi i} \oint_C\frac{dw}{A_a(w)(z-w)},\qquad D_a=\frac1{2\pi}\oint_C\frac{|dw|}{|A_a(w)|},\quad S=\sup_C|w|. \tag{3.2} $$

The integral is the sum of principal parts of $1/A_a$. Consequently $U_a$ extends entirely, equals $1$ modulo each enclosed zero's full multiplicity, and vanishes to full multiplicity at every other zero.

For an explicit inverse without an unjustified exchange of integrals, write

$$ f_w^+(x)=-ie^{iwx}\int_x^\infty e^{-iwv}p_a(v)\,dv,\qquad f_w^-(x)= ie^{iwx}\int_{-\infty}^x e^{-iwv}p_a(v)\,dv. $$

Both satisfy $(D-iw)f_w=ip_a$ and their difference is $-ie^{iwx}A_a(w)$. Thus the contour integrals of $f_w^\pm/A_a(w)$ coincide. Call their common value, including the normalization $1/(2\pi i)$, the physical function $f_C$.

To compute its transform, let $\alpha$ have multiplicity $m$. For $1\le j\le m$, the functions $(j-1)!^{-1}\partial_w^{j-1}f_w^\pm|_{w=\alpha}$ agree, because the differentiated defect vanishes. They decay at both ends. Differentiating the ODE and taking Fourier transforms gives successively $A_a(z)/(z-\alpha)^j$. Taking the finitely many residues of $f_w^+/A_a(w)$ proves $\widehat f_C=U_a$. This argument never integrates a growing individual $f_w^+$ over the whole real line. It also handles repeated zeros.

Let $U^\#(z)=\overline{U(\bar z)}$, set $H=i(U_a-U_a^\#)$, and form its even and odd projections. At $\mathcal A,\bar{\mathcal A},-\mathcal A,-\bar{\mathcal A}$ the even projection has values $i/2,-i/2,i/2,-i/2$, and the odd projection has values $i/2,-i/2,-i/2,i/2$. Elsewhere both vanish on the zero set. If $M=\sum_{\alpha\in\mathcal A}m_\alpha$, (2.2) gives energy $-M$ in each sector. The even inverse is real even; the odd inverse becomes real odd after multiplication by a unit phase. Denote either real physical selector by $h$. The multiplier $e^{az}$ need not be even: only the reflected zero set is used here.

4. Uniform contour cost without a separation hypothesis

Lemma 4.1. For $T\ge20$ and a zero $T+i\eta$, $0<\eta<1/2$, one may choose $C$ with left and right sides in $[T-1,T-1/2]$ and $[T+1/2,T+1]$, bottom in $[\eta/3,2\eta/3]$, and top at imaginary part $1$, such that

$$ 1\le M\le140\log(T+2),\quad S\le T+2,\quad \log D_a\le\epsilon T+2000\log(T+2)\log(e/\eta). \tag{4.1} $$

Proof. Write $Z(z)=\zeta(1/2+iz)$ and anchor at $z_0=T-3i/2$. In the radius-eight disk, the corresponding $s$ has $-6\le\Re s\le10$ and $T-8\le\Im s\le T+8$. Euler--Maclaurin at $N=1$ with four Bernoulli terms gives

$$ |Z(z)|\le7(T+25)^8\le(T+2)^{16}=:\mathcal M. $$

For details, the remainder has bound $|(s)_8|\sup|B_8|/[8!(\Re s+7)]$, with $\sup|B_8|<1$, $\Re s+7\ge1$, and $|s+j|\le T+25$. The two initial terms, four Bernoulli terms, and remainder are each bounded by $(T+25)^8$. The last displayed inequality follows from $(T+2)^2\ge2(T+25)$.

The Euler product gives $|Z(z_0)|\ge1/\zeta(2)>1/2$. Jensen on the radius-eight disk therefore counts at most

$$ N\le\frac{16\log(T+2)+\log2}{\log(8/7)} \le140\log(T+2) \tag{4.2} $$

zeros in radius seven, with multiplicities. Choose a zero-free circle of radius $R\in(6,7)$ and divide $Z$ by its finite disk Blaschke product $B$. The zero-free quotient $g$ has $|g|\le\mathcal M$ and $|g(z_0)|>1/2$. Harnack applied to $\log\mathcal M-\log|g|$ gives, on radius four,

$$ |g(z)|\ge2^{-5}(T+2)^{-64}. $$

Each disk factor there has modulus at least $|z-\alpha|/11$. To choose whole rectangle edges, use the elementary logarithmic average

$$ \frac1\ell\int_I\log_+\frac{11}{|t-b|}\,dt \le\log(22/\ell)+1,\qquad |I|=\ell\le1/2. \tag{4.3} $$

The centered interval maximizes this integral by symmetric decreasing rearrangement; direct integration gives the bound. Sum (4.3) over the local zeros, including multiplicities, then choose an edge coordinate at most its average. A vertical edge uses $b=\Re\alpha$ and a horizontal edge uses $b=\Im\alpha$. The same estimate then holds uniformly along that entire edge. Use intervals of length $1/2$ for the vertical edges and length $\eta/3$ for the bottom edge. The top is at distance at least $1/2$ from every nontrivial ordinate. All rectangle points lie inside radius three of $z_0$.

It follows uniformly on $C$ that

$$ |Z(z)|\ge2^{-5}(T+2)^{-64} (\eta/(66e))^N, \qquad \log(1/|Z(z)|)\le1000K, \tag{4.4} $$

where $K=\log(T+2)\log(e/\eta)$. Here $\log(66e/\eta)\le6\log(e/\eta)$ and (4.2) suffice. In particular the chosen contour avoids zeros. This argument averages logarithmic losses, rather than forcing a separate minimum gap from each zero, and therefore avoids an extra $\log\log T$ loss.

We record a completion-factor bound to make the constants effective. For $s=\sigma+it$, $-1/2\le\sigma\le1/2$, $t\ge19$, put $w=s/2$ and $y=t/2$. Binet's formula at $v=w+2$ has remainder $|R(v)|\le1/(12\Re v)\le1/21$. One way to see the bound is

$$ R(v)=\int_0^\infty e^{-vu}b(u)\,du,\quad b(u)=2\sum_{n\ge1}(u^2+4\pi^2n^2)^{-1}\in(0,1/12]. $$

The gamma recurrence, $|w(w+1)|\le2y^2$, and $|s(s-1)|/2\ge2y^2$ yield

$$ \left|\tfrac12s(s-1)\pi^{-s/2}\Gamma(s/2)\right| \ge \sqrt{2\pi}\,\pi^{-1/4}e^{-9/4-1/21} y^{5/4}e^{-\pi t/4} \ge e^{-\pi t/4}. \tag{4.5} $$

The last prefactor already exceeds $3.15$ at $y=19/2$. Combining (4.4)--(4.5), perimeter at most six, and $|\Re z-T|\le1$, gives $\log D_0\le\pi T/4+1000K+\pi/4+\log(3/\pi)$. Finally $D_a\le e^{-a(T-1)}D_0$, proving (4.1). The rectangle contains the given zero and has the required disjoint reflections. Its multiplicity is bounded by (4.2). $\square$

5. Shifted theta tails, compactification, and norm

For $x\ge0$ put $R_x=\pi q e^{2x}$. If $R_x\ge\pi$, the full theta series gives

$$ |p_a(x)|\le B_a(x):=20\pi^2e^{9x/2-R_x}. \tag{5.1} $$

Indeed the absolute summands are bounded by $(4\pi^2n^4e^{9x/2}+6\pi n^2e^{5x/2})e^{-R_xn^2}$. The sum $\sum n^4e^{-R_x(n^2-1)}$ is at most two: successive ratios are bounded by $16e^{-3\pi}<1/2$. The coefficient $8\pi^2+12\pi$ is less than $20\pi^2$. The same bound holds at $-x$.

If $R_x\ge6$, the upper-tail integral satisfies $|f_w^+(x)|\le B_a(x)/R_x$. In fact $B_a(x+u)\le B_a(x)e^{-(2R_x-9/2)u}$, while $|e^{-iwu}|\le e^u$. The ODE gives the derivative estimate. After contour integration and parity projection,

$$ |h(x)|\le2D_aB_a(|x|),\qquad |h'(x)|\le2D_a(1+S)B_a(|x|). \tag{5.2} $$

Let $L\ge2$, $\delta=e^{-2L}$, $b=L-\delta$ and $A=\pi q e^{2b}\ge17$. Define the even cubic cutoff $\chi$ to be one on $[-b,b]$, zero outside $[-L,L]$, and $1-3s^2+2s^3$ for $s=(|x|-b)/\delta\in(0,1)$. It is $C^1$ and $|\chi'|\le3/(2\delta)$. For $r=(1-\chi)h$, the elementary tail estimates are

$$ \begin{aligned} \int_b^\infty B_a^2&\le200\pi^3q^{-1}e^{7b-2A},\\ \int_b^\infty e^{2x}B_a^2&\le200\pi^3q^{-1}e^{9b-2A},\\ \int_b^\infty e^{x/2}B_a&\le20\pi q^{-1}e^{3b-A}. \end{aligned} \tag{5.3} $$

For example the logarithmic slopes of the three integrands are at most $9-4A$, $11-4A$, and $5-2A$, respectively. Integrating the resulting exponentials bounds the integrals by their initial values divided by $2A,2A,A$; this proves (5.3).

Using (5.2), both half-lines, and $|r'|^2\le2|h'|^2+2|\chi'h|^2$ gives

$$ \begin{aligned} N(r)^2&\le12000\pi^3q^{-1}D_a^2(1+S)^2e^{11L-2A},\\ \sup_{|\Im z|\le1/2}|\widehat r(z)| &\le80\pi q^{-1}D_a e^{3L-A}=:E. \end{aligned} \tag{5.4} $$

The coefficients of the first bound before $200\pi^3/q$ are $16$ from $h'$, $36$ from $\chi'h$, and $8$ from the weighted norm. Each Fourier value of $h$ at the four selected clusters has modulus $1/2$. Equation (2.2) therefore gives $|Q(h,r)|\le2ME$. Since $Q(h)=-M$,

$$ |Q(\chi h)+M| \le320\pi Mq^{-1}D_a e^{3L-A} +240000\pi^3q^{-1}D_a^2(1+S)^2e^{11L-2A}. \tag{5.5} $$

Now $A\ge\pi q e^{2L}-2\pi$, since $q\le1$ and $e^{2L}(1-e^{-2\delta})\le2$. A sufficient condition for $Q(\chi h)\le-M/2$ is

$$ D_a(1+S)q^{-1}e^{6L-\pi q e^{2L}}\le10^{-8}. \tag{5.6} $$

Indeed the relative error in (5.5) is then bounded by

$$ 320\pi e^{2\pi-6}10^{-8} +240000\pi^3e^{4\pi-2}10^{-16}<1/2. \tag{5.7} $$

We also need a global physical norm, with only polynomial loss as $a\uparrow\pi/4$. Set $K_a=\int_0^\infty e^v|p_a(v)|\,dv$. Use the upper inverse for $x\ge0$ and lower inverse for $x\le0$ to get

$$ |h(x)|\le2D_a e^{-|x|}K_a. $$

Termwise integration of the entire absolute theta series, followed by the substitution $t=\pi q n^2e^{2v}$ and a complete-gamma upper bound, gives

$$ \begin{aligned} K_a&\le\zeta(3/2)\pi^{-3/4} \bigl[2\Gamma(11/4)q^{-11/4}+3\Gamma(7/4)q^{-7/4}\bigr]\\ &\le10q^{-11/4}. \end{aligned} \tag{5.8} $$

For the last constant use $\zeta(3/2)<3$, $\Gamma(7/4)\le1$ by log-convexity, $\Gamma(11/4)=(7/4)\Gamma(7/4)$, and $\pi^{-3/4}<1/2$. Since $\int_{\mathbb R}e^{-2|x|}dx=1$,

$$ \|\chi h\|_2^2\le400D_a^2q^{-11/2},\qquad \frac{Q(\chi h)}{\|\chi h\|_2^2} \le-\frac{Mq^{11/2}}{800D_a^2}. \tag{5.9} $$

The compact $C^1$ function lies in $H_0^1(-L,L)$, and its weak second derivative is integrable. Real, parity-preserving smooth interior approximation converges in $X$, so preserves any strictly weaker negative margin in (5.9).

6. Explicit parameter choice

Under Theorem 1.1 choose

$$ \epsilon=\frac{10000K}{dT},\qquad a=\pi/4-\epsilon, \qquad q=\sin(2\epsilon). $$

The hypothesis implies $\epsilon\le\sqrt d/2\le1/2<\pi/4$; also $q\ge\epsilon$. Since $\sin(2\epsilon)\ge2\epsilon(1-2\epsilon^2/3)$,

$$ \pi q e^{2L} =\frac{(1+d)T}{2}\sin(2\epsilon) \ge\epsilon T+5000K. \tag{6.1} $$

All remaining costs in (5.6) fit well inside this excess. Specifically, Lemma 4.1 and $S\le T+2$ give

$$ \log D_a\le\epsilon T+2000K,\quad \log(1+S)\le2K,\quad \log(1/q)\le K,\quad 6L\le3K,\quad 8\log10\le4K. \tag{6.2} $$

For the third bound, $\epsilon\ge1/T$, so $q\ge1/T$. Also $K>5$. Thus (6.1)--(6.2) prove (5.6). The hypothesis itself forces $T\ge20000K$, giving $L\ge2$; $e^{-2e^{-2L}}\ge1/2$ and (6.1) give $A\ge5000K>17$. All tail conditions therefore hold.

Finally the logarithm of the reciprocal margin in (5.9) is at most

$$ \log800+\tfrac{11}{2}\log(1/q)+2\log D_a \le7K+\tfrac{11}{2}K+20000K/d+4000K <25000K/d. $$

An individual smooth interior approximation retains half this margin, which is more than the weaker $\exp(-30000K/d)$. This proves Theorem 1.1.

7. Several clusters and the counting consequence

The zero-count assertion is trivial when $N_{\rm off}=0$; in the subspace construction assume $k\ge1$. Use the common parameters $\epsilon=10000K/(dH)$, $a=\pi/4-\epsilon$, and $q=\sin(2\epsilon)$ in Theorem 1.3. For each center choose the rectangle of Lemma 4.1. The separation in (1.4) makes all its real projections disjoint. Let $M_j$ be its selected multiplicity and $D_j$ its contour cost. With $D_* =\exp(\epsilon H+2000K)$ and $S=H+2$,

$$ 1\le M_j\le140\log(H+2),\quad D_j\le D_*,\quad S_j\le S. $$

There is no need for each $T_j$ to satisfy the large-parameter condition separately: Lemma 4.1 applies at every $T_j\ge20$. Let $h_j$ be its real selector in either fixed parity. Their zero-side supports are disjoint, so $Q(h_j,h_\ell)=-M_j\delta_{j\ell}$. For real coefficients set

$$ h(c)=\sum_jc_jh_j,\qquad \|c\|_M^2=\sum_jM_jc_j^2, \qquad D(c)=\sum_jD_j|c_j|. $$

The tail bounds (5.4) apply with $D_a$ replaced by $D(c)$. The cross term in (5.5) becomes $320\pi q^{-1}D(c)(\sum M_j|c_j|)e^{3L-A}$, and its quadratic term uses $D(c)^2$. Weighted Cauchy--Schwarz gives

$$ D(c)\le D_*\sqrt k\,\|c\|_M,\qquad \sum M_j|c_j|\le\sqrt{140k\log(H+2)}\,\|c\|_M. $$

Put $F=k\sqrt{140\log(H+2)}$. Thus a sufficient simultaneous condition is

$$ F D_*(1+S)q^{-1}e^{6L-\pi q e^{2L}}\le10^{-8}. \tag{7.1} $$

The same substitution as in (5.7) bounds the linear error unchanged and multiplies its quadratic error by $k/F^2\le1$. Since separated centers give $k\le H$ and $\sqrt{140\log(H+2)}\le(H+2)^2$ for $H\ge20$, $\log F\le3K$. This adds only $3K$ to (6.2), so (7.1) holds. Consequently the entire cutoff image satisfies

$$ Q(\chi h(c))\le-\tfrac12\|c\|_M^2,\qquad \|\chi h(c)\|_2^2\le400kD_*^2q^{-11/2}\|c\|_M^2. \tag{7.2} $$

The first inequality proves injectivity. Simultaneously approximate the finite basis by smooth interior tests. Uniform convergence on its compact coefficient unit sphere permits $Q\le-\|c\|_M^2/4$ and norm squared at most $800kD_*^2q^{-11/2}\|c\|_M^2$. The smooth image is still $k$-dimensional and has quotient at most $-q^{11/2}/(3200kD_*^2)$. Its reciprocal logarithm is bounded by

$$ 2K+K+\tfrac{11}{2}K+20000K/d+4000K<25000K/d. $$

This proves the subspace assertion with exponent $30000$.

For (1.5), greedily choose one remaining ordinate in the specified multiset, then delete all remaining ordinates within real distance at most three of its real height. Chosen heights are more than three apart. Every deleted ordinate lies in the radius-seven disk anchored at $T_j-3i/2$: its squared distance is at most $9+4<49$. Equation (4.2) bounds its deleted multiplicity by $140\log(T_j+2)\le140\log(H+2)$, even at interval endpoints. If $k$ choices exhaust the multiset, then $N_{\rm off}\le140k\log(H+2)$, proving (1.5).

8. Gaussian filtering and a margin uniform in the slack

We prove Theorem 1.4 using the same contour selection, but a different zero-free multiplier. For each center $20\le T\le H$ set

$$ A_{H,T}(z)=\exp\bigl(\pi z/4-(z-T)^2/H\bigr)\Xi(z),\qquad p_{H,T}(x)=\frac1{2\pi}\int_{\mathbb R}A_{H,T}(t)e^{itx}\,dt, \qquad L_0=\frac12\log\frac H{2\pi}. \tag{8.1} $$

The filter is centered separately at each $T$, while its width uses the common maximum height $H$. The multiplier is real on the real axis and has no zeros. Hence $p_{H,T}(-x)=\overline{p_{H,T}(x)}$ and the zero multiset of $A_{H,T}$ is that of $\Xi$.

8.1 Uniform physical tails

For $y\ge2$, the functional equation writes $\Xi(t+iy)=\xi(1/2+y-it)$. Put $a=1/4+y/2$, $b=|t|/2$, and $r=|a-ib|$. Binet's formula, with remainder at most $1/(12a)$, gives

$$ |\Gamma(a-ib)|e^{\pi b/2} \le\sqrt{2\pi}\,e^{1/15}r^{a-1/2} \le3r^{y/2}. \tag{8.2} $$

Indeed the non-power exponent is $b\arctan(a/b)-a\le0$, including its limit at $b=0$; also $r\ge1$ and $a-1/2=y/2-1/4$. With $V=|t|+y+1$, $|s|,|s-1|\le V$, $r\le V/2$, and $|\zeta(1/2+y-it)|\le\zeta(1/2+y)<2$, completion yields

$$ e^{\pi t/4}|\Xi(t+iy)| \le3V^2\left(\frac V{2\pi}\right)^{y/2}. \tag{8.3} $$

Shift the inversion contour to height $y$. To justify this first at any fixed $y$, the Gaussian dominates the polynomial growth of the completed zeta function after gamma-exponential cancellation throughout the intervening finite strip; thus the vertical ends tend to zero. With $u=t-T$, use $|t|\le H+|u|$ and

$$ \left(1+\frac{|u|+y+1}{H}\right)^{y/2} \le e^{y(|u|+y+1)/(2H)},\qquad -\frac{u^2}{H}+\frac{y|u|}{2H} \le-\frac{u^2}{2H}+\frac{y^2}{8H}. $$

The Gaussian shift contributes $y^2/H$. The total loss is at most $2y^2/H$, since $y^2+y(y+1)/2+y^2/8\le2y^2$ for $y\ge2$. For $j=0,1,2$, the remaining polynomial is at most $(H+y+1+|u|)^{j+2}$. Gaussian moments and $(v+w)^m\le2^{m-1}(v^m+w^m)$ give

$$ |p_{H,T}^{(j)}(x)| \le20(H+y+1)^{j+2}\sqrt H\, e^{-y(x-L_0)+2y^2/H}. \tag{8.4} $$

For completeness, for $2\le m\le4$ and $W=H+y+1\ge H\ge20$, the normalized $m$th absolute moment under $e^{-u^2/(2H)}$ divided by $W^m$ is at most one. The largest resulting integral constant is $48/\sqrt{2\pi}<20$. When $2\le y\le H$, (8.4) is bounded by

$$ 2000H^{j+5/2}e^{-y(x-L_0)+2y^2/H}. \tag{8.5} $$

For every fixed higher derivative order the same proof has a finite order-dependent constant. Taking arbitrarily large fixed $y$ and using the reflection symmetry proves exponential integrability of every physical derivative with every fixed exponential weight.

8.2 Contour inversion and compactification

Choose each rectangle by Lemma 4.1 and define $U,D$ by (3.2), replacing $A_a$ with $A_{H,T}$. Equation (4.4), the pointwise cancellation of (4.5) by $e^{\pi\Re w/4}$, and $|\Re w-T|\le1$ imply

$$ D\le e^{1000K+1/H}\le D_*:=e^{1001K}. \tag{8.6} $$

Here $\log(T+2)\log(e/\eta_j)\le K$, and the perimeter factor $6/(2\pi)$ is less than one. The Gaussian costs at most $e^{1/H}$ on each contour, uniformly in its center.

The inversion proof of Section 3 applies with $p_{H,T}$ in place of $p_a$. Its upper and lower ODE inverses differ by $-ie^{iwx}A_{H,T}(w)$; at every enclosed zero, all derivatives of this defect below its multiplicity vanish. Taking residues before Fourier integration therefore gives the entire selector and its exact zero values. The exponential derivative bounds just proved justify the ODE tails, all transform evaluations, and (2.2). The same parity projections have energy $-M$, without any symmetry assumption on the multiplier beyond being real on the real axis.

Let $L=\tfrac12\log((1+d)H/(2\pi))$, $\delta=e^{-2L}$, $b=L-\delta$, and use the cubic cutoff of Section 5. Put

$$ y=\frac{dH}{64},\quad R=\frac{d^2H}{1024},\quad B=2000H^{5/2}e^{-R}. \tag{8.7} $$

Hypothesis (1.6) gives $K>5$, $y\ge2$, and $y\le H$. Since $\frac12\log(1+d)\ge d/4$ and $\delta\le2\pi/H$, we have $b-L_0\ge d/8$. Consequently the exponent in (8.5) at $b$ is at most $-3d^2H/2048\le-R$. For $x\ge b$, $|p_{H,T}(x)|\le B e^{-y(x-b)}$, and the same bound holds at $-x$. The ODE inverses and $S\le H+2$ now give, for $s=|x|-b\ge0$,

$$ |h(x)|\le\frac{2DB}{y-1}e^{-ys}\le2DB e^{-ys},\qquad |h'(x)|\le2DB\left(1+\frac{S}{y-1}\right)e^{-ys} \le4DHB e^{-ys}. \tag{8.8} $$

Also $|\chi'|\le3H/2$, so for $r=(1-\chi)h$, $|r'|\le7DHB e^{-ys}$. Integrate over both half-lines. The derivative contributes at most $49D^2H^2B^2/y$, and the weighted square contributes at most $4D^2B^2e^{2b}/(y-1)$. Using $e^{2b}\le H$ yields

$$ N(r)^2\le2\cdot10^8D^2H^7e^{-2R},\qquad \sup_{|\Im z|\le1/2}|\widehat r(z)| \le8000DH^3e^{-R}. \tag{8.9} $$

For the second inequality, integrate $4DB e^{b/2}/(y-1/2)$ and use $e^{b/2}\le H^{1/2}$.

8.3 Norm and simultaneous margin

The global norm is estimated on the real frequency axis. Fractional-part integration, valid for $\Re s>0$, $s\ne1$, gives

$$ \zeta(s)=\frac{s}{s-1}-s\int_1^\infty\{v\}v^{-s-1}\,dv. $$

At $s=1/2+it$ this implies $|\zeta(s)|\le2(|t|+1)$. Binet at real part $1/4$ gives $|\Gamma(1/4+it/2)|e^{\pi|t|/4}<5$. Completion therefore gives

$$ |A_{H,T}(t)|\le5(|t|+2)^3e^{-(t-T)^2/H},\qquad \|p_{H,T}\|_2^2\le400H^{13/2}. \tag{8.10} $$

For the last constant, Plancherel and $T+2\le1.1H$ bound the integral of the sixth-degree polynomial by $32\sqrt{\pi H/2}[(H+2)^6+15H^3/64] <64\sqrt{\pi/2}H^{13/2}$; multiplication by $25/(2\pi)$ is less than $400$.

The real axis lies outside every rectangle and at distance at least $\eta_0/3$ from it. Hence $|U(t)|\le3D|A_{H,T}(t)|/\eta_0$. Reflection and parity projection in $L^2$ show

$$ \|h\|_2^2\le20000D^2\eta_0^{-2}H^{13/2}. \tag{8.11} $$

This uses the entire contour selector on the whole real axis; it does not integrate an individually growing upper ODE inverse there.

For separated clusters use the notation of Section 7, in particular $\|c\|_M^2=\sum M_jc_j^2$, $D(c)=\sum D_j|c_j|$, and $F=k\sqrt{140\log(H+2)}$. Bounds (8.8)--(8.9) sum with $D(c)$; the disjoint zero-side supports still give $Q(h(c))=-\|c\|_M^2$. The same weighted Cauchy--Schwarz inequalities as in Section 7 imply that the cutoff-energy error divided by $\|c\|_M^2$ is at most

$$ 32000FD_*H^3e^{-R} +4\cdot10^9kD_*^2H^7e^{-2R}. \tag{8.12} $$

Now $R\ge9000K$, $\log F\le3K$, $\log k\le K$, and $\log H\le K$. Since $\log32000<3K$ and $\log(4\cdot10^9)<5K$, the two terms in (8.12) are bounded by $e^{1010K-R}$ and $e^{2015K-2R}$, respectively; each is less than $1/4$. It follows that

$$ Q(\chi h(c))\le-\tfrac12\|c\|_M^2,\qquad \|\chi h(c)\|_2^2\le20000kD_*^2\eta_0^{-2}H^{13/2}\|c\|_M^2. \tag{8.13} $$

As before, negativity proves injectivity. Simultaneous real parity-preserving smooth interior approximation gives the quotient bound $-\eta_0^2/(160000kD_*^2H^{13/2})$. Its reciprocal logarithm is at most

$$ \log160000+\log k+2002K+2\log(1/\eta_0) +\tfrac{13}{2}\log H<3000K. $$

This proves the weaker advertised margin $\exp(-10000K)$. The greedy counting proof in Section 7 applies unchanged. Taking $H=T$, $\eta_0=\eta$, and $d=\sqrt{10^7K/T}$ proves (1.7).

9. Interpretation, verification, and limitations

For fixed displacement and fixed positive slack $d$, the guaranteed negative margin in Theorem 1.1 is a negative power of $T$. Reducing the slack to obtain (1.3) worsens this guarantee to a stretched exponential. Gaussian filtering retains a power margin at the larger support excess (1.7). Thus a radius estimate and a numerical sensitivity estimate are different outputs. No claim of an optimal exponent or optimal constant is made.

Contraposition gives a conditional zero-exclusion statement if one already has complete all-test positivity at the specified physical radius, or an actual upper bound on the complete negative index. Neither a finite trial matrix nor a truncated real-zero sampling matrix gives that hypothesis. Our construction supplies negative witnesses under a hypothetical spectral defect; it does not remove the missing arithmetic lower-bound problem.

The proof is analytic and uses no computed zeta zero. A short CPU Arb checker certifies (5.7), the completion prefactor at $t=19$, and the finite arithmetic constants used in Sections 4--7. A second checker verifies 22 finite comparisons for the Gaussian alternative in Section 8. These finite checks do not certify the analytic theorem by themselves and do not exhibit an off-line zero. Run python check_constants.py and python check_gaussian_constants.py in the unpacked supplement with Python 3 and python-flint==0.9.0. The checkers use 160-bit Arb arithmetic, take less than a second on one CPU, and write finite_constants.json and gaussian_constants.json. The expected output, dependency list, proof ledger, and precise literature-reading scope accompany the script. The supplement's code is distributed under the MIT license.

The manuscript has received a fresh internal reconstruction of inversion, multiplicities, parity, full-theta estimates, cutoff, and norm assembly. A peer independently checked the edge-averaging mechanism and a floating Stirling scale, explicitly leaving the shifted tails, global norm, and cutoff assembly unverified. In a later focused discussion, halasz_resonator checked the Gaussian angular inequality, square completion, real-axis norm argument, and compatibility of distinct cluster centers with exact zero selection. The peer also reported numerical checks; we have not inspected their code. These focused comments do not independently verify the whole theorem. Prior-art comparison is limited to the cited sources and the inspected portions noted in the supplement; no claim to have read inaccessible material is made.

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